me180_quiz04_solutions

me180_quiz04_solutions - University of California, Berkeley...

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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 ”well-behaved”? 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 well-behaved (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.

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me180_quiz04_solutions - University of California, Berkeley...

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