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Unformatted text preview: University of California, Berkeley ME 180, Engineering Analysis Using the Finite Element Method Spring 2008 Instructor: T. Zohdi Quiz 4 : Solutions Problem 1 a) Derive the approximation functions φ e I ( ξ ) for a master element with nodes at ξ = 1, ξ = 0, and ξ = 1, and show that it meets the typical criteria for approximation functions. Solution (2 points): We derive the approximation functions based on the first typical criterion: φ e I = δ IJ , J = 1 , 2 , ··· , nen , where nen is the number of nodes in the element. Based on this, the approximation functions can be defined as φ e 1 = (0 ξ )(1 ξ ) (0 ( 1))(1 ( 1)) = ( ξ )(1 ξ ) 2 , φ e 2 = ( 1 ξ )(1 ξ ) ( 1 0)(1 0) = (1 + ξ )(1 ξ ) , φ e 3 = ( 1 ξ )(0 ξ ) ( 1 1)(0 1) = (1 + ξ )( ξ ) 2 . Now, checking the second typical criterion for shape functions yields the following: nen X I = 1 φ e I = 1 , Ω e = [ 1 , 1] = ( ξ )(1 ξ ) 2 + (1 + ξ )(1 ξ ) + ( 1 ξ )( ξ ) 2 , = ξ + ξ 2 2 + 2 2 ξ 2 2 + ξ + ξ 2 2 , = 2 2 = 1 , for all values of ξ . Thus, the shape functions meet the typical criteria. b) Write down the map x = M ( ξ ) for an element in physical space with nodes at x 1 , x 2 , and x 3 . Solution (2 points): This is fairly straightforward, as follows: x = nen X I = 1 N e I x I , = N e 1 x 1 + N e 2 x 2 + N e 3 x 3 , = ( ξ )(1 ξ ) 2 x 1 + (1 + ξ )(1 ξ ) x 2 + ( 1 ξ )( ξ ) 2 x 3 . c) A desirable property of the isoparametric mapping is that it is invertible for all points in the element. State the mathematical condition that expresses this property. Solution (2 points): Invertability of the map over the whole element would be guaranteed if J = det " ∂ x ∂ ξ # , , ∀ ξ , (for a system where the number of spatial dimensions > 1) and, in particular, we would like J to be strictly positive, to prevent elements from being overly distorted. For 1D, this reduces to J = ∂ x ∂ξ , , ∀ ξ. d) Consider an element in physical space with nodes 1, 2, and 3 at positions x 1 , x 2 , and x 3 , such that x 1 < x 2 < x 3 . What restriction must be placed on x 2 , in terms of x 1 , x 3 , for the element to be ”wellbehaved”? Solution (6 points): First, calculate an expression for J : J = det ∂ x ∂ ξ ! , = ∂ x ∂ξ , = ∂ N 1 ∂ξ x 1 + ∂ N 2 ∂ξ x 2 + ∂ N 3 ∂ξ x 3 , = 1 + 2 ξ 2 x 1 + ( 2 ξ ) x 2 + 1 + 2 ξ 2 x 3 . Next, state the problem we would like to answer. In this case, we want to know where we can place x 2 in physical space, in terms of x 1 and x 3 , so that the element is wellbehaved (i.e., J is positive, ∀ ξ ). To this end, let us first look at how J varies over the master element (i.e., how J varies with respect to ξ ): ∂ J ∂ξ = x 1 2 x 2 + x 3 . So, for fixed values of x 1 , x 2 , and x 3 , the slope is a constant, i.e., J ( ξ ) is a straight line. Also, we note) is a straight line....
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This note was uploaded on 08/01/2008 for the course ME 180 taught by Professor Zohdi during the Spring '08 term at Berkeley.
 Spring '08
 ZOHDI

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