me211solution_8 - Find each internal torque at the solid...

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Find each internal torque at the solid inner shaft and the outer tube. (FBD) 2 BC BA TT k i p i == n max 4 2(0.5) () 1 0 . 2 ksi = (0.5) 2 BC BC Tc J τ π max 44 2(1) ( ) (1 0.75 ) 2 BA BA J 1.86 ksi = Angle of Twist: 3 2(12) 0.002032 0.75 )11(10 ) 2 BA BA B TL rad JG φ = / 43 2(24) 0.044448 (0.5 )11(10 ) 2 BC BC CB rad JG = / 0.002032 0.044448 0.04648 2.66 rad CBC B φφφ =+ = + = = °
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Express the radius as a function of x () rr L r rx r x x L L + =+ = Then, the polar moment of inertia is 4 4 4 4 ( ) 22 rL rx r Jx L x LL ππ + ⎛⎞ == ⎜⎟ ⎝⎠ + Angle of Twist: 44 4 3 00 0 1 ( ) 3 ( ) 1 2 L Tdx TL dx TL TL 4 7 J xG rG φ = = ++ ∫∫ π
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Equilibrium: 3000(0.05)2 0 AB TT +− = ------ (1) Compatibility condition: // CA CB φφ = (400) (600) JG JG = 1.5 A B T = T m m ------ (2) Solving Eqs, (1) and (2) yields: 120 B TN =⋅ 180 A max 4 180(0.02) ( ) 14.3 (0.02) 2 AC Tc MPa = J τ π == max 4 120(0.02) () 9 . 5 5 (0.02) 2 CB Tc MPa = J
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Equilibrium: 800(12) 0 TS TT −−= ------(1) Compatibility condition: φφ = 44 4 (0.75)(12) (0.75)(12) (1.5 1.25 ) (0.5) 22 GG ππ = ------(2) Solving Eqn (1) and (2) 9376.42 T Tl =⋅ b i n b i n 223.58 S //// 46 800(12)(0.5)(12) 223.58(0.75)(12) 2 0.1085 6.22 rad = = (0.5 )(11)(10 ) (0.5 )(11)(10 ) CD BD AB CA φφφφ =++= × + °
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Section property: 33 11 (0.2)(0.2) (0.15)(0.15) 91.14583(10 ) 12 12 6 4 I m =− = Bending stress: My I σ = 6 6 (0.075) 30(10 ) 91.14583(10 ) M = 36458 36.5 M Nm =⋅ kNm = max 6 36458(0.1) 40.0 91.14583(10 ) Mc MPa = I ==
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Shear force diagram and Moment diagram The maximum moment is max 44.8 M kip ft =⋅ max max 3 44.8(12)(4.5) 4.42 1 (9)(9 ) 12 Mc ksi I σ == =
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Centroid: Segment A x xA 1 160(80) 80 1.024(10 6 ) 2 40(80)
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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_8 - Find each internal torque at the solid...

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