111.2EQUILIBRIUMOFADEFORMABLEBODYEXAMPLE1.1Determine the resultant internal loadings acting on the cross section at C ofthe cantilevered beam shown in Fig.1-4a.270N/mALC6m(a)Fig.1-4SOLUTIONSupport Reactions. The support reactions at A do not have to bedeterminedifsegmentCBisconsideredFree-BodyDiagram.Thefree-bodydiagram of segment CB is showninFig.1-4b.Itisimportanttokeepthedistributedloadingonthesegment540Nuntil after the section is made.Only then should this loading be replaced180N/mby a single resultantforce.Notice that theintensity ofthe distributedMloading at Cis found by proportion, ie., from Fig.1-4a,w/6m=(270N/m)/9m,w=180N/m.ThemagnitudeoftheresultantNof the distributed load is equal to the area under the loading curveAT(triangle)andacts through thecentroid of this area.Thus,F=(180N/m)(6m)=540N,whichacts(6m)=2mfromCas(b)shown in Fig.1-4b.Equations of Equilibrium.Applyingthe equations of equilibrium wehave±2=0-Nc=0Nc=0Ans.+EF,=0:Vc-540N=0Vc= 540NAns.540N135N(+ZMc = 0;-Mc-540N(2m)= 090N/ml180N/mAns.Mc=-1080N·mM.1215NThe negative sign indicates that Mc acts in the opposite direction tothat shown on the free-body diagram.Try solving this problem using-1.5m-3645N-mlimCo.5msegment AC,byfirst checkingthe support reactions atA,which aregiven(c)inFig.1-4c
1.2 Equilibrium of a Deformable Body 11 1 EXAMPLE 1.1 Determine the resultant internal loadings acting on the cross section at C of the cantilevered beam shown in Fig. 1–4a. SOLUTION Support Reactions. The support reactions at A do not have to be determined if segment CB is considered. Free-Body Diagram. The free-body diagram of segment CB is shown in Fig. 1–4b. It is important to keep the distributed loading on the segment until after the section is made. Only then should this loading be replaced by a single resultant force. Notice that the intensity of the distributed loading at C is found by proportion, i.e., from Fig. 1–4a, w>6 m = (270 N>m)>9 m, w = 180 N>m. The magnitude of the resultant of the distributed load is equal to the area under the loading curve (triangle) and acts through the centroid of this area. Thus, F = 1 2(180 N>m)(6 m) = 540 N, which acts 1 3(6 m) = 2 m from C as shown in Fig. 1–4b. Equations of Equilibrium. Applying the equations of equilibrium we have S + ΣFx = 0; -NC = 0 NC = 0 Ans. + cΣFy = 0; VC - 540 N = 0 VC = 540 N Ans. a+ΣMC = 0; -MC - 540 N(2 m) = 0 MC = -1080 N # m Ans. The negative sign indicates that MC acts in the opposite direction to that shown on the free-body diagram. Try solving this problem using segment AC, by first checking the support reactions at A, which are given in Fig. 1–4c. (a) A B C 3 m 6 m 270 N/m 2 m 4 m (b) C B 540 N 180 N/m VC MC NC VC MC NC 1.5 m 0.5 m 1 m 90 N/m 540 N 135 N (c) 1215 N 3645 N m A C 180 N/m Fig. 1–4
12CHAPTER 1STRESSEXAMPLE1.2The500-kgengine is suspended from the craneboominFig.1-5aDetermine the resultant internal loadings acting on the cross section ofthe boom at point E.1.5mSOLUTIONAE-m--m--1mSupport Reactions.We will consider segment AE of the boom, so wemust first determine the pin reactions at A.Since member CD is atwo-force member,it acts like a cable,and therefore exerts a force FcDhavingaknowndirection.Thefree-bodydiagramof theboomisshowninFig.1-5b.Applyingtheequations of equilibrium,(a)(+EMA = 0;Fcp()(2m) - [500(9.81) NJ(3m)= 0FcDFcD=12262.5N±=0Ar- (12262.5N)() = 02mA=9810NA,500(9.81) N+F=0-A,+(12262.5N)()-500(9.81)N=0(b)A,=2452.5NFree-BodyDiagram.Thefree-bodydiagram of segmentAEis showninFig.1-5c.Equations of Equilibrium.±=0MENE+9810N=09810NN,AANg=-9810N=-9.81kNAns.1m+EF = 0;-VE-2452.5N=02452.5 NVE=-2452.5N=-2.45kNAns.(c)Fig.1-5(+ME=0;Me + (2452.5N)(1m) = 0Ans.ME=-2452.5N·m=-2.45kN·m
