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Shigley’s MED, 10th edition Chapter 1 Solutions, Page 1/12 Chapter 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 ( )2 1.10 1.43 . 0.85 0.95d n Ans= = ______________________________________________________________________________ 1-10 (a) X1 + X2: ( ) ( ) 1 2 1 1 2 2 1 2 1 2 1 2 error . x x X e X e e x x X X e e Ans + = + + + = = + − + = + (b) X1 − X2: ( ) ( ) ( ) 1 2 1 1 2 2 1 2 1 2 1 2 . x x X e X e e x x X X e e Ans − = + − + = − − − = − (c) X1 X2: ( )( )1 2 1 1 2 2 1 2 1 2 1 2 2 1 1 2 1 2 1 2 2 1 1 2 1 2 . x x X e X e e x x X X X e X e e e e eX e X e X X Ans X X = + + = − = + + ≈ + = + Shigley’s MED, 10th edition Chapter 1 Solutions, Page 2/12 (d) X1/X2: 1 1 1 1 1 1 2 2 2 2 2 2 1 2 2 1 1 1 2 1 2 2 2 2 2 1 2 1 2 1 1 1 1 2 2 2 2 1 2 1 1 11 1 then 1 1 1 1 Thus, . x X e X e X x X e X e X e e e X e e e e X X e X X X X X x X X e e e Ans 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 ( ) ( )33 25 1032 1000 1.006 in .2.5d S d Ans n d σ pi = ⇒ = ⇒ = Table A-17: d = 141 in Ans. Factor of safety: ( ) ( ) ( ) 3 3 25 10 4.79 . 32 1000 1.25 S n Ans σ pi = = = ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 1 Solutions, Page 3/12 1-13 (a) Eq. (1-6) 1 1 8480 122.9 kcycles 69 k i i i x f x N = = = =∑ Eq. (1-7) 2 2 1/ 22 1 1104 600 69(122.9) 30.3 kcycles . 1 69 1 k i i i x f x N x s Ans N = − − = = = − − ∑ (b) Eq. (1-5) 115115 115 122.9 0.2607 ˆ 30.3 x x x x x x z s µ σ − − − = = = = − Interpolating from Table (A-10) 0.2600 0.3974 0.2607 x ⇒ x = 0.3971 0.2700 0.3936 NΦ(−0.2607) = 69 (0.3971) = 27.4 ≈ 27 Ans. From the data, the number of instances less than 115 kcycles is x f f x f x2 60 2 120 7200 70 1 70 4900 80 3 240 19200 90 5 450 40500 100 8 800 80000 110 12 1320 145200 120 6 720 86400 130 10 1300 169000 140 8 1120 156800 150 5 750 112500 160 2 320 51200 170 3 510 86700 180 2 360 64800 190 1 190 36100 200 0 0 0 210 1 210 44100 Σ 69 8480 1 104 600 Shigley’s MED, 10th edition Chapter 1 Solutions, Page 4/12 2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal) ____________________________________________________________________________ 1-14 x f f x f x2 174 6 1044 181656 182 9 1638 298116 190 44 8360 1588400 198 67 13266 2626668 206 53 10918 2249108 214 12 2568 549552 222 6 1332 295704 Σ 197 39126 7789204 Eq. (1-6) 1 1 39 126 198.61 kpsi 197 k i i i x f x N = = = =∑ Eq. (1-7) 2 2 1 22 1 7 789 204 197(198.61) 9.68 kpsi . 1 197 1 k i i i x f x N x s Ans N = − − = = = − − ∑ ______________________________________________________________________________ 1-15 122.9 kcycles and 30.3 kcyclesLL s= = Eq. (1-5) 10 1010 122.9 ˆ 30.3 x L x x L x z s µ σ − − − = = = Thus, x10 = 122.9 + 30.3 z10 = L10 From Table A-10, for 10 percent failure, z10 = −1.282. Thus, L10 = 122.9 + 30.3(−1.282) = 84.1 kcycles Ans. ___________________________________________________________________________ Shigley’s MED, 10th edition Chapter 1 Solutions, Page 5/12 1-16 x f fx fx2 93 19 1767 164331 95 25 2375 225625 97 38 3686 357542 99 17 1683 166617 101 12 1212 122412 103 10 1030 106090 105 5 525 55125 107 4 428 45796 109 4 436 47524 111 2 222 24642 Σ 136 13364 1315704 Eq. (1-6) 1 1 13 364 / 136 98.26471 =98.26 kpsi k i i i x f x N = = = =∑ Eq. (1-7) 2 2 1 22 1 1 315 704 136(98.26471) 4.30 kpsi 1 136 1 k i i i x f x N x s N = − − = = = − − ∑ Note, for accuracy in the calculation given above, x needs to be of more significant figures than the rounded value. For a normal distribution, from Eq. (1-5), and a yield strength exceeded by 99 percent (R = 0.99, pf = 0.01), 0.01 0.01 0.01 98.26 ˆ 4.30 x x x x x x x z s µ σ − − − = = = Solving for the yield strength gives x0.01 = 98.26 + 4.30 z0.01 From Table A-10, z0.01 = − 2.326. Thus x0.01 = 98.26 + 4.30(− 2.326) = 88.3 kpsi Ans. ______________________________________________________________________________ 1-17 Eq. (1-9): R = 1 n i i R = ∏ = 0.98(0.96)0.94 = 0.88 Overall reliability = 88 percent Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 1 Solutions, Page 6/12 1-18 Obtain the coefficients of variance for strength and stress ˆ 23.5 0.07532 312 syS S sy C S σ = = = ˆ ˆ 145 0.09667 1 500 TC T τ σ σ σ τ = = = = For R = 0.99, from Table A-10, z = − 2.326. Eq. (1-12): ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 2.326 0.07532 1 2.326 0.09667 1.3229 1.32 . 1 2.326 0.07532 S S z C z C n z C Ans σ+ − − − = − + − − − − − = = = − − From the given equation for stress, max 3 16syS T n d τ pi = = Solving for d gives 1/3 1/3 6 16 16(1500)1.3229 0.0319 m 31.9 mm .(312)10 sy T nd Ans Spi pi = = = = ______________________________________________________________________________ 1-19 Obtain the coefficients of variance for stress and strength ˆ ˆ 5 0.09231 65 PC Pσ σ σ σ σ µ = = = = ˆ ˆ 6.59 0.06901 95.5 ySS S S y C S σσ µ = = = = (a) 1.2n = Shigley’s MED, 10th edition Chapter 1 Solutions, Page 7/12 Eq. (1-11): ( )2 2 2 2 2 2 1 1.2 1 1.6127 1.2 0.06901 0.09231 d d S n z n C Cσ − − = − = − = − + + Interpolating Table A-10, 1.61 0.0537 1.6127 Φ ⇒ Φ = 0.0534 1.62 0.0526 R = 1 − 0.0534 = 0.9466 Ans. ( ) ( ) 2 2 4 65 1.24 1.020 in . / ( / 4) 4 95.5 y y y y S S d S Pn n d Ans P d P S pi σ pi pi pi = = = ⇒ = = = (b) 1.5n = ( )2 2 2 1.5 1 3.605 1.5 0.06901 0.09231 z − = − = − + 3.6 0.000159 3.605 Φ ⇒ Φ = 0.00015645 3.7 0.000108 R = 1 − 0.00015645 = 0.9998 Ans. ( ) ( ) 4 65 1.54 1.140 in . 95.5y Pnd Ans Spi pi = = = ______________________________________________________________________________ 1-20 max max 90 383 473 MPaa bσµ σ σ σ= = + = + = From footnote 9, p. 25 of text, ( ) max 1/ 22 2 2 2 1/ 2 ˆ ˆ ˆ (8.4 22.3 ) 23.83 MPa a bσ σ σ σ σ σ= + = + = max max max max max ˆ ˆ 23.83 0.0504 473 C σ σσ σ σ σ µ σ = = = = ˆ ˆ 42.7 0.0772 553 y y y y S S S S y C S σ σ µ = = = = Shigley’s MED, 10th edition Chapter 1 Solutions, Page 8/12 max 553 1.169 1.17 . 