Chapter 10 = + We see the maximum and minimum occur at C = 4 and 12 respectively where

Transcription

1 Chpter From Eq. (10-) nd (10-5) K W K = + C Plot 100(K W K )/ K W v. C for C 1 obtining 100(K W -K )/K W We ee the mximum nd minimum occur t C = nd 1 repectivel where Mximum = 1.6 % An., nd Minimum = 0.7 % An. 10- A = Sd m dim(a ucu ) = [dim (S) dim(d m )] ucu = kpi in m dim(a SI ) = [dim (S) dim(d m )] SI = MP mm m m MP mm m m ASI = Aucu = ( 5.) Aucu 6.895( 5. ) Aucu An. m kpi in For muic wire, from Tble 10-: A ucu = 01 kpi in m, m = 0.15; wht i A SI? A SI = 6.895(5.) 0.15 (01) = 15 MP mm m An. 10- Given: Muic wire, d =.5 mm, OD = 1 mm, plin ground end, N t = 1 coil. C Shigle MED, 10 th edition Chpter 10 Solution, Pge 1/1

2 () Tble 10-1: N = N t 1 = 1 1 = 1 coil D = OD d = 1.5 = 8.5 mm C = D/d = 8.5/.5 = 11. Tble 10-5: d =.5/5. = in G = 81.0(10 ) MP Eq. (10-9): (b) Tble 10-1: k = = = 1.1 N / mm An. d G 8D N L = d N t =.5(1) = 5 mm Tble 10-: m = 0.15, A = 11 MP mm m A 11 Eq. (10-1): Sut = = = 196 MP m 0.15 d.5 Tble 10-6: S = 0.5(196) = 871. MP Eq. (10-5): Eq. (10-7): K F = = = S π = = = N An. 8K D F (c) L0 = + L = + 5 = 16.8 mm An. k (d) ( L 0 ) = = 19.9 mm. Spring need to be upported. An. cr Given: Deign lod, F 1 = 10 N. Referring to Prob. 10- olution, C = 11., N = 1 coil, S = 871. MP, F = N, L 0 = 16.8 mm nd (L 0 ) cr = 19.9 mm. Eq. (10-18): C 1 C = 11. O.K. Eq. (10-19): N 15 N = 1 O.K. Eq. (10-17): ξ = F = 1 = 0.9 F 10 1 Shigle MED, 10 th edition Chpter 10 Solution, Pge /1

3 Eq. (10-0): ξ 0.15, ξ = 0.9 O. K. From Eq. (10-7) for ttic ervice 8F1 D 8(10)(8.5) 1 = K MP = = π (.5) S 871. n = = = Eq. (10-1): n 1., n = 1.9 O.K = 1 = 67 = MP / = 871. / S S / (n ) d : Not olid-fe (but w the bi of the deign). Not O.K. L 0 (L 0 ) cr : Not O.K. Deign i untifctor. Operte over rod? An Given: Oil-tempered wire, d = 0. in, D = in, N t = 1 coil, L 0 = 5 in, qured end. () Tble 10-1: (b) Tble 10-1: Tble 10-5: Eq. (10-9): L = d (N t + 1) = 0.(1 + 1) =.6 in An. N = N t = 1 = 10 coil G = 11. Mpi k = = = 8 lbf/in d G 6 8D N 8 10 F = k = k (L 0 L ) = 8(5.6) = 67. lbf An. (c) Eq. (10-1): C = D/d = /0. = 10 Eq. (10-5): K = = = Eq. (10-7): Tble 10-: 8F D = K ( 10 ) pi = π = ( 0. ) m = 0.187, A = 17 kpi in m Shigle MED, 10 th edition Chpter 10 Solution, Pge /1

4 A 17 Eq. (10-1): Sut = = = kpi m d 0. Tble 10-6: S = 0.50 S ut = 0.50(198.6) = 99. kpi S 99. n = = =.0 An Given: Oil-tempered wire, d = mm, C = 10, plin end, L 0 = 80 mm, nd t F = 50 N, = 15 mm. () k = F/ = 50/15 =. N/mm An. (b) D = Cd = 10() = 0 mm OD = D + d = 0 + = mm An. (c) From Tble 10-5, G = 77. GP Eq. (10-9): N d G = = = 11.6 coil 8kD 8. 0 Tble 10-1: N t = N = 11.6 coil An. (d) Tble 10-1: L = d (N t + 1) = ( ) = 50. mm An. (e) Tble 10-: m = 0.187, A = 1855 MP mm m A 1855 Eq. (10-1): Sut = = = 11 MP m d Tble 10-6: S = 0.50 S ut = 0.50(11) = MP = L 0 L = = 9.6 mm F = k =.(9.6) = N Eq. (10-5): K + (10) + = = = 1.15 (10) Eq. (10-7): 8F D = K MP = π = ( ) Shigle MED, 10 th edition Chpter 10 Solution, Pge /1

5 S n = = =.0 An Sttic ervice pring with: HD teel wire, d = in, OD = in, N t = 8 coil, plin nd ground end. Preliminrie Tble 10-5: A = 10 kpi in m, m = Eq. (10-1): A 10 Sut = = m d = 6. kpi Tble 10-6: S = 0.5(6.) = kpi Then, D = OD d = = 0.8 in Eq. (10-1): C = D/d = 0.8/0.08 = 10 Eq. (10-5): + (10) + K = = = 1.15 (10) Tble 10-1: N = N t 1 = 8 1 = 7 coil L = dn t = 0.08(8) = 0.6 in Eq. (10-7) For olid-fe, n = 1. : S / ( 0.08 ) 101.8( 10 ) / 1. n π F = = = lbf 8K D 8(1.15)(0.8) Eq. (10-9): k = = = 16. lbf/in 6 d G 8D N F = = = 1.1 in k 16. () L 0 = + L = = 1.78 in An. (b) Tble 10-1: p = L = = 0. in Nt 8 An. (c) From bove: F = lbf An. (d) From bove: k = 16. lbf/in An. (e) Tble 10- nd Eq. (10-1):.6D.6(0.8) ( L0) cr = = =.1 in α 0.5 Since L 0 < (L 0 ) cr, buckling i unlikel An Given: Deign lod, F 1 = 16.5 lbf. Referring to Prob olution, C = 10, N = 7 coil, S = kpi, F = lbf, = 1.1 in, L 0 = 1.78 in, nd (L 0 ) cr =.1 in. Eq. (10-18): C 1 C = 10 O.K. Shigle MED, 10 th edition Chpter 10 Solution, Pge 5/1

