{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# FEM - d3(they are above d4 f1-1 0 0.5 f5 = p 0 0-1 1 =...

This preview shows pages 1–2. Sign up to view the full content.

%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 %nodes 1&5 fixed and force applied at node 3, find nodal displacements 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 %reactions at fixed nodes 1 & 5 %subsitute above into this f1 -k 0 0 d2 f5= 0 0 -k

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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}...
View Full Document

{[ snackBarMessage ]}