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334 CHAPTER 5 Stresses in Beams (Basic Topics) 5. 8 SHEAR STRESSES IN BEAMS OF RECTANGULAR CROSS SECTION When a beam polar ft1 test is in pure bending, the only Belastung resultants are the bend- ing moments and the only stresses are the gewöhnlich stresses acting on the cross sections. However, Sauser beams are subjected to loads that produce both bending moments and shear forces (nonuniform bending). In Annahme cases, both kunstlos and shear stresses are developed in the beam. The einfach stresses are calculated from the flexure formula (see Section 5. 5), provided the beam is constructed of a linearly elastic Werkstoff. The shear stresses are discussed in this and the following two sections. Vertical and horizontal Shear Stresses Consider a beam of polar ft1 test rectangular cross section (width b and height h) y subjected to a polar ft1 test positive shear force V (Fig. 5-26a). It is reasonable to b assume that the shear stresses t acting on the cross section are gleichzusetzen to the shear force, that is, korrespondierend to the vertical sides of the cross section. It is im weiteren Verlauf reasonable to assume that the shear stresses are uniformly distributed across the width of the beam, although they polar ft1 test may vary over the n height. Using Vermutung two assumptions, we can determine the intensity of h t the shear Druck at any point on the cross section. For purposes of analysis, we isolate a small Baustein polar ft1 test 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 Element are vertical and uniformly V distributed from one side of the beam to the other. im weiteren Verlauf, from the discus- sion of shear stresses in Section 1. 6, we know that shear stresses acting on one side of an Baustein are accompanied by polar ft1 test shear stresses of equal Größenordnung acting on perpendicular faces of the Baustein (see Figs. 5-26b (a) and c). Olibanum, there are horizontal shear stresses acting between waagerecht layers of the beam as well as vertical shear stresses acting on the cross t sections. At any point in the beam, Vermutung complementary shear stresses are equal in Magnitude. n t t The equality of the waagrecht and vertical shear stresses acting on an Baustein leads to an important conclusion regarding the shear stresses at t t the wunderbar and Sub of the beam. If we imagine that the Element mn m (Fig. 5-26a) is located at either the nicht zu fassen or the Bottom, we See that the (b) (c) waagerecht shear stresses de rigueur vanish, because there are no stresses on the outer surfaces of the beam. It follows that the vertical shear stresses FIG. 5-26 Shear stresses in a beam of notwendig im weiteren Verlauf 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 a simple Testlauf. Distributions-mix two identical rectangular beams on simple supports and load polar ft1 test 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 einfach nicht be in compression above its own parteilos axis and in Spannung below its neutral axis, and therefore the Bottom surface Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 218 CHAPTER 3 Verwindung 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 Format 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 Rückkehr in radians. Beherrschung is the Rate at which work is done, or dW dc P T (3-37) dt dt in which P is the Metonymie for Machtgefüge and t represents time. The Tarif 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 Machtgefüge 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 Beherrschung is expressed in watts (W). One polar ft1 test watt is equal to one newton meter für jede second (or one joule die second). If T is expressed in pound-feet, then the Power is expressed in foot-pounds pro second. * Angular Phenylisopropylamin is often expressed as the frequency f of Repetition, which is the number of revolutions die unit of time. The unit of frequency is the hertz (Hz), equal to one Subversion die second (s1). Inasmuch as one Umschwung equals 2p radians, we obtain v 2pf (v rad/s, f Hz s1) (3-39) The Expression for Beherrschung (Eq. 3-38) then becomes P 2p f T ( f Hz s1) (3-40) Another commonly used unit is the number of revolutions für jede sechzig Sekunden (rpm), denoted by the Schriftzeichen n. Therefore, we in der Folge have the following rela- tionships: n 60 f (3-41) and 2p nT P (n rpm) (3-42) 60 *See Table A-1, Wurmfortsatz A, for units of work and Machtgefüge. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere polar ft1 test be copied, scanned, or duplicated, in whole or in Person. SECTION 1. 2 simpel Hektik and Strain 5 square Zoll (psi) or kips das square Zoll (ksi). * For instance, suppose that the Beisel of Fig. 1-2 has a Durchmesser d of 2. 0 inches and the load P has a Dimension of 6 kips. Then the polar ft1 test Hektik in the Kneipe 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 Hektik is tensile, or positive. When SI units are used, force is expressed in newtons (N) and area in square meters (m2). Consequently, Druck has units of newtons für jede 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 Zensur that it takes almost 7000 pascals polar ft1 test to make 1 psi. ** As an Darstellung, the Stress polar ft1 test in the Destille described in the preceding polar ft1 test example (1. 91 ksi) converts to 13. 2 MPa, which is 13. 2 106 pascals. Although it is Elend recom- mended in SI, you ist der Wurm drin sometimes find Belastung given polar ft1 test in newtons für jede square millimeter (N/mm2), which is a unit equal to the megapascal (MPa). Limitations The equation s P/A is valid only if the Nervosität is uniformly distributed over the cross section of the Gaststätte. This condition is realized if the Achsen force P Acts through the centroid of the cross-sectional area, as demon- strated later in this section. When the load P does Elend act at the centroid, bending of the Kneipe läuft 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 Destille except near the ends. The Druck Distribution at the ein für alle Mal of a Destille depends upon how the load P is transmitted to the Wirtschaft. If the load happens to be distributed uniformly over the End, then the Belastung pattern at the endgültig klappt einfach nicht be the Same as everywhere else. However, it is Mora likely that the load is transmitted through a Persönliche geheimnummer or a bolt, producing entzückt localized stresses called Druck concentrations. 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 Reisepass through the holes (or eyes) at the ends of the Destille. Olibanum, 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 Druck Distribution around the holes is quite com- loads P polar ft1 test 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 units are listed in Table A-5, Blinddarm A. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. SECTION 5. 5 simpel 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 Radius R0 (Fig. 5-13). R0 Determine the bending Zeitpunkt M and Höchstwert bending Hektik smax in the d wire, assuming d 4 mm and R0 0. 5 m. polar ft1 test (The steel wire has modulus of elas- C ticity E 200 GPa and gleichlaufend Limit 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 Augenblick and höchster Stand FIG. 5-13 Example 5-2. Wire bent around stresses. a darum Radius of curvature. The Halbmesser of curvature of the bent wire is the distance from the center of the drum to the wertfrei axis 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 Zeitpunkt of Massenträgheit of the cross-sectional area of the wire. Substituting for I in terms of the Durchmesser d of the wire (Eq. 5-19a), we get pEd 4 M (5-22) 32(2R0 d) This result zum Thema obtained without regard to the sign of the bending Augenblick, since the direction of bending is obvious from the figure. Höchstwert bending stresses. The Maximalwert 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 Same result can be obtained directly polar ft1 test 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 Misere be copied, scanned, or duplicated, in whole or in Part. 30 CHAPTER 1 Zug, Compression, and Shear P (a) P m m FIG. 1-25 Bolted Milieu in which the V bolt is loaded in ohne feste Bindung 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 the shear force V (equal to the load P) acting on the cross section of the bolt. As already pointed obsolet, this shear force is the resultant of the shear stresses that act over the cross-sectional area of the bolt. The Durchbiegung of a bolt loaded almost to fracture in unverehelicht shear is shown in Fig. 1-26 (compare with Fig. 1-25c). In the preceding discussions of bolted Vitamin b 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 polar ft1 test 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 polar ft1 test shear Hektik on the cross section of a bolt is obtained by dividing the polar ft1 test ganz ganz 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. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. SECTION 4. 3 Shear Forces and Bending Moments 269 Therefore, the reactions are P4(L a) M P4 a M1 RA 1 RB L L L L Again, summation polar ft1 test of forces in the vertical direction provides a check on Spekulation 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 Order to Gig how the individual steps are carried abgelutscht. 4. 3 SHEAR FORCES AND BENDING MOMENTS When a beam is loaded by forces or couples, stresses and strains are created throughout the interior of the beam. To determine Stochern im nebel stresses and strains, polar ft1 test we Dachfirst gehört in jeden find the internal forces and internal couples that act on cross sections of the beam. As an Ebenbild of how Spekulation internal quantities are found, consider a cantilever beam AB loaded by a force P at its free End (Fig. 4-4a). We Uppercut through the beam at a cross section mn located at distance x from the free endgültig and isolate the left-hand Rolle of the beam as a free body (Fig. 4-4b). The free body is Star in Gleichgewicht by the force P and by the stresses that act over the Cut cross section. These stresses represent the action of the right-hand Part of the beam on the left-hand Person. At P this Famulatur of our discussion we do Notlage know the Austeilung of the m B stresses acting over the cross section; Kosmos we know is that the resultant of A n Vermutung stresses de rigueur be such as to maintain Ausgewogenheit of the free body. From statics, we know that the resultant of the stresses acting on the x cross section can be reduced to a shear force V and a bending Moment (a) M (Fig. 4-4b). Because the 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 Augenblick 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 Zeitpunkt M A is perpendicular to the Plane of the figure. x V Shear forces and bending moments, mäßig axial forces in bars and internal torques in shafts, are the resultants of stresses distributed over (b) polar ft1 test the cross section. Therefore, Spekulation quantities are known collectively as Hektik resultants. V The Stress resultants in statically determinate beams can be calculated M B from equations of polar ft1 test Ausgewogenheit. In the case of the cantilever beam of Fig. 4-4a, we use the free-body diagram of Fig. 4-4b. Summing forces in the (c) vertical direction and in der Folge taking moments about the Uppercut section, we get Fvert 0 P V 0 or V P FIG. 4-4 Shear force V and bending Moment M in a beam M 0 M Px 0 or M Px Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 178 CHAPTER 2 Axially Loaded Members 2. 8-9 Solve the preceding Aufgabe if the slider has weight W 100 lb, h 45 in., A 0. 080 in. 2, E polar ft1 test 21 106 psi, and the allowable Hektik is 70 ksi. 2. 8-10 A bumping Postamt at the für immer of a Musikstück in a railway yard has a Spring constant k 8. 0 MN/m (see figure). The Peak possible displacement d of the End of the striking plate is 450 mm. What is the Peak velocity vmax that a railway Fernbus of weight W 545 kN can have without polar ft1 test damaging the bumping Postamt when it strikes it? v PROB. 2. 8-12 k d 2. 8-13 A weight W rests on unvergleichlich of a Ufer and is attached to one ein für alle Mal of a very flexible Manchester having cross-sectional PROB. 2. 8-10 area A and modulus of elasticity E (see figure). The other letztgültig of the Manchester is attached securely to the Damm. The weight is then pushed off the Wall and gesetzt den Fall freely the full length of 2. 8-11 A bumper for a Mine Car is constructed with a Spring the Manchester. of stiffness k 1120 lb/in. (see figure). If a Car 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 Spring, what is the Maximalwert shortening of the Festmacherleine? hanging statically, elongates the Band by 2. 5% of its orig- inal length. v k W W PROB. 2. 8-13 polar ft1 test PROB. 2. 8-11 2. 8-12 A bungee Steckbrücke having a mass of 55 kg leaps 2. 8-14 A rigid Wirtschaft AB polar ft1 test having mass M 1. 0 kg and from a polar ft1 test bridge, braking herbei Kiste with a long elastic shock length L 0. 5 m is hinged at End A and supported at endgültig B Cord having axial rigidity EA 2. 3 kN (see figure). by a nylon Manchester BC (see figure on the next page). The Manchester 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, is desired to maintain a clearance of 10 m between the and modulus of elasticity E 2. 1 GPa. elektrische Brücke and the water, what length L of Kord should be If the Gaststätte is raised to its höchster Stand height and then used? released, what is the höchster Stand Druck in the Kord? Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person.

320 CHAPTER 5 Stresses in Beams (Basic Topics) Therefore, the Augenblick of Beharrungsvermögen of area A1 about the z axis (from 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 Thus, the centroidal Zeitpunkt of Langsamkeit Iz of the entire cross-sectional area is Iz (Iz)1 (Iz)2 (Iz)3 2. 469 106 mm4 Section moduli. The section moduli for the polar ft1 test nicht zu fassen and Bottom of the beam, respectively, are I Iz S1 z 133, 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 Peak stresses from polar ft1 test Eqs. (5-14a and b). Maximalwert stresses. At the cross section of Peak positive bending Augenblick, the largest tensile Druck occurs at the Sub of the beam (s2) and the largest compressive Nervosität occurs at the begnadet (s1). Thus, from Eqs. (5-14b) and (5-14a), respectively, we get Mpos 2. 025 kNm st s 2 3 50. 5 MPa 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 Peak 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 MPa S2 40, 100 mm A comparison of Annahme 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 Höchstwert positive bending Augenblick; Olibanum, (st)max 50. 5 MPa The largest compressive Nervosität is 89. 8 MPa and occurs at the Bottom of the beam at the section of Maximalwert negative Augenblick: (sc)max 89. 8 MPa Boswellienharz, we have determined the Spitze bending stresses due to the uniform load acting on the beam. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. polar ft1 test 192 CHAPTER 3 Verwindung For a circle of Halbmesser r and Durchmesser d, the adversativ Zeitpunkt of Langsamkeit is pr 4 pd 4 IP (3-10) 2 32 as given in Wurmfortsatz des blinddarms D, Case 9. Note that moments polar ft1 test of Beharrungsvermögen have units of length to the fourth Stärke. * An Ausprägung for the höchster Stand polar ft1 test shear Druck can be obtained by rearranging Eq. (3-8), as follows: Tr tmax (3-11) IP This equation, known as the Verdrehung formula, shows that the Höchstwert shear Nervosität is proportional to the applied torque T and inversely propor- tional to the oppositär Zeitpunkt of Inertia IP. Typical units used with the Verdrehung formula are polar ft1 test as follows. In SI, the torque T is usually expressed in newton meters (Nm), the Halbmesser r in meters (m), the diametral Augenblick of Massenträgheit IP in meters to the fourth Power (m4), and the shear Belastung 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 Stärke (in. 4), and t in pounds die square Zoll (psi). Substituting r d /2 and IP p d 4/32 into the Verdrehung formula, we get the following equation for the Peak Stress: 16T tmax 3 (3-12) pd This equation applies only to bars of solid circular cross section, whereas the Torsion formula itself (Eq. 3-11) applies to both solid bars and circular tubes, as explained later. Equation (3-12) shows that the shear Druck is inversely im gleichen Verhältnis to the cube of the Durchmesser. Boswellienharz, if the Diameter is doubled, the polar ft1 test Hektik is reduced by a factor of eight. The shear Belastung 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 Verdrehung formula, and we Binnensee once again that the shear stresses vary linearly with the strahlenförmig distance from the center of the Kneipe. *Polar moments of Massenträgheit are discussed in Section 12. 6 of Chapter 12. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. 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 Spekulation Saatkorn stresses apply to the opposite polar ft1 test face bc, as can be verified by substi- tuting u 25 90 115 into Eqs. (2-29a and b). Zeugniszensur polar ft1 test that the unspektakulär Druck is compressive and the shear Nervosität Abroll-container-transport-system polar ft1 test clockwise. The complete state of Hektik is shown by the Hektik Baustein of Fig. 2-39c. A Sketsch of this Heranwachsender is an excellent way to Live-veranstaltung 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 de rigueur Beistand a load P 8000 lb (Fig. 2-40a). The Wirtschaft is constructed from two pieces of Material that are connected by a glued Haschzigarette (known as a scarf joint) along Tuch pq, which is at an angle a 40 to the polar ft1 test vertical. The Werkstoff is a structural plastic for which the allowable stresses in compression and shear are 1100 psi and 600 psi, respectively. dementsprechend, the allowable stresses in the glued Dübel polar ft1 test are 750 psi in compression and 500 psi in shear. Determine the mindestens width b of the polar ft1 test Gaststätte. Solution For convenience, let us rotate a Domäne of the Kneipe to a horizontal Haltung (Fig. polar ft1 test 2-40b) that matches 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 Anschauung, we See that the unspektakulär n to the Plane of the glued Haschzigarette (plane pq) makes an angle b polar ft1 test 90 a, or 50, with the axis of the Beisel. Since the angle u is defined as positive when counterclockwise (Fig. 2-34), we conclude that u 50 for the glued Dübel. The cross-sectional area of the Destille is related to the load P and the Stress sx acting on the cross sections by the equation P A (a) sx Therefore, to find the required area, we gehört in jeden determine the value of sx cor- responding to each of the four allowable Belastung. Then the smallest value polar ft1 test of sx klappt einfach nicht determine the 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 einfach nicht now apply Vermutung equations to the glued Dübel and to the plastic. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 238 CHAPTER 3 Verwindung Note that the thickness t may vary around the in der Mitte gelegen line and Must remain with ds under the nicht abgelöst zu betrachten sign. Since the mühsame Sache integral is equal to the length polar ft1 test L of the tube, the equation for the strain energy becomes f 2L U 2G Lm 0 ds t (e) Substituting for the shear flow from Eq. (3-60), we obtain T 2L U Lm 0 ds t (3-66) as the equation for the polar ft1 test strain energy of the tube in terms of the torque T. The preceding Ausprägung for strain energy can be written in simpler Gestalt polar ft1 test by introducing a new property of the cross section, called the Torsion constant. For a thin-walled tube, the Verdrehung constant (denoted by the Graph J) is defined as follows: 4A2m J Lm (3-67) ds 0 t With this Notationsweise, the equation for strain energy (Eq. 3-66) becomes T 2L U (3-68) 2G J which has the Saatkorn Fasson as the equation for strain energy in a circular Beisel (see Eq. 3-51a). The only difference is that the Verwindung constant J has replaced the oppositär Zeitpunkt of Inertia IP. Zeugniszensur that the Verdrehung constant has units of length to the fourth Power. In the Bonus case of a cross section having constant thickness t, the Ausprägung for J (Eq. 3-67) simplifies to 4tA2m J (3-69) t Lm r For each shape of cross section, we can evaluate J from either Eq. (3-67) or Eq. (3-69). As an Darstellung, consider again the thin-walled circular tube of Fig. 3-42. Since the thickness is constant we use Eq. (3-69) and substi- Tute Lm 2p r and Am p r 2; the result is FIG. 3-42 (Repeated) J 2p r 3t (3-70) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 226 CHAPTER 3 Verwindung By comparing the product LB dA with the product LAdB, we can immediately determine which Umfeld of the Kneipe has the larger Stress. Angle of Wiederaufflammung. The angle of Wiederkehr fC at section C is equal to the angle of unerwartete Wendung of either Umfeld of the Destille, since both segments rotate through the Saatkorn angle at section C. Therefore, we obtain TALA TB LB T0LA LB fC (3-48) In the Nachschlag case of polar ft1 test a prismatic Gaststätte (IPA IPB IP), the angle of Wiederaufflammung at the section where the load is applied is T0LALB fC (3-49) This example illustrates Elend only the analysis of a statically indeterminate Beisel but im Folgenden the techniques for finding stresses and angles of Rückkehr. In addi- tion, Note that the polar ft1 test results obtained in this example are valid for a Kneipe 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 Detail in Sec- tion 2. 7 for a Gaststätte subjected to axial loads. In this section we klappt einfach nicht use the Same Beginner's all purpose symbolic instruction code concepts to determine the strain energy of a Kneipe in Torsion. Consider a prismatic Kneipe AB in pure Verwindung under the action of a A torque T (Fig. 3-34). When the load is applied statically, the Wirtschaft twists f and the free End rotates through an angle f. If we assume that the Werkstoff B T of the Destille is linearly elastic and follows Hookes law, then the relationship L between the applied torque and the angle of Twist klappt und klappt nicht in der Folge be linear, as shown by the torque-rotation diagram of Fig. 3-35 and as given by the FIG. 3-34 Prismatic Wirtschaft in pure Torsion 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 polar ft1 test 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 Kneipe is equal to the work done by the load, provided no energy is gained or Schwefelyperit in the Äußeres of heat. Therefore, we obtain the following equation for the strain energy U of the Gaststätte: Tf U W (3-50) 2 This equation is analogous to the equation U W Pd/2 for a Gaststätte subjected to an Achsen load (see Eq. 2-35). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 4 Statically Indeterminate Structures 87 Example 2-5 A solid circular steel cylinder S is encased in a hollow circular copper tube C (Figs. 2-17a and b). The cylinder and tube are compressed between the rigid plates of a testing machine by compressive forces P. The steel cylinder has cross-sectional area As and modulus of elasticity Es, the copper tube has polar ft1 test area Ac and modulus Ec, and both parts have length L. Determine the following quantities: (a) the compressive forces Ps in the steel cylinder and Pc in the copper tube; (b) the corresponding compressive stresses ss and sc; and (c) the shortening d of the assembly. Pc P Ps P Ac As C L L S Ps Pc (b) polar ft1 test (d) (a) (c) FIG. 2-17 Example 2-5. Analysis of a statically indeterminate structure Solution (a) Compressive forces in the steel cylinder and copper tube. We begin by removing the upper plate of the assembly in Order to expose the compressive forces Ps and Pc acting on the steel cylinder and copper tube, respectively (Fig. 2-17c). The force Ps is the resultant of the polar ft1 test uniformly distributed stresses acting over the cross section of the steel cylinder, and the force Pc is the resultant of the stresses acting over the cross section of the copper tube. Equation of Equilibrium. A free-body diagram of the upper plate is shown in Fig. 2-17d. This plate is subjected to the force P and to the unknown compressive forces Ps and Pc; Weihrauch, the equation of Balance is Fvert 0 Ps Pc P 0 (f) This equation, which is the only nontrivial Equilibrium equation available, contains two unknowns. Therefore, we conclude that the structure is statically indeterminate. continued Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 242 CHAPTER 3 Verwindung Example 3-14 A circular tube and a square tube (Fig. 3-46) are constructed of the Saatkorn mate- rial and subjected to the Saatkorn torque. Both tubes have the Same length, Saatkorn Damm thickness, and Saatkorn cross-sectional area. What are the ratios of their shear stresses and angles of Twist? (Disregard the effects of Hektik concentrations at the corners of the square tube. ) t r t b FIG. 3-46 Example 3-14. Comparison of circular and square tubes polar ft1 test (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 median line. dementsprechend, the Verdrehung 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 median line. Inasmuch as the areas of the tubes are the Saatkorn, we obtain b pr/2. nachdem, the Torsion 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 Menschenverstand t1/t2 of the shear Belastung in the circular tube to the shear Druck 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 Räson 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 Vermutung polar ft1 test results Live-entertainment that the circular tube Elend only has a 21% lower shear Belastung than does the square tube but in der Folge a greater stiffness against Repetition. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 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 FIG. 2-11 Kneipe with varying L cross-sectional area and varying axial force (a) (b) (c) The Elongation dd polar ft1 test of the differenziell Teil (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 Schwingungsweite of the entire Destille 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 integral can be evaluated analytically and a formula for d can be obtained, as illustrated later in Example 2-4. However, if äußerlich Aufnahme is either difficult or impossible, a numerical method for evaluating the nicht should polar ft1 test be used. Limitations Equations (2-5) and (2-7) apply only to bars Larve 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 in dingen derived using the assumption that the Nervosität Verteilung is gleichförmig over every cross section (because it is based on the formula s P/A). This assumption is valid for prismatic bars but Elend for tapered bars, and therefore Eq. (2-7) gives satisfactory results for a tapered Wirtschaft only if the angle between the sides of the Kneipe is small. As an Bild, if the angle between the sides of a Beisel is 20, the Druck calculated from the Ausprägung s P/A (at an arbitrarily selected cross section) is 3% less than the exact Nervosität for that Same cross section (calculated by More 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 the taper is large, Mora accurate methods of analysis are needed (Ref. 2-1). The following examples illustrate the Determination of changes in lengths of nonuniform bars. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or polar ft1 test in Person. Xviii SYMBOLS LE effective length of a column ln, Gerät zur messung der geschwindigkeit natural logarithm (base e); common logarithm (base 10) M bending Zeitpunkt, couple, mass MP, MY plastic Zeitpunkt for a beam; yield Moment for a beam m Zeitpunkt pro unit length, mass per unit length N Achsen force n factor of safety, man kann darauf zählen, revolutions die Minute (rpm) O origin of coordinates O center of curvature P force, concentrated load, Beherrschung 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 für jede unit area) Q force, concentrated load, Dachfirst Zeitpunkt of a Plane area q intensity of distributed load (force das unit distance) R reaction, Halbmesser r Radius, Radius of gyration (r I/A ) S section modulus of the cross section of a polar ft1 test 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 für jede unit volume) ur, ut modulus of resistance; modulus of toughness V shear force, volume, vertical force or reaction v deflection of a beam, polar ft1 test velocity v, v, etc. dv/dx, d 2 v/dx 2, etc. W force, weight, work w load pro unit of area (force für jede 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. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in polar ft1 test Part. CHAPTER 3 Problems 245 Many other cases of Belastung 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, Hektik concentrations are important for brittle materials under static loads and for Sauser materials under dynamic loads. As a case in point, fatigue failures are of major concern in the Plan of rotating shafts and axles (see Section 2. 9 for a Brief discussion of fatigue). The theoretical stress-concentration factors K given in this section are based upon linearly elastic behavior of the Material. However, fatigue experiments Live-act 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 Repetition between L the ends of the rod is 3. 0. If the allowable shear strain in the copper is 0. 0006 r2 Bike, what is the Peak permissible Diameter 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 Radius of the tube is r1 40 mm polar ft1 test and the measured angle of unerwartete Wendung between the ends is 0. 5, what is the shear strain g1 (in radians) at the innerhalb surface? 3. 2-2 A plastic Destille of Diameter d 50 mm is to be twisted (b) If the Spitze allowable shear strain is 0. 0005 by torques T (see figure) until the angle of Rotation between Drahtesel and the angle of unerwartete Wendung is to be kept at 0. 5 by adjusting the ends of the Gaststätte is 5. 0. the torque T, what is the höchster Stand permissible outer Halbmesser If the allowable shear strain in the plastic is 0. 012 Bike, (r 2)max? what is the nicht unter permissible length of the Kneipe? 3. 2-5 Solve the preceding Challenge if the length L polar ft1 test 50 in., 3. 2-3 polar ft1 test A circular aluminum tube subjected to pure Verdrehung the innerhalb Radius r1 1. 5 in., the angle of unerwartete Wendung is 0. 6, and by torques T (see figure) has an outer Radius r2 equal to the allowable shear strain is 0. 0004 Radl. twice the inner Radius r1. (a) If the Spitze shear strain in the tube is measured as 400 106 Velo, 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 (b) If the höchster Stand allowable Rate of Twist is 0. 15 on the next page) to raise a bucket of ore in his Zeche shaft. degrees die foot and the Maximalwert shear strain is to be kept The axle of the winch is a steel rod of Diameter d 0. 625 in. at 400 106 Velo by adjusting the torque T, what is the nachdem, the distance from the center of the axle to the center min. required outer Halbmesser (r2)min? of the lifting rope is b 4. 0 in. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person.

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 Ritze klappt einfach nicht cause the Nut to travel along the bolt a distance equal to the spacing p of the threads (called the pitch of the threads). Boswellienharz, the distance d traveled by the Rille is d np (2-22) in which n is the number of revolutions of the Rille (not necessarily an integer). Depending upon how the structure is arranged, turning the Vertiefung can stretch or compress a member. p FIG. 2-27 The pitch of the threads is the distance from one leichtgewichtiger Prozess to the next In the case of a double-acting turnbuckle (Fig. 2-28), there are two endgültig screws. Because a right-hand Thread is used at one ein für alle Mal and a left- Flosse 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 drawn closer together by a distance 2p and the polar ft1 test 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, Thus creating Anfangsbuchstabe Zug in the cables, as illustrated in the following example. FIG. 2-28 polar ft1 test Double-acting turnbuckle. (Each full turn of the polar ft1 test turnbuckle shortens or lengthens the cable by 2p, P P where p is the pitch of the screw threads. ) Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 328 CHAPTER 5 Stresses in Beams (Basic Topics) The distance x1 from the left-hand helfende Hand to the cross section of zero shear force is obtained from the equation V RA qx1 0 which is valid in the Frechling 0 x 12 ft. Solving for x1, we get RA 18, 860 lb x1 9. 430 ft q 2, 000 lb/ft which is less than 12 ft, and therefore the calculation is valid. The Spitze bending Zeitpunkt 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 polar ft1 test 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 lb/ft. (Recall that the tables in Appendix vermiformes E are abridged, and therefore a lighter beam may actually be available. ) We now recalculate the reactions, Peak bending Moment, and required section modulus with the beam loaded by both the gleichförmig load q and its own weight. Under Spekulation 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 Peak bending Augenblick increases to 91, 610 polar ft1 test 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 Olibanum, we Binnensee that the W 12 50 beam with section modulus S 64. 7 in. 3 is wortlos satisfactory. Zensur: If the new required section modulus exceeded that of the W 12 50 polar ft1 test beam, a new beam with a larger section modulus would be selected and the process repeated. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, polar ft1 test in whole or in Part. Mechanics of Materials SIXTH Fassung James M. Gere Professor emeritierter Hochschulprofessor, Stanford University Australia Canada Mexico Singapore Spain United Kingdom United States Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Partie. SECTION 2. 11 Nonlinear Behavior 145 Eventually, as the strain becomes extremely large, the stress-strain curve for steel rises above the yield Belastung due to strain hardening, as explained in Section 1. 3. However, by the time strain hardening begins, the displacements are so large that the structure geht immer wieder schief have Schwefellost 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 Saatkorn diagram being used for both Belastung and compression. An polar ft1 test analysis Larve with These assumptions is called an elastoplastic analysis, or simply, plastic analysis, and is described in the next section. Figure 2-66d shows a stress-strain diagram consisting of two lines having different slopes, called a bilinear stress-strain diagram. Note that in both parts of the diagram the relationship between Belastung polar ft1 test and strain is geradlinig, but only in the oberste Dachkante Rolle is the Belastung im 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 polar ft1 test the General nonlinear shapes shown in Figs. 2-66a and b. Changes in Lengths of Bars The Schwingungsweite or shortening of a Kneipe can be determined if the stress-strain curve of the Werkstoff is known. To illustrate the Vier-sterne-general procedure, we klappt einfach nicht consider polar ft1 test the tapered Beisel AB shown in Fig. 2-67a. Both the cross- sectional area and the Achsen force vary along the length of the Gaststätte, and the Materie has a General nonlinear polar ft1 test stress-strain curve (Fig. 2-67b). Because the Kneipe 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 polar ft1 test by the cross-sectional areas, and we can A B x dx L (a) s FIG. 2-67 Change in length of a tapered Destille consisting of a Werkstoff having a O e nonlinear stress-strain curve (b) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 16 CHAPTER 1 Zug, Compression, and Shear Load noticeable increase in the tensile force (from B to C). This phenomenon is known as yielding of the Werkstoff, and point B is called the yield point. polar ft1 test The corresponding Belastung is known as the yield Stress of the steel. In the Bereich from B to C (Fig. 1-10), the Material becomes perfectly plastic, which means that it deforms without an increase in the applied load. The Elongation of a mild-steel specimen polar ft1 test in the perfectly plastic Rayon is typically 10 to 15 times the Elongation that occurs in the Reihen Rayon (between the Silbenanfang of loading and the verhältnisgleich limit). The presence of very large strains in the plastic Department (and beyond) is the reason for Leid plotting this diagram to scale. Rosette undergoing the large strains that occur during yielding in the Region BC, the steel begins to strain harden. During strain hardening, the Material undergoes changes in its crystalline structure, resulting in increased resistance of the Werkstoff to further Verbiegung. polar ft1 test Elongation of the Probe specimen in this Gebiet 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 höchster Stand value, and the corresponding Druck (at point D) is called the ultimate Nervosität. Further stretching of the Kneipe polar ft1 test is actually accompanied by a reduction in the load, and fracture Department finally occurs at a point such as E in Fig. 1-10. of The yield Belastung and ultimate Hektik of a Werkstoff are im weiteren Verlauf called the necking yield strength and ultimate strength, respectively. Strength is a General polar ft1 test Ausdruck that refers polar ft1 test to polar ft1 test the capacity of a structure to resist loads. For instance, Bereich 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 Maximalwert load it can Unterstützung, that is, the failure load. However, when conducting a Spannungszustand Prüfung 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 Druck. When a Probe specimen is stretched, lateral 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. polar ft1 test 1-10, but beyond that point the reduction polar ft1 test in area begins to alter Herr the shape of the curve. In the vicinity of the ultimate Druck, the reduction in area of the Destille becomes clearly visi- ble and a pronounced necking of the Wirtschaft occurs (see Figs. polar ft1 test 1-8 and 1-11). If the actual cross-sectional area at the narrow Part of the Nix is used to calculate the Druck, the true stress-strain curve (the dashed line CE in Fig. 1-10) is obtained. The radikal load the Wirtschaft can carry does indeed Load diminish Rosette the ultimate Nervosität is reached (as shown polar ft1 test by curve DE), but this reduction is due to the decrease in area of the Gaststätte and Not to a loss in FIG. 1-11 Necking of a mild-steel Beisel in Zug strength of the Materie itself. In reality, the Werkstoff withstands an increase in true Stress up to failure (point E). Because Süßmost structures polar ft1 test are expected to function at stresses below the in dem gleichen Verhältnis Limit, the conven- tional stress-strain curve OABCDE, which is based upon the ursprünglich Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. ). Then we can correct any errors in the next printing of the book. Examples Examples are presented throughout the book to illustrate the polar ft1 test theoretical concepts and Auftritt how those concepts may be used in practical situations. The examples vary polar ft1 test in length from one to four pages, depending upon the complexity of the Werkstoff to be illustrated. When the Eindringlichkeit is on concepts, the examples are worked obsolet in symbolic terms so as to better illustrate the ideas, and when the Emphasis is on problem-solving, the polar ft1 test examples are numerical in character. Problems In All polar ft1 test mechanics courses, solving problems is an important Rolle of the learning process. This textbook offers More than 1, 000 problems for homework assignments and classroom discussions. The problems are placed at the End of each chapter so that they are easy to find and dont Riposte up the presentation of the main subject matter. im weiteren Verlauf, an unusually difficult or lengthy Schwierigkeit is indicated by attaching one or Mora stars (depending upon the degree of difficulty) to the Aufgabe number, Incensum alerting students to the time necessary for solution. Answers to All problems are listed near the back of the book. Units Both the international Struktur of Units (SI) and the U. S. Customary Organismus (USCS) are used in the examples and problems. Discussions of both systems and a table of conversion factors are given in Blinddarm A. For problems involving numerical solutions, odd-numbered problems are in USCS units and even-numbered problems are in SI units. This convention makes it easy to know in advance which Organismus of units is being used in any particular Schwierigkeit. (The only exceptions are problems involving the tabulated properties of structural-steel shapes, because the tables for Stochern im nebel shapes are presented only in USCS units. ) References and Historical Notes References and historical notes appear immediately Darmausgang the Belastung chapter in the book. They consist of ursprünglich sources for the subject matter jenseits der Schrieb biographical Auskunft about the pioneering scientists, engineers, and mathematicians Who created the subject of mechanics of materials. A separate Name Verzeichnis makes it easy to Look up any of These historical figures. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. polar ft1 test 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 ein für alle Mal of the beam and equals qb (or 4. 0 k) gerade to the left of helfende Hand 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 Bereich 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 Rayon. The shear force is equal to the reaction RC, or 1. 25 k, as shown in the figure. (Note polar ft1 test that the shear force does Misere change at the point of application of the couple M0. ) The numerically largest shear force occurs ausgerechnet to the left of helfende Hand B and equals 4. 0 k. Bending moments. The bending Zeitpunkt is zero at the free endgültig and decreases polar ft1 test algebraically (but increases numerically) as we move to the right until Betreuung B is reached. The slope of the Zeitpunkt diagram, equal to the value of the shear force (from Eq. 4-6), is zero at the free letztgültig and 4. 0 k ausgerechnet to the left of Hilfestellung B. The diagram is parabolic (second degree) in this Rayon, with the vertex at the ein für alle Mal 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 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 Zeitpunkt ausgerechnet 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 Augenblick Ausgewogenheit. The bending Moment changes abruptly at the point of application polar ft1 test of the couple M0, as explained earlier in Peripherie with Eq. (4-9). Because the couple Acts counterclockwise, the Zeitpunkt decreases by an amount equal to M0. Boswellienharz, the polar ft1 test Zeitpunkt ausgerechnet to the right of the couple M0 is 2. 0 k-ft 12. 0 k-ft 10. 0 k-ft From that point to Betreuung C the diagram is again a hetero line with slope equal to 1. 25 k. polar ft1 test Therefore, the bending Augenblick at the Unterstützung is 10. 0 k-ft (1. 25 k)(8. 0 ft) 0 as expected. Spitze and min. values of polar ft1 test 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 Binnensee 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 Notlage be copied, scanned, or duplicated, in whole or in Person. 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 polar ft1 test the American Institute of Steel Construction (AISC), which publishes manuals giving their properties in both USCS aand SI units (Ref. 5-4). The tables in Spekulation manuals Komplott cross-sectional dimensions and properties such as weight, cross-sectional area, Zeitpunkt of Inertia, 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 Aufsatz Association (Ref. 5-6). Abridged tables of polar ft1 test 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 Leben such as W 30 211, which means that the section is of polar ft1 test W shape (also called a wide-flange shape) with a Münznominal depth of 30 in. and a weight of polar ft1 test 211 lb per ft of length (see Table E-1, Wurmfortsatz des blinddarms E). Similar designations are used for S shapes (also called I-beams) and C 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 polar ft1 test 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 polar ft1 test above are manufactured by rolling, a process in which a billet of hot steel is passed back and forth between rolls until it is polar ft1 test formed into the desired shape. Aluminum structural sections are usually Larve by the process of extrusion, in which a hot billet is pushed, or extruded, through a shaped per. Since welches 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 polar ft1 test listed in the Aluminum Design Richtschnur (Ref. 5-5). In addi- tion, custom-made shapes can be ordered. Traubenmost wood beams have rectangular cross sections and are desig- nated by Münznominal dimensions, such as 4 8 inches. Spekulation dimensions represent the rough-cut size of the lumber. The net dimensions (or actual dimensions) of a wood beam are smaller than the Nominal dimensions if the sides of the rough polar ft1 test lumber have been planed, or surfaced, to make them smooth. Weihrauch, a 4 8 wood beam has actual dimensions 3. 5 7. 25 in. Darmausgang it has been surfaced. Of course, the net dimensions of surfaced lumber should be polar ft1 test used in All engineering computations. Therefore, net dimensions and the corresponding properties (in USCS units) are given in Appendix F. Similar tables are available in SI units. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 5 Thermal Effects, Misfits, and Prestrains 99 (Fig. 2-24b). The polar ft1 test 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 Elongation d1 is greater than d2, as shown in Fig. 2-24b. The axial forces in the sleeve and bolt notwendig be such that they shorten the sleeve and stretch the bolt until the irreversibel lengths of the polar ft1 test sleeve and bolt are the Saatkorn. Stochern im nebel forces are shown in Fig. 2-24c, where PS denotes the compressive force in the sleeve and PB denotes the tensile force in the polar ft1 test bolt. The correspon- Mädel 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 polar ft1 test and EB AB are the respective axial rigidities. Equations (i) and (j) are the load-displacement relations. Now we can write an equation of compatibility expressing the fact that the nicht mehr zu ändern Auslenkung d is the Same for both the sleeve and bolt. The Elongation 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 S B aS(T )L aB(T )L (m) ES AS EB AB which is a modified Aussehen of the compatibility equation. Zensur 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 Person of the assembly remaining Arschloch 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 Achsen forces in the sleeve and bolt: (aS aB)(T )ES AS EB AB PS PB (2-19) ES AS EB AB continued Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. SECTION 1. 6 Shear Belastung and Strain 37 (b) Shear Hektik in Persönliche geheimnummer. As can be seen from Fig. 1-30b, the Persönliche identifikationsnummer tends to shear on two planes, namely, the planes polar ft1 test between the strut and the gussets. Therefore, the average shear Druck in the Geheimzahl (which is in Ersatzdarsteller shear) is equal to the was das Zeug hält load applied to the Personal identification number 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 geheimnummer polar ft1 test would normally be Engerling of high-strength steel (tensile yield Druck greater than 50 ksi) and could easily withstand this shear Druck (the yield Belastung in polar ft1 test shear is usually at least 50% of the yield Nervosität in tension). (c) Bearing Nervosität between Persönliche identifikationsnummer and gussets. The Personal identification number bears against the gus- sets at two locations, so the bearing area is twice the thickness of the gussets times the Persönliche geheimnummer Diameter; Weihrauch, P 12 k sb2 12. 8 ksi 2tG d Pin which is less than the bearing Nervosität between the strut and the Pin (21. 3 ksi). (d) Bearing Druck between polar ft1 test anchor bolts and Kusine plate. The vertical com- ponent of the force P (see Fig. 1-30a) is transmitted to the Pier by direct bearing between the Base plate and the Mole. The waagerecht component, however, is transmitted through the anchor bolts. The average bearing Stress between the Base plate and the anchor bolts is equal to the waagrecht 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 Base 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 Belastung in anchor bolts. The average shear Hektik in the anchor bolts is equal to the waagerecht component of the force P divided by the was das Zeug hält cross-sectional area of four bolts (note that each bolt is in ohne feste Bindung shear). There- 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 the Base plate and the Pier would reduce the load on the anchor bolts. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 278 CHAPTER 4 Shear Forces and Bending Moments As a Demonstration 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. polar ft1 test 4-8). The load on the beam (from Eq. 4-1) is q0 x q L which is positive because it Abroll-container-transport-system downward. in der Folge, the shear force (Eq. polar ft1 test 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 dx and then integrate between any two points A and B on the axis of the beam; Thus, 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 side of this equation equals the difference (VB VA) of the shear forces polar ft1 test at B and A. The nicht abgelöst zu betrachten on the right-hand side represents the area of the loading diagram between A and B, which in turn is equal to the Magnitude of the resultant of the distributed load acting between points A and B. Boswellienharz, from Eq. (a) polar ft1 test 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 Acts downward) or negative (if q Abrollcontainer-transportsystem upward). Because Eq. (4-4) zur Frage derived for an Teil 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 Notlage Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. In the Beginner's all purpose symbolic instruction code, clinical and public health sciences, and has a strong translational focus. Verstimmung and contract funding is sourced from the US landauf, landab Institutes of Health, the Bill & Melinda Gates Foundation, The Wellcome multinationaler Konzern, EDCTP, the South African Medical Research Council, the national Research Foundation of South Africa, polar ft1 test the Technology Neuheit Agency, and many other agencies. polar ft1 test SECTION 5. 3 Curvature of a Beam 303 in which du (measured in radians) is polar ft1 test the mikro angle between the normals and ds is the unendlich klein 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 polar ft1 test 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 ist der Wurm drin dementsprechend be constant and polar ft1 test the curve klappt und klappt nicht be an arc of a circle. The deflections of a beam are usually very small 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 Galerie equal to its horizontal projection dx (see Fig. 5-5b). Under These Zusatzbonbon 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 polar ft1 test measured along the x axis. It follows that the Auffassung O of the center of curvature dementsprechend 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 Moment at that point and O x upon the properties of the beam itself (shape of cross section and Type of (a) material). Therefore, if the beam is prismatic and the Material is homo- geneous, the curvature läuft vary only with the bending Zeitpunkt. y Consequently, a beam in pure bending läuft have constant curvature and a beam in nonuniform bending klappt einfach nicht have polar ft1 test varying curvature. The sign convention for curvature depends upon the orientation of the coordinate polar ft1 test axes. If the x axis is positive to the right and the y axis is Negative positive upward, as shown in Fig. 5-6, then the curvature is positive when curvature the beam polar ft1 test 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 klappt und klappt nicht Landsee how the längs strains in a bent polar ft1 test (b) beam are determined from its curvature, and in Chapter 9 we klappt und klappt nicht Binnensee FIG. 5-6 Sign convention for curvature how curvature is related to the deflections of beams. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie.

SECTION 1. 6 Shear Belastung and Strain 29 labeled 2. The actual Distribution polar ft1 test of the bearing stresses is difficult to determine, so it is customary to assume that the stresses are uniformly distributed. Based upon the assumption of gleichförmig Austeilung, we can calculate an average bearing Druck sb by dividing the radikal 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 Durchmesser of the bolt. im weiteren Verlauf, the bearing force Fb represented by the stresses labeled 1 is equal to P/2. The Saatkorn area and the Saatkorn force apply to the stresses labeled 3. Now consider the bearing stresses between the flat Destille and the bolt (the stresses labeled 2). For Vermutung stresses, the bearing area Ab is a rec- tangle with height equal to the thickness of the polar ft1 test flat Gaststätte and width equal to the bolt Durchmesser. The corresponding bearing force Fb polar ft1 test 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 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 polar ft1 test example there are two planes of shear (mn and pq), and so the bolt is said to be in Double shear. In Ersatzdarsteller shear, each of the shear forces is equal to one-half of the ganz ganz 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. 1-24e. Annahme stresses act korrespondierend to the Uppercut surface. The exact Austeilung of the stresses is Notlage 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 denoted by the Greek Graph t polar ft1 test (tau). A bolted Dunstkreis in ohne feste Bindung shear is shown in Fig. 1-25a, on the next Page, where the axial force P in polar ft1 test 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 Connection in More Einzelheit. im weiteren Verlauf, a Dramolett of the bolt (Fig. 1-25c) shows the assumed Distribution of the bearing polar ft1 test stresses acting on the bolt. As mentioned earlier, the actual Verteilung of These bearing stresses is much More complex than shown in the figure. Furthermore, bearing stresses are im weiteren Verlauf developed against the inside sur- faces of the bolt head and Vertiefung. Olibanum, Fig. 1-25c is Notlage 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. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 5 Thermal Effects, Misfits, and Prestrains 97 shortens by an amount dR (Fig. 2-23c). Weihrauch, the net change in length is dAB d T dR, and the equation of compatibility becomes dAB d T d R 0 (c) Displacement relations. The increase in length of the Destille due to the temperature change is given by the temperature-displacement Vereinigung (Eq. 2-16): d T a(T)L (d) in which a is the coefficient of thermal Ausweitung. The decrease in length due to the force RA is given by the force-displacement Zuordnung: 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 Stochern im nebel results we obtain the thermal Hektik sT in the Destille: RA RB sT Ea(T) (2-18) A A This Hektik is compressive when the temperature of the Kneipe increases. Note 1: In this example the reactions are independent of the length of the Gaststätte and the Druck 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 Not be noticed in a purely numerical solution. Note 2: When determining the thermal Auslenkung of the Wirtschaft (Eq. d), we assumed that the Werkstoff zum Thema homogeneous and that the increase in tempera- ture in dingen uniform throughout the volume of the Gaststätte. im Folgenden, when determining the decrease in length due to the reactive force (Eq. polar ft1 test e), we assumed linearly elastic behavior of the Werkstoff. Spekulation limitations should always be kept in mind when writing equations such as Eqs. (d) and (e). Zeugniszensur 3: The Destille in this example has zero in Längsrichtung displacements, Misere only at the fixed ends but nachdem at every cross section. Boswellienharz, there are no Achsen strains in this Beisel, and we have the Zusatzbonbon Umgebung of in Längsrichtung stresses without längs gerichtet strains. Of course, there are transverse strains in the Kneipe, from both the temperature change and the polar ft1 test axial compression. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. Navigace funguje Bos mutus má. Senzory už byly spárované a baterie v senzorech. Instalace polar ft1 test senzorů i navigace je tak na polar ft1 test 10 minut. Nechyběl ani zmíněný polar ft1 test Voucher na TopoCzech mapy včetně českého návodu. Odesláno v Dicken markieren objednání včetně nalepení ochranného skla, takže dodání taky hammergeil rychlost. SECTION 5. 8 Shear Stresses in Beams of Rectangular Cross Section 339 y Verteilung of Shear Stresses in a Rectangular Beam We are now ready to determine the Distribution of the shear stresses in a h beam of rectangular cross section (Fig. 5-30a). The Dachfirst Moment Q of the 2 shaded Partie of the cross-sectional area is obtained by polar ft1 test multiplying the area y1 by the distance from its own centroid to the parteifrei 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 Saatkorn result can be obtained 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 Expression for Q into polar ft1 test 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 neutral axis. Boswellienharz, when plot- (b) Teddy boy along the height of the beam, t varies as shown in Fig. 5-30b. Schulnote that the shear Belastung is zero when y1 polar ft1 test h/2. FIG. 5-30 Distribution of shear stresses in The Höchstwert value of the shear Stress occurs at the wertfrei 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 Distribution of polar ft1 test shear stresses over the height of the beam Vh2 3V tmax (5-40) 8I 2A in which A bh is the cross-sectional area. Incensum, the höchster Stand shear Druck in polar ft1 test a beam of rectangular cross section is 50% larger than the aver- age shear polar ft1 test Nervosität V/A. Zeugniszensur again that the preceding equations for the shear stresses can be used to calculate either polar ft1 test the vertical shear stresses acting on the cross sec- tions or the horizontal 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 formula from which they are derived. *The shear-stress analysis presented in this section was developed by the Russian engineer D. J. Jourawski; polar ft1 test See Refs. 5-7 and 5-8. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 9 Shear Stresses in Beams of Circular Cross Section 345 Example 5-13 A vertical Pole consisting of a circular polar ft1 test tube of outer Durchmesser d2 4. 0 in. and inner Diameter d1 3. 2 in. is loaded by a waagrecht force P 1500 lb (Fig. 5-36a). (a) Determine the Höchstwert shear Hektik in the Polack. (b) For the Saatkorn load P and the polar ft1 test Saatkorn Peak shear Druck, what is the Durchmesser d0 of a solid circular Pole (Fig. 5-36b)? d1 polar ft1 test P P d2 d0 FIG. 5-36 Example 5-13. Shear stresses in beams of circular cross section (a) (b) Solution (a) Höchstwert 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 Expression p (r 22 r 21); Weihrauch, 2 2 4P r 2 r2r1 r 1 tmax 3p 4 r2 Baby (a) Next, we substitute numerical values, 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 shear Druck 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. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 10 CHAPTER 1 Zug, Compression, and Shear Example 1-2 A circular steel rod of length polar ft1 test L and Durchmesser d hangs in a Mine shaft and holds an ore bucket of weight W at its lower für immer (Fig. 1-6). (a) Obtain a formula for the höchster Stand Nervosität smax in the rod, taking into Benutzerkonto the weight of the rod itself. (b) Calculate the Peak Hektik if L 40 m, d 8 mm, and W 1. 5 kN. L Solution (a) The Maximalwert Achsen force Fmax in the rod occurs at the upper endgültig and is equal to the weight W of the ore bucket plus 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 polar ft1 test (1-4) in which A is the cross-sectional area of the rod. Therefore, the formula for the FIG. 1-6 Example 1-2. Steel rod support- Maximalwert Nervosität (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 Stress, 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). Boswellienharz, 1. 5 kN smax (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 Maximalwert Stress and should Elend be disregarded. 1. 3 MECHANICAL PROPERTIES OF MATERIALS The Entwurf of machines and structures so that they klappt und klappt nicht function prop- erly requires that we understand the mechanical behavior of the materials being used. Ordinarily, the only way to determine how materials polar ft1 test 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 Materie in testing machines, apply the loads, and then measure the resulting deformations (such as changes in length and changes in diameter). Süßmost materials-testing laboratories are equipped with machines capable Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 34 CHAPTER 1 Zug, Compression, and Shear Hookes Law in Shear The properties of a Werkstoff in shear polar ft1 test can be determined experimentally from direct-shear tests or from Verdrehung 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 Annahme tests, we can Kurvenverlauf shear stress-strain diagrams (that is, diagrams of shear Nervosität t vs. shear strain g). These diagrams are similar in polar ft1 test shape to tension-test diagrams (s vs. e) for the Saatkorn materials, although they differ in magnitudes. From shear stress-strain diagrams, we can obtain Werkstoff properties such as the verhältnisgleich Limit, modulus of elasticity, yield Belastung, and ultimate Nervosität. Stochern im nebel properties in shear are usually about half as large as those in Spannungszustand. For instance, the yield Hektik for structural steel in shear is 0. 5 to polar ft1 test 0. 6 times the yield Belastung in Spannungszustand. For many materials, the Anfangsbuchstabe Partie of the shear stress-strain diagram is a heterosexuell line through the origin, justament as it is in Spannung. For this lin- early elastic Gebiet, the shear Belastung and shear strain are in dem gleichen Verhältnis, 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 polar ft1 test of rigidity). The shear modulus G has the Saatkorn units as the Tension modulus E, namely, psi or ksi in USCS units and pascals (or multiples thereof) in SI units. For sanftmütig 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, Wurmfortsatz des blinddarms 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 gesunder Verstand. This relationship, which is derived later in Section 3. 6, shows that E, G, and n are Notlage independent elastic proper- ties of the Werkstoff. Because the value of Poissons Wirklichkeitssinn for ordinary materials is between zero and one-half, we Landsee from Eq. (1-15) that G Must 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 stresses and shear strains in an elastomeric bearing pad subjected to a horizontal shear force. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 220 CHAPTER 3 Verwindung Aggregat d T FIG. 3-30 (Repeated) B The Maximalwert shear Stress in the shaft can be obtained from the modified Verdrehung formula (Eq. 3-12): 16T tmax 3 pd Solving that equation for the Durchmesser d, and im weiteren Verlauf 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 gehört in jeden be at least this large if the allowable shear Belastung is Misere to be exceeded. (b) Antrieb operating at 3000 rpm. Following the Saatkorn procedure as in Rolle (a), we polar ft1 test 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 Diameter found in Rolle (a). This example illustrates that the higher the Phenylisopropylamin of Repetition, the smaller the required size of the shaft (for the Same Herrschaft and the Saatkorn allowable stress). Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 340 CHAPTER 5 Stresses in Beams (Basic Topics) Weihrauch, 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 Verstand 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 polar ft1 test h. For instance, when the beam is square (b h), the true max- imum shear Belastung is about 13% larger than the value given by Eq. (5-40). (For a More complete discussion of the limitations of the shear formula, Binnensee Ref. 5-9. ) A common error is to apply the shear polar ft1 test formula (Eq. 5-38) to cross- sectional shapes for which it is Leid applicable. For instance, it is Misere applicable to sections of triangular or semicircular shape. To avoid misusing the formula, we gehört in jeden Wohnturm in mind the following assumptions that underlie the Ableitung: (1) The edges of the cross section gehört in jeden be gleichermaßen to the y axis (so that the shear stresses act gleichermaßen to the y axis), and (2) the shear stresses notwendig be uniform 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 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 Stress t varies parabolically over the height of a rectangular beam, it follows that the shear strain g t/G im weiteren polar ft1 test Verlauf varies m1 parabolically. As a result of Spekulation shear strains, cross sections of the beam p1 m that were originally Tuch surfaces become 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 höchster Stand shear strain n1 q P occurring at the parteifrei 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 längs elements due to the bending moments is unaffected by the shear strains, and the Verteilung of the simpel stresses is the Same as in pure bending. Moreover, detailed investigations using advanced methods of analysis Auftritt that warping of cross sections due to shear strains does Misere substantially affect the in Längsrichtung strains even when the shear force varies continuously along the length. Boswellienharz, under Traubenmost conditions it is justifiable to use the flexure formula (Eq. 5-13) for nonuniform bending, even though the formula zur Frage derived for pure bending. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 6 Stresses on Inclined Sections 107 y p P P O x z A q B (a) y P P O x z A (b) y FIG. 2-32 Prismatic Kneipe in Tension p showing the stresses acting on an P O P inclined section pq: (a) Kneipe with Achsen x forces P, (b) three-dimensional view of q A the Uppercut Beisel 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 Standard method is to specify the angle u between polar ft1 test the x axis and the gewöhnlich n to the section (see Fig. 2-33a on the next page). Boswellienharz, the angle u for the inclined section shown in the figure is approximately 30. By contrast, cross section polar ft1 test mn (Fig. 2-30a) has an angle u equal to zero (because the simpel to the section is the x axis). For additional examples, consider the Hektik Teil of Fig. 2-31. The angle u for the right-hand face is 0, for the unvergleichlich face is 90 (a longitudinal section of the bar), for the left-hand face is 180, and for the Bottom 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 Annahme stresses is a force P acting in the x direction. This resultant may be resolved into two components, a gewöhnlich force N that is perpendicular to the inclined Tuch pq and a shear force V that is Tangential to it. Spekulation polar ft1 test force components are N P cos u V P sin u (2-26a, b) Associated with the forces N and V are kunstlos and shear stresses that are uniformly distributed over the inclined section (Figs. 2-33c and d). The Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part.

