8_ch 10 Mechanical Design budynas_SM_ch10

# 8_ch 10 Mechanical Design budynas_SM_ch10 - ± = L s y ± s...

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268 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 10-12 Given: A232 (Cr-V steel), SQ&GRD ends, d = 0 . 192 in, OD = 3 in, L 0 = 9 in, N t = 8turns. Table 10-4: A = 169 kpsi · in m , m = 0 . 168 Table 10-5: G = 11 . 2(10 6 ) psi D = OD d = 3 0 . 192 = 2 . 808 in C = D / d = 2 . 808 / 0 . 192 = 14 . 625 (large) K B = 4(14 . 625) + 2 4(14 . 625) 3 = 1 . 090 Table 10-1: N a = N t 2 = 8 2 = 6 turns S ut = 169 (0 . 192) 0 . 168 = 223 . 0 kpsi Table 10-6: S sy = 0 . 50(223 . 0) = 111 . 5 kpsi k = Gd 4 8 D 3 N a = 11 . 2(10 6 )(0 . 192) 4 8(2 . 808) 3 (6) = 14 . 32 lbf/in Table 10-1: L s = dN t = 0 . 192(8) = 1 . 536 in Now F s = ky s , y s = L 0 L s = 9 1 . 536 = 7 . 464 in τ s = K B ± 8( ky s ) D π d 3 ² = 1 . 090 ± 8(14 . 32)(7 . 464)(2 . 808) π (0 . 192) 3 ² (10 3 ) = 117 . 7 kpsi (1) τ s > S sy , that is,117 . 7 > 111 . 5 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 = (111 500 / 1 . 2)( π )(0 . 192) 3 8(1 . 090)(14 . 32)(2 . 808) = 5 . 892 in L
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Unformatted text preview: ± = L s + y ± s = 1 . 536 + 5 . 892 = 7 . 428 in Wind the spring to a free length of 7.428 in. Ans. 10-13 Given: A313 (stainless steel) SQ&GRD ends, d = . 2 mm, OD = . 91 mm, L = 15 . 9 mm, N t = 40 turns. Table 10-4: A = 1867 MPa · mm m , m = . 146 Table 10-5: G = 69 . 0 GPa D = OD − d = . 91 − . 2 = . 71 mm C = D / d = . 71 / . 2 = 3 . 55 (small) K B = 4(3 . 55) + 2 4(3 . 55) − 3 = 1 . 446 N a = N t − 2 = 40 − 2 = 38 turns S ut = 1867 (0 . 2) . 146 = 2361 . 5 MPa...
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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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