SOLUTION MANUALChapter 1 Problems 1-1 through 1-6 are for student research. No standard solutions are provided. 1-7 From Fig. 1-2, cost of grinding to 0.0005 in is 270%. Cost of turning to 0.003 in is 60%. Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans. ______________________________________________________________________________ 1-8 CA = CB, 10 + 0.8 P = 60 + 0.8 P 0.005 P 2 P 2 = 50/0.005 P = 100 parts Ans. ______________________________________________________________________________ 1-9 Max. load = 1.10 P Min. area = (0.95)2A Min. strength = 0.85 S To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be ( )21.101.43 .0.85 0.95dn A = = ns) ______________________________________________________________________________ 1-10 (a) X1 + X2: ( ) (1 2 1 1 2 21 2 1 21 2error .x x X e X ee x x X Xe e Ans+ = + + += = + += + (b) X1 X2: ( )( ) ( )1 2 1 1 2 21 2 1 2 1 2 .x x X e X ee x x X X e e Ans = + += = (c) X1 X2: ( )( )1 2 1 1 2 21 2 1 2 1 2 2 1 1 21 21 2 2 1 1 21 2 .x x X e X ee x x X X X e X e e ee eX e X e X X AnsX X= + += = + +| |+ = + |\ . Chapter 1 Solutions - Rev. B, Page 1/6 (d) X1/X2: 1 1 1 1 1 12 2 2 2 2 212 2 1 1 1 2 12 2 2 2 1 2 11 1 1 1 22 2 2 1 21111 1 then 1 1 11Thus, .x X e X e Xx X e X e Xe e e X e e e22eX X e X X X Xx X X e ee Ansx X X X X| | + += = |+ +\ .| | | | | || | ++ + + | | | |+\ . \ . \ .\ .| |= |\ . X ______________________________________________________________________________ 1-11 (a) x1 = 7 = 2.645 751 311 1 X1 = 2.64 (3 correct digits) x2 = 8 = 2.828 427 124 7 X2 = 2.82 (3 correct digits) x1 + x2 = 5.474 178 435 8 e1 = x1 X1 = 0.005 751 311 1 e2 = x2 X2 = 0.008 427 124 7 e = e1 + e2 = 0.014 178 435 8 Sum = x1 + x2 = X1 + X2 + e = 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks (b) X1 = 2.65, X2 = 2.83 (3 digit significant numbers) e1 = x1 X1 = 0.004 248 688 9 e2 = x2 X2 = 0.001 572 875 3 e = e1 + e2 = 0.005 821 564 2 Sum = x1 + x2 = X1 + X2 + e = 2.65 +2.83 0.001 572 875 3 = 5.474 178 435 8 Checks ______________________________________________________________________________ 1-12 ( ) ( )3325 1016 10000.799 in .2.5dSd An dot= = = ns Table A-17: d = 78in Ans. Factor of safety: ( )( )( )337825 103.29 .16 1000Sn Aot= = = ns ______________________________________________________________________________ 1-13 Eq. (1-5): R =1niiR== 0.98(0.96)0.94 = 0.88 Overall reliability = 88 percent Ans. ______________________________________________________________________________ Chapter 1 Solutions - Rev. B, Page 2/6 1-14 a = 1.500 0.001 in b = 2.000 0.003 in c = 3.000 0.004 in d = 6.520 0.010 in (a) d a b c = w = 6.520 1.5 2 3 = 0.020 in = 0.001 + 0.003 + 0.004 +0.010 = 0.018 allt = wt w = 0.020 0.018 in Ans. (b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore, d = 6.520 + 0.008 = 6.528 in Ans. ______________________________________________________________________________ 1-15 V = xyz, and x = a A a, y = b A b, z = c A c, V abc = ( )( )( ) V a a b b c cabc bc a ac b ab c a b c b c a c a b a b c= A A A= A A A A A A A A A A A A The higher order terms in A are negligible. Thus, V bc a ac b ab c A A + A + A and, .V bc a ac b ab c a b c a b cAnsV abc a b c a b cA A + A + A A A A A A A= + + = + + For the numerical values given, ( )31.500 1.875 3.000 8.4375 in V = = ( )30.002 0.003 0.0040.00427 0.00427 8.4375 0.036 in1.500 1.875 3.000VVVA = + + = A = = V = 8.438 0.036 in3 Ans. ______________________________________________________________________________ Chapter 1 Solutions - Rev. B, Page 3/6 1-16 wmax = 0.05 in, wmin = 0.004 in 0.05 0.0040.027 in2+ = w= Thus, A w = 0.05 0.027 = 0.023 in, and then, w = 0.027 0.023 in. 0.027 0.042 1.51.569 ina b caa = =w= tw = 0.023 = tallta + 0.002 + 0.005 ta = 0.016 in Thus, a = 1.569 0.016 in Ans. ______________________________________________________________________________ 1-17 ( ) 2 3.734 2 0.139 4.012 ino iD D d = + = + = ( )all0.028 2 0.004 0.036 inoDt t = = + = Do = 4.012 0.036 in Ans. ______________________________________________________________________________ 1-18 From O-Rings, Inc. (oringsusa.com), Di = 9.19 0.13 mm, d = 2.62 0.08 mm ( ) 2 9.19 2 2.62 14.43 mmo iD D d = + = + = ( )all0.13 2 0.08 0.29 mmoDt t = = + = Do = 14.43 0.29 mm Ans. ______________________________________________________________________________ 1-19 From O-Rings, Inc. (oringsusa.com), Di = 34.52 0.30 mm, d = 3.53 0.10 mm ( ) 2 34.52 2 3.53 41.58 mmo iD D d = + = + = ( )all0.30 2 0.10 0.50 mmoDt t = = + = Do = 41.58 0.50 mm Ans. ______________________________________________________________________________ Chapter 1 Solutions - Rev. B, Page 4/6 1-20 From O-Rings, Inc. (oringsusa.com), Di = 5.237 0.035 in, d = 0.103 0.003 in ( ) 2 5.237 2 0.103 5.443 ino iD D d = + = + = ( )all0.035 2 0.003 0.041 inoDt t = = + = Do = 5.443 0.041 in Ans. ______________________________________________________________________________ 1-21 From O-Rings, Inc. (oringsusa.com), Di = 1.100 0.012 in, d = 0.210 0.005 in ( ) 2 1.100 2 0.210 1.520 ino iD D d = + = + = ( )all0.012 2 0.005 0.022 inoDt t = = + = Do = 1.520 0.022 in Ans. ______________________________________________________________________________ 1-22 From Table A-2, (a) o = 150/6.89 = 21.8 kpsi Ans. (b) F = 2 /4.45 = 0.449 kip = 449 lbf Ans. (c) M = 150/0.113 = 1330 lbf in = 1.33 kip in Ans. (d) A = 1500/ 25.42 = 2.33 in2 Ans. (e) I = 750/2.544 = 18.0 in4 Ans. (f) E = 145/6.89 = 21.0 Mpsi Ans. (g) v = 75/1.61 = 46.6 mi/h Ans. (h) V = 1000/946 = 1.06 qt Ans. ______________________________________________________________________________ 1-23 From Table A-2, (a) l = 5(0.305) = 1.53 m Ans. (b) o = 90(6.89) = 620 MPa Ans. (c) p = 25(6.89) = 172 kPa Ans. Chapter 1 Solutions - Rev. B, Page 5/6 Chapter 1 Solutions - Rev. B, Page 6/6 (d) Z =12(16.4) = 197 cm3 Ans. (e) w = 0.208(175) = 36.4 N/m Ans. (f) o = 0.001 89(25.4) = 0.0480 mm Ans. (g) v = 1200(0.0051) = 6.12 m/s Ans. (h) = 0.002 15(1) = 0.002 15 mm/mm Ans. (i) V = 1830(25.43) = 30.0 (106) mm3 Ans.

______________________________________________________________________________ 1-24 (a) o = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi Ans. (b) o = F /A = 9440/23.8 = 397 psi Ans. (c) y =Fl3/3EI = 270(31.5)3/[3(30)106(0.154)] = 0.609 in Ans. (d) u = Tl /GJ = 9740(9.85)/[11.3(106)(t /32)1.004] = 8.648(102) rad = 4.95 Ans. ______________________________________________________________________________ 1-25 (a) o =F / wt = 1000/[25(5)] = 8 MPa Ans. (b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4 Ans. (c) I =t d4/64 = t (25.4)4/64 = 20.4(103) mm4 Ans. (d) t =16T /t d 3 = 16(25)103/[t (12.7)3] = 62.2 MPa Ans. ______________________________________________________________________________ 1-26 (a) t =F /A = 2 700/[t (0.750)2/4] = 6110 psi = 6.11 kpsi Ans. (b) o = 32Fa/t d 3 = 32(180)31.5/[t (1.25)3] = 29 570 psi = 29.6 kpsi Ans. (c) Z =t (do4 di4)/(32 do) = t (1.504 1.004)/[32(1.50)] = 0.266 in3 Ans. (d) k = (d 4G)/(8D 3 N) = 0.06254(11.3)106/[8(0.760)3 32] = 1.53 lbf/in Ans. ______________________________________________________________________________ Chapter 2 2-1 From Tables A-20, A-21, A-22, and A-24c, (a) UNS G10200 HR: Sut = 380 (55) MPa (kpsi), Syt = 210 (30) Mpa (kpsi) Ans. (b) SAE 1050 CD: Sut = 690 (100) MPa (kpsi), Syt = 580 (84) Mpa (kpsi) Ans. (c) AISI 1141 Q&T at 540C (1000F): Sut = 896 (130) MPa (kpsi), Syt = 765 (111) Mpa (kpsi) Ans. (d) 2024-T4: Sut = 446 (64.8) MPa (kpsi), Syt = 296 (43.0) Mpa (kpsi) Ans. (e) Ti-6Al-4V annealed: Sut = 900 (130) MPa (kpsi), Syt = 830 (120) Mpa (kpsi) Ans. ______________________________________________________________________________ 2-2 (a) Maximize yield strength: Q&T at 425C (800F) Ans. (b)Maximize elongation: Q&T at 650C (1200F) Ans. ______________________________________________________________________________ 2-3 Conversion of kN/m3 to kg/ m3 multiply by 1(103) / 9.81 = 102 AISI 1018 CD steel: Tables A-20 and A-5 ( )( )3370 1047.4 kN m/kg .76.5 102ySAns = = 2011-T6 aluminum: Tables A-22 and A-5 ( )( )3169 1062.3 kN m/kg .26.6 102ySAns = = Ti-6Al-4V titanium: Tables A-24c and A-5 ( )( )3830 10187 kN m/kg .43.4 102ySAns = = ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the ultimate strength in tension ( )( )( )342.5 6.89 1040.7 kN m/kg70.6 102utSAns = = ______________________________________________________________________________ 2-4 AISI 1018 CD steel: Table A-5 ( ) ( )6630.0 10106 10 in .0.282EAns = = 2011-T6 aluminum: Table A-5 ( ) ( )6610.4 10106 10 in .0.098EAns = = Ti-6Al-6V titanium: Table A-5 Chapter 2 - Rev. D, Page 1/19 ( ) ( )6616.5 10103 10 in .0.160EAns = = No. 40 cast iron: Table A-5 ( ) ( )6614.5 1055.8 10 in .0.260EAns = = ______________________________________________________________________________ 2-5 22 (1 )2E GG v E vG+ = = From Table A-5 Steel: ( )( )30.0 2 11.50.304 .2 11.5v A= = ns Aluminum: ( )( )10.4 2 3.900.333 .2 3.90v A= = ns Beryllium copper: ( )( )18.0 2 7.00.286 .2 7.0v A= = ns Gray cast iron: ( )( )14.5 2 6.00.208 .2 6.0v A= = ns ______________________________________________________________________________ 2-6 (a) A0 = t (0.503)2/4, o = Pi / A0 For data in elastic range, = A l / l0 = A l / 2 For data in plastic range, 0 00 0 01 1l l A l ll l l A A= = = = On the next two pages, the data and plots are presented. Figure (a) shows the linear part of the curve from data points 1-7. Figure (b) shows data points 1-12. Figure (c) shows the complete range. Note: The exact value of A0 is used without rounding off. (b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsi Ans. From Fig. (b) the equation for the dotted offset line is found to be o = 30.5(106) 61 000 (1) The equation for the line between data points 8 and 9 is o = 7.60(105) + 42 900 (2) Chapter 2 - Rev. D, Page 2/19 Solving Eqs. (1) and (2) simultaneously yields o = 45.6 kpsi which is the 0.2 percent offset yield strength. Thus, Sy = 45.6 kpsi Ans. The ultimate strength from Figure (c) is Su = 85.6 kpsi Ans. The reduction in area is given by Eq. (2-12) is ( ) ( )000.1987 0.1077100 100 45.8 % .0.1987fA AR AnsA = = = Data Point Pi Al, Ai o 1 0 0 0 0 2 1000 0.0004 0.00020 5032 3 2000 0.0006 0.00030 10065 4 3000 0.001 0.00050 15097 5 4000 0.0013 0.00065 20130 6 7000 0.0023 0.00115 35227 7 8400 0.0028 0.00140 42272 8 8800 0.0036 0.00180 44285 9 9200 0.0089 0.00445 46298 10 8800 0.1984 0.00158 44285 11 9200 0.1978 0.00461 46298 12 9100 0.1963 0.01229 45795 13 13200 0.1924 0.03281 66428 14 15200 0.1875 0.05980 76492 15 17000 0.1563 0.27136 85551 16 16400 0.1307 0.52037 82531 17 14800 0.1077 0.84506 74479 (a) Linear range Chapter 2 - Rev. D, Page 3/19 (b) Offset yield (c) Complete range (c) The material is ductile since there is a large amount of deformation beyond yield. (d) The closest material to the values of Sy, Sut, and R is SAE 1045 HR with Sy = 45 kpsi, Sut = 82 kpsi, and R = 40 %. Ans. ______________________________________________________________________________ 2-7 To plot o true vs.c, the following equations are applied to the data. truePAo = Eq. (2-4) Chapter 2 - Rev. D, Page 4/19 00ln for 0 0.0028 inln for 0.0028 inlllAlAcc= s A s= A > where 220(0.503)0.1987 in4A t= = The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log o vs log c The curve fit gives m = 0.2306 log o0 = 5.1852 o0 = 153.2 kpsi Ans. For 20% cold work, Eq. (2-14) and Eq. (2-17) give, A = A0 (1 W) = 0.1987 (1 0.2) = 0.1590 in2 00.230600.1987ln ln 0.22310.1590Eq. (2-18): 153.2(0.2231) 108.4 kpsi .Eq. (2-19), with 85.6 from Prob. 2-6, 85.6107 kpsi .1 1 0.2myuuuAAS ASSS AnsWco c'= = == = ==' = = = ns P AL A c o true log c log otrue 0 0 0.198 713 0 0 1000 0.0004 0.198 713 0.000 2 5032.388 -3.699 01 3.701 7742000 0.0006 0.198 713 0.000 3 10 064.78 -3.522 94 4.002 8043000 0.001 0.198 713 0.000 5 15 097.17 -3.301 14 4.178 8954000 0.0013 0.198 713 0.000 65 20 129.55 -3.187 23 4.303 8347000 0.0023 0.198 713 0.001 149 35 226.72 -2.939 55 4.546 8728400 0.0028 0.198 713 0.001 399 42 272.06 -2.854 18 4.626 0538800 0.0036 0.198 4 0.001 575 44 354.84 -2.802 61 4.646 9419200 0.0089 0.197 8 0.004 604 46 511.63 -2.336 85 4.667 5629100 0.196 3 0.012 216 46 357.62 -1.913 05 4.666 12113200 0.192 4 0.032 284 68 607.07 -1.491 01 4.836 36915200 0.187 5 0.058 082 81 066.67 -1.235 96 4.908 84217000 0.156 3 0.240 083 108 765.20 -0.619 64 5.036 49 16400 0.130 7 0.418 956 125 478.20 -0.377 83 5.098 56814800 0.107 7 0.612 511 137 418.80 -0.212 89 5.138 046 Chapter 2 - Rev. D, Page 5/19 ______________________________________________________________________________ 2-8 Tangent modulus at o = 0 is ( ) ( )635000 025 10 psi0.2 10 0E oA = =A Ans. At o = 20 kpsi Chapter 2 - Rev. D, Page 6/19 ( )( )( )( ) ( )3620326 19 1014.0 10 psi1.5 1 10E= Ans. (10-3) o (kpsi) 0 0 0.20 5 0.44 10 0.80 16 1.0 19 1.5 26 2.0 32 2.8 40 3.4 46 4.0 49 5.0 54 ______________________________________________________________________________ 2-9 W = 0.20, (a) Before cold working: Annealed AISI 1018 steel. Table A-22, Sy = 32 kpsi, Su = 49.5 kpsi, o0 = 90.0 kpsi, m = 0.25, f = 1.05 After cold working: Eq. (2-16), u = m = 0.25 Eq. (2-14), 01 11.251 1 0.20iAA W= = = Eq. (2-17), 0ln ln1.25 0.223i uiAAc c = = = < Eq. (2-18), S 93% increase Ans. ( )0.25090 0.223 61.8 kpsi .my io c' = = = Ans Eq. (2-19), 49.561.9 kpsi .1 1 0.20uuSS AW' = = = ns 25% increase Ans. (b) Before: 49.51.5532uySS = = After: 61.91.0061.8uySS'= =' Ans. Lost most of its ductility ______________________________________________________________________________ 2-10 W = 0.20, (a) Before cold working: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, Su = 61.5 kpsi, o0 = 110 kpsi, m = 0.24, f = 0.85 After cold working: Eq. (2-16), u = m = 0.24 Chapter 2 - Rev. D, Page 7/19 Eq. (2-14), 01 11.251 1 0.20iAA W= = = Eq. (2-17), 0ln ln1.25 0.223i uiAAc c = = = < Eq. (2-18), 174% increase Ans. ( )0.240110 0.223 76.7 kpsi .my iS A o c' = = = ns Eq. (2-19), 61.576.9 kpsi .1 1 0.20uuSS AW' = = = ns 25% increase Ans. (b) Before: 61.52.2028uySS = = After: 76.91.0076.7uySS'= =' Ans. Lost most of its ductility ______________________________________________________________________________ 2-11 W = 0.20, (a) Before cold working: 2024-T4 aluminum alloy. Table A-22, Sy = 43.0 kpsi, Su = 64.8 kpsi, o0 = 100 kpsi, m = 0.15, f = 0.18 After cold working: Eq. (2-16), u = m = 0.15 Eq. (2-14), 01 11.251 1 0.20iAA W= = = Eq. (2-17), 0ln ln1.25 0.223iiAAfc c = = = > Material fractures. Ans. ______________________________________________________________________________ 2-12 For HB = 275, Eq. (2-21), Su = 3.4(275) = 935 MPa Ans. ______________________________________________________________________________ 2-13 Gray cast iron, HB = 200. Eq. (2-22), Su = 0.23(200) 12.5 = 33.5 kpsi Ans. From Table A-24, this is probably ASTM No. 30 Gray cast iron Ans. ______________________________________________________________________________ 2-14 Eq. (2-21), 0.5HB = 100 HB = 200 Ans. ______________________________________________________________________________ Chapter 2 - Rev. D, Page 8/19 2-15 For the data given, converting HB to Su using Eq. (2-21) HB Su (kpsi) Su2 (kpsi)230 115 13225 232 116 13456 232 116 13456 234 117 13689 235 117.5 13806.25235 117.5 13806.25235 117.5 13806.25236 118 13924 236 118 13924 239 119.5 14280.25ESu = 1172 ESu2 = 137373 1172117.2 117 kpsi .10uuSS AN= = = ns Eq. (20-8), ( )102 221137373 10 117.21.27 kpsi .1 9uu uiSS NSs AN= = = =ns ______________________________________________________________________________ 2-16 For the data given, converting HB to Su using Eq. (2-22) HB Su (kpsi) Su2 (kpsi)230 40.4 1632.16 232 40.86 1669.54 232 40.86 1669.54 234 41.32 1707.342235 41.55 1726.403235 41.55 1726.403235 41.55 1726.403236 41.78 1745.568236 41.78 1745.568239 42.47 1803.701ESu = 414.12 ESu2 = 17152.63 Chapter 2 - Rev. D, Page 9/19 414.1241.4 kpsi .10uuSS AN= = =ns Eq. (20-8), ( )102 22117152.63 10 41.41.20 .1 9uu uiSS NSs AN= = = =ns ______________________________________________________________________________ 2-17 (a) 2345.534.5 in lbf / in .2(30)Ru A = ns (b) P AL A A0 / A 1 o = P/A0 0 0 0 0 1000 0.0004 0.0002 5 032.39 2000 0.0006 0.0003 10 064.78 3000 0.0010 0.0005 15 097.17 4000 0.0013 0.000 65 20 129.55 7000 0.0023 0.001 15 35 226.72 8400 0.0028 0.0014 42 272.06 8800 0.0036 0.0018 44 285.02 9200 0.0089 0.004 45 46 297.97 9100 0.1963 0.012 291 0.012 291 45 794.73 13200 0.1924 0.032 811 0.032 811 66 427.53 15200 0.1875 0.059 802 0.059 802 76 492.30 17000 0.1563 0.271 355 0.271 355 85 550.60 16400 0.1307 0.520 373 0.520 373 82 531.17 