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Shigley’s Mechanical Engineering Design, 10th edition in SI Units 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 ______________________________________________________________________________ 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 Mechanical Engineering Design, 10th edition in SI Units 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 ( ) ( )63 172.4 1032 113 0.0052 m 5.2861 mm .2.5σ pi= ⇒ = ⇒ = =d S d Ans n d Table A-17: d = 5.5 mm Ans. Factor of safety: ( ) ( ) 6 33 172.4 10 2.82 .113 5.5 10 σ pi − = = = × S n Ans ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units 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 Mechanical Engineering Design, 10th edition in SI Units 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 1.2 6 7.2 8.64 1.25 9 11.25 14.0625 1.31 44 57.64 75.5084 1.37 67 91.79 125.7523 1.42 53 75.26 106.8692 1.48 12 17.76 26.2848 1.53 6 9.18 14.0454 Σ 197 270.08 371.1626 Eq. (1-6) 1 1 270.08 1.37 GPa 197 = = = =∑ k i i i x f x N Eq. (1-7) 2 2 1 22 1 371.16 197(1.37) 0.067 GPa . 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 Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 5/12 1-16 x f fx fx2 0.64 19 12.16 7.78 0.66 25 16.50 10.89 0.64 38 24.32 15.56 0.67 17 11.39 7.63 0.7 12 8.40 5.88 0.71 10 7.10 5.04 0.72 5 3.60 2.59 0.74 4 2.96 2.19 0.75 4 3.00 2.25 0.77 2 1.54 1.19 Σ 136 90.97 61.01 Eq. (1-6) 1 1 90.97 / 136 0.67 MPa = = = =∑ k i i i x f x N Eq. (1-7) 2 2 1 22 1 61.01 136(0.67) 0.034 MPa 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 0.67 ˆ 0.034 µ σ − − − = = = x x x x x x x z s Solving for the yield strength gives x0.01 = 0.67 + 0.034 z0.01 From Table A-10, z0.01 = − 2.326. Thus x0.01 = 0.67 + 0.034(− 2.326) = 0.60 MPa Ans. ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 6/12 1-17 _____________________________________________________________________________ 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)10sy T nd Ans Spi pi = = = = ______________________________________________________________________________ 1-19 Obtain the coefficients of variance for stress and strength ˆ ˆ 22.24 0.07692 289.13 σ σ σ σ σµ = = = = PC P Eq. (1-9) Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 7/12 ˆ ˆ 45.44 0.06901 658.45 σσ µ = = = = ySS S S y C S (a) 1.2n = Eq. (1-11): ( )2 2 2 2 2 2 1 1.2 1 1.7695 1.2 0.06901 0.07692σ − − = − = − = − + + d d S n z n C C Interpolating Table A-10, 1.76 0.0392 1.7695 Φ ⇒ Φ = 0.03844 1.77 0.0384 R = 1 − 0.03844 = 0.96156 Ans. ( ) ( ) 2 2 3 6 / ( / 4) 4 4 289.13 10 1.24 0.0259m 25.9mm . 658.45 10 pi σ pi pi pi = = = × ⇒ = = = = × y y y y S S d S n P d P Pnd Ans S (b) 1.5n = ( )2 2 2 2 2 2 1 1.5 1 3.8770 1.2 0.06901 0.07692σ − − = − = − = − + + d d S n z n C C 3.8 0.000072 3.8770 Φ ⇒ Φ = 0.00005367 3.9 0.000048 R = 1 − 0.00005367 = 0.99994633 Ans. ( ) ( ) 3 6 4 289.13 10 1.54 0.029m 29mm . 658.45 10pi pi × = = = = ×y Pnd Ans S ______________________________________________________________________________ 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= + = + = Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 8/12 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 = = = = 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 _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 9/12 1-22 ______________________________________________________________________________ 1-23 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 10/12 1-24 ______________________________________________________________________________ 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. ______________________________________________________________________________ 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 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 11/12 1-28 ______________________________________________________________________________ 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. (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. Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 1 Solutions, Page 12/12 (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 ______________________________________________________________________________ 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 Mechanical Engineering Design, 10th edition in SI Units 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 = = ⋅ ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 2/22 ______________________________________________________________________________ 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 − = = Thepercent difference from the value in Table A-5 is 0.208 0.211 0.0142 1.42 percent . 