1_ch 10 Mechanical Design budynas_SM_ch10

# 1_ch 10 Mechanical Design budynas_SM_ch10 - 2 4(10 67 − 3...

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Chapter 10 10-1 10-2 A = Sd m dim( A uscu ) = dim( S ) dim( d m ) = kpsi · in m dim( A SI ) = dim( S 1 ) dim ( d m 1 ) = MPa · mm m A SI = MPa kpsi · mm m in m A uscu = 6 . 894 757(25 . 40) m A uscu . = 6 . 895(25 . 4) m A uscu Ans. For music wire, from Table 10-4: A uscu = 201, m = 0 . 145 ; what is A SI ? A SI = 6 . 89(25 . 4) 0 . 145 (201) = 2214 MPa · mm m Ans. 10-3 Given: Music wire, d = 0 . 105 in, OD = 1 . 225 in, plain ground ends, N t = 12 coils. Table 10-1: N a = N t 1 = 12 1 = 11 L s = dN t = 0 . 105(12) = 1 . 26 in Table 10-4: A = 201, m = 0 . 145 (a) Eq. (10-14): S ut = 201 (0 . 105) 0 . 145 = 278 . 7 kpsi Table 10-6: S sy = 0 . 45(278 . 7) = 125 . 4 kpsi D = 1 . 225 0 . 105 = 1 . 120 in C = D d = 1 . 120 0 . 105 = 10 . 67 Eq. (10-6): K B = 4(10 . 67)
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Unformatted text preview: + 2 4(10 . 67) − 3 = 1 . 126 Eq. (10-3): F | S sy = π d 3 S sy 8 K B D = π (0 . 105) 3 (125 . 4)(10 3 ) 8(1 . 126)(1 . 120) = 45 . 2 lbf Eq. (10-9): k = d 4 G 8 D 3 N a = (0 . 105) 4 (11 . 75)(10 6 ) 8(1 . 120) 3 (11) = 11 . 55 lbf/in L = F | S sy k + L s = 45 . 2 11 . 55 + 1 . 26 = 5 . 17 in Ans . 1 2 " 4" 1" 1 2 " 4" 1"...
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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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