me180_quiz02_solutions

# me180_quiz02_solutions - u is speciﬁed so the second...

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University of California, Berkeley ME 180, Engineering Analysis Using the Finite Element Method Spring 2008 Instructor: T. Zohdi Quiz 2 Solutions Problem 1 Derive the weak form corresponding to the strong form d dx x 2 du dx ! = - 5 u + e x in Ω = (0 , L ) , u = 4 on Γ u = 0 , du dx = - 14 on Γ q = L . Solution: (10 points) First, create the weighted residual, and integrate it over the domain: Z Ω wR d Ω = 0 , Z Ω w " d dx x 2 du dx ! + 5 u - e x # d Ω = 0 , Z Ω w d dx x 2 du dx ! + 5 wu - we x d Ω = 0 . Next, use the product rule, i.e., d dx wx 2 du dx ! = dw dx x 2 du dx + w d dx x 2 du dx ! and subtitute its result back into the integral statement of the weighted residual: Z Ω " d dx wx 2 du dx ! - dw dx x 2 du dx # + 5 wu - we x d Ω = 0 . Next, apply the divergence theorem: Z Ω - x 2 dw dx du dx + 5 wu - we x d Ω + Z Γ=Γ u S Γ q wx 2 du dx n x d Γ = 0 , Z Ω - x 2 dw dx du dx + 5 wu - we x d Ω + Z Γ u wx 2 du dx n x d Γ + Z Γ q wx 2 du dx n x d Γ = 0 . Now, we notice that the weighing function is equal to zero on the part of the boundary where

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Unformatted text preview: u is speciﬁed, so the second integral disappears. Also, we notice that the domain Ω is a line, so d Ω = dx , and the integral over its boundary (or any part thereof) is just the evaluation at a point, so Z Ω-x 2 dw dx du dx + 5 wu-we x ! dx + wx 2 du dx n x ± ± ± ± ± Γ q = L = , where the normal to the boundary on the right hand side is n x | L = 1. Thus, the weak form is as follows: Z Ω-x 2 dw dx du dx + 5 wu-we x ! dx-14 L 2 w ( L ) = ....
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me180_quiz02_solutions - u is speciﬁed so the second...

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