14_ch 05 Mechanical Design budynas_SM_ch05

14_ch 05 Mechanical Design budynas_SM_ch05 - = 32 2(16 54 =...

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128 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design τ xy = 16 T π d 3 = 16(30) π (0 . 020 3 ) = 19 . 10(10 6 )Pa = 19 . 10 MPa σ ± = ( σ 2 x + 3 τ 2 xy ) 1 / 2 = [95 . 49 2 + 3(19 . 1) 2 ] 1 / 2 = 101 . 1MPa n = S y σ ± = 280 101 . 1 = 2 . 77 Ans. B: σ x = 4 P π d 3 = 4(8)(10 3 ) π (0 . 020 2 ) = 25 . 47(10 6 )Pa = 25 . 47 MPa τ xy = 16 T π d 3 + 4 3 V A = 16(30) π (0 . 020 3 ) + 4 3 ± 0 . 55(10 3 ) ( π/ 4)(0 . 020 2 ) ² = 21 . 43(10 6 )Pa = 21 . 43 MPa σ ± = [25 . 47 2 + 3(21 . 43 2 )] 1 / 2 = 45 . 02 MPa n = 280 45 . 02 = 6 . 22 Ans. 5-15 S y = 32 kpsi At A, M = 6(190) = 1 140 lbf · in, T = 4(190) = 760 lbf · in . σ x = 32 M π d 3 = 32(1140) π (3 / 4) 3 = 27 520 psi τ zx = 16 T π d 3 = 16(760) π (3 / 4) 3 = 9175 psi τ max = ³ ´ 27 520 2 µ 2 + 9175 2 = 16 540 psi n = S y 2 τ max
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Unformatted text preview: = 32 2(16 . 54) = . 967 Ans . MSS predicts yielding 5-16 From Prob. 4-15, σ x = 27 . 52 kpsi, τ zx = 9 . 175 kpsi . For Eq. (5-15), adjusted for coordinates, σ ± = ¶ 27 . 52 2 + 3(9 . 175) 2 · 1 / 2 = 31 . 78 kpsi n = S y σ ± = 32 31 . 78 = 1 . 01 Ans . DE predicts no yielding, but it is extremely close. Shaft size should be increased....
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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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