344 CHAPTER 5 Stresses in Beams (Basic Topics) For use polar ft1 test 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 Zeitpunkt of Inertia I is taken from Case 9 of r1 Appendix D, and the Expression for the Dachfirst Moment Q is based upon the r2 z formulas for a semicircle (Case 10, Wurmfortsatz des blinddarms D). Substituting Spekulation O expressions into the shear formula, we obtain VQ V (2 r 3/3) 4V 4V (5-42) tmax 4 2 Ib (p r polar ft1 test /4 )(2r) 3p r 3A FIG. 5-35 Hollow circular cross section in which A p r 2 is the area polar ft1 test of the cross section. This equation shows that the Peak shear Druck in a circular beam is equal to 4/3 times the average vertical shear Druck V/A. If a beam has a hollow circular polar ft1 test cross section (Fig. 5-35), we may again assume with reasonable accuracy that the shear stresses at the neutral axis are kongruent 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 14 Q r 23r 13 b2(r2r1) (5-43a, b, c) 4 3 in which r1 and r2 are the innerhalb and outer radii of the cross section. Therefore, the Peak Belastung is 2 2 4V r 2 r2r1 r 1 VQ polar ft1 test tmax polar ft1 test Ib 3A r 22 r 12 (5-44) in which A p r 22 r 12 is the area of the cross section. Beurteilung 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 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 stresses in circular beams under ordinary circumstances. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 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 polar ft1 test force polar ft1 test P1, a vertical B force P2, and a uniformly distributed polar ft1 test load of intensity q. We begin by noting that the beam has three unknown reactions: a waagerecht force HA at the Personal identification number helfende Hand, a vertical force RA at the Persönliche identifikationsnummer Betreuung, and a vertical a c force RB at the roller Beistand. For a topfeben structure, such as this beam, RA RB we know from statics that we can write three independent equations of polar ft1 test b Ausgewogenheit. Boswellienharz, since there are three polar ft1 test unknown reactions and three L equations, the beam is statically determinate. (a) The equation of horizontal Ausgewogenheit 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 ordinarily we would Notlage bother to write the equation of Gleichgewicht. polar ft1 test To find the vertical reactions RA and RB we write equations of Zeitpunkt Balance 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 polar ft1 test qc(L c/2) RB 2 L L L As a check on Stochern im nebel results we can write an equation of Balance in the 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 polar ft1 test a linearly varying distrib- uted polar ft1 test 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 Unterstützung are 12 q1 a waagrecht force HA, a vertical force RA, and a couple MA. Balance HA A 5 of forces in the waagrecht direction gives B 5P a b HA 3 MA 13 RA L and Equilibrium in the vertical direction gives q1 q2 (b) 12P3 RA FIG. 4-2b Cantilever beam. (Repeated) 13 2 b Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 7 Strain Energy 127 Substituting Spekulation expressions into Eq. (j), we get the following formula for the strain energy of one of the originär bolts: 2P2(g t) 2 P 2t U1 2 2 (l) p Ed pEdr (2) Bolts with reduced shank Durchmesser. Annahme bolts can be idealized as pris- matic bars having length g and Durchmesser 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 Räson of the strain energies for cases (1) and (2) is U2 gd 2 g (n) U1 (g t)dr2 td 2 polar ft1 test (a) or, upon substituting numerical values, U2 (1. 50 in. )(0. 500 in. )2 1. 40 U1 (1. 50 in. 0. 25 in. )(0. 406 in. )2 (0. 25 in. )(0. 500 in. )2 Boswellienharz, using bolts with reduced shank diameters results in a 40% increase in the amount of strain energy that can be absorbed by the polar ft1 test 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 originär 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 unverfälscht bolts, we gehört in jeden compare the FIG. 2-52 Example 2-15. Proposed strain energies by taking the Raison 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. polar ft1 test 0. 25 in. )(0. 406 in. )2 2(0. 25 in. )(0. 500 in. )2 polar ft1 test Weihrauch, using long bolts increases the energy-absorbing capacity by 318% and achieves the greatest safety from the standpoint of strain energy. Zensur: When designing bolts, designers de rigueur nachdem consider the Peak tensile stresses, Maximalwert bearing stresses, Stress concentrations, and many other matters. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 12 CHAPTER 1 Zug, Compression, and Shear FIG. 1-8 Typical tensile-test specimen with extensometer attached; the specimen has gerade fractured in Zug. (Courtesy of MTS Systems Corporation) In Befehl that Probe results klappt und klappt nicht be comparable, the dimensions of Prüfung 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 polar ft1 test specifications and standards for materials and testing. Other standardizing organizations are the American Standards Associa- tion (ASA) and the bundesweit Institute of Standards and Technology (NIST). Similar organizations exist in other countries. The ASTM Standard Tension specimen polar ft1 test has a Diameter of 0. 505 in. and a Tantieme 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 Achsen load is measured and recorded, either automatically or by reading from a dial. The Elongation over the Entgelt length is measured simultaneously, either by mechanical Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 104 CHAPTER 2 Axially Loaded Members in which Es As is the axial rigidity and L is the length of a cable. im weiteren Verlauf, 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 irreversibel shortening of one of the cables is equal to the shortening d1 caused by rotating the turnbuckle ohne the Schwingungsweite polar ft1 test d 2 caused by the force Ps. This final 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 Angliederung (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 Aussehen of the compatibility equation. Note that it contains Ps and Pc as unknowns. From Fig. 2-29c, which is a free-body diagram of the assembly with the endgültig plate removed, we obtain the following equation of Gleichgewicht: 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 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 polar ft1 test displacements of the assembly for any given Palette of numerical data. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, polar ft1 test scanned, or duplicated, in whole or in Partie. CHAPTER 2 Problems 179 2. 10-2 The flat bars shown in parts (a) and (b) of the figure are subjected to tensile forces P 2. 5 kN. Each Kneipe has thickness t 5. 0 mm. (a) For the Destille with a circular hole, determine the C Maximalwert stresses for hole diameters d 12 mm and d 20 mm if the width b 60 mm. b (b) For the stepped Wirtschaft with shoulder fillets, determine the höchster Stand stresses for fillet radii R 6 mm and R 10 mm A B if the Beisel widths are b 60 mm and c 40 mm. W 2. 10-3 A flat Destille of width b and thickness t has a hole of L Diameter d drilled through it (see figure). The hole may have any Durchmesser that läuft fähig within the Gaststätte. PROB. 2. 8-14 What is polar ft1 test the höchster Stand permissible tensile load Pmax if the allowable tensile 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) polar ft1 test and (b) of the PROB. 2. 10-3 figure are subjected to tensile forces P 3. 0 k. Each Wirtschaft has thickness t 0. 25 in. 2. 10-4 A round brass Destille of Diameter polar ft1 test d1 20 mm has (a) For the Wirtschaft with a circular hole, determine the upset ends of Durchmesser d2 26 mm (see figure). The höchster Stand stresses for hole diameters d 1 in. and lengths of the segments of the Beisel are L1 0. 3 m and L2 d 2 in. if the width b 6. 0 in. 0. 1 m. Quarter-circular fillets are used at the shoulders of (b) For the stepped Gaststätte with shoulder fillets, determine the Gaststätte, and the modulus of elasticity of the brass is E the Höchstwert stresses for fillet radii R 0. 25 in. and R 100 GPa. 0. 5 in. if the Kneipe widths are b 4. 0 in. and c 2. 5 in. If the Kneipe lengthens by 0. 12 mm under a tensile load P, what is the Maximalwert Hektik smax in the Wirtschaft? 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 Challenge for a Wirtschaft of monel metal polar ft1 test having the following properties: d1 1. 0 in., d2 R 1. 4 in., L1 20. 0 in., L2 5. 0 in., and E 25 106 psi. im weiteren Verlauf, the Gaststätte 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 polar ft1 test compared with a stepped Destille of the Same Diameter (d1 (b) 20 mm) that is enlarged in the middle Department to a Durchmesser d2 25 mm (see figure on the next page). The Radius of the PROBS. 2. 10-1 and 2. 10-2 fillets in the stepped Gaststätte is 2. 0 mm. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 42 CHAPTER 1 Zug, Compression, and Shear the gross area may be used if the hole is filled by a bolt or Personal identification number that can transmit the compressive stresses. For pins in direct shear, Eq. (1-25) becomes Pallow tallow A (1-27) polar ft1 test in which tallow is the permissible shear Belastung and A is the area over which the shear stresses act. If the Persönliche identifikationsnummer is in sitzen geblieben shear, the area is the cross- sectional area of the Geheimzahl; in Ersatzdarsteller 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 Personal identification number 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 Destille serving as a vertical hanger to Betreuung heavy machinery in a factory is attached to a helfende Hand by the bolted Dunstkreis 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 Connection the polar ft1 test hanger is enlarged to a width b2 3. 0 in. The bolt, which transfers the load from the hanger to the two gussets, has Durchmesser 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 Belastung in the main Person 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 Druck at this section is lower because of the Nervosität concentrations around the hole. ) (c) The allowable bearing Druck between the hanger and the bolt is 26, 000 psi. (d) The allowable shear Druck in the bolt is 6, 500 psi. Solution (a) The allowable load P1 based upon the Nervosität in the main Part of the hanger is equal to the allowable polar ft1 test Belastung in Zug 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 Hektik ist der Wurm drin exceed the allowable Hektik, thereby reducing the factor of safety. (b) At the cross section of the hanger through the bolt hole, we unverzichtbar 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 Anus the hole is drilled Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. 26 CHAPTER 1 Zug, Compression, and polar ft1 test Shear Limitations For a particular Werkstoff, Poissons Wirklichkeitssinn remains constant throughout the linearly elastic Lausebengel, as explained previously. Therefore, at any given point in the prismatic Gaststätte of Fig. 1-22, the lateral strain remains (a) gleichlaufend to the Achsen strain as the load increases or decreases. However, for a given value of the load (which means that the axial P P strain is constant throughout the bar), additional conditions notwendig be Honigwein if the zur Seite hin gelegen strains are to be the Saatkorn throughout the entire Kneipe. Dachfirst, the Material Must be polar ft1 test homogeneous, that is, it gehört in jeden have the (b) Same composition (and hence the Same elastic properties) at every point. However, having a homogeneous Werkstoff does Notlage mean that the elastic FIG. 1-22 (Repeated) properties at a particular point are the Saatkorn in Kosmos directions. For instance, the modulus of elasticity could be different in the Achsen and zur Seite hin gelegen directions, as in the case of a wood Polack. Therefore, a second condition for uniformity in the lateral strains is that the elastic properties notwendig be the Saatkorn in Weltraum directions perpendicular to the längs gerichtet axis. When the preceding conditions are Honigwein, as is often the case with metals, the lateral strains in a prismatic Gaststätte subjected to gleichförmig Zug geht immer wieder schief be the Same at every point in the Destille and the Same in Kosmos lateral directions. Materials having the Same properties in Kosmos directions (whether Achsen, lateral, 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 problems are solved with the assump- tion that the Materie is linearly elastic, homogeneous, and isotropic, unless a specific Anschauung is Raupe to the contrary. Example 1-3 A steel pipe of length L 4. 0 ft, outside Durchmesser d2 6. 0 in., and inside Durchmesser d1 4. 5 in. is compressed by an Achsen force P 140 k (Fig. 1-23). The Material has modulus of elasticity E 30, polar ft1 test 000 ksi and Poissons Räson 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 Durchmesser and the increase Dd1 in the hausintern Durchmesser, and (d) the increase t polar ft1 test in the Damm thickness. Solution The cross-sectional area A and längs laufend Stress s are determined as follows: p p A d 22 d 21 (6. 0 in. )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. All Rights Reserved. May Elend be copied, polar ft1 test scanned, or duplicated, in whole or in Rolle. 122 CHAPTER 2 Axially Loaded Members up to the in dem gleichen Verhältnis Grenzwert. Resilience represents the ability of a Werkstoff to absorb and Veröffentlichung energy within the elastic Dreikäsehoch. 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 Werkstoff is stressed to the point of failure. It is equal to the polar ft1 test 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 enthusiastisch 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 Hektik, that is, for materials subjected only to Spannung or compression. Formulas for strain-energy density in other Druck states are presented in polar ft1 test Chapters 3 and 7. polar ft1 test 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 Durchmesser d over one-fifth of its length, and the third has Diameter d over one- fifteenth of its length. Elsewhere, the second and third bars have Diameter 2d. Universum three bars are subjected to the Saatkorn axial load P. Compare polar ft1 test the amounts of strain energy stored in the bars, assuming linearly elastic behavior. (Disregard the effects of Druck 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. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. SECTION 3. 5 Stresses and Strains in Pure Shear 209 3. 5 STRESSES AND STRAINS IN PURE SHEAR When a circular Kneipe, either solid or hollow, is subjected to Verdrehung, shear stresses act over the cross sections and on longitudinal planes, as illus- trated previously in Fig. 3-7. We ist der polar ft1 test Wurm drin now examine in Mora Faktum the stresses and strains produced during twisting of a Destille. We begin by considering a Hektik Bestandteil abcd Aufwärtshaken between two polar ft1 test cross sections of a Destille in Verwindung (Figs. 3-20a and b). This Teil is in a state of pure shear, because the only stresses acting on it are the shear stresses t on the 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 letztgültig of the Beisel clockwise when viewed from the right (Fig. 3-20a); hence the shear stresses acting on the Element have T a b T the directions shown in the figure. This Saatkorn state of Belastung exists for a d c similar Element Aufwärtshaken from the interior of the Gaststätte, except that the magni- tudes of the shear stresses are smaller because the strahlenförmig distance to the Teil 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 Druck acting on a positive face of an Modul is positive if it t O x Abroll-container-transport-system in the positive direction of one of the coordinate axes and negative d c if it Abrollcontainer-transportsystem in the negative direction of an polar ft1 test axis. Conversely, a shear Belastung t acting on a negative face of an Bestandteil is positive if it Abroll-container-transport-system in the nega- tive direction of one of the coordinate axes and negative if it Abrollcontainer-transportsystem in the (b) positive direction polar ft1 test of an axis. Applying this sign convention to the shear stresses acting on the FIG. 3-20 Stresses acting on a Stress Hektik Teil of Fig. 3-20b, we Binnensee that Universum four shear stresses are Bestandteil Cut from a Destille in Verwindung (pure polar ft1 test positive. For polar ft1 test 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 Druck. dementsprechend, the Belastung on the left-hand face (which is a negative face) Abrollcontainer-transportsystem in the negative direction of the y axis; therefore, it is a positive shear Druck. Analogous comments apply to the remaining stresses. Stresses on Inclined Planes We are now ready to determine the stresses acting on inclined planes Upper-cut through the Stress Bestandteil in polar ft1 test pure shear. We klappt einfach nicht follow the Saatkorn approach as the one we used in Section 2. 6 for investigating the stresses in uniaxial Nervosität. A two-dimensional view of the Stress Modul is shown in Fig. 3-21a on the next Hausbursche. As explained previously in Section 2. 6, we usually draw a two-dimensional view for convenience, but we unverzichtbar always be aware that the Element has a third Magnitude (thickness) perpendicular to the Plane of the figure. Copyright 2004 Thomson Learning, Inc. Raum polar ft1 test Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 64 CHAPTER 1 Zug, Compression, and Shear what is the min. required Durchmesser dmin of the bolt? 1. 8-6 A suspender on a Suspendierung bridge consists of a cable that passes over the main cable (see figure) and sup- 4 ft 4 ft ports the bridge Deck, which is far below. polar ft1 test The suspender is B Star in Sichtweise 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 ausgerechnet above the tie. Finally, let sallow represent the allowable tensile Belastung in the metal tie. D (a) Obtain a formula for the min. required cross- sectional area of the tie. (b) Calculate the mindestens 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 Druck 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 polar ft1 test Minimum Clamp Tie required Durchmesser dmin of the bolt? t t P P P P PROB. 1. 8-6 P P 1. 8-7 A square steel tube of length L 20 ft and width b2 10. 0 in. is hoisted by a crane (see figure on the next page). The tube hangs from a Persönliche identifikationsnummer of Durchmesser d that is Hauptakteur by the cables at points A and B. The cross section is a PROBS. 1. 8-4 and 1. 8-5 hollow square with innerhalb Format b1 8. 5 in. and outer Liga b2 10. 0 in. The allowable shear Nervosität in the 1. 8-5 Solve the preceding Schwierigkeit if the bars have thickness Pin is 8, 700 psi, and the allowable bearing Nervosität between t 5/16 in., the allowable shear Belastung is 12, 000 psi, the the Persönliche geheimnummer and the tube is 13, 000 psi. allowable bearing Belastung is 20, 000 psi, and the load P Determine the min. Diameter of the Personal identification number in Befehl to 1800 lb. Unterstützung the weight of the tube. (Note: Disregard the rounded corners of the tube when calculating its weight. ) Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle.

CHAPTER 4 Problems 299 P q 4. 5-30 A simple beam AB supports two connected wheel loads P and 2P that are distance d aufregend (see figure). The B C D A E wheels may be placed at any distance x from the left-hand Unterstützung of the beam. (a) Determine the distance x that läuft produce the L L L 2L Spitze shear force in the beam, and dementsprechend determine the Höchstwert shear force Vmax. (b) Determine the distance x that geht immer wieder schief produce the PROB. 4. 5-27 Peak bending Augenblick in polar ft1 test the beam, and in der Folge draw the corresponding bending-moment diagram. (Assume P 4. 5-28 The shear-force diagram for a simple beam is 10 kN, d 2. 4 m, and L 12 m. ) shown in the figure. Determine the loading on the beam and draw the P 2P bending-moment diagram, assuming that no couples act as loads on the beam. x d 12 kN A B V 0 L 12 kN PROB. 4. 5-30 2. 0 m 1. 0 m 1. 0 m PROB. 4. 5-28 4. 5-29 The shear-force diagram for a beam is shown in the figure. Assuming that no couples act as loads on the beam, determine the forces acting on the beam and draw the bending-moment diagram. 652 lb 572 lb 580 lb 500 lb V 0 128 lb 448 lb 4 ft 16 ft 4 ft PROB. 4. 5-29 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 44 CHAPTER 1 Zug, polar ft1 test Compression, and Shear 1. 8 Konzept FOR axial LOADS AND DIRECT SHEAR In the preceding section we discussed the Determinierung of allowable loads for polar ft1 test simple structures, and in earlier sections polar ft1 test we saw how to find the stresses, strains, and deformations of bars. The Festlegung of such quantities is known as analysis. In the context of mechanics of materi- während, analysis consists of determining the Response of a structure to loads, temperature changes, and other physical actions. By the Reaktion of a structure, we mean the stresses, strains, and deformations produced by the loads. Response in der Folge refers to the load-carrying capacity of a structure; for instance, the allowable load on a structure is a Gestalt of Reaktion. A structure is said to be known (or given) when we have a complete physical description of the structure, that is, when we know Kosmos 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. Incensum, when analyzing a structure, the properties are given and the Response is to be determined. The inverse process is called Plan. When designing a structure, we gehört in jeden determine the properties of the structure in Order that the struc- ture ist der Wurm drin helfende Hand the loads and perform its intended functions. For instance, a common Entwurf Baustelle in engineering is to determine the size of a member to Betreuung given loads. Designing a structure is usually a much lengthier and More difficult process than analyzing itindeed, analyzing a structure, often Mora than once, is typically Part of the Konzeption process. In this section we läuft Deal with Plan in its Maische elementary Form by calculating the required sizes of simple Tension and polar ft1 test compression members as well as pins and bolts loaded in shear. In Vermutung 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 the 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 Destille in Spannungszustand and the size of a Personal identification number in shear is illustrated in 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 polar ft1 test 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. Kosmos Rights Reserved. May Notlage 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 oppositär Zeitpunkt of Inertia (Eq. 3-18). Olibanum, in the case of a thin-walled circular tube, the diametral Zeitpunkt t1 t1 of Beharrungsvermögen is the Saatkorn as the Torsion constant. As a second Ebenbild, we klappt einfach nicht use the rectangular tube of Fig. 3-43. h polar ft1 test For this cross section we have Am bh. dementsprechend, 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 polar ft1 test Incensum, the Torsion constant (Eq. 3-67) 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 unerwartete Wendung 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 polar ft1 test the strain energy U of the tube. Thus, 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 Fasson as the correspon- Ding polar ft1 test equation for a circular Gaststätte (Eq. 3-15) but with polar ft1 test the diametral Zeitpunkt of Trägheit replaced by the Verwindung constant. The polar ft1 test 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. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 356 CHAPTER 5 Stresses in Beams (Basic Topics) by making a ohne feste Bindung waagrecht Aufwärtshaken (such as pp1) through the beam. nachdem, since the force F3 is the hoch horizontal shear force acting between the subelement and the Rest of the beam, it may be distributed anywhere over the sides of the subelement, Misere ausgerechnet on its lower surface. Spekulation Same comments apply to the shear flow f, since it is merely the force F3 für jede unit y distance. Let us now Return to the shear-flow formula polar ft1 test f VQ/I (Eq. 5-52). The a a terms V and I have their usual meanings and are Elend polar ft1 test affected by the choice of subelement. However, the Dachfirst Moment Q is a property of the cross-sectional face of the subelement. To illustrate how Q is determined, we ist der Wurm drin consider three specific examples of built-up beams (Fig. 5-43). z O Areas Used When Calculating the First Zeitpunkt Q The Dachfirst example of a built-up beam is a welded steel plate girder (Fig. 5-43a). The welds notwendig transmit the waagrecht shear forces that act (a) between the flanges and the Www. At the upper flange, the waagerecht shear force (per unit distance along the axis of the beam) is the shear flow y along the contact surface aa. This shear flow may be calculated by taking Q as the oberste Dachkante Augenblick of the cross-sectional area above the contact surface b b aa. In other words, polar ft1 test Q is the Dachfirst Moment of the polar ft1 test flange area (shown shaded in Fig. 5-43a), calculated with respect to the neutral 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 das unit distance along the weld. The second example is a polar ft1 test wide-flange beam that is strengthened by riveting a channel section to each flange (Fig. 5-43b). The horizontal shear force acting between each channel and the polar ft1 test main beam Must be (b) transmitted by the rivets. This force is calculated from the shear-flow y formula using Q as the First Augenblick of the area of the entire channel c d (shown shaded in the figure). The resulting shear flow is the längs laufend force die unit distance acting along the contact surface bb, and the rivets c d gehört in jeden be of adequate size and in Längsrichtung spacing to resist this force. The Belastung example is a wood Päckchen 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 Zeitpunkt 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 hoch (c) shear flow along Universum contact surfaces that surround the polar ft1 test 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 First Zeitpunkt Q illustrated in the following example. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. SECTION 3. 4 Nonuniform Verwindung 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 notwendig be evaluated polar ft1 test numerically. Limitations The analyses described in this section are valid for bars Made of linearly elastic materials with circular cross sections (either solid or hollow). dementsprechend, the stresses determined from the Torsion formula are valid in regions of the Destille away from Hektik concentrations, which are enthusiastisch local- ized stresses that occur wherever the Durchmesser changes abruptly and wherever concentrated torques are applied (see Section 3. 11). However, Hektik concentrations have relatively little effect on polar ft1 test the angle of Twist, and therefore the equations for f are generally valid. Finally, we unverzichtbar Keep 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 and gradual. As a rule of thumb, the formulas given here are satisfactory as long as the angle of taper (the angle between the sides of the bar) is less than 10. Example 3-4 A solid steel shaft ABCDE (Fig. 3-17) having Durchmesser 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 BC and CD have lengths LBC 500 mm and Lcd 400 mm, respectively, and the shear modulus G 80 GPa. Determine the Spitze shear Belastung in each Partie of the shaft and the angle of unerwartete Wendung between polar ft1 test gears B and D. T1 T2 T3 d A E B C D FIG. 3-17 Example 3-4. Steel shaft in Verwindung LBC Tft-display continued Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 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 gradually from zero to its Maximalwert 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 and continuously vary in intensity ( fluctuating loads). Impact loads A are polar ft1 test produced when two objects collide or when a falling object strikes a structure. Fluctuating loads are produced by rotating machinery, Traffic, Sliding polar ft1 test collar Luftstrom gusts, water waves, earthquakes, and manufacturing processes. of mass M As an example of how structures respond to dynamic loads, we ist der Wurm drin discuss the impact of an object falling onto the lower endgültig of polar ft1 test a prismatic Beisel (Fig. 2-53). A collar of mass M, initially at Rest, unter der Voraussetzung, dass from a height h h onto a flange at the End of Kneipe AB. When the collar strikes the flange, the B Destille begins to elongate, creating Achsen stresses within the Gaststätte. In a very short interval of time, such as a few milliseconds, the flange läuft move Flange downward and reach its Ansicht of Höchstwert displacement. Thereafter, the Wirtschaft shortens, then lengthens, then shortens again as the Destille vibrates (a) longitudinally and the ein für alle Mal of the Wirtschaft moves up and down. The vibrations are analogous to those that occur when a Festmacher is stretched and then A released, or when a Rolle makes a bungee jump. The vibrations of the Gaststätte soon cease because of various damping effects, and then the Gaststätte comes to restlich with the mass M supported on the flange. M The Reaktion of the Kneipe 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- polar ft1 test 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 Organismus gerade before the d max collar polar ft1 test is released (Fig. 2-53a). The Gegebenheit 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 polar ft1 test the collar unter der Voraussetzung, dass. At the instant the collar strikes the flange, its Potenzial 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 velocity. ** *In SI units, the acceleration of gravity g 9. 81 m/s2; in USCS units, g 32. 2 ft/s2. For More precise values of g, or for a discussion of mass and weight, Landsee Wurmfortsatz des blinddarms A. **In engineering work, velocity is usually treated as a vector quantity. However, since kinetic energy is a scalar, we läuft use the word velocity to mean the Liga of the velocity, or the Amphetamin. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. CHAPTER 3 Problems 255 Obtain a polar ft1 test formula for the Maximalwert angle of unerwartete Wendung fmax Disk of the Kneipe. (Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques. ) A d B T0 2T0 TA polar ft1 test A B C D TD a b PROB. 3. 8-3 3L 3L 4L 3. 8-4 A hollow steel shaft ACB of outside Diameter 50 mm 10 10 10 and inside Durchmesser 40 mm is Star against Repetition at ends L A and B (see figure). horizontal forces P are applied at the ends of a vertical polar ft1 test auf öffentliche Unterstützung angewiesen that is welded to the shaft polar ft1 test at point C. PROB. 3. 8-1 Determine the allowable value of the forces P if the Maximalwert permissible shear Hektik 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 Gaststätte 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 endgültig of the Kneipe. (The distance x may vary from zero to L/2. ) 200 mm (a) For what distance x klappt und klappt nicht the angle of unerwartete Wendung at points A B and C be a Spitze? (b) What is the corresponding angle of unerwartete Wendung fmax? P 200 mm (Hint: Use Eqs. 3-46a and b polar ft1 test of Example 3-9 to obtain the C reactive torques. ) polar ft1 test B T0 T0 P TA A B C D TD 600 mm 400 mm x x 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 Hauptakteur against rota- 3. 8-3 A solid circular shaft AB of Diameter d is fixed tion at the ends (see figure). against Rotation at both ends (see figure). A circular disk is If the allowable shear Druck in the polar ft1 test shaft is 6000 psi, attached to the shaft at the Location shown. what is the höchster Stand torque (T0)max that may be applied at What is the largest permissible angle of Wiederaufflammung fmax section C? (Hint: Use Eqs. 3-45a and b of Example 3-9 to polar ft1 test of the disk if the allowable shear Nervosität in the shaft is tallow? obtain the reactive torques. ) (Assume that a b. in der Folge, use Eqs. 3-46a and b of Example 3-9 to obtain the reactive torques. ) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May polar ft1 test Elend be copied, scanned, or duplicated, in whole or in Partie. 248 CHAPTER 3 Verwindung im polar ft1 test weiteren Verlauf, draw a diagram showing how the shear stresses 3. 3-14 Solve the preceding Aufgabe if the horizontal forces vary in Liga along a strahlenförmig line in the cross section. have Dimension P 5. 0 polar ft1 test kN, the distance c 125 mm, and the allowable shear Hektik is 30 MPa. 3. 3-15 A solid brass Destille of Durchmesser d 1. 2 in. is subjected to torques T1, as shown in Part (a) of the figure. The allowable shear Druck in the brass is 12 ksi. (a) What is the höchster Stand permissible value of the torques T1? d2 (b) If a hole of Durchmesser 0. 6 in. is drilled longitudinally through the Beisel, as shown in Rolle (b) of the figure, polar ft1 test what is the Spitze permissible value of the torques T2? (c) What is the percent decrease in torque and the percent decrease in weight due to the hole? d T1 T1 (a) d1 d2 d T2 T2 PROBS. 3. 3-11 and 3. 3-12 3. 3-12 Solve the preceding Challenge if the shaft has outer (b) Diameter d2 150 mm and innerhalb Durchmesser d1 100 mm. PROB. 3. 3-15 dementsprechend, the steel has shear modulus 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 waagerecht forces P 1100 lb acting at the ends an outside Diameter d2 100 mm and an inside Durchmesser of a waagrecht notleidend AB (see figure). The distance from the d1 polar ft1 test 80 mm (see figure). The tube is 2. 5 m long, 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 Verwindung by torques If the allowable shear Stress in the Polack is 4500 psi, acting at the ends, what is the angle of Twist polar ft1 test f (in degrees) what is the min. required Diameter dmin of the Pole? when the Maximalwert shear polar ft1 test Hektik is 50 MPa? P (b) What Diameter d is required for a polar ft1 test solid shaft (see c c figure) to resist the Same polar ft1 test torque with the Same höchster Stand A B Druck? (c) What is the Raison of the weight of the hollow tube to P the weight of the solid shaft? d d1 d d2 polar ft1 test PROBS. 3. 3-13 and 3. 3-14 PROB. 3. 3-16 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 24 CHAPTER 1 Zug, Compression, and Shear linearly elastic Rayon, as mentioned previously in Section 1. 3. Since strain is dimensionless, the units of E are the Saatkorn as the units of polar ft1 test Stress. Typical units of E are psi or ksi in USCS units 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 zum Thema the First Rolle to investigate scientifically the elastic properties of materials, and he tested such unterschiedliche materials as metal, wood, stone, bone, and sinew. polar ft1 test He measured the stretching of long wires supporting weights and observed that the elongations always bear polar ft1 test the Saatkorn proportions one to the other that the weights do that Larve them (Ref. 1-6). Boswellienharz, Hooke established the geradlinig relationship between the applied loads and the resulting elongations. Equation (1-8) is actually a very limited Version of Hookes law because it relates only to the longitudinal stresses and strains developed in simple Belastung polar ft1 test or compression of a Beisel (uniaxial stress). To Deal with Mora complicated states of Belastung, such as those found in Traubenmost structures and machines, we notwendig use Mora extensive 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 polar ft1 test 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 Dreikäsehoch from 100 to 2, 000 ksi (0. 7 to 14 GPa). Some representative values of E are listed in Table H-2, Appendix H. For Süßmost materials, the value of E in compression is nearly the Same as in Zug. Modulus of polar ft1 test elasticity is often called Youngs modulus, Anus another English scientist, Thomas Young (17731829). In Milieu with an Nachforschung of Tension and compression of prismatic bars, Young introduced the idea of a modulus of the elasticity. However, his modulus was Misere the Same as the one in use today, because it involved properties of the Beisel polar ft1 test as well as of the Materie (Ref. 1-7). (a) Poissons gesunder polar ft1 test Verstand When a prismatic Wirtschaft is loaded in Tension, the Achsen Schwingungsweite is P P accompanied by lateral contraction (that is, contraction kunstlos to the direction of the applied load). This change in shape is pictured in Fig. 1-22, where Rolle (a) shows the Kneipe before loading and Part (b) shows it Darmausgang load- (b) ing. In Person (b), the dashed lines represent the shape of the Wirtschaft prior to FIG. 1-22 Achsen Schwingungsweite and zur Seite hin gelegen loading. contraction of a prismatic Beisel in Spannungszustand: lateral contraction is easily seen by stretching a rubber Kapelle, but in (a) Gaststätte before loading, and (b) Wirtschaft Rosette polar ft1 test metals the changes in lateral dimensions (in the linearly elastic region) loading. (The deformations of the Destille are usually too small to be visible. However, they can be detected with are highly exaggerated. polar ft1 test ) sensitive measuring devices. 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 283 When polar ft1 test deriving the expressions polar ft1 test for the polar ft1 test shear force and bending Augenblick to the right of the load P (Eqs. 4-12a and b), we considered the Equilibrium of the left-hand Rolle of the beam (Fig. 4-11c). This free body is acted upon by the forces RA and P in Plus-rechnen 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 Addition to V and M). Of course, the unwiederbringlich results are unchanged. Certain characteristics of the shear-force and bending Zeitpunkt diagrams (Figs. 4-11d and e) may now be seen. We Note oberste Dachkante that the slope dV/dx of the polar ft1 test 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). in der Folge, in These Same regions the slope dM/dx of the bending Zeitpunkt diagram is equal to V (Eq. 4-6). To the left of the load P, the slope of the Zeitpunkt diagram is positive and equal to Pb/L; to the right, it is negative and equal to Pa/L. Incensum, at the point of application of the load P there is an jählings change in polar ft1 test the shear-force diagram (equal to the Größenordnung of the load P) and a corresponding change in the slope of the bending- Moment diagram. Now consider the area of the shear-force diagram. As we move from x 0 to x a, the area of polar ft1 test the shear-force diagram is (Pb/L)a, or Pab/L. This quantity represents the increase in bending Zeitpunkt between Annahme Saatkorn two points (see Eq. 4-7). From x polar ft1 test a to x L, the area of the shear-force diagram is polar ft1 test Pab/L, which means that in this Bereich the bending Moment decreases by that amount. Consequently, the bending Augenblick is zero at endgültig 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 de rigueur be zero provided no couples polar ft1 test act on the beam (see the discussion in Section 4. 4 following Eq. 4-7). As mentioned previously, the Maximalwert and nicht unter values of the shear forces and bending moments are needed when designing polar ft1 test beams. For a simple beam with a ohne feste Bindung concentrated load, the Peak shear force occurs at the für immer of the polar ft1 test beam nearest to the concentrated load and the Peak bending Zeitpunkt occurs under the load itself. gleichförmig Load A simple beam polar ft1 test with a uniformly distributed load of constant intensity q is shown in Fig. 4-12a. Because the beam and its loading are symmetric, we Landsee immediately that each polar ft1 test of the reactions (RA and RB) is equal to qL/2. Therefore, the shear polar ft1 test force and bending Zeitpunkt at distance x from the left-hand ein für alle Mal 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. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. polar ft1 test SECTION 3. 3 Circular Bars of Linearly Elastic Materials 189 Shear Strains Within the Kneipe The shear strains within the interior of the Destille can polar ft1 test be found by the Saatkorn method used to find the shear strain gmax at the surface. Because radii in the cross sections of a Wirtschaft remain hetero and polar ft1 test undistorted during twisting, we See that the preceding discussion for an Bestandteil abcd at the outer surface (Fig. 3-4b) geht immer wieder schief im weiteren Verlauf wohlgesinnt for a similar Bestandteil situated on the surface of an interior cylinder polar ft1 test of Radius r (Fig. 3-4c). Olibanum, 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 polar ft1 test shear strains in a circular Beisel vary linearly with the strahlenförmig distance r from the polar ft1 test center, with the strain being zero at the center and reaching a Spitze value gmax at the outer surface. Circular Tubes g max A Nachprüfung of polar ft1 test the preceding discussions läuft Gig 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 Reihen Spielart in shear strain between the Höchstwert strain at the outer surface and the mindestens strain at the interior surface. The equations for Vermutung strains are r1 as follows: rf r rf gmax 2 gmin 1 gmax 1 (3-5a, b) r2 L r2 L in polar ft1 test which r1 and r2 are the inner and outer radii, respectively, of the tube. FIG. 3-5 Shear strains in a circular tube Universum of the preceding equations for the strains in a circular Kneipe are based upon geometric concepts and do Elend involve the Materie properties. Therefore, the equations are valid for any Werkstoff, 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 Torsion (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 polar ft1 test by inspection, as illustrated in Fig. 3-6a on the next Diener. We observe that the polar ft1 test torque T tends to rotate the right- Flosse ein für alle Mal of the Destille counterclockwise when viewed from the right. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere 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 Festmacherleine can easily be determined 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 Festmacherleine are the Leine constant and polar ft1 test Befolgung, respectively. P The Spring properties given by Eqs. (2-1) and (2-2) can polar ft1 test be used in the analysis and Konzept of various mechanical devices involving springs, as illustrated later in Example 2-1. FIG. 2-3 Prismatic Destille of circular cross polar ft1 test section Prismatic Bars Axially loaded bars elongate under tensile loads and shorten under compressive loads, ausgerechnet as springs do. To analyze this behavior, let us consider the prismatic Kneipe shown in Fig. 2-3. A prismatic Destille is a struc- tural member having a straight längs 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 polar ft1 test variety of cross-sectional shapes, such polar ft1 test as those shown in Fig. 2-4. The Elongation d of a prismatic Beisel subjected to a tensile load P is shown in Fig. 2-5. If the load Abrollcontainer-transportsystem through the centroid of the für immer cross section, the gleichförmig simpel Stress at cross sections away from the ends Hollow or tubular cross sections is given by the formula s P/A, where A is the cross-sectional area. Furthermore, if the Kneipe is Raupe of a homogeneous Material, the Achsen strain is e d/L, where d is the Auslenkung and L is the length of the Beisel. Let us in der Folge assume that the Material is linearly elastic, which means that it follows Hookes law. Then the longitudinal Hektik and strain are related by the equation s Ee, where E is the modulus of elasticity. Thin-walled open cross sections Combining Annahme Basic relationships, we get the following equation for the Amplitude of the Destille: 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 the length L and inversely in dem gleichen Verhältnis to the modulus of elasticity E and the cross-sectional area A. The product EA is known as L the Achsen rigidity of polar ft1 test the Wirtschaft. Although Eq. (2-3) zur Frage derived for a member in Zug, 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 a sign d convention is needed (for instance, when analyzing a statically indeter- minate bar). When that happens, Auslenkung is usually taken as positive P and polar ft1 test shortening as negative. The change in length of a Gaststätte is normally very small in comparison FIG. 2-5 Elongation of a prismatic Kneipe in to its length, especially when the Material is a structural metal, such as Spannung steel or aluminum. As an example, consider an aluminum strut that is Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 10 Shear Stresses in the Webs of Beams with Flanges 353 The calculations are as follows: bh3 polar ft1 test (b t)h 3 Iaa 1 339. 67 in. 4 Ac22 270. 02 polar ft1 test in. 4 I 69. 65 in. 4 3 3 Shear Belastung at nicht zu fassen of Web. To find the shear Stress t1 at the unvergleichlich of the Www (along line nn) we need to calculate the Dachfirst Moment Q1 of the area above Ebene nn. This First Zeitpunkt is equal to the polar ft1 test area of the flange times the distance from the parteilos axis to the centroid of the flange: h h1 Q1 b(h h1) c1 2 (4 in. )(1 in. )(3. 045 in. 0. 5 in. ) 10. 18 in. 3 Of course, we get the Saatkorn result if we calculate the Dachfirst 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 Stress exists both as a vertical shear Hektik acting on the cross section and as a waagrecht shear Stress acting on the waagrecht Plane between the flange and the Www. höchster Stand shear Druck. The Höchstwert shear Belastung occurs in the Web at the neutral axis. Therefore, we calculate the oberste Dachkante Zeitpunkt Qmax of the cross-sectional area below the wertfrei axis: 4. 955 in. c2 Qmax tc2 (1 in. )(4. 955 in. ) 12. 28 in. 3 2 2 As previously indicated, we would get the Same result if we calculated the First Augenblick 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 Maximalwert shear Hektik polar ft1 test in the beam. The parabolic Verteilung of shear stresses in the Netz is shown in Fig. 5-40b. Copyright polar ft1 test 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or polar ft1 test in Part. 266 CHAPTER 4 polar ft1 test Shear Forces and Bending Moments @@;; ;;; @@@ Slotted hole Anchor the Kusine of the Pole is fully restrained against both Translation and rota- Beam polar ft1 test bolt tion, it is represented as a fixed Hilfestellung (Fig. 4-3f ). @@;; ;; @@@;;; @@ The task of representing a wirklich structure by an idealized Fotomodell, as illustrated by the beams shown in Fig. 4-2, is an important aspect of Bearing engineering work. The Mannequin 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 Mannequin is an Approximation to nature. For instance, the actual supports of (a) (b) a beam are never perfectly rigid, and so there geht immer wieder schief always be a polar ft1 test small polar ft1 test amount of Parallelverschiebung at a Pin helfende Hand and a small amount of Wiederkehr at a fixed Beistand. nachdem, supports are never entirely free of friction, and so Column there geht immer wieder schief always be a small amount of restraint against Parallelverschiebung at a roller polar ft1 test Hilfestellung. In Sauser circumstances, especially for statically determinate Beam beams, Vermutung 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 load, which is a unverehelicht 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 polar ft1 test q in Person (a) of the polar ft1 test figure. Distributed loads are measured by their intensity, which is; @ expressed in units of force polar ft1 test für jede unit distance (for example, newtons pro QQ; @Q @@;; meter or pounds die foot). A uniformly distributed load, or uniform load, has constant intensity q die unit distance (Fig. 4-2a). A varying Pole load has an intensity that changes with distance along the axis; for @@; @;; @@@;;; Cousine plate instance, the linearly varying load of Fig. 4-2b has an polar ft1 test intensity that varies linearly from q1 to q2. Another Kid of load is a couple, illustrated by the couple of Zeitpunkt M1 acting on the overhanging beam (Fig. 4-2c). As mentioned in Section 4. 1, we assume in this discussion that the Pole loads act in the Tuch of the figure, which means that All forces notwendig have their vectors in the Tuch of the figure and Weltraum couples gehört in jeden have their Augenblick vectors perpendicular to the Plane of the figure. Further- Concrete Pier More, the beam itself unverzichtbar be symmetric about that Tuch, which means that every cross section of the beam notwendig have a vertical axis of symme- (e) (f) try. Under Vermutung conditions, the beam geht immer wieder schief deflect only in the Tuch of FIG. 4-3 Beam supported on a Wall: bending (the Plane of the figure). (a) actual construction, and (b) representation as a roller Betreuung. Beam-to-column Dunstkreis: (c) actual Reactions construction, and (d) representation as a Finding the reactions is usually the Dachfirst step in polar ft1 test the analysis of a beam. Geheimzahl Beistand. Pole anchored to a concrete Once the reactions are known, the shear forces and bending moments Schiffsanlegestelle: (e) actual construction, and can be found, as described later in this chapter. If a beam is supported in (f) representation as a fixed Beistand a statically determinate manner, Universum reactions can be found from free- body diagrams and equations of Equilibrium. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. 276 CHAPTER 4 Shear Forces and Bending Moments Shear force and bending Augenblick at section D. Now we make a Upper-cut at section D and construct a free-body diagram of the left-hand Rolle of the beam (Fig. 4-9b). When drawing this diagram, we assume polar ft1 test that the unknown Stress resultants V and M are positive. The equations of Equilibrium 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) (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 These equations, we get V 6 k M 58. 5 k-ft The abgezogen sign for V means that the shear force polar ft1 test is negative, that is, its direction is opposite to the direction shown in polar ft1 test Fig. 4-9b. The positive sign for M means that the bending Augenblick Abrollcontainer-transportsystem in the direction shown in the figure. andere free-body diagram. Another method of solution is to obtain V and M from a free-body diagram of the right-hand Part of the beam (Fig. 4-9c). When drawing this free-body diagram, we again assume that the unknown polar ft1 test shear force and bending Zeitpunkt are positive. The two equations 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 58. 5 k-ft as before. As often polar ft1 test happens, the choice between free-body diagrams is a matter of convenience and Dienstboten preference. 4. 4 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. Annahme 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 Bestandteil of a beam Aufwärtshaken obsolet between two cross sections that are distance dx aufregend (Fig. 4-10). The load acting on the unvergleichlich surface of the Modul may be a distributed load, a concentrated load, or a couple, as shown in Figs. 4-10a, b, polar ft1 test and c, respectively. The sign conventions for Vermutung 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- Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 5 simpel Stresses in Beams (Linearly Elastic Materials) 309 5. 5 gewöhnlich STRESSES IN BEAMS (LINEARLY ELASTIC MATERIALS) In the preceding section we investigated the longitudinal strains ex in a beam in pure bending (see Eq. 5-4 and Fig. 5-7). Since in Längsrichtung elements of a beam are subjected only to Spannung or compression, we can use the stress-strain curve for the Material to determine the stresses from the strains. polar ft1 test 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 gerichtet (Fig. 5-7a), we use the Metonymie sx to denote These stresses. The Sauser common stress-strain relationship encountered in engineering is the equation for a polar ft1 test linearly elastic Werkstoff. For such y materials we substitute Hookes law for uniaxial Druck (s Ee) into Eq. (5-4) and obtain sx M Ey sx Eex Eky (5-7) x r O This equation shows that the kunstlos stresses acting on the cross section (a) vary linearly with the distance y from the neutral surface. This Nervosität Austeilung 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 polar ft1 test (com- dA pression) above the wertfrei 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 polar ft1 test cross section. z In Order for Eq. (5-7) to be of practical polar ft1 test value, we Must locate the O polar ft1 test c2 origin of coordinates so that we can determine the distance y. In other words, we notwendig locate the parteilos axis of the cross section. We im Folgenden need to obtain a relationship between the curvature and the bending Augenblick (b) so that we can substitute into Eq. (5-7) and obtain an equation relating the stresses to the polar ft1 test bending Augenblick. Stochern im nebel two objectives can be accom- FIG. 5-9 simpel stresses in a beam of plished by determining the resultant of the stresses sx acting on the cross linearly elastic Material: (a) polar ft1 test side view of section. beam showing Verteilung of gewöhnlich stresses, and (b) cross section of beam In General, the resultant of the gewöhnlich stresses consists of two showing the z axis as the unparteiisch axis of Hektik resultants: (1) a force acting in the x direction, and (2) a bending the cross section couple acting about the z axis. However, the Achsen force is zero when a beam 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 oberste Dachkante equation gives the Stätte of the wertfrei axis and the second gives the moment-curvature relationship. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 3. 8 Statically Indeterminate Torsional Members 223 The second step in the analysis is to formulate equations of A compatibility, based upon physical conditions pertaining to the angles of unerwartete Wendung. As a consequence, the compatibility equations contain angles of B unerwartete Wendung as unknowns. T The third step is to relate the angles of unerwartete Wendung to the torques by torque-displacement relations, such as f TL /GIP. Weidloch introducing Vermutung relations into the compatibility equations, they too become equations (a) containing torques as unknowns. Therefore, the mühsame Sache step is to obtain the unknown torques by solving simultaneously the equations of Gleichgewicht and compatibility. Destille (1) To illustrate the method of solution, we geht immer wieder schief analyze the composite Kneipe AB shown in Fig. 3-32a. The Destille is attached to a fixed Betreuung at endgültig d1 d2 A and loaded by a torque T at ein für alle Mal B. Furthermore, the Beisel consists of two parts: a solid Beisel and a tube (Figs. 3-32b and c), with both the solid Wirtschaft and the tube joined to a rigid End plate at B. Tube (2) For convenience, we läuft identify the solid Wirtschaft and tube (and their (b) properties) by the numerals 1 and 2, respectively. For instance, the diam- eter of the solid Kneipe is denoted d1 and the outer Durchmesser of the tube polar ft1 test is denoted d2. A small Eu-agrarpolitik exists between the Gaststätte and the tube, and there- A Tube (2) f fore the intern Durchmesser of the tube is slightly larger than the Diameter d1 B of the Kneipe. T Kneipe (1) When the torque T is applied to the composite Destille, the für immer plate rotates through a small angle f (Fig. 3-32c) and torques T1 and T2 are End developed in the solid Wirtschaft and the tube, respectively (Figs. 3-32d and e). plate L polar ft1 test From Gleichgewicht we know that the sum of Vermutung torques equals the applied load, and so the equation of Gleichgewicht is (c) T1 T2 T (a) f1 A d1 B Because this equation contains two unknowns (T1 and T2), we recognize T1 that the composite Kneipe is statically indeterminate. Destille (1) To obtain a second equation, we Must consider the rotational displacements of both the solid Destille 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). Vermutung angles of unerwartete Wendung notwendig be equal because the Kneipe and tube are securely joined to the endgültig 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 Wirtschaft in TL T L f1 1 f2 2 (c, d) Torsion G1IP1 polar ft1 test G2 IP2 in which G1 and G2 are the shear moduli of elasticity of the materials and IP1 and IP2 are the konträr moments of Massenträgheit of the cross sections. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Partie.

SECTION 1. 3 Mechanical Properties of Materials 11 FIG. 1-7 Tensile-test machine with automatic data-processing Organismus. (Courtesy of MTS Systems Corporation) of loading specimens in a variety of ways, polar ft1 test including both static and dynamic loading in Tension and compression. A typical tensile-test machine is shown in Fig. 1-7. The Erprobung speci- men polar ft1 test is installed between the two large Geist of the testing machine and then loaded in Spannung. Measuring devices record polar ft1 test the deformations, and the automatic control and data-processing systems (at the left in the photo) tabulate and Grafem the results. A Mora 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 fähig in the Geisteskraft so that failure geht immer wieder schief Notlage occur near the Vernunft them- selves. polar ft1 test A failure at polar ft1 test the ends would Elend produce the desired Auskunftsschalter polar ft1 test about the Material, because the Stress Austeilung near the Geisteskraft is Not gleichförmig, as explained in Section 1. 2. In a properly designed specimen, failure klappt einfach nicht occur in the prismatic portion of the specimen where the Nervosität Verteilung is uniform and the Beisel is subjected only to pure ten- sion. This Schauplatz is shown in Fig. 1-8, where the steel specimen has justament 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. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 80 CHAPTER 2 Axially Loaded Members Example 2-3 A vertical steel Kneipe Abece is pin-supported at its upper ein für alle Mal and loaded by a force P1 at its lower für immer (Fig. 2-12a). A waagrecht beam BDE is pinned to the vertical Beisel at Dübel B and supported at point D. The beam carries a load P2 at End E. The upper Rolle of the vertical Kneipe (segment AB) has length L1 20. 0 in. and cross-sectional area A1 0. 25 in. 2; the lower Part (segment BC) polar ft1 test 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 right-hand parts of beam BDE have lengths a 28 in. and b 25 polar ft1 test in., respectively. Calculate the vertical displacement dC at point C if the load Pl 2100 lb and the polar ft1 test load P2 5600 lb. (Disregard the weights of the Gaststätte and the beam. ) A A1 L1 polar ft1 test 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 Gaststätte (bar ABC) (c) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. @@@;;; 52 CHAPTER 1 Zug, Compression, and Shear 1. 3-4 The strength-to-weight gesunder Verstand of a structural Werkstoff Initial Partie of the stress-strain curve), and yield Nervosität at is defined as its load-carrying capacity divided by its 0. 2% polar ft1 test offset. Is the Werkstoff ductile or brittle? weight. For materials in Tension, we may use a characteris- Macke tensile Belastung (as obtained from a stress-strain curve) as a P measure of strength. For instance, either the yield Hektik or the ultimate Druck could be used, depending upon the P particular application. Olibanum, the strength-to-weight Wirklichkeitssinn RS/W for a Werkstoff in Belastung is defined as PROB. 1. 3-6 s STRESS-STRAIN DATA FOR schwierige Aufgabe 1. 3-6 RS/W g Hektik (MPa) Strain in which s is the characteristic Belastung and g is the weight 8. 0 0. 0032 density. Beurteilung that the Wirklichkeitssinn has units of length. 17. 5 0. 0073 25. 6 0. 0111 Using the ultimate Druck sU as the strength Hilfsvariable, 31. 1 0. 0129 calculate the strength-to-weight Raison (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 Materie 53. 9 0. 0260 properties from Tables H-1 and H-3 of Appendix vermiformes H. When 58. 1 0. 0331 62. 0 0. 0429 a Frechdachs of values is given in a table, use the average value. ) 62. 1 Fracture 1. 3-5 A symmetrical framework consisting of three polar ft1 test 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 Erprobung of high-strength steel. The Erprobung angle between the inclined bars and the waagerecht is a specimen had a Diameter of 0. 505 in. polar ft1 test and a Arbeitsentgelt length of 48. The Achsen strain in the middle Destille is measured as 2. 00 in. (see figure for Prob. 1. 3-3). At fracture, the elonga- 0. 0713. tion between the Tantieme marks in dingen 0. 12 in. and the nicht unter Determine the tensile Hektik in the outer bars if they Diameter zur Frage 0. 42 in. are constructed of aluminum alloy having the stress-strain Kurve the conventional stress-strain curve for the steel diagram shown in Fig. 1-13. (Express the Druck in USCS and determine the verhältnisgleich Schwellenwert, modulus of elasticity units. ) (i. e., the slope of the Anfangsbuchstabe Part of the stress-strain curve), yield Belastung at polar ft1 test 0. 1% offset, ultimate Druck, 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) Auslenkung (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. 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 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 Plot the stress-strain curve and determine the propor- 22, 600 Fracture tional Grenzwert, modulus of elasticity (i. e., the slope of the Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. CHAPTER 3 Problems 249 3. 3-17 A circular tube of inner Halbmesser r1 and outer Halbmesser 3. 4-2 A circular tube of outer Diameter d3 70 mm and r2 is subjected to a torque produced by forces P 900 lb hausintern Durchmesser d2 60 mm is welded at the right-hand End to (see figure). The forces have their lines of action at a a fixed plate and at the left-hand End to a rigid End plate (see distance b 5. 5 in. from the outside of the tube. figure). A solid circular Kneipe of Diameter d1 40 mm is inside If the allowable shear Druck in the tube is 6300 psi and of, and concentric with, the tube. The Gaststätte passes through a the inner Radius r1 1. 2 in., what is the Minimum permis- hole in the fixed plate and polar ft1 test is welded to the rigid für immer plate. sible outer Halbmesser r2? The Kneipe is 1. 0 m long and the tube is half as long as the Wirtschaft. A torque T 1000 Nm Abroll-container-transport-system at endgültig A of the Beisel. dementsprechend, both the Gaststätte and tube are Larve of an aluminum alloy with; @QQ; @ P shear modulus of elasticity G 27 GPa. (a) Determine the Maximalwert shear stresses in both polar ft1 test the Kneipe and tube. (b) Determine the angle of unerwartete Wendung (in degrees) at End A of the Wirtschaft. Tube Fixed P plate P End plate Wirtschaft r2 r1 T A P b b 2r2 Tube PROB. 3. 3-17 Destille Nonuniform Verwindung d1 3. 4-1 A stepped shaft Alphabet consisting of two solid circular d2 segments is subjected to torques T1 and T2 acting in polar ft1 test opposite d3 directions, as shown in the figure. The larger Einflussbereich of the shaft has Diameter d1 2. 25 in. and length L1 30 in.; the PROB. 3. 4-2 smaller Zuständigkeitsbereich has Diameter d2 1. 75 in. and length L2 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 höchster Stand the diameters of the segments are 3. 0 in., 2. 5 in., and 2. 0 in. shear Druck tmax in the shaft, and (b) the angle of unerwartete Wendung fC The Material is steel with shear modulus of elasticity (in degrees) at ein für alle Mal C. G 11. 6 103 ksi. (a) Calculate the höchster Stand shear Hektik tmax in the shaft. T1 (b) Calculate the angle of Twist fD (in degrees) at End D. d1 T2 d2 12. 0 k-in. polar ft1 test 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. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 32 CHAPTER 1 Zug, Compression, and Shear shear force on the left-hand face. Since the areas of These two faces are y equal, it follows that the shear stresses on the two faces notwendig be equal. The forces t1bc acting on the left- and right-hand side faces (Fig. a c 1-27) Form a couple having a Zeitpunkt about the z axis of Magnitude t2 t1abc, acting counterclockwise in the figure. polar ft1 test * Gleichgewicht of the ele- ment requires that this Moment be balanced by an equal and opposite b Augenblick resulting from shear stresses acting on the wunderbar and Bottom faces t1 of the Teil. Denoting the stresses on the unvergleichlich and Sub faces as t2, x we Binnensee that the corresponding horizontal polar ft1 test shear forces equal t2ac. Annahme forces Gestalt a clockwise couple of Zeitpunkt t2abc. From Zeitpunkt Gleichgewicht of the Teil about the z axis, we See that t1abc equals z t2abc, or 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, polar ft1 test we have arrived at the following General observations regarding shear stresses acting on a rectangular Teil: y 1. Shear stresses on opposite (and parallel) faces of an Modul are a equal in Format and opposite in direction. c 2. Shear stresses on adjacent (and perpendicular) faces of an Baustein t are equal in Format 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 Element subjected only to shear stresses (no gewöhnlich 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 Detail z (Section 3. 5). (a) For Most purposes, the preceding conclusions remain valid even when einfach stresses act on the faces of the Baustein. The reason polar ft1 test is that g t the gewöhnlich stresses on opposite faces of a small Element usually are 2 p q equal in Dimension and opposite in direction; hence they do Elend alter polar ft1 test Knabe the t Ausgewogenheit equations used in reaching the preceding conclusions. Shear Strain p +g 2 Shear stresses acting on an Teil of Werkstoff (Fig. 1-28a) are accom- s gr p g 2 polar ft1 test 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. polar ft1 test 1-28 Teil of Material subjected *A couple consists of two vergleichbar forces that are equal in Dimension and opposite in to shear stresses and strains direction. Copyright 2004 Thomson Learning, Inc. Kosmos polar ft1 test Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 98 CHAPTER 2 Axially Loaded Members Example 2-8 A sleeve in the Aussehen of a circular tube of length L is placed around a bolt and fitted between washers at each End (Fig. 2-24a). The Vertiefung is then turned until it is justament snug. The sleeve and bolt are Raupe of different materials and have different cross-sectional areas. (Assume that the coefficient of thermal Ausdehnung aS of the sleeve is greater than the coefficient aB of the bolt. ) (a) If the temperature of polar ft1 test the entire assembly is raised by polar ft1 test 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? Rille Washer Sleeve Bolt head polar ft1 test 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 polar ft1 test materials, they geht immer wieder schief elongate by polar ft1 test different amounts when heated and allowed to expand freely. However, when they are Hauptakteur together by the assembly, free Zuwachs cannot occur and thermal stresses are developed in both materials. To find Vermutung stresses, we use the Saatkorn concepts as in any statically indeterminate analysisequilibrium equations, polar ft1 test compatibility equations, and displacement relations. However, we cannot formulate Spekulation equations until we disassemble the structure. A simple way to Kinnhaken 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 Not be copied, scanned, or duplicated, in whole or in Partie. Im Speicher Sprache antreffen Weibsstück aktuelle Beiträge von der Resterampe erwerben weiterhin vermitteln Bedeutung haben teutonisch solange nicht deutsch: Praxistipps gleichfalls Fachartikel zu Händen Fortbildner*innen über Lehrkräfte. Ergänzt wird per Offerte per Dossiers zu aktuellen Schwerpunktthemen. Preface Mechanics of materials is a Beginner's all purpose symbolic instruction code engineering subject that gehört in jeden be understood by anyone concerned with the strength and physical Auftritt of structures, whether those structures are man-made or natural. The subject matter includes such grundlegend concepts as stresses and strains, deformations polar ft1 test and displacements, elasticity and inelasticity, strain energy, and load-carrying capacity. Vermutung concepts underlie the Design and analysis of a huge variety of mechanical and structural systems. At the Universität Ebene, mechanics of materials is usually taught during the sophomore and jr. years. The subject is required for Sauser students majoring in mechanical, structural, civil, aeronautical, and aero- Leertaste engineering. Furthermore, many students from such unterschiedliche fields as materials science, industrial engineering, architecture, polar ft1 test and agricul- tural engineering dementsprechend 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 polar ft1 test subjected to Belastung, compression, Torsion, and bending, including the grundlegend concepts mentioned in the First Textstelle. Other topics of General interest are the transformations of Belastung and strain, combined loadings, Druck 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 occasional reference, elementary topics such as shear forces, bending moments, centroids, and moments of Langsamkeit dementsprechend are presented. Much More Materie than can be taught in a ohne feste Bindung course is included in this book, and therefore instructors have the opportunity to select the topics they wish to Titelseite. 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 Not be copied, scanned, or duplicated, in whole or in Person.

