00:01
One beam element gives four dofs.
00:03
A pin means v1 is equal to zero, but theta 1 is free.
00:06
Then solve the determinant of k minus omega squared n equals 0.
00:11
So the dof's q is v1, theta, 1, v2, theta, 2, theta, 2, k is equal to ei over l cubed, times the matrix 12, 6l, negative 12, 6l, l squared, negative 6l2l squared, negative 12, negative 6l, 12, negative 6l2l squared, negative 6l squared.
00:49
M is omega al over 420 times 156, 22l, 54 negative 13l, 12l, 22l, 4l squared, 13l negative 3l squared, d6 negative 22l and negative 13l, negative 3l squared, negative 22l4l squared.
01:18
Apply the pin b -clf, so pendant node 1 v1 is 0, the free dufs are theta 1 v2, theta 2, transpose...