Chapter_10_Solutions

Chapter_10_Solutions - Chapter 10 10-1 From Eqs. (10-4) and...

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Chapter 10 10-1 From Eqs. (10-4) and (10-5) 4 1 0.615 4 2 44 4 WB CC KK C 3  Plot 100( K W K B )/ K W vs. C for 4 C 12 obtaining We see the maximum and minimum occur at C = 4 and 12 respectively where Maximum = 1.36 % Ans ., and Minimum = 0.743 % Ans . ______________________________________________________________________________ 10-2 A = Sd m d i m ( A uscu ) = [dim ( S ) dim( d m )] uscu = kpsi in m d i m ( A SI ) = [dim ( S ) dim( d m )] SI = MPa mm m  SI uscu uscu uscu MPa mm 6.894757 25.4 6.895 25.4 . kpsi in m mm m AA A A A n s  For music wire, from Table 10-4: A uscu = 201 kpsi in m , m = 0.145; what is A SI ? _____________________________________________________________________________ 0-3 Given: Music wire, d = 2.5 mm, OD = 31 mm, plain ground ends, N t = 14 coils. A SI = 6.895(25.4) 0.145 (201) = 2215 MPa mm m Ans . _ 1 Chapter 10 - Rev. A, Page 1/41
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( a ) Table 10-1: N a = N 1 = 14 1 = 13 coils L s = d N t = 2.5(14) = 35 mm Table 10-4: m = 0.145, A = 2211 MPa mm m Eq. (10-14): t 0.145 2211 1936 MPa 2.5 ut m A S d  Table 10-6: S sy = 0.45(1936) = 871.2 MPa D = OD d = 31 2.5 = 28.5 mm C = D/d = 28.5/2.5 = 11.4 Eq. (10-5):    4 11.4 2 42 1.117 4 3 4 11.4 3 B C K C  Eq. (10-7):   3 3 2.5 871.2 167.9 N 8 8 1.117 28.5 sy s B dS F KD Table 10-5): d = 2.5/25.4 = 0.098 in G = 81.0(10 3 ) MPa Eq. (10-9):   43 4 3 3 2.5 81 10 1.314 N / mm 8 8 28.5 13 a dG k DN 0 167.9 35 162.8 mm . 1.314 s s F L LA k   n s ( b ) F s = 167.9 N Ans . ( c ) k = 1.314 N/mm Ans . ( d ) 0 cr 149.9 mm 0.5 L . Spring needs to be supported. Ans . 2.63 28.5 _____________________________________________________________________________ 0-4 Given: Design load, F 1 = 130 N. 4, N = 13 coils, S sy = 871.2 MPa, F s = 167.9 N, Eq. (10-19): 3 N a 15 N a = 13 O.K. _ 1 Referring to Prob. 10-3 solution, C = 11. a L 0 = 162.8 mm and ( L 0 ) cr = 149.9 mm . Eq. (10-18): 4 C 12 C = 11.4 O.K. Chapter 10 - Rev. A, Page 2/41
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Eq. (10-17): 1 167.9 1 1 0.29 130 s F F  Eq. (10-20): 0.15, 0.29 . . OK  From Eq. (10-7) for static service 1 1 33 1 8 8(130)(28.5) 1.117 674 MPa (2.5) 871.2 1.29 674 B sy FD K d S n      Eq. (10-21): n s 1 . 2, n = 1 . 29 O.K. 1 167.9 167.9 674 870.5 MPa 130 130 / 871.2 / 870.5 1 s sy s S  S sy / s ( n s ) d : Not solid-safe (but was the basis of the design). Not O.K. L 0 ( L 0 ) 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, N t = 12 coils, L 0 = 5 in, squared ends. ( a ) Table 10-1: L s = d ( N t + 1) = 0.2(12 + 1) = 2.6 in Ans . ( b ) Table 10-1: N a = N t 2 = 12 2 = 10 coils Table 10-5: G = 11.2 Mpsi Eq. (10-9):    46 4 3 3 0.2 11.2 10 28 lbf/in 8 82 10 dG k DN F s = k y s = k ( L 0 L s ) = 28(5 2.6) = 67.2 lbf Ans . ( c ) Eq. (10-1): C = D/d = 2/0.2 = 10 Eq. (10-5):   410 2 42 1.135 43 4 1 0 3 B C K C  Eq. (10-7):   3 3 3 8 67.2 2 8 1.135 48.56 10 psi 0.2 sB FD K d Chapter 10 - Rev. A, Page 3/41
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Table 10-4: m = 0.187, A = 147 kpsi in m Eq. (10-14): 0.187 147 198.6 kpsi 0.2 ut m A S d  Table 10-6: S sy = 0.50 S ut = 0.50(198.6) = 99.3 kpsi 99.3 2.04 . 48.56 sy s s S nA n s ______________________________________________________________________________ 10-6 Given: Oil-tempered wire, d = 4 mm, C = 10, plain ends, L 0 = 80 mm, and at F = 50 N, y = 15 mm. ( a ) k = F/y = 50/15 = 3.333 N/mm Ans . ( b ) D = Cd = 10(4) = 40 mm OD = D + d = 40 + 4 = 44 mm Ans . ( c ) From Table 10-5, G = 77.2 GPa Eq. (10-9):    43 4 33 4 77.2 10 11.6 coils 8 8 3.333 40 a dG N kD Table 10-1: N t = N a = 11.6 coils Ans . ( d ) Table 10-1: L s = d ( N t + 1) = 4(11.6 + 1) = 50. 4 mm Ans . ( e ) Table 10-4: m = 0.187, A = 1855 MPa mm m Eq. (10-14): 0.187 1855 1431 MPa 4 ut m A S d Table 10-6: S sy = 0.50 S ut = 0.50(1431) = 715.5 MPa y s = L 0 L s = 80 50.4 = 29.6 mm F s = k y s = 3.333(29.6) = 98.66 N Eq. (10-5): 4 2 4(10) 2 1.135 4 ( 1 0 ) 3 B C K C  Chapter 10 - Rev. A, Page 4/41
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Eq. (10-7):    3 3 8 98.66 40 8 1.135 178.2 MPa 4 s sB FD K d  715.5 4.02 .
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Chapter_10_Solutions - Chapter 10 10-1 From Eqs. (10-4) and...

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