FEM - d3 %%(they are above) d4 f1 -1 0 0 .5 f5 = p[ 0 0 -1]...

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%spring asssemblage, global stiffness by direct superposition f1 k -k 0 0 0 d1 f2 -k 2k -k 0 0 d2 f3 = 0 -k 2k -k 0 d3 f4 0 0 -k 2k -k d4 f5 0 0 0 -k k d5 f f1 same matrix as above 0 0 d2 p = d3 0 d4 f5 0 f %solve 0 2k -k 0 d2 p = -k 2k -k d3 0 0 -k 2k d4 0 %k/p[d]=[2 -1 0, -1 2 -1,0 -1 2] k=[^]; p=[0;1;0]; p d=k\p d %then solution obtained by multiplying the factor p/k as d2 .5 d3 = p/k 1 d4 .5 d %subsitute above into this f1 -k 0 0 d2 f5= 0 0 -k
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Unformatted text preview: d3 %%(they are above) d4 f1 -1 0 0 .5 f5 = p[ 0 0 -1] 1 = p {.5 .5 .5 %p=0 (at node 3) and node 5 fixed known displacement B %same as first part %solve part %0 0 2k -k 0 0 d2 0 = -k 2k -k 0 d3 0 0 -k 2k -k d4 B 1/b d2 d3 d4= [2,-1,0, -1 2 -1, 0 -1 2] {0 0 1} %use matlab now k=[2,-1,0;-1,2,-1. ..] p=[0;0;1] d=k\p %final solution obtained by multiplying each answer by b %to get f1 and f5 f1 -1 0 0 f5 = k 0 0 -1 * d2 d3 d4= kB[same matrix^]*{answer for d}= -kB*{1st and3rd solution}...
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This note was uploaded on 09/11/2011 for the course MECH 231 taught by Professor Den during the Spring '11 term at Rutgers.

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FEM - d3 %%(they are above) d4 f1 -1 0 0 .5 f5 = p[ 0 0 -1]...

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