6 Eq. (10-19): N 15 N = 7 O.K. F Eq. (10-17): ξ = 1 = 1 = 0.1 F Eq. (10-0): ξ 0.15, ξ = 0.1 not O. K., but probbl cceptble. From Eq. (10-7) for ttic ervice 8F1 D 8(16.5)(0.8) 1 = K = = ( 10 ) pi = 7.5 kpi π (0.080) S n = = = Eq. (10-1): Eq. (10-1): n 1., n = 1.7 O.K = 1 = 7.5 = 8.8 kpi n = S / = / 8.8 = 1.0 n 1., n = 1. It i olid-fe (bi of deign). O.K. Eq. (10-1) nd Tble 10-: L 0 (L 0 ) cr 1.78 in.1 in O.K Given: A8 muic wire, qured nd ground end, d = in, OD = 0.08 in, L 0 = 0.58 in, N t = 8 coil. D = OD d = = 0.01 in Eq. (10-1): C = D/d = 0.01/0.007 =.9 Eq. (10-5): + (.9) + K = = = Tble 10-1: N = N t = 8 = 6 coil (high) Tble 10-5: G = 1.0 Mpi Eq. (10-9): 6 d G ( 1.0) 10 k = = =.58 lbf/in 8D N Tble 10-1: Eq. (10-7): L = dn t = 0.007(8) = 0.66 in = L 0 L = = 0.1 in F = k =.58(0.1) = 1.05 lbf 8F 8( 1.05) 0.01 D = K ( 10 ) pi = π = (1) Tble 10-: A = 01 kpi in m, m = 0.15 Shigle MED, 10 th edition Chpter 10 Solution, Pge 6/1

7 A 01 Eq. (10-1): Sut = = = 1.7 kpi m 0.15 d Tble 10-6: S = 0.5 S ut = 0.5(1.7) = kpi > S, tht i, 5.1 > kpi, the pring i not olid-fe. Return to Eq. (1) with F = k nd = S /n, nd olve for, giving The free length hould be wound to ( S / ) ( 10 ) / 1. ( n π ) = = = 0.19 in 8K kd L 0 = L + = = 0.15 in An. Thi onl ddree the olid-fe criteri. There re dditionl problem Given: 159 phophor-bronze, qured nd ground. end, d = 0.01 in, OD = 0.18 in, L 0 = 0.50 in, N t = 16 coil. D = OD d = = 0.11 in Eq. (10-1): C = D/d = 0.11/0.01 = 8.1 Eq. (10-5): + ( 8.1) + K = = = Tble 10-1: Tble 10-5: Eq. (10-9): Tble 10-1: Eq. (10-7): N = N t = 16 = 1 coil G = 6 Mpi 6 d G 0.01 ( 6) 10 k = = = 1.89 lbf/in 8D N L = dn t = 0.01(16) = 0. in = L 0 L = = 0.76 in F = k = 1.89(0.76) = 0.8 lbf 8F 8( 0.8) 0.11 D = K ( 10 ) pi = π 0.01 = (1) Tble 10-: A = 15 kpi in m, m = 0 A 15 Eq. (10-1): Sut = = = 15 kpi m 0 d 0.01 Tble 10-6: S = 0.5 S ut = 0.5(15) = 7.5 kpi > S, tht i, 7. > 7.5 kpi, the pring i not olid-fe. Return to Eq. (1) with F = k nd = S /n, nd olve for, giving The free length hould be wound to ( S / ) 7.5 ( 10 ) /1. ( 0.01 n π ) = = = 0.9 in 8K kd Shigle MED, 10 th edition Chpter 10 Solution, Pge 7/1

8 L 0 = L + = = 0.5 in An Given: A1 tinle teel, qured nd ground end, d = in, OD = 0.50 in, L 0 = 0.68 in, N t = 11. coil. D = OD d = = 0.00 in Eq. (10-1): C = D/d = 0.00/0.050 = Eq. (10-5): + ( ) + K = = = 1.85 Tble 10-1: Tble 10-5: Eq. (10-9): Tble 10-1: Eq. (10-7): N = N t = 11. = 9. coil G = 10 Mpi 6 d G ( 10) 10 k = = = lbf/in 8D N L = dn t = 0.050(11.) = 0.56 in = L 0 L = = 0.1 in F = k = 106.1(0.1) = 1.7 lbf 8F 8( 1.7) 0. D = K ( 10 ) pi = π = Tble 10-: A = 169 kpi in m, m = 0.16 A 169 Eq. (10-1): Sut = = = 61.7 kpi m 0.16 d Tble 10-6: S = 0.5 S ut = 0.5(61.7) = 91.6 kpi S 91.6 n = = = 1.8 Spring i olid-fe (n > 1.) An Given: A7 hrd-drwn wire, qured nd ground end, d = 0.18 in, OD =.1 in, L 0 =.5 in, N t = 5.75 coil. D = OD d = = 1.97 in Eq. (10-1): C = D/d = 1.97/0.18 = 1. (high) Eq. (10-5): + ( 1.) + K = = = Tble 10-1: Tble 10-5: Eq. (10-9): Tble 10-1: N = N t = 5.75 =.75 coil G = 11. Mpi 6 d G 0.18 ( 11.) 10 k = = =.77 lbf/in 8D N L = dn t = 0.18(5.75) = in = L 0 L = = 1.69 in F = k =.77(1.69) = 9.0 lbf Shigle MED, 10 th edition Chpter 10 Solution, Pge 8/1

9 Eq. (10-7): 8F D = K ( 10 ) pi = π = ( 0.18 ) Tble 10-: A = 10 kpi in m, m = A 10 Eq. (10-1): Sut = = = 01. kpi m d 0.18 Tble 10-6: S = 0.5 S ut = 0.5(01.) = 90.6 kpi S 90.6 n = = = 1.6 Spring i olid-fe (n > 1.) An Given: A9 OQ&T teel, qured nd ground end, d = 0.18 in, OD = 0.9 in, L 0 =.86 in, N t = 1 coil. D = OD d = = 0.78 in Eq. (10-1): C = D/d = 0.78/0.18 = Eq. (10-5): + ( 5.667) + K = = = Tble 10-1: N = N t = 1 = 10 coil A9 OQ&T teel i not given in Tble From Tble A-5, for crbon teel, G = 11.5 Mpi. 6 d G 0.18 ( 11.5) 10 Eq. (10-9): k = = = lbf/in 8D N Tble 10-1: Eq. (10-7): L = dn t = 0.18(1) = in = L 0 L = = 1.0 in F = k = 109.0(1.0) = 11. lbf 8F 8( 11.) 0.78 D = K ( 10 ) pi = π 0.18 = (1) Tble 10-: A = 17 kpi in m, m = A 17 Eq. (10-1): Sut = = = 1.9 kpi m d 0.18 Tble 10-6: S = 0.50 S ut = 0.50(1.9) = kpi > S, tht i, 1.7 > kpi, the pring i not olid-fe. Return to Eq. (1) with F = k nd = S /n, nd olve for, giving The free length hould be wound to ( S / ) ( 10 ) /1. ( 0.18 n π ) = = = in 8K kd Shigle MED, 10 th edition Chpter 10 Solution, Pge 9/1