14800 0.1077 0.845 059 0.845 059 74 479.35 From the figures on the next page, ( )( ) ( )( )513 31(43 000)(0.001 5) 45 000(0.004 45 0.001 5)2145 000 76 500 (0.059 8 0.004 45)281 000 0.4 0.059 8 80 000 0.845 0.466.7 10 in lbf/in .T iiu AAns= = ++ + + + + Chapter 2 - Rev. D, Page 10/19 Chapter 2 - Rev. D, Page 11/19 2-18, 2-19 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________ 2-20 Appropriate tables: Youngs modulus and Density (Table A-5)1020 HR and CD (Table A-20), 1040 and 4140 (Table A-21), Aluminum (Table A-24), Titanium (Table A-24c) Appropriate equations: For diameter, ( )24/ 4yyF FS dFA d Sot t= = = = Weight/length = A, Cost/length = $/in = ($/lbf) Weight/length, Deflection/length = o /L = F/(AE) With F = 100 kips = 100(103) lbf, MaterialYoung'sModulus DensityYieldStrength Cost/lbf Diameter Weight/lengthCost/lengthDeflection/lengthunits Mpsi lbf/in^3 kpsi $/lbf in lbf/in $/in in/in 1020HR 30 0.282 30 $0.27 2.060 0.9400 $0.25 1.000E031020CD 30 0.282 57 $0.30 1.495 0.4947 $0.15 1.900E031040 30 0.282 80 $0.35 1.262 0.3525 $0.12 2.667E034140 30 0.282 165 $0.80 0.878 0.1709 $0.14 5.500E03Al 10.4 0.098 50 $1.10 1.596 0.1960 $0.22 4.808E03Ti 16.5 0.16 120 $7.00 1.030 0.1333 $0.93 7.273E03 The selected materials with minimum values are shaded in the table above. Ans. ______________________________________________________________________________ 2-21 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three would favor steel, cast iron, or maybe a less common ferrous material. The expectation would likely be hot-rolled steel. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 7.95 lbf, the unit weight is determined to be 3 327.95 lbf0.281 lbf/in 0.28 lbf/in[ (1 in) / 4](36 in)WAl t= = = w which agrees well with the unit weight of 0.282 lbf/in3 reported in Table A-5 for carbon steel. Nickel steel and stainless steel have similar unit weights, but surface finish and darker coloring do not favor their selection. To select a likely specification from Table Chapter 2 - Rev. D, Page 12/19 A-20, perform a Brinell hardness test, then use Eq. (2-21) to estimate an ultimate strength of . Assuming the material is hot-rolled due to the rough surface finish, appropriate choices from Table A-20 would be one of the higher carbon steels, such as hot-rolled AISI 1050, 1060, or 1080. Ans. 0.5 0.5(200) 100 kpsiu BS H = = =______________________________________________________________________________ 2-22 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous material like aluminum. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 2.90 lbf, the unit weight is determined to be 3 322.9 lbf0.103 lbf/in 0.10 lbf/in[ (1 in) / 4](36 in)WAl t= = = w which agrees reasonably well with the unit weight of 0.098 lbf/in3 reported in Table A-5 for aluminum. No other materials come close to this unit weight, so the material is likely aluminum. Ans. ______________________________________________________________________________ 2-23 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic, and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous copper-based material such as copper, brass, or bronze. To further distinguish the material, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 9 lbf, the unit weight is determined to be 3 329.0 lbf0.318 lbf/in 0.32 lbf/in[ (1 in) / 4](36 in)WAl t= = = w which agrees reasonably well with the unit weight of 0.322 lbf/in3 reported in Table A-5 for copper. Brass is not far off (0.309 lbf/in3), so the deflection test could be used to gain additional insight. From the measured deflection and utilizing the deflection equation for an end-loaded cantilever beam from Table A-9, Youngs modulus is determined to be ( )( )334100 2417.7 Mpsi3 3 (1) 64 (17 / 32)FlEIy t= = = which agrees better with the modulus for copper (17.2 Mpsi) than with brass (15.4 Mpsi). The conclusion is that the material is likely copper. Ans. ______________________________________________________________________________ 2-24 and 2-25 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________ Chapter 2 - Rev. D, Page 13/19 2-26 For strength, o = F/A = S A = F/S For mass, m = Al = (F/S) l Thus, f 3(M ) = /S , and maximize S/ (| = 1) In Fig. (2-19), draw lines parallel to S/ From the list of materials given, both aluminum alloy and high carbon heat treated steel are good candidates, having greater potential than tungsten carbide or polycarbonate. The higher strength aluminum alloys have a slightly greater potential. Other factors, such as cost or availability, may dictate which to choose. Ans. ______________________________________________________________________________ 2-27 For stiffness, k = AE/l A = kl/E For mass, m = Al = (kl/E) l =kl2 /E Thus, f 3(M) = /E , and maximize E/ (| = 1) In Fig. (2-16), draw lines parallel to E/ Chapter 2 - Rev. D, Page 14/19 From the list of materials given, tungsten carbide (WC) is best, closely followed by aluminum alloys, and then followed by high carbon heat-treated steel. They are close enough that other factors, like cost or availability, would likely dictate the best choice. Polycarbonate polymer is clearly not a good choice compared to the other candidate materials. Ans. ______________________________________________________________________________ 2-28 For strength, o = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 90 ]. The section modulus is strictly a function of the dimensions of the cross section and has the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a number. For example, for a circular cross section, C = ( )14 t . Then, for strength, Eq. (1) is 2/33/ 2Fl FlS ACA CS| |= = |\ . (2) Chapter 2 - Rev. D, Page 15/19 For mass, 2/3 2/35/32/3 Fl Fm Al l lCS C S | | | | |= = = | | \ . \ . \ ||. Thus, f 3(M) = /S 2/3, and maximize S 2/3/ (| = 2/3) In Fig. (2-19), draw lines parallel to S 2/3/ From the list of materials given, a higher strength aluminum alloy has the greatest potential, followed closely by high carbon heat-treated steel. Tungsten carbide is clearly not a good choice compared to the other candidate materials. .Ans. ______________________________________________________________________________ 2-29 Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a Chapter 2 - Rev. D, Page 16/19 constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 = cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12 (a constant). Thus, Eq. (2-27) becomes 1/ 233klACE| |= |\ . and Eq. (2-29) becomes 1/ 25/ 21/ 23km Al lC E | | | |= = | |\ . \ . Thus, minimize ( )3 1/ 2f ME= , or maximize1/ 2EM= . From Fig. (2-16) From the list of materials given, aluminum alloys are clearly the best followed by steels and tungsten carbide. Polycarbonate polymer is not a good choice compared to the other candidate materials. Ans. ______________________________________________________________________________ 2-30 For stiffness, k = AE/l A = kl/E For mass, m = Al = (kl/E) l =kl2 /E Chapter 2 - Rev. D, Page 17/19 So, f 3(M) = /E, and maximize E/ . Thus, | = 1. Ans. ______________________________________________________________________________ 2-31 For strength, o = F/A = S A = F/S For mass, m = Al = (F/S) l So, f 3(M ) = /S, and maximize S/ . Thus, | = 1. Ans. ______________________________________________________________________________ 2-32 Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 = cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12. Thus, Eq. (2-27) becomes 1/ 233klACE| |= |\ . and Eq. (2-29) becomes 1/ 25/ 21/ 23km Al lC E | | |= = | \ . \ ||. So, minimize ( )3 1/ 2f ME= , or maximize1/ 2EM= . Thus, | = 1/2. Ans. ______________________________________________________________________________ 2-33 For strength, o = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 90 ]. The section modulus is strictly a function of the dimensions of the cross section and has the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a number. For example, for a circular cross section, C =( )14 t . Then, for strength, Eq. (1) is 2/33/ 2Fl FlS ACA CS| |= = |\ . (2) For mass, 2/3 2/35/32/3 Fl Fm Al l lCS C S | | | | |= = = | | \ . \ . \ ||. So, f 3(M) = /S 2/3, and maximize S 2/3/. Thus, | = 2/3. Ans. ______________________________________________________________________________ 2-34 For stiffness, k=AE/l, or, A = kl/E. Chapter 2 - Rev. D, Page 18/19 Chapter 2 - Rev. D, Page 19/19 Thus, m = Al = (kl/E )l = kl 2 /E. Then, M = E / and | = 1. From Fig. 2-16, lines parallel to E / for ductile materials include steel, titanium, molybdenum, aluminum alloys, and composites. For strength, S = F/A, or, A = F/S. Thus, m = Al = F/Sl = Fl /S. Then, M = S/ and | = 1. From Fig. 2-19, lines parallel to S/ give for ductile materials, steel, aluminum alloys, nickel alloys, titanium, and composites. Common to both stiffness and strength are steel, titanium, aluminum alloys, and composites. Ans. 