5_ch 10 Mechanical Design budynas_SM_ch10

5_ch 10 Mechanical Design budynas_SM_ch10 - s = L − L s =...

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Chapter 10 265 Table 10-1: L s = N t d = 40(0 . 006) = 0 . 240 in Now F s = ky s where y s = L 0 L s = 0 . 390 in. Thus, τ s = K B ± 8( ky s ) D π d 3 ² = 1 . 294 ± 8(1 . 895)(0 . 39)(0 . 030) π (0 . 006) 3 ² (10 3 ) = 338 . 2 kpsi (1) τ s > S sy , that is, 338 . 2 > 189 . 9 kpsi; the spring is not solid-safe. Solving Eq. (1) for y s gives y ± s = ( τ s / n s )( π d 3 ) 8 K B kD = (189 900 / 1 . 2)( π )(0 . 006) 3 8(1 . 294)(1 . 895)(0 . 030) = 0 . 182 in Using a design factor of 1.2, L ± 0 = L s + y ± s = 0 . 240 + 0 . 182 = 0 . 422 in The spring should be wound to a free length of 0.422 in. Ans. 10-8 Given: B159 (phosphor bronze), SQ&GRD ends, d = 0 . 012 in, OD = 0 . 120 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 = 0 . 120 0 . 012 = 0 . 108 in C = D / d = 0 . 108 / 0 . 012 = 9 K B = 4(9) + 2 4(9) 3 = 1 . 152 Table 10-1: N a = N t 2 = 15 . 1 2 = 13 . 1 turns S ut = 145 0 . 012 0 = 145 kpsi Table 10-6: S sy = 0 . 35(145) = 50 . 8 kpsi k = Gd 4 8 D 3 N a = 6(10 6 )(0 . 012) 4 8(0 . 108) 3 (13 . 1) = 0 . 942 lbf/in Table 10-1: L s = dN t = 0 . 012(15 . 1) = 0 . 181 in Now F s = ky s , y
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Unformatted text preview: s = L − L s = . 81 − . 181 = . 629 in τ s = K B ± 8( ky s ) D π d 3 ² = 1 . 152 ± 8(0 . 942)(0 . 6)(0 . 108) π (0 . 012) 3 ² (10 − 3 ) = 108 . 6 kpsi (1) τ s > S sy , that is, 108 . 6 > 50 . 8 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 = (50 . 8 / 1 . 2)( π )(0 . 012) 3 (10 3 ) 8(1 . 152)(0 . 942)(0 . 108) = . 245 in L ± = L s + y ± s = . 181 + . 245 = . 426 in Wind the spring to a free length of 0.426 in. Ans....
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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