473 yS n Ans σ = = = = Eq. (1-11): ( )2 2 2 2 2 2 1 1.169 1 1.635 1.169 0.0772 0.0504 d d S n z n C Cσ − − = − = − = − + + From Table A-10, Φ(− 1.635) = 0.05105 R = 1 − 0.05105 = 0.94895 = 94.9 percent Ans. ______________________________________________________________________________ 1-21 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 allt t=∑w = 0.001 + 0.003 + 0.004 +0.010 = 0.018 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. ______________________________________________________________________________ 1-22 V = xyz, and x = a ± ∆ a, y = b ± ∆ b, z = c ± ∆ c, V abc= ( ) ( )( )V a a b b c c abc bc a ac b ab c a b c b c a c a b a b c = ± ∆ ± ∆ ± ∆ = ± ∆ ± ∆ ± ∆ ± ∆ ∆ ± ∆ ∆ ± ∆ ∆ ± ∆ ∆ ∆ The higher order terms in ∆ are negligible. Thus, V bc a ac b ab c∆ ≈ ∆ + ∆ + ∆ and, .V bc a ac b ab c a b c a b c Ans V abc a b c a b c ∆ ∆ + ∆ + ∆ ∆ ∆ ∆ ∆ ∆ ∆ ≈ = + + = + + For the numerical values given, ( ) 31.500 1.875 3.000 8.4375 inV = = Shigley’s MED, 10th edition Chapter 1 Solutions, Page 9/12 ( ) 30.002 0.003 0.004 0.004267 0.004267 8.4375 0.0360 in 1.500 1.875 3.000 V V V ∆ ≈ + + = ⇒ ∆ ≈ = V = 8.4375 ± 0.0360 in3 Ans. This answer yields 8.4735 8.4015 V ≈ in, whereas, exact is 8.473551.. 8.401551.. V = in ______________________________________________________________________________ 1-23 wmax = 0.05 in, wmin = 0.004 in 0.05 0.004 0.027 in 2 + =w = Thus, ∆ w = 0.05 − 0.027 = 0.023 in, and then, w = 0.027 ± 0.023 in. 0.027 0.042 1.5 1.569 in a b c a a − − = − − = w = tw = all t∑ ⇒ 0.023 = ta + 0.002 + 0.005 ⇒ ta = 0.016 in Thus, a = 1.569 ± 0.016 in Ans. ______________________________________________________________________________ 1-24 ( )2 3.734 2 0.139 4.012 ino iD D d= + = + = ( )all 0.028 2 0.004 0.036 inoDt t= = + =∑ Do = 4.012 ± 0.036 in Ans. ______________________________________________________________________________ 1-25 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= + = + = ( )all 0.13 2 0.08 0.29 mmoDt t= = + =∑ Do = 14.43 ± 0.29 mm Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 1 Solutions, Page 10/12 1-26 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= + = + = ( )all 0.30 2 0.10 0.50 mmoDt t= = + =∑ Do = 41.58 ± 0.50 mm Ans. ______________________________________________________________________________ 1-27 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= + = + = ( )all 0.035 2 0.003 0.041 inoDt t= = + =∑ Do = 5.443 ± 0.041 in Ans. ______________________________________________________________________________ 1-28 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= + = + = ( )all 0.012 2 0.005 0.022 inoDt t= = + =∑ Do = 1.520 ± 0.022 in Ans. ______________________________________________________________________________ 1-29 From Table A-2, (a) σ = 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. Shigley’s MED, 10th edition Chapter 1 Solutions, Page 11/12 (h) V = 1000/946 = 1.06 qt Ans. ______________________________________________________________________________ 1-30 From Table A-2, (a) l = 5(0.305) = 1.53 m Ans. (b) σ = 90(6.89) = 620 MPa Ans. (c) p = 25(6.89) = 172 kPa Ans. (d) Z =12(16.4) = 197 cm3 Ans. (e) w = 0.208(175) = 36.4 N/m Ans. (f) δ = 0.001 89(25.4) = 0.048 0 mm Ans. (g) v = 1 200(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. ______________________________________________________________________________ 1-31 (a) σ = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi Ans. (b) σ = 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) θ = Tl /GJ = 9 740(9.85)/[11.3(106)(pi /32)1.004] = 8.648(10−2) rad = 4.95° Ans. ______________________________________________________________________________ 1-32 (a) σ =F / wt = 1000/[25(5)] = 8 MPa Ans. (b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4 Ans. (c) I =pi d4/64 = pi (25.4)4/64 = 20.4(103) mm4 Ans. (d) τ =16T /pi d 3 = 16(25)103/[pi (12.7)3] = 62.2 MPa Ans. ______________________________________________________________________________ 1-33 Shigley’s MED, 10th edition Chapter 1 Solutions, Page 12/12 (a) τ =F /A = 2 700/[pi (0.750)2/4] = 6110 psi = 6.11 kpsi Ans. (b) σ = 32Fa/pi d 3 = 32(180)31.5/[pi (1.25)3] = 29 570 psi = 29.6 kpsi Ans. (c) Z =pi (do4 − di4)/(32 do) = pi (1.504 − 1.004)/[32(1.50)] = 0.266 in3 Ans. (d) k = (d 4G)/(8D 3 N) = 0.062 54(11.3)106/[8(0.760)3 32] = 1.53 lbf/in Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 2 Solutions, Page 1/22 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 540°C (1000°F): 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 425°C (800°F) Ans. (b) Maximize elongation: Q&T at 650°C (1200°F) 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 10 47.4 kN m/kg . 76.5 102 yS Ans ρ = = ⋅ 2011-T6 aluminum: Tables A-22 and A-5 ( ) ( ) 3169 10 62.3 kN m/kg . 26.6 102 yS Ans ρ = = ⋅ Ti-6Al-4V titanium: Tables A-24c and A-5 ( ) ( ) 3830 10 187 kN m/kg . 43.4 102 yS Ans ρ = = ⋅ 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 10 40.7 kN m/kg 70.6 102 utS Ans ρ = = ⋅ ______________________________________________________________________________ 2-4 AISI 1018 CD steel: Table A-5 ( ) ( ) 6 630.0 10 106 10 in . 0.282 E Ans γ = = 2011-T6 aluminum: Table A-5 ( ) ( ) 6 610.4 10 106 10 in . 0.098 E Ans γ = = Shigley’s MED, 10th edition Chapter 2 Solutions, Page 2/22 Ti-6Al-6V titanium: Table A-5 ( ) ( ) 6 616.5 10 103 10 in . 0.160 E Ans γ = = No. 40 cast iron: Table A-5 ( ) ( ) 6 614.5 10 55.8 10 in . 0.260 E Ans γ = = ______________________________________________________________________________ 2-5 22 (1 ) 2 E GG v E v G − + = ⇒ = Using values for E and G from Table A-5, Steel: ( )( ) 30.0 2 11.5 0.304 . 2 11.5 v Ans − = = The percent difference from the value in Table A-5 is 0.304 0.292 0.0411 4.11 percent . 0.292 Ans− = = Aluminum: ( )( ) 10.4 2 3.90 0.333 . 2 3.90 v Ans − = = The percent difference from the value in Table A-5 is 0 percent Ans. Beryllium copper: ( )( ) 18.0 2 7.0 0.286 . 2 7.0 v Ans − = = The percent difference from the value in Table A-5 is 0.286 0.285 0.00351 0.351 percent . 0.285 Ans− = = Gray cast iron: ( )( ) 14.5 2 6.0 0.208 . 2 6.0 v Ans − = = The percent difference from the value in Table A-5 is 0.208 0.211 0.0142 1.42 percent . 0.211 Ans− = − = − ______________________________________________________________________________ 2-6 (a) A0 = pi (0.503)2/4 = 0.1987 in2, σ = Pi / A0 Shigley’s MED, 10th edition Chapter 2 Solutions, Page 3/22 For data in elastic range, ∫ = ∆ l / l0 = ∆ l / 2 For data in plastic range, 0 0 0 0 0 1 1l l Al l l l l A −∆ = = = − = −ò 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 σ = 30.5(106)∫ − 61 000 (1) The equation for the line between data points 8 and 9 is σ = 7.60(105)∫ + 42 900 (2) Solving Eqs. (1) and (2) simultaneously yields σ = 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 ( ) ( )0 0 0.1987 0.1077100 100 45.8 % . 