- #1
urocissa
- 6
- 0
the buckling shape for clamped-free column is v(x)=1-cos(n*pi/2*L), n=1,3,5 ...
how could i use the discrete model to get the buckling mode shape matrix?
for example 3x3 matrix
form a stiffness matrix and solve the eigenvalue problem?
L=1;E=1;I=1
l=L/3 % 3-element beam
k=(E*I/l^3)*[24 0 -12 6*l 0 0;
0 8*l^2 -6*l 2*l^2 0 0;
-12 -6*l 24 0 -12 6*l;
6*l 2*l^2 0 8*l^2 -6*l 2*l^2 ;
0 0 -12 -6*l 12 -6*l ;
0 0 6*l 2*l^2 -6*l 4*l^2]
[v,d]=eig(k)
v =
0.0524 0.1142 0.0947 0.3177 -0.7286 0.5860
-0.3090 -0.6433 -0.5420 0.4409 -0.0323 -0.0386
0.1941 0.2589 0.0603 0.5004 -0.3124 -0.7373
-0.5357 -0.1915 0.7575 0.2857 0.1437 -0.0133
0.3924 0.1830 -0.0086 0.6031 0.5834 0.3290
-0.6505 0.6602 -0.3460 0.0941 0.0970 0.0550
how could i use the discrete model to get the buckling mode shape matrix?
for example 3x3 matrix
form a stiffness matrix and solve the eigenvalue problem?
L=1;E=1;I=1
l=L/3 % 3-element beam
k=(E*I/l^3)*[24 0 -12 6*l 0 0;
0 8*l^2 -6*l 2*l^2 0 0;
-12 -6*l 24 0 -12 6*l;
6*l 2*l^2 0 8*l^2 -6*l 2*l^2 ;
0 0 -12 -6*l 12 -6*l ;
0 0 6*l 2*l^2 -6*l 4*l^2]
[v,d]=eig(k)
v =
0.0524 0.1142 0.0947 0.3177 -0.7286 0.5860
-0.3090 -0.6433 -0.5420 0.4409 -0.0323 -0.0386
0.1941 0.2589 0.0603 0.5004 -0.3124 -0.7373
-0.5357 -0.1915 0.7575 0.2857 0.1437 -0.0133
0.3924 0.1830 -0.0086 0.6031 0.5834 0.3290
-0.6505 0.6602 -0.3460 0.0941 0.0970 0.0550