12 Chapter 1 Stress 1 EXAMPLE 1.2 The 500-kg engine is suspended from the crane boom in Fig. 1–5a. Determine the resultant internal loadings acting on the cross section of the boom at point E. SOLUTION Support Reactions. We will consider segment AE of the boom, so we must first determine the pin reactions at A. Since member CD is a two-force member, it acts like a cable, and therefore exerts a force FCD having a known direction. The free-body diagram of the boom is shown in Fig. 1–5b. Applying the equations of equilibrium, a+ΣMA = 0; FCD 1 3 5 2(2 m) - [500(9.81) N](3 m) = 0 FCD = 12 262.5 N S+ ΣFx = 0; Ax - (12 262.5N)1 4 5 2 = 0 Ax = 9810 N + cΣFy = 0; -Ay + (12 262.5 N)1 3 5 2 - 500(9.81) N = 0 Ay = 2452.5 N Free-Body Diagram. The free-body diagram of segment AE is shown in Fig. 1–5c. Equations of Equilibrium. S+ ΣFx = 0; NE + 9810 N = 0 NE = -9810 N = -9.81 kN Ans. + cΣFy = 0; -VE - 2452.5 N = 0 VE = -2452.5 N = -2.45 kN Ans. a+ΣME = 0; ME + (2452.5N)( 1 m) = 0 ME = -2452.5 N # m = -2.45 kN # m Ans. A 1 m 1 m 1 m 1.5 m E C B D (a) A 2 m 1 m 500(9.81) N (b) 3 4 5 Ay Ax FCD 9810 N 2452.5 N VE ME NE (c) A E 1 m Fig. 1–5
131.2EQUILIBRIUMOFADEFORMABLEBODYEXAMPLE1.3Determine theresultant internal loadings acting on the cross section at CofthebeamshowninFig.1-6a.900 1b-8ft2ft-300 Ib/ftAO5A30309FBDA,(a)(b)Fig.1-6SOLUTIONSupport Reactions.Here we will consider segment BC,but first wemustfind the force components atpin A.Thefree-bodydiagram of theentirebeamisshowninFig.1-6b.SincememberBDisatwo-forcemember,like member CD in Example 1.2, the force at B has a known direction,Fig.1-6b.We have(+ZMA=0;(900b)(2ft)-(FBDsin30°)10ft=0FBD=360lbFree-Body Diagram.Using this result, the free-body diagram ofsegment BCis shown inFig.1-6c.Equations of Equilibrium.M±F=0Nc-(360lb)cos30°=0NB/30°Nc=312lbAns.Vo360 1b+IZF, =0;(360lb)sin30°-Vc=0(c)Vc=180lbAns.(+ZMc=0;Mc-(360lb)sin30(2ft)=0Ans.Mc=360lb-ft
1.2 Equilibrium of a Deformable Body 13 1 EXAMPLE 1.3 Determine the resultant internal loadings acting on the cross section at C of the beam shown in Fig. 1–6a. SOLUTION Support Reactions. Here we will consider segment BC, but first we must find the force components at pin A. The free-body diagram of the entire beam is shown in Fig. 1–6b. Since member BD is a two-force member, like member CD in Example 1.2, the force at B has a known direction, Fig. 1–6b. We have a+ΣMA = 0; (900 lb)(2 ft) - (FBD sin 30°) 10 ft = 0 FBD = 360 lb Free-Body Diagram. Using this result, the free-body diagram of segment BC is shown in Fig. 1–6c. Equations of Equilibrium. S+ ΣFx = 0; NC - (360 lb) cos 30° = 0 NC = 312 lb Ans. + cΣFy = 0; (360 lb) sin 30° - VC = 0 VC = 180 lb Ans. a+ΣMC = 0; MC - (360 lb) sin 30°(2 ft) = 0 MC = 360 lb # ft Ans. 30 2 ft 6 ft 300 lb/ft (a) 2 ft B A C D (b) 2 ft 900 lb 8 ft A 30 Ax FBD Ay 2 ft (c) B 360 lb VC MC 30 NC Fig. 1–6
14CHAPTER 1STRESSEXAMPLE1.4Determine the resultant internal loadings acting on thecross section at B ofthe pipe shown in Fig.1-7a.End A is subjected to a vertical force of 50 N,ahorizontal force of 30 N, and a couple moment of 70 N·m.Neglect thepipe's mass.0.75mSOLUTION05mRDTheproblem can be solved byconsidering segmentAB,sowedo notneed50NtocalculatethesupportreactionsatC.1.25mFree-BodyDiagram.Thefree-bodydiagram of segmentABis shown in30NAFig.1-7b,wherethex,yzaxesareestablishedatB.Theresultantforceand70N-mmoment components at the section areassumed to act in thepositive(a)coordinatedirections andtopass through thecentroidofthecross-sectionalarea at B.Equations of Equilibrium. Applying the six scalar equations ofequilibrium,wehave*ZF =0;(FB) = 0Ans.Easn,ma,fas.(FB),=-30NEF,=0;(FB)+30N=0Ans.Ca05m(FB)-50N=0(FB),=50NAns.ZF,=0;(Ms)B(FB)r(MB) = 0:(MB)+70N·m-50N(0.5m)=0SON1.25mAns(MB)=-45N·m30NOAZ(MB) = 0;70N-m(M),+50N(1.25m)=0(b)Ans.(Mb),=-62.5N·mFig.1-7Z(MB)= 0;Ans.(MB)+(30N)(1.25)=0(Mg),=-37.5N·mNOTE:Whatdothenegativesignsfor (FB)(Mg)x,(Ms)y,and (MB)indicate?Thenormalforce Ng=I(F)/=30N,whereasthe shearforce is Vg = V(o)? + (50)? = 50 N. Also, the torsional momentisTg=I(Mg)/=62.5N·m,andthebendingmomentisMg=V(45)2+ (37.5)2=58.6N·m.*The magnitude of each moment about the X, y,or z axis is equal to themagnitude ofeach force times the perpendicular distance from the axis to the line of action of the forceThe direction of each moment is determined using the right-hand rule, with positivemoments (thumb) directed along the positive coordinate axes