0.1987 fA AR Ans A − − = = = Data Point Pi ∆l, Ai ∫ σ 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 Shigley’s MED, 10th edition Chapter 2 Solutions, Page 4/22 (a) Linear range (b) Offset yield (c) Complete range y = 3.05E+07x - 1.06E+01 0 10000 20000 30000 40000 50000 0.000 0.001 0.001 0.002 S tr e ss ( p si ) Strain Series1 Linear (Series1) 0 5000 10000 15000 20000 25000 30000 35000 40000 45000 50000 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 S tr e ss ( p si ) Strain Y 0 10000 20000 30000 40000 50000 60000 70000 80000 90000 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 S tr e ss ( p si ) Strain U Shigley’s MED, 10th edition Chapter 2 Solutions, Page 5/22 (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 σ true vs.ε, the following equations are applied to the data. true P A σ = Eq. (2-4) 0 0 ln for 0 0.0028 in (0 8400 lbf ) ln for 0.0028 in ( 8400 lbf ) l l P l A l P A ε ε = ≤ ∆ ≤ ≤ ≤ = ∆ > > where 2 2 0 (0.503) 0.1987 in 4 A pi= = 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 σ vs log ε The curve fit gives m = 0.2306 log σ0 = 5.1852 ⇒ σ0 = 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 0 0.2306 0 0.1987ln ln 0.2231 0.1590 Eq. (2-18): 153.2(0.2231) 108.4 kpsi . Eq. (2-19), with 85.6 from Prob. 2-6, 85.6 107 kpsi . 1 1 0.2 m y u u u A A S Ans S SS Ans W ε σ ε ′ = = = = = = = ′ = = = − − Shigley’s MED, 10th edition Chapter 2 Solutions, Page 6/22 P ∆l A ε σtrue log ε log σtrue 0 0 0.198 7 0 0 1 000 0.000 4 0.198 7 0.000 2 5 032.71 -3.699 3.702 2 000 0.000 6 0.198 7 0.000 3 10 065.4 -3.523 4.003 3 000 0.001 0 0.198 7 0.000 5 15 098.1 -3.301 4.179 4 000 0.001 3 0.198 7 0.000 65 20 130.9 -3.187 4.304 7 000 0.002 3 0.198 7 0.001 15 35 229 -2.940 4.547 8 400 0.002 8 0.198 7 0.001 4 42 274.8 -2.854 4.626 8 800 0.198 4 0.001 51 44 354.8 -2.821 4.647 9 200 0.197 8 0.004 54 46 511.6 -2.343 4.668 9 100 0.196 3 0.012 15 46 357.6 -1.915 4.666 13 200 0.192 4 0.032 22 68 607.1 -1.492 4.836 15 200 0.187 5 0.058 02 81 066.7 -1.236 4.909 17 000 0.156 3 0.240 02 108 765 -0.620 5.036 16 400 0.130 7 0.418 89 125 478 -0.378 5.099 14 800 0.107 7 0.612 45 137 419 -0.213 5.138 Shigley’s MED, 10th edition Chapter 2 Solutions, Page 7/22 ______________________________________________________________________________ 2-8 Tangent modulus at σ = 0 is( ) ( ) 6 3 5000 0 25 10 psi 0.2 10 0 E σ − ∆ − = ≈ = ∆ − ò Ans. At σ = 20 kpsi Shigley’s MED, 10th edition ( ) ( )(20 26 19 10 1.5 1 10 E − ≈ = − ∫ (10-3) σ (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: kpsi, σ0 = 90.0 kpsi, m After cold working: Eq. (2 Eq. (2-14), 0 1 1 0.20i A A W = = = − − Eq. (2-17), ln ln1.25 0.223i uε ε= = = < Eq. (2-18), y iS Ansσ ε′ = = = Eq. (2-19), 1 1 0.20u S Ans′ = = = − − (b) Before: 49.5 32 u y S S = = Lost most of its ductility ______________________________________________________________________________ 2-10 W = 0.20, (a) Before cold working: AISI 1212 HR steel. Table A σ0 = 110 kpsi, m = 0.24, After cold working: Eq. (2 Chapter 2 )( ) ( ) ( ) 3 6 3 26 19 10 14.0 10 psi 1.5 1 10− ≈ = Ans. ______________________________________________________________________________ ) Before cold working: Annealed AISI 1018 steel. Table A-22, Sy = 0.25, ∫f = 1.05 After cold working: Eq. (2-16), ∫u = m = 0.25 1 1 1.25 1 1 0.20A W = = = − − 0ln ln1.25 0.223i u i A A ε ε= = = < ( )0.250 90 0.223 61.8 kpsi .my iS Ansσ ε= = = 93% increase 49.5 61.9 kpsi . 1 1 0.20 uSS Ans W = = = − − 25% increase 49.5 1.55 32 = = After: 61.9 1.00 61.8 u y S S ′ = = ′ ductility. ______________________________________________________________________________ ) Before cold working: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, 0.24, ∫f = 0.85 q. (2-16), ∫u = m = 0.24 Chapter 2 Solutions, Page 8/22 ______________________________________________________________________________ y = 32 kpsi, Su = 49.5 93% increase Ans. 25% increase Ans. Ans. ______________________________________________________________________________ = 28 kpsi, Su = 61.5 kpsi, Shigley’s MED, 10th edition Chapter 2 Solutions, Page 9/22 Eq. (2-14), 0 1 1 1.25 1 1 0.20i A A W = = = − − Eq. (2-17), 0ln ln1.25 0.223i u i A A ε ε= = = < Eq. (2-18), ( )0.240 110 0.223 76.7 kpsi .my iS Ansσ ε′ = = = 174% increase Ans. Eq. (2-19), 61.5 76.9 kpsi . 1 1 0.20 u u SS Ans W ′ = = = − − 25% increase Ans. (b) Before: 61.5 2.20 28 u y S S = = After: 76.9 1.00 76.7 u y S S ′ = = ′ 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, σ0 = 100 kpsi, m = 0.15, ∫f = 0.18 After cold working: Eq. (2-16), ∫u = m = 0.15 Eq. (2-14), 0 1 1 1.25 1 1 0.20i A A W = = = − − Eq. (2-17), 0ln ln1.25 0.223i f i A A ε ε= = = > 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. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 2 Solutions, Page 10/22 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 13806.25 235 117.5 13806.25 235 117.5 13806.25 236 118 13924 236 118 13924 239 119.5 14280.25 ΣSu = 1172 ΣSu2 = 137373 Eq. (1-6) 1172 117.2 117 kpsi . 10 u u S S Ans N = = = ≈ ∑ Eq. (1-7), ( ) 10 2 2 2 1 137373 10 117.2 1.27 kpsi . 1 9u u u i S S NS s Ans N = − − = = = − ∑ ______________________________________________________________________________ 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 1707.342 235 41.55 1726.403 235 41.55 1726.403 235 41.55 1726.403 236 41.78 1745.568 236 41.78 1745.568 239 42.47 1803.701 ΣSu = 414.12 ΣSu2 = 17152.63 Shigley’s MED, 10th edition Chapter 2 Solutions, Page 11/22 Eq. (1-6) 414.12 41.4 kpsi . 10 u u S S Ans N = = = ∑ Eq. (1-7), ( ) 10 2 2 2 1 17152.63 10 41.4 1.20 . 1 9u u u i S S NS s Ans N = − − = = = − ∑ ______________________________________________________________________________ 2-17 (a) Eq. (2-9) 2 345.6 34.7 in lbf / in . 2(30)Ru Ans≈ = ⋅ (b) A0 = pi(0.5032)/4 = 0.19871 in2 P ∆L A (A0 / A) – 1 ∫ σ = P/A0 0 0 0 0 1 000 0.000 4 0.000 2 5 032. 