Whatever system you are handling with, from frquency analysis if you find one or more eigenvalues it means that you have one or more rigid body motions, so your structure is not perfectly constrained (it is not an iso-static structure).

Consider now a bar loaded with a axial compressive load. You can compute the Euler Critical Load, that is the load for which the structure goes to be unstable. But when the bar goes to be unstable, your first eigenvalue goes to be zero.

Now, if you have the stiffness matrix of a particular system, you can introduce the geometrical stiffness matrix, that is a matrix that takes into account the axial load applied to, for example, a bar.

So, you can perform an iteration in matlab, computing for each cycle, for each value of the load applied, the eigenvalues of the system as:

eig(K-Kg), where Kg is the geometrical stiffness matrix. When the first eigenvalue goes to be zero, you found the critical load that brings your structure/system to instability.

If the stiffness matrix K of your structure is singular, that is, determinant of K is zero, it admits rigid-body motion and is statically unstable. This is eqivalent to having a zero eigenvalue.

If column vectors of K or row vectors of K are not linearly independent, K is singular.

Dear Huajiang Ouyang ,

I am getting some eigenvalues as -ve and near to e-10. Hence here all values are not positive.

If you get very small eigenvalues, such as in the order of e^-10, they are effectively zero and thus the structure is effectively unstable (or effectively allows rigi-body motion).

Eigenvalues must be positive to become positive definite matrix, so we can not consider zero.

Have you solved your problem now? If not, you are welcome to show me all the eigenavlues of [K]. You can sen [K] if it is now very big and if you wish.

Please note the definition of eigenvalues must be clear. Lambda is called an eigenvalue if [K]*{x}=lambda*{x}.

Dear professor Huajiang Ouyang,

Detail explanation of my problem is given in attached file with matlab stiffness.m file in next answer

Please, go through this

Thank You

Continue

stiffness.m file is attached. Kindly run this in matlab

will try it tomorrow when I go to work.

I read through your Matlab code.

When you calculate the eigenvalues of [K] in line 35, the [K] is the stiffness matrix of the truss wihtout the external constraints at nodes 4 and 5. Therefore the whole truss is not constrained and thus allows rigid-body motion => positive SEMI-DEFINITE => zero eigenvalues

smallest eig value

ans =

-2.2141e-11

is effectively mean zero eigenvalue.

When you calculate a real-number quantity which should be zero in Matlab, it will not be an exact zero as it is a real number. Insteady it would be a very small positive or negative number.

By the way, your truss looks odd to me --- there is not a rod linking nodes 2 and 3 or linking nodes 4 and 5. I suggest that you caculate

max(eigenvalue(stiffness(activeDof,activeDof))) / min(eigenvalue(stiffness(activeDof,activeDof)))

and let me know what it is.

Yes Sir,

I am getting all eigenvalues positive.

eig(stiffness(activeDof,activeDof))

Ans is 1.0e+05 *

0.0335

0.0841

0.4216

1.3706

2.3273

3.2454

I wiil apply this to my topology code and will discuss results.

Thank You sir

This looks right now ---the unconstrained truss (allowing rigid-body motion) has zero eiegnavlues while the properly constrained truss has non-zero eigenvalues.