me211solution_9 - The weight of the man FBD of the diving...

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The weight of the man : 78(9.81) 765.18 WN == FBD of the diving board: 0; (1.5) 765.18(4) 0 B A MF =− = 2040.48 B FN , = 0; 765.18 0 yA B FF F =+ −= , 275.3 A = − Sketch of shear force diagram and moment diagram The maximum moment occurs at point B max 275.3(1.5) 1912.95 M Nm
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12-25 (continued) Section properties: 0.01(0.35)(0.02) 3(0.035)(0.01)(0.03) 0.012848 (0.35)(0.02) 3(0.01)(0.03) yA y A + == = + % The moment of inertia about the neutral axis: 32 6 1 (0.35)(0.02 ) (0.35)(0.02)(0.012848 0.01) 12 1 (0.03)(0.03 ) 0.03(0.03)(0.035 0.012848) 12 0.79925(10 ) NA I =+ ++ = Absolute maximum bending stress: max max 6 1912.95(0.05 0.012848) 88.92 0.79925(10 ) Mc MPa I σ = Maximum normal strain: 6 max max 9 88.92(10 ) 125(10 ) E ε 3 0.711(10 ) / mm mm =
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12-46 Centroid: 0.125(2)(0.25) 1.25(0.25)(2) 2.375(4)(0.25) 1.53125 (2)(0.25) (0.25)(2) (4)(0.25) yA y A ++ == = % The moment of inertia about the neutral axis: 32 3 4 11 (2)(0.25 ) 2(0.25)(1.53125 0.125) (0.25)(2 ) 0.25(2)(1.53125 1.25) 12 12 1 (4)(0.25 ) 4(0.25)(2.375 1.53125) 12 1.91471 NA I in =+ = 2 Maximum bending stress: max Mc I σ = The maximum compressive stress occurs at the top: () max 2(12)(1.53125) 19.2 1.91471 c ksi The maximum tensile stress occurs at the bottom max 2(12)(2.5 1.53125) 12.1 1.91471 t ksi
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12-51 2 2 2 b b 1 (1) 400 w = , 80 0 wl = 1 1.5 800 w = , 1 533 = Consider the half of the pin because it s symmetric.
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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_9 - The weight of the man FBD of the diving...

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