Shigley's Mechanical Design Chapter 3 Solutions
Chapter 3 Solutions- shigley mech. design 9th edition
Description:
solutions to chapter 3 problems from shigley's book
Transcript:
Chapter 3 3-1 0oM E = 18 6(100) 0BR = 33.3 lbf .BR Ans = 0yF E = 100 0o BR R + = 66.7 lbf .oR Ans = 33.3 lbf .C BR R A = = ns ______________________________________________________________________________ 3-2 Body AB: 0xF E = Ax BxR R = 0yF E = Ay ByR R = 0BM E = (10) (10) 0Ay AxR R = Ax AyR R = Body OAC: 0OM E = (10) 100(30) 0AyR = 300 lbf .AyR Ans = 0xF E = 300 lbf .Ox AxR R A = = ns 0yF E = 100 0Oy AyR R + = 200 lbf .OyR Ans = ______________________________________________________________________________ Chapter 3 - Rev. A, Page 1/100 3-3 0.81.39 kN .tan30OR Ans = = 0.81.6 kN .sin30AR Ans = = ______________________________________________________________________________ 3-4 Step 1: Find RA & RE 4.57.794 mtan3009 7.794(400cos30 )4.5(400sin30 ) 0400 N .AEEhMRR Ans= =E = == 2 20 400cos30 0 346.4 N0 400 400sin30 0 200 N346.4 200 400N .x AxAxy AyAyAF RRF RRR Ans= + == = + = = + == Step 2: Find components of RC on link 4 and RD ( )( )440 400(4.5) 7.794 1.9 0 305.4 N .0 305.4 N0 ( ) 400 NCDDx Cxy CyMRR AnsF RF R= === == = Chapter 3 - Rev. A, Page 2/100 Step 3: Find components of RC on link 2 ( )( )( )2220 305.4 346.4 0 41 N0 200 NxCxCxyCyFRRFR=+ ==== _____________________________________________________________________________________________________________________ Chapter 3 - Rev. A, Page 3/100 3-5 0CM E = 11500 300(5) 1200(9) 0 R + + = 18.2 kN . R Ans = 0yF E = 28.2 9 5 0 R + =25.8 kN . R Ans = 18.2(300) 2460 N m . M Ans = = 22460 0.8(900) 1740 N m . M Ans = = 31740 5.8(300) 0 checks! M = =_____________________________________________________________________________ 3-6 0yF E = 0500 40(6) 740 lbf . R Ans = + = 00 M E = 0500(8) 40(6)(17) 8080 lbf in . M Ans = + = 18080 740(8) 2160 lbf in . M Ans = + = 22160 240(6) 720 lbf in . M Ans = + = 31720 (240)(6) 0 checks!2M = + = ______________________________________________________________________________ Chapter 3 - Rev. A, Page 4/100 3-7 0BM E = 12.2 1(2) 1(4) 0 R + = 10.91 kN . R Ans = 0yF E = 20.91 2 4 0 R + = 26.91 kN . R Ans = 10.91(1.2) 1.09 kN m . M Ans = = 21.09 2.91(1) 4 kN m . M Ans = = 34 4(1) 0 checks! M = + = ______________________________________________________________________________ 3-8 Break at the hinge at B Beam OB: From symmetry, 1200 lbf .BR V Ans = = Beam BD: 0DM E = 2200(12) (10) 40(10)(5) 0 R + = 2440 lbf . R Ans = 0yF E = 3200 440 40(10) 0 R + + = 3160 lbf . R Ans = Chapter 3 - Rev. A, Page 5/100 1200(4) 800 lbf in . M Ans = = 2800 200(4) 0 checks at hinge M = = 3800 200(6) 400 lbf in . M Ans = = 41400 (240)(6) 320 lbf in .2M Ans = + = 51320 (160)(4) 0 checks!2M = = ______________________________________________________________________________ 3-9 1 1 11 20 0 01 21 1 11 29 300 5 1200 15009 300 5 1200 1500 (1)9 300 5 1200 1500 (2)q R x x x R xV R x x R xM R x x x R x = + = + = + 1 At x = 1500+ V = M = 0. Applying Eqs. (1) and (2), 1 2 1 29 5 0 14 R R R R + = + = 1 11500 9(1500 300) 5(1500 1200) 0 8.2 kN . R R A = =214 8.2 5.8 kN . ns R Ans = = 0 300 : 8.2 kN, 8.2 N m300 1200 : 8.2 9 0.8 kN 8.2 9( 300) 0.8 2700 N m1200 1500 : 8.2 9 5 5.8 kN 8.2 9( 300x V M xx VM x x xx VM x xs s = = s s = = = = + s s = = = ) 5( 1200) 5.8 8700 N m x x = + Plots of V and M are the same as in Prob. 3-5. ______________________________________________________________________________ Chapter 3 - Rev. A, Page 6/100 3-10 1 2 1 0 00 01 0 1 10 01 2 20 0500 8 40 14 40 20500 8 40 14 40 20 (1)500 8 20 14 20 20 (2)at 20 in, 0, Eqs. (1) and (2) giveq R x M x x x xV R M x x x xM R x M x x xx V MR += + = + = + = = =( )0 020 0 0500 40 20 14 0 740 lbf .