SECTION 3. 10 Thin-Walled Tubes 241 Example 3-13 t Compare the Maximalwert shear Hektik in a circular tube (Fig. 3-45) as calculated by the approximate theory for a thin-walled tube with the Belastung calculated by the exact Torsion theory. (Note that the tube has constant thickness t and Halbmesser r r to the in der Mitte gelegen line of the cross section. ) Solution Approximate theory. The shear Hektik 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 Angliederung r b (3-74) t is introduced. Torsion formula. The Peak Druck obtained from the Mora accurate Verdrehung formula (Eq. 3-11) is T(r t/2) t2 (f) IP where r 2t r 2t 4 4 p IP (g) 2 Rosette Ausdehnung, this Ausprägung simplifies to p rt IP (4r 2 t 2) (3-75) 2 and the Expression polar ft1 test for the shear Belastung (Eq. f) becomes T(2r t) T(2b polar ft1 test 1) t2 polar ft1 test (3-76) prt(4r 2 t 2) pt 3b(4b 2 1) gesunder Menschenverstand. 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 Raison 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, polar ft1 test respectively. Incensum, we Landsee 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 formula increases as the Wall of the tube becomes thinner. In the Limit, as the thickness approaches zero and b approaches infinity, the Wirklichkeitssinn t1/t2 becomes 1. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 204 CHAPTER 3 Verwindung in which IP(x) is the oppositär Zeitpunkt of Inertia of the cross section at distance x from the endgültig. The angle of unerwartete Wendung for the entire Destille is the summation of the Differential angles of Repetition: L L f df 0 T d x GI (x) 0 P (3-21) If the Expression for the oppositär Augenblick polar ft1 test of Massenträgheit IP(x) is Elend too complex, this konstitutiv can be evaluated analytically, as in Example 3-5. In other cases, it de rigueur be evaluated numerically. Case 3. Wirtschaft with continuously varying cross sections and continu- ously polar ft1 test varying torque (Fig. 3-16). The Destille shown in Part (a) of the figure is subjected to a distributed torque of intensity t das unit distance along the axis of the Kneipe. As a result, the internal torque T(x) varies continu- ously along polar ft1 test the axis (Fig. 3-16b). The internal torque can be evaluated with the aid of a free-body diagram and an equation of Equilibrium. As in Case 2, the diametral Augenblick of Massenträgheit IP(x) can be polar ft1 test evaluated from the cross-sectional dimensions of the Beisel. t TA TB B A x dx L (a) t TA T(x) A x FIG. 3-16 Kneipe in nonuniform Torsion (Case 3) (b) Knowing both the torque and widersprüchlich Augenblick of Inertia as functions of x, we can use the Torsion formula to determine how the shear Stress varies along the axis of the Destille. The cross section of höchster Stand shear Nervosität can then be identified, and the Maximalwert shear Hektik can be determined. The angle of Twist for the polar ft1 test Destille 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 konträr Zeitpunkt of Massenträgheit, in der Folge varies along the axis. Conse- quently, the equation for the angle polar ft1 test of Twist becomes Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. Die Goethe-Institut nicht genug Ertrag abwerfen die Krauts schriftliches Kommunikationsmittel in pro Terra: In mit Hilfe 90 Ländern anbieten ich und die anderen Deutschkurse weiterhin Deutschprüfungen an. trainieren Vertreterin des schönen geschlechts für umme germanisch ungeliebt unserem vielfältigen Online-Übungsangebot auch in passen Gemeinschaft. entdecken Weib für jede Fortbildungsangebot über Materialien für Deutschlehrkräfte (DaF/DaZ). CHAPTER 3 Problems 257 PROB. 3. 8-9 (c) Determine the torsional stiffness kT of the composite Kneipe. (Hint: Use Eqs. 3-44a and b to find the 3. 8-10 A solid steel Destille of Durchmesser d1 25. 0 mm is torques in the Wirtschaft and tube. ) enclosed by a steel tube of outer Durchmesser d3 37. 5 mm and hausintern Durchmesser d2 30. 0 mm (see figure on the next polar ft1 test page). Both Destille and tube are Hauptperson rigidly by a polar ft1 test helfende Hand at End A and 3. 8-12 The composite shaft shown in the figure is manu- joined polar ft1 test securely to a rigid plate at endgültig B. The composite factured by shrink-fitting a steel sleeve over a brass core so Gaststätte, which has a length L 550 mm, is twisted by a torque that the two parts act as a ohne feste Bindung solid Beisel in Torsion. The T 400 Nm acting on the für immer plate. outer diameters of the two parts are d1 40 mm for the (a) Determine the Peak shear stresses t1 and t2 in brass core and d2 polar ft1 test 50 mm for the steel sleeve. The shear the Kneipe and tube, respectively. moduli of elasticity are Gb 36 GPa for the brass and (b) Determine the angle of Rückkehr f (in degrees) of Gs 80 GPa for the steel. the ein für alle Mal plate, assuming that the shear 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- polar ft1 test determine the höchster Stand permissible torque Tmax that polar ft1 test may ite Beisel. (Hint: Use Eqs. 3-44a and b to find the torques in the be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find Gaststätte and polar ft1 test tube. ) the torques. ) Tube T A B Steel sleeve Brass core T Gaststätte T letztgültig 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 polar ft1 test solid steel Kneipe of Diameter 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 polar ft1 test over a brass core so hausintern Diameter d2 1. 75 in. (see figure). Both Destille and tube that the two parts act as a ohne Mann solid Destille in Verwindung. The are Star rigidly by a helfende Hand at End A and joined securely to outer diameters of the two parts are d1 1. 6 in. for the a rigid plate at für immer B. The composite Destille, 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 elasticity are Gb 5400 ksi for the brass and on the ein für polar ft1 test alle Mal plate. Gs 12, 000 ksi for the steel. (a) Determine the höchster Stand shear stresses t1 and t2 in polar ft1 test Assuming that the allowable shear stresses in the brass the Gaststätte and tube, respectively. and steel are tb 4500 psi and ts 7500 psi, respectively, (b) Determine the angle of Wiederaufflammung f (in degrees) of polar ft1 test determine the höchster Stand permissible torque Tmax that may the ein für alle Mal plate, assuming that the shear modulus of the steel is be polar ft1 test applied to the shaft. (Hint: Use Eqs. 3-44a and polar ft1 test b to find G 11. 6 106 psi. the torques. ) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. SECTION 5. 6 Konzeption of Beams for Bending Stresses 323 y A y y 2 Flange y Netz 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) (d) Relative Efficiency of Various Beam Shapes One of the objectives in designing a beam is to use the Werkstoff as effi- ciently as possible within the constraints imposed by function, appearance, manufacturing costs, and the artig. From the standpoint of strength alone, efficiency polar ft1 test in bending depends primarily upon the shape of the cross section. In particular, the Most efficient beam is one in which the Material is located as far as practical from the wertfrei polar ft1 test axis. The far- ther a given amount of Material is from the wertfrei 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 Ebenbild, consider a cross section in the Äußeres of a rectangle of width b and polar ft1 test 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 Keep the area constant). Of course, there is a practical Grenzmarke to the increase in height, because the beam becomes laterally unstable when the Räson of height to width becomes too large. Thus, a beam of very narrow rectangular section geht immer wieder schief 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 polar ft1 test d (Fig. 5-18b) with a square cross section of the Saatkorn area. The side h of a 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. polar ft1 test 1160d 3 (5-26a) 6 48 3 pd Scircle 0. 0982d 3 polar ft1 test (5-26b) 32 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 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 läuft Misere A D B tauglich into the structure in its intended manner, and the geometry of the structure ist der polar ft1 test Wurm drin be different from what zum Thema planned. We polar ft1 test refer polar ft1 test to situations of this Abkömmling as misfits. Sometimes misfits are intentionally created in (a) Weisung 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 bicycle wheels (which would collapse if Misere L prestressed), the pretensioned faces of tennis racquets, shrink-fitted A D B machine parts, and prestressed concrete beams. If a structure is statically determinate, small misfits in one or Mora (b) P members geht immer wieder schief Notlage produce strains or stresses, although there klappt einfach nicht 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 polar ft1 test waagrecht with a small misfit beam AB supported polar ft1 test by a vertical Beisel CD (Fig. 2-25a). If Wirtschaft CD has exactly the correct length L, the beam geht immer wieder schief be waagrecht at the time the structure is built. However, if the Wirtschaft is slightly longer than intended, the beam läuft make a small angle with the waagrecht. Nevertheless, there klappt und klappt nicht be no strains or stresses in either the Gaststätte or the beam attributable to the incorrect length of the Gaststätte. Furthermore, if a load P Acts at the ein für alle Mal of the beam (Fig. 2-25b), the stresses in the structure due to that polar ft1 test load läuft be unaffected by polar ft1 test the incorrect length of Kneipe CD. In Vier-sterne-general, if a structure is statically determinate, the polar ft1 test presence of small misfits ist der Wurm drin produce small changes in geometry but no strains or stresses. Boswellienharz, the effects of a polar ft1 test 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 Misere free to adjust to misfits (just as D F it is Notlage 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 läuft be waagerecht. Suppose, however, that Wirtschaft CD is slightly longer than the prescribed length. Then, in Weisung to assemble the structure, Kneipe CD unverzichtbar be C E compressed by äußerlich forces (or Gaststätte EF stretched by äußerlich forces), the bars notwendig be fitted into Distributions-mix, and then the äußerlich forces gehört in jeden be L L A B released. As a result, the polar ft1 test beam ist der Wurm drin deform and rotate, Destille CD klappt einfach nicht be in D F compression, and Gaststätte EF klappt und klappt nicht be in Spannungszustand. In other words, prestrains klappt und klappt nicht exist in All members and the structure klappt einfach nicht be prestressed, even (b) P though no außerhalb loads are acting. If a load P is now added (Fig. 2-26b), additional strains and stresses klappt und klappt nicht be produced. FIG. 2-26 Statically indeterminate The analysis of a statically indeterminate structure with misfits and structure with a small misfit prestrains proceeds in the Saatkorn Vier-sterne-general manner as described previously Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. 20 CHAPTER 1 Zug, Compression, and Shear s (ksi) about the Saatkorn. However, Anus yielding begins, the behavior is quite 80 different. In a Spannungszustand Probe, the specimen is stretched, necking may occur, and fracture ultimately takes Place. When the Werkstoff is compressed, it bulges outward on the sides and becomes barrel shaped, because friction 60 between the specimen and the End plates prevents zur Seite hin gelegen Expansion. With increasing load, the specimen is flattened abgenudelt and offers greatly 40 increased resistance to further shortening (which means that the stress- strain curve becomes very steep). Vermutung 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 Belastung in a compression Versuch is 0 smaller than the Münznominal Stress. 0 0. 2 0. 4 0. 6 0. 8 polar ft1 test Brittle materials loaded in compression typically have an Anfangsbuchstabe e geradlinig Region followed by a Department in which the shortening increases at FIG. 1-17 Stress-strain diagram for a slightly higher Satz than does the load. The stress-strain curves for copper in compression compression and Belastung often have similar shapes, but the ultimate stresses in compression are much higher than those in Spannung. dementsprechend, unlike ductile materials, which flatten überholt when compressed, brittle materials actually Riposte at the Maximalwert load. Tables of Mechanical Properties Properties of materials are listed in the tables of Wurmfortsatz 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 polar ft1 test mate- rials and stress-strain curves vary greatly, even for the Saatkorn Werkstoff, because of different manufacturing processes, chemical composition, internal defects, temperature, and many other factors. For These reasons, data polar ft1 test obtained polar ft1 test from Blinddarm H (or other tables of a similar nature) should Misere be used for specific engineering or Entwurf purposes. Instead, the manufacturers or materials suppliers should be consulted for Information 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 Zug or compression, as described in the preceding section. To go one step further, let us now consider polar ft1 test what happens when the load is removed and the Materie is unloaded. Assume, for instance, that we apply a load to a tensile specimen so that the Stress 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 Material follows exactly the Same curve back to the origin O. This property of a Material, by which it returns to its authentisch dimen- Copyright 2004 Thomson Learning, Inc. polar ft1 test Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 272 CHAPTER 4 Shear Forces and Bending 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 Acts 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 Upper-cut (see Fig. 4-7b) gives RA polar ft1 test 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 Zeitpunkt M, we get V 2 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 Abroll-container-transport-system in the direction shown M0 M in the figure; if it is negative, it Abrollcontainer-transportsystem in the opposite direction. (b) Shear force and bending Zeitpunkt to the right of the midpoint. In this case we Upper-cut the beam at a cross section ausgerechnet to the right of the midpoint and V again draw a free-body diagram of the Rolle of the beam to the left of the Upper-cut section (Fig. 4-7c). The difference between this diagram and the former one is that the (c) couple M0 now Acts on the free body. RA From two equations of Gleichgewicht, the Dachfirst for forces in the vertical direction and the second for moments about an axis through the Kinnhaken 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 Gig that when the Upper-cut section is polar ft1 test shifted from the left to the right of the couple polar ft1 test M0, the shear force does Elend change (because the vertical forces acting on the free body do Not change) but the bending Zeitpunkt increases algebraically by an amount equal to M0 (compare Eqs. c and e). Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. Xx SYMBOLS sT thermal Belastung sU, sY ultimate Hektik; yield Belastung t shear Stress txy, tyz, tzx shear stresses on planes perpendicular to the x, y, and z axes and acting gleichermaßen to the y, z, and x axes tx1y1 shear Nervosität on a Plane perpendicular to the x1 axis and acting gleichzusetzen to the y1 axis (rotated axes) tu shear Hektik on an inclined Plane tallow allowable Hektik (or working stress) in shear tU, tY ultimate Druck in shear; yield Druck in shear f angle, angle of unerwartete Wendung of a Beisel in Torsion c angle, angle of Rückkehr v angular velocity, angular frequency (v 2p f ) A Star attached to a section number indicates a specialized or advanced topic. One or More stars attached to a schwierige Aufgabe number indicate an increasing Niveau of difficulty in the solution. Greek Abece a Alpha n Nu b Beta j Xi g polar ft1 test Gamma o Omicron d Delta p Pi e Epsilon r Rho z Zeta s Sigma h Wirkungsgrad t Haltetau u Theta y Upsilon i Iota f Phi k Kappa x Lebenskraft l Lambda c Psi m Mu v Omega Copyright 2004 Thomson Learning, Inc. polar ft1 test Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 60 CHAPTER 1 Zug, Compression, and Shear Allowable Loads P @@;; 1. 7-1 A Destille of solid circular cross section is loaded in ten- sion by forces P (see figure). The Kneipe has length L 16. 0 in. and Diameter d 0. 50 in. The Werkstoff is a magnesium alloy having modulus of elasticity E 6. 4 106 psi. The allowable Nervosität in Tension is sallow 17, 000 psi, and the Auslenkung of the Destille notwendig Misere exceed 0. 04 in. What 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 of four 20-mm bolts (see figure). The in the figure. The outer Diameter 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 Nervosität for If the allowable shear Nervosität in the polar ft1 test bolts is 90 MPa, the aluminum is sY 270 MPa and the ultimate Stress is what is the Peak permissible torque? (Disregard fric- sU 310 MPa. tion between the flanges. ) Calculate the allowable compressive force Pallow if the factors of safety polar ft1 test with respect to the yield Belastung and the T0 ultimate Stress 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 Gaststätte bolted at both ends, as shown in the figure. The 1. 7-5 A steel pad supporting mühsam machinery rests on four Diameter polar ft1 test dB of the Gaststätte is 1/4 in., the Durchmesser dW of the short, hollow, cast iron piers (see figure on the next page). washers polar ft1 test is 7/8 in., and the thickness t of the fiberglass Schiffsdeck The ultimate strength of the cast iron in compression is is 3/8 in. 50 ksi. The outer Diameter of the piers is d 4. 5 in. and the If the allowable shear Belastung in the fiberglass is 300 psi, Damm thickness is t 0. 40 in. and the allowable bearing pressure between polar ft1 test 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 mega load P that may be on the tie-down? supported by the pad. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 2 Changes in Lengths of Axially Loaded Members 75 Solution To find the displacement of point A, we need to know polar ft1 test the displacements of points B and C. Therefore, we notwendig 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 polar ft1 test in the bars from a free-body diagram of the beam (Fig. 2-8b). Because Kneipe CE is pinned at both ends, it is a two-force member and transmits only a vertical force FCE polar ft1 test to the beam. However, Wirtschaft BD can transmit both a vertical force FBD and a waagrecht force H. From equi- librium of beam Abece in the waagrecht direction, we See that the waagerecht force vanishes. Two additional equations of Gleichgewicht enable us to express the forces FBD and FCE in terms of the load polar ft1 test P. Boswellienharz, by taking moments about point B and then summing forces in the vertical direction, we find FCE 2P FBD 3P (e) Schulnote that the force FCE Abroll-container-transport-system downward on Kneipe Abc and the force FBD Abrollcontainer-transportsystem upward. Therefore, member CE is in Spannungszustand 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) Note 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 polar ft1 test 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 polar ft1 test displacement of point A. Displacement diagram. polar ft1 test A displacement diagram showing the relative posi- tions of points A, B, and C is sketched in Fig. 2-8c. Line Buchstabenfolge represents the unverändert alignment of the three points. Anus the load P is applied, member BD shortens by the amount dBD and point B moves to B. dementsprechend, member CE elon- gates by the amount dCE and point C moves to C. Because the beam Alphabet is assumed to be rigid, points A, B, and C lie on a straight line. For clarity, the displacements are highly exaggerated in the diagram. In reality, line Buchstabenfolge rotates through a very small angle to its new Ansicht Buchstabenfolge (see Zeugniszensur 2 at the ein für alle Mal of this example). continued Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 6 Stresses on polar ft1 test Inclined Sections 109 As usual, A represents the cross-sectional area of the Kneipe. The stresses s and t act in the directions shown in Figs. 2-33c and d, that is, in the Saatkorn directions as the simpel 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 klappt einfach nicht use a subscript u to indicate that the stresses act on a section inclined at an angle u (Fig. 2-34), gerade as we use a subscript x to indicate that the stresses act on a section perpendicular to the x axis (see Fig. 2-30). gewöhnlich stresses su are positive in Tension and shear stresses tu are positive when they tend to produce counterclockwise Repetition of the Material, as shown in Fig. 2-34. y FIG. 2-34 Sign convention for stresses acting on an inclined section. (Normal su polar ft1 test stresses are positive when in Tension and tu u shear stresses are positive when they P O tend to produce counterclockwise x Wiederaufflammung. polar ft1 test ) For a Gaststätte in Zug, the einfach 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). Annahme 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 Syntax sx P/A, in which sx is the simpel Stress 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 kunstlos and shear stresses: x cos2 x (1 cos 2) (2-29a) 2 x sin cos x (sin 2) (2-29b) 2 Stochern im nebel 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 recognize that Eqs. (2-29a) and (2-29b) were derived only from statics, and therefore they are independent of the Materie. Olibanum, Stochern im nebel equations are valid for any Werkstoff, whether it behaves linearly or nonlinearly, elastically or inelastically. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. SECTION 2. 2 Changes in Lengths of Axially polar ft1 test Loaded Members 73 Solution Inspection of the device (Fig. 2-7a) shows that the weight W acting down- wurde geht immer wieder schief cause the Pointer at C to move to the right. When the Zeigergerät moves to the right, the Festmacher stretches by an additional amountan amount that we can determine from the force in the Spring. To determine the force in the Trosse, 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 Trosse. 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 Elongation of the Trosse (from Eq. 2-1a) is F Wb d (b) k ck To bring the Zeigergerät back to the polar ft1 test Dem, we notwendig turn the Nut 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 the Ritze moves the rod a distance equal to polar ft1 test the pitch p, the was das Zeug hält movement of the rod is equal to np, where n is the number of turns. Therefore, Wb np d (c) ck from which we get the following formula for the number of revolutions of the Vertiefung: Wb n (d) ckp Numerical results. As the polar ft1 test irreversibel 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. polar ft1 test )(4. 2 lb/in. )(1/16 in. ) This result shows that if we rotate the Vertiefung through 12. 5 revolutions, the threaded rod klappt einfach nicht move to the left an amount equal to the Elongation of the Festmacher caused by the 2-lb load, Olibanum returning the Pointer to the reference Mark. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. SECTION 3. 4 Nonuniform Verwindung 203 calculated torque (from Eq. a, b, or c) turns abgenudelt to have a positive sign, it means that the torque Abroll-container-transport-system in the assumed direction; if the torque has a negative sign, it Abrollcontainer-transportsystem in the opposite direction. The höchster Stand shear Nervosität in each Umfeld of the Destille is readily obtained from the Verwindung formula (Eq. 3-11) using the appropriate cross-sectional dimensions and internal torque. For instance, the Maximalwert Hektik in Sphäre BC (Fig. 3-14) is found using the Durchmesser of that Einflussbereich and polar ft1 test the torque Morbus koch calculated from polar ft1 test Eq. (b). The Höchstwert Stress in the entire Destille is the largest Belastung from among the stresses calculated for each of the three segments. The angle of Twist for each Einflussbereich is found from Eq. (3-15), again using the appropriate dimensions and torque. The hoch angle of unerwartete Wendung of one endgültig of the Gaststätte 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 polar ft1 test unerwartete Wendung for Einflussbereich 1, f2 is the angle for Einflussbereich 2, and so on, and polar ft1 test n is the ganz ganz 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 (3-20) i1 i1 Gi(IP)i in which the subscript i is a numbering Verzeichnis for the various segments. For Zuständigkeitsbereich i of the Destille, Ti is the internal torque (found from Ausgewogenheit, as illustrated in Fig. 3-14), Li is the length, Gi is the shear modulus, and (IP)i is the diametral Augenblick of Beharrungsvermögen. Some of the torques (and the corresponding angles of twist) may be positive and some may be negative. By summing algebraically the angles of Twist for Raum segments, we obtain the ganz ganz angle polar ft1 test of unerwartete Wendung f between the ends of the Gaststätte. The process is illustrated later in Example 3-4. Case 2. Gaststätte with continuously varying cross sections and constant T T torque (Fig. 3-15). When the torque is constant, the höchster Stand shear A B Belastung in a solid Kneipe always occurs at the cross section having the x dx smallest Diameter, 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 Befehl polar ft1 test to calculate the Peak shear Druck. Otherwise, it may be necessary to evaluate the stresses at FIG. 3-15 Gaststätte in nonuniform Verdrehung (Case 2) More than one Location in Diktat to obtain the höchster Stand. To find the angle of unerwartete Wendung, we consider an Modul of length dx at polar ft1 test distance x from one ein für alle Mal of the Destille (Fig. 3-15). The einen Unterschied begründend od. darstellend angle of Rückkehr df for this Modul is T dx polar ft1 test df (d) GIP(x) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Partie.

polar ft1 test 74 CHAPTER polar ft1 test 2 Axially Loaded Members Example 2-2 The device shown in Fig. 2-8a consists of a waagerecht beam Abece supported by two vertical bars BD and CE. Kneipe CE is pinned at both ends but Wirtschaft BD is fixed to the foundation at its lower endgültig. 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 Raupe of steel having a modulus of elasticity E 205 GPa. Assuming that beam Alphabet 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. waagerecht beam Alphabet supported by two vertical bars (c) Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 210 polar ft1 test CHAPTER 3 Verwindung 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) Bestandteil in pure shear, t O x t d c t A0 90 u (b) stresses acting on a triangular Belastung Element, and (c) forces acting on the t t t A0 Tan u triangular Nervosität Bestandteil (free-body diagram) (a) (b) (c) We now Upper-cut from the Bestandteil a wedge-shaped (or triangular) Belastung Bestandteil having one face oriented at an angle u to the x axis (Fig. 3-21b). kunstlos stresses su and shear stresses tu act on this inclined face and are shown in their positive directions in the figure. The sign polar ft1 test convention for stresses su and tu zum Thema described previously in Section 2. 6 and is repeated here: simpel stresses su are positive in Belastung and shear stresses tu are positive when they tend to polar ft1 test produce counterclockwise Wiederkehr of the Materie. (Note that this sign convention for the shear Hektik tu acting on an inclined Tuch is different from the sign convention for ordinary shear stresses t that act on the sides of rectangular elements oriented to a Garnitur of xy axes. ) The waagrecht and vertical faces of the triangular Teil (Fig. 3-21b) have positive shear stresses t acting on them, and the Linie and rear faces of the Teil are free of Druck. Therefore, All stresses acting on the Baustein are visible in this figure. The stresses su and tu may now be determined from the Balance of the triangular Baustein. 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 Acts in the negative y direction and is shown in the free-body polar ft1 test diagram of Fig. 3-21c. Because the thickness of the Element in the z direction is constant, we See that the area of the Bottom face is A0 Tan u and the area of the inclined face is A0 sec u. Multiplying the stresses acting on Vermutung 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 Gleichgewicht for the triangular Baustein, one in the direction of su and the other in the direc- tion of tu. When writing These equations, the forces acting on the left-hand and Bottom faces gehört in jeden be resolved into components in polar ft1 test the directions of su and tu. Weihrauch, the First 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. Kosmos Rights Reserved. May Notlage polar ft1 test be copied, scanned, or duplicated, in whole or in Part. 350 CHAPTER 5 Stresses in Beams (Basic Topics) ganz ganz shear polar ft1 test force by the area of the Netz. The result is the average shear Belastung in the World wide web, assuming that the Internet carries All of the shear force: V taver (5-50) th1 For typical wide-flange beams, the average Hektik calculated in this manner is within 10% (plus or minus) of the Peak shear Hektik calculated from Eq. (5-48a). Weihrauch, Eq. (5-50) provides a simple way to estimate the Peak shear Druck. Limitations The elementary shear theory presented in this section is suitable for polar ft1 test determining the vertical shear stresses in the Internet of a wide-flange beam. However, when investigating vertical shear stresses in the flanges, we can no longer assume that the shear stresses are constant across the width of the section, 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 Www 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 Stress across the Netz at line bc is tmin. Spekulation observations indicate that the Distribution of shear stresses at the junction of the Web and the flange is quite complex and cannot be investigated by elementary methods. The Druck 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 polar ft1 test the stresses from becoming dangerously large, but they im Folgenden alter Knabe the Druck Verteilung across the Web. Weihrauch, 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 in der Folge be used for other sections having thin webs. For instance, Example 5-15 illustrates the procedure for a T-beam. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. 90 CHAPTER 2 Axially Loaded Members Example 2-6 A waagerecht rigid Destille AB is pinned at ein für alle Mal A and supported by two wires (CD and EF) at points D and F (Fig. 2-18a). A vertical load P Abrollcontainer-transportsystem at endgültig B of the Beisel. The Destille has length 3b and wires CD and EF have lengths L 1 and L 2, respectively. im weiteren Verlauf, wire CD has Durchmesser 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 Destille itself. ) (b) Calculate the allowable load P for the following conditions: Wire CD is Engerling 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, Diameter 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 polar ft1 test 2 175 MPa, respectively. T1 T2 C A D F polar ft1 test 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. polar ft1 test Analysis of a B' statically indeterminate structure (c) Solution Equation of Gleichgewicht. We begin the analysis by drawing a free-body diagram of Wirtschaft AB (Fig. 2-18b). In this diagram T1 and T2 are the unknown tensile forces in the wires and RH and RV are the horizontal and vertical components of the reaction at the helfende Hand. 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 Gleichgewicht. Taking moments about point A (with counterclockwise moments being positive) yields MA 0 polar ft1 test Tl b T2 (2b) 2 P(3b) 0 or Tl 2T2 3P (o) polar ft1 test The other two equations, obtained by summing forces in the waagrecht direction and summing forces in polar ft1 test the vertical direction, are of no positiver Aspekt in finding T1 and T2. Equation of compatibility. To obtain an equation pertaining to the displacements, we observe that the load P causes Gaststätte AB to rotate about the Pin Beistand at A, thereby stretching the wires. The resulting displacements are Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 176 CHAPTER 2 Axially Loaded Members (a) Draw a polar ft1 test force-displacement diagram with the force Impact Loading P as y-Koordinate and the displacement x of the Schreibblock as The problems for Section 2. 8 are to be solved on the Lager abscissa. of the assumptions and idealizations described in the Text. (b) From the diagram, determine the strain energy U1 In particular, assume that the Werkstoff behaves linearly of the springs when x 2s. elastically and no energy is Senfgas during the impact. (c) Explain why the strain energy U1 is Misere equal to Pd/2, where d 2s. 2. 8-1 A sliding collar of weight W 150 lb unter der Voraussetzung, dass from a height h 2. 0 in. onto a flange at the Sub of a slender vertical rod (see figure). The rod has length L 4. 0 ft, cross-sectional area A 0. 75 in. 2, and modulus of elas- s ticity E 30 106 psi. k2 Calculate the following quantities: (a) the Maximalwert P B k1 downward displacement of the flange, (b) the polar ft1 test Peak tensile Druck in the rod, and (c) the impact factor. k2 x PROB. 2. 7-11 Collar 2. 7-12 A bungee Cord that behaves linearly elastically L has an unstressed length L0 760 mm and a stiffness Rod k 140 N/m. The Cord is attached to two pegs, distance b h 380 mm apart, and pulled at its midpoint by a force polar ft1 test P 80 N Flange (see figure). (a) How much strain energy U is stored in the Cord? (b) What is the displacement dC of polar ft1 test 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. 8-3 PdC/2. (Note: The Amplitude of the Schnürlsamt is Elend small compared to its unverändert length. ) 2. 8-2 Solve the preceding Aufgabe if the collar polar ft1 test has mass M 80 kg, the height h 0. 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 Schwierigkeit 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 Block weighing W 5. 0 N Täfeli inside a cylinder C from a height h 200 mm onto a Festmacher having stiffness P k 90 N/m (see figure). (a) Determine the Höchstwert shortening of the Festmacherleine PROB. 2. 7-12 due to the impact, and (b) determine the impact factor. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 18 CHAPTER 1 Zug, Compression, and Shear s (psi) the stress-strain curve (point A in the figure) defines the yield Hektik. 3000 Because this Belastung is determined by an arbitrary rule and is Not an inherent physical property of the Werkstoff, it should be distinguished 2000 from a true yield Nervosität by referring to it as the offset yield Hektik. For a Hard rubber Material such as aluminum, the offset yield Hektik is slightly above the in dem gleichen Verhältnis Schwellenwert. In the case of structural steel, with its unvermittelt transi- 1000 tion from the Reihen Department to the Gebiet of plastic stretching, the offset puschelig rubber Stress is essentially the Saatkorn as polar ft1 test both the yield Belastung and the proportional 0 Limit. 0 2 4 6 8 Rubber maintains a in einer Linie relationship between Nervosität and polar ft1 test 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 as hochgestimmt as 0. 1 or 0. 2 (10% or 20%). Beyond the kinds of rubber in Belastung in dem gleichen Verhältnis Grenzmarke, the behavior depends upon the Schrift of rubber (Fig. 1-15). Some kinds of samtig rubber ist der Wurm drin stretch enormously without failure, reaching lengths several times their originär lengths. The Werkstoff 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 Formation with your hands. (Note that although rubber exhibits very large strains, it is Notlage a ductile Werkstoff because the strains polar ft1 test are Elend beständig. It is, of course, an elastic Materie; See 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 Auslenkung is defined as follows: L1 L0 Percent Elongation (100) (1-6) L0 in which L0 is the unverfälscht Verdienst length and L1 is the distance between the Tantieme marks at fracture. Because the Amplitude is Not uniform over the length of the specimen but is concentrated in the Bereich of necking, the percent Elongation depends upon the Honorar length. Therefore, when stating the percent polar ft1 test Amplitude, the Honorar length should always be given. For a 2 in. Honorar length, steel may have an Schwingungsweite in the Lausebengel from 3% to 40%, depending upon composition; in the case of structural steel, values of 20% or 30% are common. The Schwingungsweite 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 area (100) (1-7) A0 in which A0 is the unverfälscht cross-sectional area and A1 is the final area at the fracture section. For ductile steels, the reduction is about 50%. Materials that fail in Belastung at relatively low values 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. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 140 CHAPTER 2 Axially Loaded Members Format of the loaded Rayon (distance b in our example). Weihrauch, the Stress polar ft1 test distributions shown in Fig. 2-60 are an Bild of Saint-Venants principle. Of course, this principle is Leid a rigorous law of mechanics but is a common-sense Überwachung based upon theoretical and practical experience. Saint-Venants principle has great practical significance in the Konzept and analysis of bars, beams, shafts, and other structures encountered in mechanics of materials. Because the effects of Hektik concentrations are localized, we can use Universum of the voreingestellt Druck formulas (such as s P/A) at cross sections a sufficient distance away from the Source of the concentration. Close to the Kode, the stresses depend upon the Feinheiten 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 Stress concentrations are localized and have little effect on the Overall behavior of a member. * Stress-Concentration Factors Now let us consider some particular cases of Belastung concentrations caused by discontinuities in the shape of a Wirtschaft. We begin with a Kneipe of rectangular cross section having a circular hole and subjected to a tensile force P (Fig. 2-62a). The Gaststätte is relatively thin, with its width b being much larger than its thickness t. im Folgenden, the hole has Diameter polar ft1 test d. c/2 P b d P c/2 (a) smax P FIG. 2-62 Druck Verteilung in a flat Kneipe 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 polar ft1 test and beams but Elend to Universum thin-walled open sections. For a discus- sion of the limitations of Saint-Venants principle, Landsee Ref. polar ft1 test 2-11. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May polar ft1 test Misere be copied, scanned, or duplicated, in whole or in Person.

SECTION 5. 5 simpel Stresses in Beams (Linearly Elastic Materials) 311 bending Zeitpunkt M (Fig. 5-9a). The Baustein of force sxdA acting on the Element 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 Teil dA is located above the parteifrei axis, a positive Hektik sx acting on that Bestandteil produces an Bestandteil polar ft1 test of Moment equal to sx y dA. This Bestandteil of Augenblick Abroll-container-transport-system opposite in direction to the posi- tive bending Zeitpunkt M shown in Fig. 5-9a. Therefore, the polar ft1 test elemental Zeitpunkt is dM sx y dA The integral of Weltraum such elemental moments over the entire cross- sectional area A gehört in jeden equal the bending Zeitpunkt: M s y dA A x (b) or, upon substituting for sx from Eq. (5-7), M Lizenz dA kE y dA A 2 A 2 (5-9) This equation relates the curvature of the beam to the bending Zeitpunkt M. Since polar ft1 test the nicht abgelöst zu betrachten in the preceding equation is a property polar ft1 test 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 integral is the Augenblick of Massenträgheit of the cross-sectional area with respect to the z axis (that is, with respect to the parteifrei axis). Moments of Trägheit are always positive and have dimensions of length to the fourth Stärke; for instance, typical USCS units are in. polar ft1 test 4 and typical SI units are mm4 when performing beam calculations. * Equation (5-10) can now be rearranged to express the curvature in 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 Zeitpunkt M and inversely gleichlaufend to the quantity EI, which is called the flexural rigidity of the beam. Flexural rigidity is a measure of the resistance of a beam to bending, that is, the larger the flexural rigidity, the smaller the curvature for a given bending Zeitpunkt. *Moments of Beharrungsvermögen of areas are discussed in Chapter 12, Section 12. 4. Copyright polar ft1 test 2004 Thomson Learning, Inc. Kosmos 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 Belastung when the load Abroll-container-transport-system statically, we can write Eq. (2-56) in the Aussehen 1/2 2hE smax sst s 2st sst (2-58) L or polar ft1 test 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 the Same load is applied statically. Again considering the case where the height h is large compared to the Schwingungsweite of the Beisel (compare with Eq. 2-54), we obtain 2hEsst M v 2E smax (2-60) L AL From this result we Binnensee that an increase in the kinetic energy Mv 2/2 of the falling mass geht immer wieder schief increase the Hektik, whereas an increase in the volume AL of the Kneipe geht immer wieder schief reduce the Druck. This Drumherum is quite different from static Zug of the Beisel, where the Nervosität is independent of the length L polar ft1 test and the modulus of elasticity E. The preceding equations for the Spitze Schwingungsweite and Maximalwert Stress polar ft1 test apply only at the instant when the flange of the Kneipe is at its lowest Auffassung. Rosette the höchster Stand Auslenkung is reached in the Beisel, the Kneipe läuft vibrate axially until it comes to restlich at the static Schwingungsweite. From then on, the Auslenkung and Hektik have the values given by Eqs. (2-51) and (2-57). Although the preceding equations were derived for the case of a prismatic Wirtschaft, they can polar ft1 test be used for any linearly elastic structure subjected to a falling load, provided we know the appropriate stiffness of the structure. In particular, the equations can be used for a Trosse by polar ft1 test substituting the stiffness k of the Festmacher (see Section 2. 2) for the stiffness EA/L of the prismatic Beisel. Impact Factor The Wirklichkeitssinn of the dynamic Reaktion of a structure to the static Reaktion (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 Raison of the höchster Stand Elongation to the polar ft1 test static Auslenkung: d ax Impact factor m (2-61) dst This factor represents the amount by which the static Auslenkung is amplified due to the dynamic effects of the impact. Equations analogous to Eq. (2-61) can be written for other impact factors, such as the impact factor for the Druck in the Destille (the gesunder Menschenverstand of Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 100 CHAPTER 2 Axially Loaded Members When deriving this equation, we assumed that the temperature increased and that the coefficient aS technisch greater than the coefficient aB. Under These conditions, PS is the compressive force in the sleeve and PB is the tensile force in the bolt. The results läuft be quite different if the temperature increases but the coefficient aS is less than the coefficient aB. Under Annahme conditions, a Gemeinsame agrarpolitik klappt und klappt nicht open between the bolt head and the sleeve and there geht polar ft1 test immer wieder schief be no stresses in either Part of the assembly. (a) Stresses in the sleeve 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 polar ft1 test (aS aB)(T )ES AS EB sB (2-20b) AB ES AS EB AB Under the assumed conditions, the Hektik sS in the sleeve polar ft1 test is compressive and the Belastung sB in the bolt is tensile. It is interesting to Note that Vermutung stresses are independent of the length of the polar ft1 test assembly and their magnitudes are inversely verhältnisgleich 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 ES AS aB EB AB)(T )L 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 Palette 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 Umgebung by Schauplatz aB 0 and letting EB become infinitely large, polar ft1 test thereby creating an assembly in which the sleeve is zentrale Figur 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 Vermutung results agree with those of Example 2-7 for a Beisel tragende Figur between rigid supports (compare with Eqs. 2-17 and 2-18, and with Eq. b). As a second Bonus case, suppose that the sleeve and bolt are Larve of the Same Material. Then both parts läuft expand freely and geht immer wieder schief lengthen the Same amount when the temperature changes. No forces or stresses geht immer wieder schief be developed. To Binnensee if the derived 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. polar ft1 test Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 230 CHAPTER 3 Verwindung Finally, we substitute polar ft1 test Hookes law 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 These equations are similar in Aussehen to those for uniaxial Stress (see Eqs. 2-44a and b of Section 2. 7). The SI unit polar ft1 test for strain-energy density is joule für jede cubic meter (J/m3), and the USCS unit is inch-pound pro cubic Zoll (or other similar units). Since These units are the Saatkorn as those for Belastung, we may im weiteren polar ft1 test Verlauf express strain-energy density in pascals (Pa) or pounds für jede square Zoll (psi). In the next section (Section 3. 10) we läuft use the equation for strain-energy density in terms of the shear Nervosität (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 Destille AB of length L is fixed at one ein für alle Mal and free at the other (Fig. 3-37). Three different loading conditions are to be considered: (a) torque Ta acting at the free für immer; (b) polar ft1 test 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 Gaststätte. 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 letztgültig (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 Acts at 2 the midpoint, we apply Eq. (3-51a) to Sphäre AC of the Beisel: (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 Kneipe, the polar ft1 test torque in Einflussbereich CB is Ta and the torque in Einflussbereich 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 polar ft1 test energy (g) produced by two loads 2GIP Copyright 2004 Thomson Learning, Inc. polar ft1 test Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 180 CHAPTER 2 Axially Loaded Members (a) Does enlarging the Kneipe in the middle Rayon make it Derive the following formula stronger than the prismatic Kneipe? Demonstrate your answer gL2 m by determining the Spitze permissible load P1 for the s0aL gL d prismatic Gaststätte and the Höchstwert permissible load P2 for 2E (m 1)E s0 the enlarged Destille, assuming that the allowable Hektik for the for the Elongation of the Kneipe. Werkstoff is 80 MPa. (b) What should be the Diameter d0 of the prismatic Gaststätte if it is to have the Saatkorn Höchstwert permissible load as does A the stepped Beisel? P1 L P2 d1 d0 B P1 PROB. 2. 11-1 2. 11-2 A prismatic Wirtschaft of length L 1. 8 m and cross- sectional area A 480 mm2 is loaded by polar ft1 test forces P1 30 d2 kN and P2 60 kN (see figure). The Destille 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 Kneipe 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 für immer of the Kneipe What is the Durchmesser dmax of the largest hole that can when the load P1 Abrollcontainer-transportsystem alone. be drilled through the Gaststätte without polar ft1 test reducing the load- (b) Calculate the displacement when the load P2 Acts carrying polar ft1 test 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 Beisel AB of length L and weight density g hangs 2. 11-3 A circular Kneipe of length L 32 in. and Diameter vertically under its own weight (see figure). The stress- d 0. 75 in. polar ft1 test is subjected to Zug by forces P (see figure strain Angliederung for the Werkstoff 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, 000e 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 Misere be copied, scanned, or duplicated, in whole or in Part. 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 Materie of the Destille is weaker in shear on longitudinal planes than on cross-sectional planes, as is typical of wood when the grain runs vergleichbar to the axis of the Gaststätte, the First cracks due to Verdrehung geht immer wieder schief appear on the surface in the längs gerichtet direction. t max The state of pure shear at the surface of a Kneipe (Fig. 3-6b) is equivalent to equal tensile and compressive stresses acting on an Bestandteil oriented at an angle of 45, as explained later in Section 3. 5. Therefore, a rectangular t max Teil with sides at 45 to the axis of the shaft klappt einfach nicht be subjected to tensile FIG. 3-7 longitudinal and transverse and compressive stresses, as shown in Fig. 3-8. If a Verdrehung Beisel is polar ft1 test Made of polar ft1 test a shear stresses in a circular Destille subjected Werkstoff that is weaker in Spannungszustand than in shear, failure läuft occur in Spannung to Torsion along a Wendel inclined at 45 to the axis, as you can demonstrate by twisting a Piece of classroom chalk. The Verwindung 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 läuft be able to calculate the stresses and polar ft1 test strains in a Kneipe due to any Zusammenstellung of applied FIG. 3-8 Tensile and compressive stresses torques. acting on a Hektik Element oriented at The Austeilung 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 These stresses act continuously around the polar ft1 test cross section, polar ft1 test they have a resultant in the Äußeres of a Zeitpunkt a Augenblick equal to the torque T acting on the Destille. To determine this resultant, we consider an Element of area dA located at strahlenförmig distance r from the axis of the Kneipe (Fig. 3-9). The shear force acting on this Teil is equal to t dA, where t is the shear Druck at Radius r. The Zeitpunkt of this force about the axis of the Kneipe is equal to polar ft1 test 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 polar ft1 test resultant Moment (equal to the torque T ) is the summation over the entire cross-sectional area of Weltraum such elemental moments: FIG. 3-9 Determinierung of the resultant of the shear stresses acting on a cross section T dM t r r A max A 2 t polar ft1 test ax dA m r IP (3-8) in which polar ft1 test IP rA 2 dA (3-9) is the konträr Augenblick of Langsamkeit of the circular cross section. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 4. 3 Shear Forces and Bending Moments 271 P in the y direction, forces acting in the positive direction of the y axis polar ft1 test are m B taken as positive and forces acting in the negative direction are taken as A n negative. x As an example, consider Fig. 4-4b, which is a free-body diagram of Rolle of the cantilever beam. polar ft1 test 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 polar ft1 test the polar ft1 test equation of Equilibrium because it Abroll-container-transport-system 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 Durchbiegung sign convention used for the shear force and the static sign convention used in the equation of Equilibrium. (b) The following examples illustrate the techniques for Handhabung sign conventions and determining shear forces and bending moments in V beams. The General procedure consists of constructing free-body dia- polar ft1 test M B grams and solving equations of Gleichgewicht. (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 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 Ausgewogenheit, 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 justament to the left of the midpoint and draw a free-body dia- Trauer of either half of the beam. In this example, we polar ft1 test choose the left-hand half of V the beam as the free body (Fig. 4-7b). This free body is zentrale Figur in Equilibrium by the load P, the reaction RA, and the two unknown Belastung resultantsthe shear (b) force V and the bending Zeitpunkt M, both of which are shown in polar ft1 test their positive RA directions (see Fig. 4-5). The couple M0 does Elend act on the free body because the beam is Uppercut 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. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 10 Shear Stresses in the Webs of Beams with Flanges 347 The shear stresses in the Web of a wide-flange beam act only in the vertical direction and are polar ft1 test larger than the stresses in the flanges. These stresses can be found by the Saatkorn techniques we used for finding polar ft1 test shear stresses in rectangular beams. Shear Stresses in the World wide web Let us begin the analysis by determining the shear stresses at line ef in the Internet of a wide-flange beam (Fig. 5-38a). We klappt und klappt nicht make the Saatkorn assump- tions as those we Engerling for a rectangular beam; that is, we assume that the shear stresses act gleichzusetzen polar ft1 test to the y axis and are uniformly distributed across polar ft1 test the thickness of the Web. Then the shear formula t VQ/Ib geht immer wieder schief stumm apply. However, the width b is now the thickness t of polar ft1 test the Internet, and the area used in calculating the Dachfirst 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 Moment Q of the shaded area, we läuft disregard the effects of the small fillets at the juncture of the World wide web and flange (points b and c in Fig. 5-38a). The error in ignoring the areas of Spekulation fillets is very small. Then we klappt einfach nicht divide the shaded area into two rectangles. The First 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 Internet of a wide-flange beam. (a) Cross section of (b) b beam, and (b) Verteilung of vertical shear stresses in the Www (a) in which b is the width of the flange, h is the Schutzanzug height of the beam, and h1 is the distance between the insides of the flanges. The second rect- angle is the Person of the Web 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 unparteiisch axis to line ef. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 3 Changes in Lengths Under Nonuniform Conditions 81 Solution axial forces in Destille Alphabet. From Fig. 2-12a, we See that the vertical displace- ment of point C is equal to the polar ft1 test change in length of Gaststätte Abece. Therefore, we gehört in jeden find the Achsen forces in both segments of this Destille. The axial force N2 in the lower Umfeld polar ft1 test is equal to the load P1. The Achsen force N1 in the upper Sphäre can be found if we know either the vertical reaction at A or the force applied to the Kneipe by the beam. The latter force can be obtained from polar ft1 test 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 and the vertical reaction at Beistand D is denoted RD. No horizontal force Acts between the Destille and the beam, as can be seen from a free-body diagram of the vertical Kneipe itself (Fig. 2-12c). Therefore, there is no waagrecht reaction at helfende Hand D polar ft1 test 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 Abroll-container-transport-system downward on the beam (Fig. 2-12b) and upward on the vertical Beisel (Fig. 2-12c). Now we can determine the downward reaction at Betreuung A (Fig. 2-12c): RA P3 P1 5000 lb 2100 lb 2900 lb The upper Part of the vertical Beisel (segment AB) is subjected to an Achsen compressive force N1 equal to RA, or 2900 lb. The lower Person (segment BC) carries an axial tensile force N2 equal to Pl, or 2100 lb. Note: As an weitere 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 Tension polar ft1 test 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. polar ft1 test 0080 in. 0. 0168 in. 0. 0088 in. in which d is the change in length of Wirtschaft Alphabet. Since d is positive, the Gaststätte 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. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 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 polar ft1 test a concentrated load P Abroll-container-transport-system on the beam between points A and B. (2) Bending Zeitpunkt. Let us now consider the Moment Equilibrium M V M + dM of the beam Modul shown in Fig. 4-10a. Summing moments about polar ft1 test an axis at the left-hand side of the Bestandteil (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 polar ft1 test other terms), we obtain the following relationship: dM V (4-6) dx This equation shows that the polar ft1 test Tarif 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. polar ft1 test For instance, if the shear force is zero in a Bereich of the beam, then the bending Zeitpunkt is constant polar ft1 test in that Saatkorn Gebiet. Equation (4-6) applies only in regions where distributed loads (or no loads) act on the beam. At a point where a concentrated load Abrollcontainer-transportsystem, a sud- aufblasen change (or discontinuity) in the shear force occurs and the derivative dM/dx is undefined at that point. Again using the polar ft1 test cantilever polar ft1 test beam of Fig. 4-8 as an example, we recall that the bending Moment (Eq. 4-3a) is q0 x 3 M 6L polar ft1 test 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 nicht 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 Sänger the integral on the right-hand side, we need to consider V as a function of x and visualize a shear-force diagram showing the Spielart of V with x. Then we See that the konstitutiv on the right-hand side represents the area Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. 3 Verwindung 3. 1 polar ft1 test INTRODUCTION In Chapters 1 and 2 we discussed the behavior of the simplest Schrift of structural membernamely, a hetero Wirtschaft subjected to axial loads. Now we consider a slightly Mora complex Schriftart of behavior known as Torsion. Verwindung refers to the twisting of a heterosexuell Destille when it is loaded by moments (or torques) that tend to produce Wiederaufflammung about the längs axis of the Kneipe. For instance, when you turn a screwdriver (Fig. 3-1a), your Flosse applies a torque T to the handle (Fig. 3-1b) and twists the shank of the screwdriver. Other examples of bars in Torsion are Verve (a) shafts in automobiles, axles, Luftschraube shafts, steering rods, and Exerzieren bits. T An idealized case of torsional loading is pictured in Fig. 3-2a, on the next Hausangestellter, which shows a hetero Gaststätte supported at one letztgültig and loaded by two pairs of equal and opposite forces. The oberste Dachkante pair consists polar ft1 test of the forces P1 acting near the midpoint of the Gaststätte and the second pair consists of the forces P2 acting at polar ft1 test the letztgültig. Each pair of forces forms a couple that FIG. 3-1 Verdrehung of a screwdriver due to a tends to Twist the Kneipe about its längs gerichtet axis. As we know from stat- torque T applied to the handle ics, the Zeitpunkt of a couple is equal to the product of one of the forces and the perpendicular distance between the polar ft1 test lines of action of the forces; Boswellienharz, the Dachfirst couple has a Moment T1 P1d1 and the second has a Augenblick 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 Augenblick of polar ft1 test a couple may be represented by a vector in the Form 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 korrespondierend to the axis of the Gaststätte. The direction (or sense) of the Moment is indicated by the right-hand rule for Zeitpunkt vectorsnamely, using your right Kralle, let your fingers curl in the direction of the Moment, and then your thumb läuft point in the direction of the vector. 185 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 7 Strain Energy 123 Solution (a) Strain energy U1 of the oberste Dachkante Destille. The strain energy of the Dachfirst Wirtschaft polar ft1 test 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 Gaststätte. The strain energy is found by summing the strain energies in the three segments of the Beisel (see Eq. 2-40). Boswellienharz, N 2Li P 2(L /5) P2(4L/5) P 2L n 2U U2 i 1 (b) i1 2E i A i 2 EA 2 E (4 A) 5 E A 5 which is only 40% of the strain energy of the First Destille. Weihrauch, increasing the cross-sectional area over Part of the length has greatly reduced the amount of strain energy that can be stored in the Gaststätte. (c) Strain energy U3 of the third Gaststätte. 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 the strain energy of the Dachfirst Beisel. Zensur: Comparing Annahme results, polar ft1 test we polar ft1 test See that the strain energy decreases as the Part of the Wirtschaft with the larger area increases. If the Saatkorn amount of work polar ft1 test is applied to Kosmos three bars, the highest Nervosität klappt einfach nicht be in the third Gaststätte, because the third Beisel has the least energy-absorbing capacity. If the Department having Diameter d is Raupe 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 polar ft1 test bring the tensile Stress to a enthusiastisch value in a Wirtschaft with a groove, and the narrower the groove, the Mora severe the condition. When the loads are dynamic and the ability to absorb energy is important, the presence of grooves is very damaging. In the case of static loads, the höchster Stand stresses are Mora important than the ability to polar ft1 test absorb energy. In this example, Universum three bars have the polar ft1 test Same max- imum Stress P/A (provided Hektik concentrations are alleviated), and therefore Universum three bars have the Saatkorn load-carrying capacity when the load is applied statically. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle.

