Budynas ch10

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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Chapter 10 10-1 10-2 A = Sd m dim( A uscu ) = dim( S ) dim( d m ) = kpsi · in m dim( A SI ) = dim( S 1 ) dim ( d m 1 ) = MPa · mm m A SI = MPa kpsi · mm m in m A uscu = 6 . 894 757(25 . 40) m A uscu . = 6 . 895(25 . 4) m A uscu Ans. For music wire, from Table 10-4: A uscu = 201, m = 0 . 145 ; what is A SI ? A SI = 6 . 89(25 . 4) 0 . 145 (201) = 2214 MPa · mm m Ans. 10-3 Given: Music wire, d = 0 . 105 in, OD = 1 . 225 in, plain ground ends, N t = 12 coils. Table 10-1: N a = N t 1 = 12 1 = 11 L s = dN t = 0 . 105(12) = 1 . 26 in Table 10-4: A = 201, m = 0 . 145 (a) Eq. (10-14): S ut = 201 (0 . 105) 0 . 145 = 278 . 7 kpsi Table 10-6: S sy = 0 . 45(278 . 7) = 125 . 4 kpsi D = 1 . 225 0 . 105 = 1 . 120 in C = D d = 1 . 120 0 . 105 = 10 . 67 Eq. (10-6): K B = 4(10 . 67) + 2 4(10 . 67) 3 = 1 . 126 Eq. (10-3): F | S sy = π d 3 S sy 8 K B D = π (0 . 105) 3 (125 . 4)(10 3 ) 8(1 . 126)(1 . 120) = 45 . 2 lbf Eq. (10-9): k = d 4 G 8 D 3 N a = (0 . 105) 4 (11 . 75)(10 6 ) 8(1 . 120) 3 (11) = 11 . 55 lbf/in L 0 = F | S sy k + L s = 45 . 2 11 . 55 + 1 . 26 = 5 . 17 in Ans . 1 2 " 4" 1" 1 2 " 4" 1"
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262 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design (b) F | S sy = 45 . 2 lbf Ans . (c) k = 11 . 55 lbf/in Ans . (d) ( L 0 ) cr = 2 . 63 D α = 2 . 63(1 . 120) 0 . 5 = 5 . 89 in Many designers provide ( L 0 ) cr / L 0 5 or more; therefore, plain ground ends are not often used in machinery due to buckling uncertainty. 10-4 Referring to Prob. 10-3 solution, C = 10 . 67, N a = 11, S sy = 125 . 4 kpsi, ( L 0 ) cr = 5 . 89 in and F = 45 . 2 lbf (at yield) . Eq. (10-18): 4 C 12 C = 10 . 67 O . K . Eq. (10-19): 3 N a 15 N a = 11 O . K . L 0 = 5 . 17 in, L s = 1 . 26 in y 1 = F 1 k = 30 11 . 55 = 2 . 60 in L 1 = L 0 y 1 = 5 . 17 2 . 60 = 2 . 57 in ξ = y s y 1 1 = 5 . 17 1 . 26 2 . 60 1 = 0 . 50 Eq. (10-20): ξ 0 . 15, ξ = 0 . 50 O . K . From Eq. (10-3) for static service τ 1 = K B 8 F 1 D π d 3 = 1 . 126 8(30)(1 . 120) π (0 . 105) 3 = 83 224 psi n s = S sy τ 1 = 125 . 4(10 3 ) 83 224 = 1 . 51 Eq. (10-21): n s 1 . 2, n s = 1 . 51 O . K . τ s = τ 1 45 . 2 30 = 83 224 45 . 2 30 = 125 391 psi S sy s = 125 . 4(10 3 ) / 125 391 . = 1 S sy s ( n s ) d : Not solid-safe. Not O.K. L 0 ( L 0 ) cr : 5 . 17 5 . 89 Margin could be higher, Not O.K. Design is unsatisfactory. Operate over a rod? Ans. L 0 L 1 y 1 F 1 y s L s F s
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Chapter 10 263 10-5 Static service spring with: HD steel wire, d = 2 mm, OD = 22 mm, N t = 8 . 5 turns plain and ground ends. Preliminaries Table 10-5: A = 1783 MPa · mm m , m = 0 . 190 Eq. (10-14): S ut = 1783 (2) 0 . 190 = 1563 MPa Table 10-6: S sy = 0 . 45(1563) = 703 . 4 MPa Then, D = OD d = 22 2 = 20 mm C = 20 / 2 = 10 K B = 4 C + 2 4 C 3 = 4(10) + 2 4(10) 3 = 1 . 135 N a = 8 . 5 1 = 7 . 5 turns L s = 2(8 . 5) = 17 mm Eq. (10-21): Use n s = 1 . 2 for solid-safe property. F s = π d 3 S sy / n s 8 K B D = π (2) 3 (703 . 4 / 1 . 2) 8(1 . 135)(20) (10 3 ) 3 (10 6 ) 10 3 = 81 . 12 N k = d 4 G 8 D 3 N a = (2) 4 (79 . 3) 8(20) 3 (7 . 5) (10 3 ) 4 (10 9 ) (10 3 ) 3 = 0 . 002 643(10 6 ) = 2643 N/m y s = F s k = 81 . 12 2643(10 3 ) = 30 . 69 mm (a) L 0 = y + L s = 30 . 69 + 17 = 47 . 7 mm Ans . (b) Table 10-1: p = L 0 N t = 47 . 7 8 . 5 = 5 . 61 mm Ans . (c) F s = 81 . 12 N (from above) Ans . (d) k = 2643 N/m (from above) Ans . (e) Table 10-2 and Eq. (10-13): ( L 0 ) cr = 2 . 63 D α = 2 . 63(20) 0 . 5 = 105 . 2 mm ( L 0 ) cr / L 0 = 105 . 2 / 47 . 7 = 2 . 21 This is less than 5. Operate over a rod? Plain and ground ends have a poor eccentric footprint. Ans. 10-6 Referring to Prob. 10-5 solution: C = 10, N a = 7 . 5, k = 2643 N/m, d = 2 mm, D = 20 mm, F s = 81 . 12 N and N t = 8 . 5 turns . Eq. (10-18): 4 C 12, C = 10 O . K .
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264 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design Eq. (10-19): 3 N a 15, N a = 7 . 5 O . K .
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