0.211 Ans− = − = − ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 3/22 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 4/22 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 5/22 ___________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 6/22 ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 7/22 ________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 8/22 ______________________________________________________________________________ ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 9/22 ______________________________________________________________________________ ______________________________________________________________________________ 2-12 For HB = 275, Eq. (2-21), Su = 3.4(275) = 935 MPa Ans. ______________________________________________________________________________ ______________________________________________________________________________ ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 10/22 ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 11/22 ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 12/22 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 13/22 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 14/22 2-18, 2-19 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 15/22 ________________________________________________________________________ ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 16/22 ______________________________________________________________________________ ______________________________________________________________________________ 2-24 and 2-25 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 17/22 ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 18/22 _ ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 19/22 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 20/22 _______________________________________________________________________ 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 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) Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 21/22 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 For mass, m = Alρ = (F/S) lρ So, f 3(M ) = ρ /S, and maximize S/ρ . Thus, β = 1. Ans. ______________________________________________________________________________ ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 22/22 ________________________________________________________________________ 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 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 23/22 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. 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 0.00483250790 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. Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 2 Solutions, Page 24/22 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). ______________________________________________________________________________ 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 Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 1/100 Chapter 3 ________________________________________________________________________ ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 2/100 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 = + = = − = + − = = − = + = ∑ ∑ o o Step 2: Find components of RC and RD on link 4 ( ) ( )4 4 0 400(4.5) 7.794 1.9 0 305.4 N . 0 305.4 N 0 ( ) 400 N C D D x Cx y Cy M R 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 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 3/100 Step 3: Find components of RC on link 2 ( ) ( ) ( ) 2 2 2 0 305.4 346.4 0 41 N 0 200 N x Cx Cx y Cy F R R F R = + − = = = = ∑ ∑ Ans. _____________________________________________________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units 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 = − = _____________________________________________________________________________ ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units 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 = − + = ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 6/100 ______________________________________________________________________________ 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 Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 7/100 ________________________________________________________________________ 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 Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 8/100 ________________________________________________________________________ 3-13 Solution depends upon the beam selected. ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 9/100 _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 10/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 11/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 12/100 ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 13/100 3-16 (a) 2 2 1 2 8 7 0.5 MPa 2 8 7 7.5 MPa 2 7.5 6 9.60MPa 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 − = = o 1 8.078 MPa 45 10.9 34.1 ccws Rτ φ = = = − = o o o Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 14/100 (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 − = = o 1 9.71 MPa 45 27.75 17.25 cws Rτ φ = = = − = o o o ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 15/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 16/100 ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 17/100 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 54.1 MPa 2 39.1 19.6 MPa 2 69.1 34.6 MPa 2 τ τ τ + = = = = = = Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 18/100 (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 + − = = = = = ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 19/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 20/100 ________________________________________________________________________ _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 21/100 ______________________________________________________________________________ 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 = = = = = _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 22/100 ________________________________________________________________________ ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 23/100 ____________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ 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 Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 24/100 ( ) 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 Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 25/100 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 sσ 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. _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 26/100 _______________________________________________________________________ ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 27/100 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 Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 28/100 ______________________________________________________________________________ ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 29/100 ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 30/100 _______________________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 31/100 3-41 ( ) ( )4 4 2 212 1018 mm , 12 113.1 mm64 4I A pi pi = = = = Model (c) 2 2 max 2000(6) 2000(9) 15 000 N mm 2 2 15 000(6) 1018 88.4 N/mm 88.4 MPa . 4 4 2000 23.6 N/mm 23.6 MPa . 3 3 113.1 M Mc I Ans V Ans A σ σ τ = + = ⋅ = = = = = = = = Model (d) 2 2000(12) 24 000 N mm 24 000(6) 1018 141.5 N/mm 141.5 MPa . M Mc I Ans σ σ = = ⋅ = = = = 2 max 4 4 2000 23.6 N/mm 23.6 MPa . 3 3 113.1 V Ans A τ = = = = Model (e) 2 2000(7.5) 15000 N mm 15000(6) 1018 88.4 N/mm 88.4 MPa . M Mc I Ans σ σ = = ⋅ = = = = 2 max 4 4 2000 23.6 N/mm 23.6 MPa . 3 3 113.1 V Ans A τ = = = = _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 32/100 ______________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 33/100 ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 34/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 35/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 36/100 _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 37/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 38/100 ____________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 39/100 3-46 1 0 cR F l cM Fx x a l = = ≤ ≤ ( ) 2 2 max 66 6 0 . c l FxM bh bh Fcxh x a Ans lb σ σ = = = ≤ ≤ _____________________________________________________________________________ 3-47 From Problem 3-46, 1 , 0 cR F V x a l = = ≤ ≤ max max 3 3 ( / ) 3 . 