7_ch 10 Mechanical Design budynas_SM_ch10

7_ch 10 Mechanical Design budynas_SM_ch10 - Now F s = ky s...

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Chapter 10 267 Table 10-6: S sy = 0 . 45(204 . 8) = 92 . 2 kpsi k = Gd 4 8 D 3 N a = 11 . 4(10 6 )(0 . 135) 4 8(1 . 865) 3 (3 . 25) = 22 . 45 lbf/in Table 10-1: L s = dN t = 0 . 135(5 . 25) = 0 . 709 in Now F s = ky s , y s = L 0 L s = 2 . 94 0 . 709 = 2 . 231 in τ s = K B ± 8( ky s ) D π d 3 ² = 1 . 096 ± 8(22 . 45)(2 . 231)(1 . 865) π (0 . 135) 3 ² (10 3 ) = 106 . 0 kpsi (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 )( π d 3 ) 8 K B kD = (92 200 / 1 . 2)( π )(0 . 135) 3 8(1 . 096)(22 . 45)(1 . 865) = 1 . 612 in L ± 0 = L s + y ± s = 0 . 709 + 1 . 612 = 2 . 321 in Wind the spring to a free length of 2.32 in. Ans. 10-11 Given: A229 (OQ&T steel), SQ&GRD ends, d = 0 . 144 in, OD = 1 . 0 in, L 0 = 3 . 75 in, N t = 13 turns. Table 10-4: A = 147 kpsi · in m , m = 0 . 187 Table 10-5: G = 11 . 4(10 6 ) psi D = OD d = 1 . 0 0 . 144 = 0 . 856 in C = D / d = 0 . 856 / 0 . 144 = 5 . 944 K B = 4(5 . 944) + 2 4(5 . 944) 3 = 1 . 241 Table 10-1: N a = N t 2 = 13 2 = 11 turns S ut = 147 (0 . 144) 0 . 187 = 211 . 2 kpsi Table 10-6: S sy = 0 . 50(211 . 2) = 105 . 6 kpsi k = Gd 4 8 D 3 N a = 11 . 4(10 6 )(0 . 144) 4 8(0 . 856) 3 (11) = 88 . 8 lbf/in Table 10-1: L s = dN t = 0 . 144(13) = 1 . 872 in
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Unformatted text preview: Now F s = ky s , y s = L L s = 3 . 75 1 . 872 = 1 . 878 in s = K B 8( ky s ) D d 3 = 1 . 241 8(88 . 8)(1 . 878)(0 . 856) (0 . 144) 3 (10 3 ) = 151 . 1 kpsi (1) s > S sy , that is,151 . 1 > 105 . 6 kpsi; the spring is not solid-safe. Solving Eq. (1) for y s gives y s = ( S sy / n )( d 3 ) 8 K B kD = (105 600 / 1 . 2)( )(0 . 144) 3 8(1 . 241)(88 . 8)(0 . 856) = 1 . 094 in L = L s + y s = 1 . 878 + 1 . 094 = 2 . 972 in Wind the spring to a free length 2.972 in. Ans....
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