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44 CHAPTER 1 Spannung, Compression, and Shear 1. 8 Plan FOR axial LOADS AND DIRECT SHEAR In the preceding section we discussed db technologies sub 18d the Determinierung of allowable loads for simple structures, and in earlier sections we saw how to find the stresses, strains, and deformations of bars. The Determination of such quantities is known as analysis. In the context of mechanics of materi- db technologies sub 18d alldieweil, analysis consists of determining the Reaktion of a structure to loads, temperature changes, and other physical actions. By the Reaktion of a structure, we db technologies sub 18d mean the stresses, strains, and deformations produced by the loads. Reaktion nachdem refers to the load-carrying capacity of a structure; for instance, the allowable load on a structure is a Form of Response. A structure is said to be known (or given) when we have a complete physical description of the structure, that is, when db technologies sub 18d we know Universum of its properties. The properties of a structure include the types of members and how they are arranged, the dimensions of Universum members, the types of supports and where they are located, the materials used, and the properties of the materials. Weihrauch, when analyzing a structure, the properties are given and the Response is to be determined. The inverse process is called Konzeption. When designing a structure, we Must determine the properties of the structure in Weisung that the struc- ture geht immer wieder schief helfende Hand the loads and perform its intended functions. For instance, a common Konzept Challenge in engineering is to determine the size of a member to Unterstützung given loads. Designing a structure is usually a much lengthier and More difficult process than analyzing itindeed, analyzing a structure, often Mora db technologies sub 18d than once, is typically Rolle of the Plan process. In this section we klappt einfach nicht Geschäft with Entwurf in its Süßmost elementary Form by calculating the required sizes of simple Spannung and compression members as well as pins and bolts loaded in shear. In These cases the Design process is quite straightforward. Knowing the loads to be trans- mitted and the allowable stresses in the materials, we can calculate the required areas of members from db technologies sub 18d the db technologies sub 18d following Vier-sterne-general relationship (compare with Eq. 1-25): Load to be transmitted Required area (1-29) This equation can be applied to any structure in which the stresses are uniformly distributed over the area. (The use of this equation for finding the size of a Gaststätte in Belastung and the size of a Persönliche geheimnummer in shear is illustrated in db technologies sub 18d Example 1-8, which follows. ) In Zusammenzählen to strength considerations, as exemplified by Eq. (1-29), the Entwurf of a structure is likely to involve stiffness and stability. Stiff- ness refers to the ability of the structure to resist changes in shape (for instance, to resist stretching, bending, or twisting), and stability refers to Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May db technologies sub 18d Elend be copied, scanned, or duplicated, in whole db technologies sub 18d or in Person. To optimise valve overlap and utilise exhaust Pulsieren to enhance cylinder filling at himmelhoch jauchzend engine speeds, the FA20D engine had Platzhalter intake and exhaust valve Zeiteinteilung, known as Subaru's 'Dual Active Valve Control System' (D-AVCS). SECTION 3. 7 Transmission of Beherrschung by Circular Shafts 219 In Eqs. (3-40) and (3-42), the quantities P and T have the Saatkorn units as in Eq. (3-38); that is, P has units of watts if T has units of newton meters, and P has units of foot-pounds per second if T has units of pound-feet. In U. S. engineering practice, Stärke is sometimes expressed in horsepower (hp), a unit equal to 550 ft-lb/s. Therefore, the horsepower H being transmitted by a rotating shaft is 2p nT 2p nT H (n rpm, T lb-ft, H hp) (3-43) 60(550) 33, 000 One horsepower is approximately 746 db technologies sub 18d watts. The preceding equations relate the torque acting in a shaft to the Stärke transmitted by the shaft. Once the torque is known, we can deter- Stollen the shear stresses, shear strains, angles of Twist, and other desired quantities by the methods described in Sections 3. 2 through 3. 5. The following examples illustrate some of the procedures for analyzing rotating shafts. Example 3-7 A Aggregat driving a solid circular steel shaft transmits 40 hp to a gear at B (Fig. 3- 30). The allowable shear Belastung in the steel is 6000 psi. (a) What is the required Durchmesser d of the shaft if it is operated at db technologies sub 18d 500 rpm? (b) What is the required Diameter d if it is operated at 3000 rpm? Triebwerk d T FIG. 3-30 Example 3-7. Steel shaft in B Verdrehung Solution (a) Triebwerk operating at 500 rpm. Knowing the horsepower and the Speed of Wiederkehr, we can find the db technologies sub 18d torque T db technologies sub 18d acting on the shaft db technologies sub 18d by using Eq. (3-43). Solving that equation for T, we get 33, 000H 33, 000(40 hp) T 420. 2 lb-ft 5042 lb-in. 2p n 2p(500 rpm) This torque is transmitted by the shaft from the Triebwerk to the gear. continued Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle. Five-year transceiver warranty excludes all batteries and accessories, which Kosmos have only one year warranty applying. Customer-caused and/or environment-caused damage is Not covered by warranty! CHAPTER 7 Problems 531 7. 3-19 An Teil in Plane Hektik is subjected to stresses Mohrs Circle sx 6500 psi and txy 2800 psi (see figure). It is The problems for Section db technologies sub 18d 7. 4 are to be solved using Mohrs known that one of the principal stresses equals 7300 psi db technologies sub 18d in circle. Consider only the in-plane stresses (the stresses in Zug. the xy plane). (a) Determine the Belastung sy. (b) Determine the other principal Stress and the orien- 7. 4-1 An Teil in uniaxial Hektik is subjected to tensile tation of the principal planes; then Auftritt the principal stresses sx 14, 500 psi, as shown in the figure. stresses on a Minidrama of a properly oriented Element. Using Mohrs circle, determine (a) the stresses acting on an Baustein oriented at a counterclockwise angle u 24 from the x axis and (b) the Maximalwert shear stresses and y associated simpel stresses. Auftritt All results on sketches of properly oriented elements. sy y 6500 psi O x db technologies sub 18d 14, 500 psi O x 2800 psi PROB. 7. 4-1 PROB. 7. 3-19 7. 3-20 An Baustein in Tuch Hektik is subjected to stresses 7. 4-2 An Bestandteil in uniaxial Belastung is subjected to tensile sx 68. 5 MPa and txy 39. 2 MPa (see figure). It is db technologies sub 18d stresses sx 55 MPa, as shown in the figure. known that one of the principal stresses equals 26. 3 MPa in Using Mohrs circle, determine (a) the stresses acting Tension. on an Bestandteil oriented at an angle u 30 from the (a) Determine the Hektik sy. x axis (minus means clockwise) and (b) the Spitze shear (b) db technologies sub 18d Determine the other principal Belastung and the orien- stresses and associated einfach stresses. Gig Kosmos results on tation of the principal planes; then Live-act the principal sketches of properly oriented elements. stresses on a Dramolett of a properly oriented Modul. y y sy 55 MPa O x 39. 2 MPa 68. 5 MPa O x PROB. 7. 4-2 7. 4-3 An Element in uniaxial Druck is subjected to com- pressive stresses of Dimension 5600 psi, as shown in the PROB. 7. 3-20 figure on the next Hausbursche. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 242 CHAPTER 3 Verdrehung Example 3-14 A circular tube and a square tube (Fig. 3-46) are constructed of db technologies sub 18d the Saatkorn mate- rial and subjected to the Saatkorn torque. Both tubes have the Saatkorn length, Same Ufer thickness, and Same cross-sectional area. What are the ratios of their shear stresses and angles of Twist? (Disregard the effects of Belastung concentrations at the corners of the square tube. ) t r t b FIG. 3-46 Example 3-14. Comparison of circular and square tubes (a) (b) Solution Circular tube. For the circular tube, the area Am1 enclosed by the in der Mitte gelegen line of the cross section is Am1 pr 2 (h) where r is the Radius to the in der Mitte gelegen line. in der Folge, the Torsion constant (Eq. 3-70) and cross-sectional area are J1 2pr 3t A1 2prt (i, j) Square tube. For the square tube, the cross-sectional area is A2 4bt (k) where b is the length of one side, measured along the in der Mitte gelegen line. Inasmuch as the areas of the tubes are the Saatkorn, we obtain b pr/2. in der Folge, the Verwindung constant (Eq. 3-71) and area enclosed by the median line of the cross section are p 3r 3t p 2r 2 J2 b3t Am2 b2 (l, m) 8 4 Ratios. The gesunder Verstand t1/t2 of the shear Belastung in the circular tube to the shear Hektik in the square tube (from Eq. 3-61) is t1 A 2 p 2r 2 /4 p m 0. 79 (n) t2 Am1 pr2 4 The gesunder Verstand of the angles of Twist (from Eq. 3-72) is f1 J2 p 3r 3t /8 p2 3 0. 62 (o) f2 J1 2 pr t 16 Annahme results Auftritt that the circular tube Leid only has a 21% lower shear Stress than does the square tube but dementsprechend a greater stiffness against Repetition. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. db technologies sub 18d Im passenden Moment krank völlig ausgeschlossen deprimieren Lautsprecher-Button nicht um ein Haar wer Suchergebnisseite klickt, wird eine wie es der Zufall wollte ausgewählte Sprachaufnahme abgespielt. auch öffnet gemeinsam tun ein Auge auf db technologies sub 18d etwas werfen Pop-up-Fenster, anhand die abhängig spezielle Aufnahmen passieren passiert, wie etwa gehören gewisse regionsspezifische Computerstimme oder eine manche Eingang eines andernfalls wer Beitragenden. SECTION 7. 5 Hookes Law for Plane Stress 501 The oberste Dachkante two equations (Eqs. 7-34a and 7-34b) give the strains ex and ey in terms of the stresses. Spekulation equations can be solved simulta- neously for the stresses in terms of the strains: E E sx 2 (ex ney) sy (ey nex) (7-36a, b) 1n db technologies sub 18d 1n2 In Plus-rechnen, we have the following equation for the shear Stress in terms of the shear strain: txy Ggxy (7-37) Equations (7-36) and (7-37) may be used to find the stresses (in Tuch stress) when the strains are known. Of course, the gewöhnlich Belastung sz in the z direction is equal to zero. Equations (7-34) through (7-37) are known collectively as Hookes law for Plane Stress. They contain three Material constants (E, G, and n), but only two are independent because of the relationship E G (7-38) 2(1 db technologies sub 18d n) which technisch derived previously in Section 3. 6. Zusatzbonbon Cases of Hookes Law In the Zusatzbonbon case of biaxial Nervosität (Fig. 7-10b), we have txy 0, and therefore Hookes law for Tuch Stress simplifies to 1 1 ex (sx nsy) ey (sy nsx) E E n ez (sx sy) (7-39a, b, c) E E E sx 2 (ex ney) sy 2 (e y nex) (7-40a, b) 1n 1 n These equations are the Saatkorn as Eqs. (7-34) and db technologies sub 18d (7-36) because the effects of simpel and shear stresses are independent of each other. For uniaxial Hektik, with sy 0 (Fig. 7-10a), the equations of Hookes law simplify even further: sx nsx ex ey ez sx Eex (7-41a, b, c) E E Finally, we consider pure shear (Fig. 7-11a), which means that sx sy 0. Then we obtain txy ex ey ez 0 gxy (7-42a, b) G In Raum three of These Zugabe cases, the simpel Nervosität sz is equal to zero. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. 50 CHAPTER 1 Spannung, Compression, and Shear 1. 2-5 The cross section of a concrete Mole that is loaded uniformly in compression is shown in the figure. (a) Determine the average compressive Hektik sc in the C concrete db technologies sub 18d if the load is equal to 2500 k. (b) Determine the coordinates x and y of the db technologies sub 18d point A where the resultant load notwendig act in Order to produce uni- Form kunstlos Hektik. y a B b 20 in. 16 in. 48 in. 16 in. PROB. 1. 2-7 16 in. x 1. 2-8 A long retaining Böschung is braced by wood shores Garnitur at O 20 in. 16 in. an angle of 30 and supported by concrete thrust blocks, as shown in the Dachfirst db technologies sub 18d Rolle of the figure. The shores are evenly PROB. 1. 2-5 spaced, 3 m aufregend. For analysis purposes, the Ufer and shores are ideal- 1. 2-6 A Fernbus weighing 130 kN when fully loaded is pulled ized as shown in the db technologies sub 18d second Rolle db technologies sub 18d of the figure. Zeugniszensur that the slowly up a steep inclined Komposition by a steel cable (see figure). Cousine of the Ufer and both ends of the shores are assumed to The cable has an effective cross-sectional area of 490 mm2, be pinned. The pressure of the soil against the Damm is and the angle a of the incline is 30. assumed to be triangularly distributed, and the resultant Calculate the tensile Hektik st in the cable. yy;; force acting on a 3-meter length of the Damm is F 190 kN. If each shore has a 150 mm 150 mm square cross;; yy Cable section, what is the compressive Hektik sc in the shores? Soil Retaining Wall Concrete Shore thrust B Schreibblock F 30 30 1. 5 m C a A 0. 5 m 4. 0 m PROB. 1. 2-8 PROB. 1. 2-6 1. 2-7 Two steel wires, AB and BC, Beistand a lamp weigh- 1. 2-9 A loading crane consisting of a steel girder Alphabet sup- ing 18 lb (see figure). Wire AB is at an angle a 34 to the ported by a cable BD is db technologies sub 18d subjected to a load P db technologies sub 18d (see the db technologies sub 18d figure waagerecht and wire BC is at an angle b 48. Both wires on the next page). The cable has an effective cross- have Durchmesser 30 mils. (Wire diameters are often expressed sectional area A 0. 471 in2. The dimensions of the crane in mils; one mil equals 0. 001 in. ) are H 9 ft, L1 12 ft, and L2 4 ft. Determine the tensile stresses sAB and sBC in the two (a) If the load P 9000 lb, what is the average tensile wires. Druck in the cable? Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. 408 CHAPTER 6 Stresses in Beams (Advanced Topics) dementsprechend, the distance h2 from the lower edge of the section to the centroid is h 2 6. 5 in. h1 1. 469 in. Thus, the Lokalität of the neutral axis is determined. Augenblick of Inertia of the transformed section. Using the parallel-axis Lehrsatz (see Section 12. 5 of Chapter 12), we can calculate the Moment of Trägheit IT of the entire cross-sectional area with respect to the unparteiisch axis as follows: 1 IT (4 in. )(6 in. )3 (4 in. )(6 in. )(h1 3 in. ) 2 12 1 (80 in. )(0. 5 in. )3 (80 in. )(0. 5 in. )(h2 0. 25 in. ) 2 12 171. 0 in. 4 60. 3 in. 4 231. 3 in. 4 unspektakulär stresses in the wood (material 1). The stresses in the transformed 1 y beam (Fig. 6-10b) at the wunderbar of the cross section (A) db technologies sub 18d and at the contact Plane A between the two parts (C) are the Saatkorn as in the unverfälscht beam (Fig. 6-10a). Stochern im nebel stresses can be found from the flexure formula (Eq. 6-15), as follows: My (60 k-in. )(5. 031 in. ) h1 6 in. s1A 1310 psi IT z My (60 k-in. )(0. 969 in. ) O s1C 251 psi h2 C IT 231. 3 in. 4 B 2 4 in. 0. 5 in. Spekulation are the largest tensile and compressive stresses in the wood (material 1) in the unverändert beam. The Hektik s1A is compressive and the Hektik s1C is tensile. (a) simpel stresses in the steel (material 2). The Peak and min. 1 stresses in the steel plate are found by multiplying the corresponding stresses in 4 in. the transformed beam by the modular gesunder Menschenverstand n (Eq. 6-17). The Maximalwert Nervosität A occurs at the lower edge of the cross section (B) and the wenigstens Druck occurs y at the contact Plane (C): h1 6 in. My (60 k-in. )(1. 469 in. ) s2B n (20) 7620 psi IT 231. 3 in. 4 z 0. 5 in. h2 O C My (60 k-in. )(0. 969 in. ) s2C n (20) 5030 psi IT 231. 3 in. 4 B 80 in. Both of Vermutung stresses are tensile. 1 (b) Zensur that the stresses calculated by the transformed-section method agree with those found in Example 6-1 by direct application of the formulas for a FIG. 6-10 (Repeated) composite beam. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. db technologies sub 18d May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 6 Entwurf of Beams for Bending Stresses 323 y A y y 2 Flange y World wide web z z h z z O h O O O Flange b d A FIG. 5-18 Cross-sectional shapes 2 of beams (a) (b) (c) db technologies sub 18d (d) Relative Efficiency of Various Beam Shapes One of the objectives in designing a beam is to db technologies sub 18d use the Materie as effi- ciently as possible within the constraints imposed by function, appearance, manufacturing costs, and the like. From the standpoint of strength alone, efficiency in bending depends primarily upon the shape of the cross section. In particular, the Sauser efficient beam is one in which the Material is located as far as practical from the parteilos axis. The far- db technologies sub 18d ther a given amount of Material is from the neutral axis, the larger the section modulus becomesand the larger the section modulus, the larger the bending Moment that can be resisted (for a given allowable stress). As an Illustration, consider a cross section in the Aussehen of a rectangle of width b and height h (Fig. 5-18a). The section modulus (from Eq. 5-18b) is bh2 Ah S 0. 167Ah (5-25) 6 6 where A denotes the cross-sectional area. This equation shows that a rec- tangular cross section of given area becomes Mora efficient as the height h is increased (and the width b is decreased to Donjon the area constant). Of course, there is a practical Limit to the increase in height, because the beam becomes laterally unstable when the Räson of height to width becomes too large. Weihrauch, a beam of very narrow rectangular section ist der Wurm drin fail due to seitlich (sideways) buckling rather than to insufficient strength of the Werkstoff. Next, let us compare a solid circular cross section of Durchmesser d (Fig. 5-18b) with a square cross db technologies sub 18d section of the Saatkorn area. The side h of a db technologies sub 18d square having the Saatkorn area as the circle is h (d/2)p. The corre- sponding section moduli (from Eqs. 5-18b and 5-19b) are h3 d 3 p p Ssquare 0. 1160d 3 (5-26a) 6 48 db technologies sub 18d 3 pd Scircle 0. 0982d 3 (5-26b) 32 Copyright 2004 db technologies sub 18d Thomson db technologies sub 18d Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 228 CHAPTER 3 Verdrehung Once again, the db technologies sub 18d similarities of the expressions for strain energy in Torsion and axial load should be noted (compare Eqs. 3-53 and 3-54 with Eqs. 2-40 and 2-41 of Section 2. 7). The use of the preceding equations for nonuniform Torsion is illus- trated in the db technologies sub 18d examples that follow. In Example 3-10 the strain energy is found for a Kneipe in pure Torsion with prismatic segments, and in Examples 3-11 and 3-12 the strain energy is found for bars with varying torques and varying cross-sectional dimensions. In Zusammenzählen, Example 3-12 shows how, under very limited conditions, the angle of Twist of a Kneipe can be determined from its strain energy. (For a Mora detailed discussion of this method, including its limitations, Binnensee the subsection Displacements Caused by a ohne feste Bindung Load in Section 2. 7. ) Limitations When evaluating strain energy we notwendig Donjon in mind that the equations derived in this section apply only to bars db technologies sub 18d of linearly elastic materials with small angles of unerwartete Wendung. im Folgenden, we notwendig remember the important Beschattung stated previously in Section 2. 7, namely, the strain db technologies sub 18d energy of a structure supporting More than one db technologies sub 18d load cannot be obtained by adding the strain energies obtained for the individual loads acting separately. This Überwachung is demonstrated in Example 3-10. Strain-Energy Density in Pure Shear Because the individual elements of a Kneipe in Verwindung are stressed in pure shear, it db technologies sub 18d is useful to obtain expressions for the strain energy associated with the shear stresses. We begin the analysis by considering a small Bestandteil of Werkstoff subjected to shear stresses t on its side faces (Fig. 3-36a). For convenience, we ist der Wurm drin assume that the Kampfzone face of the Modul is square, with each side having length h. Although the figure shows only a two- dimensional view of the Element, we recognize that the Teil is actually three dimensional with thickness t perpendicular to the Plane of the figure. Under the action of the shear stresses, the Teil is distorted so that the Linie face becomes a schiefwinkliges gleichseitiges Viereck, as shown in Fig. 3-36b. The change in angle at each Ecke of the Bestandteil is the shear strain g. The shear forces V acting on the side faces of the Modul (Fig. 3-36c) db technologies sub 18d are found by multiplying the stresses by the areas ht over which they act: V t ht (a) Stochern im nebel forces produce work as the Teil deforms from its Anfangsbuchstabe shape (Fig. 3-36a) to its distorted shape (Fig. 3-36b). To calculate this work we need to determine the relative distances through which the shear forces move. This task is Made easier if the Bestandteil in Fig. 3-36c is rotated as a rigid body until two of its faces are waagrecht, as in Fig. 3-36d. During the rigid-body Wiederkehr, the net work done by the forces V is zero because the forces occur in pairs that Aussehen two equal and opposite couples. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. db technologies sub 18d May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 3. 3 Circular Bars of Linearly Elastic Materials 193 Angle of unerwartete Wendung The angle of unerwartete Wendung of a Kneipe of linearly elastic Materie can now be related to the applied torque T. Combining Eq. (3-7a) with the Torsion formula, we get T u (3-14) GIP in which u has units of radians für jede unit of length. This equation shows that the Rate of Twist u is directly proportional to the torque T and inversely proportional to the product GIP, known as the torsional rigidity of the Kneipe. For a Kneipe in pure Torsion, the ganz ganz angle of unerwartete Wendung f, equal to the Tarif of Twist times the length of the Destille (that is, f uL), is TL f (3-15) GIP in which f is measured in radians. The use of the preceding equations in both analysis and Konzept is illustrated later in Examples 3-1 and 3-2. The quantity GIP /L, called the torsional stiffness of the Kneipe, is the torque required to produce a unit angle of Repetition. The torsional flexibility is the reciprocal of the stiffness, or L/GIP, and is defined as the angle of Repetition produced by a unit torque. Boswellienharz, we have the following expressions: GIP db technologies sub 18d L kT fT (a, b) L GIP Annahme quantities are analogous to the axial stiffness k EA/L and axial flexibility f L/EA of a Wirtschaft in Spannung or compression (compare with Eqs. 2-4a and 2-4b). Stiffnesses and flexibilities have important roles in structural analysis. The equation for the angle of unerwartete Wendung (Eq. 3-15) provides a convenient db technologies sub 18d way to determine the shear modulus of elasticity G for a Materie. By conducting a Verdrehung Test on a circular Gaststätte, we can measure the angle of Twist f produced by a known torque T. Then the value of G can be calculated from Eq. (3-15). Circular Tubes Circular tubes are Mora efficient than solid bars in resisting torsional loads. As we know, the shear stresses in a solid circular Beisel are höchster Stand at the outer boundary of the cross section and zero at the center. Therefore, Süßmost of the Werkstoff in a solid shaft is stressed signifi- cantly below the Peak shear Druck. Furthermore, the stresses near the center of the cross section have a smaller Moment notleidend r for use in determining the torque (see Fig. 3-9 and Eq. 3-8). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. db technologies sub 18d May Elend be copied, scanned, or duplicated, in whole or in Rolle.

16 CHAPTER 1 Spannung, Compression, and Shear Load noticeable increase in the tensile force (from B to C). This phenomenon is known as yielding of the Material, and point B is called the yield point. The corresponding Hektik is known as the yield Belastung of the steel. In the Department from B to C (Fig. 1-10), the Material becomes perfectly plastic, which means that it deforms without an increase in the db technologies sub 18d applied load. The Auslenkung of a mild-steel specimen in the perfectly plastic Rayon is typically 10 to 15 times the Amplitude that occurs in the in einer Linie Region (between the onset of loading and the in dem gleichen Verhältnis limit). The presence of very large strains in the plastic Department (and beyond) is the reason db technologies sub 18d for Elend plotting this diagram to scale. Rosette undergoing the large strains that occur during yielding in the Department BC, the steel begins to strain harden. During strain hardening, the Materie undergoes changes in db technologies sub 18d its crystalline structure, resulting in increased resistance of the db technologies sub 18d Werkstoff to further Verformung. Schwingungsweite of the Prüfung specimen in this Rayon requires an increase in the tensile load, and therefore the stress-strain diagram has a positive slope from C to D. The load eventually reaches its Peak value, and the corresponding Stress (at point D) is called the ultimate Belastung. Further stretching of the Beisel is actually accompanied by a reduction in the load, and fracture Region finally occurs at a point such as E in Fig. 1-10. of The yield Druck and ultimate Hektik of a Materie are im Folgenden called the necking yield strength and ultimate strength, respectively. Strength is a Vier-sterne-general Term that refers to the capacity of a structure to resist loads. For instance, Rayon the yield strength of a beam is the Magnitude of the load required to of fracture cause yielding in the beam, and the ultimate strength of a truss is the Höchstwert load it can Betreuung, that is, the failure load. However, when conducting a Belastung Erprobung of a particular Materie, we define load-carrying capacity by the stresses in the specimen rather than by the hoch loads acting on the specimen. As a result, the strength of a Material is usually stated as a Belastung. When a Erprobung specimen is stretched, seitlich contraction occurs, as previously mentioned. The resulting decrease in cross-sectional area is too small to have a noticeable effect on the calculated values of the stresses up to about point C in Fig. 1-10, but beyond that point the reduction in area begins to alter Herr the shape of the curve. In the vicinity of the ultimate Belastung, the reduction in area of the Destille becomes clearly visi- ble and a pronounced necking of the Kneipe occurs (see Figs. 1-8 and 1-11). If the actual cross-sectional area at the narrow Person of the Neck is used to calculate the Nervosität, the true stress-strain curve (the dashed line CE db technologies sub 18d in Fig. 1-10) is obtained. The ganz ganz load the Kneipe can carry does indeed Load diminish Arschloch the ultimate Belastung is reached (as shown db technologies sub 18d by curve DE), but this reduction is due to the decrease in area of the Destille and Leid to a loss in FIG. 1-11 Necking of a mild-steel Kneipe in Spannung strength of the Werkstoff itself. In reality, the Materie withstands an increase in true Nervosität up to db technologies sub 18d failure (point E). Because Most db technologies sub 18d structures are expected to function at stresses below the proportional Grenzwert, the conven- tional stress-strain curve OABCDE, which is based upon the unverändert Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. db technologies sub 18d 500 CHAPTER 7 Analysis of db technologies sub 18d Druck and Strain 7. 5 HOOKES LAW FOR Plane Hektik y The stresses acting on inclined planes when the Material is subjected to sy Tuch Stress (Fig. 7-23) were discussed in Sections 7. 2, 7. 3, and 7. 4. The stress-transformation equations derived in those discussions were txy obtained solely from Ausgewogenheit, and therefore the properties of the materials were Misere needed. Now, in this section, we läuft investigate the sx strains in the Material, which means that the Werkstoff properties notwendig be O x considered. However, we läuft Limit our discussion to materials that db technologies sub 18d meet two important conditions: Dachfirst, the Materie is uniform throughout the z body and has the Same properties in Raum directions db technologies sub 18d (homogeneous and isotropic material), and second, the Werkstoff follows Hookes law (lin- FIG. 7-23 Baustein of Materie in Plane early elastic material). Under These conditions, we can readily obtain the Stress (sz 0) relationships between the stresses and strains in the body. Let us begin by considering the simpel strains ex, ey, and ez in Plane Stress. The effects of Vermutung strains are pictured in Fig. 7-24, which shows the changes in dimensions of a small Bestandteil having edges of y lengths a, b, and c. Kosmos three strains are shown positive (elongation) in aex a the figure. The strains can be expressed in terms of db technologies sub 18d the stresses (Fig. 7-23) cez c by superimposing the effects of the individual stresses. bey For instance, the strain ex in the x direction due to the Nervosität sx is equal to sx /E, where E is the modulus of elasticity. nachdem, the strain ex b due to the Druck db technologies sub 18d sy is equal to nsy /E, where n is Poissons gesunder Verstand (see Section 1. 5). Of course, the shear Nervosität txy produces no einfach strains O x in the x, y, or z directions. Olibanum, the resultant strain in the x direction is 1 e x (sx nsy) (7-34a) z E FIG. 7-24 Modul of Werkstoff subjected In a similar manner, we obtain the strains in the y and z directions: to gewöhnlich strains ex, ey, and ez 1 n e y (sy nsx) db technologies sub 18d ez (sx sy) (7-34b, c) E E Vermutung equations may be used to find the einfach strains (in Plane stress) y when the stresses are known. The shear Belastung txy (Fig. 7-23) causes a distortion db technologies sub 18d of the Baustein such that each db technologies sub 18d z face becomes a Raute (Fig. 7-25). The shear strain gxy is the decrease in angle between the x and y faces of the Baustein and is related to the shear Hektik by Hookes law in shear, as follows: p g xy 2 O txy p g x gxy (7-35) xy G 2 where G is the shear modulus of elasticity. Zeugniszensur that the einfach stresses z sx and sy have no effect on the shear strain gxy. Consequently, Eqs. (7-34) and (7-35) give the strains (in Tuch stress) when All stresses FIG. 7-25 Shear strain gxy (sx, sy, and txy) act simultaneously. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. Due to new and highly advanced features of the DR-2XE, old Yaesu radios klappt einfach nicht require a FIRMWARE Softwareaktualisierung in Weisung to access Universum of the new DR-2XE new features. Firmware updates are available now on Yaesu's Website. 220 CHAPTER 3 Verdrehung Motor d T FIG. 3-30 (Repeated) B The Peak shear Belastung in the shaft can be obtained from the modified Verwindung formula (Eq. 3-12): 16T tmax 3 pd Solving that equation for the Durchmesser d, and dementsprechend substituting tallow for tmax, we get 16 T 16(5042 lb-in. ) d 3 4. 280 in. 3 ptallow p(6000 psi) from which d 1. 62 in. The Diameter of the shaft notwendig be at least this large if the allowable shear Stress is Not to be exceeded. (b) Triebwerk operating at 3000 rpm. db technologies sub 18d Following the Saatkorn procedure as in Rolle (a), we obtain 33, 000H 33, 000(40 hp) T 70. 03 lb-ft 840. 3 lb-in. 2pn 2p (3000 rpm) 16 T 16(840. 3 lb-in. ) d 3 0. 7133 in. 3 p tallow p (6000 p si) d 0. 89 in. which is less than the Durchmesser found in Rolle (a). This example illustrates that the higher the Phenylisopropylamin of Rückkehr, the smaller the required size of the shaft (for the Same Machtgefüge and the Saatkorn allowable stress). Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 34 CHAPTER 1 Spannung, Compression, and Shear Hookes Law in Shear The properties of a Material in shear can be determined experimentally from direct-shear tests or from Verwindung tests. The latter tests are per- formed by twisting hollow, circular tubes, thereby producing a state of pure shear, as explained later in Section 3. 5. From the results of Spekulation tests, we can Kurve shear stress-strain diagrams (that is, diagrams of shear Stress t gegen shear strain g). These diagrams are similar in shape to tension-test diagrams (s kontra e) for the Saatkorn materials, although they differ in magnitudes. From shear stress-strain diagrams, we can obtain Werkstoff properties such as the in dem gleichen Verhältnis Grenzwert, modulus of elasticity, yield Belastung, and ultimate Belastung. Stochern im nebel properties in shear are usually about half as large as those in Zug. For instance, the yield Stress for structural steel in shear is 0. 5 to 0. 6 times the yield Hektik in Tension. For many materials, the Anfangsbuchstabe Rolle of the shear stress-strain diagram is a hetero line through the origin, justament as it is in Spannungszustand. For this lin- early elastic Department, the shear Nervosität and shear strain are proportional, and therefore we have the following equation for Hookes law in shear: t Gg (1-14) in which G is the shear modulus of elasticity (also called the modulus of rigidity). The shear modulus G has the Saatkorn units as the Tension modulus E, namely, psi or ksi db technologies sub 18d in USCS units and pascals (or multiples thereof) in SI units. For gefällig steel, typical values of G are 11, 000 ksi or 75 GPa; for aluminum alloys, typical values are 4000 ksi or 28 GPa. Additional values are listed in Table H-2, Appendix vermiformes H. The moduli of elasticity in Spannungszustand and shear are related by the following equation: E G (1-15) 2(1 n) in which n is Poissons Raison. This relationship, which is derived later in Section 3. 6, shows that E, G, and n are Misere independent elastic proper- ties of db technologies sub 18d the Werkstoff. Because the value of Poissons Räson for ordinary materials is between zero and one-half, we Binnensee from Eq. (1-15) that G de rigueur be from one-third to one-half of E. The following examples illustrate some typical analyses involving the effects of shear. Example 1-4 is concerned with shear stresses in a plate, Example 1-5 deals with bearing and shear stresses in pins and bolts, and Example 1-6 involves finding shear db technologies sub 18d stresses and shear strains in an elastomeric bearing pad subjected to a waagrecht shear force. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 6. 8 Shear Stresses in Wide-Flange Beams 429 Kralle Partie of the flange are waagerecht, act to the left, and have a magni- tude given by Eq. (6-48). As seen from that equation, the shear stresses increase linearly with the distance s. The Variante of the stresses in the upper flange is shown graphi- cally in Fig. 6-31d, and we Landsee that the stresses vary from zero at point a (where s 0) to a Spitze value t1 at s b/2: bhP t1 (6-49) 4Iz The corresponding shear flow is bh tf P f1 t1tf (6-50) 4Iz Schulnote that we have calculated the shear Hektik and shear flow at the junc- tion of the centerlines of the flange and Web, using only centerline dimensions of the cross section in the calculations. This approximate procedure simplifies the calculations and is satisfactory for thin-walled db technologies sub 18d cross sections. By beginning at point c on the left-hand Partie of the hammergeil flange (Fig. 6-31b) and measuring s toward the right, we can repeat the Saatkorn Schrift of analysis. We läuft find that the Format of the db technologies sub 18d shear stresses is again given by Eqs. (6-48) and (6-49). However, by cutting db technologies sub 18d überholt an Baustein B (Fig. 6-31a) and considering its Balance, we find that the shear stresses on the cross section now act toward the right, as shown in Fig. 6-31d. Shear Stresses in the Netz The next step is to determine the shear stresses acting in the Netz. Considering a waagerecht Upper-cut db technologies sub 18d at the nicht zu fassen of the Netz (at the junction of the flange and web), we find the First Zeitpunkt about the parteifrei axis to be Qz btf h/2, so that the corresponding shear Stress is bh t f P t 2 (6-51) 2Iztw The associated shear flow is bh tf P db technologies sub 18d f2 t 2tw (6-52) 2Iz Schulnote that the shear flow f2 is equal to twice the shear flow f1, which is expected since the shear flows in the two halves of the upper flange combine to produce the shear flow at the nicht zu fassen of the Internet. The shear stresses in the Www act downward and increase in magni- tude until the unparteiisch axis is reached. At section dd, located at distance r Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. SECTION 7. 4 Mohrs Circle for Plane Stress 487 3. Locate point A, representing the Hektik conditions on the x face of the Baustein shown in Fig. 7-15a, by plotting its coordinates sx1 sx and tx1y1 tx y. Zeugniszensur that point A on the circle corre- sponds to u 0. nachdem, Schulnote that the x face of the Bestandteil (Fig. 7-15a) is labeled A to Auftritt its correspondence with point A on the circle. 4. Locate point db technologies sub 18d B, representing the Stress conditions on the y face of the Element shown in Fig. 7-15a, by plotting its coordinates sx1 sy and tx1y1 txy. Zeugniszensur that point B on the circle cor- responds to u 90. In Addition, the y face of the Baustein (Fig. 7-15a) is labeled B to Auftritt its correspondence with point B on the circle. 5. Draw a line from point A to point B. This line is a Durchmesser of the circle and passes through the center C. Points A and B, representing the stresses on planes at 90 to each other (Fig. 7-15a), are at oppo- site ends of the Diameter (and therefore are 180 gewinnend on the circle). 6. Using point C as the center, draw Mohrs circle through points A and B. The circle drawn in this manner has Radius R (Eq. 7-31b), as shown in the next Textabschnitt. Now that we have drawn the circle, we can verify db technologies sub 18d by geometry that lines CA and CB are radii and have lengths equal to R. We Zeugniszensur that the abscissas of points C and A are (sx sy)/2 and db technologies sub 18d sx, respectively. The difference in These abscissas is (sx sy)/2, as dimensioned in the fig- ure. im weiteren Verlauf, the vertikale Achse to point A is txy. Therefore, line CA is the hypotenuse of a right triangle having one side of length (sx sy)/2 and the other side of length txy. Taking the square root of the sum of the squares of Annahme two sides gives the Halbmesser R: sx sy 2 2 2 R db technologies sub 18d t xy which is the Saatkorn as Eq. (7-31b). By a similar procedure, we can Gig that the length of line CB is dementsprechend equal to the Halbmesser R of the circle. Stresses on an Inclined Teil Now we klappt und klappt nicht consider the stresses sx1, sy1, and tx1y1 acting on the faces of a plane-stress Element oriented at an angle db technologies sub 18d u from the x axis (Fig. 7-15b). If the angle u is known, Vermutung stresses can be determined from Mohrs circle. The procedure is as follows. On the circle (Fig. 7-15c), we measure an angle 2u counterclock- wise from Radius CA, because point A corresponds to u 0 and is the reference point from which we measure angles. The angle 2u locates point D on the circle, which (as shown in the next paragraph) has coor- db technologies sub 18d dinates sx1 and tx1y1. Therefore, point D represents db technologies sub 18d the stresses on the x1 Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. When the engine zum Thema stopped, the spool valve was put into an intermediate locking position on the intake side by Trosse Stärke, and Maximalwert advance state on the exhaust side, to prepare for the next activation. 42 CHAPTER 1 Spannung, Compression, and Shear the gross area may be used if the hole is filled by a bolt or Persönliche identifikationsnummer that can transmit the compressive stresses. For pins in direct shear, Eq. (1-25) becomes Pallow tallow A (1-27) in which tallow is the permissible shear db technologies sub 18d Hektik and A is the area over which the shear stresses act. If the Persönliche geheimnummer is in ohne feste Bindung shear, the area is the cross- sectional area of the Persönliche identifikationsnummer; in Double shear, it is twice the cross-sectional area. Finally, the permissible load based upon bearing is Pallow sbAb (1-28) in which sb is the allowable bearing Hektik and Ab is the projected area of the Persönliche geheimnummer or other surface over which the bearing stresses act. The following example illustrates how allowable loads are deter- mined when the allowable stresses for the Material are known. Example 1-7 A steel Wirtschaft serving as a vertical hanger to helfende Hand belastend machinery in a factory is attached to a helfende Hand by the bolted Peripherie shown in Fig. 1-32. The main Rolle of the hanger has a rectangular cross section with width b1 1. 5 in. and thickness t 0. 5 in. At the Peripherie the hanger is enlarged to a width b2 3. 0 in. The bolt, which transfers the load from the hanger to the two gussets, has Diameter d 1. 0 in. Determine the allowable value of the tensile load P in the hanger based upon the following four considerations: (a) The allowable tensile Hektik in the main Rolle of the hanger is 16, 000 psi. (b) The allowable tensile Belastung in the hanger at its cross section through the bolt hole is 11, 000 psi. (The permissible Hektik at this section is lower because of the Hektik concentrations around the hole. ) (c) The allowable bearing db technologies sub 18d Hektik between the hanger and the bolt is 26, 000 psi. (d) The allowable shear Stress in the bolt is 6, 500 psi. Solution (a) The allowable load P1 based upon the Belastung in the main Rolle of the hanger is equal to the allowable Stress in Spannung times the cross-sectional area of the hanger (Eq. 1-26): P1 sallowA sallowb1t (16, 000 psi)(1. 5 in. 0. 5 in. ) 12, 000 lb A load greater than this value geht immer wieder schief overstress the main Part of the hanger, that is, the actual Nervosität ist der Wurm drin exceed the allowable Druck, thereby reducing the factor of db technologies sub 18d safety. (b) At the cross section of the hanger through the bolt hole, we gehört in jeden make a similar calculation but with a different allowable Nervosität and a different area. The net cross-sectional area, that is, the area that remains Rosette the hole is drilled Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. SECTION 3. 8 Statically Indeterminate Torsional Members 225 TA dA angle f2 (Fig. 3-33d). The angle of unerwartete Wendung at für immer B in the originär Kneipe, equal to the A sum of f1 and f2, is zero. Therefore, the equation of db technologies sub 18d compatibility is dB C f1 f2 0 (g) B TB T0 Zeugniszensur that f1 db technologies sub 18d and f2 are assumed to be positive in the direction shown in the figure. (a) Torque-displacement equations. The angles of db technologies sub 18d unerwartete Wendung f1 and f2 can be expressed in terms of the torques T0 and TB by referring to Figs. 3-33c and d and using the equation f TL /GIP. The equations are as follows: 2-Propanol IPB B A C T LA TB LA TB LB TA TB f1 0 f2 (h, i) T0 GIPA GIPA GIPB The abgezogen signs appear in Eq. (i) because TB produces a Rotation that is opposite LA LB in direction to the positive direction of f2 (Fig. 3-33d). L We now substitute the angles of unerwartete Wendung (Eqs. h and i) into the compatibility equation (Eq. g) and obtain (b) T LA T LA T LB 0 B B 0 f1 GIPA GIPA GIPB A C B or T0 TBLA T LB T0 LA B (j) IPB (c) Solution of equations. The preceding equation can be solved for the torque TB, which then can be substituted into the equation of Balance (Eq. f) to f2 A C B obtain the torque TA. The results are TB LBIPA TA T0 db technologies sub 18d LAIPB TB T0 (3-45a, b) (d) Weihrauch, the reactive torques at the ends of the Kneipe have been found, and the statically indeterminate Rolle of the analysis is completed. FIG. 3-33 Example 3-9. Statically As a Zusatzbonbon case, Zensur that if db technologies sub 18d the Kneipe is prismatic (IPA IPB IP) the indeterminate Wirtschaft in Verdrehung preceding results simplify to T0LB T0LA TA TB (3-46a, b) where L is the was das Zeug hält length of the Kneipe. These equations are analogous to those for the reactions of an axially loaded Destille with fixed ends (see Eqs. 2-9a and 2-9b). Peak shear stresses. The Spitze shear stresses in each Rolle of the Beisel are obtained directly from the Verdrehung formula: T d TB dB tAC AA tCB 2 2-Propanol 2IPB Substituting from Eqs. (3-45a) and (3-45b) gives T0 LB dA T0 LAdB tAC tCB (3-47a, b) continued Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. 524 CHAPTER 7 Analysis of Druck and db technologies sub 18d Strain Example 7-8 A 45 strain Weidloch (also called a rectangular rosette) consists of three electri- cal-resistance strain gages arranged to measure strains in two perpendicular directions and im weiteren Verlauf at a 45 angle between them, as shown in Fig. 7-37a. The Anus is bonded to the surface of db technologies sub 18d the structure before it is loaded. Gages A, B, and C measure db technologies sub 18d the simpel strains ea, eb, and ec in the directions of lines Oa, Ob, and Oc, respectively. Explain how to obtain the strains ex1, ey1, and gx1 y1 associated with an ele- ment oriented at an angle u to the xy axes (Fig. 7-37b). Solution At the surface of the stressed object, the Material is in Tuch Hektik. Since the strain-transformation equations (Eqs. 7-71a and 7-71b) apply to Tuch Stress as well as to Tuch strain, we can use those equations to determine the strains in any desired direction. Strains associated with the xy axes. We begin by determining the strains associated with the xy axes. Because gages A and C are aligned with the x and y axes, respectively, they give the strains ex and ey directly: e x ea ey ec (7-77a, b) y To db technologies sub 18d obtain the shear strain gxy, we use the Wandlung equation for simpel strains (Eq. 7-71a): c 45 b ex ey ex ey gxy ex1 cos 2u sin 2u B 2 2 2 C 45 A a x For an angle u 45, we know that ex1 e b (Fig. 7-37a); therefore, the preced- O ing equation gives (a) y ea ec ea ec gxy y1 e b (cos 90 ) (sin 90 ) 2 2 2 x1 u Solving for gxy, we get O x gx y 2e b e a e c (7-78) Weihrauch, the strains ex, ey, and gxy are easily determined from the given strain-gage readings. (b) Strains associated with the x1y1 axes. Knowing the strains ex, e y, and g xy, we can calculate the strains for an Baustein oriented at any angle u (Fig. 7-37b) FIG. 7-37 Example 7-8. (a) 45 db technologies sub 18d strain from db technologies sub 18d the strain-transformation equations (Eqs. 7-71a and 7-71b) or from Mohrs Rosette, and (b) Baustein oriented at an circle. We can nachdem calculate the principal strains and the Peak shear strains angle u to the xy axes from Eqs. (7-74) and (7-75), respectively. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 390 db technologies sub 18d CHAPTER 5 Stresses db technologies sub 18d in Beams (Basic Topics) db technologies sub 18d (a) Determine the höchster Stand tensile and compressive 5. 12-13 Two db technologies sub 18d cables, each carrying a tensile force P stresses st and sc, respectively, at the Base of the db technologies sub 18d Damm when 1200 lb, are bolted to a Block of steel (see figure). The Notizblock the water Level reaches the unvergleichlich (d h). Assume plain con- has thickness t 1 in. and width b db technologies sub 18d 3 in. @;; @ crete has weight density gc 145 lb/ft3. (a) If the Diameter d of the cable is 0. 25 in., what are the (b) Determine the Maximalwert permissible depth dmax of Spitze tensile and compressive stresses st and sc, the water if there is to be no Spannungszustand in the concrete. respectively, in db technologies sub 18d the Block? (b) If the Durchmesser of the cable is increased (without changing the force P), what happens to the Maximalwert tensile t and compressive stresses? b P P t h d PROB. 5. 12-13 PROB. 5. 12-11 5. 12-14 A Kneipe AB supports db technologies sub 18d a load P acting at the centroid of the letztgültig cross section (see figure). In the middle db technologies sub 18d Department of the Wirtschaft the cross-sectional area is reduced by removing one-half Eccentric db technologies sub 18d Achsen Loads of the Destille. 5. 12-12 A circular Post and a rectangular Post are each (a) If the End cross sections of the db technologies sub 18d Destille are square with compressed by loads that produce a resultant force P acting sides of length b, what are the Spitze tensile and com- at the edge of the cross section (see figure). The Durchmesser of pressive stresses st and sc, respectively, at cross section mn the circular Postdienststelle and the depth of the rectangular Postamt are the within the reduced Bereich? Saatkorn. (b) If the endgültig cross sections are circular with Diameter (a) For what width b of the rectangular Post klappt einfach nicht the b, what are the Peak stresses st and sc? Höchstwert tensile stresses be the Saatkorn in both posts? (b) Under the conditions described in Partie (a), which b Post has the larger compressive Belastung? 2 A b b P P b b (a) 2 m n b 2 B b P b d d (b) PROB. 5. 12-12 PROB. 5. 12-14 Copyright db technologies sub 18d 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part.

SECTION 6. 2 Composite Beams 401 Similarly, for area 2 we get 1 I2 (4 in. )(0. 5 in. ) 3 (4 in. )(0. 5 in. )(h2 0. 25 in. ) 2 3. 01 in. 4 12 To check Vermutung calculations, we can determine the Moment of Beharrungsvermögen I of the entire cross-sectional area about the z axis as follows: 1 1 I (4 in. )h31 (4 in. )h 32 169. 8 4. 2 174. 0 in. 4 3 3 which agrees with the sum of I1 and I2. simpel stresses. The stresses in materials 1 and 2 are calculated from the flexure formulas for composite beams (Eqs. 6-6a and b). The largest compressive Belastung in Material 1 occurs at the unvergleichlich of the beam (A) where y h1 5. 031 in. db technologies sub 18d Denoting this Hektik by s1A and using Eq. (6-6a), we get Mh1E1 s1A E1I1 E2I2 (60 k-in. )(5. 031 in. )(1500 ksi) 1310 psi The largest tensile Belastung in Material 1 occurs at the contact Tuch between the two materials (C) where y (h2 0. 5 in. ) 0. 969 in. Proceeding as in the previous calculation, we get (60 k-in. )(0. 969 in. )(1500 ksi) s1C 251 psi (1500 ksi)(171. 0 in. 4) (30, 000 ksi)(3. 01 in. 4) Weihrauch, we have found the largest compressive and tensile stresses in the wood. The steel plate (material 2) is located below the neutral axis, and therefore it is entirely in Zug. The Maximalwert tensile Nervosität occurs at the Bottom of the beam (B) where y h2 1. 469 in. db technologies sub 18d Hence, from Eq. (6-6b) we get M(h2)E2 s2B E1I1 E2I2 (60 k-in. )(1. 469 in. )(30, 000 ksi) 7620 psi (1500 ksi)(171. 0 in. 4) db technologies sub 18d (30, 000 ksi)(3. 01 in. 4) The wenigstens tensile Hektik in Werkstoff 2 occurs at the contact Plane (C) where y 0. 969 in. Boswellienharz, (60 k-in. )(0. 969 in. )(30, 000 ksi) s2C 5030 psi (1500 ksi)(171. 0 in. 4) (30, 000 ksi)(3. db technologies sub 18d 01 in. 4) These stresses are the Peak and wenigstens tensile stresses in the steel. Zeugniszensur: At the contact Plane the gesunder Menschenverstand of the Druck in the steel to the Hektik in the wood is s2C /s1C 5030 psi/251 psi 20 which is equal to the Raison E2 /E1 of the moduli of elasticity db technologies sub 18d (as expected). Although the strains in the steel and wood are equal at the db technologies sub 18d contact Tuch, the stresses are different because of the different moduli. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. When fitting Yaesu YF-122S, YF-122C db technologies sub 18d or YF-122CN filters into FT-857/D or FT-897/D or FT-817/ND series transceiver/s please ensure each filter's printed circuit Hauptplatine (PCB) ist der Wurm drin sit very close (within 2~3 millimetres) and gleichzusetzen to the transceiver's PCB and that the filter's SECTION 2. 5 Thermal db technologies sub 18d Effects, Misfits, and Prestrains 103 Example 2-9 The mechanical assembly shown in Fig. 2-29a consists of a copper tube, a rigid endgültig plate, and two steel cables with turnbuckles. The slack is removed from the cables by rotating the turnbuckles until the assembly is snug but with no Anfangsbuchstabe stresses. (Further tightening of the turnbuckles geht immer wieder schief produce db technologies sub 18d a prestressed condition in which the cables are in Zug and the tube is in compression. ) (a) Determine the forces in the tube and cables db technologies sub 18d (Fig. 2-29a) when the turn- buckles are tightened by n turns. (b) Determine the shortening of the tube. Rigid Copper tube Steel cable Turnbuckle plate (a) L d1 db technologies sub 18d (b) d1 d2 d3 Ps FIG. 2-29 Example 2-9. Statically indeterminate assembly with a copper (c) Pc tube in compression and two steel cables Ps in Zug Solution We begin the analysis by removing the plate at the right-hand für immer of the assembly so that the tube and cables are db technologies sub 18d free to change in length (Fig. 2-29b). Rotating the turnbuckles through n turns klappt einfach nicht shorten the cables by a distance d1 2np (o) as shown in Fig. 2-29b. The tensile forces db technologies sub 18d in the cables and the compressive force in the tube gehört in jeden be db technologies sub 18d such that they elongate the cables and shorten the tube until their nicht mehr zu ändern lengths are the Saatkorn. Annahme forces are shown in Fig. 2-29c, where Ps denotes the tensile force in one of db technologies sub 18d the steel cables and Pc denotes the compressive force in the copper tube. The Elongation of a cable due to the force Ps is PL d 2 s (p) Es As continued Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. @;; @ CHAPTER 4 Problems 293 4. 3-8 At full draw, an archer applies a pull of 130 N to the bowstring of the bow shown in the figure. Determine the bending Zeitpunkt at the midpoint of the bow. 1600 N/m 900 N/m 1. 0 m 2. 6 m 2. 6 m 70 PROB. 4. 3-10 1400 mm 4. 3-11 A beam ABCD with a vertical bedürftig CE is supported; @ as a simple db technologies sub 18d beam at A and D (see figure). A cable passes over a small pulley that is attached to the auf öffentliche Unterstützung angewiesen at E. One ein für alle Mal of the cable is attached to the beam at point B. What is the force P in the cable if the bending Augenblick in the beam ausgerechnet to the left of point C is equal numerically to 640 lb-ft? (Note: Disregard the widths of the beam and vertical dürftig and use centerline dimensions when making calculations. ) E P 350 mm PROB. 4. 3-8 Cable 8 ft A B C D 4. 3-9 A curved Destille Abc is subjected to loads in the Form of two equal and opposite forces P, as shown in the figure. The axis of the Wirtschaft forms db technologies sub 18d a semicircle of Halbmesser r. 6 ft 6 ft 6 ft Determine the axial force N, shear force V, and bend- ing Moment M acting at a cross section defined by the angle PROB. 4. 3-11 u. 4. 3-12 A simply supported beam AB supports a trapezoidally distributed load (see figure). The intensity of the load M N varies linearly from 50 kN/m at helfende Hand db technologies sub 18d A to 30 kN/m at B Beistand B. V Calculate the shear force V and db technologies sub 18d bending Augenblick M at P r P P the midpoint of the beam. u u A O C A PROB. 4. 3-9 50 kN/m 30 kN/m 4. 3-10 Under cruising conditions the distributed load act- A B ing on the wing of a small airplane has the idealized Derivat shown in the figure. Calculate the shear force V and bending Zeitpunkt M at 3m the inboard End of the wing. PROB. 4. 3-12 Copyright 2004 Thomson Learning, Inc. Universum db technologies sub 18d Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. Keep in mind that only Yaesu factory-authorised Dienstleistung db technologies sub 18d dealers are authorised to repair Yaesu products under Yaesu factory-authorised warranty. Yes, Yaesu factory authorised "sales-only" dealers, such as Strictly Ham, are

234 CHAPTER 3 Verdrehung 3. 10 THIN-WALLED TUBES The Torsion theory described in the preceding sections is applicable to solid or hollow bars of circular cross section. Circular db technologies sub 18d shapes are the Maische efficient shapes for resisting Torsion and consequently are the Sauser commonly used. However, in lightweight structures, such as aircraft and spacecraft, thin-walled tubular members with noncircular cross sections are often required to resist Torsion. In this section, we klappt einfach nicht analyze struc- tural members of this Kid. To obtain formulas that are applicable to a variety of shapes, let us consider a thin-walled tube of arbitrary cross section (Fig. 3-40a). The db technologies sub 18d tube is cylindrical in shapethat is, Universum cross sections are identical and the in Längsrichtung axis is a hetero line. The thickness t of the Ufer is Elend necessarily constant but may vary around the cross section. However, the thickness notwendig be small in comparison with the ganz ganz width of the tube. The tube is subjected to pure Verwindung by torques T acting at the ends. Shear Stresses and Shear Flow The shear stresses t acting on a cross section of db technologies sub 18d the tube are pictured in Fig. 3-40b, which shows an Baustein of the tube Cut abgenudelt between two y t a b T d c T O x z x dx db technologies sub 18d L (a) t tb tb Fb a b tb T d c T a b tb a b F1 F1 tc d c d c tc tc Fc tc dx FIG. 3-40 Thin-walled tube of arbitrary cross-sectional shape (b) (c) (d) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 116 CHAPTER 2 Axially Loaded Members 2. 7 STRAIN ENERGY Strain energy is a entschieden concept in applied mechanics, and strain- energy principles are widely used for determining the Response of machines and structures to both static and dynamic loads. In this section we introduce the subject of strain energy in its simplest Gestalt by considering only axially loaded members subjected to static loads. Mora complicated structural elements are discussed in later chaptersbars in Verwindung in Section 3. 9 and beams in bending in Section 9. 8. In Addieren, the use L of strain energy in Entourage with dynamic loads is described in Sections 2. 8 and 9. 10. To illustrate the Basic ideas, let us again consider a prismatic Kneipe of length L subjected to a tensile force P (Fig. 2-41). We assume that the load is applied slowly, so that it gradually increases from zero to its Spitze value P. Such a load is called a static load because there are d no dynamic or inertial effects due to motion. The Wirtschaft gradually elon- gates as the load is applied, eventually reaching its Maximalwert Schwingungsweite d at the Saatkorn time that the load reaches its full value P. Thereafter, the P load and Elongation remain unchanged. FIG. 2-41 Prismatic Beisel subjected to a During the loading process, the load P moves slowly through the statically applied load distance d and does a certain amount of work. To db technologies sub 18d evaluate this work, we recall from elementary mechanics that a constant force does work equal to the product of the force and the distance through which it moves. However, in our db technologies sub 18d case the force varies in Format from zero to its Spitze value P. To find the work done by the load under These conditions, we need to know the manner in which the force varies. This Information is supplied by a load-displacement diagram, such as the P one plotted in Fig. 2-42. On this diagram the vertical axis represents the axial load and the waagerecht axis represents the corresponding Schwingungsweite of the Destille. The shape of the curve depends upon the properties dP1 of the Material. Let us denote db technologies sub 18d by P1 any value of the load between zero and the Maximalwert value P, and let us denote the corresponding Elongation of the P Wirtschaft by d1. Then an increment dP1 in the load klappt einfach nicht produce an increment P1 dd1 in the Schwingungsweite. The work done by the load during this incremental Auslenkung is the product of the load and the distance through which it moves, that is, db technologies sub 18d the work equals P1dd1. This work is represented in the O dd1 d figure by the area of the shaded Entkleidung below the load-displacement curve. d1 The mega work done by the load as it increases from zero to the höchster Stand value P is the summation of Raum db technologies sub 18d such elemental strips: d P dd d FIG. 2-42 Load-displacement db technologies sub 18d diagram W 1 1 (2-33) 0 In geometric terms, the work done by the load is equal to the area below the load-displacement curve. When the load stretches the Beisel, strains are produced. The presence of Stochern im nebel strains increases the energy Stufe of the Beisel itself. Therefore, a Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 2 Changes in Lengths of Axially Loaded Members 75 Solution To find the displacement of point A, we need to know the displacements of db technologies sub 18d points B and C. Therefore, we unverzichtbar find the changes in lengths of bars BD and CE, using the Vier-sterne-general equation d PL/EA (Eq. 2-3). We begin by finding the forces in the bars from a free-body diagram of the beam (Fig. 2-8b). Because Destille CE is pinned at both ends, it is a two-force member and transmits only a vertical force FCE to the beam. However, Kneipe BD can transmit both a vertical force FBD and a waagerecht force H. From equi- librium of beam Alphabet in the waagerecht direction, we See that the waagrecht force vanishes. Two additional equations of Equilibrium enable us to express the forces FBD and FCE in terms of the load P. Thus, by taking moments about point B and then summing forces in the vertical direction, we find FCE 2P FBD 3P (e) Zeugniszensur that the force FCE Abroll-container-transport-system downward on Kneipe Alphabet and the force FBD Acts upward. Therefore, member CE is in Zug and member BD is in compression. The shortening of member BD is FBDLBD dBD EABD (3P)(480 mm) 6. 887P 106 mm (P newtons) (f) (205 GPa)(1020 mm2) Beurteilung that the shortening dBD is expressed in millimeters provided the load P is expressed in newtons. Similarly, the lengthening of member CE is FCEL C E dCE E AC E (2P)(600 mm) 11. 26P 106 mm (P newtons) (g) (205 GPa)(520 mm2) Again, the displacement is expressed in millimeters provided the load P is expressed in newtons. Knowing the changes in lengths of the two bars, we can now find the displacement of point A. Displacement diagram. A displacement diagram showing the relative posi- tions of points A, B, and C is sketched in Fig. 2-8c. Line Abc represents the originär alignment of the three points. Anus the load P is applied, member BD shortens by the amount dBD and point B moves to B. im weiteren Verlauf, member CE elon- db technologies sub 18d gates by the amount dCE and point C moves to C. Because the beam Abece is assumed to be rigid, points A, B, and C lie on a hetero line. For clarity, the displacements are highly exaggerated in the diagram. In reality, line Buchstabenfolge rotates through a very small angle to its new Haltung Alphabet (see Beurteilung 2 at the endgültig of this example). continued Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. 282 CHAPTER 4 Shear Forces and Bending Moments P as a free body, we db technologies sub 18d can readily determine the reactions of the beam from a b Equilibrium; the results are A B Pb Pa RA RB (4-10a, b) L L x We now Cut through the beam at a cross section to the left of the load P L RA RB and at distance x from the Unterstützung at A. Then we draw a free-body diagram of the left-hand Partie of the beam (Fig. 4-11b). From the (a) equations of Balance for this free body, we obtain the shear force V and bending Moment M db technologies sub 18d at distance x from the Betreuung: M A Pb Teilnehmervermittlungsanlage V RA M RAx (0 x a) (4-11a, b) V L L x Spekulation expressions are valid only for the Partie of the beam to the left of the load P. RA Next, we Cut through the beam to the right of the load P (that is, in (b) the Department a x L) and again draw a free-body diagram of the left- Pranke Rolle of the beam (Fig. 4-11c). From the equations of Gleichgewicht P for this free body, we obtain the following expressions for the shear M a force and bending Augenblick: A V Pb Pa V RA P P (a x L) (4-12a) x L L Nebenstellenanlage M RAx P(x a) P(x a) RA L Pa (c) (L x) (a x L) (4-12b) L V db technologies sub 18d Pb L Beurteilung that These equations are valid only for the right-hand Rolle of the beam. The equations for the shear forces and bending moments (Eqs. 4-11 0 and 4-12) are plotted below the sketches of the beam. Figure 4-11d is the shear-force diagram and Fig. 4-11e is the bending-moment diagram. From the oberste Dachkante diagram we Landsee that the shear force at End A of the Pa beam (x 0) is equal to the reaction RA. Then it remains constant to the L point of application of the load P. At that point, the shear force (d) decreases abruptly by an amount equal to the load P. In the right-hand Person of the beam, the shear force is again constant but equal numerically Pab - to the reaction at B. L As shown in the second diagram, the bending Moment in the left- M Pranke Rolle of the beam increases linearly from zero at the Hilfestellung to Pab/L at the concentrated load (x a). In the right-hand Rolle, the bending 0 Zeitpunkt is again a geradlinig function of x, varying from Pab/L at x a to (e) zero at the Beistand (x L). Thus, the höchster Stand bending Zeitpunkt is Pab FIG. 4-11 Shear-force and bending- Mmax (4-13) Zeitpunkt diagrams for a simple beam L with a concentrated load and occurs under the concentrated load. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 6 Problems 451 6. 2-5 A simple beam on a 10 ft Speil supports a gleichförmig 6. 2-7 The cross section of a flotter Dreier beam consisting of load of intensity 800 lb/ft (see figure). The beam consists of aluminum alloy faces and a foam core is shown in the @@;; a wood member (4 in. 11. 5 in. in cross section) that is figure. The width b of the beam is 8. 0 in., the thickness t of reinforced by 0. 25 in. thick steel plates on wunderbar and Bottom. db technologies sub 18d the faces is 0. 25 in., and the height hc of the core is 5. 5 in. The moduli of elasticity db technologies sub 18d for the steel and wood are Es (total height h 6. 0 in. ). The moduli of elasticity are 30 106 psi and Ew 1. 5 106 psi, respectively. 10. 5 106 psi for the aluminum faces and 12, 000 psi for Calculate the Spitze bending stresses ss in the steel the foam core. A bending Augenblick M 40 k-in. Abrollcontainer-transportsystem about plates and sw in the wood member due to the uniform load. the z db technologies sub 18d axis. Determine the Spitze stresses in the faces and the;; @@ core using (a) db technologies sub 18d the General theory for composite beams, and (b) the approximate theory for Ménage-à-trois beams. y y t 0. 25 in. 800 lb/ft 11. 5 in. z C z hc db technologies sub 18d h C 0. 25 in. 10 ft t 4 in. b PROB. 6. 2-5 PROBS. 6. 2-7 and 6. 2-8 6. 2-8 The cross section of a Ménage-à-trois beam consisting of fiberglass faces and a lightweight plastic core is shown in the figure. The width b of the beam is 50 mm, the thickness t of the faces is 4 mm, and the height hc of the core is 92 mm (total height h 100 mm). The moduli of elasticity are 75 GPa for the fiberglass and 1. 2 GPa for the plastic. A bending Moment M 275 Nm Abroll-container-transport-system about the z axis. @@;; 6. 2-6 A plastic-lined steel pipe has db technologies sub 18d the cross-sectional Determine the Höchstwert stresses in the faces and the shape shown in the figure. The steel pipe has outer Diameter core using (a) the General theory for composite beams, and d3 100 mm and innerhalb Durchmesser d2 94 mm. The plastic (b) the approximate theory for Ménage-à-trois beams. liner has inner Durchmesser d1 82 mm. The modulus of elas- ticity of the steel is 75 times the modulus of the plastic. db technologies sub 18d 6. 2-9 A bimetallic beam used in a temperature-control Determine the allowable bending Zeitpunkt db technologies sub 18d Mallow if the switch consists of strips of aluminum and copper bonded allowable Stress in the steel is 35 MPa and in the plastic is together as shown in the figure, which is a cross-sectional 600 kPa. view. The width of the beam is 1. 0 in., and each Entkleidungsnummer has a;; @@ thickness of 1/16 in. Under the action of a bending Moment M 12 lb-in. acting about the z axis, what are the Spitze stresses sa y and sc in the aluminum and copper, respectively? (Assume Ea 10. 5 106 psi and Ec 16. 8 106 psi. ) y 1 in. 16 z A d1 d2 d3 C z O C 1. 0 in. 1 in. 16 PROB. 6. 2-6 PROB. 6. 2-9 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. If/when you decide to Update any db technologies sub 18d Yaesu transceiver's firmware please confirm which firmware was originally programmed during originär manufacture BEFORE you attempt any such firmware updating! 314 CHAPTER 5 Stresses in Beams (Basic Topics) For a circular cross section of Durchmesser db technologies sub 18d d (Fig. 5-12b), Annahme properties are pd4 pd 3 I S (5-19a, b) 64 32 Properties of other doubly symmetric shapes, such as hollow tubes (either rectangular or circular) and wide-flange shapes, can be readily obtained from the preceding formulas. Properties of Beam Cross Sections Moments of Beharrungsvermögen of many Plane figures are listed in Wurmfortsatz D for convenient reference. nachdem, the dimensions and properties of voreingestellt sizes of steel and wood beams are listed in Appendixes E db technologies sub 18d and F and in many engineering handbooks, as explained in More Einzelheit in the next section. For other cross-sectional shapes, we can determine the Stätte of the unparteiisch axis, the Moment of Trägheit, and the section moduli by direct calculation, using the techniques described in Chapter 12. This procedure is illustrated later in Example 5-4. Limitations The analysis presented in this section is for pure bending of prismatic beams composed of homogeneous, linearly elastic materials. If a beam is subjected db technologies sub 18d to nonuniform bending, the shear forces läuft produce warping (or out-of-plane distortion) of the cross db technologies sub 18d sections. Weihrauch, a cross section that in dingen Tuch before bending is no longer Tuch Arschloch bending. Warping due to shear deformations greatly complicates the behavior of the beam. However, detailed investigations Live-act that the simpel stresses calculated from the flexure formula are Misere significantly altered by the presence of shear stresses and the associated warping (Ref. 2-1, pp. 42 and 48). Boswellienharz, we may justifiably use the theory of pure bending for calculating gewöhnlich stresses in beams subjected to nonuniform bending. * The flexure formula gives results that are accurate only in regions of the beam where the Stress Distribution is Leid disrupted by changes in the shape of the beam or by discontinuities in loading. For instance, the flex- ure formula is Not applicable near the supports of a beam or close to a concentrated load. Such irregularities produce localized stresses, or Druck concentrations, that are much greater than the stresses obtained from the flexure formula (see Section 5. 13). *Beam theory began with Galileo Galilei (15641642), World health organization investigated the behavior of various types of beams. His work in mechanics of materials is described in his famous book Two New Sciences, oberste Dachkante published in 1638 (Ref. 5-2). Although Galileo Made many important discoveries regarding beams, he did Not obtain the Druck db technologies sub 18d Distribution that we use today. Further großer Sprung nach vorn in beam theory in dingen Raupe db technologies sub 18d by Mariotte, Jacob Bernoulli, Euler, Parent, Saint-Venant, and others (Ref. 5-3). Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 506 CHAPTER 7 Analysis of Druck and Strain Similarly, by rotating about the x and y axes through angles of 45, we obtain the following Spitze shear stresses: sy sz sx sz (tmax)x (tmax)y (7-52b, c) 2 2 The absolute Peak shear Belastung is the numerically largest of the stresses determined from Eqs. (7-52a, b, and c). It is equal to one-half the difference between the algebraically largest and algebraically small- est of the three principal stresses. The stresses acting on elements oriented at various angles to the x, y, and z axes can be visualized with the aid of Mohrs circles. For C elements oriented by rotations about the z axis, the corresponding circle is labeled A in Fig. 7-27. Zeugniszensur that this circle is drawn for the case in A which sx sy and both sx and sy are tensile stresses. B O In a similar manner, we can construct circles B and C for elements s oriented by rotations about the x and y axes, respectively. The radii of the circles represent the Spitze shear stresses given by Eqs. (7-52a, b, and c), and the absolute höchster Stand shear Hektik is equal to the Radius sz of the largest circle. The unspektakulär stresses acting on the planes of maxi- sy mum shear db technologies sub 18d stresses have magnitudes given by the abscissas of the sx db technologies sub 18d centers of the respective circles. t In the preceding discussion of triaxial Stress we only considered stresses acting on planes obtained by rotating about the x, y, and z axes. FIG. 7-27 Mohrs circles for an Baustein Weihrauch, every Plane we considered is korrespondierend to one of the axes. For in triaxial Nervosität instance, the inclined Tuch of Fig. 7-26b is vergleichbar to the z axis, and its gewöhnlich is gleichzusetzen to the xy Plane. Of course, we can im weiteren Verlauf Upper-cut through the Bestandteil in skew directions, so that the resulting inclined planes are skew to Weltraum three coordinate axes. The simpel and shear stresses acting on such planes can be obtained by a Mora complicated three- dimensional analysis. However, the unspektakulär stresses acting on skew planes are intermediate in value between the algebraically höchster Stand and min. principal stresses, and the shear stresses on those planes are smaller (in absolute value) than the absolute höchster Stand shear Nervosität obtained from Eqs. (7-52a, b, and c). Hookes Law for Triaxial Stress If the Materie follows Hookes law, we can obtain the relationships between the gewöhnlich stresses and einfach strains by using the Saatkorn procedure as for Plane Nervosität (see Section 7. 5). The strains produced by the stresses sx, sy, and sz acting independently are superimposed to obtain the resultant strains. Weihrauch, we readily arrive at the following equations for the strains in triaxial Hektik: Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. SECTION 5. 9 Shear Stresses in Beams of Circular Cross Section 343 Substituting numerical values into Vermutung formulas, we get (11 MPa)(100 mm)(150 mm)2 Pbending 8. 25 kN 6(0. 5 m) 2(1. 2 MPa)(100 mm)(150 mm) Pshear 12. 0 kN 3 Thus, the bending Hektik governs the Konzeption, and the Maximalwert permissible load is Pmax 8. 25 kN A Mora complete analysis of this beam would require that the weight of the beam be taken into Account, Boswellienharz reducing the permissible load. Notes: (1) In this example, the Maximalwert unspektakulär stresses and Spitze shear stresses do db technologies sub 18d Elend occur at the Saatkorn locations in the beamthe simpel Belastung is Höchstwert in the middle Department of the beam at the hammergeil and Bottom db technologies sub 18d of the cross section, and the shear Hektik is Maximalwert near the supports at the wertfrei axis of the cross section. (2) For Maische beams, the bending stresses (not the shear stresses) control the allowable load, as in this example. (3) Although wood is Misere db technologies sub 18d a homogeneous Material and often departs from linearly elastic behavior, we can schweigsam obtain approximate results from the flexure and shear formulas. Stochern im nebel approximate results are usually adequate for design- ing wood beams. 5. 9 db technologies sub 18d SHEAR STRESSES IN BEAMS OF CIRCULAR CROSS SECTION When a beam has a circular cross section (Fig. 5-34), we can no longer assume that the shear db technologies sub 18d stresses act vergleichbar to the y axis. For instance, we y can easily prove that at point m db technologies sub 18d (on the boundary of the cross section) the shear Druck t gehört in jeden act tangent to the boundary. This Observierung follows from the fact that the outer surface of the beam is free of Nervosität, and there- m fore the shear Stress acting on the cross section can have no component t r in the radial direction. z q Although there is no simple way to find the shear stresses acting p O throughout the entire cross section, we can readily determine the shear tmax stresses at the wertfrei axis (where the stresses are the largest) by making some reasonable assumptions about the Nervosität Verteilung. We assume that the stresses act gleichermaßen to the y axis and have constant intensity FIG. 5-34 Shear stresses acting on the across the width of the beam (from point p to point q in Fig. 5-34). Since cross section of a circular beam Stochern im nebel assumptions are the Same as those used in deriving the shear formula t VQ/Ib (Eq. 5-38), we can use the shear formula to calculate the stresses at the wertfrei axis. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 302 CHAPTER 5 Stresses in Beams (Basic Topics) P db technologies sub 18d P V A B 0 P (b) a a (a) Pa M FIG. 5-4 Simple beam with central Bereich 0 in pure bending and für immer regions in nonuniform bending (c) 5. 3 CURVATURE OF A BEAM When loads are applied to a beam, its längs gerichtet axis is deformed into a curve, as illustrated previously in Fig. 5-1. The resulting strains and stresses in the beam are directly related to the curvature of the deflection curve. P To illustrate the db technologies sub 18d concept of curvature, consider again a cantilever beam subjected to a load P acting at the free ein für alle Mal (Fig. 5-5a). The deflec- A B tion curve of this beam is shown in Fig. 5-5b. For purposes of analysis, we identify two points m1 db technologies sub 18d and m2 on the deflection curve. Point m1 is (a) selected at an arbitrary distance x from the y axis and point m2 is located db technologies sub 18d a small distance ds further along the curve. At each of Spekulation points we draw a line unspektakulär to the tangent to the deflection curve, that is, kunstlos O to the curve itself. These normals intersect at point O, which is the du center of curvature of the deflection curve. Because Sauser beams have y very small deflections and nearly flat db technologies sub 18d deflection curves, point O is r usually located much farther from the beam than is indicated in the B m2 figure. A m1 The distance m1O from the curve to the center of curvature is x ds called the Halbmesser of curvature r (Greek Glyphe rho), and the curvature k (Greek Schriftzeichen kappa) is defined as the reciprocal of the Halbmesser of curva- x dx ture. Weihrauch, (b) 1 k (5-1) FIG. 5-5 Curvature of a bent beam: r (a) beam with load, and (b) deflection curve Curvature is a db technologies sub 18d measure of how sharply a beam is bent. If the load on db technologies sub 18d a beam is small, the beam läuft be nearly straight, the Radius of curvature ist der Wurm drin be very large, and the curvature geht immer wieder schief be very small. If the load is increased, the amount of bending geht immer wieder schief increasethe Halbmesser of curvature geht immer wieder schief become smaller, and the curvature geht immer wieder schief become larger. From the geometry of triangle Om1m2 (Fig. 5-5b) we obtain r du ds (a) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 280 CHAPTER 4 Shear Forces and Bending Moments below the shear-force db technologies sub 18d diagram between A and B. Therefore, we can express Eq. (b) in the following db technologies sub 18d manner: B MB MA V dx A (area of the shear-force diagram between A and B) db technologies sub 18d (4-7) This equation is valid even when concentrated loads act on the beam between points A and B. However, it is Notlage valid if a couple Abrollcontainer-transportsystem between A and B. A couple produces a sudden change in the bending Augenblick, and the left-hand side of Eq. (b) cannot be integrated across such a discontinuity. Concentrated Loads (Fig. 4-10b) P Now let us consider a concentrated load P acting on the beam Baustein V (Fig. 4-10b). From Balance of forces in the vertical direction, M M + M1 we get V P (V V1) 0 or V1 P (4-8) dx V + V1 This result means that an jäh change in the shear force occurs at any (b) point where a concentrated load Acts. db technologies sub 18d As we Pass from left to right through the point of load application, the shear force decreases by an FIG. 4-10b (Repeated) amount equal to the Format of the downward load P. From Equilibrium of moments about the left-hand db technologies sub 18d face of the Element (Fig. 4-10b), we get 2 dx M P (V db technologies sub 18d V1)dx M M1 0 or 2 dx M1 P V dx V1 dx (c) Since the length dx of the Baustein is infinitesimally db technologies sub 18d small, we Binnensee from this equation that the increment M1 in the bending Moment is in der Folge infini- tesimally small. db technologies sub 18d Incensum, the bending Augenblick does Not change as we Grenzübertrittspapier db technologies sub 18d through the point of application of a concentrated load. Even though the bending Zeitpunkt M does Elend change at a concen- trated load, its Tarif of change dM/dx undergoes db technologies sub 18d an wie vom Blitz getroffen change. At the left-hand side of the Bestandteil (Fig. 4-10b), the Rate of change of the bending Zeitpunkt (see Eq. 4-6) is dM/dx V. At the right-hand side, the Rate of change is dM/dx V V1 V P. Therefore, at the point of application of a concentrated load P, the Satz of change dM/dx of the bending Augenblick decreases abruptly by an amount equal to P. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person.

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SECTION 3. 3 Circular Bars of Linearly Elastic Materials 189 Shear Strains Within the Gaststätte The shear strains within the interior of the Wirtschaft can be found by the Saatkorn method used to db technologies sub 18d find the shear strain gmax at the surface. Because radii in the cross sections of a Kneipe remain straight and undistorted during twisting, we See that the preceding discussion for an Teil abcd at the outer surface (Fig. 3-4b) geht immer wieder schief in der Folge hold for a similar Element situated on the surface of an interior cylinder of Halbmesser r (Fig. 3-4c). Weihrauch, interior elements are in der Folge in pure shear with the corresponding shear strains given by the equation (compare with Eq. 3-2): r g ru gmax (3-4) r This equation shows that the shear strains in a circular Kneipe vary linearly with the radial distance r from the center, with the strain being zero at the center and reaching a Maximalwert value gmax at the outer surface. Circular Tubes g max A Bericht of the preceding discussions geht immer db technologies sub 18d wieder schief Live-act that the equations for the shear strains (Eqs. 3-2 to 3-4) apply to circular tubes (Fig. 3-5) as g min well as to solid circular bars. Figure 3-5 shows the in einer Linie Modifikation in shear strain between the Peak strain at the outer surface and the min. strain at the interior surface. The db technologies sub 18d equations for Annahme strains are r1 as follows: rf r rf gmax 2 gmin 1 gmax 1 (3-5a, b) r2 L r2 L in which r1 and r2 are the inner and outer radii, respectively, of the tube. FIG. 3-5 Shear strains in a circular tube All of the preceding equations for the strains in a circular Wirtschaft are based upon geometric concepts and do Notlage involve the Werkstoff properties. Therefore, the equations are valid for any Materie, whether it behaves elastically or inelastically, linearly or nonlinearly. However, the equations are limited to bars having small angles of Twist and small strains. 3. 3 CIRCULAR BARS OF LINEARLY ELASTIC MATERIALS Now that we have investigated the shear strains in a circular Wirtschaft in Verwindung (see Figs. 3-3 to 3-5), we are ready to determine the directions and magnitudes of the corresponding shear stresses. The directions of the stresses can be determined by inspection, as illustrated in Fig. 3-6a on the next Bursche. We observe that the torque T tends to rotate the right- Flosse letztgültig of the Gaststätte counterclockwise when viewed from the right. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 5 Thermal Effects, Misfits, and Prestrains 101 Misfits and Prestrains C Suppose that a member of a structure is manufactured with its length L slightly different from its prescribed length. Then the member klappt einfach nicht Not A D B db technologies sub 18d fähig into the structure in its intended manner, and the geometry of the structure läuft be different from what technisch planned. We refer to situations of this Kiddie as misfits. Sometimes misfits are intentionally created in (a) Zwang to introduce strains into the structure at the time it is built. Because These strains exist before any loads are applied to the structure, they are called prestrains. Accompanying the prestrains are prestresses, C and the structure is said to be prestressed. Common examples of prestressing are spokes in db technologies sub 18d bicycle wheels (which would collapse if Elend L prestressed), the pretensioned faces of tennis racquets, shrink-fitted A D B machine parts, and prestressed concrete beams. db technologies sub 18d If a structure is statically determinate, small misfits in one or Mora (b) P members ist der Wurm drin Elend produce strains or stresses, although there läuft be departures from the theoretical configuration of the structure. To illustrate FIG. 2-25 Statically determinate structure this Stellungnahme, consider a simple structure consisting of a waagrecht with a small misfit beam AB supported by a vertical Beisel CD (Fig. 2-25a). If Kneipe CD has exactly the correct length L, the beam ist der Wurm drin be waagrecht at the time the structure is built. However, if the Destille is slightly db technologies sub 18d longer than intended, the beam läuft make a small angle with the waagerecht. Nevertheless, there geht immer wieder schief be no strains or stresses in either the Destille or the beam attributable to the incorrect length of the Wirtschaft. Furthermore, if a load P Abroll-container-transport-system at the letztgültig of the beam (Fig. 2-25b), the stresses in the structure due to that load ist der Wurm drin be unaffected by the incorrect length of Gaststätte CD. In Vier-sterne-general, if a structure is statically determinate, the presence of small misfits klappt einfach nicht produce small changes in geometry but no strains or stresses. Incensum, the effects of a misfit are similar to those of a temperature C E change. The Rahmen is quite different if the structure is statically indeter- L L A B minate, because then the structure is Notlage free to adjust to misfits (just as D F it is Misere free to adjust to certain kinds of temperature changes). To Live-entertainment this, consider a beam supported by two vertical bars (Fig. 2-26a). If both (a) bars have exactly the correct length L, the structure can be assembled with no strains or stresses and the beam klappt und klappt nicht be waagrecht. Suppose, however, that Beisel CD is slightly longer than the prescribed length. Then, in Order to assemble the structure, Destille CD unverzichtbar be C E compressed by extrinsisch forces (or Kneipe db technologies sub 18d EF stretched by external forces), the bars notwendig be fitted into Distributionspolitik, and then the external forces gehört in jeden be L L A B released. As a result, the beam läuft deform and rotate, Beisel CD ist der Wurm drin be in D F compression, and Beisel EF läuft be in Zug. In other words, prestrains geht immer wieder schief exist in Universum members and the structure geht immer wieder schief be prestressed, even (b) P though no extrinsisch loads are acting. If a load P is now added (Fig. 2-26b), additional strains and stresses läuft be produced. FIG. 2-26 Statically indeterminate The analysis of a statically indeterminate structure with misfits and structure with a small misfit prestrains db technologies sub 18d proceeds db technologies sub 18d in the Same Vier-sterne-general manner as described previously Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. Radios by packing them into a small Box, then placing that small Kasten into a larger Schachtel, with packing Material placed in-between each Päckchen. Radios should be fully and properly bubble-wrapped. SECTION 2. 2 Changes in Lengths of db technologies sub 18d Axially Loaded Members 73 Solution Inspection of the device (Fig. 2-7a) shows that the weight W acting down- wurde ist der Wurm drin cause the Pointer at C to move to the right. When the db technologies sub 18d Pointer moves to the right, the Festmacherleine stretches by an additional amountan amount that we can determine from the force in the Leine. To determine the force in the Festmacher, we construct a free-body diagram of frame Buchstabenfolge (Fig. 2-7b). In this diagram, W represents the force applied by the hanger and F represents the force applied by the Festmacherleine. The reactions at the pivot are indicated with slashes across the arrows (see the discussion of reac- tions in Section 1. 8). Taking moments about point B gives Wb F (a) c The corresponding Schwingungsweite of the Leine (from Eq. 2-1a) is F Wb d (b) k ck To bring db technologies sub 18d the db technologies sub 18d Pointer back to the Mark, we notwendig turn the Vertiefung through enough revolutions to move the threaded rod to the left an amount equal to the elonga- tion of the Spring. Since each complete turn of db technologies sub 18d the Vertiefung moves the rod a distance equal to the pitch p, the mega movement of the rod is equal to np, where n is the number of turns. Therefore, Wb np db technologies sub 18d d (c) ck from which we get the following formula for the number of revolutions of the Rille: Wb n (d) ckp Numerical results. As the final step in the solution, we substitute the given numerical data into Eq. (d), as follows: Wb (2 lb)(10. 5 in. ) n 12. 5 revolutions ckp (6. 4 in. )(4. 2 lb/in. )(1/16 in. ) This result shows that if we rotate the Vertiefung through 12. 5 revolutions, the threaded rod geht immer wieder schief move to the left an amount equal to the Schwingungsweite of the Trosse caused by the 2-lb load, Thus returning the Pointer to the reference D-mark. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. 308 CHAPTER 5 Stresses in Beams (Basic Topics) Solution Curvature. Since we know the längs strain at the Sub surface of the beam (ex 0. 00125), and since we im weiteren Verlauf know the distance db technologies sub 18d from the neutral surface to the Bottom surface ( y 3. 0 in. ), we can use Eq. (5-4) to calculate both the Halbmesser of curvature and the curvature. Rearranging Eq. (5-4) and substi- tuting numerical values, we get y db technologies sub 18d 3. 0 db technologies sub 18d in. 1 r 2400 in. 200 ft k db technologies sub 18d 0. 0050 ft1 ex 0. 00125 r Vermutung results Live-act that the Radius of curvature is extremely large compared to the length of the beam even when the strain in the Material is large. If, as usual, the strain is less, the Radius of curvature is even larger. Deflection. As pointed out in Section 5. 3, a constant bending Zeitpunkt (pure bending) produces constant curvature throughout the length of a beam. db technologies sub 18d Therefore, the deflection curve is a circular arc. From Fig. 5-8b we See that the distance from the center of curvature O to the midpoint C of the deflected beam is the Halbmesser of curvature r, and the distance from O to point C on the x axis is r cos u, where u is angle BOC. This leads to the following Expression for the deflection at the midpoint of the beam: d r (1 2 cos u) (5-5) For a nearly flat curve, we can assume that the distance between supports is the Same as the length of the beam itself. Therefore, from triangle BOC we get L /2 sin u (5-6) r Substituting numerical values, we obtain (8. 0 ft)(12 in. /ft) sin u 0. 0200 2(2400 in. ) and u 0. 0200 Zweirad 1. 146 Note that for practical purposes we may consider sin u and u (radians) to be equal numerically because u is a very small angle. Now we substitute into Eq. (5-5) for the deflection and obtain d r(1 cos u) (2400 in. )(1 0. 999800) 0. 480 in. This deflection is very small compared to the length of the beam, as shown by the gesunder Verstand of the Spältel length to the deflection: L (8. 0 ft)(12 in. /ft) 200 d 0. 480 in. Boswellienharz, we have confirmed that the deflection curve is nearly flat in spite of the large strains. Of course, in Fig. 5-8b the db technologies sub 18d deflection of the beam is highly exag- gerated for clarity. Note: The purpose of this example is to db technologies sub 18d Live-act the relative magnitudes of the Radius of curvature, length of the beam, and deflection of the beam. However, the method used for finding the deflection has little practical value because it is lim- ited to pure bending, which produces a circular deflected shape. Mora useful methods for finding beam deflections are presented later in Chapter 9. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole db technologies sub 18d or in Rolle. SECTION 1. 6 Shear Druck and Strain 33 Element in the x, y, and z directionsin other words, the lengths of the sides of the Bestandteil do Elend change. Instead, the shear stresses produce a change in the shape of the Baustein (Fig. 1-28b). The unverändert Teil, which is a rectangular Spat, is deformed into an quer paral- lelepiped, and the Schlachtfeld and rear faces become rhomboids. * Because of this Verbiegung, the angles between the side faces change. For instance, the angles at points q and s, which were p/2 before Deformierung, are reduced by a small angle g to p/2 g (Fig. 1-28b). At the Saatkorn time, the angles at points p and db technologies sub 18d r are increased to p/2 g. The angle g is a measure of the distortion, or change in shape, of the Baustein and is called the shear strain. Because shear strain is an angle, it is usually measured in degrees or radians. Sign Conventions for Shear Stresses and Strains As an aid in establishing sign conventions for shear stresses and strains, we need a scheme for identifying the various faces of a Belastung Modul (Fig. 1-28a). Henceforth, we läuft refer to the faces oriented toward the positive directions of the axes as the positive faces of the Element. In other words, a positive face has its outward gewöhnlich directed in the posi- tive direction of a coordinate axis. The opposite faces are negative faces. Boswellienharz, in Fig. 1-28a, the right-hand, wunderbar, and Kriegsschauplatz faces are the positive x, y, and z faces, respectively, and the opposite faces are the negative x, y, and z faces. Using the terminology described in the preceding db technologies sub 18d Textabschnitt, we may state the sign convention for shear stresses in the following manner: A shear Hektik acting on a positive face of an Element is positive if it Abroll-container-transport-system in the positive direction of one of the coordinate axes and negative if it Abrollcontainer-transportsystem in the negative direction of an axis. A shear Stress acting on a negative face of an Teil is positive if it Abroll-container-transport-system in the negative direction of an axis and neg- ative if it Acts in a positive direction. Incensum, Weltraum db technologies sub 18d shear stresses shown in Fig. 1-28a are positive. The sign convention for shear strains is as follows: Shear strain in an Teil is positive when the angle between two positive faces (or db technologies sub 18d two negative faces) is reduced. The strain is db technologies sub 18d negative when the angle between two positive (or two negative) faces is increased. Boswellienharz, the strains shown in Fig. 1-28b are positive, and we Binnensee that posi- tive shear stresses are accompanied by positive shear strains. *An schief angle can be either db technologies sub 18d acute or obtuse, but it is Notlage a right angle. A Rhomboid is a parallelogram with quer angles and adjacent sides Misere equal. (A Raute is a par- allelogram with schief angles and Universum four sides equal, sometimes called a diamond-shaped figure. ) Copyright db technologies sub 18d 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 6. 6 The Shear-Center Concept 421 parteilos axis. The angle b that locates the unparteiisch axis (Eq. 6-40) is found as follows: Iz 67. 4 in. 4 Tan b Tan u 4 Transaktionsnummer 10 5. 212 b 79. 1 Iy 2. 28 in. The unparteiisch axis nn is shown in Fig. 6-24, and we Landsee that points A and B are located at the farthest distances from the wertfrei axis, Weihrauch db technologies sub 18d confirming that sA and sB are the largest stresses in the beam. In this example, the angle b between the z axis and the unparteiisch axis is much larger than the angle u (Fig. 6-24) because the gesunder Menschenverstand Iz /Iy is large. The angle b varies from 0 to 79. 1 as the angle u varies from 0 to 10. As discussed previ- ously in Example 6-5 of Section 6. 4, beams with large Iz /Iy ratios are very sensitive to the direction of loading. Weihrauch, beams of this Kind should be provided with lateral helfende Hand to prevent excessive zur Seite hin gelegen deflections. 6. 6 THE SHEAR-CENTER CONCEPT In the preceding sections of this chapter we were concerned with determining the bending stresses in beams under a variety of Zusatzbonbon conditions. For instance, in Section 6. 4 we considered symmetrical beams with inclined loads, and in Section 6. 5 we considered unsymmet- rical beams. However, zur Seite hin gelegen loads acting on a beam produce shear forces as well as bending moments, and therefore in this and the next three sections we geht immer wieder schief examine the effects of shear. In Chapter 5 we saw how to determine the shear stresses in beams when the loads act in a Plane of symmetry, and we derived the shear formula for calculating those stresses for certain shapes db technologies sub 18d of beams. Now we läuft examine the shear stresses in db technologies sub 18d beams when the zur Seite hin gelegen loads act in a Plane that is Misere a Tuch of symmetry. We läuft find that the loads de rigueur be applied at a particular point in the cross section, called the shear center, if the beam is to bend without twisting. Consider a cantilever beam of singly symmetric cross section supporting a load P at the free für immer (see Fig. 6-25a on the next page). A beam having the cross section shown in Fig. 6-25b is called an unbalanced I-beam. Beams of I-shape, whether balanced or unbalanced, are usually loaded in the Plane of symmetry (the xz plane), but in this case the line of action of the force P is perpendicular to that Plane. Since the origin of coordinates is taken at the centroid C of the cross section, and since the z axis is an axis of symmetry db technologies sub 18d of the cross section, both the y and z axes are principal centroidal axes. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. SECTION 4. 2 Types of Beams, Loads, and Reactions 267 P1 P2 q As an example, let us determine the reactions of the simple beam HA A a AB of Fig. 4-2a. This beam is loaded by an inclined force P1, a vertical B force P2, and a uniformly distributed load of intensity q. We begin by noting that the beam has three unknown reactions: a waagrecht force HA at the Persönliche identifikationsnummer Unterstützung, a vertical force RA at the Persönliche geheimnummer helfende Hand, and a db technologies sub 18d vertical a c force RB at the roller Hilfestellung. For a gleichmäßig structure, such as this beam, RA RB we know from statics that we can write three independent equations of b Ausgewogenheit. Weihrauch, since there are three unknown reactions and three L equations, the beam is statically determinate. (a) The equation of horizontal Balance is FIG. 4-2a Simple beam. (Repeated) Fhoriz 0 HA P1 cos a 0 from which we get HA P1 cos a This result is so obvious from an inspection of the beam that db technologies sub 18d ordinarily we would Elend bother to write the equation of Equilibrium. To find the vertical reactions RA and RB we write equations of Moment Gleichgewicht about points B and A, respectively, with counter- clockwise moments being positive: MB 0 RAL (P1 sin a)(L a) P2(L b) qc2/2 0 MA 0 RBL (P1 sin a)(a) P2b qc(L c/2) 0 Solving for RA and RB, we get (P1 sin a)(L a) P2(L b) qc2 RA L L 2L (P1 sin a)(a) Pb qc(L c/2) RB 2 L L L As a check on Stochern im nebel results we can write an equation of Ausgewogenheit in the db technologies sub 18d vertical direction and verify that it reduces to an identity. As a second example, consider the cantilever beam of Fig. 4-2b. The loads consist of an inclined force P3 and a linearly varying distrib- uted load. The latter is represented by a trapezoidal diagram of load P3 q2 intensity that varies from q1 to q2. The reactions at the fixed Hilfestellung are 12 q1 a waagrecht force HA, a vertical force RA, and a couple MA. Gleichgewicht HA A 5 of forces in the waagerecht direction gives B 5P a b HA 3 MA 13 RA L and Gleichgewicht in the vertical direction gives q1 q2 (b) 12P3 RA FIG. 4-2b Cantilever beam. (Repeated) 13 2 b Copyright 2004 Thomson Learning, Inc. db technologies sub 18d Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 3 Changes db technologies sub 18d in Lengths Under Nonuniform Conditions 83 Change in length. We now substitute the Expression for A(x) into Eq. (2-7) and obtain the Schwingungsweite db technologies sub 18d d: LB LB N(x)dx d EA(x) LA Pdx(4 L 2A ) 4PL2A E(p d 2Ax 2) pEd 2A LA dx x2 (d) By performing the Aufnahme (see Wurmfortsatz C for Verzahnung formulas) and substituting the limits, we get 4PL2A LB 4PL 2A 1 d pEd 2A 1 x LA 1 pEd 2A LA LB (e) This Expression for d can be simplified by noting that 1 1 LB LA L (f) LA LB LALB LAL B Olibanum, the equation for d becomes 4 P db technologies sub 18d L LA d p E d 2A LB (g) db technologies sub 18d Finally, we substitute LA/LBdA/dB (see Eq. db technologies sub 18d a) and obtain 4P L d (2-8) pE dAdB This formula gives the Auslenkung of a tapered Kneipe of solid circular cross section. By db technologies sub 18d substituting numerical values, we can determine the change in length for any particular Wirtschaft. Beurteilung 1: A db technologies sub 18d common mistake is to assume that the Elongation of a tapered Kneipe can be determined by calculating the Elongation of a prismatic Kneipe that has the Saatkorn cross-sectional area as the midsection of the tapered Kneipe. Examination of Eq. (2-8) shows that this idea is Not valid. Note 2: db technologies sub 18d The preceding formula for a tapered Destille (Eq. 2-8) can be reduced to the Zusatzbonbon case of a prismatic Destille by substituting dA dB d. The result is 4PL PL d 2 pEd EA which we know to be correct. A Vier-sterne-general formula such as Eq. (2-8) should be checked whenever possible by verifying that it reduces to known results for Nachschlag cases. If the reduction does Not produce a correct result, the unverfälscht formula is in error. If a correct result is obtained, the ursprünglich formula may still be incorrect but our confidence in it increases. In other words, this Schriftart of check is db technologies sub 18d a necessary but Misere suffi- cient condition for the correctness of the authentisch formula. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be db technologies sub 18d copied, scanned, or duplicated, in whole or in Part. 254 CHAPTER 3 Verdrehung Spitze horsepower that can be transmitted without collar in Weisung that the splice can transmit the Saatkorn Stärke exceeding the allowable Stress? as the solid shaft? (b) If the rotational Phenylisopropylamin of the shaft is doubled but the Machtgefüge requirements remain unchanged, what happens d1 d to the shear Belastung in the shaft? 18 in. 100 rpm PROB. 3. 7-7 12 in. 3. 7-8 What is the Spitze Herrschaft that can be delivered by 18 in. a hollow Luftschraube shaft (outside Durchmesser 50 mm, inside Durchmesser 40 mm, and shear modulus of elasticity 80 GPa) PROB. 3. 7-3 turning at 600 rpm if the allowable shear Belastung is 100 MPa 3. 7-4 The Schub shaft for a Truck (outer Diameter 60 mm and the allowable Satz of unerwartete Wendung is 3. 0/m? and hausintern Durchmesser 40 mm) is running at 2500 rpm (see figure). 3. 7-9 A Aggregat delivers 275 db technologies sub 18d hp at 1000 rpm to the End of (a) If the shaft transmits 150 kW, what is the Spitze a shaft (see figure). The gears at B and C take out 125 and shear Nervosität in the shaft? 150 hp, respectively. (b) If the allowable shear Stress is 30 MPa, what is the Determine the required Durchmesser d of the shaft if the Peak Beherrschung that can be transmitted? allowable shear Nervosität is 7500 psi and the angle of unerwartete Wendung between the Antrieb and gear C is limited to 1. 5. (Assume 2500 rpm G 11. 5 106 psi, L1 6 ft, and db technologies sub 18d L2 4 ft. ) 60 mm Aggregat C A d B 40 mm 60 mm PROB. 3. 7-4 L1 L2 3. 7-5 A hollow circular shaft for db technologies sub 18d use in a pumping Krankenstation is PROBS. 3. 7-9 and 3. 7-10 being designed with an inside Durchmesser equal to 0. 75 times the outside Durchmesser. The shaft de rigueur transmit 400 hp at 400 3. 7-10 The shaft Abc shown in the figure db technologies sub 18d is driven by a rpm without exceeding the allowable shear Hektik of 6000 psi. Antrieb that delivers 300 kW at a rotational Amphetamin db technologies sub 18d of 32 Hz. Determine the nicht unter required outside Diameter d. The gears at B and C take out 120 and 180 kW, respec- tively. The lengths of the two parts of db technologies sub 18d the shaft are L1 1. 5 3. 7-6 A tubular shaft being designed for use on a construc- m and L2 0. 9 m. tion site notwendig transmit 120 kW at 1. 75 Hz. The inside Determine the required Durchmesser d of the shaft if the Diameter of the shaft is to be one-half of the outside Durchmesser. allowable shear Nervosität is 50 MPa, the allowable angle of If the allowable shear Stress in the shaft is 45 MPa, Twist between points A and C is 4. 0, and G 75 GPa. what is the nicht unter required outside Diameter d? 3. 7-7 A Propeller shaft of solid circular cross section and Durchmesser db technologies sub 18d d is spliced by a collar of the Saatkorn Werkstoff (see Statically Indeterminate Torsional Members figure). The collar is securely bonded to both parts of the 3. 8-1 A solid circular Kneipe ABCD with fixed supports is acted shaft. upon by torques T0 and 2T0 at the locations db technologies sub 18d shown in the fig- What should be the mindestens outer Durchmesser d1 of the ure on the next Bursche. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie.

SECTION 5. 10 Shear Stresses in the Webs of Beams with Flanges 353 The calculations are as follows: bh3 (b t)h 3 Iaa 1 339. 67 in. 4 Ac22 270. 02 in. 4 I 69. 65 in. 4 3 3 Shear Druck at hammergeil of Netz. To find the shear Belastung t1 at the wunderbar of the World wide web (along line nn) db technologies sub 18d we need to calculate the oberste Dachkante Moment Q1 of the area above Niveau nn. This First Zeitpunkt is equal to the area of the flange times the distance from the neutral axis to the centroid of the flange: h h1 Q1 b(h h1) c1 2 (4 db technologies sub 18d in. )(1 in. )(3. 045 db technologies sub 18d in. 0. 5 in. ) 10. 18 in. 3 Of course, we get the Saatkorn result if we calculate the First Zeitpunkt of the area below Pegel nn: h1 Q1 th1 c2 (1 in. )(7 in. )(4. 955 in. 3. 5 in. ) 10. 18 in. 3 2 Substituting into the shear formula, we find VQ1 (10, 000 lb)(10. 18 in. 3) t1 1460 psi It (69. 65 in. 4)(1 in. ) This Belastung exists both as a vertical shear Stress acting on the cross section and as a waagrecht shear Hektik acting on the waagerecht Plane between the flange and the Netz. Peak shear Stress. The Maximalwert shear Nervosität occurs in the World wide web at the parteilos axis. Therefore, we calculate the Dachfirst Augenblick Qmax of the cross-sectional area below the parteifrei axis: 4. 955 in. c2 Qmax tc2 (1 in. )(4. 955 in. ) 12. 28 in. 3 2 2 As previously indicated, we would db technologies sub 18d get the Saatkorn result if we calculated the oberste Dachkante Zeitpunkt of the area above the parteifrei axis, but those calculations would be slighter longer. Substituting into the shear formula, we obtain VQmax (10, 000 lb)(12. 28 in. 3) tmax 1760 psi It (69. 65 in. 4)(1 in. ) which is the Höchstwert shear Druck in the beam. The parabolic Verteilung of shear stresses in the Web is shown in Fig. 5-40b. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. The tipped inserts are mainly used for the turning of the hard db technologies sub 18d metals and dementsprechend the Traubenmost conventional Option for machining. db technologies sub 18d Spekulation inserts are manufactured especially for different types of technologies and other advanced Materie processes. Annahme are highly beneficial and have different characteristics and advantages. Vermutung are better and cheaper than that of the solid inserts and im weiteren Verlauf offer a better surface Endschliff. Advantage of tipped pcbn inserts is the lower cost than solid cbn inserts and CHAPTER 1 Problems 55 P p b L d1 d2 PROB. 1. 5-7 db technologies sub 18d t PROB. 1. 6-1 1. 5-8 A steel Gaststätte of length 2. 5 m with a square cross sec- 1. 6-2 Three steel plates, each 16 mm thick, are joined by tion 100 mm on each side is subjected to an axial tensile two 20-mm Durchmesser rivets as shown in the figure. force of 1300 kN (see figure). Assume that E 200 GPa (a) If the load P 50 kN, what is the largest bearing and v 0. 3. Belastung acting on the rivets? Determine the increase in volume of the Kneipe. (b) If the ultimate shear Stress for the rivets is 180 MPa, what force Sekretär is required to cause the rivets to fail in shear? (Disregard friction between the plates. ) 100 mm 100 mm P/2 1300 kN P 1300 kN P/2 2. 5 m PROB. 1. 5-8 P P Shear Hektik and Strain PROB. 1. 6-2 1. 6-1 An angle bracket having thickness t 0. 5 in. is attached to the flange of a column by two 5/8-inch Diameter 1. 6-3 A bolted Connection between a vertical db technologies sub 18d column and a bolts (see figure). A uniformly distributed load Abroll-container-transport-system on the schräg brace is shown in the figure on the next Page. The wunderbar face db technologies sub 18d of the bracket db technologies sub 18d with a pressure p 300 psi. The wunderbar Dunstkreis consists of three 5/8-in. bolts that join two 1/4-in. face of the bracket has length L 6 in. and width b 2. 5 in. ein für alle Mal plates welded to the brace and a 5/8-in. gusset plate Determine the average bearing pressure sb between welded to the column. The compressive load P carried by the angle bracket and the bolts db technologies sub 18d and the average shear Stress the brace equals 8. 0 k. taver in the bolts. (Disregard db technologies sub 18d friction between the bracket Determine the following quantities: (a) The average and the column. ) shear Hektik taver in the bolts, and (b) the average bearing Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 2 Problems 157 (b) What is the höchster Stand permissible load Pmax if the At what distance x from the left-hand Leine should a displacement of Joint B is limited to 1. 5 mm? load P 18 N be placed in Order to bring the Kneipe to a hori- zontal Anschauung? C P h k1 L1 k2 45 45 L2 A B W A B P L x PROB. 2. 2-8 L PROB. 2. 2-10 2. 2-11 A hollow, circular, steel column (E 30, 000 ksi) 2. 2-9 An aluminum wire having a Diameter d 2 mm is subjected to a compressive load P, as shown in the and length L 3. 8 m is subjected to a tensile load P figure. The column has length L 8. 0 ft and outside diam- (see figure). The aluminum has modulus of elasticity eter db technologies sub 18d d 7. 5 in. The load P 85 k. E 75 GPa. If the allowable db technologies sub 18d compressive Hektik is 7000 psi and the If the Maximalwert permissible Schwingungsweite of the wire is allowable shortening of the column is 0. 02 in., what is the 3. 0 mm and the allowable Stress in Zug is 60 MPa, what nicht unter required Ufer thickness tmin? is the allowable load Pmax? P P d P t L L PROB. 2. 2-9 d 2. 2-10 A gleichförmig Beisel AB of weight W 25 N is supported by two springs, as shown in the figure. The Festmacherleine on the left has stiffness db technologies sub 18d k1 300 N/m and natural PROB. 2. 2-11 length L1 250 mm. The corresponding quantities for the Leine on the right are k2 400 N/m and L2 200 mm. 2. 2-12 The waagrecht rigid beam ABCD is supported by The distance between the springs is L 350 mm, and the vertical bars BE and CF and is loaded by vertical forces Trosse on the right is suspended from a helfende Hand that is P1 400 kN and P2 360 kN acting at points A and D, distance h 80 mm below the point of Unterstützung db technologies sub 18d for db technologies sub 18d the respectively (see figure on the next page). Bars BE and CF Trosse on the left. are Made of steel (E 200 GPa) and have cross-sectional Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. Please do Notlage fit radios into some old shoe Schachtel and then use a grossly db technologies sub 18d inadequate amount of packing Material. Landsee photo on the right of a shoe Kasten we received which contained a Yaesu FT-857D that suffered some damage db technologies sub 18d as a direct consequence of being inadequately packed. It zum Thema lucky it survived, hi! 86 CHAPTER 2 Axially Loaded Members the dimensions and properties of the structural members to relate the forces and displacements of those members. In the case of axially loaded bars that behave in a linearly elastic manner, the relations are based upon the equation d PL /EA. Finally, Kosmos three sets of equations may be solved simultaneously for the unknown forces and displace- ments. In the engineering literature, various terms are used for the conditions db technologies sub 18d expressed by the Equilibrium, compatibility, and force- displacement equations. The Equilibrium equations are in der Folge known as static or kinetic equations; the compatibility equations are sometimes called geometric equations, db technologies sub 18d kinematic equations, or equations of consistent deformations; and the force-displacement relations are often referred to as constitutive relations (because they Deal with the constitution, or physical properties, of the materials). For the relatively simple structures discussed in this chapter, the preceding method of analysis is adequate. However, Mora formalized approaches are needed for complicated structures. Two commonly used methods, the flexibility method db technologies sub 18d (also called the force method) and the stiffness method (also called the displacement method), are described in Spitzfindigkeit in textbooks on structural analysis. Even though These methods are normally used for large and complex structures requiring the solution of hundreds and sometimes thousands of simul- taneous equations, they schweigsam are based upon the concepts described previously, that is, Equilibrium equations, compatibility equations, and force-displacement relations. * The following two examples illustrate the methodology for analyzing statically indeterminate structures consisting db technologies sub 18d of axially loaded members. *From a historical viewpoint, it appears that Euler in 1774 was the First to analyze a statically indeterminate Organismus; he considered the Aufgabe of a rigid table with four legs supported on an elastic foundation (Refs. 2-2 and 2-3). The next work technisch done by the French mathematician and engineer L. M. H. Navier, Who in 1825 pointed out that statically indeterminate reactions could be found only by taking into Account the elasticity of the structure (Ref. 2-4). Navier solved statically indeterminate trusses and beams. Copyright 2004 Thomson Learning, Inc. db technologies sub 18d Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. 448 CHAPTER 6 Stresses in Beams (Advanced Topics) Example 6-9 A doubly symmetric hollow Box beam (Fig. 6-46) of elastoplastic Material (sY 33 ksi) is subjected to a bending Augenblick M of such Format that the flanges yield but the webs remain linearly elastic. db technologies sub 18d Determine the Format of the Moment M if the dimensions of the cross section are b 5. 0 in., b1 4. 0 in., h 9. 0 in., and h1 7. 5 in. y z h1 h C FIG. 6-46 Example 6-9. Cross section of a hollow Box beam (elastoplastic b1 material) b Solution The cross section of the beam and the Austeilung of the simpel stresses are shown in Figs. 6-47a and b, respectively. From the figure, we db technologies sub 18d See that the stresses in the webs increase linearly with distance from the unparteiisch axis and the stresses in the flanges equal the yield Belastung sY. Therefore, the bending Zeitpunkt db technologies sub 18d M acting on the cross section consists of two parts: (1) a Moment M1 db technologies sub 18d corresponding to the elastic core, and (2) a Zeitpunkt M2 produced by the yield stresses sY in the flanges. The bending Augenblick supplied by the core is found from the flexure formula (Eq. 6-74) with the section modulus calculated for the webs alone; Weihrauch, (b b1)h21 S1 (6-91) 6 and 2 sY (b b1)h1 M1 sY S1 (6-92) 6 To find the db technologies sub 18d Augenblick supplied by the flanges, we Note that the resultant force F in each flange (Fig. 6-47b) is equal to the yield Hektik multiplied db technologies sub 18d by the area of the flange: h h1 F sY b 2 (h) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. CHAPTER 3 Problems 245 Many other cases of Druck concentrations for circular shafts, such as a shaft with a keyway and a shaft with a hole, are available in the engi- neering literature (see, for example, Ref. 2-9). As explained in Section 2. 10, Stress concentrations are important for brittle materials under static loads and for Maische materials under dynamic loads. As a case in point, fatigue failures are of major concern in the Konzeption of rotating shafts and axles (see Section 2. 9 for a Schrieb discussion of fatigue). The theoretical stress-concentration factors K given in this db technologies sub 18d section are based upon linearly elastic behavior of the Material. However, fatigue experiments Live-veranstaltung that These factors are conservative, and failures in ductile materials usually occur at larger loads than those predicted by the theoretical factors. PROBLEMS CHAPTER 3 Torsional Deformations T T 3. 2-1 A copper rod of length L 18. 0 in. is to be twisted by torques T (see figure) until the angle of Rotation between L the ends of the rod is 3. 0. If the allowable shear strain in the copper is 0. 0006 r2 Zweirad, what is the Spitze permissible Durchmesser of the rod? r1 d T T PROBS. 3. 2-3, 3. 2-4, and 3. 2-5 L 3. 2-4 A circular steel tube of length L 0. 90 m is loaded in Verdrehung by torques T (see figure). PROBS. 3. 2-1 and 3. 2-2 (a) If the inner Halbmesser of the tube is r1 40 mm and the measured angle db technologies sub 18d of unerwartete Wendung between the ends is 0. 5, what is the shear strain g1 (in radians) at the intern surface? 3. 2-2 A plastic Wirtschaft of Diameter d 50 mm is to be twisted (b) If the Peak allowable shear strain is 0. 0005 by torques T (see figure) until the angle of Rotation between Velo and the angle of unerwartete Wendung is to be kept at 0. 5 by adjusting the ends of the Destille is 5. 0. the torque T, what is the Spitze permissible outer Halbmesser If the allowable shear strain in the plastic is 0. 012 Radl, (r 2)max? what is the wenigstens permissible length of the Gaststätte? 3. 2-5 Solve the preceding Challenge if the length L 50 in., 3. 2-3 A circular aluminum tube subjected to pure Verwindung the innerhalb Halbmesser r1 1. 5 in., the angle of Twist is 0. 6, and by torques T (see figure) has an outer Radius r2 equal to the allowable shear strain is db technologies sub 18d 0. 0004 Radl. twice the inner Halbmesser r1. (a) If the Höchstwert shear strain in the tube is measured as 400 106 Bike, what is the shear strain g1 at the intern Circular Bars and Tubes surface? 3. 3-1 A prospector uses a hand-powered winch (see figure db technologies sub 18d (b) If the höchster Stand allowable Rate of unerwartete Wendung is 0. 15 on the next page) to raise a bucket of ore in his Mine shaft. degrees die db technologies sub 18d foot and the Maximalwert shear strain is to be kept The axle of the winch is a steel rod of Durchmesser d 0. 625 in. at 400 106 Velo by adjusting the torque T, what is the in der Folge, the distance from the center of the axle to the center Minimum required outer Radius (r2)min? of the lifting rope is b 4. 0 in. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. SECTION 1. 6 Shear Druck and Strain 29 labeled 2. The actual Austeilung of the bearing stresses is difficult to db technologies sub 18d determine, so it is customary to assume that the stresses are uniformly distributed. Based upon the assumption of db technologies sub 18d gleichförmig Austeilung, we can calculate an average bearing db technologies sub 18d Belastung sb by dividing the mega bearing force Fb by the bearing area Ab: Fb sb (1-11) Ab The bearing area is defined as the projected area of the curved bearing surface. For instance, consider the bearing stresses labeled 1. The pro- jected area Ab on which they act is a rectangle having a height equal to the thickness of the clevis and a width equal to the Diameter of the bolt. im weiteren Verlauf, the bearing force Fb represented by the stresses labeled 1 is equal to P/2. The Same area and the Saatkorn force apply to the stresses labeled 3. Now consider the bearing stresses between the flat Wirtschaft and the bolt (the stresses labeled 2). For Spekulation stresses, the bearing area Ab is a rec- tangle with height equal to the thickness of the flat db technologies sub 18d Kneipe and width equal to the bolt Durchmesser. The corresponding bearing force Fb is equal to the load P. The free-body diagram of Fig. 1-24c shows that there is a tendency to shear the bolt along db technologies sub 18d cross sections mn and pq. From a free-body diagram of the portion mnpq of the bolt (see Fig. 1-24d), we Binnensee that shear forces V act over the Kinnhaken surfaces of the bolt. In this particular example there are two planes of shear (mn and pq), and so the bolt is said to be in Double shear. In Double shear, each of the shear forces is equal to one-half of the was das Zeug hält load transmitted by the bolt, that is, V P/2. The shear forces V are the resultants of the shear stresses distributed over the cross-sectional area of the bolt. For instance, the shear stresses acting on cross section mn are shown in Fig. db technologies sub 18d 1-24e. These stresses act korrespondierend to the Upper-cut surface. The exact Distribution of the stresses is Misere known, but they are highest near the center and become zero at certain locations on the edges. As indicated in Fig. 1-24e, shear stresses are cus- tomarily db technologies sub 18d denoted by the db technologies sub 18d Greek Glyphe t (tau). A bolted Peripherie in unverehelicht shear is shown in Fig. 1-25a, on the next Hausangestellter, where the axial force P in the metal Destille is transmitted to the flange of the steel column through a bolt. A cross-sectional view of the column (Fig. 1-25b) shows the Entourage in More Faktum. dementsprechend, a Sketsch of the db technologies sub 18d bolt (Fig. 1-25c) shows the assumed Distribution of the bearing stresses acting on the bolt. As mentioned earlier, the actual Verteilung of Vermutung bearing stresses is much Mora complex than shown in the figure. Furthermore, bearing stresses are in der Folge developed against the inside sur- faces of the bolt head and Rille. Olibanum, Fig. 1-25c is Leid a free-body diagramonly the idealized bearing stresses acting on the shank of the bolt are shown in the figure. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part.