2 000 0.000 6 0.000 3 10 070 3 000 0.001 0 0.000 5 15 100 4 000 0.001 3 0.000 65 20 130 7 000 0.002 3 0.001 15 35 230 8 400 0.002 8 0.001 4 42 270 8 800 0.003 6 0.001 8 44 290 9 200 0.008 9 0.004 45 46 300 9 100 0.196 3 0.012 28 0.012 28 45 800 13 200 0.192 4 0.032 80 0.032 80 66 430 15 200 0.187 5 0.059 79 0.059 79 76 500 17 000 0.156 3 0.271 34 0.271 34 85 550 16 400 0.130 7 0.520 35 0.520 35 82 530 14 800 0.107 7 0.845 03 0.845 03 74 480 From the figures on the next page, ( ) ( ) ( ) ( ) 5 1 3 3 1 (43 000)(0.001 5) 45 000(0.004 45 0.001 5) 2 1 45 000 76 500 (0.059 8 0.004 45) 2 81 000 0.4 0.059 8 80 000 0.845 0.4 66.7 10 in lbf/in . T i i u A Ans = ≈ = + − + + − + − + − ≈ ⋅ ∑ Shigley’s MED, 10th edition Chapter 2 Chapter 2 Solutions, Page 12/22 Shigley’s MED, 10th edition Chapter 2 Solutions, Page 13/22 2-18, 2-19 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________ 2-20 Appropriate tables: Young’s 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, ( ) 2 4 / 4 y y F F FS d A d S σ pi pi = = = ⇒ = Weight/length = ρA, Cost/length = $/in = ($/lbf) Weight/length, Deflection/length = δ /L = F/(AE) With F = 100 kips = 100(103) lbf, Material Young's Modulus Density Yield Strength Cost/lbf Diameter Weight/ length Cost/ length Deflection/ length units Mpsi lbf/in3 kpsi $/lbf in lbf/in $/in in/in 1020 HR 30 0.282 30 0.27 2.060 0.9400 0.25 1.000E-03 1020 CD 30 0.282 57 0.30 1.495 0.4947 0.15 1.900E-03 1040 30 0.282 80 0.35 1.262 0.3525 0.12 2.667E-03 4140 30 0.282 165 0.80 0.878 0.1709 0.145.500E-03 Al 10.4 0.098 50 1.10 1.596 0.1960 0.22 4.808E-03 Ti 16.5 0.16 120 7.00 1.030 0.1333 $0.93 7.273E-03 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 2 7.95 lbf 0.281 lbf/in[ (1 in) / 4](36 in) W Al pi = = =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 A-20, perform a Brinell hardness test, then use Eq. (2-21) to estimate an ultimate strength Shigley’s MED, 10th edition Chapter 2 Solutions, Page 14/22 of 0.5 0.5(200) 100 kpsiu BS H= = = . 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. ______________________________________________________________________________ 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 2 2.9 lbf 0.103 lbf/in[ (1 in) / 4](36 in) W Al pi = = =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 2 9.0 lbf 0.318 lbf/in[ (1 in) / 4](36 in) W Al pi = = =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, Young’s modulus is determined to be ( ) ( ) 33 4 100 24 17.7 Mpsi 3 3 (1) 64 (17 / 32) FlE Iy pi = = = 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. ______________________________________________________________________________ 2-26 For strength, σ = F/A = S ⇒ A = F/S Shigley’s MED, 10th edition Chapter 2 Solutions, Page 15/22 For mass, m = Alρ = (F/S) lρ Thus, f 3(M ) = ρ /S , and maximize S/ρ (β = 1) In Fig. (2-19), draw lines parallel to S/ρ The higher strength aluminum alloys have the greatest potential, as determined by comparing each material’s bubble to the S/ρ guidelines. Ans. ______________________________________________________________________________ 2-27 For stiffness, k = AE/l ⇒ A = kl/E For mass, m = Alρ = (kl/E) lρ =kl2 ρ /E Thus, f 3(M) = ρ /E , and maximize E/ρ (β = 1) In Fig. (2-16), draw lines parallel to E/ρ Shigley’s MED, 10th edition Chapter 2 Solutions, Page 16/22 From the list of materials given, tungsten carbide (WC) is best, closely followed by aluminum alloys. 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, σ = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 104 ]. 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 pi − . Then, for strength, Eq. (1) is 2/3 3/2 Fl FlS A CA CS = ⇒ = (2) Shigley’s MED, 10th edition Chapter 2 Solutions, Page 17/22 For mass, 2/3 2/3 5/3 2/3 Fl F m Al l l CS C S ρρ ρ = = = Thus, f 3(M) = ρ /S 2/3, and maximize S 2/3/ρ (β = 2/3) In Fig. (2-19), draw lines parallel to S 2/3/ρ 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. 77, 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 (a constant). Thus, Eq. (2-27) becomes Shigley’s MED, 10th edition Chapter 2 Solutions, Page 18/22 1/23 3 klA CE = and Eq. (2-29) becomes 1/2 5/2 1/23 k m Al l C E ρρ = = Thus, minimize ( )3 1/2f M E ρ = , or maximize 1/2EM ρ = . 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 = Alρ = (kl/E) lρ =kl2 ρ /E So, f 3(M) = ρ /E, and maximize E/ρ . Thus, β = 1. Ans. ______________________________________________________________________________ 2-31 For strength, σ = F/A = S⇒ A = F/S Shigley’s MED, 10th edition Chapter 2 Solutions, Page 19/22 For mass, m = Alρ = (F/S) lρ So, f 3(M ) = ρ /S, and maximize S/ρ . Thus, β = 1. Ans. ______________________________________________________________________________ 2-32 Eq. (2-26), p. 77, 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 the circular cross section (see p.77), C = (4pi)−1. Another 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, a constant. Thus, Eq. (2-27) becomes 1/23 3 klA CE = and Eq. (2-29) becomes 1/2 5/2 1/23 k m Al l C E ρρ = = So, minimize ( )3 1/2f M E ρ = , or maximize 1/2EM ρ = . Thus, β = 1/2. Ans. ______________________________________________________________________________ 2-33 For strength, σ = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 104 ]. The section modulus is strictly a function of the dimensions of the cross section and has the units in3 (ips) or m3 (SI). The area of the cross section has the units in2 or m2. Thus, for a given cross section, Z =C (A)3/2, where C is a number. For example, for a circular cross section, Z = pid 3/(32)and the area is A = pid 2/4. This leads to C = ( ) 14 pi − . So, with Z =C (A)3/2, for strength, Eq. (1) is 2/3 3/2 Fl FlS A CA CS = ⇒ = (2) For mass, 2/3 2/3 5/3 2/3 Fl F m Al l l CS C S ρρ ρ = = = So, f 3(M) = ρ /S 2/3, and maximize S 2/3/ρ. Thus, β = 2/3. Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 2 Solutions, Page 20/22 2-34 For stiffness, k=AE/l, or, A = kl/E. Thus, m = ρAl =ρ (kl/E )l = kl 2 ρ /E. Then, M = E /ρ and β = 1. From Fig. 2-16, lines parallel to E /ρ for ductile materials include steel, titanium, molybdenum, aluminum alloys, and composites. For strength, S = F/A, or, A = F/S. Thus, m = ρAl =ρ F/Sl = Fl ρ /S. Then, M = S/ρ and β = 1. From Fig. 2-19, lines parallel to S/ρ 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. ______________________________________________________________________________ 2-35 See Prob. 1-13 solution for x = 122.9 kcycles and xs = 30.3 kcycles. Also, in that solution it is observed that the number of instances less than 115 kcycles predicted by the normal distribution is 27; whereas, the data indicates the number to be 31. From Eq. (1-4), the probability density function (PDF), with xµ = and ˆ xsσ = , is 2 21 1 1 1 122.9( ) exp exp 2 2 30.32 30.3 2xx x x xf x ss pi pi − − = − = − (1) The discrete PDF is given by f /(Nw), where N = 69 and w = 10 kcycles. From the Eq. (1) and the data of Prob. 1-13, the following plots are obtained. Shigley’s MED, 10th edition Chapter 2 Solutions, Page 21/22 Range midpoint (kcycles) Frequency Observed PDF Normal PDF x f f /(Nw) f (x) 60 2 0.002898551 0.001526493 70 1 0.001449275 0.002868043 80 3 0.004347826 0.004832507 90 5 0.007246377 0.007302224 100 8 0.011594203 0.009895407 110 12 0.017391304 0.012025636 120 6 0.008695652 0.013106245 130 10 0.014492754 0.012809861 140 8 0.011594203 0.011228104 150 5 0.007246377 0.008826008 160 2 0.002898551 0.006221829 170 3 0.004347826 0.003933396 180 2 0.002898551 0.002230043 190 1 0.001449275 0.001133847 200 0 0 0.000517001 210 1 0.001449275 0.00021141 Plots of the PDF’s are shown below. 