10 L 0 = L + = =.51 in An Given: A chrome-vndium teel, qured nd ground end, d = in, OD =.75 in, L 0 = 7.5 in, N t = 8 coil. D = OD d = =.565 in Eq. (10-1): C = D/d =.565/0.185 = 1.86 (high) Eq. (10-5): + ( 1.86 ) + K = = = Tble 10-1: Tble 10-5: Eq. (10-9): Tble 10-1: Eq. (10-7): N = N t = 8 = 6 coil G = 11. Mpi. 6 d G ( 11.) 10 k = = = 16.0 lbf/in 8D N L = dn t = 0.185(8) = 1.8 in = L 0 L = = 6.0 in F = k = 16.0(6.0) = 97.5 lbf 8F 8( 97.5).565 D = K ( 10 ) pi = π = (1) Tble 10-: A = 169 kpi in m, m = A 169 Eq. (10-1): Sut = = =. kpi m d Tble 10-6: S = 0.50 S ut = 0.50(.) = 11. kpi S 11. n = = = 1.0 Spring i not olid-fe (n < 1.) Return to Eq. (1) with F = k nd = S /n, nd olve for, giving The free length hould be wound to ( S / ) 11. ( 10 ) / 1. ( n π ) = = = in 8K kd L 0 = L + = = 6.59 in An Given: A1 tinle teel, qured nd ground end, d = 0.5 mm, OD = 0.95 mm, L 0 = 1.1 mm, N t = 8 coil. D = OD d = = 0.7 mm Eq. (10-1): C = D/d = 0.7/0.5 =.8 (low) + (.8) + Eq. (10-5): K = = = Shigle MED, 10 th edition Chpter 10 Solution, Pge 10/1

11 Tble 10-1: N = N t = 8 = 6 coil (high) Tble 10-5: Eq. (10-9): Tble 10-1: Eq. (10-7): G = 69.0(10 ) MP. d G 0.5 ( 69.0) 10 k = = =.78 N/mm 8D N L = dn t = 0.5(8) = 9.5 mm = L 0 L = =.6 mm F = k =.78(.6) = 7.09 N 8F 8( 7.09) 0.7 D = K MP = π 0.5 = (1) Tble 10- (di. le thn tble): A = 1867 MP mm m, m = 0.16 A 1867 Eq. (10-1): Sut = = = 86 MP m 0.16 d 0.5 Tble 10-6: S = 0.5 S ut = 0.5(86) = 7 MP > S, tht i, 10 > 7 MP, the pring i not olid-fe. Return to Eq. (1) with F = k nd = S /n, nd olve for, giving ( S / n ) ( 7 / 1.) π ( 0.5 ) = = = 1. mm 8KkD 8( 1.610).78( 0.7) The free length hould be wound to L 0 = L + = = 10.7 mm An. Thi onl ddree the olid-fe criteri. There re dditionl problem Given: A8 muic wire, qured nd ground end, d = 1. mm, OD = 6.5 mm, L 0 = 15.7 mm, N t = 10. coil. D = OD d = = 5. mm Eq. (10-1): C = D/d = 5./1. =.17 + (.17 ) + Eq. (10-5): K = = = Tble (10-1): N = N t = 10. = 8. coil Tble 10-5 (d = 1./5. = 0.07 in): G = 81.7(10 ) MP. d G 1. ( 81.7) 10 Eq. (10-9): k = = = 17.5 N/mm 8D N Tble 10-1: L = dn t = 1.(10.) = 1. mm = L 0 L = =.6 mm F = k = 17.5(.6) = 60.0 N Shigle MED, 10 th edition Chpter 10 Solution, Pge 11/1

12 Eq. (10-7): 8F D = K MP = π = (1) ( 1. ) Tble 10-: A = 11 MP mm m, m = 0.15 A 11 Eq. (10-1): Sut = = = 15 MP m 0.15 d 1. Tble 10-6: S = 0.5 S ut = 0.5(15) = 969 MP S 969 n = = = 1.51 Spring i olid-fe (n > 1.) An Given: A9 OQ&T teel, qured nd ground end, d =.5 mm, OD = 50.6 mm, L 0 = 75.5 mm, N t = 5.5 coil. Eq. (10-1): Eq. (10-5): Tble 10-1: D = OD d = = 7.1 mm C = D/d = 7.1/.5 = 1.6 (high) + ( 1.6 ) + K = = = N = N t = 5.5 =.5 coil A9 OQ&T teel i not given in Tble From Tble A-5, for crbon teel, G = 79.(10 ) MP. d G.5 ( 79.) 10 Eq. (10-9): k = = =.067 N/mm 8D N Tble 10-1: Eq. (10-7): L = dn t =.5(5.5) = 19.5 mm = L 0 L = = 56.5 mm F = k =.067(56.5) = 8.8 N 8F 8( 8.8) 7.1 D = K MP = π.5 = (1) Tble 10-: A = 1855 MP mm m, m = A 1855 Eq. (10-1): Sut = = = 168 MP m d.5 Tble 10-6: S = 0.50 S ut = 0.50(168) = 7 MP S 7 n = = = 1.0 Spring i not olid-fe (n < 1.) 70.8 Return to Eq. (1) with F = k nd = S /n, nd olve for, giving ( S / n ) ( 7 /1.) π (.5 ) = = = 8.96 mm 8KkD 8( 1.098).067 ( 7.1) The free length hould be wound to Shigle MED, 10 th edition Chpter 10 Solution, Pge 1/1

13 L 0 = L + = = 68. mm An Given: 159 phophor-bronze, qured nd ground end, d =.8 mm, OD = 1. mm, L 0 = 71. mm, N t = 1.8 coil. D = OD d = 1..8 = 7.6 mm Eq. (10-1): C = D/d = 7.6/.8 = 7.6 Eq. (10-5): + ( 7.6) + K = = = Tble 10-1: Tble 10-5: Eq. (10-9): Tble 10-1: Eq. (10-7): N = N t = 1.8 = 10.8 coil G = 1.(10 ) MP. d G.8 ( 1.) 10 k = = =.75 N/mm 8D N L = dn t =.8(1.8) = 8.6 mm = L 0 L = =.76 mm F = k =.75(.76) = 108. N 8F 8( 108.) 7.6 D = K MP = π.8 = (1) Tble 10- (d =.8/5. = in): A = 9 MP mm m, m = 0.06 A 9 Eq. (10-1): Sut = = = MP m 0.06 d.8 Tble 10-6: S = 0.5 S ut = 0.5(855.7) = 99.5 MP S 99.5 n = = = 1.81 Spring i olid-fe (n > 1.) An Given: A chrome-vndium teel, qured nd ground end, d =.5 mm, OD = 69. mm, L 0 = 15.6 mm, N t = 8. coil. Eq. (10-1): Eq. (10-5): Tble 10-1: Tble 10-5: Eq. (10-9): Tble 10-1: D = OD d = = 6.7 mm C = D/d = 6.7/.5 = 1.8 (high) + ( 1.8) + K = = = N = N t = 8. = 6. coil G = 77.(10 ) MP. d G.5 ( 77.) 10 k = = =.57 N/mm 8D N L = dn t =.5(8.) = 6.9 mm Shigle MED, 10 th edition Chpter 10 Solution, Pge 1/1