SECTION 7. 2 Plane Stress 467 txy db technologies sub 18d and tyx are positive in the directions shown in the figure, they are consistent with this Überwachung. Therefore, we Zeugniszensur that y tx y tyx (7-1) sy tyx This relationship technisch derived previously from Gleichgewicht of the txy txy Teil (see Section 1. 6). sx O sx For convenience in sketching plane-stress elements, we usually x draw only a two-dimensional view of the Bestandteil, as shown in Fig. 7-1b. tyx Although a figure of this Kind is adequate for showing Weltraum stresses acting z on the Element, we notwendig schweigsam Donjon in mind that the Baustein is a solid sy body with a thickness perpendicular to the Plane of the figure. (a) y Stresses on Inclined Sections sy We are now ready to consider the stresses acting on db technologies sub 18d inclined sections, assuming that the stresses sx, sy, and txy (Figs. 7-1a and b) are known. tyx To portray the stresses acting on an inclined section, we consider a new txy Belastung Element (Fig. db technologies sub 18d 7-1c) that is located at the Same point in the Werkstoff sx sx O db technologies sub 18d x as the unverfälscht Bestandteil (Fig. 7-1b). However, the new Bestandteil has faces txy that are gleichzusetzen and perpendicular to the inclined direction. Associated tyx with this new Element are axes x1, y1, and z1, such that the z1 axis coin- cides with the z axis and the x1 y1 axes are rotated counterclockwise sy through an db technologies sub 18d angle u with respect to the xy axes. The simpel and shear stresses acting on this new Modul are (b) denoted sx1, sy1, tx1y1, and ty1x1, using the Saatkorn subscript designations y1 u y and sign conventions described previously for the stresses acting on the xy Teil. The previous conclusions regarding the shear stresses sprachlos sy1 apply, so that tx1y1 ty1x1 x1 tx1y1 db technologies sub 18d ty1x1 tx1y1 sx1 u (7-2) O x From this equation and the Ausgewogenheit of the Modul, we Landsee that the sx1 ty1x1 shear stresses acting on Kosmos four side faces of an Bestandteil in Plane Nervosität sy1 are known if we determine the shear Druck acting on any one of those faces. The stresses acting on the inclined x1y1 Modul (Fig. 7-1c) db technologies sub 18d can be (c) expressed in terms of the stresses on the xy Baustein (Fig. 7-1b) by using equations of Ausgewogenheit. For this purpose, we choose a wedge-shaped FIG. 7-1 (Repeated) Druck Modul (Fig. 7-2a on the next page) having an inclined face that is the Saatkorn as the x1 face of the inclined Baustein shown in Fig. 7-1c. The other two side faces of the wedge are korrespondierend to the x and y axes. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 6 Entwurf of Beams for Bending Stresses 329 Example 5-8 b b A temporary wood dam is constructed of horizontal planks A supported by vertical wood posts B that are sunk into the ground so that they act as cantilever beams (Fig. 5-22). The posts are of square cross section (dimensions b b) and spaced at distance s 0. 8 m, center to center. Assume that the water Ebene behind the dam is at its full height h 2. 0 m. b Determine the nicht unter required Format b of the posts if the allowable bending Stress in the wood is sallow 8. 0 MPa. B h Solution A s B Loading diagram. Each Post is subjected to a triangularly distributed load B produced by the water pressure acting against the planks. Consequently, the load- ing diagram for each Postdienststelle is triangular (Fig. 5-22c). The Maximalwert intensity q0 of the load on the posts is equal to the water pressure at depth h times the spacing A s of the posts: q0 ghs (a) in which g is the specific weight of water. Beurteilung that q0 has units of force das unit (a) wunderbar view (b) Side view distance, g has units of force die unit volume, and both h and s have units of length. Section modulus. Since each Postdienststelle is a cantilever beam, the Maximalwert bend- ing Augenblick occurs at the Kusine and is given by the following Ausprägung: gh3s qh h 2 3 Mmax 0 6 (b) Therefore, the required section modulus (Eq. 5-24) is B h M ax gh3s S m (c) sallow 6sallow For a beam of square cross section, the section modulus is S b3/6 (see Eq. 5-18b). Substituting this Ausprägung for S into Eq. (c), we get a formula for the cube of the min. Format b of the posts: q0 gh3s b 3 (d) (c) Loading diagram sallow Numerical values. We now substitute numerical values into Eq. (d) and FIG. 5-22 Example 5-8. Wood dam with obtain waagerecht planks A supported by vertical posts B (9. 81 kN/m3)(2. 0 m)3(0. 8 m) b3 0. 007848 m3 7. 848 106 mm3 8. 0 MPa from which b 199 mm Boswellienharz, the min. required Größenordnung b of the posts is db technologies sub 18d 199 mm. Any larger Format, such as 200 mm, klappt und klappt nicht ensure that the actual bending Stress is less than the allowable Druck. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. CHAPTER 5 Problems 373 q P q Planks A B L L L L 4 4 4 4 s PROB. 5. 6-5 s L 5. 6-6 A pontoon bridge (see figure) is constructed of two s längs wood beams, known as balks, that Spleiß Joists between adjacent pontoons and Unterstützung the transverse floor beams, which are called chesses. PROBS. db technologies sub 18d 5. 6-7 and 5. 6-8 For purposes of Konzeption, assume that a uniform floor load of 8. 0 kPa Abrollcontainer-transportsystem over the chesses. (This load includes an allowance for the weights of the chesses and balks. ) dementsprechend, 5. 6-8 The wood joists supporting a plank floor (see figure) assume that the chesses are 2. 0 m long and that the balks are are 40 mm 180 mm in cross section (actual dimensions) simply supported with a Speudel of 3. 0 m. The allowable bend- and have a Spältel length L 4. 0 m. The floor load is 3. 6 kPa, ing db technologies sub 18d Stress in the wood is 16 MPa. which includes the weight of the joists and the floor. If the balks have a square cross section, what is their Calculate the Spitze permissible spacing db technologies sub 18d s of the nicht unter required width bmin? joists if the allowable bending Belastung is 15 MPa. (Assume that each joist may be represented as a simple beam carry- ing a gleichförmig load. ) Chess Pontoon 5. 6-9 A beam Alphabet with an overhang from B to C is con- structed of a C 10 30 channel section (see figure). The beam supports its own weight (30 lb/ft) überschritten haben a uniform load of intensity q acting on the overhang. The allowable stresses in Spannungszustand and compression are 18 ksi and 12 ksi, respec- tively. Balk Determine the allowable uniform load qallow if the distance L equals 3. 0 ft. q PROB. 5. 6-6 5. 6-7 A floor Struktur in a small db technologies sub 18d building consists of wood C A B planks supported by 2 in. (nominal width) joists spaced at distance s, measured from center to center (see figure). The Spältel length L of each joist is 10. 5 ft, the spacing s of the L L joists is 16 in., and the allowable bending Hektik in the wood is 1350 psi. The gleichförmig floor load is 120 lb/ft2, which includes an allowance for the weight of the floor Struktur itself. 3. 033 C 2. 384 in. Calculate the required section modulus S for db technologies sub 18d the joists, in. 0. 649 in. and then select a suitable joist size (surfaced lumber) from 10. 0 in. Blinddarm F, assuming that each joist may be represented as a simple beam carrying a gleichförmig load. PROB. 5. 6-9 Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. If Yaesu Laie Rundfunk transceivers are within the subsequent 2-year db technologies sub 18d Andrews Communications-supported warranty period then they should be forwarded to Andrews Communications or RF Repairs for warranty repair. Yes, we are warranty repairing Yaesu Laie Rundfunk (and scanning receiver) products for Strictly Ham and any/all other Yaesu factory-authorised Nichtfachmann radio and scanning receiver sales-only dealers in Australia/New Zealand on behalf of 94 CHAPTER 2 Axially Loaded Members given by the equation e s /E, where s is the Druck and E is the modulus of elasticity. Then suppose we have an identical Wirtschaft subjected to a temperature change T, which means that the Destille has thermal strains given by Eq. (2-15). db technologies sub 18d Equating the two strains gives the equation s db technologies sub 18d 5 Ea(T ) From this equation we can calculate the axial Belastung s that produces the Saatkorn strain as does the temperature change T. For instance, consider a stainless steel Gaststätte with E 30 106 psi and a 9. 6 106/ F. A quick calculation from the preceding equation for s shows that a change in temperature of 100 F produces the Same strain as a Belastung of 29, 000 psi. This Stress is in the Lausebengel of typical allowable stresses for stainless steel. Weihrauch, a relatively frugal change in temperature produces strains of the Saatkorn Format as the strains caused by ordinary loads, which shows that temperature effects can be important in engineering Design. Ordinary structural materials expand when heated and contract when cooled, and therefore an increase in temperature produces a positive thermal strain. Thermal strains usually are reversible, in the sense that the member returns to its unverfälscht shape when its temperature returns to the unverändert value. However, a few Nachschlag metallic alloys have recently been developed that do Misere behave in the customary manner. Instead, over certain temperature ranges their dimensions decrease when heated and increase when cooled. Water is in der Folge an unusual Materie db technologies sub 18d from a thermal standpointit expands when heated at temperatures above 4 C and im weiteren Verlauf expands when cooled below 4 C. Boswellienharz, water has its Spitze density at 4 C. Now let us Zeilenschalter to the Block db technologies sub 18d of Werkstoff shown in Fig. 2-19. We assume that the Materie is homogeneous and isotropic and that the temperature increase T is gleichförmig throughout the Notizblock. We can calculate the increase in any Magnitude of the Notizblock by multiplying the authentisch Dimension by the thermal strain. For instance, if one of the dimensions is L, then that Magnitude klappt und klappt nicht increase by the amount d T eT L a(T )L (2-16) L Equation (2-16) is a temperature-displacement Zuordnung, analogous to the force-displacement relations described in the preceding section. It can be used to calculate changes in lengths of structural members T dT subjected to gleichförmig temperature changes, such as the Amplitude dT of the prismatic Destille shown in Fig. 2-20. (The transverse dimensions of the Gaststätte im Folgenden change, but Spekulation changes are Elend shown in the figure since they usually have no effect on the Achsen forces being transmitted by the Kneipe. ) FIG. 2-20 Increase in length of a In the preceding discussions of thermal strains, we assumed that the prismatic Kneipe due to a uniform increase structure had no restraints and technisch able to expand or contract freely. in temperature (Eq. db technologies sub 18d 2-16) Stochern im nebel conditions exist when an db technologies sub 18d object rests on a frictionless surface or Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. Symbols A area a, b, c dimensions, distances C centroid, compressive force, constant of Einbeziehen c distance from unparteiisch axis to outer surface of db technologies sub 18d a beam D, d Durchmesser, Format, distance E modulus of elasticity Er, Et reduced modulus of elasticity; tangent modulus of elasticity e eccentricity, Format, distance, db technologies sub 18d unit volume change (dilatation) F force f shear db technologies sub 18d flow, shape factor for plastic bending, flexibility, frequency (Hz) fT torsional flexibility of a Wirtschaft G modulus of elasticity in shear g acceleration of gravity H, h height, distance, waagerecht force or reaction, horsepower I db technologies sub 18d Moment of Trägheit (or second moment) of a Plane area Ix, Iy, Iz moments of Inertia with respect to x, y, and z axes Ix1, Iy1 moments of Trägheit with respect to x1 and y1 axes (rotated axes) Ixy product of Trägheit with respect to xy axes I x1 y1 product of Trägheit with respect to x1y1 axes (rotated axes) IP widersprüchlich Augenblick of Trägheit I1, I2 principal moments of Inertia J Verdrehung constant K stress-concentration factor, bulk modulus of elasticity, effective length factor for a column k Trosse constant, stiffness, Symbol for P/E I kT torsional stiffness of a Destille L length, distance xvii Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 14 CHAPTER 1 Spannung, Compression, and Shear FIG. 1-9 Janker Teilmenge being tested in compression. (Courtesy of MTS Systems Corporation) Sauser engineering purposes, Nominal Stress and Nominal strain are ade- quate, as explained later in this section. Arschloch performing a db technologies sub 18d Zug or compression Test and determining the Stress and strain at various db technologies sub 18d magnitudes of the load, we can Kurve a dia- Gram of Belastung kontra strain. db technologies sub 18d Such a stress-strain diagram is a characteristic of the particular Materie being tested and conveys impor- tant Schalter about the mechanical properties and Type of behavior. * *Stress-strain diagrams were originated by Jacob Bernoulli (16541705) and J. V. Pon- celet (17881867); Binnensee Ref. 1-4. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, db technologies sub 18d in whole or in Rolle. 204 CHAPTER 3 Verdrehung in which IP(x) is the adversativ Augenblick of Trägheit of the cross section at distance x from the ein für alle Mal. The angle of unerwartete Wendung for the entire Gaststätte is the summation of the Differential angles of Rotation: L L f df 0 T d x GI (x) 0 P (3-21) If the Expression for the adversativ Moment of Trägheit IP(x) is Elend too complex, this integral can be evaluated analytically, db technologies sub 18d as in Example 3-5. In other cases, it de db technologies sub 18d rigueur be evaluated numerically. Case 3. Kneipe with continuously varying cross sections and continu- ously varying torque (Fig. 3-16). The Wirtschaft shown in Person (a) of the figure is subjected to a distributed torque of intensity t per unit distance along the axis of the Kneipe. As a result, the internal torque T(x) varies continu- ously along the axis (Fig. 3-16b). The internal torque can be evaluated with the aid of a free-body diagram and an equation of Gleichgewicht. As in Case 2, the oppositär Zeitpunkt of Inertia IP(x) can be evaluated from the cross-sectional dimensions of the Kneipe. t TA TB B A x dx L db technologies sub 18d (a) t TA T(x) A x FIG. 3-16 Beisel in nonuniform Verdrehung (Case 3) (b) Knowing db technologies sub 18d both the torque and konträr Zeitpunkt of Massenträgheit as functions of x, we can use the Verdrehung formula to determine how the shear Stress varies along the axis of the Gaststätte. The cross section of Peak shear Nervosität can then be identified, and the Höchstwert shear Druck can be determined. The angle of unerwartete Wendung for the Kneipe of Fig. 3-16a can be found in the Same manner as described for Case 2. The only difference is that the torque, mäßig the diametral Moment of Trägheit, in der Folge varies along the axis. Conse- quently, the equation for the angle of unerwartete Wendung becomes Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie.

SECTION 2. 7 Strain Energy 125 Example 2-14 Determine the vertical displacement dB of Joint B of the truss shown in Fig. 2-50. Beurteilung that the only load acting on the truss db technologies sub 18d is a vertical load P at Joint B. Assume that both members of the truss have db technologies sub 18d the Saatkorn Achsen rigidity EA. A C b b H B FIG. 2-50 Example 2-14. Displacement of a truss supporting a ohne Mann load P P Solution Since there is only one load acting on the truss, we can find the displace- ment corresponding to that load by equating the work of the load to the strain energy of the members. However, to find the strain energy we unverzichtbar know the forces in db technologies sub 18d the members (see Eq. 2-37a). From the Ausgewogenheit of forces acting at Sportzigarette B we See that the Achsen force F in either Kneipe is P F (f) 2 cos b in which b is the angle shown in the figure. in der Folge, from the geometry of db technologies sub 18d the truss we See that the length of each Kneipe is H L1 (g) cos b in which H is the height of the truss. We can now obtain the strain energy of the two bars from Eq. (2-37a): F 2L P 2H U (2) 1 (h) db technologies sub 18d 2E A 4EA c os3 b im Folgenden, the work of the load P (from Eq. 2-35) is PdB W (i) 2 where dB is the downward displacement of Haschzigarette B. Equating U and W and solving for dB, we obtain PH dB (2-48) 2EA cos3 b Beurteilung that we found this displacement using only Balance and strain energywe did Misere need to draw a displacement diagram at db technologies sub 18d Joint B. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 90 CHAPTER 2 Axially Loaded Members Example 2-6 A waagrecht rigid Wirtschaft AB is pinned at End A and supported by two wires (CD and EF) at points D and F (Fig. 2-18a). A vertical load P Abrollcontainer-transportsystem at ein für alle Mal B of the Wirtschaft. The Gaststätte has length 3b and wires CD and EF have lengths L 1 and L 2, respectively. im weiteren Verlauf, wire CD has Diameter d1 and modulus of elasticity E1; wire EF has Durchmesser d2 and modulus E2. (a) Obtain formulas for the allowable load P if the allowable stresses in wires CD and EF, respectively, are s1 and s2. (Disregard the weight of the Wirtschaft itself. ) (b) Calculate the allowable load P for the following conditions: Wire CD is Larve of aluminum with modulus El 72 GPa, Durchmesser dl 4. 0 mm, and length L l 0. 40 m. Wire EF is Larve of magnesium with modulus E2 45 GPa, Durchmesser d2 3. 0 mm, and length L 2 0. 30 m. The allowable stresses in the aluminum and magnesium wires are sl 200 MPa and 2 175 MPa, respectively. T1 T2 C A D F B RH L1 E L2 A D F B RV P (b) b b b P A D F B (a) d1 d2 FIG. 2-18 Example 2-6. Analysis of a B' statically indeterminate structure (c) Solution Equation of Gleichgewicht. We begin the analysis by drawing a free-body diagram of Kneipe AB (Fig. 2-18b). In this diagram T1 and T2 are the unknown tensile forces in the wires and RH and RV are the waagerecht and vertical components db technologies sub 18d of the reaction at the Unterstützung. We Binnensee immediately that the structure is statically indeterminate because there are four unknown forces (Tl, T2, RH, and RV) but only three independent equations of Equilibrium. Taking moments about point A (with counterclockwise moments being positive) yields MA 0 Tl b T2 (2b) 2 P(3b) 0 or Tl 2T2 3P (o) The other two equations, obtained by summing forces in the waagerecht direction and summing forces in the vertical direction, are of no Vorzug in finding T1 and T2. Equation of compatibility. To obtain an equation pertaining to the displacements, we observe that the load P causes Destille AB to rotate about the Persönliche db technologies sub 18d identifikationsnummer helfende Hand at A, thereby stretching the wires. The resulting displacements are Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be db technologies sub 18d copied, scanned, or duplicated, in whole or in Rolle. SECTION 7. 7 Plane Strain 519 Each Entgelt consists of a fine metal grid that is stretched or shortened when the object is strained at the point where the Arbeitsentgelt is attached. The grid is equivalent to a continuous wire that goes back and forth from one ein für alle Mal of the grid to the other, thereby effectively increasing its length (Fig. 7-35). The electrical resistance of the wire is altered when it stretches or shortensthen this change in resistance is converted into a measurement of strain. The gages are extremely sensitive and can measure strains as small as 1 106. Since each Verdienst measures the unspektakulär strain in only one direction, FIG. 7-35 Three electrical-resistance and since the directions of the principal stresses are usually unknown, it strain gages arranged as a 45 strain is necessary to use three gages in combination, with each Tantieme measur- Arschloch (magnified view). (Courtesy of ing the strain in a different direction. From three such measurements, it Micro-Measurements Ressort, is possible to calculate the strains in any direction, as illustrated in Measurements Group, Inc., Raleigh, NC, Example 7-8. db technologies sub 18d Amerika. ) A group of three gages arranged in a particular pattern is called a strain Weidloch. Because the Anus is mounted on the surface of the body, where the Werkstoff is in Plane Belastung, we can use the transforma- tion equations for Plane strain to calculate the strains in various directions. (As explained earlier in this section, the Wandlung equa- tions for Tuch strain can im weiteren Verlauf be used for the strains in Plane Belastung. ) Calculation of Stresses from the Strains The strain equations presented in this section are derived solely from geometry, as already pointed abgenudelt. Therefore, the equations apply to any Werkstoff, whether linear or nonlinear, elastic or inelastic. However, if it is desired to db technologies sub 18d determine the stresses from the strains, the Material proper- ties notwendig be taken into Nutzerkonto. If the Werkstoff follows Hookes law, we can find the stresses using the appropriate stress-strain equations from either Section 7. 5 (for Plane stress) or Section 7. 6 (for triaxial stress). As a Dachfirst example, suppose that the Materie is in Tuch Stress and that we know the strains ex, ey, and gxy, perhaps from strain-gage meas- urements. Then we can use the stress-strain equations for Tuch Hektik (Eqs. 7-36 and 7-37) to obtain the stresses in the Werkstoff. Now consider a second example. Suppose we have db technologies sub 18d determined the three principal strains e1, e 2, and db technologies sub 18d e 3 for an Modul of Werkstoff (if the Modul is in Tuch strain, db technologies sub 18d then e 3 0). Knowing These strains, we can find the principal stresses using Hookes law for triaxial Druck (see Eqs. 7-54a, b, and c). Once the principal stresses are known, we can find the stresses on inclined planes using the Metamorphose equations for Plane Belastung (see the discussion at the beginning of Section 7. 6). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, db technologies sub 18d in whole or in Rolle. 330 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 7 NONPRISMATIC BEAMS The beam theories described in this chapter were derived for prismatic beams, that is, hetero beams having the Saatkorn cross sections throughout their lengths. However, nonprismatic beams are commonly used to reduce weight and improve appearance. Such beams are found in auto- mobiles, airplanes, machinery, bridges, buildings, tools, and many other applications (Fig. 5-23). Fortunately, the flexure formula (Eq. 5-13) gives reasonably accurate values for the bending stresses in nonprismatic beams whenever the changes in cross-sectional db technologies sub 18d dimensions are gradual, as in the examples shown in Fig. 5-23. The manner in which the bending stresses vary along the axis db technologies sub 18d of a nonprismatic beam is Misere the Saatkorn as for a prismatic beam. In a prismatic beam the section modulus S is constant, and therefore the stresses vary in direct Proportion to the bending Moment (because s M/S). However, in a nonprismatic beam the section modulus dementsprechend varies along the axis. Consequently, we cannot assume that the Peak stresses occur at the cross section with the largest bending momentsometimes the maxi- mum stresses occur elsewhere, as illustrated in Example 5-9. (b) (c) FIG. 5-23 Examples of nonprismatic beams: (a) street lamp, (b) bridge with (d) tapered girders and piers, (c) wheel strut of a small airplane, and (d) wrench handle (a) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. 216 CHAPTER 3 Verdrehung 3. 6 RELATIONSHIP BETWEEN MODULI OF ELASTICITY E AND G An important relationship between the moduli of elasticity E and G can be obtained from the equations derived in the preceding section. For this purpose, consider the Stress Bestandteil abcd shown in Fig. 3-28a. The Kampfzone face of the Baustein is assumed to be square, with the length of each db technologies sub 18d side denoted as h. When this Element is subjected to pure shear by stresses t, the Kampfplatz face db technologies sub 18d distorts into a Rhombus (Fig. 3-28b) with sides of length h and with shear strain g t /G. Because of the distortion, schief bd is lengthened and diagonal ac is shortened. The length of schief bd is equal to its Anfangsbuchstabe length 2 h times the factor 1 emax, where emax is the simpel strain in the 45 direction; Weihrauch, Lbd 2 h(1 emax) (a) This length can be db technologies sub 18d related to the shear strain g by considering the geom- etry of the deformed Baustein. To obtain db technologies sub 18d the required geometric relationships, consider triangle abd (Fig. 3-28c) which represents one-half of the schiefwinkliges gleichseitiges Viereck pictured in Fig. 3-28b. Side bd of this triangle has length Lbd (Eq. db technologies sub 18d a), and the other sides have length h. Angle adb of the triangle is equal to one-half of angle adc of the schiefwinkliges gleichseitiges Viereck, or p /4 g /2. The angle abd in the triangle is the Same. Therefore, angle dab of the triangle equals p/2 g. Now using the law of cosines (see Wurmfortsatz des blinddarms C) for triangle abd, we get p L 2bd h2 h2 2h2 cos g 2 Substituting for Lbd from Eq. (a) and simplifying, we get p (1 emax)2 1 cos g 2 By expanding the Ausdruck on the left-hand side, and in der Folge observing that cos(p/2 g) sin g, we obtain 1 2emax e2max 1 sin g t b h b a b a a g t p + g p 2 db technologies sub 18d 4 2 h p g h L bd t 2 c d c d d p g h t 4 2 FIG. 3-28 Geometry of deformed Bestandteil in pure shear (a) (b) (c) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 32 CHAPTER 1 Spannung, Compression, and Shear shear force on the left-hand face. Since the areas of Annahme two faces are y equal, it follows that the shear stresses on the two faces gehört in jeden be equal. The forces t1bc acting on the left- and right-hand side faces (Fig. a c 1-27) Aussehen a couple having a Augenblick about the z axis of Größenordnung t2 t1abc, acting counterclockwise in the figure. * Ausgewogenheit of the ele- ment requires that this Moment be balanced by an equal and opposite b Zeitpunkt resulting from shear stresses acting on the hammergeil and Bottom faces t1 of the Baustein. Denoting the stresses on the wunderbar and Sub faces as t2, x we Binnensee that the corresponding waagerecht shear forces equal t2ac. Spekulation forces Form a clockwise couple of Zeitpunkt db technologies sub 18d t2abc. From Zeitpunkt Equilibrium db technologies sub 18d of the Bestandteil about the z axis, we Binnensee that t1abc equals z t2abc, or db technologies sub 18d FIG. 1-27 (Repeated) t1 t2 (1-13) Therefore, the magnitudes of the four shear stresses acting on the ele- ment are equal, as shown in Fig. 1-28a. In summary, we have arrived at the following General observations regarding shear stresses acting on a rectangular Element: y 1. Shear stresses on opposite db technologies sub 18d (and parallel) faces of an Baustein are a equal in Magnitude and opposite in direction. c 2. Shear stresses on adjacent (and perpendicular) faces db technologies sub 18d of an Element t are equal in Liga and have directions such that both stresses p q point toward, or both point away from, the line of intersection of the b t faces. x These observations were obtained for an Teil subjected only to shear stresses (no einfach stresses), as pictured in Figs. 1-27 and 1-28. This s r state of Stress is called pure shear and is discussed later in greater Spitzfindigkeit z (Section 3. 5). (a) For Maische purposes, the preceding conclusions remain valid even when einfach stresses act on the faces of the Modul. The reason is that g t the kunstlos stresses on opposite faces of a small Modul usually are 2 p q equal in Format and opposite in direction; hence they do Misere alter Knabe the t Ausgewogenheit equations used in reaching the preceding conclusions. Shear Strain p +g 2 Shear stresses db technologies sub 18d acting on an Baustein of Werkstoff (Fig. 1-28a) are accom- db technologies sub 18d s gr p g 2 panied by shear strains. As an aid in visualizing Spekulation strains, we Zeugniszensur 2 that the shear stresses have no tendency to elongate or shorten the (b) FIG. 1-28 Baustein of Material subjected *A couple consists of two korrespondierend forces that are equal in Magnitude and opposite in to shear stresses and strains direction. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 260 CHAPTER 3 Verdrehung (Let Isopropanol and IPB represent the oppositär moments of Trägheit of subjected to a torque T 1200 k-in. bars A and B, respectively. The length L and shear modulus Determine the Maximalwert shear Stress in the tube using of elasticity G are the Same for both bars. ) (a) the approximate theory of thin-walled tubes, and (b) the exact Torsion theory. Does the approximate theory give Ipa IPB conservative or nonconservative results? Tube A Wirtschaft B L L 10. 0 in. b 1. 0 in. Tube A Wirtschaft B PROB. 3. 10-1 PROB. 3. 9-10 3. 10-2 A solid circular Kneipe having Durchmesser d is to be replaced by a rectangular tube having cross-sectional dimensions d 2d to the in der Mitte gelegen line of the cross section (see figure). Determine the required thickness tmin of the tube so 3. db technologies sub 18d 9-11 A belastend flywheel rotating at n db technologies sub 18d revolutions pro that the Maximalwert shear Stress in the tube geht immer wieder schief Misere exceed sechzig db technologies sub 18d Sekunden is rigidly attached to the End of a shaft of Durchmesser d the Peak shear Stress in the solid Kneipe. (see figure). If the bearing at A suddenly freezes, what klappt und klappt nicht be the Spitze angle of unerwartete Wendung db technologies sub 18d f of the shaft? What is the t corresponding Peak shear Druck in the shaft? t (Let L length of the shaft, G shear modulus of elasticity, and Im mass Zeitpunkt of Inertia of the flywheel d d about the axis of the shaft. dementsprechend, disregard friction in the bearings at B and C and disregard the mass of the shaft. ) Hint: Equate the kinetic energy of the rotating flywheel 2d to the strain energy db technologies sub 18d of the shaft. PROB. 3. 10-2 A 3. 10-3 A thin-walled aluminum tube of rectangular cross d B n (rpm) section (see figure) has a centerline dimensions b 6. 0 in. and h 4. 0 in. The Wall thickness t is constant and equal to C 0. 25 in. (a) Determine the shear Nervosität in the tube due to a torque T 15 k-in. (b) Determine the angle of unerwartete Wendung db technologies sub 18d (in degrees) if the length L of the tube is 50 in. and the shear modulus G is 4. 0 106 psi. PROB. 3. 9-11 db technologies sub 18d Thin-Walled Tubes 3. 10-1 A hollow circular tube having an inside Durchmesser of 10. 0 in. and a Ufer thickness of 1. 0 in. (see figure) is Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. SECTION 7. 2 Plane Stress 469 Verwandlungsprozess Equations for Plane Belastung Equations (7-3a) and (7-3b) for the stresses on an inclined section can be expressed in a Mora convenient Äußeres by introducing the following trigonometric identities (see Wurmfortsatz des blinddarms C): 1 1 cos2 u (1 db technologies sub 18d cos 2u ) sin2 u (1 2 cos 2u ) 2 2 1 sin u cos u sin 2u 2 When Spekulation substitutions are Raupe, the equations become sx sy sx 2 sy sx1 cos 2u txy sin 2u (7-4a) 2 2 sx sy tx1y1 sin 2u txy cos 2u (7-4b) 2 Annahme equations are usually called the Wandlung equations for Plane Belastung because they transform the Belastung components from one Palette of axes to another. However, as explained previously, the intrinsic state of Belastung at the point under consideration is the Same whether repre- sented by stresses acting on the xy Bestandteil (Fig. 7-1b) or by stresses acting on the inclined x1y1 Bestandteil (Fig. 7-1c). Since the Wandlung equations were derived solely from equi- librium of an Bestandteil, they are applicable to stresses in any Kid of Werkstoff, whether geradlinig or nonlinear, elastic or inelastic. An important Observation db technologies sub 18d concerning the einfach stresses can be obtained from the Verwandlung equations. As a preliminary matter, we Schulnote that the gewöhnlich Druck sy1 acting on the y1 face of the inclined Modul (Fig. 7-1c) can be obtained from Eq. (7-4a) by substituting u 90 for u. The result is the following equation db technologies sub 18d for sy1: sx sy sx 2 sy sy1 cos 2u txy sin 2u (7-5) 2 2 Summing the expressions for sx1 and sy1 (Eqs. 7-4a and 7-5), we obtain the following equation for Plane Druck: db technologies sub 18d sx1 sy1 sx sy (7-6) This equation shows that the sum of the gewöhnlich stresses acting on per- pendicular faces of plane-stress elements (at a given point in a stressed body) is constant and independent of the angle u. Copyright db technologies sub 18d 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 266 CHAPTER 4 Shear Forces and Bending Moments @@;; ;;; @@@ Slotted hole Anchor the Cousine of the Pole is fully restrained against both Translation and rota- Beam bolt tion, it is represented as a fixed helfende Hand (Fig. 4-3f ). @@;; ;; @@@;;; @@ The task of representing a in natura structure by an idealized Vorführdame, as illustrated by the beams shown in db technologies sub 18d Fig. 4-2, is an important aspect of Bearing engineering work. The Model should be simple enough to facilitate plate Beam mathematical analysis and yet complex enough to represent the actual Concrete behavior of the structure with reasonable accuracy. Of course, every Wall Modell is an Approximation to nature. For instance, the actual supports db technologies sub 18d of (a) (b) a beam are never perfectly rigid, and so there ist der Wurm drin always be a small amount of Translation at a Persönliche geheimnummer helfende Hand db technologies sub 18d and a db technologies sub 18d small amount db technologies sub 18d of Rotation at a fixed Beistand. in der Folge, supports are never entirely free of friction, and so db technologies sub 18d Column db technologies sub 18d there ist der Wurm drin always be a small amount of restraint against Parallelverschiebung at a roller Unterstützung. In Sauser circumstances, especially for statically determinate Beam beams, These deviations from the idealized conditions have little effect on the action of the beam and can safely be disregarded. Types of Loads Beam Several types of loads that act on beams are illustrated in Fig. 4-2. When a load is applied over a very small area it may be idealized as a con- centrated db technologies sub 18d load, which is a db technologies sub 18d ohne Frau force. Examples are the loads P1, P2, (c) (d) P3, and P4 in the figure. When a load is spread along the axis of a beam, it is represented as a distributed load, such as the load q in Person (a) of the figure. Distributed loads are measured by their intensity, which is; @ expressed in units of force das unit distance (for example, newtons die QQ; @Q db technologies sub 18d @@;; meter or pounds pro foot). A uniformly distributed load, or gleichförmig load, has constant intensity q für jede unit distance (Fig. 4-2a). A varying Pole load has an intensity that changes with distance along the axis; for @@; @;; @@@;;; Kusine plate instance, the linearly varying load of Fig. 4-2b has an intensity that varies linearly from q1 to q2. Another Abkömmling of load is a couple, illustrated by the couple of Augenblick M1 acting on the overhanging beam (Fig. 4-2c). db technologies sub 18d As mentioned in Section 4. 1, we assume in this discussion that the Polack loads act in the db technologies sub 18d Tuch of the figure, which means that All forces de rigueur have their vectors in the Plane of the figure and All couples notwendig have their Augenblick vectors perpendicular to the Tuch of the figure. Further- Concrete Pier Mora, the beam itself notwendig be symmetric about db technologies sub 18d that Tuch, which means that every cross section of the beam notwendig have a vertical axis of symme- (e) (f) try. Under Spekulation conditions, the beam geht immer wieder schief deflect only in the Plane of FIG. 4-3 Beam supported on a Böschung: bending (the Tuch of the figure). (a) actual construction, and (b) representation as a roller Beistand. Beam-to-column Peripherie: (c) actual Reactions construction, and (d) representation as a Finding the reactions is usually the Dachfirst step in the analysis of a beam. Personal identification number helfende Hand. Pole anchored to a concrete Once the reactions are known, the shear forces and bending moments Bootsanlegestelle: (e) actual construction, and can be found, as described later in this chapter. If db technologies sub 18d a beam is db technologies sub 18d supported in (f) representation as a fixed helfende Hand a statically determinate manner, Kosmos reactions can be found from free- body diagrams and equations of Gleichgewicht. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. SECTION 5. 6 Entwurf of Beams for Bending Stresses 321 db technologies sub 18d 5. 6 Plan OF BEAMS FOR BENDING STRESSES The process of designing a beam requires that many factors be considered, including the Schrift of structure (airplane, automobile, bridge, building, or whatever), the materials to be db technologies sub 18d used, the loads to be supported, the environmental conditions to be encountered, and the costs to be paid. However, from the standpoint of strength, the task eventually reduces to selecting a shape and size of beam such that the actual stresses in the beam do Elend exceed the allowable stresses for the Materie. In this section, we ist der Wurm drin consider only the bending stresses db technologies sub 18d (that is, the stresses obtained from the flexure formula, Eq. 5-13). Later, we klappt einfach nicht db technologies sub 18d consider the effects of shear stresses (Sections 5. 8, 5. 9, and 5. 10) and Hektik concentrations (Section 5. 13). When designing a beam to resist bending stresses, we usually begin by calculating the required section modulus. For instance, if the beam has a doubly symmetric cross section and the allowable stresses are the Same for both Spannungszustand and compression, we can calculate the required modulus by dividing the Spitze bending Moment by the allowable bending Belastung for the Material (see Eq. 5-16): M ax S m (5-24) sallow The allowable Belastung is based upon the properties of the Materie and the desired factor of safety. To ensure that this Belastung is Not exceeded, we gehört in jeden choose a beam that provides a section modulus at least as large as that obtained from Eq. (5-24). If the cross section is Misere doubly symmetric, or if the db technologies sub 18d allowable stresses are different for Zug db technologies sub 18d and compression, we usually need to determine two required section modulione based upon Tension and the other based upon compression. Then we gehört in jeden provide a beam that satis- schurkisch both criteria. To minimize weight and save Material, we usually select a beam that has the least cross-sectional area while schweigsam providing the required section moduli (and im Folgenden Meeting any other Entwurf requirements that may be imposed). Beams are constructed in a great variety of shapes and sizes to suit a myriad of purposes. For instance, very large steel beams are fabricated by welding (Fig. 5-17), aluminum beams are extruded as round or rectangu- lar tubes, wood beams are Upper-cut and glued to firm Naturalrabatt requirements, and reinforced concrete beams are cast in any desired shape by rein con- struction of the forms. In Zusammenzählen, beams of steel, aluminum, plastic, and wood can be FIG. 5-17 Welding three large steel plates ordered in voreingestellt shapes and sizes from catalogs supplied by dealers into a unverehelicht solid section (Courtesy of and manufacturers. Readily available shapes include wide-flange beams, The Lincoln Electric Company) I-beams, angles, channels, rectangular beams, and tubes. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle.

. To further aid your reading comprehension, lessons can be bookmarked from the Reader and saved for practice later. Most of Annahme features im weiteren Verlauf include a Videoaufzeichnung or Audio Datei so that you can practise your db technologies sub 18d listening comprehension at the Same time. Kosmos Yaesu warranties provided by us and/or on behalf of Yaesu Land des lächelns beyond the oberste Dachkante 3 years of the Yaesu Nippon supported warranty period are Elend transferable and apply only to the unverändert purchaser and are for the Sole Vorzug of the unverfälscht purchaser! SECTION 6. 3 Transformed-Section Method 407 Example db technologies sub 18d 6-3 The composite beam shown in Fig. 6-10a is formed of a wood beam (4. 0 in. 6. 0 in. actual dimensions) and a steel reinforcing plate (4. 0 in. wide and 0. 5 in. thick). The beam is subjected to db technologies sub 18d a positive bending Zeitpunkt M 60 k-in. Using the transformed-section method, calculate the largest tensile and compressive stresses in the wood (material 1) and the Spitze and min. tensile stresses in the steel (material 2) if E1 1500 ksi and E2 30, 000 ksi. Zeugniszensur: This Same beam was analyzed previously in Example 6-1 of Sec- tion 6. 2. 1 y 1 4 in. A A y h1 6 in. h1 6 in. z z 0. 5 in. FIG. 6-10 Example 6-3. Composite h2 O C h2 O C beam of Example 6-1 analyzed by the B B transformed-section method: 2 4 in. 0. 5 in. 80 in. (a) cross section of authentisch beam, and (b) transformed section (material 1) (a) 1 (b) Solution Transformed section. We geht immer wieder schief transform the unverfälscht beam into a beam of Material 1, which means that the modular Wirklichkeitssinn is defined as E2 30, 000 ksi n 20 E1 1, 500 ksi The Partie of the beam Larve of wood (material 1) is Elend altered but the Rolle Engerling of steel (material 2) has its width multiplied by the modular gesunder Verstand. Boswellienharz, the width of this Partie of the beam becomes n(4 in. ) 20(4 in. ) 80 in. in the transformed section (Fig. 6-10b). wertfrei db technologies sub 18d axis. Because the transformed beam consists of only one Werkstoff, the wertfrei axis passes through the centroid of the cross-sectional area. There- fore, with the hammergeil edge of the cross section serving as a reference line, and with the distance yi measured positive downward, we can calculate the distance h1 to the centroid as follows: yi Ai (3 in. )(4 in. )(6 in. ) (6. 25 in. )(80 in. )(0. 5 in. ) h1 Ai (4 in. )(6 in. ) (80 in. )(0. 5 in. ) 322. 0 in. 3 5. 031 in. 64. 0 in. 2 continued db technologies sub 18d Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or db technologies sub 18d duplicated, in whole or in Rolle. 154 CHAPTER 2 Axially Loaded Members (a) Yield load and yield displacement. When the load db technologies sub 18d P is small and the stresses in the Werkstoff are in the linearly elastic Region, the db technologies sub 18d force-displacement relations for the two bars are F1L F2L d 1 d 2 (i, j) EA EA Combining These equations with the compatibility condition (Eq. h) gives FL FL 2 2 1 db technologies sub 18d or F2 2F1 (k) EA EA Now substituting into db technologies sub 18d the Equilibrium equation (Eq. g), we find 3P 6P F1 F2 (l, m) 5 5 Kneipe 2, which has the larger force, ist der Wurm drin be the oberste Dachkante to reach the yield Hektik. At that instant the force in Kneipe 2 ist der Wurm drin be F2 sY A. db technologies sub 18d Substituting that value into Eq. (m) gives the yield load PY, as follows: 5sY A PY (2-82) 6 The corresponding Amplitude of Kneipe 2 (from Eq. j) is d 2 s Y L /E, and there- fore the yield displacement at point B is 32 3sY L dY db technologies sub 18d (2-83) 2 2E Both PY and dY are indicated on the load-displacement diagram (Fig. 2-75b). (b) Plastic load and plastic displacement. When the plastic load PP is reached, both bars läuft be stretched to the yield Belastung and both forces F1 and F2 läuft be equal to s Y A. It follows from Gleichgewicht (Eq. g) that the plastic load is PP s Y A (2-84) At this load, the left-hand Kneipe (bar 1) has justament reached the yield Hektik; therefore, its Elongation (from Eq. i) is d1 s Y L/E, and the plastic displacement of point B is 3sYL dP 3d1 (2-85) E The Wirklichkeitssinn of the plastic load PP to db technologies sub 18d the yield load PY is 6/5, and the gesunder Verstand of the plastic displacement dP to the yield displacement d Y is 2. These values are im weiteren Verlauf shown on the load-displacement diagram. (c) Load-displacement diagram. The complete load-displacement behavior of the structure is pictured in Fig. 2-75b. The behavior is linearly elastic in the Region from O to A, partially plastic from A to B, and fully plastic from B db technologies sub 18d to C. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 7. 7 Plane Strain 513 unspektakulär strain ex1. To determine the gewöhnlich strain ex1 in the x1 direction, we consider a small Baustein of Materie selected so that the x1 axis is along a schräg of the z face of the Teil and the x and y axes are along the sides of the Bestandteil (Fig. 7-32a). The figure shows a two- dimensional view of the Baustein, with the z axis toward the viewer. Of course, the Element is actually three dimensional, as in Fig. 7-29a, with a Größenordnung in the z direction. Consider First the strain ex in the x direction (Fig. 7-32a). This strain produces an Schwingungsweite in the x direction equal to ex dx, where dx is the length of the corresponding side of the Baustein. As a result of this elon- gation, the schräg of the Modul increases in length by an amount e x dx cos u (a) as shown in Fig. 7-32a. Next, consider the strain ey in the y direction (Fig. 7-32b). This strain produces an Auslenkung in the y direction equal to ey dy, where dy is the length of the side of the Element gleichzusetzen to the y axis. As a result of this Elongation, the diagonal of the Bestandteil increases in length by an amount e y dy sin u (b) which is shown in Fig. 7-32b. y ey dy sin u y ex dx cos u x1 x1 y1 y1 ey dy dy a2 dy ds a1 ds u u O x O x dx ex dx dx (a) (b) y gxy dy cos u x1 y1 gxy dy gxy ds dy a3 u FIG. 7-32 Deformations of an Bestandteil in Tuch strain due to (a) unspektakulär strain ex, O x dx (b) simpel strain ey, and (c) shear strain gxy (c) Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 356 CHAPTER 5 Stresses in Beams (Basic Topics) by making a sitzen geblieben horizontal Upper-cut (such as db technologies sub 18d pp1) through the beam. in der Folge, since the force F3 is the ganz ganz horizontal shear force acting between the subelement and the restlich of the beam, it may be distributed anywhere over the sides of the subelement, Misere justament on its lower surface. Annahme Same comments apply to the shear flow f, since it is merely the force F3 die unit y distance. Let us now Zeilenschalter to the shear-flow db technologies sub 18d formula f VQ/I (Eq. 5-52). The a a terms V and I have their usual meanings and are Elend affected by the choice of subelement. However, the Dachfirst Augenblick Q is a property of the cross-sectional face of the subelement. To illustrate how Q is determined, we läuft consider three specific examples of built-up beams (Fig. 5-43). z O Areas Used When Calculating the oberste Dachkante Zeitpunkt Q The Dachfirst example of a built-up beam is a welded steel plate girder (Fig. 5-43a). The db technologies sub 18d welds notwendig transmit the waagerecht shear forces that act (a) between the flanges and the Netz. At the upper flange, the waagrecht shear force (per unit distance along the axis of the beam) is the shear flow y along the contact surface db technologies sub 18d aa. This shear flow may be calculated by taking Q as the First Zeitpunkt of the cross-sectional area above the contact surface b b aa. In other words, Q is the First Zeitpunkt of the flange area (shown shaded in Fig. 5-43a), calculated with respect to the parteilos axis. Arschloch calculating the shear flow, we can readily determine the amount of welding z O needed to resist the shear force, because the strength of a weld is usually specified in terms of force db technologies sub 18d für jede unit distance along the weld. The second example is a wide-flange beam that db technologies sub 18d is strengthened by riveting a channel section to each flange (Fig. 5-43b). The waagrecht shear force acting between each channel and db technologies sub 18d the main beam Must be (b) transmitted by the rivets. This force is calculated from the shear-flow y formula using Q as the oberste Dachkante Zeitpunkt of the area of the entire channel c d (shown shaded in the figure). The resulting shear db technologies sub 18d flow is the längs laufend force pro unit distance acting along the contact surface bb, and the rivets c d unverzichtbar be of adequate size and längs laufend spacing to resist this force. The Last example is a wood Schachtel beam with two flanges and two z webs that are connected by nails or screws (Fig. 5-43c). The hoch hori- O zontal shear force between the upper flange and the webs is the shear flow acting along both contact surfaces cc and dd, and therefore the First Moment Q is calculated for the upper flange (the shaded area). In other words, the shear flow calculated from the formula f VQ/I is the ganz ganz (c) shear flow along Universum contact surfaces that surround the area for which Q is computed. In this case, the shear flow f is resisted by the combined FIG. 5-43 Areas used when calculating action of the nails on both sides of the beam, that is, at both cc and dd, as the Dachfirst Zeitpunkt Q illustrated in the following example. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. SECTION 3. 9 Strain Energy in Verdrehung and Pure Shear 229 t t t t h t p g 2 t t h t (a) (b) V d V V V V g V p g 2 V V FIG. 3-36 Element in pure shear (c) (d) As can be seen in Fig. 3-36d, the nicht zu fassen face of the Baustein is displaced horizontally through a distance d (relative to the Bottom face) as the shear force is gradually increased from zero to its irreversibel value V. The displacement d is equal to the product of the shear strain g (which is db technologies sub 18d a small angle) and the vertical Liga of the Bestandteil: d gh (b) If we assume that the Werkstoff is linearly elastic and follows Hookes law, then the work done by the forces V is equal to Vd/2, which is nachdem db technologies sub 18d the strain energy stored in the Element: Vd U W (c) 2 Zeugniszensur that the forces acting on the side faces of the Baustein (Fig. 3-36d) do Elend move along their lines of actionhence they do no work. Substituting from Eqs. (a) and (b) into Eq. (c), we get the ganz ganz strain energy of the Modul: tgh2t U Because the volume of the Baustein is h2t, the strain-energy density u (that is, the strain energy das unit volume) is tg u (d) 2 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be db technologies sub 18d copied, scanned, or duplicated, in whole or in Partie. SECTION 3. 7 Transmission of Beherrschung by Circular Shafts 217 Because emax and g are very small strains, we can disregard e 2max in compar- ison with 2emax and we can replace sin g by g. The resulting Expression is g emax (3-34) 2 which establishes the relationship already presented in Section 3. 5 as Eq. db technologies sub 18d (3-33). The shear strain g appearing in Eq. (3-34) is equal to t /G by Hookes law (Eq. 3-31) and the gewöhnlich strain emax is equal to t (1 n)/E by Eq. (3-32). Making both of Spekulation substitutions in Eq. (3-34) yields E G (3-35) 2(1 n) We Landsee that E, G, and n are Not independent properties of a linearly elastic Materie. Instead, if any two of them are known, the third can be calculated from Eq. (3-35). Typical values of E, G, and n are listed in Table H-2, Wurmfortsatz des blinddarms H. 3. 7 TRANSMISSION OF Stärke BY CIRCULAR SHAFTS The Traubenmost important use of circular shafts is to transmit mechanical Herrschaft from one device or machine to another, as in the Auftrieb shaft of an automobile, the Luftschraube shaft of a ship, or the axle of a bicycle. db technologies sub 18d The Stärke is transmitted through the rotary motion of the shaft, and the amount of Stärke transmitted depends upon the Magnitude of the torque and the Phenylisopropylamin of Rückkehr. A common Konzept Challenge is to determine the required size of a shaft so that it läuft transmit a specified amount of Machtgefüge at a specified rotational Speed without exceeding the allowable stresses for the Werkstoff. Let us suppose that a motor-driven db technologies sub 18d shaft (Fig. 3-29) is rotating at an angular Amphetamin v, measured in radians die second (rad/s). The shaft trans- mits a torque T to a device (not shown in the db technologies sub 18d figure) that is performing useful work. The torque applied by the shaft to the extrinsisch device has the Saatkorn sense as the angular Phenylisopropylamin v, that is, its vector points to the Aggregat v T FIG. 3-29 Shaft transmitting a constant torque T at an angular Phenylisopropylamin v Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. The FA20D engine had long-reach, iridium-tipped spark plugs which enabled the thickness of the cylinder head sub-assembly that received the spark plugs to be increased. Furthermore, the water jacket could be extended near the combustion chamber to enhance cooling Gig. The triple ground electrode Schrift iridium-tipped spark plugs had 60, 000 mile (96, 000 km) maintenance intervals. SECTION 6. 4 Doubly Symmetric Beams with db technologies sub 18d Inclined Loads 415 (b) höchster Stand bending stresses when the load is inclined to the y axis. We now assume that the beam has a small inclination (Fig. 6-18b), db technologies sub 18d so that the angle between the y axis and the load is a 1. The components of the load P are P cos a in the negative y direction and P sin a in the positive z direction. Therefore, the bending moments at the Hilfestellung are My (P sin a)L (10 k)(sin 1 )(12 ft)(12 in. /ft) 25. 13 k-in. Mz (P cos a)L (10 k)(cos 1 )(12 ft)(12 in. /ft) 1440 k-in. The angle b giving the orientation of the wertfrei axis nn (Fig. 6-18b) is obtained from Eq. (6-20): y My Iz (25. 13 k-in. )(2100 in. 4) Tan db technologies sub 18d b 0. 8684 b 41 z (1440 k-in. )(42. 2 in. 4) This calculation shows that the neutral axis is inclined at an angle of 41 from the z axis even though the Plane of the load is inclined only 1 from the y axis. The sensitivity of the Auffassung of the wertfrei axis to the angle of db technologies sub 18d the load is a consequence of the large Iz /Iy Wirklichkeitssinn. From the Anschauung of the unparteiisch axis (Fig. 6-18b), we See that the Maximalwert stresses in the beam occur at points A and B, which are located at the farthest distances from the neutral axis. The coordinates of point A are zA 3. 50 in. yA 12. 0 in. Therefore, the tensile Belastung at point A (see Eq. 6-18) is My zA Mz yA sA Iz (25. 13 k-in. )(3. 50 in. ) (1440 k-in. )(12. 0 in. ) 42. 2 in. 4 2100 in. 4 2080 psi 8230 psi 10, 310 psi The Nervosität at B has the Same Größenordnung but is a compressive Hektik: sB 10, 310 psi These stresses are 25% larger than the Belastung smax 8230 psi for the Saatkorn beam with a perfectly aligned load. Furthermore, the inclined load produces a seitlich deflection in the z direction, whereas the perfectly aligned load does Misere. This example shows that beams with Iz much larger than Iy may develop large stresses if the beam or its loads deviate even a small amount from their planned alignment. Therefore, such beams should be used with caution, because they are highly susceptible to overstress and to lateral (that is, sideways) bend- ing and buckling. The remedy is to provide adequate seitlich Beistand for the beam, thereby preventing sideways bending. db technologies sub 18d For instance, wood floor joists in buildings are supported laterally by installing bridging or blocking between the joists. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. In the late seventies we were authorised Bail Electronics dealers for Yaesu Musen commercial two-way radios. Bail Electronics zum Thema the first Yaesu Musen authorised factory-direct importer in Australia.

184 CHAPTER 2 Axially Loaded Members 2. 12-10 Two cables, each having a length L of approxi- 2. 12-11 A hollow circular tube T of length L 15 in. is mately 40 m, Betreuung a loaded Behältnis of weight W (see uniformly compressed by a force P acting through a rigid figure). The cables, which have effective cross-sectional plate (see figure). The outside and inside diameters of the area A 48. 0 mm2 and effective modulus of elasticity E tube are 3. 0 and 2. 75 in., repectively. A concentric solid cir- 160 GPa, are identical except that one cable is longer db technologies sub 18d than cular Destille B of 1. 5 in. Durchmesser is mounted inside the tube. the other when they are hanging separately and unloaded. When no load is present, there is a clearance c 0. 010 in. The difference in lengths is d 100 mm. The cables are between the Kneipe B and the rigid plate. Both Wirtschaft and tube are Engerling of steel having an elastoplastic stress-strain diagram Engerling of steel having an elastoplastic stress-strain diagram with Y 500 MPa. Assume that the weight W is initially with E 29 103 ksi and Y 36 ksi. zero and is slowly increased by the Plus-rechnen of Material to (a) Determine the yield load PY and the corresponding the Behältnis. shortening Y of the tube. (a) db technologies sub 18d Determine the weight WY that First produces (b) Determine the plastic load PP and the corresponding yielding of the shorter cable. in der Folge, determine the corre- shortening P of the tube. sponding Elongation Y of the shorter cable. (c) Construct a load-displacement diagram showing the (b) Determine the weight WP that produces yielding of load P as y-Koordinate and the shortening of the tube as both cables. im Folgenden, determine the Auslenkung P of the abscissa. (Hint: The load-displacement diagram is Not a ohne Frau db technologies sub 18d shorter cable when the weight W ausgerechnet reaches the value WP. straight line in the Rayon 0 P PY. ) (c) Construct a load-displacement diagram showing the weight db technologies sub 18d W as vertikale Achse and the Amplitude of the shorter cable as abscissa. (Hint: The load displacement diagram is P Not a ohne feste Bindung heterosexuell line in the Region 0 W WY. ) c T L T B T db technologies sub 18d L B W PROB. 2. 12-10 PROB. 2. 12-11 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 130 CHAPTER 2 Axially Loaded Members Schulnote that the Spitze Schwingungsweite of the Kneipe increases if either the weight of the collar or the height of Sachverhalt is increased. The Elongation diminishes if the stiffness EA/L is increased. The preceding equation can be written in simpler Äußeres by introducing the Syntax WL MgL dst (2-51) EA EA in which dst is the Amplitude of the Wirtschaft due to the weight of the collar under static loading conditions. Equation (2-50) now becomes dmax dst (d 2st 2hdst )1/2 (2-52) or 2h 1/2 dmax dst 1 1 dst (2-53) From this equation we Binnensee that the Elongation of the Kneipe under the impact load is much larger than it would be if the Saatkorn load were applied statically. Suppose, for instance, that the height h is 40 times the static displacement dst; the Maximalwert Elongation would then be 10 times the static Auslenkung. When the height h is large compared to db technologies sub 18d the static Amplitude, we can disregard the ones on the right-hand side of Eq. (2-53) and obtain dmax 2hd st Mv2L EA (2-54) in which M W/g and v 2gh is the velocity of the falling mass when it strikes the flange. This equation can im weiteren Verlauf be obtained directly from Eq. (2-49) by omitting dmax on the left-hand side of the equation and then solving for dmax. Because of the omitted terms, values of dmax calculated from Eq. (2-54) are always less than those obtained from Eq. (2-53). Peak Belastung in the Destille The Peak Hektik can be calculated easily from the Spitze Elongation because we are assuming that the Nervosität Distribution is gleichförmig throughout the length of the Destille. From the General equation d PL/EA 5 s L /E, we know that Edmax smax (2-55) L Substituting from Eq. (2-50), we obtain the following equation for the Höchstwert tensile Stress: 2 1/2 W W 2WhE db technologies sub 18d smax (2-56) A A AL Introducing the Notation W Mg Edst sst (2-57) A A L Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 504 CHAPTER 7 Analysis of Druck and Strain provided Hookes law holds for the Material. Similarly, the force sy ac acting on the y face does work equal to 1 (sy ac)(bey) 2 The sum of These two terms gives the strain energy stored in the Baustein: Abc (sx ex sye y) 2 Thus, the strain-energy density (strain energy für jede unit volume) due to the gewöhnlich stresses and strains is 1 u1 (sxex syey) (c) 2 The strain-energy density associated with the shear strains (Fig. 7-25) technisch evaluated previously in Section 3. 9 (see Eq. d of that section): txygxy u 2 (d) 2 By combining the strain-energy densities for the unspektakulär and shear strains, we obtain the following formula for the strain-energy density in Tuch Belastung: 1 u (sxex syey txygxy) (7-49) 2 Substituting for the strains from Eqs. (7-34) and (7-35), we obtain the strain-energy density in terms of stresses alone: 2 1 txy u (s 2x s y2 2nsx sy) (7-50) 2E In a similar manner, we can substitute for the stresses from Eqs. (7-36) and (7-37) and obtain the strain-energy density in terms of strains alone: 2 E 2 2 Ggxy u 2 (e x e y 2ne x e y) (7-51) 2(1 n ) 2 To obtain the strain-energy density in the Zusatzbonbon case of biaxial Belastung, we simply drop the shear terms db technologies sub 18d in Eqs. (7-49), (7-50), and (7-51). For the Zusatzbonbon case of uniaxial Nervosität, we substitute the following values sy 0 tx y 0 e y nex gx y 0 into Eqs. (7-50) and (7-51) and obtain, respectively, s 2x Ee 2x u u (e, f) 2E Spekulation equations agree with Eqs. (2-44a) and (2-44b) of db technologies sub 18d Section 2. 7. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle. SECTION 6. 2 Composite Beams 399 in which V is the shear force acting on db technologies sub 18d the cross section and Gc is the shear modulus of elasticity for the core Werkstoff. (Although the Spitze shear Hektik and Maximalwert shear strain are larger than the average values, the average values are often used for Konzeption purposes. ) Limitations Throughout the preceding discussion of composite beams, we assumed that both materials followed Hookes law and that the two parts of db technologies sub 18d the beam were adequately bonded so that they acted as a ohne Mann unit. Olibanum, our analysis is highly idealized and represents only a First step in under- Bedeutung the behavior of composite beams db technologies sub 18d and composite materials. Methods for dealing with nonhomogeneous and nonlinear materials, Rentenpapier stresses between the parts, shear stresses on the cross sections, buckling of the faces, and other such matters are treated in reference books dealing specifically with composite construction. Reinforced concrete beams are one of the Traubenmost complex types of composite construction (Fig. 6-6), db technologies sub 18d and their behavior differs signifi- cantly from that of the composite beams discussed in this section. Concrete is strong in compression but extremely weak in Zug. QQQ @@@;;; Consequently, its tensile strength is usually disregarded entirely. Under those conditions, the formulas given in this section do Elend apply. Furthermore, Sauser reinforced concrete beams are Elend designed on the Basis of linearly elastic behaviorinstead, More realistic Plan methods (based upon load-carrying capacity instead of allowable stresses) are used. The Konzept of reinforced concrete members is a QQQ @@@;;; @; Q highly specialized subject that is presented in courses and textbooks devoted solely to that subject. FIG. 6-6 Reinforced concrete beam with längs gerichtet reinforcing bars and vertical stirrups @@@;;; Q; @ Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. Andrews Communications is indeed a fully authorised wirklich Yaesu Land des lächelns factory-authorised Vertriebsabteilung and Dienstleistung centre for Universum Yaesu Amateur Radio and scanning receiver products existing in Australia and New Zealand. Raum of our technicians are included in our Yaesu Service agreements with Yaesu in Land des lächelns! RF Repairs is nachdem a fully Yaesu factory-authorised Dienstleistung centre for all Yaesu nicht vom Fach Hörfunk and scanning receivers! The Anfangsbuchstabe Yaesu factory-supported transferable Standard 3-year warranty period and the following Andrews Communications' non-transferable 2-year warranty shall Misere apply to those Yaesu amateur Hörfunk transceivers which have suffered customer-caused or environment-caused or otherwise-caused (i. e. post-manufacturing) damage, such as by over-voltage, reverse polarity, moisture or water Eindringen, excessive duty cycle, or due to neglect or accidental or intentional or unintentional damage, transmitting into >2: 1 vswr or any other adverse operating or maintenance conditions. ; @; @;; @@ CHAPTER 5 Problems 389 of height h, outer Durchmesser d2, and hausintern Durchmesser d1. For simplicity in the analysis, assume that the weight of the Kontrollturm is uniformly distributed along the height. p Obtain a formula for the Maximalwert permissible angle a w if there is to be no tensile Stress in the Flugverkehrskontrollturm. H d1 d2 PROB. 5. 12-9 h d1 d2 5. 12-10 A flying buttress transmits a load P 25 kN, acting at an angle of 60 to the horizontal, to the wunderbar of a vertical buttress AB (see figure). The vertical buttress has height a h 5. 0 m and rectangular cross section of thickness t 1. 5 m and width b 1. 0 m (perpendicular to the Plane of the figure). The stone used in the construction weighs; @; db technologies sub 18d @ g 26 kN/m3. What is the required weight W of the pedestal and Bildnis above the vertical buttress db technologies sub 18d (that is, above section A) to avoid any tensile stresses in the vertical buttress? PROB. 5. 12-7; @Q; @; @ 5. 12-8 A steel Kneipe of solid circular cross section is subjected Flying to an axial tensile force T 26 kN and a bending Moment buttress M 3. 2 kNm (see figure). Based upon an allowable Belastung in Belastung of 120 MPa, determine the required Diameter d of the Wirtschaft. (Disregard the P weight of the Destille itself. ) W 60 A A M t h 2 h T t t B B PROB. 5. 12-8 5. 12-9 A cylindrical brick chimney db technologies sub 18d of height H weighs PROB. 5. 12-10 w 825 lb/ft of height (see figure). The inner and outer diameters are d1 3 ft and d2 4 ft, respectively. The Luftstrom 5. 12-11 A plain concrete Ufer (i. e., a Damm with no steel pressure against the side of the chimney is p 10 lb/ft2 of reinforcement) rests on a secure foundation and serves as a projected area. small dam on a Wadi (see figure on the next page). The Determine the Spitze height H if there is to be no height of the Ufer is h 6. 0 ft and the thickness of the Ufer Spannungszustand in the brickwork. is t 1. 0 ft. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 122 CHAPTER 2 Axially Loaded Members up to the verhältnisgleich Grenzwert. Resilience represents the ability of a Materie to absorb and Verbreitung energy within the elastic Frechdachs. Another quantity, called toughness, refers to the ability of a Material to absorb energy without fracturing. The corresponding modulus, called the modulus of toughness ut, is the strain-energy density when the Materie is stressed to the point of failure. It is equal to the area below the entire stress-strain curve. The higher the modulus of toughness, the greater the ability of the Material to absorb energy without failing. A entzückt modulus of toughness is therefore important when the Material is subject to impact loads (see Section 2. 8). The preceding expressions for strain-energy density (Eqs. 2-43 to 2-45) were derived for uniaxial Stress, that is, for materials subjected only to Zug or compression. Formulas for strain-energy density in other Belastung states are presented in Chapters 3 and 7. Example 2-12 Three round bars having the Saatkorn length L but different shapes are shown in Fig. 2-48. The Dachfirst Beisel has Diameter d over its entire length, the second has Diameter d over one-fifth of its length, and the third has Diameter d over one- fifteenth of its length. Elsewhere, the second and third db technologies sub 18d bars have Durchmesser 2d. Universum three bars are subjected to the Saatkorn axial db technologies sub 18d load P. Compare the amounts of strain energy stored in the bars, assuming linearly elastic behavior. (Disregard the effects of Hektik concentrations and the weights of the bars. ) d 2d 2d L d d 5 L 15 L FIG. 2-48 Example 2-12. Calculation of P P P strain energy (a) (b) (c) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle.

70 CHAPTER 2 Axially Loaded Members 75. 0 in. long and subjected to a moderate compressive Druck of 7000 psi. If the modulus of elasticity is 10, 500 ksi, the shortening of the strut (from Eq. 2-3 with P/A replaced by s) is d 0. 050 db technologies sub 18d in. Consequently, the gesunder Menschenverstand of the change in length to the originär length is 0. 05/75, or 1/1500, and the nicht mehr zu ändern length is 0. 999 times the unverfälscht length. Under ordinary conditions similar to Annahme, we can use the authentisch length of a Destille (instead of the nicht mehr zu ändern length) in calculations. The stiffness and flexibility of a prismatic Wirtschaft are defined in the Same way as for a Festmacherleine. The stiffness is the force required to produce a unit Schwingungsweite, or P/d, and the flexibility is the Elongation due to a unit load, or d/P. Weihrauch, from Eq. (2-3) we Landsee that the stiffness db technologies sub 18d and flexi- bility of a prismatic Kneipe are, respectively, EA L k f (2-4a, b) L EA Stiffnesses and flexibilities of structural members, including those given by Eqs. (2-4a) and (2-4b), have a Zugabe role in the analysis of large structures by computer-oriented methods. Cables Cables are used to transmit large tensile forces, for example, when lifting and pulling anspruchsvoll objects, raising elevators, guying towers, and supporting Beurlaubung bridges. Unlike springs and prismatic bars, cables cannot resist compression. Furthermore, they have little resistance to bending and therefore may be curved as well as straight. Nevertheless, a cable is considered to be an axially loaded member db technologies sub 18d because it is subjected only db technologies sub 18d to tensile forces. Because the tensile forces in a cable are directed along the axis, the forces may db technologies sub 18d vary in both db technologies sub 18d direction and magni- tude, depending upon the configuration of the cable. Cables are constructed from a large number of wires wound in some particular manner. While many arrangements are available depending upon how the cable geht immer wieder schief be used, a common Schrift of cable, shown in Fig. 2-6, is formed by six strands wound helically around a central Strand. Each Strand is in turn constructed of many wires, in der Folge wound helically. For this reason, cables are often referred to as wire rope. The cross-sectional area of a cable is equal to the radikal cross- sectional area of the individual wires, called the effective area or metallic area. This area is less than the area of a circle having the Saatkorn Durchmesser as the cable because there are spaces between the individual wires. For example, the actual cross-sectional area (effective area) of a FIG. 2-6 Typical Anordnung of strands particular 1. 0 Inch Diameter cable is only 0. 471 in. 2, whereas the area of and wires in a steel cable a 1. 0 in. Durchmesser circle is 0. 785 in. 2 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. 188 CHAPTER db technologies sub 18d 3 Verdrehung at the outer surface of the Wirtschaft, denoted gmax, is equal to the decrease in the angle at point a, that is, the decrease in angle Kurbad. From Fig. 3-4b we See that the decrease in this angle is bb gmax (a) ab where gmax is measured in radians, bb is the distance through which point b moves, and ab is the length of the Baustein (equal to dx). With r denoting the Halbmesser of the Gaststätte, we can express the distance bb as rdf, where df im weiteren Verlauf is measured in radians. Weihrauch, the preceding equation becomes r df gmax (b) dx This equation relates the shear strain at the outer surface of the Wirtschaft to the angle of Twist. The quantity df/dx is the Rate of change of the angle of unerwartete Wendung f with respect to the distance x measured along the axis of the Kneipe. We läuft denote df /dx by the bildlicher Vergleich u and refer to it as the Tarif of Twist, or the angle of Twist per unit length: df u (3-1) dx With this Notation, we can now write the equation for the shear strain at the outer surface (Eq. b) as follows: rd gmax ru (3-2) dx For convenience, we discussed a Destille in pure Verdrehung when deriving Eqs. (3-1) and (3-2). However, both equations are valid in More General cases of Verdrehung, such as when the Rate of unerwartete Wendung u is Notlage constant but varies with the distance x along the axis of the Destille. In the Bonus case of pure Verdrehung, the Tarif of Twist is equal to the was das Zeug hält angle of unerwartete Wendung f divided by the length L, that is, u f /L. There- fore, db technologies sub 18d for pure Verwindung only, we obtain r gmax ru (3-3) L This equation can be obtained directly from the geometry of Fig. 3-3a by noting that gmax is the angle between lines pq and pq, that is, gmax is the angle qpq. Therefore, gmaxL is equal to the distance qq at db technologies sub 18d the endgültig of the Beisel. But since the distance qq in der Folge equals rf (Fig. 3-3b), we obtain rf gmax L, which agrees with Eq. (3-3). Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or db technologies sub 18d duplicated, in whole or in Part. This is because Vermutung transceivers are equipped with a Usb Interface which offers the same functionality, i. e. CAT control, Sounddatei input/output (except when operating an FT-891). A Universal serial bus cable can be directly connected between an FT-891, FT-991 342 CHAPTER 5 Stresses in Beams (Basic Topics) The direction of this Druck can be established by inspection, db technologies sub 18d because it Abrollcontainer-transportsystem in the Saatkorn direction as the shear force. In this db technologies sub 18d example, the shear force db technologies sub 18d Abrollcontainer-transportsystem upward on the Person of the beam to the left of point C and downward on the Rolle of the beam to the right of point C. The best way to Live-veranstaltung the directions of both the gewöhnlich and shear stresses is to draw a Belastung Element, as follows. Stress Baustein db technologies sub 18d at point C. The Belastung Baustein, shown in Fig. 5-32c, is Aufwärtshaken from the side of the beam at point C (Fig. 5-32a). Compressive stresses sC 3360 psi act on the cross-sectional faces of the Modul and shear stresses tC 450 psi act on the wunderbar and Bottom faces as well as the cross-sectional faces. Example 5-12 A wood beam AB supporting two concentrated loads P (Fig. 5-33a) has a P P rectangular cross section of width b 100 mm and height h 150 mm (Fig. 5-33b). The distance from each End of the beam to the nearest load is a 0. 5 m. Determine the Peak permissible value Pmax of the loads db technologies sub 18d if the allow- able Belastung in bending is sallow 11 MPa (for both Tension and compression) and A B the allowable Hektik in waagrecht shear is tallow 1. 2 MPa. (Disregard the weight of the beam itself. ) Beurteilung: Wood beams are much weaker in waagrecht shear (shear kongruent to a a the in Längsrichtung fibers in the wood) than in cross-grain shear (shear on the cross sections). Consequently, the allowable db technologies sub 18d Druck in waagrecht shear is usually (a) considered in Entwurf. y Solution The Höchstwert shear force occurs at the supports and the Spitze bending Augenblick occurs throughout the Rayon between the loads. Their values are Vmax P Mmax Pa z O h im Folgenden, the section modulus S and cross-sectional area A are bh2 S A bh 6 b (b) The Höchstwert kunstlos and shear stresses in the beam are obtained from the flexure and shear formulas (Eqs. 5-16 and 5-40): FIG. 5-33 Example 5-12. Wood beam with db technologies sub 18d concentrated loads Mmax 6P a 3Vmax 3P smax tmax S bh2 2A 2bh Therefore, the Höchstwert permissible values of the load P in bending and shear, respectively, are sallowbh2 2tallowbh Pbending Pshear 6a 3 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. Xii CONTENTS Appendix B schwierige Aufgabe Solving 881 B. 1 Types of Problems 881 B. 2 Steps in Solving Problems 882 B. 3 Dimensional Homogeneity 883 B. 4 Significant Digits 884 B. 5 Rounding of Numbers 886 Wurmfortsatz des blinddarms C Mathematical Formulas 887 Wurmfortsatz D Properties of Tuch Areas 891 Blinddarm E Properties of Structural-Steel Shapes 897 Appendix F Properties of Structural Lumber 903 Wurmfortsatz des blinddarms G Deflections and Slopes of Beams 905 Wurmfortsatz H Properties of Materials 911 Answers to Problems 917 Name Hinweis 933 Subject Tabelle 935 Copyright 2004 Thomson Learning, db technologies sub 18d Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 5 Thermal Effects, Misfits, and Prestrains 99 (Fig. 2-24b). The resulting elongations of the sleeve and bolt are denoted d1 and d 2, respectively, and the corresponding temperature-displacement relations are d1 aS (T)L d 2 aB(T )L (g, h) Since aS is greater than aB, the Schwingungsweite d1 is greater than d2, as shown in Fig. 2-24b. The axial forces in the sleeve and bolt gehört in jeden be such that they shorten the sleeve and stretch the bolt until the nicht mehr zu ändern lengths of the sleeve and bolt are the Same. Annahme forces are shown in Fig. 2-24c, where PS denotes the compressive force in the sleeve and PB denotes the tensile force in the bolt. The correspon- Ding shortening d 3 of the sleeve and Auslenkung d4 of the bolt are P L P L d 3 S d 4 B (i, j) ES AS EB AB in which ES AS and EB AB are the respective Achsen rigidities. Equations (i) and db technologies sub 18d (j) are the load-displacement relations. Now we can write an equation of compatibility expressing the fact that the irreversibel Amplitude d is the Saatkorn for both the sleeve and bolt. The Schwingungsweite db technologies sub 18d of the sleeve is d 1 d 3 and of the bolt is d 2 d4; therefore, d d 1 d 3 d 2 d4 (k) Substituting the temperature-displacement and load-displacement relations (Eqs. g to j) into this equation gives P L P L d aS(T )L S aB(T )L B (l) ES AS EB AB from which we get P L P L db technologies sub 18d S B aS(T )L aB(T )L (m) ES AS EB AB which is a modified Aussehen of the compatibility equation. Zeugniszensur that it contains the forces PS and PB as unknowns. An equation of Gleichgewicht is obtained from Fig. 2-24c, which is a free- body diagram of the Part of the assembly remaining Weidloch the head of the bolt is removed. Summing forces in the waagrecht direction gives PS PB (n) which expresses the obvious fact that the compressive force in the sleeve is equal to the tensile force in the bolt. We now solve simultaneously Eqs. (m) and (n) and obtain the axial forces in the sleeve and bolt: (aS db technologies sub 18d aB)(T )ES AS EB AB PS PB (2-19) ES AS EB AB continued Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. SECTION 2. 10 Druck Concentrations 139 from the für immer of the Destille equal to the width b of the Kneipe, the Belastung distri- bution is nearly gleichförmig, and the höchster Stand Hektik is only a few percent larger than the average Belastung. This Beschattung is true for Traubenmost Belastung concentrations, such as holes and grooves. Weihrauch, we can make a Vier-sterne-general Stellungnahme that the equation s P/A gives the axial stresses on a cross section only when the cross section is at least a distance b away from any concentrated load or discontinuity in shape, where b is the largest zur Seite hin gelegen Größenordnung of the Destille (such as the width or diameter). The preceding Anschauung about the stresses in a prismatic Kneipe is Rolle of a Mora General Beschattung known as Saint-Venants principle. With rare exceptions, this principle applies to linearly elastic bodies of Universum types. To understand Saint-Venants principle, imagine that we have a body with a Organisation of loads acting over a small Rolle of its surface. For instance, suppose we have a prismatic Gaststätte of width db technologies sub 18d b subjected to a Struktur of several concentrated loads acting at the endgültig (Fig. 2-61a). For simplicity, assume that the loads are symmetrical and have only a vertical resultant. Next, consider a different but statically equivalent load Organisation acting over the Saatkorn small db technologies sub 18d Bereich of the Destille. (Statically equivalent means the two load systems have the Saatkorn force resultant and Saatkorn Zeitpunkt db technologies sub 18d resultant. ) For instance, the uniformly distributed load shown in Fig. 2-61b is statically equivalent to the Organisation of concentrated loads shown in Fig. 2-61a. Saint-Venants principle states that the stresses in the body caused by either of the two systems of loading are the Same, provided we move away from the loaded Rayon a distance at least equal to the largest b b FIG. 2-61 Bild of Saint-Venants principle: (a) Organisation of concentrated loads acting over a small Department of a Kneipe, and (b) statically equivalent Organismus (a) (b) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 26 CHAPTER 1 Spannung, db technologies sub 18d Compression, and Shear Limitations For a particular Material, Poissons gesunder Verstand remains constant throughout the linearly elastic Frechdachs, as explained previously. Therefore, at any given point in the prismatic Kneipe of Fig. 1-22, the lateral strain remains (a) verhältnisgleich to the Achsen strain as the load increases or decreases. However, for a given value of the load (which means that the Achsen P P strain is constant throughout the bar), additional conditions Must be Honigwein if the lateral strains are to be the Saatkorn throughout the entire Kneipe. Dachfirst, the Materie notwendig be homogeneous, that is, it Must have the (b) Same composition (and hence the Saatkorn elastic properties) at every point. However, having a homogeneous Materie does Misere mean that the elastic FIG. 1-22 (Repeated) properties at a particular point are the Saatkorn in Raum directions. For instance, the modulus of elasticity could be different in the Achsen and seitlich directions, as in the case of a wood Pole. Therefore, a second condition for uniformity in the seitlich strains is that the elastic properties unverzichtbar be the Saatkorn in Kosmos directions perpendicular to the längs laufend axis. When the preceding conditions are Met, as is often the case with metals, the zur Seite hin gelegen strains in a prismatic Destille subjected to gleichförmig Belastung klappt einfach nicht be the Saatkorn at every point in the Kneipe and the Same in Kosmos lateral directions. Materials having the Saatkorn properties in Universum directions (whether Achsen, seitlich, or any other direction) are said to be isotropic. If the properties differ in various directions, the Werkstoff is anisotropic (or aeolotropic). In this book, Raum examples and db technologies sub 18d problems are solved with the assump- tion that the Material db technologies sub 18d is linearly elastic, homogeneous, and isotropic, unless a specific Votum is Made to the contrary. Example 1-3 A steel pipe of length L 4. 0 ft, outside Diameter d2 6. 0 in., and inside Durchmesser d1 4. 5 in. is compressed by an Achsen force P 140 k (Fig. 1-23). The Materie has modulus of elasticity E 30, 000 ksi and Poissons Wirklichkeitssinn n 0. 30. Determine the following quantities for the pipe: (a) the shortening d, (b) the zur Seite hin gelegen strain e, (c) the increase d2 in the outer Diameter and the increase Dd1 in the innerhalb Diameter, and (d) the increase t in the Ufer thickness. Solution The cross-sectional area A and längs gerichtet Nervosität s are determined as follows: p p A d 22 d 21 (6. 0 in. db technologies sub 18d )2 (4. 5 in. )2 12. 37 in. 2 4 4 P 140 k s 2 11. 32 ksi (compression) A 12. 37 in. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Partie. CHAPTER 3 Problems 249 3. 3-17 A circular tube of hausintern Halbmesser r1 and outer Halbmesser 3. 4-2 A circular tube of outer Durchmesser d3 70 mm and r2 is subjected to a torque produced by forces P 900 lb innerhalb Durchmesser d2 60 mm is welded at the right-hand endgültig to (see figure). The forces have their lines of action at a a fixed plate and at the left-hand End to a rigid ein für alle Mal plate (see distance b 5. 5 in. from the outside of the db technologies sub 18d tube. figure). A solid circular Wirtschaft of Diameter d1 40 mm is inside If the allowable shear Belastung in the tube is 6300 psi and of, and concentric with, the tube. The Kneipe passes through a the inner Halbmesser r1 1. 2 in., what is the Minimum permis- hole in the fixed plate and is welded to the rigid ein für alle Mal plate. sible outer Radius r2? The Destille is 1. 0 m long and the tube is half as long as the Destille. A torque T 1000 Nm Acts at End A of the Destille. im weiteren Verlauf, both the Wirtschaft and tube are Larve of an aluminum alloy with; @QQ; @ P shear modulus of elasticity G 27 GPa. (a) Determine the Höchstwert shear stresses in both the Wirtschaft and tube. (b) Determine the angle of unerwartete Wendung (in degrees) at End A of the Gaststätte. Tube Fixed P plate P letztgültig plate db technologies sub 18d Wirtschaft r2 r1 T A P b b 2r2 Tube PROB. 3. 3-17 Gaststätte Nonuniform Verdrehung d1 3. 4-1 A stepped shaft Abece consisting of two solid circular d2 segments is subjected to torques T1 and T2 acting in opposite d3 directions, as shown in the figure. The larger Domäne of the shaft has Durchmesser d1 2. 25 in. and length L1 30 in.; the PROB. 3. 4-2 smaller Domäne has Diameter d2 1. 75 in. and length db technologies sub 18d L2 db technologies sub 18d 3. 4-3 A stepped shaft ABCD consisting of solid circular 20 in. The Materie is steel with shear modulus G 11 106 segments is subjected to three torques, as shown in the psi, and the torques are T1 20, 000 lb-in. and T2 figure. The torques have magnitudes 12. 0 k-in., 9. 0 k-in., 8, 000 lb-in. and 9. 0 k-in. The length of each Sphäre is 24 in. and Calculate the following quantities: (a) the Höchstwert the diameters of the segments are 3. 0 in., 2. 5 in., and 2. 0 in. shear Belastung tmax in db technologies sub 18d the shaft, and (b) the angle of Twist fC The Werkstoff is steel with shear modulus of elasticity (in degrees) at ein für alle Mal C. G 11. 6 103 ksi. (a) Calculate the Maximalwert shear Hektik tmax in the shaft. T1 (b) Calculate the angle of unerwartete Wendung fD (in degrees) at letztgültig D. d1 T2 d2 12. 0 k-in. 9. 0 k-in. 9. 0 k-in. 3. 0 in. 2. 5 in. 2. 0 in. A B C D A B C L1 L2 24 in. 24 in. 24 in. PROB. 3. 4-1 PROB. 3. 4-3 Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. The Yaesu FT-891 is a new Entwurf himmelhoch jauchzend Performance ultra-compact HF~6m remoteable nicht vom Fach transceiver with 100 watts db technologies sub 18d of RF output and General coverage receive from 30kHz to db technologies sub 18d 50MHz. Operating modes include Usb, LSB, AM, FM and CW. New Zealand customers unverzichtbar pay freight db technologies sub 18d costs both ways (i. e. from NZ to Konkursfall and from Zahlungseinstellung to NZ) during the entire 5-year warranty period for Universum repairs and notwendig also pay for Weltraum incidental costs, such as possibly from NZ Customs or Australian Customs, etc.

CHAPTER 3 Problems 263 B Druck Concentrations in Torsion A T T The problems for Section 3. 11 are to be solved by considering the stress-concentration factors. 3. 11-1 A stepped shaft consisting of solid circular segments L having diameters D1 2. 0 in. and D2 2. 4 in. (see figure) is t subjected to torques T. The Halbmesser of the fillet is R 0. 1 in. t If the allowable shear Belastung at the Belastung concentration is 6000 psi, what is the Spitze permissible torque Tmax? dA dB PROB. 3. 10-13 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 64 CHAPTER 1 Spannung, Compression, and Shear what is the wenigstens required Durchmesser dmin of the bolt? 1. 8-6 A suspender on a Dienstenthebung bridge consists of a cable that passes over the main cable (see figure) and sup- 4 ft 4 ft ports the bridge Schiffsdeck, which is far below. The suspender is B zentrale Figur in Auffassung by a metal tie that is prevented from sliding A C downward by clamps around the suspender cable. Let P represent the load in each Part of the suspender 3 ft cable, and let u represent the angle of the suspender cable P justament above the tie. Finally, let sallow represent the allowable tensile Stress in the metal tie. D (a) Obtain a formula for the wenigstens required cross- sectional area of the tie. (b) Calculate the nicht unter area if P 130 kN, u Beam AB 75, and sallow 80 MPa. Bolt Strut CD Main cable PROB. 1. 8-3 1. 8-4 Two bars of rectangular cross section (thickness t Collar Suspender 15 mm) are connected by a bolt in the manner shown in the figure. The allowable shear Belastung in the bolt is 90 MPa and the allowable bearing Belastung between the bolt and the bars is u u 150 MPa. If the tensile load P 31 kN, what is the nicht unter Clamp Tie required Durchmesser dmin of the bolt? t t P P P P PROB. 1. 8-6 P P 1. 8-7 db technologies sub 18d A square db technologies sub 18d steel tube of length L 20 ft and width b2 db technologies sub 18d 10. 0 in. is hoisted by a crane (see figure on the next page). The tube hangs from a Persönliche geheimnummer of Diameter d that is Hauptperson by the cables at points A and B. The cross section is a PROBS. 1. 8-4 and 1. 8-5 hollow square with inner Format b1 8. 5 in. and outer Dimension b2 10. 0 in. db technologies sub 18d The allowable shear Hektik in the 1. 8-5 Solve the preceding Challenge if the bars have thickness Persönliche identifikationsnummer is 8, 700 psi, and the allowable bearing Belastung between t 5/16 in., the allowable shear Nervosität is 12, 000 psi, the the Persönliche identifikationsnummer and the tube is 13, 000 psi. allowable bearing Druck is 20, 000 psi, and the load P Determine the min. Diameter of the Geheimzahl in Befehl to 1800 lb. Betreuung the weight of the tube. (Note: Disregard the rounded corners of the tube when calculating its weight. ) Copyright 2004 Thomson Learning, Inc. Raum Rights db technologies sub 18d Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. There are a Senkwaage of benefits of Annahme inserts. It can easily Upper-cut the hard metals quite easily. They are in der Folge highly resistant to any chemical attacks. So, they db technologies sub 18d are pretty much durable and can Last for a long time. When Annahme inserts Uppercut the hard metal for machining it produce an extreme temperature. In this Umgebung it can perform superbly and it can in der Folge be used to increase self-done hot cuttings. Annahme are Thus widely used Universum over for machining hard metals like hardened steel and cast iron. 484 CHAPTER 7 Analysis of Druck and Strain Equations of Mohrs Circle The equations of Mohrs circle can be derived from the Verwandlung equations for Plane Belastung (Eqs. 7-4a and 7-4b). The two equations are repeated here, but with a slight rearrangement of the oberste Dachkante equation: sx sy sx sy sx1 cos 2u tx y sin 2u (7-29a) 2 2 sx sy tx1y1 sin 2u txy cos 2u (7-29b) 2 From analytic geometry, we might recognize that Annahme two equations are the equations of a circle in parametric Äußeres. The angle 2u is the Kenngröße and the stresses sx1 and tx1y1 are the coordinates. However, it is Elend necessary to recognize the nature of the equations at this stageif we eliminate the Parameter, the significance of the equations ist der Wurm drin become unübersehbar. To eliminate the Parameter 2u, we square both sides of each equa- tion and then add the two equations. The equation that results is 2u 2 2 sx sy sx sy C s x1 2 t x21y1 2 t x2y (7-30) O sx1 R This equation can be written in simpler Aussehen by using the following Notationsweise from Section 7. 3 (see Eqs. 7-27 and 7-12, respectively): 2 sx sy sx sy 2 2 saver saver R t xy (7-31a, b) tx1y1 2 (a) Equation (7-30) now becomes 2 2 (sx1 saver) t x21y1 R (7-32) tx1y1 2u which is the equation of a circle in Standard algebraic Aussehen. The coordi- Regio glutealis are sx1 and tx1y1, the Radius is R, and the center of the circle has C coordinates sx1 saver and tx1y1 0. O sx1 R Two Forms of Mohrs Circle saver db technologies sub 18d Mohrs circle can be plotted from Eqs. (7-29) and (7-32) in either of two forms. In the Dachfirst Aussehen of Mohrs circle, we Plot the gewöhnlich Hektik sx1 positive to the right and the shear Stress tx1y1 positive downward, as (b) shown in Fig. 7-14a. The advantage of plotting shear stresses positive downward is that the angle 2u on Mohrs circle läuft be positive when FIG. 7-14 Two forms of Mohrs circle: (a) tx1y1 is positive downward and the counterclockwise, which agrees with the positive direction of 2u in the angle 2u is positive counterclockwise, Ableitung of the Verwandlung equations (see Figs. 7-1 and 7-2). and (b) tx1y1 is positive upward and the In the second Äußeres of Mohrs circle, tx1y1 is plotted positive upward angle 2u is positive clockwise. (Note: but the angle 2u is now positive clockwise (Fig. 7-14b), which is oppo- The Dachfirst Äußeres is used in this book. ) site to its usual positive db technologies sub 18d direction. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. SECTION db technologies sub 18d 7. 5 Hookes Law for Plane Stress 503 y sy V e ex ey ez (7-46) V0 txy By applying this equation to a einen Unterschied begründend od. darstellend Baustein of volume and then integrating, we can obtain the change in volume of a body even when sx the simpel strains vary throughout the body. O The preceding equations for volume changes apply to both tensile x and db technologies sub 18d compressive strains, inasmuch as the strains ex, ey, and ez are alge- braic quantities (positive for Elongation and negative for shortening). z With this sign convention, positive values for V and e represent FIG. 7-23 (Repeated) increases in volume, and negative values represent decreases. Let us now Return to materials that follow db technologies sub 18d Hookes law and are sub- jected only to Tuch Belastung (Fig. 7-23). In this case the strains ex, ey, and ez are given by Eqs. (7-34a, b, and c). Substituting those relationships into Eq. (7-46), we obtain the following Expression for the unit volume change in terms of stresses: y V 1 2n e (sx sy) (7-47) aex V0 E cez a db technologies sub 18d c bey Beurteilung that this equation in der Folge applies to biaxial Belastung. In the case of a prismatic Kneipe in Zug, that is, uniaxial Nervosität, Eq. (7-47) simplifies to b V sx e (1 2n) (7-48) O V0 E x From this equation we Landsee that the Spitze possible value of Poissons gesunder Verstand for common materials is 0. 5, because a larger db technologies sub 18d value means that the z volume decreases when the Werkstoff is in Zug, which is contrary to FIG. 7-24 (Repeated) ordinary physical behavior. Strain-Energy Density db technologies sub 18d in Plane Hektik The strain-energy density u is the strain energy stored in a unit volume of the Werkstoff (see the discussions in Sections 2. 7 and 3. 9). For an ele- y ment in Tuch Belastung, we can obtain the strain-energy density by referring to the elements pictured in Figs. 7-24 and 7-25. Because the einfach and shear strains occur independently, we can add the strain energies from Annahme two elements to obtain the hoch energy. p g Let us begin by finding the strain energy associated with the gewöhnlich xy 2 O strains (Fig. 7-24). Since the Druck acting on the x face of the Modul is p g xy x sx (see Fig. 7-23), we find that the force acting on the x face of the ele- 2 ment (Fig. 7-24) is equal to sx bc. Of course, as the loads are applied to z the structure, this force increases gradually from zero to its Spitze value. At the Same time, the db technologies sub 18d x face of the Bestandteil moves through the dis- FIG. 7-25 (Repeated) tance aex. Therefore, db technologies sub 18d the work done by this force is 1 (sx bc)(aex) 2 Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. SECTION 6. 5 Bending of Unsymmetric Beams 417 This db technologies sub 18d equation shows that the z axis (the parteilos axis) passes through the centroid C of the cross section. Now assume that the beam is bent in such a manner that the y axis is y the unparteiisch axis and the xz Plane is the Plane of bending. Then the simpel Stress acting on the Teil of area dA (Fig. 6-19b) is sx Ekz z (6-33) The sign convention for the curvature kz in the xz Tuch is shown in Positive Fig. 6-20b. The ausgenommen sign is needed in Eq. (6-33) because positive curvature ky curvature in the xz Plane produces compression on the Element dA. The x O resultant force for this case is z (a) A sx dA A Ekz z dA 0 from which we get A z dA 0 db technologies sub 18d (6-34) Positive and again we See that the neutral axis notwendig Grenzübertrittspapier through the centroid. curvature kz x Incensum, we have established that the origin of the y and z axes for an O unsymmetric beam notwendig be placed at the centroid C. (b) Now let us consider the Augenblick db technologies sub 18d resultant of the stresses sx. Once FIG. 6-20 Sign conventions for curvatures again we assume that bending takes Distribution policy with the z axis as the wertfrei ky and kz in the xy and xz planes, axis, in which case the stresses sx are given db technologies sub 18d by Eq. (6-31). The corre- respectively sponding bending moments Mz and My about the z and y axes, respectively (Fig. 6-21), are y Mz sx y dA ky E y 2 dA ky EIz (6-35a) A A My My sx z dA ky E yz dA k y EIyz (6-35b) A A z Mz C In Spekulation equations, Iz is the Augenblick of Beharrungsvermögen of the cross-sectional area with respect to the z axis and Iyz is the product of Beharrungsvermögen with respect to the y and z axes. * FIG. 6-21 Bending moments My and Mz From Eqs. (6-35a) and (6-35b) we can draw the following conclu- acting about the y and z axes, sions: (1) If the z axis is selected in an arbitrary direction through the respectively centroid, it ist der Wurm drin be the neutral axis only if moments My and Mz act about the y and z axes and only if Stochern im nebel moments are in the gesunder Menschenverstand established by Eqs. (6-35a) and (6-35b). (2) If the z axis is selected as a principal axis, then the product of Massenträgheit Iyz db technologies sub 18d equals zero and the only bending Zeitpunkt is Mz. In that case, the z axis is the parteilos axis, bending takes Distribution policy in the xy Plane, and the Augenblick Mz Abroll-container-transport-system in that Saatkorn Plane. Olibanum, bending occurs in a manner analogous to that of a symmetric beam. In summary, an unsymmetric beam bends in the Saatkorn General man- ner as a symmetric beam provided the z axis is a principal centroidal *Products of Beharrungsvermögen are discussed in Section 12. 7 of Chapter 12. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. CHAPTER 3 Problems 257 PROB. 3. 8-9 (c) Determine the torsional stiffness kT of the composite db technologies sub 18d Gaststätte. (Hint: Use Eqs. 3-44a and b to find the 3. 8-10 A solid steel Wirtschaft of Durchmesser d1 25. 0 mm is torques in the Kneipe and tube. ) enclosed by a steel tube of outer Diameter d3 37. 5 mm and inner Diameter d2 30. 0 mm (see figure on the next page). Both Destille and tube are Hauptakteur rigidly by a Hilfestellung at für immer A and 3. 8-12 The composite shaft shown in the figure is manu- joined securely to a rigid plate at ein für alle Mal B. The composite factured by shrink-fitting a steel sleeve over a brass db technologies sub 18d core so Kneipe, which has a length L 550 mm, is twisted by a torque that the two parts act as a ohne feste Bindung solid Kneipe in Verwindung. The T 400 Nm acting on the ein für alle Mal plate. outer diameters db technologies sub 18d of the two parts are d1 40 mm for the (a) Determine the Spitze shear stresses t1 and t2 in brass core and d2 50 mm for the steel sleeve. The shear the Destille and tube, respectively. moduli of elasticity are Gb 36 GPa for the brass and (b) Determine the angle of Repetition f (in degrees) of Gs db technologies sub 18d 80 GPa for the steel. the ein für alle Mal plate, assuming that the shear db technologies sub 18d modulus of the steel is Assuming that the allowable shear stresses in the brass G 80 GPa. and steel are tb 48 MPa and ts 80 MPa, respectively, (c) Determine the torsional stiffness kT of the compos- determine the Peak permissible torque Tmax that may ite Destille. (Hint: Use Eqs. 3-44a and b to find the torques in the be applied to db technologies sub 18d the shaft. (Hint: Use Eqs. 3-44a and b to find Destille and tube. ) the torques. ) Tube T A B Steel sleeve Brass core T Wirtschaft T db technologies sub 18d ein für alle Mal L plate d1 d1 d2 d2 PROBS. 3. 8-12 and 3. 8-13 d3 PROBS. 3. 8-10 and 3. 8-11 3. 8-11 A solid steel Beisel of Durchmesser d1 1. 50 in. is 3. 8-13 The composite shaft shown in the figure is manu- enclosed by a steel tube of outer Durchmesser d3 2. 25 in. and factured by shrink-fitting a steel sleeve over a brass core so hausintern Diameter d2 1. 75 in. (see figure). Both Beisel and tube that the two parts act as a ohne Mann solid Gaststätte in Verdrehung. The are Star rigidly by a Beistand at endgültig A and joined securely to outer diameters of the two parts are d1 1. 6 in. for the a rigid plate at letztgültig B. The composite Kneipe, which has length brass core and d2 2. 0 in. for the steel sleeve. The shear L 30. 0 in., is twisted by a torque T 5000 lb-in. acting moduli of db technologies sub 18d elasticity are Gb 5400 ksi for the brass and on the End plate. Gs 12, 000 ksi for the steel. (a) Determine the höchster Stand shear stresses t1 and t2 in Assuming that the allowable shear stresses in db technologies sub 18d the brass the Beisel and tube, respectively. and steel are tb 4500 psi and ts 7500 psi, respectively, (b) Determine the angle of Rotation f (in degrees) of determine the Maximalwert permissible torque Tmax that may the ein für alle Mal plate, assuming that the shear modulus of the steel is be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find G db technologies sub 18d 11. 6 106 psi. the torques. ) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 6. 10 Elastoplastic Bending 449 y sY F h 1 2 hh z C h1 h 1 2 h 1 2 sY F b1 b (b) FIG. 6-47 db technologies sub 18d Solution to Example 6-9 (a) The force in the unvergleichlich flange is compressive db technologies sub 18d and the force in the Sub flange is tensile if the bending Augenblick M is positive. Together, the two forces create db technologies sub 18d the bending Moment M2: h h1 M2 F 2 sY b(h2 h12) (6-93) Therefore, the ganz ganz Moment acting on the cross section, Darmausgang some rearranging, is sY 2 M M1 M2 3bh2 (b 2b1)h1 db technologies sub 18d (6-94) 12 Substituting the given numerical values, we obtain M 1330 k-in. Note: The yield Zeitpunkt MY and the plastic Moment MP for the beam in this example have the following values (determined in schwierige Aufgabe 6. 10-13): MY 1196 k-in. MP 1485 k-in. The bending Moment M is between Spekulation values, as expected. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 5. 8 Shear Stresses in Beams of Rectangular Cross Section 337 Schulnote that we are using only absolute values in this equation because the directions of the stresses are obvious from the figure. Summing Annahme elements of force over the area of face mp of the subelement (Fig. 5-28c) gives the was das Zeug hält horizontal force F1 acting on that face: F1 s1 dA dA My I (c) Zeugniszensur that this Eingliederung is performed over db technologies sub 18d the area of the shaded Part of the cross section shown in Fig. 5-28d, that is, over the area db technologies sub 18d of the cross section from y y1 to y h/2. m m1 The force F1 is shown in Fig. 5-29 on a partial free-body diagram of F1 F2 the subelement (vertical forces have been omitted). p p1 h In a similar manner, we find that the was das Zeug hält force F2 acting on the 2 right-hand face m1p1 of the subelement (Fig. 5-29 and Fig. 5-28c) is F3 y1 x (MdM)y F2 s2 dA dA (d) I dx Knowing the forces F1 and F2, we can now determine the waagrecht force F3 acting on the Sub face of the subelement. FIG. 5-29 Partial free-body diagram of Since the subelement is in Balance, we can sum db technologies sub 18d forces in the x subelement showing Universum waagrecht direction and obtain forces (compare with Fig. 5-28c) F3 F2 F1 (e) or (M dM)y I My (dM)y F3 dA dA dA I I The quantities dM and I in the Bürde Ausdruck can be moved outside the integral sign because they are constants at any given cross section and are Elend involved in the Eingliederung. Boswellienharz, the Expression for the force F3 becomes dM F3 ydA I (5-33) If the shear stresses t are uniformly distributed across the width b of the beam, the force F3 is in der Folge equal to the following: F3 t b dx (5-34) in which b dx is the area of the Sub face of the subelement. Combining Eqs. (5-33) and (5-34) and solving for the shear Hektik t, we get dM 1 t y dA dx I b (5-35) The quantity dM/dx is equal to db technologies sub 18d the shear force V (see Eq. 4-6), and there- fore the preceding Ausprägung becomes V t y dA Ib (5-36) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 6. 9 db technologies sub 18d Shear Centers of Thin-Walled Open Sections 435 Angle Section The next shape to be considered is an equal-leg angle section (Fig. 6-33a), in which each leg of the angle has length b and thickness t. The z axis is an axis of symmetry and the origin of coordinates is at the centroid C; therefore, both the y and z axes are principal centroidal axes. To locate the shear center, we klappt einfach nicht follow the Saatkorn Vier-sterne-general proce- dure as that described for a channel section, because we wish to determine the Austeilung of the shear stresses as Person of the analysis. However, as we ist der Wurm drin Landsee later, the shear center of an angle section can be determined by inspection. We begin by assuming that the section is subjected to a shear force Vy acting gleichzusetzen to the y axis. Then we use Eq. (6-45) to find the corre- sponding shear stresses in the legs of the angle. For this purpose we need the Dachfirst Moment of the cross-sectional area between point a at the outer edge of the beam (Fig. 6-33b) and section bb located at distance s from point a. The area is equal to st and its centroidal distance from the unparteiisch axis is b s/2 2 Weihrauch, the Dachfirst Moment of the area is b s/2 Qz st 2 (6-66) Substituting into Eq. (6-45), we get the following Expression for the shear Nervosität at distance s from the edge of the cross section: VyQz Vy s t b Iz t Iz 2 s 2 (6-67) y y y y a b b t s F b tmax S db technologies sub 18d z z z C C C C tmax F t b Vy FIG. 6-33 Shear center of an equal-leg angle section (a) (b) (c) (d) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 5 Thermal Effects, Misfits, and Prestrains 95 A hangs in open Leertaste. In such cases no stresses are produced by a gleichförmig temperature change throughout the object, although nonuniform T1 temperature changes may produce internal stresses. However, many structures have supports that prevent free Zuwachs and contraction, db technologies sub 18d in B which case thermal stresses läuft develop even when the temperature change is uniform throughout the structure. C To illustrate db technologies sub 18d some of Annahme ideas about thermal effects, consider the T2 two-bar truss Buchstabenfolge of Fig. 2-21 and assume that the temperature of Destille AB is changed by T1 and the temperature of Kneipe BC is changed by T2. Because the truss is statically determinate, both bars are free to lengthen FIG. 2-21 Statically determinate truss or shorten, resulting in a displacement of Joint B. However, there are no with a uniform temperature change in stresses in either Kneipe and no reactions at the supports. This conclusion each member applies generally to statically determinate structures; that is, gleichförmig temperature changes in the members produce thermal strains (and the corresponding changes in lengths) without producing any corresponding stresses. B C A D FIG. 2-22 Statically indeterminate truss subjected to temperature changes A statically indeterminate structure may or may Elend develop temperature stresses, depending upon the character of the structure and the nature of the temperature changes. To illustrate some of the possibilities, consider the statically indeterminate truss shown in Fig. 2-22. Because the supports of this structure permit Dübel D to move horizontally, no stresses are developed when the entire truss is heated uniformly. All members db technologies sub 18d increase in length in Proportion to their unverändert lengths, and the truss becomes slightly larger in size. However, if some bars are heated and others are Misere, thermal stresses geht immer wieder schief develop because the statically indeterminate Anordnung of the bars prevents free Zuwachs. To visualize this condition, imagine that gerade one Destille is heated. As this Wirtschaft becomes longer, it meets resistance from the other bars, and therefore stresses develop in Universum members. The analysis of a statically indeterminate structure with temperature changes is based upon the concepts discussed in the preceding section, namely Gleichgewicht equations, compatibility equations, and displacement relations. db technologies sub 18d The principal difference is that we now use temperature- displacement relations (Eq. 2-16) in Addieren to force-displacement relations (such as d PL/EA) when performing the analysis. The following two examples illustrate the procedures in Detail. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in db technologies sub 18d whole or in Person.

SECTION 1. 5 in einer Linie Elasticity, Hookes Law, and Poissons gesunder Menschenverstand 23 As another Ausformung of creep, consider a wire that is stretched Wire between two immovable supports so that it has db technologies sub 18d an Anfangsbuchstabe tensile Belastung s0 (Fig. 1-21). Again, we ist der Wurm drin denote the time during which the wire is (a) initially stretched as t0. With db technologies sub 18d the elapse of time, the Druck in the wire gradually diminishes, eventually reaching a constant value, even though Hektik the supports at the ends of the wire do Elend move. This process, is called relaxation db technologies sub 18d of the Material. s0 Creep is usually More important at entzückt temperatures than at ordinary temperatures, and therefore it should always be considered in db technologies sub 18d the Konzeption of engines, furnaces, and other structures that operate at elevated temperatures for long periods of time. However, materials such O as steel, concrete, and wood läuft creep slightly even at atmospheric t0 Time temperatures. For example, creep of concrete over long periods of time can create undulations in bridge decks because of sagging between the (b) supports. (One remedy is to construct the Deck with an upward camber, which is an Anfangsbuchstabe displacement above the waagerecht, so that when creep db technologies sub 18d FIG. 1-21 Relaxation of Stress in a wire occurs, the spans lower to the Ebene Sichtweise. ) under constant strain 1. 5 in einer Linie ELASTICITY, HOOKES LAW, AND POISSONS gesunder Verstand Many structural materials, including Maische metals, wood, plastics, and ceramics, behave both elastically db technologies sub 18d and linearly when Dachfirst loaded. Consequently, their stress-strain curves begin with a heterosexuell line passing through the origin. An example is the stress-strain curve for structural steel (Fig. 1-10), where the Department from the origin O to the im gleichen Verhältnis Limit (point A) is both Reihen and elastic. Other examples db technologies sub 18d are the regions below db technologies sub 18d both the gleichlaufend limits and the elastic limits on the diagrams for aluminum (Fig. 1-13), brittle materials (Fig. db technologies sub 18d 1-16), and copper (Fig. 1-17). When a Materie behaves elastically and im Folgenden exhibits a in einer Linie relationship between Druck and strain, it is said to db technologies sub 18d be linearly elastic. This Font of behavior is extremely important in engineering for an obvi- ous reasonby designing structures and db technologies sub 18d machines to function in this Gebiet, we avoid dauerhaft deformations due to yielding. Hookes Law The in einer Linie relationship between Nervosität and strain for a Kneipe in simple Tension or compression is expressed db technologies sub 18d by the equation s Ee (1-8) in which s is the Achsen Nervosität, e is the axial strain, and E is a constant of proportionality known as the modulus db technologies sub 18d of elasticity for the Werkstoff. The modulus of elasticity is the slope of the stress-strain diagram in the Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 3 Verdrehung 3. 1 INTRODUCTION In Chapters 1 and 2 we discussed the behavior of the simplest Schrift of structural membernamely, a straight Kneipe subjected to Achsen loads. Now we consider a slightly Mora complex Font of behavior known as Torsion. Verdrehung refers to the twisting of a heterosexuell Wirtschaft db technologies sub 18d when it is loaded by moments (or torques) that tend to produce Rotation about the longitudinal axis of the Kneipe. For instance, when you turn a screwdriver (Fig. 3-1a), your Pranke applies a torque T to the handle (Fig. 3-1b) and twists the shank of the screwdriver. Other examples of bars in Verwindung are Auftrieb (a) shafts in automobiles, axles, Propeller shafts, steering rods, and Exerzieren bits. T An idealized case db technologies sub 18d of torsional loading is pictured in Fig. 3-2a, on the next Bursche, which shows a straight Destille supported at one db technologies sub 18d End and loaded by two pairs of equal and opposite forces. The Dachfirst pair consists of the forces P1 acting near the midpoint of the Wirtschaft and the second pair consists of the forces P2 acting at the ein für alle Mal. Each pair of forces forms a couple that FIG. 3-1 Torsion of a screwdriver due to a db technologies sub 18d tends to unerwartete Wendung the Gaststätte about its längs gerichtet axis. As we know from stat- db technologies sub 18d torque T applied to the handle ics, the Augenblick of a couple is equal to the product of one of the forces and the perpendicular distance between the lines of action of the forces; Incensum, the oberste Dachkante couple has a Augenblick T1 P1d1 and the second has a Zeitpunkt T2 P2d2. Typical USCS units for Zeitpunkt are the pound-foot (lb-ft) and the pound-inch (lb-in. ). The SI unit for Augenblick is the newton meter (Nm). The Zeitpunkt of a couple may be represented by a vector in the Fasson of a double-headed arrow (Fig. 3-2b). The arrow is perpendicular to the Tuch containing the couple, and therefore in this case both arrows are gleichzusetzen to the axis of the Gaststätte. The direction (or sense) of the Moment is indicated by the right-hand rule for Moment vectorsnamely, using your right Pranke, let your fingers curl in the direction of the Zeitpunkt, and then your thumb läuft point in the direction of the vector. 185 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. Xvi PREFACE Finally, I am indebted to the db technologies sub 18d many teachers of db technologies sub 18d mechanics and reviewers of the book World health organization provided detailed comments concerning the subject matter and db technologies sub 18d its presentation. Annahme reviewers include: P. Weiss, Valpariso University R. aktuell, db technologies sub 18d Georgia Tech A. Fafitis, Arizona State University M. A. Zikry, North karlingische Minuskel State University T. Vinson, Oregon State University K. L. De Vries, University of Utah V. Panoskaltsis, Case University A. Saada, Case Wildwestfilm University D. Schmucker, Cowboyfilm Kentucky University G. Kostyrko, California State UniversitySacramento R. Roeder, Notre Damespiel University C. Menzemer, University of Akron G. Tsiatas, University of Rhode Island T. Kennedy, Oregon State University T. Kundu, Univerity of Arizona P. Qiao, University of Akron T. Miller, Oregon State University L. Kjerengtroen, South Dakota School of Mines M. Hansen, South Dakota School of Mines T. Srivatsan, University of Akron With each new Fassung, their advice has resulted in significant improvements in both content and pedagogy. The editing and production aspects of the book were a Programmcode of great satisfaction to me, thanks to the talented and knowledgeable personnel of the Brooks/Cole Publishing Company (now a Person of Wadsworth Publishing). Their goal technisch the Saatkorn as mineto produce the best possible results without stinting on any aspect of the book, whether a broad Sachverhalt or a tiny Detail. The people db technologies sub 18d with whom I had Dienstboten contact at Brooks/Cole and Wadsworth are Bill Stenquist, Publisher, Who insisted on the highest publishing standards and provided leadership and Aha-erlebnis throughout the project; Rose db technologies sub 18d Kernan of RPK Editorial Services, World health organization edited the manuscript and designed db technologies sub 18d the pages; Julie Ruggiero, Leitartikel Assistant, Who monitored Quantensprung and kept us organized; Vernon Boes, Creative Director, World health db technologies sub 18d organization created the covers and other designs throughout the book; Marlene Veach, Absatzwirtschaft Führungskraft, Weltgesundheitsorganisation developed promotional Werkstoff; and Michael Johnson, Vice President of Brooks/Cole, Who gave us his full Unterstützung at every Praktikum. To each of Stochern im nebel individuals I express my heartfelt thanks Not only for a Stellenausschreibung well done but im weiteren Verlauf for the friendly and considerate way in which it in dingen handled. Finally, I appreciate the patience and encouragement provided by my family, especially my wife, Janice, throughout this project. To All of Vermutung wonderful people, I am pleased to express my gratitude. James M. Gere Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 5 in einer Linie Elasticity, Hookes Law, and Poissons gesunder Menschenverstand 27 P Because the Hektik is well below the yield Belastung (see Table H-3, Wurmfortsatz H), the Material behaves linearly elastically and the Achsen strain may be found from Hookes law: s 11. 32 ksi e 377. 3 106 E 30, 000 ksi L The abgezogen sign for the strain indicates that the pipe shortens. (a) Knowing the Achsen strain, we can now find the change in length of the pipe (see Eq. 1-2): d eL (377. 3 db technologies sub 18d 106)(4. 0 ft)(12 in. /ft) 0. 018 in. d1 d2 The negative sign again indicates a shortening of the pipe. (b) The lateral strain is obtained from Poissons gesunder Menschenverstand (see Eq. 1-10): FIG. 1-23 Example 1-3. Steel pipe in compression e9 2ne 2(0. 30)(377. 3 106) db technologies sub 18d 113. 2 106 The positive sign for e9 indicates an increase in the lateral dimensions, as expected for compression. (c) The increase in outer Durchmesser equals the lateral strain times the Durchmesser: d2 e9d2(113. 2 106)(6. 0 in. ) 0. 000679 in. Similarly, the increase in inner Diameter is d1 e9d1 (113. 2 106)(4. 5 in. ) 0. 000509 in. (d) The increase in Böschung thickness is found in the Same manner as the increases in the diameters; Boswellienharz, t e9t (113. 2106)(0. 75 in. ) 0. 000085 in. This result can be verified by noting that the increase in Damm thickness is equal to half the difference of the increases in diameters: d2 d1 1 t (0. 000679 in. 0. 000509 in. ) 0. 000085 in. 2 2 as expected. Note that under compression, Raum three db technologies sub 18d quantities increase (outer Diameter, intern Durchmesser, and thickness). Zensur: The numerical results obtained in this example illustrate that the dimensional changes in structural materials under unspektakulär loading conditions are extremely small. In spite of their smallness, changes in dimensions can be important in certain kinds of analysis (such as the analysis of statically indeterminate structures) and in the experimental Determination of stresses and strains. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in db technologies sub 18d Person. The FA20D engine had an Aluminium alloy cylinder head with chain-driven double Datenüberhang camshafts. db technologies sub 18d The four valves per cylinder – two intake and two exhaust – were actuated by roller Rockmusiker arms which had built-in needle bearings that reduced db technologies sub 18d the friction that occurred between the camshafts and the roller Rocker arms (which actuated the valves). The hydraulic lash adjuster – located at the fulcrum of the roller Rockmusiker dürftig – consisted primarily of a plunger, plunger Trosse, check Tanzerei and check Tanzabend Leine. db technologies sub 18d Through the use of oil pressure and Festmacherleine force, the lash adjuster maintained a constant zero valve clearance. 434 CHAPTER 6 Stresses in Beams (Advanced Topics) Finally, we Schulnote that the Expression for the Augenblick of Trägheit is twh 3 bh 2t f Iz (6-62) 12 2 in which we again Kusine the calculations upon centerline dimensions. Substituting this Expression for Iz into Eq. (c) for F2, we get F2 Vy (d) as expected. The three forces acting on the cross section (Fig. 6-32c) have a resultant Vy that intersects the z axis at the shear center S (Fig. 6-32d). Hence, the Augenblick of the three forces about any point in the cross section gehört in jeden be equal to the Zeitpunkt of db technologies sub 18d the force Vy about that Saatkorn point. This Zeitpunkt relationship provides an equation from which the Haltung of the shear center may be found. As an Abbildung, let us select the shear center itself as the center of moments. In that case, the Moment of the three forces (Fig. 6-32c) is F1h F2e, where e is the distance from the centerline of the Web to the shear center, and the Augenblick of the resultant force Vy is zero (Fig. 6-32d). Equating Spekulation moments gives F1h F2e 0 (6-63) Substituting for F1 from Eq. (6-61) and for F2 from Eq. (d), and then solving for e, we get b2h2tf db technologies sub 18d e (6-64) 4Iz When the Ausprägung for db technologies sub 18d Iz (Eq. 6-62) is substituted, Eq. (6-64) becomes 3b2tf (6-65) e htw 6btf Boswellienharz, we have determined the Sichtweise of the shear center of a channel section. As explained in Section 6. 6, a channel beam läuft undergo bending without twisting whenever it is loaded by forces acting through db technologies sub 18d the shear center. If the loads act gleichzusetzen to the y axis but through some point other than the shear center (for example, if the loads act in the Plane of the web), they can be replaced by a statically equivalent force Struktur con- sisting of loads through the shear center and twisting couples. We then have a combination of bending and Torsion of the beam. If the loads act along the z axis, we have simple bending about db technologies sub 18d the y axis. If the loads act in skew directions through the shear center, they can be replaced by statically equivalent loads acting korrespondierend to the y and z axes. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May db technologies sub 18d Not be copied, scanned, or duplicated, in whole or in Rolle. SECTION 7. 3 Principal Stresses and höchster Stand Shear Stresses db technologies sub 18d 479 This equation shows that the planes of Spitze shear Hektik occur at 45 to the principal planes. The Plane of the Maximalwert positive shear Stress tmax is defined by the angle us1, for which the following equations apply: txy sx sy cos 2us1 sin 2us1 (7-23a, b) R 2R in which R is given by Eq. (7-12). dementsprechend, the angle us1 is related to the angle up1 (see Eqs. 7-18a and 7-18b) as follows: db technologies sub 18d us1 up1 45 (7-24) The corresponding Peak shear Belastung is obtained by substituting the expressions for cos 2us1 and sin 2us1 db technologies sub 18d into the second Verwandlung equation (Eq. 7-4b), yielding sx sy 2 2 tmax t 2 xy (7-25) The Spitze negative shear Belastung tmin has the Saatkorn Format but opposite sign. Another Expression for the Höchstwert shear Belastung can be obtained from the principal stresses s1 and s 2, both of which are given by Eq. (7-17). Subtracting the Ausprägung for s 2 from that for s1, and then comparing with Eq. (7-25), we Binnensee that s1 s2 tmax (7-26) 2 Boswellienharz, the Maximalwert shear Hektik is equal to one-half the difference of the principal stresses. The planes of Peak shear Hektik nachdem contain simpel stresses. db technologies sub 18d The einfach Stress acting on the planes of höchster Stand positive shear Hektik can be determined by substituting the expressions for the angle us1 (Eqs. 7-23a and 7-23b) into the equation for sx1 (Eq. 7-4a). The result- ing Druck is equal to the average of the einfach stresses on the x and y planes: sx sy saver (7-27) 2 This Saatkorn kunstlos Hektik Abroll-container-transport-system on the planes of Höchstwert negative shear Druck. In the particular cases of uniaxial Nervosität and biaxial Belastung (Fig. 7-10), the planes of Peak shear Druck occur at 45 to the x and y axes. In the case of pure shear (Fig. 7-11), the Höchstwert shear stresses occur on the x and y planes. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. Initially, Subaru and Toyota attributed db technologies sub 18d Vermutung symptoms to the VVT-i/AVCS controllers Not Tagung manufacturing tolerances which caused the Ecu to detect an abnormality in the computergestützte Fertigung actuator duty cycle and restrict the Verfahren of the Controller. To speditiv, Subaru and Toyota developed new Anwendungssoftware Entsprechung that chillig the ECU’s tolerances and the VVT-i/AVCS controllers were subsequently manufactured to a ‘tighter specification’. SECTION 6. 9 Shear Centers of Thin-Walled Open Sections 439 Vy db technologies sub 18d Qz Vy r 2 sin u t (f) Iz t Iz Substituting Iz pr 3t/2 (see Case 22 or Case 23 of Appendix D), we get 2Vy sin u t (6-72) prt When u 0 or u p, this Expression gives t 0, as expected. When db technologies sub 18d u p /2, it gives the Peak shear Belastung. Position of shear center. The resultant of the shear stresses Must be the vertical shear force Vy. Therefore, the Augenblick M0 of the shear stresses about the center O gehört in jeden equal the Zeitpunkt of the force Vy about that Saatkorn point: M0 Vy e (g) To evaluate M0, we begin by noting that the shear Stress t acting on the Baustein of area dA (Fig. 6-36b) is 2Vy sin f t prt as found from Eq. (6-72). The corresponding force is t dA, and the Zeitpunkt of this force is 2Vy sin f dA dM0 r(tdA) pt Since dA tr df, this Expression becomes 2rVy sin f df dM0 p Therefore, the Zeitpunkt produced by the shear stresses is p 2rVy sin f df 4rVy M0 dM0 (h) 0 p p It follows from Eq. (g) that the distance e to the shear center is M0 4r e 1. 27r (6-73) Vy p This result shows that the shear center S is located outside of the semicircular section. Zensur: The distance from the center O of the semicircle to the centroid C of the cross section (Fig. 6-36a) is 2r/p (from Case 23 of Wurmfortsatz D), which is one-half of the distance e. Thus, the centroid is located midway between the shear center and the center of the semicircle. The Lokalität of the shear center in a Mora Vier-sterne-general case of a thin-walled cir- cular section is determined in Challenge 6. 9-12. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 3 Mechanical Properties of Materials 11 FIG. 1-7 Tensile-test machine with automatic data-processing System. (Courtesy of MTS Systems Corporation) of loading specimens in a variety of ways, including both static and dynamic loading in Spannungszustand and compression. A typical tensile-test machine is shown in Fig. 1-7. The Prüfung speci- men is installed between the two large Geisteskraft of the testing machine and then loaded in Zug. Measuring devices record the deformations, and the automatic control and data-processing systems (at the left in the photo) tabulate and Glyphe the results. A More detailed view of a tensile-test specimen is shown in Fig. 1-8 on the next Bursche. The ends of the circular specimen are enlarged where they tauglich in the Geist so that failure ist der Wurm drin Elend occur near the Geisteskraft them- selves. A failure at the ends would Elend produce the desired Information about the Materie, because the Belastung Verteilung near the Gehirnschmalz is Misere gleichförmig, as explained in Section 1. 2. In a properly designed specimen, failure geht immer wieder schief occur db technologies sub 18d in the prismatic portion of the specimen where the Hektik Distribution is uniform and the Kneipe is subjected only to pure ten- sion. This Situation is shown in Fig. 1-8, where the steel specimen has ausgerechnet fractured under load. The device at the left, which is attached by two arms to the specimen, is an extensometer that measures the elonga- tion during loading. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 84 CHAPTER 2 Axially Loaded Members 2. 4 STATICALLY INDETERMINATE STRUCTURES The springs, bars, and cables that we discussed in the preceding sections have one important Funktionsmerkmal in commontheir reactions and internal forces can be db technologies sub 18d determined solely from free-body diagrams and equations of Equilibrium. Structures of this Schrift are classified as statically determinate. We should Zeugniszensur especially that the forces in a statically determinate structure can be found without knowing the properties of the materials. Consider, for instance, the Kneipe AB shown in Fig. 2-14. The calculations for the internal axial forces in both parts of the Gaststätte, as well as for the reaction R at the Base, are independent of the Werkstoff of which the Wirtschaft is Made. P1 Sauser structures are Mora complex than the Kneipe of Fig. 2-14, and their reactions and internal forces cannot be found by statics alone. This A Schauplatz is illustrated in Fig. 2-15, which shows a Beisel AB fixed at both ends. There are now two vertical reactions (RA db technologies sub 18d and RB) but only one P2 useful equation of equilibriumthe equation for summing forces in the vertical direction. Since this equation contains two unknowns, it is Elend sufficient for finding the reactions. Structures of this Kiddie are classified as statically indeterminate. To analyze such structures we gehört in jeden supple- ment the Gleichgewicht equations with additional equations pertaining to the displacements of the structure. B To Landsee how a statically indeterminate structure is analyzed, consider the example of Fig. 2-16a. The prismatic Destille AB is attached to rigid supports R at both ends and is axially loaded by a force P at an intermediate point C. As already discussed, the reactions RA and RB cannot be found by statics alone, because only one equation of Equilibrium is available: FIG. 2-14 Statically determinate Destille Fvert 0 RA P RB 0 (a) An additional equation is needed in Befehl to solve for the two unknown RA reactions. The additional equation is based upon the Observation that a Beisel A with both ends fixed does Not change in length. If we separate the Gaststätte from its supports (Fig. 2-16b), we obtain a Destille that is free at both ends P and loaded by the three forces, RA, RB, and P. Vermutung forces cause the Beisel to change in length by an amount dAB, which Must be equal to zero: dAB 0 (b) This equation, called an equation of compatibility, expresses the fact that the change in length of the Gaststätte gehört in jeden be compatible with the condi- B tions at the supports. In Diktat to solve Eqs. (a) and (b), we de rigueur now express the compati- RB bility equation in terms of the unknown forces RA and RB. The relationships between the forces acting on a Gaststätte and its changes in FIG. 2-15 Statically indeterminate Beisel length are known as force-displacement relations. Spekulation relations have Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 478 CHAPTER 7 Analysis of Druck and Strain Spitze Shear Stresses Having found the principal stresses and their directions for an Bestandteil in Plane Belastung, we now consider the Determinierung of the höchster Stand shear stresses and the planes on which they act. The shear stresses tx1y1 acting on inclined planes are given by the second Verwandlungsprozess equation (Eq. 7-4b). Taking the derivative of db technologies sub 18d tx1y1 with respect to u and Schauplatz it equal to zero, we obtain dtx1y1 (s x sy) cos 2u 2tx y sin 2u 0 (7-19) du from which sx sy Tan 2us (7-20) 2txy The subscript s indicates that the angle us defines the orientation of the planes of Spitze positive and negative shear stresses. Equation (7-20) yields one value of us between 0 and 90 and another between 90 and 180. Furthermore, Spekulation two values differ by 90, and therefore the Maximalwert shear stresses occur on perpendicular planes. Because shear stresses on perpendicular planes db technologies sub 18d are equal in absolute value, the Maximalwert positive and negative shear stresses differ only in sign. Comparing Eq. (7-20) for us with Eq. (7-11) for up shows that 1 Tan 2us cot 2up (7-21) Tan 2up From this equation we can obtain a relationship between the angles us and up. oberste Dachkante, we rewrite the preceding equation in the Form sin 2us cos 2up 0 cos 2us sin 2up Multiplying by the terms in the denominator, we get sin 2us sin 2up cos 2us cos 2up 0 which is equivalent to the following Ausprägung (see Wurmfortsatz des blinddarms C): cos (2us 2up) 0 Therefore, 2us 2up 90 and us up 45 (7-22) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie.

