hw1_s04_sln

# hw1_s04_sln - EMA 405 Spring 2004 Solutions for HW#1 1 The...

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Unformatted text preview: EMA 405, Spring 2004 Solutions for HW #1 1. The purpose of this exercise is to evaluate the coefficients of a linear, 2D interpolation function and to use the completed expansion to find derivative information. The computation of strains also reinforces the concept that the degree of the piecewise polynomial approximations for displacement and strain are different. x y 1 2 3 1 2 1a) With the given displacement vectors, U = at the origin, y x U ˆ 05 . ˆ 1 + = . at ( x =1, y =2), and at (3,0), we have three equations for each component of displacement, and they may be used to find the three unknowns in each linear, 2D expansion given by Cook in (1.2-1a- b). For the horizontal component, U y x U ˆ 1 . ˆ 2 − − = . y x x 3 2 1 β β β + + = , so the three equations are: 2 . ) ( ) 3 ( 1 . ) 2 ( ) 1 ( ) ( ) ( 3 2 1 3 2 1 3 2 1 − = + + = + + = + + β β β β β β β β β This can be solved using basic algebra skills, but it’s worth noting that we can write the system as a matrix equation, where the unknown coefficients ( β i ) are the components of an algebraic vector, , − = ⋅ 2 . 1 . 3 1 2 1 1 1 3 2 1 β β β so we can alternatively invert the 3x3 matrix to find the coefficients. 1/14 The solution vectors for U x and U y ( U y x y 6 5 4 β β β + + = ) are . − = − = 24 / 1 30 / 1 12 / 1 15 / 1 6 5 4 3 2 1 β β β β β β Once we know the coefficients, we also know the linear, 2D functions for the displacement components throughout the element: y x U y x U y x 24 1 30 1 12 1 15 1 + − = + − = 1b) Once we have the linear, 2D functions for each component of displacement, we can evaluate the strains: 20 1 24 1 15 1 = ∂ ∂ + ∂ ∂ = = ∂ ∂ = − = ∂ ∂ = y U x U y U x U x y xy y yy x xx γ ε ε 1c) For this example of an initial strain due to thermal, we have the node at the origin fixed with respect to both components of displacement, and the bottom of the triangle has its y-component of displacement constrained. This allows the triangle to expand freely, but there is no ambiguity as to the final position of the triangle. Using the information at the end of Section II. of the Intro to Continuum Mechanics notes with no stress (free expansion), ε=ε , T y U T x U y yy x xx ∆ = ∂ ∂ = ∆ = ∂ ∂ = α ε α ε 2/14 Integrating these two equations, and using the constraints to evaluate the constants of integration, we have y Ty U x Tx U y x 03 ....
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## This note was uploaded on 07/29/2008 for the course EMA 405 taught by Professor Witt during the Spring '04 term at University of Wisconsin.

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hw1_s04_sln - EMA 405 Spring 2004 Solutions for HW#1 1 The...

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