ME211solution_5 - Bearing stress: P 30(103 ) b = = =...

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Bearing stress: 3 22 30(10 ) 48.3 (0.06 0.053 ) 4 b P MPa A σ π == = Average shear stress: 3 30(10 ) 18.4 (0.052)(0.01) avg V MPa A τ =
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Case (a) : 0; 6 6 y FP P =− == k N 2 6 306 (0.005) 4 Pk N MPa A σ π = Case (b) : ;61 5 Fk x x 0.4 0.2 x mm => The stretched length due to the applied load is larger than the originally stretched length Hence, load is entirely supported by spring. (No force at O) 2 6 306 (0.005) 4 N MPa A =
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0.85 0.65 0; 450(12) 2 0 2 O MF + ⎛⎞ =− ⎜⎟ ⎝⎠ = 3600 lb F = A= 22 (0.85 0.65 ) 0.07854 12 in 2 π −= 3600 45.8 0.07854 avg F ksi A τ == =
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Support reactions: From FBD (a) 0; (sin 45 )(5) 1.5(7) 0 DB C MF = i p i p 2.97 BC Fk = From FBD (b) 0; (10) 1.5(7) 0 Ay MD =− = 1.05 y Dk = 0; 1.5 0 1.5 xx x FA A = = k i p k i p 0; 1.05 0 1.05 yy y A = = 22 1.5 1.05 1.831 A =+= i p Pin A is subjected to double shear. Therefore, 0.9155 2 A A F Vk == i p 2 0.9155 ;6 4 A allow A A V A d τ π di 0.441 A = n p Pin B is subjected to single shear. Therefore, 2.97 ABB C VFF k i = 2 2.97 4 A allow A B V A d 0.794 B = n
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Equation of equilibrium :
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This homework help was uploaded on 04/04/2008 for the course ME 211 taught by Professor Thouless during the Fall '08 term at University of Michigan.

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ME211solution_5 - Bearing stress: P 30(103 ) b = = =...

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