14 Chapter 1 Stress 1 EXAMPLE 1.4 *The magnitude of each moment about the x, y, or z axis is equal to the magnitude of each force times the perpendicular distance from the axis to the line of action of the force. The direction of each moment is determined using the right-hand rule, with positive moments (thumb) directed along the positive coordinate axes. Determine the resultant internal loadings acting on the cross section at B of the pipe shown in Fig. 1–7a. End A is subjected to a vertical force of 50 N, a horizontal force of 30 N, and a couple moment of 70 N # m. Neglect the pipe’s mass. SOLUTION The problem can be solved by considering segment AB, so we do not need to calculate the support reactions at C. Free-Body Diagram. The free-body diagram of segment AB is shown in Fig. 1–7b, where the x, y, z axes are established at B. The resultant force and moment components at the section are assumed to act in the positive coordinate directions and to pass through the centroid of the cross-sectional area at B. Equations of Equilibrium. Applying the six scalar equations of equilibrium, we have* ΣFx = 0; (FB)x = 0 Ans. ΣFy = 0; (FB)y + 30 N = 0 (FB)y = -30 N Ans. ΣFz = 0; (FB)z - 50 N = 0 (FB)z = 50 N Ans. Σ(MB)x = 0; (MB)x + 70 N # m - 50 N (0.5 m) = 0 (MB)x = -45 N # m Ans. Σ(MB)y = 0; (MB)y + 50 N (1.25 m) = 0 (MB)y = -62.5 N # m Ans. Σ(MB)z = 0; (MB)z + (30 N)(1.25) = 0 Ans. (MB)z = -37.5 N # m NOTE: What do the negative signs for (FB)y, (MB)x, (MB)y, and (MB)z indicate? The normal force NB = (FB)y = 30 N, whereas the shear force is VB = 2(0)2 + (50)2 = 50 N. Also, the torsional moment is TB = (MB)y = 62.5 N # m, and the bending moment is MB = 2(45)2 + (37.5)2 = 58.6 N # m. 0.75 m 50 N 1.25 m B A 0.5 m C D 70 N m (a) 30 N 1.25 m 70 N·m 30 N (b) y A 50 N 0.5 m x z B (FB)z (MB)z (MB)x (FB)x (MB)y (FB)y Fig. 1–7
151.2EQUILIBRIUMOFADEFORMABLEBODYItissuggestedthatyoutestyourselfonthesolutionstotheseexamples, bycoveringthemoverand thentryingtothink aboutwhichequilibrium equationsmustbeusedandhowtheyareapplied inorderto determinetheunknowns.Thenbeforesolvinganyoftheproblems,buildyourskillsbyfirsttryingto solvethePreliminaryProblems,whichactuallyrequirelittleornocalculations,andthendosomeoftheFundamentalProblemsgivenonthefollowingpages.Thesolutionsandanswerstoalltheseproblemsaregiveninthebackofthebook.Doingthis throughout the book will help immensely in understanding how to apply the theory, and thereby developyourproblem-solving skills.PRELIMINARYPROBLEMSPi-1.In each case,explainhowtofindthe resultantinternalloadingacting on the cross section atpointA.Draw allnecessary free-bodydiagrams, and indicate therelevantequationsofequilibrium.Donotcalculatevalues.The lettereddimensions,angles,and loadsare assumed tobeknown.(d)P(a)A(e)(b).(c)(f)
1.2 Equilibrium of a Deformable Body 15 1 It is suggested that you test yourself on the solutions to these examples, by covering them over and then trying to think about which equilibrium equations must be used and how they are applied in order to determine the unknowns. Then before solving any of the problems, build your skills by first trying to solve the Preliminary Problems, which actually require little or no calculations, and then do some of the Fundamental Problems given on the following pages. The solutions and answers to all these problems are given in the back of the book. Doing this throughout the book will help immensely in understanding how to apply the theory, and thereby develop your problem-solving skills. P1–1. In each case, explain how to find the resultant internal loading acting on the cross section at point A. Draw all necessary free-body diagrams, and indicate the relevant equations of equilibrium. Do not calculate values. The lettered dimensions, angles, and loads are assumed to be known. P B A u 2a a a (a) C D B a a w A C a P (b) B A C P u a (c) a/2 a/2 M O P B r A (d) f u (e) a a 2a A C B P u a a a a 3a a P (f) B A D C PRELIMINARY PROBLEMS