Chapter 10 1" 2 4" 1 = MPa mm. 201 = kpsi (0.105) S sy = 0.45(278.7) = kpsi D = = in. C = D d = = 10.

Transcription

1 budynas_sm_ch10.qxd 01/29/ :26 Page 261 Chapter " 4" 1" 2 1" 4" 1" 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 mmm in m A uscu = (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) (201) = 2214 MPa mm m Ans Given: Music wire, d = in, OD = 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 = (a) Eq. (10-14): S ut = Table 10-6: 201 = kpsi (0.105) S sy = 0.45(278.7) = kpsi D = = in C = D d = = Eq. (10-6): K B = 4(10.67) + 2 4(10.67) 3 = Eq. (10-3): F Ssy = πd3 S sy 8K B D = π(0.105)3 (125.4)(10 3 ) = 45.2 lbf 8(1.126)(1.120) Eq. (10-9): k = d4 G = (0.105)4 (11.75)(10 6 ) = lbf/in 8D 3 N a 8(1.120) 3 (11) L 0 = F S sy k + L s = = 5.17 in Ans

2 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design (b) F Ssy = 45.2 lbf Ans. (c) k = lbf/in Ans. (d) (L 0 ) cr = 2.63D 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 Referring to Prob solution, C = 10.67, N a = 11, S sy = kpsi, (L 0 ) cr = 5.89 in and F = 45.2 lbf (at yield). Eq. (10-18): 4 C 12 C = O.K. Eq. (10-19): 3 N a 15 N a = 11 O.K. y 1 F1 y s L 0 F s L 1 L s L 0 = 5.17 in, L s = 1.26 in y 1 = F 1 k = 30 = 2.60 in L 1 = L 0 y 1 = = 2.57 in ξ = y s = 1 = 0.50 y Eq. (10-20): ξ 0.15, ξ = 0.50 O.K. From Eq. (10-3) for static service ( ) [ ] 8F1 D 8(30)(1.120) τ 1 = K B = = psi π(0.105) 3 n s = S sy = 125.4(103 ) = 1.51 τ Eq. (10-21): n s 1.2, n s = 1.51 O.K. ( ) ( ) τ s = τ 1 = = psi S sy /τ s = 125.4(10 3 )/ =. 1 S sy /τ s (n s ) d : Not solid-safe. Not O.K. L 0 (L 0 ) cr : Margin could be higher, Not O.K. Design is unsatisfactory. Operate over a rod? Ans.

3 budynas_sm_ch10.qxd 01/29/ :26 Page 263 Chapter Static service spring with: HD steel wire, d = 2mm, OD = 22 mm, N t = 8.5 turns plain and ground ends. Preliminaries Table 10-5: A = 1783 MPa mm m, m = Eq. (10-14): S ut = 1783 = 1563 MPa (2) Table 10-6: S sy = 0.45(1563) = 703.4MPa Then, D = OD d = 22 2 = 20 mm C = 20/2 = 10 K B = 4C + 2 4C 3 = 4(10) + 2 4(10) 3 = N a = = 7.5 turns L s = 2(8.5) = 17 mm Eq. (10-21): Use n s = 1.2 for solid-safe property. [ F s = πd3 S sy /n s 8K B D = π(2)3 (703.4/1.2) (10 3 ) 3 (10 6 ] ) = N 8(1.135)(20) 10 3 [ k = d4 G = (2)4 (79.3) (10 3 ) 4 (10 9 ] ) = (10 6 ) = 2643 N/m 8D 3 N a 8(20) 3 (7.5) (10 3 ) 3 y s = F s k = (10 3 ) = mm (a) L 0 = y + L s = = 47.7mm Ans. (b) Table 10-1: p = L 0 = 47.7 = 5.61 mm Ans. N t 8.5 (c) F s = N (from above) Ans. (d) k = 2643 N/m (from above) Ans. (e) Table 10-2 and Eq. (10-13): (L 0 ) cr = 2.63D = 2.63(20) α 0.5 (L 0 ) cr /L 0 = 105.2/47.7 = 2.21 = 105.2mm This is less than 5. Operate over a rod? Plain and ground ends have a poor eccentric footprint. Ans Referring to Prob solution: C = 10, N a = 7.5, k = 2643 N/m, d = 2mm, D = 20 mm, F s = N and N t = 8.5 turns. Eq. (10-18): 4 C 12, C = 10 O.K.

4 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design Eq. (10-19): 3 N a 15, N a = 7.5 O.K. y 1 = F 1 k = (10 3 ) = 28.4mm (y) for yield = 81.12(1.2) 2643(10 3 ) = 36.8mm y s = = mm 2643(10 3 ) ξ = (y) for yield 1 = 36.8 y = Eq. (10-20): ξ 0.15, ξ = O.K. Table 10-6: S sy = 0.45S ut O.K. As-wound ( ) [ ][ 8Fs D 8(81.12)(20) 10 3 ] τ s = K B = = 586 MPa π(2) 3 (10 3 ) 3 (10 6 ) Eq. (10-21): S sy = = 1.2 O.K. (Basis for Prob solution) τ s 586 Table 10-1: L s = N t d = 8.5(2) = 17 mm L 0 = F s k + L s = = 47.7mm D = 2.63(20) = 105.2mm α 0.5 (L 0 ) cr = L = 2.21 which is less than 5. Operate over a rod? Not O.K. Plain and ground ends have a poor eccentric footprint Given: A228 (music wire), SQ&GRD ends, d = in, OD = in, L 0 = 0.63 in, N t = 40 turns. Table 10-4: A = 201 kpsi in m, m = D = OD d = = in C = D/d = 0.030/0.006 = 5 K B = 4(5) + 2 4(5) 3 = Table 10-1: N a = N t 2 = 40 2 = 38 turns 201 S ut = = kpsi (0.006) S sy = 0.45(422.1) = kpsi k = Gd4 = 12(106 )(0.006) 4 8D 3 N a 8(0.030) 3 (38) Ans. = lbf/in

