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www.konkur.in Chapter 1 Problems 1-1 through 1-6 are for student research. No standard solutions are provided. 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. ______________________________________________________________________________ 1-7 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 nd = 1.10 = 1.43 Ans. 2 0.85 ( 0.95 ) ______________________________________________________________________________ 1-10 (a) X1 + X2: x1 + x2 = X 1 + e1 + X 2 + e2 error = e = ( x1 + x2 ) − ( X 1 + X 2 ) = e1 + e2 (b) X1 − X2: Ans. x1 − x2 = X 1 + e1 − ( X 2 + e2 ) e = ( x1 − x2 ) − ( X1 − X 2 ) = e1 − e2 (c) X1 X2: Ans. x1 x2 = ( X 1 + e1 )( X 2 + e2 ) e = x1 x2 − X 1 X 2 = X 1e2 + X 2 e1 + e1e2 e e ≈ X 1e2 + X 2 e1 = X 1 X 2 1 + 2 X1 X 2 Shigley’s MED, 10th edition Ans. Chapter 1 Solutions, Page 1/12 www.konkur.in (d) X1/X2: x1 X 1 + e1 X 1 1 + e1 X 1 = = x2 X 2 + e2 X 2 1 + e2 X 2 −1 e2 e2 1 + ≈ 1− X2 X2 then 1 + e1 X 1 e1 e2 e1 e2 − ≈ 1 + 1 − ≈ 1+ X 1 X 2 X1 X 2 1 + e2 X 2 x1 X 1 X 1 e1 e2 − ≈ − Ans. x2 X 2 X 2 X 1 X 2 ______________________________________________________________________________ Thus, e= 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 ______________________________________________________________________________ 1-11 (a) S 1-12 σ = nd 3 32 (1000 ) 25 (10 ) ⇒ = 2.5 πd3 Table A-17: d = 1 14 in Factor of safety: ⇒ d = 1.006 in Ans. Ans. n= S σ = 25 (103 ) 32 (1000 ) π (1.25 ) = 4.79 Ans. 3 ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 1 Solutions, Page 2/12 www.konkur.in 1-13 (a) f x2 7200 4900 19200 40500 80000 145200 86400 169000 156800 112500 51200 86700 64800 36100 0 44100 x 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 fx 120 70 240 450 800 1320 720 1300 1120 750 320 510 360 190 0 210 Σ 69 8480 1 104 600 x = Eq. (1-6) 1 N k ∑fx i =1 i i = 8480 = 122.9 kcycles 69 Eq. (1-7) k sx = (b) ∑fx i =1 2 i i − N x2 N −1 Eq. (1-5) 1/ 2 1 104 600 − 69(122.9)2 = 69 − 1 z115 = = 30.3 kcycles Ans. x − µx x − x 115 − 122.9 = 115 = = −0.2607 σˆ x sx 30.3 Interpolating from Table (A-10) 0.2600 0.2607 0.2700 0.3974 x 0.3936 ⇒ NΦ(−0.2607) = 69 (0.3971) = 27.4 ≈ 27 x = 0.3971 Ans. From the data, the number of instances less than 115 kcycles is Shigley’s MED, 10th edition Chapter 1 Solutions, Page 3/12 www.konkur.in 2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal) ____________________________________________________________________________ 1-14 x 174 182 190 198 206 214 222 Σ Eq. (1-6) x = 6 9 44 67 53 12 6 fx 1044 1638 8360 13266 10918 2568 1332 f x2 181656 298116 1588400 2626668 2249108 549552 295704 197 39126 7789204 f 1 N k ∑fx i =1 i i k ∑fx 2 i i = 39 126 = 198.61 kpsi 197 − N x2 12 7 789 204 − 197(198.61) 2 Eq. (1-7) sx = = = 9.68 kpsi Ans. N −1 197 − 1 ______________________________________________________________________________ i =1 1-15 L = 122.9 kcycles and sL = 30.3 kcycles x − µ x x10 − L x10 − 122.9 = = sL 30.3 σˆ Eq. (1-5) z10 = 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 4/12 www.konkur.in 1-16 Eq. (1-6) x = x f 93 95 97 99 101 103 105 107 109 111 Σ 19 25 38 17 12 10 5 4 4 2 136 1 N k ∑fx i =1 i i k Eq. (1-7) ∑fx sx = i =1 2 i i fx 1767 2375 3686 1683 1212 1030 525 428 436 222 13364 fx2 164331 225625 357542 166617 122412 106090 55125 45796 47524 24642 1315704 = 13 364 / 136 = 98.26471 = 98.26 kpsi − N x2 N −1 12 1 315 704 − 136(98.26471) 2 = 136 − 1 = 4.30 kpsi 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), x − µx x −x x − 98.26 z0.01 = = 0.01 = 0.01 sx 4.30 σˆ x 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. ______________________________________________________________________________ n 1-17 Eq. (1-9): R = ∏ Ri = 0.98(0.96)0.94 = 0.88 i =1 Overall reliability = 88 percent Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 1 Solutions, Page 5/12 www.konkur.in 1-18 Obtain the coefficients of variance for strength and stress σˆ S 23.5 CS = sy = = 0.07532 Ssy 312 Cσ = 145 σˆτ σˆT = = = 0.09667 T 1 500 τ For R = 0.99, from Table A-10, z = − 2.326. Eq. (1-12): n = 1 + 1 − (1 − z 2CS2 )(1 − z 2Cσ2 ) 1 − z 2CS2 2 2 2 2 1 + 1 − 1 − ( −2.326 ) ( 0.07532 ) 1 − ( −2.326 ) ( 0.09667 ) = = 1.3229 =1.32 2 2 1 − ( −2.326 ) ( 0.07532 ) Ans. From the given equation for stress, τ max = S sy n = 16T π d3 Solving for d gives 1/ 3 16 T n 16(1500)1.3229 d = = = 0.0319 m = 31.9 mm Ans. 6 π Ssy π (312)10 ______________________________________________________________________________ 1-19 1/ 3 Obtain the coefficients of variance for stress and strength 5 σˆ σˆ Cσ = σ = P = = 0.09231 P 65 µσ CS = (a) 6.59 σˆ S σˆ S = = = 0.06901 µS Sy 95.5 y n = 1.2 Shigley’s MED, 10th edition Chapter 1 Solutions, Page 6/12 www.konkur.in n = (b) n C + Cσ 2 d 2 S 2 =− 1.2 − 1 1.2 ( 0.069012 ) + 0.092312 2 Interpolating Table A-10, 1.61 0.0537 1.6127 Φ 1.62 0.0526 ⇒ R = 1 − 0.0534 = 0.9466 Ans. Sy σ nd − 1 z = − Eq. (1-11): = Sy P / (π d / 4) 2 = π d 2S y 4P ⇒ d = = −1.6127 Φ = 0.0534 4Pn = π Sy 4 ( 65 )1.2 = 1.020 in Ans. π ( 95.5) n = 1.5 1.5 − 1 z = − 1.52 ( 0.069012 ) + 0.092312 3.6 3.605 3.7 0.000159 Φ 0.000108 R = 1 − 0.00015645 = 0.9998 ⇒ Φ = 0.00015645 Ans. 4 ( 65 )1.5 = 1.140 in Ans. π ( 95.5) 4 Pn = π Sy d = = −3.605 ______________________________________________________________________________ 1-20 µσ max = σ max = σ a + σ b = 90 + 383 = 473 MPa From footnote 9, p. 25 of text, σˆσ max ( = σˆσ2a + σˆσ2b Cσ max = CS y = σˆσ µσ σˆ S y µS y Shigley’s MED, 10th edition max = max = σˆ S ) 1/ 2 = (8.42 + 22.32 )1/ 2 = 23.83 MPa σˆσ 23.83 = = 0.0504 473 σ max max y Sy = 42.7 = 0.0772 553 Chapter 1 Solutions, Page 7/12 www.konkur.in n = Eq. (1-11): Sy σ max = 553 = 1.169 = 1.17 Ans. 473 z = − From Table A-10, nd − 1 n C + Cσ 2 d 2 S 2 =− 1.169 − 1 1.1692 ( 0.07722 ) + 0.05042 = −1.635 Φ(− 1.635) = 0.05105 R = 1 − 0.05105 = 0.94895 = 94.9 percent Ans. ______________________________________________________________________________ 1-21 (a) a = 1.500 ± 0.001 in b = 2.000 ± 0.003 in c = 3.000 ± 0.004 in d = 6.520 ± 0.010 in w = d − a − b − c = 6.520 − 1.5 − 2 − 3 = 0.020 in tw = ∑ tall = 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, V = 1.500 (1.875 ) 3.000 = 8.4375 in 3 Shigley’s MED, 10th edition Chapter 1 Solutions, Page 8/12 www.konkur.in ∆V 0.002 0.003 0.004 ≈ + + = 0.004267 V 1.500 1.875 3.000 V = 8.4375 ± 0.0360 in3 ⇒ ∆V ≈ 0.004267 ( 8.4375) = 0.0360 in 3 Ans. 8.4735 8.473551.. in, whereas, exact is V = in 8.4015 8.401551.. ______________________________________________________________________________ This answer yields V ≈ 1-23 wmax = 0.05 in, wmin = 0.004 in 0.05 + 0.004 w= = 0.027 in 2 Thus, ∆ w = 0.05 − 0.027 = 0.023 in, and then, w = 0.027 ± 0.023 in. w= a −b −c 0.027 = a − 0.042 − 1.5 a = 1.569 in tw = Thus, ∑t all ⇒ 0.023 = ta + 0.002 + 0.005 ⇒ ta = 0.016 in a = 1.569 ± 0.016 in Ans. ______________________________________________________________________________ 1-24 Do = Di + 2d = 3.734 + 2 ( 0.139 ) = 4.012 in t Do = ∑ tall = 0.028 + 2 ( 0.004 ) = 0.036 in 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 Do = Di + 2d = 9.19 + 2 ( 2.62 ) = 14.43 mm t Do = ∑ tall = 0.13 + 2 ( 0.08 ) = 0.29 mm Do = 14.43 ± 0.29 mm Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 1 Solutions, Page 9/12 www.konkur.in 1-26 From O-Rings, Inc. (oringsusa.com), Di = 34.52 ± 0.30 mm, d = 3.53 ± 0.10 mm Do = Di + 2d = 34.52 + 2 ( 3.53) = 41.58 mm t Do = ∑ tall = 0.30 + 2 ( 0.10 ) = 0.50 mm 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 Do = Di + 2d = 5.237 + 2 ( 0.103) = 5.443 in t Do = ∑ tall = 0.035 + 2 ( 0.003) = 0.041 in 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 Do = Di + 2d = 1.100 + 2 ( 0.210 ) = 1.520 in t Do = ∑ tall = 0.012 + 2 ( 0.005 ) = 0.022 in 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 (d) A = 1500/ 25.42 = 2.33 in2 (e) I = 750/2.544 = 18.0 in4 (f) E = 145/6.89 = 21.0 Mpsi (g) v = 75/1.61 = 46.6 mi/h Shigley’s MED, 10th edition Ans. Ans. Ans. Ans. Ans. Chapter 1 Solutions, Page 10/12 www.konkur.in (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 (g) v = 1 200(0.0051) = 6.12 m/s Ans. Ans. (h) ∫ = 0.002 15(1) = 0.002 15 mm/mm (i) V = 1830(25.43) = 30.0 (106) mm3 Ans. Ans. ______________________________________________________________________________ 1-31 (a) σ = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi (b) σ = F /A = 9440/23.8 = 397 psi Ans. 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)(π /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 =π d4/64 = π (25.4)4/64 = 20.4(103) mm4 Ans. (d) τ =16T /π d 3 = 16(25)103/[π (12.7)3] = 62.2 MPa Ans. ______________________________________________________________________________ 1-33 Shigley’s MED, 10th edition Chapter 1 Solutions, Page 11/12 www.konkur.in (a) τ =F /A = 2 700/[π (0.750)2/4] = 6110 psi = 6.11 kpsi Ans. (b) σ = 32Fa/π d 3 = 32(180)31.5/[π (1.25)3] = 29 570 psi = 29.6 kpsi Ans. (c) Z =π (do4 − di4)/(32 do) = π (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 1 Solutions, Page 12/12 www.konkur.in 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. ______________________________________________________________________________ Conversion of kN/m3 to kg/ m3 multiply by 1(103) / 9.81 = 102 AISI 1018 CD steel: Tables A-20 and A-5 3 S y 370 (10 ) = = 47.4 kN ⋅ m/kg Ans. ρ 76.5 (102 ) 2011-T6 aluminum: Tables A-22 and A-5 3 S y 169 (10 ) = = 62.3 kN ⋅ m/kg Ans. ρ 26.6 (102 ) Ti-6Al-4V titanium: Tables A-24c and A-5 3 S y 830 (10 ) = = 187 kN ⋅ m/kg Ans. ρ 43.4 (102 ) ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the ultimate strength in tension 3 Sut 42.5 ( 6.89 ) (10 ) = = 40.7 kN ⋅ m/kg Ans 70.6 (102 ) ρ ______________________________________________________________________________ 2-3 2-4 AISI 1018 CD steel: Table A-5 6 E 30.0 (10 ) = = 106 (106 ) in 0.282 γ 2011-T6 aluminum: Table A-5 6 E 10.4 (10 ) = = 106 (106 ) in γ 0.098 Shigley’s MED, 10th edition Ans. Ans. Chapter 2 Solutions, Page 1/22 www.konkur.in Ti-6Al-6V titanium: Table A-5 6 E 16.5 (10 ) = = 103 (106 ) in Ans. γ 0.160 No. 40 cast iron: Table A-5 6 E 14.5 (10 ) = = 55.8 (106 ) in Ans. γ 0.260 ______________________________________________________________________________ 2-5 2G (1 + v) = E ⇒ v= E − 2G 2G Using values for E and G from Table A-5, 30.0 − 2 (11.5 ) Steel: v= = 0.304 Ans. 2 (11.5 ) The percent difference from the value in Table A-5 is 0.304 − 0.292 = 0.0411 = 4.11 percent 0.292 Ans. 10.4 − 2 ( 3.90) = 0.333 Ans. 2 ( 3.90) The percent difference from the value in Table A-5 is 0 percent Ans. Aluminum: v= Beryllium copper: v= 18.0 − 2 ( 7.0 ) 2 ( 7.0 ) = 0.286 Ans. The percent difference from the value in Table A-5 is 0.286 − 0.285 = 0.00351 = 0.351 percent 0.285 v= 14.5 − 2 ( 6.0 ) = 0.208 2 ( 6.0 ) The percent difference from the value in Table A-5 is Gray cast iron: Ans. Ans. 0.208 − 0.211 = − 0.0142 = − 1.42 percent Ans. 0.211 ______________________________________________________________________________ 2-6 (a) A0 = π (0.503)2/4 = 0.1987 in2, σ = Pi / A0 Shigley’s MED, 10th edition Chapter 2 Solutions, Page 2/22 www.konkur.in For data in elastic range, ∫ = ∆ l / l0 = ∆ l / 2 A ∆l l − l0 l For data in plastic range, ò = = = −1 = 0 −1 l0 l0 l0 A On the next two pages, the data and plots are presented. Figure (a) shows the linear part of the curve from data points 1-7. Figure (b) shows data points 1-12. Figure (c) shows the complete range. Note: The exact value of A0 is used without rounding off. (b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsi Ans. From Fig. (b) the equation for the dotted offset line is found to be 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 R= A0 − Af A0 (100 ) = Data Point Pi 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 1000 2000 3000 4000 7000 8400 8800 9200 8800 9200 9100 13200 15200 17000 16400 14800 Shigley’s MED, 10th edition 0.1987 − 0.1077 (100 ) = 45.8 % 0.1987 ∫ σ 0 0 0.0004 0.0006 0.001 0.0013 0.0023 0.0028 0.0036 0.0089 0.1984 0.1978 0.1963 0.1924 0.1875 0.1563 0.1307 0.1077 0.00020 0.00030 0.00050 0.00065 0.00115 0.00140 0.00180 0.00445 0.00158 0.00461 0.01229 0.03281 0.05980 0.27136 0.52037 0.84506 0 5032 10065 15097 20130 35227 42272 44285 46298 44285 46298 45795 66428 76492 85551 82531 74479 ∆l, Ai Ans. Chapter 2 Solutions, Page 3/22 www.konkur.in 50000 y = 3.05E+07x - 1.06E+01 Stress (psi) 40000 30000 Series1 20000 Linear (Series1) 10000 0 0.000 0.001 0.001 0.002 Strain Stress (psi) (a) Linear range 50000 Y 45000 40000 35000 30000 25000 20000 15000 10000 5000 0 0.000 0.002 0.004 0.006 0.008 0.010 0.012 0.014 Strain (b) Offset yield 90000 80000 U Stress (psi) 70000 60000 50000 40000 30000 20000 10000 0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Strain (c) Complete range Shigley’s MED, 10th edition Chapter 2 Solutions, Page 4/22 www.konkur.in (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. P σ true = A Eq. (2-4) l ε = ln for 0 ≤ ∆l ≤ 0.0028 in (0 ≤ P ≤ 8 400 lbf ) l0 ε = ln for ∆l > 0.0028 in (P > 8 400 lbf ) π (0.503)2 = 0.1987 in 2 4 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 where A0 = A0 A 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 A0 0.1987 = ln = 0.2231 A 0.1590 Eq. (2-18): S y′ = σ 0ε m = 153.2(0.2231)0.2306 = 108.4 kpsi ε = ln Ans. Eq. (2-19), with Su = 85.6 from Prob. 2-6, Su′ = Su 85.6 = = 107 kpsi 1 − W 1 − 0.2 Shigley’s MED, 10th edition Ans. Chapter 2 Solutions, Page 5/22 www.konkur.in P 0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800 Shigley’s MED, 10th edition ∆l 0 0.000 4 0.000 6 0.001 0 0.001 3 0.002 3 0.002 8 A 0.198 7 0.198 7 0.198 7 0.198 7 0.198 7 0.198 7 0.198 7 0.198 4 0.197 8 0.196 3 0.192 4 0.187 5 0.156 3 0.130 7 0.107 7 ε 0 0.000 2 0.000 3 0.000 5 0.000 65 0.001 15 0.001 4 0.001 51 0.004 54 0.012 15 0.032 22 0.058 02 0.240 02 0.418 89 0.612 45 σtrue 0 5 032.71 10 065.4 15 098.1 20 130.9 35 229 42 274.8 44 354.8 46 511.6 46 357.6 68 607.1 81 066.7 108 765 125 478 137 419 log ε -3.699 -3.523 -3.301 -3.187 -2.940 -2.854 -2.821 -2.343 -1.915 -1.492 -1.236 -0.620 -0.378 -0.213 log σtrue 3.702 4.003 4.179 4.304 4.547 4.626 4.647 4.668 4.666 4.836 4.909 5.036 5.099 5.138 Chapter 2 Solutions, Page 6/22 www.konkur.in ______________________________________________________________________________ 2-8 Tangent modulus at σ = 0 is E= ∆σ 5000 − 0 ≈ = 25 106 psi −3 ∆ò 0.2 10 − 0 ( ) ( ) Ans. At σ = 20 kpsi Shigley’s MED, 10th edition Chapter 2 Solutions, Page 7/22 www.konkur.in ( 26 − 19 ) (103 ) E20 ≈ = 14.0 (106 ) psi −3 (1.5 − 1) (10 ) Ans. ∫ (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: Annealed AISI 1018 steel. Table A-22, Sy = 32 kpsi, Su = 49.5 kpsi, σ0 = 90.0 kpsi, m = 0.25, ∫f = 1.05 After cold working: Eq. (2-16), (2 ∫u = m = 0.25 A0 1 1 Eq. (2-14), = = = 1.25 Ai 1 − W 1 − 0.20 A Eq. (2-17), ε i = ln 0 = ln1.25 = 0.223 < ε u Ai Eq. (2-18), S y′ = σ 0ε im = 90 ( 0.223) Eq. (2-19), Su′ = (b) Before: Su 49.5 = = 1.55 Sy 32 0.25 = 61.8 kpsi Su 49.5 = = 61.9 kpsi 1 − W 1 − 0.20 After: Ans. 93% increase Ans. 25% increase Su′ 61.9 = = 1.00 S y′ 61.8 Ans. Ans. Ans. Lost most of its ductility. ductility ______________________________________________________________________________ 2-10 W = 0.20, (a)) Before cold working: AISI 1212 HR steel. Table A A-22, Sy = 28 kpsi, Su = 61.5 kpsi, σ0 = 110 kpsi, m = 0.24, ∫f = 0.85 After cold working: Eq. q. (2 (2-16), ∫u = m = 0.24 Shigley’s MED, 10th edition Chapter 2 Solutions, Page 8/22 www.konkur.in Eq. (2-14), Eq. (2-17), A0 1 1 = = = 1.25 Ai 1 − W 1 − 0.20 A ε i = ln 0 = ln1.25 = 0.223 < ε u Ai Eq. (2-18), S y′ = σ 0ε im = 110 ( 0.223) Eq. (2-19), Su′ = (b) Before: Su 61.5 = = 2.20 Sy 28 0.24 = 76.7 kpsi Su 61.5 = = 76.9 kpsi 1 − W 1 − 0.20 After: Ans. 174% increase Ans. 25% increase Su′ 76.9 = = 1.00 S y′ 76.7 Ans. Ans. 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 A0 1 1 Eq. (2-14), = = = 1.25 Ai 1 − W 1 − 0.20 A Eq. (2-17), Ans. ε i = ln 0 = ln1.25 = 0.223 > ε f Material fractures. Ai ______________________________________________________________________________ 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 9/22 www.konkur.in 2-15 For the data given, converting HB to Su using Eq. (2-21) HB 230 232 232 234 235 235 235 236 236 239 Su (kpsi) 115 116 116 117 117.5 117.5 117.5 118 118 119.5 ΣSu = Eq. (1-6) Su = ∑S u N = 1172 Su2 (kpsi) 13225 13456 13456 13689 13806.25 13806.25 13806.25 13924 13924 14280.25 ΣSu2 = 137373 1172 = 117.2 ≈ 117 kpsi Ans. 10 Eq. (1-7), 10 ∑S 2 u − NSu2 137373 − 10 (117.2 ) s Su = = = 1.27 kpsi Ans. N −1 9 ______________________________________________________________________________ i =1 2 2-16 For the data given, converting HB to Su using Eq. (2-22) Su (kpsi) 40.4 40.86 40.86 41.32 41.55 41.55 41.55 41.78 41.78 42.47 Su2 (kpsi) 1632.16 1669.54 1669.54 1707.342 1726.403 1726.403 1726.403 1745.568 1745.568 1803.701 ΣSu = 414.12 ΣSu2 = 17152.63 HB 230 232 232 234 235 235 235 236 236 239 Shigley’s MED, 10th edition Chapter 2 Solutions, Page 10/22 www.konkur.in Eq. (1-6) Su = ∑S = u N 414.12 = 41.4 kpsi Ans. 10 Eq. (1-7), 10 ∑S 2 u − NSu2 17152.63 − 10 ( 41.4 ) s Su = = = 1.20 Ans. N −1 9 ______________________________________________________________________________ i =1 uR ≈ 2-17 (a) Eq. (2-9) 2 45.62 = 34.7 in ⋅ lbf / in 3 2(30) Ans. (b) A0 = π(0.5032)/4 = 0.19871 in2 P ∆L 0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800 0 0.000 4 0.000 6 0.001 0 0.001 3 0.002 3 0.002 8 0.003 6 0.008 9 A (A0 / A) – 1 0.196 3 0.192 4 0.187 5 0.156 3 0.130 7 0.107 7 0.012 28 0.032 80 0.059 79 0.271 34 0.520 35 0.845 03 σ = P/A0 ∫ 0 0.000 2 0.000 3 0.000 5 0.000 65 0.001 15 0.001 4 0.001 8 0.004 45 0.012 28 0.032 80 0.059 79 0.271 34 0.520 35 0.845 03 0 5 032. 10 070 15 100 20 130 35 230 42 270 44 290 46 300 45 800 66 430 76 500 85 550 82 530 74 480 From the figures on the next page, 5 1 uT ≈ ∑ Ai = (43 000)(0.001 5) + 45 000(0.004 45 − 0.001 5) 2 i =1 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 ) ( ) ≈ 66.7 103 in ⋅ lbf/in 3 Shigley’s MED, 10th edition Ans. Chapter 2 Solutions, Page 11/22 www.konkur.in Shigley’s MED, 10th edition Chapter 2 Solutions, Page 12/22 www.konkur.in 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 A20), 1040 and 4140 (Table A-21), Aluminum (Table A-24), Titanium (Table A-24c) Appropriate equations: F F For diameter, σ = = = Sy A (π / 4 ) d 2 ⇒ d= 4F π Sy Weight/length = ρA, Cost/length = $/in = ($/lbf) Weight/length, Deflection/length = δ /L = F/(AE) With F = 100 kips = 100(103) lbf, Young's Material Modulus units Mpsi 1020 HR 1020 CD 1040 4140 Al Ti Density lbf/in3 30 30 30 30 10.4 16.5 0.282 0.282 0.282 0.282 0.098 0.16 Yield Weight/ Strength Cost/lbf Diameter length kpsi $/lbf in lbf/in 30 57 80 165 50 120 0.27 0.30 0.35 0.80 1.10 7.00 2.060 1.495 1.262 0.878 1.596 1.030 0.9400 0.4947 0.3525 0.1709 0.1960 0.1333 Cost/ Deflection/ length length $/in in/in 0.25 0.15 0.12 0.14 0.22 $0.93 1.000E-03 1.900E-03 2.667E-03 5.500E-03 4.808E-03 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 w= W 7.95 lbf = = 0.281 lbf/in 3 2 Al [π (1 in) / 4](36 in) 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 13/22 www.konkur.in of Su = 0.5H B = 0.5(200) = 100 kpsi . 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 W 2.9 lbf w= = = 0.103 lbf/in 3 Al [π (1 in)2 / 4](36 in) 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 w= W 9.0 lbf = = 0.318 lbf/in 3 2 Al [π (1 in) / 4](36 in) 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 100 ( 24 ) Fl 3 E= = = 17.7 Mpsi 3Iy 3 (π (1) 4 64 ) (17 / 32) 3 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 Shigley’s MED, 10th edition ⇒ A = F/S Chapter 2 Solutions, Page 14/22 www.konkur.in 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 15/22 www.konkur.in 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 = 4 π ) −1 . Then, for strength, Eq. (1) is Fl =S CA3/2 Shigley’s MED, 10th edition ⇒ Fl A= CS 2/3 (2) Chapter 2 Solutions, Page 16/22 www.konkur.in For mass, Thus, Fl m = Al ρ = CS 2/3 F lρ = C 2/3 ρ l 5/3 2/3 S 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 17/22 www.konkur.in 1/2 kl 3 A= 3CE and Eq. (2-29) becomes k 5/2 ρ m = Al ρ = l 1/2 3C E 1/2 Thus, minimize f 3 ( M ) = ρ E 1/2 , or maximize M = E1/2 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 18/22 www.konkur.in 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 = (4π)−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 1/2 kl 3 A= 3CE and Eq. (2-29) becomes k 5/2 ρ m = Al ρ = l 1/2 3C E 1/2 So, minimize f 3 ( M ) = ρ E1/2 . Thus, β = 1/2. Ans. ρ E1/2 ______________________________________________________________________________ 2-33 For strength, , or maximize M = 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 = πd 3/(32)and the area is A = πd 2/4. This leads to C = 4 π ) −1 . So, with Z =C (A)3/2, for strength, Eq. (1) is Fl =S CA3/2 For mass, Fl A= CS ⇒ Fl m = Al ρ = CS 2/3 F lρ = C 2/3 2/3 (2) ρ l 5/3 2/3 S 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 19/22 www.konkur.in 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 sx = 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 µ = x and σˆ = sx , is 1 x − x 2 1 x − 122.9 2 1 1 f ( x) = exp − = exp − sx 2π 2 sx 30.3 2π 2 30.3 (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 20/22 www.konkur.in Range midpoint (kcycles) Frequency x 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 Observed PDF f /(Nw) 0.002898551 0.001449275 0.004347826 0.007246377 0.011594203 0.017391304 0.008695652 0.014492754 0.011594203 0.007246377 0.002898551 0.004347826 0.002898551 0.001449275 0 0.001449275 Normal PDF f (x) 0.001526493 0.002868043 0.004832507 0.007302224 0.009895407 0.012025636 0.013106245 0.012809861 0.011228104 0.008826008 0.006221829 0.003933396 0.002230043 0.001133847 0.000517001 0.00021141 Plots of the PDF’s are shown below. 0.02 0.018 0.016 Probability Density 0.014 0.012 0.01 Normal Distribution Histogram 0.008 0.006 0.004 0.002 0 0 20 40 60 80 100 120 140 160 180 200 220 L (kcycles) Shigley’s MED, 10th edition Chapter 2 Solutions, Page 21/22 www.konkur.in 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 2 Solutions, Page 22/22 www.konkur.in Chapter 3 3-1 ΣM O = 0 18 RB − 6(100) = 0 RB = 33.3 lbf Ans. ΣFy = 0 RO + RB − 100 = 0 RO = 66.7 lbf Ans. RC = RB = 33.3 lbf Ans. ______________________________________________________________________________ 3-2 Body AB: ΣFx = 0 RAx = RBx ΣFy = 0 RAy = RBy ΣM B = 0 RAy (10) − RAx (10) = 0 RAx = RAy Body OAC: ΣM O = 0 RAy (10) − 100(30) = 0 RAy = 300 lbf Ans. ΣFx = 0 ROx = − RAx = −300 lbf ΣFy = 0 ROy + RAy − 100 = 0 ROy = −200 lbf Ans. Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 1/100 www.konkur.in 3-3 0.8 = 1.39 kN Ans. tan 30o 0.8 RA = = 1.6 kN Ans. sin 30o RO = ______________________________________________________________________________ 3-4 Step 1: Find RA & RE 4.5 = 7.794 m tan 30o ΣM A = 0 h= 9 RE − 7.794(400 cos 30o ) −4.5(400sin 30o ) = 0 RE = 400 N Ans. ∑F x =0 RAx + 400 cos 30o = 0 RAx = −346.4 N ∑F y =0 RAy + 400 − 400sin 30o = 0 RAy = −200 N RA = 346.42 + 2002 = 400 N Ans. Step 2: Find components of RC and RD on link 4 ∑M C =0 400(4.5) − ( 7.794 − 1.9 ) RD = 0 RD = 305.4 N Ans. x =0 ⇒ ( RCx )4 = 305.4 N y =0 ⇒ ( RCy ) 4 = −400 N ∑F ∑F Shigley’s MED, 10th edition Chapter 3 Solutions, Page 2/100 www.konkur.in Step 3: Find components of RC on link 2 ∑ Fx = 0 ( RCx )2 + 305.4 − 346.4 = 0 ( RCx )2 = 41 N ∑F y =0 (R ) Cy 2 = 200 N Ans. _____________________________________________________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 3/100 www.konkur.in 3-5 ΣM C = 0 −1500 R1 + 300(5) + 1200(9) = 0 R1 = 8.2 kN Ans. ΣFy = 0 8.2 − 9 − 5 + R2 = 0 R2 = 5.8 kN Ans. M 1 = 8.2(300) = 2460 N ⋅ m Ans. M 2 = 2460 − 0.8(900) = 1740 N ⋅ m Ans. M 3 = 1740 − 5.8(300) = 0 checks! _____________________________________________________________________________ 3-6 ΣFy = 0 RO = 500 + 40(6) = 740 lbf Ans. ΣM O = 0 M O = 500(8) + 40(6)(17) = 8080 lbf ⋅ in Ans. M 1 = −8080 + 740(8) = −2160 lbf ⋅ in Ans. M 2 = −2160 + 240(6) = −720 lbf ⋅ in Ans. 1 M 3 = −720 + (240)(6) = 0 checks! 2 ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 4/100 www.konkur.in 3-7 ΣM B = 0 −2.2 R1 + 1(2) − 1(4) = 0 R1 = −0.91 kN Ans. ΣFy = 0 −0.91 − 2 + R2 − 4 = 0 R2 = 6.91 kN Ans. M 1 = −0.91(1.2) = −1.09 kN ⋅ m M 2 = −1.09 − 2.91(1) = −4 kN ⋅ m M 3 = −4 + 4(1) = 0 checks! Ans. Ans. ______________________________________________________________________________ 3-8 Break at the hinge at B Beam OB: From symmetry, R1 = VB = 200 lbf Ans. Beam BD: ΣM D = 0 200(12) − R2 (10) + 40(10)(5) = 0 R2 = 440 lbf Ans. ΣFy = 0 −200 + 440 − 40(10) + R3 = 0 R3 = 160 lbf Ans. Shigley’s MED, 10th edition Chapter 3 Solutions, Page 5/100 www.konkur.in M 1 = 200(4) = 800 lbf ⋅ in Ans. M 2 = 800 − 200(4) = 0 checks at hinge M 3 = 800 − 200(6) = −400 lbf ⋅ in Ans. 1 M 4 = −400 + (240)(6) = 320 lbf ⋅ in Ans. 2 1 M 5 = 320 − (160)(4) = 0 checks! 2 ______________________________________________________________________________ 3-9 q = R1 x −1 − 9 x − 300 −1 − 5 x − 1200 −1 + R2 x − 1500 V = R1 − 9 x − 300 − 5 x − 1200 + R2 x − 1500 0 0 0 M = R1 x − 9 x − 300 − 5 x − 1200 + R2 x − 1500 1 1 −1 (1) 1 (2) At x = 1500+ V = M = 0. Applying Eqs. (1) and (2), R1 − 9 − 5 + R2 = 0 ⇒ R1 + R2 = 14 1500 R1 − 9(1500 − 300) − 5(1500 − 1200) = 0 R2 = 14 − 8.2 = 5.8 kN ⇒ R1 = 8.2 kN Ans. Ans. 0 ≤ x ≤ 300 : V = 8.2 kN, M = 8.2 x N ⋅ m 300 ≤ x ≤ 1200 : V = 8.2 − 9 = −0.8 kN M = 8.2 x − 9( x − 300) = −0.8 x + 2700 N ⋅ m 1200 ≤ x ≤ 1500 : V = 8.2 − 9 − 5 = −5.8 kN M = 8.2 x − 9( x − 300) − 5( x − 1200) = −5.8 x + 8700 N ⋅ m Plots of V and M are the same as in Prob. 3-5. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 6/100 www.konkur.in 3-10 q = RO x −1 − MO x V = RO − M O x −1 −2 − 500 x − 8 −1 − 40 x − 14 + 40 x − 20 0 − 500 x − 8 − 40 x − 14 + 40 x − 20 0 1 M = RO x − M O − 500 x − 8 − 20 x − 14 + 20 x − 20 1 2 0 1 (1) 2 (2) at x = 20+ in, V = M = 0, Eqs. (1) and (2) give RO − 500 − 40 ( 20 − 14 ) = 0 ⇒ RO (20) − M O − 500(20 − 8) − 20(20 − 14) = 0 ⇒ 2 0 ≤ x ≤ 8: RO = 740 lbf Ans. M O = 8080 lbf ⋅ in Ans. V = 740 lbf, M = 740 x − 8080 lbf ⋅ in 8 ≤ x ≤ 14 : V = 740 − 500 = 240 lbf M = 740 x − 8080 − 500( x − 8) = 240 x − 4080 lbf ⋅ in 14 ≤ x ≤ 20 : V = 740 − 500 − 40( x − 14) = −40 x + 800 lbf M = 740 x − 8080 − 500( x − 8) − 20( x − 14)2 = −20 x 2 + 800 x − 8000 lbf ⋅ in Plots of V and M are the same as in Prob. 3-6. ______________________________________________________________________________ 3-11 q = R1 x −1 − 2 x − 1.2 −1 + R2 x − 2.2 −1 − 4 x − 3.2 V = R1 − 2 x − 1.2 + R2 x − 2.2 − 4 x − 3.2 0 0 −1 0 (1) M = R1 x − 2 x − 1.2 + R2 x − 2.2 − 4 x − 3.2 at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2), 1 1 1 (2) R1 − 2 + R2 − 4 = 0 ⇒ R1 + R2 = 6 (3) 3.2 R1 − 2(2) + R2 (1) = 0 ⇒ 3.2 R1 + R2 = 4 (4) Solving Eqs. (3) and (4) simultaneously, R1 = -0.91 kN, R2 = 6.91 kN Ans. 0 ≤ x ≤ 1.2 : V = −0.91 kN, M = −0.91x kN ⋅ m 1.2 ≤ x ≤ 2.2 : V = −0.91 − 2 = −2.91 kN M = −0.91x − 2( x − 1.2) = −2.91x + 2.4 kN ⋅ m 2.2 ≤ x ≤ 3.2 : V = −0.91 − 2 + 6.91 = 4 kN M = −0.91x − 2( x −1.2) + 6.91( x − 2.2) = 4 x − 12.8 kN ⋅ m Plots of V and M are the same as in Prob. 3-7. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 7/100 www.konkur.in 3-12 q = R1 x −1 − 400 x − 4 −1 + R2 x − 10 −1 − 40 x − 10 + 40 x − 20 + R3 x − 20 0 0 V = R1 − 400 x − 4 + R2 x − 10 − 40 x − 10 + 40 x − 20 + R3 x − 20 0 0 1 1 0 M = R1 x − 400 x − 4 + R2 x − 10 − 20 x − 10 + 20 x − 20 + R3 x − 20 1 1 2 M = 0 at x = 8 in ∴ 8 R1 − 400(8 − 4) = 0 ⇒ 2 R1 = 200 lbf −1 (1) 1 (2) Ans. at x = 20+, V =M = 0. Applying Eqs. (1) and (2), 200 − 400 + R2 − 40(10) + R3 = 0 ⇒ R2 + R3 = 600 200(20) − 400(16) + R2 (10) − 20(10) 2 = 0 ⇒ R2 = 440 lbf Ans. R3 = 600 − 440 = 160 lbf 0 ≤ x ≤ 4: Ans. V = 200 lbf, M = 200 x lbf ⋅ in 4 ≤ x ≤ 10 : V = 200 − 400 = −200 lbf, M = 200 x − 400( x − 4) = −200 x + 1600 lbf ⋅ in 10 ≤ x ≤ 20 : V = 200 − 400 + 440 − 40( x − 10) = 640 − 40 x lbf M = 200 x − 400( x − 4) + 440( x − 10) − 20 ( x − 10 ) = −20 x 2 + 640 x − 4800 lbf ⋅ in 2 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, ( l − 2a ) xc = 2 2 wl l wl l M c = ( l − 2a ) − = − a 2 2 2 2 4 At reaction, M r = wa2 2 a = 2.25, l = 10 in, w = 100 lbf/in Mc = 100(10) 10 − 2.25 = 125 lbf ⋅ in 2 4 Mr = 100 ( 2.252 ) = 253 lbf ⋅ in 2 (b) Optimal occurs when M c = M r Shigley’s MED, 10th edition Ans. Chapter 3 Solutions, Page 8/100 www.konkur.in 2 wl l wa − a = ⇒ a 2 + al − 0.25l 2 = 0 2 4 2 Taking the positive root ( ) 1 l −l + l 2 + 4 ( 0.25l 2 ) = 2 − 1 = 0.207 l 2 2 for l = 10 in, w = 100 lbf, lbf a = 0.207(10) = 2.07 in M min = (100 2 ) 2.072 = 214 lbf ⋅ in a= Ans. ______________________________________________________________________________ 3-15 (a) 20 − 10 = 5 kpsi 2 20 + 10 CD = = 15 kpsi 2 C= R = 152 + 82 = 17 kpsi σ 1 = 5 + 17 = 22 kpsi σ 2 = 5 − 17 = −12 kpsi 1 8 φ p = tan −1 = 14.04o cw 2 15 τ 1 = R = 17 kpsi φs = 45o − 14.04o = 30.96o ccw (b) 9 + 16 = 12.5 kpsi 2 16 − 9 CD = = 3.5 kpsi 2 C= R = 52 + 3.52 = 6.10 kpsi σ 1 = 12.5 + 6.1 = 18.6 kpsi σ 2 = 12.5 − 6.1 = 6.4 kpsi 1 5 o φ p = tan −1 = 27.5 ccw 2 3.5 τ 1 = R = 6.10 kpsi φs = 45o − 27.5o = 17.5o cw Shigley’s MED, 10th edition Chapter 3 Solutions, Page 9/100 www.konkur.in (c) 24 + 10 = 17 kpsi 2 24 − 10 CD = = 7 kpsi 2 C= R = 7 2 + 6 2 = 9.22 kpsi σ 1 = 17 + 9.22 = 26.22 kpsi σ 2 = 17 − 9.22 = 7.78 kpsi 1 7 φ p = 90o + tan −1 = 69.7o ccw 2 6 τ 1 = R = 9.22 kpsi φs = 69.7o − 45o = 24.7o ccw (d) −12 + 22 = 5 kpsi 2 12 + 22 CD = = 17 kpsi 2 C= R = 17 2 + 122 = 20.81 kpsi σ 1 = 5 + 20.81 = 25.81 kpsi σ 2 = 5 − 20.81 = −15.81 kpsi 1 17 τ 1 = R = 20.81 kpsi φ p = 90o + tan −1 = 72.39o cw 2 12 φs = 72.39 − 45 = 27.39o cw ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 10/100 www.konkur.in 3-16 (a) −8 + 7 = −0.5 MPa 2 8+7 CD = = 7.5 MPa 2 C= R = 7.52 + 62 = 9.60 MPa σ 1 = 9.60 − 0.5 = 9.10 MPa σ 2 = −0.5 − 9.6 = −10.1 Mpa 1 7.5 o φ p = 90o + tan −1 = 70.67 cw 2 6 τ 1 = R = 9.60 MPa φs = 70.67o − 45o = 25.67o cw (b) 9−6 = 1.5 MPa 2 9+6 CD = = 7.5 MPa 2 C= R = 7.52 + 32 = 8.078 MPa σ 1 = 1.5 + 8.078 = 9.58 MPa σ 2 = 1.5 − 8.078 = −6.58 MPa 1 3 o φ p = tan −1 = 10.9 cw 2 7.5 τ 1 = R = 8.078 MPa φs = 45o − 10.9o = 34.1o ccw Shigley’s MED, 10th edition Chapter 3 Solutions, Page 11/100 www.konkur.in (c) 12 − 4 = 4 MPa 2 12 + 4 CD = = 8 MPa 2 C= R = 82 + 7 2 = 10.63 MPa σ 1 = 4 + 10.63 = 14.63 MPa σ 2 = 4 − 10.63 = −6.63 MPa 1 8 φ p = 90o + tan −1 = 69.4o ccw 2 7 τ 1 = R = 10.63 MPa φs = 69.4o − 45o = 24.4o ccw (d) 6−5 = 0.5 MPa 2 6+5 CD = = 5.5 MPa 2 C= R = 5.52 + 82 = 9.71 MPa σ 1 = 0.5 + 9.71 = 10.21 MPa σ 2 = 0.5 − 9.71 = −9.21 MPa 1 8 φ p = tan −1 = 27.75o ccw 2 5.5 τ 1 = R = 9.71 MPa φs = 45o − 27.75o = 17.25o cw ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 12/100 www.konkur.in 3-17 (a) 12 + 6 = 9 kpsi 2 12 − 6 CD = = 3 kpsi 2 C= R = 32 + 42 = 5 kpsi σ 1 = 5 + 9 = 14 kpsi σ 2 = 9 − 5 = 4 kpsi 1 4 tan −1 = 26.6o ccw 3 2 τ 1 = R = 5 kpsi φp = φs = 45o − 26.6o = 18.4o ccw (b) 30 − 10 = 10 kpsi 2 30 + 10 CD = = 20 kpsi 2 C= R = 202 + 102 = 22.36 kpsi σ 1 = 10 + 22.36 = 32.36 kpsi σ 2 = 10 − 22.36 = −12.36 kpsi 1 10 φ p = tan −1 = 13.28o ccw 2 20 τ 1 = R = 22.36 kpsi φs = 45o − 13.28o = 31.72o cw Shigley’s MED, 10th edition Chapter 3 Solutions, Page 13/100 www.konkur.in (c) −10 + 18 = 4 kpsi 2 10 + 18 CD = = 14 kpsi 2 C= R = 142 + 92 = 16.64 kpsi σ 1 = 4 + 16.64 = 20.64 kpsi σ 2 = 4 − 16.64 = −12.64 kpsi 1 14 φ p = 90o + tan −1 = 73.63o cw 2 9 τ 1 = R = 16.64 kpsi φs = 73.63 − 45 = 28.63o cw (d) 9 + 19 = 14 kpsi 2 19 − 9 CD = = 5 kpsi 2 C= R = 52 + 82 = 9.434 kpsi σ 1 = 14 + 9.43 = 23.43 kpsi σ 2 = 14 − 9.43 = 4.57 kpsi 1 5 φ p = 90o + tan −1 = 61.0o cw 2 8 τ 1 = R = 9.34 kpsi φs = 61o − 45o = 16o cw ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 14/100 www.konkur.in 3-18 (a) −80 − 30 = −55 MPa 2 80 − 30 CD = = 25 MPa 2 C= R = 252 + 202 = 32.02 MPa σ 1 = 0 MPa σ 2 = −55 + 32.02 = −22.98 = −23.0 MPa σ 3 = −55 − 32.0 = −87.0 MPa τ1 2 = 23 = 11.5 MPa, 2 τ 2 3 = 32.0 MPa, τ1 3 = 87 = 43.5 MPa 2 (b) 30 − 60 = −15 MPa 2 60 + 30 CD = = 45 MPa 2 C= R = 452 + 302 = 54.1 MPa σ 1 = −15 + 54.1 = 39.1 MPa σ 2 = 0 MPa σ 3 = −15 − 54.1 = −69.1 MPa 39.1 + 69.1 = 54.1 MPa 2 39.1 τ1 2 = = 19.6 MPa 2 69.1 = 34.6 MPa τ2 3 = 2 τ1 3 = Shigley’s MED, 10th edition Chapter 3 Solutions, Page 15/100 www.konkur.in (c) 40 + 0 = 20 MPa 2 40 − 0 CD = = 20 MPa 2 C= R = 202 + 202 = 28.3 MPa σ 1 = 20 + 28.3 = 48.3 MPa σ 2 = 20 − 28.3 = −8.3 MPa σ 3 = σ z = −30 MPa τ1 3 = 48.3 + 30 = 39.1 MPa, 2 τ 1 2 = 28.3 MPa, τ2 3 = 30 − 8.3 = 10.9 MPa 2 (d) 50 = 25 MPa 2 50 CD = = 25 MPa 2 C= R = 252 + 302 = 39.1 MPa σ 1 = 25 + 39.1 = 64.1 MPa σ 2 = 25 − 39.1 = −14.1 MPa σ 3 = σ z = −20 MPa 64.1 + 20 20 − 14.1 = 42.1 MPa, τ 1 2 = 39.1 MPa, τ 2 3 = = 2.95 MPa 2 2 ______________________________________________________________________________ τ1 3 = 3-19 (a) Since there are no shear stresses on the stress element, the stress element already represents principal stresses. σ 1 = σ x = 10 kpsi σ 2 = 0 kpsi σ 3 = σ y = −4 kpsi 10 − (−4) = 7 kpsi 2 10 τ 1 2 = = 5 kpsi 2 0 − (−4) = 2 kpsi τ2 3 = 2 τ1 3 = Shigley’s MED, 10th edition Chapter 3 Solutions, Page 16/100 www.konkur.in (b) 0 + 10 = 5 kpsi 2 10 − 0 CD = = 5 kpsi 2 C= R = 52 + 42 = 6.40 kpsi σ 1 = 5 + 6.40 = 11.40 kpsi σ 2 = 0 kpsi, σ 3 = 5 − 6.40 = −1.40 kpsi τ 1 3 = R = 6.40 kpsi, τ1 2 = 11.40 = 5.70 kpsi, 2 τ3 = 1.40 = 0.70 kpsi 2 (c) −2 − 8 = −5 kpsi 2 8−2 CD = = 3 kpsi 2 C= R = 32 + 42 = 5 kpsi σ 1 = −5 + 5 = 0 kpsi, σ 2 = 0 kpsi σ 3 = −5 − 5 = −10 kpsi τ1 3 = 10 = 5 kpsi, 2 τ 1 2 = 0 kpsi, τ 2 3 = 5 kpsi (d) 10 − 30 = −10 kpsi 2 10 + 30 CD = = 20 kpsi 2 C= R = 202 + 102 = 22.36 kpsi σ 1 = −10 + 22.36 = 12.36 kpsi σ 2 = 0 kpsi σ 3 = −10 − 22.36 = −32.36 kpsi 12.36 32.36 = 6.18 kpsi, τ 2 3 = = 16.18 kpsi kps 2 2 ______________________________________________________________________________ τ 1 3 = 22.36 kpsi, Shigley’s MED, 10th edition τ1 2 = Chapter 3 Solutions, Page 17/100 www.konkur.in 3-20 From Eq. (3-15), σ 3 − (−6 + 18 − 12)σ 2 + −6(18) + (−6)(−12) + 18(−12) − 92 − 62 − (−15)2 σ − −6(18)(−12) + 2(9)(6)(−15) − (−6)(6) 2 − 18(−15)2 − (−12)(9)2 = 0 σ 3 − 594σ + 3186 = 0 Roots are: 21.04, 5.67, –26.71 kpsi Ans. 21.04 − 5.67 = 7.69 kpsi 2 5.67 + 26.71 τ2 3 = = 16.19 kpsi 2 21.04 + 26.71 τ max = τ 1 3 = = 23.88 kpsi 2 τ1 2 = Ans. _____________________________________________________________________________ 3-21 From Eq. (3-15) ( ) σ 3 − (20 + 0 + 20)σ 2 + 20(0) + 20(20) + 0(20) − 402 − −20 2 − 02 σ ( ) ( − 20(0)(20) + 2(40) −20 2 (0) − 20 −20 2 σ 3 − 40σ 2 − 2 000σ + 48 000 = 0 Roots are: 60, 20, –40 kpsi 60 − 20 = 20 kpsi 2 20 + 40 = 30 kpsi τ2 3 = 2 60 + 40 = 50 kpsi τ max = τ 1 3 = 2 ) 2 2 − 0(0) 2 − 20(40)2 = 0 Ans. τ1 2 = Ans. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 18/100 www.konkur.in 3-22 From Eq. (3-15) 2 2 σ 3 − (10 + 40 + 40)σ 2 + 10(40) + 10(40) + 40(40) − 202 − ( −40 ) − ( −20 ) σ − 10(40)(40) + 2(20)( −40)(−20) − 10( −40) − 40(−20) − 40(20) = 0 σ 3 − 90σ 2 = 0 2 Roots are: 90, 0, 0 MPa 2 2 Ans. τ2 3 = 0 τ 1 2 = τ 1 3 = τ max = 90 = 45 MPa 2 Ans. _____________________________________________________________________________ 3-23 σ= F 15000 = = 33 950 psi = 34.0 kpsi A (π 4 ) ( 0.752 ) δ= FL L 60 = σ = 33 950 = 0.0679 in AE E 30 (106 ) δ 0.0679 = 1130 (10−6 ) = 1130µ L 60 From Table A-5, v = 0.292 ò1 = = ò2 = −vò1 = −0.292(1130) = −330 µ ∆d = ò2 d = −330 (10 −6 Ans. Ans. Ans. Ans. ) (0.75) = −248 (10 ) in −6 Ans. _____________________________________________________________________________ 3-24 σ= F 3000 = = 6790 psi = 6.79 kpsi A (π 4 ) ( 0.752 ) δ= FL L 60 = σ = 6790 = 0.0392 in AE E 10.4 (106 ) δ 0.0392 = 653 (10−6 ) = 653µ L 60 From Table A-5, v = 0.333 ò1 = = ò2 = −vò1 = −0.333(653) = −217 µ Ans. Ans. Ans. Ans. ∆d = ò2 d = −217 (10−6 ) (0.75) = −163 (10−6 ) in Ans. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 19/100 www.konkur.in 3-25 ∆d −0.0001d = = −0.0001 d d From Table A-5, v = 0.326, E = 119 GPa −ò −0.0001 ò1 = 2 = = 306.7 (10−6 ) v 0.326 FL F δ= and σ = , so AE A δE σ= = ò1 E = 306.7 (10−6 ) (119) (109 ) = 36.5 MPa L ò2 = F = σ A = 36.5 (106 ) π ( 0.03) 2 = 25 800 N = 25.8 kN Ans. 4 Sy = 70 MPa > σ , so elastic deformation assumption is valid. _____________________________________________________________________________ 3-26 δ= FL L 8(12) = σ = 20 000 = 0.185 in AE E 10.4 (106 ) Ans. _____________________________________________________________________________ 3-27 δ= FL L 3 = σ = 140 (106 ) = 0.00586 m = 5.86 mm AE E 71.7 (109 ) Ans. _____________________________________________________________________________ 3-28 δ= FL L 10(12) = σ = 15 000 = 0.173 in AE E 10.4 (106 ) Ans. _____________________________________________________________________________ 3-29 With σ z = 0, 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, σx = Eòx Eòy −v 1 1 −v −v 1 = Eòx + vEòy 1 − v2 = E (òx + vòy ) 1 − v2 Likewise, Shigley’s MED, 10th edition Chapter 3 Solutions, Page 20/100 www.konkur.in σy = E (òy + ν òx ) 1 − v2 From Table A-5, E = 207 GPa and ν = 0.292. Thus, E (òx + vòy ) σx = = 207 (109 ) 0.0019 + 0.292 ( −0.000 72 ) 1− v 1 − 0.292 9 207 (10 ) −0.000 72 + 0.292 ( 0.0019 ) 2 σy = 2 (10 ) = 382 MPa −6 Ans. (10 ) = −37.4 MPa −6 Ans. 1 − 0.292 _____________________________________________________________________________ 3-30 2 With σ z = 0, 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, Eòx Eòy σx = −v 1 1 −v −v 1 = Eòx + vEòy 1 − v2 = E (òx + vòy ) 1 − v2 Likewise, σy = E (òy + ν òx ) 1 − v2 From Table A-5, E = 71.7 GPa and ν = 0.333. Thus, E (òx + vòy ) 71.7 (109 ) 0.0019 + 0.333 ( −0.000 72 ) σx = = 10−6 ) = 134 MPa Ans. ( 2 2 1− v 1 − 0.333 9 71.7 (10 ) −0.000 72 + 0.333 ( 0.0019 ) 10−6 ) = −7.04 MPa Ans. σy = ( 2 1 − 0.333 _____________________________________________________________________________ 3-31 c (a) R1 = F l 6M σ= 2 = bh F (σ (b) m = m F ac F l 6 ac σ bh 2l Ans. F ⇒ F = bh 2 l 6ac 2 σ )( bm b )( hm h ) ( lm l1 ) 1( s)( s )2 ( s ) 2 = =s ( s )( s) ( am a )( cm c ) M max = R1a = Ans. For equal stress, the model load varies by the square of the scale factor. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 21/100 www.konkur.in 3-32 wl w l l wl 2 , M max x =l /2 = l − = 2 2 2 2 8 2 2 6M 6 wl 3Wl 4 σ bh σ= 2 = 2 = ⇒ W= Ans. 2 bh bh 8 4bh 3 l (a) (b) R1 = Wm (σ m / σ )(bm / b)(hm / h) 2 1( s )( s )2 = = = s2 W lm / l s Ans. wmlm wm s 2 = s2 ⇒ = = s Ans. wl w s 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 W3 . WT = ΣW at centroid of W’s l − x3 − d3 RA = WT l Under wheel 3, M 3 = RA x3 − W1a13 − W2 a23 = For maximum, ( l − x3 − d3 ) W T l dM 3 W = 0 = ( l − d3 − 2 x3 ) T dx3 l ( l − d3 ) = x3 − W1a13 − W2 a23 x3 = ⇒ l − d3 2 2 WT − W1a13 − W2 a23 4l This means the midpoint of d 3 intersects the midpoint of the beam. Substitute into M ⇒ M 3 For wheel i, l − di xi = , 2 Mi ( l − di ) = 4l 2 i −1 WT − ∑ W j a ji j =1 Note for wheel 1: ΣW j a ji = 0 WT = 104.4, W1 = W2 = W3 = W4 = 104.4 = 26.1 kips 4 476 (1200 − 238) 2 = 238 in, M 1 = (104.4) = 20 128 kip ⋅ in 2 4(1200) Wheel 2: d 2 = 238 − 84 = 154 in Wheel 1: d1 = Shigley’s MED, 10th edition Chapter 3 Solutions, Page 22/100 www.konkur.in M2 = (1200 − 154) 2 (104.4) − 26.1(84) = 21 605 kip ⋅ in = M max 4(1200) Ans. Check if all of the wheels are on the rail rail. (b) xmax = 600 − 77 = 523 in Ans. (c) See above sketch. (d) Inner axles _____________________________________________________________________________ 3-34 (a) Let a = total area of entire envelope Let b = area of side notch A = a − 2b = 40(2)(37.5) − 25 ( 34) = 2150 mm2 1 1 3 3 ( 40 )( 75) − ( 34 )( 25) 12 12 6 4 I = 1.36 (10 ) mm Ans. I = I a − 2Ib = Dimensions in mm. (b) Aa = 0.375(1.875) = 0.703 125 in 2 Ab = 0.375(1.75) = 0.656 25 in 2 A = 2(0.703125) + 0.656 25 = 2.0625 in 2 2(0.703 125)(0.9375) + 0.656 25(0.6875) = 0.858 in Ans. 2.0625 0.375(1.875)3 Ia = = 0.206 in 4 12 1.75(0.375)3 Ib = = 0.007 69 in 4 12 I1 = 2 0.206 + 0.703 125(0.0795) 2 + 0.00769 + 0.656 25(0.1705) 25(0.1705) 2 = 0.448 in 4 y= Ans. (c) Use two negative areas. Aa = 625 mm 2 , Ab = 5625 mm 2 , Ac = 10 000 mm 2 A = 10 000 − 5625 − 625 = 3750 mm 2 ; Shigley’s MED, 10th edition Chapter 3 Solutions, Page 23/100 www.konkur.in ya = 6.25 mm, yb = 50 mm, yc = 50 mm 10 000(50) − 5625(50) − 625(6.25) = 57.29 mm 3750 c1 = 100 − 57.29 = 42.71 mm Ans. y= Ans. 50(12.5)3 = 8138 mm 4 12 75(75)3 Ib = = 2.637 (106 ) mm 4 12 100(100)3 Ic = = 8.333 (106 ) in 4 12 2 2 I1 = 8.333 (106 ) + 10 000(7.29) 2 − 2.637 (106 ) + 5625 ( 7.29 ) − 8138 + 625 ( 57.29 − 6.25 ) Ia = I1 = 4.29 (106 ) in 4 Ans. (d) Aa = 4 ( 0.875 ) = 3.5 in 2 Ab = 2.5 ( 0.875 ) = 2.1875 in 2 A = Aa + Ab = 5.6875 in 2 2.9375 ( 3.5 ) + 1.25(2.1875) = 2.288 in Ans. 5.6875 1 1 3 2 3 2 I = (4) ( 0.875 ) + 3.5 ( 2.9375 − 2.288 ) + ( 0.875 )( 2.5 ) + 2.1875 ( 2.288 − 1.25 ) 12 12 4 I = 5.20 in Ans. _____________________________________________________________________________ y= 3-35 1 (20)(40)3 = 1.067 (105 ) mm4 12 A = 20(40) = 800 mm2 Mmax is at A. At the bottom of the section, Mc 450 000(20) σ max = = = 84.3 MPa Ans. I 1.067 (105 ) I= Due to V, τmax is between A and B at y = 0. 3 V 3 3000 = τ max = = 5.63 MPa Ans. 2 A 2 800 _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 24/100 www.konkur.in 3-36 1 (1)(2)3 = 0.6667 in 4 12 A = 1(2) = 2 in 2 I= ΣM O = 0 8 RA − 100(8)(12) = 0 RA = 1200 lbf RO = 1200 − 100(8) = 400 lbf M max is at A. At the top of the beam, σ max = Mc 3200(0.5) = = 2400 psi I 0.6667 Ans. Due to V, τ max is at A, at y = 0. 3 V 3 800 = = 600 psi Ans. 2 A 2 2 _____________________________________________________________________________ τ max = 3-37 1 (0.75)(2)3 = 0.5 in 4 12 A = (0.75)(2) = 1.5 in 2 I= ΣM A = 0 15RB − 1000(20) = 0 RB = 1333.3 lbf RA = 3000 − 1333.3 + 1000 = 2666.7 lbf M max is at B. At the top of the beam, σ max = Mc 5000(1) = = 10 000 psi I 0.5 Ans. Due to V, τ max is between B and C at y = 0. 3 V 3 1000 = = 1000 psi Ans. 2 A 2 1.5 _____________________________________________________________________________ τ max = Shigley’s MED, 10th edition Chapter 3 Solutions, Page 25/100 www.konkur.in 3-38 I= πd4 = π (50) 4 ( ) = 306.796 103 mm 4 64 64 2 πd π (50) 2 A= = = 1963 mm 2 4 4 ΣM B = 0 6(300)(150) − 200 RA = 0 RA = 1350 kN RB = 6(300) − 1350 = 450 kN M max is at A. At the top, Mc 30 000(25) = = 2.44 kN/mm 2 = 2.44 GPa I 30 6796 Due to V, τ max is at A, at y = 0. σ max = Ans. 4 V 4 750 2 = = 0.509 kN/mm = 509 MPa Ans. 3 A 3 1963 _____________________________________________________________________________ τ max = 3-39 8σ I wl 2 wl 2 c ⇒ σ max = ⇒ w = max 8 8I cl 2 4 (a) l = 48 in; Table A-8, I = 0.537 in M max = w= 8 (12 ) (103 ) ( 0.537 ) 1( 482 ) = 22.38 lbf/in Ans. (b) l = 60 in, I ≈ (1 12 )( 2 ) ( 33 ) − (1 12 )(1.625 ) ( 2.6253 ) = 2.051 in 4 w= 8 (12 ) (103 ) ( 2.051) (1.5) ( 602 ) = 36.5 lbf/in Ans. (c) l = 60 in; Table A-6, I = 2 ( 0.703 ) = 1.406 in 4 y = 0.717 in, cmax = 1.783 in 8 (12 ) (103 ) (1.406 ) w= = 21.0 lbf/in Ans. 1.783 ( 602 ) (d) l = 60 in, Table A-7, I = 2.07 in 4 w= 8 (12 ) (103 ) ( 2.07 ) 1.5 ( 602 ) = 36.8 lbf/in Ans. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 26/100 www.konkur.in 3-40 I= π ( 0.5 ) = 3.068 (10 ) in , 64 4 −3 4 A= π 4 ( 0.5 ) = 0.1963 in 2 2 Model (c) 500(0.5) 500(0.75 / 2) M = + = 218.75 lbf ⋅ in 2 2 Mc 218.75(0.25) σ= = I 3.068 (10 −3 ) σ = 17 825 psi = 17.8 kpsi τ max = Ans. 4 V 4 500 = = 3400 psi = 3.4 kpsi 3 A 3 0.1963 Ans. Model (d) M = 500(0.625) = 312.5 lbf ⋅ in σ= Mc 312.5(0.25) = I 3.068 (10−3 ) σ = 25 464 psi = 25.5 kpsi τ max = Ans. 4 V 4 500 = = 3400 psi = 3.4 kpsi 3 A 3 0.1963 Ans. Model (e) M = 500(0.4375) = 218.75 lbf ⋅ in σ= Mc 218.75(0.25) = I 3.068 (10−3 ) σ = 17 825 psi = 17.8 kpsi τ max = Ans. 4 V 4 500 = = 3400 psi = 3.4 kpsi 3 A 3 0.1963 Ans. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 27/100 www.konkur.in 3-41 I= π π 12 ) = 1018 mm , A = (12 ) = 113.1 mm ( 64 4 4 4 2 2 Model (c) 2000(6) 2000(9) M = + = 15 000 N ⋅ mm 2 2 Mc 15 000(6) = σ= I 1018 σ = 88.4 N/mm 2 = 88.4 MPa Ans. τ max = 4 V 4 2000 2 = = 23.6 N/mm = 23.6 MPa 3 A 3 113.1 Model (d) M = 2000(12) = 24 000 N ⋅ mm Mc 24 000(6) = σ= I 1018 σ = 141.5 N/mm 2 = 141.5 MPa Ans. 4 V 4 2000 2 = τ max = = 23.6 N/mm = 23.6 MPa 3 A 3 113.1 Ans. Ans. Model (e) M = 2000(7.5) = 15000 N ⋅ mm Mc 15000(6) = I 1018 σ = 88.4 N/mm 2 = 88.4 MPa Ans. 4 V 4 2000 2 τ max = = = 23.6 N/mm = 23.6 MPa 3 A 3 113.1 σ= Ans. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 28/100 www.konkur.in 3-42 (a) σ = d= (b) τ = d= (c) τ = Mc M ( d / 2 ) 32 M = = π d 4 / 64 π d 3 I 3 32M = πσ 3 32(218.75) = 0.420 in Ans. π (30 000) V V = A πd2 / 4 4V πτ = 4(500) = 0.206 in π (15000) Ans. 4V 4 V = 3 A 3 (π d 2 / 4 ) 4 4V 4 4(500) = = 0.238 in Ans. 3 πτ 3 π (15000) _____________________________________________________________________________ d= 3-43 p1 + p2 1 x − l + terms for x > l + a a p + p2 1 2 V = − F + p1 x − l − 1 x − l + terms for x > l + a 2a p1 p + p2 2 3 M = − Fx + x−l − 1 x − l + terms for x > l + a 2 6a q = −F x −1 + p1 x − l − 0 At x = (l + a ) + , V = M = 0, terms for x > l + a = 0 p1 + p2 2 2F a = 0 ⇒ p1 − p2 = 2a a 2 pa p + p2 3 6 F (l + a) − F (l + a) + 1 − 1 a =0 ⇒ 2 p1 − p2 = 2 6a a2 − F + p1a − From (1) and (2) Shigley’s MED, 10th edition p1 = 2F (3l + 2a), a2 p2 = 2F (3l + a) a2 (1) (2) (3) Chapter 3 Solutions, Page 29/100 www.konkur.in From similar triangles b a = p2 p1 + p2 ⇒ b= ap2 p1 + p2 (4) Mmax occurs where V = 0 xmax = l + a − 2b p1 p + p2 (a − 2b) 2 − 1 ( a − 2b)3 2 6a p p + p2 = − Fl − F ( a − 2b) + 1 ( a − 2b) 2 − 1 (a − 2b)3 2 6a Normally Mmax = − Fl M max = − F (l + a − 2b) + The fractional increase in the magnitude is ∆= F (a − 2b) − ( p1 2 ) (a − 2b)2 + ( p1 + p2 ) 6a (a − 2b)3 Fl (5) For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in From (3) From (4) p1 = 2(1500) 3 (1.5) + 2(1.2) = 14 375 lbf/in 1.22 p2 = 2(1500) 3(1.5) + 1.2 = 11 875 lbf/in 1.22 b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in Substituting into (5) yields ∆ = 0.036 89 or 3.7% higher than -Fl _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 30/100 www.konkur.in 3-44 300(30) 40 + 1800 = 6900 lbf 2 30 300(30) 10 R2 = − 1800 = 3900 lbf 2 30 3900 a= = 13 in 300 R1 = MB = −1800(10) = −18 000 lbf⋅in Mx = 27 in = (1/2)3900(13) = 25 350 lbf⋅in 0.5(3) + 2.5(3) = 1.5 in 6 1 I1 = (3)(13 ) = 0.25 in 4 12 1 I 2 = (1)(33 ) = 2.25 in 4 12 Applying the parallel-axis theorem, y= I z = 0.25 + 3(1.5 − 0.5) 2 + 2.25 + 3(2.5 − 1.5) 2 = 8.5 in 4 (a) −18000(−1.5) At x = 10 in, y = −1.5 in, σ x = − = −3176 psi 8.5 −18000(2.5) At x = 10 in, y = 2.5 in, σ x = − = 5294 psi 8.5 25350( −1.5) At x = 27 in, y = −1.5 in, σ x = − = 4474 psi 8.5 25350(2.5) At x = 27 in, y = 2.5 in, σ x = − = −7456 psi 8.5 Max tension = 5294 psi Ans. Max compression = −7456 psi Ans. (b) The maximum shear stress due to V is at B, at the neutral axis. Vmax = 5100 lbf Q = y′A′ = 1.25(2.5)(1) = 3.125 in 3 VQ 5100(3.125) = 1875 psi Ans. (τ max )V = = Ib 8.5(1) (c) There are three potentially critical locations for the maximum shear stress, all at x = 27 in: (i) at the top where the bending stress is maximum, (ii) at the neutral axis where Shigley’s MED, 10th edition Chapter 3 Solutions, Page 31/100 www.konkur.in the transverse shear is maximum, or (iii) in the web just above the flange where bending stress and shear stress are in their largest combination. For (i): The maximum bending stress was previously found to be − 7456 psi, and the shear stress is zero. From Mohr’s circle, σ 7456 = 3728 psi τ max = max = 2 2 For (ii): The bending stress is zero, and the transverse shear stress was found previously to be 1875 psi. Thus, τmax = 1875 psi. For (iii): The bending stress, at y = – 0.5 in, is −18000( −0.5) = −1059 psi σx = − 8.5 The transverse shear stress is Q = y′A′ = (1)(3)(1) = 3.0 in 3 τ= VQ 5100(3.0) = = 1800 psi Ib 8.5(1) From Mohr’s circle, −1059 2 + 1800 = 1876 psi 2 The critical location is at x = 27 in, at the top surface, where τmax = 3728 psi. Ans. _____________________________________________________________________________ 2 τ max = 3-45 (a) L = 10 in. Element A: My −(1000)(10)(0.5) =− 10−3 = 101.9 kpsi 4 I (π / 64)(1) VQ , Q = 0 ⇒ τA = 0 τA = Ib ( σA = − ) σ 101.9 2 τ max = A + τ A2 = + (0) = 50.9 kpsi 2 2 2 2 Ans. Element B: σB = − My , y=0 I 4r Q = y′A′ = 3π Shigley’s MED, 10th edition ⇒ σB = 0 2 3 4 ( 0.5 ) π r 4r = = = 1/ 12 in 3 2 6 6 3 Chapter 3 Solutions, Page 32/100 www.konkur.in ( ) VQ (1000)(1/12) = 10−3 = 1.698 kpsi 4 Ib (π / 64)(1) (1) τB = 0 2 τ max = + 1.6982 = 1.698 kpsi 2 Ans. Element C: σC = − My −(1000)(10)(0.25) =− 10−3 = 50.93 kpsi 4 I (π / 64)(1) ( r r r y1 y1 y1 ) ) ( Q = ∫ ydA = ∫ y (2 x ) dy = ∫ y 2 r 2 − y 2 dy ( 2 = − r 2 − y2 3 ( ) ) 3/ 2 r y1 ( 2 = − r2 − r2 3 ) 3/2 ( − r 2 − y12 ) 3/2 2 2 r − y12 3 For C, y1 = r /2 =0.25 in = Q= 3/ 2 3/2 2 0.52 − 0.252 ) = 0.05413 in3 ( 3 b = 2 x = 2 r 2 − y12 = 2 0.52 − 0.252 = 0.866 in τC = VQ (1000)(0.05413) = (10−3 ) = 1.273 kpsi Ib (π / 64)(1) 4 (0.866) τ max 50.93 2 = + (1.273) = 25.50 kpsi 2 2 Ans. (b) Neglecting transverse shear stress: Element A: Since the transverse shear stress at point A is zero, there is no change. τ max = 50.9 kpsi Ans. % error = 0% Ans. Element B: Since the only stress at point B is transverse shear stress, neglecting the transverse shear stress ignores the entire stress. 2 0 τ max = = 0 psi Ans. 2 1.698 − 0 % error = * (100) = 100% Ans. 1.698 Shigley’s MED, 10th edition Chapter 3 Solutions, Page 33/100 www.konkur.in Element C: 2 50.93 = 25.47 kpsi 2 τ max = Ans. 25.50 − 25.47 % error = *(100) = 0.12% Ans. 25.50 (c) Repeating the process with different beam lengths produces the results in the table. L = 10 in A B C L = 4 in A B C L = 1 in A B C L = 0.1in A B C Bending stress, σ (kpsi) Transverse shear stress, τ (kpsi) Max shear stress, τmax (kpsi) Max shear stress, neglecting τ, τmax (kpsi) % error 102 0 50.9 0 1.70 1.27 50.9 1.70 25.50 50.9 0 25.47 0 100 0.12 40.7 0 20.4 0 1.70 1.27 20.4 1.70 10.26 20.4 0 10.19 0 100 0.77 10.2 0 5.09 0 1.70 1.27 5.09 1.70 2.85 5.09 0 2.55 0 100 10.6 1.02 0 0.509 0 1.70 1.27 0.509 1.70 1.30 0.509 0 0.255 0 100 80.4 Discussion: The transverse shear stress is only significant in determining the critical stress element as the length of the cantilever beam becomes smaller. As this length decreases, bending stress reduces greatly and transverse shear stress stays the same. This causes the critical element location to go from being at point A, on the surface, to point B, in the center. The maximum shear stress is on the outer surface at point A for all cases except L = 0.1 in, where it is at point B at the center. When the critical stress element is at point A, there is no error from neglecting transverse shear stress, since it is zero at that location. Neglecting the transverse shear stress has extreme significance at the stress element at the center at point B, but that location is probably only of practical significance for very short beam lengths. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 34/100 www.konkur.in 3-46 c F l c M = Fx 0 ≤ x ≤ a l 6 M 6 ( c l ) Fx σ= 2= bh bh 2 R1 = h= 6 Fcx lbσ max 0≤ x≤a Ans. _____________________________________________________________________________ 3-47 c From Problem 3-46, R1 = F = V , 0 ≤ x ≤ a l 3 V 3 (c / l ) F 3 Fc τ max = = ⇒ h= 2 bh 2 bh 2 lbτ max Ans. 6 Fcx . lbσ max Sub in x = e and equate to h above. 3 Fc 6 Fce = 2 lbτ max lbσ max From Problem 3-46, h( x) = 3 Fcσ max Ans. 2 8 lbτ max _____________________________________________________________________________ e= 3-48 (a) x-z plane ΣM O = 0 = 1.5(0.5) + 2(1.5) sin(30o )(2.25) − R2 z (3) R2 z = 1.375 kN Ans. ΣFz = 0 = R1z − 1.5 − 2(1.5) sin(30o ) + 1.375 R1z = 1.625 kN Ans. x-y plane ΣM O = 0 = −2(1.5) cos(30o )(2.25) + R2 y (3) R2 y = 1.949 kN Ans. ΣFy = 0 = R1 y − 2(1.5) cos(30o ) + 1.949 R1 y = 0.6491 kN Shigley’s MED, 10th edition Ans. Chapter 3 Solutions, Page 35/100 www.konkur.in (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, Vz = x′ − 0.125 M y = ∫ Vz dx′ = ( x′ ) 2 2 − 0.125x′ + C At x′ = 0, M y = C = −0.9375 ⇒ M y = 0.5 ( x′ ) − 0.125 x′ + 0.9375 2 1.949 x′ + 0.6491 = −1.732 x′ + 0.6491 1.125 −1.732 2 Mz = ( x′ ) + 0.6491x′ + C 2 2 At x′ = 0, M z = C = 0.9737 ⇒ M z = −0.8662 ( x′ ) − 0.125 x′ − 0.9375 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. Vy = − x (m) 0 0.5− 1.5 1.625 1.875 3− Vz (kN) –1.625 –1.625 –0.1250 0 0.2500 1.375 Shigley’s MED, 10th edition Vy (kN) 0.6491 0.6491 0.6491 0.4327 0 –1.949 My Mz V (kN) (kN⋅m) (kN⋅m) 1.750 0 0 1.750 –0.8125 0.3246 0.6610 0.9375 0.9737 0.4327 –0.9453 1.041 0.2500 –0.9141 1.095 2.385 0 0 M (kN⋅m) 0 0.8749 1.352 1.406 1.427 0 Chapter 3 Solutions, Page 36/100 www.konkur.in (d) The bending stress is obtained from Eq. (3-27), −M z yA M y zA σx = + Iz Iy 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. 40(75)3 34(25)3 Iz = − = 1.36(106 ) mm 4 = 1.36(10−6 ) m 4 12 12 3 25(40) 25(6)3 Iy = 2 = 2.67(105 ) mm 4 = 2.67(10 −7 ) m 4 + 12 12 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. _____________________________________________________________________________ 3-49 (a) x-z plane 3 600 ΣM O = 0 = (1000)(4) − (10) + M Oy 5 2 M Oy = 1842.6 lbf ⋅ in Ans. 3 600 ΣFz = 0 = ROz − (1000) + 5 2 ROz = 175.7 lbf Ans. x-y plane 4 600 ΣM O = 0 = − (1000)(4) − (10) + M Oz 5 2 M Oz = 7442.5 lbf ⋅ in Ans. 4 600 ΣFy = 0 = ROy − (1000) − 5 2 ROy = 1224.3 lbf Ans. Shigley’s MED, 10th edition Chapter 3 Solutions, Page 37/100 www.konkur.in (b) ( (c) 1/2 V ( x ) = Vy ( x ) 2 + Vz ( x) 2 M ( x ) = M y ( x ) 2 + M z ( x ) 2 x (m) 0 4− 10− Vz (kN) –175.7 –175.7 424.3 1/2 Vy (kN) 1224.3 1224.3 424.3 V (kN) 1237 1237 600 My (kN⋅m) Mz (kN⋅m) M (kN⋅m) –1842.6 –7442.6 7667 –2545.4 –2545.4 3600 0 0 0 (d) The maximum tensile bending stress will be at the outer corner of the cross section in the positive y, negative z quadrant, where y = 1.5 in and z = –1 in. 2(3)3 (1.625)(2.625)3 Iz = − = 2.051 in 4 12 12 3 3(2) (2.625)(1.625)3 Iy = − = 1.601 in 4 12 12 At x = 0, using Eq. (3-27), M y M z σx = − z + y Iz Iy ( −7442.6)(1.5) ( −1842.6)( −1) σx = − + = 6594 psi 2.051 1.601 Check at x = 4 in, ( −2545.4)(1.5) ( −2545.4)( −1) + = 2706 psi σx = − 2.051 1.601 The critical location is at x = 0, where σx = 6594 psi. Ans. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 38/100 www.konkur.in 3-50 (a) The area within the wall median line, Am, is Square: Am = (b − t ) 2 . From Eq. (3-45) (3 Tsq = 2 Amtτ all = 2(b − t ) 2 tτ all Round: Am = π (b − t ) 2 / 4 Trd = 2π (b − t ) 2 tτ all / 4 Ratio of Torques Tsq 2(b − t ) 2 tτ all 4 = = = 1.27 Ans. 2 Trd π (b − t ) tτ all / 2 π (b) Twist per unit length from Eq. (3-46) is θ1 = TLm 2 A tτ L L τ L = m all2 m = all m = C m 2 4GAmt 4GAmt 2G Am Am Square: θsq = C Round: θrd = C 4(b − t ) (b − t )2 π (b − t ) π (b − t ) / 4 2 =C 4(b − t ) (b − t )2 θ sq = 1 . Twists wists are the same. Ans. θ rd _____________________________________________________________________________ 3-51 (a) The area enclosed by the section median line is Am = (1 − 0.0625)2 = 0.8789 in2 and the length of the section median line is Lm = 4(1 − 0.0625) = 3.75 in. From Eq. (3-45), T = 2 Amtτ = 2(0.8789)(0.0625)(12 000) = 1318 lbf ⋅ in From Eq. (3-46), θ = θ1l = TLml 4GAm2 t = (1318)(3.75) ( 36 ) ( ) 4 (11.5 ) 10 (0.8789) ( 0.0625 ) 6 2 Ans. = 0.0801 rad = 4.59o Ans. (b) The radius at the median line is rm = 0.125 + (0.5)(0.0625) = 0.15625 in. The area enclosed by the section median line is Am = (1 − 0.0625)2 – 4(0.15625)2 + 4(π π /4)(0.15625)2 = 0.8579 Shigley’s MED, 10th edition Chapter 3 Solutions, Page 39/100 www.konkur.in in2. The length of the section median line is Lm = 4[1 – 0.0625 – 2(0.15625)] + 2π(0.15625) = 3.482 in. From Eq. (3-45), T = 2 Amtτ = 2(0.8579)(0.0625)(12 000) = 1287 lbf ⋅ in Ans. From Eq. (3-46), (1287)(3.482) ( 36 ) TLm l = = 0.0762 rad = 4.37o Ans. θ = θ1l = 2 6 2 4GAm t 4 (11.5 ) 10 (0.8579) ( 0.0625 ) ( ) _____________________________________________________________________________ 3-52 θ1 = 3Ti Ti = ⇒ GLi ci3 T = T1 + T2 + T3 = θ1G 3 θ1GLi ci3 3 3 ∑ Li ci3 Ans. i =1 From Eq. (3-47), τ = Gθ1c G and θ1 are constant, therefore the largest shear stress occurs when c is a maximum. τ max = Gθ1cmax Ans. _____________________________________________________________________________ 3-53 (b) Solve part (b) first since the angle of twist per unit length is needed for part (a). τ max = τ allow = 12 ( 6.89 ) = 82.7 MPa θ1 = τ max Gcmax = ( ) = 0.348 rad/m 79.3 (10 ) (0.003) 82.7 106 9 (a) T1 = T2 = T3 = θ1GL1c13 3 θ 2GL2 c23 3 = ( ) 0.348(79.3) 109 (0.020)(0.0023 ) = θ3GL3c33 Ans. 3 0.348(79.3) (109 ) (0.030)(0.0033 ) 3 9 0.348(79.3) (10 ) (0)(03 ) = 1.47 N ⋅ m Ans. = 7.45 N ⋅ m Ans. = = 0 Ans. 3 3 T = T1 + T2 + T3 = 1.47 + 7.45 + 0 = 8.92 N ⋅ m Ans. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 40/100 www.konkur.in 3-54 (b) Solve part (b) first since the angle of twist per unit length is needed for part (a). τ 12 000 θ1 = max = = 8.35 10−3 rad/in Ans. Gcmax 11.5 106 (0.125) ( ( ) (a) T1 = T2 = θ1GL1c13 3 = θ 2GL2 c23 ) ( 8.35) (10−3 ) (11.5) (106 ) ( 0.75) ( 0.06253 ) 3 −3 ( 8.35) (10 ) (11.5) (106 ) (1) ( 0.1253 ) = = 5.86 lbf ⋅ in Ans. = 62.52 lbf ⋅ in Ans. 3 −3 6 3 θ GL c3 ( 8.35) (10 ) (11.5) (10 ) ( 0.625 ) ( 0.0625 ) T3 = 3 3 3 = = 4.88 lbf ⋅ in Ans. 3 3 T = T1 + T2 + T3 = 5.86 + 62.52 + 4.88 = 73.3 lbf ⋅ in Ans. _____________________________________________________________________________ 3 3-55 (b) Solve part (b) first since the angle of twist per unit length is needed for part (a). τ max = τ allow = 12 ( 6.89 ) = 82.7 MPa θ1 = τ max Gcmax = ( ) = 0.348 rad/m 79.3 (10 ) (0.003) 82.7 106 9 (a) T1 = T2 = T3 = θ1GL1c13 3 θ 2GL2 c23 3 = ( ) 0.348(79.3) 109 (0.020)(0.0023 ) = θ3GL3c33 Ans. 3 0.348(79.3) (109 ) (0.030)(0.0033 ) 3 9 0.348(79.3) (10 ) (0.025)(0.0023 ) = 1.47 N ⋅ m = 7.45 N ⋅ m Ans. Ans. = = 1.84 N ⋅ m Ans. 3 3 T = T1 + T2 + T3 = 1.47 + 7.45 + 1.84 = 10.8 N ⋅ m Ans. _____________________________________________________________________________ 3-56 (a) From Eq. (3-40), with two 2-mm strips, ( 80 ) (106 ) ( 0.030 ) ( 0.0022 ) T= = = 3.08 N ⋅ m 3 + 1.8 / (b / c) 3 + 1.8 / ( 0.030 / 0.002 ) τ max bc 2 Tmax = 2(3.08) = 6.16 N ⋅ m Shigley’s MED, 10th edition Ans. Chapter 3 Solutions, Page 41/100 www.konkur.in 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 α≈ 3 + 1.8 / (30 / 2) From Eq. (3-41), Tl 3.08(0.3) θ= = = 0.151 rad 3 β bc G 0.321( 0.030 ) 0.0023 ( 79.3) 109 ( kt = T θ = 6.16 = 40.8 N ⋅ m 0.151 ) ( ) Ans. Ans. From Eq. (3-40), with a single 4-mm strip, (80 ) (106 ) ( 0.030) ( 0.0042 ) Tmax = = = 11.9 N ⋅ m 3 + 1.8 / (b / c) 3 + 1.8 / ( 0.030 / 0.004 ) τ max bc 2 Ans. 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 β= 8−6 From Eq. (3-41) Tl 11.9(0.3) = = 0.0769 rad θ= 3 β bc G 0.305 ( 0.030 ) 0.0043 ( 79.3) 109 ( kt = T θ = ) 11.9 = 155 N ⋅ m 0.0769 ( ) Ans. Ans. (b) From Eq. (3-47), with two 2-mm strips, ( ) ( ) 2 6 Lc 2τ ( 0.030 ) 0.002 ( 80 ) 10 T= = = 3.20 N ⋅ m 3 3 Tmax = 2(3.20) = 6.40 N ⋅ m Ans. 3Tl 3(3.20)(0.3) θ= 3 = = 0.151 rad Lc G ( 0.030 ) 0.0023 ( 79.3) 109 ( ) ( ) kt = T θ = 6.40 0.151 = 42.4 N ⋅ m Ans. Ans. From Eq. (3-47), with a single 4-mm strip, Tmax ( ) ( ) 2 6 Lc 2τ ( 0.030 ) 0.004 ( 80 ) 10 = = = 12.8 N ⋅ m 3 3 Shigley’s MED, 10th edition Ans. Chapter 3 Solutions, Page 42/100 www.konkur.in θ= 3Tl 3(12.8)(0.3) = = 0.0757 rad 3 Lc G ( 0.030 ) 0.0043 ( 79.3) 109 ( ) ( ) kt = T θ = 12.8 0.0757 = 169 N ⋅ m Ans. Ans. 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). H (40000) T = 9.55 = 9.55 = 152.8 N ⋅ m n 2500 Tr 16T τ max = = 3 J πd 13 16 152.8 ( ) = 0.0223 m = 22.3 mm = 6 π ( 70 ) 10 H (40000) = 1528 N ⋅ m (b) T = 9.55 = 9.55 n 250 13 16T d = πτ max ( ) Ans. 13 16(1528) d= = 0.0481 m = 48.1 mm Ans. π ( 70 ) 106 _____________________________________________________________________________ ( ) 3-58 (a) Obtain the torque from the given power and speed using Eq. (3-42). 63025H 63025(50) T= = = 1261 lbf ⋅ in n 2500 Tr 16T τ max = = 3 J πd 16T 16 (1261) d = = = 0.685 in πτ max π (20 000) 63025H 63025(50) = = 12610 lbf ⋅ in (b) T = n 250 13 13 Ans. 13 16(12 610) d= = 1.48 in Ans. π (20 000) _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 43/100 www.konkur.in 3-59 τ max 16T = π d3 ( 50 ) (106 ) π ( 0.033 ) = = 265 N ⋅ m 16 16 Tn 265(2000) = = 55.5 103 W = 55.5 kW Ans. Eq. (3-44), H = 9.55 9.55 _____________________________________________________________________________ 3-60 π π 16T τ = 3 ⇒ T = τ d 3 = (110 ) 106 0.0203 = 173 N ⋅ m 16 16 πd π π 0.0204 ( 79.3) 109 15 π d 4Gθ Tl 180 ⇒ l= = θ= JG 32T 32(173) l = 1.89 m Ans. _____________________________________________________________________________ ⇒ T= τ maxπ d 3 ( ) ( )( ( 3-61 ) ) ( ) 16T π π ⇒ T = τ d 3 = ( 30 000 ) 0.753 = 2485 lbf ⋅ in 3 16 16 πd Tl 32Tl 32(2485)(24) = = = 0.167 rad = 9.57o Ans. θ= 4 4 6 JG π d G π 0.75 (11.5 ) 10 ( τ= ( ) ) ( ) _____________________________________________________________________________ 3-62 (a) Tsolid = Jτ max π do4τ max = r 16d o Thollow = Jτ max π (d o4 − di4 )τ max = r 16d o ( ) ( ) 364 Tsolid − Thollow di4 % ∆T = (100%) = 4 (100%) = (100%) = 65.6% Tsolid do 404 (b) Wsolid = kd o2 , ( Whollow = k d o 2 − di 2 ) Ans. ( ) ( ) 362 Wsolid − Whollow di2 %∆W = (100%) = 2 (100%) = (100%) = 81.0% Wsolid do 402 Ans. _____________________________________________________________________________ 3-63 4 4 Jτ max π d 4τ max Jτ max π d − ( xd ) τ max (a) Tsolid = = Thollow = = r 16d r 16d 4 T −T ( xd ) %∆T = solid hollow (100%) = (100%) = x 4 (100%) Ans. 4 Tsolid d Shigley’s MED, 10th edition Chapter 3 Solutions, Page 44/100 www.konkur.in ( Whollow = k d 2 − ( xd ) (b) Wsolid = kd 2 2 ) W − Whollow ( xd ) (100%) = x 2 (100%) %∆W = solid (100%) = Wsolid d2 2 Ans. Plot %∆T and %∆W versus x. Percent Reduction in Torque and Weight Percent Reduction 100 80 60 Weight 40 Torque 20 difference 0 0 0.2 0.4 0.6 0.8 1 x The value of greatest difference in percent reduction of weight and torque is 25% and occurs at x = 2 2 . _____________________________________________________________________________ 3-64 Tc (a) τ = J ( )= ⇒ 120 10 ( ) 4200 ( d 2 ) 6 (π 32 ) d 4 − ( 0.70d ) 4 = ( ) 2.8149 104 d 3 13 2.8149 104 = 6.17 10−2 m = 61.7 mm d = 120(106 ) 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) 4200 ( d i 2 ) 4200 ( 0.050 2 ) Tc = = = 30.8 MPa Ans. 4 J (π 32 ) d 4 − ( d ) (π 32 ) ( 0.080 )4 − ( 0.050 )4 i _____________________________________________________________________________ τ= Shigley’s MED, 10th edition Chapter 3 Solutions, Page 45/100 www.konkur.in 3-65 T = 9.55 H (1500) = 9.55 = 1433 N ⋅ m n 10 13 16 1433 ( ) = 0.045 m = 45 mm 16T 16T τ= 3 ⇒ dC = = π dC πτ π 80 106 ( ) From Table A-17, select 50 mm. Ans. 16 ( 2 )(1433) (a) τ start = = 117 (106 ) Pa = 117 MPa Ans. 3 π ( 0.050 ) 13 ( ) (b) Design activity _____________________________________________________________________________ 3-66 T= 63 025 H 63 025(1) = = 7880 lbf ⋅ in n 8 16T τ= 3 π dC 13 ⇒ 16T dC = πτ 16 ( 7880 ) = π (15 000 ) 13 = 1.39 in From Table A-17, select 1.40 in. Ans. _____________________________________________________________________________ 3-67 For a square cross section with side length b, and a circular section with diameter d, π π Asquare = Acircular ⇒ b 2 = d 2 ⇒ b= d 4 2 From Eq. (3-40) with b = c, (τ max )square = 3 T 1.8 T 1.8 T 2 T 3+ (4.8) = 6.896 3 = 3 3+ = 3 2 bc b/c b 1 d π d For the circular cross section, 16T T (τ max )circular = 3 = 5.093 3 d πd T (τ max )square 6.896 d 3 = = 1.354 (τ max )circular 5.093 T d3 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), Tl Tl Tl Tl θsquare = = = = 11.50 4 4 3 4 β bc G β b G d G π 0.141 d G 2 Shigley’s MED, 10th edition Chapter 3 Solutions, Page 46/100 www.konkur.in For the circular cross section, Tl Tl Tl θ rd = = = 10.19 4 4 GJ G (π d 32 ) d G Tl θ sq 11.50 d 4G = = 1.129 θ rd 10.19 Tl 4 d G The angle of twist in the square cross section is 12.9% greater. Ans. _____________________________________________________________________________ 3-68 (a) T1 = 0.15T2 ∑ T = 0 = (500 − 75)(4) − (T 2 1700 − 4.25T2 = 0 ⇒ T1 = 0.15 ( 400 ) = 60 lbf (b) ∑M O − T1 )( 5 ) = 1700 − (T2 − 0.15T2 )( 5 ) T2 = 400 lbf Ans. Ans. = 0 = −575(10) + 460(28) − RC (40) RC = 178.25 lbf ∑F =0=R O Ans. + 575 − 460 + 178.25 RO = −293.25 lbf Ans. (c) (d) The maximum bending moment is at x = 10 in, and is M = 2932.5 lbf·in. 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 Shigley’s MED, 10th edition Chapter 3 Solutions, Page 47/100 www.konkur.in from A to B is T = (500 − 75)(4) = 1700 lbf·in. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Mc 32 M 32 ( 2932.5 ) = = = 15 294 psi = 15.3 kpsi Ans. I πd3 π (1.25)3 Tr 16T 16(1700) τ= = 3= = 4433 psi = 4.43 kpsi Ans. π (1.25)3 J πd σ= (e) σx 2 15.3 2 σ 15.3 σ1, σ 2 = ± x + (τ xy ) = ± + ( 4.43) 2 2 2 2 Ans. σ 1 = 16.5 kpsi 2 σ 2 = −1.19 kpsi 2 Ans. 2 2 σ 15.3 τ max = x + (τ xy ) = Ans. + ( 4.43) = 8.84 kpsi 2 2 _____________________________________________________________________________ 3-69 (a) T2 = 0.15T1 2 2 ∑ T = 0 = (1800 − 270 ) (200) + (T 2 306 (103 ) − 106.25T1 = 0 ⇒ − T1 ) (125) = 306 (103 ) + 125 ( 0.15T1 − T1 ) T1 = 2880 N Ans. T2 = 0.15 ( 2880 ) = 432 N Ans. (b) ∑M O = 0 = 3312(230) + RC (510) − 2070(810) RC = 1794 N Ans. ∑F y = 0 = RO + 3312 + 1794 − 2070 RO = −3036 N Ans. (c) Shigley’s MED, 10th edition Chapter 3 Solutions, Page 48/100 www.konkur.in (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, Mc 32 M 32 ( 698.3 ) = = = 263 (103 ) Pa = 263 MPa Ans. 3 3 I πd π (0.030) Tr 16T 16(306) τ= = 3= = 57.7 106 Pa = 57.7MPa Ans. 3 π (0.030) J πd σ= ( ) (e) 2 263 2 σ 263 ± x + (τ xy ) = ± + ( 57.7 ) 2 2 2 2 Ans. σ 1 = 275 MPa σ1 , σ 2 = σx 2 σ 2 = −12.1 MPa 2 Ans. 2 2 σ 263 τ max = x + (τ xy ) = + ( 57.7 ) = 144 MPa 2 2 2 2 Ans. _____________________________________________________________________________ 3-70 (a) T2 = 0.15T1 ∑ T = 0 = ( 300 − 50 ) (4) + (T − T1 ) (3) = 1000 + ( 0.15T1 − T1 ) (3) 1000 − 2.55T1 = 0 T1 = 392.16 lbf Ans. 2 ⇒ T2 = 0.15 ( 392.16 ) = 58.82 lbf Ans. (b) ∑M Oy = 0 = −450.98(16) − RC z (22) RC z = −327.99 lbf Ans. ∑F z = 0 = RO z + 450.98 − 327.99 RO z = −122.99 lbf Ans. ∑M Oz = 0 =350(8) + RC y (22) RC y = −127.27 lbf Ans. ∑F y = 0 = RO y + 350 − 127.27 RO y = −222.73 lbf Ans. Shigley’s MED, 10th edition Chapter 3 Solutions, Page 49/100 www.konkur.in (c) (d) Combine the bending moments from both planes at A and B to find the critical location. M A = (983.92) 2 + (−1781.84) 2 = 2035 lbf ⋅ in M B = (1967.84) 2 + (−763.65) 2 = 2111 lbf ⋅ in The critical location is at B. The torque transmitted through the shaft from A to B is T = (300 − 50)(4) = 1000 lbf·in. For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Mc 32 M 32 ( 2111) = = = 21 502 psi = 21.5 kpsi Ans. πd3 π (1)3 I Tr 16T 16(1000) = 5093 psi = 5.09 kpsi Ans. τ= = 3= J πd π (1)3 σ= (e) σx 2 21.5 2 σ 21.5 ± x + (τ xy ) = ± σ1 , σ 2 = + ( 5.09 ) 2 2 2 2 σ 1 = 22.6 kpsi Ans. 2 σ 2 = −1.14 kpsi 2 Ans. 2 2 σ 21.5 Ans. τ max = x + (τ xy ) = + ( 5.09 ) = 11.9 kpsi 2 2 _____________________________________________________________________________ 2 Shigley’s MED, 10th edition 2 Chapter 3 Solutions, Page 50/100 www.konkur.in 3-71 (a) T2 = 0.15T1 ∑ T = 0 = ( 300 − 45 ) (125) + (T 2 31 875 − 127.5T1 = 0 ⇒ − T1 ) (150) = 31 875 + ( 0.15T1 − T1 ) (150) T1 = 250 N ⋅ mm Ans. T2 = 0.15 ( 250 ) = 37.5 N ⋅ mm Ans. (b) ∑M Oy = 0 = 345sin 45o (300) − 287.5(700) − RC z (850) RC z = −150.7 N ∑F z = 0 = RO z − 345 cos 45o + 287.5 − 150.7 RO z = 107.2 N ∑M Oz y Ans. = 0 = 345sin 45o (300) + RC y (850) RC y = −86.10 N ∑F Ans. Ans. = 0 = RO y + 345 cos 45o − 86.10 RO y = −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 MED, 10th edition Chapter 3 Solutions, Page 51/100 www.konkur.in ( −47.37 ) + ( −32.16) M= 2 2 = 57.26 N ⋅ m 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, Mc 32M 32 ( 57.26 ) = = = 72.9 (106 ) Pa = 72.9 MPa Ans. σ= 3 3 πd π (0.020) I Tr 16T 16(31.88) τ= = 3= = 20.3 (106 ) Pa = 20.3 MPa Ans. 3 π (0.020) J πd (e) 2 72.9 2 σ 72.9 ± x + (τ xy ) = ± + ( 20.3) 2 2 2 2 Ans. σ 1 = 78.2 MPa σ1 , σ 2 = σx 2 σ 2 = −5.27 MPa 2 Ans. 2 2 σ 72.9 τ max = x + (τ xy ) = + ( 20.3) = 41.7 MPa 2 2 2 2 Ans. _____________________________________________________________________________ 3-72 (a) ∑ T = 0 = − 300(cos 20º )(10) + FB (cos 20º )(4) FB = 750 lbf Ans. (b) ∑M Oz = 0 = 300(cos 20º )(16) − 750(sin 20º )(39) +RC y (30) RC y = 183.1 lbf Ans. ∑F y = 0 =RO y + 300(cos 20º ) + 183.1 − 750(sin 20º ) RO y = −208.5 lbf Ans. ∑M Oy = 0 = 300(sin 20º )(16) − RC z (30) − 750(cos 20º )(39) RC z = −861.5 lbf Ans. ∑F z = 0 =RO z − 300(sin 20º ) − 861.5 + 750(cos 20º ) RO z = 259.3 lbf Ans. Shigley’s MED, 10th edition Chapter 3 Solutions, Page 52/100 www.konkur.in (c) (d) Combine the bending moments from both planes at A and C to find the critical location. M A = (−3336) 2 + (−4149) 2 = 5324 lbf ⋅ in M C = (−2308) 2 + (−6343)2 = 6750 lbf ⋅ in The critical location is at C. The torque transmitted through the shaft from A to B is T = 300 cos ( 20º )(10 ) = 2819 lbf ⋅ in . For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Mc 32 M 32 ( 6750 ) = = = 35 203 psi = 35.2 kpsi Ans. I πd3 π (1.25)3 Tr 16T 16(2819) = 7351 psi = 7.35 kpsi Ans. τ= = 3= J πd π (1.25)3 σ= (e) σx 2 35.2 2 σ 35.2 σ1, σ 2 = ± x + (τ xy ) = ± + ( 7.35 ) 2 2 2 2 Ans. σ 1 = 36.7 kpsi 2 σ 2 = −1.47 kpsi 2 Ans. 2 2 σ 35.2 Ans. τ max = x + (τ xy ) = + ( 7.35 ) = 19.1 kpsi 2 2 _____________________________________________________________________________ 2 Shigley’s MED, 10th edition 2 Chapter 3 Solutions, Page 53/100 www.konkur.in 3-73 (a) ∑ T = 0 = −11 000(cos 20º )(300) + F (cos 25º )(150) B FB = 22 810 N (b) ∑M Oz = 0 = −11 000(sin 20º )(400) − 22 810(sin 25º )(750) +RC y (1050) RC y = 8319 N ∑F y Ans. = 0 =RO y − 11000(sin 20º ) − 22 810sin(25º ) + 8319 RO y = 5083 N ∑M Ans. Oy Ans. = 0 =11 000(cos 20º )(400) − 22 810(cos 25º )(750) − RC z (1050) RC z = −10 830 N Ans. ∑F z = 0 =RO z − 11 000(cos 20º ) + 22 810(cos 25º ) − 10 830 RO z = 494 N 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, M= ( 2496 ) + ( 3249 ) 2 2 = 4097 N ⋅ m The torque transmitted through the shaft from A to B is T = 11 000 cos ( 20º )( 0.3) = 3101 N ⋅ m . For a stress element on the outer surface where the bending stress and the torsional stress are both maximum, Shigley’s MED, 10th edition Chapter 3 Solutions, Page 54/100 www.konkur.in Mc 32 M 32 ( 4097 ) = = = 333.9 (10 6 ) Pa = 333.9 MPa Ans. I π d 3 π (0.050)3 Tr 16T 16(3101) = 126.3 106 Pa = 126.3 MPa Ans. τ= = 3= J πd π (0.050)3 σ= ( ) (e) σx 2 333.9 2 σ 333.9 σ1 , σ 2 = ± x + (τ xy ) = ± + (126.3) 2 2 2 2 Ans. σ 1 = 376 MPa 2 σ 2 = −42.4 MPa 2 Ans. 2 2 σ 333.9 τ max = x + (τ xy ) = Ans. + (126.3) = 209 MPa 2 2 _____________________________________________________________________________ 3-74 (a) ( ΣM D ) z = 6.13Cx − 3.8(92.8) − 3.88(362.8) = 0 2 C x = 287.2 lbf 2 Ans. ( ΣM C ) z = 6.13Dx + 2.33(92.8) − 3.88(362.8) = 0 Dx = 194.4 lbf 3.8 (808) = 500.9 lbf 6.13 2.33 = 0 ⇒ Dz = (808) = 307.1 lbf 6.13 ( ΣM D ) x = 0 ( ΣM C ) x Ans. ⇒ Cz = Ans. Ans. (b) For DQC, let x′, y′, z′ correspond to the original − y, x, z axes. Shigley’s MED, 10th edition Chapter 3 Solutions, Page 55/100 www.konkur.in (c) The critical stress element is just to the right of Q, where the bending moment in both planes is a maximum, and where the torsional and axial loads exist. T = 808(3.88) = 3135 lbf ⋅ in M = 669.22 + 11672 = 1345 lbf ⋅ in 16T 16(3135) τ= 3= = 11 070 psi Ans. πd π (1.133 ) σb = ± 32 M 32(1345) =± = ± 9495 psi 3 πd π 1.133 Ans. σa = − F 362.8 =− = −362 psi A (π / 4) 1.132 Ans. ( ) ( ) (d) The critical stress element will be where the bending stress and axial stress are both in compression. σ max = −9495 − 362 = −9857 psi −9857 2 + 11 070 = 12 118 psi = 12.1 kpsi 2 2 τ max = Ans. −9857 −9857 2 ± + 11 070 2 2 σ 1 = 7189 psi = 7.19 kpsi Ans. 2 σ1,σ 2 = σ 2 = −17 046 psi = −17.0 kpsi Ans. _____________________________________________________________________________ 3-75 (a) ( ΣM D ) z = 0 6.13Cx − 3.8(46.6) − 3.88(140) = 0 Cx = 117.5 lbf Ans. ( ΣM C ) z = 0 −6.13Dx − 2.33(46.6) + 3.88(140) = 0 Dx = 70.9 lbf 3.8 (406) = 251.7 lbf 6.13 2.33 = 0 ⇒ Dz = (406) = 154.3 lbf 6.13 ( ΣM D ) x = 0 ( ΣM C ) x Ans. ⇒ Cz = Shigley’s MED, 10th edition Ans. Ans. Chapter 3 Solutions, Page 56/100 www.konkur.in (b) For DQC, let x′, y′, z′ correspond to the original − y, x, z axes. (c) The critical stress element is just to the right of Q, where the bending moment in both planes is a maximum, and where the torsional and axial loads exist. T = 406(3.88) = 1575 lbf ⋅ in M = 273.82 + 586.32 = 647.1 lbf ⋅ in 16T 16(1575) τ= 3= = 8021 psi Ans. πd π (13 ) σb = ± 32 M 32(647.1) =± = ± 6591 psi 3 πd π 13 σa = − F 140 =− = −178.3 psi A (π / 4) 12 ( ) ( ) Ans. Ans. (d) The critical stress element will be where the bending stress and axial stress are both in compression. σ max = −6591 − 178.3 = −6769 psi −6769 2 = + 8021 = 8706 psi = 8.71 kpsi 2 2 τ max Ans. −6769 −6769 2 ± σ1 ,σ 2 = + 8021 2 2 σ 1 = 5321 psi = 5.32 kpsi Ans. 2 Shigley’s MED, 10th edition Chapter 3 Solutions, Page 57/100 www.konkur.in σ 2 = −12 090 psi = −12.1 kpsi Ans. _____________________________________________________________________________ 3-76 ( ΣM B ) z = −5.62(362.8) + 1.3(92.8) + 3 Ay = 0 Ay = 639.4 lbf Ans. ( ΣM A ) z = −2.62(362.8) + 1.3(92.8) + 3By = 0 By = 276.6 lbf 5.62 (808) = 1513.7 lbf 3 2.62 = 0 ⇒ Bz = (808) = 705.7 lbf 3 ( ΣM B ) y = 0 ( ΣM A ) y Ans. ⇒ Az = Ans. Ans. (b) (c) The critical stress element is just to the left of A, where the bending moment in both planes is a maximum, and where the torsional and axial loads exist. Shigley’s MED, 10th edition Chapter 3 Solutions, Page 58/100 www.konkur.in T = 808(1.3) = 1050 lbf ⋅ in 16(1050) = 7847 psi Ans. τ= π 0.883 ( ) M = (829.8) 2 + (2117) 2 = 2274 lbf ⋅ in 32 M 32(2274) σb = ± =± = ±33 990 psi 3 πd π 0.883 ( σa = − ) F 92.8 =− = −153 psi A (π / 4) 0.882 ( ) Ans. Ans. (d) The critical stress will occur when the bending stress and axial stress are both in compression. σ max = −33 990 − 153 = −34 143 psi −34 143 2 = + 7847 = 18 789 psi = 18.8 kpsi 2 2 τ max Ans. −34143 −34143 2 ± + 7847 2 2 σ 1 = 1717 psi = 1.72 kpsi Ans. 2 σ1,σ 2 = σ 2 = −35 860 psi = −35.9 kpsi Ans. _____________________________________________________________________________ 3-77 Ft = T 100 = = 1600 N c / 2 0.125 / 2 Fn = 1600 tan 20 = 582.4 N TC = Ft ( b 2 ) = 1600 ( 0.250 2 ) = 200 N ⋅ m P= TC 200 = = 2667 N ( a 2 ) ( 0.150 2 ) ∑ ( M A )z = 0 450 RDy − 582.4(325) − 2667(75) = 0 RDy = 865.1 N ∑ ( M A ) y = 0 = −450 RDz + 1600(325) ⇒ RDz = 1156 N ∑ Fy = 0 = RAy + 865.1 − 582.4 − 2667 ⇒ RAy = 2384 N ∑ Fz = 0 = RAz + 1156 − 1600 ⇒ RAz = 444 N Shigley’s MED, 10th edition Chapter 3 Solutions, Page 59/100 www.konkur.in 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. M B = AB RA2 y + RA2z = 0.075 23842 + 4442 = 181.9 N ⋅ m M C = CD RD2 y + RD2 z = 0.125 865.12 + 11562 = 180.5 N ⋅ m The stresses at B and C are almost identical, but the maximum stresses occur at B. 32 M B 32(181.9) = = 68.6 106 Pa = 68.6 MPa σB = 3 3 πd π 0.030 ( τB = ) Ans. ( ) ( ) 16TB 16(200) = = 37.7 106 Pa = 37.7 MPa 3 3 πd π 0.030 σ max = ( σB ) 68.6 σ 68.6 2 + B + τ B2 = + + 37.7 = 85.3 MPa 2 2 2 2 2 2 σ 68.6 2 τ max = B + τ B2 = + 37.7 = 51.0 MPa 2 2 2 Ans. 2 Ans. _____________________________________________________________________________ 3-78 Ft = T 100 = = 1600 N c / 2 0.125 / 2 Fn = 1600 tan 20 = 582.4 N TC = Ft ( b 2 ) = 1600 ( 0.250 2 ) = 200 N ⋅ m P= TC 200 = = 2667 N ( a 2 ) ( 0.150 2 ) ⇒ RDy = 420.6 N ∑ ( M A ) z = 0 = 450 RDy − 582.4(325) ∑ ( M A ) y = 0 = −450 RDz + 1600(325) − 2667(75) ⇒ RDz = 711.1 N ∑ Fy = 0 = RAy + 420.6 − 582.4 ∑ Fz = 0 = RAz + 711.1 − 1600 + 2667 Shigley’s MED, 10th edition ⇒ RAy = 161.8 N ⇒ RAz = −1778 N Chapter 3 Solutions, Page 60/100 www.konkur.in 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. M B = AB RA2 y + RA2z = 0.075 161.82 + ( −1778 ) = 133.9 N ⋅ m 2 M C = CD RD2 y + RD2 z = 0.125 420.62 + 711.12 = 103.3 N ⋅ m The maximum stresses occur at B. Ans. 32M B 32(133.9) = = 50.5 106 Pa = 50.5 MPa σB = 3 3 πd π 0.030 ( τB = ) ( ) ( ) 16TB 16(200) = = 37.7 106 Pa = 37.7 MPa 3 3 πd π 0.030 σ max = ( σB ) 50.5 σ 50.5 2 + B + τ B2 = + + 37.7 = 70.6 MPa 2 2 2 2 2 2 σ 50.5 2 τ max = B + τ B2 = + 37.7 = 45.4 MPa 2 2 2 Ans. 2 Ans. _____________________________________________________________________________ 3-79 T 900 = = 180 lbf c / 2 10 / 2 Fn = 180 tan 20 = 65.5 lbf Ft = TC = Ft ( b 2 ) = 180 ( 5 2 ) = 450 lbf ⋅ in P= TC 450 = = 150 lbf ( a 2) ( 6 2) ∑ ( M A ) z = 0 = 20 RDy − 65.5(14) − 150(4) ⇒ RDy = 75.9 lbf ⇒ RDz = 126 lbf ∑ ( M A ) y = 0 = −20 RDz + 180(14) ∑ Fy = 0 = RAy + 75.9 − 65.5 − 150 ∑ Fz = 0 = RAz + 126 − 180 Shigley’s MED, 10th edition ⇒ RAy = 140 lbf ⇒ RAz = 54.0 lbf Chapter 3 Solutions, Page 61/100 www.konkur.in 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. M B = AB RA2 y + RA2z = 4 1402 + 542 = 600 lbf ⋅ in M C = CD RD2 y + RD2 z = 6 75.92 + 1262 = 883 lbf ⋅ in The maximum stresses occur at C. σC = 32 M C τC = 16TC πd πd σ max 3 3 = = 32(883) ( π 1.3753 16(450) ( π 1.3753 σC ) ) Ans. = 3460 psi = 882 psi 3460 σ 3460 2 = + C + τ C2 = + + 882 = 3670 psi 2 2 2 2 2 2 Ans. σ 3460 2 τ max = C + τ C2 = + 882 = 1940 psi Ans. 2 2 _____________________________________________________________________________ 3-80 (a) Rod AB experiences constant torsion throughout its length, and maximum bending moment at the wall. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be at the wall, at either the top (compression) or the bottom (tension) on the y axis. We will select the bottom element for this analysis. (b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces. Mc M ( d / 2 ) 32 M 32 ( 8 )( 200 ) = = = = 16 297 psi = 16.3 kpsi σx = 3 I π d 4 / 64 π d 3 π (1) 2 τ xz = 2 Tr T ( d / 2 ) 16T 16 ( 5 )( 200 ) = = = = 5093 psi = 5.09 kpsi 3 J π d 4 / 32 π d 3 π (1) Shigley’s MED, 10th edition Chapter 3 Solutions, Page 62/100 www.konkur.in (c) 16.3 2 2 σ 16.3 ± x + (τ xz ) = ± + ( 5.09 ) 2 2 2 2 σ 1 = 17.8 kpsi Ans. σ1, σ 2 = σx 2 σ 2 = −1.46 kpsi 2 Ans. 2 2 σ 16.3 τ max = x + (τ xz ) = + ( 5.09 ) = 9.61 kpsi 2 2 2 2 Ans. _____________________________________________________________________________ 3-81 (a) Rod AB experiences constant torsion throughout its length, and maximum bending moments at the wall in both planes of bending. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface at the wall, with its critical location determined by the plane of the combined bending moments. My = – (100)(8) = – 800 lbf·in Mz = (175)(8) = 1400 lbf·in M tot = M y2 + M z2 = ( −800 ) 2 + 14002 = 1612 lbf ⋅ in My −1 800 = tan = 29.7º 1400 Mz The combined bending moment vector is at an angle of 29.7º CCW from the z axis. The critical bending stress location, and thus the critical stress element, will be ±90º from this vector, as shown. There are two equally critical stress elements, one in tension (119.7º CCW from the z axis) and the other in compression (60.3º CW from the z axis). We’ll continue the analysis with the element in tension. (b) Transverse shear is zero at the critical stress elements on the outer surfaces. M c M ( d / 2 ) 32M tot 32 (1612 ) = = = 16 420 psi = 16.4 kpsi σ x = tot = tot 4 3 π d / 64 πd3 I π (1) θ = tan −1 τ= Tr T ( d / 2 ) 16T 16 ( 5 )(175 ) = = = = 4456 psi = 4.46 kpsi 3 J π d 4 / 32 π d 3 π (1) Shigley’s MED, 10th edition Chapter 3 Solutions, Page 63/100 www.konkur.in (c) 16.4 2 σ 16.4 ± x +τ 2 = ± + ( 4.46 ) 2 2 2 2 σ 1 = 17.5 kpsi Ans. σ1, σ 2 = σx 2 σ 2 = −1.13 kpsi 2 Ans. 2 σ 16.4 τ max = x + τ 2 = + ( 4.46 ) = 9.33 kpsi 2 2 2 2 Ans. _____________________________________________________________________________ 3-82 (a) Rod AB experiences constant torsion and constant axial tension throughout its length, and maximum bending moments at the wall from both planes of bending. Both torsional shear stress and bending stress will be maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface at the wall, with its critical location determined by the plane of the combined bending moments. My = – (100)(8) – (75)(5) = – 1175 lbf·in Mz = (–200)(8) = –1600 lbf·in M tot = M y2 + M z2 = ( −1175 ) + ( −1600 ) 2 2 = 1985 lbf ⋅ in My −1 1175 = tan = 36.3º 1600 Mz The combined bending moment vector is at an angle of 36.3º CW from the negative z axis. The critical bending stress location will be ±90º from this vector, as shown. Since there is an axial stress in tension, the critical stress element will be where the bending is also in tension. The critical stress element is therefore on the outer surface at the wall, at an angle of 36.3º CW from the y axis. (b) Transverse shear is zero at the critical stress element on the outer surface. M c M ( d / 2 ) 32M tot 32 (1985 ) = = = 20 220 psi = 20.2 kpsi σ x ,bend = tot = tot 4 3 I π d / 64 πd3 π (1) θ = tan −1 σ x ,axial = Fx Fx 75 = = = 95.5 psi = 0.1 kpsi , which is essentially negligible 2 A π d / 4 π (1) 2 / 4 σ x = σ x ,axial + σ x ,bend = 20 220 + 95.5 = 20 316 psi = 20.3 kpsi τ= Tr 16T 16 ( 5 )( 200 ) = = = 5093 psi = 5.09 kpsi 3 J πd3 π (1) Shigley’s MED, 10th edition Chapter 3 Solutions, Page 64/100 www.konkur.in (c) σx 20.3 2 σ 20.3 σ1, σ 2 = ± x +τ 2 = ± + ( 5.09 ) 2 2 2 2 Ans. σ 1 = 21.5 kpsi 2 σ 2 = −1.20 kpsi 2 Ans. 2 σ 20.3 τ max = x + τ 2 = + ( 5.09 ) = 11.4 kpsi 2 2 2 2 Ans. _____________________________________________________________________________ 3-83 T = (2)(200) = 400 lbf·in The maximum shear stress due to torsion occurs in the middle of the longest side of the rectangular cross section. From the table on p. 116, with b/c = 1.5/0.25 = 6, α = 0.299. From Eq. (3-40), T 400 = = 14 270 psi = 14.3 kpsi Ans. τ max = 2 α bc ( 0.299 )(1.5 )( 0.25 )2 ____________________________________________________________________________ 3-84 (a) The cross section at A will experience bending, torsion, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be at either the top (compression) or the bottom (tension) on the y axis. We’ll select the bottom element for this analysis. (b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces. Mc M ( d / 2 ) 32 M 32 (11)( 250 ) σx = = = = = 28 011 psi = 28.0 kpsi 3 π d 4 / 64 π d 3 I π (1) τ xz = Tr T ( d / 2 ) 16T 16 (12 )( 250 ) = = = = 15 279 psi = 15.3 kpsi 3 J π d 4 / 32 π d 3 π (1) Shigley’s MED, 10th edition Chapter 3 Solutions, Page 65/100 www.konkur.in (c) 28.0 2 2 σ 28.0 ± x + (τ xz ) = ± + (15.3) 2 2 2 2 σ 1 = 34.7 kpsi Ans. σ1 , σ 2 = σx 2 σ 2 = −6.7 kpsi 2 Ans. 2 2 σ 28.0 τ max = x + (τ xz ) = + (15.3) = 20.7 kpsi 2 2 2 2 Ans. ____________________________________________________________________________ 3-85 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) = 3600 lbf·in Mz = (250)(11) = 2750 lbf·in M tot = M y2 + M z2 = ( 3600 ) + ( 2750 ) 2 M z My θ = tan −1 2 = 4530 lbf ⋅ in 2750 = tan −1 = 37.4º 3600 The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 37.4º CCW from the z axis. (b) M c M ( d / 2 ) 32 M tot 32 ( 4530 ) σ x ,bend = tot = tot 4 = = = 46 142 psi = 46.1 kpsi 3 π d / 64 πd3 I π (1) σ x ,axial = Fx Fx 300 = = = 382 psi = 0.382 kpsi 2 A π d / 4 π (1)2 / 4 σ x = σ x ,axial + σ x ,bend = 46 142 + 382 = 46 524 psi = 46.5 kpsi τ= Tr 16T 16 (12 )( 250 ) = = = 15 279 psi = 15.3 kpsi 3 J πd3 π (1) Shigley’s MED, 10th edition Chapter 3 Solutions, Page 66/100 www.konkur.in (c) σx 46.5 2 σ 46.5 σ1, σ 2 = ± x +τ 2 = ± + (15.3) 2 2 2 2 Ans. σ 1 = 51.1 kpsi 2 σ 2 = −4.58 kpsi 2 Ans. 2 σ 46.5 τ max = x + τ 2 = + (15.3) = 27.8 kpsi 2 2 2 2 Ans. ____________________________________________________________________________ 3-86 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) – (–100)(11) = 4700 lbf·in Mz = (250)(11) = 2750 lbf·in M tot = M y2 + M z2 = ( 4700 ) + ( 2750 ) 2 M z My θ = tan −1 2 = 5445 lbf ⋅ in 2750 = tan −1 = 30.3º 4700 The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 30.3º CCW from the z axis. (b) M c M ( d / 2 ) 32 M tot 32 ( 5445 ) = = = 55 462 psi = 55.5 kpsi σ x ,bend = tot = tot 4 3 I π d / 64 πd3 π (1) Shigley’s MED, 10th edition Chapter 3 Solutions, Page 67/100 www.konkur.in σ x ,axial = Fx Fx 300 = = = 382 psi = 0.382 kpsi 2 A π d / 4 π (1)2 / 4 σ x = σ x ,axial + σ x ,bend = 55 462 + 382 = 55 844 psi = 55.8 kpsi τ= Tr 16T 16 (12 )( 250 ) = = = 15 279 psi = 15.3 kpsi 3 J πd3 π (1) (c) σx 55.8 2 σ 55.8 ± x +τ 2 = ± σ1 , σ 2 = + (15.3) 2 2 2 2 σ 1 = 59.7 kpsi Ans. 2 σ 2 = −3.92 kpsi 2 Ans. 2 σ 55.8 τ max = x + τ 2 = + (15.3) = 31.8 kpsi 2 2 2 2 Ans. ____________________________________________________________________________ 3-87 (a) The cross section at A will experience bending, torsion, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface, where the stress concentration will also be applicable. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be at either the top (compression) or the bottom (tension) on the y axis. We’ll select the bottom element for this analysis. (b) Transverse shear is zero at the critical stress elements on the top and bottom surfaces. r / d = 0.125 /1 = 0.125 D / d = 1.5 /1 = 1.5 Fig. A-15-8 K t ,torsion = 1.39 K t ,bend = 1.59 σ x = K t ,bend Fig. A-15-9 32 (11)( 250 ) Mc 32 M = K t ,bend = (1.59) = 44 538 psi = 44.5 kpsi 3 I πd3 π (1) τ xz = K t ,torsion 16 (12 )( 250 ) Tr 16T = K t ,torsion = (1.39) = 21 238 psi = 21.2 kpsi 3 3 J πd π (1) Shigley’s MED, 10th edition Chapter 3 Solutions, Page 68/100 www.konkur.in (c) 44.5 2 2 σ 44.5 ± x + (τ xz ) = ± + ( 21.2 ) 2 2 2 2 Ans. σ 1 = 53.0 kpsi σ1, σ 2 = σx 2 σ 2 = −8.48 kpsi 2 Ans. 2 2 σ 44.5 τ max = x + (τ xz ) = + ( 21.2 ) = 30.7 kpsi 2 2 2 2 Ans. ____________________________________________________________________________ 3-88 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface, where the stress concentration will also be applicable. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) = 3600 lbf·in Mz = (250)(11) = 2750 lbf·in M tot = M y2 + M z2 = ( 3600 ) + ( 2750 ) 2 M z My θ = tan −1 2 = 4530 lbf ⋅ in 2750 = tan −1 = 37.4º 3600 The combined bending moment vector is at an angle of 37.4º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 37.4º CCW from the z axis. (b) r / d = 0.125 /1 = 0.125 D / d = 1.5 /1 = 1.5 Fig. A-15-7 K t , axial = 1.75 K t ,torsion = 1.39 Fig. A-15-8 K t ,bend = 1.59 Fig. A-15-9 Shigley’s MED, 10th edition Chapter 3 Solutions, Page 69/100 www.konkur.in σ x ,bend = K t ,bend 32 ( 4530 ) Mc 32 M = K t ,bend = (1.59) = 73 366 psi = 73.4 kpsi 3 3 I πd π (1) σ x ,axial = K t ,axial Fx 300 = (1.75 ) = 668 psi = 0.668 kpsi 2 A π (1) / 4 σ x = σ x ,axial + σ x ,bend = 73 366 + 668 = 74 034 psi = 74.0 kpsi τ = K t ,torsion 16 (12 )( 250 ) Tr 16T = K t ,torsion = (1.39) = 21 238 psi = 21.2 kpsi 3 3 J πd π (1) (c) 74.0 2 σ 74.0 ± x +τ 2 = ± + ( 21.2 ) 2 2 2 2 Ans. σ 1 = 79.6 kpsi σ1 , σ 2 = σx 2 σ 2 = −5.64 kpsi 2 Ans. 2 σ 74.0 τ max = x + τ 2 = + ( 21.2 ) = 42.6 kpsi 2 2 2 2 Ans. ____________________________________________________________________________ 3-89 (a) The cross section at A will experience bending, torsion, axial, and transverse shear. Both torsional shear stress and bending stress will be a maximum on the outer surface, where the stress concentration is also applicable. The transverse shear will be very small compared to bending and torsion, due to the reasonably high length to diameter ratio, so it will not dominate the determination of the critical location. The critical stress element will be on the outer surface, with its critical location determined by the plane of the combined bending moments. My = (300)(12) – (–100)(11) = 4700 lbf·in Mz = (250)(11) = 2750 lbf·in M tot = M y2 + M z2 = ( 4700 ) + ( 2750 ) 2 M z My θ = tan −1 Shigley’s MED, 10th edition 2 = 5445 lbf ⋅ in 2750 = tan −1 = 30.3º 4700 Chapter 3 Solutions, Page 70/100 www.konkur.in The combined bending moment vector is at an angle of 30.3º CCW from the y axis. The critical bending stress location will be 90º CCW from this vector, where the tensile bending stress is additive with the tensile axial stress. The critical stress element is therefore on the outer surface, at an angle of 30.3º CCW from the z axis. (b) r / d = 0.125 /1 = 0.125 D / d = 1.5 /1 = 1.5 Fig. A-15-7 K t , axial = 1.75 K t ,torsion = 1.39 Fig. A-15-8 K t ,bend = 1.59 Fig. A-15-9 σ x ,bend = K t ,bend 32 ( 5445 ) Mc 32 M = K t ,bend = (1.59) = 88185 psi = 88.2 kpsi 3 3 I πd π (1) σ x ,axial = K t ,axial Fx 300 = (1.75 ) = 668 psi = 0.668 kpsi 2 A π (1) / 4 σ x = σ x ,axial + σ x ,bend = 88 185 + 668 = 88 853 psi = 88.9 kpsi τ = K t ,torsion 16 (12 )( 250 ) Tr 16T = K t ,torsion = (1.39) = 21 238 psi = 21.2 kpsi 3 3 πd J π (1) (c) σx 88.9 2 σ 88.9 σ1 , σ 2 = ± x +τ 2 = ± + ( 21.2 ) 2 2 2 2 Ans. σ 1 = 93.7 kpsi 2 σ 2 = −4.80 kpsi 2 Ans. 2 σ 88.9 τ max = x + τ 2 = Ans. + ( 21.2 ) = 49.2 kpsi 2 2 ____________________________________________________________________________ 2 3-90 2 (a) M = F(p / 4), c = p / 4, I = bh3 / 12, b = π dr nt, h = p / 2 Shigley’s MED, 10th edition Chapter 3 Solutions, Page 71/100 www.konkur.in F ( p / 4 ) ( p / 4 ) Mc Fp 2 σb = ± =± =± 3 I bh3 /12 16 (π d r nt )( p / 2 ) / 12 6F Ans. π d r nt p F F 4F (b) σ a = − = − 2 =− 2 A π dr / 4 π dr σb = ± Ans. Tr T ( d r / 2 ) 16T = = Ans. π d r4 / 32 π d r3 J (c) The bending stress causes compression in the x direction. The axial stress causes compression in the y direction. The torsional stress shears across the y face in the negative z direction. τt = (d) Analyze the stress element from part (c) using the equations developed in parts (a) and (b). d r = d − p = 1.5 − 0.25 = 1.25 in σ x = σb = − σ y = σa = − 6 (1500 ) 6F =− = − 4584 psi = − 4.584 kpsi π d r nt p π (1.25 )( 2 )( 0.25 ) 4 (1500 ) 4F =− = − 1222 psi = − 1.222 kpsi 2 π dr π (1.252 ) τ yz = −τ t = − 16 ( 235 ) 16T =− = − 612.8 psi = − 0.6128 kpsi 3 π dr π (1.253 ) Use Eq. (3-15) for the three-dimensional stress element. 2 2 σ 3 − ( −4.584 − 1.222 ) σ 2 + ( −4.584 )( −1.222 ) − ( −0.6128 ) σ − − ( −4.584 )( −0.6128 ) = 0 σ + 5.806σ + 5.226σ − 1.721 = 0 3 2 The roots are at 0.2543, – 4.584, and –1.476. Thus, the ordered principal stresses are σ1 = 0.2543 kpsi, σ2 = –1.476 kpsi, and σ3 = – 4.584 kpsi. Ans. From Eq. (3-16), the principal shear stresses are Shigley’s MED, 10th edition Chapter 3 Solutions, Page 72/100 www.konkur.in 0.2543 − ( −1.476 ) = 0.8652 kpsi Ans. 2 2 σ − σ 3 ( −1.476 ) − ( −4.584 ) τ 2/3 = 2 = = 1.554 kpsi Ans. 2 2 0.2543 − ( −4.584 ) σ −σ = 2.419 kpsi Ans. τ 1/3 = 1 3 = 2 2 ____________________________________________________________________________ τ 1/2 = 3-91 σ1 − σ 2 = As shown in Fig. 3-32, the maximum stresses occur at the inside fiber where r = ri. Therefore, from Eq. (3-50) ri2 pi ro2 1 + ro2 − ri2 ri2 r2 + r2 = pi o2 i2 Ans. r −r o i r2 p r2 σ r ,max = 2i i 2 1 − o2 = − pi Ans. ro − ri ri ______________________________________________________________________________ σ t ,max = 3-92 If pi = 0, Eq. (3-49) becomes − po ro2 − ri 2 ro2 po / r 2 σt = ro2 − ri 2 po ro2 ri 2 1 + ro2 − ri 2 r 2 The maximum tangential stress occurs at r = ri. So =− σ t ,max = − 2 po ro2 Ans. ro2 − ri 2 For σr, we have − po ro2 − ri 2 ro2 po / r 2 σr = ro2 − ri 2 po ro2 ri 2 − 1 ro2 − ri 2 r 2 So σr = 0 at r = ri. Thus at r = ro p r2 r2 − r2 σ r ,max = 2 o o 2 i 2 o = − po Ans. ro − ri ro ______________________________________________________________________________ = Shigley’s MED, 10th edition Chapter 3 Solutions, Page 73/100 www.konkur.in 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 (σ t )av = σ 1 = σ 2 = p (π / 4 ) di2 π di t = pdi 4t Ans. The radial stress on the inner surface of the shell is, σ3 = − p Ans. ______________________________________________________________________________ 3-94 σt > σl > σr τmax = (σt − σr)/2 at r = ri τ max ro2 pi 1 ri 2 pi ro2 ri 2 pi ro2 = 2 2 1 + 2 − 2 2 1 − 2 = 2 2 2 ro − ri ri ro − ri ri ro − ri ro2 − ri 2 32 − 2.752 = (10 000) = 1597 psi Ans. τ max ro 2 32 ______________________________________________________________________________ ⇒ 3-95 pi = σt > σl > σr τmax = (σt − σr)/2 at r = ri ri 2 pi ro2 ro2 pi 1 ri 2 pi ro2 ri 2 pi ro2 τ max = 2 2 1 + 2 − 2 2 1 − 2 = 2 2 2 = 2 2 2 ro − ri ri ro − ri ri ro − ri ri ro − ri ⇒ ri = ro (τ max − pi ) (25 − 4)106 = 100 = 91.7 mm 25 (106 ) τ max t = ro − ri = 100 − 91.7 = 8.3 mm Ans. ______________________________________________________________________________ 3-96 σt > σl > σr τmax = (σt − σr)/2 at r = ri 1 ri 2 pi ro2 1 + 2 ro2 − ri 2 ri 2 τ max = ri 2 pi ro2 − 2 2 1 − 2 ro − ri ri ri 2 pi ro2 = 2 2 2 ro − ri ri ro2 pi = 2 2 ro − ri 42 (500) = 2 = 4129 psi Ans. 4 − 3.752 ______________________________________________________________________________ 3-97 From Eq. (3-49) with pi = 0, Shigley’s MED, 10th edition Chapter 3 Solutions, Page 74/100 www.konkur.in σt = − ro2 po ri 2 1 + ro2 − ri 2 r 2 ro2 po ri 2 σ r = − 2 2 1 − 2 ro − ri r σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro ro2 po ri 2 ri 2 po 1 ro2 po ri 2 ro2 po ri 2 1 − + 1 + = = 2 ro2 − ri 2 ro 2 ro2 − ri 2 ro 2 ro2 − ri 2 ro 2 ro2 − ri 2 τ max = − ro2 − ri 2 32 − 2.752 τ = (10 000) = 1900 psi Ans. max ri 2 2.752 ______________________________________________________________________________ ⇒ 3-98 po = From Eq. (3-49) with pi = 0, r2 p r2 σ t = − 2o o 2 1 + i 2 ro − ri r σr = − ro2 po ri 2 1 − ro2 − ri 2 r 2 σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro 1 ro2 po ri 2 ro2 po ri 2 ro2 po ri 2 ri 2 po 1 − + 1 + = = 2 ro2 − ri 2 ro 2 ro2 − ri 2 ro 2 ro2 − ri 2 ro 2 ro2 − ri 2 τ max = − ⇒ ri = ro τ max (τ max + po ) = 100 25 (106 ) ( 25 + 4 )106 = 92.8 mm t = ro − ri = 100 − 92.8 = 7.2 mm Ans. ______________________________________________________________________________ 3-99 From Eq. (3-49) with pi = 0, ro2 po ri 2 σ t = − 2 2 1 + 2 ro − ri r σr = − ro2 po ri 2 1 − ro2 − ri 2 r 2 σt > σl > σr, and since σt and σr are negative, τmax = (σr − σt)/2 at r = ro Shigley’s MED, 10th edition Chapter 3 Solutions, Page 75/100 www.konkur.in τ max ro2 po ri 2 ri 2 po 1 ro2 po ri 2 ro2 po ri 2 = − 2 2 1 − 2 + 2 2 1 + 2 = 2 2 2 = 2 2 2 ro − ri ro ro − ri ro ro − ri ro ro − ri 3.752 (500) = 3629 psi Ans. 42 − 3.752 ______________________________________________________________________________ = 3-100 From Table A-20, Sy=490 MPa From Eq. (3-49) with pi = 0, σt = − ro2 po ri 2 1 + ro2 − ri 2 r 2 Maximum will occur at r = ri [0.8(−490)] ( 25 − 19 ) σ (r − r ) 2ro2 po σ t ,max = − 2 2 ⇒ po = − t ,max o2 i = − = 82.8 MPa Ans. ro − ri 2ro 2(252 ) ______________________________________________________________________________ 2 2 2 2 3-101 From Table A-20, Sy = 71 kpsi From Eq. (3-49) with pi = 0, σt = − ro2 po ri 2 1 + ro2 − ri 2 r 2 Maximum will occur at r = ri σ t ,max ( ro − ri ) [ 0.8(−71)] (1 − 0.75 ) 2r 2 p =− = 12.4 kpsi Ans. σ t ,max = − 2 o o2 ⇒ po = − 2 ro − ri 2ro 2(12 ) ______________________________________________________________________________ 2 2 2 2 3-102 From Table A-20, Sy=490 MPa From Eq. (3-50) ri 2 pi ro2 σ t = 2 2 1 + 2 ro − ri r Maximum will occur at r = ri σ t ,max ⇒ 2 2 ri 2 pi ro2 pi ( ro + ri ) = 2 2 1 + 2 = ro − ri ri ro2 − ri 2 pi = σ t ,max (ro2 − ri 2 ) 0.8(490)] (252 − 192 ) [ = = 105 MPa Ans. ro2 + ri 2 (252 + 192 ) ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 76/100 www.konkur.in 3-103 From Table A-20, Sy =71 MPa From Eq. (3-50) σt = ri 2 pi ro2 1 + ro2 − ri 2 r 2 Maximum will occur at r = ri r 2 p r 2 p (r 2 + r 2 ) σ t ,max = 2i i 2 1 + o2 = i 2o 2i ro − ri ri ro − ri ⇒ pi = σ t ,max (ro2 − ri 2 ) = [ 0.8(71)] (12 − 0.752 ) = 15.9 ksi Ans. ro2 + ri 2 (12 + 0.752 ) ______________________________________________________________________________ 3-104 The longitudinal stress will be due to the weight of the vessel above the maximum stress point. From Table A-5, the unit weight of steel is γs = 0.282 lbf/in3. The area of the wall is Awall = (π /4)(3602 − 358.52) = 846. 