HW08_Solution - EML 4507 FEA & DESGN Fall 2010 HW08...

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Unformatted text preview: EML 4507 FEA & DESGN Fall 2010 HW08 Solution 1. Repeat Example 4.2 with the approximate deflection in the following form: 2 3 4 1 2 3 ( ) v x c x c x c x = + + . Compare the deflection curve with the exact solution. Solution: The given form of approximate deflection satisfies the essential boundary conditions. The second derivative of deflection becomes: v"=2 c 1 + 6 c 2 x + 12 c 3 x 2 . Using the expression for strain energy in a beam given in Eq. (4.17) we obtain ( ) 2 2 1 2 3 2 6 12 2 L EI U c c x c x dx = + + ∫ The above strain energy is differentiated with respect to unknown coefficients c 1 and c 2 : ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 1 2 3 1 2 3 1 2 2 3 4 1 2 3 1 2 3 2 2 2 3 4 5 2 6 12 1 2 3 1 2 3 3 4 5 3 2 2 6 12 2 2 3 4 6 2 6 12 6 2 3 12 2 6 12 12 L L L U EI c c x c x dx EI Lc L c c L c U EI c c x c x xdx EI L c L c c L c U EI c c x c x x dx EI L c L c c L c ∂ = + + = + + ∂ ∂ = + + = + + ∂ ∂ = + + = + + ∂ ∫ ∫ ∫ The potential energy of the external forces can be derived as follows: 2 3 4 1 2 3 2 3 4 2 3 1 2 3 1 2 3 3 4 2 3 2 1 2 5 4 3 3 ( )( ) ( ) (2 3 4 ) 2 3 3 4 4 5 L V p c x c x c x dx F c L c L c L C c L c L c L p L p L c FL CL c FL CL p L c FL CL = -- + +- + +- + + =-- +-- +-- ∫ Then, the stationary condition of the total potential energy becomes the following matrix equation: 2 3 3 2 1 1 3 2 3 4 4 3 2 1 2 4 5 4 3 3 4 5 1 144 3 5 5 4 6 8 2 6 12 18 3 4 8 18 L L L p L FL CL c EI L L L c p L FL CL c p L FL CL L L L - + +...
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This note was uploaded on 06/07/2011 for the course EML 4507 taught by Professor Kim during the Fall '11 term at University of Florida.

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HW08_Solution - EML 4507 FEA & DESGN Fall 2010 HW08...

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