5 budynas_sm_ch10.qxd 01/29/ :26 Page 265 Chapter Table 10-1: L s = N t d = 40(0.006) = in Now F s = ky s where y s = L 0 L s = in. Thus, [ ] [ ] 8(kys )D 8(1.895)(0.39)(0.030) τ s = K B = (10 3 ) = kpsi π(0.006) 3 (1) τ s > S sy, that is, > kpsi; the spring is not solid-safe. Solving Eq. (1) for y s gives y s = (τ s/n s )( ) 8K B kd Using a design factor of 1.2, = ( /1.2)(π)(0.006)3 8(1.294)(1.895)(0.030) L 0 = L s + y s = = in The spring should be wound to a free length of in. Ans. = in 10-8 Given: B159 (phosphor bronze), SQ&GRD ends, d = in, OD = in, L 0 = 0.81 in, N t = 15.1 turns. Table 10-4: A = 145 kpsi in m, m = 0 Table 10-5: G = 6 Mpsi D = OD d = = in C = D/d = 0.108/0.012 = 9 K B = 4(9) + 2 4(9) 3 = Table 10-1: N a = N t 2 = = 13.1 turns S ut = 145 = 145 kpsi Table 10-6: S sy = 0.35(145) = 50.8 kpsi k = Gd4 = 6(106 )(0.012) 4 = lbf/in 8D 3 N a 8(0.108) 3 (13.1) Table 10-1: L s = dn t = 0.012(15.1) = in Now F s = ky s, y s = L 0 L s = = in [ ] [ ] 8(kys )D 8(0.942)(0.6)(0.108) τ s = K B = (10 3 ) = kpsi π(0.012) 3 (1) τ s > S sy, that is, > 50.8 kpsi; the spring is not solid safe. Solving Eq. (1) for y s gives y s = (S sy/n) 8K B kd = (50.8/1.2)(π)(0.012)3 (10 3 ) = in 8(1.152)(0.942)(0.108) L 0 = L s + y s = = in Wind the spring to a free length of in. Ans.

6 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design 10-9 Given: A313 (stainless steel), SQ&GRD ends, d = in, OD = in, L 0 = 0.75 in, N t = 10.4 turns. Table 10-4: A = 169 kpsi in m, m = Table 10-5: G = 10(10 6 ) psi D = OD d = = in C = D/d = 0.200/0.040 = 5 K B = 4(5) + 2 4(5) 3 = Table 10-6: N a = N t 2 = = 8.4 turns 169 S ut = = kpsi (0.040) Table 10-13: S sy = 0.35(270.4) = 94.6 kpsi k = Gd4 = 10(106 )(0.040) 4 = lbf/in 8D 3 N a 8(0.2) 3 (8.4) Table 10-6: L s = dn t = 0.040(10.4) = in Now F s = ky s, y s = L 0 L s = = in [ ] [ ] 8(kys )D 8(47.62)(0.334)(0.2) τ s = K B = (10 3 ) = kpsi π(0.040) 3 (1) τ s > S sy, that is, > 94.6 kpsi; the spring is not solid-safe. Solving Eq. (1) for y s gives y s = (S sy/n)( ) 8K B kd L 0 = L s + y s Wind the spring to a free length in. = (94600/1.2)(π)(0.040)3 8(1.294)(47.62)(0.2) = = in Ans. = in Given: A227 (hard drawn steel), d = in, OD = 2.0in, L 0 = 2.94 in, N t = 5.25 turns. Table 10-4: A = 140 kpsi in m, m = Table 10-5: G = 11.4(10 6 ) psi D = OD d = = in C = D/d = 1.865/0.135 = K B = 4(13.81) + 2 4(13.81) 3 = N a = N t 2 = = 3.25 turns S ut = 140 = kpsi (0.135) 0.190

7 budynas_sm_ch10.qxd 01/29/ :26 Page 267 Chapter Table 10-6: S sy = 0.45(204.8) = 92.2 kpsi k = Gd4 = 11.4(106 )(0.135) 4 = lbf/in 8D 3 N a 8(1.865) 3 (3.25) Table 10-1: L s = dn t = 0.135(5.25) = in Now F s = ky s, y s = L 0 L s = = in [ ] [ ] 8(kys )D 8(22.45)(2.231)(1.865) τ s = K B = (10 3 ) = kpsi π(0.135) 3 (1) τ s > S sy, that is, 106 > 92.2 kpsi; the spring is not solid-safe. Solving Eq. (1) for y s gives y s = (S sy/n)( ) 8K B kd L 0 = L s + y s = (92200/1.2)(π)(0.135)3 8(1.096)(22.45)(1.865) = = in Wind the spring to a free length of 2.32 in. Ans. = in Given: A229 (OQ&T steel), SQ&GRD ends, d = in, OD = 1.0 in, L 0 = 3.75 in, N t = 13 turns. Table 10-4: A = 147 kpsi in m, m = Table 10-5: G = 11.4(10 6 ) psi D = OD d = = in C = D/d = 0.856/0.144 = K B = 4(5.944) + 2 4(5.944) 3 = Table 10-1: N a = N t 2 = 13 2 = 11 turns S ut = 147 = kpsi (0.144) Table 10-6: S sy = 0.50(211.2) = kpsi k = Gd4 = 11.4(106 )(0.144) 4 = 88.8 lbf/in 8D 3 N a 8(0.856) 3 (11) Table 10-1: L s = dn t = 0.144(13) = in Now F s = ky s, y s = L 0 L s = = in [ ] [ ] 8(kys )D 8(88.8)(1.878)(0.856) τ s = K B = (10 3 ) = kpsi π(0.144) 3 (1) τ s > S sy, that is,151.1 > kpsi; the spring is not solid-safe. Solving Eq. (1) for y s gives y s = (S sy/n)( ) ( /1.2)(π)(0.144)3 = = in 8K B kd 8(1.241)(88.8)(0.856) L 0 = L s + y s = = in Wind the spring to a free length in. Ans.

