Mechanics of Materials, 10th Edition Solution Manual

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Solution11–1.The shaft is supported by a smooth thrust bearing atBanda journal bearing atC. Determine the resultant internalloadings acting on the cross section atE.AEDBC4 ft400 lb800 lb4 ft4 ft4 ftSolutionSupport Reactions:We will only need to computeCyby writing the momentequation of equilibrium aboutBwith reference to the free-body diagram of theentire shaft, Fig.a.a+ΣMB=0;Cy(8)+400(4)-800(12)=0Cy=1000 lbInternal Loadings:Using the result forCy,sectionDEof the shaft will be considered.Referring to the free-body diagram, Fig.b,S+ΣFx=0;NE=0Ans.+ cΣFy=0;VE+1000-800=0VE=-200 lbAns.a+ΣME=0; 1000(4)-800(8)-ME=0ME=-2400 lb#ft=-2.40 kip#ftAns.The negative signs indicates thatVEandMEact in the opposite sense to that shownon the free-body diagram.Ans:NE=0,VE=-200 lb,ME=-2.40 kip#ft

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Solution21–2.Determine the resultant internal normal and shear force inthe member at (a) sectiona–aand (b) sectionb–b, each ofwhich passes through the centroidA. The 500-lb load isapplied along the centroidal axis of the member.30°Ababa500 lb500 lbSolution(a)+SΣFx=0;Na-500=0Na=500 lbAns.+ TΣFy=0;Va=0Ans.(b)R+ΣFx=0;Nb-500 cos 30°=0Nb=433 lbAns.+QΣFy=0;Vb-500 sin 30°=0Vb=250 lbAns.Ans:(a)Na=500 lb,Va=0,(b)Nb=433 lb,Vb=250 lb

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Solution31–3.Determine the resultant internal loadings acting on sectionb–bthrough the centroidCon the beam.30°6 ft3 ft60°bb900 lb/ftCABSolutionSupport Reaction:a+ΣMA=0;NB(9 sin 30°)-12(900)(9)(3)=0NB=2700 lbEquations of Equilibrium:For sectionb–bS+ΣFx=0;Vb-b+12 (300)(3) sin 30°-2700=0Vb-b=2475 lb=2.475 kipAns.+ cΣFy=0;Nb-b-12 (300)(3) cos 30°=0Nb-b=389.7 lb=0.390 kipAns.a+ΣMC=0;2700(3 sin 30°)-12 (300)(3)(1)-Mb-b=0Mb-b=3600 lb#ft=3.60 kip#ftAns.Ans:Vb-b=2.475 kip,Nb-b=0.390 kip,Mb-b=3.60 kip#ft

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41–4.The shaft is supported by a smooth thrust bearing atAanda smooth journal bearing atB. Determine the resultantinternal loadings acting on the cross section atC.ADBC900 N1.5 m600 N/m1.5 m1 m1 m1 mSolutionSupport Reactions:We will only need to computeByby writing the momentequation of equilibrium aboutAwith reference to the free-body diagram of theentire shaft, Fig.a.a+ΣMA=0;By(4.5)-600(2)(2)-900(6)=0By=1733.33 NInternal Loadings:Using the result ofBy, sectionCDof the shaft will beconsidered. Referring to the free-body diagram of this part, Fig.b,S+ΣFx=0;NC=0Ans.+ cΣFy=0;VC-600(1)+1733.33-900=0VC=-233 NAns.a+ΣMC=0;1733.33(2.5)-600(1)(0.5)-900(4)-MC=0MC=433 N#mAns.The negative sign indicates thatVCacts in the opposite sense to that shown on thefree-body diagram.*Ans:NC=0,VC=-233 N,MC=433 N#m

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51–5.Determine the resultant internal loadings acting on thecross section at pointB.AC12 ft3 ft60 lb/ftBSolutionS+ΣFx=0;NB=0Ans.+ cΣFy=0;VB-12 (48)(12)=0VB=288 lbAns.a+ΣMB=0;-MB-12 (48)(12)(4)=0MB=-1152 lb#ft=-1.15 kip#ftAns.Ans:NB=0VB=288 lb,MB=-1.15 kip#ft,

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61–6.Determine the resultant internal loadings on the crosssection at pointD.2 mD1.25 kN/m1.5 m1 mEBCAF0.5 m 0.5 m 0.5 mSolutionSupport Reactions:MemberBCis the two force member.a+ΣMA=0;45FBC(1.5)-1.875(0.75)=0FBC=1.1719 kN+ cΣFy=0;Ay+45(1.1719)-1.875=0Ay=0.9375 kNS+ΣFx=0;35 (1.1719)-Ax=0Ax=0.7031 kNEquations of Equilibrium:For pointDS+ΣFx=0;ND-0.7031=0ND=0.703 kNAns.+ cΣFy=0;0.9375-0.625-VD=0VD=0.3125 kNAns.a+ΣMD=0;MD+0.625(0.25)-0.9375(0.5)=0MD=0.3125 kN#mAns.Ans:ND=0.703 kN,VD=0.3125 kN,MD=0.3125 kN#m

