EN0175-22

EN0175-22 - EN0175 11 / 21/ 06 Principle of minimum...

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EN0175 11 / 21/ 06 Principle of minimum potential energy (continued) The potential energy of a system is = S i i V i i V S u t V u f V w V d d d Principle of minimum potential energy states that for all kinematically admissible , the actual displacement field minimizes . i u V Example 1: g ρ x L We have shown in the beginning of the semester that the exact solution is: () x L x E g u = 2 . Now we discuss how to solve the same problem by using the principle of minimum potential energy. Procedure: 1) Pick any displacement such that ( ) ( ) 0 0 = = L u u . 2) Minimize for the chosen parameters. V An obvious choice is () ( ) ( ) x f x L x x u = since this satisfies the clamped displacement boundary conditions for any . We can assume x f ( ) x f to be a polynomial function. N N x C x C x C C x f + + + + = L 2 2 1 0 To minimize the potential energy , we take N C C C C V , , , , 2 1 0 L 0 0 = C V , 0 1 = C V , …, 0 = N C V 1
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EN0175 11 / 21/ 06 This gives algebraic equations to determine 1 + N 1 + N parameters . N C C C C , , , , 2 1 0 L For the present example, take the simplest form that ( ) 0 C x f = . () ( ) x L x C x u = 0 () x L C x u 2 0 = = ε x L EC E 2 0 = = σ 2 2 0 2 2 1 2 1 x L EC w = = σε The potential energy of the system is ( ) 0 2 0 3 0 0 0 2 2 0 0 0 0 6 2 1 d d 2 2 1 d d gC EC L A A x x L x gC A x x L EC A x x L x gC A x w V L L L L ρ = = = To minimize , V E g C g EC C V 2 0 2 0 0 0 0 = = = Therefore ( x L x ) E g u = 2 , which is identical to the exact solution for this problem. Suppose x C C x f 1 0 + = is assumed, we will get a function of potential energy in terms of and , 0 C 1 C ( ) 1 0 , C C V Minimizing the potential energy requires that 0 0 = C V , 0 1
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This note was uploaded on 01/10/2010 for the course EN 0175 taught by Professor Huajiangao during the Spring '06 term at Brown.

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EN0175-22 - EN0175 11 / 21/ 06 Principle of minimum...

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