8 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design Given: A232 (Cr-V steel), SQ&GRD ends, d = in, OD = 3 in, L 0 = 9 in, N t = 8turns. Table 10-4: A = 169 kpsi in m, m = Table 10-5: G = 11.2(10 6 ) psi D = OD d = = in C = D/d = 2.808/0.192 = (large) K B = 4(14.625) + 2 4(14.625) 3 = Table 10-1: N a = N t 2 = 8 2 = 6 turns S ut = 169 = kpsi (0.192) Table 10-6: S sy = 0.50(223.0) = kpsi k = Gd4 = 11.2(106 )(0.192) 4 = lbf/in 8D 3 N a 8(2.808) 3 (6) Table 10-1: L s = dn t = 0.192(8) = in Now F s = ky s, y s = L 0 L s = = in [ ] [ ] 8(kys )D 8(14.32)(7.464)(2.808) τ s = K B = (10 3 ) = kpsi π(0.192) 3 (1) τ s > S sy, that is,117.7 > kpsi; the spring is not solid safe. Solving Eq. (1) for y s gives y s = (S sy/n)( ) 8K B kd L 0 = L s + y s = ( /1.2)(π)(0.192)3 8(1.090)(14.32)(2.808) = = in Wind the spring to a free length of in. Ans. = in Given: A313 (stainless steel) SQ&GRD ends, d = 0.2mm, OD = 0.91 mm, L 0 = 15.9mm, N t = 40 turns. Table 10-4: A = 1867 MPa mm m, m = Table 10-5: G = 69.0GPa D = OD d = = 0.71 mm C = D/d = 0.71/0.2 = 3.55 (small) K B = 4(3.55) + 2 4(3.55) 3 = N a = N t 2 = 40 2 = 38 turns S ut = 1867 = MPa (0.2) 0.146

9 budynas_sm_ch10.qxd 01/29/ :26 Page 269 Chapter Table 10-6: S sy = 0.35(2361.5) = 826.5MPa [ k = d4 G = (0.2)4 (69.0) (10 3 ) 4 (10 9 ] ) 8D 3 N a 8(0.71) 3 (38) (10 3 ) 3 = (10 3 )(10 6 ) = N/m or N/mm L s = dn t = 0.2(40) = 8mm F s = ky s y s = L 0 L s = = 7.9 [ ] [ ][ 8(kys )D 8(1.0147)(7.9)(0.71) 10 3 (10 3 )(10 3 ] ) τ s = K B = π(0.2) 3 (10 3 ) 3 = 2620(1) = 2620 MPa (1) τ s > S sy, that is,2620 > MPa; the spring is not solid safe. Solve Eq. (1) for y s giving y s = (S sy/n)( ) 8K B kd L 0 = L s + y s = (826.5/1.2)(π)(0.2)3 8(1.446)(1.0147)(0.71) = = mm = 2.08 mm Wind the spring to a free length of mm. This only addresses the solid-safe criteria. There are additional problems. Ans Given: A228 (music wire), SQ&GRD ends, d = 1mm, OD = 6.10 mm, L 0 = 19.1mm, N t = 10.4 turns. Table 10-4: A = 2211 MPa mm m, m = Table 10-5: G = 81.7GPa D = OD d = = 5.1mm C = D/d = 5.1/1 = 5.1 N a = N t 2 = = 8.4 turns K B = 4(5.1) + 2 4(5.1) 3 = S ut = 2211 = 2211 MPa (1) Table 10-6: S sy = 0.45(2211) = 995 MPa [ k = d4 G = (1)4 (81.7) (10 3 ) 4 (10 9 ] ) = (10 6 ) 8D 3 N a 8(5.1) 3 (8.4) (10 3 ) 3 = 9165 N/m or N/mm L s = dn t = 1(10.4) = 10.4mm F s = ky s

