16_ch 05 Mechanical Design budynas_SM_ch05

16_ch 05 Mechanical Design budynas_SM_ch05 - = . 016 19...

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130 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-18 For a thin walled cylinder made of AISI 1018 steel, S y = 54 kpsi, S ut = 64 kpsi . The state of stress is σ t = pd 4 t = p (8) 4(0 . 05) = 40 p , σ l = pd 8 t = 20 p , σ r =− p These three are all principal stresses. Therefore, σ ± = 1 2 [( σ 1 σ 2 ) 2 + ( σ 2 σ 3 ) 2 + ( σ 3 σ 1 ) 2 ] 1 / 2 = 1 2 [(40 p 20 p ) 2 + (20 p + p ) 2 + ( p 40 p ) 2 ] = 35 . 51 p = 54 p = 1 . 52 kpsi (for yield) Ans . For rupture, 35 . 51 p . = 64 p . = 1 . 80 kpsi Ans . 5-19 For hot-forged AISI steel w = 0 . 282 lbf/in 3 , S y = 30 kpsi and ν = 0 . 292. Then ρ = w/ g = 0.282 / 386 lbf · s 2 /in ; r i = 3in ; r o = 5in ; r 2 i = 9 ; r 2 o = 25 ; 3 + ν = 3 . 292 ; 1 + 3 ν = 1.876. Eq. (3-55) for r = r i becomes σ t = ρω 2 ± 3 + ν 8 ²³ 2 r 2 o + r 2 i ± 1 1 + 3 ν 3 + ν ²´ Rearranging and substituting the above values: S y ω 2 = 0 . 282 386 ± 3 . 292 8 ²³ 50 + 9 ± 1 1 . 876 3 . 292 ²´
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Unformatted text preview: = . 016 19 Setting the tangential stress equal to the yield stress, = 30 000 . 016 19 1 / 2 = 1361 rad/s or n = 60 / 2 = 60(1361) / (2 ) = 13 000 rev/min Now check the stresses at r = ( r o r i ) 1 / 2 , or r = [5(3)] 1 / 2 = 3 . 873 in r = 2 3 + 8 ( r o r i ) 2 = . 282 2 386 3 . 292 8 (5 3) 2 = . 001 203 2 Applying Eq. (3-55) for t t = 2 . 282 386 3 . 292 8 9 + 25 + 9(25) 15 1 . 876(15) 3 . 292 = . 012 16 2...
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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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