Solution_hw4 - 5.1 P A a B C L = 6 5 V AB = P VBC = Pa L V P Pa/L x Us = 5.7 L V 2 x 2 AG 0 dx = 3 5 AG P 2 dx 0 P2a 2 2 0 L L dx = 3 P2a 5 AGL(L a

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5.1 α == 6 5 VPV AB BC Pa L , = C B P a L A U dx P dx dx L a s V AG L AG Pa L L AGL x + = ∫∫ 2 22 2 2 2 0 3 5 2 00 3 5 [] + ( ) Pa/L V x P 5.7 Introduce a fictitious horizontal force Q at end B. Vertical reactions at supports are P/2. W e h a v e MR R Q BC P =− + (c o s ) ( s i n ) 1 2 θ MM CA BC = T h u s δθ π B EI BC M Q BC = 2 0 2 d + 2 2 0 2 1 EI P RR Q R [ ( cos ) ( sin ) ][ sin ] θθ R d Setting Q=0 and integrating: δ B PR EI = 3 2
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5.15 W e h a v e ] ) cos( 1 [ = L x m a v m π ( a ) = ) cos( ) ( " 2 L x m L m a v m and referring to Example 5.12: Ud x a EI d v dx L EI m m L L m == = ⋅⋅⋅⋅ 2 2 0 2 24 2 246 2 2 () ,,, ( b ) Also WP v P a x m m L =− = = ,, , 2 2 2610 ( c ) In Eq.(c), m=4, 8, , do not appear, because they give zero deflection at x=L/2.
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This note was uploaded on 03/24/2010 for the course ME 031 taught by Professor Kim during the Winter '10 term at Korea Advanced Institute of Science and Technology.

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Solution_hw4 - 5.1 P A a B C L = 6 5 V AB = P VBC = Pa L V P Pa/L x Us = 5.7 L V 2 x 2 AG 0 dx = 3 5 AG P 2 dx 0 P2a 2 2 0 L L dx = 3 P2a 5 AGL(L a

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