45_ch 03 Mechanical Design budynas_SM_ch03

# 45_ch 03 Mechanical Design budynas_SM_ch03 - Ans. 3-60 ρ =...

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58 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design ( σ t ) max = 0 . 282 386 ± 2 π (7200) 60 ² 2 ³ 3 + 0 . 292 8 ´ × ± 0 . 375 2 + 5 2 + (0 . 375 2 )(5 2 ) 0 . 375 2 1 + 3(0 . 292) 3 + 0 . 292 (0 . 375 2 ) ² = 8556 psi τ max = 8556 2 = 4278 psi Ans. Radial stress: σ r = k µ r 2 i + r 2 o r 2 i r 2 o r 2 r 2 Maxima: d σ r dr = k µ 2 r 2 i r 2 o r 3 2 r = 0 r = r i r o = · 0 . 375(5) = 1 . 3693 in ( σ r ) max = 0 . 282 386 ± 2 π (7200) 60 ² 2 ³ 3 + 0 . 292 8 ´± 0 . 375 2 + 5 2 0 . 375 2 (5 2 ) 1 . 3693 2 1 . 3693 2 ² = 3656 psi Ans. 3-59 ω = 2 π (2069) / 60 = 216 . 7 rad/s, ρ = 3320 kg/m 3 , ν = 0 . 24, r i = 0 . 0125 m, r o = 0 . 15 m ; use Eq. (3-55) σ t = 3320(216 . 7) 2 ³ 3 + 0 . 24 8 ´± (0 . 0125) 2 + (0 . 15) 2 + (0 . 15) 2 1 + 3(0 . 24) 3 + 0 . 24 (0 . 0125) 2 ² (10) 6 = 2.85 MPa
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Unformatted text preview: Ans. 3-60 ρ = (6 / 16) 386(1 / 16)( π/ 4)(6 2 − 1 2 ) = 5 . 655(10 − 4 ) lbf · s 2 /in 4 τ max is at bore and equals σ t 2 Eq. (3-55) ( σ t ) max = 5 . 655(10 − 4 ) ± 2 π (10 000) 60 ² 2 ³ 3 + . 20 8 ´± . 5 2 + 3 2 + 3 2 − 1 + 3(0 . 20) 3 + . 20 (0 . 5) 2 ² = 4496 psi τ max = 4496 2 = 2248 psi Ans....
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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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