0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02 0 20 40 60 80 100 120 140 160 180 200 220 Normal Distribution Histogram L (kcycles) P ro b a b il it y D e n si ty Shigley’s MED, 10th edition Chapter 2 Solutions, Page 22/22 It can be seen that the data is not perfectly normal and is skewed to the left indicating that the number of instances below 115 kcycles for the data (31) would be higher than the hypothetical normal distribution (27). Shigley’s MED, 10th edition Chapter 3 Solutions, Page 1/100 Chapter 3 3-1 0OMΣ = 18 6(100) 0BR − = 33.3 lbf .BR Ans= 0yFΣ = 100 0O BR R+ − = 66.7 lbf .OR Ans= 33.3 lbf .C BR R Ans= = ______________________________________________________________________________ 3-2 Body AB: 0xFΣ = Ax BxR R= 0yFΣ = Ay ByR R= 0BMΣ = (10) (10) 0Ay AxR R− = Ax AyR R= Body OAC: 0OMΣ = (10) 100(30) 0AyR − = 300 lbf .AyR Ans= 0xFΣ = 300 lbf .Ox AxR R Ans= − = − 0yFΣ = 100 0Oy AyR R+ − = 200 lbf .OyR Ans= − ______________________________________________________________________________ Shigley’s MED, 10th edition 3-3 0.8 1.39 kN . tan 30O R Ans= = o 0.8 1.6 kN . sin 30A R Ans= = o ______________________________________________________________________________ 3-4 Step 1: Find RA & RE 2 2 0 400cos30 0 346.4 N 0 400 400sin 30 0 200 N 346.4 200 400 N . x Ax Ax y Ay Ay A F R R F R R R Ans = + = = − = + − = = − = + = ∑ ∑ Step 2: Find components of ( ( )4 4 0 400(4.5) 7.794 1.9 0 305.4 N . 0 305.4 N 0 ( ) 400 N C D x Cx y Cy M R Ans F R F R = − − = = = ⇒ = ⇒ ∑ ∑ ∑ 4.5 7.794 m tan 30 0 9 7.794(400cos30 ) 4.5(400sin 30 ) 0 400 N . A E E h M R R Ans = = Σ = − − = = o o o Chapter 3 1.39 kN .R Ans 1.6 kN .R Ans ______________________________________________________________________________ 0 400cos30 0 346.4 N 0 400 400sin 30 0 200 N 346.4 200 400 N .R Ans = + = = + − = = + = o o Step 2: Find components of RC and RD on link 4 ) 4 4 400(4.5) 7.794 1.9 0 305.4 N . 0 305.4 N 0 ( ) 400 N DR R Ans − − = = = − 9 7.794(400cos30 ) 4.5(400sin 30 ) 0− = o Chapter 3 Solutions, Page 2/100 ______________________________________________________________________________ Shigley’s MED, 10th edition Step 3: Find components of RC ( ) ( ) ( ) 2 2 2 0 305.4 346.4 0 41 N 0 200 N x Cx Cx y Cy F R R F R = + − = = = = ∑ ∑ _____________________________________________________________________________________________________________________ Chapter 3 C on link 2305.4 346.4 0+ − = Ans. _____________________________________________________________________________________________________________________ Chapter 3 Solutions, Page 3/100 _____________________________________________________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 4/100 3-5 0CMΣ = 11500 300(5) 1200(9) 0R− + + = 1 8.2 kN .R Ans= 0yFΣ = 28.2 9 5 0R− − + = 2 5.8 kN .R Ans= 1 8.2(300) 2460 N m .M Ans= = ⋅ 2 2460 0.8(900) 1740 N m .M Ans= − = ⋅ 3 1740 5.8(300) 0 checks!M = − = _____________________________________________________________________________ 3-6 0yFΣ = 500 40(6) 740 lbf .OR Ans= + = 0OMΣ = 500(8) 40(6)(17) 8080 lbf in .OM Ans= + = ⋅ 1 8080 740(8) 2160 lbf in .M Ans= − + = − ⋅ 2 2160 240(6) 720 lbf in .M Ans= − + = − ⋅ 3 1720 (240)(6) 0 checks! 2 M = − + = ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 5/100 3-7 0BMΣ = 12.2 1(2) 1(4) 0R− + − = 1 0.91 kN .R Ans= − 0yFΣ = 20.91 2 4 0R− − + − = 2 6.91 kN .R Ans= 1 0.91(1.2) 1.09 kN m .M Ans= − = − ⋅ 2 1.09 2.91(1) 4 kN m .M Ans= − − = − ⋅ 3 4 4(1) 0 checks!M = − + = ______________________________________________________________________________ 3-8 Break at the hinge at B Beam OB: From symmetry, 1 200 lbf .BR V Ans= = Beam BD: 0DMΣ = 2200(12) (10) 40(10)(5) 0R− + = 2 440 lbf .R Ans= 0yFΣ = 3200 440 40(10) 0R− + − + = 3 160 lbf .R Ans= Shigley’s MED, 10th edition Chapter 3 Solutions, Page 6/100 1 200(4) 800 lbf in .M Ans= = ⋅ 2 800 200(4) 0 checks at hingeM = − = 3 800 200(6) 400 lbf in .M Ans= − = − ⋅ 4 1400 (240)(6) 320 lbf in . 2 M Ans= − + = ⋅ 5 1320 (160)(4) 0 checks! 2 M = − = ______________________________________________________________________________ 3-9 1 1 1 1 1 2 0 0 0 1 2 1 1 1 1 2 9 300 5 1200 1500 9 300 5 1200 1500 (1) 9 300 5 1200 1500 (2) q R x x x R x V R x x R x M R x x x R x − − − − = − − − − + − = − − − − + − = − − − − + − At x = 1500+ V = M = 0. Applying Eqs. (1) and (2), 1 2 1 29 5 0 14R R R R− − + = ⇒ + = 1 11500 9(1500 300) 5(1500 1200) 0 8.2 kN .R R Ans− − − − = ⇒ = 2 14 8.2 5.8 kN .R Ans= − = 0 300 : 8.2 kN, 8.2 N m 300 1200 : 8.2 9 0.8 kN 8.2 9( 300) 0.8 2700 N m 1200 1500 : 8.2 9 5 5.8 kN 8.2 9( 300 x V M x x V M x x x x V M x x ≤ ≤ = = ⋅ ≤ ≤ = − = − = − − = − + ⋅ ≤ ≤ = − − = − = − − ) 5( 1200) 5.8 8700 N mx x− − = − + ⋅ Plots of V and M are the same as in Prob. 3-5. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 7/100 3-10 1 2 1 0 0 1 0 1 1 1 2 2 500 8 40 14 40 20 500 8 40 14 40 20 (1) 500 8 20 14 20 20 (2) at 20 in, 0, Eqs. (1) and (2) give O O O O O O q R x M x x x x V R M x x x x M R x M x x x x V M R − − − − + = − − − − − + − = − − − − − + − = − − − − − + − = = = ( ) 2 500 40 20 14 0 740 lbf . (20) 500(20 8) 20(20 14) 0 8080 lbf in . O O O O O R Ans R M M Ans − − − = ⇒ = − − − − − = ⇒ = ⋅ 0 8 : 740 lbf, 740 8080 lbf in 8 14 : 740 500 240 lbf 740 8080 500( 8) 240 4080 lbf in 14 20 : 740 500 40( 14) 40 800 lbf 740 8080 x V M x x V M x x x x V x x M x ≤ ≤ = = − ⋅ ≤ ≤ = − = = − − − = − ⋅ ≤ ≤ = − − − = − + = − − 2 2500( 8) 20( 14) 20 800 8000 lbf inx x x x− − − = − + − ⋅ Plots of V and M are the same as in Prob. 3-6. ______________________________________________________________________________ 3-11 1 1 1 1 1 2 0 0 0 1 2 1 1 1 1 2 2 1.2 2.2 4 3.2 2 1.2 2.2 4 3.2 (1) 2 1.2 2.2 4 3.2 (2) q R x x R x x V R x R x x M R x x R x x − − − − = − − + − − − = − − + − − − = − − + − − − at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2), 1 2 1 2 1 2 1 2 2 4 0 6 (3) 3.2 2(2) (1) 0 3.2 4 (4) R R R R R R R R − + − = ⇒ + = − + = ⇒ + = Solving Eqs. (3) and (4) simultaneously, R1 = -0.91 kN, R2 = 6.91 kN Ans. 0 1.2 : 0.91 kN, 0.91 kN m 1.2 2.2 : 0.91 2 2.91 kN 0.91 2( 1.2) 2.91 2.4 kN m 2.2 3.2 : 0.91 2 6.91 4 kN 0.91 2( x V M x x V M x x x x V M x x ≤ ≤ = − = − ⋅ ≤ ≤ = − − = − = − − − = − + ⋅ ≤ ≤ = − − + = = − − −1.2) 6.91( 2.2) 4 12.8 kN mx x+ − = − ⋅ Plots of V and M are the same as in Prob. 3-7. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 8/100 3-12 1 1 1 0 0 1 1 2 3 0 0 1 1 0 1 2 3 1 1 2 2 1 1 2 3 1 400 4 10 40 10 40 20 20 400 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 x V R x R x x x R x M R x x R x x x R x M x R − − − − = − − + − − − + − + − = − − + − − − + − + − = − − + − − − + − + − = = ∴ − 18 4) 0 200 lbf .R Ans− = ⇒ = at x = 20+, V =M = 0. Applying Eqs. (1) and (2), 2 3 2 3 2 2 2 200 400 40(10) 0 600 200(20) 400(16) (10) 20(10) 0 440 lbf . R R R R R R Ans − + − + = ⇒ + = − + − = ⇒ = 3 600 440 160 lbf .R Ans= − = 0 4 : 200 lbf, 200 lbf in 4 10 : 200 400 200 lbf, 200 400( 4) 200 1600 lbf in 10 20 : 200 400 440 40( 10) 640 40 lbf 200 400( 4) x V M x x V M x x x x V x x M x x ≤ ≤ = = ⋅ ≤ ≤ = − = − = − − = − + ⋅ ≤ ≤ = − + − − = − = − − + ( )2 2440( 10) 20 10 20 640 4800 lbf inx x x 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, ( ) ( ) 2 2 2 2 2 2 2 2 4 c c l a x l l lM l a a − = = − − = − w wl At reaction, 2 2 r M a= w a = 2.25, l = 10 in, w = 100 lbf/in ( )2 100(10) 10 2.25 125 lbf in 2 4 100 2.25 253 lbf in . 2 c r M M Ans = − = ⋅ = = ⋅ (b) Optimal occurs when c rM M= Shigley’s MED, 10th edition 2 2 4 2 l l a a a al l − = ⇒ w w Taking the positive root (2 21 4 0.25 2 1 0.207 .2 2a l l l l Ans = − + + = − = for l = 10 in, w = 100 lbf ( )min 100 2 2.07 214 lbfinM = = ⋅ ______________________________________________________________________________ 3-15 (a) 20 10 5 kpsi 2 C −= = 20 10 15 kpsi 2 CD += = 2 215 8 17 kpsiR = + = 1 5 17 22 kpsiσ = + = 2 5 17 12 kpsiσ = − = − 11 8tan 14.04 cw 2 15p φ − = = 1 17 kpsi 45 14.04 30.96 ccws Rτ φ = = = − = o o o (b) 9 16 12.5 kpsi 2 C += = 16 9 3.5 kpsi 2 CD −= = 2 25 3.5 6.10 kpsiR = + = 1 12.5 6.1 18.6 kpsiσ = + = 2 12.5 6.1 6.4 kpsiσ = − = 11 5tan 27.5 ccw 2 3.5p φ − = = 1 6.10 kpsi 45 27.5 17.5 cws Rτ φ = = = − = o o o Chapter 3 2 20.25 0a a al l⇒ + − = ) ( )2 24 0.25 2 1 0.207 .2 2la l l l l Ans = − + + = − = = 100 lbf, a = 0.207(10) = 2.07 in 2100 2 2.07 214 lbf in= = ⋅ ______________________________________________________________________________ 15 kpsi 15 8 17 kpsi 5 17 12 kpsi tan 14.04 cwo 45 14.04 30.96 ccwo o o 12.5 kpsi 3.5 kpsi 5 3.5 6.10 kpsi 12.5 6.1 18.6 kpsi 12.5 6.1 6.4 kpsi tan 27.5 ccwo 45 27.5 17.5 cwo o o Chapter 3 Solutions, Page 9/100 ______________________________________________________________________________ Shigley’s MED, 10th edition (c) 2 2 1 2 24 10 17 kpsi 2 24 10 7 kpsi 2 7 6 9.22 kpsi 17 9.22 26.22 kpsi 17 9.22 7.78 kpsi C CD R σ σ + = = − = = = + = = + = = − = 11 790 tan 69.7 ccw 2 6p φ − = + = o o 1 9.22 kpsi 69.7 45 24.7 ccws Rτ φ = = = − = o o o (d) 2 2 1 2 12 22 5 kpsi 2 12 22 17 kpsi 2 17 12 20.81 kpsi 5 20.81 25.81 kpsi 5 20.81 15.81 kpsi C CD R σ σ − + = = + = = = + = = + = = − = − 11 1790 tan 72.39 cw 2 12p φ − = + = o o 1 20.81 kpsi 72.39 45 27.39 cws Rτ φ = = = − = ______________________________________________________________________________ Chapter 3 17 kpsi 7 kpsi 7 6 9.22 kpsi 17 9.22 26.22 kpsi 17 9.22 7.78 kpsi 90 tan 69.7 ccw = + = o o 69.7 45 24.7 ccwo o o 5 kpsi 17 kpsi 17 12 20.81 kpsi 5 20.81 25.81 kpsi 5 20.81 15.81 kpsi 1 1790 tan 72.39 cw 2 12 = + = o o 72.39 45 27.39 cwo ______________________________________________________________________________ Chapter 3 Solutions, Page 10/100 ______________________________________________________________________________ Shigley’s MED, 10th edition 3-16 (a) 2 2 1 2 8 7 0.5 MPa 2 8 7 7.5 MPa 2 7.5 6 9.60 MPa 9.60 0.5 9.10 MPa 0.5 9.6 10.1 Mpa C CD R σ σ − + = = − + = = = + = = − = = − − = − 11 7.590 tan 70.67 cw 2 6p φ − = + = o o 1 9.60 MPa 70.67 45 25.67 cws Rτ φ = = = − = o o o (b) 2 2 1 2 9 6 1.5 MPa 2 9 6 7.5 MPa 2 7.5 3 8.078 MPa 1.5 8.078 9.58 MPa 1.5 8.078 6.58 MPa C CD R σ σ − = = + = = = + = = + = = − = − 11 3tan 10.9 cw 2 7.5p φ − = = 1 8.078 MPa 45 10.9 34.1 ccws Rτ φ = = = − = o o o Chapter 3 0.5 MPa 7.5 MPa 7.5 6 9.60 MPa 9.60 0.5 9.10 MPa 0.5 9.6 10.1 Mpa 90 tan 70.67 cwo o 70.67 45 25.67 cwo o o 7.5 MPa 7.5 3 8.078 MPa 1.5 8.078 9.58 MPa 1.5 8.078 6.58 MPa tan 10.9 cwo 45 10.9 34.1 ccwo o o Chapter 3 Solutions, Page 11/100 Shigley’s MED, 10th edition (c) 2 2 1 2 12 4 4 MPa 2 12 4 8 MPa 2 8 7 10.63 MPa 4 10.63 14.63 MPa 4 10.63 6.63 MPa C CD R σ σ − = = + = = = + = = + = = − = − 11 890 tan 69.4 ccw 2 7p φ − = + = o o 1 10.63 MPa 69.4 45 24.4 ccws Rτ φ = = = − = o o o (d) 2 2 1 2 6 5 0.5 MPa 2 6 5 5.5 MPa 2 5.5 8 9.71 MPa 0.5 9.71 10.21 MPa 0.5 9.71 9.21 MPa C CD R σ σ − = = + = = = + = = + = = − = − 11 8tan 27.75 ccw 2 5.5p φ − = = 1 9.71 MPa 45 27.75 17.25 cws Rτ φ = = = − = o o o ______________________________________________________________________________ Chapter 3 8 7 10.63 MPa 4 10.63 14.63 MPa 4 10.63 6.63 MPa 90 tan 69.4 ccw = + = o o 69.4 45 24.4 ccwo o o 5.5 MPa 5.5 8 9.71 MPa 0.5 9.71 10.21 MPa 0.5 9.71 9.21 MPa tan 27.75 ccwo 45 27.75 17.25 cwo o o ______________________________________________________________________________ Chapter 3 Solutions, Page 12/100 ______________________________________________________________________________ Shigley’s MED, 10th edition 3-17 (a) 2 2 1 2 12 6 9 kpsi 2 12 6 3 kpsi 2 3 4 5 kpsi 5 9 14 kpsi 9 5 4 kpsi C CD R σ σ + = = − = = = + = = + = = − = 11 4tan 26.6 ccw 2 3p φ − = = 1 5 kpsi 45 26.6 18.4 ccws Rτ φ = = = − = o o o (b) 2 2 1 2 30 10 10 kpsi 2 30 10 20 kpsi 2 20 10 22.36 kpsi 10 22.36 32.36 kpsi 10 22.36 12.36 kpsi C CD R σ σ − = = + = = = + = = + = = − = − 11 10tan 13.28 ccw 2 20p φ − = = 1 22.36 kpsi 45 13.28 31.72 cws Rτ φ = = = − = o o o Chapter 3 3 4 5 kpsi tan 26.6 ccwo 45 26.6 18.4 ccwo o o 20 kpsi 20 10 22.36 kpsi 10 22.36 32.36 kpsi 10 22.36 12.36 kpsi tan 13.28 ccwo 45 13.28 31.72 cwo o o Chapter 3 Solutions, Page 13/100 Shigley’s MED, 10th edition (c) 2 2 1 2 10 18 4 kpsi 2 10 18 14 kpsi 2 14 9 16.64 kpsi 4 16.64 20.64 kpsi 4 16.64 12.64 kpsi C CD R σ σ − + = = + = = = + = = + = = − = − 11 1490 tan 73.63 cw 2 9p φ − = + = o o 1 16.64 kpsi 73.63 45 28.63 cws Rτ φ = = = − = (d) 2 2 1 2 9 19 14 kpsi 2 19 9 5 kpsi 2 5 8 9.434 kpsi 14 9.43 23.43 kpsi 14 9.43 4.57 kpsi C CD R σ σ + = = − = = = + = = + = = − = 