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19_ch 03 Mechanical Design budynas_SM_ch03

# 19_ch 03 Mechanical Design budynas_SM_ch03 - x 1 − ν 2...

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32 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design 3-13 σ = F A = 2000 ( π/ 4)(0 . 5 2 ) = 10 190 psi = 10 . 19 kpsi Ans. δ = FL AE = σ L E = 10 190 72 30(10 6 ) = 0 . 024 46 in Ans. 1 = δ L = 0 . 024 46 72 = 340(10 6 ) = 340 µ Ans. From Table A-5, ν = 0 . 292 2 = − ν 1 = − 0 . 292(340) = − 99 . 3 µ Ans. d = 2 d = − 99 . 3(10 6 )(0 . 5) = − 49 . 6(10 6 ) in Ans. 3-14 From Table A-5, E = 71 . 7 GPa δ = σ L E = 135(10 6 ) 3 71 . 7(10 9 ) = 5 . 65(10 3 ) m = 5 . 65 mm Ans. 3-15 With σ z = 0, solve the first two equations of Eq. (3-19) simultaneously. Place E on the left- hand side of both equations, and using Cramer’s rule, σ x = E x ν E y 1 1 ν ν 1 = E x + ν E y 1 ν 2 = E ( x + ν y ) 1 ν 2 Likewise, σ y = E ( y + ν x ) 1 ν
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Unformatted text preview: x ) 1 − ν 2 From Table A-5, E = 207 GPa and ν = . 292 . Thus, σ x = E ( ± x + ν± y ) 1 − ν 2 = 207(10 9 )[0 . 0021 + . 292( − . 000 67)] 1 − . 292 2 (10 − 6 ) = 431 MPa Ans. σ y = 207(10 9 )[ − . 000 67 + . 292(0 . 0021)] 1 − . 292 2 (10 − 6 ) = − 12 . 9 MPa Ans. 3-16 The engineer has assumed the stress to be uniform. That is, ² F t = − F cos θ + τ A = ⇒ τ = F A cos θ When failure occurs in shear S su = F A cos θ ± ² t ³ F...
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