2 2 2 V c l F Fch Ans bh bh lb τ τ = = ⇒ = From Problem 3-46, max 6( ) Fcxh x lbσ = . Sub in x = e and equate to h above. max max max 2 max 3 6 2 3 . 8 Fc Fce lb lb Fc e Ans lb τ σ σ τ = = _____________________________________________________________________________ 3-48 (a) x-z plane 20 1.5(0.5) 2(1.5)sin(30 )(2.25) (3)O zM RΣ = = + −o 2 1.375 kN .zR Ans= 10 1.5 2(1.5)sin(30 ) 1.375z zF RΣ = = − − +o 1 1.625 kN .zR Ans= x-y plane 20 2(1.5)cos(30 )(2.25) (3)O yM RΣ = = − +o 2 1.949 kN .yR Ans= 10 2(1.5) cos(30 ) 1.949y yF RΣ = = − +o 1 0.6491 kN .yR Ans= Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 40/100 (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, ( ) ( ) 2 2 0.125 0.125 2 At 0, 0.9375 0.5 0.125 0.9375 z y z y y V x x M V dx x C x M C M x x ′= − ′ ′ ′= = − + ′ ′ ′= = = − ⇒ = − + ∫ ( ) ( ) 2 2 1.949 0.6491 1.732 0.6491 1.125 1.732 0.6491 2 At 0, 0.9737 0.8662 0.125 0.9375 y z z z V x x M x x C x 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 kN·m. The table below shows values at key locations on the shear and bending moment diagrams. Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 41/100 x (m) Vz (kN) Vy (kN) V (kN) My (kN⋅m) Mz (kN⋅m) M (kN⋅m) 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 (d) The bending stress is obtained from Eq. (3-27), y Az A x z y M zM y I I σ − = + 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 3 6 4 6 440(75) 34(25) 1.36(10 ) mm 1.36(10 ) m 12 12z I −= − = = 3 3 5 4 7 425(40) 25(6)2 2.67(10 ) mm 2.67(10 ) m 12 12y I − = + = = 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 kN·m, Mz = 1.075 kN·m, and σx = 100.1 MPa. Ans. _____________________________________________________________________________Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 42/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 43/100 ______________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 44/100 ___________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 45/100 ___________________________________________________________________________ 3-52 3 1 1 3 3 3 i i i i i i T GL cT GL c θθ = ⇒ = 3 31 1 2 3 1 . 3 i ii GT T T T L c Ansθ = = + + = ∑ From Eq. (3-47), τ = Gθ1c G and θ1 are constant, therefore the largest shear stress occurs when c is a maximum. max 1 max .G c Ansτ θ= _____________________________________________________________________________ 3-53 (b) Solve part (b) first since the angle of twist per unit length is needed for part (a). ( )max allow 12 6.89 82.7 MPaτ τ= = = ( ) ( ) 6 max 1 9 max 82.7 10 0.348 rad/m . 79.3 10 (0.003) Ans Gc τθ = = = (a) Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 46/100 ( )9 331 1 1 1 0.348(79.3) 10 (0.020)(0.002 ) 1.47 N m . 3 3 GL cT Ansθ= = = ⋅ ( ) ( ) 9 33 2 2 2 2 9 33 3 3 3 3 1 2 3 0.348(79.3) 10 (0.030)(0.003 ) 7.45 N m . 3 3 0.348(79.3) 10 (0)(0 ) 0 . 3 3 1.47 7.45 0 8.92 N m . GL cT Ans GL cT Ans T T T T Ans θ θ = = = ⋅ = = = = + + = + + = ⋅ _____________________________________________________________________________ _____________________________________________________________________________ 3-55 (b) Solve part (b) first since the angle of twist per unit length is needed for part (a). ( )max allow 12 6.89 82.7 MPaτ τ= = = ( ) ( ) 6 max 1 9 max 82.7 10 0.348 rad/m . 79.3 10 (0.003) Ans Gc τθ = = = (a) ( )9 331 1 1 1 0.348(79.3) 10 (0.020)(0.002 ) 1.47 N m . 3 3 GL cT Ansθ= = = ⋅ ( ) ( ) 9 33 2 2 2 2 9 33 3 3 3 3 1 2 3 0.348(79.3) 10 (0.030)(0.003 ) 7.45 N m . 3 3 0.348(79.3) 10 (0.025)(0.002 ) 1.84 N m . 3 3 1.47 7.45 1.84 10.8 N m . GL cT Ans GL cT Ans T T T T Ans θ θ = = = ⋅ = = = ⋅ = + + = + + = ⋅ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 47/100 _____________________________________________________________________________ 3-56 (a) From Eq. (3-40), with two 2-mm strips, ( )( )( )( ) ( ) 6 22 max max 80 10 0.030 0.002 3.08 N m 3 1.8 / ( / ) 3 1.8 / 0.030 / 0.002 2(3.08) 6.16 N m . bcT b c T Ans τ = = = ⋅ + + = = ⋅ From the table on p. 116, with b/c = 30/2 = 15, α = β and has a value between 0.313 and 0.333. From Eq. (3-40), 1 0.321 3 1.8 / (30 / 2)α ≈ =+ From Eq. (3-41), ( )( )( )( )3 3 9 3.08(0.3) 0.151 rad . 0.321 0.030 0.002 79.3 10 6.16 40.8 N m . 0.151t Tl Ans bc G Tk Ans θ β θ = = = = = = ⋅ From Eq. (3-40), with a single 4-mm strip, ( ) ( )( )( ) ( ) 6 22 max max 80 10 0.030 0.004 11.9 N m . 