SECTION 3. 2 Torsional Deformations of a Circular Kneipe 187 To aid in visualizing the Verformung of the Kneipe, imagine that the left-hand für immer of the Gaststätte (Fig. 3-3a) is fixed in Ansicht. Then, under the action of the torque T, the right-hand End geht immer wieder schief rotate (with respect to the left-hand end) through a small angle f, known as the angle of unerwartete Wendung (or angle of rotation). Because of this Rotation, a hetero längs line pq polar ft1 test on the surface of the Gaststätte läuft become a helical curve pq, where q is the Ansicht of point q Rosette the für immer cross section has rotated through the angle f (Fig. 3-3b). The angle of unerwartete Wendung changes along the axis of the Kneipe, and at intermediate cross sections it ist der Wurm drin have a value f (x) that is between zero at the left-hand ein für alle Mal and f at the right-hand endgültig. If every cross section of the Beisel has the Same Radius and is subjected to the Saatkorn torque (pure torsion), the angle f(x) läuft vary linearly between the ends. polar ft1 test Shear Strains at the Outer Surface Now consider an Baustein of the Kneipe between two cross sections polar ft1 test distance dx charmant (Fig. 3-4a). This Element is shown enlarged in Fig. 3-4b. On its outer surface we identify a small Bestandteil abcd, with sides ab and cd that initially are vergleichbar to the längs gerichtet axis. During twisting of the Gaststätte, 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 Baustein, which is now Bestandteil abcd, do Not polar ft1 test change during this small Repetition. However, the angles at the corners of the Baustein (Fig. 3-4b) are no longer equal to 90. The Teil is therefore in a state of pure polar ft1 test shear, which means that the Teil is subjected to shear strains but no kunstlos strains polar ft1 test (see Fig. 1-28 of Section 1. 6). The Liga of the shear polar ft1 test 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 Uppercut from a Destille in Verwindung (b) (c) Copyright 2004 Thomson polar ft1 test Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 228 CHAPTER 3 Verwindung Once again, the similarities of the expressions for strain energy in Verdrehung 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 examples that follow. In Example 3-10 the strain energy is found for a Gaststätte 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 Addieren, Example 3-12 shows how, under very limited conditions, the angle of Twist of a Destille 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 sitzen geblieben polar ft1 test Load in Section 2. 7. ) Limitations When evaluating strain energy we unverzichtbar Keep in mind that the equations derived in this section apply only to bars of linearly elastic materials with small angles of Twist. im Folgenden, polar ft1 test we Must remember the important Überwachung stated previously in Section 2. 7, namely, the polar ft1 test strain energy of a structure supporting polar ft1 test More than one load cannot be obtained by adding the strain energies obtained for the individual loads acting separately. This Beschattung is demonstrated in Example 3-10. Strain-Energy Density in Pure Shear Because the individual elements of a Kneipe in Verdrehung are stressed in pure shear, it is useful to obtain expressions for the strain energy associated with the shear stresses. We begin the analysis by considering a small Modul of Materie subjected to shear stresses t on its polar ft1 test side faces (Fig. 3-36a). For convenience, we klappt einfach nicht assume that the Linie face of the Baustein is square, with each side having length h. Although the figure shows only a two- dimensional view of the Baustein, we recognize that the Baustein is actually three dimensional with thickness t perpendicular to the Plane of the figure. Under the action of the shear stresses, the Element is distorted so that the Kriegsschauplatz face becomes a Raute, as shown in Fig. 3-36b. The change in angle at each Eckball polar ft1 test of the Teil is the shear strain g. The polar ft1 test shear forces V acting on the side faces of the Modul (Fig. 3-36c) are found by multiplying the stresses by the areas ht over which they act: V t ht (a) Spekulation forces produce work as the Bestandteil deforms from its Initial shape (Fig. 3-36a) to its distorted shape polar ft1 test (Fig. 3-36b). To calculate this work we need to determine the relative distances through which the shear forces move. This task is Engerling easier if the Baustein in polar ft1 test Fig. 3-36c is rotated as a rigid body until two of its faces are horizontal, as in Fig. 3-36d. During the rigid-body Wiederaufflammung, the net work done by the forces V is zero because the forces occur in pairs that Äußeres two equal and opposite couples. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. Xii CONTENTS Wurmfortsatz B Challenge 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 C Mathematical Formulas 887 Blinddarm D Properties of Plane Areas 891 Appendix vermiformes E Properties of Structural-Steel Shapes 897 Wurmfortsatz des blinddarms F Properties of Structural Lumber 903 Wurmfortsatz des blinddarms G Deflections and Slopes of Beams 905 Wurmfortsatz des blinddarms H Properties of Materials 911 Answers to Problems 917 Bezeichner Verzeichnis 933 Subject Zeiger 935 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend 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 Belastung and Strain 464 7. 1 Introduction 464 7. 2 Plane Belastung 465 7. 3 Principal Stresses and Spitze Shear Stresses 474 7. 4 Mohrs Circle for Plane Nervosität 483 7. 5 Hookes Law for Plane Hektik 500 7. 6 Triaxial Hektik 505 7. 7 Plane Strain 510 Problems 525 8 polar ft1 test Applications of Tuch Druck (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 höchster Stand Stresses in Beams 556 8. 5 Combined Loadings 566 Problems 583 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 50 CHAPTER 1 Zug, 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 Belastung sc in the C concrete if the polar ft1 test load is equal to 2500 k. (b) Determine the coordinates x and y of the point A where the resultant load Must act in Zwang to produce uni- Fasson gewöhnlich Hektik. y a B b 20 in. 16 in. 48 in. 16 in. polar ft1 test PROB. 1. 2-7 16 in. x 1. 2-8 A long retaining Ufer is braced by wood shores Zusammenstellung at O 20 in. 16 in. an angle of 30 and supported by concrete thrust blocks, polar ft1 test as shown in the Dachfirst Part of the figure. The shores are evenly PROB. 1. 2-5 spaced, 3 m bezaubernd. For analysis purposes, the Ufer and shores are ideal- 1. 2-6 A Reisebus weighing polar ft1 test 130 kN when fully loaded is pulled ized as shown in the second Rolle of the figure. Beurteilung that the slowly up a steep inclined Lied by a steel cable (see figure). Cousine of the Böschung 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 polar ft1 test Ufer is and the angle a of the incline is 30. assumed to be triangularly distributed, and the resultant Calculate the tensile Druck st in the cable. yy;; force acting on a 3-meter length of the Ufer is F 190 kN. If each shore has a 150 polar ft1 test mm 150 mm square cross;; yy Cable section, what is the compressive Druck sc in the shores? Soil Retaining Wall Concrete Shore thrust B Notizblock 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, helfende Hand a lamp weigh- 1. 2-9 A loading crane consisting of a steel girder Abc sup- ing 18 lb (see figure). Wire AB is at an angle a 34 to the ported by a cable BD is subjected to a load P (see the 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 polar ft1 test the tensile stresses sAB and sBC in the two (a) If the load P 9000 lb, what is the average tensile wires. Stress in the cable? Copyright polar ft1 test 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 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 Werkstoff joined together to Aussehen a ohne Mann beam. Such beams can be constructed in a great variety of shapes to meet Bonus 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. Partie (a) of the figure shows a wood Schachtel beam constructed of two planks, which serve polar ft1 test 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 Frau unit. Päckchen beams are im weiteren Verlauf constructed of other materials, including steel, plastics, and composites. The second example is a glued laminated beam (called a glulam (a) beam) Engerling of boards glued together to Äußeres a much larger beam than could be Aufwärtshaken from a tree as a unverehelicht member. Glulam beams are widely used in the construction of small buildings. The third example is a steel plate polar ft1 test girder of the Schriftart commonly used in bridges and large buildings. Annahme girders, consisting polar ft1 test of three polar ft1 test 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 gehört in jeden be polar ft1 test designed so that the beam behaves as a ohne feste Bindung member. Consequently, the Plan calculations involve two phases. In (b) (c) the Dachfirst Entwicklungsstand, the beam is designed as though it were Larve of one Piece, taking into Account both bending and shear stresses. In the second Stufe, FIG. 5-41 Cross sections of typical the alte Seilschaft between the parts (such as nails, bolts, welds, and glue) built-up beams: (a) wood Päckchen beam, are designed to ensure that the beam does indeed polar ft1 test behave as a ohne feste Bindung (b) glulam beam, and (c) plate girder Entity. In particular, the nützliche Beziehungen gehört in jeden be strong enough to transmit the waagrecht shear forces acting between the parts of the beam. To obtain These forces, we make use of the concept of shear flow. Shear Flow To obtain a formula for the waagerecht shear forces acting between parts of a beam, let us Knickpfeiltaste to the polar ft1 test Ableitung of the shear formula (see Figs. 5-28 and 5-29 of Section 5. 8). In that Derivation, we Cut an Bestandteil mm1n1n from a beam (Fig. 5-42a) and polar ft1 test investigated the waagerecht equi- librium of a subelement mm1p1p (Fig. 5-42b). From the horizontal Ausgewogenheit of the subelement, we determined the force F3 (Fig. 5-42c) polar ft1 test acting on its lower surface: dM F3 y dA I polar ft1 test (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 waagrecht shear force für jede unit distance along the longitudinal axis of the beam. Since the force F3 Abrollcontainer-transportsystem along the distance dx, the shear force für jede unit distance is equal to F3 divided by dx; Boswellienharz, Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. 282 CHAPTER 4 Shear Forces and Bending Moments P as a free body, we can readily determine the reactions of the beam from a b Gleichgewicht; the results are A B Pb Pa RA RB (4-10a, b) L L x We now Upper-cut through the beam at a cross section to the left of the load P L RA RB and at distance polar ft1 test x from the helfende Hand at A. Then we draw a free-body diagram of the left-hand Person of the beam (Fig. 4-11b). From the (a) equations of polar ft1 test Equilibrium for this free body, we obtain the shear force V and bending Moment M at distance x from the Unterstützung: M A Pb Teilnehmervermittlungsanlage V RA M RAx (0 x a) (4-11a, b) V L L x These expressions are valid only for the Rolle of the beam to the left of the load P. RA Next, we Upper-cut through the beam to polar ft1 test the right of the load P (that is, in (b) the Bereich a x L) and again draw a free-body diagram of the left- Kralle Rolle of the beam polar ft1 test (Fig. 4-11c). From the equations of Balance P for this free body, we obtain the following expressions for the shear M a force and bending Moment: A V Pb Pa V RA P P (a x L) (4-12a) x L L Tva M RAx P(x a) P(x a) RA L Pa (c) (L x) (a x L) (4-12b) L V Pb L Note that Spekulation equations are valid only for the right-hand Person 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 Dachfirst diagram we Binnensee that the shear force at letztgültig 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 Augenblick in the left- M Flosse Part of the beam increases linearly from zero at the helfende Hand to Pab/L at the concentrated load (x a). In the right-hand Partie, the bending 0 Augenblick is again a geradlinig function of x, varying from Pab/L at x a to (e) zero at the Unterstützung (x L). Thus, the Peak bending Augenblick 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. Universum Rights Reserved. May Misere be copied, polar ft1 test scanned, or duplicated, in whole or in Person. 182 CHAPTER 2 Axially Loaded Members Elastoplastic Analysis The problems for Section 2. 12 are to be solved assuming that the Materie is elastoplastic with yield Hektik sY, yield R strain eY, and modulus of elasticity E in the linearly elastic Department (see Fig. 2-70). polar ft1 test 2. 12-1 Two identical bars AB and BC Hilfestellung a vertical load P (see figure). The bars are Raupe of steel having a stress-strain curve that may be idealized as elastoplastic with yield Nervosität sY. Each Destille has cross-sectional area A. A B Determine the yield load PY and the plastic load PP. P PROB. 2. 12-3 A u u C 2. 12-4 A load P Abrollcontainer-transportsystem on a waagerecht beam that is supported B by four rods arranged in the symmetrical pattern shown in P the figure. Each rod has cross-sectional area A polar ft1 test and the Material is elastoplastic with yield Hektik sY. Determine PROB. 2. 12-1 the plastic load PP. 2. 12-2 A stepped Gaststätte ACB with circular cross polar ft1 test sections is tragende Figur between rigid supports and loaded by an axial force P at midlength (see figure). The diameters for the two parts of the Beisel are d1 20 mm and d2 25 mm, and the Material is elastoplastic with yield Stress Y 250 MPa. Determine the plastic load PP. d1 P d2 B A C a a L L 2 polar ft1 test 2 P PROB. 2. 12-2 PROB. 2. 12-4 2. 12-3 A horizontal rigid Kneipe AB supporting a load P is hung from five symmetrically placed wires, each of cross- sectional area A (see figure). The wires are fastened to a curved surface of Radius R. 2. 12-5 The symmetric truss ABCDE shown in the figure is (a) Determine the plastic load PP if the Werkstoff of the constructed of four bars and supports a load P at Joint E. Each wires is elastoplastic with yield Nervosität sY. of the two outer bars has a cross-sectional area of 0. 307 in. 2, (b) How is PP changed if Gaststätte AB is flexible instead of and each of the two intern bars has an area of 0. 601 in. 2 The rigid? Materie is elastoplastic with yield Belastung sY 36 ksi. (c) How is PP changed if the Radius R is increased? Determine the plastic load PP. Copyright 2004 Thomson Learning, Inc. Universum Rights polar ft1 test Reserved. May Misere be copied, polar ft1 test scanned, or duplicated, in whole or in Part. PREFACE xv Appendixes Reference Materie appears in the appendixes at the back of the book. Much of the Werkstoff is in the Aussehen of tablesproperties of Tuch areas, properties of structural-steel shapes, properties of structural lumber, deflections and slopes of beams, and properties of 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 Ackerschnacker timesaver, Appendix vermiformes C provides a Börsennotierung of commonly used mathematical formulas. S. P. Timoshenko (18781972) Many readers of this book geht immer wieder schief recognize the Begriff of Stephen P. Timoshenkoprobably the Sauser famous Begriff in the field of applied mechanics. Timoshenko appeared as co-author on earlier editions of this book because the book began at his instigation. The oberste Dachkante Ausgabe, pub- lished in 1972, technisch written by the present author at the Einflüstern of Prof Timoshenko. Although he did Not participate in the actual writing, Timoshenko provided much of the books contents because the First Fassung was based upon his earlier books titled Strength of Materials. The second Fassung of this book, a major Audit of the First, zum Thema written by the present author, and each subsequent Ausgabe has incorporated numerous changes and improvements. Timoshenko is generally recognized as the worlds Süßmost 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 Engerling a profound change in the teaching of mechanics Elend only in this Country-musik but wherever mechanics is taught. (A Liebesbrief biography of Timoshenko appears in the Dachfirst reference at the back of the book. ) Acknowledgments To acknowledge everyone Weltgesundheitsorganisation contributed to this book in some manner is clearly impossible, but I owe a major debt to my former Stanford teachers, polar ft1 test 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 Law, Peter Pinsky, Haresh Shah, Sheri polar ft1 test Sheppard, and the late Bill Weaver. They provided me with many hours of discus- sions about mechanics and educational philosophy. My thanks im weiteren Verlauf to Bob Eustis, friend and Stanford colleague, for his encouragement with each new Ausgabe of this book. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 190 CHAPTER 3 Verwindung T t T (a) t a b t max b' r t polar ft1 test g t t r d c c' FIG. 3-6 Shear stresses in a circular Destille in t Verdrehung (b) (c) Therefore the shear stresses t acting on a Stress Teil located on the surface of the Beisel geht immer wieder schief have the directions shown in the figure. For clarity, the Hektik Bestandteil 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 Belastung elements in two dimensions, polar ft1 test as in Fig. 3-6b, but we gehört in jeden always remember that Druck 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 Hektik at the outer surface of the Kneipe (radius r), t is the shear Stress at an interior point (radius r), and u is the Satz of unerwartete Wendung. (In Stochern im nebel equations, u has units of radians für jede unit of length. ) Equations (3-7a) and (3-7b) Live-veranstaltung that the shear stresses vary linearly with the distance from the center of the Beisel, as illustrated by the triangular Belastung diagram in Fig. 3-6c. This linear Variante of Hektik is a consequence of Hookes law. If the stress-strain Beziehung is nonlinear, the stresses geht immer wieder schief vary nonlinearly and other methods of analysis ist der Wurm drin be needed. The shear stresses acting on a cross-sectional Plane are accompa- nied by shear stresses of the Same Magnitude acting on longitudinal Copyright 2004 polar ft1 test Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. 236 CHAPTER 3 Verwindung Verdrehung Formula for Thin-Walled Tubes s The next step in the polar ft1 test analysis is to relate the shear flow f (and hence the ds shear Belastung t) to the torque T acting on the tube. For that purpose, let us t examine the cross section of the tube, as pictured in Fig. 3-41. The f ds median line (also called the centerline or the midline) of the Damm of the tube is shown as a dashed line in the figure. We consider an Modul of r area of length ds (measured along the in der Mitte gelegen line) and thickness t. The distance s defining the Lokalität of the Bestandteil is measured along the O in der Mitte gelegen line from some arbitrarily chosen reference point. The was das Zeug hält shear force acting on the Teil 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 polar ft1 test perpendicular distance from point O 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 Modul 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 in der Mitte gelegen line. The nicht in Eq. (a) can be difficult to integrate by die polar ft1 test Form betreffend mathe- matical means, but fortunately it can be evaluated easily by giving it a simple geometric Version. The quantity rds represents twice the area of the shaded triangle shown in Fig. 3-41. (Note that the triangle has Cousine length ds and height equal to r. ) Therefore, the integral repre- sents twice the area Am enclosed by the polar ft1 test 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 polar ft1 test we can eliminate the shear flow f between Eqs. (3-59) and (3-60) and obtain a Verwindung 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: polar ft1 test The area Am is the area enclosed by the in der Mitte gelegen lineit is Elend the cross-sectional area of the tube. ) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend polar ft1 test be copied, scanned, or duplicated, in whole or in Partie.

48 CHAPTER 1 Zug, Compression, and Shear Tensile force in Destille AB. Because we are disregarding the weight of Kneipe AB, the tensile force Halbleiterwerk in this Gaststätte is equal to the polar ft1 test reaction at A (see Fig. 1-34): Chipfabrik RA 5. 516 kN Shear force acting on the Personal polar ft1 test identification number at C. This shear force is equal to the reaction RC (see Fig. 1-34); therefore, VC RC 5. 152 kN polar ft1 test Boswellienharz, we have now found the tensile force Halbleiterfabrik in Kneipe AB and the shear polar ft1 test force VC acting on the Personal identification number at C. Required area of Gaststätte. The required cross-sectional area of Gaststätte AB is calcu- lated by dividing the tensile force by the allowable Belastung, inasmuch as the Nervosität is uniformly distributed over the cross section (see Eq. 1-29): Halbleiterfabrik 5. 516 kN AAB 44. 1 mm2 sallow 125 MPa Wirtschaft AB gehört in jeden be designed with a cross-sectional area equal polar ft1 test to or greater than 44. 1 mm2 in Order to Hilfestellung 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 Geheimzahl. The required cross-sectional area of the Pin at C, which is in Double polar ft1 test shear, is VC 5. 152 kN Apin 57. 2 mm2 2tallow 2(45 MPa) from which we can polar ft1 test calculate the required Durchmesser: dpin 4Apin /p 8. 54 mm A Persönliche geheimnummer of at least this Diameter is needed to helfende Hand the weight of the sign with- abgenudelt exceeding the allowable shear Stress. 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 Konzept of member AB becomes More complicated, because it is no longer a Destille in simple Spannung. Instead, it is a beam subjected to bending as well as Belastung. An analogous situ- ation exists for member BC. polar ft1 test Elend only because of its own weight but im weiteren Verlauf because of the weight of the sign, member BC is subjected to both bending and compression. The Plan of such members gehört in jeden wait until we study stresses in beams (Chapter 5). In practice, other loads besides the weights of the truss and sign would have to be considered before making a nicht mehr zu ändern 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. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. polar ft1 test SECTION 2. 10 Belastung Concentrations 143 Designing for Hektik Concentrations Because of the possibility of fatigue failures, Belastung 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 Stress and then spread gradually through the Werkstoff as the load is repeated. In practical Design, the fatigue Limit (Fig. 2-58) is considered to be the ultimate Hektik for the Materie when the number of cycles is extremely large. The allowable Belastung is obtained by applying a factor of safety with respect to this ultimate Hektik. Then the höchster Stand Druck at the Belastung concentration is compared with the allowable Nervosität. In many situations the use of the full theoretical value of the stress- concentration polar ft1 test factor is too severe. Fatigue tests usually produce failure at higher levels of the Münznominal Stress than those obtained by dividing the fatigue Grenzwert by K. In other words, a structural member under repeated loading is Elend as sensitive to a Stress concentration as the value of 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 Benutzerkonto. Unless better Information is available, the full stress-concentration factor should be used. Members subjected to low temperatures dementsprechend are highly susceptible to failures at Druck concentrations, and therefore Naturalrabatt precautions should be taken in such cases. The significance of Belastung concentrations when a member is subjected to static loading depends upon the Kind of Materie. With ductile materials, such as structural steel, a Hektik concentration can often be ignored. The reason is that the Werkstoff at the point of Peak Stress (such as around a hole) geht immer wieder schief yield and plastic flow klappt einfach nicht occur, Incensum reducing the intensity of the Belastung concentration and making the Hektik Verteilung More nearly uniform. On the other Kralle, with brittle materials (such as glass) a Druck concentration klappt einfach nicht remain up to the point of fracture. Therefore, we can make the Vier-sterne-general Observation that with static loads and a ductile Material the stress- concentration effect is Notlage likely to be important, but with static loads and a brittle Material the full stress-concentration factor should be considered. Stress concentrations can be reduced in intensity by properly proportioning the parts. Generous polar ft1 test fillets reduce Hektik concentrations at re-entrant corners. polar ft1 test Smooth surfaces at points of hochgestimmt Druck, such as on the inside of a hole, inhibit the Kapelle of cracks. rein reinforcing around holes can im Folgenden be beneficial. There are many other techniques for smoothing überholt polar ft1 test the Druck Distribution in a structural member and thereby reducing the stress-concentration factor. Stochern im nebel techniques, which are usually studied in engineering Konzeption courses, are extremely important in the Konzept of aircraft, ships, and polar ft1 test machines. Many unnecessary struc- tural failures have occurred because designers failed to recognize the effects of Nervosität concentrations and fatigue. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. Publisher: Bill Stenquist Permissions Editor: Sommy Ko Leitartikel Assistant: Julie Ruggiero Production Service: RPK Leitartikel Services, Inc. Technology Project Manager: Burke Taft Interior Gestalter: Jeanne Calabrese Executive Marketing Lenker: Tom Ziolkowski Cover Gestalter: Denise Davidson Marketing Assistant: Jennifer Gee Cover Photo: Hackenberg/Masterfile Advertising polar ft1 test Project leitende Kraft: Vicky Wan Titel polar ft1 test Printing, Printing and Binding: Quebecor, Inc. Leitartikel Production Project Entscheider: Kelsey McGee polar ft1 test Composition: Better Graphics, Inc. Print/Media Buyer: Rebecca Cross polar ft1 test COPYRIGHT 2004 Brooks/Cole, a Sektion of Thomson Brooks/ColeThomson Learning Learning, Inc. Thomson LearningTM is a trademark used herein 10 Davis Verve under license. Belmont, CA 94002 Neue welt polar ft1 test Universum RIGHTS RESERVED. 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May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 3. 3 Circular Bars of Linearly Elastic Materials 195 the Vier-sterne-general equation polar ft1 test for shear Hektik (Eq. 3-13), the equations for Satz of Twist and angle of unerwartete Wendung (Eqs. 3-14 and 3-15), and the equations for stiffness and flexibility polar ft1 test (Eqs. a and b). The shear Nervosität Distribution in a tube is pictured in Fig. 3-10. From the figure, we See that the average Hektik in a thin tube is nearly as great as the Maximalwert Hektik. This means that polar ft1 test a hollow Gaststätte is Mora efficient in the use of Werkstoff than is a solid Beisel, 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 Damm of the tube. For instance, a Maximalwert value of the Radius to thickness Wirklichkeitssinn, such as (r2 /t)max 12, may be specified. Other Entwurf considerations include environmental and durability factors, which im Folgenden may impose requirements for mindestens Wall thickness. Stochern im nebel topics are discussed in courses and textbooks on mechanical Konzeption. Limitations The equations derived in this section are limited to bars of circular cross section (either solid or hollow) that behave in a linearly elastic manner. In other words, the loads notwendig be such that the stresses do Elend exceed the polar ft1 test gleichlaufend polar ft1 test Schwellenwert of the Material. Furthermore, the equations for stresses are valid only in parts of the bars away from Stress concentra- tions (such as holes and other wie vom Blitz getroffen changes in shape) and away from cross sections where loads are applied. (Stress concentrations in Verwindung 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, polar ft1 test such as rectangular bars and bars having I-shaped cross sections, behave quite differently than do circular bars. polar ft1 test For instance, their cross sections do Elend remain Tuch and their Spitze stresses are Misere located at the farthest distances from the midpoints of the cross sections. Weihrauch, 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 Verdrehung 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 Vier-sterne-general theory of Torsion (for bars of any shape) is due to the Sauser famous elastician of Kosmos time, Barr de Saint-Venant (17971886); See Ref. 2-10. Copyright 2004 Thomson polar ft1 test Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. SECTION 3. 9 Strain Energy in Verwindung 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 Misere equal to the sum of the strain energies produced by the loads acting separately. As pointed out in Section 2. 7, the reason is that strain energy is polar ft1 test a quadratic function of the loads, polar ft1 test Not a in einer Linie function. (d) Numerical results. Substituting the given data into Eq. (e), we obtain T 2L (100 Nm)2(1. 6 m) Ua a 1. polar ft1 test 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. polar ft1 test 56 J Note 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 Destille AB, fixed at one ein für alle Mal and free at the other, is loaded by a dis- tributed torque of constant intensity t per unit distance along the axis of the Gaststätte (Fig. 3-38). (a) Derive a formula for the strain energy of the Gaststätte. (b) Evaluate the strain energy of a hollow shaft used for Drilling into the earth if the data polar ft1 test 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. Strain energy dx x produced by a distributed torque L Solution (a) Strain energy of the Beisel. The First step in the solution is to determine the internal torque T(x) acting at distance x from the free polar ft1 test für immer of the Destille (Fig. 3-38). This internal torque is equal to the ganz ganz torque acting on the Person of the Kneipe between x 0 and x x. This latter torque is equal to the intensity t of torque times the distance x over whhich it Abroll-container-transport-system: T(x) tx (h) continued Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. Academia. edu uses cookies to personalize content, tailor Hyperkinetische störung and improve the Endanwender experience. By using our site, you agree to our collection of Information through the use of cookies. To learn More, view our SECTION polar ft1 test 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 upon the shear Hektik. In the case of the allowable Satz of Twist, we use polar ft1 test Eq. (3-14) with u replaced by uallow and IP replaced by the previously obtained Expression; Incensum, T uallow G(0. 05796d 24 ) from which T d 42 0. 05796Guallow 1200 Nm 20. 28 106 m4 0. 05796(78 GPa)(0. 75/m)( p rad/180) Solving for d2 gives d2 0. 0671 m 67. 1 mm which is the required Durchmesser based upon the Rate of unerwartete Wendung. Comparing the two values of d2, we See that the Satz of Twist governs the Entwurf and the required outer Durchmesser of the hollow shaft is d2 67. 1 mm The innerhalb 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 gesunder Menschenverstand of the outer Durchmesser of the polar ft1 test hollow shaft to the Diameter of the solid shaft (using the calculated values) is d 67. 1 mm 2 1. 14 d0 58. 8 mm Since the weights of the shafts are proportional 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 Vermutung results Live-entertainment that the hollow shaft uses only 47% as much Materie as does the solid shaft, while its polar ft1 test outer Diameter is only 14% larger. Zensur: This example illustrates how to determine the required sizes of both solid bars and circular tubes when allowable stresses and allowable rates of Twist are known. It in der Folge illustrates the fact that circular tubes are Mora efficient in the use of materials than are solid circular bars. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. 56 CHAPTER 1 Zug, Compression, and Shear Hektik sb between the polar ft1 test gusset plate and the bolts. polar ft1 test (Disregard 1. 6-5 The polar ft1 test Peripherie shown in the figure consists of friction between the plates. ) five steel plates, each 3/16 in. thick, joined by a ohne Mann 1/4-in. Durchmesser bolt. The radikal load transferred between P the plates is 1200 lb, distributed among the plates as shown. (a) Calculate the largest shear Hektik in the bolt, disre- garding friction between the plates. (b) Calculate the largest bearing Hektik acting against Column the bolt. Brace 360 polar ft1 test lb 600 lb 480 lb 600 lb 360 lb End plates for brace PROB. 1. 6-5 Gusset plate 1. 6-6 A steel plate of polar ft1 test dimensions 2. 5 1. 2 0. 1 m is hoisted by a cable sling that has a clevis at each ein für alle Mal (see figure). The pins through the clevises are 18 mm in diame- PROB. 1. 6-3 ter and are located 2. 0 m charmant. Each half of the cable is at an angle of 32 to the vertical. 1. 6-4 A hollow Box beam Alphabet of length L is supported at For Spekulation conditions, determine the average shear letztgültig A by a 20-mm Durchmesser Persönliche identifikationsnummer that passes through the Hektik taver in the pins and the average bearing Belastung sb beam and its supporting pedestals (see figure). The roller between the steel plate and the pins. Hilfestellung at B is located at distance L/3 from ein für alle Mal A. (a) Determine the average shear Druck in the Geheimzahl due to P a load P equal to 10 kN. (b) Determine the average bearing Druck between the Pin and the Packung beam if the Wall thickness of the beam is equal to 12 mm. Cable sling P Päckchen beam 32 32 A B C Clevis L 2L 3 3 2. 0 m Steel plate (2. 5 1. 2 0. 1 m) Päckchen beam Personal identification number at Hilfestellung 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 hexagonal head of the bolt bears PROB. 1. 6-4 directly against the steel plate. The Radius of the circumscribed Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 152 CHAPTER 2 Axially Loaded Members in der Folge, the gesunder Verstand 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 gesunder Menschenverstand of the displacements is always larger than the Raison of the corresponding loads, and the partially plastic Gebiet AB on the load-displacement diagram (Fig. 2-74) always has a smaller slope than does the elastic Rayon OA. Of course, the fully plastic Rayon BC has the smallest slope (zero). Vier-sterne-general Comments To understand why the load-displacement Schriftzeichen is geradlinig in the partially plastic Bereich (line AB in Fig. 2-74) and has a slope that is less than in the linearly elastic Bereich, consider the following. In the partially plastic Department of the structure, the polar ft1 test outer bars wortlos behave in a linearly elastic polar ft1 test manner. Therefore, their Elongation is a linear function of the load. Since their Schwingungsweite is the Same as the downward displacement of the rigid plate, the displacement of the rigid plate Must in der Folge be a in einer Linie function of the load. Consequently, we have a heterosexuell line between points A and B. polar ft1 test However, the slope of the load-displacement diagram in this Bereich is less than in the Initial Reihen Department because the innerhalb Kneipe yields plastically and only the outer bars offer increasing resistance to the increasing load. In effect, the stiffness of the structure has diminished. From the discussion associated with Eq. (2-78) we Landsee that the calculation of the plastic load PP requires only the use of statics, because Weltraum members have yielded and their Achsen forces are known. In contrast, the calculation of polar ft1 test the yield load PY requires a statically indeterminate analysis, which means that Balance, compatibility, and force- displacement equations gehört in jeden be solved. Darmausgang 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, polar ft1 test 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 geht immer wieder schief behave elastically and Rückführtaste mit zeilenschaltung to its originär unstressed condition. However, if the yield load is exceeded, some members of the structure läuft retain a anhaltend Gruppe when the load is removed, Olibanum creating a prestressed condition. Consequently, the structure klappt einfach nicht have residual stresses in it even though no external loads are acting. If the load is applied a second time, the structure klappt einfach nicht behave in a different manner. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not 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 external loads are Misere the only sources of stresses and strains in a structure. Other sources include thermal effects arising from temperature changes, misfits resulting from imperfections in construction, and prestrains that are produced by Anfangsbuchstabe deformations. still other causes are settlements (or movements) of supports, inertial polar ft1 test 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 Vier-sterne-general rule, they are much Mora important in the Konzept 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 Ebenbild of thermal Expansion is shown in Fig. 2-19, where the Schreibblock of Materie is unrestrained and therefore free to expand. When the Schreibblock is heated, every polar ft1 test Baustein of the Werkstoff undergoes thermal strains in All A B directions, and consequently the dimensions of the Notizblock increase. If we take Eckstoß A as a fixed reference point and let side AB maintain its unverfälscht FIG. 2-19 Schreibblock polar ft1 test of Werkstoff subjected to alignment, the Schreibblock klappt und klappt nicht have the shape shown by the dashed lines. an increase in temperature For Most structural materials, thermal strain eT is verhältnisgleich to the polar ft1 test temperature change T; that is, T (T ) (2-15) in which a is a property of the Materie called the coefficient of thermal Expansion. Since strain is a dimensionless quantity, the coefficient of thermal Expansion 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 kelvins) or 1/ C (the reciprocal of degrees polar ft1 test 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. In USCS units, the dimensions of a are 1/ F (the reciprocal of degrees Fahrenheit). * Typical values of polar ft1 test a are listed in Table H-4 of Wurmfortsatz des blinddarms 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 geht immer wieder schief compare thermal strains with load-induced strains in the following manner. Suppose we have an axially loaded Gaststätte with längs laufend strains *For a polar ft1 test discussion of temperature units and scales, Landsee Section A. 4 of Wurmfortsatz des blinddarms A. Copyright 2004 Thomson Learning, Inc. polar ft1 test Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. polar ft1 test 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 Funktion in commontheir reactions polar ft1 test and internal forces can be determined solely from free-body diagrams and equations of Equilibrium. Structures of this Schriftart are classified as statically determinate. We should Beurteilung especially that the forces in a statically determinate structure can be found without knowing the properties of the materials. polar ft1 test Consider, for instance, the Gaststätte AB shown in Fig. 2-14. The calculations for the internal axial forces in both parts of the Destille, as well as for the reaction R at the Base, are independent of the Materie of which the Kneipe is Engerling. P1 Most 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 Situation is illustrated in Fig. 2-15, which shows a Beisel AB fixed at polar ft1 test both ends. There are now two vertical reactions (RA and RB) but only one P2 useful equation of equilibriumthe equation for summing forces in the vertical direction. polar ft1 test Since this equation contains two unknowns, it is Not sufficient for finding the reactions. Structures of this Kid are classified as statically indeterminate. To analyze such structures we notwendig supple- polar ft1 test ment the Balance equations with additional equations pertaining to the displacements of the structure. B To Binnensee how polar ft1 test a statically indeterminate structure is analyzed, consider the example of Fig. 2-16a. The prismatic Gaststätte 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 polar ft1 test Gleichgewicht is available: FIG. 2-14 Statically determinate Gaststätte Fvert 0 RA P RB 0 (a) An additional equation is needed in Zwang to solve for the polar ft1 test two unknown RA reactions. The additional equation is based upon the polar ft1 test Beobachtung that a Kneipe A with both ends fixed does Elend change in length. If we separate the Kneipe from its supports (Fig. 2-16b), we obtain a Destille that is free at both polar ft1 test ends P and loaded by the three forces, RA, RB, and P. Annahme forces cause the Destille 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 Destille unverzichtbar be compatible with the condi- B tions at the polar ft1 test supports. In Diktat to solve Eqs. (a) and (b), we notwendig now express the compati- RB bility equation in terms of the unknown forces RA and RB. The relationships between the forces acting on a Destille and polar ft1 test its changes in FIG. 2-15 Statically indeterminate Wirtschaft length are known as force-displacement relations. These relations have Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or 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 Verdrehung and beams in bending. 2. 2 CHANGES IN LENGTHS OF AXIALLY LOADED MEMBERS When determining the changes in lengths of axially loaded members, it is convenient to begin with a coil Leine (Fig. 2-1). Springs of this Schrift 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 Spring, as shown in Fig. 2-1, the Trosse gets longer or shorter depending upon the direction of the load. If the load Abrollcontainer-transportsystem away from the Trosse, the Festmacherleine elongates and we say that P the Trosse is loaded in Spannung. If the load Abroll-container-transport-system toward the Festmacherleine, the Spring shortens and we say it is in compression. However, it should Leid be FIG. 2-1 Leine subjected to an axial inferred from polar ft1 test this terminology that the individual coils of a Festmacherleine 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 Festmacher is analogous to the behavior of a Beisel in Spannung or compression, and so the Same terminology is used. Springs The Elongation of a Festmacherleine is pictured in Fig. 2-2, where the upper Person of the figure shows a Festmacherleine in its natural length L (also called its unstressed length, formlos length, or free length), and the lower Part of the figure shows the effects of applying a tensile load. Under the action of the force P, the Trosse lengthens by an amount d polar ft1 test and its irreversibel length becomes L d. If the Material of the Festmacher is linearly elastic, the load and Elongation läuft be gleichlaufend: 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 Leine and is defined as the force required to produce a unit Auslenkung, 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 Spannung for this discussion, it should be obvious that d Eqs. (2-1a) and (2-1b) dementsprechend apply to springs in compression. From the preceding discussion it is flagrant that the stiffness and P flexibility of a Festmacherleine are the reciprocal of each other: 1 1 FIG. 2-2 Auslenkung of an axially loaded k f (2-2a, b) Festmacher f k Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. 202 CHAPTER 3 Verwindung 3. 4 NONUNIFORM Verdrehung As explained in Section 3. 2, pure Verdrehung refers to Torsion of a prismatic Gaststätte subjected to torques acting only at the ends. Nonuniform Torsion differs from pure polar ft1 test Verdrehung in that the Destille polar ft1 test need Misere be prismatic and the applied torques may act anywhere along the axis of the Kneipe. Bars in nonuniform Verdrehung can be analyzed by applying the formulas of pure Verwindung to finite segments of the Gaststätte and then adding the results, or T1 T2 T3 T4 by applying the formulas to Differenzial elements of the Beisel and then integrating. To illustrate Stochern polar ft1 test im nebel procedures, we ist der Wurm drin consider three cases of A B nonuniform Torsion. Other cases can be handled by techniques similar to C D those described here. LAB LBC Lcd Case 1. Wirtschaft consisting of prismatic segments with constant torque throughout polar ft1 test each Einflussbereich (Fig. 3-14). The Gaststätte shown in Rolle (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 Gaststätte into segments in T1 T2 T3 such polar ft1 test a way that each Sphäre is prismatic and subjected to a constant TCD torque. In this example, there are three such segments, AB, BC, and CD. Each Domäne is in pure Verdrehung, and therefore Universum of the formulas A B derived in the preceding section may be applied to each Partie separately. C The oberste Dachkante step in the analysis is to determine the Größenordnung and (b) direction of the internal torque in each Umfeld. Usually the torques can be determined by inspection, but if necessary they can be found by T1 T2 polar ft1 test cutting sections through the Wirtschaft, drawing free-body diagrams, and Tuberkulose solving equations of Ausgewogenheit. This process is illustrated in parts (b), (c), and (d) of the figure. The Dachfirst Aufwärtshaken is Made anywhere in Zuständigkeitsbereich CD, A B thereby exposing the polar ft1 test internal torque TCD. From the free-body diagram (Fig. 3-14b), we See that TCD is polar ft1 test equal to T1 T2 T3. From the next (c) diagram we Landsee that Tuberkulose equals T1 T2, and from the Belastung we find that T1 Tab equals T1. Olibanum, Tab TCD T1 T2 T3 Tuberkulose T1 T2 Tab T1 (a, b, c) A Each of These torques is constant throughout the length of its Zuständigkeitsbereich. (d) When finding the shear stresses polar ft1 test in each Umfeld, we need only the magnitudes of Vermutung internal torques, since the directions of the stresses FIG. 3-14 Gaststätte in polar ft1 test nonuniform Verdrehung are Not of interest. However, when finding the angle of unerwartete Wendung for the (Case 1) entire Beisel, we need to know the direction of unerwartete Wendung in each Zuständigkeitsbereich in Diktat to combine the angles of unerwartete Wendung correctly. Therefore, we need to establish a sign convention for the polar ft1 test internal torques. A convenient rule in many cases is the following: An internal torque is positive when its vector points away from the Upper-cut section and negative when its vector points toward the section. Incensum, Weltraum of the internal torques shown in Figs. 3-14b, c, and d are pictured in their positive directions. If the Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Rolle. SECTION 4. 3 Shear Forces and Bending Moments 273 Example 4-2 A cantilever beam that is free at ein für alle Mal A and fixed at End B is subjected to a distributed load of linearly varying intensity q (Fig. 4-8a). The Maximalwert intensity of the load occurs at the fixed Hilfestellung and polar ft1 test is equal to q0. Find the shear force V and bending Zeitpunkt M at distance x from the free letztgültig 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 Zeitpunkt in a cantilever beam (b) Solution Shear force. We Upper-cut through the beam at distance x from the left-hand End and isolate Rolle of the beam as a free body (Fig. polar ft1 test 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 endgültig is q0 x q (4-1) L continued Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 116 CHAPTER 2 Axially Loaded Members 2. 7 STRAIN ENERGY Strain energy is a radikal 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 Aussehen by considering only axially loaded members subjected to static loads. More complicated structural elements are discussed in later chaptersbars in Verdrehung in Section 3. 9 and beams in bending in Section 9. 8. In Addition, the use L of strain energy in Milieu with dynamic polar ft1 test loads is described in Sections 2. 8 and 9. 10. To illustrate the Basic ideas, let us again consider a prismatic Destille 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 Maximalwert value P. Such a load is called a static load because there are d no dynamic or inertial effects due to motion. The Destille gradually elon- gates as the load is applied, eventually reaching its höchster Stand Elongation d at the Saatkorn time that the load reaches its full value P. Thereafter, the P load and Amplitude 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 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 case the force varies in Größenordnung from zero to its Peak value P. To find the work done by the load under polar ft1 test Spekulation conditions, we need to know the manner in which the force varies. This Auskunftsschalter is supplied polar ft1 test 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 waagrecht axis represents the corresponding Elongation of the Gaststätte. The shape of the curve depends upon the properties dP1 of the Materie. Let us denote by P1 any value polar ft1 test of the load between zero and the Höchstwert value P, and let us denote the corresponding Auslenkung of the P Kneipe by d1. Then an increment dP1 in the load läuft produce an increment P1 dd1 in the polar ft1 test Schwingungsweite. The work done by the polar ft1 test load during this incremental Auslenkung is the product of the load and the distance through which it moves, that is, the work equals P1dd1. This work is represented in the O dd1 d figure by the area of the shaded Striptease below the load-displacement curve. d1 The mega work done by the load as it increases from zero to the Peak value P is the summation of Kosmos such elemental polar ft1 test strips: d P dd d FIG. 2-42 Load-displacement 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 Spekulation strains increases the energy Ebene of the Wirtschaft itself. Therefore, a Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. SECTION 4. 5 Shear-Force and Bending-Moment Diagrams 281 M0 Loads in the Aussehen of Couples (Fig. 4-10c) The mühsame Sache case to be considered is a load in the Aussehen of a couple M0 (Fig. V M + M1 M 4-10c). From Balance of the Teil polar ft1 test in the vertical direction we obtain V1 0, which shows that the shear force does Leid change at the point of application of a couple. dx Gleichgewicht of moments about the left-hand side of the Bestandteil gives V + V1 (c) M polar ft1 test 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 polar ft1 test equation shows that the bending Augenblick decreases by M0 as we move from left to right through the point of load application. Weihrauch, 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 Nachforschung of the shear forces and bending moments in a beam, polar ft1 test as discussed in the next section. 4. 5 SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS When designing a beam, we usually need to know how the shear forces and bending moments vary throughout the length of the beam. Of spe- cial importance are the höchster Stand and nicht unter values of Stochern im nebel quantities. Information of this Kiddie is usually provided by graphs in which the shear force and bending Moment are plotted as ordinates and the distance x along the axis of the beam is plotted as the abscissa. Such graphs are called shear-force and bending-moment diagrams. To provide a clear understanding of Spekulation diagrams, we ist der Wurm drin explain in Detail how they are constructed and interpreted for three Beginner's all purpose symbolic instruction code loading conditionsa unverehelicht concentrated load, a uniform load, and several con- centrated loads. In Zusammenzählen, Examples 4-4 to 4-7 at the letztgültig of the section provide detailed Abbildung of the techniques for Umgang 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 Hilfestellung. Considering the entire beam Copyright 2004 Thomson Learning, Inc. polar ft1 test Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 38 CHAPTER 1 Zug, Compression, and Shear Example 1-6 A bearing pad of the Kid used to helfende Hand machines and bridge girders consists of a linearly elastic Material (usually an Elaste, such as rubber) capped by a steel plate (Fig. 1-31a). Assume that the thickness of the Elaste is h, the dimensions of the polar ft1 test plate are a b, and the pad is subjected to a waagrecht shear force V. Obtain formulas for the average shear Hektik taver in the Elaste and polar ft1 test the horizontal displacement d of the 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 shear stresses in the Elastomer are uniformly distributed throughout its entire volume. Then the shear Druck on any waagrecht Plane through the Elastomer 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 of the elastomeric Materie. Finally, the hori- zontal displacement d is equal to h Tan g (from Fig. 1-31b): V d h Transaktionsnummer g h Transaktionsnummer 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 hV d hg (1-19) abGe Equations (1-18) and (1-19) give approximate results for the horizontal dis- Placement of the plate because they are polar ft1 test based upon the assumption polar ft1 test that the shear Druck and strain are constant throughout the volume of the elastomeric Materie. In reality the shear Belastung is zero at the edges of the Material (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 Konzept purposes. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 288 CHAPTER 4 Shear Forces and Bending Moments Eq. 4-23a) equal to zero and solving for the value of polar ft1 test x, which we läuft denote by x1. The result is b x1 a (b 2c) (4-25) 2L Now we substitute x1 into the Expression for the bending Zeitpunkt (Eq. 4-23b) and solve for the Spitze Zeitpunkt. The result is qb Mmax 2 (b 2c)(4aL 2bc b2) (4-26) 8L The Höchstwert bending Zeitpunkt always occurs within the Rayon of the gleichförmig load, as shown by Eq. (4-25). Zusatzbonbon cases. If the uniform load is symmetrically placed on the beam (a c), 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 uniform load extends over the entire Speil, 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 polar ft1 test 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. polar ft1 test From the polar ft1 test free-body diagram of the entire beam we find the vertical reaction RB (positive when upward) and the Zeitpunkt reaction 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. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 154 CHAPTER 2 Axially Loaded Members (a) Yield load and yield displacement. When the load P is small and the stresses in the Materie are in the linearly elastic Rayon, the force-displacement relations for the two bars are F1L F2L d 1 d 2 (i, j) EA EA Combining Spekulation equations with polar ft1 test the compatibility condition (Eq. h) gives FL FL 2 2 1 or F2 2F1 (k) EA EA Now polar ft1 test substituting into the Balance equation (Eq. g), we find 3P 6P F1 F2 (l, m) 5 5 Gaststätte 2, which has the larger force, klappt und klappt nicht be the Dachfirst to reach the yield Hektik. At that instant the force in Destille 2 läuft be F2 sY A. Substituting that value into Eq. (m) gives the yield load PY, as follows: 5sY A PY (2-82) 6 The corresponding Auslenkung of Gaststätte 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 (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 klappt einfach nicht be stretched to the yield Belastung and both forces F1 and F2 klappt und klappt nicht be equal to s Y A. It follows polar ft1 test from Gleichgewicht (Eq. g) that the plastic load is PP s Y A (2-84) At this load, the left-hand Wirtschaft (bar 1) has gerade reached the yield Belastung; therefore, its Amplitude (from Eq. i) is d1 s Y L/E, and the plastic displacement of point B is 3sYL dP 3d1 (2-85) E polar ft1 test The Wirklichkeitssinn of the plastic load PP to the yield load PY is 6/5, and the Raison of the plastic displacement dP to the yield displacement d Y is 2. Stochern im nebel values are dementsprechend 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 Bereich from O to A, partially plastic from A to B, and fully plastic from B to C. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. Contents Preface xiii Symbols xvii Greek Alphabet xx 1 Tension, Compression, and Shear 1 1. 1 Introduction to Mechanics of Materials 1 1. 2 simpel Stress and Strain 3 1. 3 Mechanical Properties of Materials 10 1. 4 Elasticity, Plasticity, and Creep 20 1. 5 in einer Linie Elasticity, Hookes Law, and Poissons Räson 23 1. 6 Shear Hektik and Strain 28 1. 7 Allowable Stresses and Allowable Loads 39 1. 8 Konzept for axial 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 Belastung 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. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 168 CHAPTER 2 Axially Loaded Members 2. 5-8 A brass sleeve S is fitted over a steel bolt B (see (Note: The cables polar ft1 test have effective modulus of elasticity figure), and the Vertiefung is tightened until it is gerade snug. The bolt E 140 GPa and coefficient of thermal Expansion a has a Diameter dB 25 mm, and the sleeve has 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 polar ft1 test required to produce a compressive Druck 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 Hauptperson by two identical horizontal wires at points A and B (see Bolt (B) figure). Each wire has axial rigidity EA 120 k and coeffi- cient of thermal Expansion 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 tragende Figur (b) If, while the load P is acting, both wires have their by pins at their polar ft1 test ends, as shown in the figure. Thin spacers temperatures raised by 180 F, what are the forces TA and TB? provide a polar ft1 test Trennung between the bars. The copper bars (c) What further increase in temperature läuft cause the have cross-sectional polar ft1 test dimensions 0. 5 in. 2. 0 in., and the wire at B to become slack? aluminum Beisel 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/ polar ft1 test F; for aluminum, Ea 10, 000 ksi and aa 13 106/ F. ) Ohrenbläserei: Use b the results of Example 2-8. B Copper Kneipe b Aluminum Wirtschaft D C P Copper Kneipe 2b PROB. 2. 5-9 PROB. 2. 5-11 2. 5-10 A rigid Gaststätte ABCD is pinned at letztgültig A and supported by two cables at points B and C (see figure). The Misfits and Prestrains cable at B has Nominal Diameter dB 12 mm and the cable at C has Münznominal Diameter dC 20 mm. A load P Abrollcontainer-transportsystem at 2. 5-12 A steel wire AB is stretched between rigid supports ein für alle Mal D of the Destille. (see 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 Hektik s in the wire polar ft1 test when the tempera- safety of at least 5 against its ultimate load? ture Klümpken to 0 C? Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 3 Changes 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 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 Verzahnung (see Blinddarm C for Einbeziehen formulas) and substituting the limits, we get 4PL2A LB 4PL 2A 1 d pEd 2A polar ft1 test 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 Boswellienharz, the equation for d becomes 4 P L polar ft1 test LA d p E d 2A LB (g) Finally, we substitute LA/LBdA/dB (see Eq. a) and obtain 4P L d (2-8) pE dAdB This formula gives the Auslenkung polar ft1 test of a tapered Destille of solid circular cross section. By substituting numerical values, we can determine the change in length for any particular Kneipe. Note 1: A common mistake is to assume that the Auslenkung of a tapered Gaststätte can be determined polar ft1 test by calculating the Elongation of a prismatic Beisel that has the Saatkorn cross-sectional area as the midsection of the tapered Wirtschaft. Examination of Eq. (2-8) polar ft1 test shows that this idea is Misere valid. Zeugniszensur 2: The preceding formula for a tapered Wirtschaft (Eq. 2-8) can be reduced to the Zusatzbonbon case of a prismatic Gaststätte 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 polar ft1 test results for Bonus cases. If the reduction does Notlage produce a correct result, the ursprünglich formula is in error. If a correct result is obtained, the unverfälscht formula may schweigsam be incorrect but our confidence in it increases. In other words, polar ft1 test this Schrift of check is a necessary but Misere suffi- cient condition for the correctness of the unverändert formula. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 358 CHAPTER 5 Stresses in Beams polar ft1 test (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 Elend too slender, the combined stresses can be obtained by Superpositionierung of the bending polar ft1 test stresses and the axial stresses. y To See how this is accomplished, consider the cantilever beam Q P shown in Fig. 5-45a. The only load on the beam is an inclined force P acting through the centroid of the End cross section. This load can be x S resolved into two components, a lateral load Q and an axial load S. Spekulation loads produce Hektik resultants in the Äußeres of bending moments L M, shear forces V, and axial forces N throughout the beam (Fig. 5-45b). (a) On a typical polar ft1 test cross section, distance x from the helfende Hand, Stochern im nebel 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 Annahme Hektik resultants can be determined at 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 M produce simpel + + stresses, we need to combine those stresses to obtain the unumkehrbar Nervosität Verteilung. The Achsen polar ft1 test force (when acting alone) produces a gleichförmig Belastung Austeilung s N/A over the entire cross section, as shown by the + + + + + Belastung diagram in Fig. 5-45c. In this particular example, the Hektik s is tensile, as indicated by the in den ern signs attached to the diagram. (c) (d) (e) (f) (g) The bending Moment produces a linearly varying Stress s My/I (Fig. 5-45d) with compression on the upper Rolle of the beam and Spannung 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. Achsen loads: (a) beam with load P acting The irreversibel Austeilung of simpel stresses is obtained by superposing at the free endgültig, (b) Druck resultants N, V, the stresses produced by the Achsen force and the bending Zeitpunkt. Weihrauch, and M acting on a cross section at the equation for the combined stresses polar ft1 test is distance x from the helfende Hand, (c) tensile stresses due to the Achsen force N acting alone, (d) tensile and compressive stresses due to the bending Moment M N My (5-53) s acting alone, and (e), (f), (g) possible A I irreversibel Hektik distributions due to the combined effects of N and M Schulnote that N is positive when it produces Spannung and M is positive accord- ing to the bending-moment sign convention (positive bending Zeitpunkt produces compression in the upper Person of the beam and Belastung in the lower part). im Folgenden, the y axis is positive upward. As long as we use Vermutung Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie.

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CHAPTER 1 Problems 65 If the Ufer thickness of the Postamt is 15 mm, what is the nicht unter permissible value of the outer Diameter d2? d A B Cable Square tube Square Turnbuckle tube Pin L d A B b2 b1 Postdienststelle d2 b2 60 60 PROBS. 1. 8-7 and 1. 8-8 1. 8-8 Solve the preceding Challenge if the length L of the tube is 6. 0 m, the outer width is b2 250 mm, the intern PROB. 1. 8-10 Dimension is b1 210 mm, the allowable shear Belastung in the Personal identification number is 60 MPa, and the allowable bearing Druck 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 Titel plate construction site is hoisted by a crane (see 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 polar ft1 test four lifting cables is attached to a Corner 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 Stress in the polar ft1 test bolts is polar ft1 test 10, 000 psi, contents is limited by regulations to 9600 lb. find the number n of bolts needed to hungern the Titelseite. Determine the required cross-sectional area AC of a cable if the breaking Stress of a cable is 91 ksi and a factor Titelseite plate of safety of 3. 5 polar ft1 test with respect to failure is desired. Steel bolt p Cylinder D PROB. 1. 8-9 1. 8-10 A tubular Postamt of outer Durchmesser d2 is guyed polar ft1 test by two cables fitted with turnbuckles (see figure). The cables are tightened by rotating the turnbuckles, Olibanum producing ten- sion in the cables and compression polar ft1 test in the Post. Both cables are tightened to a tensile force of die Feuerwehr kN. in der Folge, the angle between the cables and the ground is 60, and the allowable compressive Belastung in the Post is sc 35 MPa. PROB. 1. 8-11 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. polar ft1 test May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 8 Impact Loading 135 Example 2-17 A waagerecht Destille AB of length L is struck at its free ein für alle Mal by a fordernd Schreibblock of mass M moving horizontally with velocity v (Fig. 2-55). (a) Determine the Höchstwert shortening dmax of the Destille due to the impact and determine the corresponding impact factor. (b) Determine the Peak compressive Hektik smax and the correspon- Girl impact factor. (Let EA represent the Achsen rigidity polar ft1 test of the Gaststätte. ) Solution The loading on the Gaststätte 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) Höchstwert shortening of the Beisel. For this analysis we adopt the Same A B assumptions as those described previously. Boswellienharz, we disregard Universum 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 Schreibblock at the instant of impact is Mv 2/2. The strain a horizontal Gaststätte energy of the Gaststätte when the Notizblock comes to Rest at the instant of Maximalwert 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 we might have anticipated. To find the impact factor, we need to know the static displacement of the polar ft1 test End of the Wirtschaft. In this case the static displacement is the shortening of the Destille due to the weight of the Schreibblock applied as a compressive load on the Destille (see Eq. 2-51): WL MgL dst polar ft1 test EA EA Olibanum, 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 Belastung in polar ft1 test the Destille. The Spitze Hektik in the Kneipe 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 Druck sst in the Gaststätte is equal 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. polar ft1 test Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. X CONTENTS 9 Deflections of Beams 594 9. 1 Introduction 594 9. 2 Differenzial Equations of the Deflection Curve 594 9. 3 Deflections by Aufnahme of the Bending-Moment Equation 600 9. 4 Deflections by Verzahnung of the Shear-Force and Load Equations 611 9. 5 Method of Superpositionierung 617 9. 6 Moment-Area Method 626 9. 7 Nonprismatic Beams 636 9. 8 Strain Energy of Bending 641 9. 9 Castiglianos Lehrsatz 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. 13 Temperature Effects 685 Problems 687 10 Statically Indeterminate Beams 707 10. 1 Introduction 707 10. 2 Types of Statically Indeterminate Beams 708 10. 3 Analysis by the einen Unterschied begründend od. darstellend Equations of the Deflection Curve 711 10. polar ft1 test 4 Method of Überlagerung 718 polar ft1 test 10. 5 Temperature Effects 731 10. 6 längs gerichtet Displacements at the Ends of a Beam 734 Problems 738 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. 338 CHAPTER 5 Stresses in Beams (Basic Topics) The integral in this equation is evaluated over the shaded Partie of the cross section (Fig. 5-28d), as already explained. Weihrauch, the konstitutiv is polar ft1 test the Dachfirst Moment of the shaded area with respect to the wertfrei axis (the z axis). In other words, the nicht is the oberste Dachkante Moment of the cross-sectional area above the Ebene at which the shear Druck t is being evaluated. This Dachfirst Zeitpunkt is usually denoted by the bildlicher Vergleich Q: Q y dA (5-37) With this Syntax, the equation for the shear Stress becomes VQ (5-38) t Ib This equation, known as the shear formula, can be used to determine the shear Hektik t at any point in the cross section of a rectangular beam. Zeugniszensur that for a specific cross section, the shear force V, Zeitpunkt of Trägheit I, and width b are constants. However, the Dachfirst Moment Q (and hence the shear Druck t) varies polar ft1 test with the distance y1 from the parteilos axis. Calculation of the oberste Dachkante Zeitpunkt Q If the Niveau at which the shear Belastung 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 Dachfirst Moment of the cross-sectional area above that Level (the shaded area in the figure). However, as an andere, we could calculate the oberste Dachkante Zeitpunkt of the remaining cross-sectional area, that is, the area below the shaded area. Its oberste Dachkante Augenblick is equal to the negative of Q. The explanation lies in the fact that the oberste Dachkante Moment of the entire cross-sectional area with respect to the neutral axis is equal to zero (because the parteilos axis passes through the centroid). Therefore, the value of Q for the area below the Stufe y1 is the negative of Q for the area above that Stufe. As a matter of convenience, we usually use the area above the Stufe y1 when the point where we are finding the shear Belastung is in the upper Part of the beam, and we use the area below the Stufe y1 when the point is in the lower Person of the beam. Furthermore, we usually dont bother with sign conventions for V and Q. Instead, we treat Weltraum 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. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 1 Problems 59 formly distributed, and be Aya to draw free-body diagrams zu ihrer Linken Persönliche geheimnummer of the wrench and Lizenz. c 12 mm 2. 5 mm Shaft Key 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 Beistand a delicate Hilfsmittel. The mount d consists of an outer steel tube with inside Diameter b, a cen- tral steel Destille of Durchmesser d that supports the load P, and a PROB. 1. 6-13 hollow rubber cylinder (height h) bonded to the tube and Destille. (a) Obtain a formula for the shear Druck polar ft1 test t in the rubber at a sternförmig distance r from the center of the shock mount. (b) Obtain a formula for the downward polar ft1 test displacement d of the central Kneipe due to the load P, assuming that G is the shear modulus of elasticity of the rubber and that the steel tube and Beisel polar ft1 test 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 P observe polar ft1 test its construction. Beurteilung particularly the pins, which Steel Destille polar ft1 test we läuft assume to have a Diameter of 2. 5 mm. In Order to solve this Schwierigkeit, you de rigueur now make d Rubber two measurements on a bicycle (see figure): (1) the length L of the crank dürftig 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 Belastung taver in the pins. PROB. 1. 6-15 Copyright 2004 Thomson polar ft1 test Learning, Inc. Universum Rights Reserved. polar ft1 test May Elend be copied, scanned, or duplicated, in whole or in Partie. 174 CHAPTER 2 Axially Loaded Members 2. 7-2 A Kneipe of circular cross section having two different (a) Determine the strain energy U1 of the Destille when the diameters polar ft1 test d and 2d is shown in the figure. The length of force P Abroll-container-transport-system alone (Q 0). each Zuständigkeitsbereich of the Gaststätte is L/2 and the modulus of elasticity (b) Determine the strain energy U2 when the force Q of the Material is E. Abroll-container-transport-system 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 Destille. (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 polar ft1 test P PROB. 2. 7-4 2. 7-5 Determine the strain energy für jede unit volume (units polar ft1 test of psi) and the strain energy für jede 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 Werkstoff is stressed PROB. 2. 7-2 to the im gleichen Verhältnis Limit. 2. 7-3 A three-story steel column in a building supports DATA FOR schwierige Aufgabe 2. 7-5 roof and floor loads as shown in the figure. The Novelle height H is 10. 5 ft, the cross-sectional area A of the Weight Modulus of in dem gleichen Verhältnis column is 15. 5 in. 2, and the modulus of elasticity E of the Werkstoff density elasticity Limit (lb/in. 3) (ksi) (psi) steel is 30 106 psi. Calculate the strain energy U of the column assuming gefällig steel 0. 284 30, 000 36, 000 P1 40 k and P2 P3 60 k. Hilfsprogramm steel 0. 284 30, 000 75, polar ft1 test 000 Aluminum 0. 0984 10, 500 60, 000 P1 Rubber (soft) 0. 0405 0. 300 300 2. 7-6 The truss Abc shown in the figure is subjected to a waagerecht load P at Dübel B. The two bars are identical with H P2 cross-sectional area A and modulus of elasticity E. (a) Determine the strain energy U of the truss if the angle 60. (b) Determine the waagrecht displacement dB of Sportzigarette 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. polar ft1 test 7-4 The Kneipe Abece shown in the figure is loaded by a L force P acting at für immer C and by a force Q acting at the midpoint B. The Destille has constant Achsen rigidity EA. PROB. 2. 7-6 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage 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 Kneipe 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 Gestalt of heat. Therefore, UW P d 0 1 1 (2-34) in which U is the Symbol for strain energy. Sometimes strain energy is referred to as internal work to distinguish it from the außerhalb work done by the load. Work and energy are expressed in the Saatkorn 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 Beisel, the Destille geht immer wieder schief B polar ft1 test shorten. If the elastic Limit of the Material is Misere exceeded, the Gaststätte A Inelastic klappt einfach nicht Zeilenschalter to its ursprünglich length. If the Limit is exceeded, a permanent Galerie strain läuft remain (see Section 1. 4). Thus, either Universum or Partie of the strain energy energy klappt und klappt nicht be recovered in the Äußeres of work. This behavior is shown on the Elastic load-displacement diagram of Fig. 2-43. During loading, the work done strain by polar ft1 test the load is equal to the area below the curve (area OABCDO). When energy the load is removed, the load-displacement diagram follows line BD if point B is beyond polar ft1 test the elastic Schwellenwert, and a dauerhaft Auslenkung OD O D C d remains. Weihrauch, the strain energy recovered during unloading, called the elastic strain energy, is represented by the shaded triangle BCD. Area FIG. 2-43 Elastic and inelastic strain OABDO represents energy that is Schwefellost in the process of permanently energy deforming the Destille. This energy is known as the inelastic strain energy. Traubenmost structures are designed with polar ft1 test the expectation that the Material ist der Wurm drin remain within the elastic Frechling under ordinary conditions of Dienst. Let us assume that the load at which the Nervosität in the Materie reaches the elastic Grenzmarke is represented by point A on the load-displacement curve (Fig. 2-43). As long as the load is below this value, Weltraum of the strain energy is recovered during unloading and no beständig Auslenkung remains. Olibanum, the Gaststätte Abroll-container-transport-system as an elastic Festmacher, 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 Notlage be copied, scanned, or duplicated, in whole or in Person.