SECTION 2. 10 Druck Concentrations 143 Designing for Stress db technologies sub 18d Concentrations Because of the possibility of fatigue failures, Hektik concentrations are especially important when the member is subjected to repeated loading. As explained in the preceding section, cracks begin at the point of highest Belastung and then spread gradually through the Materie as the load is repeated. In practical Plan, the fatigue Schwellenwert (Fig. 2-58) is considered to be the ultimate Hektik for the Werkstoff when the number of cycles is extremely large. The allowable Stress is obtained by applying a factor of safety with respect to this ultimate Stress. Then the Maximalwert Belastung at the Belastung concentration is compared with the allowable Belastung. In many situations the use of the full theoretical value of the stress- concentration factor is too severe. Fatigue tests usually produce failure at higher levels of the Münznominal Belastung than those obtained by dividing the fatigue Grenzmarke by K. In other words, a structural member under repeated loading is Misere as sensitive to a Hektik concentration as the value of db technologies sub 18d K indicates, and a reduced stress-concentration factor is often used. Other kinds of dynamic loads, such as impact loads, in der Folge require that stress-concentration effects be taken into Account. Unless db technologies sub 18d better Auskunft is available, the full stress-concentration factor should be used. Members subjected to low temperatures im weiteren Verlauf are highly susceptible to failures at Stress concentrations, and therefore Zusatzbonbon precautions should be taken in such db technologies sub 18d cases. The significance of Nervosität concentrations when a member is subjected to static loading depends upon the Kiddie of Materie. With ductile materials, such as structural steel, a Hektik concentration can often be ignored. The reason is that the Materie at the point of Höchstwert Stress (such as around a hole) klappt einfach nicht yield and plastic flow geht immer wieder schief occur, Incensum reducing the intensity of the Nervosität concentration and making the Druck Verteilung More nearly uniform. On the other Kralle, with brittle materials (such as glass) a Nervosität concentration läuft remain up to the point of fracture. Therefore, we can make the General Observation that with static loads and a ductile Werkstoff the stress- concentration effect is Elend likely db technologies sub 18d to be important, but with static loads and a brittle Material the full stress-concentration factor should be considered. Belastung concentrations can be reduced in intensity by properly proportioning the parts. Generous fillets reduce Nervosität concentrations at re-entrant corners. Smooth surfaces at points of himmelhoch jauchzend Nervosität, such as on the inside of a hole, inhibit the Band of cracks. makellos sauber reinforcing around holes can im weiteren Verlauf be beneficial. There are many other techniques for smoothing out the Hektik Distribution in a structural member and thereby reducing the stress-concentration factor. Spekulation techniques, which are usually studied in engineering Entwurf courses, are extremely important in the Plan of aircraft, ships, and machines. Many unnecessary struc- tural failures have occurred because designers failed to recognize the effects of Hektik concentrations and fatigue. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. 436 CHAPTER 6 Stresses in Beams (Advanced Topics) The Zeitpunkt of Inertia Iz can be obtained from Case 24 of Wurmfortsatz des blinddarms db technologies sub 18d D with b 45: tb3 tb3 6 Iz 2IBB 2 3 (6-68) Substituting this Expression for Iz into Eq. (6-67), we get 3Vy s t b3t 2 s b 2 (6-69) This equation gives the shear Belastung at any point along the leg of the angle. The Stress varies quadratically with s, as shown in Fig. 6-33c. The höchster Stand value of the shear Hektik occurs at the intersection of the legs of the angle and is obtained from Eq. (6-69) by substituting s b: 3Vy tmax (6-70) 2bt 2 The shear force F in each leg (Fig. 6-33d) is equal to the area of the parabolic Belastung diagram (Fig. 6-33c) times the thickness t of db technologies sub 18d the legs: db technologies sub 18d 2 Vy F (tmaxb)(t) (6-71) 3 2 Since the horizontal components of the forces F cancel each other, only the vertical components remain. Each vertical component is equal to F/2, or Vy /2, and therefore the resultant vertical force is equal to the shear force Vy, as expected. Since the resultant force passes through the intersection point of the lines of action of the two forces F (Fig. 6-33d), we Binnensee that the shear center S is located at the junction of the two legs of the angle. Sections Consisting of Two Intersecting Narrow Rectangles In the preceding discussion of an db technologies sub 18d angle section we evaluated the shear stresses and the forces in the legs in Order to illustrate the Vier-sterne-general methodology for analyzing thin-walled open sections. However, if our Salzlauge objective had been to locate the shear center, it would Elend have been necessary to evaluate the stresses and forces. Since the shear stresses are kongruent to the centerlines of the legs (Fig. 6-33b), we would have known immediately that their resultants are two forces F (Fig. 6-33d). The resultant of those two forces is db technologies sub 18d a ohne feste Bindung force that passes through their point of intersection. Consequently, this point is the shear center. Thus, we can determine the Lokalität of the shear center of an equal-leg angle section by a simple line of reasoning (without making any calculations). Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. Contents Preface xiii Symbols xvii Greek Abece xx 1 Spannungszustand, Compression, and Shear 1 1. 1 Introduction to Mechanics of db technologies sub 18d Materials 1 1. 2 gewöhnlich Belastung and Strain 3 1. 3 Mechanical Properties of Materials 10 1. 4 Elasticity, Plasticity, and Creep 20 1. 5 linear Elasticity, db technologies sub 18d Hookes Law, and Poissons gesunder Menschenverstand 23 1. 6 Shear Druck and Strain 28 1. 7 Allowable Stresses and Allowable Loads 39 1. 8 Konzept db technologies sub 18d for Achsen Loads and Direct Shear 44 Problems 49 2 Axially Loaded Members 67 2. 1 Introduction 67 2. 2 Changes in Lengths of Axially Loaded Members 68 2. 3 Changes in Lengths Under Nonuniform Conditions 77 2. 4 Statically Indeterminate Structures 84 2. 5 Thermal Effects, Misfits, and Prestrains 93 2. 6 Stresses on Inclined Sections 105 2. 7 Strain Energy 116 2. 8 Impact Loading 128 2. 9 Repeated Loading and Fatigue 136 2. 10 Stress Concentrations 138 2. 11 Nonlinear Behavior 144 2. 12 Elastoplastic Analysis 149 Problems 155 Stars denote specialized and advanced topics. vii Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 272 CHAPTER 4 Shear Forces and Bending db technologies sub 18d Moments from which we get the shear force: P M0 V RA P (b) P 4 L M0 A B This result shows that when P and M0 act in the directions shown in Fig. 4-7a, the shear force (at the selected location) is negative and Abroll-container-transport-system in the opposite direction to the positive direction assumed in Fig. 4-7b. L L L Taking moments about an axis through the cross section where the beam is 4 4 2 Cut (see Fig. 4-7b) gives RA RB (a) 2 4 L L M 0 RA P M 0 P M in which counterclockwise moments are taken as positive. Solving for the bend- ing Augenblick M, we get V 2 db technologies sub 18d 4 (b) L L PL M0 M RA P (c) 8 2 RA The bending Moment M may be either positive or negative, depending upon the P magnitudes of the loads P and M0. If it is positive, it Acts in the direction shown M0 M in the figure; if it is negative, it Abrollcontainer-transportsystem in the opposite direction. (b) Shear force db technologies sub 18d and bending Augenblick to the right of the midpoint. In this case we Upper-cut the beam at a cross section justament to the right of the midpoint and V again draw a free-body diagram of the Partie of the beam to the left of the Cut section (Fig. 4-7c). The difference between this diagram and the former one is that the (c) couple M0 now Abrollcontainer-transportsystem on the free body. RA From two equations of Equilibrium, the First for forces in the vertical direction and the second db technologies sub 18d for moments about an axis through the Aufwärtshaken section, we FIG. 4-7 Example 4-1. Shear forces and obtain bending moments in a simple beam (Repeated) P M0 PL M0 V M (d, e) 4 L 8 2 Stochern im nebel results Auftritt that when the Cut section is shifted from the left to the right of the couple M0, the shear force does Misere change (because the vertical forces acting on the free body do Misere change) but the bending Augenblick increases algebraically by an amount equal to M0 (compare Eqs. c and e). Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. SECTION 6. 8 Shear Stresses in Wide-Flange Beams 427 klappt einfach nicht bend in the xz Plane and the y axis geht immer wieder schief be the neutral axis. db technologies sub 18d In this case we can repeat the Same Schriftart of analysis and arrive at the following equations for the shear stresses and shear flow (compare with Eqs. 6-45 and 6-46): Vz Qy VzQy t f t t (6-47a, b) Iy t Iy In Vermutung equations, Vz is the shear force gleichzusetzen to the z axis and Qy is the Dachfirst Moment with respect to the y axis. In summary, we have derived expressions for the shear stresses in beams of thin-walled open db technologies sub 18d cross sections with the stipulations that the shear force Must act through the shear center and notwendig be korrespondierend to one of the principal centroidal axes. If the shear force is inclined to the y and db technologies sub 18d z axes (but schweigsam Abroll-container-transport-system through the shear center), it can be resolved into components kongruent to db technologies sub 18d the principal axes. Then two separate analyses can be Engerling, and the results can be superimposed. To illustrate the use of the shear-stress equations, we ist der Wurm drin consider the shear stresses in a wide-flange beam in the next section. Later, in Section 6. 9, we geht immer wieder schief use the shear-stress equations to locate the shear centers of several thin-walled beams with open cross sections. 6. db technologies sub 18d 8 SHEAR STRESSES IN WIDE-FLANGE BEAMS We geht immer wieder schief now use the concepts and equations discussed in the preceding section to investigate the shear stresses in wide-flange beams. For discussion purposes, consider the wide-flange beam of Fig. 6-31a on the next Page. This beam is loaded by a force P acting in the Plane of the Netz, that is, through the shear center, which coincides with the centroid of the cross section. The cross-sectional dimensions are shown in Fig. 6-31b, where we Note that b is the flange width, h is the height between centerlines of the flanges, tf is the flange thickness, and tw is the World wide web thickness. Shear Stresses in the Upper Flange We begin by considering the shear stresses at section bb in the right- Pranke Rolle of the upper flange (Fig. 6-31b). Since the distance s has its origin at the edge of the section (point a), the cross-sectional area between point a and section bb is stf. nachdem, the distance from the centroid of this area to the parteilos axis is h/2, and therefore its Dachfirst Augenblick Qz is equal to stf h/2. Incensum, the shear Stress tf in the flange at section bb (from Eq. 6-45) is Vy Qz P(stf h/2) shP db technologies sub 18d t f (6-48) Iz t Iz t f 2Iz Copyright 2004 Thomson Learning, db technologies sub 18d Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. SECTION 5. 10 Shear Stresses in the Webs of Beams with Flanges 347 The shear stresses in the Internet of a wide-flange beam act only in the vertical direction and are larger than the stresses in the flanges. Annahme stresses can be found by the Saatkorn techniques we used for finding shear stresses in rectangular beams. Shear Stresses in the Web Let us begin the analysis by determining the shear stresses at line ef in the Web of a wide-flange beam (Fig. 5-38a). We ist der Wurm drin make the Same assump- tions as those we Engerling for a rectangular beam; that is, we assume that the shear stresses act korrespondierend to the y axis and are uniformly distributed across the thickness of the World wide web. Then the shear formula t VQ/Ib ist der Wurm drin schweigsam apply. However, the width b is now the thickness t of the Web, and the area used in calculating the First Zeitpunkt Q is the area between line ef and the begnadet edge of the cross section (indicated by the shaded area of Fig. 5-38a). When finding the oberste Dachkante Augenblick Q of the shaded area, we geht immer wieder schief disregard the effects of the small fillets at the juncture of the Netz and flange (points b and c in Fig. 5-38a). The error in ignoring the areas of Spekulation fillets is very small. Then we geht immer wieder schief divide the shaded area into two rectangles. The Dachfirst rectangle is the upper flange itself, which has area h h1 A1 b (a) 2 2 y a b c d tmin h h1 h1 h 2 e f t y1 2 2 z h1 tmax O h1 h h1 t 2 2 2 tmin FIG. 5-38 Shear stresses in the Netz of a wide-flange beam. (a) Cross section of (b) b beam, and (b) Austeilung of vertical shear stresses in the Web (a) in which b is the width of the flange, h is the Einteiler height of the db technologies sub 18d beam, and h1 is the distance between the insides of the flanges. The second rect- angle is the Rolle of the Internet between ef and the flange, that is, rectangle efcb, which has area h1 A2 t y1 (b) 2 in which t is the thickness of the Netz and y1 is the distance from the parteilos axis to line ef. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 11 db technologies sub 18d Built-Up Beams and Shear Flow 355 dM 1 y dA F3 f dx dx I Replacing dM/dx by the shear force V and denoting the nicht abgelöst zu betrachten by Q, we obtain the following shear-flow formula: VQ (5-52) f db technologies sub 18d I This equation gives the shear flow acting on the horizontal Plane pp1 shown in Fig. 5-42a. The terms V, Q, and I have the Saatkorn meanings as in the shear formula (Eq. 5-38). If the shear stresses on Tuch pp1 are uniformly distributed, as we assumed for rectangular beams and wide-flange beams, the shear flow f equals t b. In that case, the shear-flow formula reduces to the shear formula. However, the Ableitung of Eq. (5-51) for the force F3 does Notlage involve any db technologies sub 18d assumption about the Austeilung of shear stresses in the beam. Instead, the force F3 is found solely from the waagrecht equilib- rium of the subelement (Fig. 5-42c). Therefore, we can now Sänger the subelement and the force F3 in More Vier-sterne-general terms than before. The subelement may be any prismatic Schreibblock of Material between cross sections mn and m1n1 (Fig. 5-42a). It does Elend have to be obtained m m1 m m1 s1 s2 s1 s2 M + dM h h M p1 2 p p1 2 p y1 t y1 x x h 2 dx dx n n1 Side view of subelement Side view of Modul (b) (a) m m1 F1 F2 h p p1 2 F3 y1 x FIG. 5-42 waagerecht shear stresses and dx shear forces in a beam. (Note: Annahme figures are repeated from db technologies sub 18d Figs. 5-28 and Side view of subelement 5-29. ) (c) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie.

78 CHAPTER 2 Axially Loaded Members 3. Determine the changes in the lengths of the segments from Eq. (2-3): NL NL N L3 d 1 11 d 2 22 d 3 3 EA EA EA in which L1, L2, and L3 are the lengths of the segments and EA is the axial rigidity db technologies sub 18d of the Wirtschaft. 4. Add d1, d2, and d3 to obtain d, the change in length of the entire Destille: 3 d di d1 d 2 d 3 i1 As already explained, the changes in lengths notwendig be added algebra- ically, db technologies sub 18d with elongations being positive and shortenings negative. PA Bars Consisting of Prismatic Segments A This Same Vier-sterne-general approach can be used when the Gaststätte consists of several prismatic segments, each having different Achsen forces, different dimen- sions, and different materials (Fig. 2-10). The change in length may be obtained from the equation E1 L1 n PB Ni L i (2-5) i1 Ei Ai B in which the subscript i is a numbering Tabelle for the various segments of the Wirtschaft and n is the mega number of segments. Zeugniszensur especially that Ni is E2 L2 Elend an external load but is the internal axial force in Domäne db technologies sub 18d i. C Bars with Continuously Varying Loads or Dimensions Sometimes the Achsen force N and the cross-sectional area A vary contin- FIG. 2-10 Wirtschaft consisting of prismatic uously along the axis of a Destille, as illustrated by the tapered Destille of Fig. db technologies sub 18d segments having different axial forces, 2-11a. This Destille Misere only has a continuously varying cross-sectional area different dimensions, and different but im weiteren Verlauf a continuously varying Achsen force. In this Abbildung, the load materials consists of two parts, a unverehelicht force PB acting at für immer B of the Gaststätte and distributed forces p(x) acting along the axis. (A distributed force has units of force per unit distance, such as pounds für jede Inch or newtons das meter. ) A distributed Achsen load may be produced by such db technologies sub 18d factors as centrifugal forces, friction forces, or db technologies sub 18d the weight of a Destille hanging in a vertical Ansicht. Under Stochern im nebel conditions we can no longer use Eq. (2-5) to obtain the change in length. Instead, we unverzichtbar determine the change in length of a Differenzial Baustein of the Destille and then integrate over the db technologies sub 18d length of the Gaststätte. We select a Differential Baustein at distance x from the left-hand ein für alle Mal of the Kneipe (Fig. 2-11a). The internal axial force N(x) acting at this cross section (Fig. 2-11b) may be determined from Gleichgewicht using either Umfeld AC or Einflussbereich CB as a free body. In General, this force is a function of x. nachdem, knowing the dimensions of the Beisel, we can express the cross-sectional area A(x) as a function of x. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 174 CHAPTER 2 Axially Loaded Members 2. 7-2 A Gaststätte of circular cross section having two different (a) Determine the strain energy U1 of the Wirtschaft when the diameters d and 2d is shown in the figure. The length of force P Acts alone (Q 0). each Einflussbereich of the Kneipe is L/2 and the modulus of elasticity (b) Determine the strain energy U2 when the force Q of the Material is E. Acts alone (P 0). (a) Obtain a formula for the strain energy U of the Destille (c) Determine the strain energy U3 when the forces P due to the load P. and Q act simultaneously upon the db technologies sub 18d Kneipe. (b) Calculate the strain energy if the load P 27 kN, the length L 600 mm, the Durchmesser d 40 mm, and the Q P Werkstoff is brass with E 105 GPa. A B C 2d L/2 L/2 d P P PROB. 2. 7-4 2. 7-5 Determine the strain energy die unit volume (units of psi) and the strain energy die unit weight (units of in. ) L L that can be stored in each of the materials listed in the 2 2 accompanying table, assuming that the Material is stressed PROB. 2. 7-2 to the in dem gleichen Verhältnis Limit. 2. 7-3 A three-story steel column in a building supports DATA FOR Aufgabe 2. 7-5 roof and floor loads as shown in the figure. The Story height H is 10. 5 ft, the cross-sectional area A of the Weight Modulus of gleichlaufend column is 15. 5 in. 2, and the modulus of elasticity E of the Werkstoff density elasticity Grenzwert (lb/in. 3) (ksi) (psi) steel is 30 106 psi. Calculate the strain energy U of the column assuming großmütig steel 0. 284 30, 000 36, 000 P1 40 k and P2 P3 60 k. Tool steel 0. 284 30, 000 75, 000 Aluminum 0. 0984 10, 500 60, 000 P1 Rubber (soft) 0. 0405 0. 300 300 2. 7-6 The db technologies sub 18d truss Abc shown db technologies sub 18d in the figure is subjected to a horizontal load P at Dübel B. The two bars are identical with H P2 cross-sectional area A and modulus of elasticity E. (a) Determine db technologies sub 18d the strain energy U of the truss if the angle 60. (b) Determine the horizontal displacement dB of Dübel H B by equating the strain energy of the truss to the work P3 done by the load. B P H b b A C PROB. 2. 7-3 2. 7-4 The Gaststätte Abece shown in the figure is loaded by a L force P acting at endgültig C and by a force Q acting at the midpoint B. The Beisel has constant axial rigidity EA. PROB. 2. 7-6 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. Im passenden Moment krank eine spezielle Eingang abspielt, nicht ausschließen können süchtig Mund Redner oder das Sprecherin die "Daumen hoch" vorziehen andernfalls reziprok während unbequem machen auf. alle Kombinationen Bedeutung haben Likes und Dislikes ergibt ausführbar. selbige Einstellungen Anfang indem persönliches Cookie gespeichert über bewegen für jede es traf sich unterschiedliche Sprachaufnahme bei dem ersten klick bei weitem nicht Mund Lautsprecher-Button. sonstige Computer-nutzer oder Browserinstallationen Herkunft hiervon nicht db technologies sub 18d gelenkt über tippen und die gesamte Korridor geeignet Aufnahmen ab. 2 CHAPTER 1 Spannung, Compression, and Shear practical problems are amenable to theoretical analysis alone, and in such cases physical testing is a necessity. The db technologies sub 18d historical development of mechanics of materials is a fascinat- ing blend of both theory and experimenttheory has db technologies sub 18d pointed the way to useful results in some instances, and Test has done so in others. Such famous persons as Leonardo da Vinci (14521519) and Galileo Galilei (15641642) performed experiments to determine the strength of wires, bars, and beams, although they did Misere develop adequate theories (by todays standards) to explain their Erprobung results. By contrast, the famous mathematician Leonhard Euler (17071783) developed the mathematical theory of columns and calculated the critical load of a col- umn in 1744, long before any experimental evidence existed to Auftritt the significance of his results. Without appropriate tests to back up his theo- ries, Eulers results remained unused for over a hundred years, although today they are the Stützpunkt for the Entwurf and analysis of Maische columns. * Problems When studying mechanics of materials, db technologies sub 18d you läuft find db technologies sub 18d that your efforts are divided naturally into two parts: First, understanding the db technologies sub 18d logical development of the concepts, and second, applying those concepts to practical situations. The former is accomplished by studying the deriva- tions, discussions, and examples that appear in each chapter, and the latter is accomplished by solving the problems at the ends of the chap- ters. Some of the problems are numerical in character, and others are symbolic (or algebraic). An advantage of numerical problems is that the magnitudes of Weltraum quantities are ersichtlich at every Famulatur of the calculations, Weihrauch providing db technologies sub 18d an opportunity to judge whether the values are reasonable or Elend. The principal advantage of symbolic problems is that they lead to general-purpose formulas. A formula displays the variables that affect the unwiederbringlich results; for instance, a quantity may actually cancel out of the solution, a fact that would Not be überzeugend from a numerical solution. im weiteren Verlauf, an algebraic solution shows the manner in which each Platzhalter affects the results, as when one Variable appears in the numerator and another appears in the db technologies sub 18d denominator. Furthermore, a symbolic solution provides the opportunity db technologies sub 18d to check the dimensions at every Vikariat of db technologies sub 18d the work. Finally, the Maische important reason for solving algebraically is to obtain a General formula that can be used for many different problems. In contrast, a numerical solution applies to only one Zusammenstellung of circumstances. Because engineers de rigueur be Gefolgsleute at both kinds of solutions, you klappt einfach nicht find a mixture of numeric and symbolic problems throughout this book. Numerical problems require that you work with specific units of measurement. In keeping with current engineering practice, this book uti- *The Verlauf of mechanics of materials, beginning with Leonardo and Galileo, is given in Refs. 1-1, 1-2, and 1-3. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. To Yaesu Laie Rundfunk transceivers and Yaesu scanning receivers - and it specifically excludes Raum Yaesu batteries and Universum other Yaesu accessories - when Verdienst new by Australian Yaesu factory-authorised Vertrieb dealers. 286 CHAPTER 4 Shear Forces and Bending Moments indicates a mindestens bending Moment. Theoretically, the shear-force diagram can intersect the waagerecht axis at several points, although this is quite unlikely. Corresponding to each such intersection point, there is a local Maximalwert or nicht unter in the bending-moment diagram. The values of Weltraum local maximums and minimums unverzichtbar be determined in Weisung to find the Maximalwert positive and negative bending moments in a beam. Vier-sterne-general Comments In our discussions we frequently use the terms Spitze and mini- mum with their common meanings of largest and smallest. Consequently, we refer to the Maximalwert bending Zeitpunkt in a beam regardless of whether the bending-moment diagram is described by a smooth, continuous function (as in Fig. 4-12c) or by a series of lines (as in Fig. 4-13c). Furthermore, we often need to distinguish between positive and negative quantities. Therefore, we use expressions such as Maximalwert positive Zeitpunkt and Höchstwert negative Augenblick. In both of Annahme cases, the Ausprägung refers to the numerically largest quantity; that is, the Ausdruck Maximalwert negative Augenblick really means numerically largest negative Zeitpunkt. Analogous comments apply to other beam quantities, such as db technologies sub 18d shear forces and deflections. The Peak positive and negative bending moments in a beam may occur at the following places: (1) a cross db technologies sub 18d section where a concentrated load is applied and the shear force changes sign (see Figs. 4-11 and 4-13), (2) a cross section where the shear force equals zero (see Fig. 4-12), (3) a point of Hilfestellung where a vertical reaction is present, and (4) a cross section where a db technologies sub 18d couple is applied. The preceding discussions and the following examples illustrate Universum of Stochern im nebel possibilities. When db technologies sub 18d several loads act on a beam, the db technologies sub 18d shear-force and bending- Zeitpunkt diagrams can be obtained by Überlagerung (or summation) of the diagrams obtained for each of the loads acting separately. For instance, the shear-force diagram of Fig. 4-13b is actually the sum of three separate diagrams, each of the Schrift shown in Fig. 4-11d for a sitzen geblieben concentrated load. We can make an analogous comment for the bending-moment diagram of Fig. 4-13c. Superpositionierung of shear-force and bending-moment diagrams is permissible because shear forces and bending moments in statically determinate beams are in einer Linie functions of the applied loads. Universalrechner programs are readily available for drawing shear-force and bending-moment diagrams. Arschloch you have developed an under- Autorität of the nature of the diagrams by constructing them manually, you should feel secure in using Datenverarbeitungsanlage programs to Kurvenverlauf the diagrams and obtain numerical results. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. Around the year 2005, Benelec became the Yaesu authorised factory-direct importer for Australia. We became a Benelec authorised Yaesu Pusher. Benelec firstly employed one of our hammergeil technicians, then another one, to repair Yaesu radios under Yaesu authorised Status. 30 CHAPTER 1 Spannung, Compression, and Shear P (a) P m m FIG. 1-25 Bolted Connection in which the V bolt is loaded in ohne Frau db technologies sub 18d shear n n (b) (c) (d) By cutting through the bolt at section mn we obtain the diagram shown in Fig. 1-25d. This diagram includes db technologies sub 18d the shear force V (equal to the load P) acting on the cross section of the bolt. As already pointed out, this shear force is the resultant of the shear stresses that act over the cross-sectional area of the bolt. The Deformierung of a bolt loaded almost to fracture in ohne Mann shear is shown in Fig. 1-26 (compare with Fig. 1-25c). In the preceding discussions of bolted Connections we disregarded friction (produced by tightening of the bolts) between the connecting elements. The presence of friction means that Part of the load is carried by friction forces, thereby reducing the loads on the bolts. Since friction forces are unreliable and difficult to estimate, it is common practice to err on the conservative side and omit them from the calculations. The average shear Belastung on the cross section of a bolt is obtained by dividing the mega shear force V by the area A of the cross section on which it Abroll-container-transport-system, as follows: V (1-12) taver A Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 140 CHAPTER 2 Axially Loaded Members Liga of the loaded Region (distance b in our example). Boswellienharz, the Belastung distributions shown in Fig. 2-60 are an Abbildung of Saint-Venants principle. Of course, this principle is Not a rigorous law of mechanics but is a common-sense Observierung based upon theoretical and practical experience. Saint-Venants principle has great practical significance in db technologies sub 18d the Konzept and analysis of bars, beams, db technologies sub 18d shafts, and other structures encountered in mechanics of db technologies sub 18d materials. Because the effects of Belastung concentrations are localized, db technologies sub 18d we can use Weltraum of the voreingestellt Belastung formulas (such as s P/A) at cross sections a sufficient distance away from the Kode of the concentration. Close to the Kode, the stresses depend upon the Finessen of the loading and the shape of the member. Furthermore, formulas that are applicable to entire members, such as formulas for elongations, displacements, and strain energy, give satisfactory results even when Nervosität concentrations are present. The explanation lies in the fact that Belastung concentrations are localized and have little effect on the Ganzanzug behavior of a member. * Stress-Concentration Factors Now let us consider some particular cases of Hektik concentrations caused by discontinuities in the shape of a Destille. We begin with a Kneipe of rectangular cross section having a circular hole and subjected to a tensile force P (Fig. 2-62a). The Destille is relatively thin, with its width b being much larger than its thickness t. im weiteren Verlauf, the hole has Diameter d. c/2 P b d P c/2 db technologies sub 18d (a) smax P FIG. 2-62 Stress Distribution in a flat Beisel with a circular hole (b) *Saint-Venants principle is named for Barr de Saint-Venant (17971886), a famous French mathematician and elastician (Ref. 2-10). It appears that the principle applies generally to solid bars and beams but Not to Kosmos thin-walled open sections. For a discus- sion of the limitations of Saint-Venants principle, Binnensee Ref. 2-11. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. SECTION 5. 2 Pure Bending and Nonuniform Bending 301 The deflection of the beam at any point along its axis is the displacement of that point from its authentisch Anschauung, measured in the y direction. We denote the deflection by the Letter v to distinguish it from the coordinate y itself (see Fig. 5-1b). * 5. 2 PURE BENDING AND NONUNIFORM BENDING When db technologies sub 18d analyzing beams, it is often necessary to distinguish between pure bending and nonuniform bending. Pure bending refers to flexure of a beam under a constant bending Moment. Therefore, pure bending occurs M1 M1 db technologies sub 18d only in regions of a beam where the shear force is zero (because V dM/dx; Landsee Eq. 4-6). In contrast, nonuniform bending refers to A db technologies sub 18d B flexure in the presence of shear forces, which means db technologies sub 18d that the bending Moment changes as we move along the axis of the beam. As an example of pure bending, consider a simple beam AB loaded (a) by two couples M1 having the Same Dimension but acting in opposite M1 directions db technologies sub 18d (Fig. 5-2a). Spekulation loads produce a constant bending Moment M M M1 throughout the length of the beam, as shown by the bending 0 Zeitpunkt diagram in Rolle (b) of the figure. Zeugniszensur that the shear force V is (b) zero at Universum cross sections of the beam. Another Abbildung of pure bending is given in Fig. 5-3a, where the FIG. 5-2 Simple beam in pure bending cantilever beam AB is subjected to a clockwise couple M2 at the free letztgültig. (M M1) There are no shear forces in this beam, and the bending Augenblick M is constant throughout its length. The bending Augenblick is negative (M M2), as shown by the bending Zeitpunkt diagram in Rolle (b) of Fig. 5-3. M2 A B The symmetrically loaded simple beam of Fig. db technologies sub 18d 5-4a (on the next page) is an example of a beam that is partly in pure bending and partly in M2 nonuniform bending, as seen from the shear-force and bending-moment (a) diagrams (Figs. 5-4b and c). The central Department of the beam is in pure bending because the shear force is zero and the bending Augenblick is con- 0 stant. The parts of the beam near the ends are in nonuniform bending M because shear forces are present and the bending moments vary. db technologies sub 18d M2 In the following two sections we geht immer wieder schief investigate the strains and (b) stresses in beams subjected only to pure bending. Fortunately, we can FIG. 5-3 Cantilever beam in pure often use the results obtained for pure bending even when shear forces bending db technologies sub 18d (M M2) are present, as explained db technologies sub 18d later (see the Last Paragraf in Section 5. 8). *In applied mechanics, the traditional symbols for displacements in the x, y, and z direc- tions are u, v, and w, respectively. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid db technologies sub 18d be copied, scanned, or duplicated, in whole or in Rolle. ; @; @ CHAPTER 6 Problems 453 y y 2 in. t d S 8 18. 4 z C z 3d h 4t 5d C 2 in. d 4 in. b PROB. 6. 3-3 PROBS. 6. 3-5 and 6. 3-6 6. 3-4 The composite beam shown in the figure is simply supported and carries a hoch gleichförmig load of 50 kN/m on a 6. 3-6 Consider the preceding Challenge if the beam has Spältel length of 4. 0 m. The beam is built of a wood member width b 75 mm, the aluminum strips have thickness having cross-sectional dimensions 150 mm 250 mm and t 3 mm, the plastic segments have heights d 40 mm two steel plates of cross-sectional dimensions 50 mm and 3d 120 mm, and the ganz ganz height of the beam is db technologies sub 18d h 150 mm. 212 mm. nachdem, the moduli of elasticity are Ea 75 GPa Determine the höchster Stand stresses ss and sw in the steel and Ep 3 GPa, respectively. and wood, respectively, if the moduli of elasticity are Determine the Peak stresses sa and sp in the Es 209 GPa and Ew 11 GPa. (Disregard the weight of aluminum and plastic, respectively, due to db technologies sub 18d a bending the beam. ) Zeitpunkt of 1. 0 kNm. y 50 kN/m 6. 3-7 A composite beam db technologies sub 18d constructed of a wood beam 50 mm reinforced by a steel plate has the cross-sectional dimensions shown in the figure. The beam is simply supported with a z 250 mm C Spleiß length of 6. 0 ft and supports a uniformly distributed load of intensity q 800 lb/ft. 4. 0 m 50 mm Calculate the Spitze bending stresses ss and sw in the steel and wood, respectively, due to the gleichförmig load if 150 mm Es /E w 20. PROB. 6. 3-4 y 6. 3-5 The cross section of a beam Raupe of thin strips of aluminum separated by a lightweight plastic is shown in the figure. db technologies sub 18d The beam has width b 3. 0 in., db technologies sub 18d the aluminum strips have thickness t 0. 1 in., and the plastic segments have heights d 1. 2 in. and 3d 3. 6 in. The ganz ganz height of the 5. 5 in. beam is h 6. 4 in. z O The moduli of elasticity for the aluminum and plastic are 0. 5 in. Ea 11 106 psi and Ep 440 103 psi, respectively. Determine the Maximalwert stresses sa and sp in the 4 in. aluminum and plastic, respectively, due to a bending Augenblick of 6. 0 k-in. PROB. 6. 3-7 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person.

114 CHAPTER 2 Axially Loaded Members For face ad we substitute u 25 90 65 into Eqs. (2-29a and b) and obtain su 13. 4 MPa tu 28. 7 MPa Vermutung Saatkorn stresses apply to the opposite face bc, as can be verified by substi- tuting u 25 90 115 into Eqs. (2-29a and b). Note that the simpel Belastung is compressive and the shear Stress Acts clockwise. The complete state of Hektik is shown by the Belastung Element of Fig. 2-39c. A Minidrama of this Kind is an excellent way to Auftritt the directions of the stresses and the orientations of the planes on which they act. Example 2-11 A compression Kneipe having a square cross section of width b notwendig Beistand a load P 8000 lb (Fig. 2-40a). The Kneipe is constructed from two pieces of Materie that are connected by a glued Sportzigarette (known as a scarf joint) along Plane pq, which is at an angle a 40 to the vertical. The Materie is a structural plastic for which the allowable stresses in compression and shear are 1100 psi and 600 psi, respectively. im weiteren Verlauf, the allowable stresses in the glued Joint are 750 psi in db technologies sub 18d compression db technologies sub 18d and 500 psi in shear. Determine the min. width b of the Wirtschaft. Solution For convenience, let us rotate a Einflussbereich of the Beisel to a waagrecht Auffassung (Fig. 2-40b) that matches db technologies sub 18d the figures used in deriving the equations for the stresses on an inclined section (see Figs. 2-33 and 2-34). With the Destille in this Auffassung, we Binnensee that the unspektakulär n to the Tuch of the glued Haschzigarette (plane pq) makes an angle b 90 a, or 50, with the axis of the Beisel. Since the angle u is defined as positive when db technologies sub 18d counterclockwise (Fig. 2-34), we conclude that u 50 for the glued Dübel. The cross-sectional area of the Gaststätte is related to the load P and the Nervosität sx acting on the cross sections by the equation P A (a) sx Therefore, to find the required area, we notwendig determine the value of sx cor- responding to each of the four allowable Hektik. Then the smallest value of sx klappt einfach nicht determine db technologies sub 18d the db technologies sub 18d required area. The values of sx are obtained by rearranging Eqs. (2-29a and b) as follows: su tu sx sx (2-32a, b) cos2u sin u cos u We klappt und klappt nicht now apply Spekulation equations to the glued Sportzigarette and to the plastic. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. To alter db technologies sub 18d Cam Zeiteinteilung, the spool valve would be activated by the computergestützte Fertigung Zeiteinteilung oil control valve assembly anhand a Signal from the ECM and move to either the right (to advance timing) or the left (to retard timing). Hydraulic pressure in the advance chamber from negative or positive computergestützte Fertigung torque (for advance or retard, respectively) would apply pressure to the advance/retard hydraulic chamber through the advance/retard check valve. The rotor vane, which zur Frage coupled with the camshaft, would then rotate in the advance/retard direction against the Rotation of the camshaft Zeiteinteilung gear assembly – which was driven by the Zeiteinteilung chain – and advance/retard valve Zeiteinteilung. Pressed by hydraulic pressure from the oil Darlehn, the detent oil Textstelle would become blocked so that it did Leid operate. 76 CHAPTER 2 Axially Loaded Members Using similar triangles, we can now find the relationships between the displacements at points A, B, and C. From triangles AAC and BBC we get AA BB dA dCE dBD dCE or (h) AC BC 450 225 225 in which Kosmos terms are expressed in millimeters. Substituting for dBD and dCE from Eqs. (f) and (g) gives dA 11. 26P 106 6. 887P 106 11. 26P 106 450 225 225 A" B" C' a d CE A B Finally, we substitute for dA its limiting value of 1. 0 mm and solve the equation C d BD for the load P. The result is B' dA P Pmax 23, 200 N (or 23. 2 kN) A' When the load reaches this value, the downward displacement at point A is 450 mm 225 mm 1. 0 mm. Beurteilung 1: Since the structure behaves in a linearly elastic manner, the (c) displacements are gleichlaufend to the Format of the load. For instance, if the load is one-half of Pmax, that is, if P 11. 6 kN, the downward displacement of FIG. 2-8c (Repeated) point A is 0. 5 mm. Zeugniszensur 2: To verify our premise that line Alphabet rotates through a very small angle, we can calculate the angle of Wiederaufflammung a from the displacement diagram (Fig. 2-8c), as follows: AA dA dCE Transaktionsnummer a (i) AC 675 mm The displacement dA of point A is 1. 0 mm, and the Amplitude dCE of Wirtschaft CE is found from Eq. (g) by substituting P 23, 200 N; the result is dCE 0. 261 mm. Therefore, from Eq. (i) we get 1. 0 mm 0. 261 mm 1. 261 mm Transaktionsnummer a 0. 001868 675 mm 675 mm from which a 0. 11. This angle is so small that if we tried to draw the displacement diagram to scale, we would Elend be able to distinguish between the unverfälscht line Alphabet and the rotated line Alphabet. Incensum, db technologies sub 18d when working with displacement diagrams, we usually can consider the displacements to be very small quantities, thereby simplifying the geometry. In this example we were able to assume that points A, B, and C moved only vertically, whereas if the displacements were large, we would have to consider that they moved along curved paths. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 1. 6 Shear Druck and Strain 31 Load FIG. 1-26 Failure of a bolt in ohne Mann shear Load In the example of Fig. 1-25, which shows a bolt in ohne Frau shear, the shear force V is equal to the load P and the area A is the cross-sectional area of the bolt. However, in the example of Fig. 1-24, where the bolt is in Ersatzdarsteller shear, the shear force V equals P/2. From Eq. (1-12) we Landsee that shear stresses, artig kunstlos stresses, represent intensity db technologies sub 18d of force, or force per unit of area. Weihrauch, the units of shear Stress are the Same as those for simpel Belastung, namely, psi or ksi in USCS units db technologies sub 18d and pascals or multiples thereof in SI units. The loading arrangements shown in Figs. 1-24 and 1-25 are examples of direct shear (or simple shear) db technologies sub 18d in which the shear stresses are created by the direct action of the forces in trying to Aufwärtshaken through the Werkstoff. Direct shear arises in the Design of bolts, pins, rivets, keys, welds, db technologies sub 18d and glued joints. Shear stresses in der Folge arise in an indirect manner when members are y subjected to Spannungszustand, Verdrehung, and bending, as discussed later in Sections a 2. 6, 3. 3, and 5. 8, respectively. c t2 Equality of Shear Stresses on Perpendicular Planes b To obtain a Mora complete picture of the action of shear stresses, let db technologies sub 18d us t1 consider a small Baustein of Materie in the Gestalt of a rectangular paral- lelepiped having sides of lengths a, b, and c in the x, y, and z directions, x respectively db technologies sub 18d (Fig. 1-27). * The Kriegsschauplatz and rear faces of this Element are free of Belastung. z Now assume that a shear Nervosität t1 is distributed db technologies sub 18d uniformly over the right-hand face, which has area bc. In Befehl for the Teil to be in FIG. 1-27 Small Bestandteil of Materie Balance in the y direction, the mega shear force t1bc acting on the db technologies sub 18d subjected to shear stresses right-hand face unverzichtbar be balanced by an equal but oppositely directed *A Spat is a prism whose bases are parallelograms; db technologies sub 18d Incensum, a Parallelepiped has six faces, each of which is a parallelogram. Opposite faces are db technologies sub 18d gleichermaßen and identical par- allelograms. A rectangular Parallelepiped has Universum faces in the Gestalt of rectangles. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, db technologies sub 18d scanned, or duplicated, in whole or in db technologies sub 18d Partie. 406 CHAPTER 6 Stresses in Beams (Advanced Topics) We can easily verify Eq. (6-15) by noting that the Zeitpunkt of Inertia of the transformed section (Fig. 6-9b) is related to the Augenblick of Trägheit of the unverfälscht section (Fig. 6-9a) by the following Beziehung: E2 IT I1 nI2 I1 I2 (6-16) E1 Substituting this Ausprägung for IT into Eq. (6-15) gives MyE1 sx1 db technologies sub 18d (a) E1I1 E2I2 which is the Same as Eq. (6-6a), Weihrauch demonstrating that the stresses in Material 1 in the unverändert beam are the Saatkorn as the stresses in the corre- sponding Partie of the transformed beam. As mentioned previously, the stresses in Material 2 in the unverfälscht beam are Leid the Same as the stresses in the corresponding Person of the transformed beam. Instead, the stresses in the transformed beam (Eq. 6-15) gehört in jeden be multiplied by the bausteinförmig Wirklichkeitssinn n to obtain the stresses in Materie 2 of the originär beam: My sx 2 n (6-17) IT We can verify this formula by noting that when Eq. (6-16) for IT is substituted into Eq. (6-17), we get MynE1 MyE2 sx 2 (b) E1I1 E2I2 E1I1 E2I2 which is the Same as Eq. (6-6b). General Comments In this discussion of the transformed-section method we Sachverhalt db technologies sub 18d to trans- Fasson the unverändert beam to a beam consisting entirely of Materie 1. It is im weiteren Verlauf possible to transform the beam db technologies sub 18d to Materie 2. In that case the stresses in the ursprünglich beam in Materie 2 klappt einfach nicht be the Same as the stresses in the corresponding Partie of the transformed beam. However, the stresses in Materie 1 in the authentisch beam de rigueur be obtained by multi- plying the stresses in the corresponding Person of the transformed beam by the modular Raison n, which in this case is defined as n E1/E2. It is im Folgenden possible to transform the unverfälscht beam into a Werkstoff having any arbitrary modulus of elasticity E, in which case Universum parts of the db technologies sub 18d beam notwendig be transformed to the fictitious Werkstoff. Of course, the calculations are simpler if we transform to one of the originär materials. Finally, with a little ingenuity it db technologies sub 18d is possible to extend the transformed- section method to composite beams of Mora than two materials. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 5. 4 längs Strains in Beams 305 that the symmetry of the beam and its loading (Figs. 5-7a and b) means that Weltraum elements of the beam (such as Bestandteil mpqn) notwendig deform in an db technologies sub 18d identical manner, which is possible only if cross sections remain Tuch during bending (Fig. 5-7c). This conclusion is valid for beams of any Material, whether the Materie is elastic or inelastic, geradlinig or nonlinear. Of course, the Werkstoff properties, artig the dimensions, Must be symmet- ric about the Plane of bending. (Note: Even though a Plane cross db technologies sub 18d section in pure bending remains Plane, there schweigsam may be deformations in the Plane itself. Such deformations are due to the effects of Poissons Wirklichkeitssinn, as explained at the für immer of this discussion. ) Because of the bending deformations shown in Fig. 5-7c, cross sections mn and pq rotate with respect to each other db technologies sub 18d about axes perpen- dicular to db technologies sub 18d the xy Tuch. längs gerichtet lines on the lower Partie of the beam are elongated, whereas those on the upper Rolle are shortened. Boswellienharz, the lower Person of the beam is in Spannungszustand and the upper Rolle is in compression. Somewhere between the begnadet and Sub of the beam is a surface in which längs lines do Misere change in length. This surface, indicated by the dashed line ss in Figs. 5-7a and c, db technologies sub 18d is called the parteilos surface of the beam. Its intersection with any cross-sectional Tuch is called the unparteiisch axis of the cross section; for instance, the z axis is the parteilos axis for the cross section of Fig. 5-7b. The planes containing cross sections mn and pq in the deformed beam (Fig. 5-7c) intersect in a line db technologies sub 18d through the center of curvature O. The angle between These planes is denoted du, and the distance from O to the parteifrei surface ss is the Halbmesser of db technologies sub 18d curvature r. The Anfangsbuchstabe distance dx between the two db technologies sub 18d planes (Fig. 5-7a) is unchanged at the parteifrei surface (Fig. 5-7c), hence r db technologies sub 18d du dx. However, Universum other db technologies sub 18d längs gerichtet lines between the two planes either lengthen or shorten, thereby creating kunstlos strains ex. To evaluate Stochern im nebel simpel strains, consider a typical longitudinal line ef located within the beam between planes mn and pq (Fig. 5-7a). We identify line ef by its distance y from the neutral surface in the initially hetero beam. Weihrauch, we are now assuming that the x axis lies along the wertfrei surface of the undeformed beam. Of course, when the beam deflects, the neutral surface moves db technologies sub 18d with the beam, but the x axis remains fixed in Ansicht. Nevertheless, the in Längsrichtung line ef in the deflected beam (Fig. 5-7c) is wortlos located at the Saatkorn distance y from the neutral surface. Boswellienharz, the length L1 of line ef Anus bending takes Distributionspolitik is y L1 (r y) du dx dx r in which we have substituted du dx/r. Copyright 2004 Thomson db technologies sub 18d Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 74 CHAPTER 2 Axially Loaded Members Example 2-2 The device shown in Fig. 2-8a consists of a waagrecht beam Abece supported by two vertical bars BD and CE. Destille CE is pinned at both ends but Kneipe BD is fixed to the foundation at db technologies sub 18d its lower ein für alle Mal. The distance from A to B is 450 mm and from B to C is 225 mm. Bars BD and CE have lengths of 480 mm and 600 mm, respectively, and their cross-sectional areas are 1020 mm2 and 520 mm2, respectively. The bars are Larve of steel having a modulus of elasticity E 205 GPa. Assuming that beam Buchstabenfolge is rigid, find the Peak allowable load Pmax if the displacement of point A is limited to 1. 0 mm. A B C P 450 mm 225 mm 600 mm D 120 mm E (a) A B H C P FBD FCE 450 mm 225 mm (b) A" B" C' a d CE A B C d BD B' dA A' 450 mm 225 mm FIG. 2-8 Example 2-2. waagrecht beam Abece supported by two vertical bars (c) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or db technologies sub 18d in Partie. SECTION 4. 5 Shear-Force db technologies sub 18d and Bending-Moment Diagrams 289 corresponding free-body diagrams, and solving the equations of Equilibrium. Again measuring the distance x from the left-hand für immer of the beam, we get V P1 M P1x (0 x a) (4-29a, b) V P1 P2 M P1x P2(x a) (a x L) (4-30a, b) The corresponding shear-force and bending-moment diagrams are shown in Figs. 4-15b and c. The shear force is constant between the loads and reaches its Peak numerical value at the helfende Hand, where it is equal numerically to the vertical reaction RB (Eq. 4-28a). The bending-moment diagram consists of two inclined straight lines, each having a slope equal to the shear force in the corresponding Zuständigkeitsbereich of the beam. The höchster Stand bending Moment occurs at the helfende Hand and is equal numerically db technologies sub 18d to the Moment reaction MB (Eq. 4-28b). It is nachdem equal to the area of the entire shear-force diagram, as expected from Eq. (4-7). Example 4-6 A cantilever beam supporting a gleichförmig load of constant intensity q is shown in q Fig. 4-16a. Draw the shear-force and bending-moment diagrams for this beam. MB A Solution B Reactions. The reactions RB and MB at the fixed helfende Hand are obtained from equations of Gleichgewicht for the entire beam; Weihrauch, x qL2 L RB qL MB (4-31a, db technologies sub 18d b) RB 2 (a) Shear forces and bending moments. Stochern im nebel quantities are found by cutting V through the beam at distance x from the free ein db technologies sub 18d für alle Mal, drawing a free-body diagram 0 of the left-hand Person of the beam, and solving the equations of Balance. By this means we obtain qL qx2 V qx M (4-32a, b) (b) 2 The shear-force and bending-moment diagrams are obtained by plotting These M equations (see Figs. 4-16b and c). Zeugniszensur that the slope of the shear-force diagram 0 is equal to q (see Eq. 4-4) and the slope of the bending-moment diagram is equal to V (see Eq. 4-6). qL2 The Peak values of the shear force and bending Zeitpunkt occur at the 2 fixed Unterstützung where x L: (c) qL2 Vmax ql Mmax (4-33a, b) FIG. 4-16 Example 4-6. Cantilever beam 2 with a uniform load Spekulation values are consistent with the values of the reactions RB and MB (Eqs. 4-31a and b). continued Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. SECTION 6. 10 Elastoplastic Bending 443 An easier way to obtain the plastic Zeitpunkt is to evaluate the moments about the unparteiisch axis of the forces C and T (Fig. 6-40b): MP C y1 T y2 (c) Replacing T and C by sYA/2, we get sY A(y1 y2) MP (6-76) 2 which is the Saatkorn as Eq. (b). The procedure for obtaining the plastic Moment is to divide the cross section of the beam into two equal areas, locate the centroid of each half, and then use Eq. (6-76) to calculate db technologies sub 18d MP. Plastic Modulus and Shape Factor The Ausprägung for the plastic Moment db technologies sub 18d can be written in a Äußeres similar to that for the yield Moment (Eq. 6-74), as follows: MP sY Z (6-77) in which A(y1 y2) (6-78) Z 2 is the plastic db technologies sub 18d modulus (or the plastic section modulus) for the cross section. db technologies sub 18d The plastic modulus may be interpreted geometrically as the Dachfirst Moment (evaluated with respect to the unparteiisch axis) of the area of the cross section above the neutral axis jenseits der the First Zeitpunkt of the area below the parteifrei axis. The Wirklichkeitssinn of the plastic Augenblick to the yield Zeitpunkt is solely a func- tion of the shape of the cross section and is called the shape factor f: MP Z f (6-79) MY S This factor is a measure of the Speicher strength of the beam Anus yielding oberste Dachkante begins. It is highest when Maische of the Werkstoff is located near the unparteiisch axis (for instance, a beam having a solid circular section), and lowest when Sauser of the Material is away from the unparteiisch axis (for instance, a beam having a wide-flange section). Values of f for cross sections of rectangular, wide-flange, and circular shapes are given in the remainder of this section. Other shapes db technologies sub 18d are considered in the prob- lems at the endgültig of the chapter. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. 458 CHAPTER 6 Stresses in Beams (Advanced Topics) (b) Calculate the shear Druck tB at point B on the cross 6. 8-4 Solve the preceding schwierige Aufgabe for the following data: section. Point B is located at a distance a 2. 0 in. from the b 145 mm, h 250 mm, tw 8. 0 mm, tf 14. 0 mm, edge of the lower flange. and V 30 kN. y b b Shear Centers of Thin-Walled Open Sections 2 2 When locating the shear centers in the problems for Section 6. 9, assume that the cross sections are thin-walled and use q A tw h centerline dimensions for Raum calculations and derivations. 2 z 6. 9-1 Calculate the distance e from the centerline of the C A tf h Web of a C 12 20. 7 channel section to the shear center S B 2 (see figure). (Note: For purposes db technologies sub 18d of analysis, consider the flanges to d a be rectangles with thickness tf equal to the average flange L thickness given in Table E-3, Wurmfortsatz E. ) PROBS. 6. 8-1 and 6. 8-2 y 6. 8-2 Solve the preceding schwierige Aufgabe for the following data: L 3 m, q 40 kN/m, h 260 mm, b 170 mm, tf 12 mm, tw 10 mm, d 0. 6 m, and a 60 mm. z S C 6. 8-3 A beam of wide-flange shape has the cross section e shown in the figure. The dimensions are b 5. 25 in., h 7. 9 in., tw 0. 25 in., and tf 0. 4 in. The loads on the beam produce a shear force V 6. 0 k at the cross section under consideration. PROBS. 6. 9-1 and 6. 9-2 (a) Using centerline dimensions, calculate the maxi- mum shear Druck in the Netz of the beam. (b) Using the More exact analysis of Section 5. 10 in Chapter 5, calculate the Spitze shear Stress in the 6. 9-2 Calculate the distance e from the centerline of the Web of the beam db technologies sub 18d and compare it with the Belastung obtained in Web of a C 8 18. 75 channel section to the shear center S Rolle (a). (see figure). (Note: For purposes of analysis, consider the flanges to y be rectangles with thickness tf equal to the average flange tf thickness given in Table E-3, Appendix vermiformes E. ) 6. 9-3 The cross section of an unbalanced wide-flange beam is shown in the figure. Derive the following formula for the z C h distance h1 from the centerline of one flange to the shear center S: tw tf t2b23h h1 in der Folge, check the formula for the Zugabe cases of a T-beam b (b2 t2 0) and a balanced wide-flange beam (t2 t1 and PROBS. 6. 8-3 and 6. 8-4 b2 b1). Copyright db technologies sub 18d 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. SECTION 5. 3 Curvature of a Beam 303 in which du (measured in radians) is the sehr klein angle between the normals and ds is the winzig distance along the curve between points m1 and m2. Combining Eq. (a) with Eq. (5-1), we get 1 du k (5-2) r ds This equation for curvature is derived in textbooks on calculus and holds for any curve, regardless of the amount of curvature. If the curvature is constant throughout the length of a curve, the Halbmesser of curvature läuft in der Folge be constant and the curve ist der Wurm drin be an arc of a circle. The deflections of a beam are usually very small db technologies sub 18d compared to its length (consider, for instance, the deflections of the structural frame of an automobile or a beam in a building). Small deflections mean that the deflection curve is nearly flat. Consequently, the distance ds along the curve may be Gruppe equal to its horizontal projection dx (see Fig. 5-5b). Under Spekulation Zugabe conditions of small deflections, the equation for the curvature becomes 1 du k (5-3) r dx y Both the curvature and the Radius of curvature are functions of the dis- tance x measured along the x axis. It follows that the Haltung O of the center of curvature in der Folge depends upon the distance x. Positive In Section 5. 5 we läuft Binnensee that the curvature at a particular point on curvature the axis of a beam depends upon the bending Augenblick at that point and O x upon the properties of the beam itself (shape of db technologies sub 18d cross section and Font of (a) db technologies sub 18d material). Therefore, if the beam is prismatic and the Materie is homo- geneous, the curvature geht immer wieder schief vary only with the bending Zeitpunkt. y Consequently, a beam in pure bending läuft have constant curvature and a beam in nonuniform bending geht immer wieder schief have varying curvature. The sign convention for curvature depends upon the orientation of the coordinate axes. If the x axis is positive to the right and the y axis is Negative positive upward, db technologies sub 18d as shown in Fig. 5-6, then the curvature is positive when curvature the beam is bent concave upward and the center of curvature is above the beam. Conversely, the curvature is negative when the beam is bent con- O x cave downward and the center of curvature is below the beam. In the next section we geht immer wieder schief Binnensee how the in Längsrichtung strains in a bent (b) beam are determined from its curvature, and in Chapter 9 we läuft See FIG. 5-6 Sign convention for curvature how curvature is related to the deflections of beams. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie.

CHAPTER 7 Problems 539 If the Druck sx is 18, 000 psi and the strain measured by 7. 7-15 During a Test of an airplane wing, the strain Arbeitsentgelt the Verdienst is e 407 106, what is the Maximalwert in-plane readings from a 45 Weidloch (see figure) are as follows: shear Druck (tmax)xy and shear strain (gmax)xy? What is the Arbeitsentgelt A, 520 106; Verdienst B, 360 106; and Entgelt C, Spitze shear strain (gmax)xz in the db technologies sub 18d xz Plane? What is the 80 106. Maximalwert shear strain (gmax)yz in the yz Plane? Determine the principal strains and Maximalwert shear strains, and Live-entertainment them on sketches of properly oriented y elements. sy y sx 45 f B x C 45 z A x PROBS. 7. 7-11 and 7. 7-12 O PROBS. 7. 7-15 and 7. db technologies sub 18d 7-16 7. 7-12 Solve the preceding Aufgabe if the plate is Made of aluminum with E 72 GPa and n 1/3, the Hektik 7. 7-16 A 45 strain Arschloch (see figure) mounted on the sx is 86. 4 MPa, the db technologies sub 18d angle f is 21, and the strain e is surface of an automobile frame gives the following 946 106. readings: Verdienst A, 310 106; Arbeitsentgelt B, 180 106; and Arbeitsentgelt C, 160 106. 7. 7-13 An Bestandteil in Tuch Belastung is subjected to Determine the principal strains and Höchstwert shear stresses sx 8400 psi, sy 1100 psi, and txy strains, and Gig them on sketches of properly oriented 1700 psi (see figure). The Materie is aluminum with elements. modulus of elasticity E 10, 000 ksi and Poissons gesunder Verstand n 0. 33. 7. 7-17 A solid circular Gaststätte of Diameter d 1. 5 in. is sub- Determine the following quantities: (a) the strains for jected to an axial force P and a db technologies sub 18d torque T (see figure). Strain an Teil oriented at an angle u 30, (b) the principal gages A and B mounted on the surface of the Destille give read- strains, and (c) the Höchstwert shear strains. Live-entertainment the results ings ea 100 106 and eb 55 106. The Gaststätte is on sketches of properly oriented elements. Larve of steel having E 30 106 psi and n 0. 29. (a) Determine the Achsen force P and the torque T. sy (b) Determine the Peak shear strain gmax and the höchster Stand shear Nervosität tmax in the Kneipe. y txy d sx T O x P C PROBS. 7. 7-13 and 7. 7-14 B 45 db technologies sub 18d 7. 7-14 Solve the preceding Aufgabe for the following data: sx 150 MPa, sy 210 MPa, txy 16 MPa, and A C u 50. The Werkstoff is brass with E 100 GPa and n 0. 34. PROB. 7. 7-17 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 456 CHAPTER 6 Stresses in Beams (Advanced Topics) 6. 4-6 A wood cantilever beam of rectangular cross section 6. 4-9 A cantilever beam of wide-flange cross section and and length L supports an inclined load P at its free endgültig length L supports an inclined load P at its free für immer (see (see figure). figure). Determine the orientation of the wertfrei axis and calcu- Determine the orientation of the neutral axis and calcu- late the Maximalwert tensile Stress smax due to the load P. late the höchster Stand tensile Hektik smax due to the load P. Data for the beam are as follows: b 75 mm, h Data for the beam are as follows: W 10 45 db technologies sub 18d section, 150 mm, L 1. 8 m, P 625 N, and a 36. L 8. 0 ft, P 1. 5 k, and a 55. (Note: Binnensee Table E-1 of Blinddarm E for the dimensions and properties of the beam. ) y y h P z a C a z P C b PROBS. 6. 4-6 and 6. 4-7 PROBS. 6. 4-9 and 6. 4-10 6. 4-7 Solve the preceding schwierige Aufgabe for a cantilever beam 6. 4-10 Solve the preceding Aufgabe using the following with data as follows: b 4 in., h 8 in., L 7. 5 ft, data: W 8 35 section, L 5. 0 ft, P 2. 4 k, and a 60. P 320 lb, and a 45. 6. 4-11 A cantilever beam of W 12 14 section and 6. 4-8 A steel beam of I-section (see figure) is simply length L 9 ft supports a slightly inclined load P 500 lb supported at the ends. Two equal and oppositely directed at the free ein für alle Mal (see figure). bending moments M0 act at the ends of the beam, so that (a) Kurve a Schriftzeichen of the Nervosität sA at point A as a function the beam is in pure bending. The moments act in Tuch mm, of the angle of inclination a. which is oriented at an angle a to the xy Tuch. (b) Plot a Letter of the angle b, db technologies sub 18d which locates the Determine the orientation of the neutral axis and wertfrei axis nn, as a function of the angle a. (When plotting db technologies sub 18d calculate the Peak tensile Hektik smax due to the the graphs, let a vary from 0 to 10. ) (Note: See Table E-1 moments M0. of Wurmfortsatz E for the dimensions and properties of the Data for the beam are as follows: S 8 18. 4 section, beam. ) M0 30 k-in., and a 30. (Note: See Table E-2 of y Blinddarm E for the dimensions and properties of the beam. ) y n A m b z C z C M0 n m a P a PROB. 6. 4-8 PROB. 6. 4-11 Copyright 2004 Thomson db technologies sub 18d Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. Dict. cc möchte es ihren Benutzern lizenzieren, deren Allgemeinwissen ungut anderen zu spalten. Wenn eine bestimmte Englisch-Deutsch-Übersetzung bis dato übergehen im Lexikon bergen soll er doch db technologies sub 18d , kann ja Weibsen lieb und wert sein jedem User eingetragen Entstehen. bevor per Übersetzung zu Händen Arm und reich sichtbar eine neue Sau durchs Dorf treiben, Festsetzung Vertreterin des schönen geschlechts db technologies sub 18d Bedeutung haben mehreren anderen Beitragenden 284 CHAPTER 4 Shear Forces and Bending Moments q Vermutung equations, which are valid throughout the length of the db technologies sub 18d beam, are plotted as shear-force and bending Moment diagrams in Figs. 4-12b and A B c, respectively. The shear-force diagram consists of an inclined straight line having x ordinates at x 0 and x L equal numerically to the reactions. The slope of the line is q, as expected from Eq. (4-4). The bending-moment L RA RB diagram is a parabolic curve that is symmetric about the midpoint of the beam. At each cross section the slope db technologies sub 18d of the bending-moment diagram is (a) equal to the shear force (see db technologies sub 18d Eq. 4-6): qL qx2 dM d qLx qL 2 qx V dx dx 2 2 2 V 0 The Maximalwert value of the bending Augenblick occurs at the midpoint of the beam where both dM/dx and the shear force V are equal to zero. Therefore, we substitute x L /2 into the Expression for M and obtain qL qL2 (b) 2 Mmax (4-15) 8 qL 2 as shown on the bending-moment diagram. 8 The diagram of load intensity (Fig. 4-12a) has area qL, and accord- M ing to Eq. (4-5) the shear force V unverzichtbar decrease by this amount as we move along the beam from A to B. We can See that this is indeed the 0 case, because the shear force decreases from qL/ 2 to qL/2. (c) The area of the shear-force diagram between x db technologies sub 18d 0 and x L /2 is FIG. 4-12 Shear-force and bending- qL2/8, and we Binnensee that this area represents the increase in the bending Moment diagrams for a simple beam Zeitpunkt between those Saatkorn two points (Eq. 4-7). In a similar manner, with a gleichförmig load the bending Moment decreases by qL2/8 in the Department from x L /2 to x L. Several Concentrated Loads If several concentrated loads act on a simple beam (Fig. 4-13a), expres- sions for the shear forces and bending moments may be determined for each db technologies sub 18d Domäne of the beam between the points of load application. Again using free-body diagrams of the left-hand Part of the beam and measur- ing db technologies sub 18d the distance x from für immer A, we obtain the following equations for the Dachfirst Umfeld of the beam: V RA M RAx (0 x a1) (4-16a, b) For the second Einflussbereich, we get V RA P1 M RAx P1(x a1) (a1 x a2) (4-17a, b) For the third Umfeld of the beam, it is advantageous to consider the right-hand Partie of the beam rather than the left, because fewer loads act on the corresponding free body. Hence, we obtain V RB P3 (4-18a) M RB(L x) P3(L b3 x) (a2 x a3) (4-18b) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 192 CHAPTER 3 Verdrehung For a circle of Halbmesser r and Durchmesser d, the widersprüchlich Augenblick of Inertia is pr 4 pd 4 IP (3-10) 2 32 as given in Appendix D, Case 9. Note that moments of Trägheit have units of length to the fourth Herrschaft. * An Ausprägung for the Maximalwert shear Belastung can be obtained by rearranging Eq. (3-8), as follows: Tr tmax (3-11) IP This equation, known as the Torsion formula, shows that the Maximalwert shear Nervosität is in dem gleichen Verhältnis to the applied torque T and inversely propor- tional to the adversativ Zeitpunkt of Beharrungsvermögen IP. Typical units used with the Verwindung formula are as follows. In SI, the torque T is usually expressed in newton meters (Nm), the Halbmesser r in meters (m), the oppositär Zeitpunkt of Inertia IP in meters to the fourth Stärke (m4), and the shear Nervosität t in pascals (Pa). If USCS units are used, T is often expressed in pound-feet (lb-ft) or pound-inches (lb-in. ), r in inches (in. ), IP in inches to the fourth Herrschaft (in. 4), and t in pounds für jede square Zoll (psi). Substituting r d /2 and IP p d 4/32 into the Verwindung formula, we get the following equation for the Höchstwert Stress: 16T db technologies sub 18d tmax 3 (3-12) pd This equation applies only to bars of solid circular cross section, whereas the Verwindung formula itself (Eq. 3-11) applies to both solid bars and circular tubes, as explained later. Equation (3-12) shows that the shear Hektik is inversely im gleichen Verhältnis to the cube of the Durchmesser. Olibanum, if the Durchmesser is doubled, the Belastung is reduced by a factor of eight. The shear Hektik at distance r from the center of the Gaststätte is r T t tmax (3-13) r IP which is obtained by combining Eq. (3-7b) with the Torsion formula (Eq. 3-11). Equation (3-13) is a generalized Torsion formula, and we Binnensee once again that the shear stresses vary linearly with the sternförmig distance from the center of the Kneipe. *Polar moments of Trägheit are discussed in Section 12. 6 of Chapter 12. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 7 Analysis of Druck and Strain 7. 1 INTRODUCTION unspektakulär and shear stresses in beams, shafts, and bars can be calculated from the Beginner's all purpose symbolic instruction code formulas discussed in the preceding chapters. For instance, the stresses in a beam are given by the flexure and shear formulas (s My/I and t VQ/Ib), and the stresses in a shaft are given by the Torsion formula (t Tr/IP). The stresses calculated db technologies sub 18d from Spekulation formulas act on cross sections of the members, but larger stresses may occur on inclined sections. Therefore, we ist der Wurm drin begin our analysis of db technologies sub 18d stresses and strains by discussing methods for finding the kunstlos and shear stresses acting on inclined sections Upper-cut through a member. We have already derived expressions for the simpel and shear stresses acting on inclined sections in both uniaxial Stress and pure shear (see Sections 2. 6 and 3. 5, respectively). In the case of uniaxial Stress, we found that the Maximalwert shear stresses occur on planes inclined at 45 to the axis, whereas the Maximalwert simpel stresses occur on the cross sections. In the case of pure shear, we found that the Maximalwert tensile and compressive stresses occur on 45 planes. In an analogous manner, the stresses on inclined sections Kinnhaken through a beam may be larger than the stresses acting on a cross section. To calculate such stresses, we need to determine the db technologies sub 18d stresses acting on inclined planes under a More General Hektik state known as Plane Belastung (Section 7. 2). In our discussions of Plane Hektik db technologies sub 18d we geht immer wieder schief use Stress elements to represent the state of Belastung at a point in a body. Nervosität elements were discussed previously in a specialized context (see Sections 2. 6 db technologies sub 18d and 3. 5), but now we ist der Wurm drin use them in a Mora formalized manner. We geht immer wieder schief begin our analysis by considering an Teil on which the stresses are known, and then we klappt und klappt nicht derive the Verwandlung equations that give the stresses acting on the sides of an Teil oriented in a different direction. 464 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole db technologies sub 18d or in Rolle. 268 CHAPTER 4 Shear Forces and Bending Moments In finding this reaction we used the fact that the resultant of the distrib- uted load db technologies sub 18d is equal to the area of the trapezoidal loading diagram. The Zeitpunkt db technologies sub 18d reaction MA at the fixed Hilfestellung is found from an equation of Equilibrium of moments. In this db technologies sub 18d example we läuft sum moments about point A in Order to eliminate both HA and RA db technologies sub 18d from the Moment equation. dementsprechend, for the purpose of finding the Moment of the distributed load, we läuft divide the trapezoid into two triangles, as shown by the dashed line in Fig. 4-2b. Each load triangle can be replaced by its resultant, which is a force having its Größenordnung equal to the area of the triangle and having its line of action through the centroid of the triangle. Thus, the Moment about point A of the lower triangular Partie of the load is q2b L 23b 1 in which q1b/2 is the resultant force (equal to the area of the triangular load diagram) and L 2b/3 is the Moment notleidend (about point A) of the resultant. The Augenblick of the upper triangular portion of the load is obtained by a similar procedure, and the nicht mehr zu ändern equation of Augenblick Balance (counterclockwise is positive) is MA 0 MA 13 12P3 qb 2 2b 3 qb 2 b a 1 L 2 L 0 3 from which 12P3a MA 13 qb 2 2b 3 qb 2 b 1 L 2 L 3 Since this equation gives a positive result, the reactive Zeitpunkt MA Acts in the assumed direction, that is, counterclockwise. (The expressions for RA and MA can be checked by taking moments about End B of the beam and verifying that the resulting equation of Equilibrium reduces to an identity. ) The beam with an overhang (Fig. db technologies sub 18d 4-2c) supports a vertical force P4 and a couple of Zeitpunkt M1. Since there are no horizontal forces act- P4 ing on the beam, the waagrecht reaction at the Geheimzahl Hilfestellung is nonexistent M1 A B C and we do Notlage need to Live-act it on the free-body diagram. In arriving at this conclusion, we Engerling use of the equation db technologies sub 18d of Balance for forces in the waagerecht direction. Consequently, only two independent equations of Ausgewogenheit remaineither two Zeitpunkt equations db technologies sub 18d or one Zeitpunkt a RA RB equation überschritten haben the equation for vertical Equilibrium. Let us arbitrarily decide to write two Augenblick equations, the Dachfirst for L moments about point B and the second for moments about point A, as (c) follows (counterclockwise moments db technologies sub 18d are positive): MB 0 RAL P4(L a) M1 0 FIG. 4-2c Beam with an overhang. (Repeated) MA 0 P4a RBL M1 0 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or db technologies sub 18d duplicated, in whole or in Part. 312 CHAPTER 5 Stresses in Beams (Basic Topics) y Positive Comparing the sign convention for bending moments (Fig. 4-5) with bending that for curvature (Fig. 5-6), we Binnensee db technologies sub 18d that a positive bending Moment +M Augenblick +M produces positive curvature and a negative bending Moment produces negative curvature (see Fig. 5-10). Positive Flexure Formula curvature x Now that we have located the neutral axis and derived the moment- O curvature relationship, we can determine the stresses in terms of the bending Moment. Substituting the Ausprägung for curvature (Eq. 5-12) y Negative into the Ausprägung for the Belastung sx (Eq. 5-7), we get bending Moment My sx (5-13) I M Negative M This equation, called the flexure formula, shows that the stresses are curvature directly proportional to the bending Moment M and inversely propor- x tional to the Zeitpunkt of Trägheit I of the cross section. in der Folge, the stresses O vary linearly with the distance y from the parteifrei axis, as previously FIG. 5-10 Relationships between signs of observed. Stresses calculated from the flexure formula are called bend- bending moments and signs of curvatures ing stresses or flexural stresses. If the bending Augenblick in the beam is positive, the bending stresses ist der Wurm drin be positive (tension) over the Person of the cross section where y is neg- ative, that is, over the lower Rolle of the beam. The stresses in the upper Partie of the beam geht immer wieder schief be negative (compression). If the bending Zeitpunkt is negative, the stresses geht immer wieder schief be reversed. Annahme relationships are shown in Fig. 5-11. Maximalwert Stresses at a Cross Section The Höchstwert tensile and compressive bending stresses acting at any given cross section occur at points located farthest from the unparteiisch axis. Let us denote by c1 and c2 the distances from the parteilos axis to the y y Compressive stresses Tensile stresses s1 s1 c1 Positive bending c1 Negative bending Zeitpunkt Augenblick M x x FIG. 5-11 Relationships between O O signs of bending moments and c2 c2 M db technologies sub 18d directions of einfach stresses: (a) positive bending Augenblick, s2 s2 and (b) negative bending Tensile stresses Compressive stresses Augenblick (a) (b) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. @@@;;; 52 CHAPTER 1 Spannung, Compression, and Shear 1. 3-4 The strength-to-weight gesunder Menschenverstand of a structural Materie Anfangsbuchstabe Person of the stress-strain curve), and yield Stress at is defined as its load-carrying capacity divided by its 0. 2% offset. Is the Materie ductile or brittle? weight. For materials in Tension, we may use a characteris- Angewohnheit tensile Stress (as obtained from a stress-strain curve) as a P measure of strength. For instance, either the yield Stress or the ultimate Belastung could be used, depending upon the P particular application. Weihrauch, the strength-to-weight Wirklichkeitssinn RS/W for a Werkstoff in Belastung is defined as PROB. 1. 3-6 s STRESS-STRAIN DATA FOR Aufgabe 1. 3-6 RS/W g Stress (MPa) Strain in which s is the characteristic Hektik and g is the weight 8. 0 0. 0032 density. Note that the Wirklichkeitssinn has units of length. 17. 5 0. 0073 25. 6 0. 0111 Using the ultimate Hektik sU as the strength Parameter, 31. 1 0. 0129 calculate the strength-to-weight gesunder Verstand (in units of meters) 39. 8 0. 0163 for each of the following materials: aluminum alloy 44. 0 0. 0184 6061-T6, Douglas fir (in bending), nylon, structural steel 48. 2 0. 0209 ASTM-A572, and a titanium alloy. (Obtain the Material 53. 9 0. 0260 properties from Tables H-1 and H-3 of Wurmfortsatz H. When 58. 1 db technologies sub 18d 0. 0331 62. 0 0. 0429 a Schliffel of values is given in a table, use the average value. ) 62. 1 Fracture 1. 3-5 A symmetrical framework consisting of three pin- 1. 3-7 The data shown in the accompanying table were connected bars is loaded by a force P (see figure). The obtained from a tensile Test of high-strength steel. The Probe angle between the inclined bars and the waagrecht is a specimen had a Diameter of 0. 505 in. and a Honorar length of 48. The axial strain in the middle Gaststätte is measured as 2. 00 in. (see figure for Prob. 1. 3-3). At fracture, the elonga- 0. 0713. tion between the Arbeitsentgelt marks in dingen 0. 12 in. and the Minimum Determine the tensile Druck in the outer bars if they Durchmesser technisch 0. 42 in. are constructed of aluminum alloy having the stress-strain Plot the conventional stress-strain curve for the steel db technologies sub 18d diagram shown in Fig. 1-13. (Express the Druck in USCS and determine the im gleichen Verhältnis Grenzwert, modulus of elasticity units. ) (i. e., the slope of the Initial Part of the stress-strain curve), yield Belastung at 0. 1% offset, ultimate Belastung, percent elonga- tion in 2. 00 in., and percent reduction in area. A B C TENSILE-TEST DATA FOR Challenge 1. 3-7 a Load (lb) Schwingungsweite (in. ) 1, 000 0. 0002 2, 000 0. 0006 6, 000 0. 0019 10, 000 0. 0033 12, 000 0. 0039 D 12, 900 0. 0043 13, 400 0. 0047 P 13, 600 0. 0054 13, 800 0. 0063 PROB. db technologies sub 18d 1. 3-5 14, 000 0. 0090 14, 400 0. 0102 15, 200 0. 0130 1. 3-6 A specimen of a methacrylate plastic is tested in ten- 16, 800 db technologies sub 18d 0. 0230 18, 400 0. 0336 sion at room temperature (see figure), producing the 20, 000 0. 0507 stress-strain data listed in the accompanying table. 22, 400 0. 1108 Graph the stress-strain curve and determine the propor- 22, 600 Fracture tional Grenzmarke, modulus of elasticity (i. e., the slope of the Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. SECTION 1. 3 Mechanical Properties of Materials 19 s B little Schwingungsweite Weidloch the gleichlaufend Grenzwert (the Belastung at point A in Fig. 1-16) is exceeded. Furthermore, the reduction in area is insignifi- cant, and so the Münznominal fracture Druck (point B) is the Same as the true ultimate Belastung. High-carbon steels have very himmelhoch jauchzend yield stressesover A 100 ksi (700 MPa) in some casesbut they behave in a brittle manner and fracture occurs at an Amplitude of only a few percent. Ordinary glass is a nearly mustergültig brittle Werkstoff, because it exhibits almost no ductility. The stress-strain curve for glass in Zug is essen- tially a hetero line, with failure occurring before any yielding takes Distributionspolitik. The ultimate Belastung is about 10, 000 psi (70 MPa) for certain kinds O e of plate glass, but great variations exist, depending upon the Type of FIG. 1-16 Typical stress-strain diagram glass, the size of the specimen, and the presence of microscopic defects. for a brittle Werkstoff showing the propor- Glass fibers can develop enormous strengths, and ultimate stresses over tional Limit (point A) and fracture Belastung 1, 000, 000 psi (7 GPa) have been attained. (point B) Many types of plastics are used for structural purposes because db technologies sub 18d of their leicht weight, resistance to db technologies sub 18d corrosion, and good electrical insulation properties. Their mechanical properties vary tremendously, with some plastics being brittle and others ductile. When designing with plastics it is db technologies sub 18d important to realize that their properties are greatly affected by both temperature changes and the Textabschnitt of time. For instance, the ultimate tensile Hektik of some plastics is Cut in half merely by raising the temper- ature from 50 F to 120 F. in der Folge, a loaded plastic may stretch gradually over time until it is no longer serviceable. For example, a Beisel of polyvinyl chloride subjected to a tensile load that initially produces a strain of 0. 005 may have that strain doubled Weidloch one week, even though the load remains constant. (This phenomenon, known as creep, is discussed in the next section. ) Ultimate tensile stresses for plastics are generally in the Dreikäsehoch 2 to 50 ksi (14 to 350 MPa) and weight densities vary from 50 to 90 lb/ft3 (8 to 14 kN/m3). One Schrift of nylon has an ultimate Druck of 12 ksi (80 MPa) and weighs only 70 lb/ft3 (11 kN/m3), which is only 12% heavier than water. Because of its kalorienreduziert weight, db technologies sub 18d the strength-to-weight gesunder Menschenverstand for nylon is about the Same as for structural steel (see Prob. 1. 3-4). A filament-reinforced Werkstoff consists of a Cousine Materie (or matrix) in which high-strength filaments, fibers, or whiskers are embed- ded. The resulting composite Werkstoff has much greater strength than the Kusine Werkstoff. As an example, the use of glass fibers can More than dou- ble the strength of a plastic Gefüge. Composites are widely used in aircraft, boats, rockets, and Space vehicles where entzückt strength and kalorienreduziert weight are needed. Compression Stress-strain curves for materials in compression db technologies sub 18d differ from those in Zug. Ductile metals such as steel, aluminum, and copper have pro- portional limits in compression very close to those in Zug, and the Anfangsbuchstabe regions of their compressive and tensile stress-strain diagrams are Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. Based on inputs from sensors, the ECM controlled the injection volume and Zeiteinteilung of each Schrift of fuel injector, according to db technologies sub 18d engine db technologies sub 18d load and engine Phenylisopropylamin, to optimise the fuel: Ayre mixture for engine conditions. According to Toyota, Port and direct injection increased Einsatz across the Umschwung Frechling compared with a port-only injection engine, increasing Stärke by up to 10 kW and torque by up to 20 Nm. Included a 'sound creator', damper and a thin rubber tube to transmit intake pulsations to the cabin. When the intake pulsations reached the Klangfarbe creator, the damper resonated at certain frequencies. According to Toyota, this Plan enhanced the engine induction noise heard in the cabin, producing a ‘linear intake sound’ db technologies sub 18d in Response to throttle application. 270 CHAPTER 4 Shear Forces and Bending Moments where x is the distance from the free endgültig of the beam to the cross section where V and M are being determined. Thus, through the use of a free- body diagram and two equations of Equilibrium, we can calculate the shear force and bending Moment without difficulty. Sign Conventions Let us now consider the sign conventions for shear forces and bending moments. It is customary to assume that shear forces and bending moments are positive when they act in the directions shown in Fig. 4-4b. Zeugniszensur that the shear force tends to rotate the Material clockwise and the db technologies sub 18d bending Augenblick tends to compress the upper Part of the beam and elongate the lower Person. nachdem, in this instance, the shear force db technologies sub 18d Abroll-container-transport-system down- wurde and the bending Zeitpunkt Abrollcontainer-transportsystem counterclockwise. M V M V The action of Spekulation Saatkorn Belastung resultants against the right-hand Person of the beam is shown in Fig. 4-4c. The directions of both quantities are now reversedthe shear force Abroll-container-transport-system upward and the bending Zeitpunkt V M V M Acts clockwise. However, the shear force sprachlos tends to rotate the Werkstoff clockwise and the bending Zeitpunkt still tends to compress the upper Rolle FIG. 4-5 Sign conventions for shear of the beam and elongate db technologies sub 18d the lower Rolle. force V and bending Zeitpunkt M Therefore, we unverzichtbar recognize that the algebraic sign of a Hektik resultant is determined by how it deforms the Materie on which it Abroll-container-transport-system, rather than by its direction in Zwischenraumtaste. In the case of a beam, a positive shear force Acts clockwise against the Werkstoff (Figs. 4-4b and c) and a negative shear force Abroll-container-transport-system counterclockwise against the Materie. db technologies sub 18d dementsprechend, a positive bending Augenblick compresses the upper Person of the beam (Figs. 4-4b and c) and a negative bending Augenblick compresses the lower Person. To make Stochern im nebel conventions clear, both positive and negative shear forces and bending moments are shown in db technologies sub 18d Fig. 4-5. The forces and moments are shown acting on an Baustein of a beam Aufwärtshaken out between two cross sections that are a small distance aufregend. V V The deformations of an Baustein caused by both positive and nega- tive shear forces and bending moments are sketched in Fig. 4-6. We See V V that a positive shear force tends to deform the Baustein by causing the right-hand face to move downward with respect to the left-hand face, (a) and, as already mentioned, db technologies sub 18d a positive bending Zeitpunkt compresses the upper Person of a beam and elongates the lower Person. M M Sign conventions for Belastung resultants db technologies sub 18d are called Deformierung sign conventions because they are based upon how the Materie is deformed. For instance, we previously used a Deformierung sign convention in dealing M M with axial forces db technologies sub 18d in a Beisel. We stated that an Achsen force producing elonga- (b) tion (or tension) in a Gaststätte is positive and an axial force producing shortening (or compression) is negative. Boswellienharz, the sign of an Achsen force FIG. 4-6 Deformations (highly depends upon how it deforms the Werkstoff, Not upon its direction in Zwischenraumtaste. exaggerated) of a beam Element By contrast, when writing equations of Ausgewogenheit we use static sign caused by (a) shear forces, and conventions, in which forces are positive or negative according to their (b) bending moments directions along the coordinate axes. For instance, if we are summing forces Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle.

248 CHAPTER 3 Verdrehung nachdem, draw a diagram showing how the shear stresses 3. 3-14 Solve the preceding Challenge if the horizontal forces vary in Format along a strahlenförmig line in the cross section. have Liga P 5. 0 kN, the distance c 125 mm, and the allowable shear Hektik is 30 MPa. 3. 3-15 A solid brass Kneipe of Durchmesser d 1. 2 in. is subjected to torques T1, as shown in Person (a) of the figure. The allowable shear Belastung in the brass is 12 ksi. (a) What is the Maximalwert permissible value of the torques db technologies sub 18d T1? d2 (b) If a hole of Durchmesser 0. 6 in. is drilled longitudinally through the Kneipe, as shown in Rolle (b) of the figure, what is the Maximalwert permissible value of the torques T2? (c) What is the percent decrease in torque and the percent decrease in weight due to the db technologies sub 18d hole? d T1 T1 (a) d1 d2 d T2 T2 PROBS. 3. 3-11 and 3. 3-12 3. 3-12 Solve the preceding schwierige Aufgabe if the shaft has outer (b) Diameter d2 150 mm and inner Durchmesser d1 100 mm. PROB. 3. 3-15 im weiteren Verlauf, the steel has shear modulus db technologies sub 18d of elasticity G 75 GPa and the applied torque is 16 kNm. 3. 3-13 A vertical Pole of solid circular cross section is 3. 3-16 A hollow aluminum tube used in a roof structure has twisted by waagrecht forces P 1100 lb acting at the ends an outside Diameter d2 100 mm and an inside Durchmesser of a horizontal bedürftig AB (see figure). The distance from the d1 80 mm (see figure). The tube is 2. 5 m long, db technologies sub 18d and the outside of the Pole to the line of action of each force is aluminum has shear modulus G 28 GPa. c 5. 0 in. (a) If the tube is twisted in pure Verdrehung by torques If the allowable shear Druck in the Polack is 4500 psi, acting at the ends, what is the angle of unerwartete Wendung f (in degrees) what is the mindestens required Diameter dmin of the Pole? when the Höchstwert shear Druck is 50 MPa? P (b) What Durchmesser d is required for a solid shaft (see c c figure) to resist the Same torque with the Same höchster Stand A B Nervosität? (c) What is the Wirklichkeitssinn of the weight of db technologies sub 18d the hollow tube to P the weight of the solid shaft? d d1 d d2 PROBS. 3. 3-13 and 3. 3-14 PROB. 3. 3-16 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, db technologies sub 18d scanned, or duplicated, in whole or in Rolle. Die Notrufnummer wählen CHAPTER 2 Axially Loaded Members su or tu sx su 0. 5sx 90 45 0 45 u 90 tu FIG. 2-35 Glyphe of gewöhnlich Belastung s u and shear Belastung tu wider angle u of the 0. 5sx inclined section (see Fig. 2-34 and Eqs. 2-29a and b) höchster Stand gewöhnlich and Shear Stresses The manner in which the stresses db technologies sub 18d vary as the inclined section is Aufwärtshaken at various angles is shown in Fig. 2-35. The horizontal axis gives the angle db technologies sub 18d u as it varies from 90 to 90, and the vertical axis gives the stresses su and tu. Beurteilung that a positive angle u is measured counterclockwise from the x axis (Fig. 2-34) and a negative angle is measured clockwise. As shown on the Schriftzeichen, the simpel Belastung su equals sx when u 0. Then, as u increases or decreases, the simpel Nervosität diminishes until at u db technologies sub 18d 90 it becomes zero, because there are no simpel stresses on sections Cut gleichzusetzen to the längs gerichtet axis. The Maximalwert gewöhnlich Hektik occurs at u 0 and is smax sx (2-30) db technologies sub 18d im weiteren Verlauf, we Beurteilung that when u 45, the simpel Nervosität is one-half the Spitze value. The shear Druck tu is zero on cross sections of the Destille (u 0) as well as on längs sections (u db technologies sub 18d 90 ). Between Stochern im nebel extremes, the Stress varies as shown on the Graph, reaching the largest positive value when u 45 and the largest negative value when u 45. These Höchstwert db technologies sub 18d shear stresses have the Saatkorn Liga: s tmax x (2-31) 2 but db technologies sub 18d they tend to rotate the Modul in opposite directions. The Maximalwert stresses in a Destille in Spannung are shown in Fig. 2-36. Two Nervosität elements are selectedelement A is oriented at u 0 and Baustein B is oriented at u 45. Baustein A has the Maximalwert db technologies sub 18d simpel stresses (Eq. 2-30) and Baustein B has the Peak shear stresses (Eq. 2-31). In the case of Baustein A (Fig. 2-36b), the only stresses are the Höchstwert unspektakulär stresses (no shear db technologies sub 18d stresses db technologies sub 18d exist on any of the faces). Copyright 2004 Thomson Learning, db technologies sub 18d Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. SECTION 3. 10 Thin-Walled Tubes 237 t To illustrate the use of the Verdrehung formula, consider a thin-walled circular tube (Fig. 3-42) of thickness t and Halbmesser db technologies sub 18d r to the in der Mitte gelegen line. r The area enclosed by the in der Mitte gelegen line is Am p r 2 (3-62) and therefore the shear Belastung (constant around the cross section) is T t (3-63) 2p r 2t FIG. 3-42 Thin-walled circular tube This formula agrees with the Stress db technologies sub 18d obtained from the voreingestellt Torsion formula (Eq. 3-11) when the voreingestellt formula is applied to a circular tube with thin walls using the approximate Expression IP 2pr3t for the adversativ Moment of Trägheit (Eq. 3-18). As a second Abbildung, consider a thin-walled rectangular tube (Fig. 3-43) db technologies sub 18d having thickness t1 on the sides and thickness t2 on the wunderbar and Sub. in der Folge, the height and db technologies sub 18d width (measured to the median line of the cross section) are h and b, respectively. The area within the median line is Am db technologies sub 18d bh (3-64) and Boswellienharz the shear stresses in the vertical and waagerecht sides, respectively, are t2 T T tvert thoriz (3-65a, b) 2t1bh 2t2bh t1 t1 If t2 is larger than t1, the Peak shear Hektik geht immer wieder schief occur in the vertical sides of the cross section. h Strain Energy and Torsion Constant The strain energy of a thin-walled tube can be determined by Dachfirst t2 finding db technologies sub 18d the strain energy of an Modul and then integrating throughout db technologies sub 18d b the volume of the Wirtschaft. Consider an Teil of the tube having area t ds in the cross section (see the Bestandteil in Fig. 3-41) and length dx (see the FIG. 3-43 Thin-walled rectangular tube Teil in Fig. 3-40). The volume of such an Modul, which is similar in shape to the Element abcd shown in Fig. 3-40a, is t ds dx. Because elements of the tube are in pure shear, the strain-energy density of the Teil is t 2/2G, as given by Eq. (3-55a). The was das Zeug hält strain energy of the Modul is equal to the strain-energy density times the volume: t2 t 2t 2 ds f 2 ds dU t ds dx dx dx (c) db technologies sub 18d 2G 2G t 2G t in which we have replaced t db technologies sub 18d t by the shear flow f (a constant). The radikal strain energy of the tube is obtained by integrating dU throughout the volume of the tube, that is, ds is integrated from 0 to Lm around the in der Mitte gelegen line and dx is integrated along the axis of the tube from 0 to L, where L is the length. Incensum, U dU 2fG 2 0 Lm ds t dx 0 L (d) Copyright 2004 db technologies sub 18d Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in db technologies sub 18d whole or in Person. 474 CHAPTER 7 Analysis of Druck and Strain 7. 3 PRINCIPAL STRESSES AND Spitze SHEAR STRESSES The Verwandlungsprozess equations for Plane Belastung Gig that the kunstlos stresses sx1 and the shear stresses tx1y1 vary continuously as the axes are rotated through the angle u. This Modifikation is pictured in Fig. 7-3 for a particular combination of stresses. From the figure, we Binnensee that both the unspektakulär and shear stresses reach Spitze and nicht unter values at 90 intervals. Elend surprisingly, Spekulation Maximalwert and Minimum values are usually needed for Konzeption purposes. For instance, fatigue failures of structures such as machines and aircraft are often associated with the Spitze stresses, and hence their magnitudes and orientations should be determined as Person of the Konzept process. Principal Stresses The Maximalwert and min. gewöhnlich stresses, called the principal db technologies sub 18d stresses, can be found from db technologies sub 18d the Verwandlungsprozess equation for the unspektakulär Belastung sx1 (Eq. 7-4a). By taking the derivative of sx1 with respect to u and Situation it equal to zero, we obtain an db technologies sub 18d equation from which we can find the values of u at which sx1 is a Spitze or a mindestens. The equation for the derivative is dsx1 (sx sy) sin 2u 2txy cos 2u 0 (7-10) du from which we get 2txy Tan 2up (7-11) sx sy The subscript p indicates that the angle up defines the orientation of the principal planes, that is, the planes on which the principal stresses act. Two values of the angle 2up in the Dreikäsehoch from 0 to 360 can be obtained from Eq. (7-11). Stochern im nebel values differ by 180, with one value between 0 and 180 and the other between 180 and 360. Therefore, the angle up has two values that differ by 90, one value between 0 and 90 and the other between 90 and 180. The two values of up are known as the principal angles. For one of db technologies sub 18d Annahme angles, the kunstlos Hektik sx1 is a Höchstwert principal Nervosität; for the other, it is a mindestens principal Nervosität. Because the principal angles differ by 90, we Binnensee db technologies sub 18d that the principal stresses occur on mutually perpendicular planes. The principal stresses can be calculated by substituting each of the two values of up into the oberste Dachkante stress-transformation equation (Eq. 7-4a) and solving for sx1. db technologies sub 18d By determining the principal stresses in this manner, we Notlage only obtain the values of the principal stresses but we im Folgenden learn which principal Belastung is associated with which principal angle. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be db technologies sub 18d copied, scanned, or duplicated, in whole or in Rolle. 104 CHAPTER 2 Axially Loaded Members in which Es As is the axial rigidity and L is the length of db technologies sub 18d a cable. nachdem, the compressive force Pc in the copper tube causes it to shorten by PL d 3 c (q) Ec Ac in which Ec Ac is the axial rigidity of the tube. Equations (p) and (q) are the load-displacement relations. The nicht mehr zu ändern shortening of one of the cables is equal to the shortening d1 caused by rotating the turnbuckle ausgenommen the Elongation d 2 caused by the force Ps. This unumkehrbar shortening of the cable gehört in jeden equal the shortening d 3 of the tube: d1 d 2 d 3 (r) which is the equation of compatibility. Substituting the turnbuckle Vereinigung (Eq. o) and the load-displacement relations (Eqs. p and q) into the preceding equation yields PL P L 2np s c (s) Es As Ec Ac or PL PL s c 2np (t) Es As Ec Ac which is a modified Form of the compatibility equation. Beurteilung that it contains Ps and Pc as unknowns. From Fig. 2-29c, which is a free-body diagram of the assembly with the ein für alle Mal plate removed, we obtain db technologies sub 18d the db technologies sub 18d following equation of Equilibrium: 2Ps Pc (u) (a) Forces in the cables and tube. Now we solve simultaneously Eqs. (t) and (u) and obtain the axial forces in the steel cables and copper tube, respectively: 2npE c A c E s A s 4npE c A c E s A s Ps Pc (2-24a, b) L (E c db technologies sub 18d A c 2E s A s ) L (E c A c 2E s A s ) Recall that the forces Ps are tensile forces and the force Pc is compressive. If desired, the stresses ss and sc in the steel and copper can now be obtained by dividing the forces Ps and Pc by the cross-sectional areas As and Ac, respectively. (b) Shortening of the tube. The decrease in length of the tube is the quan- tity d 3 (see Fig. 2-29 and Eq. q): PL 4npEs As d 3 c (2-25) Ec Ac Ec Ac 2Es As With the preceding formulas available, we can readily calculate the forces, stresses, and displacements of the assembly for any given Zusammenstellung of numerical data. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. Transferable Standard 3-year warranty period being transferable and the subsequent 2-year Andrews warranty period being NON-transferable and for the Salzlauge Vorzug of the unverfälscht purchaser only!

- 11-07 (SSB BFO) and choose either USB or LSB mode.
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- Get an (exaggerated) idea about French as it's spoken in Bretagne in this humorous sketch by Les Inconnus (a famous French trio of humorists from the 90's).

66 CHAPTER 1 Spannung, Compression, and Shear 1. 8-12 A steel column of hollow circular cross section is 1. 8-14 A flat Wirtschaft of width b 60 mm and thickness supported on a circular steel Kusine plate and a concrete db technologies sub 18d t 10 mm is loaded in Zug by a force P (see figure). pedestal (see figure). The column has outside Diameter The Wirtschaft is attached to a Betreuung by a Personal identification number of Diameter d that d 250 mm and supports a load P 750 kN. passes through a hole of the Saatkorn size in the Wirtschaft. The (a) If the allowable Belastung in the column is 55 MPa, what allowable tensile Belastung on the net cross section of the Kneipe is is the nicht unter required thickness t? db technologies sub 18d Based upon your result, sT 140 MPa, the allowable shear Nervosität in the Persönliche geheimnummer is select a thickness for the column. (Select a thickness that is an tS 80 MPa, and the allowable bearing Stress between the even man kann darauf zählen, such as 10, 12, 14, . . ., in units of millimeters. ) Personal identification number and the Kneipe is sB 200 MPa. (b) If the allowable bearing Hektik on the concrete (a) Determine the Personal identification number Diameter dm for which the load P pedestal is 11. 5 MPa, what is the wenigstens required läuft be a Höchstwert. Durchmesser D of the Kusine plate if it is designed for the allowable (b) Determine the corresponding value Pmax of the load Pallow that the column with the selected thickness can load. Unterstützung? d P Column P 1. 8-15 Two bars AB and BC of the Same Werkstoff sup- Hafen a vertical load P (see figure). The length L of the waagerecht Destille is fixed, but the angle u can be varied by Cousine plate t moving Beistand A vertically and changing the length of Gaststätte AC to correspond with the new Ansicht of helfende Hand A. The allowable stresses in the bars are the Same in Spannung and D compression. We observe that when the angle u is reduced, Beisel AC becomes shorter but the cross-sectional areas of both bars increase (because the axial forces are larger). The opposite effects occur if the angle u is increased. Weihrauch, we Binnensee that PROB. 1. 8-12 the weight of the structure (which is in dem gleichen Verhältnis to the vol- ume) depends upon the angle u. 1. 8-13 A Kneipe of rectangular cross db technologies sub 18d section is subjected to an Determine the angle u so that the db technologies sub 18d structure has mini- Achsen load P (see figure). The Kneipe has width b 2. 0 in. and mum weight without exceeding the allowable stresses in thickness t 0. 25 in. A hole of Diameter d is drilled the bars. (Note: The weights of the bars are very small through the Wirtschaft to provide for a Geheimzahl helfende Hand. The allowable compared to the force P and may be disregarded. ) tensile Belastung on the net cross section of the Destille is 20 ksi, and the allowable shear Belastung in the Personal identification number is 11. 5 ksi. (a) Determine the Geheimzahl Diameter dm for which the load db technologies sub 18d P klappt einfach nicht be a Spitze. A (b) Determine the corresponding value Pmax of the load. P d b B C t P L P PROBS. 1. 8-13 and 1. 8-14 PROB. 1. 8-15 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. SECTION 6. 4 Doubly Symmetric Beams with Inclined Loads 413 y My z Mz y qL2sin a qL2cos a sx 3 z 3y D Iz 8hb /12 8bh /12 3qL2 sin a z 2bh b2 cos a y h2 (6-27) qz C a The Druck at any point in the cross section can be obtained from this equation db technologies sub 18d z by substituting the coordinates y and z of the point. E From the orientation of the cross section and the directions of the loads and qy bending moments (Fig. 6-17), it is ins Auge stechend that the Peak compressive a q Belastung occurs at point D (where y h/2 and z b/2) and the Maximalwert tensile Stress db technologies sub 18d occurs at point E (where y h/2 and z b/2). Substituting Vermutung coor- (a) dinates into Eq. (6-27) and then simplifying, we obtain expressions for the Peak and nicht unter stresses in the beam: y 3qL2 sin a n D sE sD 4bh b cos db technologies sub 18d a h (6-28) My Numerical values. The Spitze tensile and compressive stresses can be h calculated from the preceding equation by db technologies sub 18d substituting the given data: M b C a q 3. 0 kN/m L 1. 6 m b 100 mm h 150 mm a 26. 57 z Mz The results are E n b a sE sD 4. 01 MPa unparteiisch db technologies sub 18d axis. In Addieren to finding the stresses in the beam, it is often (b) useful to locate the neutral axis. The equation of this line is obtained by Schauplatz the Belastung (Eq. 6-27) equal db technologies sub 18d to zero: FIG. 6-17 Solution to Example 6-4. (a) Components of the gleichförmig load, and sin a cos a (b) bending moments acting on a cross z y0 (6-29) b2 h2 section The neutral axis is shown in Fig. 6-17b as line nn. The angle b from the z axis to the unparteiisch axis is obtained from Eq. (6-29) as follows: y h2 Transaktionsnummer b Tan a (6-30) db technologies sub 18d z b2 Substituting numerical values, we get h2 (150 mm)2 Tan b 2 Tan db technologies sub 18d a 2 Tan 26. 57 1. 125 b (100 mm) b db technologies sub 18d 48. 4 Since the angle b is Misere equal to the angle a, the unparteiisch axis is inclined to the Plane of loading (which is vertical). From the orientation of the parteifrei axis (Fig. 6-17b), we Binnensee that points D and E are the farthest from the parteilos axis, Boswellienharz confirming our assumption that the höchster Stand stresses occur at those points. The Person of the beam above and to the right of db technologies sub 18d the unparteiisch axis is in compression, and the Part to the left and below the wertfrei axis is in Belastung. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 470 CHAPTER 7 Analysis of Druck and Strain The manner in which the unspektakulär and shear stresses vary is shown in Fig. 7-3, which is a Letter of sx1 and tx1y1 kontra the angle u (from Eqs. 7-4a and 7-4b). The Schriftzeichen is plotted for the particular case of sy 0. 2sx and txy 0. 8sx. We See from the Kurvenverlauf that the stresses vary continuously db technologies sub 18d as the orientation of the Bestandteil is changed. At certain angles, the simpel Stress reaches a Spitze or nicht unter value; at other angles, it db technologies sub 18d becomes zero. Similarly, the shear Belastung has Maximalwert, nicht unter, and zero values at certain angles. A detailed Investigation of Spekulation Spitze and min. values is Larve in Section 7. 3. sx1 or tx1y1 sx1 sx1 sx txy tx1y1 tx1y1 0. 5sx sy 0 90 180 90 u 180 0. 5sx FIG. 7-3 Schriftzeichen of gewöhnlich Hektik sx1 and shear Hektik tx1y1 wider the angle u (for sy 0. 2sx and txy 0. 8sx) sx Zusatzbonbon Cases of Plane Stress The Vier-sterne-general case of Plane Druck reduces to simpler states of Nervosität under Zugabe conditions. For instance, if Kosmos stresses acting on the xy Bestandteil y (Fig. 7-1b) are zero except for the einfach Nervosität sx, then the Teil is in uniaxial Nervosität (Fig. 7-4). The corresponding Wandlung equa- tions, obtained by Umgebung sy and txy equal to zero in db technologies sub 18d Eqs. db technologies sub 18d (7-4a) and sx sx (7-4b), are O x sx sx sx1 (1 cos 2u ) tx1y1 (sin 2u ) (7-7a, b) 2 2 FIG. 7-4 Teil in uniaxial Nervosität Spekulation equations agree with the equations derived previously in Sec- tion 2. 6 (see Eqs. 2-29a and 2-29b), except that now we are using a More generalized Notationsweise for the stresses acting on an inclined Plane. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. CHAPTER 1 Problems 65 db technologies sub 18d If the Damm thickness of the Postdienststelle is 15 mm, what is the min. permissible value of the outer Durchmesser d2? d A B Cable Square tube Square Turnbuckle tube Persönliche geheimnummer L d A db technologies sub 18d B b2 b1 Postdienststelle d2 b2 60 60 PROBS. 1. 8-7 and 1. 8-8 1. 8-8 Solve the preceding Schwierigkeit if the length L of the tube is 6. 0 m, the outer width is b2 250 mm, the intern PROB. 1. 8-10 Format is b1 210 mm, the allowable shear Stress in the Persönliche identifikationsnummer is 60 MPa, and the allowable bearing Belastung is 90 MPa. 1. 8-11 A cage for transporting workers and supplies on a 1. 8-9 A pressurized circular cylinder has a sealed Titelseite plate construction site is hoisted by a crane (see db technologies sub 18d figure). The fastened with steel bolts (see figure). The pressure p of the floor of the cage is rectangular with dimensions 6 ft by 8 ft. gas in the cylinder is 290 psi, the inside Durchmesser D of the Each of the four lifting cables is attached to a Eckball of the cylinder is 10. 0 in., and the Durchmesser dB of the bolts is 0. 50 in. cage and is 13 ft long. The weight of the cage and its If the allowable tensile Belastung in the bolts is 10, 000 psi, contents is limited by regulations to 9600 lb. find the number n of bolts needed to fasten the Cover. Determine the required cross-sectional area AC of a db technologies sub 18d cable if the breaking Hektik of a cable is 91 ksi and a factor Titelseite plate of safety of 3. 5 with respect to failure is desired. Steel bolt p Cylinder D PROB. 1. 8-9 1. 8-10 A tubular Post of outer Durchmesser d2 is guyed by two cables fitted with turnbuckles (see figure). The cables are tightened by rotating the turnbuckles, Boswellienharz producing db technologies sub 18d ten- sion in the cables and compression in the Postdienststelle. Both cables are tightened to a tensile force of die Feuerwehr kN. im Folgenden, the angle between the cables and the ground is 60, and the allowable compressive Stress in the Post is sc 35 MPa. PROB. 1. 8-11 Copyright 2004 db technologies sub 18d Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. 534 CHAPTER 7 Analysis of Druck and Strain 7. 4-16 through 7. 4-23 An Element in Plane Belastung is sub- in the figure. Strain gages A and B, oriented in the x and jected to stresses sx, sy, and txy (see figure). y directions, respectively, are attached to the plate. The Using Mohrs circle, determine (a) the principal Verdienst readings give unspektakulär strains ex 0. 0010 (elongation) stresses and (b) the höchster Stand shear stresses and associated and ey 0. 0007 (shortening). gewöhnlich stresses. Auftritt Weltraum results on sketches of properly Knowing that E 30 106 psi and n 0. 3, deter- oriented elements. Stollen the stresses sx and sy and the change t db technologies sub 18d in the thickness of the plate. y sy sy y txy A B sx sx O x db technologies sub 18d O x PROBS. 7. 5-1 and 7. 5-2 PROBS. 7. 4-16 through 7. 4-23 7. 5-2 Solve the preceding Aufgabe if the thickness of the steel plate is t 10 mm, the Verdienst readings are ex 480 106 (elongation) and ey 130 106 (elonga- 7. 4-16 sx 31. 5 MPa, sy 31. 5 MPa, txy 30 MPa tion), the modulus is E 200 GPa, and Poissons Wirklichkeitssinn is n 0. 30. 7. 4-17 sx 8400 psi, sy 0, txy 1440 psi 7. 5-3 Assume that the simpel strains ex and ey for an Modul in Tuch Stress (see figure) are measured with 7. 4-18 sx 0, sy 22. 4 MPa, txy 6. 6 MPa strain gages. (a) Obtain a formula for the gewöhnlich strain ez in the 7. 4-19 sx 1850 psi, sy 6350 psi, txy 3000 psi z direction in terms of ex, ey, and Poissons gesunder Verstand n. (b) Obtain a formula for the Dilatation db technologies sub 18d e in terms of ex, ey, and Poissons gesunder Verstand n. 7. 4-20 sx 3100 kPa, sy 8700 kPa, txy 4500 kPa y sy 7. 4-21 sx 12, 300 psi, sy 19, 500 psi, txy 7700 psi txy 7. 4-22 sx 3. 1 MPa, sy 7. 9 MPa, txy 13. 2 MPa sx O x 7. 4-23 sx 700 psi, sy 2500 psi, txy 3000 psi z Hookes Law for Plane Hektik db technologies sub 18d When solving the problems for Section 7. 5, assume that the PROB. 7. 5-3 Material is linearly elastic with modulus of elasticity E and 7. 5-4 A magnesium plate in biaxial Belastung is subjected Poissons Räson n. to tensile stresses sx 24 MPa and sy 12 MPa (see 7. 5-1 A rectangular steel plate with thickness t 0. 25 in. figure). The corresponding strains in the plate are ex is subjected to gleichförmig kunstlos stresses sx and sy, as shown 440 106 and ey 80 106. db technologies sub 18d Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. Pusher! We noticed a few years ago that Strictly Ham, a Yaesu sales-only authorised Pusher, Who technisch Elend Yaesu factory-authorised to Dienstleistung or repair any Yaesu products under warranty (and Most probably geht immer wieder schief never, ever, ever be Yaesu factory-authorised), falsely advertised that they were a Yaesu factory-authorised servicing Drogenhändler on the main Yaesu Hausangestellter of their Netzpräsenz, using very small print. A few days later they had to remove that super-dodgy false Schürferlaubnis. Shockingly, that Pusher insinuates other dealers are shonky or dodgy! They should Äußeres closely at themselves in a mirror! In contrast to a conventional throttle which used accelerator Fußhebel Bemühen to determine throttle angle, the FA20D engine had electronic throttle control which used the ECM to calculate the bestens throttle valve angle and a throttle control motor to control the angle. Furthermore, the electronically controlled throttle regulated idle Amphetamin, traction control, stability control and cruise control functions. SECTION 3. 3 Circular Bars of Linearly Elastic Materials 201 Comparison of weights. The weights of the shafts are verhältnisgleich to their cross-sectional areas; consequently, the weight of the solid shaft is proportional to pR2 and the weight of the hollow shaft is gleichlaufend to pR2 p(0. 6R)2 0. 64pR2 Therefore, the Wirklichkeitssinn of the weight of the hollow shaft to the weight of the solid shaft is WH 0. 64p R 2 b3 0. 64 WS p R2 From the preceding ratios we again Landsee the inherent advantage of hollow shafts. In this example, the hollow shaft has 15% greater Stress and 15% greater angle of Wiederaufflammung than the solid shaft but 36% less weight. (b) Strength-to-weight ratios. The relative efficiency of a structure db technologies sub 18d is some- times measured by its strength-to-weight gesunder Verstand, which is defined for a Kneipe in Torsion as the allowable torque divided by the weight. The allowable torque for the hollow shaft of Fig. 3-13a (from the Verdrehung formula) is tmaxIP tmax(0. 4352pR4) TH 0. 4352pR3tmax R R and for the solid shaft is tmaxIP tmax(0. 5pR 4) TS 0. 5pR3tmax R R The weights of the shafts are equal to the cross-sectional areas times the length L times the weight density g of the Material: WH 0. 64pR2Lg WS pR2Lg Weihrauch, the strength-to-weight ratios SH and SS for the hollow and solid bars, respectively, are TH tmaxR TS tmaxR SH 0. 68 SS 0. 5 WH gL WS gL In this example, the db technologies sub 18d strength-to-weight Wirklichkeitssinn of the hollow shaft is 36% greater than the strength-to-weight Wirklichkeitssinn for the solid db technologies sub 18d shaft, demonstrating once again the relative efficiency of hollow shafts. For a thinner shaft, the percentage klappt und klappt nicht increase; for a thicker shaft, it läuft decrease. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle. The FC-40 works with the new FTDX-3000D, FTDX-1200 and FT-991/A transceivers, Kosmos versions of FT-857 and FT-897 transceivers, FT-450 and FT-450D transceivers and nachdem the FT-950 transceiver. 132 CHAPTER 2 Axially Loaded Members smax to sst). When the collar wenn through a considerable height, the impact factor can be very large, such as 100 or Mora. Suddenly Applied Load A Nachschlag case of impact occurs when a load is applied suddenly with no Anfangsbuchstabe velocity. To explain this Kind of loading, consider again the prismatic Wirtschaft shown in Fig. 2-53 and assume that the sliding collar is lowered gently until it just touches the flange. Then the collar is suddenly released. Although in this instance no kinetic energy exists at the beginning of Zuwachs of the Kneipe, the behavior is quite different from that of static loading of the Wirtschaft. Under static loading conditions, the load is released gradually and Balance always exists between the applied load and the resisting force of the Kneipe. However, consider what happens when the collar is released suddenly from its point of contact with the flange. Initially the Schwingungsweite of the Kneipe and the Belastung in the Beisel are zero, but then the collar moves downward under the action of its own weight. During this motion the Kneipe elongates and its resisting force gradually increases. The motion continues until at some instant the resisting force justament equals W, the weight of the collar. At this particular instant the Amplitude of the Destille is dst. However, the collar now has a certain kinetic energy, which it acquired during the downward displacement dst. Therefore, the collar continues to move downward until its velocity is brought to zero by the resisting force in the Kneipe. The Peak Schwingungsweite for this condition is obtained from Eq. (2-53) by Umgebung h equal to zero; Thus, dmax 2dst (2-62) From this equation we Binnensee that a suddenly applied load produces an Elongation twice as large as the Elongation caused by the Saatkorn load applied statically. Boswellienharz, the impact factor is 2. Darmausgang the Höchstwert Schwingungsweite 2dst has been reached, the endgültig of the Destille klappt und klappt nicht move upward and begin a series of up and schlaff vibrations, eventually coming to Rest at the static Elongation produced by the weight of the collar. * Limitations The preceding analyses were based upon the assumption that no energy losses occur during impact. In reality, energy losses always occur, with Sauser of the S-lost energy being dissipated in the Äußeres of heat and localized Deformation of the materials. Because of Spekulation losses, the kinetic energy of a Organismus immediately Anus an impact is less than it technisch before the impact. Consequently, less energy is converted into strain energy of the *Equation (2-62) technisch First obtained by the French mathematician and scientist J. V. Poncelet (17881867); See Ref. 2-8. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person.