14 Eq. (10-7): = L 0 L = = mm F = k =.57(178.7) = 1. N 8F 8( 1.) 6.7 D = K MP = π.5 = (1) Tble 10-: A = 005 MP mm m, m = A 005 Eq. (10-1): Sut = = = 1557 MP m d.5 Tble 10-6: S = 0.50 S ut = 0.50(1557) = 779 MP > S, tht i, 8 > 779 MP, the pring i not olid-fe. Return to Eq. (1) with F = k nd = S /n, nd olve for, giving ( S / n ) ( 779 /1.) π (.5 ) = = = 19.5 mm 8K kd The free length hould be wound to L 0 = L + = = 176. mm An. Thi onl ddree the olid-fe criteri. There re dditionl problem Given: A7 HD teel. From the figure: L 0 =.75 in, OD = in, nd d = 0.15 in. Thu D = OD d = 0.15 = in () counting, N t = 1.5 coil. Since the end re qured long 1/ turn on ech end, N p = = 1 turn An. =.75 / 1 = 0.96 in An. The olid tck i 1 wire dimeter L = 1(0.15) = in An. (b) From Tble 10-5, G = 11. Mpi k 0.15 (11.) 10 = = = 6 d G 8D N (1) 6.08 lbf/in An. (c) F = k(l 0 - L ) = 6.08( ) = 18. lbf An. (d) C = D/d = 1.865/0.15 = 1.81 Shigle MED, 10 th edition Chpter 10 Solution, Pge 1/1

15 (1.81) + K = = (1.81) 8F D 8(18.)(1.865) = K = = 8.5 ( 10 ) pi = 8.5 kpi An. π ( 0.15 ) 10-1 For the wire dimeter nlzed, G = Mpi per Tble Ue qured nd ground end. The following i pred-heet tud uing Fig. 10- for prt () nd (b). For N, k = F mx / = 0/ = 10 lbf/in. For, F = F = 0(1 + ξ) = 0( ) = lbf. () Spring over Rod (b) Spring in Hole Source Prmeter Vlue Source Prmeter Vlue d d ID OD D D Eq. (10-1) C Eq. (10-1) C Eq. (10-9) N Eq. (10-9) N Tble 10-1 N t Tble 10-1 N t Tble 10-1 L Tble 10-1 L L L L L Eq. (10-1) (L 0 ) cr Eq. (10-1) (L 0 ) cr Tble 10- A Tble 10- A Tble 10- m Tble 10- m Eq. (10-1) S ut Eq. (10-1) S ut Tble 10-6 S Tble 10-6 S Eq. (10-5) K Eq. (10-5) K Eq. (10-7) Eq. (10-7) Eq. (10-) n Eq. (10-) n Eq. (10-) fom Eq. (10-) fom For n 1., the optiml ize i d = in for both ce. 10- In Prob. 10-1, there i n dvntge of firt electing d one cn elect from the vilble ize (Tble A-8). Selecting C firt require clcultion of d where then ize mut be elected from Tble A-8. Conider prt () of the problem. It i required tht ID = D d = in. (1) From Eq. (10-1), D = Cd. Subtituting thi into the firt eqution ield d = C 1 () Strting with C = 10, from Eq. () we find tht d = in. From Tble A-8, the cloet dimeter i d = in. Subtituting thi bck into Eq. (1) give D = in, with C = 0.890/0.090 = 9.889, which re cceptble. From thi point the olution i the me Prob For prt (b), ue Shigle MED, 10 th edition Chpter 10 Solution, Pge 15/1

16 OD = D + d = in. () nd, d = C 1 () () Spring over rod (b) Spring in Hole Source Prmeter Vlue Source Prmeter Vlue C C Eq. () d Eq. () d Tble A-8 d Tble A-8 d Eq. (1) D Eq. () D Eq. (10-1) C Eq. (10-1) C Eq. (10-9) N Eq. (10-9) N Tble 10-1 N t Tble 10-1 N t 1.86 Tble 10-1 L Tble 10-1 L L L L L 0.77 Eq. (10-1) (L 0 ) cr Eq. (10-1) (L 0 ) cr.550 Tble 10- A Tble 10- A Tble 10- m Tble 10- m 0.15 Eq. (10-1) S ut Eq. (10-1) S ut 87.6 Tble 10-6 S Tble 10-6 S 19.1 Eq. (10-5) K Eq. (10-5) K 1.15 Eq. (10-7) Eq. (10-7) 9.6 n = S / n n = S / n 1.81 Eq. (10-) fom Eq. (10-) fom Agin, for n 1., the optiml ize i = in. Although thi pproch ued le itertion thn in Prob. 10-1, thi w due to the initil vlue picked nd not the pproch. 10- One pproch i to elect A7 HD teel for it low cot. Tr L 0 = 8 mm, then for = = 10.5 mm when F = 5 N. The pring rte i k = F/ = 5/10.5 =.86 N/mm. For clernce of 1.5 mm with crew, ID = = 11.5 mm. Strting with d = mm, D = ID + d = = 1.5 mm C = D/d = 1.5/ = 6.65 (cceptble) Tble 10-5 (d = /5. = in): G = 79. GP d G (79.)10 Eq. (10-9): N = = = 15.9 coil 8kD 8(.86)1.5 Aume qured nd cloed. Shigle MED, 10 th edition Chpter 10 Solution, Pge 16/1

17 Tble 10-1: N t = N + = = 17.9 coil L = dn t = (17.9) =5.8 mm = L 0 L = = 1. mm F = k =.86(1.) = 5.9 N Eq. (10-5): + ( 6.65) + K = = = 1.1 ( 6.65) Eq. (10-7): 8F D 8(5.9)1.5 = K = 1.1 = 67.5 MP π ( ) Tble 10-: A = 178 MP mm m, m = Eq. (10-1): A 178 Sut = = = m MP d Tble 10-6: S = 0.5S ut = 0.5(156) = 70. MP n S 70. = = =.6 > 1. O. K No other dimeter in the given rnge work. So pecif A7-7 HD teel, d = mm, D = 1.5 mm, ID = 11.5 mm, OD = 15.5 mm, qured nd cloed, N t = 17.9 coil, N = 15.9 coil, k =.86 N/mm, L = 5.8 mm, nd L 0 = 8 mm. An. 10- Select A7 HD teel for it low cot. Tr L 0 = 8 mm, then for = = 10.5 mm when F = 5 N. The pring rte i k = F/ = 5/10.5 =.86 N/mm. For clernce of 1.5 mm with crew, ID = = 11.5 mm. D d = 11.5 (1) nd, D =Cd () Strting with C = 8, give D = 8d. Subtitute into Eq. (1) reulting in d = mm. Selecting the neret dimeter in the given rnge, d = 1.6 mm. From thi point, the clcultion re hown in the third column of the predheet output hown. We ee tht for d = 1.6 mm, the pring i not olid fe. Iterting on C we find tht C = 6.5 provide cceptble reult with the pecifiction A7-7 HD teel, d = mm, D = 1.5 mm, ID = 11.5 mm, OD = 15.5 mm, qured nd cloed, N t = 17.9 coil, N = 15.9 coil, k =.86 N/mm, L = 5.8 mm, nd L 0 = 8 mm. An. Shigle MED, 10 th edition Chpter 10 Solution, Pge 17/1