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 kpsi5 20.81 15.81 kpsiCCDRoo += =+= == + == + == = 11 1790 tan 72.39 cw2 12p| ( | |= + = | (\ . Chapter 3 - Rev. A, Page 10/100 120.81 kpsi 72.39 45 27.39 cwsR t| = == = ______________________________________________________________________________ Chapter 3 - Rev. A, Page 11/100 3-16 (a) 2 2128 70.5 MPa28 77.5 MPa27.5 6 9.60 MPa9.60 0.5 9.10 MPa0.5 9.6 10.1 MpaCCDRoo += = += == + == == = 11 7.590 tan 70.67 cw2 6p| ( | |= + = | (\ . 19.60 MPa70.67 45 25.67 cwsR t| = == = (b) 2 2129 61.5 MPa29 67.5 MPa27.5 3 8.078 MPa1.5 8.078 9.58 MPa1.5 8.078 6.58 MPaCCDRoo= =+= == + == + == = 11 3tan 10.9 cw2 7.5p| | |= = |\ . 18.078 MPa45 10.9 34.1 ccwsR t| = == = Chapter 3 - Rev. A, Page 12/100 (c) 2 21212 44 MPa212 48 MPa28 7 10.63 MPa4 10.63 14.63 MPa4 10.63 6.63 MPaCCDRoo= =+= == + == + == = 11 890 tan 69.4 ccw2 7p| ( | |= + = | (\ . 110.63 MPa69.4 45 24.4 ccwsR t| = == = (d) 2 2126 50.5 MPa26 55.5 MPa25.5 8 9.71 MPa0.5 9.71 10.21 MPa0.5 9.71 9.21 MPaCCDRoo= =+= == + == + == = 11 8tan 27.75 ccw2 5.5p| | |= = |\ . 19.71 MPa45 27.75 17.25 cwsR t| = == = ______________________________________________________________________________ Chapter 3 - Rev. A, Page 13/100 3-17 (a) 2 21212 69 kpsi212 63 kpsi23 4 5 kpsi5 9 14 kpsi9 5 4 kpsiCCDRoo+= == == + == + == = 11 4tan 26.6 ccw2 3p| | |= = |\ . 15 kpsi45 26.6 18.4 ccwsR t| = == = (b) 2 21230 1010 kpsi230 1020 kpsi220 10 22.36 kpsi10 22.36 32.36 kpsi10 22.36 12.36 kpsiCCDRoo= =+= == + == + == = 11 10tan 13.28 ccw2 20p| | |= = |\ . 122.36 kpsi45 13.28 31.72 cwsR t| = == = Chapter 3 - Rev. A, Page 14/100 (c) 2 21210 184 kpsi210 1814 kpsi214 9 16.64 kpsi4 16.64 20.64 kpsi4 16.64 12.64 kpsiCCDRoo += =+= == + == + == = 11 1490 tan 73.63 cw2 9p| ( | |= + = | (\ . 116.64 kpsi 73.63 45 28.63 cwsR t| = == = (d) 2 2129 1914 kpsi219 95 kpsi25 8 9.434 kpsi14 9.43 23.43 kpsi14 9.43 4.57 kpsiCCDRoo+= == == + == + == = 11 590 tan 61.0 cw2 8p| ( | |= + = | (\ . 19.34 kpsi61 45 16 cwsR t| = == = ______________________________________________________________________________ Chapter 3 - Rev. A, Page 15/100 3-18 (a) 2 212380 3055 MPa280 3025 MPa225 20 32.02 MPa0 MPa55 32.02 22.98 23.0 MPa55 32.0 87.0 MPaCCDRooo = = = == + === + = = = = 1 2 2 3 1 323 8711.5 MPa, 32.0 MPa, 43.5 MPa2 2t t t = = = = = (b) 2 212330 6015 MPa260 3045 MPa245 30 54.1 MPa15 54.1 39.1 MPa0 MPa15 54.1 69.1 MPaCCDRooo= = += == + == + === = 1 31 22 339.1 69.154.1 MPa239.119.6 MPa269.134.6 MPa2ttt+= == == = Chapter 3 - Rev. A, Page 16/100 (c) 2 212340 020 MPa240 020 MPa220 20 28.3 MPa20 28.3 48.3 MPa20 28.3 8.3 MPa30 MPazCCDRooo o+= == == + == + == = = = 1 3 1 2 2 348.3 30 30 8.339.1 MPa, 28.3 MPa, 10.9 MPa2 2t t t+ = = = = = (d) 2 21235025 MPa25025 MPa225 30 39.1 MPa25 39.1 64.1 MPa25 39.1 14.1 MPa20 MPazCCDRooo o= == == + == + == = = = 1 3 1 2 2 364.1 20 20 14.142.1 MPa, 39.1 MPa, 2.95 MPa2 2t t t+ = = = = = ______________________________________________________________________________ 3-19 (a) Since there are no shear stresses on the stress element, the stress element already represents principal stresses. 12310 kpsi0 kpsi4 kpsixyo ooo o= === = 1 31 22 310 ( 4)7 kpsi2105 kpsi20 ( 4)2 kpsi2ttt = == = = = Chapter 3 - Rev. A, Page 17/100 (b) 2 212 30 105 kpsi210 05 kpsi25 4 6.40 kpsi5 6.40 11.40 kpsi0 kpsi, 5 6.40 1.40 kpsiCCDRoo o+= == == + == + == = = 1 3 1 2 311.40 1.406.40 kpsi, 5.70 kpsi, 0.70 kpsi2 2R t t t = = = = = = (c) 2 21 232 85 kpsi28 23 kpsi23 4 5 kpsi5 5 0 kpsi, 0 kpsi5 5 10 kpsiCCDRo oo = = = == + == + = == = 1 3 1 2 2 3105 kpsi, 0 kpsi, 5 kpsi2t t t = = = = (d) 2 212310 3010 kpsi210 3020 kpsi220 10 22.36 kpsi10 22.36 12.36 kpsi0 kpsi10 22.36 32.36 kpsiCCDRooo= = += == + == + === = 1 3 1 2 2 312.36 32.3622.36 kpsi, 6.18 kpsi, 16.18 kpsi2 2t t t = = = = = ______________________________________________________________________________ Chapter 3 - Rev. A, Page 18/100 3-20 From Eq. (3-15), 3 2 2 22 2 23( 6 18 12) 6(18) ( 6)( 12) 18( 12) 9 6 ( 15) 6(18)( 12) 2(9)(6)( 15) ( 6)(6) 18( 15) ( 12)(9) 0594 3186 02o o oo o ( + + + + ( + = + = Roots are: 21.04, 5.67, 26.71 kpsi Ans. 1 22 3max 1 321.04 5.677.69 kpsi25.67 26.7116.19 kpsi221.04 26.71 23.88 kpsi .2 Ansttt t= =+= =+= = = _____________________________________________________________________________ 3-21 From Eq. (3-15) ( )( ) ( )23 2 222 23 2(20 0 20) 20(0) 20(20) 0(20) 40 20 2 0 20(0)(20) 2(40) 20 2 (0) 20 20 2 0(0) 20(40) 040 2 000 48 000 02o o oo o o ( + + + + + ( ( + ( + == Roots are: 60, 20, 40 kpsi Ans. 1 22 3max 1 360 2020 kpsi220 4030 kpsi260 4050 kpsi .2 Ansttt t= =+= =+= = = _____________________________________________________________________________ Chapter 3 - Rev. A, Page 19/100 3-22 From Eq. (3-15) ( ) ( )2 23 2 22 2 23 2(10 40 40) 10(40) 10(40) 40(40) 20 40 2010(40)(40) 2(20)( 40)( 20) 10( 40) 40( 20) 40(20) 090 0o o oo o ( + + + + + ( + = = Roots are: 90, 0, 0 MPa Ans. 2 31 2 1 3 max09045 MPa .2 Anstt t t== = = = _____________________________________________________________________________ 3-23 ( )( )( )( )26611500033 950 psi 34.0 kpsi .4 0.756033 950 0.0679 in .30 100.06791130 10 1130 .60FAnsAFL LAnsAE EAnsLoto oo = = = == = = == = = = From Table A-5, v = 0.292 ( ) ( )2 16 620.292(1130) 330 .330 10 (0.75) 248 10 in .v Ad d Annss = = = A = = = _____________________________________________________________________________ 3-24 ( )( )( )( )266130006790 psi 6.79 kpsi .4 0.75606790 0.0392 in .10.4 100.0392653 10 653 .60FAnsAFL LAnsAE EAnsLoto oo = = = == = = == = = = From Table A-5, v = 0.333 ( ) ( )2 16 620.333(653) 217 .217 10 (0.75) 163 10 in .v Ansd d Ans = = = A = = = Chapter 3 - Rev. A, Page 20/100 _____________________________________________________________________________ 3-25 20.00010.0001d dd dA = = = From Table A-5, v = 0.326, E = 119 GPa ( )( ) ( )( ) ( )6 216 91260.0001306.7 100.326 and , so= 306.7 10 (119) 10 36.5 MPa0.0336.5 10 25 800 N 25.8 kN .4vFL FAE AEELF A Ano oooto = = == == = == = = =s Sy = 70 MPa > o , so elastic deformation assumption is valid. _____________________________________________________________________________ 3-26 ( )68(12)20 000 0.185 in .10.4 10FL LAnsAE Eo o = = = = _____________________________________________________________________________ 3-27 ( ) ( )693140 10 0.00586 m 5.86 mm .71.7 10FL LAnsAE Eo o = = = = = _____________________________________________________________________________ 3-28 ( )610(12)15 000 0.173 in .10.4 10FL LAnsAE Eo o = = = = _____________________________________________________________________________ 3-29 With 0,zo = solve the first two equations of Eq. (3-19) simulatenously. Place E on the left-hand side of both equations, and using Cramers rule, ( )2 211 1 11xy x x yxE vE E E vEv v vvo+ += = = yv Likewise, Chapter 3 - Rev. A, Page 21/100 ( )21y xyEvvo += From Table A-5, E = 207 GPa and = 0.292. Thus, ( ) ( ) ( )( )( ) ( )( )962 2962207 10 0.0019 0.292 0.000 7210 382 MPa .1 1 0.292207 10 0.000 72 0.292 0.001910 37.4 MPa .1 0.292x yxyE vAnsvAnsoo+ ( + = = = + ( = = _____________________________________________________________________________ 3-30 With 0,zo = solve the first two equations of Eq. (3-19) simulatenously. Place E on the left-hand side of both equations, and using Cramers rule, ( )2 211 1 11xy x x yxE vE E E vEv v vvo+ += = = yv Likewise, ( )21y xyEvvo += From Table A-5, E = 71.7 GPa and = 0.333. Thus, ( ) ( ) ( )( )( ) ( )( )962 296271.7 10 0.0019 0.333 0.000 7210 134 MPa .1 1 0.33371.7 10 0.000 72 0.333 0.001910 7.04 MPa .1 0.333x yxyE vAnsvAnsoo+ ( + = = = + ( = = _____________________________________________________________________________ 3-31 (a) 1 max 1 c acR F M Ra Fl l= = = 22 26 66M ac bh lF Fbh bh l acoo = = = Ans. (b) ( )( )( ) ( )( )( )221 21( )( ) ( ) .