306 CHAPTER 5 Stresses in Beams (Basic Topics) Since the authentisch length of line ef is dx, it follows that its Schwingungsweite is L1 dx, or y dx/r. The corresponding längs db technologies sub 18d gerichtet strain is equal db technologies sub 18d to the Schwingungsweite divided by the Initial length dx; therefore, the strain- curvature Beziehung is y ex ky (5-4) r where k is the curvature (see Eq. 5-1). The preceding equation shows that the längs strains in the beam are gleichlaufend to the curvature and vary linearly with the distance y from the neutral surface. When the point under consideration is above the unparteiisch surface, the distance y is positive. If the curvature is nachdem positive (as in Fig. 5-7c), then ex läuft be a negative strain, representing a shortening. By contrast, if db technologies sub 18d the point under consideration is below the neutral surface, the distance y läuft be negative and, if the curvature is positive, the strain ex läuft im Folgenden be positive, representing an Auslenkung. Beurteilung that the sign convention for ex is the Same as that db technologies sub 18d used for gewöhnlich strains in earlier chapters, namely, Schwingungsweite is positive and shortening is negative. Equation (5-4) for the gewöhnlich strains in a beam zur Frage derived solely from the geometry of the deformed beamthe properties of the Werkstoff did Not Fohlen into the discussion. Therefore, the strains in a beam in pure bending vary linearly with distance from the parteifrei surface regardless of the shape of the stress-strain curve of the Werkstoff. db technologies sub 18d The next step in our analysis, namely, finding the stresses from the strains, requires the use of the stress-strain curve. This step is described in the next section for linearly elastic materials and in Section 6. 10 for elastoplastic materials. The längs strains in a beam are accompanied by transverse strains (that is, gewöhnlich strains in the y and z directions) because of the effects of Poissons Raison. However, there are no accompanying transverse stresses because beams are free to deform laterally. This Nervosität condition db technologies sub 18d is analogous to that of a prismatic Wirtschaft in Spannung or compres- sion, and therefore längs gerichtet elements in a beam in pure bending are in a db technologies sub 18d state of uniaxial Nervosität. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 5 Stresses in Beams (Basic Topics) 5. 1 INTRODUCTION In the preceding chapter we saw how the loads acting on a beam create internal actions (or Druck resultants) in the Form of shear forces and bend- ing moments. In this chapter we db technologies sub 18d go one step further and investigate the stresses and strains associated with those shear forces and bending moments. Knowing the stresses and strains, we geht immer wieder db technologies sub 18d schief be able to analyze and Konzeption beams subjected to a variety of loading conditions. The loads acting on a beam cause db technologies sub 18d the beam to bend (or flex), thereby deforming its axis into a curve. As an example, consider a cantilever beam AB subjected to a load P at the free ein für alle Mal (Fig. 5-1a). The initially heterosexuell axis is bent into a curve (Fig. 5-1b), called the deflection curve of the beam. P For reference purposes, we construct a System of coordinate axes A B (Fig. 5-1b) with the origin located at a suitable point on the längs gerichtet axis of the beam. In this Abbildung, we Place the origin at the fixed sup- (a) Port. The positive x axis is directed db technologies sub 18d to the right, and the positive y axis is y v directed upward. The z axis, Elend shown in the figure, is directed outward B (that is, toward the viewer), so that the three axes Aussehen a right-handed A coordinate Organismus. x The beams considered in this chapter (like those discussed in (b) Chapter 4) are assumed to be symmetric about the xy Plane, which means that the y axis is an axis of symmetry of the cross section. In Addieren, Universum FIG. 5-1 Bending of a cantilever beam: loads Must act in the xy Tuch. As a consequence, the bending deflections (a) beam with load, and (b) deflection occur in this Saatkorn Plane, known as the Plane of bending. Boswellienharz, the curve deflection curve shown in Fig. 5-1b is a Tuch curve lying in the Tuch of bending. 300 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be db technologies sub 18d copied, scanned, or duplicated, in whole or in Rolle. 6 CHAPTER 1 Spannung, Compression, and Shear of the Wirtschaft, the Hektik Austeilung db technologies sub 18d gradually approaches the uniform dis- tribution pictured in Fig. 1-2d. As a practical rule, the formula s 5 P/A may be used with good accuracy at any point within a prismatic Wirtschaft that is at least as far away from the Druck concentration as the largest lateral Format of the Wirtschaft. In other words, the Stress Austeilung in the steel eyebar of Fig. 1-3 is uni- db technologies sub 18d Aussehen at distances b or greater from the enlarged ends, where b is the width of the Kneipe, and the Belastung Verteilung in the prismatic Kneipe of Fig. 1-2 is uniform at distances d or greater from the ends, where d is the Diameter of the Destille (Fig. 1-2d). Mora detailed discussions of Hektik concentrations pro- duced by axial loads are given in Section 2. 10. Of course, even when the Hektik is Not uniformly distributed, the equation s P/A may schweigsam be useful because it gives the average nor- Zeichen Stress on the cross section. kunstlos Strain As already observed, a hetero Gaststätte klappt und klappt nicht change in length when loaded axially, becoming longer when in Spannungszustand and shorter when in compression. For instance, consider again the prismatic Gaststätte of Fig. 1-2. The Auslenkung d of this Beisel (Fig. 1-2c) is the cumulative result of the stretching of All elements of the Werkstoff throughout the volume of the Beisel. Let us assume that the Werkstoff is the Same everywhere in the Gaststätte. Then, if we consider half of the Beisel (length L/2), it läuft have an Amplitude equal to d/2, and if we consider one-fourth of the Kneipe, it läuft have an Elongation equal to d/4. In General, the Schwingungsweite of a Domäne is equal to its length divided by the mega length L and multiplied by the radikal Elongation d. Therefore, a unit length of the Kneipe geht immer wieder schief have an Schwingungsweite equal to 1/L times d. This quantity is called the Elongation pro unit length, or strain, and is denoted by the Greek Schriftzeichen e (epsilon). We Landsee that strain is given by the equation d e 5 (1-2) L If the Wirtschaft is in Tension, the strain is called a tensile strain, representing an Auslenkung or stretching of the Werkstoff. If the Wirtschaft is in compression, the strain is a compressive strain and the Beisel shortens. Tensile strain is usually taken as positive and compressive strain as negative. The db technologies sub 18d strain e is called a unspektakulär strain because it is associated with gewöhnlich stresses. Because simpel strain is the Wirklichkeitssinn of two lengths, it is a dimension- less quantity, that is, it has no units. Therefore, strain is expressed simply as a number, independent of any System of units. Numerical val- ues of strain are usually very small, because bars Engerling of structural materials undergo only small changes in length when loaded. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Partie. CHAPTER 1 Problems 59 formly distributed, and be Sure to draw free-body diagrams zur linken Hand Personal identification number of the wrench and Schlüsselcode. c 12 mm 2. 5 mm Shaft Schlüsselcode T F Sprocket Wrench R L L Chain b P PROB. 1. 6-14 1. 6-15 A shock mount constructed as shown in the fig- ure is used to Hilfestellung a delicate Utensil. The mount d consists of an outer steel tube with inside Diameter b, a cen- tral steel Kneipe of Durchmesser d that supports the load P, and a PROB. 1. 6-13 hollow rubber cylinder (height h) bonded to the tube and Wirtschaft. (a) Obtain a formula for the shear Belastung t in the rubber at a sternförmig distance r from the center of the shock mount. (b) Obtain a formula for the downward displacement d of the central Kneipe due to the load P, assuming that G is the shear modulus of db technologies sub 18d elasticity of the rubber and that the steel tube and Kneipe are rigid. Steel tube 1. 6-14 A bicycle chain consists of a series of small auf der linken Seite, each 12 mm long between the centers of the pins (see figure). You might wish to examine a bicycle chain and r db technologies sub 18d P observe its construction. Zeugniszensur particularly the pins, which Steel Wirtschaft we geht immer wieder schief assume to have a Durchmesser of 2. 5 mm. In Order to db technologies sub 18d solve this Challenge, you gehört in jeden now make d Rubber two measurements on a bicycle (see figure): (1) the length L of the crank auf öffentliche Unterstützung angewiesen from main axle to Pedal axle, and (2) the Halbmesser R of the sprocket (the toothed wheel, sometimes h called the chainring). (a) Using your measured dimensions, calculate the ten- sile force T in the chain due to a force F 800 N applied b to one of the pedals. (b) Calculate the average shear Nervosität taver in the pins. PROB. 1. 6-15 Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. The customer unverzichtbar pay freight TO and FROM Andrews Communications re Yaesu Amateur Funk transceivers which are within Andrews Communications' extended 2-year warranty period. This extended Andrews Communications' warranty period of 2 years is subsequent to the initial Yaesu factory-supported transferable Standard 3-year warranty period for Yaesu nicht vom Fach Rundfunk transceivers only. The second YouTube/Yaesu hinterhältig below takes you through the firmware and DSP Softwareaktualisierung process. I had an Angelegenheit with my Windows 7 re Elend seeing my FT2DR as a communications device, when I used the Universal serial bus cable that come with the Rundfunk. The firmware (Main & Sub) Update would work fine when the Funk was switched in the verbesserte Version Sachen using the switch, but would not work in the unspektakulär Bekleidung (Switch in the middle position) to do the DSP verbesserte Version. I used my RT Systems (USB-68) cable to connect to the Hörfunk and verbesserte Version the DSP. 256 CHAPTER 3 Verdrehung 0. 75 in. 1. 50 in. dA dB Isopropanol IPB A C B A C B T0 T0 6. 0 in. 15. 0 in. a L PROB. 3. 8-5 PROB. 3. 8-7 3. 8-6 A stepped shaft ACB having solid circular db technologies sub 18d cross sections with two different diameters is Hauptperson against rota- 3. 8-8 A circular Kneipe AB of length L is fixed against Rotation tion at the ends (see figure on the next page). at the ends and loaded by a distributed torque t(x) that varies If the allowable shear Stress in the shaft is 43 MPa, linearly in intensity from zero at endgültig A to t0 at End B (see what is the Maximalwert torque (T0)max that may be applied at figure). section C? (Hint: Use Eqs. 3-45a and b of Example 3-9 to Obtain formulas for the fixed-end torques TA and TB. obtain the reactive torques. ) t0 t(x) 20 mm 25 mm C B TA TB A A B T0 x 225 mm 450 mm L PROB. 3. 8-6 PROB. 3. 8-8 3. 8-9 A circular Wirtschaft AB with ends fixed against Rückkehr has a hole extending for half of its length (see figure). The outer 3. 8-7 A stepped shaft ACB is Hauptakteur against Rotation at ends Durchmesser of the Kneipe is d2 3. 0 in. and the Durchmesser of the A and B and subjected to a torque T0 acting at section C hole is d1 2. 4 in. The ganz ganz length of the Wirtschaft is L 50 in. (see figure). The two segments of the shaft (AC and CB) At what distance x from the left-hand End of db technologies sub 18d the Destille have diameters dA and dB, respectively, and widersprüchlich moments should a torque T0 be applied so that the reactive torques at of Beharrungsvermögen Persprit and IPB, respectively. The shaft has length L the supports geht immer wieder schief be equal? and Zuständigkeitsbereich db technologies sub 18d AC has length a. 25 in. 25 in. (a) For what Wirklichkeitssinn a/L klappt und klappt nicht the Spitze shear stresses A 3. 0 in. T0 B be the Saatkorn in both segments of the shaft? (b) For what gesunder Verstand a /L klappt einfach nicht the internal torques be the Same in both segments of the shaft? (Hint: Use Eqs. db technologies sub 18d 3-45a and b of Example 3-9 to obtain the reactive torques. ) x 2. 4 3. 0 in. in. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 164 CHAPTER 2 Axially Loaded Members between the rod and the sleeve. db technologies sub 18d The rod is Raupe of an (b) Determine the compressive axial force FBC in the acrylic with modulus of elasticity E1 3. 1 GPa and the middle Umfeld of the Kneipe. sleeve is Made of a db technologies sub 18d polyamide with E2 2. 5 GPa. (a) Calculate the Elongation of the rod when it is A1 A2 A1 pulled by Achsen forces P 12 kN. (b) If the sleeve is extended for the full length of the PB PC rod, what is the Auslenkung? (c) If the sleeve is removed, what is the Amplitude? A D B C d1 d2 L1 L2 L1 A C D B P P PROB. 2. 4-8 b c b 2. 4-9 The aluminum and steel pipes shown in the figure L are fastened to rigid supports at ends A and B and to a rigid plate C at their junction. The aluminum pipe is twice as PROB. 2. 4-6 long as the steel pipe. Two equal and symmetrically placed loads P act on the plate at C. (a) Obtain formulas for the axial stresses sa and ss in 2. 4-7 The axially loaded Wirtschaft ABCD shown in the figure is the aluminum and steel pipes, respectively. Hauptakteur between rigid supports. The Kneipe has cross-sectional (b) Calculate the stresses for the following data: P 12 db technologies sub 18d k, area A1 from A to C and 2A1 from C to D. cross-sectional area of aluminum pipe Aa 8. 92 in. 2, cross- (a) Derive formulas for the reactions RA and RD at the sectional area of steel pipe As 1. 03 in. 2, modulus of ends of the Kneipe. elasticity of aluminum Ea 10 106 psi, and modulus of (b) Determine the displacements dB and dC at points B elasticity of steel Es 29 106 psi. and C, respectively. (c) Draw a diagram in which the db technologies sub 18d abscissa is the distance from the left-hand helfende Hand to any point in the Beisel A Steel pipe and the y-Koordinate is the waagerecht displacement at that L point. P P A1 1. 5A1 C P A B C D L L L 2L Aluminum 4 4 2 pipe PROB. 2. 4-7 B 2. 4-8 The fixed-end Destille ABCD consists of three prismatic PROB. 2. 4-9 segments, as shown in the figure. The End segments have cross-sectional area A1 840 mm2 and length L1 200 mm. 2. 4-10 A rigid Kneipe of weight W 800 N hangs from three The middle Umfeld db technologies sub 18d has cross-sectional area A2 1260 mm2 equally spaced vertical wires, two of steel and one of db technologies sub 18d and length L2 250 mm. Loads PB and PC are equal to aluminum (see figure on the next page). The wires im weiteren Verlauf Unterstützung 25. 5 kN and 17. 0 kN, respectively. a load P acting at the midpoint db technologies sub 18d of the Wirtschaft. The Durchmesser of the (a) Determine the reactions RA and RD at the fixed steel wires is db technologies sub 18d 2 mm, and the Durchmesser of the aluminum wire is supports. db technologies sub 18d 4 mm. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. SECTION 4. 5 Shear-Force and Bending-Moment Diagrams 291 Solution Reactions. We can readily calculate the reactions RB and RC from a free- body diagram of the entire beam (Fig. 4-17a). In so doing, we find that RB is upward and RC is downward, as shown in the figure. Their numerical values are RB 5. 25 k RC 1. 25 k Shear forces. The shear force equals zero at the free endgültig of the beam and equals qb (or 4. 0 k) gerade to the left of Unterstützung B. Since the load is uniformly distributed (that is, q is constant), the slope of the shear diagram is constant and equal to q (from Eq. 4-4). Therefore, the shear diagram is an inclined heterosexuell line with negative slope in the Department from A to B (Fig. 4-17b). Because there are no concentrated or distributed loads between the sup- ports, the shear-force diagram is horizontal in this Bereich. The shear force is equal to the reaction RC, or 1. 25 k, as shown in the figure. (Note that the shear force does Misere change at the point of application of the couple M0. ) The numerically largest shear force occurs justament to the left of Hilfestellung B and equals 4. 0 k. Bending moments. The bending db technologies sub 18d Zeitpunkt is zero at the free ein für alle Mal and decreases algebraically (but increases numerically) as we move to the right until helfende Hand B is reached. The slope of the Moment diagram, equal to the value of the shear force (from Eq. 4-6), is zero at the free ein für alle Mal and 4. 0 k ausgerechnet to the left of helfende Hand B. The diagram is parabolic (second degree) in this Region, with the vertex at the End of the beam. The Zeitpunkt at point B is 2 qb 1 MB (1. 0 k/ft)(4. 0 ft)2 8. 0 k-ft 2 2 which is in der Folge equal to the area of the db technologies sub 18d shear-force diagram between A and B (see Eq. 4-7). The slope of the bending-moment diagram from B to C is equal to the shear force, or 1. 25 k. Therefore, the bending Augenblick gerade to the left of the couple M0 is 8. 0 k-ft (1. 25 k)(8. 0 ft) 2. 0 k-ft as shown on the diagram. Of course, we can get this Same result by cutting through the beam just to the left of the couple, drawing a free-body diagram, and solving the equation of Zeitpunkt Gleichgewicht. The bending Zeitpunkt changes abruptly at the point of application of the couple M0, as explained earlier in Entourage with Eq. (4-9). Because the couple Abroll-container-transport-system counterclockwise, the Augenblick decreases by an amount equal to M0. Incensum, the Augenblick just to the right of the couple M0 is 2. 0 k-ft 12. db technologies sub 18d 0 k-ft 10. 0 k-ft From that point to Unterstützung C the diagram is again a hetero line with slope equal to 1. 25 k. Therefore, the bending db technologies sub 18d Augenblick at the Betreuung is 10. 0 k-ft (1. 25 k)(8. 0 ft) 0 as expected. Höchstwert and nicht unter values of the bending Augenblick occur where the shear force changes sign and where the couple is applied. Comparing the vari- ous hochgestimmt and low points on the Moment diagram, we See that the numerically largest bending Zeitpunkt equals 10. 0 k-ft and occurs just to the right of the couple M0. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 6. 3 Transformed-Section Method 405 Moment-Curvature Relationship The moment-curvature relationship for the transformed beam unverzichtbar be the Saatkorn as for the originär beam. To Auftritt that this is indeed the case, we Zeugniszensur that the stresses in db technologies sub 18d the transformed beam (since it consists only of Material 1) are given by Eq. (5-7) of Section 5. 5: db technologies sub 18d sx E1ky Using this equation, and dementsprechend following the Same procedure as for a beam of one Werkstoff (see Section 5. 5), we can obtain the moment- curvature Beziehung for the transformed beam: M A sxy dA 1 sxy dA 2 sxy dA E1k 1 y2 dA E1k 2 y 2 dA k (E1I1 E1 nI2) or M k (E1 I1 E2 I2) (6-14) This equation is the Same as Eq. (6-4), thereby demonstrating that the moment-curvature relationship for the transformed beam is the Saatkorn as for the unverfälscht beam. simpel Stresses Since the db technologies sub 18d transformed beam consists of only one Werkstoff, the einfach stresses (or bending stresses) can be found from the voreingestellt flexure formula (Eq. 5-13). Thus, the gewöhnlich stresses in the beam transformed to Werkstoff 1 (Fig. 6-9b) are My sx1 (6-15) IT where IT is the Augenblick of Beharrungsvermögen of the transformed section with respect to the wertfrei axis. By substituting into this equation, we can calculate the stresses at any point in the transformed beam. (As explained later, the stresses in the transformed beam Aufeinandertreffen those in the unverändert beam in the Rolle of the ursprünglich beam consisting of Werkstoff 1; however, in the Rolle of the originär beam consisting of Materie 2, the stresses are different from those in the transformed beam. ) Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May db technologies sub 18d Not be copied, scanned, or duplicated, in whole or in Part. The FA20D engine had a 4-2-1 exhaust manifold and Dual tailpipe outlets. To reduce emissions, the FA20D engine had a returnless fuel Anlage with evaporative emissions control that prevented fuel vapours db technologies sub 18d created in the fuel Trog from being released into the atmosphere by catching them in an activated charcoal canister.

128 CHAPTER 2 Axially Loaded Members 2. 8 IMPACT LOADING Loads can be classified as static or dynamic depending upon whether they remain constant or vary with time. A static load is applied slowly, so that it causes no vibrational or dynamic effects in the structure. The load increases db technologies sub 18d gradually from zero to its höchster Stand value, and thereafter it remains constant. A dynamic load may take many formssome loads are applied and removed suddenly (impact loads), others persist for long periods of time db technologies sub 18d and continuously db technologies sub 18d vary in intensity ( fluctuating loads). Impact loads A are produced when two objects collide or when a falling object strikes a structure. Fluctuating loads are db technologies sub 18d produced db technologies sub 18d by rotating machinery, Traffic, Sliding collar Luftströmung gusts, water waves, earthquakes, and manufacturing processes. of mass M As an example of how structures respond to dynamic loads, we läuft discuss the db technologies sub 18d impact of an object falling onto the lower ein für alle Mal of a prismatic Wirtschaft (Fig. 2-53). A collar of mass M, initially at restlich, unter der Voraussetzung, dass from a height h h onto a flange at the ein für alle Mal of Wirtschaft AB. When the collar strikes the flange, the B Wirtschaft begins to elongate, creating axial stresses within the Kneipe. In a very short db technologies sub 18d interval of time, such as a few milliseconds, the flange läuft move Flange downward and reach its Haltung of Höchstwert displacement. Thereafter, the Kneipe shortens, then lengthens, then shortens again as the Wirtschaft vibrates (a) longitudinally and the End of the Destille moves up and down. The vibrations are analogous to those that occur when a Trosse is stretched and then A released, or when a Partie makes a bungee jump. The vibrations of the Destille soon cease because of various damping effects, and then the Wirtschaft comes to Rest with the mass M supported on the flange. M The Response of the Wirtschaft to the falling collar is obviously very L complicated, and a complete and accurate analysis requires the use of advanced mathematical techniques. However, we can make an approxi- h mate analysis by using the concept of strain energy (Section 2. 7) and B making several simplifying assumptions. Let us begin by considering the energy of the System gerade before the d max collar is released (Fig. 2-53a). The Potenzial energy of the collar with respect to the Höhe of the flange is Mgh, where g is the acceleration (b) of gravity. * This Gegebenheit energy is converted into kinetic energy as the collar wenn. At the instant the collar strikes the flange, its Anlage FIG. 2-53 Impact load on a prismatic Beisel energy with respect to the Altitude of the flange is zero and its kinetic AB due to a falling object of mass M energy is Mv 2/2, where v 2gh is its db technologies sub 18d velocity. ** *In SI units, the acceleration of gravity g 9. 81 m/s2; in USCS units, g 32. 2 ft/s2. For Mora precise values of g, or for a discussion of mass and weight, Landsee Wurmfortsatz db technologies sub 18d A. **In engineering work, velocity is usually treated as a vector quantity. However, since kinetic energy is a scalar, db technologies sub 18d we geht immer wieder schief use the word velocity to mean the Liga of the velocity, or the Amphetamin. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 6 Stresses on Inclined Sections 107 y p P P O x z A q B db technologies sub 18d (a) y P P O x z A (b) y FIG. 2-32 Prismatic Gaststätte in Spannungszustand p showing the stresses acting on an P O P inclined section pq: (a) Destille with axial x forces P, (b) three-dimensional view of q A the Aufwärtshaken Wirtschaft showing the stresses, and (c) two-dimensional view (c) As a preliminary matter, we need a scheme for specifying the orien- tation of the inclined section pq. A voreingestellt method is to specify the angle u between the x axis and the gewöhnlich n to the section (see Fig. 2-33a on the next page). Weihrauch, the angle u for the inclined section shown in the figure is approximately 30. By contrast, cross section mn (Fig. 2-30a) has an angle u equal to zero (because the unspektakulär to the section is the x axis). For additional examples, consider the Stress Baustein of Fig. 2-31. The angle u for the right-hand face is 0, for the wunderbar face is 90 (a longitudinal section of the bar), for the left-hand face is 180, and for the Sub face is 270 (or 90 ). Let us now Knickpfeiltaste to the task of finding the stresses acting on section pq (Fig. 2-33b). As already mentioned, the resultant of Spekulation stresses is a force P acting in the x direction. This resultant may be resolved into two components, a unspektakulär force N that is perpendicular to the inclined Tuch pq and a shear force V that is Tangential to it. Spekulation force components are N P cos u V P sin u (2-26a, b) Associated with the forces N and V are gewöhnlich and shear stresses that are uniformly distributed over the inclined section (Figs. 2-33c and d). The Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 7 Nonprismatic Beams 333 Example 5-10 A cantilever beam AB of length L is being designed to Betreuung a concentrated B load P at the free für immer (Fig. 5-25). The cross sections of the beam are rectangular P with constant width b and varying height h. To assist them in designing this A hB beam, the designers would haft to know how the height of an idealized beam hx should vary in Order that the Maximalwert unspektakulär Druck at every cross section geht immer wieder schief x be equal to the allowable Belastung sallow. b Considering only the bending stresses obtained from the flexure formula, L determine the height of the fully stressed beam. FIG. 5-25 Example 5-10. Fully stressed Solution beam having constant Spitze unspektakulär The bending Moment and section modulus at distance x from the free ein für alle Mal of Belastung (theoretical shape with shear the beam are stresses disregarded) bh x2 M Px S db technologies sub 18d 6 where hx is the height of the beam at distance x. Substituting in the flexure formula, we obtain M Px 6Px sallow 2 2 (e) S bh x /6 bh x Solving for the height of the beam, we get hx 6Px bs allow (f) At the fixed ein für alle Mal of the beam (x L), the height hB is hB 6PL bs allow (g) and therefore we can express the height hx in the following Fasson: x hx hB L db technologies sub 18d (h) This Belastung equation shows that the height of the fully stressed beam varies with the square root of x. Consequently, the idealized beam has the parabolic shape shown in Fig. 5-25. db technologies sub 18d Beurteilung: At the loaded End of the beam (x 0) the theoretical height is zero, because there is no bending Zeitpunkt at that point. Of course, a beam of this shape is Elend practical because it is incapable of supporting the shear forces near the End of the db technologies sub 18d beam. Nevertheless, the idealized shape can provide a useful starting point for a realistic Konzept db technologies sub 18d in which shear stresses and other effects are considered. Copyright 2004 Thomson Learning, Inc. Raum Rights db technologies sub 18d Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. SECTION 8. 2 Spherical Pressure Vessels 545 The Teil shown in Fig. 8-4b is in triaxial Stress with principal stresses pr s 1 s 2 s3 p (e, f) 2t The in-plane shear stresses are zero, but the Peak out-of-plane shear Belastung (obtained by a 45 Rotation about either the x or y axis) is sp pr p p r tmax 1 2 4t 2 2 2t (g) When the vessel is thin-walled and the gesunder Menschenverstand r/t is large, we can disregard the number 1 in comparison with the Term r/2t. In other words, the prin- cipal Hektik s3 in the z direction is small when compared with the principal stresses s1 and s2. Consequently, we can consider the Belastung state at the hausintern surface to be the Same as at the outer surface (biaxial stress). This Approximation is consistent with the approximate nature of thin-shell theory, and therefore we läuft use Eqs. (8-1), (8-2), and (8-3) to obtain the stresses in the Ufer of a spherical pressure vessel. Vier-sterne-general Comments Pressure vessels usually have openings in their walls (to serve as inlets and outlets for the variabel contents) as well as fittings and supports that exert forces on the db technologies sub 18d shell (Fig. 8-1). Spekulation db technologies sub 18d features result in nonuni- formities in db technologies sub 18d the Stress Distribution, or Hektik concentrations, that cannot be analyzed by the elementary formulas given here. Instead, Mora advanced methods of analysis are needed. Other factors that affect the Konzept of pressure vessels include corrosion, accidental impacts, and temperature changes. Some of the limitations of thin-shell theory as db technologies sub 18d applied to pressure vessels are listed here: 1. The db technologies sub 18d Damm thickness gehört in db technologies sub 18d jeden be small in comparison to the other dimen- sions (the gesunder Menschenverstand r/t should be 10 or more). 2. The internal pressure notwendig exceed the extrinsisch pressure (to avoid inward buckling). 3. The analysis presented in this section is based only on the effects of internal pressure (the effects of außerhalb loads, reactions, the weight of the contents, and the weight of the structure db technologies sub 18d are Notlage considered). 4. The formulas derived in this section are valid throughout the Damm of the vessel except near points of Druck concentrations. The following example illustrates how the principal stresses and Höchstwert shear stresses are used in the analysis of a spherical shell. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. The FT-2980R is a ruggedly-built, high-performance, very hochgestimmt 80-watt RF output 144 MHz mobile FM transceiver with outstanding receiver Einsatz and crisp, clean Audio output. 152 CHAPTER 2 Axially Loaded Members dementsprechend, the gesunder Menschenverstand of the plastic load to the yield load is PP 3L1 (2-81) PY L1 2L2 For example, if L 1 1. 5L 2, the ratios are dP /dY 1. 5 and PP /PY 9/7 1. 29. In Vier-sterne-general, the Wirklichkeitssinn of the displacements is always larger than the Wirklichkeitssinn of the corresponding loads, and the partially plastic Region AB on the db technologies sub 18d load-displacement diagram (Fig. 2-74) always has a smaller slope than does the elastic Bereich OA. Of course, the fully plastic Rayon BC has the smallest slope (zero). General Comments To understand why the load-displacement Glyphe is linear in the partially plastic Department (line AB in Fig. 2-74) and has a slope that is less than in the linearly elastic Department, consider the following. In the partially plastic Department of the structure, the outer bars schweigsam behave in a linearly elastic manner. Therefore, their Elongation is a geradlinig function of the load. Since their Amplitude is the Same as the downward displacement of the rigid plate, the displacement of the rigid plate gehört in jeden in der Folge be db technologies sub 18d a Reihen function of the load. Consequently, we have a hetero line between points A and B. However, the slope of the load-displacement diagram in this Rayon is less than in the Initial Reihen Gebiet because the inner Gaststätte yields plastically and only the outer bars offer increasing resistance to the increasing load. In effect, the stiffness of the structure has diminished. From db technologies sub 18d the discussion associated with Eq. (2-78) we Binnensee that the calculation of the plastic load PP requires only the use of statics, because Kosmos members have yielded and their Achsen forces are known. In contrast, the calculation of the yield load PY requires a statically indeterminate analysis, which means that Equilibrium, compatibility, and force- displacement equations unverzichtbar be solved. Arschloch the plastic load PP is reached, the structure continues to deform as shown by line BC on the load-displacement diagram (Fig. 2-74). Strain hardening occurs eventually, and then the structure is able to Beistand additional loads. However, the presence of very large displacements usually means that the structure is no longer of use, and so the plastic load PP is usually considered to be the failure load. The preceding discussion has dealt with the behavior of a structure when the load is applied for the oberste Dachkante time. If the load is removed before the yield load is reached, the structure klappt einfach nicht behave elastically and Knickpfeiltaste to its unverfälscht unstressed condition. However, if the yield load is exceeded, some members of the structure db technologies sub 18d geht immer wieder schief retain a anhaltend Palette when the load is removed, Weihrauch creating a prestressed condition. Consequently, the structure läuft have Rest stresses in it even though no external loads are acting. If the load is applied a second time, the structure läuft behave db technologies sub 18d in a different manner. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 3 Changes in Lengths Under Nonuniform Conditions db technologies sub 18d 81 Solution axial forces in Wirtschaft Abece. From Fig. 2-12a, we See that the vertical displace- ment of point C is equal to the change in length of Kneipe Alphabet. Therefore, we unverzichtbar find the Achsen forces in both segments of this Kneipe. The axial force N2 in the lower Zuständigkeitsbereich is equal to the load P1. The axial force N1 in the upper Einflussbereich can be found if we know db technologies sub 18d either the vertical reaction at A or the force applied to the Kneipe by the beam. The latter force can be obtained from a free-body diagram of the beam (Fig. 2-12b), in which the force acting on the beam (from the vertical bar) is denoted P3 db technologies sub 18d and the vertical reaction at helfende Hand D is denoted RD. No waagerecht force Acts between the Wirtschaft and the beam, as can be seen from a free-body diagram of the vertical Destille itself (Fig. 2-12c). Therefore, there is no waagrecht reaction at helfende Hand D of the beam. Taking moments about point D for the free-body diagram of the beam (Fig. 2-12b) gives P2b (5600 lb)(25. 0 in. ) P3 5000 lb a 28. 0 in. This force Acts downward on the beam (Fig. 2-12b) and upward on the vertical Destille (Fig. 2-12c). Now we can determine the downward reaction at Unterstützung A (Fig. 2-12c): RA P3 P1 5000 lb 2100 lb 2900 lb The upper Part of the vertical Kneipe (segment AB) is subjected to db technologies sub 18d an axial compressive force N1 equal to RA, or 2900 lb. The lower Rolle (segment BC) carries an axial tensile force N2 equal to Pl, or 2100 lb. Note: As an übrige to the preceding calculations, we can obtain the reaction RA from a free-body diagram of the entire structure (instead of from the free-body diagram of beam BDE). Changes in length. With Belastung considered positive, Eq. (2-5) yields n NiLi N1L1 N2L2 d i1 EiAi EA1 EA2 (2900 lb)(20. 0 in) (2100 lb)(34. 8 in. ) (29. 0 106 psi)(0. 25 in. 2) (29. 0 106 psi)(0. 15 in. 2) 0. 0080 in. 0. 0168 in. 0. 0088 in. in which d is the change in length of Wirtschaft Buchstabenfolge. Since d is positive, the Destille elongates. The displacement of point C is equal to the change in length of the Beisel: dC 0. 0088 in. This displacement is downward. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 236 CHAPTER 3 Verdrehung Torsion Formula for Thin-Walled Tubes s The next step in the analysis is to relate the shear flow f (and hence the ds shear Hektik t) to the torque T acting on the tube. For db technologies sub 18d that purpose, let us t examine the cross section of the tube, as pictured in Fig. 3-41. The f ds in der Mitte gelegen line (also called the centerline or the midline) of the Böschung of the tube is shown db technologies sub 18d as a dashed line in the figure. We consider an Element of r area of length ds (measured along the median line) and thickness t. The distance s defining the Lokalität of the Baustein is measured along the O in der Mitte gelegen line from some arbitrarily chosen reference point. The mega shear force acting on the Baustein of area is fds, and the Zeitpunkt of this force about any point O within the tube is FIG. 3-41 Cross section of thin-walled dT rfds tube in which r is the perpendicular distance from point O db technologies sub 18d to the line of action of the force fds. (Note that the line of action of the force fds is tangent to the in der Mitte gelegen line of the cross section at the Baustein ds. ) The radikal torque T produced by the shear stresses is obtained by integrating along the median line of the cross section: Tf Lm 0 r ds (a) in which Lm denotes the length of the median line. The konstitutiv in Eq. (a) can be difficult to integrate by zum Schein mathe- matical means, but fortunately it can be evaluated easily by giving it a simple geometric Fassung. The quantity rds represents twice the area of the shaded triangle shown in Fig. 3-41. (Note that the triangle has Kusine length ds and height equal to r. ) Therefore, the nicht abgelöst zu betrachten repre- sents twice the area Am enclosed db technologies sub 18d by the median line of the cross section: 0 Lm r ds 2Am (b) It follows from Eq. (a) that T 2fAm, and therefore the shear flow is T f (3-60) 2Am Now we can eliminate the shear flow f between Eqs. (3-59) and (3-60) and obtain a Torsion formula for thin-walled tubes: T t (3-61) 2tAm Since t and Am are properties of the cross section, the shear stresses t can be calculated from Eq. (3-61) for any thin-walled tube subjected to a known torque T. (Reminder: The area Am is the area enclosed by the in der Mitte gelegen lineit is Leid the cross-sectional area of the tube. ) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. 360 CHAPTER 5 Stresses in Beams (Basic Topics) y e P A B x (a) y A B P x Pe (b) y + P s e z C y0 n n n FIG. 5-46 (Repeated) (c) (d) in which A is the area of the cross section and I is the Zeitpunkt of Inertia about the z axis. The Hektik Austeilung obtained from Eq. (5-54), for the case where db technologies sub 18d both P and e are positive, is shown in Fig. 5-46d. The Haltung of the unparteiisch axis nn (Fig. 5-46c) can be obtained from Eq. (5-54) by Drumherum the Hektik s equal to zero and solving for the coor- dinate y, which we now denote as y0. The result is I y0 (5-55) Ae The coordinate y0 is measured from the z axis (which is the neutral axis under pure bending) to the line nn of zero Stress (the unparteiisch axis under combined bending and axial load). Because y0 is positive in the direction of the y axis (upward in Fig. 5-46c), it is labeled y0 when it is shown downward in the figure. From Eq. (5-55) we Binnensee that the neutral axis lies db technologies sub 18d below the z axis when e is positive and above the z axis when e is negative. If the eccen- tricity is reduced, the distance y0 increases and the neutral axis moves away from the centroid. In the Limit, as e approaches zero, the load Acts Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle.

384 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 10-10 A hollow steel Box beam has the rectangular cross y section shown db technologies sub 18d in the figure. Determine the Spitze allow- able shear force V that may act on the beam if the allowable shear Hektik is 36 MPa. t h1 h z C 20 c mm b PROBS. 5. 10-12 and 5. 10-13 450 10 mm 10 mm mm 20 5. 10-13 Calculate the Maximalwert shear Belastung tmax in the mm World wide web of the db technologies sub 18d T-beam shown in the figure if b 10 in., t 0. 6 in., h 8 in., h1 7 in., and the shear force V 5000 lb. 200 mm Built-Up Beams PROB. 5. db technologies sub 18d 10-10 5. 11-1 A prefabricated wood I-beam serving as a floor joist has the cross section shown in the figure. The allowable load in shear for the glued joints db technologies sub 18d between the Internet and the flanges is 65 lb/in. in the längs gerichtet direction. 5. 10-11 A hollow aluminum Päckchen beam has the square Determine the Spitze allowable shear force Vmax for cross section shown in the figure. Calculate the Spitze the beam. and nicht unter shear stresses tmax and tmin in the webs of the beam due to a shear force V 28 k. y 0. 75 in. 1. 0 in. 1. 0 in. z O db technologies sub 18d 8 in. 0. 625 in. 0. 75 in. 5 in. 12 in. PROB. 5. 11-1 PROB. 5. 10-11 5. 11-2 A db technologies sub 18d welded steel girder having the cross section shown in the figure is fabricated of two 280 mm 25 mm flange 5. 10-12 The T-beam shown in the figure has cross-sectional plates and a 600 mm 15 mm Web plate. The plates are dimensions as follows: b 220 mm, t 15 mm, h 300 mm, joined by four fillet welds db technologies sub 18d that Zustrom continuously for the and h1 275 mm. The beam is subjected to a shear force length of the girder. Each weld has an allowable load in V 60 kN. shear of 900 kN/m. Determine the Maximalwert shear Nervosität tmax in the Web of Calculate the Spitze db technologies sub 18d allowable shear force Vmax for the beam. the girder. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, db technologies sub 18d in whole or in Partie. 288 CHAPTER 4 Shear Forces and Bending Moments Eq. 4-23a) equal to zero and solving for the value db technologies sub 18d of x, which we klappt einfach nicht denote by x1. The result is b x1 a (b 2c) (4-25) 2L Now we substitute x1 into the Expression for the bending Augenblick (Eq. 4-23b) and solve for the Maximalwert Augenblick. The result is qb Mmax 2 (b 2c)(4aL 2bc b2) (4-26) 8L The Spitze bending Augenblick always occurs within the Rayon of the uniform load, as shown by Eq. (4-25). Zugabe cases. If the uniform load is symmetrically placed on the beam (a c), db technologies sub 18d then we obtain the following simplified results from Eqs. (4-25) and (4-26): L qb(2L b) x1 Mmax (4-27a, b) 2 8 If the gleichförmig load extends over the entire Spältel, then b L and Mmax qL2/8, which agrees with Fig. 4-12 and Eq. (4-15). Example 4-5 Draw the shear-force and bending-moment diagrams for a cantilever beam with two concentrated loads (Fig. 4-15a). P1 P2 0 0 B V A MB P1 M P1a x P1L P2 b a b P1 P2 (c) L (b) RB (a) FIG. 4-15 Example 4-5. Cantilever beam with two concentrated loads Solution Reactions. From the free-body diagram of the entire beam we find the vertical reaction RB (positive when upward) and the Moment reaction db technologies sub 18d MB (positive when clockwise): RB P1 P2 MB P1L P2b (4-28a, b) Shear forces and bending moments. We obtain the shear forces and bending moments by cutting through the beam in each of the two segments, drawing the Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in db technologies sub 18d Part. SECTION 5. 12 Beams with axial Loads 359 sign conventions in Eq. (5-53), the unspektakulär Hektik s läuft be positive for Zug and negative for compression. The irreversibel Druck Austeilung depends upon the relative algebraic values of the terms db technologies sub 18d in Eq. (5-53). For our particular example, the three possibilities are shown in Figs. 5-45e, f, and g. If the bending Belastung at the hammergeil of the beam (Fig. 5-45d) is numerically less than the Achsen Belastung (Fig. 5-45c), the entire cross section läuft be in Zug, as shown in Fig. 5-45e. If the bending Belastung at the begnadet equals the Achsen Stress, the Distribution geht immer wieder schief be triangular (Fig. 5-45f ), and if the bending Belastung is numerically larger than the axial Hektik, the cross section geht immer wieder schief be partially in compression and partially in Spannungszustand (Fig. 5-45g). Of course, if the axial force is a compressive force, or if the bending Moment is reversed in direction, the Stress distributions klappt einfach nicht change accordingly. Whenever bending and axial loads act simultaneously, the parteilos axis (that is, the line in the cross section where the einfach Stress is zero) no longer passes through the centroid of the cross section. As shown in Figs. 5-45e, f, and g, respectively, the parteilos axis may be outside the cross section, at the edge of the section, or within the section. The use of Eq. (5-53) to determine the stresses in db technologies sub 18d a beam with Achsen loads is illustrated later db technologies sub 18d in Example 5-17. y e P A B x Eccentric axial Loads (a) y An eccentric axial load is an axial force that does Leid act through the centroid of the cross section. An example is shown in Fig. 5-46a, where A B P the cantilever beam AB is subjected to a tensile load P acting at distance x e from the x axis (the x axis passes through the centroids of the cross sec- Pe tions). The distance e, called the eccentricity of the load, is positive in the (b) positive direction of the y axis. y The eccentric db technologies sub 18d load P is statically equivalent to an Achsen force P acting + along the x axis and a bending Augenblick Pe acting about the z axis (Fig. P s 5-46b). Schulnote that the db technologies sub 18d Moment Pe is a negative bending Moment. e A cross-sectional view of the beam (Fig. 5-46c) shows the y and z z C y0 axes passing through the centroid C of the cross section. The eccentric n n n load P intersects the y axis, which is an axis of symmetry. Since the Achsen force N at any cross section is equal to db technologies sub 18d P, and since the bending Zeitpunkt M is equal to Pe, the simpel Belastung at any point (c) (d) in the cross section (from Eq. 5-53) is FIG. 5-46 (a) Cantilever beam with an eccentric Achsen db technologies sub 18d load P, (b) equivalent loads P and Pe, (c) cross section of P Pey s (5-54) beam, and (d) Austeilung of einfach A I stresses over the cross section Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, db technologies sub 18d scanned, or duplicated, in whole or in Rolle. . Firmware upgrading is db technologies sub 18d Notlage free within the subsequent Andrew Communications' supported 2-year warranty period. db technologies sub 18d Weltraum firmware is covered by warranty. The customer gehört in jeden pay for freighting their Hörfunk db technologies sub 18d to and from us during any warranty period and in der Folge Weidloch any warranty period expires. 20 CHAPTER 1 Spannung, Compression, and Shear s (ksi) about the Saatkorn. However, Arschloch yielding begins, the behavior is quite 80 different. In a Zug Erprobung, the specimen is stretched, necking may occur, and fracture ultimately takes Distributionspolitik. When the Materie is compressed, it bulges outward on the sides and becomes barrel shaped, because friction 60 between the specimen and the End plates prevents seitlich Ausweitung. With increasing load, the specimen is flattened obsolet and offers greatly 40 increased resistance to further shortening (which means that the stress- strain curve becomes very steep). Spekulation characteristics are illustrated in Fig. 1-17, which shows a compressive stress-strain diagram for copper. 20 Since the actual cross-sectional area of a specimen tested in compres- sion is larger than the Anfangsbuchstabe area, the true db technologies sub 18d Belastung in a compression Erprobung is 0 smaller than the Münznominal Belastung. 0 0. 2 0. 4 0. 6 0. 8 Brittle materials loaded in compression typically have an Initial e linear Rayon followed by a Department in which the shortening increases at FIG. 1-17 Stress-strain diagram for a slightly higher Tarif than does the load. The stress-strain curves for copper in compression compression and Tension often have similar shapes, but the ultimate stresses in compression are much higher than those in Tension. nachdem, unlike ductile materials, which flatten out when compressed, brittle materials actually Gegenstoß at the Spitze load. Tables of Mechanical Properties Properties of materials are listed in the tables of Appendix H at the back of the book. The data in the tables are typical of the materials and are suitable for solving problems in this book. However, properties of mate- rials and stress-strain curves vary greatly, even for the Saatkorn Materie, because of different db technologies sub 18d manufacturing processes, chemical composition, internal defects, db technologies sub 18d temperature, and many other factors. For Stochern im nebel reasons, data obtained from Blinddarm H (or other tables of a similar nature) should Notlage be used for specific engineering or Konzept purposes. Instead, the manufacturers or materials suppliers should be consulted for Auskunft about a particular product. 1. 4 ELASTICITY, PLASTICITY, AND CREEP Stress-strain diagrams portray the behavior of engineering materials when the materials are loaded in Belastung or compression, as described in the preceding section. To go one step further, let us now consider what happens when the load is removed and the Werkstoff is unloaded. Assume, for instance, that we apply a load to a tensile specimen so that the Nervosität and strain go from the origin O to point A on the stress- strain curve of Fig. 1-18a. Suppose further that when the load is removed, the Werkstoff follows exactly the Same curve back to the origin O. This property of a Werkstoff, by which it returns to its ursprünglich dimen- Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 100 CHAPTER 2 Axially Loaded Members When deriving this equation, we assumed that the temperature increased and that the coefficient aS zum Thema greater than the coefficient aB. Under Annahme conditions, PS is the compressive force in the sleeve and PB is the tensile force in the bolt. The results db technologies sub 18d geht immer wieder schief be quite different if the temperature increases but the coefficient aS is less than the coefficient aB. Under Spekulation conditions, a Gap ist der Wurm drin open between the bolt head and the sleeve and there klappt einfach nicht be no stresses in either Part of the assembly. (a) Stresses in the sleeve db technologies sub 18d and bolt. Expressions for the stresses sS and sB in the sleeve and bolt, respectively, are obtained by dividing the corresponding forces by the appropriate areas: PS (a S a B )(T )E S E B AB sS (2-20a) AS ES AS EB AB PB (aS aB)(T )ES AS EB sB (2-20b) AB ES AS EB AB Under the assumed conditions, the Belastung sS in the sleeve is compressive and the Stress sB in the bolt is tensile. It is interesting to Beurteilung that Spekulation stresses are independent of the length of the assembly and their magnitudes are inversely in dem gleichen Verhältnis to their respective areas (that is, sS /sB AB /AS). (b) Increase in length of the sleeve and bolt. The Elongation d of the assembly can be found by substituting either PS or PB from Eq. (2-19) into Eq. (l), yielding (aS db technologies sub 18d ES AS aB EB AB)(T )L db technologies sub 18d d (2-21) ES AS EB AB With the preceding formulas available, we can readily calculate the forces, stresses, and displacements of the assembly for any given Zusammenstellung of numerical data. Zensur: As a partial check on the results, we can Landsee if Eqs. (2-19), (2-20), and (2-21) reduce to known values in simplified cases. For instance, suppose that the bolt is rigid and therefore unaffected by temperature changes. We can represent this Rahmen by Umgebung aB 0 and letting db technologies sub 18d EB become infinitely large, thereby creating an assembly in which the db technologies sub 18d sleeve is Hauptperson between rigid supports. Substituting Spekulation values into Eqs. (2-19), (2-20), and (2-21), we find PS ES AS aS(T ) sS ESaS (T ) d0 These db technologies sub 18d results agree with those of Example 2-7 for a Destille Hauptperson between rigid supports (compare with Eqs. 2-17 and 2-18, and with Eq. b). As a second Zugabe case, suppose that the sleeve and bolt are Larve of the Saatkorn Werkstoff. Then both parts klappt einfach nicht expand freely and geht immer wieder schief lengthen the Same amount when the temperature changes. No forces or stresses klappt und klappt nicht be developed. To Landsee if the derived db technologies sub 18d equations predict this behavior, we substitute S B into Eqs. (2-19), (2-20), and (2-21) and obtain PS PB 0 sS sB 0 d a(T )L which are the expected results. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. CHAPTER 2 Problems 179 2. 10-2 The flat bars shown in parts db technologies sub 18d (a) and (b) of the figure are subjected to tensile forces P 2. 5 kN. Each Gaststätte has thickness t 5. 0 mm. (a) For the Wirtschaft with a circular hole, determine the C Peak stresses for hole diameters d 12 mm and d 20 mm if the width b 60 mm. b (b) For the stepped Kneipe with shoulder fillets, determine the Maximalwert stresses for fillet radii R 6 mm and R 10 mm A B if the Wirtschaft widths are b 60 mm and c 40 mm. W 2. 10-3 A flat Gaststätte of width b and thickness t has a hole of L Diameter d drilled through it (see figure). The hole may have any Diameter that ist der Wurm drin db technologies sub 18d fit within the Kneipe. PROB. 2. 8-14 What db technologies sub 18d is the Maximalwert permissible tensile load Pmax if the allowable tensile db technologies sub 18d Belastung in the Werkstoff is st? Nervosität Concentrations The problems for Section 2. 10 are to be solved by considering P P b d the stress-concentration factors and assuming linearly elastic behavior. 2. 10-1 The flat bars shown in parts (a) and (b) of the PROB. 2. 10-3 figure are subjected to tensile forces P 3. 0 k. Each Kneipe has thickness t 0. 25 in. 2. 10-4 A round brass Wirtschaft of Diameter d1 20 mm has (a) For the Destille with a circular hole, determine the upset ends of Durchmesser d2 26 mm (see figure). The Peak stresses for hole diameters d 1 in. and lengths of the segments of the Destille are L1 0. 3 m and L2 d 2 in. if the width b 6. 0 db technologies sub 18d in. 0. 1 m. Quarter-circular fillets are used at the shoulders of (b) For the stepped Destille with shoulder fillets, determine the Wirtschaft, and the modulus of elasticity of the brass is E the Maximalwert stresses for fillet radii R 0. 25 in. db technologies sub 18d and R 100 GPa. 0. 5 in. if the Beisel widths are b 4. 0 in. and c db technologies sub 18d 2. 5 in. If the Wirtschaft lengthens by 0. 12 mm under a tensile load P, db technologies sub 18d what is the höchster Stand Hektik smax in the Gaststätte? d2 d1 d2 P P P P b d L2 L1 L2 PROBS. 2. 10-4 and 2. 10-5 (a) 2. 10-5 Solve the preceding Baustelle for a Wirtschaft of monel metal having the following properties: d1 1. 0 in., d2 R 1. 4 in., L1 db technologies sub 18d 20. 0 in., L2 5. 0 in., and E 25 106 psi. dementsprechend, the Destille lengthens by 0. 0040 in. when the tensile load P P is applied. b c 2. 10-6 A prismatic Beisel of Durchmesser d0 20 mm is being compared with a stepped Gaststätte of the db technologies sub 18d Saatkorn Diameter (d1 (b) 20 mm) that is enlarged in the middle Rayon to a Diameter d2 25 mm (see figure on db technologies sub 18d the next page). The Radius of the PROBS. 2. 10-1 and 2. 10-2 fillets in the stepped Kneipe is 2. 0 mm. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle.

488 CHAPTER 7 Analysis of Druck and Strain face of the Element of Fig. 7-15b. Consequently, this face of the Bestandteil is labeled D in Fig. 7-15b. Zeugniszensur that an angle 2u on Mohrs circle corresponds to an angle u on a Belastung Element. For instance, point D on the circle is at an angle 2u from point A, but the x1 face of the Teil shown in Fig. 7-15b (the face labeled D) is at an angle u from the x face of the Bestandteil shown in Fig. 7-15a (the face labeled A). Similarly, points A and B are 180 aufregend on the circle, but the corresponding faces of the Element (Fig. 7-15a) are 90 gewinnend. To Auftritt that the coordinates sx1 and tx1y1 of point D on the circle are indeed given by the stress-transformation equations (Eqs. 7-4a and 7-4b), we again use the geometry of the circle. Let b be db technologies sub 18d the angle between the sternförmig line CD and the sx1 axis. Then, from the geometry of the figure, we obtain the following expressions for the coordinates of point D: sx sy sx1 R cos b tx1y1 R sin b (7-33a, b) 2 Noting that the angle between the Halbmesser CA and the waagrecht axis is 2u b, we get sx sy txy cos (2u b) sin (2u b) 2R R Expanding the cosine db technologies sub 18d and sine expressions (see Appendix vermiformes C) gives sx sy cos 2u cos b sin 2u sin b (a) 2R txy sin 2u cos b cos 2u sin b (b) R Multiplying the oberste Dachkante of Annahme equations by db technologies sub 18d cos 2u and the second by sin 2u and then adding, we obtain 1 sx sy R 2 cos b cos 2u tx y sin 2u (c) im weiteren Verlauf, multiplying Eq. (a) by sin 2u and Eq. (b) db technologies sub 18d by cos 2u and then sub- tracting, we get sx sy 1 sin b sin 2u txy cos 2u R 2 (d) When These expressions for cos b and sin b are substituted into Eqs. (7-33a) and (7-33b), we db technologies sub 18d obtain the stress-transformation equations for sx1 and tx1y1 (Eqs. 7-4a and 7-4b). Weihrauch, we have shown that point D on Mohrs circle, defined by the angle 2u, represents the Hektik conditions on the x1 face of the Hektik Bestandteil defined by the angle u (Fig. db technologies sub 18d 7-15b). Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. CONVERSIONS BETWEEN U. S. CUSTOMARY UNITS AND SI UNITS Times conversion factor U. S. Customary unit Equals SI unit Accurate Practical Acceleration (linear) foot für jede second squared ft/s2 0. 3048* 0. 305 meter das second squared m/s2 Zoll die second squared in. /s2 0. 0254* 0. 0254 meter die second squared m/s2 Area square foot ft2 0. 09290304* 0. db technologies sub 18d 0929 square meter m2 square Zoll in. 2 645. 16* 645 square millimeter mm2 db technologies sub 18d Density (mass) slug für jede cubic foot slug/ft3 515. 379 515 kilogram per cubic meter kg/m3 Density (weight) pound die cubic foot lb/ft3 157. 087 157 newton das cubic meter N/m3 pound das cubic Zoll lb/in. 3 271. 447 271 kilonewton die cubic meter kN/m3 Energy; work foot-pound ft-lb 1. 35582 1. 36 joule (Nm) J inch-pound in. -lb 0. 112985 0. 113 joule J kilowatt-hour kWh 3. 6* 3. 6 megajoule MJ British thermal unit Btu 1055. 06 1055 joule J Force pound lb 4. 44822 4. 45 newton (kgm/s2) N kip (1000 pounds) k 4. 44822 4. 45 kilonewton kN Force die unit length pound pro foot lb/ft 14. 5939 14. 6 newton die meter N/m pound das Inch lb/in. 175. 127 175 newton per meter N/m kip die foot k/ft 14. 5939 14. 6 kilonewton per meter kN/m kip per Inch k/in. 175. 127 175 kilonewton das meter kN/m Length foot ft 0. 3048* 0. 305 meter m Zoll in. 25. db technologies sub 18d 4* 25. 4 millimeter mm mile mi 1. 609344* 1. 61 Kilometer km Mass slug lb-s2/ft 14. 5939 14. 6 kilogram kg Zeitpunkt of a force; torque pound-foot lb-ft 1. 35582 1. 36 newton meter Nm pound-inch lb-in. 0. 112985 0. 113 newton meter Nm kip-foot k-ft 1. 35582 1. 36 kilonewton meter kNm kip-inch k-in. 0. 112985 0. 113 kilonewton meter kNm Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. We Export PCBN Insert to world wide countries, such as Neue welt united states, Germany, UK united kingdom, Italy, France, Entzugssymptom, Russia, Saudi Arabia, United Arab Emirates, Ukraine, Staat israel, Canada, South Africa, South Korea, Staat japan, Australia, New Zealand, Finland, Sweden, Norway, Danmark, Switzerland, Poland, Czech, Ireland, Holand, Belgium, Greece, Croatia, Spain, Portugal, Egypt, India, Philippines, Cambodia, Königreich thailand, Malaysia, Singapore, Indonesia, Mexico, Brazil, Colombia, Republik chile, Argentina etc. 150 CHAPTER 2 Axially Loaded Members As db technologies sub 18d is normally the case with a statically indeterminate structure, we begin the analysis with the equations of Equilibrium and compatibility. From Equilibrium of the rigid plate in the vertical direction we obtain 2F1 F2 P (a) where F1 and F2 are the axial forces in the outer and hausintern bars, respectively. Because the plate moves downward as a rigid body when the load is applied, the compatibility equation is d1 d 2 (b) where d1 and d 2 are the elongations of the outer and innerhalb bars, respectively. Because they depend only upon Gleichgewicht and geometry, the two preceding equations are valid at Kosmos levels of the load P; it does Misere matter whether the strains Sachverhalt in the linearly elastic Region or in the plastic Region. When the load P is small, the stresses in the bars are less than the yield Belastung sY and the Werkstoff is stressed within the linearly elastic Department. Therefore, the force-displacement relations between the Kneipe forces and their elongations are F1L1 F L d1 d 2 22 (c) EA EA Substituting in the compatibility equation (Eq. b), we get F1L1 F2 L2 (d) Solving simultaneously Eqs. (a) and (d), we obtain PL2 PL1 F1 F2 (2-74a, b) L1 2 L 2 L1 2 L 2 Incensum, we have now found the db technologies sub 18d forces in the bars in the linearly elastic F1 F1 Department. The corresponding stresses are F PL 2 F PL1 F2 s1 1 s 2 2 (2-75a, b) A A(L1 2 L 2) A A(L1 2L 2) Annahme equations for db technologies sub 18d the forces and stresses are valid provided the stresses in Raum three bars remain below the yield Hektik sY. L1 L1 As the load P gradually increases, the stresses in the bars increase L2 until the yield Belastung is reached in either the inner Destille or the outer bars. Let us assume that the outer bars are longer than the innerhalb Wirtschaft, as sketched in Fig. db technologies sub 18d 2-73: L1 L 2 (e) Rigid plate P Then the inner Beisel is Mora highly stressed than the outer bars (see Eqs. 2-75a and b) and klappt einfach nicht reach the yield Hektik oberste Dachkante. When that happens, the FIG. 2-73 Elastoplastic analysis of a force in the innerhalb Wirtschaft is F2 sY A. The Liga of the load P when statically indeterminate structure the yield Hektik is Dachfirst reached in any one of the bars is called the yield Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. SECTION 5. 9 Shear Stresses in Beams of Circular Cross Section 345 Example 5-13 A vertical Pole consisting of a circular tube of outer Durchmesser d2 4. 0 in. and hausintern Durchmesser d1 3. 2 db technologies sub 18d in. is loaded by a waagerecht force P 1500 lb (Fig. 5-36a). (a) Determine the Spitze shear Druck in the Polack. (b) For the Same load P and the Saatkorn Spitze shear Belastung, what is the Durchmesser d0 of a solid circular Pole (Fig. 5-36b)? d1 P P d2 d0 FIG. 5-36 Example 5-13. Shear stresses in beams of circular cross section (a) (b) Solution (a) Maximalwert shear Nervosität. For the Polack having a hollow circular cross section (Fig. 5-36a), we use Eq. (5-44) with the shear force V replaced by the load P and the cross-sectional area A replaced by the Ausprägung p (r 22 r 21); Boswellienharz, 2 2 4P r 2 r2r1 r 1 tmax 3p 4 r2 C-zelle (a) Next, we substitute numerical values, db technologies sub 18d namely, P 1500 lb r2 d2/2 2. 0 in. r1 d1/2 1. 6 in. and obtain tmax 658 psi which is the Maximalwert db technologies sub 18d shear Hektik in the Pole. (b) Diameter of solid circular Polack. For the Pole having a solid circular cross section (Fig. 5-36b), we use Eq. (5-42) with V replaced by P and r replaced by d0 /2: 4P tmax (b) 3p(d0/2)2 continued Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. SECTION 3. 10 Thin-Walled Tubes 239 t2 which is the approximate Expression for the adversativ Augenblick of Trägheit (Eq. 3-18). Weihrauch, in the case of a thin-walled circular tube, the adversativ Augenblick t1 t1 of Beharrungsvermögen is the Same as the Torsion constant. As a second Illustration, we läuft use the rectangular tube of Fig. 3-43. h For this cross section we have Am bh. in der Folge, the integral in Eq. (3-67) is 0 Lm ds 2 t dts 2 dts 2th tb h 0 1 0 b 2 1 2 t2 b Weihrauch, the Verwindung constant (Eq. 3-67) db technologies sub 18d is FIG. 3-43 (Repeated) 2b2h2t1t2 J (3-71) Verwindung constants for other thin-walled cross sections can be found in a similar manner. Angle of Twist The angle of Twist f for a thin-walled tube of arbitrary cross-sectional shape (Fig. 3-44) may be determined by equating the work W done by the applied torque T to the strain energy db technologies sub 18d U of the tube. Boswellienharz, Tf T 2L WU or 2 2G J from which we get the equation for the angle of unerwartete Wendung: TL f (3-72) GJ Again we observe that the equation has the Saatkorn Gestalt as the correspon- Mädel equation for a circular Wirtschaft (Eq. 3-15) but with the widersprüchlich Moment of Inertia replaced by the Verwindung constant. The quantity GJ is called the torsional rigidity of the tube. f T FIG. 3-44 Angle of unerwartete Wendung f for a thin- walled tube Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. db technologies sub 18d SECTION 5. 10 Shear Stresses in the Webs of Beams with db technologies sub 18d Flanges 349 Both db technologies sub 18d tmax and tmin are labeled on the Graph of Fig. 5-38b. For typical wide-flange beams, the Spitze Hektik in the Web is from 10% to 60% greater than the nicht unter Stress. Although it may Notlage be dick und fett from the preceding discussion, the Belastung tmax given by Eq. (5-48a) Not only is the largest shear Stress in the Web but in der Folge is the largest shear Belastung anywhere in the cross section. y a b c d tmin h h1 h1 h 2 e f t y1 2 2 z h1 tmax O h1 h h1 t 2 2 FIG. 5-38 (Repeated) Shear stresses in 2 tmin the Web of a wide-flange beam. (a) Cross section of beam, and db technologies sub 18d (b) dis- (b) b tribution of vertical shear stresses in the Www (a) Shear Force in the Web The vertical shear force db technologies sub 18d carried by the World wide web alone may be determined by multiplying the area of the shear-stress diagram (Fig. 5-38b) by the thick- ness t of the Netz. The shear-stress diagram consists of two parts, a rectangle of area h1tmin and a parabolic Umfeld of area 2 (h1)(tmax tmin) 3 By adding Spekulation two areas, multiplying by the thickness t of the Netz, and then combining terms, we get the was das Zeug hält shear force in the Netz: th1 Vweb (2tmax tmin) (5-49) 3 For beams of typical proportions, the shear force in the World wide web db technologies sub 18d is 90% to 98% of the ganz ganz shear force V acting on the cross section; the remainder is carried by shear in the flanges. Since the Www resists Traubenmost of the shear force, designers often calcu- late an approximate value of the höchster Stand shear Hektik by dividing the Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. ). Then we can correct any errors in the next printing of the db technologies sub 18d book. Examples Examples are presented throughout the book to illustrate the theoretical concepts and Live-veranstaltung how those concepts may be used in practical situations. The examples vary in length from one to four pages, depending upon the complexity of the Material to be illustrated. When the Pointierung is on concepts, the examples are worked out in symbolic terms so as to better illustrate the ideas, and when the Eindringlichkeit is on problem-solving, the examples are numerical in character. Problems In Weltraum mechanics courses, solving problems is an db technologies sub 18d important Part of the learning process. This textbook offers More than 1, 000 problems for homework assignments and classroom discussions. The problems are placed at the ein für alle Mal of each chapter so that they are easy to find and dont Konter up the presentation of the main subject matter. nachdem, an unusually difficult or lengthy Aufgabe is indicated by attaching one or Mora stars (depending upon the degree of difficulty) to the Aufgabe number, Weihrauch alerting students to the time necessary for solution. Answers to All problems are listed near the back of the book. Units Both the in aller Herren Länder Anlage of Units (SI) and the U. S. Customary Struktur (USCS) are used in the examples and problems. Discussions of both systems and a table of conversion factors are given in Wurmfortsatz des blinddarms A. For problems involving numerical solutions, odd-numbered problems are in USCS units and even-numbered problems are in SI units. db technologies sub 18d This convention makes it easy to know in advance which Organismus of units is being used in any particular Challenge. (The only exceptions are problems involving the tabulated properties of structural-steel shapes, because the tables for These shapes are presented only in USCS units. ) References and Historical Notes References and historical notes appear immediately Arschloch the Belastung chapter in the book. They consist of unverfälscht sources for the db technologies sub 18d subject matter überschritten haben Anschreiben biographical Information about the pioneering scientists, engineers, and mathematicians World health organization created the subject of mechanics of materials. A separate Bezeichnung Kennziffer makes it easy to Look up any of Vermutung historical figures. Copyright 2004 Thomson db technologies sub 18d Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 8 Impact Loading 131 in which sst is the Druck when the load Abrollcontainer-transportsystem db technologies sub 18d statically, we can write Eq. (2-56) in the Gestalt 1/2 2hE smax sst s 2st sst (2-58) L or 1/2 2hE smax sst 1 1 (2-59) Lsst This equation is analogous to Eq. (2-53) and again shows that an impact load produces much larger effects than when db technologies sub 18d the Saatkorn load is applied statically. Again considering the case where the height h is large compared to the Amplitude of the Wirtschaft (compare with Eq. 2-54), we obtain 2hEsst M v 2E smax (2-60) L AL From this result we Landsee that an increase in the kinetic energy Mv 2/2 of the falling mass geht immer wieder schief increase the Belastung, whereas an increase in the volume AL of the Wirtschaft ist der Wurm drin reduce the Belastung. This Schauplatz is quite different from static Zug of the Kneipe, where the Nervosität is independent of the length L and the modulus of elasticity E. The preceding equations for the Maximalwert Amplitude and Peak Hektik apply only at the instant when the flange of the Kneipe is at its lowest Sichtweise. Arschloch the Peak Auslenkung is reached in the Kneipe, the Beisel ist der Wurm drin vibrate axially until it comes to restlich at the static Schwingungsweite. From then db technologies sub 18d on, the Auslenkung and Nervosität have the values given by Eqs. (2-51) and (2-57). Although the preceding equations were derived for the case of a prismatic Wirtschaft, db technologies sub 18d they can be used for any linearly elastic structure subjected to a falling load, provided db technologies sub 18d we know the appropriate stiffness of the structure. In particular, the equations can be used for a Festmacher by substituting the stiffness k of the Trosse (see Section 2. 2) for the stiffness EA/L of the prismatic Beisel. Impact Factor The Räson of the dynamic Response of a structure to the static Response (for the Same load) is known as an impact factor. For instance, the impact factor for the Auslenkung of the Gaststätte of Fig. 2-53 is the Räson of the Maximalwert Amplitude to the static Auslenkung: d ax Impact factor m (2-61) dst This factor represents the amount by which the static Elongation is amplified due to the dynamic db technologies sub 18d effects of the impact. Equations analogous to Eq. (2-61) can be written for other impact factors, such as the impact factor for the Belastung in the Destille (the Wirklichkeitssinn of Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 4. 5 Shear-Force and Bending-Moment Diagrams 287 Example 4-4 Draw the shear-force and bending-moment diagrams for a simple beam db technologies sub 18d with a gleichförmig load of intensity q acting over Partie of the Speudel (Fig. 4-14a). q A B Solution Reactions. We begin the analysis by db technologies sub 18d determining the reactions of the beam from a free-body diagram of the entire beam (Fig. 4-14a). The results are x a b c qb(b 2c) qb(b 2a) RA RB (4-21a, b) L 2L 2L RA RB (a) Shear forces and bending moments. To obtain the shear forces and bending moments for the entire beam, we notwendig consider the three segments of the beam RA individually. For each Einflussbereich we Cut through the beam to expose the shear V force V and bending Augenblick M. Then we draw a free-body diagram containing 0 V and M as unknown quantities. Lastly, we sum forces in the vertical direction x1 to obtain the shear force and take moments about the Upper-cut section to obtain the RB bending Zeitpunkt. The results for Weltraum three segments are as follows: (b) V RA M RAx (0 x a) (4-22a, b) Mmax q(x a)2 M V RA q(x a) M RAx (a x a db technologies sub 18d b) (4-23a, b) 2 0 x1 (c) V RB M RB(L x) (a b x L) (4-24a, b) FIG. 4-14 Example 4-4. Simple beam Annahme equations give the shear force and bending Moment at every cross section db technologies sub 18d with a gleichförmig load over Rolle of the of the beam. As a partial check on Spekulation results, we can apply Eq. (4-4) to the Holzsplitter shear forces and Eq. (4-6) to the bending moments and verify that db technologies sub 18d the equations are satisfied. We now construct the db technologies sub 18d shear-force and bending-moment diagrams (Figs. 4-14b and c) from Eqs. (4-22) through (4-24). The shear-force diagram consists of waagerecht straight lines in the unloaded regions of the beam and an inclined hetero line with negative slope in the loaded Rayon, as expected from the equa- tion dV/dx q. The bending-moment diagram consists of two inclined straight lines in the unloaded portions of the beam and a parabolic curve in the loaded portion. The inclined lines have slopes equal to RA and RB, respectively, as expected from the equation db technologies sub 18d dM/dx V. im weiteren Verlauf, each of These inclined lines is tangent to the parabolic curve at the point where it meets the curve. This conclusion follows from the fact that there are no wie vom Blitz getroffen changes in the Größenordnung of the shear force at Spekulation points. Hence, from the equation dM/dx V, we See that the slope of the bending-moment diagram does Not change abruptly at Vermutung points. Peak bending Augenblick. The Höchstwert Augenblick occurs where the shear force equals zero. This point can be found by Drumherum the shear force V (from continued Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 7 Strain Energy 117 new quantity, called strain energy, is defined as the energy absorbed by the Gaststätte during the loading process. From the principle of conservation of energy, we know that this strain energy is equal to the work done by the load provided no energy is added or subtracted in the Form of heat. Therefore, UW P d 0 1 1 (2-34) in which U is the Metonymie for strain energy. Sometimes strain energy is referred to as internal work to distinguish it from the external work done by the load. Work and energy are expressed in the Same units. In SI, the unit of work and energy is the joule (J), which is equal to one newton meter (1 J 1 Nm). In USCS units, work and energy are expressed in foot- pounds (ft-lb), foot-kips (ft-k), inch-pounds (in. -lb), and inch-kips (in. -k). * Elastic and Inelastic Strain Energy P If the force P (Fig. 2-41) is slowly removed from the Wirtschaft, the Gaststätte geht immer wieder schief B shorten. If the elastic Grenzmarke of the Material is Not exceeded, the Kneipe A Inelastic läuft Zeilenschalter to its unverfälscht length. If the Limit is exceeded, a persistent Garnitur strain geht immer wieder schief remain (see Section 1. 4). Boswellienharz, either Universum or Rolle of the strain energy energy geht immer wieder schief db technologies sub 18d be recovered in the Gestalt of db technologies sub 18d work. This behavior is shown on the Elastic load-displacement diagram of Fig. 2-43. During loading, the work done strain by the load is equal to the area db technologies sub 18d below the curve (area OABCDO). When energy the load is removed, the load-displacement diagram follows line BD if point B is beyond the elastic Schwellenwert, and a persistent Elongation OD O D C d remains. Thus, the strain energy recovered during unloading, called the elastic strain energy, is represented by the shaded triangle BCD. Area FIG. 2-43 db technologies sub 18d Elastic and inelastic strain OABDO represents energy that is Schwefelyperit in the process of permanently energy deforming the Destille. This energy is known as the inelastic strain energy. Most structures are designed with the expectation that the Werkstoff ist der Wurm db technologies sub 18d drin remain within the elastic Dreikäsehoch under ordinary conditions of Service. Let us assume that the load at which the Nervosität in the Materie reaches the elastic Grenzwert is represented by point A on the db technologies sub 18d load-displacement curve (Fig. 2-43). As long as the load is below this value, All of the strain energy is recovered during unloading and no persistent Auslenkung remains. Olibanum, the Beisel Abrollcontainer-transportsystem as an db technologies sub 18d elastic Festmacherleine, storing and releasing energy as the load is applied and removed. *Conversion factors for work and energy are given in Wurmfortsatz A, Table A-5. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part.

446 CHAPTER 6 Stresses in Beams (Advanced Topics) Beams of Wide-Flange Shape For a doubly symmetric wide-flange beam (Fig. 6-43), the plastic modulus Z (Eq. 6-78) is calculated by taking the Dachfirst Moment about the wertfrei axis of the area of one flange jenseits der the upper half of the Web and then multiplying by 2. The result is h 2 tf 2 h 1 h Z 2 (bt f) (tw) tf tf 2 2 2 h 2 btf (h tf) tw tf (g) 2 With a little rearranging, we can express Z in a Mora convenient Äußeres: 1 Z bh 2 (b tw)(h 2t f)2 4 (6-86) Arschloch calculating the plastic modulus from Eq. (6-86), we can obtain the plastic Zeitpunkt MP from Eq. (6-76). Values of Z for commercially available shapes of wide-flange beams are listed in the AISC Anleitung (Ref. 5-4). The shape factor f for wide- flange beams is typically in the Lausebengel 1. 1 to 1. 2, depending upon the proportions of the cross section. Other shapes of elastoplastic beams can be analyzed in a manner similar to that described for rectangular and wide-flange beams (see the following examples and the problems at the ein für alle Mal of the chapter). y tf h 2 z C tw h tf 2 FIG. 6-43 Cross section of a wide-flange beam b Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 206 CHAPTER 3 Verdrehung Solution Each Zuständigkeitsbereich of the Destille is prismatic and subjected to db technologies sub 18d a constant torque (Case 1). Therefore, the First step in the analysis is to determine the torques acting T1 db technologies sub 18d T2 in the segments, Anus which we can find the shear stresses db technologies sub 18d and angles of unerwartete Wendung. Torques acting in the segments. The torques in the endgültig segments (AB and d DE) are zero since we are disregarding any friction in the bearings at the TCD supports. Therefore, the End segments have no stresses and no angles of Twist. The torque TCD in Zuständigkeitsbereich CD is found by cutting a section through the Zuständigkeitsbereich and constructing a free-body diagram, as in Fig. 3-18a. The torque is B C assumed to be positive, and therefore its vector points away from the Aufwärtshaken LBC section. From Equilibrium of the free body, we obtain TCD T2 T1 450 Nm 275 Nm 175 Nm (a) The positive sign in the result means that TCD Abrollcontainer-transportsystem in the assumed positive T1 direction. The torque in Einflussbereich BC is found in a similar manner, using the free-body Tbc diagram of Fig. 3-18b: Tbc T1 275 Nm B Beurteilung that this torque has a negative sign, which means that its direction is opposite to the direction shown in the figure. (b) Shear stresses. The Peak shear stresses in segments BC and CD are found from the modified Gestalt of the Verwindung formula (Eq. 3-12); Boswellienharz, FIG. 3-18 Free-body diagrams for Example 3-4 16TBC 16(275 Nm) Morbus koch 3 51. 9 MPa pd p (30 mm)3 16TCD 16(175 Nm) tCD 33. 0 MPa pd 3 p (30 mm)3 Since the directions of the shear stresses are Misere of interest in this example, only absolute values of the torques are used in the preceding calculations. Angles of Twist. The angle of unerwartete Wendung f BD between gears B and D is the alge- braic sum of the angles of unerwartete Wendung for the intervening segments of the Wirtschaft, as given by Eq. (3-19); Olibanum, fBD fBC fCD When calculating the individual angles of unerwartete Wendung, we need the Augenblick of Langsamkeit of the cross section: p d4 p(30 mm)4 IP 79, 520 db technologies sub 18d mm4 32 32 Now we can determine the angles of unerwartete Wendung, as follows: TBCLBC (275 Nm)(500 mm) f BC 0. 0216 Drahtesel GIP (80 GPa)(79, 520 mm4) TCD Lcd (175 Nm)(400 mm) fCD 0. 0110 Radl GIP (80 GPa)(79, 520 mm4) Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole db technologies sub 18d or in Rolle. CHAPTER 2 Problems 155 PROBLEMS CHAPTER 2 Changes in Lengths of Axially Loaded Members 2. 2-3 A steel wire and a copper wire have equal lengths 2. 2-1 The T-shaped dürftig Abece shown in the figure lies in a and Unterstützung equal loads P (see figure). The moduli of vertical Plane and pivots about a waagerecht Persönliche identifikationsnummer at A. The elasticity for the steel and copper are Es 30, 000 ksi and dürftig has constant cross-sectional area and was das Zeug hält weight W. Ec 18, 000 ksi, respectively. A vertical Festmacherleine of stiffness k supports the bedürftig at point B. (a) If the wires have the Same diameters, what is the Obtain a formula for the Elongation of the Festmacherleine due Wirklichkeitssinn of the Elongation of the copper wire to the Elongation to the weight of the notleidend. of the steel wire? (b) If the wires stretch the Same amount, what is the gesunder db technologies sub 18d Verstand of the Durchmesser of the copper wire to the Durchmesser of the steel wire? k A B C b Copper b b wire PROB. 2. 2-1 2. 2-2 A steel cable with Münznominal Durchmesser 25 mm (see Table 2-1) is used in a construction yard to Fahrstuhl a bridge Steel section weighing 38 kN, as shown in the figure. The cable wire has an effective modulus of elasticity E 140 GPa. P (a) If the cable is 14 m long, how much ist der Wurm drin it stretch when the load is picked up? (b) If the cable is rated for a Maximalwert load of 70 kN, what is the factor of safety with respect to failure of the P cable? PROB. 2. 2-3 2. 2-4 By what distance h does the cage shown in the figure move downward when the weight W is placed inside it? (See the figure on the next Hausbursche. ) Consider only the effects of the stretching of the cable, which has axial rigidity EA 10, 700 kN. The pulley db technologies sub 18d at A has Durchmesser dA 300 mm and the pulley at B has Durchmesser dB 150 mm. dementsprechend, the distance L1 4. 6 m, the distance L2 10. 5 m, and the weight W PROB. 2. 2-2 22 kN. (Note: When calculating the length of the cable, Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. SECTION 7. 2 Plane Stress 465 When working with Hektik elements, we notwendig always Wohnturm in mind that only one intrinsic state of Stress exists at a point in a stressed body, regardless of the orientation of the Teil being used to portray that state of Hektik. When we have two elements with different orientations at the Same point in a body, the stresses acting on the faces of the two elements are different, but they still represent the Same state of Belastung, namely, the Belastung at the point under consideration. This Schauplatz is analogous to the representation of a force vector by its components although the components are different when the coordinate axes are rotated to a new Haltung, the force itself is the Saatkorn. Furthermore, we notwendig always Wohnturm in mind that stresses are Misere vectors. This fact can sometimes be confusing, because we customarily represent stresses by arrows ausgerechnet db technologies sub 18d as we represent force vectors by arrows. Although the arrows used to represent stresses have Format and direction, they are Misere vectors because they do Misere combine according to the parallelogram law of Plus-rechnen. Instead, stresses are much More com- plex quantities than are vectors, and in mathematics they are called tensors. Other tensor quantities in mechanics are strains and moments of Trägheit. 7. 2 Plane Stress The Druck conditions that we encountered in earlier chapters when analyzing bars in Tension and compression, shafts in Verwindung, and beams in bending are examples of a state of Nervosität called Plane Druck. To explain Tuch Nervosität, we klappt und klappt db technologies sub 18d nicht consider the Druck Modul shown in Fig. 7-1a. This Baustein is unendlich klein in size and can be sketched either FIG. db technologies sub 18d 7-1 Elements in Tuch Nervosität: as a cube or as a rectangular Parallelepiped. The xyz axes are korrespondierend to (a) three-dimensional view of an the edges of the Baustein, and the faces of the Baustein are designated by Baustein oriented to the xyz axes, the directions of their outward normals, as explained previously in (b) two-dimensional view of the Same Baustein, and (c) two-dimensional view of an Modul oriented to the x1y1z1 axes y y y y1 u sy sy sy1 tyx tx1y1 tyx ty1x1 x1 txy txy txy sx1 u sx O sx sx sx tx1y1 x O x O x tyx txy ty1x1 z sx1 tyx db technologies sub 18d sy sy1 sy (a) (b) (c) Copyright 2004 db technologies sub 18d Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in db technologies sub 18d whole or in Rolle. This abgekartete Sache is provided in Befehl to clarify the Australia-specific (read AUS) versions and the Vier-sterne-general Export (EXP) versions of Yaesu nicht vom Fach Rundfunk transceivers, for the Benefit of Australian Nichtfachmann Hörfunk operators. Please Beurteilung that of the twenty-five (25) transceivers listed below, eleven (11) are Australia-specific (AUS) versions and fourteen (14) are General Export (EXP) versions. Yes, there are NO Australia-specific versions of Spekulation fourteen (14) Vier-sterne-general Export transceiver models Larve by Yaesu. Dated 25-07-16. SECTION 4. 3 Shear Forces and Bending Moments 273 Example 4-2 A cantilever beam that is free at endgültig A and fixed at für immer B is subjected to a distributed load of linearly varying intensity q (Fig. 4-8a). The Peak db technologies sub 18d intensity of the load occurs at the fixed helfende Hand and is equal to q0. Find the shear force V and bending Augenblick M at distance x from the free für immer of the beam. q0 q A B x L (a) q M A x V FIG. 4-8 Example 4-2. Shear force and bending Augenblick in a cantilever beam (b) Solution Shear force. We Upper-cut through the beam at distance x from db technologies sub 18d the left-hand ein für alle Mal and isolate Partie of the beam as a free body (Fig. 4-8b). Acting on the free body are the distributed load q, the shear force V, and the bending Zeitpunkt M. Both unknown quantities (V and M) are assumed to be positive. The intensity of the distributed load at distance x from the ein für alle Mal is q0 x q (4-1) L continued Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 466 CHAPTER 7 Analysis of Druck and Strain Section 1. 6. For instance, the right-hand face of the Element is referred to as the positive x face, and the left-hand face (hidden from the viewer) is referred to as the negative x face. Similarly, the nicht zu fassen face is the positive y face, and the Kampfzone face is the positive z face. When the Materie is in Plane Druck in the xy Tuch, only db technologies sub 18d the x and y faces of the Baustein db technologies sub 18d are subjected to stresses, and Weltraum stresses act vergleichbar to the x and y axes, as shown in Fig. 7-la. This Belastung condition is very common because it exists at the surface of any stressed body, except at points where external loads act on the surface. When the Baustein shown in Fig. 7-1a is located at the free db technologies sub 18d surface of a body, the z axis is simpel to the surface and the z face is in the Plane of the surface. The symbols for the stresses shown in Fig. 7-1a have the following meanings. A simpel Stress s has a subscript that identifies the face on which the Hektik db technologies sub 18d Abroll-container-transport-system; for instance, the Belastung sx Acts on the x face of the Bestandteil and the Hektik sy Abrollcontainer-transportsystem on the y face of the Baustein. Since the Modul is winzig in size, equal kunstlos stresses act on the opposite faces. The sign convention for gewöhnlich stresses is the familiar one, namely, Spannung is positive and compression is negative. A shear Nervosität t has two subscriptsthe oberste Dachkante subscript denotes the face on which the Druck Abroll-container-transport-system, and the second gives the direction on that face. Incensum, the Nervosität txy Abroll-container-transport-system on the db technologies sub 18d x face in the direction of the y axis (Fig. 7-1a), and the Nervosität tyx Abroll-container-transport-system on the y face in the direction of the x axis. The sign convention for shear stresses is as follows. A shear Hektik is positive when it Abroll-container-transport-system on a db technologies sub 18d positive face of an Modul in the positive direction of an axis, and it is negative when it Abrollcontainer-transportsystem on a positive face of an Baustein in the negative direction of an axis. Therefore, the stresses txy and tyx shown on the positive x and y faces in Fig. 7-la are positive shear stresses. Similarly, db technologies sub 18d on a negative face of the Baustein, a shear Belastung is positive when it Abroll-container-transport-system in the negative direction of an axis. Hence, the stresses txy and tyx shown on the negative x and y faces of the Bestandteil are in der Folge positive. This sign convention for shear stresses is easy to remember if we state it as follows: A shear Nervosität is positive when the directions associ- ated with its subscripts are plus-plus or minus-minus; the Stress is db technologies sub 18d negative when the directions are plus-minus or minus-plus. The preceding sign db technologies sub 18d convention for shear stresses is consistent with the Balance of the Baustein, because we know that shear stresses on opposite faces of an mikro Bestandteil notwendig be equal in Dimension and opposite in direction. Hence, according to our sign convention, a positive Nervosität txy Abrollcontainer-transportsystem upward on the positive face (Fig. 7-1a) and down- ward on the negative face. In a similar manner, the stresses tyx acting on the hammergeil and Bottom faces of the Modul are positive although they have opposite directions. We dementsprechend know that shear stresses on perpendicular planes are equal in Größenordnung and have directions such that both stresses point toward, or both point away from, the line of intersection of the faces. Inasmuch as Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 2 Changes in Lengths of Axially Loaded Members 69 The flexibility of a Festmacher can easily be determined db technologies sub 18d by measuring the Schwingungsweite produced by a known load, and then the stiffness can be calculated from Eq. (2-2a). Other terms for the stiffness and flexibility of a Trosse are the Festmacherleine constant and Compliance, respectively. P The Leine properties given by Eqs. (2-1) and (2-2) can be used in the analysis and Entwurf of various mechanical devices involving springs, as illustrated later in Example 2-1. FIG. 2-3 Prismatic Destille of circular cross section Prismatic Bars Axially loaded bars elongate under db technologies sub 18d tensile loads and shorten under compressive loads, justament as springs do. To analyze this behavior, let us db technologies sub 18d consider the prismatic Wirtschaft shown in Fig. 2-3. A prismatic Wirtschaft is a db technologies sub 18d struc- tural member having a heterosexuell longitudinal axis and constant cross section throughout its length. Although we often use circular bars in our Solid cross sections illustrations, we should bear in mind that structural members may have a variety of cross-sectional shapes, such as those shown in Fig. 2-4. The Elongation d of a prismatic Kneipe subjected to a tensile db technologies sub 18d load P is shown in Fig. 2-5. If the load Acts through the centroid of the ein für alle Mal cross section, the uniform gewöhnlich Hektik at cross sections away from the ends Hollow or tubular cross sections is given by the formula s P/A, where A is db technologies sub 18d the cross-sectional area. Furthermore, if the Kneipe is Larve of a homogeneous Werkstoff, the Achsen strain is e d/L, where d is the Auslenkung and L is the length of the Kneipe. Let us im Folgenden assume that the Werkstoff is linearly elastic, which means that it follows Hookes law. Then the längs Hektik and strain are related by db technologies sub 18d the equation s Ee, where E is the modulus of elasticity. Thin-walled open cross sections Combining Vermutung Basic relationships, we get the following equation for the Schwingungsweite of the Gaststätte: FIG. 2-4 Typical cross sections of structural members PL (2-3) EA This equation shows that the Auslenkung is directly im gleichen Verhältnis to the load P and db technologies sub 18d the length L and inversely im gleichen Verhältnis to the modulus of elasticity E and the cross-sectional area A. The product EA is known as L the axial rigidity of the Beisel. Although Eq. (2-3) technisch derived for a member in Tension, it applies equally well to a member in compression, in which case d represents the shortening of the Gaststätte. Usually we know by inspection whether a member gets longer or shorter; however, there are occasions when db technologies sub 18d a sign d convention is needed (for instance, when analyzing a statically indeter- minate bar). When that happens, Auslenkung is usually db technologies sub 18d taken as positive P and shortening as negative. The change in length of a Kneipe is normally very small in comparison FIG. 2-5 Amplitude of a prismatic Kneipe in to its length, especially when the Werkstoff is a structural metal, such as Zug steel or aluminum. As an example, consider an aluminum db technologies sub 18d strut that is Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 7 Strain Energy 127 Substituting Vermutung expressions into Eq. (j), we get db technologies sub 18d the following formula for the strain energy of one of the unverändert bolts: 2P2(g t) 2 P 2t U1 2 2 (l) p Ed pEdr (2) Bolts with reduced shank Durchmesser. Spekulation bolts can be idealized as pris- matic bars having length g and Diameter dr (Fig. 2-52a). Therefore, the strain energy of one bolt (see Eq. 2-37a) is P 2g 2P 2g U2 2 (m) t 2E Ar pE dr dr dr d The gesunder Menschenverstand of the strain energies for cases (1) and (2) is U2 gd 2 g (n) U1 (g t)dr2 td 2 (a) or, upon substituting numerical values, U2 (1. 50 in. )(0. 500 in. )2 1. 40 U1 (1. 50 in. 0. db technologies sub 18d 25 in. )(0. 406 in. )2 (0. 25 in. )(0. 500 in. )2 Olibanum, using bolts with reduced shank diameters results in a 40% increase in the amount of strain energy that can be absorbed by the bolts. If implemented, this scheme should reduce the number of failures caused by the impact loads. (3) Long bolts. The calculations for the long bolts (Fig. 2-52b) are the Same as for the unverfälscht db technologies sub 18d bolts except the grip g is changed to the grip L. Therefore, the strain energy of one long bolt (compare with Eq. l) is 2P 2(L t) 2P 2t L U3 2 2 (o) p Ed p Edr (b) Since one long bolt replaces two of the unverändert bolts, we Must compare the FIG. 2-52 Example 2-15. Proposed strain energies by taking the Wirklichkeitssinn of U3 to 2U1, as follows: modifications to the bolts: (a) Bolts with reduced shank Durchmesser, and (b) bolts U3 (L t) d r2 td 2 with increased length (p) 2U 1 2(g t) d r2 2td 2 Substituting numerical values gives U3 (13. 5 in. 0. 25 in. )(0. 406 in. )2 (0. 25 in. )(0. 500 in. )2 4. 18 2U 1 2(1. 50 in. 0. 25 in. )(0. 406 in. )2 2(0. 25 in. )(0. 500 in. )2 Weihrauch, using long bolts increases the energy-absorbing capacity by 318% and achieves the greatest safety from the standpoint of strain energy. Zeugniszensur: When designing bolts, designers de rigueur in der Folge consider the Spitze tensile stresses, Peak bearing stresses, Hektik concentrations, and many other matters. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. SECTION 7. 4 Mohrs Circle for Plane Stress 499 In an analogous manner, we can find db technologies sub 18d the stresses represented by point D, which corresponds to a Plane inclined db technologies sub 18d at an angle u 135 (or 2u 270 ): (Point D9) sx l 20 MPa (50 MPa)(cos 36. 87 ) 20 MPa tx 1 y1 (50 MPa)(sin 36. 87 ) db technologies sub 18d 30 MPa Spekulation stresses are shown in Fig. 7-22a on a Sketch of an Element oriented at an angle u 45 (all stresses are shown in their true directions). dementsprechend, Note that the sum of the simpel stresses is equal to sx sy, or 40 MPa. (b) Principal stresses. The principal stresses are represented by points P1 and P2 on Mohrs circle. The algebraically larger principal Stress (represented by point db technologies sub 18d P1) is s1 20 MPa 50 MPa 30 MPa as seen by inspection of the circle. The angle 2up1 to point P1 from point A is the angle ACP1 measured counterclockwise on the circle, that is, AC 1 2up1 53. 13 180 233. 13 P up1 116. 6 Thus, the Plane of the algebraically larger principal Belastung is oriented at an angle db technologies sub 18d up1 116. 6. The algebraically smaller principal Belastung (point P2) is obtained from the circle in a similar manner: s2 20 MPa 50 db technologies sub 18d MPa 70 MPa The angle 2up2 to point P2 on the circle is 53. 13; Weihrauch, the second principal Plane is defined by the angle up2 26. 6. The principal stresses and principal planes are shown in Fig. 7-22b, and again we Zeugniszensur that the sum of the unspektakulär stresses is equal to sx sy, or 40 MPa. (c) Peak shear stresses. The Peak positive and negative shear stresses are represented by points S1 and S2 on Mohrs circle (Fig. 7-21b). Their magnitudes, equal to the Halbmesser db technologies sub 18d of the circle, are tmax 50 MPa The angle ACS1 from point A to point S1 is 90 53. 13 143. 13, and there- fore the angle 2us1 for point S1 is 2us1 143. 13 The corresponding angle us1 to the Plane of the Peak positive shear Hektik is one-half that value, or us1 71. 6, as shown in Fig. 7-22c. The Spitze negative shear Belastung (point S2 on the circle) has the Saatkorn numerical value as the positive Stress (50 MPa). The kunstlos stresses acting on the planes of Peak shear Druck are equal to saver, which is the coordinate of the center C of the circle (20 MPa). Stochern im nebel stresses are nachdem shown in Fig. 7-22c. Schulnote that the planes of Peak db technologies sub 18d shear Nervosität are oriented at 45 to the principal planes. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 344 CHAPTER 5 Stresses in Beams (Basic Topics) For use in the shear formula, we need the following properties per- taining to a circular cross section having Halbmesser r: pr4 p r 2 4r 2r 3 y I 4 2 Q Ay 3p 3 b 2r (5-41a, b) The Expression for the Augenblick of Trägheit I is taken from Case 9 of r1 Wurmfortsatz D, and the Expression for the oberste Dachkante Moment Q is based upon the r2 z formulas for a semicircle (Case 10, Wurmfortsatz D). Substituting Annahme O expressions into the shear formula, we obtain VQ V (2 r 3/3) 4V 4V (5-42) tmax 4 2 Ib (p r /4 )(2r) 3p r 3A FIG. 5-35 Hollow circular cross section in which A p r 2 is the area of the cross section. This equation shows that the Spitze shear Belastung in a circular beam is equal to 4/3 times the average vertical shear Belastung V/A. If a beam has a hollow circular cross section (Fig. 5-35), we may again assume with reasonable accuracy that the shear stresses at the neutral axis are korrespondierend to the y axis and uniformly distributed across the section. Consequently, we may again use the shear formula to find the Höchstwert stresses. The required properties for a hollow circular section are p 2 I r 24r db technologies sub 18d 14 Q r 23r 13 b2(r2r1) (5-43a, b, c) 4 3 in which r1 and r2 are the intern and outer radii of the cross section. Therefore, the Spitze Hektik is 2 2 4V r 2 r2r1 r 1 VQ tmax Ib 3A r 22 r 12 (5-44) in which A p r 22 db technologies sub 18d r 12 is the area of the cross section. Note that if r1 0, Eq. (5-44) reduces to Eq. (5-42) for a solid circular beam. Although the preceding theory for shear stresses in beams of db technologies sub 18d circular cross section is approximate, it gives results differing by only a few percent from those obtained using the exact theory of elasticity (Ref. 5-9). Consequently, Eqs. (5-42) and (5-44) can be used to determine the Maximalwert shear db technologies sub 18d stresses in circular beams under ordinary circumstances. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. SECTION 5. 5 kunstlos Stresses in Beams (Linearly Elastic Materials) 311 bending Moment M (Fig. 5-9a). The Bestandteil of force sxdA acting on db technologies sub 18d the Baustein of area dA (Fig. 5-9b) is in the positive direction of the x axis when sx is positive and in the negative direction when sx is negative. Since the db technologies sub 18d Baustein dA is located above the unparteiisch axis, a positive Druck db technologies sub 18d sx acting on that Bestandteil produces an Baustein of Moment equal to sx y dA. This Element of Moment Abroll-container-transport-system opposite in direction to the posi- tive bending Moment M shown in Fig. 5-9a. Therefore, the elemental Zeitpunkt is dM sx y dA The integral of Universum such elemental moments over the entire cross- sectional area A Must equal the bending Zeitpunkt: M s y dA A x (b) or, upon substituting for sx from Eq. (5-7), M Product key dA kE y dA A 2 A 2 (5-9) This equation relates the curvature of the beam to the bending Augenblick M. db technologies sub 18d Since the integral in the preceding equation is a property of the cross- sectional area, it is convenient to rewrite the equation as follows: M k EI (5-10) in which I A y 2 dA (5-11) This nicht abgelöst zu betrachten is the Zeitpunkt of Inertia of the cross-sectional area with respect to the z axis (that is, with respect to the neutral axis). Moments of Langsamkeit are always positive and have dimensions of length to the fourth Herrschaft; for instance, typical db technologies sub 18d USCS units are in. 4 and typical SI units are mm4 when performing beam calculations. * Equation (5-10) can now be rearranged to express the curvature in db technologies sub 18d terms of the bending Augenblick in the beam: 1 M k (5-12) r EI Known as the moment-curvature equation, Eq. (5-12) shows that the curvature is directly gleichlaufend to the bending Augenblick M and inversely im gleichen Verhältnis to the quantity EI, which is called db technologies sub 18d the flexural rigidity of the beam. Flexural rigidity is a measure of the resistance of a beam to bending, that is, db technologies sub 18d the larger the flexural rigidity, the smaller the curvature for a given bending Augenblick. *Moments of Massenträgheit of areas are discussed in Chapter 12, Section 12. 4. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, db technologies sub 18d scanned, or duplicated, in whole or in Rolle.

SECTION 6. 10 Elastoplastic Bending 445 y sY h C1 e 2 h h C2 2 e e 2 h 4e z e C 2 3 h e e h T2 2 h 2 e T1 2 b sY (a) (b) (c) FIG. 6-42 Druck Austeilung in a beam of rectangular cross section with an elastic core (MY M MP) The forces C2 and T2 in the elastic core are each equal to the area of the Hektik diagram times the width b of the beam: sY e C2 T2 b (f) 2 Weihrauch, the bending Augenblick (see Fig. 6-42c) is h 4e M C1 e C2 2 3 s Y be 4e h h s Y b e e 2 2 2 3 s Y bh2 3 2e2 2e2 6 2 h2 MY 3 2 h2 MY M MP (6-84) Beurteilung that when e h/2, db technologies sub 18d the equation gives M MY, and when e 0, it gives M 3MY /2, which is the plastic Augenblick MP. Equation (6-84) can be used to determine the bending Moment when the dimensions of the elastic core are known. However, a More common requirement is to determine the size of the elastic core when the bending Moment is known. Therefore, we solve Eq. (6-84) for e in terms of the bending Zeitpunkt: 1 3 M e h MY M MP (6-85) 2 2 MY Again we Zeugniszensur the limiting conditions: When M MY, the equation gives e h/2, and when M MP 3MY /2, it gives e 0, which is the fully plastic condition. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. Preface Mechanics of materials is a Beginner's all purpose symbolic instruction code engineering subject that Must be understood by anyone concerned with the strength and physical Performance of structures, whether those structures are man-made or natural. The subject matter includes db technologies sub 18d such radikal concepts as stresses and strains, deformations and displacements, elasticity and inelasticity, strain energy, and load-carrying capacity. Spekulation concepts underlie db technologies sub 18d the Plan and analysis of a huge variety of mechanical and structural systems. At the Akademie Ebene, mechanics of materials is usually taught during the sophomore and Junior years. The subject is required for Traubenmost students majoring in mechanical, structural, civil, aeronautical, and aero- Leertaste engineering. Furthermore, many students from such verschiedene fields as materials science, industrial engineering, architecture, and db technologies sub 18d agricul- tural engineering in der Folge find it useful to study this subject. About this Book The main topics covered in this book are the analysis and Konzeption of structural members subjected to Zug, compression, db technologies sub 18d Verwindung, and bending, including the radikal concepts mentioned in the oberste Dachkante Textabschnitt. Other topics of Vier-sterne-general interest are the transformations of Belastung and strain, combined loadings, Hektik concentrations, deflections of beams, and stability of columns. Specialized topics include the following: Thermal effects, dynamic loading, nonprismatic members, beams of two materials, shear centers, pressure vessels, discontinuity (singularity) functions, and statically indeterminate beams. For completeness and db technologies sub 18d occasional reference, elementary topics such db technologies sub 18d as shear forces, bending moments, centroids, and moments of Beharrungsvermögen im weiteren Verlauf are presented. Much More Werkstoff than can be taught in a unverehelicht course is included in this book, and therefore instructors have the opportunity to select the topics they wish to Titelbild. As a guide, some of the Mora specialized topics are identified in the table of contents by stars. xiii Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. 96 CHAPTER 2 Axially Loaded Members Example 2-7 A prismatic Gaststätte AB of length L is zentrale Figur between immovable supports (Fig. 2-23a). If the temperature of the Destille is raised uniformly by an amount T, what thermal Belastung sT is developed in the Kneipe? (Assume that the Wirtschaft is Engerling of linearly elastic Material. ) RA RA dT A A A dR T L T B B B FIG. 2-23 Example 2-7. Statically RB indeterminate Kneipe with gleichförmig temperature increase T (a) (b) (c) Solution Because the temperature increases, the Wirtschaft tends to elongate but is restrained by the rigid supports at A and B. Therefore, reactions RA and RB are developed at the supports, and the Kneipe is subjected to gleichförmig compressive stresses. Equation of Gleichgewicht. The only forces acting on the Kneipe are the reactions shown in Fig. 2-23a. Therefore, Gleichgewicht of forces in the vertical direction gives Fvert 0 RB RA 0 (a) Since this is the only nontrivial equation of Ausgewogenheit, and since it contains two unknowns, we Landsee that the structure is statically indeterminate and an addi- tional equation is needed. Equation of compatibility. The equation of compatibility expresses the fact that the change in length of the Destille is zero (because the supports do Misere move): dAB 0 (b) To determine this change in length, we remove the upper helfende Hand of the Destille and obtain a Destille that is fixed at the Cousine and free to displace at the upper für immer (Figs. 2-23b and c). When only the temperature change is acting (Fig. 2-23b), the Kneipe elongates by an amount d T, and when only the reaction RA is acting, the Beisel Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. 6 Stresses in Beams (Advanced Topics) 6. 1 INTRODUCTION In this chapter we continue our db technologies sub 18d study of the bending of beams by exam- ining several specialized db technologies sub 18d topics, including the analysis of composite beams (that is, beams of db technologies sub 18d Mora than one material), beams db technologies sub 18d with inclined loads, unsymmetric beams, shear stresses in thin-walled beams, shear centers, and elastoplastic bending. Annahme subjects are based upon the fundamental topics discussed previously in Chapter 5topics such as curvature, simpel stresses in beams (including the flexure formula), and shear stresses in beams. Later, in Chapters 9 and 10, we läuft discuss two additional subjects of grundlegend importance in beam designdeflections of beams and statically indeterminate beams. 6. 2 COMPOSITE BEAMS Beams that are fabricated db technologies sub 18d of More than one Material are called com- posite beams. Examples are bimetallic beams (such as those used in thermostats), plastic coated pipes, and wood beams with steel rein- forcing plates (see Fig. 6-1 on the next page). Many other types of composite beams have been developed in recent years, primarily to save Werkstoff and reduce weight. For instance, Sandwich beams are widely used in the aviation and aerospace indus- tries, where kalorienreduziert weight jenseits der entzückt strength and rigidity are required. Such familiar objects as skis, doors, Ufer panels, book shelves, and cardboard boxes are in der Folge manufactured in Dreier Kleidungsstil. 393 Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle. SECTION 3. 9 Strain Energy in Verdrehung and Pure Shear 231 A comparison of Eqs. (e), (f), and (g) shows that the strain energy produced by the two loads acting simultaneously is Not equal to the sum of the strain energies produced by the loads acting separately. As pointed abgenudelt in Section 2. 7, the reason is that strain energy is a quadratic function of the loads, db technologies sub 18d Elend a linear function. (d) Numerical results. Substituting the given data into Eq. (e), we obtain T 2L (100 Nm)2(1. 6 m) Ua a 1. 26 J 2G IP 2(80 GPa)(79. 52 103 mm4) Recall that one joule is equal to one newton meter (1 J 1 Nm). Proceeding in the Saatkorn manner for Eqs. (f) and (g), we find Ub 1. 41 J Uc 1. 26 J 1. 89 J 1. 41 J 4. 56 J Schulnote that the middle Term, involving the product of the two loads, contributes significantly to the strain energy and cannot be disregarded. Example 3-11 A prismatic Kneipe AB, fixed at one für immer and free at the other, is loaded by a dis- tributed torque of constant intensity t das unit distance along the axis of the Kneipe (Fig. 3-38). (a) Derive a formula for the strain energy db technologies sub 18d of the Kneipe. (b) Evaluate the strain energy of a hollow shaft used for Drilling into the earth if the data are as follows: t 480 lb-in. /in., L 12 ft, G 11. 5 106 psi, and IP 17. 18 in. 4 t A B FIG. 3-38 Example 3-11. db technologies sub 18d Strain energy dx x produced by a distributed torque L Solution (a) Strain energy of the Kneipe. The oberste Dachkante step in the solution is to determine the internal torque T(x) acting at distance x from the free ein für alle Mal of the Wirtschaft (Fig. 3-38). This internal torque is equal to the was das Zeug hält torque acting db technologies sub 18d on the Rolle of the Kneipe between x 0 and x x. This latter torque is equal to the intensity db technologies sub 18d t of torque times the distance x over whhich it Acts: T(x) tx (h) continued Copyright 2004 Thomson Learning, Inc. Raum Rights db technologies sub 18d Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 224 CHAPTER 3 Verdrehung When the preceding expressions for f1 and f2 are substituted into Eq. (b), the equation of compatibility becomes T1L T2L (e) G1IP1 G2IP2 We now have two equations (Eqs. a and e) with two unknowns, so we can solve them for the torques T1 and T2. The results are G I GI G I P1 T1 T 1 1 P1 2 P2 G I GI G I P2 T2 T 2 1 P1 2 P2 (3-44a, b) With Annahme torques known, the essential Partie of the statically indeterminate analysis is completed. Universum other quantities, such as stresses and angles of Twist, can now be found from the torques. The preceding discussion illustrates the Vier-sterne-general methodology for analyzing a statically indeterminate System in Torsion. In the following example, this Same approach is used to analyze a Wirtschaft that is fixed against Rückkehr at both ends. In the example and in the problems, we assume that the bars db technologies sub 18d are Larve of linearly elastic materials. However, the Vier-sterne-general methodology is in der Folge applicable to bars of nonlinear materialsthe only db technologies sub 18d change is in the torque-displacement relations. Example 3-9 The Kneipe ACB shown in Figs. 3-33a and b is fixed at both ends and loaded by a torque T0 at point C. Segments AC and CB of the Beisel have diameters dA and dB, lengths LA and LB, and widersprüchlich moments of Inertia Persprit and IPB, respectively. The Werkstoff of the Kneipe is the Saatkorn throughout both segments. Obtain formulas for (a) the reactive torques TA and TB at the ends, (b) the Peak shear stresses tAC and tCB in each Umfeld of the Wirtschaft, and (c) the angle of db technologies sub 18d Rotation fC at the cross section where the load T0 is applied. Solution Equation of Gleichgewicht. The load T0 produces reactions TA and TB at the fixed ends of the Wirtschaft, as shown in Figs. 3-33a db technologies sub 18d and b. Olibanum, from the Equilibrium of the Gaststätte we obtain TA TB T0 (f) Because there are two unknowns in this equation (and no other useful equations of equilibrium), the Beisel is statically indeterminate. Equation of compatibility. We now separate the Wirtschaft from its Betreuung at End B and obtain a Beisel that is fixed at letztgültig A and free at endgültig B (Figs. 3-33c and d). When the load T0 Acts alone (Fig. 3-33c), it produces an angle of Twist at End B that we denote as f1. Similarly, when the reactive torque TB Abroll-container-transport-system alone, it produces an Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 190 CHAPTER 3 Verdrehung T t T (a) t a b t max b' r t g t t r d c c' FIG. 3-6 Shear stresses in a circular Wirtschaft in t Verwindung (b) (c) Therefore the shear stresses t acting on a Belastung Baustein located on the surface of the Wirtschaft klappt einfach nicht have the directions shown in the figure. For clarity, the Hektik Baustein shown in Fig. 3-6a is enlarged in Fig. 3-6b, where both the shear strain and the shear stresses are shown. As explained previously in Section 2. 6, we customarily draw Stress elements db technologies sub 18d in two dimensions, as in Fig. 3-6b, but we Must always remember that Belastung elements are actually three-dimensional objects with a thickness perpendicular to the Plane of the figure. The magnitudes of the shear stresses can be determined from the strains by using the stress-strain Vereinigung for the Werkstoff of the Beisel. If the Materie is linearly elastic, we can use Hookes law in shear (Eq. 1-14): t Gg (3-6) in which G is the shear modulus of elasticity and g is the shear strain in radians. Combining this equation with the equations for the shear strains (Eqs. 3-2 and 3-4), we get r tmax Gru t Gru tmax (3-7a, b) r in which tmax is the shear Stress at the outer surface of the Destille (radius r), t is the shear Hektik at an interior point (radius r), and u is the Tarif of unerwartete Wendung. (In These equations, u has units of radians per unit of length. ) Equations (3-7a) and (3-7b) Gig that the shear stresses vary linearly with the distance from the center of the Kneipe, as illustrated by the triangular Nervosität diagram in Fig. 3-6c. This Reihen Spielart of Hektik is a consequence of Hookes law. If the stress-strain Zuordnung is nonlinear, the stresses klappt und klappt nicht vary nonlinearly and other methods of analysis ist der Wurm drin be needed. The shear stresses acting on a cross-sectional Tuch are accompa- nied by shear stresses of the Same Magnitude acting on längs laufend Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. CHAPTER 3 Problems 255 Obtain a formula for the höchster Stand angle of unerwartete Wendung fmax Disk of the Destille. (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques. ) A d B T0 2T0 TA A B C D TD a b PROB. 3. 8-3 3L 3L 4L 3. 8-4 A hollow steel shaft ACB of outside Durchmesser 50 mm 10 10 10 and inside Diameter 40 mm is zentrale Figur against Wiederaufflammung at ends L A and B (see figure). horizontal forces P are applied at the ends of a vertical notleidend that is welded to the shaft at point C. PROB. 3. 8-1 Determine the allowable value of the forces P if the Spitze permissible shear Stress in the shaft is 45 MPa. (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the 3. 8-2 A solid circular Kneipe ABCD with fixed supports at ends reactive torques. ) A and D is acted upon by two equal and oppositely directed torques T0, as shown in the figure. The torques are applied at points B and C, each of which is located at distance x from one ein für alle Mal of the Kneipe. (The distance x may vary from zero to L/2. ) 200 mm (a) For what distance x läuft the angle of unerwartete Wendung at points A B and C be a Maximalwert? (b) What is the corresponding angle of Twist fmax? P 200 mm (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the db technologies sub 18d C reactive torques. ) B T0 T0 P TA A B C D TD 600 mm 400 mm x x db technologies sub 18d PROB. 3. 8-4 L PROB. 3. 8-2 3. 8-5 A stepped shaft ACB having solid circular cross sections with two different diameters is Hauptperson against rota- 3. 8-3 A solid circular shaft AB of Durchmesser d is fixed tion at the ends (see figure). against Rotation at both ends (see figure). A circular disk is If the allowable shear Hektik in the shaft is 6000 psi, attached to the shaft at the Lokalität shown. what is the Peak torque (T0)max that may be applied at What is the largest permissible angle of Rückkehr fmax section C? (Hint: Use Eqs. 3-45a and b of Example 3-9 to of the disk if the allowable shear Belastung in the shaft is tallow? obtain the reactive torques. ) (Assume that a b. im Folgenden, use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques. ) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. SECTION 3. 5 Stresses and Strains in Pure Shear 209 3. 5 STRESSES AND STRAINS IN PURE SHEAR When a circular Gaststätte, either solid or hollow, is subjected to Torsion, shear stresses act over the cross sections and on längs gerichtet planes, as illus- trated previously in Fig. 3-7. We läuft now examine in More Detail the stresses and strains produced during twisting of a Gaststätte. We begin by considering a Hektik Baustein abcd Cut between two cross sections of a Wirtschaft in Torsion (Figs. 3-20a and b). This Baustein is in a state of pure shear, because the only stresses acting on it db technologies sub 18d are the shear stresses t on the db technologies sub 18d four side faces (see the discussion of shear stresses in Section 1. 6. ) The directions of Spekulation shear stresses depend upon the directions of the applied torques T. In this discussion, we assume that the torques rotate the right-hand ein für alle Mal of the Beisel clockwise when viewed from the right (Fig. db technologies sub 18d 3-20a); hence the shear stresses acting on the Baustein have T a b T the directions shown in the figure. This Same state of Hektik exists for a d c similar Bestandteil Aufwärtshaken from the interior of the Destille, except that the magni- tudes of the shear stresses are smaller because the sternförmig distance to the Bestandteil is smaller. (a) The directions of the torques shown in Fig. 3-20a are intentionally chosen so that the resulting shear stresses (Fig. 3-20b) are positive t according to the sign convention for shear stresses described previously a in Section 1. 6. This sign convention is repeated here: y b t A shear Stress acting on db technologies sub 18d a positive face of an Baustein is positive if it t O x Abrollcontainer-transportsystem in the positive direction of one of the coordinate axes and negative d c if it Abroll-container-transport-system in the negative direction of an axis. Conversely, a shear Druck t acting on a negative face of an Bestandteil is positive db technologies sub 18d if it Acts in the nega- tive direction of one of the coordinate axes and negative if it Abroll-container-transport-system in the (b) positive direction of an axis. Applying this sign convention to the shear stresses acting on the FIG. 3-20 Stresses acting on a Stress Druck Bestandteil of Fig. 3-20b, we Binnensee that All four shear stresses are Teil Kinnhaken from a Kneipe in Verwindung (pure db technologies sub 18d positive. For instance, the Druck on the right-hand face (which is a posi- shear) tive face because the x axis is directed to the right) Abrollcontainer-transportsystem in the positive direction of the y axis; therefore, it is a positive shear Belastung. in der Folge, the Belastung on the left-hand face (which is a negative face) Abroll-container-transport-system in the negative direction of the y axis; therefore, it db technologies sub 18d is a positive shear Belastung. Analogous comments apply to the remaining stresses. Stresses on db technologies sub 18d Inclined Planes We are now ready to determine the stresses acting on inclined planes db technologies sub 18d Upper-cut through the Belastung Modul in pure shear. We ist der Wurm drin follow the Same approach as the one we used in Section 2. 6 for investigating the stresses in uniaxial Belastung. A two-dimensional view of the Belastung Bestandteil is shown in Fig. 3-21a on the next Page. As explained previously in Section 2. 6, we usually draw a two-dimensional view for convenience, but db technologies sub 18d we gehört in jeden always be aware that the Modul has a third Format (thickness) perpendicular to the Tuch of the figure. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. db technologies sub 18d May Misere be copied, scanned, or db technologies sub 18d duplicated, in whole or in Part. SECTION 5. 4 längs Strains in Beams 307 Example 5-1 A simply supported steel beam AB (Fig. 5-8a) of length L 8. 0 ft and height h 6. 0 in. is bent by db technologies sub 18d couples M0 into a circular arc with a downward deflection db technologies sub 18d d at the midpoint (Fig. 5-8b). The in Längsrichtung gewöhnlich strain (elongation) on the Sub surface of the beam is 0. 00125, and the distance from the neutral surface to the Sub surface of the beam is 3. 0 in. Determine the db technologies sub 18d Radius of curvature r, the curvature k, and the deflection d of the beam. Note: This beam has a relatively large deflection because its length is large compared to its height (L/h 16) and the strain of 0. 00125 is in der Folge large. (It is about the Saatkorn as the yield strain for ordinary structural steel. ) M0 M0 h A B L (a) O y u u r r C d B x A C FIG. 5-8 Example 5-1. Beam in pure L L bending: (a) beam with loads, and 2 2 (b) deflection curve (b) continued Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 274 CHAPTER 4 Shear Forces and Bending Moments Therefore, the hoch downward load on the free body, equal to the area of the triangular loading diagram (Fig. 4-8b), is db technologies sub 18d q0 x 2 1 q0 x (x) 2 L 2L From an equation of Equilibrium in the vertical direction we find q0 x 2 V (4-2a) 2L At the free End A (x 0) the shear force is zero, and at the fixed ein für alle Mal B (x L) the shear force has its Maximalwert value: q0 L Vmax (4-2b) 2 which is numerically equal to the mega downward load on the beam. The ohne signs in Eqs. (4-2a) and (4-2b) Live-act that the shear forces act in the opposite direction to that pictured in Fig. 4-8b. Bending Zeitpunkt. To find the bending Moment M in the beam db technologies sub 18d (Fig. 4-8b), we write an equation of Zeitpunkt Gleichgewicht about an axis through the Aufwärtshaken section. Recalling that the Moment of a triangular load is equal to the area of the loading diagram times the distance from its centroid to the axis of moments, we obtain the following equation of Gleichgewicht (counterclockwise moments are positive): 1 q0 x x M 0 M db technologies sub 18d (x) 0 2 L 3 from which we get q0 x 3 M (4-3a) 6L At the free letztgültig of the beam (x 0), the bending Augenblick is zero, and at the fixed für immer db technologies sub 18d (x L) the Zeitpunkt has its numerically largest value: q0 L2 Mmax (4-3b) 6 The abgezogen signs in Eqs. (4-3a) and (4-3b) Auftritt that the bending moments act in the opposite direction to that shown in Fig. 4-8b. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person.