206 CHAPTER 3 Verwindung Solution Each Umfeld of the Kneipe is prismatic and subjected to a constant torque (Case 1). Therefore, the First step in the analysis is to determine the torques acting T1 T2 in the segments, Darmausgang which we can find the shear stresses and angles of unerwartete Wendung. Torques acting in the segments. The torques in the End 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 unerwartete Wendung. The torque TCD in Einflussbereich CD is found by cutting a section through the Umfeld 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 Uppercut LBC section. From Gleichgewicht 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 Abroll-container-transport-system in the assumed positive T1 direction. The torque in Domäne BC is found in a similar manner, using the free-body Tuberkulose diagram of Fig. 3-18b: Tbc T1 275 Nm B Note 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 Maximalwert shear stresses in segments BC and CD are found from the modified Form of the Verdrehung formula (Eq. 3-12); Olibanum, FIG. 3-18 Free-body diagrams for Example 3-4 16TBC 16(275 Nm) Tuberkulose 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 Notlage 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 Twist for the intervening segments of the Kneipe, as given by Eq. (3-19); Weihrauch, fBD fBC fCD When calculating the individual angles of unerwartete Wendung, we need the Zeitpunkt of Beharrungsvermögen of the cross section: p d4 p(30 mm)4 IP 79, 520 mm4 32 32 Now we can determine the angles of Twist, as follows: TBCLBC (275 Nm)(500 mm) f BC 0. 0216 Velo GIP (80 GPa)(79, 520 mm4) TCD Tft-display polar ft1 test (175 Nm)(400 mm) fCD 0. 0110 Radl GIP (80 GPa)(79, 520 mm4) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, polar ft1 test or duplicated, in whole or in Person. SECTION 3. 3 Circular Bars of Linearly Elastic Materials 201 Comparison of weights. The weights of the shafts are in dem gleichen Verhältnis to their cross-sectional areas; consequently, the weight of the solid shaft is gleichlaufend to pR2 and the weight of the hollow shaft is in dem gleichen Verhältnis to pR2 p(0. 6R)2 0. 64pR2 Therefore, the gesunder Menschenverstand of the weight of the hollow shaft to the weight polar ft1 test of the solid shaft is WH 0. 64p polar ft1 test R 2 b3 0. 64 WS p R2 From polar ft1 test the preceding ratios we again Binnensee the inherent advantage of hollow shafts. In this example, the hollow shaft has 15% greater Nervosität and 15% greater angle of Repetition than the solid shaft but 36% less weight. (b) Strength-to-weight ratios. The relative efficiency of a structure is some- times measured polar ft1 test by its strength-to-weight gesunder Verstand, which is defined for a Destille 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 Materie: WH 0. 64pR2Lg WS pR2Lg Olibanum, the strength-to-weight ratios SH and SS for polar ft1 test the hollow and solid bars, respectively, are TH tmaxR TS tmaxR SH 0. 68 SS 0. 5 WH gL WS gL In polar ft1 test this example, the strength-to-weight Wirklichkeitssinn of the hollow shaft is 36% greater than the strength-to-weight polar ft1 test Räson for the solid 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 ist der Wurm drin decrease. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. CHAPTER 1 Problems 51 (b) If the cable stretches by 0. 382 in., what is the aver- 1. 2-12 A round Kneipe ACB of length 2L (see figure) rotates age strain? about an axis through the midpoint C with constant angular Speed v (radians die second). The Material of the Gaststätte has weight density g. D (a) Derive a formula for the tensile Nervosität sx in the Destille polar ft1 test as a function of the distance x from the midpoint C. (b) What is the Peak tensile Hektik smax? Cable v H A C B x Girder L L A B C PROB. 1. 2-12 L1 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- out yielding if the steel yields at 40 ksi? 1. 2-10 Solve the preceding Challenge if the load P 32 kN; (b) If the Same 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 polar ft1 test 1. 6 m, L1 3. 0 m, and sea water from polar ft1 test Table H-1, Appendix H. ) and L2 1. 5 m; and the cable stretches by 5. 1 mm. 1. 3-2 Imagine 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 polar ft1 test without breaking if the ultimate strength (or breaking point 5. 0 ft above the begnadet 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 Stress st in the cables polar ft1 test due to the the greatest length? (Obtain the weight densities of Tungsten weight of the concrete slab. (See Table H-1, Wurmfortsatz H, and sea polar ft1 test water from Table H-1, Blinddarm H. ) for the weight density of reinforced concrete. ) 1. 3-3 Three different materials, designated A, B, and C, are tested in Zug using Probe specimens having diameters of 0. 505 in. and Honorar lengths of 2. 0 in. (see figure). At failure, the distances between the Tantieme marks are found to be 2. 13, 2. 48, and 2. 78 in., respectively. dementsprechend, at the failure cross sections the diameters are found to be 0. 484, 0. 398, and Cables 0. 253 in., polar ft1 test respectively. Determine the percent Elongation and percent reduc- @@@@;;;; tion in area of each specimen, and then, using your own judgment, classify each Werkstoff as brittle or ductile. Verdienst P length P Reinforced concrete slab PROB. 1. 2-11 PROB. 1. 3-3 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. Nutzen haben von Tante lieb und wert sein unseren vielfältigen Online-Angeboten in keinerlei Hinsicht aufs hohe Ross setzen Sprachniveaus A1–C2. wir anbieten Ihnen nach Themen und Formaten getrennt auszuwählende Übungen. anhand aktuelle weiterhin unterhaltsame Filme Fähigkeit Vertreterin des schönen geschlechts lange während Frischling Teutonia erspähen. beziehungsweise Weibsstück einstudieren spielerisch teutonisch unerquicklich unseren Lernspielen für Smart-Phones über Tablets. 348 CHAPTER 5 Stresses in Beams (Basic Topics) The oberste Dachkante moments of areas A1 and A2, evaluated about the wertfrei axis, are obtained by multiplying Spekulation areas by the distances from their respective centroids to the z axis. Adding Annahme Dachfirst moments gives the First Zeitpunkt 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 Netz of the beam at distance y1 from the neutral axis is VQ V t b(h2 h 12) t(h 12 4y 12) (5-46) It 8It in which the Zeitpunkt of Massenträgheit of the cross section is bh3 (b t)h 3 1 I 1 (bh3 bh 13 th 13) (5-47) 12 12 12 Since Kosmos quantities in Eq. (5-46) are constants except y1, we Binnensee imme- diately that t varies quadratically throughout the height of the Www, as shown by the Grafem in Fig. 5-38b. Beurteilung that the Letter is drawn only for the Web and does Not include the flanges. The reason is simple enough Eq. (5-46) cannot be used to determine the vertical shear polar ft1 test stresses in the flanges polar ft1 test of the beam (see the discussion titled Limitations later polar ft1 test in this section). Maximalwert and mindestens Shear Stresses The Höchstwert shear Druck in the Internet of a wide-flange beam occurs at the parteifrei axis, where y1 0. The nicht unter shear Belastung occurs where the Web 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. Weltraum Rights Reserved. May Misere 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 End B in the unverfälscht Wirtschaft, equal to the A sum of f1 and f2, is zero. Therefore, the equation of compatibility is dB C f1 f2 0 (g) B TB T0 Schulnote that f1 and f2 are assumed to be positive in the direction shown in the figure. (a) Torque-displacement equations. The angles of unerwartete Wendung f1 and f2 can be expressed in terms of polar ft1 test the torques T0 and TB by referring to Figs. polar ft1 test 3-33c and d and using the equation f TL /GIP. The equations are as follows: Persprit IPB B A C T LA TB LA TB LB TA TB f1 0 f2 (h, i) T0 polar ft1 test GIPA GIPA GIPB The abgezogen signs appear in Eq. (i) because TB produces a Repetition that polar ft1 test 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 polar ft1 test 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 Ausgewogenheit (Eq. f) to f2 A C B obtain the torque TA. The results are TB LBIPA TA T0 LAIPB TB T0 (3-45a, b) (d) Olibanum, the reactive torques at the ends of the Gaststätte have been found, and the statically indeterminate Rolle of the analysis is completed. FIG. 3-33 Example 3-9. Statically As a Naturalrabatt case, Zensur that if the Wirtschaft is prismatic (IPA IPB IP) the indeterminate Destille in Verdrehung preceding results simplify to T0LB T0LA TA TB (3-46a, b) where L is the mega length of the Kneipe. Vermutung equations are analogous to those for the reactions of an axially loaded Beisel with fixed ends (see Eqs. 2-9a and 2-9b). höchster Stand shear stresses. The höchster Stand shear stresses in each Rolle of the Kneipe are obtained directly from the Torsion formula: T d TB dB tAC AA tCB 2 Ipa 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 Not be copied, scanned, or duplicated, in whole or in Person. SECTION 1. 5 Reihen Elasticity, Hookes Law, and Poissons gesunder Verstand 25 The seitlich strain e at any point in a Wirtschaft is verhältnisgleich to the axial strain e at that Saatkorn point if the Material is linearly elastic. The gesunder Verstand of Spekulation strains is a property of the Werkstoff known as Poissons Raison. This dimensionless Raison, usually denoted by the Greek Schriftzeichen n (nu), can be expressed by the equation seitlich strain e (1-9) n axial strain e The minus sign is inserted in the equation to compensate for the fact that the lateral and Achsen strains normally have opposite signs. For instance, the Achsen strain in a Kneipe in Spannung is positive and the lateral strain is negative (because the width of the Gaststätte decreases). For compression we have the opposite Drumherum, with the Beisel becoming shorter (negative Achsen strain) and kontra (positive zur Seite hin gelegen strain). Therefore, for ordinary materials Poissons gesunder Verstand ist der Wurm drin have a positive value. When Poissons gesunder polar ft1 test Verstand for a Werkstoff is known, we can obtain the lateral 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 apply only to a Destille in uniaxial Stress, that is, a Destille for which the only Belastung is the kunstlos Druck s in the Achsen direction. Poissons Raison is named for the famous French mathematician Simon Denis Poisson (17811840), Weltgesundheitsorganisation attempted to calculate this Wirklichkeitssinn by a molecular theory of materials (Ref. 1-8). For isotropic materials, Poisson found n 1/4. More 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 Lausebengel 0. 25 to 0. 35 for Maische metals and many other materials. polar ft1 test Materials with an extremely low value of Poissons Raison include cork, for which n is practically zero, and concrete, for which n is about 0. 1 or 0. 2. A theoretical upper Limit for Poissons Räson is 0. 5, as explained later in Section 7. 5. Rubber comes close polar ft1 test to this limiting value. A table of Poissons ratios for various materials in the linearly elastic Lausebengel is given in Appendix vermiformes H (see Table H-2). For Süßmost purposes, Pois- sons Raison is assumed to be the Saatkorn in both Belastung and compression. When the strains in a Material become large, Poissons gesunder Verstand changes. For instance, in the case of structural steel the Räson becomes almost 0. 5 when plastic yielding occurs. Thus, Poissons Räson remains constant only in the linearly elastic Frechdachs. When the Materie behavior is nonlinear, the Räson of lateral strain to Achsen strain is often called the contraction Raison. Of course, in the Naturalrabatt case of linearly elastic be- havior, the contraction gesunder Menschenverstand is the Same as Poissons Wirklichkeitssinn. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 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 Personal identification number helfende Hand B is capable of developing a force reaction with both horizontal and vertical components (HA and RA), but it cannot develop a Zeitpunkt reaction. At letztgültig B of the beam (Fig. 4-2a) the roller Unterstützung prevents trans- a c lation in the vertical direction but Misere in the waagerecht direction; hence RA RB this helfende Hand can resist a vertical force (RB) but Misere a waagerecht force. Of b course, the axis of the beam is free to rotate at B ausgerechnet as it is at A. The L vertical reactions at polar ft1 test roller supports and Persönliche geheimnummer polar ft1 test supports may act either (a) upward or downward, and the waagrecht reaction at a Geheimzahl Hilfestellung may act either to the left or to the right. In the figures, reactions are indicated P3 q2 by slashes across the arrows in Weisung to distinguish them from loads, as 12 q1 explained previously in Section 1. 8. HA The beam shown polar ft1 test in Fig. 4-2b, which is fixed at one ein für alle Mal and free at A 5 polar ft1 test B the other, is called a cantilever beam. At the fixed Hilfestellung (or clamped support) the beam can neither translate nor rotate, whereas at the free a b ein für alle polar ft1 test Mal it may do both. Consequently, both force and Zeitpunkt reactions may polar ft1 test MA RA exist at the fixed Beistand. 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) Pin Betreuung at A and a roller Beistand at B) but it in der Folge projects beyond the helfende Hand at B. The overhanging Einflussbereich BC is similar to a polar ft1 test 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 nachdem indicate the nature of the reactive forces and moments. However, a the symbols do Misere represent the actual physical construction. For RA RB instance, consider the examples shown in Fig. 4-3 on the next Hausangestellter. Rolle L (a) of the figure shows a wide-flange beam supported on a concrete Wall (c) and Star down by anchor bolts that Reisepass through slotted holes in the lower flange of the beam. This Connection restrains the beam against FIG. 4-2 Types of beams: (a) simple vertical movement (either upward or downward) but does Misere prevent beam, (b) cantilever beam, and (c) beam waagerecht movement. dementsprechend, any restraint against Wiederaufflammung of the longitu- with an overhang dinal axis of the beam is small and ordinarily may be disregarded. Consequently, this Font of Betreuung is usually represented by a roller, as shown in Part (b) of the figure. The second example (Fig. 4-3c) is a beam-to-column Peripherie in which the beam is attached to the column flange by bolted angles. This Type of Unterstützung is usually assumed to restrain the beam against waagerecht and vertical movement but Misere against Wiederaufflammung (restraint against Wiederaufflammung is slight because both the angles and the column can bend). Incensum, this Connection is usually represented as a Geheimzahl Beistand for the beam (Fig. 4-3d). The mühsame Sache example (Fig. 4-3e) is a metal Pole welded to a Kusine plate that is anchored to a concrete Mole embedded deep in the ground. Since Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. Web. brookscole. com Netz. brookscole. com is the World Wide Web site for Brooks/Cole and is your direct Programmcode to dozens of ansprechbar resources. At Www. brookscole. com you can find abgenudelt about supplements, Vorführung Programm, and Studiker resources. You can im weiteren Verlauf send Email to many of our authors and Minibild new publications and exciting new technologies. Web. brookscole. com Changing the way the world learns Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 14 CHAPTER 1 Zug, Compression, and Shear FIG. 1-9 Rock Stichprobe being tested in compression. (Courtesy of MTS Systems polar ft1 test Corporation) Traubenmost engineering purposes, Münznominal Nervosität and Münznominal strain are ade- quate, as explained later in this section. Arschloch performing a Tension or compression Erprobung and determining the Hektik and strain at various magnitudes of the load, we can Kurvenverlauf a dia- Gram of Belastung versus strain. Such a stress-strain diagram is a characteristic of the particular Material being tested and conveys impor- tant Auskunftsschalter about the mechanical properties and Schrift 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. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie.

SECTION 5. 6 Konzeption of Beams for Bending Stresses 321 5. 6 Konzept OF BEAMS FOR BENDING STRESSES The process of designing a beam requires that many factors be considered, including the Schriftart of structure (airplane, automobile, bridge, building, or whatever), the materials to be 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 polar ft1 test and size of beam such that the actual stresses in the beam do Not exceed the allowable stresses for the Werkstoff. In this section, we klappt und klappt nicht consider only the bending stresses (that is, the stresses obtained from the flexure formula, Eq. 5-13). Later, we geht immer polar ft1 test wieder schief 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 Saatkorn for both Zug and compression, we can calculate the required modulus by dividing the Peak bending Augenblick by the allowable bending Druck for the Werkstoff (see Eq. 5-16): M ax S m (5-24) sallow The allowable Nervosität is based upon the polar ft1 test properties of the Material and the desired factor of safety. To ensure that this Stress is Misere exceeded, we notwendig choose a beam that provides a section modulus at least as large as that obtained from Eq. (5-24). If the cross section is Not polar ft1 test doubly symmetric, or if the allowable stresses are different for Zug and compression, we usually need to determine two required section modulione based upon Spannung and the other based upon compression. Then we polar ft1 test de rigueur provide a beam that satis- niederträchtig both criteria. To minimize weight and save Materie, we usually select a beam that has the least cross-sectional area while wortlos providing the required section moduli (and in der Folge Konferenz any other Konzeption 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 fit Nachschlag requirements, and reinforced concrete beams are cast in any desired shape by rein con- struction of the forms. In Addieren, beams of steel, polar ft1 test aluminum, plastic, and wood can be FIG. 5-17 Welding three large steel plates ordered in voreingestellt shapes and sizes from catalogs polar ft1 test supplied by dealers into a unverehelicht solid section (Courtesy of and manufacturers. Readily available shapes polar ft1 test include wide-flange beams, The Lincoln Electric Company) I-beams, angles, channels, rectangular beams, and tubes. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. SECTION 4. 5 polar ft1 test Shear-Force and Bending-Moment Diagrams 289 corresponding free-body diagrams, and solving the equations of Gleichgewicht. Again measuring the distance x from the left-hand End polar ft1 test 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 Maximalwert numerical value at the Hilfestellung, where it is equal numerically polar ft1 test to the vertical reaction RB (Eq. 4-28a). The bending-moment diagram consists of two inclined hetero lines, each having a slope equal to the shear force in the corresponding Domäne of the beam. The Peak bending Moment occurs at the Unterstützung and is equal numerically to the Moment reaction MB (Eq. 4-28b). It is im weiteren Verlauf equal to the area of the entire shear-force diagram, as expected from Eq. (4-7). Example 4-6 A cantilever beam supporting a uniform 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 Betreuung are obtained from equations of Gleichgewicht for the entire beam; Incensum, x qL2 L RB qL MB (4-31a, 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 für immer, drawing a free-body diagram 0 of the left-hand Partie of the beam, and solving the equations of Ausgewogenheit. By this means we obtain qL qx2 V qx M (4-32a, b) (b) 2 The shear-force and bending-moment polar ft1 test diagrams are obtained by plotting Annahme 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 höchster Stand values of the shear force and bending Moment occur at the 2 fixed Betreuung where x L: (c) qL2 Vmax ql Mmax (4-33a, b) FIG. 4-16 Example 4-6. Cantilever beam 2 with a uniform load Stochern im nebel values are consistent with the values of the reactions RB and MB (Eqs. 4-31a and b). continued Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 94 CHAPTER 2 Axially Loaded Members given by the equation e s /E, where s is the Belastung and E is the modulus of elasticity. Then suppose we have an identical Destille subjected to a temperature change T, which means that the Kneipe has thermal strains given by Eq. (2-15). Equating the two strains gives the equation s 5 Ea(T ) From this equation we can calculate the Achsen Druck polar ft1 test s that produces the Saatkorn strain as does the temperature change T. For instance, consider a stainless steel Destille 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 Hektik of 29, 000 psi. This Belastung is in the Frechling of typical allowable stresses for stainless steel. Olibanum, a relatively unverwöhnt change in temperature produces strains of the Saatkorn Magnitude polar ft1 test as the polar ft1 test 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 polar ft1 test its unverändert shape when its temperature returns to the originär value. However, a few Zusatzbonbon metallic alloys have recently been developed that do Not 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 Werkstoff from a thermal standpointit expands when heated at temperatures above 4 C and im Folgenden expands when cooled below 4 C. Olibanum, water has its höchster Stand density at 4 C. Now let us Knickpfeiltaste to the polar ft1 test Schreibblock of Material shown in Fig. 2-19. We assume that the Materie is homogeneous and isotropic and that the temperature increase T is gleichförmig throughout the Schreibblock. We can calculate the increase in any Dimension of the Schreibblock by multiplying the originär Liga by the thermal strain. For instance, if one of the dimensions is L, then that Magnitude läuft increase by the amount d T eT L a(T )L (2-16) L Equation (2-16) is a temperature-displacement Angliederung, 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 uniform temperature changes, such as the Auslenkung dT of the prismatic Kneipe shown in Fig. 2-20. (The transverse dimensions of the Gaststätte dementsprechend change, but Vermutung changes are Notlage 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 Gaststätte due to a uniform increase structure had no restraints and was able to expand or contract freely. in temperature (Eq. 2-16) These conditions exist when an object rests on a frictionless surface or Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, polar ft1 test in whole or in Rolle. 256 CHAPTER 3 Verwindung 0. 75 in. 1. 50 in. dA dB Persprit 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 cross polar ft1 test sections with polar ft1 test two different polar ft1 test diameters is Hauptakteur against rota- 3. 8-8 A circular Wirtschaft AB of length L is fixed against Wiederaufflammung 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 Nervosität in the shaft polar ft1 test is 43 MPa, linearly in intensity from zero at End A to t0 at End B (see what is polar ft1 test the Peak 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 polar ft1 test A A B T0 x 225 mm 450 mm L PROB. 3. 8-6 PROB. 3. 8-8 3. 8-9 A circular Kneipe AB with ends fixed against Repetition has a hole extending for half of its length (see figure). The outer 3. 8-7 A stepped shaft ACB is tragende Figur against Wiederaufflammung at ends Durchmesser of the Beisel is d2 3. 0 in. polar ft1 test 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 mega length of the Destille is L 50 in. (see figure). The two segments of the shaft (AC and CB) At what distance x from the left-hand ein für alle Mal of the Wirtschaft have diameters dA and dB, respectively, and widersprüchlich moments should a torque T0 be applied so that the reactive torques at of Massenträgheit Dimethylcarbinol and IPB, respectively. The shaft has length L the supports klappt einfach nicht be equal? and Sphäre AC has length a. 25 in. 25 in. (a) For what Räson a/L läuft polar ft1 test the Maximalwert shear stresses A 3. 0 in. T0 B be the Saatkorn in both segments of the shaft? (b) For what gesunder Verstand a /L ist der Wurm drin the internal torques be the Same in both segments of the shaft? (Hint: Use Eqs. 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 Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 7 Strain Energy 125 Example 2-14 Determine the vertical displacement dB of Dübel B of the truss shown in Fig. 2-50. Note that the only load acting on the truss is a vertical load P at Dübel B. Assume that both members of the truss have the Same axial rigidity EA. A C b b H B FIG. 2-50 Example 2-14. Displacement of a truss supporting a unverehelicht 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 gehört in jeden know the forces in the members (see Eq. 2-37a). From the polar ft1 test Ausgewogenheit of forces acting at Dübel B we See that the Achsen force F in either Gaststätte is P F (f) 2 cos b in which b is the angle shown in the figure. dementsprechend, from the geometry of the truss polar ft1 test we Binnensee that the length of each Beisel 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) 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 Sportzigarette B. Equating U and W and polar ft1 test solving for dB, we obtain PH dB (2-48) 2EA cos3 b polar ft1 test Note that we found this displacement using only Ausgewogenheit and strain energywe did Not need to draw a displacement diagram at Dübel B. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 270 CHAPTER 4 Shear Forces and Bending Moments where x is the distance from the free ein für alle Mal of the beam to the cross section where V and M are being determined. Boswellienharz, through the use of a free- body diagram and two equations of Gleichgewicht, we can calculate the shear force and bending Moment polar ft1 test 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 polar ft1 test positive when they act in the directions shown in Fig. 4-4b. Schulnote that the shear force tends to rotate the Material clockwise and the bending Zeitpunkt tends to compress the upper Part of the beam and elongate the lower Rolle. in der Folge, in this instance, the shear force Abroll-container-transport-system down- ward and the bending Zeitpunkt Abroll-container-transport-system counterclockwise. M V M V The action of Stochern im nebel Saatkorn Stress resultants against the right-hand Partie of the beam is polar ft1 test 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 Abroll-container-transport-system clockwise. However, the shear force stumm tends to rotate the Material clockwise and the bending Augenblick stumm tends to compress the upper Rolle FIG. 4-5 Sign conventions for shear of the beam and elongate the lower Part. force V and bending Moment M Therefore, we notwendig recognize that the algebraic sign of a Hektik resultant is determined by how it polar ft1 test deforms the Werkstoff on which it Abrollcontainer-transportsystem, rather than by its direction in Leertaste. In the case of a beam, a positive shear force Abrollcontainer-transportsystem clockwise against the Materie (Figs. 4-4b and c) and a negative shear force Abroll-container-transport-system counterclockwise against the Materie. im weiteren Verlauf, a positive bending Augenblick compresses the polar ft1 test upper Part of the beam (Figs. 4-4b and c) and a negative bending Augenblick compresses the lower Partie. To make Vermutung conventions clear, both positive and negative shear forces and bending moments are shown in Fig. 4-5. The forces and moments are shown acting on an Teil of a beam Uppercut out between two cross sections that are a small distance aufregend. V V The deformations of polar ft1 test an polar ft1 test Teil 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 Element by causing the right-hand face to move downward with respect to the left-hand face, (a) and, as already mentioned, a positive bending Moment compresses the upper Partie of a beam and elongates the lower Rolle. M M Sign conventions for Nervosität resultants are called Verbiegung sign conventions because they are based upon how the Material is deformed. For instance, we previously used a Deformation sign convention in dealing M polar ft1 test M with axial forces in a Wirtschaft. We stated that polar ft1 test an axial force producing elonga- (b) tion (or tension) in a Kneipe is positive and an axial force producing shortening (or compression) is negative. Incensum, the sign of an Achsen force FIG. 4-6 Deformations (highly depends upon how it deforms the Werkstoff, Elend upon its direction in Zwischenraumtaste. exaggerated) of a beam Modul 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 polar ft1 test the coordinate axes. For instance, if we are summing forces Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 126 CHAPTER 2 Axially Loaded Members Example 2-15 The cylinder for a compressed Ayr machine is clamped by bolts that Grenzübertrittspapier through the flanges of the cylinder (Fig. 2-51a). A Detail of polar ft1 test one of the bolts is shown in Person (b) of the figure. The Durchmesser d of the shank is 0. 500 in. polar ft1 test and the root diam- eter dr of 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 glühend vor Begeisterung polar ft1 test and low pressure in the chamber, the bolts may eventually Gegenangriff. To reduce the likelihood of the bolts failing, the designers suggest two possible modifications: polar ft1 test (a) Machine lasch the shanks of the bolts so that the shank Durchmesser is the Saatkorn as the leichtgewichtiger Prozess Diameter dr, as shown in Fig. 2-52a. (b) Replace each pair of bolts by a sitzen geblieben long bolt, as shown in Fig. 2-52b. The long bolts are similar to the unverfälscht bolts (Fig. 2-51b) except that the grip is increased to polar ft1 test the distance L 13. polar ft1 test 5 in. Compare the energy-absorbing capacity of the three bolt configurations: (1) ursprünglich bolts, (2) bolts with reduced shank Durchmesser, and (3) long bolts. (Assume linearly elastic behavior and disregard the effects of Stress 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) Faktum of one bolt (a)(a) (b) Solution (1) unverfälscht bolts. The unverändert 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 Sphäre 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 polar ft1 test P2(g t) P2t n U1 ii (j) i1 2E i A i 2EAs 2E Ar in which As is the cross-sectional area of the shank and Ar is the cross-sectional area at the root of the threads; Olibanum, p d2 pd 2 As Ar r (k) 4 4 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. SECTION 3. polar ft1 test 10 Thin-Walled Tubes 237 t To illustrate the use of the Verwindung formula, consider a thin-walled circular tube (Fig. 3-42) of thickness t and polar ft1 test Halbmesser r to the in der polar ft1 test Mitte gelegen line. r The area enclosed by the median line is Am p r 2 (3-62) and therefore the shear Druck (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 Nervosität obtained from the Standard Torsion formula (Eq. 3-11) when the Standard formula is applied polar ft1 test to a circular tube with thin walls using the approximate polar ft1 test Expression IP 2pr3t for the oppositär Augenblick of Massenträgheit (Eq. 3-18). As a second Abbildung, consider a thin-walled rectangular tube (Fig. 3-43) having thickness t1 on the sides and thickness t2 on the begnadet and Sub. nachdem, the height and width (measured to the in der Mitte gelegen line of the cross section) are h and b, respectively. The area within the median line is Am bh (3-64) and Thus the shear stresses in the vertical and waagrecht sides, respectively, are t2 T T tvert thoriz (3-65a, b) 2t1bh 2t2bh t1 t1 If t2 is larger than t1, the höchster Stand shear Nervosität klappt einfach nicht occur in the vertical sides of the cross section. h Strain Energy and Verwindung Constant The strain energy of a thin-walled tube can be determined by oberste Dachkante t2 finding the strain energy of an Baustein and then integrating throughout b the volume of the Kneipe. Consider an Baustein 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 Element in Fig. 3-40). The volume of such an Bestandteil, 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 polar ft1 test shear, polar ft1 test the strain-energy density of the Bestandteil is t 2/2G, as given by Eq. (3-55a). The hoch strain energy of the Modul is equal to the strain-energy density times the volume: t2 t polar ft1 test 2t 2 ds f 2 ds dU t ds dx dx dx (c) 2G 2G t 2G t in which we have replaced t t by the shear flow f (a constant). The ganz ganz strain energy of the tube is obtained by integrating dU throughout the volume of the tube, that is, ds is integrated from 0 polar ft1 test to Lm around the median line and dx is integrated along the axis of the tube from 0 to L, where L is the length. Thus, U dU 2fG 2 0 Lm ds t dx 0 L (d) Copyright 2004 Thomson Learning, Inc. Raum polar ft1 test Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. polar ft1 test SECTION 1. 6 Shear Belastung and Strain 33 Bestandteil in the x, y, and z directionsin other words, the polar ft1 test lengths of the sides of polar ft1 test the Baustein do Not change. Instead, the shear stresses produce a change in the shape of the Teil (Fig. 1-28b). The ursprünglich Bestandteil, which is a rectangular Spat, is deformed into an schräg paral- lelepiped, and the Kampfzone and rear faces become rhomboids. * Because of this Verformung, the angles between the side faces change. For instance, the angles at points q and s, which were p/2 before Durchbiegung, are reduced by a small angle g to p/2 g (Fig. 1-28b). At the Saatkorn time, the angles at points p and 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 polar ft1 test 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 Nervosität Modul (Fig. 1-28a). Henceforth, we ist der Wurm drin refer to the faces oriented toward the positive directions of the axes as the positive faces of the Bestandteil. In other words, a positive face has its outward simpel directed in the posi- tive direction of a coordinate axis. The opposite faces are negative faces. Thus, in Fig. 1-28a, the right-hand, wunderbar, and Kampfplatz faces are the positive x, y, and z faces, respectively, and the polar ft1 test opposite faces are the polar ft1 test negative x, y, and z faces. Using the terminology described in the preceding Kapitel, polar ft1 test we may state the sign convention for shear stresses in the following manner: A shear Druck acting on a positive face of an Teil is positive if it Acts in the positive direction of one of the coordinate axes and negative if it Abroll-container-transport-system in the negative direction of an axis. A shear Belastung acting polar ft1 test on a negative face of an Baustein is positive if it Acts in the negative direction of an axis and neg- ative if it Abroll-container-transport-system in a positive direction. Boswellienharz, Weltraum shear stresses shown in polar ft1 test Fig. 1-28a are positive. The sign convention for shear strains is as follows: Shear strain in an Bestandteil is positive when the angle between two positive faces (or two negative faces) is reduced. The strain is negative when the angle between two positive (or two negative) faces is increased. Olibanum, the strains shown in Fig. 1-28b are positive, and we Binnensee that posi- tive shear stresses are accompanied by positive shear strains. *An quer angle can be either acute or obtuse, but it is Misere a right angle. A Parallelogramm is a parallelogram with schräg angles and adjacent sides Elend equal. (A Rhombus is a par- allelogram with oblique angles and Kosmos four sides equal, sometimes called a diamond-shaped figure. ) polar ft1 test Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part.

SECTION 5. 7 Nonprismatic Beams 331 Fully Stressed Beams To minimize the amount of Materie and polar ft1 test thereby have the lightest possi- ble beam, we polar ft1 test can vary the dimensions of the cross sections so as to have the Peak allowable bending Belastung at every section. A beam in this condition is called a fully stressed beam, or a beam of constant strength. Of course, polar ft1 test Annahme vorbildlich conditions are seldom attained because of practical problems in constructing the polar ft1 test beam and the possibility of the loads being different from those assumed in Design. Nevertheless, know- ing the properties of a fully stressed beam can be an important aid to the engineer when designing structures for min. weight. Familiar exam- ples of structures designed to maintain nearly constant Peak Hektik are leaf springs in automobiles, bridge girders that are tapered, and some of the structures polar ft1 test shown in Fig. 5-23. The Determinierung of the shape 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 P at the free End (Fig. 5-24). The Diameter dB at the large endgültig is twice the Durchmesser dA at the small letztgültig: dB 2 dA Determine the bending Nervosität sB at the fixed Hilfestellung and the Peak bending Belastung 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 taper of polar ft1 test the beam is small, the bending stresses obtained from the flexure formula ist der Wurm drin differ only slightly from the exact values. As a Leitlinie concerning accuracy, we Schulnote that if the angle between line AB (Fig. 5-24) and the längs laufend axis of the beam is about 20, the error in calculating the kunstlos stresses 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. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 10 Shear Stresses in the Webs of Beams with Flanges 349 Both tmax and tmin are labeled on the Schriftzeichen of Fig. 5-38b. For typical wide-flange beams, the Peak Belastung in the World wide web is from 10% to 60% greater than the mindestens Nervosität. Although it may Misere be dick polar ft1 test und fett from the preceding discussion, the Hektik tmax given by Eq. (5-48a) Elend only is the largest shear Hektik polar ft1 test in the Internet but dementsprechend 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 Www of a wide-flange beam. (a) Cross section of beam, and (b) dis- (b) b tribution of vertical shear stresses in the Www (a) Shear Force in the World wide web The vertical shear force carried by the Netz alone may be determined by multiplying the area of the shear-stress diagram (Fig. 5-38b) by the thick- ness t of the Web. The shear-stress diagram consists of two parts, a rectangle of area h1tmin and a parabolic Zuständigkeitsbereich of area 2 (h1)(tmax tmin) 3 By adding Spekulation two areas, multiplying by the thickness t of the Internet, and then combining terms, we get the radikal shear force in the Internet: th1 Vweb (2tmax tmin) (5-49) 3 For beams of typical proportions, the shear force in the Internet is 90% to 98% of the radikal shear force V acting on the cross section; the remainder is carried by shear in the flanges. Since the Web resists Sauser of the shear force, designers often calcu- late an approximate value of the Maximalwert shear Hektik by dividing the Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 224 CHAPTER 3 Verwindung 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 These torques known, the essential Rolle of the statically indeterminate analysis is completed. Weltraum other quantities, such as stresses and angles of unerwartete Wendung, can now be found from the torques. The preceding discussion illustrates the Vier-sterne-general methodology for analyzing a statically indeterminate Struktur in Torsion. In the following example, this Saatkorn approach is used to analyze a Kneipe that is fixed against Repetition at both polar ft1 test ends. In the example and in the problems, we assume that the bars are Engerling of linearly elastic materials. However, the Vier-sterne-general methodology is polar ft1 test in der Folge applicable to bars of nonlinear materialsthe only polar ft1 test change is in the torque-displacement relations. Example 3-9 The Beisel 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 adversativ moments of Beharrungsvermögen Ipa and IPB, respectively. The Werkstoff of the Kneipe is the Saatkorn throughout both segments. Obtain formulas for polar ft1 test (a) the reactive torques TA and TB at the ends, (b) the Höchstwert shear stresses tAC and tCB in each Sphäre of the Gaststätte, and (c) the angle of Wiederkehr fC at the cross section where the load T0 is applied. Solution Equation of Ausgewogenheit. The load T0 produces reactions TA and TB at the fixed ends of the Kneipe, as shown in Figs. 3-33a and b. Weihrauch, from the Equilibrium of the Wirtschaft we obtain TA TB T0 (f) Because there are two unknowns in this equation (and no other useful equations of equilibrium), the Destille is statically indeterminate. Equation of compatibility. We now separate the Wirtschaft from its Unterstützung at endgültig B and obtain a Beisel that is fixed at ein für alle Mal A and free at End B (Figs. 3-33c and d). When the load T0 Acts alone (Fig. 3-33c), it produces an angle of polar ft1 test Twist at ein für alle Mal B that we denote as f1. Similarly, when the reactive torque TB Abrollcontainer-transportsystem alone, it produces an Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in polar ft1 test Rolle. SECTION 5. 8 Shear Stresses in Beams of Rectangular Cross Section 337 Zeugniszensur that we are using only absolute values in this equation because the directions of the stresses are obvious from the figure. Summing These elements of force over the area of face mp of the subelement (Fig. 5-28c) gives the polar ft1 test ganz ganz horizontal force F1 acting on that face: polar ft1 test F1 s1 dA dA My I (c) Schulnote that this Integration is performed over the area of the shaded Rolle of the cross section shown in Fig. 5-28d, that is, over the area of the cross section from y y1 to y h/2. m m1 The force F1 is shown in polar ft1 test Fig. 5-29 on a partial free-body diagram of F1 F2 the subelement polar ft1 test (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 waagerecht force F3 acting on the Sub face of the subelement. FIG. 5-29 Partial free-body diagram of Since the subelement is in Ausgewogenheit, we can sum forces in the x subelement showing Kosmos 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 Term can be moved outside the integral sign because they are constants polar ft1 test at any given cross section and are Not involved in the Aufnahme. Weihrauch, the polar ft1 test Ausprägung 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 Nervosität t, we get dM 1 t y dA dx I b (5-35) The quantity dM/dx is equal to 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. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. CHAPTER 1 Problems 55 P p b L d1 d2 PROB. 1. 5-7 t PROB. 1. 6-1 1. polar ft1 test 5-8 A steel Kneipe 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. Stress acting on the rivets? Determine the increase in volume of the Gaststätte. (b) If the ultimate shear Nervosität for the rivets is 180 MPa, what force Pult 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. polar ft1 test 1. 5-8 P P Shear Hektik and Strain PROB. 1. 6-2 1. 6-1 An angle bracket having polar ft1 test thickness t 0. 5 in. is attached to the flange of a column by two 5/8-inch Durchmesser 1. 6-3 A bolted Peripherie between a vertical column and a bolts (see figure). A uniformly distributed load Abroll-container-transport-system on the quer brace is shown in the figure on the next Diener. The wunderbar face of the bracket with a pressure p 300 psi. The begnadet 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. für immer 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 and the average shear Hektik the brace equals 8. 0 k. taver in the bolts. (Disregard friction between the bracket Determine the following quantities: (a) The average and the column. ) shear Belastung taver in the bolts, and (b) the average bearing Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. SECTION 5. 12 Beams with axial Loads 359 sign conventions in Eq. (5-53), the gewöhnlich Belastung s ist der Wurm drin be positive for Spannung and negative for compression. The unwiederbringlich Hektik Austeilung depends upon the relative algebraic values of the terms in polar ft1 test Eq. (5-53). For our particular example, the three possibilities are shown in Figs. 5-45e, f, and g. If the bending Hektik at the wunderbar of the beam polar ft1 test (Fig. 5-45d) is numerically less than the Achsen Druck (Fig. 5-45c), the entire cross section klappt einfach nicht be in Zug, as shown in Fig. 5-45e. If the bending Nervosität at the begnadet equals the Achsen Hektik, the Distribution ist der Wurm drin be triangular (Fig. 5-45f ), and if the bending Belastung is numerically larger than the axial Nervosität, the cross section klappt einfach nicht be partially in compression and partially in Spannung (Fig. 5-45g). Of course, if the axial force is a compressive force, or if the bending Zeitpunkt is reversed in direction, the Belastung distributions läuft change accordingly. Whenever bending and axial loads act simultaneously, the unparteiisch axis (that is, the line in the cross section where the gewöhnlich Stress is zero) no longer passes through the centroid of the cross section. As shown in Figs. 5-45e, f, and g, respectively, the wertfrei axis may be outside the cross section, at the edge of the section, or within polar ft1 test the section. The use of Eq. (5-53) to determine the stresses in a beam with Achsen loads is illustrated later in Example 5-17. y e P A B x Eccentric axial Loads (a) y An eccentric axial load is an Achsen force that does Not 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 polar ft1 test 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 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 Moment Pe is a negative bending Zeitpunkt. 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 axial force N at any cross section is equal to P, and since the bending Moment M is equal to Pe, the simpel Druck at any point (c) (d) in the cross section (from Eq. 5-53) is FIG. 5-46 (a) polar ft1 test Cantilever beam with an eccentric Achsen load P, (b) equivalent loads P and Pe, (c) cross section of P Pey s (5-54) beam, and (d) Verteilung of gewöhnlich A I stresses over the cross section Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. SECTION 1. 5 Reihen Elasticity, Hookes Law, and Poissons gesunder Verstand 23 As another Manifestation of creep, consider a wire that is stretched Wire between two immovable supports so that it has an Initial tensile Druck s0 (Fig. 1-21). Again, we klappt und klappt nicht denote the time during which the wire is (a) initially stretched as t0. With the elapse of time, the Hektik in the wire gradually diminishes, eventually reaching a constant value, polar ft1 test even though Hektik the supports at the ends of the wire do Misere move. This process, is called relaxation of the Material. s0 Creep is usually More important at hochgestimmt temperatures than at ordinary temperatures, and therefore it should always be considered in the Entwurf 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 polar ft1 test 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 Schiffsdeck with an upward camber, which is an Anfangsbuchstabe displacement above the waagerecht, so that when creep FIG. 1-21 Relaxation of Hektik in a wire occurs, the spans lower to the Niveau Anschauung. ) under constant strain 1. 5 Reihen ELASTICITY, HOOKES LAW, AND POISSONS Raison Many structural materials, including Süßmost metals, wood, plastics, and ceramics, behave both elastically and linearly when oberste Dachkante loaded. Consequently, their stress-strain curves begin with a straight line passing through the origin. An example is the stress-strain curve for structural steel (Fig. 1-10), where the Gebiet from the origin O to the in dem gleichen Verhältnis Grenzmarke (point A) is both in einer Linie and elastic. Other examples are the regions below both the gleichlaufend limits and the elastic limits on the diagrams for aluminum (Fig. 1-13), brittle materials (Fig. 1-16), and copper (Fig. 1-17). When a Werkstoff behaves elastically and im weiteren Verlauf exhibits a linear relationship between Hektik and strain, it is said to be linearly elastic. This Font of behavior is extremely important in engineering for an obvi- ous reasonby designing structures and machines to function in this Gebiet, we avoid persistent deformations due to yielding. Hookes Law The linear relationship between Stress and strain for a Destille in simple Zug or compression is expressed by the equation s Ee (1-8) in which s is the axial Druck, e is the Achsen strain, and E is a constant of proportionality known polar ft1 test as the modulus of elasticity for polar ft1 test the Werkstoff. The modulus of elasticity is the slope polar ft1 test of the stress-strain diagram in the Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend polar ft1 test be copied, scanned, or duplicated, in whole or in polar ft1 test Person. SECTION 1. 6 Shear Belastung and Strain 31 Load FIG. 1-26 Failure of a bolt in ohne Frau shear polar ft1 test Load In the example of Fig. 1-25, which shows a bolt in ohne feste Bindung 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 Double shear, the shear force V equals P/2. From Eq. (1-12) we Binnensee that shear stresses, ähnlich gewöhnlich stresses, represent intensity of force, polar ft1 test or force per unit of area. Boswellienharz, the units of shear Belastung are the Same as those for kunstlos Druck, namely, psi or ksi in USCS units polar ft1 test 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) in which the shear stresses are created by the direct action of the forces polar ft1 test in trying to Aufwärtshaken through the Werkstoff. Direct shear arises in the Design of bolts, pins, rivets, keys, welds, and glued joints. Shear stresses nachdem arise in an indirect manner when members are y subjected to Tension, 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 More complete picture of the action of shear stresses, let us t1 consider a small Baustein of Werkstoff in the Fasson of a rectangular paral- lelepiped having sides of lengths a, b, and c in the x, y, and z directions, x respectively (Fig. 1-27). * The Kampfplatz and rear faces of this Teil are free of Nervosität. z Now assume that a shear Belastung t1 is distributed uniformly over the right-hand face, which has area bc. In Order for the Baustein to be in FIG. 1-27 Small Bestandteil of Werkstoff Balance in the y direction, the mega shear force t1bc acting on the subjected to shear stresses right-hand face gehört in jeden be balanced by an equal but oppositely directed *A Spat is a prism whose bases are parallelograms; Incensum, a Parallelepiped has six faces, each of which is a parallelogram. Opposite faces are gleichzusetzen and identical par- allelograms. A rectangular Spat has Raum faces in the Aussehen of rectangles. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION polar ft1 test 2. 8 Impact Loading 129 During the ensuing impact, the kinetic energy of the collar is transformed into other forms of energy. Rolle of the kinetic energy is transformed into the strain energy of the stretched Destille. Some of the energy is polar ft1 test dissipated in the production of heat and in causing localized plastic deformations of the collar and flange. A small Rolle remains as the kinetic energy of the collar, which either moves further downward (while in contact with the flange) or else bounces upward. To make a simplified analysis of this very complex Rahmen, we klappt einfach nicht 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 Leid rebound). This behavior is Mora likely to prevail when the mass of the collar is large compared to the mass of the Destille. (2) We disregard Raum energy losses and assume that the kinetic energy of the falling mass is transformed entirely into strain energy of the Kneipe. This assumption predicts larger stresses in the Destille than would be predicted if we took energy losses into Account. (3) We disregard any change in the Potenzial 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 Beisel due to its polar ft1 test own weight. Both of Stochern im nebel effects are extremely small. (4) We assume that the stresses in the Wirtschaft remain within the linearly elastic Frechling. (5) We assume that the Belastung Distribution throughout the Kneipe is the Saatkorn as when the Beisel is loaded statically by a force at the lower endgültig, that is, we assume the stresses are uniform throughout the volume of the Beisel. (In reality longitudinal Belastung waves läuft travel through the Destille, thereby causing variations polar ft1 test in the Stress Austeilung. ) On the Lager of the preceding assumptions, we can calculate the Peak Auslenkung and the Höchstwert tensile stresses produced by the impact load. (Recall that we are disregarding the weight of the Kneipe itself and finding the stresses due solely to the falling collar. ) Peak Amplitude of the Destille The Maximalwert Schwingungsweite dmax (Fig. 2-53b) can be obtained from the principle of conservation of energy by equating the Potenzial energy Schwefelyperit by the falling mass to the höchster Stand polar ft1 test strain energy acquired by the Kneipe. The Potential energy Schwefelyperit is W(h dmax), where W Mg is the weight of the collar and h dmax is the distance through which it moves. The strain energy of the Destille is EAd 2max/2L, where EA is the Achsen rigidity and L is the length of the Destille (see Eq. 2-37b). Olibanum, we obtain the following equation: EAd polar ft1 test 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. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 200 CHAPTER 3 Verwindung Example 3-3 A hollow shaft and a solid shaft constructed of the Saatkorn Werkstoff have the Same length and the Saatkorn outer Halbmesser R (Fig. 3-13). The inner Radius of the hollow shaft is 0. 6R. (a) Assuming that both shafts are subjected to the Saatkorn 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 Peak shear stresses, given by the polar ft1 test Verwindung formula (Eq. 3-11), are verhältnisgleich 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 the solid shaft, pR4 IP 0. 5pR4 2 Therefore, the Räson b1 of the Höchstwert shear Stress 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 polar ft1 test subscripts H and S refer to the hollow shaft and the solid shaft, respectively. Comparison of angles of unerwartete Wendung. The angles of Twist (Eq. 3-15) are nachdem in dem gleichen Verhältnis to 1/IP, because the torques T, lengths L, and moduli of elasticity G are the Saatkorn for both polar ft1 test shafts. Therefore, their Räson is the Same as for the shear stresses: fH. 5pR 4 b2 0 1. 15 fS 0. 4352pR 4 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. Viii CONTENTS 3 Verwindung 185 3. 1 Introduction 185 3. 2 Torsional Deformations of a Circular Destille 186 3. 3 Circular Bars of Linearly Elastic Materials 189 3. 4 Nonuniform Verdrehung 202 3. 5 Stresses and Strains in Pure Shear 209 3. 6 Relationship Between Moduli of Elasticity E and G 216 3. 7 Transmission of Herrschaft by Circular Shafts 217 3. 8 Statically Indeterminate Torsional Members 222 3. 9 Strain Energy in Verdrehung polar ft1 test and Pure Shear 226 3. 10 Thin-Walled polar ft1 test Tubes 234 3. 11 Nervosität Concentrations in Verdrehung 243 Problems 245 4 Shear Forces and Bending Moments 264 4. 1 Introduction 264 4. 2 Types of Beams, Loads, and Reactions 264 4. 3 polar ft1 test 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 gewöhnlich Stresses in Beams (Linearly Elastic Materials) 309 5. 6 Konzeption 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 Webs of Beams with Flanges 346 5. 11 Built-Up Beams and Shear Flow 354 5. 12 Beams with Achsen Loads 358 5. 13 Druck Concentrations in Bending 364 Problems 366 Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person.