5 in2 The volume of the wall and dome are Vwall = Awall h = 846.5 (720) = 609.5 (103) in3 Vdome = (2π /3)(1803 − 179.253) = 152.0 (103) in3 The weight of the structure on the wall area at the tank bottom is W = γs Vtotal = 0.282(609.5 +152.0) (103) = 214.7(103) lbf 214.7 (103 ) W =− = −254 psi σl = − Awall 846.5 The maximum pressure will occur at the bottom of the tank, pi = γwater h. From Eq. (3-50) with r = ri ro2 + ri 2 ri 2 pi ro2 1 + = p i 2 2 ro2 − ri 2 ri 2 ro − ri 1 ft 2 1802 + 179.252 = 62.4(55) = 5708 5710 psi Ans. 2 2 2 144 in 180 − 179.25 1 ft 2 r2 p r2 σ r = 2i i 2 1 − o2 = − pi = −62.4(55) = −23.8 psi Ans. 2 ro − ri ri 144 in Note: These stresses are very idealized as the floor of the tank will restrict the values calculated. σt = Since σ1 ≥ σ2 ≥ σ3, σ1 = σt = 5708 psi, σ2 = σr = − 24 psi andσ3 = σl = − 254 psi. From Eq. (3-16), Shigley’s MED, 10th edition Chapter 3 Solutions, Page 77/100 www.konkur.in 5708 + 254 = 2981 ≈ 2980 psi 2 5708 + 24 τ1 2 = = 2866 ≈ 2870 psi Ans. 2 −24 + 254 τ2 3 = = 115 psi 2 ______________________________________________________________________________ τ1 3 = 3-105 Stresses from additional pressure are, Eq. (3-51), 50 (179.252 ) = 5963 psi (σ l )50psi = 2 180 − 179.252 (σr)50 psi = − 50 psi Eq. (3-50) 1802 + 179.252 = 11 975 psi (σ t )50psi = 50 2 180 − 179.252 Adding these to the stresses found in Prob. 3-104 gives σt = 5708 + 11 975 = 17683 psi = 17.7 kpsi Ans. σr = − 23.8 − 50 = − 73.8 psi Ans. σl = − 254 + 5963 = 5709 psi Ans. Note: These stresses are very idealized as the floor of the tank will restrict the values calculated. From Eq. (3-16) 17 683 + 73.8 τ1 3 = = 8879 psi 2 17 683 − 5709 τ1 2 = = 5987 psi Ans. 2 5709 + 23.8 = 2866 psi τ2 3 = 2 ______________________________________________________________________________ 3-106 Since σt and σr are both positive and σt > σr τ max = (σ t ) max 2 From Eq. (3-55), σt is maximum at r = ri = 0.3125 in. The term 3 +ν ρω 8 2 Shigley’s MED, 10th edition 0.282 2π ( 5000 ) 3 + 0.292 = = 82.42 lbf/in 60 8 386 2 Chapter 3 Solutions, Page 78/100 www.konkur.in (σ t )max τ max = ( )( ) 0.31252 2.752 1 + 3(0.292) 2 2 2 = 82.42 0.3125 + 2.75 + − 0.3125 3 + 0.292 0.31252 = 1260 psi 1260 = 630 psi 2 ( ) Ans. Radial stress: 2 2 ri2 ro2 σ r = k ri + ro − 2 − r 2 r Maxima: r 2r 2 dσ r = k 2 i 3o − 2r = 0 ⇒ r = ri ro = 0.3125(2.75) = 0.927 in dr r 0.31252 2.752 2 2 2 − 0.927 (σ r )max = 82.42 0.3125 + 2.75 − 0.927 2 = 490 psi Ans. ______________________________________________________________________________ ( ) 3-107 ω = 2π (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 3 + 0.24 2 2 1 + 3(0.24) ( 0.01)2 (10)−6 ( 0.01) + (0.125) + (0.125) − 3 + 0.24 8 = 1.85 MPa Ans. ______________________________________________________________________________ σ t = 3320(209.4)2 3-108 ω = 2π (12 000)/60 = 1256.6 rad/s, ( 5 / 16 ) ρ= = 6.749 10−4 lbf ⋅ s 2 / in 4 2 2 386 (1 16 )(π 4 ) 5 − 0.75 ( ) ( ) The maximum shear stress occurs at bore where τmax = σt /2. From Eq. (3-55) 1 + 3(0.20) 2 3 + 0.20 2 2 2 (σ t )max = 6.749(10−4 ) (1256.6 ) (0.375)2 0.375 + 2.5 + 2.5 − 3 + 0.20 8 = 5360 psi τmax = 5360 / 2 = 2680 psi Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 79/100 www.konkur.in 3-109 ω = 2π (3500)/60 = 366.5 rad/s, mass of blade = m = ρV = (0.282 / 386) [1.25(30)(0.125)] = 3.425(10−3) lbf⋅s2/in F = (m/2) ω 2r = [3.425(10−3)/2]( 366.52)(7.5) = 1725 lbf Anom = (1.25 − 0.5)(1/8) = 0.093 75 in2 σnom = F/ Anom = 1725/0.093 75 = 18 400 psi Ans. Note: Stress concentration Fig. A-15-1 gives Kt = 2.25 which increases σmax and fatigue. ______________________________________________________________________________ 3-110 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57), 207(109 )δ (0.052 − 0.0252 )(0.0252 − 0) −9 3 10 = 3.105(10 )δ 3 2 2(0.025) (0.05 − 0) where p is in MPa and δ is in mm. ( p= ) (1) Maximum interference, 1 δ max = [50.042 − 50.000] = 0.021 mm Ans. 2 Minimum interference, 1 δ min = [50.026 − 50.025] = 0.0005 mm Ans. 2 From Eq. (1) pmax = 3.105(103)(0.021) = 65.2 MPa Ans. pmin = 3.105(103)(0.0005) = 1.55 MPa Ans. ______________________________________________________________________________ 3-111 ν = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in Eq. (3-57), 30(106 )δ (22 − 12 )(12 − 0) 7 p= = 1.125(10 )δ 2(13 ) (22 − 0) (1) where p is in psi and δ is in inches. Maximum interference, 1 δ max = [2.0016 − 2.0000] = 0.0008 in 2 Minimum interference, Shigley’s MED, 10th edition Ans. Chapter 3 Solutions, Page 80/100 www.konkur.in 1 2 δ min = [2.0010 − 2.0010] = 0 Ans. From Eq. (1), pmax = 1.125(107)(0.0008) = 9 000 psi Ans. pmin = 1.125(107)(0) = 0 Ans. ______________________________________________________________________________ 3-112 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57), 207(109 )δ p= 2(0.025)3 (0.052 − 0.0252 )(0.0252 − 0) −9 3 10 = 3.105(10 )δ 2 (0.05 − 0) where p is in MPa and δ is in mm. ( ) (1) Maximum interference, 1 δ max = [50.059 − 50.000] = 0.0295 mm Ans. 2 Minimum interference, 1 δ min = [50.043 − 50.025] = 0.009 mm Ans. 2 From Eq. (1) pmax = 3.105(103)(0.0295) = 91.6 MPa Ans. pmin = 3.105(103)(0.009) = 27.9 MPa Ans. ______________________________________________________________________________ 3-113 ν = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in Eq. (3-57), 30(106 )δ (22 − 12 )(12 − 0) 7 p= = 1.125(10 )δ 3 2 2(1 ) (2 − 0) (1) where p is in psi and δ is in inches. Maximum interference, 1 δ max = [2.0023 − 2.0000] = 0.00115 in 2 Minimum interference, 1 2 δ min = [2.0017 − 2.0010] = 0.00035 Ans. Ans. From Eq. (1), Shigley’s MED, 10th edition Chapter 3 Solutions, Page 81/100 www.konkur.in pmax = 1.125(107)(0.00115) = 12 940 psi pmin = 1.125(107)(0.00035) = 3 938 Ans. Ans. ______________________________________________________________________________ 3-114 ν = 0.292, E = 207 GPa, ri = 0, R = 25 mm, ro = 50 mm Eq. (3-57), 207(109 )δ (0.052 − 0.0252 )(0.0252 − 0) −9 3 10 = 3.105(10 )δ 3 2 2(0.025) (0.05 − 0) where p is in MPa and δ is in mm. ( p= ) (1) Maximum interference, 1 δ max = [50.086 − 50.000] = 0.043 mm Ans. 2 Minimum interference, 1 δ min = [50.070 − 50.025] = 0.0225 mm Ans. 2 From Eq. (1) pmax = 3.105(103)(0.043) = 134 MPa Ans. pmin = 3.105(103)(0.0225) = 69.9 MPa Ans. ______________________________________________________________________________ 3-115 ν = 0.292, E = 30 Mpsi, ri = 0, R = 1 in, ro = 2 in Eq. (3-57), 30(106 )δ (22 − 12 )(12 − 0) 7 p= = 1.125(10 )δ 3 2 2(1 ) (2 − 0) (1) where p is in psi and δ is in inches. Maximum interference, 1 δ max = [2.0034 − 2.0000] = 0.0017 in 2 Minimum interference, 1 2 δ min = [2.0028 − 2.0010] = 0.0009 Ans. Ans. From Eq. (1), pmax = 1.125(107)(0.0017) = 19 130 psi Ans. pmin = 1.125(107)(0.0009) = 10 130 Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 82/100 www.konkur.in 3-116 From Table A-5, Ei = Eo = 30 Mpsi, νi = νo = 0.292. ri = 0, R = 1 in, ro = 1.5 in 1 The radial interference is δ = ( 2.002 − 2.000 ) = 0.001 in Ans. 2 Eq. (3-57), 2 2 2 2 6 2 2 2 Eδ ro − R R − ri 30 10 0.001 1.5 − 1 1 − 0 = p= 3 3 2 2R ro2 − ri 2 2 1 1.5 − 0 = 8333 psi = 8.33 kpsi Ans. ( )( ) ( ) ( ) ( ( )( ) ) The tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. R2 + r 2 12 + 02 (σ t )i r = R = − p 2 i 2 = −(8333) 2 2 = −8333 psi = −8.33 kpsi Ans. R − ri 1 −0 ro2 + R 2 1.52 + 12 (σ t )o r = R = p 2 = (8333) 2 2 = 21 670 psi = 21.7 kpsi Ans. ro − R 2 1.5 − 1 ______________________________________________________________________________ 3-117 From Table A-5, Ei = 30 Mpsi, Eo =14.5 Mpsi, νi = 0.292, νo = 0.211. ri = 0, R = 1 in, ro = 1.5 in 1 The radial interference is δ = ( 2.002 − 2.000 ) = 0.001 in Ans. 2 Eq. (3-56), p= p= δ 1 r +R 1 R +ν o + 2 Ei Eo r − R 2 o 2 o 2 R 2 + ri 2 2 2 −ν i R − ri 0.001 1.52 + 12 12 + 02 1 1 1 + 0.211 + − 0.292 2 2 2 2 6 6 1 − 0 30 10 14.5 10 1.5 − 1 ( ) = 4599 psi Ans. ( ) The tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. R2 + r 2 12 + 02 (σ t )i r = R = − p 2 i2 = −(4599) 2 2 = −4599 psi Ans. R − ri 1 −0 ro2 + R 2 1.52 + 12 (σ t )o r = R = p 2 = (4599) 2 2 = 11 960 psi Ans. ro − R 2 1.5 − 1 ______________________________________________________________________________ 3-118 From Table A-5, Ei = Eo = 30 Mpsi, νi = νo = 0.292. ri = 0, R = 0.5 in, ro = 1 in Shigley’s MED, 10th edition Chapter 3 Solutions, Page 83/100 www.konkur.in The minimum and maximum radial interferences are 1 δ min = (1.002 − 1.002 ) = 0.000 in Ans. 2 1 δ max = (1.003 − 1.001) = 0.001 in Ans. 2 Since the minimum interference is zero, the minimum pressure and tangential stresses are zero. Ans. The maximum pressure is obtained from Eq. (3-57). ( )( 2 2 2 2 Eδ ro − R R − ri p= 3 2R ro2 − ri 2 ( ) ( ) ( ) )( 30 106 0.001 12 − 0.52 0.52 − 0 p= 2 0.53 12 − 0 ( ) ) = 22 500 psi Ans The maximum tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. R2 + r 2 0.52 + 02 (σ t )i r = R = − p 2 i 2 = −(22 500) 2 2 = −22 500 psi Ans. R − ri 0.5 − 0 ro2 + R 2 12 + 0.52 = (22 500) = 37 500 psi Ans. r =R ro2 − R 2 12 − 0.52 ______________________________________________________________________________ (σ t )o =p 3-119 From Table A-5, Ei = 10.4 Mpsi, Eo =30 Mpsi, νi = 0.333, νo = 0.292. ri = 0, R = 1 in, ro = 1.5 in The minimum and maximum radial interferences are 1 δ min = [2.003 − 2.002] = 0.0005 in Ans. 2 1 δ max = [2.006 − 2.000] = 0.003 in Ans. 2 Eq. (3-56), Shigley’s MED, 10th edition Chapter 3 Solutions, Page 84/100 www.konkur.in p= p= δ 1 r +R 1 R 2 + ri 2 R + ν 2 2 −ν i o + 2 Ei R − ri Eo r − R 2 o 2 o 2 δ 1 1.52 + 12 12 + 02 1 1 + 0.292 + − 0.333 2 2 2 2 6 6 30 10 1.5 − 1 10.4 10 1 − 0 ( ) p = 6.229 (10 ) δ psi p = 6.229 (10 ) δ p = 6.229 (10 ) δ ( ) 6 6 min min 6 max max Ans. = 6.229 (10 6 ) ( 0.0005 ) = 3114.6 psi = 3.11 kpsi = 6.229 (106 ) ( 0.003) = 18 687 psi = 18.7 kpsi Ans. Ans. The tangential stresses at the interface for the inner and outer members are given by Eqs. (3-58) and (3-59), respectively. Minimum interference: R2 + r 2 12 + 02 (σ t )i min = − pmin 2 i 2 = −(3.11) 2 2 = −3.11 kpsi Ans. R − ri 1 −0 (σ t )o min = pmin ro2 + R 2 1.52 + 12 = (3.11) = 8.09 kpsi ro2 − R 2 1.52 − 12 Ans. Maximum interference: R2 + r 2 12 + 02 (σ t )i max = − pmax 2 i 2 = −(18.7) 2 2 = −18.7 kpsi Ans. R − ri 1 −0 ro2 + R 2 1.52 + 12 = (18.7) = 48.6 kpsi Ans. max ro2 − R 2 1.52 − 12 ______________________________________________________________________________ (σ t )o = pmax 3-120 d = 20 mm, ri = 37.5 mm, ro = 57.5 mm From Table 3-4, for R = 10 mm, rc = 37.5 + 10 = 47.5 mm rn = ( 102 2 47.5 − 47.52 − 102 ) = 46.96772 mm e = rc − rn = 47.5 − 46.96772 = 0.53228 mm ci = rn − ri = 46.9677 − 37.5 = 9.4677 mm co = ro − rn = 57.5 − 46.9677 = 10.5323 mm A = π d 2 / 4 = π (20) 2 / 4 = 314.16 mm 2 M = Frc = 4000(47.5) = 190 000 N ⋅ mm Using Eq. (3-65) for the bending stress, and combining with the axial stress, Shigley’s MED, 10th edition Chapter 3 Solutions, Page 85/100 www.konkur.in F Mci 4000 190 000(9.4677) + = + = 300 MPa A Aeri 314.16 314.16(0.53228)(37.5) σi = Ans. F Mco 4000 190 000(10.5323) − = − = −195 MPa Ans. A Aero 314.16 314.16(0.53228)(57.5) ______________________________________________________________________________ σo = 3-121 d = 0.75 in, ri = 1.25 in, ro = 2.0 in From Table 3-4, for R = 0.375 in, rc = 1.25 + 0.375 = 1.625 in rn = 0.3752 ( 2 1.625 − 1.6252 − 0.3752 ) = 1.60307 in e = rc − rn = 1.625 − 1.60307 = 0.02193 in ci = rn − ri = 1.60307 − 1.25 = 0.35307 in co = ro − rn = 2.0 − 1.60307 = 0.39693 in A = π d 2 / 4 = π (0.75) 2 / 4 = 0.44179 in 2 M = Frc = 750(1.625) = 1218.8 lbf ⋅ in Using Eq. (3-65) for the bending stress, and combining with the axial stress, σi = F Mci 750 1218.8(0.35307) + = + = 37 230 psi = 37.2 kpsi A Aeri 0.44179 0.44179(0.02193)(1.25) Ans. F Mco 750 1218.8(0.39693) − = − = −23 269 psi = −23.3 kpsi Ans. A Aero 0.44179 0.44179(0.02193)(2.0) ______________________________________________________________________________ σo = 3-122 d = 6 mm, ri = 10 mm, ro = 16 mm From Table 3-4, for R = 3 mm, rc = 10 + 3 = 13 mm rn = ( 32 2 13 − 132 − 32 ) = 12.82456 mm e = rc − rn = 13 − 12.82456 = 0.17544 mm ci = rn − ri = 12.82456 − 10 = 2.82456 mm co = ro − rn = 16 − 12.82456 = 3.17544 mm A = π d 2 / 4 = π (6) 2 / 4 = 28.2743 mm 2 M = Frc = 300(13) = 3900 N ⋅ mm Using Eq. (3-65) for the bending stress, and combining with the axial stress, Shigley’s MED, 10th edition Chapter 3 Solutions, Page 86/100 www.konkur.in F Mci 300 3900(2.82456) + = + = 233 MPa A Aeri 28.2743 28.2743(0.17544)(10) σi = Ans. F Mco 300 3900(3.17544) − = − = −145 MPa Ans. A Aero 28.2743 28.2743(0.17544)(16) ______________________________________________________________________________ σo = 3-123 d = 6 mm, ri = 10 mm, ro = 16 mm From Table 3-4, for R = 3 mm, rc = 10 + 3 = 13 mm rn = ( 32 2 13 − 132 − 32 ) = 12.82456 mm e = rc − rn = 13 − 12.82456 = 0.17544 mm ci = rn − ri = 12.82456 − 10 = 2.82456 mm co = ro − rn = 16 − 12.82456 = 3.17544 mm A = π d 2 / 4 = π (6) 2 / 4 = 28.2743 mm 2 The angle θ of the line of radius centers is R+d /2 −1 10 + 6 / 2 o θ = sin −1 = sin = 30 R + d + R 10 + 6 + 10 M = F ( R + d / 2 ) sin θ = 300 (10 + 6 / 2 ) sin 30o = 1950 N ⋅ mm Using Eq. (3-65) for the bending stress, and combining with the axial stress, F sin θ Mci 300 sin 30o 1950(2.82456) + = + = 116 MPa A Aeri 28.2743 28.2743(0.17544)(10) σi = Ans. F sin θ Mco 300 sin 30o 1950(3.17544) − = − = −72.7 MPa Ans. A Aero 28.2743 28.2743(0.17544)(16) Note that the shear stress due to the shear force is zero at the surface. ______________________________________________________________________________ σo = 3-124 d = 0.25 in, ri = 0.5 in, ro = 0.75 in From Table 3-4, for R = 0.125 in, rc = 0.5 + 0.125 = 0.625 in rn = ( 0.1252 2 0.625 − 0.6252 − 0.1252 ) = 0.618686 in e = rc − rn = 0.625 − 0.618686 = 0.006314 in ci = rn − ri = 0.618686 − 0.5 = 0.118686 in co = ro − rn = 0.75 − 0.618686 = 0.131314 in Shigley’s MED, 10th edition Chapter 3 Solutions, Page 87/100 www.konkur.in A = π d 2 / 4 = π (0.25)2 / 4 = 0.049087 in 2 M = Frc = 75(0.625) = 46.875 lbf ⋅ in Using Eq. (3-65) for the bending stress, and combining with the axial stress, σi = F Mci 75 46.875(0.118686) + = + = 37 428 psi = 37.4 kpsi A Aeri 0.049087 0.049087(0.006314)(0.5) Ans. F Mco 75 46.875(0.131314) − = − = −24 952 psi = −25.0 kpsi Ans. A Aero 0.049087 0.049087(0.006314)(0.75) ______________________________________________________________________________ σo = 3-125 d = 0.25 in, ri = 0.5 in, ro = 0.75 in From Table 3-4, for R = 0.125 in, rc = 0.5 + 0.125 = 0.625 in rn = ( 0.1252 2 0.625 − 0.6252 − 0.1252 ) = 0.618686 in e = rc − rn = 0.625 − 0.618686 = 0.006314 in ci = rn − ri = 0.618686 − 0.5 = 0.118686 in co = ro − rn = 0.75 − 0.618686 = 0.131314 in A = π d 2 / 4 = π (0.25)2 / 4 = 0.049087 in 2 The angle θ of the line of radius centers is R+d /2 −1 0.5 + 0.25 / 2 o θ = sin −1 = sin = 30 R+d +R 0.5 + 0.25 + 0.5 M = F ( R + d / 2 ) sin θ = 75 ( 0.5 + 0.25 / 2 ) sin 30o = 23.44 lbf ⋅ in Using Eq. (3-65) for the bending stress, and combining with the axial stress, σi = F sin θ Mci 75sin 30o 23.44(0.118686) + = + = 18 716 psi = 18.7 kpsi A Aeri 0.049087 0.049087(0.006314)(0.5) Ans. F sin θ Mco 75sin 30o 23.44(0.131314) − = − = −12 478 psi = −12.5 kpsi Ans. σo = A Aero 0.049087 0.049087(0.006314)(0.75) Note that the shear stress due to the shear force is zero at the surface. ______________________________________________________________________________ 3-126 (a) σ = ± [3(4)][0.5(0.1094)] = ±8021 psi = ±8.02 kpsi Mc =± I (0.75) ( 0.10943 ) /12 Ans. (b) ri = 0.125 in, ro = ri + h = 0.125 + 0.1094 = 0.2344 in From Table 3-4, Shigley’s MED, 10th edition Chapter 3 Solutions, Page 88/100 www.konkur.in rc = 0.125 + (0.5)(0.1094) = 0.1797 in 0.1094 = 0.174006 in ln(0.2344 / 0.125) e = rc − rn = 0.1797 − 0.174006 = 0.005694 in rn = ci = rn − ri = 0.174006 − 0.125 = 0.049006 in co = ro − rn = 0.2344 − 0.174006 = 0.060394 in A = bh = 0.75(0.1094) = 0.08205 in 2 M = −3(4) = −12 lbf ⋅ in The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using Eq. (3-65), σi = Mci −12(0.049006) = = −10 070 psi = −10.1 kpsi Aeri 0.08205(0.005694)(0.125) σo = − Mco −12(0.060394) =− = 6618 psi = 6.62 kpsi Aero 0.08205(0.005694)(0.2344) σ i −10.1 = = 1.26 σ −8.02 σ 6.62 Ko = o = = 0.825 σ 8.02 (c) K i = Ans. Ans. Ans. Ans. ______________________________________________________________________________ 3-127 (a) σ = ± [3(4)][0.5(0.1406)] = ±4856 psi = ±4.86 kpsi Mc =± I (0.75) ( 0.14063 ) / 12 Ans. (b) ri = 0.125 in, ro = ri + h = 0.125 + 0.1406 = 0.2656 in From Table 3-4, rc = 0.125 + (0.5)(0.1406) = 0.1953 in 0.1406 = 0.186552 in ln(0.2656 / 0.125) e = rc − rn = 0.1953 − 0.186552 = 0.008748 in rn = ci = rn − ri = 0.186552 − 0.125 = 0.061552 in co = ro − rn = 0.2656 − 0.186552 = 0.079048 in A = bh = 0.75(0.1406) = 0.10545 in 2 M = −3(4) = −12 lbf ⋅ in The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using Eq. (3-65), Shigley’s MED, 10th edition Chapter 3 Solutions, Page 89/100 www.konkur.in σi = Mci −12(0.061552) = = −6406 psi = −6.41 kpsi Aeri 0.10545(0.008748)(0.125) σo = − Ans. Mco −12(0.079048) =− = 3872 psi = 3.87 kpsi Aero 0.10545(0.008748)(0.2656) Ans. σ i −6.41 = = 1.32 Ans. σ −4.86 σ 3.87 Ko = o = = 0.80 Ans. σ 4.86 ______________________________________________________________________________ (c) K i = 3-128 (a) σ = ± [3(4)][0.5(0.1094)] = ±8021 psi = ±8.02 kpsi Mc =± I (0.75) ( 0.10943 ) /12 Ans. (b) ri = 0.25 in, ro = ri + h = 0.25 + 0.1094 = 0.3594 in From Table 3-4, rc = 0.25 + (0.5)(0.1094) = 0.3047 in 0.1094 = 0.301398 in ln(0.3594 / 0.25) e = rc − rn = 0.3047 − 0.301398 = 0.003302 in rn = ci = rn − ri = 0.301398 − 0.25 = 0.051398 in co = ro − rn = 0.3594 − 0.301398 = 0.058002 in A = bh = 0.75(0.1094) = 0.08205 in 2 M = −3(4) = −12 lbf ⋅ in The negative sign on the bending moment is due to the sign convention shown in Fig. 3-34. Using Eq. (3-65), σi = Mci −12(0.051398) = = −9106 psi = −9.11 kpsi Aeri 0.08205(0.003302)(0.25) σo = − Ans. Mco −12(0.058002) =− = 7148 psi = 7.15 kpsi Aero 0.08205(0.003302)(0.3594) Ans. σ i −9.11 = = 1.14 Ans. σ −8.02 σ 7.15 Ko = o = = 0.89 Ans. σ 8.02 ______________________________________________________________________________ (c) K i = 3-129 ri = 25 mm, ro = ri + h = 25 + 87 = 112 mm, rc = 25 + 87/2 = 68.5 mm The radius of the neutral axis is found from Eq. (3-63), given below. Shigley’s MED, 10th edition Chapter 3 Solutions, Page 90/100 www.konkur.in rn = A ∫ ( dA / r ) For a rectangular area with constant width b, the denominator is ro bdr r ∫ri r = b ln roi Applying this equation over each of the four rectangular areas, dA 45 54.5 92 112 = 9 ln + 31 ln + 31 ln + 9 ln = 16.3769 r 45 25 82.5 92 A = 2 [ 20(9) + 31(9.5)] = 949 mm2 ∫ rn = A ∫ ( dA / r ) = 949 = 57.9475 mm 16.3769 e = rc − rn = 68.5 − 57.9475 = 10.5525 mm ci = rn − ri = 57.9475 − 25 = 32.9475 mm co = ro − rn = 112 − 57.9475 = 54.0525 mm M = 150F2 = 150(3.2) = 480 kN∙mm We need to find the forces transmitted through the section in order to determine the axial stress. It is not immediately obvious which plane should be used for resolving the axial versus shear directions. It is convenient to use the plane containing the reaction force at the bushing, which assumes its contribution resolves entirely into shear force. To find the angle of this plane, find the resultant of F1 and F2. Fx = F1x + F2 x = 2.4 cos 60o + 3.2 cos 0o = 4.40 kN Fy = F1 y + F2 y = 2.4sin 60o + 3.2sin 0o = 2.08 kN F = ( 4.402 + 2.082 ) 12 = 4.87 kN This is the pin force on the lever which acts in a direction θ = tan −1 Fy Fx = tan −1 2.08 = 25.3o 4.40 On the surface 25.3° from the horizontal, find the internal forces in the tangential and normal directions. Resolving F1 into components, Ft = 2.4 cos ( 60o − 25.3o ) = 1.97 kN Fn = 2.4sin ( 60o − 25.3o ) = 1.37 kN The transverse shear stress is zero at the inner and outer surfaces. Using Eq. (3-65) for the bending stress, and combining with the axial stress due to Fn, Shigley’s MED, 10th edition Chapter 3 Solutions, Page 91/100 www.konkur.in σi = Fn Mci 1370 ( 3200 )(150 ) (32.9475) + = + = 64.6 MPa A Aeri 949 949(10.5525)(25) Ans. Fn Mco 1370 ( 3200 )(150 ) (54.0525) − = − = −21.7 MPa Ans. A Aero 949 949(10.5525)(112) ______________________________________________________________________________ σo = 3-130 ri = 2 in, ro = ri + h = 2 + 4 = 6 in, rc = 2 + 0.5(4) = 4 in A = (6 − 2 − 0.75)(0.75) = 2.4375 in 2 Similar to Prob. 3-129, dA 3.625 6 = 0.75 ln + 0.75 ln = 0.682 920 in r 2 4.375 A 2.4375 rn = = = 3.56923 in 0.682 920 ( dA / r ) ∫ ∫ e = rc − rn = 4 − 3.56923 = 0.43077 in ci = rn − ri = 3.56923 − 2 = 1.56923 in co = ro − rn = 6 − 3.56923 = 2.43077 in M = Frc = 6000(4) = 24 000 lbf ⋅ in Using Eq. (3-65) for the bending stress, and combining with the axial stress, σi = F Mci 6000 24 000(1.56923) + = + = 20 396 psi = 20.4 kpsi A Aeri 2.4375 2.4375(0.43077)(2) Ans. F Mco 6000 24 000(2.43077) − = − = −6 799 psi = −6.80 kpsi Ans. A Aero 2.4375 2.4375(0.43077)(6) ______________________________________________________________________________ σo = 3-131 ri = 12 in, ro = ri + h = 12 + 3 = 15 in, rc = 12 + 3/2 = 13.5 in π π I = a 3b = (1.53 )(0.75) = 1.988 in 4 4 4 A = π ab = π (1.5)(0.75) = 3.534 M = 20(3 + 1.5) = 90 kip ⋅ in Since the radius is large compared to the cross section, assume Eq. 3-67 is applicable for the bending stress. Combining the bending stress and the axial stress, F Mci rc 20 90(1.5)(13.5) + = + = 82.1 kpsi Ans. A Iri 3.534 (1.988)(12) F Mc r 20 90(1.5)(13.5) − = −55.5 kpsi Ans. σo = − o c = A Iro 3.534 1.988(15) σi = ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 92/100 www.konkur.in 3-132 ri = 1.25 in, ro = ri + h = 1.25 + 0.5 + 1 + 0.5 = 3.25 in rc = (ri + ro) / 2 = (1.25 + 3.25)/2 = 2.25 in Ans. ro dA For outer rectangle, ∫ = b ln ri r A r2 , A = π r2 = For circle, O 2 2 ∫ ( dA r ) 2 rc − rc − r O O dA ∴ ∫ = 2π ( rc − rc2 − r 2 ) r O Combine the integrals subtracting the circle from the rectangle ) ( ( ) dA 3.25 = 1.25ln − 2π 2.25 − 2.252 − 0.52 = 0.840 904 in r 1.25 A = 1.25(2) − π (0.52 ) = 1.714 60 in 2 Ans. A 1.714 60 rn = = = 2.0390 in Ans. ∑ ∫ (dA / r ) 0.840 904 Σ∫ e = rc − rn = 2.25 − 2.0390 = 0.2110 in Ans. ci = rn − ri = 2.0390 − 1.25 = 0.7890 in co = ro − rn = 3.25 − 2.0390 = 1.2110 in M = 2000(4.5 + 1.25 + 0.5 + 0.5) = 13 500 lbf ⋅ in F Mci 2000 13 500(0.7890) = + = 20 720 psi = 20.7 kpsi Ans. σi = + A Aeri 1.7146 1.7146(0.2110)(1.25) F Mco 2000 13 500(1.2110) σo = − = − = −12 738 psi = −12.7 kpsi Ans. A Aero 1.7146 1.7146(0.2110)(3.25) ______________________________________________________________________________ 3-133 From Eq. (3-68), 2 (1 −ν 2 ) E 1 3 3 = F 2 (1 d ) 8 13 a = KF 1 3 Use ν = 0.292, F in newtons, E in N/mm2 and d in mm, then 1/3 3 [(1 − 0.2922 ) / 207 000] K = 1 / 30 8 = 0.03685 From Eq. (3-69), pmax = 3F 3F 3F 1/3 3F 1/3 = = = = 352 F 1/3 MPa 2 1/3 2 2 2 2π a 2π ( KF ) 2π K 2π (0.03685) Shigley’s MED, 10th edition Chapter 3 Solutions, Page 93/100 www.konkur.in From Eq. (3-71), the maximum principal stress occurs on the surface where z = 0, and is equal to – pmax. Ans. σ max = σ z = − pmax = −352 F 1/3 MPa From Fig. 3-37, Ans. τ max = 0.3 pmax = 106 F 1/3 MPa ______________________________________________________________________________ 3-134 From Eq. (3-68), 1 −ν 12 ) E1 + (1 −ν 22 ) E2 ( 3 F a=3 1 d1 + 1 d 2 8 2 2 3 (10 ) (1 − 0.292 ) ( 207 000 ) + (1 − 0.333 ) ( 71 700 ) a= = 0.0990 mm 1 25 + 1 40 8 3 From Eq. (3-69), 3 (10 ) 3F pmax = = = 487.2 MPa 2 2π a 2π ( 0.09902 ) From Fig. 3-37, the maximum shear stress occurs at a depth of z = 0.48 a. z = 0.48a = 0.48 ( 0.0990 ) = 0.0475 mm Ans. The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48. σ 1 = σ 2 = −487.2 1 − 0.48 tan −1 (1/ 0.48) (1 + 0.333) − σ3 = 1 = −101.3 MPa 2 (1 + 0.482 ) −487.2 = −396.0 MPa 1 + 0.482 From Eq. (3-72), ( −101.3) − ( −396.0 ) = 147.4 MPa Ans. σ −σ τ max = 1 3 = 2 2 Note that if a closer examination of the applicability of the depth assumption from Fig. 337 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow for calculating and plotting the stresses versus the depth for specific values of ν. For ν = 0.333 for aluminum, the maximum shear stress occurs at a depth of z = 0.492a with τmax = 0.3025 pmax. Shigley’s MED, 10th edition Chapter 3 Solutions, Page 94/100 www.konkur.in This gives τmax = 0.3025 pmax = (0.3025)(487.2) = 147.38 MPa. Even though the depth assumption was a little off, it did not have significant effect on the the maximum shear stress. ______________________________________________________________________________ 3-135 From the solution to Prob. 3-134, a = 0.0990 mm and pmax = 487.2 MPa. Assuming applicability of Fig. 3-37, the maximum shear stress occurs at a depth of z = 0.48 a = 0.0475 mm. Ans. The principal stresses are obtained from Eqs. (3-70) and (3-71) at a depth of z/a = 0.48. σ 1 = σ 2 = −487.2 1 − 0.48 tan −1 (1/ 0.48) (1 + 0.292 ) − σ3 = 1 = −92.09 MPa 2 (1 + 0.482 ) −487.2 = −396.0 MPa 1 + 0.482 From Eq. (3-72), σ −σ ( −92.09 ) − ( −396.0 ) = 152.0 MPa Ans. τ max = 1 3 = 2 2 Note that if a closer examination of the applicability of the depth assumption from Fig. 337 is desired, implementing Eqs. (3-70), (3-71), and (3-72) on a spreadsheet will allow for calculating and plotting the stresses versus the depth for specific values of ν. For ν = 0.292 for steel, the maximum shear stress occurs at a depth of z = 0.478a with τmax = 0.3119 pmax. ______________________________________________________________________________ 3-136 From Eq. (3-68), 2 3F 2 (1 −ν ) E a=3 8 1 d1 + 1 d 2 2 3 ( 20 ) 2 (1 − 0.292 ) ( 207 000 ) a=3 = 0.1258 mm 1 30 + 1 ∞ 8 From Eq. (3-69), 3 ( 20 ) 3F pmax = = = 603.4 MPa 2 2π a 2π ( 0.12582 ) From Fig. 3-37, the maximum shear stress occurs at a depth of z = 0.48a = 0.48 ( 0.1258) = 0.0604 mm Ans. Also from Fig. 3-37, the maximum shear stress is Shigley’s MED, 10th edition Chapter 3 Solutions, Page 95/100 www.konkur.in τ max = 0.3 pmax = 0.3(603.4) = 181 MPa Ans. ______________________________________________________________________________ 3-137 Aluminum Plate-Ball interface: From Eq. (3-68), 2 2 3F (1 −ν 1 ) E1 + (1 −ν 2 ) E2 a= 1 d1 + 1 d 2 8 3 2 6 2 6 3F (1 − 0.292 ) ( 30 ) (10 ) + (1 − 0.333 ) (10.4 ) (10 ) a= = 3.517 (10−3 ) F 1/3 in 8 1 1 + 1 ∞ 3 From Eq. (3-69), 3F 3F pmax = = = 3.860 104 F 1/3 psi 2 2 −3 1/3 2π a 2π 3.517 10 F By examination of Eqs. (3-70), (3-71), and (3-72), it can be seen that the only difference in the maximum shear stress for the plate and the ball will be due to poisson’s ratio in Eq. (3-70). The larger poisson’s ratio will create the greater maximum shear stress, so the aluminum plate will be the critical element in this interface. Applying the equations for the aluminum plate, ( ( ) ) σ 1 = −3.86 (104 ) F 1/3 1 − 0.48 tan −1 (1/ 0.48) (1 + 0.333) − σ3 = −3.86 (104 ) F 1/3 1 + 0.48 From Eq. (3-72), τ max = σ1 − σ 3 2 1 1/3 = −8025 F psi 2 (1 + 0.482 ) = −3.137 (104 ) F 1/3 psi ( −8025F ) − ( −3.137 (10 ) F ) = = 1.167 (10 ) F 2 1/3 4 1/3 4 1/3 2 Comparing this stress to the allowable stress, and solving for F, psi 3 20 000 = 5.03 lbf F = 4 1.167 (10 ) Table-Ball interface: From Eq. (3-68), 3F (1 − 0.292 a= 8 3 2 ) ( 30 ) (10 ) + (1 − 0.211 ) (14.5) (10 ) 6 1 1+1 ∞ 2 6 = 3.306 (10−3 ) F 1/3 in From Eq. (3-69), Shigley’s MED, 10th edition Chapter 3 Solutions, Page 96/100 www.konkur.in ( ) 3F 3F = = 4.369 104 F 1/3 psi 2 2 2π a 2π 3.306 10−3 F 1/3 The steel ball has a higher poisson’s ratio than the cast iron table, so it will dominate. pmax = ( ) σ 1 = −4.369 (104 ) F 1/3 1 − 0.48 tan −1 (1/ 0.48) (1 + 0.292 ) − σ3 = −4.369 (104 ) F 1/3 1 + 0.482 From Eq. (3-72), τ max = σ1 − σ 3 1 = −8258 F 1/3 psi 2 2 (1 + 0.48 ) = −3.551(104 ) F 1/3 psi ( −8258F ) − ( −3.551(10 ) F ) = = 1.363 (10 ) F 2 1/3 4 1/3 4 1/3 2 Comparing this stress to the allowable stress, and solving for F, psi 3 20 000 = 3.16 lbf F = 4 1.363 (10 ) The steel ball is critical, with F = 3.16 lbf. Ans. ______________________________________________________________________________ 3-138 v1 = 0.333, E1 = 10.4 Mpsi, l = 2 in, d1 = 1.25 in, v2 = 0.211, E2 = 14.5 Mpsi, d2 = –12 in. From Eq. 3-73, with b = KcF1/2 2 (1 − 0.3332 ) 10.4 (106 ) + (1 − 0.2112 ) 14.5 (106 ) Kc = π (2) 1/1.25 + 1/ ( −12 ) 12 = 2.593 (10−4 ) By examination of Eqs. (3-75), (3-76), and (3-77), it can be seen that the only difference in the maximum shear stress for the two materials will be due to poisson’s ratio in Eq. (375). The larger poisson’s ratio will create the greater maximum shear stress, so the aluminum roller will be the critical element in this interface. Instead of applying these equations, we will assume the poisson’s ratio for aluminum of 0.333 is close enough to 0.3 to make Fig. 3-39 applicable. τ max = 0.3 pmax pmax = 4000 = 13 300 psi 0.3 From Eq. (3-74), pmax = 2F / (πbl ), so we have Shigley’s MED, 10th edition Chapter 3 Solutions, Page 97/100 www.konkur.in pmax = 2F 2F 1 2 = π lK c F 1 2 π lK c So, π lK c pmax F = 2 2 π (2)(2.593) (10−4 ) (13 300) = 2 = 117.4 lbf Ans. ______________________________________________________________________________ 2 3-139 v = 0.292, E = 30 Mpsi, l = 0.75 in, d1 = 2(0.47) = 0.94 in, d2 = 2(0.62) = 1.24 in. Eq. (3-73): 2(40) 2 (1 − 0.2922 ) 30 (106 ) b= π (0.75) 1/ 0.94 + 1/ 1.24 12 Eq. (3-74): pmax = = 1.052 (10−3 ) in 2 ( 40 ) 2F == = 32 275 psi = 32.3 kpsi π bl π (1.052 ) (10−3 ) ( 0.75 ) Ans. From Fig. 3-39, τ max = 0.3 pmax = 0.3 ( 32 275) = 9682.5 psi = 9.68 kpsi Ans. ______________________________________________________________________________ 3-140 Use Eqs. (3-73) through (3-77). 1/2 2 F (1 −ν 12 ) / E1 + (1 − v22 ) / E2 b= (1 / d1 ) + (1 / d 2 ) πl 1/2 2(600) (1 − 0.2922 ) / (30(106 )) + (1 − 0.2922 ) / (30(106 )) = 1/ 5 + 1 / ∞ π (2) b = 0.007 631 in 2F 2(600) = = 25 028 psi π bl π (0.007 631)(2) z2 z σ x = −2ν pmax 1 + 2 − = −2 ( 0.292 )( 25 028 ) b b = −7102 psi = −7.10 kpsi Ans. pmax = Shigley’s MED, 10th edition ( 1 + 0.7862 − 0.786 ) Chapter 3 Solutions, Page 98/100 www.konkur.in z2 2 1+ 2 2 b − 2 z = −25 028 1 + 2 ( 0.786 ) − 2 ( 0.786 ) σ y = − pmax 1 + 0.7862 b z2 ( ) 1+ 2 b = −4 646 psi = −4.65 kpsi Ans. − pmax −25 028 = = −19 677 psi = −19.7 kpsi Ans. σz = 2 z 1 + 0.7862 1+ 2 b σ − σ z −4 646 − ( −19 677 ) τ max = y = = 7 516 psi = 7.52 kpsi Ans. 2 2 ______________________________________________________________________________ 3-141 Use Eqs. (3-73) through (3-77). 2 F (1 −ν 12 ) E1 + (1 −ν 2 2 ) E2 b= πl 1/ d1 + 1/ d 2 12 2(2000) (1 − 0.2922 ) 207 (103 ) + (1 − 0.2112 ) 100 (103 ) = π (40) 1/ 150 + 1/ ∞ b = 0.2583 mm 12 pmax = 2F 2(2000) = = 123.2 MPa π bl π (0.2583)(40) ( z2 z σ x = −2ν pmax 1 + 2 − = −2 ( 0.292 )(123.2 ) 1 + 0.7862 − 0.786 b b = −35.0 MPa Ans. z2 1 + 2 0.7862 1+ 2 2 ( ) − 2 0.786 z b − 2 = −123.2 σ y = − pmax ( ) 1 + 0.7862 b z2 ( ) 1+ 2 b = −22.9 MPa Ans. − pmax −123.2 σz = = = −96.9 MPa Ans. 2 z 1 + 0.7862 1+ 2 b σ y − σ z −22.9 − ( −96.9 ) = = 37.0 MPa Ans. τ max = 2 2 ) ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 99/100 www.konkur.in 3-142 Use Eqs. (3-73) through (3-77). 2 F (1 −ν 12 ) E1 + (1 −ν 2 2 ) E2 b= πl 1/ d1 + 1/ d 2 12 2(250) (1 − 0.2112 ) 14.5 (106 ) + (1 − 0.2112 ) 14.5 (106 ) = π (1.25) 1/ 3 + 1/ ∞ b = 0.007 095 in 12 2F 2(250) = = 17 946 psi π bl π (0.007 095)(1.25) z2 z σ x = −2ν pmax 1 + 2 − = −2 ( 0.211)(17 946 ) b b = −3 680 psi = −3.68 kpsi Ans. pmax = ( 1 + 0.7862 − 0.786 ) z2 1 + 2 0.7862 1+ 2 2 ( ) z b −2 = −17 946 − 2 ( 0.786 ) σ y = − pmax 2 2 b z 1 + ( 0.786 ) 1 + b2 = −3 332 psi = −3.33 kpsi Ans. − pmax −17 946 σz = = = −14 109 psi = −14.1 kpsi Ans. 2 z 1 + 0.7862 1+ 2 b σ y − σ z −3 332 − ( −14 109 ) τ max = = = 5 389 psi = 5.39 kpsi Ans. 2 2 ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 3 Solutions, Page 100/100 www.konkur.in Chapter 4 For a torsion bar, kT = T/θ = Fl/θ, and so θ = Fl/kT. For a cantilever, kl = F/δ ,δ = F/kl. For the assembly, k = F/y, or, y = F/k = lθ + δ Thus F Fl 2 F y= = + k kT kl Solving for k kk 1 k= 2 = 2l T Ans. l 1 kl l + kT + kT kl ______________________________________________________________________________ 4-1 For a torsion bar, kT = T/θ = Fl/θ, and so θ = Fl/kT. For each cantilever, kl = F/δl, δl = F/kl, and,δL = F/kL. For the assembly, k = F/y, or, y = F/k = lθ + δl +δL. Thus F Fl 2 F F y= = + + k kT kl k L Solving for k k L kl kT 1 k= 2 = Ans. 2 l 1 1 kl k Ll + kT k L + kT kl + + kT kl k L ______________________________________________________________________________ 4-2 4-3 (a) For a torsion bar, k =T/θ =GJ/l. Two springs in parallel, with J =πdi 4/32, and d1 = d2 = d, J G J G π d4 d4 k = 1 + 2 = G 1 + 2 x l − x 32 x l − x π 1 1 Gd 4 + 32 x l−x Deflection equation, = θ= Ans. (1) T1 x T2 ( l − x ) = JG JG T2 ( l − x ) (2) x From statics, T1 + T2 = T = 1500. Substitute Eq. (2) results in T1 = Shigley’s MED, 10th edition Chapter 4 Solutions, Page 1/80 www.konkur.in l−x T2 + T2 = 1500 x T2 = 1500 ⇒ x l Ans. (3) l−x Ans. (4) l 1 π 1 3 (b) From Eq. (1), k = ( 0.54 )11.5 (106 ) + = 28.2 (10 ) lbf ⋅ in/rad 32 5 10 − 5 10 − 5 From Eq. (4), T1 = 1500 = 750 lbf ⋅ in Ans. 10 5 From Eq. (3), T2 = 1500 = 750 lbf ⋅ in Ans. 10 16Ti 16 (1500 ) From either section, τ = = = 30.6 (103 ) psi = 30.6 kpsi Ans. π di3 π ( 0.53 ) T1 = 1500 Substitute into Eq. (2) resulting in Ans. ______________________________________________________________________________ 4-4 Deflection to be the same as Prob. 4-3 where T1 = 750 lbf⋅in, l1 = l / 2 = 5 in, and d1 = 0.5 in θ1=θ2=θ T1 ( 4 ) π 32 Or, 4 1 d G = T2 ( 6 ) π 32 = π 4 2 d G 32 T1 = 15 (103 ) d14 ( 0.5 ) G 4 ⇒ 4T1 6T2 = 4 = 60 (103 ) d14 d2 (1) (2) T2 = 10 (103 ) d 24 Equal stress, τ 1 = τ 2 ⇒ 750 ( 5 ) (3) 16T1 16T2 T T = ⇒ 13 = 23 3 3 d1 d 2 π d1 π d 2 (4) Divide Eq. (4) by the first two equations of Eq.(1) results in T1 T2 3 d1 d3 = 2 ⇒ d 2 = 1.5d1 (5) 4T1 6T2 d14 d 24 Statics, T1 + T2 = 1500 (6) Substitute in Eqs. (2) and (3), with Eq. (5) gives 15 (103 ) d14 + 10 (103 ) (1.5d1 ) = 1500 4 Solving for d1 and substituting it back into Eq. (5) gives d1 = 0.388 8 in, d2 = 0.583 2 in Ans. Shigley’s MED, 10th edition Chapter 4 Solutions, Page 2/80 www.konkur.in From Eqs. (2) and (3), T1 = 15(103)(0.388 8)4 = 343 lbf⋅in T2 = 10(103)(0.583 2)4 = 1 157 lbf⋅in Deflection of T is θ1 = Spring constant is k= The stress in d1 is τ1 = Ans. Ans. 343 ( 4 ) T1l1 = = 0.053 18 rad J1G (π / 32 ) ( 0.388 84 )11.5 (106 ) T θ1 = 1500 = 28.2 (103 ) lbf ⋅ in 0.053 18 Ans. 16 ( 343) 16T1 = = 29.7 (103 ) psi = 29.7 kpsi 3 3 π d1 π ( 0.388 8 ) Ans. 16 (1 157 ) 16T2 = = 29.7 (103 ) psi = 29.7 kpsi Ans. 3 π d 2 π ( 0.583 2 )3 ______________________________________________________________________________ The stress in d1 is 4-5 τ2 = (a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the taper be α where tan α = (r2 − r1)/2. Thus, the radius in the taper as a function of x is r = r1 + x tan α, and the area is A = π (r1 + x tan α)2. The deflection of the tapered portion is l l l F F dx F 1 dx = =− δ =∫ 2 ∫ AE π E 0 ( r1 + x tan α ) π E ( r1 + x tan α ) tan α 0 = 1 1 F 1 1 F − = − π E r1 tan α tan α ( r1 + l tan α ) π E tan α r1 r2 = F r2 − r1 F l tan α Fl = = π E tan α r1r2 π E tan α r1r2 π r1r2 E = 4 Fl π d1d 2 E 0 Ans. (b) For section 1, δ1 = Fl 4 Fl 4(1000)(2) = = = 3.40(10−4 ) in 2 2 6 AE π d1 E π (0.5 )(30)(10 ) For the tapered section, 4 Fl 4 1000(2) δ= = = 2.26(10−4 ) in π d1d 2 E π (0.5)(0.75)(30)(106 ) For section 2, Shigley’s MED, 10th edition Ans. Ans. Chapter 4 Solutions, Page 3/80 www.konkur.in Fl 4 Fl 4(1000)(2) = = = 1.51(10−4 ) in Ans. 2 2 6 AE π d1 E π (0.75 )(30)(10 ) ______________________________________________________________________________ δ2 = 4-6 (a) Let the radii of the straight sections be r1 = d1 /2 and r2 = d2 /2. Let the angle of the taper be α where tan α = (r2 − r1)/2. Thus, the radius in the taper as a function of x is r = r1 + x tan α, and the polar second area moment is J = (π /2) (r1 + x tan α)4. The angular deflection of the tapered portion is l l l T 2T dx 1 2T 1 =− θ = ∫ dx = 4 ∫ GJ 3 π G ( r1 + x tan α )3 tan α π G 0 ( r1 + x tan α ) 0 = 0 2 2 T 1 1 T 1 1 − = 3 − 3 3π G r1 tan α tan α ( r1 + l tan α ) 3π G tan α r13 r23 2 2 2 T r23 − r13 2 T l r23 − r13 2 Tl ( r1 + r1r2 + r2 ) = = = 3π G tan α r13 r23 3π G r2 − r1 r13r23 3π G r13r23 2 2 32 Tl ( d1 + d1d 2 + d 2 ) = 3π G d13d 23 Ans. (b) The deflections, in degrees, are: Section 1, Tl GJ θ1 = 32(1500)(2) 180 180 32Tl 180 = = = 2.44 deg 4 4 6 π π d1 G π π (0.5 )11.5(10 ) π Ans. Tapered section, θ= = 32 Tl (d12 + d1d 2 + d 2 2 ) 180 3π Gd13 d 23 π 2 2 32 (1500)(2) 0.5 + (0.5)(0.75) + 0.75 180 = 1.14 deg 3π 11.5(106 )(0.53 )(.753 ) π Ans. Section 2, Tl 180 32Tl 180 32(1500)(2) 180 Ans. = = = 0.481 deg 4 GJ π π d 2 G π π (0.754 )11.5(106 ) π ______________________________________________________________________________ θ2 = 4-7 The area and the elastic modulus remain constant. However, the force changes with respect to x. From Table A-5, the unit weight of steel is γ = 0.282 lbf/in3 and the elastic modulus is E = 30 Mpsi. Starting from the top of the cable (i.e. x = 0, at the top). F = γ(A)(l−x) Shigley’s MED, 10th edition Chapter 4 Solutions, Page 4/80 www.konkur.in γ γ l 2 0.282 [500(12) ] Fdx w l 1 = ∫ (l − x)dx = lx − x 2 = = = 0.169 in AE E 0 E 2 0 2E 2(30)106 2 l δc = ∫ l o From the weight at the bottom of the cable, 4(5000) [500(12)] Wl 4Wl = = = 5.093 in 2 AE π d E π (0.52 )30(106 ) Ans. δ = δ c + δW = 0.169 + 5.093 = 5.262 in δW = The percentage of total elongation due to the cable’s own weight 0.169 (100) = 3.21% Ans. 5.262 ______________________________________________________________________________ 4-8 ΣFy = 0 = R 1 − F ⇒ R 1 = F ΣMA = 0 = M1 − Fa ⇒ M1 = Fa VAB = F, MAB =F (x − a ), VBC = MBC = 0 Section AB: 1 F x2 θ AB = F x − a dx = − ax ( ) + C1 EI ∫ EI 2 θAB = 0 at x = 0 ⇒ C1 = 0 y AB = F x2 F x3 x2 − ax dx = − a + C2 EI ∫ 2 EI 6 2 (1) (2) yAB = 0 at x = 0 ⇒ C2 = 0, and Fx 2 y AB = Ans. ( x − 3a ) 6 EI Section BC: θ BC = 1 EI ∫ ( 0 ) dx = 0 + C 3 From Eq. (1), at x = a (with C1 = 0), θ = θ BC = − F a2 Fa 2 − a ( a ) = − = C3. Thus, EI 2 2 EI Fa 2 2 EI Shigley’s MED, 10th edition Chapter 4 Solutions, Page 5/80 www.konkur.in yBC = − Fa 2 Fa 2 dx = − x + C4 2 EI ∫ 2 EI (3) From Eq. (2), at x = a (with C2 = 0), y = − Fa 2 Fa 3 a + C4 = − 2 EI 3EI yBC C4 = ⇒ F a3 a2 Fa3 . Thus, from Eq. (3) −a = − EI 6 2 3EI Fa 3 6 EI Fa 2 Fa 3 Fa 2 =− x+ = ( a − 3x ) 2 EI 6 EI 6 EI Substitute into Eq. (3), obtaining Ans. The maximum deflection occurs at x= l, Fa 2 ymax = Ans. ( a − 3l ) 6 EI ______________________________________________________________________________ 4-9 ΣMC = 0 = F (l /2) − R1 l ⇒ R1 = F /2 ΣFy = 0 = F /2 + R 2 − F ⇒ R 2 = F /2 Break at 0 ≤ x ≤ l /2: VAB = R 1 = F /2, MAB = R 1 x = Fx /2 Break at l /2 ≤ x ≤ l : VBC = R 1 − F = − R 2 = − F /2, MBC = R 1 x − F ( x − l / 2) = F (l − x) /2 Section AB: θ AB = 1 EI Fx F x2 dx = + C1 ∫2 EI 4 2 l F 2 +C = 0 From symmetry, θAB = 0 at x = l /2 ⇒ 1 4 EI θ AB = F x 2 Fl 2 F − = 4 x2 − l 2 ) ( EI 4 16 EI 16 EI Shigley’s MED, 10th edition ⇒ C1 = − Fl 2 . Thus, 16 EI (1) Chapter 4 Solutions, Page 6/80 www.konkur.in y AB = F 16 EI yAB = 0 at x = 0 y AB = 2 2 ∫ ( 4 x − l ) dx = ⇒ Fx ( 4 x 2 − 3l 2 ) 48 EI F 4 x3 2 − l x + C2 16 EI 3 C2 = 0, and, (2) yBC is not given, because with symmetry, Eq. (2) can be used in this region. The maximum deflection occurs at x =l /2, ymax l F 2 Fl 3 2 l = 4 − 3l 2 = − 48 EI 2 48 EI Ans. ______________________________________________________________________________ 4-10 From Table A-6, for each angle, I1-1 = 207 cm4. Thus, I = 2(207) (104) = 4.14(106) mm4 From Table A-9, use beam 2 with F = 2500 N, a = 2000 mm, and l = 3000 mm; and beam 3 with w = 1 N/mm and l = 3000 mm. Fa 2 w l4 (a − 3l ) − 6 EI 8 EI 2500(2000) 2 (1)(3000) 4 = [ 2000 − 3(3000)] − 6(207)103 (4.14)106 8(207)(103 )(4.14)(106 ) = −25.4 mm Ans. ymax = M O = − Fa − ( wl 2 / 2) = − 2500(2000) − [1(30002)/2] = − 9.5(106) N⋅mm From Table A-6, from centroid to upper surface is y = 29 mm. From centroid to bottom surface is y = 29.0 − 100= − 71 mm. The maximum stress is compressive at the bottom of the beam at the wall. This stress is My −9.5(106 )(−71) =− = −163 MPa Ans. I 4.14(106 ) ______________________________________________________________________________ σ max = − Shigley’s MED, 10th edition Chapter 4 Solutions, Page 7/80 www.konkur.in 4-11 14 10 (450) + (300) = 465 lbf 20 20 6 10 RC = (450) + (300) = 285 lbf 20 20 RO = M1 = 465(6)12 = 33.48(103) lbf⋅in M2 = 33.48(103) +15(4)12 = 34.20(103) lbf⋅in M max 34.2 ⇒ 15 = Z = 2.28 in 3 Z Z For deflections, use beams 5 and 6 of Table A-9 2 F1a[l − (l / 2)] l l F2l 3 2 y x =10ft = + a − 2l − 6 EIl 2 48 EI 2 σ max = −0.5 = 450(72)(120) 300(2403 ) 2 2 2 120 + 72 − 240 − ( ) 48(30)(106 ) I 6(30)(106 ) I (240) I = 12.60 in 4 ⇒ I / 2 = 6.30 in 4 Select two 5 in-6.7 lbf/ft channels from Table A-7, I = 2(7.49) = 14.98 in4, Z =2(3.00) = 6.00 in3 12.60 1 ymidspan = − = −0.421 in 14.98 2 34.2 σ max = = 5.70 kpsi 6.00 ______________________________________________________________________________ 4-12 I= π (1.54 ) = 0.2485 in 4 64 From Table A-9 by superposition of beams 6 and 7, at x = a = 15 in, with b = 24 in and l = 39 in Fba 2 wa y= [a + b 2 − l 2 ] + (2la 2 − a 3 − l 3 ) 6 EIl 24 EI yA = 340(24)15 152 + 242 − 392 6(30)106 (0.2485)39 (150 / 12)(15) 2(39)(152 ) − 153 − 393 = −0.0978 in + 6 24(30)10 (0.2485) Ans. At x = l /2 = 19.5 in 2 2 3 Fa[l − (l / 2)] l l w(l / 2) l l 2 3 y= + a − 2l + 2l − − l 6 EIl 2 24 EI 2 2 2 Shigley’s MED, 10th edition Chapter 4 Solutions, Page 8/80 www.konkur.in y= 340(15)(19.5) 19.52 + 152 − 392 6 6(30)(10 )(0.2485)(39) (150 /12)(19.5) 2(39)(19.52 ) − 19.53 − 393 = −0.1027 in + 24(30)(106 )(0.2485) Ans. −0.1027 + 0.0978 (100) = 5.01% Ans. −0.0978 ______________________________________________________________________________ % difference = 4-13 I = ( ) 1 (6)(323 ) = 16.384 103 mm 4 12 From Table A-9-10, beam 10 Fa 2 yC = − (l + a ) 3EI Fax 2 y AB = l − x2 ) ( 6 EIl dy AB Fa 2 = (l − 3x 2 ) dx 6 EIl dy AB = θA dx Fal 2 Fal θA = = 6 EIl 6 EI Fa 2l yO = −θ A a = − 6 EI At x = 0, With both loads, Fa 2l Fa 2 − (l + a ) 6 EI 3EI Fa 2 400(3002 ) =− (3l + 2a) = − [3(500) + 2(300)] = −3.72 mm 6 EI 6(207)103 (16.384)103 At midspan, 2 2 Fa (l / 2) 2 l 3 Fal 2 3 400(300)(5002 ) yE = = = 1.11 mm l − = 6 EIl 24 207 (103 )16.384 (103 ) 2 24 EI yO = − Ans. Ans. _____________________________________________________________________________ π 4-14 I = (24 − 1.54 ) = 0.5369 in 4 64 From Table A-5, E = 10.4 Mpsi From Table A-9, beams 1 and 2, by superposition Shigley’s MED, 10th edition Chapter 4 Solutions, Page 9/80 www.konkur.in −200 [ 4(12)] 300 [ 2(12) ] F l 3 F a2 yB = − B + A (a − 3l ) = + [ 2(12) − 3(4)(12)] 6 3EI 6 EI 3(10.4)10 (0.5369) 6(10.4)106 (0.5369) yB = −1.94 in Ans. ______________________________________________________________________________ 3 2 4-15 From Table A-7, I = 2(1.85) = 3.70 in4 From Table A-5, E = 30.0 Mpsi From Table A-9, beams 1 and 3, by superposition [5 + 2(5 / 12)] (60 ) = −0.182 in Ans. Fl 3 ( w + wc )l 4 150(603 ) − =− − 6 3EI 8 EI 3(30)10 (3.70) 8(30)106 (3.70) ______________________________________________________________________________ π 4 4-16 I = d 64 From Table A-5, E = 207(103 ) MPa From Table A-9, beams 5 and 9, with FC = FA = F, by superposition F l3 Fa 1 − FB l 3 + 2 Fa (4a 2 − 3l 2 ) yB = − B + (4a 2 − 3l 2 ) ⇒ I = 48 EI 24 EI 48 EyB 4 yA = − I= { } 1 −550(10003 ) + 2 ( 375) (250) 4(2502 ) − 3(10002 ) 3 48(207)10 ( −2 ) ( ) = 53.624 103 mm 4 d= 4 64 π I = 4 64 π (53.624)103 = 32.3 mm Ans. ______________________________________________________________________________ 4-17 From Table A-9, beams 8 (region BC for this beam with a = 0) and 10 (with a = a), by superposition yBC ( ) ( ) MA 3 Fax 2 x − 3lx 2 + 2l 2 x + l − x2 6 EIl 6 EIl 1 = M A x 3 − 3lx 2 + 2l 2 x + Fax l 2 − x 2 Ans. 6 EIl d M F (x − l) = A x3 − 3lx 2 + 2l 2 x ( x − l ) + [( x − l )2 − a(3x − l )] 6 EI x =l dx 6 EIl y AB = ( ) ( ) ( ) M Al F (x − l) (x − l) + [( x − l )2 − a (3x − l )] 6 EI 6 EI (x − l) = − M Al + F ( x − l ) 2 − a (3 x − l ) Ans. 6 EI ______________________________________________________________________________ =− { Shigley’s MED, 10th edition } Chapter 4 Solutions, Page 10/80 www.konkur.in 4-18 Note to the instructor: Beams with discontinuous loading are better solved using singularity functions. This eliminates matching the slopes and displacements at the discontinuity as is done in this solution. a wa ∑ M C = 0 = R1l − wa l − a + 2 ⇒ R1 = 2l ( 2l − a ) Ans. wa wa 2 F = 0 = 2 l − a + R − w a ⇒ R = Ans. ( ) 2 ∑ y 2 2l 2l wa w VAB = R1 − wx = ( 2l − a ) − wx = 2l ( a − x ) − a 2 Ans. 2l 2l 2 wa VBC = − R2 = − Ans. 2l w x2 M AB = ∫ VAB dx = 2l ax − − a 2 x + C1 2l 2 wx 2al − a 2 − lx Ans. M AB = 0 at x = 0 ∴ C1 = 0 ⇒ M AB = 2l wa 2 wa 2 M BC = ∫ VBC dx = ∫ − dx = − x + C2 2l 2l wa 2 wa 2 M BC = 0 at x = l ∴ C2 = ⇒ M BC = (l − x) Ans. 2 2l M 1 wx 1 w 2 1 2 2 1 3 2al − a 2 − lx ) dx = θ AB = ∫ AB dx = ( alx − a x − lx + C3 ∫ EI EI 2l EI 2l 2 3 1 w 2 1 2 2 1 3 alx − a x − lx + C3 dx ∫ EI 2l 2 3 1 w 1 3 1 2 3 1 4 = alx − a x − lx + C3 x + C4 EI 2l 3 6 12 y AB = ∫ θ AB dx = y AB = 0 at x = 0 ∴ C4 = 0 θ BC = ∫ M BC 1 dx = EI EI wa 2 1 wa 2 1 2 ( l − x ) dx = lx − x + C5 ∫ 2l EI 2l 2 θ AB = θ BC at x = a ∴ 1 EI w 2 1 4 1 3 1 wa 2 1 2 wa 3 ala − a − la + C = la − a + C ⇒ C = + C5 3 5 3 2l 2 3 2 6 EI 2l Shigley’s MED, 10th edition (1) Chapter 4 Solutions, Page 11/80 www.konkur.in 1 wa 2 1 2 1 wa 2 1 2 1 3 lx − x + C dx = 5 lx − x + C5 x + C6 ∫ EI 2l 2 EI 2l 2 6 2 2 wa l = 0 at x = l ∴ C6 = − − C5l 6 1 wa 2 1 2 1 3 1 3 = lx − x − l + C5 ( x − l ) EI 2l 2 6 3 yBC = ∫ θ BC dx = yBC yBC y AB = yBC at x = a ∴ w 1 1 5 1 4 wa 2 1 2 1 3 1 3 3 ala − a − la + C3a = la − a − l + C5 (a − l ) 2l 3 6 12 2l 2 6 3 C3 a = wa 2 3la 2 − 4l 3 ) + C5 (a − l ) ( 24l Substituting (1) into (2) yields C5 = (2) wa 2 −a 2 − 4l 2 ) . Substituting this back into (2) gives ( 24l wa 2 4al − a 2 − 4l 2 ) . Thus, ( 24l w y AB = ( 4alx3 − 2a 2 x3 − lx 4 + 4a3lx − a 4 x − 4a 2l 2 x ) 24 EIl wx 2 ⇒ y AB = 2ax 2 (2l − a ) − lx 3 − a 2 ( 2l − a ) Ans. 24 EIl w yBC = Ans. ( 6a 2lx 2 − 2a 2 x3 − a 4 x − 4a 2l 2 x + a 4l ) 24 EIl This result is sufficient for yBC. However, this can be shown to be equivalent to w w yBC = 4alx 3 − 2a 2 x 3 − lx 4 − 4a 2l 2 x + 4a 3lx − a 4 x ) + ( x − a)4 ( 24 EIl 24 EI w yBC = y AB + ( x − a)4 Ans. 24 EI by expanding this or by solving the problem using singularity functions. ______________________________________________________________________________ C3 = 4-19 The beam can be broken up into a uniform load w downward from points A to C and a uniform load w upward from points A to B. Using the results of Prob. 4-18, with b = a for A to C and a = a for A to B, results in wx wx 2 2 y AB = 2bx 2 (2l − b) − lx 3 − b 2 ( 2l − b ) − 2ax 2 (2l − a) − lx 3 − a 2 ( 2l − a ) 24 EIl 24 EIl wx 2 2 = 2bx 2 (2l − b) − b 2 ( 2l − b ) − 2ax 2 (2l − a) + a 2 ( 2l − a ) Ans. 24 EIl w 2 yBC = 2bx 3 (2l − b) − lx 4 − b 2 x ( 2l − b ) 24 EIl − ( 4alx 3 − 2a 2 x 3 − lx 4 − 4a 2l 2 x + 4a 3lx − a 4 x ) − l ( x − a) 4 Ans. Shigley’s MED, 10th edition Chapter 4 Solutions, Page 12/80 www.konkur.in w 4blx3 − 2b 2 x3 − lx 4 − 4b 2l 2 x + 4b3lx − b 4 x + l ( x − b)4 24 EIl w 4alx3 − 2a 2 x3 − lx 4 − 4a 2l 2 x + 4a 3lx − a 4 x + l ( x − a )4 − 24 EIl w ( x − b) 4 − ( x − a ) 4 + y AB = Ans. 24 EI ______________________________________________________________________________ yCD = 4-20 Note to the instructor: See the note in the solution for Problem 4-18. wa 2 wa ∑ Fy = 0 = RB − 2l − wa ⇒ RB = 2l ( 2l + a ) Ans. For region BC, isolate right-hand element of length (l + a − x) wa 2 VAB = − RA = − , VBC = w ( l + a − x ) Ans. 2l wa 2 w 2 M AB = − RA x = − x, M BC = − ( l + a − x ) Ans. 2l 2 wa 2 2 EIθ AB = ∫ M AB dx = − x + C1 4l wa 2 3 EIy AB = − x + C1 x + C2 12l wa 2 3 yAB = 0 at x = 0 ⇒ C2 = 0 ∴ EIy AB = − x + C1 x 12l w a 2l yAB = 0 at x = l ⇒ C1 = ∴ 12 w a 2 3 w a 2l wa2 x 2 wa 2 x 2 2 EIy AB = − x + x= l − x ⇒ y = l − x 2 ) Ans. ( ) ( AB 12l 12 12l 12 EIl w 3 EIθ BC = ∫ M BC dx = − ( l + a − x ) + C3 6 w 4 EIy BC = − ( l + a − x ) + C3 x + C4 24 wa 4 wa 4 yBC = 0 at x = l ⇒ − (1) + C3l + C4 = 0 ⇒ C4 = − C3l 24 24 wa 2l wa 2l wa 3 wa 2 θAB = θBC at x = l ⇒ − + = + C3 ⇒ C3 = − (l + a ) 4 12 6 6 wa 2 2 a + 4l ( l + a ) . Substitute back into yBC Substitute C3 into Eq. (1) gives C4 = 24 1 w wa 2 wa 4 wa 2l 4 yBC = − (l + a − x ) − x (l + a ) + + ( l + a ) EI 24 6 24 6 =− w 4 ( l + a − x ) − 4a 2 ( l − x )( l + a ) − a 4 24 EI Shigley’s MED, 10th edition Ans. Chapter 4 Solutions, Page 13/80 www.konkur.in 4-21 Table A-9, beam 7, w l 100(10) R1 = R2 = = = 500 lbf ↑ 2 2 wx 100 x 2(10) x 2 − x 3 − 103 y AB = 2lx 2 − x 3 − l 3 ) = ( 6 24 EI 24 ( 30 )10 ( 0.05 ) = 2.7778 (10−6 ) x ( 20 x 2 − x 3 − 1000 ) Slope: θ AB = w wl 3 2 3 3 6 l l − 4 l − l = ( ) 24EI x =l 24 EI 100 (103 ) w l3 (x −l) = (x −l) = ( x − 10 ) = 2.7778 (10−3 ) ( x − 10 ) x =l 24 EI 24(30)106 (0.05) At x = l, θ AB yBC = θ AB d y AB w = 6lx 2 − 4 x 3 − l 3 ) ( dx 24 EI = From Prob. 4-20, 2 wa 2 100 ( 4 ) RA = = = 80 lbf ↓ 2l 2(10) RB = 100 ( 4 ) wa ( 2l + a ) = [ 2(10) + 4] = 480 lbf ↑ 2l 2(10) 100 ( 42 ) x wa 2 x 2 2 l − x = 102 − x 2 ) = 8.8889 (10−6 ) x (100 − x 2 ) ( ) ( 6 12 EIl 12 ( 30 )10 ( 0.05 ) w 4 =− ( l + a − x ) − 4a 2 ( l − x )( l + a ) − a 4 24 EI 100 (10 + 4 − x )4 − 4 ( 42 ) (10 − x )(10 + 4 ) − 44 =− 6 24 ( 30 )10 ( 0.05 ) y AB = yBC = −2.7778 (10−6 ) (14 − x ) + 896 x − 9216 Superposition, RA = 500 − 80 = 420 lbf ↑ RB = 500 + 480 = 980 lbf ↑ 4 y AB = 2.7778 (10 −6 ) x ( 20 x 2 − x 3 − 1000 ) + 8.8889 (10 −6 ) x (100 − x 2 ) Ans. Ans. 4 yBC = 2.7778 (10−3 ) ( x − 10 ) − 2.7778 (10−6 ) (14 − x ) + 896 x − 9216 Ans. The deflection equations can be simplified further. However, they are sufficient for plotting. Using a spreadsheet, x y 0 0.5 1 1.5 2 2.5 3 3.5 0.000000 -0.000939 -0.001845 -0.002690 -0.003449 -0.004102 -0.004632 -0.005027 x y 4 4.5 5 5.5 6 6.5 7 7.5 -0.005280 -0.005387 -0.005347 -0.005167 -0.004853 -0.004421 -0.003885 -0.003268 Shigley’s MED, 10th edition Chapter 4 Solutions, Page 14/80 www.konkur.in x y x y 8 8.5 9 9.5 10 10.5 11 11.5 -0.002596 -0.001897 -0.001205 -0.000559 0.000000 0.000439 0.000775 0.001036 12 0.001244 12.5 0.001419 13 13.5 14 0.001575 0.001722 0.001867 0.002 Beam Deflection, Prob. 4-21 0.001 0.000 -0.001 y (in) -0.002 -0.003 -0.004 -0.005 -0.006 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 x (in) ______________________________________________________________________________ 4-22 (a) Useful relations k= F 48 EI = 3 y l 3 kl 3 1800 ( 36 ) I= = = 0.05832 in 4 48 E 48(30)106 From I = bh 3/12, and b = 10 h, then I = 5 h 4/6, or, h= 4 6 I 4 6(0.05832) = = 0.514 in 5 5 h is close to 1/2 in and 9/16 in, while b is close to 5.14 in. Changing the height drastically changes the spring rate, so changing the base will make finding a close solution easier. Trial and error was applied to find the combination of values from Table A-17 that yielded the closet desired spring rate. h (in) 1/2 1/2 1/2 9/16 9/16 Shigley’s MED, 10th edition b (in) 5 5½ 5¾ 5 4 b/h 10 11 11.5 8.89 7.11 k (lbf/in) 1608 1768 1849 2289 1831 Chapter 4 Solutions, Page 15/80 www.konkur.in h = ½ in, b = 5 ½ in should be selected because it results in a close spring rate and b/h is still reasonably close to 10. (b) I = 5.5(0.5)3 / 12 = 0.05729 in 4 Mc ( Fl / 4)c 4σ I 4(60)103 (0.05729) σ= = ⇒ F= = = 1528 lbf I I lc ( 36 ) (0.25) (1528) ( 36 ) Fl 3 y= = = 0.864 in Ans. 48EI 48(30)106 (0.05729) ______________________________________________________________________________ 3 4-23 From the solutions to Prob. 3-68, T1 = 60 lbf and T2 = 400 lbf I= πd4 64 = π (1.25)4 64 = 0.1198 in 4 From Table A-9, beam 6, Fb x Fb x z A = 1 1 ( x 2 + b12 − l 2 ) + 2 2 ( x 2 + b2 2 − l 2 ) 6 EIl 6 EIl x =10in (−575)(30)(10) = (102 + 302 − 402 ) 6(30)106 (0.1198)(40) 460(12)(10) + 102 + 122 − 402 ) = 0.0332 in ( 6 6(30)10 (0.1198)(40) Ans. dz d Fb x Fb x = − 1 1 ( x 2 + b12 − l 2 ) + 2 2 ( x 2 + b2 2 − l 2 ) 6 EIl dx x =10in x =10in dx 6 EIl Fb Fb = − 1 1 (3 x 2 + b12 − l 2 ) + 2 2 (3 x 2 + b2 2 − l 2 ) 6 EIl 6 EIl x =10in (θ A ) y = − =− (575)(30) 3 (102 ) + 302 − 402 6(30)106 (0.1198)(40) −460(12) 3 (102 ) + 122 − 402 − 6 6(30)10 (0.1198)(40) = 6.02(10−4 ) rad Ans. ______________________________________________________________________________ 4-24 From the solutions to Prob. 3-69, T1 = 2880 N and T2 = 432 N I= πd4 64 = π (30)4 Shigley’s MED, 10th edition 64 ( ) = 39.76 103 mm 4 Chapter 4 Solutions, Page 16/80 www.konkur.in The load in between the supports supplies an angle to the overhanging end of the beam. That angle is found by taking the derivative of the deflection from that load. From Table A-9, beams 6 (subscript 1) and 10 (subscript 2), y A = θ BC C ( a2 ) beam 6 + ( y A )beam10 (1) d F a ( l − x ) 2 Fa = 1 1 x + a12 − 2lx ) = 1 1 ( 6lx − 3 x 2 − a12 − 2l 2 ) ( x =l dx 6 EIl x =l 6 EIl Fa = 1 1 ( l 2 − a12 ) 6 EIl Equation (1) is thus θ BC C F1a1 2 Fa2 l − a12 ) a2 − 2 2 (l + a2 ) ( 6 EIl 3EI −3312(230) 2070(3002 ) 2 2 = 510 − 230 300 − ( ) ( 510 + 300 ) ( ) 6(207)103 (39.76)103 (510) 3(207)103 (39.76)103 = −7.99 mm Ans. yA = The slope at A, relative to the z axis is (θ A ) z = F1a1 2 d F (x − l) ( x − l )2 − a2 (3x − l ) (l − a12 ) + 2 6 EIl x =l + a2 dx 6 EI F1a1 2 F l − a12 ) + 2 3( x − l ) 2 − 3a2 ( x − l ) − a2 (3 x − l ) ( x = l + a2 6 EIl 6 EI Fa F = 1 1 (l 2 − a12 ) − 2 ( 3a2 2 + 2la2 ) 6 EIl 6 EI −3312(230) = (5102 − 2302 ) 6(207)103 (39.76)103 (510) 2070 3(3002 ) + 2(510)(300) − 3 3 6(207)10 (39.76)10 = −0.0304 rad Ans. ______________________________________________________________________________ = 4-25 From the solutions to Prob. 3-70, T1 = 392.16 lbf and T2 = 58.82 lbf I= πd4 = 64 From Table A-9, beam 6, Shigley’s MED, 10th edition π (1) 4 64 = 0.049 09 in 4 Chapter 4 Solutions, Page 17/80 www.konkur.in ( −350)(14)(8) Fb x y A = 1 1 ( x 2 + b12 − l 2 ) = (82 + 142 − 222 ) = 0.0452 in Ans. 6 6 EIl x =8in 6(30)10 (0.049 09)(22) ( −450.98)(6)(8) F b x z A = 2 2 ( x 2 + b2 2 − l 2 ) = (82 + 62 − 222 ) = 0.0428 in Ans. 6 6 EIl x =8in 6(30)10 (0.049 09)(22) The displacement magnitude is δ = y A2 + z A2 = 0.04522 + 0.04282 = 0.0622 in Ans. d y d Fb x Fb = 1 1 ( x 2 + b12 − l 2 ) = 1 1 (3a12 + b12 − l 2 ) x = a1 6 EIl d x x = a1 dx 6 EIl (θ A ) z = = (−350)(14) 3 ( 82 ) + 142 − 222 = 0.00242 rad 6 6(30)10 (0.04909)(22) Ans. dz d −F b x Fb = − 2 2 ( x 2 + b22 − l 2 ) = 2 2 ( 3a12 + b22 − l 2 ) x = a1 6 EIl dx 6 EIl d x x = a1 (θ A ) y = − = (450.98)(6) 3 ( 82 ) + 62 − 222 = −0.00356 rad 6 6(30)10 (0.04909)(22) Ans. The slope magnitude is Θ A = 0.002422 + ( −0.00356 ) = 0.00430 rad Ans. 2 ______________________________________________________________________________ 4-26 From the solutions to Prob. 3-71, T1 = 250 N and T2 = 37.5 N I= πd4 64 = π (20) 4 64 = 7 854 mm 4 ( −345sin 45o ) (550)(300) 3002 + 5502 − 8502 F1 y b1 x 2 yA = ( x + b12 − l 2 ) = ( ) 6(207)103 (7 854)(850) 6 EIl x =300mm = 1.60 mm Ans. Fb x F b x z A = 1z 1 ( x 2 + b12 − l 2 ) + 2 2 ( x 2 + b2 2 − l 2 ) 6 EIl 6 EIl x =300mm ( 345cos 45 ) (550)(300) o = ( 300 6(207)10 (7 854)(850) 3 + 2 + 5502 − 8502 ) −287.5(150)(300) ( 3002 + 1502 − 8502 ) = −0.650 mm 6(207)103 (7 854)(850) Ans. The displacement magnitude is δ = y A2 + z A2 = 1.602 + ( −0.650 ) = 1.73 mm 2 Shigley’s MED, 10th edition Ans. Chapter 4 Solutions, Page 18/80 www.konkur.in d F b x F b d y = 1 y 1 ( x 2 + b12 − l 2 ) = 1 y 1 (3a12 + b12 − l 2 ) d x x = a1 dx 6 EIl x = a1 6 EIl (θ A ) z = = − ( 345sin 45o ) (550) 3 ( 3002 ) + 5502 − 8502 = 0.00243 rad 6(207)10 (7 854)(850) Ans. 3 dz d F b x Fb x = − 1z 1 ( x 2 + b12 − l 2 ) + 2 2 ( x 2 + b2 2 − l 2 ) 6 EIl x = a1 dx 6 EIl d x x = a1 (θ A ) y = − F1z b1 Fb 3a12 + b12 − l 2 ) − 2 2 ( 3a12 + b22 − l 2 ) ( 6 EIl 6 EIl o ( 345cos 45 ) (550) 3 3002 + 5502 − 8502 =− ( ) 6(207)103 (7 854)(850) =− − −287.5(150) 3 ( 3002 ) + 1502 − 8502 = 1.91 ⋅10−4 rad 6(207)103 (7 854)(850) Ans. The slope magnitude is Θ A = 0.002432 + 0.0001912 = 0.00244 rad Ans. ______________________________________________________________________________ 4-27 From the solutions to Prob. 3-72, FB = 750 lbf I= πd4 64 = π (1.25)4 64 = 0.1198 in 4 From Table A-9, beams 6 (subscript 1) and 10 (subscript 2) F2 y a2 x 2 F1 y b1 x 2 yA = x + b12 − l 2 ) + l − x2 ) ( ( 6 EIl 6 EIl x =16in ( −300 cos 20 ) (14)(16) o = 6 6(30)10 (0.119 8)(30) = 0.0805 in Ans. (162 + 142 − 302 ) + ( 750sin 20 ) (9)(16) o ( 30 6(30)10 (0.119 8)(30) 2 6 − 162 ) F ax F b x z A = 1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( l 2 − x 2 ) 6 EIl 6 EIl x =16in ( 300sin 20 ) (14)(16) o = 6(30)106 (0.119 8)(30) = −0.1169 in (16 2 + 14 − 30 2 2 ) ( −750 cos 20 ) (9)(16) + o 6(30)106 (0.119 8)(30) ( 30 2 − 162 ) Ans. The displacement magnitude is δ = y A2 + z A2 = 0.08052 + ( −0.1169 ) = 0.142 in 2 Shigley’s MED, 10th edition Ans. Chapter 4 Solutions, Page 19/80 www.konkur.in d F b x F ax d y = 1 y 1 ( x 2 + b12 − l 2 ) + 2 y 2 ( l 2 − x 2 ) 6 EIl d x x = a1 dx 6 EIl x = a1 (θ A ) z = ( 3a + b − l ) + 6EIl ( l − 3a ) 6 EIl ( −300 cos 20 ) (14) 3 16 + 14 − 30 = ( ) 6(30)10 (0.119 8)(30) = F1 y b1 2 1 2 1 2 F2 y a2 2 2 1 o 2 2 2 6 ( 750sin 20 ) (9) o + 302 − 3 (162 ) = 8.06 (10−5 ) rad 6(30)10 (0.119 8)(30) 6 Ans. dz d F b x F ax = − 1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( l 2 − x 2 ) 6 EIl x = a1 dx 6 EIl d x x = a1 (θ A ) y = − F1z b1 F a 3a12 + b12 − l 2 ) − 2 z 2 ( l 2 − 3a12 ) ( 6 EIl 6 EIl o 300sin 20 ) (14) −750 cos 20o ) (9) ( ( 2 2 2 302 − 3 (162 ) =− 3 (16 ) + 14 − 30 − 6 6 6(30)10 (0.119 8)(30) 6(30)10 (0.119 8)(30) =− = 0.00115 rad Ans. The slope magnitude is Θ A = 8.06 (10−5 ) + 0.001152 = 0.00115 rad Ans. ______________________________________________________________________________ 2 4-28 From the solutions to Prob. 3-73, FB = 22.8 (103) N 4 π d 4 π ( 50 ) I= = = 306.8 (103 ) mm 4 64 64 From Table A-9, beam 6, F bx F b x y A = 1 y 1 ( x 2 + b12 − l 2 ) + 2 y 2 ( x 2 + b2 2 − l 2 ) 6 EIl 6 EIl x = 400mm 11(103 ) sin 20o (650)(400) = ( 4002 + 6502 − 10502 ) 6(207)103 (306.8)103 (1050) 22.8 (103 ) sin 25o (300)(400) + ( 4002 + 3002 − 10502 ) 6(207)103 (306.8)103 (1050) = −3.735 mm Ans. Shigley’s MED, 10th edition Chapter 4 Solutions, Page 20/80 www.konkur.in F bx F b x z A = 1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( x 2 + b2 2 − l 2 ) 6 EIl 6 EIl x = 400mm 11(103 ) cos 20o (650)(400) = ( 4002 + 6502 − 10502 ) 6(207)103 (306.8)103 (1050) −22.8 (103 ) cos 25o (300)(400) + ( 4002 + 3002 − 10502 ) = 1.791 mm 6(207)103 (306.8)103 (1050) The displacement magnitude is δ = y A2 + z A2 = ( −3.735) 2 + 1.7912 = 4.14 mm Ans. Ans. d y d F b x F bx = 1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( x 2 + b2 2 − l 2 ) 6 EIl x = a1 d x x = a1 dx 6 EIl (θ A ) z = ( 3a + b − l ) + 6 EIl ( 3a + b − l ) 6 EIl 11(10 ) sin 20 (650) 3 ( 400 ) + 650 = 6(207)10 (306.8)10 (1050) = F1 y b1 2 1 3 2 1 F2 y b2 2 2 1 2 2 o 2 3 2 3 2 − 10502 22.8 (103 ) sin 25o (300) 3 ( 4002 ) + 3002 − 10502 + 6(207)103 (306.8)103 (1050) = −0.00507 rad Ans. dz d F b x F bx = − 1z 1 ( x 2 + b12 − l 2 ) + 2 z 2 ( x 2 + b2 2 − l 2 ) 6 EIl x = a1 dx 6 EIl d x x = a1 (θ A ) y = − F1z b1 F b 3a12 + b12 − l 2 ) − 2 z 2 ( 3a12 + b22 − l 2 ) ( 6 EIl 6 EIl 3 o 11(10 ) cos 20 (650) 3 ( 4002 ) + 6502 − 10502 =− 6(207)103 (306.8)103 (1050) =− −22.8 (103 ) cos 25o (300) 3 ( 4002 ) + 3002 − 10502 − 6(207)103 (306.8)103 (1050) = −0.00489 rad Ans. The slope magnitude is Θ A = ( −0.00507 ) + ( −0.00489 ) 2 2 = 0.00704 rad Ans. ______________________________________________________________________________ 4-29 From the solutions to Prob. 3-68, T1 = 60 lbf and T2 = 400 lbf , and Prob. 4-23, I = 0.119 8 in4. From Table A-9, beam 6, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 21/80 www.konkur.in dz d F1z b1 x 2 F bx x + b12 − l 2 ) + 2 z 2 ( x 2 + b22 − l 2 ) ( = − 6 EIl x =0 dx 6 EIl d x x =0 F b F b −575(30) = − 1z 1 ( b12 − l 2 ) − 2 z 2 ( b2 2 − l 2 ) = − ( 302 − 402 ) 6 EIl 6 EIl 6(30)106 (0.119 8)(40) 460(12) − 122 − 402 ) = −0.00468 rad Ans. ( 6 6(30)10 (0.119 8)(40) (θO ) y = − F a (l − x ) 2 dz d F1z a1 ( l − x ) 2 x + a12 − 2lx ) + 2 z 2 x + a22 − 2lx ) ( ( = − 6 EIl dx 6 EIl d x x =l x =l F a F a = − 1z 1 ( 6lx − 2l 2 − 3 x 2 − a12 ) + 2 z 2 ( 6lx − 2l 2 − 3 x 2 − a22 ) 6 EIl 6 EIl x =l (θC ) y = − F1z a1 2 F a l − a12 ) − 2 z 2 ( l 2 − a22 ) ( 6 EIl 6 EIl 2 −575(10) ( 40 − 102 ) 460(28) ( 402 − 282 ) =− − = −0.00219 rad Ans. 6(30)106 (0.119 8)(40) 6(30)106 (0.119 8)(40) ______________________________________________________________________________ =− 4-30 From the solutions to Prob. 3-69, T1 = 2 880 N and T2 = 432 N, and Prob. 4-24, I = 39.76 (103) mm4. From Table A-9, beams 6 and 10 d y d Fb x Fa x (θO ) z = = 1 1 ( x 2 + b12 − l 2 ) + 2 2 (l 2 − x 2 ) 6 EIl x =0 d x x = 0 dx 6 EIl Fa Fb Fal Fb = 1 1 (3 x 2 + b12 − l 2 ) + 2 2 (l 2 − 3x 2 ) = 1 1 (b12 − l 2 ) + 2 2 6 EIl 6 EI 6 EIl x = 0 6 EIl −3 312(280) 2 070(300)(510) = 2802 − 5102 ) + ( 3 3 6(207)10 (39.76)10 (510) 6(207)103 (39.76)103 = 0.0131 rad Ans. d y d F1a1 (l − x) 2 Fa x ( x + a12 − 2lx) + 2 2 (l 2 − x 2 ) = 6 EIl x =l d x x =l dx 6 EIl (θC ) z = Fa Fa Fal Fa = 1 1 (6lx − 2l 2 − 3x 2 − a12 ) + 2 2 (l 2 − 3x 2 ) = 1 1 (l 2 − a12 ) − 2 2 6 EIl 3EI 6 EIl x =l 6 EIl −3 312(230) 2 070(300)(510) = (5102 − 2302 ) − 3 3 6(207)10 (39.76)10 (510) 3(207)103 (39.76)103 = −0.0191 rad Ans. ______________________________________________________________________________ 4-31 From the solutions to Prob. 3-70, T1 = 392.19 lbf and T2 = 58.82 lbf , and Prob. 4-25, I = 0.0490 9 in4. From Table A-9, beam 6 Shigley’s MED, 10th edition Chapter 4 Solutions, Page 22/80 www.konkur.in d F1 y b1 x 2 F1 y b1 2 2 d y x + b12 − l 2 ) = (b1 − l ) ( = d x x = 0 dx 6 EIl x = 0 6 EIl −350(14) = 142 − 222 ) = 0.00726 rad Ans. ( 6 6(30)10 (0.04909)(22) (θO ) z = dz d F2 z b2 x 2 F b x + b22 − l 2 ) = − 2 z 2 ( b22 − l 2 ) ( = − 6 EIl x =0 dx 6 EIl d x x =0 −450.98(6) =− 62 − 222 ) ( 6 6(30)10 (0.04909)(22) (θO ) y = − = −0.00624 rad Ans. The slope magnitude is ΘO = 0.007262 + ( −0.00624 ) = 0.00957 rad Ans. 2 d y d F1 y a1 (l − x) 2 x + a12 − 2lx ) ( = d x x =l dx 6 EIl x =l F a F a = 1 y 1 ( 6lx − 2l 2 − 3 x 2 − a12 ) = 1 y 1 (l 2 − a12 ) 6 EIl x =l 6 EIl −350(8) = ( 222 − 82 ) = −0.00605 rad Ans. 6(30)106 (0.0491)(22) (θC ) z = dz d F2 z a2 (l − x) 2 x + a22 − 2lx ) ( = − x =l dx 6 EIl d x x =l (θC ) y = − F a F a = − 2 z 2 ( 6lx − 2l 2 − 3 x 2 − a22 ) = − 2 z 2 ( l 2 − a22 ) 6 EIl 6 EIl x =l −450.98(16) =− ( 222 − 162 ) = 0.00846 rad 6(30)106 (0.04909)(22) The slope magnitude is ΘC = ( −0.00605) 2 Ans. + 0.008462 = 0.0104 rad Ans. ______________________________________________________________________________ 4-32 From the solutions to Prob. 3-71, T1 =250 N and T1 =37.5 N, and Prob. 4-26, I = 7 854 mm4. From Table A-9, beam 6 F b d y d F1 y b1 x 2 x + b12 − l 2 ) = 1 y 1 (b12 − l 2 ) ( = d x x = 0 dx 6 EIl x = 0 6 EIl (θO ) z = −345sin 45o (550) = ( 5502 − 8502 ) = 0.00680 rad 6(207)103 (7 854)(850) Shigley’s MED, 10th edition Ans. Chapter 4 Solutions, Page 23/80 www.konkur.in dz d F1z b1 x 2 F bx x + b12 − l 2 ) + 2 z 2 ( x 2 + b22 − l 2 ) ( = − 6 EIl x =0 dx 6 EIl d x x=0 (θO ) y = − 345cos 45o (550) F1z b1 2 2 F2 z b2 2 2 b − l − b − l = − 5502 − 8502 ) ( ) ( ) ( 1 2 3 6 EIl 6 EIl 6(207)10 (7 854)(850) −287.5(150) − (1502 − 8502 ) = 0.00316 rad Ans. 6(207)103 (7 854)(850) =− The slope magnitude is ΘO = 0.006802 + 0.003162 = 0.00750 rad Ans. d F1 y a1 (l − x) 2 F1 y a1 d y x + a12 − 2lx ) = 6lx − 2l 2 − 3 x 2 − a12 ) ( ( = d x x =l dx 6 EIl x =l 6 EIl x =l (θC ) z = −345sin 45o (300) = (l − a ) = 8502 − 3002 ) = −0.00558 rad ( 3 6 EIl 6(207)10 (7 854)(850) F1 y a1 2 2 1 Ans. dz d F1z a1 (l − x) 2 F a (l − x) 2 x + a12 − 2lx ) + 2 z 2 x + a22 − 2lx ) ( ( = − 6 EIl x =l dx 6 EIl d x x =l (θC ) y = − 345cos 45o (300) F1z a1 2 F2 z a2 2 2 2 =− l − a1 ) − l − a2 ) = − 8502 − 3002 ) ( ( ( 3 6 EIl 6 EIl 6(207)10 (7 854)(850) −287.5(700) − 8502 − 7002 ) = 6.04 (10−5 ) rad Ans. ( 3 6(207)10 (7 854)(850) The slope magnitude is ΘC = ( −0.00558) 2 + 6.04 (10−5 ) = 0.00558 rad Ans. 2 ________________________________________________________________________ 4-33 From the solutions to Prob. 3-72, FB = 750 lbf, and Prob. 4-27, I = 0.119 8 in4. From Table A-9, beams 6 and 10 d F b x F ax d y (θO ) z = = 1 y 1 ( x 2 + b12 − l 2 ) + 2 y 2 ( l 2 − x 2 ) 6 EIl d x x = 0 dx 6 EIl x = 0 F2 y a2 2 F1 y b1 2 2 F2 y a2l F1 y b1 = 3 x 2 + b12 − l 2 ) + l − 3x 2 ) = ( ( ( b1 − l ) + 6EI 6 EIl 6 EIl x =0 6 EIl −300 cos 20o (14) 750sin 20o (9)(30) 2 2 = (14 − 30 ) + 6(30)106 (0.119 8) = 0.00751 rad 6(30)106 (0.119 8)(30) Shigley’s MED, 10th edition Ans. Chapter 4 Solutions, Page 24/80 www.konkur.in dz d F1z b1 x 2 F ax x + b12 − l 2 ) + 2 z 2 ( l 2 − x 2 ) ( = − 6 EIl x =0 dx 6 EIl d x x =0 (θO ) y = − F a F b F al F b = − 1z 1 ( 3 x 2 + b12 − l 2 ) + 2 z 2 ( l 2 − 3x 2 ) = − 1z 1 ( b12 − l 2 ) − 2 z 2 6 EIl 6 EIl 6 EI 6 EIl x=0 300sin 20o (14) −750 cos 20o (9)(30) 2 2 =− (14 − 30 ) − 6(30)106 (0.119 8) = 0.0104 rad 6(30)106 (0.119 8)(30) Ans. The slope magnitude is ΘO = 0.007512 + 0.0104 2 = 0.0128 rad Ans. d F1 y a1 (l − x) 2 F2 y a2 x 2 dy x + a12 − 2lx ) + l − x 2 ) ( ( = 6 EIl dx x =l dx 6 EIl x =l F2 y a2 2 F1 y a1 2 F2 y a2l F1 y a1 = 6lx − 2l 2 − 3x 2 − a12 ) + l − 3x 2 ) = (l − a12 ) − ( ( 6 EIl 3EI 6 EIl x =l 6 EIl (θC ) z = −300 cos 20o (16) 750sin 20o (9)(30) 2 2 = 30 − 16 − ( ) 3(30)106 (0.119 8) = −0.0109 rad 6(30)106 (0.119 8)(30) Ans. dz d F1z a1 (l − x) 2 F ax x + a12 − 2lx ) + 2 z 2 ( l 2 − x 2 ) ( = − 6 EIl x =l dx 6 EIl d x x =l (θC ) y = − F a F a F al F a = − 1z 1 ( 6lx − 2l 2 − 3 x 2 − a12 ) + 2 z 2 ( l 2 − 3 x 2 ) = − 1z 1 ( l 2 − a12 ) + 2 z 2 6 EIl 6 EIl 3EI 6 EIl x =l 300sin 20o (16) −750 cos 20o (9)(30) 2 2 =− ( 30 − 16 ) + 3(30)106 (0.119 8) = −0.0193 rad 6(30)106 (0.119 8)(30) ( −0.0109) + ( −0.0193) The slope magnitude is ΘC = 2 2 Ans. = 0.0222 rad Ans. ______________________________________________________________________________ 4-34 From the solutions to Prob. 3-73, FB = 22.8 kN, and Prob. 4-28, I = 306.8 (103) mm4. From Table A-9, beam 6 d F1 y b1 x 2 F bx d y x + b12 − l 2 ) + 2 y 2 ( x 2 + b22 − l 2 ) ( = 6 EIl d x x = 0 dx 6 EIl x = 0 (θO ) z = 11(103 ) sin 20o (650) = b −l )+ b −l ) = 6502 − 10502 ) ( ( ( 3 3 6 EIl 6 EIl 6(207)10 (306.8)10 (1050) F1 y b1 2 1 2 F2 y b2 2 2 2 22.8 (103 ) sin 25o (300) + 3002 − 10502 ) = −0.0115 rad ( 3 3 6(207)10 (306.8)10 (1050) Shigley’s MED, 10th edition Ans. Chapter 4 Solutions, Page 25/80 www.konkur.in dz d F1z b1 x 2 F bx x + b12 − l 2 ) + 2 z 2 ( x 2 + b22 − l 2 ) ( = − 6 EIl x =0 dx 6 EIl d x x =0 F b F b = − 1z 1 ( b12 − l 2 ) − 2 z 2 ( b22 − l 2 ) 6 EIl 6 EIl 11(103 ) cos 20o (650) =− 6502 − 10502 ) ( 3 3 6(207)10 (306.8)10 (1050) (θO ) y = − −22.8 (103 ) cos 25o (300) − 3002 − 10502 ) = −0.00427 rad ( 3 3 6(207)10 (306.8)10 (1050) ( −0.0115) + ( −0.00427 ) The slope magnitude is ΘO = 2 2 Ans. = 0.0123 rad Ans. d F1 y a1 (l − x) 2 F2 y a2 (l − x) 2 d y x + a12 − 2lx ) + x + a22 − 2lx ) ( ( = 6 EIl d x x =l dx 6 EIl x =l (θC ) z = F a F a = 1 y 1 (6lx − 2l 2 − 3 x 2 − a12 ) + 2 y 2 ( 6lx − 2l 2 − 3x 2 − a22 ) 6 EIl 6 EIl x =l 11(103 ) sin 20o (400) = l −a )+ l −a ) = 10502 − 4002 ) ( ( ( 3 3 6 EIl 6 EIl 6(207)10 (306.8)10 (1050) F1 y a1 2 2 1 F2 y a2 2 2 2 22.8 (103 ) sin 25o (750) + 10502 − 7502 ) = 0.0133 rad ( 3 3 6(207)10 (306.8)10 (1050) Ans. dz d F1z a1 (l − x) 2 F a (l − x) 2 x + a12 − 2lx ) + 2 z 2 x + a22 − 2lx ) ( ( = − 6 EIl x =l dx 6 EIl d x x =l (θC ) y = − F a F a = − 1z 1 ( 6lx − 2l 2 − 3x 2 − a12 ) + 2 z 2 ( 6lx − 2l 2 − 3 x 2 − a22 ) 6 EIl 6 EIl x =l 11(103 ) cos 20o (400) F1z a1 2 F2 z a2 2 2 2 =− ( l − a1 ) − 6EIl ( l − a2 ) = − 6(207)10 (10502 − 4002 ) 3 6 EIl (306.8)103 (1050) −22.8 (103 ) cos 25o (750) − (10502 − 7502 ) = 0.0112 rad 6(207)103 (306.8)103 (1050) Ans. The slope magnitude is ΘC = 0.01332 + 0.0112 2 = 0.0174 rad Ans. ______________________________________________________________________________ 4-35 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad. In Prob. 4-29, I = 0.119 8 in4, and it was found that the greater angle occurs at the bearing at O where (θO)y = − 0.00468 rad. Sinceθ is inversely proportional to I, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 26/80 www.konkur.in θ new Inew = θ old Iold ⇒ 4 Inew = π d new /64 = θ old Iold /θ new 1/4 or, d new 64 θ old = I old π θ new The absolute sign is used as the old slope may be negative. 1/4 64 −0.00468 d new = 0.119 8 = 1.82 in Ans. π 0.00105 ______________________________________________________________________________ 4-36 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad. In Prob. 4-30, I = 39.76 (103) mm4, and it was found that the greater angle occurs at the bearing at C where (θC)y = − 0.0191 rad. See the solution to Prob. 4-35 for the development of the equation 1/4 d new 64 θ old = I old π θ new 1/4 64 −0.0191 d new = 39.76 (103 ) = 62.0 mm Ans. π 0.00105 ______________________________________________________________________________ 4-37 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad. In Prob. 4-31, I = 0.0491 in4, and the maximum slope is ΘC = 0.0104 rad. See the solution to Prob. 4-35 for the development of the equation 1/4 d new 64 θ old = I old π θ new 1/4 64 0.0104 d new = 0.0491 = 1.77 in Ans. π 0.00105 ______________________________________________________________________________ 4-38 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad. In Prob. 4-32, I = 7 854 mm4, and the maximum slope is ΘO = 0.00750 rad. See the solution to Prob. 4-35 for the development of the equation Shigley’s MED, 10th edition Chapter 4 Solutions, Page 27/80 www.konkur.in 1/4 d new 64 θ old = I old π θ new 1/4 64 0.00750 d new = 7 854 = 32.7 mm Ans. π 0.00105 ______________________________________________________________________________ 4-39 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad. In Prob. 4-33, I = 0.119 8 in4, and the maximum slope Θ = 0.0222 rad. See the solution to Prob. 4-35 for the development of the equation 1/4 d new 64 θ old = I old π θ new 1/4 64 0.0222 d new = 0.119 8 = 2.68 in Ans. π 0.00105 ______________________________________________________________________________ 4-40 The required new slope in radians is θ new = 0.06(π /180) = 0.00105 rad. In Prob. 4-34, I = 306.8 (103) mm4, and the maximum slope is ΘC = 0.0174 rad. See the solution to Prob. 4-35 for the development of the equation 1/4 d new 64 θ old = I old π θ new 1/4 64 0.0174 d new = 306.8 (103 ) = 100.9 mm Ans. π 0.00105 ______________________________________________________________________________ 4-41 IAB = π 14/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = (0.25)(1.5)3/12 = 0.07031 in4, ICD = π (3/4)4/64 = 0.01553 in4. For Eq. (3-41), p. 116, b/c = 1.5/0.25 = 6 ⇒ β = 0.299. The deflection can be broken down into several parts 1. The vertical deflection of B due to force and moment acting on B (y1). 2. The vertical deflection due to the slope at B, θB1, due to the force and moment acting on B (y2 = CD θB1 = 2θB1). Shigley’s MED, 10th edition Chapter 4 Solutions, Page 28/80 www.konkur.in 3. The vertical deflection due to the rotation at B, θB2, due to the torsion acting at B (y3 = BC θB1 = 5θB1). 4. The vertical deflection of C due to the force acting on C (y4). 5. The rotation at C, θC, due to the torsion acting at C (y3 = CD θC = 2θC). 6. The vertical deflection of D due to the force acting on D (y5). 1. From Table A-9, beams 1 and 4 with F = − 200 lbf and MB = 2(200) = 400 lbf⋅in −200 ( 63 ) 400 ( 62 ) y1 = − + = 0.01467 in 3 ( 30 )106 ( 0.04909 ) 2 ( 30 )106 ( 0.04909 ) 2. From Table A-9, beams 1 and 4 2 M x2 M x Fx d Fx θ B1 = ( x − 3l ) + B = ( 3x − 6l ) + B 2 EI x =l 6 EI EI x =l dx 6 EI 6 l = − ( −200 )( 6 ) + 2 ( 400 ) = 0.004074 rad [ − Fl + 2M B ] = 6 2 EI 2 ( 30 )10 ( 0.04909 ) y 2 = 2(0.004072) = 0.00815 in 3. The torsion at B is TB = 5(200) = 1000 lbf⋅in. From Eq. (4-5) 1000 ( 6 ) TL = 0.005314 rad = 6 JG AB 0.09818 (11.5 )10 θB2 = y 3 = 5(0.005314) = 0.02657 in 4. For bending of BC, from Table A-9, beam 1 y4 = − −200 ( 53 ) 3 ( 30 )106 ( 0.07031) = 0.00395 in 5. For twist of BC, from Eq. (3-41), p. 116, with T = 2(200) = 400 lbf⋅in θC = 400 ( 5 ) 0.299 (1.5 ) 0.253 (11.5 )106 = 0.02482 rad y 5 = 2(0.02482) = 0.04964 in 6. For bending of CD, from Table A-9, beam 1 y6 = − −200 ( 23 ) 3 ( 30 )106 ( 0.01553) Shigley’s MED, 10th edition = 0.00114 in Chapter 4 Solutions, Page 29/80 www.konkur.in Summing the deflections results in 6 yD = ∑ yi = 0.01467 + 0.00815 + 0.02657 + 0.00395 + 0.04964 + 0.00114 = 0.1041 in Ans. i =1 This problem is solved more easily using Castigliano’s theorem. See Prob. 4-71. ______________________________________________________________________________ 4-42 The deflection of D in the x direction due to Fz is from: 1. The deflection due to the slope at B, θB1, due to the force and moment acting on B (x1 = BC θB1 = 5θB1). 2. The deflection due to the moment acting on C (x2). 1. For AB, IAB = π 14/64 = 0.04909 in4. From Table A-9, beams 1 and 4 2 M B x 2 M x d Fx Fx x − 3 l + ( ) ( 3x − 6l ) + B = 2 EI x =l 6 EI EI x =l dx 6 EI θ B1 = 6 l = − (100 )( 6 ) + 2 ( −200 ) = −0.002037 rad [ − Fl + 2M B ] = 6 2 EI 2 ( 30 )10 ( 0.04909 ) x 1 = 5(− 0.002037) = − 0.01019 in 2. For BC, IBC = (1.5)(0.25)3/12 = 0.001953 in4. From Table A-9, beam 4 x2 = 2 ( −100 ) 5 MCl2 = = −0.04267 in 2 EI 2 ( 30 )106 ( 0.001953) The deflection of D in the x direction due to Fx is from: 3. The elongation of AB due to the tension. For AB, the area is A = π 12/4 = 0.7854 in2 −150 ( 6 ) Fl x3 = = −3.82 (10−5 ) in = 6 AE AB 0.7854 ( 30 )10 4. The deflection due to the slope at B, θB2, due to the moment acting on B (x1 = BC θB2 = 5θB2). With IAB = 0.04907 in4, θB2 = 5 ( −150 ) 6 M Bl = = −0.003056 rad EI 30 (106 ) 0.04909 Shigley’s MED, 10th edition Chapter 4 Solutions, Page 30/80 www.konkur.in x4 = 5(− 0.003056) = − 0.01528 in 5. The deflection at C due to the bending force acting on C. With IBC = 0.001953 in4 150 ( 53 ) Fl 3 x5 = − = − = −0.10667 in 3 ( 30 )106 ( 0.001953) 3EI BC 6. The elongation of CD due to the tension. For CD, the area is A = π (0.752)/4 = 0.4418 in2 −150 ( 2 ) Fl x6 = = −2.26 (10−5 ) in = 6 AE 0.4418 30 10 ( ) CD Summing the deflections results in xD = ∑ xi = −0.01019 − 0.04267 − 3.82 (10−5 ) 6 i =1 − 0.01528 − 0.10667 − 2.26 (10−5 ) = −0.1749 in Ans. ______________________________________________________________________________ 4-43 JOA = JBC = π (1.54)/32 = 0.4970 in4, JAB = π (14)/32 = 0.09817 in4, IAB = π (14)/64 = 0.04909 in4, and ICD = π (0.754)/64 = 0.01553 in4. T lOA l AB lBC Tl Tl Tl + + θ = + + = GJ OA GJ AB GJ BC G J OA J AB J BC 250(12) 2 9 2 + + = 0.0260 rad 6 11.5(10 ) 0.4970 0.09817 0.4970 Simplified Tl 250(12)(13) = θs = GJ 11.5 (106 ) ( 0.09817 ) = Ans. θ s = 0.0345 rad Ans. Simplified is 0.0345/0.0260 = 1.33 times greater Ans. yD = Fy lOC 3 3EI AB + θ s ( lCD ) + Fy lCD 3 3EI CD = 250 (133 ) 3(30)106 ( 0.04909 ) + 0.0345(12) + 250 (123 ) 3(30)106 ( 0.01553) yD = 0.847 in Ans. ______________________________________________________________________________ 4-44 Reverse the deflection equation of beam 7 of Table A-9. Using units in lbf, inches Shigley’s MED, 10th edition Chapter 4 Solutions, Page 31/80 www.konkur.in y=− ( 3000 / 12 ) x 2 25 x 2 − x3 − 25 12 3 wx 2lx 2 − x3 − l 3 ) = − ( ) ( ( ) 24 EI 24 ( 30 )106 ( 485 ) { = 7.159 (10−10 ) x 27 (106 ) − 600 x 2 + x3 } Ans. The maximum height occurs at x = 25(12)/2 = 150 in ymax = 7.159 (10−10 )150 27 (106 ) − 600 (1502 ) + 1503 = 1.812 in Ans. ______________________________________________________________________________ 4-45 From Table A-9, beam 6, Fbx 2 yL = x + b2 − l 2 ) ( 6 EIl Fb 3 yL = x + b2 x − l 2 x ) ( 6 EIl dyL Fb = ( 3x 2 + b2 − l 2 ) dx 6 EIl dyL dx Let ξ = dyL dx = Fb ( b 2 − l 2 ) 6 EIl x =0 and set I = x =0 dL = π d L4 64 . Thus, 32 Fb ( b 2 − l 2 ) 1/4 Ans. 3π Elξ For the other end view, observe beam 6 of Table A-9 from the back of the page, noting that a and b interchange as do x and –x dR = 32 Fa ( l 2 − a 2 ) 3π Elξ 1/4 Ans. For a uniform diameter shaft the necessary diameter is the larger of d L and d R . ______________________________________________________________________________ 4-46 The maximum slope will occur at the left bearing. Incorporating a design factor into the solution for d L of Prob. 4-45, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 32/80 www.konkur.in 32nFb ( l 2 − b 2 ) d= 3π Elξ 1/4 d= 4 32(1.28)(3000)(200) ( 3002 − 2002 ) 3π (207)103 (300)(0.001) d = 38.1 mm Ans. π ( 38.14 ) = 103.4 (103 ) mm 4 64 From Table A-9, beam 6, the maximum deflection will occur in BC where dyBC /dx = 0 I= d Fa ( l − x ) 2 x + a 2 − 2lx ) = 0 ⇒ 3 x 2 − 6lx + ( a 2 + 2l 2 ) = 0 ( dx 6 EIl 3x 2 − 6 ( 300 ) x + 1002 + 2 ( 3002 ) = 0 ⇒ x 2 − 600 x + 63333 = 0 x= 1 600 ± 6002 − 4(1)63 333 = 463.3, 136.7 mm 2 x = 136.7 mm is acceptable. Fa ( l − x ) 2 ymax = x + a 2 − 2lx ) ( 6 EIl x =136.7 mm 3 (103 )100 ( 300 − 136.7 ) 136.7 2 + 1002 − 2 ( 300 )136.7 = −0.0678 mm Ans. 6 ( 207 )10 (103.4 )10 ( 300 ) ______________________________________________________________________________ = 3 3 4-47 I = π (1.254)/64 = 0.1198 in4. From Table A-9, beam 6 F1a1 (l − x) 2 F b x ( x + a12 − 2lx) + 2 2 ( x 2 + b2 2 − l 2 ) 6 EIl 6 EIl 2 2 δ= 2 150(5)(20 − 8) 2 2 = (8 + 5 − 2(20)(8) ) 6 6(30)10 ( 0.1198 ) (20) 250(10)(8) 2 2 2 + 8 + 10 − 20 ( ) 6 6(30)10 ( 0.1198 ) (20) 2 1/2 = 0.0120 in Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 4 Solutions, Page 33/80 www.konkur.in 4-48 I = π (1.254)/64 = 0.1198 in4. For both forces use beam 6 of Table A-9. For F1 = 150 lbf: 0≤x≤5 150 (15 ) x Fb x y = 1 1 ( x 2 + b12 − l 2 ) = ( x 2 + 152 − 202 ) 6 EIl 6 ( 30 )106 ( 0.1198 )( 20 ) = 5.217 (10−6 ) x ( x 2 − 175 ) 5 ≤ x ≤ 20 y= (1) F1a1 ( l − x ) 2 150 ( 5 )( 20 − x ) x 2 + 52 − 2 ( 20 ) x x + a12 − 2lx ) = ( 6 EIl 6 ( 30 )106 ( 0.1198 )( 20 ) = 1.739 (10−6 ) ( 20 − x ) ( x 2 − 40 x + 25 ) (2) For F2 = 250 lbf: 0 ≤ x ≤ 10 250 (10 ) x Fb x z = 2 2 ( x 2 + b22 − l 2 ) = x 2 + 102 − 202 ) ( 6 6 EIl 6 ( 30 )10 ( 0.1198 )( 20 ) = 5.797 (10−6 ) x ( x 2 − 300 ) (3) 10 ≤ x ≤ 20 F a (l − x ) 2 250 (10 )( 20 − x ) x 2 + 102 − 2 ( 20 ) x z= 2 2 x + a22 − 2lx ) = ( 6 EIl 6 ( 30 )106 ( 0.1198 )( 20 ) = 5.797 (10−6 ) ( 20 − x ) ( x 2 − 40 x + 100 ) (4) Plot Eqs. (1) to (4) for each 0.1 in using a spreadsheet. There are 201 data points, too numerous to tabulate here but the plot is shown below, where the maximum deflection of δ = 0.01255 in occurs at x = 9.9 in. Ans. 0.015 0.01 0.005 y (in) 0 Displacement (in) z (in) -0.005 Total (in) -0.01 -0.015 0 5 10 15 20 x (in) ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 4 Solutions, Page 34/80 www.konkur.in 4-49 The larger slope will occur at the left end. From Table A-9, beam 8 MBx 2 ( x + 3a 2 − 6al + 2l 2 ) 6 EIl dy AB M B = (3x 2 + 3a 2 − 6al + 2l 2 ) dx 6 EIl y AB = With I = π d 4/64, the slope at the left bearing is dy AB dx = θA = x=0 MB (3a 2 − 6al + 2l 2 ) 4 6 E (π d / 64 ) l Solving for d 32 M B 32(1000) 3(42 ) − 6(4)(10) + 2 (102 ) d=4 3a 2 − 6al + 2l 2 ) = 4 ( 3π Eθ Al 3π (30)106 (0.002)(10) = 0.461 in Ans. ______________________________________________________________________________ 4-50 From Table A-5, E = 10.4 Mpsi ΣMO = 0 = 18 FBC − 6(100) ⇒ FBC = 33.33 lbf The cross sectional area of rod BC is A = π (0.52)/4 = 0.1963 in2. The deflection at point B will be equal to the elongation of the rod BC. 33.33(12) FL yB = = 6.79 (10−5 ) in Ans. = 6 AE BC ( 0.1963) 30 (10 ) ______________________________________________________________________________ 4-51 ΣMO = 0 = 6 FAC − 11(100) ⇒ FAC = 183.3 lbf The deflection at point A in the negative y direction is equal to the elongation of the rod AC. From Table A-5, Es = 30 Mpsi. 183.3 (12 ) FL yA = − = −3.735 (10−4 ) in =− 2 6 AE AC π ( 0.5 ) / 4 30 (10 ) By similar triangles the deflection at B due to the elongation of the rod AC is y A yB1 = 6 18 ⇒ yB1 = 3 y A = 3( −3.735)10−4 = −0.00112 in From Table A-5, Ea = 10.4 Mpsi The bar can then be treated as a simply supported beam with an overhang AB. From Table A-9, beam 10, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 35/80 www.konkur.in dyBC Fa 2 d F (x − l) Fa 2 2 yB 2 = BD (l + a ) = 7 − (l + a ) − ( x − l ) − a (3 x − l ) x =l + a 3EI dx 6 EI dx x =l + a 3EI F Fa 2 7 Fa Fa 2 3( x − l ) 2 − 3a( x − l ) − a (3 x − l ) |x =l + a − =7 (l + a ) = − (2l + 3a ) − (l + a ) 6 EI 3EI 6 EI 3EI 100 ( 52 ) 7 (100 ) 5 =− (6 + 5) [ 2(6) + 3(5)] − 6(10.4)106 ( 0.25(23 ) / 12 ) 3(10.4)106 ( 0.25(23 ) /12 ) ( ) = −0.01438 in yB = yB1 + yB2 = − 0.00112 − 0.01438 = − 0.0155 in Ans. ______________________________________________________________________________ 4-52 From Table A-5, E = 207 GPa, and G = 79.3 GPa. Fl AB 3 FlOC l AB 2 Fl AC l AB 2 Fl AB 3 Tl Tl yB = l + l + = + + AB AB 3EI AB G (π dOC 4 / 32 ) G (π d AC 4 / 32 ) 3E (π d34 / 64 ) GJ OC GJ AC = l 32 Fl AB 2 lOC 2l AB + AC 4 + 4 4 π GdOC Gd AC 3Ed AB The spring rate is k = F/ |yB|. Thus 32l 2 k = AB π lOC l AC 2l AB + + 4 4 4 GdOC Gd AC 3Ed AB −1 −1 32 ( 2002 ) 2 ( 200 ) 200 200 = + + 3 4 3 4 3 4 π 79.3 (10 )18 79.3 (10 )12 3 ( 207 )10 ( 8 ) = 8.10 N/mm Ans. _____________________________________________________________________________ 4-53 For the beam deflection, use beam 5 of Table A-9. F R1 = R2 = 2 F F δ1 = , and δ 2 = 2k1 2k 2 y AB = −δ1 + δ1 − δ 2 l x+ Fx (4 x 2 − 3l 3 ) 48EI 1 k2 − k1 x y AB = F − + x+ (4 x 2 − 3l 3 ) 48 EI 2k1 2k1k2l Shigley’s MED, 10th edition Ans. Chapter 4 Solutions, Page 36/80 www.konkur.in For BC, since Table A-9 does not have an equation (because of symmetry) an equation will need to be developed as the problem is no longer symmetric. This can be done easily using beam 6 of Table A-9 with a = l /2 F ( l / 2 )( l − x ) 2 l 2 − F Fk2 − Fk1 yBC = + x+ x + − 2lx 2k1 2k1k 2l EIl 4 1 k 2 − k1 ( l − x ) 4 x 2 + l 2 − 8lx Ans. = F − + x+ ( ) 48 EI 2k1 2k1k2l ______________________________________________________________________________ 4-54 Fa F , and R2 = (l + a ) l l Fa F , and δ 2 = (l + a) δ1 = lk1 lk2 R1 = y AB = −δ1 + δ1 − δ 2 l x+ Fax 2 (l − x 2 ) 6 EIl a x ax 2 y AB = F − + k a − k1 ( l + a ) + (l − x 2 ) Ans. 2 2 6 EIl k1l k1k2l δ −δ F (x − l) ( x − l )2 − a(3 x − l ) yBC = −δ1 + 1 2 x + l 6 EI a x (x − l) ( x − l ) 2 − a (3x − l ) yBC = F − + Ans. k a − k1 ( l + a ) + 2 2 6 EI k1l k1k2l ______________________________________________________________________________ 4-55 Let the load be at x ≥ l/2. The maximum deflection will be in Section AB (Table A-9, beam 6) y AB = Fbx 2 ( x + b2 − l 2 ) 6 EIl dy AB Fb = ( 3x 2 + b 2 − l 2 ) = 0 dx 6 EIl x= ⇒ l 2 − b2 l2 , xmax = = 0.577l 3 3 3x 2 + b2 − l 2 = 0 Ans. For x ≤ l/2, xmin = l − 0.577l = 0.423l Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 4 Solutions, Page 37/80 www.konkur.in 4-56 M O = 1(3000)(1500) + 2500(2000) = 9.5 (106 ) N·mm RO = 1(3000) + 2500 = 5 500 N From Prob. 4-10, I = 4.14(106 ) mm 4 x2 1 − 2500 x - 2 000 2 dy x3 2 EI = −9.5 (106 ) x + 2 750 x 2 − − 1250 x − 2000 + C1 dx 6 M = −9.5 (106 ) + 5500 x − dy = 0 at x = 0 ∴ C1 = 0 dx dy x3 2 EI = −9.5 (106 ) x + 2 750 x 2 − − 1250 x − 2000 dx 6 x4 3 EIy = −4.75 (106 ) x 2 + 916.67 x 3 − − 416.67 x − 2000 + C2 24 y = 0 at x = 0 ∴ C2 = 0 , and therefore y=− 1 3 114 (106 ) x 2 − 22 (103 ) x 3 + x 4 + 10 (103 ) x − 2000 24 EI yB = − 1 114 (106 ) 30002 − 22 (103 ) 30003 3 6 24 ( 207 )10 ( 4.14 )10 3 +30004 + 10 (103 ) ( 3000 − 2000 ) = −25.4 mm Ans. MO = 9.5 (106) N⋅m. The maximum stress is compressive at the bottom of the beam where y = 29.0 − 100 = − 71 mm −9.5 (106 ) (−71) My σ max = − =− = −163 (106 ) Pa = −163MPa Ans. I 4.14(106 ) The solutions are the same as Prob. 4-10. ______________________________________________________________________________ 4-57 See Prob. 4-11 for reactions: RO = 465 lbf and RC = 285 lbf. Using lbf and inch units M = 465 x − 450 ⟨ x − 72⟩1 − 300 ⟨x − 120⟩1 Shigley’s MED, 10th edition Chapter 4 Solutions, Page 38/80 www.konkur.in dy 2 2 = 232.5 x 2 − 225 x − 72 − 150 x − 120 + C1 dx EIy = 77.5 x3 − 75 ⟨x − 72⟩ 3 − 50 ⟨x − 120⟩3 − C1x EI y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = 240 in 0 = 77.5(2403) − 75(240 −72)3 − 50(240 − 120)3 + C1 x ⇒ and, EIy = 77.5 x3 − 75 ⟨x − 72⟩ 3 − 50 ⟨x − 120⟩3 −2.622(106) x C1 = − 2.622(106) lbf⋅in2 Substituting y = − 0.5 in at x = 120 in gives 30(106) I (− 0.5) = 77.5 (1203) − 75(120 − 72)3 − 50(120 − 120)3 −2.622(106)(120) I = 12.60 in4 Select two 5 in × 6.7 lbf/ft channels; from Table A-7, I = 2(7.49) = 14.98 in4 12.60 1 Ans. − = −0.421 in 14.98 2 The maximum moment occurs at x = 120 in where Mmax = 34.2(103) lbf⋅in ymidspan = Mc 34.2(103 )(2.5) = = 5 710 psi O.K. I 14.98 The solutions are the same as Prob. 4-17. ______________________________________________________________________________ σ max = 4-58 I = π (1.54)/64 = 0.2485 in4, and w = 150/12 = 12.5 lbf/in. 1 24 RO = (12.5 ) 39 + (340) = 453.0 lbf 2 39 12.5 2 1 M = 453.0 x − x − 340 x − 15 2 dy 12.5 3 2 EI = 226.5 x 2 − x − 170 x − 15 + C1 dx 6 3 3 EIy = 75.5 x − 0.5208 x 4 − 56.67 x − 15 + C1 x + C2 y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = 39 in y= ⇒ C1 = −6.385(10 4 ) lbf ⋅ in 2 Thus, 1 3 75.5 x 3 − 0.5208 x 4 − 56.67 x − 15 − 6.385 (104 ) x EI Evaluating at x = 15 in, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 39/80 www.konkur.in 1 75.5 (153 ) − 0.5208 (154 ) − 56.67 (15 − 15 )3 − 6.385 (104 ) (15) 30(10 )(0.2485) = −0.0978 in Ans. yA = 6 1 75.5 (19.53 ) − 0.5208 (19.54 ) − 56.67 (19.5 − 15 )3 − 6.385 (104 ) (19.5) 30(10 )(0.2485) = − 0.1027 in Ans. ymidspan = 6 5 % difference Ans. The solutions are the same as Prob. 4-12. ______________________________________________________________________________ 3 (14 )100 7 (14 )100 4-59 I = 0.05 in4, RA = = 420 lbf ↑ and RB = = 980 lbf ↑ 10 10 M = 420 x − 50 x2 + 980 ⟨ x − 10 ⟩1 EI dy 2 = 210 x 2 − 16.667 x 3 + 490 x − 10 + C1 dx EIy = 70 x 3 − 4.167 x 4 + 163.3 x − 10 + C1 x + C2 3 y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = 10 in ⇒ C1 = − 2 833 lbf⋅in2. Thus, y= 1 70 x 3 − 4.167 x 4 + 163.3 x − 10 3 − 2833 x 6 30 (10 ) 0.05 3 = 6.667 (10−7 ) 70 x 3 − 4.167 x 4 + 163.3 x − 10 − 2833 x Ans. The tabular results and plot are exactly the same as Prob. 4-21. ______________________________________________________________________________ 4-60 RA = RB = 400 N, and I = 6(323) /12 = 16 384 mm4. First half of beam, M = − 400 x + 400 ⟨ x − 300 ⟩1 dy 2 EI = −200 x 2 + 200 x − 300 + C1 dx From symmetry, dy/dx = 0 at x = 550 mm ⇒ ⇒ 0 = − 200(5502) + 200(550 – 300) 2 + C1 C1 = 48(106) N·mm2 EIy = − 66.67 x3 + 66.67 ⟨ x − 300 ⟩3 + 48(106) x + C2 Shigley’s MED, 10th edition Chapter 4 Solutions, Page 40/80 www.konkur.in y = 0 at x = 300 mm C2 = − 12.60(109) N·mm3. ⇒ The term (EI)−1 = [207(103)16 384] −1 = 2.949 (10−10 ) Thus y = 2.949 (10−10) [− 66.67 x3 + 66.67 ⟨ x − 300 ⟩3 + 48(106) x − 12.60(109)] yO = − 3.72 mm Ans. y|x = 550 mm =2.949 (10−10) [− 66.67 (5503) + 66.67 (550 − 300)3 + 48(106) 550 − 12.60(109)] = 1.11 mm Ans. The solutions are the same as Prob. 4-13. ______________________________________________________________________________ 4-61 1 ( M A − Fa ) l 1 ∑ M A = 0 = M A + R2l − F (l + a) ⇒ R2 = l ( Fl + Fa − M A ) ∑M B = 0 = R1l + Fa − M A M = R1 x − M A + R2 x − l ⇒ R1 = 1 dy 1 1 2 = R1 x 2 − M A x + R2 x − l + C1 dx 2 2 1 1 1 3 EIy = R1 x 3 − M A x 2 + R2 x − l + C1 x + C2 6 2 6 EI y = 0 at x = 0 ⇒ C2 = 0 y = 0 at x = l ⇒ 1 1 C1 = − R1l 2 + M Al . Thus, 6 2 EIy = y= 1 1 1 1 3 1 R1 x 3 − M A x 2 + R2 x − l + − R1l 2 + M Al x 6 2 6 2 6 1 3 ( M A − Fa ) x3 − 3M A x 2l + ( Fl + Fa − M A ) x − l + ( Fal 2 + 2M Al 2 ) x 6 EIl Ans. In regions, 1 ( M A − Fa ) x 3 − 3M A x 2l + ( Fal 2 + 2M Al 2 ) x 6 EIl x M A ( x 2 − 3lx + 2l 2 ) + Fa ( l 2 − x 2 ) = Ans. 6 EIl y AB = Shigley’s MED, 10th edition Chapter 4 Solutions, Page 41/80 www.konkur.in 1 3 M A − Fa ) x3 − 3M A x 2l + ( Fl + Fa − M A )( x − l ) + ( Fal 2 + 2 M Al 2 ) x ( 6 EIl 1 3 3 = M A x3 − 3 x 2l − ( x − l ) + 2 xl 2 + F −ax3 + ( l + a )( x − l ) + axl 2 6 EIl yBC = { 1 = {−M ( x − l ) l + Fl ( x − l ) ( x − l ) − a (3x − l )} 6 EIl ( x − l ) −M l + F x − l − a 3x − l = Ans. ) ( ) } { ( 6 EI } 2 2 A 2 A The solutions reduce to the same as Prob. 4-17. ______________________________________________________________________________ w (b − a ) 1 4-62 ∑ M D = 0 = R1l − w ( b − a ) l − b + ( b − a ) ⇒ R1 = ( 2l − b − a ) 2 2l w w 2 2 x−a + x−b 2 2 dy 1 w w 3 3 EI = R1 x 2 − x−a + x − b + C1 dx 2 6 6 1 w w 4 4 3 EIy = R1 x − x−a + x − b + C1 x + C2 6 24 24 M = R1 x − y = 0 at x = 0 y = 0 at x = l ⇒ C2 = 0 1 1 w w 4 4 C1 = − R1l 3 − ( l − a ) + ( l − b ) l 6 24 24 y= 1 1 w (b − a ) w w 4 ( 2l − b − a ) x3 − x − a + x − b EI 6 2l 24 24 4 1 1 w (b − a ) w w 4 4 −x ( 2l − b − a ) l 3 − ( l − a ) + ( l − b ) l 6 2l 24 24 = { w 4 2 ( b − a )( 2l − b − a ) x3 − l x − a + l x − b 24 EIl 4 } 4 4 − x 2 ( b − a )( 2l − b − a ) l 2 − ( l − a ) + ( l − b ) Ans. The above answer is sufficient. In regions, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 42/80 www.konkur.in } w 4 4 2 ( b − a )( 2l − b − a ) x 3 − x 2 ( b − a )( 2l − b − a ) l 2 − ( l − a ) + ( l − b ) { 24 EIl wx 4 4 = 2 ( b − a )( 2l − b − a ) x 2 − 2 ( b − a )( 2l − b − a ) l 2 + ( l − a ) − ( l − b ) 24 EIl y AB = yBC = { w 4 2 ( b − a )( 2l − b − a ) x 3 − l ( x − a ) 24 EIl } 4 4 − x 2 ( b − a )( 2l − b − a ) l 2 − ( l − a ) + ( l − b ) yCD = { w 4 4 2 ( b − a )( 2l − b − a ) x 3 − l ( x − a ) + l ( x − b ) 24 EIl } 4 4 − x 2 ( b − a )( 2l − b − a ) l 2 − ( l − a ) + ( l − b ) These equations can be shown to be equivalent to the results found in Prob. 4-19. ______________________________________________________________________________ 4-63 I1 = π (1.3754)/64 = 0.1755 in4, I2 = π (1.754)/64 = 0.4604 in4, R1 = 0.5(180)(10) = 900 lbf Since the loading and geometry are symmetric, we will only write the equations for the first half of the beam 2 For 0 ≤ x ≤ 8 in M = 900 x − 90 x − 3 At x = 3, M = 2700 lbf⋅in Writing an equation for M / I, as seen in the figure, the magnitude and slope reduce since I 2 > I 1. To reduce the magnitude at x = 3 in, we add the term, − 2700(1/I 1 − 1/ I 2)⟨ x − 3 ⟩0. The slope of 900 at x = 3 in is also reduced. We account for this with a ramp function, ⟨ x − 3⟩1 . Thus, 1 1 1 1 M 900 x 90 0 1 = − 2700 − x − 3 − 900 − x − 3 − x −3 I I1 I2 I1 I 2 I1 I 2 = 5128 x − 9520 x − 3 − 3173 x − 3 − 195.5 x − 3 0 E 1 2 2 dy 1 2 3 = 2564 x 2 − 9520 x − 3 − 1587 x − 3 − 65.17 x − 3 + C1 dx Boundary Condition: Shigley’s MED, 10th edition dy = 0 at x = 8 in dx Chapter 4 Solutions, Page 43/80 www.konkur.in 0 = 2564 ( 8 ) − 9520 ( 8 − 3 ) − 1587 ( 8 − 3 ) − 65.17 ( 8 − 3) + C1 ⇒ 2 2 3 C1 = − 68.67 (103) lbf/in2 Ey = 854.7 x 3 − 4760 x − 3 − 529 x − 3 − 16.29 x − 3 − 68.67(103 ) x + C2 2 ⇒ y = 0 at x = 0 3 4 C2 = 0 Thus, for 0 ≤ x ≤ 8 in 1 2 3 4 y= 854.7 x 3 − 4760 x − 3 − 529 x − 3 − 16.29 x − 3 − 68.7(103 ) x 6 30(10 ) Ans. Using a spreadsheet, the following graph represents the deflection equation found above Beam Deflection 0 0 1 2 3 4 5 6 7 8 -0.002 -0.004 y (in) -0.006 -0.008 -0.01 -0.012 x (in) The maximum is ymax = −0.0102 in at x = 8 in Ans. ______________________________________________________________________________ 4-64 The force and moment reactions at the left support are F and Fl respectively. The bending moment equation is M = Fx − Fl Plots for M and M /I are shown. M /I can be expressed using singularity functions Shigley’s MED, 10th edition Chapter 4 Solutions, Page 44/80 www.konkur.in M F Fl Fl l = x− − x− I 2 I1 2 I1 4 I1 2 0 F l + x− 2 I1 2 1 where the step down and increase in slope at x = l /2 are given by the last two terms. Integrate dy F 2 Fl Fl l E = x − x− x− dx 4 I1 2 I1 4 I1 2 dy/dx = 0 at x = 0 ⇒ C1 = 0 1 F l + x− 4 I1 2 2 + C1 2 3 F 3 Fl 2 Fl l F l Ey = x − x − x− + x− + C2 12 I1 4 I1 8I1 2 12 I1 2 y = 0 at x = 0 ⇒ C2 = 0 2 3 F 3 l l 2 y= +2 x− 2 x − 6lx − 3l x − 24 EI1 2 2 3 2 F l 5Fl 3 l y x =l /2 = 2 − 6 l − 3 l (0) + 2(0) = − Ans. 24 EI1 2 96 EI1 2 2 3 F l 3Fl 3 3 l 2 y x =l = Ans. 2 ( l ) − 6l ( l ) − 3l l − + 2 x − = − 24 EI1 2 16 EI1 2 The answers are identical to Ex. 4-10. ______________________________________________________________________________ 4-65 Place a dummy force, Q, at the center. The reaction, R1 = wl / 2 + Q / 2 wx2 ∂M x wl Q M = + x− = 2 ∂Q 2 2 2 Integrating for half the beam and doubling the results ymax 1 = 2 EI l /2 ∫ 0 ∂M M ∂Q 2 dx = Q = 0 EI l /2 ∫ 0 wl wx 2 x x − dx 2 2 2 Note, after differentiating with respect to Q, it can be set to zero ymax w = 2 EI l /2 ∫ 0 l /2 w x 3l x 4 5w x ( l − x )dx = − = 2 EI 3 4 0 384 EI 2 Ans. ______________________________________________________________________________ 4-66 Place a fictitious force Q pointing downwards at the end. Use the variable x originating at the free end and positive to the left Shigley’s MED, 10th edition Chapter 4 Solutions, Page 45/80 www.konkur.in ∂M = −x ∂Q l l 1 l ∂M 1 wx 2 w = ∫M dx = − − x dx = x3dx ( ) ∫ ∫ EI Q EI 2 2 EI δ Q =0 0 0 0 M = −Qx − ymax wx 2 2 wl 4 Ans. 8 EI ______________________________________________________________________________ = 4-67 From Table A-7, I1-1 = 1.85 in4. Thus, I = 2(1.85) = 3.70 in4 First treat the end force as a variable, F. Adding weight of channels of 2(5)/12 = 0.833 lbf/in. Using the variable x as shown in the figure M = −F x − 5.833 2 x = − F x − 2.917 x 2 2 ∂M = −x ∂F δA = 1 EI ∫ 60 0 M ∂M 1 dx= ∂F EI ∫ 60 0 ( F x + 2.917 x 2 )( x ) d x = 60 1 ( Fx 3 / 3 + 2.917 x 4 / 4) 0 EI (150)(603 ) / 3 + (2.917)(604 ) / 4 = 0.182 in in the direction of the 150 lbf force 30(106 )(3.70) ∴ y A = − 0.182 in Ans. ______________________________________________________________________________ = 4-68 The energy includes torsion in AC, torsion in CO, and bending in AB. Neglecting transverse shear in AB M = Fx, ∂M =x ∂F In AC and CO, ∂T T = Fl AB , = l AB ∂F The total energy is T 2l T 2l U = + + 2GJ AC 2GJ CO Shigley’s MED, 10th edition l AB ∫ 0 M2 dx 2 EI AB Chapter 4 Solutions, Page 46/80 www.konkur.in The deflection at the tip is δ= δ= = k= ∂U Tl AC ∂T TlCO ∂T = + + ∂F GJ AC ∂F GJ CO ∂F l AB ∫ 0 Tl l Tl l M ∂M 1 dx = AC AB + CO AB + EI 3 ∂F GJ AC GJ CO EI AB l AB ∫ Fx dx 2 0 2 2 3 3 Tl AC l AB TlCO l AB Fl AB Fl AC l AB FlCO l AB Fl AB + + = + + 4 4 4 GJ AC GJ CO 3EI AB G (π d AC / 32 ) G (π d CO / 32 ) 3E (π d AB / 64 ) 2 l AC 32 Fl AB l 2l AB + CO4 + 4 4 π Gd AC GdCO 3Ed AB F δ = π 2 32l AB l AC l 2l AB + CO4 + 4 4 Gd AC GdCO 3Ed AB −1 −1 2 ( 200 ) 200 200 = 8.10 N/mm Ans. = + + 32 ( 2002 ) 79.3 (103 )184 79.3 (103 )124 3 ( 207 )103 ( 84 ) ______________________________________________________________________________ π 4-69 I1 = π (1.3754)/64 = 0.1755 in4, I2 = π (1.754)/64 = 0.4604 in4 Place a fictitious force Q pointing downwards at the midspan of the beam, x = 8 in R1 = 1 1 (10)180 + Q = 900 + 0.5Q 2 2 For 0 ≤ x ≤ 3 in M = ( 900 + 0.5Q ) x ∂M = 0.5 x ∂Q ∂M = 0.5 x ∂Q For 3 ≤ x ≤ 13 in M = ( 900 + 0.5Q ) x − 90( x − 3)2 By symmetry it is equivalent to use twice the integral from 0 to 8 3 8 8 M ∂M 1 1 2 2 dx = 900 x dx + 900 x − 90 ( x − 3) x dx δ = 2∫ ∫ ∫ EI 2 3 0 EI ∂Q Q =0 EI1 0 3 300 x 3 1 = + EI1 0 EI 2 8 1 4 9 2 3 3 300 x − 90( 4 x − 2 x + 2 x ) 3 120.2 (103 ) 8100 1 8100 3 3 = + 145.5 (10 ) − 25.31(10 ) = + EI1 EI 2 30 (106 ) 0.1755 30 (106 ) 0.4604 = 0.0102 in Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 4 Solutions, Page 47/80 www.konkur.in 4-70 I = π (0.54)/64 = 3.068 (10−3) in4, J = 2 I = 6.136 (10−3) in4, A =π (0.52)/4 = 0.1963 in2. Consider x to be in the direction of OA, y vertically upward, and z in the direction of AB. Resolve the force F into components in the x and y directions obtaining 0.6 F in the horizontal direction and 0.8 F in the negative vertical direction. The 0.6 F force creates strain energy in the form of bending in AB and OA, and tension in OA. The 0.8 F force creates strain energy in the form of bending in AB and OA, and torsion in OA. Use the dummy variable x to originate at the end where the loads are applied on each segment, 0.6 F: AB OA ∂M = 0.6 x ∂F M = 0.6 F x M = 4.2 F Fa = 0.6 F ∂M = 4.2 ∂F ∂Fa = 0.6 ∂F 0.8 F: AB M = 0.8 F x ∂M = 0.8 x ∂F OA M = 0.8 F x ∂M = 0.8 x ∂F ∂T = 5.6 ∂F Once the derivatives are taken the value of F = 15 lbf can be substituted in. The deflection of B in the direction of F is* T = 5.6 F (δ B ) F ∂U Fa L ∂Fa TL ∂T 1 ∂M = + + M dx ∑ ∫ ∂F AE OA ∂F JG OA ∂F EI ∂F 0.6 (15 )15 5.6 (15 )15 = 0.6 ) + ( 5.6 ) 6 ( 0.1963 ( 30 )10 6.136 (10−3 )11.5 (106 ) = 7 15 15 ( 4.22 ) 15 2 + 0.6 x d x + dx+ ( ) 30 (106 ) 3.068 (10−3 ) ∫0 30 (106 ) 3.068 (10−3 ) ∫0 7 15 15 15 2 2 + ( 0.8 x ) d x + ( 0.8 x ) d x 6 −3 ∫ 6 −3 ∫ 30 (10 ) 3.068 (10 ) 0 30 (10 ) 3.068 (10 ) 0 = 1.38 (10−5 ) + 0.1000 + 6.71(10−3 ) + 0.0431 + 0.0119 + 0.1173 = 0.279 in Shigley’s MED, 10th edition Ans. Chapter 4 Solutions, Page 48/80 www.konkur.in *Note. Be careful, this is not the actual deflection of point B. For this, dummy forces must be placed on B in the x, y, and z directions. Determine the energy due to each, take derivatives, and then substitute the values of Fx = 9 lbf, Fy = − 12 lbf, and Fz = 0. This can be done separately and then added by vector addition. The actual deflections of B are found to be δB = 0.0831 i − 0.2862 j − 0.00770 k in From this, the deflection of B in the direction of F is (δ B ) F = 0.6 ( 0.0831) + 0.8 ( 0.2862 ) = 0.279 in which agrees with our result. ______________________________________________________________________________ 4-71 Strain energy. AB: Bending and torsion, BC: Bending and torsion, CD: Bending. IAB = π (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, IBC = 0.25(1.53)/12 = 0.07031 in4, ICD = π (0.754)/64 = 0.01553 in4. For the torsion of bar BC, Eq. (3-41) is in the form of θ =TL/(JG), where the equivalent of J is Jeq = βbc 3. With b/c = 1.5/0.25 = 6, JBC = βbc 3 = 0.299(1.5)0.253 = 7.008 (10−3) in4. Use the dummy variable x to originate at the end where the loads are applied on each segment, ∂M AB: Bending M = F x + 2 F = x +2 ∂F ∂T Torsion T = 5 F =5 ∂F ∂M BC: Bending M = F x =x ∂F ∂T Torsion T = 2 F =2 ∂F ∂M CD: Bending M = F x =x ∂F ∂U Tl ∂T 1 ∂M =∑ +∑ ∫M dx ∂F JG ∂F EI ∂F 6 5F ( 6) 2F (5) 1 2 = 5 + 2 + F ( x + 2) d x 6 ( ) ∫ −3 6 6 0.09818 (11.5 )10 7.008 (10 )11.5 (10 ) 30 (10 ) 0.04909 0 δD = 5 + 2 1 1 F x 2d x + F x 2d x ∫ ∫ 6 6 30 (10 ) 0.07031 0 30 (10 ) 0.01553 0 = 1.329 (10−4 ) F + 2.482 (10−4 ) F + 1.141(10−4 ) F + 1.98 (10−5 ) F + 5.72 (10−6 ) F = 5.207 (10−4 ) F = 5.207 (10−4 ) 200 = 0.104 in Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 4 Solutions, Page 49/80 www.konkur.in 4-72 AAB = π (12)/4 = 0.7854 in2, IAB = π (14)/64 = 0.04909 in4, IBC = 1.5 (0.253)/12 = 1.953 (10−3) in4, ACD = π (0.752)/4 = 0.4418 in2, IAB = π (0.754)/64 = 0.01553 in4. For (δD )x let F = Fx = − 150 lbf and Fz = − 100 lbf . Use the dummy variable x to originate at the end where the loads are applied on each segment, ∂M y CD: M y = Fz x =0 ∂F ∂Fa Fa = F =1 ∂F ∂M y BC: M y = F x + 2 Fz =x ∂F ∂Fa Fa = Fz =0 ∂F ∂M y AB: M y = 5F + 2 Fz + Fz x =5 ∂F ∂Fa Fa = F =1 ∂F 5 ∂Fa ∂U FL 1 + (δ D ) x = = ( F x + 2 Fz ) x d x ∂F AE CD ∂F EI BC ∫0 + = 1 EI AB 6 ∫ ( 5F + 2 F z 0 FL ∂Fa + Fz x )( 5 ) d x + AE AB ∂F F ( 2) 1 3 F 1 + ( 5) + Fz ( 52 ) 6 ( ) 6 −3 0.4418 ( 30 )10 30 (10 )1.953 (10 ) 3 + F ( 6) 1 F 25 F ( 6 ) + 10 Fz ( 6 ) + z ( 62 ) 5 + 1 6 ( ) 2 30 (10 ) 0.04909 0.7854 ( 30 )10 6 = 1.509 (10−7 ) F + 7.112 (10−4 ) F + 4.267 (10−4 ) Fz + 1.019 (10−4 ) F + 1.019 (10−4 ) Fz + 2.546 (10−7 ) F = 8.135 (10−4 ) F + 5.286 (10−4 ) Fz Substituting F = Fx = − 150 lbf and Fz = − 100 lbf gives (δ D ) x = 8.135 (10−4 ) ( −150 ) + 5.286 (10−4 ) ( −100 ) = −0.1749 in Ans. ______________________________________________________________________________ 4-73 IOA = IBC = π (1.54)/64 = 0.2485 in4, JOA = JBC = 2 IOA = 0.4970 in4, IAB = π (14)/64 = 0.04909 in4, JAB = 2 IAB = 0.09818 in4, ICD = π (0.754)/64 = 0.01553 in4 Let Fy = F, and use the dummy variable x to originate at the end where the loads are applied on each segment, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 50/80 www.konkur.in OC: M =Fx ∂M = x, ∂F DC: M =Fx ∂M =x ∂F T = 12 F ∂T = 12 ∂F ∂U 1 ∂M TL ∂T = ∑ +∑ ∫M dx ∂F EI ∂F JG OC ∂F The terms involving the torsion and bending moments in OC must be split up because of the changing second-area moments. (δ D ) y = (δ D ) y = 2 12 F ( 4 ) 12 F ( 9 ) 1 12 + 12 + F x 2d x ( ) ( ) 0.4970 (11.5 )106 0.09818 (11.5 )106 30 (106 ) 0.2485 ∫0 11 + 13 12 1 1 1 F x 2d x + F x 2d x + F x 2d x ∫ ∫ 6 6 6 30 (10 ) 0.04909 2 30 (10 ) 0.2485 11 30 (10 ) 0.01553 ∫0 = 1.008 (10−4 ) F + 1.148 (10−3 ) F + 3.58 (10−7 ) F + 2.994 (10−4 ) F + 3.872 (10−5 ) F + 1.2363 (10−3 ) F = 2.824 (10−3 ) F = 2.824 (10−3 ) 250 = 0.706 in Ans. For the simplified shaft OC, (δ B ) y = 13 12 12 F (13) 1 1 2 12 + F x d x + F x 2d x ( ) ∫ ∫ 6 6 0.09818 (11.5 )106 30 (10 ) 0.04909 0 30 (10 ) 0.01553 0 = 1.6580 (10−3 ) F + 4.973 (10−4 ) F + 1.2363 (10−3 ) F = 3.392 (10−3 ) F = 3.392 (10−3 ) 250 = 0.848 in Ans. Simplified is 0.848/0.706 = 1.20 times greater Ans. ______________________________________________________________________________ 4-74 Place a dummy force Q pointing downwards at point B. The reaction at C is RC = Q + (6/18)100 = Q + 33.33 This is the axial force in member BC. Isolating the beam, we find that the moment is not a function of Q, and thus does not contribute to the strain energy. Thus, only energy in the member BC needs to be considered. Let the axial force in BC be F, where Shigley’s MED, 10th edition Chapter 4 Solutions, Page 51/80 www.konkur.in F = Q + 33.33 ∂F =1 ∂Q FL ∂F ( 0 + 33.33)12 = (1) = 6.79 (10−5 ) in Ans. = 2 6 AE ∂ Q BC Q =0 π ( 0.5 ) / 4 30 (10 ) Q =0 ______________________________________________________________________________ δB = 4-75 ∂U ∂Q IOB = 0.25(23)/12 = 0.1667 in4 AAC = π (0.52)/4 = 0.1963 in2 ΣMO = 0 = 6 RC − 11(100) − 18 Q RC = 3Q + 183.3 ΣMA = 0 = 6 RO − 5(100) − 12 Q ⇒ RO = 2Q + 83.33 Bending in OB. BD: Bending in BD is only due to Q which when set to zero after differentiation gives no contribution. AD: Using the variable x as shown in the figure above ∂M = − (7 + x ) ∂Q OA: Using the variable x as shown in the figure above M = −100 x − Q ( 7 + x ) M = − ( 2Q + 83.33) x Axial in AC: F = 3Q + 183.3 Shigley’s MED, 10th edition ∂M = −2 x ∂Q ∂F =3 ∂Q Chapter 4 Solutions, Page 52/80 www.konkur.in ∂U FL ∂F ∂M 1 δB = = + ∑ M ∂Q dx ∂Q Q = 0 AE ∂Q Q = 0 EI Q = 0 = 183.3 (12 ) 1 3 + 6 ( ) 0.1963 ( 30 )10 EI 5 ∫ (100 x )( 7 + x ) d x + ∫ 2 (83.33) x dx 6 2 0 0 5 6 2 = 1.121(10 ) + 100 x 7 + x d x + 166.7 x dx ( ) ∫0 10.4 (106 ) 0.1667 ∫0 1 −3 = 1.121(10−3 ) + 5.768 (10−7 ) 100 (129.2 ) + 166.7 ( 72 ) = 0.0155 in Ans. ______________________________________________________________________________ 4-76 There is no bending in AB. Using the variableθ, rotating counterclockwise from B M = PR sin θ Fr = P cos θ Fθ = P sin θ ∂MFθ = 2 PR sin 2 θ ∂P ∂M = R sin θ ∂P ∂Fr = cos θ ∂P ∂Fθ = sin θ ∂P A = 6(4) = 24 mm 2 , ro = 40 + 12 (6) = 43 mm, ri = 40 − 12 (6) = 37 mm, From Table 3-4, p. 135, for a rectangular cross section 6 rn = = 39.92489 mm ln(43 / 37) From Eq. (4-33), the eccentricity is e = R − rn =40 − 39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 From Eq. (4-38) π π π π M ∂M 1 ∂ ( MFθ ) 2 2 Fθ R ∂Fθ 2 2 CFr R ∂Fr δ =∫ θ θ θ d + d − d + ∫ ∫ ∫ 0 AeE 0 AE 0 AE 0 ∂P AG ∂P ∂P ∂P dθ 2 π π π P ( R sin θ ) 2 PR ( sin θ ) 2 2 PR sin θ 2 CPR ( cos θ ) =∫ dθ + ∫ dθ − ∫ dθ + ∫ dθ 0 0 0 0 AeE AE AE AG π PR R π (10)(40) 40 EC (207 ⋅103 )(1.2) = + 1 − 2 + = + 1 − 2 + 4 AE e G 4(24)(207 ⋅103 ) 0.07511 79.3 ⋅103 Ans. δ = 0.0338 mm ______________________________________________________________________________ π 2 2 2 2 Shigley’s MED, 10th edition Chapter 4 Solutions, Page 53/80 www.konkur.in 4-77 Place a dummy force Q pointing downwards at point A. Bending in AB is only due to Q which when set to zero after differentiation gives no contribution. For section BC use the variableθ, rotating counterclockwise from B ∂M M = PR sin θ + Q ( R + R sin θ ) = R (1 + sin θ ) ∂Q ∂Fr Fr = ( P + Q ) cos θ = cos θ ∂Q ∂Fθ Fθ = ( P + Q ) sin θ = sin θ ∂Q MFθ = PR sin θ + QR (1 + sin θ ) ( P + Q ) sin θ ∂MFθ = PR sin 2 θ + + PR sin θ (1 + sin θ ) + 2QR sin θ (1 + sin θ ) ∂Q But after differentiation, we can set Q = 0. Thus, ∂MFθ = PR sin θ (1 + 2sin θ ) ∂Q A = 6(4) = 24 mm 2 , ro = 40 + 12 (6) = 43 mm, ri = 40 − 12 (6) = 37 mm, From Table 3-4, p.135, for a rectangular cross section 6 rn = = 39.92489 mm ln(43 / 37) From Eq. (4-33), the eccentricity is e = R − rn =40 − 39.92489 = 0.07511 mm From Table A-5, E = 207(103) MPa, G = 79.3(103) MPa From Table 4-1, C = 1.2 From Eq. (4-38), π π π 1 ∂ ( MFθ ) 2 F R ∂F 2 2 CFr R ∂Fr dθ + ∫ θ θ dθ − ∫ dθ + ∫ dθ 0 0 AE 0 AE 0 ∂Q AG ∂Q ∂Q PR 2 π2 PR π2 2 PR π2 = sin θ (1 + sin θ ) dθ + sin θ dθ − sin θ (1 + 2sin θ ) dθ ∫ ∫ AeE 0 AE 0 AE ∫0 CPR π2 + cos 2 θ dθ ∫ 0 AG 2 π CE π PR π PR π PR π CPR PR π R = + 1 + − + 2 + = + 1 − 2 + 4 G 4 AeE 4 AE 4 AE 4 AG AE 4 e δ =∫ π 2 M ∂M AeE ∂Q 10 ( 40 ) = 24 ( 207 )103 3 π π 1.2 ( 207 )10 40 + 1 −2+ 4 79.3 (103 ) 4 0.07511 = 0.0766 mm Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 4 Solutions, Page 54/80 www.konkur.in 4-78 A = 3(2.25) −2.25(1.5) = 3.375 in2 (1 + 1.5)(3)(2.25) − (1 + 0.75 + 1.125)(1.5)(2.25) R= = 2.125 in 3.375 Section is equivalent to the “T” section of Table 3-4, p. 135, 2.25(0.75) + 0.75(2.25) = 1.7960 in 2.25ln[(1 + 0.75) /1] + 0.75ln[(1 + 3) / (1 + 0.75)] e = R − rn = 2.125 − 1.7960 = 0.329 in rn = For the straight section 1 I z = (2.25) ( 33 ) + 2.25(3)(1.5 − 1.125)2 12 2 1 2.25 3 − (1.5) ( 2.25 ) + 1.5(2.25) 0.75 + − 1.125 2 12 = 2.689 in 4 For 0 ≤ x ≤ 4 in M = − Fx ∂M = − x, ∂F V =F ∂V =1 ∂F For θ ≤ π /2 Fr = F cos θ ∂Fr = cos θ , ∂F Fθ = F sin θ ∂Fθ = sin θ ∂F ∂M = (4 + 2.125sin θ ) ∂F ∂MFθ MFθ = F (4 + 2.125sin θ ) F sin θ = 2 F (4 + 2.365sin θ ) sin θ ∂F M = F (4 + 2.125sin θ ) Use Eqs. (4-31) and (4-24) (with C = 1) for the straight part, and Eq. (4-38) for the curved part, integrating from 0 to π/2, and double the results Shigley’s MED, 10th edition Chapter 4 Solutions, Page 55/80 www.konkur.in δ= π /2 2 1 4 2 F (4)(1) (4 + 2.125sin θ ) 2 Fx dx + + F dθ E I ∫0 3.375(G / E ) ∫0 3.375(0.329) π /2 2 F (4 + 2.125sin θ ) sin θ F sin 2 θ (2.125) dθ − ∫ dθ 0 0 3.375 3.375 2 π /2 (1) F cos θ (2.125) +∫ dθ 0 3.375(G / E ) +∫ π /2 Substitute I = 2.689 in4, F = 6700 lbf, E = 30 (106) psi, G = 11.5 (106) psi 2 ( 6700 ) 43 4 1 π π + + 16 + 17(1) + 4.516 δ= 6 30 (10 ) 3 ( 2.689 ) 3.375(11.5 / 30) 3.375(0.329) 2 4 + 2.125 π 2 2.125 π π 4 (1) + 2.125 + − 3.375 4 3.375 4 3.375 (11.5 / 30 ) 4 = 0.0226 in Ans. ______________________________________________________________________________ 4-79 Since R/h = 35/4.5 = 7.78 use Eq. (4-38), integrate from 0 to π , and double the results ∂M M = FR (1 − cosθ ) = R (1 − cosθ ) ∂F ∂Fr Fr = F sin θ = sin θ ∂F ∂Fθ Fθ = F cos θ = cos θ ∂F MFθ = F 2 Rcosθ (1 − cosθ ) ∂ ( MFθ ) = 2 FRcosθ (1 − cosθ ) ∂F From Eq. (4-38), FR 2 π FR π (1 − cos θ ) 2 dθ + cos 2 θ dθ δ = 2 ∫ ∫ 0 0 AE AeE − = 2 FR π 1.2 FR π 2 cos θ (1 − cos θ ) dθ + sin θ dθ ∫ ∫ 0 0 AE AG 2 FR 3π R 3π E + + 0.6π AE 2 e 2 G A = 4.5(3) = 13.5 mm2, E = 207 (103) N/mm2, G = 79.3 (103) N/mm2, and from Table 3-4, p. 135, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 56/80 www.konkur.in h 4.5 = = 34.95173 mm ro 37.25 ln ln 32.75 ri and e = R − rn = 35 − 34.95173 = 0.04827 mm. Thus, 2 F ( 35 ) 3π 35 3π 207 δ= + + 0.6π = 0.08583F 3 13.5 ( 207 )10 2 0.04827 2 79.3 1 where F is in N. For δ = 1 mm, F = = 11.65 N Ans. 0.08583 Note: The first term in the equation for δ dominates and this is from the bending moment. Try Eq. (4-41), and compare the results. ______________________________________________________________________________ rn = 4-80 R/h = 20 > 10 so Eq. (4-41) can be used to determine deflections. Consider the horizontal reaction to be applied at B and subject to the constraint (δ B ) H = 0. M= ∂M = − R sin θ ∂H FR (1 − cos θ ) − HR sin θ 2 0 <θ < π 2 By symmetry, we may consider only half of the wire form and use twice the strain energy Eq. (4-41) then becomes, (δ B ) H = ∫ π /2 0 ∂U 2 ≈ ∂H EI π /2 ∂M ∫ M ∂H Rdθ = 0 0 FR 2 (1 − cos θ ) − HR sin θ (− R sin θ ) R dθ = 0 π F F F 30 − + +H =0 ⇒ H = = = 9.55 N π π 2 4 4 Ans. The reaction at A is the same where H goes to the left. Substituting H into the moment equation we get, M= FR [π (1 − cos θ ) − 2 sin θ ] 2π Shigley’s MED, 10th edition ∂M R = [π (1 − cos θ ) − 2sin θ ] ∂F 2π 0 <θ < π 2 Chapter 4 Solutions, Page 57/80 www.konkur.in ∂U 2 ∂M 2 π /2 FR 2 ≈∫ M Rd = [π (1 − cos θ ) − 2sin θ ]2 R dθ θ ∂P EI ∂F EI ∫0 4π 2 FR 3 π /2 2 = (π + π 2 cos 2 θ + 4sin 2 θ − 2π 2 cos θ − 4π sin θ + 4π sin θ cos θ ) dθ 2 ∫ 0 2π EI FR 3 2 π π π π + π 2 + 4 − 2π 2 − 4π + 2π = 2 2π EI 2 4 4 δP = (3π 2 − 8π − 4) FR 3 (3π 2 − 8π − 4) (30)(403 ) = = 0.224 mm Ans. 8π EI 8π 207 (103 ) π ( 24 ) / 64 ______________________________________________________________________________ = 4-81 The radius is sufficiently large compared to the wire diameter to use Eq. (4-41) for the curved beam portion. The shear and axial components will be negligible compared to bending. Place a fictitious force Q pointing to the left at point A. ∂M M = PR sin θ + Q( R sin θ + l ) = R sin θ + l ∂Q Note that the strain energy in the straight portion is zero since there is no real force in that section. From Eq. (4-41), π /2 δ = ∫ 0 PR 2 = EI 1 ∂M 1 M Rdθ = EI ∂Q Q = 0 EI ∫ π /2 0 ∫ π /2 0 PR sin θ ( R sin θ + l )Rdθ PR 2 π 1(52 ) π ( R sin θ + l sin θ ) dθ = EI 4 R + l = (5) + 4 6 4 30 (10 ) π ( 0.125 ) / 64 4 2 = 0.551 in Ans. ______________________________________________________________________________ 4-82 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. ∂M AB =x ∂P Straight portion: M AB = Px Curved portion: M BC = P [ R (1 − cos θ ) + l ] ∂M BC = [ R (1 − cos θ ) + l ] ∂P From Eq. (4-41) with the addition of the bending strain energy in the straight portion of the wire, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 58/80 www.konkur.in π /2 1 ∂M BC 1 ∂M AB M BC Rdθ M AB dx + ∫0 EI ∂P EI ∂P P l 2 PR π /2 2 = x dx + R(1 − cos θ ) + l ] dθ [ ∫ ∫ EI 0 EI 0 3 Pl PR π /2 2 R (1 − 2 cos θ + cos 2 θ ) + 2 Rl (1 − cos θ ) + l 2 dθ = + ∫ 0 3EI EI Pl 3 PR π /2 2 R cos 2 θ − ( 2 R 2 + 2 Rl ) cos θ + ( R + l )2 dθ = + ∫ 0 3EI EI Pl 3 PR π 2 π = + R − ( 2 R 2 + 2 Rl ) + ( R + l ) 2 3EI EI 4 2 δ =∫ l 0 l3 π 3 π 2 2 3 + 4 R − R ( 2 R + 2 Rl ) + 2 R ( R + l ) = P EI = 43 π 3 1 π 2 2 + (5 ) − 5 2(5 ) + 2(5)(4) + ( 5 )( 5 + 4 ) 6 4 2 30 (10 ) π ( 0.125 ) / 64 3 4 = 0.850 in Ans. ______________________________________________________________________________ 4-83 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. Place a dummy force, Q, at A vertically downward. The only load in the straight section is the axial force, Q. Since this will be zero, there is no contribution. In the curved section ∂M = R (1 − cos θ ) ∂Q M = PR sin θ + QR (1 − cos θ ) From Eq. (4-41) π /2 1 ∂M 1 δ = ∫ M Rdθ = 0 EI ∂Q Q =0 EI = = PR 3 EI π /2 ∫ ∫0 ( sin θ − sin θ cos θ ) dθ = 1( 53 ) 2 ( 30 )106 π ( 0.1254 ) / 64 π /2 0 PR sin θ R (1 − cos θ ) Rdθ PR3 1 PR3 1 − = EI 2 2 EI = 0.174 in Ans. ______________________________________________________________________________ 4-84 Both the radius and the length are sufficiently large to use Eq. (4-41) for the curved beam portion and to neglect transverse shear stress for the straight portion. Shigley’s MED, 10th edition Chapter 4 Solutions, Page 59/80 www.konkur.in Place a dummy force, Q, at A vertically downward. The load in the straight section is the axial force, Q, whereas the bending moment is only a function of P and is not a function of Q. When setting Q = 0, there is no axial or bending contribution. In the curved section M = P R (1 − cos θ ) + l − QR sin θ ∂M = − R sin θ ∂Q From Eq. (4-41) π /2 δ = ∫ 0 =− =− 1 ∂M M EI ∂Q PR 2 EI 1 Rdθ = Q =0 EI π /2 ∫ π /2 0 P R (1 − cos θ ) + l ( − R sin θ ) Rdθ ∫ ( R sin θ − R sin θ cos θ + l sin θ ) dθ = − 0 PR 2 1 PR 2 R + l − R = − ( R + 2l ) EI 2 2 EI 1( 52 ) 5 + 2 ( 4 ) = −0.452 in 2 ( 30 )106 π ( 0.1254 ) / 64 Since the deflection is negative, δ is in the opposite direction of Q. Thus the deflection is δ = 0.452 in ↑ Ans. ______________________________________________________________________________ 4-85 Consider the force of the mass to be F, where F = 9.81(1) = 9.81 N. The load in AB is tension ∂FAB FAB = F =1 ∂F For the curved section, the radius is sufficiently large to use Eq. (4-41). There is no bending in section DE. For section BCD, let θ be counterclockwise originating at D ∂M M = FR sin θ = R sin θ 0 ≤θ ≤π ∂F Using Eqs. (4-29) and (4-41) 3 π 1 π FR ∂M Fl Fl ∂FAB +∫ M Rdθ = 1) + ∫ sin 2 θ dθ ( 0 0 EI ∂F AE EI AE AB ∂F δ = π ( 403 ) Fl π FR 3 F l π R 3 9.81 80 = + = + + = AE 2 EI E A 2 I 207 (103 ) π ( 22 ) / 4 2 π ( 24 ) / 64 = 6.067 mm Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 4 Solutions, Page 60/80 www.konkur.in 4-86 AOA = 2(0.25) = 0.5 in2, IOAB = 0.25(23)/12 = 0.1667 in4, IAC = π (0.54)/64 = 3.068 (10-3) in4 Applying a force F at point B, using statics (AC is a two-force member), the reaction forces at O and C are as shown. ∂FOA OA: Axial FOA = 3F =3 ∂F ∂M OA = −2 x ∂F Bending M OA = −2 Fx AB: Bending ∂M AB = −x ∂F M AB = − F x AC: Isolating the upper curved section ∂M AC = 3R ( sin θ + cos θ − 1) ∂F 10 20 1 1 Fl ∂FOA 2 δ = + 4 Fx dx + F x 2d x ∫ ∫ ( EI )OAB 0 AE OA ∂F ( EI )OAB 0 M AC = 3FR ( sin θ + cos θ − 1) + = 9 FR 3 ( EI ) AC π /2 ∫ ( sin θ + cosθ − 1) 2 dθ 0 4 F (103 ) F ( 203 ) 3F (10 ) 3 + + ( ) 0.5 (10.4 )106 3 (10.4 )106 ( 0.1667 ) 3 (10.4 )106 ( 0.1667 ) + 9 F (103 ) 30 (10 ) 3.068 (10 6 π /2 −3 ( sin )∫ 2 θ + 2sin θ cos θ − 2sin θ + cos2 θ − 2 cos θ + 1) dθ 0 π π π = 1.731(10−5 ) F + 7.691(10−4 ) F + 1.538 (10−3 ) F + 0.09778 F + 1 − 2 + − 2 + 4 2 4 = 0.0162 F = 0.0162 (100 ) = 1.62 in Ans. _____________________________________________________________________________ 4-87 AOA = 2(0.25) = 0.5 in2, IOAB = 0.25(23)/12 = 0.1667 in4, IAC = π (0.54)/64 = 3.068 (10-3) in4 Applying a vertical dummy force, Q, at A, from statics the reactions are as shown. The dummy force is transmitted through section OA and member AC. Shigley’s MED, 10th edition Chapter 4 Solutions, Page 61/80 www.konkur.in ∂FOA =1 ∂Q OA: FOA = 3F + Q M AC = ( 3F + Q ) R sin θ − ( 3F + Q ) R (1 − cos θ ) AC: ∂M AC = R ( sin θ + cos θ − 1) ∂Q Fl ∂FOA 1 π / 2 ∂M AC Rdθ δ = + ∫ M AC ∂Q AE OA ∂Q EI AC 0 Q=0 3FlOA 3FR 3 = + ( AE )OA ( EI ) AC = 3 (100 )10 10.4 (10 ) 0.5 6 + π /2 ∫ ( sin θ + cos θ − 1) 2 dθ 0 3 (100 )103 π π π + 1 − 2 + − 2 + = 0.462 in 4 2 30 (10 ) 3.068 (10 ) 4 6 −3 Ans. ______________________________________________________________________________ 4-88 I = π (64)/64 = 63.62 mm4 0≤θ ≤π/2 ∂M = R sin θ ∂F ∂T T = FR (1 − cos θ ) = R(1 − cos θ ) ∂F According to Castigliano’s theorem, a positive ∂ U/∂ F will yield a deflection of A in the negative y direction. Thus the deflection in the positive y direction is M = FR sin θ (δ A ) y = − ∂U 1 = − ∂F EI ∫ π /2 0 F ( R sin θ ) 2 R dθ + 1 GJ ∫ π /2 0 F [ R (1 − cos θ )]2 R dθ Integrating and substituting J = 2 I and G = E / 2 (1 + ν ) π FR 3 3π + (1 + ) − 2 = − 4 − 8 + (3 − 8) ν π π ν [ ] 4 4 EI 4 (250)(80)3 = −[4π − 8 + (3π − 8)(0.29)] = − 12.5 mm Ans. 4(200)103 ( 63.62 ) ______________________________________________________________________________ (δ A ) y = − FR 3 EI 4-89 The force applied to the copper and steel wire assembly is Fc + Fs = 400 lbf (1) Since the deflections are equal, δ c = δ s Shigley’s MED, 10th edition Chapter 4 Solutions, Page 62/80 www.konkur.in Fl Fl = AE c AE s Fc l Fs l = 2 6 3(π / 4)(0.1019) (17.2)10 (π / 4)(0.1055) 2 (30)106 Yields, Fc = 1.6046 Fs . Substituting this into Eq. (1) gives 1.604 Fs + Fs = 2.6046Fs = 400 ⇒ Fs = 153.6 lbf Fc = 1.6046 Fs = 246.5 lbf F 246.5 σc = c = = 10 075 psi = 10.1 kpsi Ans. Ac 3(π / 4)(0.1019) 2 F 153.6 σs = s = = 17 571 psi = 17.6 kpsi Ans. As (π / 4)(0.10552 ) Fl 153.6(100)(12) = 0.703 in δ = = 2 6 AE s (π / 4)(0.1055) (30)10 Ans. ______________________________________________________________________________ 4-90 (a) Bolt stress σ b = 0.75(65) = 48.8 kpsi Ans. π Fb = 6σ b Ab = 6(48.8) (0.52 ) = 57.5 kips 4 Fb 57.43 Cylinder stress = −13.9 kpsi Ans. σc = − = Ac (π / 4)(5.52 − 52 ) (b) Force from pressure π D2 π (52 ) P= p= (500) = 9817 lbf = 9.82 kip 4 4 ΣFx = 0 Total bolt force Pb + Pc = 9.82 (1) Since δ c = δ b , Pc l Pbl = 2 2 (π / 4)(5.5 − 5 ) E 6(π / 4)(0.52 ) E Pc = 3.5 Pb (2) Substituting this into Eq. (1) Pb + 3.5 Pb = 4.5 Pb = 9.82 ⇒ Pb = 2.182 kip. From Eq. (2), Pc = 7.638 kip Using the results of (a) above, the total bolt and cylinder stresses are 2.182 = 50.7 kpsi Ans. σ b = 48.8 + 6(π / 4)(0.52 ) Shigley’s MED, 10th edition Chapter 4 Solutions, Page 63/80 www.konkur.in 7.638 = −12.0 kpsi Ans. (π / 4)(5.52 − 52 ) ______________________________________________________________________________ σ c = −13.9 + 4-91 Tc + Ts = T θc = θs (1) Tc l Tl = s ( JG )c ( JG )s ⇒ Substitute this into Eq. (1) ( JG )c T +T = T ( JG )s s s ⇒ Tc = ⇒ Ts = ( JG )c T ( JG )s s (2) ( JG )s T ( JG )s + ( JG )c The percentage of the total torque carried by the shell is % Torque = 100 ( JG ) s ( JG )s + ( JG )c Ans. ______________________________________________________________________________ 4-92 RO + RB = W (1) δOA = δAB Fl Fl = AE OA AE AB 400 RO 600 RB 3 = ⇒ RO = RB AE AE 2 Substitute this unto Eq. (1) 3 RB + RB = 4 2 ⇒ RB = 1.6 kN (2) Ans. 3 RO = 1.6 = 2.4 kN Ans. 2 2 400(400) Fl = 0.0223 mm Ans. δA = = 3 AE OA 10(60)(71.7)(10 ) ______________________________________________________________________________ From Eq. (2) 4-93 See figure in Prob. 4-92 solution. Procedure 1: 1. Let RB be the redundant reaction. Shigley’s MED, 10th edition Chapter 4 Solutions, Page 64/80 www.konkur.in ⇒ RO = 4 000 − RB 2. Statics. RO + RB = 4 000 N 3. Deflection of point B. δ B = (1) RB ( 600 ) ( RB − 4000 )( 400 ) + =0 AE AE (2) 4. From Eq. (2), AE cancels and RB = 1 600 N Ans. and from Eq. (1), RO = 4 000 − 1 600 = 2 400 N Ans. 2 400(400) Fl = 0.0223 mm Ans. = 3 AE OA 10(60)(71.7)(10 ) ______________________________________________________________________________ δA = 4-94 (a) Without the right-hand wall, the deflection of point C would be δC = ∑ 5 (103 ) 8 2 (103 ) 5 Fl = + AE (π / 4 ) 0.752 (10.4 )106 ( π / 4 ) 0.52 (10.4 )106 = 0.01360 in > 0.005 in ∴ Hits wall Ans. (b) Let RC be the reaction of the wall at C acting to the left (←). Thus, the deflection of point C is now 5 (103 ) − RC 8 2 (103 ) − RC 5 + δC = 2 6 2 / 4 0.75 10.4 10 / 4 0.5 10.4 π π ( ) ( ) ( ) ( )106 = 0.01360 − 4 RC 5 8 + 2 = 0.005 6 2 π (10.4 )10 0.75 0.5 or, 0.01360 − 4.190 (10 −6 ) RC = 0.005 ⇒ RC = 2 053 lbf = 2.05 kip ← Ans. Σ Fx = 5 000 + RA − 2 053 = 0 ⇒ RA = −2 947 lbf, ⇒ RA = 2.95 kip ← Ans. Deflection. AB is 2 947 lbf in tension. Thus RA ( 8 ) 2947 ( 8 ) = 5.13 (10−3 ) in → Ans. AAB E (π / 4 ) 0.752 (10.4 )106 ______________________________________________________________________________ δ B = δ AB = = 4-95 Since θOA = θAB, TOA (4) TAB (6) = JG JG Shigley’s MED, 10th edition ⇒ 3 TOA = TAB 2 (1) Chapter 4 Solutions, Page 65/80 www.konkur.in Statics. TOA + TAB = 200 (2) Substitute Eq. (1) into Eq. (2), 3 5 TAB + TAB = TAB = 200 ⇒ 2 2 From Eq. (1) TAB = 80 lbf ⋅ in Ans. 3 3 TOA = TAB = 80 = 120 lbf ⋅ in Ans. 2 2 80 ( 6 ) 180 TL θA = = 0.3900 = 4 6 JG AB ( π / 32 ) 0.5 (11.5 )10 π τ max = 16T πd3 ⇒ τ AB = 16 ( 80 ) π ( 0.53 ) 16 (120 ) τ OA = π ( 0.53 ) Ans. = 4890 psi = 4.89 kpsi = 3260 psi = 3.26 kpsi Ans. Ans. ______________________________________________________________________________ 4-96 Since θOA = θAB, TOA (4) TAB (6) = 4 (π / 32 ) 0.5 G (π / 32 ) 0.754 G Statics. TOA + TAB = 200 ⇒ TOA = 0.2963 TAB (1) (2) Substitute Eq. (1) into Eq. (2), 0.2963TAB + TAB = 1.2963TAB = 200 From Eq. (1) TAB = 154.3 lbf ⋅ in ⇒ TOA = 0.2963TAB = 0.2963 (154.3) = 45.7 lbf ⋅ in θA = 154.3 ( 6 ) 180 (π / 32 ) 0.75 (11.5)10 π τ max = 4 16T πd3 τ AB = ⇒ 6 τ OA = = 0.1480 16 ( 45.7 ) π ( 0.53 ) Ans. Ans. Ans. = 1862 psi =1.86 kpsi Ans. 16 (154.3) = 1862 psi =1.86 kpsi Ans. π ( 0.753 ) ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 4 Solutions, Page 66/80 www.konkur.in 4-97 Procedure 1 1. Arbitrarily, choose RC as a redundant reaction. 