(20) 500(20 8) 20(20 14) 0 8080 lbf in .R AnsR M M = = = = Ans 0 8: 740 lbf, 740 8080 lbf in8 14: 740 500 240 lbf 740 8080 500( 8) 240 4080 lbf in14 20: 740 500 40( 14) 40 800 lbf 740 8080x V M xx VM x x xx V x xM xs s = = s s = == = s s = = += 2 2500( 8) 20( 14) 20 800 8000 lbf in x x x x = + Plots of V and M are the same as in Prob. 3-6. ______________________________________________________________________________ 3-11 1 1 1 11 20 0 01 21 1 11 22 1.2 2.2 4 3.22 1.2 2.2 4 3.2 (1)2 1.2 2.2 4 3.2 (2)q R x x R x xV R x R x xM R x x R x x = + = + = + at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2), Solving Eqs. (3) and (4) simultaneously, 1 2 1 21 2 1 22 4 0 6 (3)3.2 2(2) (1) 0 3.2 4 (4)R R R RR R R R + = + = + = + = R1 = -0.91 kN, R2 = 6.91 kN Ans. 0 1.2: 0.91 kN, 0.91 kN m1.2 2.2: 0.91 2 2.91 kN 0.91 2( 1.2) 2.91 2.4 kN m2.2 3.2: 0.91 2 6.91 4 kN 0.91 2(x V M xx VM x x xx VM x xs s = = s s = = = = + s s = + == 1.2) 6.91( 2.2) 4 12.8 kN m x x + = Plots of V and M are the same as in Prob. 3-7. ______________________________________________________________________________ Chapter 3 - Rev. A, Page 7/100 3-12 1 1 1 0 0 11 2 30 0 1 1 01 2 31 1 2 2 11 2 31400 4 10 40 10 40 20 20400 4 10 40 10 40 20 20 (1)400 4 10 20 10 20 20 20 (2)0 at 8 in 8 400(q R x x R x x x R xV R x R x x x R xM R x x R x x x R xM x R = + + + = + + + = + + + = = 18 4) 0 200 lbf . R Ans = = at x = 20+, V =M = 0. Applying Eqs. (1) and (2), 2 3 2 322 2200 400 40(10) 0 600200(20) 400(16) (10) 20(10) 0 440 lbf . R R R RR R A + + = + = + = =3600 440 160 lbf .nsR Ans = =0 4: 200 lbf, 200 lbf in4 10: 200 400 200 lbf, 200 400( 4) 200 1600 lbf in10 20: 200 400 440 40( 10) 640 40 lbf 200 400( 4)x V M xx VM x x xx V x xM x xs s = = s s = = = = + s s = + = = + ( )22440( 10) 20 10 20 640 x x x 4800 lbf in x = + Plots of V and M are the same as in Prob. 3-8. ______________________________________________________________________________ 3-13 Solution depends upon the beam selected. ______________________________________________________________________________ 3-14 (a) Moment at center, ( )( )22222 2 2 2 4ccl axl l lM l a a= (| | | |= = ( | |\ . \ . ( w wl At reaction, 22rM a = w a = 2.25, l = 10 in, w = 100 lbf/in ( )2100(10) 102.25 125 lbf in2 4100 2.25253 lbf in .2crMM Ans| |= = |\ .= = (b) Optimal occurs when c rM M = Chapter 3 - Rev. A, Page 8/100 22 20.25 02 4 2l l aa a al l| | = + = |\ .w w Taking the positive root ( ) ( )2 214 0.25 2 1 0.207 .2 2la l l l l A (= + + = = ( ns for l = 10 in, w = 100 lbf, a = 0.207(10) = 2.07 in ( )2min100 2 2.07 214 lbf in M = = ______________________________________________________________________________ 3-15 (a) 20 105 kpsi2C = = 20 1015 kpsi2CD += = 2 215 8 17 kpsi R = + = 15 17 22 kpsi o = + = 25 17 12 kpsi o = = 11 8tan 14.04 cw2 15p| | |= = |\ . 117 kpsi45 14.04 30.96 ccwsR t| = == = (b) 9 1612.5 kpsi2C += = 16 93.5 kpsi2CD = = 2 25 3.5 6.10 kpsi R = + = 112.5 6.1 18.6 kpsi o = + = 212.5 6.1 6.4 kpsi o = = 11 5tan 27.5 ccw2 3.5p| | |= = |\ . 16.10 kpsi45 27.5 17.5 cwsR t| = == = Chapter 3 - Rev. A, Page 9/100 (c) 2 21224 1017 kpsi224 107 kpsi27 6 9.22 kpsi17 9.22 26.22 kpsi17 9.22 7.78 kpsiCCDRoo+= == == + == + == = 11 790 tan 69.7 ccw2 6p| ( | |= + = | (\ . 19.22 kpsi69.7 45 24.7 ccwsR t| = == = (d) 2 21212 225 kpsi212 2217 kpsi217 12 20.81 kpsi5 20.81 25.81 kp
Shigley's Mechanical Design Chapter 3 Solutions
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