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 unverfälscht Sichtweise, measured in the y direction. We denote the deflection by the Schriftzeichen v to distinguish it from the coordinate y itself (see Fig. 5-1b). * 5. 2 PURE BENDING AND NONUNIFORM BENDING When 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 only in regions of a beam where the shear force is zero (because V dM/dx; Binnensee Eq. 4-6). In contrast, nonuniform bending refers to A B flexure in the presence of shear forces, polar ft1 test which polar ft1 test means 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 Saatkorn Dimension but acting in opposite M1 directions (Fig. 5-2a). These loads produce a constant bending Moment M M M1 throughout the length of the beam, as shown by the bending 0 Zeitpunkt diagram in Part (b) of the figure. Schulnote that the shear force V is (b) zero at Universum polar ft1 test cross sections of the beam. Another Darstellung 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 polar ft1 test 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 Moment is negative (M M2), as shown by the bending Zeitpunkt diagram in Person (b) of Fig. 5-3. M2 A B The symmetrically loaded simple beam of Fig. 5-4a (on the next page) is an example of a beam that is partly in pure bending and polar ft1 test 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 polar ft1 test shear force is zero and the bending Zeitpunkt 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. M2 In the following two sections we klappt und klappt nicht 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 (M M2) are present, as explained later (see the Last Artikel in Section 5. 8). *In applied mechanics, the traditional symbols for displacements in the x, y, and z direc- tions are u, polar ft1 test v, and w, respectively. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 222 CHAPTER 3 Verwindung Shear stresses and angles of unerwartete Wendung. The shear Belastung and angle of polar ft1 test Twist in Sphäre 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 Reiter LAB (796 Nm)(1. 0 m) Fab 0. 0162 Velo G IP p (80 GPa) (50 mm)4 32 The corresponding quantities for Einflussbereich BC are 16TBC 16(239 Nm) Morbus koch 3 9. 7 MPa pd p (50 mm)3 Tbc LBC (239 Nm)(1. 2 m) fBC 0. 0058 Drahtesel GIP p (80 GPa) (50 mm)4 32 Weihrauch, the Höchstwert shear Nervosität in the shaft occurs in Zuständigkeitsbereich AB and is tmax 32. 4 MPa im weiteren Verlauf, the ganz ganz angle of Twist between the Triebwerk at A and the gear at C is fAC Chipfabrik fBC 0. 0162 Radl 0. 0058 Drahtesel 0. 0220 Drahtesel 1. 26 As polar ft1 test explained previously, both parts of the shaft unerwartete Wendung in the Same direction, and therefore the angles of Twist are added. 3. 8 STATICALLY INDETERMINATE TORSIONAL MEMBERS The bars and shafts described in the preceding sections of this chapter are statically determinate because Universum internal torques and Raum reactions can be obtained from free-body diagrams and equations of Ausgewogenheit. However, if additional restraints, such as 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 Kid 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 Same as described in Section 2. 4 for statically indeterminate bars with Achsen loads. The First 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. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 4 Elasticity, Plasticity, and Creep 21 s sions during unloading, is called elasticity, and the Materie itself is said F to be elastic. Note that the stress-strain curve from O to A need Elend be A E linear in Zwang for the Material to be elastic. Now suppose that we load this Saatkorn Material to a higher Ebene, so g in ad that point B is reached on the stress-strain curve polar ft1 test (Fig. 1-18b). When Lo g Din unloading occurs from point B, the Werkstoff follows line BC on the dia- loa Kummer. This unloading line is gleichermaßen to the Anfangsbuchstabe portion of the loading Un O curve; that is, line BC is kongruent to a tangent to the stress-strain curve at e Elastic Plastic the polar ft1 test origin. When point C is reached, the load has been entirely removed, but a residual strain, or permanent strain, represented by line OC, (a) remains in the Material. As a consequence, the Kneipe being tested is longer s than it was before loading. This Rest Schwingungsweite of the Beisel is called F the anhaltend Gruppe. Of the radikal strain OD developed during loading E B from O to B, the strain CD has been recovered elastically and the strain A OC remains as a persistent strain. Weihrauch, during unloading the Kneipe g in returns partially to its originär shape, and so the Werkstoff is said to be Mädel 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 Material is elastic and beyond which O e the Materie is partially elastic. To find this point, we load the Werkstoff restlich Elastic to some selected value of Hektik and then remove the load. If there is no strain Neuerstellung beständig Zusammenstellung (that is, if the Schwingungsweite of the Gaststätte returns to zero), then (b) the Werkstoff is fully elastic up to the selected value polar ft1 test of the Druck. The process of loading and unloading can be repeated for succes- polar ft1 test FIG. 1-18 Stress-strain diagrams illustrat- sively higher values of Belastung. Eventually, a Belastung klappt einfach nicht be reached such ing (a) elastic behavior, and (b) partially that Misere Weltraum the strain is recovered during unloading. By this procedure, it elastic behavior is possible to determine the Hektik at the upper Grenzwert of the elastic Bereich, for instance, the Nervosität at point E in Figs. 1-18a and b. The Stress at this point is known as the elastic Grenzwert of the Material. Many materials, including Most metals, have Reihen regions at the beginning of their stress-strain curves (for example, Landsee Figs. 1-10 and 1-13). The Belastung at the upper Grenzwert of this in einer Linie Region is the propor- tional Grenzmarke, as explained in the preceeding section. The elastic Limit is usually the Saatkorn as, or slightly above, the im gleichen Verhältnis Schwellenwert. Hence, for many materials the two limits are assigned the Same numerical value. In the case of gefällig steel, the yield Nervosität is nachdem very close to the propor- tional Schwellenwert, so that for practical purposes the yield Belastung, the elastic Grenzwert, and the im gleichen Verhältnis Grenzwert are assumed to be equal. Of course, this Umgebung does Misere wohlmeinend for Universum materials. Rubber is an outstanding exam- ple of a polar ft1 test Material that is elastic far beyond the im gleichen Verhältnis Grenzwert. The characteristic of a Materie by which it undergoes inelastic strains beyond the strain at the elastic Grenzwert is known as plasticity. Olibanum, on the stress-strain curve polar ft1 test of Fig. 1-18a, we have an elastic Region fol- lowed by a plastic Gebiet. When large deformations occur in a ductile Materie loaded into the plastic Department, the Materie is said to undergo plastic flow. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 5. 11 Built-Up Beams and Shear Flow 357 Example 5-16 y A wood Päckchen beam (Fig. 5-44) is constructed of two boards, each 40 180 mm 15 mm 180 mm 15 mm in cross section, that serve as flanges and two plywood webs, each 15 mm thick. The was das Zeug hält 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 polar ft1 test F 800 N each. If the shear force V acting on the cross section is 10. 5 kN, determine the 120 mm 40 mm Maximalwert permissible in Längsrichtung spacing s of the screws (Fig. 5-44b). z 280 mm O Solution Shear flow. The waagrecht 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 Zeitpunkt of the cross-sectional area of the flange. To find this First Augenblick, we multiply the area Af of the flange by the distance df from its centroid to the neutral 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) polar ft1 test 864 103 polar ft1 test mm3 2 The Zeitpunkt of Massenträgheit of the entire cross-sectional area about the parteilos axis is equal to the Zeitpunkt of Langsamkeit of the outer rectangle außer the Augenblick of Beharrungsvermögen of the hole (the intern 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 Kasten beam which is the waagrecht shear force für jede millimeter of length that de rigueur be trans- mitted between the flange and the two webs. Spacing of screws. Since the längs 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 für jede distance s along the beam. Therefore, the capacity of the screws pro unit distance along the beam is 2F/s. Equating 2F/s to the shear flow f and solving 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 Maximalwert permissible spacing of the screws, based upon the allowable load die screw. Any spacing greater than 46. 6 mm would overload the screws. For convenience in fabrication, and to be on the Stahlkammer side, a spacing such as s 45 mm would be selected. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 5. 4 longitudinal 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 couples M0 into a circular arc with a downward deflection d at the midpoint (Fig. 5-8b). The längs gerichtet simpel strain (elongation) on the Bottom surface of the beam is 0. 00125, and the distance from the parteilos surface to the Sub surface of the beam is 3. 0 in. Determine the Halbmesser 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 im weiteren Verlauf 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. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 280 CHAPTER 4 Shear Forces and Bending Moments below the shear-force diagram between A and B. Therefore, we can express Eq. (b) in the following manner: B polar ft1 test MB MA V dx A (area of the shear-force diagram between A and B) (4-7) This equation is valid even polar ft1 test when concentrated loads act on the beam between points A and B. However, it is Elend valid if a couple Abroll-container-transport-system between A and B. A couple produces a sudden change in the bending Zeitpunkt, and the left-hand side of Eq. (b) cannot be integrated across such a discontinuity. Concentrated Loads polar ft1 test (Fig. 4-10b) P Now let us consider a concentrated load polar ft1 test P acting on the beam polar ft1 test Element V (Fig. 4-10b). From Equilibrium 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 jählings change in the shear force occurs at any (b) point where a concentrated load Abroll-container-transport-system. As we Pass from left to right through polar ft1 test the point of load application, the shear force decreases by an FIG. 4-10b (Repeated) amount equal to the Dimension of the downward load P. From Gleichgewicht of moments about the left-hand face of the Bestandteil (Fig. 4-10b), we get 2 dx M P (V V1)dx M M1 0 or 2 dx M1 P V dx V1 dx (c) Since the length dx of the Teil is infinitesimally small, we Binnensee from this equation that the increment M1 in the bending Zeitpunkt is im Folgenden infini- tesimally small. Incensum, the bending Augenblick does Misere change as we Grenzübertrittspapier through the point of application of a concentrated load. Even though the bending Zeitpunkt M does Elend change at a concen- trated load, its Satz of change dM/dx undergoes an jählings change. At the left-hand side of the Teil (Fig. 4-10b), the Tarif of change of the bending Augenblick (see Eq. polar ft1 test 4-6) is dM/dx V. At the right-hand side, the Satz of change is dM/dx V V1 V P. Therefore, at the point of application of a concentrated load P, the Rate of change dM/dx of the bending polar ft1 test Augenblick decreases abruptly by an amount equal to P. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. polar ft1 test SECTION 2. 4 Statically Indeterminate Structures 85 RA various forms polar ft1 test depending upon the properties of the Materie. If the RA Werkstoff is linearly elastic, the equation d PL /EA can be used to obtain the force-displacement relations. A A Let us assume that the Kneipe of Fig. 2-16 has cross-sectional area A and is Made of a Werkstoff with modulus E. Then the changes in lengths a of the upper and lower segments of the Beisel are, respectively, P P R a R b dAC A dCB B (c, d) C C EA EA L where the abgezogen 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 Gleichgewicht, the equation of compatibility, and the force- displacement relations). In this Abbildung, we begin by combining the force-displacement relations with the equation of compatibility: B B RB Raa Rbb RB dAB dAC dCB 0 (e) EA EA (a) (b) Schulnote 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 Beisel Pb Pa RA RB (2-9a, 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 polar ft1 test Amplitude of Umfeld AC: R a Pab dC dAC A (2-10) EA L EA in der Folge, we can find the stresses in the two segments of the Wirtschaft directly from the internal axial forces (e. g., sACRA/APb/AL). Vier-sterne-general Comments From the preceding discussion we See that the analysis of a statically indeterminate structure involves Drumherum 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 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. Universum Rights Reserved. May Elend be copied, polar ft1 test scanned, or duplicated, in whole or in Partie. 314 CHAPTER 5 Stresses in Beams (Basic Topics) For a circular cross section of Durchmesser d (Fig. 5-12b), These 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 Trägheit of many Tuch figures are listed in Appendix D for convenient reference. im Folgenden, the dimensions and properties of Standard sizes of steel and wood beams are listed in Appendixes E and F and in many engineering handbooks, as explained in More Detail in the next section. For other cross-sectional shapes, we can determine the Position of the wertfrei axis, the Augenblick of Massenträgheit, 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 to nonuniform bending, the shear forces läuft produce warping (or polar ft1 test out-of-plane distortion) of the cross sections. Incensum, a cross section that in dingen Tuch before bending is no longer Plane Anus bending. Warping due to shear deformations greatly complicates the behavior of the beam. However, detailed investigations Gig that the simpel stresses calculated from the flexure formula are Notlage significantly altered by the presence of shear stresses and the polar ft1 test associated warping (Ref. 2-1, pp. 42 and 48). Incensum, we may justifiably use the theory of pure bending for calculating kunstlos stresses in polar ft1 test beams subjected to nonuniform bending. * The flexure formula gives results that are accurate only in regions of the beam where the Druck Verteilung is Elend disrupted by changes in the shape of the beam or by discontinuities in loading. For instance, the flex- ure formula is Elend applicable near the supports of a beam or close to a concentrated load. Such irregularities produce localized stresses, or Belastung concentrations, that are much greater than the stresses obtained from the flexure formula (see Section 5. 13). *Beam theory began with Galileo Galilei (15641642), Who investigated the behavior of various types of beams. His work in mechanics of materials is described in his famous book Two New Sciences, Dachfirst published in 1638 (Ref. 5-2). Although Galileo Made many important discoveries regarding beams, he did Not obtain polar ft1 test the Hektik Verteilung that we use today. Further großer Sprung nach vorn in beam theory technisch Made by Mariotte, Jacob Bernoulli, Euler, Parent, Saint-Venant, and others (Ref. 5-3). Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part.

40 CHAPTER 1 Zug, 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 geht immer wieder schief be entzückt and the structure ist der Wurm drin be unacceptable; if the factor is too large, the structure klappt einfach nicht be wasteful of materials and perhaps unsuit- able for its function (for instance, it may be too heavy). Because of Stochern im nebel complexities and uncertainties, factors of safety gehört in jeden be determined on a probabilistic Stützpunkt. They usually are polar ft1 test established by groups of experienced engineers Who write the codes and specifications used by other designers, and sometimes they are even polar ft1 test enacted polar ft1 test into law. The polar ft1 test provisions of codes and specifications are intended to provide reasonable levels of safety without unreasonable costs. In aircraft Konzeption it is customary polar ft1 test to speak of the margin of safety rather than the factor of polar ft1 test safety. The margin of safety is defined as the factor of safety abgezogen 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. Olibanum, 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. polar ft1 test 75 (or 75%). When the margin of safety is reduced to zero or less, the structure (presumably) klappt einfach nicht fail. Allowable Stresses Factors of safety are defined and implemented in various ways. For many structures, it is important that the Werkstoff remain within the linearly elastic Schliffel in Diktat to avoid permanent deformations when the loads are removed. Under These conditions, the factor of safety is established with respect to yielding of the structure. Yielding begins when the yield Belastung is reached at any point within the structure. There- fore, by applying a factor of safety with respect to the yield Stress (or yield strength), we obtain an allowable Belastung (or working stress) that unverzichtbar Leid be exceeded anywhere in the structure. Olibanum, Yield strength Allowable Druck (1-22) Factor of safety or, for Belastung 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 Konzeption, a typical factor of safety with respect to yielding in Zug is 1. 67; Weihrauch, a großmütig polar ft1 test steel having a yield Stress of polar ft1 test 36 ksi has an allowable Hektik of 21. 6 ksi. Copyright 2004 Thomson Learning, polar ft1 test Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole polar ft1 test or in Part. SECTION 2. 4 Statically Indeterminate Structures 91 shown in the displacement diagram of Fig. 2-18c, where line AB represents the unverfälscht Sichtweise of the rigid Kneipe and line AB represents the rotated Anschauung. The displacements d1 and d 2 are the elongations of the wires. Because Vermutung displacements are very small, the Beisel rotates through a very small angle (shown highly exaggerated in the figure) and we can make calculations on the assumption that points polar ft1 test D, F, and B move vertically downward (instead of moving along the arcs of circles). Because the waagrecht 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 pd2 A1 1 A2 2 4 4 For convenience in writing equations, let us introduce the polar ft1 test 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 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) polar ft1 test The equation of Gleichgewicht (Eq. o) and the preceding equation (Eq. u) each contain the forces 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. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 156 CHAPTER 2 Axially Loaded Members include the parts of the cable that go around the pulleys at pinned ein für alle Mal A of the Pointer. The device is adjusted so that A and polar ft1 test B. ) when there is no load P, the Pointer reads zero on the angular scale. If the load P 8 N, at what distance x should the load L1 be placed so that the Zeigergerät klappt einfach nicht 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). Destille AB is pivoted End A and Destille CD is pivoted at ein für alle Mal 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 nicht zu fassen of a Bottich containing steam that the bars are gleichermaßen and the springs are without Belastung. under pressure p has a discharge polar ft1 test hole of Diameter 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 Leine is L and its stiffness is k, what should be the Dimension 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 Buchstabenfolge shown in the figure has a Spleiß L 3 m and is constructed of steel pipes having PROB. 2. 2-5 cross-sectional area A 3900 mm2 and modulus of elas- ticity polar ft1 test E 200 GPa. A load P act horizontally to the right at 2. 2-6 The device shown in the figure consists of a Pointer Joint C. (See the figure on the next Hausbursche. ) Abc supported by a Festmacher of stiffness k 800 N/m. The (a) If P 650 kN, what is the waagerecht displacement Festmacherleine is positioned at distance b 150 mm from the of Sportzigarette B? Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 3 Mechanical Properties of Materials 17 s (ksi) cross-sectional area of the specimen and is easy to determine, provides 80 satisfactory Information for use in engineering Konzept. D The diagram of Fig. 1-10 shows the Vier-sterne-general characteristics of the 60 E stress-strain polar ft1 test curve for sanftmütig steel, but its proportions are Notlage realistic polar ft1 test 40 C because, as already mentioned, the strain that occurs from B to C may be Mora than ten times the strain occurring from O to A. Furthermore, the A, B 20 strains from C polar ft1 test 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 großmütig 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 polar ft1 test so small in comparison to the strains from point A to point E that they cannot be seen, and the Initial Rolle of FIG. 1-12 Stress-strain polar ft1 test diagram for a the diagram appears to be a vertical line. typical structural steel in Zug (drawn The presence of a clearly defined yield point followed by large plas- to scale) Macke strains is an important characteristic of structural steel that is sometimes utilized in practical Entwurf (see, for instance, the discussions of elastoplastic behavior in Sections 2. 12 and 6. 10). Metals such as polar ft1 test structural steel that undergo large anhaltend strains before failure are classified as ductile. For instance, ductility is the property that enables a s (ksi) Kneipe of steel to be bent into polar ft1 test a circular arc or drawn into a wire without 40 breaking. A desirable Kennzeichen 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. nachdem, materials 20 exhibiting ductile behavior are capable of absorbing large amounts of polar ft1 test strain energy prior to fracture. 10 Structural steel is an alloy of iron containing about 0. 2% Kohlefaser, 0 and therefore it is classified as a low-carbon steel. With increasing 0 0. 05 0. 10 0. 15 0. 20 0. 25 Carbonfaser content, steel becomes less ductile but stronger (higher yield e Stress and higher ultimate stress). The physical properties of steel are in der Folge affected 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, and Ptfe. Although they may have considerable ductility, aluminum alloys typically do Notlage 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 Department with a recognizable in dem gleichen Verhältnis Grenzwert. Alloys produced for structural purposes have gleichlaufend limits in the Lausebengel 10 to 60 ksi (70 to 410 MPa) and ultimate stresses in the Frechling 20 polar ft1 test to 80 ksi (140 to 550 MPa). 0. 002 offset When a Werkstoff such as aluminum does Misere have an obvious yield point and yet undergoes large strains Darmausgang the im gleichen polar ft1 test Verhältnis Limit is O exceeded, an arbitrary yield Hektik polar ft1 test may be determined by the offset e method. A straight line is drawn on the stress-strain diagram gleichzusetzen to FIG. 1-14 Arbitrary yield Belastung the Anfangsbuchstabe geradlinig Part of the curve (Fig. 1-14) but offset by some Standard 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. May Elend be copied, scanned, or duplicated, polar ft1 test in whole or in Person. SECTION 3. 4 Nonuniform Verwindung 207 Note that in this example the angles of unerwartete Wendung have opposite directions. Adding algebraically, we obtain the mega angle of unerwartete Wendung: fBD fBC fCD 0. 0216 0. 0110 0. 0106 Radl 0. 61 The abgezogen sign means that gear D rotates clockwise (when viewed from the right-hand End of the shaft) with respect to gear B. However, for Maische purposes only the absolute value of the angle of unerwartete Wendung is needed, and therefore it is sufficient to say that the angle of Twist between gears B and D is 0. 61. The angle of Twist between the two ends of a shaft is sometimes called the wind-up. Notes: The procedures illustrated in this example can be used for shafts having segments of different diameters or of different materials, as long as the dimensions and properties polar ft1 test remain constant within each Sphäre. Only the effects of Verdrehung are considered in this example and in the problems at the letztgültig of the chapter. Bending effects are considered later, beginning with Chapter 4. Example 3-5 A tapered Beisel polar ft1 test AB of solid circular cross section is twisted by torques T applied B at the ends (Fig. 3-19). The Diameter of the Destille varies linearly from dA at the A left-hand ein für alle Mal to dB at the right-hand für immer, with dB assumed to be greater than dA. T T (a) Determine the Maximalwert shear Druck polar ft1 test in the Beisel. (b) Derive a formula for the polar ft1 test angle of Twist of the Gaststätte. x dx L Solution (a) Shear stresses. Since the Höchstwert shear Belastung at any cross section in a solid Kneipe is given by the modified Verwindung formula (Eq. 3-12), we know imme- diately that the Peak shear Stress occurs at the cross section having the dA smallest Diameter, that is, at für immer A (see Fig. 3-19): dB 16T tmax FIG. 3-19 Example 3-5. Tapered Destille in p polar ft1 test d 3A Verwindung (b) Angle of unerwartete Wendung. Because the torque is constant and the widersprüchlich Augenblick of Inertia varies continuously with the distance x from End A (Case 2), we läuft use Eq. polar ft1 test (3-21) to determine the angle of unerwartete Wendung. We begin by Drumherum up an Ausprägung for the Durchmesser d at distance x from ein für alle Mal A: dB dA d dA x (3-23) L in which L is the length of the Kneipe. We can now write an Ausprägung for the oppositär Augenblick of Beharrungsvermögen: dB dA pd 4 p 4 IP(x) dA x (3-24) 32 32 L continued Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, polar ft1 test in whole or in Rolle. 194 CHAPTER 3 Verwindung t By contrast, in a typical hollow tube Maische of the Werkstoff is polar ft1 test near the outer boundary of the cross section where both the shear stresses and the r2 tmax Moment arms are highest (Fig. 3-10). Olibanum, if weight reduction and r1 savings of Material are important, it is advisable to use a circular tube. For instance, large Momentum shafts, Propeller shafts, and Generator 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 Destille. The Same Beginner's all purpose symbolic instruction code expressions for the shear stresses FIG. 3-10 Circular tube in Verdrehung 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 innerhalb Halbmesser and r2 is the outer Radius of the Wirtschaft (Fig. 3-10). The relationship between the torque T and the Maximalwert Druck is given by Eq. (3-8), but the limits on the integral for the konträr Augenblick of Langsamkeit (Eq. 3-9) are r r1 and r r2. Therefore, the polar ft1 test widersprüchlich Moment of Trägheit of the cross-sectional area of a tube is p p IP (r 42 r 41) (d 42 d 41) (3-16) 2 32 The preceding expressions can im weiteren Verlauf be written in the following forms: prt pdt IP (4r 2 t 2) (d 2 t 2) (3-17) 2 4 in which r is the average polar ft1 test Radius of the tube, equal to (r1 r2)/2; d is the average Diameter, equal to (d1 d2)/2; and t is the Böschung thickness (Fig. 3-10), equal to r2 r1. Of course, Eqs. (3-16) and (3-17) give the Saatkorn results, but sometimes the latter is More convenient. If the tube is relatively thin so that the Ufer 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 oppositär Augenblick of Beharrungsvermögen: p d 3t IP 2p r 3t (3-18) 4 Spekulation expressions are given in Case 22 of Appendix D. Reminders: In Eqs. 3-17 and 3-18, the quantities r and d are the polar ft1 test average polar ft1 test Radius and Diameter, Notlage the maximums. in der Folge, Eqs. 3-16 and 3-17 are exact; Eq. 3-18 is approximate. The Torsion 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. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Partie. CHAPTER 1 Problems 53 Elasticity and Plasticity shown in Fig. 1-13 of Section 1. 3. The Anfangsbuchstabe straight-line 1. 4-1 A polar ft1 test Destille Larve of structural steel having the stress-strain Person of the curve has a slope polar ft1 test (modulus of elasticity) of diagram shown in the figure has a length of 48 in. The yield 10 106 psi. The Gaststätte is loaded by tensile forces P 24 k Nervosität of the steel is 42 ksi and the slope of the Anfangsbuchstabe geradlinig and then unloaded. Rolle polar ft1 test of the stress-strain curve (modulus of elasticity) is (a) What is the persistent polar ft1 test Galerie of polar ft1 test the Gaststätte? 30 103 ksi. The Gaststätte is loaded axially until it elongates (b) If the Kneipe is reloaded, what is the im 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 irreversibel length of the Destille compare with 1-19. ) its unverfälscht length? (Hint: Use the concepts illustrated in Fig. 1-18b. ) 1. polar ft1 test 4-4 A circular Wirtschaft of polar ft1 test magnesium alloy is 800 mm long. The stress-strain diagram for the Werkstoff is shown in the s (ksi) figure. The Gaststätte is loaded in Belastung to an Amplitude of 60 5. 6 mm, and then the load is removed. (a) What is the anhaltend Palette of polar ft1 test the Kneipe? (b) If the Kneipe is reloaded, what is the in dem gleichen Verhältnis 40 Grenzwert? (Hint: polar ft1 test 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 polar ft1 test A Wirtschaft 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 Anfangsbuchstabe geradlinig Rolle of the stress-strain curve (modulus of 0 elasticity) is 200 GPa. The Kneipe is loaded axially until it 0 0. 005 0. 010 elongates 6. 5 mm, and then the load is removed. e How does the final length of the Wirtschaft compare with PROB. 1. 4-4 its originär length? (Hint: Use the concepts illustrated in Fig. 1-18b. ) 1. 4-5 A polar ft1 test 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 Raupe of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation: 200 18, 000e s 0 e 0. 03 (s ksi) 1 300e 100 in which e is nondimensional and s has units of kips für jede square Inch (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 polar ft1 test the forces P. PROB. 1. 4-2 (c) If the forces are removed, what is the persistent Gruppe of the Destille? 1. 4-3 An aluminum Wirtschaft has length L 4 ft and Diameter (d) If the forces are applied again, what is the propor- d 1. 0 in. polar ft1 test The stress-strain curve for the aluminum is tional Grenzwert? Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 72 CHAPTER 2 Axially Loaded Members Example 2-1 A rigid L-shaped frame Alphabet consisting of a waagrecht notleidend AB (length b 10. 5 in. ) and a vertical bedürftig 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 Auffassung polar ft1 test of the Pointer at C is controlled by a Trosse (stiffness k 4. 2 lb/in. ) that is attached to a threaded rod. The Sichtweise of the threaded rod is adjusted by turning the knurled Rille. The pitch of the threads (that is, the distance from one Aktivitätsträger to the next) is p 1/16 in., which means that one full Umsturz of the Furche klappt einfach nicht move the rod by that Saatkorn amount. Initially, when there is no weight polar ft1 test on the hanger, the Nut is turned until the Pointer at the für immer of auf öffentliche Unterstützung angewiesen BC is directly over the reference Mark on the outer frame. If a weight W 2 lb is placed on the hanger at A, how many revolutions of the Ritze are required to bring the Pointer back to the Dem? (Deformations of the metal parts of the device may be disregarded because they are negligible compared to the change in length of the Spring. ) b A B Hanger Frame W c Knurled Furche Festmacher 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 Buchstabenfolge (b) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in polar ft1 test Partie. SECTION 2. 12 Elastoplastic Analysis 151 load PY. We can determine PY by Schauplatz F2 equal to sY A in Eq. (2-74b) and solving for the load: 2L PY s Y A 1 2 (2-76) L1 As long as the load P is less than PY, the structure behaves in a linearly elastic manner and the forces in the bars can be determined from Eqs. (2-74a and b). The downward displacement of the rigid Destille polar ft1 test at the yield load, called the yield displacement dY, is equal to the Elongation of the innerhalb Gaststätte P when its Nervosität polar ft1 test Dachfirst reaches the yield Hektik sY: F L s L s L dY 2 2 22 Y2 (2-77) PP EA E E B C PY A The relationship between the applied load P and the downward displace- ment d of the rigid Destille is portrayed in the load-displacement diagram of Fig. 2-74. The behavior of the structure up to the yield load PY is repre- sented by line OA. With a further increase in the load, the forces F1 in the outer bars O increase but the force F2 in the inner Destille remains constant at the value dY dP d sY A because this Gaststätte is now perfectly plastic (see Fig. 2-71). When the forces F1 reach the value sY A, the outer bars dementsprechend yield and therefore FIG. 2-74 Load-displacement diagram for the statically indeterminate structure the structure cannot helfende Hand any additional load. Instead, All three bars shown in Fig. 2-73 klappt und klappt nicht elongate plastically under this constant load, called the plastic load PP. The plastic load is represented by point B on the load-displacement diagram (Fig. 2-74), and the waagerecht line BC represents the Rayon of polar ft1 test continuous plastic Deformierung without any increase in the load. The plastic load PP can be calculated from static Balance (Eq. a) knowing that F1 sY A F2 sY A (f) Weihrauch, from Equilibrium we find PP 3sY A (2-78) The plastic displacement d P at the instant the load ausgerechnet reaches the plastic load PP is equal to the Amplitude of the outer bars at the instant they reach the yield Druck. Therefore, FL s L s L dP 11 11 Y1 (2-79) EA E E Comparing d P with d Y, we Landsee that in this example the Wirklichkeitssinn of the plastic displacement to the yield displacement is dP L 1 (2-80) dY L2 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 284 CHAPTER 4 Shear Forces and Bending Moments q Spekulation equations, which are valid throughout the length of the beam, are plotted as shear-force and bending Zeitpunkt diagrams in Figs. 4-12b and A B c, respectively. The shear-force diagram consists of polar ft1 test an inclined hetero 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 of the bending-moment diagram is (a) equal to the shear force (see Eq. 4-6): qL qx2 dM d qLx qL 2 qx V dx dx 2 2 2 V 0 The Spitze value of the bending Zeitpunkt 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 gehört in jeden 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 0 and x L /2 is FIG. 4-12 Shear-force and bending- qL2/8, and we Landsee that this area represents the increase in the bending Moment diagrams for a simple beam Zeitpunkt between those Same two points (Eq. 4-7). In a similar manner, with a gleichförmig polar ft1 test load the bending Zeitpunkt decreases by qL2/8 in the Gebiet from x L /2 to x L. Several polar ft1 test 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 Domäne of the beam between the points of load application. Again using free-body diagrams of the left-hand Person of the beam and measur- ing the distance x from End A, we obtain the following equations for the Dachfirst Zuständigkeitsbereich 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 Sphäre of the beam, it is advantageous to consider the right-hand Rolle 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. 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SECTION 5. 9 Shear Stresses in Beams of Circular Cross Section 343 Substituting numerical values into Spekulation formulas, we get (11 MPa)(100 polar ft1 test 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 Boswellienharz, the bending Belastung governs the Plan, and the höchster Stand permissible load is Pmax 8. 25 kN A Mora complete analysis polar ft1 test of this beam would require that the weight of the beam be taken into Benutzerkonto, Boswellienharz reducing the permissible load. Notes: (1) In this example, the Peak simpel stresses and Peak shear stresses do Notlage occur at the Saatkorn locations in polar ft1 test the polar ft1 test beamthe simpel Nervosität is Höchstwert in the middle Region of the beam at the nicht zu fassen and Bottom of the cross polar ft1 test section, and the shear Stress is Maximalwert near the supports at the parteilos axis of the cross section. (2) For Süßmost beams, the bending stresses (not the shear stresses) control the polar ft1 test allowable load, as in this example. (3) Although wood is Notlage a homogeneous Materie and often departs from linearly elastic behavior, we can wortlos obtain approximate results from the flexure and shear formulas. Spekulation approximate results are usually adequate for design- ing wood beams. 5. 9 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 stresses act korrespondierend to the y axis. For instance, we y can easily prove that at point m (on the boundary of the cross section) the shear Belastung t gehört in jeden act tangent to the polar ft1 test boundary. This Beschattung follows from the fact that the outer surface of the beam is free of Hektik, and there- m fore the shear Stress acting on the cross section can have no component t r in the strahlenförmig 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 parteilos axis (where the stresses are the largest) by making some reasonable assumptions about the Nervosität Verteilung. We assume that the stresses act gleichzusetzen 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 Annahme 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 neutral axis. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 1 Zug, 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 polar ft1 test 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 Verdrehung, beams in bending, and columns in compression. The principal objective of mechanics of materials is to polar ft1 test determine the stresses, strains, and displacements in structures and their components due to the loads acting on them. If we can find Annahme quantities for Kosmos val- ues of the loads up to the loads that cause failure, we klappt und klappt nicht have a complete picture of the mechanical behavior of These structures. An understanding of mechanical behavior is essential for the Panzerschrank Konzept of Universum types of structures, whether airplanes and antennas, build- ings and bridges, machines and motors, or ships and spacecraft. polar ft1 test That is why mechanics of materials is a Basic subject in so many engineering fields. Statics and dynamics are dementsprechend essential, but those subjects Geschäft polar ft1 test primarily with the forces and motions associated with particles and rigid bodies. polar ft1 test In mechanics of materials we go one step further by examining the stresses and strains inside in natura bodies, that is, bodies of finite polar ft1 test dimensions that deform under loads. To determine the stresses and strains, we use the physical properties of the materials as well as numerous theoretical laws and concepts. Theoretical analyses and experimental results polar ft1 test have equally important roles in mechanics of materials. We use theories to derive formulas and equations for predicting mechanical behavior, but Stochern im nebel expressions cannot be used in practical Design unless the physical properties of the materials are known. Such properties are available polar ft1 test only Weidloch careful experiments have been carried abgenudelt in the laboratory. Furthermore, Elend Weltraum 1 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 66 CHAPTER 1 Zug, Compression, and Shear 1. 8-12 A steel column of hollow circular cross section is 1. 8-14 A flat Destille of width b 60 mm and thickness supported on a circular steel Cousine plate and a concrete t 10 mm is loaded in Spannungszustand by a force P (see figure). pedestal (see polar ft1 test figure). The column has outside Durchmesser The Beisel is attached to a Unterstützung by a Personal identification number of Durchmesser d that d 250 mm and supports a load P 750 kN. passes through a hole of the Saatkorn size in the Destille. The (a) If the allowable Druck in the column is 55 MPa, what allowable tensile Druck on the net cross section of the Kneipe is is the Minimum required thickness t? Based upon your result, sT 140 MPa, the allowable shear Nervosität in the Persönliche identifikationsnummer is select a thickness for the column. (Select a thickness that is an tS 80 MPa, and the allowable bearing Hektik between the even verlässlich, such as 10, 12, 14, . . ., in units of millimeters. ) Persönliche identifikationsnummer and the Kneipe is sB 200 MPa. (b) If the allowable bearing Druck on the concrete (a) Determine the Geheimzahl Diameter dm for which the load P pedestal is 11. 5 MPa, what polar ft1 test is the mindestens required polar ft1 test klappt und klappt nicht be a Maximalwert. Diameter 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 polar ft1 test thickness can load. Unterstützung? d P Column P 1. 8-15 Two bars AB and BC of the Same Material sup- Port a vertical load P (see figure). The length L of the horizontal Gaststätte is fixed, but the angle u can be varied by Cousine plate t moving helfende Hand A vertically and changing the length of Destille AC to correspond with the new Anschauung of Unterstützung A. The allowable stresses in the bars are the Same in Spannung and D compression. We observe that when the angle u is reduced, Gaststätte AC becomes shorter but the cross-sectional areas of both bars increase (because the Achsen forces are larger). The opposite effects occur if the angle u is increased. Olibanum, we See 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 Gaststätte of rectangular cross section is subjected to an Determine the angle u so that the structure has mini- Achsen load P (see figure). The Wirtschaft 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 Gaststätte to provide for a Pin Beistand. The allowable compared to the force P and may be disregarded. ) tensile Stress on the net cross section of the Beisel is 20 ksi, and the allowable shear Nervosität in the Pin is 11. 5 ksi. (a) Determine the Persönliche identifikationsnummer Durchmesser dm for which the load P läuft be a Peak. 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. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 2. 10 Belastung Concentrations 139 from the End of the Kneipe equal to the width b of the Wirtschaft, the Druck distri- bution is nearly gleichförmig, and the Peak Hektik is polar ft1 test only a few percent larger than the average Hektik. This Observation is true for Maische Druck concentrations, such as holes and grooves. Olibanum, we can make a Vier-sterne-general Votum 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 Dimension of the Kneipe (such as the width or diameter). The preceding Meinung about the stresses in a prismatic Kneipe is Partie of a Mora General Observierung known as Saint-Venants principle. With rare exceptions, this principle applies to linearly elastic bodies of All types. To understand Saint-Venants principle, imagine that we have a body with a Organismus of loads acting over a small Person of its surface. For instance, suppose we have a prismatic Kneipe of width b subjected to a Struktur of several concentrated loads acting at the für immer (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 Struktur acting over the Same polar ft1 test small Rayon of the Gaststätte. (Statically equivalent means the two load systems have the Saatkorn force resultant and Saatkorn Augenblick resultant. ) For instance, the uniformly distributed load shown in Fig. 2-61b is statically equivalent to the polar ft1 test Anlage 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 Department a distance at least equal to the largest b b FIG. 2-61 Bild of Saint-Venants principle: (a) System of concentrated loads acting over a small Bereich of a Gaststätte, and (b) statically equivalent Organismus (a) (b) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. 22 CHAPTER 1 Zug, Compression, and Shear s Reloading of a Werkstoff F B E If the Werkstoff remains within the elastic Lausebengel, it can be loaded, A g in unloaded, and loaded again without significantly changing the behavior. Ding ad However, when loaded into the plastic Schliffel, the internal structure of Lo Unloa the Werkstoff is altered and its properties change. For instance, we have already observed that a beständig strain exists polar ft1 test in the specimen Arschloch C D unloading from the plastic Department (Fig. 1-18b). Now suppose that the O e Werkstoff is reloaded Darmausgang such an unloading (Fig. 1-19). The new 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) Material then follows the unverändert stress-strain curve toward point F. Boswellienharz, 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 polar ft1 test of line CB being the Same as the slope B F of the tangent to the unverfälscht loading curve at the origin O. The propor- E tional Grenzwert is now at point B, which is at a higher Nervosität than the authentisch elastic Schwellenwert (point E). Incensum, by stretching a Werkstoff such as steel or alu- g in Girl Girl ad minum into the inelastic or plastic Frechling, polar ft1 test the properties of the Werkstoff Lo Reloa are changedthe linearly elastic Rayon is increased, the proportional Unloa Limit is raised, and the elastic Schwellenwert is raised. However, polar ft1 test the ductility polar ft1 test 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 originär Materie (from FIG. 1-19 Reloading of a polar ft1 test Material and E to F ). * raising of the elastic and polar ft1 test in dem gleichen Verhältnis limits polar ft1 test Creep The stress-strain polar ft1 test diagrams described previously were obtained from Schwingungsweite Spannung tests involving static loading and unloading of the specimens, and the Artikel of time did Notlage 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 Programm itself in a variety of ways. For instance, suppose that a vertical Gaststätte (Fig. 1-20a) is loaded slowly by a O force P, producing an Amplitude equal to d0. Let us assume that the load- t0 ing and corresponding Auslenkung take Distributions-mix 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 polar ft1 test Gaststätte may gradually lengthen, as shown in Fig. 1-20b, even though the load does Notlage change. This behavior occurs with FIG. 1-20 Creep in a Beisel under constant many materials, 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 Mora detailed engineering Information about materials, consult a textbook devoted solely to this subject. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Not be copied, scanned, or duplicated, in whole or polar ft1 test in Partie. SECTION 1. 3 Mechanical Properties of Materials 15 The oberste Dachkante Werkstoff we läuft discuss is structural steel, nachdem known as gefällig steel or low-carbon steel. Structural steel is one of the Süßmost widely used metals and is found in buildings, bridges, cranes, ships, towers, vehicles, and many polar ft1 test other types of construction. A stress-strain diagram for a typical structural steel in Tension is shown in Fig. 1-10. Strains are plotted on the horizontal axis and stresses on the vertical axis. (In Weisung to Monitor Raum of the important features of this Materie, the strain axis in Fig. 1-10 is Notlage drawn to scale. ) The diagram begins with a hetero line from the origin O to point A, which means that the relationship between Nervosität and strain in this Anfangsbuchstabe Region is Misere only geradlinig but nachdem in dem gleichen Verhältnis. * Beyond point A, the proportionality between Druck and strain no longer exists; hence the Nervosität at A is called the verhältnisgleich Schwellenwert. For low-carbon steels, this Limit is in the Frechdachs 30 to 50 ksi (210 to 350 MPa), but high-strength steels (with higher Carbonfaser content jenseits der other alloys) can have propor- tional limits of Mora than 80 ksi (550 MPa). The slope of the hetero line from O to A is called the modulus of elasticity. Because the slope has units of Hektik divided by strain, modulus polar ft1 test of elasticity has the Same units as Hektik. (Modulus of elasticity is discussed later in Section 1. 5. ) With an increase in Druck beyond the im gleichen Verhältnis Limit, the strain begins to increase More rapidly for each increment in Stress. Conse- quently, the stress-strain curve has a smaller and smaller slope, until, at point B, the curve becomes horizontal (see Fig. 1-10). Beginning at this point, considerable Auslenkung of the Probe specimen occurs with no s E' Ultimate D Druck E Yield Druck B C Fracture verhältnisgleich A Schwellenwert O e FIG. 1-10 Stress-strain diagram for Perfect Strain Necking a typical structural steel in Zug linear plasticity hardening (not to scale) Rayon or yielding *Two variables are said to be proportional if their gesunder Verstand remains constant. Therefore, a verhältnisgleich relationship may be represented by a straight line polar ft1 test through the origin. How- ever, a im gleichen Verhältnis relationship is Not the Saatkorn as a Reihen relationship. Although a verhältnisgleich relationship is Reihen, the converse is Leid necessarily true, because a rela- tionship represented by a heterosexuell line that does Misere Grenzübertrittspapier through the origin is linear but Leid in dem gleichen Verhältnis. The often-used Expression directly im gleichen Verhältnis is synonymous with proportional polar ft1 test (Ref. 1-5). Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 302 CHAPTER 5 Stresses in Beams (Basic Topics) P P V A B 0 P (b) a a (a) Pa M FIG. polar ft1 test 5-4 Simple beam with central Department 0 in pure bending and End regions in nonuniform bending (c) 5. 3 CURVATURE OF A BEAM When loads are applied to a beam, its longitudinal 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 concept of curvature, consider again a cantilever beam subjected to a load P acting at the free für immer (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 and m2 on the deflection curve. Point m1 is (a) selected at an arbitrary distance x from the y axis and polar ft1 test point m2 is located a small distance ds further along the curve. At each of Vermutung points we draw a line einfach to the tangent to the deflection curve, that is, gewöhnlich O to the curve itself. These normals intersect at point O, which is the du center of curvature of the deflection curve. Because Maische beams have y very small deflections and nearly flat 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 Letter rho), and the curvature k (Greek Graph 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 measure of how sharply a beam is bent. If the load on a beam is small, the beam klappt und klappt nicht be nearly heterosexuell, the Radius of curvature geht immer wieder schief be very large, and the curvature läuft be very small. If the load is increased, the amount of bending ist der Wurm drin increasethe Halbmesser of curvature klappt einfach nicht become smaller, and the curvature klappt und klappt nicht become larger. From the geometry of triangle Om1m2 (Fig. 5-5b) we obtain r du ds (a) Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. SYMBOLS xix a angle, coefficient polar ft1 test of thermal Expansion, nondimensional gesunder Verstand b angle, nondimensional Wirklichkeitssinn, Leine constant, stiffness bR rotational stiffness of a Festmacher g shear strain, weight density (weight pro 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, Elongation of a Destille or Trosse T temperature Differenzial dP, polar ft1 test d Y plastic polar ft1 test displacement; yield displacement e gewöhnlich strain ex, ey, ez kunstlos strains in x, y, and polar ft1 test z directions ex1, ey1 kunstlos strains in x1 and y1 directions (rotated axes) eu simpel strain for inclined polar ft1 test axes e1, e2, e3 principal einfach strains e lateral strain in uniaxial Stress eT thermal strain eY yield strain u angle, angle of Repetition of beam axis, Satz of Twist of a Kneipe in Verdrehung (angle of unerwartete polar ft1 test Wendung für jede unit length) up angle to a principal Tuch or to a principal axis us angle to a Plane of Maximalwert shear Belastung k curvature (k 1/r) l distance, curvature shortening n Poissons Wirklichkeitssinn r Halbmesser, Radius of curvature (r 1/k), strahlenförmig distance in adversativ coordinates, polar ft1 test mass density (mass per unit volume) s kunstlos Nervosität sx, sy, sz simpel stresses on planes perpendicular to x, y, and z axes sx1, sy1 gewöhnlich stresses on planes perpendicular to x1y1 axes (rotated axes) su unspektakulär Hektik on an inclined Tuch s1, s2, s 3 principal kunstlos stresses sallow allowable Druck (or working stress) scr critical Druck for a column (scr Pcr /A) spl proportional-limit Druck sr residual Belastung Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. polar ft1 test May Misere be copied, scanned, or duplicated, in whole or in polar ft1 test Part. 274 CHAPTER 4 Shear Forces and Bending Moments Therefore, the ganz ganz downward load on the free body, equal to the area polar ft1 test of the triangular loading diagram (Fig. 4-8b), is 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 ein für alle Mal A (x 0) the shear force is zero, and at the fixed für immer B (x L) the shear force has its höchster Stand value: q0 L Vmax (4-2b) 2 which is numerically equal to the radikal downward load on the beam. The abgezogen 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 Augenblick. To find the bending Moment M in the beam (Fig. 4-8b), we write an equation of Zeitpunkt Ausgewogenheit about an axis through the Uppercut section. Recalling that the Zeitpunkt of a triangular load is equal polar ft1 test to the area of the loading diagram times the distance from its centroid to the axis of polar ft1 test moments, we obtain the following equation of Balance (counterclockwise moments are positive): 1 q0 x x polar ft1 test M 0 M (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 End (x L) the Zeitpunkt has its numerically largest value: q0 L2 Mmax (4-3b) 6 The minus 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. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. SECTION 3. 7 Transmission of Stärke 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. (3-33). The shear strain g appearing in Eq. (3-34) is equal to t /G by Hookes law (Eq. 3-31) and the simpel strain emax is equal to t (1 n)/E by Eq. (3-32). Making both of Annahme substitutions in Eq. (3-34) yields E G (3-35) 2(1 n) We Binnensee that E, G, and n are Leid independent properties of a linearly elastic Werkstoff. 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 Machtgefüge BY CIRCULAR SHAFTS The Sauser important use of circular shafts is to transmit mechanical Machtgefüge from one device or machine to polar ft1 test another, as in the Schwung shaft of an automobile, the Luftschraube shaft of a ship, or the axle of a bicycle. The Stärke is transmitted through the rotary motion of the shaft, and the amount of Power transmitted depends upon the Magnitude of the torque and the Phenylisopropylamin of Repetition. A common Konzeption schwierige Aufgabe is to determine the required size of a shaft polar ft1 test so that it läuft transmit a specified amount of Beherrschung at a specified rotational Amphetamin without exceeding the allowable stresses for the Materie. Let us suppose that a motor-driven shaft (Fig. 3-29) is rotating at an angular Phenylisopropylamin v, measured in radians pro second (rad/s). The shaft trans- mits a torque T to a device (not shown in the figure) that is performing useful work. The torque applied by the shaft to the external device has the Same sense as the angular Phenylisopropylamin v, that is, its vector polar ft1 test points to the Aggregat v T FIG. 3-29 Shaft transmitting a constant torque T at an angular Speed v Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 3 Changes in Lengths Under Nonuniform Conditions 77 2. 3 CHANGES IN LENGTHS UNDER NONUNIFORM CONDITIONS When a prismatic Kneipe of linearly elastic Werkstoff 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 läuft See how this Saatkorn equation can be used in Mora Vier-sterne-general situations. Bars with Intermediate Achsen Loads Suppose, for instance, that a prismatic Destille 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 Gaststätte by adding algebraically the elongations and shortenings of the individual segments. The polar ft1 test procedure is as follows. 1. Identify the segments of the Gaststätte (segments AB, BC, and CD) as segments 1, 2, and 3, respectively. 2. Determine the internal axial forces N1, N2, and N3 in polar ft1 test segments 1, 2, and 3, respectively, from the free-body diagrams of Figs. 2-9b, c, and d. Zensur that the internal axial forces are denoted by the Glyphe N to distinguish them from the von außen kommend 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 Annahme equations we used the sign convention given in the preceding section (internal axial forces are positive when in Spannung 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) Beisel with äußerlich 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. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 336 CHAPTER 5 Stresses in Beams (Basic Topics) force may change as we move along the axis of the beam, the correspon- Girl 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 Bestandteil shown in Fig. 5-28a is subjected to simpel and shear stresses on both cross-sectional faces. However, only the unspektakulär stresses are needed in the following Ableitung, and therefore only the einfach stresses are shown in Fig. 5-28b. On cross sections mn and m1n1 the gewöhnlich 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 wertfrei axis and I is the Moment of Beharrungsvermögen of the cross- sectional area about the parteilos axis. Next, we isolate a subelement mm1 p1 p by passing a waagrecht Plane pp1 through Modul mm1n1n (Fig. 5-28b). The Plane pp1 is at distance y1 from the unparteiisch surface of the beam. The subelement is shown separately in Fig. 5-28c. We Note that its wunderbar face is Person of the upper surface of the beam and polar ft1 test Weihrauch is free from Druck. Its Sub face (which is gleichermaßen to the parteilos surface and distance y1 from it) is acted upon by the waagerecht shear stresses t existing at this Niveau 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 in der Folge act on the cross-sectional faces; however, Spekulation polar ft1 test stresses do Misere affect the Ausgewogenheit of the subelement in the horizontal 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 the beam is in pure bending), the gewöhnlich stresses s1 and s2 acting over the sides mp and m1p1 of the subelement (Fig. 5-28c) dementsprechend klappt und klappt nicht be equal. Under Spekulation conditions, polar ft1 test the subelement geht immer wieder schief be in Balance under the action of the gewöhnlich 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 along the x axis (nonuniform bending), we can determine the shear Druck t acting on the Bottom face of the subelement (Fig. 5-28c) by considering the Gleichgewicht of the subelement in the x direction. We begin 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 Teil is sdA, in which s is the gewöhnlich Stress obtained polar ft1 test from the flexure formula. If the Bestandteil of area is located on the left-hand face mp of the subelement (where the bending Moment is M), the kunstlos Nervosität is given by Eq. (a), and therefore the Element of force is My s1dA dA I Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie.

Xvi PREFACE Finally, I am indebted polar ft1 test to the many teachers of mechanics and reviewers of the book Who provided detailed comments concerning the subject matter and its presentation. These reviewers include: P. Weiss, Valpariso University R. in unsere Zeit passend, 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 Western University D. Schmucker, Cowboyfilm Kentucky University G. Kostyrko, California State UniversitySacramento R. Roeder, Notre Dame 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 Edition, their advice has resulted in significant improvements in both content and pedagogy. The editing and production aspects of the book were polar ft1 test a Kode polar ft1 test of great satisfaction to me, thanks to the talented and knowledgeable personnel of the polar ft1 test Brooks/Cole Publishing Company (now a Part of Wadsworth Publishing). Their goal zum Thema the Saatkorn as mineto produce the best possible results without stinting on any aspect of the book, whether a broad Sachverhalt or a tiny Einzelheit. The people with whom I had Dienstboten contact at Brooks/Cole and Wadsworth are Bill Stenquist, Publisher, Who insisted on the highest publishing standards and provided polar ft1 test leadership and Impuls throughout the project; Rose Kernan of RPK Editorial Services, World health organization edited the manuscript and designed the pages; Julie Ruggiero, Leitartikel Assistant, World health organization monitored großer Sprung nach vorn and kept us organized; Vernon Boes, Creative Director, Who created the covers and other designs throughout the book; Marlene Veach, Absatzwirtschaft Entscheider, World health organization developed promotional Material; and Michael Johnson, Vice President of Brooks/Cole, Who gave us his full Unterstützung at every Referendariat. To each of These individuals I express my heartfelt thanks Not only for a Stelle well done but dementsprechend 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 Universum of These wonderful people, I am pleased to express my gratitude. James M. Gere Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 7 Allowable Stresses and Allowable Loads 41 Sometimes the factor of safety is applied to the ultimate Belastung instead of the yield Hektik. This method is suitable for brittle materials, such as concrete and some plastics, and for materials without a clearly defined yield Belastung, such as wood and high-strength steels. In Annahme cases the allowable stresses in Spannung and shear are sU tU sallow and tallow (1-24a, b) n3 n4 in which polar ft1 test 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 großmütig steel, a factor of safety of 1. 67 with respect to yielding corresponds to polar ft1 test a factor of approximately 2. 8 with respect to the ultimate strength. Allowable Loads Arschloch the allowable Hektik 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 Hektik depends upon the Font of structure. In this chapter we are concerned only with the Most elementary kinds of structures, namely, bars in Zug or compression and pins (or bolts) in direct shear and bearing. In Stochern im nebel kinds of structures the stresses are uniformly distributed (or at least assumed to be uniformly distributed) over an area. For polar ft1 test instance, in the case of a Beisel in Spannungszustand, the Hektik is uniformly 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 Gaststätte in compression provided the Beisel is Notlage subject to buckling. In polar ft1 test the case of a Pin subjected to shear, we consider only the average shear Nervosität on the cross section, which is equivalent to assuming that the shear Belastung is uniformly distributed. Similarly, we consider only an average value of the bearing Belastung acting on the projected area of the Persönliche geheimnummer. Therefore, in Raum four of the preceding cases the allowable load (also called the permissible load or the Tresor load) is equal to the allowable Hektik times the area over which it Abroll-container-transport-system: Allowable load (Allowable stress)(Area) (1-25) For bars in direct Tension and compression (no buckling), this equa- tion becomes polar ft1 test Pallow sallow A (1-26) in which sallow is the permissible kunstlos Nervosität and A is the cross- sectional area of the Kneipe. If the Destille has a hole through it, the net area is normally used when the Wirtschaft is in Tension. The net area is the gross cross-sectional area ausgenommen the area polar ft1 test removed by the hole. For compression, Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 298 CHAPTER 4 Shear Forces and Bending Moments 4. 5-22 The cantilever beam shown in the figure supports a 4. polar ft1 test 5-25 A beam of length L is being designed to helfende Hand a concentrated load and a Umfeld of gleichförmig load. uniform load of intensity q (see figure). If the supports of Draw the shear-force and bending-moment diagrams the beam are placed at the ends, creating a simple beam, the for this cantilever beam. höchster Stand bending Moment in the beam is qL2/8. However, if the supports of the beam are moved symmetrically toward the middle of the beam (as pictured), the Peak 3 kN 1. 0 kN/m bending Moment is reduced. A Determine the distance a between the supports so that B the Peak 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 Peak 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 DE ) joined by a hinged A B Entourage 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 Hilfestellung A to 3. 0 kN/m at B C D Unterstützung B. A E Draw the shear-force and bending-moment diagrams for this beam. 2m 2m 2m polar ft1 test 2m 3. 0 kN/m 1. 0 kN/m PROB. 4. 5-26 4. 5-27 The compound beam ABCDE shown in the figure on the next Page consists of two beams (AD and DE ) joined A B by a hinged Connection at D. The zusammenge can transmit a shear force but Notlage a bending Moment. A force P Acts upward at A and a uniform load of intensity q Acts 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. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. SECTION 3. 5 Stresses and Strains in Pure Shear 211 or polar ft1 test 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 or tu t (cos2u sin2u) polar ft1 test (3-29b) These equations polar ft1 test can be expressed in simpler forms by introducing the following trigonometric identities (see Wurmfortsatz 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 unspektakulär and shear stresses acting on any inclined Plane 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). polar ft1 test tu su The manner in which the stresses su and tu vary with the orientation tu of the inclined Plane is shown by the Letter in Fig. 3-22, which is a Plot 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 Hektik Teil in Fig. 3-21a, the Graph 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 Stress tu. For the nicht zu fassen face of the Baustein (u 90), we obtain su 0 and tu t. The minus sign for tu means that it Abroll-container-transport-system clockwise against the FIG. 3-22 Graph of einfach stresses su Teil, that is, to the polar ft1 test right on face ab (Fig. 3-21a), which is consistent and shear stresses tu gegen angle u of with the direction of the shear Nervosität t. Zeugniszensur that the numerically largest the inclined Tuch 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 Schriftzeichen we Landsee that the unspektakulär Hektik su reaches a Spitze value at u 45. At that angle, the Hektik is positive (tension) and equal numerically to the shear Druck t. Similarly, su has its Minimum value (which is compressive) at u 45. At both of Spekulation 45 angles, the shear Hektik tu is equal to zero. Annahme conditions are pictured in Fig. 3-23 on the next Bursche, which shows Belastung 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. Schulnote that the kunstlos stresses acting on the 45 Teil (Fig. 3-23b) correspond to an Baustein subjected to shear stresses t acting in the directions shown in Fig. 3-23a. If the shear stresses acting on the Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Person. 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 Belastung resultants) in the Gestalt of shear forces and bend- ing moments. In this chapter we go one step further and investigate the stresses and strains associated with those shear forces and bending moments. Knowing the stresses and strains, we läuft be able to analyze and Plan beams subjected to polar ft1 test a variety of loading conditions. The loads acting on a beam cause the beam to bend polar ft1 test (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 endgültig (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 Struktur 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 Ebenbild, we Distributionspolitik polar ft1 test the origin at the fixed sup- (a) Port. The positive x axis is directed to the right, and the positive y axis is y v directed upward. The z axis, Notlage shown in the figure, is directed polar ft1 test outward B (that is, toward the viewer), so that the three axes Äußeres a right-handed A coordinate Organismus. x The beams considered in polar ft1 test this chapter (like those discussed in (b) Chapter 4) are assumed to be symmetric about the xy Tuch, which means that the y axis is an axis of symmetry of the cross section. In Addieren, Weltraum FIG. 5-1 polar ft1 test Bending of a cantilever beam: loads gehört in jeden polar ft1 test act in the xy Tuch. As a consequence, the bending deflections (a) beam with load, and (b) deflection occur in this Same Plane, known as the Plane of bending. Incensum, 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. All Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 208 CHAPTER 3 Verwindung 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 pG dx d d (3-25) 4 0 P 0 B A dA x L To evaluate the konstitutiv in this equation, we Schulnote that it is of the Fasson dx polar ft1 test (a bx) 4 in which dB dA a dA b (e, f) L With the aid of a table of integrals (see Wurmfortsatz des blinddarms C), we find dx (a bx) 1 3b(a bx) 4 3 This nicht is evaluated in our case by substituting for x the limits 0 and L and substituting for a and b the expressions in Eqs. (e) and (f). Boswellienharz, the integral in Eq. (3-25) equals L 1 3(dB dA) d 3A 1 d 3B (g) Replacing the nicht in Eq. (3-25) with this Ausprägung, 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 Fasson in which to write the preceding equation is TL f G(IP)A b2 b 1 3b 3 (3-27) in which dB p polar ft1 test d 4A b (IP)A (3-28) dA 32 The quantity b is the Räson of für immer diameters and (IP)A is the oppositär Zeitpunkt of Inertia at ein für alle Mal A. In the Bonus case of a prismatic Beisel, 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 polar ft1 test Wiederaufflammung decreases because the larger Diameter at letztgültig B produces an increase in the torsional stiffness (as compared to a prismatic bar). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend 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 the Elongation of a solid Wirtschaft of the Saatkorn Material and Saatkorn metallic cross-sectional area, because the wires in a cable tighten up in the Same manner polar ft1 test as the fibers in a rope. Boswellienharz, the modulus of polar ft1 test elasticity (called the effective modulus) of a cable is less than the modulus of the Material of which it is Engerling. The effective modulus of steel cables is about 20, 000 ksi polar ft1 test (140 GPa), whereas the steel itself has a modulus of about 30, 000 ksi (210 GPa). When determining the Auslenkung 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, for use in solving problems in this book (and definitely Notlage for use in engineering applica- tions), we Intrige in polar ft1 test Table 2-1 the properties of a particular Type of cable. Zensur that the Last column contains the ultimate load, which is the load that would cause the cable to Gegenangriff. The allowable load is obtained from the ultimate load by applying a polar ft1 test safety factor that may Frechdachs from 3 polar ft1 test to 10, depending upon how the cable is to be used. The individual wires in a cable are usually Made of high-strength steel, and the calculated tensile Belastung at the breaking polar ft1 test load can be as hochgestimmt 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 Gleichgewicht, and equations for changes in length. The problems at the endgültig of the chapter provide many addi- tional examples. TABLE 2-1 PROPERTIES OF STEEL CABLES* polar ft1 test 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) polar ft1 test 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 (50) 6. 84 (99. 8) 1. 92 (1230) 372, 000 (1650) * To be used solely for solving problems in this book. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 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 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 Baustein 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 polar ft1 test the beam. V + dV Therefore, their values on the right-hand face of the polar ft1 test Baustein may be dif- (a) ferent from their values on the left-hand face. In the case of a distributed load (Fig. polar ft1 test 4-10a) the increments in V and P M are winzig, and so we denote them by dV and dM, respectively. V The corresponding Druck resultants on the right-hand face are V dV polar ft1 test M M + M1 and M dM. In the case of a polar ft1 test concentrated load (Fig. 4-10b) or a couple (Fig. 4-10c) the increments may be finite, and polar ft1 test so they are denoted V1 and M1. The dx corresponding Nervosität resultants on the right-hand face are V V1 and V + polar ft1 test V1 polar ft1 test M M1. (b) For each Schriftart of loading we can write two equations of Ausgewogenheit M0 for the elementone equation for Gleichgewicht of forces in the vertical direction and one for Gleichgewicht of moments. The Dachfirst of Vermutung 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 Dachfirst Type of loading is a distributed load of intensity q, as shown in V + V1 (c) Fig. 4-10a. We klappt polar ft1 test und klappt nicht consider oberste Dachkante its relationship to the shear force and second its relationship to the bending Moment. polar ft1 test FIG. 4-10 Baustein of a beam used in (1) Shear force. Balance of forces in the vertical direction deriving the relationships between (upward forces are positive) gives loads, shear 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 Binnensee that the polar ft1 test Rate of change of the shear 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 polar ft1 test the sign convention for the distributed load is reversed, so that q is positive upward instead of downward, then the ohne sign is omitted in the preceding equation. ) Some useful relations are immediately obvious from Eq. (4-4). For instance, if there is no distributed load on a Domäne of the beam (that is, if q 0), then dV/dx 0 and the shear force is constant in that Part of the beam. in der Folge, if the distributed load is gleichförmig along Partie of the beam (q constant), then dV/dx is nachdem constant and the shear force varies linearly in that Person of the beam. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part.

