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A typical nonlinear FE problem can be given by:

\begin{equation} \mathbf{R} = \mathbf{K}\rho + \mathbf{F}_{NL}(\rho), \end{equation}

where R is the vector of applied forces, K is the material or linear stiffness matrix, $F_{NL}$ is the nonlinear force function and $\rho$ is the vector of displacements.

The tangent stiffness matrix is:

\begin{equation} K_T = K + \dfrac{\partial F_{NL}}{\partial \rho}(\rho) \end{equation}

If an incremental procedure is applied, $\dfrac{\partial F_{NL}}{\partial \rho}$ will need to be updated at load application step, since it is a function of the deformed configuration.

But why does the material stiffness matrix $K$ not need to be updated? $K$ is formed by rotating local stiffness matrices and assembling them into a global frame of reference. Surely when the geometry becomes deformed $K$ will need updating also, even if it is the linear stiffness matrix?

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Depends on what you mean by "nonlinear ". Elastic plastic material constitutive model? Large deflections? Large rotations? Follower forces? Element failure? Contact elements? In some situations you will update everything, in other situations you will have a linear matrix which does not get updated and another one that does.

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  • $\begingroup$ I was thinking of large deflections but small strains. $\endgroup$ Commented Dec 9, 2017 at 6:50
  • $\begingroup$ That is for a 2D beam such that $\varepsilon_{xx} = \dfrac{\partial u}{\partial x} + \dfrac{1}{2} \left( \dfrac{\partial v}{\partial x} \right)^2$. $\endgroup$ Commented Dec 9, 2017 at 7:18
  • $\begingroup$ For small strain large displ., the approach you mentioned is the one I am familiar with. The linear part of K does not need to be updated. Instead you can use coordinate transforms. i.e. the element shape is basically the same, just moved around. The rigid body motion does not contribute to the strain energy. So just subtract it out. For large strain it would be different. This manual for NASTRAN might help. See the description starting at section 5.2 docs.plm.automation.siemens.com/data_services/resources/… $\endgroup$
    – Daniel K
    Commented Dec 9, 2017 at 21:43

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