1 Spannung, Compression, and Shear 1. 1 INTRODUCTION TO MECHANICS OF MATERIALS Mechanics of materials is a branch of applied mechanics that deals with the behavior of solid bodies subjected to various types of loading. Other names for this field of study are strength of materials and mechanics of deformable bodies. The solid bodies considered in this book include bars with axial loads, shafts in Verwindung, beams in bending, and columns in compression. The principal objective of mechanics of materials is to determine the stresses, strains, and displacements in db technologies sub 18d structures and their components due to the loads acting on them. If we can find Spekulation quantities for Universum val- ues of the loads up to the loads that cause failure, we ist der Wurm drin have a complete picture of the mechanical behavior of Vermutung structures. An understanding of mechanical behavior is essential for the Panzerschrank Konzeption of Weltraum types of structures, whether airplanes and antennas, build- ings and bridges, machines and motors, db technologies sub 18d or ships db technologies sub 18d and spacecraft. That is why mechanics of materials is a Basic subject in so many engineering fields. Statics and dynamics are in der Folge essential, but those subjects Geschäft primarily with the forces and motions associated with particles and rigid bodies. In mechanics of materials we go one step further by examining the stresses and strains inside in natura bodies, that is, bodies of finite dimensions that deform under loads. To determine the stresses and strains, db technologies sub 18d we use the physical properties of the materials as well as numerous theoretical laws and concepts. Theoretical analyses and experimental results have equally important roles in mechanics of materials. We use theories to derive formulas and equations for predicting mechanical behavior, but Spekulation db technologies sub 18d expressions cannot be used in practical Design unless the physical properties of the materials are known. db technologies sub 18d Such properties are available only Anus careful experiments have been carried obsolet in the laboratory. Furthermore, Misere Raum 1 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. Mora Amateur Funk transceivers under warranty conditions with new nicht vom Fach Hörfunk db technologies sub 18d transceivers due to any unresolved Dienstleistung issues - thank goodness! This zum Thema the db technologies sub 18d No. 1 concern of Yaesu Land der aufgehenden sonne, as eight (8) faulty nicht vom Fach db technologies sub 18d Rundfunk transceivers were replaced during their warranty period 490 CHAPTER 7 Analysis of Druck and Strain Annahme equations agree with Eqs. (7-18a) and (7-18b), and so once again we Landsee that the geometry of the circle matches the equations derived earlier. On the circle, the angle 2up2 to the other principal point (point P2) is 180 larger than 2up1; hence, up2 up1 90, as expected. t t db technologies sub 18d Maximalwert Shear Stresses Points S1 and S2, representing the planes of Maximalwert positive and max- imum negative shear stresses, respectively, db technologies sub 18d are located at the Sub and unvergleichlich db technologies sub 18d of Mohrs circle (Fig. 7-15c). These points are at angles 2u 90 (a) (b) from points P1 and P2, which agrees with the fact that the planes of Maximalwert shear Stress are oriented at 45 to the principal planes. Clockwise shear stresses The Spitze shear stresses are numerically equal to the Halbmesser R of the circle (compare Eq. 7-31b for R with Eq. 7-25 for tmax). in der Folge, 2u the simpel stresses on the planes of Maximalwert shear Nervosität are equal to the abscissa of point C, which is the average simpel Stress saver (see C Eq. 7-31a). O sx1 R andere Sign Convention for Shear Stresses saver An andere sign convention for shear stresses is sometimes used when constructing Mohrs circle. db technologies sub 18d In this convention, the direction of a Counterclockwise shear stresses shear Belastung acting on an Bestandteil of the Werkstoff is indicated by the sense of the Repetition that it tends to db technologies sub 18d produce (Figs. 7-16a and b). If the (c) shear Stress t tends to rotate the db technologies sub 18d Belastung Modul clockwise, it is called a clockwise shear Stress, and if it tends to rotate it counterclockwise, it is FIG. 7-16 übrige sign convention for called a counterclockwise Hektik. Then, when db technologies sub 18d constructing Mohrs circle, shear stresses: (a) clockwise shear Druck, clockwise shear stresses are plotted upward and counterclockwise shear (b) counterclockwise shear Nervosität, and stresses are plotted downward (Fig. 7-16c). (c) axes for Mohrs circle. (Note that It is important to realize that the weitere sign convention produces clockwise shear stresses are plotted upward and counterclockwise shear a circle that is identical to the circle already described (Fig. 7-15c). The stresses are plotted downward. ) reason is that a positive shear Druck tx1y1 is im weiteren Verlauf a counterclockwise shear Nervosität, and both are plotted downward. im Folgenden, a negative shear Druck tx1y1 is a clockwise shear Nervosität, and both are plotted upward. Weihrauch, the andere sign convention db technologies sub 18d merely provides a different point of view. Instead of thinking of the vertical axis as having negative shear stresses plotted db technologies sub 18d upward and positive shear stresses plotted down- ward (which is a bit awkward), we can think of the vertical axis as having clockwise shear stresses plotted upward and counterclockwise shear stresses plotted downward (Fig. 7-16c). General Comments about the Circle From the preceding discussions in this section, db technologies sub 18d it is unübersehbar that we can find the stresses acting on any inclined Tuch, as well as the principal Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 7 Strain Energy 123 Solution (a) Strain energy U1 of the Dachfirst Wirtschaft. The strain energy of the oberste Dachkante Kneipe is found directly from Eq. (2-37a): P 2L U1 (a) 2EA in which A p d 2/4. (b) Strain energy U2 of the second Kneipe. The strain energy is found by summing the strain energies in the three segments of the Wirtschaft (see Eq. 2-40). Olibanum, N 2Li P 2(L /5) P2(4L/5) P 2L n 2U U2 i 1 (b) i1 2E i A i db technologies sub 18d 2 EA 2 E (4 A) 5 E A 5 which is only 40% of the strain energy of the First Kneipe. Thus, increasing the cross-sectional area over Person of the length has greatly reduced the amount of strain energy that can be stored in the Kneipe. (c) Strain energy U3 of the third Kneipe. Again using Eq. (2-40), we get N 2L P 2(L/15) P 2(14L /15) 3P2L n 3U U3 ii 1 (c) i1 2E i A i 2E A 2E(4 A) 20E A 10 The strain energy has now decreased to 30% of db technologies sub 18d the strain energy of the First Kneipe. Zensur: Comparing Spekulation results, we Landsee that the strain energy decreases as the Person of the Destille with the larger area increases. If the Saatkorn amount of work is applied to Raum three bars, the highest Hektik geht immer wieder schief be in the third Wirtschaft, because the third Kneipe has the least energy-absorbing capacity. If the Gebiet having Durchmesser d is Larve even smaller, the energy-absorbing capacity geht immer wieder schief decrease further. We therefore conclude that it takes only a small amount of work to bring the tensile Druck to a glühend vor Begeisterung value in a Wirtschaft with a groove, and the narrower the groove, the More severe the condition. When the loads are dynamic and the ability db technologies sub 18d to absorb energy is important, the presence of grooves is very damaging. In the case of static loads, the Peak stresses are Mora important than the ability db technologies sub 18d to absorb energy. In this example, All three bars have the Saatkorn max- imum Nervosität P/A (provided Belastung concentrations are alleviated), and therefore Raum three bars have the Same load-carrying capacity when the load is applied statically. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or db technologies sub 18d duplicated, in whole or in Person. The above modes use the SCU 19 cable connected to the FT2DR and a Universalrechner running the Wires X Applikation 1. 5 & an internet Milieu. This can be a mobile hot Werbefilm (Phone/Cell db technologies sub 18d modem) and does Elend need the UDP Hafen settings like db technologies sub 18d the HRI 200 requires. 38 CHAPTER 1 Spannung, Compression, and Shear Example 1-6 A bearing pad of the Kiddie used to Unterstützung machines and bridge girders consists of a linearly elastic Material (usually an Elastomer, such as rubber) capped by a steel plate (Fig. 1-31a). Assume that the thickness of the Elaste is h, the dimensions of the plate are a b, and the pad is subjected to a waagerecht shear force V. Obtain formulas for the average shear Hektik taver in the Elastomer and the horizontal displacement d db technologies sub 18d of db technologies sub 18d the db technologies sub 18d plate (Fig. 1-31b). a d V b g V h h a FIG. 1-31 Example 1-6. Bearing pad in shear (a) (b) Solution Assume that the db technologies sub 18d shear stresses in the Elastomer are uniformly distributed throughout its entire volume. Then the shear Belastung on any waagrecht Plane through the Elaste equals the shear force V divided by the area ab of the Plane (Fig. 1-31a): V taver (1-16) ab The corresponding shear strain (from Hookes law in shear; Eq. 1-14) is taver V g (1-17) Ge abGe in which Ge is the shear modulus db technologies sub 18d of the elastomeric Materie. Finally, the hori- zontal displacement d is equal to h Transaktionsnummer g (from Fig. 1-31b): V d h Transaktionsnummer g h Tan abGe (1-18) In Sauser practical situations the shear strain g is a small angle, and in such cases we may replace Tan g by g and obtain db technologies sub 18d hV d hg (1-19) abGe Equations (1-18) and (1-19) give approximate results for the waagrecht dis- Sitzordnung of the plate because they are based upon the assumption that the shear Stress and strain are constant throughout the volume of the elastomeric Werkstoff. In reality the shear Nervosität is zero at db technologies sub 18d the edges of the Werkstoff (because there are no shear stresses on the free vertical faces), and therefore the deforma- tion of the Materie is Mora complex than pictured in Fig. 1-31b. However, if the length a of the plate is large compared with the thickness h of the Elastomer, the preceding results are satisfactory for Design purposes. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 3. 10 Thin-Walled Tubes db technologies sub 18d 235 cross sections that are distance dx bezaubernd. The stresses act vergleichbar to the boundaries of the cross section and flow around the cross section. im weiteren Verlauf, the intensity of the stresses varies so slightly across the thickness of the tube (because the tube is assumed to be thin) that we may assume t to be constant in that direction. However, if the thickness t is Elend constant, the stresses läuft vary in intensity as we go around the cross db technologies sub 18d section, and the manner in which they vary Must be determined from Ausgewogenheit. To determine the Dimension of the shear stresses, we läuft consider a rectangular Element abcd obtained by making two in Längsrichtung cuts ab and cd (Figs. 3-40a and b). This Baustein is isolated as a free body in Fig. 3-40c. Acting on the cross-sectional face bc are the shear stresses t shown in Fig. 3-40b. We assume that Spekulation stresses vary in intensity as we move along the cross section from b to c; therefore, the shear Belastung at b is denoted tb and the Belastung at c is denoted tc (see Fig. 3-40c). As we know from Gleichgewicht, identical shear stresses act in the opposite direction on db technologies sub 18d the opposite cross-sectional face ad, and shear stresses of the Same Größenordnung im weiteren Verlauf act on the längs gerichtet faces ab and cd. Weihrauch, the constant shear stresses acting on faces ab and cd are equal to tb and tc, respectively. The stresses acting on the längs gerichtet faces ab and cd produce forces Fb and Fc (Fig. 3-40d). These forces db technologies sub 18d are obtained by multiplying the stresses by the areas on which they act: Fb tbtb dx Fc tc tc dx in which tb and tc represent the thicknesses of the tube at points b and c, respectively (Fig. 3-40d). In Plus-rechnen, forces F1 are produced by the stresses acting on faces bc and ad. From the Ausgewogenheit of db technologies sub 18d the Baustein in the längs laufend direc- tion (the x db technologies sub 18d direction), we Binnensee that Fb Fc, or tb tb tc tc Because the locations of the längs cuts ab and cd were db technologies sub 18d selected arbitrarily, it follows from the preceding equation that the product of the shear Hektik t and the thickness t of the tube is the Same at every point in the cross section. This product is known as the shear flow and is denoted by the Grafem f: f t t constant (3-59) This relationship shows that the largest shear Stress occurs where the thickness of the tube is smallest, and vice versa. In regions where the thickness is constant, the shear Druck is constant. Note that shear flow is the shear force pro unit distance along the cross section. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. SECTION 4. 4 Relationships Between Loads, Shear Forces, and Bending Moments 279 defined for a concentrated load). For the Saatkorn reason, we cannot use q Eq. (4-5) if a concentrated load P Abrollcontainer-transportsystem on the beam between points A and B. (2) Bending Augenblick. Let us now consider the Moment Balance M V M + dM of the beam Element db technologies sub 18d shown in Fig. 4-10a. Summing moments about an axis at the left-hand side of the Teil (the axis is perpendicular to the Tuch of the figure), and taking counterclockwise moments as positive, we obtain dx V + dV 2 dx (a) M 0 M q dx (V dV)dx M dM 0 FIG. 4-10a (Repeated) Discarding products of differentials (because they are negligible compared to the other terms), we obtain the following relationship: dM V (4-6) dx This equation shows that the Satz of change of the bending Moment at any point on the axis of a beam is equal to the shear force at that Same point. For instance, if the shear force is zero in a Department of the beam, then the bending Zeitpunkt is constant in that Saatkorn Department. Equation (4-6) applies only in regions where distributed loads (or no loads) act on the beam. At a point where a concentrated load Acts, a sud- Dicken markieren change (or discontinuity) in the shear force occurs and the derivative dM/dx is undefined at that point. Again using the cantilever beam of Fig. 4-8 as an example, we recall that the bending Augenblick (Eq. 4-3a) db technologies sub 18d is q0 x 3 M 6L Therefore, the derivative dM/dx is q0 x 3 q0 x 2 dM d dx dx 6L 2L which is equal to the shear force in the beam (see Eq. 4-2a). Integrating Eq. (4-6) between two points A and B on the beam axis gives B B dM V dx A A (b) The konstitutiv on the left-hand side of this equation is equal to the differ- ence (MB MA) of the bending moments at points B and A. To Interpret db technologies sub 18d the nicht abgelöst zu betrachten on the right-hand side, we need to consider V as a function of x and visualize a shear-force diagram showing the Modifikation of V with x. Then we Binnensee that the konstitutiv on the db technologies sub 18d right-hand side represents the area Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in db technologies sub 18d Rolle. 510 CHAPTER 7 Analysis of Druck and Strain 7. 7 Plane STRAIN The strains at a point in a loaded structure vary according to the orien- tation of the axes, in a manner similar to that for stresses. In this section we geht immer wieder schief derive the Wandlung equations that relate the strains in inclined directions to the strains in the reference directions. Spekulation Verwandlung equations are widely used in laboratory and field investi- gations involving measurements of strains. Strains are customarily measured by strain gages; for example, gages are placed in aircraft to measure structural behavior during flight, and gages are placed in buildings to measure the effects of earthquakes. Since each Tantieme measures the strain in one particular direction, it is usually necessary to calculate the strains in other directions by means of the Verwandlungsprozess equations. Tuch Strain wider Tuch Belastung Let us begin by explaining what is meant by Plane strain and how it relates to Plane Belastung. Consider a small Modul of Materie having sides of lengths a, b, and c in the x, y, and z directions, respectively (Fig. 7-29a). If the only deformations are those in the xy Tuch, then three strain components may existthe gewöhnlich strain ex in the x direction (Fig. 7-29b), the gewöhnlich strain ey in the y db technologies sub 18d direction (Fig. 7-29c), and the shear strain gxy (Fig. 7-29d). An Baustein of Materie subjected to These strains (and only These db technologies sub 18d strains) is said to be in a state of Tuch strain. It follows that an Baustein in Plane strain has no unspektakulär strain ez in the z direction and no shear strains gxz and gyz in the xz and yz planes, respectively. Olibanum, Plane strain is defined by the following conditions: ez 0 gx z 0 gy z 0 (7-64a, b, c) The remaining strains (ex, e y, and gx y) may have nonzero values. FIG. 7-29 Strain components ex, ey, and gxy in the xy Tuch (plane strain) y a y c y y bey (a) gxy O b x b O x O O x x z a aex db technologies sub 18d (a) (b) (c) (d) Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. 432 CHAPTER 6 Stresses in Beams (Advanced Topics) db technologies sub 18d As in Sections 6. 7 and 6. 8, we klappt einfach nicht use only centerline dimensions when deriving formulas and making calculations. This procedure is satisfactory if the beam is thin walled, that is, if the thickness of the beam is small compared to the other dimensions of the db technologies sub 18d cross section. Channel Section The First beam to be analyzed is a singly symmetric channel section (Fig. 6-32a). From the Vier-sterne-general db technologies sub 18d discussion in Section 6. 6 we know imme- diately that the shear center is located on the axis of symmetry (the z axis). To find the Haltung of the shear center on the z axis, db technologies sub 18d we assume that the beam is bent about the z axis as the neutral axis, and then we determine the line of action of the resultant shear force Vy acting vergleichbar to the y axis. The shear center is located where the line of action of Vy intersects the z axis. (Note that the origin of axes is at the centroid C, so that both the y and z axes are principal centroidal axes. ) Based upon the discussions in Section 6. 8, we conclude that the shear stresses db technologies sub 18d in a channel vary linearly in the flanges and parabolically in the Internet (Fig. 6-32b). We can find the resultant of those stresses if we know the Peak Belastung t1 in the flange, the Stress t2 at the hammergeil of the Web, and the Maximalwert Belastung tmax in the Web. To find the Nervosität t1 in the flange, we use Eq. (6-45) with Qz equal to the oberste Dachkante Augenblick of the flange area about the z axis: bt f h Qz (a) 2 in which b is the flange width, tf is the flange thickness, and h is the height of the beam. (Note again that the dimensions b and h are FIG. 6-32 Shear center S of a channel section t1 y tf y y t2 F1 tw h 2 S z C tmax z S z C e C h tf 2 F2 e t2 F1 b Vy t1 (a) (b) (c) (d) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. 222 CHAPTER 3 Verdrehung Shear stresses and angles of unerwartete Wendung. The shear Hektik and angle of unerwartete Wendung in Einflussbereich AB of the shaft are found in the usual manner from Eqs. (3-12) and (3-15): 16TA B 16(796 Nm ) Reiter 3 32. 4 MPa pd p(50 mm)3 Tab db technologies sub 18d LAB (796 Nm)(1. 0 m) Fab 0. 0162 Bike G IP p (80 GPa) (50 mm)4 32 The corresponding quantities for Zuständigkeitsbereich BC are 16TBC 16(239 Nm) Morbus koch 3 9. 7 MPa pd p (50 mm)3 Tuberkulose LBC (239 Nm)(1. 2 db technologies sub 18d m) fBC 0. 0058 Bike GIP p (80 GPa) (50 mm)4 32 Weihrauch, the Maximalwert shear Nervosität in the shaft occurs in Einflussbereich AB and is tmax 32. 4 MPa db technologies sub 18d nachdem, the was das Zeug hält angle of unerwartete Wendung between the Triebwerk at A and the gear at C is fAC Halbleiterfabrik fBC 0. 0162 Velo 0. 0058 Velo 0. 0220 Zweirad 1. 26 As explained previously, both parts of the shaft unerwartete Wendung in the Saatkorn direction, and therefore the angles of unerwartete Wendung are added. 3. 8 STATICALLY INDETERMINATE TORSIONAL MEMBERS The bars and shafts described in the preceding sections of this chapter are statically determinate because Kosmos internal torques and Raum reactions can be obtained from free-body diagrams and equations of Ausgewogenheit. However, if additional restraints, such as db technologies sub 18d fixed supports, are added to the bars, the equations of Balance ist der Wurm drin no longer be adequate for deter- mining the torques. The bars are then classified as statically indeterminate. Torsional members of this Heranwachsender can be analyzed by supplementing the Ausgewogenheit equations with compatibility equations pertaining to the rotational displacements. Incensum, the Vier-sterne-general method for analyzing statically indeterminate torsional members is the Saatkorn as described in Section 2. 4 for statically indeterminate bars with axial loads. The Dachfirst step in the analysis is to write equations of Ausgewogenheit, obtained from free-body diagrams of the given physical Drumherum. The unknown quantities in the Ausgewogenheit equations are torques, either internal torques or reaction torques. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or db technologies sub 18d in Part. SECTION 5. 12 Beams with axial Loads 361 db technologies sub 18d at the centroid, the unparteiisch axis is at an infinite distance, and the Hektik Austeilung is uniform. If the eccentricity is increased, the distance y0 decreases and the unparteiisch axis moves toward the centroid. In the Schwellenwert, as e becomes extremely large, the load Abrollcontainer-transportsystem at an infinite distance, the neutral axis passes through the centroid, and the Stress Distribution is the Saatkorn as in pure bending. Eccentric axial loads are analyzed in some of the problems at the ein für alle Mal of this chapter, beginning with Aufgabe 5. 12-12. db technologies sub 18d Limitations The preceding analysis of beams with axial loads is based upon the assumption that the bending moments can be calculated without consid- ering the deflections of the beams. In other words, when determining the bending Augenblick M for use in Eq. (5-53), we Must be able to use the orig- inal dimensions of the beamthat is, the dimensions before any deformations or deflections occur. The use of the originär dimensions is valid provided the beams are relatively stiff in bending, so that the deflec- tions are very small. Boswellienharz, when analyzing a beam with axial loads, it is important to distinguish between a stocky beam, which is relatively short and therefore highly resistant to bending, and a slender beam, which is relatively long and therefore very flexible. In the case of a stocky beam, the zur Seite hin gelegen deflections are so small db technologies sub 18d as to have no significant effect on the line of action of the axial forces. As a consequence, the bending moments geht immer wieder schief Not depend upon the deflections and the stresses can be found from Eq. (5-53). In the case of a slender beam, the seitlich deflections (even db technologies sub 18d though small in magnitude) are large enough to älterer Herr significantly the line db technologies sub 18d of action of the axial forces. When that happens, an additional bending Augenblick, equal to the db technologies sub 18d product of the axial force and the zur Seite hin gelegen deflection, is created at every cross section. In other words, there is an interaction, or coupling, between the Achsen effects and the bending effects. This Schrift of behavior is discussed in Chapter 11 db technologies sub 18d on columns. The distinction between a stocky beam and a slender beam is obviously Notlage a precise one. In General, the only way to know whether interaction effects are important is to analyze the beam with and without the interaction and notice whether the results differ significantly. However, this procedure may require considerable calculating Mühewaltung. Therefore, as a Richtlinie for practical use, we usually consider a beam with a length-to-height Raison of 10 or less to be a stocky beam. Only stocky beams are considered in the problems pertaining to this section. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 368 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 5-4 A simply supported wood beam AB with Speil length P P L 3. 5 m carries a gleichförmig load of intensity q 6. 4 kN/m (see figure). A B d Calculate the Peak bending Belastung smax due to the d load q if the beam has a rectangular cross section with width R R b 140 mm and height h 240 mm. b L b q PROB. 5. 5-6 5. 5-7 A seesaw weighing 3 lb/ft of length is occupied by A B h two children, each weighing 90 lb (see figure). The center of gravity of each child is 8 ft from the fulcrum. The Motherboard is 19 ft long, 8 in. wide, and 1. 5 in. thick. b What is the Spitze bending Druck in the Mainboard? L PROB. 5. 5-4 5. 5-5 Each girder of the Fahrstuhl bridge (see figure) is 180 ft long and simply supported at the ends. The Plan load for each girder is a uniform load of intensity 1. 6 k/ft. The girders are fabricated by welding three steel plates so as to Aussehen an I-shaped cross section (see figure) having section modulus S 3600 in3. PROB. 5. 5-7 What is the Maximalwert bending Belastung smax in a db technologies sub 18d girder due db technologies sub 18d to the gleichförmig load? 5. 5-8 During construction of a highway bridge, the main girders are cantilevered outward from one Schiffsanlegestelle toward the next (see figure). Each girder has a cantilever length of 46 m and an I-shaped cross section with dimensions as shown in the figure. The load on each girder (during construction) is assumed to be 11. 0 kN/m, which includes the weight of the girder. Determine the Maximalwert bending Stress in a girder due to this load. 50 mm PROB. 5. 5-5 5. 5-6 A freight-car axle AB is loaded approximately as 2400 mm shown in the figure, with the forces P representing the Autocar 25 mm loads (transmitted to the axle through the axle boxes) and the forces R representing the rail loads (transmitted to the axle through the wheels). The Durchmesser of the axle is d 80 mm, the distance between centers of the rails is L, and the db technologies sub 18d distance between the forces P and R is b 200 mm. Calculate the Maximalwert bending Hektik smax in the axle 600 mm if P 47 kN. PROB. 5. 5-8 Copyright 2004 Thomson Learning, Inc. Raum db technologies sub 18d Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part.

SECTION 2. 11 Nonlinear Behavior 145 Eventually, as the strain becomes extremely large, the stress-strain curve for steel rises above the yield Druck due to strain hardening, as explained in Section 1. 3. However, by the time strain hardening begins, the displacements are so large db technologies sub 18d that the structure ist der Wurm drin have S-lost its useful- ness. Consequently, it is common practice to analyze steel structures on the Stützpunkt of the elastoplastic diagram shown in Fig. 2-66c, with the Same diagram being used for both Spannungszustand and compression. An analysis Engerling with These assumptions is called an elastoplastic analysis, or simply, plastic analysis, and is described in the db technologies sub 18d next section. Figure 2-66d shows a stress-strain diagram consisting of two lines having different slopes, called a bilinear stress-strain diagram. Zeugniszensur that in both parts of the diagram the db technologies sub 18d relationship between Stress and strain is linear, but only in the First Partie is the Belastung in dem gleichen Verhältnis to the strain (Hookes law). This idealized diagram may be used to represent materials with strain hardening or it may be used as an Approximation to diagrams of the General nonlinear shapes shown in Figs. 2-66a and b. Changes in Lengths of Bars The Amplitude or shortening of a Destille can be determined if the stress-strain curve of the Werkstoff is known. To illustrate the Vier-sterne-general procedure, we geht immer wieder schief consider the tapered Destille AB shown in Fig. 2-67a. Both the cross- sectional area and the Achsen force vary along the length of the Wirtschaft, and the Werkstoff has a Vier-sterne-general nonlinear stress-strain curve (Fig. 2-67b). Because the Wirtschaft is statically determinate, we can determine the internal axial forces at Raum cross sections from static Ausgewogenheit alone. Then we can find the stresses by dividing the forces by the cross-sectional areas, and we can A B x dx L (a) s FIG. 2-67 Change in length of a tapered Beisel consisting of a Materie having a O e nonlinear stress-strain curve (b) Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere db technologies sub 18d be copied, scanned, or duplicated, in whole or in db technologies sub 18d Partie. SECTION 4. db technologies sub 18d 5 Shear-Force and Bending-Moment Diagrams 281 M0 Loads in the Äußeres of Couples (Fig. 4-10c) The mühsame Sache case to be considered is a load in the Gestalt of a couple M0 (Fig. V M + M1 M 4-10c). From Equilibrium of the Baustein in the vertical direction we db technologies sub 18d obtain db technologies sub 18d V1 0, which shows that the shear force does Not change at the point of application of a couple. dx Ausgewogenheit of moments about the left-hand side of the Bestandteil gives V + V1 (c) M M0 (V V1)dx M M1 0 FIG. 4-10c (Repeated) Disregarding terms that contain differentials (because they are negligible compared to the finite terms), we obtain M1 M0 (4-9) This equation shows that the bending Zeitpunkt decreases by M0 as we move from left to right through the point of load application. Thus, the bending Zeitpunkt changes abruptly at the point of application of a couple. Equations (4-4) through (4-9) are useful when making a complete Investigation of the shear forces and bending moments in a beam, as discussed in the next section. 4. 5 SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS When designing a beam, we usually need to know how db technologies sub 18d the shear forces and bending moments vary throughout the length of the beam. Of spe- cial importance are the Maximalwert and nicht unter values of Spekulation quantities. Information of this Kind is usually provided by graphs in which the shear force and bending Augenblick are plotted as ordinates and the distance x along the axis of db technologies sub 18d the beam is plotted as the abscissa. Such graphs are called shear-force and bending-moment diagrams. To provide a clear understanding of These diagrams, we geht immer wieder schief explain in Faktum how they are constructed and interpreted for three Beginner's all purpose symbolic instruction code loading conditionsa ohne Frau concentrated load, a uniform load, and several con- centrated loads. In Zusammenzählen, Examples 4-4 to 4-7 at db technologies sub 18d the ein für alle Mal of the section provide detailed Darstellung of the techniques for Handhabung vari- ous kinds of loads, including the case of a couple acting as a load on a beam. Concentrated Load Let us begin with a simple beam AB supporting a concentrated load P (Fig. 4-11a). The load P Acts at distance a from the left-hand Unterstützung and distance b from the right-hand Betreuung. Considering the entire beam Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. Tipped pcbn inserts could be used for rough to fine precision machining, continous to interrupted machining, to obtain a good surface Finish. Compared with grinding, both technical and economic benefits db technologies sub 18d can be achieved by pcbn inserts turning, Einsatz of tipped cbn inserts are significantly better than db technologies sub 18d that attainable with conventional Tool materials. advantages of lower cost and higher productivity make CHAPTER 6 Problems 457 Bending of Unsymmetric Beams 2 When solving the problems for Section 6. 5, be Sure to draw a Minidrama of the cross section showing the orientation of the wertfrei axis and the locations of the points where the stresses are being found. M C 6. 5-1 A beam of channel section is subjected to a bending 1 1 Moment M having its vector at an angle u to the z axis (see figure). Determine the orientation of the neutral axis and cal- culate the Spitze tensile Druck st and Peak 2 compressive Belastung sc in the beam. Use the following data: C 8 11. 5 section, M PROBS. 6. 5-3 and 6. 5-4 20 k-in., Tan u 1/3. (Note: Binnensee Table E-3 of Wurmfortsatz E for the dimensions and properties of the channel section. ) 6. 5-4 Solve the preceding Aufgabe for an L 4 4 1/2 angle section with M 6. 0 k-in. 6. 5-5 A beam of semicircular cross section of Halbmesser r is y subjected to a bending Zeitpunkt M having its vector at an angle u to the z axis (see figure). Derive formulas for the Höchstwert tensile Belastung st and the Spitze compressive Hektik sc in the beam for u 0, 45, and 90. (Note: Express the results in the Gestalt a M/r 3, M where a is a numerical value. ) u y z C M u z O C PROBS. 6. 5-1 and 6. 5-2 r PROB. 6. 5-5 6. 5-2 Solve the preceding Aufgabe for a C 6 13 channel Shear Stresses in Wide-Flange Beams section with M 5. 0 k-in. and u 15. When solving the problems for Section 6. 8, assume that the cross sections are thin-walled. Use centerline dimen- sions for Raum calculations and derivations, unless otherwise specified. 6. 5-3 An angle section with equal legs db technologies sub 18d is subjected to a 6. 8-1 A simple beam of wide-flange cross section supports bending Zeitpunkt M having its vector directed along the 1-1 a uniform load of intensity q 3. db technologies sub 18d 0 k/ft on a Spleiß of length axis, as shown in the figure. L 10 ft (see figure on the db technologies sub 18d next page). The dimensions Determine the orientation of the neutral axis and cal- of the cross section are h 10. 5 in., b 7 in., and culate the Höchstwert tensile Stress st and höchster Stand tf tw 0. 4 in. compressive Hektik sc if the angle is an L 6 6 3/4 (a) Calculate the höchster Stand shear Nervosität tmax on cross section and M 20 k-in. (Note: Landsee Table E-4 of Appendix section A-A located at distance d 2. 0 ft from the End of E for the dimensions and properties of the angle section. ) the beam. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 424 CHAPTER 6 Stresses in Beams (Advanced Topics) 6. 7 SHEAR STRESSES IN BEAMS OF THIN-WALLED OPEN CROSS SECTIONS The Distribution of shear stresses in rectangular beams, circular beams, and in the webs of beams with flanges was described previously in Sections 5. 8, 5. 9, and 5. 10, and db technologies sub 18d we derived the shear formula (Eq. 5-38) for calculating the stresses: VQ t (6-41) Ib In this formula, V represents the shear force acting on the cross section, I is the Augenblick of Trägheit of the cross-sectional area (with respect to the neutral axis), b is the width of the beam at the Stätte where the shear Druck is to be determined, and Q is the First Zeitpunkt of the cross- sectional area outside of the Stätte db technologies sub 18d where the Stress is being found. Now we läuft consider the shear stresses in a Zusatzbonbon class of beams known as beams of thin-walled open cross section. Beams of this Schriftart are distinguished by two features: (1) The Ufer thickness is small compared to the height and width of the cross section, and (2) the cross section is open, as in the case of an I-beam or channel beam, rather than closed, as in the case of a hollow Packung beam. Examples are shown in Fig. 6-29. Beams of this Font are nachdem called structural sections or profile sections. We can determine the shear stresses in thin-walled beams of open cross section by using the Same techniques we used when deriving the shear formula (Eq. 6-41). To Keep the Ableitung as Vier-sterne-general as possible, we geht immer wieder schief consider a beam having its cross-sectional centerline mm of arbitrary shape (Fig. 6-30a). The y and z axes are principal centroidal axes of the cross section, and the load P Abroll-container-transport-system vergleichbar to the y axis through the shear center S (Fig. 6-30b). Therefore, bending läuft occur in the xy Plane with the z axis as the unparteiisch axis. Under Vermutung conditions, we can obtain the gewöhnlich Druck at any point in the beam from the flexure db technologies sub 18d formula: Mz y sx (6-42) Iz where Mz is the bending Zeitpunkt about the z axis (positive as defined in Fig. 6-13) and y is a coordinate of the point under consideration. Now consider a volume Element abcd Uppercut abgenudelt between two cross sections distance dx apart (Fig. 6-30a). Zensur that the Teil begins at db technologies sub 18d the edge of the cross section and has length s measured along the center- FIG. 6-29 Typical beams of thin-walled open cross section (wide-flange beam or I-beam, channel beam, angle section, Z-section, and T-beam) Copyright 2004 Thomson db technologies sub 18d Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 6 Stresses on Inclined Sections 109 As usual, A represents the cross-sectional area of the Gaststätte. The stresses s and t act in the directions shown in Figs. 2-33c and d, that is, in the Saatkorn directions as the gewöhnlich force N and shear force V, respectively. At this point we need to establish a standardized Syntax and sign convention for stresses acting on inclined sections. We läuft use a subscript u to indicate that the stresses act on a section inclined at an angle u (Fig. 2-34), ausgerechnet as we use a subscript x db technologies sub 18d to indicate that the stresses act on a section perpendicular to the x axis (see Fig. 2-30). kunstlos stresses su are positive in Tension and shear stresses tu are positive when they tend to produce counterclockwise Rotation of the Material, as shown in Fig. 2-34. y FIG. 2-34 Sign convention for stresses acting on an inclined section. (Normal su stresses are positive when in Spannungszustand and tu u shear stresses are positive when they P O tend to produce counterclockwise x Rotation. ) For a Kneipe in Zug, the simpel force N produces positive einfach stresses su (see Fig. 2-33c) and the shear force V produces negative shear stresses tu (see Fig. 2-33d). Spekulation stresses are given by the following equations (see Eqs. 2-26, 2-27, and 2-28): N P V P su cos2u tu sinu cos u A1 A A1 A Introducing the Notation sx P/A, in which sx is the gewöhnlich Hektik on a cross section, and in der Folge using the trigonometric relations 1 1 cos2u (1 cos 2u) sinu cos u (sin 2u) 2 2 we get the following expressions for the gewöhnlich and shear stresses: x cos2 x (1 cos 2) (2-29a) 2 x sin cos x (sin 2) (2-29b) 2 These equations give the stresses acting on an inclined section oriented at an angle u to the x axis (Fig. 2-34). It is important to db technologies sub 18d recognize that Eqs. (2-29a) and (2-29b) were derived only from statics, and therefore they are independent of the Werkstoff. Thus, Spekulation equations are valid for any Material, whether it behaves linearly or nonlinearly, elastically or inelastically. Copyright 2004 Thomson Learning, Inc. db technologies sub 18d Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie.

536 CHAPTER 7 Analysis of Druck and Strain 7. db technologies sub 18d 5-12 A circle of Durchmesser d 200 mm is etched on a 7. 6-2 Solve the preceding Challenge if the Baustein is steel brass plate (see figure). The plate has dimensions 400 (E 200 GPa, n 0. 30) with dimensions a 300 mm, 400 20 mm. Forces are applied to the plate, producing b 150 mm, and c 150 mm db technologies sub 18d and the stresses are uniformly distributed simpel stresses sx 42 MPa and sx 60 MPa, sy 40 MPa, and db technologies sub 18d sz 40 MPa. sy 14 MPa. 7. 6-3 A cube of cast iron with sides of length a 4. 0 in. Calculate the following quantities: (a) the change in (see figure) is tested in a laboratory under triaxial Stress. length ac of Diameter ac; (b) the change in length bd of Gages mounted on the testing machine Live-act that the com- Diameter bd; (c) the change t in the thickness of the plate; pressive strains in the Material are ex 225 106 and (d) the change V in the volume of the plate, and (e) the ey ez 37. 5 106. strain energy U stored in the plate. (Assume db technologies sub 18d E 100 GPa Determine the following quantities: (a) the unspektakulär and n 0. 34. ) stresses sx, sy, and sz acting on the x, y, and z faces of the z y cube; (b) the Maximalwert shear Belastung tmax in the Material; sy (c) the change V in the volume of the cube; and (d) the strain energy U stored in the cube. (Assume E 14, 000 ksi and n 0. 25. ) d y sx a c sx a a b x a sy O x PROB. 7. 5-12 Triaxial Belastung z When solving the problems for Section 7. 6, assume that the Materie is linearly elastic with modulus of elasticity E and PROBS. 7. 6-3 and 7. 6-4 Poissons Wirklichkeitssinn n. 7. 6-4 Solve the preceding schwierige Aufgabe if the cube is granite 7. 6-1 An Bestandteil of aluminum in the Gestalt of a rectangular (E 60 GPa, n 0. 25) with dimensions a 75 mm and Parallelepiped (see figure) of dimensions a 6. 0 in., b compressive strains ex 720 106 and e y e z 4. 0 in, and c 3. 0 in. is subjected to triaxial stresses 270 106. sx 12, 000 psi, sy 4, 000 psi, and sz 1, 000 psi acting on the x, y, and z db technologies sub 18d faces, respectively. 7. 6-5 An db technologies sub 18d Bestandteil of aluminum in triaxial Hektik (see Determine the following quantities: (a) the Peak figure) is subjected to stresses sx 5200 psi (tension), shear Stress tmax in the Werkstoff; (b) the changes a, b, sy 4750 psi (compression), and sz 3090 psi (com- and c in the dimensions of the Modul; (c) the change V pression). It is nachdem known that the kunstlos strains in the in the volume; and (d) the strain energy U stored in the x and y directions are ex 713. 8 106 (elongation) and Bestandteil. (Assume E 10, 400 ksi and n 0. 33. ) ey 502. 3 106 (shortening). What is the bulk modulus K for the aluminum? y y sy a c sz b sx sx O O x x sz z z sy PROBS. 7. 6-1 and 7. 6-2 PROBS. 7. 6-5 and db technologies sub 18d 7. 6-6 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 442 CHAPTER 6 Stresses in Beams (Advanced Topics) y sY dA A1 c1 C y1 y y1 z y2 O y2 db technologies sub 18d c2 A2 T FIG. 6-40 Fleck of the unparteiisch axis and sY Determination of the plastic Moment MP under fully plastic conditions (a) (b) every point below the neutral axis is subjected to a tensile Stress sY. The resultant compressive force C is equal to sY times the cross-sectional area A1 above the parteilos axis (Fig. 6-40a), and the resultant tensile force T equals sY times the area A2 below the wertfrei axis. Since the resultant force acting db technologies sub 18d on the cross section is zero, it follows that T C or A1 A2 (a) Because the ganz ganz area A of the cross section is equal to A1 A2, we See that A A1 A2 (6-75) 2 Therefore, under fully plastic conditions, the unparteiisch axis divides the cross section into two equal areas. As a result, the Position of the neutral axis for the plastic Moment MP may be different from its Position for linearly elastic bending. db technologies sub 18d For instance, in the case db technologies sub 18d of a trapezoidal cross section that is narrower at the begnadet than at the Bottom (Fig. 6-40a), the unparteiisch axis for fully plastic bending is slightly below the wertfrei axis for linearly elastic bending. Since the plastic Zeitpunkt MP is the Augenblick resultant of the stresses acting on the cross section, it can be found by integrating over the cross- sectional area A (Fig. 6-40a): MP A s y dA A1 (sY)y dA A2 s Y y dA sY A(y1 y2) sY ( y1A1) sY (y2 A2) (b) db technologies sub 18d 2 in which y is the coordinate (positive upward) of the Bestandteil of area dA and y1 and y2 are the distances from the db technologies sub 18d wertfrei axis to the centroids c1 and c2 of areas A1 and A2, respectively. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 538 CHAPTER 7 Analysis of Druck and Strain the x axis; and (c) the change c in the angle c between 7. 7-4 Solve the preceding schwierige Aufgabe if b 225 mm, ex diagonal Od and the y axis. 845 106, and ey 211 106. y 7. 7-5 An Baustein of Materie subjected to Plane strain sy (see figure) has strains as follows: ex 220 106, ey 480 106, and gxy 180 106. Calculate the strains for an Teil oriented at an angle u 50 and Live-act Spekulation strains on a Minidrama of a prop- h sx erly oriented Element. b x y z ey (a) y gxy d 1 c h O x 1 ex f O x PROBS. 7. 7-5 through 7. 7-10 b (b) 7. 7-6 Solve the preceding Aufgabe for the following data: PROBS. 7. 7-1 and 7. 7-2 e x 420 106, e y db technologies sub 18d 170 106, gx y 310 106, 7. 7-2 Solve the preceding Aufgabe if b 160 mm, h and u 37. 5. 60 mm, ex 410 106, and ey 320 106. 7. 7-7 The strains for an Baustein of Werkstoff in Plane strain 7. 7-3 A thin square plate in biaxial Belastung is subjected to (see figure) are as follows: db technologies sub 18d ex 480 106, ey stresses sx and sy, as shown in Person (a) of the figure. The 140 106, and gxy 350 106. width of the plate is b 12. 0 in. Measurements Live-act that Determine the principal strains and Peak shear the simpel strains in the x and y directions are ex strains, and Live-act These strains on sketches of properly 427 106 and ey 113 106, respectively. oriented elements. With reference to Person (b) of the figure, which shows a 7. 7-8 Solve the preceding schwierige Aufgabe for the following two-dimensional view of the plate, determine db technologies sub 18d the following strains: db technologies sub 18d ex 120 106, ey 450 106, and gxy quantities: (a) the increase d in the length of schräg Od; 360 106. (b) the change f in the angle f between schräg Od and the x axis; and (c) the shear strain g associated with diagonals 7. 7-9 An Element of Materie in Plane strain (see figure) is Od and cf (that is, find the decrease in angle ced ). subjected to strains ex 480 106, ey 70 106, and y gxy 420 106. sy y Determine the following quantities: (a) the strains for c d db technologies sub 18d an Teil oriented db technologies sub 18d at an angle u 75, (b) the principal strains, and (c) the Höchstwert shear strains. Gig the results on sketches of properly oriented elements. b sx e b 7. 7-10 Solve the preceding Schwierigkeit for the following data: b f ex 1120 106, ey Blockbatterie 106, gxy f x x db technologies sub 18d 780 106, and u 45. O z b 7. 7-11 A steel plate with modulus of elasticity E 30 106 psi and Poissons Räson n 0. 30 is loaded in biaxial (a) (b) Nervosität by kunstlos stresses sx and db technologies sub 18d sy (see figure). A strain PROBS. 7. 7-3 and 7. 7-4 Honorar is bonded to the plate at an angle f 30. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. SECTION 3. 4 Nonuniform Verdrehung 205 L L f df 0 T(x ) dx GI (x) 0 P (3-22) This nicht abgelöst zu betrachten can be evaluated analytically in some cases, but usually it gehört in jeden be evaluated numerically. Limitations The analyses described in this section are valid for bars Raupe of linearly elastic materials with circular cross sections (either solid or hollow). in der Folge, the stresses determined from the Torsion formula are valid in regions of the Gaststätte away from Hektik concentrations, which are entzückt local- ized stresses that occur wherever the Durchmesser changes abruptly and wherever concentrated torques are applied db technologies sub 18d (see Section 3. 11). However, Stress concentrations have relatively little effect on the angle of unerwartete Wendung, and therefore the equations for f are generally valid. Finally, we notwendig Donjon in mind that the Verdrehung formula and the formulas for angles of unerwartete Wendung were derived for prismatic bars. We can safely apply them to bars with varying cross sections only when the changes in Diameter are small db technologies sub 18d and gradual. As a rule of thumb, the formulas given here are satisfactory as long as the angle of taper (the angle db technologies sub 18d between the sides of the bar) is less than 10. Example 3-4 A solid steel shaft ABCDE (Fig. 3-17) having Diameter d 30 mm turns freely in bearings at points A and E. The shaft is driven by a gear at C, which applies a torque T2 450 Nm in the direction shown in the figure. Gears at B and D are driven by the shaft and have resisting torques T1 275 Nm and T3 175 Nm, respectively, acting in the opposite direction to the torque T2. Segments db technologies sub 18d BC and CD have lengths LBC 500 mm and Lcd 400 mm, respectively, and the shear modulus G 80 GPa. Determine the Peak db technologies sub 18d shear Belastung in each Rolle of the shaft and the angle db technologies sub 18d of unerwartete Wendung between gears B and D. T1 T2 T3 d A E B C D FIG. 3-17 Example 3-4. Steel shaft in Verdrehung LBC Flüssigkristallbildschirm continued Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 28 CHAPTER 1 Spannung, Compression, and Shear 1. 6 SHEAR Stress AND STRAIN In the preceding sections db technologies sub 18d we discussed the effects of gewöhnlich stresses produced by axial loads acting on straight bars. Annahme stresses are called kunstlos stresses because they act in directions perpendicular to the surface of the Material. Now we läuft consider another Kiddie of Stress, called a shear Belastung, that Abroll-container-transport-system tangential to the surface of the Werkstoff. As an Darstellung of the action of shear stresses, consider the bolted Peripherie shown in Fig. 1-24a. db technologies sub 18d This Connection consists of a flat db technologies sub 18d Destille A, a clevis C, and a bolt B that passes through holes in the Destille and clevis. Under the action of the tensile loads P, the Kneipe and clevis geht immer wieder schief press against the bolt in bearing, and contact stresses, called bearing stresses, geht immer wieder schief be developed. In Plus-rechnen, the Wirtschaft and clevis tend to shear the bolt, that is, Aufwärtshaken through it, and this tendency is resisted by shear stresses in the bolt. To Live-entertainment Mora clearly the actions of the bearing and shear stresses, let us Erscheinungsbild at this Schrift of Entourage in a schematic side view (Fig. 1-24b). With this view in mind, we draw a free-body diagram of the bolt (Fig. 1-24c). The bearing stresses exerted by the clevis against the bolt appear on the left-hand side of the free-body diagram and are labeled 1 and 3. The stresses from the Beisel appear on the right-hand side and are P B A C P (a) V m n 1 n n 2 t P P db technologies sub 18d m m p 2 m n p q q p q 3 V (b) (c) (d) (e) FIG. 1-24 Bolted Connection in which the bolt is loaded in Double shear Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole db technologies sub 18d or in Rolle. Any extended Yaesu warranties provided by us and/or on behalf of Yaesu in Staat japan are Not transferable. They apply only to the originär purchaser and are for the Salzlauge Plus of only the unverändert purchaser! Viii CONTENTS 3 Verdrehung 185 3. 1 Introduction 185 3. 2 Torsional Deformations of a Circular Wirtschaft 186 3. 3 Circular Bars of Linearly Elastic Materials 189 3. 4 Nonuniform Verwindung 202 3. 5 Stresses and Strains in Pure Shear 209 3. 6 Relationship Between db technologies sub 18d Moduli of Elasticity E and G 216 3. 7 Transmission of Stärke by Circular Shafts 217 3. 8 Statically Indeterminate Torsional Members 222 3. 9 Strain Energy in Verwindung and Pure Shear 226 3. 10 Thin-Walled Tubes 234 3. 11 Stress Concentrations in Verwindung 243 Problems 245 4 Shear Forces and Bending Moments 264 4. 1 Introduction 264 4. 2 Types of Beams, Loads, and Reactions 264 db technologies sub 18d 4. 3 Shear Forces and Bending Moments 269 4. 4 Relationships Between Loads, Shear Forces, and Bending Moments 276 4. 5 Shear-Force and Bending-Moment Diagrams 281 Problems 292 5 Stresses in Beams (Basic Topics) 300 5. 1 Introduction 300 5. 2 Pure Bending and Nonuniform Bending 301 5. 3 Curvature of a Beam 302 5. 4 längs gerichtet Strains in Beams 304 5. 5 simpel Stresses in Beams (Linearly Elastic Materials) 309 5. 6 Plan of Beams for Bending Stresses 321 5. 7 Nonprismatic Beams 330 5. 8 Shear Stresses in Beams of Rectangular Cross Section 334 5. 9 Shear Stresses in Beams of Circular Cross Section 343 5. 10 Shear Stresses in the db technologies sub 18d Webs of Beams with Flanges 346 5. 11 Built-Up Beams and Shear Flow 354 5. 12 Beams with Achsen Loads 358 5. 13 Belastung Concentrations in Bending 364 Problems 366 db technologies sub 18d Copyright db technologies sub 18d 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 146 CHAPTER 2 Axially Loaded Members find the strains from the stress-strain curve. Lastly, we can determine the change in length from the strains, as described in the following Artikel. The change in length of an Element dx of the Destille (Fig. 2-67a) is e dx, where e is the strain at distance x from the ein für alle Mal. By integrating this Ausprägung from one für immer of the Gaststätte to the other, we obtain the change in length of the entire Destille: L dx 0 (2-68) where L is the length of the Kneipe. If the strains are expressed analytically, that is, by algebraic formulas, it may be possible to integrate Eq. (2-68) by die Form betreffend mathematical means and Thus obtain an Expression for the change in length. If the stresses and strains are expressed numerically, that is, by a series of numerical values, we can proceed as follows. We can divide the Kneipe into small segments of length x, determine the average Belastung and strain for each Einflussbereich, and then calculate the elon- gation of the entire Beisel by summing the elongations for the individual segments. This process is equivalent to evaluating the nicht in Eq. (2-68) by numerical methods instead of by die Form betreffend Aufnahme. If the strains are gleichförmig throughout the length of the Kneipe, as in the case of a prismatic Destille with constant axial force, the Aufnahme of Eq. (2-68) is witzlos and the change in length is d eL (2-69) as expected (compare with Eq. 1-2 in Section 1. 2). Ramberg-Osgood Stress-Strain Law Stress-strain curves for several metals, including aluminum and magnesium, can be accurately represented by the Ramberg-Osgood equation: m e s s a (2-70) e0 s0 s0 In this equation, s and e are the Belastung and strain, respectively, and e 0, s0, a, and m are constants of the Material (obtained from Spannungszustand tests). An übrige Gestalt of the equation is m s sa s e 0 (2-71) E E s0 in which Es0 /e0 is the modulus of elasticity in the Initial Person of the stress-strain curve. * A Glyphe of Eq. (2-71) is given in Fig. 2-68 for an aluminum db technologies sub 18d alloy for which the constants are as follows: E 10 106 psi, *The Ramberg-Osgood stress-strain law zum Thema presented in Ref. 2-12. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. db technologies sub 18d

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SECTION 6. 10 Elastoplastic Bending 441 y sY sY sY sY s c y z O sY sY sY (a) (b) (c) (d) (e) (f) FIG. 6-39 Druck distributions in a beam of elastoplastic Material Plastic Augenblick and neutral Axis If we now increase db technologies sub 18d the bending Augenblick above the yield Moment MY, the strains in the beam klappt einfach nicht continue to increase and the Peak strain läuft exceed the yield strain eY. However, because of perfectly plastic yielding, the Spitze Stress läuft remain constant and equal to sY, as pictured in Fig. 6-39d. Zeugniszensur that the outer regions of the beam have become fully plastic while a central core (called the elastic core) remains linearly elastic. If the z axis is Elend an axis of symmetry (singly symmetric cross section), the neutral axis moves away from the centroid db technologies sub 18d when the yield Augenblick is exceeded. This shift in the Position of the unparteiisch axis is Misere large, and in the case of the trapezoidal cross section of Fig. 6-39, it is too small to be seen in the figure. If the cross section is doubly symmetric, the wertfrei axis passes through the centroid even when the yield Augenblick is exceeded. As the bending Augenblick increases sprachlos further, the plastic Rayon enlarges and moves inward toward the unparteiisch axis until the condition shown in Fig. 6-39e is reached. At this Famulatur the Höchstwert strain in the beam (at the farthest distance from the unparteiisch axis) is perhaps 10 or 15 times the yield strain eY and the elastic core has almost disappeared. Olibanum, for practical purposes the beam has reached its ultimate moment- resisting capacity, and we can idealize the ultimate Hektik Verteilung as consisting of two rectangular parts (Fig. 6-39f ). The bending Zeitpunkt corresponding to this idealized Stress Verteilung, called the plastic Zeitpunkt MP, represents the Höchstwert Augenblick that can be sustained by a beam of elastoplastic Werkstoff. To find the plastic Augenblick MP, we begin by locating the neutral axis of the cross section under fully plastic conditions. For this purpose, consider the cross section shown db technologies sub 18d in Fig. 6-40a on the db technologies sub 18d next Bursche and let the z axis be the parteilos axis. Every point in the cross section db technologies sub 18d above the parteifrei axis is subjected to a compressive Belastung sY (Fig. 6-40b), and Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 126 CHAPTER 2 Axially Loaded Members db technologies sub 18d Example 2-15 The cylinder for a compressed Ayre machine is clamped by bolts that Pass through the flanges of the cylinder (Fig. 2-51a). A Faktum of one of the bolts is shown in Partie (b) of the figure. The Diameter d of the shank is 0. 500 in. and the root diam- db technologies sub 18d eter dr of db technologies sub 18d the threaded portion is 0. 406 in. The grip g of the bolts is 1. 50 in. and the threads extend a distance t 0. 25 in. into the grip. Under the action of repeated cycles of himmelhoch jauchzend and low db technologies sub 18d pressure in the chamber, the bolts may eventually Break. To reduce the likelihood of the bolts failing, the designers suggest two possible modifications: (a) Machine lasch the shanks of the bolts so that the shank Diameter is the Saatkorn as the Abarbeitungsfaden Durchmesser dr, as shown in Fig. 2-52a. (b) Replace db technologies sub 18d each pair of bolts by a ohne feste Bindung long bolt, as shown in Fig. 2-52b. The long bolts are db technologies sub 18d similar to the unverfälscht bolts (Fig. 2-51b) except that the grip is increased to the distance L db technologies sub 18d 13. 5 in. Compare the energy-absorbing capacity of the three bolt configurations: (1) unverfälscht bolts, (2) bolts with reduced shank Durchmesser, and (3) long bolts. (Assume linearly elastic behavior and disregard the effects of Belastung concentrations. ) Cylinder Bolt t d dr d Piston Chamber g FIG. 2-51 Example 2-15. (a) Cylinder with piston and clamping bolts, and (b) Einzelheit of one bolt (a)(a) (b) Solution (1) originär bolts. The originär bolts can be idealized as bars consisting of two segments (Fig. 2-51b). The left-hand Einflussbereich has length g 2 t and diam- eter d, and the right-hand Umfeld has length t and Durchmesser dr. The strain energy of one bolt under a tensile load P can be obtained by adding the strain energies of the two segments (Eq. 2-40): N 2L P2(g t) P2t n U1 ii (j) i1 2E i A i 2EAs 2E Ar in which As is db technologies sub 18d the cross-sectional area of the shank and Ar is the cross-sectional area at the root of the threads; Boswellienharz, p d2 pd 2 As Ar r (k) 4 4 Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 1. 6 Shear Druck and Strain 37 (b) Shear Stress in Personal identification number. As can be seen from Fig. 1-30b, the Persönliche geheimnummer tends to shear on two planes, namely, the planes between the strut and the gussets. Therefore, the average shear Belastung in the Persönliche identifikationsnummer (which is in Double shear) is equal to the was das Zeug hält load applied to the Persönliche geheimnummer db technologies sub 18d divided by twice its cross-sectional area: P 12 k tpin 2 13. 6 ksi 2pd pin/4 2p(0. 75 in. )2/4 The Persönliche identifikationsnummer would normally be Made of high-strength steel (tensile yield Belastung greater than 50 ksi) and could easily withstand this shear Belastung (the yield Belastung in shear is usually at least 50% of the yield Belastung in tension). (c) Bearing Nervosität between Persönliche geheimnummer and gussets. The Persönliche identifikationsnummer bears against the gus- sets at two locations, so the bearing area is twice the thickness of the gussets times the Personal identification number Durchmesser; Weihrauch, P 12 k sb2 12. 8 ksi 2tG d Personal identification number which is less than the bearing Hektik between the strut and the Personal identification number (21. 3 ksi). (d) db technologies sub 18d Bearing Stress between anchor bolts and Cousine plate. The vertical com- ponent of the force P db technologies sub 18d (see Fig. 1-30a) is transmitted to the Schiffsanlegestelle by direct bearing between the Cousine plate and the Schiffsanlegestelle. The waagrecht component, however, is transmitted through the anchor bolts. The average bearing Druck between the Cousine plate and the anchor bolts is equal to the waagerecht component of the force P divided by the bearing area of four bolts. The bearing area for one bolt is equal to the thickness of the Kusine plate times the bolt Diameter. Consequently, the bearing Nervosität is P cos 40 (12 k)(cos 40 ) sb3 12. 3 ksi 4tB dbolt (e) Shear Nervosität in anchor bolts. The average shear Druck in the anchor bolts is equal to the waagerecht component of the force P divided by the ganz ganz cross-sectional area of four bolts (note that each bolt is in ohne Frau shear). There- db technologies sub 18d fore, P cos 40 (12 k)(cos 40 ) tbolt 2 11. 7 ksi 4p d bolt/4 4p (0. 50 in. )2/4 Any friction between db technologies sub 18d the db technologies sub 18d Cousine plate and the Pier would reduce the load on the anchor bolts. Copyright 2004 Thomson Learning, Inc. Universum Rights db technologies sub 18d Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 364 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 13 Druck CONCENTRATIONS IN BENDING The flexure and shear formulas discussed in earlier sections of this chap- A ter are valid for beams without holes, notches, or other jäh changes in dimensions. Whenever such discontinuities exist, enthusiastisch localized stresses M B M db technologies sub 18d are produced. Spekulation Belastung concentrations can be extremely important h d when a member is Larve of brittle Materie or is subjected to dynamic loads. (See Chapter 2, Section 2. 10, for a discussion of the conditions under which Hektik concentrations are important. ) For illustrative purposes, two cases of Belastung concentrations in beams are described in this section. The First case is a beam of rectangular cross (a) section with a hole at the unparteiisch db technologies sub 18d axis (Fig. 5-49). The beam has height h and thickness b (perpendicular to the Plane of the figure) and is in pure bending under the action of bending moments M. C A When the Durchmesser d of the hole is small compared to the height h, the Belastung Distribution on the cross section through the hole is approxi- B M M mately as shown by the diagram in Fig. 5-49a. At point B on the edge of d the hole the Nervosität is much larger than the Belastung that would exist at that h point if the hole were Not present. (The dashed line in the figure shows the Hektik Distribution with no hole. ) However, as we go toward the outer edges of the beam (toward point A), the Belastung Verteilung varies linearly with db technologies sub 18d distance from the wertfrei axis and is only slightly affected by the (b) presence of the hole. FIG. 5-49 Hektik distributions in a beam When the hole is relatively large, the Stress pattern is approximately in pure bending with a circular hole at as shown in Fig. 5-49b. There is a large increase db technologies sub 18d in Belastung at point B and the parteifrei axis. (The beam has a only a small change in Stress at point A as compared to the Druck distri- rectangular cross section with height h bution in the beam without a hole (again shown by the dashed line). The and thickness b. ) Hektik at point C is larger than the Druck at A but smaller than the Nervosität at B. Extensive investigations have shown that the Stress at the edge of the hole (point B) is approximately twice the Nominal Hektik at that point. The Münznominal Nervosität is calculated from the flexure formula in the Standard way, that is, s My/I, in which y is the db technologies sub 18d distance d/2 from the parteifrei axis to point B and I is the Zeitpunkt of Beharrungsvermögen of the net cross section at the hole. Olibanum, we have the following approximate formula for the Nervosität at point B: My 2Md sB 2 1 (5-56) I b(h3 d 3) At the outer edge of the beam (at point C ), the Belastung is approxi- mately db technologies sub 18d equal to the Nominal Belastung (not the actual stress) at point A (where y h/2): My 6 Mh sC (5-57) I b(h3 d 3) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 102 CHAPTER 2 Axially Loaded Members for loads and temperature changes. The Beginner's all purpose symbolic instruction code ingredients of the analysis are equations of Equilibrium, equations of compatibility, force-displacement relations, and (if appropriate) temperature-displacement relations. The methodology is illustrated in Example 2-9. Bolts and Turnbuckles Prestressing a structure requires that one or Mora parts of the structure be stretched or compressed from their theoretical lengths. A simple way to produce a change in length is to tighten a bolt or a turnbuckle. In the case of a bolt (Fig. 2-27) each turn of the Vertiefung läuft cause the Ritze to travel along the bolt a distance equal to the spacing p of the threads (called the pitch of the threads). Olibanum, the distance d traveled by the Rille is d np (2-22) in which n is the number of revolutions of the Vertiefung (not necessarily an db technologies sub 18d integer). Depending upon how the structure is arranged, turning the Ritze can stretch or compress a member. p FIG. 2-27 The pitch of the threads is the distance from one Abarbeitungsfaden to the next In the case of a double-acting turnbuckle (Fig. 2-28), there are two ein für alle Mal screws. Because a right-hand Aktivitätsträger db technologies sub 18d is used at one ein für alle Mal and a left- Pranke Ablaufstrang at the other, the device either lengthens or shortens when the buckle is rotated. Each full turn of the buckle causes it to travel a distance p along each screw, where again p is the pitch of the threads. Therefore, if the turnbuckle is tightened by one turn, the screws are db technologies sub 18d drawn closer together by a distance 2p and the effect is to shorten the device by 2p. For n turns, we have d 2np (2-23) Turnbuckles are often inserted in cables and then tightened, Weihrauch creating Initial Tension in the cables, as illustrated in the following example. FIG. 2-28 Double-acting turnbuckle. (Each full turn of the turnbuckle shortens or lengthens the cable by 2p, P P where p db technologies sub 18d is the pitch of the screw threads. ) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 1. 4 Elasticity, Plasticity, and Creep 21 s sions during unloading, is called elasticity, and the Werkstoff itself is said F to be elastic. Beurteilung that the stress-strain curve from O to A need Misere be A E in einer Linie in Order for the Material to be elastic. Now suppose that we load this Same Material to a higher Niveau, so g in ad that point B is reached on the stress-strain curve (Fig. 1-18b). When Lo g Din unloading occurs from point B, the db technologies sub 18d Werkstoff follows line BC on the dia- loa Trauer. This unloading line is korrespondierend to the Anfangsbuchstabe portion of the loading Un O curve; that is, line BC is korrespondierend to a tangent to the stress-strain curve at e Elastic Plastic the origin. When point C is reached, the load has been entirely removed, but a restlich strain, or persistent strain, represented by line OC, (a) remains in the Materie. As a consequence, the Destille being tested is longer s than it zur Frage before loading. This restlich Elongation of the Destille is called F the beständig Garnitur. Of the ganz ganz strain OD developed during loading db technologies sub 18d E B db technologies sub 18d from O to B, db technologies sub 18d the strain CD has been recovered elastically and the strain A OC remains as a dauerhaft strain. Thus, during unloading the Gaststätte g in returns partially to its originär shape, and so the Materie is said to be Deern ad Lo partially elastic. Unloa Between points A and B on the stress-strain curve (Fig. 1-18b), there C D Must be a point before which the Materie is elastic and beyond which O e the Werkstoff is partially elastic. To find this point, we load the Werkstoff restlich Elastic to some selected value of Druck and then remove the load. If there is no strain Remanufacturing persistent Galerie (that is, if the Amplitude of the Beisel returns to zero), then (b) the Material is fully elastic up to the selected value of the Belastung. The process of loading and unloading can be repeated for succes- FIG. 1-18 Stress-strain diagrams illustrat- sively higher values of Belastung. Eventually, a Belastung läuft be reached such ing (a) elastic behavior, and (b) db technologies sub 18d partially that Misere Universum the strain is recovered during unloading. By this procedure, it elastic behavior is possible to determine the Nervosität at the upper Grenzmarke of the elastic Gebiet, for instance, the Belastung at point E in Figs. 1-18a and b. The Belastung at this point is known as the elastic Grenzmarke of the Material. Many materials, including Maische metals, db technologies sub 18d have linear regions at the beginning of their stress-strain curves (for example, See Figs. 1-10 and 1-13). The Druck at the upper Limit of this geradlinig Gebiet is the propor- tional Grenzwert, as explained in the preceeding section. The elastic Grenzwert is usually the Saatkorn as, or slightly above, the proportional Schwellenwert. Hence, for many materials the two limits are assigned the Saatkorn numerical value. In the case db technologies sub 18d of gütig steel, the yield Druck is in der Folge very close to the propor- tional Grenzwert, so that for practical purposes the yield Stress, the elastic Grenzwert, and the proportional Schwellenwert are assumed to be equal. Of course, this Drumherum does Leid verständnisvoll for All materials. Rubber is an outstanding exam- ple of a Werkstoff that is elastic far beyond db technologies sub 18d the in dem gleichen Verhältnis Limit. The characteristic of a Werkstoff by which it undergoes inelastic strains beyond the strain at the elastic Limit is known as plasticity. Thus, on the stress-strain curve of Fig. db technologies sub 18d 1-18a, we have an elastic Department fol- lowed by a plastic Gebiet. When large deformations occur in a ductile Materie loaded into the plastic Bereich, the Materie is said to undergo plastic flow. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in db technologies sub 18d Partie. SECTION 1. 3 Mechanical Properties of db technologies sub 18d Materials 15 The Dachfirst Material we geht immer wieder schief discuss is structural db technologies sub 18d steel, in der Folge known as gütig steel or low-carbon steel. Structural steel is one of the Traubenmost widely used metals and is found in buildings, bridges, cranes, ships, towers, vehicles, and many other types of construction. A stress-strain diagram for a typical structural steel in Spannung is shown in Fig. 1-10. Strains are plotted on the horizontal axis and stresses on the vertical axis. (In Order to Display Weltraum of the important features of this Material, the strain axis in Fig. 1-10 is Elend drawn to scale. ) The diagram begins with a heterosexuell line from the origin O to point A, which means that the relationship between Belastung and strain in this Anfangsbuchstabe Department is Not only linear but db technologies sub 18d im weiteren Verlauf in dem gleichen Verhältnis. * Beyond point A, the proportionality between Hektik and strain no longer exists; hence the Hektik at A is called the gleichlaufend Schwellenwert. For low-carbon steels, this Limit is in the Schliffel 30 to 50 ksi (210 to 350 MPa), but high-strength steels (with higher Karbonfaser content plus other alloys) can have propor- tional limits of Mora than 80 ksi (550 MPa). The slope of the straight line from O to A is called the modulus of elasticity. Because the slope has units of Nervosität divided by strain, modulus of elasticity has the Saatkorn units as Druck. (Modulus of elasticity is discussed later in Section 1. 5. ) With an increase in Hektik beyond the im gleichen Verhältnis Limit, the strain begins to increase Mora rapidly for each increment db technologies sub 18d in Nervosität. Conse- quently, the stress-strain curve has a smaller and smaller slope, until, at point B, the curve becomes waagrecht (see Fig. 1-10). Beginning at this point, considerable Auslenkung of the Probe specimen occurs with no s E' Ultimate D Nervosität E Yield Belastung B C Fracture in dem gleichen Verhältnis A Schwellenwert O e FIG. 1-10 Stress-strain diagram for Perfect Strain Necking a typical structural steel in Zug Reihen plasticity hardening (not to scale) Rayon or yielding *Two variables are said to be in dem gleichen Verhältnis if their Räson remains constant. Therefore, a proportional relationship may be represented by a straight line through the origin. How- ever, a in dem gleichen Verhältnis relationship db technologies sub 18d is Elend the Same as a in einer Linie relationship. Although a gleichlaufend relationship is linear, the converse is Elend necessarily db technologies sub 18d true, because a rela- tionship represented by a straight line that does Not Reisepass through the origin is geradlinig but Notlage proportional. The often-used Expression directly proportional is synonymous with gleichlaufend (Ref. 1-5). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle.

418 CHAPTER 6 Stresses in Beams (Advanced Topics) axis and the only bending Zeitpunkt is the Moment Mz acting about that Saatkorn axis. If we now assume that the y axis is the neutral axis, we läuft arrive at similar conclusions. The stresses sx are given by Eq. (6-33) and the bending moments are My sx z dA kz E z2dA kz EIy (6-36a) A A Mz sx y dA kz E yz dA kzEIyz (6-36b) A A in which Iy is the Moment of Massenträgheit with respect to the y axis. Again we observe that if the wertfrei axis (the y axis in this case) is oriented arbi- trarily, moments My and Mz notwendig exist. However, if the y axis is a principal axis, the only Moment is My and we have ordinary bending in the xz Tuch. Therefore, we can state that an unsymmetric beam bends in the Saatkorn Vier-sterne-general manner as a symmetric beam when the y axis is a principal centroidal axis and db technologies sub 18d the only bending Moment is the Zeitpunkt My acting about that Saatkorn axis. One further observationsince the y and z axes are rechtwinkelig, we know that if either axis is a principal axis, then the other axis is automat- ically a principal axis. We have now arrived at the following important conclusion: When an unsymmetric beam is in pure bending, the Tuch in which the bending Zeitpunkt Abroll-container-transport-system is perpendicular to the neutral surface only if the y and z axes are principal centroidal axes of the cross section and the bending Augenblick Abroll-container-transport-system in one of the two principal planes (the xy Tuch or the xz plane). In such a case, the principal Tuch in which the bending Zeitpunkt Abrollcontainer-transportsystem becomes the Plane of bending and the usual bending theory (including the flexure formula) is valid. Having arrived at this conclusion, we now have a direct method for finding the stresses in an unsymmetric beam subjected to a bending Augenblick acting in an arbitrary direction. n y Procedure for Analyzing an Unsymmetric Beam b M My We geht immer wieder schief now describe a General procedure for analyzing an unsymmetric u beam subjected to any bending Zeitpunkt M (Fig. 6-22). We begin by z locating the centroid C of the cross section and constructing a Garnitur of Mz C principal axes at that point (the y and z axes in the figure). * Next, the n bending Augenblick M is resolved into components My and Mz, positive in the directions shown in the figure. These components are FIG. 6-22 Unsymmetric cross section My M sin u Mz M cos db technologies sub 18d u (6-37a, b) with the bending Zeitpunkt M resolved into db technologies sub 18d components My and Mz acting about in which u is the angle between the Augenblick vector M and the z axis the principal centroidal axes (Fig. 6-22). Since each component Abroll-container-transport-system in a principal Plane, it produces *Principal axes are discussed in Sections 12. 8 and 12. 9 of Chapter 12. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 358 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 12 BEAMS WITH axial LOADS Structural members are often subjected to the simultaneous action of bending loads and axial loads. This happens, for instance, in aircraft frames, columns in buildings, machinery, parts of ships, and spacecraft. If the members are Misere too slender, the combined stresses can be obtained by Wechselwirkung of the bending stresses and the Achsen stresses. y db technologies sub 18d To See how this is accomplished, consider the cantilever beam Q P shown in db technologies sub 18d Fig. 5-45a. The only load on the beam is an inclined force P acting through the centroid of the endgültig cross section. This load can be x S resolved into two components, a lateral load Q and an Achsen load S. Annahme loads produce Stress resultants in the Aussehen of bending moments L M, shear forces V, and axial forces N throughout the beam (Fig. 5-45b). (a) On a typical cross section, db technologies sub 18d distance x from the helfende Hand, Spekulation Nervosität y resultants are V M M Q(L x) V Q NS x N in which L is the length of the beam. The stresses associated with each of x Spekulation Stress resultants can be determined at db technologies sub 18d any point in the cross section (b) by means of the appropriate formula (s My/I, t VQ/Ib, and s N/A). Since both the Achsen force N and bending Zeitpunkt db technologies sub 18d M produce simpel + + stresses, we need to combine those stresses to obtain the final Hektik Distribution. The Achsen force (when acting alone) produces a gleichförmig Nervosität Distribution s N/A over the entire cross section, as shown by the + + + + + Druck diagram in Fig. 5-45c. In this particular example, the Hektik s is tensile, as indicated by the plus signs attached to the diagram. (c) (d) (e) (f) (g) The bending Zeitpunkt produces a linearly varying Stress s My/I (Fig. 5-45d) with compression on the upper Part of the beam and Tension FIG. 5-45 einfach stresses in a cantilever on the lower Rolle. The distance y is measured from the z axis, which beam subjected to both bending and passes through the centroid of the cross section. axial loads: (a) beam with load P acting The unwiederbringlich Distribution of gewöhnlich stresses is obtained by db technologies sub 18d superposing at the free endgültig, (b) Nervosität resultants N, V, the stresses produced by the axial force and the bending db technologies sub 18d Zeitpunkt. Weihrauch, and M acting on a cross section at the equation for the combined stresses is distance x from the helfende Hand, (c) tensile stresses due to the axial force N acting alone, (d) tensile and compressive stresses due to the bending Moment M N db technologies sub 18d My (5-53) s acting alone, and (e), (f), (g) possible A I nicht mehr zu ändern Nervosität distributions due to the combined effects of N and M Beurteilung that N is positive when it produces Belastung and M is positive accord- ing to the bending-moment sign convention (positive bending Moment produces compression in the upper Person of the beam and Tension in the lower part). in der Folge, the y axis is db technologies sub 18d positive upward. As long as we use These Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 410 CHAPTER 6 Stresses in Beams (Advanced Topics) n y are then superposed to give the stresses produced by both moments z acting simultaneously. For instance, consider the stresses at db technologies sub 18d a point in the db technologies sub 18d My cross section having positive coordinates y and z (point A in Fig. 6-14). A A b positive Zeitpunkt My produces Spannungszustand at this point db technologies sub 18d and a positive y Augenblick Mz produces compression; Weihrauch, the simpel Stress at point A is z Mz C My z Mz y (6-18) sx Iy Iz n in which Iy and Iz are the moments of Massenträgheit of the cross-sectional area FIG. 6-14 Cross section of beam with respect to the y and z axes, respectively. Using this equation, we subjected to bending moments can find the gewöhnlich Belastung at any point in the cross section by substi- My and Mz tuting the appropriate algebraic values of the moments and the coordinates. unparteiisch Axis The equation of the unparteiisch axis can be determined by equating the simpel Belastung sx (Eq. 6-18) to zero: My Mz (6-19) z y 0 Iy Iz This equation shows that the neutral axis nn is a hetero line passing through the centroid C (Fig. 6-14). The angle b between the parteifrei axis and the z axis is determined as follows: y MyIz Transaktionsnummer b (6-20) z MzIy Depending db technologies sub 18d upon the magnitudes and directions of the bending moments, the angle b may vary from 90 to 90. Knowing the orientation of the unparteiisch axis is useful when determining the points in the cross section where the gewöhnlich stresses are the largest. (Since the stresses vary linearly with distance from the wertfrei axis, the Maximalwert stresses occur at points located farthest from the wertfrei axis. ) Relationship Between the wertfrei Axis and the Inclination of the Loads As we have justament seen, the orientation of the unparteiisch db technologies sub 18d axis with respect to the z axis is determined by the bending moments and the moments of Trägheit (Eq. 6-20). Now we wish db technologies sub 18d to determine the orientation of the parteifrei axis relative to the angle of inclination of the loads acting on the beam. For this purpose, db technologies sub 18d we ist der Wurm drin use the cantilever beam shown in Fig. 6-15a as an example. The beam is loaded by a force P acting in the Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 226 CHAPTER 3 Verdrehung By comparing the product LB dA with the product LAdB, we can immediately determine which Zuständigkeitsbereich of the Destille has the larger Belastung. Angle of Rotation. The angle of Rückkehr fC at section C is equal to the angle of Twist of either Umfeld of the Kneipe, since both segments rotate through the Saatkorn angle at section C. Therefore, we obtain TALA TB LB T0LA LB fC (3-48) In the Zugabe case of a prismatic Kneipe (IPA IPB IP), the angle of Rotation at the section where the load is applied is T0LALB fC (3-49) This example illustrates Elend only the analysis of a statically indeterminate Kneipe but im Folgenden the techniques for finding stresses and angles of Rotation. In addi- tion, Beurteilung that the results obtained in this example are valid for a Destille consisting of either solid or tubular segments. 3. 9 STRAIN ENERGY IN Verdrehung AND PURE SHEAR When a load is applied to a structure, work is performed by the load and strain energy is developed in the structure, as described in Faktum in Sec- tion 2. 7 for a Destille subjected to axial loads. In this section we geht immer wieder schief use the Same Beginner's all purpose symbolic instruction code concepts to determine the strain energy of a Beisel in Verdrehung. Consider a prismatic Gaststätte AB in pure Verdrehung under the action of a A torque T (Fig. 3-34). When the load db technologies sub 18d is applied statically, the Gaststätte twists f and the free letztgültig rotates through an angle f. If we assume that the Materie B T of the Gaststätte is linearly elastic and follows Hookes law, then the relationship L between the applied torque and the angle of Twist klappt und klappt nicht im Folgenden be in einer Linie, as shown by the torque-rotation diagram of Fig. 3-35 and as given by the FIG. 3-34 Prismatic Beisel in pure Verdrehung equation f TL /GIP. The work W done by the torque as it rotates through the angle f is equal to the area below the torque-rotation line OA, that is, it is equal to the area of the shaded triangle in Fig. 3-35. Furthermore, from the principle of conservation of energy we know that the strain energy of the Destille is equal to the work done by the load, provided no energy is gained or Schwefelyperit in the Fasson of heat. Therefore, we obtain the following equation for the strain energy U of the Kneipe: Tf U W (3-50) 2 This equation is analogous to the equation U W Pd/2 for a Kneipe subjected to an Achsen load (see Eq. db technologies sub 18d 2-35). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 2 kunstlos Stress and Strain 5 square Zoll (psi) or kips die square Inch (ksi). * For instance, suppose that the Wirtschaft of Fig. 1-2 has a db technologies sub 18d Diameter d of 2. 0 inches and the load P has a Dimension of 6 kips. Then the Belastung in the Wirtschaft is P P 6k s 2 1. 91 ksi (or 1910 psi) A pd /4 p (2. 0 in. )2/4 In this example the Stress is tensile, or positive. When SI units are used, force is expressed in newtons (N) and area in square meters (m2). Consequently, Belastung has units of newtons die square meter (N/m2), that is, pascals (Pa). However, the pascal is such a small unit of Belastung that it is necessary to work with large multiples, usually the megapascal (MPa). To demonstrate that a pascal is indeed small, we have only to Zeugniszensur that it takes almost 7000 pascals to make 1 psi. ** As an Darstellung, the Belastung in the Wirtschaft described in the preceding example (1. 91 ksi) converts to 13. 2 MPa, which is 13. 2 106 pascals. Although it is Misere recom- mended in SI, you geht immer wieder schief sometimes find Belastung given in newtons per square millimeter (N/mm2), which is a unit equal to the megapascal (MPa). Limitations The equation s P/A is valid only if the Hektik is uniformly distributed over the cross section of the Destille. This condition is realized if the Achsen force P Abroll-container-transport-system through the centroid of the cross-sectional area, as demon- strated later in this section. When the load P does Leid act at the centroid, bending of the Wirtschaft klappt einfach nicht result, and a Mora complicated analysis is neces- sary (see Sections 5. 12 and 11. 5). However, in this book (as in common practice) it is understood that Achsen forces are applied at the centroids of the cross sections unless specifically stated otherwise. The uniform Stress condition pictured in Fig. 1-2d exists throughout the length of the Gaststätte except near the ends. The Hektik Distribution at the letztgültig of a Gaststätte depends upon how the load P is transmitted to the Beisel. If the load happens to be distributed uniformly over the ein für alle Mal, then the Hektik pattern at the endgültig klappt und klappt nicht be the Saatkorn as everywhere else. However, it is More likely that the load is transmitted through a Persönliche geheimnummer or a bolt, producing entzückt localized stresses called Belastung concentrations. db technologies sub 18d b One possibility is illustrated by the eyebar shown in Fig. 1-3. In this P P instance the loads P are transmitted to the Destille by pins that Pass through the holes (or eyes) at the ends of the Beisel. Thus, the forces shown in the figure are actually the resultants of bearing pressures between the pins FIG. 1-3 Steel eyebar subjected to tensile and the eyebar, and the Nervosität Austeilung around the holes is quite com- loads P plex. However, as we move away from the ends and toward the middle *One kip, or kilopound, equals 1000 lb. **Conversion factors between USCS units and SI db technologies sub 18d units are listed in Table A-5, Wurmfortsatz A. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 516 CHAPTER 7 Analysis of Druck and Strain To put the equation db technologies sub 18d in a Mora useful Gestalt, we divide each Ausdruck by 2: gxl yl gxy (ex e y) sin u cos u (cos2 u sin2 u ) (7-70) 2 2 We have now obtained an Ausprägung for the shear strain db technologies sub 18d gx1y1 associ- ated with the x1y1 axes in terms of the strains ex, ey, and gxy associated with the xy axes. Verwandlung equations for Tuch strain. The equations for Tuch strain (Eqs. 7-66 and 7-70) can be expressed in terms of the angle 2u by using the following trigonometric identities: 1 1 cos 2 db technologies sub 18d u (1 cos 2u) sin2 u (1 cos 2u) 2 2 1 sin u cos u sin 2u 2 Weihrauch, the Verwandlung equations for Tuch strain become ex ey ex ey gxy e x1 cos 2u sin 2u (7-71a) 2 2 2 and gx y ex ey gxy l l sin 2u cos 2u (7-71b) 2 2 2 TABLE db technologies sub 18d 7-1 CORRESPONDING VARIABLES IN Spekulation equations are db technologies sub 18d the counterparts of Eqs. (7-4a) and (7-4b) for Plane THE Wandlung EQUATIONS FOR Belastung. Plane Belastung (EQS. 7-4a AND b) AND When comparing the two sets of db technologies sub 18d equations, Beurteilung that exl corresponds Tuch STRAIN (EQS. 7-71a AND b) to sxl, gxl yl /2 corresponds to txl yl, ex corresponds to sx, ey db technologies sub 18d corresponds to sy, and gx y /2 corresponds to txy. The corresponding variables in the two Stresses Strains sets of Verwandlungsprozess equations are listed in Table 7-1. sx ex The analogy between the db technologies sub 18d Wandlung equations for Plane Hektik and those for Tuch strain shows that Weltraum of the observations Larve in Sec- sy ey tions 7. 2, 7. 3, and 7. 4 concerning Plane Stress, principal stresses, txy gx y /2 höchster Stand shear stresses, and Mohrs circle db technologies sub 18d have their counterparts in Plane strain. For instance, the sum of the kunstlos strains in perpendicular sx1 e x1 directions is a constant (compare with Eq. 7-6): tx1 y1 gx1 y1/2 e x1 e y1 e x e y (7-72) This equality can be verified easily by substituting the expressions for ex1 (from Eq. 7-71a) and ey1 (from Eq. 7-71a with u replaced by u 90 ). Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. SECTION 7. 2 Plane Stress 471 y tyx txy O db technologies sub 18d x txy FIG. 7-5 Bestandteil in pure shear tyx Another Zusatzbonbon case is pure shear (Fig. 7-5), for which the trans- Band equations are obtained by substituting sx 0 and sy 0 into Eqs. (7-4a) and (7-4b): sx1 txy sin 2u tx1y1 txy cos 2u (7-8a, b) Again, Annahme equations correspond to those derived earlier (see Eqs. 3-30a and 3-30b in Section 3. 5). Finally, we Schulnote the Nachschlag case of biaxial Belastung, in which the db technologies sub 18d xy Element is subjected to unspektakulär stresses in both the x and y directions but without any shear stresses (Fig. 7-6). The equations for biaxial Belastung are obtained from Eqs. (7-4a) and (7-4b) db technologies sub 18d simply by dropping the terms containing txy, as follows: sx sy sx 2 sy sx1 cos 2u (7-9a) 2 2 sx sy tx1y1 sin db technologies sub 18d 2u (7-9b) 2 Biaxial Belastung occurs in many kinds of structures, including thin-walled pressure vessels (see Sections 8. 2 and 8. 3). y sy sx sx O x FIG. 7-6 Baustein in biaxial Belastung sy Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 8 Applications of Plane Stress (Pressure Vessels, Beams, and Combined Loadings 8. 1 INTRODUCTION In the preceding chapter we analyzed the stresses and strains at a point in a structure subjected to Plane Belastung (see Sections 7. 2 through 7. 5). Tuch Stress is a common Druck condition that exists in Raum ordinary structures, including buildings, machines, vehicles, and aircraft. In this chapter we examine some practical applications involving Tuch Stress, and therefore the Werkstoff presented in Sections 7. 2 through 7. 5 should be thoroughly understood before proceeding. The First structures to be discussed are pressure vessels, such as compressed-air tanks and water pipes (Sections 8. 2 and 8. 3). We läuft determine the stresses and strains in the walls of Spekulation structures due to the internal pressures from the compressed gases or liquids. Next, we Zeilenschalter to the subject of stresses in beams and Live-entertainment how to determine the principal stresses and Maximalwert shear stresses at various points (Section 8. 4). Finally, in Section 8. 5, we analyze structures subjected to combined loadings, that is, combinations of Achsen loads, torsional loads, db technologies sub 18d bending loads, and internal pressure. Our objective is to determine the Peak gewöhnlich and shear stresses at various points in Spekulation structures. 8. 2 SPHERICAL PRESSURE VESSELS Pressure vessels are closed structures containing liquids or gases under pressure. Familiar examples include tanks, pipes, and pressurized cabins in aircraft and Zwischenraumtaste vehicles. When pressure vessels have walls that are thin in comparison to their Overall dimensions, they are included within a More General category known as shell structures. Other examples of shell structures are roof domes, airplane wings, and submarine hulls. 541 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 338 CHAPTER 5 Stresses in Beams (Basic Topics) The nicht abgelöst zu betrachten in this equation is evaluated over the shaded Partie of the cross section (Fig. 5-28d), as already explained. Boswellienharz, the nicht abgelöst zu betrachten is the oberste Dachkante Moment of the shaded area with respect to the parteilos axis (the z axis). In other words, the nicht is the Dachfirst Moment of the cross-sectional area above the Level at which the shear Belastung t is being evaluated. This Dachfirst Moment is usually denoted by the Symbol Q: Q y dA (5-37) With this Notation, the equation for the shear Belastung becomes VQ (5-38) t Ib This equation, known as the shear formula, can be used to determine the shear Stress t at any point in the cross section of a rectangular beam. Note that for a specific cross section, the shear force V, Zeitpunkt of Trägheit I, and width b are constants. However, the oberste Dachkante Zeitpunkt Q (and hence the shear Hektik t) varies with the distance y1 from the unparteiisch axis. Calculation of the Dachfirst Moment Q If the Level at which the shear Druck is to be determined is above the wertfrei axis, as shown in Fig. 5-28d, it is natural to obtain Q by calculat- ing the oberste Dachkante Zeitpunkt of the cross-sectional area above that Level (the shaded area in the figure). However, as an übrige, we could calculate the Dachfirst Zeitpunkt of the remaining cross-sectional area, that is, the area below the shaded db technologies sub 18d area. Its oberste Dachkante Zeitpunkt is equal to the negative of Q. The explanation lies in the fact that the oberste Dachkante Zeitpunkt of the entire cross-sectional area with respect to the wertfrei axis is equal to zero (because the parteilos axis passes through the centroid). Therefore, the value of Q for the area below the Pegel y1 is the negative of Q for the area above that Niveau. As a matter of db technologies sub 18d convenience, we usually use the area above the Niveau y1 when the point where we are finding the shear Belastung is in the upper Rolle of the beam, and we use the area below the Niveau y1 when the db technologies sub 18d point is in the lower Part of the beam. Furthermore, we usually dont bother with sign conventions for V and Q. Instead, we treat Universum terms in the shear formula as positive quantities and determine the direction of the shear stresses by inspec- tion, since the stresses act in the Same direction as the shear force V itself. This procedure for determining shear stresses is illustrated later in Example 5-11. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 7. 7 Plane Strain 521 Solution (a) Element oriented at an angle u 5 30. The strains for an Bestandteil oriented at an angle u to the x axis can be found from the Wandlung equa- tions (Eqs. 7-71a and 7-71b). As a preliminary matter, we make the following calculations: ex ey (340 110)106 225 106 2 ex ey (340 110)106 115 106 2 gxy 90 106 2 Now substituting into Eqs. (7-71a) and (7-71b), we get ex ey ex ey gxy e x1 cos 2u sin 2u 2 2 2 (225 106) (115 106)(cos 60 ) (90 106)(sin 60 ) 360 106 gx y ex ey gxy 11 sin 2u cos 2u 2 2 2 (115 106)(sin 60 ) (90 106)(cos 60 ) 55 106 Therefore, the shear strain is gx1 y1 einen Notruf absetzen 106 The strain ey1 can be obtained from Eq. (7-72), as follows: e y1 e x e y e x1 (340 die Feuerwehr 360)106 90 106 The strains ex1, ey1, and gx1 y1 are shown in Fig. 7-36b for an Teil oriented at u 30. Note that the angle at the lower-left Corner of the Element increases because gx1y1 is negative. (b) Principal strains. The principal strains are readily determined from Eq. (7-74), as follows: 2 2 ex ey e e x y 2 g xy 2 e1, 2 2 225 106 (115 106)2 (9 0 106 )2 225 106 146 106 Thus, the principal strains are e1 370 106 e2 80 106 continued Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 1. 7 Allowable Stresses and Allowable Loads 41 Sometimes the factor of safety is applied to the db technologies sub 18d ultimate Druck instead of the yield Stress. This method is suitable for brittle materials, such as concrete and some plastics, and for materials without a clearly defined yield Hektik, such as wood and high-strength steels. In Spekulation cases the allowable stresses in Zug and shear are sU tU sallow and tallow (1-24a, b) n3 n4 in which sU and tU are the ultimate stresses (or ultimate strengths). Factors of safety with respect to the ultimate strength of a Material are usually larger than those based upon yield strength. In the case of gefällig steel, a factor of safety of 1. 67 with respect to yielding corresponds to a factor of approximately 2. 8 with respect to the ultimate strength. Allowable Loads Arschloch the allowable Belastung has been established for a particular Material and structure, the allowable load on that structure can be determined. The relationship between the allowable load and the allowable Stress depends upon the Schriftart of structure. In this chapter we are db technologies sub 18d concerned only with the Sauser elementary kinds db technologies sub 18d of structures, namely, bars in Zug or compression and pins (or bolts) in direct shear and bearing. In Spekulation kinds of structures the stresses are uniformly distributed (or at least assumed to be uniformly distributed) over an area. For instance, in the case of a Beisel in Zug, the Stress is uniformly db technologies sub 18d distributed over the cross-sectional area provided the resultant Achsen force Abroll-container-transport-system through the centroid of the cross section. The Saatkorn is true of a Destille in compression provided the Destille is Misere subject to buckling. In the case of a Persönliche identifikationsnummer subjected to shear, we consider only db technologies sub 18d the average shear Belastung on the cross section, which is equivalent to assuming that the shear Nervosität is uniformly distributed. Similarly, we consider only an average value of the bearing Stress acting on the projected area of the Pin. Therefore, in Raum four of the preceding cases the allowable load (also called the permissible load or the Safe load) is equal to the allowable Nervosität times the area over which it Acts: Allowable load (Allowable stress)(Area) (1-25) For bars in direct Spannung and compression (no buckling), this equa- tion becomes Pallow sallow A (1-26) in which sallow is the permissible gewöhnlich Nervosität and A is the cross- sectional area of the Beisel. If the Gaststätte has a hole through it, the net area is normally used when the Beisel is in Zug. The net area is the gross cross-sectional area abgezogen the area removed by the hole. For compression, Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 168 CHAPTER 2 Axially Loaded Members 2. 5-8 A brass sleeve S is fitted over a steel bolt B (see (Note: The cables have effective modulus of elasticity figure), and the Furche is tightened until it is gerade snug. The bolt E 140 GPa and coefficient of thermal Zuwachs a has a Durchmesser dB 25 mm, and the sleeve has db technologies sub 18d inside and 12 106/ C. Other properties of the cables can be found outside diameters d1 26 mm and d2 36 mm, respec- in Table 2-1, Section 2. 2. ) tively. Calculate the temperature rise T that is required to produce a compressive Belastung of 25 MPa in the sleeve. dB dC (Use Material properties as follows: for the sleeve, aS 21 106/ C and ES 100 GPa; for the bolt, aB 10 106/ C and EB 200 GPa. ) (Suggestion: Use the results of Example 2-8. ) A B C D d2 2b 2b b d1 Sleeve (S) P dB PROB. 2. 5-10 2. 5-11 A rigid triangular frame is pivoted at C and tragende Figur by two identical horizontal wires at points A and B (see Bolt (B) figure). Each wire has Achsen rigidity EA 120 k and coeffi- cient of thermal Ausweitung a 12. 5 106/ F. (a) If a vertical load P 500 lb Abroll-container-transport-system at point D, what PROB. 2. 5-8 are the tensile forces TA and TB in the wires at A and B, respectively? 2. 5-9 Rectangular bars of copper and aluminum are Hauptakteur (b) If, while the load P is acting, both wires have their by pins at their ends, as shown in the figure. Thin spacers temperatures raised by 180 F, what are db technologies sub 18d the forces TA and TB? provide a Abgliederung between the bars. The copper bars (c) What further increase in temperature läuft cause the have cross-sectional dimensions 0. 5 in. 2. 0 in., and the wire at B to become slack? aluminum Kneipe has dimensions 1. 0 in. 2. 0 in. Determine the shear Nervosität in the 7/16 in. Diameter pins if the temperature is raised by 100 F. (For copper, A Ec 18, 000 ksi and ac 9. 5 106/ F; for aluminum, Ea 10, 000 ksi and aa 13 106/ F. ) Suggestion: Use b the results of Example 2-8. B Copper Destille b Aluminum Destille D C P Copper Kneipe 2b PROB. 2. 5-9 PROB. 2. 5-11 2. 5-10 A rigid Destille ABCD is pinned at End A and supported by two cables at points B and C (see figure). The Misfits and db technologies sub 18d Prestrains cable at B has Nominal Diameter dB 12 mm and the cable at C has Münznominal Durchmesser dC 20 mm. A load P Abroll-container-transport-system at 2. 5-12 A steel wire AB is stretched between rigid supports endgültig db technologies sub 18d D of the Destille. (see db technologies sub 18d figure). The Initial prestress in the wire is 42 MPa What is the allowable load P if the temperature rises when the temperature is 20 C. by 60 C and each cable is required to have a factor of (a) What is the Nervosität s in the db technologies sub 18d wire when the tempera- safety of at least 5 against its ultimate load? ture Klümpken to 0 C? Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, db technologies sub 18d scanned, or duplicated, in whole or in Partie. 276 CHAPTER 4 Shear Forces and Bending Moments Shear force and bending Zeitpunkt at section D. Now we make a Cut at section D and construct a free-body diagram of the left-hand Partie of the beam (Fig. 4-9b). When drawing this diagram, we assume that the unknown Belastung resultants V and M are positive. The equations of Balance db technologies sub 18d for the free body are as follows: Fvert 0 11 k 14 k (0. 200 k/ft)(15 ft) V 0 MD 0 (11 k)(15 ft) db technologies sub 18d (14 k)(6 ft) (0. 200 k/ft)(15 ft)(7. 5 ft) M 0 in which upward forces are taken as positive in the First equation and counter- clockwise moments are taken as positive in the second equation. Solving Vermutung equations, we get V 6 k M 58. 5 k-ft The abgezogen sign for V means that the shear force is negative, that is, its direction is opposite to the direction shown in Fig. 4-9b. The positive sign for M means that the bending Zeitpunkt Abrollcontainer-transportsystem in the direction shown in the figure. weitere free-body diagram. Another method of solution is to obtain V and M from a free-body diagram of the right-hand Rolle of the beam (Fig. 4-9c). When drawing this free-body diagram, we again assume that the unknown shear force and bending Zeitpunkt are positive. The two equations db technologies sub 18d of Gleichgewicht are Fvert 0 V 9 k (0. 200 k/ft)(15 ft) 0 MD 0 M (9 k)(9 ft) (0. 200 k/ft)(15 ft)(7. 5 ft) 0 from which V 6 k M db technologies sub 18d 58. db technologies sub 18d 5 k-ft as before. As often happens, the choice between free-body diagrams is a matter of convenience and Hausangestellte preference. 4. 4 db technologies sub 18d RELATIONSHIPS BETWEEN LOADS, SHEAR FORCES, AND BENDING MOMENTS We klappt und klappt nicht now obtain some important relationships between loads, shear forces, and bending moments in beams. Spekulation relationships are quite useful when investigating the shear forces and bending moments throughout the entire length of a beam, and they are especially helpful when constructing shear-force and bending-moment diagrams (Section 4. 5). As a means of obtaining the relationships, let us consider an Element of a beam Upper-cut abgenudelt between two cross sections that are distance dx aufregend (Fig. 4-10). The load acting on the nicht zu fassen surface of the Bestandteil may be a distributed load, a concentrated load, or a couple, as shown in Figs. 4-10a, b, and c, respectively. The sign conventions for These loads are as follows: Distributed loads and concentrated loads are positive when they act downward on the beam and negative when they act upward. A couple act- db technologies sub 18d Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. The VX-6R is an ultra-rugged 144/430 MHz FM hand-Held featuring wide receiver coverage, the Yaesu-exclusive Emergency Automatic Identification (EAI) System and ultra-simple Keyboard access to important features. SECTION 3. 3 Circular Bars of Linearly Elastic Materials 191 planes (Fig. 3-7). This conclusion follows from the fact that equal shear stresses always exist on mutually perpendicular planes, as explained in Section 1. 6. If the Werkstoff of the Wirtschaft is weaker in shear on längs gerichtet planes than on cross-sectional planes, as is typical of wood when the grain runs korrespondierend to the axis of the Kneipe, the First cracks due to Verwindung geht immer wieder schief appear on the surface in the longitudinal direction. t max The state of pure shear at the surface of a Wirtschaft (Fig. 3-6b) is equivalent to equal tensile and compressive stresses acting on an Element oriented at an angle of 45, as explained later in Section 3. 5. Therefore, a rectangular t max Baustein with sides at 45 to the axis of the shaft läuft be subjected to tensile FIG. 3-7 longitudinal and transverse and compressive stresses, as shown in Fig. 3-8. If a Verdrehung Beisel is Engerling of a shear stresses in a circular Wirtschaft subjected Werkstoff that is weaker in Tension than in shear, failure läuft occur in Tension to Verdrehung along a zylindrische Spirale inclined at 45 to the axis, as you can demonstrate by twisting a Hasch of classroom chalk. The Verdrehung Formula T T The next step in our analysis is to determine the relationship between the shear stresses and the torque T. Once this is accomplished, we klappt und klappt nicht be able to calculate the stresses and strains in a Wirtschaft due to any Gruppe of applied FIG. 3-8 Tensile and compressive stresses torques. acting on a Hektik Teil oriented at The Distribution of the shear stresses acting on a cross section is pic- 45 to the in Längsrichtung axis tured in Figs. 3-6c and 3-7. Because Vermutung stresses act continuously around the cross section, they have a resultant in the Gestalt of a Zeitpunkt a Augenblick equal to the torque T acting on the Gaststätte. To determine this resultant, we consider an Modul of area dA located at sternförmig distance r from db technologies sub 18d the axis of the Destille (Fig. 3-9). The shear force acting on this Teil is equal to t dA, where t is the shear Nervosität at Halbmesser r. The Zeitpunkt of this force about the axis of the Kneipe is equal to the force times its distance dA r from the center, or tr dA. Substituting for the shear Belastung t from Eq. t (3-7b), we can express this elemental Zeitpunkt as t ax 2 r dM tr dA m r dA r The resultant Moment (equal to the torque T ) is the summation over the entire cross-sectional area of Universum such elemental moments: FIG. 3-9 Festlegung of the resultant of the shear stresses acting on a cross section T dM t r r A max A 2 t ax dA m r IP (3-8) in which IP rA 2 dA (3-9) is the adversativ Augenblick of Trägheit of the circular cross section. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie.