( )( )m m m m mm mb b h h l l F s s ss AnsF a a c c s so o= = =3-32 For equal stress, the model load varies by the square of the scale factor. _____________________________________________________________________________ Chapter 3 - Rev. A, Page 22/100 21 max/ 2,2 2 2 2x ll l lR M l= | |= = |\ .w w 8l= ww(a) 2 22 2 26 6 3 4 .8 4 3M l Wl bhW Abh bh bh loo = = = =w ns (b) 2 22( / )( / )( / ) 1( )( )./m m m mmW b b h h s ss AnW l l so o= = = s22 .m m ml ss sl s= = =w ww w Ans For equal stress, the model load w varies linearly with the scale factor. _____ _____________ -33 (a) Can solve by iteration or derive _ __________________________________________________________ 3equations for the general case. Find maximum moment under wheel 3W . W W = E at centroid of Ws T3 3dA Tl xR Wl= Under wheel 3, ( )3 33 3 1 13 2 23 3 1 13 2 23 A Tl x dM R x = Wa W a W x Wa W al = For maximum, ( )3 33 3 330 22TdM l d Wl d x xdx l = = = Substitute into ( )233 1 14 Tl d3 2 23M M W Wal = W aintersects the midpoint of the beam. For wheel i, This means the midpoint of 3d( )21 il dl d1,2 4 iiT j jiji ix M W W al == = Note for wheel 1: 0j jiW a E = 1 2 3 4104.4104.4, 26.1 kips4TW W W W W = = = = = = Wheel 1: 21 1476 (1200 238)238 in, (104.4) 20128 kip in2 4(1200)d M = = = = Wheel 2: 238 84 154 in d = = 2Chapter 3 - Rev. A, Page 23/100 22 max(1200 154)(104.4) 26.1(84) 21605 kip in .4(1200)M M A= = = ns Check if all of the wheels are on the rail. (b) max600 77 523 in . x Ans = = (c) See above sketch. (d) Inner axles _____________________________________________________________________________ 3-34 (a) Let a = total area of entire envelope Let b = area of side notch ( )( ) ( ) ( )( )( )23 36 42 40(3)(25) 25 34 2150 mm1 12 40 75 34 2512 121.36 10 mm .a bA a bI I II Ans= = == = = Dimensions in mm. (b) 2220.375(1.875) 0.703 125 in0.375(1.75) 0.656 25 in2(0.703125) 0.656 25 2.0625 inabAAA= == == + = 34342 212(0.703 125)(0.9375) 0.656 25(0.6875)0.858 in .2.06250.375(1.875)0.206 in121.75(0.375)0.007 69 in122 0.206 0.703 125(0.0795) 0.00769 0.65625(0.1705) 0.448 in .aby AII4nsI Ans+= == == = ( ( = + + + = (c) Use two negative areas. 2 22625 mm , 5625 mm , 10 000 mm10 000 5625 625 3750 mm ;a b cA A AA = = == =2 Chapter 3 - Rev. A, Page 24/100 ( )( )13436 436 46.25 mm, 50 mm, 50 mm10 000(50) 5625(50) 625(6.25)57.29 mm .3750100 57.29 42.71 mm .50(12.5)8138 mm1275(75)2.637 10 mm12100(100)8.333 10 in12a b cabcy y yy Ansc AnsIII= = = = == == == == = ( ) ( ) ( ) ( )( )2 26 2 616 418.333 10 10000(7.29) 2.637 10 5625 7.29 8138 625 57.29 6.254.29 10 in .II Ans ( ( (= + + + = (d) ( )( )( )2224 0.875 3.5 in2.5 0.875 2.1875 in5.6875 in2.9375 3.5 1.25(2.1875)2.288 in .5.6875aba bAAA A Ay Ans= == == + =+= =( ) ( ) ( )( ) (3 2 341 1(4) 0.875 3.5 2.9375 2.288 0.875 2.5 2.1875 2.288 1.2512 125.20 in .II Ans= + + + =)2 _____________________________________________________________________________ 3-35 ( )3 521(20)(40) 1.067 10 mm1220(40) 800 mmIA= == =4 Mmax is at A. At the bottom of the section, ( )max5450 000(20)84.3 MPa .1.067 10McAnsIo = = = Due to V, tmax is between A and B at y = 0. max3 3 30005.63 MPa .2 2 800VAnsAt | |= = = |\ . _____________________________________________________________________________ Chapter 3 - Rev. A, Page 25/100 3-36 3 41(1)(2) 0.6667 in12I = = 21(2) 2 in A = = 0oM E = 8 100(8)(12) 0AR = 1200 lbfAR = 1200 100(8) 400 lbfoR = = is at A. At the top of the beam, maxM max3200(0.5)2400 psi .0.6667McAnsIo = = = Due to V, maxt is at A, at y = 0. max3 3 800600 psi .2 2 2VAnsAt | |= = = |\ . _____________________________________________________________________________ 3-37 3 41(0.75)(2) 0.5 in12I = = 2(0.75)(2) 1.5 in A = = 0AM E = 15 1000(20) 0BR = 1333.3 lbfBR = 3000 1333.3 1000 2666.7 lbfAR = + = is at B. At the top of the beam, maxM max5000(1)10000 psi .0.5McAnsIo = = = Due to V, maxt is between B and C at y = 0. max3 3 10001000 psi .2 2 1.5VAnsAt | |= = = |\ . _____________________________________________________________________________ Chapter 3 - Rev. A, Page 26/100 3-38 ( )4 43 4(50)306.796 10 mm64 64dI t t= = = 2 22(50)1963 mm4 4dA t t= = = 0BM E = 6(300)(150) 200 0AR = 1350 kNAR = 6(300) 1350 450 kNBR = = maxM is at A. At the top, max McIo = =Due to V, maxt is at A, at y = 0. 2max4 4 7500.509 kN/mm 509 MPa .3 3 1963VAnsAt | |= = = = |\ . _____________________________________________________________________________ 3-39 2 2maxmax max 28 8 8 I l l cMI cloo = = =w ww (a) 448 in; Table A-8, 0.537 in l I = = ( ) ( )( )( )328 12 10 0.53722.38 lbf/in .1 48 Ans = = w (b) ( )( )( ) ( )( )( )3 360 in, 1 12 2 3 1 12 1.625 2.625 2.051 in l I = 4= ( )( )( )( )( )328 12 10 2.05136.5 lbf/in .1.5 60 Ans = = w (c) ( )460 in; Table A-6, 2 0.703 1.406 in l I = = = y = 0.717 in, cmax = 1.783 in ( )( )( )( )328 12 10 1.40621.0 lbf/in .1.783 60 Ans = = w (d) 460 in, Table A-7, 2.07 in l I = = ( ) ( )( )( )328 12 10 2.0736.8 lbf/in .1.5 60 Ans = = w _____________________________________________________________________________ Chapter 3 - Rev. A, Page 27/100 3-40 ( ) ( ) ( )4 3 4 2 20.5 3.068 10 in , 0.5 0.1963 in64 4I At t= = = = Model (c) ( )3max500(0.5) 500(0.75 / 2)218.75 lbf in2 2218.75(0.25)3.068 1017 825 psi 17.8 kpsi .4 4 5003400 psi 3.4 kpsi .3 3 0.1963MMcIAnsVAnsAoot= + = = == == = = = Model (d) ( )3500(0.625) 312.5 lbf in312.5(0.25)3.068 1025 464 psi 25.5 kpsi .MMcIAnsoo= = = == = max4 4 5003400 psi 3.4 kpsi .3 3 0.1963VAnsAt = = = = Model (e) ( )3max500(0.4375) 218.75 lbf in218.75(0.25)3.068 1017 825 psi 17.8 kpsi .4 4 5003400 psi 3.4 kpsi .3 3 0.1963MMcIAnsVAnsAoot= = = == == = = = _____________________________________________________________________________ 3-41 Chapter 3 - Rev. A, Page 28/100 ( ) ( )4 4 212 1018 mm , 12 113.1 mm64 4I A2t t= = = = Model (c) 22max2000(6) 2000(9)15 000 N mm2 215 000(6)101888.4 N/mm 88.4 MPa .4 4 200023.6 N/mm 23.6 MPa .3 3 113.1MMcIAnsVAnsAoot= + = = == =| |= = = = |\ . Model (d) 22000(12) 24 000 N mm24 000(6)1018141.5 N/mm 141.5 MPa .MMcIAnsoo= = = == = 2max4 4 200023.6 N/mm 23.6 MPa .3 3 113.1VAnsAt | |= = = = |\ . Model (e) 22000(7.5) 15000 N mm15000(6)101888.4 N/mm 88.4 MPa .MMcIAnsoo= = = == = 2max4 4 200023.6 N/mm 23.6 MPa .3 3 113.1VAnsAt | |= = = = |\ . _____________________________________________________________________________ ( )4 3/ 232/ 64M dMc MI d dot t= = = 3-42 (a)Chapter 3 - Rev. A, Page 29/100 3332 32(218.75)0.420 in .(30 000)Md Ato t= = = ns (b) 2/ 4V VA dtt= = 4 4( 500)0.206 in .(15000)Vd Anstt t= = = (c) ( )24 43 3 / 4V VA dtt= = 4 4 4 4(500)0.238 in .3 3 (15000)Vd Att t= = = ns______________ __________________ ______________________________ _____________ _ _3-43 1 0 11 211 21 212 31 1 2terms for terms for 2 terms for 2 6p pq F x p x l x l x l aap pV F p x l x l x l aap p pM Fx x l x l x l aa += + + > ++= + + > ++= + + > + terms for x > l + a = 0 At x ( ) , 0, l a V M+= + = = 2 1 21 1 22 Fp p =23 1 1 21 2 20 (1)26 ( )( ) 0 2 (2)2 6p pF p a aa ap a p p F l aF l a a p pa a+ + =+ + + + = = From (1) and (2) 1 2 2 22 2(3 2 ), (3 ) (3)F Fp l a p l aa a= + = + From similar triang les 22 1 2 1 2 (4)ap b abp p p p p= =+ + Chapter 3 - Rev. A, Page 30/100 Mmax occurs where V = 0 max2 x l a b = + 2 3 1 1 2max2 3 1 1 2( 2 ) ( 2 ) ( 2 )2 6( 2 ) ( 2 ) ( 2 )2 6p p pM F l a b a b a bap p pFl F a b a b a ba+= + + += + Normally Mmax = Fl The fractional increase in the magnitude is ( ) ( )2 31 22 ( 2 ) 6 ( 2 ) (5)a b p p a a b + + ( For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in (3) 1( 2 ) F a b p FlA =( )1 22(1500)3 1.5 2(1.2) 14 375 lbf/in1.2p = + ( = ( )2 22(1500)3 1.5 1.2 11 875 lbf/in1.2p = + ( = (4) b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in Substituting into (5) yields _____________________________________________________________________________ -44 A = 0.036 89 or 3.7% higher than -Fl 3Chapter 3 - Rev. A, Page 31/100 12300(30)R =401800 6900 lbf2 30300(30) 101800 3900 lbf2 30390013 in300Ra+ == == = MB = 1800(10) = 18 000 lbfin x = 27 in = (1/2)3900(13) = 25 350 lbfin = = MB = 1800(10) = 18 000 lbfin x = 27 in = (1/2)3900(13) = 25 350 lbfin MM 3 413 420.5(3) 2.5(3)1.5 