MEM427_HW_4_Solution

MEM427_HW_4_Solution - MEM427 HWAssignment#4Solutions...

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MEM427 HW Assignment #4 Solutions Fall 2010 1 Homework Assignment #4(a) A simply supported beam is subjected to a uniformly distributed load of intensity q as shown. Two approximate solutions,   L x x C v 1 1 and L x C v sin 2 2 have been proposed. Determine the values of constants 1 C and 2 C for these two approximate solutions, respectively, by using the principle of minimum total potential energy. Compare the deflections ( v ), bending moments ( M ) and shear forces ( V ) of the approximate solutions with those of the exact solution. The exact solutions are:  3 2 3 2 24 x Lx L EI qx v ex , 2 2 2 qx qLx M ex , qx qL V ex 2 Solution: For  L x x C v 1 1 , we have 1 2 1 2 2 C dx v d . The total potential energy is   1 3 2 1 0 1 2 1 0 1 2 2 1 2 1 6 2 2 2 ) ( ) ( 2 C qL EILC dx q L x x C C EI dx x p x v dx v d EI L L Constant 1 C can be obtained by minimizing the total potential energy EI qL C qL EILC C 24 0 6 4 2 1 3 1 1 1 Thus, the approximate solution is  L x x EI qL v 24 2 1 The corresponding bending moment and shear force are 12 2 2 1 2 1 qL dx v d EI M 0 3 1 3 1 dx v d EI V For L x C v sin 2 2 , we have L x C L dx v d sin 2 2 2 2 2 . The total potential energy is 2 2 2 3 4 0 2 2 4 2 2 0 2 2 2 2 2 2 2 4 sin sin 2 ) ( ) ( 2 C qL C L EI dx q L x C L x L EI C dx x p x v dx v d EI L L Constant 2 C can be obtained by minimizing the total potential energy
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This note was uploaded on 11/14/2010 for the course MEM 427 taught by Professor Tein-mintan during the Fall '10 term at Drexel.

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MEM427_HW_4_Solution - MEM427 HWAssignment#4Solutions...

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