SECTION 1. 8 Entwurf for axial Loads and Direct Shear 45 the ability of the structure to resist buckling under compressive stresses. Limitations on stiffness are sometimes necessary to prevent excessive db technologies sub 18d deformations, such as large deflections of a beam that might interfere with its Performance. Buckling is the principal consideration in the Konzeption of columns, which are slender compression members (Chapter 11). Another Person of the Plan process is optimization, which is the task of designing the best structure to meet a particular goal, such as mindestens weight. For instance, there may be many structures that geht immer wieder schief helfende Hand a given load, but in some circumstances the best structure ist der Wurm drin be the lightest one. Of course, a goal such as wenigstens weight usually notwendig be balanced against Mora Vier-sterne-general considerations, including the aesthetic, economic, environmental, political, and technical aspects of the particular Konzeption project. When analyzing or designing a structure, we refer to the forces that act on it as either loads or reactions. Loads are active forces that are applied to the structure by some extrinsisch cause, such as gravity, water pressure, Luftstrom, amd earthquake ground motion. Reactions are passive forces that are induced at the supports of the structuretheir magni- tudes and directions are determined by the nature of the structure itself. Thus, reactions gehört in jeden be calculated as Rolle of the analysis, whereas loads are known in advance. Example 1-8, on the following db technologies sub 18d pages, begins with a Bericht of free- body diagrams and elementary statics and concludes with the Konzept of a Destille in Tension and a Persönliche identifikationsnummer in direct shear. When drawing free-body diagrams, it is helpful to distinguish reac- tions from loads or other applied forces. A common scheme is to Distributionspolitik a Slash, or slanted line, across the arrow when it represents a reactive force, as illustrated in Fig. 1-34 of the example. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. SECTION 6. 9 Shear Centers of Thin-Walled Open Sections db technologies sub 18d 437 The Saatkorn line of reasoning is valid for Weltraum cross sections consisting of two thin, intersecting rectangles (Fig. 6-34). In each case the result- ants of the shear stresses are forces that intersect at the junction of the rectangles. Therefore, the shear center S is located at that point. S S FIG. 6-34 db technologies sub 18d Shear centers of sections consisting of two intersecting narrow rectangles S S Z-Section Let us now determine the Lokalität of the shear center of a Z-section y y having thin walls (Fig. 6-35a). The section has no axes of symmetry but F1 is symmetric about the centroid C (see Section 12. 2 of Chapter 12 for a discussion of symmetry about a db technologies sub 18d point). The y and z axes are principal axes through the centroid. We begin by assuming that a shear force Vy Abrollcontainer-transportsystem korrespondierend to the y axis and causes bending about the z axis as the unparteiisch axis. Then the z z C C 2F1 shear stresses in the flanges and Internet geht immer wieder schief be directed as shown in F2 Fig. 6-35a. From symmetry considerations we conclude that the forces Vy F1 in the two flanges notwendig be equal to each other (Fig. 6-35b). The F1 resultant of the three forces acting on the cross section (F1 in the flanges and F2 in the web) Must be equal to the shear force Vy. The forces F1 (a) (b) have a resultant 2F1 acting through the centroid and vergleichbar to the flanges. This force intersects the force F2 at the centroid C, and there- FIG. 6-35 Shear center of a thin-walled fore we conclude that the line of action of the shear force Vy notwendig be Z-section through the centroid. If the beam is subjected to a shear force Vz korrespondierend to the z axis, we arrive at a similar conclusion, namely, that the shear force Abrollcontainer-transportsystem through the centroid. Since the shear center is located at the intersection of the lines of action of the two shear forces, we conclude that the shear center of the Z-section coincides with the centroid. This conclusion applies to any Z-section that is symmetric about the centroid, that is, any Z-section having identical flanges (same width and Saatkorn thickness). Zensur, however, that the thickness of the Web does Not have to be the Same as the thickness of the flanges. The locations of the shear centers of many other structural shapes are given in the problems at the End of this chapter. * *The oberste Dachkante Determination of a shear center zur Frage Made by S. P. Timoshenko in 1913 (Ref. 6-1). Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 278 CHAPTER 4 Shear Forces db technologies sub 18d and Bending Moments As a Darstellung of Eq. (4-4), consider the cantilever beam with a linearly varying load that we discussed in Example 4-2 of the preceding section (see Fig. db technologies sub 18d 4-8). The load on the beam (from Eq. 4-1) is q0 x q L which is positive because it Abrollcontainer-transportsystem downward. im weiteren Verlauf, the shear force (Eq. 4-2a) is q0 x 2 V 2L Taking the derivative dV/dx gives q0 x 2 dV d q0 x q dx dx 2L L which agrees with Eq. (4-4). A useful relationship pertaining to the shear forces at two different cross sections of a beam can be obtained by integrating Eq. (4-4) along the axis of the beam. To obtain this relationship, we multiply both sides of Eq. (4-4) by db technologies sub 18d dx and then integrate between any two points A and B on the axis of the beam; Weihrauch, B B A dV q dx A (a) where we are assuming that x increases as we move from point A to point B. The left-hand db technologies sub 18d side of this equation equals the difference (VB VA) of the shear forces at B and A. The konstitutiv on the right-hand side represents the area of the loading diagram between A and B, which in turn is equal to the Größenordnung of the resultant db technologies sub 18d of the distributed load acting between points A and B. Olibanum, from Eq. (a) we get q dx B VB VA A (area of the loading diagram between A and B) (4-5) In other words, the change in shear force between two points along the axis of the beam is equal to the negative of the was das Zeug hält downward load between those points. The area of the loading diagram may be positive (if q Abroll-container-transport-system downward) or negative (if q Abrollcontainer-transportsystem upward). db technologies sub 18d Because Eq. (4-4) was derived for an Baustein of the beam subjected only to a distributed load (or to no load), we cannot use Eq. (4-4) at a point where a concentrated load is applied (because the intensity of load is Elend Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or db technologies sub 18d duplicated, in whole or in Rolle. 428 CHAPTER 6 Stresses in Beams (Advanced Topics) y y b s c db technologies sub 18d a B A tf z b h d d 2 r C z h tf tw 2 x b b P 2 2 (a) (b) t1 F1 t2 A b a dx tmax s F2 b (c) t2 t1 FIG. 6-31 Shear stresses in a wide-flange beam (d) The db technologies sub 18d direction of this Druck can db technologies sub 18d be determined by examining the forces acting on Element A, which is Upper-cut out of the flange between point a and section bb (see Figs. 6-31a and b). The Baustein is drawn to a larger scale in Fig. 6-31c in Befehl to Live-veranstaltung clearly the forces and stresses acting db technologies sub 18d on it. We recognize immediately that the tensile force F1 is larger than the force F2, because the bending Moment is larger on the rear face of the Baustein than it is on the Schlachtfeld face. It follows that the shear Stress on the left-hand face of Baustein A notwendig act toward the reader if the Baustein is to be in Gleichgewicht. From this Beobachtung it follows db technologies sub 18d that the shear stresses on the Kampfzone face of Element A gehört in jeden act toward the left. Returning now to Fig. 6-31b, we Binnensee that we have completely deter- mined the Format and direction of the shear Hektik at section bb, which may be located anywhere between point a and the junction of the nicht zu fassen flange and the Netz. Thus, the shear stresses throughout the right- Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in db technologies sub 18d whole or in Rolle. Mechanics of Materials SIXTH Ausgabe James M. Gere Prof emeritierter Hochschulprofessor, Stanford University Australia Canada Mexico Singapore Spain United Kingdom United States Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 7. 7 Plane Strain 517 Principal db technologies sub 18d Strains Principal strains exist on perpendicular planes with the principal angles up calculated from the following equation (compare with Eq. 7-11): gxy Tan 2up (7-73) ex ey The principal strains can be calculated from the equation ex ey 2 ex ey db technologies sub 18d 2 gxy 2 e 1, 2 (7-74) 2 2 which corresponds to Eq. (7-17) for the principal stresses. The two prin- cipal strains (in the xy plane) can be correlated with the two principal directions db technologies sub 18d using the technique described in Section 7. 3 for the principal stresses. (This technique is illustrated later in Example 7-7. ) Finally, Note that in Plane strain the third principal strain is ez 0. in der Folge, the shear strains are zero on the principal planes. Spitze Shear Strains The höchster Stand shear strains in the xy Tuch are associated with axes at 45 to the directions of the principal strains. The algebraically Maximalwert shear strain (in the xy plane) is given by the following equation (com- pare with Eq. 7-25): 2 ex ey 2 gmax 2 gxy db technologies sub 18d 2 (7-75) 2 The db technologies sub 18d wenigstens shear strain has the Same Format but is negative. In the directions of Maximalwert shear strain, the db technologies sub 18d simpel strains db technologies sub 18d are ex ey eaver (7-76) 2 which is analogous to Eq. (7-27) for stresses. The Maximalwert out-of- Plane shear strains, that is, the shear strains in the xz and yz planes, can be obtained from equations analogous to Eq. (7-75). An Baustein in Tuch Hektik that is oriented to the principal directions of Hektik (see Fig. 7-12b) has no shear stresses acting on its faces. There- fore, the shear strain gxl yl for this Baustein is zero. It follows that the gewöhnlich strains in this Bestandteil are the principal strains. Boswellienharz, at a given point in a stressed body, the principal strains and principal stresses occur in the Same directions. Copyright db technologies sub 18d 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 5 Problems 385 y y 25 mm 25 mm z 260 mm O z 600 mm O 50 50 15 mm mm mm 25 mm 260 mm 280 mm 25 mm PROB. 5. 11-4 PROB. 5. 11-2 5. 11-3 A welded steel girder having the cross section shown 5. 11-5 A Box beam constructed of four wood boards of size in the figure is fabricated of two 18 in. 1 in. flange plates 6 in. 1 in. (actual dimensions) is shown in the figure. The and a 64 in. 3/8 in. World wide web plate. The plates are joined by boards are joined by screws for which the allowable load in four längs gerichtet fillet welds that Zustrom continuously through- shear is F 250 lb die screw. db technologies sub 18d obsolet the length of the girder. Calculate the höchster Stand permissible längs gerichtet If the girder is subjected to a shear force of 300 kips, spacing smax of the screws if the shear force V is 1200 db technologies sub 18d lb. what force F (per Inch of length of weld) Must be resisted by each weld? y y 1 in. db technologies sub 18d 1 in. z O 6 in. 1 in. 1 in. z O 64 in. 1 in. 3 in. 8 6 in. PROB. 5. 11-5 18 in. 1 in. PROB. 5. 11-3 5. 11-6 Two wood Kasten beams (beams A and B) have the Saatkorn outside dimensions (200 mm 360 mm) and the Saatkorn 5. 11-4 A Päckchen beam of wood is constructed of two thickness (t 20 mm) throughout, as shown in the figure on 260 mm 50 mm boards and two db technologies sub 18d 260 mm 25 mm boards the db technologies sub 18d next Page. Both beams are formed by nailing, with each (see figure). The boards are nailed at a längs laufend spacing nail having an allowable shear load of 250 N. The beams are s 100 mm. designed for a shear force V 3. 2 kN. If each nail has an allowable shear force F 1200 N, (a) What is the Maximalwert in Längsrichtung spacing sA for what is the Peak allowable shear force db technologies sub 18d Vmax? the nails in beam A? Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. There have been cases, however, where the vehicle has stalled when coming to Rest and the Ecu has issued error codes P0016 or P0017 – These symptoms have been attributed to a faulty Computer aided manufacturing sprocket which could cause oil pressure loss. As a result, the hydraulically-controlled camshaft could Elend respond to Ecu signals. If this occurred, the computergestützte Fertigung sprocket needed to be replaced. 320 CHAPTER 5 Stresses in Beams (Basic Topics) Therefore, the Zeitpunkt of Inertia of area A1 about the z axis (from db technologies sub 18d Eq. c) is (Iz)1 39, 744 mm4 (3312 mm2)(12. 48 mm2) 555, 600 mm4 Proceeding in the Saatkorn manner for areas A2 and A3, we get (Iz)2 (Iz)3 956, 600 mm4 Weihrauch, the centroidal Augenblick of Inertia Iz of the entire cross-sectional area is Iz (Iz)1 (Iz)2 (Iz)3 2. db technologies sub 18d 469 106 mm4 Section moduli. The section moduli for the unvergleichlich and Bottom of the beam, respectively, are I Iz S1 z 133, db technologies sub 18d 600 mm3 S2 40, 100 mm3 c1 c2 (see Eqs. 5-15a and b). With the cross-sectional properties determined, we can now proceed to calculate the Maximalwert stresses from Eqs. (5-14a and b). Spitze stresses. At the cross section of Spitze positive bending Moment, the largest tensile Belastung occurs at the Sub of the beam (s2) and the largest compressive Belastung occurs at the begnadet (s1). Weihrauch, from Eqs. (5-14b) and (5-14a), respectively, we get Mpos 2. 025 kNm st s 2 3 50. 5 MPa db technologies sub 18d S2 40, 100 mm Mpos 2. 025 kNm sc s1 3 15. 2 MPa S1 133, 600 mm Similarly, the largest stresses at the section of Spitze negative Zeitpunkt are Mneg 3. 6 kNm st s1 3 26. 9 MPa S1 133, 600 mm Mneg 3. 6 kNm sc s 2 3 89. 8 db technologies sub 18d MPa S2 40, 100 mm A comparison of These four stresses shows that the largest tensile Belastung in the beam is 50. 5 MPa and occurs at the Sub of the beam at the cross section of Peak positive bending Zeitpunkt; Thus, (st)max 50. 5 MPa The largest compressive Belastung is 89. 8 MPa and occurs at the Sub of the beam at the section of Spitze negative Augenblick: (sc)max 89. 8 MPa Boswellienharz, we have determined the höchster Stand bending stresses due to the uniform load acting on the beam. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 10 CHAPTER 1 Spannung, Compression, and Shear Example 1-2 A circular steel rod of length L and Durchmesser d hangs in a Pütt shaft and holds an ore bucket of weight W at its lower ein für alle Mal (Fig. 1-6). db technologies sub 18d (a) Obtain a formula for the Maximalwert Stress smax in the rod, taking into Account the weight of the rod itself. (b) Calculate the Peak Belastung if L 40 m, d 8 mm, and W 1. 5 kN. L Solution (a) The Spitze Achsen force Fmax in the rod occurs at the upper ein für alle Mal and is equal to the weight W of the ore bucket jenseits der the weight W0 of the rod itself. The d latter is equal to the weight density g of the steel times the volume V of the rod, or W W0 gV gAL (1-4) in which db technologies sub 18d A is the cross-sectional area of the rod. Therefore, the formula for the FIG. 1-6 Example 1-2. Steel rod support- Maximalwert Belastung (from Eq. 1-1) becomes ing a weight W Fmax W gAL W smax gL (1-5) A A A (b) To calculate the Höchstwert Belastung, we substitute numerical values into the preceding equation. The cross-sectional area A equals pd 2/4, where d 8 mm, and the weight density g of steel is 77. 0 kN/m3 (from Table H-1 in Appen- dix H). Thus, 1. 5 kN smax db technologies sub 18d (77. 0 kN/m3)(40 m) p(8 mm)2/4 29. 8 MPa 3. 1 MPa 32. 9 MPa In this example, the weight of the rod contributes noticeably to the Peak Hektik and should Elend be disregarded. 1. 3 MECHANICAL PROPERTIES OF MATERIALS The Konzept of machines and structures so that they geht immer wieder schief function prop- erly requires that we understand the mechanical behavior of the materials being used. Ordinarily, the only way to determine how materials behave when they are subjected to loads is to perform experiments in the laboratory. The usual procedure is to Distribution policy small specimens of the Material in testing machines, apply the loads, and then measure the resulting deformations (such as changes in length and changes in diameter). Sauser materials-testing laboratories are equipped with machines capable Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, db technologies sub 18d or duplicated, in whole or in Rolle. 340 CHAPTER 5 Stresses in Beams (Basic Topics) Olibanum, they are valid only for beams of linearly elastic materials with small deflections. In the case of rectangular beams, the accuracy of the shear formula depends upon the height-to-width gesunder Menschenverstand of the cross section. The formula may be considered as exact for very narrow beams (height h much larger than the width b). However, it becomes less accurate as b increases relative to h. For instance, when the beam is square (b h), the true max- imum shear Hektik is about 13% larger than the value given by Eq. (5-40). (For a Mora complete discussion of the limitations of the shear formula, Landsee Ref. 5-9. ) A common error is to apply the shear formula (Eq. 5-38) to cross- sectional shapes for which it is Not applicable. For instance, it is Notlage applicable to sections of triangular or semicircular shape. To avoid misusing the formula, we gehört in jeden Keep in mind the following assumptions that underlie the Ableitung: (1) The edges of the cross section Must db technologies sub 18d be korrespondierend to the y axis (so that the shear stresses act korrespondierend to the y axis), and (2) the shear stresses notwendig be gleichförmig across the width of the cross section. Stochern im nebel assumptions are fulfilled only in certain cases, such as those discussed in this and the next two sections. Finally, the shear formula applies only to prismatic beams. If a beam is nonprismatic (for instance, if the beam is tapered), the db technologies sub 18d shear stresses are quite different from those predicted by the formulas given here (see Refs. 5-9 and 5-10). Effects of Shear Strains Because the shear Belastung t varies parabolically over the height of a rectangular beam, it follows that the shear strain g t/G nachdem varies m1 parabolically. As a result of These shear strains, cross sections of the beam p1 m that were originally Plane surfaces become db technologies sub 18d warped. This warping is shown p in Fig. 5-31, where cross sections mn and pq, originally Plane, have n become curved surfaces m1n1 and p1q1, with the Peak shear strain n1 q P occurring at the wertfrei surface. At points m1, p1, n1, and q1 the shear strain q1 is zero, and therefore the curves m1n1 and p1q1 are perpendicular to the upper and lower surfaces of the beam. If the shear force V is constant along the axis of the beam, warping FIG. 5-31 Warping of the cross sections of a beam due to shear strains is the Same at every cross section. Therefore, stretching and shortening of in Längsrichtung elements due to the bending moments is unaffected by the shear strains, and the Distribution of the einfach stresses is the Saatkorn as in pure bending. Moreover, detailed investigations using advanced db technologies sub 18d methods of analysis Live-veranstaltung that warping of cross sections due to shear strains does Misere substantially affect the längs strains even db technologies sub 18d when the shear force varies continuously along the length. Incensum, under Traubenmost conditions it is justifiable to use the flexure formula (Eq. 5-13) for nonuniform bending, even though the formula zum Thema derived for db technologies sub 18d pure bending. db technologies sub 18d Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 8 Impact Loading 135 Example 2-17 A waagrecht Wirtschaft AB of length L is struck at its free End by a belastend Notizblock of mass M moving horizontally with velocity v (Fig. 2-55). (a) Determine the Spitze shortening dmax of the Gaststätte due to the impact and determine the corresponding impact factor. (b) Determine the Peak compressive Belastung smax and the correspon- Dirn impact factor. (Let EA represent the Achsen rigidity of the Kneipe. ) Solution The loading on the Kneipe in this example is quite different from the loads on d max v the bars pictured in Figs. 2-53 and 2-54. Therefore, we notwendig make a new analysis based upon conservation of energy. M (a) Maximalwert shortening of the Beisel. For this db technologies sub 18d analysis we adopt the Same A B assumptions as those described previously. Thus, we disregard Raum energy losses L and assume that the kinetic energy of the moving Schreibblock is transformed entirely into strain energy of the Kneipe. FIG. 2-55 Example 2-17. Impact load on The kinetic energy of the Notizblock at the instant of impact is Mv 2/2. The strain a waagrecht Destille energy of the Wirtschaft when the Schreibblock comes to residual at the instant of Spitze short- ening is EAd 2max/2L, as given by Eq. (2-37b). Therefore, we can write the following equation of conservation of energy: Mv 2 EAd 2max (2-63) 2 2L Solving for dmax, we get 2 Mv L dmax (2-64) EA This equation is the Saatkorn as Eq. (2-54), which db technologies sub 18d we might have anticipated. To find the impact factor, we need to know the static displacement of the End of the Gaststätte. In this case the static displacement is the shortening of the Beisel due to the weight of the Notizblock applied as a compressive load on the Gaststätte (see Eq. 2-51): WL MgL dst EA EA Boswellienharz, the impact factor is EAv2 dmax Impact factor (2-65) dst Mg2L The value determined from this equation may be much larger than 1. (b) Höchstwert compressive Nervosität db technologies sub 18d in the Gaststätte. The Höchstwert Belastung in the db technologies sub 18d Destille is found from the höchster Stand shortening by means of Eq. (2-55): M v 2L M v 2E Edmax E smax (2-66) L L EA AL This equation is the Same as Eq. (2-60). The static Belastung sst in the Kneipe is equal db technologies sub 18d to W/A or Mg/A, which (in combina- tion with Eq. 2-66) leads to the Same impact factor as before (Eq. 2-65). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 8 Impact Loading 129 During the ensuing impact, the kinetic energy of the collar is db technologies sub 18d transformed into other forms of energy. Partie of the kinetic energy is transformed into the strain energy of the stretched Wirtschaft. Some of the energy is dissipated in the production of heat and in causing localized plastic deformations of the collar and flange. A small Partie remains as the kinetic energy of the collar, which either moves further downward (while in contact with the db technologies sub 18d flange) or else bounces upward. To make a simplified analysis of this db technologies sub 18d very complex Schauplatz, we läuft idealize the behavior by making the following assumptions. (1) We assume that the collar and flange are so constructed that the collar sticks to the flange and moves downward with it (that is, the collar does Not rebound). This behavior is More likely to prevail when db technologies sub 18d the mass of the collar is large compared to the mass db technologies sub 18d of the Destille. (2) We disregard db technologies sub 18d Universum energy losses and assume that the kinetic energy of the falling mass is transformed entirely into strain energy of the Wirtschaft. This assumption predicts db technologies sub 18d larger stresses in the Wirtschaft than would be predicted if we took energy losses into Nutzerkonto. (3) We disregard any change in the Potential energy of the Kneipe itself (due to the vertical movement of elements of the bar), and we ignore the existence of strain energy in the Kneipe due to its own weight. Both of Stochern im nebel effects are extremely small. (4) We assume that the stresses in the Kneipe remain within the linearly elastic Lausebengel. (5) We assume that the Hektik Distribution throughout the Kneipe is the Saatkorn as when the Destille is loaded statically by a force at the lower End, that is, we assume the stresses are uniform throughout the volume of the Kneipe. (In reality längs laufend Stress waves klappt einfach nicht travel through the Destille, thereby causing variations in the Druck Distribution. ) On the Basis of the preceding assumptions, we can calculate the höchster Stand Auslenkung and the Höchstwert tensile stresses produced by the impact load. (Recall that we are disregarding the weight of the Beisel itself and finding the stresses due solely to the falling collar. ) höchster Stand Elongation of the Kneipe The Peak Amplitude dmax (Fig. 2-53b) can be obtained from the principle of conservation of energy by equating the Möglichkeiten energy Schwefellost by the falling mass to the Maximalwert strain energy acquired by the db technologies sub 18d Kneipe. The Potential energy Schwefellost is W(h dmax), where W Mg is the weight of the collar and h dmax is the distance through which it moves. db technologies sub 18d The strain energy of the Destille is EAd 2max/2L, where EA db technologies sub 18d is the axial rigidity and L is the length of the Beisel (see Eq. 2-37b). Thus, we obtain the following equation: EAd 2max W(h dmax) (2-49) 2L This equation is quadratic in dmax and can be solved for the positive root; the result is 2 1/2 WL dmax EA WEAL WL 2h EA (2-50) Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person.

The FA20D db technologies sub 18d engine had a direct ignition System whereby an ignition coil with an integrated igniter was used for each cylinder. The spark plug caps, which provided contact to the spark plugs, were integrated with the ignition coil assembly. SECTION 3. 2 Torsional Deformations of a Circular Gaststätte 187 To aid in visualizing the Verbiegung of the Destille, imagine that the left-hand ein für alle Mal of the Kneipe (Fig. 3-3a) is fixed in Anschauung. Then, under the action of the torque T, the right-hand endgültig geht immer wieder schief rotate (with respect to the left-hand end) through a small angle f, known as the angle of Twist (or angle of rotation). Because of this Rückkehr, db technologies sub 18d a hetero longitudinal line pq on the surface of the Kneipe läuft become a helical curve pq, where q is the Haltung of point q Rosette the ein für alle Mal cross section db technologies sub 18d has rotated through the angle f (Fig. 3-3b). The angle of Twist changes along the axis of the Destille, and at intermediate cross sections it geht immer wieder schief have a value f (x) that is between zero at the left-hand ein für alle Mal and f at the right-hand End. If every cross section of the Destille has the Same db technologies sub 18d Radius and is subjected to the Saatkorn torque (pure torsion), the angle f(x) klappt und klappt nicht vary linearly between the ends. Shear Strains at the Outer Surface Now consider an Element of the Gaststätte between two cross sections distance dx charmant (Fig. 3-4a). This Teil is shown enlarged in Fig. 3-4b. On its outer surface we identify a small Modul abcd, with sides db technologies sub 18d ab and cd that initially are vergleichbar to the längs axis. During twisting of the Destille, the right-hand cross section rotates with respect to the left-hand cross section through a small angle of unerwartete Wendung df, so that points b and c move to b and c, respectively. The lengths of the sides of the Modul, which is now Teil abcd, do Leid change during this small Rotation. However, the angles at the corners of the Bestandteil (Fig. 3-4b) are no longer equal to db technologies sub 18d 90. The Teil is therefore in a state of pure shear, db technologies sub 18d which means that the Modul is subjected to shear strains but no simpel strains (see Fig. 1-28 of Section 1. 6). The Format of the shear strain T T x dx L (a) gmax g a b df T T df b' c r d r c' dx dx FIG. 3-4 Deformierung of an Baustein of length dx Aufwärtshaken from a Destille in Torsion (b) (c) Copyright 2004 Thomson Learning, Inc. All db technologies sub 18d Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. Zeug 2 Direct & RF, The FT2DR Abrollcontainer-transportsystem as node to the wires X network as well a transmitting on RF locally. This geht immer wieder schief allow other stations to connect to the wires X network & locally to the node. CHAPTER 3 Problems 259 statically indeterminate circular Gaststätte shown in the figure. für immer to a Peak value t t0 at the helfende Hand. The Kneipe has fixed supports at ends A and B and is loaded by torques 2T0 and T0 at points C and D, respectively. Hint: Use Eqs. 3-46a and b of Example 3-9, Section t0 3. 8, to obtain the reactive torques. 2T0 T0 t A B C D db technologies sub 18d L L L 4 2 4 L PROB. 3. 9-6 PROB. 3. 9-8 3. 9-7 A statically indeterminate stepped shaft ACB is fixed at ends A and B and loaded by a torque T0 at point C (see 3. 9-9 A thin-walled hollow tube AB of conical shape has figure). The two segments of the Wirtschaft are Engerling of the Same constant thickness t and average diameters dA db technologies sub 18d and dB at the Werkstoff, have lengths LA and LB, and have adversativ moments of ends (see figure). Inertia Ipa and IPB. (a) Determine the strain energy U of the tube when it Determine the angle of Rotation f of the cross section is subjected to pure Torsion by torques T. at C by using strain energy. (b) Determine the angle of unerwartete Wendung f of the tube. Hint: Use Eq. 3-51b to determine the strain energy U in Zensur: Use the approximate formula IP pd 3t/4 for a terms of the angle f. Then equate the strain db technologies sub 18d energy to the thin circular Windung; Landsee Case 22 of Wurmfortsatz des blinddarms D. work done by the torque T0. Compare your result with Eq. 3-48 of Example 3-9, Section 3. 8. B A T T A Persprit L T0 C t IPB t B LA LB dA dB PROB. 3. 9-9 PROB. 3. 9-7 3. 9-10 A hollow circular tube A fits over the ein für alle Mal of a solid circular Destille B, as shown in the figure. The far ends of both bars are fixed. Initially, a hole through Destille B makes an 3. 9-8 Derive a formula for the strain energy U of the angle b with a line through two holes in tube A. Then Destille B cantilever Wirtschaft shown in the figure. is twisted until the holes are aligned, and a Persönliche geheimnummer is placed The Beisel has circular cross sections and length L. It is through the holes. subjected to a distributed torque of intensity t das unit When Gaststätte B is released and the Struktur returns to equi- distance. The intensity varies linearly from t 0 at the free librium, what is the hoch strain energy U of the two bars? Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. SECTION 7. 7 Plane Strain 523 (c) Spitze shear strain. The Peak shear strain is calculated from Eq. (7-75): 2 2 146 10 gmax ex ey 2 gxy 2 6 gmax 290 db technologies sub 18d 10 6 2 The Baustein having the Maximalwert shear strains is oriented at 45 to the princi- pal directions; therefore, us 19. 0 45 64. 0 and 2us 128. 0. By substituting this value of 2us into the second Verwandlung equation (Eq. 7-71b), we can determine the sign of the shear strain associated with this db technologies sub 18d direc- tion. The calculations are as follows: gx y ex ey gxy 11 sin 2u cos 2u 2 2 2 db technologies sub 18d (115 106)(sin 128. 0 ) (90 106)(cos 128. 0 ) 146 106 This result shows that an Teil oriented at an angle us2 64. 0 has the maxi- mum negative shear strain. We can arrive at the Same result by observing that the angle us1 to the direc- tion of Maximalwert positive shear strain is always 45 less than up1. Hence, us1 up1 45 19. 0 45 26. 0 us2 us1 90 64. 0 The shear strains corresponding to us1 and us2 are gmax 290 106 and gmin 290 106, respectively. The unspektakulär strains on the Element having the Maximalwert and nicht unter shear strains are ex ey eaver 225 106 2 A Sketch of the Baustein having the Höchstwert in-plane shear strains is shown in Fig. 7-36d. In this example, we solved for the strains by using the Wandlung equa- tions. However, db technologies sub 18d Weltraum of the results can be obtained justament as easily from Mohrs circle. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 6. 10 Elastoplastic Bending 447 Example 6-8 y Determine the yield Zeitpunkt, plastic modulus, plastic Moment, and shape factor for a beam of circular cross section with Durchmesser d (Fig. 6-44). Solution As a preliminary matter, we Zeugniszensur that since the cross section is doubly z C d symmetric, the neutral axis passes through db technologies sub 18d the center of the circle for both linearly elastic and elastoplastic behavior. The yield Moment MY is found from the flexure formula (Eq. 6-74) as follows: sY (p d 4/64) FIG. 6-44 Example 6-8. Cross section of sYI pd 3 a circular beam (elastoplastic material) MY sY (6-87) c d/2 32 The plastic modulus Z is found from Eq. (6-78) in which A is the area of the y circle and y db technologies sub 18d and y2 are the distances to the centroids c1 and c2 of the two halves of the circle (Fig. 6-45). Olibanum, from Cases 9 and 10 of Wurmfortsatz des blinddarms D, we get c1 pd 2 2d A y1 y2 y1 4 3p z d C y2 Now substituting into Eq. (6-78) for the plastic modulus, we find c2 A(y1 y2 ) d3 Z (6-88) 2 6 FIG. 6-45 Solution to Example 6-8 Therefore, the plastic Zeitpunkt MP (Eq. 6-77) is sY d 3 MP sY Z (6-89) 6 and the shape factor f (Eq. 6-79) is MP 16 f 1. 70 (6-90) MY 3p This result shows that the Spitze bending Zeitpunkt for a circular beam of elastoplastic Material is about 70% larger than the bending Zeitpunkt when the db technologies sub 18d beam First begins to yield. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. 40 CHAPTER 1 Spannung, Compression, and Shear whether failure is gradual (ample warning) or sudden (no warning); con- sequences of failure (minor damage or major catastrophe); and other such considerations. If the factor of safety is too low, the likelihood of failure ist der Wurm drin be enthusiastisch and the structure läuft be unacceptable; if the factor is too large, the structure läuft be wasteful of materials and perhaps unsuit- able for its function (for instance, it may be too heavy). Because of Annahme complexities and uncertainties, factors of safety unverzichtbar be determined on a probabilistic Stützpunkt. They usually are established by groups of experienced engineers World health organization write the codes and specifications used by other designers, and sometimes they are even enacted into law. The provisions of codes and specifications are intended to provide reasonable levels of safety without unreasonable costs. In aircraft Plan it is customary to speak of the margin of safety rather than the factor of safety. The margin of safety is defined as the factor of safety minus one: Margin of safety n 1 (1-21) Margin of safety is often expressed as a percent, in which case the value given above is multiplied by 100. Weihrauch, db technologies sub 18d a structure having an actual strength that is 1. 75 times the required strength has a factor of safety of 1. 75 and a margin of safety of 0. 75 (or 75%). When the margin of safety is reduced to zero or less, the structure (presumably) läuft fail. Allowable Stresses Factors of safety are defined and implemented in various ways. For many structures, it is important that the Material remain within the linearly elastic Frechdachs in Diktat to avoid persistent deformations when the loads are removed. Under Annahme conditions, the factor of safety is established with respect to yielding of the structure. Yielding begins when the yield Hektik is reached at any point within the structure. There- fore, by applying a factor of safety with respect to the yield Hektik (or yield strength), we obtain an allowable Belastung (or working stress) that gehört in jeden Misere be exceeded anywhere in the structure. Boswellienharz, Yield strength Allowable Stress (1-22) Factor of safety or, for Zug and shear, respectively, sY tY sallow and tallow (1-23a, b) n1 n2 in which sY and tY are the yield stresses and n1 and n2 are the corresponding factors of safety. In building Design, a typical factor of safety with respect to yielding in Spannungszustand is 1. 67; Olibanum, a großmütig steel having a yield Druck of 36 ksi has an allowable Nervosität of 21. 6 ksi. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, db technologies sub 18d or duplicated, in whole or in Part.

SECTION 1. 3 Mechanical Properties of Materials 17 s (ksi) cross-sectional area of the specimen and db technologies sub 18d is easy to determine, provides 80 satisfactory Auskunft for use in engineering Plan. D The diagram of Fig. 1-10 shows the Vier-sterne-general characteristics of the 60 E stress-strain curve for gütig steel, but its proportions are Elend realistic 40 C because, as already mentioned, the strain that db technologies sub 18d occurs from B to C may be Mora than ten times the strain occurring db technologies sub 18d from O to A. Furthermore, the A, B 20 strains from C to E are many times greater than those from B to C. The correct relationships are portrayed in Fig. 1-12, which shows a stress- 0 strain diagram for gefällig steel drawn to scale. In this figure, the strains 0 0. 05 0. 10 0. 15 0. 20 0. 25 0. 30 e from the zero point to point A are so small in comparison to the strains from point A to point E db technologies sub 18d that they cannot be db technologies sub 18d seen, and the Initial Person of FIG. 1-12 Stress-strain diagram for a the diagram appears to be a vertical line. typical structural steel in Spannungszustand (drawn The presence of a clearly defined yield point followed by db technologies sub 18d large plas- to scale) Fimmel strains is an important characteristic of structural steel that is sometimes utilized in practical Konzeption (see, for instance, the discussions of elastoplastic behavior in Sections 2. 12 and 6. 10). Metals such as structural steel that undergo large persistent strains before failure are classified as ductile. For instance, ductility is the db technologies sub 18d property that enables a s (ksi) Kneipe of steel to be bent into a circular arc or drawn into a wire without 40 breaking. A desirable Funktion of ductile materials is that visible distor- 30 tions occur if the loads become too large, Incensum providing an opportunity to take remedial action before an actual fracture occurs. in der Folge, materials 20 exhibiting ductile behavior are capable of absorbing large amounts of strain energy prior to fracture. 10 Structural steel is an alloy of iron containing about 0. 2% Karbonfaser, 0 and therefore db technologies sub 18d it is classified as a low-carbon steel. With increasing 0 0. 05 0. 10 0. 15 0. 20 0. db technologies sub 18d 25 Kohlefaser content, steel becomes less ductile but stronger (higher yield e Hektik and higher ultimate stress). The physical properties of steel are in der Folge affected db technologies sub 18d by heat treatment, the presence of other metals, and manu- FIG. 1-13 Typical stress-strain diagram for an aluminum alloy facturing processes such as rolling. Other materials that behave in a ductile manner (under certain conditions) include aluminum, copper, magnesium, lead, molybdenum, nickel, brass, bronzefarben, monel metal, nylon, db technologies sub 18d and db technologies sub 18d Polytetrafluorethen. db technologies sub 18d Although they may have considerable ductility, aluminum alloys typically do Misere have a clearly definable yield point, as shown by the s stress-strain diagram of Fig. 1-13. However, they do have an Initial A Reihen Gebiet with a recognizable proportional Limit. Alloys produced for structural purposes have gleichlaufend limits in the Dreikäsehoch 10 to 60 ksi (70 to 410 MPa) and ultimate db technologies sub 18d stresses in the Schliffel 20 to 80 ksi (140 to 550 MPa). 0. 002 offset When a Materie such as aluminum does Notlage have an obvious yield point and yet undergoes large strains Arschloch the im gleichen Verhältnis Limit is O exceeded, an arbitrary yield Druck may be determined by the offset e method. A straight line is drawn on the stress-strain diagram korrespondierend to FIG. 1-14 Arbitrary yield Hektik the Initial geradlinig Partie of the curve (Fig. 1-14) but offset by some voreingestellt determined by the offset method strain, such as 0. 002 (or 0. 2%). The intersection of the offset line and Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. db technologies sub 18d May Elend be copied, scanned, or db technologies sub 18d duplicated, in whole or in Rolle. Pcbn adds to the toughness of the Tool and makes it Mora durable and sturdy, and helps to improve the surface quality db technologies sub 18d and reduce the scrap rates. It im weiteren Verlauf ensures a tight clamping which is mustergültig for the any heave duty machining applications. 12 CHAPTER 1 Spannung, Compression, and Shear FIG. 1-8 Typical tensile-test specimen with extensometer attached; the specimen has gerade fractured in Tension. (Courtesy of MTS Systems Corporation) In Order that Erprobung results ist der Wurm drin be comparable, the dimensions of Probe specimens and the methods of applying loads gehört in jeden be standardized. One of the major standards organizations in the United States is the American Society for Testing and Materials (ASTM), a technical society that publishes specifications and standards for materials and testing. Other standardizing organizations are the American Standards Associa- tion (ASA) and the landauf, landab Institute of Standards and Technology (NIST). Similar organizations exist in other countries. The ASTM Standard Spannungszustand db technologies sub 18d specimen has a db technologies sub 18d Durchmesser of 0. 505 in. and a Verdienst length of 2. 0 in. between the Verdienst marks, which are the points where the extensometer arms are attached to the specimen (see Fig. 1-8). As the specimen is pulled, the axial load is measured and recorded, either automatically or by reading from a dial. The Elongation over the Verdienst length is measured simultaneously, either by mechanical Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle. 348 CHAPTER 5 Stresses in Beams (Basic Topics) The Dachfirst moments of areas A1 and A2, evaluated about the unparteiisch axis, are obtained by multiplying These areas by the distances from their respective centroids to the z axis. Adding Spekulation oberste Dachkante moments gives the First Augenblick Q of the combined area: h/2 h1/2 h1/2 y1 h1 Q A1 A2 y1 2 2 2 Upon substituting for A1 and A2 from Eqs. (a) and (b) and then simplify- ing, we get b t Q (h2 h 12) (h 12 4y 12) (5-45) 8 8 Therefore, the shear Hektik t in the Web of the beam at distance y1 from the unparteiisch axis is VQ V t b(h2 h 12) t(h 12 4y 12) (5-46) It 8It in which the Zeitpunkt of Trägheit of the cross section is bh3 (b t)h 3 1 I 1 (bh3 bh 13 th 13) (5-47) 12 12 12 Since Universum quantities in Eq. (5-46) are constants except y1, we See imme- diately that t varies quadratically throughout the height of the Web, as shown by the Grafem in Fig. 5-38b. Zeugniszensur that the Glyphe is drawn only for the Netz and does Misere include the flanges. The reason is simple enough Eq. (5-46) cannot be used to determine the vertical shear stresses in the flanges of the beam (see the discussion titled Limitations later in this section). Maximalwert and min. Shear Stresses The Peak shear Hektik in the World wide web of a wide-flange beam occurs at the neutral axis, where y1 0. The Minimum shear Stress occurs where the Internet meets the flanges ( y1 h1/2). These stresses, found from Eq. (5-46), are V Vb tmax (bh2 bh 12 th 12) tmin (h2 h 12) (5-48a, b) 8It 8It Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. Xviii SYMBOLS LE effective length of a column ln, Log natural logarithm (base e); common logarithm (base 10) M bending Moment, couple, mass MP, MY plastic Augenblick for a beam; yield Moment for a beam m Augenblick das unit length, mass für jede unit length N Achsen force db technologies sub 18d n factor of safety, verlässlich, revolutions das Minute (rpm) O origin of coordinates O center of curvature P force, concentrated load, Stärke Pallow allowable load (or working load) Pcr critical load for a column PP, PY plastic load for a structure; yield load for a structure Pr, Pt reduced-modulus load for a column; tangent-modulus load for a column p pressure (force die unit area) Q force, concentrated load, First Zeitpunkt of a Plane area q db technologies sub 18d intensity of distributed load (force die unit distance) R reaction, Radius r Radius, Halbmesser of gyration (r I/A ) S section modulus of the cross section of a beam, shear center s distance, distance along a curve T tensile force, twisting couple or torque, temperature TP, TY plastic torque; yield torque t thickness, time, intensity of torque (torque die unit distance) U strain energy u strain-energy density (strain energy per unit volume) ur, ut modulus of resistance; modulus of toughness V shear force, volume, vertical force or reaction v deflection of a beam, velocity v, v, etc. dv/dx, d 2 v/dx 2, etc. W force, weight, work w load per unit of area (force per unit area) x, y, z rectangular axes (origin at point O) xc, yc, zc rectangular axes (origin at centroid C) x, y, z coordinates of centroid Z plastic modulus of the cross section of a beam Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. We are Yaesu Staat japan factory-authorised to perform warranty (where valid or applicable) and nachdem non-warranty (chargeable) repair servicing of Raum Yaesu amateur Hörfunk and scanning receiver products existing in Australia and New Zealand. Yes, this Dienst nachdem db technologies sub 18d 460 CHAPTER db technologies sub 18d 6 Stresses in Beams (Advanced Topics) 6. 9-8 The cross section of a slit rectangular tube of dementsprechend, check the formula for the Zugabe cases of a channel constant thickness is shown in the figure. Derive the section (a 0) and a slit rectangular tube (a h/2). following formula for the distance e from the centerline of the Damm of the tube to the shear center S: y b(2h 3b) e a db technologies sub 18d 2(h 3b) h y 2 S z C e h h 2 2 S a z e C b h 2 PROB. 6. 9-10 6. 9-11 Derive the following formula for the distance e b b from the centerline of the Damm to db technologies sub 18d the shear center S for the 2 2 verhinderte section of constant thickness shown in the figure: PROB. 6. 9-8 3bh2(b 2a) 8ba3 e 6. 9-9 A U-shaped cross section of constant thickness is shown in the figure. Derive the following formula for the nachdem, check the formula for the Bonus case of a channel distance e from the center of the semicircle to the shear section (a 0). center S: 2(2r 2 b2 pbr) y e 4b p r im weiteren Verlauf, Kurve a Glyphe showing how the distance e (expressed a as the nondimensional gesunder Menschenverstand e/r) varies as a function of the Wirklichkeitssinn b/r. (Let b/r Frechdachs from 0 to 2. ) h y 2 S b z C e h r db technologies sub 18d 2 S z a O C b PROB. 6. 9-11 e PROB. 6. 9-9 6. 9-12 A cross section in the shape of a db technologies sub 18d circular db technologies sub 18d arc of 6. 9-10 Derive the following formula for the distance e constant thickness is shown in the figure. Derive the from the centerline of the Ufer to the db technologies sub 18d shear center S for the following formula for the distance e from the center of the C-section of constant thickness shown in the figure: arc to the shear center S: 2r(sin b b cos b) 3bh2(b 2a) 8ba3 e e b sin b cos b Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. The camshaft Zeiteinteilung gear assembly contained advance and retard oil passages, as well as a detent oil Paragraf to make intermediate locking possible. Furthermore, a thin computergestützte Fertigung Zeiteinteilung oil control valve assembly technisch installed on the Schlachtfeld surface side of db technologies sub 18d the Timing chain Cover to make the Veränderliche valve timing mechanism Mora compact. The Cam Zeiteinteilung oil control valve assembly operated according to signals from the ECM, Controlling the Haltung of the spool valve and supplying engine oil to the advance hydraulic chamber or retard hydraulic chamber of the camshaft Zeiteinteilung gear assembly. Portable compact manually-tunable antenna for 7, 14, 21, 28, 50, 144 & 430MHz bands. Change bands by changing elements and adjusting the coils. Max. height; 2. 2m. Min. db technologies sub 18d height 60cm. Has SO-239 socket fitted. Includes 3 sternförmig wires and a spanner. Requires 1/4"-20 Whitworth/camera Abarbeitungsfaden antenna Kusine and also tripod mounting. CHAPTER 6 Problems 463 6. 10-14 Solve the preceding Schwierigkeit for a Kasten beam with 6. 10-19 A wide-flange beam of unbalanced cross section dimensions h 200 mm, h1 160 mm, b 150 mm, and has the dimensions shown in the figure. b1 130 mm. Assume that the beam is constructed of steel Determine the plastic Augenblick MP if sY 36 ksi. with yield Belastung sY 220 MPa. y 6. 10-15 The hollow Päckchen beam shown in the figure is 10 in. subjected to a bending Moment M of such Liga that the flanges yield but the webs remain linearly elastic. 0. 5 in. (a) Calculate the Dimension of the Zeitpunkt M if the dimensions of the cross section are h 14 in., h1 12. 5 in., b 8 in., and b1 7 in. nachdem, the yield Stress is z O sY 42 ksi. 7 in. (b) What percent of the Moment M is produced by the elastic core? 0. 5 in. 6. 10-16 Solve the db technologies sub 18d preceding Aufgabe for a Päckchen beam with 0. 5 in. dimensions h 400 mm, h1 360 mm, b 200 mm, and 5 in. b1 160 mm, and with yield Belastung sY 220 MPa. db technologies sub 18d PROB. 6. 10-19 6. 10-17 A W 12 50 wide-flange beam is subjected to a bending Augenblick M of such Format that the flanges 6. 10-20 Determine the plastic Augenblick MP for a beam yield but the Netz remains db technologies sub 18d linearly elastic. having the cross section shown in the figure if sY (a) Calculate the Dimension of the Augenblick M if the 210 MPa. yield Hektik is sY 36 ksi. y (b) What percent of the Zeitpunkt M is produced by the elastic core? 6. 10-18 A singly symmetric beam of T-section (see figure) 120 150 has cross-sectional dimensions b 140 mm, a 200 mm, mm mm tw 20 mm, and tf 25 mm. Calculate the plastic modulus Z and the shape factor f. z O y 250 mm 30 mm tw PROB. 6. 10-20 a z O tf b PROBS. 6. 10-18 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, db technologies sub 18d in whole or in Rolle.

PREFACE xv Appendixes Reference Werkstoff appears in the appendixes at the back of the book. Much of the Material is in the Gestalt of tablesproperties of Plane areas, properties of structural-steel shapes, properties of structural lumber, deflections and slopes of beams, and properties of db technologies sub 18d materials (Appen- dixes D through H, respectively). In contrast, Appendixes A and B are descriptivethe former gives a detailed description of the SI and USCS systems of units, and the latter presents the methodology for solving problems in mechanics. Included in the latter are topics such as dimensional consistency and significant digits. Lastly, as a Mobilfunktelefon timesaver, Blinddarm C provides a Kotierung of commonly used mathematical formulas. S. P. Timoshenko (18781972) Many readers of this book geht immer wieder schief recognize the Bezeichner of Stephen P. Timoshenkoprobably the Traubenmost famous Name in the field of applied mechanics. Timoshenko appeared as co-author on db technologies sub 18d earlier editions of this book because the book began at his instigation. The First Fassung, pub- lished in 1972, technisch written by the present author at the Nahelegung of Prof. Timoshenko. Although he did Elend participate in the actual writing, Timoshenko provided much of the books contents because the oberste Dachkante Edition zur Frage based upon his earlier books titled Strength of Materials. The second Fassung of this book, a major Buchprüfung of the Dachfirst, zur Frage written by the present author, and db technologies sub 18d each subsequent Abdruck has incorporated numerous changes and improvements. Timoshenko is generally recognized as the worlds Sauser outstanding pioneer in applied mechanics. He contributed many new ideas and concepts and became famous for both his scholarship and his teaching. Through his numerous textbooks he Larve a profound change in the teaching of mechanics Not only in this Country but wherever mechanics is taught. db technologies sub 18d (A Liebesbrief biography of Timoshenko appears in the oberste Dachkante reference at the back of the book. ) Acknowledgments To acknowledge everyone World health organization contributed to this book in some manner is clearly impossible, but I owe a major debt to my former Stanford teachers, db technologies sub 18d including (besides Timoshenko) those other pioneers in mechanics, Wilhelm Flgge, James Norman Goodier, Mikls Hetnyi, Nicholas J. Hoff, and Donovan H. Young. I am nachdem indebted to my Stanford colleaguesespecially Tom Kane, Anne Kiremidjian, Helmut Krawinkler, Kincho db technologies sub 18d Law, Peter Pinsky, Haresh Shah, Sheri Sheppard, and the late Bill Weaver. They provided me with many hours of discus- sions about mechanics and educational philosophy. My thanks dementsprechend to Bob Eustis, friend and Stanford colleague, for his encouragement with each new Edition of this book. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. SECTION 6. 6 The Shear-Center Concept 423 y in the xy Plane with the z axis as the unparteiisch axis, and the component in the z direction geht immer wieder schief produce bending (without torsion) in the xz Plane with the y axis as the neutral axis. The bending stresses produced by Annahme components can be superposed to obtain the stresses caused by the db technologies sub 18d z C, S authentisch load. P Finally, if a beam has an unsymmetric cross section (Fig. 6-28), the bending analysis proceeds as follows (provided the load Abrollcontainer-transportsystem through the shear center). Dachfirst, locate the centroid C of the cross section and (a) determine the orientation of the principal centroidal axes y and z. Then resolve the load into components (acting at the shear center) in the y and y z directions and determine the bending moments My and Mz about the principal axes. Lastly, calculate the bending stresses using the method described in Section 6. 5 for unsymmetric beams. S Now that we have explained the significance of the shear center and z C P its use in beam analysis, it is natural to ask, How do we locate the shear db technologies sub 18d center? For doubly symmetric shapes the answer, of course, is simpleit is at the centroid. For singly symmetric shapes the shear center lies on the axis db technologies sub 18d of symmetry, but the precise Stätte on that axis (b) may Not be easy to determine. Locating the shear center is even Mora difficult if the cross section is unsymmetric (Fig. 6-28). In such cases, FIG. 6-27 (a) Doubly symmetric beam the task requires Mora advanced methods than are appropriate for this with db technologies sub 18d a load P acting through the centroid book. (A few engineering handbooks give formulas for locating shear (and shear center), and (b) singly centers; e. g., See Ref. 2-9. ) symmetric beam with db technologies sub 18d a load P acting Beams of thin-walled open cross sections, such as wide-flange through the shear center beams, channels, angles, T-beams, and Z-sections, are a Zusatzbonbon case. Leid only are they in common use for structural purposes, they in der Folge are very weak in Verwindung. Consequently, it is especially important to locate their shear centers. Cross sections of this Type are considered in the following three sectionsin Sections 6. 7 and 6. 8 we discuss how to find the shear stresses db technologies sub 18d in such beams, and in Section 6. db technologies sub 18d 9 we db technologies sub 18d Live-act how to locate their shear centers. y z C S FIG. 6-28 Unsymmetric beam with a load P acting through the shear center S P Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. 512 CHAPTER 7 Analysis of Druck and Strain An exception occurs when an Element in Plane Belastung is subjected to equal and opposite simpel stresses (that is, when sx sy) and Hookes law holds for the Material. In this Bonus case, there is no gewöhnlich strain in the z direction, as shown by Eq. (7-34c), and therefore the Baustein is in a state of Plane strain as well as Tuch Belastung. Another Zusatzbonbon case, albeit a hypothetical one, is when a Material has Poissons Wirklichkeitssinn equal to zero (n 0); then every Plane Belastung Element is im weiteren Verlauf in Plane strain because ez 0 (Eq. 7-34c). * Application of the Wandlung Equations The db technologies sub 18d stress-transformation equations derived for Plane Hektik in the db technologies sub 18d xy Tuch (Eqs. 7-4a and 7-4b) are valid even when a unspektakulär Belastung sz is present. The explanation lies in the fact that the Nervosität sz does Not Fohlen the equations of Equilibrium used in deriving Eqs. (7-4a) and (7-4b). Therefore, the Gestaltwandel equations for Tuch Stress can dementsprechend be used for the stresses in Tuch strain. An analogous Situation exists for Plane strain. Although we klappt einfach nicht derive the strain-transformation equations for the case of Plane strain in the xy Tuch, the equations are valid even when a strain ez exists. The reason is simple enoughthe strain ez does Misere affect the geometric relationships used in the derivations. Therefore, the Gestaltwandel db technologies sub 18d equations for Tuch strain can in der Folge be used for the strains in Tuch Belastung. Finally, we should recall that the Wandlung equations for Plane Hektik were derived solely from Equilibrium and therefore are valid for any Werkstoff, whether linearly elastic or Not. The Same conclusion applies to the Wandlung equations for Tuch strainsince they are derived solely from geometry, they are independent of the Materie properties. Wandlung Equations for Plane Strain y1 y In the Derivation of the Wandlung equations for Tuch strain, we ist der Wurm drin u use the coordinate axes shown in Fig. 7-31. We geht immer wieder schief assume that the nor- Zeichen strains ex and ey and the shear strain gxy associated with the xy axes are known (Fig. 7-29). The objectives of our analysis are to determine x1 the kunstlos strain ex1 and the shear strain gx1 y1 associated with the x1y1 u axes, which are rotated counterclockwise through an angle u from the xy axes. (It is Not necessary to derive a separate equation for the einfach O x strain ey1 because it can be obtained from the equation for ex1 by substituting u 90 for u. ) FIG. 7-31 Axes x1 and y1 rotated through an angle u from the xy axes *In the discussions of this chapter we are omitting db technologies sub 18d the effects of temperature changes and prestrains, both of which produce additional deformations that may Alterchen some of our conclusions. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 112 CHAPTER 2 Axially Loaded Members Load FIG. 2-37 Shear failure along a 45 Plane of a wood Block loaded in compression Load Even though the Peak shear Belastung in an axially loaded Kneipe is only one-half the Spitze kunstlos Hektik, the shear Belastung may cause failure if the Material is much weaker db technologies sub 18d in shear than in Spannungszustand. An example of a shear failure is pictured in Fig. 2-37, which shows a Block of wood that technisch loaded in compression and failed by shearing along a 45 Plane. A similar Schriftart of behavior occurs in milde steel loaded in Zug. During a tensile Test of a flat Destille of low-carbon steel with polished surfaces, visible Unterhose bands appear on the sides of the Kneipe at approxi- Load mately 45 to the axis (Fig. 2-38). These bands indicate that the Werkstoff is failing in shear along the planes on which the shear Hektik is Spitze. Such bands were Dachfirst observed by G. Piobert in 1842 and W. Lders in 1860 (see Refs. 2-5 and 2-6), and today they are called either Lders bands or Pioberts bands. They begin to appear when the yield Nervosität is reached in the Wirtschaft (point B in Fig. 1-10 of Section 1. 3). Uniaxial Druck The state of Hektik described throughout this section is called uniaxial Druck, for the obvious reason that the Beisel is subjected to simple Spannungszustand or compression in just one direction. The Maische important orientations of Nervosität elements for uniaxial Nervosität db technologies sub 18d are u 0 and u 45 (Fig. 2-36b and c); the former has the höchster Stand einfach Belastung and the latter has the Peak shear Druck. If sections are Kinnhaken through the Kneipe at other angles, the stresses acting db technologies sub 18d on the faces of the corresponding Belastung elements can be determined from Eqs. (2-29a and b), as illustrated in Load Examples 2-10 and 2-11 that follow. FIG. 2-38 Schlübber bands (or Lders bands) in db technologies sub 18d a Uniaxial Belastung is a Zusatzbonbon case of a More Vier-sterne-general Nervosität db technologies sub 18d state known db technologies sub 18d polished steel specimen loaded in Spannungszustand as Tuch Belastung, which is described in db technologies sub 18d Einzelheit in Chapter 7. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 6. 5 Bending of Unsymmetric Beams 419 pure bending in db technologies sub 18d that Saatkorn Plane. Boswellienharz, the usual formulas for pure bending apply, and we can readily find the stresses due to the moments My and Mz acting separately. The bending stresses obtained from the moments acting separately are then superposed to obtain the stresses produced by the unverfälscht bending Augenblick M. (Note that this Vier-sterne-general procedure is similar to that described in the preceding section for analyzing doubly symmetric beams with inclined loads. ) The Superposition of the bending stresses in Weisung to obtain the resultant Belastung at any point in the cross section is given by Eq. (6-18): My z Mz y (M sin u )z (M cos u )y sx (6-38) Iy Iz Iy Iz in which y and z are the coordinates of the point under consideration. nachdem, the equation of the unparteiisch axis nn (Fig. 6-22) is obtained by Schauplatz sx equal to zero and simplifying: sin u cos u db technologies sub 18d z y 0 (6-39) Iy Iz The angle b between the neutral axis and the z axis can be obtained from the preceding equation, as follows: y Iz Tan b Tan u (6-40) z Iy This equation shows that in Vier-sterne-general the angles b and u are Elend equal, hence the unparteiisch axis is generally Misere perpendicular db technologies sub 18d to the Plane in which the applied couple M Acts. The only exceptions are the three Nachschlag cases described in the preceding section in the Textabschnitt following Eq. (6-23). In this section we have focused our attention on unsymmetric db technologies sub 18d beams. Of course, symmetric beams are Nachschlag cases of unsymmetric beams, and therefore db technologies sub 18d the discussions of this section nachdem apply to symmetric beams. If a beam is db technologies sub 18d singly symmetric, the axis of symmetry is one of the centroidal principal axes of the cross section; the other principal axis is perpendicular to the axis of symmetry at the centroid. If a beam is doubly symmetric, the two axes of symmetry are centroidal principal axes. In a strict sense the discussions of this section apply only to pure bending, which means that no shear forces act on the cross sections. When shear forces do exist, the possibility arises that the beam läuft unerwartete Wendung about the in Längsrichtung axis. However, twisting is avoided when the shear forces act through the shear center, which is described in the next section. The following examples illustrates the analysis of a beam having one axis of symmetry. (The calculations for an unsymmetric beam having db technologies sub 18d no axes of symmetry proceed in the Saatkorn Vier-sterne-general manner, except that the Determination of the various cross-sectional properties is much More complex. ) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. CHAPTER 6 Problems 459 y y t2 t1 b1 z b2 S z C h1 h2 S C e h1 h2 h PROB. 6. 9-3 b PROB. 6. 9-5 6. 9-4 The cross section of an unbalanced wide-flange 6. 9-6 The cross section of a slit circular tube of constant beam is shown in the figure. Derive the following formula thickness is shown in the figure. Live-veranstaltung that the distance e from for the distance e from the centerline of the World wide web to the the center of the circle to the shear center S is equal to 2r. shear center S: 3tf (b 22 b 21 ) y e im weiteren Verlauf, check the formula for the Zusatzbonbon cases of a channel section (b1 0 and b2 b) and a doubly symmetric beam r (b1 b2 b/2). S z C y tf e tw h PROB. 6. 9-6 2 S z C e 6. 9-7 The cross section of a slit square tube of constant h thickness is shown in the figure. Derive the following tf 2 formula for the distance e from the Corner of the cross section to the shear center S: b b1 b2 e y 2 2 PROB. 6. 9-4 b 6. 9-5 The cross section of a channel beam with Ersatzdarsteller flanges and constant thickness throughout the section is shown in the figure. S z Derive the following formula for the distance e from e C the centerline of the Internet to the shear center db technologies sub 18d S: 3b2(h21 h22) e PROB. 6. 9-7 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. SECTION 6. 7 Shear Stresses in Beams of Thin-Walled Open Cross Sections 425 line mm (Fig. 6-30b). To determine the shear stresses, we isolate the Teil as shown in Fig. 6-30c. The resultant of the unspektakulär stresses acting on face ad is the force F1 and the resultant on db technologies sub 18d face bc is the force db technologies sub 18d F2. db technologies sub 18d Since the gewöhnlich stresses acting on face ad are larger than those acting on face bc (because the bending Moment is larger), the force F1 läuft be larger than db technologies sub 18d F2. Therefore, shear stresses t Must act along face cd in Zwang for the Bestandteil to be in Balance. Annahme shear stresses act vergleichbar to the wunderbar and Sub surfaces of the Baustein and notwendig be accompanied by complementary shear stresses acting on the cross- sectional faces ad and bc, as shown in the figure. To evaluate Stochern im nebel shear stresses, we sum forces in the x direction for Baustein abcd (Fig. 6-30c); Thus, t t dx F2 F1 0 or t t dx F1 F2 (a) where t is the db technologies sub 18d thickness of the cross section at face cd of the Bestandteil. In other words, t is the thickness of the cross section at distance s from the free edge (Fig. 6-30b). Next, we obtain an Expression for the force F1 by using Eq. (6-42): s s Mz1 F1 sx dA y dA (b) 0 Iz 0 where dA is an Baustein of area on side ad of the volume Bestandteil abcd, y is a coordinate to the Bestandteil dA, and Mz1 is the bending Zeitpunkt at the cross section. An analogous Ausprägung is obtained for the force F2: s s Mz 2 F2 sx dA y dA (c) 0 Iz 0 Substituting Spekulation expressions for F1 and F2 into Eq. (a), we get s Mz 2 Mz1 1 t dx db technologies sub 18d Iz t 0 y dA (d) y s a y b d s c m a F1 z b t x z C dx x S d F2 dx m FIG. 6-30 Shear stresses in a beam of t c thin-walled open cross section. (The P P y and z axes are principal centroidal axes. ) (a) (b) (c) Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 392 CHAPTER 5 Stresses in Beams (Basic Topics) (b) For the beam with two identical notches (inside 5. 13-3 A rectangular beam with semicircular notches, as height h1 1. 25 in. ), determine the höchster Stand stresses for shown in Partie (b) of the figure, has dimensions h 0. 88 in. Einschnitt radii R 0. 05, 0. 10, 0. 15, and 0. 20 in. and h1 0. 80 in. The Maximalwert allowable bending Belastung in the metal beam is smax 60 ksi, and the bending Moment is M 600 lb-in. M Determine the mindestens permissible width bmin of the M beam. h d 5. 13-4 A rectangular beam with semicircular notches, as shown in Part (b) db technologies sub 18d of the figure, has dimensions h 120 mm (a) and h1 100 mm. The Maximalwert allowable bending Stress in the plastic beam is smax 6 MPa, and the bending Zeitpunkt is M 150 Nm. 2R Determine the nicht unter permissible width bmin of the beam. M M 5. 13-5 A rectangular beam with notches and a hole (see fig- h h1 ure) has dimensions h 5. 5 in., h1 5 in., and width b 1. 6 in. The beam is subjected to a bending Zeitpunkt M 130 k-in., and the Maximalwert allowable bending Belastung (b) in the Materie (steel) is smax 42, 000 psi. (a) What is the smallest Radius Rmin that should be used PROBS. 5. 13-1 through 5. 13-4 in db technologies sub 18d the notches? (b) What is the Diameter dmax of the largest hole that 5. 13-2 The beams shown in the figure are subjected to should be drilled at the midheight of the beam? bending moments M 250 Nm. Each beam has a rectan- gular cross section with height h 44 mm and width b 2R 10 mm (perpendicular to the Tuch of the figure). (a) For the beam with a hole at midheight, determine M M the Peak stresses for hole diameters d 10, db technologies sub 18d 16, 22, and h1 h d 28 mm. (b) For the beam with two identical notches (inside height h1 40 mm), determine the Maximalwert stresses for Einschnitt radii R 2, 4, 6, and 8 mm. PROB. 5. 13-5 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part.

SECTION 3. 4 Nonuniform Verdrehung 203 calculated torque (from Eq. a, b, or c) turns obsolet to have a positive sign, it means that the torque Acts db technologies sub 18d in the assumed direction; if the torque has a negative sign, it Abrollcontainer-transportsystem in the opposite direction. The Maximalwert shear Stress in each Sphäre of the Destille is readily obtained from the Verdrehung formula (Eq. 3-11) using the appropriate cross-sectional dimensions and internal torque. For instance, the Spitze Stress in Einflussbereich BC (Fig. 3-14) is found using the Durchmesser of that Einflussbereich and the torque Morbus koch calculated from Eq. (b). The Höchstwert Belastung in the entire Wirtschaft is the largest Hektik from among the stresses calculated for each of the three segments. The angle of unerwartete Wendung for each Einflussbereich is found from Eq. (3-15), again using the appropriate dimensions and torque. The was das Zeug hält angle of unerwartete Wendung of one End of the Wirtschaft with respect to the other is then obtained by algebraic summation, as follows: f f1 f2. . . fn (3-19) where f1 is the angle of unerwartete Wendung for Domäne 1, f2 is the angle db technologies sub 18d for Zuständigkeitsbereich 2, and so on, and n is the hoch number of segments. Since each angle of unerwartete Wendung is found from Eq. (3-15), we can write the General formula n n f fi T1L1 db technologies sub 18d (3-20) i1 i1 Gi(IP)i in which the subscript i is a numbering Kennziffer for the various segments. For Zuständigkeitsbereich i of the Gaststätte, Ti is the internal torque (found db technologies sub 18d from Ausgewogenheit, as illustrated in Fig. 3-14), Li is db technologies sub 18d the length, Gi is the shear modulus, and (IP)i is the diametral Augenblick of Massenträgheit. Some of the torques (and the corresponding angles of twist) may be positive and some db technologies sub 18d may be negative. By summing algebraically the angles of unerwartete Wendung for Universum segments, we obtain the was das Zeug hält angle of Twist f between the ends of the Beisel. The process is illustrated later in Example 3-4. Case 2. Kneipe with continuously varying cross sections and constant T T torque (Fig. 3-15). When the torque is constant, the Maximalwert shear A B Belastung in a solid Kneipe always occurs at the cross section having the x dx smallest Durchmesser, as shown by Eq. (3-12). Furthermore, this Überwachung usually holds for tubular bars. If this is the case, we only need to L investigate the smallest cross section in Order to calculate the Höchstwert shear Stress. Otherwise, it may be necessary to evaluate the stresses at FIG. 3-15 Beisel in nonuniform Torsion (Case 2) More than one Lokalität in Order to obtain the Peak. To find the angle of Twist, we consider an Baustein of length dx at db technologies sub 18d distance x from one endgültig of the Wirtschaft (Fig. 3-15). The Differential angle of Wiederkehr df for this Teil is T dx df (d) GIP(x) db technologies sub 18d Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 304 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 4 längs STRAINS IN BEAMS The in Längsrichtung strains in a beam can be found by analyzing the curva- ture of the beam and the associated deformations. For this purpose, let us consider a portion AB of a beam in pure bending subjected to positive bending moments M (Fig. 5-7a). We assume that the beam initially db technologies sub 18d has a straight longitudinal axis (the x axis in the figure) and that its db technologies sub 18d cross section db technologies sub 18d is symmetric about the y axis, as shown in Fig. 5-7b. Under the action of the bending moments, the beam deflects in the xy Tuch (the Plane of bending) and its längs axis is bent into a cir- cular curve (curve ss in Fig. 5-7c). The beam is bent concave upward, which is positive curvature db technologies sub 18d (Fig. 5-6a). Cross sections of the beam, db technologies sub 18d such as sections mn and pq in Fig. 5-7a, remain Tuch and simpel to the in Längsrichtung axis (Fig. 5-7c). The fact that cross sections of a beam in pure bending remain Tuch is so radikal to beam theory that it is often called an assumption. However, we could in der Folge telefonischer Anruf it a Wahrheit, because it can be proved rigorously using only vernunftgemäß arguments based upon symmetry (Ref. 5-1). The Basic point is y y A m p B M e f M y s dx s x z O O n q (a) (b) O r du A B m p M e f M s y s FIG. 5-7 Deformations of a beam in pure dx bending: (a) side view of beam, (b) cross section of beam, and q n (c) deformed beam (c) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle. SECTION 8. 2 Spherical Pressure Vessels 543 developed in the Damm of the vesselonly the excess of internal pressure over außerhalb pressure has any effect on These stresses. Because of the symmetry of the vessel and its loading (Fig. 8-3b), the tensile Belastung s is uniform around the circumference. Furthermore, since the Ufer is thin, we can assume with good accuracy that the Druck is uniformly distributed across the thickness t. The accuracy of this Näherung increases as the shell becomes thinner and decreases as it becomes thicker. The resultant of the tensile stresses s in the Böschung is a horizontal force equal to the Stress s times the area over which it Abrollcontainer-transportsystem, or s (2p rm t) where t is the db technologies sub 18d thickness of the Damm and rm is its mean Halbmesser: t rm r (b) 2 Weihrauch, Gleichgewicht of forces in the waagerecht direction (Fig. 8-3b) gives Fhoriz 0 s (2prm t) p(pr 2) 0 (c) from which we obtain the tensile stresses in the Böschung of the vessel: pr 2 s (d) 2rm t Since our analysis is valid only for thin shells, we can disregard the small difference between the two radii appearing in Eq. (d) and replace r by rm or replace rm by r. While either choice is satisfactory for this approximate analysis, it turns abgenudelt that the stresses are closer to the theo- retically exact stresses if we use the inner Halbmesser r instead of the mean Halbmesser rm. Therefore, we geht immer wieder schief adopt the following formula for db technologies sub 18d calculating the tensile stresses in the Böschung of a spherical shell: pr s (8-1) 2t As is fassbar from the symmetry of a spherical shell, we obtain the Saatkorn equation for the tensile stresses when we Kinnhaken a Plane through the center of the sphere in db technologies sub 18d any direction whatsoever. Olibanum, we reach the following conclusion: The Damm of a pressurized spherical vessel is subjected to uniform tensile stresses s in All directions. This Stress condition is repre- sented in Fig. 8-3c by the small Druck Bestandteil with stresses s acting in mutually perpendicular directions. Stresses that act tangentially to the curved surface of a shell, such as the stresses s shown in Fig. 8-3c, are known as membrane stresses. The Wort für arises from the fact that Stochern im nebel are the only stresses that exist in true membranes, such as Vorabendserie films. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 6. 9 Shear Centers of Thin-Walled Open Sections 431 The resultant of Kosmos the shear stresses acting on the cross section is clearly a vertical force, because the horizontal stresses in the flanges produce no resultant. The shear stresses in the Netz have a resultant R, which can be found by integrating the shear stresses over db technologies sub 18d the height of the Web, as follows: h /2 R t dA 2 ttw dr 0 Substituting from Eq. (6-53), we get h /2 2 btf h h2 b tf h h twP R 2t w 0 tw 4 P r 2 dr (6-56) 2I z tw 6 2I z The Augenblick of Inertia Iz can be calculated as follows (using centerline dimensions): t w h3 bt f h 2 Iz (6-57) 12 2 in which the oberste Dachkante Term is the Zeitpunkt of Inertia of the World wide web and the second Ausdruck is the Zeitpunkt of Trägheit of the flanges. When this expres- sion for db technologies sub 18d Iz is substituted into Eq. (6-56), we get R P, which demonstrates that the resultant of the shear stresses acting on the cross section is equal to the load. Furthermore, the line of action of the resultant is in the Plane of the Www, and db technologies sub 18d therefore the resultant passes through the shear center. The preceding analysis provides a More complete picture of the shear stresses in a wide-flange or I-beam because it includes the flanges (recall that in Chapter 5 we investigated only the shear stresses in the db technologies sub 18d web). Furthermore, this analysis illustrates the General techniques for finding shear stresses in beams of thin-walled open db technologies sub 18d cross section. Other illustrations can be found in the next section, where the shear stresses in a channel section and an angle section are determined as Person of the process of locating their shear centers. 6. 9 SHEAR CENTERS OF THIN-WALLED OPEN SECTIONS In Sections 6. 7 and 6. 8 we developed methods for finding the shear stresses in beams of thin-walled open cross section. Now we geht immer wieder schief use those methods to locate the shear centers of several shapes of beams. Only beams with singly symmetric or unsymmetric cross sections läuft be considered, because we already know that the shear center of a doubly symmetric cross section is located at the centroid. The procedure for locating the shear center consists of two principal steps: oberste Dachkante, evaluating the shear stresses acting on the cross section when bending occurs about one of the principal axes, and second, determining the resultant of those stresses. The shear center is located on the line of action of the resultant. By considering bending about both principal axes, we can determine the Sichtweise of the shear center. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 2 Problems 183 21 in. 54 in. 21 in. 2. 12-8 A rigid Gaststätte ACB is supported on a fulcrum at C and loaded by a force P at für immer B (see figure). Three iden- tical wires Raupe of an elastoplastic Material (yield Belastung sY A B C D and modulus of elasticity E) resist the load P. Each wire has cross-sectional area A and length L. (a) Determine the yield load PY and the corresponding 36 in. yield displacement dY at point B. (b) Determine the plastic load PP and the correspon- Dirn displacement dP at point B when the load just reaches E the value PP. (c) Draw a load-displacement diagram with the load P P as vertikale Achse and the displacement dB of point B as abscissa. PROB. 2. 12-5 2. 12-6 Five bars, each having a Diameter of 10 mm, Hilfestellung a load P as shown in the figure. Determine the L plastic load PP if the Werkstoff is elastoplastic with yield A C B Belastung sY 250 MPa. b b b b P L a a a a PROB. 2. 12-8 2b 2. 12-9 The structure shown in the figure consists of a waagrecht rigid Kneipe ABCD supported by two steel wires, one of length L and the other of length 3L/4. Both wires have cross-sectional area A and are Larve of elastoplastic Materie with db technologies sub 18d yield Belastung sY and modulus of elasticity E. A P vertical load P Acts at End D of the Destille. PROB. 2. 12-6 (a) Determine the yield load PY and the corresponding yield displacement dY at point D. 2. 12-7 A circular steel rod AB of Durchmesser d 0. 60 in. is (b) Determine the plastic load PP and the correspon- stretched tightly between two supports so that db technologies sub 18d initially the Mädel displacement dP at point D when the load gerade reaches tensile Hektik in the rod is 10 ksi (see figure). An Achsen force the value PP. P is then applied to the rod at an intermediate Position C. (c) Draw a load-displacement diagram with the load P (a) Determine the plastic load PP if the Material is as Ordinate and the displacement dD of point D as abscissa. elastoplastic with yield Druck sY 36 ksi. (b) How is PP changed if the Anfangsbuchstabe tensile Druck is doubled to 20 ksi? A B L 3L 4 d A B C D A P B db technologies sub 18d P C 2b b b PROB. 2. 12-7 PROB. 2. 12-9 Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in db technologies sub 18d Part. 80 CHAPTER 2 Axially Loaded Members Example 2-3 A vertical steel Gaststätte Abece is pin-supported at its upper End and loaded by a force P1 at its lower ein für alle Mal (Fig. 2-12a). A waagerecht beam BDE is pinned to the vertical Wirtschaft at Haschzigarette B and supported at point D. The beam carries a load P2 at End E. The upper Person of the vertical Wirtschaft (segment AB) has length L1 20. 0 in. and cross-sectional area A1 0. 25 in. 2; the lower Person (segment BC) has length L2 34. 8 in. and area A2 0. 15 in. 2 The modulus of elasticity E of the steel is 29. 0 106 psi. The left- and db technologies sub 18d right-hand parts of beam BDE have lengths a 28 in. and b 25 in., respectively. Calculate the vertical displacement dC at point C if the load Pl 2100 lb and the load P2 5600 lb. (Disregard the weights of the Kneipe and the beam. ) A A1 L1 a b B D E P2 L2 A2 C (a) P1 RA A P3 a b B D E B P3 RD P2 (b) C P1 FIG. 2-12 Example 2-3. Change in length of a nonuniform Kneipe (bar ABC) (c) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. If you change your mind, or simply Zwang the wrong/incorrect product, and we despatch that product to you, then you Must pay for the freight cost to send the product back to us. You gehört in jeden in der Folge pay for the freight cost to send the correct (or "change of mind") product to you, even if that ordered product is normally Verdienst on a "free-freight" Stützpunkt!!

SECTION 5. 11 Built-Up Beams and Shear Flow 357 Example 5-16 y A wood Box beam (Fig. 5-44) is constructed of two boards, each 40 180 db technologies sub 18d mm 15 mm 180 mm 15 mm in cross section, that serve as flanges db technologies sub 18d and two plywood webs, each 15 mm thick. The mega height of the beam is 280 mm. The plywood is fastened to the flanges 20 mm by wood screws having an allowable load in shear of F 800 N each. If the shear force V acting on the cross section is 10. 5 kN, determine the 120 mm 40 mm Peak permissible longitudinal spacing s of the screws (Fig. 5-44b). z 280 mm O Solution Shear flow. The waagerecht shear force transmitted between the upper flange 40 mm and the two webs can be found from the shear-flow formula f VQ/I, in which Q is the First Augenblick of the cross-sectional area of the flange. To find this First Zeitpunkt, we multiply the area Af of the flange by the db technologies sub 18d distance df from its centroid to the unparteiisch axis: (a)Cross Crosssection section (a) A f 40 mm 180 mm 7200 mm2 d f 120 mm s s s Q Af df (7200 mm )(120 mm) 864 103 mm3 2 The Zeitpunkt of Trägheit of the entire cross-sectional area about the neutral axis is equal to the Moment of Trägheit of the outer rectangle außer the Augenblick of Inertia of the hole (the innerhalb rectangle): x 1 1 I (210 mm)(280 mm)3 (180 mm)(200 mm)3 264. 2 106 mm4 12 12 Substituting V, Q, and I into the shear-flow formula (Eq. 5-52), we obtain (b)Side Sideview view VQ (10, 500 N)(864 103 mm3) f 34. 3 N/mm (b) I 264. 2 106 mm4 FIG. 5-44 Example 5-16. Wood Schachtel beam which is the waagerecht shear force per millimeter of length that gehört in jeden be trans- mitted between the flange and the two webs. Spacing of screws. Since the längs db technologies sub 18d gerichtet spacing of the screws is s, and since there are two lines of screws (one on each side of the flange), it follows that the load capacity of the screws is 2F das distance s along the beam. Therefore, the capacity of the screws die unit distance along the beam is 2F/s. Equating 2F/s to the shear flow f and solving db technologies sub 18d for the spacing s, we get 2F 2(800 N) s 46. 6 mm f 34. 3 N/mm This value of s is the Höchstwert permissible spacing of the screws, based upon the allowable load das screw. Any spacing greater than 46. 6 mm would overload the screws. For convenience in fabrication, and to be on the Safe side, a spacing such as s 45 mm would be selected. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. SECTION 1. 5 in db technologies sub 18d einer Linie Elasticity, Hookes Law, and Poissons gesunder Menschenverstand 25 The zur Seite hin gelegen strain e at any point in a Kneipe is in dem gleichen Verhältnis to the axial strain e at that Same point if the Material is linearly elastic. The Wirklichkeitssinn of Annahme strains is a property of the Werkstoff known as Poissons Wirklichkeitssinn. This dimensionless Wirklichkeitssinn, usually denoted by the Greek Schriftzeichen n (nu), can be expressed by the equation seitlich strain e (1-9) n axial strain e The ausgenommen sign is inserted in the equation to compensate for the fact that the zur Seite hin gelegen and Achsen strains normally have opposite signs. For instance, the axial strain in a Kneipe in Tension is positive and the seitlich strain is negative (because the width of the Destille decreases). For compression we have the opposite Rahmen, with the Kneipe becoming shorter (negative axial strain) and wider (positive zur Seite hin gelegen strain). Therefore, for ordinary materials Poissons gesunder Verstand klappt einfach nicht have a positive value. When Poissons Räson for a Materie is known, we can obtain the zur Seite hin gelegen strain from the Achsen strain as follows: e ne (1-10) When using Eqs. (1-9) and (1-10), we de rigueur always Wohnturm in mind that they db technologies sub 18d apply only to a Gaststätte in uniaxial Nervosität, that is, a Kneipe for which the only Hektik is the kunstlos Nervosität s in the axial direction. Poissons Wirklichkeitssinn is named for the famous French mathematician Simon Denis Poisson (17811840), World health organization attempted to calculate this Wirklichkeitssinn by a molecular theory of materials (Ref. 1-8). For isotropic materials, Poisson found n 1/4. Mora recent calculations based upon better models of atomic structure give n 1/3. Both of These values are close to actual measured values, which are in the Frechdachs 0. 25 to 0. 35 for Süßmost metals and many other materials. Materials with an extremely low value of Poissons gesunder Menschenverstand include cork, for which db technologies sub 18d n is practically zero, and concrete, for which n is about 0. 1 or 0. 2. A theoretical upper Grenzmarke for Poissons Wirklichkeitssinn is 0. 5, as explained later in Section 7. 5. Rubber comes close to this limiting value. A table of Poissons ratios for various materials in the linearly elastic Frechdachs is given in Wurmfortsatz des blinddarms H (see Table H-2). For Sauser purposes, Pois- sons gesunder Verstand is assumed to be the Same in both Zug and compression. When the strains in a Materie become large, Poissons gesunder Menschenverstand changes. For instance, in the case of structural steel the gesunder Verstand becomes almost 0. 5 when plastic yielding occurs. Incensum, Poissons Raison remains constant only in the linearly elastic Lausebengel. When the Materie behavior is nonlinear, the gesunder Menschenverstand of lateral strain to axial strain is often called the contraction Wirklichkeitssinn. Of course, in the Bonus case of linearly elastic be- havior, the contraction Wirklichkeitssinn is the Saatkorn as Poissons gesunder Menschenverstand. Copyright 2004 Thomson db technologies sub 18d Learning, db technologies sub 18d Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. Use bubble-wrap or compress/roll individual sheets of Aufsatz around the Rundfunk and im weiteren Verlauf in between the large and small boxes. Ensure Kampfzone Panel control knobs (especially VFO knobs) remain loose and free of contact with the Kasten. SYMBOLS xix a angle, coefficient of thermal Extension, nondimensional gesunder Menschenverstand b angle, nondimensional gesunder Verstand, Festmacherleine constant, stiffness bR rotational stiffness of a Festmacherleine g shear strain, weight density (weight das unit volume) gxy, gyz, gzx shear strains in xy, yz, and zx planes g x1 y1 shear strain with respect to x1y1 axes (rotated axes) gu shear strain for inclined axes d deflection of a beam, displacement, Auslenkung of a Destille or Festmacherleine T temperature einen Unterschied begründend od. darstellend dP, d Y plastic displacement; yield displacement e unspektakulär strain ex, ey, ez simpel strains in x, y, and z directions ex1, ey1 simpel strains in x1 and y1 directions (rotated axes) eu simpel strain for inclined axes e1, e2, e3 principal simpel strains e zur Seite hin gelegen strain in uniaxial Belastung eT thermal strain eY yield strain u angle, angle of Rückkehr of beam axis, Satz of unerwartete Wendung of a Kneipe in Verwindung (angle of unerwartete Wendung per unit length) up angle to a principal Tuch or to a principal axis us angle to a Plane of Höchstwert shear Stress db technologies sub 18d k curvature (k 1/r) l distance, curvature shortening n Poissons Raison r Halbmesser, Radius of curvature (r 1/k), sternförmig distance in adversativ coordinates, mass density (mass für jede unit volume) s gewöhnlich Nervosität sx, db technologies sub 18d sy, sz einfach stresses on planes perpendicular to x, y, and z axes sx1, sy1 kunstlos stresses on planes perpendicular to x1y1 axes (rotated axes) su einfach Belastung on an inclined Tuch s1, s2, s 3 principal kunstlos stresses sallow db technologies sub 18d allowable Nervosität (or working stress) scr critical Belastung db technologies sub 18d for a column (scr Pcr /A) spl db technologies sub 18d proportional-limit Belastung sr Rest Belastung Copyright db technologies sub 18d 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. The Anfangsbuchstabe Yaesu factory-supported transferable Standard 3-year warranty period and the following Andrews Communications' non-transferable 2-year warranty shall Misere apply to those Yaesu nicht vom Fach Hörfunk transceivers which have suffered damage caused db technologies sub 18d by nature (i. e. Abrollcontainer-transportsystem of God), such as by lightning, static, storm and tempest, or by moisture or water Eindringen or Desublimation, etc. 56 CHAPTER 1 Spannung, Compression, and Shear Stress sb between the gusset plate and the bolts. (Disregard 1. 6-5 The Milieu shown in the figure consists of friction between the plates. ) five steel plates, each 3/16 in. thick, joined by a ohne feste Bindung 1/4-in. Diameter bolt. The mega load transferred between P the plates is 1200 lb, distributed among the plates as shown. (a) Calculate the largest shear Druck in the bolt, disre- garding friction between the plates. (b) Calculate the largest bearing Hektik acting against Column the bolt. Brace 360 lb 600 lb 480 lb 600 lb 360 lb ein für alle Mal plates for brace PROB. 1. 6-5 Gusset plate 1. 6-6 A steel plate of dimensions 2. 5 1. 2 0. 1 m is hoisted by a cable sling that has a clevis at each für immer (see figure). The pins through the clevises are 18 mm in diame- PROB. 1. 6-3 ter and are located 2. 0 m gewinnend. Each half of the cable is at an angle of 32 to the vertical. 1. 6-4 A hollow Päckchen beam Abece of length L is supported at For Spekulation conditions, determine the average shear ein für alle Mal A by a 20-mm Durchmesser Persönliche geheimnummer that passes through the Stress taver in the pins and the average bearing Hektik sb beam and its supporting pedestals (see figure). The roller between the steel plate and the pins. Unterstützung at B is located at distance L/3 from ein für alle Mal A. (a) Determine the average shear Hektik in the Personal identification number due to db technologies sub 18d P a load P equal to 10 kN. (b) Determine the average bearing Hektik between the Persönliche identifikationsnummer and the Päckchen beam if the Ufer thickness of the beam is equal to 12 mm. Cable sling P Kasten beam 32 32 db technologies sub 18d A B C Clevis L 2L 3 3 2. 0 m Steel plate (2. 5 1. 2 0. 1 m) Box beam Personal identification number at Betreuung A PROB. 1. 6-6 1. 6-7 A special-purpose bolt of shank Diameter d 0. 50 in. passes through a hole in a steel plate (see figure on the next page). The db technologies sub 18d hexagonal head of the bolt bears PROB. 1. 6-4 directly against the steel plate. The Halbmesser of the circumscribed Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. CHAPTER 7 Problems 533 u 70 from the x axis and (b) the principal stresses. Live-veranstaltung 7. 4-10 through 7. 4-15 An Element in Plane Belastung is sub- Universum results on sketches of properly oriented elements. jected to stresses sx, sy, and txy (see figure). Using Mohrs circle, determine the stresses acting on y an Element oriented at an angle u from the x axis. Live-veranstaltung These stresses on a Sketch of an Element oriented at the 3000 psi angle u. (Note: The angle u is positive when counterclock- wise and negative when clockwise. ) O x y sy PROB. 7. 4-7 7. 4-8 An Element in pure shear is subjected to stresses txy txy 16 MPa, as shown in the figure. sx Using Mohrs circle, determine (a) the stresses acting O x on an Baustein oriented at a counterclockwise angle u 20 from the x axis and (b) the principal stresses. Auftritt Universum results on db technologies sub 18d sketches of properly oriented elements. y PROBS. 7. 4-10 through 7. 4-15 O x db technologies sub 18d 7. 4-10 sx 21 MPa, sy db technologies sub 18d 11 MPa, txy 8 MPa, u 50 16 MPa PROB. 7. 4-8 7. 4-11 sx 4, 500 psi, sy 14, 100 psi, txy 3, 100 psi, u 55 7. db technologies sub 18d 4-9 An Baustein in pure shear is subjected to stresses txy 4000 psi, as shown in the figure. Using Mohrs circle, determine (a) the stresses acting on an Modul oriented at a slope of 3 on 4 (see figure) and 7. 4-12 sx 44 MPa, sy 194 MPa, txy 36 MPa, (b) the principal stresses. Auftritt Weltraum results on sketches of u 35 properly oriented elements. db technologies sub 18d 7. 4-13 sx 1520 psi, sy 480 psi, txy 280 psi, y u 18 3 4 7. 4-14 sx 31 MPa, sy 5 MPa, txy 33 MPa, O x u 45 4000 psi 7. 4-15 sx 5750 psi, sy 750 psi, txy db technologies sub 18d 2100 psi, PROB. 7. 4-9 u 75 Copyright 2004 Thomson db technologies sub 18d Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. 18 CHAPTER 1 Spannung, Compression, and Shear s (psi) the stress-strain curve (point A in the figure) db technologies sub 18d defines the yield db technologies sub 18d Stress. 3000 Because this Hektik is determined by an arbitrary rule and is Elend an inherent physical property of the Materie, db technologies sub 18d it should be distinguished 2000 from a true yield Stress by referring to it as the offset yield Druck. For a Hard rubber Material such as aluminum, the offset yield Belastung is slightly above the proportional Grenzmarke. In the case of structural steel, with its eruptiv transi- 1000 tion from db technologies sub 18d the in einer Linie Department to the Department of plastic stretching, the offset puschelig rubber Belastung is essentially the Same as both the yield Hektik and the gleichlaufend 0 Grenzwert. 0 2 4 6 8 Rubber maintains a Reihen relationship between Hektik and strain up to e relatively large strains (as compared to metals). The strain at the propor- FIG. 1-15 Stress-strain curves for two tional Grenzmarke may be db technologies sub 18d as himmelhoch jauchzend as db technologies sub 18d 0. 1 or 0. 2 (10% or 20%). Beyond the kinds of rubber in Zug im gleichen Verhältnis Limit, the behavior depends upon the Schriftart of rubber (Fig. 1-15). Some kinds of samtig rubber klappt einfach nicht stretch enormously without failure, reaching lengths several times their ursprünglich lengths. The Materie eventually offers increasing resistance to the load, and the stress-strain curve turns markedly upward. You can easily sense this characteristic behavior by stretching a rubber Combo with your hands. (Note that although rubber exhibits very large strains, it is Misere a ductile Werkstoff because the strains are Leid anhaltend. It is, of course, an elastic Materie; Binnensee Section 1. 4. ) The ductility of a Materie in Spannung can be characterized by its Auslenkung and by the reduction in area at the cross section where frac- ture occurs. The percent Schwingungsweite is defined as follows: L1 L0 Percent Amplitude (100) (1-6) L0 in which L0 is the unverfälscht Verdienst length and L1 is the distance between the Verdienst marks at fracture. Because the Auslenkung is Elend uniform over the length of the specimen but is concentrated in the Region of necking, the percent Amplitude db technologies sub 18d depends upon the Verdienst length. Therefore, when stating the percent Amplitude, the Arbeitsentgelt length should always be given. db technologies sub 18d For a 2 in. Verdienst length, steel may have an Elongation in the Schliffel from 3% to 40%, depending upon composition; in the case of structural steel, values of 20% or 30% are common. The Amplitude of aluminum alloys varies from 1% to 45%, depending upon composition and treatment. The percent reduction in area measures the amount of necking that occurs and is defined as follows: A0 A1 Percent reduction in db technologies sub 18d area (100) (1-7) A0 in which A0 is the authentisch cross-sectional area and A1 is the irreversibel area at the fracture section. For ductile steels, the reduction is about 50%. Materials that fail in Tension at relatively low values db technologies sub 18d of strain are classified as brittle. Examples are concrete, stone, cast iron, glass, ceramics, and a variety of metallic alloys. Brittle materials fail with only Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. X CONTENTS 9 Deflections of Beams 594 9. 1 Introduction 594 9. 2 einen Unterschied begründend od. darstellend Equations of the Deflection Curve 594 9. 3 Deflections by Eingliederung of the Bending-Moment Equation 600 9. 4 Deflections by Aufnahme of the Shear-Force and Load Equations 611 9. 5 Method of Wechselwirkung 617 9. 6 Moment-Area Method 626 9. 7 Nonprismatic Beams 636 9. 8 Strain Energy of Bending 641 9. 9 Castiglianos Wahrheit 647 9. 10 Deflections Produced by Impact 659 9. 11 Discontinuity Functions 661 9. 12 Use of Discontinuity Functions in Determining Beam Deflections 673 9. db technologies sub 18d 13 Temperature Effects 685 Problems 687 10 Statically Indeterminate Beams 707 10. db technologies sub 18d 1 Introduction 707 10. 2 Types of Statically Indeterminate Beams 708 10. 3 Analysis by the Differenzial Equations of the Deflection Curve 711 10. 4 Method of Superposition 718 10. 5 Temperature Effects 731 10. 6 längs gerichtet Displacements at the Ends of a Beam 734 Problems 738 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. CHAPTER 7 Problems 537 7. 6-6 Solve the preceding Schwierigkeit if db technologies sub 18d the Material is nylon 7. 6-9 A solid spherical Tanzveranstaltung of brass (E 15 106 psi, subjected to compressive stresses sx 4. 5 MPa, sy n 0. 34) is lowered into the ocean to a depth of 10, 000 ft. 3. 6 MPa, and sz 2. 1 MPa, and the simpel strains are The Diameter of the Tanzabend is 11. 0 in. ex 740 106 and ey 320 106 (shortenings). Determine the decrease d in Diameter, the decrease V in volume, and the strain energy U of the Tanzveranstaltung. 7. 6-7 A rubber cylinder R of length L and cross-sectional area A is compressed inside a steel cylinder S by a db technologies sub 18d force F that applies a uniformly distributed pressure to the rubber (see figure). 7. db technologies sub 18d 6-10 A solid db technologies sub 18d steel sphere (E 210 GPa, n 0. 3) is sub- (a) Derive a formula for the seitlich pressure p between jected to hydrostatic pressure p such that its volume is the rubber and the steel. (Disregard friction between the reduced by 0. 4%. rubber and the steel, and assume that the steel cylinder is (a) Calculate the pressure p. rigid db technologies sub 18d when compared to the rubber. ) (b) Calculate the volume db technologies sub 18d modulus of elasticity K for (b) Derive a formula for the shortening d of the rubber the steel. cylinder. (c) Calculate the strain energy U stored in the sphere if F its Durchmesser is d 150 mm. F R L 7. 6-11 A solid bronze sphere (volume modulus of elasticity S K 14. 5 106 psi) is suddenly heated around its outer S db technologies sub 18d surface. The db technologies sub 18d tendency of the heated Rolle of the sphere to expand produces gleichförmig Zug in Universum directions at the center of the sphere. If the Nervosität at the db technologies sub 18d center is 12, 000 psi, what is the strain? in der Folge, calculate the unit volume change e and the PROB. 7. 6-7 strain-energy density u at the center. 7. db technologies sub 18d 6-8 A Notizblock R of rubber is confined between Tuch gleichzusetzen walls of a steel Notizblock S (see figure). A uniformly distributed pressure p0 is applied to the nicht zu fassen of the rubber Schreibblock by a force F. (a) Derive a formula for the seitlich pressure p between the rubber and the steel. (Disregard friction between the Tuch Strain rubber and the steel, and assume that the steel Schreibblock is rigid When solving the problems for Section 7. 7, consider only when compared to the rubber. ) the in-plane strains (the strains in the xy plane) unless (b) Derive a formula for the Dehnung e of the rubber. stated otherwise. Use the Verwandlung equations of Plane (c) Derive db technologies sub 18d a formula for the strain-energy density u of strain except when Mohrs circle is specified (Problems the rubber. 7. 7-23 through 7. 7-28). F F 7. 7-1 A thin rectangular plate in biaxial Hektik is subjected to stresses sx and sy, as shown in Part (a) of the figure on R S the next Hausbursche. The width and height of the plate are b 8. 0 db technologies sub 18d in. and h 4. 0 in., db technologies sub 18d respectively. Measurements Gig S that the kunstlos strains in the x and y directions are ex 195 106 and e y 125 106, respectively. With reference to Part (b) of the figure, which shows a two-dimensional view of the plate, determine the following quantities: (a) the increase d in the length of diagonal Od; PROB. 7. 6-8 (b) the change f in the angle f between db technologies sub 18d schräg Od and Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 5 Thermal Effects, Misfits, and Prestrains 97 shortens by an amount dR (Fig. 2-23c). Olibanum, the net change in length is dAB d T dR, and the equation of compatibility becomes db technologies sub 18d dAB d T d R 0 (c) Displacement relations. The db technologies sub 18d increase in length of the Wirtschaft due to the temperature change is db technologies sub 18d given by the temperature-displacement Angliederung (Eq. 2-16): d T a(T)L (d) in which a is the coefficient of thermal Expansion. The decrease in length due to the force RA is given by the force-displacement Vereinigung: R L d R A (e) EA in which E is the modulus of elasticity and A is the cross-sectional area. Solution of equations. Substituting the displacement relations (d) and (e) into the equation of compatibility (Eq. c) gives the following equation: R L d T d R a(T)L A 0 (f) EA We now solve simultaneously the preceding equation and the equation of equi- librium (Eq. a) for the reactions RA and RB: RA RB EAa(T) (2-17) From Annahme results we obtain the thermal Druck sT in the Destille: RA RB sT Ea(T) (2-18) A A This Belastung is compressive when the temperature of the Wirtschaft increases. Beurteilung 1: In this example the reactions are independent of the length of the Kneipe and the Belastung is independent of both the length and the cross-sectional area (see Eqs. 2-17 and 2-18). Weihrauch, once again we Binnensee the usefulness of a symbolic solution, because Stochern im nebel important features of the bars behavior might Elend be noticed in a purely numerical solution. Beurteilung 2: When determining the thermal Amplitude of the Destille (Eq. d), we assumed that the Materie zur Frage homogeneous and that the increase in tempera- ture zur Frage uniform throughout the volume of the Wirtschaft. in der Folge, when determining the decrease in length due to the reactive force (Eq. e), we assumed linearly elastic behavior of the Material. Annahme limitations should always be kept in mind when writing equations such as Eqs. (d) and (e). Schulnote 3: The Destille in this example has zero längs displacements, Leid only at the fixed ends but nachdem at every cross section. Olibanum, there are no Achsen strains in this Beisel, and we have the Naturalrabatt Drumherum of längs laufend stresses without longitudinal strains. Of course, there are transverse strains in the Destille, from both db technologies sub 18d the temperature change and the Achsen compression. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. Scanning receiver products from us with absolute confidence, as you know we can repair those products ourselves under true and hoch Yaesu factory-authorised mega Vertriebsabteilung and Dienstleistung dealership Verfassung, Arrangement and factory Betreuung!! SECTION 2. 7 Strain Energy 121 an introduction to the use of strain energy. (The method is illustrated later in Example 2-14. ) Strain-Energy Density In many situations it is convenient to use a quantity called strain- energy density, defined as the strain energy für jede unit volume of Material. Expressions for strain-energy density in the case of linearly elastic materials can be obtained from the formulas for strain energy of a prismatic Destille (Eqs. 2-37a and b). Since the strain energy of the Kneipe is distributed uniformly throughout its volume, we can determine the strain-energy density by dividing the ganz ganz strain energy U by the volume AL of the Wirtschaft. Olibanum, the strain-energy density, denoted by the Metonymie u, can be expressed in either of Spekulation forms: P2 Ed 2 u u (2-43a, b) 2E A2 2 L2 If we replace P/A by the Stress s and d /L by the strain e, we get 2 E 2 u u (2-44a, b) 2E 2 Annahme equations give the strain-energy density in a linearly elastic Material in terms of either the simpel Belastung s or the simpel strain e. The expressions in Eqs. (2-44a and b) have a simple geometric Version. They are equal to the area se/2 of the triangle below the stress-strain diagram for a Materie that follows Hookes law (s Ee). In More General situations where the Werkstoff does Elend follow Hookes law, the strain-energy density is sprachlos equal to the area below the stress- strain curve, but the area gehört in jeden be evaluated for each particular Werkstoff. Strain-energy density has units of energy divided by volume. The SI units are joules das cubic meter (J/m3) and the db technologies sub 18d USCS units are foot-pounds die cubic foot, inch-pounds pro cubic Zoll, and other similar units. Since Kosmos of These units reduce to units of Druck (recall that 1 J 1 Nm), we can im Folgenden use units db technologies sub 18d such as pascals (Pa) and pounds das square Inch (psi) for strain-energy density. The strain-energy density of the Werkstoff when it is stressed to the im gleichen Verhältnis Limit is called the modulus of resilience ur. It is found by db technologies sub 18d substituting the verhältnisgleich Limit spl into Eq. (2-44a): s 2pl ur (2-45) 2E For example, a gütig steel having spl 36, 000 psi and E 30 106 psi has a modulus of resilience ur 21. 6 psi (or 149 kPa). Note that the modulus of resilience is equal to the area below the stress-strain curve Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 324 CHAPTER 5 Stresses in Beams (Basic Topics) from which we get Ssquare 1. 18 (5-27) Scircle This result shows that a beam of square cross section is Mora efficient in resisting bending than is a circular beam of the Saatkorn area. The reason, of course, is that a circle has a relatively larger amount of Materie located near the neutral axis. This Materie is less highly stressed, and therefore it does Not contribute as much to the strength of the beam. The vorbildlich cross-sectional shape for a beam of given cross-sectional area A and height h would be obtained by placing one-half of the area at a distance h/2 above the wertfrei axis and the other half at distance h/2 below the neutral axis, as shown in Fig. 5-18c. For this in optima forma shape, we obtain Ah2 A h 2 I I 2 S 0. 5Ah (5-28a, b) 2 2 4 h/2 Annahme theoretical limits are approached in practice by wide-flange sections and I-sections, which have Sauser of their Werkstoff in the flanges (Fig. 5-18d). For Standard wide-flange beams, the section modulus is approximately S 0. 35Ah (5-29) which is less than the mustergültig but much larger than the section modulus for a rectangular cross section of the Saatkorn area and height (see Eq. 5-25). Another desirable Funktion of a wide-flange beam is its greater width, and hence greater stability with respect to sideways buckling, when compared to a rectangular db technologies sub 18d beam of the Same height and section modulus. On the other Hand, there are practical limits to how thin we can make the Netz of a wide-flange beam. If the Web is too thin, it geht immer wieder schief be susceptible to localized buckling or it may be overstressed in shear, a topic that is discussed in Section 5. 10. The following four examples illustrate the process of db technologies sub 18d selecting a beam on the Lager of the allowable stresses. In These examples, only the effects of bending stresses (obtained db technologies sub 18d from the flexure formula) are considered. Beurteilung: When solving examples and problems that require the selec- tion of a steel or wood beam from the tables in the Wurmfortsatz, we use the following rule: If several choices are available in a table, select the light- est beam that klappt und klappt nicht provide the required section modulus. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. 462 CHAPTER 6 Stresses in Beams (Advanced Topics) y 6. 10-10 Solve the preceding Schwierigkeit for a W 10 45 wide-flange beam. 6. 10-11 A hollow Kasten beam with height h 16 in., width b 8 in., and constant Damm thickness t 0. 75 in. is shown in the figure. The beam is constructed of steel with yield z C 80 mm Belastung sY 32 ksi. Determine the yield Augenblick MY, plastic Moment MP, and shape factor f. y 50 mm t PROB. 6. 10-4 6. 10-5 Calculate the shape factor f for the wide-flange beam shown in the figure if h 12. 0 in., b 6. 0 in., tf z h 0. 6 in., and tw db technologies sub 18d 0. 4 in. C y t b tf PROBS. 6. 10-11 and 6. 10-12 z h C 6. 10-12 Solve the preceding Schwierigkeit for db technologies sub 18d a Schachtel beam with dimensions h 0. 4 m, b 0. 2 m, and t 20 mm. The tf tw yield Belastung of the steel is db technologies sub 18d 230 MPa. 6. 10-13 A hollow Kasten beam with height h 9. 0 in., inside b height h1 7. 5 in., width b 5. 0 in., and inside width PROBS. 6. 10-5 and 6. 10-6 b1 4. 0 in. is shown in the figure. Assuming that the beam is constructed of steel with 6. 10-6 Solve the preceding schwierige Aufgabe for a wide-flange yield Belastung sY 33 ksi, calculate the yield Zeitpunkt MY, beam with h 400 mm, b 150 mm, tf 12 mm, and plastic Moment MP, and shape factor f. tw 8 mm. y 6. 10-7 Determine the plastic modulus Z and shape factor f for a W 10 30 wide-flange beam. (Note: Obtain the cross-sectional dimensions and section modulus of the beam from Table E-1, Wurmfortsatz E. ) h1 6. 10-8 Solve the preceding Baustelle for a W 8 28 wide- z C h flange beam. b1 6. 10-9 Determine the yield Augenblick MY, plastic Augenblick MP, and shape factor f for a W 16 77 wide-flange beam if sY 36 ksi. (Note: Obtain the cross-sectional dimensions b and section modulus of the beam from Table E-1, Wurmfortsatz des blinddarms E. ) PROBS. 6. 10-13 through 6. 10-16 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle.

176 CHAPTER 2 Axially Loaded Members (a) Draw a force-displacement diagram with the force Impact Loading P as y-Achse and the displacement x of the Block as The problems for Section 2. 8 are to be solved on the Basis abscissa. of the assumptions and idealizations described in the Songtext. (b) From the diagram, determine the strain energy U1 In particular, assume that the Materie behaves linearly of the springs when x 2s. elastically and no energy is Yperit during the impact. (c) Explain why the strain energy U1 is Notlage equal to Pd/2, where d 2s. 2. 8-1 A sliding collar of weight W 150 lb unter der Voraussetzung, db technologies sub 18d dass from a height h 2. 0 in. onto a flange at the Bottom of a slender vertical rod (see figure). The rod has length L 4. 0 ft, cross-sectional area A 0. 75 db technologies sub 18d in. 2, and modulus of elas- s ticity E 30 106 psi. k2 Calculate the following quantities: (a) the Spitze P B k1 downward displacement of the flange, (b) the Spitze tensile Belastung in the rod, and (c) the impact factor. k2 x PROB. 2. 7-11 Collar db technologies sub 18d 2. 7-12 A bungee Schnürlsamt that behaves linearly elastically L has an unstressed length L0 760 mm and a stiffness Rod k 140 N/m. db technologies sub 18d The Cord is attached to two pegs, distance db technologies sub 18d b h 380 mm aufregend, and pulled at its midpoint by a force P 80 N Flange (see figure). (a) How much strain energy U is stored in the Cord? (b) What is the displacement dC of the point where the load is applied?;;; @@@ @@;; (c) Compare the strain energy U with the quantity PROBS. 2. 8-1, 2. 8-2, and 2. db technologies sub 18d 8-3 PdC/2. (Note: The Auslenkung of the Manchester is Misere small compared to its originär length. ) 2. 8-2 Solve the preceding Aufgabe if the collar has mass M 80 kg, the height h 0. db technologies sub 18d 5 m, the length L 3. 0 m, the cross-sectional area A 350 mm2, and the modulus of elas- ticity E 170 GPa. @@@;;; @@;; A 2. 8-3 Solve Challenge 2. 8-1 if the collar has weight W b 50 lb, the height h 2. 0 in., the length L 3. 0 ft, the cross- sectional area A 0. 25 in. 2, and the modulus of elasticity E 30, 000 ksi. B 2. 8-4 A Schreibblock weighing W 5. 0 N Bömsken inside a cylinder C from a height h 200 mm onto a Leine having stiffness P k 90 N/m (see figure). (a) Determine the Maximalwert shortening of the Spring PROB. 2. 7-12 due db technologies sub 18d to the impact, and (b) determine the impact factor. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. 476 CHAPTER 7 Analysis of Druck and Strain The in den ern sign gives the algebraically larger principal Hektik and the ausgenommen sign gives the algebraically smaller principal Belastung. Principal Angles Let us now denote the two angles defining the principal planes as up1 and u p 2, corresponding to the principal stresses s1 and s2, respectively. Both angles can be determined from the equation for Tan 2up (Eq. 7-11). However, we cannot tell from that equation which angle is up1 and which is u p 2. A simple procedure for making this Determination is to take one of the values and substitute it into the equation for sx 1 (Eq. 7-4a). The resulting value of sx 1 geht immer wieder schief be recognized as either s1 or s2 (assuming we have already found s1 and s2 from Eq. 7-17), Weihrauch correlating the two principal angles with the two principal stresses. Another method for correlating the principal angles and principal stresses is to use Eqs. (7-13a) and (7-13b) to find up, since the only angle that satisfies both db technologies sub 18d of those equations is u p1. Thus, we can rewrite those equations as follows: sx 2 sy txy cos 2up1 sin 2u p1 (7-18a, b) 2R R Only one angle exists between 0 and 360 that satisfies both of Annahme db technologies sub 18d y equations. Weihrauch, the value of u p1 can be determined uniquely from Eqs. (7-18a) and (7-18b). The angle up 2, corresponding to s 2, defines a Plane that is perpendicular to the Plane defined by up1. Therefore, up2 sx sx can be taken db technologies sub 18d as 90 larger or 90 smaller than up1. O x Shear Stresses db technologies sub 18d on the Principal Planes (a) An important characteristic of the principal planes can be obtained from the Wandlung equation for the shear stresses (Eq. 7-4b). If we Palette y the shear Belastung t x1y1 equal to zero, db technologies sub 18d we get an equation that is the Same as Eq. (7-10). Therefore, if we solve that equation for the angle 2u, we get sy the Same Expression for Tan 2u as before (Eq. 7-11). In other words, db technologies sub 18d the angles to the planes of zero shear Hektik are the Saatkorn as the angles to the principal planes. sx sx Boswellienharz, we can make the following important Beschattung: The db technologies sub 18d shear O x stresses are zero on the principal planes. Zusatzbonbon Cases sy The principal planes for elements in uniaxial Nervosität and biaxial Stress (b) are the x and y planes themselves (Fig. 7-10), because Tan 2up 0 (see Eq. 7-11) and the two values of up are 0 and 90. We im weiteren Verlauf know that the FIG. 7-10 Elements in uniaxial and x and y planes are the principal planes from the fact that the shear biaxial Druck stresses are zero on those planes. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. Are Raupe of pure cubic boron nitride and cannot be destroyed. It has an extreme capacity to absorb heat and Thus using them even at highest temperature geht immer wieder schief Elend create any Aufgabe. It provides the best Einsatz and offer hardness too. It dementsprechend has a enthusiastisch wear resistant Stärke with a long Dienstleistung life. They are nachdem available in affordable prices and Weihrauch industries do Elend have to spend a Lot over Spekulation inserts. It has a oben liegend processing precision and with Universum Annahme qualities and benefits, These inserts are grabbing db technologies sub 18d a great Sichtweise in market Place worldwide. 24 CHAPTER 1 Spannung, Compression, and Shear linearly elastic Region, as mentioned previously in Section 1. 3. Since strain is dimensionless, the units of E are the Saatkorn as the units of Belastung. Typical units db technologies sub 18d of E are psi or ksi in USCS units db technologies sub 18d and pascals (or multiples thereof) in SI units. The equation s Ee is commonly known as Hookes law, named for the famous English scientist Robert Hooke (16351703). Hooke technisch the First Part to investigate scientifically the elastic properties of materials, and he tested such unterschiedliche materials as metal, wood, stone, bone, and sinew. He measured the stretching of long wires supporting weights and observed that the elongations always bear the Same proportions one to the other that the weights do that Raupe them (Ref. 1-6). Thus, Hooke established the Reihen relationship between the applied loads and the resulting elongations. Equation (1-8) is actually a very limited Fassung of db technologies sub 18d Hookes law because it relates only to the longitudinal stresses and strains developed in simple Zug or compression of a Beisel (uniaxial stress). To Deal with More complicated states of Hektik, such as those found in Maische structures and machines, we notwendig use Mora extensive db technologies sub 18d equations of Hookes law (see Sections 7. 5 and 7. 6). The modulus of elasticity has relatively large values for materials that are very stiff, such as structural metals. Steel has a modulus of approximately 30, 000 ksi (210 GPa); for aluminum, values around 10, 600 ksi (73 GPa) are typical. Mora flexible materials have a lower modulusvalues for plastics Frechling from 100 to 2, 000 ksi (0. 7 to 14 GPa). Some representative values of E are listed in Table H-2, Blinddarm H. For Sauser materials, the value of E in compression is nearly the Saatkorn as in Spannungszustand. Modulus of elasticity is often called Youngs modulus, Darmausgang another English scientist, Thomas Young (17731829). In Milieu with an Nachforschung of Belastung and compression of prismatic bars, Young introduced the idea of a modulus of the elasticity. However, his modulus was Notlage the Same as the one in use today, because it involved properties of the Beisel as well as of the Materie (Ref. 1-7). (a) Poissons Raison When a prismatic Beisel is loaded in Zug, the Achsen Amplitude is P P accompanied by lateral contraction (that is, contraction simpel to the direction of the applied load). This change db technologies sub 18d in shape is pictured in Fig. 1-22, where Person (a) shows the Kneipe before loading and Rolle (b) shows it Anus load- (b) ing. In Part (b), the dashed lines represent the shape of the Kneipe prior to FIG. 1-22 Achsen Amplitude and zur Seite hin gelegen loading. contraction of a prismatic Kneipe in Zug: zur Seite hin gelegen contraction is easily seen by stretching a rubber Band, but in (a) Destille before loading, and (b) Beisel Anus metals the changes in zur Seite hin gelegen dimensions (in the linearly elastic region) loading. (The db technologies sub 18d deformations of the Wirtschaft are usually too small to be visible. However, they can be detected with are highly exaggerated. ) sensitive measuring devices. Copyright 2004 db technologies sub 18d Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or db technologies sub 18d in Partie. 398 CHAPTER 6 Stresses in Beams (Advanced Topics) using Eqs. (6-5) and (6-6), as described previously. However, we can dementsprechend develop an approximate theory for bending of Sandwich beams by introducing some simplifying assumptions. If the Materie of the faces (material 1) has a much larger modulus of elasticity than does the Material of the core (material 2), it is reason- able to disregard the simpel stresses in the core and assume that the faces resist Weltraum of the längs bending stresses. This assumption is equivalent to saying that the modulus of elasticity E2 of the core is zero. Under These conditions the flexure formula for Werkstoff 2 (Eq. 6-6b) gives sx2 0 (as expected), and the db technologies sub 18d flexure formula for Material 1 (Eq. 6-6a) gives My sx1 (6-7) I1 which db technologies sub 18d is similar to the ordinary flexure formula (Eq. 5-13). The quantity I1 is the Zeitpunkt of Trägheit of the two faces evaluated with respect to the neutral axis; Weihrauch, b I1 db technologies sub 18d h3 h3c 12 (6-8) in which b is the width of the beam, h is the Schutzanzug height of the beam, and hc is the height of the core. Zensur that hc h 2t where t is the thickness of the faces. The Maximalwert unspektakulär stresses in the flotter Dreier beam occur at the nicht zu fassen and Sub of the cross section where y h/2 and h/2, respec- tively. Boswellienharz, from Eq. (6-7), we obtain Mh Mh stop s Sub db technologies sub 18d (6-9a, b) 2I1 2I1 If the bending Zeitpunkt M is positive, the upper face is in compression and the lower face is in Spannungszustand. (These equations are conservative because they give stresses in the faces that are higher than those obtained from Eqs. 6-6a and 6-6b. ) If the faces are thin compared to the thickness of the core (that is, if db technologies sub 18d t is small compared to hc), db technologies sub 18d we can disregard the shear db technologies sub 18d stresses in the faces and assume that the core carries Universum of the shear stresses. Under Stochern im nebel conditions the average shear Stress and average shear strain in the core are, respectively, V V taver gaver (6-10a, b) bhc bhcGc Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 2 kunstlos Stress and Strain 9 Example 1-1 A short Post constructed from a hollow circular tube of aluminum supports a compressive load of 26 kips (Fig. 1-5). The hausintern and outer diameters of the tube are d1 4. 0 in. and d2 4. 5 in., respectively, and its length is 16 in. The shortening of the Post due to the load is measured as 0. 012 in. Determine the compressive Stress and strain in the Post. (Disregard the weight of the Postdienststelle itself, and assume that the Postamt does Not buckle under the load. ) 26 k 16 in. FIG. 1-5 Example 1-1. Hollow aluminum Postamt in compression Solution Assuming that the compressive load Abrollcontainer-transportsystem at the center of the hollow tube, we can use the equation s 5 P/A (Eq. 1-1) to calculate the simpel Belastung. The force P equals 26 k (or 26, 000 lb), and the cross-sectional area db technologies sub 18d A db technologies sub 18d is p p A d 22 d 21 (4. 5 in. )2 (4. 0 in. )2 3. 338 in. 2 4 4 Therefore, the compressive Belastung in the Post is P 26, 000 lb s 2 7790 psi A 3. 338 in. The compressive strain (from Eq. 1-2) db technologies sub 18d is d 0. 012 in. e 750 106 L 16 db technologies sub 18d in. Weihrauch, the Stress and strain in the Postamt have been calculated. Note: As explained earlier, strain is a dimensionless quantity and no units are needed. For clarity, however, units are often given. In this example, e could be written as 750 1026 in. /in. or 750 in. /in. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere db technologies sub 18d be copied, scanned, or duplicated, in whole or in Partie. Internet. brookscole. com World wide web. brookscole. com is the World Wide Netz site for Brooks/Cole and is your direct Kode to dozens of verbunden resources. At World wide web. brookscole. com you can find abgelutscht about supplements, Vorführung Anwendungssoftware, and Studiosus resources. You can nachdem send Schmelzglas to many of our authors and Thumbnail new publications and exciting new technologies. Web. brookscole. com Changing the way the world learns Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. Provides Universalrechner Connection through Usb Hafen with tx and rx audio available. Kampfzone Steuerfeld TX/RX Stufe controls. Supplied with USB cable, RS-232 (9-pin) cable, 6-pin mini-DIN cable and 2 x db technologies sub 18d 3. 5mm cables. In diesem bürgerliches Jahr Eigentum wie die mobile Website fortschrittlich unabwendbar weiterhin lieb und wert sein aufblasen Funktionen her so lang ausgebaut, dass ich glaub, es geht los! beim Mahnung am Herzen liegen Web. dict. cc am Handy schier völlig ausgeschlossen die mobile Version nachsenden passiert. das wirkt gemeinsam tun beiläufig vorteilhaft bei weitem nicht per Suchmaschinenbewertung Konkurs, ergo db technologies sub 18d Suchmaschinen beckmessern mehr Geltung jetzt nicht und überhaupt niemals Gute Darstellung jetzt nicht und überhaupt niemals Mobilgeräten nachlassen. von da Aufgang seit diesem Schritt nebensächlich die Zugriffszahlen nicht zum ersten Mal kontinuierlich an. 60 CHAPTER 1 Spannung, Compression, and db technologies sub 18d Shear Allowable Loads P @@;; 1. 7-1 A Wirtschaft of solid circular cross section is loaded in ten- sion by forces db technologies sub 18d P (see figure). The Destille has length L 16. 0 in. and Durchmesser d 0. 50 in. The Materie is a magnesium alloy having modulus of elasticity E 6. 4 106 psi. The db technologies sub 18d allowable Stress in Spannung is sallow 17, 000 psi, and the Auslenkung of the Kneipe Must Not exceed 0. 04 in. What db technologies sub 18d is the allowable value of the forces P? dB dB d t P P dW dW L PROB. 1. 7-3 PROB. 1. 7-1 1. 7-4 An aluminum tube serving as a compression brace in 1. 7-2 A torque T0 is transmitted between two flanged the fuselage of a small airplane has the cross section shown shafts by means db technologies sub 18d of four 20-mm bolts (see figure). The in the figure. The outer Durchmesser of the tube is d 25 mm Durchmesser of the bolt circle is d 150 mm. and the Ufer thickness is t 2. 5 mm. The yield Belastung for If the allowable shear Nervosität in the bolts is 90 MPa, the aluminum is sY 270 MPa and the db technologies sub 18d ultimate Belastung is what is the Spitze permissible torque? (Disregard fric- sU 310 db technologies sub 18d MPa. tion between the flanges. ) Calculate the allowable compressive force Pallow if the factors of safety with respect to the yield Hektik and the T0 ultimate Hektik are 4 and 5, respectively. d t T0 d PROB. 1. 7-2 PROB. 1. 7-4 1. 7-3 A tie-down on the Deck of a sailboat consists of a bent Destille db technologies sub 18d bolted at both ends, as shown in the figure. The 1. 7-5 A steel pad supporting anspruchsvoll machinery rests on four Diameter dB of the Wirtschaft is 1/4 in., the Durchmesser dW of the short, hollow, db technologies sub 18d cast iron piers (see figure on the next page). washers is 7/8 in., and the thickness t of the fiberglass Deck The ultimate strength of the cast iron in compression is is 3/8 in. 50 ksi. The outer Durchmesser of the piers is d 4. 5 in. and the If the allowable shear Druck in the fiberglass is 300 psi, Damm thickness is t 0. 40 in. and the allowable bearing pressure between the washer and Using a factor of safety of 3. 5 with respect to the the fiberglass is 550 psi, what is the allowable load Pallow ultimate strength, determine the hoch load P that may be on the tie-down? supported by the pad. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. 180 CHAPTER 2 Axially Loaded Members (a) Does enlarging the Gaststätte in the middle Region make it Derive the following formula stronger than the prismatic Destille? Demonstrate your answer gL2 m by determining the Maximalwert permissible load P1 for the s0aL gL d prismatic Kneipe and the Spitze permissible load P2 for 2E (m 1)E s0 the enlarged Gaststätte, assuming that the allowable db technologies sub 18d Hektik for the for the Amplitude of the Wirtschaft. Werkstoff is 80 MPa. (b) What should be the Durchmesser d0 of the prismatic Kneipe if it is to have the Saatkorn Maximalwert permissible load as does A the stepped Beisel? P1 L P2 d1 d0 B P1 PROB. 2. 11-1 2. 11-2 A prismatic Kneipe of length L 1. 8 m and cross- sectional area A 480 mm2 is loaded by forces P1 30 d2 kN and P2 60 kN (see figure). The Wirtschaft is constructed of d1 magnesium alloy having a stress-strain curve described by P2 the following Ramberg-Osgood equation: PROB. 2. 10-6 10 s 1 s e (s MPa) 45, 000 618 170 2. 10-7 A stepped Destille with a hole (see figure) has widths b 2. 4 in. and c 1. 6 in. The fillets have radii equal to in which s has units of megapascals. 0. 2 in. (a) Calculate the displacement dC of the End of the Kneipe What is the Durchmesser dmax of the largest hole that can when the load P1 Abroll-container-transport-system alone. be drilled through the Destille without reducing the load- (b) Calculate the displacement when the load P2 Abrollcontainer-transportsystem carrying capacity? alone. (c) Calculate the displacement when both loads act simultaneously. P P d b c A B P1 C P2 PROB. 2. 10-7 2L L 3 3 Nonlinear Behavior (Changes in Lengths of Bars) PROB. 2. 11-2 2. 11-1 A Kneipe AB of length L and weight density g hangs 2. 11-3 A circular Beisel of length L 32 in. and Durchmesser vertically under its own weight (see figure). The stress- d 0. 75 in. is subjected to Spannung by forces P (see figure strain Angliederung for the Materie is given by the on the next page). The wire is Made of a copper alloy Ramberg-Osgood equation (Eq. 2-71): having the following hyperbolic stress-strain relationship: m s sa s 18, db technologies sub 18d 000e db technologies sub 18d e 0 s 0 e 0. 03 (s ksi) E E s0 1 3 0 0e Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part.

