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Shigley s Mechanical Engineering Design Solution Manual

Nov 21, 2014

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Chapter 1 Problems 1-1 through 1-4 are for student research.1-5(a) Point vehiclesQ =carshour =vx =42.1v v20.324Seek stationary point maximumdQdv= 0 =42.1 2v0.324 v* = 21.05 mphQ* =42.1(21.05) 21.0520.324= 1368 cars/h Ans.(b)Q =vx +l =

0.324v(42.1) v2 + lv

1Maximize Q with l = 10/5280 miv Q22.18 1221.43122.19 1221.43322.20 1221.435 22.21 1221.43522.22 1221.434% loss of throughput =1368 12211221= 12% Ans.(c) % increase in speed 22.2 21.0521.05= 5.5%Modest change in optimal speed Ans. xl2l2vxvbudy_sm_ch01.qxd11/21/200615:23Page 12 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design1-6 This and the following problem may be the students rst experience with a gure of merit. Formulate fom to reect larger gure of merit for larger merit. Use a maximization optimization algorithm. When one gets into computer implementa-tion and answers are not known, minimizing instead of maximizing is the largest errorone can make.

FV = F1sin W = 0

FH = F1cos F2 = 0From whichF1 = W/sin F2 = W cos /sin fom = $ = (volume).= (l1A1+l2A2)A1 =F1S=WS sin , l2 =l1cos A2 =

F2S

= W cos S sin fom =

l2cos WS sin +l2W cos S sin

= Wl2S

1 +cos2cos sin

Set leading constant to unityfom0 20 5.8630 4.0440 3.2245 3.0050 2.8754.736 2.82860 2.886Checksecondderivativetoseeifamaximum, minimum, orpointofinectionhasbeenfound. Or, evaluate fom on either side of*. * = 54.736Ans.fom* = 2.828Alternative:dd

1 +cos2cos sin

= 0And solve resulting tran-scendental for *.budy_sm_ch01.qxd11/21/200615:23Page 2Chapter 1 31-7(a) x1+ x2 = X1+e1+ X2+e2error = e = (x1+ x2) ( X1+ X2)= e1+e2Ans.(b) x1 x2 = X1+e1( X2+e2)e = (x1 x2) ( X1 X2) = e1e2Ans.(c) x1x2 = ( X1+e1)( X2+e2)e = x1x2 X1X2 = X1e2+ X2e1+e1e2.= X1e2+ X2e1 = X1X2