2. Statics. ΣFx = 0, 12(103) − 6(103) − RO − RC = 0 RO = 6(103) − RC (1) 3. The deflection of point C. 12(103 ) − 6(103 ) − RC (20) 6(103 ) + RC (10) R (15) − − C =0 δC = AE AE AE 4. The deflection equation simplifies to − 45 RC + 60(103) = 0 ⇒ RC = 1 333 lbf = 1.33 kip Ans. RO = 6(103) − 1 333 = 4 667 lbf = 4.67 kip Ans. From Eq. (1), FAB = FB + RC = 6 +1.333 = 7.333 kips compression FAB −7.333 = = −14.7 kpsi Ans. A (0.5)(1) Deflection of A. Since OA is in tension, R l 4 667(20) δ A = δ OA = O OA = = 0.00622 in Ans. AE (0.5)(1)(30)106 ______________________________________________________________________________ σ AB = 4-98 Procedure 1 1. Choose RB as redundant reaction. 2. Statics. RC = wl − RB (1) 1 w l 2 − RB ( l − a ) (2) 2 3. Deflection equation for point B. Superposition of beams 2 and 3 of Table A-9, MC = R (l − a ) w (l − a ) 4l ( l − a ) − ( l − a )2 − 6l 2 = 0 yB = B + 3EI 24 EI 3 2 4. Solving for RB. w 2 2 RB = 6l − 4l ( l − a ) + ( l − a ) 8(l − a ) = w 3l 2 + 2al + a 2 ) ( 8 (l − a ) Ans. Substituting this into Eqs. (1) and (2) gives Shigley’s MED, 10th edition Chapter 4 Solutions, Page 67/80 www.konkur.in RC = wl − RB = w 5l 2 − 10al − a 2 ) ( 8 (l − a ) Ans. 1 2 w wl − RB ( l − a ) = ( l 2 − 2al − a 2 ) Ans. 2 8 ______________________________________________________________________________ MC = 4-99 See figure in Prob. 4-98 solution. Procedure 1 1. Choose RB as redundant reaction. 2. Statics. RC = wl − RB (1) 1 M C = wl 2 − RB ( l − a ) (2) 2 3. Deflection equation for point B. Let the variable x start at point A and to the right. Using singularity functions, the bending moment as a function of x is 1 M = − w x 2 + RB x − a 2 1 ∂M = x−a ∂RB 1 ∂U 1 ∂M = M dx ∫ ∂RB EI 0 ∂RB l yB = l l 1 1 1 1 2 = − w x 2 ( 0 ) dx + − w x + RB ( x − a ) ( x − a ) dx = 0 ∫ ∫ EI 0 2 EI a 2 or, 1 1 a 3 3 R − w ( l 4 − a 4 ) − ( l 3 − a 3 ) + B ( l − a ) − ( a − a ) = 0 2 4 3 3 Solving for RB gives RB = w 8 (l − a ) 3 w 2 2 3 ( l 4 − a 4 ) − 4a ( l 3 − a 3 ) = 8 ( l − a ) ( 3l + 2al + a ) Ans. From Eqs. (1) and (2) RC = wl − RB = w 5l 2 − 10al − a 2 ) ( 8 (l − a ) Ans. 1 2 w wl − RB ( l − a ) = ( l 2 − 2al − a 2 ) Ans. 2 8 ______________________________________________________________________________ MC = Shigley’s MED, 10th edition Chapter 4 Solutions, Page 68/80 www.konkur.in 4-100 Note: When setting up the equations for this problem, no rounding of numbers was made in the calculations. It turns out that the deflection equation is very sensitive to rounding. Procedure 2 1. Statics. R1 + R2 = wl 1 2 wl 2 2. Bending moment equation. R2l + M 1 = (1) (2) 1 M = R1 x − w x 2 − M 1 2 dy 1 1 EI = R1 x 2 − w x 3 − M 1 x + C1 dx 2 6 1 1 1 EIy = R1 x3 − w x 4 − M 1 x 2 + C1 x + C2 6 24 2 (3) (4) EI = 30(106)(0.85) = 25.5(106) lbf⋅in2. 3. Boundary condition 1. At x = 0, y = − R1/k1 = − R1/[1.5(106)]. Substitute into Eq. (4) with value of EI yields C2 = − 17 R1. Boundary condition 2. At x = 0, dy /dx = − M1/k2 = − M1/[2.5(106)]. Substitute into Eq. (3) with value of EI yields C1 = − 10.2 M1. Boundary condition 3. At x = l, y = − R2/k3 = − R1/[2.0(106)]. Substitute into Eq. (4) with value of EI yields 1 3 1 1 R1l − wl 4 − M 1l 2 − 10.2 M 1l − 17 R1 (5) 6 24 2 Equations (1), (2), and (5), written in matrix form with w = 500/12 lbf/in and l = 24 in, are −12.75R2 = 1 0 1 24 1 0 2287 12.75 −532.8 R1 1 3 R2 = 12 (10 ) M 576 1 Solving, the simultaneous equations yields R1 = 554.59 lbf, R2 = 445.41.59 lbf, M1 = 1310.1 lbf⋅in Ans. For the deflection at x = l /2 = 12 in, Eq. (4) gives Shigley’s MED, 10th edition Chapter 4 Solutions, Page 69/80 www.konkur.in y x =12in = 1 1 500 4 1 1 12 − (1310.1)122 ( 554.59 )123 − 6 24 12 2 25.5 (10 ) 6 −10.2 (1310.1)12 − 17 ( 554.59 ) = −5.51(10−3 ) in Ans. ______________________________________________________________________________ 4-101 Cable area, A = π 4 (0.52 ) = 0.1963 in 2 Procedure 2 1. Statics. RA + FBE + FDF = 5(103) 3 FDF + FBE = 10(103) (1) (2) 2. Bending moment equation. M = RA x + FBE x − 16 − 5000 x − 32 1 1 dy 1 1 2 2 = RA x 2 + FBE x − 16 − 2500 x − 32 + C1 dx 2 2 1 1 2500 3 3 EIy = RA x 3 + FBE x − 16 − x − 32 + C1 x + C2 6 6 3 EI 3. B.C. 1: At x = 0, y = 0 ⇒ (3) (4) C2 = 0 B.C. 2: At x = 16 in, FBE (38) Fl yB = − = −6.453(10−6 ) FBE =− 6 AE 0.1963(30)10 BE Substituting into Eq. (4) and evaluating at x = 16 in 1 EIyB = 30(106 )(1.2)(−6.453)(10−6 ) FBE = RA (163 ) + C1 (16) 6 Simplifying gives 682.7 RA + 232.3 FBE + 16 C1 = 0 (5) B.C. 2: At x = 48 in, FDF (38) Fl yD = − = −6.453(10−6 ) FDF =− 6 AE 0.1963(30)10 DF Substituting into Eq. (4) and evaluating at x = 48 in, 1 1 2500 RA ( 483 ) + FBE (48 − 16)3 − (48 − 32)3 + 48C1 6 6 3 18 432 RA + 5 461 FBE + 232.3 FDF + 48 C1 = 3.413(106) EIyD = −232.3FDF = Simplifying gives Shigley’s MED, 10th edition (6) Chapter 4 Solutions, Page 70/80 www.konkur.in Equations (1), (2), (5) and (6) in matrix form are 1 1 0 RA 5000 1 1 3 0 FBE 10 000 0 = 0 682.7 232.3 0 16 FDF 6 3.413 10 C 18 432 5 461 232.3 48 ( ) 1 Solve simultaneously or use software. The results are RA = − 970.5 lbf, FBE = 3956 lbf, FDF = 2015 lbf, and C1 = − 16 020 lbf⋅in2. 3956 2015 = 20.2 kpsi, σ DF = = 10.3 kpsi Ans. σ BE = 0.1963 0.1963 EI = 30(106)(1.2) = 36(106) lbf⋅in2 y= = 1 2500 3 3 970.5 3 3956 − x + x − 16 − x − 32 − 16 020 x 6 6 6 3 36 10 ( ) −161.8 x + 659.3 x − 16 ( )( 1 36 106 B: x = 16 in, 3 3 − 833.3 x − 32 − 16 020 x 3 ) yB = 1 −161.8 (163 ) − 16 020 (16 ) = −0.0255 in 36 (106 ) yC = 1 −161.8 ( 323 ) + 659.3 ( 32 − 16 )3 − 16 020 ( 32 ) 6 36 (10 ) Ans. C: x = 32 in, = −0.0865 in Ans. D: x = 48 in, yD = ( ) 1 −161.8 483 + 659.3 ( 48 − 16 )3 − 833.3 ( 48 − 32 )3 − 16 020 ( 48 ) 6 36 10 ( ) = −0.0131 in Ans. ______________________________________________________________________________ 4-102 Beam: EI = 207(103)21(103) = 4.347(109) N⋅mm2. Rods: A = (π /4)82 = 50.27 mm2. Procedure 2 1. Statics. Shigley’s MED, 10th edition Chapter 4 Solutions, Page 71/80 www.konkur.in RC + FBE − FDF = 2 000 (1) RC + 2FBE = 6 000 (2) 2. Bending moment equation. M = − 2 000 x + FBE ⟨x − 75 ⟩1 + RC ⟨x − 150 ⟩1 dy 1 1 2 2 = −1000 x 2 + FBE x − 75 + RC x − 150 + C1 dx 2 2 1000 3 1 1 3 3 EIy = − x + FBE x − 75 + RC x − 150 + C1 x + C2 3 6 6 EI (3) (4) 3. B.C 1. At x = 75 mm, FBE ( 50 ) Fl yB = − = −4.805 (10−6 ) FBE =− 3 50.27 ( 207 )10 AE BE Substituting into Eq. (4) at x = 75 mm, 4.347 (109 ) −4.805 (10−6 ) FBE = − 1000 ( 753 ) + C1 ( 75) + C2 3 Simplifying gives 20.89 (103 ) FBE + 75C1 + C2 = 140.6 (10 6 ) (5) B.C 2. At x = 150 mm, y = 0. From Eq. (4), − or, 1000 1 3 1503 ) + FBE (150 − 75 ) + C1 (150 ) + C2 = 0 ( 3 6 70.31(103 ) FBE + 150C1 + C2 = 1.125 (109 ) (6) B.C 3. At x = 225 mm, FDF ( 65 ) Fl yD = = 6.246 (10−6 ) FDF = 3 AE DF 50.27 ( 207 )10 Substituting into Eq. (4) at x = 225 mm, Shigley’s MED, 10th edition Chapter 4 Solutions, Page 72/80 www.konkur.in 4.347 (109 ) 6.246 (10−6 ) FDF = − 1000 1 3 2253 ) + FBE ( 225 − 75 ) ( 3 6 1 3 + RC ( 225 − 150 ) + C1 ( 225 ) + C2 6 Simplifying gives 70.31(103 ) RC + 562.5 (103 ) FBE − 27.15 (103 ) FDF + 225C1 + C2 = 3.797 (109 ) (7) Equations (1), (2), (5), (6), and (7) in matrix form are 2 (103 ) 1 1 −1 0 0 RC 3 1 2 0 0 0 6 10 ( ) FBE 3 6 0 20.89 (10 ) 0 75 1 F DF = 140.6 (10 ) 3 0 70.31(10 ) 0 150 1 C1 1.125 109 ( ) 70.31(103 ) 562.5 (103 ) −27.15 (103 ) 225 1 C2 9 3.797 (10 ) Solve simultaneously or use software. The results are RC = − 2378 N, FBE = 4189 N, FDF = − 189.2 N Ans. and C1 = 1.036 (107) N⋅mm2, C2 = − 7.243 (108) N⋅mm3. The bolt stresses are σBE = 4189/50.27 = 83.3 MPa, σDF = − 189/50.27= − 3.8 MPa Ans. The deflections are From Eq. (4) yA = 1 −7.243 (108 ) = −0.167 mm 9 4.347 (10 ) Ans. For points B and D use the axial deflection equations*. 4189 ( 50 ) Fl yB = − = −0.0201 mm =− 50.27 ( 207 )103 AE BE Ans. −189 ( 65 ) Fl yD = = −1.18 (10−3 ) mm Ans. = 3 AE 50.27 207 10 ( ) DF *Note. The terms in Eq. (4) are quite large, and due to rounding are not very accurate for calculating the very small deflections, especially for point D. ______________________________________________________________________________ 4-103 (a) The cross section at A does not rotate. Thus, for a single quadrant we have Shigley’s MED, 10th edition Chapter 4 Solutions, Page 73/80 www.konkur.in ∂U =0 ∂M A The bending moment at an angle θ to the x axis is FR ∂M M = MA − =1 (1 − cos θ ) 2 ∂M A The rotation at A is π /2 ∂U 1 ∂M = M Rdθ = 0 θA = ∫ ∂M A EI 0 ∂M A Thus, 1 EI π /2 ∫ M A − 0 FR (1 − cos θ ) (1) Rdθ = 0 2 ⇒ FR π FR =0 MA − + 2 2 2 or, FR 2 1 − 2 π Substituting this into the equation for M gives FR 2 M= cos θ − (1) π 2 The maximum occurs at B where θ = π /2 MA = M max = M B = − FR Ans. π (b) Assume B is supported on a knife edge. The deflection of point D is ∂ U/∂ F. We will deal with the quarter-ring segment and multiply the results by 4. From Eq. (1) ∂M R 2 = cos θ − ∂F 2 π Thus, ∂U 4 = δD = ∂F EI π /2 ∫ 0 ∂M FR3 M Rdθ = ∂F EI π /2 ∫ 0 2 2 FR 3 π 2 cos θ − dθ = − EI 4 π π 3 FR π 2 − 8 ) Ans. ( 4π EI ______________________________________________________________________________ = 4-104 Pcr = I= Pcr = Cπ 2 EI l2 π (D 64 4 −d4) = π D4 64 (1 − K ) 4 where K = d D Cπ E π D 1 − K 4 ) ( 2 l 64 2 Shigley’s MED, 10th edition 4 Chapter 4 Solutions, Page 74/80 www.konkur.in 1/4 64 Pcr l 2 D= 3 Ans. 4 π CE (1 − K ) ______________________________________________________________________________ 4-105 A = π D 2 (1 − K 2 ) , π I= D 4 (1 − K 4 ) = π D 4 (1 − K 2 )(1 + K 2 ) , where K = d / D. 4 64 64 The radius of gyration, k, is given by I D2 k2 = = (1 + K 2 ) A 16 From Eq. (4-46) S y2l 2 S y2l 2 Pcr = Sy − 2 2 = Sy − 2 2 4π k CE 4π ( D / 16 )(1 + K 2 ) CE (π / 4 ) D 2 (1 − K 2 ) 4 Pcr = π D (1 − K 2 π D (1 − K 2 )S 2 y 2 )S y − = 4 Pcr + 4 S y2l 2π D 2 (1 − K 2 ) π 2 D 2 (1 + K 2 ) CE 4S y2l 2 (1 − K 2 ) π (1 + K 2 ) CE 4 S y2l 2 (1 − K 2 ) 4 Pcr D= + 2 2 2 π S y (1 − K ) π (1 + K ) CEπ (1 − K ) S y 1/2 1/2 S yl 2 Pcr = 2 + Ans. 2 2 2 π S y (1 − K ) π CE (1 + K ) ______________________________________________________________________________ 0.9 4-106 (a) ΣM A = 0, (0.75)(800) − 0.92 + 0.52 FBO (0.5) = 0 ⇒ FBO = 1373 N Using nd = 4, design for Fcr = nd FBO = 4(1373) = 5492 N l = 0.92 + 0.52 = 1.03 m, S y = 165 MPa In-plane: 1/2 1/2 bh3 / 12 I k = = A bh l 1.03 = = 142.7 k 0.007218 = 0.2887 h = 0.2887(0.025) = 0.007 218 m, C = 1.0 1/2 2 9 l 2π (207)(10 ) = 6 k 1 165(10 ) Shigley’s MED, 10th edition = 157.4 Chapter 4 Solutions, Page 75/80 www.konkur.in Since (l / k )1 > (l / k ) use Johnson formula. Try 25 mm x 12 mm, 2 165 (106 ) 1 6 Pcr = 0.025(0.012) 165 (10 ) − (142.7) = 29.1 kN 9 2π 1(207)10 This is significantly greater than the design load of 5492 N found earlier. Check out-ofplane. Out-of-plane: k = 0.2887(0.012) = 0.003 464 in, l 1.03 = = 297.3 k 0.003 464 Since (l / k )1 < (l / k ) use Euler equation. Pcr = 0.025(0.012) C = 1.2 1.2π 2 ( 207 )109 = 8321 N 297.32 This is greater than the design load of 5492 N found earlier. It is also significantly less than the in-plane Pcr found earlier, so the out-of-plane condition will dominate. Iterate the process to find the minimum h that gives Pcr greater than the design load. With h = 0.010, Pcr = 4815 N (too small) h = 0.011, Pcr = 6409 N (acceptable) Use 25 mm x 11 mm. If standard size is preferred, use 25 mm x 12 mm. Ans. ( ) P 1373 =− = −10.4 106 Pa = −10.4 MPa dh 0.012(0.011) No, bearing stress is not significant. Ans. ______________________________________________________________________________ (b) σ b = − 4-107 This is an open-ended design problem with no one distinct solution. ______________________________________________________________________________ F = 1500(π /4)22 = 4712 lbf. From Table A-20, Sy = 37.5 kpsi Pcr = nd F = 2.5(4712) = 11 780 lbf 4-108 (a) Assume Euler with C = 1 1/4 64 (11790 ) 502 ⇒ = 3 6 π (1) 30 (10 ) Use d = 1.25 in. The radius of gyration, k = ( I / A)1/2 = d /4 = 0.3125 in π P l2 I = d = cr 2 64 Cπ E 4 Shigley’s MED, 10th edition 1/4 64 P l 2 d = 3 cr π CE = 1.193 in Chapter 4 Solutions, Page 76/80 www.konkur.in l 50 = = 160 k 0.3125 2π 2 (1)30 (106 ) = 126 = 37.5 (103 ) 2 6 4 π ( 30 )10 (π / 64 )1.25 Pcr = = 14194 lbf 502 1/2 1/2 2 l 2π CE = k 1 S y Since 14 194 lbf > 11 780 lbf, d = 1.25 in is satisfactory. ∴ use Euler Ans. 1/4 64 (11780 )162 = 0.675 in, so use d = 0.750 in d= 3 6 π (1) 30 (10 ) k = 0.750/4 = 0.1875 in l 16 = = 85.33 use Johnson k 0.1875 (b) 2 37.5 (103 ) 1 3 Pcr = ( 0.750 ) 37.5 (10 ) − 85.33 = 12 748 lbf 6 π 4 2 1 30 10 ( ) Use d = 0.75 in. π 2 (c) 14194 = 3.01 Ans. 4 712 12 748 n(b ) = = 2.71 Ans. 4 712 ______________________________________________________________________________ n( a ) = 4-109 From Table A-20, Sy = 180 MPa 4F sinθ = 2 943 735.8 sin θ In range of operation, F is maximum when θ = 15° 735.8 Fmax = = 2843 N per bar sin15o F= Pcr = ndFmax = 3.50 (2 843) = 9 951 N l = 350 mm, h = 30 mm Shigley’s MED, 10th edition Chapter 4 Solutions, Page 77/80 www.konkur.in Try b = 5 mm. Out of plane, k = b / 12 = 5/ 12 = 1.443 mm l 350 = = 242.6 k 1.443 2 9 l 2π (1.4 ) 207 (10 ) = 180 (106 ) k 1 1/2 Pcr = A Cπ 2 E (l / k ) 2 = 5(30) = 178.3 1.4π 2 ( 207 )103 ( 242.6 ) 2 ∴ use Euler = 7 290 N Too low. Try b = 6 mm. k = 6/ 12 = 1.732 mm l 350 = = 202.1 k 1.732 1.4π 2 ( 207 )103 Cπ 2 E Pcr = A = 6(30) = 12 605 N 2 2 (l / k ) ( 202.1) O.K. Use 25 × 6 mm bars Ans. The factor of safety is 12 605 = 4.43 Ans. 2843 ______________________________________________________________________________ n= 4-110 P = 1 500 + 9 000 = 10 500 lbf Ans. ΣMA = 10 500 (4.5/2) − 9 000 (4.5) +M = 0 M = 16 874 lbf⋅in e = M / P = 16 874/10 500 = 1.607 in Ans. From Table A-8, A = 2.160 in2, and I = 2.059 in4. The stresses are determined using Eq. (4-55) k2 = I 2.059 = = 0.953 in 2 A 2.160 P ec 10500 1.607 ( 3 / 2 ) 1 + = −17157 psi = −17.16 kpsi Ans. 1 + 2 = − A k 2.160 0.953 ______________________________________________________________________________ σc = − 4-111 This is a design problem which has no single distinct solution. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 4 Solutions, Page 78/80 www.konkur.in 4-112 Loss of potential energy of weight = W (h +δ ) 1 Increase in potential energy of spring = kδ 2 2 1 W (h +δ ) = kδ 2 2 2 W 2 W or, δ 2 − δ− h = 0 . W = 30 lbf, k = 100 lbf/in, h = 2 in yields k k δ 2 − 0.6 δ − 1.2 = 0 Taking the positive root (see discussion on p. 206 of text) 1 δ max = 0.6 + (−0.6) 2 + 4(1.2) = 1.436 in 2 Ans. Fmax = k δ max = 100 (1.436) = 143.6 lbf Ans. ______________________________________________________________________________ 4-113 The drop of weight W1 converts potential energy, W1 h, to kinetic energy 1 W1 2 v1 . 2 g Equating these provides the velocity of W1 at impact with W2. 1 W1 2 (1) v1 ⇒ v1 = 2 gh 2 g Since the collision is inelastic, momentum is conserved. That is, (m1 + m2) v2 = m1 v1, where v2 is the velocity of W1 + W2 after impact. Thus W1h = W1 + W2 W v2 = 1 v1 g g ⇒ v2 = W1 W1 v1 = 2 gh W1 + W2 W1 + W2 (2) The kinetic and potential energies of W1 + W2 are then converted to potential energy of the spring. Thus, 1 W1 + W2 2 1 v2 + (W1 + W2 ) δ = kδ 2 2 g 2 Substituting in Eq. (1) and rearranging results in W1 + W2 W12 h (3) δ −2 =0 k W1 + W2 k Solving for the positive root (see discussion on p. 206) δ2 −2 2 1 W1 + W2 W12 h W + W2 + 4 1 + 8 2 k k W1 + W2 k δ = 2 Shigley’s MED, 10th edition (4) Chapter 4 Solutions, Page 79/80 www.konkur.in W1 = 40 N, W2 = 400 N, h = 200 mm, k = 32 kN/m = 32 N/mm. 2 1 40 + 400 402 200 40 + 400 = 29.06 mm + 4 + 8 32 40 + 400 32 32 δ = 2 2 Ans. Fmax = kδ = 32(29.06) = 930 N Ans. ______________________________________________________________________________ 1 4-114 The initial potential energy of the k1 spring is Vi = k1a 2 . The movement of the weight 2 1 1 2 W the distance y gives a final potential of Vf = k1 ( a − y ) + k 2 y 2 . Equating the two 2 2 energies give 1 2 1 1 2 k1a = k1 ( a − y ) + k2 y 2 2 2 2 Simplifying gives ( k1 + k2 ) y 2 − 2ak1 y = 0 2k1a . Without damping the weight will vibrate between k1 + k2 2k1a these two limits. The maximum displacement is thus y max = Ans. k1 + k2 With W = 5 lbf, k1 = 10 lbf/in, k2 = 20 lbf/in, and a = 0.25 in This has two roots, y = 0, 2 ( 0.25 )10 = 0.1667 in Ans. 10 + 20 ______________________________________________________________________________ ymax = Shigley’s MED, 10th edition Chapter 4 Solutions, Page 80/80 www.konkur.in Chapter 5 5-1 Sy = 350 MPa. MSS: σ1 − σ3 = Sy /n DE: n= ⇒ σ ′ = (σ A2 − σ Aσ B + σ B2 ) 1/2 n= Sy (σ 1 − σ 3 ) = (σ x2 − σ xσ y + σ y2 + 3τ xy2 ) 1/2 Sy σ′ (a) MSS: σ1 = 100 MPa,σ2 = 100 MPa,σ3 = 0 n= DE: (b) MSS: Ans. σ ′ = (1002 − 100(100) + 1002 )1/2 = 100 MPa, 350 = 3.5 100 − 0 350 = 3.5 100 Ans. 350 = 4.04 86.6 Ans. n= σ1 = 100 MPa,σ2 = 50 MPa,σ3 = 0 n= DE: 350 = 3.5 100 − 0 Ans. σ ′ = (1002 − 100(50) + 502 )1/2 = 86.6 MPa, n= 2 100 100 2 ± + (−75) = 140, − 40 MPa 2 2 σ 1 = 140, σ 2 = 0, σ 3 = −40 MPa 350 MSS: n = = 1.94 Ans. 140 − (−40) (c) σ A , σ B = DE: σ ′ = 1002 + 3 ( −752 ) 1/2 = 164 MPa, n= 350 = 2.13 164 Ans. −50 − 75 −50 + 75 2 (d) σ A , σ B = ± + (−50) = −11.0, −114.0 MPa 2 2 σ 1 = 0, σ 2 = −11.0, σ 3 = −114.0 MPa 350 MSS: n= = 3.07 Ans. 0 − (−114.0) DE: σ ′ = [(−50) 2 − (−50)(−75) + (−75) 2 + 3(−50) 2 ]1/ 2 = 109.0 MPa 350 n= = 3.21 Ans. 109.0 2 100 + 20 2 100 − 20 (e) σ A , σ B = ± + ( −20 ) = 104.7, 15.3 MPa 2 2 2 Shigley’s MED, 10th edition Chapter 5 Solutions, Page 1/52 www.konkur.in σ 1 = 104.7, σ 2 = 15.3, σ 3 = 0 MPa MSS: n= 350 = 3.34 104.7 − 0 Ans. 1/2 2 σ ′ = 1002 − 100(20) + 202 + 3 ( −20 ) = 98.0 MPa 350 n= = 3.57 Ans. 98.0 ______________________________________________________________________________ DE: 5-2 Sy = 350 MPa. MSS: σ1 − σ3 = Sy /n DE: n= ⇒ σ ′ = (σ A2 − σ Aσ B + σ B2 ) (a) MSS: DE: (b) MSS: 1/2 = Sy (σ 1 − σ 3 ) Sy n ⇒ n= σ 1 = 100 MPa, σ 3 = 0 ⇒ n = n= Sy σ′ 350 = 3.5 100 − 0 350 = 3.5 [100 − (100)(100) + 1002 ]1/2 2 σ 1 = 100 , σ 3 = −100 MPa ⇒ Ans. 350 = 1.75 100 − (−100) Ans. 350 = 2.02 Ans. 1/2 1002 − (100)(−100) + ( −100 )2 350 (c) MSS: = 3.5 Ans. σ 1 = 100 MPa, σ 3 = 0 ⇒ n = 100 − 0 350 DE: n= = 4.04 Ans. 1/2 2 100 − (100)(50) + 502 350 (d) MSS: = 2.33 Ans. σ 1 = 100, σ 3 = −50 MPa ⇒ n = 100 − (−50) 350 DE: n= = 2.65 Ans. 1/2 1002 − (100)(−50) + ( −50 )2 350 (e) MSS: σ 1 = 0, σ 3 = −100 MPa ⇒ n = = 3.5 Ans. 0 − (−100) 350 DE: n= = 4.04 Ans. 1/2 ( −50 ) 2 − (−50)(−100) + ( −100 ) 2 ______________________________________________________________________________ DE: 5-3 n= n= Ans. From Table A-20, Sy = 37.5 kpsi Shigley’s MED, 10th edition Chapter 5 Solutions, Page 2/52 www.konkur.in MSS: σ1 − σ3 = Sy /n n= ⇒ σ ′ = (σ A2 − σ Aσ B + σ B2 ) 1/2 DE: n= Sy (σ 1 − σ 3 ) = (σ x2 − σ xσ y + σ y2 + 3τ xy2 ) 1/2 Sy σ′ σ 1 = 25 kpsi, σ 3 = 0 (a) MSS: n= DE: ⇒ 37.5 1/2 252 − (25)(15) + 152 n= = 1.72 σ 1 = 15 kpsi, σ 3 = −15 ⇒ (b) MSS: n= DE: 37.5 = 1.5 25 − 0 n= 37.5 1/2 152 − (15)(−15) + ( −15 )2 Ans. Ans. 37.5 = 1.25 15 − (−15) = 1.44 Ans. Ans. 2 20 20 (c) σ A , σ B = ± + (−10) 2 = 24.1, − 4.1 kpsi 2 2 σ 1 = 24.1, σ 2 = 0, σ 3 = −4.1 kpsi 37.5 MSS: n= = 1.33 Ans. 24.1 − (−4.1) σ ′ = 202 + 3 ( −102 ) DE: 1/2 = 26.5 kpsi ⇒ n= 37.5 = 1.42 26.5 Ans. −12 + 15 −12 − 15 2 ± + (−9) = 17.7, −14.7 kpsi 2 2 σ 1 = 17.7, σ 2 = 0, σ 3 = −14.7 kpsi 37.5 MSS: n= = 1.16 Ans. 17.7 − (−14.7) 2 (d) σ A , σ B = 1/ 2 σ ′ = (−12) 2 − (−12)(15) + 152 + 3(−9)2 DE: n= 37.5 = 1.33 28.1 = 28.1 kpsi Ans. −24 − 24 2 −24 + 24 ± + ( −15 ) = −9, − 39 kpsi 2 2 σ 1 = 0, σ 2 = −9, σ 3 = −39 kpsi 37.5 MSS: n= = 0.96 Ans. 0 − ( −39 ) 2 (e) σ A , σ B = Shigley’s MED, 10th edition Chapter 5 Solutions, Page 3/52 www.konkur.in 2 2 2 σ ′ = ( −24 ) − ( −24 )( −24 ) + ( −24 ) + 3 ( −15 ) 1/ 2 = 35.4 kpsi 37.5 n= = 1.06 Ans. 35.4 ______________________________________________________________________________ DE: 5-4 From Table A-20, Sy = 47 kpsi. MSS: σ1 − σ3 = Sy /n n= ⇒ σ ′ = (σ A2 − σ Aσ B + σ B2 ) Sy (σ 1 − σ 3 ) Sy σ′ n 47 (a) MSS: = 1.57 Ans. σ 1 = 30 kpsi, σ 3 = 0 ⇒ n = 30 − 0 47 DE: n= = 1.57 Ans. 2 [30 − (30)(30) + 302 ]1/2 47 (b) MSS: = 0.78 Ans. σ 1 = 30 , σ 3 = −30 kpsi ⇒ n = 30 − (−30) 47 DE: n= = 0.90 Ans. 1/2 302 − (30)(−30) + ( −30 )2 47 (c) MSS: σ 1 = 30 kpsi, σ 3 = 0 ⇒ n = = 1.57 Ans. 30 − 0 47 DE: n= = 1.81 Ans. 1/2 2 30 − (30)(15) + 152 47 (d) MSS: = 1.57 Ans. σ 1 = 0, σ 3 = −30 kpsi ⇒ n = 0 − (−30) 47 DE: n= = 1.81 Ans. 1/2 ( −30 )2 − (−30)(−15) + ( −15 )2 47 (e) MSS: σ 1 = 10, σ 3 = −50 kpsi ⇒ n = = 0.78 Ans. 10 − (−50) 47 DE: n= = 0.84 Ans. 1/2 2 ( −50 ) − (−50)(10) + 102 ______________________________________________________________________________ DE: Shigley’s MED, 10th edition 1/2 = Sy ⇒ n= Chapter 5 Solutions, Page 4/52 www.konkur.in 5-5 Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. (a) MSS and DE: n= OB 4.95" = = 3.51 Ans. OA 1.41" (b) MSS: n= OD 3.91" = = 3.49 Ans. OC 1.12" DE: n= OE 4.51" = = 4.03 Ans. OC 1.12" (c) MSS: n= OG 2.50" = = 2.00 Ans. OF 1.25" DE: n = OH 2.86" = = 2.29 Ans. OF 1.25" (d) MSS: n = OJ 3.51" OK 3.65" = = 3.05 Ans. , DE: n = = = 3.17 Ans. OI 1.15" OI 1.15" OM 3.54" ON 3.77" = = 3.34 Ans. , DE: n = = = 3.56 Ans. OL 1.06" OL 1.06" ______________________________________________________________________________ (e) MSS: n = Shigley’s MED, 10th edition Chapter 5 Solutions, Page 5/52 www.konkur.in 5-6 Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. (a) σA = 25 kpsi, σB = 15 kpsi MSS: n= OB 4.37" = = 1.50 OA 2.92" Ans. DE: OC 5.02" = = 1.72 Ans. OA 2.92" (b) σA = 15 kpsi, σB = − 15 kpsi n= MSS: n= OE 2.66" = = 1.25 OD 2.12" Ans. n= OF 3.05" = = 1.44 OD 2.12" Ans. DE: 2 20 20 (c) σ A , σ B = ± + (−10) 2 = 24.1, − 4.1 kpsi 2 2 MSS: n = OH 3.25" = = 1.34 OG 2.43" Ans. DE: n = OI 3.46" = = 1.42 OG 2.43" Ans. −12 + 15 −12 − 15 2 (d) σ A , σ B = ± + (−9) = 17.7, −14.7 MPa 2 2 2 MSS: n = OK 2.67" = = 1.16 OJ 2.30" Ans. DE: n = OL 3.06" = = 1.33 OJ 2.30" Ans. −24 − 24 2 −24 + 24 ± + ( −15 ) = −9, − 39 kpsi 2 2 2 (e) σ A , σ B = ON 3.85" OP 4.23" = = 0.96 Ans. DE: n = = = 1.06 Ans. OM 4.00" OM 4.00" ______________________________________________________________________________ MSS: n = Shigley’s MED, 10th edition Chapter 5 Solutions, Page 6/52 www.konkur.in 5-7 Sy = 295 MPa, σA = 75 MPa, σB = − 35 MPa, (a) n = (σ Sy 2 A − σ Aσ B + σ ) 2 1/2 B = 295 1/2 752 − 75 ( −35 ) + ( −35 )2 = 3.03 Ans. (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n= OB 2.50" = = 3.01 Ans. OA 0.83" ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 7/52 www.konkur.in 5-8 Sy = 295 MPa, σA = 30 MPa, σB = − 100 MPa, (a) n = (σ Sy 2 A − σ Aσ B + σ ) 2 1/2 B = 295 1/2 302 − 30 ( −100 ) + ( −100 )2 = 2.50 Ans. (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n= OB 2.50" = = 3.01 Ans. OA 0.83" ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 8/52 www.konkur.in 2 5-9 100 100 2 Sy = 295 MPa, σ A , σ B = ± + (−25) = 105.9, − 5.9 MPa 2 2 Sy 295 (a) n = = = 2.71 1/2 2 2 1/2 (σ A − σ Aσ B + σ B ) 105.92 − 105.9 ( −5.9) + ( −5.9 )2 Ans. (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n= OB 2.87" = = 2.71 Ans. OA 1.06" ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 9/52 www.konkur.in −30 − 65 −30 + 65 2 Sy = 295 MPa, σ A , σ B = ± + 40 = −3.8, −91.2 MPa 2 2 Sy 295 (a) n = = = 3.30 1/2 2 2 1/2 2 (σ A − σ Aσ B + σ B ) ( −3.8) − ( −3.8)( −91.2 ) + ( −91.2 )2 2 5-10 Ans. (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n= OB 3.00" = = 3.33 Ans. OA 0.90" ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 10/52 www.konkur.in −80 + 30 2 −80 − 30 Sy = 295 MPa, σ A , σ B = ± + ( −10 ) = 30.9, −80.9 MPa 2 2 Sy 295 (a) n = = = 2.95 Ans. 1/2 2 2 1/2 (σ A − σ Aσ B + σ B ) 30.92 − 30.9 ( −80.9 ) + ( −80.9 )2 2 5-11 (b) Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n= OB 2.55" = = 2.93 Ans. OA 0.87" ______________________________________________________________________________ 5-12 Syt = 60 kpsi, Syc = 75 kpsi. Eq. (5-26) for yield is σ σ n= 1 − 3 S yt S yc (a) σ 1 = 25 kpsi, σ 3 = 0 (b) σ 1 = 15, σ 3 = −15 kpsi Shigley’s MED, 10th edition −1 −1 25 0 n = − = 2.40 60 75 ⇒ Ans. −1 ⇒ 15 −15 n= − = 2.22 60 75 Ans. Chapter 5 Solutions, Page 11/52 www.konkur.in 2 20 20 (c) σ A , σ B = ± + (−10) 2 = 24.1, − 4.1 kpsi , 2 2 σ 1 = 24.1, σ 2 = 0, σ 3 = −4.1 kpsi ⇒ −1 24.1 −4.1 n= − = 2.19 75 60 Ans. −12 + 15 −12 − 15 2 (d) σ A , σ B = ± + (−9) = 17.7, −14.7 kpsi 2 2 2 σ 1 = 17.7, σ 2 = 0, σ 3 = −14.7 kpsi −1 ⇒ 17.7 −14.7 n= − = 2.04 75 60 Ans. −24 − 24 2 −24 + 24 ± + ( −15 ) = −9, − 39 kpsi 2 2 2 (e) σ A , σ B = −1 0 −39 n= − Ans. = 1.92 60 75 ______________________________________________________________________________ σ 1 = 0, σ 2 = −9, σ 3 = −39 kpsi ⇒ 5-13 Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. (a) σ A = 25, σ B = 15 kpsi n= OB 3.49" = = 2.39 Ans. OA 1.46" (b) σ A = 15, σ B = −15 kpsi n= OD 2.36" = = 2.23 Ans. OC 1.06" (c) 2 σ A, σ B = n= 20 20 ± + ( −10) 2 = 24.1, − 4.1 kpsi 2 2 OF 2.67" = = 2.19 Ans. OE 1.22" (d) −12 + 15 −12 − 15 2 ± + (−9) = 17.7, −14.7 kpsi 2 2 2 σ A ,σ B = Shigley’s MED, 10th edition Chapter 5 Solutions, Page 12/52 www.konkur.in n= OH 2.34" = = 2.03 Ans. OG 1.15" (e) −24 − 24 2 −24 + 24 ± σ A ,σ B = + ( −15 ) = −9, − 39 kpsi 2 2 OJ 3.85" n= = = 1.93 Ans. OI 2.00" ______________________________________________________________________________ 2 5-14 Since εf > 0.05, and Syt ≠ Syc, the Coulomb-Mohr theory for ductile materials will be used. (a) From Eq. (5-26), −1 −1 σ σ 150 −50 n= 1 − 3 = − = 1.23 Ans. S yt S yc 235 285 (b) Plots for Problems 5-14 to 5-18 are found here. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. n= OB 1.94" = = 1.23 Ans. OA 1.58" ______________________________________________________________________________ 5-15 (a) From Eq. (5-26), −1 −1 σ1 σ 3 50 −150 n= − − = = 1.35 Ans. S 235 285 yt S yc Shigley’s MED, 10th edition Chapter 5 Solutions, Page 13/52 www.konkur.in (b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OD 2.14" = = 1.35 Ans. OC 1.58" ______________________________________________________________________________ n= 2 5-16 125 125 2 ± σ A, σ B = + (−75) = 160, − 35 kpsi 2 2 (a) From Eq. (5-26), −1 −1 σ1 σ 3 160 −35 n= − = − = 1.24 Ans. S yt S yc 235 285 (b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OF 2.04" = = 1.24 Ans. OE 1.64" ______________________________________________________________________________ n= −80 − 125 −80 + 125 2 ± + 50 = −47.7, − 157.3 kpsi 2 2 (a) From Eq. (5-26), 2 5-17 σ A, σ B = −1 −1 σ σ −157.3 0 n= 1 − 3 = − = 1.81 Ans. S S 235 285 yt yc (b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OH 2.99" = = 1.82 Ans. OG 1.64" ______________________________________________________________________________ n= 125 + 80 125 − 80 2 ± + (−75) = 180.8, 24.2 kpsi 2 2 (a) From Eq. (5-26), 2 5-18 σ A, σ B = Shigley’s MED, 10th edition Chapter 5 Solutions, Page 14/52 www.konkur.in −1 −1 σ σ 0 180.8 n= 1 − 3 = − = 1.30 Ans. S S 235 285 yt yc (b) The plot for this problem is found on the page for Prob. 5-14. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OJ 2.37" = = 1.30 Ans. OI 1.83" ______________________________________________________________________________ n= 5-19 Sut = 30 kpsi, Suc = 90 kpsi BCM: Eqs. (5-31), p. 250 MM: see Eqs. (5-32), p. 250 (a) σA = 25 kpsi, σB = 15 kpsi BCM : Eq. (5-31a), n = MM: Eq. (5-32a), n = Sut σA Sut = = σA (b) σA = 15 kpsi, σB = −15 kpsi, 30 = 1.2 Ans. 25 30 = 1.2 Ans. 25 −1 15 −15 BCM: Eq. (5-31a), n = − = 1.5 Ans. 30 90 MM: σA ≥ 0 ≥ σB, and |σB /σA | ≤ 1, Eq. (5-32a), n = Sut σA = 30 = 2.0 Ans. 15 = 30 = 1.24 Ans. 24.14 2 (c) σ A , σ B = 20 20 ± + (−10) 2 = 24.14, − 4.14 kpsi 2 2 −1 24.14 −4.14 BCM: Eq. (5-31b), n = − = 1.18 Ans. 90 30 MM: σA ≥ 0 ≥ σB, and |σB /σA | ≤ 1, Eq. (5-32a), n = Sut σA −15 + 10 −15 − 10 2 (d) σ A , σ B = ± + (−15) = 17.03, − 22.03 kpsi 2 2 2 Shigley’s MED, 10th edition Chapter 5 Solutions, Page 15/52 www.konkur.in −1 17.03 −22.03 BCM: Eq. (5-31b), n = − = 1.23 Ans. 90 30 MM: σA ≥ 0 ≥ σB, and |σB /σA | ≥ 1, Eq. (5-32b), −1 −1 ( 90 − 30 )17.03 −22.03 ( S − Sut ) σ A σ B n = uc − − = 1.60 Ans. = Suc Sut Suc 90 30 90 ( ) −20 − 20 −20 + 20 2 (e) σ A , σ B = ± + (−15) = −5, − 35 kpsi 2 2 2 BCM: Eq. (5-31c), n = − Suc =− Suc =− σB 90 = 2.57 Ans. −35 90 = 2.57 Ans. −35 σB ______________________________________________________________________________ MM: Eq. (5-32c), n = − 5-20 Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. (a) σA = 25, σB = 15 kpsi BCM & MM: OB 1.74" n= = = 1.19 Ans. OA 1.46" (b) σA = 15, σB = − 15 kpsi OC 1.59" = = 1.5 Ans. OD 1.06" OE 2.12" MM: n = = = 2.0 Ans. OC 1.06" BCM: n = 2 20 20 (c) σ A , σ B = ± + (−10) 2 2 2 = 24.14, − 4.14 kpsi OG 1.44" BCM: n = = = 1.18 Ans. OF 1.22" OH 1.52" MM: n = = = 1.25 Ans. OF 1.22" Shigley’s MED, 10th edition Chapter 5 Solutions, Page 16/52 www.konkur.in −15 + 10 −15 − 10 2 (d) σ A , σ B = ± + (−15) = 17.03, − 22.03 kpsi 2 2 OJ 1.72" BCM: n = = = 1.24 Ans. OI 1.39" OK 2.24" MM: n = = = 1.61 Ans. OI 1.39" 2 −20 − 20 −20 + 20 2 (e) σ A , σ B = ± + ( −15) = −5, − 35 kpsi 2 2 OM 4.55" BCM and MM: n = = = 2.57 Ans. OL 1.77" ______________________________________________________________________________ 2 5-21 From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi BCM: Eqs. (5-31), MM: Eqs. (5-32) (a) σA = 15, σB = 10 kpsi. Sut 31 = 2.07 Ans. σ A 15 S 31 MM: Eq. (5-32a), n = ut = = 2.07 Ans. σ A 15 (b), (c) The plot is shown below is for Probs. 5-21 to 5-25. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. BCM: Eq. (5-31a), n = = BCM and MM: n= OB 1.86" = = 2.07 Ans. OA 0.90" ______________________________________________________________________________ 5-22 Sut = 31 kpsi, Suc = 109 kpsi Shigley’s MED, 10th edition Chapter 5 Solutions, Page 17/52 www.konkur.in BCM: Eq. (5-31), MM: Eqs. (5-32) (a) σA = 15, σB = − 50 kpsi, |σB /σA | > 1 −1 15 −50 BCM: Eq. (5-31b), n = − = 1.06 Ans. 31 109 −1 −1 (109 − 31)15 −50 ( S − Sut ) σ A σ B MM: Eq. (3-32b), n = uc − − = 1.24 Ans. = Suc Sut Suc 109 109 ( 31) (b), (c) The plot is shown in the solution to Prob. 5-21. OD 2.78" = = 1.07 Ans. OC 2.61" OE 3.25" MM: n = = = 1.25 Ans. OC 2.61" ______________________________________________________________________________ BCM: n = 5-23 From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi BCM: Eq. (5-31), MM: Eqs. (5-32) 2 15 15 σ A , σ B = ± + (−10) 2 = 20, − 5 kpsi 2 2 −1 20 −5 (a) BCM: Eq. (5-32b), n = − = 1.45 Ans. 31 109 S 31 MM: Eq. (5-32a), n = ut = = 1.55 Ans. σ A 20 (b), (c) The plot is shown in the solution to Prob. 5-21. OG 1.48" = = 1.44 Ans. OF 1.03" OH 1.60" MM: n = = = 1.55 Ans. OF 1.03" ______________________________________________________________________________ BCM: n = 5-24 From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi BCM: Eq. (5-31), MM: Eqs. (5-32) −10 − 25 −10 + 25 2 σ A, σ B = ± + (−10) = −5, − 30 kpsi 2 2 2 Shigley’s MED, 10th edition Chapter 5 Solutions, Page 18/52 www.konkur.in Suc 109 = 3.63 Ans. σ B −30 S 109 MM: Eq. (5-32c), n = − uc = − = 3.63 Ans. σB −30 (b), (c) The plot is shown in the solution to Prob. 5-21. (a) BCM: Eq. (5-31c), n = − − OJ 5.53" = = 3.64 Ans. OI 1.52" ______________________________________________________________________________ BCM and MM: n = 5-25 From Table A-24, Sut = 31 kpsi, Suc = 109 kpsi BCM: Eq. (5-31), MM: Eqs. (5-32) −35 + 13 −35 − 13 2 ± σ A, σ B = + (−10) = 15, − 37 kpsi 2 2 2 −1 15 −37 (a) BCM: Eq. (5-31b), n = − = 1.21 Ans. 31 109 −1 −1 (109 − 31)15 −37 ( S − Sut ) σ A σ B MM: Eq. (5-32b), n = uc − − = 1.46 Ans. = S S S 109 31 109 ( ) uc ut uc (b), (c) The plot is shown in the solution to Prob. 5-21. OL 2.42" = = 1.21 Ans. OK 2.00" OM 2.91" MM: n = = = 1.46 Ans. OK 2.00" ______________________________________________________________________________ BCM: n = 5-26 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), (a) σA = 15, σB = − 10 kpsi. −1 15 −10 BCM: Eq. (5-31b), n = − = 1.42 Ans. 36 35 Shigley’s MED, 10th edition Chapter 5 Solutions, Page 19/52 www.konkur.in (b) The plot is shown below is for Probs. 5-26 to 5-30. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. OB 1.28" n= = = 1.42 Ans. OA 0.90" ______________________________________________________________________________ 5-27 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), (a) σA = 15, σB = − 15 kpsi. −1 10 −15 BCM: Eq. (5-31b), n = − = 1.42 Ans. 36 35 (b) The plot is shown in the solution to Prob. 5-26. OD 1.28" = = 1.42 Ans. OC 0.90" ______________________________________________________________________________ n= 5-28 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), 2 12 12 (a) σ A , σ B = ± + (−8)2 = 16, − 4 kpsi 2 2 −1 16 −4 BCM: Eq. (5-31b), n = − = 1.79 Ans. 36 35 (b) The plot is shown in the solution to Prob. 5-26. Shigley’s MED, 10th edition Chapter 5 Solutions, Page 20/52 www.konkur.in OF 1.47" = = 1.79 Ans. OE 0.82" ______________________________________________________________________________ n= 5-29 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), −10 − 15 −10 + 15 2 (a) σ A , σ B = ± + 10 = −2.2, − 22.8 kpsi 2 2 2 BCM: Eq. (5-31c), n = − 35 = 1.54 Ans. −22.8 (b) The plot is shown in the solution to Prob. 5-26. OH 1.76" = = 1.53 Ans. OG 1.15" ______________________________________________________________________________ n= 5-30 Sut = 36 kpsi, Suc = 35 kpsi BCM: Eq. (5-31), 15 + 8 2 15 − 8 (a) σ A , σ B = ± + ( −8 ) = 20.2, 2.8 kpsi 2 2 2 BCM: Eq. (5-31a), n = 36 = 1.78 Ans. 20.2 (b) The plot is shown in the solution to Prob. 5-26. OJ 1.82" = = 1.78 Ans. OI 1.02" ______________________________________________________________________________ n= 5-31 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32). For this problem, MM reduces to the MNS theory. S 36 (a) σA = 15, σB = − 10 kpsi. Eq. (5-32a), n = ut = = 2.4 Ans. σ A 15 (b) The plot on the next page is for Probs. 5-31 to 5-35. Note: The drawing in this manual may not be to the scale of original drawing. The measurements were taken from the original drawing. Shigley’s MED, 10th edition Chapter 5 Solutions, Page 21/52 www.konkur.in n= OB 2.16" = = 2.43 Ans. OA 0.90" ______________________________________________________________________________ 5-32 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the MNS theory. (a) σA = 10, σB = − 15 kpsi. Eq. (3-32b) is not valid and must use Eq, (3-32c), S 35 n = − uc = − = 2.33 Ans. −15 σB (b) The plot is shown in the solution to Prob. 5-31. OD 2.10" n= = = 2.33 Ans. OC 0.90" ______________________________________________________________________________ 5-33 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the MNS theory. 2 12 12 (a) σ A , σ B = ± + (−8)2 = 16, − 4 kpsi 2 2 S 36 n = ut = = 2.25 Ans. σ A 16 (b) The plot is shown in the solution to Prob. 5-31. OF 1.86" = = 2.27 Ans. OE 0.82" ______________________________________________________________________________ n= 5-34 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the MNS theory. Shigley’s MED, 10th edition Chapter 5 Solutions, Page 22/52 www.konkur.in −10 − 15 −10 + 15 2 (a) σ A , σ B = ± + 10 = −2.2, − 22.8 kpsi 2 2 Suc 35 n=− =− = 1.54 Ans. −22.8 σB (b) The plot is shown in the solution to Prob. 5-31. 2 OH 1.76" = = 1.53 Ans. OG 1.15" ______________________________________________________________________________ n= 5-35 Sut = 36 kpsi, Suc = 35 kpsi. MM: Use Eq. (5-32).For this problem, MM reduces to the MNS theory. 15 + 8 2 15 − 8 (a) σ A , σ B = ± + ( −8 ) = 20.2, 2.8 kpsi 2 2 S 36 n = ut = = 1.78 Ans. σ A 20.2 (b) The plot is shown in the solution to Prob. 5-31. 2 OJ 1.82" = = 1.78 Ans. OI 1.02" ______________________________________________________________________________ n= 5-36 Given: AISI 1006 CD steel, F = 0.55 kN, P = 4.0 kN, and T = 25 N⋅m. From Table A-20, Sy =280 MPa. Apply the DE theory to stress elements A and B 4 ( 4 )103 4P A: σx = 2 = = 22.6 (106 ) Pa = 22.6 MPa 2 πd π ( 0.015 ) 3 16 ( 25 ) 16T 4 V 4 0.55 (10 ) = 41.9 (106 ) Pa = 41.9 MPa τ xy = 3 + = + π d 3 A π ( 0.0153 ) 3 (π / 4 ) 0.0152 σ ′ = 22.62 + 3 ( 41.92 ) n= B: 280 = 3.68 76.0 1/2 = 76.0 MPa Ans. 3 3 32 Fl 4 P 32 ( 0.55 )10 ( 0.1) 4 ( 4 )10 σx = + = + = 189 (106 ) Pa = 189 MPa 3 2 3 2 πd πd π ( 0.015 ) π ( 0.015 ) τ xy = 16 ( 25 ) 16T = = 37.7 (106 ) Pa = 37.7 MPa 3 3 πd π ( 0.015 ) Shigley’s MED, 10th edition Chapter 5 Solutions, Page 23/52 www.konkur.in σ ′ = (σ x2 + 3τ xy2 ) 1/2 = 1892 + 3 ( 37.72 ) 1/2 = 200 MPa S y 280 = = 1.4 Ans. σ ′ 200 ______________________________________________________________________________ n= 5-37 From Prob. 3-44, the critical location is at the top of the beam at x = 27 in from the left end, where there is only a bending stress of σ = − 7 456 psi. Thus, σ′ = 7 456 psi and (Sy)min = nσ′ = 2(7 456) = 14 912 psi Choose (Sy)min = 15 kpsi Ans. ______________________________________________________________________________ 5-38 From Table A-20 for 1020 CD steel, Sy = 57 kpsi. From Eq. (3-42), p.116, T= 63 025H n (1) where n is the shaft speed in rev/min. From Eq. (5-3), for the MSS theory, τ max = Sy 2nd = 16T πd3 (2) where nd is the design factor. Substituting Eq. (1) into Eq. (2) and solving for d gives 32 ( 63 025 ) Hnd d = nπ S y 1/3 (3) Substituting H = 20 hp, nd = 3, n = 1750 rev/min, and Sy = 57(103) psi results in 32 ( 63 025 ) 20 ( 3) d min = = 0.728 in Ans. 3 1750π ( 57 )10 ______________________________________________________________________________ 1/3 5-39 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-68, in the plane of analysis σ1 = 16.5 kpsi, σ2 = − 1.19 kpsi, and τmax = 8.84 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 16.5 kpsi, σ2 = 0, and σ3 = − 1.19 kpsi Shigley’s MED, 10th edition Chapter 5 Solutions, Page 24/52 www.konkur.in MSS: From Eq. (5-3), n= Sy σ1 − σ 3 = 54 = 3.05 16.5 − ( −1.19 ) Ans. Note: Whenever the two principal stresses of a plane stress state are of opposite sign, the maximum shear stress found in the analysis is the true maximum shear stress. Thus, the factor of safety could have been found from Sy 54 = 3.05 Ans. 2τ max 2 ( 8.84 ) DE: The von Mises stress can be found from the principal stresses or from the stresses found in part (d) of Prob. 3-68. That is, n= = Eqs. (5-13) and (5-19) S Sy 54 n= y = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 16.52 − 16.5 ( −1.19 ) + ( −1.19 ) 2 = 3.15 Ans. or, Eqs. (5-15) and (5-19) using the results of part (d) of Prob. 3-68 Sy Sy 54 = = 1/2 1/2 2 2 2 σ ′ (σ + 3τ ) 15.3 + 3 ( 4.432 ) = 3.15 Ans. ______________________________________________________________________________ n= 5-40 From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-69, in the plane of analysis σ1 = 275 MPa, σ2 = − 12.1 MPa, and τmax = 144 MPa The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 275 MPa, σ2 = 0, and σ3 = − 12.1 MPa MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 370 = 1.29 275 − ( −12.1) Ans. DE: From Eqs. (5-13) and (5-19) S Sy 370 n= y = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 2752 − 275 ( −12.1) + ( −12.1)2 = 1.32 Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 25/52 www.konkur.in 5-41 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-70, in the plane of analysis σ1 = 22.6 kpsi, σ2 = − 1.14 kpsi, and τmax = 11.9 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 22.6 kpsi, σ2 = 0, and σ3 = − 1.14 kpsi MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 54 = 2.27 22.6 − ( −1.14 ) Ans. DE: From Eqs. (5-13) and (5-19) Sy Sy 54 n= = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 22.62 − 22.6 ( −1.14 ) + ( −1.14 ) 2 = 2.33 Ans. ______________________________________________________________________________ 5-42 From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-71, in the plane of analysis σ1 = 78.2 MPa, σ2 = − 5.27 MPa, and τmax = 41.7 MPa The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 78.2 MPa, σ2 = 0, and σ3 = − 5.27 MPa MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 370 = 4.43 78.2 − ( −5.27 ) Ans. DE: From Eqs. (5-13) and (5-19) S Sy 370 n= y = = 1/2 1/2 2 2 σ ′ (σ A − σ Aσ B + σ B ) 78.22 − 78.2 ( −5.27 ) + ( −5.27 ) 2 = 4.57 Ans. ______________________________________________________________________________ 5-43 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-72, in the plane of analysis σ1 = 36.7 kpsi, σ2 = − 1.47 kpsi, and τmax = 19.1 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 36.7 kpsi, σ2 = 0, and σ3 = − 1.47 kpsi Shigley’s MED, 10th edition Chapter 5 Solutions, Page 26/52 www.konkur.in MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 54 = 1.41 36.7 − ( −1.47 ) Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54 n= y = = 1/2 1/2 2 2 σ ′ (σ A − σ Aσ B + σ B ) 36.7 2 − 36.7 ( −1.47 ) + ( −1.47 ) 2 = 1.44 Ans. ______________________________________________________________________________ 5-44 From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-73, in the plane of analysis σ1 = 376 MPa, σ2 = − 42.4 MPa, and τmax = 209 MPa The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 376 MPa, σ2 = 0, and σ3 = − 42.4 MPa MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 370 = 0.88 376 − ( −42.4 ) Ans. DE: From Eqs. (5-13) and (5-19) S Sy 370 n= y = = 1/2 1/2 2 2 σ ′ (σ A − σ Aσ B + σ B ) 3762 − 376 ( −42.4 ) + ( −42.4 ) 2 = 0.93 Ans. ______________________________________________________________________________ 5-45 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-74, in the plane of analysis σ1 = 7.19 kpsi, σ2 = − 17.0 kpsi, and τmax = 12.1 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 7.19 kpsi, σ2 = 0, and σ3 = − 17.0 kpsi MSS: From Eq. (5-3), n = Sy = 54 = 2.23 7.19 − ( −17.0 ) Ans. σ1 − σ 3 DE: From Eqs. (5-13) and (5-19) S Sy 54 n= y = = 1/2 1/2 2 2 σ ′ (σ A − σ Aσ B + σ B ) 7.192 − 7.19 ( −17.0 ) + ( −17.0 ) 2 = 2.51 Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 27/52 www.konkur.in 5-46 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-76, in the plane of analysis σ1 = 1.72 kpsi, σ2 = − 35.9 kpsi, and τmax = 18.8 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 1.72 kpsi, σ2 = 0, and σ3 = − 35.9 kpsi MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 54 = 1.44 1.72 − ( −35.9 ) Ans. DE: From Eqs. (5-13) and (5-19) Sy Sy 54 n= = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 1.722 − 1.72 ( −35.9 ) + ( −35.9 )2 = 1.47 Ans. ______________________________________________________________________________ 5-47 From Table A-20, Sy = 370 MPa. From the solution of Prob. 3-77, Bending: σB = 68.6 MPa, Torsion: τB = 37.7 MPa For a plane stress analysis it was found that τmax = 51.0 MPa. With combined bending and torsion, the plane stress analysis yields the true τmax. S 370 MSS: From Eq. (5-3), n = y = = 3.63 Ans. 2τ max 2 ( 51.0 ) DE: From Eqs. (5-15) and (5-19) Sy Sy 370 n= = = 1/2 1/2 σ ′ (σ B2 + 3τ B2 ) 68.62 + 3 ( 37.7 2 ) = 3.91 Ans. ______________________________________________________________________________ 5-48 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-79, Bending: σC = 3460 psi, Torsion: τC = 882 kpsi For a plane stress analysis it was found that τmax = 1940 psi. With combined bending and torsion, the plane stress analysis yields the true τmax. 54 (103 ) Sy MSS: From Eq. (5-3), n = = = 13.9 Ans. 2τ max 2 (1940 ) Shigley’s MED, 10th edition Chapter 5 Solutions, Page 28/52 www.konkur.in DE: From Eqs. (5-15) and (5-19) 54 (103 ) Sy Sy n= = = σ ′ (σ C2 + 3τ C2 )1/2 34602 + 3 ( 8822 ) 1/2 = 14.3 Ans. ______________________________________________________________________________ 5-49 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-80, in the plane of analysis σ1 = 17.8 kpsi, σ2 = − 1.46 kpsi, and τmax = 9.61 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 17.8 kpsi, σ2 = 0, and σ3 = − 1.46 kpsi MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 54 = 2.80 17.8 − ( −1.46 ) Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54 n= y = = 1/2 1/2 2 2 σ ′ (σ A − σ Aσ B + σ B ) 17.82 − 17.8 ( −1.46 ) + ( −1.46 ) 2 = 2.91 Ans. ______________________________________________________________________________ 5-50 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-81, in the plane of analysis σ1 = 17.5 kpsi, σ2 = − 1.13 kpsi, and τmax = 9.33 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 17.5 kpsi, σ2 = 0, and σ3 = − 1.13 kpsi MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 54 = 2.90 17.5 − ( −1.13) Ans. DE: From Eqs. (5-13) and (5-19) S Sy 54 n= y = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 17.52 − 17.5 ( −1.13) + ( −1.13) 2 = 2.98 Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 29/52 www.konkur.in 5-51 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-82, in the plane of analysis σ1 = 21.5 kpsi, σ2 = − 1.20 kpsi, and τmax = 11.4 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 21.5 kpsi, σ2 = 0, and σ3 = − 1.20 kpsi MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 54 = 2.38 21.5 − ( −1.20 ) Ans. DE: From Eqs. (5-13) and (5-19) Sy Sy 54 n= = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 21.52 − 21.5 ( −1.20 ) + ( −1.20 ) 2 = 2.44 Ans. ______________________________________________________________________________ 5-52 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-83, the concern was failure due to twisting of the flat bar where it was found that τmax = 14.3 kpsi in the middle of the longest side of the rectangular cross section. The bar is also in bending, but the bending stress is zero where τmax exists. S 54 MSS: From Eq. (5-3), n = y = = 1.89 Ans. 2τ max 2 (14.3) DE: From Eqs. (5-15) and (5-19) S Sy 54 n= y = = = 2.18 Ans. 1/2 1/2 2 σ ′ ( 3τ max ) 3 (14.32 ) ______________________________________________________________________________ 5-53 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-84, in the plane of analysis σ1 = 34.7 kpsi, σ2 = − 6.7 kpsi, and τmax = 20.7 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 34.7 kpsi, σ2 = 0, and σ3 = − 6.7 kpsi MSS: From Eq. (5-3), n = Shigley’s MED, 10th edition Sy σ1 − σ 3 = 54 = 1.30 34.7 − ( −6.7 ) Ans. Chapter 5 Solutions, Page 30/52 www.konkur.in DE: From Eqs. (5-13) and (5-19) Sy Sy 54 n= = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 34.7 2 − 34.7 ( −6.7 ) + ( −6.7 ) 2 = 1.40 Ans. ______________________________________________________________________________ 5-54 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-85, in the plane of analysis σ1 = 51.1 kpsi, σ2 = − 4.58 kpsi, and τmax = 27.8 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 51.1 kpsi, σ2 = 0, and σ3 = − 4.58 kpsi MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 54 = 0.97 51.1 − ( −4.58 ) Ans. DE: From Eqs. (5-13) and (5-19) Sy Sy 54 n= = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 51.12 − 51.1( −4.58 ) + ( −4.58 ) 2 = 1.01 Ans. ______________________________________________________________________________ 5-55 From Table A-20, Sy = 54 kpsi. From the solution of Prob. 3-86, in the plane of analysis σ1 = 59.7 kpsi, σ2 = − 3.92 kpsi, and τmax = 31.8 kpsi The state of stress is plane stress. Thus, the three-dimensional principal stresses are σ1 = 59.7 kpsi, σ2 = 0, and σ3 = − 3.92 kpsi MSS: From Eq. (5-3), n = Sy σ1 − σ 3 = 54 = 0.85 59.7 − ( −3.92 ) Ans. DE: From Eqs. (5-13) and (5-19) Sy Sy 54 n= = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 59.7 2 − 59.7 ( −3.92 ) + ( −3.92 ) 2 = 0.87 Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 31/52 www.konkur.in 5-56 For Prob. 3-84, from Prob. 5-53 solution, with 1018 CD, DE theory yields, n = 1.40. From Table A-21, for 4140 Q&T @400°F, Sy = 238 kpsi. From Prob. (3-87) solution which considered stress concentrations for Prob. 3-84 σ1 = 53.0 kpsi, σ2 = − 8.48 kpsi, and τmax = 30.7 kpsi DE: From Eqs. (5-13) and (5-19) S Sy 238 n= y = = 1/2 1/2 σ ′ (σ A2 − σ Aσ B + σ B2 ) 53.02 − 53.0 ( −8.48 ) + ( −8.48 ) 2 = 4.12 Ans. Using the 4140 versus the 1018 CD, the factor of safety increases by a factor of 4.12/1.40 = 2.94. Ans. ______________________________________________________________________________ 5-57 Design Decisions Required: • • • • Material and condition Design factor Failure model Diameter of pin Using F = 416 lbf from Ex. 5-3, σ max = 32M πd3 1 32M 3 d = πσ max Decision 1: Select the same material and condition of Ex. 5-3 (AISI 1035 steel, Sy = 81 kpsi) Decision 2: Since we prefer the pin to yield, set nd a little larger than 1. Further explanation will follow. Decision 3: Use the Distortion Energy static failure theory. Decision 4: Initially set nd = 1 S S σ max = y = y = 81 000 psi nd 1 1 32(416)(15) 3 d = = 0.922 in π (81 000) Shigley’s MED, 10th edition Chapter 5 Solutions, Page 32/52 www.konkur.in Choose preferred size of d = 1.000 in π (1)3 (81 000) F= n= 32(15) = 530 lbf 530 = 1.27 416 Set design factor to nd = 1.27 Adequacy Assessment: σ max = Sy nd = 81 000 = 63 800 psi 1.27 1 32(416)(15) 3 d = = 1.00 in (OK) π (63 800) F= π (1)3 (81 000) = 530 lbf 32(15) 530 n= = 1.27 (OK) 416 ______________________________________________________________________________ 5-58 From Table A-20, for a thin walled cylinder made of AISI 1020 CD steel, Syt = 57 kpsi, Sut = 68 kpsi. Since r/t = 7.5/0.0625 = 120 > 10, the shell can be considered thin-wall. From the solution of Prob. 3-93 the principal stresses are σ1 = σ 2 = pd p (15) = = 60 p, 4t 4(0.0625) σ3 = − p From Eq. (5-12) 1 2 1 = 2 σ′ = 1/ 2 (σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 1/ 2 (60 p − 60 p ) 2 + (60 p + p ) 2 + (− p − 60 p) 2 = 61 p For yield, σ′ = Sy ⇒ 61p = 57 (103) ⇒ p = 934 psi Ans. For rupture, 61 p = 68 ⇒ p = 1.11 kpsi Ans. ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 33/52 www.konkur.in 5-59 For AISI 1020 HR steel, from Tables A-5 and A-20, w = 0.282 lbf/in3, Sy = 30 kpsi, and ν = 0.292. Then, ρ = w/g = 0.282/386 lbf⋅s2/in. For the problem, ri = 3 in, and ro = 5 in. Substituting into Eqs. (3-55), p. 129, gives σt = 9 ( 25 ) 1 + 3 ( 0.292 ) 2 0.282 2 3 + 0.292 ω r 9 + 25 + 2 − 386 8 r 3 + 0.292 225 = 3.006 (10−4 ) ω 2 34 + 2 − 0.5699r 2 = F ( r ) ω 2 r 225 σ r = 3.006 (10−4 ) ω 2 34 − 2 − r 2 = G ( r ) ω 2 r (1) (2) For the distortion-energy theory, the von Mises stress will be σ ′ = (σ t2 − σ tσ r + σ 22 ) 1/2 = ω 2 F 2 (r ) − F (r )G (r ) + G 2 (r ) 1/2 (3) Although it was noted that the maximum radial stress occurs at r = (rori )1/2 we are more interested as to where the von Mises stress is a maximum. One could take the derivative of Eq. (3) and set it to zero to find where the maximum occurs. However, it is much easier to plot σ′/ω 2 for 3 ≤ r ≤ 5 in. Plotting Eqs. (1) through (3) results in 70 60 50 σ'/ω2 40 tan 30 radial 20 von Mises 10 0 3 3.5 4 4.5 5 r (in) It can be seen that there is no maxima, and the greatest value of the von Mises stress is the tangential stress at r = ri. Substituting r = 3 in into Eq. (1) and setting σ′ = Sy gives 30 (103 ) ω= 3.006 10−4 34 + 225 − 0.5699 32 ( ) ( ) 32 Shigley’s MED, 10th edition 1/2 = 1361 rad/s Chapter 5 Solutions, Page 34/52 www.konkur.in 60ω 60(1361) = = 13 000 rev/min Ans. 2π 2π ________________________________________________________________________ n= 5-60 Since r/t = 1.75/0.065 = 26.9 > 10, we can use thin-walled equations. From Eqs. (3-53) and (3-54), p. 128, di = 3.5 − 2(0.065) = 3.37 in p(d i + t ) 2t 500(3.37 + 0.065) = 13 212 psi = 13.2 kpsi σt = 2(0.065) pd 500(3.37) = 6481 psi = 6.48 kpsi σl = i = 4t 4(0.065) σ r = − pi = −500 psi = −0.5 kpsi σt = These are all principal stresses, thus, from Eq. (5-12), { } 1 2 2 2 1/2 (13.2 − 6.48 ) + [6.48 − (−0.5)] + ( −0.5 − 13.2 ) 2 = 11.87 kpsi Sy 46 n= = ′ σ 11.87 n = 3.88 Ans. ________________________________________________________________________ σ′ = 5-61 From Table A-20, S y = 320 MPa With pi = 0, Eqs. (3-49), p. 127, are b2 r2 p r2 σ t = − 2o o 2 1 + i 2 = c 1 + 2 ro − ri r r (1) b2 ro2 po ri 2 σ r = − 2 2 1 − 2 = c 1 − 2 ro − ri r r For the distortion-energy theory, the von Mises stress is σ ′ = (σ − σ tσ r + σ 2 t ) 2 1/2 r b 2 2 b 2 b 2 b 2 2 = c 1 + 2 − 1 + 2 1 − 2 + 1 − 2 r r r r 1/2 1/2 b4 = c 1 + 3 4 r Shigley’s MED, 10th edition Chapter 5 Solutions, Page 35/52 www.konkur.in We see that the maximum von Mises stress occurs where r is a minimum at r = ri. Here, σr = 0 and thus σ′ = − σt . Setting − σt = Sy = 320 MPa at r = 0.1 m in Eq. (1) results in 2 ( 0.15 ) po 2r 2 p −σ t r = r = 2 o o2 = = 3.6 po = 320 ⇒ po = 88.9 MPa Ans. i ro − ri 0.152 − 0.12 ________________________________________________________________________ 2 5-62 From Table A-24, Sut = 31 kpsi for grade 30 cast iron. From Table A-5, ν = 0.211 and w = 0.260 lbf/in3. In Prob. 5-59, it was determined that the maximum stress was the tangential stress at the inner radius, where the radial stress is zero. Thus at the inner radius, Eq. (3-55), p. 129, gives 1 + 3v 2 0.260 2 3.211 1 + 3(0.211) 2 3+ v 2 2 2 2 ri = 3 σ t = ρω 2 ω 2ro + ri − 2 (5 ) + 3 − 3 + v 386 3.211 8 8 ( ) = 0.01471ω 2 = 31 103 ⇒ 1452 rad/ sec n = 60(1452)/(2π ) = 13 870 rev/min Ans. ________________________________________________________________________ 5-63 From Table A-20, for AISI 1035 CD, Sy = 67 kpsi. From force and bending-moment equations, the ground reaction forces are found in two planes as shown. The maximum bending moment will be at B or C. Check which is larger. In the xy plane, M B = 223(8) = 1784 lbf ⋅ in and M C = 127(6) = 762 lbf ⋅ in. In the xz plane, M B = 123(8) = 984 lbf ⋅ in and M C = 328(6) = 1968 lbf ⋅ in. Shigley’s MED, 10th edition Chapter 5 Solutions, Page 36/52 www.konkur.in 1 M B = [(1784)2 + (984) 2 ] 2 = 2037 lbf ⋅ in 1 M C = [(762) 2 + (1968) 2 ] 2 = 2110 lbf ⋅ in So point C governs. The torque transmitted between B and C is T = (300 − 50)(4) = 1000 lbf·in. The stresses are τ xz = 16T 16(1000) 5093 = = 3 psi d πd3 πd3 σx = 32M C 32(2110) 21 492 = = psi πd3 πd3 d3 For combined bending and torsion, the maximum shear stress is found from 1/2 τ max σ x 2 2 = + τ xz 2 1/ 2 21.49 2 5.09 2 = + 3 3 2d d = 11.89 kpsi d3 Max Shear Stress theory is chosen as a conservative failure theory. From Eq. (5-3) Sy 11.89 67 = ⇒ d = 0.892 in Ans. 3 2n d 2 ( 2) ________________________________________________________________________ τ max = 5-64 = As in Prob. 5-63, we will assume this to be a statics problem. Since the proportions are unchanged, the bearing reactions will be the same as in Prob. 5-63 and the bending moment will still be a maximum at point C. Thus xy plane: M C = 127(3) = 381 lbf ⋅ in xz plane: M C = 328(3) = 984 lbf ⋅ in So 1/ 2 M max = (381) 2 + (984) 2 = 1055 lbf ⋅ in 32 M C 32(1055) 10 746 10.75 = = psi = 3 kpsi σx = 3 3 3 πd πd d d Since the torsional stress is unchanged, τ xz = 5.09 kpsi d3 For combined bending and torsion, the maximum shear stress is found from Shigley’s MED, 10th edition Chapter 5 Solutions, Page 37/52 www.konkur.in 1/ 2 τ max σ 2 = x + τ xz2 2 1/ 2 10.75 2 5.09 2 = + 3 3 2d d = 7.40 kpsi d3 Using the MSS theory, as was used in Prob. 5-63, gives S 7.40 67 ⇒ d = 0.762 in Ans. τ max = y = 3 = 2n d 2 ( 2) ________________________________________________________________________ 5-65 For AISI 1018 HR, Table A-20 gives Sy = 32 kpsi. Transverse shear stress is a maximum at the neutral axis, and zero at the outer radius. Bending stress is a maximum at the outer radius, and zero at the neutral axis. Model (c): From Prob. 3-40, at outer radius, σ ′ = σ = 17.8 kpsi Sy 32 = = 1.80 σ ′ 17.8 At neutral axis, n= σ ′ = 3τ 2 = 3 ( 3.4 ) = 5.89 kpsi 2 Sy 32 = = 5.43 σ ′ 5.89 The bending stress at the outer radius dominates. n = 1.80 n= Ans. Model (d): Assume the bending stress at the outer radius will dominate, as in model (c). From Prob. 3-40, σ ′ = σ = 25.5 kpsi n= Sy 32 = = 1.25 σ ′ 25.5 Ans. Model (e): From Prob. 3-40, σ ′ = σ = 17.8 kpsi Sy 32 = = 1.80 Ans. σ ′ 17.8 Model (d) is the most conservative, thus safest, and requires the least modeling time. Model (c) is probably the most accurate, but model (e) yields the same results with less modeling effort. ________________________________________________________________________ n= 5-66 For AISI 1018 HR, from Table A-20, Sy = 32 kpsi. Model (d) yields the largest bending moment, so designing to it is the most conservative approach. The bending moment is M = 312.5 lbf⋅in. For this case, the principal stresses are Shigley’s MED, 10th edition Chapter 5 Solutions, Page 38/52 www.konkur.in σ1 = 32M , σ2 = σ3 = 0 πd3 Using a conservative yielding failure theory use the MSS theory and Eq. (5-3) σ1 − σ 3 = Sy S 32 M = y 3 n πd ⇒ n 1/3 ⇒ 32 Mn d = π S y 32 ( 312.5 ) 2.5 11 Thus, d = in Ans. = 0.629 in ∴ Use d = 3 16 π ( 32 )10 ________________________________________________________________________ 1/3 5-67 When the ring is set, the hoop tension in the ring is equal to the screw tension. σt = ri 2 pi ro2 1 + ro2 − ri 2 r 2 The differential hoop tension dF at r for the ring of width w, is dF = wσ t dr . Integration yields F=∫ ro ri wr 2 p wσ t dr = 2 i 2i ro − ri ro ∫ ro ri ro2 wri 2 pi ro2 1 + dr = r − = wri pi (1) 2 2 2 r r − r r o i ri The screw equation is Fi = T 0.2d (2) From Eqs. (1) and (2) pi = F T = wri 0.2d wri dFx = fpi ri dθ Fx = ∫ 2π 0 = fpi wri dθ = 2π f T 0.2d 2π f Tw ri ∫ dθ 0.2d wri 0 Ans. ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 39/52 www.konkur.in 5-68 T = 20 N⋅m, Sy = 450 MPa (a) From Prob. 5-67, T = 0.2 Fi d T 20 Fi = = = 16.7 (103 ) N = 16.7 kN 0.2d 0.2 6 (10−3 ) (b) From Prob. 5-67, F =wri pi ( ) Ans. 16.7 103 Fi F pi = = = = 111.3 106 Pa = 111.3 MPa − 3 − 3 wri wri 12 10 ( 25 / 2 ) 10 ( ) ( ( ( ) ) pi ri 2 + ro2 ri 2 pi ro2 σ t = 2 2 1 + = ro − ri r r =r ro2 − ri 2 (c) Ans. ) i = ( 111.3 0.01252 + 0.0252 0.0252 − 0.01252 ) = 185.5 MPa Ans. σ r = − pi = −111.3 MPa (d) τ max = σ1 − σ 3 = σt −σr 2 2 185.5 − (−111.3) = = 148.4 MPa 2 σ ' = (σ A2 − σ Aσ B + σ B2 )1/2 Ans. = 185.52 − (185.5)( −111.3) + ( −111.3) 2 = 259.7 MPa 1/2 Ans. (e) Maximum Shear Stress Theory n= Sy 2τ max = 450 = 1.52 2 (148.4 ) Ans. Distortion Energy theory Sy 450 = = 1.73 Ans. σ ′ 259.7 ________________________________________________________________________ n= Shigley’s MED, 10th edition Chapter 5 Solutions, Page 40/52 www.konkur.in 5-69 The moment about the center caused by the force F is F re where re is the effective radius. This is balanced by the moment about the center caused by the tangential (hoop) stress. For the ring of width w ro Fre = ∫ rσ t w dr ri ro2 r + ∫ri r dr w pi ri 2 ro2 − ri 2 r re = + ro2 ln o 2 2 ri F ( ro − ri ) 2 = w pi ri 2 ro2 − ri 2 ro From Prob. 5-67, F = wri pi. Therefore, re = ri ro2 − ri 2 r + ro2 ln o 2 2 ro − ri 2 ri For the conditions of Prob. 5-67, ri = 12.5 mm and ro = 25 mm 12.5 252 − 12.52 25 re = 2 + 252 ln Ans. = 17.8 mm 2 25 − 12.5 2 12.5 ________________________________________________________________________ 5-70 (a) The nominal radial interference is δnom = (2.002 − 2.001) /2 = 0.0005 in. From Eq. (3-57), p. 130, 2 2 2 2 Eδ ( ro − R )( R − ri ) p= 3 2R ro2 − ri 2 30 (106 ) 0.0005 (1.52 − 12 )(12 − 0.6252 ) = 3072 psi Ans. = 1.52 − 0.6252 2 (13 ) Inner member: pi = 0, po = p = 3072 psi. At fit surface r =R = 1 in, Eq. (3-49), p. 127, σt = − p 12 + 0.6252 R 2 + ri 2 = − 3072 = −7 010 psi 2 2 R 2 − ri 2 1 − 0.625 σr = − p = − 3072 psi Shigley’s MED, 10th edition Chapter 5 Solutions, Page 41/52 www.konkur.in Eq. (5-13) σ ′ = (σ A2 − σ Aσ B + σ A2 ) 1/2 1/2 2 = ( −7010 ) − ( −7010 )( −3072 ) + ( −3072 ) = 6086 psi Ans. Outer member: pi = p = 3072 psi, po = 0. At fit surface r =R = 1 in, Eq. (3-49), p. 127, σt = p 1.52 + 12 ro2 + R 2 = 3072 2 2 = 7 987 psi ro2 − R 2 1.5 − 1 σr = − p = − 3072 psi Eq. (5-13) σ ′ = (σ A2 − σ Aσ B + σ A2 ) 1/2 = 7987 2 − 7987 ( −3072 ) + ( −3072 ) 1/2 = 9888 psi Ans. (b) For a solid inner tube, 30 (106 ) 0.0005 (1.52 − 12 )(12 ) = 4167 psi Ans. p= 1.52 2 (13 ) Inner member: σt = σr = − p = − 4167 psi 1/2 2 2 σ ′ = ( −4167 ) − ( −4167 )( −4167 ) + ( −4167 ) = 4167 psi Ans. Outer member: pi = p = 4167 psi, po = 0. At fit surface r =R = 1 in, Eq. (3-49), p. 127, σt = p 1.52 + 12 ro2 + R 2 = 4167 2 2 = 10834 psi ro2 − R 2 1.5 − 1 σr = − p = − 4167 psi Eq. (5-13) σ ′ = (σ A2 − σ Aσ B + σ A2 ) 1/2 = 10 8342 − 10 834 ( −4167 ) + ( −4167 ) 1/2 = 13 410 psi Ans. ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 42/52 www.konkur.in 5-71 From Table A-5, E = 207 (103) MPa. The nominal radial interference is δnom = (40 − 39.98) /2 = 0.01 mm. From Eq. (3-57), p. 130, 2 2 2 2 Eδ ( ro − R )( R − ri ) p= 3 2R ro2 − ri 2 207 (103 ) 0.01 ( 32.52 − 202 )( 202 − 102 ) = 26.64 MPa Ans. = 32.52 − 102 2 ( 203 ) Inner member: pi = 0, po = p = 26.64 MPa. At fit surface r =R = 20 mm, Eq. (3-49), p. 127, σt = − p 202 + 102 R 2 + ri 2 = − 26.64 = −44.40 MPa 2 2 R 2 − ri 2 20 − 10 σr = − p = − 26.64 MPa Eq. (5-13) σ ′ = (σ A2 − σ Aσ B + σ A2 ) 1/2 1/2 = ( −44.40 ) − ( −44.40 )( −26.64 ) + ( −26.64 ) 2 = 38.71 MPa Ans. Outer member: pi = p = 26.64 MPa, po = 0. At fit surface r =R = 20 mm, Eq. (3-49), p. 127, σt = p 32.52 + 202 ro2 + R 2 = 26.64 = 59.12 MPa 2 2 ro2 − R 2 32.5 − 20 σr = − p = − 26.64 MPa Eq. (5-13) σ ′ = (σ A2 − σ Aσ B + σ A2 ) 1/2 = 59.122 − 59.12 ( −26.64 ) + ( −26.64 ) 1/2 = 76.03 MPa Ans. ________________________________________________________________________ 5-72 From Table A-5, E = 207 (103) MPa. The nominal radial interference is δnom = (40.008 − 39.972) /2 = 0.018 mm. From Eq. (3-57), p. 130, Shigley’s MED, 10th edition Chapter 5 Solutions, Page 43/52 www.konkur.in 2 2 2 2 Eδ ( ro − R )( R − ri ) p= 2R3 ro2 − ri 2 207 (103 ) 0.018 ( 32.52 − 202 )( 202 − 102 ) = 47.94 MPa Ans. = 32.52 − 102 2 ( 203 ) Inner member: pi = 0, po = p = 47.94 MPa. At fit surface r =R = 20 mm, Eq. (3-49), p. 127, σt = − p 202 + 102 R 2 + ri 2 = − 47.94 = −79.90 MPa 2 2 R 2 − ri 2 20 − 10 σr = − p = − 47.94 MPa Eq. (5-13) σ ′ = (σ A2 − σ Aσ B + σ A2 ) 1/2 1/2 = ( −79.90 ) − ( −79.90 )( −47.94 ) + ( −47.94 ) 2 = 69.66 MPa Ans. Outer member: pi = p = 47.94 MPa, po = 0. At fit surface r =R = 20 mm, Eq. (3-49), p. 127, σt = p 32.52 + 202 ro2 + R 2 = 47.94 = 106.4 MPa 2 2 ro2 − R 2 32.5 − 20 σr = − p = − 47.94 MPa Eq. (5-13) σ ′ = (σ A2 − σ Aσ B + σ A2 ) 1/2 = 106.42 − 106.4 ( −47.94 ) + ( −47.94 ) 1/2 = 136.8 MPa Ans. ________________________________________________________________________ 5-73 From Table A-5, for carbon steel, Es = 30 kpsi, and νs = 0.292. While for Eci = 14.5 Mpsi, and νci = 0.211. For ASTM grade 20 cast iron, from Table A-24, Sut = 22 kpsi. For midrange values, δ = (2.001 − 2.0002)/2 = 0.0004 in. Eq. (3-50), p. 127, Shigley’s MED, 10th edition Chapter 5 Solutions, Page 44/52 www.konkur.in δ p= 1 r +R 1 R 2 + ri 2 ν R + 2 2 −ν i o + 2 Ei R − ri Eo r − R 0.0004 = = 2613 psi 22 + 12 12 1 1 1 + 0.211 + − 0.292 2 6 2 6 2 14.5 (10 ) 2 − 1 30 (10 ) 1 2 o 2 o 2 At fit surface, with pi = p =2613 psi, and po = 0, from Eq. (3-50), p. 127 σt = p 22 + 12 ro2 + R 2 = 2613 2 2 = 4355 psi ro2 − R 2 2 −1 σr = − p = − 2613 psi From Modified-Mohr theory, Eq. (5-32a), since σA > 0 > σB and | σB /σA | <1, Sut 22 = 5.05 Ans. σ A 4.355 ________________________________________________________________________ n= 5-74 = E = 207 GPa Eq. (3-57), p. 130, can be written in terms of diameters, p= Eδ d 2D3 ( ) 3 (d o2 − D 2 )( D 2 − d i2 ) 207 10 (0.062) (502 − 452 )(452 − 402 ) = 3 (d o2 − di2 ) (502 − 402 ) 2 ( 45 ) p = 15.80 MPa Outer member: From Eq. (3-50), Outer radius: (σ t )o = 452 (15.80) (2) = 134.7 MPa 502 − 452 (σ r ) o = 0 452 (15.80) 502 1 + = 150.5 MPa 502 − 452 452 = −15.80 MPa Inner radius: (σ t )i = (σ r )i Bending (no slipping): Shigley’s MED, 10th edition I = (π /64)(504 − 404) = 181.1 (103) mm4 Chapter 5 Solutions, Page 45/52 www.konkur.in At ro : (σ x ) o = ± Mc 675(0.05 / 2) =± = ±93.2(106 ) Pa = ±93.2 MPa −9 I 181.1 10 At ri : (σ x )i = ± 675(0.045 / 2) = ±83.9 106 Pa = ±83.9 MPa −9 181.1 10 ( ( ) ( ) ) Torsion: J = 2I = 362.2 (103) mm4 At ro : (τ ) = Tc 900(0.05 / 2) = = 62.1 106 Pa = 62.1 MPa −9 J 362.2 10 At ri : (τ ) = 900(0.045 / 2) = 55.9 106 Pa = 55.9 MPa −9 362.2 10 xy o xy i ( ( ( ) ) ( ) ) Outer radius, is plane stress. Since the tangential stress is positive the von Mises stress will be a maximum with a negative bending stress. That is, σ x = −93.2 MPa, σ y = 134.7 MPa, τ xy = 62.1 MPa σ ′ = (σ x2 − σ xσ y + σ y2 + 3τ xy2 ) 1/2 Eq. (5-15) 2 2 = ( −93.2 ) − ( −93.2 )(134.7 ) + 134.7 2 + 3 ( 62.1) S y 415 no = = = 1.84 Ans. σ ′ 226 1/2 = 226 MPa Inner radius, 3D state of stress From Eq. (5-14) with τyz = τzx = 0 and σx = + 83.9 MPa 1/2 1 (83.9 − 150.5)2 + (150.5 + 15.8) 2 + (−15.8 − 83.9) 2 + 6(55.9) 2 = 174 MPa σ′ = 2 With σx = − 83.9 MPa 1/2 1 (−83.9 − 150.5) 2 + (150.5 + 15.8) 2 + (−15.8 + 83.9) 2 + 6(55.9) 2 = 230 MPa σ′ = 2 Shigley’s MED, 10th edition Chapter 5 Solutions, Page 46/52 www.konkur.in S y 415 = = 1.80 Ans. σ ′ 230 ________________________________________________________________________ 5-75 From the solution of Prob. 5-74, p = 15.80 MPa ni = Inner member: From Eq. (3-50), Outer radius: (σ t )o = − ro2 + ri 2 452 + 402 p = − (15.80) = −134.8 MPa o ro2 − ri 2 452 − 402 (σ r )o = − p = −15.80 MPa Inner radius: (σ t )i ( ) 2 452 2ro2 = − 2 2 po = − 2 (15.80) = −150.6 MPa ro − ri 45 − 402 (σ r )i = 0 Bending (no slipping): I = (π /64)(504 − 404) = 181.1 (103) mm4 Mc 675(0.045 / 2) = ±83.9(106 ) Pa = ±83.9 MPa At ro : (σ x )o = ± = ± −9 I 181.1 10 ( At ri : (σ x ) i = ± ) ( ) 675(0.040 / 2) = ±74.5 106 Pa = ±74.5 MPa 181.1 10−9 ( ) Torsion: J = 2I = 362.2 (103) mm4 At ro : (τ ) = Tc 900(0.045 / 2) = = 55.9 106 Pa = 55.9 MPa −9 J 362.2 10 At ri : (τ ) = 900(0.040 / 2) = 49.7 106 Pa = 49.7 MPa 362.2 10−9 xy o xy i ( ( ) ( ) ) ( ) Outer radius, 3D state of stress Shigley’s MED, 10th edition Chapter 5 Solutions, Page 47/52 www.konkur.in From Eq. (5-14) with τyz = τzx = 0 and σx = + 83.9 MPa 1/2 1 (83.9 + 134.8) 2 + (−134.8 + 15.8) 2 + (−15.8 − 83.9) 2 + 6(55.9) 2 = 213 MPa σ′ = 2 With σx = − 83.9 MPa 1/2 1 (−83.9 + 134.8) 2 + (−134.8 + 15.8) 2 + (−15.8 + 83.9) 2 + 6(55.9) 2 = 142 MPa σ′ = 2 S y 415 no = = = 1.95 Ans. σ ′ 213 Inner radius, plane stress. Worst case is when σx is positive σ x = 74.5 MPa, σ y = −150.6 MPa, τ xy = 49.7 MPa Eq. (5-15) 1/2 σ ′ = (σ x2 − σ xσ y + σ y2 + 3τ xy2 ) 2 1/ 2 = 74.52 − 74.5 ( −150.6 ) + ( −150.6 ) + 3 ( 49.7 ) = 216 MPa S y 415 ni = = = 1.92 Ans. σ ′ 216 ______________________________________________________________________________ 2 5-76 For AISI 1040 HR, from Table A-20, Sy = 290 MPa. From Prob. 3-110, pmax = 65.2 MPa. From Eq. (3-50) at the inner radius R of the outer member, r 2 + R2 502 + 252 = 65.2 = 108.7 MPa σ t = p o2 ro − R 2 502 − 252 σ r = − p = −65.2 MPa These are principal stresses. From Eq. (5-13) σ o′ = (σ t2 − σ tσ r + σ r2 ) 1/ 2 2 = 108.7 2 − 108.7 ( −65.2 ) + ( −65.2 ) 1/ 2 = 152.2 MPa Sy 290 = = 1.91 Ans. σ o′ 152.2 ________________________________________________________________________ n= 5-77 For AISI 1040 HR, from Table A-20, Sy = 42 kpsi. From Prob. 3-111, pmax = 9 kpsi. From Eq. (3-50) at the inner radius R of the outer member, Shigley’s MED, 10th edition Chapter 5 Solutions, Page 48/52 www.konkur.in ro2 + R 2 22 + 12 = 9 = 15 kpsi ro2 − R 2 22 − 12 σ r = − p = −9 kpsi These are principal stresses. From Eq. (5-13) σt = p σ o′ = (σ t2 − σ tσ r + σ r2 ) 1/2 2 = 152 − 15( −9) + ( −9 ) 1/2 = 21 kpsi S y 42 = = 2 Ans. σ o′ 21 ________________________________________________________________________ n= 5-78 For AISI 1040 HR, from Table A-20, Sy = 290 MPa. From Prob. 3-111, pmax = 91.6 MPa. From Eq. (3-50) at the inner radius R of the outer member, r 2 + R2 502 + 252 σ t = p o2 = 91.6 = 152.7 MPa ro − R 2 502 − 252 σ r = − p = −91.6 MPa These are principal stresses. From Eq. (5-13) σ o′ = (σ t2 − σ tσ r + σ r2 ) 1/ 2 2 = 152.7 2 − 152.7( −91.6) + ( −91.6 ) 1/2 = 213.8 MPa Sy 290 = = 1.36 Ans. σ o′ 213.8 ________________________________________________________________________ n= 5-79 For AISI 1040 HR, from Table A-20, Sy = 42 kpsi. From Prob. 3-111, pmax = 12.94 kpsi. From Eq. (3-50) at the inner radius R of the outer member, r 2 + R2 22 + 12 σ t = p o2 = 12.94 = 21.57 kpsi ro − R 2 22 − 12 σ r = − p = −12.94 kpsi These are principal stresses. From Eq. (5-13) σ o′ = (σ t2 − σ tσ r + σ r2 ) 1/2 2 = 21.57 2 − 21.57( −12.94) + ( −12.94 ) 1/2 = 30.20 kpsi Sy 42 = = 1.39 Ans. σ o′ 30.2 ________________________________________________________________________ n= Shigley’s MED, 10th edition Chapter 5 Solutions, Page 49/52 www.konkur.in 5-80 For AISI 1040 HR, from Table A-20, Sy = 290 MPa. From Prob. 3-111, pmax = 134 MPa. From Eq. (3-50) at the inner radius R of the outer member, r 2 + R2 502 + 252 σ t = p o2 = 134 2 = 223.3 MPa ro − R 2 50 − 252 σ r = − p = −134 MPa These are principal stresses. From Eq. (5-13) σ o′ = (σ t2 − σ tσ r + σ r2 ) 1/2 2 = 223.32 − 223.3( −134) + ( −134 ) 1/2 = 312.6 MPa Sy 290 = = 0.93 Ans. σ o′ 312.6 ________________________________________________________________________ n= 5-81 For AISI 1040 HR, from Table A-20, Sy = 42 kpsi. From Prob. 3-111, pmax = 19.13 kpsi. From Eq. (3-50) at the inner radius R of the outer member, r 2 + R2 22 + 12 σ t = p o2 = 19.13 = 31.88 kpsi ro − R 2 22 − 12 σ r = − p = −19.13 kpsi These are principal stresses. From Eq. (5-13) σ o′ = (σ t2 − σ tσ r + σ r2 ) 1/ 2 2 = 31.882 − 31.88( −19.13) + ( −19.13 ) 1/2 = 44.63 kpsi Sy 42 = = 0.94 Ans. σ o′ 44.63 ________________________________________________________________________ n= 5-82 θ K θ θ 1 K 3θ σ p = 2 I cos ± I sin cos sin 2 2 2π r 2 2 2 2π r 2 1/2 2 θ θ 3θ KI + sin cos cos 2 2 2 2π r Shigley’s MED, 10th edition Chapter 5 Solutions, Page 50/52 www.konkur.in 1/2 KI θ 2 θ 2θ 2 3θ 2θ 2θ 2 3θ = + sin cos cos cos ± sin cos sin 2 2 2 2 2 2 2 2π r KI KI θ θ θ θ θ = cos 1 ± sin cos ± sin cos = 2 2 2 2 2 2π r 2π r Plane stress: The third principal stress is zero and θ θ θ θ KI KI cos 1 + sin , σ 2 = cos 1 − sin , σ 3 = 0 Ans. σ1 = 2 2 2 2 2π r 2π r Plane strain: Equations for σ1 andσ2 are still valid,. However, KI θ cos Ans. 2 2π r ________________________________________________________________________ σ 3 = ν (σ 1 + σ 2 ) = 2ν 5-83 For θ = 0 and plane strain, the principal stress equations of Prob. 5-82 give σ1 = σ 2 = (a) DE: Eq. (5-18) or, KI , σ 3 = 2ν 2π r KI = 2νσ 1 2π r 1 2 2 2 1/2 σ 1 − σ 1 ) + (σ 1 − 2νσ 1 ) + ( 2νσ 1 − σ 1 ) = S y ( 2 σ1 − 2νσ1 = Sy 1 For ν = , 3 1 1 − 2 3 σ 1 = S y (a) MSS: Eq. (5-3) , with n =1 ⇒ σ 1 = 3S y σ1 − σ3 = Sy ν= 1 3 ⇒ ⇒ Ans. σ1 − 2νσ1 = Sy σ 1 = 3S y Ans. 2 3 σ 3 = σ1 − S y = 2S y ⇒ σ 3 = σ1 Radius of largest circle 1 2 σ R = σ1 − σ1 = 1 2 3 6 ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 51/52 www.konkur.in 5-84 Given: a = 16 mm, KIc = 80 MPa⋅ m and S y = 950 MPa (a) Ignoring stress concentration F = SyA =950(100 − 16)(12) = 958(103) N = 958 kN Ans. (b) From Fig. 5-26: h/b = 1, a/b = 16/100 = 0.16, β = 1.3 Eq. (5-37) K I = βσ π a 80 = 1.3 F π (16)10−3 100(12) F = 329.4(103) N = 329.4 kN Ans. ________________________________________________________________________ 5-85 Given: a = 0.5 in, KIc = 72 kpsi⋅ in and Sy = 170 kpsi, Sut = 192 kpsi ro = 14/2 = 7 in, ri = (14 − 2)/2 = 6 in a 0.5 = = 0.5, ro − ri 7 − 6 Fig. 5-30: β B 2.4 Eq. (5-37): K Ic = βσ π a ri 6 = = 0.857 ro 7 ⇒ 72 = 2.4σ π ( 0.5 ) ⇒ σ = 23.9 kpsi Eq. (3-50), p. 127, at r = ro = 7 in: r2 p 62 p Ans. σ t = 2i i 2 ( 2 ) ⇒ 23.9 = 2 i 2 ( 2 ) ⇒ pi = 4.315 kpsi ro − ri 7 −6 ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 5 Solutions, Page 52/52 www.konkur.in Chapter 6 6-1 Eq. (2-21): Eq. (6-8): Table 6-2: Eq. (6-19): Sut = 3.4 H B = 3.4(300) = 1020 MPa Se′ = 0.5Sut = 0.5(1020) = 510 MPa a = 1.58, b = −0.085 ka = aSutb = 1.58(1020) −0.085 = 0.877 Eq. (6-20): kb = 1.24d −0.107 = 1.24(10) −0.107 = 0.969 Eq. (6-18): Se = ka kb Se′ = (0.877)(0.969)(510) = 433 MPa Ans. ______________________________________________________________________________ 6-2 (a) Table A-20: Eq. (6-8): Sut = 80 kpsi Se′ = 0.5(80) = 40 kpsi Ans. (b) Table A-20: Eq. (6-8): Sut = 90 kpsi Se′ = 0.5(90) = 45 kpsi Ans. (c) Aluminum has no endurance limit. Ans. (d) Eq. (6-8): Sut > 200 kpsi, Se′ = 100 kpsi Ans. ______________________________________________________________________________ 6-3 Sut = 120 kpsi, σ rev = 70 kpsi Fig. 6-18: f = 0.82 Eq. (6-8): Se′ = Se = 0.5(120) = 60 kpsi Eq. (6-14): ( f Sut ) 2 [ 0.82(120) ] a= = = 161.4 kpsi Se 60 Eq. (6-15): f Sut 1 b = − log 3 Se 2 1 0.82(120) = − log = −0.0716 3 60 1 σ 70 −0.0716 Eq. (6-16): N = rev = = 116 700 cycles Ans. 161.4 a ______________________________________________________________________________ 1/ b 6-4 Sut = 1600 MPa, σ rev = 900 MPa Fig. 6-18: Sut = 1600 MPa = 232 kpsi. Off the graph, so estimate f = 0.77. Eq. (6-8): Sut > 1400 MPa, so Se = 700 MPa Eq. (6-14): ( f Sut )2 [ 0.77(1600)] a= = = 2168.3 MPa Se 700 2 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 1/58 www.konkur.in Eq. (6-15): f Sut 1 b = − log 3 Se 1 0.77(1600) = − log = −0.081838 3 700 1 σ 900 −0.081838 Eq. (6-16): N = rev = = 46 400 cycles Ans. 2168.3 a ______________________________________________________________________________ 1/ b 6-5 Sut = 230 kpsi, N = 150 000 cycles Fig. 6-18, point is off the graph, so estimate: f = 0.77 Eq. (6-8): Sut > 200 kpsi, so Se′ = Se = 100 kpsi Eq. (6-14): ( f Sut ) 2 [ 0.77(230) ] a= = = 313.6 kpsi Se 100 Eq. (6-15): f Sut 1 b = − log 3 Se Eq. (6-13): S f = aN b = 313.6(150 000)−0.08274 = 117.0 kpsi 2 1 0.77(230) = − log = −0.08274 3 100 Ans. ______________________________________________________________________________ 6-6 Sut = 1100 MPa = 160 kpsi Fig. 6-18: f = 0.79 Eq. (6-8): Se′ = Se = 0.5(1100) = 550 MPa Eq. (6-14): ( f Sut ) 2 [ 0.79(1100) ] a= = = 1373 MPa Se 550 Eq. (6-15): f Sut 1 1 0.79(1100) b = − log = − log = −0.06622 3 3 550 Se Eq. (6-13): S f = aN b = 1373(150 000)−0.06622 = 624 MPa 2 Ans. ______________________________________________________________________________ 6-7 Sut = 150 kpsi, S yt = 135 kpsi, N = 500 cycles Fig. 6-18: f = 0.798 From Fig. 6-10, we note that below 103 cycles on the S-N diagram constitutes the lowcycle region, in which Eq. (6-17) is applicable. Shigley’s MED, 10th edition Chapter 6 Solutions, Page 2/58 www.konkur.in Eq. (6-17): log( 0.798) /3 S f = Sut N ( log f ) /3 = 150 ( 500 ) = 122 kpsi Ans. The testing should be done at a completely reversed stress of 122 kpsi, which is below the yield strength, so it is possible. Ans. ______________________________________________________________________________ 6-8 The general equation for a line on a log Sf - log N scale is Sf = aNb, which is Eq. (6-13). By taking the log of both sides, we can get the equation of the line in slope-intercept form. log S f = b log N + log a Substitute the two known points to solve for unknowns a and b. Substituting point (1, Sut), log Sut = b log(1) + log a From which a = Sut . Substituting point (103 , f Sut ) and a = Sut log f Sut = b log103 + log Sut From which b = (1/ 3) log f ∴ S f = Sut N (log f )/3 1 ≤ N ≤ 103 ______________________________________________________________________________ 6-9 Read from graph: (103 , 90 ) and (106 , 50). From S = aN b log S1 = log a + b log N1 log S2 = log a + b log N 2 From which log a = log S1 log N 2 − log S2 log N1 log N 2 / N1 log 90 log106 − log 50 log103 log106 / 103 = 2.2095 = a = 10log a = 102.2095 = 162.0 kpsi log 50 / 90 b= = −0.0851 3 ( S f ) ax = 162 N −0.0851 103 ≤ N ≤ 106 in kpsi Shigley’s MED, 10th edition Ans. Chapter 6 Solutions, Page 3/58 www.konkur.in Check: ( S f ) ax 3 = 162(103 )−0.0851 = 90 kpsi 10 ( S f ) ax 6 = 162(106 ) −0.0851 = 50 kpsi 10 The end points agree. ______________________________________________________________________________ 6-10 d = 1.5 in, Sut = 110 kpsi Eq. (6-8): Se′ = 0.5(110) = 55 kpsi Table 6-2: a = 2.70, b = − 0.265 Eq. (6-19): ka = aSut b = 2.70(110) −0.265 = 0.777 Eq. (6-20): kb = 0.879 d −0.107= 0.879(1.5) −0.107 =0.842 Eq. (6-18): Se = kakb Se′ = 0.777(0.842)(55) = 36.0 kpsi Ans. ______________________________________________________________________________ 6-11 For AISI 4340 as-forged steel, Eq. (6-8): Table 6-2: Eq. (6-19): Se = 100 kpsi a = 39.9, b = − 0.995 ka = 39.9(260)−0.995 = 0.158 Eq. (6-20): 0.75 kb = 0.30 −0.107 = 0.907 Each of the other modifying factors is unity. Se = 0.158(0.907)(100) = 14.3 kpsi Ans. For AISI 1040: Se′ = 0.5(113) = 56.5 kpsi ka = 39.9(113)−0.995 = 0.362 kb = 0.907 (same as 4340) Each of the other modifying factors is unity Se = 0.362(0.907)(56.5) = 18.6 kpsi Ans. Not only is AISI 1040 steel a contender, it has a superior endurance strength. Ans. ______________________________________________________________________________ 6-12 D = 1 in, d = 0.8 in, T = 1800 lbf⋅in, f = 0.9, and from Table A-20 for AISI 1020 CD, Sut = 68 kpsi, and Sy = 57 kpsi. Shigley’s MED, 10th edition Chapter 6 Solutions, Page 4/58 www.konkur.in (a) Fig. A-15-15: r 0.1 D 1 = = 0.125, = = 1.25, K ts = 1.40 d 0.8 d 0.8 Get the notch sensitivity either from Fig. 6-21, or from the curve-fit Eqs. (6-34) and (6-35b). Using the equations, a = 0.190 − 2.51(10−3 ) ( 68 ) + 1.35 (10−5 ) ( 68) − 2.67 (10−8 )( 683 ) = 0.07335 2 1 qs = a r 1+ Eq. (6-32): = 1 = 0.812 0.07335 1+ 0.1 Kfs = 1 + qs (Kts − 1) = 1 + 0.812(1.40 − 1) = 1.32 For a purely reversing torque of T = 1800 lbf⋅in, τ a = K fs Tr K fs 16T 1.32(16)(1800) = = = 23 635 psi = 23.6 kpsi J πd3 π (0.8)3 Eq. (6-8): Se′ = 0.5(68) = 34 kpsi Eq. (6-19): ka = 2.70(68)−0.265 = 0.883 Eq. (6-20): kb = 0.879(0.8)−0.107 = 0.900 Eq. (6-26): kc = 0.59 Eq. (6-18) (labeling for shear): Sse = 0.883(0.900)(0.59)(34) = 15.9 kpsi For purely reversing torsion, use Eq. (6-54) for the ultimate strength in shear. Eq. (6-54): Ssu = 0.67 Sut = 0.67(68) = 45.6 kpsi Adjusting the fatigue strength equations for shear, Eq. (6-14): Eq. (6-15): ( f Ssu ) a= S se 2 [0.9(45.6)] = 2 15.9 = 105.9 kpsi f S su 1 1 0.9(45.6) b = − log = − log = −0.137 27 3 3 15.9 S se 1 τ a b 1 23.3 −0.137 27 Eq. (6-16): N = = = 61.7 103 cycles Ans. a 105.9 (b) For an operating temperature of 750o F, the temperature modification factor, from Table 6-4 is kd = 0.90. Shigley’s MED, 10th edition ( ) Chapter 6 Solutions, Page 5/58 www.konkur.in Sse = 0.883(0.900)(0.59)(0.9)(34) = 14.3 kpsi ( f Ssu ) a= 2 S se [0.9(45.6)] = 2 = 117.8 kpsi 14.3 f S su 1 1 0.9(45.6) b = − log = − log = −0.152 62 3 3 14.3 S se 1 1 τ b 23.3 −0.152 62 N = a = = 40.9 103 cycles Ans. a 117.8 ______________________________________________________________________________ 6-13 ( ) L = 0.6 m, Fa = 2 kN, n = 1.5, N = 104 cycles, Sut = 770 MPa, S y = 420 MPa (Table A-20) First evaluate the fatigue strength. Se′ = 0.5(770) = 385 MPa ka = 57.7(770) −0.718 = 0.488 Since the size is not yet known, assume a typical value of kb = 0.85 and check later. All other modifiers are equal to one. Eq. (6-18): Se = 0.488(0.85)(385) = 160 MPa In kpsi, Sut = 770/6.89 = 112 kpsi Fig. 6-18: f = 0.83 Eq. (6-14): a= Eq. (6-15): Eq. (6-13): ( f Sut ) Se 2 [0.83(770)] 2 = 160 = 2553 MPa f Sut 1 1 0.83(770) b = − log = − log = −0.2005 3 3 160 Se S f = aN b = 2553(104 )−0.2005 = 403 MPa Now evaluate the stress. M max = (2000 N)(0.6 m) = 1200 N ⋅ m σ a = σ max = Mc M ( b / 2 ) 6 M 6 (1200 ) 7200 = = = = 3 Pa, with b in m. I b(b3 ) /12 b3 b3 b Compare strength to stress and solve for the necessary b. 6 S f 403 (10 ) n= = = 1.5 σ a 7200 / b3 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 6/58 www.konkur.in b = 0.0299 m Select b = 30 mm. Since the size factor was guessed, go back and check it now. 1/ 2 Eq. (6-25): d e = 0.808 ( hb ) = 0.808b = 0.808 ( 30 ) = 24.24 mm −0.107 24.2 Eq. (6-20): kb = = 0.88 7.62 Our guess of 0.85 was slightly conservative, so we will accept the result of b = 30 mm. Ans. Checking yield, σ max = 7200 10−6 ) = 267 MPa 3 ( 0.030 Sy 420 = 1.57 σ max 267 ______________________________________________________________________________ ny = 6-14 = Given: w =2.5 in, t = 3/8 in, d = 0.5 in, nd = 2. From Table A-20, for AISI 1020 CD, Sut = 68 kpsi and Sy = 57 kpsi. Eq. (6-8): Se′ = 0.5(68) = 34 kpsi Table 6-2: Eq. (6-21): Eq. (6-26): ka = 2.70(68) −0.265 = 0.88 kb = 1 (axial loading) kc = 0.85 Eq. (6-18): Se = 0.88(1)(0.85)(34) = 25.4 kpsi Table A-15-1: d / w = 0.5 / 2.5 = 0.2, K t = 2.5 Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). The relatively large radius is off the graph of Fig. 6-20, so we will assume the curves continue according to the same trend and use the equations to estimate the notch sensitivity. a = 0.246 − 3.08 (10−3 ) ( 68 ) + 1.51(10−5 ) ( 68) − 2.67 (10−8 )( 683 ) = 0.09799 2 1 = 0.836 a 1 + 0.09799 1+ 0.25 r K f = 1 + q ( K t − 1) = 1 + 0.836(2.5 − 1) = 2.25 q= Eq. (6-32): 1 σa = K f Shigley’s MED, 10th edition = Fa 2.25 Fa = = 3Fa A (3 / 8)(2.5 − 0.5) Chapter 6 Solutions, Page 7/58 www.konkur.in Since a finite life was not mentioned, we’ll assume infinite life is desired, so the completely reversed stress must stay below the endurance limit. Se 25.4 =2 σ a 3Fa Fa = 4.23 kips Ans. ______________________________________________________________________________ nf = 6-15 = Given: D = 2 in, d = 1.8 in, r = 0.1 in, M max = 25 000 lbf ⋅ in, M min = 0. From Table A-20, for AISI 1095 HR, Sut = 120 kpsi and Sy = 66 kpsi. Eq. (6-8): Se′ = 0.5Sut = 0.5 (120 ) = 60 kpsi ka = aSutb = 2.70(120) −0.265 = 0.76 Eq. (6-19): Eq. (6-24): d e = 0.370d = 0.370(1.8) = 0.666 in Eq. (6-20): kb = 0.879d e −0.107 = 0.879(0.666) −0.107 = 0.92 Eq. (6-26): kc = 1 Eq. (6-18): Se = ka kb kc Se′ = (0.76)(0.92)(1)(60) = 42.0 kpsi Fig. A-15-14: D / d = 2 /1.8 = 1.11, r / d = 0.1 /1.8 = 0.056 ⇒ K t = 2.1 Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). Using the equations, a = 0.246 − 3.08 (10−3 ) (120 ) + 1.51(10−5 ) (120 ) − 2.67 (10−8 )(1203 ) = 0.04770 2 1 q= 1+ a r = 1 = 0.87 0.04770 1+ 0.1 K f = 1 + q ( K t − 1) = 1 + 0.87(2.1 − 1) = 1.96 Eq. (6-32): I = (π / 64)d 4 = (π / 64)(1.8)4 = 0.5153 in 4 Mc 25 000(1.8 / 2) = = 43 664 psi = 43.7 kpsi I 0.5153 =0 σ max = σ min Eq. (6-36): Shigley’s MED, 10th edition σm = K f σ max + σ min 2 = (1.96 ) ( 43.7 + 0 ) = 42.8 kpsi 2 Chapter 6 Solutions, Page 8/58 www.konkur.in σa = K f σ max − σ min 2 = (1.96 ) ( 43.7 − 0 ) 2 = 42.8 kpsi 1 σ a σ m 42.8 42.8 = + = + nf Se Sut 42.0 120 Eq. (6-46): n f = 0.73 Ans. A factor of safety less than unity indicates a finite life. Check for yielding. It is not necessary to include the stress concentration for static yielding of a ductile material. Sy 66 = 1.51 Ans. σ max 43.7 ______________________________________________________________________________ ny = 6-16 = From a free-body diagram analysis, the bearing reaction forces are found to be 2.1 kN at the left bearing and 3.9 kN at the right bearing. The critical location will be at the shoulder fillet between the 35 mm and the 50 mm diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. The bending moment at this point is M = 2.1(200) = 420 kN∙mm. With a rotating shaft, the bending stress will be completely reversed. σ rev = Mc 420 (35 / 2) = = 0.09978 kN/mm 2 = 99.8 MPa 4 I (π / 64)(35) This stress is far below the yield strength of 390 MPa, so yielding is not predicted. Find the stress concentration factor for the fatigue analysis. Fig. A-15-9: r/d = 3/35 = 0.086, D/d = 50/35 = 1.43, Kt =1.7 Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). Using the equations, with Sut = 470 MPa = 68.2 kpsi and r = 3 mm = 0.118 in, a = 0.246 − 3.08 (10−3 ) ( 68.2 ) + 1.51(10−5 ) ( 68.2 ) − 2.67 (10−8 ) ( 68.2 ) = 0.09771 2 Eq. (6-32): 1 = 0.78 0.09771 a 1+ 1+ 0.118 r K f = 1 + q ( K t − 1) = 1 + 0.78(1.7 − 1) = 1.55 Eq. (6-8): S e' = 0.5 Sut = 0.5(470) = 235 MPa Eq. (6-19): ka = aSutb = 4.51(470) −0.265 = 0.88 Eq. (6-24): kb = 1.24d −0.107 = 1.24(35) −0.107 = 0.85 q= Shigley’s MED, 10th edition 1 3 = Chapter 6 Solutions, Page 9/58 www.konkur.in Eq. (6-26): kc = 1 Eq. (6-18): S e = k a kb kc S e' = (0.88)(0.85)(1)(235) = 176 MPa Se 176 = 1.14 Infinite life is predicted. Ans. K f σ rev 1.55 ( 99.8 ) ______________________________________________________________________________ nf = 6-17 = From a free-body diagram analysis, the bearing reaction forces are found to be RA = 2000 lbf and RB = 1500 lbf. The shearforce and bending-moment diagrams are shown. The critical location will be at the shoulder fillet between the 1-5/8 in and the 1-7/8 in diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. M = 16 000 – 500 (2.5) = 14 750 lbf ∙ in With a rotating shaft, the bending stress will be completely reversed. Mc 14 750(1.625 / 2) = = 35.0 kpsi I (π / 64)(1.625) 4 This stress is far below the yield strength of 71 kpsi, so yielding is not predicted. σ rev = Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.95 Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). Using the equations, a = 0.246 − 3.08 (10−3 ) ( 85) + 1.51(10−5 ) ( 85 ) − 2.67 (10−8 ) ( 85 ) = 0.07690 2 1 q= = 1 = 0.76 . 0.07690 1+ 0.0625 Eq. (6-32): a r K f = 1 + q ( K t − 1) = 1 + 0.76(1.95 − 1) = 1.72 Eq. (6-8): S e' = 0.5Sut = 0.5(85) = 42.5 kpsi Eq. (6-19): ka = aSutb = 2.70(85) −0.265 = 0.832 Eq. (6-20): Eq. (6-26): kb = 0.879 d −0.107 = 0.879(1.625) −0.107 = 0.835 1+ 3 kc = 1 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 10/58 www.konkur.in Eq. (6-18): S e = ka kb kc S e' = (0.832)(0.835)(1)(42.5) = 29.5 kpsi Se 29.5 = 0.49 Ans. K f σ rev 1.72 ( 35.0 ) Infinite life is not predicted. Use the S-N diagram to estimate the life. nf = = Fig. 6-18: f = 0.867 Eq. (6-14): Eq. (6-15): ( f Sut ) a= Se 2 [0.867(85)] = 2 29.5 f Sut 1 b = − log 3 Se = 184.1 1 0.867(85) = − log = −0.1325 3 29.5 1 1 K f σ rev b (1.72)(35.0) −0.1325 Eq. (6-16): N = = 4611 cycles = a 184.1 N = 4600 cycles Ans. ______________________________________________________________________________ 6-18 From a free-body diagram analysis, the bearing reaction forces are found to be RA = 1600 lbf and RB = 2000 lbf. The shear-force and bending-moment diagrams are shown. The critical location will be at the shoulder fillet between the 1-5/8 in and the 1-7/8 in diameters, where the bending moment is large, the diameter is smaller, and the stress concentration exists. M = 12 800 + 400 (2.5) = 13 800 lbf ∙ in With a rotating shaft, the bending stress will be completely reversed. Mc 13 800(1.625 / 2) = = 32.8 kpsi σ rev = I (π / 64)(1.625)4 This stress is far below the yield strength of 71 kpsi, so yielding is not predicted. Fig. A-15-9: r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt =1.95 Get the notch sensitivity either from Fig. 6-20, or from the curve-fit Eqs. (6-34) and (6-35a). Usingthe equations, Shigley’s MED, 10th edition Chapter 6 Solutions, Page 11/58 www.konkur.in a = 0.246 − 3.08 (10−3 ) ( 85 ) + 1.51(10−5 ) ( 85 ) − 2.67 (10−8 ) ( 85 ) = 0.07690 2 1 3 Eq. (6-32): 1 = 0.76 a 1 + 0.07690 1+ 0.0625 r K f = 1 + q ( K t − 1) = 1 + 0.76(1.95 − 1) = 1.72 Eq. (6-8): S e' = 0.5Sut = 0.5(85) = 42.5 kpsi Eq. (6-19): ka = aSutb = 2.70(85) −0.265 = 0.832 Eq. (6-20): kb = 0.879 d −0.107 = 0.879(1.625) −0.107 = 0.835 Eq. (6-26): kc = 1 Eq. (6-18): S e = ka kb kc S e' = (0.832)(0.835)(1)(42.5) = 29.5 kpsi q= Se = 29.5 = 0.52 Ans. K f σ rev 1.72 ( 32.8 ) Infinite life is not predicted. Use the S-N diagram to estimate the life. Fig. 6-18: f = 0.867 nf = Eq. (6-14): Eq. (6-15): ( f Sut ) a= Se 2 = [0.867(85)] = 2 29.5 = 184.1 f Sut 1 1 0.867(85) b = − log = − log = −0.1325 3 3 29.5 Se 1 1 K f σ rev b (1.72)(32.8) −0.1325 Eq. (6-16): N = = 7527 cycles = a 184.1 N = 7500 cycles Ans. ______________________________________________________________________________ 6-19 Table A-20: Sut = 120 kpsi, S y = 66 kpsi N = (950 rev/min)(10 hr)(60 min/hr) = 570 000 cycles One approach is to guess a diameter and solve the problem as an iterative analysis problem. Alternatively, we can estimate the few modifying parameters that are dependent on the diameter and solve the stress equation for the diameter, then iterate to check the estimates. We’ll use the second approach since it should require only one iteration, since the estimates on the modifying parameters should be pretty close. First, we will evaluate the stress. From a free-body diagram analysis, the reaction forces at the bearings are R1 = 2 kips and R2 = 6 kips. The critical stress location is in the middle of the span at the shoulder, where the bending moment is high, the shaft diameter is smaller, and a stress concentration factor exists. If the critical location is not obvious, Shigley’s MED, 10th edition Chapter 6 Solutions, Page 12/58 www.konkur.in prepare a complete bending moment diagram and evaluate at any potentially critical locations. Evaluating at the critical shoulder, M = 2 kip (10 in ) = 20 kip ⋅ in σ rev = Mc M ( d / 2 ) 32 M 32 ( 20 ) 203.7 = = = = kpsi I d3 π d 4 / 64 π d 3 πd3 Now we will get the notch sensitivity and stress concentration factor. The notch sensitivity depends on the fillet radius, which depends on the unknown diameter. For now, let us estimate a value of q = 0.85 from observation of Fig. 6-20, and check it later. Fig. A-15-9: D / d = 1.4d / d = 1.4, Eq. (6-32): r / d = 0.1d / d = 0.1, K t = 1.65 K f = 1 + q ( K t − 1) = 1 + 0.85(1.65 − 1) = 1.55 Now, evaluate the fatigue strength. Se' = 0.5(120) = 60 kpsi ka = 2.70(120) −0.265 = 0.76 Since the diameter is not yet known, assume a typical value of kb = 0.85 and check later. All other modifiers are equal to one. Se = (0.76)(0.85)(60) = 38.8 kpsi Determine the desired fatigue strength from the S-N diagram. Fig. 6-18: f = 0.82 Eq. (6-14): ( f Sut ) a= 2 Se [0.82(120)] = 38.8 2 = 249.6 Eq. (6-15): f Sut 1 1 0.82(120) b = − log = − log = −0.1347 3 3 38.8 Se Eq. (6-13): S f = aN b = 249.6(570 000) −0.1347 = 41.9 kpsi Compare strength to stress and solve for the necessary d. nf = Sf K f σ rev = 41.9 = 1.6 (1.55 ) ( 203.7 / d 3 ) d = 2.29 in Shigley’s MED, 10th edition Chapter 6 Solutions, Page 13/58 www.konkur.in Since the size factor and notch sensitivity were guessed, go back and check them now. Eq. (6-20): kb = 0.91d −0.157 = 0.91( 2.29 ) −0.157 = 0.80 From Fig. 6-20 with r = d/10 = 0.229 in, we are off the graph, but it appears our guess for q of 0.85 is low. Assuming the trend of the graph continues, we’ll choose q = 0.91 and iterate the problem with the new values of kb and q. Intermediate results are Se = 36.5 kpsi, Sf = 39.6 kpsi, and Kf = 1.59. This gives nf = Sf K f σ rev = 39.6 = 1.6 (1.59 ) ( 203.7 / d 3 ) d = 2.36 in Ans. A quick check of kb and q show that our estimates are still reasonable for this diameter. ______________________________________________________________________________ 6-20 S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ m = 15 kpsi, σ a = 25 kpsi, σ m = τ a = 0 Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/2 σ m′ = (σ m2 + 3τ m2 ) 1/2 1/ 2 = 25.00 kpsi 1/2 = 25.98 kpsi 2 = 252 + 3 ( 0 ) 2 = 02 + 3 (15 ) 2 2 ′ = (σ max σ max + 3τ max ) 1/2 ( ) 1/2 2 2 = (σ a + σ m ) + 3 (τ a + τ m ) 1/2 = 252 + 3 152 = 36.06 kpsi Sy 60 ny = = = 1.66 Ans. ′ 36.06 σ max (a) Modified Goodman, Table 6-6 nf = 1 = 1.05 (25.00 / 40) + (25.98 / 80) Ans. (b) Gerber, Table 6-7 2 2 2(25.98)(40) 1 80 25.00 nf = = 1.31 −1 + 1 + 2 25.98 40 80(25.00) (c) ASME-Elliptic, Table 6-8 Ans. 1 = 1.32 Ans. (25.00 / 40) + (25.98 / 60) 2 ______________________________________________________________________________ nf = Shigley’s MED, 10th edition 2 Chapter 6 Solutions, Page 14/58 www.konkur.in 6-21 S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ m = 20 kpsi, σ a = 10 kpsi, σ m = τ a = 0 Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/ 2 σ m′ = (σ m2 + 3τ m2 ) 1/2 1/2 2 = 102 + 3 ( 0 ) = 10.00 kpsi 1/2 2 = 02 + 3 ( 20 ) 2 2 ′ = (σ max σ max + 3τ max ) 1/2 ( ) = 34.64 kpsi 1/2 2 2 = (σ a + σ m ) + 3 (τ a + τ m ) 1/2 = 102 + 3 202 = 36.06 kpsi Sy 60 ny = = = 1.66 Ans. ′ σ max 36.06 (a) Modified Goodman, Table 6-6 1 nf = = 1.46 (10.00 / 40) + (34.64 / 80) (b) Gerber, Table 6-7 Ans. 2 2 2(34.64)(40) 1 80 10.00 nf = = 1.74 −1 + 1 + 2 34.64 40 80(10.00) Ans. (c) ASME-Elliptic, Table 6-8 1 = 1.59 Ans. (10.00 / 40) + (34.64 / 60) 2 ______________________________________________________________________________ nf = 6-22 2 S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ a = 10 kpsi, τ m = 15 kpsi, σ a = 12 kpsi, σ m = 0 Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/2 1/2 2 = 122 + 3 (10 ) σ m′ = (σ m2 + 3τ m2 ) 1/2 1/ 2 2 = 02 + 3 (15 ) 2 2 ′ = (σ max + 3τ max σ max ) 1/2 = 21.07 kpsi = 25.98 kpsi 1/2 2 2 = (σ a + σ m ) + 3 (τ a + τ m ) 2 1/2 = (12 + 0 ) + 3 (10 + 15 ) = 44.93 kpsi Sy 60 ny = = = 1.34 Ans. ′ σ max 44.93 2 (a) Modified Goodman, Table 6-6 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 15/58 www.konkur.in nf = 1 = 1.17 (21.07 / 40) + (25.98 / 80) Ans. (b) Gerber, Table 6-7 2 2 2(25.98)(40) 1 80 21.07 nf = = 1.47 −1 + 1 + 2 25.98 40 80(21.07) (c) ASME-Elliptic, Table 6-8 Ans. 1 = 1.47 Ans. (21.07 / 40) + (25.98 / 60) 2 ______________________________________________________________________________ nf = 6-23 2 S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ a = 30 kpsi, σ m = σ a = τ a = 0 Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/ 2 σ m′ = (σ m2 + 3τ m2 ) 1/ 2 1/2 2 = 02 + 3 ( 30 ) = 51.96 kpsi = 0 kpsi 2 2 ′ = (σ max + 3τ max σ max ) 1/2 1/2 2 2 = (σ a + σ m ) + 3 (τ a + τ m ) 1/2 2 = 3 ( 30 ) = 51.96 kpsi Sy 60 ny = = = 1.15 Ans. ′ σ max 51.96 (a) Modified Goodman, Table 6-6 nf = 1 = 0.77 (51.96 / 40) Ans. (b) Gerber criterion of Table 6-7 is only valid for σm > 0; therefore use Eq. (6-47). σ′ S 40 nf a = 1 ⇒ nf = e = = 0.77 Ans. Se σ a′ 51.96 (c) ASME-Elliptic, Table 6-8 nf = 1 = 0.77 (51.96 / 40) 2 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. Since σ'm = 0, the stress state is completely reversed and the S-N diagram is applicable for σ'a. Fig. 6-18: f = 0.875 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 16/58 www.konkur.in Eq. (6-14): ( f Sut ) 2 [ 0.875(80) ] a= = = 122.5 Se 40 Eq. (6-15): f Sut 1 1 0.875(80) b = − log = − log = −0.08101 3 3 40 Se 2 1 σ 51.96 −0.08101 Eq. (6-16): N = rev = = 39 600 cycles Ans. 122.5 a ______________________________________________________________________________ 1/ b 6-24 S e = 40 kpsi, S y = 60 kpsi, Sut = 80 kpsi, τ a = 15 kpsi, σ m = 15 kpsi, τ m = σ a = 0 Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/ 2 1/2 2 = 02 + 3 (15 ) σ m′ = (σ m2 + 3τ m2 ) = 25.98 kpsi 1/2 2 = 152 + 3 ( 0 ) 1/2 2 2 ′ = (σ max + 3τ max σ max ) 1/2 = 15.00 kpsi 1/2 2 2 = (σ a + σ m ) + 3 (τ a + τ m ) 2 1/2 = (15 ) + 3 (15 ) = 30.00 kpsi Sy 60 ny = = = 2.00 Ans. ′ 30 σ max 2 (a) Modified Goodman, Table 6-6 nf = 1 = 1.19 (25.98 / 40) + (15.00 / 80) Ans. (b) Gerber, Table 6-7 2 2 2(15.00)(40) 1 80 25.98 nf = = 1.43 −1 + 1 + 2 15.00 40 80(25.98) (c) ASME-Elliptic, Table 6-8 Ans. 1 = 1.44 Ans. (25.98 / 40) + (15.00 / 60) 2 ______________________________________________________________________________ nf = 6-25 2 Given: Fmax = 28 kN, Fmin = −28 kN . From Table A-20, for AISI 1040 CD, Sut = 590 MPa, S y = 490 MPa, Check for yielding Shigley’s MED, 10th edition Chapter 6 Solutions, Page 17/58 www.konkur.in σ max = Fmax 28 000 = = 147.4 N/mm 2 = 147.4 MPa A 10(25 − 6) Sy 490 = 3.32 Ans. σ max 147.4 Determine the fatigue factor of safety based on infinite life ny = = Eq. (6-8): S e' = 0.5(590) = 295 MPa Eq. (6-19): ka = aSutb = 4.51(590) −0.265 = 0.832 Eq. (6-21): Eq. (6-26): kb = 1 (axial) kc = 0.85 Eq. (6-18): S e = ka kb kc S e' = (0.832)(1)(0.85)(295) = 208.6 MPa Fig. 6-20: q = 0.83 Fig. A-15-1: d / w = 0.24, K t = 2.44 K f = 1 + q ( K t − 1) = 1 + 0.83(2.44 − 1) = 2.20 σa = K f 28 000 − ( −28 000 ) Fmax − Fmin = 2.2 = 324.2 MPa 2A 2(10)(25 − 6) Fmax + Fmin =0 2A Note, since σm = 0, the stress is completely reversing, and σm = K f Se 208.6 = 0.64 Ans. σ a 324.2 Since infinite life is not predicted, estimate the life from the S-N diagram. With σm = 0, the stress state is completely reversed, and the S-N diagram is applicable for σa. nf = = Sut = 590/6.89 = 85.6 kpsi Fig. 6-18: f = 0.87 Eq. (6-14): ( f Sut )2 [ 0.87(590)] a= = = 1263 Se 208.6 Eq. (6-15): f Sut 1 b = − log 3 Se 2 1 0.87(590) = − log = −0.1304 3 208.6 1 σ 324.2 −0.1304 Eq. (6-16): N = rev = = 33 812 cycles 1263 a N = 34 000 cycles Ans. ________________________________________________________________________ 1/ b Shigley’s MED, 10th edition Chapter 6 Solutions, Page 18/58 www.konkur.in 6-26 Sut = 590 MPa, S y = 490 MPa, Fmax = 28 kN, Fmin = 12 kN Check for yielding Fmax 28 000 = = 147.4 N/mm 2 = 147.4 MPa A 10(25 − 6) Sy 490 ny = = = 3.32 Ans. σ max 147.4 σ max = Determine the fatigue factor of safety based on infinite life. From Prob. 6-25: S e = 208.6 MPa, K f = 2.2 σa = K f 28 000 − (12 000 ) Fmax − Fmin = 2.2 = 92.63 MPa 2A 2(10)(25 − 6) σm = K f 28 000 + 12 000 Fmax + Fmin = 2.2 = 231.6 MPa 2A 2(10)(25 − 6) Modified Goodman criteria: 1 σ a σ m 92.63 231.6 = + = + n f Se Sut 208.6 590 n f = 1.20 Ans. Gerber criteria: 2 2 2σ m Se 1 Sut σ a nf = −1 + 1 + 2 σ m Se Sutσ a 2 2 2(231.6)(208.6) 1 590 92.63 = −1 + 1 + 2 231.6 208.6 590(92.63) n f = 1.49 Ans. ASME-Elliptic criteria: nf = 1 1 = 2 2 (σ a / Se ) + (σ m / S y ) (92.63 / 208.6) + (231.6 / 490) 2 2 = 1.54 Ans. The results are consistent with Fig. 6-27, where for a mean stress that is about half of the yield strength, the Modified Goodman line should predict failure significantly before the other two. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 6 Solutions, Page 19/58 www.konkur.in 6-27 Sut = 590 MPa, S y = 490 MPa (a) Fmax = 28 kN, Fmin = 0 kN Check for yielding σ max = ny = Fmax 28 000 = = 147.4 N/mm 2 = 147.4 MPa A 10(25 − 6) Sy σ max = 490 = 3.32 147.4 Ans. From Prob. 6-25: S e = 208.6 MPa, K f = 2.2 σa = K f Fmax − Fmin 28 000 − 0 = 2.2 = 162.1 MPa 2A 2(10)(25 − 6) 28 000 + 0 Fmax + Fmin = 2.2 = 162.1 MPa 2A 2(10)(25 − 6) For the modified Goodman criteria, σm = K f 1 σ a σ m 162.1 162.1 = + = + n f Se Sut 208.6 590 n f = 0.95 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress. For the modified Goodman criteria, see Ex. 6-12. σ rev = σa 162.1 = = 223.5 MPa 1 − (σ m / Sut ) 1 − (162.1 / 590) Fig. 6-18: f = 0.87 Eq. (6-14): ( f Sut )2 [ 0.87(590)] a= = = 1263 Se 208.6 Eq. (6-15): f Sut 1 b = − log 3 Se Eq. (6-16): σ N = rev a 2 1/ b 1 0.87(590) = − log = −0.1304 3 208.6 1 223.5 −0.1304 = = 586 000 cycles 1263 Ans. (b) Fmax = 28 kN, Fmin = 12 kN The maximum load is the same as in part (a), so σ max = 147.4 MPa Shigley’s MED, 10th edition Chapter 6 Solutions, Page 20/58 www.konkur.in n y = 3.32 Ans. Factor of safety based on infinite life: Fmax − Fmin 28 000 − 12 000 = 2.2 = 92.63 MPa 2A 2(10)(25 − 6) σa = K f 28 000 + 12 000 Fmax + Fmin = 2.2 = 231.6 MPa 2A 2(10)(25 − 6) 1 σ a σ m 92.63 231.6 = + = + n f Se Sut 208.6 590 σm = K f n f = 1.20 Ans. (c) Fmax = 12 kN, Fmin = −28 kN The compressive load is the largest, so check it for yielding. σ min = ny = Fmin −28 000 = = −147.4 MPa A 10(25 − 6) S yc = σ min −490 = 3.32 −147.4 Ans. Factor of safety based on infinite life: σa = K f 12 000 − ( −28 000 ) Fmax − Fmin = 2.2 = 231.6 MPa 2A 2(10)(25 − 6) σm = K f 12 000 + ( −28 000 ) Fmax + Fmin = 2.2 = −92.63 MPa 2A 2(10)(25 − 6) Se 208.6 = 0.90 Ans. σ a 231.6 Since infinite life is not predicted, estimate a life from the S-N diagram. For a negative mean stress, we shall assume the equivalent completely reversed stress is the same as the actual alternating stress. Get a and b from part (a). For σm < 0, nf = = 1 σ 231.6 −0.1304 Eq. (6-16): N = rev = = 446 000 cycles Ans. 1263 a ______________________________________________________________________________ 1/ b 6-28 Eq. (2-21): Sut = 0.5(400) = 200 kpsi Eq. (6-8): S e' = 0.5(200) = 100 kpsi Shigley’s MED, 10th edition Chapter 6 Solutions, Page 21/58 www.konkur.in Eq. (6-19): Eq. (6-25): ka = aSutb = 14.4(200) −0.718 = 0.321 Eq. (6-20): kb = 0.879d e −0.107 = 0.879(0.1388) −0.107 = 1.09 d e = 0.37 d = 0.37(0.375) = 0.1388 in Since we have used the equivalent diameter method to get the size factor, and in doing so introduced greater uncertainties, we will choose not to use a size factor greater than one. Let kb = 1. Eq. (6-18): Se = (0.321)(1)(100) = 32.1 kpsi 40 − 20 40 + 20 Fa = = 10 lb Fm = = 30 lb 2 2 32M a 32(10)(12) σa = = = 23.18 kpsi π d3 π (0.375)3 32 M m 32(30)(12) = = 69.54 kpsi σm = πd3 π (0.375)3 (a) Modified Goodman criterion 1 σ a σ m 23.18 69.54 = + = + nf Se Sut 32.1 200 n f = 0.94 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). σa 23.18 = = 35.54 kpsi 1 − (σ m / Sut ) 1 − (69.54 / 200) f = 0.775 σ rev = Fig. 6-18: Eq. (6-14): ( f Sut )2 [ 0.775(200) ] a= = = 748.4 Se 32.1 Eq. (6-15): f Sut 1 b = − log 3 Se Eq. (6-16): σ N = rev a 2 1/ b 1 0.775(200) = − log = −0.228 3 32.1 1 35.54 −0.228 = = 637 000 cycles 748.4 Ans. (b) Gerber criterion, Table 6-7 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 22/58 www.konkur.in 2 2 2σ m Se 1 Sut σ a nf = −1 + 1 + 2 σ m Se Sutσ a 2 2 2(69.54)(32.1) 1 200 23.18 = −1 + 1 + 2 69.54 32.1 200(23.18) = 1.16 Infinite life is predicted Ans. ______________________________________________________________________________ 6-29 E = 207.0 GPa 1 (a) I = (20)(43 ) = 106.7 mm 4 12 Fl 3 3EIy y= ⇒ F= 3 3EI l 9 3(207)(10 )(106.7)(10−12 )(2)(10−3 ) Fmin = = 48.3 N 1403 (10−9 ) Fmax = 3(207)(109 )(106.7)(10 −12 )(6)(10 −3 ) = 144.9 N 1403 (10 −9 ) Ans. Ans. (b) Get the fatigue strength information. Eq. (2-21): Sut = =3.4HB = 3.4(490) = 1666 MPa From problem statement: Sy = 0.9Sut = 0.9(1666) = 1499 MPa Eq. (6-8): Se′ = 700 MPa Eq. (6-19): Eq. (6-25): Eq. (6-20): Eq. (6-18): ka = 1.58(1666)-0.085 = 0.84 de = 0.808[20(4)]1/2 = 7.23 mm kb = 1.24(7.23)-0.107 = 1.00 Se = 0.84(1)(700) = 588 MPa This is a relatively thick curved beam, so use the method in Sect. 3-18 to find the stresses. The maximum bending moment will be to the centroid of the section as shown. M = 142F N∙mm, A = 4(20) = 80 mm2, h = 4 mm, ri = 4 mm, ro = ri + h = 8 mm, rc = ri + h/2 = 6 mm Table 3-4: rn = Shigley’s MED, 10th edition h 4 = = 5.7708 mm ln( ro / ri ) ln(8 / 4) Chapter 6 Solutions, Page 23/58 www.konkur.in e = rc − rn = 6 − 5.7708 = 0.2292 mm ci = rn − ri = 5.7708 − 4 = 1.7708 mm co = ro − rn = 8 − 5.7708 = 2.2292 mm Get the stresses at the inner and outer surfaces from Eq. (3-65) with the axial stresses added. The signs have been set to account for tension and compression as appropriate. σi = − σo = Mci F (142 F )(1.7708) F − =− − = −3.441F MPa Aeri A 80(0.2292)(4) 80 Mco F (142 F )(2.2292) F − = − = 2.145 F MPa Aero A 80(0.2292)(8) 80 (σ i ) min = −3.441(144.9) = −498.6 MPa (σ i ) max = −3.441(48.3) = −166.2 MPa (σ o ) min = 2.145(48.3) = 103.6 MPa (σ o ) max = 2.145(144.9) = 310.8 MPa (σ i ) a = (σ i )m = −166.2 − ( −498.6 ) = 166.2 MPa 2 −166.2 + ( −498.6 ) = −332.4 MPa 2 310.8 − 103.6 (σ o ) a = = 103.6 MPa 2 310.8 + 103.6 (σ o ) m = = 207.2 MPa 2 To check for yielding, we note that the largest stress is –498.6 MPa (compression) on the inner radius. This is considerably less than the estimated yield strength of 1499 MPa, so yielding is not predicted. Check for fatigue on both inner and outer radii since one has a compressive mean stress and the other has a tensile mean stress. Inner radius: S 588 Since σm < 0, n f = e = = 3.54 σ a 166.2 Outer radius: Since σm > 0, using the Modified Goodman line, σ σ 103.6 207.2 1/ n f = a + m = + Se Sut 588 1666 n f = 3.33 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 24/58 www.konkur.in Infinite life is predicted at both inner and outer radii. The outer radius is critical, with a fatigue factor of safety of nf = 3.33. Ans. ______________________________________________________________________________ 6-30 From Table A-20, for AISI 1018 CD, Sut = 64 kpsi, S y = 54 kpsi Eq. (6-8): S e' = 0.5(64) = 32 kpsi Eq. (6-19): Eq. (6-20): Eq. (6-26): Eq. (6-18): ka = 2.70(64) −0.265 = 0.897 kb = 1 (axial) kc = 0.85 Se = (0.897)(1)(0.85)(32) = 24.4 kpsi Fillet: Fig. A-15-5: D / d = 3.5 / 3 = 1.17, r / d = 0.25 / 3 = 0.083, K t = 1.85 Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the graph. q = 0.85 K f = 1 + q ( K t − 1) = 1 + 0.85(1.85 − 1) = 1.72 σ max = Fmax 5 = = 3.33 kpsi w2 h 3.0(0.5) σ min = −16 = −10.67 kpsi 3.0(0.5) σa = K f σ max − σ min 2 σ max + σ min 2 σm = K f ny = Sy σ min = = 1.72 3.33 − (−10.67) = 12.0 kpsi 2 3.33 + (−10.67) = −6.31 kpsi = 1.72 2 54 = 5.06 −10.67 ∴ Does not yield. Since the midrange stress is negative, nf = Se σa = 24.4 = 2.03 12.0 Hole: Fig. A-15-1: d / w1 = 0.4 / 3.5 = 0.11 ∴ K t = 2.68 Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the graph, q = 0.85 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 25/58 www.konkur.in K f = 1 + 0.85(2.68 − 1) = 2.43 σ max = Fmax 5 = = 3.226 kpsi h ( w1 − d ) 0.5(3.5 − 0.4) σ min = Fmin −16 = = −10.32 kpsi h ( w1 − d ) 0.5(3.5 − 0.4) σa = K f σ max − σ min 2 σ max + σ min 2 σm = K f ny = Sy σ min = = 2.43 3.226 − (−10.32) = 16.5 kpsi 2 3.226 + (−10.32) = −8.62 kpsi = 2.43 2 54 = 5.23 −10.32 ∴ does not yield Since the midrange stress is negative, nf = Se σa = 24.4 = 1.48 16.5 Thus the design is controlled by the threat of fatigue at the hole with a minimum factor of safety of n f = 1.48. Ans. ______________________________________________________________________________ 6-31 Sut = 64 kpsi, S y = 54 kpsi Eq. (6-8): S e' = 0.5(64) = 32 kpsi Eq. (6-19): Eq. (6-20): Eq. (6-26): Eq. (6-18): ka = 2.70(64) −0.265 = 0.897 kb = 1 (axial) kc = 0.85 Se = (0.897)(1)(0.85)(32) = 24.4 kpsi Fillet: Fig. A-15-5: D / d = 2.5 / 1.5 = 1.67, r / d = 0.25 /1.5 = 0.17, K t ≈ 2.1 Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the graph. q = 0.85 K f = 1 + q ( K t − 1) = 1 + 0.85(2.1 − 1) = 1.94 σ max = Fmax 16 = = 21.3 kpsi w2 h 1.5(0.5) σ min = −4 = −5.33 kpsi 1.5(0.5) Shigley’s MED, 10th edition Chapter 6 Solutions, Page 26/58 www.konkur.in σa = K f σ max − σ min 2 = 1.94 21.3 − (−5.33) = 25.8 kpsi 2 σ max + σ min 21.3 + (−5.33) = 1.94 = 15.5 kpsi 2 2 σm = K f ny = Sy σ max = 54 = 2.54 21.3 ∴ Does not yield. Using Modified Goodman criteria, 1 σ a σ m 25.8 15.5 = + = + nf Se Sut 24.4 64 n f = 0.77 Hole: Fig. A-15-1: d / w1 = 0.4 / 2.5 = 0.16 ∴ K t = 2.55 Use Fig. 6-20 or Eqs. (6-34) and (6-35a) for q. Estimate a little high since it is off the graph. q = 0.85 K f = 1 + 0.85(2.55 − 1) = 2.32 σ max = Fmax 16 = = 15.2 kpsi h ( w1 − d ) 0.5(2.5 − 0.4) σ min = Fmin −4 = = −3.81 kpsi h ( w1 − d ) 0.5(2.5 − 0.4) σ max − σ min 15.2 − (−3.81) = 2.32 = 22.1 kpsi 2 2 σ + σ min 15.2 + (−3.81) σ m = K f max = 13.2 kpsi = 2.32 2 2 Sy 54 ny = = = 3.55 ∴ Does not yield. σ max 15.2 σa = K f Using Modified Goodman criteria 1 σ a σ m 22.1 13.2 = + = + nf Se Sut 24.4 64 n f = 0.90 Thus the design is controlled by the threat of fatigue at the fillet with a minimum factor of safety of n f = 0.77 Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 6 Solutions, Page 27/58 www.konkur.in 6-32 Sut = 64 kpsi, S y = 54 kpsi From Prob. 6-30, the fatigue factor of safety at the hole is nf = 1.48. To match this at the fillet, S S 24.4 nf = e ⇒ σ a = e = = 16.5 kpsi σa n f 1.48 where Se is unchanged from Prob. 6-30. The only aspect of σa that is affected by the fillet radius is the fatigue stress concentration factor. Obtaining σa in terms of Kf, σa = K f σ max − σ min 2 = Kf 3.33 − ( −10.67) = 7.00 K f 2 Equating to the desired stress, and solving for Kf, σ a = 7.00 K f = 16.5 ⇒ K f = 2.36 Assume since we are expecting to get a smaller fillet radius than the original, that q will be back on the graph of Fig. 6-20, so we will estimate q = 0.8. K f = 1 + 0.80( K t − 1) = 2.36 ⇒ K t = 2.7 From Fig. A-15-5, with D / d = 3.5/3 = 1.17 and Kt = 2.6, find r / d. Choosing r / d = 0.03, and with d = w2 = 3.0, r = 0.03w2 = 0.03 ( 3.0 ) = 0.09 in At this small radius, our estimate for q is too high. From Fig. 6-20, with r = 0.09, q should be about 0.75. Iterating, we get Kt = 2.8. This is at a difficult range on Fig. A-155 to read the graph with any confidence, but we’ll estimate r / d = 0.02, giving r = 0.06 in. This is a very rough estimate, but it clearly demonstrates that the fillet radius can be relatively sharp to match the fatigue factor of safety of the hole. Ans. ______________________________________________________________________________ 6-33 S y = 60 kpsi, Sut = 110 kpsi Inner fiber where rc = 3 / 4 in 3 3 ro = + = 0.84375 4 16(2) 3 3 ri = − = 0.65625 4 32 Table 3-4, p. 135, rn = Shigley’s MED, 10th edition h 3 /16 = = 0.74608 in ro 0.84375 ln ln 0.65625 ri Chapter 6 Solutions, Page 28/58 www.konkur.in e = rc − rn = 0.75 − 0.74608 = 0.00392 in ci = rn − ri = 0.74608 − 0.65625 = 0.08983 3 3 A = = 0.035156 in 2 16 16 Eq. (3-65), p. 133, σi = Mci −T (0.08983) = = −993.3T Aeri (0.035156)(0.00392)(0.65625) where T is in lbf∙in and σ i is in psi. 1 2 σ a = 496.7T σ m = (−993.3)T = −496.7T Eq. (6-8): Se' = 0.5 (110 ) = 55 kpsi Eq. (6-19): ka = 2.70(110) −0.265 = 0.777 Eq. (6-25): d e = 0.808 ( 3 / 16 )( 3 / 16 ) Eq. (6-20): kb = 0.879 ( 0.1515 ) Eq. (6-19): Se = (0.777)(1)(55) = 42.7 kpsi 1/ 2 −0.107 = 0.1515 in = 1.08 (round to 1) For a compressive midrange component, σ a = S e / n f . Thus, 42.7 3 T = 28.7 lbf ⋅ in 0.4967T = Outer fiber where rc = 2.5 in 3 = 2.59375 32 3 ri = 2.5 − = 2.40625 32 3 / 16 rn = = 2.49883 2.59375 ln 2.40625 e = 2.5 − 2.49883 = 0.00117 in co = 2.59375 − 2.49883 = 0.09492 in Mco T (0.09492) = = 889.7T psi σo = Aero (0.035156)(0.00117)(2.59375) ro = 2.5 + 1 2 σ m = σ a = (889.7T ) = 444.9T psi Shigley’s MED, 10th edition Chapter 6 Solutions, Page 29/58 www.konkur.in (a) Using Eq. (6-46), for modified Goodman, we have σa σm 1 + = Se Sut n 0.4449T 0.4449T 1 + = 42.7 110 3 T = 23.0 lbf ⋅ in Ans. (b) Gerber, Eq. (6-47), at the outer fiber, 2 nσ a nσ m + =1 Se Sut 2 3(0.4449T ) 3(0.4449T ) + =1 42.7 110 T = 28.2 lbf ⋅ in Ans. (c) To guard against yield, use T of part (b) and the inner stress. Sy 60 = 2.14 Ans. σ i 0.9933(28.2) ______________________________________________________________________________ ny = 6-34 = From Prob. 6-33, S e = 42.7 kpsi, S y = 60 kpsi, and Sut = 110 kpsi (a) Assuming the beam is straight, σ max = Goodman: Mc M ( h / 2 ) 6 M 6T = = 2 = = 910.2T 3 I bh /12 bh (3 /16)3 0.4551T 0.4551T 1 + = 42.7 110 3 T = 22.5 lbf ⋅ in Ans. 2 (b) Gerber: 3(0.4551T ) 3(0.4551T ) + =1 42.7 110 T = 27.6 lbf ⋅ in Ans. Sy 60 = 2.39 Ans. σ max 0.9102(27.6) ______________________________________________________________________________ (c) ny = = Shigley’s MED, 10th edition Chapter 6 Solutions, Page 30/58 www.konkur.in 6-35 K f ,bend = 1.4, K f ,axial = 1.1, K f ,tors = 2.0, S y = 300 MPa, Sut = 400 MPa, Se = 200 MPa Bending: σ m = 0, σ a = 60 MPa σ m = 20 MPa, σ a = 0 τ m = 25 MPa, τ a = 25 MPa Axial: Torsion: Eqs. (6-55) and (6-56): σ a′ = [1.4(60) + 0] σ m′ = [0 + 1.1(20)] 2 2 + 3[ 2.0(25) ] = 120.6 MPa 2 + 3[ 2.0(25)] = 89.35 MPa 2 Using Modified Goodman, Eq. (6-46), 1 σ a′ σ m′ 120.6 89.35 = + = + n f Se Sut 200 400 n f = 1.21 Ans. ′ = σ a′ + σ m′ , Check for yielding, using the conservative σ max Sy 300 = = 1.43 Ans. ′ ′ σ a + σ m 120.6 + 89.35 ______________________________________________________________________________ ny = 6-36 K f ,bend = 1.4, K f ,tors = 2.0, S y = 300 MPa, Sut = 400 MPa, Se = 200 MPa Bending: σ max = 150 MPa, σ min = −40 MPa, σ m = 55 MPa, σ a = 95 MPa Torsion: τ m = 90 MPa, τ a = 9 MPa Eqs. (6-55) and (6-56): σ a′ = [1.4(95)] σ m′ = [1.4(55)] 2 2 + 3[ 2.0(9)] = 136.6 MPa 2 + 3[ 2.0(90) ] = 321.1 MPa 2 Using Modified Goodman, 1 σ a′ σ m′ 136.6 321.1 = + = + n f Se Sut 200 400 n f = 0.67 Ans. ′ = σ a′ + σ m′ , Check for yielding, using the conservative σ max ny = Shigley’s MED, 10th edition Sy σ a′ + σ m′ = 300 = 0.66 136.6 + 321.1 Ans. Chapter 6 Solutions, Page 31/58 www.konkur.in Since the conservative yield check indicates yielding, we will check more carefully with ′ obtained directly from the maximum stresses, using the distortion energy failure σ max theory, without stress concentrations. Note that this is exactly the method used for static failure in Ch. 5. ′ = σ max (σ max ) + 3 (τ max ) = 2 2 (150 ) 2 + 3 ( 90 + 9 ) = 227.8 MPa 2 Sy 300 = = 1.32 Ans. ′ σ max 227.8 Since yielding is not predicted, and infinite life is not predicted, we would like to estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). ny = σ a′ 136.6 = = 692.5 MPa 1 − (σ m′ / Sut ) 1 − (321.1 / 400) This stress is much higher than the ultimate strength, rendering it impractical for the S-N diagram. We must conclude that the stresses from the combination loading, when increased by the stress concentration factors, produce such a high midrange stress that the equivalent completely reversed stress method is not practical to use. Without testing, we are unable to predict a life. ______________________________________________________________________________ σ rev = 6-37 Table A-20: S ut = 64 kpsi, S y = 54 kpsi From Prob. 3-68, the critical stress element experiences σ = 15.3 kpsi and τ = 4.43 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving σa = 15.3 kpsi, σm = 0 kpsi, τa = 0 kpsi, τm = 4.43 kpsi. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/2 σ m′ = (σ m2 + 3τ m2 ) 1/ 2 2 = 15.32 + 3 ( 0 ) 1/2 2 = 0 2 + 3 ( 4.43) 2 2 ′ = (σ max + 3τ max σ max ) 1/2 = 15.3 kpsi 1/2 = 7.67 kpsi 2 = 15.32 + 3 ( 4.43 ) 1/ 2 = 17.11 kpsi Check for yielding, using the distortion energy failure theory. Sy 54 ny = = = 3.16 ′ 17.11 σ max Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5 ( 64 ) = 32 kpsi Eq. (6-19): ka = 2.70(64) −0.265 = 0.90 Eq. (6-20): kb = 0.879(1.25) −0.107 = 0.86 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 32/58 www.konkur.in Eq. (6-18): Se = 0.90(0.86)(32) = 24.8 kpsi Using Modified Goodman, 1 σ a′ σ m′ 15.3 7.67 = + = + nf Se Sut 24.8 64 n f = 1.36 Ans. ______________________________________________________________________________ 6-38 Table A-20: S ut = 440 MPa, S y = 370 MPa From Prob. 3-69, the critical stress element experiences σ = 263 MPa and τ = 57.7 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving σa = 263 MPa, σm = 0, τa = 0 MPa, τm = 57.7 MPa. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/2 σ m′ = (σ m2 + 3τ m2 ) 1/ 2 1/2 2 = 2632 + 3 ( 0 ) = 263 MPa 1/ 2 2 = 02 + 3 ( 57.7 ) 2 2 ′ = (σ max σ max + 3τ max ) 1/ 2 = 99.9 MPa 1/ 2 2 = 2632 + 3 ( 57.7 ) = 281 MPa Check for yielding, using the distortion energy failure theory. Sy 370 = = 1.32 ′ σ max 281 Obtain the modifying factors and endurance limit. ny = Eq. (6-8): Se′ = 0.5 ( 440 ) = 220 MPa Eq. (6-19): ka = 4.51(440) −0.265 = 0.90 Eq. (6-20): Eq. (6-18): kb = 1.24(30) −0.107 = 0.86 Se = 0.90(0.86)(220) = 170 MPa Using Modified Goodman, 1 σ a′ σ m′ 263 99.9 = + = + nf Se Sut 170 440 n f = 0.56 Infinite life is not predicted. Ans. ______________________________________________________________________________ 6-39 Table A-20: S ut = 64 kpsi, S y = 54 kpsi From Prob. 3-70, the critical stress element experiences σ = 21.5 kpsi and τ = 5.09 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving Shigley’s MED, 10th edition Chapter 6 Solutions, Page 33/58 www.konkur.in σa = 21.5 kpsi, σm = 0 kpsi, τa = 0 kpsi, τm = 5.09 kpsi. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/2 σ m′ = (σ m2 + 3τ m2 ) 1/2 2 = 21.52 + 3 ( 0 ) 1/2 2 = 0 2 + 3 ( 5.09 ) 2 2 ′ = (σ max σ max + 3τ max ) 1/ 2 1/ 2 = 21.5 kpsi = 8.82 kpsi 2 = 21.52 + 3 ( 5.09 ) 1/2 = 23.24 kpsi Check for yielding, using the distortion energy failure theory. ny = Sy 54 = = 2.32 ′ σ max 23.24 Obtain the modifying factors and endurance limit. ka = 2.70(64) −0.265 = 0.90 kb = 0.879(1) −0.107 = 0.88 Se = 0.90(0.88)(0.5)(64) = 25.3 kpsi Using Modified Goodman, 1 σ a′ σ m′ 21.5 8.82 = + = + n f Se Sut 25.3 64 n f = 1.01 Ans. ______________________________________________________________________________ 6-40 Table A-20: S ut = 440 MPa, S y = 370 MPa From Prob. 3-71, the critical stress element experiences σ = 72.9 MPa and τ = 20.3 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving σa = 72.9 MPa, σm = 0 MPa, τa = 0 MPa, τm = 20.3 MPa. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/2 σ m′ = (σ m2 + 3τ m2 ) 1/ 2 2 = 72.92 + 3 ( 0 ) 1/ 2 2 = 02 + 3 ( 20.3) 2 2 ′ = (σ max σ max + 3τ max ) 1/2 1/2 = 72.9 MPa = 35.2 MPa 2 = 72.92 + 3 ( 20.3) 1/2 = 80.9 MPa Check for yielding, using the distortion energy failure theory. Sy 370 ny = = = 4.57 ′ σ max 80.9 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 34/58 www.konkur.in Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5 ( 440 ) = 220 MPa Eq. (6-19): ka = 4.51(440) −0.265 = 0.90 Eq. (6-20): Eq. (6-18): kb = 1.24(20) −0.107 = 0.90 Se = 0.90(0.90)(220) = 178.2 MPa Using Modified Goodman, 1 σ a′ σ m′ 72.9 35.2 = + = + nf Se Sut 178.2 440 n f = 2.04 Ans. ______________________________________________________________________________ 6-41 Table A-20: S ut = 64 kpsi, S y = 54 kpsi From Prob. 3-72, the critical stress element experiences σ = 35.2 kpsi and τ = 7.35 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving σa = 35.2 kpsi, σm = 0 kpsi, τa = 0 kpsi, τm = 7.35 kpsi. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/2 σ m′ = (σ m2 + 3τ m2 ) 1/2 2 = 35.2 2 + 3 ( 0 ) 1/2 2 = 0 2 + 3 ( 7.35 ) 2 2 ′ = (σ max + 3τ max σ max ) 1/2 1/2 = 35.2 kpsi = 12.7 kpsi 2 = 35.2 2 + 3 ( 7.35 ) 1/ 2 = 37.4 kpsi Check for yielding, using the distortion energy failure theory. ny = Sy 54 = = 1.44 ′ 37.4 σ max Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): ka = 2.70(64) −0.265 = 0.90 Eq. (6-20): kb = 0.879(1.25) −0.107 = 0.86 Eq. (6-18): Se = 0.90(0.86)(32) = 24.8 kpsi Using Modified Goodman, Shigley’s MED, 10th edition Chapter 6 Solutions, Page 35/58 www.konkur.in 1 σ a′ σ m′ 35.2 12.7 = + = + nf Se Sut 24.8 64 n f = 0.62 Infinite life is not predicted. Ans. ______________________________________________________________________________ 6-42 Table A-20: S ut = 440 MPa, S y = 370 MPa From Prob. 3-73, the critical stress element experiences σ = 333.9 MPa and τ = 126.3 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving σa = 333.9 MPa, σm = 0 MPa, τa = 0 MPa, τm = 126.3 MPa. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/ 2 σ m′ = (σ m2 + 3τ m2 ) 1/2 2 = 333.9 2 + 3 ( 0 ) 1/2 = 333.9 MPa 1/2 = 218.8 MPa 2 = 0 2 + 3 (126.3) 2 2 ′ = (σ max + 3τ max σ max ) 1/2 2 = 333.9 2 + 3 (126.3 ) 1/ 2 = 399.2 MPa Check for yielding, using the distortion energy failure theory. ny = Sy 370 = = 0.93 ′ σ max 399.2 The sample fails by yielding, infinite life is not predicted. Ans. The fatigue analysis will be continued only to obtain the requested fatigue factor of safety, though the yielding failure will dictate the life. Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5(440) = 220 MPa Eq. (6-19): ka = 4.51(440) −0.265 = 0.90 Eq. (6-20): Eq. (6-18): kb = 1.24(50) −0.107 = 0.82 Se = 0.90(0.82)(220) = 162.4 MPa Using Modified Goodman, 1 σ a′ σ m′ 333.9 218.8 = + = + nf Se Sut 162.4 440 n f = 0.39 Infinite life is not predicted. Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 6 Solutions, Page 36/58 www.konkur.in 6-43 Table A-20: Sut = 64 kpsi, S y = 54 kpsi From Prob. 3-74, the critical stress element experiences completely reversed bending stress due to the rotation, and steady torsional and axial stresses. σ a ,bend = 9.495 kpsi, σ a ,axial = 0 kpsi, τ a = 0 kpsi, σ m,bend = 0 kpsi σ m,axial = −0.362 kpsi τ m = 11.07 kpsi Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/ 2 σ m′ = (σ m2 + 3τ m2 ) 1/2 2 2 = ( 9.495 ) + 3 ( 0 ) 1/ 2 = 9.495 kpsi 2 2 = ( −0.362 ) + 3 (11.07 ) 2 2 ′ = (σ max σ max + 3τ max ) 1/2 1/2 = 19.18 kpsi 2 2 = ( −9.495 − 0.362 ) + 3 (11.07 ) 1/2 = 21.56 kpsi Check for yielding, using the distortion energy failure theory. Sy 54 ny = = = 2.50 ′ σ max 21.56 Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): ka = 2.70(64) −0.265 = 0.90 Eq. (6-20): Eq. (6-18): kb = 0.879(1.13) −0.107 = 0.87 Se = 0.90(0.87)(32) = 25.1 kpsi Using Modified Goodman, 1 σ a′ σ m′ 9.495 19.18 = + = + nf Se Sut 25.1 64 n f = 1.47 Ans. ______________________________________________________________________________ 6-44 Table A-20: S ut = 64 kpsi, S y = 54 kpsi From Prob. 3-76, the critical stress element experiences completely reversed bending stress due to the rotation, and steady torsional and axial stresses. Shigley’s MED, 10th edition Chapter 6 Solutions, Page 37/58 www.konkur.in σ a ,bend = 33.99 kpsi, σ a ,axial = 0 kpsi, τ a = 0 kpsi, σ m,bend = 0 kpsi σ m,axial = −0.153 kpsi τ m = 7.847 kpsi Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/ 2 σ m′ = (σ m2 + 3τ m2 ) 1/2 2 2 = ( 33.99 ) + 3 ( 0 ) 1/2 = 33.99 kpsi 2 2 = ( −0.153) + 3 ( 7.847 ) 2 2 ′ = (σ max σ max + 3τ max ) 1/2 1/2 = 13.59 kpsi 2 2 = ( −33.99 − 0.153 ) + 3 ( 7.847 ) 1/2 = 36.75 kpsi Check for yielding, using the distortion energy failure theory. ny = Sy 54 = = 1.47 ′ σ max 36.75 Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): ka = 2.70(64) −0.265 = 0.90 Eq. (6-20): kb = 0.879(0.88) −0.107 = 0.89 Eq. (6-18): Se = 0.90(0.89)(32) = 25.6 kpsi Using Modified Goodman, 1 σ a′ σ m′ 33.99 13.59 = + = + nf Se Sut 25.6 64 n f = 0.65 Infinite life is not predicted. Ans. ______________________________________________________________________________ 6-45 Table A-20: S ut = 440 MPa, S y = 370 MPa From Prob. 3-77, the critical stress element experiences σ = 68.6 MPa and τ = 37.7 MPa. The bending is completely reversed due to the rotation, and the torsion is steady, giving σa = 68.6 MPa, σm = 0 MPa, τa = 0 MPa, τm = 37.7 MPa. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/2 σ m′ = (σ m2 + 3τ m2 ) 1/2 2 = 68.6 2 + 3 ( 0 ) 2 = 0 2 + 3 ( 37.7 ) 2 2 ′ = (σ max σ max + 3τ max ) 1/2 Shigley’s MED, 10th edition 1/ 2 1/2 = 68.6 MPa = 65.3 MPa 2 = 68.6 2 + 3 ( 37.7 ) 1/2 = 94.7 MPa Chapter 6 Solutions, Page 38/58 www.konkur.in Check for yielding, using the distortion energy failure theory. ny = Sy 370 = = 3.91 ′ 94.7 σ max Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5(440) = 220 MPa Eq. (6-19): ka = 4.51(440) −0.265 = 0.90 Eq. (6-20): Eq. (6-18): kb = 1.24(30) −0.107 = 0.86 Se = 0.90(0.86)(220) = 170 MPa Using Modified Goodman, 1 σ a′ σ m′ 68.6 65.3 = + = + nf Se Sut 170 440 n f = 1.81 Ans. ______________________________________________________________________________ 6-46 Table A-20: Sut = 64 kpsi, S y = 54 kpsi From Prob. 3-79, the critical stress element experiences σ = 3.46 kpsi and τ = 0.882 kpsi. The bending is completely reversed due to the rotation, and the torsion is steady, giving σa = 3.46 kpsi, σm = 0, τa = 0 kpsi, τm = 0.882 kpsi. Obtain von Mises stresses for the alternating, mid-range, and maximum stresses. σ a′ = (σ a2 + 3τ a2 ) 1/ 2 σ m′ = (σ m2 + 3τ m2 ) 1/ 2 1/2 2 = 3.462 + 3 ( 0 ) 2 = 02 + 3 ( 0.882 ) 2 2 ′ = (σ max σ max + 3τ max ) 1/ 2 = 3.46 kpsi 1/2 = 1.53 kpsi 1/ 2 2 = 3.46 2 + 3 ( 0.882 ) = 3.78 kpsi Check for yielding, using the distortion energy failure theory. ny = Sy 54 = = 14.3 ′ 3.78 σ max Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): ka = 2.70(64) −0.265 = 0.90 Eq. (6-20): kb = 0.879(1.375) −0.107 = 0.85 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 39/58 www.konkur.in Eq. (6-18): Se = 0.90(0.85)(32) = 24.5 kpsi Using Modified Goodman, 1 σ a′ σ m′ 3.46 1.53 = + = + nf Se Sut 24.5 64 n f = 6.06 Ans. ______________________________________________________________________________ 6-47 Table A-20: Sut = 64 kpsi, S y = 54 kpsi From Prob. 3-80, the critical stress element experiences σ = 16.3 kpsi and τ = 5.09 kpsi. Since the load is applied and released repeatedly, this gives σmax = 16.3 kpsi, σmin = 0 kpsi, τmax = 5.09 kpsi, τmin = 0 kpsi. Consequently, σm = σa = 8.15 kpsi, τm = τa = 2.55 kpsi. For bending, from Eqs. (6-34) and (6-35a), a = 0.246 − 3.08 (10−3 ) ( 64 ) + 1.51(10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.10373 2 1 1 = 0.75 a 1 + 0.10373 1+ 0.1 r K f = 1 + q ( K t − 1) = 1 + 0.75(1.5 − 1) = 1.38 q= Eq. (6-32): 3 = For torsion, from Eqs. (6-34) and (6-35b), a = 0.190 − 2.51(10−3 ) ( 64 ) + 1.35 (10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.07800 2 1 1 = 0.80 0.07800 a 1+ 1+ 0.1 r K fs = 1 + qs ( K ts − 1) = 1 + 0.80(2.1 − 1) = 1.88 q= Eq. (6-32): 3 = Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56). { σ a′ = (1.38 )( 8.15 ) + 3 (1.88 )( 2.55 ) 2 } 2 1/2 = 13.98 kpsi σ m′ = σ a′ = 13.98 kpsi ′ = σ a′ + σ m′ , Check for yielding, using the conservative σ max Sy 54 ny = = = 1.93 σ a′ + σ m′ 13.98 + 13.98 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 40/58 www.konkur.in Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): ka = aSutb = 2.70(64) −0.265 = 0.90 Eq. (6-24): de = 0.370d = 0.370 (1) = 0.370 in Eq. (6-20): kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.98 Eq. (6-18): Se = (0.90)(0.98)(32) = 28.2 kpsi Using Modified Goodman, 1 σ a′ σ m′ 13.98 13.98 = + = + nf Se Sut 28.2 64 n f = 1.40 Ans. ______________________________________________________________________________ 6-48 Table A-20: Sut = 64 kpsi, S y = 54 kpsi From Prob. 3-81, the critical stress element experiences σ = 16.4 kpsi and τ = 4.46 kpsi. Since the load is applied and released repeatedly, this gives σmax = 16.4 kpsi, σmin = 0 kpsi, τmax = 4.46 kpsi, τmin = 0 kpsi. Consequently, σm = σa = 8.20 kpsi, τm = τa = 2.23 kpsi. For bending, from Eqs. (6-34) and (6-35a), a = 0.246 − 3.08 (10−3 ) ( 64 ) + 1.51(10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.10373 2 1 = 0.75 a 1 + 0.10373 1+ 0.1 r K f = 1 + q ( K t − 1) = 1 + 0.75(1.5 − 1) = 1.38 q= Eq. (6-32): 1 3 = For torsion, from Eqs. (6-34) and (6-35b), a = 0.190 − 2.51(10−3 ) ( 64 ) + 1.35 (10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.07800 2 1 = 0.80 a 1 + 0.07800 1+ 0.1 r K fs = 1 + qs ( K ts − 1) = 1 + 0.80(2.1 − 1) = 1.88 q= Eq. (6-32): 1 3 = Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56). Shigley’s MED, 10th edition Chapter 6 Solutions, Page 41/58 www.konkur.in { σ a′ = (1.38 )( 8.20 ) + 3 (1.88 )( 2.23) 2 } 2 1/2 = 13.45 kpsi σ m′ = σ a′ = 13.45 kpsi ′ = σ a′ + σ m′ , Check for yielding, using the conservative σ max Sy 54 ny = = = 2.01 σ a′ + σ m′ 13.45 + 13.45 Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): Eq. (6-24): ka = aSutb = 2.70(64) −0.265 = 0.90 Eq. (6-20): Eq. (6-18): kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.98 d e = 0.370d = 0.370(1) = 0.370 in Se = (0.90)(0.98)(32) = 28.2 kpsi Using Modified Goodman, 1 σ a′ σ m′ 13.45 13.45 = + = + nf S e Sut 28.2 64 n f = 1.46 Ans. ______________________________________________________________________________ 6-49 Table A-20: Sut = 64 kpsi, S y = 54 kpsi From Prob. 3-82, the critical stress element experiences repeatedly applied bending, axial, and torsional stresses of σx,bend = 20.2 kpsi, σx,axial = 0.1 kpsi, and τ = 5.09 kpsi.. Since the axial stress is practically negligible compared to the bending stress, we will simply combine the two and not treat the axial stress separately for stress concentration factor and load factor. This gives σmax = 20.3 kpsi, σmin = 0 kpsi, τmax = 5.09 kpsi, τmin = 0 kpsi. Consequently, σm = σa = 10.15 kpsi, τm = τa = 2.55 kpsi. For bending, from Eqs. (6-34) and (6-35a), a = 0.246 − 3.08 (10−3 ) ( 64 ) + 1.51(10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.10373 2 1 = 0.75 a 1 + 0.10373 1+ 0.1 r K f = 1 + q ( K t − 1) = 1 + 0.75(1.5 − 1) = 1.38 q= Eq. (6-32): 1 3 = For torsion, from Eqs. (6-34) and (6-35b), Shigley’s MED, 10th edition Chapter 6 Solutions, Page 42/58 www.konkur.in a = 0.190 − 2.51(10−3 ) ( 64 ) + 1.35 (10−5 ) ( 64 ) − 2.67 (10−8 ) ( 64 ) = 0.07800 2 1 1 = 0.80 a 1 + 0.07800 1+ 0.1 r K fs = 1 + qs ( K ts − 1) = 1 + 0.80(2.1 − 1) = 1.88 q= Eq. (6-32): 3 = Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56). { σ a′ = (1.38 )(10.15 ) + 3 (1.88 )( 2.55 ) 2 } 2 1/2 = 16.28 kpsi σ m′ = σ a′ = 16.28 kpsi ′ = σ a′ + σ m′ , Check for yielding, using the conservative σ max Sy 54 ny = = = 1.66 σ a′ + σ m′ 16.28 + 16.28 Obtain the modifying factors and endurance limit. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): Eq. (6-24): ka = aSutb = 2.70(64) −0.265 = 0.90 Eq. (6-20): Eq. (6-18): d e = 0.370d = 0.370(1) = 0.370 in kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.98 Se = (0.90)(0.98)(32) = 28.2 kpsi Using Modified Goodman, 1 σ a′ σ m′ 16.28 16.28 = + = + nf S e Sut 28.2 64 n f = 1.20 Ans. ____________________________________________________________________________ 6-50 Table A-20: Sut = 64 kpsi, S y = 54 kpsi From Prob. 3-83, the critical stress element on the neutral axis in the middle of the longest side of the rectangular cross section experiences a repeatedly applied shear stress of τmax = 14.3 kpsi, τmin = 0 kpsi. Thus, τm = τa = 7.15 kpsi. Since the stress is entirely shear, it is convenient to check for yielding using the standard Maximum Shear Stress theory. S / 2 54 / 2 ny = y = = 1.89 τ max 14.3 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 43/58 www.konkur.in Find the modifiers and endurance limit. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): ka = aSutb = 2.70(64) −0.265 = 0.90 The size factor for a rectangular cross section loaded in torsion is not readily available. Following the procedure on p. 297, we need an equivalent diameter based on the 95 percent stress area. However, the stress situation in this case is nonlinear, as described on p. 116. Noting that the maximum stress occurs at the middle of the longest side, or with a radius from the center of the cross section equal to half of the shortest side, we will simply choose an equivalent diameter equal to the length of the shortest side. de = 0.25 in Eq. (6-20): kb = 0.879de −0.107 = 0.879(0.25)−0.107 = 1.02 We will round down to kb = 1. Eq. (6-26): Eq. (6-18): kc = 0.59 S se = 0.9(1)(0.59)(32) = 17.0 kpsi Since the stress is entirely shear, we choose to use a load factor kc = 0.59, and convert the ultimate strength to a shear value rather than using the combination loading method of Sec. 6-14. From Eq. (6-54), Ssu = 0.67Su = 0.67 (64) = 42.9 kpsi. Using Modified Goodman, 1 1 = = 1.70 Ans. (τ a / S se ) + (τ m / S su ) (7.15 / 17.0) + (7.15 / 42.9) ______________________________________________________________________________ nf = 6-51 Table A-20: Sut = 64 kpsi, S y = 54 kpsi From Prob. 3-84, the critical stress element experiences σ = 28.0 kpsi and τ = 15.3 kpsi. Since the load is applied and released repeatedly, this gives σmax = 28.0 kpsi, σmin = 0 kpsi, τmax = 15.3 kpsi, τmin = 0 kpsi. Consequently, σm = σa = 14.0 kpsi, τm = τa = 7.65 kpsi. From Table A-15-8 and A-15-9, D / d = 1.5 /1 = 1.5, r / d = 0.125 /1 = 0.125 K t ,bend = 1.60, K t ,tors = 1.39 Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: Eq. (6-32): Shigley’s MED, 10th edition qbend = 0.78, qtors = 0.82 Chapter 6 Solutions, Page 44/58 www.konkur.in K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.78 (1.60 − 1) = 1.47 K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.82 (1.39 − 1) = 1.32 Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56). { σ a′ = (1.47 )(14.0 ) + 3 (1.32 )( 7.65 ) 2 } 2 1/2 = 27.0 kpsi σ m′ = σ a′ = 27.0 kpsi ′ = σ a′ + σ m′ , Check for yielding, using the conservative σ max Sy 54 ny = = = 1.00 σ a′ + σ m′ 27.0 + 27.0 Since stress concentrations are included in this quick yield check, the low factor of safety is acceptable. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): ka = aSutb = 2.70(64) −0.265 = 0.897 Eq. (6-24): Eq. (6-20): Eq. (6-18): de = 0.370d = 0.370 (1) = 0.370 in kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.978 Se = (0.897)(0.978)(0.5)(64) = 28.1 kpsi Using Modified Goodman, 1 σ a′ σ m′ 27.0 27.0 = + = + nf Se Sut 28.1 64 n f = 0.72 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). σ rev = σ a′ 27.0 = = 46.7 kpsi 1 − (σ m′ / Sut ) 1 − (27.0 / 64) Fig. 6-18: f = 0.9 Eq. (6-14): ( f Sut ) 2 [ 0.9(64) ] a= = = 118.07 Se 28.1 Eq. (6-15): f Sut 1 b = − log 3 Se 2 Shigley’s MED, 10th edition 1 0.9(64) = − log = −0.1039 3 28.1 Chapter 6 Solutions, Page 45/58 www.konkur.in 1 σ 46.7 −0.1039 Eq. (6-16): N = rev = = 7534 cycles 7500 cycles Ans. 118.07 a ______________________________________________________________________________ 1/ b 6-52 Table A-20: Sut = 64 kpsi, S y = 54 kpsi From Prob. 3-85, the critical stress element experiences σx,bend = 46.1 kpsi, σx,axial = 0.382 kpsi and τ = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to demonstrate the process. Since the load is applied and released repeatedly, this gives σmax,bend = 46.1 kpsi, σmin,bend = 0 kpsi, σmax,axial = 0.382 kpsi, σmin,axial = 0 kpsi, τmax = 15.3 kpsi, τmin = 0 kpsi. Consequently, σm,bend = σa,bend = 23.05 kpsi, σm,axial = σa,axial = 0.191 kpsi, τm = τa = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9, D / d = 1.5 /1 = 1.5, r / d = 0.125 /1 = 0.125 K t ,bend = 1.60, K t ,tors = 1.39, K t ,axial = 1.75 Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = qaxial =0.78, qtors = 0.82 Eq. (6-32): K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.78 (1.60 − 1) = 1.47 K f ,axial = 1 + qaxial ( Kt ,axial − 1) = 1 + 0.78 (1.75 − 1) = 1.59 K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.82 (1.39 − 1) = 1.32 Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56). 1/ 2 2 0.191) 2 ( σ a′ = (1.47 )( 23.05 ) + (1.59 ) + 3 1.32 7.65 ( )( ) 0.85 { σ m′ = (1.47 )( 23.05 ) + (1.59 )( 0.191) + 3 (1.32 )( 7.65 ) 2 } 2 1/2 = 38.45 kpsi = 38.40 kpsi ′ = σ a′ + σ m′ , Check for yielding, using the conservative σ max Sy 54 ny = = = 0.70 σ a′ + σ m′ 38.45 + 38.40 Since the conservative yield check indicates yielding, we will check more carefully with ′ obtained directly from the maximum stresses, using the distortion energy with σ max failure theory, without stress concentrations. Note that this is exactly the method used for static failure in Ch. 5. Shigley’s MED, 10th edition Chapter 6 Solutions, Page 46/58 www.konkur.in ′ = σ max ny = (σ + σ max,axial ) + 3 (τ max ) = 2 max,bend Sy 54 = = 1.01 ′ 53.5 σ max 2 ( 46.1 + 0.382 ) 2 + 3 (15.3) = 53.5 kpsi 2 Ans. This shows that yielding is imminent, and further analysis of fatigue life should not be interpreted as a guarantee of more than one cycle of life. Eq. (6-8): Se′ = 0.5(64) = 32 kpsi Eq. (6-19): ka = aSutb = 2.70(64) −0.265 = 0.897 Eq. (6-24): Eq. (6-20): Eq. (6-18): de = 0.370d = 0.370 (1) = 0.370 in kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.978 Se = (0.897)(0.978)(0.5)(64) = 28.1 kpsi Using Modified Goodman, 1 σ a′ σ m′ 38.45 38.40 = + = + nf Se Sut 28.1 64 n f = 0.51 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). σ rev = σ a′ 38.45 = = 96.1 kpsi 1 − (σ m′ / Sut ) 1 − (38.40 / 64) This stress is much higher than the ultimate strength, rendering it impractical for the S-N diagram. We must conclude that the fluctuating stresses from the combination loading, when increased by the stress concentration factors, are so far from the Goodman line that the equivalent completely reversed stress method is not practical to use. Without testing, we are unable to predict a life. ______________________________________________________________________________ 6-53 Table A-20: Sut = 64 kpsi, S y = 54 kpsi From Prob. 3-86, the critical stress element experiences σx,bend = 55.5 kpsi, σx,axial = 0.382 kpsi and τ = 15.3 kpsi. The axial load is practically negligible, but we’ll include it to demonstrate the process. Since the load is applied and released repeatedly, this gives σmax,bend = 55.5 kpsi, σmin,bend = 0 kpsi, σmax,axial = 0.382 kpsi, σmin,axial = 0 kpsi, τmax = 15.3 kpsi, τmin = 0 kpsi. Consequently, σm,bend = σa,bend = 27.75 kpsi, σm,axial = σa,axial = 0.191 kpsi, τm = τa = 7.65 kpsi. From Table A-15-7, A-15-8 and A-15-9, Shigley’s MED, 10th edition Chapter 6 Solutions, Page 47/58 www.konkur.in D / d = 1.5 /1 = 1.5, K t ,bend = 1.60, r / d = 0.125 /1 = 0.125 K t ,tors = 1.39, K t ,axial = 1.75 Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = qaxial =0.78, qtors = 0.82 Eq. (6-32): K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.78 (1.60 − 1) = 1.47 K f ,axial = 1 + qaxial ( Kt ,axial − 1) = 1 + 0.78 (1.75 − 1) = 1.59 K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.82 (1.39 − 1) = 1.32 Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56). 1/2 2 0.191) 2 ( σ a′ = (1.47 )( 27.75 ) + (1.59 ) + 3 1.32 7.65 )( ) ( 0.85 { σ m′ = (1.47 )( 27.75 ) + (1.59 )( 0.191) + 3 (1.32 )( 7.65 ) 2 } 2 1/ 2 = 44.71 kpsi = 44.66 kpsi Since these stresses are relatively high compared to the yield strength, we will go ahead and check for yielding using the distortion energy failure theory. ′ = σ max ny = (σ + σ max,axial ) + 3 (τ max ) = 2 max,bend Sy 54 = = 0.87 ′ σ max 61.8 2 ( 55.5 + 0.382 ) 2 + 3 (15.3) = 61.8 kpsi 2 Ans. This shows that yielding is predicted. Further analysis of fatigue life is just to be able to report the fatigue factor of safety, though the life will be dictated by the static yielding failure, i.e. N = 1/2 cycle. Ans. Eq. (6-8): Se′ = 0.5 ( 64 ) = 32 kpsi Eq. (6-19): ka = aSutb = 2.70(64) −0.265 = 0.897 Eq. (6-24): Eq. (6-20): Eq. (6-18): de = 0.370d = 0.370 (1) = 0.370 in kb = 0.879d e −0.107 = 0.879(0.370) −0.107 = 0.978 Se = (0.897)(0.978)(0.5)(64) = 28.1 kpsi Using Modified Goodman, 1 σ a′ σ m′ 44.71 44.66 = + = + nf Se Sut 28.1 64 n f = 0.44 Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 6 Solutions, Page 48/58 www.konkur.in 6-54 From Table A-20, for AISI 1040 CD, Sut = 85 kpsi and Sy = 71 kpsi. From the solution to Prob. 6-17 we find the completely reversed stress at the critical shoulder fillet to be σrev = 35.0 kpsi, producing σa = 35.0 kpsi and σm = 0 kpsi. This problem adds a steady torque which creates torsional stresses of τm = Tr 2500 (1.625 / 2 ) = = 2967 psi = 2.97 kpsi, τ a = 0 kpsi J π (1.6254 ) / 32 From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt,bend =1.95, Kt,tors =1.60 Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = 0.76, qtors = 0.81 Eq. (6-32): K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.76 (1.95 − 1) = 1.72 K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.81(1.60 − 1) = 1.49 Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56). { } σ ′ = {(1.72 )( 0 ) + 3 (1.49 )( 2.97 ) } σ a′ = (1.72 )( 35.0 ) + 3 (1.49 )( 0 ) 2 1/2 2 2 1/2 2 m = 60.2 kpsi = 7.66 kpsi ′ = σ a′ + σ m′ , Check for yielding, using the conservative σ max Sy 71 ny = = = 1.05 σ a′ + σ m′ 60.2 + 7.66 From the solution to Prob. 6-17, Se = 29.5 kpsi. Using Modified Goodman, 1 σ a′ σ m′ 60.2 7.66 = + = + nf Se Sut 29.5 85 n f = 0.47 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). σ rev = Fig. 6-18: σ a′ 60.2 = = 66.2 kpsi 1 − (σ m′ / Sut ) 1 − (7.66 / 85) f = 0.867 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 49/58 www.konkur.in Eq. (6-14): ( f Sut ) a= 2 Se [0.867(85)] = 2 29.5 Eq. (6-15): f Sut 1 b = − log 3 Se Eq. (6-16): σ N = rev a 1/ b = 184.1 1 0.867(85) = − log = −0.1325 3 29.5 1 66.2 −0.1325 = = 2251 cycles 184.1 N = 2300 cycles Ans. ______________________________________________________________________________ 6-55 From the solution to Prob. 6-18 we find the completely reversed stress at the critical shoulder fillet to be σrev = 32.8 kpsi, producing σa = 32.8 kpsi and σm = 0 kpsi. This problem adds a steady torque which creates torsional stresses of τm = Tr 2200 (1.625 / 2 ) = = 2611 psi = 2.61 kpsi, J π (1.6254 ) / 32 τ a = 0 kpsi From Table A-15-8 and A-15-9, r/d = 0.0625/1.625 = 0.04, D/d = 1.875/1.625 = 1.15, Kt,bend =1.95, Kt,tors =1.60 Eqs. (6-34) and (6-35), or Figs. 6-20 and 6-21: qbend = 0.76, qtors = 0.81 Eq. (6-32): K f ,bend = 1 + qbend ( Kt ,bend − 1) = 1 + 0.76 (1.95 − 1) = 1.72 K f ,tors = 1 + qtors ( Kt ,tors − 1) = 1 + 0.81(1.60 − 1) = 1.49 Obtain von Mises stresses for the alternating and mid-range stresses from Eqs. (6-55) and (6-56). { } σ ′ = {(1.72 )( 0 ) + 3 (1.49 )( 2.61) } σ a′ = (1.72 )( 32.8 ) + 3 (1.49 )( 0 ) 2 1/2 2 2 1/2 2 m = 56.4 kpsi = 6.74 kpsi ′ = σ a′ + σ m′ , Check for yielding, using the conservative σ max Sy 71 ny = = = 1.12 σ a′ + σ m′ 56.4 + 6.74 From the solution to Prob. 6-18, Se = 29.5 kpsi. Using Modified Goodman, 1 σ a′ σ m′ 56.4 6.74 = + = + nf Se Sut 29.5 85 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 50/58 www.konkur.in n f = 0.50 Ans. Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). σ rev = σ a′ 56.4 = = 61.3 kpsi 1 − (σ m′ / Sut ) 1 − (6.74 / 85) Fig. 6-18: f = 0.867 Eq. (6-14): ( f Sut ) a= Se 2 [0.867(85)] = 29.5 f Sut 1 b = − log 3 Se Eq. (6-15): σ Eq. (6-16): N = rev a 1/ b 2 = 184.1 1 0.867(85) = − log = −0.1325 3 29.5 1 61.3 −0.1325 = = 4022 cycles 184.1 N = 4000 cycles Ans. ______________________________________________________________________________ 6-56 Sut = 55 kpsi, S y = 30 kpsi, K ts = 1.6, L = 2 ft, Fmin = 150 lbf , Fmax = 500 lbf Eqs. (6-34) and (6-35b), or Fig. 6-21: qs = 0.80 Eq. (6-32): K fs = 1 + qs ( Kts − 1) = 1 + 0.80 (1.6 − 1) = 1.48 Tmax = 500(2) = 1000 lbf ⋅ in, Tmin = 150(2) = 300 lbf ⋅ in τ max = 16 K fsTmax τ min = 16 K fsTmin τm = πd 3 πd τ max + τ min 3 2 τ −τ τ a = max min 2 = 16(1.48)(1000) = 11 251 psi = 11.25 kpsi π (0.875)3 16(1.48)(300) = 3375 psi = 3.38 kpsi π (0.875)3 11.25 + 3.38 = = 7.32 kpsi 2 11.25 − 3.38 = = 3.94 kpsi 2 = Since the stress is entirely shear, it is convenient to check for yielding using the standard Maximum Shear Stress theory. S / 2 30 / 2 ny = y = = 1.33 τ max 11.25 Find the modifiers and endurance limit. Shigley’s MED, 10th edition Chapter 6 Solutions, Page 51/58 www.konkur.in Eq. (6-8): Se′ = 0.5(55) = 27.5 kpsi Eq. (6-19): ka = 14.4(55) −0.718 = 0.81 Eq. (6-24): d e = 0.370(0.875) = 0.324 in Eq. (6-20): kb = 0.879(0.324) −0.107 = 0.99 Eq. (6-26): kc = 0.59 Eq. (6-18): S se = 0.81(0.99)(0.59)(27.5) = 13.0 kpsi Since the stress is entirely shear, we will use a load factor kc = 0.59, and convert the ultimate strength to a shear value rather than using the combination loading method of Sec. 6-14. From Eq. (6-54), Ssu = 0.67Su = 0.67 (55) = 36.9 kpsi. (a) Modified Goodman, Table 6-6 nf = 1 1 = = 1.99 (τ a / S se ) + (τ m / S su ) (3.94 / 13.0) + (7.32 / 36.9) Ans. (b) Gerber, Table 6-7 2 2 2τ m S se 1 S su τ a nf = −1 + 1 + 2 τ m S se S suτ a 2 2 2(7.32)(13.0) 1 36.9 3.94 = −1 + 1 + 2 7.32 13.0 36.9(3.94) n f = 2.49 Ans. ______________________________________________________________________________ 6-57 Sut = 145 kpsi, S y = 120 kpsi From Eqs. (6-34) and (6-35a), or Fig. 6-20, with a notch radius of 0.1 in, q = 0.9. Thus, with Kt = 3 from the problem statement, K f = 1 + q ( K t − 1) = 1 + 0.9(3 − 1) = 2.80 4 P −2.80(4)( P) = = −2.476 P πd2 π (1.2) 2 1 σ m = −σ a = (−2.476 P ) = −1.238 P 2 f P ( D + d ) 0.3P ( 6 + 1.2 ) Tmax = = = 0.54 P 4 4 σ max = − K f From Eqs. (6-34) and (6-35b), or Fig. 6-21, with a notch radius of 0.1 in, qs = 0.92. Thus, with Kts = 1.8 from the problem statement, Shigley’s MED, 10th edition Chapter 6 Solutions, Page 52/58 www.konkur.in K fs = 1 + qs ( K ts − 1) = 1 + 0.92(1.8 − 1) = 1.74 16 K fsT 16(1.74)(0.54 P) = 2.769 P πd π (1.2)3 τ 2.769 P τ a = τ m = max = = 1.385 P 2 2 Eqs. (6-55) and (6-56): τ max = 3 = σ a′ = [(σ a / 0.85)2 + 3τ a2 ]1/2 = [(1.238 P / 0.85) 2 + 3(1.385P )2 ]1/2 = 2.81P σ m′ = [σ m 2 + 3τ m2 ]1/2 = [(−1.238 P) 2 + 3(1.385P )2 ]1/2 = 2.70 P Eq. (6-8): Se′ = 0.5(145) = 72.5 kpsi Eq. (6-19): ka = 2.70(145)−0.265 = 0.722 Eq. (6-20): Eq. (6-18): kb = 0.879(1.2) −0.107 = 0.862 Se = (0.722)(0.862)(72.5) = 45.12 kpsi Modified Goodman: 1 σ a′ σ m′ 2.81P 2.70 P 1 = + = + = n f Se Sut 45.12 145 3 P = 4.12 kips Ans. Sy 120 = 5.29 Ans. σ a′ + σ m′ (2.81)(4.12) + (2.70)(4.12) ______________________________________________________________________________ Yield (conservative): n y = 6-58 = From Prob. 6-57, K f = 2.80, K f s = 1.74, S e = 45.12 kpsi 4 Pmax 4(18) = −2.80 = −44.56 kpsi 2 πd π (1.22 ) 4P 4(4.5) σ min = − K f min2 = −2.80 = −11.14 kpsi πd π (1.2) 2 D+d 6 + 1.2 Tmax = f Pmax = 0.3(18) = 9.72 kip ⋅ in 4 4 D+d 6 + 1.2 Tmin = f Pmin = 0.3(4.5) = 2.43 kip ⋅ in 4 4 16Tmax 16(9.72) τ max = K f s = 1.74 = 49.85 kpsi 3 πd π (1.2)3 16Tmin 16(2.43) τ min = K f s = 1.74 = 12.46 kpsi 3 πd π (1.2)3 −44.56 − ( −11.14) σa = = 16.71 kpsi 2 σ max = − K f Shigley’s MED, 10th edition Chapter 6 Solutions, Page 53/58 www.konkur.in −44.56 + ( −11.14) = −27.85 kpsi 2 49.85 − 12.46 τa = = 18.70 kpsi 2 49.85 + 12.46 τm = = 31.16 kpsi 2 σm = Eqs. (6-55) and (6-56): σ a′ = [(σ a / 0.85)2 + 3τ a2 ]1/2 = [(16.71/ 0.85)2 + 3(18.70) 2 ]1/2 = 37.89 kpsi σ m′ = [σ m 2 + 3τ m2 ]1/2 = [(−27.85)2 + 3(31.16) 2 ]1/ 2 = 60.73 kpsi Modified Goodman: 1 σ a′ σ m′ 37.89 60.73 = + = + n f Se Sut 45.12 145 nf = 0.79 Since infinite life is not predicted, estimate a life from the S-N diagram. First, find an equivalent completely reversed stress (See Ex. 6-12). σ rev = σ a′ 37.89 = = 65.2 kpsi 1 − (σ m′ / Sut ) 1 − (60.73 /145) Fig. 6-18: f = 0.8 Eq. (6-14): ( f Sut ) a= 2 Se [0.8(145)] = 2 45.12 Eq. (6-15): f Sut 1 b = − log 3 Se Eq. (6-16): σ N = rev a 1/ b = 298.2 1 0.8(145) = − log = −0.1367 3 45.12 1 65.2 −0.1367 = = 67 607 cycles 298.2 N = 67 600 cycles Ans. ______________________________________________________________________________ 6-59 For AISI 1020 CD, From Table A-20, Sy = 390 MPa, Sut = 470 MPa. Given: Se = 175 MPa. 360 + 160 360 − 160 First Loading: = 260 MPa, = 100 MPa (σ m )1 = (σ a )1 = 2 2 Shigley’s MED, 10th edition Chapter 6 Solutions, Page 54/58 www.konkur.in Goodman: (σ a )1 1 − (σ m )1 / Sut 2 0.9 ( 470 ) (σ a )e1 = a= = 100 = 223.8 MPa > Se ∴ finite life 1 − 260 / 470 = 1022.5 MPa 175 0.9 ( 470 ) 1 b = − log = −0.127 767 3 175 −1/0.127 767 223.8 N = = 145 920 cycles 1022.5 320 + ( −200 ) 320 − ( −200 ) Second loading: (σ m ) 2 = = 60 MPa, = 260 MPa (σ a ) 2 = 2 2 (σ a ) e 2 = (a) Miner’s method: n1 n2 + =1 N1 N 2 260 = 298.0 MPa 1 − 60 / 470 298.0 N2 = 1022.5 ⇒ −1/0.127767 = 15 520 cycles 80 000 n2 + =1 145 920 15 520 ⇒ n2 = 7000 cycles Ans. (b) Manson’s method: The number of cycles remaining after the first loading Nremaining =145 920 − 80 000 = 65 920 cycles Two data points: 0.9(470) MPa, 103 cycles 223.8 MPa, 65 920 cycles a2 (10 ) 0.9 ( 470 ) = b 223.8 a2 ( 65 920 ) 2 3 b2 1.8901 = ( 0.015170 ) 2 b log1.8901 = −0.151 997 log 0.015170 223.8 a2 = = 1208.7 MPa −0.151 997 ( 65 920 ) b2 = 1/ −0.151 997 298.0 n2 = = 10 000 cycles Ans. 1208.7 ______________________________________________________________________________ 6-60 Given: Se = 50 kpsi, Sut = 140 kpsi, f =0.8. Using Miner’s method, Shigley’s MED, 10th edition Chapter 6 Solutions, Page 55/58 www.konkur.in 0.8 (140 ) a= = 250.88 kpsi 50 0.8 (140 ) 1 b = − log = −0.116 749 3 50 1/ −0.116 749 95 N1 = = 4100 cycles σ 1 = 95 kpsi, 250.88 2 1/ −0.116 749 σ 2 = 80 kpsi, 80 N2 = 250.88 σ 3 = 65 kpsi, 65 N3 = 250.88 = 17 850 cycles 1/ −0.116 749 = 105 700 cycles 0.2 N 0.5 N 0.3 N + + = 1 ⇒ N = 12 600 cycles Ans. 4100 17 850 105 700 ______________________________________________________________________________ 6-61 Given: Sut = 530 MPa, Se = 210 MPa, and f = 0.9. (a) Miner’s method 0.9 ( 530 ) a= = 1083.47 MPa 210 0.9 ( 530 ) 1 b = − log = −0.118 766 3 210 2 1/ −0.118 766 350 1083.47 σ 1 = 350 MPa, N1 = = 13 550 cycles 1/ −0.118 766 260 σ 2 = 260 MPa, N 2 = 1083.47 = 165 600 cycles 1/ −0.118 766 225 = 559 400 cycles σ 3 = 225 MPa, N 3 = 1083.47 n1 n2 n3 + + =1 N1 N 2 N 3 n3 5000 50 000 + + = 184 100 cycles Ans. 13 550 165 600 559 400 (b) Manson’s method: The life remaining after the first series of cycling is NR1 = 13 550 − 5000 = 8550 cycles. The two data points required to define S e′,1 are [0.9(530), 103] and (350, 8550). Shigley’s MED, 10th edition Chapter 6 Solutions, Page 56/58 www.konkur.in 3 0.9 ( 530 ) a2 (10 ) = b 350 a2 ( 8550 ) 2 b2 b2 = a2 = ⇒ log (1.362 9 ) log ( 0.116 96 ) 350 ( 8550 ) −0.144 280 1.3629 = ( 0.11696 ) 2 b = −0.144 280 = 1292.3 MPa −1/0.144 280 260 N2 = = 67 090 cycles 1292.3 N R 2 = 67 090 − 50 000 = 17 090 cycles 0.9 ( 530 ) 260 b3 = = a3 (103 ) b3 a3 (17 090 ) b3 ⇒ 1.834 6 = ( 0.058 514 ) 2 b log (1.834 6 ) 260 = −0.213 785, a3 = = 2088.7 MPa −0.213 785 log ( 0.058 514 ) (17 090 ) −1/0.213 785 225 N3 = = 33 610 cycles Ans. 2088.7 ______________________________________________________________________________ 6-62 Given: Se = 45 kpsi, Sut = 85 kpsi, f = 0.86, and σa = 35 kpsi and σm = 30 kpsi for 12 (103) cycles. σa 35 Gerber equivalent reversing stress: σ rev = = = 39.98 kpsi 2 2 1 − (σ m / Sut ) 1 − ( 30 / 85 ) (a) Miner’s method: σrev < Se. According to the method, this means that the endurance limit has not been reduced and the new endurance limit is S e′ = 45 kpsi. Ans. (b) Manson’s method: Again, σrev < Se. According to the method, this means that the material has not been damaged and the endurance limit has not been reduced. Thus, the new endurance limit is S e′ = 45 kpsi. Ans. ______________________________________________________________________________ 6-63 Given: Se = 45 kpsi, Sut = 85 kpsi, f = 0.86, and σa = 35 kpsi and σm = 30 kpsi for 12 (103) cycles. σa 35 = = 54.09 kpsi Goodman equivalent reversing stress: σ rev = 1 − (σ m / Sut ) 1 − ( 30 / 85 ) Initial cycling Shigley’s MED, 10th edition Chapter 6 Solutions, Page 57/58 www.konkur.in 0.86 ( 85 ) a= = 116.00 kpsi 45 0.86 ( 85 ) 1 b = − log = −0.070 235 3 45 2 σ 1 = 54.09 kpsi, 1/ −0.070 235 54.09 N1 = 116.00 = 52 190 cycles (a) Miner’s method (see discussion on p. 333): The number of remaining cycles at 54.09 kpsi is Nremaining = 52 190 − 12 000 = 40 190 cycles. The new coefficients are b′ = b, and a′ =Sf /Nb = 54.09/(40 190) − 0.070 235 = 113.89 kpsi. The new endurance limit is Se′,1 = a′N eb′ = 113.89 (106 ) − 0.070 235 = 43.2 kpsi Ans. (b) Manson’s method (see discussion on p. 334): The number of remaining cycles at 54.09 kpsi is Nremaining = 52 190 − 12 000 = 40 190 cycles. At 103 cycles, Sf = 0.86(85) = 73.1 kpsi. The new coefficients are b′ = [log(73.1/54.09)]/log(103/40 190) = − 0.081 540 and a′ = σ1/ (Nremaining) b′ = 54.09/(40 190) − 0.081 540 = 128.39 kpsi. The new endurance limit is Se′,1 = a′N eb′ = 128.39 (106 ) − 0.081 540 = 41.6 kpsi Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 6 Solutions, Page 58/58 www.konkur.in Chapter 7 7-1 (a) DE-Gerber, Eq. (7-10): A = 4 ( K f M a ) + 3 ( K fsTa ) = 4 [ (2.2)(70) ] + 3 [ (1.8)(45) ] = 338.4 N ⋅ m 2 2 2 2 B = 4 ( K f M m ) + 3 ( K fsTm ) = 4 [ (2.2)(55) ] + 3 [ (1.8)(35) ] = 265.5 N ⋅ m 2 2 2 6 8(2)(338.4) 2(265.5) ( 210 ) 10 d = 1+ 1+ 6 6 π ( 210 ) 10 338.4 ( 700 ) 10 d = 25.85 (10−3) m = 25.85 mm Ans. 2 ( ) ( ) ( ) 2 1/2 1/3 (b) DE-elliptic, Eq. (7-12) can be shown to be 1/3 2 2 338.4 ) 265.5 ) ( ( 16(2) = + 2 2 ( 210 ) 106 ( 560 ) 106 π −3 d = 25.77 (10 ) m = 25.77 mm Ans. 16n d = π 1/3 A B + Se2 S y2 2 2 ( ) ( ) (c) DE-Soderberg, Eq. (7-14) can be shown to be 16(2) 338.4 16n A B 265.5 d= + + = 6 π 210 10 560 106 π Se S y d = 27.70 (10−3) m = 27.70 mm Ans. 1/3 ( ) ( ) 1/3 (d) DE-Goodman: Eq. (7-8) can be shown to be 1/3 16(2) 338.4 265.5 = + 6 π 210 106 700 10 d = 27.27 (10−3) m = 27.27 mm Ans. ________________________________________________________________________ Criterion d (mm) Compared to DE-Gerber DE-Gerber 25.85 DE-Elliptic 25.77 0.31% Lower Less conservative DE-Soderberg 27.70 7.2% Higher More conservative DE-Goodman 27.27 5.5% Higher More conservative ______________________________________________________________________________ 16n A B d= + π Se Sut 7-2 1/3 ( ) ( ) This problem has to be done by successive trials, since Se is a function of shaft size. The material is SAE 2340 for which Sut = 175 kpsi, Sy = 160 kpsi, and HB ≥ 370. Shigley’s MED, 10th edition Chapter 7 Solutions, Page 1/45 www.konkur.in Eq. (6-19), p. 295: ka = 2.70(175) −0.265 = 0.69 Trial #1: Choose dr = 0.75 in Eq. (6-20), p. 296: kb = 0.879(0.75) −0.107 = 0.91 Eq. (6-8), p.290: Eq. (6-18), p. 295: Se′ = 0.5Sut = 0.5 (175 ) = 87.5 kpsi Se = 0.69 (0.91)(87.5) = 54.9 kpsi d r = d − 2r = 0.75D − 2 D / 20 = 0.65 D d 0.75 D= r = = 1.15 in 0.65 0.65 D 1.15 r= = = 0.058 in 20 20 Fig. A-15-14: d = d r + 2r = 0.75 + 2(0.058) = 0.808 in d 0.808 = = 1.08 dr 0.75 r 0.058 = = 0.077 dr 0.75 Kt = 1.9 Fig. 6-20, p. 296: r = 0.058 in, q = 0.90 Eq. (6-32), p. 303: Kf = 1 + 0.90 (1.9 – 1) = 1.81 Fig. A-15-15: Kts = 1.5 Fig. 6-21, p. 304: r = 0.058 in, qs = 0.92 Eq. (6-32), p. 303: Kfs = 1 + 0.92 (1.5 – 1) = 1.46 We select the DE-ASME Elliptic failure criteria, Eq. (7-12), with d as dr, and Mm = Ta = 0, 1/3 2 2 1/2 1.46(400) 16(2.5) 1.81(600) + 3 dr = 4 3 160 103 π 54.9 10 dr = 0.799 in ( ) ( ) Trial #2: Choose dr = 0.799 in. kb = 0.879(0.799) −0.107 = 0.90 Se = 0.69 (0.90)(0.5)(175) = 54.3 kpsi d 0.799 D= r = = 1.23 in 0.65 0.65 r = D / 20 = 1.23/20 = 0.062 in Shigley’s MED, 10th edition Chapter 7 Solutions, Page 2/45 www.konkur.in Figs. A-15-14 and A-15-15: d = d r + 2r = 0.799 + 2(0.062) = 0.923 in d 0.923 = = 1.16 d r 0.799 r 0.062 = = 0.078 d r 0.799 With these ratios only slightly different from the previous iteration, we are at the limit of readability of the figures. We will keep the same values as before. K t = 1.9, K ts = 1.5, q = 0.90, qs = 0.92 ∴ K f = 1.81, K fs = 1.46 Using Eq. (7-12) produces dr = 0.802 in. Further iteration produces no change. With dr = 0.802 in, 0.802 D= = 1.23 in 0.65 d = 0.75(1.23) = 0.92 in A look at a bearing catalog finds that the next available bore diameter is 0.9375 in. In nominal sizes, we select d = 0.94 in, D = 1.25 in, r = 0.0625 in Ans. ______________________________________________________________________________ 7-3 F cos 20°(d / 2) = TA, F = 2 TA / ( d cos 20°) = 2(340) / (0.150 cos 20°) = 4824 N. The maximum bending moment will be at point C, with MC = 4824(0.100) = 482.4 N∙m. Due to the rotation, the bending is completely reversed, while the torsion is constant. Thus, Ma = 482.4 N∙m, Tm = 340 N∙m, Mm = Ta = 0. For sharp fillet radii at the shoulders, from Table 7-1, Kt = 2.7, and Kts = 2.2. Examining Figs. 6-20 and 6-21 (pp. 303 and 304 respectively) with Sut = 560 MPa, conservatively estimate q = 0.8 and qs = 0.9. These estimates can be checked once a specific fillet radius is determined. Eq. (6-32): K f = 1 + 0.8(2.7 − 1) = 2.4 K fs = 1 + 0.9(2.2 − 1) = 2.1 (a) We will choose to include fatigue stress concentration factors even for the static analysis to avoid localized yielding. 1/ 2 Eq. (7-15): ′ σ max Shigley’s MED, 10th edition 2 32 K f M a 2 16 K fsTm = + 3 3 3 π d π d Chapter 7 Solutions, Page 3/45 www.konkur.in Eq. (7-16): n= Sy π d 3S y 2 2 −1/2 = 4 ( K f M a ) + 3 ( K fsTm ) ′ σ max 16 Solving for d, 1/3 16n 1/2 4( K f M a ) 2 + 3( K fsTa )2 d = π S y 16(2.5) = π ( 420 ) 106 ( ){ 4 [ (2.4)(482.4)] + 3[ (2.1)(340) ] d = 0.0430 m = 43.0 mm (b) } 2 1/2 2 1/3 Ans. ka = 4.51(560) −0.265 = 0.84 Assume kb = 0.85 for now. Check later once a diameter is known. Se = 0.84(0.85)(0.5)(560) = 200 MPa Selecting the DE-ASME Elliptic criteria, use Eq. (7-12) with M m = Ta = 0. 2 2.1(340) 16(2.5) 2.4(482.4) + 3 d = 4 6 420 106 π 200 10 = 0.0534 m = 53.4 mm ( ) ( ) 2 1/2 1/3 With this diameter, we can refine our estimates for kb and q. Eq. (6-20): kb = 1.51d −0.157 = 1.51( 53.4 ) −0.157 = 0.81 Assuming a sharp fillet radius, from Table 7-1, r = 0.02d = 0.02 (53.4) = 1.07 mm. Fig. (6-20): Fig. (6-21): q = 0.72 qs = 0.77 Iterating with these new estimates, Eq. (6-32): Eq. (6-18): Eq. (7-12): Kf = 1 + 0.72 (2.7 – 1) = 2.2 Kfs = 1 + 0.77 (2.2 – 1) = 1.9 Se = 0.84(0.81)(0.5)(560) = 191 MPa d = 53 mm Ans. Further iteration does not change the results. _____________________________________________________________________________ Shigley’s MED, 10th edition Chapter 7 Solutions, Page 4/45 www.konkur.in 7-4 We have a design task of identifying bending moment and torsion diagrams which are preliminary to an industrial roller shaft design. Let point C represent the center of the span of the roller. FCy = 30(8) = 240 lbf FCz = 0.4(240) = 96 lbf T = FCz (2) = 96(2) = 192 lbf ⋅ in T 192 FBz = = = 128 lbf 1.5 1.5 FBy = FBz tan 20o = 128 tan 20o = 46.6 lbf (a) xy-plane ΣM O = 240(5.75) − FAy (11.5) − 46.6(14.25) = 0 240(5.75) − 46.6(14.25) FAy = = 62.3 lbf 11.5 ΣM A = FOy (11.5) − 46.6(2.75) − 240(5.75) = 0 240(5.75) + 46.6(2.75) FOy = = 131.1 lbf 11.5 Bending moment diagram: Shigley’s MED, 10th edition Chapter 7 Solutions, Page 5/45 www.konkur.in xz-plane ΣM O = 0 = 96(5.75) − FAz (11.5) + 128(14.25) 96(5.75) + 128(14.25) FAz = = 206.6 lbf 11.5 ΣM A = 0 = FOz (11.5) + 128(2.75) − 96(5.75) 96(5.75) − 128(2.75) FOz = = 17.4 lbf 11.5 Bending moment diagram: M C = 100 2 + ( −754) 2 = 761 lbf ⋅ in M A = ( −128) 2 + ( −352) 2 = 375 lbf ⋅ in Torque: The torque is constant from C to B, with a magnitude previously obtained of 192 lbf∙in. (b) xy-plane M xy = −131.1x + 15 x − 1.75 − 15 x − 9.75 − 62.3 x − 11.5 2 Shigley’s MED, 10th edition 2 1 Chapter 7 Solutions, Page 6/45 www.konkur.in Bending moment diagram: Mmax = –516 lbf ∙ in and occurs at 6.12 in. M C = 131.1(5.75) − 15(5.75 − 1.75) 2 = 514 lbf ⋅ in This is reduced from 754 lbf ∙ in found in part (a). The maximum occurs at x = 6.12 in rather than C, but it is close enough. xz-plane M xz = 17.4 x − 6 x − 1.75 + 6 x − 9.75 + 206.6 x − 11.5 2 2 1 Bending moment diagram: Let M net = M xy2 + M xz2 Plot Mnet(x), 1.75 ≤ x ≤ 11.5 in Mmax = 516 lbf ∙ in at x = 6.25 in Torque: The torque rises from 0 to 192 lbf∙in linearly across the roller, then is constant to B. Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 7 Solutions, Page 7/45 www.konkur.in 7-5 This is a design problem, which can have many acceptable designs. See the solution for Prob. 7-17 for an example of the design process. ______________________________________________________________________________ 7-6 If students have access to finite element or beam analysis software, have them model the shaft to check deflections. If not, solve a simpler version of shaft for deflection. The 1 in diameter sections will not affect the deflection results much, so model the 1 in diameter as 1.25 in. Also, ignore the step in AB. From Prob. 7-4, integrate Mxy and Mxz. xy plane, with dy/dx = y' ( ) 131.1 2 62.3 3 3 2 x + 5 x − 1.75 − 5 x − 9.75 − x − 11.5 + C1 2 2 131.1 3 5 5 62.3 4 4 3 EIy = − x + x − 1.75 − x − 9.75 − x − 11.5 + C1 x + C2 6 4 4 6 EIy ′ = − (1) ( ) y = 0 at x = 0 From (1), ⇒ C2 = 0 y = 0 at x = 11.5 ⇒ C1 = 1908.4 lbf ⋅ in 3 x = 0: EIy' = 1908.4 x = 11.5: EIy' = –2153.1 xz plane (treating z ↑ + ) ( ) 17.4 2 206.6 3 3 2 x − 2 x − 1.75 + 2 x − 9.75 + x − 11.5 + C3 2 2 17.4 3 1 1 206.6 4 4 3 EIz = x ) − x − 1.75 + x − 9.75 + x − 11.5 + C3 x + C4 ( 6 2 2 6 EIz ′ = z = 0 at x = 0 From (2), ⇒ (2) C4 = 0 z = 0 at x = 11.5 ⇒ C3 = 8.975 lbf ⋅ in 3 x = 0: EIz' = 8.975 x = 11.5: EIz' = –683.5 Shigley’s MED, 10th edition Chapter 7 Solutions, Page 8/45 www.konkur.in At O: EIθ = 1908.42 + 8.9752 = 1908.4 lbf ⋅ in 3 At A: EIθ = ( −2153.1) 2 + ( −683.5) 2 = 2259.0 lbf ⋅ in 3 (dictates size) θ= 2259 = 0.000 628 rad 30 10 (π / 64 ) 1.254 n= 0.001 = 1.59 0.000 628 ( ) ( 6 ) At gear mesh, B xy plane With I = I1 in section OCA, y′A = −2153.1/ EI1 Since y'B/A is a cantilever, from Table A-9-1, with I = I 2 in section AB Fx( x − 2l ) 46.6 yB′ / A = = (2.75)[2.75 − 2(2.75)] = −176.2 / EI 2 2 EI 2 2 EI 2 2153.1 176.2 ∴ y′B = y′A + y′B / A = − − 6 4 6 30 10 (π / 64 ) 1.25 30 10 (π / 64 ) 0.8754 ( ) ( ) ( ) ( ) = –0.000 803 rad (magnitude greater than 0.0005 rad) xz plane 683.5 z ′A = − , EI1 z ′B = − z B′ / A = − ( 128 2.752 2 EI 2 ) = − 484 EI 2 484 683.5 − = −0.000 751 rad 30 10 (π / 64 ) 1.254 30 106 (π / 64 ) 0.8754 ( ) 6 ( ) ( ) ( ) θ B = (−0.000 803) 2 + (−0.000 751) 2 = 0.00110 rad Crowned teeth must be used. Shigley’s MED, 10th edition Chapter 7 Solutions, Page 9/45 www.konkur.in Finite element results: θO = 5.47(10−4 ) rad Error in simplified model 3.0% θ A = 7.09(10−4 ) rad 11.4% 0.0% θ B = 1.10(10 ) rad The simplified model yielded reasonable results. −3 Sut = 72 kpsi, S y = 39.5 kpsi Strength At the shoulder at A, x = 10.75 in. From Prob. 7-4, M xy = −209.3 lbf ⋅ in, M xz = −293.0 lbf ⋅ in, T = 192 lbf ⋅ in M = ( −209.3) 2 + ( −293) 2 = 360.0 lbf ⋅ in Se′ = 0.5(72) = 36 kpsi ka = 2.70(72) −0.265 − 0.869 −0.107 1 kb = = 0.879 0.3 kc = k d = k e = k f = 1 Fig. A-15-8: Fig. A-15-9: Fig. 6-20: Fig. 6-21: Eq. (6-32): Se = 0.869(0.879)(36) = 27.5 kpsi D / d = 1.25, r / d = 0.03 Kts = 1.8 Kt = 2.3 q = 0.65 qs = 0.70 K f = 1 + 0.65(2.3 − 1) = 1.85 K fs = 1 + 0.70(1.8 − 1) = 1.56 Using DE-ASME Elliptic, Eq. (7-11) with M m = Ta = 0, 1/2 1 16 = n π 13 ( ) 2 1.85(360) 2 1.56(192) 4 + 3 39 500 27 500 n = 3.91 Perform a similar analysis at the profile keyway under the gear. The main problem with the design is the undersized shaft overhang with excessive slope at the gear. The use of crowned-teeth in the gears will eliminate this problem. ______________________________________________________________________________ 7-7 through 7-16 These are design problems, which can have many acceptable designs. See the solution for Prob. 7-17 for an example of the design process. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 7 Solutions, Page 10/45 www.konkur.in 7-17 (a) One possible shaft layout is shown in part (e). Both bearings and the gear will be located against shoulders. The gear and the motor will transmit the torque through the keys. The bearings can be lightly pressed onto the shaft. The left bearing will locate the shaft in the housing, while the right bearing will float in the housing. (b) From summing moments around the shaft axis, the tangential transmitted load through the gear will be Wt = T / (d / 2) = 2500 / (4 / 2) = 1250 lbf The radial component of gear force is related by the pressure angle. Wr = Wt tan φ = 1250 tan 20o = 455 lbf ( W = Wr2 + Wt 2 ) 1/ 2 ( = 4552 + 12502 ) 1/2 = 1330 lbf Reactions RA and RB , and the load W are all in the same plane. From force and moment balance, RA = 1330(2 / 11) = 242 lbf RB = 1330(9 / 11) = 1088 lbf M max = RA (9) = 242(9) = 2178 lbf ⋅ in Shear force, bending moment, and torque diagrams can now be obtained. Shigley’s MED, 10th edition Chapter 7 Solutions, Page 11/45 www.konkur.in (c) Potential critical locations occur at each stress concentration (shoulders and keyways). To be thorough, the stress at each potentially critical location should be evaluated. For now, we will choose the most likely critical location, by observation of the loading situation, to be in the keyway for the gear. At this point there is a large stress concentration, a large bending moment, and the torque is present. The other locations either have small bending moments, or no torque. The stress concentration for the keyway is highest at the ends. For simplicity, and to be conservative, we will use the maximum bending moment, even though it will have dropped off a little at the end of the keyway. (d) At the gear keyway, approximately 9 in from the left end of the shaft, the bending is completely reversed and the torque is steady. M a = 2178 lbf ⋅ in Tm = 2500 lbf ⋅ in M m = Ta = 0 From Table 7-1, p. 365, estimate stress concentrations for the end-milled keyseat to be Kt = 2.14 and Kts = 3.0. For the relatively low strength steel specified (AISI 1020 CD), roughly estimate notch sensitivities of q = 0.75 and qs = 0.80, obtained by observation of Figs. 6-20 and 6-21, assuming a typical radius at the bottom of the keyseat of r / d = 0.02, and a shaft diameter of up to 3 inches. Eq. (6-32): K f = 1 + 0.75(2.14 − 1) = 1.9 K fs = 1 + 0.8(3.0 − 1) = 2.6 Eq. (6-19): ka = 2.70(68) −0.265 = 0.883 For estimating kb , guess d = 2 in. Eq. (6-20) kb = (2 / 0.3) −0.107 = 0.816 Eq. (6-18) Se = 0.883(0.816)(0.5)(68) = 24.5 kpsi Selecting the DE-Goodman criteria for a conservative first design, 1/3 Eq. (7-8): 2 1/2 2 1/ 2 4 K M 3 K T 16n ( f a ) ( fs m ) d= + Se Sut π 1/3 1/ 2 2 1/2 3 ( 2.6 ⋅ 2500 )2 4 1.9 ⋅ 2178 ( ) 16(1.5) + d= π 24 500 68 000 d = 1.57 in Shigley’s MED, 10th edition Ans. Chapter 7 Solutions, Page 12/45 www.konkur.in With this diameter, the estimates for notch sensitivity and size factor were conservative, but close enough for a first iteration until deflections are checked. Check yielding with this diameter. 1/ 2 Eq. (7-15): ′ σ max 2 32 K f M a 2 16 K fsTm = + 3 3 3 π d π d 1/2 2 32(1.9)(2178) 2 16(2.6)(2500) ′ = σ max + 3 3 3 π (1.57) π (1.57) ′ = 57 /18.4 = 3.1 Ans. n y = S y / σ max = 18389 psi = 18.4 kpsi (e) Now estimate other diameters to provide typical shoulder supports for the gear and bearings (p. 364). Also, estimate the gear and bearing widths. (f) Entering this shaft geometry into beam analysis software (or Finite Element software), the following deflections are determined: Left bearing slope: 0.000 532 rad Right bearing slope: − 0.000 850 rad Gear slope: − 0.000 545 rad Right end of shaft slope: − 0.000 850 rad Gear deflection: − 0.001 45 in Right end of shaft deflection: 0.005 10 in Comparing these deflections to the recommendations in Table 7-2, everything is within typical range except the gear slope is a little high for an uncrowned gear. (g) To use a non-crowned gear, the gear slope is recommended to be less than 0.0005 rad. Since all other deflections are acceptable, we will target an increase in diameter only for the long section between the left bearing and the gear. Increasing this diameter from the proposed 1.56 in to 1.75 in, produces a gear slope of − 0.000 401 rad. All other deflections are improved as well. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 7 Solutions, Page 13/45 www.konkur.in 7-18 (a) Use the distortion-energy elliptic failure locus. The torque and moment loadings on the shaft are shown in the solution to Prob. 7-17. Candidate critical locations for strength: • Left seat keyway • Right bearing shoulder • Right keyway Table A-20 for 1030 HR: Sut = 68 kpsi, S y = 37.5 kpsi, H B = 137 Eq. (6-8): Se′ = 0.5(68) = 34.0 kpsi Eq. (6-19): ka = 2.70(68) −0.265 = 0.883 kc = k d = ke = 1 Left keyway See Table 7-1 for keyway stress concentration factors, Kt = 2.14 Profile keyway Kts = 3.0 For an end-mill profile keyway cutter of 0.010 in radius, estimate notch sensitivities. Fig. 6-20: Fig. 6-21: Eq. (6-32): q = 0.51 qs = 0.57 K fs = 1 + qs ( K ts − 1) = 1 + 0.57(3.0 − 1) = 2.1 K f = 1 + 0.51(2.14 − 1) = 1.6 −0.107 Eq. (6-20): Eq. (6-18): 1.875 kb = = 0.822 0.30 Se = 0.883(0.822)(34.0) = 24.7 kpsi 1 Eq. (7-11): 2 2 1.6(2178) 2 2.1(2500) 1 16 = 4 + 3 n f π (1.8753 ) 24 700 37 500 nf = 3.5 Ans. Right bearing shoulder The text does not give minimum and maximum shoulder diameters for 03-series bearings (roller). Use D = 1.75 in. r 0.030 D 1.75 = = 0.019, = = 1.11 d 1.574 d 1.574 Fig. A-15-9: Fig. A-15-8: Shigley’s MED, 10th edition Kt = 2.4 K ts = 1.6 Chapter 7 Solutions, Page 14/45 www.konkur.in Fig. 6-20: Fig. 6-21: Eq. (6-32): q = 0.65 qs = 0.70 K f = 1 + 0.65(2.4 − 1) = 1.91 K fs = 1 + 0.70(1.6 − 1) = 1.42 0.453 M = 2178 = 493 lbf ⋅ in 2 1/ 2 Eq. (7-11): 2 1.91(493) 2 1.42(2500) 1 16 = 4 + 3 n f π (1.5743 ) 24 700 37 500 nf = 4.2 Ans. Right keyway Use the same stress concentration factors as for the left keyway. There is no bending moment, thus Eq. (7-11) reduces to: 1 16 3K fsTm 16 3(2.1)(2500) = = nf π d 3S y π (1.53 ) (37 500) nf = 2.7 Ans. Yielding Check for yielding at the left keyway, where the completely reversed bending is maximum, and the steady torque is present. Using Eq. (7-15), with Mm = Ta = 0, 1/2 ′ σ max 2 32 K f M a 2 16 K fsTm = + 3 3 3 π d π d 1/ 2 2 32 1.6 2178 2 16 ( 2.1)( 2500 ) ( )( ) + 3 = π (1.875 )3 π (1.875 )3 = 8791 psi = 8.79 kpsi S 37.5 Ans. ny = y = = 4.3 ′ σ max 8.79 Check in smaller diameter at right end of shaft where only steady torsion exists. 1/2 ′ σ max 16 K fsTm 2 = 3 3 π d 1/2 16 2.1 2500 2 ( )( ) = 3 3 π (1.5 ) = 13 722 psi = 13.7 kpsi Shigley’s MED, 10th edition Chapter 7 Solutions, Page 15/45 www.konkur.in ny = Sy 37.5 = = 2.7 ′ σ max 13.7 Ans. (b) One could take pains to model this shaft exactly, using finite element software. However, for the bearings and the gear, the shaft is basically of uniform diameter, 1.875 in. The reductions in diameter at the bearings will change the results insignificantly. Use E = 30 Mpsi for steel. To the left of the load, from Table A-9, case 6, θ AB = d y AB Fb 1449(2)(3 x 2 + 2 2 − 112 ) = (3 x 2 + b 2 − l 2 ) = dx 6 EIl 6(30)(106 )(π / 64)(1.8754 )(11) = 2.4124(10 −6 )(3 x 2 − 117) At x = 0 in: θ = −2.823(10−4 ) rad θ = 3.040(10−4 ) rad At x = 9 in: To the right of the load, from Table A-9, case 6, θ BC = ( d y BC Fa = −3 x 2 + 6 xl − 2l 2 − a 2 dx 6 EIl ) At x = l = 11 in: Fa 2 1449(9)(112 − 92 ) 2 l −a = = 4.342(10−4 ) rad θ= 6 4 6 EIl 6(30)(10 )(π / 64)(1.875 )(11) ( ) Obtain allowable slopes from Table 7-2. Left bearing: n fs = Allowable slope 0.001 = = 3.5 Actual slope 0.000 282 3 n fs = 0.0008 = 1.8 0.000 434 2 Ans. Right bearing: Ans. Gear mesh slope: Table 7-2 recommends a minimum relative slope of 0.0005 rad. While we don’t know the slope on the next shaft, we know that it will need to have a larger diameter and be stiffer. At the moment we can say 0.0005 = 1.6 Ans. 0.000 304 ______________________________________________________________________________ n fs < Shigley’s MED, 10th edition Chapter 7 Solutions, Page 16/45 www.konkur.in 7-19 The most likely critical locations for fatigue are at locations where the bending moment is high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-72, indicate decreasing moments in both planes to the left of A and to the right of C, with combined values at A and C of MA = 5324 lbf∙in and MC = 6750 lbf∙in. The torque is constant between A and B, with T = 2819 lbf∙in. The most likely critical locations are at the stress concentrations near A and C. The two shoulders near A can be eliminated since the shoulders near C have the same geometry but a higher bending moment. We will consider the following potentially critical locations: • keyway at A • shoulder to the left of C • shoulder to the right of C Table A-20: Eq. (6-8): Sut = 64 kpsi, Sy = 54 kpsi Se′ = 0.5(64) = 32.0 kpsi Eq. (6-19): ka = 2.70(64) −0.265 = 0.897 kc = k d = ke = 1 Keyway at A Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 365), with d = 1.75 in, r = 0.02d = 0.035 in. Table 7-1: Kt = 2.14, Kts = 3.0 Fig. 6-20: q = 0.65 Fig. 6-21: qs = 0.71 Eq. (6-32): K f = 1 + q ( K t − 1) = 1 + 0.65(2.14 − 1) = 1.7 K fs = 1 + qs ( K ts − 1) = 1 + 0.71(3.0 − 1) = 2.4 −0.107 Eq. (6-20): Eq. (6-18): 1.75 kb = = 0.828 0.30 Se = 0.897(0.828)(32) = 23.8 kpsi Shigley’s MED, 10th edition Chapter 7 Solutions, Page 17/45 www.konkur.in We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. Using Eq. (7-9) with Mm = Ta = 0, A = 4 ( K f M a ) = 4 (1.7 )( 5324 ) = 18102 lbf ⋅ in = 18.10 kip ⋅ in 2 2 B = 3 ( K fsTm ) = 3 ( 2.4 )( 2819 ) = 11 718 lbf ⋅ in = 11.72 kip ⋅ in 2 1 8A = n π d 3 Se 2 2 1/2 2 BSe 1 + 1 + ASut 2 1/ 2 2 (11.72 )( 23.8 ) = 1 + 1 + π (1.753 ) ( 23.8 ) (18.10 )( 64 ) n = 1.3 8 (18.10 ) Shoulder to the left of C r / d = 0.0625 / 1.75 = 0.036, D / d = 2.5 / 1.75 = 1.43 Fig. A-15-9: Fig. A-15-8: Fig. 6-20: Fig. 6-21: Eq. (6-32): Kt = 2.2 Kts = 1.8 q = 0.71 qs = 0.76 K f = 1 + q ( K t − 1) = 1 + 0.71(2.2 − 1) = 1.9 K fs = 1 + qs ( K ts − 1) = 1 + 0.76(1.8 − 1) = 1.6 −0.107 Eq. (6-20): Eq. (6-18): 1.75 kb = = 0.828 0.30 Se = 0.897(0.828)(32) = 23.8 kpsi For convenience, we will use the full value of the bending moment at C, even though it will be slightly less at the shoulder. Using Eq. (7-9) with Mm = Ta = 0, A = 4 ( K f M a ) = 4 (1.9 )( 6750 ) = 25 650 lbf ⋅ in = 25.65 kip ⋅ in 2 2 B = 3 ( K fsTm ) = 3 (1.6 )( 2819 ) = 7812 lbf ⋅ in = 7.812 kip ⋅ in 2 1 8A = n π d 3 Se 2 2 1/2 2 BS e 1 + 1 + AS ut 2 1/2 2 ( 7.812 )( 23.8 ) = 1 + 1 + π (1.753 ) ( 23.8 ) ( 25.65 )( 64 ) 8 ( 25.65 ) Shigley’s MED, 10th edition Chapter 7 Solutions, Page 18/45 www.konkur.in n = 0.96 Shoulder to the right of C r / d = 0.0625 / 1.3 = 0.048, D / d = 1.75 / 1.3 = 1.35 Fig. A-15-9: Fig. A-15-8: Fig. 6-20: Fig. 6-21: Eq. (6-32): Kt = 2.0 Kts = 1.7 q = 0.71 qs = 0.76 K f = 1 + q ( K t − 1) = 1 + 0.71(2.0 − 1) = 1.7 K fs = 1 + qs ( K ts − 1) = 1 + 0.76(1.7 − 1) = 1.5 −0.107 1.3 Eq. (6-20): kb = = 0.855 0.30 Eq. (6-18): Se = 0.897(0.855)(32) = 24.5 kpsi For convenience, we will use the full value of the bending moment at C, even though it will be slightly less at the shoulder. Using Eq. (7-9) with Mm = Ta = 0, A = 4 ( K f M a ) = 4 (1.7 )( 6750 ) = 22 950 lbf ⋅ in = 22.95 kip ⋅ in 2 2 B = 3 ( K fsTm ) = 3 (1.5 )( 2819 ) = 7324 lbf ⋅ in = 7.324 kip ⋅ in 2 1 8A = n π d 3 Se 2 2 1/2 2 BSe 1 + 1 + ASut 2 1/2 2 ( 7.324 )( 24.5 ) = 1 + 1 + π (1.33 ) ( 24.5 ) ( 22.95 )( 64 ) n = 0.45 8 ( 22.95 ) The critical location is at the shoulder to the right of C, where n = 0.45 and finite life is predicted. Ans. Though not explicitly called for in the problem statement, a static check for yielding is especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with Mm = Ta = 0, 1/ 2 ′ σ max 2 32 K f M a 2 16 K fsTm = + 3 3 3 π d π d 1/ 2 2 2 16 (1.5 )( 2819 ) 32 1.7 6750 ( )( ) + 3 = 3 3 π π 1.3 1.3 ( ) ( ) Shigley’s MED, 10th edition = 55 845 psi = 55.8 kpsi Chapter 7 Solutions, Page 19/45 www.konkur.in n= Sy 54 = = 0.97 ′ σ max 55.8 This indicates localized yielding is predicted at the stress-concentration, though after localized cold-working it may not be a problem. The finite fatigue life is still likely to be the failure mode that will dictate whether this shaft is acceptable. It is interesting to note the impact of stress concentration on the acceptability of the proposed design. This problem is linked with several previous problems (see Table 1-2, p. 34) in which the shaft was considered to have a constant diameter of 1.25 in. In each of the previous problems, the 1.25 in diameter was more than adequate for deflection, static, and fatigue considerations. In this problem, even though practically the entire shaft has diameters larger than 1.25 in, the stress concentrations significantly reduce the anticipated fatigue life. ______________________________________________________________________________ 7-20 For a shaft with significantly varying diameters over its length, we will choose to use shaft analysis software or finite element software to calculate the deflections. Entering the geometry from the shaft as defined in Prob. 7-19, and the loading as defined in Prob. 3-72, the following deflection magnitudes are determined: Location Left bearing O Right bearing C Left Gear A Right Gear B Slope Deflection (rad) (in) 0.00640 0.00000 0.00434 0.00000 0.00260 0.04839 0.01078 0.07517 Comparing these values to the recommended limits in Table 7-2, we find that they are all out of the desired range. This is not unexpected since the stress analysis of Prob. 7-19 also indicated the shaft is undersized for infinite life. The slope at the right gear is the most excessive, so we will attempt to increase all diameters to bring it into compliance. Using Eq. (7-18) at the right gear, d new nd ( dy / dx )old = d old ( slope )all 1/4 (1)(0.01078) = 0.0005 1/4 = 2.15 Multiplying all diameters by 2.15, we obtain the following deflections: Location Left bearing O Right bearing C Left Gear A Right Gear B Shigley’s MED, 10th edition Slope (rad) 0.00030 0.00020 0.00012 0.00050 Deflection (in) 0.00000 0.00000 0.00225 0.00350 Chapter 7 Solutions, Page 20/45 www.konkur.in This brings the slope at the right gear just to the limit for an uncrowned gear, and all other slopes well below the recommended limits. For the gear deflections, the values are below recommended limits as long as the diametral pitch is less than 20. ______________________________________________________________________________ 7-21 The most likely critical locations for fatigue are at locations where the bending moment is high, the cross section is small, stress concentration exists, and torque exists. The twoplane bending moment diagrams, shown in the solution to Prob. 3-73, indicate both planes have a maximum bending moment at B. At this location, the combined bending moment from both planes is M = 4097 N∙m, and the torque is T = 3101 N∙m. The shoulder to the right of B will be eliminated since its diameter is only slightly smaller, and there is no torque. Comparing the shoulder to the left of B with the keyway at B, the primary difference between the two is the stress concentration, since they both have essentially the same bending moment, torque, and size. We will check the stress concentration factors for both to determine which is critical. Table A-20: Sut = 440 MPa, Sy = 370 MPa Keyway at A Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 365), with d = 50 mm, r = 0.02d = 1 mm. Table 7-1: Fig. 6-20: Fig. 6-21: Eq. (6-32): Kt = 2.14, Kts = 3.0 q = 0.66 qs = 0.72 K f = 1 + q ( K t − 1) = 1 + 0.66(2.14 − 1) = 1.8 K fs = 1 + qs ( K ts − 1) = 1 + 0.72(3.0 − 1) = 2.4 Shoulder to the left of B r / d = 2 / 50 = 0.04, D / d = 75 / 50 = 1.5 Fig. A-15-9: Kt = 2.2 Shigley’s MED, 10th edition Chapter 7 Solutions, Page 21/45 www.konkur.in Fig. A-15-8: Fig. 6-20: Fig. 6-21: Eq. (6-32): Kts = 1.8 q = 0.73 qs = 0.78 K f = 1 + q ( K t − 1) = 1 + 0.73(2.2 − 1) = 1.9 K fs = 1 + qs ( K ts − 1) = 1 + 0.78(1.8 − 1) = 1.6 Examination of the stress concentration factors indicates the keyway will be the critical location. Eq. (6-8): Se′ = 0.5(440) = 220 MPa Eq. (6-19): ka = 4.51(440) −0.265 = 0.899 −0.107 Eq. (6-20): Eq. (6-18): 50 kb = = 0.818 7.62 kc = k d = ke = 1 Se = 0.899(0.818)(220) = 162 MPa We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. Using Eq. (7-9) with Mm = Ta = 0, A = 4 ( K f M a ) = 4 (1.8 )( 4097 ) = 14 750 N ⋅ m 2 2 B = 3 ( K fsTm ) = 3 ( 2.4 )( 3101) = 12 890 N ⋅ m 2 2 2 1/2 1 8 A 2 BSe = 1 + 1 + n π d 3 Se ASut 2 1/2 6 2 12 890 162 10 ( )( ) ( ) 8 (14 750 ) = 1 + 1 + 3 6 6 π ( 0.050 ) (162 ) (10 ) (14 750 )( 440 ) (10 ) n = 0.25 Infinite life is not predicted. Ans. Though not explicitly called for in the problem statement, a static check for yielding is especially warranted with such a low fatigue factor of safety. Using Eq. (7-15), with Mm = Ta = 0, 1/2 ′ σ max 2 32 K f M a 2 16 K fsTm = + 3 3 3 π d π d 1/ 2 2 32 1.8 4097 2 16 ( 2.4 )( 3101) ( )( ) + 3 = π ( 0.050 )3 π ( 0.050 )3 Shigley’s MED, 10th edition ( ) = 7.98 108 Pa = 798 MPa Chapter 7 Solutions, Page 22/45 www.konkur.in n= Sy 370 = = 0.46 ′ σ max 798 This indicates localized yielding is predicted at the stress-concentration. Even without the stress concentration effects, the static factor of safety turns out to be 0.93. Static failure is predicted, rendering this proposed shaft design unacceptable. This problem is linked with several previous problems (see Table 1-2, p. 34) in which the shaft was considered to have a constant diameter of 50 mm. The results here are consistent with the previous problems, in which the 50 mm diameter was found to slightly undersized for static, and significantly undersized for fatigue. Though in the current problem much of the shaft has larger than 50 mm diameter, the added contribution of stress concentration limits the fatigue life. ______________________________________________________________________________ 7-22 For a shaft with significantly varying diameters over its length, we will choose to use shaft analysis software or finite element software to calculate the deflections. Entering the geometry from the shaft as defined in Prob. 7-21, and the loading as defined in Prob. 3-73, the following deflection magnitudes are determined: Location Slope (rad) 0.01445 0.01843 0.00358 0.00366 Left bearing O Right bearing C Left Gear A Right Gear B Deflection (mm) 0.000 0.000 3.761 3.676 Comparing these values to the recommended limits in Table 7-2, we find that they are all well out of the desired range. This is not unexpected since the stress analysis in Prob. 7-21 also indicated the shaft is undersized for infinite life. The transverse deflection at the left gear is the most excessive, so we will attempt to increase all diameters to bring it into compliance. Using Eq. (7-17) at the left gear, assuming from Table 7-2 an allowable deflection of yall = 0.01 in = 0.254 mm, d new nd yold = d old yall 1/4 (1)(3.761) = 0.254 1/4 = 1.96 Multiplying all diameters by 2, we obtain the following deflections: Location Left bearing O Right bearing C Left Gear A Right Gear B Shigley’s MED, 10th edition Slope (rad) 0.00090 0.00115 0.00022 0.00023 Deflection (mm) 0.000 0.000 0.235 0.230 Chapter 7 Solutions, Page 23/45 www.konkur.in This brings the deflection at the gears just within the limit for a spur gear (assuming P < 10 teeth/in), and all other deflections well below the recommended limits. ______________________________________________________________________________ 7-23 (a) Label the approximate locations of the effective centers of the bearings as A and B, the fan as C, and the gear as D, with axial dimensions as shown. Since there is only one gear, we can combine the radial and tangential gear forces into a single resultant force with an accompanying torque, and handle the statics problem in a single plane. From statics, the resultant reactions at the bearings can be found to be RA = 209.9 lbf and RB = 464.5 lbf. The bending moment and torque diagrams are shown, with the maximum bending moment at D of MD = 209.9(6.98) = 1459 lbf∙in and a torque transmitted from D to C of T = 633 (8/2) = 2532 lbf∙in. Due to the shaft rotation, the bending stress on any stress element will be completely reversed, while the torsional stress will be steady. Since we do not have any information about the fan, we will ignore any axial load that it would introduce. It would not likely contribute much compared to the bending anyway. Potentially critical locations are identified as follows: • Keyway at C, where the torque is high, the diameter is small, and the keyway creates a stress concentration. Shigley’s MED, 10th edition Chapter 7 Solutions, Page 24/45 www.konkur.in • • • • Keyway at D, where the bending moment is maximum, the torque is high, and the keyway creates a stress concentration. Groove at E, where the diameter is smaller than at D, the bending moment is still high, and the groove creates a stress concentration. There is no torque here, though. Shoulder at F, where the diameter is smaller than at D or E, the bending moment is still moderate, and the shoulder creates a stress concentration. There is no torque here, though. The shoulder to the left of D can be eliminated since the change in diameter is very slight, so that the stress concentration will undoubtedly be much less than at D. Table A-20: Eq. (6-8): Sut = 68 kpsi, Sy = 57 kpsi Se′ = 0.5(68) = 34.0 kpsi Eq. (6-19): ka = 2.70(68) −0.265 = 0.883 Keyway at C Since there is only steady torsion here, only a static check needs to be performed. We’ll use the maximum shear stress theory. τ= Eq. (5-3): Tr 2532 (1.00 / 2 ) = = 12.9 kpsi J π (1.004 ) / 32 ny = Sy / 2 τ = 57 / 2 = 2.21 12.9 Keyway at D Assuming r / d = 0.02 for typical end-milled keyway cutter (p. 365), with d = 1.75 in, r = 0.02d = 0.035 in. Table 7-1: Fig. 6-20: Fig. 6-21: Eq. (6-32): Kt = 2.14, Kts = 3.0 q = 0.66 qs = 0.72 K f = 1 + q ( K t − 1) = 1 + 0.66(2.14 − 1) = 1.8 K fs = 1 + qs ( K ts − 1) = 1 + 0.72(3.0 − 1) = 2.4 −0.107 Eq. (6-20): Eq. (6-18): 1.75 kb = = 0.828 0.30 Se = 0.883(0.828)(34.0) = 24.9 kpsi We will choose the DE-Gerber criteria since this is an analysis problem in which we would like to evaluate typical expectations. Using Eq. (7-9) with Mm = Ta = 0, Shigley’s MED, 10th edition Chapter 7 Solutions, Page 25/45 www.konkur.in A = 4 ( K f M a ) = 4 (1.8 )(1459 ) = 5252 lbf ⋅ in = 5.252 kip ⋅ in 2 2 B = 3 ( K fsTm ) = 3 ( 2.4 )( 2532 ) = 10 525 lbf ⋅ in = 10.53 kip ⋅ in 2 1 8A = n π d 3 Se 2 2 1/2 2 BSe 1 + 1 + ASut 2 1/ 2 2 (10.53)( 24.9 ) = 1 + 1 + π (1.753 ) ( 24.9 ) ( 5.252 )( 68 ) n = 3.59 Ans. 8 ( 5.252 ) Groove at E We will assume Figs. A-15-14 is applicable since the 2 in diameter to the right of the groove is relatively narrow and will likely not allow the stress flow to fully develop. (See Fig.7-9 for the stress flow concept.) r / d = 0.1 / 1.55 = 0.065, D / d = 1.75 / 1.55 = 1.13 Fig. A-15-14: Kt = 2.1 Fig. 6-20: q = 0.76 Eq. (6-32): K f = 1 + q ( K t − 1) = 1 + 0.76(2.1 − 1) = 1.8 −0.107 Eq. (6-20): Eq. (6-18): 1.55 kb = = 0.839 0.30 Se = 0.883(0.839)(34) = 25.2 kpsi Using Eq. (7-9) with Mm = Ta = Tm = 0, A = 4 ( K f M a ) = 4 (1.8 )(1115 ) = 4122 lbf ⋅ in = 4.122 kip ⋅ in 2 2 B=0 2 1/2 2 BS e 1 + 1 + AS ut 8 ( 4.122 ) 2 1/2 = 1 + 1 + ( 0 ) 3 π (1.55 ) ( 25.2 ) 1 8A = n π d 3 Se { n = 4.47 } Ans. Shoulder at F Fig. A-15-9: r / d = 0.125 / 1.40 = 0.089, Kt = 1.7 Shigley’s MED, 10th edition D / d = 2.0 / 1.40 = 1.43 Chapter 7 Solutions, Page 26/45 www.konkur.in Fig. 6-20: Eq. (6-32): q = 0.78 K f = 1 + q ( K t − 1) = 1 + 0.78(1.7 − 1) = 1.5 −0.107 Eq. (6-20): Eq. (6-18): 1.40 kb = = 0.848 0.30 Se = 0.883(0.848)(34) = 25.5 kpsi Using Eq. (7-9) with Mm = Ta = Tm = 0, A = 4 ( K f M a ) = 4 (1.5 )( 845 ) = 2535 lbf ⋅ in = 2.535 kip ⋅ in 2 2 B=0 2 1/2 1 8 A 2 BSe = 1 + 1 + n π d 3 Se ASut 1/2 8 ( 2.535 ) 1 + ( 0 ) 2 = 1 + π (1.403 ) ( 25.5 ) { n = 5.42 } Ans. (b) The deflection will not be much affected by the details of fillet radii, grooves, and keyways, so these can be ignored. Also, the slight diameter changes, as well as the narrow 2.0 in diameter section, can be neglected. We will model the shaft with the following three sections: Section Diameter Length (in) (in) 1 1.00 2.90 2 1.70 7.77 3 1.40 2.20 The deflection problem can readily (though tediously) be solved with singularity functions. For examples, see Ex. 4-7, p. 173, or the solution to Prob. 7-24. Alternatively, shaft analysis software or finite element software may be used. Using any of the methods, the results should be as follows: Location Left bearing A Right bearing B Fan C Gear D Shigley’s MED, 10th edition Slope Deflection (rad) (in) 0.000290 0.000000 0.000400 0.000000 0.000290 0.000404 0.000146 0.000928 Chapter 7 Solutions, Page 27/45 www.konkur.in Comparing these values to the recommended limits in Table 7-2, we find that they are all within the recommended range. ______________________________________________________________________________ 7-24 Shaft analysis software or finite element software can be utilized if available. Here we will demonstrate how the problem can be simplified and solved using singularity functions. Deflection: First we will ignore the steps near the bearings where the bending moments are low. Thus let the 30 mm dia. be 35 mm. Secondly, the 55 mm dia. is very thin, 10 mm. The full bending stresses will not develop at the outer fibers so full stiffness will not develop either. Thus, ignore this step and let the diameter be 45 mm. Statics: Left support: R1 = 7(315 − 140) / 315 = 3.889 kN Right support: R2 = 7(140) / 315 = 3.111 kN Determine the bending moment at each step. x(mm) 0 40 100 140 210 275 315 M(N · m) 0 155.56 388.89 544.44 326.67 124.44 0 I35 = (π/64)(0.0354) = 7.366(10-8) m4, I40 = 1.257(10-7) m4, I45 = 2.013(10-7) m4 Plot M/I as a function of x. x(m) 0 0.04 0.04 0.1 0.1 0.14 0.14 0.21 0.21 0.275 0.275 0.315 Shigley’s MED, 10th edition M/I (109 N/m3) 0 2.112 1.2375 3.094 1.932 2.705 2.705 1.623 2.6 0.99 1.6894 0 Step Slope 52.8 ∆Slope –0.8745 30.942 –21.86 –1.162 19.325 –11.617 0 –15.457 –34.78 0.977 -24.769 -9.312 0.6994 -42.235 -17.47 Chapter 7 Solutions, Page 28/45 www.konkur.in The steps and the change of slopes are evaluated in the table. From these, the function M/I can be generated: 0 1 0 M / I = 52.8 x − 0.8745 x − 0.04 − 21.86 x − 0.04 − 1.162 x − 0.1 1 1 0 −11.617 x − 0.1 − 34.78 x − 0.14 + 0.977 x − 0.21 −9.312 x − 0.21 + 0.6994 x − 0.275 1 0 1 − 17.47 x − 0.275 109 Integrate twice: E dy 1 2 1 = 26.4 x 2 − 0.8745 x − 0.04 − 10.93 x − 0.04 − 1.162 x − 0.1 dx 2 2 1 −5.81 x − 0.1 − 17.39 x − 0.14 + 0.977 x − 0.21 + C1 109 (1) 2 − 0.581 x − 0.1 −4.655 x − 0.21 + 0.6994 x − 0.275 − 8.735 x − 0.275 2 1 2 3 Ey = 8.8x3 − 0.4373 x − 0.04 − 3.643 x − 0.04 3 3 −1.937 x − 0.1 − 5.797 x − 0.14 + 0.4885 x − 0.21 −1.552 x − 0.21 + 0.3497 x − 0.275 3 2 2 2 − 2.912 x − 0.275 3 + C1x + C2 109 Boundary conditions: y = 0 at x = 0 yields C2 = 0; y = 0 at x = 0.315 m yields C1 = –0.295 25 N/m2. Equation (1) with C1 = –0.295 25 provides the slopes at the bearings and gear. The following table gives the results in the second column. The third column gives the results from a similar finite element model. The fourth column gives the results of a full model which models the 35 and 55 mm diameter steps. x (mm) 0 140 315 Shigley’s MED, 10th edition θ (rad) –0.001 4260 –0.000 1466 0.001 3120 F.E. Model –0.001 4270 –0.000 1467 0.001 3280 Full F.E. Model –0.001 4160 –0.000 1646 0.001 3150 Chapter 7 Solutions, Page 29/45 www.konkur.in The main discrepancy between the results is at the gear location (x = 140 mm). The larger value in the full model is caused by the stiffer 55 mm diameter step. As was stated earlier, this step is not as stiff as modeling implicates, so the exact answer is somewhere between the full model and the simplified model which in any event is a small value. As expected, modeling the 30 mm dia. as 35 mm does not affect the results much. It can be seen that the allowable slopes at the bearings are exceeded. Thus, either the load has to be reduced or the shaft “beefed” up. If the allowable slope is 0.001 rad, then the maximum load should be Fmax = (0.001/0.001 426)7 = 4.91 kN. With a design factor this would be reduced further. To increase the stiffness of the shaft, apply Eq. (7-18) to the most offending deflection (at x = 0) to determine a multiplier to be used for all diameters. d new nd ( dy / dx )old = d old ( slope )all 1/4 = (1)(0.0014260) 0.001 1/4 = 1.093 Form a table: Old d, mm 20.00 30.00 35.00 40.00 45.00 55.00 New ideal d, mm 21.86 32.79 38.26 43.72 49.19 60.12 Rounded up d, mm 22.00 34.00 40.00 44.00 50.00 62.00 Repeating the full finite element model results in x = 0: θ = –9.30 × 10-4 rad x = 140 mm: θ = –1.09 × 10-4 rad x = 315 mm: θ = 8.65 × 10-4 rad This is well within our goal. Have the students try a goal of 0.0005 rad at the gears. Strength: Due to stress concentrations and reduced shaft diameters, there are a number of locations to look at. A table of nominal stresses is given below. Note that torsion is only to the right of the 7 kN load. Using σ = 32M/(πd3) and τ = 16T/(πd3), x (mm) σ (MPa) τ (MPa) σ ′(MPa) 0 15 40 100 110 140 210 275 300 330 0 22.0 37.0 61.9 47.8 60.9 52.0 39.6 17.6 0 0 0 0 0 0 6 8.5 12.7 20.2 68.1 0 22.0 37.0 61.9 47.8 61.8 53.1 45.3 39.2 118.0 Table A-20 for AISI 1020 CD steel: Sut = 470 MPa, Sy = 390 MPa At x = 210 mm: Eq. (6-19): ka = 4.51(470) −0.265 = 0.883 Shigley’s MED, 10th edition Chapter 7 Solutions, Page 30/45 www.konkur.in Eq. (6-20): Eq. (6-18): Fig. A-15-8: Fig. A-15-9: Fig. 6-20: Fig. 6-21: Eq. (6-32): kb = (40 / 7.62) −0.107 = 0.837 Se = 0.883 (0.837)(0.5)(470) = 174 MPa D / d = 45 / 40 = 1.125, r / d = 2 / 40 = 0.05 Kts = 1.4 Kt = 1.9 q = 0.75 qs = 0.79 Kf = 1 + 0.75(1.9 –1) = 1.68 Kf s = 1 + 0.79(1.4 – 1) = 1.32 Choosing DE-ASME Elliptic to inherently include the yield check, from Eq. (7-11), with Mm = Ta = 0, 1/2 2 2 1.32(107) 1 16 1.68(326.67) +3 = 4 6 n π 0.043 174 106 390 10 n = 1.98 ( ) ( ) ( ) At x = 330 mm: The von Mises stress is the highest but it comes from the steady torque only. Fig. A-15-9: Fig. 6-21: Eq. (6-32): Eq. (7-11): D / d = 30 / 20 = 1.5, r / d = 2 / 20 = 0.1 Kts = 1.42 qs = 0.79 Kf s = 1 + 0.79(1.42 – 1) = 1.33 1 16 = n π 0.023 ( () 3 ) 1.33(107) 390 (10 ) 6 n = 2.49 Note that since there is only a steady torque, Eq. (7-11) reduces to essentially the equivalent of the distortion energy failure theory. Check the other locations. If worse-case is at x = 210 mm, the changes discussed for the slope criterion will improve the strength issue. ______________________________________________________________________________ 7-25 and 7-26 With these design tasks each student will travel different paths and almost all details will differ. The important points are • The student gets a blank piece of paper, a statement of function, and some constraints – explicit and implied. At this point in the course, this is a good experience. • It is a good preparation for the capstone design course. Shigley’s MED, 10th edition Chapter 7 Solutions, Page 31/45 www.konkur.in • The adequacy of their design must be demonstrated and possibly include a designer’s notebook. • Many of the fundaments of the course, based on this text and this course, are useful. The student will find them useful and notice that he/she is doing it. • Don’t let the students create a time sink for themselves. Tell them how far you want them to go. ______________________________________________________________________________ 7-27 This task was once given as a final exam problem. This problem is a learning experience. Following the task statement, the following guidance was added. • • • Take the first half hour, resisting the temptation of putting pencil to paper, and decide what the problem really is. Take another twenty minutes to list several possible remedies. Pick one, and show your instructor how you would implement it. The students’ initial reaction is that he/she does not know much from the problem statement. Then, slowly the realization sets in that they do know some important things that the designer did not. They knew how it failed, where it failed, and that the design wasn’t good enough; it was close, though. Also, a fix at the bearing seat lead-in could transfer the problem to the shoulder fillet, and the problem may not be solved. To many students’ credit, they chose to keep the shaft geometry, and selected a new material to realize about twice the Brinell hardness. ______________________________________________________________________________ 7-28 In Eq. (7-22) set πd4 πd2 I= , A= 64 4 to obtain π d gE ω = l 4 γ 2 (1) or d= 4l 2ω γ π gE 2 (2) (a) From Eq. (1) and Table A-5 9 π 0.025 9.81(207)(10 ) = 883 rad/s 76.5 103 0.6 4 2 ω = Shigley’s MED, 10th edition ( ) Ans. Chapter 7 Solutions, Page 32/45 www.konkur.in (b) From Eq. (1), we observe that the critical speed is linearly proportional to the diameter. Thus, to double the critical speed, we should double the diameter to d = 50 mm. Ans. (c) From Eq. (2), lω = π2 d gE 4 l γ Since d / l is the same regardless of the scale, lω = constant = 0.6(883) = 529.8 529.8 = 1766 rad/s Ans. ω= 0.3 Thus the first critical speed doubles. ______________________________________________________________________________ 7-29 From Prob. 7-28, ω = 883 rad/s A = 4.909 (10 −4 ) m 2 , I = 1.917 (10 −8 ) m 4 , E = 207(109 ) Pa, ( ) γ = 7.65 (104 ) N/m 3 ( ) w = Aγ l = 4.909 10 −4 7.65 10 4 (0.6) = 22.53 N One element: Eq. (7-24): ( 0.3(0.3) 0.62 − 0.32 − 0.32 ) ( ) ( ) (0.6) y = w δ = 22.53(1.134) (10 ) = 2.555 (10 ) m y = 6.528 (10 ) Σw y = 22.53(2.555) (10 ) = 5.756 (10 ) Σw y = 22.53(6.528) (10 ) = 1.471(10 ) δ11 = ( ) 9 6(207) 10 (1.917) 10 = 1.134 10−6 m/N −8 −6 1 −5 1 11 −10 2 1 −5 −10 2 ω1 = g −4 −8 ( ( ) ) 5.756 10−4 Σw y = 9.81 = 620 rad/s Σw y 2 1.471 10−8 (30% low) Two elements: Shigley’s MED, 10th edition Chapter 7 Solutions, Page 33/45 www.konkur.in δ11 = δ 22 = δ12 = δ 21 = ( ) = 6.379 10 m/N ( ) (0.6) 0.45(0.15) 0.62 − 0.452 − 0.152 ( ) ( 9 6(207) 10 (1.917) 10 −8 ) −7 0.15(0.15)(0.62 − 0.152 − 0.152 ) = 4.961 10−7 m/N 9 −8 6(207) 10 (1.917) 10 (0.6) ( ) ( ( ) ( ) ( ) ) ( ) ( ) y1 = y2 = w1δ11 + w2δ12 = 11.265(6.379) 10 −7 + 11.265(4.961) 10 −7 = 1.277 10−5 m ( y = y = 1.632 10 2 1 2 2 −10 ) ( ) Σw y = 2(11.265)(1.277) 10 −5 = 2.877 10 −4 ( ) ( Σw y 2 = 2(11.265)(1.632) 10 −10 = 3.677 10 −9 ( ( 2.877 10−4 ω1 = 9.81 3.677 10−9 ) = 876 rad/s ) ) (0.8% low) Three elements: δ11 = δ 33 = ( 0.5(0.1) 0.62 − 0.52 − 0.12 ( ) ( 9 6(207) 10 (1.917) 10 ( 0.3(0.3) 0.62 − 0.32 − 0.32 δ 22 = ( ) ( 9 6(207) 10 (1.917) 10 δ12 = δ 32 = ( −8 −8 ) ) (0.6) ) (0.6) ( ) ( 6(207) 10 (1.917) 10 ( 0.1(0.1) 0.62 − 0.12 − 0.12 −8 ) ( ) = 3.500 10 −7 m/N ( ) = 1.134 10−6 m/N 0.3(0.1) 0.62 − 0.32 − 0.12 9 ) ) ) (0.6) ( ) = 5.460 10−7 m/N ( ) ( ) ( ) y = 7.51 3.500 (10 ) + 5.460 (10 ) + 2.380 (10 ) = 8.516 (10 ) y = 7.51 5.460 (10 ) + 1.134 (10 ) + 5.460 (10 ) = 1.672 (10 ) y = 7.51 2.380 (10 ) + 5.460 (10 ) + 3.500 (10 ) = 8.516 (10 ) Σw y = 7.51 8.516 (10 ) + 1.672 (10 ) + 8.516 (10 ) = 2.535 (10 ) Σw y = 7.51{8.516 (10 ) + 1.672 (10 ) + 8.516 (10 ) } = 3.189 (10 ) δ13 = 9 6(207) 10 (1.917) 10 −8 (0.6) = 2.380 10−7 m/N −7 −7 −7 −6 −7 −6 −7 −5 −7 −7 −7 −6 1 2 3 −6 2 Shigley’s MED, 10th edition −5 −6 2 −6 −5 2 −4 −6 2 −9 Chapter 7 Solutions, Page 34/45 www.konkur.in ( ( 2.535 10−4 ω1 = 9.81 3.189 10−9 ) = 883 rad/s ) The result is the same as in Prob. 7-28. The point was to show that convergence is rapid using a static deflection beam equation. The method works because: • If a deflection curve is chosen which meets the boundary conditions of momentfree and deflection-free ends, as in this problem, the strain energy is not very sensitive to the equation used. • Since the static bending equation is available, and meets the moment-free and deflection-free ends, it works. ______________________________________________________________________________ 7-30 (a) For two bodies, Eq. (7-26) is (m1δ11 − 1/ ω 2 ) m2δ12 m1δ 21 (m2δ 22 − 1/ ω 2 ) =0 Expanding the determinant yields, 2 1 1 2 − (m1δ11 + m2δ 22 ) 2 + m1m2 (δ11δ 22 − δ12δ 21 ) = 0 ω ω1 (1) Eq. (1) has two roots 1 / ω12 and 1 / ω22 . Thus 1 1 1 1 2 − 2 2 − 2 = 0 ω ω1 ω ω2 or, 2 2 1 1 1 1 1 1 2 + 2 + 2 + 2 2 = 0 ω ω1 ω2 ω ω1 ω2 (2) Equate the third terms of Eqs. (1) and (2), which must be identical. 