33_ch 04 Mechanical Design budynas_SM_ch04

33_ch 04 Mechanical Design budynas_SM_ch04 -...

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102 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design δ Q = U Q ± ± ± ± Q = 0 = 1 EI ² π 0 ( PR sin θ ) R (1 cos θ ) Rd θ =− 2 3 Deflection is upward and equals 2( 3 / ) Ans. 4-64 Equation (4-28) becomes U = 2 ² π 0 M 2 θ 2 R / h > 10 where M = FR cos θ ) and M F = R cos θ ) δ = U F = 2 ² π 0 M M F θ = 2 ² π 0 3 cos θ ) 2 d θ = 3 π 3 Since I = bh 3 / 12 = 4(6) 3 / 12 = 72 mm 4 and R = 81 / 2 = 40 . 5mm, we have δ = 3 π (40 . 5) 3 F 131(72) = 66 . 4 F mm Ans. where F is in kN. 4-65 M Px , M P x 0 x l M = Pl + cos
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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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