38_ch 04 Mechanical Design budynas_SM_ch04

38_ch 04 Mechanical Design budynas_SM_ch04 -...

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Unformatted text preview: budynas_SM_ch04.qxd 11/28/2006 20:50 FIRST PAGES Page 107 107 Chapter 4 But ∂ M /∂ M A = 1. Therefore π /2 π /2 M dθ = 0 MA − 0 FR (1 − cos θ ) d θ = 0 2 Since this term is zero, we have MA = 2 FR 1− 2 π Substituting into Eq. (1) M= 2 FR cos θ − 2 π The maximum occurs at B where θ = π/2. It is MB = − 4-72 Ans. For one quadrant M= δ= 2 FR cos θ − 2 π ∂U =4 ∂F F R3 = EI = 4-73 FR π F R3 EI π /2 0 π /2 0 ; ∂M 2 R cos θ − = ∂F 2 π M ∂M R dθ EI ∂F 2 cos θ − π π 2 − 4 π 2 dθ Ans. Cπ2E I Pcr = l2 π π D4 4 4 I = (D − d ) = (1 − K 4 ) 64 64 Pcr = C π 2 E π D4 (1 − K 4 ) 2 l 64 64 Pcr l 2 D= π 3 C E (1 − K 4 ) 1/4 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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