1 1 ω ω 2 1 2 2 = m1m2 (δ11δ 22 − δ12δ 21 ) ⇒ 1 ω22 = ω12 m1m2 (δ11δ 22 − δ12δ 21 ) and it follows that Shigley’s MED, 10th edition Chapter 7 Solutions, Page 35/45 www.konkur.in ω2 = g2 1 ω1 w1w2 (δ11δ 22 − δ12δ 21 ) Ans. (b) In Ex. 7-5, part (b), the first critical speed of the two-disk shaft (w1 = 35 lbf, w 2 = 55 lbf) is ω1 = 124.8 rad/s. From part (a), using influence coefficients, 1 3862 = 466 rad/s Ans. 124.8 35(55) 2.061(3.534) − 2.2342 10−8 ______________________________________________________________________________ ω2 = 7-31 ( ) In Eq. (7-22), for ω1, the term I / A appears. For a hollow uniform diameter shaft, ω1 ∝ ( ( ) ) ( )( ) 2 2 2 2 π do4 − di4 / 64 I 1 do + di do − di 1 = = = d o2 + di2 2 2 2 2 A 16 d − d 4 π d o − di / 4 o i This means that when a solid shaft is hollowed out, the critical speed increases beyond that of the solid shaft of the same size. Ans. By how much? (1/ 4) d o2 + di2 (1/ 4) d o2 d = 1+ i do 2 The possible values of di are 0 ≤ di ≤ d o , so the range of the critical speeds is ω1 1 + 0 to about ω1 1 + 1 or from ω1 to 2 ω1 . For the specific case where the inner diameter is half of the outer diameter, the ratio of the critical speeds is 2 1 do di Ans. 1 + = 1 + 2 = 1.12 d d o o ______________________________________________________________________________ 2 7-32 All steps will be modeled using singularity functions with a spreadsheet. Programming Shigley’s MED, 10th edition Chapter 7 Solutions, Page 36/45 www.konkur.in both loads will enable the user to first set the left load to 1, the right load to 0 and calculate δ11 and δ21. Then set the left load to 0 and the right to 1 to get δ12 and δ22. The spreadsheet shows the δ11 and δ21 calculation. A table for M / I vs. x is easy to make. First, draw the bending-moment diagram as shown with the data. x M 0 0 1 2 3 0.875 1.75 1.625 4 1.5 5 6 7 1.375 1.25 1.125 8 1 M (lbf-in) 2 1.5 1 0.5 0 0 2 4 6 8 10 12 14 16 x (in) x M 9 10 11 0.875 0.75 0.625 12 0.5 13 14 15 0.375 0.25 0.125 16 0 The second-area moments are: 0 ≤ x ≤ 1 in and 15 ≤ x ≤ 16 in, I1 = π ( 2 4 ) / 64 = 0.7854 in 4 1 ≤ x ≤ 9 in , I 2 = π ( 2.4724 ) / 64 = 1.833 in 4 9 ≤ x ≤ 15 in , I 3 = π ( 2.7634 ) / 64 = 2.861 in 4 Divide M by I at the key points x = 0, 1, 2, 9, 14, 15, and 16 in and plot x M/I x M/I 0 0 1 1 2 2 3 4 5 6 7 8 1.1141 0.4774 0.9547 0.9547 0.8865 0.8183 0.7501 0.6819 0.6137 0.5456 9 9 10 11 12 13 14 14 15 15 0.4774 0.3058 0.2621 0.2185 0.1748 0.1311 0.0874 0.0874 0.0437 0.1592 Shigley’s MED, 10th edition 16 0 Chapter 7 Solutions, Page 37/45 www.konkur.in 1.2 M/I (lbf/in3) 1 0.8 0.6 0.4 0.2 0 0 2 4 6 8 10 12 14 16 x (in) From this diagram, one can see where changes in value (steps) and slope occur. Using a spreadsheet, one can form a table of these changes. An example of a step is, at x = 1 in, M/I goes from 0.875/0.7854 = 1.1141 lbf/in3 to 0.875/1.833 = 0.4774 lbf/in3, a step change of 0.4774 − 1.1141 = − 0.6367 lbf/in3. A slope change also occurs at at x = 1 in. The slope for 0 ≤ x ≤ 1 in is 1.1141/1 = 1.1141 lbf/in2, which changes to (0.9547 − 0.4774)/1 = 0.4774 lbf/in2, a change of 0.4774 − 1.1141 = − 0.6367 lbf/in2. Following this approach, a table is made of all the changes. The table shown indicates the column letters and row numbers for the spreadsheet. A x 1a 1b 2 2 9a 9b 14 14 15a 15b 16 1 2 3 4 5 6 7 8 9 10 11 12 B M 0.875 0.875 1.75 1.75 0.875 0.875 0.25 0.25 0.125 0.125 0 C M/I 1.114085 0.477358 0.954716 0.954716 0.477358 0.305854 0.087387 0.087387 0.043693 0.159155 0.000000 D step 0.000000 -0.636727 0.000000 0.000000 0.000000 -0.171504 0.000000 0.000000 0.000000 0.115461 0.000000 E Slope 1.114085 0.477358 0.477358 -0.068194 -0.068194 -0.043693 -0.043693 -0.043693 -0.043693 -0.159155 -0.159155 F ∆ Slope 0.000000 -0.636727 0.000000 -0.545552 0.000000 0.024501 0.000000 0.000000 0.000000 -0.115461 0.000000 The equation for M / I in terms of the spreadsheet cell locations is: M / I = E2 ( x) + D3 x − 1 + F3 x − 1 + F5 x − 2 0 1 1 + D7 x − 9 + F7 x − 9 + D11 x − 15 + F11 x − 15 0 Shigley’s MED, 10th edition 1 0 1 Chapter 7 Solutions, Page 38/45 www.konkur.in Integrating twice gives the equation for Ey. Assume the shaft is steel. Boundary conditions y = 0 at x = 0 and at x = 16 inches provide integration constants (C1 = − 4.906 lbf/in and C2 = 0). Substitution back into the deflection equation at x = 2 and 14 in provides the δ ’s. The results are: δ11 = 2.917(10–7) and δ12 = 1.627(10–7). Repeat for F1 = 0 and F2 = 1, resulting in δ21 = 1.627(10–7) and δ22 = 2.231(10–7). This can be verified by finite element analysis. y1 y2 y12 ∑ wy = 18(2.917)(10−7 ) + 32(1.627)(10−7 ) = 1.046(10−5 ) = 18(1.627)(10−7 ) + 32(2.231)(10−7 ) = 1.007(10−5 ) = 1.093(10−10 ), y22 = 1.014(10−10 ) = 5.105(10−4 ), ∑ w y 2 = 5.212(10−9 ) Neglecting the shaft, Eq. (7-23) gives ω1 = 386 5.105(10−4 ) = 6149 rad/s or 58 720 rev/min 5.212(10−9 ) Ans. Without the loads, we will model the shaft using 2 elements, one between 0 ≤ x ≤ 9 in, and one between 0 ≤ x ≤ 16 in. As an approximation, we will place their weights at x = 9/2 = 4.5 in, and x = 9 + (16 − 9)/2 = 12.5 in. From Table A-5, the weight density of steel is γ = 0.282 lbf/in3. The weight of the left element is π π w1 = γ ∑ d 2l = 0.282 22 (1) + 2.4722 ( 8 ) = 11.7 lbf 4 4 The right element is π w2 == 0.282 2.7632 ( 6 ) + 2 2 (1) = 11.0 lbf 4 The spreadsheet can be easily modified to give δ11 = 9.605 (10−7 ) , δ12 = δ 21 = 5.718 (10−7 ) , δ 22 = 5.472 (10−7 ) y1 = 1.753 (10 −5 ) , y2 = 1.271(10 −5 ) y12 = 3.072 (10−10 ) , y22 = 1.615 (10−10 ) ∑ w y =3.449 (10 ) , ∑ w y −4 Shigley’s MED, 10th edition 2 =5.371 (10−9 ) Chapter 7 Solutions, Page 39/45 www.konkur.in 3.449 (10−4 ) = 4980 rad/s ω1 = 386 −9 5.371(10 ) A finite element model of the exact shaft gives ω1 = 5340 rad/s. The simple model is 6.8% low. Combination: Using Dunkerley’s equation, Eq. (7-32): 1 1 1 + ⇒ ω1 3870 rad/s Ans. 2 6149 49802 ω ______________________________________________________________________________ 2 1 7-33 ≈ We must not let the basis of the stress concentration factor, as presented, impose a viewpoint on the designer. Table A-16 shows Kts as a decreasing monotonic as a function of a/D. All is not what it seems. Let us change the basis for data presentation to the full section rather than the net section. τ = K tsτ 0 = K ts′τ 0′ K ts = 32T 32T = K ts′ 3 3 π AD πD Therefore K ts′ = K ts A Form a table: K ts′ has the following attributes: • It exhibits a minimum; • It changes little over a wide range; • Its minimum is a stationary point minimum at a / D B 0.100; Shigley’s MED, 10th edition Chapter 7 Solutions, Page 40/45 www.konkur.in • Our knowledge of the minima location is 0.075 ≤ ( a / D ) ≤ 0.125 We can form a design rule: In torsion, the pin diameter should be about 1/10 of the shaft diameter, for greatest shaft capacity, that is, a ≈ 0.10 D. Ans. However, it is not catastrophic if one forgets the rule. ______________________________________________________________________________ 7-34 From the solution to Prob. 3-72, the torque to be transmitted through the key from the gear to the shaft is T = 2819 lbf∙in. From Prob. 7-19, the nominal shaft diameter supporting the gear is 1.00 in. From Table 7-6, a 0.25 in square key is appropriate for a 1.00 in shaft diameter. The force applied to the key is F= T 2819 = = 5638 lbf r 1.00 / 2 Selecting 1020 CD steel for the key, with Sy = 57 kpsi, and using the distortion-energy theory, Ssy = 0.577 Sy = (0.577)(57) = 32.9 kpsi. Failure by shear across the key: F F = A tl S S n = sy = sy τ F / tl τ= ⇒ l= 1.1( 5638 ) nF = = 0.754 in tS sy 0.25 ( 32 900 ) Failure by crushing: σ= S Sy F F = n= y = σ 2 F / ( tl ) A (t / 2) l ⇒ l= 2 Fn 2 ( 5638 )(1.1) = = 0.870 in tS y 0.25 ( 57 ) (103 ) Select ¼-in square key, 7/8 in long, 1020 CD steel. Ans. ______________________________________________________________________________ 7-35 From the solution to Prob. 3-73, the torque to be transmitted through the key from the gear to the shaft is T = 3101 N∙m. From Prob. 7-21, the nominal shaft diameter supporting the gear is 50 mm. To determine an appropriate key size for the shaft diameter, we can either convert to inches and use Table 7-6, or we can look up standard metric key sizes from the internet or a machine design handbook. It turns out that the recommended metric key for a 50 mm shaft is 14 x 9 mm. Since the problem statement specifies a square key, we will use a 14 x 14 mm key. For comparison, using Table 7-6 as a guide, for d = 50 mm = 1.97 in, a 0.5 in square key is appropriate. This is equivalent to 12.7 mm. A 14 x 14 mm size is conservative, but reasonable after rounding up to standard sizes. Shigley’s MED, 10th edition Chapter 7 Solutions, Page 41/45 www.konkur.in The force applied to the key is F= T 3101 = = 124 (103 ) N r 0.050 / 2 Selecting 1020 CD steel for the key, with Sy = 390 MPa, and using the distortion-energy theory, Ssy = 0.577 Sy = 0.577(390) = 225 MPa. Failure by shear across the key: τ= F F = A tl n= S sy τ = S sy F / ( tl ) ⇒ 1.1(124 ) (103 ) nF l= = = 0.0433 m = 43.3 mm tS sy ( 0.014 )( 225 ) (106 ) Failure by crushing: n= Sy σ σ= F F = A (t / 2) l = Sy 2 F / ( tl ) ⇒ 2 (124 ) (103 ) (1.1) 2 Fn l= = = 0.0500 m = 50.0 mm tS y ( 0.014 )( 390 ) (106 ) Select 14 mm square key, 50 mm long, 1020 CD steel. Ans. ______________________________________________________________________________ 7-36 Choose basic size D = d = 15 mm. From Table 7-9, a locational clearance fit is designated as 15H7/h6. From Table A-11, the tolerance grades are ∆D = 0.018 mm and ∆d = 0.011 mm. From Table A-12, the fundamental deviation is δF = 0 mm. Hole: Eq. (7-36): Shaft: Eq. (7-37): 7-37 Dmax = D + ∆D = 15 + 0.018 = 15.018 mm Dmin = D = 15.000 mm Ans. Ans. dmax = d + δF = 15.000 + 0 = 15.000 mm Ans. dmin = d + δF – ∆d = 15.000 + 0 – 0.011 = 14.989 mm Ans. Choose basic size D = d = 1.75 in. From Table 7-9, a medium drive fit is designated as H7/s6. From Table A-13, the tolerance grades are ∆D = 0.0010 in and ∆d = 0.0006 in. From Table A-14, the fundamental deviation is δF = 0.0017 in. Hole: Eq. (7-36): Dmax = D + ∆D = 1.75 + 0.0010 = 1.7510 in Shigley’s MED, 10th edition Ans. Chapter 7 Solutions, Page 42/45 www.konkur.in Dmin = D = 1.7500 in Ans. Shaft: Eq. (7-38): dmin = d + δF = 1.75 + 0.0017 = 1.7517 in Ans. dmax = d + δF + ∆d = 1.75 + 0.0017 + 0.0006 = 1.7523 in Ans. ______________________________________________________________________________ 7-38 Choose basic size D = d = 45 mm. From Table 7-9, a sliding fit is designated as H7/g6. From Table A-11, the tolerance grades are ∆D = 0.025 mm and ∆d = 0.016 mm. From Table A-12, the fundamental deviation is δF = –0.009 mm. Hole: Eq. (7-36): Dmax = D + ∆D = 45 + 0.025 = 45.025 mm Dmin = D = 45.000 mm Ans. Ans. Shaft: Eq. (7-37): dmax = d + δF = 45.000 + (–0.009) = 44.991 mm Ans. dmin = d + δF – ∆d = 45.000 + (–0.009) – 0.016 = 44.975 mm Ans. ______________________________________________________________________________ 7-39 Choose basic size D = d = 1.250 in. From Table 7-9, a close running fit is designated as H8/f7. From Table A-13, the tolerance grades are ∆D = 0.0015 in and ∆d = 0.0010 in. From Table A-14, the fundamental deviation is δF = –0.0010 in. Hole: Eq. (7-36): Dmax = D + ∆D = 1.250 + 0.0015 = 1.2515 in Dmin = D = 1.2500 in Ans. Ans. Shaft: Eq. (7-37): dmax = d + δF = 1.250 + (–0.0010) = 1.2490 in Ans. dmin = d + δF – ∆d = 1.250 + (–0.0010) – 0.0010 = 1.2480 in Ans. ______________________________________________________________________________ 7-40 Choose basic size D = d = 35 mm. From Table 7-9, a locational interference fit is designated as H7/p6. From Table A-11, the tolerance grades are ∆D = 0.025 mm and ∆d = 0.016 mm. From Table A-12, the fundamental deviation is δF = 0.026 mm. Hole: Eq. (7-36): Dmax = D + ∆D = 35 + 0.025 = 35.025 mm Dmin = D = 35.000 mm The bearing bore specifications are within the hole specifications for a locational interference fit. Now find the necessary shaft sizes. Shaft: Eq. (7-38): dmin = d + δF = 35 + 0.026 = 35.026 mm Shigley’s MED, 10th edition Ans. Chapter 7 Solutions, Page 43/45 www.konkur.in dmax = d + δF + ∆d = 35 + 0.026 + 0.016 = 35.042 mm Ans. ______________________________________________________________________________ 7-41 Choose basic size D = d = 1.5 in. From Table 7-9, a locational interference fit is designated as H7/p6. From Table A-13, the tolerance grades are ∆D = 0.0010 in and ∆d = 0.0006 in. From Table A-14, the fundamental deviation is δF = 0.0010 in. Hole: Eq. (7-36): Dmax = D + ∆D = 1.5000 + 0.0010 = 1.5010 in Dmin = D = 1.5000 in The bearing bore specifications exactly match the requirements for a locational interference fit. Now check the shaft. Shaft: Eq. (7-38): dmin = d + δF = 1.5000 + 0.0010 = 1.5010 in dmax = d + δF + ∆d = 1.5000 + 0.0010 + 0.0006 = 1.5016 in The shaft diameter of 1.5020 in is greater than the maximum allowable diameter of 1.5016 in, and therefore does not meet the specifications for the locational interference fit. Ans. ______________________________________________________________________________ 7-42 (a) Basic size is D = d = 35 mm. Table 7-9: H7/s6 is specified for medium drive fit. Table A-11: Tolerance grades are ∆D = 0.025 mm and ∆d = 0.016 mm. Table A-12: Fundamental deviation is δ F = +0.043 mm. Eq. (7-36): Dmax = D + ∆D = 35 + 0.025 = 35.025 mm Dmin = D = 35.000 mm Eq. (7-38): dmin = d + δF = 35 + 0.043 = 35.043 mm Ans. dmax = d + δF + ∆d = 35 + 0.043 + 0.016 = 35.059 mm Ans. (b) Eq. (7-42): Eq. (7-43): Eq. (7-40): δ min = d min − Dmax = 35.043 − 35.025 = 0.018 mm δ max = d max − Dmin = 35.059 − 35.000 = 0.059 mm pmax ( ( ) ( ) Shigley’s MED, 10th edition )( ) 2 2 2 2 Eδ max d o − d d − di = 2d 3 d o2 − di2 207 109 ( 0.059 ) 602 − 352 352 − 0 = 602 − 0 2 353 ( )( ) = 115 MPa Ans. Chapter 7 Solutions, Page 44/45 www.konkur.in pmin ( )( ) 2 2 2 2 Eδ min d o − d d − di = 2d 3 d o2 − di2 207 109 ( 0.018) 602 − 352 352 − 0 = 602 − 0 2 353 ( ) ( ) ( )( ) = 35.1 MPa Ans. (c) For the shaft: Eq. (7-44): σ t ,shaft = − p = −115 MPa Eq. (7-46): σ r ,shaft = − p = −115 MPa Eq. (5-13): σ ′ = (σ 12 − σ 1σ 2 + σ 22 ) 1/2 1/ 2 = (−115) 2 − (−115)(−115) + (−115) 2 n = S y / σ ′ = 390 /115 = 3.4 = 115 MPa Ans. For the hub: 602 + 352 do2 + d 2 = 115 = 234 MPa 2 2 do2 − d 2 60 − 35 Eq. (7-45): σ t ,hub = p Eq. (7-46): σ r ,hub = − p = −115 MPa Eq. (5-13): σ ′ = (σ 12 − σ 1σ 2 + σ 22 ) 1/2 = (234) 2 − (234)( −115) + ( −115) 2 1/ 2 = 308 MPa n = S y / σ ′ = 600 / 308 = 1.9 Ans. (d) A value for the static coefficient of friction for steel to steel can be obtained online or from a physics textbook as approximately f = 0.8. Eq. (7-49) T = (π / 2) f pmin ld 2 ( ) = (π / 2)(0.8)(35.1) 106 (0.050)(0.035) 2 = 2700 N ⋅ m Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 7 Solutions, Page 45/45 www.konkur.in Chapter 8 ______________________________________________________________________________ Notes to the Instructor. 1. For Problems 8-33 33 through 88-44 and 8-51 through 8-58. There is a statement preceding Prob. 8-33 33 which covers a change which will be corrected in the second printing of the 10th edition. 2. For Probs. 8-41 to 8-44.. These problems, as well as many others in this chapter are best implemented using a spreadsheet. ______________________________________________________________________________ ____________________________________________________________________________ 8-1 (a) Thread depth= 2.55 mm Ans. Width = 2.5 mm Ans. dm = 25 - 1.25 - 1.25 25 = 22.5 22 mm dr = 25 - 5 = 20 mm l = p = 5 mm Ans. (b) Thread depth = 2.5 mm Ans. Width at pitch line = 2.5 mm Ans. dm = 22.5 mm dr = 20 mm l = p = 5 mm Ans. ______________________________________________________________________________ 8-2 From Table 8-1, d r = d − 1.226 869 p d m = d − 0.649 519 p d − 1.226 869 p + d − 0.649 519 p d = = d − 0.938 194 p 2 πd 2 π At = = (d − 0.938 194 p)2 Ans. 4 4 ______________________________________________________________________________ 8-3 From Eq. (c) of Sec. 8--2, tan λ + f PR = F 1 − f tan λ Shigley’s MED, 10th edition Chap. 8 Solutions, Page 1/70 www.konkur.in PRd m Fd m tan λ + f = 2 2 1 − f tan λ T Fl / (2π ) 1 − f tan λ 1 − f tan λ e= 0 = = tan λ TR Fd m / 2 tan λ + f tan λ + f TR = Ans. Using f = 0.08, 08, form a table and plot the efficiency curve. e λ, deg. 0 0 0 0.678 20 0.796 30 0.838 40 0.8517 45 0.8519 ______________________________________________________________________________ 8-4 Given F = 5 kN, l = 5 mm, and dm = d − p/2 = 25 − 5/2 = 22.5 mm, the torque required to raise the load is found using Eqs. (8-1) (8 and (8-6) TR = 5 ( 22.5 ) 5 + π ( 0.09 ) 22.5 5 ( 0.06 ) 45 = 15.85 N ⋅ m + π 2 22.5 − 0.09 5 2 ( ) ( ) Ans. The torque required to lower the load, from Eqs. (8-2) (8 and (8-6) is TL = 5 ( 22.5 ) π ( 0.09 ) 22.5 − 5 5 ( 0.06 ) 45 = 7.83 N ⋅ m + 2 2 π ( 22.5 ) + 0.09 ( 5 ) Ans. Since TL is positive, the thread is self self-locking. From Eq.(8-4) 4) the efficiency is 5 ( 5) = 0.251 Ans. 2π (15.85 ) ______________________________________________________________________________ e= 8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom segment of the screwss must be in compression. Whereas, Where tension specimens and their the grips must be in tension. Both screws must be of the same-hand same hand threads. ______________________________________________________________________________ Shigley’s MED, 10th edition Chap. 8 Solutions, Page 2/70 www.konkur.in 8-6 Screws rotate at an angular rate of n= 1720 = 28.67 rev/min 60 (a) The lead is 0.25 in, so the linear speed of the press head is V = 28.67(0.25) = 7.17 in/min Ans. (b) F = 2500 lbf/screw d m = 2 − 0.25 / 2 = 1.875 in sec α = 1 / cos(29o / 2) = 1.033 Eq. (8-5): TR = 2500(1.875) 0.25 + π (0.05)(1.875)(1.033) = 221.0 lbf · in 2 π (1.875) − 0.05(0.25)(1.033) Eq. (8-6): Tc = 2500(0.08)(3.5 / 2) = 350 lbf · in Ttotal = 350 + 221.0 = 571 lbf · in/screw 571(2) Tmotor = = 20.04 lbf · in 60(0.95) Tn 20.04(1720) H = = = 0.547 hp Ans. 63 025 63 025 ______________________________________________________________________________ 8-7 L = 3.5 in, T = 3.5 F, M = (L − 3/8) F = (3.5 − 0.375) F = 3.125 F, Sy = 41 kpsi. 32M 32(3.125) F = = 41 000 σ = Sy = 3 πd π (0.1875)3 F = 8.49 lbf T = 3.5(8.49) = 29.7 lbf ∙ in Ans. (b) Eq. (8-5), 2α = 60° , l = 1/10 = 0.1 in, f = 0.15, sec α = 1.155, p = 0.1 in 3 d m = − 0.649 519 ( 0.1) = 0.6850 in 4 Fclamp (0.6850) 0.1 + π (0.15)(0.6850)(1.155) TR = 2 π (0.6850) − 0.15(0.1)(1.155) TR = 0.075 86Fclamp TR 29.7 Fclamp = = = 392 lbf Ans. 0.075 86 0.075 86 Shigley’s MED, 10th edition Chap. 8 Solutions, Page 3/70 www.konkur.in (c) The column has one end fixed and the other end pivoted. Base the decision on the mean diameter column. Input: C = 1.2, D = 0.685 in, A = π(0.6852)/4 = 0.369 in2, Sy = 41 kpsi, E = 30(106) psi, L = 6 in, k = D/4 =0.171 25 in, L/k = 35.04. From Eq. (4-45), 2π 2 (1.2 ) 30 (106 ) = 131.7 = 41 000 From Eq. (4-46), the limiting clamping force for buckling is 1/2 1/2 2 l 2π CE = k 1 S y 2 Sy l 1 Fclamp = Pc r = A S y − 2π k CE 2 41(103 ) 1 3 = 0.369 41(10 ) − 35.04 = 14.6 (103 ) lbf 6 2π 1.2 ( 30 )10 Ans (d) This is a subject for class discussion. ______________________________________________________________________________ 8-8 T = 8(3.5) = 28 lbf ⋅ in dm = 3 1 − = 0.6667 in 4 12 l = 1 = 0.1667 in, 6 α = 290 = 14.50, 2 sec 14.50 = 1.033 From Eqs. (8-5) and (8-6) Ttotal = 0.6667 F 2 0.1667 + π ( 0.15 )( 0.6667 )(1.033) 0.15 (1) F = 0.1542 F + 2 π ( 0.6667 ) − 0.15 ( 0.1667 )(1.033) 28 = 182 lbf Ans. 0.1542 _____________________________________________________________________________ F= 8-9 dm = 1.5 − 0.25/2 = 1.375 in, l = 2(0.25) = 0.5 in From Eq. (8-1) and Eq. (8-6), Shigley’s MED, 10th edition Chap. 8 Solutions, Page 4/70 www.konkur.in 2.2 (103 ) (1.375) 0.5 + π (0.10)(1.375) 2.2 (103 ) (0.15)(2.25) TR = π (1.375) − 0.10(0.5) + 2 2 = 330 + 371 = 701 lbf · in Since n = V/l = 2/0.5 = 4 rev/s = 240 rev/min so the power is 701( 240 ) Tn = = 2.67 hp Ans. 63 025 63 025 ______________________________________________________________________________ H= 8-10 dm = 40 − 4 = 36 mm, l = p = 8 mm From Eqs. (8-1) and (8-6) T = = ω = H = T = F = 36 F 8 + π (0.14)(36) 0.09(100) F + 2 π (36) − 0.14(8) 2 (3.831 + 4.5) F = 8.33F N · m (F in kN) 2π n = 2π (1) = 2π rad/s Tω H 3000 = = 477 N · m 2π ω 477 = 57.3 kN Ans. 8.33 Fl 57.3(8) = = 0.153 Ans. 2π T 2π (477) ______________________________________________________________________________ e= 8-11 (a) Table A-31, nut height H = 12.8 mm. L ≥ l + H = 2(15) + 12.8 = 42.8 mm. Rounding up, L = 45 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm From Table 8-7, ld = L − LT = 45− 34 = 11 mm, lt = l − ld = 2(15) − 11 = 19 mm, Ad = π (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17) kb = 153.9 (115 ) 207 Ad At E = = 874.6 MN/m Ad lt + At ld 153.9 (19 ) + 115 (11) Ans. (c) From Eq. (8-22), with l = 2(15) = 30 mm Shigley’s MED, 10th edition Chap. 8 Solutions, Page 5/70 www.konkur.in km = 0.5774π ( 207 )14 0.5774π Ed = = 3 116.5 MN/m 0.5774l + 0.5d 0.5774 ( 30 ) + 0.5 (14 ) 2 ln 5 2 ln 5 0.5774l + 2.5d 0.5774 ( 30 ) + 2.5 (14 ) Ans. 8-12 (a) Table A-31, nut height H = 12.8 mm. Table A-33, washer thickness t = 3.5 mm. Thus, the grip is l = 2(15) + 3.5 = 33.5 mm. L ≥ l + H = 33.5 + 12.8 = 46.3 mm. Rounding up L = 50 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm From Table 8-7, ld = L − LT = 50− 34 = 16 mm, lt = l − ld = 33.5 − 16 = 17.5 mm, Ad = π (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17) kb = 153.9 (115 ) 207 Ad At E = = 808.2 MN/m Ad lt + At ld 153.9 (17.5 ) + 115 (16 ) Ans. (c) From Eq. (8-22) km = 0.5774π ( 207 )14 0.5774π Ed = = 2 969 MN/m 0.5774l + 0.5d 0.5774 ( 33.5 ) + 0.5 (14 ) 2 ln 5 2 ln 5 0.5774l + 2.5d 0.5774 ( 33.5 ) + 2.5 (14 ) Ans. ______________________________________________________________________________ 8-13 (a) Table 8-7, l = h + d /2 = 15 + 14/2 = 22 mm. L ≥ h + 1.5d = 36 mm. Rounding up L = 40 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(14) +6 = 34 mm Shigley’s MED, 10th edition Chap. 8 Solutions, Page 6/70 www.konkur.in From Table 8-7, ld = L − LT = 40− 34 = 6 mm, lt = l − ld = 22 − 6 = 16 mm Ad = π (142) / 4 = 153.9 mm2. From Table 8-1, At = 115 mm2. From Eq. (8-17) kb = 153.9 (115 ) 207 Ad At E = = 1 162.2 MN/m Ad lt + At ld 153.9 (16 ) + 115 ( 6 ) Ans. (c) From Eq. (8-22), with l = 22 mm km = 0.5774π ( 207 )14 0.5774π Ed = = 3 624.4 MN/m 0.5774l + 0.5d 0.5774 ( 22 ) + 0.5 (14 ) 2 ln 5 2 ln 5 0.5774l + 2.5d 0.5774 ( 22 ) + 2.5 (14 ) Ans. ______________________________________________________________________________ 8-14 (a) From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 2 + 1 + 7/16 = 3 7/16 in. Rounding up, L = 3.5 in Ans. (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L − LT = 3.5 − 1.25 = 2.25 in, lt = l − ld = 3 − 2.25 = 0.75 in Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) kb = 0.1963 ( 0.1419 ) 30 Ad At E = = 1.79 Mlbf/in Ad lt + At ld 0.1963 ( 0.75 ) + 0.1419 ( 2.25 ) Shigley’s MED, 10th edition Ans. Chap. 8 Solutions, Page 7/70 www.konkur.in (c) Top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20) k1 = 0.5774π ( 30 ) 0.5 1.155 (1.5 ) + 0.75 − 0.5 ( 0.75 + 0.5 ) ln 1.155 (1.5 ) + 0.75 + 0.5 ( 0.75 − 0.5 ) = 22.65 Mlbf/in Lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.75 + 2(1) tan 30° = 1.905 in, E = 30 Mpsi. Eq. (8-20) ⇒ k2 = 210.7 Mlbf/in Cast iron: t = 1 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20) ⇒ k3 = 12.27 Mlbf/in From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3)−1 = (1/22.65 + 1/210.7 + 1/12.27)−1 = 7.67 Mlbf/in Ans. 8-15 (a) From Table A-32, the washer thickness is 0.095 in. Thus, l = 2 + 1 + 2(0.095) = 3.19 in. From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 3.19 + 7/16 = 3.63 in. Rounding up, L = 3.75 in Ans. (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in From Table 8-7, ld = L − LT = 3.75 − 1.25 = 2.5 in, lt = l − ld = 3.19 − 2.5 = 0.69 in Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) Shigley’s MED, 10th edition Chap. 8 Solutions, Page 8/70 www.konkur.in kb = 0.1963 ( 0.1419 ) 30 Ad At E = = 1.705 Mlbf/in Ad lt + At ld 0.1963 ( 0.69 ) + 0.1419 ( 2.5 ) Ans. (c) Each steel washer frustum: t = 0.095 in, d = 0.531 in (Table A-32), D = 0.75 in, E = 30 Mpsi. From Eq. (8-20) k1 = 0.5774π ( 30 ) 0.531 1.155 ( 0.095 ) + 0.75 − 0.531 ( 0.75 + 0.531) ln 1.155 ( 0.095 ) + 0.75 + 0.531 ( 0.75 − 0.531) = 89.20 Mlbf/in Top plate, top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30° = 0.860 in, E = 30 Mpsi. Eq. (8-20) ⇒ k2 = 28.99 Mlbf/in Top plate, lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.860 + 2(1) tan 30° = 2.015 in, E = 30 Mpsi. Eq. (8-20) ⇒ k3 = 234.08 Mlbf/in Cast iron: t = 1 in, d = 0.5 in, D = 0.75 + 2(0.095) tan 30° = 0.860 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20) ⇒ k4 = 15.99 Mlbf/in From Eq. (8-18) km = (2/k1 + 1/k2 +1/k3+1/k4)−1 = (2/89.20 + 1/28.99 + 1/234.08 + 1/15.99)−1 = 8.08 Mlbf/in Ans. ______________________________________________________________________________ 8-16 (a) From Table 8-7, l = h + d /2 = 2 + 0.5/2 = 2.25 in. L ≥ h + 1.5 d = 2 + 1.5(0.5) = 2.75 in Ans. (b) From Table 8-7, LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in ld = L − LT = 2.75 − 1.25 = 1.5 in, lt = l − ld = 2.25 − 1.5 = 0.75 in Shigley’s MED, 10th edition Chap. 8 Solutions, Page 9/70 www.konkur.in Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) kb = 0.1963 ( 0.1419 ) 30 Ad At E = = 2.321 Mlbf/in Ad lt + At ld 0.1963 ( 0.75 ) + 0.1419 (1.5 ) Ans. (c) Top steel frustum: t = 1.125 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20) k1 = 0.5774π ( 30 ) 0.5 1.155 (1.125 ) + 0.75 − 0.5 ( 0.75 + 0.5 ) ln 1.155 (1.125 ) + 0.75 + 0.5 ( 0.75 − 0.5 ) = 24.48 Mlbf/in Lower steel frustum: t = 0.875 in, d = 0.5 in, D = 0.75 + 2(0.25) tan 30° = 1.039 in, E = 30 Mpsi. Eq. (8-20) ⇒ k2 = 49.36 Mlbf/in Cast iron: t = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20) ⇒ k3 = 23.49 Mlbf/in From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3)−1 = (1/24.48 + 1/49.36 + 1/23.49)−1 = 9.645 Mlbf/in Ans. ______________________________________________________________________________ 8-17 a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in. Rounding up, L = 4.75 in Ans. (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in Shigley’s MED, 10th edition Chap. 8 Solutions, Page 10/70 www.konkur.in From Table 8-7, ld = L − LT = 4.75 − 1.25 = 3.5 in, lt = l − ld = 4.19 − 3.5 = 0.69 in Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) kb = 0.1963 ( 0.1419 ) 30 Ad At E = = 1.322 Mlbf/in Ad lt + At ld 0.1963 ( 0.69 ) + 0.1419 ( 3.5 ) Ans. (c) Upper and lower halves are the same. For the upper half, Steel frustum: t = 0.095 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi. From Eq. (8-20) k1 = 0.5774π ( 30 ) 0.531 1.155 ( 0.095 ) + 0.75 − 0.531 ( 0.75 + 0.531) ln 1.155 ( 0.095 ) + 0.75 + 0.531 ( 0.75 − 0.531) = 89.20 Mlbf/in Aluminum: t = 2 in, d = 0.5 in, D =0.75 + 2(0.095) tan 30° = 0.860 in, and E = 10.3 Mpsi. Eq. (8-20) ⇒ k2 = 9.24 Mlbf/in For the top half, km′ = (1/k1 + 1/k2)−1 = (1/89.20 + 1/9.24)−1 = 8.373 Mlbf/in Since the bottom half is the same, the overall stiffness is given by km = (1/ km′ + 1/ km′ )−1 = km′ /2 = 8.373/2 = 4.19 Mlbf/in Ans ______________________________________________________________________________ 8-18 (a) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in. Rounding up, L = 4.75 in Ans. (b) From Eq. (8-13), LT = 2d + 1/4 = 2(0.5) + 0.25 = 1.25 in Shigley’s MED, 10th edition Chap. 8 Solutions, Page 11/70 www.konkur.in From Table 8-7, ld = L − LT = 4.75 − 1.25 = 3.5 in, lt = l − ld = 4.19 − 3.5 = 0.69 in Ad = π (0.52)/4 = 0.1963 in2. From Table 8-2, At = 0.1419 in2. From Eq. (8-17) kb = 0.1963 ( 0.1419 ) 30 Ad At E = = 1.322 Mlbf/in Ad lt + At ld 0.1963 ( 0.69 ) + 0.1419 ( 3.5 ) Ans. (c) Upper aluminum frustum: t = [4 + 2(0.095)] /2 = 2.095 in, d = 0.5 in, D = 0.75 in, and E = 10.3 Mpsi. From Eq. (8-20) 0.5774π (10.3) 0.5 k1 = = 7.23 Mlbf/in 1.155 ( 2.095 ) + 0.75 − 0.5 ( 0.75 + 0.5 ) ln 1.155 ( 2.095 ) + 0.75 + 0.5 ( 0.75 − 0.5 ) Lower aluminum frustum: t = 4 − 2.095 = 1.905 in, d = 0.5 in, D = 0.75 +4(0.095) tan 30° = 0.969 in, and E = 10.3 Mpsi. Eq. (8-20) ⇒ k2 = 11.34 Mlbf/in Steel washers frustum: t = 2(0.095) = 0.190 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi. Eq. (8-20) ⇒ k3 = 53.91 Mlbf/in From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3)−1 = (1/7.23 + 1/11.34 + 1/53.91)−1 = 4.08 Mlbf/in Ans. ______________________________________________________________________________ 8-19 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 50 + 8.4 = 58.4 mm. Rounding up, L = 60 mm Shigley’s MED, 10th edition Ans. Chap. 8 Solutions, Page 12/70 www.konkur.in (b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L − LT = 60 − 26 = 34 mm, lt = l − l = 50 − 34 = 16 mm. Ad = π (102) / 4 = 78.54 mm2. From Table 8-1, At = 58 mm2. From Eq. (8-17) 78.54 ( 58.0 ) 207 Ad At E kb = = = 292.1 MN/m Ans. Ad lt + At ld 78.54 (16 ) + 58.0 ( 34 ) (c) Upper and lower frustums are the same. For the upper half, Aluminum: t = 10 mm, d = 10 mm, D = 15 mm, and from Table 8-8, E = 71 GPa. From Eq. (8-20) 0.5774π ( 71)10 k1 = = 1576 MN/m 1.155 (10 ) + 15 − 10 (15 + 10 ) ln 1.155 (10 ) + 15 + 10 (15 − 10 ) Steel: t = 15 mm, d = 10 mm, D = 15 + 2(10) tan 30° = 26.55 mm, and E = 207 GPa. From Eq. (8-20) k2 = 0.5774π ( 207 )10 1.155 (15 ) + 26.55 − 10 ( 26.55 + 10 ) ln 1.155 (15 ) + 26.55 + 10 ( 26.55 − 10 ) = 11 440 MN/m For the top half, km′ = (1/k1 + 1/k2)−1 = (1/1576 + 1/11 440)−1 = 1385 MN/m Since the bottom half is the same, the overall stiffness is given by km = (1/ km′ + 1/ km′ )−1 = km′ /2 = 1385/2 = 692.5 MN/m Shigley’s MED, 10th edition Ans. Chap. 8 Solutions, Page 13/70 www.konkur.in 8-20 (a) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 60 + 8.4 = 68.4 mm. Rounding up, L = 70 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L − LT = 70 − 26 = 44 mm, lt = l − ld = 60 − 44 = 16 mm. Ad = π (102) / 4 = 78.54 mm2. From Table 8-1, At = 58 mm2. From Eq. (8-17) kb = 78.54 ( 58.0 ) 207 Ad At E = = 247.6 MN/m Ad lt + At ld 78.54 (16 ) + 58.0 ( 44 ) Ans. (c) Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq. (8-20) 0.5774π (10.3) 71 k1 = = 1576 MN/m 1.155 ( 2.095 ) + 15 − 10 (15 + 10 ) ln 1.155 ( 2.095 ) + 15 + 10 (15 − 10 ) Lower aluminum frustum: t = 20 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq. (8-20) ⇒ k2 = 1 201 MN/m Top steel frustum: t = 20 mm, d = 10 mm, D = 15 + 2(10) tan 30° = 26.55 mm, and E = 207 GPa. Eq. (8-20) ⇒ k3 = 9 781 MN/m Shigley’s MED, 10th edition Chap. 8 Solutions, Page 14/70 www.konkur.in Lower steel frustum: t = 10 mm, d = 10 mm, D = 15 + 2(20) tan 30° = 38.09 mm, and E = 207 GPa. Eq. (8-20) ⇒ k4 = 29 070 MN/m From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3+1/k4)−1 = (1/1 576 + 1/1 201 + 1/9 781 +1/29 070)−1 = 623.5 MN/m Ans. ______________________________________________________________________________ 8-21 (a) From Table 8-7, l = h + d /2 = 10 + 30 + 10/2 = 45 mm. L ≥ h + 1.5 d = 10 + 30 + 1.5(10) = 55 mm Ans. (b) From Eq. (8-14), LT = 2d + 6 = 2(10) + 6 = 26 mm, ld = L − LT = 55 − 26 = 29 mm, lt = l − ld = 45 − 29 = 16 mm. Ad = π (102) / 4 = 78.54 mm2. From Table 8-1, At = 58 mm2. From Eq. (8-17) 78.54 ( 58.0 ) 207 Ad At E kb = = = 320.9 MN/m Ans. Ad lt + At ld 78.54 (16 ) + 58.0 ( 29 ) (c) Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq. (8-20) 0.5774π (10.3) 71 k1 = = 1576 MN/m 1.155 ( 2.095 ) + 15 − 10 (15 + 10 ) ln 1.155 ( 2.095 ) + 15 + 10 (15 − 10 ) Lower aluminum frustum: t = 5 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq. (8-20) ⇒ k2 = 2 300 MN/m Top steel frustum: t = 12.5 mm, d = 10 mm, D = 15 + 2(10) tan 30° = 26.55 mm, and E = 207 GPa. Eq. (8-20) ⇒ k3 = 12 759 MN/m Shigley’s MED, 10th edition Chap. 8 Solutions, Page 15/70 www.konkur.in Lower steel frustum: t = 17.5 mm, d = 10 mm, D = 15 + 2(5) tan 30° = 20.77 mm, and E = 207 GPa. Eq. (8-20) ⇒ k4 = 6 806 MN/m From Eq. (8-18) km = (1/k1 + 1/k2 +1/k3+1/k4)−1 = (1/1 576 + 1/2 300 + 1/12 759 +1/6 806)−1 = 772.4 MN/m Ans. ______________________________________________________________________________ kb 8-22 Equation (f ), p. 428: C = kb + k m Ad At E Eq. (8-17): kb = Ad lt + At ld 0.5774π ( 207 ) d km = Eq. (8-22): 0.5774 ( 40 ) + 0.5d 2 ln 5 0.5774 ( 40 ) + 2.5d See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are mm, mm2, MN/m): d 10 12 14 16 20 24 30 d 10 12 14 16 20 24 30 l 40 40 40 40 40 40 40 At 58 84.3 115 157 245 353 561 ld 24 25 21 17 14 11 4 lt 16 15 19 23 26 29 36 Ad 78.53982 113.0973 153.938 201.0619 314.1593 452.3893 706.8583 kb 356.0129 518.8172 686.2578 895.9182 1373.719 1944.24 2964.343 H 8.4 10.8 12.8 14.8 18 21.5 25.6 L> 48.4 50.8 52.8 54.8 58 61.5 65.6 km 1751.566 2235.192 2761.721 3330.796 4595.515 6027.684 8487.533 L 50 55 55 55 60 65 70 LT 26 30 34 38 46 54 66 C 0.16892 0.188386 0.199032 0.211966 0.230133 0.243886 0.258852 Use a M14 × 2 bolt, with length 55 mm. Ans. _____________________________________________________________________________ kb 8-23 Equation (f ), p. 428: C = kb + k m Shigley’s MED, 10th edition Chap. 8 Solutions, Page 16/70 www.konkur.in Eq. (8-17): kb = Ad At E Ad lt + At ld For upper frustum, Eq. (8-20), with D = 1.5 d and t = 1.5 in: k1 = 0.5774π ( 30 ) d 1.155 (1.5 ) + 0.5d ( 2.5d ) ln 1.155 (1.5 ) + 2.5d ( 0.5d ) = 0.5774π ( 30 ) d (1.733 + 0.5d ) ln 5 (1.733 + 2.5d ) Lower steel frustum, with D = 1.5d + 2(1) tan 30° = 1.5d + 1.155, and t = 0.5 in: 0.5774π ( 30 ) d k2 = (1.733 + 0.5d )( 2.5d + 1.155 ) ln (1.733 + 2.5d )( 0.5d + 1.155 ) For cast iron frustum, let E = 14. 5 Mpsi, and D = 1.5 d, and t = 1 in: 0.5774π (14.5 ) d k3 = (1.155 + 0.5d ) ln 5 (1.155 + 2.5d ) Overall, km = (1/k1 +1/k2 +1/k3)−1 See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are in, in2, Mlbf/in): Shigley’s MED, 10th edition Chap. 8 Solutions, Page 17/70 www.konkur.in d 0.375 0.4375 0.5 0.5625 0.625 0.75 0.875 At 0.0775 0.1063 0.1419 0.182 0.226 0.334 0.462 d 0.375 0.4375 0.5 0.5625 0.625 0.75 0.875 ld 2.5 2.375 2.25 2.125 2.25 2 1.75 Ad H L> 0.110447 0.328125 3.328125 0.15033 0.375 3.375 0.19635 0.4375 3.4375 0.248505 0.484375 3.484375 0.306796 0.546875 3.546875 0.441786 0.640625 3.640625 0.60132 0.75 3.75 lt 0.5 0.625 0.75 0.875 0.75 1 1.25 L 3.5 3.5 3.5 3.5 3.75 3.75 3.75 LT 1 1.125 1.25 1.375 1.5 1.75 2 l 3 3 3 3 3 3 3 kb k1 k2 k3 km C 1.031389 15.94599 178.7801 8.461979 5.362481 0.161309 1.383882 19.21506 194.465 10.30557 6.484256 0.175884 1.791626 22.65332 210.6084 12.26874 7.668728 0.189383 2.245705 26.25931 227.2109 14.35052 8.915294 0.20121 2.816255 30.03179 244.2728 16.55009 10.22344 0.215976 3.988786 38.07191 279.7762 21.29991 13.02271 0.234476 5.341985 46.7663 317.1203 26.51374 16.06359 0.24956 Use a 169 −12 UNC × 3.5 in long bolt Ans. ______________________________________________________________________________ 8-24 Equation (f ), p. 428: Eq. (8-17): kb = Shigley’s MED, 10th edition C= kb kb + k m Ad At E Ad lt + At ld Chap. 8 Solutions, Page 18/70 www.konkur.in Top frustum, Eq. (8-20), with E = 10.3Mpsi, D = 1.5 d, and t = l /2: 0.5774π (10.3) d k1 = 1.155 l / 2 + 0.5d ln 5 1.155 l / 2 + 2.5d Middle frustum, with E = 10.3 Mpsi, D = 1.5d + 2(l − 0.5) tan 30°, and t = 0.5 − l /2 0.5774π (10.3) d k2 = 1.155 ( 0.5 − 0.5l ) + 0.5d + 2 ( l − 0.5 ) tan 300 2.5d + 2 ( l − 0.5 ) tan 300 ln 1.155 ( 0.5 − 0.5l ) + 2.5d + 2 ( l − 0.5 ) tan 300 0.5d + 2 ( l − 0.5 ) tan 300 { { } } Lower frustum, with E = 30Mpsi, D = 1.5 d, t = l − 0.5 0.5774π ( 30 ) d k3 = 1.155 ( l − 0.5 ) + 0.5d ln 5 1.155 ( l − 0.5 ) + 2.5d See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are in, in2, Mlbf/in) Size 1 2 3 4 5 6 8 10 d 0.073 0.086 0.099 0.112 0.125 0.138 0.164 0.19 At 0.00263 0.0037 0.00487 0.00604 0.00796 0.00909 0.014 0.0175 Ad 0.004185 0.005809 0.007698 0.009852 0.012272 0.014957 0.021124 0.028353 L> 0.6095 0.629 0.6485 0.668 0.6875 0.707 0.746 0.785 L 0.75 0.75 0.75 0.75 0.75 0.75 0.75 1 LT 0.396 0.422 0.448 0.474 0.5 0.526 0.578 0.63 l 0.5365 0.543 0.5495 0.556 0.5625 0.569 0.582 0.595 ld 0.354 0.328 0.302 0.276 0.25 0.224 0.172 0.37 Size 1 2 3 4 5 6 8 10 d 0.073 0.086 0.099 0.112 0.125 0.138 0.164 0.19 lt 0.1825 0.215 0.2475 0.28 0.3125 0.345 0.41 0.225 kb 0.194841 0.261839 0.333134 0.403377 0.503097 0.566787 0.801537 1.15799 k1 1.084468 1.321595 1.570439 1.830494 2.101297 2.382414 2.974009 3.602349 k2 1.954599 2.449694 2.993366 3.587564 4.234381 4.936066 6.513824 8.342138 k3 7.09432 8.357692 9.621064 10.88444 12.14781 13.41118 15.93792 18.46467 km 0.635049 0.778497 0.930427 1.090613 1.258846 1.434931 1.809923 2.214214 C 0.23478 0.251687 0.263647 0.27 0.285535 0.28315 0.306931 0.343393 Use a 2−56 UNC × 0.75 in long bolt. Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chap. 8 Solutions, Page 19/70 www.konkur.in 8-25 For half of joint, Eq. (8-20): t = 20 mm, d = 14 mm, D = 21 mm, and E = 207 GPa k1 = 0.5774π ( 207 )14 1.155 ( 20 ) + 21 − 14 ( 21 + 14 ) ln 1.155 ( 20 ) + 21 + 14 ( 21 − 14 ) = 5523 MN/m km = (1/k1 + 1/k1)−1 = k1/2 = 5523/2 = 2762 MN/m Ans. From Eq. (8-22) with l = 40 mm km = 0.5774π ( 207 )14 0.5774 ( 40 ) + 0.5 (14 ) 2 ln 5 0.5774 ( 40 ) + 2.5 (14 ) = 2762 MN/m Ans. which agrees with the earlier calculation. For Eq. (8-23), from Table 8-8, A = 0.787 15, B = 0.628 73 km = 207(14)(0.78 715) exp [0.628 73(14)/40] = 2843 MN/m Ans. This is 2.9% higher than the earlier calculations. ______________________________________________________________________________ 8-26 (a) Grip, l = 10 in. Nut height, H = 41/64 in (Table A-31). L ≥ l + H = 10 + 41/64 = 10.641 in. Let L = 10.75 in. Table 8-7, LT = 2d + 0.5 = 2(0.75) + 0.5 = 2 in, ld = L − LT = 10.75 − 2 = 8.75 in, lt = l − ld = 10 − 8.75 = 1.25 in Ad = π(0.752)/4 = 0.4418 in2, At = 0.373 in2 (Table 8-2) Eq. (8-17), 0.4418 ( 0.373) 30 Ad At E kb = = = 1.296 Mlbf/in Ans. Ad lt + At ld 0.4418 (1.25 ) + 0.373 ( 8.75 ) Eq. (4-4), p. 163, 2 2 Am Em (π / 4 ) (1.125 − 0.75 ) 30 km = = = 1.657 Mlbf/in Ans. l 10 Eq. (f), p. 428, C = kb/(kb + km) = 1.296/(1.296 + 1.657) = 0.439 Shigley’s MED, 10th edition Ans. Chap. 8 Solutions, Page 20/70 www.konkur.in (b) Let: Nt = no. of turns, p = pitch of thread (in), N = no. of threads per in = 1/p. Then, δ = δb + δm = Nt p = Nt / N (1) But, δb = Fi / kb, and, δm = Fi / km. Substituting these into Eq. (1) and solving for Fi gives Fi = kb k m N t kb + k m N ( 2) 1.296 (1.657 )106 1/ 3 = 15 150 lbf Ans. 1.296 + 1.657 16 ______________________________________________________________________________ = 8-27 Proof for the turn-of-nut equation is given in the solution of Prob. 8-26, Eq. (2), where Nt = θ / 360°. The relationship between the turn-of-nut method and the torque-wrench method is as follows. k + km Nt = b Fi N kb k m T = KFd i (turn-of-nut) (torque-wrench) Eliminate Fi k + km NT θ Nt = b = Ans. 360o kbkm Kd ______________________________________________________________________________ 8-28 (a) From Ex. 8-4, Fi = 14.4 kip, kb = 5.21(106) lbf/in, km = 8.95(106) lbf/in Eq. (8-27): T = kFid = 0.2(14.4)(103)(5/8) = 1800 lbf · in Ans. From Prob. 8-27, Shigley’s MED, 10th edition Chap. 8 Solutions, Page 21/70 www.konkur.in 5.21 + 8.95 k + km Nt = b (14.4)(103 )11 Fi N = 6 5.21( 8.95 )10 kb k m = 0.0481 turns = 17.3o Ans. Bolt group is (1.5) / (5/8) = 2.4 diameters. Answer is much lower than RB&W recommendations. ______________________________________________________________________________ 8-29 C = kb / (kb + km) = 3/(3+12) = 0.2, P = Ptotal/ N = 80/6 = 13.33 kips/bolt Table 8-2, At = 0.141 9 in2; Table 8-9, Sp = 120 kpsi; Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(0.141 9)(120) = 12.77 kips (a) From Eq. (8-28), the factor of safety for yielding is S p At np = 120 ( 0.141 9 ) = CP + Fi 0.2 (13.33) + 12.77 (b) From Eq. (8-29), the overload factor is nL = S p At − Fi CP = = 1.10 120 ( 0.141 9 ) − 12.77 0.2 (13.33) Ans. = 1.60 Ans. (c) From Eq. (803), the joint separation factor of safety is Fi 12.77 = = 1.20 Ans. P (1 − C ) 13.33 (1 − 0.2 ) ______________________________________________________________________________ n0 = 1/2 − 13 UNC Grade 8 bolt, K = 0.20 (a) Proof strength, Table 8-9, Sp = 120 kpsi Table 8-2, At = 0.141 9 in2 Maximum, Fi = Sp At = 120(0.141 9) = 17.0 kips Ans. (b) From Prob. 8-29, C = 0.2, P = 13.33 kips Joint separation, Eq. (8-30) with n0 = 1 Minimum Fi = P (1 − C) = 13.33(1 − 0.2) = 10.66 kips Ans. (c) Fi = (17.0 + 10.66)/2 = 13.8 kips Eq. (8-27), T = KFi d = 0.2(13.8)103(0.5)/12 = 115 lbf ⋅ ft Ans. ______________________________________________________________________________ 8-30 8-31 (a) Table 8-1, At = 20.1 mm2. Table 8-11, Sp = 380 MPa. Eq. (8-31), Fi = 0.75 Fp = 0.75 At Sp = 0.75(20.1)380(10−3) = 5.73 kN Eq. (f ), p. 428, Shigley’s MED, 10th edition C= kb 1 = = 0.278 kb + k m 1 + 2.6 Chap. 8 Solutions, Page 22/70 www.konkur.in Eq. (8-28) with np = 1, −3 S p At − Fi 0.25S p At 0.25 ( 20.1) 380 (10 ) P= = = = 6.869 kN C C 0.278 Ptotal = NP = 8(6.869) = 55.0 kN Ans. (b) Eq. (8-30) with n0 = 1, F 5.73 P= i = = 7.94 kN 1 − C 1 − 0.278 Ptotal = NP = 8(7.94) = 63.5 kN Ans. Bolt stress would exceed proof strength ______________________________________________________________________________ 8-32 (a) Table 8-2, At = 0.141 9 in2. Table 8-9, Sp = 120 kpsi. Eq. (8-31), Fi = 0.75 Fp = 0.75 At Sp = 0.75(0.141 9)120 = 12.77 kips Eq. (f ), p. 428, C= kb 4 = = 0.25 kb + k m 4 + 12 Eq. (8-28) with np = 1, S A − Fi 0.25 NS p At Ptotal = N p t = C C 80 ( 0.25 ) P C N = total = = 4.70 0.25S p At 0.25 (120 ) 0.141 9 Round to N = 5 bolts Ans. (b) Eq. (8-30) with n0 = 1, F Ptotal = N i 1− C P (1 − C ) 80 (1 − 0.25 ) N = total = = 4.70 Fi 12.77 Round to N = 5 bolts Ans. Shigley’s MED, 10th edition Chap. 8 Solutions, Page 23/70 www.konkur.in ______________________________________________________________________________ Problems 8-33 through 8-44 and 8-51 through 8-58. Note to the instructor: In previous editions of this solutions manual, in Probs. 8-33 through 8-36 the cylinder diameter was mistakenly used, instead of the effective gasket diameter, in determining the external force from the cylinder pressure. To correct this, the second printing of the 10th edition changes the problem statement so that the old value of the cylinder diameter is used for the effective gasket diameter. Consequently, the solutions to these problems, as well as several others that reference these problems remain numerically the same as before, with the diameter being redefined in the problem statements. ______________________________________________________________________________ 8-33 Bolts: From Table A-31, the nut height is H = 10.8 mm. L ≥ l +H = 40 + 10.8 = 50.8 mm. Round up to L = 55 mm Ans. Eq. (8-14): LT = 2d + 6 = 2(12) + 6 = 30 mm Table 8-7: ld = L − LT = 55 − 30 = 25 mm, lt = l −ld = 40 − 25 = 15 mm Ad =π (122)/4 = 113.1 mm2, Table 8-1: At = 84.3 mm2 Eq. (8-17): kb = 113.1( 84.3) 207 Ad At E = = 518.8 MN/m Ad lt + At ld 113.1(15 ) + 84.3 ( 25 ) Members: Steel cyl. head: t = 20 mm, d = 12 mm, D = 18 mm, E = 207 GPa. Eq. (8-20), k1 = 0.5774π ( 207 )12 1.155 ( 20 ) + 18 − 12 (18 + 12 ) ln 1.155 ( 20 ) + 18 + 12 (18 − 12 ) = 4470 MN/m Cast iron: t = 20 mm, d = 12 mm, D = 18 mm, E = 100 GPa (from Table 8-8). The only difference from k1 is the material k2 = (100/207)(4470) = 2159 MN/m Eq. (8-18): km = (1/4470 + 1/2159)−1 = 1456 MN/m C = kb / (kb + km) = 518.8/(518.8+1456) = 0.263 Table 8-11: Sp = 650 MPa Shigley’s MED, 10th edition Chap. 8 Solutions, Page 24/70 www.konkur.in For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(84.3)(650)10−3 = 41.1 kN The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 100 mm. The external load per bolt is P = Ptotal /N. Thus P = [6π (1002)/4](10−3)/10 = 4.712 kN/bolt Yielding factor of safety, Eq. (8-28): 650 ( 84.3)10−3 np = = = 1.29 CP + Fi 0.263 ( 4.712 ) + 41.10 S p At Ans. Overload factor of safety, Eq. (8-29): nL = S p At − Fi CP = 650 ( 84.3)10−3 − 41.10 = 11.1 0.263 ( 4.712 ) Ans. Separation factor of safety, Eq. (8-30): Fi 41.10 = = 11.8 Ans. P (1 − C ) 4.712 (1 − 0.263) ______________________________________________________________________________ n0 = 8-34 Instructors: See note preceding Prob. 8-33. Bolts: Grip, l = 1/2 + 5/8 = 1.125 in. From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 1.125 + 7/16 = 1.563 in. Round up to L = 1.75 in Ans. Eq. (8-13): LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in Table 8-7: ld = L − LT = 1.75 − 1.25 = 0.5 in, lt = l −ld = 1.125 − 0.5 = 0.625 in Ad = π (0.52)/4 = 0.196 3 in2, Table 8-2: At = 0.141 9 in2 Eq. (8-17): kb = 0.196 3 ( 0.141 9 ) 30 Ad At E = = 4.316 Mlbf/in Ad lt + At ld 0.196 3 ( 0.625 ) + 0.141 9 ( 0.5 ) Members: Steel cyl. head: t = 0.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. Eq. (8-20), Shigley’s MED, 10th edition Chap. 8 Solutions, Page 25/70 www.konkur.in k1 = 0.5774π ( 30 ) 0.5 1.155 ( 0.5 ) + 0.75 − 0.5 ( 0.75 + 0.5 ) ln 1.155 ( 0.5 ) + 0.75 + 0.5 ( 0.75 − 0.5 ) = 33.30 Mlbf/in Cast iron: Has two frusta. Midpoint of complete joint is at (1/2 + 5/8)/2 = 0.5625 in. Upper frustum, t = 0.5625 −0.5 = 0.0625 in, d = 0.5 in, D = 0.75 + 2(0.5) tan 30° = 1.327 in, E = 14.5 Mpsi (from Table 8-8) Eq. (8-20) ⇒ k2 = 292.7 Mlbf/in Lower frustum, t = 0.5625 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi Eq. (8-20) ⇒ k3 = 15.26 Mlbf/in Eq. (8-18): km = (1/33.30 + 1/292.7 + 1/15.26)−1 = 10.10 Mlbf/in C = kb / (kb + km) = 4.316/(4.316+10.10) = 0.299 Table 8-9: Sp = 85 kpsi For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.141 9)(85) = 9.05 kips The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 3.5 in. The external load per bolt is P = Ptotal /N. Thus, P = [1 500π (3.52)/4](10−3)/10 = 1.443 kips/bolt Yielding factor of safety, Eq. (8-28): S A 85 ( 0.141 9 ) np = p t = = 1.27 CP + Fi 0.299 (1.443) + 9.05 Ans. Overload factor of safety, Eq. (8-29): nL = S p At − Fi CP = 85 ( 0.141 9 ) − 9.05 0.299 (1.443) = 6.98 Ans. Separation factor of safety, Eq. (8-30): Fi 9.05 = = 8.95 Ans. P (1 − C ) 1.443 (1 − 0.299 ) ______________________________________________________________________________ n0 = Shigley’s MED, 10th edition Chap. 8 Solutions, Page 26/70 www.konkur.in 8-35 Instructors: See note preceding Prob. 8-33. Bolts: Grip: l = 20 + 25 = 45 mm. From Table A-31, the nut height is H = 8.4 mm. L ≥ l +H = 45 + 8.4 = 53.4 mm. Round up to L = 55 mm Ans. Eq. (8-14): LT = 2d + 6 = 2(10) + 6 = 26 mm Table 8-7: ld = L − LT = 55 − 26 = 29 mm, lt = l −ld = 45 − 29 = 16 mm Ad =π (102)/4 = 78.5 mm2, Table 8-1: At = 58.0 mm2 Eq. (8-17): kb = 78.5 ( 58.0 ) 207 Ad At E = = 320.8 MN/m Ad lt + At ld 78.5 (16 ) + 58.0 ( 29 ) Members: Steel cyl. head: t = 20 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20), k1 = 0.5774π ( 207 )10 1.155 ( 20 ) + 15 − 10 (15 + 10 ) ln 1.155 ( 20 ) + 15 + 10 (15 − 10 ) = 3503 MN/m Cast iron: Has two frusta. Midpoint of complete joint is at (20 + 25)/2 = 22.5 mm Upper frustum, t = 22.5 − 20 = 2.5 mm, d = 10 mm, D = 15 + 2(20) tan 30° = 38.09 mm, E = 100 GPa (from Table 8-8), Eq. (8-20) ⇒ k2 = 45 880 MN/m Lower frustum, t = 22.5 mm, d = 10 mm, D = 15 mm, E = 100 GPa Eq. (8-20) ⇒ k3 = 1632 MN/m Eq. (8-18): km = (1/3503 + 1/45 880 + 1/1632)−1 = 1087 MN/m C = kb / (kb + km) = 320.8/(320.8+1087) = 0.228 Table 8-11: Sp = 830 MPa For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(58.0)(830)10−3 = 36.1 kN The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 0.8 m. The external load per bolt is P = Ptotal /N. Thus, P = [550π (0.82)/4]/36 = 7.679 kN/bolt Shigley’s MED, 10th edition Chap. 8 Solutions, Page 27/70 www.konkur.in Yielding factor of safety, Eq. (8-28): np = S p At CP + Fi = 830 ( 58.0 )10−3 = 1.27 0.228 ( 7.679 ) + 36.1 Ans. Overload factor of safety, Eq. (8-29): nL = S p At − Fi CP = 830 ( 58.0 )10−3 − 36.1 0.228 ( 7.679 ) = 6.88 Ans. Separation factor of safety, Eq. (8-30): Fi 36.1 = = 6.09 Ans. P (1 − C ) 7.679 (1 − 0.228 ) ______________________________________________________________________________ n0 = 8-36 Instructors: See note preceding Prob. 8-33. Bolts: Grip, l = 3/8 + 1/2 = 0.875 in. From Table A-31, the nut height is H = 3/8 in. L ≥ l + H = 0.875 + 3/8 = 1.25 in. Let L = 1.25 in Ans. Eq. (8-13): LT = 2d + 0.25 = 2(7/16) + 0.25 = 1.125 in Table 8-7: ld = L − LT = 1.25 − 1.125 = 0.125 in, lt = l −ld = 0.875 − 0.125 = 0.75 in Ad = π (7/16)2/4 = 0.150 3 in2, Table 8-2: At = 0.106 3 in2 Eq. (8-17), kb = 0.150 3 ( 0.106 3) 30 Ad At E = = 3.804 Mlbf/in Ad lt + At ld 0.150 3 ( 0.75 ) + 0.106 3 ( 0.125 ) Members: Steel cyl. head: t = 0.375 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi. Eq. (8-20), k1 = 0.5774π ( 30 ) 0.4375 1.155 ( 0.375 ) + 0.65625 − 0.4375 ( 0.65625 + 0.4375 ) ln 1.155 ( 0.375 ) + 0.65625 + 0.4375 ( 0.65625 − 0.4375 ) = 31.40 Mlbf/in Cast iron: Has two frusta. Midpoint of complete joint is at (3/8 + 1/2)/2 = 0.4375 in. Shigley’s MED, 10th edition Chap. 8 Solutions, Page 28/70 www.konkur.in Upper frustum, t = 0.4375 −0.375 = 0.0625 in, d = 0.4375 in, D = 0.65625 + 2(0.375) tan 30° = 1.089 in, E = 14.5 Mpsi (from Table 8-8) Eq. (8-20) ⇒ k2 = 195.5 Mlbf/in Lower frustum, t = 0.4375 in, d = 0.4375 in, D = 0.65625 in, E = 14.5 Mpsi Eq. (8-20) ⇒ k3 = 14.08 Mlbf/in Eq. (8-18): km = (1/31.40 + 1/195.5 + 1/14.08)−1 = 9.261 Mlbf/in C = kb / (kb + km) = 3.804/(3.804 + 9.261) = 0.291 Table 8-9: Sp = 120 kpsi For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.106 3)(120) = 9.57 kips The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 3.25 in. The external load per bolt is P = Ptotal /N. Thus, P = [1 200π (3.252)/4](10−3)/8 = 1.244 kips/bolt Yielding factor of safety, Eq. (8-28): S p At 120 ( 0.106 3) np = = = 1.28 Ans. CP + Fi 0.291(1.244 ) + 9.57 Overload factor of safety, Eq. (8-29): nL = S p At − Fi CP = 120 ( 0.106 3) − 9.57 = 8.80 Ans. 0.291(1.244 ) Separation factor of safety, Eq. (8-30): n0 = Fi 9.57 = = 10.9 P (1 − C ) 1.244 (1 − 0.291) Ans. ______________________________________________________________________________ 8-37 Instructors: See note preceding Prob. 8-33. From Table 8-7, h = t1 = 20 mm For t2 > d, l = h + d /2 = 20 + 12/2 = 26 mm L ≥ h + 1.5 d = 20 + 1.5(12) = 38 mm. Round up to L = 40 mm LT = 2d + 6 = 2(12) + 6 = 30 mm Shigley’s MED, 10th edition Chap. 8 Solutions, Page 29/70 www.konkur.in ld = L − LT = 40 − 20 = 10 mm lt = l − ld = 26 − 10 = 16 mm From Table 8-1, At = 84.3 mm2. Ad = π (122)/4 = 113.1 mm2 Eq. (8-17), 113.1( 84.3) 207 Ad At E kb = = = 744.0 MN/m Ad lt + At ld 113.1(16 ) + 84.3 (10 ) Similar to Fig. 8-21, we have three frusta. Top frusta, steel: t = l / 2 = 13 mm, d = 12 mm, D = 18 mm, E = 207 GPa. Eq. (8-20) k1 = 0.5774π ( 207 )12 1.155 (13) + 18 − 12 (18 + 12 ) ln 1.155 (13) + 18 + 12 (18 − 12 ) = 5 316 MN/m Middle frusta, steel: t = 20 − 13 = 7 mm, d = 12 mm, D = 18 + 2(13 − 7) tan 30° = 24.93 mm, E = 207 GPa. Eq. (8-20) ⇒ k2 = 15 660 MN/m Lower frusta, cast iron: t = 26 − 20 = 6 mm, d = 12 mm, D = 18 mm, E = 100 GPa (see Table 8-8). Eq. (8-20) ⇒ k3 = 3 887 MN/m Eq. (8-18), km = (1/5 316 + 1/15 660 + 1/3 887)−1 = 1 964 MN/m C = kb / (kb + km) = 744.0/(744.0 + 1 964) = 0.275 Table 8-11: Sp = 650 MPa. For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(84.3)(650)10−3 = 41.1 kN The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 100 mm. The external load per bolt is P = Ptotal /N. Thus P = [6π (1002)/4](10−3)/10 = 4.712 kN/bolt Yielding factor of safety, Eq. (8-28) np = S p At CP + Fi = 650 ( 84.3)10−3 0.275 ( 4.712 ) + 41.1 = 1.29 Ans. Overload factor of safety, Eq. (8-29) nL = Shigley’s MED, 10th edition S p At − Fi CP = 650 ( 84.3)10−3 − 41.1 = 10.7 0.275 ( 4.712 ) Ans. Chap. 8 Solutions, Page 30/70 www.konkur.in Separation factor of safety, Eq. (8-30) Fi 41.1 = = 12.0 Ans. P (1 − C ) 4.712 (1 − 0.275 ) ______________________________________________________________________________ n0 = 8-38 Instructors: See note preceding Prob. 8-33. From Table 8-7, h = t1 = 0.5 in For t2 > d, l = h + d /2 = 0.5 + 0.5/2 = 0.75 in L ≥ h + 1.5 d = 0.5 + 1.5(0.5) = 1.25 in. Let L = 1.25 in LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in. All threaded. From Table 8-1, At = 0.141 9 in2. The bolt stiffness is kb = At E / l = 0.141 9(30)/0.75 = 5.676 Mlbf/in Similar to Fig. 8-21, we have three frusta. Top frusta, steel: t = l / 2 = 0.375 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi 0.5774π ( 30 ) 0.5 k1 = = 38.45 Mlbf/in 1.155 ( 0.375 ) + 0.75 − 0.5 ( 0.75 + 0.5 ) ln 1.155 ( 0.375 ) + 0.75 + 0.5 ( 0.75 − 0.5 ) Middle frusta, steel: t = 0.5 − 0.375 = 0.125 in, d = 0.5 in, D = 0.75 + 2(0.75 − 0.5) tan 30° = 1.039 in, E = 30 Mpsi. Eq. (8-20) ⇒ k2 = 184.3 Mlbf/in Lower frusta, cast iron: t = 0.75 − 0.5 = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi. Eq. (8-20) ⇒ k3 = 23.49 Mlbf/in Eq. (8-18), km = (1/38.45 + 1/184.3 + 1/23.49)−1 = 13.51 Mlbf/in C = kb / (kb + km) = 5.676 / (5.676 + 13.51) = 0.296 Table 8-9, Sp = 85 kpsi. For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.141 9)(85) = 9.05 kips The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 3.5 in. The external load per bolt is P = Ptotal /N. Thus P = [1 500π (3.52)/4](10−3)/10 = 1.443 kips/bolt Yielding factor of safety, Eq. (8-28) np = S p At CP + Fi = 85 ( 0.141 9 ) = 1.27 0.296 (1.443) + 9.05 Ans. Overload factor of safety, Eq. (8-29) Shigley’s MED, 10th edition Chap. 8 Solutions, Page 31/70 www.konkur.in nL = S p At − Fi CP = 85 ( 0.141 9 ) − 9.05 0.296 (1.443) = 7.05 Ans. Separation factor of safety, Eq. (8-30) Fi 9.05 = = 8.91 Ans. P (1 − C ) 1.443 (1 − 0.296 ) ______________________________________________________________________________ n0 = 8-39 Instructors: See note preceding Prob. 8-33. From Table 8-7, h = t1 = 20 mm For t2 > d, l = h + d /2 = 20 + 10/2 = 25 mm L ≥ h + 1.5 d = 20 + 1.5(10) = 35 mm. Let L = 35 mm LT = 2d + 6 = 2(10) + 6 = 26 mm ld = L − LT = 35 − 26 = 9 mm lt = l − ld = 25 − 9 = 16 mm From Table 8-1, At = 58.0 mm2. Ad = π (102)/4 = 78.5 mm2 Eq. (8-17), 78.5 ( 58.0 ) 207 Ad At E kb = = = 530.1 MN/m Ad lt + At ld 78.5 (16 ) + 58.0 ( 9 ) Similar to Fig. 8-21, we have three frusta. Top frusta, steel: t = l / 2 = 12.5 mm, d = 10 mm, D = 15 mm, E = 207 GPa. Eq. (8-20) 0.5774π ( 207 )10 k1 = = 4 163 MN/m 1.155 (12.5 ) + 15 − 10 (15 + 10 ) ln 1.155 (12.5 ) + 15 + 10 (15 − 10 ) Middle frusta, steel: t = 20 − 12.5 = 7.5 mm, d = 10 mm, D = 15 + 2(12.5 − 7.5) tan 30° = 20.77 mm, E = 207 GPa. Eq. (8-20) ⇒ k2 = 10 975 MN/m Lower frusta, cast iron: t = 25 − 20 = 5 mm, d = 10 mm, D = 15 mm, E = 100 GPa (see Table 8-8). Eq. (8-20) ⇒ k3 = 3 239 MN/m Eq. (8-18), km = (1/4 163 + 1/10 975 + 1/3 239)−1 = 1 562 MN/m C = kb / (kb + km) = 530.1/(530.1 + 1 562) = 0.253 Table 8-11: Sp = 830 MPa. For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(58.0)(830)10−3 = 36.1 kN The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 0.8 m. The external load per bolt is P = Ptotal /N. Thus Shigley’s MED, 10th edition Chap. 8 Solutions, Page 32/70 www.konkur.in P = [550π (0.82)/4]/36 = 7.679 kN/bolt Yielding factor of safety, Eq. (8-28) np = S p At CP + Fi = 830 ( 58.0 )10−3 = 1.27 0.253 ( 7.679 ) + 36.1 Ans. Overload factor of safety, Eq. (8-29) nL = S p At − Fi CP = 830 ( 58.0 )10−3 − 36.1 = 6.20 0.253 ( 7.679 ) Ans. Separation factor of safety, Eq. (8-30) Fi 36.1 = = 6.29 Ans. P (1 − C ) 7.679 (1 − 0.253) ______________________________________________________________________________ n0 = 8-40 Instructors: See note preceding Prob. 8-33. From Table 8-7, h = t1 = 0.375 in For t2 > d, l = h + d /2 = 0.375 + 0.4375/2 = 0.59375 in L ≥ h + 1.5 d = 0.375 + 1.5(0.4375) = 1.031 in. Round up to L = 1.25 in LT = 2d + 0.25 = 2(0.4375) + 0.25 = 1.125 in ld = L − LT = 1.25 − 1.125 = 0.125 lt = l − ld = 0.59375 − 0.125 = 0.46875 in Ad = π (7/16)2/4 = 0.150 3 in2, Table 8-2: At = 0.106 3 in2 Eq. (8-17), 0.150 3 ( 0.106 3) 30 Ad At E = = 5.724 Mlbf/in Ad lt + At ld 0.150 3 ( 0.46875 ) + 0.106 3 ( 0.125 ) Similar to Fig. 8-21, we have three frusta. Top frusta, steel: t = l / 2 = 0.296875 in, d = 0.4375 in, D = 0.65625 in, E = 30 Mpsi kb = k1 = 0.5774π ( 30 ) 0.4375 1.155 ( 0.296875 ) + 0.656255 − 0.4375 ( 0.75 + 0.656255 ) ln 1.155 ( 0.296875 ) + 0.75 + 0.656255 ( 0.75 − 0.656255 ) = 35.52 Mlbf/in Middle frusta, steel: t = 0.375 − 0.296875 = 0.078125 in, d = 0.4375 in, D = 0.65625 + 2(0.59375 − 0.375) tan 30° = 0.9088 in, E = 30 Mpsi. Eq. (8-20) ⇒ k2 = 215.8 Mlbf/in Lower frusta, cast iron: t = 0.59375 − 0.375 = 0.21875 in, d = 0.4375 in, D = 0.65625 in, E = 14.5 Mpsi. Eq. (8-20) ⇒ k3 = 20.55 Mlbf/in Shigley’s MED, 10th edition Chap. 8 Solutions, Page 33/70 www.konkur.in Eq. (8-18), km = (1/35.52 + 1/215.8 + 1/20.55)−1 = 12.28 Mlbf/in C = kb / (kb + km) = 5.724/(5.724 + 12.28) = 0.318 Table 8-9, Sp = 120 kpsi. For a non-permanent connection, using Eqs. (8-31) and (8-32) Fi = 0.75 At Sp = 0.75(0.106 3)(120) = 9.57 kips The total external load is Ptotal = pg Ag, where Ag is the effective area of the cylinder, based on the effective sealing diameter of 3.25 in. The external load per bolt is P = Ptotal /N. Thus P = [1 200π (3.252)/4](10−3)/8 = 1.244 kips/bolt Yielding factor of safety, Eq. (8-28) np = S p At CP + Fi = 120 ( 0.106 3) = 1.28 0.318 (1.244 ) + 9.57 Ans. Overload factor of safety, Eq. (8-29) nL = S p At − Fi CP = 120 ( 0.106 3) − 9.57 0.318 (1.244 ) = 8.05 Ans. Separation factor of safety, Eq. (8-30) Fi 9.57 = = 11.3 Ans. P (1 − C ) 1.244 (1 − 0.318 ) ______________________________________________________________________________ n0 = 8-41 Instructors: See note preceding Prob. 8-33. This is a design problem and there is no closed-form solution path or a unique solution. What is presented here is one possible iterative approach. We will demonstrate this with an example. 1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on members, and combining using Eq. (8-18), yields km = 1 141 MN/m (see Prob. 8-33 for method of calculation. 2. Look up the nut height in Table A-31. For the example, H = 8.4 mm. From this, L is rounded up from the calculation of l + H = 40 + 8.4 = 48.4 mm to 50 mm. Next, calculations are made for LT = 2(10) + 6 = 26 mm, ld = 50 − 26 = 24 mm, lt = 40 − 24 = 16 mm. From step 1, Ad =π (102)/4 = 78.54 mm2. Next, from Table 8-1, At = 78.54 mm2. From Eq. (8-17), kb = 356 MN/m. Finally, from Eq. (f), p. 428, C = 0.238. Shigley’s MED, 10th edition Chap. 8 Solutions, Page 34/70 www.konkur.in 3. From Prob. 8-33, the bolt circle diameter is E = 200 mm. Substituting this for Db in Eq. (8-34), the number of bolts are N= π Db = π ( 200 ) 4d 4 (10 ) Rounding this up gives N = 16. = 15.7 4. Next, select a grade bolt. Based on the solution to Prob. 8-33, the strength of ISO 9.8 was so high to give very large factors of safety for overload and separation. Try ISO 4.6 with Sp = 225 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi = 9.79 kN. 5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-33, pg = 6 MPa, and Ag = π (1002)/4. This gives P = 3.39 kN/bolt. 6. Using Eqs. (8-28) to (8-30) yield np = 1.23, nL = 4.05, and n0 = 3.79. Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables used from the text. The results for four bolt sizes are shown below. The dimension of each term is consistent with the example given above. d 8 10 12 14 km 854 1141 1456 1950 H 6.8 8.4 10.8 12.8 L 50 50 55 55 LT 22 26 30 34 ld 28 24 25 21 lt 12 16 15 19 d 8 10 12 14 C 0.215 0.238 0.263 0.276 N 20 16 13* 12 Sp 225 225 225 225 Fi 6.18 9.79 14.23 19.41 P 2.71 3.39 4.17 4.52 np 1.22 1.23 1.24 1.25 Ad At kb 50.26 36.6 233.9 78.54 58 356 113.1 84.3 518.8 153.9 115 686.3 nL 3.53 4.05 4.33 5.19 n0 2.90 3.79 4.63 5.94 *Rounded down from13.08997, so spacing is slightly greater than four diameters. Any one of the solutions is acceptable. A decision-maker might be cost such as N × cost/bolt, and/or N × cost per hole, etc. ________________________________________________________________________ Shigley’s MED, 10th edition Chap. 8 Solutions, Page 35/70 www.konkur.in 8-42 Instructors: See note preceding Prob. 8-33. This is a design problem and there is no closed-form solution path or a unique solution. What is presented here is one possible iterative approach. We will demonstrate this with an example. 1. Select the diameter, d. For this example, let d = 0.5 in. Using Eq. (8-20) on three frusta (see Prob. 8-34 solution), and combining using Eq. (8-19), yields km = 10.10 Mlbf/in. 2. Look up the nut height in Table A-31. For the example, H = 0.4375 in. From this, L is rounded up from the calculation of l + H = 1.125 + 0.4375 = 1.5625 in to 1.75 in. Next, calculations are made for LT = 2(0.5) + 0.25 = 1.25 in, ld = 1.75 − 1.25 = 0.5 in, lt = 1.125 − 0.5 = 0.625 in. From step 1, Ad =π (0.52)/4 = 0.1963 in2. Next, from Table 8-1, At = 0.141 9 in2. From Eq. (8-17), kb = 4.316 Mlbf/in. Finally, from Eq. (f), p. 428, C = 0.299. 3. From Prob. 8-34, the bolt circle diameter is E = 6 in. Substituting this for Db in Eq. (834), for the number of bolts N= π Db = π ( 6) 4d 4 ( 0.5 ) Rounding this up gives N = 10. = 9.425 4. Next, select a grade bolt. Based on the solution to Prob. 8-34, the strength of SAE grade 5 was adequate. Use this with Sp = 85 kpsi. From Eqs. (8-31) and (8-32) for a nonpermanent connection, Fi = 9.046 kips. 5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-34, pg = 1 500 psi, and Ag = π (3.52)/4 . This gives P = 1.660 kips/bolt. 6. Using Eqs. (8-28) to (8-30) yield np = 1.26, nL = 6.07, and n0 = 7.78. d 0.375 0.4375 0.5 0.5625 km 6.75 9.17 10.10 11.98 H 0.3281 0.375 0.4375 0.4844 d 0.375 0.4375 0.5 0.5625 C 0.261 0.273 0.299 0.308 N 13 11 10 9 L LT ld 1.5 1 0.5 1.5 1.125 0.375 1.75 1.25 0.5 1.75 1.375 0.375 Sp 85 85 85 85 Fi 4.941 6.777 9.046 11.6 P 1.277 1.509 1.660 1.844 lt 0.625 0.75 0.625 0.75 np 1.25 1.26 1.26 1.27 Ad At 0.1104 0.0775 0.1503 0.1063 0.1963 0.1419 0.2485 0.182 nL 4.95 5.48 6.07 6.81 kb 2.383 3.141 4.316 5.329 n0 5.24 6.18 7.78 9.09 Any one of the solutions is acceptable. A decision-maker might be cost such as N × cost/bolt, and/or N × cost per hole, etc. ________________________________________________________________________ Shigley’s MED, 10th edition Chap. 8 Solutions, Page 36/70 www.konkur.in 8-43 Instructors: See note preceding Prob. 8-33. This is a design problem and there is no closed-form solution path or a unique solution. What is presented here is one possible iterative approach. We will demonstrate this with an example. 1. Select the diameter, d. For this example, let d = 10 mm. Using Eq. (8-20) on three frusta (see Prob. 8-35 solution), and combining using Eq. (8-19), yields km = 1 087 MN/m. 2. Look up the nut height in Table A-31. For the example, H = 8.4 mm. From this, L is rounded up from the calculation of l + H = 45 + 8.4 = 53.4 mm to 55 mm. Next, calculations are made for LT = 2(10) + 6 = 26 mm, ld = 55 − 26 = 29 mm, lt = 45 − 29 = 16 mm. From step 1, Ad =π (102)/4 = 78.54 mm2. Next, from Table 8-1, At = 58.0 mm2. From Eq. (8-17), kb = 320.9 MN/m. Finally, from Eq. (f), p. 428, C = 0.228. 3. From Prob. 8-35, the bolt circle diameter is E = 1000 mm. Substituting this for Db in Eq. (8-34), for the number of bolts N= π Db = π (1000 ) = 78.5 4d 4 (10 ) Rounding this up gives N = 79. A rather large number, since the bolt circle diameter, E is so large. Try larger bolts. 4. Next, select a grade bolt. Based on the solution to Prob. 8-35, the strength of ISO 9.8 was so high to give very large factors of safety for overload and separation. Try ISO 5.8 with Sp = 380 MPa. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi = 16.53 kN. 5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-35, pg = 0.550 MPa, and Ag = π (8002)/4 . This gives P = 4.024 kN/bolt. 6. Using Eqs. (8-28) to (8-30) yield np = 1.26, nL = 6.01, and n0 = 5.32. Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables used from the text. The results for three bolt sizes are shown below. The dimension of each term is consistent with the example given above. d 10 20 36 km 1087 3055 6725 H 8.4 18 31 L 55 65 80 d 10 20 36 C 0.228 0.308 0.361 N 79 40 22 Sp 380 380 380 Shigley’s MED, 10th edition LT 26 46 78 ld 29 19 2 lt 16 26 43 Fi P np 16.53 4.024 1.26 69.83 7.948 1.29 232.8 14.45 1.3 Ad 78.54 314.2 1018 At 58 245 817 nL 6.01 9.5 14.9 n0 5.32 12.7 25.2 kb 320.9 1242 3791 Chap. 8 Solutions, Page 37/70 www.konkur.in A large range is presented here. Any one of the solutions is acceptable. A decision-maker might be cost such as N × cost/bolt, and/or N × cost per hole, etc. ________________________________________________________________________ 8-44 Instructors: See note preceding Prob. 8-33. This is a design problem and there is no closed-form solution path or a unique solution. What is presented here is one possible iterative approach. We will demonstrate this with an example. 1. Select the diameter, d. For this example, let d = 0.375 in. Using Eq. (8-20) on three frusta (see Prob. 8-36 solution), and combining using Eq. (8-19), yields km = 7.42 Mlbf/in. 2. Look up the nut height in Table A-31. For the example, H = 0.3281 in. From this, L ≥ l + H = 0.875 + 0.3281 = 1.2031 in. Rounding up, L = 1.25. Next, calculations are made for LT = 2(0.375) + 0.25 = 1 in, ld = 1.25 − 1 = 0.25 in, lt = 0.875 − 0.25 = 0.625 in. From step 1, Ad =π (0.3752)/4 = 0.1104 in2. Next, from Table 8-1, At = 0.0775 in2. From Eq. (8-17), kb = 2.905 Mlbf/in. Finally, from Eq. (f), p. 428, C = 0.263. 3. From Prob. 8-36, the bolt circle diameter is E = 6 in. Substituting this for Db in Eq. (834), for the number of bolts N= π Db = π (6) 4d 4 ( 0.375 ) Rounding this up gives N = 13. = 12.6 4. Next, select a grade bolt. Based on the solution to Prob. 8-36, the strength of SAE grade 8 seemed high for overload and separation. Try SAE grade 5 with Sp = 85 kpsi. From Eqs. (8-31) and (8-32) for a non-permanent connection, Fi = 4.941 kips. 5. The external load requirement per bolt is P = 1.15 pg Ag/N, where from Prob 8-34, pg = 1 200 psi, and Ag = π (3.252)/4. This gives P = 0.881 kips/bolt. 6. Using Eqs. (8-28) to (8-30) yield np = 1.27, nL = 6.65, and n0 = 7.81. Steps 1 - 6 can be easily implemented on a spreadsheet with lookup tables for the tables used from the text. For this solution we only looked at one bolt size, 83 − 16 , but evaluated changing the bolt grade. The results for four bolt grades are shown below. The dimension of each term is consistent with the example given above. d km H L 0.375 7.42 0.3281 1.25 Shigley’s MED, 10th edition LT 1 ld lt Ad At kb 0.25 0.625 0.1104 0.0775 2.905 Chap. 8 Solutions, Page 38/70 www.konkur.in d 0.375 0.375 0.375 0.375 C 0.281 0.281 0.281 0.281 N 13 13 13 13 SAE grade 1 2 4 5 Sp 33 55 65 85 Fi 1.918 3.197 3.778 4.941 P 0.881 0.881 0.881 0.881 np 1.18 1.24 1.25 1.27 nL 2.58 4.30 5.08 6.65 n0 3.03 5.05 5.97 7.81 Note that changing the bolt grade only affects Sp, Fi , np, nL, and n0. Any one of the solutions is acceptable, especially the lowest grade bolt. ________________________________________________________________________ 8-45 (a) Fb′ = RFb′,max sin θ Half of the external moment is contributed by the line load in the interval 0 ≤ θ ≤ π π M = ∫ Fb′R 2 sin θ dθ = 0 2 π M = Fb′,max R 2 2 2 π 0 Fb′,max R 2 sin 2 θ dθ M π R2 from which Fb′,max = Fmax = ∫ φ2 ∫φ Fb′R sin θ dθ = 1 M φ2 M R sin θ dθ = (cos φ1 - cos φ2 ) 2 ∫φ πR 1 πR Noting φ1 = 75°, φ2 = 105°, Fmax = 12 000 (cos 75o - cos105o ) = 494 lbf π (8 / 2) Fmax = Fb′,max R∆φ = (b) Fmax = Ans. M 2M 2π ( R) = 2 N RN πR 2(12 000) = 500 lbf (8 / 2)(12) Ans. (c) F = Fmax sin θ M = 2 Fmax R [(1) sin2 90° + 2 sin2 60° + 2 sin2 30° + (1) sin2 (0)] = 6FmaxR from which, Fmax = Shigley’s MED, 10th edition M 12 000 = = 500 lbf 6R 6(8 / 2) Ans. Chap. 8 Solutions, Page 39/70 www.konkur.in The simple general equation resulted from part (b) 2M RN ________________________________________________________________________ Fmax = 8-46 (a) From Table 8-11, Sp = 600 MPa. From Table 8-1, At = 353 mm2. Eq. (8-31): Table 8-15: Fi = 0.9 At S p = 0.9 ( 353 )( 600 ) (10 −3 ) = 190.6 kN K = 0.18 Eq. (8-27): T = K Fi d = 0.18(190.6)(24) = 823 N⋅m Ans. (b) Washers: t = 4.6 mm, d = 24 mm, D = 1.5(24) = 36 mm, E = 207 GPa. Eq. (8-20), k1 = 0.5774π ( 207 ) 24 1.155 ( 4.6 ) + 36 − 24 ( 36 + 24 ) ln 1.155 ( 4.6 ) + 36 + 24 ( 36 − 24 ) = 31 990 MN/m Cast iron: t = 20 mm, d = 24 mm, D = 36 + 2(4.6) tan 30° = 41.31 mm, E = 135 GPa. Eq. (8-20) ⇒ k2 = 10 785 MN/m Steel joist: t = 20 mm, d = 24 mm, D = 41.31 mm, E = 207 GPa. Eq. (8-20) ⇒ k3 = 16 537 MN/m Eq. (8-18): km = (2 / 31 990 + 1 / 10 785 +1 / 16 537)−1 = 4 636 MN/m Bolt: l = 2(4.6) + 2(20) = 49.2 mm. Nut, Table A-31, H = 21.5 mm. L > 49.2 + 21.5 = 70.7 mm. From Table A-17, use L = 80 mm. From Eq. (8-14) LT = 2(24) + 6 = 54 mm, ld = 80 − 54 = 26 mm, lt = 49.2 − 26 = 23.2 mm From Table (8-1), At = 353 mm2, Ad = π (242) / 4 = 452.4 mm2 Eq. (8-17): kb = 452.4 ( 353) 207 Ad At E = = 1680 MN/m Ad lt + At ld 452.4 ( 23.2 ) + 353 ( 26 ) C = kb / (kb + km) = 1680 / (1680 + 4636) = 0.266, Sp = 600 MPa, Fi = 190.6 kN, P = Ptotal / N = 18/4 = 4.5 kN Shigley’s MED, 10th edition Chap. 8 Solutions, Page 40/70 www.konkur.in Yield: From Eq. (8-28) np = S p At CP + Fi = 600 ( 353)10−3 = 1.10 Ans. 0.266 ( 4.5 ) + 190.6 Load factor: From Eq. (8-29) nL = S p At − Fi CP = 600 ( 353)10−3 − 190.6 = 17.7 0.266 ( 4.5 ) Ans. Separation: From Eq. (8-30) n0 = Fi 190.6 = = 57.7 P (1 − C ) 4.5 (1 − 0.266 ) Ans. As was stated in the text, bolts are typically preloaded such that the yielding factor of safety is not much greater than unity which is the case for this problem. However, the other load factors indicate that the bolts are oversized for the external load. ______________________________________________________________________________ 8-47 (a) ISO M 20 × 2.5 grade 8.8 coarse pitch bolts, lubricated. Table 8-2, At = 245 mm2 Table 8-11, Sp = 600 MPa Fi = 0.90 At Sp = 0.90(245)600(10−3) = 132.3 kN Table 8-15, K = 0.18 Eq. (8-27), T = KFi d = 0.18(132.3)20 = 476 N ⋅ m Ans. (b) Table A-31, H = 18 mm, L ≥ LG + H = 48 + 18 = 66 mm. Round up to L = 80 mm per Table A-17. LT = 2d + 6 = 2(20) + 6 = 46 mm ld = L - LT = 80 − 46 = 34 mm lt = l - ld = 48 − 34 = 14 mm Ad = π (202) /4 = 314.2 mm2, kb = Ad At E 314.2(245)(207) = = 1251.9 MN/m Ad lt + Alt d 314.2(14) + 245(34) Members: Since all members are steel use Eq. (8-22) with E = 207 MPa, l = 48 mm, d = 20mm Shigley’s MED, 10th edition Chap. 8 Solutions, Page 41/70 www.konkur.in km = 0.5774π ( 207 ) 20 0.5774π Ed = = 4236 MN/m 0.5774l + 0.5d 0.5774 ( 48 ) + 0.5 ( 20 ) 2 ln 5 2 ln 5 0.5774l + 2.5d 0.5774 ( 48 ) + 2.5 ( 20 ) kb 1251.9 = = 0.228 kb + km 1251.9 + 4236 P = Ptotal / N = 40/2 = 20 kN, C= Yield: From Eq. (8-28) S p At 600 ( 245 )10−3 np = = = 1.07 Ans. CP + Fi 0.228 ( 20 ) + 132.3 Load factor: From Eq. (8-29) nL = S p At − Fi CP 600 ( 245 )10 −3 − 132.3 = = 3.22 0.228 ( 20 ) Ans. Separation: From Eq. (8-30) Fi 132.3 = = 8.57 Ans. P (1 − C ) 20 (1 − 0.228 ) ______________________________________________________________________________ n0 = 8-48 From Prob. 8-29 solution, Pmax =13.33 kips, C = 0.2, Fi = 12.77 kips, At = 0.141 9 in2 F 12.77 σi = i = = 90.0 kpsi At 0.141 9 CP 0.2 (13.33) Eq. (8-39), σa = = = 9.39 kpsi 2 At 2 ( 0.141 9 ) Eq. (8-41), σ m = σ a + σ i = 9.39 + 90.0 = 99.39 kpsi (a) Goodman Eq. (8-45) for grade 8 bolts, Se = 23.2 kpsi (Table 8-17), Sut = 150 kpsi (Table 8-9) S ( S − σ i ) 23.2 (150 − 90.0 ) n f = e ut = = 0.856 Ans. σ a ( Sut + Se ) 9.39 (150 + 23.2 ) (b) Gerber Eq. (8-46) 1 nf = Sut Sut2 + 4 Se ( Se + σ i ) − Sut2 − 2σ i Se 2σ a Se = 1 150 1502 + 4 ( 23.2 )( 23.2 + 90.0 ) − 1502 − 2 ( 90.0 ) 23.2 = 1.32 2 ( 9.39 ) 23.2 Ans. (c) ASME-elliptic Eq. (8-47) with Sp = 120 kpsi (Table 8-9) Shigley’s MED, 10th edition Chap. 8 Solutions, Page 42/70 www.konkur.in nf = = ( Se S p S p2 + Se2 − σ i2 − σ i Se 2 2 σ a ( S p + Se ) ) 23.2 120 1202 + 23.22 − 902 − 90 ( 23.2 ) = 1.30 9.39 (1202 + 23.22 ) Ans. ______________________________________________________________________________ 8-49 Attention to the Instructor. Part (d) requires the determination of the endurance strength, Se, of a class 5.8 bolt. Table 8-17 does not provide this and the student will be required to estimate it by other means [see the solution of part (d)]. Per bolt, Pbmax = 60/8 = 7.5 kN, Pbmin = 20/8 = 2.5 kN kb 1 C= = = 0.278 kb + k m 1 + 2.6 (a) Table 8-1, At = 20.1 mm2; Table 8-11, Sp = 380 MPa Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(20.1)380(10−3) = 5.73 kN S A 380 ( 20.1)10−3 np = p t = = 0.98 Ans. Yield, Eq. (8-28), CP + Fi 0.278 ( 7.5) + 5.73 (b) Overload, Eq. (8-29), nL = (c) Separation, Eq. (8-30), n0 = (d) Goodman, Eq. (8-35), Eq. (8-36), σ m = CP = 380 ( 20.1)10−3 − 5.73 0.278 ( 7.5 ) Fi 5.73 = = 1.06 P (1 − C ) 7.5 (1 − 0.278) σa = C ( Pb max + Pb min ) 2 At S p At − Fi + C ( Pb max − Pb min ) 2 At = = 0.915 Ans. Ans. 0.278 ( 7.5 − 2.5 )103 2 ( 20.1) 3 Fi 0.278 ( 7.5 + 2.5 )10 5.73 (10 = + At 2 ( 20.1) 20.1 3 = 34.6 MPa ) = 354.2 MPa Table 8-11, Sut = 520 MPa, σi = Fi /At = 5.73(103)/20.1 = 285 MPa We have a problem for Se. Table 8-17 does not list Se for class 5.8 bolts. Here, we will estimate Se using the methods of Chapter 6. Estimate S e′ from the, Eq. (6-8), p. 290, Se′ = 0.5Sut = 0.5 ( 520 ) = 260 MPa . Table 6-2, p. 296, Eq. (6-19), p. 295, a = 4.51, b = − 0.265 ka = aSutb = 4.51( 520 −0.265 ) = 0.860 Eq. (6-21), p. 296, kb = 1 Eq. (6-26), p. 298, kc = 0.85 The fatigue stress-concentration factor, from Table 8-16, is Kf = 2.2. For simple axial loading and infinite-life it is acceptable to reduce the endurance limit by Kf and use the nominal stresses in the stress/strength/design factor equations. Thus, Eq. (6-18), p. 295, Se = ka kb kc S e′ / Kf = 0.86(1)0.85(260) / 2.2 = 86.4 MPa Shigley’s MED, 10th edition Chap. 8 Solutions, Page 43/70 www.konkur.in Eq. (8-38), nf = Se ( Sut − σ i ) Sutσ a + Se (σ m − σ i ) = 86.4 ( 520 − 285 ) 520 ( 34.6 ) + 86.4 ( 354.2 − 285 ) = 0.847 Ans. It is obvious from the various answers obtained, the bolted assembly is undersized. This can be rectified by a one or more of the following: more bolts, larger bolts, higher class bolts. ______________________________________________________________________________ 8-50 Per bolt, Pbmax = Pmax /N = 80/10 = 8 kips, Pbmin = Pmin /N = 20/10 = 2 kips C = kb / (kb + km) = 4/(4 + 12) = 0.25 (a) Table 8-2, At = 0.141 9 in2, Table 8-9, Sp = 120 kpsi and Sut = 150 kpsi Table 8-17, Se = 23.2 kpsi Eqs. (8-31) and (8-32), Fi = 0.75 At Sp ⇒ σi = Fi /At = 0.75 Sp = 0.75(120) =90 kpsi Eq. (8-35), σa = Eq. (8-36), σm = Eq. (8-38), nf = C ( Pb max − Pb min ) 2 At C ( Pb max + Pb min ) 2 At Se ( Sut − σ i ) = = 0.25 ( 8 − 2 ) 2 ( 0.141 9 ) +σi = = 5.29 kpsi 0.25 ( 8 + 2 ) 2 ( 0.141 9 ) + 90 = 98.81 kpsi 23.2 (150 − 90 ) = 1.39 Ans. Sutσ a + Se (σ m − σ i ) 150 ( 5.29 ) + 23.2 ( 98.81 − 90 ) ______________________________________________________________________________ 8-51 Instructors: See note preceding Prob. 8-33. From Prob. 8-33, C = 0.263, Pmax = 4.712 kN / bolt, Fi = 41.1 kN, Sp = 650 MPa, and At = 84.3 mm2 σi = 0.75 Sp = 0.75(650) = 487.5 MPa 3 CP 0.263 ( 4.712 )10 σa = = = 7.350 MPa Eq. (8-39): 2 At 2 ( 84.3) Eq. (8-40) σm = CP Fi + = 7.350 + 487.5 = 494.9 MPa 2 At At (a) Goodman: From Table 8-11, Sut = 900 MPa, and from Table 8-17, Se = 140 MPa S ( S − σ i ) 140 ( 900 − 487.5 ) Eq. (8-45): n f = e ut = = 7.55 Ans. σ a ( Sut + Se ) 7.350 ( 900 + 140 ) (b) Gerber: Eq. (8-46): Shigley’s MED, 10th edition Chap. 8 Solutions, Page 44/70 www.konkur.in nf = = 1 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S ut ut e e i ut i e 2σ a Se 1 900 9002 + 4 (140 )(140 + 487.5 ) − 9002 − 2 ( 487.5 )(140 ) 2 ( 7.350 )140 = 11.4 Ans. (c) ASME-elliptic: Eq. (8-47): nf = = ( Se S p S p2 + Se2 − σ i2 − σ i Se σ a ( S p2 + Se2 ) ) 140 650 650 2 + 1402 − 487.52 − 487.5 (140 ) = 9.73 2 2 7.350 ( 650 + 140 ) Ans. ______________________________________________________________________________ 8-52 Instructors: See note preceding Prob. 8-33. From Prob. 8-34, C = 0.299, Pmax = 1.443 kips/bolt, Fi = 9.05 kips, Sp = 85 kpsi, and At = 0.141 9 in2 σ i = 0.75S p = 0.75 (85) = 63.75 kpsi Eq. (8-37): σa = CP 0.299 (1.443) = = 1.520 kpsi 2 At 2 ( 0.141 9 ) Eq. (8-38) σm = CP + σ i = 1.520 + 63.75 = 65.27 kpsi 2 At (a) Goodman: From Table 8-9, Sut = 120 kpsi, and from Table 8-17, Se = 18.8 kpsi S ( S − σ i ) 18.8 (120 − 63.75 ) Eq. (8-45): n f = e ut = = 5.01 Ans. σ a ( Sut + Se ) 1.520 (120 + 18.8 ) (b) Gerber: Eq. (8-46): Shigley’s MED, 10th edition Chap. 8 Solutions, Page 45/70 www.konkur.in nf = = 1 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S ut ut e e i ut i e 2σ a Se 1 120 1202 + 4 (18.6 )(18.6 + 63.75 ) − 1202 − 2 ( 63.75 )(18.6 ) 2 (1.520 )18.6 = 7.45 Ans. (c) ASME-elliptic: Eq. (8-47): nf = = ( Se S p S p2 + S e2 − σ i2 − σ i Se σ a ( S p2 + Se2 ) ) 18.6 85 852 + 18.62 − 63.752 − 63.75 (18.6 ) = 6.22 2 2 1.520 ( 85 + 18.6 ) Ans. ______________________________________________________________________________ 8-53 Instructors: See note preceding Prob. 8-33. From Prob. 8-35, C = 0.228, Pmax = 7.679 kN/bolt, Fi = 36.1 kN, Sp = 830 MPa, and At = 58.0 mm2 σi = 0.75 Sp = 0.75(830) = 622.5 MPa 3 CP 0.228 ( 7.679 )10 σa = Eq. (8-37): = = 15.09 MPa 2 At 2 ( 58.0 ) CP + σ i = 15.09 + 622.5 = 637.6 MPa 2 At (a) Goodman: From Table 8-11, Sut = 1040 MPa, and from Table 8-17, Se = 162 MPa S ( S − σ i ) 162 (1040 − 622.5 ) Eq. (8-45): n f = e ut = = 3.73 Ans. σ a ( Sut + Se ) 15.09 (1040 + 162 ) Eq. (8-38) Shigley’s MED, 10th edition σm = Chap. 8 Solutions, Page 46/70 www.konkur.in (b) Gerber: Eq. (8-46): 1 nf = Sut Sut2 + 4 Se ( Se + σ i ) − Sut2 − 2σ i Se 2σ a Se = 1 1040 10402 + 4 (162 )(162 + 622.5 ) − 10402 − 2 ( 622.5 )(162 ) 2 (15.09 )162 = 5.74 Ans. (c) ASME-elliptic: Eq. (8-47): nf = = ( Se S p S p2 + Se2 − σ i2 − σ i Se 2 2 σ a ( S p + Se ) ) 162 830 8302 + 1622 − 622.52 − 622.5 (162 ) = 5.62 2 2 15.09 ( 830 + 162 ) Ans. ______________________________________________________________________________ 8-54 Instructors: See note preceding Prob. 8-33. From Prob. 8-36, C = 0.291, Pmax = 1.244 kips/bolt, Fi = 9.57 kips, Sp = 120 kpsi, and At = 0.106 3 in2 σ i = 0.75S p = 0.75 (120 ) = 90 kpsi Eq. (8-37): σa = CP 0.291(1.244 ) = = 1.703 kpsi 2 At 2 ( 0.106 3) Eq. (8-38) σm = CP + σ i = 1.703 + 90 = 91.70 kpsi 2 At (a) Goodman: From Table 8-9, Sut = 150 kpsi, and from Table 8-17, Se = 23.2 kpsi S (S −σi ) 23.2 (150 − 90 ) Eq. (8-45): n f = e ut = = 4.72 Ans. σ a ( Sut + Se ) 1.703 (150 + 23.2 ) (b) Gerber: Eq. (8-46): Shigley’s MED, 10th edition Chap. 8 Solutions, Page 47/70 www.konkur.in nf = = 1 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S ut ut e e i ut i e 2σ a Se 1 150 1502 + 4 ( 23.2 )( 23.2 + 90 ) − 1502 − 2 ( 90 )( 23.2 ) 2 (1.703) 23.2 = 7.28 Ans. (c) ASME-elliptic: Eq. (8-47): nf = = ( Se S p S p2 + S e2 − σ i2 − σ i Se σ a ( S p2 + Se2 ) ) 23.2 120 1202 + 23.22 − 902 − 90 ( 23.2 ) = 7.24 2 2 1.703 (120 + 18.6 ) Ans. ______________________________________________________________________________ 8-55 Instructors: See note preceding Prob. 8-33. From Prob. 8-51, C = 0.263, Se = 140 MPa, Sut = 900 MPa, At = 84.4 mm2, σi = 487.5 MPa, and Pmax = 4.712 kN. Pmin = Pmax / 2 = 4.712/2 = 2.356 kN σa = Eq. (8-35): Eq. (8-36): σm = = C ( Pmax − Pmin ) 2 At = 0.263 ( 4.712 − 2.356 )103 2 ( 84.3) = 3.675 MPa C ( Pmax + Pmin ) +σi 2 At 0.263 ( 4.712 + 2.356 )103 + 487.5 = 498.5 MPa 2 ( 84.3) Eq. (8-38): Se ( Sut − σ i ) 140 ( 900 − 487.5 ) = = 11.9 Ans. Sutσ a + Se (σ m − σ i ) 900 ( 3.675 ) + 140 ( 498.5 − 487.5 ) ______________________________________________________________________________ nf = Shigley’s MED, 10th edition Chap. 8 Solutions, Page 48/70 www.konkur.in 8-56 Instructors: See note preceding Prob. 8-33. From Prob. 8-52, C = 0.299, Se = 18.8 kpsi, Sut = 120 kpsi, At = 0.141 9 in2, σi = 63.75 kpsi, and Pmax = 1.443 kips Pmin = Pmax / 2 = 1.443/2 = 0.722 kips σa = Eq. (8-35): Eq. (8-36): σm = = C ( Pmax − Pmin ) 0.299 (1.443 − 0.722 ) = = 0.760 kpsi 2 At 2 ( 0.141 9 ) C ( Pmax + Pmin ) +σi 2 At 0.299 (1.443 + 0.722 ) + 63.75 = 66.03 kpsi 2 ( 0.141 9 ) Eq. (8-38): nf = Se ( Sut − σ i ) = 18.8 (120 − 63.75 ) = 7.89 Ans. Sutσ a + Se (σ m − σ i ) 120 ( 0.760 ) + 18.8 ( 66.03 − 63.75 ) ______________________________________________________________________________ 8-57 Instructors: See note preceding Prob. 8-33. From Prob. 8-53, C = 0.228, Se = 162 MPa, Sut = 1040 MPa, At = 58.0 mm2, σi = 622.5 MPa, and Pmax = 7.679 kN. Pmin = Pmax / 2 = 7.679/2 = 3.840 kN σa = Eq. (8-35): Eq. (8-36): σm = C ( Pmax − Pmin ) 2 At = 0.228 ( 7.679 − 3.840 )103 2 ( 58.0 ) = 7.546 MPa C ( Pmax + Pmin ) +σi 2 At 0.228 ( 7.679 + 3.840 )103 = + 622.5 = 645.1 MPa 2 ( 58.0 ) Eq. (8-38): Shigley’s MED, 10th edition Chap. 8 Solutions, Page 49/70 www.konkur.in Se ( Sut − σ i ) 162 (1040 − 622.5 ) = = 5.88 Ans. Sutσ a + Se (σ m − σ i ) 1040 ( 7.546 ) + 162 ( 645.1 − 622.5 ) ______________________________________________________________________________ nf = 8-58 Instructors: See note preceding Prob. 8-33. From Prob. 8-54, C = 0.291, Se = 23.2 kpsi, Sut = 150 kpsi, At = 0.106 3 in2, σi = 90 kpsi, and Pmax = 1.244 kips Pmin = Pmax / 2 = 1.244/2 = 0.622 kips σa = Eq. (8-35): Eq. (8-36): σm = = C ( Pmax − Pmin ) 2 At = 0.291(1.244 − 0.622 ) 2 ( 0.106 3) = 0.851 kpsi C ( Pmax + Pmin ) +σi 2 At 0.291(1.244 + 0.622 ) + 90 = 92.55 kpsi 2 ( 0.106 3) Eq. (8-38): Se ( Sut − σ i ) 23.2 (150 − 90 ) = = 7.45 Ans. Sutσ a + Se (σ m − σ i ) 150 ( 0.851) + 23.2 ( 92.55 − 90 ) ______________________________________________________________________________ nf = 8-59 Let the repeatedly-applied load be designated as P. From Table A-22, Sut = 93.7 kpsi. Referring to the Figure of Prob. 3-122, the following notation will be used for the radii of Section AA. ri = 1.5 in, ro = 2.5 in, rc = 2.0 in From Table 3-4, p. 135, with R = 0.5 in R2 0.52 rn = = = 1.968 246 in 2 rc − rc2 − R 2 2 2 − 22 − 0.52 e co ci A = = = = ( ) ( ) rc − rn = 2.0 − 1.968 246 = 0.031 754 in ro - rn = 2.5 − 1.968 246 = 0.531 754 in rn - ri = 1.968 246 − 1.5 = 0.468 246 in π (12 ) / 4 = 0.7854 in 2 If P is the maximum load Shigley’s MED, 10th edition Chap. 8 Solutions, Page 50/70 www.konkur.in M = Prc = 2 P P rc P 2(0.468) σ i = 1 + c i = 1 + = 26.29P A eri 0.785 4 0.031 754(1.5) 26.294 P σ = 13.15P σa = σm = i = 2 2 (a) Eye: Section AA, Table 6-2, p. 296, a = 14.4 kpsi, b = − 0.718 Eq. (6-19), p. 295, ka = 14.4(93.7) −0.718 = 0.553 Eq. (6-23), p. 297, de = 0.370 d Eq. (6-20), p. 296, 0.37 kb = 0.30 −0.107 = 0.978 Eq. (6-26), p. 298, kc = 0.85 Eq. (6-8), p. 290, Se′ = 0.5Sut = 0.5 ( 93.7 ) = 46.85 kpsi Eq. (6-18) p. 295, Se = 0.553(0.978)0.85(46.85) = 21.5 kpsi From Table 6-7, p. 315, for Gerber 2 2 2σ m Se 1 Sut σ a nf = −1 + 1 + 2 σ m Se Sutσ a With σm = σa, 2 2 2Se 1 1 Sut2 93.7 2 2(21.5) 1.557 nf = −1 + 1 + = − 1 + 1 + = 2 σ a Se P 93.7 Sut 2 13.15P(21.5) where P is in kips. Thread: Die cut. Table 8-17 gives Se = 18.6 kpsi for rolled threads. Use Table 8-16 to find Se for die cut threads Se = 18.6(3.0/3.8) = 14.7 kpsi Table 8-2, At = 0.663 in2, σ = P/At = P /0.663 = 1.51 P, σa = σm =σ /2 = 0.755 P From Table 6-7, Gerber 2 2 2S e 1 1 Sut2 93.7 2 2(14.7) 19.01 −1 + 1 + nf = −1 + 1 + = = 2 σ a Se Sut 2 0.755P(14.7) P 93.7 Shigley’s MED, 10th edition Chap. 8 Solutions, Page 51/70 www.konkur.in Comparing 1910/P with 19 200/P, we conclude that the eye is weaker in fatigue. Ans. (b) Strengthening steps can include heat treatment, cold forming, cross section change (a round is a poor cross section for a curved bar in bending because the bulk of the material is located where the stress is small). Ans. (c) For nf = 2 P = 1.557 (103 ) = 779 lbf, max. load Ans. 2 ______________________________________________________________________________ 8-60 Member, Eq. (8-22) with E =16 Mpsi, d = 0.75 in, and l = 1.5 in km = 0.5774π (16 ) 0.75 0.5774π Ed = = 13.32 Mlbf/in 0.5774l + 0.5d 0.5774 (1.5 ) + 0.5 ( 0.75 ) 2 ln 5 2 ln 5 0.5774l + 2.5d 0.5774 (1.5 ) + 2.5 ( 0.75 ) Bolt, Eq. (8-13), LT = 2d + 0.25 = 2(0.75) + 0.25 = 1.75 in l = 1.5 in ld = L − LT = 2.5 − 1.75 = 0.75 in lt = l − ld = 1.5 − 0.75 = 0.75 in Table 8-2, At = 0.373 in2 Ad = π(0.752)/4 = 0.442 in2 Eq. (8-17), kb = 0.442 ( 0.373) 30 Ad At E = = 8.09 Mlbf/in Ad lt + At ld 0.442 ( 0.75 ) + 0.373 ( 0.75 ) C= kb 8.09 = = 0.378 kb + k m 8.09 + 13.32 Eq. (8-35), σa = C ( Pmax − Pmin ) 2 At = 0.378 ( 6 − 4 ) 2 ( 0.373) = 1.013 kpsi Eq.(8-36), Shigley’s MED, 10th edition Chap. 8 Solutions, Page 52/70 www.konkur.in σm = C ( Pmax + Pmin ) Fi 0.378 ( 6 + 4 ) 25 + = + = 72.09 kpsi 2 At At 2 ( 0.373) 0.373 (a) From Table 8-9, Sp = 85 kpsi, and Eq. (8-51), the yielding factor of safety is np = Sp σm +σa = 85 = 1.16 72.09 + 1.013 Ans. (b) From Eq. (8-29), the overload factor of safety is nL = S p At − Fi CPmax = 85 ( 0.373) − 25 0.378 ( 6 ) = 2.96 Ans. (c) From Eq. (8-30), the factor of safety based on joint separation is n0 = Fi 25 = = 6.70 Pmax (1 − C ) 6 (1 − 0.378 ) Ans. (d) From Table 8-17, Se = 18.6 kpsi; Table 8-9, Sut = 120 kps; the preload stress is σi = Fi / At = 25/0.373 = 67.0 kpsi; and from Eq. (8-38) nf = Se ( Sut − σ i ) = 18.6 (120 − 67.0 ) = 4.56 Ans. Sutσ a + Se (σ m − σ i ) 120 (1.013) + 18.6 ( 72.09 − 67.0 ) ______________________________________________________________________________ 8-61 (a) Table 8-2, At = 0.1419 in2 Table 8-9, Sp = 120 kpsi, Sut = 150 kpsi Table 8-17, Se = 23.2 kpsi Eqs. (8-31) and (8-32), σi = 0.75 Sp = 0.75(120) = 90 kpsi kb 4 = = 0.2 kb + k m 4 + 16 CP 0.2 P σa = = = 0.705P kpsi 2 At 2(0.141 9) C = Eq. (8-45) for the Goodman criterion, S (S − σi ) 23.2(150 − 90) 11.4 n f = e ut = = = 2 σ a ( Sut + Se ) 0.705P(150 + 23.2) P ⇒ P = 5.70 kips Ans. (b) Fi = 0.75At Sp = 0.75(0.141 9)120 = 12.77 kips Yield, Eq. (8-28), Shigley’s MED, 10th edition Chap. 8 Solutions, Page 53/70 www.konkur.in np = S p At = 120 ( 0.141 9 ) = 1.22 0.2 ( 5.70 ) + 12.77 CP + Fi Load factor, Eq. (8-29), S A - Fi 120(0.141 9) − 12.77 nL = p t = = 3.74 CP 0.2(5.70) Ans. Ans. Separation load factor, Eq. (8-30) Fi 12.77 = = 2.80 Ans. P(1 - C ) 5.70(1 − 0.2) ______________________________________________________________________________ n0 = 8-62 Table 8-2, At = 0.969 in2 (coarse), At = 1.073 in2 (fine) Table 8-9, Sp = 74 kpsi, Sut = 105 kpsi Table 8-17, Se = 16.3 kpsi Coarse thread, Fi = 0.75 At Sp = 0.75(0.969)74 = 53.78 kips σi = 0.75 Sp = 0.75(74) = 55.5 kpsi σa = CP 0.30 P = = 0.155P kpsi 2 At 2(0.969) Gerber, Eq. (8-46), nf = 1 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S ut ut e e i ut i e 2σ a Se 1 105 1052 + 4 (16.3)(16.3 + 55.5 ) − 1052 − 2 ( 55.5 )16.3 = 64.28 2 ( 0.155 P )16.3 P With nf =2, = P = 64.28 = 32.14 kip 2 Ans. Fine thread, Fi = 0.75 At Sp = 0.75(1.073)74 = 59.55kips σi = 0.75 Sp = 0.75(74) = 55.5 kpsi σa = CP 0.32 P = = 0.149 P kpsi 2 At 2(1.073) The only thing that changes in Eq. (8-46) is σa. Thus, 0.155 64.28 66.87 nf = = = 2 ⇒ P = 33.43 kips 0.149 P P Shigley’s MED, 10th edition Ans. Chap. 8 Solutions, Page 54/70 www.konkur.in Percent improvement, 33.43 − 32.14 (100) 4% Ans. 32.14 ______________________________________________________________________________ 8-63 For an M 30 × 3.5 ISO 8.8 bolt with P = 65 kN/bolt and C = 0.28 Table 8-1, Table 8-11, Table 8-17, At = 561 mm2 Sp = 600 MPa, Sut = 830 MPa Se = 129 MPa Eq. (8-31), Fi = 0.75Fp = 0.75 At Sp = 0.75(5610600(10−3) = 252.45 kN σi = 0.75 Sp = 0.75(600) = 450 MPa CP 0.28 ( 65)10 = = 16.22 MPa σa = 2 At 2 ( 561) 3 Eq. (8-39), Gerber, Eq. (8-46), nf = = 1 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S ut ut e e i ut i e 2σ a Se 1 830 8302 + 4 (129 )(129 + 450 ) − 8302 − 2 ( 450 )129 2 (16.22 )129 = 4.75 Ans. The yielding factor of safety, from Eq. (8-28) is np = S p At CP + Fi = 600 ( 561)10−3 0.28 ( 65 ) + 252.45 = 1.24 Ans. From Eq. (8-29), the load factor is nL = S p At − Fi CP 600 ( 561)10−3 − 252.45 = = 4.62 0.28 ( 65) Ans. The separation factor, from Eq. (8-30) is Fi 252.45 = = 5.39 Ans. P (1 − C ) 65 (1 − 0.28 ) ______________________________________________________________________________ n0 = Shigley’s MED, 10th edition Chap. 8 Solutions, Page 55/70 www.konkur.in At = 0.077 5 in2 Sp = 85 kpsi, Sut = 120 kpsi Se = 18.6 kpsi 8-64 (a) Table 8-2, Table 8-9, Table 8-17, Unthreaded grip, kb = Am = Ad E π (0.375) 2 (30) = = 0.245 Mlbf/in per bolt l 4(13.5) π [( D + 2t ) 2 - D 2 ] = π (4.752 - 42 ) = 5.154 in 2 4 4 Am E 5.154(30) 1 km = = = 2.148 Mlbf/in/bolt. l 12 6 (b) Ans. Ans. Fi = 0.75 At Sp = 0.75(0.0775)(85) = 4.94 kip σ i = 0.75S p = 0.75(85) = 63.75 kpsi 2000 π 2 P = pA = (4) = 4189 lbf/bolt 6 4 kb 0.245 C = = = 0.102 kb + k m 0.245 + 2.148 CP 0.102(4.189) = = 2.77 kpsi σa = 2 At 2(0.0775) From Eq. (8-46) for Gerber fatigue criterion, nf = = 1 S S 2 + 4 S ( S + σ ) − S 2 − 2σ S ut ut e e i ut i e 2σ a Se 1 120 1202 + 4 (18.6 )(18.6 + 63.75 ) − 1202 − 2 ( 63.75 )18.6 = 4.09 2 ( 2.77 )18.6 Ans. (c) Pressure causing joint separation from Eq. (8-30) Fi =1 P(1 − C ) Fi 4.94 P= = = 5.50 kip 1 − C 1 − 0.102 P 5.50 p = = 6 = 2.63 kpsi Ans. A π (42 ) / 4 ______________________________________________________________________________ n0 = 8-65 From the solution of Prob. 8-64, At = 0.077 5 in2, Sut = 120 kpsi, Se = 18.6 kpsi, C = 0.102, σi = 63.75 kpsi Pmax = pmaxA = 2 π (42)/4 = 25.13 kpsi, Pmin = pminA = 1.2 π (42)/4 = 15.08 kpsi, Shigley’s MED, 10th edition Chap. 8 Solutions, Page 56/70 www.konkur.in Eq. (8-35), σa = Eq. (8-36), σm = C ( Pmax − Pmin ) 0.102 ( 25.13 − 15.08 ) = = 6.61 kpsi 2 At 2 ( 0.077 5 ) C ( Pmax + Pmin ) 2 At +σi = 0.102 ( 25.13 + 15.08 ) 2 ( 0.077 5 ) + 63.75 = 90.21 kpsi Eq. (8-38), Se ( Sut − σ i ) 18.6 (120 − 63.75 ) nf = = = 0.814 Sutσ a + Se (σ m − σ i ) 120 ( 6.61) + 18.6 ( 90.21 − 63.75 ) Ans. This predicts a fatigue failure. ______________________________________________________________________________ 8-66 Members: Sy = 57 kpsi, Ssy = 0.577(57) = 32.89 kpsi. Bolts: SAE grade 5, Sy = 92 kpsi, Ssy = 0.577(92) = 53.08 kpsi Shear in bolts, π (0.252 ) As = 2 = 0.0982 in 2 4 AS 0.0982(53.08) Fs = s sy = = 2.61 kips n 2 Bearing on bolts, Ab = 2(0.25)0.25 = 0.125 in2 AS 0.125(92) Fb = b yc = = 5.75 kips n 2 Bearing on member, 0.125(57) Fb = = 3.56 kips 2 Tension of members, At = (1.25 − 0.25)(0.25) = 0.25 in2 0.25(57) = 7.13 kip 2 F = min(2.61, 5.75, 3.56, 7.13) = 2.61 kip Ft = Ans. The shear in the bolts controls the design. ______________________________________________________________________________ 8-67 Members, Table A-20, Sy = 42 kpsi Bolts, Table 8-9, Sy = 130 kpsi, Ssy = 0.577(130) = 75.01 kpsi Shear of bolts, Shigley’s MED, 10th edition Chap. 8 Solutions, Page 57/70 www.konkur.in π ( 5 /16 )2 As = 2 = 0.1534 in 2 4 τ= Fs 5 = = 32.6 kpsi As 0.1534 n= S sy τ = 75.01 = 2.30 32.6 Ans. Bearing on bolts, Ab = 2(0.25)(5/16) = 0.1563 in2 5 = −32.0 kpsi σb = − 0.1563 S 130 n= y = = 4.06 Ans. σ b 32.0 Bearing on members, n= Tension of members, Sy σb = 42 = 1.31 32 Ans. At = [2.375 − 2(5/16)](1/4) = 0.4375 in2 σt = 5 = 11.4 kpsi 0.4375 Sy 42 = 3.68 Ans. σ t 11.4 ______________________________________________________________________________ n= = 8-68 Members: Table A-20, Sy = 490 MPa, Ssy = 0.577(490) = 282.7 MPa Bolts: Table 8-11, ISO class 5.8, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Shear in bolts, π (202 ) 2 As = 2 = 628.3 mm 4 AS 628.3(242.3)10−3 Fs = s sy = = 60.9 kN n 2.5 Bearing on bolts, Ab = 2(20)20 = 800 mm2 Shigley’s MED, 10th edition Chap. 8 Solutions, Page 58/70 www.konkur.in Fb = Ab S yc n = 800(420)10−3 = 134 kN 2.5 Bearing on member, Fb = 800(490)10−3 = 157 kN 2.5 Tension of members, At = (80 − 20)(20) = 1 200 mm2 1 200(490)10−3 Ft = = 235 kN 2.5 F = min(60.9, 134, 157, 235) = 60.9 kN Ans. The shear in the bolts controls the design. ______________________________________________________________________________ 8-69 Members: Table A-20, Sy = 320 MPa Bolts: Table 8-11, ISO class 5.8, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa Shear of bolts, As = π (202)/4 = 314.2 mm2 90 (103 ) = 95.48 MPa τs = 3 ( 314.2 ) S 242.3 n = sy = = 2.54 Ans. 95.48 τs Bearing on bolt, Ab = 3(20)15 = 900 mm2 90 (103 ) σb = − = −100 MPa 900 S 420 n= y = = 4.2 Ans. 100 σb Bearing on members, S 320 n= y = = 3.2 Ans. 100 σb Tension on members, 90 (103 ) F σt = = = 46.15 MPa A 15[190 − 3 ( 20 )] S 320 n= y = = 6.93 Ans. 46.15 σt ______________________________________________________________________________ Shigley’s MED, 10th edition Chap. 8 Solutions, Page 59/70 www.konkur.in 8-70 Members: Sy = 57 kpsi Bolts: Sy = 100 kpsi, Ssy = 0.577(100) = 57.7 kpsi Shear of bolts, π (1/ 4 ) 2 A = 3 = 0.1473 in 2 4 F 5 τs = = = 33.94 kpsi As 0.1473 n= S sy τs = 57.7 = 1.70 33.94 Ans. Bearing on bolts, Ab = 3(1/4)(5/16) = 0.2344 in2 σb = − n= Sy σb F 5 =− = −21.3 kpsi Ab 0.2344 = 100 = 4.69 21.3 Ans. Bearing on members, Ab = 0.2344 in2 (From bearing on bolts calculation) σb = − 21.3 kpsi (From bearing on bolts calculation) n= Sy σb = 57 = 2.68 21.3 Ans. Tension in members, failure across two bolts, At = 5 2.375 − 2 (1 / 4 ) = 0.5859 in 2 16 F 5 = = 8.534 kpsi At 0.5859 S 57 n= y = = 6.68 Ans. σ t 8.534 σt = ______________________________________________________________________________ Shigley’s MED, 10th edition Chap. 8 Solutions, Page 60/70 www.konkur.in 8-71 By symmetry, the reactions at each support is 1.6 kN. The free-body diagram for the left member is ∑M ∑M B =0 1.6(250) − 50 RA = 0 ⇒ RA = 8 kN A =0 200(1.6) − 50 RB = 0 ⇒ RB = 6.4 kN Members: Table A-20, Sy = 370 MPa Bolts: Table 8-11, Sy = 420 MPa, Ssy = 0.577(420) = 242.3 MPa π Bolt shear, As = (122 ) = 113.1 mm 2 4 F 8(103 ) τ = max = = 70.73 MPa As 113.1 S 242.3 n = sy = = 3.43 τ 70.73 Ab = td = 10(12) = 120 mm2 8(103 ) = −66.67 MPa σb = − 120 S 370 n= y = = 5.55 66.67 σb Bearing on member, Strength of member. The bending moments at the hole locations are: in the left member at A, MA = 1.6(200) = 320 N · m. In the right member at B, MB = 8(50) = 400 N · m. The bending moment is greater at B 1 I B = [10(503 ) − 10(123 )] = 102.7(103 ) mm 4 12 M c 400(25) (103 ) = 97.37 MPa σB = A = IA 102.7(103 ) S 370 n= y = = 3.80 σ A 97.37 At the center, call it point C, MC = 1.6(350) = 560 N · m Shigley’s MED, 10th edition Chap. 8 Solutions, Page 61/70 www.konkur.in 1 (10)(503 ) = 104.2(103 ) mm 4 12 M c 560(25) σC = C = (103 ) = 134.4 MPa IC 104.2(103 ) S 370 n= y = = 2.75 < 3.80 more critical at C σ C 134.4 n = min(3.04, 3.80, 2.75) = 2.72 Ans. ______________________________________________________________________________ IC = 8-72 The free-body diagram of the bracket, assuming the upper bolt takes all the shear and tensile load is Fs = 2500 lbf P= 2500 ( 3) 7 = 1071 lbf Table A-31, H = 7/16 = 0.4375 in. Grip, l = 2(1/2) = 1 in. L ≥ l + H = 1.4375 in. Use 1.5 in bolts. Eq. (8-13), LT = 2d + 0.25 = 2(0.5) + 0.25 = 1.25 in Table 8-7, ld = L − LT = 1.5 − 1.25 = 0.25 in lt = l − ld = 1 − 0.25 = 0.75 in Table 8-2, At = 0.141 9 in2 Ad = π (0.52) /4 = 0.196 3 in2 0.196 3 ( 0.141 9 ) 30 Ad At E Eq. (8-17), kb = = = 4.574 Mlbf/in Ad lt + At ld 0.196 3 ( 0.75 ) + 0.141 9 ( 0.25 ) Eq. (8-22), 0.5774π ( 30 ) 0.5 0.5774π Ed km = = = 16.65 Mlbf/in 0.5774 l + 0.5d 0.5774 (1) + 0.5 ( 0.5 ) 2 ln 5 2 ln 5 0.5774 l + 2.5d 0.5774 (1) + 2.5 ( 0.5 ) Shigley’s MED, 10th edition Chap. 8 Solutions, Page 62/70 www.konkur.in kb 4.574 = = 0.216 kb + km 4.574 + 16.65 Table 8-9, Sp = 65 kpsi Eqs. (8-31) and (8-32), Fi = 0.75 At Sp = 0.75(0.141 9)65 = 6.918 kips σi = 0.75 Sp = 0.75(65) = 48.75 kips C= CP + Fi 0.216 (1.071) + 6.918 = = 50.38 kpsi At 0.141 9 F 3 τs ≈ s = = 21.14 kpsi At 0.141 9 Eq. (a), p. 432, σ b = Direct shear, von Mises stress, Eq. (5-15), p. 237 σ ′ = (σ b2 + 3τ s2 ) 1/2 = 50.382 + 3 ( 21.142 ) 1/2 = 62.3 kpsi Sp 65 = = 1.04 Ans. σ ′ 62.3 ______________________________________________________________________________ n= 8-73 2P(200) = 14(50) 14(50) P = = 1.75 kN per bolt 2(200) Fs = 7 kN/bolt S p = 380 MPa At = 245 mm 2 , Ad = π (202 ) = 314.2 mm 2 4 Fi = 0.75(245)(380)(10−3 ) = 69.83 kN σ i = 0.75 ( 380 ) = 285 MPa CP + Fi 0.25(1.75) + 69.83 3 = (10 ) = 287 MPa At 245 F 7(103 ) τ = s = = 22.3 MPa Ad 314.2 σ ′ = [287 2 + 3(22.32 )]1/ 2 = 290 MPa σb = Sp 380 = = 1.31 Ans. σ ′ 290 ______________________________________________________________________________ n= Shigley’s MED, 10th edition Chap. 8 Solutions, Page 63/70 www.konkur.in 8-74 Using the result of Prob. 5-67 for lubricated assembly (replace 0.2 with 0.18 per Table 8-15) 2π f T Fx = 0.18d With a design factor of nd gives T = 0.18nd Fx d 0.18(3)(1000)d = = 716d 2π f 2π (0.12) or T/d = 716. Also, T = K (0.75S p At ) d = 0.18(0.75)(85 000) At = 11 475 At Form a table Size 1 4 - 28 5 16 - 24 3 8 − 24 At T/d = 11 475At n 0.0364 417.70 1.75 0.058 665.55 2.8 0.0878 1007.50 4.23 where the factor of safety in the last column of the table comes from n= Select a 3" 8 2π f (T / d ) 2π (0.12)(T / d ) = = 0.0042(T / d ) 0.18Fx 0.18(1000) - 24 UNF cap screw. The setting is given by T = (11 475At )d = 1007.5(0.375) = 378 lbf · in Given the coarse scale on a torque wrench, specify a torque wrench setting of 400 lbf · in. Check the factor of safety 2π f T 2π (0.12)(400) = = 4.47 0.18Fx d 0.18(1000)(0.375) ______________________________________________________________________________ n= Shigley’s MED, 10th edition Chap. 8 Solutions, Page 64/70 www.konkur.in 8-75 Bolts, from Table 8-11, Sy = 420 MPa Channel, From Table A-20, Sy = 170 MPa. From Table A-7, t = 6.4 mm Cantilever, from Table A-20, Sy = 190 MPa F′A = F′B = F′C = F / 3 M = (50 + 26 + 125) F = 201 F FA′′ = FC′′ = 201F = 2.01 F 2 ( 50 ) 1 (1) FC = FC′ + FC′′ = + 2.01 F = 2.343F 3 Shear on Bolts: The shoulder bolt shear area, As = π(102) / 4 = 78.54 mm2 Max. force, Ssy = 0.577(420) = 242.3 KPa τ max = FC S sy = As n From Eq. (1), FC = 2.343 F. Thus F= S sy As 242.3 78.54 −3 = 10 = 4.06 kN n 2.343 2.0 2.343 Bearing on bolt: The bearing area is Ab = td = 6.4(10) = 64 mm2. Similar to shear F= S y Ab 420 64 −3 = 10 = 5.74 kN n 2.343 2.0 2.343 Bearing on channel: Ab = 64 mm2, Sy = 170 MPa. Shigley’s MED, 10th edition Chap. 8 Solutions, Page 65/70 www.konkur.in F= S y Ab 170 64 −3 = 10 = 2.32 kN n 2.343 2.0 2.343 Bearing on cantilever:: Ab = 12(10) = 120 mm2, Sy = 190 MPa. F= S y Ab 190 120 −3 = 10 = 4.87 kN n 2.343 2.0 2.343 Bending of cantilever:: At C I= 1 (12 ) ( 503 − 103 ) = 1.24 (105 ) mm 4 12 σ max = Sy n = Mc 151Fc = I I ⇒ F= Sy I n 151c 5 190 1.24 (10 ) −3 10 = 3.12 kN F= 2.0 151( 25 ) So F = 2.32 kN based on bearing on channel. Ans. ______________________________________________________________________________ 8-76 Bolts, from Table 8-11, 11, Sy = 420 MPa Bracket, from Table A--20, Sy = 210 MPa 12 = 4 kN; M = 12(200) = 2400 N · m 3 2400 FA′′ = FB′′ = = 37.5 kN 64 FA = FB = (4) 2 + (37.5)2 = 37.7 kN FO = 4 kN F′ = Bolt shear: The shoulder bolt shear area, As = π(122) / 4 = 113.1 mm2 Ssy = 0.577(420) = 242.3 KPa 37.7(10)3 = 333 MPa 113 S 242.3 n = sy = = 0.728 333 τ τ = Shigley’s MED, 10th edition Ans. Chap. 8 Solutions, Page 66/70 www.konkur.in Bearing on bolts: Ab = 12(8) = 96 mm 2 37.7(10)3 = −393 MPa σb = − 96 S 420 n = yc = = 1.07 Ans. 393 σb Bearing on member: σ b = −393 MPa S 210 n = yc = = 0.534 393 σb Ans. Bending stress in plate: bd 3 bh3 bd 3 I = − − 2 + a 2bd 12 12 12 3 3 3 8(12) 8(136) 8(12) = − − 2 + (32) 2 (8)(12) 12 12 12 6 4 = 1.48(10) mm Ans. Mc 2400(68) = (10)3 = 110 MPa σ = 6 I 1.48(10) Sy 210 n= = = 1.91 Ans. σ 110 Failure is predicted for bolt shear and bearing on member. ______________________________________________________________________________ _____________________________________________________________________________ Shigley’s MED, 10th edition Chap. 8 Solutions, Page 67/70 www.konkur.in 8-77 FA = 1208 − 125 = 1083 lbf, Bolt shear: 3625 FA′′ = FB′′ = = 1208 lbf 3 FB = 1208 + 125 = 1333 lbf As = (π / 4)(0.3752) = 0.1104 in2 τ max = Fmax 1333 = = 12 070 psi As 0.1104 From Table 8-10, Sy = 100 kpsi, Ssy = 0.577(100) = 57.7 kpsi n= S sy τ max = 57.7 = 4.78 12.07 Ans. Bearing on bolt: Bearing area is Ab = td = 0.375 (0.375) = 0.1406 in2. σb = − n= Sy σb F 1333 =− = −9 481 psi Ab 0.1406 = 100 = 10.55 9.481 Ans. Bearing on member: From Table A-20, Sy = 54 kpsi. Bearing stress same as bolt n= Sy σb = 54 = 5.70 9.481 Ans. Bending of member: At B, M = 250(13) = 3250 lbf⋅in I= Shigley’s MED, 10th edition 3 1 3 3 3 4 2 − = 0.2484 in 12 8 8 Chap. 8 Solutions, Page 68/70 www.konkur.in Mc 3250 (1) = = 13 080 psi I 0.2484 σ= Sy 54 = 4.13 Ans. σ 13.08 ______________________________________________________________________________ n= = 8-78 The direct shear load per bolt is F′ = 2000/6 = 333.3 lbf. The moment is taken only by the four outside bolts. This moment is M = 2000(5) = 10 000 lbf · in. 10 000 Thus F ′′ = = 1000 lbf and the resultant bolt load is 2(5) F = (333.3) 2 + (1000) 2 = 1054 lbf Bolt strength, Table 8-9, Sy = 100 kpsi; Channel and Plate strength, Sy = 42 kpsi Shear of bolt: As = π (0.5)2/4 = 0.1963 in2 n= Ssy τ = (0.577)(100) = 10.7 1.054 / 0.1963 Ans. Bearing on bolt: Channel thickness is t = 3/16 in, Ab = 0.5(3/16) = 0.09375 in2 n= Bearing on channel: Bearing on plate: n= 100 = 8.89 1.054 / 0.09375 42 = 3.74 1.054 / 0.09375 Ab = 0.5(0.25) = 0.125 in2 n= 42 = 4.98 1.054 / 0.125 Ans. Ans. Ans. Strength of plate: I = Shigley’s MED, 10th edition 0.25(7.5)3 0.25(0.5)3 − 12 12 3 0.25(0.5) − 2 + ( 0.25 )( 0.5 ) (2.5) 2 = 7.219 in 4 12 Chap. 8 Solutions, Page 69/70 www.konkur.in M = 5000 lbf · in per plate Mc 5000(3.75) σ = = = 2597 psi I 7.219 42 n= = 16.2 Ans. 2.597 ______________________________________________________________________________ 8-79 to 8-81 Specifying bolts, screws, dowels and rivets is the way a student learns about such components. However, choosing an array a priori is based on experience. Here is a chance for students to build some experience. Shigley’s MED, 10th edition Chap. 8 Solutions, Page 70/70 www.konkur.in Chapter 9 Figure for Probs. 9-1 to 9-4 9-1 Given, b = 50 mm, d = 50 mm, h = 5 mm, τallow = 140 MPa. F = 0.707 hlτallow = 0.707(5)[2(50)](140)(10−3) = 49.5 kN Ans. ______________________________________________________________________________ 9-2 Given, b = 2 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi. F = 0.707 hlτallow = 0.707(5/16)[2(2)](25) = 22.1 kip Ans. ______________________________________________________________________________ 9-3 Given, b = 50 mm, d = 30 mm, h = 5 mm, τallow = 140 MPa. F = 0.707 hlτallow = 0.707(5)[2(50)](140)(10−3) = 49.5 kN Ans. ______________________________________________________________________________ 9-4 Given, b = 4 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi. F = 0.707 hlτallow = 0.707(5/16)[2(4)](25) = 44.2 kip Ans. ______________________________________________________________________________ 9-5 Prob. 9-1 with E7010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in = 2.923(4.45/25.4) = 0.512 kN/mm F = f l = 0.512[2(50)] = 51.2 kN Ans. ______________________________________________________________________________ 9-6 Prob. 9-2 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in Shigley’s MED, 10th edition Chapter 9 Solutions, Page 1/35 www.konkur.in F = f l = 4.64[2(2)] = 18.6 kip Ans. ______________________________________________________________________________ 9-7 Prob. 9-3 with E7010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in = 2.923(4.45/25.4) = 0.512 kN/mm F = f l = 0.512[2(50)] = 51.2 kN Ans. ______________________________________________________________________________ 9-8 Prob. 9-4 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64[2(4)] = 37.1 kip Ans. ______________________________________________________________________________ 9-9 Table A-20: 1018 CD: Sut = 440 MPa, Sy = 370 MPa 1018 HR: Sut = 400 MPa, Sy = 220 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: τ all = min(0.30Sut , 0.40S y ) = min[0.30(400), 0.40(220)] = min(120, 88) = 88 MPa for both materials. Eq. (9-3): F = 0.707hlτall = 0.707(5)[2(50)](88)(10−3) = 31.1 kN Ans. ______________________________________________________________________________ 9-10 Table A-20: 1020 CD: Sut = 68 kpsi, Sy = 57 kpsi 1020 HR: Sut = 55 kpsi, Sy = 30 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: τ all = min(0.30Sut , 0.40S y ) = min[0.30(55), 0.40(30)] = min(16.5, 12.0) = 12.0 kpsi for both materials. Eq. (9-3): F = 0.707hlτall = 0.707(5/16)[2(2)](12.0) = 10.6 kip Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 9 Solutions, Page 2/35 www.konkur.in 9-11 Table A-20: 1035 HR: Sut = 500 MPa, Sy = 270 MPa 1035 CD: Sut = 550 MPa, Sy = 460 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: τ all = min(0.30Sut , 0.40S y ) = min[0.30(500), 0.40(270)] = min(150, 108) = 108 MPa for both materials. Eq. (9-3): F = 0.707hlτall = 0.707(5)[2(50)](108)(10−3) = 38.2 kN Ans. ______________________________________________________________________________ Table A-20: 1035 HR: Sut = 72 kpsi, Sy = 39.5 kpsi 1020 CD: Sut = 68 kpsi, Sy = 57 kpsi, 1020 HR: Sut = 55 kpsi, Sy = 30 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: τ all = min(0.30Sut , 0.40S y ) = min[0.30(55), 0.40(30)] = min(16.5, 12.0) = 12.0 kpsi for both materials. Eq. (9-3): F = 0.707hlτall = 0.707(5/16)[2(4)](12.0) = 21.2 kip Ans. ______________________________________________________________________________ 9-12 9-13 2 (100 ) (103 ) 2F Eq. (9-3): = = 141 MPa Ans. τ= hl 5 2 ( 50 + 50 ) ______________________________________________________________________________ 9-14 Eq. (9-3): τ= 2 ( 40 ) 2F = = 22.6 kpsi hl ( 5 /16 ) 2 ( 2 + 2 ) Ans. ______________________________________________________________________________ 2 (100 ) (103 ) 2F 9-15 Eq. (9-3): τ = = = 177 MPa Ans. hl 5 2 ( 50 + 30 ) ______________________________________________________________________________ 9-16 Eq. (9-3): τ = 2 ( 40 ) 2F = = 15.1 kpsi hl ( 5 / 16 ) 2 ( 4 + 2 ) Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 9 Solutions, Page 3/35 www.konkur.in 9-17 b = d =50 mm, c = 150 mm, h = 5 mm, and τallow = 140 MPa. (a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kN and τ in MPa): F (103 ) V = 2.829 F τ ′y = = A 1.414 ( 5)( 50 ) Secondary shear, Table 9-1: Ju = d ( 3b 2 + d 2 ) 6 = 50 3 ( 502 ) + 502 6 = 83.33 (103 ) mm3 J = 0.707 h Ju = 0.707(5)(83.33)(103) = 294.6(103) mm4 τ x′′ = τ ′′y = Mry J = 175 F (103 ) ( 25 ) 294.6 (103 ) = 14.85 F τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 14.852 + ( 2.829 + 14.85 ) = 23.1F (1) 2 F= τ allow 23.1 = 2 140 = 6.06 kN Ans. 23.1 (b) For E7010 from Table 9-6, τallow = 21 kpsi = 21(6.89) = 145 MPa 1020 HR bar: Sut = 380 MPa, Sy = 210 MPa 1015 HR support: Sut = 340 MPa, Sy = 190 MPa Table 9-3, E7010 Electrode: Sut = 482 MPa, Sy = 393 MPa The support controls the design. Table 9-4: τallow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76) = 76 MPa The allowable load, from Eq. (1) is τ allow 76 = 3.29 kN Ans. 23.1 23.1 ______________________________________________________________________________ F= 9-18 = b = d =2 in, c = 6 in, h = 5/16 in, and τallow = 25 kpsi. Shigley’s MED, 10th edition Chapter 9 Solutions, Page 4/35 www.konkur.in (a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and τ in kpsi): V F = = 1.132 F A 1.414 ( 5 /16 )( 2 ) Secondary shear, Table 9-1: 2 2 d ( 3b 2 + d 2 ) 2 3 ( 2 ) + 2 Ju = = = 5.333 in 3 6 6 τ ′y = J = 0.707 h Ju = 0.707(5/16)(5.333) = 1.178 in4 τ x′′ = τ ′′y = Mry J = 7 F (1) 1.178 = 5.942 F τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 5.9422 + (1.132 + 5.942 ) = 9.24 F (1) 2 F= τ allow 9.24 = 2 25 = 2.71 kip Ans. 9.24 (b) For E7010 from Table 9-6, τallow = 21 kpsi 1020 HR bar: Sut = 55 kpsi, Sy = 30 kpsi 1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi The support controls the design. Table 9-4: τallow = min(0.30Sut, 0.40Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11) = 11 kpsi The allowable load, from Eq. (1) is τ allow 11 = 1.19 kip Ans. 9.24 9.24 ______________________________________________________________________________ F= 9-19 = b =50 mm, c = 150 mm, d = 30 mm, h = 5 mm, and τallow = 140 MPa. (a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kN and τ in MPa): Shigley’s MED, 10th edition Chapter 9 Solutions, Page 5/35 www.konkur.in F (103 ) V τ ′y = = = 2.829 F A 1.414 ( 5)( 50 ) Secondary shear, Table 9-1: Ju = d ( 3b 2 + d 2 ) 6 = 50 3 ( 302 ) + 502 6 = 43.33 (103 ) mm3 J = 0.707 h Ju = 0.707(5)(43.33)(103) = 153.2(103) mm4 τ x′′ = Mry J = 175F (103 ) (15 ) 153.2 (103 ) = 17.13F Mrx 175F (10 ) ( 25 ) = = 28.55 F J 153.2 (103 ) 3 τ ′′y = τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 17.132 + ( 2.829 + 28.55 ) = 35.8 F (1) 2 F= τ allow 35.8 = 2 140 = 3.91 kN Ans. 35.8 (b) For E7010 from Table 9-6, τallow = 21 kpsi = 21(6.89) = 145 MPa 1020 HR bar: Sut = 380 MPa, Sy = 210 MPa 1015 HR support: Sut = 340 MPa, Sy = 190 MPa Table 9-3, E7010 Electrode: Sut = 482 MPa, Sy = 393 MPa The support controls the design. Table 9-4: τallow = min(0.30Sut, 0.40Sy ) =min[0.30(340), 0.40(190) = min(102, 76) = 76 MPa The allowable load, from Eq. (1) is τ allow 76 = 2.12 kN Ans. 35.8 35.8 ______________________________________________________________________________ F= 9-20 = b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, and τallow = 25 kpsi. Shigley’s MED, 10th edition Chapter 9 Solutions, Page 6/35 www.konkur.in (a) Primary shear, Table 9-1(Note: b and d are interchanged between problem figure and table figure. Note, also, F in kip and τ in kpsi): V F = = 0.5658F A 1.414 ( 5 /16 )( 4 ) Secondary shear, Table 9-1: τ ′y = Ju = d ( 3b2 + d 2 ) = 6 4 3 ( 22 ) + 42 6 = 18.67 in 3 J = 0.707 h Ju = 0.707(5/16)(18.67) = 4.125 in4 τ x′′ = τ ′′y = Mry J = 8 F (1) 4.125 = 1.939 F Mrx 8 F ( 2 ) = = 3.879 F J 4.125 τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 1.9392 + ( 0.5658 + 3.879 ) = 4.85F (1) 2 F= τ allow 4.85 = 2 25 = 5.15 kip Ans. 4.85 (b) For E7010 from Table 9-6, τallow = 21 kpsi 1020 HR bar: Sut = 55 kpsi, Sy = 30 kpsi 1015 HR support: Sut = 50 kpsi, Sy = 27.5 kpsi Table 9-3, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi The support controls the design. Table 9-4: τallow = min(0.30Sut, 0.40Sy ) =min[0.30(50), 0.40(27.5) = min(15, 11) = 11 kpsi The allowable load, from Eq. (1) is τ allow 11 = 2.27 kip Ans. 4.85 4.85 ______________________________________________________________________________ F= Shigley’s MED, 10th edition = Chapter 9 Solutions, Page 7/35 www.konkur.in 9-21 Given, b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, τallow = 140 MPa. Primary shear (F in kN, τ in MPa, A in mm2): F (103 ) V τ ′y = = = 1.414 F A 1.414 ( 5 )( 50 + 50 ) Secondary shear: Table 9-1: (b + d ) Ju = 3 = ( 50 + 50 ) 3 = 166.7 (103 ) mm3 6 6 J = 0.707 h Ju = 0.707(5)166.7(103) = 589.2(103) mm4 τ x′′ = τ ′′y = Mry J = 175 F (103 ) (25) 589.2 (103 ) = 7.425F Maximum shear: τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 7.4252 + (1.414 + 7.425 ) = 11.54 F 2 τ allow 2 140 = 12.1 kN Ans. 11.54 11.54 ______________________________________________________________________________ F= 9-22 = Given, b = 2 in, c = 6 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi. Primary shear: τ ′y = V F = = 0.5658 F A 1.414 ( 5 /16 )( 2 + 2 ) Secondary shear: Table 9-1: Ju (b + d ) = 3 ( 2 + 2) = 3 = 10.67 in 3 6 6 J = 0.707 h Ju = 0.707(5/16)10.67 = 2.357 in4 τ x′′ = τ ′′y = Mry J = 7 F (1) = 2.970 F 2.357 Maximum shear: τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 2.9702 + ( 0.566 + 2.970 ) = 4.618 F 2 τ allow 2 25 = 5.41 kip Ans. 4.618 4.618 ______________________________________________________________________________ F= 9-23 = Given, b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, τallow = 140 MPa. Shigley’s MED, 10th edition Chapter 9 Solutions, Page 8/35 www.konkur.in Primary shear (F in kN, τ in MPa, A in mm2): F (103 ) V = 1.768 F τ ′y = = A 1.414 ( 5 )( 50 + 30 ) Secondary shear: Table 9-1: Ju = (b + d ) 3 = ( 50 + 30 ) 3 = 85.33 (103 ) mm3 6 6 J = 0.707 h Ju = 0.707(5)85.33(103) = 301.6(103) mm4 τ x′′ = τ ′′y = Mry J = 175F (103 ) (15) 301.6 (103 ) = 8.704 F 3 Mrx 175F (10 ) (25) = = 14.51F J 301.6 (103 ) Maximum shear: τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 8.7042 + (1.768 + 14.51) = 18.46 F 2 τ allow 2 140 = 7.58 kN Ans. 18.46 18.46 ______________________________________________________________________________ F= 9-24 = Given, b = 4 in, c = 6 in, d = 2 in, h = 5/16 in, τallow = 25 kpsi. Primary shear: τ ′y = V F = = 0.3772 F A 1.414 ( 5 /16 )( 4 + 2 ) Secondary shear: Table 9-1: (b + d ) = 3 ( 4 + 2) = 3 = 36 in 3 6 6 J = 0.707 h Ju = 0.707(5/16)36 = 7.954 in4 Ju τ x′′ = Mry τ ′′y = Mrx 8 F (2) = = 2.012 F J 7.954 J = 8 F (1) = 1.006 F 7.954 Maximum shear: τ max = τ x′′2 + (τ ′y + τ ′′y ) = F 1.0062 + ( 0.3772 + 2.012 ) = 2.592 F 2 Shigley’s MED, 10th edition 2 Chapter 9 Solutions, Page 9/35 www.konkur.in τ allow 25 = 9.65 kip Ans. 2.592 2.592 ______________________________________________________________________________ F= 9-25 = Given, b = 50 mm, d = 50 mm, h = 5 mm, E6010 electrode. A = 0.707(5)(50 +50 + 50) = 530.3 mm2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa. Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 272(320)−0.995 = 0.875 kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-8) and (6-18): Se = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa Electrode endurance: E6010, Table 9-3, Sut = 427 MPa Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 272(427)−0.995 = 0.657 As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.657(1)(0.59)(1)(0.5)(427) = 82.8 MPa The members and electrode are basically of equal strength. We will use Se = 82.6 MPa. For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5) τ A 82.6 ( 530.3) F = allow = = 16.2 (103 ) N = 16.2 kN Ans. K fs 2.7 ______________________________________________________________________________ 9-26 Given, b = 2 in, d = 2 in, h = 5/16 in, E6010 electrode. A = 0.707(5/16)(2 +2 + 2) = 1.326 in2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi. Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 39.9(47)−0.995 = 0.865 kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-8) and (6-18): Se = 0.865(1)(0.59)(1)(0.5)(47) = 12.0 kpsi Electrode endurance: E6010, Table 9-3, Sut = 62 kpsi Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: Shigley’s MED, 10th edition ka = 39.9(62)−0.995 = 0.657 Chapter 9 Solutions, Page 10/35 www.konkur.in As before, kb = 1 (uniform shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.657(1)(0.59)(1)(0.5)(62) = 12.0 kpsi Thus the members and electrode are of equal strength. For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5) τ A 12.0 (1.326 ) F = allow = = 5.89 kip Ans. K fs 2.7 ______________________________________________________________________________ 9-27 Given, b = 50 mm, d = 30 mm, h = 5 mm, E7010 electrode. A = 0.707(5)(50 +50 + 30) = 459.6 mm2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 320 MPa. Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 272(320)−0.995 = 0.875 kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Eqs. (6-8) and (6-18): Se = 0.875(1)(0.59)(1)(0.5)(320) = 82.6 MPa Electrode endurance: E6010, Table 9-3, Sut = 482 MPa Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 272(482)−0.995 = 0.582 As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.582(1)(0.59)(1)(0.5)(482) = 82.7 MPa The members and electrode are basically of equal strength. We will use Se =82.6 MPa. For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5) τ A 82.6 ( 459.6 ) F = allow = = 14.1(103 ) N = 14.1 kN Ans. K fs 2.7 ______________________________________________________________________________ 9-28 Given, b = 4 in, d = 2 in, h = 5/16 in, E7010 electrode. A = 0.707(5/16)(4 +4 + 2) = 2.209 in2 Member endurance limit: From Table A-20 for AISI 1010 HR, Sut = 47 kpsi. Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 39.9(47)−0.995 = 0.865 kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Shigley’s MED, 10th edition Chapter 9 Solutions, Page 11/35 www.konkur.in Eqs. (6-8) and (6-18): Se = 0.865(1)(0.59)(1)(0.5)(47) = 12.0 kpsi Electrode endurance: E7010, Table 9-3, Sut = 70 kpsi Eq. 6-19 and Table 6-2, pp. 295 and 296, respectively: ka = 39.9(70)−0.995 = 0.582 As before, kb = 1 (direct shear), kc = 0.59 (torsion, shear), kd = 1 Se = 0.582(1)(0.59)(1)(0.5)(70) = 12.0 kpsi Thus the members and electrode are of equal strength. For a factor of safety of 1, and with Kfs = 2.7 (Table 9-5) τ A 12.0 ( 2.209 ) F = allow = = 9.82 kip Ans. K fs 2.7 ______________________________________________________________________________ 9-29 τ′ = 0 (why?) Primary shear: Secondary shear: Table 9-1: Ju = 2π r3 = 2π (1.5)3 = 21.21 in3 J = 0.707 h Ju = 0.707(1/4)(21.21) = 3.749 in4 2 welds: Mr 8 F (1.5 ) = = 1.600 F 2 J 2 ( 3.749 ) τ ′′ = τ ′′ = τ allow ⇒ 1.600 F = 20 ⇒ F = 12.5 kip Ans. ______________________________________________________________________________ 9-30 l = 2 + 4 + 4 = 10 in 2 (1) + 4 ( 0 ) + 4 ( 2 ) = 1 in 10 2 ( 4) + 4 ( 2) + 4 ( 0) y= = 1.6 in 10 x= M = FR = F(10 − 1) = 9 F r1 = (1 − 1) + ( 4 − 1.6) r3 = ( 2 − 1) 2 Shigley’s MED, 10th edition 2 2 = 2.4 in, r2 = 12 + ( 2 − 1.6 ) = 1.077 in 2 + 1.62 = 1.887 in Chapter 9 Solutions, Page 12/35 www.konkur.in 1 ( 0.707 )( 5 /16 ) ( 23 ) = 0.1473 in 4 12 1 = J G3 = ( 0.707 )( 5 / 16 ) ( 43 ) = 1.178 in 4 12 J G1 = J G2 3 ( J = ∑ J i + Ai rG2i i =1 ) = 0.1473 + 0.707 ( 5 /16 )( 2 ) ( 2.42 ) + 1.178 + 0.707 ( 5 / 16 )( 4 ) (1.077 2 ) + 1.178 + 0.707 ( 5 / 16 )( 4 ) (1.887 2 ) = 9.220 in 4 1.6 o = 28.07 4 −1 α = tan −1 r = 1.62 + ( 4 − 1) = 3.4 in 2 Primary shear (τ in kpsi, F in kip) : τ′ = Secondary shear: τ ′′ = V F = = 0.4526 F A 0.707 ( 5 /16 )(10 ) Mr 9 F ( 3.4 ) = = 3.319 F J 9.220 τ max = ( 3.319F sin 28.07 ) + ( 3.319F cos 28.07 o 2 o + 0.4526 F ) 2 = 3.724 F τmax = τallow ⇒ 3.724 F = 25 ⇒ F = 6.71 kip Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 9 Solutions, Page 13/35 www.konkur.in 9-31 l = 30 + 50 + 50 = 130 mm 30 (15 ) + 50 ( 0 ) + 50 ( 25 ) = 13.08 mm 130 30 ( 50 ) + 50 ( 25 ) + 50 ( 0 ) y= = 21.15 mm 130 x= M = FR = F(200 − 13.08) = 186.92 F (M in N⋅m, F in kN) r1 = (15 − 13.08) + ( 50 − 21.15) r3 = ( 25 − 13.08) 2 2 = 28.92 mm, r2 = 13.082 + ( 25 − 21.15) = 13.63 mm 2 2 + 21.152 = 24.28 mm 1 ( 0.707 )( 5 ) ( 303 ) = 7.954 (103 ) mm 4 12 1 = J G3 = ( 0.707 )( 5 ) ( 503 ) = 36.82 (103 ) mm 4 12 J G1 = J G2 3 ( J = ∑ J i + Ai rG2i i =1 ) = 7.954 (103 ) + 0.707 ( 5 )( 30 ) ( 28.922 ) + 36.82 (103 ) + 0.707 ( 5 )( 50 ) (13.632 ) + 36.82 (103 ) + 0.707 ( 5 )( 50 ) ( 24.282 ) = 307.3 (103 ) mm 4 21.15 o = 29.81 50 − 13.08 α = tan −1 r = 21.152 + ( 50 − 13.08) = 42.55 mm 2 Primary shear (τ in MPa, F in kN) : F (103 ) V τ′ = = = 2.176 F A 0.707 ( 5 )(130 ) Secondary shear: 3 Mr 186.92 F (10 ) ( 42.55 ) τ ′′ = = = 25.88 F J 307.3 (103 ) Shigley’s MED, 10th edition Chapter 9 Solutions, Page 14/35 www.konkur.in τ max = ( 25.88F sin 29.81 ) + ( 25.88F cos 29.81 o 2 o + 2.176 F ) 2 = 27.79 F τmax = τallow ⇒ 27.79 F = 140 ⇒ F = 5.04 kN Ans. ______________________________________________________________________________ 9-32 Weld Pattern 1. Figure of merit a2 J a 3 /12 a 2 fom′ = u = = = 0.0833 lh ah 12h h a ( 3a 2 + a 2 ) 5 a2 a2 = = 0.3333 3h h 2. fom′ = 3. ( 2a ) − 6a 2 a 2 = 5a 2 = 0.2083 a 2 fom′ = 12 ( a + a ) 2ah 24h h 4. fom′ = 5. ( 2a ) fom′ = 6 ( 2a ) h Rank______ 1 4 4 a2 1 8a 3 + 6 a 3 + a 3 a4 − = 0.3056 3ah 12 2a + a h 2 3 a2 1 8a 3 = = 0.3333 6h 4a 24ah h 1 2π ( a / 2 ) a2 a3 6. 3 fom′ = = = 0.25 4ah π ah h ______________________________________________________________________________ 3 Shigley’s MED, 10th edition Chapter 9 Solutions, Page 15/35 www.konkur.in 9-33 Weld Pattern 1. 2. 3. 4.* 5. & 7. Figure of merit 3 a2 I u ( a /12 ) fom′ = = = 0.0833 lh ah h a3 / 6 ) ( a2 ′ fom = = 0.0833 2ah h 2 ( aa / 2 ) = 0.25 a 2 fom′ = 2ah h a 2 / 12 ) ( 6a + a ) 7 a 2 ( a2 fom′ = = = 0.1944 3ah 36h h a a2 a x= , y= = 2 a + 2a 3 2 3 2a a a3 a Iu = − 2a 2 + ( a + 2a ) = 3 3 3 3 3 a2 I u ( a / 3) 1 a 2 = = = 0.1111 lh 3ah 9 h h 2 ( a / 6 ) ( 3a + a ) = 1 a 2 = 0.1667 a 2 fom′ = 4ah 6 h h fom′ = 6. & 8. π ( a / 2) a2 a2 fom′ = = = 0.125 π ah 8h h Rank______ 6 6 1 2 5 3 3 9. 4 *Note. Because this section is not symmetric with the vertical axis, out-of-plane deflection may occur unless special precautions are taken. See the topic of “shear center” in books with more advanced treatments of mechanics of materials. ______________________________________________________________________________ 9-34 Attachment and member (1018 HR), Sy = 220 MPa and Sut = 400 MPa. The member and attachment are weak compared to the properties of the lowest electrode. Decision Specify the E6010 electrode Controlling property, Table 9-4: τall = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa For a static load, the parallel and transverse fillets are the same. Let the length of a bead be l = 75 mm, and n be the number of beads. Shigley’s MED, 10th edition Chapter 9 Solutions, Page 16/35 www.konkur.in τ= F = τ all n ( 0.707 ) hl nh = 100 (103 ) F = = 21.43 0.707lτ all 0.707 ( 75 )( 88 ) where h is in millimeters. Make a table Number of beads, n 1 2 3 4 Decision Leg size, h (mm) 21.43 10.71 7.14 5.36 → 6 mm Specify h = 6 mm on all four sides. Weldment specification: Pattern: All-around square, four beads each side, 75 mm long Electrode: E6010 Leg size: h = 6 mm ______________________________________________________________________________ 9-35 Decision: Choose a parallel fillet weldment pattern. By so-doing, we’ve chosen an optimal pattern (see Prob. 9-32) and have thus reduced a synthesis problem to an analysis problem: Table 9-1, case 2, rotated 90°: A = 1.414hd = 1.414(h)(75) = 106.05h mm2 Primary shear τ ′y = 12 (103 ) 113.2 V = = A 106.05h h Secondary shear: d (3b 2 + d 2 ) 6 75[3(752 ) + 752 ] = = 281.3 (103 ) mm3 6 J = 0.707(h)(281.3) (103 ) = 198.8 (103 ) h mm 4 Ju = With α = 45°, Shigley’s MED, 10th edition Chapter 9 Solutions, Page 17/35 www.konkur.in 3 Mry 12 (10 ) (187.5)(37.5) 424.4 Mr cos 45o τ ′′x = = = = = τ ′′y J J h 198.8 (103 ) h τ max = τ ′′x 2 + (τ ′y + τ ′′y ) = 2 1 684.9 424.42 + (113.2 + 424.4) 2 = h h Attachment and member (1018 HR): Sy = 220 MPa, Sut = 400 MPa Decision: Use E60XX electrode which is stronger τ all = min[0.3(400), 0.4(220)] = 88 MPa τ max = τ all = h= 684.9 = 88 MPa h 684.9 = 7.78 mm 88 Decision: Specify 8 mm leg size Weldment Specifications: Pattern: Parallel horizontal fillet welds Electrode: E6010 Type: Fillet Length of each bead: 75 mm Leg size: 8 mm ______________________________________________________________________________ 9-36 Problem 9-35 solves the problem using parallel horizontal fillet welds, each 75 mm long obtaining a leg size rounded up to 8 mm. For this problem, since the width of the plate is fixed and the length has not been determined, we will explore reducing the leg size by using two vertical beads 75 mm long and two horizontal beads such that the beads have a leg size of 6 mm. Decision: Use a rectangular weld bead pattern with a leg size of 6 mm (case 5 of Table 9-1 with b unknown and d = 75 mm). Materials: Attachment and member (1018 HR): Sy = 220 MPa, Sut = 400 MPa From Table 9-4, AISC welding code, τall = min[0.3(400), 0.4(220)] = min(120, 88) = 88 MPa Select a stronger electrode material from Table 9-3. Decision: Specify E6010 Solving for b: In Prob. 9-35, every term was linear in the unknown h. This made solving for h relatively easy. In this problem, the terms will not be linear in b, and so we will use an iterative solution with a spreadsheet. Throat area and other properties from Table 9-1: A = 1.414(6)(b + 75) = 8.484(b + 75) Shigley’s MED, 10th edition (1) Chapter 9 Solutions, Page 18/35 www.konkur.in Ju ( b + 75 ) = 6 3 , J = 0.707 (6) Ju = 0.707(b +75)3 (2) Primary shear (τ in MPa, h in mm): 12 (103 ) V = τ ′y = A A (3) Secondary shear (See Prob. 9-35 solution for the definition of α) : Mr τ ′′ = J 3 Mry 12 (10 ) (150 + b / 2 ) (37.5) Mr τ ′′x = τ ′′ cos α = cos α = = 3 J J 0.707 ( b + 75 ) 3 Mr Mrx 12 (10 ) (150 + b / 2 ) (b / 2) τ ′′y = τ ′′ sin α = sin α = = 3 J J 0.707 ( b + 75 ) τ max = τ ′y 2 + (τ ′′x + τ ′′y ) 2 (4) (5) (6) Enter Eqs. (1) to (6) into a spreadsheet and iterate for various values of b. A portion of the spreadsheet is shown below. 2 4 b (mm) A (mm ) J (mm ) 41 984.144 1103553.5 42 992.628 1132340.4 43 1001.112 1161623.6 44 1009.596 1191407.4 τ'y (Mpa) 12.19334 12.08912 11.98667 11.88594 τ"y (Mpa) τ"x (Mpa) 69.5254 67.9566 66.43718 64.96518 38.00722 38.05569 38.09065 38.11291 τmax (Mpa) 90.12492 88.63156 87.18485 < 88 Mpa 85.7828 We see that b ≥ 43 mm meets the strength goal. Weldment Specifications: Pattern: Horizontal parallel weld tracks 43 mm long, vertical parallel weld tracks 75 mm long Electrode: E6010 Leg size: 6 mm ______________________________________________________________________________ 9-37 Materials: Member and attachment (1018 HR): Table 9-4: Shigley’s MED, 10th edition S y = 32 kpsi, Sut = 58 kpsi τ all = min[0.3(58), 0.4(32)] = 12.8 kpsi Chapter 9 Solutions, Page 19/35 www.konkur.in Decision: Use E6010 electrode. From Table 9-3: S y = 50 kpsi, Sut = 62 kpsi, τ all = min[0.3(62), 0.4(50)] = 20 kpsi Decision: Since 1018 HR is weaker than the E6010 electrode, use τ all = 12.8 kpsi Decision: Use an all-around square weld bead track. l1 = 6 + a = 6 + 6.25 = 12.25 in Throat area and other properties from Table 9-1: A = 1.414 h(b + d ) = 1.414( h )(6 + 6) = 16.97 h Primary shear ( ) 3 V F 20 10 1179 = psi τ ′y = = = A A 16.97h h Secondary shear (b + d )3 (6 + 6)3 = = 288 in 3 6 6 J = 0.707 h(288) = 203.6h in 4 Ju = τ x′′ = τ ′′y = τ max Mry J = ( ) 20 103 (6.25 + 3)(3) = 2726 psi h 203.6h 1 4762 = τ x′′2 + (τ ′y + τ ′′y ) 2 = 27262 + (1179 + 2726) 2 = psi h h Relate stress to strength τ max = τ all ⇒ ( ) 4762 = 12.8 103 h ⇒ h= 4762 = 0.372 in 12.8 103 ( ) Decision: Specify 3 / 8 in leg size Specifications: Pattern: All-around square weld bead track Electrode: E6010 Type of weld: Fillet Weld bead length: 24 in Leg size: 3 / 8 in Attachment length: 12.25 in ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 9 Solutions, Page 20/35 www.konkur.in 9-38 This is a good analysis task to test a student’s understanding. (1) Solicit information related to a priori decisions. (2) Solicit design variables b and d. (3) Find h and round and output all parameters on a single screen. Allow return to Step 1 or Step 2. (4) When the iteration is complete, the final display can be the bulk of your adequacy assessment. Such a program can teach too. ______________________________________________________________________________ 9-39 The objective of this design task is to have the students teach themselves that the weld patterns of Table 9-2 can be added or subtracted to obtain the properties of a contemplated weld pattern. The instructor can control the level of complication. We have left the presentation of the drawing to you. Here is one possibility. Study the problem’s opportunities, and then present this (or your sketch) with the problem assignment. Use b1 as the design variable. Express properties as a function of b1. From Table 9-3, case 3: A = 1.414h(b − b1 ) bd 2 b1d 2 (b − b1 )d 2 − = 2 2 2 I = 0.707hI u V F τ′ = = A 1.414h(b − b1 ) Mc Fa (d / 2) = τ ′′ = I 0.707 hI u Iu = Parametric study Let a = 10 in, b = 8 in, d = 8 in, b1 = 2 in, τ all = 12.8 kpsi, l = 2(8 − 2) = 12 in Shigley’s MED, 10th edition Chapter 9 Solutions, Page 21/35 www.konkur.in A = 1.414h(8 − 2) = 8.48h in 2 I u = (8 − 2)(82 / 2) = 192 in 3 I = 0.707(h)(192) = 135.7h in 4 10 000 1179 τ′ = = psi 8.48h h 10 000(10)(8 / 2) 2948 τ ′′ = = psi 135.7h h 1 3175 τ max = 11792 + 29482 = = 12 800 psi h h from which h = 0.248 in. Do not round off the leg size – something to learn. Iu 192 = = 64.5 in hl 0.248(12) A = 8.48(0.248) = 2.10 in 2 fom ' = I = 135.7(0.248) = 33.65 in 4 h2 0.2482 vol = l = 12 = 0.369 in 3 2 2 I 33.65 eff = = = 91.2 in vol 0.369 1179 = 4754 psi τ′ = 0.248 2948 τ ′′ = = 11 887 psi 0.248 3175 = 12 800 psi τ max = 0.248 Now consider the case of uninterrupted welds, b1 = 0 A = 1.414( h )(8 − 0) = 11.31h I u = (8 − 0)(82 / 2) = 256 in 3 I = 0.707(256)h = 181h in 4 10 000 884 = τ′ = 11.31h h 10 000(10)(8 / 2) 2210 = τ ′′ = 181h h 1 2380 τ max = 8842 + 22102 = = τ all h h τ 2380 h = max = = 0.186 in τ all 12 800 Do not round off h. Shigley’s MED, 10th edition Chapter 9 Solutions, Page 22/35 www.konkur.in A = 11.31(0.186) = 2.10 in 2 I = 181(0.186) = 33.67 in 4 884 0.1862 = 4753 psi, vol = 16 = 0.277 in 3 0.186 2 2210 τ ′′ = = 11 882 psi 0.186 I 256 fom ' = u = = 86.0 in hl 0.186(16) I 33.67 eff = 2 = = 121.7 in (h / 2)l (0.1862 / 2)16 τ′ = Conclusions: To meet allowable stress limitations, I and A do not change, nor do τ and σ. To meet the shortened bead length, h is increased proportionately. However, volume of bead laid down increases as h2. The uninterrupted bead is superior. In this example, we did not round h and as a result we learned something. Our measures of merit are also sensitive to rounding. When the design decision is made, rounding to the next larger standard weld fillet size will decrease the merit. Had the weld bead gone around the corners, the situation would change. Here is a follow up task analyzing an alternative weld pattern. ______________________________________________________________________________ 9-40 From Table 9-2 For the box A = 1.414h(b + d ) Subtracting b1 from b and d1 from d A = 1.414h ( b − b1 + d − d1 ) Iu = d2 d3 b d2 1 1 (3b + d ) − 1 − 1 = ( b − b1 ) d 2 + d 3 − d13 6 6 2 2 6 ( ) l = 2(b − b1 + d − d1 ) fom = I u / hl ______________________________________________________________________________ Length of bead Shigley’s MED, 10th edition Chapter 9 Solutions, Page 23/35 www.konkur.in 9-41 Computer programs will vary. ______________________________________________________________________________ 9-42 τall = 12 kpsi. Use Fig. 9-17(a) for general geometry, but employ beads and then beads. Horizontal parallel weld bead pattern b = 3 in, d = 6 in Table 9-2: A = 1.414hb = 1.414(h)(3) = 4.24h in 2 bd 2 3(6)2 = = 54 in 3 2 2 I = 0.707 hI u = 0.707( h)(54) = 38.2h in 4 10 2.358 τ′ = = kpsi 4.24h h Mc 10(10)(6 / 2) 7.853 τ ′′ = = = kpsi I 38.2h h Iu = τ max = τ ′2 + τ ′′2 = 1 8.199 2.3582 + 7.8532 = kpsi h h Equate the maximum and allowable shear stresses. τ max = τ all = 8.199 = 12 h from which h = 0.683 in. It follows that I = 38.2(0.683) = 26.1 in 4 The volume of the weld metal is vol = h 2l (0.683) 2 (3 + 3) = = 1.40 in 3 2 2 The effectiveness, (eff)H, is (eff) H = Shigley’s MED, 10th edition I 26.1 = = 18.6 in vol 1.4 Ans. Chapter 9 Solutions, Page 24/35 www.konkur.in Vertical parallel weld beads b = 3 in d = 6 in From Table 9-2, case 2 A = 1.414hd = 1.414(h)(6) = 8.48h in 2 d 3 63 Iu = = = 72 in 3 6 6 I = 0.707 hI u = 0.707(h)(72) = 50.9h 10 1.179 τ′ = = psi 8.48h h Mc 10(10)(6 / 2) 5.894 = = psi τ ′′ = I 50.9h h 1 6.011 1.1792 + 5.8942 = kpsi τ max = τ ′2 + τ ′′2 = h h Equating τ max to τ all gives h = 0.501 in. It follows that I = 50.9(0.501) = 25.5 in 4 h 2l 0.5012 = (6 + 6) = 1.51 in 3 2 2 I 25.5 (eff ) V = = = 16.7 in vol 1.51 vol = Ans. ______________________________________________________________________________ 9-43 F = 0, T = 15 kip⋅in. Table 9-1: Ju = 2π r 3 = 2π (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4 Tr 15 (1) = = 13.5 kpsi Ans. J 1.111 ______________________________________________________________________________ τ max = 9-44 F = 2 kip, T = 0. Table 9-2: A = 1.414 π h r = 1.414 π (1/4)(1) = 1.111 in2 Iu = π r 3 = π (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4 Shigley’s MED, 10th edition Chapter 9 Solutions, Page 25/35 www.konkur.in τ′ = V 2 = = 1.80 kpsi A 1.111 τ ′′ = Mr 2 ( 6 )(1) = = 21.6 kpsi I 0.5553 τ max = (τ′ 2 + τ′′ 2)1/2 = (1.802 + 21.62)1/2 = 21.7 kpsi Ans. ______________________________________________________________________________ 9-45 F = 2 kip, T = 15 kip⋅in. Bending: Table 9-2: A = 1.414 π h r = 1.414 π (1/4)(1) = 1.111 in2 Iu = π r 3 = π (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4 τ′ = V 2 = = 1.80 kpsi A 1.111 (τ ′′) M = Mr 2 ( 6 )(1) = = 21.6 kpsi I 0.5553 Torsion: Table 9-1: Ju = 2π r 3 = 2π (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4 (τ ′′)T = Tr 15 (1) = = 13.5 kpsi J 1.111 τ max = τ ′2 + (τ ′′ )M + (τ ′′ )T = 1.802 + 21.62 + 13.52 = 25.5 kpsi 2 2 Ans. ______________________________________________________________________________ 9-46 F = 2 kip, T = 15 kip⋅in. Bending: Table 9-2: A = 1.414 π h r = 1.414 π h (1) = 4.442h in2 Iu = π r 3 = π (1)3 = 3.142 in3, I = 0.707 h (3.142) = 2.221h in4 τ′ = Shigley’s MED, 10th edition V 2 0.4502 = = kpsi A 4.442h h Chapter 9 Solutions, Page 26/35 www.konkur.in (τ ′′ )M Mr 2 ( 6 )(1) 5.403 = = kpsi I 2.221h h = Torsion: Ju = 2π r 3 = 2π (1)3 = 6.283 in3, J = 0.707 h (6.283) = 4.442 in4 Table 9-1: (τ ′′)T = Tr 15 (1) 3.377 = = kpsi J 4.442h h τ max = τ ′ + (τ ′′ ) M + (τ ′′ )T 2 2 τ max = τ all 2 2 6.387 = 20 h ⇒ 2 2 6.387 0.4502 5.403 3.377 = kpsi + + = h h h h Should specify a 83 in weld. ⇒ h = 0.319 in Ans. Ans. ______________________________________________________________________________ 9-47 h = 9 mm, d = 200 mm, b = 25 mm From Table 9-2, case 2: A = 1.414(9)(200) = 2.545(103) mm2 d 3 2003 = = 1.333 106 mm3 6 6 I = 0.707h Iu = 0.707(9)(1.333)(106) = 8.484(106) mm4 ( ) Iu = ( ) 25 103 F = 9.82 MPa τ′ = = A 2.545(103 ) M = 25(150) = 3750 N⋅m τ ′′ = Mc 3750(100) = (103 ) = 44.20 MPa I 8.484(106 ) τ max = τ ′2 + τ ′′2 = 9.822 + 44.202 = 45.3 MPa Ans. ______________________________________________________________________________ 9-48 h = 0.25 in, b = 2.5 in, d = 5 in. Table 9-2, case 5: Shigley’s MED, 10th edition A = 0.707h (b +2d) = 0.707(0.25)[2.5 + 2(5)] = 2.209 in2 Chapter 9 Solutions, Page 27/35 www.konkur.in y= d2 52 = = 2 in b + 2d 2.5 + 2 ( 5 ) 2d 3 − 2d 2 y + ( b + 2d ) y 2 3 2 ( 53 ) = − 2 ( 52 ) ( 2 ) + 2.5 + 2 ( 5 ) ( 22 ) = 33.33 in 3 3 Iu = I = 0.707 h Iu = 0.707(1/4)(33.33) = 5.891 in4 Primary shear: τ′ = F 2 = = 0.905 kpsi A 2.209 Secondary shear (the critical location is at the bottom of the bracket): y = 5 − 2 = 3 in τ ′′ = My 2 ( 5 )( 3) = = 5.093 kpsi I 5.891 τ max = τ ′2 + τ ′′2 = 0.9052 + 5.0932 = 5.173 kpsi τ all 18 = = 3.48 Ans. τ max 5.173 ______________________________________________________________________________ n= 9-49 The largest possible weld size is 1/16 in. This is a small weld and thus difficult to accomplish. The bracket’s load-carrying capability is not known. There are geometry problems associated with sheet metal folding, load-placement and location of the center of twist. This is not available to us. We will identify the strongest possible weldment. Use a rectangular, weld-all-around pattern – Table 9-2, case 6: Shigley’s MED, 10th edition Chapter 9 Solutions, Page 28/35 www.konkur.in A = 1.414 h(b + d ) = 1.414(1 / 16)(1 + 7.5) = 0.7512 in 2 x = b / 2 = 0.5 in y = d / 2 = 7.5 / 2 = 3.75 in d2 7.52 Iu = (3b + d ) = [3(1) + 7.5] = 98.44 in 3 6 6 I = 0.707hI u = 0.707(1 / 16)(98.44) = 4.350 in 4 M = (3.75 + 0.5)W = 4.25W V W τ′ = = = 1.331W A 0.7512 Mc 4.25W (7.5 / 2) = = 3.664W τ ′′ = I 4.350 τ max = τ ′2 + τ ′′2 = W 1.3312 + 3.6642 = 3.90W Material properties: The allowable stress given is low. Let’s demonstrate that. For the 1020 CD bracket, bracket use HR properties of Sy = 30 kpsi and Sut = 55. The 1030 HR support, Sy = 37.5 kpsi and Sut = 68. The E6010 electrode has strengths of Sy = 50 and Sut = 62 kpsi. Allowable stresses: 1020 HR: τall = min[0.3(55), 0.4(30)] = min(16.5, 12) = 12 kpsi 1020 HR: τall = min[0.3(68), 0.4(37.5)] = min(20.4, 20.4, 15) = 15 kpsi E6010: 18 kpsi τall = min[0.3(62), 0.4(50)] = min(18.6, 20) = 18.6 Since Table 9-66 gives 18.0 kpsi as the allowable shear stress, use this lower value. Therefore, the allowable shear stress is τall = min(14.4, 12, 18.0) = 12 kpsi However, the allowable stress in the problem statement is 1.5 kpsi which is low from the weldment perspective. The load associated with this strength is τ max = τ all = 3.90W = 1500 W = 1500 = 385 lbf 3.90 If the welding can be accomplished (1/16 leg size is a small weld), tthe weld strength is 12 000 psi and the load associated with this strength is W = 12 000/3.90 = 3077 lbf. Can the bracket carry such a load? There are geometry problems associated with sheet metal folding. Load placement is important and the center of twist has not been identified. Also, the load-carrying load capability of the top bend is unknown. Shigley’s MED, 10th edition Chapter 9 Solutions, Page 29/35 www.konkur.in These uncertainties may require the use of a different weld pattern. Our solution provides the best weldment and thus insight for comparing a welded joint to one o which employs screw fasteners. ______________________________________________________________________________ 9-50 F FB FBx FBy = 100 lbf , τ all = 3 kpsi = 100(16 / 3) = 533.3 lbf = −533.3cos 60o = −266.7 lbf = −533.3cos 30o = −462 lbf It follows that RAy = 562 lbf and RAx = 266.7 lbf, RA = 622 lbf M = 100(16) = 1600 lbf · in The OD of the tubes is 1 in. From Table 9-1, 9 case 6: A = 2 [1.414(π hr ) ] = 2(1.414)(π h)(1 / 2) = 4.442h in 2 J u = 2π r 3 = 2π (1 / 2)3 = 0.7854 in 3 J = 2(0.707)hJ u = 1.414(0.7854)h = 1.111h in 4 The weld only carries the torsional load between the handle and tube A. Consequently, the primary shear in the weld is zero, and the maximum shear stress is comprised entirely of the secondary shear. Shigley’s MED, 10th edition Chapter 9 Solutions, Page 30/35 www.konkur.in Tc Mc 1600(0.5) 720.1 = = = psi J J 1.111h h 720.1 = = 3000 h τ max = τ ′′ = τ max = τ all h= 720.1 = 0.240 → 1 / 4 in 3000 Decision: Use 1/4 in fillet welds Ans. ______________________________________________________________________________ 9-51 For the pattern in bending shown, find the centroid G of the weld group. x= 75 ( 6 )(150 ) + 325 ( 9 )(150 ) ( 6 )(150 ) + ( 9 )(150 ) I 6 mm = 2 ( I G + Ax 2 ) = 225 mm 6 mm 0.707 ( 6 ) (1503 ) 2 = 2 + 0.707 ( 6 )(150 )( 225 − 75 ) = 31.02 (106 ) mm 4 12 I 9 mm 0.707 ( 9 ) (1503 ) 2 = 2 + 0.707 ( 9 )(150 )(175 − 75 ) = 22.67 (106 ) mm4 12 I = I 6 mm + I 9 mm = (31.02 + 22.67)(106) = 53.69(106) mm4 The critical location is at B. With τ in MPa, and F in kN F (103 ) V τ′ = = = 0.3143F A 2 0.707 ( 6 + 9 )(150 ) 3 Mc 200 F (10 ) ( 225 ) = = 0.8381F τ ′′ = I 53.69 (106 ) Shigley’s MED, 10th edition Chapter 9 Solutions, Page 31/35 www.konkur.in τ max = τ ′2 + τ ′′2 = F 0.31432 + 0.83812 = 0.8951F Materials: 1015 HR (Table A-20): Sy = 190 MPa, E6010 Electrode(Table 9-3): Sy = 345 MPa τall = 0.577(190) = 109.6 MPa Eq. (5-21), p. 239 τ all / n 109.6 / 2 = 61.2 kN Ans. 0.8951 0.8951 ______________________________________________________________________________ F= 9-52 = In the textbook, Fig. Problem 9-52b is a free-body diagram of the bracket. Forces and moments that act on the welds are equal, but of opposite sense. (a) M = 1200(0.366) = 439 lbf · in Ans. (b) Fy = 1200 sin 30° = 600 lbf Ans. (c) Fx = 1200 cos 30° = 1039 lbf Ans. (d) From Table 9-2, case 6: A = 1.414(0.25)(0.25 + 2.5) = 0.972 in 2 d2 2.52 Iu = (3b + d ) = [3(0.25) + 2.5] = 3.39 in 3 6 6 The second area moment about an axis through G and parallel to z is I = 0.707hI u = 0.707(0.25)(3.39) = 0.599 in 4 Ans. (e) Refer to Fig. Problem 9-52b. The shear stress due to Fy is τ1 = Fy A = 600 = 617 psi 0.972 The shear stress along the throat due to Fx is τ2 = Fx 1039 = = 1069 psi A 0.972 The resultant of τ1 and τ2 is in the throat plane τ ′ = τ12 + τ 22 = 6172 + 10692 = 1234 psi The bending of the throat gives τ ′′ = Shigley’s MED, 10th edition Mc 439(1.25) = = 916 psi I 0.599 Chapter 9 Solutions, Page 32/35 www.konkur.in The maximum shear stress is τ max = τ ′2 + τ ′′2 = 12342 + 9162 = 1537 psi (f) Materials: 1018 HR Member: E6010 Electrode: n= S sy τ max Ans. Sy = 32 kpsi, Sut = 58 kpsi (Table A-20) Sy = 50 kpsi (Table 9-3) = 0.577 S y τ max = 0.577(32) = 12.0 1.537 Ans. (g) Bending in the attachment near the base. The cross-sectional area is approximately equal to bh. A1 bh = 0.25(2.5) = 0.625 in 2 F 1039 = 1662 psi τ xy = x = A1 0.625 I bd 2 0.25(2.5) 2 = = = 0.260 in 3 c 6 6 At location A, F M σy = y + A1 I /c 600 439 σy = + = 2648 psi 0.625 0.260 The von Mises stress σ ′ is σ ′ = σ y2 + 3τ xy2 = 26482 + 3(1662) 2 = 3912 psi Thus, the factor of safety is, S 32 n= y = = 8.18 σ ′ 3.912 Ans. The clip on the mooring line bears against the side of the 1/2-in hole. If the clip fills the hole F −1200 = = −9600 psi σ = td 0.25(0.50) S 32(103 ) n=− y =− = 3.33 Ans. σ′ −9600 Further investigation of this situation requires more detail than is included in the task statement. (h) In shear fatigue, the weakest constituent of the weld melt is 1018 HR with Sut = 58 kpsi, Eq. (6-8), p. 290, gives Shigley’s MED, 10th edition Chapter 9 Solutions, Page 33/35 www.konkur.in Se′ = 0.5Sut = 0.5(58) = 29.0 kpsi ka = 14.4(58)-0.718 = 0.780 Eq. (6-19), p. 295: For uniform shear stress on the throat, assume kb = 1. Eq.(6-26), p. 298: kc = 0.59 From Eq. (6-18), p. 295, the endurance strength in shear is Sse = 0.780(1)(0.59)(29.0) = 13.35 kpsi From Table 9-5, the shear stress-concentration factor is Kf s = 2.7. The loading is repeatedly-applied 1.537 τ τ a = τ m = K f s max = 2.7 = 2.07 kpsi 2 2 Table 6-7, p. 315: Gerber factor of safety nf, adjusted for shear, with Ssu = 0.67Sut 2 1 S su τ a −1 + nf = 2 τ m S se 2τ S 1 + m se S suτ a 2 2 2(2.07)(13.35) 1 0.67(58) 2.07 = −1 + 1 + = 5.83 Ans. 2 2.07 13.35 0.67(58)(2.07) Attachment metal should be checked for bending fatigue. ______________________________________________________________________________ 9-53 (a) Use b = d = 4 in. Since h = 5/8 in, the primary shear is τ′ = F = 0.2829 F 1.414(5 / 8)(4) The secondary shear calculations, for a moment arm of 14 in give 4[3(42 ) + 42 ] = 42.67 in 3 6 J = 0.707hJ u = 0.707(5 / 8)42.67 = 18.85 in 4 Mry 14F (2) τ ′′x = τ ′′y = = = 1.485F J 18.85 Ju = Thus, the maximum shear and allowable load are: Shigley’s MED, 10th edition Chapter 9 Solutions, Page 34/35 www.konkur.in τ max = F 1.4852 + (0.2829 + 1.485) 2 = 2.309F 25 τ all F = = = 10.8 kip Ans. 2.309 2.309 The load for part (a) has increased by a factor of 10.8/2.71 = 3.99 Ans. (b) From Prob. 9-18b, τall = 11 kpsi Fall = τ all 2.309 = 11 = 4.76 kip 2.309 The allowable load in part (b) has increased by a factor of 4.76/1.19 = 4 Ans. ______________________________________________________________________________ 9-54 Purchase the hook having the design shown in Fig. Problem 9-54b. Referring to text Fig. 9-29a, this design reduces peel stresses. ______________________________________________________________________________ 9-55 (a) l/2 1 l / 2 Pω cosh(ω x) A dx = A1 ∫ cosh(ω x) dx = 1 sinh(ω x) ∫ − l / 2 − l / 2 ω l 4b sinh(ωl / 2) −l / 2 A1 A1 = [sinh(ωl / 2) − sinh(−ωl / 2)] = [sinh(ωl / 2) − (− sinh(ωl / 2))] l/2 τ = ω = (b) (c) 2 A1 sinh(ωl / 2) ω Pω P = [2sinh(ωl / 2)] = 4bl sinh(ωl / 2) 2bl ω Pω cosh(ω l / 2) Pω τ (l / 2) = = Ans. 4b sinh(ω l / 2) 4b tanh(ω l / 2) K = Pω τ (l / 2) ωl / 2 2bl = = 4b tanh(ωl / 2) P tanh(ωl / 2) τ Ans. Ans. For computer programming, it can be useful to express the hyperbolic tangent in terms of exponentials: ωl exp(ωl / 2) − exp(−ωl / 2) K = Ans. 2 exp(ωl / 2) + exp(−ωl / 2) ______________________________________________________________________________ 9-56 This is a computer programming exercise. All programs will vary. Shigley’s MED, 10th edition Chapter 9 Solutions, Page 35/35 www.konkur.in Chapter 10 10-1 From Eqs. (10-4) and (10-5) KW − K B = 4C − 1 0.615 4C + 2 + − 4C − 4 C 4C − 3 Plot 100(KW − KB)/ KW vs. C for 4 ≤ C ≤ 12 obtaining 1.4 1.3 1.2 1.1 100(KW-KB)/KW 1 0.9 0.8 0.7 4 6 8 10 12 C We see the maximum and minimum occur at C = 4 and 12 respectively where Maximum = 1.36 % Ans., and Minimum = 0.743 % Ans. ______________________________________________________________________________ A = Sdm dim(Auscu) = [dim (S) dim(d m)]uscu = kpsi⋅inm 10-2 dim(ASI) = [dim (S) dim(d m)]SI = MPa⋅mmm ASI = MPa mm m m ⋅ m Auscu = 6.894 757 ( 25.4 ) Auscu kpsi in 6.895 ( 25.4 ) Auscu m Ans. For music wire, from Table 10-4: Auscu = 201 kpsi⋅inm, m = 0.145; what is ASI? ASI = 6.895(25.4)0.145 (201) = 2215 MPa⋅mmm Ans. ______________________________________________________________________________ 10-3 Given: Music wire, d = 2.5 mm, OD = 31 mm, plain ground ends, Nt = 14 coils. Shigley’s MED, 10th edition Chapter 10 Solutions, Page 1/41 www.konkur.in (a) Table 10-1: Na = Nt − 1 = 14 − 1 = 13 coils D = OD − d = 31− 2.5 = 28.5 mm C = D/d = 28.5/2.5 = 11.4 ⇒ G = 81.0(103) MPa Table 10-5: d = 2.5/25.4 = 0.098 in Eq. (10-9): 2.54 ( 81)103 d 4G k= = = 1.314 N / mm 8 D 3 N a 8 ( 28.53 )13 (b) Table 10-1: Ls = d Nt = 2.5(14) = 35 mm A = 2211 MPa⋅mmm Table 10-4: m = 0.145, Eq. (10-14): Sut = Table 10-6: Ssy = 0.45(1936) = 871.2 MPa Eq. (10-5): KB = Eq. (10-7): Fs = L0 = (c) Ans. A 2211 = = 1936 MPa m d 2.50.145 4C + 2 4 (11.4 ) + 2 = = 1.117 4C − 3 4 (11.4 ) − 3 π d 3 S sy 8K B D = π ( 2.53 ) 871.2 8 (1.117 ) 28.5 = 167.9 N Fs 167.9 + Ls = + 35 = 162.8 mm k 1.314 Ans. Ans. 2.63 ( 28.5 ) = 149.9 mm . Spring needs to be supported. Ans. 0.5 ______________________________________________________________________________ (d) 10-4 ( L0 )cr = Given: Design load, F1 = 130 N. Referring to Prob. 10-3 solution, C = 11.4, Na = 13 coils, Ssy = 871.2 MPa, Fs = 167.9 N, L0 = 162.8 mm and (L0)cr = 149.9 mm. Eq. (10-18): 4 ≤ C ≤ 12 C = 11.4 O.K. Eq. (10-19): 3 ≤ Na ≤ 15 Na = 13 O.K. F 167.9 Eq. (10-17): ξ = s − 1 = − 1 = 0.29 F1 130 Shigley’s MED, 10th edition Chapter 10 Solutions, Page 2/41 www.konkur.in Eq. (10-20): ξ ≥ 0.15, ξ = 0.29 O.K . From Eq. (10-7) for static service 8(130)(28.5) 8F1D = 1.117 = 674 MPa 3 π (2.5)3 πd S 871.2 n = sy = = 1.29 674 τ1 τ1 = K B Eq. (10-21): ns ≥ 1.2, n = 1.29 O.K. 167.9 167.9 = 674 = 870.5 MPa 130 130 S sy / τ s = 871.2 / 870.5 1 τ s = τ1 Ssy/τs ≥ (ns )d : Not solid-safe (but was the basis of the design). Not O.K. L0 ≤ (L0)cr: 162.8 ≥ 149.9 Not O.K. Design is unsatisfactory. Operate over a rod? Ans. ______________________________________________________________________________ 10-5 Given: Oil-tempered wire, d = 0.2 in, D = 2 in, Nt = 12 coils, L0 = 5 in, squared ends. (a) Table 10-1: Ls = d (Nt + 1) = 0.2(12 + 1) = 2.6 in Ans. (b) Table 10-1: Table 10-5: Na = Nt − 2 = 12 − 2 = 10 coils G = 11.2 Mpsi Eq. (10-9): k= 0.24 (11.2 )106 d 4G = = 28 lbf/in 8D 3 N a 8 ( 23 )10 Fs = k ys = k (L0 − Ls ) = 28(5 − 2.6) = 67.2 lbf (c) Eq. (10-1): C = D/d = 2/0.2 = 10 4C + 2 4 (10 ) + 2 = = 1.135 4C − 3 4 (10 ) − 3 Eq. (10-5): KB = Eq. (10-7): τ s = KB Table 10-4: m = 0.187, A = 147 kpsi⋅inm Shigley’s MED, 10th edition Ans. 8 ( 67.2 ) 2 8 Fs D = 1.135 = 48.56 (103 ) psi 3 πd π ( 0.23 ) Chapter 10 Solutions, Page 3/41 www.konkur.in A 147 = = 198.6 kpsi m d 0.20.187 Eq. (10-14): Sut = Table 10-6: Ssy = 0.50 Sut = 0.50(198.6) = 99.3 kpsi S sy 99.3 = 2.04 Ans. τ s 48.56 ______________________________________________________________________________ ns = 10-6 = Given: Oil-tempered wire, d = 4 mm, C = 10, plain ends, L0 = 80 mm, and at F = 50 N, y = 15 mm. (a) k = F/y = 50/15 = 3.333 N/mm (b) D = Cd = 10(4) = 40 mm Ans. OD = D + d = 40 + 4 = 44 mm Ans. (c) From Table 10-5, G = 77.2 GPa 4 3 d 4G 4 ( 77.2 )10 = = 11.6 coils 8kD3 8 ( 3.333) 403 Eq. (10-9): Na = Table 10-1: Nt = Na = 11.6 coils Ans. (d) Table 10-1: Ls = d (Nt + 1) = 4(11.6 + 1) = 50.4 mm (e) Table 10-4: m = 0.187, A = 1855 MPa⋅mmm Ans. A 1855 = = 1431 MPa d m 40.187 Eq. (10-14): Sut = Table 10-6: Ssy = 0.50 Sut = 0.50(1431) = 715.5 MPa ys = L0 − Ls = 80 − 50.4 = 29.6 mm Fs = k ys = 3.333(29.6) = 98.66 N 4C + 2 4(10) + 2 = = 1.135 4C − 3 4(10) − 3 Eq. (10-5): KB = Eq. (10-7): τ s = KB Shigley’s MED, 10th edition 8 ( 98.66 ) 40 8 Fs D = 1.135 = 178.2 MPa 3 πd π ( 43 ) Chapter 10 Solutions, Page 4/41 www.konkur.in S sy 715.5 = 4.02 Ans. τ s 178.2 ______________________________________________________________________________ ns = 10-7 = Static service spring with: HD steel wire, d = 0.080 in, OD = 0.880 in, Nt = 8 coils, plain and ground ends. Preliminaries Table 10-5: A = 140 kpsi · inm, m = 0.190 A 140 Eq. (10-14): Sut = m = = 226.2 kpsi d 0.0800.190 Table 10-6: Ssy = 0.45(226.2) = 101.8 kpsi Then, D = OD − d = 0.880 − 0.080 = 0.8 in Eq. (10-1): C = D/d = 0.8/0.08 = 10 4C + 2 4(10) + 2 Eq. (10-5): KB = = = 1.135 4C − 3 4(10) − 3 Table 10-1: Na = Nt − 1 = 8 − 1 = 7 coils Ls = dNt = 0.08(8) = 0.64 in Eq. (10-7) For solid-safe, ns = 1.2 : 3 3 π d 3S sy / ns π ( 0.08 ) 101.8 (10 ) / 1.2 Fs = = = 18.78 lbf 8K B D 8(1.135)(0.8) Eq. (10-9): k = 0.084 (11.5 )106 d 4G = = 16.43 lbf/in 8D 3 N a 8 ( 0.83 ) 7 ys = Fs 18.78 = = 1.14 in k 16.43 (a) L0 = ys + Ls = 1.14 + 0.64 = 1.78 in Ans. L 1.78 (b) Table 10-1: p = 0 = = 0.223 in Ans. Nt 8 (c) From above: Fs = 18.78 lbf Ans. (d) From above: k = 16.43 lbf/in Ans. 2.63D 2.63(0.8) (e) Table 10-2 and Eq. (10-13): ( L0 )c r = = = 4.21 in α 0.5 Since L0 < (L0)cr, buckling is unlikely Ans. ______________________________________________________________________________ 10-8 Given: Design load, F1 = 16.5 lbf. Referring to Prob. 10-7 solution, C = 10, Na = 7 coils, Ssy = 101.8 kpsi, Fs = 18.78 lbf, ys = 1.14 in, L0 = 1.78 in, and (L0)cr = 4.21 in. Eq. (10-18): 4 ≤ C ≤ 12 Shigley’s MED, 10th edition C = 10 O.K. Chapter 10 Solutions, Page 5/41 www.konkur.in Eq. (10-19): 3 ≤ Na ≤ 15 Na = 7 O.K. Fs 18.78 −1 = − 1 = 0.14 F1 16.5 Eq. (10-20): ξ ≥ 0.15, ξ = 0.14 not O.K . , but probably acceptable. From Eq. (10-7) for static service Eq. (10-17): ξ = 8F D = 74.5 (103 ) psi = 74.5 kpsi τ 1 = K B 1 3 = 1.135 3 π (0.080) πd n= S sy τ1 = 8(16.5)(0.8) 101.8 = 1.37 74.5 Eq. (10-21): ns ≥ 1.2, n = 1.37 O.K. Eq. (10-21): 18.78 18.78 = 74.5 = 84.8 kpsi 16.5 16.5 ns = S sy / τ s = 101.8 / 84.8 = 1.20 ns ≥ 1.2, ns = 1.2 It is solid-safe (basis of design). O.K. τ s = τ1 Eq. (10-13) and Table 10-2: L0 ≤ (L0)cr 1.78 in ≤ 4.21 in O.K. ______________________________________________________________________________ 10-9 Given: A228 music wire, squared and ground ends, d = 0.007 in, OD = 0.038 in, L0 = 0.58 in, Nt = 38 coils. Eq. (10-1): Eq. (10-5): Table 10-1: Table 10-5: Eq. (10-9): Table 10-1: Eq. (10-7): Table 10-4: D = OD − d = 0.038 − 0.007 = 0.031 in C = D/d = 0.031/0.007 = 4.429 4C + 2 4 ( 4.429 ) + 2 KB = = = 1.340 4C − 3 4 ( 4.429 ) − 3 Na = Nt − 2 = 38 − 2 = 36 coils (high) G = 12.0 Mpsi 0.007 4 (12.0 )106 d 4G k= = = 3.358 lbf/in 8D 3 N a 8 ( 0.0313 ) 36 Ls = dNt = 0.007(38) = 0.266 in ys = L0 − Ls = 0.58 − 0.266 = 0.314 in Fs = kys = 3.358(0.314) = 1.054 lbf 8 (1.054 ) 0.031 8F D = 325.1(103 ) psi τ s = K B s 3 = 1.340 3 πd π ( 0.007 ) (1) A = 201 kpsi⋅inm, m = 0.145 Shigley’s MED, 10th edition Chapter 10 Solutions, Page 6/41 www.konkur.in Eq. (10-14): Table 10-6: A 201 = = 412.7 kpsi m d 0.007 0.145 Ssy = 0.45 Sut = 0.45(412.7) = 185.7 kpsi Sut = τs > Ssy, that is, 325.1 > 185.7 kpsi, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and τs = Ssy /ns, and solve for ys, giving ( Ssy / ns ) π d 3 185.7 (103 ) / 1.2 π ( 0.0073 ) ys = = 8 K B kD The free length should be wound to 8 (1.340 ) 3.358 ( 0.031) L0 = Ls + ys = 0.266 + 0.149 = 0.415 in = 0.149 in Ans. This only addresses the solid-safe criteria. There are additional problems. ______________________________________________________________________________ 10-10 Given: B159 phosphor-bronze, squared and ground. ends, d = 0.014 in, OD = 0.128 in, L0 = 0.50 in, Nt = 16 coils. Eq. (10-1): Eq. (10-5): Table 10-1: Table 10-5: Eq. (10-9): Table 10-1: Eq. (10-7): Table 10-4: Eq. (10-14): Table 10-6: D = OD − d = 0.128 − 0.014 = 0.114 in C = D/d = 0.114/0.014 = 8.143 4C + 2 4 ( 8.143) + 2 KB = = = 1.169 4C − 3 4 ( 8.143) − 3 Na = Nt − 2 = 16 − 2 = 14 coils G = 6 Mpsi 0.0144 ( 6 )106 d 4G k= = = 1.389 lbf/in 8D 3 N a 8 ( 0.1143 )14 Ls = dNt = 0.014(16) = 0.224 in ys = L0 − Ls = 0.50 − 0.224 = 0.276 in Fs = kys = 1.389(0.276) = 0.3834 lbf 8 ( 0.3834 ) 0.114 8F D = 47.42 (103 ) psi τ s = K B s 3 = 1.169 πd π ( 0.0143 ) (1) A = 145 kpsi⋅inm, m = 0 A 145 Sut = m = = 145 kpsi d 0.0140 Ssy = 0.35 Sut = 0.35(135) = 47.25 kpsi τs > Ssy, that is, 47.42 > 47.25 kpsi, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and τs = Ssy /ns, and solve for ys, giving ( Ssy / ns ) π d 3 47.25 (103 ) /1.2 π ( 0.0143 ) ys = = 8 K B kD The free length should be wound to Shigley’s MED, 10th edition 8 (1.169 )1.389 ( 0.114 ) = 0.229 in Chapter 10 Solutions, Page 7/41 www.konkur.in L0 = Ls + ys = 0.224 + 0.229 = 0.453 in Ans. ______________________________________________________________________________ 10-11 Given: A313 stainless steel, squared and ground ends, d = 0.050 in, OD = 0.250 in, L0 = 0.68 in, Nt = 11.2 coils. Eq. (10-1): Eq. (10-5): Table 10-1: Table 10-5: Eq. (10-9): Table 10-1: Eq. (10-7): Table 10-4: Eq. (10-14): Table 10-6: D = OD − d = 0.250 − 0.050 = 0.200 in C = D/d = 0.200/0.050 = 4 4C + 2 4 ( 4 ) + 2 KB = = = 1.385 4C − 3 4 ( 4 ) − 3 Na = Nt − 2 = 11.2 − 2 = 9.2 coils G = 10 Mpsi 0.0504 (10 )106 d 4G k= = = 106.1 lbf/in 8D 3 N a 8 ( 0.23 ) 9.2 Ls = dNt = 0.050(11.2) = 0.56 in ys = L0 − Ls = 0.68 − 0.56 = 0.12 in Fs = kys = 106.1(0.12) = 12.73 lbf 8 (12.73) 0.2 8F D τ s = K B s 3 = 1.385 = 71.8 (103 ) psi 3 πd π ( 0.050 ) A = 169 kpsi⋅inm, m = 0.146 A 169 Sut = m = = 261.7 kpsi d 0.0500.146 Ssy = 0.35 Sut = 0.35(261.7) = 91.6 kpsi S sy 91.6 = 1.28 Spring is solid-safe (ns > 1.2) Ans. τ s 71.8 ______________________________________________________________________________ ns = = 10-12 Given: A227 hard-drawn wire, squared and ground ends, d = 0.148 in, OD = 2.12 in, L0 = 2.5 in, Nt = 5.75 coils. Eq. (10-1): Eq. (10-5): Table 10-1: Table 10-5: Eq. (10-9): Table 10-1: D = OD − d = 2.12 − 0.148 = 1.972 in C = D/d = 1.972/0.148 = 13.32 (high) 4 13.32 + 2 ( ) = 1.099 4C + 2 KB = = 4C − 3 4 (13.32 ) − 3 Na = Nt − 2 = 5.75 − 2 = 3.75 coils G = 11.4 Mpsi 0.1484 (11.4 )106 d 4G k= = = 23.77 lbf/in 8D 3 N a 8 (1.9723 ) 3.75 Ls = dNt = 0.148(5.75) = 0.851 in ys = L0 − Ls = 2.5 − 0.851 = 1.649 in Fs = kys = 23.77(1.649) = 39.20 lbf Shigley’s MED, 10th edition Chapter 10 Solutions, Page 8/41 www.konkur.in 8 ( 39.20 )1.972 8Fs D = 1.099 = 66.7 (103 ) psi 3 3 πd π ( 0.148 ) Eq. (10-7): τ s = KB Table 10-4: A = 140 kpsi⋅inm, m = 0.190 A 140 Sut = m = = 201.3 kpsi d 0.1480.190 Ssy = 0.35 Sut = 0.45(201.3) = 90.6 kpsi Eq. (10-14): Table 10-6: S sy 90.6 Ans. = 1.36 Spring is solid-safe (ns > 1.2) τ s 66.7 ______________________________________________________________________________ ns = = 10-13 Given: A229 OQ&T steel, squared and ground ends, d = 0.138 in, OD = 0.92 in, L0 = 2.86 in, Nt = 12 coils. D = OD − d = 0.92 − 0.138 = 0.782 in Eq. (10-5): C = D/d = 0.782/0.138 = 5.667 4C + 2 4 ( 5.667 ) + 2 KB = = = 1.254 4C − 3 4 ( 5.667 ) − 3 Table 10-1: Na = Nt − 2 = 12 − 2 = 10 coils Eq. (10-1): A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels, G = 11.5 Mpsi. 0.1384 (11.5 )106 d 4G Eq. (10-9): k= = = 109.0 lbf/in 8D3 N a 8 ( 0.7823 )10 Table 10-1: Eq. (10-7): Table 10-4: Eq. (10-14): Table 10-6: Ls = dNt = 0.138(12) = 1.656 in ys = L0 − Ls = 2.86 − 1.656 = 1.204 in Fs = kys = 109.0(1.204) = 131.2 lbf 8 (131.2 ) 0.782 8F D τ s = K B s 3 = 1.254 = 124.7 (103 ) psi 3 πd π ( 0.138 ) (1) A = 147 kpsi⋅inm, m = 0.187 A 147 Sut = m = = 212.9 kpsi d 0.1380.187 Ssy = 0.50 Sut = 0.50(212.9) = 106.5 kpsi τs > Ssy, that is, 124.7 > 106.5 kpsi, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and τs = Ssy /ns, and solve for ys, giving ( Ssy / ns ) π d 3 106.5 (103 ) /1.2 π ( 0.1383 ) ys = 8K B kD = 8 (1.254 )109.0 ( 0.782 ) = 0.857 in The free length should be wound to Shigley’s MED, 10th edition Chapter 10 Solutions, Page 9/41 www.konkur.in L0 = Ls + ys = 1.656 + 0.857 = 2.51 in Ans. ______________________________________________________________________________ 10-14 Given: A232 chrome-vanadium steel, squared and ground ends, d = 0.185 in, OD = 2.75 in, L0 = 7.5 in, Nt = 8 coils. Eq. (10-1): Eq. (10-5): D = OD − d = 2.75 − 0.185 = 2.565 in C = D/d = 2.565/0.185 = 13.86 (high) 4C + 2 4 (13.86 ) + 2 KB = = = 1.095 4C − 3 4 (13.86 ) − 3 Table 10-1: Na = Nt − 2 = 8 − 2 = 6 coils Table 10-5: G = 11.2 Mpsi. 0.1854 (11.2 )106 d 4G k= = = 16.20 lbf/in 8D3 N a 8 ( 2.5653 ) 6 Eq. (10-9): Table 10-1: Eq. (10-7): Ls = dNt = 0.185(8) = 1.48 in ys = L0 − Ls = 7.5 − 1.48 = 6.02 in Fs = kys = 16.20(6.02) = 97.5 lbf 8 ( 97.5 ) 2.565 8F D = 110.1(103 ) psi τ s = K B s 3 = 1.095 3 πd π ( 0.185 ) (1) A = 169 kpsi⋅inm, m = 0.168 A 169 Eq. (10-14): Sut = m = = 224.4 kpsi d 0.1850.168 Table 10-6: Ssy = 0.50 Sut = 0.50(224.4) = 112.2 kpsi S 112.2 ns = sy = = 1.02 Spring is not solid-safe (ns < 1.2) τ s 110.1 Return to Eq. (1) with Fs = kys and τs = Ssy /ns, and solve for ys, giving 3 3 S sy / ns ) π d 3 112.2 (10 ) / 1.2 π ( 0.185 ) ( ys = = = 5.109 in 8K B kD 8 (1.095 )16.20 ( 2.565 ) The free length should be wound to Table 10-4: L0 = Ls + ys = 1.48 + 5.109 = 6.59 in Ans. ______________________________________________________________________________ 10-15 Given: A313 stainless steel, squared and ground ends, d = 0.25 mm, OD = 0.95 mm, L0 = 12.1 mm, Nt = 38 coils. D = OD − d = 0.95 − 0.25 = 0.7 mm Eq. (10-1): C = D/d = 0.7/0.25 = 2.8 (low) 4C + 2 4 ( 2.8 ) + 2 Eq. (10-5): KB = = = 1.610 4C − 3 4 ( 2.8 ) − 3 Shigley’s MED, 10th edition Chapter 10 Solutions, Page 10/41 www.konkur.in Table 10-1: Na = Nt − 2 = 38 − 2 = 36 coils Table 10-5: G = 69.0(103) MPa. 0.254 ( 69.0 )103 d 4G k= = = 2.728 N/mm 8D3 N a 8 ( 0.73 ) 36 Eq. (10-9): Table 10-1: Eq. (10-7): (high) Ls = dNt = 0.25(38) = 9.5 mm ys = L0 − Ls = 12.1 − 9.5 = 2.6 mm Fs = kys = 2.728(2.6) = 7.093 N 8 ( 7.093) 0.7 8F D τ s = K B s 3 = 1.610 = 1303 MPa πd π ( 0.253 ) (1) Table 10-4 (dia. less than table): A = 1867 MPa⋅mmm, m = 0.146 A 1867 Eq. (10-14): Sut = m = = 2286 MPa d 0.250.146 Table 10-6: Ssy = 0.35 Sut = 0.35(2286) = 734 MPa τs > Ssy, that is, 1303 > 734 MPa, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and τs = Ssy /ns, and solve for ys, giving ( S sy / ns ) π d 3 ( 734 / 1.2 ) π ( 0.253 ) ys = = = 1.22 mm 8 K B kD 8 (1.610 ) 2.728 ( 0.7 ) The free length should be wound to L0 = Ls + ys = 9.5 + 1.22 = 10.72 mm Ans. This only addresses the solid-safe criteria. There are additional problems. ______________________________________________________________________________ 10-16 Given: A228 music wire, squared and ground ends, d = 1.2 mm, OD = 6.5 mm, L0 = 15.7 mm, Nt = 10.2 coils. D = OD − d = 6.5 − 1.2 = 5.3 mm Eq. (10-1): C = D/d = 5.3/1.2 = 4.417 4C + 2 4 ( 4.417 ) + 2 Eq. (10-5): KB = = = 1.368 4C − 3 4 ( 4.417 ) − 3 Table (10-1): Na = Nt − 2 = 10.2 − 2 = 8.2 coils Table 10-5 (d = 1.2/25.4 = 0.0472 in): G = 81.7(103) MPa. 1.24 ( 81.7 )103 d 4G Eq. (10-9): k= = = 17.35 N/mm 8D3 N a 8 ( 5.33 ) 8.2 Table 10-1: Ls = dNt = 1.2(10.2) = 12.24 mm ys = L0 − Ls = 15.7 − 12.24 = 3.46 mm Fs = kys = 17.35(3.46) = 60.03 N Shigley’s MED, 10th edition Chapter 10 Solutions, Page 11/41 www.konkur.in 8 ( 60.03) 5.3 8 Fs D = 1.368 = 641.4 MPa 3 πd π (1.23 ) Eq. (10-7): τ s = KB Table 10-4: A = 2211 MPa⋅mmm, m = 0.145 A 2211 Sut = m = 0.145 = 2153 MPa d 1.2 Ssy = 0.45 Sut = 0.45(2153) = 969 MPa Eq. (10-14): Table 10-6: S sy (1) 969 = 1.51 Spring is solid-safe (ns > 1.2) Ans. τ s 641.4 ______________________________________________________________________________ ns = = 10-17 Given: A229 OQ&T steel, squared and ground ends, d = 3.5 mm, OD = 50.6 mm, L0 = 75.5 mm, Nt = 5.5 coils. Eq. (10-1): Eq. (10-5): Table 10-1: D = OD − d = 50.6 − 3.5 = 47.1 mm C = D/d = 47.1/3.5 = 13.46 (high) 4C + 2 4 (13.46 ) + 2 KB = = = 1.098 4C − 3 4 (13.46 ) − 3 Na = Nt − 2 = 5.5 − 2 = 3.5 coils A229 OQ&T steel is not given in Table 10-5. From Table A-5, for carbon steels, G = 79.3(103) MPa. 3.54 ( 79.3)103 d 4G Eq. (10-9): k= = = 4.067 N/mm 8D3 N a 8 ( 47.13 ) 3.5 Table 10-1: Eq. (10-7): Ls = dNt = 3.5(5.5) = 19.25 mm ys = L0 − Ls = 75.5 − 19.25 = 56.25 mm Fs = kys = 4.067(56.25) = 228.8 N 8 ( 228.8 ) 47.1 8F D τ s = K B s 3 = 1.098 = 702.8 MPa πd π ( 3.53 ) (1) A = 1855 MPa⋅mmm, m = 0.187 A 1855 Eq. (10-14): Sut = m = = 1468 MPa d 3.50.187 Table 10-6: Ssy = 0.50 Sut = 0.50(1468) = 734 MPa S sy 734 ns = = = 1.04 Spring is not solid-safe (ns < 1.2) τ s 702.8 Return to Eq. (1) with Fs = kys and τs = Ssy /ns, and solve for ys, giving ( S sy / ns ) π d 3 = ( 734 / 1.2 ) π ( 3.53 ) = 48.96 mm ys = 8 K B kD 8 (1.098 ) 4.067 ( 47.1) The free length should be wound to Table 10-4: Shigley’s MED, 10th edition Chapter 10 Solutions, Page 12/41 www.konkur.in L0 = Ls + ys = 19.25 + 48.96 = 68.2 mm Ans. ______________________________________________________________________________ 10-18 Given: B159 phosphor-bronze, squared and ground ends, d = 3.8 mm, OD = 31.4 mm, L0 = 71.4 mm, Nt = 12.8 coils. Eq. (10-1): Eq. (10-5): D = OD − d = 31.4 − 3.8 = 27.6 mm C = D/d = 27.6/3.8 = 7.263 4C + 2 4 ( 7.263) + 2 KB = = = 1.192 4C − 3 4 ( 7.263) − 3 Table 10-1: Na = Nt − 2 = 12.8 − 2 = 10.8 coils Table 10-5: G = 41.4(103) MPa. 3.84 ( 41.4 )103 d 4G k= = = 4.752 N/mm 8 D 3 N a 8 ( 27.63 )10.8 Eq. (10-9): Table 10-1: Eq. (10-7): Ls = dNt = 3.8(12.8) = 48.64 mm ys = L0 − Ls = 71.4 − 48.64 = 22.76 mm Fs = kys = 4.752(22.76) = 108.2 N 8 (108.2 ) 27.6 8F D = 165.2 MPa τ s = K B s 3 = 1.192 πd π ( 3.83 ) (1) Table 10-4 (d = 3.8/25.4 = 0.150 in): A = 932 MPa⋅mmm, m = 0.064 A 932 Eq. (10-14): Sut = m = = 855.7 MPa d 3.80.064 Table 10-6: Ssy = 0.35 Sut = 0.35(855.7) = 299.5 MPa S 299.5 ns = sy = = 1.81 Spring is solid-safe (ns > 1.2) Ans. τ s 165.2 ______________________________________________________________________________ 10-19 Given: A232 chrome-vanadium steel, squared and ground ends, d = 4.5 mm, OD = 69.2 mm, L0 = 215.6 mm, Nt = 8.2 coils. Eq. (10-1): Eq. (10-5): D = OD − d = 69.2 − 4.5 = 64.7 mm C = D/d = 64.7/4.5 = 14.38 (high) 4C + 2 4 (14.38 ) + 2 KB = = = 1.092 4C − 3 4 (14.38 ) − 3 Table 10-1: Na = Nt − 2 = 8.2 − 2 = 6.2 coils Table 10-5: G = 77.2(103) MPa. 4.54 ( 77.2 )103 d 4G k= = = 2.357 N/mm 8D3 N a 8 ( 64.73 ) 6.2 Eq. (10-9): Table 10-1: Ls = dNt = 4.5(8.2) = 36.9 mm Shigley’s MED, 10th edition Chapter 10 Solutions, Page 13/41 www.konkur.in Eq. (10-7): Table 10-4: Eq. (10-14): Table 10-6: ys = L0 − Ls = 215.6 − 36.9 = 178.7 mm Fs = kys = 2.357(178.7) = 421.2 N 8 ( 421.2 ) 64.7 8F D = 832 MPa τ s = K B s 3 = 1.092 πd π ( 4.53 ) (1) A = 2005 MPa⋅mmm, m = 0.168 A 2005 Sut = m = = 1557 MPa d 4.50.168 Ssy = 0.50 Sut = 0.50(1557) = 779 MPa τs > Ssy, that is, 832 > 779 MPa, the spring is not solid-safe. Return to Eq. (1) with Fs = kys and τs = Ssy /ns, and solve for ys, giving ( S sy / ns ) π d 3 ( 779 / 1.2 ) π ( 4.53 ) ys = = = 139.5 mm 8 K B kD 8 (1.092 ) 2.357 ( 64.7 ) The free length should be wound to L0 = Ls + ys = 36.9 + 139.5 = 176.4 mm Ans. This only addresses the solid-safe criteria. There are additional problems. ______________________________________________________________________________ 10-20 Given: A227 HD steel. From the figure: L0 = 4.75 in, OD = 2 in, and d = 0.135 in. Thus D = OD − d = 2 − 0.135 = 1.865 in (a) By counting, Nt = 12.5 coils. Since the ends are squared along 1/4 turn on each end, N a = 12.5 − 0.5 = 12 turns Ans. p = 4.75 / 12 = 0.396 in Ans. The solid stack is 13 wire diameters Ls = 13(0.135) = 1.755 in Ans. (b) From Table 10-5, G = 11.4 Mpsi 0.1354 (11.4) (106 ) d 4G k = = = 6.08 lbf/in 8D 3 N a 8 (1.8653 ) (12) (c) Fs = k(L0 - Ls ) = 6.08(4.75 − 1.755) = 18.2 lbf Ans. Ans. (d) C = D/d = 1.865/0.135 = 13.81 Shigley’s MED, 10th edition Chapter 10 Solutions, Page 14/41 www.konkur.in 4(13.81) + 2 = 1.096 4(13.81) − 3 8F D 8(18.2)(1.865) τ s = K B s 3 = 1.096 = 38.5 (103 ) psi = 38.5 kpsi 3 πd π ( 0.135 ) KB = Ans. ______________________________________________________________________________ 10-21 For the wire diameter analyzed, G = 11.75 Mpsi per Table 10-5. Use squared and ground ends. The following is a spread-sheet study using Fig. 10-3 for parts (a) and (b). For Na, k = Fmax /y = 20/2 = 10 lbf/in. For τs, F = Fs = 20(1 + ξ) = 20(1 + 0.15) = 23 lbf. Source Eq. (10-1) Eq. (10-9) Table 10-1 Table 10-1 1.15y + Ls Eq. (10-13) Table 10-4 Table 10-4 Eq. (10-14) Table 10-6 Eq. (10-5) Eq. (10-7) Eq. (10-3) Eq. (10-22) (a) Spring over a Rod (b) Spring in a Hole Parameter Values Source Parameter Values d 0.075 0.080 0.085 d 0.075 0.080 ID 0.800 0.800 0.800 OD 0.950 0.950 D 0.875 0.880 0.885 D 0.875 0.870 C 11.667 11.000 10.412 Eq. (10-1) C 11.667 10.875 Na 6.937 8.828 11.061 Eq. (10-9) Na 6.937 9.136 Nt 8.937 10.828 13.061 Table 10-1 Nt 8.937 11.136 Ls 0.670 0.866 1.110 Table 10-1 Ls 0.670 0.891 L0 2.970 3.166 3.410 1.15y + Ls L0 2.970 3.191 (L0)cr 4.603 4.629 4.655 Eq. (10-13) (L0)cr 4.603 4.576 A 201.000 201.000 201.000 Table 10-4 A 201.000 201.000 m 0.145 0.145 0.145 Table 10-4 m 0.145 0.145 Sut 292.626 289.900 287.363 Eq. (10-14) Sut 292.626 289.900 Ssy 131.681 130.455 129.313 Table 10-6 Ssy 131.681 130.455 KB 1.115 1.122 1.129 Eq. (10-5) KB 1.115 1.123 135.335 112.948 95.293 Eq. (10-7) 135.335 111.787 τs τs ns 0.973 1.155 1.357 Eq. (10-3) ns 0.973 1.167 fom −0.282 −0.391 −0.536 Eq. (10-22) fom −0.282 −0.398 For ns ≥ 1.2, the optimal size is d = 0.085 in for both cases. 0.085 0.950 0.865 10.176 11.846 13.846 1.177 3.477 4.550 201.000 0.145 287.363 129.313 1.133 93.434 1.384 −0.555 ______________________________________________________________________________ 10-22 In Prob. 10-21, there is an advantage of first selecting d as one can select from the available sizes (Table A-28). Selecting C first requires a calculation of d where then a size must be selected from Table A-28. Consider part (a) of the problem. It is required that ID = D − d = 0.800 in. (1) From Eq. (10-1), D = Cd. Substituting this into the first equation yields d= 0.800 C −1 (2) Starting with C = 10, from Eq. (2) we find that d = 0.089 in. From Table A-28, the closest diameter is d = 0.090 in. Substituting this back into Eq. (1) gives D = 0.890 in, with C = 0.890/0.090 = 9.889, which are acceptable. From this point the solution is the same as Prob. 10-21. For part (b), use Shigley’s MED, 10th edition Chapter 10 Solutions, Page 15/41 www.konkur.in OD = D + d = 0.950 in. d= and, (3) 0.800 C −1 (a) Spring over a rod Parameter Values C 10.000 10.5 Eq. (2) d 0.089 0.084 Table A-28 d 0.090 0.085 Eq. (1) D 0.890 0.885 Eq. (10-1) C 9.889 10.412 Eq. (10-9) Na 13.669 11.061 Table 10-1 Nt 15.669 13.061 Table 10-1 Ls 1.410 1.110 1.15y + Ls L0 3.710 3.410 Eq. (10-13) (L0)cr 4.681 4.655 Table 10-4 A 201.000 201.000 Table 10-4 m 0.145 0.145 Eq. (10-14) Sut 284.991 287.363 Table 10-6 Ssy 128.246 129.313 Eq. (10-5) KB 1.135 1.128 τs Eq. (10-7) 81.167 95.223 ns 1.580 1.358 ns = Ssy/τs Eq. (10-22) fom -0.725 -0.536 Source (4) (b) Spring in a Hole Source Parameter Values C 10.000 Eq. (4) d 0.086 Table A-28 d 0.085 Eq. (3) D 0.865 Eq. (10-1) C 10.176 Eq. (10-9) Na 11.846 Table 10-1 Nt 13.846 Table 10-1 Ls 1.177 1.15y + Ls L0 3.477 Eq. (10-13) (L0)cr 4.550 Table 10-4 A 201.000 Table 10-4 m 0.145 Eq. (10-14) Sut 287.363 Table 10-6 Ssy 129.313 Eq. (10-5) KB 1.135 τs Eq. (10-7) 93.643 ns 1.381 ns = Ssy/τs Eq. (10-22) fom -0.555 Again, for ns ≥ 1.2, the optimal size is = 0.085 in. Although this approach used less iterations than in Prob. 10-21, this was due to the initial values picked and not the approach. ______________________________________________________________________________ 10-23 One approach is to select A227 HD steel for its low cost. Try L0 = 48 mm, then for y = 48 − 37.5 = 10.5 mm when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286 N/mm. For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm. Starting with d = 2 mm, D = ID + d = 11.25 + 2 = 13.25 mm C = D/d = 13.25/2 = 6.625 (acceptable) Table 10-5 (d = 2/25.4 = 0.0787 in): G = 79.3 GPa d 4G 24 (79.3)103 Eq. (10-9): Na = = = 15.9 coils 8kD 3 8(4.286)13.253 Assume squared and closed. Shigley’s MED, 10th edition Chapter 10 Solutions, Page 16/41 www.konkur.in Table 10-1: Eq. (10-5): Nt = Na + 2 = 15.9 + 2 = 17.9 coils Ls = dNt = 2(17.9) =35.8 mm ys = L0 − Ls = 48 − 35.8 = 12.2 mm Fs = kys = 4.286(12.2) = 52.29 N 4C + 2 4 ( 6.625 ) + 2 KB = = = 1.213 4C − 3 4 ( 6.625 ) − 3 8(52.29)13.25 8Fs D = 1.213 = 267.5 MPa πd3 π ( 23 ) Eq. (10-7): τ s = KB Table 10-4: A = 1783 MPa · mmm, m = 0.190 A 1783 Sut = m = 0.190 = 1563 MPa d 2 Ssy = 0.45Sut = 0.45(1563) = 703.3 MPa Eq. (10-14): Table 10-6: ns = S sy τs = 703.3 = 2.63 >1.2 267.5 O.K . No other diameters in the given range work. So specify A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L0 = 48 mm. Ans. ______________________________________________________________________________ 10-24 Select A227 HD steel for its low cost. Try L0 = 48 mm, then for y = 48 − 37.5 = 10.5 mm when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286 N/mm. For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm. and, D − d = 11.25 (1) D =Cd (2) Starting with C = 8, gives D = 8d. Substitute into Eq. (1) resulting in d = 1.607 mm. Selecting the nearest diameter in the given range, d = 1.6 mm. From this point, the calculations are shown in the third column of the spreadsheet output shown. We see that for d = 1.6 mm, the spring is not solid safe. Iterating on C we find that C = 6.5 provides acceptable results with the specifications A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L0 = 48 mm. Ans. Shigley’s MED, 10th edition Chapter 10 Solutions, Page 17/41 www.konkur.in Source Eq. (2) Table A-28 Eq. (1) Eq. (10-1) Eq. (10-9) Table 10-1 Table 10-1 L0 − Ls Fs = kys Table 10-4 Table 10-4 Eq. (10-14) Table 10-6 Eq. (10-5) Eq. (10-7) ns = Ssy/τs C d d D C Na Nt Ls ys Fs A m Sut Ssy KB τs ns Parameter Values 8.000 7 6.500 1.607 1.875 2.045 1.600 1.800 2.000 12.850 13.050 13.250 8.031 7.250 6.625 7.206 10.924 15.908 9.206 12.924 17.908 14.730 23.264 35.815 33.270 24.736 12.185 142.594 106.020 52.224 1783.000 1783.000 1783.000 0.190 0.190 0.190 1630.679 1594.592 1562.988 733.806 717.566 703.345 1.172 1.200 1.217 1335.568 724.943 268.145 0.549 0.990 2.623 The only difference between selecting C first rather than d as was done in Prob. 10-23, is that once d is calculated, the closest wire size must be selected. Iterating on d uses available wire sizes from the beginning. ______________________________________________________________________________ 10-25 A stock spring catalog may have over two hundred pages of compression springs with up to 80 springs per page listed. • Students should be made aware that such catalogs exist. • Many springs are selected from catalogs rather than designed. • The wire size you want may not be listed. • Catalogs may also be available on disk or the web through search routines. For example, disks are available from Century Spring at 1 - (800) - 237 - 5225 www.centuryspring.com • It is better to familiarize yourself with vendor resources rather than invent them yourself. • Sample catalog pages can be given to students for study. ______________________________________________________________________________ 10-26 Given: ID = 0.6 in, C = 10, L0 = 5 in, Ls = 5 − 3 = 2 in, sq. & grd ends, unpeened, HD A227 wire. (a) With ID = D − d = 0.6 in and C = D/d = 10 ⇒10 d − d = 0.6 ⇒ d = 0.0667 in Ans., and D = 0.667 in. (b) Table 10-1: Ls = dNt = 2 in ⇒ Nt = 2/0.0667 = 30 coils Ans. (c) Table 10-1: Shigley’s MED, 10th edition Na = Nt − 2 = 30 − 2 = 28 coils Chapter 10 Solutions, Page 18/41 www.konkur.in Table 10-5: G = 11.5 Mpsi 0.0667 4 (11.5 )106 d 4G k= = = 3.424 lbf/in 8D3 N a 8 ( 0.6673 ) 28 Eq. (10-9): (d) Table 10-4: Eq. (10-14): Ans. A = 140 kpsi⋅inm, m = 0.190 A 140 Sut = m = = 234.2 kpsi d 0.0667 0.190 Table 10-6: Ssy = 0.45 Sut = 0.45 (234.2) = 105.4 kpsi Eq. (10-5): Fs = kys = 3.424(3) = 10.27 lbf 4C + 2 4 (10 ) + 2 KB = = = 1.135 4C − 3 4 (10 ) − 3 Eq. (10-7): τ s = KB 8 (10.27 ) 0.667 8 Fs D = 1.135 3 πd π ( 0.06673 ) = 66.72 (103 ) psi = 66.72 kpsi S sy 105.4 = 1.58 Ans. τ s 66.72 (e) τa = τm = 0.5τs = 0.5(66.72) = 33.36 kpsi, r = τa / τm = 1. Using the Gerber fatigue failure criterion with Zimmerli data, ns = Eq. (10-30): = Ssu = 0.67 Sut = 0.67(234.2) = 156.9 kpsi The Gerber ordinate intercept for the Zimmerli data is S sa 35 S se = = = 39.9 kpsi 2 2 1 − ( S sm / S su ) 1 − ( 55 / 156.9 ) Table 6-7, p. 315, 2 2 S se r 2 S su2 S sa = −1 + 1 + 2 S se rS su 2 12 (156.92 ) 2 ( 39.9 ) = −1 + 1 + = 37.61 kpsi 2 ( 39.9 ) 1(156.9 ) S sa 37.61 = 1.13 Ans. τ a 33.36 ______________________________________________________________________________ nf = = 10-27 Given: OD ≤ 0.9 in, C = 8, L0 = 3 in, Ls = 1 in, ys = 3 − 1 = 2 in, sq. ends, unpeened, music wire. (a) Try OD = D + d = 0.9 in, C = D/d = 8 ⇒ D = 8d ⇒ 9d = 0.9 ⇒ d = 0.1 Ans. Shigley’s MED, 10th edition Chapter 10 Solutions, Page 19/41 www.konkur.in D = 8(0.1) = 0.8 in (b) Table 10-1: Ls = d (Nt + 1) ⇒ Nt = Ls / d − 1 = 1/0.1− 1 = 9 coils Ans. Table 10-1: (c) Table 10-5: Eq. (10-9): (d) Na = Nt − 2 = 9 − 2 = 7 coils G = 11.75 Mpsi k= 0.14 (11.75 )106 d 4G = = 40.98 lbf/in 8D 3 N a 8 ( 0.83 ) 7 Ans. Fs = kys = 40.98(2) = 81.96 lbf 4C + 2 4 ( 8 ) + 2 = = 1.172 4C − 3 4 ( 8 ) − 3 Eq. (10-5): KB = Eq. (10-7): τ s = KB Table 10-4: A = 201 kpsi⋅inm, m = 0.145 Eq. (10-14): Sut = Table 10-6: Ssy = 0.45 Sut = 0.45(280.7) = 126.3 kpsi ns = 8 ( 81.96 ) 0.8 8Fs D = 1.172 = 195.7 (103 ) psi = 195.7 kpsi 3 3 πd π ( 0.1 ) A 201 = = 280.7 kpsi m d 0.10.145 S sy τs = 126.3 = 0.645 195.7 Ans. (e) τa = τm = τs /2 = 195.7/2 = 97.85 kpsi. Using the Gerber fatigue failure criterion with Zimmerli data, Eq. (10-30): Ssu = 0.67 Sut = 0.67(280.7) = 188.1 kpsi The Gerber ordinate intercept for the Zimmerli data is S sa 35 S se = = = 38.3 kpsi 2 2 1 − ( S sm / S su ) 1 − ( 55 /188.1) Table 6-7, p. 315, 2 2 S se r 2 S su2 S sa = −1 + 1 + 2 S se rS su 2 12 (188.12 ) 2 ( 38.3) = −1 + 1 + = 36.83 kpsi 2 ( 38.3) 1(188.1) Shigley’s MED, 10th edition Chapter 10 Solutions, Page 20/41 www.konkur.in nf = S sa τa = 36.83 = 0.376 97.85 Ans. Obviously, the spring is severely under designed and will fail statically and in fatigue. Increasing C would improve matters. Try C = 12. This yields ns = 1.83 and nf = 1.00. ______________________________________________________________________________ 10-28 Given: Fmax = 300 lbf, Fmin = 150 lbf, ∆y = 1 in, OD = 2.1 − 0.2 = 1.9 in, C = 7, unpeened, squared & ground, oil-tempered wire. (a) D = OD − d = 1.9 − d (1) C = D/d = 7 ⇒ D = 7d (2) Substitute Eq. (2) into (1) 7d = 1.9 − d ⇒ d = 1.9/8 = 0.2375 in (b) From Eq. (2): D = 7d = 7(0.2375) = 1.663 in (c) k= (d) Table 10-5: G = 11.6 Mpsi Ans. ∆F 300 − 150 = = 150 lbf/in ∆y 1 Ans. 4 6 d 4G 0.2375 (11.6 )10 = = 6.69 coils 8D 3k 8 (1.6633 )150 Eq. (10-9): Na = Table 10-1: Nt = Na + 2 = 8.69 coils Ans. Table 10-6: A = 147 kpsi⋅inm, m = 0.187 A 147 Sut = m = = 192.3 kpsi d 0.23750.187 Ssy = 0.5 Sut = 0.5(192.3) = 96.15 kpsi Eq. (10-5): KB = Eq. (10-7): τ s = KB (e) Table 10-4: Eq. (10-14): Shigley’s MED, 10th edition Ans. 4C + 2 4 ( 7 ) + 2 = = 1.2 4C − 3 4 ( 7 ) − 3 8 Fs D = S sy π d3 Chapter 10 Solutions, Page 21/41 www.konkur.in Fs = π d 3 S sy 8K B D = π ( 0.23753 ) 96.15 (103 ) 8 (1.2 )1.663 = 253.5 lbf ys = Fs / k = 253.5/150 = 1.69 in Table 10-1: Ls = Nt d = 8.46(0.2375) = 2.01 in L0 = Ls + ys = 2.01 + 1.69 = 3.70 in Ans. ______________________________________________________________________________ 10-29 For a coil radius given by: R = R1 + R2 − R1 θ 2π N The torsion of a section is T = PR where dL = R dθ ∂U 1 ∂T 1 2π N 3 = T dL = PR dθ ∂P GJ ∫ ∂P GJ ∫0 3 P 2π N R2 − R1 = θ dθ R1 + GJ ∫0 2π N δP = 2π N 4 P 1 2π N R2 − R1 = R + θ 1 GJ 4 R2 − R1 2π N 0 π PN π PN = R24 − R14 ) = ( R1 + R2 ) ( R12 + R22 ) ( 2GJ ( R2 − R1) 2GJ π 4 16 PN J = d ∴ δp = ( R1 + R2 ) ( R12 + R22 ) 4 32 Gd P d 4G k = = Ans. δ P 16 N ( R1 + R2 ) ( R12 + R22 ) ______________________________________________________________________________ 10-30 Given: Fmin = 4 lbf, Fmax = 18 lbf, k = 9.5 lbf/in, OD ≤ 2.5 in, nf = 1.5. For a food service machinery application select A313 Stainless wire. Table 10-5: G = 10(106) psi Note that for 0.013 ≤ d ≤ 0.10 in A = 169, m = 0.146 0.10 < d ≤ 0.20 in A = 128, m = 0.263 18 − 4 18 + 4 Fa = = 7 lbf , Fm = = 11 lbf , r = 7 / 11 2 2 169 Try, d = 0.080 in, Sut = = 244.4 kpsi (0.08)0.146 Shigley’s MED, 10th edition Chapter 10 Solutions, Page 22/41 www.konkur.in Ssu = 0.67Sut = 163.7 kpsi, Ssy = 0.35Sut = 85.5 kpsi Try unpeened using Zimmerli’s endurance data: Ssa = 35 kpsi, Ssm = 55 kpsi S sa 35 Gerber: S se = = = 39.5 kpsi 2 1 − (S sm / S su ) 1 − (55 / 163.7) 2 2 (7 / 11) 2 (163.7)2 2(39.5) S sa = −1 + 1 + = 35.0 kpsi 2(39.5) (7 / 11)(163.7) α = S sa / n f = 35.0 / 1.5 = 23.3 kpsi β = 8(7) −3 8Fa (10−3 ) = (10 ) = 2.785 kpsi 2 2 πd π (0.08 ) 2 C = D= KB = τa = nf = 2(23.3) − 2.785 2(23.3) − 2.785 3(23.3) + − = 6.97 4(2.785) 4(2.785) 4(2.785) Cd = 6.97(0.08) = 0.558 in 4C + 2 4(6.97) + 2 = = 1.201 4C − 3 4(6.97) − 3 8(7)(0.558) −3 8F D K B a 3 = 1.201 (10 ) = 23.3 kpsi 3 πd π (0.08 ) 35 / 23.3 = 1.50 checks Gd 4 10(106 )(0.08) 4 = = 31.02 coils 8kD 3 8(9.5)(0.558)3 = 31.02 + 2 = 33 coils, Ls = dN t = 0.08(33) = 2.64 in = Fmax / k = 18 / 9.5 = 1.895 in = (1 + ξ ) ymax = (1 + 0.15)(1.895) = 2.179 in = 2.64 + 2.179 = 4.819 in D 2.63(0.558) = 2.63 = = 2.935 in α 0.5 = 1.15(18 / 7)τ a = 1.15(18 / 7)(23.3) = 68.9 kpsi = S sy / τ s = 85.5 / 68.9 = 1.24 Na = Nt ymax ys L0 ( L0 )cr τs ns f = kg = π d DN aγ 2 2 9.5(386) = 109 Hz π (0.08 )(0.558)(31.02)(0.283) 2 2 These steps are easily implemented on a spreadsheet, as shown below, for different diameters. Shigley’s MED, 10th edition Chapter 10 Solutions, Page 23/41 www.konkur.in d1 d 0.080 m 0.146 A 169.000 Sut 244.363 Ssu 163.723 Ssy 85.527 Sse 39.452 Ssa 35.000 23.333 α 2.785 β C 6.977 D 0.558 KB 1.201 23.333 τa nf 1.500 Na 30.993 Nt 32.993 LS 2.639 ys 2.179 L0 4.818 (L0)cr 2.936 τs 69.000 ns 1.240 f,(Hz) 108.895 d2 0.0915 0.146 169.000 239.618 160.544 83.866 39.654 35.000 23.333 2.129 9.603 0.879 1.141 23.333 1.500 13.594 15.594 1.427 2.179 3.606 4.622 69.000 1.215 114.578 d3 0.1055 0.263 128 231.257 154.942 80.940 40.046 35.000 23.333 1.602 13.244 1.397 1.100 23.333 1.500 5.975 7.975 0.841 2.179 3.020 7.350 69.000 1.173 118.863 d4 0.1205 0.263 128 223.311 149.618 78.159 40.469 35.000 23.333 1.228 17.702 2.133 1.074 23.333 1.500 2.858 4.858 0.585 2.179 2.764 11.220 69.000 1.133 121.775 The shaded areas depict conditions outside the recommended design conditions. Thus, one spring is satisfactory. The specifications are: A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, L0 = 3.606 in, and Nt = 15.59 turns Ans. ______________________________________________________________________________ 10-31 The steps are the same as in Prob. 10-30 except that the Gerber-Zimmerli criterion is replaced with Goodman-Zimmerli: S se = S sa 1 − ( S sm S su ) The problem then proceeds as in Prob. 10-30. The results for the wire sizes are shown Shigley’s MED, 10th edition Chapter 10 Solutions, Page 24/41 www.konkur.in below (see solution to Prob. 10-30 for additional details). d m A Sut Ssu Ssy Sse Ssa α β C D d1 0.080 0.146 169.000 244.363 163.723 85.527 52.706 43.513 29.008 2.785 9.052 0.724 d2 0.0915 0.146 169.000 239.618 160.544 83.866 53.239 43.560 29.040 2.129 12.309 1.126 Iteration of d for the first trial d3 d4 d1 d2 d3 d4 0.1055 0.1205 d 0.080 0.0915 0.1055 0.1205 0.263 0.263 KB 1.151 1.108 1.078 1.058 128.000 128.000 τa 29.008 29.040 29.090 29.127 231.257 223.311 nf 1.500 1.500 1.500 1.500 154.942 149.618 Na 14.191 6.456 2.899 1.404 80.940 78.159 Nt 16.191 8.456 4.899 3.404 54.261 55.345 Ls 1.295 0.774 0.517 0.410 43.634 43.691 ymax 2.875 2.875 2.875 2.875 29.090 29.127 L0 4.170 3.649 3.392 3.285 1.602 1.228 (L0)cr 3.809 5.924 9.354 14.219 16.856 22.433 τs 85.782 85.876 86.022 86.133 1.778 2.703 ns 0.997 0.977 0.941 0.907 f (Hz) 140.040 145.559 149.938 152.966 Without checking all of the design conditions, it is obvious that none of the wire sizes satisfy ns ≥ 1.2. Also, the Gerber line is closer to the yield line than the Goodman. Setting nf = 1.5 for Goodman makes it impossible to reach the yield line (ns < 1) . The table below uses nf = 2. d1 d2 d 0.080 0.0915 m 0.146 0.146 A 169.000 169.000 Sut 244.363 239.618 Ssu 163.723 160.544 Ssy 85.527 83.866 Sse 52.706 53.239 Ssa 43.513 43.560 21.756 21.780 α 2.785 2.129 β C 6.395 8.864 D 0.512 0.811 Iteration of d for the second trial d3 d4 d1 0.1055 0.1205 d 0.080 0.263 0.263 KB 1.221 128.000 128.000 τa 21.756 231.257 223.311 nf 2.000 154.942 149.618 Na 40.243 80.940 78.159 Nt 42.243 54.261 55.345 Ls 3.379 43.634 43.691 ymax 2.875 21.817 21.845 L0 6.254 1.602 1.228 (L0)cr 2.691 12.292 16.485 τs 64.336 1.297 1.986 ns 1.329 f (Hz) 98.936 d2 d3 d4 0.0915 0.1055 0.1205 1.154 1.108 1.079 21.780 21.817 21.845 2.000 2.000 2.000 17.286 7.475 3.539 19.286 9.475 5.539 1.765 1.000 0.667 2.875 2.875 2.875 4.640 3.875 3.542 4.266 6.821 10.449 64.407 64.517 64.600 1.302 1.255 1.210 104.827 109.340 112.409 The satisfactory spring has design specifications of: A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.811 + 0.092 = 0.903 in, L0 = 4.640 in, and .Nt = 19.3 turns. Ans. ______________________________________________________________________________ 10-32 This is the same as Prob. 10-30 since Ssa = 35 kpsi. Therefore, the specifications are: A313 stainless wire, unpeened, squared and ground, d = 0.0915 in, OD = 0.879 + 0.092 = Shigley’s MED, 10th edition Chapter 10 Solutions, Page 25/41 www.konkur.in 0.971 in, L0 = 3.606 in, and Nt = 15.59 turns Ans. ______________________________________________________________________________ 10-33 For the Gerber-Zimmerli fatigue-failure criterion, Ssu = 0.67Sut , S sa S se = , 1 − (S sm / S su ) 2 S sa r 2S su2 = −1 + 1 + 2S se 2 2S se rS su The equation for Ssa is the basic difference. The last 2 columns of diameters of Ex. 10-5 are presented below with additional calculations. d Sut Ssu Sse Ssy Ssa 0.105 0.112 d 0.105 0.112 278.691 276.096 Na 8.915 6.190 186.723 184.984 Ls 1.146 0.917 38.325 38.394 L0 3.446 3.217 125.411 124.243 (L0)cr 6.630 8.160 34.658 34.652 KB 1.111 1.095 23.105 23.101 23.105 23.101 τa α 1.732 1.523 nf 1.500 1.500 β C 12.004 13.851 τs 70.855 70.844 D 1.260 1.551 ns 1.770 1.754 ID 1.155 1.439 fn 105.433 106.922 OD 1.365 1.663 fom −0.973 −1.022 There are only slight changes in the results. ______________________________________________________________________________ 10-34 As in Prob. 10-34, the basic change is Ssa. S sa For Goodman, S se = 1 − ( S sm / S su ) Recalculate Ssa with rS se S su S sa = rS su + S se Calculations for the last 2 diameters of Ex. 10-5 are given below. Shigley’s MED, 10th edition Chapter 10 Solutions, Page 26/41 www.konkur.in d Sut Ssu Sse Ssy Ssa 0.105 278.691 186.723 49.614 125.411 34.386 22.924 α 1.732 β C 11.899 D 1.249 ID 1.144 OD 1.354 0.112 276.096 184.984 49.810 124.243 34.380 22.920 1.523 13.732 1.538 1.426 1.650 d 0.105 0.112 Na 9.153 6.353 Ls 1.171 0.936 L0 3.471 3.236 (L0)cr 6.572 8.090 KB 1.112 1.096 22.924 22.920 τa nf 1.500 1.500 70.301 70.289 τs ns 1.784 1.768 fn 104.509 106.000 fom −0.986 −1.034 There are only slight differences in the results. ______________________________________________________________________________ 10-35 Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi · inm , m = 0.190, rel cost = 1. 140 Try d = 0.067 in, Sut = = 234.0 kpsi (0.067)0.190 Table 10-6: Ssy = 0.45Sut = 105.3 kpsi Table 10-7: Sy = 0.75Sut = 175.5 kpsi Eq. (10-34) with D/d = C and C1 = C S F σ A = max2 [( K ) A (16C ) + 4] = y πd ny π d 2S y 4C 2 − C − 1 (16C ) + 4 = 4C (C − 1) n y Fmax π d 2S y 4C − C − 1 = (C − 1) − 1 4n F y max 2 π d 2S y 1 1 π d Sy C 2 − 1 + − 1 C + − 2 = 0 4 4n y Fmax 4 4n y Fmax 2 2 2 π d 2S y π d 2S y 1 π d Sy C = ± − + 2 take positive root 16n F 2 16n y Fmax 4 n F y max y max 2 3 1 π (0.067 )(175.5)(10 ) = 2 16(1.5)(18) + Shigley’s MED, 10th edition 2 π (0.067) 2 (175.5)(103 ) π (0.067) 2 (175.5)(103 ) − + 2 = 4.590 16(1.5)(18) 4(1.5)(18) Chapter 10 Solutions, Page 27/41 www.konkur.in D = Cd = 4.59 ( 0.067 ) = 0.3075 in Fi = π d 3τ i 8D = πd3 33 500 C − 3 ± 1000 4 − 8D exp(0.105C ) 6.5 Use the lowest Fi in the preferred range. This results in the best fom. Fi = π (0.067)3 33 500 4.590 − 3 − 1000 4 − = 6.505 lbf 8(0.3075) exp[0.105(4.590)] 6.5 For simplicity, we will round up to the next integer or half integer. Therefore, use Fi = 7 lbf 18 − 7 k = = 22 lbf/in 0.5 d 4G (0.067) 4 (11.5)(106 ) Na = = = 45.28 turns 8kD3 8(22)(0.3075)3 G 11.5 Nb = N a − = 45.28 − = 44.88 turns E 28.6 L0 = (2C − 1 + N b )d = [2(4.590) − 1 + 44.88](0.067) = 3.555 in L18 lbf = 3.555 + 0.5 = 4.055 in Body: K B = τ max (n y ) body r2 (K )B τB (n y ) B fom 4C + 2 4(4.590) + 2 = = 1.326 4C − 3 4(4.590) − 3 8K B Fmax D 8(1.326)(18)(0.3075) −3 = = (10 ) = 62.1 kpsi π d3 π (0.067)3 S 105.3 = sy = = 1.70 62.1 τ max 2r 2(0.134) = 2d = 2(0.067) = 0.134 in, C2 = 2 = =4 d 0.067 4C2 − 1 4(4) − 1 = = = 1.25 4C2 − 4 4(4) − 4 8(18)(0.3075) −3 8F D = ( K ) B max3 = 1.25 (10 ) = 58.58 kpsi 3 πd π (0.067) S 105.3 = sy = = 1.80 58.58 τB π 2d 2 ( Nb + 2) D π 2 (0.067) 2 (44.88 + 2)(0.3075) = −(1) =− = −0.160 4 4 Several diameters, evaluated using a spreadsheet, are shown below. Shigley’s MED, 10th edition Chapter 10 Solutions, Page 28/41 www.konkur.in d 0.067 0.072 0.076 0.081 0.085 0.09 0.095 0.104 Sut 233.977 230.799 228.441 225.692 223.634 221.219 218.958 215.224 Ssy 105.290 103.860 102.798 101.561 100.635 99.548 98.531 96.851 Sy 175.483 173.100 171.331 169.269 167.726 165.914 164.218 161.418 C 4.589 5.412 6.099 6.993 7.738 8.708 9.721 11.650 D 0.307 0.390 0.463 0.566 0.658 0.784 0.923 1.212 Fi (calc) 6.505 5.773 5.257 4.675 4.251 3.764 3.320 2.621 Fi (rd) 7.0 6.0 5.5 5.0 4.5 4.0 3.5 3.0 k 22.000 24.000 25.000 26.000 27.000 28.000 29.000 30.000 Na 45.29 27.20 19.27 13.10 9.77 7.00 5.13 3.15 Nb 44.89 26.80 18.86 12.69 9.36 6.59 4.72 2.75 L0 3.556 2.637 2.285 2.080 2.026 2.071 2.201 2.605 L18 lbf 4.056 3.137 2.785 2.580 2.526 2.571 2.701 3.105 KB 1.326 1.268 1.234 1.200 1.179 1.157 1.139 1.115 62.118 60.686 59.707 58.636 57.875 57.019 56.249 55.031 τmax (ny)body 1.695 1.711 1.722 1.732 1.739 1.746 1.752 1.760 58.576 59.820 60.495 61.067 61.367 61.598 61.712 61.712 τB (ny)B 1.797 1.736 1.699 1.663 1.640 1.616 1.597 1.569 (ny)A 1.500 1.500 1.500 1.500 1.500 1.500 1.500 1.500 fom -0.138 −0.135 −0.133 −0.135 −0.138 −0.154 −0.160 −0.144 Except for the 0.067 in wire, all springs satisfy the requirements of length and number of coils. The 0.085 in wire has the highest fom. ______________________________________________________________________________ 10-36 Given: Nb = 84 coils, Fi = 16 lbf, OQ&T steel, OD = 1.5 in, d = 0.162 in. D = OD − d = 1.5 − 0.162 = 1.338 in (a) Eq. (10-39): L0 = 2(D − d) + (Nb + 1)d = 2(1.338 − 0.162) + (84 + 1)(0.162) = 16.12 in 2d + L0 = 2(0.162) + 16.12 = 16.45 in overall D 1.338 (b) C = = = 8.26 d 0.162 4C + 2 4(8.26) + 2 KB = = = 1.166 4C − 3 4(8.26) − 3 8(16)(1.338) 8F D τ i = K B i 3 = 1.166 = 14 950 psi π (0.162)3 πd (c) From Table 10-5 use: G = 11.4(106) psi and E = 28.5(106) psi Ans. or G 11.4 = 84 + = 84.4 turns E 28.5 d 4G (0.162)4 (11.4)(106 ) k = = = 4.855 lbf/in 8D3 N a 8(1.338)3 (84.4) Ans. Na = Nb + Shigley’s MED, 10th edition Ans. Chapter 10 Solutions, Page 29/41 www.konkur.in A = 147 psi · inm , m = 0.187 147 Sut = = 207.1 kpsi (0.162)0.187 S y = 0.75(207.1) = 155.3 kpsi S sy = 0.50(207.1) = 103.5 kpsi (d) Table 10-4: Body π d 3S sy F = π KBD π (0.162)3 (103.5)(103 ) = 8(1.166)(1.338) = 110.8 lbf Torsional stress on hook point B 2r2 2(0.25 + 0.162 / 2) = = 4.086 d 0.162 4C2 − 1 4(4.086) − 1 (K )B = = = 1.243 4C2 − 4 4(4.086) − 4 π (0.162)3 (103.5)(103 ) F = = 103.9 lbf 8(1.243)(1.338) C2 = Normal stress on hook point A 2r1 1.338 = = 8.26 d 0.162 4C12 − C1 − 1 4(8.26) 2 − 8.26 − 1 (K ) A = = = 1.099 4C1(C1 − 1) 4(8.26)(8.26 − 1) 4 16( K ) A D S yt = σ = F + 3 π d 2 πd 155.3(103 ) F = = 85.8 lbf [16(1.099)(1.338)] / π (0.162)3 + 4 / π (0.162)2 C1 = = min(110.8, 103.9, 85.8) = 85.8 lbf { } Ans. (e) Eq. (10-48): F − Fi 85.8 − 16 = = 14.4 in Ans. k 4.855 ______________________________________________________________________________ y = 10-37 Fmin = 9 lbf, Fmax = 18 lbf 18 − 9 18 + 9 Fa = = 4.5 lbf , Fm = = 13.5 lbf 2 2 Shigley’s MED, 10th edition Chapter 10 Solutions, Page 30/41 www.konkur.in 0.013 ≤ d ≤ 0.1 A = 169 kpsi · inm , 0.1 ≤ d ≤ 0.2 A = 128 kpsi · inm , E = 28 Mpsi, G = 10 Gpsi Try d = 0.081 in and refer to the discussion following Ex. 10-7 169 Sut = = 243.9 kpsi (0.081)0.146 S su = 0.67Sut = 163.4 kpsi S sy = 0.35Sut = 85.4 kpsi S y = 0.55Sut = 134.2 kpsi A313 stainless: m = 0.146 m = 0.263 Table 10-8: Sr = 0.45Sut = 109.8 kpsi Sr / 2 109.8 / 2 Se = = = 57.8 kpsi 2 1 − [Sr / (2Sut )] 1 − [(109.8 / 2) / 243.9]2 r = Fa / Fm = 4.5 / 13.5 = 0.333 r 2 Sut2 Table 6-7: Sa = −1 + 2 Se (0.333) 2 (243.92 ) Sa = −1 + 1 + 2(57.8) 2 2S e 1+ rSut 2 2(57.8) 0.333(243.9) = 42.2 kpsi Hook bending 16C 4 Sa S (σ a ) A = Fa ( K ) A + = = a 2 2 2 πd π d (n f ) A S 4.5 (4C 2 − C − 1)16C + 4 = a 2 4C (C − 1) 2 πd This equation reduces to a quadratic in C (see Prob. 10-35). The useable root for C is 2 2 2 d S d S π π a a C = 0.5 + − 144 144 π (0.081)2 (42.2)(103 ) = 0.5 + 144 = 4.91 Shigley’s MED, 10th edition π d 2Sa 36 + 2 2 π (0.081) 2 (42.2)(103 ) π (0.081)2 (42.2)(103 ) − + 2 144 36 Chapter 10 Solutions, Page 31/41 www.konkur.in D = Cd = 0.398 in C − 3 π d 3τ i π d 3 33 500 Fi = = ± 1000 4 − 8D 8D exp(0.105C ) 6.5 Use the lowest Fi in the preferred range. 33 500 4.91 − 3 π (0.081)3 Fi = − 1000 4 − 8(0.398) exp[0.105(4.91)] 6.5 = 8.55 lbf For simplicity we will round up to next 1/4 integer. Fi = 8.75 lbf 18 − 9 k = = 36 lbf/in 0.25 d 4G (0.081) 4 (10)(106 ) Na = = = 23.7 turns 8kD3 8(36)(0.398)3 G 10 Nb = N a − = 23.7 − = 23.3 turns E 28 L0 = (2C − 1 + N b )d = [2(4.91) − 1 + 23.3](0.081) = 2.602 in Lmax = L0 + ( Fmax − Fi ) / k = 2.602 + (18 − 8.75) / 36 = 2.859 in 4.5(4) 4C 2 − C − 1 (σ a ) A = + 1 2 πd C − 1 18(10-3 ) 4(4.912 ) − 4.91 − 1 + 1 = 21.1 kpsi 2 π (0.081 ) 4.91 − 1 Sa 42.2 (n f ) A = = = 2 checks (σ a ) A 21.1 = Body: 4C + 2 4(4.91) + 2 = = 1.300 4C − 3 4(4.91) − 3 8(1.300)(4.5)(0.398) −3 (10 ) = 11.16 kpsi τa = π (0.081)3 F 13.5 τm = m τa = (11.16) = 33.47 kpsi Fa 4.5 KB = The repeating allowable stress from Table 10-8 is Ssr = 0.30Sut = 0.30(243.9) = 73.17 kpsi The Gerber intercept is given by Eq. (10-42) as S se = Shigley’s MED, 10th edition 73.17 / 2 = 38.5 kpsi 1 − [(73.17 / 2) / 163.4]2 Chapter 10 Solutions, Page 32/41 www.konkur.in From Table 6-7, 2 2 2(33.47)(38.5) 1 163.4 11.16 (n f )body = −1 + 1 + = 2.53 2 33.47 38.5 163.4(11.16) Let r2 = 2d = 2(0.081) = 0.162 2r 4(4) − 1 C2 = 2 = 4, ( K ) B = = 1.25 d 4(4) − 4 (K )B 1.25 (τ a ) B = (11.16) = 10.73 kpsi τa = KB 1.30 (K )B 1.25 τm = (τ m ) B = (33.47) = 32.18 kpsi KB 1.30 Table 10-8: (Ssr )B = 0.28Sut = 0.28(243.9) = 68.3 kpsi 68.3 / 2 (S se ) B = = 35.7 kpsi 1 − [(68.3 / 2) / 163.4]2 2 2 2(32.18)(35.7) 1 163.4 10.73 (n f ) B = −1 + 1 + = 2.51 2 32.18 35.7 163.4(10.73) Yield Bending: 4 Fmax (4C 2 − C − 1) (σ A ) max = + 1 2 πd C −1 4(18) 4(4.91) 2 − 4.91 − 1 + 1 (10-3 ) = 84.4 kpsi 2 π (0.081 ) 4.91 − 1 134.2 (n y ) A = = 1.59 84.4 = Body: τ i = ( Fi / Fa )τ a = (8.75 / 4.5)(11.16) = 21.7 kpsi r = τ a /(τ m − τ i ) = 11.16 / (33.47 − 21.7) = 0.948 r 0.948 (S sy − τ i ) = (85.4 − 21.7) = 31.0 kpsi r +1 0.948 + 1 (S ) 31.0 = sa y = = 2.78 τa 11.16 (S sa ) y = (n y ) body Hook shear: Ssy = 0.3Sut = 0.3(243.9) = 73.2 kpsi τ max = (τ a ) B + (τ m ) B = 10.73 + 32.18 = 42.9 kpsi 73.2 (n y ) B = = 1.71 42.9 7.6π 2d 2 ( N b + 2) D 7.6π 2 (0.081)2 (23.3 + 2)(0.398) fom = − =− = −1.239 4 4 Shigley’s MED, 10th edition Chapter 10 Solutions, Page 33/41 www.konkur.in A tabulation of several wire sizes follow d Sut Ssu Sr Se Sa C D OD Fi (calc) Fi (rd) k Na Nb L0 L18 lbf (σa)A (nf)A KB (τa)body (τm)body Ssr Sse (nf)body (K)B (τa)B (τm)B (Ssr)B (Sse)B (nf)B Sy (σA)max (ny)A τi r (Ssy)body (Ssa)y (ny)body (Ssy)B (τB)max (ny)B fom 0.081 243.920 163.427 109.764 57.809 42.136 4.903 0.397 0.478 8.572 8.75 36.000 23.86 23.50 2.617 2.874 21.068 2.000 1.301 11.141 33.424 73.176 38.519 2.531 1.250 10.705 32.114 68.298 35.708 2.519 134.156 84.273 1.592 21.663 0.945 85.372 30.958 2.779 73.176 42.819 1.709 −1.246 0.085 242.210 162.281 108.994 57.403 41.841 5.484 0.466 0.551 7.874 9.75 36.000 17.90 17.54 2.338 2.567 20.920 2.000 1.264 10.994 32.982 72.663 38.249 2.547 1.250 10.872 32.615 67.819 35.458 2.463 133.215 83.682 1.592 23.820 1.157 84.773 32.688 2.973 72.663 43.486 1.671 −1.234 0.092 239.427 160.416 107.742 56.744 41.360 6.547 0.602 0.694 6.798 10.75 36.000 11.38 11.02 2.127 2.328 20.680 2.000 1.216 10.775 32.326 71.828 37.809 2.569 1.250 11.080 33.240 67.040 35.050 2.388 131.685 82.720 1.592 25.741 1.444 83.800 34.302 3.183 71.828 44.321 1.621 −1.245 0.098 237.229 158.943 106.753 56.223 40.980 7.510 0.736 0.834 5.987 11.75 36.000 8.03 7.68 2.126 2.300 20.490 2.000 1.185 10.617 31.852 71.169 37.462 2.583 1.250 11.200 33.601 66.424 34.728 2.341 130.476 81.961 1.592 27.723 1.942 83.030 36.507 3.438 71.169 44.801 1.589 −1.283 0.105 234.851 157.350 105.683 55.659 40.570 8.693 0.913 1.018 5.141 12.75 36.000 5.55 5.19 2.266 2.412 20.285 2.000 1.157 10.457 31.372 70.455 37.087 2.596 1.250 11.294 33.883 65.758 34.380 2.298 129.168 81.139 1.592 29.629 2.906 82.198 39.109 3.740 70.455 45.177 1.560 −1.357 0.12 230.317 154.312 103.643 54.585 39.786 11.451 1.374 1.494 3.637 13.75 36.000 2.77 2.42 2.918 3.036 19.893 2.000 1.117 10.177 30.532 69.095 36.371 2.616 1.250 11.391 34.173 64.489 33.717 2.235 126.674 79.573 1.592 31.097 4.703 80.611 40.832 4.012 69.095 45.564 1.516 −1.639 optimal fom Shigley’s MED, 10th edition Chapter 10 Solutions, Page 34/41 www.konkur.in The shaded areas show the conditions not satisfied. ______________________________________________________________________________ 10-38 For the hook, M = FR sinθ, ∂M/∂F ∂F = R sinθ δF = 1 EI ∫ π /2 0 F ( R sin θ ) R dθ = 2 π FR 3 2 EI The total deflection of the body and the two hooks π FR 3 8FD3 N b 8FD 3 N b π F ( D / 2)3 + 2 = + d 4G d 4G E (π / 64)(d 4 ) 2 EI 8FD 3 G 8FD 3 N a = 4 Nb + = dG E d 4G G ∴ N a = Nb + Q.E.D. E ______________________________________________________________________________ δ = 10-39 Table 10-5 (d = 4 mm = 0.1575 in): E = 196.5 GPa Table 10-4 for A227: Eq. (10-14): Eq. (10-57): A = 1783 MPa · mmm, m = 0.190 A 1783 Sut = m = 0.190 = 1370 MPa d 4 Sy = σall = 0.78 Sut = 0.78(1370) = 1069 MPa D = OD − d = 32 − 4 = 28 mm C = D/d = 28/4 = 7 Eq. (10-43): 2 4C 2 − C − 1 4 ( 7 ) − 7 − 1 Ki = = = 1.119 4C (C − 1) 4(7)(7 − 1) 32Fr π d3 At yield, Fr = My , σ = Sy. Thus, Eq. (10-44): σ = Ki My = Shigley’s MED, 10th edition π d 3S y π ( 43 )1069 (10−3 ) = = 6.00 N · m 32Ki 32(1.119) Chapter 10 Solutions Solutions, Page 35/41 www.konkur.in Count the turns when M = 0 N = 2.5 − My k= where from Eq. (10-51): k d 4E 10.8 DN Thus, N = 2.5 − My 4 d E / (10.8DN ) Solving for N gives N = = 2.5 1 + [10.8DM y / (d 4 E )] 2.5 { } 1 + [10.8(28)(6.00) ] / 44 (196.5) This means (2.5 − 2.413)(360°) or 31.3° from closed. = 2.413 turns Ans. Treating the hand force as in the middle of the grip, 87.5 r = 112.5 − 87.5 + = 68.75 mm 2 6.00 (103 ) My Fmax = = = 87.3 N Ans. r 68.75 ______________________________________________________________________________ 10-40 The spring material and condition are unknown. Given d = 0.081 in and OD = 0.500, (a) D = 0.500 − 0.081 = 0.419 in Using E = 28.6 Mpsi for an estimate k′ = d 4E (0.081)4 (28.6)(106 ) = = 24.7 lbf · in/turn 10.8DN 10.8(0.419)(11) for each spring. The moment corresponding to a force of 8 lbf Fr = (8/2)(3.3125) = 13.25 lbf · in/spring The fraction windup turn is n = Fr 13.25 = = 0.536 turns k′ 24.7 The arm swings through an arc of slightly less than 180°, say 165°. This uses up 165/360 or 0.458 turns. So n = 0.536 − 0.458 = 0.078 turns are left (or Shigley’s MED, 10th edition Chapter 10 Solutions, Page 36/41 www.konkur.in 0.078(360°) = 28.11° ). The original configuration of the spring was Ans. (b) D 0.419 = = 5.17 d 0.081 4C 2 − C − 1 4(5.17) 2 − 5.17 − 1 Ki = = = 1.168 4C ( C − 1) 4(5.17)(5.17 − 1) C = σ = Ki 32(13.25) 32M = 1.168 = 297 (103 ) psi = 297 kpsi 3 3 (0.081) πd π Ans. To achieve this stress level, the spring had to have set removed. ______________________________________________________________________________ 10-41 (a) Consider half and double results Straight section: M = 3FR, ∂M = 3R ∂F Upper 180° section: M = F [ R + R(1 − cos φ )] ∂M = FR(2 − cos φ ), = R(2 − cos φ ) ∂F Lower section: M = FR sin θ, ∂M = R sin θ ∂F Considering bending only: Shigley’s MED, 10th edition Chapter 10 Solutions Solutions, Page 37/41 www.konkur.in π ∂U 2 l/2 2 = 9 FR dx + FR 2 (2 − cos φ ) 2 R dφ + ∫ ∫ 0 0 ∂F EI π 2F 9 2 π π = R l + R 3 4π − 4sin φ 0 + + R 3 EI 2 2 4 2 2 2FR 19π 9 FR = R + l = (19π R + 18l ) EI 4 2 2 EI δ = ∫ π /2 0 F ( R sin θ ) 2 R dθ The spring rate is k= F δ = 2 EI R (19π R + 18 l ) Ans. 2 (b) Given: A227 HD wire, d = 2 mm, R = 6 mm, and l = 25 mm. Table 10-5 (d = 2 mm = 0.0787 in): k= E = 197.2 GPa 2 (197.2 )109 π ( 0.0024 ) / ( 64 ) 0.0062 19π ( 0.006 ) + 18 ( 0.025 ) = 10.65 (103 ) N/m = 10.65 N/mm Ans. (c) The maximum stress will occur at the bottom of the top hook where the bendingmoment is 3FR and the axial fore is F. Using curved beam theory for bending, σi = Eq. (3-65), p. 133: Axial: σa = Combining, Mci 3FRci = 2 Aeri (π d / 4 ) e ( R − d / 2 ) F F = A πd2 / 4 σ max = σ i + σ a = F= 4F πd2 3Rci + 1 = S y e ( R − d / 2) π d 2Sy 3Rci 4 + 1 e ( R − d / 2) (1) Ans. For the clip in part (b), Eq. (10-14) and Table 10-4: Eq. (10-57): Shigley’s MED, 10th edition Sut = A/dm = 1783/20.190 = 1563 MPa Sy = 0.78 Sut = 0.78(1563) = 1219 MPa Chapter 10 Solutions, Page 38/41 www.konkur.in Table 3-4, p. 135: rn = 12 ( 2 6 − 6 −1 2 2 ) = 5.95804 mm e = rc − rn = 6 − 5.95804 = 0.04196 mm ci = rn − (R − d /2) = 5.95804 − (6 − 2/2) = 0.95804 mm Eq. (1): F= π ( 0.0022 )1219 (106 ) = 46.0 N Ans. 3 ( 6 ) 0.95804 4 + 1 0.04196 ( 6 − 1) ______________________________________________________________________________ 10-42 (a) ∂M = −x ∂F M = − Fx, M = Fl + FR (1 − cos θ ) , { 0≤ x≤l ∂M = l + R (1 − cos θ ) ∂F 0 ≤θ ≤π / 2 π /2 2 1 l − Fx ( − x ) dx + F l + R (1 − cos θ ) Rdθ ∫ ∫ 0 0 EI F = 4l 3 + 3R 2π l 2 + 4 ( π − 2 ) l R + ( 3π − 8 ) R 2 12 EI δF = { } } The spring rate is k= F δF = 12 EI 4l + 3R 2π l + 4 (π − 2 ) l R + ( 3π − 8 ) R 2 3 2 Ans. (b) Given: A313 stainless wire, d = 0.063 in, R = 0.625 in, and l = 0.5 in. Table 10-5: Shigley’s MED, 10th edition E = 28 Mpsi Chapter 10 Solutions, Page 39/41 www.konkur.in I= k= π 64 d4 = π ( 0.063 ) = 7.733 (10 ) in 64 −7 4 4 12 ( 28 )106 ( 7.733)10−7 4 ( 0.53 ) + 3 ( 0.625 ) 2π ( 0.52 ) + 4 (π − 2 ) 0.5 ( 0.625 ) + ( 3π − 8 ) ( 0.6252 ) = 36.3 lbf/in (c) Table 10-4: Ans. A = 169 kpsi⋅inm, m = 0.146 Eq. (10-14): Sut = A/ d m = 169/0.0630.146 = 253.0 kpsi Eq. (10-57): Sy = 0.61 Sut = 0.61(253.0) = 154.4 kpsi One can use curved beam theory as in the solution for Prob. 10-41. However, the equations developed in Sec. 10-12 are equally valid. C = D/d = 2(0.625 + 0.063/2)/0.063 = 20.8 Eq. (10-43): 2 4C 2 − C − 1 4 ( 20.8 ) − 20.8 − 1 Ki = = = 1.037 4C ( C − 1) 4 ( 20.8 )( 20.8 − 1) Eq. (10-44), setting σ = Sy: Ki Solving for F yields 32 Fr = Sy πd3 ⇒ F = 3.25 lbf 1.037 32 F ( 0.5 + 0.625 ) π ( 0.063 3 ) = 154.4 (103 ) Ans. Try solving part (c) of this problem using curved beam theory. You should obtain the same answer. ______________________________________________________________________________ 10-43 (a) M = − Fx σ= M Fx Fx = = 2 I / c I / c bh / 6 Constant stress, bh 2 Fx = ⇒ σ 6 Shigley’s MED, 10th edition h= 6 Fx bσ (1) Ans. Chapter 10 Solutions, Page 40/41 www.konkur.in At x = l, 6 Fl bσ ho = (b) ⇒ h = ho x / l Ans. M = − Fx, ∂ M /∂ F =− x l y=∫ 0 = l l M ( ∂M / ∂F ) − Fx ( − x ) 1 12 Fl 3/ 2 1/2 dx = ∫ 1 3 dx = x dx EI E 0 12 bho ( x / l )3/2 bho3 E ∫0 2 12 Fl 3/2 3/2 8 Fl 3 l = 3 3 bho3 E bho E F bho3 E = 3 Ans. y 8l ______________________________________________________________________________ k= 10-44 Computer programs will vary. ______________________________________________________________________________ 10-45 Computer programs will vary. Shigley’s MED, 10th edition Chapter 10 Solutions, Page 41/41 www.konkur.in Chapter 11 11-1 For the deep-groove 02-series ball bearing with R = 0.90, the design life xD, in multiples of rating life, is L 60L D nD 60 ( 25000 ) 350 xD = D = = = 525 Ans. LR L10 106 The design radial load is FD = 1.2 ( 2.5) = 3.0 kN 1/3 Eq. (11-9): 525 C10 = 3.0 1/1.483 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.9 ) C10 = 24.3 kN Table 11-2: Ans. Choose an 02-35 mm bearing with C10 = 25.5 kN. Ans. 1.483 3 525 ( 3 / 25.5 ) − 0.02 Eq. (11-21): R = exp − Ans. = 0.920 4.459 − 0.02 ______________________________________________________________________________ 11-2 For the angular-contact 02-series ball bearing as described, the rating life multiple is L 60L D nD 60 ( 40 000 ) 520 xD = D = = = 1248 LR L10 106 The design radial load is FD = 1.4 ( 725 ) = 1015 lbf = 4.52 kN Eq. (11-9): 1/3 1248 C10 = 1015 1/1.483 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.9 ) = 10 930 lbf = 48.6 kN Table 11-2: Select an 02-60 mm bearing with C10 = 55.9 kN. Ans. 1.483 3 1248 ( 4.52 / 55.9 ) − 0.02 Eq. (11-21): R = exp − Ans. = 0.945 4.439 ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 11 Solutions, Page 1/28 www.konkur.in 11-3 For the straight-roller 03-series bearing selection, xD = 1248 rating lives from Prob. 11-2 solution. FD = 1.4 ( 2235) = 3129 lbf = 13.92 kN 1248 C10 = 13.92 1 Table 11-3: 3/10 = 118 kN Select an 03-60 mm bearing with C10 = 123 kN. Ans. 1.483 10/3 1248 (13.92 / 123) − 0.02 R = exp − Eq. (11-21): = 0.917 Ans. 4.459 − 0.02 ______________________________________________________________________________ 11-4 The combined reliability of the two bearings selected in Probs. 11-2 and 11-3 is R = ( 0.945)( 0.917 ) = 0.867 Ans. We can choose a reliability goal of 0.90 = 0.95 for each bearing. We make the selections, find the existing reliabilities, multiply them together, and observe that the reliability goal is exceeded due to the roundup of capacity upon table entry. Another possibility is to use the reliability of one bearing, say R1. Then set the reliability goal of the second as R2 = 0.90 R1 or vice versa. This gives three pairs of selections to compare in terms of cost, geometry implications, etc. ______________________________________________________________________________ 11-5 Establish a reliability goal of contact ball bearing, 0.90 = 0.95 for each bearing. For an 02-series angular 1/3 1248 C10 = 1015 1/1.483 0.02 + 4.439 ln (1/ 0.95 ) = 12822 lbf = 57.1 kN Select an 02-65 mm angular-contact bearing with C10 = 63.7 kN. 1.483 3 1248 ( 4.52 / 63.7 ) − 0.02 RA = exp − = 0.962 4.439 Shigley’s MED, 10th edition Chapter 11 Solutions, Page 2/28 www.konkur.in For an 03-series straight roller bearing, 3/10 1248 C10 = 13.92 1/1.483 0.02 + 4.439 ln (1/ 0.95 ) = 136.5 kN Select an 03-65 mm straight-roller bearing with C10 = 138 kN. 1.483 10/3 1248 (13.92 /138 ) − 0.02 RB = exp − = 0.953 4.439 The overall reliability is R = (0.962)(0.953) = 0.917, which exceeds the goal. ______________________________________________________________________________ 11-6 For the straight cylindrical roller bearing specified with a service factor of 1, R = 0.95 and FR = 20 kN. L 60L D nD 60 ( 8000 ) 950 xD = D = = = 456 LR L10 106 3/10 456 C10 = 20 1/1.483 0.02 + 4.439 ln (1/ 0.95 ) = 145 kN Ans. ______________________________________________________________________________ 11-7 Both bearings need to be rated in terms of the same catalog rating system in order to compare them. Using a rating life of one million revolutions, both bearings can be rated in terms of a Basic Load Rating. 1/ a Eq. (11-3): L C A = FA A LR = 8.96 kN 1/ a L n 60 = FA A A LR ( 3000 )( 500 )( 60 ) = 2.0 106 1/3 Bearing B already is rated at one million revolutions, so CB = 7.0 kN. Since CA > CB, bearing A can carry the larger load. Ans. ______________________________________________________________________________ 11-8 FD = 2 kN, LD = 109 rev, R = 0.90 1/ a 1/3 L 109 Eq. (11-3): C10 = FD D = 2 6 = 20 kN Ans. 10 LR ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 11 Solutions, Page 3/28 www.konkur.in 11-9 FD = 800 lbf, ΛD = 12 000 hours, nD = 350 rev/min, R = 0.90 12 000 ( 350 )( 60 ) L n 60 Eq. (11-3): C10 = FD D D = 800 = 5050 lbf Ans 106 LR ______________________________________________________________________________ 1/3 1/ a 11-10 FD = 4 kN, ΛD = 8 000 hours, nD = 500 rev/min, R = 0.90 8 000 ( 500 )( 60 ) L n 60 Eq. (11-3): C10 = FD D D = 4 = 24.9 kN Ans 6 10 LR ______________________________________________________________________________ 1/ a 1/3 11-11 FD = 650 lbf, nD = 400 rev/min, R = 0.95 ΛD = (5 years)(40 h/week)(52 week/year) = 10 400 hours Assume an application factor of one. The multiple of rating life is xD = LD (10 400 )( 400 )( 60 ) = = 249.6 LR 106 1/3 249.6 Eq. (11-9): C10 = (1)( 650 ) 1/1.483 0.02 + 4.439 ln (1/ 0.95 ) = 4800 lbf Ans. ______________________________________________________________________________ 11-12 FD = 9 kN, LD = 108 rev, R = 0.99 Assume an application factor of one. The multiple of rating life is xD = LD 108 = = 100 LR 106 1/3 100 Eq. (11-9): C10 = (1)( 9 ) 1/1.483 0.02 + 4.439 ln (1/ 0.99 ) = 69.2 kN Ans. ______________________________________________________________________________ 11-13 FD = 11 kips, ΛD = 20 000 hours, nD = 200 rev/min, R = 0.99 Assume an application factor of one. Use the Weibull parameters for Manufacturer 2 in Table 11-6. The multiple of rating life is Shigley’s MED, 10th edition Chapter 11 Solutions, Page 4/28 www.konkur.in xD = LD ( 20 000 )( 200 )( 60 ) = = 240 LR 106 1/3 240 C10 = (1)(11) Eq. (11-9): 1/1.483 0.02 + 4.439 ln (1/ 0.99 ) = 113 kips Ans. ______________________________________________________________________________ 11-14 From the solution to Prob. 3-68, the ground reaction force carried by the bearing at C is RC = FD = 178 lbf. Use the Weibull parameters for Manufacturer 2 in Table 11-6. xD = LD 15000 (1200 )( 60 ) = = 1080 LR 106 1/ a Eq. (11-10): xD C10 = a f FD 1/ b x0 + (θ − x0 )(1 − RD ) 1/3 1080 C10 = 1.2 (178 ) 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.95 ) = 2590 lbf Ans. ______________________________________________________________________________ 11-15 From the solution to Prob. 3-69, the ground reaction force carried by the bearing at C is RC = FD = 1.794 kN. Use the Weibull parameters for Manufacturer 2 in Table 11-6. xD = LD 15000 (1200 )( 60 ) = = 1080 LR 106 1/ a Eq. (11-10): xD C10 = a f FD 1/ b x0 + (θ − x0 )(1 − RD ) 1/3 1080 C10 = 1.2 (1.794 ) 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.95 ) = 26.1 kN Ans. ______________________________________________________________________________ 11-16 From the solution to Prob. 3-70, RCz = –327.99 lbf, RCy = –127.27 lbf 1/2 2 2 RC = FD = ( −327.99 ) + ( −127.27 ) = 351.8 lbf Use the Weibull parameters for Manufacturer 2 in Table 11-6. 15000 (1200 )( 60 ) L xD = D = = 1080 LR 106 Shigley’s MED, 10th edition Chapter 11 Solutions, Page 5/28 www.konkur.in 1/ a Eq. (11-10): xD C10 = a f FD 1/ b xo + (θ − xo )(1 − RD ) 1/3 1080 C10 = 1.2 ( 351.8 ) 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.95 ) = 5110 lbf Ans. ______________________________________________________________________________ 11-17 From the solution to Prob. 3-71, RCz = –150.7 N, RCy = –86.10 N 1/ 2 2 2 RC = FD = ( −150.7 ) + ( −86.10 ) = 173.6 N Use the Weibull parameters for Manufacturer 2 in Table 11-6. xD = LD 15000 (1200 )( 60 ) = = 1080 LR 106 1/ a Eq. (11-10): xD C10 = a f FD 1/ b x0 + (θ − x0 )(1 − RD ) 1/3 1080 C10 = 1.2 (173.6 ) 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.95 ) = 2520 N Ans. ______________________________________________________________________________ 11-18 From the solution to Prob. 3-77, RAz = 444 N, RAy = 2384 N ( RA = FD = 4442 + 23842 ) 1/2 = 2425 N = 2.425 kN Use the Weibull parameters for Manufacturer 2 in Table 11-6. The design speed is equal to the speed of shaft AD, d 125 nD = F ni = (191) = 95.5 rev/min dC 250 xD = LD 12 000 ( 95.5 )( 60 ) = = 68.76 LR 106 1/ a Eq. (11-10): xD C10 = a f FD 1/ b x0 + (θ − x0 )(1 − RD ) 1/3 68.76 C10 = (1)( 2.425 ) 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.95 ) = 11.7 kN Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 11 Solutions, Page 6/28 www.konkur.in 11-19 From the solution to Prob. 3-79, RAz = 54.0 lbf, RAy = 140 lbf ( RA = FD = 54.02 + 1402 ) 1/ 2 = 150.1 lbf Use the Weibull parameters for Manufacturer 2 in Table 11-6. The design speed is equal to the speed of shaft AD, d 10 nD = F ni = ( 280 ) = 560 rev/min dC 5 xD = LD 14 000 ( 560 )( 60 ) = = 470.4 LR 106 1/ a Eq. (11-10): xD C10 = a f FD 1/ b x0 + (θ − x0 )(1 − RD ) 470.4 C10 = (1)(150.1) 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.98 ) 3/10 = 1320 lbf Ans. ______________________________________________________________________________ 11-20 (a) Fa = 3 kN, Fr = 7 kN, nD = 500 rev/min, V = 1.2 From Table 11-2, with a 65 mm bore, C0 = 34.0 kN. Fa / C0 = 3 / 34 = 0.088 From Table 11-1, 0.28 ≤ e ≤ 3.0. Fa 3 = = 0.357 VFr (1.2 )( 7 ) Since this is greater than e, interpolating Table 11-1 with Fa / C0 = 0.088, we obtain X2 = 0.56 and Y2 = 1.53. Eq. (11-12): Fe = X iVFr + Yi Fa = ( 0.56 )(1.2 )( 7 ) + (1.53)( 3) = 9.29 kN Fe > Fr so use Fe. Ans. (b) Use Eq. (11-10) to determine the necessary rated load the bearing should have to carry the equivalent radial load for the desired life and reliability. Use the Weibull parameters for Manufacturer 2 in Table 11-6. 