18 Source Prmeter Vlue C Eq. () d Tble A-8 d Eq. (1) D Eq. (10-1) C Eq. (10-9) N Tble 10-1 N t Tble 10-1 L L 0 L F = k F Tble 10- A Tble 10- m Eq. (10-1) S ut Tble 10-6 S Eq. (10-5) K Eq. (10-7) n = S / n The onl difference between electing C firt rther thn d w done in Prob. 10-, i tht once d i clculted, the cloet wire ize mut be elected. Iterting on d ue vilble wire ize from the beginning A tock pring ctlog m hve over two hundred pge of compreion pring with up to 80 pring per pge lited. Student hould be mde wre tht uch ctlog exit. Mn pring re elected from ctlog rther thn deigned. The wire ize ou wnt m not be lited. Ctlog m lo be vilble on dik or the web through erch routine. For exmple, dik re vilble from Centur Spring t 1 - (800) It i better to fmilirize ourelf with vendor reource rther thn invent them ourelf. Smple ctlog pge cn be given to tudent for tud Given: ID = 0.6 in, C = 10, L 0 = 5 in, L = 5 = in, q. & grd end, unpeened, HD A7 wire. () With ID = D d = 0.6 in nd C = D/d = d d = 0.6 d = in An., nd D = in. (b) Tble 10-1: L = dn t = in N t = / = 0 coil An. (c) Tble 10-1: N = N t = 0 = 8 coil Shigle MED, 10 th edition Chpter 10 Solution, Pge 18/1

19 Tble 10-5: G = 11.5 Mpi Eq. (10-9): 6 d G ( 11.5) 10 k = = =. lbf/in 8D N An. (d) Tble 10-: A = 10 kpi in m, m = A 10 Eq. (10-1): Sut = = =. kpi m d Tble 10-6: S = 0.5 S ut = 0.5 (.) = 105. kpi Eq. (10-5): Eq. (10-7): F = k =.() = 10.7 lbf + ( 10) + K = = = F D = K = 1.15 π ( ) = pi = 66.7 kpi S 105. n = = = 1.58 An (e) = m = 0.5 = 0.5(66.7) =.6 kpi, r = / m = 1. Uing the Gerber ftigue filure criterion with Zimmerli dt, Eq. (10-0): S u = 0.67 S ut = 0.67(.) = kpi The Gerber ordinte intercept for the Zimmerli dt i S 5 Se = = = 9.9 kpi 1 S / S 1 55 / Tble 6-7, p. 15, r S u S e S = S e rs u m u ( 9.9) = = 7.61 kpi ( 9.9) 1( 156.9) S 7.61 n f = = = 1.1 An Given: OD 0.9 in, C = 8, L 0 = in, L = 1 in, = 1 = in, q. end, unpeened, muic wire. () Tr OD = D + d = 0.9 in, C = D/d = 8 D = 8d 9d = 0.9 d = 0.1 An. Shigle MED, 10 th edition Chpter 10 Solution, Pge 19/1

20 D = 8(0.1) = 0.8 in (b) Tble 10-1: L = d (N t + 1) N t = L / d 1 = 1/0.1 1 = 9 coil An. Tble 10-1: (c) Tble 10-5: Eq. (10-9): (d) N = N t = 9 = 7 coil G = Mpi k = = = 0.98 lbf/in An. 6 d G 8D N F = k = 0.98() = lbf Eq. (10-5): K = = = Eq. (10-7): 8F D = K ( 10 ) pi kpi = π = = ( 0.1 ) Tble 10-: A = 01 kpi in m, m = 0.15 A 01 Eq. (10-1): Sut = = = 80.7 kpi m 0.15 d 0.1 Tble 10-6: S = 0.5 S ut = 0.5(80.7) = 16. kpi n S 16. = = = 0.65 An (e) = m = / = 195.7/ = kpi. Uing the Gerber ftigue filure criterion with Zimmerli dt, Eq. (10-0): S u = 0.67 S ut = 0.67(80.7) = kpi The Gerber ordinte intercept for the Zimmerli dt i S 5 Se = = = 8. kpi 1 S / S 1 55 /188.1 Tble 6-7, p. 15, r S u S e S = S e rs u m u ( 8.) = = 6.8 kpi ( 8.) 1( 188.1) Shigle MED, 10 th edition Chpter 10 Solution, Pge 0/1

21 n f S 6.8 = = = 0.76 An Obvioul, the pring i everel under deigned nd will fil tticll nd in ftigue. Increing C would improve mtter. Tr C = 1. Thi ield n = 1.8 nd n f = Given: F mx = 00 lbf, F min = 150 lbf, = 1 in, OD =.1 0. = 1.9 in, C = 7, unpeened, qured & ground, oil-tempered wire. () D = OD d = 1.9 d (1) Subtitute Eq. () into (1) C = D/d = 7 D = 7d () 7d = 1.9 d d = 1.9/8 = 0.75 in An. (b) From Eq. (): D = 7d = 7(0.75) = 1.66 in An. (c) (d) Tble 10-5: F k = = = 150 lbf/in An. 1 G = 11.6 Mpi Eq. (10-9): N = = = 6.69 coil d G 6 8D k Tble 10-1: N t = N + = 8.69 coil An. (e) Tble 10-: A = 17 kpi in m, m = Eq. (10-1): A 17 Sut = = m d 0.75 = 19. kpi Tble 10-6: S = 0.5 S ut = 0.5(19.) = kpi Eq. (10-5): K = = = F Eq. (10-7): D = K S = Shigle MED, 10 th edition Chpter 10 Solution, Pge 1/1

