2_ch 04 Mechanical Design budynas_SM_ch04

# 2_ch 04 Mechanical Design budynas_SM_ch04 -...

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Chapter 4 71 Thus T 1 = Kd 4 x θ ; T 2 = Kd 4 θ l x If x = l / 2, then T 1 = T 2 . If x < l / 2, then T 1 > T 2 Using τ = 16 T d 3 and θ = 32 Tl / ( G π d 4 ) gives T = π d 3 τ 16 and so θ all = 32 l G π d 4 · π d 3 τ 16 = 2 l τ all Gd Thus, if x < l / 2, the allowable twist is θ all = 2 x τ all Gd Ans. Since k = Kd 4 ± 1 x + 1 l x ² = π Gd 4 32 ± 1 x + 1 l x ² Ans. Then the maximum torque is found to be T max = π d 3 x τ all 16 ± 1 x + 1 l x ² Ans. 4-4 Both legs have the same twist angle. From Prob. 4-3, for equal shear, d is linear in x . Thus, d 1 = 0 . 2 d 2 Ans. k = π G 32 ³ ( 0 . 2 d 2 ) 4 0 . 2 l + d 4 2 0 . 8 l ´ = π G 32 l µ 1 . 258 d 4 2 Ans. θ all = 2(0
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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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