11 590 tan 61.0 cw 2 8p φ − = + = o o 1 9.34 kpsi 61 45 16 cws Rτ φ = = = − = o o o ______________________________________________________________________________ Chapter 3 4 kpsi 14 kpsi 14 9 16.64 kpsi 4 16.64 20.64 kpsi 4 16.64 12.64 kpsi 1 1490 tan 73.63 cw = + = o o 16.64 kpsi 73.63 45 28.63 cwo 5 8 9.434 kpsi 14 9.43 23.43 kpsi 14 9.43 4.57 kpsi 90 tan 61.0 cw = + = o o 61 45 16 cw ______________________________________________________________________________ Chapter 3 Solutions, Page 14/100 ______________________________________________________________________________ Shigley’s MED, 10th edition 3-18 (a) 2 2 1 2 3 80 30 55 MPa 2 80 30 25 MPa 2 25 20 32.02 MPa 0 MPa 55 32.02 22.98 23.0 MPa 55 32.0 87.0 MPa C CD R σ σ σ − − = = − − = = = + = = = − + = − = − = − − = − 1 2 2 3 1 3 23 8711.5 MPa, 32.0 MPa, 43.5 MPa 2 2 τ τ τ= = = = = (b) 2 2 1 2 3 30 60 15 MPa 2 60 30 45 MPa 2 45 30 54.1 MPa 15 54.1 39.1 MPa 0 MPa 15 54.1 69.1 MPa C CD R σ σ σ − = = − + = = = + = = − + = = = − − = − 1 3 1 2 2 3 39.1 69.1 54.1MPa 2 39.1 19.6 MPa 2 69.1 34.6 MPa 2 τ τ τ + = = = = = = Chapter 3 55 MPa 25 MPa 25 20 32.02 MPa 55 32.02 22.98 23.0 MPa 55 32.0 87.0 MPa = − + = − = − 1 2 2 3 1 3 23 8711.5 MPa, 32.0 MPa, 43.5 MPa 2 2 τ τ τ= = = = = 15 MPa 45 MPa 45 30 54.1 MPa 15 54.1 39.1 MPa 15 54.1 69.1 MPa 54.1 MPa 19.6 MPa 34.6 MPa Chapter 3 Solutions, Page 15/100 Shigley’s MED, 10th edition (c) 2 2 1 2 3 40 0 20 MPa 2 40 0 20 MPa 2 20 20 28.3 MPa 20 28.3 48.3 MPa 20 28.3 8.3 MPa 30 MPaz C CD R σ σ σ σ + = = − = = = + = = + = = − = − = = − 1 3 1 2 2 3 48.3 30 30 8.339.1 MPa, 28.3 MPa, 10.9 MPa 2 2 τ τ τ + − = = = = = (d) 2 2 1 2 3 50 25 MPa 2 50 25 MPa 2 25 30 39.1 MPa 25 39.1 64.1 MPa 25 39.1 14.1 MPa 20 MPaz C CD R σ σ σ σ = = = = = + = = + = = − = − = = − 1 3 1 2 2 3 64.1 20 20 14.142.1 MPa, 39.1 MPa, 2.95 MPa 2 2 τ τ τ + − = = = = = ______________________________________________________________________________ 3-19 (a) Since there are no shear stresses on the stress element, the stress element already represents principal stresses. 1 2 3 10 kpsi 0 kpsi 4 kpsi x y σ σ σ σ σ = = = = = − 1 3 1 2 2 3 10 ( 4) 7 kpsi 2 10 5 kpsi 2 0 ( 4) 2 kpsi 2 τ τ τ − − = = = = − − = = Chapter 3 20 MPa 20 20 28.3 MPa 20 28.3 48.3 MPa 20 28.3 8.3 MPa 1 3 1 2 2 3 48.3 30 30 8.339.1 MPa, 28.3 MPa, 10.9 MPa 2 2 τ τ τ + − = = = = = 25 30 39.1 MPa 25 39.1 64.1 MPa 25 39.1 14.1 MPa 1 3 1 2 2 3 64.1 20 20 14.142.1 MPa, 39.1 MPa, 2.95 MPa 2 2 τ τ τ + − = = = = = ______________________________________________________________________________ Since there are no shear stresses on the stress element, the stress element already represents principal stresses. 7 kpsi 2 kpsi Chapter 3 Solutions, Page 16/100 39.1 MPa, 28.3 MPa, 10.9 MPa 42.1 MPa, 39.1 MPa, 2.95 MPa ______________________________________________________________________________ Shigley’s MED, 10th edition (b) 2 2 1 2 3 0 10 5 kpsi 2 10 0 5 kpsi 2 5 4 6.40 kpsi 5 6.40 11.40 kpsi 0 kpsi, 5 6.40 1.40 kpsi C CD R σ σ σ + = = − = = = + = = + = = = − = − 1 3 1 2 36.40 kpsi, 5.70 kpsi, 0.70 kpsiRτ τ τ= = = = = = (c) 2 2 1 2 3 2 8 5 kpsi 2 8 2 3 kpsi 2 3 4 5 kpsi 5 5 0 kpsi, 0 kpsi 5 5 10 kpsi C CD R σ σ σ − − = = − − = = = + = = − + = = = − − = − 1 3 1 2 2 3 10 5 kpsi, 0 kpsi, 5 kpsi 2 τ τ τ= = = = (d) 2 2 1 2 3 10 30 10 kpsi 2 10 30 20 kpsi 2 20 10 22.36 kpsi 10 22.36 12.36 kpsi 0 kpsi 10 22.36 32.36 kpsi C CD R σ σ σ − = = − + = = = + = = − + = = = − − = − 1 3 1 2 2 322.36 kpsi, 6.18 kpsi, 16.18 kpsτ τ τ= = = = = ______________________________________________________________________________ Chapter 3 5 4 6.40 kpsi 5 6.40 11.40 kpsi 0 kpsi, 5 6.40 1.40 kpsi= = − = − 1 3 1 2 3 11.40 1.406.40 kpsi, 5.70 kpsi, 0.70 kpsi 2 2 τ τ τ= = = = = = 1 2 3 4 5 kpsi 5 5 0 kpsi, 0 kpsi 5 5 10 kpsi σ σ= − + = = 1 3 1 2 2 35 kpsi, 0 kpsi, 5 kpsiτ τ τ= = = = 10 kpsi 20 kpsi 20 10 22.36 kpsi 10 22.36 12.36 kpsi 10 22.36 32.36 kpsi 1 3 1 2 2 3 12.36 32.3622.36 kpsi, 6.18 kpsi, 16.18 kps 2 2 τ τ τ= = = = = ______________________________________________________________________________ Chapter 3 Solutions, Page 17/100 6.40 kpsi, 5.70 kpsi, 0.70 kpsi 22.36 kpsi, 6.18 kpsi, 16.18 kpsi ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 18/100 3-20 From Eq. (3-15), 3 2 2 2 2 2 2 2 3 ( 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) 0 594 3186 0 σ σ σ σ σ − − + − + − + − − + − − − − − − − − + − − − − − − − = − + = Roots are: 21.04, 5.67, –26.71 kpsi Ans. 1 2 2 3 max 1 3 21.04 5.67 7.69 kpsi 2 5.67 26.71 16.19 kpsi 2 21.04 26.71 23.88 kpsi . 2 Ans τ τ τ τ − = = + = = + = = = _____________________________________________________________________________ 3-21 From Eq. (3-15) ( ) ( ) ( ) 23 2 2 2 2 2 2 3 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) 0 40 2 000 48 000 0 σ σ σ σ σ σ − + + + + + − − − − − + − − − − − = − − + = Roots are: 60, 20, –40 kpsi Ans. 1 2 2 3 max 1 3 60 20 20 kpsi 2 20 40 30 kpsi 2 60 40 50 kpsi . 2 Ans τ τ τ τ − = = + = = + = = = _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 19/100 3-22 From Eq. (3-15) ( ) ( )2 23 2 2 2 2 2 3 2 (10 40 40) 10(40) 10(40) 40(40) 20 40 20 10(40)(40) 2(20)( 40)( 20) 10( 40) 40( 20) 40(20) 0 90 0 σ σ σ σ σ − + + + + + − − − − − − + − − − − − − − = − = Roots are: 90, 0, 0 MPa Ans. 2 3 1 2 1 3 max 0 90 45 MPa . 2 Ans τ τ τ τ = = = = = _____________________________________________________________________________ 3-23 ( )( ) ( ) ( ) 2 6 6 1 15000 33 950 psi 34.0 kpsi . 4 0.75 6033 950 0.0679 in . 30 10 0.0679 1130 10 1130 . 60 F Ans A FL L Ans AE E Ans L σ pi δ σ δ µ− = = = = = = = = = = = =ò From Table A-5, v = 0.292 ( ) ( ) 2 1 6 6 2 0.292(1130) 330 . 330 10 (0.75) 248 10 in . v Ans d d Ans µ − − = − = − = − ∆ = = − = − ò ò ò _____________________________________________________________________________ 3-24 ( )( ) ( ) ( ) 2 6 6 1 3000 6790 psi 6.79 kpsi . 4 0.75 606790 0.0392 in . 10.4 10 0.0392 653 10 653 . 60 F Ans A FL L Ans AE E Ans L σ pi δ σ δ µ− = = = = = = = = = = = =ò From Table A-5, v = 0.333 ( ) ( ) 2 1 6 6 2 0.333(653) 217 . 217 10 (0.75) 163 10 in . v Ans d d Ans µ − − = − = − = − ∆ = = − = − ò ò ò _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 20/100 3-25 2 0.0001 0.0001d d d d ∆ − = = = −ò From Table A-5, v = 0.326, E = 119 GPa ( ) ( ) ( ) ( ) ( ) 62 1 6 9 1 2 6 0.0001 306.7 10 0.326 and , so = 306.7 10 (119) 10 36.5 MPa 0.03 36.5 10 25 800 N 25.8 kN . 