Why the tangent stiffness matrix needs to be positive

[Rocketeer3k]

I am working on a crack propagation problem using cohesive elements with Abaqus Implicit Newton-Raphson algorithm. It says in the manual that viscous regularization should be applied to the cohesive elements to help with convergence. The viscosity guarantees that the tangent stiffness matrix is positive for sufficiently small increments.

I do not understand the last sentence above and would be very grateful if somebody could explain that to me.

I am good with FEM but I do not understand the mathematical process in depth. So please, why does the tangent stiffness matrix need to be positive, why does this happen at small increments and how does viscosity help?

 

[burningDiesel]

Short answer: it needs to be postive so that it can be reversed. K*u = f, so if you want the displacements you need to know K^-1
(tangent is because you are using newton raphson, but it's basicly the same thing just at some derivatives at some locations). How viscosity helps I can't say without looking it up
Long answer: a FEM book

 

[Rocketeer3k]

sdebock, thx for the answer. My knowledge of linear algebra is too far gone to understand a FEM book.

so a matrix to be inversed cannot be negative?

 

[burningDiesel]

I was a bit too fast and incomplete I guess.
A matrix with a 0 eigenvalue can not be inverted. If the matrix is positive definite it has only positive eigenvalues and hence, it can be inverted. You can think of a zero eigenvalue as a rigid body translation or rotation, i.e. a movement (u) without applying any forces (f).
If your matrix is positive definite you can also (in most cases) use decomposition to solve Ku=f, not needing the (expensive) inversion of K.

Negative eigenvalues on the other hand are often symptoms of instabilities. One example is buckling of a structure. So if you perform a buckling analysis, negative eigenvalues are actually expected! Another example is material instability.
These indefinite matrices (both positive & negative eigenvalues) can still be inverted, but Ku=f will not always have a unique solution anymore.

As far as I know, negative matrices (with only negative eigenvalues) never occur in (normal) finite elements.