56_ch 03 Mechanical Design budynas_SM_ch03

56_ch 03 Mechanical Design budynas_SM_ch03 -...

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Chapter 3 69 3-86 From Table A-5: ν 1 = 0 . 211, ν 2 = 0 . 292, E 1 = 14 . 5 ( 10 6 ) psi, E 2 = 30 ( 10 6 ) psi, d 1 = 6 in, d 2 =∞ , l = 2 in (a) Eq. (3-73): b = ± 2(800) π (2) (1 0 . 211 2 ) / 14 . 5(10 6 ) + (1 0 . 292 2 ) / [30(10 6 )] 1 / 6 + 1 / = 0 . 012 135 in p max = 2(800) π (0 . 012 135)(2) = 20 984 psi For z = 0 in, σ x 1 =− 2 ν 1 p max =− 2(0 . 211)20 984 =− 8855 psi in wheel σ x 2 =− 2(0 . 292)20 984 =− 12 254 psi In plate
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