# How does rigid body rotation affect resuts in a simulation?

I have been running a finite-element simulation where, for the sake of convergence, I chose to stay in small deformation hypothesis (in comsol, include non-linear geometry box not checked for who that may concern).

Now, my simulation is made-up so that geometry actually undergoes very large deformations.

Under comsol, I've read that rotation matrix is computed in a different manner (some kind of linear version).

My question is : in a finite-element simulation with a presence of a rigid body, do the rotation matrix of that rigid body is computed AFTER the displacement field computation or BEFORE ?

In other terms, does the fact that the rotation matrix is linearized affects the displacement of the rigid body ?

• I don't understand your question. You first said that you are using a linear model but then you said that you have large deformations. Can you clarify this? Can you also provide what is the equation/problem you are solving? – nicoguaro Jul 13 '16 at 20:07
• If you expect that your model undergoes finite rotations and large deformations, result obtained with a linear model (which is valid only under the infinitesimal rotation and deformation hypothesis) will be grossly inaccurate. – Stefano M Jul 13 '16 at 20:22
• Well all I have is a rigid body moving into a viscous low-young modulus cast. So the low young modulus cast deform largely, and the viscous assumption helps it converge. I know it is way inaccurate, but that's the side point of my question : I need to know if the rotation matrix helps compute the displacement, or if the rotation matrix is computed after the displacement have been computed ? – Blue_Elephant Jul 13 '16 at 21:07

You question is unclear. However, rotations in the context of linear mechanics are often confusing. Therefore a response is probably needed.

1) I will assume you have a deformable body and a rigid body in your simulation, i.e., there are two objects that may interact. 2) I will assume the deformable body is subject to a torque while the rigid body has a constant rotational velocity.

In a typical FE simulation under these conditions, the rigid body is treated as a node at which the Euler equations are solved. If contact is expected then some extra computations are done. That means that rotations are taken care of reasonably accurately.

If the deformable body is linear elastic (or small-strain elastic-plastic/viscoelastic), rotations are typically accounted for by using an objective stress rate (such as the Jaumann rate). More accurate results are obtained by unrotating the stress and rate of deformation using the deformation gradient, updating the stress, and then rotating back.

The same approach can be used for nonlinear materials, even at large strains. However, a more convenient approach for "simple" material models is to use a large deformation strain measure and use related constitutive models.

Due to the Euler's rotation theorem a rigid body in 3D space has 3 rotational degrees of freedom, (plus 3 displacement degrees of freedom for a total of 6 degrees of freedom.). This typically means that the rotation matrix (which has 9 components) is not the primary unknown but, when needed, is computed from the rigid body's rotational DOFs.

Several different choices for the three rotation parameters are possible, see "Rotation formalisms in three dimensions". In general the dependence between the rotation parameters and the rotation matrix is non linear, but when $\sin\theta \approx \theta$, simplified expression can be obtained. E.g., for a small $\theta$ rotation about the $z$ axis, we have \begin{equation} \begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \approx \begin{bmatrix} 1 & -\theta & 0 \\ \theta & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{equation} As you can see the "linearised" rotation matrix is correct up to $\theta^2$ terms, so in general it is not possible to use small rotation theory when $\theta$ is not small with respect to unity.

As pointed out by Biswajit Banerjee in his answer, formulations with finite rotations and small/infinitesimal strains are possible, but I would not call them "fully linear" because the dependence of the rotation matrix from the rotation parameters is still non-linear.

• So you are actually saying that rotation matrix is computed after the displacement field ? Which means, my displacement does not depend on the rotation matrix ? – Blue_Elephant Jul 14 '16 at 10:34
• In FEM typically a rigid body is described by a single node with 6 DOFs (3 displacement and 3 "rotation" components, collectively called generalised displacements). The displacement field of points belonging to the rigid body can be recovered, via the rotation matrix, from these 6 DOFs. To be more precise, the displacement field is a non-linear function of the generalised nodal displacements. Note that the rotation matrix shows up in the (generalised) force balance equations of the rigid body: changes to the rotation matrix affect also the computed (generalised) nodal displacements. – Stefano M Jul 14 '16 at 12:28