17_ch 16 Mechanical Design budynas_SM_ch16

# 17_ch 16 Mechanical Design budynas_SM_ch16 - in Â s 2 I x =...

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412 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design The bearing and shear stress estimates are σ b = 2 . 15(10 3 ) 10(22 . 5 13) =− 22 . 6MPa Ans. τ = 2 . 15(10 3 ) 10[0 . 25 π (17 . 75) 2 ] = 0 . 869 MPa Ans. 16-22 ω 1 = 2 π n / 60 = 2 π (1800) / 60 = 188 . 5 rad/s ω 2 = 0 From Eq. (16-51), I 1 I 2 I 1 + I 2 = Tt 1 ω 1 ω 2 = 320(8 . 3) 188 . 5 0 = 14 . 09 N · m · s 2 Eq. (16-52): E = 14 . 09 ± 188 . 5 2 2 ² (10 3 ) = 250 kJ Eq. (16-55): ± T = E C p m = 250(10 3 ) 500(18) = 27 . 8 C Ans. 16-23 n = n 1 + n 2 2 = 260 + 240 2 = 250 rev/min C s = 260 240 250 = 0 . 08 Ans. ω = 2 π (250) / 60 = 26 . 18 rad/s I = E 2 E 1 C s ω 2 = 5000(12) 0 . 08(26 . 18) 2 = 1094 lbf ·
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Unformatted text preview: in Â· s 2 I x = m 8 ( d 2 o + d 2 i ) = W 8 g ( d 2 o + d 2 i ) W = 8 gI d 2 o + d 2 i = 8(386)(1094) 60 2 + 56 2 = 502 lbf w = . 260 lbf/in 3 for cast iron V = W w = 502 . 260 = 1931 in 3 Also, V = Ï€ t 4 ( d 2 o âˆ’ d 2 i ) = Ï€ t 4 ( 60 2 âˆ’ 56 2 ) = 364 t in 3 Equating the expressions for volume and solving for t , t = 1931 364 = 5 . 3 in Ans....
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