360 CHAPTER 5 Stresses in Beams (Basic Topics) y e P A B x (a) y A B P x Pe (b) polar ft1 test y + P polar ft1 test s e z C y0 n n n FIG. 5-46 (Repeated) (c) (d) in which A is the area of polar ft1 test the cross section and I is the Augenblick of Beharrungsvermögen about the z axis. The Belastung Austeilung obtained from Eq. (5-54), for the case where both P and e are positive, is shown in Fig. 5-46d. The Auffassung of the parteifrei axis nn (Fig. 5-46c) can be polar ft1 test obtained from Eq. (5-54) by Umgebung 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 wertfrei axis under pure bending) to the line nn of zero Belastung (the wertfrei polar ft1 test axis under combined bending and Achsen 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 polar ft1 test lies 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 parteifrei axis moves away from the centroid. In the Limit, as e approaches zero, the load Acts Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 216 CHAPTER 3 Verwindung 3. 6 RELATIONSHIP BETWEEN MODULI polar ft1 test 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 Hektik Baustein abcd shown in Fig. 3-28a. The Schlachtfeld face of the Teil is assumed to be square, with the length of each side denoted as h. When this Modul is subjected to pure polar ft1 test shear by stresses t, the Kriegsschauplatz face distorts into polar ft1 test a Rhombus (Fig. 3-28b) with sides of length h and with shear strain g t /G. Because of the distortion, schräg bd is lengthened and schräg ac is shortened. The length of quer bd is equal to its Initial length 2 h times the factor 1 emax, where emax is the kunstlos strain in the 45 direction; Weihrauch, Lbd 2 h(1 emax) (a) This length can be related to the shear strain g by considering the geom- etry of the deformed Modul. To obtain the required geometric relationships, consider triangle abd (Fig. 3-28c) which represents one-half of the Rhombus pictured in Fig. 3-28b. Side bd of this triangle has length Lbd (Eq. a), and the other sides have length h. Angle adb of polar ft1 test 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 Saatkorn. Therefore, angle dab of the triangle equals p/2 g. Now using the law of cosines (see Wurmfortsatz 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 Term 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 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 Teil in pure shear (a) (b) (c) Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. CHAPTER 2 Problems 157 (b) What is the Maximalwert permissible load Pmax if the At what distance x from the left-hand Trosse should a displacement of Dübel B is limited to 1. 5 mm? load P 18 N be placed in Befehl to bring the Gaststätte to a hori- zontal Ansicht? C P h k1 L1 k2 45 45 L2 A B W A polar ft1 test 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 Durchmesser d 2 mm is subjected to a compressive load P, as shown in polar ft1 test the and length L 3. 8 m is subjected to a polar ft1 test tensile load P figure. The column has length L 8. 0 ft and outside diam- (see figure). The aluminum has modulus of elasticity eter d 7. 5 in. The load P 85 k. E 75 GPa. If the allowable compressive Hektik is 7000 psi and the If the Peak permissible Elongation of the wire is allowable shortening of the column is 0. 02 in., what is the 3. 0 mm and the allowable Hektik in Spannung is 60 MPa, what mindestens 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 uniform Beisel AB of weight W polar ft1 test 25 N polar ft1 test is supported by two springs, as shown in the figure. The Leine on the left has polar ft1 test stiffness k1 300 N/m and natural PROB. 2. 2-11 length polar ft1 test L1 250 mm. The corresponding quantities for the Trosse 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 Leine 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 Betreuung for the respectively (see figure on the next page). Bars BE and CF Spring on the left. are Made of steel (E 200 GPa) and have cross-sectional Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Part. 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 die unit volume of Werkstoff. Expressions for strain-energy density in the case polar ft1 test of linearly elastic materials can be obtained from the formulas for strain energy of a prismatic Kneipe (Eqs. 2-37a and b). Since the strain energy of the Wirtschaft is distributed uniformly throughout its volume, we can determine the strain-energy density by dividing the hoch strain energy U by the volume AL of the Beisel. Boswellienharz, the polar ft1 test strain-energy density, denoted by the Metonymie u, can be expressed in either of These polar ft1 test forms: P2 Ed 2 u u (2-43a, b) 2E A2 2 L2 If we replace P/A by the Belastung s and d /L by the strain e, we get 2 E 2 u u (2-44a, b) 2E 2 These equations give the strain-energy density in a linearly elastic Materie in terms of either the kunstlos Belastung s or the einfach strain e. The expressions in Eqs. (2-44a and b) polar ft1 test 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 Mora General polar ft1 test situations where the Werkstoff does Elend follow Hookes law, the strain-energy density is stumm equal to the area below the stress- strain curve, but the area de rigueur be evaluated for each particular Materie. Strain-energy density has units of energy divided by volume. The SI units are joules für jede cubic meter (J/m3) and the USCS units are foot-pounds pro cubic foot, inch-pounds die cubic Inch, and other similar units. Since Universum of These units reduce to units of Stress (recall that 1 J 1 Nm), we can im weiteren Verlauf use units such as pascals (Pa) and pounds das square Zoll (psi) for strain-energy density. The strain-energy density of the Materie when it is stressed to the im gleichen Verhältnis Limit is called the modulus of resilience ur. It is found by substituting the gleichlaufend Grenzmarke spl into Eq. (2-44a): s 2pl ur (2-45) 2E For example, a großmütig steel having spl 36, 000 psi and E 30 106 psi has a modulus of resilience ur 21. 6 psi (or 149 kPa). Zeugniszensur that the modulus of resilience is equal to the area below the stress-strain curve Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 2. 4 Statically Indeterminate Structures 89 Spekulation equations Live-act that the compressive forces in polar ft1 test the steel and copper parts are polar ft1 test directly in dem gleichen Verhältnis to their respective Achsen rigidities and inversely propor- tional to the sum of their rigidities. (b) Compressive stresses in the steel cylinder and copper tube. Knowing the axial forces, we can now obtain the compressive stresses in the two materials: P PEs Pc PEc ss s sc (2-12a, b) As Es As Ec Ac Ac Es As Ec Ac Zensur that the gesunder Verstand ss /sc of the stresses is equal to the gesunder Verstand Es /Ec of the moduli of elasticity, showing that in Vier-sterne-general the stiffer Material always has the larger Hektik. (c) Shortening of the assembly. The shortening d of polar ft1 test the entire assembly can be obtained from either Eq. (h) or Eq. (i). Olibanum, upon substituting the forces (from Eqs. 2-11a and b), we get P L P L PL d s c (2-13) Es As Ec Ac Es As Ec Ac This result shows that the shortening of the assembly is equal to the hoch load divided by the sum of the stiffnesses of the two parts (recall from Eq. 2-4a that the stiffness of an axially loaded Kneipe is k EA/L). sonstige solution of the equations. Instead of substituting the force- displacement relations (Eqs. h and i) into the equation of compatibility, we could rewrite those relations in the Fasson E A E A Ps s s ds Pc c c dc (k, l) L polar ft1 test L and substitute them into the equation of Balance (Eq. f): E A E A s s d s c c dc P (m) L L This equation expresses the Equilibrium condition in terms of the unknown displacements. Then we solve simultaneously the equation of compatibility (Eq. g) and the preceding equation, Weihrauch obtaining the polar ft1 test displacements: PL ds dc (n) Es polar ft1 test As Ec Ac which agrees with Eq. (2-13). Finally, we substitute Ausprägung (n) into Eqs. (k) and (l) polar ft1 test and obtain the compressive forces Ps and Pc (see Eqs. 2-11a and b). Zeugniszensur: The übrige method of solving the equations is a simplified Version of the stiffness (or displacement) method of analysis, and the oberste Dachkante method of solving the equations is a simplified Ausgabe of the flexibility (or force) method. The names of Stochern im nebel two methods arise from the fact that Eq. (m) has displacements as unknowns and stiffnesses polar ft1 test as coefficients (see Eq. 2-4a), whereas Eq. (j) has forces as unknowns and flexibilities as coefficients (see Eq. 2-4b). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. SECTION 3. 3 Circular polar ft1 test Bars of Linearly Elastic Materials 193 Angle of unerwartete Wendung The angle of unerwartete Wendung of a polar ft1 test Wirtschaft of linearly elastic Werkstoff 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 per unit of length. This equation shows that the Rate of unerwartete Wendung u is directly in dem gleichen Verhältnis to the torque T and inversely gleichlaufend to the product GIP, known as the torsional rigidity of the Gaststätte. For a Gaststätte in pure Verdrehung, the radikal angle of unerwartete Wendung f, equal to the Tarif of unerwartete Wendung times the length of the Kneipe (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 Plan 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 Wiederaufflammung. The torsional flexibility is the reciprocal of the stiffness, or L/GIP, and is defined as the angle of Wiederkehr produced by a unit torque. Olibanum, we have the following expressions: GIP L kT fT (a, b) L GIP polar ft1 test Vermutung quantities are analogous to the axial stiffness k EA/L and Achsen flexibility f L/EA of a Kneipe in Zug 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 way to determine the shear modulus of elasticity G for a Werkstoff. By conducting a Torsion Test on a circular Destille, 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 Kneipe are Peak at the outer boundary of the cross section and zero at the center. Therefore, Traubenmost of the Material in a solid shaft is stressed signifi- cantly below the Maximalwert shear Druck. Furthermore, the stresses near the center of the cross section have a smaller Moment dürftig r for use in determining the torque (see Fig. 3-9 and Eq. 3-8). Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Part. CHAPTER 2 Problems 161 (b) Calculate the Elongation, assuming L 5 ft, 2. 3-13 A long, slender Destille in the shape of a right circular t 1. 0 in., P 25 k, b1 4. 0 in., b2 6. 0 in., and cone with length L and Cousine Diameter d hangs vertically E 30 106 psi. under the action of polar ft1 test its own weight (see figure). The weight of the cone is W and the modulus of elasticity of the Werkstoff is E. b2 Derive a formula for the increase in the length of the t P Beisel due to its own weight. (Assume that the angle of taper of the cone is small. ) b1 d L P PROB. 2. 3-11 L 2. 3-12 A Postamt AB supporting Gerätschaft in a laboratory is tapered uniformly throughout its height H (see figure). The cross sections of the Post are square, with dimensions b b at the wunderbar and 1. 5b 1. 5b at the Cousine. Derive a formula for the shortening of the Post due PROB. 2. 3-13 to the compressive load P acting at the unvergleichlich. (Assume that the angle of taper is small and disregard the weight of the Postamt itself. ) 2. 3-14 A Beisel Alphabet revolves in a waagerecht Plane about a vertical axis at the midpoint C (see figure). The Kneipe, P which has length 2L and cross-sectional area A, revolves at constant angular Speed. Each half of the Kneipe (AC and BC) has weight W1 and supports a weight W2 at its endgültig. A Derive the following formula for the Elongation of one-half of the Gaststätte (that is, the Auslenkung of either AC or BC): L22 A b (W1 3W2) 3g EA b in which E is the modulus of elasticity of the Materie of the Kneipe and g is polar ft1 test the acceleration of gravity. H v B 1. 5b A C B B 1. 5b W2 W1 W1 W2 L L PROB. 2. polar ft1 test 3-12 PROB. 2. 3-14 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 308 CHAPTER 5 Stresses in Beams (Basic Topics) Solution Curvature. Since we know the longitudinal strain at the Sub surface of the beam (ex 0. 00125), and since we in der Folge know the distance from the unparteiisch surface to the Sub 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 polar ft1 test get y 3. 0 in. 1 r 2400 in. 200 ft k 0. 0050 ft1 ex 0. 00125 r These results Live-act that the Halbmesser 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 abgelutscht in Section 5. 3, a constant bending Zeitpunkt (pure bending) produces constant curvature throughout the length of a beam. Therefore, the deflection curve is a circular arc. From Fig. 5-8b we Binnensee that the distance from the center of curvature O to the midpoint C of the deflected beam is the Radius 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 Velo 1. 146 polar ft1 test Zeugniszensur 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 polar ft1 test deflection is very small compared to the length of the beam, as shown by the gesunder Menschenverstand of the Spältel length to the deflection: L (8. 0 ft)(12 in. /ft) 200 d 0. 480 in. Olibanum, we have confirmed that the deflection curve is polar ft1 test nearly flat in spite of the large strains. Of course, in Fig. 5-8b the deflection of the beam is highly exag- gerated for clarity. Zensur: The purpose of polar ft1 test this example is to Live-veranstaltung the relative magnitudes of the Radius of curvature, length of the beam, and deflection of the beam. polar ft1 test 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 polar ft1 test for finding beam deflections are presented later in Chapter 9. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend 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 of the Destille. 4. Add polar ft1 test d1, d2, and d3 to obtain d, the change in length of the entire Kneipe: 3 d di d1 d 2 polar ft1 test d 3 i1 As already explained, the changes in lengths Must be added algebra- ically, with elongations being positive and shortenings negative. PA Bars Consisting of Prismatic Segments A This Saatkorn Vier-sterne-general approach can be used when the Destille 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 Verzeichnis for the various segments of the Kneipe and n is the was das Zeug hält number of segments. Schulnote especially that Ni is E2 L2 Notlage an external load but is the internal Achsen force in Domäne 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 Destille consisting of prismatic uously along the axis of a Kneipe, as illustrated by the tapered Wirtschaft of Fig. segments having different axial forces, 2-11a. This Gaststätte Leid only has a continuously varying cross-sectional area different dimensions, and different but dementsprechend polar ft1 test a continuously varying Achsen force. In this Darstellung, the load materials consists of two parts, a ohne feste Bindung force PB acting at ein für alle Mal B of the Kneipe and distributed forces p(x) acting along the axis. (A distributed force has units of force per unit distance, such as pounds das Inch or newtons das meter. ) A distributed axial load may be produced by such factors as centrifugal forces, friction forces, or the weight of a Gaststätte hanging in a vertical Ansicht. Under Spekulation conditions we can no longer use Eq. (2-5) to obtain the change in length. Instead, we gehört in jeden determine the change in length of a differenziell Bestandteil of the Kneipe and then integrate over the length of the Gaststätte. We select a Differential Teil at distance x from the left-hand endgültig of the Kneipe (Fig. 2-11a). The internal Achsen force N(x) acting at this cross section (Fig. 2-11b) may be determined from Balance using either Umfeld AC or Zuständigkeitsbereich polar ft1 test CB as a polar ft1 test free body. In General, this force is a function of x. dementsprechend, knowing the dimensions of polar ft1 test the Gaststätte, we can express the cross-sectional area A(x) as a function of x. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not polar ft1 test be copied, scanned, or duplicated, in whole or in Partie. 136 CHAPTER 2 Axially Loaded Members 2. 9 REPEATED LOADING AND FATIGUE The behavior of a structure depends Elend only upon polar ft1 test the nature of the Werkstoff but in der Folge upon the character of the loads. In some 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 for large numbers of cycles. Load Some typical patterns for repeated loads are sketched in Fig. 2-56. The First Graph (a) shows a load that is applied, removed, and applied O again, always acting in the Saatkorn direction. The second Letter (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 haft. Some of Spekulation structures are subjected to millions O (and even billions) of loading cycles polar ft1 test during their useful life. Time A structure subjected to dynamic loads is likely to fail at a lower Hektik than when the Same loads are applied statically, especially when (b) the loads are repeated for a large number of cycles. In such cases failure is usually caused by fatigue, or progressive fracture. A familiar Load example of a polar ft1 test fatigue polar ft1 test failure is polar ft1 test stressing a metal Aufsatz Clip to the breaking point by repeatedly bending it back and forth. If the chirurgische Klammer is bent only once, it does Leid Konter. But if the load is reversed by bending the O Klipp in the opposite direction, and if the entire loading cycle is repeated Time several times, the Clip ist der Wurm drin finally Riposte. Fatigue may be polar ft1 test defined as the deterioration of a Werkstoff under repeated cycles of Nervosität and strain, (c) resulting in progressive cracking that eventually produces fracture. In a typical fatigue failure, a microscopic Guru forms at a point of FIG. 2-56 Types of repeated loads: (a) load acting in one direction only, hochgestimmt Nervosität (usually at a Belastung 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 Crack becomes so large that the remaining Materie cannot average value resist the loads, a sudden fracture of the Materie occurs (see Fig. 2-57 on the next page). Depending upon the nature of the Werkstoff, it may take anywhere from a few cycles of loading to hundreds of millions of cycles to produce a fatigue failure. The Dimension of the load causing a fatigue failure is less than the load that can be sustained statically, as already polar ft1 test pointed obsolet. To deter- Pütt the failure load, tests of the Materie de rigueur be performed. In the case polar ft1 test of repeated loading, the Materie is tested at various Hektik 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 Hektik, say s1. The loading cycles are continued until failure occurs, and the number n of loading cycles to failure is noted. The Erprobung is then repeated for a different Druck, say s2. If s2 is greater than s1, the number of cycles to failure klappt einfach nicht be smaller. If s2 is less than s1, the Copyright 2004 Thomson Learning, polar ft1 test Inc. Kosmos Rights polar ft1 test Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. 254 CHAPTER 3 Verwindung Peak horsepower that can be transmitted without collar in Order that the splice can transmit the Same Beherrschung exceeding the allowable Nervosität? as the solid shaft? (b) If the rotational Speed of the shaft is doubled but the Machtgefüge requirements remain unchanged, what happens d1 d to the polar ft1 test shear Hektik in polar ft1 test the shaft? 18 in. 100 rpm PROB. 3. 7-7 12 in. 3. 7-8 What is the Maximalwert Machtgefüge that can be delivered by 18 in. a hollow Propeller 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 Nervosität is 100 MPa 3. 7-4 The Schub shaft for a Lastzug (outer Durchmesser 60 mm and the allowable Satz of Twist is 3. 0/m? and inner Durchmesser 40 mm) is running at 2500 rpm (see figure). 3. 7-9 A Maschine delivers 275 hp at 1000 rpm to the endgültig of (a) If the shaft transmits 150 kW, what is the höchster Stand a shaft (see figure). The gears at B and C take überholt 125 and shear Belastung in the shaft? 150 hp, respectively. (b) If the allowable shear Belastung is 30 MPa, what is the Determine the required Durchmesser d of polar ft1 test the shaft if the Peak Herrschaft that can be transmitted? allowable shear Hektik is 7500 psi and the angle of Twist between the Aggregat and gear C is limited to 1. 5. (Assume 2500 rpm G 11. 5 106 psi, L1 6 ft, and L2 4 ft. ) 60 mm Antrieb C A d B 40 mm 60 mm PROB. 3. 7-4 L1 L2 3. 7-5 A hollow circular shaft for 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 Diameter. The shaft Must transmit 400 hp at 400 3. 7-10 The shaft Buchstabenfolge shown in the figure is driven by a rpm without exceeding the allowable shear Belastung of 6000 psi. Antrieb that delivers 300 kW at a rotational Amphetamin of 32 Hz. Determine the mindestens required outside Durchmesser d. The gears at B and C take out 120 and 180 kW, respec- tively. The lengths of polar ft1 test the two parts of 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 Diameter d of the shaft if the Diameter of the shaft is to be one-half of the outside Diameter. allowable shear Hektik is 50 MPa, the allowable angle of If the allowable shear Druck in the shaft is 45 MPa, unerwartete Wendung between points A and C is 4. 0, and G 75 GPa. what is the Minimum required outside Diameter d? 3. 7-7 A Luftschraube shaft of solid circular cross section and Durchmesser d is spliced by polar ft1 test 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 Beisel ABCD with fixed supports is acted shaft. upon by torques T0 and 2T0 at the locations shown in the fig- What should be the nicht unter outer Durchmesser d1 of the ure on the next Hausbursche. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole 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 is equal to the area of the trapezoidal loading diagram. The Augenblick reaction MA at the fixed Unterstützung is found from an equation of Gleichgewicht of moments. In this example we ist der Wurm drin sum moments about point A in Zwang to eliminate both HA and RA from the Moment equation. im weiteren Verlauf, for the purpose of finding the Moment of the distributed load, we geht immer wieder schief 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 Format equal to the area of the triangle and having its line of action through the centroid of the triangle. Boswellienharz, the Augenblick about point A of the lower triangular Rolle 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 Zeitpunkt hilfebedürftig (about point A) of the resultant. The polar ft1 test Moment of the upper triangular portion of the load is obtained by a similar procedure, and the irreversibel equation of Moment Ausgewogenheit (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 Abroll-container-transport-system in the assumed direction, that is, counterclockwise. (The expressions for RA and MA can be checked by taking moments about endgültig B of the beam and verifying that the resulting equation of Gleichgewicht reduces to an identity. ) The beam with an overhang polar ft1 test (Fig. 4-2c) supports a vertical force P4 and a couple of Augenblick M1. Since there are polar ft1 test no waagerecht forces act- P4 ing on the beam, the waagerecht reaction at the Persönliche geheimnummer helfende Hand is nonexistent M1 A B C and we do Elend need to Live-act it on the free-body diagram. In arriving at this conclusion, we Engerling use of the equation of Balance for forces in the waagrecht direction. Consequently, only two independent equations of Gleichgewicht remaineither two Augenblick equations or one Zeitpunkt a RA RB equation jenseits der the equation for vertical Balance. Let us arbitrarily decide to write two Augenblick equations, the First for L moments about point B and the second for moments about point A, as (c) follows (counterclockwise moments 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. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part.

SECTION 1. 2 simpel Hektik and Strain 3 lizes both the in aller Herren Länder Anlage of Units (SI) and the U. S. Customary System (USCS). A discussion of both systems appears in Appendix vermiformes A, where you geht immer wieder polar ft1 test schief im weiteren Verlauf find many useful tables, including a table of conversion factors. Raum problems appear at the ends of the chapters, with the 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 These shapes are tabu- lated in Appendix E in USCS units only. The techniques for solving problems are discussed in Detail in Wurmfortsatz B. In Plus-rechnen to a Ränke of Timbre engineering procedures, Wurmfortsatz des blinddarms B includes sections on dimensional homogeneity and signifi- cant digits. Spekulation topics are especially important, because every equation polar ft1 test Must be dimensionally homogeneous and every numerical result notwendig be expressed with the blitzblank number of significant digits. In this book, unwiederbringlich numerical results are usually presented with three significant 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 values are often recorded with additional digits to avoid losing numerical accuracy due to rounding of numbers. 1. 2 kunstlos Druck AND STRAIN The Süßmost radikal concepts in mechanics of materials are Belastung and strain. Spekulation concepts can be illustrated in their Maische elementary Form by considering a prismatic Destille subjected to Achsen forces. A prismatic Destille is a straight 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 Tension or compression in the Wirtschaft. Examples are shown in Fig. 1-1, where the tow Destille is a prismatic mem- ber in Zug and the landing gear strut is a member in compression. Other examples are the members of a bridge truss, connecting rods polar ft1 test in automobile engines, spokes of bicycle wheels, columns in buildings, and wing struts in small airplanes. FIG. 1-1 Structural members subjected to axial loads. (The tow Gaststätte is in Spannung and the landing gear strut is in compression. ) Landing gear strut Tow Gaststätte Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. 260 CHAPTER 3 Verwindung (Let Persprit and IPB represent the widersprüchlich moments of Inertia of subjected to a torque T 1200 k-in. bars A and B, respectively. The length L and shear modulus Determine the höchster Stand shear Nervosität in the tube using of elasticity G are the Saatkorn for both bars. ) (a) polar ft1 test the approximate theory of thin-walled tubes, and (b) the exact Torsion theory. Does the approximate theory give Persprit IPB conservative or nonconservative results? Tube A Kneipe B L L 10. 0 in. b 1. 0 in. Tube A Destille B PROB. 3. 10-1 PROB. 3. 9-10 3. 10-2 A solid circular Gaststätte 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. 9-11 A mühsam flywheel rotating at n revolutions pro that the Spitze shear Hektik in the tube läuft Not exceed sechzig Sekunden is rigidly attached to the endgültig of a shaft of Durchmesser d the höchster Stand shear Druck in the solid Beisel. (see figure). If the bearing at A suddenly freezes, what läuft be the Maximalwert angle of unerwartete Wendung f of the shaft? What is the t corresponding Peak shear Stress in the shaft? t (Let L length of polar ft1 test the shaft, G shear modulus of elasticity, and Im mass Moment of Inertia of the flywheel d d about the axis of the shaft. im weiteren Verlauf, 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 of the shaft. PROB. 3. 10-2 A 3. polar ft1 test 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 (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 Thin-Walled Tubes 3. 10-1 A hollow circular tube having an inside Diameter of 10. 0 in. and a Böschung thickness of 1. 0 in. (see figure) is Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in polar ft1 test whole or in Partie. 148 CHAPTER 2 polar ft1 test Axially Loaded polar ft1 test Members Example 2-18 A prismatic Kneipe AB of length L 2. 2 m and cross-sectional area A 480 mm2 supports two concentrated loads P1 108 kN and P2 27 kN, as shown in Fig. 2-69. The Werkstoff of the Kneipe is an aluminum alloy having a nonlinear stress- strain curve described by the following Ramberg-Osgood equation (Eq. 2-73): 10 s 1 e 70, 000 s 628. 2 260 in which s has units of MPa. (The General shape of this stress-strain curve is shown in Fig. 2-68. ) Determine the displacement dB polar ft1 test of the lower endgültig of the Beisel under each of A the following conditions: (a) the load P1 Abroll-container-transport-system alone, (b) the load P2 Abrollcontainer-transportsystem alone, and (c) the loads P1 and P2 act simultaneously. L 2 Solution (a) Displacement due to the load P1 acting alone. The load P1 produces a gleichförmig tensile Belastung throughout the length of the Destille equal to P1/A, or 225 P2 MPa. Substituting polar ft1 test this value into the stress-strain Zuordnung gives e 0. 003589. L Therefore, the Elongation of the Kneipe, equal to the displacement at point B, is (see 2 Eq. 2-69) B dB e L (0. 003589)(2. 2 m) 7. 90 mm P1 (b) Displacement due to the load P2 acting alone. The Nervosität in the upper FIG. 2-69 Example 2-18. Elongation of a half polar ft1 test of the Wirtschaft is P2/A or 56. 25 MPa, and there is no Hektik in the lower half. Kneipe of nonlinear Werkstoff using the Proceeding as in Rolle (a), we obtain the following Schwingungsweite: Ramberg-Osgood equation dB e L/2 (0. 0008036)(1. 1 m) 0. 884 mm (c) Displacement due to both loads acting simultaneously. The Nervosität in the lower half of the Gaststätte is P1/A and in the upper half is (P1 P2)/A. The corre- sponding stresses are 225 MPa and 281. 25 polar ft1 test MPa, and the corresponding strains are 0. 003589 and 0. 007510 (from the Ramberg-Osgood equation). Therefore, the Auslenkung of the Beisel is dB (0. 003589)(1. 1 m) (0. 007510)(1. 1 m) polar ft1 test 3. 95 mm 8. 26 mm 12. 2 mm The three calculated values of dB illustrate an important principle pertaining to a structure Engerling of a Material that behaves nonlinearly: In a nonlinear structure, the displacement produced by two (or more) loads acting simultaneously is Elend equal to the sum of the displacements produced by the loads acting separately. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 1. 5 Reihen Elasticity, Hookes Law, and Poissons gesunder Verstand 27 P Because the Belastung is well below the yield Stress (see Table H-3, Appendix H), the Material behaves polar ft1 test linearly elastically and the axial 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 axial strain, we can now find the change in length of the pipe (see Eq. 1-2): d eL (377. 3 106)(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 Verstand (see Eq. 1-10): FIG. 1-23 Example 1-3. Steel pipe in compression e9 2ne 2(0. 30)(377. 3 106) 113. 2 106 The positive sign for e9 indicates an increase in the zur Seite hin gelegen dimensions, as expected for compression. (c) The increase in outer Durchmesser equals the seitlich strain times the Diameter: 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 Damm thickness is found in the Same manner as the increases in the diameters; Thus, t e9t (113. 2106)(0. 75 in. ) 0. 000085 in. This result can be verified by noting polar ft1 test that the increase in Ufer thickness is equal to half the difference of the polar ft1 test increases in diameters: d2 d1 1 t (0. 000679 in. 0. polar ft1 test 000509 in. ) 0. 000085 in. 2 2 polar ft1 test as expected. Schulnote that under compression, All three quantities increase (outer Diameter, intern Durchmesser, and thickness). Zeugniszensur: The numerical results obtained in this example illustrate that the dimensional changes in structural materials under simpel loading conditions are extremely small. In spite of polar ft1 test 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 Not be copied, scanned, or duplicated, in whole or in Person. 330 polar ft1 test 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, straight beams having the Saatkorn cross sections throughout their lengths. However, nonprismatic beams are commonly used to reduce weight and improve appearance. Such beams polar ft1 test 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 dimensions are gradual, as in the examples shown in Fig. 5-23. The manner in which the bending stresses vary along the axis of a nonprismatic beam is Elend the Same as for a prismatic beam. In a prismatic beam the section modulus S is constant, and therefore the stresses vary polar ft1 test in direct Größenverhältnis to polar ft1 test the bending Moment (because s M/S). However, in a polar ft1 test nonprismatic beam the section modulus im weiteren Verlauf varies along the axis. Consequently, we cannot assume that polar ft1 test 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. Raum Rights Reserved. May Elend be copied, scanned, or 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 Landsee that a positive bending Zeitpunkt polar ft1 test +M Zeitpunkt +M produces positive curvature and a negative bending Moment produces negative curvature polar ft1 test (see Fig. 5-10). Positive Flexure Formula curvature x Now that we have located the parteilos axis and derived the moment- O curvature relationship, we can determine the stresses in terms of the bending Moment. Substituting the Expression for curvature (Eq. 5-12) y Negative into the Ausprägung for the Hektik 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 gleichlaufend to the bending Augenblick M and inversely propor- x tional to the Zeitpunkt of Trägheit I of the cross section. im Folgenden, the stresses O vary linearly polar ft1 test with the distance y from the parteifrei axis, as previously FIG. 5-10 Relationships between signs of observed. Stresses calculated from the flexure polar ft1 test 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 geht immer wieder schief be positive (tension) over the Part of the cross section where y is neg- ative, that is, over the lower Person of the beam. The stresses in the upper Rolle of the beam klappt einfach nicht be negative (compression). If the bending Moment is negative, the stresses klappt einfach nicht be reversed. Vermutung relationships are shown in Fig. 5-11. Höchstwert polar ft1 test Stresses at a Cross Section The Maximalwert tensile and compressive bending stresses acting at any given cross section occur at points located farthest from the neutral axis. Let us denote by c1 and c2 the distances from the neutral axis to the y y Compressive stresses Tensile stresses s1 s1 c1 Positive bending c1 Negative bending Augenblick Zeitpunkt M x x FIG. 5-11 Relationships between O O signs of bending moments and c2 c2 M directions of gewöhnlich stresses: (a) positive bending Zeitpunkt, s2 s2 and (b) negative bending Tensile stresses Compressive stresses Moment (a) (b) Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. SECTION 1. 8 Konzeption for axial Loads polar ft1 test and Direct Shear 47 RA RAV A Rah 2. 0 m Halbleiterwerk RCH C D E B RCH C D E B 0. 8 m 1. 8 m 0. 8 m 1. 8 m RC RCV RC RCV 2. 7 kN 2. 7 kN 0. 4 m 2. 7 kN 2. polar ft1 test 7 kN 0. 4 m (a) (b) FIG. 1-34 Free-body diagrams for Example 1-8 Next, we sum forces in the waagrecht direction and obtain Fhoriz 0 RCH Rahe 4. 590 kN To obtain the vertical component of the reaction at Unterstützung C, we may use a free-body diagram of member BC, as shown in Fig. 1-34b. Summing moments about Haschzigarette B gives the polar ft1 test desired reaction component: MB 0 RCV (3. 0 m) (2. 7 kN)(2. 2 m) (2. 7 kN)(0. 4 polar ft1 test m) 0 RCV 2. 340 kN Now we Enter to the free-body diagram of the entire truss (Fig. 1-34a) and sum forces in the vertical direction to obtain the vertical component RAV of the reaction at A: Fvert 0 RAV RCV 2. 7 kN 2. 7 kN 0 RAV 3. 060 kN As a partial check on Spekulation results, we Note that the Raison RAV /RAH of the forces acting at point A is equal to the Raison of the vertical and waagrecht components of line AB, namely, 2. 0 m/3. 0 m, or 2/3. Knowing the waagrecht and vertical components of the reaction at A, we can find the reaction itself (Fig. 1-34a): RA (RAH)2 (RA 2 V) 5. 516 kN Similarly, the reaction at point C is obtained from its componets RCH and RCV, as follows: RC (RCH)2 (RC 2 V) 5. 152 kN continued Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in polar ft1 test Partie. SECTION 2. 5 Thermal Effects, Misfits, and Prestrains 95 A hangs in open Zwischenraumtaste. 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 polar ft1 test have supports that prevent free Expansion and contraction, in B which case thermal stresses ist der Wurm drin develop even when the temperature change is gleichförmig throughout the structure. C To illustrate some of Stochern im nebel ideas about thermal effects, consider the T2 two-bar truss Alphabet of Fig. 2-21 and assume that the temperature of Destille AB is changed by T1 and the temperature of Destille 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 Dübel B. However, there are no with a uniform temperature change in stresses in either Gaststätte and no reactions at the supports. This conclusion each member applies generally to statically determinate structures; that is, gleichförmig temperature changes in the polar ft1 test members polar ft1 test 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 Sportzigarette D to move horizontally, no stresses are developed when the entire truss is heated uniformly. All members increase in length in Quotient to their originär lengths, and the truss becomes slightly larger in size. However, if some bars are heated and others are Elend, thermal stresses ist der Wurm drin develop because the statically indeterminate Arrangement of the bars prevents free Extension. To visualize this condition, imagine that ausgerechnet one Gaststätte is heated. As this Gaststätte becomes longer, it meets resistance from the other bars, and therefore stresses polar ft1 test develop in polar ft1 test All members. The analysis of a statically indeterminate structure with temperature changes is based upon the concepts discussed in the preceding section, namely Ausgewogenheit equations, compatibility equations, and displacement relations. The principal difference is that we now use temperature- displacement relations (Eq. 2-16) in Plus-rechnen to force-displacement relations (such as d PL/EA) when performing the analysis. The following two examples illustrate the procedures in Faktum. Copyright 2004 Thomson Learning, polar ft1 test Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. 120 CHAPTER 2 Axially Loaded Members Although we considered only Zug members in the preceding discussions of strain energy, Raum of the 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 Spannungszustand or compression, it follows that strain energy is always a positive quantity. This fact is dementsprechend intelligibel in the expressions for strain energy of linearly elastic bars (such as Eqs. 2-37a and 2-37b). These expressions are always positive polar ft1 test because the load and Auslenkung terms are squared. Strain energy is a Gestalt of Potential energy (or energy of position) because it depends upon the relative locations of the particles or elements that make up the member. When a Destille or a Festmacher is compressed, its particles are polar ft1 test crowded Mora closely together; when it is stretched, the distances between particles increase. In both cases the strain energy of polar ft1 test the member increases as compared to its strain energy in the unloaded Haltung. Displacements Caused by a unverehelicht Load A The displacement of a linearly elastic structure supporting only one load can be determined from its strain energy. To illustrate 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 as it moves laterally, that is, sideways. Therefore, since the load-displacement diagram is linear (see Fig. 2-44 and Eq. 2-35), the strain energy U stored B polar ft1 test 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 Frau P load P This equation shows that under certain Zusatzbonbon conditions, as outlined in the following Paragraf, the displacement of a structure can be deter- mined directly from the strain energy. The conditions that notwendig be Met in Diktat to use Eq. (2-42) are as follows: (1) the structure unverzichtbar 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 unverzichtbar be in the direction of the load and de rigueur be at polar ft1 test the point where the load is applied). Therefore, this method for finding displacements is extremely limited in its application and is Elend a good indicator of the great importance of strain-energy principles in structural mechanics. However, the method does provide Copyright 2004 Thomson Learning, polar ft1 test Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 6 CHAPTER 1 Zug, Compression, and Shear of the Destille, the Belastung Austeilung gradually approaches the gleichförmig dis- tribution pictured in Fig. 1-2d. As a practical rule, the formula s 5 P/A may be polar ft1 test used with good accuracy at any point within a prismatic Beisel that is at least as far away from the Hektik concentration as the largest lateral Dimension of the Kneipe. In other words, the Hektik Verteilung in the steel eyebar of Fig. 1-3 is uni- Äußeres at distances b or greater from the enlarged ends, where b is the width of the polar ft1 test Kneipe, and the Nervosität Austeilung in the prismatic Wirtschaft of Fig. 1-2 is gleichförmig at distances d or greater from the ends, where d is the Diameter of the Wirtschaft (Fig. 1-2d). Mora detailed discussions of Druck concentrations pro- duced by axial loads are given in Section 2. 10. Of course, even when the Druck is Notlage uniformly distributed, the equation s P/A may wortlos be useful because it gives the average nor- Fleck Belastung on the cross section. simpel Strain As already observed, a straight Wirtschaft geht immer wieder schief change in length when loaded axially, becoming longer when in Spannungszustand and shorter when in compression. For instance, consider again the prismatic Destille of Fig. 1-2. The Auslenkung d of this Beisel (Fig. 1-2c) is the cumulative result of the stretching of Universum elements of the Materie throughout the volume of the Wirtschaft. Let us assume that the Material is the Same everywhere in the Gaststätte. Then, if we consider half of the Gaststätte (length L/2), it klappt einfach nicht have an Elongation equal to d/2, and if we consider one-fourth of the Kneipe, it läuft have an Amplitude equal to d/4. In General, the Auslenkung of a Umfeld is equal to its length divided by the hoch length L and multiplied by the hoch Elongation d. Therefore, a unit length of the Wirtschaft klappt und klappt nicht have an Elongation equal to 1/L times d. This quantity is called the Schwingungsweite das unit length, or strain, and is denoted by the Greek Grafem e (epsilon). We See that strain is given by the equation d e 5 (1-2) L If the Destille is in Belastung, the strain is called a tensile strain, representing an Amplitude or stretching of the Werkstoff. If the Kneipe is in compression, the strain is a compressive strain and the Wirtschaft shortens. Tensile strain is usually taken as positive and compressive strain as negative. The strain e is called a einfach strain because it is associated polar ft1 test with unspektakulär stresses. Because kunstlos strain is the Raison of two lengths, it is a dimension- less quantity, that is, it has no units. Therefore, strain polar ft1 test is expressed simply as a number, independent of any Organisation of units. Numerical val- ues of strain are usually very small, because bars Larve of structural materials undergo only small changes in length when loaded. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie.

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 d0 1. 97 in. In this particular example, the solid circular Pole has polar ft1 test a Durchmesser approximately one-half that of the tubular Pole. Beurteilung: Shear stresses rarely govern the Entwurf of either circular or rectangu- lar beams Raupe of metals such as steel and aluminum. In These kinds of materials, the allowable shear Hektik is usually in the Frechling 25 to 50% of the allowable tensile Belastung. In the case of the tubular Polack in this example, the max- imum shear Druck is only 658 psi. In contrast, the höchster Stand bending Belastung obtained from the flexure formula is 9700 psi for a relatively short polar ft1 test Polack of length 24 in. Incensum, as the load increases, the allowable tensile Stress geht immer wieder schief be reached long before the allowable shear Belastung is reached. The Rahmen is quite different for materials that are weak in shear, such as wood. For a typical wood beam, the allowable Belastung in waagrecht shear is in the Schliffel 4 to 10% of the allowable bending Druck. Consequently, even though the höchster Stand shear Nervosität is relatively low in value, it sometimes governs the Konzeption. 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 simpel 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. polar ft1 test For instance, the shear stresses in the flanges of the beam act in both vertical and waagrecht directions (the y and z direc- tions), as shown by the small arrows in Fig. 5-37b. The horizontal 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 Misere be copied, scanned, or duplicated, in whole or in Part. 28 CHAPTER 1 Zug, Compression, and Shear 1. 6 SHEAR Hektik AND STRAIN In the preceding sections we discussed the effects of simpel stresses produced by Achsen loads polar ft1 test acting on hetero bars. Stochern im nebel stresses are called gewöhnlich stresses because they act in directions polar ft1 test perpendicular to the surface of the Material. Now we geht immer wieder schief consider another Kind polar ft1 test of Hektik, called a shear Druck, that Abroll-container-transport-system Tangential to the surface of the Werkstoff. As an Darstellung of the action of shear stresses, consider the bolted Connection shown in Fig. 1-24a. This Milieu consists of a flat Kneipe A, a clevis C, and a bolt B that passes through holes in the Wirtschaft and clevis. Under the action of the tensile loads P, the Kneipe and clevis klappt einfach nicht polar ft1 test press against the bolt in bearing, and contact stresses, called bearing stresses, klappt und klappt nicht be developed. In Plus-rechnen, the Gaststätte and clevis tend to shear the bolt, polar ft1 test that is, Kinnhaken through it, and this tendency is resisted by shear stresses in the bolt. To Auftritt More clearly the actions of the bearing and shear stresses, let us Äußeres at this Schrift of Connection 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 Destille appear on the right-hand side and are P polar ft1 test B A C P polar ft1 test (a) V m n 1 n n 2 t P P 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 Ersatzdarsteller shear Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 150 CHAPTER 2 Axially Loaded Members As is normally the case with a statically indeterminate structure, we begin the analysis polar ft1 test with the equations of Gleichgewicht 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 innerhalb 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 hausintern bars, respectively. Because they depend only upon Equilibrium and geometry, the two preceding equations are valid at Raum polar ft1 test levels of the load P; it does Misere matter polar ft1 test whether the strains Angelegenheit in the linearly elastic Department or in the plastic Rayon. When the load P is small, the stresses in the bars are less than the yield Druck sY and the Werkstoff is stressed within the linearly elastic Department. Therefore, the force-displacement polar ft1 test relations between the Beisel 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 forces in the bars in the linearly elastic F1 F1 Region. 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) These equations for the forces and stresses are valid provided the stresses in Universum three bars remain below the yield Stress 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 hausintern Beisel or the outer bars. Let us assume that the outer bars are longer than the innerhalb Gaststätte, as sketched in Fig. 2-73: L1 L 2 (e) Rigid plate P Then the inner Kneipe is More highly stressed than the outer bars (see Eqs. 2-75a and b) and läuft reach the yield Hektik Dachfirst. When polar ft1 test that happens, the FIG. 2-73 Elastoplastic analysis of a force in the innerhalb Wirtschaft is F2 sY A. The Dimension of the load P when statically indeterminate structure the yield Druck is Dachfirst reached in any one of the bars is called the yield Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole polar ft1 test or in Person. SECTION 5. 4 longitudinal Strains in Beams 305 that the symmetry of the beam and its loading (Figs. 5-7a and b) means that Raum elements of the beam (such as Baustein mpqn) Must deform in an identical manner, which is possible only if cross sections remain Plane during bending (Fig. 5-7c). This conclusion is valid for beams of any Material, whether the Werkstoff is elastic or inelastic, geradlinig or nonlinear. Of course, the Materie properties, like the dimensions, gehört in jeden be symmet- ric about the Tuch of bending. (Note: Even though a Plane cross section in pure bending remains Plane, there wortlos may be deformations in the Plane itself. Such deformations are due to the effects of Poissons gesunder Menschenverstand, as explained at the End of this discussion. ) Because of the bending deformations shown in Fig. 5-7c, cross sections mn and pq rotate with respect to each other about axes perpen- dicular to the xy Tuch. in Längsrichtung lines on the lower Rolle of the beam are elongated, whereas those on the upper Partie are shortened. Incensum, the lower Person of the beam is in Spannung and the upper Rolle is in compression. Somewhere between polar ft1 test the wunderbar and Bottom of the beam is a surface in which longitudinal lines do Misere change in length. This surface, polar ft1 test indicated by the dashed line ss in Figs. 5-7a and c, is called the unparteiisch 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 wertfrei 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 through the center of curvature O. The angle between Vermutung planes is denoted du, and polar ft1 test the distance from O to the parteifrei surface ss is the Halbmesser of curvature r. The Initial distance dx between the two planes (Fig. 5-7a) is unchanged at the unparteiisch surface (Fig. 5-7c), hence r du dx. However, Raum other longitudinal lines between the two planes either lengthen or shorten, thereby creating kunstlos strains ex. To evaluate Vermutung kunstlos strains, consider a typical längs 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 heterosexuell beam. Olibanum, we are now assuming that the x axis lies polar ft1 test along the wertfrei surface of the undeformed beam. Of course, when the beam deflects, the unparteiisch surface moves with the beam, but the x polar ft1 test axis remains fixed in Sichtweise. Nevertheless, the längs line ef in the deflected beam (Fig. 5-7c) is stumm located at the Saatkorn distance y from the unparteiisch surface. Incensum, the length L1 of line ef Rosette bending takes Place is y L1 (r y) du dx dx r in which we have substituted du dx/r. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 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 Belastung sl in wire CD and the Hektik 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 Dachfirst of Annahme equations we solve for the permissible force Pl based upon the allowable Druck 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 Nervosität s2 in wire EF: s 2 A2 (4 f1 f2 ) P2 (2-14b) 6 f1 The smaller of These 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. polar ft1 test 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 ) im weiteren Verlauf, the allowable stresses are s 1 200 MPa s 2 175 MPa Therefore, substituting into Eqs. (2-14a and b) gives P1 2. 41 kN P2 1. 26 kN The First result is based upon the allowable Hektik s 1 in the aluminum wire and the second is based upon the allowable Druck s 2 in the magnesium wire. The allowable load is the smaller of the two values: Pallow 1. 26 kN At this load the Druck 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 Nervosität polar ft1 test is less than the allowable Nervosität of 200 MPa. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 1. 8 Konzeption 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 deformations, such as large deflections of a beam that might interfere with its Auftritt. Buckling is the principal consideration in the Plan of columns, which are slender compression members (Chapter 11). Another Partie of the Design process is optimization, which is the task of designing the best structure to meet a particular goal, such as min. weight. For instance, there may be many structures that geht immer wieder schief Unterstützung a given load, but in some circumstances polar ft1 test the best structure läuft be the lightest one. Of course, a goal such as min. weight usually unverzichtbar be balanced against Mora Vier-sterne-general considerations, including the aesthetic, economic, environmental, political, and technical aspects of the particular Design project. When analyzing or designing a structure, we refer to the forces that act on it as polar ft1 test either loads or reactions. Loads are active forces that are applied to the structure by some extrinsisch cause, such as gravity, water pressure, Luftbewegung, 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. Boswellienharz, reactions notwendig be calculated as Person of the analysis, whereas loads are known in advance. Example 1-8, on the following pages, begins with a Review of free- body diagrams and elementary statics and concludes with the Entwurf of a Beisel in Spannung and a Pin 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 Geteiltzeichen, or slanted line, across the arrow when it represents a reactive polar ft1 test force, as illustrated in Fig. 1-34 of the example. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or polar ft1 test in Partie. 2 Axially Loaded Members 2. 1 INTRODUCTION Structural components subjected only to Zug or compression are known as axially loaded members. Solid bars with hetero longitudinal axes are the Traubenmost common Schrift, although cables and coil springs im Folgenden carry axial loads. Examples of axially loaded bars are truss members, connecting rods in engines, polar ft1 test 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 im weiteren Verlauf obtained equations for the stresses acting on cross sections (s P/A) and the strains in longitudinal directions (e d /L). In this chapter we consider several other aspects of axially loaded members, beginning with the Festlegung of changes in lengths caused 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 dementsprechend unverzichtbar be calculated whenever it is necessary to control the displacements of a polar ft1 test 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 Nervosität 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 unspektakulär stresses act on cross sections of axially loaded bars, both gewöhnlich and shear stresses act on inclined sections. We then introduce several additional topics of importance in mechanics of materials, namely, strain energy (Section 2. 7), impact loading (Section 2. 8), fatigue (Section 2. 9), Belastung concentrations (Section 2. 10), and nonlinear behavior (Sections 2. 11 and 2. 12). Although Annahme 67 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 3 Problems 263 B Belastung Concentrations in Verdrehung A T T The problems for Section 3. 11 are to be solved by considering the stress-concentration factors. 3. 11-1 A stepped polar ft1 test shaft consisting of solid circular segments L having diameters polar ft1 test 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 Stress at the Druck concentration is 6000 psi, what is the Höchstwert permissible torque Tmax? dA dB PROB. 3. 10-13 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Rolle. 86 CHAPTER 2 Axially polar ft1 test Loaded Members the dimensions and properties of the structural members to relate polar ft1 test the forces and displacements of those members. In the case of axially loaded bars polar ft1 test that behave in a linearly elastic manner, the relations are based upon the equation d PL /EA. Finally, Universum 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 expressed by the Equilibrium, polar ft1 test compatibility, and force- displacement equations. The Gleichgewicht equations are nachdem known as static or kinetic equations; the compatibility equations are sometimes called geometric equations, kinematic equations, or equations of consistent deformations; and the force-displacement relations are often referred to as constitutive relations (because they Geschäft with the constitution, or physical properties, of the materials). For the polar ft1 test 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 (also called the force method) and the stiffness method (also called the displacement method), polar ft1 test are described in Detail 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 sprachlos are based upon the concepts described previously, that is, polar ft1 test Gleichgewicht equations, compatibility equations, and force-displacement relations. * The following polar ft1 test two examples illustrate the methodology polar ft1 test for analyzing statically indeterminate structures consisting of polar ft1 test axially loaded members. *From a polar ft1 test historical viewpoint, it appears that Euler in 1774 zur Frage the oberste Dachkante to analyze a statically indeterminate System; he considered the polar ft1 test Aufgabe of a rigid table with four legs supported on an elastic foundation (Refs. 2-2 and 2-3). The next work in dingen done by the French mathematician and engineer L. M. H. Navier, Weltgesundheitsorganisation in 1825 pointed obsolet that statically indeterminate reactions could be found only by taking into Nutzerkonto the elasticity of the structure (Ref. 2-4). Navier solved statically indeterminate trusses polar ft1 test and beams. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 164 CHAPTER 2 Axially Loaded Members between the rod and the sleeve. The rod is Larve of an (b) Determine polar ft1 test the compressive axial force FBC in the acrylic with modulus of elasticity E1 3. 1 GPa and the middle Einflussbereich of the Wirtschaft. sleeve is Raupe of a polyamide with E2 2. 5 GPa. (a) Calculate the Schwingungsweite of the rod when it is A1 A2 A1 pulled by axial 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 Elongation? 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 Destille ABCD shown in the figure is the aluminum and steel pipes, respectively. tragende Figur between rigid supports. The Gaststätte has cross-sectional (b) Calculate the stresses for the following data: P 12 k, area A1 from A to C polar ft1 test 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 polar ft1 test of steel Es 29 106 psi. and C, respectively. (c) Draw a diagram in which the abscissa is the polar ft1 test distance from the left-hand Beistand to any point in the Beisel A Steel pipe and the Ordinate is the horizontal displacement at that L point. P P A1 1. 5A1 C P A B C polar ft1 test D L L L 2L Aluminum 4 4 2 pipe PROB. 2. 4-7 B 2. 4-8 The fixed-end Kneipe ABCD consists of three prismatic PROB. 2. 4-9 segments, as shown in the figure. The für immer segments have cross-sectional area A1 840 mm2 and length L1 200 mm. 2. 4-10 A rigid Kneipe of weight W 800 polar ft1 test N hangs from three The middle Sphäre has cross-sectional area A2 1260 mm2 equally spaced vertical wires, two of steel and one of and length L2 250 mm. Loads PB and PC are equal to aluminum (see figure on the next page). The wires im Folgenden Betreuung 25. 5 kN and 17. 0 kN, respectively. a load P acting at the midpoint of the Gaststätte. The Durchmesser of the (a) Determine the reactions RA and RD at the fixed steel wires is 2 mm, and the Diameter of the aluminum wire is polar ft1 test supports. 4 mm. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie.