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71–7.Determine the resultant internal loadings at cross sectionsat pointsEandFon the assembly.2 mD0.5 m 0.5 m1.25 kN/m0.5 m1.5 m1 mEBCAFSolutionSupport Reactions:MemberBCis the two-force member.a+ΣMA=0;45FBC(1.5)-1.875(0.75)=0FBC=1.1719 kN+ cΣFy=0;Ay+45(1.1719)-1.875=0Ay=0.9375 kNS+ΣFx=0;35 (1.1719)-Ax=0Ax=0.7031 kNEquations of Equilibrium:For pointF+ bΣFx′=0;NF-1.1719=0NF=1.17 kNAns.a +ΣFy′=0;VF=0Ans.a+ΣMF=0;MF=0Ans.Equations of Equilibrium:For pointE+dΣFx=0;NE-35 (1.1719)=0NE=0.703 kNAns.+ cΣFy=0;VE-0.625+45 (1.1719)=0VE=-0.3125 kNAns.a+ΣME=0;-ME-0.625(0.25)+45 (1.1719)(0.5)=0ME=0.3125 kN#mAns.Negative sign indicates thatVEacts in the opposite direction to that shown on FBD.Ans:NF=1.17 kN,VF=0,MF=0,NE=0.703 kN,VE=-0.3125 kN,ME=0.3125 kN#m

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8*1–8.The beam supports the distributed load shown. Determinethe resultant internal loadings acting on the cross section atpointC. Assume the reactions at the supportsAandBarevertical.1.5 m3 mDCAB4 kN/m1.5 mSolutionSupport Reactions:Referring to the FBD of the entire beam, Fig.a,a+ΣMA=0;By(6)-12(4)(6)(2)=0By=4.00 kNInternal Loadings:Referring to the FBD of the right segment of the beam sectionedthroughC, Fig.b,S+ΣFx=0;NC=0Ans.+ cΣFy=0;VC+4.00-12(3)(4.5)=0VC=2.75 kNAns.a+ΣMC=0;4.00(4.5)-12(3)(4.5)(1.5)-MC=0MC=7.875 kN#mAns.Ans:NC=0,VC=2.75 kN,MC=7.875 kN#m

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91–9.The beam supports the distributed load shown. Determinethe resultant internal loadings acting on the cross section atpointD. Assume the reactions at the supportsAandBarevertical.1.5 m3 mDCAB4 kN/m1.5 mSolutionSupport Reactions:Referring to the FBD of the entire beam, Fig.a,a+ΣMA=0;By(6)-12(4)(6)(2)=0By=4.00 kNInternal Loadings:Referring to the FBD of the right segment of the beam sectionedthroughD, Fig.b,S+ΣFx=0;ND=0Ans.+ cΣFy=0;VD+4.00-12(1.00)(1.5)=0VD=-3.25 kNAns.a+ΣMD=0;4.00(1.5)-12(1.00)(1.5)(0.5)-MD=0MD=5.625 kN#mAns.The negative sign indicates thatVDacts in the sense opposite to that shown onthe FBD.Ans:ND=0,VD=-3.25 kN,MD=5.625 kN#m

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101–10.The boomDFof the jib crane and the columnDEhave auniform weight of 50 lb/ft. If the supported load is 300 lb,determine the resultant internal loadings in the crane oncross sections at pointsA,B, andC.5 ft7 ftCDFEBA300 lb2 ft8 ft3 ftSolutionEquations of Equilibrium:For pointAd+ΣFx=0;NA=0Ans.+ cΣFy=0;VA-150-300=0VA=450 lbAns.a+ΣMA=0;-MA-150(1.5)-300(3)=0MA=-1125 lb#ft=-1.125 kip#ftAns.Negative sign indicates thatMAacts in the opposite direction to that shown on FBD.Equations of Equilibrium:For pointBd+ΣFx=0;NB=0Ans.+ cΣFy=0;VB-550-300=0VB=850 lbAns.a+ΣMB=0;-MB-550(5.5)-300(11)=0MB=-6325 lb#ft=-6.325 kip#ftAns.Negative sign indicates thatMBacts in the opposite direction to that shown on FBD.Equations of Equilibrium:For pointCd+ΣFx=0;VC=0Ans.+ cΣFy=0;-NC-250-650-300=0NC=-1200 lb=-1.20 kipAns.a+ΣMC=0;-MC-650(6.5)-300(13)=0MC=-8125 lb#ft=-8.125 kip#ftAns.Negative signs indicate thatNCandMCact in the opposite direction to that shownon FBD.Ans:NA=0,VA=450 lb,MA=-1.125 kip#ft,NB=0,VB=850 lb,MB=-6.325 kipft,#VC=0,NC=-1.20 kip,MC=-8.125 kip#ft