4 v FL F AE A E E L F A Ans δ σ δ σ pi σ − − − − = = = = = = = = = = = = òò ò Sy = 70 MPa > σ , so elastic deformation assumption is valid. _____________________________________________________________________________ 3-26 ( )6 8(12)20 000 0.185 in . 10.4 10 FL L Ans AE E δ σ= = = = _____________________________________________________________________________ 3-27 ( ) ( ) 6 9 3140 10 0.00586 m 5.86 mm . 71.7 10 FL L Ans AE E δ σ= = = = = _____________________________________________________________________________3-28 ( )6 10(12)15 000 0.173 in . 10.4 10 FL L Ans AE E δ σ= = = = _____________________________________________________________________________ 3-29 With 0,zσ = solve the first two equations of Eq. (3-19) simulatenously. Place E on the left-hand side of both equations, and using Cramer’s rule, ( ) 2 2 1 1 1 1 1 x y x yx y x E v E E vE vE v v v v σ − ++ = = = − − − − ò ò ò òò ò Likewise, Shigley’s MED, 10th edition Chapter 3 Solutions, Page 21/100 ( ) 21 y x y E v ν σ + = − ò ò From Table A-5, E = 207 GPa and ν = 0.292. Thus, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 9 6 2 2 9 6 2 207 10 0.0019 0.292 0.000 72 10 382 MPa . 1 1 0.292 207 10 0.000 72 0.292 0.0019 10 37.4 MPa . 1 0.292 x y x y E v Ans v Ans σ σ − − + − + = = = − − − + = = − − ò ò _____________________________________________________________________________ 3-30 With 0,zσ = solve the first two equations of Eq. (3-19) simulatenously. Place E on the left-hand side of both equations, and using Cramer’s rule, ( ) 2 2 1 1 1 1 1 x y x yx y x E v E E vE vE v v v v σ − ++ = = = − − − − ò ò ò òò ò Likewise, ( ) 21 y x y E v ν σ + = − ò ò From Table A-5, E = 71.7 GPa and ν = 0.333. Thus, ( ) ( ) ( ) ( ) ( ) ( ) ( ) 9 6 2 2 9 6 2 71.7 10 0.0019 0.333 0.000 72 10 134 MPa . 1 1 0.333 71.7 10 0.000 72 0.333 0.0019 10 7.04 MPa . 1 0.333 x y x y E v Ans v Ans σ σ − − + − + = = = − − − + = = − − ò ò _____________________________________________________________________________ 3-31 (a) 1 max 1 c acR F M R a F l l = = = 2 2 2 6 6 6 M ac bh lF F bh bh l ac σ σ = = ⇒ = Ans. (b) ( )( )( ) ( )( )( ) 2 2 1 21( )( ) ( ) .( )( ) m m m mm m m b b h h l lF s s s s Ans F a a c c s s σ σ = = = For equal stress, the model load varies by the square of the scale factor. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 22/100 3-32 (a) 2 1 max /2,2 2 2 2 8x l l l l lR M l = = = − = w w w 2 2 2 2 2 6 6 3 4 . 8 4 3 M l Wl bhW Ans bh bh bh l σ σ = = = ⇒ = w (b) 2 2 2( / )( / )( / ) 1( )( ) . / m m m m m W b b h h s s s Ans W l l s σ σ = = = 2 2 . m m ml ss s Ans l s = ⇒ = = w w w w For equal stress, the model load w varies linearly with the scale factor. _____________________________________________________________________________ 3-33 (a) Can solve by iteration or derive equations for the general case. Find maximum moment under wheel 3W . TW W= Σ at centroid of W’s 3 3 A T l x dR W l − − = Under wheel 3, ( )3 3 3 3 1 13 2 23 3 1 13 2 23A T l x d M R x W a W a W x W a W a l − − = − − = − − For maximum, ( )3 33 3 3 3 0 2 2 TdM l dWl d x x dx l − = = − − ⇒ = Substitute into ( ) 2 3 3 1 13 2 234 T l d M M W W a W a l − ⇒ = − − This means the midpoint of 3d intersects the midpoint of the beam. For wheel i, ( ) 2 1 1 , 2 4 i ii i i T j ji j l dl d x M W W a l − = − − = = −∑ Note for wheel 1: 0j jiW aΣ = 1 2 3 4 104.4104.4, 26.1 kips 4T W W W W W= = = = = = Wheel 1: 2 1 1 476 (1200 238)238 in, (104.4) 20128 kip in 2 4(1200)d M − = = = = ⋅ Wheel 2: 2 238 84 154 ind = − = Shigley’s MED, 10th edition 2 2 max (1200 154) 4(1200)M M Ans − = − = ⋅ = Check if all of the wheels are on the rail (b) max 600 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 ( ( )6 4 2 40(2)(37.5) 25 34 2150 mm 1 12 40 75 34 25 12 12 1.36 10 mm . a b A a b I I I I Ans = − = − = = − = − = (b) 0.375(1.875) 0.703 125 in 0.375(1.75) 0.656 25 in 2(0.703125) 0.656 25 2.0625 in a b A A A = = = = = + = (c) Use two negative areas. 2 2 2625 mm , 5625 mm , 10 000 mm 10 000 5625 625 3750 mm ; a b cA A A A = = = = − − = 3 3 1 2(0.703 125)(0.9375) 0.656 25(0.6875) 2.0625 0.375(1.875) 0.206 in 12 1.75(0.375) 0.007 69 in 12 2 0.206 0.703 125(0.0795) 0.00769 0.65625(0.1 a b y Ans I I I Ans = = = = = = = + + + = Chapter 3 2 2 max (1200 154) (104.4) 26.1(84) 21605 kip in .M M Ans= − = ⋅ = wheels are on the rail. 600 77 523 in .x Ans= − = See above sketch. _____________________________________________________________________________ of entire envelope = area of side notch ( ) )( ) ( )( ) 2 3 3 6 4 2 40(2)(37.5) 25 34 2150 mm 1 12 40 75 34 25 12 12 1.36 10 mm .I Ans = − = − = = − = − Dimensions in mm. 2 2 2 0.375(1.875) 0.703 125 in 0.375(1.75) 0.656 25 in 2(0.703125) 0.656 25 2.0625 in= + = Use two negative areas. 2 2 2 2 625 mm , 5625 mm , 10 000 mm 10 000 5625 625 3750 mm ; a b cA A A= = = = − − = 4 4 2 2 4 2(0.703 125)(0.9375) 0.656 25(0.6875) 0.858 in . 2.0625 0.206 in 0.007 69 in 2 0.206 0.703 125(0.0795) 0.00769 0.65625(0.1705) 0.448 in . y Ans I Ans + = = = + + + = Chapter 3 Solutions, Page 23/100 (104.4) 26.1(84) 21605 kip in .M M Ans _____________________________________________________________________________ Dimensions in mm. 2 2 4705) 0.448 in .I Ans = + + + = Shigley’s MED, 10th edition Chapter 3 Solutions, Page 24/100 ( ) ( ) 1 3 4 3 6 4 3 6 4 6.25 mm, 50 mm, 50 mm 10 000(50) 5625(50) 625(6.25) 57.29 mm . 3750 100 57.29 42.71 mm . 50(12.5) 8138 mm 12 75(75) 2.637 10 mm 12 100(100) 8.333 10 in 12 a b c a b c y y y y Ans c Ans I I I = = = − − = = = − = = = = = = = ( ) ( ) ( ) ( ) ( ) 2 26 2 6 1 6 4 1 8.333 10 10000(7.29) 2.637 10 5625 7.29 8138 625 57.29 6.25 4.29 10 in . I I Ans = + − + − + − = (d) ( ) ( ) ( ) 2 2 2 4 0.875 3.5 in 2.5 0.875 2.1875 in 5.6875 in 2.9375 3.5 1.25(2.1875) 2.288 in . 5.6875 a b a b A A A A A y Ans = = = = = + = + = = ( ) ( ) ( )( ) ( )3 2 3 2 4 1 1(4) 0.875 3.5 2.9375 2.288 0.875 2.5 2.1875 2.288 1.25 12 12 5.20 in . I I Ans = + − + + − = _____________________________________________________________________________ 3-35 ( )3 5 4 2 1 (20)(40) 1.067 10 mm 12 20(40) 800 mm I A = = = = Mmax is at A. At the bottom of the section, ( )max 5 450 000(20) 84.3 MPa . 1.067 10 Mc Ans I σ = = = Due to V, τmax is between A and B at y = 0. max 3 3 3000 5.63 MPa . 2 2 800 V Ans A τ = = = _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 25/100 3-36 3 41 (1)(2) 0.6667 in 12 I = = 21(2) 2 inA = = 0OMΣ = 8 100(8)(12) 0AR − = 1200
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