10 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design y s = L 0 L s = = 8.7mm [ ] [ ] 8(kys )D 8(9.165)(8.7)(5.1) τ s = K B = = 1333 MPa π(1) 3 (1) τ s > S sy, that is, 1333 > 995 MPa; the spring is not solid safe. Solve Eq. (1) for y s giving y s = (S sy/n)( ) 8K B kd L 0 = L s + y s = (995/1.2)(π)(1)3 8(1.287)(9.165)(5.1) = = mm Wind the spring to a free length of mm. Ans. = 5.43 mm Given: A229 (OQ&T spring steel), SQ&GRD ends, d = 3.4 mm, OD = 50.8 mm, L 0 = 74.6 mm, N t = Table 10-4: A = 1855 MPa mm m, m = Table 10-5: Table 10-6: G = 77.2GPa D = OD d = = 47.4mm C = D/d = 47.4/3.4 = (large) N a = N t 2 = = 3.25 turns K B = 4(13.94) + 2 4(13.94) 3 = S ut = 1855 = 1476 MPa (3.4) S sy = 0.50(1476) = 737.8MPa k = d4 G = (3.4)4 (77.2) 8D 3 N a 8(47.4) 3 (3.25) = 3750 N/m or N/mm L s = dn t = 3.4(5.25) = [ (10 3 ) 4 (10 9 ] ) = (10 6 ) (10 3 ) 3 F s = ky s y s = L 0 L s = = mm [ ] 8(kys )D τ s = K B [ ] 8(3.750)(56.75)(47.4) = = 720.2MPa π(3.4) 3 (1) τ s < S sy, that is, < MPa

11 budynas_sm_ch10.qxd 01/29/ :26 Page 271 Chapter The spring is solid safe. With n s = 1.2, y s = (S sy/n)( ) 8K B kd L 0 = L s + y s = (737.8/1.2)(π)(3.4)3 8(1.095)(3.75)(47.4) = = mm = mm Wind the spring to a free length of mm. Ans Given: B159 (phosphor bronze), SQ&GRD ends, d = 3.7 mm, OD = 25.4 mm, L 0 = 95.3mm, N t = 13 turns. Table 10-4: A = 932 MPa mm m, m = Table 10-5: G = 41.4GPa D = OD d = = 21.7mm C = D/d = 21.7/3.7 = K B = 4(5.865) + 2 4(5.865) 3 = N a = N t 2 = 13 2 = 11 turns S ut = 932 = 857.1MPa (3.7) Table 10-6: S sy = 0.35(857.1) = 300 MPa [ k = d4 G = (3.7)4 (41.4) (10 3 ) 4 (10 9 ] ) = (10 6 ) 8D 3 N a 8(21.7) 3 (11) (10 3 ) 3 = 8629 N/m or N/mm L s = dn t = 3.7(13) = 48.1mm F s = ky s y s = L 0 L s = = 47.2mm [ ] 8(kys )D τ s = K B [ ] 8(8.629)(47.2)(21.7) = = 553 MPa π(3.7) 3 (1) τ s > S sy, that is, 553 > 300 MPa; the spring is not solid-safe. Solving Eq. (1) for y s gives y s = (S sy/n)( ) 8K B kd L 0 = L s + y s = (300/1.2)(π)(3.7)3 8(1.244)(8.629)(21.7) = = mm = mm Wind the spring to a free length of mm. Ans.

12 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design Given: A232 (Cr-V steel), SQ&GRD ends, d = 4.3mm, OD = 76.2mm, L 0 = 228.6mm, N t = 8 turns. Table 10-4: A = 2005 MPa mm m, m = Table 10-5: G = 77.2GPa D = OD d = = 71.9mm C = D/d = 71.9/4.3 = (large) K B = 4(16.72) + 2 4(16.72) 3 = N a = N t 2 = 8 2 = 6 turns S ut = 2005 = 1569 MPa (4.3) Table 10-6: S sy = 0.50(1569) = 784.5MPa [ k = d4 G = (4.3)4 (77.2) (10 3 ) 4 (10 9 ] ) = (10 6 ) 8D 3 N a 8(71.9) 3 (6) (10 3 ) 3 = 1479 N/m or N/mm L s = dn t = 4.3(8) = 34.4mm F s = ky s y s = L 0 L s = = 194.2mm [ ] [ ] 8(kys )D 8(1.479)(194.2)(71.9) τ s = K B = = 713.0MPa π(4.3) 3 (1) τ s < S sy, that is, < 784.5; the spring is solid safe. With n s = 1.2 Eq. (1) becomes y s = (S sy/n)( ) 8K B kd = (784.5/1.2)(π)(4.3)3 8(1.078)(1.479)(71.9) = 178.1mm L 0 = L s + y s = = 212.5mm Wind the spring to a free length of L 0 = mm. Ans For the wire diameter analyzed, G = Mpsi per Table Use squared and ground ends. The following is a spread-sheet study using Fig for parts (a) and (b). For N a, k = 20/2 = 10 lbf/in.

13 budynas_sm_ch10.qxd 01/29/ :26 Page 273 Chapter (a) Spring over a Rod (b) Spring in a Hole Source Parameter Values Source Parameter Values d d D D ID ID OD OD Eq. (10-2) C Eq. (10-2) C Eq. (10-9) N a Eq. (10-9) N a Table 10-1 N t Table 10-1 N t Table 10-1 L s Table 10-1 L s y + L s L y + L s L Eq. (10-13) (L 0 ) cr Eq. (10-13) (L 0 ) cr Table 10-4 A Table 10-4 A Table 10-4 m Table 10-4 m Eq. (10-14) S ut Eq. (10-14) S ut Table 10-6 S sy Table 10-6 S sy Eq. (10-6) K B Eq. (10-6) K B Eq. (10-3) n s Eq. (10-3) n s Eq. (10-22) fom Eq. (10-22) fom For n s 1.2, the optimal size is d = in for both cases From the figure: L 0 = 120 mm, OD = 50 mm, and d = 3.4mm. Thus D = OD d = = 46.6mm (a) By counting, N t = 12.5 turns. Since the ends are squared along 1/4 turn on each end, N a = = 12 turns Ans. p = 120/12 = 10 mm Ans. The solid stack is 13 diameters across the top and 12 across the bottom. L s = 13(3.4) = 44.2mm Ans. (b) d = 3.4/25.4 = in and from Table 10-5, G = 78.6GPa k = d4 G = (3.4)4 (78.6)(10 9 ) (10 3 ) = 1080 N/m Ans. 8D 3 N a 8(46.6) 3 (12) (c) F s = k(l 0 L s ) = 1080( )(10 3 ) = 81.9N (d) C = D/d = 46.6/3.4 = K B = 4(13.71) + 2 4(13.71) 3 = Ans. τ s = 8K B F s D = 8(1.096)(81.9)(46.6) π(3.4) 3 = 271 MPa Ans One approach is to select A HD steel for its low cost. Then, for y 1 3/8 at F 1 = 10 lbf, k 10/ = lbf/in. Try d = in #14 gauge