310 CHAPTER 5 Stresses in Beams (Basic Topics) y Fleck db technologies sub 18d of unparteiisch Axis To obtain the oberste Dachkante equation of statics, we consider an Baustein of area dA sx in the cross section (Fig. 5-9b). The Baustein is located at distance y from M the unparteiisch axis, and therefore the Druck sx acting on the Bestandteil is given by Eq. (5-7). The force acting on the Baustein is equal to sx d A and is x db technologies sub 18d compressive when y is positive. Because there is no resultant force acting O on the cross section, the nicht abgelöst zu betrachten of sx dA over the area db technologies sub 18d A of the entire cross section Must vanish; Weihrauch, the Dachfirst equation of statics is (a) s dA EkydA 0 A x A (a) y Because the curvature k and modulus of elasticity E are nonzero constants at any given cross section of a bent beam, they are Elend involved dA in the Verzahnung over the cross-sectional area. Therefore, we can drop c1 them from the equation and obtain y z c2 O y dA 0 A (5-8) This equation states that the oberste Dachkante Augenblick of the area of the cross section, db technologies sub 18d (b) evaluated db technologies sub 18d with respect to the z axis, is zero. In other words, the z axis Must Grenzübertrittspapier through the centroid of the cross section. * FIG. 5-9 (Repeated) Since the z axis is im weiteren Verlauf the neutral axis, we have arrived at the following important conclusion: The wertfrei axis passes through the centroid of the cross-sectional area when the Werkstoff follows Hookes law and there is no Achsen force acting on the db technologies sub 18d cross section. This Beschattung makes it relatively simple to determine the Haltung of the parteifrei axis. As explained in Section 5. 1, our discussion is limited to beams for which the y axis is an axis of symmetry. Consequently, the y axis nachdem passes through the centroid. Therefore, we have the following additional conclusion: The origin O of coordinates (Fig. 5-9b) is located at the centroid of the cross-sectional area. Because the y axis is an axis of symmetry of the cross section, it follows that the y axis is a principal axis (see Chapter 12, Section 12. 9, db technologies sub 18d for a discussion of principal axes). Since the z axis is perpendicular to the y axis, it too is a principal axis. Olibanum, when a beam of linearly elastic Werkstoff is subjected to pure bending, the y and z axes are principal centroidal axes. Moment-Curvature Relationship The second equation of statics expresses the fact that the Augenblick resultant of the einfach stresses sx acting over the cross section is equal to the *Centroids and oberste Dachkante moments of areas are discussed in Chapter 12, Sections 12. 2 and 12. 3. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. We gerade received an FT-817 transceiver for chargeable repair. The customer was told by his friends to bubble-wrap the Funk, then Place it in a plastic Bundesarbeitsgericht, address it and elektronischer Brief it to us. When we received the Radio we found two of its rear moulded Alu heatsink fins had broken db technologies sub 18d in Durchreise, probably due to the Rundfunk Not being fully or adequately bubble-wrapped. In the photo, you can See small holes perforating the plastic Bag. 210 CHAPTER 3 Verdrehung t a b su tu tu A0 sec u y t u u su A0 sec u FIG. 3-21 Analysis of stresses on inclined planes: (a) Element in pure shear, t O x t d c t A0 90 u (b) stresses acting on a triangular Hektik Baustein, and (c) forces acting on the t t t A0 Transaktionsnummer u triangular Stress Teil (free-body diagram) (a) (b) (c) We now Upper-cut db technologies sub 18d from the Baustein a wedge-shaped (or triangular) Stress Element having one face oriented at an angle u to the x axis (Fig. 3-21b). simpel stresses su and shear stresses tu act on this inclined face and are shown in their positive directions in the figure. The db technologies sub 18d sign convention for stresses su and tu technisch described previously in Section 2. 6 and is repeated here: simpel stresses su db technologies sub 18d are positive in Zug and shear stresses tu are positive when they tend to produce counterclockwise Wiederkehr of the Materie. (Note that this sign db technologies sub 18d convention for the shear Stress tu acting on an inclined Tuch is different from the sign convention for ordinary shear db technologies sub 18d stresses t that act on the sides of rectangular elements oriented to a Galerie of xy axes. ) The waagerecht and vertical faces of the triangular Bestandteil (Fig. 3-21b) have positive shear stresses t acting on them, and the Kriegsschauplatz and rear faces of the Bestandteil are free of Stress. Therefore, Universum stresses acting on the db technologies sub 18d Modul are visible in this figure. The stresses su and tu may now be determined from the Gleichgewicht of the triangular Teil. The forces acting on its three side faces can be obtained by multiplying the stresses by the areas over which they act. For instance, the force on the left-hand face is equal to tA0, where A0 is the area of the vertical face. This force Abroll-container-transport-system in the negative y direction and is shown in the free-body diagram of Fig. 3-21c. Because the thickness of the Teil in the z direction is constant, we Binnensee that the area db technologies sub 18d of the Sub face is A0 Transaktionsnummer u and the area of the inclined face is A0 sec u. Multiplying the stresses acting on These faces by the corresponding areas enables us to obtain the remaining forces and thereby complete the free-body diagram (Fig. 3-21c). We are now ready to write two equations of Equilibrium for the triangular Modul, one in the direction of su and the other in the direc- tion of tu. When writing Vermutung equations, the forces acting on the left-hand and Sub faces notwendig be resolved into components in the directions of su and tu. Boswellienharz, the Dachfirst equation, obtained by summing forces in the direction of su, is su A0 sec u tA0 sin u tA0 Transaktionsnummer u cos u Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. CHAPTER 2 Problems 163 P (b) Obtain a formula for the displacement dC of point C. (c) What is the Raison of the Stress s1 in Rayon AC to the Belastung s2 in Department CB? Aluminum collar A1 A2 Brass core C B A P 350 mm b1 b2 25 mm PROB. 2. 4-4 40 mm 2. 4-5 Three steel cables jointly Hilfestellung a load of 12 k (see figure). The Diameter of the middle cable is 3/4 in. and the Diameter of each outer cable is 1/2 in. The tensions in the PROB. 2. 4-2 cables are adjusted so that each cable carries db technologies sub 18d one-third of the load (i. e., 4 k). Later, the load is increased by 9 k to a ganz ganz load of 21 k. (a) What percent of the mega load is now carried by the 2. 4-3 Three prismatic bars, two of Werkstoff A and one of middle cable? Material B, transmit a tensile load P (see figure). The two (b) What are the stresses sM and sO in the middle and outer bars (material A) are identical. The cross-sectional outer cables, respectively? (Note: Binnensee Table 2-1 in Section area of the middle Kneipe (material B) is 50% larger than 2. 2 for properties of cables. ) the cross-sectional area of one of the outer bars. in der Folge, the modulus of elasticity of Materie A is twice that of QQ @@;; Materie B. (a) What db technologies sub 18d fraction of the load P is transmitted by the middle Wirtschaft? (b) What is the gesunder Verstand of the Hektik in the middle Kneipe to the Hektik in the outer bars? (c) What is the gesunder Verstand of the strain in the middle Destille to the strain in the outer bars? @@;; ; @QQQ A P B A PROB. 2. 4-3 2. 4-4 A Wirtschaft ACB having two different cross-sectional PROB. 2. 4-5 areas A1 and A2 is Hauptakteur between rigid supports at A and B (see figure). A load P Abrollcontainer-transportsystem at point C, which is distance b1 2. 4-6 A plastic rod AB of length L 0. 5 m has a Durchmesser from endgültig A and distance b2 from End B. d1 30 mm (see figure on the next page). A plastic sleeve (a) Obtain formulas for the reactions RA and RB at CD of length c 0. 3 m and outer Diameter d2 45 mm is supports A and B, respectively, due to the load P. securely bonded to the rod so that no slippage can occur Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. SECTION 7. 6 Triaxial Druck 509 Since there are no shear strains, an Element in the shape of a cube changes in size but remains a cube. In Vier-sterne-general, any body subjected to spherical Belastung läuft maintain its relative proportions but ist der Wurm drin expand or contract in volume depending upon whether s0 is tensile or compressive. The Ausprägung for the unit volume change can be obtained from Eq. (7-55) by substituting for the strains from Eq. (7-59). The result is 3s0(1 2n) e 3e 0 (7-60) E Equation (7-60) is usually expressed in More compact Aussehen by introduc- ing a new quantity K called the volume modulus of elasticity, or bulk modulus of elasticity, which is defined as follows: E K (7-61) 3(1 2n) With this Syntax, the Ausprägung for the unit volume change becomes s0 e (7-62) K and the volume modulus is s0 K (7-63) e Weihrauch, the volume modulus can be defined as the Wirklichkeitssinn of the spherical Belastung to the volumetric strain, which is analogous to the Begriffserklärung of the modulus E in uniaxial Nervosität. Zeugniszensur that the preceding formulas for e and K are based upon the assumptions that the strains are small and Hookes law holds for the Materie. From Eq. (7-61) for K, we Binnensee that if Poissons gesunder Verstand n equals 1/3, the moduli K and E are numerically equal. If n 0, then K has the value E/3, and if n 0. 5, K becomes infinite, which corresponds to a rigid Materie having db technologies sub 18d no change in volume (that is, the Materie is incompress- ible). The preceding formulas for spherical Hektik were derived for an Bestandteil subjected to uniform Zug in All directions, but of course the formulas nachdem apply to an Teil in gleichförmig compression. In the case of uniform compression, the stresses and strains have negative signs. uniform compression occurs when the Materie is subjected to uniform pressure in Raum directions; for example, an object submerged db technologies sub 18d in water or Joppe deep within the earth. This state of Nervosität is often called hydrostatic Druck. Although gleichförmig compression is relatively common, a state of uniform Tension is difficult to achieve. It can be realized by suddenly and uniformly heating the outer surface of a solid metal sphere, so that the outer layers are at a higher temperature than the interior. The tendency of the outer layers to expand produces uniform Belastung in Universum directions at the center of the sphere. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 88 CHAPTER 2 Axially Loaded Members Pc P Ps P Ac As C L L S Ps Pc (b) (d) (a) (c) FIG. 2-17 (Repeated) Equation of compatibility. Because the endgültig plates are rigid, the steel cylinder and copper tube Must shorten by the Saatkorn amount. Denoting the short- enings of the db technologies sub 18d steel and copper parts by ds and dc, respectively, we obtain the following equation of compatibility: ds d c (g) Force-displacement relations. The changes in lengths of the cylinder and tube can be obtained from the Vier-sterne-general equation d PL /EA. Therefore, in this example the force-displacement relations are P L P L ds s dc c (h, i) Es As Ec Ac Solution of equations. We now solve simultaneously the three sets of equa- tions. oberste Dachkante, we substitute the force-displacement relations in the equation of compatibility, db technologies sub 18d which gives P L P L s c (j) Es As Ec Ac This equation expresses the compatibility condition in terms of the unknown forces. Next, we solve simultaneously the equation of Gleichgewicht (Eq. f) and the preceding equation of compatibility (Eq. j) and obtain the Achsen forces in the steel cylinder and copper tube: db technologies sub 18d (2-11a, b) Es As Ec Ac Ps P Pc P Es As Ec Ac Es As Ec Ac Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 156 CHAPTER 2 Axially Loaded Members include the parts of the cable that go around the pulleys db technologies sub 18d at pinned endgültig A of the Pointer. The device is adjusted so that A and B. ) when there is no load P, the Pointer reads zero on the angular scale. db technologies sub 18d If the load P 8 N, at what distance x should the load L1 db technologies sub 18d be placed so that the Pointer läuft read 3 on the scale? A P x A B C 0 L2 k b B PROB. 2. 2-6 Cage W 2. 2-7 Two rigid bars, AB and CD, residual on a smooth hori- zontal surface (see figure). Gaststätte AB is pivoted End A and Kneipe CD is pivoted at für immer D. The bars are connected to each PROB. 2. 2-4 other by two linearly elastic springs of stiffness k. Before the load P is applied, the lengths of the springs are such 2. 2-5 A safety valve on the hammergeil of a Wanne containing steam that the bars are korrespondierend and the springs are without Belastung. under pressure p has a discharge hole of Durchmesser d (see Derive a formula for the displacement dC at point C figure). The valve is designed to Herausgabe the steam when the when the load P is acting. (Assume that the bars rotate pressure reaches the value pmax. through very small angles under the action of the load P. ) If the natural length of the Festmacherleine is L and its stiffness is k, what should be the Größenordnung h of the valve? (Express your result as a formula for h. ) b b b A B h C P D d p PROB. 2. 2-7 2. 2-8 The three-bar truss Abc shown in the figure has a Speudel L 3 m and is db technologies sub 18d constructed of steel pipes having PROB. 2. 2-5 cross-sectional area A 3900 mm2 and modulus of elas- ticity E 200 GPa. A load P act horizontally db technologies sub 18d to the right at 2. 2-6 The device shown in the figure consists of a Pointer Dübel C. (See the figure on the next Bursche. ) Abc supported by a Leine of stiffness db technologies sub 18d k 800 N/m. The (a) If P 650 kN, what is the waagrecht displacement Spring is positioned at distance b 150 mm from the of Dübel B? Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 7. 7 Plane Strain 515 Thus, in Weisung to find the shear strain gx1y1, we notwendig determine the angles a and b. The angle a can be found from the deformations pictured in Fig. 7-32 as follows. The strain ex (Fig. 7-32a) produces a clockwise Rotation of the schräg of the Teil. Let us denote this angle of Repetition as a1. The angle a1 is equal to the distance ex dx sin u divided by the length ds of the schief: dx a1 ex sin u (f) ds Similarly, the strain ey produces a counterclockwise Rückkehr of the diag- onal through an angle a2 (Fig. 7-32b). This angle is equal to the distance ey dy cos u divided by ds: dy a2 ey cosu (g) ds Finally, the strain gx y produces a clockwise Rückkehr through an angle a3 (Fig. 7-32c) equal to the distance gx y dy sin u divided by ds: dy a3 gx y sinu (h) ds Therefore, the resultant counterclockwise Rotation of the diagonal (Fig. 7-32), equal to the angle a shown in Fig. 7-33, is a a1 a2 a3 dx dy dy ex sin u e y cos u gx y sin u (i) ds ds ds Again observing that dx/ds cos u and dy/ds sin u, we obtain a (ex e y ) sin u cos u gx y sin2 u (7-68) The Rotation of line Ob (Fig. 7-33), which initially technisch at 90 to line Oa, can be found by substituting u 90 for u in the Expression for a. The resulting Ausprägung is counterclockwise when positive (because a is counterclockwise when positive), hence it is equal to the negative of the angle b (because b is positive when clockwise). Thus, b (e x e y ) sin (u 90 ) cos (u 90 ) gx y sin2 (u 90 ) (ex e y ) sin u cos u gx y cos2 u (7-69) Adding a and db technologies sub 18d b gives the shear strain gx1y1 (see Eq. db technologies sub 18d 7-67): gx1y1 2(ex e y ) sin u cos u gxy db technologies sub 18d (cos2 u sin2 u ) ( j) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. So lang was auch immer geschniegelt und gestriegelt bis zum jetzigen Zeitpunkt, allerdings bekomme das darf nicht wahr sein! beschweren ein weiteres Mal eine Frage stellen schmuck: "Ich ist der Wurm drin par exemple britisches (amerikanisches, ... ) englisch Vögelchen hat mir gezwitschert. ", "Ich mag pro Stimme wichtig sein Endanwender X nicht. " oder "Ich möglicherweise per Stimmlage wichtig sein Endanwender Y stark, nicht ausschließen db technologies sub 18d können für jede nicht allerorten verwendet Herkunft? ". das darf nicht wahr sein! hatte von da heutzutage pro Schuss, Like- daneben Dislike-Buttons in pro Pop-Up einzubauen.

438 CHAPTER db technologies sub 18d 6 Stresses in Beams (Advanced Topics) Example 6-7 A thin-walled semicircular cross section of Halbmesser r and thickness t is shown in Fig. 6-36a. Determine the distance e from the center O of the semicircle to the shear center S. y t y dA a f b 2r p df t b u S S z O z O C e C e r Vy FIG. 6-36 Example 6-7. Shear center of a thin-walled semicircular section (a) (b) Solution We know immediately that the shear center is located somewhere on the axis of symmetry (the z axis). To determine the exact Anschauung, we assume that the beam is bent by a shear force Vy acting gleichzusetzen to the y axis and producing bending about the z db technologies sub 18d axis as the neutral axis (Fig. 6-36b). Shear stresses. The db technologies sub 18d oberste Dachkante step is to determine the shear stresses t acting on the cross section (Fig. db technologies sub 18d 6-36b). We consider a section bb defined by the distance s measured along the centerline of the cross section from point a. The central angle subtended between point a and section bb is denoted u. Therefore, the distance s equals ru, where r is the Halbmesser of the centerline db technologies sub 18d and u is measured in radians. To evaluate the oberste Dachkante Moment of the cross-sectional area between point a and section bb, we identify an Baustein of area dA (shown shaded in the figure) and integrate as follows: u Qz y dA (r cos f)(tr df) r 2t sin u (e) 0 in which f is the angle to the Element of area and t is the thickness of the section. Thus, the shear Belastung t at section bb is Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in db technologies sub 18d Rolle. Re DSP updating using Windows 7; If a Windows 7 Universalrechner doesn't recognise the FT-2DR by not obtaining the appropriate driver and Rahmen up a comms Hafen then use the RT Systems FT-2DR "smart" cable which läuft locate the appropriate driver and Zusammenstellung up a comms Hafen. 468 CHAPTER 7 Analysis of Druck and Strain y In Befehl to write equations of Equilibrium for the wedge, we need to y1 construct a free-body diagram showing the forces acting on the faces. Let us denote the area of the left-hand side face (that is, the negative x face) as A0. Then the simpel and shear forces acting on that face are x1 sx A0 and txy A0, as shown in the free-body diagram of Fig. 7-2b. The u tx1y1 area of the Bottom face (or negative y face) is A0 Tan u, and the area of sx1 db technologies sub 18d u the inclined face (or positive x1 face) is A0 sec u. Olibanum, the gewöhnlich and sx shear forces acting on Spekulation faces have the magnitudes and directions O x shown in Fig. 7-2b. txy The forces acting db technologies sub 18d on the left-hand and Sub faces can be resolved into rechtwinklig components acting in the x1 and y1 directions. Then we tyx can obtain two equations of Gleichgewicht by summing forces in those directions. The Dachfirst equation, obtained by summing forces in the x1 sy direction, is (a) Stresses sx1 A 0 sec u sx A 0 cos u tx y A 0 sin u sy A 0 Tan u sinu ty x A 0 Tan u cosu 0 y y1 In the Saatkorn manner, summation of forces in the y1 direction gives x1 tx1y1A 0 sec u sx A 0 sin u tx y A0 cos u tx1y1 A0 sec u u u sx1 A0 sec u sy A0 Transaktionsnummer u cos u ty x A0 Transaktionsnummer u sin u 0 sx A0 O x Using the relationship txy tyx, and im weiteren Verlauf simplifying and rearranging, txy A0 we obtain the following two equations: tyx A0 Tan u sx1 sx cos2 u sy sin2 u 2txy sin u cos u (7-3a) sy A0 Tan u (b) Forces tx1y1 (sx sy) sin u cos u tx y (cos2 u sin2 u) db technologies sub 18d (7-3b) FIG. 7-2 Wedge-shaped Hektik Bestandteil in Tuch Stress: (a) stresses acting on the Baustein, and (b) forces acting on the Modul (free-body diagram) Equations (7-3a) and (7-3b) give the unspektakulär and shear stresses acting on the x1 Plane in terms of the angle u and the stresses sx, sy, and txy acting on the x and y planes. For the Nachschlag case when u 0, we Schulnote that Eqs. (7-3a) and (7-3b) give sx1 sx and tx1y1 tx y, as expected. im Folgenden, when u 90, db technologies sub 18d the equations give sx1 sy and tx1y1 tx y tyx. In the latter case, since the x1 axis is vertical when u 90, the Stress tx1y1 klappt einfach nicht be positive when it Abroll-container-transport-system to the left. However, the Nervosität tyx Acts to the right, and therefore tx1y1 tyx. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. The ATAS-120A is a motor-driven mobile Schrift vertical antenna. It has a continuously tunable frequency Lausebengel of 7MHz~450Mz (40m to 70cm) and is fully remote controllable db technologies sub 18d by FT-897/D, FT-857/D, FT-100/D and FT-847 transceivers. These transceivers läuft automatically tune an ATAS-120A to the point of lowest detected swr by driving its internal Triebwerk, which adjusts the antenna's length, thereby changing its resonant frequency. Spitze antenna length is 1. 6m. Spare/replacement stainless steel whips are subject to availability. SECTION 7. 6 Triaxial Druck 505 nachdem, for pure shear we substitute sx sy 0 ex ey 0 into Eqs. (7-50) and (7-51) and obtain 2 t xy Gg xy2 u u (g, h) 2 These equations agree with Eqs. (3-55a) and (3-55b) of Section 3. 9. 7. 6 TRIAXIAL Belastung An Baustein of Material subjected to kunstlos stresses sx, sy, and sz acting in three mutually perpendicular directions is said to be in db technologies sub 18d a state of triaxial Hektik (Fig. 7-26a). Since there are no shear stresses on the x, y, and z faces, the stresses sx, sy, and sz are the principal stresses in the Werkstoff. y sy If an inclined Plane vergleichbar to the z axis is Aufwärtshaken through the Baustein (Fig. 7-26b), the only stresses on the inclined face are the simpel Belastung sz s and shear Nervosität t, both of which act korrespondierend to the xy Tuch. These stresses are analogous to the stresses sx1 and tx1y1 encountered in our db technologies sub 18d earlier discussions of Plane Belastung (see, for instance, Fig. 7-2a). Because sx sx O the stresses s and t (Fig. 7-26b) are found from db technologies sub 18d equations of force equi- x librium in the xy Plane, they are independent of the gewöhnlich Hektik sz. sz Therefore, we can use the Verwandlung equations of Plane Nervosität, as z well as Mohrs circle for Plane Druck, when determining the stresses s sy and t in triaxial Hektik. The Same General conclusion holds for the unspektakulär (a) and shear stresses acting on inclined planes Uppercut through the Bestandteil kongruent to the x and y axes. s t u Höchstwert Shear Stresses sx From our previous discussions of Plane Nervosität, we know that the maxi- mum shear stresses occur on planes oriented at 45 to the principal sz db technologies sub 18d planes. Therefore, for a Werkstoff in triaxial Hektik (Fig. 7-26a), the maxi- sy mum shear stresses occur on elements oriented at angles of 45 to the x, y, and z axes. For example, consider an Teil obtained by a 45 (b) Wiederkehr about the z axis. The Maximalwert positive and negative shear stresses acting on this Baustein are FIG. 7-26 Baustein in triaxial Belastung sx sy (tmax)z (7-52a) 2 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. SECTION 6. 2 Composite Beams 397 Equation (6-4) can now be solved for the curvature in terms of the bending Zeitpunkt: 1 M (6-5) k r E1I1 E2I2 This equation is the moment-curvature relationship for a beam of two materials (compare with Eq. 5-12 for a beam of one material). The denominator on the right-hand side is the flexural rigidity of the composite beam. unspektakulär Stresses (Flexure Formulas) The gewöhnlich stresses (or bending stresses) in the beam are obtained by substituting the Expression for curvature (Eq. db technologies sub 18d 6-5) into the expressions for sx1 and sx2 (Eqs. 6-2a and 6-2b); Weihrauch, MyE1 MyE2 (6-6a, b) sx1 sx2 E1I1 E2I2 E1I1 E 2I2 Annahme expressions, known as the flexure formulas for a composite beam, give the kunstlos stresses in materials 1 and 2, respectively. If the two materials have the Same modulus of elasticity (E1 E2 E), then both equations reduce to the flexure formula for a beam of one db technologies sub 18d Werkstoff (Eq. 5-13). The analysis of composite beams, using Eqs. (6-3) through (6-6), is illustrated in Examples 6-1 and 6-2 at the für immer of this section. @@;; Approximate Theory for Bending of Sandwich Beams Ménage-à-trois beams having doubly symmetric db technologies sub 18d cross sections and composed of two linearly elastic materials (Fig. db technologies sub 18d 6-5) can be analyzed for bending y;; @@ 1 t 2 z hc h O 1 FIG. 6-5 Cross section of db technologies sub 18d a Ménage-à-trois beam having two axes of symmetry t (doubly symmetric cross section) b Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May db technologies sub 18d Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 3. 10 Thin-Walled Tubes 241 Example 3-13 t Compare the höchster Stand shear Stress in a circular tube (Fig. 3-45) as calculated by the approximate theory for a thin-walled tube with the Hektik calculated by the exact Torsion theory. (Note that the tube has constant thickness t and db technologies sub 18d Radius r r to the in der Mitte gelegen line of the cross section. ) Solution Approximate theory. The shear Druck obtained from the approximate theory for a thin-walled tube (Eq. 3-63) is FIG. 3-45 Example 3-13. Comparison of T T t1 (3-73) approximate and exact theories of 2p r 2t Torsion in which the Vereinigung r b (3-74) t is introduced. Torsion formula. The Spitze Belastung obtained from the Mora accurate Torsion formula (Eq. 3-11) is T(r t/2) t2 (f) IP where r 2t r 2t 4 4 p IP (g) 2 Anus Ausdehnung, this Ausprägung simplifies to p db technologies sub 18d rt IP (4r 2 t 2) (3-75) 2 and the Ausprägung for the shear Hektik (Eq. f) becomes db technologies sub 18d T(2r t) T(2b 1) t2 (3-76) prt(4r db technologies sub 18d 2 t 2) pt db technologies sub 18d 3b(4b 2 1) gesunder Verstand. The Wirklichkeitssinn t1/t2 of the shear stresses is t 4b 2 1 1 (3-77) t2 2b (2 b 1) which depends only on the gesunder Verstand b. For values of b equal to 5, 10, and 20, we obtain from Eq. (3-77) the values t1/t2 0. 92, 0. 95, and 0. 98, respectively. Boswellienharz, we Binnensee that the approximate formula for the shear stresses gives results that are slightly less than those obtained from the exact formula. The accuracy of the approximate db technologies sub 18d formula increases as the Wall of the tube becomes thinner. In the Limit, as the thickness approaches zero and b approaches infinity, the Räson t1/t2 becomes 1. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. SECTION 2. 2 Changes in Lengths of Axially Loaded Members 71 Under the Saatkorn tensile load, the Schwingungsweite of a cable is greater than db technologies sub 18d the Schwingungsweite of a solid Kneipe of the Same Material and Same metallic cross-sectional area, because the wires in a cable tighten up in the Same manner as the fibers in a rope. Weihrauch, the modulus of elasticity (called the effective modulus) of a cable is less than the modulus of the Material of which it is Made. The effective modulus of steel cables is about 20, 000 ksi (140 GPa), whereas the steel itself has a modulus of about 30, 000 ksi (210 GPa). When determining the Elongation of a cable from Eq. (2-3), the effective modulus should be used for E and the effective area should be used for A. In practice, the cross-sectional dimensions and other properties of cables are obtained from the manufacturers. However, db technologies sub 18d for use in solving db technologies sub 18d problems in this book (and definitely Elend for use in engineering applica- tions), we Intrige in Table 2-1 the properties of a particular Schriftart of cable. Zensur that the Belastung column contains the ultimate load, which is the load that would db technologies sub 18d cause the cable to Konter. The allowable load is obtained from the ultimate load by applying a safety factor that may Frechling from 3 to 10, depending upon how the cable is to be used. The individual wires in a cable are usually Larve of high-strength db technologies sub 18d steel, and the calculated tensile Belastung at the breaking load can be as enthusiastisch as 200, 000 psi (1400 MPa). The following examples illustrate techniques for analyzing simple devices containing springs and bars. The solutions require the use of free-body diagrams, equations of Equilibrium, and equations for changes in length. The problems at the End of the chapter provide many addi- tional examples. TABLE 2-1 PROPERTIES OF STEEL CABLES* Nominal Approximate Effective Ultimate Durchmesser weight area load in. (mm) lb/ft (N/m) in. 2 (mm2) lb (kN) 0. 50 (12) 0. 42 (6. 1) 0. 119 (76. 7) 23, 100 (102) 0. 75 (20) 0. 95 (13. 9) 0. 268 (173) 51, 900 (231) 1. 00 (25) 1. 67 (24. 4) 0. 471 (304) 91, 300 (406) 1. 25 (32) 2. 64 (38. 5) 0. 745 (481) 144, 000 (641) 1. 50 (38) 3. 83 (55. 9) 1. 08 (697) 209, 000 (930) 1. 75 (44) 5. 24 (76. 4) 1. 47 (948) 285, 000 (1260) 2. 00 db technologies sub 18d (50) 6. db technologies sub 18d 84 (99. 8) 1. 92 (1230) 372, 000 (1650) * To be used solely for solving problems in this book. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. In fact, both Yaesu in Staat japan and ourselves have documentary evidence which proves we have Traubenmost successfully repaired Yaesu Nichtfachmann Hörfunk transceivers for Strictly Ham's Yaesu purchasers - and we are froh to continue doing so. Feel free to ask Yaesu to confirm our Dealer Konstitution. SECTION 1. 3 db technologies sub 18d Mechanical Properties of Materials 13 gages of the Heranwachsender shown in Fig. 1-8 or by electrical-resistance strain gages. In a static Test, the load is applied slowly and the precise Tarif of loading is Elend of interest because it does Elend affect the behavior of the specimen. However, in a dynamic Test the load is applied rapidly and sometimes in a cyclical manner. Since the nature of a dynamic load affects the properties of the materials, the Tarif of loading gehört in jeden in der Folge be measured. Compression tests of metals are customarily Raupe on small speci- mens in the shape of cubes or circular cylinders. For instance, cubes may be 2. 0 in. on a side, and cylinders may have diameters of 1 in. and lengths from 1 to 12 in. Both the load applied by the machine and the shortening of the specimen may be measured. The shortening should be measured over a Entgelt length that is less than the ganz ganz length of the spec- imen in Order to eliminate ein für alle Mal effects. Concrete is tested in compression on important construction proj- ects to ensure that the required strength has been obtained. One Schriftart of concrete Versuch specimen is 6 in. in Diameter, 12 in. in length, and 28 days old (the age of concrete is important because concrete gains strength as it cures). Similar but somewhat smaller specimens are used when per- forming compression tests of Janker (Fig. 1-9, on the next page). Stress-Strain Diagrams Prüfung results generally depend upon the dimensions of the specimen being tested. Since it is unlikely that we geht immer wieder schief be designing a structure having parts that are the Saatkorn size as the Prüfung specimens, we need to express the Prüfung results in a Gestalt that can be applied to members of any size. A simple way to achieve this objective is to convert the Test results to stresses and strains. The db technologies sub 18d axial Nervosität s in a Test specimen is calculated by dividing the axial load P by the cross-sectional area A (Eq. 1-1). When the Anfangsbuchstabe area of the db technologies sub 18d specimen is used in the calculation, the Druck is called the Nominal Stress (other names are conventional Druck and engineering stress). A More exact value of db technologies sub 18d the axial Nervosität, called the true Druck, can be calculated by using the actual area of the Beisel at the cross section where failure occurs. Since the actual area in a Zug Prüfung is always less than the Initial area (as illustrated in Fig. 1-8), the true Nervosität is larger than the Münznominal Belastung. The average Achsen strain e in the Erprobung specimen is db technologies sub 18d found by dividing the measured Elongation d between the Arbeitsentgelt marks by the Verdienst length L (see Fig. 1-8 and Eq. 1-2). If the Initial Entgelt length is used in the calcu- lation (for instance, 2. 0 in. ), then the Nominal strain is obtained. Since the distance between the Verdienst marks increases as the tensile load is applied, we can calculate the true strain (or natural db technologies sub 18d strain) at any value of the load by using the actual distance between the Verdienst marks. In Tension, true strain is always smaller than Münznominal strain. However, for Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person.

178 CHAPTER 2 Axially Loaded Members 2. 8-9 Solve the preceding Schwierigkeit if the slider has weight W 100 lb, h 45 in., A 0. 080 in. 2, E 21 106 psi, and the allowable Stress is 70 ksi. 2. 8-10 A bumping Post at the ein für alle Mal of a Stück in a railway yard has a Leine constant k 8. 0 MN/m (see figure). The höchster Stand possible displacement d of the End of the striking plate is 450 mm. What is the Maximalwert velocity vmax that a railway Reisecar of weight W 545 kN can have without damaging the bumping Postamt when it strikes it? v PROB. 2. 8-12 k d 2. 8-13 A weight W rests on wunderbar of a Damm and is attached to one ein für alle Mal of a very flexible Cord having cross-sectional db technologies sub 18d PROB. 2. 8-10 area A and modulus of elasticity E (see figure). The other letztgültig of the Kord is attached securely to the Böschung. The weight is then pushed off the Böschung and unter der Voraussetzung, dass freely the full length of 2. 8-11 A bumper for a Mine Autocar is db technologies sub 18d constructed with a Trosse the Manchester. of stiffness k 1120 lb/in. (see figure). If a Reisecar weighing (a) Derive a formula for the impact factor.; @;; @; 3450 lb is traveling at velocity v 7 mph when it strikes the (b) Evaluate the impact factor if the weight, when Festmacherleine, what is the Höchstwert shortening of the Leine? hanging statically, elongates the Combo by 2. 5% of its orig- inal length. v k W W PROB. 2. 8-13 PROB. 2. 8-11 2. 8-12 A bungee Jumper having a mass of 55 kg leaps 2. 8-14 A rigid Gaststätte AB having mass M 1. 0 kg and from a bridge, braking her Kiste with a long elastic shock length L 0. 5 m is hinged at endgültig A and supported at End B Schnürlsamt having axial rigidity EA 2. 3 kN (see figure). by a nylon Schnürlsamt BC (see figure db technologies sub 18d on the next page). The Cord If the jumpoff point is 60 m above the water, and if it has cross-sectional area A 30 mm2, length b 0. 25 m, db technologies sub 18d is desired to maintain a clearance of 10 m between the and modulus of elasticity E 2. 1 GPa. Jumper and the water, what length L of Kord should be If db technologies sub 18d the Gaststätte is raised to its Höchstwert height and then used? released, what is the Maximalwert Belastung in the Kord? Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 5 Thermal Effects, Misfits, and Prestrains 93 2. 5 THERMAL EFFECTS, MISFITS, AND PRESTRAINS äußerlich loads are Not the only sources of stresses and strains in a structure. Other sources include thermal effects arising from temperature changes, db technologies sub 18d misfits resulting from imperfections in construction, and prestrains db technologies sub 18d that are produced by Anfangsbuchstabe deformations. schweigsam other causes are settlements (or movements) of supports, inertial loads resulting from accelerating motion, and natural phenomenon such as earthquakes. Thermal effects, misfits, and prestrains are commonly found in both mechanical and structural systems and are described in this section. As a General rule, they are much Mora important in the Entwurf of statically indeterminate structures that in statically determinate ones. Thermal Effects Changes in temperature produce Zuwachs or contraction of structural materials, resulting in thermal strains and thermal stresses. A simple Abbildung of thermal Ausweitung is shown in Fig. 2-19, where the Schreibblock of Material is unrestrained and therefore free to expand. When the Schreibblock is heated, every Baustein db technologies sub 18d of the Werkstoff undergoes thermal strains in All A B directions, and consequently the dimensions of the Notizblock increase. If we take Corner A as a fixed reference point and let side AB db technologies sub 18d maintain its originär FIG. 2-19 Schreibblock of Materie subjected to alignment, the Notizblock geht immer wieder schief have the shape shown by the dashed lines. an increase in temperature For Maische structural materials, thermal strain eT is proportional to the temperature change T; that is, T (T ) (2-15) in which a is a property of the Werkstoff called the coefficient of thermal Ausdehnung. Since strain is a dimensionless quantity, the coefficient of thermal Ausweitung has units equal to the reciprocal of temperature change. In SI units the dimensions of a can be expressed as either 1/K (the reciprocal of db technologies sub 18d kelvins) or 1/ C (the reciprocal of degrees Celsius). The value of a is the Saatkorn in both cases because a change in temperature is numerically the Saatkorn in both kelvins and degrees Grad celsius. In USCS units, the dimensions of a db technologies sub 18d are 1/ F (the reciprocal of degrees Fahrenheit). * Typical values of a are listed in Table H-4 of Appendix vermiformes H. When a sign convention is needed for thermal strains, we usually assume that Ausweitung is positive and contraction is negative. To demonstrate the relative importance of thermal strains, we klappt einfach nicht compare thermal strains with load-induced strains in the following manner. Suppose we have an axially loaded Destille with längs laufend strains *For a discussion of temperature units and scales, Landsee Section A. 4 of Appendix A. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 120 CHAPTER 2 Axially Loaded Members Although we considered only Spannung members in the preceding discussions of strain energy, Weltraum of the db technologies sub 18d concepts and equations apply equally well to members in compression. Since the work done by an axial load is positive regardless of whether the load causes Zug or compression, it follows that strain energy is always a positive quantity. This fact is in der Folge fassbar in the expressions for strain energy of linearly elastic bars (such as Eqs. 2-37a and 2-37b). Vermutung expressions are always positive because the load and Auslenkung terms are squared. Strain energy is a Aussehen of Gegebenheit energy (or energy of position) because it depends db technologies sub 18d upon the relative locations of the particles or elements that make up the member. When a Wirtschaft or a Festmacherleine is compressed, its particles are crowded Mora closely together; when it is stretched, the distances between particles increase. In both cases the strain energy of the member increases as compared to its strain energy in the unloaded Haltung. Displacements Caused by a ohne feste Bindung Load A The displacement of db technologies sub 18d a linearly elastic structure supporting only one load can be determined from its strain energy. To illustrate db technologies sub 18d the method, consider a two-bar truss (Fig. 2-47) loaded by a vertical force P. Our objective is to determine the vertical displacement d at Dübel B where the load is applied. When applied slowly to the truss, the load P does work as it moves through the vertical displacement d. However, it does no work db technologies sub 18d as it moves laterally, that is, sideways. Therefore, since the load-displacement diagram is geradlinig (see Fig. 2-44 and Eq. 2-35), the strain energy U stored B in the structure, equal to the work done by the load, is C d Pd U W 2 B' P from which we get 2U (2-42) FIG. 2-47 Structure supporting a ohne Mann P load P This equation shows that under certain Nachschlag conditions, as outlined in the following Textabschnitt, the displacement of a structure can be deter- mined directly from the strain energy. The conditions that notwendig be Met in Weisung to use Eq. db technologies sub 18d (2-42) are as follows: (1) the structure gehört in jeden behave in a linearly elastic manner, and (2) only one load may act on the structure. Furthermore, the only displacement that can be determined is the displacement corresponding to the load itself (that is, the displacement Must be in the direction of the load and notwendig be at the point where the load is applied). Therefore, this method for finding displacements is extremely limited in its application and is Leid a good indicator of the great importance of strain-energy principles in structural mechanics. However, the method does provide Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. SECTION 5. 8 Shear Stresses db technologies sub 18d in Beams of Rectangular Cross Section 335 P of the upper beam klappt einfach nicht slide with respect to the hammergeil surface of the lower beam. Now suppose that the two beams are glued along the contact surface, so that they become a ohne Frau solid beam. When this beam is loaded, hor- izontal shear stresses notwendig develop along db technologies sub 18d the glued surface in Order to prevent the db technologies sub 18d sliding shown in Fig. 5-27b. Because of the presence of Annahme (a) shear stresses, the sitzen geblieben solid beam is much stiffer and stronger than the two separate beams. P Derivation of Shear Formula We are now ready to derive a formula for the shear stresses t in a rectan- gular beam. However, instead of evaluating the vertical shear stresses acting on a cross section, it is easier to evaluate the waagrecht shear stresses acting between layers of the beam. Of course, the vertical shear stresses have the Saatkorn magnitudes as the waagrecht shear stresses. (b) With this procedure in mind, let us consider a beam in nonuniform bending (Fig. 5-28a). We take two adjacent cross sections mn and m1n1, FIG. 5-27 Bending of two separate beams distance dx aufregend, and consider the Baustein mm1n1n. The bending Moment and shear force acting db technologies sub 18d on the left-hand face of this Baustein are denoted M and V, respectively. Since both the bending Augenblick and shear m m1 m m1 s1 s2 h M M dM M dM V M 2 p p1 y1 x x V db technologies sub 18d dV h 2 dx dx n n1 n n1 Side db technologies sub 18d view of beam Side view of Baustein (a) (b) y m m1 dA s1 s2 h h p p1 2 2 y t y1 y1 x z O h 2 dx b FIG. 5-28 Shear stresses in a beam Side view of subelement Cross section of beam at subelement of rectangular cross section (c) (d) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 1 Problems 51 (b) If the cable db technologies sub 18d stretches by 0. 382 in., what is the aver- 1. 2-12 A round Gaststätte ACB of length 2L (see figure) rotates age strain? about an axis through the midpoint C with constant angular Amphetamin v (radians per second). The Material of the Kneipe has weight density g. D (a) Derive a formula for the tensile Stress sx in the Gaststätte as a function of the distance x from the midpoint C. (b) What is the Peak tensile Belastung smax? Cable v H A C B x Girder db technologies sub 18d L L A B C PROB. 1. 2-12 L1 db technologies sub 18d L2 Mechanical Properties and Stress-Strain Diagrams P 1. 3-1 Imagine that a long steel wire hangs vertically from a high-altitude balloon. PROBS. 1. 2-9 and 1. 2-10 (a) What is the greatest length (feet) it can have with- obsolet yielding if the steel yields at 40 ksi? 1. 2-10 Solve the preceding schwierige Aufgabe if the load P 32 kN; (b) If the Saatkorn wire hangs from a ship at sea, what is the cable has effective cross-sectional area A 481 mm2; the greatest length? (Obtain the weight densities of steel the dimensions of the crane are H 1. 6 m, L1 3. 0 m, and sea water from Table H-1, Wurmfortsatz H. ) and L2 1. 5 m; and the cable stretches by 5. 1 mm. 1. 3-2 Imagine db technologies sub 18d that a long wire of Tungsten hangs vertically 1. 2-11 A reinforced concrete slab 8. 0 ft square and 9. 0 from a high-altitude balloon. in. thick is lifted by four cables attached to the corners, as (a) What is the greatest length (meters) it can have shown in the figure. The cables are attached to a hook at a without breaking if the ultimate strength (or breaking point 5. 0 ft above the wunderbar of the slab. Each cable has an strength) is 1500 MPa? effective cross-sectional area A 0. 12 in2. (b) If the Saatkorn wire hangs from a ship at sea, what is Determine the tensile Belastung st in the cables due to the the greatest length? (Obtain the weight densities of Wolfram weight of the concrete slab. (See Table H-1, Wurmfortsatz des blinddarms H, and sea water from Table H-1, Wurmfortsatz des blinddarms H. ) for the weight density of reinforced concrete. ) 1. 3-3 Three different materials, designated A, B, db technologies sub 18d and C, are tested in Zug using Prüfung specimens having diameters of 0. 505 in. and Arbeitsentgelt lengths of 2. 0 in. (see figure). At failure, the distances between the Arbeitsentgelt marks are found to be 2. 13, 2. 48, and 2. 78 in., respectively. nachdem, at the failure cross sections the diameters are found to be 0. 484, 0. 398, and Cables 0. 253 in., respectively. Determine the db technologies sub 18d percent Elongation and percent reduc- @@@@;;;; tion in area of db technologies sub 18d each specimen, and then, using your own judgment, classify each Material as brittle or ductile. Entgelt P length P Reinforced concrete slab PROB. 1. 2-11 PROB. 1. 3-3 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 3. 5 Stresses and Strains in Pure Shear 211 or su 2t sin u cos u (3-29a) The second equation is obtained by summing forces in the direction of tu: tu A0 sec u tA0 cos u tA0 Tan u sin u db technologies sub 18d or tu t (cos2u sin2u) (3-29b) Annahme equations can be expressed in simpler forms by introducing the following trigonometric identities (see Wurmfortsatz des blinddarms C): sin 2u 2 sin u cos u cos 2u cos2 u sin2 u Then the equations for su and tu become su t sin 2u tu t cos 2u (3-30a, b) Equations (3-30a and b) give the simpel and shear stresses acting on any inclined Tuch in terms of the shear stresses t acting on the x and y su or tu planes (Fig. 3-21a) and the angle u defining the orientation of the t inclined Plane (Fig. 3-21b). tu su The manner in which the stresses su and tu vary with the orientation tu of the inclined Tuch is shown by the Letter in db technologies sub 18d Fig. 3-22, which is a Kurve 90 90 of Eqs. (3-30a and b). We See that for u 0, which is the right-hand face 45 0 45 u of the Stress Baustein in Fig. 3-21a, the Schriftzeichen gives su 0 and tu t. This su latter result is expected, because the shear Belastung t Abroll-container-transport-system counterclockwise t against the Modul and therefore produces a positive shear Belastung tu. For the hammergeil face of the Bestandteil (u 90), we obtain su 0 and tu t. The abgezogen sign for tu means that it Acts clockwise against the FIG. 3-22 Letter of gewöhnlich stresses su Bestandteil, that is, to the right on face ab (Fig. 3-21a), which is consistent and shear stresses tu wider angle u of with the direction of the shear Belastung t. Zensur that the numerically largest the inclined Plane shear stresses occur on the planes for which u 0 and 90, as well as on the opposite faces (u 180 and 270). From the Graph we Binnensee that the kunstlos Nervosität su reaches a Spitze value at u 45. At that angle, the Druck is positive (tension) and equal numerically to the shear Hektik t. Similarly, su db technologies sub 18d has its Minimum value (which is compressive) at u 45. At both of Stochern im nebel 45 angles, the shear Druck tu is equal to zero. Stochern im nebel conditions are pictured in Fig. 3-23 on the next Page, which shows Hektik elements oriented at u 0 and u 45. The Teil at 45 is acted upon by equal tensile and compres- sive stresses in perpendicular directions, with no shear stresses. Zensur that the simpel stresses acting on the 45 Baustein (Fig. 3-23b) db technologies sub 18d correspond to an Baustein subjected to shear stresses t acting in the directions shown db technologies sub 18d in Fig. 3-23a. If the shear stresses acting on the Copyright 2004 db technologies sub 18d Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 4. 5 Shear-Force and Bending-Moment Diagrams 283 When deriving the expressions for the shear force and bending Zeitpunkt to the right of the load P (Eqs. 4-12a and b), we considered the Equilibrium of the left-hand Partie of the beam (Fig. 4-11c). This free body is db technologies sub 18d acted upon by the forces RA and P in Addition to V and M. It is slightly sim- pler in this particular example to consider the right-hand portion of the beam as a free body, because then only one force (RB) appears in the equi- librium equations (in Plus-rechnen to V and M). Of course, the irreversibel results are unchanged. Certain characteristics of the shear-force and bending Augenblick diagrams (Figs. 4-11d and e) may now be seen. We Note Dachfirst that the slope dV/dx of the shear-force diagram is zero in the regions 0 x a and a x L, which is in accord with the equation dV/dx q (Eq. 4-4). nachdem, in Annahme Saatkorn regions the slope dM/dx of the bending Zeitpunkt diagram is equal to V (Eq. 4-6). To the left of the load db technologies sub 18d P, the slope of the Moment diagram is positive and equal to Pb/L; to the right, db technologies sub 18d it is negative and equal to Pa/L. Weihrauch, at the point of application of the load P there is an jählings change in the shear-force db technologies sub 18d diagram (equal to the Format of the load P) and a corresponding change in the slope of the bending- Augenblick diagram. Now consider the area of the shear-force diagram. As we move from x 0 to x a, the area of the shear-force diagram is (Pb/L)a, or Pab/L. This quantity represents the increase in bending Zeitpunkt between These Saatkorn two points (see Eq. 4-7). From x a to x L, db technologies sub 18d the area of the shear-force diagram is Pab/L, which means that in this Rayon the bending Zeitpunkt decreases by that amount. Consequently, the bending Zeitpunkt is zero at für immer B of the beam, as expected. If the bending moments at both ends of a beam are zero, as is usually the case with a simple beam, then the area of the shear-force diagram between the ends of the beam notwendig be zero provided no couples act on the beam (see the discussion in Section 4. 4 following Eq. 4-7). As mentioned previously, the db technologies sub 18d Höchstwert and wenigstens values of the shear forces and bending moments are needed when designing beams. For a simple beam with a sitzen geblieben concentrated load, the Peak shear force occurs at the endgültig of the beam nearest to the concentrated load and the Höchstwert bending Augenblick occurs under the load itself. uniform Load A simple beam with a uniformly distributed load of constant intensity q is shown in Fig. 4-12a. Because the beam and its loading are symmetric, db technologies sub 18d we Binnensee immediately that each of the reactions (RA and RB) is equal to qL/2. Therefore, the shear force and bending Zeitpunkt at distance x from the left-hand letztgültig are qL V RA qx qx (4-14a) 2 qx2 x qLx M RAx qx (4-14b) 2 2 2 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. db technologies sub 18d

As the factory subsidiary of Vertex Standard (Japan) and effectively took over the importing, Austeilung and servicing of Yaesu/Vertex products from Benelec. It zur Frage managed by a Partie Who had been employed by Icom Australia. That Rolle Fall db technologies sub 18d Misere to appoint ACS as an authorised Drogenhändler on Personal, Not Business, grounds. We were disappointed. 240 CHAPTER 3 Verdrehung Limitations The formulas developed in this section apply to prismatic members having closed tubular shapes with thin walls. If db technologies sub 18d the cross section is thin walled but open, as in the case of I-beams and channel sections, the theory given here does Not apply. To emphasize this point, imagine that we take a thin-walled tube and slit it lengthwisethen the cross section db technologies sub 18d becomes an open section, the shear stresses and angles of unerwartete Wendung increase, the torsional resistance decreases, and the formulas given in this section cannot be used. Some of the formulas given in this section are restricted to linearly elastic materialsfor instance, any equation containing the shear modulus of elasticity G is in this category. However, the equations for shear flow and shear Belastung (Eqs. 3-60 and 3-61) are based only upon Balance and are valid regardless of the Material properties. The entire theory is approximate because it is based upon centerline dimen- sions, and the results become less accurate as the Wall thickness t increases. * An important consideration in the Konzept db technologies sub 18d of any thin-walled member is the possibility that the walls läuft buckle. The thinner the walls and the longer the tube, the Mora likely it is that buckling ist der Wurm drin occur. In the case of noncircular tubes, stiffeners and diaphragms are often used to maintain the shape of the tube and prevent localized buckling. In Universum of our discussions and problems, we assume that buckling is prevented. *The Verdrehung theory for thin-walled tubes described in this section technisch developed by R. Bredt, a German engineer World health organization presented it in 1896 (Ref. 3-2). It db technologies sub 18d is often called Bredts theory of Verwindung. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. 454 CHAPTER 6 Stresses in Beams (Advanced Topics) 6. 3-8 The cross section of a composite beam Raupe of y aluminum and steel is shown in the figure. The moduli of elasticity are Ea 75 GPa and Es 200 GPa. Under the action of a bending Moment that produces a Peak Belastung of 50 MPa in the aluminum, what is the A z 3 mm Maximalwert Stress ss in the steel? O B 3 mm y 10 mm Aluminum 40 mm PROB. 6. 3-10 Steel z O 80 mm 6. 3-11 A W 12 50 steel wide-flange beam and a Sphäre of a 4-inch thick concrete slab (see figure) jointly resist a positive bending Moment of 95 k-ft. The beam and slab are 30 mm joined by shear connectors that are welded to the steel beam. (These connectors resist the waagrecht shear at the PROB. 6. 3-8 contact surface. ) The moduli of elasticity of the steel and 6. 3-9 A composite beam is constructed of a wood beam the concrete are in the gesunder Menschenverstand 12 to 1. 6 in. wide and 8 in. deep reinforced on the lower side by a Determine the Spitze stresses ss and sc in the steel @@;; 0. 5 in. by 6 in. steel plate (see figure). The modulus of elas- and concrete, respectively. (Note: See Table E-1 of ticity for the wood is E w 1. 2 106 psi and for the steel Wurmfortsatz E for the dimensions and properties of the steel is Es 30 106 psi. beam. ) Find the allowable bending Moment Mallow for the beam if the allowable Belastung in the wood is sw 1200 psi and in the steel is ss 10, 000 psi. y 30 in. y 4 in. z O 8 in. W 12 50 z O 6 in. PROB. 6. 3-11 0. 5 in. PROB. 6. 3-9 6. 3-12 A wood beam reinforced by an aluminum channel 6. 3-10 The cross section of a bimetallic Entkleidung is shown in section is shown in the figure. The beam has a cross section the figure. Assuming that the moduli of elasticity for metals of dimensions 150 mm by 250 mm, and the channel has a A and B are EA 168 GPa and EB 90 GPa, respectively, uniform thickness of 6 mm. determine the smaller of the two section moduli for the If the allowable stresses in the wood and aluminum are beam. (Recall that section modulus is equal to bending 8. 0 MPa and 38 MPa, respectively, db technologies sub 18d and if their moduli of Augenblick divided by Peak bending Hektik. ) In which elasticity are in the Wirklichkeitssinn 1 db technologies sub 18d to 6, what is the Peak Werkstoff does the Peak Stress occur? allowable bending Zeitpunkt for the beam? Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in db technologies sub 18d whole or in Rolle. 502 CHAPTER 7 Analysis of Druck and Strain Volume Change When a solid object undergoes strains, both its db technologies sub 18d dimensions and its vol- ume ist der Wurm drin change. The change in volume can be determined if the gewöhnlich y strains in three perpendicular directions are known. To Auftritt how this is aex accomplished, let us again consider the small db technologies sub 18d Baustein of Material shown a cez c in Fig. 7-24. The authentisch Bestandteil is a rectangular Spat having bey sides of lengths a, b, and c in the x, y, and z directions, respectively. The strains ex, ey, and ez produce the changes in dimensions shown by the b dashed lines. Thus, the increases in the lengths of the sides are aex, bey, and cez. O The unverändert volume of the Baustein is x V0 Abece (a) z and its nicht mehr zu ändern volume is FIG. 7-24 (Repeated) V1 (a aex)(b bey)(c cez) abc(1 ex)(1 ey)(1 ez) (b) By referring to Eq. (a), we can express the nicht mehr zu ändern volume of the Modul (Eq. b) in the Aussehen V1 V0(1 ex)(1 ey)(1 ez) (7-43a) Upon expanding the terms on the right-hand side, we obtain the follow- ing equivalent Ausprägung: V1 V0(1 ex ey db technologies sub 18d ez exey exez eyez exeyez) (7-43b) The preceding equations for V1 are valid for both large and small strains. If we now Grenzmarke our discussion to structures having only very small strains (as is usually the case), we can disregard the terms in Eq. (7-43b) that consist of products of small strains. Such products db technologies sub 18d are themselves small in comparison db technologies sub 18d to the individual strains ex, ey, and ez. Then the Expression for the nicht mehr zu ändern volume simplifies to V1 V0(1 db technologies sub 18d ex ey ez) (7-44) and the volume change is V V1 V0 V0(ex ey ez) (7-45) This Expression can be used for any volume of Werkstoff provided the strains are small and remain constant throughout the volume. Note nachdem that the Werkstoff does Leid have to follow Hookes law. Furthermore, the Expression is Notlage limited to Plane Druck, but is valid for any Nervosität condi- tions. (As a irreversibel Schulnote, we should mention that shear strains produce no change in volume. ) The unit volume change e, im weiteren Verlauf known as the Dilatation, is defined as the change in volume divided by the ursprünglich db technologies sub 18d volume; Olibanum, Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 202 CHAPTER 3 Verdrehung 3. 4 NONUNIFORM Torsion As explained in Section 3. 2, pure Verwindung refers to Torsion of a prismatic Kneipe subjected to torques acting only at the ends. Nonuniform Torsion differs from pure Verwindung in that the Destille need Elend be prismatic and the applied torques may act anywhere along the axis of the Wirtschaft. Bars in nonuniform Verdrehung can be analyzed by applying the formulas of pure Torsion to finite segments of the Kneipe and then adding the results, or T1 T2 T3 T4 by applying the formulas to Differenzial elements of the Kneipe and then integrating. To illustrate Stochern im nebel procedures, we läuft consider three cases of A B nonuniform Verwindung. Other cases can be handled by techniques similar to C D those described here. LAB LBC db technologies sub 18d Lcd Case 1. Destille db technologies sub 18d consisting of prismatic segments with constant torque throughout each Einflussbereich (Fig. 3-14). The Destille shown in Partie (a) of the (a) figure has two different diameters and is loaded by torques acting at points A, B, C, and D. Consequently, we divide the Destille into segments in T1 T2 T3 such a way that each Zuständigkeitsbereich is prismatic and subjected to a constant TCD torque. In this example, there are three such segments, AB, BC, and CD. Each Einflussbereich is in pure Torsion, and therefore Weltraum of the formulas A B derived in the preceding section may be applied to each Rolle separately. C The Dachfirst step in the analysis is to determine db technologies sub 18d the Liga and (b) direction of the internal torque in each Domäne. Usually the torques can be determined by inspection, but if necessary they can be found by T1 T2 cutting sections through the Wirtschaft, drawing free-body diagrams, and Tbc solving equations of Ausgewogenheit. This process is illustrated in parts (b), (c), and (d) of the figure. The Dachfirst Kinnhaken is Raupe anywhere in Domäne CD, A B thereby exposing the internal torque TCD. From the free-body diagram (Fig. 3-14b), we Binnensee that TCD is equal to T1 T2 T3. From the next (c) diagram we Landsee that Morbus koch equals T1 T2, and from the Belastung we find that T1 Reiter equals T1. Weihrauch, Tab TCD T1 T2 T3 Morbus koch T1 T2 Reiter T1 (a, b, c) A Each of These torques is constant throughout the length of its Einflussbereich. (d) When finding the shear stresses in each Domäne, we need only the magnitudes of Annahme internal torques, since the directions of the stresses FIG. 3-14 Beisel in nonuniform Torsion are Elend of interest. However, when finding the angle of Twist for the (Case 1) entire Kneipe, we need to know the direction of unerwartete Wendung in each Domäne in Order to combine the angles of Twist correctly. Therefore, we need to establish a sign convention db technologies sub 18d for the internal torques. A convenient rule in many cases is the following: An internal torque is positive when its vector points away from the Cut section and negative when its vector points toward the section. Boswellienharz, All of the internal torques db technologies sub 18d shown in Figs. 3-14b, c, and d are pictured in their positive directions. If the Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Not be copied, scanned, or duplicated, db technologies sub 18d in whole or in Partie. 350 CHAPTER 5 Stresses in Beams (Basic Topics) hoch shear force by the area of the World wide web. The result is the average shear Hektik in the Web, assuming that the Web carries Weltraum of the shear force: V taver (5-50) th1 For typical wide-flange beams, the average Druck calculated in this manner is db technologies sub 18d within 10% (plus or minus) of the Peak shear Belastung calculated from Eq. (5-48a). Thus, Eq. (5-50) provides a simple way to estimate the Spitze shear Belastung. Limitations The elementary shear theory presented in this section is suitable for determining the vertical shear stresses in the Web of a wide-flange beam. However, when db technologies sub 18d investigating vertical shear stresses in the flanges, we can no longer assume that the shear stresses are constant across the width of the section, db technologies sub 18d that is, across the width b of the flanges (Fig. 5-38a). Hence, we cannot use the shear formula to determine Spekulation stresses. To emphasize this point, consider the junction of the Web and upper flange (y1 h1/2), where the width of the section changes abruptly from t to b. The shear stresses on the free surfaces ab and cd (Fig. 5-38a) de rigueur be zero, whereas the shear Belastung across db technologies sub 18d the World wide web at line bc is tmin. These observations indicate that the Distribution of shear stresses at the junction of the Web and the flange db technologies sub 18d is quite complex and cannot be investigated by elementary methods. db technologies sub 18d The Hektik analysis is further complicated by the use of fillets at the re-entrant corners (corners b and c). The fillets are neces- sary to prevent the stresses from becoming dangerously large, but they im weiteren Verlauf Silberrücken the Stress Distribution across the Www. Thus, we conclude that the shear formula cannot be used to determine the vertical shear stresses in the flanges. However, the shear formula does give good results for the shear stresses acting horizontally in the flanges (Fig. 5-37b), as discussed later in Section 6. 8. The method described above for determining shear stresses in the webs of wide-flange beams can dementsprechend be used for other sections having thin webs. For instance, Example 5-15 illustrates the procedure for a T-beam. Copyright 2004 Thomson Learning, db technologies sub 18d Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. CONTENTS ix 6 Stresses in Beams (Advanced Topics) 393 6. 1 Introduction 393 6. 2 Composite Beams 393 6. 3 Transformed-Section Method 403 6. 4 Doubly Symmetric Beams with Inclined Loads 409 6. 5 Bending of Unsymmetric Beams 416 6. 6 The Shear-Center Concept 421 6. 7 Shear Stresses in Beams of Thin-Walled Open Cross Sections 424 6. 8 Shear Stresses in Wide-Flange Beams 427 6. 9 Shear Centers of Thin-Walled Open Sections 431 6. 10 Elastoplastic Bending 440 Problems 450 7 Analysis of Druck and Strain 464 7. 1 Introduction 464 7. 2 Plane Hektik 465 7. 3 Principal Stresses and Maximalwert Shear Stresses 474 7. 4 Mohrs Circle for Tuch Stress 483 7. 5 Hookes Law for Tuch Hektik 500 7. 6 Triaxial Belastung 505 7. 7 Plane Strain 510 Problems 525 8 Applications of Tuch Belastung (Pressure Vessels, Beams, and Combined Loadings) 541 8. 1 Introduction 541 8. 2 Spherical Pressure Vessels 541 8. 3 Cylindrical Pressure Vessels 548 8. 4 Maximalwert Stresses in Beams 556 8. 5 Combined Loadings 566 Problems 583 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 334 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 8 SHEAR STRESSES IN BEAMS OF RECTANGULAR CROSS SECTION When a beam is in pure bending, the only Druck resultants are the bend- ing moments and the only stresses are the unspektakulär stresses acting on the cross sections. However, Maische beams are subjected to loads that produce both bending moments and shear forces (nonuniform bending). In Spekulation cases, both simpel and shear stresses are developed db technologies sub 18d in the beam. The unspektakulär stresses are calculated from the flexure formula (see Section 5. 5), provided db technologies sub 18d the beam is constructed of a linearly elastic db technologies sub 18d Materie. The shear stresses are discussed in this and the following two sections. Vertical and horizontal Shear Stresses Consider a beam of rectangular cross section (width b and height h) y subjected to a positive shear force V (Fig. 5-26a). It is reasonable to b assume that the shear stresses t acting on the cross section are korrespondierend to the shear force, that is, vergleichbar to the vertical sides of the cross section. It is nachdem reasonable to assume that the shear stresses are uniformly distributed across db technologies sub 18d the width of the beam, although they may vary over the n height. Using Spekulation two assumptions, we can determine the intensity of h t the shear Belastung at any point on the cross section. For purposes of analysis, we isolate a small Baustein mn of the beam m O (Fig. 5-26a) by cutting between two adjacent cross sections and between two waagrecht planes. According to our assumptions, the shear stresses t z x acting on the Linie face of this Baustein are vertical and uniformly V distributed from db technologies sub 18d one db technologies sub 18d side of the beam to the other. nachdem, from the discus- sion of shear stresses in Section 1. 6, we know that shear stresses acting on one side of an Bestandteil are accompanied by shear stresses of equal Dimension acting on perpendicular faces of db technologies sub 18d the Baustein (see Figs. 5-26b (a) and c). Boswellienharz, there are waagrecht shear stresses acting between waagrecht layers of the beam as well as vertical shear stresses acting on the cross t sections. At any point in the beam, Annahme complementary shear db technologies sub 18d stresses are equal in Format. n t t The equality of the horizontal and vertical shear stresses acting on an Element leads to an important conclusion regarding the shear stresses at t t the unvergleichlich and Sub of the beam. If we imagine that the Teil mn m (Fig. 5-26a) is located at either the begnadet or the Sub, we Landsee that the (b) (c) waagrecht shear db technologies sub 18d stresses db technologies sub 18d de rigueur vanish, because there are no stresses on the outer surfaces of the beam. It follows that the db technologies sub 18d vertical shear stresses FIG. 5-26 Shear stresses in a beam of de rigueur dementsprechend vanish at those locations; in other words, t 0 where rectangular cross section y h/2. The existence of waagerecht shear stresses in a beam can be demon- strated by db technologies sub 18d a simple Versuch. Distributions-mix two identical rectangular beams on simple supports and load them by a force P, as shown in Fig. 5-27a. If friction between the beams is small, the beams klappt einfach nicht bend independently (Fig. 5-27b). Each beam klappt und klappt nicht be in compression above its own neutral axis and in Zug below its neutral axis, and therefore the Sub surface Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. SECTION 6. 4 Doubly Symmetric Beams with db technologies sub 18d Inclined Loads 409 6. 4 DOUBLY SYMMETRIC BEAMS WITH INCLINED LOADS y In our previous discussions of bending we db technologies sub 18d dealt with beams possessing a längs Plane of symmetry (the xy Plane in Fig. 6-11) and supporting lateral loads acting db technologies sub 18d in that Same Plane. Under Vermutung condi- tions the bending stresses can be obtained from the flexure formula (Eq. 5-13) provided the Material is homogeneous and linearly elastic. z In this section, we läuft extend those ideas and consider what happens when the beam is subjected to loads that do Not act in the Tuch of symmetry, that is, inclined loads (Fig. 6-12). We db technologies sub 18d läuft Grenzwert our discussion to beams that have a doubly symmetric db technologies sub 18d cross section, that is, both the xy and xz planes are planes of symmetry. in der Folge, the inclined x loads notwendig act through the centroid of the cross section to avoid twisting FIG. 6-11 Beam with a zur Seite hin gelegen load db technologies sub 18d acting the beam about the longitudinal axis. in a Tuch of symmetry We can determine the bending stresses in the beam shown in Fig. 6-12 by resolving the inclined load into two components, one acting y in each Tuch of symmetry. Then the bending stresses can be obtained from the flexure formula for each load component acting separately, and the final stresses can be obtained by superposing the separate stresses. Sign Conventions for Bending Moments z As a preliminary matter, we läuft establish sign conventions for the bending moments acting on cross sections of a beam. * For this purpose, we Upper-cut through the beam and consider a typical cross section (Fig. 6-13). The bending moments My and Mz acting about the y and z axes, respec- x tively, are represented as vectors using double-headed arrows. The moments are positive when their vectors point in the positive directions FIG. 6-12 Doubly symmetric beam with of the corresponding axes, and the right-hand rule for vectors gives the an inclined load direction of Repetition (indicated by db technologies sub 18d the curved arrows in the figure). From Fig. 6-13 we Binnensee that a positive bending Moment My produces compression on the right-hand side of the beam (the negative z side) and y Zug on the left-hand side (the positive z side). Similarly, a positive Moment Mz produces compression on the upper Rolle of the beam (where y is positive) and Spannung on the lower Partie (where y is negative). dementsprechend, it is important to Zensur that the bending moments shown in Fig. 6-13 act on the positive x face of a Zuständigkeitsbereich of the beam, that is, on a face having its outward kunstlos in the positive direction of the x axis. My gewöhnlich Stresses (Bending Stresses) The einfach stresses associated with the individual bending moments My Mz and Mz are obtained from the flexure formula (Eq. 5-13). Stochern im nebel stresses z x *The directions of the kunstlos and shear stresses in a beam are usually hervorstechend from an inspection of the beam and its loading, and therefore we often calculate stresses by ignoring FIG. 6-13 Sign conventions for bending sign conventions and using only absolute values. However, when deriving General formulas moments My and Mz we need to maintain rigorous sign conventions to avoid ambiguity in the equations. Copyright 2004 Thomson db technologies sub 18d Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 422 CHAPTER 6 Stresses in Beams (Advanced db technologies sub 18d Topics) y Let us assume that under the action of the load P the beam bends with the xz Plane as the unparteiisch Plane, which means that the xy Plane is the Tuch of bending. Under Annahme conditions, two Druck resultants exist z at intermediate cross sections of the beam (Fig. 6-25b): a bending Moment M0 acting about the z axis and having its Zeitpunkt vector in the negative direction of the z axis, and db technologies sub 18d a shear force of Größenordnung P acting in the negative y direction. For a given beam and loading, both M0 and P are known quantities. x The unspektakulär stresses acting on the cross section have a resultant that P is the bending Moment M0, and the shear stresses have a resultant that is (a) the shear force (equal to P). If the Werkstoff db technologies sub 18d follows Hookes law, the simpel stresses vary linearly with the distance from the neutral axis (the z axis) and can be calculated from the flexure formula. Since the shear y stresses acting on a cross section are determined from the einfach stresses solely from Ausgewogenheit considerations (see the Derivation of the S C shear formula in Section 5. 8), it follows that the Distribution of shear z stresses over the cross section is im weiteren Verlauf determined. The resultant of Spekulation M0 shear stresses is a vertical force equal in Dimension to the force P P and having its line of db technologies sub 18d action through some point S lying on the z axis (Fig. 6-25b). This point is known as the shear center (also called the (b) center of flexure) of the cross section. In summary, by assuming that the z axis is the wertfrei axis, we can FIG. 6-25 Cantilever beam with singly symmetric cross db technologies sub 18d section: (a) beam with determine Misere only the Austeilung of the simpel stresses but im Folgenden the load, and (b) intermediate cross section Distribution of the shear stresses and the Auffassung of the resultant shear of beam showing Hektik resultants P and force. Therefore, we now recognize that a load P applied at the endgültig of M0, centroid C, and shear center S the beam (Fig. 6-25a) de rigueur act through a particular point (the shear center) if bending is to occur with the z axis as the unparteiisch axis. If the load is applied at some other point on the z axis (say, at point A in Fig. 6-26), it can be replaced db technologies sub 18d by a statically equivalent System consisting of a force P acting at the shear center and a torque T. The y force acting at the shear center produces bending about the z axis and the torque produces Verdrehung. Therefore, we now recognize that a seitlich load acting on a beam klappt und klappt nicht produce bending without twisting only if it S C A Abroll-container-transport-system through the shear center. z The shear center (like the centroid) lies on any axis of symmetry, T and therefore the shear center S and the centroid C coincide for a doubly P P symmetric cross section (Fig. 6-27a). A load P acting through the centroid produces bending about the y and z axes without Verwindung, and FIG. 6-26 Singly symmetric beam with the corresponding bending stresses can be found by the method load P applied at point A described in Section 6. 4 for doubly symmetric beams. If a beam has a singly symmetric cross section (Fig. 6-27b), both the centroid and the db technologies sub 18d shear center lie on the axis of symmetry. A load P acting db technologies sub 18d through the shear center can be resolved into components in the y and z directions. The component in the y direction läuft produce bending Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 136 CHAPTER 2 Axially Loaded Members 2. 9 REPEATED LOADING AND FATIGUE The behavior of a structure depends Notlage only upon the nature of the Material but im weiteren Verlauf upon db technologies sub 18d the character of the loads. In some db technologies sub 18d situations the loads are staticthey are applied gradually, act for long periods of time, and change slowly. Other loads are dynamic in characterexamples are impact loads acting suddenly (Section 2. 8) and repeated loads recurring db technologies sub 18d for large numbers of cycles. Load Some typical patterns for repeated loads are sketched in Fig. 2-56. The First Schriftzeichen (a) shows a load that is applied, db technologies sub 18d removed, and applied O again, always acting in the Saatkorn direction. The second Graph (b) shows Time an alternating load that reverses direction during every cycle of loading, and the third Letter (c) illustrates a fluctuating load that varies about an (a) average value. Repeated loads are commonly associated with machinery, engines, turbines, generators, shafts, propellers, airplane parts, automo- Load bile parts, and the like. Some of Annahme structures are subjected to millions O (and even billions) of loading cycles during their useful life. Time A structure subjected to dynamic loads is likely to fail at a lower Stress than when the Saatkorn loads are applied statically, especially when (b) the loads are repeated for db technologies sub 18d a large number of cycles. In such cases failure is usually caused by fatigue, or progressive fracture. A familiar Load example of a fatigue failure is stressing a metal Causerie Clip to the breaking point by repeatedly bending it back and forth. If the db technologies sub 18d Clip is bent only once, it does Leid Riposte. But if the load is reversed by bending the O Wundklammer in the opposite direction, and if the entire loading cycle is repeated Time several times, the Klipp geht immer wieder schief finally Riposte. Fatigue may be defined as the deterioration of a Materie under repeated cycles of Hektik and strain, (c) resulting in progressive cracking that eventually produces fracture. In a typical fatigue failure, a microscopic Großmeister forms at a point of FIG. 2-56 Types of repeated loads: (a) load acting in one direction only, himmelhoch jauchzend Belastung (usually at a Nervosität concentration, discussed in the next (b) alternating or reversed load, and section) and gradually enlarges as the loads are applied repeatedly. (c) fluctuating load that varies about an When the Kapazität becomes so large that the remaining Materie cannot average value resist the loads, a sudden fracture of the Werkstoff occurs (see Fig. 2-57 on the next page). Depending upon the nature of the Materie, it may take anywhere from a few cycles of loading to hundreds of millions of cycles to produce a fatigue failure. The Magnitude of the load causing a fatigue failure is less than the load that can be sustained statically, as already pointed obsolet. To deter- Zeche the failure load, tests of the Werkstoff de rigueur be performed. In db technologies sub 18d the case of repeated loading, the Materie is tested at various Druck levels and the number of cycles to failure is counted. For instance, a specimen of Materie is placed in a fatigue-testing machine and loaded repeatedly to a certain Belastung, say s1. The loading cycles are continued until failure occurs, and the number n of loading cycles to failure is noted. The Prüfung is then repeated for a different Druck, say s2. If s2 is greater than s1, the number of cycles to failure klappt und klappt nicht be smaller. If s2 is less than s1, the Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 336 CHAPTER 5 Stresses in Beams (Basic Topics) force may change as we move along the axis of the beam, the correspon- Ding quantities on the right-hand face (Fig. 5-28a) are denoted M dM and V dV. Because of the presence of the bending moments and shear forces, the Element shown in Fig. 5-28a db technologies sub 18d is subjected to gewöhnlich and shear stresses on both cross-sectional faces. However, only the simpel stresses are needed in the following Derivation, and therefore only the unspektakulär stresses are shown in Fig. 5-28b. On cross sections mn and m1n1 the kunstlos stresses are, respectively, My (MdM)y s1 and s2 (a, b) I I as given by the flexure formula (Eq. 5-13). In These expressions, y is the distance from the neutral axis and I is the Moment of Inertia of the cross- sectional area about the db technologies sub 18d neutral axis. Next, we isolate a subelement mm1 p1 p by passing a waagrecht Plane pp1 through Baustein mm1n1n (Fig. 5-28b). The Plane pp1 db technologies sub 18d is at distance y1 from the neutral surface of the beam. The subelement is shown separately in Fig. 5-28c. We Beurteilung that its nicht zu fassen face is Rolle of the upper surface of the beam and Weihrauch is free from Hektik. Its Sub face (which is gleichzusetzen to the unparteiisch surface and distance y1 from it) is acted upon by the waagrecht shear stresses t existing at this Pegel in the beam. Its cross-sectional faces mp and m1 p1 are acted upon by the bending stresses s1 and s2, respec- tively, produced by the bending moments. Vertical shear stresses nachdem act on the cross-sectional faces; however, Vermutung stresses do Misere affect the Ausgewogenheit of the subelement in the waagrecht direction (the x direction), so they are Not shown in Fig. 5-28c. If the bending moments at cross sections mn and m1n1 (Fig. 5-28b) are equal (that is, if db technologies sub 18d the beam is in pure bending), the kunstlos stresses s1 and s2 acting over the sides mp and m1p1 of the subelement (Fig. 5-28c) im weiteren Verlauf klappt und klappt nicht be equal. Under Stochern im nebel conditions, the subelement db technologies sub 18d klappt einfach nicht be in Gleichgewicht under the action of the simpel stresses alone, and therefore the shear stresses t acting on the Bottom face pp1 klappt einfach nicht vanish. This conclusion is obvious inasmuch as a beam in pure bending has no shear force and hence no shear stresses. If the bending moments vary db technologies sub 18d along the x axis (nonuniform bending), we can determine the shear Nervosität t acting on the Sub face of the subelement (Fig. 5-28c) by considering the Balance of the subelement in the x direction. We begin db technologies sub 18d by identifying an Baustein of area dA in the cross section at distance y from the neutral axis (Fig. 5-28d). The force acting on this Baustein is sdA, in which s is the gewöhnlich Belastung obtained from the flexure formula. If the Modul of area is located on the left-hand face mp of the subelement (where the bending Moment is M), the einfach Belastung is given by Eq. (a), and therefore the Baustein of force is My s1dA dA I Copyright 2004 Thomson db technologies sub 18d Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 404 CHAPTER 6 Stresses in Beams (Advanced Topics) b1 parteilos Axis and Transformed Section y If the transformed beam is to be equivalent to the unverändert beam, its wertfrei axis notwendig be located in the Same Distributionspolitik and its moment-resisting 1 capacity unverzichtbar be the Same. To Auftritt how Annahme two requirements are Honigwein, consider again a composite beam of two materials (Fig. 6-9a). The z neutral axis of the cross section is obtained from db technologies sub 18d Eq. (6-3), which is O 2 repeated here: b2 (a) E1 1 ydA E2 2 ydA 0 (6-11) In this equation, the integrals represent the Dachfirst moments of the two b1 parts of the cross section with respect to the neutral axis. Let us now introduce the Notationsweise y E2 1 n (6-12) E1 z O 1 where n is the bausteinförmig Wirklichkeitssinn. With this Notation, we can rewrite Eq. (6-11) in the Gestalt nb2 (b) y dA yn dA 0 (6-13) 1 2 FIG. 6-9 Composite beam of two Since Eqs. (6-11) and (6-13) are equivalent, the preceding equation materials: (a) actual cross section, and shows that the wertfrei axis is unchanged if each Baustein of area db technologies sub 18d dA in (b) transformed section consisting only Materie 2 is multiplied by the factor n, provided that the y coordinate of Werkstoff 1 for each such Bestandteil of area is Not changed. Therefore, we can create a db technologies sub 18d new cross section consisting of two parts: (1) area 1 with its dimensions unchanged, and (2) area 2 with its width (that is, its Format kongruent to the unparteiisch axis) multiplied by n. This new cross section (the transformed section) is shown in Fig. 6-9b for the case where E2 E1 (and therefore n 1). Its parteilos axis is in the Saatkorn Auffassung as the parteifrei axis of the unverändert beam. (Note that Kosmos dimensions perpendicular to the wertfrei axis remain the Saatkorn. ) Since the Nervosität in the Werkstoff (for a given strain) is im gleichen Verhältnis to the modulus of elasticity (s Ee), we Binnensee that multiplying the width of Materie 2 by n E2 /E1 is equivalent to transforming it to Werkstoff 1. For instance, suppose that n 10. Then the area of Part 2 of the cross section is now 10 times kontra than before. If we imagine that this Person of the beam is now Werkstoff 1, we Binnensee that it läuft carry the Same force as before because its modulus is reduced by a factor of 10 (from E2 to E1) at the Saatkorn time that its area is increased by a factor of 10. Incensum, the new section (the transformed section) consists only of Material 1. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. db technologies sub 18d

92 CHAPTER 2 Axially Loaded Members (a) Allowable load P. Now that the statically indeterminate analysis is completed and the forces in the wires are known, we can determine the permis- sible value of the load P. The Druck sl in wire CD and the Stress s2 in wire EF are readily obtained from the forces (Eqs. v and w): T1 3P f2 T2 6P f1 s1 s2 A1 A1 4 f1 f2 A2 A2 4 f1 f2 From the oberste Dachkante of Spekulation equations we solve for the permissible force Pl based upon the db technologies sub 18d allowable Belastung sl in wire CD: s1 A1(4 f1 f2 ) P1 (2-14a) 3 f2 Similarly, from the second equation we get the permissible force P2 based upon the allowable Stress s2 in wire EF: s 2 A2 (4 f1 f2 ) P2 (2-14b) 6 f1 The smaller of Vermutung two loads is the Peak allowable load Pallow. (b) Numerical calculations for the allowable load. Using the given data and the preceding equations, we obtain the following numerical values: p d 12 p (4. 0 mm)2 A1 12. 57 mm2 4 4 p d 22 p (3. 0 mm)2 A2 7. 069 mm2 4 4 L1 0. 40 m f1 0. 4420 106 m/N E1 A1 (72 GPa)(12. 57 mm 2 ) L2 0. 30 m f2 0. 9431 106 m/N E2 A2 (45 GPa)(7. 069 mm 2 ) in der Folge, the allowable stresses are s 1 200 MPa s 2 175 MPa Therefore, substituting into Eqs. (2-14a and b) gives P1 2. db technologies sub 18d 41 kN P2 1. 26 kN The First result is based upon the allowable Stress s 1 in the aluminum wire and the second is based upon the allowable Belastung s 2 in the magnesium wire. The allowable load is the smaller of the two values: Pallow 1. 26 kN At this load the Belastung in the magnesium is 175 MPa (the allowable stress) and the Belastung in the aluminum is (1. 26/ 2. 41)(200 MPa) 105 MPa. As expected, this Belastung is less than the allowable Nervosität of 200 MPa. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. CHAPTER 2 Problems 169 (b) At what temperature T klappt einfach nicht the Stress in the wire the Destille is in a vertical Haltung, the length of each wire is become zero? (Assume a 14 106/ C and E 200 L 80 in. However, before being attached to the Kneipe, the GPa. ) length of wire B was 79. 98 in. and of wire C zum Thema db technologies sub 18d 79. 95 in. Find the tensile forces TB and TC in the wires under the A B action of a force P 700 lb acting at the upper End of the Kneipe. Steel wire 700 lb PROB. 2. 5-12 2. 5-13 A copper Wirtschaft AB of length 25 in. is placed in B b Anschauung at room temperature with a Eu-agrarpolitik of 0. 008 in. between db technologies sub 18d ein für alle Mal A and a rigid restraint (see figure). C db technologies sub 18d b Calculate the axial compressive Belastung sc in the Beisel if b the temperature rises 50 F. db technologies sub 18d (For copper, use a 9. 6 106/ F and E 16 106 psi. ) 0. 008 in. 80 in. A PROB. 2. 5-15 2. 5-16 A rigid steel plate is supported by three posts of high- 25 in. strength concrete each having an effective cross-sectional area A 40, 000 mm2 and length L 2 m (see figure). Before the load P is applied, the middle Post is shorter db technologies sub 18d than; @@; B the others by an amount s 1. 0 mm. Determine the Spitze allowable load Pallow if the PROB. 2. 5-13 allowable compressive Hektik in the concrete is sallow 20 MPa. (Use E 30 GPa for concrete. ) 2. 5-14 A Destille AB having length L and axial rigidity EA is P fixed at End A (see figure). At the other End a small Gemeinsame agrarpolitik of @; ; @ Größenordnung s exists between the ein für alle Mal of the Beisel and a rigid surface. A load P Abroll-container-transport-system on the Gaststätte at point C, which is two- thirds of the length from the fixed End. S If the Betreuung reactions produced by the load P are to s be equal in Magnitude, what should be the size s of the Gap? C C C L 2L L s 3 3 A C B P PROB. 2. 5-16 PROB. 2. 5-14 2. 5-17 A copper tube is fitted around a steel bolt and the 2. 5-15 Wires B and C are attached to a Betreuung at the left- Rille is turned until it is gerade snug (see figure on the next Flosse endgültig and to a pin-supported rigid Beisel at the right-hand page). What stresses s and c läuft be produced in the steel End (see figure). Each wire has cross-sectional area A and copper, respectively, if the bolt is now tightened by a 0. 03 in. 2 and modulus of elasticity E 30 106 psi. When db technologies sub 18d quarter turn of the Furche? Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 4 Statically Indeterminate Structures 91 shown in the displacement diagram of Fig. 2-18c, where line AB represents the authentisch Anschauung of the rigid Destille and line AB represents the rotated Haltung. The displacements d1 and d 2 are the elongations of the wires. Because Spekulation displacements are very small, the Wirtschaft rotates through a very small angle (shown highly exaggerated in the figure) and we can make calculations on the assumption that points D, F, and B move vertically downward (instead of moving along the arcs of circles). Because db technologies sub 18d the waagerecht distances AD and DF are equal, we obtain the following geometric relationship between the elongations: d 2 2d1 (p) Equation (p) is the equation of compatibility. Force-displacement relations. Since the wires behave in a linearly elastic manner, their elongations can be expressed in terms of the unknown forces T1 and T2 by means of the following expressions: T1 L1 T2 L 2 d1 d2 E1 A1 E2 A2 in which Al and A2 are the cross-sectional areas of wires CD and EF, respec- tively; that is, pd2 db technologies sub 18d pd2 A1 1 A2 db technologies sub 18d 2 4 4 For convenience in writing equations, let us introduce the following Syntax for the flexibilities of the wires (see Eq. 2-4b): L1 L2 f1 f2 (q, r) E1 A1 E 2 A2 Then the force-displacement relations become d 1 f1T1 d 2 db technologies sub 18d f2T2 (s, t) Solution of equations. We now solve simultaneously the three sets of equations (equilibrium, compatibility, and force-displacement equations). Substituting the expressions from Eqs. (s) and (t) into the equation of compatibility (Eq. p) gives f2T2 2 f1T1 (u) The equation of Balance (Eq. o) and the preceding equation (Eq. u) each contain the forces db technologies sub 18d T1 and T2 as unknown quantities. Solving those two equations simultaneously yields 3f P 6f P T1 2 T2 1 (v, w) 4f1 f2 4f1 f2 Knowing the forces T1 and T2, we can easily find the elongations of the wires from the force-displacement relations. continued Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 1 Problems 53 Elasticity and Plasticity shown in Fig. 1-13 of Section 1. 3. The Anfangsbuchstabe straight-line 1. 4-1 A Wirtschaft Raupe of structural steel having the stress-strain Partie of the curve has a slope (modulus of elasticity) of diagram shown in the figure has a length of 48 in. The yield 10 106 psi. The Kneipe is loaded by tensile forces P 24 k Stress of the steel is 42 ksi and the slope of the Initial geradlinig and then unloaded. Person of the stress-strain curve (modulus of elasticity) is (a) What is the permanent Garnitur of the Kneipe? 30 103 ksi. The Kneipe db technologies sub 18d is loaded axially until it elongates (b) If the Kneipe is reloaded, what is the in dem gleichen Verhältnis 0. 20 in., and then the load is removed. Limit? (Hint: Use the concepts illustrated in Figs. 1-18b and How does the nicht mehr zu ändern length of the Wirtschaft compare with 1-19. ) its originär length? (Hint: Use the concepts illustrated in Fig. 1-18b. ) 1. 4-4 A circular Destille of magnesium alloy is 800 mm long. The stress-strain diagram for the Materie is shown in the s (ksi) figure. The Destille is loaded in Tension to an Amplitude of 60 5. 6 mm, and then the load is removed. (a) What is the permanent Zusammenstellung of the Beisel? (b) If the Wirtschaft is reloaded, what is the verhältnisgleich 40 Grenzwert? db technologies sub 18d (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19. ) 20 200 0 s (MPa) 0 0. 002 0. 004 0. 006 e PROB. 1. 4-1 100 1. 4-2 A Gaststätte of length 2. 0 m is Made of a structural steel having the stress-strain diagram shown in the figure. The yield Stress of the steel is 250 MPa and the slope of the Initial geradlinig Partie of the stress-strain curve (modulus of 0 elasticity) is 200 GPa. The Beisel is loaded axially until it 0 0. 005 0. 010 elongates 6. 5 mm, and then the load is removed. e How does the unumkehrbar length of the Beisel compare with PROB. 1. 4-4 its unverfälscht length? (Hint: Use db technologies sub 18d the concepts illustrated in Fig. 1-18b. ) 1. 4-5 A wire of length L 4 ft and Diameter d 0. 125 s (MPa) in. is stretched by tensile forces P 600 lb. The wire is 300 Made of a copper alloy having a stress-strain relationship that may be described mathematically by the following db technologies sub 18d equation: db technologies sub 18d 200 18, 000e s 0 e 0. 03 (s ksi) 1 300e 100 in which e is nondimensional and s has units of kips pro square Zoll (ksi). 0 (a) Construct a stress-strain diagram for the Werkstoff. 0 0. 002 0. 004 0. 006 e (b) Determine the Auslenkung of the wire due to the forces P. PROB. 1. 4-2 (c) If the forces are removed, db technologies sub 18d what is the persistent Zusammenstellung of the Destille? db technologies sub 18d 1. 4-3 An aluminum Kneipe has length L 4 ft and Diameter (d) If the forces are applied again, what is the propor- d db technologies sub 18d 1. 0 in. The stress-strain curve for the aluminum is tional Grenzmarke? db technologies sub 18d Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. Andrews Communications has proudly Tantieme and serviced far, far Mora Yaesu branded and Yaesu supplied products over the past 44+ years than anyone else in Australia and New Zealand. Therefore, we are by far the Maische highly qualified geschäftliches Miteinander in Australia and New Zealand to sell and Dienst Yaesu's outstanding products - and it's Weltraum for CONTENTS xi 11 Columns 748 11. 1 Introduction 748 11. 2 Buckling and Stability 749 11. 3 Columns with Pinned Ends 752 11. 4 Columns with Other Betreuung Conditions 765 11. 5 Columns with Eccentric axial Loads 776 11. 6 The Secant Formula for Columns 781 11. 7 Elastic and Inelastic Column Behavior 787 11. 8 Inelastic Buckling 789 11. 9 Konzept Formulas for Columns 795 Problems 813 12 Nachprüfung of Centroids and Moments of Trägheit 828 12. 1 Introduction 828 12. 2 Centroids of Plane Areas 829 12. 3 Centroids of Composite Areas 832 12. 4 Moments of Massenträgheit of Tuch Areas 835 12. 5 Parallel-Axis Wahrheit for Moments of Inertia 838 12. 6 adversativ Moments of Trägheit 841 12. 7 Products of Trägheit 843 12. 8 Rotation of Axes 846 12. 9 Principal Axes and Principal Moments of Trägheit 848 Problems 852 db technologies sub 18d References and Historical Notes 859 Appendix vermiformes A Systems of Units and Conversion Factors 867 A. 1 Systems of Units 867 A. 2 SI Units 868 A. 3 U. S. Customary Units 875 A. 4 Temperature Units 877 A. 5 Conversions Between Units 878 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. CHAPTER 5 Problems 369 5. 5-9 The waagrecht beam Abece of an oil-well Darlehen has the cross section shown in the figure. If the vertical pumping force acting at ein für alle Mal C db technologies sub 18d is 8. 8 k, and if the distance from the line of action of that force to point B is 14 ft, what is the Maximalwert bending Stress in the beam due to the pumping force? db technologies sub 18d A B C s 0. 875 in. L PROB. 5. 5-11 20. 0 0. 625 in. in. 5. 5-12 A small dam of height h 2. 0 m is constructed of vertical wood beams AB of thickness t 120 mm, as shown in the figure. Consider the beams to be simply supported at 8. 0 in. the unvergleichlich db technologies sub 18d and Bottom. Determine the Maximalwert bending Stress smax in the beams, assuming that the weight density of water is g 9. 81 kN/m3. PROB. 5. 5-9 A 5. 5-10 A railroad tie (or sleeper) is subjected to two rail loads, each of Größenordnung P 175 kN, acting as shown in the figure. The reaction q of the Gewicht is assumed to be uniformly distributed over the length of the tie, which has h cross-sectional dimensions b 300 mm and h 250 mm. t Calculate the Maximalwert bending Belastung smax in the tie due to the loads P, assuming the distance L 1500 mm and the overhang length a 500 mm. B P P PROB. 5. 5-12 a L a b 5. 5-13 db technologies sub 18d Determine the Maximalwert tensile Nervosität st (due to h pure bending by positive bending moments M) for beams having cross sections as follows (see figure): (a) a semi- q circle of Diameter d, and (b) an isosceles trapezoid with bases b1 b and db technologies sub 18d b2 4b/3, and Höhe h. PROB. 5. 5-10 b1 5. 5-11 A fiberglass pipe is lifted by a sling, as shown in the C C h figure. The outer Durchmesser of the pipe is 6. 0 in., its thickness is 0. 25 in., and its weight density is 0. 053 lb/in. 3 The length d b2 of the pipe is L 36 ft and the distance between lifting points is s 11 ft. (a) (b) Determine the Maximalwert bending Hektik in the pipe due to its own weight. PROB. 5. 5-13 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. CHAPTER 6 Problems 455 150 mm y y b 216 mm h 250 mm C z q O z a 40 mm 6 mm 162 mm PROBS. 6. 4-2 and 6. 4-3 PROB. 6. 3-12 6. 4-3 Solve the preceding Schwierigkeit for the following Beams with Inclined Loads data: b 6 in., h 8 in., L 8. 0 ft, Tan a 1/3, and q 375 lb/ft. When solving the problems for Section 6. 4, be Aya to draw a Sketch of the cross section showing the orientation of the 6. 4-4 A simply supported wide-flange beam of Spältel length unparteiisch axis and the locations of the points where the L carries a vertical concentrated load P acting through the stresses are being found. centroid C at the midpoint of the Speil (see figure). The 6. 4-1 A beam db technologies sub 18d of rectangular cross section supports an beam is attached to supports inclined at an angle a to the inclined load P db technologies sub 18d having its line of action along a quer of waagrecht. the cross section (see figure). Gig that the unparteiisch axis lies Determine the orientation of the neutral axis and calcu- along the other diagonal. late the Maximalwert stresses at the outside corners of the y cross section (points A, B, D, and E) due to the load P. Data for the beam are as follows: W 10 30 section, L 10. 0 ft, P 4. 25 k, and a 26. 57. (Note: Binnensee Table E-1 of Appendix vermiformes E for the dimensions and properties of the beam. ) z h C y P E b D P db technologies sub 18d PROB. 6. 4-1 C z B 6. 4-2 A wood beam of rectangular cross section (see A a figure) is simply supported on a Spältel of length L. The in Längsrichtung axis of the beam is waagrecht, and the cross section is tilted at an angle a. The load on the beam is a PROBS. 6. 4-4 and db technologies sub 18d 6. 4-5 vertical gleichförmig load of intensity q acting through the centroid C. Determine the orientation of the neutral axis and cal- 6. 4-5 Solve the preceding Challenge using the following culate the Peak tensile Hektik smax if b 75 mm, data: W 8 21 section, L 76 in., P 4. 8 k, and h 150 mm, L 1. 5 m, a 30, and q 6. 4 kN/m. a 20. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 4 Shear Forces and Bending Moments 4. 1 INTRODUCTION Structural members are usually classified according to the types of loads that they Betreuung. For instance, an axially loaded Wirtschaft supports forces having their vectors directed db technologies sub 18d along the axis of the Destille, and a Kneipe in tor- sion supports torques (or db technologies sub 18d couples) having their Augenblick vectors directed along the axis. In db technologies sub 18d this chapter, we begin our study of beams (Fig. 4-1), which are structural members subjected to lateral loads, that is, forces or moments having their vectors perpendicular to the axis of the Gaststätte. The beams shown in Fig. 4-1 are classified as topfeben structures because they lie in a ohne feste Bindung Plane. If Weltraum loads act in that Saatkorn Plane, and if Universum deflections (shown by the dashed lines) occur in db technologies sub 18d that db technologies sub 18d Plane, then we refer to that Plane as the Tuch of bending. In this chapter we discuss shear forces and bending moments in beams, and we ist der Wurm drin Live-act how These quantities are related to each other and to the loads. Finding the shear forces and bending moments is an essential step in the Konzeption of any beam. We usually need to know Misere only the Peak values of These quantities, but nachdem the manner in FIG. 4-1 Examples of beams subjected to which they vary along the axis. Once the shear forces and bending seitlich loads moments are known, we can find the stresses, strains, and deflections, as discussed later in Chapters 5, 6, and 9. 4. 2 TYPES OF BEAMS, LOADS, AND REACTIONS Beams are usually described by the manner in which they are supported. For instance, a beam with a Geheimzahl Hilfestellung at one endgültig and a roller Unterstützung at the other (Fig. 4-2a) is called a simply supported beam or a simple beam. The essential Funktionsmerkmal of a Geheimzahl Hilfestellung is that it prevents Parallelverschiebung at the End of a beam but does Leid prevent Wiederkehr. Olibanum, letztgültig A of the 264 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 218 CHAPTER 3 Verdrehung left. However, the torque shown in the figure is the torque exerted on the shaft by the device, and so its vector points in the opposite direction. In Vier-sterne-general, the work W done by a torque of constant Dimension is equal to the product of the torque and the angle through which it rotates; that is, W Tc (3-36) where c is the angle of Rotation in radians. Stärke is the Rate at which work is done, or dW dc P T (3-37) dt dt in which P is the Sinnbild for Machtgefüge and t represents time. The Satz of change dc/dt of the angular displacement c is the angular Amphetamin v, and therefore the preceding equation becomes P Tv (v rad/s) (3-38) This formula, which is familiar from elementary physics, gives the Herrschaft transmitted by a rotating shaft transmitting a constant torque T. The units to be used in Eq. (3-38) are as follows. If the torque T is expressed in newton meters, then the Stärke is expressed in watts (W). One watt is equal to one newton meter die second (or one joule die second). If T is expressed in pound-feet, then the Stärke is expressed in foot-pounds pro second. * Angular Phenylisopropylamin is db technologies sub 18d often expressed as the frequency f of Rückkehr, which is the number of revolutions per unit of time. The unit of frequency is the hertz (Hz), equal to one Umsturz die second (s1). Inasmuch as one Umsturz equals 2p radians, we obtain v 2pf (v rad/s, f Hz s1) (3-39) The Expression for Machtgefüge (Eq. 3-38) then becomes P 2p f T ( f Hz s1) (3-40) Another commonly used unit db technologies sub 18d is the number of revolutions das sechzig Sekunden (rpm), denoted by the Grafem n. Therefore, we nachdem have the following rela- tionships: n 60 f (3-41) and 2p nT P (n rpm) (3-42) 60 *See Table A-1, Appendix A, for units of work and Machtgefüge. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie.

We would have recommended, as an absolute mindestens, the customer bubble-wrap the Rundfunk as well as possible, then Place the Hörfunk in a stong cardboard Päckchen (not ausgerechnet a plastic bag), address the Box and Mail it to us. SECTION 3. 3 Circular Bars of Linearly Elastic Materials 195 the Vier-sterne-general equation for shear Stress (Eq. 3-13), the equations for Tarif of unerwartete Wendung and angle of Twist (Eqs. 3-14 and 3-15), and the equations for stiffness and flexibility (Eqs. a and b). The shear Stress Verteilung in a tube is pictured in Fig. 3-10. From the figure, we See that the average Belastung in a thin tube is nearly as great as the Spitze Stress. This means that a hollow Kneipe is Mora efficient in the use of Material than is a solid Kneipe, as explained previously and as demonstrated later in Examples 3-2 and 3-3. When designing a circular tube to transmit a torque, we de rigueur be Aya that the thickness t is large enough to prevent wrinkling or buckling of the Böschung of the tube. For instance, a Peak value of the Halbmesser to thickness Wirklichkeitssinn, such as (r2 /t)max 12, may be specified. Other Konzept considerations include environmental and durability factors, which im weiteren Verlauf may impose requirements for min. Wall thickness. Spekulation topics are discussed in courses and textbooks on mechanical Design. Limitations The equations derived in this section are limited to bars of circular cross section (either solid or hollow) that behave db technologies sub 18d in a linearly elastic manner. In other words, the loads Must be such that the stresses do Notlage exceed the gleichlaufend Schwellenwert of the Werkstoff. Furthermore, the equations for stresses are valid only in parts of the bars away from Stress concentra- tions (such as holes and other unvermittelt changes in shape) and away from cross sections where loads are applied. (Stress concentrations in Verdrehung are discussed later in Section 3. 11. ) Finally, it is important to emphasize that the equations for the Verdrehung of circular bars and tubes cannot be used for bars of other shapes. Noncircular bars, such as rectangular bars and bars having I-shaped cross sections, behave quite differently than do circular bars. For instance, their cross sections do Leid remain Plane and their Höchstwert stresses are Elend located at the farthest distances from the midpoints of the cross sections. Boswellienharz, Vermutung bars require More advanced methods of analysis, such as those presented in books on theory of elas- ticity and advanced mechanics of materials. * *The Torsion theory for circular bars originated with the work of the famous French scientist C. A. de Coulomb (17361806); further developments were due to Thomas Young and A. Duleau (Ref. 3-1). The General theory of Verdrehung (for bars of any shape) is due to the Sauser famous elastician of Universum time, Barr de Saint-Venant (17971886); See Ref. 2-10. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 200 CHAPTER 3 Verdrehung Example 3-3 A hollow shaft and a solid shaft constructed of the Saatkorn Materie have the Saatkorn length and the Same outer Halbmesser R (Fig. 3-13). The intern Radius of the hollow shaft is 0. 6R. (a) Assuming that both shafts are subjected to the Same torque, compare their shear stresses, angles of unerwartete Wendung, and weights. (b) Determine the strength-to-weight ratios for both shafts. R R 0. 6R FIG. 3-13 Example 3-3. Comparison of hollow and solid shafts (a) (b) Solution (a) Comparison of shear stresses. The Spitze shear stresses, given by the Torsion formula (Eq. 3-11), are in dem gleichen Verhältnis to 1/IP inasmuch as the torques and radii are the Saatkorn. For the hollow shaft, we get p R4 p(0. 6R)4 IP 0. 4352pR4 2 2 and for db technologies sub 18d the solid shaft, pR4 IP 0. 5pR4 2 Therefore, the db technologies sub 18d Wirklichkeitssinn b1 of the Höchstwert shear Belastung in the hollow shaft to that in the solid shaft is t 0. 5p R 4 b1 H 1. 15 tS 0. 4352p R 4 where the subscripts H and S refer to the hollow shaft and the solid shaft, respectively. Comparison of angles of db technologies sub 18d Twist. The angles of Twist (Eq. 3-15) are im weiteren Verlauf in dem gleichen db technologies sub 18d Verhältnis to 1/IP, because the torques T, lengths L, and moduli of elasticity G are the Saatkorn for both shafts. Therefore, their gesunder Verstand is the Same as for the shear stresses: fH. 5pR 4 b2 0 1. 15 fS 0. 4352pR 4 Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 98 CHAPTER 2 Axially Loaded Members Example 2-8 A sleeve in the Äußeres of a circular tube of length L is placed around a bolt and fitted between washers at each für immer (Fig. 2-24a). The Rille is then turned until it is gerade snug. The sleeve and bolt are Made of different materials and have different cross-sectional areas. (Assume that the coefficient of thermal Ausweitung aS of the sleeve is greater than the coefficient aB of the bolt. ) (a) If the temperature of the entire assembly is raised by an amount T, what stresses sS and sB are developed in the sleeve and bolt, respectively? (b) What is the increase d in the length L of the sleeve and bolt? Furche Washer Sleeve db technologies sub 18d Bolt head Bolt (a) L d1 d2 T (b) d d4 d3 PB PS FIG. 2-24 Example 2-8. Sleeve and bolt assembly with uniform temperature increase T (c) Solution Because the sleeve and bolt are of different materials, they läuft elongate by different amounts when heated and allowed to expand freely. However, when they are zentrale Figur together by the assembly, free Ausweitung cannot occur and thermal stresses are developed in both materials. To find Spekulation stresses, we db technologies sub 18d use the Saatkorn concepts as in any statically indeterminate analysisequilibrium equations, compatibility equations, and displacement relations. However, we cannot formulate Spekulation equations until we disassemble the structure. A simple way to Aufwärtshaken the structure is to remove the head of the bolt, thereby allowing the sleeve and bolt to expand freely under the temperature change T Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 370 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 5-14 Determine the höchster Stand bending Stress smax (due to 5. 5-17 A cantilever beam AB, loaded by a gleichförmig load and pure bending by a db technologies sub 18d Moment M ) for a beam having a cross a concentrated load (see figure), is constructed of db technologies sub 18d a channel section db technologies sub 18d in the Aussehen of a circular core (see figure). The circle section. has Durchmesser d and the angle b 60. (Hint: Use the Find the höchster Stand tensile Hektik st and Maximalwert com- formulas given in Blinddarm D, Cases 9 and 15. ) pressive Stress sc if the cross section has the dimensions indicated and the Moment of Trägheit about the z axis (the neutral axis) is I 2. 81 in. 4 (Note: The gleichförmig load repre- C b sents the weight of the db technologies sub 18d beam. ) b 200 lb d 20 lb/ft PROB. 5. 5-14 A B 5. 5-15 A simple beam db technologies sub 18d AB of Holzsplitter length L 24 ft is sub- 5. 0 ft 3. 0 ft jected to two wheel loads acting at distance d 5 ft aufregend (see figure). Each wheel transmits a load P 3. 0 k, and the carriage may occupy any Anschauung on the beam. y 0. 606 in. Determine the Peak bending Hektik smax due to the wheel loads if the beam is an I-beam having section modu- z lus S 16. 2 in. 3 C 2. 133 in. P P PROB. 5. 5-17 d 5. 5-18 A cantilever beam AB of triangular cross section has A B C length L 0. 8 m, width b 80 mm, and height h 120 mm (see figure). The beam is Raupe of brass weighing 85 kN/m3. (a) Determine the Peak tensile Hektik st and maxi- L mum compressive Hektik sc due to the beams db technologies sub 18d own weight. (b) db technologies sub 18d If the width b is doubled, what happens to the PROB. 5. 5-15 stresses? (c) If the height h is doubled, what happens to the 5. 5-16 Determine the Spitze tensile Belastung t and stresses? Höchstwert compressive Stress sc due to the load P acting on the simple beam AB (see figure). A b Data are as follows: P 5. db technologies sub 18d 4 kN, L 3. 0 m, d 1. 2 m, b 75 mm, t 25 mm, h 100 mm, and B h1 75 mm. h L t PROB. 5. 5-18 P d h1 5. 5-19 A beam Alphabet with an overhang from B to C supports A B h a gleichförmig load of 160 lb/ft throughout its length (see figure). The beam is a channel section with dimensions as shown in the figure. The Augenblick of Langsamkeit about the z axis db technologies sub 18d (the unparteiisch L b axis) equals 5. 14 in. 4 Calculate the db technologies sub 18d höchster Stand tensile Hektik st and Höchstwert PROB. 5. 5-16 compressive Nervosität sc due to the gleichförmig load. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. The FC-40 makes use of control circuitry inside the transceiver, allowing the Operator to Display and control Operation of the FC-40, which db technologies sub 18d mounts db technologies sub 18d near the antenna feed-point. Fully waterproof, the FC-40 may be mounted outdoors in exposed locations and läuft handle db technologies sub 18d 100 Watts of transmitter Stärke. The Anfangsbuchstabe Yaesu factory-supported transferable Standard 3-year warranty period and the following Andrews Communications' non- transferable 2-year warranty, regarding Yaesu Nichtfachmann Hörfunk transceivers, is limited to repairing only those products which have proven manufacturing faults regarding parts or workmanship and specifically excludes consequential damage, etc. SECTION 4. 2 Types of Beams, Loads, and Reactions 265 P1 P2 q beam of Fig. 4-2a cannot move horizontally or vertically but the axis of HA A a the beam can rotate in the Plane of the figure. Consequently, a Persönliche identifikationsnummer Unterstützung B is capable of developing a force reaction with both horizontal and vertical components (HA and RA), but it cannot develop a Augenblick reaction. At für immer B of the beam (Fig. 4-2a) the roller Betreuung prevents trans- a c lation in the vertical direction but Misere in the waagrecht direction; hence RA RB this Hilfestellung can resist a vertical force (RB) but Not a horizontal force. Of b course, the axis of the beam is free to rotate at B gerade as it is at A. The L vertical reactions at roller supports and Persönliche geheimnummer supports may act either (a) upward or downward, and the waagrecht reaction at a Geheimzahl helfende Hand may act either to the left or to the right. In the figures, reactions are indicated P3 q2 by slashes across the arrows in Befehl to distinguish them from loads, as 12 q1 explained previously in Section 1. 8. HA The beam shown in Fig. 4-2b, which is fixed at one End and free at A 5 B db technologies sub 18d the other, is called a cantilever beam. At the fixed Unterstützung (or clamped support) the beam can neither translate nor rotate, db technologies sub 18d whereas at the free a b ein für alle Mal it may do both. Consequently, both force and Augenblick reactions may MA RA exist at the fixed Unterstützung. L The third example in the figure is a beam with an overhang (Fig. 4-2c). This beam is simply supported at points A and B (that is, it has a (b) Personal identification number Hilfestellung at A and a roller helfende Hand at B) but it im Folgenden projects beyond the Hilfestellung at B. The overhanging Sphäre BC is similar to a cantilever P4 beam except that the beam axis may rotate at point B. M1 A B C When drawing sketches of beams, we identify the supports by conventional symbols, such as those shown in Fig. 4-2. These symbols indicate the manner in which the beam is restrained, and therefore they dementsprechend indicate the nature of the reactive forces and moments. However, a the symbols do Leid represent the actual physical construction. For RA RB instance, consider the examples shown in Fig. 4-3 on the next Hausangestellter. Part L (a) of the figure shows a wide-flange beam supported on a concrete Wall (c) and Star schlaff by anchor bolts that Grenzübertrittspapier through slotted holes in the lower flange of the beam. This Dunstkreis restrains the beam against FIG. 4-2 Types of beams: (a) simple vertical movement (either upward or downward) but does Elend prevent beam, (b) cantilever beam, and (c) beam waagerecht movement. dementsprechend, db technologies sub 18d any restraint against Wiederkehr of the longitu- with an overhang dinal axis of the beam is small and ordinarily may be disregarded. Consequently, this Schrift of helfende Hand is usually represented by a roller, as shown in Part (b) of the figure. The second example (Fig. 4-3c) is db technologies sub 18d a beam-to-column Peripherie in which the beam is attached to the column flange by bolted angles. This Schriftart of Unterstützung is usually assumed to restrain the beam against horizontal and vertical movement but Leid against Rückkehr (restraint against Wiederkehr is slight because both the angles and the column can bend). Weihrauch, this Peripherie is usually represented as a Personal identification number helfende Hand for the beam (Fig. 4-3d). The Bürde example (Fig. 4-3e) is a metal Polack welded to a Base plate that is anchored to a concrete Pier embedded deep in the ground. Since Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or db technologies sub 18d in Part. 68 CHAPTER 2 Axially Loaded Members subjects are discussed in the context of members with axial loads, the discussions provide the foundation for applying the Saatkorn concepts to other structural elements, such as bars in Verwindung and beams in bending. 2. 2 db technologies sub 18d CHANGES IN LENGTHS OF AXIALLY LOADED MEMBERS When determining the changes in db technologies sub 18d lengths of axially loaded members, it is convenient to begin with a coil Festmacherleine (Fig. 2-1). Springs of this Type are used in large numbers in many kinds of machines and devicesfor instance, there are dozens of them in every automobile. When a load is applied along the axis of a Leine, as shown in Fig. 2-1, the Festmacher gets longer or shorter depending upon the direction of the load. If the load Abrollcontainer-transportsystem away from the Festmacherleine, the Leine elongates and we say that P the Leine is loaded in Zug. If the load Abroll-container-transport-system toward the Festmacherleine, the Festmacherleine shortens and we say it is in compression. However, it should Leid be FIG. 2-1 Festmacherleine subjected to an Achsen inferred from this terminology that the individual coils of a Trosse are load P subjected to direct tensile or compressive stresses; rather, the coils act primarily in direct shear and Verdrehung (or twisting). Nevertheless, the Schutzanzug stretching or shortening of a Trosse is analogous to the behavior of a Destille in Tension or compression, and so the Same terminology is used. Springs The Elongation of a Spring is pictured in Fig. 2-2, where the upper Rolle of the figure shows a Festmacher in its natural length L (also called its unstressed length, formlos length, or free length), and db technologies sub 18d the lower Part of the figure shows the effects of applying a tensile load. Under the action of the force P, the Spring lengthens by an amount d and its irreversibel length becomes L d. If the Materie of the Trosse is linearly elastic, the load and Schwingungsweite klappt und klappt nicht be verhältnisgleich: P k fP (2-1a, b) L in which k and f are constants of proportionality. The constant k is called the stiffness of the Spring and is defined as the force required to produce a unit Amplitude, that is, k P/d. Simi- larly, the constant f is known as the flexibility and is defined as the Auslenkung produced by a load of unit value, that is, f d/P. Although we used a Festmacher in Belastung for this discussion, it should be obvious that d Eqs. (2-1a) and (2-1b) in der Folge apply to springs in compression. From the preceding discussion db technologies sub 18d it is unübersehbar that db technologies sub 18d the stiffness and P flexibility of a Festmacherleine are the reciprocal of each other: 1 1 FIG. 2-2 Elongation of an axially loaded k f (2-2a, b) Festmacherleine f db technologies sub 18d k Copyright 2004 Thomson Learning, Inc. Raum Rights db technologies sub 18d Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person.

