# Thread: Solving non linear Matrix

1. ## Solving non linear Matrix

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:

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

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:
$y(0)=0 , y(L)=0, y'(0)=0 , y'(L) = 0$

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:

$P = \frac{4\pi^2 \cdot EI}{L} \cdot n$ 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:

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

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

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

I computed the determinant of the above and got:

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

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

Any guidance is appreciated.

2. ## Re: Solving non linear Matrix

If Ax= 0 then there are two possibilities:
1) A is invertible. In that case, multiplying both sides by the inverse, x= A^(-1)0= 0. x= 0 is the only solution- there is no "non-trivial" solution.

2) A is not invertible. In that case, the determinate of A must be 0. And there are an infinite number of solutions. In fact, the set of solutions form a subspace. The best way to find that subspace is to row reduce the matrix. You will wind up with quantity in the last row that must be set to 0 to be equal to the right hand side (which will always be the 0 vector) that will give the conditions on k.

3. ## Re: Solving non linear Matrix

Thanks that makes more sense, yes there will be an infinite set of solutions! Can you recommend a numerical solution to solve for kL ?

And I don't really see how I can row reduce when I have trig functions and constants, it would get really messy

4. ## Re: Solving non linear Matrix

Hey Halls, I have solved the problem!

Please ignore my final equation in post #1, it is incorrect.

I re evaluated the determinant, simplified and got a final equation of:

$2 = x\sin x + 2\cos x$

Where x = kL

The thing is that I had to use an online graphing calculator to determine that the solution was:

$x=2\pi \cdot n \to P = \frac{\pi^2\cdot 4 \cdot EI}{L^2}$

I was wondering, how would I determine this solution numerically?

For exam/test situation do you recommend I use Newtons method?