22 F S π = = = 5.5 lbf 8K D = F / k = 5.5/150 = 1.69 in Tble 10-1: L = N t d = 8.6(0.75) =.01 in L 0 = L + = =.70 in An For coil rdiu given b: R = R + 1 R R1 π N θ The torion of ection i T = PR where dl = R dθ U 1 T 1 δ P = = = P GJ P GJ P π N R R1 = R 0 1 θ dθ GJ + π N π N T dl PR d 0 π N π 1 R1 θ 1 π 0 P 1 N R R = + GJ R R N π PN π PN = = + + GJ ( R R1) GJ π 16PN J = d δ p = ( R 1 + R) ( R1 + R ) Gd P d G k = = An. δ 16 N( R + R ) R + R P ( R R1 ) ( R1 R) ( R1 R ) Given: F min = lbf, F mx = 18 lbf, k = 9.5 lbf/in, OD.5 in, n f = 1.5. For food ervice mchiner ppliction elect A1 Stinle wire. Tble 10-5: G = 10(10 6 ) pi Note tht for 0.01 d 0.10 in A = 169, m = < d 0.0 in A = 18, m = F = = 7 lbf, Fm = = 11 lbf, r = 7 / 11 Tr, d = in, 169 S ut = 0.16 (0.08) =. kpi θ Shigle MED, 10 th edition Chpter 10 Solution, Pge /1

23 S u = 0.67S ut = 16.7 kpi, S = 0.5S ut = 85.5 kpi Tr unpeened uing Zimmerli endurnce dt: S = 5 kpi, S m = 55 kpi S 5 Gerber: S = e 9.5 kpi 1 ( S / S ) = 1 (55 / 16.7) = m u (7 / 11) (16.7) (9.5) S = kpi (9.5) (7 / 11)(16.7) = α = S / n = 5.0 / 1.5 =. kpi f 8F 8(7) β = (10 ) = (10 ).785 kpi = π (0.08 ) (.).785 (.).785 (.) C = + (.785) 6.97 (.785) = (.785) D = Cd = 6.97(0.08) = in + (6.97) + K = = = 1.01 (6.97) 8F D 8(7)(0.558) = K 1.01 (10 ). kpi = π (0.08 ) = n = 5 /. = 1.50 check f 6 Gd 10(10 )(0.08) N = = = 1.0 coil 8kD 8(9.5)(0.558) Nt = = coil, L = dnt = 0.08() =.6 in mx = Fmx / k = 18 / 9.5 = in = (1 + ξ) mx = ( )(1.895) =.179 in L0 = =.819 in D.6(0.558) ( L0) cr =.6 = =.95 in α 0.5 = 1.15(18 / 7) = 1.15(18 / 7)(.) = 68.9 kpi n = S / = 85.5 / 68.9 = 1. f kg 9.5(86) = = = 109 Hz DN γ π (0.08 )(0.558)(1.0)(0.8) Thee tep re eil implemented on predheet, hown below, for different dimeter. Shigle MED, 10 th edition Chpter 10 Solution, Pge /1

24 d 1 d d d d m A S ut S u S S e S α.... β C D K n f N N t L S L (L 0 ) cr n f,(hz) The hded re depict condition outide the recommended deign condition. Thu, one pring i tifctor. The pecifiction re: A1 tinle wire, unpeened, qured nd ground, d = in, OD = = in, L 0 =.606 in, nd N t = turn An The tep re the me in Prob except tht the Gerber-Zimmerli criterion i replced with Goodmn-Zimmerli: S e = 1 S ( S S ) m u The problem then proceed in Prob The reult for the wire ize re hown Shigle MED, 10 th edition Chpter 10 Solution, Pge /1

25 below (ee olution to Prob for dditionl detil). Itertion of d for the firt tril d 1 d d d d 1 d d d d d m K A S ut n f S u N S N t S e L S mx α L β (L 0 ) cr C D n f (Hz) Without checking ll of the deign condition, it i obviou tht none of the wire ize tif n 1.. Alo, the Gerber line i cloer to the ield line thn the Goodmn. Setting n f = 1.5 for Goodmn mke it impoible to rech the ield line (n < 1). The tble below ue n f =. Itertion of d for the econd tril d 1 d d d d 1 d d d d d m K A S ut n f S u N S N t S e L S mx α L β (L 0 ) cr C D n f (Hz) The tifctor pring h deign pecifiction of: A1 tinle wire, unpeened, qured nd ground, d = in, OD = = 0.90 in, L 0 =.60 in, nd.n t = 19. turn. An. 10- Thi i the me Prob ince S = 5 kpi. Therefore, the pecifiction re: A1 tinle wire, unpeened, qured nd ground, d = in, OD = = Shigle MED, 10 th edition Chpter 10 Solution, Pge 5/1

26 0.971 in, L 0 =.606 in, nd N t = turn An. 10- For the Gerber-Zimmerli ftigue-filure criterion, S u = 0.67S ut, S e S r S u S e =, S 1 1 = ( Sm / Su) S e rs u The eqution for S i the bic difference. The lt column of dimeter of Ex re preented below with dditionl clcultion. d d S ut N S u L S e L S (L 0 ) cr S K α β n f C D n ID f n OD fom There re onl light chnge in the reult. 10- A in Prob. 10-, the bic chnge i S. S For Goodmn, Se = 1 ( Sm / Su) Reclculte S with rsesu S = rs + S u Clcultion for the lt dimeter of Ex re given below. e Shigle MED, 10 th edition Chpter 10 Solution, Pge 6/1

27 d d S ut N S u L S e L S (L 0 ) cr S K α β n f C D n ID f n OD fom There re onl light difference in the reult Ue: E = 8.6 Mpi, G = 11.5 Mpi, A = 10 kpi in m, m = 0.190, rel cot = 1. Tr d = in, 10 S ut = (0.067) =.0 kpi Tble 10-6: S = 0.5S ut = 105. kpi Tble 10-7: S = 0.75S ut = kpi Eq. (10-) with D/d = C nd C 1 = C F S mx σ A = [( K) (16 ) ] A C + = n C C C( C 1) 1 (16 C ) d S π + = n F mx S C 1 = ( C 1) 1 nf mx 1 S 1 S C C + = 0 nf mx nf mx 1 S S S C = ± + tke poitive root 16nF mx 16nF mx nf mx 1 π (0.067 )(175.5)(10 ) = 16(1.5)(18) π (0.067) (175.5)(10 ) π (0.067) (175.5)(10 ) + 16(1.5)(18) + =.590 ( 1.5)(18) Shigle MED, 10 th edition Chpter 10 Solution, Pge 7/1

28 D F i = Cd = = in i 500 C = = D 8D ± exp(0.105 C) 6.5 Ue the lowet F i in the preferred rnge. Thi reult in the bet fom. F i π (0.067) = 1000 = lbf 8(0.075) exp[0.105(.590)] 6.5 For implicit, we will round up to the next integer or hlf integer. Therefore, ue F i = 7 lbf 18 7 k = = lbf/in d G (0.067) (11.5)(10 ) N = = = 5.8 turn 8kD 8()(0.075) G 11.5 Nb = N = 5.8 =.88 turn E 8.6 L0 = (C 1 + Nb) d = [(.590) ](0.067) =.555 in L = =.055 in 18 lbf + (.590) + od: K = = = 1.6 (.590) 8 K Fmx D 8(1.6)(18)(0.075) mx (10 = = ) = 6.1 kpi π (0.067) S 105. ( n) bod = = = 1.70 mx 6.1 r (0.1) r = d = (0.067) = 0.1 in, C = = = d () 1 ( K) = = = 1.5 () 8Fmx D 8(18)(0.075) = ( K) = 1.5 (10 ) kpi π (0.067) = S 105. ( n) = = = ( Nb + ) D π (0.067) (.88 + )(0.075) fom = (1) = = Severl dimeter, evluted uing predheet, re hown below. Shigle MED, 10 th edition Chpter 10 Solution, Pge 8/1