in61(3)(1 ) 0.25 in 121(1)(3 ) 2.25 in 12yII+= == == = Applying the parallel-axis theorem, (a)20.25 3(1.5 0.5) 2.25 3zI ( = + + + 2 4 (2.5 1.5) 8.5 in ( =18000( 1.5)At 10 in, 1.5 in, 3176 psi8.518000(2.5)At 10 in, 2.5 in, 5294 psi8.525350( 1.5)At 27 in, 1.5 in, 4474 psi8.5At 27 in, 2.5 in, xxxxx yx yx yx yoooo = = = = = = = == = = == = =25350(2.5)7456 psi8.5 = Max tension 5294 psi .Max compression 7456 psi .AnsAns== aximum shear stress due to V is at B, at the neutral axis. (b) The m max5100 lbf V =( )3max1.25(2.5)(1) 3.125 in5100(3.125)1875 psi .8.5(1)VQ y AVQAnsIbt' ' = = == = = (c) There are three potentially critical locations for the maximum shear stress, all at x = 27 in: (i) at the top where the bending stress is maximum, (ii) at the neutral axis where Chapter 3 - Rev. A, Page 32/100 the transverse shear is maximum, or (iii) in the web just above the flange where bending stress and shear stress are in their largest combination. For (i): The maximum bending stress was previously found to be 7456 psi, and the shear stress is zero. From Mohrs circle, maxmax74563728 psi 2 2ot = = = For (ii): The bending stress is zero, and the transverse shear stress was found previously to be 1875 psi. Thus, tmax = 1875 psi. For (iii): The bending stress at y = 0.5 in is 18000( 0.5)1059 psi8.5xo = = The transverse shear stress is 3(1)(3)(1) 3.0 in5100(3.0)1800 psi 8.5(1)Q y AVQIbt' ' = = == = = From Mohrs circle, 22max10591800 1876 psi 2t | |= + = |\ . The critical location is at x = 27 in, at the top surface, where tmax = 3728 psi. Ans. _____________________________________________________________________________ 3-45 (a) L = 10 in. Element A: ( )34(1000)(10)(0.5)10 101.9 kpsi( / 64)(1)A MyIot = = = , 0A AVQQ 0Ibt t = = = 2 22 2max101.9(0) 50.9 kpsi .2 2AA Ansot t| | | |= + = + = | | \ . \ . Element B: , 0 0B BMyyIo o = = = ( )32 334 0.54 41/ 12 in3 2 6 6r r rQ y A tt | || |' ' = = = = = | |\ .\ . Chapter 3 - Rev. A, Page 33/100 ( )34(1000)(1/ 12)10 1.698 kpsi( / 64)(1) (1)B VQIbtt = = = 22max01.698 1.698 kpsi .2 Ans t | |= + = |\ . Element C: ( )34(1000)(10)(0.25)10 50.93 kpsi( / 64)(1)C MyIot = = = ( )( ) ( ) ( )( )2 21 1 13/ 2 3/ 2 3/ 22 2 2 2 2 2113/ 22 21(2 ) 22 23 323r r ry y yryQ ydA y x dy y r y dyr y r r r yr y= = = (= = ( = } } } For C, y1 = r /2 =0.25 in ( )3/ 22 220.5 0.25 0.054133Q = = in3 2 2 2 212 2 2 0.5 0.25 0.866 in b x r y = = = = ( )34(1000)(0.05413)10 1.273 kpsi( / 64)(1) (0.866)C VQIbtt = = = 22max50.93(1.273) 25.50 kpsi .2 Ans t | |= + = |\ . (b) Neglecting transverse shear stress: Element A: Since the transverse shear stress at point A is zero, there is no change. max50.9 kpsi . Ans t = % error 0% . Ans = Element B: Since the only stress at point B is transverse shear stress, neglecting the transverse shear stress ignores the entire stress. 2max00 psi .2 Ans t | |= = |\ . 1.698 0% error *(100) 100% .1.698 Ans | |= = |\ . Chapter 3 - Rev. A, Page 34/100 Element C: 2max50.9325.47 kpsi .2 Ans t | |= = |\ . 25.50 25.47% error *(100) 0.12% .25.50 Ans| | = = |\ . (c) Repeating the process with different beam lengths produces the results in the table. Bending stress, o (kpsi) Transverse shear stress, t (kpsi) Max shear stress, tmax (kpsi) Max shear stress, neglecting t, tmax (kpsi) % error L = 10 in A 102 0 50.9 50.9 0 B 0 1.70 1.70 0 100 C 50.9 1.27 25.50 25.47 0.12 L = 4 in A 40.7 0 20.4 20.4 0 B 0 1.70 1.70 0 100 C 20.4 1.27 10.26 10.19 0.77 L = 1 in A 10.2 0 5.09 5.09 0 B 0 1.70 1.70 0 100 C 5.09 1.27 2.85 2.55 10.6 L = 0.1in A 1.02 0 0.509 0.509 0 B 0 1.70 1.70 0 100 C 0.509 1.27 1.30 0.255 80.4 Discussion: The transverse shear stress is only significant in determining the critical stress element as the length of the cantilever beam becomes smaller. As this length decreases, bending stress reduces greatly and transverse shear stress stays the same. This causes the critical element location to go from being at point A, on the surface, to point B, in the center. The maximum shear stress is on the outer surface at point A for all cases except L = 0.1 in, where it is at point B at the center. When the critical stress element is at point A, there is no error from neglecting transverse shear stress, since it is zero at that location. Neglecting the transverse shear stress has extreme significance at the stress element at the center at point B, but that location is probably only of practical significance for very short beam lengths. _____________________________________________________________________________ Chapter 3 - Rev. A, Page 35/100 3-46 1 0cR FlcM Fx x al== s s ( )2 2max666 0 .c l FxMbh bhFcxh xlboo= == s s a Ans _____________________________________________________________________________ 3-47 From Problem 3-46, 1, 0cR F V x al= = s s maxmax3 3 ( / ) 3 .2 2 2V c l F Fch Abh bh lbtt= = = ns From Problem 3-46, max6( ) Fcxh x . lbo=Sub in x = e and equate to h above. max maxmax2max3 623 .8Fc Fcelb lbFce Albt oot== ns _____________________________________________________________________________ 3-48 (a) x-z plane 20 1.5(0.5) 2(1.5) sin(30 )(2.25) (3)O zM R E = = + 21.375 kN .zR Ans = 10 1.5 2(1.5) sin(30 ) 1.375z zF R E = = + 11.625 kN .zR Ans = x-y plane 20 2(1.5) cos(30 )(2.25) (3)O yM R E = = + 21.949 kN .yR Ans = 10 2(1.5) cos(30 ) 1.949y yF R E = = + 10.6491 kN .yR Ans = Chapter 3 - Rev. A, Page 36/100 (b) (c) The transverse shear and bending moments for most points of interest can readily be taken straight from the diagrams. For 1.5 < x < 3, the bending moment equations are parabolic, and are obtained by integrating the linear expressions for shear. For convenience, use a coordinate shift of x' = x 1.5. Then, for 0 < x' < 1.5, ( )( )220.1250.1252At 0, 0.9375 0.5 0.125 0.9375zy zy yV xxM V dx x Cx M C M x x' = '' ' = = +' ' = = = = +}' ( )( )221.9490.6491 1.732 0.64911.1251.7320.64912At 0, 0.9737 0.8662 0.125 0.9375yzz zV x xM x x Cx M C M x x' ' = + = + ' ' = + +' ' = = = = ' By programming these bending moment equations, we can find My, Mz, and their vector combination at any point along the beam. The maximum combined bending moment is found to be at x = 1.79 m, where M = 1.433 kNm. The table below shows values at key locations on the shear and bending moment diagrams. x (m) Vz (kN) Vy (kN) V (kN) My (kNm) Mz (kNm) M (kNm) 0 1.625 0.6491 1.750 0 0 0 0.5 1.625 0.6491 1.750 0.8125 0.3246 0.8749 1.5 0.1250 0.6491 0.6610 0.9375 0.9737 1.352 1.625 0 0.4327 0.4327 0.9453 1.041 1.406 1.875 0.2500 0 0.2500 0.9141 1.095 1.427 3 1.375 1.949 2.385 0 0 0 Chapter 3 - Rev. A, Page 37/100 (d) The bending stress is obtained from Eq. (3-27), y Az Axz yM zM yI Io = + The maximum tensile bending stress will be at point A in the cross section of Prob. 3-34 (a), where distances from the neutral axes for both bending moments will be maximum. At A, for Mz, yA = 37.5 mm, and for My, zA = 20 mm. 3 36 4 640(75) 34(25)1.36(10 ) mm 1.36(10 ) m12 12zI = = =4 3 35 4 725(40) 25(6)2 2.67(10 ) mm 2.67(10 ) m12 12yI (= + = = ( 4 It is apparent the maximum bending moment, and thus the maximum stress, will be in the parabolic section of the bending moment diagrams. Programming Eq. (3-27) with the bending moment equations previously derived, the maximum tensile bending stress is found at x = 1.77 m, where My = 0.9408 kNm, Mz = 1.075 kNm, and ox = 100.1 MPa. Ans. _____________________________________________________________________________ 3-49 (a) x-z plane 3 6000 (1000)(4) (10)5 2O OyM M E = = + 1842.6 lbf in .OyM Ans = 3 60 (1000)5 2z OzF R E = = +00 175.7 lbf .OzR Ans = x-y plane 4 6000 (1000)(4) (10)5 2O OzM M E = = + 7442.5 lbf in .OzM Ans = 4 60 (1000)5 2y OyF R E = = 00 1224.3 lbf .OyR Ans = Chapter 3 - Rev. A, Page 38/100 (b) ( (c) 1/ 22 2( ) ( ) ( )y zV x V x V x ( = + 1/ 22 2( ) ( ) ( )y zM x M x M x ( = + x (m) Vz (kN) Vy (kN) V (kN) My (kNm) Mz (kNm) M (kNm) 0 175.7 1224.3 1237 1842.6 7442.6 7667 4 175.7 1224.3 1237 2545.4 2545.4 3600 10 424.3 424.3 600 0 0 0 (d) The maximum tensile bending stress will be at the outer corner of the cross section in the positive y, negative z quadrant, where y = 1.5 in and z = 1 in. 3 342(3) (1.625)(2.625)2.051 in12 12zI = = 3 343(2) (2.625)(1.625)1.601 in12 12yI = = At x = 0, using Eq. (3-27), yzxz yM zM yI Io = + ( 