CramX Logo
RegisterLog in
Mechanics of Materials, 10th Edition Solution Manual - Document preview page 1

Mechanics of Materials, 10th Edition Solution Manual - Page 1

Document preview content for Mechanics of Materials, 10th Edition Solution Manual

Mechanics of Materials, 10th Edition Solution Manual

Solve your textbook questions faster and more accurately with Mechanics of Materials, 10th Edition Solution Manual, a reliable guide for every chapter.

Finn Hill
Contributor
4.4
0
12 months ago
Preview (16 of 1569 Pages)
100%
Log in to unlock
Page 1 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 1 preview imageSolution11–1.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
Page 2 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 2 preview image
Page 3 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 3 preview imageSolution21–2.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
Page 4 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 4 preview imageSolution31–3.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
Page 5 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 5 preview image41–4.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
Page 6 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 6 preview image51–5.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,
Page 7 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 7 preview image61–6.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
Page 8 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 8 preview image71–7.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
Page 9 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 9 preview image8*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
Page 10 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 10 preview image91–9.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
Page 11 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 11 preview image101–10.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
Page 12 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 12 preview image111–11.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
Page 13 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 13 preview image121–12.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
Page 14 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 14 preview image131–13.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
Page 15 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 15 preview image141–14.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
Page 16 of 16
Mechanics of Materials, 10th Edition Solution Manual - Page 16 preview image151–15.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
Preview Mode

This document has 1569 pages. Sign in to access the full document!

Study Now!

X-Copilot AI
Unlimited Access
Secure Payment
Instant Access
24/7 Support
Document Chat

Document Details

Subject
Mechanical Engineering
Textbook
Mechanics Of Materials by Russell C...

Related Documents

View all
Solved Assignments

A system is a group of parts that work together to

Textbook Guides

Solution Manual for Introduction to Robotics: Mechanics and Control, 4th Edition

4.2
00
Textbook Guides

Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

4.6
00
Textbook Guides

Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition

5.0
00
Textbook Guides

Solution Manual for Fluid Mechanics for Engineers, 2017 Edition

4.4
00
Textbook Guides

Solution Manual For Applied Strength Of Materials, 5th Edition

4.2
00
Textbook Guides

Solution Manual for Machine Elements in Mechanical Design, 6th Edition

4.7
00
Textbook Guides

Mechanical Vibrations: Theory and Applications 1st Edition Solution Manual

4.8
00
Textbook Guides

Solution Manual For System Dynamics, 4th Edition

4.7
00
Textbook Guides

Solution Manual For Design Of Machine Elements, 8th Edition

4.5
00