29 d S ut S S C D F i (clc) F i (rd) k N N b L L 18 lbf K mx (n ) bod (n ) (n ) A fom Except for the in wire, ll pring tif the requirement of length nd number of coil. The in wire h the highet fom Given: N b = 8 coil, F i = 16 lbf, OQ&T teel, OD = 1.5 in, d = 0.16 in. D = OD d = = 1.8 in () Eq. (10-9): L 0 = (D d) + (N b + 1)d = ( ) + (8 + 1)(0.16) = 16.1 in An. or d + L 0 = (0.16) = 16.5 in overll D 1.8 (b) C = = = 8.6 d (8.6) + K = = = (8.6) 8Fi D 8(16)(1.8) i = K = = pi An. π (0.16) (c) From Tble 10-5 ue: G = 11.(10 6 ) pi nd E = 8.5(10 6 ) pi k G 11. N = Nb + = 8 + = 8. turn E 8.5 = 6 d G (0.16) (11.)(10 ) = =.855 lbf/in An. 8D N 8(1.8) (8.) Shigle MED, 10 th edition Chpter 10 Solution, Pge 9/1

30 (d) Tble 10-: A = 17 pi in m, m = Sut = (0.16) = 07.1 kpi S S = 0.75(07.1) = 155. kpi = 0.50(07.1) = 10.5 kpi od F πd S = πkd π(0.16) (10.5)(10 ) = = lbf 8(1.166)(1.8) Torionl tre on hook point K C r ( / ) d (.086) 1 = = =.086 = = = 1. (.086) π (0.16) (10.5)(10 ) F = = 10.9 lbf 8(1.)(1.8) Norml tre on hook point A r1 1.8 C1 = = = 8.6 d C1 1 (8.6) ( K) A = = = C1( C1 1) (8.6)(8.6 1) 16( K) AD S t = σ = F (10 ) F = = [ 16(1.099)(1.8) ] / π (0.16) + { / π (0.16) } = min(110.8, 10.9, 85. 8) = 85.8 lbf An lbf (e) Eq. (10-8): F F i = = = 1. in An. k F min = 9 lbf, F mx = 18 lbf F = =.5 lbf, Fm = = 1.5 lbf Shigle MED, 10 th edition Chpter 10 Solution, Pge 0/1

31 A1 tinle: 0.01 d 0.1 A = 169 kpi in m, m = d 0. A = 18 kpi in m, m = 0.6 E = 8 Mpi, G = 10 Gpi Tr d = in nd refer to the dicuion following Ex Sut = 0.16 (0.081) =.9 kpi Su = 0.67Sut = 16. kpi S = 0.5Sut = 85. kpi S = 0.55S = 1. kpi ut Tble 10-8: S r = 0.5S ut = kpi Sr / / Se = = = 1 [ Sr / ( Sut)] 1 [(109.8 / ) /.9] r = F / F =.5 / 1.5 = 0. m 57.8 kpi r S ut S e Tble 6-7: S = S e rs ut (0.) (.9 ) (57.8) S = =. kpi (57.8) 0.(.9) Hook bending 16C S S ( σ ) A = F ( K) A + = = π.5 ( 1)16 d n f A C C C S + = C( C 1) Thi eqution reduce to qudrtic in C (ee Prob. 10-5). The ueble root for C i S S S C = π (0.081) (.)(10 ) π (0.081) (.)(10 ) π (0.081) (.)(10 ) = =.91 Shigle MED, 10 th edition Chpter 10 Solution, Pge 1/1

32 D = Cd = 0.98 in i 500 C Fi = = D 8D ± exp(0.105 C) 6.5 Ue the lowet F i in the preferred rnge. π (0.081) F i = (0.98) exp[0.105(.91)] 6.5 = 8.55 lbf For implicit we will round up to next 1/ integer. Fi = 8.75 lbf 18 9 k = = 6 lbf/in d G (0.081) (10)(10 ) N = = =.7 turn 8kD 8(6)(0.98) G 10 Nb = N =.7 =. turn E 8 L0 = (C 1 + Nb) d = [(.91) 1 +.](0.081) =.60 in L = L + ( F F ) / k =.60 + ( ) / 6 =. 859 in mx 0 mx i.5() C 1 A ( σ ) = + 1 C 1-18(10 ) (.91 ).91 1 = kpi π (0.081 ) = S. ( n f ) A = = = check ( σ ) 1.1 A od: K m + (.91) + = = = 1.00 (.91) 8(1.00)(.5)(0.98) (10 = ) = kpi π (0.081) Fm 1.5 = = (11.16) =.7 kpi F.5 The repeting llowble tre from Tble 10-8 i S r = 0.0S ut = 0.0(.9) = 7.17 kpi The Gerber intercept i given b Eq. (10-) 7.17 / S e = = 1 [(7.17 / ) / 16.] 8.5 kpi Shigle MED, 10 th edition Chpter 10 Solution, Pge /1

33 From Tble 6-7, (.7)(8.5) ( n f ) bod = = (11.16) Let r = d = (0.081) = 0.16 r () 1 C = =, ( K) = = 1.5 d () ( K) 1.5 ( ) = = (11.16) = 10.7 kpi K 1.0 ( K) 1.5 ( m) = m = (.7) =.18 kpi K 1.0 Tble 10-8: (S r ) = 0.8S ut = 0.8(.9) = 68. kpi 68. / ( Se) = = 5.7 kpi 1 [(68. / ) / 16.] (.18)(5.7) ( n f ) = = (10.7) Yield ending: F mx ( C 1) ( σ A) mx = 1 + C 1 (18) (.91) = 1 (10 ) 8. kpi π (0.081 ) = 1. ( n) A = = od: i = ( Fi / F ) = (8.75 /.5)(11.16) = 1.7 kpi r = /( m i) = / (.7 1.7) = 0.98 r 0.98 ( S) = ( S i) = ( ) = 1.0 kpi r ( S) 1.0 ( n) bod = = = Hook her: S = 0.Sut = 0.(.9) = 7. kpi mx = ( ) + ( m) = =.9 kpi 7. ( n) = = ( Nb + ) D 7.6 π (0.081) (. + )(0.98) fom = = = 1.9 Shigle MED, 10 th edition Chpter 10 Solution, Pge /1

34 A tbultion of everl wire ize follow d S ut S u S r S e S C D OD F i (clc) F i (rd) k N N b L L 18 lbf (σ ) A (n f ) A K ( ) bod ( m ) bod S r S e (n f ) bod (K) ( ) ( m ) (S r ) (S e ) (n f ) S (σ A ) mx (n ) A i r (S ) bod (S ) (n ) bod (S ) ( ) mx (n ) fom optiml fom Shigle MED, 10 th edition Chpter 10 Solution, Pge /1

