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ME211solution_11 - FBD of the sectioned part From...

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FBD of the sectioned part From equilibrium equations 150 V N = , 0 N = , 150(0.08) 12 M Nm = = oint A: P 4 12(0.01) 15.3 (0.01) 4 A Mc MPa I σ π = = = 0 A τ = oint B: P 0 B σ = ( ) 2 4 1 4(0.01) 150 0.01 ' ' 2 3 0.637 0.01 (0.02) 4 B VQ VA y MPa It It π π τ π = = = =
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FBD Point A: 3 4 10.5(10 )(0.05) 107 (0.05) 4 A Mc MPa I σ π = = = 3 4 3(10 )(0.05) 15.3 (0.05)
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