20_ch 05 Mechanical Design budynas_SM_ch05

20_ch 05 Mechanical Design budynas_SM_ch05 -...

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134 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design M A = 20 ± 193 . 7 2 + 233 . 5 2 = 6068 lbf · in M B = 10 ± 246 . 3 2 + 676 . 6 2 = 7200 lbf · in (maximum) σ x = 32(7200) π d 3 = 73 340 d 3 τ xy = 16(3383) π d 3 = 17 230 d 3 σ ± = ( σ 2 x + 3 τ 2 ) 1 / 2 = S y n ² ³ 73 340 d 3 ´ 2 + 3 ³ 17 230 d 3 ´ 2 µ 1 / 2 = 79 180 d 3 = 60 000 3 . 5 d = 1 . 665 in so use a standard diameter size of 1.75 in Ans. 5-26 From Prob. 5-25, τ max = ² ³ σ x 2 ´ 2 + τ 2 µ 1 / 2 = S y 2 n ² ³ 73 340 2 d 3 ´ 2 + ³ 17 230 d 3 ´ 2 µ 1 / 2 = 40 516 d 3 = 60 000 2(3 . 5) d = 1 . 678 in so use 1 . 75 in Ans. 5-27 T = (270 50)(0 . 150) = 33 N · m, S y = 370 MPa ( T 1 0 . 15 T 1 )(0 . 125) = 33 T 1 = 310 .
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