35 The hded re howw the condition not tified For the hook, M = FR inθ, M/ FF = R inθ 1 π / π FR δ F = F ( Rin EI θ ) R dθ = 0 EI The totl deflection of the bod nd the two hook 8FD N δ = d b π FR 8 FD Nb π F( D / ) + = + G EI d G E ( π / 6)( d ) 8FD G 8FD N = N b + = d G E d G G N = Nb + Q.E.D. E 10-9 Tble 10-5 (d = mm = in): E = GP Tble 10- for A7: Eq. (10-1): Eq. (10-57): A = 178 MP mm m, m = A 178 Sut = = = 170 MP m d S = σ ll = 0.78 S ut = 0.78(170) = 1069 MP D = OD d = = 8 mm Eq. (10-): C = D/d = 8/ = 7 K i C = = = C( C 1) (7)(7 1) Fr Eq. (10-): σ = Ki At ield, Fr = M, σ = S. Thu, M S π = = = K (1.119) i 6.00 N m Shigle MED, 10 th edition Chpter 10 Solution, Pge 5/1

36 Count the turn when M = 0 M N =.5 k d E where from Eq. (10-51): k = 10.8DN Thu, M N =.5 d E / (10.8 DN) Solving for N give.5 N = 1 + [10.8 DM / ( d E)].5 = = 1 + {[ 10.8(8)(6.00) ] / (196.5) }.1 turn Thi men (.5.1)(60 ) or 1. from cloed. An. Treting the hnd force in the middle of the grip, 87.5 r = = mm M 6.00( 10 ) Fmx = = = 87. N An. r The pring mteril nd condition re unknown. Given d = in nd OD = 0.500, () D = = 0.19 in Uing E = 8.6 Mpi for n etimte k 6 d E (0.081) (8.6)(10 ) = = = 10.8DN 10.8(0.19)(11).7 lbf in/turn for ech pring. The moment correponding to force of 8 lbf The frction windup turn i Fr = (8/)(.15) = 1.5 lbf in/pring n Fr 1.5 = = = k turn The rm wing through n rc of lightl le thn 180, 165. Thi ue up 165/60 or 0.58 turn. So n = = turn re left (or Shigle MED, 10 th edition Chpter 10 Solution, Pge 6/1

37 0.078(60 ) = 8.1 ). The originl configurtion of the pring w An. (b) D 0.19 C = = = 5.17 d C 1 (5.17) Ki = = = C 1 (5.17)(5.17 1) ( ) M (1.5) σ = Ki = d ( 10 ) pi 97 kp π = = π (0.081) pi An. To chieve thi tre level, the pring hd to hve et removed () Conider hlf nd double reult Stright ection: M = FR, M F = R Upper 180 ection: M = F[ R + R(1 co φ)] M = FR( co φ), = R( co φ) F Lower ection: M = FR in θ, Conidering bending onl: M F = R inθ Shigle MED, 10 th edition Chpter 10 Solution, Pge 7/1

38 U l / π π / δ = = 9 FR dx + FR ( co φ) R dφ + F( Rin θ) R dθ F EI F 9 π π π = R l R in R 0 EI + π φ + + FR 19π 9 FR = R + l = (19π R + 18 l) EI EI The pring rte i F k = = δ EI (19 18 ) R π R + l An. (b) Given: A7 HD wire, d = mm, R = 6 mm, nd l = 5 mm. Tble 10-5 (d = mm = in): E = 197. GP 9 π / 6 k = ( 10 ) N/m N/mm An π ( 0.006) + 18( 0.05) = = (c) The mximum tre will occur t the bottom of the top hook where the bendingmoment i FR nd the xil fore i F. Uing curved bem theor for bending, Eq. (-65), p. 1: σ Mc FRci Aer d e R d i i = = i ( π / ) ( / ) F F Axil: σ = = A / Combining, F Rc σ = σ + σ = + 1 = S e( R d / ) i mx i d S π F = Rci + 1 e( R d / ) (1) An. For the clip in prt (b), Eq. (10-1) nd Tble 10-: S ut = A/d m = 178/ = 156 MP Eq. (10-57): S = 0.78 S ut = 0.78(156) = 119 MP Shigle MED, 10 th edition Chpter 10 Solution, Pge 8/1

39 Tble -, p. 15: 1 r n = = mm ( ) e = r c r n = = mm c i = r n (R d /) = (6 /) = mm Eq. (1): 6 π ( 0.00 ) 119( 10 ) F = = 6.0 N An. ( 6) ( 6 1) 10- () The pring rte i M M = Fx, = x 0 x l F M M = Fl + FR 1 co θ, = l + R 1 co θ 0 θ π / F l π / { Fx( x) dx 0 F 1 co 0 l R Rd } { l R π l ( π ) l R ( π 8) R } 1 δ F = + + θ θ EI F = EI F 1EI k = = δ F l + R π l + ( π ) l R + ( π 8) R An. (b) Given: A1 tinle wire, d = 0.06 in, R = 0.65 in, nd l = 0.5 in. Tble 10-5: E = 8 Mpi Shigle MED, 10 th edition Chpter 10 Solution, Pge 9/1

40 π π I = d = ( 0.06 ) = 7.7( 10 ) in k = = 6. lbf/in An. + π + ( π ) + ( π ) (c) Tble 10-: A = 169 kpi in m, m = 0.16 Eq. (10-1): Eq. (10-57): S ut = A/ d m = 169/ = 5.0 kpi S = 0.61 S ut = 0.61(5.0) = 15. kpi One cn ue curved bem theor in the olution for Prob However, the eqution developed in Sec re equll vlid. C = D/d = ( /)/0.06 = 0.8 Eq. (10-): K i ( ) C = = = 1.07 C ( ) Eq. (10-), etting σ = S : ( + ) Fr F Ki = S = π Solving for F ield F =.5 lbf An. ( 0.06 ) Tr olving prt (c) of thi problem uing curved bem theor. You hould obtin the me nwer. 10- () M = Fx M Fx Fx σ = = = I c I c bh / / / 6 Contnt tre, bh Fx 6Fx 6 = h = σ bσ (1) An. Shigle MED, 10 th edition Chpter 10 Solution, Pge 0/1

41 At x = l, 6Fl ho = h = ho bσ x / l An. (b) M = Fx, M / F = x ( ) ( ) 1 bh ( x / l) M M / F 1 Fx x 1Fl l l / l 1/ = dx = dx x dx / EI E = bh o oe 0 1Fl 8Fl = = bh E / / l o bho E F bho E k = = An. 8l 10- Computer progrm will vr Computer progrm will vr. Shigle MED, 10 th edition Chpter 10 Solution, Pge 1/1