24_ch 10 Mechanical Design budynas_SM_ch10

# 24_ch 10 Mechanical Design budynas_SM_ch10 - 134 2 84 4 = 1...

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284 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The repeating allowable stress from Table 7-8 is S sr = 0 . 30 S ut = 0 . 30(243 . 9) = 73 . 17 kpsi The Gerber intercept is S se = 73 . 17 / 2 1 [(73 . 17 / 2) / 163 . 4] 2 = 38 . 5 kpsi From Table 6-7, ( n f ) body = 1 2 ± 163 . 4 33 . 47 ² 2 ± 11 . 16 38 . 5 ² 1 + ³ 1 + ´ 2(33 . 47)(38 . 5) 163 . 4(11 . 16) µ 2 = 2 . 53 Let r 2 = 2 d = 2(0 . 081) = 0 . 162 C 2 = 2 r 2 d = 4, ( K ) B = 4(4) 1 4(4) 4 = 1 . 25 ( τ a ) B = ( K ) B K B τ a = 1 . 25 1 . 30 (11 . 16) = 10 . 73 kpsi ( τ m ) B = ( K ) B K B τ m = 1 . 25 1 . 30 (33 . 47) = 32 . 18 kpsi Table 10-8: ( S sr ) B = 0 . 28 S ut = 0 . 28(243 . 9) = 68 . 3 kpsi ( S se ) B = 68 . 3 / 2 1 [(68 . 3 / 2) / 163 . 4] 2 = 35 . 7 kpsi ( n f ) B = 1 2 ± 163 . 4 32 . 18 ² 2 ± 10 . 73 35 . 7 ² 1 + ³ 1 + ´ 2(32 . 18)(35 . 7) 163 . 4(10 . 73) µ 2 = 2 . 51 Yield Bending: ( σ A ) max = 4 F max π d 2 ´ (4 C 2 C 1) C 1 + 1 µ = 4(18) π (0 . 081 2 ) ´ 4(4 . 91) 2 4 . 91 1 4 . 91 1 + 1 µ (10 3 ) = 84 . 4 kpsi ( n y ) A =
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Unformatted text preview: 134 . 2 84 . 4 = 1 . 59 Body: τ i = ( F i / F a ) τ a = (8 . 75 / 4 . 5)(11 . 16) = 21 . 7 kpsi r = τ a / ( τ m − τ i ) = 11 . 16 / (33 . 47 − 21 . 7) = . 948 ( S sa ) y = r r + 1 ( S sy − τ i ) = . 948 . 948 + 1 (85 . 4 − 21 . 7) = 31 . 0 kpsi ( n y ) body = ( S sa ) y τ a = 31 . 11 . 16 = 2 . 78 Hook shear: S sy = . 3 S ut = . 3(243 . 9) = 73 . 2 kpsi τ max = ( τ a ) B + ( τ m ) B = 10 . 73 + 32 . 18 = 42 . 9 kpsi...
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