Hello,

I am working on the derivation of ideal buckling of columns under fixed supports at both ends.

The general equation that describes the deflection of the column is:

$\displaystyle y(x) = A \cos(kx) + B \sin (kx) + Cx + D ; k = \sqrt{ \frac{P}{EI}} $

Where A,B,C,D are constants, k is also a constant ( E = modulus of elasticity and I is second moment of inertia of the member)

I have four boundary conditions for a fixed-fixed support that are:

$\displaystyle y(0)=0 , y(L)=0, y'(0)=0 , y'(L) = 0$

where L is the full length of the column. These boundary condition simply mean that the deflection at each support is 0 and the slope of the deflection curve is also 0.

The purpose of this exercise is to determine the critical axial load P that will cause buckling.

Ideally I want to solve for k, then isolate for P which is gives critical axial load. The solution to this is:

$\displaystyle P = \frac{4\pi^2 \cdot EI}{L} \cdot n$ where n is any natural number, which determines the shape of the deflected shape. For simplicity we will take n=1.

I have generated this matrix to solve:

$\displaystyle \begin{bmatrix} 1 & 0 & 0 & 1 \\ \cos(kL) & \sin(kL) & L & 1\\ 0 & k & 1 & 0 \\ -k\sin(kL) & \cos(kL) & 1 & 0\\ \end{bmatrix} \cdot \begin{bmatrix} A \\B\\C\\D \end{bmatrix} = \{0\}$

I want the non trivial solution, that is the 4x4 matrix = 0.

I have forgotten how to solve such matrices...

However, in class the prof briefly mentioned that we use Cramer's method to solve

$\displaystyle \begin{bmatrix} 1 & 0 & 0 & 1 \\ \cos(kL) & \sin(kL) & L & 1\\ 0 & k & 1 & 0 \\ -k\sin(kL) & \cos(kL) & 1 & 0\\ \end{bmatrix} = \{0\}$

This included finding the determinant of the matrix until you get a numerical equation.

I computed the determinant of the above and got:

$\displaystyle 2\cos(kl) \cdot ( \cos(kl) - 1 ) + kl \sin (kl) = 0$

If the above is true, how would you explicitly solve for k?

Any guidance is appreciated.