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21_ch 10 Mechanical Design budynas_SM_ch10

21_ch 10 Mechanical Design budynas_SM_ch10 - 25 162 2 162 =...

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Chapter 10 281 10-29 Given: N b = 84 coils, F i = 16 lbf, OQ&T steel, OD = 1.5 in, d = 0 . 162 in. D = 1 . 5 0 . 162 = 1 . 338 in (a) Eq. (10-39): L 0 = 2( D d ) + ( N b + 1) d = 2(1 . 338 0 . 162) + (84 + 1)(0 . 162) = 16 . 12 in Ans . or 2 d + L 0 = 2(0 . 162) + 16 . 12 = 16 . 45 in overall. (b) C = D d = 1 . 338 0 . 162 = 8 . 26 K B = 4(8 . 26) + 2 4(8 . 26) 3 = 1 . 166 τ i = K B 8 F i D π d 3 = 1 . 166 8(16)(1 . 338) π (0 . 162) 3 = 14 950 psi Ans . (c) From Table 10-5 use: G = 11 . 4(10 6 ) psi and E = 28 . 5(10 6 ) psi N a = N b + G E = 84 + 11 . 4 28 . 5 = 84 . 4 turns k = d 4 G 8 D 3 N a = (0 . 162) 4 (11 . 4)(10 6 ) 8(1 . 338) 3 (84 . 4) = 4 . 855 lbf/in Ans . (d) Table 10-4: A = 147 psi · in m , m = 0 . 187 S ut = 147 (0 . 162) 0 . 187 = 207 . 1 kpsi S y = 0 . 75(207 . 1) = 155 . 3 kpsi S sy = 0 . 50(207 . 1) = 103 . 5 kpsi Body F = π d 3 S sy π K B D = π (0 . 162) 3 (103 . 5)(10 3 ) 8(1 . 166)(1 . 338) = 110 . 8 lbf Torsional stress on hook point B C 2 = 2 r 2 d = 2(0 .
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Unformatted text preview: . 25 + . 162 / 2) . 162 = 4 . 086 ( K ) B = 4 C 2 − 1 4 C 2 − 4 = 4(4 . 086) − 1 4(4 . 086) − 4 = 1 . 243 F = π (0 . 162) 3 (103 . 5)(10 3 ) 8(1 . 243)(1 . 338) = 103 . 9 lbf Normal stress on hook point A C 1 = 2 r 1 d = 1 . 338 . 162 = 8 . 26 ( K ) A = 4 C 2 1 − C 1 − 1 4 C 1 ( C 1 − 1) = 4(8 . 26) 2 − 8 . 26 − 1 4(8 . 26)(8 . 26 − 1) = 1 . 099...
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