hw3_s04_sln - EMA 405 Spring 2004 Solutions for HW#3 1 The...

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EMA 405, Spring 2004 Solutions for HW #3 1. The objective of this exercise is to calculate the work-equivalent loads for a surface pressure acting on a Q8 planar element. The calculations will make use of the formula for work- equivalent loads, () ) , ( ˆ ) , ( y x N y x P dS i i = n F where the unit vector is the outward surface normal, which is illustrated in Fig. 1.1 for the top surface, and N i are the interpolation functions. Since the applied pressure is on the top of the element only, and since only N 3 , N 4 and N 7 are nonzero along the top surface, we know immediately that F 1 = F 2 = F 5 = F 6 = F 8 = 0 . Furthermore, since P is uniform and the element is symmetric about x =0, we expect that F 4 = F 3 . Thus, there are only two work-equivalent load computations to do. x y n 1 2 3 4 5 6 7 8 Figure 1.1. Sketch of the Q8 element and the unit outward normal for the top surface. 1/9
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() () () () () [] () 3 ˆ 3 2 2 ˆ 1 2 1 ˆ 1 1 1 1 1 4 1 ˆ ) , ( ˆ ) , ( 1 1 3 2 1 1 1 1 1 3 3 bP x x bP x x dx bP y x y x dx bP y x N y x P dS y y y y y n F = + = + = + + = = () () () () 3 4 ˆ 1 1 1 2 1 ˆ ) , ( ˆ ) , ( 1 1 1 7 7 bP y x x dx bP y x N y x P dS y y y n F = + + = = The work equivalent load vector at the middle node has a magnitude of 2/3 of the total force on the top of the element (2 bP in this case) directed against the surface normal. At the corners, the work equivalent loads have a magnitude of 1/6 of the total force. This seems somewhat odd in that the pressure is uniform, but the work equivalent loads are very nonuniform. This peculiarity results from the different interpolation functions with N 7 having a much larger weighting over the top surface than N 3 and N 4 . 2.1 Introduction The test considered here will be used to verify whether the work-equivalent loads computed in problem 1 provide the expected response for a uniform pressure. 2.2 Preliminary analysis While we have the results of problem 1 for the work-equivalent loads, we need to decide on some specifics for our test computation. I’ll use two Q8 elements side-by-side, each having a length of 2 m and a height of 2 m. The analysis type will be “plane stress,” so the default base dimension of 1 m is assumed. To avoid making all of the computations come out to unity, I’ll set E =4 Pa and ν =0.3.
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along the bottom, and U x =0 at one point), the response to a surface pressure is comparable to a spring element with k =(4 Pa)(4 m 2 )/(2 m)=8 N/m. With a uniform pressure of 1 Pa and the top surface area of 4 m 2 , the net force is 4 N, so we expect a uniform displacement of 0.5 m downward for the top surface. This should give uniform ε yy =-0.25, xx =0.075 and σ yy =-1 Pa. All other stresses and strains should be zero. 2.3.
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hw3_s04_sln - EMA 405 Spring 2004 Solutions for HW#3 1 The...

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