me180_quiz04_solutions

me180_quiz04_solutions - University of California Berkeley...

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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:
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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 # , 0 , 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 , 0 , 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 .
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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 that J is continuous in the variable ξ . So, lets look at the three cases: The slope equals zero: This corresponds to x 2 = x 1 + 1 2 ( x 3 - x 1 ). In this case, we only need to verify that J is positive, for any value of ξ . Indeed: J ( ξ = 0) = x 3 - x 1 2 .
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