e1X1+e2X2

Ans.(d)x1x2=X1+e1X2+e2=X1X2

1 +e1/X11 +e2/X2

1 +e2X2

1.= 1 e2X2and

1 +e1X1

1 e2X2

.= 1 +e1X1e2X2e =x1x2 X1X2.=X1X2

e1X1e2X2

Ans.1-8(a) x1 =5 = 2.236 067 977 5X1 = 2.23 3-correct digitsx2 =6 = 2.449 487 742 78X2 = 2.44 3-correct digitsx1+ x2 =5 +6 = 4.685 557 720 28e1 = x1 X1 =5 2.23 = 0.006 067 977 5e2 = x2 X2 =6 2.44 = 0.009 489 742 78e = e1+e2 =5 2.23 +6 2.44 = 0.015 557 720 28Sum = x1+ x2 = X1+ X2+e= 2.23 +2.44 +0.015 557 720 28= 4.685 557 720 28 (Checks) Ans.(b) X1 = 2.24, X2 = 2.45e1 =5 2.24 = 0.003 932 022 50e2 =6 2.45 = 0.000 510 257 22e = e1+e2 = 0.004 442 279 72Sum = X1+ X2+e= 2.24 +2.45 +(0.004 442 279 72)= 4.685 557 720 28 Ans. budy_sm_ch01.qxd11/21/200615:23Page 34 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design1-9(a) = 20(6.89) = 137.8 MPa(b) F = 350(4.45) = 1558 N = 1.558 kN(c) M= 1200 lbf in (0.113) = 135.6 N m(d) A = 2.4(645) = 1548 mm2(e) I = 17.4 in4(2.54)4= 724.2 cm4(f) A = 3.6(1.610)2= 9.332 km2(g) E = 21(1000)(6.89) = 144.69(103) MPa = 144.7 GPa(h) v = 45 mi/h (1.61) = 72.45 km/h(i) V = 60 in3(2.54)3= 983.2 cm3= 0.983 liter1-10(a) l = 1.5/0.305 = 4.918 ft = 59.02 in(b) = 600/6.89 = 86.96 kpsi(c) p = 160/6.89 = 23.22 psi(d) Z =1.84(105)/(25.4)3= 11.23 in3(e) w = 38.1/175 = 0.218 lbf/in(f) = 0.05/25.4 = 0.00197 in(g) v = 6.12/0.0051 = 1200 ft /min(h)= 0.0021 in/in(i) V =30/(0.254)3= 1831 in31-11(a) =20015.3 = 13.1 MPa(b) =42(103)6(102)2 = 70(106) N/m2= 70 MPa(c) y =1200(800)3(103)33(207)109(64)103(103)4 = 1.546(102) m = 15.5 mm(d) =1100(250)(103)79.3(109)(/32)(25)4(103)4= 9.043(102) rad = 5.181-12(a) =60020(6) = 5 MPa(b) I =1128(24)3= 9216 mm4(c) I =64324(101)4= 5.147 cm4(d) =16(16)(253)(103)3 = 5.215(106) N/m2= 5.215 MPabudy_sm_ch01.qxd11/21/200615:23Page 4Chapter 1 51-13(a) =120(103)(/4)(202) = 382 MPa(b) =32(800)(800)(103)(32)3(103)3= 198.9(106) N/m2= 198.9 MPa(c) Z =32(36)(364264) = 3334 mm3(d) k =(1.6)4(103)4(79.3)(109)8(19.2)3(103)3(32)= 286.8 N/mbudy_sm_ch01.qxd11/21/200615:23Page 5FIRST PAGES2-1 From Table A-20Sut = 470 MPa (68 kpsi), Sy = 390 MPa (57 kpsi) Ans.2-2 From Table A-20Sut = 620 MPa (90 kpsi), Sy = 340 MPa (49.5 kpsi) Ans.2-3 Comparison of yield strengths:Sutof G10500 HR is 620470 = 1.32 times larger than SAE1020 CD Ans.Sytof SAE1020 CD is 390340 = 1.15 times larger than G10500 HR Ans.From Table A-20, the ductilities (reduction in areas) show,SAE1020 CD is 4035 = 1.14 times larger than G10500 Ans. The stiffness values of these materials are identical Ans.Table A-20 TableA-5SutSyDuctility Stiffness MPa (kpsi) MPa (kpsi) R% GPa (Mpsi)SAE1020 CD 470(68) 390(57) 40 207(30)UNS10500 HR 620(90) 340(495) 35 207(30)2-4 From Table A-211040 Q&TSy = 593 (86) MPa (kpsi) at 205C (400F) Ans.2-5 From Table A-211040 Q&T R = 65% at 650C (1200F) Ans.2-6 Using Table A-5, the specic strengths are:UNS G10350 HR steel:SyW =39.5(103)0.282= 1.40(105) in Ans.2024 T4 aluminum:SyW =43(103)0.098= 4.39(105) in Ans.Ti-6Al-4V titanium:SyW =140(103)0.16= 8.75(105) in Ans.ASTM 30 gray cast iron has no yield strength. Ans.Chapter 2budynas_SM_ch02.qxd11/22/200616:28Page 6FIRST PAGESChapter 2 72-7 The specic moduli are:UNS G10350 HR steel:EW =30(106)0.282= 1.06(108) in Ans.2024 T4 aluminum:EW =10.3(106)0.098= 1.05(108) in Ans.Ti-6Al-4V titanium:EW =16.5(106)0.16= 1.03(108) in Ans.Gray cast iron:EW =14.5(106)0.26= 5.58(107) in Ans.2-8 2G(1 + ) = E =E 2G2GFrom Table A-5Steel: =30 2(11.5)2(11.5)= 0.304 Ans.Aluminum: =10.4 2(3.90)2(3.90)= 0.333 Ans.Beryllium copper: =18 2(7)2(7)= 0.286 Ans.Gray cast iron: =14.5 2(6)2(6)= 0.208 Ans.2-9 0100 0.0020.10.0040.20.0060.30.0080.40.0100.50.0120.60.0140.70.0160.8(Lower curve)(Upper curve)20304050Stress PA0 kpsiStrain, 607080EYUSu 85.5 kpsi Ans.E 900.003 30 000 kpsi Ans.Sy 45.5 kpsi Ans.R (100) 45.8% Ans.A0 AFA00.1987 0.10770.1987 ll0l l0l0ll0 1AA0 1budynas_SM_ch02.qxd11/22/200616:28Page 7FIRST PAGES8 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-10 To plot truevs. , the following equations are applied to the data.A0 =(0.503)24= 0.1987 in2Eq. (2-4) = lnll0for 0 L 0.0028 in = ln A0Afor L> 0.0028 intrue =PAThe results are summarized in the table below and plotted on the next page.The last 5 points of data are used to plot log vs log The curve t gives m = 0.2306log 0 = 5.1852 0 = 153.2 kpsiAns.For 20% cold work, Eq. (2-10) and Eq. (2-13) give,A = A0(1 W) = 0.1987(1 0.2) = 0.1590 in2 = ln A0A= ln0.19870.1590 = 0.2231Eq. (2-14):S

y = 0m= 153.2(0.2231)0.2306= 108.4 kpsi Ans.Eq. (2-15), with Su = 85.5 kpsi from Prob. 2-9,S