3 1.8 / ( / ) 3 1.8 / 0.030 / 0.004 bcT Ans b c τ = = = ⋅ + + Interpolating from the table on p. 116, with b/c = 30/4 = 7.5, 7.5 6 (0.307 0.299) 0.299 0.305 8 6 β −= − + = − From Eq. (3-41) ( )( )( )( )3 3 9 11.9(0.3) 0.0769 rad . 0.305 0.030 0.004 79.3 10 11.9 155 N m . 0.0769t Tl Ans bc G Tk Ans θ β θ = = = = = = ⋅ (b) From Eq. (3-47), with two 2-mm strips, ( )( )( )( )2 62 max 0.030 0.002 80 10 3.20 N m 3 3 2(3.20) 6.40 N m . LcT T Ans τ = = = ⋅ = = ⋅ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 48/100 ( )( )( )( )3 3 9 3 3(3.20)(0.3) 0.151 rad . 0.030 0.002 79.3 10 6.40 0.151 42.4 N m .t Tl Ans Lc G k T Ans θ θ = = = = = = ⋅ From Eq. (3-47), with a single 4-mm strip, ( ) ( ) ( ) ( )2 62 max 0.030 0.004 80 10 12.8 N m . 3 3 LcT Ansτ= = = ⋅ ( )( )( )( )3 3 9 3 3(12.8)(0.3) 0.0757 rad . 0.030 0.004 79.3 10 12.8 0.0757 169 N m .t Tl Ans Lc G k T Ans θ θ = = = = = = ⋅ 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 m 2500 HT n = = = ⋅ max 3 16Tr T J d τ pi = = ( ) ( ) ( ) 1 31 3 6 max 16 152.816 0.0223 m 22.3 mm . 70 10 Td Ans piτ pi = = = = (b) (40000)9.55 9.55 1528 N m 250 HT n = = = ⋅ ( )( ) 1 3 6 16(1528) 0.0481 m 48.1 mm . 70 10 d Ans pi = = = _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 49/100 ___________________________________________________________________________ 3-59 ( )( ) ( )6 33 max max 3 50 10 0.0316 265 N m 16 16 dT T d piτ pi τ pi = ⇒ = = = ⋅ Eq. (3-44), ( )3265(2000) 55.5 10 W 55.5 kW .9.55 9.55TnH Ans= = = = _____________________________________________________________________________ 3-60 ( )( )( ) ( )( ) ( ) 3 6 3 3 4 9 4 16 110 10 0.020 173 N m 16 16 0.020 79.3 10 15 180 32 32(173) 1.89 m . T T d d Tl d Gl JG T l Ans pi pi τ τ pi pi pi pi θθ = ⇒ = = = ⋅ = ⇒ = = = _____________________________________________________________________________ ___________________________________________________________________________ 3-62 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 50/100 (a) 4 4 4 max max max max solid hollow ( ) 16 16 o o i o o J d J d d T T r d r d τ pi τ τ pi τ− = = = = ( ) ( ) 44 solid hollow 4 4 solid 36 % (100%) (100%) (100%) 65.6% . 40 i o T T dT Ans T d −∆ = = = = (b) ( )2 2 2solid hollow, o o iW kd W k d d= = − ( ) ( ) 22 solid hollow 2 2 solid 36 % (100%) (100%) (100%) 81.0% . 40 i o W W dW Ans W d −∆ = = = = _____________________________________________________________________________ 3-63 (a) ( )444 max max max max solid hollow 16 16 d xdJ d JT T r d r d pi ττ pi τ τ − = = = = 4 4solid hollow 4 soli ( )% (100%) (100%) (100%) . d T T xdT x Ans T d −∆ = = = (b) ( )( )22 2solid hollow W kd W k d xd= = − ( )2 2solid hollow 2 solid % (100%) (100%) (100%) .xdW WW x Ans W d −∆ = = = Plot %∆T and %∆W versus x. The value of greatest difference in percent reduction of weight and torque is 25% andoccurs at 2 2x = . _____________________________________________________________________________ 0 20 40 60 80 100 0 0,2 0,4 0,6 0,8 1 P e rc e n t R e d u ct io n x Percent Reduction in Torque and Weight Weight Torque difference Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 51/100 3-64 (a) ( ) ( )( ) ( ) ( )46 344 2.8149 104200 2 120 10 32 0.70 dTc J dd d τ pi = ⇒ = = − ( ) ( ) 1 34 2 6 2.8149 10 6.17 10 m 61.7 mm 120(10 ) d − = = = From Table A-17, the next preferred size is d = 80 mm. Ans. di = 0.7d = 56 mm. The next preferred size smaller is di = 50 mm Ans. (b) ( ) ( ) ( ) ( ) ( ) ( ) ( )4 4 44 4200 2 4200 0.050 2 30.8 MPa . 32 0.080 0.05032 i i dTc Ans J d d τ pipi = = = = −− _____________________________________________________________________________ 3-65 (1500)9.55 9.55 1433 N m 10 HT n = = = ⋅ ( ) ( )( ) 1 3 1 3 3 6 16 143316 16 = 0.045 m 45 mm 80 10CC T Td d τ piτpi pi = ⇒ = = = From Table A-17, select 50 mm. Ans. (a) ( )( )( ) ( ) 6 start 3 16 2 1433 117 10 Pa 117 MPa . 0.050 Ansτ pi = = = (b) Design activity _____________________________________________________________________________ ___________________________________________________________________________ 3-67 For a square cross section with side length b, and a circular section with diameter d, 2 2 square circular 4 2 A A b d b dpi pi= ⇒ = ⇒ = Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 52/100 From Eq. (3-40) with b = c, ( ) 3 max 2 3 3 3square 1.8 1.8 23 3 (4.8) 6.896 / 1 T T T T bc b c b d d τ pi = + = + = = For the circular cross section, ( )max 3 3circular 16 5.093T Td dτ pi= = ( ) ( ) 3max square max circular 3 6.896 1.354 5.093 T d T d τ τ = = The shear stress in the