Discrete Model for Buckling Mode Shapes of a Clamped-Free Column

[urocissa]

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

 

[urocissa]

how to separate the translation mode and rotation mode?

 

[AlephZero]

First you do a stress analysis for a unit applied load (i.e. the axial load on the column).

Then you formulate the geometric stiffness (or stress stiffness) matrix Kg for the stress distribution.

Then you do an eigensolution with the elastic stiffness Ke and Kg:

det(Ke + b Kg) = 0

where b is the "buckling factor", i.e. the scale factor from your unit load to the buckling load.

A textbook on the FE method will tell you how to set up Kg. (I had a quick look on the web but I couldn't find anything online)

If one of the eigenvectors of Ke is the same shape as the buckling mode, that is just a coincidence.

 

[urocissa]

thank you~