14 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design For a clearance of 0.05 in: ID = (7/16) = in; OD = = in D = = in C = /0.08 = G = 11.5 Mpsi N a = d4 G 8kD = (0.08)4 (11.5)(10 6 ) = 12.0 turns 3 8(26.67)(0.5675) 3 N t = = 14 turns, L s = dn t = 0.08(14) = 1.12 in O.K. L 0 = in, y s = = in F s = ky s = 26.67(0.755) = lbf K B = 4(7.094) + 2 4(7.094) 3 = ( ) [ ] 8Fs D 8(20.14)(0.5675) τ s = K B = = psi π(0.08) 3 Table 10-4: A = 140 kpsi in m, m = S sy = 0.45 = kpsi (0.080) n = = 1.50 > 1.2 O.K τ 1 = F 1 τ s = 10 (68.05) = kpsi, F s n 1 = = 3.01 > O.K. There is much latitude for reducing the amount of material. Iterate on y 1 using a spread sheet. The final results are: y 1 = 0.32 in, k = lbf/in, N a = 10.3 turns, N t = 12.3 turns, L s = in, L 0 = in, y s = in, F s = 26.1 lbf, K B = 1.197, τ s = kpsi, n s = 1.15, and n 1 = ID = in, OD = in, d = in Try other sizes and/or materials A stock spring catalog may have over two hundred pages of compression springs with up to 80 springs per page listed. Students should be 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) Itisbetter to familiarize yourself with vendor resources rather than invent them yourself. Sample catalog pages can be given to students for study.

15 budynas_sm_ch10.qxd 01/29/ :26 Page 275 Chapter For a coil radius given by: R = R 1 + R 2 R 1 2π N θ The torsion of a section is T = PR where dl = Rdθ δ p = U P = 1 T T GJ P dl = 1 2π N PR 3 dθ GJ 0 = P 2π N ( R 1 + R ) 2 R 3 1 GJ 0 2π N θ dθ = P ( )( ) [ ( 1 2π N R 1 + R ) ] 2 R 4 2π N 1 GJ 4 R 2 R 1 2π N θ 0 π PN ( = R 4 2GJ(R 2 R 1 ) 2 R1) 4 π PN = 2GJ (R 1 + R 2 ) ( R1 2 + ) R2 2 J = π 32 d4 δ p = 16PN Gd (R R 2 ) ( R1 2 + ) R2 2 k = P d 4 G = δ p 16N(R 1 + R 2 ) ( R1 2 + ) Ans. R For a food service machinery application select A313 Stainless wire. G = 10(10 6 ) psi Note that for d 0.10 in A = 169, m = < d 0.20 in A = 128, m = F a = 18 4 = 7 lbf, F m = = 11 lbf, 2 2 r = 7/11 Try d = in, 169 S ut = = kpsi (0.08) S su = 0.67S ut = kpsi, S sy = 0.35S ut = 85.5 kpsi Try unpeened using Zimmerli s endurance data: S sa = 35 kpsi, S sm = 55 kpsi Gerber: S se = S sa 1 (S sm /S su ) = 35 2 S sa = (7/11)2 (163.7) 2 2(39.5) α = S sa /n f = 35.0/1.5 = 23.3 kpsi β = 8F [ a 8(7) πd 2 (10 3 ) = π( ) 2(23.3) C = + 4(2.785) 4(2.785) D = Cd = 6.97(0.08) = in = 39.5 kpsi 1 (55/163.7) 2 [ ] 1 + 2(39.5) = 35.0 kpsi (7/11)(163.7) ] (10 3 ) = kpsi [2(23.3) ] 2 3(23.3) 4(2.785) = 6.97

16 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design K B = 4(6.97) + 2 4(6.97) 3 = ( ) [ ] 8Fa D 8(7)(0.558) τ a = K B = (10 3 ) = 23.3 kpsi π( ) n f = 35/23.3 = 1.50 checks N a = Gd4 8kD = 10(106 )(0.08) 4 = turns 3 8(9.5)(0.558) 3 N t = = 33 turns, L s = dn t = 0.08(33) = 2.64 in y max = F max /k = 18/9.5 = in, y s = (1 + ξ)y max = ( )(1.895) = in L 0 = = in (L 0 ) cr = 2.63 D α = 2.63(0.558) = in 0.5 τ s = 1.15(18/7)τ a = 1.15(18/7)(23.3) = 68.9 kpsi n s = S sy /τ s = 85.5/68.9 = 1.24 kg f = π 2 d 2 DN a γ = 9.5(386) = 109 Hz π 2 ( )(0.558)(31.02)(0.283) These steps are easily implemented on a spreadsheet, as shown below, for different diameters. d 1 d 2 d 3 d 4 d m A S ut S su S sy S se S sa α β C D K B τ a n f N a N t L s y s L (L 0 ) cr τ s n s f (Hz)

