6_ch 10 Mechanical Design budynas_SM_ch10

# 6_ch 10 Mechanical Design budynas_SM_ch10 - + . 161 = . 577...

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266 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 10-9 Given: A313 (stainless steel), SQ&GRD ends, d = 0 . 040 in, OD = 0 . 240 in, L 0 = 0 . 75 in, N t = 10 . 4 turns. Table 10-4: A = 169 kpsi · in m , m = 0 . 146 Table 10-5: G = 10(10 6 ) psi D = OD d = 0 . 240 0 . 040 = 0 . 200 in C = D / d = 0 . 200 / 0 . 040 = 5 K B = 4(5) + 2 4(5) 3 = 1 . 294 Table 10-6: N a = N t 2 = 10 . 4 2 = 8 . 4 turns S ut = 169 (0 . 040) 0 . 146 = 270 . 4 kpsi Table 10-13: S sy = 0 . 35(270 . 4) = 94 . 6 kpsi k = Gd 4 8 D 3 N a = 10(10 6 )(0 . 040) 4 8(0 . 2) 3 (8 . 4) = 47 . 62 lbf/in Table 10-6: L s = dN t = 0 . 040(10 . 4) = 0 . 416 in Now F s = ky s , y s = L 0 L s = 0 . 75 0 . 416 = 0 . 334 in τ s = K B ± 8( ky s ) D π d 3 ² = 1 . 294 ± 8(47 . 62)(0 . 334)(0 . 2) π (0 . 040) 3 ² (10 3 ) = 163 . 8 kpsi (1) τ s > S sy , that is, 163 . 8 > 94 . 6 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 = (94 600 / 1 . 2)( π )(0 . 040) 3 8(1 . 294)(47 . 62)(0 . 2) = 0 . 161 in L ± 0 = L s + y ± s = 0 . 416
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Unformatted text preview: + . 161 = . 577 in Wind the spring to a free length 0.577 in. Ans. 10-10 Given: A227 (hard drawn steel), d = . 135 in, OD = 2 . 0 in, L = 2 . 94 in, N t = 5 . 25 turns. Table 10-4: A = 140 kpsi · in m , m = . 190 Table 10-5: G = 11 . 4(10 6 ) psi D = OD − d = 2 − . 135 = 1 . 865 in C = D / d = 1 . 865 / . 135 = 13 . 81 K B = 4(13 . 81) + 2 4(13 . 81) − 3 = 1 . 096 N a = N t − 2 = 5 . 25 − 2 = 3 . 25 turns S ut = 140 (0 . 135) . 190 = 204 . 8 kpsi...
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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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