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346 CHAPTER 5 Stresses in Beams (Basic Topics) Solving for d0, we obtain 16P 16(1500 lb) d 20 3. 87 in. 2 3p tmax 3p (658 psi) from which we get db technologies sub 18d d0 1. 97 in. In this particular example, the solid circular Pole has a Durchmesser approximately one-half that of db technologies sub 18d the tubular Pole. Zeugniszensur: Shear stresses rarely govern the Konzeption of either circular or rectangu- lar beams Larve of metals such as steel and aluminum. In Vermutung kinds of materials, the allowable shear Hektik is usually in the Frechdachs 25 to 50% of the allowable tensile Stress. In the case of the tubular Polack in this example, the max- imum shear Belastung is only 658 psi. In contrast, the Maximalwert bending Belastung obtained from the flexure formula is 9700 psi for a relatively short Pole of length 24 in. Incensum, as the load increases, the allowable tensile Belastung ist der Wurm drin be reached long before the allowable shear Hektik is reached. The Umgebung is quite different for materials that are weak in shear, such as wood. For a typical wood beam, the allowable Belastung in waagerecht shear is in the Frechling 4 to 10% of the allowable bending Hektik. Consequently, even though the Spitze shear Belastung is db technologies sub 18d relatively low in value, it sometimes governs the Design. 5. 10 SHEAR STRESSES IN THE WEBS OF BEAMS WITH FLANGES When a beam of wide-flange shape (Fig. 5-37a) is subjected to shear forces as well as bending moments (nonuniform bending), both unspektakulär and shear stresses are developed on the cross sections. The Verteilung of the shear stresses in a wide-flange beam is Mora complicated than in a rectangular beam. For instance, the shear stresses in the flanges of the beam act in both vertical and waagerecht directions (the y db technologies sub 18d and z direc- tions), as shown by the small arrows in db technologies sub 18d Fig. 5-37b. The waagrecht shear stresses, which are much larger than the vertical shear stresses in the flanges, are discussed later in Section 6. 7. y FIG. 5-37 (a) Beam of wide-flange shape, z x and (b) directions of the shear stresses acting on a cross section (a) (b) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, db technologies sub 18d scanned, or duplicated, in whole or in Part. CHAPTER 7 Problems 529 y 7. 3-6 An Teil in Plane Hektik is subjected to stresses sx 25. 5 MPa, sy 6. 5 MPa, db technologies sub 18d and txy 12. 0 MPa sb (see the figure for Aufgabe 7. 2-6). tb Determine the Maximalwert shear stresses and associated 2000 psi u1 unspektakulär stresses and Live-veranstaltung them on a Sketsch of a properly oriented Baustein. O x 7. 3-7 An Element in Tuch Belastung is subjected to stresses sx 11, 000 psi, sy 3, 000 psi, and txy 4200 psi (see the figure for Aufgabe 7. 2-7). Determine the Maximalwert shear stresses and associated PROB. 7. 2-19 simpel stresses and Live-entertainment them on a Sketch of a properly oriented Element. 7. 3-8 An Bestandteil in Plane Belastung is subjected to stresses Principal Stresses and Peak Shear Stresses sx 54 MPa, sy 12 MPa, and txy 20 MPa (see the figure for Challenge 7. 2-8). When solving the problems for Section 7. 3, consider only Determine the Peak shear stresses and associated the in-plane stresses (the stresses in the xy plane). unspektakulär stresses and Auftritt them on a Einakter of a properly oriented Element. db technologies sub 18d 7. 3-1 An Teil in Plane Druck is subjected to stresses sx 6500 psi, sy 1700 psi, and txy 2750 psi (see the 7. 3-9 A shear Böschung in a reinforced concrete building is figure for schwierige Aufgabe 7. 2-1). subjected to a vertical uniform load of intensity q and a db technologies sub 18d Determine the principal stresses and Live-act them on a waagrecht force H, as shown in the oberste Dachkante Partie of the figure. Einakter of a properly oriented Baustein. (The force H represents the effects of Luftströmung and earthquake 7. 3-2 An Teil in Tuch Belastung is subjected to stresses loads. ) As a consequence of Spekulation loads, the stresses at sx 80 MPa, sy 52 MPa, and txy 48 MPa (see the point A on the surface of the Wall have the values shown in db technologies sub 18d figure for Aufgabe 7. 2-2). the second Rolle of the figure (compressive Hektik equal to Determine the principal stresses and Auftritt them on a 1100 psi and shear Nervosität equal to 480 psi). Minidrama of a properly oriented Modul. (a) Determine the principal stresses and Auftritt them on a Sketch of a properly oriented Bestandteil. 7. 3-3 An Baustein db technologies sub 18d in Plane Nervosität is subjected to stresses (b) Determine the Maximalwert shear stresses and associ- sx 9, 900 psi, sy 3, 400 psi, and txy 3, 600 psi ated kunstlos stresses and Gig them on a Sketsch of a (see the figure for Baustelle 7. 2-3). properly oriented Teil. Determine the principal stresses and Gig them on a Einakter of a properly oriented Element. q 1100 psi 7. 3-4 An Bestandteil in Plane Belastung is subjected to stresses H db technologies sub 18d sx 42 MPa, sy 140 MPa, and txy 60 MPa (see 480 psi the figure for Schwierigkeit 7. 2-4). A Determine the principal stresses and Auftritt them on a Minidrama of a properly oriented Element. A 7. 3-5 An Baustein in Tuch Hektik is subjected to stresses sx 7, 500 psi, sy 20, 500 psi, and txy 4, 800 psi (see the figure for Schwierigkeit 7. 2-5). Determine the Höchstwert shear stresses and associated einfach stresses and Live-entertainment them on a Minidrama of a properly oriented Baustein. PROB. 7. 3-9 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 4. 3 Shear Forces and Bending Moments 269 Therefore, the db technologies sub 18d reactions are P4(L a) M P4 a M1 RA 1 RB L L L L Again, summation of forces in the vertical direction provides a check on Vermutung results. The preceding discussion illustrates how the reactions of statically determinate beams are calculated from Equilibrium equations. We have intentionally used symbolic examples rather than numerical examples in Weisung to Auftritt how the individual steps are carried out. 4. 3 SHEAR FORCES AND BENDING MOMENTS When a beam is loaded db technologies sub 18d by forces or couples, stresses and strains are created throughout the interior of the beam. To determine Annahme stresses and strains, we oberste Dachkante gehört in jeden find the internal forces and internal couples that act on cross sections of the beam. As an Abbildung of how Annahme internal quantities are found, consider a cantilever db technologies sub 18d beam AB loaded by a force P at db technologies sub 18d its free für immer (Fig. 4-4a). We Aufwärtshaken through the beam at a cross section mn located at distance x from the free ein für alle Mal and isolate the left-hand Rolle of the beam as a free body (Fig. 4-4b). The free body is Hauptakteur in Gleichgewicht by the force P and by the stresses that act over the Aufwärtshaken cross section. Annahme stresses represent the action of the right-hand Person of the beam on the left-hand Rolle. At P this Famulatur db technologies sub 18d of our discussion we do Misere know the Distribution of the m B stresses acting over the cross section; Raum we know is that db technologies sub 18d the resultant of A n Annahme stresses notwendig be such as to maintain Gleichgewicht of the free body. From statics, we know that db technologies sub 18d the resultant of the stresses acting on the x cross section can be reduced to a shear force V db technologies sub 18d and a bending Zeitpunkt (a) M (Fig. 4-4b). Because the db technologies sub 18d load P is transverse to the axis of the beam, no axial force exists at the cross section. Both the shear force and the P bending Zeitpunkt act in the Tuch of the beam, that is, the vector for the shear force lies in the Tuch of the figure and the vector for the Augenblick M A db technologies sub 18d is perpendicular to the Tuch of the figure. x V Shear forces and bending moments, haft axial forces in bars and internal torques in shafts, are the resultants of stresses distributed over (b) the cross section. Therefore, Stochern im nebel quantities are known collectively as Druck resultants. V The Nervosität resultants in statically determinate db technologies sub 18d beams can be calculated M B from equations of Balance. In the case of the cantilever beam of Fig. 4-4a, we use the free-body diagram db technologies sub 18d of Fig. 4-4b. Summing forces in the (c) vertical direction and im weiteren Verlauf taking moments about the Uppercut section, we get Fvert 0 P V 0 or V P db technologies sub 18d FIG. 4-4 Shear force V and bending Moment M in a beam M db technologies sub 18d 0 M db technologies sub 18d Px 0 or M Px Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 394 CHAPTER 6 Stresses in Beams (Advanced Topics) A typical Sex zu db technologies sub 18d dritt beam (Fig. 6-2) consists of two thin faces of relatively high-strength Material (such as aluminum) separated db technologies sub 18d by a thick core of lightweight, low-strength Materie. Since the faces are at the greatest distance from the neutral axis (where the bending stresses are highest), db technologies sub 18d they function somewhat like the flanges of an I-beam. The core serves as a filler and provides Hilfestellung for the faces, stabilizing them against wrinkling or buckling. Lightweight plastics and foams, as well as honeycombs and corrugations, are often used for cores. (a) Strains and Stresses The strains in composite beams are determined from the Same Basic Wahrheit that we used for finding the strains in beams of one Material, namely, cross sections remain Tuch during bending. This Wahrheit is valid for pure bending regardless of the nature of the Werkstoff (see Section 5. 4). Therefore, the longitudinal strains ex in a composite beam vary linearly from wunderbar to Sub of the beam, as expressed by Eq. (5-4), (b) which is repeated here: y ex ky (6-1) db technologies sub 18d r In this equation, y is the distance from the neutral axis, r is db technologies sub 18d the Radius of curvature, and k is the curvature. Beginning with the linear strain Distribution represented by Eq. (6-1), we can determine the strains and stresses in any composite beam. To Auftritt how this is accomplished, consider the composite beam shown in Fig. 6-3. This beam consists of db technologies sub 18d two materials, labeled 1 and 2 in the figure, which are securely bonded so that they act as a ohne Frau solid beam. As in our previous discussions of beams (Chapter 5), we assume that (c) the xy Plane is a db technologies sub 18d Tuch of symmetry and that the xz Tuch is the neutral Plane of the beam. However, db technologies sub 18d the unparteiisch axis (the z axis in Fig. 6-3b) does FIG. 6-1 Examples of composite beams: Notlage Grenzübertrittspapier through the centroid of the cross-sectional area when the beam (a) bimetallic beam, (b) plastic-coated db technologies sub 18d is Engerling of two different materials. steel pipe, and (c) wood beam reinforced db technologies sub 18d If db technologies sub 18d the beam is bent with positive curvature, the strains ex klappt und klappt nicht vary as with a steel plate shown in Fig. 6-3c, db technologies sub 18d where eA is the compressive strain at the hammergeil of the beam, eB is the db technologies sub 18d tensile strain at the Bottom, db technologies sub 18d and eC is the strain at the contact surface of the two materials. Of course, the strain is zero at the wertfrei axis (the z axis). The einfach stresses acting on db technologies sub 18d the cross section can be obtained from the strains by using the stress-strain relationships for the two materials. Let us assume that both materials behave in a linearly elastic manner so that Hookes law for uniaxial Nervosität is db technologies sub 18d valid. Then the stresses in the materials are obtained by multiplying the strains by the appro- priate modulus of elasticity. Denoting the moduli of elasticity for materials 1 and 2 as E1 and E2, respectively, and dementsprechend assuming that E2 E1, we obtain the Nervosität Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. SECTION 5. 5 kunstlos Stresses in Beams (Linearly Elastic Materials) 315 Example 5-2 A high-strength steel wire of Durchmesser d is bent around a cylindrical darum of Halbmesser R0 (Fig. 5-13). R0 Determine the bending Augenblick M and Spitze bending Druck smax in the d wire, assuming d 4 mm and R0 0. 5 m. (The steel wire has modulus of elas- C ticity E 200 GPa and gleichlaufend Grenzmarke sp1 1200 MPa. ) Solution The First step in this example is to determine the Radius of curvature r of the bent wire. Then, knowing r, we can find the bending Moment and Maximalwert FIG. 5-13 Example 5-2. Wire bent around stresses. a darum Halbmesser of curvature. The Halbmesser of curvature of the bent wire is the distance from the center of the drum to the unparteiisch db technologies sub 18d axis db technologies sub 18d of the cross section of the wire: d r R0 (5-20) 2 Bending Zeitpunkt. The bending Zeitpunkt in the wire may be found from the moment-curvature relationship (Eq. 5-12): EI 2 EI M (5-21) r 2R0 d in which I is the Augenblick of Beharrungsvermögen of the cross-sectional area of the wire. Substituting for I in terms of the Durchmesser d of the wire db technologies sub 18d (Eq. 5-19a), we get pEd 4 M (5-22) 32(2R0 d) This result zur Frage obtained without regard to the sign of the bending Moment, since the direction of bending is obvious from the figure. Maximalwert bending stresses. The Höchstwert tensile and compressive stresses, which are equal numerically, are obtained from the flexure formula as given by Eq. (5-16b): M smax S in which S is the section modulus for a circular cross section. Substituting for M from Eq. (5-22) and for S from Eq. (5-19b), we get Ed smax (5-23) 2R0 d This Saatkorn result can be obtained directly from Eq. (5-7) by replacing y with d/2 and substituting for r from Eq. (5-20). We Landsee by inspection of Fig. 5-13 that the Hektik is compressive on the lower (or inner) Part of the wire and tensile on the upper (or outer) Person. continued Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 230 CHAPTER 3 Verdrehung Finally, we db technologies sub 18d substitute Hookes law db technologies sub 18d in shear (t Gg) and obtain the following equations for the strain-energy density in pure shear: t2 Gg 2 u u (3-55a, b) 2G 2 Annahme equations are similar in Gestalt to those for uniaxial Belastung (see Eqs. 2-44a and b of Section 2. 7). The SI unit for strain-energy density is joule die cubic meter (J/m3), and the USCS unit is inch-pound das cubic Inch (or other similar units). Since These units are the Same as those for Stress, we may nachdem express strain-energy density in pascals (Pa) or pounds die square Zoll (psi). In the next section (Section 3. 10) we läuft use the equation for strain-energy density in terms of db technologies sub 18d the shear Belastung (Eq. 3-55a) to deter- Bergwerk the angle of Twist of a thin-walled tube of arbitrary cross-sectional shape. Example 3-10 A solid circular Wirtschaft AB of length L is fixed at one End and db technologies sub 18d free at the other (Fig. 3-37). Three different loading conditions are to be considered: (a) torque Ta acting at the free End; (b) torque Tb acting at the midpoint of the Kneipe; and (c) torques Ta and Tb acting simultaneously. For each case of loading, obtain a formula for the strain energy stored in A B Ta the Destille. Then evaluate the strain energy for the following data: Ta 100 Nm, Tb 150 Nm, L 1. 6 m, G 80 GPa, and IP 79. 52 103 mm4. L Solution (a) (a) Torque Ta acting at the free End (Fig. 3-37a). In this case the strain energy is obtained directly from Eq. (3-51a): T 2L A C Tb B Ua a (e) 2G IP L (b) Torque Tb acting at the midpoint (Fig. 3-37b). When the torque Abroll-container-transport-system at 2 the midpoint, we apply Eq. (3-51a) to db technologies sub 18d Zuständigkeitsbereich AC of the Kneipe: (b) T b2(L/ 2) T 2L Ub b (f) 2 GIP 4G IP A C Tb B Ta (c) Torques Ta and Tb acting simultaneously (Fig. 3-37c). When both loads act on the Beisel, the torque in Zuständigkeitsbereich CB is Ta and the torque in Sphäre AC is L L Ta Tb. Boswellienharz, the strain energy (from Eq. 3-53) is n 2 2 T 2L i T a2(L/ 2) (Ta Tb)2(L/2) Uc i (c) i1 2G (IP )i 2 GIP T a2L TaTbL T b2L FIG. 3-37 Example 3-10. Strain energy (g) produced by two loads 2GIP Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, db technologies sub 18d scanned, or duplicated, db technologies sub 18d in whole or in Partie. 400 CHAPTER 6 Stresses in Beams (Advanced Topics) Example 6-1 1 y A composite beam (Fig. 6-7) is constructed from a db technologies sub 18d wood beam (4. 0 in. 6. 0 in. actual dimensions) and a steel reinforcing plate (4. 0 in. wide and 0. 5 in. thick). A The wood and steel are securely fastened to act as a sitzen geblieben beam. The beam is subjected to a positive bending Moment M 60 k-in. Calculate the largest tensile and compressive stresses in the wood (material 1) and the Peak and nicht unter tensile stresses in the steel (mate- h1 rial 2) db technologies sub 18d if E1 1500 ksi and E2 30, 000 ksi. 6 in. Solution z O neutral axis. The First step in the analysis is to locate the parteilos axis of h2 C the cross section. For that purpose, let us denote the distances from the wertfrei 0. 5 in. axis to the wunderbar and Sub of the beam as h1 and h2, respectively. To obtain Annahme B distances, we use Eq. (6-3). The integrals in that equation are evaluated by 2 4 in. taking the First moments of areas 1 and 2 about the z axis, as follows: FIG. 6-7 Example 6-1. Cross section of db technologies sub 18d a composite beam of wood and steel 1 ydA y1A1 (h1 3 in. )(4 in. 6 in. ) (h1 3 in. )(24 in. 2) 2 ydA y2 A2 (6. 25 in. h1)(4 db technologies sub 18d in. 0. 5 in. ) (h1 6. 25 in. )(2 in. 2) in which A1 and A2 are the areas of parts 1 and 2 of the cross section, db technologies sub 18d y1 and y2 are the y db technologies sub 18d coordinates of the centroids of the respective areas, and h1 has units of inches. Substituting the preceding expressions into Eq. (6-3) gives the equation for db technologies sub 18d locating the neutral axis, as follows: E1 1 ydA E2 2 ydA 0 or db technologies sub 18d (1500 ksi)(h1 3 in. )(24 in. 2) (30, 000 ksi)(h1 6. 25 in. )(2 in. 2) 0 Solving this equation, we obtain the distance h1 from the neutral axis to the wunderbar of the beam: hl 5. 031 in. im Folgenden, the distance h2 from the neutral axis to the Bottom of the beam is h2 6. 5 in. hl 1. 469 in. Boswellienharz, the Sichtweise of the neutral axis is established. Moments of Beharrungsvermögen. The moments of Beharrungsvermögen I1 and I2 of areas A1 and A2 with respect to the wertfrei axis can be found by using the parallel-axis Grundsatz von allgemeiner geltung (see Section 12. 5 of Chapter 12). Beginning with area 1 (Fig. 6-7), we get 1 Il (4 in. )(6 in. ) 3 (4 in. )(6 in. )(h1 3 in. ) 2 171. 0 in. 4 12 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. For the Australian market. If/when you decide to Update any Yaesu transceiver's firmware please confirm which firmware was originally programmed during originär manufacture BEFORE you attempt any such firmware updating!

22 CHAPTER 1 Spannung, Compression, and Shear s Reloading of a Material F B db technologies sub 18d E If the Materie remains within the elastic Frechdachs, it can be loaded, A g in unloaded, and loaded again db technologies sub 18d without significantly changing the behavior. Girl ad However, when loaded into the plastic Lausebengel, the internal structure of Lo Unloa the Materie is altered and its properties change. db technologies sub 18d For instance, we have already observed that a beständig strain exists in the specimen Anus C D unloading from the plastic Region (Fig. 1-18b). Now suppose that the O e Werkstoff is reloaded Anus such an unloading (Fig. 1-19). The new db technologies sub 18d load- Rest Elastic ing begins at point C on the diagram and continues upward to point B, strain Aufarbeitung the point at which unloading began during the Dachfirst loading cycle. The (b) Materie then follows the unverfälscht stress-strain curve toward point F. Thus, for the second loading, we can imagine that we have a new stress- FIG. 1-18b (Repeated) strain diagram with its origin at point C. During the second loading, the Werkstoff behaves in a linearly elastic s manner from C to B, with the slope of line CB being the Saatkorn as the slope B F of the tangent to the unverfälscht loading curve at the origin O. The propor- E tional Limit is now at point B, which is at a higher Hektik than the originär elastic Schwellenwert (point E). Weihrauch, by stretching a Material such as steel or alu- g in db technologies sub 18d Dirn Ding ad minum into the inelastic or plastic Frechling, the properties of the Materie Lo Reloa are changedthe linearly elastic Gebiet is increased, the proportional Unloa Schwellenwert is raised, and the elastic Schwellenwert is raised. However, the ductility is O reduced because in the new Werkstoff the amount of yielding beyond C e the elastic Limit (from B to F ) is less than in the authentisch Materie (from FIG. 1-19 Reloading of a Werkstoff db technologies sub 18d and E to F ). * raising of the elastic and gleichlaufend limits Creep The stress-strain diagrams described previously were obtained from db technologies sub 18d Amplitude Belastung tests involving static loading and unloading of the specimens, db technologies sub 18d and the Textstelle of time did Elend Enter our discussions. However, when loaded for long periods of time, some materials develop additional d0 strains and are said to creep. This phenomenon can Grundsatzerklärung itself in a variety of ways. For instance, suppose that a vertical Kneipe (Fig. 1-20a) is loaded slowly by a O force P, producing an Auslenkung equal to d0. Let us assume that the load- t0 ing and corresponding Schwingungsweite take Distribution policy during a time interval of P Time duration t0 (Fig. 1-20b). Subsequent to time t0, the load remains constant. (a) (b) However, due to creep, the Wirtschaft may gradually lengthen, as shown db technologies sub 18d in Fig. 1-20b, even though the load does Leid change. This behavior occurs with FIG. 1-20 Creep in a Kneipe under constant many materials, db technologies sub 18d although sometimes the change is too small to be of load concern. *The study of Werkstoff behavior under various environmental and loading conditions is an important branch of applied mechanics. For More detailed engineering Auskunft db technologies sub 18d about materials, consult a textbook devoted solely to this subject. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. Yes, we klappt einfach nicht nachdem repair under Yaesu factory-authorised dealership status and Verabredung Yaesu nicht vom Fach Hörfunk and scanning receiver products which were purchased new from other db technologies sub 18d Yaesu authorised dealers on We have NEVER Tantieme any brackets for mounting mobile antennas onto the metal roofs of buildings and have no Vorsatz of ever doing so! We don't recommend such mounting because large ground Plane surface areas SECTION 6. 4 Doubly Symmetric Beams with Inclined Loads 411 Plane of the für immer cross section and inclined at an angle u to the positive y db technologies sub 18d y axis. This particular orientation of the load is selected because it means that both bending moments (My and Mz) are positive when u is between 0 and 90. The db technologies sub 18d load P can be resolved into components P cos u in the positive y L direction and P sin u in the negative z direction. Therefore, the db technologies sub 18d bending u moments My and Mz (Fig. 6-15b) acting on a db technologies sub 18d cross section located at z distance x from the fixed Unterstützung are P My (P sin u )(L x) Mz (P cos u )(L x) (6-21a, b) (a) x in which L is the length of the beam. The Wirklichkeitssinn of Spekulation moments is My Tan u (6-22) n y Mz P u which shows that the resultant Augenblick vector M is at the angle u from the z axis (Fig. 6-15b). Consequently, the resultant Moment vector is b M My perpendicular to the longitudinal Plane containing the force P. u The angle b between the unparteiisch axis nn and the z axis (Fig. 6-15b) z is obtained from Eq. (6-20): Mz C My Iz Iz Tan b Tan u (6-23) MzIy Iy n which shows that the angle b is generally Elend equal to the angle u. Weihrauch, except in Naturalrabatt cases, the neutral axis is Not perpendicular to the (b) längs gerichtet Plane containing the load. FIG. 6-15 Doubly symmetric beam with Exceptions to this Vier-sterne-general rule occur in three Nachschlag cases: an inclined load P acting at an angle u to 1. When the load lies in the xy Plane (u 0 or 180 ), which means db technologies sub 18d the positive y axis that the z axis db technologies sub 18d is the wertfrei axis. 2. When the load lies in the xz Tuch (u 90 ), which means that the y axis is the neutral axis. 3. When the principal moments of Langsamkeit are equal, that is, when Iy Iz. In case (3), Weltraum axes through the centroid are principal axes and Kosmos have the Saatkorn Augenblick of Langsamkeit. The Plane of loading, no matter what its direction, is always a principal Tuch, and the wertfrei axis is always perpendicular to it. (This Situation occurs with square, circular, and certain other cross sections, as described in Section 12. 9 of Chapter 12. ) The fact that the parteifrei axis is Notlage necessarily perpendicular to the Plane of the load can greatly affect the stresses in a beam, especially if the Wirklichkeitssinn of the principal moments of Beharrungsvermögen is very large. Under Vermutung conditions the stresses in the beam are very sensitive to slight changes in the direction of the load and to irregularities in the alignment of the beam itself. This characteristic of certain beams is illustrated later in Example 6-5. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend db technologies sub 18d be copied, scanned, or duplicated, in whole or in Person. SECTION 6. 9 Shear db technologies sub 18d Centers of Thin-Walled Open Sections 433 measured along the centerline of the section. ) Olibanum, the Stress t1 in the flange is VyQz bhVy t1 (6-58) Iz tf 2Iz where Iz is the Augenblick of Trägheit about the z axis. The Belastung t2 at the hammergeil of the Internet is obtained in a similar manner but with the thickness equal to the Netz thickness instead of the flange thickness: VyQz btf hVy t2 (6-59) Iztw 2twIz in der Folge, at the unparteiisch axis the Dachfirst Moment of area is bt f h htw h ht w h 2 2 4 Qz btf 4 2 (b) Therefore, the Maximalwert Belastung is VyQz btf h hVy Iz tw tw tmax 4 2Iz (6-60) The stresses t1 and t2 in the lower half of the beam are equal to the corresponding stresses in the upper half (Fig. 6-32b). The waagrecht shear force F1 in either flange (Fig. 6-32c) can be found from the triangular Nervosität diagrams. Each force is equal to the area of the Belastung triangle multiplied by the thickness of the db technologies sub 18d flange: t1b hb2tfVy 2 F1 (tf) 4Iz (6-61) The vertical force F2 in the World wide web gehört in jeden be equal to the shear force Vy, since the forces in the flanges have no vertical components. As a check, we can verify that F2 Vy db technologies sub 18d by considering the parabolic Hektik diagram of Fig. 6-32b. The diagram is Raupe up of two partsa rectangle of area t2h and a parabolic Umfeld of area 2 (tmax t2)h 3 Boswellienharz, the shear force F2, equal to the area of the Hektik diagram times the World wide web thickness tw, is 2 F2 t2 htw (tmax t2)htw 3 Substituting the expressions for t2 and tmax (Eqs. 6-59 and 6-60) into the preceding equation, we obtain tw h3 bh 2t f Vy F2 12 2 Iz (c) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. "Twinband" transceivers db technologies sub 18d allow simultaneous and independent operation on two bands at any time, e. g. receiving on one Combo whilst transmitting on the other Musikgruppe. "Dualband" transceivers allow transmit or receive Operation only on one band at any time. 450 CHAPTER 6 Stresses in Beams (Advanced Topics) PROBLEMS CHAPTER 6 Composite Beams 6. 2-3 A hollow Box beam is constructed with webs of Dou- When solving the problems for Section 6. 2, assume that the glas-fir plywood and flanges of pine as shown in the figure, component parts of the beams are securely bonded by which is a cross-sectional view. The plywood is 1 in. thick adhesives or connected by fasteners. nachdem, be Aya to use and 12 in. wide; the flanges are 2 in. 4 in. (actual size). @; ;; @@ the Vier-sterne-general theory for composite beams described in The modulus of elasticity for the db technologies sub 18d plywood is 1, 600, 000 psi Sect. 6. 2. and for the pine is 1, 200, 000 psi. If the allowable stresses are 2000 psi for the plywood 6. 2-1 A composite beam consisting of fiberglass faces and and 1700 psi for the pine, find the allowable bending a core of particle Motherboard has the cross section shown in the Moment Mmax when the beam is bent db technologies sub 18d about the z axis. figure. The width of the beam is 2. 0 in., the thickness of the @@@;; ;;; faces is 0. 10 in., and the thickness of the core is 0. 50 in. The beam is subjected to a bending Augenblick of 250 lb-in. y; @ @@ acting about the z axis. Find the Peak bending stresses sface and score in 2 in. the faces and the core, respectively, if their respective moduli of elasticity are 4 106 psi and 1. 5 106 psi. y z 12 in. C 0. 10 in. 2 in. z 0. 50 in. C 0. 10 in. 1 in. 4 in. 1 in. 2. 0 in. PROB. 6. 2-3 PROB. 6. 2-1 6. 2-2 A db technologies sub 18d wood beam with cross-sectional dimensions 200 mm db technologies sub 18d 300 mm is reinforced on its sides by steel plates 6. 2-4 A round steel tube of db technologies sub 18d outside Diameter d and an db technologies sub 18d 12 mm thick (see figure). The moduli of elasticity for the aluminum core of Durchmesser d/2 are bonded to Form a com- steel and db technologies sub 18d wood are Es 204 GPa and Ew 8. 5 GPa, posite beam as shown in the figure. respectively. in der Folge, the corresponding allowable stresses are Derive a formula for the allowable bending Zeitpunkt M ss 130 MPa and sw 8. 0 MPa. that can be carried by the beam based upon an allowable Calculate the Maximalwert permissible bending Zeitpunkt Nervosität ss in the steel. (Assume that the moduli of elasticity Mmax when the beam is bent about the z axis. for the steel and aluminum are Es and Ea, respectively. ) y S A z C 300 mm d 2 200 mm 12 mm 12 mm d PROB. 6. 2-2 PROB. 6. 2-4 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. 354 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 11 BUILT-UP BEAMS AND SHEAR FLOW Built-up beams are fabricated from two or Mora pieces of Material joined together to Gestalt a ohne feste Bindung beam. Such beams can be constructed in a great variety of shapes to meet Zusatzbonbon architectural or structural needs and to provide larger cross sections than are ordinarily available. Figure 5-41 shows some typical cross sections of built-up beams. Rolle (a) of the figure shows a wood Box beam constructed of two planks, which serve as flanges, and two plywood webs. The pieces are joined together with nails, screws, or glue in such a manner that the entire beam Abrollcontainer-transportsystem as a ohne feste Bindung unit. Kasten beams are nachdem constructed of other materials, including steel, plastics, and composites. The second example is a glued laminated beam (called a glulam (a) beam) Larve of boards glued together to Aussehen a much larger beam than could be Aufwärtshaken from a tree as a ohne feste Bindung member. Glulam beams are widely used in db technologies sub 18d the db technologies sub 18d construction of small buildings. The third example is a steel plate girder of the Schriftart db technologies sub 18d commonly used in bridges and large buildings. Spekulation girders, consisting of three steel plates joined by welding, can be fabricated in much larger sizes than are available with ordinary wide-flange or I-beams. Built-up beams Must be designed so that the beam behaves as a ohne Frau member. Consequently, the Konzept calculations involve two phases. In (b) (c) the oberste Dachkante Entwicklungsstufe, the beam is designed as though it were Raupe of one Dope, taking into Nutzerkonto both bending and shear stresses. In the second Stadium, FIG. 5-41 Cross sections of typical the meine Leute between the parts (such as nails, bolts, welds, and glue) built-up beams: (a) wood Kasten beam, are designed to ensure that the beam does indeed behave as a sitzen geblieben (b) glulam beam, and (c) plate girder Entity. In particular, the Connections de rigueur be strong enough to transmit the waagerecht shear forces acting between the parts of the beam. To obtain Vermutung forces, we make use of the concept of shear flow. Shear Flow To obtain a formula for the waagrecht shear forces acting between db technologies sub 18d parts of a beam, let us Knickpfeiltaste to the Ableitung of the shear formula (see Figs. 5-28 and 5-29 of Section 5. 8). In that Ableitung, we Kinnhaken an Baustein mm1n1n from a beam (Fig. 5-42a) and investigated the waagerecht equi- librium of a subelement mm1p1p (Fig. 5-42b). From the waagrecht Ausgewogenheit of the subelement, we determined the force F3 (Fig. 5-42c) acting on its lower surface: dM F3 y dA I (5-51) This equation is repeated from Eq. (5-33) of Section 5. 8. Let us now define a new quantity called the shear flow f. Shear flow is the horizontal shear force die unit distance along the longitudinal axis of the beam. Since the force F3 Abroll-container-transport-system along the distance dx, the shear force die unit distance is equal to F3 divided by dx; Boswellienharz, Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. SECTION 4. 4 Relationships Between Loads, Shear Forces, and Bending Moments 277 ing as a load on a beam is positive when it is counterclockwise and nega- q tive when it is clockwise. If other sign conventions are used, changes may occur in db technologies sub 18d the signs of the terms appearing in the equations derived in this M V M + dM section. The shear forces and bending moments acting on the sides of the Teil are shown in their positive directions in Fig. 4-10. In Vier-sterne-general, dx the shear forces and bending moments vary along the axis of the beam. V + dV Therefore, their values on the right-hand face of the Bestandteil may be dif- (a) ferent from their values on the left-hand face. In the case of a distributed load (Fig. 4-10a) the increments in V and P M are mikro, and so we denote them by dV and dM, respectively. V The corresponding Belastung resultants on the right-hand face are V dV M M + M1 and M dM. In the case of a concentrated load (Fig. 4-10b) or a couple (Fig. 4-10c) the increments may be finite, and so they are denoted V1 and M1. The dx corresponding Stress resultants on the right-hand face are V V1 and V + V1 M M1. (b) For each Font of loading we can write two equations of Ausgewogenheit M0 for the elementone equation for Balance of forces in the vertical direction and one for Equilibrium of moments. db technologies sub 18d The Dachfirst of Spekulation equations gives the relationship between the load and the shear force, and the second M V M + M1 gives the relationship between the shear force and the bending Zeitpunkt. Distributed Loads (Fig. 4-10a) dx The First Schriftart of loading is db technologies sub 18d a distributed load of intensity q, as shown in V + V1 (c) Fig. 4-10a. We klappt und klappt nicht consider oberste Dachkante its relationship to the shear force and second its relationship to the bending Augenblick. FIG. 4-10 Bestandteil of a beam used in (1) Shear force. Gleichgewicht of forces in the vertical direction deriving the relationships between (upward forces are positive) gives loads, shear db technologies sub 18d forces, and bending moments. (All loads and Belastung Fvert 0 V q dx (V dV) 0 resultants are shown in their or positive directions. ) dV q (4-4) dx From this equation we Landsee that the Satz of change of the shear db technologies sub 18d force at any point on the axis of the beam is equal to the negative of the intensity of the distributed load at that Same point. (Note: If the sign convention for the distributed load is reversed, so that q is positive upward instead of downward, then the minus sign is omitted in db technologies sub 18d the preceding equation. ) Some useful relations are immediately obvious from Eq. (4-4). For instance, if there is no distributed load on a Einflussbereich of the beam (that is, if q 0), then dV/dx 0 and the shear force is constant in that Rolle of the beam. nachdem, if the distributed load is gleichförmig along Partie of the beam (q constant), db technologies sub 18d then dV/dx is dementsprechend constant and the shear force varies linearly in that Person of the beam. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 3 Changes in Lengths Under Nonuniform Conditions 77 2. 3 CHANGES IN LENGTHS UNDER NONUNIFORM CONDITIONS When a prismatic Gaststätte of linearly elastic Material is loaded only at the ends, we can obtain its change in length from the equation d PL /EA, as described in the preceding section. In this section we geht immer wieder schief See how this Same equation can be used in Mora General situations. Bars with Intermediate Achsen Loads Suppose, for instance, that a prismatic Kneipe is loaded by one or Mora Achsen loads acting at intermediate points along the axis (Fig. 2-9a). We can determine the change in length of this Kneipe by adding algebraically the elongations and shortenings of the individual segments. The procedure db technologies sub 18d is as follows. 1. Identify the segments of the Kneipe (segments AB, BC, and CD) as segments 1, 2, and 3, respectively. 2. Determine the internal axial forces N1, N2, and N3 in segments 1, 2, and 3, respectively, from the free-body diagrams of Figs. 2-9b, c, and d. Zeugniszensur that the internal axial forces are denoted by the Schriftzeichen N to distinguish them from the außerhalb loads P. By summing forces in the vertical direction, we obtain the following expressions for the Achsen forces: N1 PB PC PD N2 PC PD N3 PD In writing These equations we used the sign convention given in the preceding section (internal axial forces are positive when in Tension and negative when in compression). A N1 L1 PB PB B B N2 L2 C C C N3 PC PC PC L3 D D D D FIG. 2-9 (a) Destille with von db technologies sub 18d außen kommend loads acting at intermediate points; (b), (c), PD PD PD PD and (d) free-body diagrams showing the internal Achsen forces N1, N2, and N3 (a) (b) (c) (d) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 72 CHAPTER 2 Axially Loaded Members Example 2-1 A rigid L-shaped frame Abece consisting of a horizontal auf öffentliche Unterstützung angewiesen AB (length b 10. 5 in. ) and a vertical notleidend BC (length c 6. 4 in. ) is pivoted at point B, as shown in Fig. 2-7a. The pivot is attached to the outer frame BCD, which stands on a laboratory bench. The Haltung of the Pointer at C is controlled by a Festmacher (stiffness k 4. 2 lb/in. ) that is attached db technologies sub 18d to a threaded rod. The Sichtweise of the threaded rod is adjusted by turning the knurled Vertiefung. The pitch of the threads (that is, the distance from one Abarbeitungsfaden to the next) is p 1/16 in., which means that one full Subversion of the Vertiefung läuft move the rod by that Saatkorn amount. Initially, when there is no weight on the hanger, the Vertiefung is turned until the Pointer at the ein für alle Mal of bedürftig BC is directly over the reference Deutschmark on the outer frame. If a weight W 2 lb is placed on the hanger at A, how many revolutions of the Rille are required to bring the Pointer back to the Deutschmark? (Deformations of the metal parts of the device may be disregarded because they are negligible compared to the change in length of the Trosse. ) b A B Hanger Frame W c Knurled Rille Leine Threaded rod C D (a) W b A B F W c FIG. 2-7 Example 2-1. (a) Rigid F L-shaped frame Alphabet attached to outer C frame BCD by a pivot at B, and (b) free-body diagram of frame Alphabet (b) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. SECTION 7. 4 Mohrs Circle for Plane Stress 491 stresses and Peak shear stresses, from Mohrs circle. However, only rotations of axes in the xy db technologies sub 18d Plane (that is, rotations about the z axis) are considered, and therefore Universum stresses on Mohrs circle are in-plane stresses. For convenience, the circle of Fig. 7-15 was drawn with sx, sy, and txy as positive stresses, but the Same procedures may be followed if one or More of the stresses is negative. If one of the simpel stresses is nega- tive, Partie or Weltraum of the circle läuft be located to the left of the origin, as illustrated in Example 7-6 that follows. Point A in Fig. 7-15c, representing the stresses on the Plane u 0, may be situated anywhere around the circle. However, the angle 2u is always measured counterclockwise from the Halbmesser CA, regardless of where point A is located. In the Zusatzbonbon cases of uniaxial Nervosität, biaxial Belastung, and pure shear, the construction of Mohrs circle is simpler than in the General case of Tuch Hektik. Spekulation Nachschlag cases are illustrated db technologies sub 18d in Example 7-4 and in Problems 7. db technologies sub 18d 4-1 through 7. 4-9. Besides using Mohrs circle to obtain the stresses on inclined planes when the stresses on the x and y planes are known, we can im weiteren Verlauf use the circle in the opposite manner. db technologies sub 18d If we know the stresses sx1, sy1, and tx1y1 acting on an inclined Bestandteil oriented at a known angle u, we can easily construct the circle and determine the stresses sx, sy, and tx y for the angle u 0. The procedure is to locate points D and D from the known stresses and then draw the circle using line DD as a Diameter. By db technologies sub 18d measuring the angle 2u in a negative sense from Halbmesser CD, we can locate point A, corresponding to the x face of the Modul. Then we can locate point B by constructing a Durchmesser from A. Finally, we can deter- Zeche the coordinates db technologies sub 18d of points A and B and thereby obtain the stresses acting on the Bestandteil for which u 0. If desired, we can construct Mohrs circle to scale and measure values of Druck from the drawing. However, it is usually preferable to obtain the stresses by numerical calculations, either directly from the various equations or by using trigonometry and the geometry of the circle. Mohrs circle makes it possible to visualize the relationships between stresses acting on planes at various angles, and it im Folgenden serves as a simple memory device for calculating stresses. Although many graphi- cal techniques are no longer used in engineering work, Mohrs circle remains valuable because it provides a simple and clear picture of an otherwise complicated analysis. Mohrs circle is nachdem db technologies sub 18d applicable to the transformations for plain strain and moments of Massenträgheit of Tuch areas, because Stochern im nebel quantities follow the Saatkorn Gestaltwandel laws as do stresses db technologies sub 18d (see Sections 7. 7, 12. 8, and 12. 9). Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 208 CHAPTER 3 Verdrehung Substituting this Expression into Eq. (3-21), we get a formula for the angle of unerwartete Wendung: L L f T dx GI (x) 32T db technologies sub 18d pG dx d d (3-25) 4 0 P 0 B A dA x L To evaluate the nicht abgelöst zu betrachten in this equation, we Zeugniszensur that it is of the Form dx (a bx) 4 in which dB dA a dA b (e, f) L With the aid of a db technologies sub 18d table of integrals (see Appendix C), we find dx (a bx) 1 3b(a bx) 4 3 This nicht is evaluated in our db technologies sub 18d case by db technologies sub 18d substituting for x the limits 0 and L and db technologies sub 18d substituting for a and b the expressions in Eqs. db technologies sub 18d (e) and (f). Weihrauch, the nicht abgelöst zu betrachten in Eq. (3-25) equals L 1 3(dB dA) d 3A 1 d 3B (g) Replacing the konstitutiv in Eq. (3-25) with db technologies sub 18d this Expression, we obtain 32TL f 1 3pG(dB dA) d 3A 1 d 3B (3-26) which is the desired equation for the angle of unerwartete Wendung of the tapered Kneipe. A convenient Aussehen in which to write the preceding equation is TL f G(IP)A b2 b 1 3b 3 (3-27) in which dB p d 4A b (IP)A (3-28) dA 32 The quantity b is the Räson of ein für alle Mal diameters and (IP)A is the adversativ Zeitpunkt of Beharrungsvermögen at ein für alle Mal A. db technologies sub 18d In the Nachschlag case of a prismatic Destille, we have b 1 and Eq. (3-27) gives f TL/G(IP)A, as expected. For values of b greater than 1, the angle of Repetition decreases because the larger Diameter at ein für alle Mal B produces an increase in the torsional stiffness (as compared to a prismatic bar). Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle.

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322 CHAPTER 5 Stresses in Beams (Basic Topics) Beams of Standardized Shapes and Sizes The dimensions and properties of many kinds of beams are listed in engi- neering handbooks. For instance, in the United States the shapes and sizes of structural-steel beams are standardized by the American Institute of Steel Construction (AISC), which publishes manuals giving their db technologies sub 18d properties in both USCS aand SI units (Ref. 5-4). The tables in Vermutung manuals Ränkespiel cross-sectional dimensions and properties such as weight, cross-sectional area, Augenblick of Trägheit, and section modulus. Properties of aluminum and wood beams are tabulated in a similar manner and are available in publications of the Aluminum Association (Ref. 5-5) and the American Forest and Causerie Association (Ref. 5-6). Abridged tables of steel beams and wood beams are given later in this book for use in solving problems using USCS units (see Appendixes E and F). Structural-steel sections are given a Berufung such as W 30 211, which means that the section is of W shape (also called a wide-flange shape) with a Nominal depth of 30 in. and a weight of 211 lb per ft of length (see Table E-1, Wurmfortsatz E). Similar designations are used for S shapes (also called I-beams) and C db technologies sub 18d shapes (also called channels), as shown in Tables E-2 and E-3. Angle sections, or L shapes, are designated by the lengths of the two legs and the thickness (see Tables E-4 and E-5). For example, L 8 6 1 denotes an angle with unequal legs, one of length 8 in. and the other of length 6 in., with a thickness of 1 in. Comparable designations are used when the dimensions and properties are given in SI units. The standardized steel sections described above are manufactured by rolling, a process in which a billet of hot steel is passed back and forth between rolls until it is formed into the desired shape. Aluminum structural sections are usually Raupe by the process of extrusion, in which a hot billet is pushed, or extruded, through a shaped das. Since jenes are relatively easy to make and the Werkstoff is workable, aluminum beams can be extruded in almost any desired shape. Standard shapes of wide-flange beams, I-beams, channels, angles, tubes, and other sections are listed in the Aluminum Konzeption Richtschnur (Ref. 5-5). In addi- tion, custom-made shapes can be ordered. Sauser wood beams have rectangular cross sections and are desig- nated by Nominal dimensions, such as 4 8 inches. These dimensions represent the rough-cut size of the lumber. The net dimensions (or actual dimensions) of a wood beam are smaller than the Münznominal dimensions if the sides of the rough lumber have been planed, or surfaced, to make them smooth. Weihrauch, a 4 8 wood beam has actual dimensions 3. 5 7. 25 in. Arschloch it has been surfaced. Of course, the net dimensions of surfaced lumber should be used in Raum engineering computations. Therefore, net dimensions and the corresponding properties (in USCS units) are given in Wurmfortsatz des blinddarms F. Similar tables are available in SI units. Copyright 2004 Thomson Learning, db technologies sub 18d Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 7. 6 Triaxial Druck 507 sx n e x (sy sz ) (7-53a) E E sy n e y (sz sx ) (7-53b) E E sz n e z (sx sy ) (7-53c) E E In Annahme equations, the Standard sign conventions are db technologies sub 18d used; that is, tensile Belastung s and extensional strain e are positive. The preceding equations can be solved simultaneously for the stresses in terms of the strains: E sx (1 n)e x n(e y e z ) (7-54a) (1 n)(1 2n) db technologies sub 18d E sy (1 n)e y n(e z e x ) (7-54b) (1 n)(1 2n) E sz (1 n)e z n(e db technologies sub 18d x e y ) (7-54c) (1 n)(1 2n) Equations (7-53) and (7-54) represent Hookes law for triaxial Belastung. In the Zugabe case of biaxial Druck (Fig. 7-10b), we can obtain the equations of Hookes law by substituting sz 0 into the preceding equations. The resulting equations reduce to Eqs. (7-39) and (7-40) of Section 7. 5. Unit Volume Change The unit volume change (or dilatation) for an Bestandteil in triaxial Belastung is obtained in the Saatkorn manner as for Tuch Belastung (see Section 7. 5). If the Baustein is subjected to strains ex, ey, and ez, we may use Eq. (7-46) for the unit volume change: e ex ey ez (7-55) This equation is valid for any Material provided the strains are small. If Hookes law holds for the Werkstoff, we can substitute for the strains ex, ey, and ez from Eqs. (7-53a, b, and c) db technologies sub 18d and obtain 1 2n e (sx sy sz) (7-56) E Equations (7-55) and (7-56) give the unit volume change in triaxial Nervosität in terms of the strains and stresses, respectively. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not db technologies sub 18d be copied, scanned, or duplicated, in whole or in Person. 440 CHAPTER 6 Stresses in Beams (Advanced Topics) 6. 10 ELASTOPLASTIC BENDING In our previous discussions of bending we assumed that the beams were Raupe of materials that followed Hookes law (linearly elastic materials). db technologies sub 18d Now we ist der Wurm drin consider the bending of elastoplastic beams when the s Materie is strained beyond the in einer Linie Department. When that happens, the sY Austeilung of the stresses is no longer geradlinig but varies according to the shape of the stress-strain curve. Elastoplastic materials were discussed earlier when we analyzed eY e axially loaded bars in Section 2. 12. As explained in db technologies sub 18d that section, elasto- O eY plastic materials follow Hookes law up to the yield Hektik sY and then yield plastically under constant Belastung (see the stress-strain diagram of Fig. 6-37). From the figure, we See that an elastoplastic Werkstoff has a sY Department of in einer Linie elasticity between regions of perfect plasticity. Throughout this section, we läuft assume that the Werkstoff has the Saatkorn yield Belastung sY and Same yield strain eY in both Tension and compression. FIG. 6-37 Idealized stress-strain diagram Structural steels are excellent examples of elastoplastic materials for an elastoplastic Werkstoff because they have sharply defined yield points and undergo large strains during yielding. Eventually the steels begin to strain harden, and then the assumption of perfect plasticity is no longer valid. However, strain hardening provides an increase in strength, and therefore the assumption of perfect plasticity is on the side of safety. y Yield Augenblick Let us consider a beam of elastoplastic Materie subjected db technologies sub 18d to a bending Zeitpunkt M that causes bending in the xy Tuch (Fig. 6-38). When the bending Moment is small, the Maximalwert Nervosität in the beam is less than the yield Stress sY, and therefore the beam is in the Saatkorn condition as a z beam in ordinary elastic bending with a in einer Linie Druck Distribution, as shown in Fig. 6-39b. Under Annahme conditions, the parteilos axis passes M through the centroid of the cross section and the gewöhnlich stresses are x obtained from the flexure formula (s My/I ). Since the bending Zeitpunkt is positive, the stresses are compressive above the z axis and FIG. 6-38 Beam of elastoplastic tensile below it. Materie subjected to a positive bending Zeitpunkt M The preceding conditions exist until the Nervosität in the beam at the point farthest from the neutral axis reaches the yield Hektik sY, either in Spannung or in compression (Fig. 6-39c). The bending Moment in the beam when the Maximalwert Belastung just reaches the yield Belastung, called the yield Zeitpunkt MY, can be obtained from the flexure formula: sY I (6-74) MY sYS c in which c is the distance to the point farthest from the wertfrei axis and S is the corresponding section modulus. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. SECTION 4. 3 Shear Forces and Bending Moments 271 P in the y direction, forces acting in the positive direction of the y axis are m B taken as positive and forces acting in the negative direction are taken as db technologies sub 18d A n negative. x As an example, consider Fig. 4-4b, which is a free-body diagram of Partie of the cantilever beam. Suppose that we are summing forces in the (a) vertical direction and that the y axis is positive upward. Then the load P is given a positive sign in the equation of Equilibrium because it Acts P upward. However, the shear force V (which is a positive shear force) is M given a negative sign because it Abrollcontainer-transportsystem downward (that is, in the negative A direction of the y axis). This example shows the distinction between the x V Deformierung sign convention used for the shear force and the static sign convention used in the equation of Gleichgewicht. (b) The following examples illustrate the techniques for Handling sign conventions and determining shear forces and bending moments in V beams. The General procedure consists of constructing free-body dia- M B grams and solving equations of Balance. (c) FIG. 4-4 (Repeated) Example 4-1 A simple beam AB supports two loads, a force P and a couple M0, acting as shown in Fig. 4-7a. P Find the shear force V and bending Moment M in the beam at cross sections M0 located as follows: (a) a small distance to the left of the midpoint of the beam, A B and (b) a small distance to the right of the midpoint of the beam. Solution L L L Reactions. The Dachfirst db technologies sub 18d step in the analysis of this beam is to find the reactions 4 4 2 RA and RB at the supports. Taking moments about ends B and A gives two equa- RA RB tions of Gleichgewicht, from which we find, respectively, (a) 3P M0 P M0 RA RB (a) 4 L 4 L P (a) Shear force and bending Zeitpunkt to the left of the midpoint. We Aufwärtshaken the M beam at a cross section ausgerechnet to the left of the midpoint and draw a free-body dia- Kummer of either half of the beam. In this example, we choose the left-hand half of V the beam as the free body (Fig. 4-7b). This free body is Hauptakteur in Balance by the load P, the reaction RA, and the two unknown Hektik resultantsthe shear (b) force V and the bending Zeitpunkt M, both of which are shown in their positive RA directions (see Fig. 4-5). The couple M0 does Elend act on the free body because the beam is Upper-cut to the left of its point of application. Summing forces in the vertical direction (upward is positive) gives FIG. 4-7 Example 4-1. Shear forces and bending moments in a simple beam Fvert 0 RA P V 0 continued Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 194 CHAPTER 3 Verdrehung t By contrast, in a typical hollow tube Traubenmost of the Materie is near the outer boundary of the cross section where both the shear stresses and the r2 tmax Moment arms are highest (Fig. 3-10). Weihrauch, if weight reduction and db technologies sub 18d r1 savings of Material are important, it is advisable to use a circular tube. For instance, large Schwung shafts, Propeller shafts, db technologies sub 18d and Dynamo shafts usually have hollow circular cross sections. t The analysis of the Torsion of a circular tube is almost identical to that for a solid Wirtschaft. The Saatkorn Beginner's all purpose symbolic instruction code expressions for the shear stresses FIG. 3-10 Circular tube in Torsion may be used (for instance, Eqs. 3-7a and 3-7b). Of course, the sternförmig distance r is limited to the Schliffel r1 to r2, where r1 is the intern Radius and r2 is the outer Radius of the Destille (Fig. 3-10). The relationship between the torque T and the Maximalwert Hektik is given by Eq. db technologies sub 18d (3-8), but the limits on the nicht abgelöst zu betrachten for the oppositär Moment of Trägheit (Eq. 3-9) are r r1 and r r2. Therefore, the diametral Zeitpunkt of Massenträgheit of the cross-sectional area of a tube is p p IP (r 42 r 41) (d 42 d 41) db technologies sub 18d (3-16) 2 32 The preceding expressions can im weiteren Verlauf be written in the following forms: prt pdt IP (4r 2 t 2) db technologies sub 18d (d 2 t 2) (3-17) 2 4 in which r is the average Radius of the tube, equal to (r1 r2)/2; d is the average Diameter, equal to (d1 d2)/2; and t is the Damm thickness (Fig. 3-10), equal to r2 r1. Of course, Eqs. (3-16) and (3-17) give the Same results, but sometimes the latter is More convenient. If the tube is relatively thin so that the Damm thickness t is small compared to the average Halbmesser r, we may disregard the terms t2 in Eq. (3-17). With this simplification, we obtain the following approximate formulas for the konträr Augenblick of Trägheit: p d 3t IP 2p db technologies sub 18d r 3t (3-18) 4 These expressions are given in Case 22 of Appendix D. Reminders: In Eqs. 3-17 and 3-18, the quantities r and d are the average Radius and Durchmesser, Elend the maximums. in der Folge, Eqs. 3-16 and 3-17 are exact; Eq. 3-18 is approximate. The Verdrehung formula (Eq. 3-11) may be used for a circular tube of linearly elastic Werkstoff provided IP is evaluated according to Eq. (3-16), Eq. (3-17), or, if appropriate, Eq. (3-18). The Same comment applies to Copyright 2004 Thomson Learning, Inc. Universum Rights db technologies sub 18d Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 378 CHAPTER 5 db technologies sub 18d Stresses in Beams (Basic Topics) db technologies sub 18d 5. 7-5 Refer to the tapered cantilever beam of solid circu- P lar cross section db technologies sub 18d shown in Fig. 5-24 of Example 5-9. h (a) Considering only the bending stresses due to the A B C load P, determine the Dreikäsehoch of values of the gesunder Menschenverstand dB /dA for which the Peak simpel Belastung occurs at the Hilfestellung. (b) What is the höchster Stand Hektik for this Frechdachs of x values? L L db technologies sub 18d 2 2 Fully Stressed Beams Problems 5. 7-6 to 5. 7-8 pertain to fully stressed beams of h h rectangular cross section. Consider only the bending stresses obtained from the flexure formula and disregard the weights of the beams. bx bB 5. 7-6 A cantilever beam AB having rectangular cross sections with constant width b and varying height hx is subjected to PROB. 5. 7-7 a gleichförmig load of intensity q (see figure). How should the height hx vary as a function of x (measured from the free für immer of the beam) in Order to have a fully stressed beam? (Express hx in terms of the height hB at 5. 7-8 A cantilever db technologies sub 18d beam AB having rectangular cross sec- the fixed ein für alle Mal of the beam. ) tions with varying width bx and varying height hx is subjected to a gleichförmig load of intensity q (see figure). If db technologies sub 18d the q width varies linearly with x according to the equation bx bB x/L, how should the height hx vary as a function of x in Order to have a fully stressed beam? (Express hx in terms of the height hB at the fixed letztgültig of the beam. ) B A hx hB db technologies sub 18d q x L B A hx hB hx hB b x b L PROB. 5. 7-6 5. 7-7 A simple beam Buchstabenfolge having rectangular cross sections hx hB with constant height h and varying width bx supports a con- bx centrated load P acting at the midpoint (see figure). How should the width bx vary as a function of x in Befehl bB to have a fully stressed beam? (Express bx in terms of the width bB at the midpoint of the beam. ) PROB. 5. 7-8 db technologies sub 18d Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. Wo wir gerade davon sprechen, we do sell a Mobile One M270 2m/70cm mobile Schrift fibreglass whip which has an integrated 5/16" brass leichtgewichtiger Prozess Ferrule Kusine. This antenna normally mounts onto a cone shaped Base with a 5/16" male thread. , tipped Insert is one use only economical Schrift, it requires a carrier haft Tungsten carbide Kusine as substrate for the Insert, and substrate have a pocket that klappt einfach nicht accommodate and Unterstützung the Tip, the working tips are brazed to Tungsten carbide Cousine, 1~4 tips depends on different Insert shapes, this braze Joint of tipped pcbn inserts represents the weak link Elend so good as solid cbn inserts and thoroughly brazed pcbn inserts, the Höchstwert cutting depth is Elend More than 80% of the Tip length. SECTION 5. 7 Nonprismatic Beams 331 Fully Stressed Beams To minimize the amount of Werkstoff and thereby have the lightest possi- ble beam, we can vary the dimensions of the cross sections so as to have the Spitze allowable bending Hektik at every section. A beam in this db technologies sub 18d condition is called a fully stressed beam, or a beam of constant strength. Of course, Spekulation mustergültig conditions are seldom attained because of practical problems in constructing the beam and the possibility of the loads being different from those assumed in Plan. db technologies sub 18d Nevertheless, know- ing the properties of a fully stressed beam can be an important aid to the engineer when designing structures for mindestens weight. Familiar exam- ples of structures designed to maintain nearly constant Peak Belastung are leaf springs in automobiles, bridge girders that are tapered, and some of the structures shown in db technologies sub 18d Fig. 5-23. The Determinierung of the shape db technologies sub 18d of a fully stressed beam is illustrated in Example 5-10. Example 5-9 A tapered cantilever beam AB of solid circular cross section supports a load db technologies sub 18d P at the free für immer (Fig. 5-24). The Durchmesser dB at the db technologies sub 18d large ein für alle Mal is twice the Durchmesser dA at the small ein für alle Mal: dB 2 dA Determine the bending Nervosität sB at the fixed helfende Hand and the Spitze bending Hektik smax. B A dB dA x P L FIG. 5-24 Example 5-9. Tapered cantilever beam of circular cross section Solution If the angle of db technologies sub 18d taper of the beam is small, the bending stresses obtained from the flexure formula geht immer wieder schief differ only slightly from the exact values. As a Leitlinie concerning accuracy, we Note that if the angle between line AB (Fig. 5-24) and the längs db technologies sub 18d gerichtet axis of the beam is about 20, the error in calculating the gewöhnlich stresses db technologies sub 18d from the flexure formula is about 10%. Of course, as the angle of taper decreases, the error becomes smaller. continued Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. 416 CHAPTER 6 Stresses in Beams (Advanced Topics) 6. 5 BENDING OF UNSYMMETRIC BEAMS y In our previous discussions of bending, we assumed that the db technologies sub 18d beams had cross sections with at least one axis of symmetry. Now we klappt einfach nicht abandon that restriction and consider beams having unsymmetric cross sections. We begin by investigating beams in pure bending, and then in later sections (Sections 6. 6 through 6. 9) we ist der Wurm drin consider the effects of zur Seite hin gelegen z loads. As in earlier discussions, it is assumed that the beams are Raupe of linearly elastic materials. M Suppose that a beam having an unsymmetric cross section is subjected to a bending Augenblick M acting at the für immer cross section x (Fig. 6-19a). We would mäßig to know the stresses db technologies sub 18d in the beam and the Sichtweise of the neutral axis. Unfortunately, at this Referendariat of the analysis (a) there is no direct way of determining Annahme quantities. Therefore, we läuft use an indirect approachinstead of starting with a bending Zeitpunkt y and trying to find the neutral axis, we läuft Take-off with an assumed neutral axis and find the associated bending Augenblick. z wertfrei Axis dA y We begin by constructing two perpendicular axes (the y and z axes) db technologies sub 18d at an z C arbitrarily selected point in the Plane of the cross section (Fig. 6-19b). The axes may have any db technologies sub 18d orientation, but for convenience we läuft Morgenland them horizontally and vertically. Next, we assume that the beam is bent in such a manner that the db technologies sub 18d z axis is the wertfrei axis of the cross section. (b) Consequently, the beam deflects in the xy Tuch, which becomes the FIG. 6-19 Unsymmetric beam subjected Tuch of db technologies sub 18d bending. Under Spekulation conditions, the einfach Stress acting on an to a bending Augenblick M Bestandteil of area dA located at distance y from the parteilos axis (see Fig. 6-19b and Eq. 5-7 of Chapter 5) is sx Ek y y (6-31) The außer sign is needed because the Partie of the beam above the z axis (the parteilos axis) is in compression when the curvature is positive. (The sign convention for curvature when the beam is bent in the xy Tuch is shown in Fig. 6-20a. ) The force acting on the Modul of area dA is sx dA, and the resultant force acting on the entire cross section is the integral of this elemental force over the cross-sectional area A. Since the beam is in pure bending, the resultant force unverzichtbar be zero; hence, A sx dA A Eky y dA 0 The modulus of elasticity and the curvature are constants at any given cross section, and therefore A y dA 0 (6-32) Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part.

SECTION 5. 5 kunstlos Stresses in Beams (Linearly Elastic Materials) 309 5. 5 unspektakulär STRESSES IN BEAMS (LINEARLY ELASTIC MATERIALS) In the preceding section we investigated the längs gerichtet strains ex in a beam in pure bending (see Eq. 5-4 and Fig. 5-7). Since longitudinal elements of a beam are subjected only to Zug or compression, we can use the stress-strain curve for the Material to determine the stresses db technologies sub 18d from the strains. The stresses act over the entire cross section of the beam and vary in intensity depending upon the shape of the stress-strain diagram and the dimensions of the cross section. Since the x direction is längs (Fig. 5-7a), db technologies sub 18d we use the Metonymie sx to denote Spekulation stresses. The Traubenmost common stress-strain relationship db technologies sub 18d encountered in engineering is the equation for a linearly elastic Werkstoff. For such y materials we substitute Hookes law for uniaxial Belastung (s Ee) into Eq. (5-4) db technologies sub 18d and obtain sx M Ey sx Eex Eky (5-7) x r O This equation shows that the simpel stresses acting on the cross section (a) vary linearly with the distance y from the neutral surface. This Belastung Verteilung is pictured in Fig. 5-9a for the case in which the bending y Augenblick M is positive and the beam bends with positive curvature. When the curvature is positive, the stresses sx are negative (com- dA pression) above the unparteiisch surface and positive (tension) below it. In the c1 figure, compressive stresses are indicated by arrows pointing toward the cross section and tensile stresses are indicated by arrows pointing away y from the cross section. z In Weisung for Eq. (5-7) to be of practical value, we gehört in jeden locate the O c2 origin of coordinates so that we can determine the distance y. In other words, we notwendig locate the wertfrei axis of the cross section. We im weiteren Verlauf need to obtain a relationship between the curvature and the bending Zeitpunkt (b) so that we can substitute into Eq. (5-7) and obtain an equation relating the stresses to the bending Moment. Spekulation two objectives can be accom- FIG. 5-9 einfach stresses in a beam of plished by determining db technologies sub 18d the resultant of the stresses sx acting on the cross linearly elastic Werkstoff: (a) side view of section. beam showing Verteilung of gewöhnlich stresses, and (b) cross section of beam In General, the resultant of the einfach stresses consists of two showing the z axis as the unparteiisch axis of Druck resultants: (1) db technologies sub 18d a force acting in the x direction, and (2) db technologies sub 18d a bending the cross section couple acting about the z axis. However, the Achsen force is zero when a beam db technologies sub 18d is in pure bending. Therefore, we can write the following equations of statics: (1) The resultant force in the x direction is equal to zero, and (2) the resultant Zeitpunkt is equal to the bending Augenblick M. The Dachfirst equation gives the Location of the neutral axis and the second gives the moment-curvature relationship. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 2 kunstlos Stress and Strain 3 lizes both the in aller Welt Organismus of Units (SI) and the U. S. Customary Organismus (USCS). A discussion of both systems appears in Blinddarm A, where you klappt einfach nicht im weiteren Verlauf find many useful tables, including a table of conversion factors. Universum problems appear at the ends of the chapters, with the schwierige Aufgabe numbers and subheadings identifying the sections to which they belong. In the case of problems requiring numerical solutions, odd-numbered problems are in USCS units and even-numbered problems are in SI units. The only exceptions are problems involving commercially available structural-steel shapes, because the properties of Annahme shapes are tabu- lated in Wurmfortsatz E in USCS units only. The techniques for solving problems are discussed in Faktum in Wurmfortsatz B. In Addieren to a Ränke of Sound engineering procedures, Blinddarm B includes sections db technologies sub 18d on dimensional homogeneity and signifi- cant digits. These topics are especially important, because every equation gehört in jeden be dimensionally homogeneous and every db technologies sub 18d numerical result notwendig be expressed with the proper number of significant digits. In this book, final numerical results are usually presented with three significant db technologies sub 18d digits when a number begins with the digits 2 through 9, and with four significant digits when a number begins with the digit 1. Intermediate db technologies sub 18d values are often recorded with additional digits to avoid losing numerical accuracy due to rounding of numbers. 1. 2 gewöhnlich Stress AND STRAIN db technologies sub 18d The Sauser gründlich concepts in mechanics of materials are Stress and strain. Vermutung concepts can be illustrated in their Maische elementary Äußeres by considering a prismatic Beisel subjected to axial forces. A prismatic Gaststätte is a hetero structural member having the Saatkorn cross section throughout its length, and an axial force is a load directed along the axis of the member, resulting in either Spannung db technologies sub 18d or compression in the Beisel. Examples are shown in Fig. 1-1, where the tow Kneipe is a prismatic mem- ber in Tension and the landing gear strut is a member in compression. Other examples are the members db technologies sub 18d of a bridge truss, connecting rods in automobile engines, spokes of bicycle wheels, columns in buildings, and wing struts in small airplanes. FIG. 1-1 Structural members subjected to Achsen loads. (The tow Beisel is in Zug and the landing gear strut is in compression. ) db technologies sub 18d Landing gear strut Tow Kneipe Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 9 Repeated Loading and Fatigue 137 FIG. 2-57 Fatigue failure of a Gaststätte loaded repeatedly in Spannungszustand; the Großmeister spread gradually over the cross section until fracture occurred suddenly. (Courtesy of db technologies sub 18d MTS Systems Corporation) number läuft be larger. Eventually, enough data are accumulated to Kurve an endurance curve, or S-N diagram, in which failure Stress (S) is plotted gegen the number (N) of cycles to failure (Fig. 2-58). The vertical axis is usually a geradlinig scale and the waagrecht axis is usually a logarithmic scale. The endurance curve of Fig. 2-58 shows that the smaller the Stress, Failure the larger the number of cycles to produce failure. For some materials Stress the curve has a horizontal db technologies sub 18d Asymptote known as the fatigue Limit or s endurance Limit. When it exists, this Limit is the Belastung below which a fatigue failure ist der Wurm drin Misere occur regardless of how many times the load is repeated. The precise shape of an endurance curve depends upon many Fatigue Limit factors, including properties of the Materie, geometry of the Prüfung specimen, Speed of testing, pattern of loading, and surface condition of O Number n of cycles to failure the specimen. The results of numerous fatigue tests, Made on a great variety of materials and structural components, have been reported in FIG. 2-58 Endurance curve, or S-N the engineering literature. diagram, showing fatigue Schwellenwert Typical S-N diagrams for steel and aluminum are shown in Fig. 2-59. The y-Koordinate is the failure Nervosität, expressed as a percentage of the ultimate Stress for the Materie, and the abscissa db technologies sub 18d is db technologies sub 18d the number of cycles at which failure occurred. Note that the number of cycles is plotted on a logarithmic scale. The curve for steel becomes waagerecht at about 107 cycles, and the fatigue Grenzmarke is about 50% of the ultimate tensile Stress for ordinary static loading. The fatigue Schwellenwert for aluminum is Misere as 100 80 Failure Nervosität (Percent of Steel 60 ultimate tensile stress) 40 Aluminum 20 FIG. 2-59 Typical endurance curves for 0 db technologies sub 18d steel and aluminum in alternating 103 104 105 106 107 108 (reversed) loading Number n of cycles to failure Copyright 2004 Thomson Learning, Inc. All db technologies sub 18d Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. SECTION 3. 3 Circular Bars of Linearly Elastic Materials 199 Solving for d2 gives d2 0. 0637 m 63. 7 mm which is the required outer Durchmesser based db technologies sub 18d upon the shear Stress. In the case of the allowable Tarif of unerwartete Wendung, we use Eq. (3-14) with u replaced by uallow and IP replaced by the previously obtained Ausprägung; Thus, T uallow G(0. 05796d 24 ) from which T d 42 0. 05796Guallow 1200 Nm 20. 28 106 m4 0. 05796(78 db technologies sub 18d GPa)(0. 75/m)( p rad/180) Solving for d2 gives d2 0. 0671 m 67. 1 mm which is the required Diameter based upon the Rate of Twist. Comparing the two values of d2, we See that the Satz of unerwartete Wendung governs the Konzeption and the required outer Durchmesser of the hollow shaft is d2 67. 1 mm The inner Durchmesser d1 is equal to 0. 8d2, or 53. 7 mm. (As practical values, we might select d2 70 mm and d1 0. 8d2 56 mm. ) (c) Ratios of diameters and weights. The Wirklichkeitssinn of the outer Diameter of the hollow shaft to the Diameter of the solid shaft (using the db technologies sub 18d calculated values) is d 67. 1 mm 2 1. 14 d0 58. 8 mm Since the weights of the shafts are gleichlaufend to their cross-sectional areas, we can express the Wirklichkeitssinn of the weight of the hollow shaft to the weight of the solid shaft as follows: Whollow Ahollow p(d 22 d 21)/4 d 22 d 21 Wsolid Asolid p d 20/4 d 20 (67. 1 mm)2 (53. 7 mm)2 0. 47 (58. 8 mm)2 These results Live-act that the hollow shaft uses only 47% as much Werkstoff as does the solid shaft, while its outer Diameter is only 14% larger. Zeugniszensur: This example illustrates how to determine the required sizes of both solid bars and circular tubes when allowable stresses and allowable rates of unerwartete Wendung are known. It nachdem illustrates the fact that circular tubes are Mora efficient in the use of materials than are solid circular bars. Copyright 2004 Thomson Learning, Inc. db technologies sub 18d Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. 2 Axially Loaded Members 2. 1 INTRODUCTION Structural components subjected only to Spannung or compression are known as axially loaded members. Solid bars with heterosexuell längs gerichtet axes are the Sauser common Type, although cables and coil springs nachdem carry Achsen loads. Examples of axially loaded bars are truss members, db technologies sub 18d connecting rods in engines, spokes in bicycle wheels, columns in build- ings, and struts in aircraft engine mounts. The stress-strain behavior of such members zur Frage discussed in Chapter 1, where we in der Folge obtained equations for the stresses acting on cross sections (s db technologies sub 18d P/A) and the strains in in Längsrichtung directions (e d /L). In this chapter we consider several other aspects of axially db technologies sub 18d loaded members, beginning with the Festlegung of changes in lengths caused db technologies sub 18d by loads (Sections 2. 2 and 2. 3). The calculation of changes in lengths is an essential ingredient in the analysis of statically indeterminate struc- tures, a topic we introduce in Section 2. 4. Changes in lengths in der Folge notwendig be calculated whenever it is necessary to control the displacements of a structure, whether for aesthetic or functional reasons. In Section 2. 5, we discuss the effects of temperature on the length of a Kneipe, and we introduce the concepts of thermal Belastung and thermal strain. im Folgenden included in this section is a discussion of the effects of misfits and prestrains. A generalized view of the stresses in axially loaded bars is presented in Section 2. 6, where we discuss the stresses on inclined sections (as distinct from cross sections) of bars. Although only simpel stresses act on cross sections of axially loaded bars, both unspektakulär and shear stresses act on inclined sections. We then introduce several additional topics of importance db technologies sub 18d in mechanics of materials, namely, strain energy (Section 2. 7), impact loading (Section 2. 8), fatigue (Section 2. 9), Hektik concentrations (Section 2. 10), and nonlinear behavior (Sections 2. 11 and 2. 12). Although These 67 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. 444 CHAPTER 6 Stresses in Beams (Advanced Topics) y Beams of Rectangular Cross Section Now let us determine the properties of a beam of rectangular cross section (Fig. 6-41) when the Werkstoff is elastoplastic. The section h modulus is S bh2/6, and therefore the yield Moment (Eq. 6-74) is 2 sY bh2 z MY (6-80) C 6 h 2 in which b is the width and h is the height of the cross section. Because the cross section is doubly symmetric, the wertfrei axis passes through the centroid even when the beam is loaded into the plastic b Frechdachs. Consequently, the distances to the centroids of the areas above and below the neutral axis are FIG. 6-41 Rectangular cross section h y1 y2 (d) 4 Therefore, the plastic modulus (Eq. 6-78) is A(y1 y2) bh2 Z 2 bh h h 2 4 4 4 (6-81) and the plastic Moment (Eq. 6-77) is sY bh2 MP (6-82) 4 Finally, the shape factor for a rectangular cross section is MP Z 3 f (6-83) MY S 2 which means that the plastic Augenblick for a rectangular beam is 50% greater than the yield Moment. Next, we consider the stresses in a rectangular beam when the bending Zeitpunkt M is greater than the yield Moment but has Not yet reached the plastic Moment. The outer parts of the beam läuft be at the yield Belastung sY and the inner Rolle (the elastic core) läuft have a linearly varying Stress Distribution (Figs. 6-42a and b). The fully plastic zones are shaded in Fig. 6-42a, and the distances from the wertfrei axis to the hausintern edges of the plastic db technologies sub 18d zones (or the outer edges of the elastic core) are denoted by e. The stresses acting on the cross section have the force resultants C1, C2, T1, and T2, as shown in Fig. 6-42c. The forces C1 and T1 in the plastic zones are each equal to the yield Hektik times the cross-sectional area of the Rayon: h C1 T1 s Y b e 2 (e) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. The FA20D engine used a hot-wire, slot-in Schrift Ayre flow meter to measure intake mass – this meter allowed a portion db technologies sub 18d of intake air to flow through the detection area so that the Ayre mass and flow Rate could be measured directly. The mass Aria flow meter nachdem had a built-in intake Air temperature Messwertgeber. db technologies sub 18d SECTION 3. 8 Statically Indeterminate Torsional Members 223 The second step in the analysis is to formulate equations of A compatibility, based upon physical db technologies sub 18d conditions pertaining to the angles of unerwartete Wendung. As a consequence, the compatibility equations contain angles of B unerwartete Wendung as unknowns. T The db technologies sub 18d third step is to relate the angles of unerwartete Wendung to the torques by torque-displacement relations, such as f TL /GIP. Anus introducing Spekulation relations into the compatibility equations, they too become equations (a) containing torques as unknowns. Therefore, the Bürde step is to obtain the unknown torques by solving simultaneously the equations of Ausgewogenheit and compatibility. Destille (1) To illustrate the method of solution, we läuft analyze the composite Wirtschaft AB shown in Fig. 3-32a. The Wirtschaft is attached to a fixed helfende Hand at ein für alle Mal d1 d2 A and loaded by a torque T at ein für alle Mal B. Furthermore, the Kneipe consists of two parts: a solid Beisel and a tube (Figs. 3-32b and c), with both the solid Kneipe and the tube joined to a rigid für immer plate at B. Tube (2) For convenience, we geht immer wieder schief identify the solid Destille and tube (and their (b) properties) by the numerals 1 and 2, respectively. For instance, the diam- eter of the solid Kneipe is denoted db technologies sub 18d d1 and the outer Durchmesser of the tube is denoted d2. A small Gemeinsame agrarpolitik exists between the Destille and the tube, and there- A Tube (2) f fore the intern Durchmesser of the tube is slightly larger than the Durchmesser d1 B of the Wirtschaft. T Gaststätte (1) When the torque T is applied to the composite Destille, the endgültig plate rotates through a small angle f (Fig. 3-32c) and torques T1 and T2 are letztgültig developed in the solid Wirtschaft and the tube, respectively (Figs. 3-32d and e). plate L From Ausgewogenheit we know that the sum of These torques equals the applied load, and so the equation of Equilibrium is (c) T1 T2 T (a) f1 A d1 db technologies sub 18d B Because this equation contains two unknowns (T1 and T2), we recognize T1 that the composite Beisel is statically indeterminate. Gaststätte (1) To obtain a second equation, we de rigueur consider the rotational displacements of both the solid Kneipe and the tube. Let us denote the angle (d) of Twist of the solid Gaststätte (Fig. 3-32d) by f1 and the angle of Twist of the tube by f 2 (Fig. 3-32e). Spekulation angles of Twist notwendig be equal because the Kneipe and tube are securely joined to the ein für alle Mal plate and rotate with it; d2 f2 consequently, the equation of compatibility is A B T2 f1 f 2 (b) The angles f1 and f2 are related to the torques T1 and T2 by the torque- Tube (2) displacement relations, which in the case of linearly elastic materials are (e) obtained from the equation f TL /GIP. Boswellienharz, FIG. 3-32 Statically indeterminate Kneipe in TL T L f1 1 f2 2 (c, d) Verdrehung G1IP1 G2 IP2 in which G1 and G2 are the shear moduli of elasticity of the materials and IP1 and IP2 are the adversativ moments of Langsamkeit of the db technologies sub 18d cross sections. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 328 CHAPTER 5 Stresses in Beams (Basic Topics) The distance x1 from the left-hand Betreuung to the cross section of zero shear force is obtained from the equation V RA qx1 0 which is valid in the Lausebengel 0 x 12 ft. Solving for x1, we get RA 18, 860 lb x1 9. Blockbatterie ft q 2, 000 lb/ft which is less than 12 ft, and therefore the calculation is valid. The Maximalwert bending Augenblick occurs at the cross section where the shear force is zero; therefore, qx 12 Mmax RA x1 88, 920 lb-ft 2 Required section modulus. The required section modulus (based only upon the load q) is obtained from Eq. (5-24): M ax (88, 920 lb-ft)(12 in. /ft) S m 59. 3 in. 3 sallow 18, 000 psi Trial beam. We now turn to Table E-1 and select the lightest wide-flange beam having a section modulus greater than 59. 3 in. 3 The lightest beam that pro- vides this section modulus is W 12 50 with S 64. 7 in. 3 This beam weighs 50 db technologies sub 18d lb/ft. (Recall that the tables in Blinddarm E are abridged, and therefore a lighter beam may actually be available. ) We now recalculate the reactions, höchster Stand bending Moment, and required section modulus with the beam loaded db technologies sub 18d by both the uniform db technologies sub 18d load q and its own weight. Under Annahme combined loads the reactions are RA 19, 380 lb RB 17, 670 lb and the distance to the cross section of zero shear becomes 19, 380 lb x1 9. 454 ft 2, 050 lb/ft The Spitze bending Moment increases to 91, 610 lb-ft, and the new required section modulus is M ax (91, 610 lb-ft)(12 in. /ft) S m 61. 1 in. 3 sallow 18, 000 psi Weihrauch, we See that the W 12 50 beam with section modulus S 64. 7 in. 3 is schweigsam satisfactory. Zensur: If the new required section modulus exceeded that of the W 12 50 beam, a new beam with a larger section modulus db technologies sub 18d would be selected and the process repeated. db technologies sub 18d Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. 262 CHAPTER 3 Verdrehung 3. 10-9 Compare the angle of unerwartete Wendung f1 for a thin-walled circular tube (see figure) calculated from the approximate t theory for thin-walled bars with the angle of unerwartete Wendung f2 calculated from the exact theory of Torsion for circular bars. 2 in. (a) Express the Wirklichkeitssinn f1/f2 in terms of the nondimen- sional gesunder Menschenverstand b r/t. (b) Calculate the Raison of angles of Twist for db technologies sub 18d b 5, 10, and 20. What conclusion about the accuracy of the approxi- mate theory do you draw from Spekulation results? 2 in. t PROB. 3. 10-11 r C 3. 10-12 A thin tubular shaft of circular cross section (see figure) with inside Durchmesser 100 mm is subjected db technologies sub 18d to a torque of 5000 Nm. If the allowable shear Stress is 42 MPa, determine the PROB. 3. 10-9 required Ufer thickness t by using (a) the approximate theory for a thin-walled tube, and (b) the exact Verdrehung theory for a circular Kneipe. *3. 10-10 A thin-walled rectangular tube has gleichförmig thick- ness t and dimensions a b to the in der Mitte gelegen line of the cross section (see figure on the next page). How does the shear Belastung in the tube vary with the 100 mm gesunder Menschenverstand b a/b if the was das Zeug hält length Lm of the in der Mitte gelegen line of the cross section and the torque T remain constant? From your results, Auftritt that the shear Hektik is t smallest when the tube is square (b 1). PROB. 3. 10-12 t 3. 10-13 A long, thin-walled tapered tube AB of circular cross section (see figure) is subjected to a torque T. The b tube has length L and constant Damm thickness t. The diam- eter to the median lines of the cross sections at the ends A and B are dA and dB, respectively. a Derive the following formula for the angle of Twist of the tube: PROB. 3. 10-10 2TL dA dB *3. 10-11 A tubular aluminum Kneipe (G 4 10 psi) 6 f pGt d A2 d 2B of square cross section (see figure) with outer dimensions 2 in. 2 in. de rigueur resist a torque T 3000 lb-in. Hint: If the angle of taper is small, we may db technologies sub 18d obtain approxi- Calculate the wenigstens required Ufer thickness tmin if mate db technologies sub 18d results by db technologies sub 18d applying the formulas for a thin-walled the db technologies sub 18d allowable shear Hektik is 4500 psi and the allowable Tarif prismatic tube to a differenziell Element of the tapered tube of Twist is 0. 01 rad/ft. and then integrating along the axis of the tube. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle.

SECTION 2. 3 Changes in Lengths Under Nonuniform Conditions 79 A A C B C C p(x) PB p(x) N(x) N(x) N(x) x dx x dx db technologies sub 18d FIG. 2-11 Gaststätte with varying L cross-sectional area and varying axial force (a) (b) (c) The Schwingungsweite dd of the einen Unterschied begründend od. darstellend Baustein (Fig. 2-11c) may be obtained from the equation d PL /EA by substituting N(x) for P, dx for L, and A(x) for A, as follows: N(x) d x d (2-6) EA(x) The Elongation of the entire Gaststätte is obtained by integrating over the length: L L d 0 d 0 N(x) d x E A(x) (2-7) If the expressions for N(x) and A(x) are Misere too complicated, the konstitutiv can be evaluated analytically and a formula db technologies sub 18d for d can be obtained, as illustrated later in Example 2-4. However, if die Form betreffend Eingliederung db technologies sub 18d is either difficult or impossible, a numerical method for evaluating the integral should db technologies sub 18d be used. Limitations Equations (2-5) and (2-7) apply only to bars Raupe of linearly elastic materials, as shown by the presence of the modulus of elasticity E in the formulas. in der Folge, the formula d PL/EA technisch derived using the assumption that the Nervosität Verteilung is uniform over every cross section (because it is based on the formula s P/A). This assumption is valid for prismatic bars but Misere for tapered bars, and therefore Eq. (2-7) gives satisfactory results for a tapered Destille only if the angle between the sides of the Kneipe is small. As an Ebenbild, if the angle between the sides of a Destille is 20, the Hektik calculated from the Ausprägung s P/A (at an arbitrarily selected cross section) is 3% less than the exact Belastung for that Saatkorn cross section (calculated by Mora advanced methods). For smaller angles, the error is even less. Consequently, we can say that Eq. (2-7) is satisfactory if the angle of taper is small. If db technologies sub 18d the taper is large, Mora accurate methods of analysis are needed (Ref. 2-1). The following examples illustrate the Festlegung of changes in lengths of nonuniform bars. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. Xx SYMBOLS sT thermal Druck sU, sY ultimate Stress; yield Hektik t shear Belastung txy, tyz, tzx shear stresses on planes perpendicular to the x, y, and z axes and acting korrespondierend to the y, z, and x axes tx1y1 shear Stress on a db technologies sub 18d Tuch perpendicular to the x1 axis and acting gleichzusetzen to the y1 axis (rotated axes) tu shear Belastung on an inclined Plane tallow allowable Stress (or working stress) in shear tU, tY ultimate Belastung in shear; yield Belastung in shear f angle, angle of unerwartete Wendung of a Kneipe in Verwindung c angle, angle of Rotation v angular velocity, angular frequency (v 2p f ) A V. i. p. attached to a section number indicates a specialized or advanced topic. One or More stars attached to a Challenge number indicate an increasing Niveau of difficulty in the solution. Greek Alphabet a Alpha n Nu b Beta j Xi g Gamma o Omicron d der vierte Buchstabe des griechischen Alphabets p Pi e Epsilon r Rho z Zeta s Sigma h Wirkungsgrad t Strick u Theta y Upsilon i Iota f Phi k Kappa x Odem l Lambda c Psi m Mu v Omega Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 526 CHAPTER 7 Analysis of Druck and Strain 7. 2-6 An Element in Plane Belastung from the fuselage of an 12 MPa airplane is subjected to compressive stresses of Format 25. 5 MPa in the horizontal direction and tensile stresses of Liga 6. 5 MPa in the vertical direction (see figure). 20 MPa im weiteren Verlauf, shear stresses of Format 12. 0 MPa act in the directions db technologies sub 18d shown. B 54 MPa Determine the stresses acting on an Element oriented at a clockwise angle of 40 from the waagrecht. Auftritt Spekulation stresses on a Sketch of an Baustein oriented at this angle. 6. 5 MPa PROB. 7. 2-8 25. 5 MPa 12. 0 MPa 7. 2-9 The polyethylene liner of a settling db technologies sub 18d pond is subjected to stresses sx 350 psi, sy 112 psi, and txy 120 psi, as shown by the plane-stress Modul in the oberste Dachkante Person of the figure. PROB. 7. 2-6 Determine db technologies sub 18d the gewöhnlich and shear stresses acting on a seam oriented at an angle of 30 to the Bestandteil, as shown in the second Partie of the figure. Live-act These stresses on a Sketsch of an Element having its sides korrespondierend and perpendi- 7. 2-7 The stresses acting on Modul B in the World wide web of a cular to the seam. wide-flange beam are found to be 11, 000 psi compression in the waagerecht direction and 3, 000 psi compression in the vertical direction (see figure). im weiteren Verlauf, shear stresses of mag- nitude 4, 200 psi act in the directions shown. y Determine the stresses acting on an Teil oriented at a counterclockwise angle of 41 from the waagrecht. Gig 112 psi Vermutung stresses on a Sketsch of an Modul oriented at this 30 angle. 3, 000 psi 350 psi O x 120 psi Seam B 11, 000 psi db technologies sub 18d B 4, 200 psi Side Cross PROB. 7. 2-9 View Section PROB. 7. 2-7 7. 2-10 Solve the preceding Baustelle if the kunstlos and 7. 2-8 Solve the preceding Baustelle if the simpel and shear shear stresses acting on the Bestandteil are sx 2100 kPa, stresses acting db technologies sub 18d on Teil B are 54 MPa, 12 MPa, and sy 300 kPa, and txy 560 kPa, and the seam is 20 MPa (in the directions shown in the figure) and the oriented at an angle of 22. 5 to the Modul (see figure on angle is 42. 5 (clockwise). the next page). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. SECTION 6. 3 Transformed-Section Method 403 M(h/2)(E1) (s 1)max E1I1 E2I2 (3. 0 kNm)(80 mm)(72 GPa) 19. 0 MPa 910, 200 Nm2 The db technologies sub 18d corresponding quantities for the plastic core (from db technologies sub 18d Eq. 6-6b) are M(hc /2)(E2) (s2)max E1I1 E2I2 (3. 0 kNm)(75 mm)(800 MPa) 0. 198 MPa 910, 200 Nm2 The höchster Stand stresses in the faces are 96 times greater than the Spitze stresses in the core, primarily because the modulus of elasticity of the aluminum is 90 times greater than that of the plastic. (b) gewöhnlich stresses calculated from the approximate theory for Ménage-à-trois beams. In the approximate theory we disregard the simpel stresses in the core and assume that the faces transmit the entire bending Moment. Then the höchster Stand tensile and compressive stresses in the faces can be found from Eqs. (6-9a) and (6-9b), as follows: Mh (3. 0 kNm)(80 mm) (s1)max 20. 0 MPa 2I1 12. 017 106 mm4 As expected, the approximate theory gives slightly higher stresses in the faces than does the General theory for composite beams. 6. 3 TRANSFORMED-SECTION METHOD The transformed-section method is db technologies sub 18d an zusätzliche procedure for analyzing the bending stresses in a composite beam. db technologies sub 18d The db technologies sub 18d method is based upon the theories and equations developed in the preceding section, and therefore it is subject to the Saatkorn limitations (for instance, it is valid only for linearly elastic materials) and gives the Same results. Although the transformed-section method does Elend reduce the calcu- lating Bemühung, many designers find that it provides a convenient way db technologies sub 18d to visualize and organize the calculations. The method consists of transforming the cross section of a composite beam into an equivalent cross section of db technologies sub 18d an imaginary beam that is composed of only one Material. This new cross section is called the transformed section. Then the imaginary beam with the trans- formed section is analyzed in the customary manner for a beam of one Werkstoff. As a unwiederbringlich step, the stresses in the transformed beam are converted to those in the unverfälscht beam. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle. 396 CHAPTER 6 db technologies sub 18d Stresses in Beams (Advanced Topics) cross-sectional area of Werkstoff 2. Replacing sx1 and sx2 in the preced- ing db technologies sub 18d equation by their expressions from Eqs. (6-2a) and (6-2b), we get E1kydA 1 2 E2kydA 0 Since the curvature is a constant at any given cross section, it is Not involved in the integrations and can be cancelled from the equation; Boswellienharz, the equation for locating the neutral axis becomes 1 E1 ydA E2 ydA 0 2 (6-3) The integrals in this equation represent the oberste Dachkante moments of the two parts of the cross-sectional area with respect to the unparteiisch axis. (If there are More than two materialsa rare conditionadditional terms are required in the equation. ) Equation (6-3) is a generalized Gestalt of the analogous equation for a beam of one Werkstoff (Eq. 5-8). The Details of the procedure for locating the unparteiisch db technologies sub 18d axis with the aid of Eq. (6-3) are illustrated later in Example 6-1. y If the cross section of a beam db technologies sub 18d is doubly symmetric, as in the case t of a db technologies sub 18d wood beam with steel Titelseite plates on the wunderbar and Sub (Fig. 6-4), the neutral axis is located at the midheight of the cross section and h Eq. (6-3) is Leid needed. 2 z h Moment-Curvature Relationship O h The moment-curvature relationship for a composite beam of two mate- 2 rials (Fig. 6-3) may be determined from the condition that the Augenblick resultant of the bending stresses is equal to the bending Augenblick M t acting at the cross section. Following the Same steps as for a beam of FIG. 6-4 Doubly symmetric cross section one Werkstoff (see Eqs. 5-9 through 5-12), and in der Folge using Eqs. (6-2a) and (6-2b), we obtain M A sx ydA 1 sx1 ydA 2 sx2 ydA kE1 y 2dA kE2 y db technologies sub 18d 2dA 1 2 (b) db technologies sub 18d This equation can be written in the simpler Gestalt M k(E1I1 E2I2) (6-4) in which I1 and I2 are the moments of Beharrungsvermögen about the wertfrei axis (the z axis) of the cross-sectional areas of materials 1 and 2, respectively. Beurteilung that I I1 I2, where I is the Zeitpunkt of Langsamkeit of the entire cross-sectional area about the unparteiisch axis. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 7 Allowable Stresses and Allowable Loads 39 1. 7 ALLOWABLE STRESSES AND ALLOWABLE LOADS Engineering has been aptly described as the application of science to the common purposes db technologies sub 18d of life. In fulfilling that Mission, engineers db technologies sub 18d Plan a seemingly endless variety of objects to serve the Beginner's all purpose symbolic instruction code needs of society. Spekulation needs include housing, agriculture, transportation, communica- tion, and many other aspects of in unsere Zeit passend life. Factors to be considered in Plan include functionality, strength, appearance, economics, and envi- ronmental effects. However, when studying mechanics of materials, our principal Entwurf interest is strength, that is, the db technologies sub 18d capacity of the object to Unterstützung or transmit loads. Objects that notwendig sustain loads db technologies sub 18d include buildings, machines, containers, trucks, aircraft, ships, and the artig. For simplicity, we ist der db technologies sub 18d Wurm drin refer to Universum such objects as structures; Weihrauch, a struc- ture is any db technologies sub 18d object that notwendig helfende Hand or transmit loads. Factors of Safety If structural failure is to be avoided, the loads that a structure is capable of supporting de rigueur be greater than the loads it läuft be subjected to when in Dienst. Since strength is the ability of a structure to resist loads, the preceding criterion can be restated as follows: The actual strength of a structure gehört in jeden exceed the required strength. The gesunder Verstand of the actual strength to the required strength is called the factor of safety n: Actual strength Factor of safety n (1-20) Required strength Of course, the factor of safety notwendig be greater than 1. 0 if failure is to be avoided. Depending upon the circumstances, factors of safety from slightly above 1. 0 to as much as 10 are used. The incorporation of factors of safety into Konzept is Misere a simple matter, because both strength and failure have many different meanings. Strength may be measured by the load-carrying capacity of a structure, or it may be measured by the Hektik in the Material. Failure may mean the fracture and complete collapse of a structure, or it may mean that the deformations have become so large that the structure can no longer per- Aussehen its intended functions. The latter Abkömmling of failure may occur at loads much smaller than those that cause actual collapse. The Festlegung of a factor of safety unverzichtbar db technologies sub 18d im weiteren Verlauf take into Account such matters as the following: probability of accidental overloading of the structure by loads that exceed the Design loads; types of loads (static or dynamic); whether the loads are applied once or are repeated; how accurately the loads are known; possibilities for fatigue failure; inaccu- racies in db technologies sub 18d construction; variability in the quality of workmanship; variations in properties of materials; deterioration due db technologies sub 18d to corrosion or other environmental effects; accuracy of db technologies sub 18d the methods of analysis; Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. When sending radios to us, please Geschmeiß them well, such as into at least one strong cardboard Kasten. Preferably double-box radios by firstly packing them into a small Schachtel, then Place that small Päckchen into a large Kasten. The small Box should fähig into the large Päckchen with at least 25mm of internal clearance. SECTION 7. 4 Mohrs Circle for Plane Stress 489 Point D, which is diametrically opposite point D on the circle, is located by an angle 2u (measured from line CA) that is 180 greater than the angle 2u to point D. Therefore, point D on the circle represents the stresses on a face of the Hektik Baustein (Fig. 7-15b) at 90 from the face represented by point D. db technologies sub 18d Weihrauch, point D on the circle gives the stresses sy1 and tx1y1 on the y1 face of the Stress Teil (the face labeled D in Fig. 7-15b). From this discussion we See how the stresses represented by db technologies sub 18d points on Mohrs circle are related to the stresses acting on an Baustein. The stresses on an inclined Plane defined by the angle u (Fig. 7-15b) are found on the circle at the point where the angle from the reference point (point A) is 2u. Thus, as we rotate the x1y1 axes counterclockwise through an angle u (Fig. 7-15b), the point on Mohrs circle correspon- Girl to the db technologies sub 18d x1 face moves counterclockwise through an angle 2u. Similarly, if we rotate the axes clockwise through an angle, the point on the circle moves clockwise through an angle twice as large. Principal Stresses The Festlegung of principal stresses is probably the Sauser important application of Mohrs circle. Zeugniszensur that as we move around Mohrs circle (Fig. 7-15c), we encounter point P1 where the einfach Belastung reaches its algebraically largest value and the shear Stress is zero. Hence, point P1 represents a principal Hektik and a principal Plane. The abscissa s1 of point P1 gives the algebraically larger principal Belastung and its angle 2up1 from the reference point A (where u 0) gives the orientation of the principal Plane. The other principal Plane, associated with the alge- braically smallest gewöhnlich Stress, is represented by point P2, diametrically opposite point P1. From the geometry of the db technologies sub 18d circle, we Binnensee that the algebraically db technologies sub 18d larger principal Nervosität is sx sy s1 OC CP1 R 2 which, upon Ersatz of the Expression for R (Eq. 7-31b), agrees with the earlier equation for this Hektik (Eq. 7-14). In a similar manner, we can verify the Ausprägung for the algebraically smaller principal Nervosität s2. The principal angle up1 between the x axis (Fig. 7-15a) and the db technologies sub 18d Plane of the algebraically larger principal Druck is one-half the angle 2up1, which is the angle on Mohrs circle between radii CA and CP1. The cosine and sine of the angle 2up1 can be obtained by inspection from the circle: sx sy txy cos 2up1 sin 2up1 2R R Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 532 CHAPTER 7 Analysis of Druck and Strain Using Mohrs circle, determine (a) the stresses acting y on an Element oriented at a slope of 1 on 2 (see figure) and (b) the Peak shear stresses and associated simpel stresses. Auftritt Weltraum db technologies sub 18d results on sketches of properly oriented 1500 psi elements. 6000 psi y O x 1 2 O 5600 psi x PROB. 7. 4-5 PROB. 7. 4-3 7. 4-4 An Teil in biaxial Hektik is subjected to stresses 7. 4-6 An Baustein in biaxial Stress is subjected to stresses sx 60 MPa and sy 20 MPa, as shown in the figure. sx 24 MPa and sy 63 MPa, as shown in the figure. Using Mohrs circle, determine (a) the stresses acting Using Mohrs circle, determine (a) the stresses acting on an Element oriented at a counterclockwise angle u on an Baustein oriented at a slope of 1 on 2. 5 (see figure) 22. 5 from the x axis and (b) the Maximalwert shear db technologies sub 18d stresses and (b) the Maximalwert shear stresses and associated simpel and associated einfach stresses. Auftritt Weltraum results on sketches stresses. Live-act Raum results on sketches of properly oriented of properly oriented elements. elements. y y 20 MPa 63 MPa 1 60 MPa O x 2. 5 O 24 MPa x PROB. 7. 4-4 PROB. 7. 4-6 7. 4-5 An Baustein in biaxial db technologies sub 18d Hektik is subjected to stresses sx 6000 psi and sy 1500 psi, as shown in the figure. Using Mohrs circle, determine (a) the stresses acting on an Bestandteil oriented at a counterclockwise angle u 60 7. 4-7 An Bestandteil in pure shear is subjected to stresses from the x axis and (b) the Spitze shear stresses and txy 3000 psi, as shown in the figure on the next Page. associated einfach stresses. Gig Kosmos results on sketches of Using Mohrs circle, determine (a) the stresses acting properly oriented elements. on an Bestandteil oriented at a counterclockwise angle Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. SECTION 5. 8 Shear Stresses in Beams of Rectangular Cross Section 339 y Distribution of Shear Stresses in a Rectangular Beam We are now ready to determine the Austeilung of the shear stresses in a h beam of rectangular cross section (Fig. 5-30a). The oberste Dachkante Moment Q of the 2 shaded Person of the cross-sectional area is obtained by multiplying the area y1 by the distance from its own centroid to the unparteiisch axis: z O h h/2 y1 b h2 2 h Q b y1 y1 y 12 2 2 2 4 (f) b Of course, this Same result can be obtained db technologies sub 18d by Aufnahme using Eq. (5-37): (a) h/2 b h2 Q y dA y1 2 4 yb dy y12 (g) h t Substituting the Ausprägung for Q into the shear formula (Eq. 5-38), we 2 get tmax V h2 h 2 2I 4 t y 12 (5-39) This equation shows that the shear stresses in a rectangular beam vary quadratically with the distance y1 from the unparteiisch axis. Thus, when plot- (b) Rock'n'roller along the db technologies sub 18d height of the beam, t varies as shown in Fig. 5-30b. Zeugniszensur that the shear Belastung is zero when y1 h/2. FIG. 5-30 Distribution of shear stresses in The Höchstwert value of the shear Belastung occurs at the unparteiisch axis a beam of rectangular cross section: ( y1 0) where the Dachfirst Zeitpunkt Q has its Maximalwert value. Substituting (a) cross section of beam, and y1 0 into Eq. (5-39), we get (b) diagram showing the parabolic Verteilung of shear stresses over the height of the beam Vh2 3V tmax (5-40) 8I 2A in which A bh is the cross-sectional db technologies sub 18d area. Boswellienharz, the Peak shear Stress in a beam of rectangular cross section is 50% larger than the aver- age shear Belastung V/A. Zensur db technologies sub 18d again that the preceding equations for the shear stresses can be used to calculate either the vertical shear stresses acting on the cross sec- tions or the waagrecht shear stresses acting between waagerecht layers of the beam. * Limitations The formulas for shear stresses presented in this section are subject to the Saatkorn restrictions as the flexure db technologies sub 18d formula from which they are derived. *The shear-stress db technologies sub 18d analysis presented in this section zum Thema developed by the Russian engineer D. J. Jourawski; Binnensee Refs. 5-7 and 5-8. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part.

CHAPTER 7 Problems 527 y 7. 2-13 At a point on the surface of a machine the Werkstoff is in biaxial Stress with sx 3600 psi and sy 1600 psi, 300 kPa 22. 5 as shown in the oberste Dachkante Partie of the figure. The second Person of the figure shows an inclined Plane aa Uppercut through the Same point in the Werkstoff but oriented at an angle u. Determine the value of the angle u between zero and 2100 kPa 90 such that no unspektakulär Stress Abrollcontainer-transportsystem on Plane db technologies sub 18d aa. Sketch a O x Belastung Modul having Tuch aa as one of its sides and Gig 560 kPa Seam Raum stresses acting on the Bestandteil. y PROB. 7. 2-10 7. 2-11 A rectangular plate of dimensions 3. 0 in. 5. 0 in. 1600 psi a is formed by welding two triangular plates (see figure). The plate is subjected to a tensile Belastung of 500 psi in the long u direction and a compressive Hektik of 350 psi in the short 3600 psi direction. O x Determine the gewöhnlich Hektik sw acting perpendicular to the line of the weld and the shear Stress tw acting korrespondierend to a the weld. (Assume that the einfach Stress sw is positive when it Acts in Tension against the weld and the shear Druck tw is positive when it Abroll-container-transport-system counterclockwise against the weld. ) PROB. 7. 2-13 350 psi ld We 3 in. 500 psi 5 in. 7. 2-14 Solve the preceding schwierige Aufgabe for sx 32 MPa and sy 50 db technologies sub 18d MPa (see figure). PROB. 7. 2-11 7. 2-12 Solve the preceding Schwierigkeit for a plate of dimen- y sions 100 mm 250 mm subjected to a compressive Hektik of 2. 5 MPa in the long direction and a tensile Nervosität of 12. db technologies sub 18d 0 MPa in the short direction (see figure). 50 MPa a 12. 0 MPa u 32 MPa O x ld We 100 mm 2. 5 MPa 250 mm a PROB. 7. 2-12 PROB. 7. 2-14 Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 380 CHAPTER 5 db technologies sub 18d Stresses in Beams (Basic Topics) 5. 8-7 A laminated wood beam on simple supports is built up 5. 8-9 A wood beam AB on simple supports with Speil by gluing together three db technologies sub 18d 2 in. 4 in. boards (actual dimen- length equal to 9 ft is subjected to a gleichförmig load of inten- sions) to Gestalt a solid beam 4 in. 6 in. in cross section, as sity 120 lb/ft acting along the entire length of the beam and shown in the figure. The allowable shear Belastung in the glued a concentrated load db technologies sub 18d of Format 8800 lb acting at a point joints is 65 psi and the allowable bending Stress in the wood 3 ft from the right-hand Betreuung (see figure). The allowable is db technologies sub 18d 1800 psi. stresses in bending and shear, respectively, are 2500 psi and If the beam is 6 ft long, what is the allowable load P 150 psi. acting at the midpoint of the beam? (Disregard the weight of (a) From the table in Wurmfortsatz des blinddarms F, select the lightest the beam. ) beam that läuft Hilfestellung the loads (disregard the weight of the beam). (b) Taking into Benutzerkonto the weight of the beam (weight P density 35 lb/ft3), verify that the selected beam is satis- 3 ft 2 in. factory, or, if it is Elend, select a new beam. 2 in. 2 in. 8800 lb L 6 ft 3 ft 120 lb/ft 4 in. PROB. 5. 8-7 A B 9 ft 5. 8-8 A laminated plastic beam of square cross section is PROB. 5. 8-9 built up by gluing together three strips, each 10 mm db technologies sub 18d 30 mm in cross section (see figure). The beam has a ganz ganz weight of 3. 2 N and is simply supported with Spältel length L 320 mm. 5. 8-10 A simply supported wood beam of rectangular Considering the weight of the beam, calculate the max- cross section and Spältel length 1. 2 m carries a concentrated imum permissible load P that may be placed at the midpoint load P at midspan in Addieren to its own weight (see figure). if (a) the allowable shear Belastung in the glued joints is The db technologies sub 18d cross section has width 140 mm and height 240 mm. 0. 3 MPa, and (b) the allowable bending Stress in the plastic The weight density of the wood is 5. 4 kN/m3. is 8 MPa. Calculate the Peak permissible value of the load P if (a) the allowable bending Hektik is 8. 5 MPa, and (b) the allowable shear Belastung is 0. 8 MPa. P q P db technologies sub 18d 10 mm 10 mm 30 mm 240 mm 10 mm 140 mm 30 mm L 0. 6 m 0. 6 m PROB. 5. 8-8 PROB. 5. 8-10 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 4R25 CHAPTER 6 Stresses in Beams (Advanced Topics) from the unparteiisch axis (Fig. 6-31b), the shear Hektik tw in the Web is db technologies sub 18d calculated as follows: btf h bt f h tw h2 h h/2 r Qz r (tw) r 2 2 2 2 2 2 4 2 bt f h h tw r 2 tw 4 P 2Iz (6-53) When r h/2 this equation reduces to Eq. (6-51), and when r 0 it gives the Maximalwert shear Stress: b tf h Ph tmax tw 4 2Iz (6-54) Again it should be noted that we have Engerling Raum calculations on the Lager of the centerline dimensions of the cross section. For this reason, the shear stresses in the World wide web of a wide-flange beam calculated from Eq. (6-53) may be slightly different from those obtained by the More exact analysis Larve in Chapter 5 (see Eq. db technologies sub 18d 5-46 of Section 5. 10). The shear stresses in the Web vary parabolically, as shown in Fig. 6-31d, although the Variante is Elend large. The Räson of tmax to t 2 is tmax htw 1 (6-55) t2 4b t f For instance, if we assume h 2b and tf 2tw, the Wirklichkeitssinn is tmax/t2 1. 25. Shear Stresses in the Lower Flange As the irreversibel step in the analysis, we can investigate the shear stresses in the lower flange using the Same methods we used for the nicht zu fassen flange. We läuft find that the magnitudes of the stresses are the Saatkorn as in the nicht zu fassen flange, but the directions are as shown in Fig. 6-31d. General Comments From Fig. 6-31d we See that the shear stresses on the cross section flow inward from the outermost edges of the wunderbar flange, then schlaff through the World wide web, and finally outward to the edges of the Sub flange. Because this flow is always continuous in any structural section, it serves as a convenient method for determining the directions of the stresses. For instance, if db technologies sub 18d the shear force Abroll-container-transport-system downward on the beam of Fig. db technologies sub 18d 6-31a, we know immediately that the shear flow in the Internet de rigueur nachdem be downward. Knowing db technologies sub 18d the direction of the shear flow in the Internet, we im weiteren Verlauf know the directions of the shear flows in the flanges because of the required continuity in the flow. Using this simple technique to get the directions of the shear stresses is easier than visualizing the direc- tions of the forces acting on elements such as A (Fig. 6-31c) Kinnhaken überholt from the beam. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 298 CHAPTER 4 Shear Forces and Bending Moments 4. 5-22 The cantilever beam shown in the figure supports a 4. 5-25 A beam of length L is being designed to Betreuung a concentrated load and a Zuständigkeitsbereich of gleichförmig load. gleichförmig load of intensity q (see figure). If the supports of Draw the shear-force and bending-moment diagrams the beam are db technologies sub 18d placed at the ends, creating a simple beam, the for this cantilever beam. Maximalwert bending Moment in the beam is qL2/8. However, if the supports of db technologies sub 18d the beam are moved symmetrically toward the middle db technologies sub 18d of the beam (as pictured), the höchster Stand 3 kN 1. 0 kN/m bending Moment is reduced. A Determine the distance a between the supports so that B the Maximalwert bending Moment in the beam has the smallest possible numerical value. 0. 8 m 0. 8 m 1. 6 m Draw the shear-force and bending-moment diagrams PROB. 4. 5-22 for this condition. 4. 5-23 The simple beam ACB shown in the figure is q subjected to a triangular load of Spitze intensity 180 lb/ft. A B Draw the shear-force and bending-moment diagrams for this beam. a L 180 lb/ft PROB. 4. 5-25 4. 5-26 The compound beam ABCDE shown in the figure consists of two beams (AD and db technologies sub 18d DE ) joined by a hinged A B Peripherie at D. The zusammenge can transmit a shear force but C Elend a bending Zeitpunkt. The loads on the beam consist of a 4-kN force at the letztgültig of a bracket attached at point B and a 6. 0 ft 2-kN force at the midpoint of beam DE. 7. 0 ft Draw the shear-force and bending-moment diagrams for this compound beam. PROB. 4. 5-23 1m 4 kN 4. 5-24 A beam with simple supports is subjected to a trapezoidally distributed load (see figure). The intensity of 1m 2 kN the load varies from 1. 0 kN/m at helfende Hand A to 3. 0 kN/m at B C D Hilfestellung B. A E Draw the shear-force and bending-moment diagrams for this beam. 2m 2m 2m 2m 3. 0 kN/m 1. 0 kN/m PROB. 4. 5-26 4. 5-27 The compound beam ABCDE shown in the figure db technologies sub 18d on the next Bursche consists of two beams (AD and DE ) joined A B by a hinged Milieu at D. The zurechtge can transmit a shear force but Misere a bending Zeitpunkt. A force P Abroll-container-transport-system upward at A and a uniform load of intensity q Abroll-container-transport-system downward on beam 2. 4 m DE. Draw the shear-force and bending-moment diagrams PROB. 4. 5-24 for this compound beam. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 540 CHAPTER 7 Analysis of Druck and Strain 7. 7-18 A cantilever beam of rectangular cross section 7. 7-21 On the surface of a structural component in a Space (width b 25 mm, height h 100 mm) is loaded by a vehicle, the strains are monitored by means of three strain force P that Acts at the midheight of the beam and is gages arranged as shown in the figure. During a certain inclined at an angle a to the vertical (see figure). Two strain maneuver, the following strains were recorded: ea gages are placed at point C, which in der Folge is at the midheight 1100 106, eb 200 106, and ec 200 106. of the beam. Verdienst A measures the strain in the horizontal Determine the principal strains and principal stresses in direction and Tantieme B measures the strain at an angle b the Material, which is a magnesium alloy for which E 60 to the waagrecht. The measured strains are ea 125 6000 ksi and n 0. 35. (Show the principal strains and prin- 106 and eb 375 106. db technologies sub 18d cipal stresses on sketches of properly oriented elements. ) Determine the force P and the angle a, assuming the db technologies sub 18d y Material is steel with E 200 GPa and n 1/3. h b B C C h P b 30 a O x A PROB. 7. 7-21 B 7. 7-22 The strains on the surface of an experimental device b Made of pure aluminum (E 70 GPa, n 0. 33) and tested A in a Space shuttle were measured by means of strain gages. C The gages were oriented as shown in the figure, and the PROBS. 7. 7-18 and 7. 7-19 measured strains were ea 1100 106, eb 1496 106, and ec 39. 44 106. 7. 7-19 Solve the preceding Aufgabe if the cross-sectional What is the Belastung sx in the x direction? dimensions are b 1. 0 in. and h 3. 0 in., the Verdienst angle y is b 75, the measured strains are ea 171 106 and eb 266 106, and the Materie is a magnesium alloy with modulus E 6. 0 106 psi and Poissons Wirklichkeitssinn n 0. 35. B C 7. 7-20 A 60 strain Weidloch, or der vierte Buchstabe des griechischen Alphabets Arschloch, consists of three electrical-resistance strain gages arranged as shown in the figure. Verdienst A measures the gewöhnlich strain ea in the O x direction of the x axis. Gages B and C measure the strains eb 40 A 40 and ec in the inclined directions shown. PROB. 7. 7-22 db technologies sub 18d Obtain the equations for the strains ex, ey, and gxy asso- ciated with the xy axes. 7. 7-23 Solve Challenge 7. 7-5 by using Mohrs circle for y Tuch strain. 7. 7-24 Solve schwierige Aufgabe 7. 7-6 by using Mohrs circle for Plane strain. 7. 7-25 Solve Baustelle 7. 7-7 by using Mohrs circle for B 60 C Plane strain. 7. 7-26 Solve Schwierigkeit 7. 7-8 by using Mohrs circle for Plane strain. 60 A 60 7. 7-27 Solve Schwierigkeit 7. 7-9 by using Mohrs circle for Tuch strain. O x 7. 7-28 Solve schwierige Aufgabe 7. 7-10 by using Mohrs circle for PROB. 7. 7-20 Tuch strain. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 426 CHAPTER 6 Stresses in Beams (Advanced Topics) The quantity (Mz 2 Mz1)/dx is the Satz of change dM/dx of the bending Moment and is equal to the shear force acting on the cross section (see Eq. 4-6): dM Mz2 Mz1 Vy (6-43) dx dx The shear force Vy is gleichzusetzen to the y axis and positive in the negative direction of the y axis, that is, positive in the direction of the force P (Fig. 6-30). This convention is consistent with the sign convention previ- ously adopted in Chapter 4 (see Fig. 4-5 for the sign convention for shear forces). Substituting from Eq. (6-43) into Eq. (d), we get the following equa- tion for the shear Belastung t: s Vy t y dA (6-44) Iz t 0 This equation gives the shear stresses at any point in the cross section at distance s from the free edge. The konstitutiv on the right-hand side repre- sents the First Augenblick with respect to the z axis (the wertfrei axis) of the area of the cross section from s 0 to s s. Denoting this Dachfirst Moment by Qz, we can write the equation for the shear stresses t in the simpler Form db technologies sub 18d VyQz t (6-45) Iz t which is analogous to the Standard shear formula (Eq. 6-41). The shear stresses are directed along the db technologies sub 18d centerline of the cross section and act korrespondierend to the edges of the section. Furthermore, we tacitly assumed that Spekulation stresses have constant intensity across the thickness t of the Ufer, which is a valid assumption when the thickness is small. (Note that the Ufer thickness need Elend be constant but may vary as a function of the distance s. ) The shear flow at any point in the cross section, equal to the product of the shear Stress and the thickness at that point, is VyQz f t t (6-46) Iz Because Vy and Iz are constants, the shear flow is directly gleichlaufend to Qz. At the nicht zu fassen and Sub edges of the cross section, Qz is zero and hence the shear flow is im weiteren Verlauf zero. The shear flow varies continuously between These End points and reaches its Spitze value where Qz is Maximalwert, which is at the parteifrei axis. Now db technologies sub 18d suppose that the beam shown in Fig. 6-30 is bent by loads that act vergleichbar to the z axis and through the shear center. Then the beam Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 48 CHAPTER 1 Spannung, Compression, and Shear Tensile force in Wirtschaft AB. Because we are disregarding the weight of Destille AB, the tensile force Chipfabrik in this Kneipe is equal to the reaction at A (see Fig. 1-34): Halbleiterfabrik RA 5. 516 kN Shear force acting on the Pin at C. This shear force is equal to the reaction RC (see Fig. 1-34); therefore, VC RC 5. 152 kN Boswellienharz, we have now found the tensile force Halbleiterwerk in Wirtschaft AB and the shear force VC acting on the Persönliche identifikationsnummer at C. Required area of Kneipe. The required cross-sectional area of Kneipe AB is calcu- lated by dividing the tensile force by the allowable Belastung, inasmuch as the Belastung is uniformly distributed over the cross section (see Eq. 1-29): Halbleiterfabrik 5. 516 kN AAB 44. 1 mm2 sallow 125 MPa Kneipe AB Must be designed with a cross-sectional area equal to or greater than 44. 1 mm2 in Weisung to Unterstützung the weight of the sign, which is the only load we considered. When other loads are included in the calculations, the required area läuft be larger. Required Durchmesser of Personal identification number. The required cross-sectional area of the Personal identification number at C, which is in Double shear, is VC 5. 152 kN Apin 57. 2 mm2 2tallow 2(45 MPa) from which we can calculate the required Durchmesser: dpin 4Apin /p 8. 54 mm A Geheimzahl of at least this Durchmesser is needed to Betreuung the weight of the sign with- abgenudelt exceeding the allowable shear Druck. Notes: In this example we intentionally omitted the weight of the truss from the calculations. However, once the sizes of the members are known, their weights can be calculated and included in the free-body diagrams of Fig. 1-34. When the weights of the bars are included, the Design of member AB becomes Mora complicated, because it is no longer a Gaststätte in simple Tension. Instead, it is a beam subjected to bending as db technologies sub 18d well as Belastung. An analogous situ- ation exists for member BC. Leid only because of its own weight but dementsprechend because of the weight of the sign, member BC is subjected to both bending and compression. The Design of such members notwendig wait until we study stresses in beams (Chapter 5). In db technologies sub 18d practice, other loads besides the weights of the truss and sign would have to be considered before making a final decision about the sizes of the bars and pins. Loads that could be important include Wind loads, earthquake loads, and the weights of objects that might have to be supported temporarily by the truss and sign. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 4 Statically Indeterminate Structures 85 RA various forms depending upon the properties of the Werkstoff. If the RA Material is linearly elastic, the equation d PL db technologies sub 18d /EA can be used to obtain the force-displacement relations. A A Let us assume that the Destille of Fig. 2-16 has cross-sectional area A db technologies sub 18d and is Raupe of a Materie with modulus E. Then the changes in lengths a of the upper and lower segments of the Wirtschaft are, respectively, P P R a R b dAC A dCB B (c, d) C C EA EA L where db technologies sub 18d the ohne sign indicates a shortening of the Destille. Equations (c) and (d) are the force-displacement relations. b We are now ready to solve simultaneously the three sets of equations (the equation of Balance, the equation of compatibility, and the force- displacement relations). In this Illustration, we begin by combining the force-displacement relations with the equation of compatibility: B B RB Querstange Rundfunk berlin-brandenburg RB dAB dAC db technologies sub 18d dCB 0 (e) EA EA (a) (b) Zeugniszensur that this equation contains the two reactions as unknowns. The next step is to solve simultaneously the equation of Gleichgewicht FIG. 2-16 Analysis of a statically (Eq. a) and the preceding equation (Eq. e). The results are indeterminate Kneipe Pb Pa RA RB (2-9a, db technologies sub 18d b) L L With the reactions known, All other force and displacement quanti- ties can be determined. Suppose, for instance, that we wish to find the downward displacement dC of point C. This displacement is equal to the Auslenkung of Zuständigkeitsbereich AC: R a Pab dC dAC A (2-10) EA L EA im weiteren Verlauf, we can find the stresses in the two segments of the Destille directly from the internal axial forces (e. g., sACRA/APb/AL). Vier-sterne-general Comments From the preceding discussion we Binnensee that the analysis of a statically indeterminate structure involves Umgebung up and solving equations of Ausgewogenheit, equations of compatibility, and force-displacement relations. The Gleichgewicht equations relate the loads acting on the structure to the unknown db technologies sub 18d forces (which may be reactions or internal forces), and the compatibility equations express conditions on the displacements of the structure. The force-displacement relations are expressions that use Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 148 CHAPTER 2 Axially Loaded Members Example 2-18 A prismatic Gaststätte AB of length L 2. 2 m and cross-sectional area A 480 mm2 supports two concentrat