u =Su1 W =85.51 0.2 = 106.9 kpsi Ans.P L A truelog log true0 0 0.198 713 0 01000 0.0004 0.198 713 0.0002 5032.388 3.69901 3.7017742000 0.0006 0.198 713 0.0003 10064.78 3.52294 4.0028043000 0.0010 0.198 713 0.0005 15097.17 3.30114 4.1788954000 0.0013 0.198 713 0.00065 20129.55 3.18723 4.3038347000 0.0023 0.198 713 0.001149 35226.72 2.93955 4.5468728400 0.0028 0.198 713 0.001399 42272.06 2.85418 4.6260538800 0.0036 0.1984 0.001575 44354.84 2.80261 4.6469419200 0.0089 0.1978 0.004604 46511.63 2.33685 4.6675629100 0.1963 0.012216 46357.62 1.91305 4.66612113200 0.1924 0.032284 68607.07 1.49101 4.83636915200 0.1875 0.058082 81066.67 1.23596 4.90884217000 0.1563 0.240083 108765.2 0.61964 5.0364916400 0.1307 0.418956 125478.2 0.37783 5.09856814800 0.1077 0.612511 137418.8 0.21289 5.138046budynas_SM_ch02.qxd11/22/200616:28Page 8FIRST PAGESChapter 2 92-11Tangent modulus at = 0 isE0 =.=5000 00.2(103) 0 = 25(106) psiAt = 20 kpsiE20.=(26 19)(103)(1.5 1)(103) = 14.0(106) psi Ans.(103) (kpsi)0 00.20 50.44 100.80 161.0 191.5 262.0 322.8 403.4 464.0 495.0 54log log y 0.2306x 5.18524.84.955.15.21.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0truetrue (psi)0200004000060000800001000001200001400001600000 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (103)(Sy)0.001 35 kpsi Ans. (kpsi)01020304050600 1 2 3 4 5budynas_SM_ch02.qxd11/22/200616:28Page 9FIRST PAGES10 Solutions Manual Instructors Solution Manual to Accompany Mechanical Engineering Design2-12 Since |o| = |i|

lnR + hR + N

=

lnRR + N

=

ln R + NR

R + hR + N =R + NR( R + N)2= R( R + h)From which, N2+2RN Rh = 0The roots are: N = R

1

1 +hR

1/2

The + sign being signicant,N = R

1 +hR

1/21

Ans.Substitute for N ino = lnR + hR + NGives 0 = lnR + hR + R

1 +hR

1/2 R= ln

1 +hR

1/2Ans.These constitute a useful pair of equations in cold-forming situations, allowing the surfacestrains to be found so that cold-working strength enhancement can be estimated.2-13From Table A-22AISI 1212 Sy = 28.0 kpsi , f = 106 kpsi, Sut = 61.5 kpsi0 = 110 kpsi, m = 0.24, f = 0.85From Eq. (2-12) u = m = 0.24Eq. (2-10)A0A

i=11 W =11 0.2 = 1.25Eq. (2-13) i = ln 1.25 = 0.2231 i< uEq. (2-14) S

y = 0mi= 110(0.2231)0.24= 76.7 kpsi Ans.Eq. (2-15) S

u =Su1 W =61.51 0.2 = 76.9 kpsi Ans.2-14For HB = 250,Eq. (2-17) Su = 0.495 (250) = 124 kpsi= 3.41 (250) = 853 MPaAns.budynas_SM_ch02.qxd11/22/200616:28Page 10FIRST PAGESChapter 2 112-15For the data given,

HB = 2530

H2B = 640 226HB =253010= 253 HB =

640 226 (2530)2/109= 3.887Eq. (2-17)Su = 0.495(253) = 125.2 kpsi Ans. su = 0.495(3.887) = 1.92 kpsi Ans.2-16 From Prob. 2-15,HB = 253 and HB = 3.887Eq. (2-18)Su = 0.23(253) 12.5 = 45.7 kpsi Ans. su = 0.23(3.887) = 0.894 kpsi Ans.2-17 (a) uR.=45.522(30) = 34.5 in lbf/in3Ans.(b)P L A A0/A 1 = P/A00 0 0 01 000 0.0004 0.0002 5032.392 000 0.0006 0.0003 10064.783 000 0.0010 0.0005 15097.174 000 0.0013 0.000 65 20129.557 000 0.0023 0.00115 35226.728 400 0.0028 0.0014 42272.068 800 0.0036 0.0018 44285.029 200 0.0089 0.004 45 46297.979 100 0.1963 0.012 291 0.012 291 45794.7313 200 0.1924 0.032 811 0.032 811 66427.5315 200 0.1875 0.059 802 0.059 802 76492.3017 000 0.1563 0.271 355 0.271 355 85550.6016 400 0.1307 0.520 373 0.520 373 82531.1714 800 0.1077 0.845059 0.845 059 74479.35budynas_SM_ch02.qxd11/22/200616:28Page 11FIRST PAGES12 Solutions Manual Instructors Solution Manual to Accompany