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came10pt25

# came10pt25 - W U − = ∏ ∵ 2 2 c L c a P L a EI − =...

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Question.10.25 A simply supported beam carries a load P at a distance c away from its left end. Obtain the beam deflection at the point where P is applied .Use the Rayleigh-Ritz method. Assume a deflection curve of the form, v = a.x(L-x), where a is to be determined. Solution . In accordance with the Rayleigh-Ritz method, W U = Where, = Potential energy U = Strain energy W = Change in potential = dV U U . 0 = dV I E y M U . . 2 . 2 2 2 ∫∫ = I E dx dA y M U 2 2 2 2 ). . .( ………..Since, dV = dA.dx But, x v I E M 2 2 . = x P c L y

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dx x v EI U L . ) 2 2 ( . 2 0 2 = ) ( . x L x a v = ……………….assumed deflection curve. ) 2 ( x L a dx dv = a x v 2 2 2 = dx a EI U L . ) 2 ( . 2 0 2 = L a EI U . . 2 2 = And, since ) .( . . . c L c a p v P W c = = Hence, we have
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Unformatted text preview: W U − = ∏ ∵ ) .( . . . 2 2 c L c a P L a EI − + = ∏ ∴ As the potential energy must be minimum at equilibrium, = ∂ ∏ ∂ a ) .( . ) 2 ( 2 = − − ∴ c L c P a EIL EIL c L c P a 4 ) .( . − − = ∴ Now, we substitute this constant back in the assumed deflection curve equation, so that we will be able to calculate the deflection at any section of the beam. ) .( . 4 ) .( . x L x EIL c L c P v − − − = Now, at the desired section, x = c ……(distance from the left end.) EIL c L c P v c 4 ) ( . 2 2 − − = ∴...
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