# Eigenvalue formulation to find the principal stresses, directions

1. Jan 7, 2009

### aznkid310

1. The problem statement, all variables and given/known data

3.) Stress analysis at a critical point in a machine member gives the three-dimensional state of stress in MPa as the following:

y =

[ 105 0 0
0 -140 210
0 210 350 ]
Using the eigenvalue formulation to find the principal stresses (eigenvalues) and principal directions (eigenvectors). Show these stresses on a properly oriented element. Use the three-dimensional Mohr’s circle to obtain the maximum shear stress and show this on a properly oriented element. Repeat this problem using MATLAB.

2. Relevant equations

I can do all except finding the directions. I'm not sure how to find that. Do i just plug back my eigenvalues into mhy homogeneous eqn?

3. The attempt at a solution

skipping some steps:

det( 105 - s 0 0
0 -140 - s 210 = [0]
0 210 350 - s )

Solving for S (using characteristic eqn): S = 105, 427.68, -217.68
These are my eigenvalues.

In a previous step, my homogeneous eqn was:

[ 105 - s 0 0
0 -140 - s 210 * [l;m;n] = [0]
0 210 350 - s ]

If i plug my S values back in, i dont know how to solve.

2. Jan 8, 2009

### nvn

As you mentioned, the equation for, e.g., your third eigenvalue, S3 = -217.684056, is (A - S3*I)*X = 0, where I = identity matrix, A is your given matrix under item 1, and X = eigenvector column vector (x1,x2,x3). Therefore, solve for x1, x2, and x3 in the following. The equation (A - S3*I)*X = 0, for your third eigenvalue, is as follows.

[322.684056 0.0 0.0]
[0.0 77.684056 210.0]*(x1,x2,x3) = (0,0,0)
[0.0 210.0 567.684056]

First, you need to get the above system of equations in row-reduced form. If you are not familiar with row-reduced form, look it up in your linear algebra text book. In row-reduced form, the above system of equations becomes as follows.

[1.0 0.0 0.0]
[0.0 1.0 2.7032574]*(x1,x2,x3) = (0,0,0)
[0.0 0.0 0.00000]

Therefore, e.g., let x2 = t. Thus, the second equation (row) shown above, 0*x1 + x2 + 2.7032574*x3 = 0, becomes 0 + t + 2.7032574*x3 = 0; and solving for x3 gives x3 = -0.369924*t. The first equation (row) shown above, x1 + 0*x2 + 0*x3 = 0, becomes x1 + 0*t + 0*t = 0; and solving for x1 gives x1 = 0*t. Normalizing the current results, we have (0.0,1.0,-0.369924)*t/sqrt(0.0^2 + 1.0^2 + 0.369924^2), for t not equal to zero. Therefore, the eigenvector corresponding to your third eigenvalue (S3 = -217.684056) is X = (0.0,0.937885,-0.346946).

3. Jan 8, 2009

### aznkid310

ah that rings a bell. So i also have to do thv same thing for my other two eigenvalues too?

4. Jan 8, 2009

Yes.