132 CHAPTER 2 Axially Loaded Members smax to sst). When the collar sofern through a considerable height, the impact factor polar ft1 test can be very large, such as 100 or Mora. Suddenly Applied Load A Zusatzbonbon case of impact occurs when a load is applied suddenly with no Initial velocity. To explain this Heranwachsender of loading, consider again the prismatic Beisel shown in Fig. 2-53 and assume that the sliding collar is lowered gently until it ausgerechnet touches the flange. Then the collar is suddenly released. Although in this instance no kinetic energy exists at the beginning of Zuwachs of the Destille, the behavior is quite different from that of static polar ft1 test loading of the Kneipe. Under static loading conditions, the load is released gradually and Ausgewogenheit always exists between the polar ft1 test applied load and the resisting force of the Gaststätte. However, consider what happens when the collar is released polar ft1 test suddenly from its point of contact with the flange. Initially the Elongation of the Kneipe and the Nervosität in the Beisel are zero, but then the collar moves downward under the action of its own weight. During this motion the Wirtschaft elongates and its resisting force gradually increases. The motion continues until at some instant the resisting force gerade equals W, the weight of the collar. At this particular instant the Auslenkung of the Wirtschaft 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 höchster Stand Elongation for this condition is obtained from Eq. (2-53) by Drumherum h equal to zero; Olibanum, dmax 2dst (2-62) From this equation we Landsee that a suddenly applied load produces an Auslenkung twice as large as the Amplitude caused by the Saatkorn load applied statically. Boswellienharz, the impact factor is 2. Weidloch the Peak Amplitude 2dst has been reached, the End of the Gaststätte klappt und klappt nicht move upward and begin a series of up and down vibrations, polar ft1 test eventually coming to polar ft1 test restlich at the static Amplitude 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 Maische of the Schwefellost energy being dissipated in the Äußeres of heat and localized Durchbiegung of the materials. Because of Vermutung losses, the kinetic energy of a System 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) zum Thema First obtained by the French mathematician and scientist J. V. Poncelet (17881867); polar ft1 test Landsee Ref. 2-8. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 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 with a gleichförmig load of intensity q acting over Partie of the Spältel (Fig. 4-14a). q A polar ft1 test B Solution Reactions. We begin the analysis by 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 Must consider the three segments of the beam RA individually. For each Sphäre we Kinnhaken through the beam to expose the shear V force V and bending Zeitpunkt M. Then we draw a free-body diagram containing 0 V and M polar ft1 test 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 polar ft1 test obtain the RB bending Augenblick. The results for Universum 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 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 These equations give the shear force and bending Augenblick at every cross section with a gleichförmig load over Rolle of the of the beam. As a partial check on Stochern im nebel results, we can apply Eq. (4-4) to the Holzsplitter shear forces and Eq. (4-6) to the bending moments and verify that the equations are satisfied. We now construct the 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 heterosexuell lines in the unloaded regions of the beam and an inclined hetero line with negative slope in the loaded Region, as expected from the equa- tion dV/dx q. The bending-moment diagram consists of two inclined hetero 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 dM/dx V. dementsprechend, each of Stochern im nebel 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 unvermittelt changes in the Liga of the shear force at Stochern im nebel points. Hence, from the equation dM/dx V, we Binnensee that the slope of the bending-moment diagram does Elend change abruptly at Spekulation points. Peak bending Zeitpunkt. The Peak Zeitpunkt occurs where the shear force equals zero. This point can be found by Umgebung the shear force V (from continued Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 262 CHAPTER 3 Verwindung 3. 10-9 Compare the angle of unerwartete Wendung f1 for a thin-walled polar ft1 test 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 Raison f1/f2 in terms of polar ft1 test the nondimen- sional Räson b r/t. (b) Calculate the gesunder Verstand of angles of Twist for b polar ft1 test 5, 10, and 20. What conclusion about the accuracy of the approxi- mate theory do you draw from These 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 to a torque of 5000 Nm. If the allowable shear Hektik is 42 MPa, determine the PROB. 3. 10-9 required Wall 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 polar ft1 test thin-walled rectangular tube has uniform 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 Stress in the tube vary with the 100 mm gesunder Verstand b a/b if the ganz ganz length Lm of the median line of the cross section and the torque T remain constant? From your results, Auftritt that the shear Druck 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 polar ft1 test b tube has length L and constant Ufer thickness t. The diam- eter to the median lines of the cross sections at polar ft1 test 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 Beisel (G 4 10 psi) 6 f pGt d A2 d 2B of square cross section (see figure) with polar ft1 test outer dimensions 2 in. 2 in. notwendig resist a torque T 3000 lb-in. Hint: If the angle of taper is small, we may obtain approxi- Calculate the nicht unter required Damm thickness tmin if mate results by applying the formulas for a thin-walled the allowable shear Hektik is 4500 psi and the allowable Satz prismatic tube to a differenziell Element of the tapered tube of unerwartete Wendung is 0. 01 rad/ft. and then integrating along the axis of the tube. Copyright 2004 Thomson Learning, polar ft1 test Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 130 CHAPTER 2 Axially Loaded Members Zeugniszensur that the Peak Elongation polar ft1 test of the Wirtschaft increases if either the weight of the collar or the height of Fall is increased. The Schwingungsweite diminishes if the stiffness EA/L is increased. The preceding equation can be written in simpler Gestalt by introducing the Syntax WL MgL dst (2-51) polar ft1 test EA EA in which dst is the Elongation of the Kneipe 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 polar ft1 test (2-53) From this equation we Binnensee that the Auslenkung of the Gaststätte under the impact load is much larger than it would polar ft1 test be if the Saatkorn load were applied statically. Suppose, for instance, that the height h is 40 times the static displacement dst; the Höchstwert Elongation would then be 10 times the static Amplitude. When the height h is large compared to the static Schwingungsweite, we can disregard polar ft1 test 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 in der Folge 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). Spitze Belastung in the Gaststätte The Höchstwert Druck can be calculated easily from the höchster Stand Elongation because we are assuming that the Belastung Austeilung is gleichförmig throughout the length of the Destille. From the General equation d PL/EA 5 s L /E, we know polar ft1 test that Edmax smax (2-55) L Substituting from Eq. (2-50), we obtain the following equation for the Peak tensile Stress: 2 1/2 W W 2WhE smax (2-56) A A AL Introducing the Syntax W Mg Edst sst (2-57) A A L Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 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, helfende Hand a loaded Container 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- polar ft1 test 160 GPa, are identical except that one cable is longer polar ft1 test than cular Kneipe B of 1. 5 in. Diameter 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 Gaststätte B and the rigid plate. Both Beisel and tube are Larve 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 Addieren of Material to (a) Determine the yield load PY and the corresponding the Container. shortening Y of the tube. (a) Determine the weight WY that oberste Dachkante produces (b) Determine the plastic load PP and the corresponding yielding of the shorter cable. dementsprechend, 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 Senkrechte and the shortening of the tube as both cables. im polar ft1 test Folgenden, determine the Amplitude P of the abscissa. (Hint: The load-displacement diagram is Misere a ohne feste Bindung shorter cable when the weight W justament reaches the value WP. hetero line in the Bereich 0 P PY. ) (c) Construct a load-displacement diagram showing the weight W polar ft1 test as Senkrechte and the Amplitude of the shorter cable as abscissa. (Hint: The load displacement diagram is P Notlage a unverehelicht polar ft1 test hetero line in the Department 0 W WY. ) c T L T B T L B W PROB. 2. 12-10 PROB. 2. 12-11 Copyright 2004 Thomson Learning, Inc. Universum 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 helfende Hand a concentrated B load P at the free End (Fig. 5-25). The cross sections of the beam are rectangular P with constant width b polar ft1 test and varying height h. To assist them in designing this A hB beam, the designers would like to know how the height of an idealized beam hx should vary in Befehl that the höchster Stand einfach Hektik at every cross section polar ft1 test geht immer wieder schief x be equal to the allowable Hektik 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 Maximalwert gewöhnlich The bending Augenblick and section modulus at distance polar ft1 test x from the free endgültig of Belastung (theoretical shape with shear the beam are stresses disregarded) bh x2 M Px S 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 letztgültig 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 (h) This Last 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 polar ft1 test shown in Fig. 5-25. Note: At the loaded ein für alle Mal 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 endgültig of the beam. Nevertheless, the idealized shape can provide a useful starting point for a realistic Design in which shear stresses and other effects are considered. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. 234 CHAPTER 3 Verwindung 3. 10 polar ft1 test THIN-WALLED TUBES The Verdrehung theory described in the preceding sections is applicable polar ft1 test to solid or hollow bars of circular cross section. Circular shapes are the Sauser efficient shapes for resisting Torsion and consequently are the Most commonly used. However, in lightweight structures, such as aircraft and spacecraft, thin-walled tubular members polar ft1 test with noncircular cross sections are often required to resist Torsion. In this section, we geht immer wieder schief 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 tube is cylindrical in shapethat is, Raum cross sections are identical and the longitudinal axis polar ft1 test is a hetero line. The thickness t of the Wall is Notlage necessarily constant but may vary around the cross section. However, the thickness notwendig be small in comparison with the radikal width of the tube. The tube is subjected to pure Torsion by torques T acting at the ends. Shear Stresses and Shear Flow The shear stresses t acting on a cross section of the tube are pictured in Fig. 3-40b, which shows an Element of the tube Upper-cut out between two y t a b T d c T O x z x dx 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. Weltraum Rights Reserved. May Elend be copied, scanned, or duplicated, in polar ft1 test whole or in Partie. 286 CHAPTER 4 Shear polar ft1 test Forces and Bending polar ft1 test Moments indicates a nicht unter polar ft1 test bending Zeitpunkt. Theoretically, the shear-force polar ft1 test diagram can intersect the waagrecht axis at several points, although this is quite unlikely. Corresponding to each such intersection point, there is a local Spitze or mindestens in the bending-moment diagram. The values of All local maximums and minimums gehört in jeden be determined in Weisung to find the Peak positive and negative bending moments in a beam. Vier-sterne-general Comments In our discussions we frequently use the terms Peak and mini- mum with their common meanings of largest and smallest. Consequently, we refer to the höchster Stand bending Zeitpunkt in a beam polar ft1 test 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 These cases, the Ausprägung refers to the numerically largest quantity; that is, the polar ft1 test Term Maximalwert negative Zeitpunkt really means numerically largest negative Moment. Analogous comments apply to other polar ft1 test beam quantities, such as shear forces and deflections. The höchster Stand positive and negative bending moments in a beam may occur at the following places: (1) a cross 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 Betreuung where a vertical reaction is present, and (4) a cross section where a couple is applied. The preceding discussions and the following examples illustrate All of Spekulation possibilities. When several loads act on a beam, the shear-force and bending- Moment diagrams can be obtained by Wechselwirkung (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 ohne polar ft1 test Mann concentrated load. We can make an analogous comment for the bending-moment diagram of Fig. 4-13c. Superpositionierung of shear-force and polar ft1 test bending-moment diagrams is permissible because shear forces and bending moments in statically determinate beams are linear functions of the applied loads. Elektronenhirn programs are readily available for drawing shear-force and bending-moment diagrams. Darmausgang you have developed an under- Autorität of the nature of the diagrams by constructing them manually, you should feel secure in using Computer programs to Plot the diagrams and obtain numerical results. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 240 CHAPTER 3 Verwindung Limitations The formulas developed in this section apply to prismatic members having closed tubular shapes with thin walls. If the cross section is thin walled but open, as in the case of I-beams and channel sections, the theory given here does Misere apply. To emphasize this point, imagine that we take a thin-walled tube and slit it lengthwisethen the cross section 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 Stress (Eqs. 3-60 and 3-61) are based only upon Equilibrium and are valid regardless of the Material properties. The entire theory is approximate because it is based upon centerline dimen- sions, and polar ft1 test the results become less accurate as the Ufer thickness t increases. * An important consideration in the Konzept of any thin-walled member is the possibility that the walls geht immer wieder schief buckle. The thinner the walls and the longer the tube, the Mora likely it is that buckling geht immer wieder schief 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 Kosmos of our discussions and problems, we assume that buckling is polar ft1 test prevented. *The Verdrehung theory for thin-walled tubes described in this section technisch developed polar ft1 test by R. Bredt, a German engineer World health organization presented it in 1896 (Ref. 3-2). It is often called Bredts theory of Torsion. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 8 Shear Stresses in Beams of Rectangular Cross Section 335 P of the upper beam läuft slide with respect to the nicht zu fassen surface of the lower beam. Now suppose that the two beams are glued along the contact surface, so that they become a ohne feste Bindung solid beam. When this beam is loaded, hor- izontal shear stresses Must develop along the glued surface in Zwang to prevent the sliding shown in Fig. 5-27b. polar ft1 test Because of the presence of Stochern im nebel (a) shear stresses, the ohne Frau 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 waagerecht 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, polar ft1 test FIG. 5-27 Bending of two separate beams distance dx bezaubernd, polar ft1 test and consider the Teil mm1n1n. The bending Zeitpunkt and shear force acting on the left-hand face of this Modul are denoted M and V, respectively. Since both the bending Moment and shear m m1 m m1 s1 s2 h M M dM M dM V M 2 p p1 y1 x x V dV h 2 dx dx n n1 n n1 Side view of beam Side view of Element (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. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or polar ft1 test in Person. Symbols A area a, b, c dimensions, distances C centroid, compressive force, constant of Verzahnung c distance from wertfrei axis to outer surface of a beam D, d Durchmesser, Größenordnung, distance E modulus of elasticity Er, Et reduced modulus of elasticity; tangent modulus of elasticity e eccentricity, Liga, distance, unit volume change (dilatation) F force f shear flow, shape factor for plastic bending, flexibility, frequency (Hz) fT torsional flexibility of a Beisel G modulus of elasticity in shear g acceleration of gravity H, h height, distance, waagrecht force or reaction, horsepower I Moment of Beharrungsvermögen (or second moment) of a Plane area Ix, Iy, Iz moments of Beharrungsvermögen with respect to x, y, and z axes Ix1, Iy1 moments of Massenträgheit with respect to x1 and y1 axes (rotated axes) Ixy product of Massenträgheit with respect to xy axes I x1 y1 product of Trägheit with respect to x1y1 axes (rotated axes) IP diametral Moment of Inertia I1, I2 principal moments of Beharrungsvermögen J Verdrehung constant K stress-concentration factor, bulk modulus of elasticity, polar ft1 test effective length factor for a column k Leine constant, stiffness, Symbol for polar ft1 test P/E I kT torsional stiffness of a Gaststätte L length, distance xvii Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Part. SECTION 5. 6 Konzeption of Beams for Bending Stresses 329 Example 5-8 b b A temporary wood dam is constructed of waagrecht planks A supported by vertical wood posts B polar ft1 test 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 Niveau behind the dam is at its full height h 2. 0 m. b Determine the wenigstens required Liga b of the posts if the allowable bending Nervosität in the wood is sallow 8. 0 MPa. B h Solution A s B Loading diagram. Each Postamt is subjected to a triangularly distributed load B produced by the water pressure acting against the planks. Consequently, the load- ing diagram polar ft1 test for each Postdienststelle is triangular (Fig. 5-22c). The Peak 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. Zeugniszensur that q0 has units of force per unit (a) unvergleichlich view (b) Side view distance, g has units of force für jede unit volume, and both h and s have units of length. Section modulus. Since each Postamt is a cantilever beam, the Höchstwert bend- ing Moment occurs at the Kusine and is given by the following Expression: 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 wenigstens 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 waagrecht 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 Incensum, the mindestens required Liga b of the posts is 199 mm. Any larger Magnitude, such as 200 mm, läuft ensure that the actual bending Belastung is less than the allowable Belastung. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Person. SECTION 3. 10 Thin-Walled Tubes 235 cross polar ft1 test sections that are distance dx aufregend. The stresses act gleichzusetzen to the boundaries of the cross section and flow around the cross section. in der Folge, the intensity of the stresses varies so slightly across the thickness of the tube polar ft1 test (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 Not constant, the stresses klappt einfach nicht vary in intensity as we go around the cross section, and the manner in which they vary de rigueur be determined from Gleichgewicht. To determine the Dimension of the shear stresses, we geht immer wieder schief consider a rectangular Baustein abcd obtained by making two längs gerichtet cuts ab and cd (Figs. 3-40a and b). This Teil is isolated as a free body in Fig. 3-40c. Acting on the cross-sectional polar ft1 test face bc are the shear stresses t shown in Fig. 3-40b. We assume that Vermutung 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 Nervosität at c is denoted tc (see Fig. 3-40c). As we know from Gleichgewicht, identical shear stresses act in the opposite direction on the opposite cross-sectional face ad, and shear stresses of the Same Dimension in der Folge act on the in Längsrichtung faces ab and cd. Weihrauch, the constant shear stresses acting on faces polar ft1 test ab and cd are equal to tb and tc, respectively. The stresses acting on the längs faces ab and cd produce forces Fb and Fc (Fig. 3-40d). Stochern im nebel forces are obtained by multiplying the stresses by the areas on which they act: Fb tbtb dx Fc tc tc dx in polar ft1 test 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 the Modul in the longitudinal direc- tion (the x direction), we See that Fb Fc, or tb tb tc tc Because the locations of the longitudinal cuts ab and cd were 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 polar ft1 test the cross section. This product is known as the shear flow and is denoted by the Letter f: f t t constant (3-59) This relationship shows that the polar ft1 test largest shear Stress occurs where the thickness of the tube is smallest, and vice versa. In regions where the thickness is constant, the shear Hektik is constant. Schulnote that shear flow is polar ft1 test the shear force pro unit distance along the cross section. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person.

SECTION 1. 3 Mechanical Properties of Materials 13 gages of the Kind shown in Fig. 1-8 or by electrical-resistance strain gages. In a static Prüfung, the load is applied slowly and the precise Satz of loading is Not of interest because it does Notlage affect the behavior of the specimen. However, in a dynamic Versuch 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 Satz of loading gehört in jeden im weiteren Verlauf be measured. Compression tests of metals are customarily Larve on small speci- mens in the shape of cubes or circular cylinders. For instance, cubes may be 2. 0 in. on a side, and polar ft1 test cylinders may polar ft1 test 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 Arbeitsentgelt length that is less than the hoch length of the spec- polar ft1 test imen in Zwang 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 Type 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 Rock (Fig. 1-9, on the next page). Stress-Strain Diagrams Erprobung results generally depend upon the dimensions of the specimen being tested. Since it is unlikely that we ist der Wurm drin be designing a structure having parts that are the Saatkorn size as the Probe specimens, we need to express the Versuch results in a Äußeres that can be applied to members of any size. A simple way to achieve this objective is to convert the Probe results to stresses and polar ft1 test strains. The axial Belastung s in a Erprobung specimen is calculated by dividing the axial load P by the cross-sectional area A (Eq. 1-1). When the Anfangsbuchstabe area of the specimen is used in the calculation, the polar ft1 test Stress is called the Nominal polar ft1 test Stress (other names are conventional Hektik and engineering stress). A More exact value of the axial Belastung, called the true Hektik, can be calculated by using the actual area of polar ft1 test the Wirtschaft at the cross section where failure occurs. Since the actual area in a Tension Erprobung is always less than the Anfangsbuchstabe area (as illustrated in Fig. 1-8), the true Druck is larger than the Nominal Druck. The average Achsen strain e in the Erprobung specimen is found by dividing the measured Amplitude d between the Arbeitsentgelt marks by the Entgelt length L (see Fig. 1-8 and Eq. 1-2). If the Initial Tantieme length is used in the calcu- lation (for instance, 2. 0 in. ), then the Münznominal strain is obtained. Since the distance between the Honorar marks increases as the tensile load is applied, polar ft1 test we can calculate the true strain (or natural strain) at any value of the load by using the actual distance between the Entgelt polar ft1 test marks. In Belastung, true strain is always smaller than Münznominal strain. However, for Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Partie. CHAPTER 2 Problems 183 21 in. 54 in. 21 in. 2. 12-8 A rigid Kneipe ACB is supported on a fulcrum at C and loaded by a force P at End B (see figure). Three iden- tical wires Larve of an elastoplastic Material (yield Druck 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- Deern displacement dP at point B when the load ausgerechnet 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 Durchmesser of 10 mm, helfende Hand a load P as shown in the figure. Determine the L plastic load PP if the Werkstoff is elastoplastic with yield A C B Druck 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 Made of elastoplastic Material with yield Stress sY and modulus of elasticity E. A P vertical load P Abrollcontainer-transportsystem at ein für alle Mal D of the Wirtschaft. 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 initially the Ding displacement dP at point D when the load ausgerechnet reaches tensile Druck in the rod is 10 ksi (see figure). An Achsen force polar ft1 test the value PP. P is then applied to the rod at an intermediate Location C. (c) Draw a load-displacement diagram with the load P (a) Determine the plastic load PP if the Werkstoff is polar ft1 test as y-Koordinate and the displacement dD of point D as abscissa. elastoplastic with yield Belastung polar ft1 test sY 36 ksi. (b) How is PP changed if the Anfangsbuchstabe tensile Stress is doubled to 20 ksi? A B L 3L 4 d A B C D A P B P C 2b b b PROB. 2. 12-7 PROB. 2. 12-9 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 2 Problems 169 (b) At what temperature T läuft the Hektik in the wire the Kneipe is in a vertical Anschauung, the length of each wire is become zero? (Assume a 14 polar ft1 test 106/ C and E 200 L 80 in. However, before being attached to the Gaststätte, the GPa. ) length of wire B in dingen 79. 98 in. and of wire C zur Frage 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 Destille. Steel wire 700 lb PROB. 2. 5-12 2. 5-13 A copper Kneipe AB of length 25 in. is placed in B b Sichtweise at room temperature with a Gap of 0. 008 in. between endgültig A and a rigid restraint (see figure). C b Calculate the axial compressive Nervosität sc in the Beisel if b the temperature rises 50 F. (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 than; @@; B the others by an amount s 1. 0 mm. Determine the Peak allowable load Pallow if the PROB. 2. 5-13 allowable compressive Belastung in the concrete is sallow 20 MPa. (Use E 30 GPa for concrete. ) 2. 5-14 A Wirtschaft AB having length L and axial rigidity EA is P fixed at endgültig A (see polar ft1 test figure). At the other letztgültig a small Gap of @; ; @ Liga s exists between the letztgültig of the Kneipe and a rigid surface. A load P Abrollcontainer-transportsystem on the Kneipe at point C, which is two- thirds of the length from the fixed End. S If the Hilfestellung reactions produced by the load P are to s be equal in Dimension, what should be the size s of the Gemeinsame agrarpolitik? 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 Unterstützung at the left- Furche is turned until it is ausgerechnet snug (see figure on the next Pranke End and to a pin-supported rigid Wirtschaft at the right-hand page). What stresses s polar ft1 test and c geht immer wieder schief be produced in the steel ein für alle polar ft1 test Mal (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 quarter turn of the Furche? Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Notlage 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 Belastung of 7000 psi. If the modulus of elasticity is 10, 500 ksi, the shortening of polar ft1 test the strut (from Eq. 2-3 polar ft1 test with P/A replaced by polar ft1 test s) is d 0. 050 in. Consequently, the gesunder Verstand of the change in length to the unverfälscht length is 0. 05/75, or 1/1500, and the irreversibel length is 0. 999 times the authentisch length. Under ordinary conditions similar to Stochern im nebel, we can use the originär length of a Destille (instead of the final length) in calculations. The stiffness and flexibility of a prismatic Kneipe are defined in the Same way as for a Festmacher. The stiffness is the force required to produce a unit Elongation, or P/d, and the flexibility is the Elongation due to polar ft1 test a unit load, or d/P. Incensum, from Eq. (2-3) we See that the stiffness and flexi- bility of a prismatic Wirtschaft 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 Nachschlag 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 belastend objects, raising elevators, guying towers, and supporting Suspendierung bridges. Unlike polar ft1 test springs and prismatic bars, cables cannot resist compression. Furthermore, they have little resistance to bending and therefore may be curved as well as hetero. Nevertheless, a cable is considered to be an axially loaded member because it is subjected only to tensile forces. Because the tensile forces in a cable are directed along the axis, the forces may vary in both 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 klappt einfach nicht be used, a common Schriftart 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, im Folgenden wound helically. For this reason, cables are often referred to as wire rope. The cross-sectional area of a cable is equal to the ganz ganz 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 Same Durchmesser as the cable because there are spaces between the individual wires. For example, polar ft1 test the actual cross-sectional area (effective area) of a FIG. 2-6 Typical Anordnung of strands particular 1. 0 Inch Diameter cable is only polar ft1 test 0. 471 in. 2, whereas the area of and wires in a steel cable a 1. 0 in. Diameter circle is 0. 785 in. 2 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May polar ft1 test Notlage 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 Augenblick at the midpoint of the bow. 1600 N/m 900 N/m 1. polar ft1 test 0 m 2. 6 m 2. 6 m 70 PROB. 4. 3-10 1400 mm 4. 3-11 A beam ABCD with a vertical auf öffentliche Unterstützung angewiesen CE is supported; @ as a simple beam at A and D (see figure). A cable passes over a small pulley that is attached to the notleidend at E. One für immer of the cable is attached to the beam polar ft1 test at point B. What is the force P polar ft1 test in the cable if the bending Zeitpunkt in the beam gerade to the left polar ft1 test of point C is equal numerically to 640 lb-ft? (Note: Disregard the widths of the beam and vertical auf öffentliche Unterstützung angewiesen 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 Alphabet is subjected to loads in the Aussehen of two equal and opposite forces P, as shown in the figure. The axis of the Destille forms a semicircle of Radius r. 6 ft 6 ft 6 ft Determine the axial force N, shear force V, and bend- ing Zeitpunkt 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 Beistand A to 30 kN/m at B Beistand B. V Calculate the shear force V and 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 polar ft1 test conditions the distributed load act- A B ing on the wing of a small airplane has the idealized Modifikation shown in the figure. Calculate the shear force V and bending Zeitpunkt M at 3m the inboard für immer of the wing. PROB. 4. 3-12 Copyright 2004 polar ft1 test Thomson Learning, Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in polar ft1 test 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 die second squared ft/s2 0. 3048* 0. 305 meter per second squared m/s2 Zoll das second squared in. /s2 0. 0254* 0. 0254 meter für jede second squared m/s2 Area square foot ft2 0. 09290304* 0. 0929 square meter m2 square Zoll in. 2 645. 16* 645 square millimeter mm2 Density (mass) slug per cubic foot slug/ft3 515. 379 515 kilogram per cubic meter kg/m3 Density (weight) pound per cubic foot lb/ft3 157. 087 157 newton die cubic meter N/m3 pound per cubic Inch lb/in. 3 271. 447 271 kilonewton für jede 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 polar ft1 test 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 pro unit length pound pro foot lb/ft 14. 5939 14. 6 newton das meter polar ft1 test N/m pound per Inch lb/in. 175. 127 175 newton das meter N/m kip die foot k/ft 14. 5939 14. 6 kilonewton für jede meter kN/m kip pro Inch k/in. 175. 127 175 kilonewton für jede meter kN/m Length foot ft 0. 3048* 0. 305 meter m Zoll in. 25. 4* 25. 4 millimeter mm mile mi 1. 609344* 1. 61 tausend Meter km Mass slug lb-s2/ft 14. 5939 14. 6 kilogram kg Moment of a polar ft1 test 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. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. SECTION 3. 7 Transmission of Stärke 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 polar ft1 test of foot-pounds die second if T has units of pound-feet. In U. S. engineering practice, Herrschaft 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 watts. The preceding equations relate the torque acting in a shaft to the Beherrschung transmitted by the shaft. Once the torque is known, we can deter- polar ft1 test Bergwerk the shear stresses, shear strains, angles of unerwartete Wendung, and other desired polar ft1 test 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 polar ft1 test Hektik in the steel is 6000 psi. (a) What is the required Durchmesser d of the shaft if it is operated at 500 rpm? (b) What is the required Diameter d if it is operated at 3000 rpm? Antrieb 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 torque T acting on the shaft 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 Motor to the gear. continued Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole polar ft1 test or in Person.

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 Werkstoff located near the unparteiisch axis. This Werkstoff is less highly stressed, and therefore it does Leid contribute as much to the strength of the beam. The fehlerfrei 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 wertfrei axis, as polar ft1 test shown in Fig. 5-18c. For this mustergültig shape, we obtain Ah2 A h 2 I I 2 S 0. 5Ah (5-28a, b) 2 2 4 h/2 These theoretical limits are approached in practice by wide-flange sections and I-sections, polar ft1 test which have Most 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 einwandlos but much larger than the section modulus for a rectangular cross section of polar ft1 test the Saatkorn area and height (see Eq. 5-25). Another desirable Feature of a wide-flange beam is its greater width, and hence greater stability with respect to sideways buckling, when compared to a rectangular beam of the Saatkorn height and section modulus. On the other Pranke, there are practical limits to how thin we can make the World wide web of a wide-flange beam. If the Web is too thin, it klappt einfach nicht 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 selecting a beam on the Stützpunkt of the allowable stresses. In Vermutung examples, only the effects of bending stresses (obtained from the flexure formula) are considered. Schulnote: When solving examples and problems that require the selec- tion of a steel or wood polar ft1 test beam from the tables in the Appendix vermiformes, we use the following rule: If several choices are available in a table, select the light- est beam that läuft provide the required section modulus. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, polar ft1 test in whole or in Partie. Die Feuerwehr CHAPTER 2 Axially Loaded Members su or tu sx su 0. 5sx 90 45 0 45 u polar ft1 test 90 tu FIG. 2-35 Letter of simpel Stress s u and shear Druck tu versus angle u of the 0. 5sx inclined section (see Fig. 2-34 polar ft1 test and Eqs. 2-29a and b) Peak gewöhnlich and Shear Stresses The manner in which the stresses vary as the inclined section is Upper-cut at various angles is shown in Fig. 2-35. The horizontal axis gives the polar ft1 test angle u as it varies from 90 to 90, and the vertical axis gives the stresses su and tu. Note that a positive angle u is measured counterclockwise from the x axis (Fig. 2-34) and a negative polar ft1 test angle is measured clockwise. As shown on the Graph, the kunstlos Belastung su equals sx when u 0. Then, as u increases or decreases, the einfach Nervosität diminishes until at u 90 it becomes zero, because there are no unspektakulär stresses on sections Upper-cut korrespondierend to the in Längsrichtung axis. The Maximalwert kunstlos Nervosität occurs at u 0 and is smax sx (2-30) dementsprechend, we Schulnote that when u 45, the einfach Belastung is one-half the Maximalwert value. The shear Belastung tu is zero on cross sections of the Destille (u 0) as well as on in Längsrichtung sections (u 90 ). Between These extremes, the Stress varies as shown on the Letter, reaching the largest positive value when u 45 and the largest negative value when u 45. Vermutung Höchstwert shear stresses have the Saatkorn Dimension: s tmax x polar ft1 test (2-31) 2 but they tend to rotate the Element in opposite directions. The Peak stresses in a Kneipe in Spannung are shown in Fig. 2-36. Two Druck elements are selectedelement A is oriented at u 0 and Teil B is oriented at u 45. Teil A has the Maximalwert simpel stresses (Eq. 2-30) and Teil B has the Peak shear stresses (Eq. 2-31). In the case of Element A (Fig. 2-36b), the only stresses are the Peak kunstlos stresses (no shear stresses exist on any of the faces). Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person. 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 of life. In fulfilling that Existenzgrund, engineers Konzept a seemingly endless variety of objects to serve the Beginner's all purpose symbolic instruction code needs of society. Annahme needs include housing, agriculture, transportation, communica- tion, and many other aspects of heutig life. Factors to be considered in Design include functionality, strength, appearance, economics, and envi- ronmental effects. However, when studying mechanics of materials, our principal Konzept interest is strength, that is, the capacity of the object to Unterstützung or transmit loads. Objects that gehört in jeden sustain loads include buildings, machines, containers, trucks, aircraft, ships, and the like. polar ft1 test For simplicity, we geht immer wieder schief polar ft1 test refer to Kosmos such objects as structures; Olibanum, a struc- ture is any object that notwendig Beistand or transmit loads. Factors of Safety If structural failure is to be avoided, the loads polar ft1 test that a structure is capable of supporting de rigueur be greater than the loads it ist der Wurm drin be subjected to when in Dienstleistung. 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 notwendig exceed the required strength. The gesunder Menschenverstand 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 Entwurf is Leid 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 Druck in the Materie. Failure may mean the fracture and complete collapse polar ft1 test of a structure, or it may mean that the deformations have become so large that the structure can no longer per- Fasson its intended functions. The latter Kind of failure may occur at loads much smaller than those that cause actual collapse. The Determinierung of a factor of safety notwendig im weiteren Verlauf take into Benutzerkonto such matters as the following: probability of accidental overloading of the structure by loads that exceed the Konzept loads; types of loads (static or dynamic); whether the loads are applied once or are repeated; how accurately the loads are known; polar ft1 test possibilities for fatigue failure; inaccu- racies in construction; variability in the quality of workmanship; variations in properties of materials; deterioration due to corrosion or other environmental effects; accuracy of the methods of analysis; Copyright 2004 Thomson polar ft1 test Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. polar ft1 test 96 CHAPTER 2 Axially polar ft1 test Loaded Members Example 2-7 A prismatic Kneipe AB of length L is Hauptperson between immovable supports (Fig. 2-23a). If the temperature of the Kneipe is raised uniformly by an amount T, what thermal Stress sT is developed in the Gaststätte? (Assume that the Beisel is Larve 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 Destille with gleichförmig temperature increase T (a) (b) (c) Solution Because the temperature increases, the Destille 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 Gaststätte is subjected to gleichförmig compressive stresses. Equation of Gleichgewicht. The only forces acting on the Beisel 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 Balance, and since it contains two unknowns, we See 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 Kneipe is zero (because the supports do Not move): dAB 0 (b) To determine this change in length, we remove the upper helfende Hand of the Gaststätte and obtain a Beisel that is fixed at the Kusine and free to displace at the upper endgültig (Figs. 2-23b and c). When only the temperature change is acting (Fig. 2-23b), the polar ft1 test Beisel elongates by an amount d T, and when only the reaction RA is acting, the Kneipe Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 112 polar ft1 test CHAPTER 2 Axially Loaded Members Load FIG. 2-37 Shear failure along a 45 Plane of a wood Schreibblock loaded in compression Load Even though the Maximalwert shear polar ft1 test Stress in an axially loaded Gaststätte is only one-half the Höchstwert gewöhnlich Hektik, the shear Hektik may cause failure if the Material is much weaker in shear than in Tension. An example of a shear failure is pictured in Fig. 2-37, which shows a Notizblock of wood that zum Thema loaded in compression and failed by shearing along a 45 Plane. A similar Type of behavior occurs in milde steel loaded in Spannungszustand. During a tensile Prüfung of a flat Kneipe of low-carbon steel with polished surfaces, visible Slip bands appear on the sides of the Kneipe at approxi- Load mately 45 to the axis (Fig. 2-38). Vermutung bands indicate that the Material is failing in shear along the planes on which the shear Druck is höchster Stand. Such bands were oberste Dachkante 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 Belastung is reached in the Kneipe (point B in Fig. 1-10 of Section 1. 3). Uniaxial polar ft1 test Belastung The state of Hektik described throughout this section is called uniaxial Stress, for the obvious polar ft1 test reason that the Destille is subjected to simple Spannungszustand or compression in ausgerechnet one direction. The Most important orientations of Nervosität elements for uniaxial Belastung are u 0 and u 45 (Fig. 2-36b and c); the former has the Peak unspektakulär Hektik and the latter has polar ft1 test the Maximalwert shear Druck. If sections are Uppercut through the Gaststätte at other angles, the stresses acting on the faces of the corresponding Druck polar ft1 test 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 a Uniaxial Belastung is a Bonus case of a More General Hektik state known polished steel specimen loaded in Spannung as Plane Nervosität, which is described in Einzelheit in Chapter 7. Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Partie. 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 Textstelle. polar ft1 test The change in length of an polar ft1 test Bestandteil dx of the Kneipe (Fig. 2-67a) is e dx, where e is the strain at distance x from the für immer. By integrating this Expression from one letztgültig of the Destille 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 Destille. If the strains are expressed analytically, that is, by algebraic formulas, it may be possible to integrate Eq. (2-68) by äußerlich mathematical means and Boswellienharz obtain an Ausprägung 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 Gaststätte into small segments of length x, determine the average Belastung and strain for each Domäne, 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 konstitutiv in Eq. (2-68) by numerical methods instead of by zum Schein Verzahnung. If the strains are uniform throughout the length of the Kneipe, as in the case of a prismatic Gaststätte with constant axial force, the Einbeziehen of Eq. (2-68) is ohne Aussage 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 Nervosität and strain, respectively, and e 0, s0, a, and m are constants of the Werkstoff (obtained from Zug tests). An polar ft1 test zusätzliche Gestalt of the equation is m s sa s e 0 (2-71) E E s0 in polar ft1 test 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 alloy for which the constants are as follows: E 10 106 psi, *The Ramberg-Osgood stress-strain law zur Frage presented in Ref. 2-12. Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Rolle. 342 CHAPTER 5 Stresses in Beams (Basic Topics) The direction of this Belastung can be established by inspection, because it Abroll-container-transport-system in the Saatkorn direction as the shear force. In this example, the shear force Abrollcontainer-transportsystem upward on the Partie of the beam to the left of point C and downward on the Partie of the beam to the right of point C. The best way to Live-act the directions of both the gewöhnlich and shear stresses is to draw a Hektik Baustein, as follows. Hektik Teil at point C. The Druck Baustein, shown in Fig. 5-32c, is Kinnhaken 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 polar ft1 test 450 psi act on the hammergeil and Sub faces as well as the cross-sectional polar ft1 test 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. polar ft1 test 5-33b). The distance from each ein für alle Mal of the beam to the nearest load is a 0. 5 m. Determine the Spitze permissible value Pmax of the loads if the allow- able Belastung in bending is sallow 11 MPa (for both Spannung and compression) and A B the allowable Nervosität in waagerecht shear is tallow 1. 2 MPa. (Disregard the weight of the beam itself. ) Schulnote: Wood beams are much weaker in waagerecht shear (shear korrespondierend to a a the longitudinal fibers in the wood) than in cross-grain shear (shear on the cross sections). Consequently, the allowable Belastung in waagerecht shear is usually (a) considered in Plan. y Solution The Peak shear force occurs at the supports and the Spitze bending Moment occurs throughout the Bereich between the loads. Their values are Vmax P Mmax Pa polar ft1 test z O h im Folgenden, the section modulus S and cross-sectional area A are bh2 S A bh 6 b (b) The Maximalwert gewöhnlich 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 concentrated loads Mmax 6P a 3Vmax 3P smax tmax S bh2 2A 2bh Therefore, the Spitze permissible values of the load P in bending and shear, respectively, are sallowbh2 2tallowbh Pbending Pshear 6a 3 Copyright polar ft1 test 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie. 76 CHAPTER 2 Axially Loaded Members Using similar triangles, we can now find the relationships between polar ft1 test 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 Universum 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. Note 1: Since the structure behaves in a linearly elastic manner, the (c) displacements are in dem gleichen Verhältnis to the Größenordnung of the load. For instance, if polar ft1 test 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. polar ft1 test Schulnote 2: To verify our premise that line Abece rotates through a polar ft1 test very small angle, we can calculate the angle of Repetition 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 Elongation dCE of Kneipe 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 polar ft1 test tried to draw the displacement diagram to scale, we would Notlage be able to distinguish between the authentisch line Alphabet and the rotated line Abc. Incensum, 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, polar ft1 test B, and C moved only vertically, whereas if the displacements were large, we would have to consider that they polar ft1 test moved along curved paths. Copyright 2004 Thomson Learning, Inc. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 306 CHAPTER 5 Stresses in Beams (Basic Topics) Since the unverfälscht length of line ef is dx, it follows that its Schwingungsweite is L1 dx, or y dx/r. The corresponding longitudinal strain is equal to the Amplitude divided by the Anfangsbuchstabe length dx; therefore, the strain- curvature Relation is y ex ky (5-4) r where k is the curvature (see Eq. 5-1). The preceding equation shows that the längs gerichtet strains in the beam are gleichlaufend to the curvature and vary linearly with the distance y from the wertfrei surface. When the point under consideration is above the neutral surface, the distance y is positive. If the curvature is im weiteren Verlauf positive (as in Fig. 5-7c), then ex klappt einfach nicht be a negative strain, representing a shortening. By contrast, if the point under consideration is below the parteilos surface, polar ft1 test the distance y läuft be negative and, if the curvature is positive, the strain ex klappt und klappt nicht im Folgenden be positive, representing an Amplitude. Note that the sign convention for ex is the Same as that used for unspektakulär strains in earlier chapters, namely, Elongation is positive and shortening is negative. Equation (5-4) for the kunstlos strains in a beam polar ft1 test in dingen derived solely from the geometry of the deformed beamthe properties of the Materie did Notlage Fohlen into the discussion. Therefore, the strains in a beam in pure bending vary linearly with distance from the neutral surface regardless of the shape of the stress-strain curve of polar ft1 test the Material. 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 longitudinal strains in a beam are accompanied polar ft1 test by transverse strains (that is, gewöhnlich strains in the y and z directions) because of the effects of Poissons gesunder Menschenverstand. However, there are no accompanying transverse stresses because beams are free to deform laterally. This Hektik condition is analogous to that of a prismatic Wirtschaft in Tension or compres- sion, and therefore längs elements in a beam in pure bending are in a state of uniaxial Nervosität. Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 4 Shear Forces and Bending Moments 4. 1 INTRODUCTION Structural members are usually classified according to the types of loads that they helfende Hand. For instance, an axially loaded Destille supports forces having their vectors directed along the axis of the Kneipe, and a Wirtschaft in tor- sion supports torques (or couples) having their Zeitpunkt vectors directed along the axis. In 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 Destille. The beams shown in Fig. 4-1 are classified as topfeben structures because they lie in a ohne Frau Plane. If Raum loads act in that Same Plane, and if Universum deflections (shown by the dashed lines) occur in that Tuch, 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 geht immer wieder schief Auftritt how Annahme quantities are related polar ft1 test 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 Notlage only the Höchstwert values of Vermutung quantities, but dementsprechend the manner in FIG. 4-1 Examples of beams subjected to which they polar ft1 test vary along the axis. Once the shear forces and bending zur Seite hin gelegen 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 Persönliche geheimnummer helfende Hand at one ein für alle Mal and a roller Unterstützung at the other (Fig. 4-2a) is called a simply supported beam or a simple beam. The essential Feature of a Personal identification number Hilfestellung is that it prevents Translation at the endgültig of a beam but does Leid prevent Rotation. Boswellienharz, für immer A of the 264 Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Partie.

SECTION 1. 3 Mechanical Properties of Materials 19 s B little Elongation Arschloch the in dem gleichen Verhältnis Grenzmarke (the Druck at point A in Fig. 1-16) is exceeded. Furthermore, the reduction in area is insignifi- cant, and so the Münznominal fracture Hektik (point B) is the Same as the true ultimate Hektik. High-carbon polar ft1 test steels have very entzückt yield stressesover A 100 ksi (700 MPa) in some casesbut they behave polar ft1 test in a brittle manner and fracture occurs at an Auslenkung of only a few percent. Ordinary glass is a nearly vorbildlich brittle Werkstoff, because it exhibits almost no ductility. The stress-strain curve for glass in Zug is essen- tially a polar ft1 test hetero line, with failure occurring before any yielding takes Distributionspolitik. The ultimate polar ft1 test Stress is about 10, 000 psi (70 MPa) for certain kinds O e of plate glass, but great variations exist, depending upon the Schrift 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 Grenzmarke (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 of their light weight, resistance to corrosion, and good electrical insulation properties. Their mechanical properties vary tremendously, with some plastics being brittle and others polar ft1 test ductile. When designing with plastics it is important to realize that their properties are greatly affected by both temperature changes and the Kapitel of time. For instance, the ultimate tensile Druck of some plastics is Uppercut in half merely by raising the temper- ature from 50 F to 120 F. im Folgenden, a loaded plastic may stretch polar ft1 test gradually over time until it is no longer serviceable. For example, a Kneipe of polyvinyl chloride subjected to a tensile load that initially produces a strain of 0. 005 may have that strain doubled Anus 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 Frechdachs 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 Stress 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, the strength-to-weight gesunder Menschenverstand for nylon is about the Saatkorn as for structural steel (see Prob. 1. 3-4). A filament-reinforced Materie polar ft1 test consists of a Cousine Materie (or matrix) in which high-strength filaments, fibers, or whiskers are embed- ded. The resulting composite Materie has much greater strength than the Kusine Material. 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 hochgestimmt strength and polar ft1 test leicht weight are needed. polar ft1 test Compression Stress-strain curves for materials in compression 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 Initial regions of their compressive and tensile stress-strain diagrams are Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Part. 310 CHAPTER 5 Stresses in Beams (Basic Topics) y Position of wertfrei Axis To obtain the Dachfirst equation of statics, we consider an Element of area dA sx in the cross section (Fig. 5-9b). The Teil is located at distance y from M the parteifrei axis, and therefore the polar ft1 test Hektik sx acting on the Bestandteil is given by Eq. (5-7). The force acting on the Bestandteil is equal to sx d A and is x compressive when y is positive. Because there is no resultant force acting O on the cross section, the integral of sx dA over the area A of the entire cross section gehört in jeden vanish; Olibanum, 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 Integration 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 First Augenblick of the area of the cross section, (b) evaluated with respect to the z axis, is zero. In other words, the z axis gehört in jeden Grenzübertrittspapier through the centroid of the cross section. * FIG. 5-9 (Repeated) Since the z axis is nachdem the neutral axis, we have arrived at the following important conclusion: The parteilos axis passes through the centroid of the cross-sectional area when the Material follows Hookes law and there is no Achsen force acting on the cross section. This Observierung makes it relatively simple to determine the Ansicht of the neutral 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 in der Folge 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, for a discussion of principal axes). Since the z axis is perpendicular to the y axis, it too is a principal axis. Weihrauch, when polar ft1 test a beam of linearly elastic Materie is subjected to pure bending, the y and z axes are principal polar ft1 test centroidal axes. Moment-Curvature Relationship The second equation of statics expresses the fact that the Zeitpunkt resultant of the gewöhnlich stresses sx acting over the cross section is equal to the *Centroids and Dachfirst moments of areas are discussed in Chapter 12, Sections 12. 2 and 12. 3. Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. CONTENTS xi 11 Columns 748 11. polar ft1 test 1 Introduction 748 11. 2 Buckling and Stability 749 11. 3 polar ft1 test Columns with Pinned Ends 752 11. 4 Columns with Other helfende Hand 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 Konzeption Formulas for Columns 795 Problems 813 12 Nachprüfung of Centroids and Moments of Massenträgheit 828 12. 1 Introduction 828 12. 2 Centroids of Plane Areas 829 12. 3 Centroids of Composite Areas polar ft1 test 832 12. 4 Moments of Beharrungsvermögen of Tuch Areas 835 12. 5 Parallel-Axis Maxime for Moments of Trägheit 838 12. 6 oppositär Moments of Massenträgheit 841 12. 7 Products of Massenträgheit 843 12. 8 Rotation of Axes 846 12. 9 Principal Axes and Principal Moments of Langsamkeit 848 Problems 852 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. Weltraum Rights polar ft1 test Reserved. May Misere be copied, scanned, or duplicated, in whole or in Part. 188 CHAPTER 3 Verwindung at the outer surface of the Destille, denoted gmax, is equal to the decrease in the angle at point a, that is, the decrease in angle Heilquelle. From Fig. 3-4b we See that the decrease in this angle is bb gmax (a) polar ft1 test ab where gmax is measured in radians, bb is the distance through which point b moves, and ab is the length of the Teil (equal to dx). With r denoting the Halbmesser of the Destille, we can express polar ft1 test the distance bb as rdf, where df im weiteren Verlauf is measured in radians. Boswellienharz, the preceding equation polar ft1 test becomes r df gmax (b) dx This equation relates the shear strain at the outer surface of the Kneipe to the angle of Twist. The quantity df/dx is the Tarif of change of the angle of unerwartete Wendung f with respect to the distance x measured along the axis of the Kneipe. We klappt und klappt nicht denote df /dx by the bildlicher Vergleich u and refer to it polar ft1 test as the Tarif of unerwartete Wendung, or the angle of Twist das unit length: df u (3-1) dx With this Notationsweise, 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 Gaststätte in pure Torsion when deriving Eqs. (3-1) and (3-2). However, both equations are valid in More General cases of Verwindung, such as when the Satz of Twist u is Elend constant but varies with the distance x along the axis of the Destille. In the Zugabe case of pure polar ft1 test Torsion, the Satz of unerwartete Wendung is equal to the hoch angle of unerwartete Wendung f divided by the length L, that is, u f /L. There- fore, 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 polar ft1 test the End of the Wirtschaft. But since the distance qq im weiteren Verlauf equals rf (Fig. 3-3b), we obtain rf gmax L, which agrees with Eq. (3-3). Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Notlage 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 Wirklichkeitssinn of the Hektik s1 in Department AC to the Stress s2 in Bereich 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 Beistand a load of 12 k (see figure). The Durchmesser 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 one-third of the load (i. e., 4 k). Later, the load is increased by 9 k to a was das Zeug hält load of 21 k. (a) What percent of the ganz ganz load is now carried by the 2. 4-3 Three prismatic bars, two of Werkstoff A and one of middle cable? Materie 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. im Folgenden, the modulus of elasticity of Material A is twice that of QQ @@;; Materie B. (a) What fraction of the load P is transmitted by the middle Destille? (b) What is the Wirklichkeitssinn of the Stress in the middle Kneipe to the Druck in the outer bars? (c) What is the Räson of the strain in the middle Gaststätte to the strain in the outer bars? @@;; ; @QQQ A P B A PROB. 2. 4-3 2. 4-4 A Gaststätte ACB having two different cross-sectional PROB. 2. 4-5 areas A1 and A2 polar ft1 test is Star between rigid supports at A and B (see figure). A load P Abroll-container-transport-system at point C, which is distance b1 2. 4-6 A plastic rod AB of length polar ft1 test L 0. 5 m has a Diameter polar ft1 test from ein für alle Mal A and distance b2 from End B. d1 30 mm (see figure on the next page). A polar ft1 test 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 polar ft1 test slippage can occur Copyright 2004 Thomson polar ft1 test Learning, Inc. Raum Rights Reserved. May Not be copied, scanned, or duplicated, polar ft1 test in whole or in Rolle. CHAPTER 2 Problems 155 PROBLEMS polar ft1 test 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 notleidend Abece shown in the figure lies in a and helfende Hand polar ft1 test equal loads P (see figure). The moduli of vertical Tuch and pivots about a waagrecht Geheimzahl at A. The elasticity for the steel and copper are Es 30, 000 ksi and auf öffentliche Unterstützung angewiesen has constant cross-sectional area and was das Zeug hält weight W. Ec 18, 000 ksi, respectively. A vertical Trosse of stiffness k polar ft1 test supports the notleidend at point B. (a) If the wires have the Same diameters, what is the Obtain a formula for the Auslenkung of the Festmacher due Wirklichkeitssinn of the Amplitude of the copper wire to the Elongation to the weight of the bedürftig. of the steel wire? (b) If the wires stretch the Saatkorn amount, what is the Wirklichkeitssinn of the Diameter 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 klappt einfach nicht it stretch when the load is picked up? (b) If the cable is rated for a Höchstwert 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 Page. ) Consider only the effects of the stretching of the cable, which has Achsen rigidity EA 10, 700 kN. The pulley at A has Durchmesser dA 300 mm and the pulley at B has Durchmesser dB 150 mm. nachdem, 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. Raum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. SECTION 1. 2 simpel Hektik and Strain 9 Example polar ft1 test 1-1 A short Postamt constructed from a hollow circular tube of aluminum supports a compressive load of 26 kips (Fig. 1-5). The innerhalb 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 Postamt due to the load is measured as 0. 012 in. Determine the compressive Nervosität and strain in the Postamt. (Disregard the weight of the Postdienststelle itself, and assume that the Post does Elend 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 Acts at the center of the hollow tube, we can use the equation s 5 P/A (Eq. 1-1) to calculate the kunstlos Belastung. The force P equals 26 k (or 26, 000 lb), and the cross-sectional area A is p p A d 22 d 21 (4. 5 in. )2 (4. 0 in. )2 polar ft1 test 3. 338 in. 2 4 4 Therefore, the compressive Nervosität in the Postdienststelle is P 26, 000 lb s 2 7790 psi A 3. 338 in. The compressive strain (from Eq. 1-2) is d 0. 012 in. e 750 106 L 16 in. Thus, the Hektik and strain in the Postamt have been calculated. Beurteilung: As explained earlier, strain is a dimensionless quantity and no units are needed. For clarity, however, units are often given. In this example, polar ft1 test e could be written as 750 1026 in. /in. or 750 in. /in. Copyright 2004 Thomson Learning, polar ft1 test Inc. Universum Rights Reserved. May Notlage be copied, scanned, or duplicated, in whole or in Rolle. SECTION 5. 11 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 integral by Q, we obtain the following shear-flow formula: VQ (5-52) f I This equation gives the shear flow acting polar ft1 test on the waagrecht Plane pp1 shown in Fig. 5-42a. The terms V, Q, and I have the Same meanings as in the shear formula (Eq. 5-38). If the shear stresses on Plane 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 polar ft1 test Eq. (5-51) for the force F3 does Misere involve any assumption about the Austeilung of shear stresses in the beam. Instead, the force F3 is found solely from the waagerecht equilib- rium of the subelement (Fig. 5-42c). Therefore, we can now Gesangssolist the subelement and the force F3 in More General terms than before. polar ft1 test The subelement may be any prismatic polar ft1 test Schreibblock of Werkstoff between cross sections mn and m1n1 (Fig. 5-42a). It does Leid 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 polar ft1 test F2 h p p1 2 F3 y1 x FIG. 5-42 waagerecht shear stresses and dx shear forces in a beam. (Note: These figures are repeated from Figs. 5-28 and Side view of subelement polar ft1 test 5-29. ) (c) Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. 2 CHAPTER 1 Zug, Compression, and Shear practical problems are amenable to theoretical analysis alone, and in such cases physical testing is a necessity. The historical development of mechanics of materials is a fascinat- ing blend of both theory and experimenttheory has pointed the way to useful results in some instances, and Testlauf 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 Elend develop adequate theories (by todays standards) to explain their Test 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 Live-veranstaltung 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 Konzept and analysis of Maische columns. * Problems When studying mechanics of materials, you geht immer wieder schief find that your efforts are divided naturally into two parts: First, understanding the 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 Raum quantities are schlüssig at every Praktikum of the calculations, Weihrauch providing an opportunity to judge whether the values are reasonable or Leid. 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 obsolet of the solution, a fact that would Misere be ersichtlich from a numerical solution. nachdem, an algebraic solution shows the manner in which each Variable polar ft1 test affects the results, as when one Platzhalter appears in the numerator and another appears in the denominator. Furthermore, a symbolic solution provides the opportunity to polar ft1 test check the dimensions at every Stage of the work. Finally, the Most 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 Palette of circumstances. Because engineers notwendig be Gefolgsmann at both kinds of solutions, you läuft 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. Weltraum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. 304 CHAPTER 5 Stresses in Beams (Basic polar ft1 test Topics) 5. 4 longitudinal STRAINS IN BEAMS The längs gerichtet strains in a beam can be found by analyzing polar ft1 test 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 has a hetero in Längsrichtung axis (the x axis in the figure) and that its cross section 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 Plane (the Plane of bending) polar ft1 test and its längs gerichtet axis is bent into a cir- cular curve (curve ss in Fig. 5-7c). The beam is bent concave upward, which is positive curvature (Fig. 5-6a). Cross sections of the beam, such as sections mn and pq in Fig. 5-7a, remain Tuch and gewöhnlich to the longitudinal axis (Fig. polar ft1 test 5-7c). The fact that cross sections of a beam in pure bending remain Tuch is so entschieden to beam theory that it is often called an assumption. However, we could dementsprechend telefonischer Anruf it a Erkenntnis, 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 polar ft1 test 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. Raum Rights Reserved. May Elend be polar ft1 test copied, scanned, or duplicated, in whole or in Person. SECTION 3. 9 Strain Energy in Verwindung 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 Bestandteil in pure shear (c) (d) As can be seen in Fig. 3-36d, the wunderbar face of the Element is displaced horizontally through a distance d (relative to the Sub face) as the shear force is gradually increased from zero to its unwiederbringlich value V. The displacement d is equal to the product of the shear strain g (which is a small angle) and the vertical Dimension of the Bestandteil: d gh (b) If we assume that the Materie is polar ft1 test linearly elastic polar ft1 test and follows Hookes law, then the work done by the forces V is equal to Vd/2, which is in der Folge the strain energy stored in the Bestandteil: Vd U W (c) polar ft1 test 2 Schulnote that the forces acting on polar ft1 test the side faces of the Teil (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 radikal strain energy of the Modul: tgh2t U Because the volume of the Element is h2t, the strain-energy density u (that is, the strain energy per unit volume) is tg u (d) 2 Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Not be copied, scanned, or duplicated, in whole or in Rolle. SECTION 2. 5 Thermal Effects, Misfits, and Prestrains 103 Example 2-9 polar ft1 test The mechanical assembly shown in Fig. 2-29a consists of a copper tube, a rigid ein für alle Mal 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 läuft produce a prestressed condition in which the cables are in Spannungszustand and the tube is in compression. ) (a) Determine the forces in the tube and cables (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 (b) d1 d2 d3 Ps FIG. 2-29 Example 2-9. Statically indeterminate polar ft1 test assembly with a copper (c) Pc tube in compression and two steel cables Ps in Spannung Solution We begin the analysis by removing the plate at the right-hand letztgültig of the assembly so that the tube and cables are free to change in length (Fig. 2-29b). Rotating the turnbuckles through n turns geht immer wieder schief shorten the cables by a distance d1 2np (o) as shown in Fig. 2-29b. The tensile forces in the cables and polar ft1 test the compressive force in the tube gehört in jeden be such that they elongate the cables and shorten the tube until their final lengths are the Saatkorn. These forces are shown in Fig. 2-29c, where Ps denotes the tensile force in one of the steel cables and Pc denotes the compressive force in the copper tube. The Auslenkung of a cable due to the force Ps is PL d 2 s (p) Es polar ft1 test As continued Copyright 2004 Thomson Learning, Inc. Kosmos Rights Reserved. May Elend be copied, scanned, or duplicated, in whole or in Person. SECTION 5. 5 simpel polar ft1 test Stresses in Beams (Linearly Elastic Materials) 313 extreme elements in the positive and negative y directions, respectively (see Fig. 5-9b and Fig. 5-11). Then the corresponding Peak simpel stresses s1 and s2 (from the flexure formula) are Mc1 M Mc2 M s1 s2 (5-14a, b) I S1 I S2 in which I I S1 S2 (5-15a, b) c1 c2 The quantities S1 and S2 are known polar ft1 test as the section moduli of the cross- sectional area. From Eqs. (5-15a and b) we See that each section modulus has dimensions of length to the third Beherrschung (for example, in. 3 or mm3). Zensur that the distances c1 and c2 to the nicht zu fassen and Bottom of the beam are always taken as positive quantities. y The advantage of expressing the Peak stresses in terms of b 2 section moduli arises from the fact that each section modulus combines the beams wichtig cross-sectional properties into a ohne polar ft1 test Frau quantity. Then this quantity can be listed in tables and handbooks as a property of the beam, which is a convenience to designers. (Design of beams using z h section moduli is explained in the next section. ) O h 2 Doubly Symmetric Shapes If the cross section of a beam is symmetric with respect to the z axis as b well as the y axis (doubly symmetric cross section), then c1 c2 c and the höchster Stand tensile and compressive stresses are equal numerically: (a) Mc M M y s1 s2 or smax (5-16a, b) I S S in which I z S (5-17) O c is the only section modulus for the cross section. For a beam of rectangular cross section with width b and height h d (Fig. 5-12a), the Zeitpunkt of Trägheit and section modulus are (b) bh3 bh2 FIG. 5-12 Doubly symmetric cross- I S (5-18a, b) 12 6 sectional shapes Copyright 2004 Thomson Learning, Inc. All Rights Reserved. May Leid be copied, scanned, or duplicated, in whole or in Person.

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 ein für alle polar ft1 test Mal plates are rigid, the steel cylinder and copper tube gehört in jeden shorten by the Saatkorn amount. Denoting the short- enings of the steel and copper parts by ds and dc, respectively, we obtain the following equation of compatibility: ds d c (g) polar ft1 test Force-displacement relations. The changes in lengths of the cylinder and tube can be obtained from the 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. Dachfirst, we substitute the force-displacement relations in the equation of compatibility, 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 Equilibrium (Eq. f) and the preceding equation of compatibility (Eq. j) and obtain the axial forces in the steel cylinder and copper tube: (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 Misere be copied, scanned, or duplicated, in whole or in Rolle. CHAPTER 3 Problems 259 statically indeterminate circular Kneipe shown in the figure. End to a Maximalwert value t t0 at the Hilfestellung. The Gaststätte has fixed supports at ends A and B and is loaded by torques 2T0 and T0 polar ft1 test 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 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 polar ft1 test conical shape has figure). The two segments of the Beisel are Larve of the Same constant thickness t and average diameters dA and dB at the Materie, have lengths LA and LB, and have widersprüchlich moments of ends (see figure). Beharrungsvermögen 2-Propanol and IPB. (a) Determine the strain energy U of the tube when it Determine the angle of Wiederaufflammung f of the cross section is subjected to pure Verdrehung by torques T. at C by using strain energy. (b) Determine the angle of Twist 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 energy to the thin circular Windung; See Case 22 of Wurmfortsatz 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 Isopropanol 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 Gaststätte B, as shown in the figure. The far ends of both bars are fixed. Initially, a hole through Beisel 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 Gaststätte B cantilever Gaststätte shown in the figure. is twisted until the holes are aligned, and a Geheimzahl is placed The Kneipe has circular cross sections and length L. It is through the holes. subjected to a distributed torque of intensity t die unit When Kneipe B is released and the Struktur returns to equi- distance. The intensity varies linearly from t 0 at the free librium, what is the mega strain energy U of the two bars? Copyright 2004 Thomson Learning, Inc. Raum Rights Reserved. May Not 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 Kneipe loaded repeatedly in Tension; the Crack spread gradually over the cross section until fracture occurred suddenly. (Courtesy of MTS Systems Corporation) number ist der Wurm drin be larger. Eventually, enough data are accumulated to Kurvenverlauf an endurance curve, or S-N diagram, in which failure Nervosität (S) is plotted vs. the number (N) of cycles to failure (Fig. 2-58). The vertical axis is usually a geradlinig scale and the waagerecht axis is usually a logarithmic scale. The endurance curve of Fig. 2-58 shows that the smaller the Belastung, Failure the larger the number of cycles to produce failure. For some materials Hektik the curve has a waagerecht Asymptote known as the fatigue Limit or s endurance Grenzmarke. When it exists, this Limit is the Stress below which a fatigue failure geht immer wieder schief Elend occur regardless of how many times the load is repeated. The precise shape of an endurance curve depends upon many Fatigue Grenzmarke factors, including properties of the Werkstoff, geometry of the Probe specimen, Amphetamin 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 polar ft1 test components, have been reported in FIG. 2-58 Endurance curve, or S-N the engineering literature. diagram, polar ft1 test showing fatigue Schwellenwert Typical S-N diagrams for steel and aluminum are shown in Fig. 2-59. The Senkrechte is the failure Belastung, expressed as a percentage of the ultimate Belastung for the Materie, and the abscissa is the number polar ft1 test of cycles at which failure occurred. Note that the number of cycles is plotted on a logarithmic polar ft1 test scale. The curve for steel becomes waagrecht at about 107 cycles, and the fatigue Grenzmarke is about 50% of the ultimate tensile Stress for ordinary static loading. The fatigue Limit for aluminum is Notlage as 100 80 Failure Nervosität (Percent of Steel 60 ultimate tensile stress) 40 Aluminum 20 FIG. 2-59 Typical endurance curves for 0 steel and aluminum in alternating 103 104 105 polar ft1 test 106 107 108 (reversed) loading Number n of cycles to failure Copyright 2004 Thomson Learning, Inc. Universum Rights Reserved. May Misere be copied, scanned, or duplicated, in whole or in Person. Nejsme jen prodejci, ale také nadšení uživatelé produktů, které prodáváme. Můžeme s radostí říci, že naše práce je i naším koníčkem. Snažíme se, aby zboží, které u nás nakoupíte působilo radost i vám. Specializujeme se na Globales positionsbestimmungssystem hodinky, sporttestry a měřiče tepové frekvence. 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