square cross section is 35.4% greater. Ans. (b) For the square cross section, from the table on p. 116, β = 0.141. From Eq. (3-41), square 43 4 411.50 0.141 2 Tl Tl Tl Tl bc G b G d G d G θ β β pi = = = = For the circular cross section, ( ) 44 10.1932rd Tl Tl Tl GJ d GG d θ pi = = = 4 4 11.50 1.129 10.19 sq rd Tl d G Tl d G θ θ = = The angle of twist in the square cross section is 12.9% greater. Ans. _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 53/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 54/100 _________________________________________________________________________ 3-69 (a) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 3 2 1 1 1 3 1 1 2 0.15 0 1800 270 (200) (125) 306 10 125 0.15 306 10 106.25 0 2880 N . 0.15 2880 432 N . T T T T T T T T T Ans T Ans = = = − + − = + − − = ⇒ = = = ∑ (b) 0 3312(230) (510) 2070(810) 1794 N . 0 3312 1794 2070 3036 N . O C C y O O M R R Ans F R R Ans = = + − = = = + + − = − ∑ ∑ (c) Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 55/100 (d) The maximum bending moment is at x = 230 mm, and is M = –698.3 N·m. 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 N·m. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, ( ) ( )33 332 698.332 263 10 Pa 263 MPa .(0.030) Mc M Ans I d σ pi pi = = = = = ( )63 316 16(306) 57.7 10 Pa 57.7MPa .(0.030) Tr T Ans J d τ pi pi = = = = = (e) ( ) ( ) ( ) ( ) 2 2 2 2 1 2 1 2 2 2 2 2 max 263 263 , 57.7 2 2 2 2 275 MPa . 12.1 MPa . 263 57.7 144 MPa . 2 2 x x xy x xy Ans Ans Ans σ σ σ σ τ σ σ σ τ τ = ± + = ± + = = − = + = + = _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 56/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 57/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 58/100 3-71 (a) ( ) ( ) ( ) ( ) 2 1 2 1 1 1 1 1 2 0.15 0 300 45 (125) (150) 31 875 0.15 (150) 31 875 127.5 0 250 N mm . 0.15 250 37.5 N mm . T T T T T T T T T Ans T Ans = = = − + − = + − − = ⇒ = ⋅ = = ⋅ ∑ (b) o o o o 0 345sin 45 (300) 287.5(700) (850) 150.7 N . 0 345cos 45 287.5 150.7 107.2 N . 0 345sin 45 (300) (850) 86.10 N . 0 345cos 45 86.10 O y C z C z z O z O z O z C y C y y O y O y M R R Ans F R R Ans M R R Ans F R R = = − − = − = = − + − = = = + = − = = + − ∑ ∑ ∑ ∑ 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, Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 59/100 ( ) ( )2 247.37 32.16 57.26 N mM = − + − = ⋅ The torque transmitted through the shaft from A to B is T = (300 − 45)(0.125) = 31.88 N·m. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, ( ) ( )63 332 57.2632 72.9 10 Pa 72.9 MPa .(0.020) Mc M Ans I d σ pi pi = = = = = ( )63 316 16(31.88) 20.3 10 Pa 20.3 MPa .(0.020) Tr T Ans J d τ pi pi = = = = = (e) ( ) ( ) ( ) ( ) 2 2 2 2 1 2 1 2 2 2 2 2 max 72.9 72.9 , 20.3 2 2 2 2 78.2 MPa . 5.27 MPa . 72.9 20.3 41.7 MPa . 2 2 x x xy x xy Ans Ans Ans σ σ σ σ τ σ σ σ τ τ = ± + = ± + = = − = + = + = _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 60/100 _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 61/100 3-73 (a) 0 11000(cos 20º )(300) (cos 25º )(150) 22 810 N . B B T F F Ans = = − + = ∑ (b) 0 11000(sin 20º )(400) 22 810(sin 25º )(750) (1050) 8319 N . O z C y C y M R R Ans = = − − + = ∑ 0 11000(sin 20º ) 22 810sin(25º ) 8319 5083 N . 0 11 000(cos 20º )(400) 22 810(cos 25º )(750) (1050) 10 830 N . 0 11 000(cos 20º ) 22 810(cos25º ) 10 830 494 N . y O y O y O y C z C z z O z O z F R R Ans M R R Ans F R R 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 mM = + = ⋅ The torque transmitted through the shaft from A to B is ( )( )11000cos 20º 0.3 3101 N mT = = ⋅ . For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 62/100 ( ) ( )63 332 409732 333.9 10 Pa 333.9 MPa .(0.050) Mc M Ans I d σ pi pi = = = = = ( )63 316 16(3101) 126.3 10 Pa 126.3 MPa .(0.050) Tr T Ans J d τ pi pi = = = = = (e) ( ) ( ) ( ) ( ) 2 2 2 2 1 2 1 2 2 2 2 2 max 333.9 333.9 , 126.3 2 2 2 2 376 MPa . 42.4 MPa . 333.9 126.3 209 MPa . 