7442.6)(1.5) ( 1842.6)( 1)6594 psi2.051 1.601xo = + = Check at x = 4 in, ( 2545.4)(1.5) ( 2545.4)( 1)2706 psi2.051 1.601xo = + = The critical location is at x = 0, where ox = 6594 psi. Ans. _____________________________________________________________________________ Chapter 3 - Rev. A, Page 39/100 3-50 The area within the wall median line, Am, is Square: 2( )mA b t = . From Eq. (3-45) 2sq all all2 2( )mT A t b t t t t = = Round: 2( ) /mA b t t = 42rd all2 ( ) / 4 T b t t t t = Ratio of Torques 2sqall2rd all2( ) 41.27( ) / 2T b t tT b t t tt t t= = = Twist per unit length from Eq. (3-46) is all all1 2 224 4 2m m m mm m mTL A t L L LCGA t GA t G A Ammt tu = = = = Square: sq24( )( )b tCb tu = Round: rd2 2( ) 4(( ) / 4 ( )b t b tC Cb t b t) tut = = Ratio equals 1. Twists are the same. _____________________________________________________________________________ 3-51 (a) The area enclosed by the section median line is Am = (1 0.0625)2 = 0.8789 in2 and the length of the section median line is Lm = 4(1 0.0625) = 3.75 in. From Eq. (3-45), 2 2(0.8789)(0.0625)(12 000) 1318 lbf in .mT A t Ans t = = = From Eq. (3-46), ( )( )( ) ( )126 2(1318)(3.75) 360.0801 rad 4.59 .44 11.5 10 (0.8789) 0.0625mmTL ll AGA tu u = = = = = ns (b) The radius at the median line is rm = 0.125 + (0.5)(0.0625) = 0.15625 in. The area enclosed by the section median line is Am = (1 0.0625)2 4(0.15625)2 + 4( /4)(0.15625)2 = 0.8579 in2. The length of the section median line is Lm = 4[1 0.0625 2(0.15625)] + 2(0.15625) = 3.482 in. Chapter 3 - Rev. A, Page 40/100 From Eq. (3-45), 2 2(0.8579)(0.0625)(12 000) 1287 lbf in .mT A t Ans t = = = From Eq. (3-46), ( )( )( ) ( )126 2(1287)(3.482) 360.0762 rad 4.37 .44 11.5 10 (0.8579) 0.0625mmTL ll AGA tu u = = = = = ns _____________________________________________________________________________ 3-52 31133 3i iii iT GTGL c uu = = iL c 33 11 2 31 .3 i iiGT T T T L c Ansu== + + = From Eq. (3-47), t = Gu1c G and u1 are constant, therefore the largest shear stress occurs when c is a maximum. max 1 max . G c Ans t u = _____________________________________________________________________________ 3-53 (b) Solve part (b) first since the twist is needed for part (a). ( )max allow12 6.89 82.7 MPa t t = = = ( )( )6max19max82.7 100.348 rad/m .79.3 10 (0.003) AnsGctu = = = (a) ( )9 331 1 110.348(79.3) 10 (0.020)(0.002 )1.47 N m .3 3GL cT Au= = = ns ( )( )9 332 2 229 333 3 331 2 30.348(79.3) 10 (0.030)(0.003 )7.45 N m .3 30.348(79.3) 10 (0)(0 )0 .3 31.47 7.45 0 8.92 N m .GL cT AGL cT AT T T T Ansuu= = = = = == + + = + + = nsns _____________________________________________________________________________ Chapter 3 - Rev. A, Page 41/100 3-54 (b) Solve part (b) first since the twist is needed for part (a). ( ) ( )3 max16max120008.35 10 rad/in .11.5 10 (0.125) AnsGctu = = = (a) ( ) ( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )( )3 6 331 1 113 6 332 2 223 6 333 3 331 2 38.35 10 11.5 10 0.75 0.06255.86 lbf in .3 38.35 10 11.5 10 1 0.12562.52 lbf in .3 38.35 10 11.5 10 0.625 0.06254.88 lbf in .3 35.86 62.52 4GL cT AGL cT AGL cT AT T T Tuuu= = = = = = = = = = + + = + + .88 73.3 lbf in . Ans = nsnsns _____________________________________________________________________________ 3-55 (b) Solve part (b) first since the twist is needed for part (a). ( )max allow12 6.89 82.7 MPa t t = = = ( )( )6max19max82.7 100.348 rad/m .79.3 10 (0.003) AnsGctu = = = (a) ( )9 331 1 110.348(79.3) 10 (0.020)(0.002 )1.47 N m .3 3GL cT Au= = = ns ( )( )9 332 2 229 333 3 331 2 30.348(79.3) 10 (0.030)(0.003 )7.45 N m .3 30.348(79.3) 10 (0.025)(0.002 )1.84 N m .3 31.47 7.45 1.84 10.8 N m .GL cT AGL cT AT T T T Ansuu= = = = = = = + + = + + = nsns _____________________________________________________________________________ 3-56 (a) From Eq. (3-40), with two 2-mm strips, ( )( )( )( )( )6 22maxmax80 10 0.030 0.0023.08 N m3 1.8 / ( / ) 3 1.8 / 0.030 / 0.0022(3.08) 6.16 N m .bcTb cT Anst= = =+ += = Chapter 3 - Rev. A, Page 42/100 From the table on p. 102, with b/c = 30/2 = 15, o = | and has a value between 0.313 and 0.333. From Eq. (3-40), 10.3213 1.8 / (30 / 2)o =+ From Eq. (3-41), ( )( )( )( )33 93.08(0.3)0.151 rad .0.321 0.030 0.002 79.3 106.1640.8 N m .0.151tTlAnsbc GTk Ansu|u= = == = = From Eq. (3-40), with a single 4-mm strip, ( )( )( )( )( )6 22maxmax80 10 0.030 0.00411.9 N m .3 1.8/ ( / ) 3 1.8/ 0.030 / 0.004bcT Ab ct= = = + + ns Interpolating from the table on p. 102, with b/c = 30/4 = 7.5, 7.5 6(0.307 0.299) 0.299 0.3058 6| = + = From Eq. (3-41) ( )( )( )( )33 911.9(0.3)0.0769 rad .0.305 0.030 0.004 79.3 1011.9155 N m .0.0769tTlAnsbc GTk Ansu|u= = == = = (b) From Eq. (3-47), with two 2-mm strips, ( )( )( )( )2 62max0.030 0.002 80 103.20 N m3 32(3.20) 6.40 N m .LcTT Anst= = = = = ( )( )( )( )33 93 3(3.20)(0.3)0.151 rad .0.030 0.002 79.3 106.40 0.151 42.4 N m .tTlAnsLc Gk T Ansuu= = == = = From Eq. (3-47), with a single 4-mm strip, ( )( )( )( )2 62max0.030 0.004 80 1012.8 N m .3 3LcT At= = = ns Chapter 3 - Rev. A, Page 43/100 ( )( )( )( )33 93 3(12.8)(0.3)0.0757 rad .0.030 0.004 79.3 1012.8 0.0757 169 N m .tTlAnsLc Gk T Ansuu= = == = = The results for the spring constants when using Eq. (3-47) are slightly larger than when using Eq. (3-40) and Eq. (3-41) because the strips are not infinitesimally thin (i.e. b/c does not equal infinity). The spring constants when considering one solid strip are significantly larger (almost four times larger) than when considering two thin strips because two thin strips would be able to slip along the center plane. _____________________________________________________________________________ 3-57 (a) Obtain the torque from the given power and speed using Eq. (3-44). (40000)9.55 9.55 152.8 N m2500HTn= = = max316 Tr TJ dtt= = ( )( )( )1 31 36max16 152.8160.0223 m 22.3 mm .70 10Td Att t (| | (= = = = | (\ . ns (b) (40000)9.55 9.55 1528 N m250HTn= = = ( )( )1 3616(1528)0.0481 m 48.1 mm .70 10d At ( (= = = ( ns _____________________________________________________________________________ 3-58 (a) Obtain the torque from the given power and speed using Eq. (3-42). 63025 63025(50)1261 lbf in2500HTn= = = max316 Tr TJ dtt= = ( )1 3 1 3max16 1261160.685 in .(20000)Td Att t ( | |= = = | (\ . ns (b) 63025 63025(50)12610 lbf in250HTn= = = 1 316(12610)1.48 in .(20000)d At (= = ( ns _____________________________________________________________________________ Chapter 3 - Rev. A, Page 44/100 3-59 ( )( ) ( )6 33maxmax350 10 0.0316 265 N m16 16d TTdtt ttt= = = = Eq. (3-44), ( )3265(2000)55.5 10 W 55.5 kW .9.55 9.55TnH A = = = = ns _____________________________________________________________________________ 3-60 ( )( )( )( )( )( )3 6 334 9416 110 10 0.020 173 N m16 160.020 79.3 10 15180 32 32(173)1.89 m .TT ddTl d GlJG Tl Anst tt ttttt uu= = = =| | |\ .= = == _____________________________________________________________________________ 3-61 ( )( )( )( )( )3 3344 616 30 000 0.75 2485 lbf in16 1632 32(2485)(24)0.167 rad 9.57 .0.75 11.5 10TT ddTl TlAnsJG d Gt tt ttut t= = = = = = = = = _____________________________________________________________________________ 3-62 (a) 4 4max max max maxsolid hollow( ) 16 16o oo oJ d J d dT Tr d r dt t t t t t = = = =4i ( )( )44solid hollow44solid36% (100%) (100%) (100%) 65.6% .40ioT T dT AT dA = = = = ns (b) ( )2 2solid hollow, o oW kd W k d d = =2i ( )( )22solid hollow22solid36% (100%) (100%) (100%) 81.0% .40ioW W dW AW dA = = = = ns _____________________________________________________________________________ 3-63 (a) ( )444maxmax max maxsolid hollow 16 16d xdJ d JT Tr d r dt tt t t t ( = = = = 44 solid hollow4soli( )% (100%) (100%) (100%) .dT T xdT xT dA = = = Ans Chapter 3 - Rev. A, Page 45/100 (b) ( )( )22 2solid hollow W kd W k d xd = = ( )22 solid hollow2solid% (100%) (100%) (100%) .xd W WW xW dA = = = Ans Plot %AT and %AW versus x. The value of greatest difference in percent reduction of weight and torque is 25% and occurs at 2 2 x = . _____________________________________________________________________________ 3-64 (a) ( ) ( )( ) ( ) ( )463 442.8149 104200 2 120 1032 0.70dTcJ dd dtt= = = ( ( ) ( )1 34262.8149 106.17 10 m 61.7 mm120(10 )d | | |= = = |\ . dFrom Table A-17, the next preferred size is d = 80 mm. Ans. i = 0.7d = 56 