17 budynas_sm_ch10.qxd 01/29/ :26 Page 277 Chapter The shaded areas depict conditions outside the recommended design conditions. Thus, one spring is satisfactory A313, as wound, unpeened, squared and ground, d = in, OD = = in, N t = turns The steps are the same as in Prob except that the Gerber-Zimmerli criterion is replaced with Goodman-Zimmerli: S sa S se = 1 (S sm /S su ) The problem then proceeds as in Prob The results for the wire sizes are shown below (see solution to Prob for additional details). Iteration of d for the first trial d 1 d 2 d 3 d 4 d 1 d 2 d 3 d 4 d d m K B A τ a S ut n f S su N a S sy N t S se L s S sa y s α L β (L 0 ) cr C τ s D n s f (Hz) Without checking all of the design conditions, it is obvious that none of the wire sizes satisfy n s 1.2. Also, the Gerber line is closer to the yield line than the Goodman. Setting n f = 1.5 for Goodman makes it impossible to reach the yield line (n s < 1). The table below uses n f = 2. Iteration of d for the second trial d 1 d 2 d 3 d 4 d 1 d 2 d 3 d 4 d d m K B A τ a S ut n f S su N a S sy N t S se L s S sa y s α L β (L 0 ) cr C τ s D n s f (Hz) The satisfactory spring has design specifications of: A313, as wound, unpeened, squared and ground, d = in, OD = = in, N t = 19.3 turns.

18 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design This is the same as Prob since S se = S sa = 35 kpsi. Therefore, design the spring using: A313, as wound, un-peened, squared and ground, d = in, OD = in, N t = turns For the Gerber fatigue-failure criterion, S su = 0.67S ut, S sa S se = 1 (S sm /S su ), S 2 sa = r 2 Ssu 2 ( ) 2 2Sse S se rs su The equation for S sa is the basic difference. The last 2 columns of diameters of Ex are presented below with additional calculations. d = d = d = d = S ut N a S su L s S se L S sy (L 0 ) cr S sa K B α τ a β n f C τ s D n s ID f n OD fom There are only slight changes in the results As in Prob , the basic change is S sa. S sa For Goodman, S se = 1 (S sm /S su ) Recalculate S sa with S sa = rs ses su rs su + S se Calculations for the last 2 diameters of Ex are given below. d = d = d = d = S ut N a S su L s S se L S sy (L 0 ) cr S sa K B α τ a β n f C τ s D n s ID f n OD fom There are only slight differences in the results.

19 budynas_sm_ch10.qxd 01/29/ :26 Page 279 Chapter Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi in m, m = 0.190, rel cost = Try d = in, S ut = = kpsi (0.067) Table 10-6: S sy = 0.45S ut = kpsi Table 10-7: S y = 0.75S ut = kpsi Eq. (10-34) with D/d = C and C 1 = C σ A = F max πd [(K ) A(16C) + 4] = S y 2 n y 4C 2 C 1 4C(C 1) (16C) + 4 = πd2 S y n y F max ( πd 4C 2 2 ) S y C 1 = (C 1) 1 4n y F max C 2 1 ( ) 1 + πd2 S y 1 C + 1 ( πd 2 ) S y 2 = 0 4 4n y F max 4 4n y F max ( C = 1 ) πd2 S y πd2 S 2 y ± πd2 S y + 2 take positive root 2 16n y F max 16n y F max 4n y F max = 1 { π( )(175.5)(10 3 ) 2 16(1.5)(18) [π(0.067)2 (175.5)( ] ) 2 π(0.067)2 (175.5)(10 3 ) (1.5)(18) 4(1.5)(18) = D = Cd = in [ ( F i = πd3 τ i 8D = πd D exp(0.105c) ± C 3 )] 6.5 Use the lowest F i in the preferred range. This results in the best fom. { ( F i = π(0.067) (0.3075) exp[0.105(4.590)] )} = lbf 6.5 For simplicity, we will round up to the next integer or half integer; therefore, use F i = 7 lbf k = 18 7 = 22 lbf/in 0.5 N a = d4 G 8kD = (0.067)4 (11.5)(10 6 ) = turns 3 8(22)(0.3075) 3 N b = N a G 11.5 = = turns E 28.6 L 0 = (2C 1 + N b )d = [2(4.590) ](0.067) = in L 18 lbf = = in

20 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design Body: K B = 4C + 2 4C 3 = 4(4.590) + 2 4(4.590) 3 = τ max = 8K B F max D = 8(1.326)(18)(0.3075) (10 3 ) = 62.1 kpsi π(0.067) 3 (n y ) body = S sy τ max = = 1.70 r 2 = 2d = 2(0.067) = in, C 2 = 2r 2 d = 2(0.134) = 4 (K ) B = 4C 2 1 4C 2 4 = 4(4) 1 4(4) 4 = 1.25 τ B = (K ) B [ 8Fmax D (n y ) B = S sy τ B = = 1.80 fom = (1) π 2 d 2 (N b + 2)D 4 ] [ 8(18)(0.3075) = 1.25 π(0.067) 3 ] (10 3 ) = kpsi = π 2 (0.067) 2 ( )(0.3075) 4 Several diameters, evaluated using a spreadsheet, are shown below. = d: S ut S sy S y C D F i (calc) F i (rd) k N a N b L L 18 lbf K B τ max (n y ) body τ B (n y ) B (n y ) A fom Except for the in wire, all springs satisfy the requirements of length and number of coils. The in wire has the highest fom.