2 2 x x xy x xy Ans Ans Ans σ σ σ σ τ σ σ σ τ τ = ± + = ± + = = − = + = + = _____________________________________________________________________________ __ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 63/100 ______________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 64/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 65/100 _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 66/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 67/100 _________________________________________________________________________ 3-77 100 1600 N / 2 0.125 / 2t TF c = = = ( ) ( ) ( ) ( ) 1600 tan 20 582.4 N 2 1600 0.250 2 200 N m 200 2667 N 2 0.150 2 n C t C F T F b T P a = = = = = ⋅ = = = ( ) 0 450 582.4(325) 2667(75) 0 865.1 N A z Dy Dy M R R = − − = = ∑ ( ) 0 450 1600(325) 1156 NA Dz DzyM R R= = − + ⇒ =∑ 0 865.1 582.4 2667 2384 Ny Ay AyF R R= = + − − ⇒ =∑ 0 1156 1600 444 Nz Az AzF R R= = + − ⇒ =∑ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 68/100 AB The maximum bending moment will either be at B or C. If this is not obvious, sketch the shear and bending moment diagrams. We will directly obtain the combined moments from each plane. 2 2 2 2 2 2 2 2 0.075 2384 444 181.9 N m 0.125 865.1 1156 180.5 N m y z y z B A A C D D M AB R R M CD R R = + = + = ⋅ = + = + = ⋅ The stresses at B and C are almost identical, but the maximum stresses occur at B. Ans. ( ) ( ) ( ) ( ) 6 3 3 6 3 3 32 32(181.9) 68.6 10 Pa 68.6 MPa 0.030 16 16(200) 37.7 10 Pa 37.7 MPa 0.030 B B B B M d T d σ pi pi τ pi pi = = = = = = = = 2 2 2 2 max 68.6 68.6 37.7 85.3 MPa . 2 2 2 2 B B B Ans σ σ σ τ = + + = + + = 2 2 2 2 max 68.6 37.7 51.0 MPa . 2 2 B B Ans σ τ τ = + = + = _____________________________________________________________________________ 3-78 100 1600 N / 2 0.125 / 2t TF c = = = ( ) ( ) ( ) ( ) 1600 tan 20 582.4 N 2 1600 0.250 2 200 N m 200 2667 N 2 0.150 2 n C t C F T F b T P a = = = = = ⋅ = = = ( ) ( ) 0 450 582.4(325) 420.6 N 0 450 1600(325) 2667(75) 711.1 N 0 420.6 582.4 161.8 N 0 711.1 1600 2667 A Dy Dyz A Dz Dzy y Ay Ay z Az M R R M R R F R R F R = = − ⇒ = = = − + − ⇒ = = = + − ⇒ = = = + − + ∑ ∑ ∑ ∑ 1778 NAzR⇒ = − Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 69/100 The maximum bending moment will either be at B or C. If this is not obvious, sketch shear and bending moment diagrams. We will directly obtain the combined moments from each plane. ( )22 2 2 2 2 2 2 0.075 161.8 1778 133.9 N m 0.125 420.6 711.1 103.3 N m y z y z B A A C D D M AB R R M CD R R = + = + − = ⋅ = + = + = ⋅ The maximum stresses occur at B. Ans. ( ) ( ) ( ) ( ) 6 3 3 6 3 3 32 32(133.9) 50.5 10 Pa 50.5 MPa 0.030 16 16(200) 37.7 10 Pa 37.7 MPa 0.030 B B B B M d T d σ pi pi τ pi pi = = = = = = = = 2 2 2 2 max 50.5 50.5 37.7 70.6 MPa . 2 2 2 2 B B B Ans σ σ σ τ = + + = + + = 2 2 2 2 max 50.5 37.7 45.4 MPa . 2 2 B B Ans σ τ τ = + = + = _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 70/100 _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 71/100 _____________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 72/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 73/100 _______________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 74/100 __________________________________________________________________________ __________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 75/100 ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 76/100 _______________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 77/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 78/100 ______________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 79/100 ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 80/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 81/100 _____________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 82/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions,Page 83/100 ______________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 84/100 Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 85/100 __________________________________________________________________________ 3-91 As shown in Fig. 3-32, the maximum stresses occur at the inside fiber where r = ri. Therefore, from Eq. (3-50) 2 2 ,max 2 2 2 2 2 2 2 2 2 ,max 2 2 2 1 . 1 . i i o t o i i o i i o i i i o r i o i i r p r r r r r r p Ans r r r p r p Ans r r r σ σ = + − + = − = − = − − ______________________________________________________________________________ 3-92 If pi = 0, Eq. (3-49) becomes 2 2 2 2 2 2 2 2 2 2 2 / 1 o o i o o t o i o o i o i p r r r p r r r p r r r r r σ − − = − = − + − The maximum tangential stress occurs at r = ri. So 2 ,max 2 2 2 . o o t o i p r Ans r r σ = − − For σr, we have 2 2 2 2 2 2 2 2 2 2 2 / 1 o o i o o r o i o o i o i p r r r p r r r p r r r r r σ − − = − = − − So σr = 0 at r = ri. Thus at r = ro Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 86/100 2 2 2 ,max 2 2 2 . o o i o r o o i o p r r r p Ans r r r σ − = = − − ______________________________________________________________________________ 3-93 The force due to the pressure on half of the sphere is resisted by the stress that is distributed around the center plane of the sphere. All planes are the same, so ( ) 2 av 1 2 / 4( ) . 