mm. The next preferred size smaller is di = 50 mm Ans. (b) ( )( ) ( ) ( )( ) ( ) ( )4 4 444200 2 4200 0.050 230.8 MPa .32 0.080 0.050 32 iidTcAnsJ d dtt t= = = = ( ( _____________________________________________________________________________ Chapter 3 - Rev. A, Page 46/100 3-65 (1500)9.55 9.55 1433 N m10HTn= = = ( )( )( )1 31 33616 143316 16 = 0.045 m 45 mm 80 10CCT Tddttt t t (| | (= = = = | (\ . From Table A-17, select 50 mm. Ans. (a) ( )( )( ) ( )6start316 2 1433117 10 Pa 117 MPa .0.050 Ans tt= = = (b) Design activity _____________________________________________________________________________ 3-66 ( )( )1 31 3363 025 63 025(1)7880 lbf in816 788016 16 = 1.39 in 15 000CCHTnT Tddttt t t= = = (| |= = = ( |\ . From Table A-17, select 1.40 in. Ans. _____________________________________________________________________________ 3-67 For a square cross section with side length b, and a circular section with diameter d, 2 2square circular 4 2A A b d b dt t= = = From Eq. (3-40) with b = c, ( )3max 2 3 3 square1.8 1.8 23 3 (4.8) 6.896/ 1T T Tbc b c b d dtt| | | | | |= + = + = = | | |\ . \ . \ .3T For the circular cross section, ( )max 3 3 circular165.093T Td dtt= = ( )( )3maxsquaremaxcircular36.8961.3545.093TdTdtt = = The shear stress in the square cross section is 35.4% greater. Ans. (b) For the square cross section, from the table on p. 102, = 0.141. From Eq. (3-41), Chapter 3 - Rev. A, Page 47/100 square 4 3 411.500.1412Tl Tl Tl Tlbc G b G d Gd Gu| | t= = = =| | |\ .4 For the circular cross section, ( )4 410.1932rd Tl Tl TlGJ d G G dut= = = 4411.501.12910.19sqrdTld GTld Guu = = The angle of twist in the square cross section is 12.9% greater. Ans. _____________________________________________________________________________ 3-68 (a) ( )( ) ( )(( )1 22 1 2 22 210.150 (500 75)(4) 5 1700 0.15 51700 4.25 0 400 lbf .0.15 400 60 lbf .T TT T T TT T AnsT Ans== = = = == = ) Ts(b) 0 575(10) 460(28) (40)178.25 178 lbf .0 575 460 178.25293.25 lbf .O CCOOM RR AnF RR Ans= = + == = + += (c) Chapter 3 - Rev. A, Page 48/100 (d) The maximum bending moment is at x = 10 in, and is M = 2932.5 lbfin. Since the shaft rotates, each stress element will experience both positive and negative bending stress as it moves from tension to compression. The torque transmitted through the shaft from A to B is T = (500 75)(4) = 1700 lbfin. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, ( )3 332 2932.53215 294 psi = 15.3 kpsi .(1.25)Mc MAnsI dot t= = = = 3 316 16(1700)4433 psi = 4.43 kpsi .(1.25)Tr TAnsJ dtt t= = = = (e) ( ) ( )( ) ( )2 2221 2122 222max15.3 15.3, 4.432 2 2 216.5 kpsi .1.19 kpsi .15.34.43 8.84 kpsi .2 2x xxyxxyAnsAnsAnso oo o tooot t| | | |= + = + | | \ . \ .== | | | |= + = + = | | \ . \ . _____________________________________________________________________________ 3-69 (a) ( ) ( ) ( ) ( )( )( )2 132 1 1 131 120.150 1800 270 (200) (125) 306 10 125 0.15306 10 106.25 0 2880 N .0.15 2880 432 N .T TT T TT T AnsT Ans== = + = + = == = T T(b) 0 3312(230) (510) 2070(810)1794 N .0 3312 1794 20703036 N .O CCy OOM RR AnsF RR Ans= = + == = + + = (c) Chapter 3 - Rev. A, Page 49/100 (d) The maximum bending moment is at x = 230 mm, and is M = 698.3 Nm. Since the shaft rotates, each stress element will experience both positive and negative bending stress as it moves from tension to compression. The torque transmitted through the shaft from A to B is T = (1800 270)(0.200) = 306 Nm. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, ( ) ( )33 332 698.332263 10 Pa 263 MPa .(0.030)Mc MAnsI dot t= = = = = ( )63 316 16(306)57.7 10 Pa 57.7MPa .(0.030)Tr TAnsJ dtt t= = = = = (e) ( ) ( )( ) ( )2 2221 2122 222max263 263, 57.72 2 2 2275 MPa .12.1 MPa .26357.7 144 MPa .2 2x xxyxxyAnsAnsAnso oo o tooot t| | | |= + = + | | \ . \ .== | | | |= + = + = | | \ . \ . _____________________________________________________________________________ 3-70 (a) ( ) ( ) ( )( )2 12 1 1 11 120.150 300 50 (4) (3) 1000 0.15 (3)1000 2.55 0 392.16 lbf .0.15 392.16 58.82 lbf .T TT T TT T AnsT Ans== = + = + = == = T T (b) Chapter 3 - Rev. A, Page 50/100 0 450.98(16) (22)327.99 lbf .0 450.98 327.99122.99 lbf .0 350(8) (22)127.27 lbf .0 350 127.27222.73 lbf .O y C zC zz OzOzOz C yC yy O yO yM RR AnsF RR AnsM RR AnsF RR Ans= = = = = + = = = += = = + = Chapter 3 - Rev. A, Page 51/100 (c) (d) Combine the bending moments from both planes at A and B to find the critical location. 2 22 2(983.92) ( 1781.84) 2035 lbf in(1967.84) ( 763.65) 2111 lbf inABMM= + == + = The critical location is at B. The torque transmitted through the shaft from A to B is T = (300 50)(4) = 1000 lbfin. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, ( )3 332 21113221502 psi = 21.5 kpsi .(1)Mc MAnsI dot t= = = = 3 316 16(1000)5093 psi = 5.09 kpsi .(1)Tr TAnsJ dtt t= = = = (e) ( ) ( )( ) ( )2 2221 2122 222max21.5 21.5, 5.092 2 2 222.6 kpsi .1.14 kpsi .21.55.09 11.9 kpsi .2 2x xxyxxyAnsAnsAnso oo o tooot t| | | |= + = + | | \ . \ .== | | | |= + = + = | | \ . \ . _____________________________________________________________________________ Chapter 3 - Rev. A, Page 52/100 3-71 (a) ( ) ( ) ( )( )2 12 1 1 11 120.150 300 45 (125) (150) 31 875 0.15 (150)31 875 127.5 0 250 N mm .0.15 250 37.5 N mm .T TT T T TT T AnsT Ans== = + = + = = = = T(b) oooo0 345sin 45 (300) 287.5(700) (850)150.7 N .0 345cos 45 287.5 150.7107.2 N .0 345sin 45 (300) (850)86.10 N .0 345cos 45 86.10O y C zC zz OzOzOz C yC yy O yO yM RR AnsF RR AnsM RR AnsF RR= = = = = + == = += = = + 157.9 N . Ans = (c) (d) From the bending moment diagrams, it is clear that the critical location is at A where both planes have the maximum bending moment. Combining the bending moments from the two planes, ( ) ( )2 247.37 32.16 57.26 N m M = + = Chapter 3 - Rev. A, Page 53/100 The torque transmitted through the shaft from A to B is T = (300 45)(0.125) = 31.88 Nm. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, ( ) ( )63 332 57.263272.9 10 Pa 72.9 MPa .(0.020)Mc MAnsI dot t= = = = = ( )63 316 16(31.88)20.3 10 Pa 20.3 MPa .(0.020)Tr TAnsJ dtt t= = = = = (e) ( ) ( )( ) ( )2 2221 2122 222max72.9 72.9, 20.32 2 2 278.2 MPa .5.27 MPa .72.920.3 41.7 MPa .2 2x xxyxxyAnsAnsAnso oo o tooot t| | | |= + = + | | \ . \ .== | | | |= + = + = | | \ . \ . _____________________________________________________________________________ 3-72 (a) 0 300(cos 20 )(10) (cos 20 )(4)750 lbf . BBT FF Ans= = +=(b) 0 300(cos 20 )(16) 750(sin 20 )(39) (30)183 lbf .0 300(cos 20 ) 183 750(sin 20 )208 lbf .0 300(sin 20 )(16) (30) 750(cos 20 )(39)861 lbf .0 300Oz C yC yy O yO yO y C zC zz OzM RR AnsF RR AnsM RR AnsF R= = +== = + + = = = = = = (sin 20 ) 861 750(cos 20 )259 lbf .OzR Ans +=Chapter 3 - Rev. A, Page 54/100 (c) (d) Combine the bending moments from both planes at A and C to find the critical location. 2 22 2( 3336) ( 4149) 5324 lbf in( 2308) ( 6343) 6750 lbf inACMM= + = = + = The critical location is at C. The torque transmitted through the shaft from A to B is . For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, ( )( ) 300cos 20 10 2819 lbf in T = = ( )3 332 67503235 203 psi = 35.2 kpsi .(1.25)Mc MAnsI dot t= = = = 3 316 16(2819)7351 psi = 7.35 kpsi .(1.25)Tr TAnsJ dtt t= = = = (e) ( ) ( )( ) ( )2 2221 2122 222max35.2 35.2, 7.352 2 2 236.7 kpsi .1.47 kpsi .35.27.35 19.1 kpsi .2 2x xxyxxyAnsAnsAnso oo o tooot t| | | |= + = + | | \ . \ .== | | | |= + = + = | | \ . \ . _____________________________________________________________________________ Chapter 3 - Rev. A, Page 55/100 3-73 (a) 0 11000(cos 20 )(300) (cos 25 )(150)22 810 N . BBT FF Ans= = +=(b) 0 11000(sin 20 )(400) 22 810(sin 25 )(750) (1050)8319 N .Oz C yC yM RR Ans= = += 0 11000(sin 20 ) 22 810sin(25 ) 83195083 N .0 11000(cos 20 )(400) 22 810(cos 25 )(750) (1050)10 830 N .0 11000(cos 20 ) 22 810(cos 25 ) 10 830494 N .y O yO yO y C zC zz OzOzF RR AnsM RR AnsF RR Ans= = +== = = = = + =(c) (d) From the bending moment diagrams, it is clear that the critical location is at B where both planes have the maximum bending moment. Combining the bending moments from the two planes, ( ) ( )2 22496 3249 4097 N m M = + = The torque transmitted through the shaft from A to B is . ( )( ) 11000cos 20 0.3 3101 N m T = =For a stress element on the outer surface where the bendi