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111–11.Determine the resultant internal loadings acting on thecross sections at pointsDandEof the frame.4 ftAD2 ftCB75 lb/ft1 ft150 lb1 ftE30°2 ft1 ftG1 ftFSolutionMemberAG:a+ΣMA=0;45FBC(3)-75(4)(5)-150 cos 30°(7)=0;FBC=1003.89 lba+ΣMB=0;Ay(3)-75(4)(2)-150 cos 30°(4)=0;Ay=373.20 lbS+ΣFx=0;Ax-35(1003.89)+150 sin 30°=0;Ax=527.33 lbFor pointD:S+ΣFx=0;ND+527.33=0ND=-527 lbAns.+ cΣFy=0;-373.20-VD=0VD=-373 lbAns.a+ΣMD=0;MD+373.20(1)=0MD=-373 lb#ftAns.For pointE:S+ΣFx=0;150 sin 30°-NE=0NE=75.0 lbAns.+ cΣFy=0;VE-75(3)-150 cos 30°=0VE=355 lbAns.a+ΣME=0;-ME-75(3)(1.5)-150 cos 30°(3)=0;ME=-727 lb#ftAns.Ans:ND=-527 lb,VD=-373 lb,MD=-373 lbft,#NE=75.0 lb,VE=355 lb,ME=-727 lb#ft

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121–12.Determine the resultant internal loadings acting on thecross sections at pointsFandGof the frame.4 ftAD2 ftCB75 lb/ft1 ft150 lb1 ftE30°2 ft1 ftG1 ftFSolutionMemberAG:a+ΣMA=0;45FBF(3)-300(5)-150 cos 30°(7)=0FBF=1003.9 lbFor pointF:+QΣFx′=0;VF=0Ans.a +ΣFy′=0;NF-1003.9=0NF=1004 lbAns.a+ΣMF=0;MF=0Ans.For pointG:d+ΣFx=0;NG-150 sin 30°=0NG=75.0 lbAns.+ cΣFy=0;VG-75(1)-150 cos 30°=0VG=205 lbAns.a+ΣMG=0;-MG-75(1)(0.5)-150 cos 30°(1)=0MG=-167 lb#ftAns.*Ans:VF=0,NF=1004 lb,MF=0,NG=75.0 lb,VG=205 lb,MG=-167 lb#ft

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131–13.The blade of the hacksaw is subjected to a pretension forceofF=100 N. Determine the resultant internal loadingsacting on sectiona–athat passes through pointD.ABCDFFabba30°225 mm150 mmESolutionInternal Loadings:Referring to the free-body diagram of the section of the hacksawshown in Fig.a,d+ΣFx=0;Na-a+100=0Na-a=-100 NAns.+ cΣFy=0;Va-a=0Ans.a+ΣMD=0;-Ma-a-100(0.15)=0Ma-a=-15 Nm#Ans.The negative sign indicates thatNa-aandMa-aact in the opposite sense to thatshown on the free-body diagram.Ans:Na-a=-100 N,Va-a=0,Ma-a=-15 N#m

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141–14.The blade of the hacksaw is subjected to a pretension forceofF=100 N. Determine the resultant internal loadingsacting on sectionb–bthat passes through pointD.ABCDFFabba30°225 mm150 mmESolutionInternal Loadings:Referring to the free-body diagram of the section of the hacksawshown in Fig.a,ΣFx′=0;Nb-b+100 cos 30°=0Nb-b=-86.6 NAns.ΣFy′=0;Vb-b-100 sin 30°=0Vb-b=50 NAns.a+ΣMD=0;-Mb-b-100(0.15)=0Mb-b=-15 N#mAns.The negative sign indicates thatNb–bandMb–bact in the opposite sense to thatshown on the free-body diagram.Ans:Nb-b=-86.6 N,Vb-b=50 N,Mb-b=-15 N#m

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151–15.The beam supports the triangular distributed load shown.Determine the resultant internal loadings on the crosssection at pointC. Assume the reactions at the supportsAandBare vertical.6 ft6 ftCAB4.5 ft800 lb/ft6 ft4.5 ftEDSolutionSupport Reactions:Referring to the FBD of the entire beam, Fig.a,a+ΣMB=0;12(0.8)(18)(6)-12(0.8)(9)(3)-Ay(18)=0Ay=1.80 kipInternal Loadings:Referring to the FBD of the left beam segment sectioned throughpointC, Fig.b,S+ΣFx=0;NC=0Ans.+ cΣFy=0;1.80-12(0.5333)(12)-VC=0VC=-1.40 kipAns.a+ΣMC=0;MC+12(0.5333)(12)(4)-1.80(12)=0MC=8.80 kip#ftAns.The negative sign indicates thatVCacts in the sense opposite to that shown onthe FBD.Ans:NC=0,VC=-1.40 kip,MC=8.80 kip#ft

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