solution_hw3[0]

# solution_hw3[0] - 4.3 Solution F C 120 F B 180 TCD = F 012...

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4.3 Solution TF N CD == (. ) 012 500 m k N m o r F = 4167 . TN AB = = 4167 018 750 .( . ) ( a ) rad GJ TL AB 028 . 0 4 9 ) 0225 . 0 ( 2 ) 10 ( 79 ) 2 . 1 ( 750 = = = π φ rad CD 077 . 0 4 9 ) 0175 . 0 ( 2 ) 10 79 ( ) 8 . 1 ( 500 = = × T h u s φφ DC D A B o rad =+ = = 15 0119 6 82 .. . ( b ) MPa c T AB AB 92 . 41 3 3 ) 0225 . 0 ( ) 750 ( 2 2 = = = τ 180 120 B C F F

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4.11 Solution Due to symmetry: B A L w B A M M R R = = = , 4 0 W e h a v e , f o r 02 ≤≤ xL : MR xM M AA wx L wLx A L =−−=−− 0 3 00 3 34 3 Therefore EIv M A L " =− + 3 43 EIv M x c A L ' + + 0 2 0 4 81 2 1 Boundary conditions: 0 : 0 ) 0 ( ' 1 = = = c v θ 96 5 2 2 0 : 0 ) ( ' L w A L M v = = H e n c e 2 60 24 192 5 5 0 3 0 2 2 0 c EIv L x w Lx w x L w + + = Boundary condition: vc () : 2 = 0 = Thus, we obtain ) 2 0 ( ) 16 40 25 ( 3 2 3 960 2 0 L x for x x L L v LEI x w + = and = = EI L w L v v 3840 7 2 max 4 0 ) (
4.15 Solution P=mg T a b l e A . 9 ( C a s e 6 ) . 075 2 . P 125 2 . P L=2 1.25=a b=0.75 B A x δ st Pbx LEI Lbx m m =− = = ×× 6 222 20 9 81 0 75 25 6 2 210 10 0 06 0 08 12 93 2 0 75 0 0535 () ( . . ) . ( . )( . )(1.
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solution_hw3[0] - 4.3 Solution F C 120 F B 180 TCD = F 012...

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