10 000 ( 500 )( 60 ) L xD = D = = 300 LR 106 Shigley’s MED, 10th edition Chapter 11 Solutions, Page 7/28 www.konkur.in 1/ a Eq. (11-10): xD C10 = a f FD 1/ b x0 + (θ − x0 )(1 − RD ) 1/3 300 C10 = (1)( 9.29 ) 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.95 ) = 73.4 kN From Table 11-2, the 65 mm bearing is rated for 55.9 kN, which is less than the necessary rating to meet the specifications. This bearing should not be expected to meet the load, life, and reliability goals. Ans. ______________________________________________________________________________ 11-21 (a) Fa = 2 kN, Fr = 5 kN, nD = 400 rev/min, V = 1 From Table 11-2, 30 mm bore, C10 = 19.5 kN, C0 = 10.0 kN Fa / C0 = 2 / 10 = 0.2 From Table 11-1, 0.34 ≤ e ≤ 0.38. Fa 2 = = 0.4 VFr (1)( 5 ) Since this is greater than e, interpolating Table 11-1, with Fa / C0 = 0.2, we obtain X2 = 0.56 and Y2 = 1.27. Eq. (11-12): Fe = X iVFr + Yi Fa = ( 0.56 )(1)( 5) + (1.27 )( 2 ) = 5.34 kN Ans. Fe > Fr so use Fe. (b) Solve Eq. (11-10) for xD. C xD = 10 a f FD a 1/ b x0 + (θ − x0 )(1 − RD ) 3 19.5 0.02 + ( 4.459 − 0.02 )(1 − 0.99 )1/1.483 xD = (1)( 5.34 ) xD = 10.66 xD = LD = Shigley’s MED, 10th edition LD LD nD ( 60 ) = LR 106 ( ) = 10.66 (10 ) = 444 h xD 106 nD ( 60 ) 6 ( 400 )( 60 ) Ans. Chapter 11 Solutions, Page 8/28 www.konkur.in ______________________________________________________________________________ 11-22 Fr = 8 kN, R = 0.9, LD = 109 rev 1/ a Eq. (11-3): L C10 = FD D LR 1/3 109 = 8 6 10 = 80 kN From Table 11-2, select the 85 mm bore. Ans. ______________________________________________________________________________ 11-23 Fr = 8 kN, Fa = 2 kN, V = 1, R = 0.99 Use the Weibull parameters for Manufacturer 2 in Table 11-6. xD = LD 10 000 ( 400 )( 60 ) = = 240 LR 106 First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-12): Fe = 0.56 (1)( 8) + 1.63 ( 2 ) = 7.74 kN Fe < Fr, so just use Fr as the design load. Eq. (11-10): xD C10 = a f FD 1/ b xo + (θ − xo )(1 − RD ) 1/ a 1/3 240 C10 = (1)( 8 ) = 82.5 kN 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.99 ) From Table 11-2, try 85 mm bore with C10 = 83.2 kN, C0 = 53.0 kN Iterate the previous process: Fa / C0 = 2 / 53 = 0.038 0.22 ≤ e ≤ 0.24 Fa 2 = = 0.25 > e VFr 1( 8) Interpolate Table 11-1 with Fa / C0 = 0.038 to obtain X2 = 0.56 and Y2 = 1.89. Table 11-1: Eq. (11-12): Fe = 0.56(1)8 + 1.89(2) = 8.26 > Fr 1/3 Eq. (11-10): Table 11-2: Iterate again: 240 C10 = (1)( 8.26 ) = 85.2 kN 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.99 ) Move up to the 90 mm bore with C10 = 95.6 kN, C0 = 62.0 kN. Shigley’s MED, 10th edition Chapter 11 Solutions, Page 9/28 www.konkur.in Fa / C0 = 2 / 62 = 0.032 Again, 0.22 ≤ e ≤ 0.24 Fa 2 = = 0.25 > e VFr 1( 8 ) Interpolate Table 11-1 with Fa / C0 = 0.032 to obtain X2 = 0.56 and Y2 = 1.95. Table 11-1: Eq. (11-12): Fe = 0.56(1)8 + 1.95(2) = 8.38 > Fr 1/3 240 Eq. (11-10): C10 = (1)( 8.38 ) = 86.4 kN 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.99 ) The 90 mm bore is acceptable. Ans. ______________________________________________________________________________ 11-24 Fr = 8 kN, Fa = 3 kN, V = 1.2, R = 0.9, LD = 108 rev First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-12): Fe = 0.56 (1.2 )( 8) + 1.63 ( 3) = 10.3 kN Fe > Fr 1/ a Eq. (11-3): L C10 = Fe D LR 1/3 108 = 10.3 6 10 = 47.8 kN From Table 11-2, try 60 mm with C10 = 47.5 kN, C0 = 28.0 kN Iterate the previous process: Fa / C0 = 3 / 28 = 0.107 0.28 ≤ e ≤ 0.30 Fa 3 = = 0.313 > e VFr 1.2 ( 8) Interpolate Table 11-1 with Fa / C0 = 0.107 to obtain X2 = 0.56 and Y2 = 1.46 Table 11-1: Eq. (11-12): Fe = 0.56 (1.2 )( 8) + 1.46 ( 3) = 9.76 kN > Fr 1/3 108 C10 = 9.76 6 = 45.3 kN 10 From Table 11-2, we have converged on the 60 mm bearing. Ans. ______________________________________________________________________________ 11-25 Fr = 10 kN, Fa = 5 kN, V = 1, R = 0.95 Use the Weibull parameters for Manufacturer 2 in Table 11-6. Eq. (11-3): Shigley’s MED, 10th edition Chapter 11 Solutions, Page 10/28 www.konkur.in xD = LD 12 000 ( 300 )( 60 ) = = 216 LR 106 First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-12): Fe = 0.56 (1)(10 ) + 1.63 ( 5) = 13.75 kN Fe > Fr, so use Fe as the design load. Eq. (11-10): xD C10 = a f FD 1/ b x0 + (θ − x0 )(1 − RD ) 1/ a 1/3 216 C10 = (1)(13.75 ) 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.95 ) = 97.4 kN From Table 11-2, try 95 mm bore with C10 = 108 kN, C0 = 69.5 kN Iterate the previous process: Fa / C0 = 5 / 69.5 = 0.072 Table 11-1: 0.27 ≤ e ≤ 0.28 Fa 5 = = 0.5 > e VFr 1(10 ) Interpolate Table 11-1 with Fa / C0 = 0.072 to obtain X2 = 0.56 and Y2 = 1.62 B 1.63 Since this is where we started, we will converge back to the same bearing. The 95 mm bore meets the requirements. Ans. ______________________________________________________________________________ 11-26 Fr = 9 kN, Fa = 3 kN, V = 1.2, R = 0.99 Use the Weibull parameters for Manufacturer 2 in Table 11-6. xD = LD 108 = = 100 LR 106 First guess: Choose from middle of Table 11-1, X = 0.56, Y = 1.63 Eq. (11-12): Fe = 0.56 (1.2 )( 9 ) + 1.63 ( 3) = 10.9 kN Fe > Fr, so use Fe as the design load. 1/ a Eq. (11-10): xD C10 = a f FD 1/ b x0 + (θ − x0 )(1 − RD ) Shigley’s MED, 10th edition Chapter 11 Solutions, Page 11/28 www.konkur.in 1/3 100 C10 = (1)(10.9 ) 1/1.483 0.02 + ( 4.459 − 0.02 )(1 − 0.99 ) = 83.9 kN From Table 11-2, try 90 mm bore with C10 = 95.6 kN, C0 = 62.0 kN. Try this bearing. Iterate the previous process: Fa / C0 = 3 / 62 = 0.048 0.24 ≤ e ≤ 0.26 Fa 3 = = 0.278 > e VFr 1.2 ( 9 ) Interpolate Table 11-1 with Fa / C0 = 0.048 to obtain X2 = 0.56 and Y2 = 1.79 Table 11-1: Eq. (11-12): Fe = 0.56 (1.2 )( 9 ) + 1.79 ( 3) = 11.4 kN > Fr 11.4 83.9 = 87.7 kN 10.9 From Table 11-2, this converges back to the same bearing. The 90 mm bore meets the requirements. Ans. ______________________________________________________________________________ C10 = 11-27 (a) nD = 1200 rev/min, LD = 15 kh, R = 0.95, a f = 1.2 From Prob. 3-72, RCy = 183.1 lbf, RCz = –861.5 lbf. 1/2 2 RC = FD = 183.12 + ( −861.5 ) = 881 lbf 15000 (1200 )( 60 ) L xD = D = = 1080 LR 106 1/3 Eq. (11-10): 1080 C10 = 1.2 ( 881) 1/1.483 0.02 + 4.439 (1 − 0.95 ) = 12800 lbf = 12.8 kips Ans. (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ 11-28 (a) nD = 1200 rev/min, LD = 15 kh, R = 0.95, a f = 1.2 From Prob. 3-72, ROy = –208.5 lbf, ROz = 259.3 lbf. 2 RC = FD = 259.32 + ( −208.5 ) Shigley’s MED, 10th edition 1/2 = 333 lbf Chapter 11 Solutions, Page 12/28 www.konkur.in xD = LD 15000 (1200 )( 60 ) = = 1080 LR 106 1/3 1080 Eq. (11-10): C10 = 1.2 ( 333) 1/1.483 0.02 + 4.439 (1 − 0.95 ) = 4837 lbf = 4.84 kips Ans. (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ 11-29 (a) nD = 900 rev/min, LD = 12 kh, R = 0.98, a f = 1.2 From Prob. 3-73, RCy = 8.319 kN, RCz = –10.830 kN. 2 1/2 RC = FD = 8.319 2 + ( −10.830 ) = 13.7 kN 12 000 900 60 ( )( ) = 648 L xD = D = LR 106 1/3 648 Eq. (11-10): C10 = 1.2 (13.7 ) = 204 kN Ans. 1/1.483 0.02 + 4.439 (1 − 0.98 ) (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________________________________________________ 11-30 (a) nD = 900 rev/min, LD = 12 kh, R = 0.98, a f = 1.2 From Prob. 3-73, ROy = 5083 N, ROz = 494 N. ( RC = FD = 50832 + 4942 xD = ) 1/2 = 5106 N = 5.1 kN LD 12 000 ( 900 )( 60 ) = = 648 LR 106 1/3 648 Eq. (11-10): C10 = 1.2 ( 5.1) = 76.1 kN Ans. 1/1.483 0.02 + 4.439 (1 − 0.98 ) (b) Results will vary depending on the specific bearing manufacturer selected. A general engineering components search site such as www.globalspec.com might be useful as a starting point. ______________________________________ ________________________________ ________ 11-31 Assume concentrated forces as shown. Shigley’s MED, 10th edition Chapter 11 Solutions, Page 13/28 www.konkur.in Pz = 8 ( 28) = 224 lbf Py = 8 ( 35) = 280 lbf T = 224 ( 2 ) = 448 lbf ⋅ in ΣT x = −448 + 1.5 F cos 20o = 0 448 F= = 318 lbf 1.5 ( 0.940 ) ΣM Oz = 5.75Py + 11.5RAy − 14.25 F sin 20o = 0 5.75 ( 280 ) + 11.5 RAy − 14.25 ( 318 )( 0.342 ) = 0 RAy = −5.24 lbf ΣM Oy = −5.75 Pz − 11.5 RAz − 14.25 F cos 20o = 0 −5.75 ( 224 ) − 11.5 RAz − 14.25 ( 318 )( 0.940 ) = 0 1/2 2 2 RA = ( −482 ) + ( −5.24 ) z z z o ΣF = RO + Pz + RA + F cos 20 = 0 RAz = −482 lbf; = 482 lbf ROz + 224 − 482 + 318 ( 0.940 ) = 0 ROz = −40.9 lbf ΣF y = ROy + Py + RAy − F sin 20o = 0 ROy + 280 − 5.24 − 318 ( 0.342 ) = 0 ROy = −166 lbf 1/2 2 2 RO = ( −40.9 ) + ( −166 ) = 171 lbf So the reaction at A governs. Reliability Goal: 0.92 = 0.96 FD = 1.2 ( 482 ) = 578 lbf xD = 35 000 ( 350 )( 60 ) / 106 = 735 1/3 735 C10 = 578 1/1.483 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.96 ) = 6431 lbf = 28.6 kN From Table 11-2, a 40 mm bore angular contact bearing is sufficient with a rating of 31.9 kN. Ans. Shigley’s MED, 10th edition Chapter 11 Solutions, Page 14/28 www.konkur.in ______________________________________________________________________________ 11-32 For a combined reliability goal of 0.95, use 0.95 = 0.975 for the individual bearings. xD = 40 000 ( 420 )( 60 ) 106 = 1008 The resultants of the given forces are RO = [(–387)2 + 4672]1/2 = 607 lbf RB = [3162 + (–1615)2]1/2 = 1646 lbf At O: 1/3 Eq. (11-9): 1008 C10 = 1.2 ( 607 ) 1/1.483 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.975 ) = 9978 lbf = 44.4 kN From Table 11-2, select an 02-55 mm angular-contact ball bearing with a basic load rating of 46.2 kN. Ans. At B: 3/10 Eq. (11-9): 1008 C10 = 1.2 (1646 ) 1/1.483 0.02 + ( 4.459 − 0.02 ) ln (1/ 0.975 ) = 20827 lbf = 92.7 kN From Table 11-3, select an 02-75 mm or 03-55 mm cylindrical roller. ______________________________________ ________________________________ ________ Ans. 11-33 The reliability of the individual bearings is R = 0.98 = 0.9899 From statics, Shigley’s MED, 10th edition Chapter 11 Solutions, Page 15/28 www.konkur.in T = (270 − 50) = (P1 − P2)125 = (P1 − 0.15 P1)125 P1 = 310.6 N, P2 = 0.15 (310.6) = 46.6 N P1 + P2 = 357.2 N FAy = 357.2 sin 45o = 252.6 N = FAz ∑M ∑F ∑M ∑F = 850REy + 300(252.6) = 0 ⇒ REy = −89.2 N z O y = 252.6 − 89.2 + ROy = 0 ⇒ ROy = −163.4 N = −850REz − 700(320) + 300(252.6) = 0 ⇒ REz = −174.4 N y O z = −174.4 + 320 − 252.6 + ROz = 0 ⇒ ROz = 107 N RO = ( −163.4 ) 2 RE = ( −89.2 ) + ( −174.4 ) = 196 N 2 + 107 2 = 195 N 2 The radial loads are nearly the same at O and E. We can use the same bearing at both locations. xD = 60 000 (1500 )( 60 ) 106 = 5400 1/3 Eq. (11-9): 5400 C10 = 1( 0.196 ) 1/1.483 0.02 + 4.439 ln (1/ 0.9899 ) = 5.7 kN From Table 11-2, select an 02-12 mm deep-groove ball bearing with a basic load rating of 6.89 kN. Ans. ______________________________________________________________________________ 11-34 R = 0.96 = 0.980 T = 12(240 cos 20o ) = 2706 lbf ⋅ in F= 2706 = 498 lbf 6 cos 25o In xy-plane: ΣM Oz = −16(82.1) − 30(210) + 42 RCy = 0 RCy = 181 lbf Shigley’s MED, 10th edition Chapter 11 Solutions, Page 16/28 www.konkur.in ROy = 82.1 + 210 − 181 = 111.1 lbf In xz-plane: ΣM Oy = 16(226) − 30(451) − 42 RCz = 0 RCz = −236 lbf ROz = 226 − 451 + 236 = 11 lbf ( = (181 RO = 111.12 + 112 RC xD = 2 + 2362 ) ) 1/ 2 = 112 lbf Ans. 1/ 2 = 297 lbf Ans. 50 000 ( 300 )( 60 ) 106 = 900 1/3 ( C10 )O 900 = 1.2 (112 ) 1/1.483 0.02 + 4.439 ln (1/ 0.980 ) = 1860 lbf = 8.28 kN 1/3 ( C10 )C 900 = 1.2 ( 297 ) 1/1.483 0.02 + 4.439 ln (1/ 0.980 ) = 4932 lbf = 21.9 kN Bearing at O: Choose a deep-groove 02-17 mm. Ans. Bearing at C: Choose a deep-groove 02-35 mm. Ans. ______________________________________________________________________________ 11-35 Shafts subjected to thrust can be constrained by bearings, one of which supports the thrust. The shaft floats within the endplay of the second (roller) bearing. Since the thrust force here is larger than any radial load, the bearing absorbing the thrust (bearing A) is heavily loaded compared to bearing B. Bearing B is thus likely to be oversized and may not contribute measurably to the chance of failure. If this is the case, we may be able to obtain the desired combined reliability with bearing A having a reliability near 0.99 and bearing B having a reliability near 1. This would allow for bearing A to have a lower capacity than if it needed to achieve a reliability of 0.99 . To determine if this is the case, we will start with bearing B. Bearing B (straight roller bearing) 30 000 ( 500 )( 60 ) xD = = 900 106 ( Fr = 36 2 + 67 2 ) 1/ 2 = 76.1 lbf = 0.339 kN Try a reliability of 1 to see if it is readily obtainable with the available bearings. Shigley’s MED, 10th edition Chapter 11 Solutions, Page 17/28 www.konkur.in 3/10 Eq. (11-9): 900 C10 = 1.2 ( 0.339 ) 1/1.483 0.02 + 4.439 ln (1/ 1.0 ) = 10.1 kN The smallest capacity bearing from Table 11-3 has a rated capacity of 16.8 kN. Therefore, we select the 02-25 mm straight cylindrical roller bearing. Ans. Bearing at A (angular-contact ball) With a reliability of 1 for bearing B, we can achieve the combined reliability goal of 0.99 if bearing A has a reliability of 0.99. ( Fr = 36 2 + 212 2 ) 1/ 2 = 215 lbf = 0.957 kN Fa = 555 lbf = 2.47 kN Trial #1: Tentatively select an 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN. Fa 2.47 = = 0.0392 C0 63.0 30 000 ( 500 )( 60 ) xD = = 900 106 Table 11-1: Interpolating, X2 = 0.56, Y2 = 1.88 Eq. (11-12): Fe = 0.56 ( 0.957 ) + 1.88 ( 2.47 ) = 5.18 kN 1/3 Eq. (11-9): 900 C10 = 1.2 ( 5.18 ) 1/1.483 0.02 + 4.439 ln (1/ 0.99 ) = 99.54 kN > 90.4 kN Trial #2: Tentatively select a 02-90 mm angular-contact ball with C10 = 106 kN and C0 = 73.5 kN. Fa 2.47 = = 0.0336 C0 73.5 Table 11-1: Interpolating, X2 = 0.56, Y2 = 1.93 Fe = 0.56 ( 0.957 ) + 1.93 ( 2.47 ) = 5.30 kN 1/3 900 C10 = 1.2 ( 5.30 ) 1/1.483 0.02 + 4.439 ln (1/ 0.99 ) Shigley’s MED, 10th edition = 102 kN < 106 kN O.K. Chapter 11 Solutions, Page 18/28 www.konkur.in Select an 02-90 mm angular-contact ball bearing. Ans. ______________________________________________________________________________ 11-36 We have some data. Let’s estimate parameters b and θ from it. In Fig. 11-5, we will use line AB. In this case, B is to the right of A. ( x )1 = For F = 18 kN, 115 ( 2000 )( 60 ) 106 = 13.8 This establishes point 1 on the R = 0.90 line. The R = 0.20 locus is above and parallel to the R = 0.90 locus. For the two-parameter Weibull distribution, x0 = 0 and points A and B are related by [see Eq. (11-8)]: x A = θ ln (1/ 0.90 ) 1/ b (1) xB = θ ln (1/ 0.20 ) 1/ b and xB/xA is in the same ratio as 600/115. Eliminating θ, b= ln ln (1/ 0.20 ) / ln (1/ 0.90 ) ln ( 600 /115 ) = 1.65 Ans. Solving for θ in Eq. (1), θ= Shigley’s MED, 10th edition xA ln (1/ RA ) 1/1.65 = 1 ln (1/ 0.90 ) 1/1.65 = 3.91 Ans. Chapter 11 Solutions, Page 19/28 www.konkur.in Therefore, for the data at hand, x 1.65 R = exp − 3.91 Check R at point B: xB = (600/115) = 5.217 5.217 1.65 R = exp − = 0.20 3.91 Note also, for point 2 on the R = 0.20 line, log ( 5.217 ) − log (1) = log ( xm )2 − log (13.8) ( xm )2 = 72 ______________________________________________________________________________ 11-37 This problem is rich in useful variations. Here is one. Decision: Make straight roller bearings identical on a given shaft. Use a reliability goal of (0.99)1/6 = 0.9983. Shaft a ( = ( 502 FAr = 239 2 + 1112 FBr 2 ) 1/2 = 264 lbf = 1.175 kN ) 1/2 + 10752 = 1186 lbf = 5.28 kN Thus the bearing at B controls. xD = 10 000 (1200 )( 60 ) 106 = 720 0.02 + 4.439 ln (1/ 0.9983) 1/1.483 720 C10 = 1.2 ( 5.28 ) 0.080 26 Select an 02-80 mm with C10 = 106 kN. Shaft b FCr = ( 8742 + 22742 ) 1/2 ( FDr = 3932 + 657 2 ) 1/2 = 0.080 26 0.3 = 97.2 kN Ans. = 2436 lbf = 766 lbf or or 10.84 kN 3.41 kN The bearing at C controls. Shigley’s MED, 10th edition Chapter 11 Solutions, Page 20/28 www.konkur.in xD = 10 000 ( 240 )( 60 ) 106 = 144 144 C10 = 1.2 (10.84 ) 0.080 26 0.3 Select an 02-90 mm with C10 = 142 kN. Shaft c ( = ( 417 FEr = 11132 + 23852 FFr + 8952 2 ) 1/2 ) 1/ 2 = 123 kN Ans. = 2632 lbf = 987 lbf or or 11.71 kN 4.39 kN The bearing at E controls. xD = 10 000 ( 80 )( 60 ) 106 = 48 48 C10 = 1.2 (11.71) 0.080 26 0.3 = 95.7 kN Select an 02-80 mm with C10 = 106 kN. Ans. ______________________________________________________________________________ 11-38 Express Eq. (11-1) as F1a L1 = C10a L10 = K For a ball bearing, a = 3 and for an 02-30 mm angular contact bearing, C10 = 20.3 kN. ( ) ( ) K = ( 20.3) 106 = 8.365 109 3 At a load of 18 kN, life L1 is given by: ( ) 9 K 8.365 10 = = 1.434 106 rev a 3 F1 18 For a load of 30 kN, life L2 is: L1 = L2 = ( ) ( ) = 0.310 10 rev ( ) 30 8.365 109 6 3 In this case, Eq. (6-57) – the Palmgren-Miner cycle-ratio summation rule – can be expressed as Shigley’s MED, 10th edition Chapter 11 Solutions, Page 21/28 www.konkur.in l1 l2 + =1 L1 L2 Substituting, 200 000 l2 + =1 6 1.434 10 0.310 106 l2 ( ) ( ) = 0.267 (10 ) rev Ans. 6 ______________________________________________________________________________ 11-39 Total life in revolutions Let: l = total turns f1 = fraction of turns at F1 f2 = fraction of turns at F2 From the solution of Prob. 11-38, L1 = 1.434(106) rev and L2 = 0.310(106) rev. Palmgren-Miner rule: l1 l2 fl f l + = 1 + 2 =1 L1 L2 L1 L2 from which l= 1 f1 / L1 + f 2 / L2 l= {0.40 / 1.434 (10 )} + {0.60 / 0.310 (10 )} = 451 585 rev 1 6 6 Ans. Total life in loading cycles 4 min at 2000 rev/min = 8000 rev/cycle 6 min at 2000 rev/min = 12 000 rev/cycle Total rev/cycle = 8000 + 12 000 = 20 000 451 585 rev = 22.58 cycles 20 000 rev/cycle Shigley’s MED, 10th edition Ans. Chapter 11 Solutions, Page 22/28 www.konkur.in Total life in hours min 22.58 cycles 10 = 3.76 h Ans. cycle 60 min/h ______________________________________________________________________________ FrA = 560 lbf FrB = 1095 lbf Fae = 200 lbf 11-40 xD = LD 40 000 ( 400 )( 60 ) = = 10.67 LR 90 106 ( ) R = 0.90 = 0.949 Eq. (11-18): Eq. (11-18): 0.47 FrA 0.47 ( 560 ) = = 175.5 lbf KA 1.5 0.47 FrB 0.47 (1095 ) FiB = = = 343.1 lbf KB 1.5 FiA = FiA ≤ ? ≥ ( FiB + Fae ) 175.5 lbf ≤ ( 343.1 + 200 ) = 543.1 lbf, so Eq. (11-19) applies. We will size bearing B first since its induced load will affect bearing A, but is not itself affected by the induced load from bearing A [see Eq. (11-19)]. From Eq. (11-19b), FeB = FrB = 1095 lbf. Eq. (11-10): FRB 10.67 = 1.4 (1095 ) 1/1.5 4.48 (1 − 0.949 ) 3/10 = 3607 lbf Ans. Select cone 32305, cup 32305, with 0.9843 in bore, and rated at 3910 lbf with K = 1.95. Ans. With bearing B selected, we re-evaluate the induced load from bearing B using the actual value for K. 0.47 FrB 0.47 (1095 ) Eq. (11-18): FiB = = = 263.9 lbf KB 1.95 Find the equivalent radial load for bearing A from Eq. (11-19), which still applies. Eq. (11-19a): FeA = 0.4 FrA + K A ( FiB + Fae ) FeA = 0.4 ( 560 ) + 1.5 ( 263.9 + 200 ) = 920 lbf Shigley’s MED, 10th edition Chapter 11 Solutions, Page 23/28 www.konkur.in FeA > FrA Eq. (11-10): 10.67 FRA = 1.4 ( 920 ) 4.48 (1 − 0.949 )1/1.5 3/10 = 3030 lbf Tentatively select cone M86643, cup M86610, with 1 in bore, and rated at 3250 lbf with K = 1.07. Iterating with the new value for K, we get FeA = 702 lbf and FrA = 2312 lbf. Ans. By using a bearing with a lower K, the rated load decreased significantly, providing a higher than requested reliability. Further examination with different combinations of bearing choices could yield additional acceptable solutions. ______________________________________________________________________________ 11-41 The thrust load on shaft CD is from the axial component of the force transmitted through the bevel gear, and is directed toward bearing C. By observation of Fig. 11-14, direct mounted bearings would allow bearing C to carry the thrust load. Ans. From the solution to Prob. 3-74, the axial thrust load is Fae = 362.8 lbf, and the bearing radial forces are FCx = 287.2 lbf, FCz = 500.9 lbf, FDx = 194.4 lbf, and FDz = 307.1 lbf. Thus, the radial forces are FrC = 287.2 2 + 500.9 2 = 577 lbf FrD = 194.42 + 307.12 = 363 lbf The induced loads are 0.47 FrC 0.47 ( 577 ) Eq. (11-18): FiC = = = 181 lbf KC 1.5 0.47 FrD 0.47 ( 363) Eq. (11-18): FiD = = = 114 lbf KD 1.5 Check the condition on whether to apply Eq. (11-19) or Eq. (11-20), where bearings C and D are substituted, respectively, for labels A and B in the equations. FiC ≤ ? ≥ FiD + Fae 181 lbf < 114 + 362.8 = 476.8 lbf, so Eq.(11-19) applies Eq. (11-19a): FeC = 0.4 FrC + KC ( FiD + Fae ) = 0.4 ( 577 ) + 1.5 (114 + 362.8 ) = 946 lbf > FrC , so use FeC Assume for tapered roller bearings that the specifications for Manufacturer 1 in Table 116 are applicable. Shigley’s MED, 10th edition Chapter 11 Solutions, Page 24/28 www.konkur.in xD = LD 108 = = 1.11 LR 90 (106 ) R = 0.90 = 0.949 1.11 Eq. (11-10): FRC = 1( 946 ) 1/1.5 4.48 (1 − 0.949 ) Eq. (11-19b): FeD = FrD = 363 lbf 3/10 = 1130 lbf Ans. 3/10 1.11 = 433 lbf Ans. Eq. (11-10): FRD = 1( 363) 4.48 (1 − 0.949 )1/1.5 ______________________________________________________________________________ 11-42 The thrust load on shaft AB is from the axial component of the force transmitted through the bevel gear, and is directed to the right. By observation of Fig. 11-14, indirect mounted bearings would allow bearing A to carry the thrust load. Ans. From the solution to Prob. 3-76, the axial thrust load is Fae = 92.8 lbf, and the bearing radial forces are FAy = 639.4 lbf, FAz = 1513.7 lbf, FBy = 276.6 lbf, and FBz = 705.7 lbf. Thus, the radial forces are FrA = 639.4 2 + 1513.7 2 = 1643 lbf FrB = 276.62 + 705.7 2 = 758 lbf The induced loads are 0.47 FrA 0.47 (1643) Eq. (11-18): FiA = = = 515 lbf KA 1.5 0.47 FrB 0.47 ( 758 ) Eq. (11-18): FiB = = = 238 lbf KB 1.5 Check the condition on whether to apply Eq. (11-19) or Eq. (11-20). FiA ≤ ? ≥ FiB + Fae 515 lbf > 238 + 92.8 = 330.8 lbf, so Eq.(11-20) applies Notice that the induced load from bearing A is sufficiently large to cause a net axial force to the left, which must be supported by bearing B. Eq. (11-20a): FeB = 0.4 FrB + K B ( FiA − Fae ) = 0.4 ( 758 ) + 1.5 ( 515 − 92.8) = 937 lbf > FrB, so use FeB Assume for tapered roller bearings that the specifications for Manufacturer 1 in Table 11-6 are applicable. Shigley’s MED, 10th edition Chapter 11 Solutions, Page 25/28 www.konkur.in 6 LD 500 (10 ) xD = = = 5.56 LR 90 (106 ) R = 0.90 = 0.949 5.56 Eq. (11-10): FRB = 1( 937 ) 4.48 (1 − 0.949 )1/1.5 Eq. (11-19b): FeA = FrA = 1643 lbf 3/10 = 1810 lbf Ans. 3/10 5.56 = 3180 lbf Ans. Eq. (11-10): FRA = 1(1643) 4.48 (1 − 0.949 )1/1.5 ______________________________________________________________________________ 11-43 The lower bearing is compressed by the axial load, so it is designated as bearing A. FrA = 25 kN FrB = 12 kN Fae = 5 kN 0.47 FrA 0.47 ( 25 ) = = 7.83 kN KA 1.5 0.47 FrB 0.47 (12 ) Eq. (11-18): FiB = = = 3.76 kN KB 1.5 Check the condition on whether to apply Eq. (11-19) or Eq. (11-20) Eq. (11-18): FiA = FiA ≤ ? ≥ FiB + Fae 7.83 kN < 3.76 + 5 = 8.76 kN, so Eq.(11-19) applies Eq. (11-19a): FeA = 0.4 FrA + K A ( FiB + Fae ) = 0.4 ( 25 ) + 1.5 ( 3.76 + 5 ) = 23.1 kN < FrA, so use FrA 60 min 8 hr 5 day 52 weeks LD = ( 250 rev/min ) ( 5 yrs ) yr hr day week ( ) = 156 106 rev Assume for tapered roller bearings that the specifications for Manufacturer 1 in Table 11-6 are applicable. Eq. (11-3): L FRA = a f FD D LR 3/10 ( ) ( ) 156 106 = 1.2 ( 25 ) 90 106 3/10 = 35.4 kN Ans. Eq. (11-19b): FeB = FrB = 12 kN Shigley’s MED, 10th edition Chapter 11 Solutions, Page 26/28 www.konkur.in 3/10 156 FRB = 1.2 (12 ) = 17.0 kN Ans. 90 ______________________________________________________________________________ Eq. (11-3): 11-44 The left bearing is compressed by the axial load, so it is properly designated as bearing A. FrA = 875 lbf FrB = 625 lbf Fae = 250 lbf Assume K = 1.5 for each bearing for the first iteration. Obtain the induced loads. Eq. (11-18): Eq. (11-18): 0.47 FrA 0.47 ( 875 ) = = 274 lbf KA 1.5 0.47 FrB 0.47 ( 625 ) FiB = = = 196 lbf KB 1.5 FiA = Check the condition on whether to apply Eq. (11-19) or Eq. (11-20). FiA ≤ ? ≥ FiB + Fae 274 lbf < 196 + 250 lbf, so Eq.(11-19) applies We will size bearing B first since its induced load will affect bearing A, but it is not affected by the induced load from bearing A [see Eq. (11-19)]. From Eq. (11-19b), FeB = FrB = 625 lbf. Eq. (11-3): L FRB = a f FD D LR 3/10 90 000 (150 )( 60 ) = 1( 625 ) 90 106 3/10 ( ) FRB = 1208 lbf Select cone 07100, cup 07196, with 1 in bore, and rated at 1570 lbf with K = 1.45. Ans. With bearing B selected, we re-evaluate the induced load from bearing B using the actual value for K. 0.47 FrB 0.47 ( 625 ) Eq. (11-18): FiB = = = 203 lbf KB 1.45 Find the equivalent radial load for bearing A from Eq. (11-19), which still applies. Shigley’s MED, 10th edition Chapter 11 Solutions, Page 27/28 www.konkur.in Eq. (11-19a): FeA = 0.4 FrA + K A ( FiB + Fae ) = 0.4 ( 875) + 1.5 ( 203 + 250 ) = 1030 lbf FeA > FrA Eq. (11-3): L FRA = a f FD D LR 3/10 90 000 (150 )( 60 ) = 1(1030 ) 90 106 3/10 ( ) FRA = 1990 lbf Any of the bearings with 1-1/8 in bore are more than adequate. Select cone 15590, cup 15520, rated at 2480 lbf with K = 1.69. Iterating with the new value for K, we get FeA = 1120 lbf and FrA = 2160 lbf. The selected bearing is still adequate. Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 11 Solutions, Page 28/28 www.konkur.in Chapter 12 12-1 Given: dmax = 25 mm, bmin = 25.03 mm, l/d = 1/2, W = 1.2 kN, µ = 55 mPa⋅s, and N = 1100 rev/min. b − d max 25.03 − 25 cmin = min = = 0.015 mm 2 2 r 25/2 = 12.5 mm r/c = 12.5/0.015 = 833.3 N = 1100/60 = 18.33 rev/s P = W/ (ld) = 1200/ [12.5(25)] = 3.84 N/mm2 = 3.84 MPa Eq. (12-7): 2 55 (10−3 )18.33 r µN = 0.182 S = = 833.32 6 c P 3.84 (10 ) Fig. 12-16: h0 /c = 0.3 ⇒ h0 = 0.3(0.015) = 0.0045 mm Fig. 12-18: f r/c = 5.4 ⇒ f = 5.4/833.3 = 0.006 48 Ans. T =f Wr = 0.006 48(1200)12.5(10−3) = 0.0972 N⋅m Hloss = 2π TN = 2π (0.0972)18.33 = 11.2 W Fig. 12-19: Q/(rcNl) = 5.1 ⇒ Ans. Q = 5.1(12.5)0.015(18.33)12.5 = 219 mm3/s Fig. 12-20: Qs /Q = 0.81 ⇒ Qs = 0.81(219) = 177 mm3/s Ans. ______________________________________________________________________________ 12-2 Given: dmax = 32 mm, bmin = 32.05 mm, l = 64 mm, W = 1.75 kN, µ = 55 mPa⋅s, and N = 900 rev/min. b − d max 32.05 − 32 cmin = min = = 0.025 mm 2 2 r ≈ 32/2 = 16 mm r/c = 16/0.025 = 640 N = 900/60 = 15 rev/s P = W/ (ld) = 1750/ [32(64)] = 0.854 MPa Shigley’s MED, 10th edition Chapter 12 Solutions, Page 1/26 www.konkur.in l/d = 64/32 = 2 Eq. (12-7): 2 55 (10−3 )15 r µN 2 = 0.797 S = = 640 c P 0.854 Eq. (12-16), Figs. 12-16, 12-19, and 12-21 l/d y∞ y1 y1/2 y1/4 yl/d h0/c 2 0.98 0.83 0.61 0.36 0.92 P/pmax Q/rcNl 2 2 0.84 3.1 0.54 3.45 0.45 4.2 0.31 5.08 0.65 3.20 h0 = 0.92 c = 0.92(0.025) = 0.023 mm Ans. pmax = P / 0.65 = 0.854/0.65 = 1.31 MPa Ans. Q = 3.20 rcNl = 3.20(16)0.025(15)64 = 1.23 (103) mm3/s Ans. ______________________________________________________________________________ 12-3 Given: dmax = 3.000 in, bmin = 3.005 in, l = 1.5 in, W = 800 lbf, N = 600 rev/min, and SAE 10 and SAE 40 at 150°F. b − d max 3.005 − 3.000 cmin = min = = 0.0025 in 2 2 r ≈ 3.000 / 2 = 1.500 in l / d = 1.5 / 3 = 0.5 r / c = 1.5 / 0.0025 = 600 N = 600 / 60 = 10 rev/s W 800 P= = = 177.78 psi ld 1.5(3) Fig. 12-12: SAE 10 at 150°F, µ ′ = 1.75 µ reyn 1.75(10− 6 )(10) r µN S = = 6002 = 0.0354 c P 177.78 2 Figs. 12-16 and 12-21: h0/c = 0.11 and P/pmax = 0.21 h0 = 0.11(0.0025) = 0.000 275 in Ans. pmax = 177.78 / 0.21 = 847 psi Ans. Fig. 12-12: SAE 40 at 150°F, µ ′ = 4.5 µ reyn Shigley’s MED, 10th edition Chapter 12 Solutions, Page 2/26 www.konkur.in 4.5 S = 0.0354 = 0.0910 1.75 h0 / c = 0.19, P / pmax = 0.275 h0 = 0.19(0.0025) = 0.000 475 in Ans. pmax = 177.78 / 0.275 = 646 psi Ans. ______________________________________________________________________________ 12-4 Given: dmax = 3.250 in, bmin = 3.256 in, l = 3 in, W = 800 lbf, and N = 1000 rev/min. b − d max 3.256 − 3.250 cmin = min = = 0.003 2 2 r ≈ 3.250 / 2 = 1.625 in l / d = 3 / 3.250 = 0.923 r / c = 1.625 / 0.003 = 542 N = 1000 / 60 = 16.67 rev/s W 800 P= = = 82.05 psi ld 3(3.25) Fig. 12-12: SAE 20W at 150°F, µ′ = 2.40 µ reyn Note to instructors: Some students may obtain a higher value of viscosity (2.85) from Fig. 12-14. The value from Fig. 12-12 is used here as the preferred value since this figure is specifically for single-viscosity oils. −6 r µN 2 2.40(10 )(16.67) S = = 542 = 0.143 82.05 c P 2 From Eq. (12-16), and Figs. 12-16 and 12-21: l/d y∞ y1 y1/2 y1/4 yl/d h0/c 0.923 0.82 0.44 0.26 0.14 0.42 P/pmax 0.923 0.83 0.44 0.31 0.21 0.42 h0 = 0.42c = 0.42(0.003) = 0.001 26 in P 82.05 pmax = = = 195 psi Ans. 0.42 0.42 Ans. Fig. 12-14: SAE 20W-40 at 150°F, µ′ = 4.4 µ reyn S = 5422 4.4(10− 6 )(16.67) = 0.263 82.05 From Eq. (12-16), and Figs. 12-16 and 12-21: Shigley’s MED, 10th edition Chapter 12 Solutions, Page 3/26 www.konkur.in l/d y∞ y1 y1/2 y1/4 yl/d h0/c 0.923 0.91 0.6 0.38 0.2 0.58 P/pmax 0.923 0.83 0.48 0.35 0.24 0.46 h0 = 0.58c = 0.58(0.003) = 0.001 74 in Ans. P 82.05 pmax = = = 178 psi Ans. 0.46 0.46 ______________________________________________________________________________ 12-5 Given: dmax = 2.000 in, bmin = 2.0024 in, l = 1 in, W = 600 lbf, N = 800 rev/min, and SAE 20 at 130°F. b − d max 2.0024 − 2 cmin = min = = 0.0012 in 2 2 d 2 r ≈ = = 1 in, l / d = 1 / 2 = 0.50 2 2 r / c = 1 / 0.0012 = 833 N = 800 / 60 = 13.33 rev/s W 600 P = = = 300 psi ld 2(1) Fig. 12-12: SAE 20 at 130°F, µ ′ = 3.75 µ reyn 3.75(10− 6 )(13.3) r µN S = = 8332 = 0.115 300 c P 2 From Figs. 12-16, 12-18 and 12-19: h0 / c = 0.23, r f / c = 3.8, Q / (rcNl ) = 5.3 h0 = 0.23(0.0012) = 0.000 276 in Ans. 3.8 f = = 0.004 56 833 The power loss due to friction is 2π f WrN 2π (0.004 56)(600)(1)(13.33) = 778(12) 778(12) = 0.0245 Btu/s Ans. Q = 5.3rcNl = 5.3(1)(0.0012)(13.33)(1) = 0.0848 in 3 / s Ans. ______________________________________________________________________________ H = Shigley’s MED, 10th edition Chapter 12 Solutions, Page 4/26 www.konkur.in 12-6 Given: dmax = 25 mm, bmin = 25.04 mm, l/d = 1, W = 1.25 kN, µ = 50 mPa⋅s, and N = 1200 rev/min. b − d max 25.04 − 25 cmin = min = = 0.02 mm 2 2 r ≈ d / 2 = 25 / 2 = 12.5 mm, l / d = 1 r / c = 12.5 / 0.02 = 625 N = 1200 / 60 = 20 rev/s W 1250 P = = = 2 MPa ld 252 50(10− 3 )(20) r µN For µ = 50 mPa · s, S = = 6252 = 0.195 6 c P 2(10 ) From Figs. 12-16, 12-18 and 12-20: 2 h0 / c = 0.52, f r / c = 4.5, Qs / Q = 0.57 h0 = 0.52(0.02) = 0.0104 mm Ans. 4.5 f = = 0.0072 625 T = f Wr = 0.0072(1.25)(12.5) = 0.1125 N · m The power loss due to friction is H = 2πT N = 2π (0.1125)(20) = 14.14 W Ans. Qs = 0.57Q The side flow is 57% of Q Ans. ______________________________________________________________________________ 12-7 Given: dmax = 1.25 in, bmin = 1.252 in, l = 2 in, W = 620 lbf, µ′ = 8.5 µ reyn, and N = 1120 rev/min. b − d max 1.252 − 1.25 cmin = min = = 0.001 in 2 2 r = d / 2 = 1.25 / 2 = 0.625 in r / c = 0.625 / 0.001 = 625 N = 1120 / 60 = 18.67 rev/s W 620 P = = = 248 psi ld 1.25(2) 8.5(10− 6 )(18.67) r µN S = = 6252 = 0.250 248 c P l / d = 2 / 1.25 = 1.6 2 From Eq. (12-16), and Figs. 12-16, 12-18, and 12-19 Shigley’s MED, 10th edition Chapter 12 Solutions, Page 5/26 www.konkur.in l/d y∞ y1 y1/2 y1/4 yl/d 1.6 1.6 1.6 0.9 4.5 3 0.58 5.3 3.98 0.36 6.5 4.97 0.185 8 5.6 0.69 4.92 3.59 h0 = 0.69 c = 0.69(0.001) =0.000 69 in Ans. h0/c fr/c Q/rcNl f = 4.92/(r/c) = 4.92/625 = 0.007 87 Ans. Q = 1.6 rcNl = 1.6(0.625) 0.001(18.57) 2 = 0.0833 in3/s Ans. ______________________________________________________________________________ 12-8 Given: dmax = 75.00 mm, bmin = 75.10 mm, l = 36 mm, W = 2 kN, N = 720 rev/min, and SAE 20 and SAE 40 at 60°C. bmin − d max 75.10 − 75 = = 0.05 mm 2 2 l / d = 36 / 75 = 0.48 0.5 (close enough) r = d / 2 = 75 / 2 = 37.5 mm r / c = 37.5 / 0.05 = 750 N = 720 / 60 = 12 rev/s W 2000 P = = = 0.741 MPa ld 75(36) cmin = Fig. 12-13: SAE 20 at 60°C, µ = 18.5 mPa · s 18.5(10− 3 )(12) r µN S = = 7502 = 0.169 6 c P 0.741(10 ) 2 From Figures 12-16, 12-18 and 12-21: h0 / c = 0.29, f r / c = 5.1, P / pmax = 0.315 h0 = 0.29(0.05) = 0.0145 mm Ans. f = 5.1 / 750 = 0.0068 T = f Wr = 0.0068(2)(37.5) = 0.51 N · m The heat loss rate equals the rate of work on the film Hloss = 2πT N = 2π(0.51)(12) = 38.5 W pmax = 0.741/0.315 = 2.35 MPa Ans. Ans. Fig. 12-13: SAE 40 at 60°C, µ = 37 mPa · s Shigley’s MED, 10th edition Chapter 12 Solutions, Page 6/26 www.konkur.in S = 0.169(37)/18.5 = 0.338 From Figures 12-16, 12-18 and 12-21: h0 / c = 0.42, f r / c = 8.5, P / pmax = 0.38 h0 = 0.42(0.05) = 0.021 mm Ans. f = 8.5 / 750 = 0.0113 T = f Wr = 0.0113(2)(37.5) = 0.85 N · m H loss = 2π TN = 2π (0.85)(12) = 64 W Ans. pmax = 0.741 / 0.38 = 1.95 MPa Ans. _____________________________________________________________________________ 12-9 Given: dmax = 56.00 mm, bmin = 56.05 mm, l = 28 mm, W = 2.4 kN, N = 900 rev/min, and SAE 40 at 65°C. b − d max 56.05 − 56 cmin = min = = 0.025 mm 2 2 r = d / 2 = 56 / 2 = 28 mm r / c = 28 / 0.025 = 1120 l / d = 28 / 56 = 0.5, N = 900 / 60 = 15 rev/s 2400 P = = 1.53 MPa 28(56) Fig. 12-13: SAE 40 at 65°C, µ = 30 mPa · s 2 30(10− 3 )(15) r µN S = = 11202 = 0.369 6 c P 1.53(10 ) From Figures 12-16, 12-18, 12-19 and 12-20: h0 / c = 0.44, f r / c = 8.5, Qs / Q = 0.71, Q / (rcNl ) = 4.85 h0 = 0.44(0.025) = 0.011 mm Ans. f = 8.5 / 1000 = 0.007 59 T = f Wr = 0.007 59(2.4)(28) = 0.51 N · m H = 2π TN = 2π (0.51)(15) = 48.1 W Ans. Q = 4.85 rcNl = 4.85(28)(0.025)(15)(28) = 1426 mm3 /s Qs = 0.71(1426) = 1012 mm3 /s Ans. _____________________________________________________________________________ 12-10 Consider the bearings as specified by minimum f : d −+t0d , b−+0tb maximum W: d −′+td0 , b−+0tb Shigley’s MED, 10th edition Chapter 12 Solutions, Page 7/26 www.konkur.in and differing only in d and d′ . Preliminaries: l /d =1 P = W / (ld ) = 700 / (1.252 ) = 448 psi N = 3600 / 60 = 60 rev/s Fig. 12-16: minimum f : maximum W: S ≈ 0.08 S ≈ 0.20 Fig. 12-12: µ = 1.38(10−6) reyn µN/P = 1.38(10−6)(60/448) = 0.185(10−6) Eq. (12-7): r = c S µN / P For minimum f : r 0.08 = = 658 c 0.185(10− 6 ) c = 0.625 / 658 = 0.000 950 If this is cmin, 0.001 in b − d = 2(0.001) = 0.002 in The median clearance is c = cmin + t d + tb t + tb = 0.001 + d 2 2 and the clearance range for this bearing is t + tb ∆c = d 2 which is a function only of the tolerances. For maximum W: r 0.2 = = 1040 c 0.185(10− 6 ) c = 0.625 / 1040 = 0.000 600 0.0005 in If this is cmin Shigley’s MED, 10th edition Chapter 12 Solutions, Page 8/26 www.konkur.in b − d ′ = 2cmin = 2(0.0005) = 0.001 in t + tb t + tb c = cmin + d = 0.0005 + d 2 2 t d + tb ∆c = 2 The difference (mean) in clearance between the two clearance ranges, crange, is td + tb t + tb − 0.0005 + d 2 2 = 0.0005 in crange = 0.001 + For the minimum f bearing b − d = 0.002 in or d = b − 0.002 in For the maximum W bearing d′ = b − 0.001 in For the same b, tb and td, we need to change the journal diameter by 0.001 in. d ′ − d = b − 0.001 − (b − 0.002) = 0.001 in Increasing d of the minimum friction bearing by 0.001 in, defines d′ of the maximum load bearing. Thus, the clearance range provides for bearing dimensions which are attainable in manufacturing. Ans. _____________________________________________________________________________ 12-11 Given: SAE 40, N = 10 rev/s, Ts = 140°F, l/d = 1, d = 3.000 in, b = 3.003 in, W = 675 lbf. b − d max 3.003 − 3 cmin = min = = 0.0015 in 2 2 r = d / 2 = 3 / 2 = 1.5 in r / c = 1.5 / 0.0015 = 1000 W 675 P = = = 75 psi ld 3(3) Trial #1: From Figure 12-12 for T = 160°F, µ = 3.5 µ reyn, ∆T = 2(160 − 140) = 40°F 3.5(10− 6 )(10) r µN S = = 10002 = 0.4667 75 c P 2 Shigley’s MED, 10th edition Chapter 12 Solutions, Page 9/26 www.konkur.in From Fig. 12-24, 9.70∆T = 0.349 109 + 6.009 40(0.4667) + 0.047 467(0.4667) 2 = 3.16 P P 75 ∆T = 3.16 = 3.16 = 24.4°F 9.70 9.70 Discrepancy = 40 − 24.4 = 15.6°F Trial #2: T = 150°F, µ = 4.5 µ reyn, ∆T = 2(150 − 140) = 20° F 4.5 (10− 6 )10 2 S = 1000 = 0.6 75 From Fig. 12-24, 9.70∆T = 0.349 109 + 6.009 40(0.6) + 0.047 467(0.6) 2 = 3.97 P P 75 ∆T = 3.97 = 3.97 = 30.7°F 9.70 9.70 Discrepancy = 20 − 30.7 = − 10.7°F Trial #3: T = 154°F, µ = 4 µ reyn, ∆T = 2(154 − 140) = 28°F 4 (10− 6 )10 S = 10002 = 0.533 75 From Fig. 12-24, 9.70∆T = 0.349 109 + 6.009 40(0.533) + 0.047 467(0.533) 2 = 3.57 P P 75 ∆T = 3.57 = 3.57 = 27.6°F 9.70 9.70 Discrepancy = 28 − 27.6 = 0.4°F O.K. Tav = 140 +28/2 = 154°F Ans. T1 = Tav − ∆T / 2 = 154 − (28 / 2) = 140°F T2 = Tav + ∆T / 2 = 154 + (28 / 2) = 168°F S = 0.4 From Figures 12-16, 12-18, to 12-20: Shigley’s MED, 10th edition Chapter 12 Solutions, Page 10/26 www.konkur.in h0 fr Q Qs = 0.75, = 11, = 3.6, = 0.33 c c rcN l Q h0 = 0.75(0.0015) = 0.001 13 in Ans. 11 f = = 0.011 1000 T = f Wr = 0.0075(3)(40) = 0.9 N · m 2π ( 0.011) 675 (1.5 )10 2π f WrN H loss = = = 0.075 Btu/s 778 (12 ) 778 (12 ) Ans. Q = 3.6rcN l = 3.6(1.5)0.0015(10)3 = 0.243 in 3 /s Ans. 3 Qs = 0.33(0.243) = 0.0802 in /s Ans. _____________________________________________________________________________ 12-12 Given: d = 2.5 in, b = 2.504 in, cmin = 0.002 in, W = 1200 lbf, SAE = 20, Ts = 110°F, N = 1120 rev/min, and l = 2.5 in. P = W/(ld) = 1200/(2.5)2 = 192 psi, N = 1120/60 = 18.67 rev/s For a trial film temperature, let Tf = 150°F Table 12-1: Eq. (12-7): µ′ = 0.0136 exp[1271.6/(150 + 95)] = 2.441 µ reyn −6 2 2 r µ N 2.5 / 2 2.441(10 )18.67 S = = = 0.0927 192 c P 0.002 Fig. 12-24: 192 0.349 109 + 6.009 40 ( 0.0927 ) + 0.047 467 ( 0.0927 2 ) 9.70 = 17.9°F ∆T = ∆T 17.9 = 110 + = 119.0°F 2 2 T f − Tav = 150 − 119.0 = 31.0°F Tav = Ts + which is not 0.1 or less, therefore try averaging for the new trial film temperature, let (T f ) new = 150 + 119.0 = 134.5°F 2 Proceed with additional trials using a spreadsheet (table also shows the first trial) Shigley’s MED, 10th edition Chapter 12 Solutions, Page 11/26 www.konkur.in Trial Tf µ' S ∆T Tav Tf −Tav New Tf 150.0 134.5 127.9 125.3 124.3 124.0 123.8 2.441 3.466 4.084 4.369 4.485 4.521 4.545 0.0927 0.1317 0.1551 0.1659 0.1704 0.1717 0.1726 17.9 22.6 25.4 26.7 27.2 27.4 27.5 119.0 121.3 122.7 123.3 123.6 123.7 123.7 31.0 13.2 5.2 2.0 0.7 0.3 0.1 134.5 127.9 125.3 124.3 124.0 123.8 123.8 Note that the convergence begins rapidly. There are ways to speed this, but at this point they would only add complexity. (a) µ ′ = 4.545(10−6 ), S = 0.1726 From Fig. 12-16: h0 = 0.482, h0 = 0.482(0.002) = 0.000 964 in c From Fig. 12-17: φ = 56° Ans. (b) e = c − h0 = 0.002 − 0.000 964 = 0.001 04 in Ans. f r (c) From Fig. 12-18: = 4.10, f = 4.10(0.002 /1.25) = 0.006 56 c Ans. (d) T = f Wr = 0.006 56(1200)(1.25) = 9.84 lbf · in H = (e) From Fig. 12-19: 2π T N 2π (9.84)(1120 / 60) = = 0.124 Btu/s 778(12) 778(12) Ans. Q = 4.16 rcNl 1120 3 Q = 4.16(1.25)(0.002) (2.5) = 0.485 in /s 60 Q From Fig. 12-20: s = 0.6, Qs = 0.6(0.485) = 0.291 in 3/s Q Ans. Ans. W / ( ld ) 1200 / 2.52 P = 0.45, pmax = = = 427 psi pmax 0.45 0.45 From Fig. 12-22: φ pmax = 16° Ans. (f) From Fig. 12-21: Shigley’s MED, 10th edition Ans. Chapter 12 Solutions, Page 12/26 www.konkur.in (g) From Fig. 12-22: φ p0 = 82° Ans. (h) From the trial table, Tf = 123.8°F Ans. (i) With ∆T = 27.5°F from the trial table, Ts + ∆T = 110 + 27.5 = 137.5°F Ans. _____________________________________________________________________________ 12-13 Given: d = 1.250 in, td = 0.001 in, b = 1.252 in, tb = 0.003 in, l = 1.25 in, W = 250 lbf, N = 1750 rev/min, SAE 10 lubricant, sump temperature Ts = 120°F. P = W/(ld) = 250/1.252 = 160 psi, N = 1750/60 = 29.17 rev/s For the clearance, c = 0.002 ± 0.001 in. Thus, cmin = 0.001 in, cmedian = 0.002 in, and cmax = 0.003 in. For cmin = 0.001 in, start with a trial film temperature of Tf = 135°F Table 12-1: Eq. (12-7): µ′ = 0.0158 exp[1157.5/(135 + 95)] = 2.423 µ reyn −6 2 2 r µ N 1.25 / 2 2.423 (10 ) 29.17 S = = = 0.1725 160 c P 0.001 Fig. 12-24: 160 0.349 109 + 6.009 40 ( 0.1725 ) + 0.047 467 ( 0.17252 ) 9.70 = 22.9°F ∆T = ∆T 22.9 = 120 + = 131.4°F 2 2 T f − Tav = 135 − 131.4 = 3.6°F Tav = Ts + which is not 0.1 or less, therefore try averaging for the new trial film temperature, let 135 + 131.4 = 133.2°F 2 Proceed with additional trials using a spreadsheet (table also shows the first trial) (T f ) new = Trial Tf µ' S ∆T Tav Tf−Tav New Tf 135.0 133.2 132.5 132.2 132.1 2.423 2.521 2.560 2.578 2.583 0.1725 0.1795 0.1823 0.1836 0.1840 22.9 23.6 23.9 24.0 24.0 131.4 131.8 131.9 132.0 132.0 3.6 1.4 0.6 0.2 0.1 133.2 132.5 132.2 132.1 132.1 Shigley’s MED, 10th edition Chapter 12 Solutions, Page 13/26 www.konkur.in With Tf = 132.1°F, ∆T = 24.0°F, µ′ = 2.583 µ reyn, S = 0.1840, Tmax = Ts + ∆T = 120 + 24.0 = 144.0°F Fig. 12-16: h0/c = 0.50, h0 = 0.50(0.001) = 0.000 50 in ∫ = 1 − h0/c = 1 − 0.50 = 0.05 in Fig. 12-18: r f /c = 4.25, f = 4.25/(0.625/0.001) = 0.006 8 Fig. 12-19: Q/(rcNl) = 4.13, Q = 4.13(0.625)0.001(29.17)1.25 = 0.0941 in3/s Fig. 12-20: Qs/Q = 0.58, Qs = 0.58(0.0941) = 0.0546 in3/s The above can be repeated for cmedian = 0.002 in, and cmax = 0.003 in. The results are shown below. cmin 0.001 in cmedian 0.002 in cmax 0.003 in 132.1 2.583 0.184 24.0 125.6 3.002 0.0534 11.1 124.1 3.112 0.0246 8.2 h0/c 144.0 0.5 131.1 0.23 128.2 0.125 h0 0.00050 0.00069 0.00038 ∫ 0.50 4.25 0.0068 4.13 0.0941 0.77 1.8 0.0058 4.55 0.207 0.88 1.22 0.0059 4.7 0.321 0.58 0.82 0.90 0.0546 0.170 0.289 Tf µ′ S ∆Τ Tmax fr/c f Q/(rcNl) Q Qs /Q Qs _____________________________________________________________________________ 12-14 Computer programs will vary. _____________________________________________________________________________ 12-15 In a step-by-step fashion, we are building a skill for natural circulation bearings. • Given the average film temperature, establish the bearing properties. • Given a sump temperature, find the average film temperature, then, establish the Shigley’s MED, 10th edition Chapter 12 Solutions, Page 14/26 www.konkur.in bearing properties. • Now we acknowledge the environmental temperature’s role in establishing the sump temperature. Sec. 12-9 and Ex. 12-5 address this problem. Given: dmax = 2.500 in, bmin = 2.504 in, l/d = 1, N = 1120 rev/min, SAE 20 lubricant, W = 300 lbf, A = 60 in2, T∞ = 70°F, and α = 1. 600 lbf load with minimal clearance: We will start by using W = 600 lbf (nd = 2). The task is to iteratively find the average film temperature, Tf , which makes Hgen and Hloss equal. b − d max 2.504 − 2.500 cmin = min = = 0.002 in 2 2 N = 1120/60 = 18.67 rev/s P = W 600 = = 96 psi ld 2.52 −6 2 2 r µ N 1.25 µ ′ (10 )18.67 S = = = 0.0760µ ′ 96 c P 0.002 µ′ = 0.0136 exp[1271.6/(Tf + 95)] Table 12-1: 2545 fr f r 2545 WNc ( 600 )18.67 ( 0.002 ) = 1050 c c 1050 fr = 54.3 c H gen = H loss = 2.7 ( 60 / 144 ) h CR A T f − T∞ ) = ( (T f − 70) 1+1 α +1 = 0.5625 (T f − 70 ) Start with trial values of Tf of 220 and 240°F. Trial Tf 220 240 µ′ 0.770 0.605 S 0.059 0.046 f r/c 1.9 1.7 Hgen 103.2 92.3 Hloss 84.4 95.6 As a linear approximation, let Hgen = mTf + b. Substituting the two sets of values of Tf and Hgen we find that Hgen = − 0.545 Tf +223.1. Setting this equal to Hloss and solving for Tf gives Tf = 237°F. Trial Tf 237 Shigley’s MED, 10th edition µ′ 0.627 S 0.048 f r/c 1.73 Hgen 93.9 Hloss 94.0 Chapter 12 Solutions, Page 15/26 www.konkur.in which is satisfactory. Table 12-16: h0/c = 0.21, h0 = 0.21 (0.002) = 000 42 in Fig. 12-24: ∆T = 96 0.349 109 + 6.009 4 ( 0.048 ) + 0.047 467 ( 0.0482 ) 9.7 = 6.31° F T1 = Ts = Tf − ∆T = 237 − 6.31/2 = 233.8°F Tmax = T1 + ∆T = 233.8 + 6.31 = 240.1°F Trumpler’s design criteria: 0.002 + 0.000 04d = 0.002 + 0.000 04(2.5) = 0.000 30 in < h0 O.K. Tmax = 240.1°F < 250°F O.K. Wst 300 = = 48 psi < 300 psi ld 2.52 nd = 2 (assessed at W = 600 lbf) O.K . O.K. We see that the design passes Trumpler’s criteria and is deemed acceptable. For an operating load of W = 300 lbf, it can be shown that Tf = 219.3°F, µ′ = 0.78, S = 0.118, f r/c = 3.09, Hgen = Hloss = 84 Btu/h, h0 = , ∆T = 10.5°F, T1 = 224.6°F, and Tmax = 235.1°F. _____________________________________________________________________________ +0.005 12-16 Given: d = 3.500 +−0.000 0.001 in, b = 3.505 −0.000 in , SAE 30, Ts = 120°F, ps = 50 psi, N = 2000/60 = 33.33 rev/s, W = 4600 lbf, bearing length = 2 in, groove width = 0.250 in, and Hloss ≤ 5000 Btu/hr. b − d max 3.505 − 3.500 cmin = min = = 0.0025 in 2 2 r = d/ 2 = 3.500/2 = 1.750 in r / c = 1.750/0.0025 = 700 l′ = (2 − 0.25)/2 = 0.875 in l′ / d = 0.875/3.500 = 0.25 W 4600 P = = = 751 psi 4rl′ 4 (1.750 ) 0.875 Shigley’s MED, 10th edition Chapter 12 Solutions, Page 16/26 www.konkur.in Trial #1: Choose (Tf )1 = 150°F. From Table 12-1, µ′ = 0.0141 exp[1360.0/(150 + 95)] = 3.63 µ reyn 3.63(10−6 )(33.33) r µN S = = 7002 = 0.0789 751 c P From Figs. 12-16 and 12-18: ∫ = 0.9, f r/ c = 3.6 2 From Eq. (12-24), ∆T = = 0.0123( f r / c)SW 2 (1 + 1.5 ò2 ) ps r 4 0.0123 ( 3.6 ) 0.0789 ( 46002 ) 1 + 1.5(0.9) 2 50 (1.7504 ) = 71.2°F Tav = Ts + ∆T / 2 = 120 + 71.2/2 = 155.6°F Trial #2: Choose (Tf )2 = 160°F. From Table 12-1 µ′ = 0.0141 exp[1360.0/(160 + 95)] = 2.92 µ reyn 2.92 S = 0.0789 = 0.0635 3.63 From Figs. 12-16 and 12-18: ∫ = 0.915, f r/ c =3 0.0123 ( 3) 0.0635 ( 46002 ) ∆T = = 46.9°F 1 + 1.5 ( 0.9152 ) 50 (1.7504 ) Tav = 120 + 46.9/2 = 143.5°F Trial #3: Thus, the plot gives (Tf )3 = 152.5°F. From Table 12-1 µ′ = 0.0141 exp[1360.0/(152.5 + 95)] = 3.43 µ reyn Shigley’s MED, 10th edition Chapter 12 Solutions, Page 17/26 www.konkur.in 3.43 S = 0.0789 = 0.0746 3.63 From Figs. 12-16 and 12-18: ∫ = 0.905, f r/ c =3.4 ∆T = Tav 0.0123 ( 3.4 ) 0.0746 ( 46002 ) 1 + 1.5 ( 0.9052 ) 50 (1.7504 ) = 120 + 63.2/2 = 151.6°F Result is close. Choose T f = = 63.2°F 152.5 + 151.6 = 152.1°F 2 Try 152°F µ′ = 0.0141 exp[1360.0/(152 + 95)] = 3.47 µ reyn Table 12-1: 3.47 S = 0.0789 = 0.0754 3.63 f r h0 = 3.4, ò = 0.902, = 0.098 c c 0.0123 ( 3.4 ) 0.0754 ( 46002 ) ∆T = = 64.1°F 1 + 1.5 ( 0.9022 ) 50 (1.7504 ) Tav = 120 + 64.1 / 2 = 152.1°F O.K. h0 = 0.098(0.0025) = 0.000 245 in Tmax = Ts + ∆T = 120 + 64.1 = 184.1°F Eq. (12-22): π ( 50 )1.750 ( 0.00253 ) π ps rc3 2 Qs = (1 + 1.5ò ) = 3 ( 3.47 )10− 6 ( 0.875) 1 + 1.5 ( 0.9022 ) 3µ l′ = 1.047 in 3 /s Hloss = ρ CpQs ∆T = 0.0311(0.42)1.047(64.1) = 0.877 Btu/s = 0.877(602) = 3160 Btu/h O.K. Trumpler’s design criteria: 0.0002 + 0.000 04(3.5) = 0.000 34 in > 0.000 245 Not O.K. Tmax = 184.1°F < 250°F O.K. Pst = 751 psi > 300 psi Not O.K. n = 1, as done Not O.K. _____________________________________________________________________________ +0.010 12-17 Given: d = 50.00 +−0.00 0.05 mm, b = 50.084 −0.000 mm , SAE 30, Ts = 55°C, ps = 200 kPa, N = 2880/60 = 48 rev/s, W = 10 kN, bearing length = 55 mm, groove width = 5 mm, and Shigley’s MED, 10th edition Chapter 12 Solutions, Page 18/26 www.konkur.in Hloss ≤ 300 W. bmin − d max 50.084 − 50 = = 0.042 mm 2 2 r = d/ 2 = 50/2 = 25 mm cmin = r / c = 25/0.042 = 595 l′ = (55 − 5)/2 = 25 mm l′ / d = 25/50 = 0.5 10 (103 ) W P = = = 4 MPa 4rl′ 4 ( 25 ) 25 Trial #1: Choose (Tf )1 = 79°C. From Fig. 12-13, µ = 13 mPa · s. 13(10−3 )(48) r µN S = = 5952 = 0.0552 6 c P 4(10 ) From Figs. 12-16 and 12-18: ∫ = 0.85, f r/ c = 2.3 From Eq. (12-25), 978(106 ) ( f r / c)SW 2 ∆T = 1 + 1.5 ò2 ps r 4 978(106 ) 2.3(0.0552)(10 2 ) = = 76.3°C 1 + 1.5(0.85) 2 200(25) 4 2 Tav = Ts + ∆T / 2 = 55 + 76.3/2 = 93.2°C Trial #2: Choose (Tf )2 = 100°C. From Fig. 12-13, µ = 7 mPa · s. 7 S = 0.0552 = 0.0297 13 From Figs. 12-16 and 12-18: ∫ = 0.90, f r/ c =1.6 6 978(10 ) 1.6(0.0297)(102 ) ∆T = = 26.9°C 1 + 1.5(0.9)2 200(25) 4 Tav = 55 + 26.9/2 = 68.5°C Shigley’s MED, 10th edition Chapter 12 Solutions, Page 19/26 www.konkur.in Trial #3: Thus, the plot gives (Tf )3 = 85.5°C. From Fig. 12-13, µ = 10.5 mPa · s. 10.5 S = 0.0552 = 0.0446 13 From Figs. 12-16 and 12-18: ∫ = 0.87, f r/ c =2.2 ∆T = 978(106 ) 2.2(0.0457)(102 ) = 58.9°C 1 + 1.5(0.87 2 ) 200(25) 4 Tav = 55 + 58.9/2 = 84.5°C Result is close. Choose T f = Fig. 12-13: 85.5 + 84.5 = 85°C 2 µ = 10.5 mPa · s 10.5 S = 0.0552 = 0.0446 13 f r h0 ò = 0.87, = 2.2, = 0.13 c c 978(106 ) 2.2(0.0457)(102 ) ∆T = = 58.9°C 1 + 1.5(0.87 2 ) 200(254 ) Tav = 55 + 58.9 / 2 = 84.5°C O.K. or 138°F From Eq. (12-22) h0 = 0.13(0.042) = 0.005 46 mm or 0.000 215 in Tmax = Ts + ∆T = 55 + 58.9 = 113.9°C or 237°F π ( 200 ) 25 ( 0.0423 ) Qs = (1 + 1.5 ò ) = 1 + 1.5 ( 0.87 ) −6 3µl′ 3 (10.5 )10 ( 25 ) = 3156 mm3 /s = 3156 ( 25.4− 3 ) = 0.193 in 3 /s 2 π ps rc3 2 Hloss = ρ CpQs ∆T = 0.0311(0.42)0.193(138) = 0.348 Btu/s = 1.05(0.348) = 0.365 kW = 365 W not O.K. Trumpler’s design criteria: 0.0002 + 0.000 04(50/25.4) = 0.000 279 in > h0 Not O.K. Tmax = 237°F O.K. Pst = 4000 kPa or 581 psi > 300 psi Not O.K. n = 1, as done Not O.K. _____________________________________________________________________________ 12-18 So far, we have performed elements of the design task. Now let’s do it more completely. The values of the unilateral tolerances, tb and td , reflect the routine capabilities of the Shigley’s MED, 10th edition Chapter 12 Solutions, Page 20/26 www.konkur.in bushing vendor and the in-house capabilities. While the designer has to live with these, his approach should not depend on them. They can be incorporated later. First we shall find the minimum size of the journal which satisfies Trumpler’s constraint of Pst ≤ 300 psi. W Pst = ≤ 300 2dl ′ W W ≤ 300 ⇒ d ≥ 2 2d l ′ / d 600(l ′ / d ) d min = 900 = 1.73 in 2(300)(0.5) In this problem we will take journal diameter as the nominal value and the bushing bore as a variable. In the next problem, we will take the bushing bore as nominal and the journal diameter as free. To determine where the constraints are, we will set tb = td = 0, and thereby shrink the design window to a point. We set d = 2.000 in b = d + 2cmin = d + 2c nd = 2 (This makes Trumpler’s nd ≤ 2 tight) and construct a table. c b d Tf * Tmax h0 Pst Tmax n 0.0010 0.0011 0.0012 0.0013 0.0014 0.0015 0.0016 0.0017 0.0018 0.0019 0.0020 2.0020 2.0022 2.0024 2.0026 2.0028 2.0030 2.0032 2.0034 2.0036 2.0038 2.0040 2 2 2 2 2 2 2 2 2 2 2 215.50 206.75 198.50 191.40 185.23 179.80 175.00 171.13 166.92 163.50 160.40 312.0 293.0 277.0 262.8 250.4 239.6 230.1 220.3 213.9 206.9 200.6 × × × × × × × × √ × √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ √ fom −5.74 −6.06 −6.37 −6.66 −6.94 −7.20 −7.45 −7.65 −7.91 −8.12 −8.32 *Sample calculation for the first entry of this column. Iteration yields: T f = 215.5°F With T f = 215.5°F , from Table 12-1 Shigley’s MED, 10th edition Chapter 12 Solutions, Page 21/26 www.konkur.in µ = 0.0136(10− 6 ) exp[1271.6 / (215.5 + 95)] = 0.817(10− 6 ) reyn 900 N = 3000 / 60 = 50 rev/s, P = = 225 psi 4 2 −6 1 0.817(10 )(50) S = = 0.182 225 0.001 From Figs. 12-16 and 12-18: Eq. (12–24): ∫ = 0.7, f r/c = 5.5 0.0123(5.5)(0.182)(9002 ) = 191.6°F [1 + 1.5(0.7 2 )](30)(14 ) 191.6°F Tav = 120°F + = 215.8°F ≈ 215.5°F 2 ∆TF = For the nominal 2-in bearing, the various clearances show that we have been in contact with the recurving of (h0)min. The figure of merit (the parasitic friction torque plus the pumping torque negated) is best at c = 0.0018 in. For the nominal 2-in bearing, we will place the top of the design window at cmin = 0.002 in, and b = d + 2(0.002) = 2.004 in. At this point, add the b and d unilateral tolerances: d = 2.000 +−0.000 b = 2.004 +−0.003 0.001 in, 0.000 in Now we can check the performance at cmin , c , and cmax . Of immediate interest is the fom of the median clearance assembly, − 9.82, as compared to any other satisfactory bearing ensemble. If a nominal 1.875 in bearing is possible, construct another table with tb = 0 and td = 0. c 0.0020 0.0030 0.0035 0.0040 0.0050 0.0055 0.0060 b 1.879 1.881 1.882 1.883 1.885 1.886 1.887 d 1.875 1.875 1.875 1.875 1.875 1.875 1.875 Tf 157.2 138.6 133.5 130.0 125.7 124.4 123.4 Tmax h0 Pst Tmax n fom √ √ − 7.36 194.30 × √ √ √ √ √ − 8.64 157.10 √ √ − 9.05 147.10 √ √ √ √ √ √ − 9.32 140.10 √ √ − 9.59 131.45 √ √ √ √ − 9.63 128.80 √ √ √ √ √ − 9.64 126.80 × The range of clearance is 0.0030 < c < 0.0055 in. That is enough room to fit in our design window. d = 1.875+−0.000 0.001 in, b = 1.881+−0.003 0.000 in The ensemble median assembly has a fom = − 9.31. We just had room to fit in a design window based upon the (h0)min constraint. Further Shigley’s MED, 10th edition Chapter 12 Solutions, Page 22/26 www.konkur.in reduction in nominal diameter will preclude any smaller bearings. A table constructed for a d = 1.750 in journal will prove this. We choose the nominal 1.875-in bearing ensemble because it has the largest figure of merit. Ans. _____________________________________________________________________________ 12-19 This is the same as Prob. 12-18 but uses design variables of nominal bushing bore b and radial clearance c. The approach is similar to that of Prob. 12-18 and the tables will change slightly. In the table for a nominal b = 1.875 in, note that at c = 0.003 in the constraints are “loose.” Set b = 1.875 in d = 1.875 − 2(0.003) = 1.869 in For the ensemble b = 1.875+−0.003 0.001 in, d = 1.869 +−0.000 0.001 in Analyze at cmin = 0.003 in, c = 0.004 in and cmax = 0.005 in At cmin = 0.003 in: T f = 138.4°F, µ′ = 3.160 µ reyn, S = 0.0297, H loss = 1035 Btu/h and the Trumpler conditions are met. At c = 0.004 in: T f = 130°F, µ′ = 3.872 µ reyn, S = 0.0205, Hloss = 1106 Btu/h, fom = −9.246 and the Trumpler conditions are O.K. At cmax = 0.005 in: T f = 125.68°F, µ′ = 4.325µ reyn, S = 0.014 66, Hloss = 1129 Btu/h and the Trumpler conditions are O.K. The ensemble figure of merit is slightly better; this bearing is slightly smaller. The lubricant cooler has sufficient capacity. _____________________________________________________________________________ 12-20 Table 12-1: µ (µ reyn) = µ 0 (106) exp [b / (T + 95)] b and T in °F The conversion from µ reyn to mPa⋅s is given on p. 612. For a temperature of C degrees Celsius, T = 1.8 C + 32. Substituting into the above equation gives µ (mPa⋅s) = 6.89 µ 0 (106) exp [b / (1.8 C + 32+ 95)] = 6.89 µ 0 (106) exp [b / (1.8 C + 127)] Ans. For SAE 50 oil at 70°C, from Table 12-1, µ 0 = 0.0170 (10−6) reyn, and b = 1509.6°F. From the equation, µ = 6.89(0.0170) 10−6(106) exp {1509.6/[1.8(70) + 127]} Shigley’s MED, 10th edition Chapter 12 Solutions, Page 23/26 www.konkur.in = 45.7 mPa⋅s Ans. From Fig. 12-13, µ = 39 mPa⋅s Ans. The figure gives a value of about 15 % lower than the equation. _____________________________________________________________________________ 12-21 Originally d = 2.000 +−0.000 b = 2.005+−0.003 0.001 in, 0.000 in Doubled, d = 4.000 +−0.000 b = 4.010 +−0.006 0.002 in, 0.000 in The radial load quadrupled to 3600 lbf when the analyses for parts (a) and (b) were carried out. Some of the results are: Part c µ′ S Tf f r/c Qs h0 /c ∫ (a) (b) 0.007 0.0035 3.416 3.416 0.0310 0.0310 135.1 135.1 0.1612 0.1612 6.56 0.870 0.1032 0.1032 0.897 0.897 Hloss 9898 1237 h0 0.000 722 0.000 361 Trumpler h0 0.000 360 0.000 280 f 0.005 67 0.005 67 The side flow Qs differs because there is a c3 term and consequently an 8-fold increase. Hloss is related by a 9898/1237 or an 8-fold increase. The existing h0 is related by a 2-fold increase. Trumpler’s (h0)min is related by a 1.286-fold increase. _____________________________________________________________________________ 12-22 Given: Oiles SP 500 alloy brass bushing, L = 0.75 in, D = 0.75 in, T∞ = 70°F, F = 400 lbf, N = 250 rev/min, and w = 0.004 in. Table 12-8: K = 0.6(10−10) in3⋅min/(lbf⋅ft⋅h) P = F/ (DL) = 400/ [0.75(0.75)] = 711 psi V = πDN/ 12 = π (0.75)250/12 = 49.1 ft/min From Table 12-10, interpolation gives V 33 49.1 100 f1 1.3 f1 1.8 = > f1 = 1.42 Table 12-11: f 2 = 1.0 Shigley’s MED, 10th edition Chapter 12 Solutions, Page 24/26 www.konkur.in Table 12-12: PVmax = 46 700 psi⋅ft/min, Pmax = 3560 psi, Vmax = 100 ft/min Pmax = 4 F 4 400 = = 905 psi < 3560 psi O.K . π DL π 0.752 PV = 711 (49.1) = 34 910 psi⋅ft/min < 46 700 psi⋅ft/min O.K. Eq. (12-32) can be written as w = f1 f 2 K 4 F Vt π DL Solving for t, t = π DLw 4 f1 f 2 KVF = π ( 0.75 ) 0.75 ( 0.004 ) 4 (1.42 )1.0 ( 0.6 )10−10 ( 49.1) 400 = 1056 h = 1056 ( 60 ) = 63 400 min Cycles = Nt = 250 (63 400) = 15.9 (106) cycles Ans. _____________________________________________________________________________ 12-23 Given: Oiles SP 500 alloy brass bushing, wmax = 0.002 in for 1000 h, N = 200 rev/min, F = 100 lbf, hCR = 2.7 Btu/(h∙ft2∙oF), Tmax = 300 oF, f s = 0.03, and nd = 2. Using Eq. (12-38) with nd F for F, fs = 0.03 from Table 12-9, and hCR = 2.7 Btu/(h ⋅ ft2 ⋅ οF), gives L≥ 720 f s nd FN 720(0.03)2(100)200 = = 1.79 in 778(2.7)(300 − 70) J h CR (T f − T∞ ) From Table 12-13, the smallest available bushing has an ID = 1 in, OD = 1 83 in, and L = 2 in. With L/D = 2/1 = 2, this is inside of the recommendations of Eq. (12-33). Thus, for the first trial, try the bushing with ID = 1 in, OD = 1 83 in, and L = 2 in. Thus, Eq. (12-31): Pmax = P= Shigley’s MED, 10th edition 4 nd F 4 2(100) = = 127.3psi < 3560 psi (OK) π DL π 1(2) nd F 2(100) = = 100 psi DL 1(2) Chapter 12 Solutions, Page 25/26 www.konkur.in Eq. (12-29): V = π DN 12 = π (1) 200 12 = 52.4 ft/min < 100 ft/min (OK) PV = 100(52.4) = 5240 psi ⋅ ft/min < 46 700 psi ⋅ ft/min (OK) From Table 12-10, interpolation gives V 33 52.4 100 f1 1.3 f1 1.8 = > f1 = 1.445 Eq. (12-32), with Tables 12-8 and 12-10: −11 f1 f 2 Knd FN t 1.445(1) 6 (10 ) 2 (100) 200 (1000) w= = = 0.000 578in < 0.001 in (OK) 3L 3(2) Answer Select ID = 1 in, OD = 1 83 in, and L = 2 in. Shigley’s MED, 10th edition Chapter 12 Solutions, Page 26/26 www.konkur.in Chapter 13 d P = 17 / 8 = 2.125 in N 1120 dG = 2 d P = ( 2.125) = 4.375 in N3 544 13-1 NG = PdG = 8 ( 4.375) = 35 teeth Ans. nG = 1600 (15 / 60 ) = 400 rev/min p = π m = 3π mm Ans. Ans. C = ( 2.125 + 4.375) / 2 = 3.25 in Ans. ______________________________________________________________________________ 13-2 C = 3 (15 + 60 ) 2 = 112.5 mm Ans. ______________________________________________________________________________ NG = 16 ( 4 ) = 64 teeth 13-3 Ans. dG = N G m = 64 ( 6 ) = 384 mm d P = N P m = 16 ( 6 ) = 96 mm Ans. Ans. C = ( 384 + 96 ) / 2 = 240 mm Ans. ______________________________________________________________________________ 13-4 Mesh: a = 1/ P = 1/ 3 = 0.3333 in Ans. b = 1.25 / P = 1.25 / 3 = 0.4167 in Ans. c = b − a = 0.0834 in Ans. p = π / P = π / 3 = 1.047 in Ans. t = p / 2 = 1.047 / 2 = 0.523 in Ans. Pinion Base-Circle: d1 = N1 / P = 21/ 3 = 7 in d1b = 7 cos 20o = 6.578 in Gear Base-Circle: Ans. d 2 = N 2 / P = 28 / 3 = 9.333 in d 2b = 9.333cos 20o = 8.770 in Base pitch: Ans. pb = pc cos φ = (π / 3) cos 20o = 0.984 in Ans. Contact Ratio: mc = Lab / pb = 1.53 / 0.984 = 1.55 Ans. See the following figure for a drawing of the gears and the arc lengths. Shigley’s MED, 10th edition Chapter 13 Solutions, Page 1/36 www.konkur.in ______________________________________________________________________________ 13-5 1/2 (a) 14 / 6 2 32 / 6 2 AO = + 2 2 (b) γ = tan −1 (14 / 32 ) = 23.63o Γ = tan −1 ( 32 /14 ) = 66.37 (c) d P = 14 / 6 = 2.333 in dG = 32 / 6 = 5.333 in Shigley’s MED, 10th edition o = 2.910 in Ans. Ans. Ans. Ans. Ans. Chapter 13 Solutions, Page 2/36 www.konkur.in (d) From Table 13-3, 0.3AO = 0.3(2.910) = 0.873 in and 10/P = 10/6 = 1.67 0.873 < 1.67 ∴ F = 0.873 in Ans. ______________________________________________________________________________ 13-6 (a) pn = π / Pn = π / 4 = 0.7854 in pt = pn / cosψ = 0.7854 / cos 30o = 0.9069 in p x = pt / tanψ = 0.9069 / tan 30o = 1.571 in pnb = pn cos φn = 0.7854 cos 25o = 0.7380 in (b) Eq. (13-7): (c) Pt = Pn cosψ = 4 cos 30o = 3.464 teeth/in Ans. φt = tan −1 ( tan φn / cosψ ) = tan −1 (tan 25o / cos30o ) = 28.3o Ans. (d) Table 13-4: a = 1/ 4 = 0.250 in Ans. b = 1.25 / 4 = 0.3125 in Ans. 20 dP = = 5.774 in Ans. 4 cos 30o 36 dG = = 10.39 in Ans. 4 cos 30o ______________________________________________________________________________ 13-7 N P = 19 teeth, N G = 57 teeth, φn = 20o , mn = 2.5 mm (a) pn = π mn = π ( 2.5) = 7.854 mm Ans. pn 7.854 = = 9.069 mm Ans. cosψ cos 30o pt 9.069 px = = = 15.71 mm Ans. tanψ tan 30o mn 2.5 mt = = = 2.887 mm Ans. cosψ cos 30o pt = (b) Shigley’s MED, 10th edition Chapter 13 Solutions, Page 3/36 www.konkur.in tan 20o = 22.80o o cos 30 φt = tan −1 (c) a = mn = 2.5 mm Ans. Ans. b = 1.25mn = 1.25 ( 2.5) = 3.125 mm dP = Ans. N = Nmt = 19 ( 2.887 ) =54.85 mm Pt Ans. dG = 57 ( 2.887 ) = 164.6 mm Ans. ______________________________________________________________________________ 13-8 (a) Using Eq. (13-11) with k = 1, φ = 20º, and m = 2, NP = = ( 2k m + m 2 + (1 + 2m ) sin 2 φ (1 + 2m ) sin 2 φ {( 2) + 1 + 2 ( 2 ) sin ( 20 ) 2 (1) 2 o ( 2) 2 ) ( )} = 14.16 teeth + 1 + 2 ( 2 ) sin 2 20o Round up for the minimum integer number of teeth. NP = 15 teeth Ans. (b) Repeating (a) with m = 3, NP = 14.98 teeth. Rounding up, NP = 15 teeth. Ans. (c) Repeating (a) with m = 4, NP = 15.44 teeth. Rounding up, NP = 16 teeth. Ans. (d) Repeating (a) with m = 5, NP = 15.74 teeth. Rounding up, NP = 16 teeth. Ans. Alternatively, a useful table can be generated to determine the largest gear that can mesh with a specified pinion, and thus also the maximum gear ratio with a specified pinion. The Max NG column was generated using Eq. (13-12) with k = 1, φ = 20º, and rounding up to the next integer. Min NP 13 14 15 16 17 18 Max NG 16 26 45 101 1309 unlimited Max m = Max NG / Min NP 1.23 1.86 3.00 6.31 77.00 unlimited With this table, we can readily see that gear ratios up to 3 can be obtained with a minimum NP of 15 teeth, and gear ratios up to 6.31 can be obtained with a minimum NP of 16 teeth. This is consistent with the results previously obtained. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 13 Solutions, Page 4/36 www.konkur.in 13-9 Repeating the process shown in the solution to Prob. 13-8, except with φ = 25º, we obtain the following results. (a) For m = 2, NP = 9.43 teeth. Rounding up, NP = 10 teeth. Ans. (b) For m = 3, NP = 9.92 teeth. Rounding up, NP = 10 teeth. Ans. (c) For m = 4, NP = 10.20 teeth. Rounding up, NP = 11 teeth. Ans. (d) For m = 5, NP = 10.38 teeth. Rounding up, NP = 11 teeth. Ans. For convenient reference, we will also generate the table from Eq. (13-12) for φ = 25º. Min NP Max NG Max m = Max NG / Min NP 9 13 1.44 10 32 3.20 11 249 22.64 12 unlimited unlimited ______________________________________________________________________________ 13-10 (a) The smallest pinion tooth count that will run with itself is found from Eq. (13-10). NP ≥ ≥ ( 2k 1 + 1 + 3sin 2 φ 2 3sin φ 2 (1) (1 + 3sin 20 2 ≥ 12.32 o ) 1 + 3sin 2 20o → 13 teeth ) Ans. (b) The smallest pinion that will mesh with a gear ratio of mG = 2.5, from Eq. (13-11) is NP ≥ ≥ ( 2k m + m 2 + (1 + 2m ) sin 2 φ 2 (1 + 2m ) sin φ 2 (1) {2.5 + 1 + 2 ( 2.5 ) sin 20 2 ≥ 14.64 → 2.52 + 1 + 2 ( 2.5 ) sin 2 20o o 15 teeth ) } Ans. The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-12) is NG ≤ ≤ N P2 sin 2 φ − 4k 2 4k − 2 N P sin 2 φ 152 sin 2 20o − 4 (1) 2 4 (1) − 2 (15 ) sin 2 20o ≤ 45.49 → 45 teeth Ans. (c) The smallest pinion that will mesh with a rack, from Eq. (13-13), Shigley’s MED, 10th edition Chapter 13 Solutions, Page 5/36 www.konkur.in 2 (1) 2k = sin 2 φ sin 2 20o ≥ 17.097 → 18 teeth Ans. ______________________________________________________________________________ NP ≥ 13-11 φn = 20o ,ψ = 30o From Eq. (13-19), φt = tan −1 ( tan 20o / cos 30o ) = 22.80o (a) The smallest pinion tooth count that will run with itself, from Eq. (13-21) is NP ≥ ≥ ( 2k cosψ 1 + 1 + 3sin 2 φt 2 3sin φt 2 (1) cos 30o (1 + 3sin 22.80 2 o ) 1 + 3sin 2 22.80o ≥ 8.48 → 9 teeth ) Ans. (b) The smallest pinion that will mesh with a gear ratio of m = 2.5, from Eq. (13-22) is 2 (1) cos 30o NP ≥ 2.5 + 2.52 + 1 + 2 ( 2.5 ) sin 2 22.80o 1 + 2 ( 2.5 ) sin 2 22.80o ≥ 9.95 → 10 teeth Ans. { } The largest gear-tooth count possible to mesh with this pinion, from Eq. (13-23) is NG ≤ ≤ N P2 sin 2 φt − 4k 2 cos 2 ψ 4k cosψ − 2 N P sin 2 φt 102 sin 2 22.80o − 4 (1) cos 2 30o 4 (1) cos 2 30o − 2 ( 20 ) sin 2 22.80o ≤ 26.08 → 26 teeth Ans. (c) The smallest pinion that will mesh with a rack, from Eq. (13-24) is o 2k cosψ 2 (1) cos 30 NP ≥ = sin 2 φt sin 2 22.80o ≥ 11.53 → 12 teeth Ans. ______________________________________________________________________________ 13-12 From Eq. (13-19), Shigley’s MED, 10th edition tan φn cosψ φt = tan −1 o −1 tan 20 = tan = 22.796o o cos 30 Chapter 13 Solutions, Page 6/36 www.konkur.in Program Eq. (13-23) on a computer using a spreadsheet or code, and increment NP. The first value of NP that can be doubled is NP = 10 teeth, where NG ≤ 26.01 teeth. So NG = 20 teeth will work. Higher tooth counts will work also, for example 11:22, 12:24, etc. Use NP = 10 teeth, NG = 20 teeth Ans. Note that the given diametral pitch (tooth size) is not relevant to the interference problem. ______________________________________________________________________________ o tan φn −1 tan 20 φt = tan 13-13 From Eq. (13-19), = 27.236o = tan o cosψ cos 45 Program Eq. (13-23) on a computer using a spreadsheet or code, and increment NP. The first value of NP that can be doubled is NP = 6 teeth, where NG ≤ 17.6 teeth. So NG = 12 teeth will work. Higher tooth counts will work also, for example 7:14, 8:16, etc. −1 Use NP = 6 teeth, NG = 12 teeth Ans. ______________________________________________________________________________ 13-14 The smallest pinion that will operate with a rack without interference is given by Eq. (1313). 2k NP = sin 2 φ Setting k = 1 for full depth teeth, NP = 9 teeth, and solving for φ, 2 (1) 2k = sin −1 = 28.126o Ans. NP 9 ______________________________________________________________________________ φ = sin −1 13-15 (a) (b) Eq. (13-3): pn = π mn = 3π mm Eq. (13-16): pt = pn / cosψ = 3π / cos 25o = 10.40 mm Eq. (13-17): p x = pt / tanψ = 10.40 / tan 25 = 22.30 mm Eq. (13-3): mt = pt / π = 10.40 / π = 3.310 mm Eq. (13-19): φt = tan −1 Shigley’s MED, 10th edition Ans. o Ans. Ans. Ans. tan φn tan 20o = tan −1 = 21.88o cosψ cos 25o Ans. Chapter 13 Solutions, Page 7/36 www.konkur.in (c) Eq. (13-2): dp = mt Np = 3.310 (18) = 59.58 mm Ans. Eq. (13-2): dG = mt NG = 3.310 (32) = 105.92 mm Ans. ______________________________________________________________________________ 13-16 (a) Sketches of the figures are shown to determine the axial forces by inspection. The axial force of gear 2 on shaft a is in the negative z-direction. The axial force of gear 3 on shaft b is in the positive z-direction. Ans. The axial force of gear 4 on shaft b is in the positive z-direction. The axial force of gear 5 on shaft c is in the negative z-direction. Ans. 12 16 ( 700 ) = +77.78 rev/min ccw 48 36 (b) nc = n5 = (c) d P 2 = 12 / 12 cos 30o = 1.155 in dG 3 Ans. ( ) = 48 / (12 cos 30 ) = 4.619 in o 1.155 + 4.619 = 2.887 in Ans. 2 = 16 / ( 8 cos 25o ) = 2.207 in Cab = dP4 ( ) d G 5 = 36 / 8 cos 25o = 4.965 in Cbc = 3.586 in Ans. ______________________________________________________________________________ 20 8 20 4 = 40 17 60 51 4 nd = ( 600 ) = 47.06 rev/min cw Ans. 51 ______________________________________________________________________________ 13-17 e= 6 18 20 3 3 = 10 38 48 36 304 3 n9 = (1200 ) = 11.84 rev/min cw Ans. 304 ______________________________________________________________________________ 13-18 e= Shigley’s MED, 10th edition Chapter 13 Solutions, Page 8/36 www.konkur.in 13-19 (a) Ans. cw about x, as viewed from the positive x axis o d P = 12 / ( 8 cos 23 ) = 1.630 in nc = (b) 12 1 ⋅ ( 540 ) = 162 rev/min 40 1 ( ) d G = 40 / 8 cos 23o = 5.432 in d P + dG = 3.531 in 2 Ans. N 32 = = 8 in Ans. P 4 ______________________________________________________________________________ (c) d= 13-20 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 45. For an exact ratio, we will choose to factor the train value into integers, such that N2 / N3 = 9 N4 / N5 = 5 (1) (2) Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the in-line requirement of the compound reverted configuration is N2 + N3 = N4 + N5 (3) With three equations and four unknowns, one free choice is available. It is necessary that all of the unknowns be integers. We will use a normalized approach to find the minimum free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity, thus N3 = 1. From (1), N2 = 9. From (3), N2 + N3 = 9 + 1 = 10 = N4 + N5 Substituting N4 = 5 N5 from (2) gives 10 = 5 N5 + N5 = 6 N5 N5 = 10 / 6 = 5 / 3 To eliminate this fraction, we need to multiply the original free choice by a multiple of 3. In addition, the smallest gear needs to have sufficient teeth to avoid interference. From Eq. (13-11) with k = 1, φ = 20°, and m = 9, the minimum number of teeth on the pinion to avoid interference is 17. Therefore, the smallest multiple of 3 greater than 17 is 18. Setting N3 = 18 and repeating the solution of equations (1), (2), and (3) yields Shigley’s MED, 10th edition Chapter 13 Solutions, Page 9/36 www.konkur.in N2 = 162 teeth N3 = 18 teeth N4 = 150 teeth N5 = 30 teeth Ans. ______________________________________________________________________________ 13-21 The solution to Prob. 13-20 applies up to the point of determining the minimum number of teeth to avoid interference. From Eq. (13-11), with k = 1, φ = 25°, and m = 9, the minimum number of teeth on the pinion to avoid interference is 11. Therefore, the smallest multiple of 3 greater than 11 is 12. Setting N3 = 12 and repeating the solution of equations (1), (2), and (3) of Prob. 13-20 yields N2 = 108 teeth N3 = 12 teeth N4 = 100 teeth N5 = 20 teeth Ans. ______________________________________________________________________________ 13-22 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 30. For an exact ratio, we will choose to factor the train value into integers, such that N2 / N3 = 6 N4 / N5 = 5 (1) (2) Assuming a constant diametral pitch in both stages, the geometry condition to satisfy the in-line requirement of the compound reverted configuration is N2 + N3 = N4 + N5 (3) With three equations and four unknowns, one free choice is available. It is necessary that all of the unknowns be integers. We will use a normalized approach to find the minimum free choice to guarantee integers; that is, set the smallest gear of the largest stage to unity, thus N3 = 1. From (1), N2 = 6. From (3), N2 + N3 = 6 + 1 = 7 = N4 + N5 Substituting N4 = 5 N5 from (2) gives 7 = 5 N5 + N5 = 6 N5 N5 = 7 / 6 To eliminate this fraction, we need to multiply the original free choice by a multiple of 6. In addition, the smallest gear needs to have sufficient teeth to avoid interference. From Eq. (13-11) with k = 1, φ = 20°, and m = 6, the minimum number of teeth on the pinion to avoid interference is 16. Therefore, the smallest multiple of 6 greater than 16 is 18. Setting N3 = 18 and repeating the solution of equations (1), (2), and (3) yields Shigley’s MED, 10th edition Chapter 13 Solutions, Page 10/36 www.konkur.in N2 = 108 teeth N3 = 18 teeth N4 = 105 teeth N5 = 21 teeth Ans. ______________________________________________________________________________ 13-23 Applying Eq. (13-30), e = (N2 / N3) (N4 / N5) = 45. For an approximate ratio, we will choose to factor the train value into two equal stages, such that N 2 / N3 = N 4 / N5 = 45 If we choose identical pinions such that interference is avoided, both stages will be identical and the in-line geometry condition will automatically be satisfied. From Eq. (13-11) with k = 1, φ = 20°, and m = 45 , the minimum number of teeth on the pinions to avoid interference is 17. Setting N3 = N5 = 17, we get N 2 = N 4 = 17 45 = 114.04 teeth Rounding to the nearest integer, we obtain N2 = N4 = 114 teeth N3 = N5 = 17 teeth Ans. Checking, the overall train value is e = (114 / 17) (114 / 17) = 44.97. ______________________________________________________________________________ 13-24 H = 25 hp, ωi = 2500 rev/min Let ωo = 300 rev/min for minimal gear ratio to minimize gear size. ωo 300 1 N N = = = 2 4 ωi 2500 8.333 N 3 N 5 Let N2 N4 1 1 = = = N3 N5 8.333 2.887 From Eq. (13-11) with k = 1, φ = 20°, and m = 2.887, the minimum number of teeth on the pinions to avoid interference is 15. Let N2 = N4 = 15 teeth N3 = N5 = 2.887(15) = 43.31 teeth Try N3 = N5 = 43 teeth. Shigley’s MED, 10th edition Chapter 13 Solutions, Page 11/36 www.konkur.in 15 15 ωo = ( 2500 ) = 304.2 43 43 Too big. Try N3 = N5 = 44. 15 15 ωo = ( 2500 ) = 290.55 rev/min 44 44 N2 = N4 = 15 teeth, N3 = N5 = 44 teeth Ans. ______________________________________________________________________________ 13-25 (a) The planet gears act as keys and the wheel speeds are the same as that of the ring gear. Thus, nA = n3 = 900 (16 / 48 ) = 300 rev/min Ans. (b) nF = n5 = 0, nL = n6 , e = −1 n − 300 −1 = 6 0 − 300 300 = n6 − 300 n6 = 600 rev/min Ans. (c) The wheel spins freely on icy surfaces, leaving no traction for the other wheel. The car is stalled. Ans. ______________________________________________________________________________ 13-26 (a) The motive power is divided equally among four wheels instead of two. (b) Locking the center differential causes 50 percent of the power to be applied to the rear wheels and 50 percent to the front wheels. If one of the rear wheels rests on a slippery surface such as ice, the other rear wheel has no traction. But the front wheels still provide traction, and so you have two-wheel drive. However, if the rear differential is locked, you have 3-wheel drive because the rear-wheel power is now distributed 50-50. ______________________________________________________________________________ 13-27 Let gear 2 be first, then nF = n2 = 0. Let gear 6 be last, then nL = n6 = –12 rev/min. 20 16 16 nL − nA = = 30 34 51 nF − nA 16 ( 0 − n A ) = −12 − nA 51 e= Shigley’s MED, 10th edition Chapter 13 Solutions, Page 12/36 www.konkur.in −12 = −17.49 rev/min 35 / 51 Ans. (negative indicates cw as viewed from the bottom of the figure) nA = ______________________________________________________________________________ 13-28 Let gear 2 be first, then nF = n2 = 0 rev/min. Let gear 6 be last, then nL = n6 = 85 rev/min. 20 16 16 nL − nA = = 30 34 51 nF − nA 16 ( 0 − n A ) = ( 85 − nA ) 51 16 − nA + nA = 85 51 16 nA 1 − = 85 51 85 nA = = 123.9 rev/min 16 1− 51 e= The positive sign indicates the same direction as n6. ∴ nA = 123.9 rev/min ccw Ans. ______________________________________________________________________________ 13-29 The geometry condition is d5 / 2 = d 2 / 2 + d3 + d 4 . Since all the gears are meshed, they will all have the same diametral pitch. Applying d = N / P, N 5 / (2 P) = N 2 / (2 P) + N 3 / P + N 4 / P N5 = N 2 + 2 N3 + 2 N 4 = 12 + 2 (16 ) + 2 (12 ) = 68 teeth Ans. Let gear 2 be first, nF = n2 = 320 rev/min. Let gear 5 be last, nL = n5 = 0 rev/min. e= 12 16 12 3 nL − nA = = 16 12 68 17 nF − nA 320 − nA = nA = − Shigley’s MED, 10th edition 17 ( 0 − nA ) 3 3 ( 320 ) = −68.57 rev/min 14 Chapter 13 Solutions, Page 13/36 www.konkur.in The negative sign indicates opposite of n2. ∴ nA = 68.57 rev/min cw Ans. ______________________________________________________________________________ 13-30 Let nF = n2, then nL = n7 = 0. e=− nL − n5 20 16 36 = −0.5217 = 16 30 46 nF − n5 0 − n5 = −0.5217 10 − n5 −0.5217 (10 − n5 ) = −n5 −5.217 + 0.5217 n5 + n5 = 0 n5 (1 + 0.5217 ) = 5.217 5.217 1.5217 n5 = nb = 3.428 turns in same direction Ans. ______________________________________________________________________________ n5 = 13-31 (a) ω = 2π n / 60 H = T ω = 2π Tn / 60 (T in N ⋅ m, H in W) ( ) 60 H 103 So T= So F32t = 2π n = 9550 H / n (H in kW, n in rev/min) 9550 ( 75 ) Ta = = 398 N ⋅ m 1800 mN 2 5 (17 ) r2 = = = 42.5 mm 2 2 Shigley’s MED, 10th edition Ta 398 = = 9.36 kN r2 42.5 Chapter 13 Solutions, Page 14/36 www.konkur.in F3b = − Fb 3 = 2 ( 9.36 ) = 18.73 kN in the positive x-direction. Ans. mN 4 5 ( 51) = = 127.5 mm 2 2 Tc 4 = 9.36 (127.5) = 1193 N ⋅ m ccw r4 = (b) ∴T4c = 1193 N ⋅ m cw Ans. Note: The solution is independent of the pressure angle. ______________________________________________________________________________ N N 13-32 d= = P 6 d 2 = 4 in, d 4 = 4 in, d 5 = 6 in, d 6 = 24 in 24 24 36 e = − − + =1/ 6 24 36 144 nF = n2 = 1000 rev/min nL = n6 = 0 n − nA 0 − nA 1 e= L = = nF − nA 1000 − nA 6 nA = −200 rev/min Noting that power equals torque times angular velocity, the input torque is T2 = 550 lbf ⋅ ft/s 60 s 1 rev 12 in H 25 hp = = 1576 lbf ⋅ in n2 1000 rev/min hp min 2π rad ft For 100 percent gear efficiency, the output power equals the input power, so Tarm = H 25 hp 550 lbf ⋅ ft/s 60 s 1 rev 12 in = = 7878 lbf ⋅ in nA 200 rev/min hp min 2π rad ft Next, we’ll confirm the output torque as we work through the force analysis and complete the free body diagrams. Gear 2 1576 = 788 lbf 2 F32r = 788 tan 20o = 287 lbf Wt = Shigley’s MED, 10th edition Chapter 13 Solutions, Page 15/36 www.konkur.in Gear 4 FA4 = 2W t = 2 ( 788) = 1576 lbf Gear 5 Arm Tout = 1576 ( 9 ) − 1576 ( 4 ) = 7880 lbf ⋅ in Ans. ______________________________________________________________________________ 13-33 Given: m = 12 mm, nP = 1800 rev/min cw, N2 = 18T, N3 = 32T, N4 = 18T, N5 = 48T Pitch Diameters: d2 = 18(12) = 216 mm, d3 = 32(12) = 384 mm, d4 = 18(12) = 216 mm, d5 = 48(12) = 576 mm Gear 2 From Eq. (13-36), 60 000 (150 ) 60 000 H Wt = = = 7.368 kN π dn π ( 216 )(1800 ) d 216 Ta 2 = Wt 2 = 7.368 = 795.7 N ⋅ m 2 2 W r = 7.368 tan 20o = 2.682 kN Shigley’s MED, 10th edition Chapter 13 Solutions, Page 16/36 www.konkur.in Gears 3 and 4 ( 384) 216 Wt = 7.368 2 2 t W = 13.10 kN W r = 13.10 tan 20o = 4.768 kN Ans. ______________________________________________________________________________ 13-34 Given: P = 5 teeth/in, N2 = 18T, N3 = 45T, φn = 20o , H = 32 hp, n2 = 1800 rev/min Gear 2 Tin = 63025 ( 32 ) 1800 = 1120 lbf ⋅ in 18 = 3.600 in 5 45 dG = = 9.000 in 5 1120 W32t = = 622 lbf 3.6 / 2 W32r = 622 tan 20o = 226 lbf dP = Fat2 = W32t = 622 lbf, ( Fa 2 = 622 2 + 226 2 ) 1/2 Far2 = W32r = 226 lbf = 662 lbf Each bearing on shaft a has the same radial load of RA = RB = 662/2 = 331 lbf. Gear 3 W23t = W32t = 622 lbf W23r = W32r = 226 lbf Fb 3 = Fb 2 = 662 lbf RC = RD = 662 / 2 = 331 lbf Shigley’s MED, 10th edition Chapter 13 Solutions, Page 17/36 www.konkur.in Each bearing on shaft b has the same radial load which is equal to the radial load of bearings A and B. Thus, all four bearings have the same radial load of 331 lbf. Ans. ______________________________________________________________________________ 13-35 Given: P = 4 teeth/in, φn = 20o , NP = 20T, n2 = 900 rev/min N P 20 = = 5.000 in P 4 63025 ( 30 )( 2 ) Tin = = 4202 lbf ⋅ in 900 W32t = Tin / ( d 2 / 2 ) = 4202 / ( 5 / 2 ) = 1681 lbf d2 = W32r = 1681 tan 20o = 612 lbf The motor mount resists the equivalent forces and torque. The radial force due to torque is Fr = 4202 = 150 lbf 14 ( 2 ) Forces reverse with rotational sense as torque reverses. The compressive loads at A and D are absorbed by the base plate, not the bolts. For W32t , the tensions in C and D are Shigley’s MED, 10th edition Chapter 13 Solutions, Page 18/36 www.konkur.in ΣM AB = 0 1681( 4.875 + 15.25) − 2 F (15.25) = 0 F = 1109 lbf If W32t reverses, 15.25 in changes to 13.25 in, 4.815 in changes to 2.875 in, and the forces change direction. For A and B, 1681( 2.875) − 2 F1 (13.25 ) = 0 ⇒ F1 = 182.4 lbf For W32r , M = 612 ( 4.875 + 11.25 / 2 ) = 6426 lbf ⋅ in a= (14 / 2 ) + (11.25 / 2 ) F2 = 6426 = 179 lbf 4 ( 8.98) 2 2 = 8.98 in At C and D, the shear forces are: FS 1 = 153 + 179 ( 5.625 / 8.98 ) + 179 ( 7 / 8.98 ) 2 2 At A and B, the shear forces are: Shigley’s MED, 10th edition Chapter 13 Solutions, Page 19/36 www.konkur.in FS 2 = 153 − 179 ( 5.625 / 8.98 ) + 179 ( 7 / 8.98 ) = 145 lbf 2 2 The shear forces are independent of the rotational sense. The bolt tensions and the shear forces for cw rotation are, For ccw rotation, ______________________________________________________________________________ 13-36 (a) N2 = N4 = 15 teeth, P= N d ⇒ d= N3 = N5 = 44 teeth N P 15 = 2.5 in Ans. 6 44 d3 = d5 = = 7.33 in Ans. 6 d2 = d4 = (b) (c) Vi = V2 = V3 = π d 2 n2 π ( 2.5 )( 2500 ) = 1636 ft/min Ans. 12 π d n π ( 2.5) ( 2500 )(15 / 44 ) Vo = V4 = V5 = 4 4 = = 558 ft/min 12 12 Input gears: Wti = 33000 12 = H 33000 ( 25 ) = = 504.3 lbf = 504 lbf Vi 1636 Wri = Wti tan φ = 504.3 tan 20o = 184 lbf W 504.3 Wi = ti = = 537 lbf Ans. cos φ cos 20o Shigley’s MED, 10th edition Ans. Ans. Ans. Chapter 13 Solutions, Page 20/36 www.konkur.in Output gears: H 33000 ( 25 ) Wto = 33000 = = 1478 lbf Vo 558 Ans. Wro = Wto tan φ = 1478 tan 20o = 538 lbf Ans. Wto 1478 Wo = = = 1573 lbf Ans. o cos 20 cos 20o (d) d 2.5 Ti = Wti 2 = 504.3 = 630 lbf ⋅ in 2 2 2 Ans. 2 44 44 To = Ti = 630 = 5420 lbf ⋅ in Ans. 15 15 ______________________________________________________________________________ (e) 13-37 H = 35 hp, ni = 1200 rev/min, φ =20o N 2 = N 4 = 16 teeth, N 3 = N 5 = 48 teeth, P = 10 teeth/in N 16 (a) nintermediate = n3 = n4 = 2 ni = (1200 ) = 400 rev/min N3 48 (b) no = N2 N4 16 16 ni = (1200 ) = 133.3 rev/min N3 N5 48 48 P= N d ⇒ d= Ans. Ans. N P 16 = 1.6 in Ans. 10 48 d3 = d5 = = 4.8 in Ans. 10 π d n π (1.6 )(1200 ) Vi = V2 = V3 = 2 2 = = 502.7 ft/min Ans. 12 12 π d n π (1.6 )( 400 ) Vo = V4 = V5 = 4 4 = = 167.6 ft/min Ans. 12 12 d2 = d4 = (c) Wti = 33000 H 33000 ( 35 ) = = 2298 lbf lbf Vi 502.7 Wri = Wti tan φ = 2298 tan 20o = 836.4 lbf W 2298 Wi = ti = = 2445 lbf Ans. cos φ cos 20o Shigley’s MED, 10th edition Ans. Ans. Chapter 13 Solutions, Page 21/36 www.konkur.in H 33000 ( 35 ) = = 6891 lbf Vo 167.6 Wto = 33000 Ans. Wro = Wto tan φ = 6891tan 20o = 2508 lbf Ans. Wto 6891 Wo = = = 7333 lbf Ans. o cos 20 cos 20o (d) d 1.6 Ti = Wti 2 = 2298 = 1838 lbf ⋅ in 2 2 2 Ans. 2 48 48 (e) To = Ti = 1838 = 16 540 lbf ⋅ in Ans. 16 16 ______________________________________________________________________________ ω 2 13-38 (a) For o = , from Eq. (13-11), with m = 2, k = 1, φ = 20o ωi 1 NP = So N P min (b) P= 2 (1) 1 + 2 ( 2 ) sin 2 20o = 15 Ans. {2 + N 15 = = 1.875 teeth/in d 8 } 22 + 1 + 2 ( 2 ) sin 2 20o = 14.16 Ans. (c) To transmit the same power with no change in pitch diameters, the speed and transmitted force must remain the same. For A, with φ = 20°, WtA = FA cos20° = 300 cos20° = 281.9 lbf For A, with φ = 25°, same transmitted load, FA = WtA/cos25° = 281.9/cos25° = 311.0 lbf Ans. Summing the torque about the shaft axis, d d WtA A = WtB B 2 2 ( d / 2 ) = W d A = 281.9 20 = 704.75 lbf WtB = WtA A ( ) ( d B / 2 ) tA d B 8 WtB 704.75 FB = = = 777.6 lbf Ans. o cos 25 cos 25o ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 13 Solutions, Page 22/36 www.konkur.in 13-39 (a) For ωo 5 = , from Eq. (13-11), with m = 5, k = 1, φ = 20o ωi 1 2 (1) NP = 5 + 52 + 1 + 2 ( 5) sin 2 25o = 10.4 2 o 1 + 2 ( 5) sin 25 { So N P min = 11 (b) m= } Ans. d 300 = = 27.3 mm/tooth N 11 Ans. (c) To transmit the same power with no change in pitch diameters, the speed and transmitted force must remain the same. For A, with φ = 20°, WtA = FA cos20° = 11 cos20° = 10.33 kN For A, with φ = 25°, same transmitted load, FA = WtA/cos25° = 10.33 / cos 25° = 11.40 kN Ans. Summing the torque about the shaft axis, d d WtA A = WtB B 2 2 d ( d / 2) 600 WtB = WtA A = WtA A = (11.40 ) = 22.80 kN d 300 ( dB / 2) B WtB 22.80 FB = = = 25.16 kN Ans. cos 25o cos 25o ______________________________________________________________________________ 13-40 (a) Using Eq. (13-11) with k = 1, φ = 20º, and m = 2, NP = = ( 2k m + m 2 + (1 + 2m ) sin 2 φ 2 1 + 2 m sin φ ( ) {( 2) + 1 + 2 ( 2 ) sin ( 20 ) 2 (1) 2 o ( 2) 2 ) ( )} = 14.16 teeth + 1 + 2 ( 2 ) sin 2 20o Round up for the minimum integer number of teeth. NF = 15 teeth, NC = 30 teeth Shigley’s MED, 10th edition Ans. Chapter 13 Solutions, Page 23/36 www.konkur.in d 125 = = 8.33 mm/tooth Ans. N 15 H 2 kW 1000 W rev 60 s T= = = 100 N ⋅ m ω 191 rev/min kW 2π rad min m= (b) (c) (d) From Eq. (13-36), Wt = 60 000 ( 2 ) 60 000 H = = 1.60 kN = 1600 N π dn π (125 )(191) Ans. Or, we could have obtained Wt directly from the torque and radius, Wt = T 100 = = 1600 N d / 2 0.125 / 2 Wr = Wt tan φ = 1600 tan 20o = 583 N Ans. Wt 1600 W= = = 1700 N Ans. cos φ cos 20o ______________________________________________________________________________ 13-41 (a) Using Eq. (13-11) with k = 1, φ = 20º, and m = 2, NP = = ( 2k m + m 2 + (1 + 2m ) sin 2 φ 2 (1 + 2m ) sin φ {( 2) + 1 + 2 ( 2 ) sin ( 20 ) 2 (1) 2 o ( 2) 2 ) ( )} = 14.16 teeth + 1 + 2 ( 2 ) sin 2 20o Round up for the minimum integer number of teeth. NC = 15 teeth, NF = 30 teeth (b) P= (c) T= (d) N 30 = = 3 teeth/in d 10 Ans. Ans. 550 lbf ⋅ ft/s 12 in rev 60 s 1 hp hp ω 70 rev/min ft 2π rad min T = 900 lbf ⋅ in Ans. H = From Eqs. (13-34) and (13-35), Shigley’s MED, 10th edition Chapter 13 Solutions, Page 24/36 www.konkur.in V= π dn π (10 )( 70 ) = 183.3 ft/min 12 H 33000 (1) Wt = 33000 = = 180 lbf Ans. V 183.3 Wr = Wt tan φ = 180 tan 20o = 65.5 lbf Ans. W 180 W= t = = 192 lbf Ans. cos φ cos 20o ______________________________________________________________________________ N d 1.30 o 13-42 (a) Eq. (13-14): γ = tan −1 P = tan −1 P = tan −1 Ans. = 18.5 3.88 NG dG (b) Eq. (13-34): (c) Eq. (13-35): 12 = V= π dn 12 = π ( 2 )(1.30 )( 600 ) 12 = 408.4 ft/min Eq. (13-38): H 10 = 33 000 = 808 lbf V 408.4 Wr = Wt tan φ cos γ = 808 tan 20o cos18.5o = 279 lbf Eq. (13-38): Wa = Wt tan φ sin γ = 808 tan 20o sin18.5o = 93.3 lbf Wt = 33 000 Ans. Ans. Ans. Ans. The tangential and axial forces agree with Prob. 3-74, but the radial force given in Prob. 3-74 is shown here to be incorrect. Ans. ______________________________________________________________________________ 13-43 Tin = 63 025H / n = 63025 ( 2.5) / 240 = 656.5 lbf ⋅ in W t = T / r = 656.5 / 2 = 328.3 lbf γ = tan −1 ( 2 / 4 ) = 26.565o Γ = tan −1 ( 4 / 2 ) = 63.435o ( ) a = 2 + 1.5 cos 26.565o / 2 = 2.67 in W r = 328.3 tan 20o cos 26.565o = 106.9 lbf W a = 328.3 tan 20o sin 26.565o = 53.4 lbf W = 106.9i – 53.4j + 328.3k lbf RAG = –2i + 5.17j, RAB = 2.5j ΣM 4 = R AG × W + R AB × FB + T = 0 Solving gives R AB × FB = 2.5 FBz i − 2.5 FBx k R AG × W = 1697i + 656.6 j − 445.9k Shigley’s MED, 10th edition Chapter 13 Solutions, Page 25/36 www.konkur.in So, (1697i + 656.6 j − 445.9k ) + ( 2.5FBz i − 2.5FBx k + Tj) = 0 FBz = −1697 / 2.5 = −678.8 lbf T = −656.6 lbf ⋅ in FBx = −445.9 / 2.5 = −178.4 lbf So, FB = −178.4i −678.8k lbf Ans. 2 1/2 FB = ( −678.8 ) + ( −178.4 ) 2 = 702 lbf FA = – (FB + W) = – (–178.8i – 678.8k + 106.9i – 53.4j + 328.3k) = 71.5i + 53.4j + 350.5k Ans. ( FA (radial) = 71.52 + 350.52 ) 1/2 = 358 lbf FA (thrust) = 53.4 lbf ______________________________________________________________________________ 13-44 d 2 = 18 / 10 = 1.8 in, d3 = 30 / 10 = 3.0 in d2 / 2 −1 0.9 o = tan = 30.96 1.5 d3 / 2 γ = tan −1 Γ = 180o − γ = 59.04o 9 DE = + 0.5 cos 59.04o = 0.8197 in 16 W t = 25 lbf W r = 25 tan 20o cos 59.04o = 4.681 lbf W a = 25 tan 20o sin 59.04o = 7.803 lbf W = –4.681i – 7.803j +25k RDG = 0.8197j + 1.25i RDC = –0.625j ΣM D = R DG × W + R DC × FC + T = 0 R DG × W = 20.49i − 31.25 j − 5.917k R DC × FC = −0.625 FCz i + 0.625 FCx k ( 20.49i − 31.25 j − 5.917k ) + ( −0.625 FCz i + 0.625 FCxk ) + Tj = 0 T = 31.25 lbf ⋅ in Ans. FC = 9.47i + 32.8k lbf Shigley’s MED, 10th edition Ans. Chapter 13 Solutions, Page 26/36 www.konkur.in ( FC = 9.47 2 + 32.82 ΣF = 0 ) 1/2 = 34.1 lbf Ans. FD = −4.79i + 7.80 j − 57.8k lbf 1/2 2 2 FD (radial) = ( −4.79 ) + ( −57.8) = 58.0 lbf Ans. a FD (thrust) = W = 7.80 lbf Ans. ______________________________________________________________________________ 13-45 NOTE: The shaft forces exerted on the gears are not shown in the figures above. Pt = Pn cosψ = 5 cos 30o = 4.330 teeth/in o tan φn −1 tan 20 φt = tan = tan = 22.80o o cosψ cos 30 18 dP = = 4.157 in 4.330 −1 The forces on the shafts will be equal to the forces transmitted to the gears through the meshing teeth. Pinion (Gear 2) W r = W t tan φt = 800 tan 22.80o = 336 lbf W a = W t tanψ = 800 tan 30o = 462 lbf W = −336i − 462 j + 800k lbf Ans. 1/2 W = ( −336 ) + ( −462 ) + 8002 2 Gear 3 2 W = 336i + 462 j − 800k lbf W = 983 lbf Ans. 32 dG = = 7.390 in 4.330 Shigley’s MED, 10th edition = 983 lbf Ans. Ans. Chapter 13 Solutions, Page 27/36 www.konkur.in TG = W t r = 800 ( 7.390 ) = 5912 lbf ⋅ in ______________________________________________________________________________ 13-46 φt = tan −1 tan φn tan 20o = tan −1 = 22.80o o cosψ cos 30 Pinion (Gear 2) W r = W t tan φt = 800 tan 22.80o = 336 lbf W a = W t tanψ = 800 tan 30o = 462 lbf W = −336i − 462 j − 800k lbf Ans. 1/2 2 2 2 W = ( −336 ) + ( −462 ) + ( −800 ) = 983 lbf Ans. Idler (Gear 3) From the diagram for the idler, noting that the radial and axial forces from gears 2 and 4 cancel each other, the force acting on the shaft is W = +1600k lbf Ans. Output gear (Gear 4) W = 336i − 462 j − 800k lbf 2 1/2 2 2 W = ( −336 ) + ( −462 ) + ( −800 ) Ans. = 983 lbf Ans. NOTE: For simplicity, the above figures only show the gear contact forces. Also, notice that the idler shaft reaction contains a couple tending to turn the shaft endover-end. Also the idler teeth are bent both ways. Idlers are more severely loaded than other gears, belying their name. Thus, be cautious. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 13 Solutions, Page 28/36 www.konkur.in 13-47 Gear 3: Pt = Pn cosψ = 7 cos 30o = 6.062 teeth/in tan 20o = 0.4203, φt = 22.8o o cos 30 54 d3 = = 8.908 in 6.062 W t = 500 lbf W a = 500 tan 30o = 288.7 lbf W r = 500 tan 22.8o = 210.2 lbf tan φt = W3 = 210.2i − 288.7 j − 500k lbf Ans. Gear 4: 14 d4 = = 2.309 in 6.062 8.908 W t = 500 = 1929 lbf 2.309 W a = 1929 tan 30o = 1114 lbf W r = 1929 tan 22.8o = 811 lbf W4 = −811i + 1114 j − 1929k lbf Ans. ______________________________________________________________________________ 13-48 Pt = 6 cos 30o = 5.196 teeth/in 42 d3 = = 8.083 in 5.196 φt = 22.8o 16 d2 = = 3.079 in 5.196 63025 ( 25 ) T2 = = 916 lbf ⋅ in 1720 Shigley’s MED, 10th edition Chapter 13 Solutions, Page 29/36 www.konkur.in T 916 = = 595 lbf r 3.079 / 2 W a = 595 tan 30o = 344 lbf W r = 595 tan 22.8o = 250 lbf Wt = W = 344i + 250 j + 595k lbf R DC = 6i, R DG = 3i − 4.04 j ΣM D = R DC × FC + R DG × W + T = 0 (1) R DG × W = −2404i − 1785 j + 2140k R DC × FC = −6 FCz j + 6 FCy k Substituting and solving Eq. (1) gives T = 2404i lbf ⋅ in FCz = −297.5 lbf FCy = −365.7 lbf ΣF = FD + FC + W = 0 Substituting and solving gives FCx = −344 lbf FDy = 106.7 lbf FDz = −297.5 lbf FC = −344i − 356.7 j − 297.5k lbf Ans. FD = 106.7 j − 297.5k lbf Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 13 Solutions, Page 30/36 www.konkur.in 13-49 Since the transverse pressure angle is specified, we will assume the given module is also in terms of the transverse orientation. d 2 = mN 2 = 4 (16 ) = 64 mm d3 = mN3 = 4 ( 36 ) = 144 mm d 4 = mN 4 = 4 ( 28) = 112 mm 6 kW 1000 W rev 60 s = 35.81 N ⋅ m ω 1600 rev/min kW 2π rad min T 35.81 Wt = = = 1119 N d 2 / 2 0.064 / 2 T= H = Shigley’s MED, 10th edition Chapter 13 Solutions, Page 31/36 www.konkur.in W r = W t tan φt = 1119 tan 20o = 407.3 N W a = W t tanψ = 1119 tan15o = 299.8 N F2 a = −1119i − 407.3 j − 299.8k N Ans. F3b = (1119 − 407.3) i − (1119 − 407.3) j = 711.7i − 711.7 j N Ans. F4 c = 407.3i + 1119 j + 299.8k N Ans. ______________________________________________________________________________ 13-50 N 14 36 = = 2.021 in, d3 = = 5.196 in o Pn cosψ 8 cos 30 8cos 30o 15 45 d4 = = 3.106 in, d5 = = 9.317 in o 5cos15 5cos15o d2 = For gears 2 and 3: φt = tan −1 ( tan φn / cosψ ) = tan −1 ( tan 20o / cos 30o ) = 22.8o For gears 4 and 5: φt = tan −1 ( tan 20o / cos15o ) = 20.6o F23t = T2 / r2 = 1200 / ( 2.021/ 2 ) = 1188 lbf 5.196 = 1987 lbf 3.106 F23r = F23t tan φt = 1188 tan 22.8o = 499 lbf F54t = 1188 F54r = 1986 tan 20.6o = 746 lbf F23a = F23t tanψ = 1188 tan 30o = 686 lbf F54a = 1986 tan15o = 532 lbf Next, designate the points of action on gears 4 and 3, respectively, as points G and H, as shown. Position vectors are Shigley’s MED, 10th edition Chapter 13 Solutions, Page 32/36 www.konkur.in R CG = 1.553j − 3k R CH = −2.598 j − 6.5k R CD = −8.5k Force vectors are F54 = −1986i − 748 j + 532k F23 = −1188i + 500 j − 686k FC = FCx i + FCy j FD = FDx i + FDy j + FDz k Now, a summation of moments about bearing C gives ΣM C = R CG × F54 + R CH × F23 + R CD × FD = 0 The terms for this equation are found to be R CG × F54 = −1412i + 5961j + 3086k R CH × F23 = 5026i + 7722 j − 3086k R CD × FD = 8.5 FDy i − 8.5 FDx j When these terms are placed back into the moment equation, the k terms, representing the shaft torque, cancel. The i and j terms give 3614 = −425 lbf 8.5 13683 FDx = = 1610 lbf 8.5 Next, we sum the forces to zero. FDy = − ΣF = FC + F54 + F23 + FD = 0 Substituting, gives ( F i + F j) + ( −1987i − 746 j + 532k ) + ( −1188i + 499 j − 686k ) + (1610i − 425 j + F k ) = 0 x C y C z D Solving gives FCx = 1987 + 1188 − 1610 = 1565 lbf FCy = 746 − 499 + 425 = 672 lbf FDz = −532 + 686 = 154 lbf So, FC = 1565i + 672j lbf Ans. FD = 1610i − 425j + 154k lbf Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 13 Solutions, Page 33/36 www.konkur.in VW = 13-51 WW t π dW nW = π ( 0.100 )( 600 ) 60 60 H 2000 = = = 637 N π VW = π m/s L = px NW = 25 (1) = 25 mm λ = tan −1 = tan −1 L π dW 25 = 4.550o lead angle π (100 ) WW t W= cos φn sin λ + f cos λ V π VS = W = = 3.152 m/s cos λ cos 4.550o In ft/min: VS = 3.28(3.152) = 10.33 ft/s = 620 ft/min Use f = 0.043 from curve A of Fig. 13-42. Then, from the first of Eq. (13-43) W= 637 = 5323 N cos14.5 sin 4.55o + 0.043cos 4.55o o ( ) W y = W sin φn = 5323sin14.5o = 1333 N ( ) W z = 5323 cos14.5o cos 4.55o − 0.043sin 4.55o = 5119 N The force acting against the worm is W = −637i + 1333 j + 5119k N Thus, A is the thrust bearing. Ans. R AG = −0.05 j − 0.10k m, R AB = −0.20k m ΣM A = R AG × W + R AB × FB + T = 0 R AG × W = −122.6i + 63.7 j − 31.85k N ⋅ m R AB × FB = 0.2 FBy i − 0.2 FBx j Substituting and solving gives T = 31.85 N ⋅ m Ans. FBx = 318.5 N, FBy = 613 N So FB = 318.5i + 613 j N Shigley’s MED, 10th edition Ans. Chapter 13 Solutions, Page 34/36 www.konkur.in 1/2 2 2 FB = ( 613) + ( 318.5 ) ΣF = FA + W + R B = 0 Or = 691 N radial FA = − ( W + FB ) = − ( −637i + 1333j + 5119k + 318.5i + 613j) = 318.5i − 1946 j − 5119k Ans. FAr = 318.5i − 1946 j N Radial 2 1/ 2 FAr = ( 318.5) + ( −1946 ) = 1972 N a Thrust FA = −5119 N ______________________________________________________________________________ 2 13-52 From Prob. 13-51, WG = 637i − 1333 j − 5119k N pt = px So dG = N G px π = 48 ( 25 ) π = 382 mm Bearing D takes the thrust load. ΣM D = R DG × WG + R DC × FC + T = 0 R DG = −0.0725i + 0.191j m R DC = −0.1075i m R DG × WG = −977.7i − 371.1j − 25.02k N ⋅ m R DC × FC = 0.1075 FCz j − 0.1075 FCy k N ⋅ m Putting it together and solving, T = 977.7 N ⋅ m Ans. FC = −233 j + 3450k N, ΣF = FC + WG + FD = 0 FC = 3460 N Ans. FD = − ( FC + WG ) = −637i + 1566 j + 1669k N Ans. Radial FDr = 1566 j + 1669k N Or ( FDr = 1566 2 + 1669 2 ) 1/2 = 2289 N (total radial) FDt = −637 i N (thrust) ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 13 Solutions, Page 35/36 www.konkur.in 13-53 VW = π (1.5 )( 600 ) = 235.7 ft/min 12 33000 ( 0.75 ) W x = WW t = = 105.0 lbf 235.7 π pt = px = = 0.3927 in 8 L = 0.3927 ( 2 ) = 0.7854 in 0.7854 = 9.46o π (1.5) 105.0 W= = 515.3 lbf o cos 20 sin 9.46o + 0.05 cos 9.46o W y = 515.3sin 20o = 176.2 lbf W z = 515.3 cos 20o ( cos 9.46o ) − 0.05sin 9.46o = 473.4 lbf λ = tan −1 So W = 105i + 176 j + 473k lbf Ans. T = 105 ( 0.75) = 78.8 lbf ⋅ in Ans. ______________________________________________________________________________ 13-54 Computer programs will vary. Shigley’s MED, 10th edition Chapter 13 Solutions, Page 36/36 www.konkur.in Chapter 14 d = 14-1 N 22 = = 3.667 in P 6 Table 14-2: Y = 0.331 π dn π (3.667)(1200) Eq. (13-34): V = = = 1152 ft/min 12 12 1200 + 1152 Eq. (14-4b): K v = = 1.96 1200 H 15 Eq. (13-35) : W t = 33 000 = 33 000 = 429.7 lbf V 1152 K W t P 1.96(429.7)(6) Eq. (14-7): = = 7633 psi = 7.63 kpsi Ans. σ = v FY 2(0.331) ________________________________________________________________________ N 18 = = 1.8 in P 10 Table 14-2: Y = 0.309 π dn π (1.8)(600) Eq. (13-34): V = = = 282.7 ft/min 12 12 1200 + 282.7 Eq. (14-4b): K v = = 1.236 1200 H 2 Eq. (13-35) : W t = 33 000 = 33 000 = 233.5 lbf V 282.7 K W t P 1.236(233.5)(10) Eq. (14-7): = = 9340 psi = 9.34 kpsi Ans. σ = v FY 1.0(0.309) ________________________________________________________________________ 14-2 d = 14-3 d = mN = 1.25(18) = 22.5 mm Y = 0.309 π dn π (22.5)(10−3 )(1800) V = = = 2.121 m/s 60 60 6.1 + 2.121 Kv = = 1.348 6.1 60 000 H 60 000(0.5) Wt = = = 0.2358 kN = 235.8 N π dn π (22.5)(1800) Table 14-2: Eq. (14-6b): Eq. (13-36): K vW t 1.348(235.8) = = 68.6 MPa Ans. FmY 12(1.25)(0.309) ________________________________________________________________________ Eq. (14-8): σ = Shigley’s MED, 10th edition Chapter 14 Solutions, Page 1/39 www.konkur.in 14-4 Table 14-2: Eq. (14-6b): Eq. (13-36): d = mN = 8(16) = 128 mm Y = 0.296 π dn π (128)(10−3 )(150) V = = = 1.0053 m/s 60 60 6.1 + 1.0053 Kv = = 1.165 6.1 60 000 H 60 000(6) Wt = = = 5.968 kN = 5968 N π dn π (128)(150) K vW t 1.165(5968) = = 32.6 MPa Ans. FmY 90(8)(0.296) ________________________________________________________________________ Eq. (14-8): 14-5 Table 14-2: Eq. (14-6b): Eq. (13-36): Eq. (14-8): σ = d = mN = 1(16) = 16 mm Y = 0.296 π dn π (16)(10−3 )(400) V = = = 0.335 m/s 60 60 6.1 + 0.335 Kv = = 1.055 6.1 60 000H 60 000(0.15) Wt = = = 0.4476 kN = 447.6 N π dn π (16)(400) F = K vW t 1.055(447.6) = = 10.6 mm σ mY 150(1)(0.296) From Table 13-2, use F = 11 mm or 12 mm, depending on availability. Ans. ________________________________________________________________________ 14-6 Table 14-2: Eq. (14-6b): Eq. (13-36): Eq. (14-8): d = mN = 2(20) = 40 mm Y = 0.322 π dn π (40)(10−3 )(200) V = = = 0.419 m/s 60 60 6.1 + 0.419 Kv = = 1.069 6.1 60 000 H 60 000(0.5) Wt = = = 1.194 kN = 1194 N π dn π (40)(200) F = K vW t 1.069(1194) = = 26.4 mm σ mY 75(2.0)(0.322) From Table 13-2, use F = 28 mm. Ans. ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 14 Solutions, Page 2/39 www.konkur.in 14-7 N 24 = = 4.8 in P 5 Table 14-2: Y = 0.337 π dn π (4.8)(50) Eq. (13-34): V = = = 62.83 ft/min 12 12 1200 + 62.83 Eq. (14-4b): K v = = 1.052 1200 H 6 Eq. (13-35) : W t = 33 000 = 33 000 = 3151 lbf V 62.83 K W t P 1.052(3151)(5) Eq. (14-7): F = v = = 2.46 in 20(103 )(0.337) σY d = Use F = 2.5 in Ans. ________________________________________________________________________ N 16 = = 4.0 in P 4 Table 14-2: Y = 0.296 π dn π (4.0)(400) Eq. (13-34): V = = = 418.9 ft/min 12 12 1200 + 418.9 Eq. (14-4b): K v = = 1.349 1200 H 20 Eq. (13-35) : W t = 33 000 = 33 000 = 1575.6 lbf V 418.9 K W t P 1.349(1575.6)(4) Eq. (14-7): F = v = = 2.39 in 12(103 )(0.296) σY Use F = 2.5 in Ans. ________________________________________________________________________ d = 14-8 14-9 Try P = 8 which gives d = 18/8 = 2.25 in and Y = 0.309. Eq. (13-34): Eq. (14-4b): Eq. (13-35): Eq. (14-7): V = π dn π (2.25)(600) = = 353.4 ft/min 12 12 1200 + 353.4 = 1.295 1200 H 2.5 W t = 33 000 = 33 000 = 233.4 lbf V 353.4 K W t P 1.295(233.4)(8) F = v = = 0.783 in 10(103 )(0.309) σY Kv = Using coarse integer pitches from Table 13-2, the following table is formed. Shigley’s MED, 10th edition Chapter 14 Solutions, Page 3/39 www.konkur.in d V Wt F Kv 9.000 1413.717 2.178 58.356 0.082 6.000 942.478 1.785 87.535 0.152 4.500 706.858 1.589 116.713 0.240 3.000 471.239 1.393 175.069 0.473 2.250 353.429 1.295 233.426 0.782 1.800 282.743 1.236 291.782 1.167 1.500 235.619 1.196 350.139 1.627 1.125 176.715 1.147 466.852 2.773 P 2 3 4 6 8 10 12 16 Other considerations may dictate the selection. Good candidates are P = 8 (F = 7/8 in) and P =10 (F = 1.25 in). Ans. ________________________________________________________________________ 14-10 Try m = 2 mm which gives d = 2(18) = 36 mm and Y = 0.309. V = π dn π (36)(10−3 )(900) = 1.696 m/s 60 60 6.1 + 1.696 Eq. (14-6b): K v = = 1.278 6.1 60 000H 60 000(1.5) Eq. (13-36): W t = = = 0.884 kN = 884 N π dn π (36)(900) 1.278(884) Eq. (14-8): F = = 24.4 mm 75(2)(0.309) Using the preferred module sizes from Table 13-2: m 1.00 1.25 1.50 2.00 3.00 4.00 5.00 6.00 8.00 10.00 12.00 16.00 20.00 25.00 32.00 40.00 50.00 Shigley’s MED, 10th edition = d 18.0 22.5 27.0 36.0 54.0 72.0 90.0 108.0 144.0 180.0 216.0 288.0 360.0 450.0 576.0 720.0 900.0 V 0.848 1.060 1.272 1.696 2.545 3.393 4.241 5.089 6.786 8.482 10.179 13.572 16.965 21.206 27.143 33.929 42.412 Kv Wt F 1.139 1768.388 86.917 1.174 1414.711 57.324 1.209 1178.926 40.987 1.278 884.194 24.382 1.417 589.463 12.015 1.556 442.097 7.422 1.695 353.678 5.174 1.834 294.731 3.888 2.112 221.049 2.519 2.391 176.839 1.824 2.669 147.366 1.414 3.225 110.524 0.961 3.781 88.419 0.721 4.476 70.736 0.547 5.450 55.262 0.406 6.562 44.210 0.313 7.953 35.368 0.243 Chapter 14 Solutions, Page 4/39 www.konkur.in Other design considerations may dictate the size selection. For the present design, m = 2 mm (F = 25 mm) is a good selection. Ans. ________________________________________________________________________ 14-11 NP 20 N 50 = = 2.5 in, d G = G = = 6.25 in P 8 P 8 π (2.5)(1200) V = = 785.4 ft/min 12 1200 + 785.4 Kv = = 1.655 1200 H 12 W t = 33 000 = 33 000 = 504.2 lbf V 785.4 dP = Eq. (14-4b): Eq. (13-36): C p = 2100 psi [Note: Using Eq. (14-13) can result in wide variation in Table 14-8: Cp due to wide variation in cast iron properties.] 2.5sin 20° = 0.4275 in, 2 Eq. (14-12): r1 = r2 = Eq. (14-14): K vW t 1 1 σ C = −C p + F cos φ r1 r2 6.25 sin 20° = 1.069 in 2 1/ 2 1/ 2 1.655(504.2) 1 1 = −2100 + 1.5 cos 20° 0.4275 1.069 = −92.5(103 ) psi = −92.5 kpsi Ans. ________________________________________________________________________ 14-12 16 48 = 1.333 in, dG = = 4 in 12 12 π (1.333)(700) V = = 244.3 ft/min 12 dP = Eq. (14-4b): Eq. (13-36): 1200 + 244.3 = 1.204 1200 H 1.5 Wt = 33 000 = 33 000 = 202.6 lbf V 244.3 C p = 2100 psi [Note: Using Eq. (14-13) can result in wide variation Kv = Table 14-8: in Cp due to wide variation in cast iron properties.] Eq. (14-12): r1 = 1.333sin 20° = 0.228 in, 2 r2 = 4 sin 20° = 0.684 in 2 Eq. (14-14): Shigley’s MED, 10th edition Chapter 14 Solutions, Page 5/39 www.konkur.in 1.204(202.6) 1 1 σ C = −2100 + F cos 20° 0.228 0.684 1/ 2 = −100(103 ) 2 2100 1.204(202.6) 1 1 F = + = 0.669 in 3 100(10 ) cos 20° 0.228 0.684 Use F = 0.75 in Ans. ________________________________________________________________________ d p = 5(24) = 120 mm, dG = 5(48) = 240 mm 14-13 V= π (120)(10−3 )(50) = 0.3142 m/s 60 3.05 + 0.3142 Eq. (14-6a): K v = = 1.103 3.05 60 000H 60(103 ) H Wt = = = 3.183H π dn π (120)(50) where H is in kW and Wt is in kN Table 14-8: C p = 163 MPa [Note: Using Eq. (14-13) can result in wide variation in Cp due to wide variation in cast iron properties]. 120 sin 20° 240 sin 20° Eq. (14-12): r1 = = 20.52 mm, r2 = = 41.04 mm 2 2 1.103(3.183) (103 ) H 1 1 −690 = −163 + 60 cos 20o 20.52 41.04 1/ 2 Eq. (14-14): H = 3.94 kW Ans. ________________________________________________________________________ 14-14 Eq. (14-6a): d P = 4(20) = 80 mm, dG = 4(32) = 128 mm π (80)(10−3 )(1000) V = = 4.189 m/s 60 3.05 + 4.189 Kv = = 2.373 3.05 60(10)(103 ) Wt = = 2.387 kN = 2387 N π (80)(1000) C p = 163 MPa [Note: Using Eq. (14-13) can result in wide variation in Table 14-8: Cp due to wide variation in cast iron properties.] Shigley’s MED, 10th edition Chapter 14 Solutions, Page 6/39 www.konkur.in Eq. (14-12): r1 = 80 sin 20° = 13.68 mm, 2 r2 = 128sin 20° = 21.89 mm 2 1/ 2 2.373(2387) 1 1 Eq. (14-14): σ C = −163 + = −617 MPa Ans. 50 cos 20° 13.68 21.89 ________________________________________________________________________ 14-15 The pinion controls the design. Bending YP = 0.303, YG = 0.359 17 30 = 1.417 in, dG = = 2.500 in 12 12 π d P n π (1.417)(525) V = = = 194.8 ft/min 12 12 1200 + 194.8 Eq. (14-4b): K v = = 1.162 1200 Eq. (6-8), p. 290: Se′ = 0.5(76) = 38.0 kpsi Eq. (6-19), p. 295: ka = 2.70(76)–0.265 = 0.857 2.25 2.25 l = = = 0.1875 in Pd 12 3Y 3(0.303) Eq. (14-3): x= P = = 0.0379 in 2P 2(12) dP = Eq. (b), p. 729: t = Eq. (6-25), p. 297: 4lx = 4(0.1875)(0.0379) = 0.1686 in de = 0.808 hb = 0.808 0.875(0.1686) = 0.310 in −0.107 0.310 kb = = 0.996 0.3 kc = kd = ke = 1 Account for one-way bending with kf = 1.66. (See Ex. 14-2.) Eq. (6-20), p. 296: Eq. (6-18), p. 295: Se = 0.857(0.996)(1)(1)(1)(1.66)(38.0) = 53.84 kpsi For stress concentration, find the radius of the root fillet (See Ex. 14-2). rf = 0.300 0.300 = = 0.025 in P 12 From Fig. A-15-6, r r 0.025 = f = = 0.148 d t 0.1686 Approximate D/d = ∞ with D/d = 3; from Fig. A-15-6, Kt = 1.68. From Fig. 6-20, with Sut = 76 kpsi and r = 0.025 in, q = 0.62. Shigley’s MED, 10th edition Chapter 14 Solutions, Page 7/39 www.konkur.in Eq. (6-32): Kf = 1 + 0.62 (1.68 – 1) = 1.42 Se 53.84 = = 16.85 psi σ all = K f nd 1.42(2.25) FYPσ all 0.875(0.303)(16 850) Wt = = = 320.4 lbf K v Pd 1.162(12) W tV 320.4(194.8) H = = = 1.89 hp Ans. 33 000 33 000 Wear ν 1 = ν 2 = 0.292, E1 = E2 = 30(106) psi 1/ 2 Eq. (14-13): Eq. (14-12): 1 Cp = 2 1 − 0.292 2π 30 (106 ) = 2285 psi dP 1.417 sin φ = sin 20° = 0.242 in 2 2 d 2.500 r2 = G sin φ = sin 20° = 0.428 in 2 2 1 1 1 1 + = + = 6.469 in −1 r1 r2 0.242 0.428 r1 = (SC )108 = 0.4H B − 10 kpsi = [0.4(149) − 10](103 ) = 49 600 psi Eq. (6-68), p. 337: From the discussion and equation developed on the bottom of p. 337, ( S ) 8 −49 600 = −33 067 psi σ C ,all = − C 10 = n 2.25 −33 067 0.875cos 20° Wt = = 22.6 lbf 2285 1.162(6.469) W tV 22.6(194.8) H = = = 0.133 hp Ans. 33 000 33 000 Rating power (pinion controls): 2 Eq. (14-14): H1 = 1.89 hp H2 = 0.133 hp Hall = (min 1.89, 0.133) = 0.133 hp Ans. ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 14 Solutions, Page 8/39 www.konkur.in 14-16 See Prob. 14-15 solution for equation numbers. Pinion controls: YP = 0.322, YG = 0.447 Bending dP = 20/3 = 6.667 in, dG = 100/3 = 33.333 in V = π d P n / 12 = π (6.667)(870) / 12 = 1519 ft/min K v = (1200 + 1519) / 1200 = 2.266 Se′ = 0.5(113) = 56.5 kpsi ka = 2.70(113) −0.265 = 0.771 l = 2.25 / Pd = 2.25 / 3 = 0.75 in x = 3(0.322) / [2(3)] = 0.161 in t = 4(0.75)(0.161) = 0.695 in d e = 0.808 2.5(0.695) = 1.065 in kb = (1.065 / 0.30) −0.107 = 0.873 kc = k d = k e = 1 kf = 1.66 (See Ex. 14-2.) Se = 0.771(0.873)(1)(1)(1)(1.66)(56.5) = 63.1 kpsi rf = 0.300 / 3 = 0.100 in r r 0.100 = f = = 0.144 d t 0.695 Kt = 1.75, q = 0.85, Kf = 1.64 Se 63.1 σ all = = = 25.7 kpsi K f nd 1.64(1.5) FYPσ all 2.5(0.322)(25 700) Wt = = = 3043 lbf K v Pd 2.266(3) H = W tV / 33 000 = 3043(1519) / 33 000 = 140 hp Ans. Wear 1/ 2 Eq. (14-13): Eq. (14-12): 1 Cp = 2 1 − 0.292 2π 30 (106 ) = 2285 psi r1 = (6.667/2) sin 20° = 1.140 in r2 = (33.333/2) sin 20° = 5.700 in Eq. (6-68), p. 337: σ C,all Shigley’s MED, 10th edition SC = [0.4(262) – 10](103) = 94 800 psi = −SC / nd = −94 800 / 1.5 = −77 400 psi Chapter 14 Solutions, Page 9/39 www.konkur.in 2 σ F cos φ 1 W = C ,all Cp K v 1 / r1 + 1 / r2 2 1 −77 400 2.5cos 20° = 2285 2.266 1 / 1.140 + 1 / 5.700 = 1130 lbf W tV 1130(1519) H = = = 52.0 hp Ans. 33 000 33 000 For 108 cycles (revolutions of the pinion), the power based on wear is 52.0 hp. Rating power (pinion controls): t H1 = 140 hp H 2 = 52.0 hp H rated = min(140, 52.0) = 52.0 hp Ans. ________________________________________________________________________ 14-17 See Prob. 14-15 solution for equation numbers. Given: φ = 20°, n = 1145 rev/min, m = 6 mm, F = 75 mm, NP = 16 milled teeth, NG = 30T, Sut = 900 MPa, HB = 260, nd = 3, YP = 0.296, and YG = 0.359. Pinion bending d P = mN P = 6(16) = 96 mm dG = 6(30) = 180 mm π d P n π (96)(10−3 )(1145) V = = = 5.76 m/s 60 (60) 6.1 + 5.76 Kv = = 1.944 6.1 Se′ = 0.5(900) = 450 MPa ka = 4.51(900) −0.265 = 0.744 l = 2.25m = 2.25(6) = 13.5 mm x = 3Ym / 2 = 3(0.296)6 / 2 = 2.664 mm t = 4lx = 4(13.5)(2.664) = 12.0 mm de = 0.808 75(12.0) = 24.23 mm −0.107 24.23 kb = = 0.884 7.62 kc = k d = ke = 1 kf = 1.66 (See Ex. 14-2) Se = 0.744(0.884)(1)(1)(1)(1.66)(450) = 491.3 MPa rf = 0.300m = 0.300(6) = 1.8 mm r/d = rf /t = 1.8/12 = 0.15, Kt = 1.68, q = 0.86, Kf = 1.58 Shigley’s MED, 10th edition Chapter 14 Solutions, Page 10/39 www.konkur.in σ all = Se 491.3 = = 239.2 MPa K f nd 1.58 (1.3) Eq. (14-8): Wt = FYmσ all 75(0.296)(6)(239.2) = = 16 390 N Kv 1.944 Eq. (13-36): H = W tπ dn 16.39π (96)(1145) = = 94.3 kW 60 000 60 000 Ans. Wear: Pinion and gear Eq. (14-12): r1 = (96/2) sin 20° = 16.42 mm r2 = (180/2) sin 20° = 30.78 mm 1/ 2 1 Eq. (14-13): C p = = 190 MPa 2 1 − 0.292 2π 207 (103 ) Eq. (6-68), p. 337: SC = 6.89[0.4(260) – 10] = 647.7 MPa −647.7 = −568 MPa σ C ,all = −SC / nd = 1.3 Eq. (14-14): σ W = C ,all Cp 2 t F cos φ 1 K v 1 / r1 + 1 / r2 o 1 −568 75cos 20 = = 3469 N 190 1.944 1 / 16.42 + 1 / 30.78 W tπ dn 3.469π (96)(1145) H = = = 20.0 kW 60 000 60 000 2 Eq. (13-36): Thus, wear controls the gearset power rating; H = 20.0 kW. Ans. ________________________________________________________________________ 14-18 NP = 17 teeth, NG = 51 teeth N 17 dP = = = 2.833 in P 6 51 dG = = 8.500 in 6 V = π d P n / 12 = π (2.833)(1120) / 12 = 830.7 ft/min Eq. (14-4b): Kv = (1200 + 830.7)/1200 = 1.692 Shigley’s MED, 10th edition Chapter 14 Solutions, Page 11/39 www.konkur.in σ all = Sy nd = 90 000 = 45 000 psi 2 Table 14-2: YP = 0.303, YG = 0.410 Eq. (14-7): Wt = Eq. (13-35): H = FYPσ all 2(0.303)(45 000) = = 2686 lbf KvP 1.692(6) W tV 2686(830.7) = = 67.6 hp 33 000 33 000 Based on yielding in bending, the power is 67.6 hp. (a) Pinion fatigue Bending Eq. (2-21), p. 52: Sut = 0.5 HB = 0.5(232) = 116 kpsi Eq. (6-8), p. 290: Se′ = 0.5Sut = 0.5(116) = 58 kpsi Eq. (6-19), p. 295: ka = 2.70(116)−0.265 = 0.766 Table 13-1, p. 688: l = Eq. (14-3): x = 1 1.25 2.25 2.25 + = = = 0.375 in Pd Pd Pd 6 3YP 3(0.303) = = 0.0758 in 2P 2(6) Eq. (b), p. 729: t = 4lx = 4(0.375)(0.0758) = 0.337 in Eq. (6-25), p. 297: de = 0.808 F t = 0.808 2(0.337) = 0.663 in 0.663 Eq. (6-20), p. 296: kb = 0.30 kc = kd = ke = 1 −0.107 = 0.919 Account for one-way bending with kf = 1.66. (See Ex. 14-2.) Eq. (6-18): Se = 0.766(0.919)(1)(1)(1)(1.66)(58) = 67.8 kpsi For stress concentration, find the radius of the root fillet (See Ex. 14-2). 0.300 0.300 rf = = = 0.050 in P 6 r r 0.05 Fig. A-15-6: = f = = 0.148 d t 0.338 Estimate D/d = ∞ by setting D/d = 3, Kt = 1.68. Shigley’s MED, 10th edition Chapter 14 Solutions, Page 12/39 www.konkur.in Fig. 6-20, p. 303: q = 0.86 Eq. (6-32), p. 303: K f = 1 + (0.86)(1.68 − 1) = 1.58 Se 67.8 = = 21.5 kpsi K f nd 1.58(2) FYPσ all 2(0.303)(21 500) Wt = = = 1283 lbf K v Pd 1.692(6) σ all = H = W tV 1283(830.7) = = 32.3 hp 33 000 33 000 Ans. (b) Pinion fatigue Wear 1/ 2 Eq. (14-13): Eq. (14-12): Eq. (6-68): 1 Cp = 2 6 2π[(1 - 0.292 ) / 30(10 )] = 2285 psi dP 2.833 sin φ = sin 20 o = 0.485 in 2 2 dG 8.500 r2 = sin φ = sin 20 o = 1.454 in 2 2 1 1 1 1 + = 2.750 in + = r1 r2 0.485 1.454 r1 = (SC )108 = 0.4H B − 10 kpsi In terms of gear notation σC = [0.4(232) – 10]103 = 82 800 psi We will introduce the design factor of nd = 2 and because it is a contact stress apply it to the load Wt by dividing by nd = 2 . (See p. 337.) σ C ,all = − σc 2 =− 82 800 = −58 548 psi 2 Solve Eq. (14-14) for Wt: o −58 548 2 cos 20 Wt = = 265 lbf 2285 1.692(2.750) W tV 265(830.7) H all = = = 6.67 hp Ans. 33 000 33 000 For 108 cycles (turns of pinion), the allowable power is 6.67 hp. 2 Shigley’s MED, 10th edition Chapter 14 Solutions, Page 13/39 www.konkur.in (c) Gear fatigue due to bending and wear Bending x= Eq. (14-3): Eq. (b), p. 729: t = 3YG 3(0.4103) = = 0.1026 in 2P 2(6) 4 lx = 4(0.375)(0.1026) = 0.392 in ± Eq. (6-25): de = 0.808 F t = 0.808 2(0.392) = 0.715 in −0.107 0.715 Eq. (6-20): kb = = 0.911 0.30 kc = kd = ke = 1 kf = 1.66. (See Ex. 14-2.) Eq. (6-18): Se = 0.766(0.911)(1)(1)(1)(1.66)(58) = 67.2 kpsi r r 0.050 = f = = 0.128 d t 0.392 Approximate D/d = ∞ by setting D/d = 3 for Fig. A-15-6; Kt = 1.80. Fig. 6-20: q = 0.82 Eq. (6-32): K f = 1 + (0.82)(1.80 − 1) = 1.66 Se 67.2 = = 20.2 kpsi K f nd 1.66(2) FYPσ all 2(0.4103)(20 200) Wt = = = 1633 lbf K v Pd 1.692(6) σ all = W tV 1633(830.7) H all = = = 41.1 hp 33 000 33 000 The gear is thus stronger than the pinion in bending. Ans. Wear Since the material of the pinion and the gear are the same, and the contact stresses are the same, the allowable power transmission of both is the same. Thus, Hall = 6.67 hp for 108 revolutions of each. As yet, we have no way to establish SC for 108/3 revolutions. (d) Pinion bending: H1 = 32.3 hp Pinion wear: H2 = 6.67 hp Gear bending: H3 = 41.1 hp Gear wear: H4 = 6.67 hp Power rating of the gear set is thus Hrated = min(32.3, 6.67, 41.1, 6.67) = 6.67 hp Ans. ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 14 Solutions, Page 14/39 www.konkur.in 14-19 dP = 16/6 = 2.667 in, dG = 48/6 = 8 in π (2.667)(300) V = = 209.4 ft/min 12 33 000(5) Wt = = 787.8 lbf 209.4 Assuming uniform loading, Ko = 1. Eq. (14-28): Qv = 6, B = 0.25(12 − 6)2 / 3 = 0.8255 A = 50 + 56(1 − 0.8255) = 59.77 0.8255 59.77 + 209.4 Eq. (14-27): K v = 59.77 Table 14-2: YP = 0.296, YG = 0.4056 From Eq. (a), Sec. 14-10 with F = 2 in 2 0.296 ( K s ) P = 1.192 6 = 1.196 0.0535 = 1.088 0.0535 2 0.4056 ( K s )G = 1.192 = 1.097 6 From Eq. (14-30) with Cmc = 1 2 Cp f = − 0.0375 + 0.0125(2) = 0.0625 10(2.667) C p m = 1, Cma = 0.093 (Fig. 14 - 11), Ce = 1 K m = 1 + 1[0.0625(1) + 0.093(1)] = 1.156 Assuming constant thickness of the gears → KB = 1 mG = NG/NP = 48/16 = 3 With N (pinion) = 108 cycles and N (gear) = 108/3, Fig. 14-14 provides the relations: (YN ) P = 1.3558(108 ) −0.0178 = 0.977 (YN )G = 1.3558(108 / 3)−0.0178 = 0.996 Fig. 14-6: J P = 0.27, Table 14-10: KR = 0.85 KT = C f = 1 Eq. (14-23): Table 14-8: I = JG 0.38 cos 20o sin 20o 3 = 0.1205 2(1) 3 + 1 C p = 2300 psi Shigley’s MED, 10th edition Chapter 14 Solutions, Page 15/39 www.konkur.in Strength: Grade 1 steel with HBP = HBG = 200 Fig. 14-2: (St)P = (St)G = 77.3(200) + 12 800 = 28 260 psi Fig. 14-5: (Sc)P = (Sc)G = 322(200) + 29 100 = 93 500 psi Fig. 14-15: (ZN )P = 1.4488(108)–0.023 = 0.948 (ZN )G = 1.4488(108/3)–0.023 = 0.973 Sec. 14-12: HBP/HBG = 1 ∴ CH = 1 Pinion tooth bending Eq. (14-15): (σ ) P = W t K o K v K s Pd K m K B F J 6 (1.156)(1) = 787.8(1)(1.196)(1.088) 2 0.27 = 13 170 psi Ans. Eq. (14-41): S Y / ( KT K R ) (S F ) P = t N σ 28 260(0.977) / [(1)(0.85)] = = 2.47 13 170 Ans. Gear tooth bending Eq. (14-15): Eq. (14-41): 6 (1.156)(1) (σ )G = 787.8(1)(1.196)(1.097) = 9433 psi 2 0.38 28 260(0.996) / [(1)(0.85)] ( S F )G = = 3.51 Ans. 9433 Ans. Pinion tooth wear 1/ 2 Eq. (14-16): K C (σ c ) P = C p W t K o K v K s m f dP F I P 1/ 2 1.156 1 = 2300 787.8(1)(1.196)(1.088) 2.667(2) 0.1205 = 98 760 psi Ans. Eq. (14-42): S Z /( KT K R ) 93 500(0.948) /[(1)(0.85)] (S H ) P = c N = 1.06 = σc 98 760 P Shigley’s MED, 10th edition Ans. Chapter 14 Solutions, Page 16/39 www.konkur.in Gear tooth wear 1/ 2 (K ) 1.097 (σ c )G = s G (σ c ) P = (98 760) = 99 170 psi 1.088 (K s )P 93 500(0.973)(1) /[(1)(0.85)] ( S H )G = = 1.08 Ans. 99 170 1/ 2 Ans. The hardness of the pinion and the gear should be increased. ________________________________________________________________________ 14-20 dP = 2.5(20) = 50 mm, dG = 2.5(36) = 90 mm π d PnP π (50)(10−3 )(100) V = = = 0.2618 m/s 60 60 60(120) Wt = = 458.4 N π (50)(10−3 )(100) With no specific information given to indicate otherwise, assume KB = Ko = Yθ = ZR = 1 Eq. (14-28): Eq. (14-27): Table 14-2: Qv = 6, B = 0.25(12 – 6)2/3 = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77 59.77 + 200(0.2618) Kv = 59.77 YP = 0.322, YG = 0.3775 0.8255 = 1.099 Similar to Eq. (a) of Sec. 14-10 but for SI units: Ks = ( 1 = 0.8433 mF Y kb ) 0.0535 ( K s ) P = 0.8433 2.5(18) 0.322 0.0535 ( K s )G = 0.8433 2.5(18) 0.3775 Cmc = Ce = C pm = 1 = 1.003 use 1 0.0535 =1.007 use 1 18 − 0.025 = 0.011 10(50) = 0.247 + 0.0167(0.709) − 0.765(10−4 )(0.7092 ) = 0.259 F = 18 / 25.4 = 0.709 in, C pf = Cma K H = 1 + 1[0.011(1) + 0.259(1)] = 1.27 Fig. 14-14: (YN )P = 1.3558(108)–0.0178 = 0.977 (YN )G = 1.3558(108/1.8)–0.0178 = 0.987 Fig. 14-6: (YJ )P = 0.33, (YJ )G = 0.38 Shigley’s MED, 10th edition Chapter 14 Solutions, Page 17/39 www.konkur.in Eq. (14-38): Table 14-8: YZ = 0.658 – 0.0759 ln(1 – 0.95) = 0.885 cos 20o sin 20o 1.8 ZI = = 0.103 2(1) 1.8 + 1 Z E = 191 MPa Strength Grade 1 steel, HBP = HBG = 200 Fig. 14-2: Fig. 14-5: Fig. 14-15: (St)P = (St)G = 0.533(200) + 88.3 = 194.9 MPa (Sc)P = (Sc)G = 2.22(200) + 200 = 644 MPa (ZN )P = 1.4488(108)–0.023 = 0.948 (Z N )G = 1.4488(108 / 1.8) −0.023 = 0.961 Fig. 14-12: H BP / H BG = 1 Eq. (14-23): ∴ ZW = CH = 1 Pinion tooth bending Eq. (14-15): 1 KH KB (σ ) P = W t K o K v K s bmt YJ P 1 1.27(1) = 458.4(1)(1.099)(1) = 43.08 MPa 18(2.5) 0.33 S Y 194.9 0.977 Eq. (14-41) for SI: ( S F ) P = t N = = 4.99 43.08 1(0.885) σ Yθ YZ P Ans. Ans. Gear tooth bending 1 1.27(1) (σ )G = 458.4(1)(1.099)(1) = 37.42 MPa 18(2.5) 0.38 194.9 0.987 ( S F )G = = 5.81 Ans. 37.42 1(0.885) Ans. Pinion tooth wear Eq. (14-16): K Z (σ c ) P = Z E W t K o K v K s H R d w1b Z I P 1.27 1 = 191 458.4(1)(1.099)(1) = 501.8 MPa 50(18) 0.103 S Z Z 644 0.948(1) Eq. (14-42) for SI: ( S H ) P = c N W = = 1.37 Ans. 501.8 1(0.885) σ c Yθ YZ P Ans. Gear tooth wear 1/ 2 (K ) (σ c )G = s G (K s )P Shigley’s MED, 10th edition 1/ 2 1 (σ c ) P = 1 (501.8) = 501.8 MPa Ans. Chapter 14 Solutions, Page 18/39 www.konkur.in 644 0.961(1) = 1.39 Ans. 501.8 1(0.885) ________________________________________________________________________ ( S H )G = 14-21 Pt = Pn cosψ = 6 cos 30° = 5.196 teeth/in 16 48 dP = = 3.079 in, dG = (3.079) = 9.238 in 5.196 16 π (3.079)(300) V = = 241.8 ft/min 12 0.8255 59.77 + 241.8 33 000(5) t W = = 682.3 lbf , K v = = 1.210 241.8 59.77 From Prob. 14-19: YP = 0.296, YG = 0.4056 ( K s ) P = 1.088, ( K s )G = 1.097, K B = 1 mG = 3, (YN ) P = 0.977, (YN )G = 0.996, K R = 0.85 (St ) P = (St )G = 28 260 psi, CH = 1, (Sc ) P = (Sc )G = 93 500 psi (Z N ) P = 0.948, (Z N )G = 0.973, C p = 2300 psi The pressure angle is: tan 20° φt = tan −1 = 22.80° cos 30° 3.079 cos 22.8° = 1.419 in, 2 a = 1 / Pn = 1 / 6 = 0.167 in (rb ) P = (rb )G = 3(rb ) P = 4.258 in Eq. (14-25): 1/ 2 2 2 3.079 9.238 2 Z = + 0.167 − 1.419 + + 0.167 − 4.2582 2 2 3.079 9.238 − + sin 22.8° 2 2 = 0.9479 + 2.1852 − 2.3865 = 0.7466 Conditions O.K . for use p N = pn cos φ n = Eq. (14-21): mN = Shigley’s MED, 10th edition π 6 1/ 2 cos 20° = 0.4920 in pN 0.492 = = 0.6937 0.95Z 0.95(0.7466) Chapter 14 Solutions, Page 19/39 www.konkur.in Eq. (14-23): sin 22.8° cos 22.8° 3 I = 3 + 1 = 0.193 2(0.6937) Fig. 14-7: J P′ 0.45, J G′ 0.54 Fig. 14-8: Corrections are 0.94 and 0.98. J P = 0.45(0.94) = 0.423, J G = 0.54(0.98) = 0.529 2 Cmc = 1, C pf = − 0.0375 + 0.0125(2) = 0.0525 10(3.079) C pm = 1, Cma = 0.093, Ce = 1 K m = 1 + (1)[0.0525(1) + 0.093(1)] = 1.146 Pinion tooth bending 5.196 1.146(1) (σ ) P = 682.3(1)(1.21)(1.088) = 6323 psi 2 0.423 28 260(0.977) / [1(0.85)] (S F ) P = = 5.14 Ans. 6323 Ans. Gear tooth bending 5.196 1.146(1) (σ )G = 682.3(1)(1.21)(1.097) = 5097 psi 2 0.529 28 260(0.996) / [1(0.85)] ( S F )G = = 6.50 Ans. 5097 Ans. Pinion tooth wear 1/ 2 1.146 1 (σ c ) P = 2300 682.3(1)(1.21)(1.088) 3.078(2) 0.193 93 500(0.948) / [(1)(0.85)] (S H ) P = = 1.54 Ans. 67 700 = 67 700 psi Ans. Gear tooth wear 1/ 2 1.097 (σ c )G = (67 700) = 67 980 psi Ans. 1.088 93 500(0.973) /[(1)(0.85)] ( S H )G = = 1.57 Ans. 67 980 ________________________________________________________________________ 14-22 Given: R = 0.99 at 108 cycles, HB = 232 through-hardening Grade 1, core and case, both gears. NP = 17T, NG = 51T, Table 14-2: YP = 0.303, YG = 0.4103 Shigley’s MED, 10th edition Chapter 14 Solutions, Page 20/39 www.konkur.in Fig. 14-6: JP = 0.292, JG = 0.396 dP = NP / P = 17 / 6 = 2.833 in, dG = 51 / 6 = 8.500 in. Pinion bending From Fig. 14-2: 0.99 Fig. 14-14: (St )107 = 77.3H B + 12 800 = 77.3(232) + 12 800 = 30 734 psi YN = 1.6831(108)–0.0323 = 0.928 V = π d P n / 12 = π (2.833)(1120 / 12) = 830.7 ft/min KT = K R = 1, S F = 2, S H = 2 30 734(0.928) σ all = = 14 261 psi 2(1)(1) Qv = 5, B = 0.25(12 − 5)2 / 3 = 0.9148 A = 50 + 56(1 − 0.9148) = 54.77 54.77 + 830.7 K v = 54.77 0.9148 = 1.472 0.0535 2 0.303 K s = 1.192 = 1.089 ⇒ use 1 6 K m = Cm f = 1 + Cmc (C p f C p m + CmaCe ) Cmc = 1 F − 0.0375 + 0.0125F 10d 2 = − 0.0375 + 0.0125(2) = 0.0581 10(2.833) =1 C pf = C pm Eq. (14-15): Cma = 0.127 + 0.0158(2) − 0.093(10−4 )(22 ) = 0.1586 Ce = 1 K m = 1 + 1[0.0581(1) + 0.1586(1)] = 1.217 KB = 1 FJ Pσ all Wt = K o K v K s Pd K m K B 2(0.292)(14 261) = = 775 lbf 1(1.472)(1)(6)(1.217)(1) W tV 775(830.7) H = = = 19.5 hp 33 000 33 000 Pinion wear Shigley’s MED, 10th edition Chapter 14 Solutions, Page 21/39 www.konkur.in Fig. 14-15: ZN = 2.466N–0.056 = 2.466(108)–0.056 = 0.879 mG = 51 / 17 = 3 Eq. (14-23): I = Fig. 14-5: (Sc )107 = 322H B + 29 100 = 322(232) + 29 100 = 103 804 psi 103 804(0.879) = 64 519 psi σ c,all = 2(1)(1) cos 20o sin 20o 3 = 1.205, 2 3 + 1 CH = 1 0.99 2 Eq. (14-16): σ Fd P I W = c,all C p K o K v K s K mC f 2 64 519 2(2.833)(0.1205) = 2300 1(1.472)(1)(1.2167)(1) = 300 lbf W tV 300(830.7) H = = = 7.55 hp 33 000 33 000 t The pinion controls, therefore Hrated = 7.55 hp Ans. ________________________________________________________________________ 14-23 l = 2.25/ Pd, t = 4 lx = x = 3Y / 2Pd 2.25 3Y 3.674 4 Y = Pd Pd 2Pd 3.674 F Y de = 0.808 F t = 0.808 F Y = 1.5487 Pd Pd −0.107 −0.0535 1.5487 F Y / P F Y d kb = = 0.8389 0.30 Pd 0.0535 F Y 1 Ks = = 1.192 Ans. kb P d ________________________________________________________________________ 14-24 YP = 0.331, YG = 0.422, JP = 0.345, JG = 0.410, Ko = 1.25. The service conditions are adequately described by Ko. Set SF = SH = 1. dP = 22 / 4 = 5.500 in dG = 60 / 4 = 15.000 in Shigley’s MED, 10th edition Chapter 14 Solutions, Page 22/39 www.konkur.in V = Pinion bending 0.99 π (5.5)(1145) 12 = 1649 ft/min (St )107 = 77.3H B + 12 800 = 77.3(250) + 12 800 = 32 125 psi YN = 1.6831[3(109 )]−0.0323 = 0.832 Eq. (14-17): 32 125(0.832) = 26 728 psi 1(1)(1) B = 0.25(12 − 6)2 / 3 = 0.8255 A = 50 + 56(1 − 0.8255) = 59.77 (σ all ) P = 0.8255 59.77 + 1649 K v = = 1.534 59.77 K s = 1, Cm = 1 F Cmc = − 0.0375 + 0.0125 F 10d 3.25 = − 0.0375 + 0.0125(3.25) = 0.0622 10(5.5) Cma = 0.127 + 0.0158(3.25) − 0.093(10−4 )(3.252 ) = 0.178 Ce = 1 K m = Cm f = 1 + (1)[0.0622(1) + 0.178(1)] = 1.240 K B = 1, Eq. (14-15): KT = 1 26 728(3.25)(0.345) = 3151 lbf 1.25(1.534)(1)(4)(1.240) 3151(1649) H1 = = 157.5 hp 33 000 W1t = Gear bending By similar reasoning, W2t = 3861 lbf and H 2 = 192.9 hp Pinion wear mG = 60 / 22 = 2.727 cos 20o sin 20o 2.727 = 0.1176 2 1 + 2.727 0.99 ( Sc )107 = 322(250) + 29 100 = 109 600 psi I = (Z N ) P = 2.466[3(109 )]−0.056 = 0.727 (Z N )G = 2.466[3(109 ) / 2.727]−0.056 = 0.769 109 600(0.727) (σ c,all ) P = = 79 679 psi 1(1)(1) Shigley’s MED, 10th edition Chapter 14 Solutions, Page 23/39 www.konkur.in 2 σ Fd P I W = c,all C p K o K v K s K mC f t 3 2 79 679 3.25(5.5)(0.1176) = = 1061 lbf 2300 1.25(1.534)(1)(1.24)(1) 1061(1649) H3 = = 53.0 hp 33 000 Gear wear Similarly, W4t = 1182 lbf , H 4 = 59.0 hp Rating H rated = min( H1, H 2 , H 3, H 4 ) = min(157.5, 192.9, 53, 59) = 53 hp Ans. Note differing capacities. Can these be equalized? ________________________________________________________________________ 14-25 From Prob. 14-24: W1t = 3151 lbf , W2t = 3861 lbf , W3t = 1061 lbf , W4t = 1182 lbf 33 000 Ko H 33 000(1.25)(40) Wt = = = 1000 lbf V 1649 Pinion bending: The factor of safety, based on load and stress, is (S F ) P = W1t 3151 = = 3.15 1000 1000 Ans. Gear bending based on load and stress W2t 3861 ( S F )G = = = 3.86 1000 1000 Ans. Pinion wear based on load: based on stress: W3t 1061 = = 1.06 1000 1000 (S H ) P = 1.06 = 1.03 n3 = Ans. Gear wear based on load: Shigley’s MED, 10th edition W4t 1182 n4 = = = 1.18 1000 1000 Chapter 14 Solutions, Page 24/39 www.konkur.in (SH )G = 1.18 = 1.09 based on stress: Ans. Factors of safety are used to assess the relative threat of loss of function 3.15, 3.86, 1.06, 1.18 where the threat is from pinion wear. By comparison, the AGMA safety factors (SF)P, (SF)G, (SH)P, (SH)G are 3.15, 3.86, 1.03, 1.09 or 3.15, 3.86, 1.061/2, 1.181/2 and the threat is again from pinion wear. Depending on the magnitude of the numbers, using SF and SH as defined by AGMA, does not necessarily lead to the same conclusion concerning threat. Therefore be cautious. ________________________________________________________________________ 14-26 Solution summary from Prob. 14-24: n = 1145 rev/min, Ko = 1.25, Grade 1 materials, NP = 22T, NG = 60T, mG = 2.727, YP = 0.331,YG = 0.422, JP = 0.345, JG = 0.410, Pd = 4T /in, F = 3.25 in, Qv = 6, (Nc)P = 3(109), R = 0.99, Km = 1.240, KT = 1, KB = 1, dP = 5.500 in, dG = 15.000 in, V = 1649 ft/min, Kv = 1.534, (Ks )P = (Ks )G = 1, (YN )P = 0.832, (YN )G = 0.859, KR = 1 Pinion HB: 250 core, 390 case Gear HB: 250 core, 390 case Bending (σ all ) P (σ all )G W1t W2t = = = = 26 728 psi 27 546 psi 3151 lbf , 3861 lbf , (St ) P = 32 125 psi (St )G = 32 125 psi H1 = 157.5 hp H 2 = 192.9 hp Wear φ = 20o , I = 0.1176, (Z N ) P = 0.727 (Z N )G = 0.769, CP = 2300 psi (Sc ) P = Sc = 322(390) + 29 100 = 154 680 psi 154 680(0.727) (σ c,all ) P = = 112 450 psi 1(1)(1) 154 680(0.769) (σ c,all )G = = 118 950 psi 1(1)(1) 2 112 450 W3t = (1061) = 2113 lbf , 79 679 H3 = 2113(1649) = 105.6 hp 33 000 2 118 950 W = (1182) = 2354 lbf , 109 600(0.769) t 4 Shigley’s MED, 10th edition H4 = 2354(1649) = 117.6 hp 33 000 Chapter 14 Solutions, Page 25/39 www.konkur.in Rated power Hrated = min(157.5, 192.9, 105.6, 117.6) = 105.6 hp Ans. Prob. 14-24: Hrated = min(157.5, 192.9, 53.0, 59.0) = 53 hp The rated power approximately doubled. ________________________________________________________________________ 14-27 The gear and the pinion are 9310 grade 1, carburized and case-hardened to obtain Brinell 285 core and Brinell 580–600 case. Table 14-3: 0.99 (St )107 = 55 000 psi Modification of St by (YN )P = 0.832 produces (σ all ) P = 45 657 psi, Similarly for (YN )G = 0.859 (σ all )G = 47 161 psi, W = 4569 lbf , W = 5668 lbf , t 1 t 2 and H1 = 228 hp H 2 = 283 hp From Table 14-8, C p = 2300 psi. Also, from Table 14-6: 0.99 (Sc )107 = 180 000 psi Modification of Sc by YN produces (σ c,all ) P = 130 525 psi (σ c,all )G = 138 069 psi and W3t = 2489 lbf , W4t = 2767 lbf, H3 = 124.3 hp H 4 = 138.2 hp Rating Hrated = min(228, 283, 124, 138) = 124 hp Ans. ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 14 Solutions, Page 26/39 www.konkur.in 14-28 Grade 2, 9310 carburized and case-hardened to 285 core and 580 case in Prob. 14-27. Summary: Table 14-3: 0.99 (St )107 = 65 000 psi (σ all ) P = 53 959 psi (σ all )G = 55 736 psi and it follows that W1t = 5400 lbf , W2t = 6699 lbf , H1 = 270 hp H 2 = 335 hp From Table 14-8, C p = 2300 psi. Also, from Table 14-6: Sc = 225 000 psi (σ c,all ) P = 181 285 psi (σ c,all )G = 191 762 psi Consequently, W3t = 4801 lbf , W4t = 5337 lbf , H 3 = 240 hp H 4 = 267 hp Rating Hrated = min(270, 335, 240, 267) = 240 hp. Ans. ________________________________________________________________________ 14-29 Given: n = 1145 rev/min, Ko = 1.25, NP = 22T, NG = 60T, mG = 2.727, dP = 2.75 in, dG = 7.5 in, YP = 0.331,YG = 0.422, JP = 0.335, JG = 0.405, P = 8T /in, F = 1.625 in, HB = 250, case and core, both gears. Cm = 1, F/dP = 0.0591, Cf = 0.0419, Cpm = 1, Cma = 0.152, Ce = 1, Km = 1.1942, KT = 1, KB = 1, Ks = 1,V = 824 ft/min, (YN )P = 0.8318, (YN )G = 0.859, KR = 1, I = 0.117 58 (St )107 = 32 125 psi (σ all ) P = 26 668 psi (σ all )G = 27 546 psi 0.99 and it follows that W1t = 879.3 lbf , W2t = 1098 lbf , Shigley’s MED, 10th edition H1 = 21.97 hp H 2 = 27.4 hp Chapter 14 Solutions, Page 27/39 www.konkur.in For wear W3t = 304 lbf , W4t = 340 lbf , H 3 = 7.59 hp H 4 = 8.50 hp Rating Hrated = min(21.97, 27.4, 7.59, 8.50) = 7.59 hp In Prob. 14-24, Hrated = 53 hp. Thus, 7.59 1 = 0.1432 = , 53.0 6.98 not 1 8 Ans. The transmitted load rating is t Wrated = min(879.3, 1098, 304, 340) = 304 lbf In Prob. 14-24 t Wrated = 1061 lbf Thus 304 1 1 = 0.2865 = , not Ans. 1061 3.49 4 ________________________________________________________________________ 14-30 SP = SH = 1, Bending Table 14-4: Pd = 4, 0.99 JP = 0.345, JG = 0.410, Ko = 1.25 (St )107 = 13 000 psi 13 000(1) = 13 000 psi 1(1)(1) 13 000(3.25)(0.345) σ all FJ P = = = 1533 lbf K o K v K s Pd K m K B 1.25(1.534)(1)(4)(1.24)(1) 1533(1649) = = 76.6 hp 33 000 = W1t J G / J P = 1533(0.410) / 0.345 = 1822 lbf = H1J G / J P = 76.6(0.410) / 0.345 = 91.0 hp (σ all ) P = (σ all )G = W1t H1 W2t H2 Wear Table 14-8: Table 14-7: C p = 1960 psi 0.99 (Sc )107 = 75 000 psi = (σ c,all ) P = (σ c,all )G Shigley’s MED, 10th edition Chapter 14 Solutions, Page 28/39 www.konkur.in 2 t 3 W W3t W4t H4 (σ ) Fd p I = c,all P C p K o K v K s K mC f 2 75 000 3.25(5.5)(0.1176) = = 1295 lbf 1960 1.25(1.534)(1)(1.24)(1) = W3t = 1295 lbf 1295(1649) = H3 = = 64.7 hp 33 000 Rating Hrated = min(76.7, 94.4, 64.7, 64.7) = 64.7 hp Ans. Notice that the balance between bending and wear power is improved due to CI’s more favorable Sc/St ratio. Also note that the life is 107 pinion revolutions which is (1/300) of 3(109). Longer life goals require power de-rating. ________________________________________________________________________ 14-31 From Table A-24a, Eav = 11.8(106) Mpsi For φ = 14.5° and HB = 156 SC = 1.4(81) = 51 693 psi 2sin14.5° / [11.8(106 )] For φ = 20° 1.4(112) = 52 008 psi 2sin 20° / [11.8(106 )] SC = 0.32(156) = 49.9 kpsi The first two calculations were approximately 4 percent higher. ________________________________________________________________________ SC = 14-32 Programs will vary. ________________________________________________________________________ 14-33 (YN ) P = 0.977, (YN )G = 0.996 (St ) P = (St )G = 82.3(250) + 12 150 = 32 725 psi 32 725(0.977) (σ all ) P = = 37 615 psi 1(0.85) 37 615(1.5)(0.423) W1t = = 1558 lbf 1(1.404)(1.043)(8.66)(1.208)(1) 1558(925) H1 = = 43.7 hp 33 000 Shigley’s MED, 10th edition Chapter 14 Solutions, Page 29/39 www.konkur.in 32 725(0.996) = 38 346 psi 1(0.85) 38 346(1.5)(0.5346) W2t = = 2007 lbf 1(1.404)(1.043)(8.66)(1.208)(1) 2007(925) H2 = = 56.3 hp 33 000 (Z N ) P = 0.948, (Z N )G = 0.973 (σ all )G = Table 14-6: 0.99 (Sc )107 = 150 000 psi 0.948(1) (σ c,allow ) P = 150 000 = 167 294 psi 1(0.85) 2 167 294 1.963(1.5)(0.195) t W3 = = 2074 lbf 2300 1(1.404)(1.043) 2074(925) H3 = = 58.1 hp 33 000 0.973 (σ c ,allow )G = (167 294) = 171 706 psi 0.948 2 171 706 1.963(1.5)(0.195) t W4 = = 2167 lbf 2300 1(1.404)(1.052) 2167(925) H4 = = 60.7 hp 33 000 H rated = min(43.7, 56.3, 58.1, 60.7) = 43.7 hp Ans. Pinion bending is controlling. ________________________________________________________________________ (YN ) P = 1.6831(108 ) −0.0323 = 0.928 14-34 (YN )G = 1.6831(108 / 3.059) −0.0323 = 0.962 Table 14-3: St = 55 000 psi 55 000(0.928) (σ all ) P = = 60 047 psi 1(0.85) 60 047(1.5)(0.423) W1t = = 2487 lbf 1(1.404)(1.043)(8.66)(1.208)(1) 2487(925) H1 = = 69.7 hp 33 000 0.962 (σ all )G = (60 047) = 62 247 psi 0.928 Shigley’s MED, 10th edition Chapter 14 Solutions, Page 30/39 www.konkur.in 62 247 0.5346 (2487) = 3258 lbf 60 047 0.423 3258 H2 = (69.7) = 91.3 hp 2487 W2t = Table 14-6: Sc = 180 000 psi (Z N ) P = 2.466(108 )−0.056 = 0.8790 (Z N )G = 2.466(108 / 3.059)−0.056 = 0.9358 180 000(0.8790) (σ c,all ) P = = 186 141 psi 1(0.85) 2 186 141 1.963(1.5)(0.195) W3t = = 2568 lbf 2300 1(1.404)(1.043) 2568(925) H3 = = 72.0 hp 33 000 0.9358 (σ c ,all )G = (186 141) = 198 169 psi 0.8790 2 198 169 1.043 W4t = (2568) = 2886 lbf 186 141 1.052 2886(925) H4 = = 80.9 hp 33 000 H rated = min(69.7, 91.3, 72, 80.9) = 69.7 hp Ans. Pinion bending controlling ________________________________________________________________________ 14-35 (YN)P = 0.928, (YN )G = 0.962 Table 14-3: (See Prob. 14-34) St = 65 000 psi 65 000(0.928) (σ all ) P = = 70 965 psi 1(0.85) 70 965(1.5)(0.423) W1t = = 2939 lbf 1(1.404)(1.043)(8.66)(1.208) 2939(925) H1 = = 82.4 hp 33 000 65 000(0.962) (σ all )G = = 73 565 psi 1(0.85) 73 565 0.5346 W2t = (2939) = 3850 lbf 70 965 0.423 3850 H2 = (82.4) = 108 hp 2939 Shigley’s MED, 10th edition Chapter 14 Solutions, Page 31/39 www.konkur.in Table 14-6: Sc = 225 000 psi (Z N ) P = 0.8790, (Z N )G = 0.9358 225 000(0.879) (σ c,all ) P = = 232 676 psi 1(0.85) 2 232 676 1.963(1.5)(0.195) W3t = = 4013 lbf 2300 1(1.404)(1.043) 4013(925) H3 = = 112.5 hp 33 000 0.9358 (σ c ,all )G = (232 676) = 247 711 psi 0.8790 2 247 711 1.043 W4t = (4013) = 4509 lbf 232 676 1.052 4509(925) H4 = = 126 hp 33 000 H rated = min(82.4, 108, 112.5, 126) = 82.4 hp Ans. The bending of the pinion is the controlling factor. ________________________________________________________________________ P = 2 teeth/in, d = 8 in, N = dP = 8 (2) = 16 teeth π π F = 4 p = 4 = 4 = 2π P 2 ∑ M x = 0 = 10(300) cos 20o − 4FB cos20o FB = 750 lbf W t = FB cos 20o = 750cos 20o = 705 lbf n = 2400 / 2 = 1200 rev/min π dn π (8)(1200) V = = = 2513 ft/min 12 12 14-36 We will obtain all of the needed factors, roughly in the order presented in the textbook. Fig. 14-2: St = 102(300) + 16 400 = 47 000 psi Fig. 14-5: Fig. 14-6: Sc = 349(300) + 34 300 = 139 000 psi J = 0.27 cos 20o sin 20o 2 I = = 0.107 2(1) 2 + 1 Eq. (14-23): Table 14-8: C p = 2300 psi Assume a typical quality number of 6. Eq. (14-28): B = 0.25(12 − Qv )2 / 3 = 0.25(12 − 6)2/ 3 = 0.8255 Shigley’s MED, 10th edition Chapter 14 Solutions, Page 32/39 www.konkur.in A = 50 + 56(1 − B) = 50 + 56(1 − 0.8255) = 59.77 B Eq. (14-27): A+ V 59.77 + 2513 K v = = A 59.77 0.8255 = 1.65 To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296. From Eq. (a), Sec. 14-10, 0.0535 0.0535 F Y 2π 0.296 K s = 1.192 = 1.192 = 1.23 P 2 The load distribution factor is applicable for straddle-mounted gears, which is not the case here since the gear is mounted outboard of the bearings. Lacking anything better, we will use the load distribution factor as a rough estimate. Eq. (14-31): Eq. (14-32): Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30): Cmc = 1 (uncrowned teeth) 2π Cp f = − 0.0375 + 0.0125(2π ) = 0.1196 10(8) Cpm = 1.1 Cma = 0.23 (commercial enclosed gear unit) Ce = 1 K m = 1 + 1[0.1196(1.1) + 0.23(1)] = 1.36 For the stress-cycle factors, we need the desired number of load cycles. Fig. 14-14: Fig. 14-15: Eq. 14-38: N = 15 000 h (1200 rev/min)(60 min/h) = 1.1 (109) rev YN = 0.9 ZN = 0.8 K R = 0.658 − 0.0759 ln (1 − R ) = 0.658 − 0.0759ln (1 − 0.95) = 0.885 With no specific information given to indicate otherwise, assume Ko = KB = KT = Cf = 1 Tooth bending Eq. (14-15): Eq. (14-41): Pd K m K B F J 2 = 705(1)(1.65)(1.23) 2π σ = W t KoKvKs (1.36)(1) = 2294 psi 0.27 S Y / ( KT K R ) SF = t N σ 47 000(0.9) / [(1)(0.885)] = = 20.8 2294 Shigley’s MED, 10th edition Ans. Chapter 14 Solutions, Page 33/39 www.konkur.in Tooth wear 1/ 2 Eq. (14-16): K C σ c = C p W t Ko Kv Ks m f dPF I 1/ 2 1.36 1 = 2300 705(1)(1.65)(1.23) 8(2π ) 0.107 = 43 750 psi Since gear B is a pinion, CH is not used in Eq. (14-42) (see p. 753), where Sc Z N / ( KT K R ) SH = σc 139 000(0.8) / [(1)(0.885)] = = 2.9 Ans 43 750 ________________________________________________________________________ m = 18.75 mm/tooth, d = 300 mm N = d/m = 300 / 18.75 = 16 teeth F = b = 4 p = 4 (π m ) = 4π (18.75) = 236 mm 14-37 ∑M = 0 = 300(11) cos 20o − 150 FB cos25o FB = 22.81 kN W t = FB cos 25o = 22.81cos 25o = 20.67 kN n = 1800 / 2 = 900 rev/min π dn π (0.300)(900) V = = = 14.14 m/s 60 60 We will obtain all of the needed factors, roughly in the order presented in the textbook. Fig. 14-2: Fig. 14-5: Fig. 14-6: Eq. (14-23): Table 14-8: x St = 0.703(300) + 113 = 324 MPa Sc = 2.41(300) + 237 = 960 MPa J = YJ = 0.27 cos 20o sin 20o 5 I = ZI = = 0.134 2(1) 5 + 1 Z E = 191 MPa Assume a typical quality number of 6. Eq. (14-28): B = 0.25(12 − Qv ) 2/ 3 = 0.25(12 − 6)2 / 3 = 0.8255 A = 50 + 56(1 − B) = 50 + 56(1 − 0.8255) = 59.77 Eq. (14-27): A + 200V K v = A B 59.77 + 200(14.14) = 59.77 0.8255 = 1.69 To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.296. Shigley’s MED, 10th edition Chapter 14 Solutions, Page 34/39 www.konkur.in Similar to Eq. (a) of Sec. 14-10 but for SI units: Ks = ( 1 = 0.8433 mF Y kb ) 0.0535 K s = 0.8433 18.75(236) 0.296 0.0535 = 1.28 Convert the diameter and facewidth to inches for use in the load-distribution factor equations. d = 300/25.4 = 11.81 in, F = 236/25.4 = 9.29 in Eq. (14-31): Eq. (14-32): Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30): Cmc = 1 (uncrowned teeth) 9.29 C pf = − 0.0375 + 0.0125(9.29) = 0.1573 10(11.81) Cpm = 1.1 Cma = 0.27 (commercial enclosed gear unit) Ce = 1 K m = K H = 1 + 1[0.1573(1.1) + 0.27(1)] = 1.44 For the stress-cycle factors, we need the desired number of load cycles. N = 12 000 h (900 rev/min)(60 min/h) = 6.48 (108) rev Fig. 14-14: Fig. 14-15: Eq. 14-38: YN = 0.9 ZN = 0.85 K R = 0.658 − 0.0759 ln (1 − R ) = 0.658 − 0.0759ln (1 − 0.98) = 0.955 With no specific information given to indicate otherwise, assume Ko = KB = KT = ZR = 1. Tooth bending Eq. (14-15): σ = W t KoKvKs 1 KH KB bmt YJ (1.44)(1) 1 = 20 670(1)(1.69)(1.28) = 53.9 MPa 236(18.75) 0.27 Eq. (14-41): S Y / ( KT K R ) SF = t N σ 324(0.9) / [(1)(0.955)] = = 5.66 53.9 Ans. Tooth wear 1/ 2 Eq. (14-16): K Z σ c = Z E W t KoKvKs H R d w1b Z I Shigley’s MED, 10th edition Chapter 14 Solutions, Page 35/39 www.konkur.in 1/ 2 1.44 1 = 191 20 670(1)(1.69)(1.28) 300(236) 0.134 = 498 MPa Since gear B is a pinion, CH is not used in Eq. (14-42) (see p. 757), where SH = Sc Z N / ( KT K R ) σc 960(0.85) / [(1)(0.955)] = = 1.72 Ans 498 ________________________________________________________________________ 14-38 From the solution to Prob. 13-40, n = 191 rev/min, Wt = 1600 N, d = 125 mm, N = 15 teeth, m = 8.33 mm/tooth. F = b = 4 p = 4 (π m ) = 4π ( 8.33) = 105 mm V = π dn 60 = π (0.125)(191) = 1.25 m/s 60 We will obtain all of the needed factors, roughly in the order presented in the textbook. Table 14-3: Table 14-6: Fig. 14-6: Eq. (14-23): Table 14-8: St = 65 kpsi = 448 MPa Sc = 225 kpsi = 1550 MPa J = YJ = 0.25 cos 20o sin 20o 2 I = ZI = = 0.107 2(1) 2 + 1 Z E = 191 MPa Assume a typical quality number of 6. Eq. (14-28): Eq. (14-27): B = 0.25(12 − Qv )2 / 3 = 0.25(12 − 6)2/ 3 = 0.8255 A = 50 + 56(1 − B) = 50 + 56(1 − 0.8255) = 59.77 A + 200V K v = A B 59.77 + 200(1.25) = 59.77 0.8255 = 1.21 To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290. Similar to Eq. (a) of Sec. 14-10 but for SI units: Ks = ( 1 = 0.8433 mF Y kb ) 0.0535 0.0535 K s = 0.8433 8.33(105) 0.290 = 1.17 Convert the diameter and facewidth to inches for use in the load-distribution factor Shigley’s MED, 10th edition Chapter 14 Solutions, Page 36/39 www.konkur.in equations. d = 125/25.4 = 4.92 in, F = 105/25.4 = 4.13 in Eq. (14-31): Cmc = 1 (uncrowned teeth) 4.13 Eq. (14-32): C pf = − 0.0375 + 0.0125(4.13) = 0.0981 10(4.92) Eq. (14-33): Cpm = 1 Fig. 14-11: Cma = 0.32 (open gearing) Eq. (14-35): Ce = 1 Eq. (14-30): K m = K H = 1 + 1[0.0981(1) + 0.32(1)] = 1.42 For the stress-cycle factors, we need the desired number of load cycles. N = 12 000 h (191 rev/min)(60 min/h) = 1.4 (108) rev Fig. 14-14: YN = 0.95 Fig. 14-15: ZN = 0.88 Eq. 14-38: K R = 0.658 − 0.0759 ln (1 − R ) = 0.658 − 0.0759ln (1 − 0.95) = 0.885 With no specific information given to indicate otherwise, assume Ko = KB = KT = ZR = 1. Tooth bending Eq. (14-15): σ = W t KoKvKs 1 KH KB bmt YJ (1.42)(1) 1 = 1600(1)(1.21)(1.17) = 14.7 MPa 105(8.33) 0.25 Since gear is a pinion, CH is not used in Eq. (14-42) (see p. 753), where S Y / ( KT K R ) SF = t N σ 448(0.95) / [(1)(0.885)] = = 32.7 14.7 Ans. Tooth wear 1/ 2 Eq. (14-16): K Z σ c = Z E W t KoKvKs H R d w1b Z I 1/ 2 1.42 1 = 191 1600(1)(1.21)(1.17) 125(105) 0.107 = 289 MPa S Z / ( KT K R ) SH = c N σc 1550(0.88) / [(1)(0.885)] = = 5.33 Ans 289 ________________________________________________________________________ Eq. (14-42): Shigley’s MED, 10th edition Chapter 14 Solutions, Page 37/39 www.konkur.in 14-39 From the solution to Prob. 13-41, n = 2(70) = 140 rev/min, Wt = 180 lbf, d = 5 in N = 15 teeth, P = 3 teeth/in. π π F = 4 p = 4 = 4 = 4.2 in P 3 π dn π (5)(140) V = = = 183.3 ft/min 12 12 We will obtain all of the needed factors, roughly in the order presented in the textbook. Table 14-3: Table 14-6: Fig. 14-6: Eq. (14-23): Table 14-8: St = 65 kpsi Sc = 225 kpsi J = 0.25 cos 20o sin 20o 2 I = = 0.107 2(1) 2 + 1 C p = 2300 psi Assume a typical quality number of 6. Eq. (14-28): B = 0.25(12 − Qv )2 / 3 = 0.25(12 − 6)2/ 3 = 0.8255 A = 50 + 56(1 − B) = 50 + 56(1 − 0.8255) = 59.77 B Eq. (14-27): A+ V 59.77 + 183.3 K v = = A 59.77 0.8255 = 1.18 To estimate a size factor, get the Lewis Form Factor from Table 14-2, Y = 0.290. From Eq. (a), Sec. 14-10, 0.0535 Eq. (14-31): Eq. (14-32): Eq. (14-33): Fig. 14-11: Eq. (14-35): Eq. (14-30): F Y 4.2 0.290 K s = 1.192 = 1.192 3 P Cmc = 1 (uncrowned teeth) 4.2 C pf = − 0.0375 + 0.0125(4.2) = 0.099 10(5) Cpm = 1 Cma = 0.32 (Open gearing) Ce = 1 K m = 1 + 1[0.099(1) + 0.32(1)] = 1.42 0.0535 = 1.17 For the stress-cycle factors, we need the desired number of load cycles. N = 14 000 h (140 rev/min)(60 min/h) = 1.2 (108) rev Fig. 14-14: YN = 0.95 Fig. 14-15: ZN = 0.88 K R = 0.658 − 0.0759 ln (1 − R ) = 0.658 − 0.0759ln (1 − 0.98) = 0.955 Eq. 14-38: With no specific information given to indicate otherwise, assume Ko = KB = KT = Cf = 1. Shigley’s MED, 10th edition Chapter 14 Solutions, Page 38/39 www.konkur.in Tooth bending Eq. (14-15): Eq. (14-41): Pd K m K B F J 3 (1.42)(1) = 180(1)(1.18)(1.17) = 1010 psi 4.2 0.25 σ = W t KoKvKs S Y / ( KT K R ) SF = t N σ 65 000(0.95) / [(1)(0.955)] = = 64.0 1010 Ans. Tooth wear 1/ 2 Eq. (14-16): K C σ c = C p W t Ko Kv Ks m f dPF I 1/ 2 1.42 1 = 2300 180(1)(1.18)(1.17) 5(4.2) 0.107 = 28 800 psi Since gear B is a pinion, CH is not used in Eq. (14-42) (see p. 753), where S Z / ( KT K R ) SH = c N σc 225 000(0.88) / [(1)(0.955)] = = 7.28 Ans 28 800 ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 14 Solutions, Page 39/39 www.konkur.in Chapter 15 15-1 Given: Uncrowned, through-hardened 300 Brinell core and case, Grade 1, NC = 109 rev of pinion at R = 0.999, NP = 20 teeth, NG = 60 teeth, Qv = 6, Pd = 6 teeth/in, normal pressure angle = 20°, shaft angle = 90°, np = 900 rev/min, F = 1.25 in, SF = SH = 1, Ko = 1. Fig. 15-7: JP = 0.249, JG = 0.216 Mesh dP = 20/6 = 3.333 in, dG = 60/6 = 10.000 in Eq. (15-7): vt = π(3.333)(900/12) = 785.3 ft/min Eq. (15-6): B = 0.25(12 – 6)2/3 = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77 Eq. (15-5): 59.77 + 785.3 K v = 59.77 Eq. (15-8): vt,max = [59.77 + (6 – 3)]2 = 3940 ft/min 0.8255 = 1.374 Since 785.3 < 3904, Kv = 1.374 is valid. The size factor for bending is: Eq. (15-10): Ks = 0.4867 + 0.2132 / 6 = 0.5222 For one gear straddle-mounted, the load-distribution factor is: Eq. (15-11): Km = 1.10 + 0.0036 (1.25)2 = 1.106 For KL, the more conservative “critical” equation will be used here. Though it is acceptable, according to AGMA, to use the less conservative “general” equation for general applications. Eq. (15-15): (KL)P = 1.6831(109)–0.0323 = 0.862 (KL)G = 1.6831(109 / 3)–0.0323 = 0.893 Eq. (15-14): (CL)P = 3.4822(109)–0.0602 = 1 (CL)G = 3.4822(109 / 3)–0.0602 = 1.069 Eq. (15-19): KR = 0.50 – 0.25 log(1 – 0.999) = 1.25 (or Table 15-3) CR = K R = 1.25 = 1.118 Shigley’s MED, 10th edition Chapter 15 Solutions, Page 1/20 www.konkur.in Bending St = sat = 44(300) + 2100 = 15 300 psi Fig. 15-13: 0.99 Eq. (15-4): (σ all ) P = sw t = Eq. (15-3): Eq. (15-4): sa t K L = 15 300(0.862) = 10 551 psi 1(1)(1.25) S F KT K R (σ all ) P FK x J P WPt = Pd K o K v K s K m 10 551(1.25)(1)(0.249) = = 690 lbf 6(1)(1.374)(0.5222)(1.106) 690(785.3) H1 = = 16.4 hp 33 000 15 300(0.893) = 10 930 psi 1(1)(1.25) 10 930(1.25)(1)(0.216) WGt = = 620 lbf 6(1)(1.374)(0.5222)(1.106) 620(785.3) H2 = = 14.8 hp Ans. 33 000 (σ all )G = The gear controls the bending rating. ________________________________________________________________________ 15-2 Refer to Prob. 15-1 for the gearset specifications. Wear Fig. 15-12: sac = 341(300) + 23 620 = 125 920 psi For the pinion, CH = 1. From Prob. 15-1, CR = 1.118. Thus, from Eq. (15-2): sac (CL ) P CH S H KT C R 125 920(1)(1) (σ c,all ) P = = 112 630 psi 1(1)(1.118) (σ c,all ) P = For the gear, from Eq. (15-16), B1 = 0.008 98(300 / 300) − 0.008 29 = 0.000 69 CH = 1 + 0.000 69(3 − 1) = 1.001 38 From Prob. 15-1, (CL)G = 1.0685. Equation (15-2) thus gives Shigley’s MED, 10th edition Chapter 15 Solutions, Page 2/20 www.konkur.in sac (CL )G CH S H KT C R 125 920(1.0685)(1.001 38) (σ c,all )G = = 120 511 psi 1(1)(1.118) (σ c,all )G = For steel: C p = 2290 psi Eq. (15-9): Cs = 0.125(1.25) + 0.4375 = 0.593 75 Fig. 15-6: I = 0.083 Eq. (15-12): Cxc = 2 Eq. (15-1): (σ ) Fd P I W = c ,all P C p K o K v K mCsC xc 2 t P 2 1.25(3.333)(0.083) 112 630 = 2290 1(1.374)(1.106)(0.5937)(2) = 464 lbf 464(785.3) H3 = = 11.0 hp 33 000 2 1.25(3.333)(0.083) 120 511 WGt = 2290 1(1.374)(1.106)(0.593 75)(2) = 531 lbf 531(785.3) H4 = = 12.6 hp 33 000 The pinion controls wear: H = 11.0 hp Ans. The power rating of the mesh, considering the power ratings found in Prob. 15-1, is H = min(16.4, 14.8, 11.0, 12.6) = 11.0 hp Ans. ________________________________________________________________________ 15-3 AGMA 2003-B97 does not fully address cast iron gears. However, approximate comparisons can be useful. This problem is similar to Prob. 15-1, but not identical. We will organize the method. A follow-up could consist of completing Probs. 15-1 and 15-2 with identical pinions, and cast iron gears. Given: Uncrowned, straight teeth, Pd = 6 teeth/in, NP = 30 teeth, NG = 60 teeth, ASTM 30 cast iron, material Grade 1, shaft angle 90°, F = 1.25, nP = 900 rev/min, φn = 20°, one gear straddle-mounted, Ko = 1, SF = 2, S H = 2. Fig. 15-7: JP = 0.268, JG = 0.228 Shigley’s MED, 10th edition Chapter 15 Solutions, Page 3/20 www.konkur.in Mesh dP = 30/6 = 5.000 in, dG = 60/6 = 10.000 in vt = π (5)(900 / 12) = 1178 ft/min Set NL = 107 cycles for the pinion. For R = 0.99, Table 15-7: Table 15-5: Eq. (15-4): sat = 4500 psi sac = 50 000 psi sat K L 4500(1) swt = = = 2250 psi S F KT K R 2(1)(1) The velocity factor Kv represents stress augmentation due to mislocation of tooth profiles along the pitch surface and the resulting “falling” of teeth into engagement. Equation (5-67) shows that the induced bending moment in a cantilever (tooth) varies directly with E of the tooth material. If only the material varies (cast iron vs. steel) in the same geometry, I is the same. From the Lewis equation of Section 14-1, M K W tP σ = = v I /c FY We expect the ratio σCI/σsteel to be ( K v )CI ECI σ CI = = Esteel σ steel ( K v )steel In the case of ASTM class 30, from Table A-24(a) (ECI)av = (13 + 16.2)/2 = 14.7 kpsi ( K v )CI = Then, 14.7 ( K v )steel = 0.7( K v )steel 30 Our modeling is rough, but it convinces us that (Kv)CI < (Kv)steel, but we are not sure of the value of (Kv)CI. We will use Kv for steel as a basis for a conservative rating. Eq. (15-6): B = 0.25(12 – 6)2/3 = 0.8255 A = 50 + 56(1 – 0.8255) = 59.77 Eq. (15-5): 59.77 + 1178 K v = 59.77 Pinion bending 0.8255 = 1.454 (σall)P = swt = 2250 psi From Prob. 15-1, Kx = 1, Km = 1.106, Ks = 0.5222 Shigley’s MED, 10th edition Chapter 15 Solutions, Page 4/20 www.konkur.in Eq. (15-3): (σ all ) P FK x J P Pd K o K v K s K m 2250(1.25)(1)(0.268) = = 149.6 lbf 6(1)(1.454)(0.5222)(1.106) 149.6(1178) H1 = = 5.34 hp 33 000 WPt = Gear bending JG 0.228 = 149.6 = 127.3 lbf 0.268 JP 127.3(1178) H2 = = 4.54 hp 33 000 WGt = WPt The gear controls in bending fatigue. H = 4.54 hp Ans. ________________________________________________________________________ 15-4 Continuing Prob. 15-3, Table 15-5: sac = 50 000 psi 50 000 sw t = σ c,all = = 35 355 psi 2 2 Eq. (15-1): σ Fd P I W = c,all C p K o K v K mCsCxc Fig. 15-6: I = 0.86 t From Probs. 15-1 and 15-2: Cs = 0.593 75, Ks = 0.5222, Km = 1.106, Cxc = 2 From Table 14-8: C p = 1960 psi 2 Thus, 1.25(5.000)(0.086) 35 355 Wt = = 91.6 lbf 1960 1(1.454)(1.106)(0.59375)(2) 91.6(1178) Ans. H3 = H 4 = = 3.27 hp 33 000 Rating Side note: Based on results of Probs. 15-3 and 15-4, H = min(5.34, 4.54, 3.27, 3.27) = 3.27 hp The mesh is weakest in wear fatigue. ________________________________________________________________________ Shigley’s MED, 10th edition Chapter 15 Solutions, Page 5/20 www.konkur.in 15-5 Uncrowned, through-hardened to 180 Brinell (core and case), Grade 1, 109 rev of pinion at R = 0.999, NP = z1 = 22 teeth, NG = z2 = 24 teeth, Qv = 5, met = 4 mm, φn = 20°, shaft angle 90°, n1 = 1800 rev/min, SF = 1, S H = S F = 1, F = b = 25 mm, Ko = KA = KT = Kθ = 1 and C p = 190 MPa . Fig. 15-7: JP = YJ1 = 0.23, JG = YJ2 = 0.205 Mesh dP = de1 = mz1 = 4(22) = 88 mm, Eq. (15-7): vet = 5.236(10–5)(88)(1800) = 8.29 m/s Eq. (15-6): B = 0.25(12 – 5)2/3 = 0.9148 A = 50 + 56(1 – 0.9148) = 54.77 dG = met z2 = 4(24) = 96 mm 0.9148 Eq. (15-5): Eq. (15-10): 54.77 + 200(8.29) Kv = = 1.663 54.77 Ks = Yx = 0.4867 + 0.008 339(4) = 0.520 Eq. (15-11): with Kmb = 1 (both straddle-mounted), Km = KHβ = 1 + 5.6(10–6)(252) = 1.0035 From Fig. 15-8, (CL ) P = (Z NT ) P = 3.4822(109 ) −0.0602 = 1.00 (CL )G = (Z NT )G = 3.4822[109 (22 / 24)]−0.0602 = 1.0054 Eq. (15-12): Cxc = Zxc = 2 (uncrowned) Eq. (15-19): KR = YZ = 0.50 – 0.25 log (1 – 0.999) = 1.25 CR = Z Z = YZ = 1.25 = 1.118 From Fig. 15-10, CH = Zw = 1 Eq. (15-9): Zx = 0.004 92(25) + 0.4375 = 0.560 Wear of Pinion Fig. 15-12: σH lim = 2.35HB + 162.89 = 2.35(180) + 162.89 = 585.9 MPa Fig. 15-6: I = ZI = 0.066 Eq. (15-2): (σ H ) P = Shigley’s MED, 10th edition (σ H ) (Z NT ) P ZW S H Kθ Z Z lim P Chapter 15 Solutions, Page 6/20 www.konkur.in = 585.9(1)(1) = 524.1 MPa 1(1)(1.118) 2 σ bd e1Z I Eq. (15-1): W = H C p 1000K A K v K H β Z x Z xc The constant 1000 expresses Wt in kN. t P 2 25(88)(0.066) 524.1 W = = 0.591 kN 190 1000(1)(1.663)(1.0035)(0.56)(2) t π dnW π (88)(1800)(0.591) 1 H3 = = = 4.90 kW 60 000 60 000 t P Eq. (13-36): Wear of Gear σH lim = 585.9 MPa (σ H )G = 585.9(1.0054) = 526.9 MPa 1(1)(1.118) (σ H )G 526.9 = 0.591 = 0.594 kN (σ H ) P 524.1 π (88)(1800)(0.594) H4 = = 4.93 kW 60 000 WGt = WPt Thus in wear, the pinion controls the power rating; H = 4.90 kW Ans. We will rate the gear set after solving Prob. 15-6. ________________________________________________________________________ 15-6 Refer to Prob. 15-5 for terms not defined below. Bending of Pinion ( K L ) P = (YNT ) P = 1.6831(109 ) −0.0323 = 0.862 ( K L )G = (YNT )G = 1.6831[109 (22 / 24)]−0.0323 = 0.864 Fig. 15-13: σF lim = 0.30HB + 14.48 = 0.30(180) + 14.48 = 68.5 MPa Eq. (15-13): Kx = Yβ = 1 From Prob. 15-5: YZ = 1.25, vet = 8.29 m/s, K A = 1, K v = 1.663, Kθ = 1, Yx = 0.52, K H β = 1.0035, YJ 1 = 0.23 Shigley’s MED, 10th edition Chapter 15 Solutions, Page 7/20 www.konkur.in Eq. (5-4): (σ F ) P = Eq. (5-3): WPt = σ F limYNT = 68.5(0.862) = 47.2 MPa 1(1)(1.25) S F Kθ YZ (σ F ) P bmetYβ YJ 1 1000 K A K vYx K H β 47.2(25)(4)(1)(0.23) = 1.25 kN 1000(1)(1.663)(0.52)(1.0035) π ( 88 )(1800 )(1.25 ) H1 = = 10.37 kW 60 000 Bending of Gear = σ F lim = 68.5 MPa 68.5(0.864) = 47.3 MPa 1(1)(1.25) 47.3(25)(4)(1)(0.205) WGt = = 1.12 kN 1000(1)(1.663)(0.52)(1.0035) π ( 88 )(1800 )(1.12 ) H2 = = 9.29 kW 60 000 Rating of mesh is Hrating = min(10.37, 9.29, 4.90, 4.93) = 4.90 kW Ans. with pinion wear controlling. ________________________________________________________________________ (σ F )G = 15-7 (a) σ σ (S F ) P = all = (S F )G = all σ P σ G ( sat K L / KT K R ) P ( sat K L / KT K R )G = t (W Pd K o K v K s K m / FK x J ) P (W Pd K o K v K s K m / FK x J )G t All terms cancel except for sat , KL , and J, (sat)P(KL)P JP = (sat )G(KL)G JG From which (sat )G = (sat ) P ( K L ) P J P J = (sat ) P P mGβ ( K L )G J G JG where β = – 0.0178 or β = – 0.0323 as appropriate. This equation is the same as Eq. (14-44). Ans. (b) In bending Shigley’s MED, 10th edition Chapter 15 Solutions, Page 8/20 www.konkur.in σ sat K L FK x J FK x J W t = all = S F Pd K o K v K s K m 11 S F KT K R Pd K o K v K s K m 11 (1) In wear 1/ 2 sacCLCU W t K o K v K mCsCxc = Cp Fd P I S H KT CR 22 22 Squaring and solving for Wt gives s 2 C 2C 2 Fd P I W t = 2 ac 2L 2H 2 S H KT CRCP 22 K o K v K mCsC xc 22 (2) Equating the right-hand sides of Eqs. (1) and (2) and canceling terms, and recognizing that CR = K R and PddP = NP, we obtain (sac )22 = S H2 (sat )11( K L )11 K x J11KT CsCxc SF CH2 N P K s I Cp (CL )22 For equal Wt in bending and wear S H2 = SF ( SF ) 2 SF =1 So we get (sac )G = Cp (CL )G CH (sat ) P ( K L ) P J P K x KT CsCxc N P IK s Ans. (c) σ σ (S H ) P = (S H )G = c,all = c,all σc P σc G Substituting in the right-hand equality gives [sacCL / (CR KT )]P C W K K K C C / ( Fd I ) o v m s xc P p P t = [sacCLCH / (CR KT )]G C W t K K K C C / ( Fd I ) o v m s xc P p G Denominators cancel, leaving (sac)P(CL)P = (sac)G(CL)GCH Solving for (sac)P gives, Shigley’s MED, 10th edition Chapter 15 Solutions, Page 9/20 www.konkur.in ( sac ) P = ( sac )G (CL )G CH (CL ) P (1) From Eq. (15-14), ( C L ) P = 3.4822 N L−0.0602 and ( C L )G = 3.4822 ( N L / mG ) Thus, −0.0602 Ans. C H = ( sac )G mG0.0602C H ( sac ) P = ( sac )G (1 mG ) −0.0602 . This equation is the transpose of Eq. (14-45). ________________________________________________________________________ 15-8 Given (HB)11 = 300 Brinell Eq. (15-23): (sat )P = 44(300) + 2100 = 15 300 psi J P −0.0323 0.249 −0.0323 mG = 15 300 ) = 17 023 psi (3 JG 0.216 17 023 − 2100 ( H B ) 21 = = 339 Brinell Ans. 44 2290 15 300(0.862)(0.249)(1)(0.593 25)(2) (sac )G = 1.0685(1) 20(0.086)(0.5222) = 141 160 psi 141 160 − 23 600 ( H B ) 22 = = 345 Brinell Ans. 341 (sac ) P = (sac )G mG0.0602CH ≈ 141 160(30.0602 ) (1) = 150 811 psi (sat )G = (sat ) P ( H B )12 = 150 811 − 23 600 = 373 Brinell 341 Ans. Core Case Pinion (HB)11 = 300 (HB)12 = 373 Ans. Gear (HB)21 = 339 (HB)22 = 345 _______________________________________________________________________ 15-9 Pinion core (sat ) P = 44 ( 300 ) + 2 100 = 15 300 psi (σ all ) P = Wt = Shigley’s MED, 10th edition 15 300 ( 0.862 ) = 10 551 psi 1(1)(1.25 ) 10 551(1.25 )( 0.249 ) = 689.7 lbf 6 (1)(1.374 )( 0.5222 )(1.106 ) Chapter 15 Solutions, Page 10/20 www.konkur.in Gear core (sat )G = 44(339) + 2100 = 17 016 psi 17 016(0.893) (σ all )G = = 12 156 psi 1(1)(1.25) 12 156(1.25)(0.216) Wt = = 689.3 lbf 6(1)(1.374)(0.5222)(1.106) Pinion case (sac ) P = 341(373) + 23 620 = 150 813 psi 150 813(1) (σ c,all ) P = = 134 895 psi 1(1)(1.118) 134 895 Wt = 2290 2 1.25(3.333)(0.086) 1(1.374)(1.106)(0.593 75)(2) = 689.0 lbf Gear case (sac )G = 341(345) + 23 620 = 141 265 psi 141 265(1.0685)(1) (σ c,all )G = = 135 010 psi 1(1)(1.118) 2 1.25(3.333)(0.086) 135 010 Wt = = 690.1 lbf 2290 1(1.1374)(1.106)(0.593 75)(2) All four transmitted loads are essentially equal, so the equations developed within Prob. 15-7 are effective. Ans. ________________________________________________________________________ 15-10 The catalog rating is 5.2 hp at 1200 rev/min for a straight bevel gearset. Also given: NP = 20 teeth, NG = 40 teeth, φn = 20°, F = 0.71 in, JP = 0.241, JG = 0.201, Pd = 10 teeth/in, through-hardened to 300 Brinell-General Industrial Service, and Qv = 5 uncrowned. Mesh d P = 20 / 10 = 2.000 in, dG = 40 / 10 = 4.000 in π d P nP π (2)(1200) vt = = = 628.3 ft/min 12 12 K o = 1, S F = 1, S H = 1 Eq. (15-6): B = 0.25(12 – 5)2/3 = 0.9148 A = 50 + 56(1 – 0.9148) = 54.77 0.9148 Eq. (15-5): Eq. (15-10): 54.77 + 628.3 K v = = 1.412 54.77 Ks = 0.4867 + 0.2132/10 = 0.508 Shigley’s MED, 10th edition Chapter 15 Solutions, Page 11/20 www.konkur.in Eq. (15-11): Km = 1.25 + 0.0036(0.71)2 = 1.252, where Kmb = 1.25 Eq. (15-15): (KL)P = 1.3558(3∙106)–0.0178 = 1.040 (KL)G = 1.3558 (3∙106/2)–0.0178 = 1.053 Eq. (15-14): (CL)P = 3.4822(3∙106)–0.0602 = 1.419 (CL)G = 3.4822(3∙106/2)–0.0602 = 1.479 Eq. (15-19): KR = 0.50 – 0.25 log(1 – 0.99) = 1.0 CR = K R = 1.0 Bending Pinion: Eq. (15-23): (sat )P = 44(300) + 2100 = 15 300 psi Eq. (15-4): (swt ) P = Eq. (15-3): Wt = Gear: (sat )G = 15 300 psi 15 300(1.053) ( sw t ) G = = 16 111 psi 1(1)(1) 16 111(0.71)(1)(0.201) Wt = = 256.0 lbf 10(1)(1.412)(0.508)(1.252) 256.0(628.3) H2 = = 4.9 hp 33 000 Eq. (15-4): Eq. (15-3): 15 300(1.040) = 15 912 psi 1(1)(1) (sw t ) P FK x J P Pd K o K v K s K m 15 912(0.71)(1)(0.241) = = 303.2 lbf 10(1)(1.412)(0.508)(1.252) 303.2(628.3) H1 = = 5.8 hp 33 000 Wear Pinion: (CH )G = 1, I = 0.078, C p = 2290 psi, Cs = 0.125(0.71) + 0.4375 = 0.526 25 Eq. (15-22): Eq. (15-2): Cxc = 2 (sac)P = 341(300) + 23 620 = 125 920 psi 125 920(1.419)(1) (σ c,all ) P = = 178 680 psi 1(1)(1) Shigley’s MED, 10th edition Chapter 15 Solutions, Page 12/20 www.konkur.in 2 Eq. (15-1): (σ ) Fd P I W = c,all P C p K o K v K mCsC xc 2 0.71(2.000)(0.078) 178 680 = 2290 1(1.412)(1.252)(0.526 25)(2) = 362.4 lbf t H3 = 362.4(628.3) = 6.9 hp 33 000 Gear: (sac )G = 125 920 psi 125 920(1.479)(1) (σ c,all ) = = 186 236 psi 1(1)(1) 2 0.71(2.000)(0.078) 186 236 W = = 393.7 lbf 2290 1(1.412)(1.252)(0.526 25)(2) 393.7(628.3) H4 = = 7.5 hp 33 000 t Rating: H = min(5.8, 4.9, 6.9, 7.5) = 4.9 hp Pinion wear controls the power rating. The catalog rating of 5.2 hp is slightly higher than predicted here, but seems reasonable. Ans. ________________________________________________________________________ 15-11 From Ex. 15-1, the core hardness of both the pinion and gear is 180 Brinell. So (HB)11 and (HB)21 are 180 Brinell and the bending stress numbers are: (sat ) P = 44(180) + 2100 = 10 020 psi (sat )G = 10 020 psi The contact strength of the gear case, based upon the equation derived in Prob. 15-7, is Cp S H2 (sat ) P ( K L ) P K x J P KT CsCxc (sac )G = (CL )G CH S F N P IK s Substituting (sat)P from above and the values of the remaining terms from Ex. 15-1, Shigley’s MED, 10th edition Chapter 15 Solutions, Page 13/20 www.konkur.in 2290 1.52 10 020(1)(1)(0.216)(1)(0.575)(2) 1.32(1) 1.5 25(0.065)(0.529) = 114 331 psi 114 331 − 23 620 ( H B ) 22 = = 266 Brinell 341 ( sac )G = The pinion contact strength is found using the relation from Prob. 15-7: (sac ) P = (sac )G mG0.0602CH = 114 331(1)0.0602 (1) = 114 331 psi 114 331 − 23 600 ( H B )12 = = 266 Brinell 341 Core Case Pinion 180 266 Gear 180 266 Realization of hardnesses The response of students to this part of the question would be a function of the extent to which heat-treatment procedures were covered in their materials and manufacturing prerequisites, and how quantitative it was. The most important thing is to have the student think about it. The instructor can comment in class when students’ curiosity is heightened. Options that will surface may include: (a) Select a through-hardening steel which will meet or exceed core hardness in the hot-rolled condition, then heat-treating to gain the additional 86 points of Brinell hardness by bath-quenching, then tempering, then generating the teeth in the blank. (b) Flame or induction hardening are possibilities. (c) The hardness goal for the case is sufficiently modest that carburizing and case hardening may be too costly. In this case the material selection will be different. (d)The initial step in a nitriding process brings the core hardness to 33–38 Rockwell C-scale (about 300–350 Brinell), which is too much. ________________________________________________________________________ 15-12 Computer programs will vary. ________________________________________________________________________ 15-13 A design program would ask the user to make the a priori decisions, as indicated in Sec. 15-5, p. 798, of the text. The decision set can be organized as follows: Shigley’s MED, 10th edition Chapter 15 Solutions, Page 14/20 www.konkur.in A priori decisions: • Function: H, Ko, rpm, mG, temp., NL, R • Design factor: nd (SF = nd , S H = nd ) • Tooth system: Involute, Straight Teeth, Crowning, φn • Straddling: Kmb • Tooth count: NP (NG = mGNP) Design decisions: • Pitch and Face: Pd , F • Quality number: Qv • Pinion hardness: (HB)1, (HB)3 • Gear hardness: (HB)2, (HB)4 First, gather all of the equations one needs, then arrange them before coding. Find the required hardnesses, express the consequences of the chosen hardnesses, and allow for revisions as appropriate. Shigley’s MED, 10th edition Chapter 15 Solutions, Page 15/20 www.konkur.in Pinion Bending Load-induced stress (Allowable stress) Tabulated strength Associated hardness W t PK o K v K m K s st = = s11 FK x J P ( sat ) P = Factor of safety W t PK o K v K m K s st = = s21 FK x J G ( sat )G = s21S F KT K R ( K L )G Pinion Wear Gear Wear 1/ 2 W K o K vCsC xc Fd P I σ c = Cp t ( sac ) P = = s12 s22 = s12 s12 S H KT CR (CL ) P (CH ) P ( sac )G = s22 S H KT CR (CL )G (CH )G (sat ) P − 2100 44 Bhn = (sat ) P − 5980 48 (sat )G − 2100 44 Bhn = (sat )G − 5980 48 (sac ) P − 23 620 341 Bhn = (sac ) P − 29 560 363.6 (sac ) P − 23 620 341 Bhn = (sac ) P − 29 560 363.6 (HB)11 (HB)21 (HB)12 (HB)22 44( H B )11 + 2100 (sat1) P = 48( H B )11 + 5980 44( H B ) 21 + 2100 (sat1)G = 48( H B )21 + 5980 341( H B )12 + 23 620 ( sac1) P = 363.6( H B )12 + 29 560 341( H B ) 22 + 23 620 (sac1)G = 363.6( H B ) 22 + 29 560 σ (s ) ( K ) n11 = all = at1 P L P s11KT K R σ (s ) ( K ) n21 = at1 G L G s21KT K R (s ) (C ) (C ) n12 = ac1 P L P H P s12 KT CR Chosen hardness New tabulated strength s11S F KT K R (K L )P Gear Bending Note: S F = nd , S H = Shigley’s MED, 10th edition SF Chapter 15 Solutions, Page 16/20 2 n22 (s ) (C ) (C ) = ac1 G L G H G s22 KT CR 2 www.konkur.in 15-14 NW = 1, NG = 56, Pt = 8 teeth/in, d = 1.5 in, Ho = 1hp, φn = 20°, ta = 70°F, Ka = 1.25, nd = 1, Fe = 2 in, A = 850 in2 (a) mG = NG/NW = 56, dG = NG/Pt = 56/8 = 7.0 in px = π / 8 = 0.3927 in, C = 1.5 + 7 = 8.5 in Eq. (15-39): a = px / π = 0.3927 / π = 0.125 in Eq. (15-40): b = 0.3683 px = 0.1446 in Eq. (15-41): ht = 0.6866 px = 0.2696 in Eq. (15-42): do = 1.5 + 2(0.125) = 1.75 in Eq. (15-43): dr = 3 – 2(0.1446) = 2.711 in Eq. (15-44): Dt = 7 + 2(0.125) = 7.25 in Eq. (15-45): Dr = 7 – 2(0.1446) = 6.711 in Eq. (15-46): c = 0.1446 – 0.125 = 0.0196 in Eq. (15-47): ( FW ) max = 2 2 ( 7 )( 0.125 ) = 2.646 in Eq. (13-27): VW = π (1.5)(1725 /12) = 677.4 ft/min π (7)(1725 / 56) VG = = 56.45 ft/min 12 L = px NW = 0.3927 in Eq. (13-28): 0.3927 o λ = tan −1 = 4.764 π (1.5) Pn = pn = Eq. (15-62): (b) Eq. (15-38): Pt 8 = = 8.028 cos λ cos 4.764° π = 0.3913 in Pn π (1.5)(1725) Vs = = 679.8 ft/min 12 cos 4.764° f = 0.103exp −0.110(679.8) 0.450 + 0.012 = 0.0250 Eq. (15-54): cos φn − f tan λ cos 20° − 0.0250 tan 4.764° e= = = 0.7563 cos φn + f cot λ cos 20° + 0.0250 cot 4.764° Shigley’s MED, 10th edition Ans. Chapter 15 Solutions, Page 17/20 www.konkur.in Eq. (15-58): Eq. (15-57): 33 000nd H o K a 33 000(1)(1)(1.25) = = 966 lbf VG e 56.45(0.7563) cos φn sin λ + f cos λ WWt = WGt cos φn cos λ − f sin λ WGt = Ans. cos 20° sin 4.764° + 0.025cos 4.764° = 966 cos 20° cos 4.764° − 0.025sin 4.764° = 106.4 lbf Ans. (c) Eq. (15-33): Cs = 1190 – 477 log 7.0 = 787 Eq. (15-36): Cm = 0.0107 −56 2 + 56(56) + 5145 = 0.767 Eq. (15-37): Cv = 0.659 exp[−0.0011(679.8)] = 0.312 Eq. (15-38): (Wt)all = 787(7)0.8(2)(0.767)(0.312) = 1787 lbf Since WGt < (W t )all , the mesh will survive at least 25 000 h. Eq. (15-61): Eq. (15-63): 0.025(966) = −29.5 lbf 0.025sin 4.764° − cos 20° cos 4.764° 29.5(679.8) Hf = = 0.608 hp 33 000 106.4(677.4) HW = = 2.18 hp 33 000 966(56.45) HG = = 1.65 hp 33 000 Wf = The mesh is sufficient Ans. Pn = Pt / cos λ = 8 / cos 4.764o = 8.028 pn = π / 8.028 = 0.3913 in 966 = 39 500 psi σG = 0.3913(0.5)(0.125) The stress is high. At the rated horsepower, σG = Shigley’s MED, 10th edition 1 39 500 = 23 940 psi 1.65 acceptable Chapter 15 Solutions, Page 18/20 www.konkur.in (d) Eq. (15-52): Amin = 43.2(8.5)1.7 = 1642 in2 < 1700 in2 Eq. (15-49): Hloss = 33 000(1 – 0.7563)(2.18) = 17 530 ft · lbf/min Assuming a fan exists on the worm shaft, 1725 + 0.13 = 0.568 ft · lbf/(min · in 2 · o F) 3939 17 530 Eq. (15-51): ts = 70 + = 88.2o F Ans. 0.568(1700) ________________________________________________________________________ Eq. (15-50): h CR = Shigley’s MED, 10th edition Chapter 15 Solutions, Page 19/20 www.konkur.in 15-15 to 15-22 Problem statement values of 25 hp, 1125 rev/min, mG = 10, Ka = 1.25, nd = 1.1, φn = 20°, ta = 70°F are not referenced in the table. The first four parameters listed in the table were selected as design decisions. px dW FG A HW HG Hf NW NG KW Cs Cm Cv VG WGt t W W f e (Pt)G Pn C-to-C ts L λ σG dG 15-15 1.75 3.60 2.40 2000 15-16 1.75 3.60 1.68 2000 15-17 1.75 3.60 1.43 2000 15-18 1.75 3.60 1.69 2000 15-19 1.75 3.60 2.40 2000 15-20 1.75 4.10 2.25 2000 38.2 36.2 1.87 3 30 38.2 36.2 1.47 3 30 38.2 36.2 1.97 3 30 38.2 36.2 1.97 3 30 125 38.2 36.2 1.97 3 30 80 607 0.759 0.236 492 854 0.759 0.236 492 1000 0.759 0.236 492 492 2430 2430 2430 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 5103 16.71 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 7290 16.71 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 8565 16.71 Shigley’s MED, 10th edition 38.0 36.1 1.85 3 30 50 15-21 1.75 3.60 2.4 2500 FAN 41.2 37.7 3.59 3 30 115 15-22 1.75 3.60 2.4 2600 FAN 41.2 37.7 3.59 3 30 185 492 563 492 492 2430 2430 2120 2524 2524 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 7247 16.71 1189 0.0193 0.948 1.795 1.979 10.156 177 5.25 24.9 5103 16.71 1038 0.0183 0.951 1.571 1.732 11.6 171 6.0 24.98 4158 19.099 1284 0.034 0.913 1.795 1.979 10.156 179.6 5.25 24.9 5301 16.7 1284 0.034 0.913 1.795 1.979 10.156 179.6 5.25 24.9 5301 16.71 Chapter 15 Solutions, Page 20/20 www.konkur.in Chapter 16 16-1 Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, θ1 = 0°, From which, sinθa = sin90° = 1. θ2 = 120°, and θa = 90°. Eq. (16-2): 0.28 pa (0.040)(0.150) 120° ∫0° sin θ (0.150 − 0.125 cos θ ) dθ 1 = 2.993 (10− 4 ) pa N · m Mf = Eq. (16-3): pa (0.040)(0.150)(0.125) 120° 2 sin θ dθ = 9.478 (10−4 ) pa N · m ∫ 0 ° 1 MN = c = 2(0.125 cos 30°) = 0.2165 m Eq. (16-4): F = 9.478 (10− 4 ) pa − 2.993 (10− 4 ) pa 0.2165 = 2.995 (10− 3 ) pa pa = F/ [2.995(10−3)] = 2200/ [2.995(10−3)] = 734.5(103) Pa for cw rotation Eq. (16-7): 2200 = 9.478 (10− 4 ) pa + 2.993 (10 − 4 ) pa 0.2165 pa = 381.9(103) Pa for ccw rotation A maximum pressure of 734.5 kPa occurs on the RH shoe for cw rotation. (b) RH shoe: Eq. (16-6): 0.28(734.5)103 (0.040)0.1502 (cos 0o − cos120o ) TR = = 277.6 N · m 1 LH shoe: 381.9 TL = 277.6 = 144.4 N · m Ans. 734.5 Ttotal = 277.6 + 144.4 = 422 N · m Ans. Shigley’s MED, 10th edition Ans. Ans. Chapter 16 Solutions, Page 1/27 www.konkur.in (c) RH shoe: Fx = 22 2200 sin 30° = 1100 N, Eqs. (16-8): 1 A = sin 2 θ 2 0o Eqs. (16-9): Rx = 2π / 3 rad 120o Ry = = 0.375, θ 1 B = − sin 2θ 2 4 0 734.5 (103 ) 0.040(0.150) 1 3 734.5 (10 ) 0.04(0.150) R = [( −1007 ) LH shoe: Fy = 2200 cos 30° = 1905 N 2 = 1.264 [0.375 − 0.28(1.264)] − 1100 = −1007 N [1.264 + 0.28(0.375)] − 1905 = 4128 N 1 + 41282 ]1/ 2 = 4249 N Ans. Fx = 1100 110 N, Fy = 1905 N Eqs. (16-10): Rx = 381.9 (103 ) 0.040(0.150) [0.375 + 0.28(1.264)] − 1100 = 570 N 1 381.9 (103 ) 0.040(0.150) Ry = [1.264 − 0.28(0.375)] − 1905 = 751 N 1 1/ 2 R = ( 597 2 + 7512 ) = 959 N Ans. ______________________________________________________________________________ 16-2 Given: r = 300/2 = 150 mm, a = R = 125 mm, b = 40 mm, f = 0.28, F = 2.2 kN, θ1 = 15°, θ2 = 105°, and θa = 90°. 90 From which, sinθa = sin90° = 1. Eq. (16-2): 0.28 pa (0.040)(0.150) 105° Mf = 0.125cos cos θ ) dθ = 2.177 (10− 4 ) pa ∫15° sin θ (0.150 − 0.125 1 Shigley’s MED, 10th edition Chapter 16 Solutions, Page 2/27 www.konkur.in MN = Eq. (16-3): pa (0.040)(0.150)(0.125) 105° 2 sin θ dθ = 7.765 (10− 4 ) pa ∫ 15 ° 1 c = 2(0.125) cos 30° = 0.2165 m F = Eq. (16-4): 7.765 (10− 4 ) pa − 2.177 (10− 4 ) pa 0.2165 = 2.581(10− 3 ) pa pa = 2200/ [2.581(10− 3)] = 852.4 (103) Pa = 852.4 kPa on RH shoe for cw rotation RH shoe: TR = Eq. (16-6): LH shoe: 2200 = Ans. 0.28(852.4)103 (0.040)(0.1502 )(cos15° − cos105°) = 263 N · m 1 7.765 (10 − 4 ) pa + 2.177 (10 − 4 ) pa pa = 479.1(10 3 ) 0.2165 Pa = 479.1 kPa on LH shoe for cw rotation Ans. 0.28(479.1)103 (0.040)(0.1502 )(cos15° − cos105°) = 148 N · m 1 = 263 + 148 = 411 N · m Ans. TL = Ttotal Comparing this result with that of Prob. 16-1, a 2.6% reduction in torque is obtained by using 25% less braking material. ______________________________________________________________________________ 16-3 Given: θ1 = 0°, θ2 = 120°, θa = 90°, sin θa = 1, a = R = 3.5 in, b = 1.25 in, f = 0.30, F = 225 lbf, r = 11/2 = 5.5 in, counter-clockwise rotation. LH shoe: Eq. (16-2), with θ1 = 0: θ f pabr 2 f pabr a Mf = sin θ ( r − a cos θ ) dθ = r (1 − cos θ 2 ) − sin 2 θ 2 ∫ sin θ a θ1 sin θ a 2 0.30 pa (1.25)5.5 3.5 2 5.5(1 − cos120o ) − sin 120° 1 2 = 14.31 pa lbf · in Eq. (16-3), with θ1 = 0: θ pabra 2 2 p bra θ 2 1 MN = sin θ dθ = a − sin 2θ 2 ∫ sin θ a θ1 sin θ a 2 4 = pa (1.25)5.5(3.5) 120° π 1 − sin 2(120°) 1 2 180° 4 = 30.41 pa lbf · in = Shigley’s MED, 10th edition Chapter 16 Solutions, Page 3/27 www.konkur.in 180o − θ 2 o c = 2r cos = 2(5.5) cos 30 = 9.526 in 2 30.41 pa − 14.31 pa F = 225 = = 1.690 pa 9.526 pa = 225 / 1.690 = 133.1 psi Eq. (16-6): f pabr 2 (cos θ1 − cos θ 2 ) 0.30(133.1)1.25(5.52 ) TL = = [1 − (−0.5)] sin θ a 1 = 2265 lbf · in = 2.265 kip · in Ans. RH shoe: 30.41 pa + 14.31 pa = 4.694 pa 9.526 pa = 225 / 4.694 = 47.93 psi 47.93 TR = 2265 = 816 lbf ·in = 0.816 kip·in 133.1 F = 225 = Ttotal = 2.27 + 0.82 = 3.09 kip ⋅ in Ans. ______________________________________________________________________________ 16-4 (a) Given: θ1 = 10°, θ2 = 75°, θa = 75°, pa = 106 Pa, f = 0.24, b = 0.075 m (shoe width), a = 0.150 m, r = 0.200 m, d = 0.050 m, c = 0.165 m. Some of the terms needed are evaluated here: θ 2 θ2 θ2 θ 1 A = r ∫ sin θ dθ − a ∫ sin θ cos θ dθ = r [ − cos θ ]θ2 − a sin 2 θ 1 θ1 θ1 2 θ1 75° 1 = 200 [ − cos θ ]10° − 150 sin 2 θ = 77.5 mm 2 10° 75π /180 rad θ2 θ 1 B = ∫ sin 2 θ dθ = − sin 2θ = 0.528 θ1 2 4 10π /180 rad 75° C = θ2 ∫θ sin θ cos θ dθ = 0.4514 1 Now converting to Pascals and meters, we have from Eq. (16-2), Mf = 0.24 (106 ) (0.075)(0.200) f pabr A= (0.0775) = 289 N · m sin θ a sin 75° Shigley’s MED, 10th edition Chapter 16 Solutions, Page 4/27 www.konkur.in From Eq. (16-3), MN = pabra 106 (0.075)(0.200)(0.150) B = (0.528) = 1230 N · m sin θ a sin 75° Finally, using Eq. (16-4), we have F = MN − M f c = 1230 − 289 = 5.70 kN 165 Ans. (b) Use Eq. (16-6) for the primary shoe. T = = fpabr 2 (cos θ1 − cos θ 2 ) sin θ a 0.24 (106 ) (0.075)(0.200) 2 (cos 10° − cos 75°) sin 75° = 541 N · m For the secondary shoe, we must first find pa. Substituting 1230 289 pa and M f = 6 pa into Eq. (16 - 7), 6 10 10 6 (1230 / 10 ) pa + (289 / 106 ) pa 5.70 = , solving gives pa = 619 (103 ) Pa 165 MN = Then T = 0.24 619 (103 ) 0.075 ( 0.2002 ) ( cos 10° − cos 75° ) sin 75° = 335 N · m so the braking capacity is Ttotal = 2(541) + 2(335) = 1750 N · m Ans. (c) Primary shoes: pabr ( C − f B ) − Fx sin θ a 106 (0.075)0.200 = [0.4514 − 0.24(0.528)](10− 3 ) − 5.70 = −0.658 kN sin 75° p br Ry = a ( B + f C ) − Fy sin θ a 106 (0.075)0.200 = [0.528 + 0.24(0.4514)] (10− 3 ) − 0 = 9.88 kN sin 75° Rx = Shigley’s MED, 10th edition Chapter 16 Solutions, Page 5/27 www.konkur.in Secondary shoes: Rx = = pabr (C + f B) − Fx sin θ a 0.619 (106 ) 0.075(0.200) sin 75° = −0.143 kN p br Ry = a ( B − f C ) − Fy sin θ a = 0.619 (106 ) 0.075(0.200) = 4.03 kN sin 75° [0.4514 + 0.24(0.528)] (10− 3 ) − 5.70 [0.528 − 0.24(0.4514)] (10− 3 ) − 0 Note from figure that +y + for secondary shoe is opposite to +y for primary shoe. Combining horizontal and vertical components, RH = −0.658 − 0.143 = −0.801 kN RV = 9.88 − 4.03 = 5.85 kN R = (−0.801)2 + 5.852 = 5.90 kN Ans. ______________________________________________________________________________ 16-5 Given: Face width b = 1.25 in, F = 90 lbf, f = 0.25. Preliminaries: θ1 = 45° − tan−1(6/8) = 8.13°, θ2 = 98.13°, θa = 90°, a = (62 + 82)1/2 = 10 in Eq. (16-2): θ Mf = f pabr 2 0.25 pa (1.25)6 sin θ ( r − a cos θ ) dθ = ∫ sin θ a θ1 1 98.13° ∫ sin θ ( 6 − 10 cos θ ) dθ 8.13° = 3.728 pa lbf · in Eq. (16-3): θ MN = pabra 2 2 p (1.25)6(10) sin θ dθ = a ∫ sin θ a θ1 1 98.13° ∫ sin 2 θ dθ 8.13° = 69.405 pa lbf · in Eq. (16-4): Using Fc = MN − Mf , we obtain 90(20) = (69.405 − 3.728) pa Shigley’s MED, 10th edition ⇒ pa = 27.4 psi Ans. Chapter 16 Solutions, Page 6/27 www.konkur.in Eq. (16-6): fpabr 2 ( cos θ1 − cos θ 2 ) 0.25(27.4)1.25 ( 6 ) ( cos8.13° − cos 98.13° ) = sin θ a 1 = 348.7 lbf · in Ans. ______________________________________________________________________________ 2 T = 16-6 For +3σˆ f : f = f + 3σˆ f = 0.25 + 3(0.025) = 0.325 From Prob. 16-5, with f = 0.25, M f = 3.728 pa. Thus, M f = (0.325/0.25) 3.728 pa = 4.846 pa. From Prob. 16-5, M N = 69.405 pa. Eq. (16-4): Using Fc = MN − Mf , we obtain 90(20) = (69.405 − 4.846) pa ⇒ pa = 27.88 psi Ans. From Prob. 16-5, pa = 27.4 psi and T = 348.7 lbf⋅in. Thus, 0.325 27.88 T = 348.7 = 461.3 lbf ·in 0.25 27.4 Ans. Similarly, for −3σˆ f : f = f − 3σˆ f = 0.25 − 3(0.025) = 0.175 M f = (0.175 / 0.25) 3.728 pa = 2.610 pa 90(20) = (69.405 − 2.610) pa ⇒ pa = 26.95 psi 0.175 26.95 T = 348.7 = 240.1 lbf · in Ans. 0.25 27.4 ______________________________________________________________________________ 16-7 Preliminaries: θ2 = 180° − 30° − tan−1(3/12) = 136°, θ1 = 20° − tan−1(3/12) = 6°, θa = 90°, sinθa = 1, a = (32 + 122)1/2 = 12.37 in, r = 10 in, f = 0.30, b = 2 in, pa = 150 psi. Eq. (16-2): Mf = 0.30(150)(2)(10) 136o ∫6° sin θ (10 − 12.37 cos θ ) dθ = 12 800 lbf · in sin 90° Eq. (16-3): MN = 150(2)(10)(12.37) 136° 2 ∫6° sin θ dθ = 53 300 lbf · in sin 90° LH shoe: cL = 12 + 12 + 4 = 28 in Now note that Mf is cw and MN is ccw. Thus, Shigley’s MED, 10th edition Chapter 16 Solutions, Page 7/27 www.konkur.in Eq. (16-6): FL = 53 300 − 12 800 = 1446 lbf 28 TL = 0.30(150)(2)(10) 2 (cos 6° − cos136°) = 15 420 lbf · in sin 90° RH shoe: M N = 53 300 pa = 355.3 pa , 150 M f = 12 800 pa = 85.3 pa 150 On this shoe, both MN and Mf are ccw. Also, cR = (24 − 2 tan 14°) cos 14° = 22.8 in Fact = FL sin14° = 361 lbf Ans. FR = FL / cos14° = 1491 lbf Thus, 1491 = Then, TR = 355.3 + 85.3 pa ⇒ pa = 77.2 psi 22.8 0.30(77.2)(2)(10)2 (cos 6° − cos136°) = 7940 lbf · in sin 90° Ttotal = 15 420 + 7940 = 23 400 lbf · in Ans. ______________________________________________________________________________ 16-8 θ2 M f = 2∫ ( fdN )(a′ cos θ − r ) 0 where dN = pbr dθ θ2 = 2 fpbr ∫ (a′ cos θ − r ) dθ = 0 0 From which θ2 θ2 0 0 a′∫ cos θ dθ = r ∫ dθ rθ 2 r (60°)(π / 180) a′ = = = 1.209r sin θ 2 sin 60° Ans. Eq. (16-15): Shigley’s MED, 10th edition Chapter 16 Solutions, Page 8/27 www.konkur.in a = 4r sin 60° = 1.170r 2(60)(π / 180) + sin[2(60)] Ans. a differs with a ′ by 100(1.170 −1.209)/1.209 = − 3.23 % Ans. ______________________________________________________________________________ 16-9 (a) Counter-clockwise clockwise rotation, θ2 = π / 4 rad, r = 13.5/2 = 6.75 75 in Eq. (16-15): 4r sin θ 2 4(6.75) sin(π / 4) a = = = 7.426 in 2θ 2 + sin 2θ 2 2π / 4 + sin(2π / 4) e = 2a = 2(7.426) = 14.85 in Ans. (b) 85) = 11 11.4° α = tan−1(3/14.85) ∑M ∑F x R = 0 = 3F x − 6.375P = 0 = −F x + R x ⇒ F x = 2.125P ⇒ R x = F x = 2.125P F y = F x tan11.4o = 0.428P ∑ Fy = −P − F y + R y R y = P + 0.428P = 1.428P Left shoe lever. ∑ M R = 0 = 7.78S x − 15.28F x 15.28 Sx = (2.125P) = 4.174 P 7.78 S y = f S x = 0.30(4.174 P) = 1.252P ∑ Fy = 0 = R y + S y + F y R y = − F y − S y = −0.428P − 1.252P = −1.68P ∑ Fx = 0 = R x − S x + F x R x = S x − F x = 4.174 P − 2.125P = 2.049 P Shigley’s MED, 10th edition Chapter 16 Solutions, Page 9/27 www.konkur.in (c) The direction of brake pulley rotation affects the sense of Sy, which has no effect on the brake shoe lever moment and hence, no effect on Sx or the brake torque. The brake shoe levers carry identical bending moments but the left lever carries a tension while the right carries compression (column loading). The right lever is designed andd used as a left lever, producing interchangeable levers (identical levers). But do not infer from these identical loadings. ______________________________________________________________________________ 16-10 r = 13.5/2 = 6.75 in, b = 6 in, θ2 = 45° = π / 4 rad. From Table 16-33 for a rigid, molded non-asbestos non lining use a conservative estimate of pa = 100 psi, f = 0.33.. Equation (16-16) 16) gives the horizontal brake hinge pin reaction which corresponds to Sx in Prob. 16-9. Thus, p br 100(6)6.75 N = S x = a ( 2θ 2 + sin 2θ 2 ) = 2 (π / 4 ) + sin 2 ( 45° ) 2 2 = 5206 lbf { } which, from Prob. 6-99 is 4.174 P. Therefore, 4.174 P = 5206 ⇒ P = 1250 lbf = 1.25 kip Ans. Ans Applying Eq. (16-18) 18) for two shoes, where from Prob. 16-9, a = 7.426 in T = 2a f N = 2(7.426)0.33(5206) = 25 520 lbf · in = 25.52 kip · in Ans. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 16 Solutions, Page 10/27 www.konkur.in 16-11 Given: D = 350 mm, b = 100 mm, pa = 620 kPa, f = 0.30, φ = 270°. Eq. (16-22): p bD 620(0.100)0.350 P1 = a = = 10.85 kN Ans. 2 2 f φ = 0.30(270°)(π / 180°) = 1.414 Eq. (16-19): P2 = P1 exp(− f φ ) = 10.85 exp(− 1.414) = 2.64 kN Ans. T = ( P1 − P2 )( D / 2) = (10.85 − 2.64)(0.350 / 2) = 1.437 kN · m Ans. ______________________________________________________________________________ 16-12 Given: D = 12 in, Eq. (16-22): f = 0.28, pa = b = 3.25 in, φ = 270°, P1 = 1800 lbf. 2 P1 2(1800) = = 92.3 psi bD 3.25(12) Ans. f φ = 0.28(270o )(π / 180o ) = 1.319 P2 = P1 exp(− f φ ) = 1800 exp(−1.319) = 481 lbf T = ( P1 − P2 )( D / 2) = (1800 − 481)(12 / 2) = 7910 lbf · in = 7.91 kip · in Ans. ______________________________________________________________________________ 16-13 Σ MO = 0 = 100 P2 − 325 F ⇒ P2 = 325(300)/100 = 975 N Shigley’s MED, 10th edition Ans. Chapter 16 Solutions, Page 11/27 www.konkur.in 100 α = cos −1 = 51.32° 160 φ = 270° − 51.32° = 218.7° f φ = 0.30(218.7) (π / 180° ) = 1.145 P1 = P2 exp( f φ ) = 975exp(1.145) = 3064 N T = ( P1 − P2 ) ( D / 2) = (3064 − 975)(200 / 2) = 209 (103 ) N · mm = 209 N · m Ans. Ans. ______________________________________________________________________________ 16-14 (a) D = 16 in, b = 3 in n = 200 rev/min f = 0.20, pa = 70 psi Eq. (16-22): p bD 70(3)(16) P1 = a = = 1680 lbf 2 2 f φ = 0.20(3π / 2) = 0.942 Eq. (16-14): Shigley’s MED, 10th edition P2 = P1 exp(− f φ ) = 1680 exp(−0.942) = 655 lbf D 16 T = ( P1 − P2 ) = (1680 − 655) 2 2 = 8200 lbf · in Ans. Tn 8200(200) H = = = 26.0 hp Ans. 63 025 63 025 3P 3(1680) P = 1 = = 504 lbf Ans. 10 10 Chapter 16 Solutions, Page 12/27 www.konkur.in (b) Force of belt on the drum: R = (16802 + 6552)1/2 = 1803 lbf Force of shaft on the drum: 1680 and 655 lbf TP1 = 1680(8) = 13 440 lbf · in TP2 = 655(8) = 5240 lbf · in Net torque on drum due to brake band: T = TP1 − TP2 = 13 440 − 5240 = 8200 lbf · in The radial load on the bearing pair is 1803 lbf. If the bearing is straddle mounted with the drum at center span, the bearing radial load is 1803/2 1803 2 = 901 lbf. lbf Shigley’s MED, 10th edition Chapter 16 Solutions, Page 13/27 www.konkur.in (c) Eq. (16-21): 2P bD 2 P1 2(1680) = = = 70 psi 3(16) 3(16) p = p θ = 0° Ans. 2 P2 2(655) = = 27.3 psi Ans. 3(16) 3(16) ______________________________________________________________________________ p θ = 270° = 16-15 Given: φ = 270°, b = 2.125 in, f = 0.20, T =150 lbf · ft, D = 8.25 in, c2 = 2.25 in (see figure). Notice that the pivoting rocker is not located on the vertical centerline of the drum. (a) To have the band tighten for ccw rotation, it is necessary to have c1 < c2 . When friction is fully developed, P1 / P2 = exp( f φ ) = exp[0.2(3π / 2)] = 2.566 If friction is not fully developed, P1/P2 ≤ exp( f φ ) To help visualize what is going on let’s add a force W parallel to P1, at a lever arm of c3. Now sum moments about the rocker pivot. ∑M = 0 = c3W + c1P1 − c2 P2 From which c2 P2 − c1P1 c3 The device is self locking for ccw rotation if W is no longer needed, that is, W ≤ 0. It follows from the equation above W = P1 c ≥ 2 P2 c1 When friction is fully developed 2.566 = 2.25 / c1 2.25 c1 = = 0.877 in 2.566 When P1/P2 is less than 2.566, friction is not fully developed. Suppose P1/P2 = 2.25, then Shigley’s MED, 10th edition Chapter 16 Solutions, Page 14/27 www.konkur.in c1 = 2.25 = 1 in 2.25 We don’t want to be at the point of slip, and we need the band to tighten. c2 ≤ c1 ≤ c2 P1 / P2 When the developed friction is very small, P1/P2 → 1 and c1 → c2 Ans. (b) Rocker has c1 = 1 in P1 c 2.25 = 2 = = 2.25 P2 c1 1 ln( P1 / P2 ) ln 2.25 f = = = 0.172 φ 3π / 2 Friction is not fully developed, no slip. T = ( P1 − P2 ) P D D = P2 1 − 1 2 P2 2 Solve for P2 2T 2(150)(12) = = 349 lbf [( P1 / P2 ) − 1]D (2.25 − 1)(8.25) P1 = 2.25P2 = 2.25(349) = 785 lbf 2P 2(785) p = 1 = = 89.6 psi Ans. bD 2.125(8.25) P2 = (c) The torque ratio is 150(12)/100 or 18-fold. 349 P2 = = 19.4 lbf 18 P1 = 2.25P2 = 2.25(19.4) = 43.6 lbf 89.6 p = = 4.98 psi Ans. 18 Comment: As the torque opposed by the locked brake increases, P2 and P1 increase (although ratio is still 2.25), then p follows. The brake can self-destruct. Protection could be provided by a shear key. ______________________________________________________________________________ 16-16 Given: OD = 250 mm, ID = 175 mm, f = 0.30, F = 4 kN. Shigley’s MED, 10th edition Chapter 16 Solutions, Page 15/27 www.konkur.in (a) From Eq. (16-23), 2 ( 4000 ) 2F pa = = = 0.194 N/mm2 = 194 kPa Ans. π d ( D − d ) π (175)(250 − 175) Eq. (16-25): Ff 4000(0.30) T = (D + d ) = (250 + 175)10− 3 = 127.5 N · m Ans. 4 4 (b) From Eq. (16-26), 4F 4(4000) = = 0.159 N/mm 2 = 159 kPa 2 2 2 2 π ( D − d ) π (250 − 175 ) pa = Eq. (16-27): T = π f pa ( D3 − d 3 ) = π (0.30)159 (103 )( 2503 − 1753 )(10− 3 ) Ans. 3 12 12 = 128 N · m Ans. ______________________________________________________________________________ 16-17 Given: OD = 6.5 in, ID = 4 in, f = 0.24, pa = 120 psi. (a) Eq. (16-23): π pa d π (120)(4) F = (D − d ) = (6.5 − 4) = 1885 lbf Ans. 2 2 Eq. (16-24) with N sliding planes: π fpa d 2 π (0.24)(120)(4) T = (D − d 2) N = (6.52 − 42 )(6) 8 8 = 7125 lbf · in Ans. (b) T = π (0.24)(120d ) 8 (6.52 − d 2 )(6) d, in T, lbf · in 2 5191 3 6769 4 7125 Ans. 5 5853 6 2545 (c) The torque-diameter curve exhibits a stationary point maximum in the range of diameter d. The clutch has nearly optimal proportions. ______________________________________________________________________________ 16-18 (a) Eq. (16-24) with N sliding planes: Shigley’s MED, 10th edition Chapter 16 Solutions, Page 16/27 www.konkur.in π f pa d ( D 2 − d 2 ) N T = 8 = π f pa N 8 (D d − d ) 2 3 Differentiating with respect to d and equating to zero gives dT π f pa N 2 = D − 3d 2 ) = 0 ( dd 8 D d* = Ans. 3 π f pa N d 2T 3π f pa N = −6 d =− d 2 dd 8 4 which is negative for all positive d. We have a stationary point maximum. (b) d* = 6.5 = 3.75 in 3 Eq. (16-24): T* = ( Ans. ) π (0.24)(120) 6.5 / 3 2 2 6.5 − 6.5 / 3 (6) = 7173 lbf · in 8 ( ) (c) The table indicates a maximum within the range: 3 ≤ d ≤ 5 in d (d) Consider: 0.45 = = 0.80 D Multiply through by D, 0.45D ≤ d ≤ 0.80D 0.45(6.5) ≤ d ≤ 0.80(6.5) 2.925 ≤ d ≤ 5.2 in * 1 d = 0.577 = d * /D = 3 D which lies within the common range of clutches. Yes. Ans. ______________________________________________________________________________ 16-19 Given: d = 11 in, l = 2.25 in, T = 1800 lbf · in, D = 12 in, f = 0.28. 0.5 = 12.53° 2.25 α = tan −1 Shigley’s MED, 10th edition Chapter 16 Solutions, Page 17/27 www.konkur.in Uniform wear Eq. (16-45): π f pa d 2 D − d2) ( 8sin α π (0.28) pa (11) 2 1800 = 12 − 112 ) = 128.2 pa ( 8sin12.53° 1800 pa = = 14.04 psi Ans. 128.2 T = Eq. (16-44): π pa d F = 2 (D − d ) = π (14.04)11 2 (12 − 11) = 243 lbf Ans. Uniform pressure Eq. (16-48): π f pa ( D3 − d 3 ) 12sin α π (0.28) pa 1800 = 123 − 113 ) = 134.1 pa ( 12sin12.53° 1800 pa = = 13.42 psi Ans. 134.1 T = Eq. (16-47): π pa π (13.42) (122 − 112 ) = 242 lbf Ans. 4 4 ______________________________________________________________________________ F = (D 2 − d 2 ) = 16-20 Uniform wear Eq. (16-34): Eq. (16-33): Thus, 1 (θ 2 − θ1) f pa ri ( ro2 − ri 2 ) 2 F = (θ2 − θ1) pari (ro − ri) T = (1 / 2)(θ 2 − θ1) f pa ri ( ro2 − ri 2 ) T = f FD f (θ 2 − θ1) pa ri (ro − ri )( D) r + ri D / 2 + d / 2 1 d = o = = 1 + O.K . 2D 2D 4 D Ans. Uniform pressure Eq. (16-38): Shigley’s MED, 10th edition T = 1 (θ 2 − θ1 ) f pa ( ro3 − ri3 ) 3 Chapter 16 Solutions, Page 18/27 www.konkur.in F = Eq. (16-37): 1 (θ 2 − θ1) pa ( ro2 − ri 2 ) 2 Thus, (1 / 3)(θ 2 − θ1) f pa ( ro3 − ri3 ) T 2 ( D / 2)3 − (d / 2)3 = = f FD (1 / 2) f (θ 2 − θ1) pa ( ro2 − ri 2 ) D 3 ( D / 2) 2 − (d / 2) 2 D 2( D / 2)3 1 − (d / D)3 1 1 − (d / D ) 3 = = O.K . Ans. 3( D / 2) 2 1 − (d / D) 2 D 3 1 − (d / D) 2 ______________________________________________________________________________ 16-21 ω = 2π n / 60 = 2π 500 / 60 = 52.4 rad/s T = H ω = 2(103 ) = 38.2 N · m 52.4 Key: T 38.2 = = 3.18 kN r 12 Average shear stress in key is 3.18(103 ) τ = = 13.2 MPa Ans. 6(40) Average bearing stress is F 3.18(103 ) σb = − =− = −26.5 MPa Ab 3(40) Let one jaw carry the entire load. F = Ans. 1 26 45 + = 17.75 mm 2 2 2 T 38.2 F = = = 2.15 kN rav 17.75 rav = The bearing and shear stress estimates are σb = −2.15 (103 ) = −22.6 MPa Ans. 10(22.5 − 13) 2.15(103 ) τ = = 0.869 MPa Ans. 10 0.25π (17.75) 2 ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 16 Solutions, Page 19/27 www.konkur.in 16-22 ω1 = 2π n / 60 = 2π (1600) / 60 = 167.6 rad/s ω2 = 0 From Eq. (16-51), I1I 2 Tt1 2800(8) = = = 133.7 lbf · in · s 2 ω1 − ω2 167.6 − 0 I1 + I 2 Eq. (16-52): E = I1I 2 133.7 2 (167.6 − 0) 2 = 1.877 (106 ) lbf ⋅ in (ω1 − ω2 ) = 2 ( I1 + I 2 ) 2 H = E / 9336 = 1.877(106) / 9336 = 201 Btu In Btu, Eq. (16-53): Eq. (16-54): H 201 = = 41.9°F Ans. C pW 0.12(40) ______________________________________________________________________________ ∆T = 16-23 n1 + n2 260 + 240 = = 250 rev/min 2 2 Cs = (ω 2 − ω 1) / ω = (n2 − n1) / n = (260 − 240) / 250 = 0.08 n= Eq. (16-62): Ans. ω = 2π (250) / 60 = 26.18 rad/s From Eq. (16-64): 6.75 (103 ) E2 − E1 I = = = 123.1 N · m · s 2 2 2 Csω 0.08(26.18) I = m 2 d o + di2 ) ( 8 ⇒ m= 8I 8(123.1) = = 233.9 kg 2 d + di 1.52 + 1.4 2 Table A-5, cast iron unit weight = 70.6 kN/m3 Volume: 2 o ⇒ ρ = 70.6(103) / 9.81 = 7197 kg / m3. V = m / ρ = 233.9 / 7197 = 0.0325 m3 V = π t ( d o2 − d i2 ) / 4 = π t (1.52 − 1.4 2 ) / 4 = 0.2278t Equating the expressions for volume and solving for t, 0.0325 t = = 0.143 m = 143 mm Ans. 0.2278 ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 16 Solutions, Page 20/27 www.konkur.in 16-24 (a) The useful work performed in one revolution of the crank shaft is U = 320 (103) 200 (10−3) 0.15 = 9.6 (103) J Accounting for friction, the total work done in one revolution is U = 9.6(103) / (1 − 0.20) = 12.0(103) J Since 15% of the crank shaft stroke accounts for 7.5% of a crank shaft revolution, the energy fluctuation is E2 − E1 = 9.6(103) − 12.0(103)(0.075) = 8.70(103) J (b) For the flywheel, Since Eq. (16-64): Ans. n = 6(90) = 540 rev/min 2π n 2π (540) ω = = = 56.5 rad/s 60 60 Cs = 0.10 E −E 8.70(103 ) I = 2 21 = = 27.25 N · m · s 2 2 Csω 0.10(56.5) Assuming all the mass is concentrated at the effective diameter, d, md 2 4 4I 4(27.25) m= 2 = = 75.7 kg Ans. d 1.22 ______________________________________________________________________________ I = mr 2 = 16-25 Use Ex. 16-6 and Table 16-6 data for one cylinder of a 3-cylinder engine. Cs = 0.30 n = 2400 rev/min or 251 rad/s 3(3368) Tm = = 804 lbf · in Ans. 4π E2 − E1 = 3(3531) = 10 590 in · lbf E −E 10 590 I = 2 21 = = 0.560 in · lbf · s 2 Ans. 2 Csω 0.30(251 ) ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 16 Solutions, Page 21/27 www.konkur.in 16-26 (a) (1) (T2 )1 = − F21rP = − T2 T rP = 2 rG −n Ans. Equivalent energy (2) (1 / 2) I 2ω22 = (1 / 2)( I 2 )1 (ω12 ) ( I 2 )1 = ω22 I I = 22 2 2 ω1 n Ans. 2 2 2 I G rG mG rG rG = = = n4 IP rP mP rP rP (3) From (2) ( I 2 )1 = (b) I e = I M + I P + n 2 I P + IL n2 IG n4I P = = n2I P n2 n2 Ans. Ans. ______________________________________________________________________________ 16-27 (a) Reflect IL, IG2 to the center shaft Shigley’s MED, 10th edition Chapter 16 Solutions, Page 22/27 www.konkur.in Reflect the center shaft to the motor shaft Ie = IM + I P + n2I P + I P m2 I + 2 I P + 2L 2 2 n n mn (b) For R = constant = nm, I e = I M + I P + n 2 I P + (c) For R = 10, Ans. IP R2I P I + + L2 2 4 n n R Ans. ∂I e 2(1) 4(102 )(1) = 0 + 0 + 2n(1) − 3 − +0=0 ∂n n n5 n6 − n2 − 200 = 0 From which n* = 2.430 Ans. 10 m* = = 4.115 2.430 Ans. Notice that n*and m** are independent of IL. ______________________________________________________________________________ 16-28 From Prob. 16-27, IP R2I P I + + L2 2 4 n n R 1 100(1) 100 2 = 10 + 1 + n (1) + 2 + + 2 n n4 10 1 100 = 12 + n 2 + 2 + 4 n n Ie = I M + I P + n2I P + Shigley’s MED, 10th edition Chapter 16 Solutions, Page 23/27 www.konkur.in Optimizing the partitioning of a double reduction lowered the gear--train inertia to 20.9/112 = 0.187, 187, or to 19% of that of a single reduction. This includes the two additional gears. ______________________________________________________________________________ 16-29 Figure 16-29 applies, t2 = 10 s, t1 = 0.5 s t2 - t1 10 − 0.5 = = 19 t1 0.5 The load torque, as seen by the motor shaft (Rule 1, Prob. 16 16-26), 26), is TL = 1300(12) = 1560 lbf · in 10 The rated motor torque Tr is Tr = 63 025(3) = 168.07 lbf · in 1125 For Eqs. (16-65): 2π (1125) = 117.81 rad/s 60 2π (1200) = 125.66 rad/s ωs = 60 −Tr 168.07 a = =− = −21.41 lbf ⋅ in ⋅ s/rad 125.66 − 117.81 ω s − ωr Trωs 168.07(125.66) b= = = 2690.4 lbf · in ωs − ωr 125.66 − 117.81 ωr = Shigley’s MED, 10th edition Chapter 16 Solutions, Page 24/27 www.konkur.in The linear portion of the squirrel-cage motor characteristic can now be expressed as TM = −21.41ω + 2690.4 lbf · in Eq. (16-68): 19 1560 − 168.07 T2 = 168.07 1560 − T2 One root is 168.07 which is for infinite time. The root for 10 s is desired. Use a successive substitution method T2 0.00 19.30 24.40 26.00 26.50 New T2 19.30 24.40 26.00 26.50 26.67 Continue until convergence to T2 = 26.771 lbf ⋅ in Eq. (16-69): a ( t2 − t1 ) −21.41(10 − 0.5) = = 110.72 lbf · in · s 2 ln (T2 / Tr ) ln(26.771 / 168.07) T −b ω = a T − b 26.771 − 2690.4 ωmax = 2 = = 124.41 rad/s Ans. a −21.41 ωmin = 117.81 rad/s Ans. 124.41 + 117.81 ω = = 121.11 rad/s 2 124.41 − 117.81 ωmax − ωmin Cs = = = 0.0545 Ans. (ωmax + ωmin ) / 2 (124.41 + 117.81) / 2 1 1 E1 = I ωr2 = (110.72)(117.81) 2 = 768 352 in · lbf 2 2 1 2 1 E2 = I ω2 = (110.72)(124.41) 2 = 856 854 in · lbf 2 2 ∆E = E2 − E1 = 856 854 − 768 352 = 88 502 in · lbf I = Eq. (16-64): ∆E = Cs I ω 2 = 0.0545(110.72)(121.11) 2 = 88 508 in · lbf, close enough Ans. During the punch Shigley’s MED, 10th edition Chapter 16 Solutions, Page 25/27 www.konkur.in 63 025H n TLω (60 / 2π ) 1560(121.11)(60 / 2π ) H = = = 28.6 hp 63 025 63 025 T = The gear train has to be sized for 28.6 hp under shock conditions since the flywheel is on the motor shaft. From Table A-18, ( ( ) m 2 W 2 d o + di2 = d o + d i2 8 8g 8 gI 8(386)(110.72) W = 2 = 2 d o + di d o2 + di2 I = ) If a mean diameter of the flywheel rim of 30 in is acceptable, try a rim thickness of 4 in d i = 30 − (4 / 2) = 28 in d o = 30 + (4 / 2) = 32 in 8(386)(110.72) W = = 189.1 lbf 322 + 282 Rim volume V is given by V = πl (d 4 2 o − di2 ) = πl 4 (322 − 282 ) = 188.5l where l is the rim width as shown in Table A-18. The specific weight of cast iron is γ = 0.260 lbf / in3, therefore the volume of cast iron is V = W γ = 189.1 = 727.3 in 3 0.260 Equating the volumes, 188.5 l = 727.3 727.3 l = = 3.86 in wide 188.5 Proportions can be varied. ______________________________________________________________________________ 16-30 Prob. 16-29 solution has I for the motor shaft flywheel as I = 110.72 lbf · in · s2 Shigley’s MED, 10th edition Chapter 16 Solutions, Page 26/27 www.konkur.in A flywheel located on the crank shaft needs an inertia of 102 I (Prob. 16-26, rule 2) I = 102(110.72) = 11 072 lbf · in · s2 A 100-fold inertia increase. On the other hand, the gear train has to transmit 3 hp under shock conditions. Stating the problem is most of the solution. Satisfy yourself that on the crankshaft: TL = 1300(12) = 15 600 lbf · in Tr = 10(168.07) = 1680.7 lbf · in ωr = 117.81 / 10 = 11.781 rad/s ωs = 125.66 / 10 = 12.566 rad/s a = −21.41(100) = −2141 lbf · in · s/rad b = 2690.35(10) = 26903.5 lbf · in TM = −2141ωc + 26 903.5 lbf · in 19 15 600 − 1680.5 T2 = 1680.6 15 600 − T2 The root is 10(26.67) = 266.7 lbf · in ω = 121.11 / 10 = 12.111 rad/s Cs = 0.0549 (same) ωmax = 121.11 / 10 = 12.111 rad/s Ans. ωmin = 117.81 / 10 = 11.781 rad/s Ans. E1, E2, ∆E and peak power are the same. From Table A-18 W = 6 8 gI 8(386)(11 072) 34.19 (10 ) = = d o2 + di2 d o2 + di2 d o2 + di2 Scaling will affect do and di , but the gear ratio changed I. Scale up the flywheel in the Prob. 16-29 solution by a factor of 2.5. Thickness becomes 4(2.5) = 10 in. d = 30(2.5) = 75 in d o = 75 + (10 / 2) = 80 in di = 75 − (10 / 2) = 70 in Shigley’s MED, 10th edition Chapter 16 Solutions, Page 27/27 www.konkur.in W = 34.19 (106 ) = 3026 lbf 802 + 702 W 3026 V = = = 11 638 in 3 0.260 γ V = π l (802 − 702 ) = 1178 l 4 11 638 l = = 9.88 in 1178 Proportions can be varied. The weight has increased 3026/189.1 or about 16-fold while the moment of inertia I increased 100-fold. The gear train transmits a steady 3 hp. But the motor armature has its inertia magnified 100-fold, and during the punch there are deceleration stresses in the train. With no motor armature information, we cannot comment. ______________________________________________________________________________ 16-31 This can be the basis for a class discussion. Shigley’s MED, 10th edition Chapter 16 Solutions, Page 28/27 www.konkur.in Chapter 17 17-1 Begin with Eq. (17-10), F1 = Fc + Fi 2 exp( f θ ) exp( f θ ) − 1 Introduce Eq. (17-9): exp( f θ ) + 1 2 exp( f θ ) 2T exp( f θ ) F1 = Fc + d = Fc + d exp( f θ ) − 1 exp( f θ ) − 1 exp( f θ ) + 1 exp( f θ ) F1 = Fc + ∆F exp( f θ ) − 1 exp( f θ ) Now add and subtract Fc exp( f θ ) − 1 exp( f θ ) exp( f θ ) exp( f θ ) F1 = Fc + Fc + ∆F − Fc exp( f θ ) − 1 exp( f θ ) − 1 exp( f θ ) − 1 exp( f θ ) exp( f θ ) = ( Fc + ∆F ) + Fc − Fc exp( f θ ) − 1 exp( f θ ) − 1 exp( f θ ) Fc = ( Fc + ∆F ) − exp( f θ ) − 1 exp( f θ ) − 1 ( F + ∆F ) exp( f θ ) − Fc = c Q.E.D. exp( f θ ) − 1 From Ex. 17-2: θd = 3.037 rad, ∆F = 664 lbf, exp( fθ ) = exp[0.80(3.037)] = 11.35, and Fc = 73.4 lbf. (73.4 + 664)11.35 − 73.4 = 802 lbf (11.35 − 1) F2 = F1 − ∆F = 802 − 664 = 138 lbf 802 + 138 Fi = − 73.4 = 396.6 lbf 2 1 F1 − Fc 1 802 − 73.4 f′ = ln ln = 0.80 Ans. = θ d F2 − Fc 3.037 138 − 73.4 ______________________________________________________________________________ F1 = 17-2 Given: F-1 Polyamide, b = 6 in, d = 2 in with n = 1750 rev/min, Hnom = 2 hp, C = 9(12) = 108 in, velocity ratio = 0.5, Ks = 1.25, nd = 1 Shigley’s MED, 10th edition Chapter 17 Solutions, Page 1/39 www.konkur.in V = π d n / 12 = π (2)(1750) / 12 = 916.3 ft/min Eq. (17-1): Table 17-2: D = d / vel ratio = 2 / 0.5 = 4 in 4−2 D−d θ d = π − 2sin −1 = π − 2sin −1 = 3.123 rad 2C 2(108) t = 0.05 in, dmin = 1.0 in, Fa = 35 lbf/in, γ = 0.035 lbf/in3, f = 0.5 w = 12 γbt = 12(0.035)6(0.05) = 0.126 lbf/ft 2 (a) Eq. (e), p. 877: 2 w V 0.126 916.3 Fc = = = 0.913 lbf g 60 32.17 60 Ans. 63 025H nom K s nd 63 025(2)(1.25)(1) = = 90.0 lbf · in n 1750 2T 2(90.0) ∆F = ( F1 )a − F2 = = = 90.0 lbf d 2 T = Table 17-4: Cp = 0.70 Eq. (17-12): (F1)a = bFaCpCv = 6(35)(0.70)(1) = 147 lbf F2 = (F1)a − [(F1)a − F2] = 147 − 90 = 57 lbf Ans. Ans. Do not use Eq. (17-9) because we do not yet know f ′ ( F1 )a + F2 147 + 57 − 0.913 = 101.1 lbf 2 2 Using Eq. (17-7) solved for f ′ (see step 8, p.888), 1 ( F1) a − Fc 1 147 − 0.913 f′ = ln ln = 0.307 = θ d F2 − Fc 3.123 57 − 0.913 Eq. (i), p. 878: Fi = − Fc = Ans. The friction is thus underdeveloped. (b) The transmitted horsepower is, with ∆F = (F1)a − F2 = 90 lbf, (∆F )V 90(916.3) Eq. (j), p. 879: H = = = 2.5 hp Ans. 33 000 33 000 H 2.5 nf s = = =1 H nom K s 2(1.25) Eq. (17-1): Eq. (17-2): Shigley’s MED, 10th edition 4−2 D−d = π + 2sin −1 = 3.160 rad 2C 2(108) L = [4C2 − (D − d)2]1/2 + (DθD + dθd)/2 θ D = π + 2sin −1 Chapter 17 Solutions, Page 2/39 www.konkur.in = [4(108)2 − (4 − 2)2]1/2 + [4(3.160) + 2(3.123)]/2 = 225.4 in dip = (c) Eq. (17-13): 3C 2 w 3(108 / 12) 2 (0.126) = = 0.151 in 2Fi 2(101.1) Ans. Ans. Comment: The solution of the problem is finished; however, a note concerning the design is presented here. The friction is under-developed. Narrowing the belt width to 5 in (if size is available) will increase f ′ . The limit of narrowing is bmin = 4.680 in, whence w = 0.0983 lbf/ft Fc = 0.713 lbf T = 90 lbf · in (same) ∆F = ( F1) a − F2 = 90 lbf Fi = 68.9 lbf ( F1) a = 114.7 lbf F2 = 24.7 lbf f ′ = f = 0.50 dip = 0.173 in Longer life can be obtained with a 6-inch wide belt by reducing Fi to attain f ′ = 0.50. Prob. 17-8 develops an equation we can use here (∆F + Fc ) exp( f θ ) − Fc exp( f θ ) − 1 F2 = F1 − ∆F F + F2 Fi = 1 − Fc 2 1 F1 − Fc f′ = ln θ d F2 − Fc F1 = dip = 3C 2w 2 Fi which in this case, θd = 3.123 rad, exp(f θ) = exp[0.5(3.123)] = 4.766, w = 0.126 lbf/ft, ∆F = 90.0 lbf, Fc = 0.913 lbf, and gives ( 0.913 + 90 ) 4.766 − 0.913 = 114.8 lbf F1 = 4.766 − 1 F2 = 114.8 − 90 = 24.8 lbf Fi = (114.8 + 24.8)/ 2 − 0.913 = 68.9 lbf f′ = Shigley’s MED, 10th edition 1 114.8 − 0.913 ln = 0.50 3.123 24.8 − 0.913 Chapter 17 Solutions, Page 3/39 www.konkur.in 3 (108 / 12 ) 0.126 dip = = 0.222 in 2(68.9) 2 So, reducing Fi from 101.1 lbf to 68.9 lbf will bring the undeveloped friction up to 0.50, with a corresponding dip of 0.222 in. Having reduced F1 and F2, the endurance of the belt is improved. Power, service factor and design factor have remained intact. ______________________________________________________________________________ 17-3 Double the dimensions of Prob. 17-2. In Prob. 17-2, F-1 Polyamide was used with a thickness of 0.05 in. With what is available in Table 17-2 we will select the Polyamide A-2 belt with a thickness of 0.11 in. Also, let b = 12 in, d = 4 in with n = 1750 rev/min, Hnom = 2 hp, C = 18(12) = 216 in, velocity ratio = 0.5, Ks = 1.25, nd = 1. V = π d n / 12 = π (4)(1750) / 12 = 1833 ft/min Eq. (17-1): Table 17-2: D = d / vel ratio = 4 / 0.5 = 8 in 8−4 D−d θ d = π − 2sin −1 = π − 2sin −1 = 3.123 rad 2C 2(216) t = 0.11 in, dmin = 2.4 in, Fa = 60 lbf/in, γ = 0.037 lbf/in3, f = 0.8 w = 12 γbt = 12(0.037)12(0.11) = 0.586 lbf/ft 2 Fc = (a) Eq. (e), p. 877: 2 w V 0.586 1833 = = 17.0 lbf g 60 32.17 60 Ans. 63 025H nom K s nd 63 025(2)(1.25)(1) = = 90.0 lbf · in n 1750 2T 2(90.0) ∆F = ( F1 )a − F2 = = = 45.0 lbf d 4 T = Table 17-4: Cp = 0.73 Eq. (17-12): (F1)a = bFaCpCv = 12(60)(0.73)(1) = 525.6 lbf F2 = (F1)a − [(F1)a − F2] = 525.6 − 45 = 480.6 lbf Fi = Eq. (i), p. 878: ( F1 )a 2 + F2 − Fc = Ans. Ans. 525.6 + 480.6 − 17.0 = 486.1 lbf 2 Ans. Eq. (17-9): f′ = Shigley’s MED, 10th edition ( F ) − Fc 1 525.6 − 17.0 ln 1 a ln = 0.0297 = θ d F2 − Fc 3.123 480.6 − 17.0 1 Chapter 17 Solutions, Page 4/39 www.konkur.in The friction is thus un underdeveloped. (b) The transmitted horsepower is, with ∆F = (F1)a − F2 = 45 lbf, (∆F )V 45(1833) = = 2.5 hp 33 000 33 000 H 2.5 = = =1 H nom K s 2(1.25) H = nf s θ D = π + 2sin −1 Eq. (17-1): Ans. 8−4 D−d = π + 2sin −1 = 3.160 rad 2C 2(216) L = [4C2 − (D − d)2]1/2 + (DθD + dθd)/2 Eq. (17-2): = [4( [4(216)2 − (8 − 4)2]1/2 + [8(3.160) + 4(3.123)]/2 (3.123)]/2 = 450.9 in Ans. 3C 2 w 3(216 / 12) 2 (0.586) = = 0.586 in Ans. 2Fi 2(486.1) ______________________________________________________________________________ dip = (c) Eq. (17-13): 17-4 As a design task, the decision set on p. 8885 is useful. A priori decisions: • Function: Hnom = 60 hp, n = 380 rev/min, C = 192 in, Ks = 1.1 • Design factor: nd = 1 • Initial tension: Catenary • Belt material.. Table 17 17-2: Polyamide A-3, Fa = 100 lbf/in, γ = 0.042 lbf/in3, f = 0.8 • Drive geometry: d = D = 48 in • Belt thickness: t = 00.13 in Design ign variable: Belt width. Use a method of trials. Initially Initially, choose b = 6 in V = π dn = π (48)(380) = 4775 ft/min 12 12 w = 12γ bt = 12(0.042)(6)(0.13) = 0.393 lbf/ft wV 2 0.393(4775 / 60) 2 Fc = = = 77.4 lbf g 32.17 63 025H nom K s nd 63 025(60)(1.1)(1) T = = = 10 946 lbf · in n 380 Shigley’s MED, 10th edition Chapter 17 Solutions, Page 5/39 www.konkur.in 2T 2(10 946) = = 456.1 lbf d 48 F1 = ( F1)a = bFaC pCv = 6(100)(1)(1) = 600 lbf F2 = F1 − ∆F = 600 − 456.1 = 143.9 lbf ∆F = Transmitted power H ∆F (V ) = 33 000 F + F2 Fi = 1 − 2 1 F − f′ = ln 1 θ d F2 − H = Eq. (17-2): 456.1(4775) = 66 hp 33 000 600 + 143.9 Fc = − 77.4 = 294.6 lbf 2 Fc 1 600 − 77.4 = ln = 0.656 Fc π 143.9 − 77.4 L = [4(192)2 − (48 − 48)2]1/2 + [48(π) + 48(π)] / 2 = 534.8 in Friction is not fully developed, so bmin is just a little smaller than 6 in (5.7 in). Not having a figure of merit, we choose the most narrow belt available (6 in). We can improve the design by reducing the initial tension, which reduces F1 and F2, thereby increasing belt life (see the result of Prob. 17-8). This will bring f ′ to 0.80 F1 = ( ∆F + Fc ) exp ( f θ ) − Fc exp ( f θ ) − 1 exp ( f θ ) = exp(0.80π ) = 12.345 Therefore (456.1 + 77.4)(12.345) − 77.4 = 573.7 lbf 12.345 − 1 F2 = F1 − ∆F = 573.7 − 456.1 = 117.6 lbf F + F2 573.7 + 117.6 Fi = 1 − Fc = − 77.4 = 268.3 lbf 2 2 F1 = These are small reductions since f ′ is close to f , but improvements nevertheless. 1 F1 − Fc 1 573.7 − 77.4 f′ = ln = ln = 0.80 θ d F2 − Fc π 117.6 − 77.4 3C 2w 3(192 / 12)2 (0.393) = = 0.562 in 2 Fi 2(268.3) ______________________________________________________________________________ dip = Shigley’s MED, 10th edition Chapter 17 Solutions, Page 6/39 www.konkur.in 17-5 From the last equation given in the problem statement, exp ( f φ ) = 1 1 − {2T / [d (a0 − a2 )b]} 2T 1 − exp ( f φ ) = 1 d (a0 − a2 )b 2T exp ( f φ ) = exp ( f φ ) − 1 d (a0 − a2 )b b= 1 2T a0 − a2 d exp ( f φ ) exp ( f φ ) − 1 But 2T/d = 33 000Hd/V. Thus, exp ( f φ ) Q.E.D. exp ( f φ ) − 1 ______________________________________________________________________________ b= 17-6 1 33 000 H d a0 − a2 V Refer to Ex. 17-1 on p. 882 for the values used below. (a) The maximum torque prior to slip is, T = 63 025H nom K s nd 63 025(15)(1.25)(1.1) = = 742.8 lbf · in n 1750 Ans. The corresponding initial tension, from Eq. (17-9), is, Fi = T exp( f θ ) + 1 742.8 11.17 + 1 = = 148.1 lbf d exp( f θ ) − 1 6 11.17 − 1 Ans. (b) See Prob. 17-4 statement. The final relation can be written bmin = = 33 000H a exp ( f θ ) 1 2 FaC pCv − (12γ t / 32.174)(V / 60) V [exp ( f θ ) − 1] 33 000(20.6)(11.17) 1 2 100(0.7)(1) − {[12(0.042)(0.13)] / 32.174} (2749 / 60) 2749(11.17 − 1) = 4.13 in Ans. This is the minimum belt width since the belt is at the point of slip. The design must Shigley’s MED, 10th edition Chapter 17 Solutions, Page 7/39 www.konkur.in round up to an available width. Eq. (17-1): D −d 18 − 6 −1 θ d = π − 2sin −1 = π − 2sin 2C 2(96) = 3.016 511 rad D − d −1 18 − 6 θ D = π + 2sin −1 = π + 2sin 2C 2(96) = 3.266 674 rad Eq. (17-2): L = [4(96)2 − (18 − 6) 2 ]1/ 2 + = 230.074 in 1 [18(3.266 674) + 6(3.016 511)] 2 Ans. 2T 2(742.8) = = 247.6 lbf d 6 ( F1 )a = bFaC pCv = F1 = 4.13(100)(0.70)(1) = 289.1 lbf F2 = F1 − ∆F = 289.1 − 247.6 = 41.5 lbf w = 12γ bt = 12(0.042)4.13(0.130) = 0.271 lbf/ft ∆F = (c) 2 2 w V 0.271 2749 Fc = = = 17.7 lbf g 60 32.17 60 F + F2 289.1 + 41.5 Fi = 1 − Fc = − 17.7 = 147.6 lbf 2 2 Transmitted belt power H ∆F (V ) 247.6(2749) = = 20.6 hp 33 000 33 000 H 20.6 n fs = = = 1.1 H nom K s 15(1.25) H = Dip: 2 3C 2 w 3(96 / 12) ( 0.271) dip = = = 0.176 in 2Fi 2(147.6) (d) If you only change the belt width, the parameters in the following table change as shown. Shigley’s MED, 10th edition Chapter 17 Solutions, Page 8/39 www.konkur.in b w Fc (F1)a F2 Fi f′ dip Ex. 17-1 This Problem 6.00 4.13 0.393 0.271 25.6 17.7 420 289 172.4 41.5 270.6 147.6 0.33* 0.80** 0.139 0.176 *Friction underdeveloped **Friction fully developed ______________________________________________________________________________ 17-7 The transmitted power is the same. Fc Fi (F1)a F2 Ha nfs f′ dip n-Fold b = 6 in b = 12 in Change 25.65 51.3 2 270.35 664.9 2.46 420 840 2 172.4 592.4 3.44 20.62 20.62 1 1.1 1.1 1 0.139 0.125 0.90 0.328 0.114 0.34 If we relax Fi to develop full friction (f = 0.80) and obtain longer life, then n-Fold b = 6 in b = 12 in Change Fc 25.6 51.3 2 Fi 148.1 148.1 1 F1 297.6 323.2 1.09 F2 50 75.6 1.51 0.80 0.80 1 f′ dip 0.255 0.503 2 ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 17 Solutions, Page 9/39 www.konkur.in 17-8 Find the resultant of F1 and F2: D−d 2C D−d sin α = 2C 2 1D − d cos α 1 − 2 2C α = sin −1 2 1 D − d R = F1 cos α + F2 cos α = ( F1 + F2 ) 1 − 2 2C D−d R y = F1 sin α − F2 sin α = ( F1 − F2 ) Ans. 2C x Ans. From Ex. 17-2, d = 16 in, D = 36 in, C = 16(12) = 192 in, F1 = 940 lbf, F2 = 276 lbf 36 − 16 = 2.9855o 2(192) α = sin −1 2 1 36 − 16 R = (940 + 276) 1 − = 1214.4 lbf 2 2(192) 36 − 16 R y = (940 − 276) = 34.6 lbf 2(192) d 16 T = ( F1 − F2 ) = (940 − 276) = 5312 lbf · in 2 2 ______________________________________________________________________________ x 17-9 This is a good class project. Form four groups, each with a belt to design. Once each group agrees internally, all four should report their designs including the forces and torques on the line shaft. If you give them the pulley locations, they could design the line shaft. ______________________________________________________________________________ Shigley’s MED, 10th edition Chapter 17 Solutions, Page 10/39 www.konkur.in 17-10 If you have the students implement a computer program, the design problem selections may differ, and the students will be able to explore them. For Ks = 1.25, nd = 1.1, d = 14 in and D = 28 in, a polyamide A-5 belt, 8 inches wide, will do (bmin = 6.58 in) ______________________________________________________________________________ 17-11 An efficiency of less than unity lowers the output for a given input. Since the object of the drive is the output, the efficiency must be incorporated such that the belt’s capacity is increased. The design power would thus be expressed as H nom K s nd Ans. eff ______________________________________________________________________________ Hd = 17-12 Some perspective on the size of Fc can be obtained from w V 12γ bt V Fc = = g 60 g 60 2 2 An approximate comparison of non-metal and metal belts is presented in the table below. γ, lbf/in b, in t, in 3 Non-metal 0.04 5.00 0.20 Metal 0.280 1.000 0.005 The ratio w / wm is w 12(0.04)(5)(0.2) = wm 12(0.28)(1)(0.005) 29 The second contribution to Fc is the belt peripheral velocity which tends to be low in metal belts used in instrument, printer, plotter and similar drives. The velocity ratio squared influences any Fc / (Fc)m ratio. It is common for engineers to treat Fc as negligible compared to other tensions in the belting problem. However, when developing a computer code, one should include Fc. ______________________________________________________________________________ 17-13 Eq. (17-8): exp( f θ ) − 1 exp( f θ ) − 1 F1 exp( f θ ) exp( f θ ) Assuming negligible centrifugal force and setting F1 = ab from step 3, p. 889, ∆F = F1 − F2 = ( F1 − Fc ) bmin = Shigley’s MED, 10th edition ∆F exp( f θ ) a exp( f θ ) − 1 (1) Chapter 17 Solutions, Page 11/39 www.konkur.in (∆F )V 33 000 33 000 H nom K s nd ∆F = V H d = H nom K s nd = Also, 1 33 000 H d exp( f θ ) Ans. exp( f θ ) − 1 a V ______________________________________________________________________________ Substituting into Eq. (1), bmin = 17-14 The decision set for the friction metal flat-belt drive is: A priori decisions • Function: Hnom = 1 hp, n = 1750 rev/min, VR = 2 , C 15 in, Ks = 1.2 , Np = 106 belt passes. • Design factor: nd = 1.05 • Belt material and properties: 301/302 stainless steel Table 17-8: Sy = 175 kpsi, E = 28 Mpsi, ν = 0.285 • Drive geometry: d = 2 in, D = 4 in • Belt thickness: t = 0.003 in Design variables: • Belt width, b • Belt loop periphery Preliminaries H d = H nom K s nd = 1(1.2)(1.05) = 1.26 hp 63 025(1.26) T = = 45.38 lbf · in 1750 A 15 in center-to-center distance corresponds to a belt loop periphery of 39.5 in. The 40 in loop available corresponds to a 15.254 in center distance. 4−2 θ d = π − 2sin −1 = 3.010 rad 2(15.254) 4−2 θ D = π + 2sin −1 = 3.273 rad 2(15.274) For full friction development Shigley’s MED, 10th edition Chapter 17 Solutions, Page 12/39 www.konkur.in exp( f θ d ) = exp[0.35(3.010)] = 2.868 π dn π (2)(1750) V = = = 916.3 ft/s 12 12 S y = 175 kpsi Eq. (17-15): S y = 14.17 (106 ) N p−0.407 = 14.17 (106 )(106 ) −0.407 = 51.212 (103 ) psi From selection step 3, p. 889, Et 28(106 )(0.003) 3 a = S f − t = 51.212(10 ) − (0.003) (1 − ν 2 )d (1 − 0.2852 )(2) = 16.50 lbf/in of belt width ( F1)a = ab = 16.50b For full friction development, from Prob. 17-13, bmin = ∆F exp( f θ d ) a exp( f θ d ) − 1 ∆F = 2T 2(45.38) = = 45.38 lbf d 2 bmin = 45.38 2.868 = 4.23 in 16.50 2.868 − 1 So Decision #1: b = 4.5 in F1 = ( F1) a = ab = 16.5(4.5) = 74.25 lbf F2 = F1 − ∆F = 74.25 − 45.38 = 28.87 lbf F + F2 74.25 + 28.87 Fi = 1 = = 51.56 lbf 2 2 Existing friction F 1 74.25 ln 1 = ln = 0.314 θ d F2 3.010 28.87 (∆F )V 45.38(916.3) Ht = = = 1.26 hp 33 000 33 000 Ht 1.26 n fs = = = 1.05 H nom K s 1(1.2) f′ = 1 This is a non-trivial point. The methodology preserved the factor of safety corresponding to nd = 1.1 even as we rounded bmin up to b. Shigley’s MED, 10th edition Chapter 17 Solutions, Page 13/39 www.konkur.in Decision #2 was taken care of with the adjustment of the center-to-center distance to accommodate the belt loop. Use Eq. (17-2) as is and solve for C to assist in this. Remember to subsequently recalculate θd and θD . ______________________________________________________________________________ 17-15 Decision set: A priori decisions • Function: Hnom = 5 hp, N = 1125 rev/min, VR = 3, C 20 in, Ks = 1.25, Np = 106 belt passes • Design factor: nd = 1.1 • Belt material: BeCu, Sy = 170 kpsi, E = 17 Mpsi, ν = 0.220 • Belt geometry: d = 3 in, D = 9 in • Belt thickness: t = 0.003 in Design decisions • Belt loop periphery • Belt width b Preliminaries: H d = H nom K s nd = 5(1.25)(1.1) = 6.875 hp 63 025(6.875) T = = 385.2 lbf · in 1125 Decision #1: Choose a 60-in belt loop with a center-to-center distance of 20.3 in. 9−3 θ d = π − 2sin −1 = 2.845 rad 2(20.3) 9−3 θ D = π + 2sin −1 = 3.438 rad 2(20.3) For full friction development: exp( f θ d ) = exp[0.32(2.845)] = 2.485 π dn π (3)(1125) V = = = 883.6 ft/min 12 12 S f = 56.67 kpsi From selection step 3, p. 889, Shigley’s MED, 10th edition Chapter 17 Solutions, Page 14/39 www.konkur.in Et 17(106 )(0.003) 3 a = S f − t = 56.67(10 ) − (0.003) = 116.4 lbf/in (1 − ν 2 )d (1 − 0.222 )(3) 2T 2(385.2) ∆F = = = 256.8 lbf d 3 ∆F exp( f θ d ) 256.8 2.485 bmin = = = 3.69 in a exp( f θ d ) − 1 116.4 2.485 − 1 Decision #2: b = 4 in F1 = ( F1) a = ab = 116.4(4) = 465.6 lbf F2 = F1 − ∆F = 465.6 − 256.8 = 208.8 lbf F + F2 465.6 + 208.8 Fi = 1 = = 337.3 lbf 2 2 Existing friction F 1 465.6 ln 1 = ln = 0.282 θ d F2 2.845 208.8 (∆F )V 256.8(883.6) H = = = 6.88 hp 33 000 33 000 H 6.88 n fs = = = 1.1 5(1.25) 5(1.25) f′ = 1 Fi can be reduced only to the point at which f ′ = f = 0.32. From Eq. (17-9) Fi = T d exp( f θ d ) + 1 385.2 2.485 + 1 = = 301.3 lbf 3 2.485 − 1 exp( f θ d ) − 1 Eq. (17-10): 2 exp( f θ d ) 2(2.485) F1 = Fi = 429.7 lbf = 301.3 2.485 + 1 exp( f θ d ) + 1 F2 = F1 − ∆F = 429.7 − 256.8 = 172.9 lbf and f ′ = f = 0.32 ______________________________________________________________________________ 17-16 This solution is the result of a series of five design tasks involving different belt thicknesses. The results are to be compared as a matter of perspective. These design tasks are accomplished in the same manner as in Probs. 17-14 and 17-15 solutions. The details will not be presented here, but the table is provided as a means of learning. Shigley’s MED, 10th edition Chapter 17 Solutions, Page 15/39 www.konkur.in Five groups of students could each be assigned a belt thickness. You can form a table from their results or use the table given here. 0.002 4.000 20.300 109.700 3.000 9.000 310.600 439.000 182.200 1.100 60.000 0.309 301.200 429.600 172.800 0.320 b CD a d D Fi F1 F2 nf s L f′ Fi F1 F2 f 0.003 3.500 20.300 131.900 3.000 9.000 333.300 461.700 209.000 1.100 60.000 0.285 301.200 429.600 172.800 0.320 t, in 0.005 4.000 20.300 110.900 3.000 9.000 315.200 443.600 186.800 1.100 60.000 0.304 301.200 429.600 172.800 0.320 0.008 1.500 18.700 194.900 5.000 15.000 215.300 292.300 138.200 1.100 70.000 0.288 195.700 272.700 118.700 0.320 0.010 1.500 20.200 221.800 6.000 18.000 268.500 332.700 204.300 1.100 80.000 0.192 166.600 230.800 102.400 0.320 The first three thicknesses result in the same adjusted Fi, F1 and F2 (why?). We have no figure of merit, but the costs of the belt and pulleys are about the same for these three thicknesses. Since the same power is transmitted and the belts are widening, belt forces are lessening. ______________________________________________________________________________ 17-17 This is a design task. The decision variables would be belt length and belt section, which could be combined into one, such as B90. The number of belts is not an issue. We have no figure of merit, which is not practical in a text for this application. It is suggested that you gather sheave dimensions and costs and V-belt costs from a principal vendor and construct a figure of merit based on the costs. Here is one trial. Preliminaries: For a single V-belt drive with Hnom = 3 hp, n = 3100 rev/min, D = 12 in, and d = 6.2 in, choose a B90 belt, Ks = 1.3 and nd = 1. From Table 17-10, select a circumference of 90 in. From Table 17-11, add 1.8 in giving Lp = 90 + 1.8 = 91.8 in Eq. (17-16b): π C = 0.25 91.8 − (12 + 6.2) + 2 = 31.47 in Shigley’s MED, 10th edition 2 π 2 91.8 − (12 + 6.2) − 2(12 − 6.2) 2 Chapter 17 Solutions, Page 16/39 www.konkur.in 12 − 6.2 θ d = π − 2sin -1 = 2.9570 rad 2(31.47) exp( f θ d ) = exp[0.5123(2.9570)] = 4.5489 π dn π (6.2)(3100) V = 12 = 12 = 5031.8 ft/min Table 17-13: Angleθ = θ d 180° π 180° = (2.957 rad) = 169.42° π The footnote regression equation of Table 17-13 gives K1 without interpolation: K1 = 0.143 543 + 0.007 468(169.42°) − 0.000 015 052(169.42°)2 = 0.9767 The design power is Hd = HnomKsnd = 3(1.3)(1) = 3.9 hp From Table 17-14 for B90, K2 = 1. From Table 17-12 take a marginal entry of Htab = 4, although extrapolation would give a slightly lower Htab. Eq. (17-17): Ha = K1K2Htab = 0.9767(1)(4) = 3.91 hp The allowable ∆Fa is given by ∆Fa = 63 025H a 63 025(3.91) = = 25.6 lbf n(d / 2) 3100(6.2 / 2) The allowable torque Ta is Ta = ∆Fa d 25.6(6.2) = = 79.4 lbf · in 2 2 From Table 17-16, Kc = 0.965. Thus, Eq. (17-21) gives, 2 2 V 5031.8 Fc = K c = 0.965 = 24.4 lbf 1000 1000 At incipient slip, Eq. (17-9) provides: T Fi = d exp( f θ ) + 1 79.4 4.5489 + 1 = 20.0 lbf = exp( f θ ) − 1 6.2 4.5489 − 1 Eq. (17-10): Shigley’s MED, 10th edition Chapter 17 Solutions, Page 17/39 www.konkur.in 2 exp( f θ ) 2(4.5489) F1 = Fc + Fi = 24.4 + 20 = 57.2 lbf 4.5489 + 1 exp( f θ ) + 1 Thus, F2 = F1 − ∆Fa = 57.2 − 25.6 = 31.6 lbf Eq. (17-26): n fs = H a Nb (3.91)(1) = = 1.003 Hd 3.9 Ans. If we had extrapolated for Htab, the factor of safety would have been slightly less than one. Life Use Table 17-16 to find equivalent tensions T1 and T2 . Kb = 57.2 + d K T2 = F1 + ( Fb ) 2 = F1 + b = 57.2 + D T1 = F1 + ( Fb )1 = F1 + 576 = 150.1 lbf 6.2 576 = 105.2 lbf 12 From Table 17-17, K = 1193, b = 10.926, and from Eq. (17-27), the number of belt passes is: −1 K − b K − b N P = + T1 T2 −1 1193 −10.926 1193 −10.926 9 = + = 6.72(10 ) passes 105.2 150.1 From Eq. (17-28) for NP > 109, 109 (91.8) 720V 720(5031.8) t > 25 340 h Ans. t = N P Lp > Suppose nf s was too small. Compare these results with a 2-belt solution. H tab = 4 hp/belt, Ta = 39.6 lbf · in/belt, ∆Fa = 12.8 lbf/belt, H a = 3.91 hp/belt NH Nb H a 2(3.91) n fs = b a = = = 2.0 Hd H nom K s 3(1.3) Also, F1 = 40.8 lbf/belt, Shigley’s MED, 10th edition F2 = 28.0 lbf/belt Chapter 17 Solutions, Page 18/39 www.konkur.in Fi = 9.99 lbf/belt, Fc = 24.4 lbf/belt ( Fb )1 = 92.9 lbf/belt, ( Fb )2 = 48 lbf/belt T1 = 133.7 lbf/belt, T2 = 88.8 lbf/belt 10 N P = 2.39(10 ) passes, t > 605 600 h Initial tension of the drive: (Fi)drive = NbFi = 2(9.99) = 20 lbf ______________________________________________________________________________ 17-18 Given: two B85 V-belts with d = 5.4 in, D = 16 in,