4 i i t i p d pd Ans d t t pi σ σ σ pi = = = = The radial stress on the inner surface of the shell is, σ3 = − p Ans. ______________________________________________________________________________ ________________________________________________________________________ 3-95 σt > σl > σr τmax = (σt − σr)/2 at r = ri ( ) 2 2 2 2 2 2 2 max 2 2 2 2 2 2 2 2 2 2 2 6 max 6 max 1 1 1 2 ( ) (25 4)10100 91.7 mm 25 10 100 91.7 8.3 mm . i i o i i o i i o o i o i i o i i o i i o i i i o o i r p r r p r r p r r p r r r r r r r r r r r p r r t r r Ans τ τ τ = + − − = = − − − − − − ⇒ = = = = − = − = ______________________________________________________________________________ ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 87/100 ___________________________________________________________________ 3-98 From Eq. (3-49) with pi = 0, 2 2 2 2 2 2 2 2 2 2 1 1 o o i t o i o o i r o i r p r r r r r p r r r r σ σ = − + − = − − − σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro ( ) ( ) 2 2 2 2 2 2 2 max 2 2 2 2 2 2 2 2 2 2 2 6 max 6 max 1 1 1 2 25 10 100 92.8 mm( ) 25 4 10 100 92.8 7.2 mm . o o i o o i o o i i o o i o o i o o i o o i i o o o i r p r r p r r p r r p r r r r r r r r r r r r r p t r r Ans τ τ τ = − − + + = = − − − − ⇒ = = = + + = − = − = ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 88/100 ______________________________________________________________________ 3-100 From Table A-20, Sy=490 MPa From Eq. (3-49) with pi = 0, 2 2 2 2 21 o o i t o i r p r r r r σ = − + − Maximum will occur at r = ri [ ]( )2 22 22 ,max ,max 2 2 2 2 0.8( 490) 25 19( )2 82.8 MPa . 2 2(25 ) t o io o t o o i o r rr p p Ans r r r σ σ − − − = − ⇒ = − = − = − ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 89/100 ________________________________________________________________________ 3-102 From Table A-20, Sy=490 MPa From Eq. (3-50) 2 2 2 2 21 i i o t o i r p r r r r σ = + − Maximum will occur at r = ri ( ) [ ] 2 22 2 ,max 2 2 2 2 2 2 2 2 2 ,max 2 2 2 2 1 ( ) 0.8(490) (25 19 ) 105 MPa .(25 19 ) i o ii i o t o i i o i t o i i o i p r rr p r r r r r r r r p Ans r r σ σ + = + = − − − − ⇒ = = = + + ______________________________________________________________________________ ________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 90/100 ______________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 91/100 __ ________________________________________________________________________ __ __ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 92/100 _________________________________________________________________ 3-107 ω = 2pi (2000)/60 = 209.4 rad/s, ρ = 3320 20 kg/m3, ν = 0.24, ri = 0.01 m, ro = 0.125 m Using Eq. (3-55) ( ) ( )2 22 2 2 63 0.24 1 3(0.24)3320(209.4) 0.01 (0.125) (0.125) 0.01 (10) 8 3 0.24 1.85 MPa . t Ans σ − + + = + + − + = ______________________________________________________________________________ Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 93/100 ________________________________________________________________________ ________________________________________________________________________ 3-110 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57), ( )9 2 2 2 9 33 2207(10 ) (0.05 0.025 )(0.025 0) 10 3.105(10 ) (1)2(0.025) (0.05 0)p δ δ− − −= = − where p is in MPa and δ is in mm. Maximum interference, max 1 [50.042 50.000] 0.021 mm . 2 Ansδ = − = Minimum interference, min 1 [50.026 50.025] 0.0005 mm . 2 Ansδ = − = From Eq. (1) pmax = 3.105(103)(0.021) = 65.2 MPa Ans. Shigley’s Mechanical Engineering Design, 10th edition in SI Units Chapter 3 Solutions, Page 94/100 pmin = 3.105(103)(0.0005) = 1.55 MPa Ans. ______________________________________________________________________________ ________________________________________________________________________ 3-112 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57), ( )9 2 2 2 9 33 2207(10 ) (0.05 0.025 )(0.025 0) 10 3.105(10 ) (1)2(0.025) (0.05 0)p δ δ− − −= = − where p is in MPa and δ is in mm. Maximum interference, max 1 [50.059 50.000] 0.0295 mm . 2 Ansδ = − = Minimum interference, min 1 [50.043 50.025] 0.009 mm . 2 Ansδ = − =
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