21 budynas_sm_ch10.qxd 01/29/ :26 Page 281 Chapter Given: N b = 84 coils, F i = 16 lbf, OQ&T steel, OD = 1.5 in, d = in. D = =1.338 in (a) Eq. (10-39): L 0 = 2(D d) + (N b + 1)d = 2( ) + (84 + 1)(0.162) = in Ans. or 2d + L 0 = 2(0.162) = in overall. (b) C = D d = = 8.26 K B = 4(8.26) + 2 4(8.26) 3 = [ ] [ ] 8Fi D 8(16)(1.338) τ i = K B = = psi Ans. π(0.162) 3 (c) From Table 10-5 use: G = 11.4(10 6 ) psi and E = 28.5(10 6 ) psi N a = N b + G 11.4 = 84 + = 84.4 turns E 28.5 k = d4 G = (0.162)4 (11.4)(10 6 ) = lbf/in Ans. 8D 3 N a 8(1.338) 3 (84.4) (d) Table 10-4: A = 147 psi in m, m = S ut = = kpsi (0.162) S y = 0.75(207.1) = kpsi S sy = 0.50(207.1) = kpsi Body F = πd3 S sy π K B D = π(0.162)3 (103.5)(10 3 ) = lbf 8(1.166)(1.338) Torsional stress on hook point B C 2 = 2r 2 d = 2( /2) = (K ) B = 4C 2 1 4C 2 4 = 4(4.086) 1 4(4.086) 4 = F = π(0.162)3 (103.5)(10 3 ) = lbf 8(1.243)(1.338) Normal stress on hook point A C 1 = 2r 1 d = = 8.26 (K ) A = 4C2 1 C 1 1 = 4(8.26) = C 1 (C 1 1) 4(8.26)(8.26 1)

22 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design S yt = σ = F [ 16(K ) A D + 4 ] πd (10 3 ) F = = 85.8 lbf [16(1.099)(1.338)]/[π(0.162) 3 ] +{4/[π(0.162) 2 ]} = min(110.8, 103.9, 85.8) = 85.8 lbf Ans. (e) Eq. (10-48): y = F F i k = = 14.4in Ans F min = 9 lbf, F max = 18 lbf F a = = 4.5 lbf, F m = = 13.5 lbf A313 stainless: d 0.1 A = 169 kpsi in m, m = d 0.2 A = 128 kpsi in m, m = E = 28 Mpsi, G = 10 Gpsi Try d = in and refer to the discussion following Ex S ut = = kpsi (0.081) S su = 0.67S ut = kpsi S sy = 0.35S ut = 85.4 kpsi S y = 0.55S ut = kpsi Table 10-8: Table 6-7: S r = 0.45S ut = kpsi S r /2 S e = 1 [S r /(2S ut )] = 109.8/2 = 57.8 kpsi 2 1 [(109.8/2)/243.9] 2 r = F a /F m = 4.5/13.5 = S a = r 2 Sut 2 ( ) 2 2Se S e rs ut S a = (0.333)2 ( ) 2(57.8) 1 + [ 1 + 2(57.8) 0.333(243.9) ] 2 = 42.2 kpsi Hook bending [ 16C (σ a ) A = F a (K ) A πd + 4 ] = S a = S a 2 πd 2 (n f ) A 2 [ 4.5 (4C 2 ] C 1)16C + 4 = S a πd 2 4C(C 1) 2 This equation reduces to a quadratic in C see Prob

23 budynas_sm_ch10.qxd 01/29/ :26 Page 283 Chapter C = 0.5 The useable root for C is πd2 S a π(0.081) 2 (42.2)(10 3 ) = (πd2 ) S 2 a πd2 S a [π(0.081)2 (42.2)( ] ) 2 π(0.081)2 (42.2)(10 3 ) = 4.91 D = Cd = in [ F i = πd3 τ i 8D = πd D exp(0.105c) ± 1000 Use the lowest F i in the preferred range. { F i = π(0.081) (0.398) exp[0.105(4.91)] 1000 = 8.55 lbf For simplicity we will round up to next 1/4 integer. F i = 8.75 lbf k = 18 9 = 36 lbf/in 0.25 N a = d4 G 8kD = (0.081)4 (10)(10 6 ) = 23.7 turns 3 8(36)(0.398) 3 ( ( 4 C N b = N a G 10 = 23.7 = 23.3 turns E 28 L 0 = (2C 1 + N b )d = [2(4.91) ](0.081) = in L max = L 0 + (F max F i )/k = ( )/36 = in (σ a ) A = 4.5(4) ( 4C 2 ) C πd 2 C 1 [ = 18(10 3 ) 4( ] ) = 21.1 kpsi π( ) (n f ) A = S a (σ a ) A = = 2 checks Body: K B = 4C + 2 4C 3 = 4(4.91) + 2 4(4.91) 3 = τ a = 8(1.300)(4.5)(0.398) (10 3 ) = kpsi π(0.081) 3 τ m = F m τ a = 13.5 (11.16) = kpsi F a 4.5 )] )}

