# 6-2 - ex2 =[0 10 0 ey2 =[0 15 20 ep =[1 0.1 E = 3e7 nu =...

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CHAP 6 Finite Elements for Plane Solids 305 2. Solve Example 6.2 using one of finite element programs in Appendix. Solution: We will solve the problem using MATLAB Toolbox in Appendix D. The program list is shown below: Nodal Displacements: 0 0 -0.0021 -0.0445 0.0189 -0.0273 0 0 Element strains: et1 = -0.0013 0.0017 -0.0032 et2 = 0.0019 0 -0.0027 Element stresses: es1 = 1.0e+004 *[-2.4709 4.4406 -3.7063] es2 = 1.0e+004 *[ 6.2354 1.8706 -3.1469] The figure below shows the undeformed and deformed shapes of the elements. The deformation is magnified by 50. Edof=[ 1 1 2 3 4 5 6; 2 1 2 5 6 7 8]; ex1 = [0 10 10]; ey1 = [0 5 15];

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Unformatted text preview: ex2 = [0 10 0]; ey2 = [0 15 20]; ep = [1 0.1]; E = 3e7; nu = 0.3; D = hooke(1, E, nu); Ke1 = plante(ex1, ey1, ep, D); Ke2 = plante(ex2, ey2, ep, D); K=zeros(8); F=zeros(8,1); F(4)=-50000;F(5)=50000; K=assem(Edof(1,:),K,Ke1); K=assem(Edof(2,:),K,Ke2); bc=[1 0;2 0;7 0;8 0]; U=solveq(K,F,bc); ed1 = extract(Edof(1,:),U); ed2 = extract(Edof(2,:),U); [es1, et1]=plants(ex1,ey1,ep,D,ed1) [es2, et2]=plants(ex2,ey2,ep,D,ed2) eldraw2(ex1,ey1,[2 1 0]); eldraw2(ex2,ey2,[2 1 0]); eldisp2(ex1,ey1,ed1,[1 1 0],50); eldisp2(ex2,ey2,ed2,[1 1 0],50); 306 Finite Element Analysis and Design...
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## This note was uploaded on 08/22/2011 for the course EML 4500 taught by Professor Staff during the Fall '08 term at University of Florida.

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6-2 - ex2 =[0 10 0 ey2 =[0 15 20 ep =[1 0.1 E = 3e7 nu =...

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