24 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design The repeating allowable stress from Table 7-8 is S sr = 0.30S ut = 0.30(243.9) = kpsi The Gerber intercept is 73.17/2 S se = = 38.5 kpsi 1 [(73.17/2)/163.4] 2 From Table 6-7, (n f ) body = 1 ( ) ( ) [ ] 2(33.47)(38.5) (11.16) = 2.53 Let r 2 = 2d = 2(0.081) = C 2 = 2r 2 d = 4, (K ) B = 4(4) 1 4(4) 4 = 1.25 (τ a ) B = (K ) B τ a = 1.25 (11.16) = kpsi K B 1.30 (τ m ) B = (K ) B τ m = 1.25 (33.47) = kpsi K B 1.30 Table 10-8: (S sr ) B = 0.28S ut = 0.28(243.9) = 68.3 kpsi Yield Bending: (S se ) B = (n f ) B = 1 2 (σ A ) max = 4F max πd /2 = 35.7 kpsi 1 [(68.3/2)/163.4] 2 ( ) ( ) = 4(18) π( ) [ (4C 2 ] C 1) + 1 C 1 [ 4(4.91) [ ] 2(32.18)(35.7) (10.73) = 2.51 ] + 1 (10 3 ) = 84.4 kpsi (n y ) A = = 1.59 Body: τ i = (F i /F a )τ a = (8.75/4.5)(11.16) = 21.7 kpsi r = τ a /(τ m τ i ) = 11.16/( ) = (S sa ) y = r r + 1 (S sy τ i ) = ( ) = 31.0 kpsi (n y ) body = (S sa) y = 31.0 τ a = 2.78 Hook shear: S sy = 0.3S ut = 0.3(243.9) = 73.2 kpsi τ max = (τ a ) B + (τ m ) B = = 42.9 kpsi

25 budynas_sm_ch10.qxd 01/29/ :26 Page 285 Chapter (n y ) B = = 1.71 fom = 7.6π 2 d 2 (N b + 2)D = 7.6π 2 (0.081) 2 ( )(0.398) 4 4 A tabulation of several wire sizes follow = d S ut S su S r S e S a C D OD F i (calc) F i (rd) k N a N b L L 18 lbf (σ a ) A (n f ) A K B (τ a ) body (τ m ) body S sr S se (n f ) body (K ) B (τ a ) B (τ m ) B (S sr ) B (S se ) B (n f ) B S y (σ A ) max (n y ) A τ i r (S sy ) body (S sa ) y (n y ) body (S sy ) B (τ B ) max (n y ) B fom optimal fom The shaded areas show the conditions not satisfied.

26 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design For the hook, F R D 2 M = FRsin θ, δ F = 1 EI π/2 The total deflection of the body and the two hooks δ = 8FD3 N b d 4 G + 2π FR 3 2 EI ( = 8FD3 N d 4 b + G ) G E N a = N b + G E QED 0 M/ F = R sin θ FR 2 sin 2 Rdθ = π 2 PR 3 EI = 8FD3 N b d 4 G + π F(D/2)3 E(π/64)(d 4 ) = 8FD3 N a d 4 G Table 10-4 for A227: Table 10-5: Eq. (10-57): From A = 140 kpsi in m, m = E = 28.5(10 6 ) psi S ut = 140 = kpsi (0.162) S y = σ all = 0.78(197.8) = kpsi D = = in C = D/d = 1.088/0.162 = 6.72 K i = 4C2 C 1 4C(C 1) σ = K i 32M Solving for M for the yield condition, = 4(6.72) (6.72)(6.72 1) = M y = πd3 S y = π(0.162)3 ( ) = 57.2 lbf in 32K i 32(1.125) Count the turns when M = 0 M y N = 2.5 d 4 E/(10.8DN) from which N = = [10.8DM y /(d 4 E)] 2.5 = turns 1 +{[10.8(1.088)(57.2)]/[(0.162) 4 (28.5)(10 6 )]}

27 budynas_sm_ch10.qxd 01/29/ :26 Page 287 Chapter This means ( )(360 ) or 29.9 from closed. Treating the hand force as in the middle of the grip r = = 2.75 in 2 F = M y r = 57.2 = 20.8 lbf Ans The spring material and condition are unknown. Given d = in and OD = 0.500, (a) D = = in Using E = 28.6 Mpsi for an estimate k = d4 E 10.8DN = (0.081)4 (28.6)(10 6 ) = 24.7 lbf in/turn 10.8(0.419)(11) for each spring. The moment corresponding to a force of 8 lbf The fraction windup turn is Fr = (8/2)(3.3125) = lbf in/spring n = Fr k = = turns 24.7 The arm swings through an arc of slightly less than 180, say 165. This uses up 165/360 or turns. So n = = turns are left (or 0.078(360 ) = 28.1 ). The original configuration of the spring was 28.1 Ans. (b) C = = 5.17 K i = 4(5.17) = (5.17)(5.17 1) 32M σ = K i [ ] 32(13.25) = = psi Ans. π(0.081) 3 To achieve this stress level, the spring had to have set removed Consider half and double results Straight section: L 2 3FR F M = 3FR, M P = 3R

28 budynas_sm_ch10.qxd 01/29/ :26 Page Solutions Manual Instructor s Solution Manual to Accompany Mechanical Engineering Design Upper 180 section: R F Lower section: M = F[R + R(1 cos φ)] = FR(2 cos φ), M P M = FRsin θ M P = R sin θ = R(2 cos φ) Considering bending only: δ = 2 [ L/2 π 9FR 2 dx + FR 2 (2 cos φ) 2 Rdφ + EI 0 0 = 2F [ ( 9 EI 2 R2 L + R 3 4π 4 sin φ π + π ) ( )] π + R π/2 0 ] F(R sin θ) 2 Rdθ = 2FR2 EI ( 19π 4 R + 9 ) 2 L = FR2 (19π R + 18L) 2EI Ans Computer programs will vary Computer programs will vary.