# Lagrange Multipliers in Multi-body Finite Element Code

I'm working on a Multibody dynamics code using the finite element method to simulate the behaviour of flexible beams (using this paper if anyone is interested/ it is relevant). I'd like to model joints, and as I only need spherical friction free joints I believe this should be easy using Lagrange multipliers.

My understanding is that simple spherical joints are done by adding some extra rows and columns to the tangent stiffness matrix $K_t$ and extra rows to the residual force vector $g$ as below. Lets say we have 6 degrees of freedom at each node, $n$ nodes, and $p$ joints. For this example $p$ is 2 - The first joint links node $i$ to node $j$ and the second node $k$ to node $m$. The initial geometry places the linked nodes in the same place. The vector $\delta p$ is the incremental change in node position/ rotation and $F_J$ is the nodal forces required to keep the nodes coincident.

$$\begin{bmatrix} K_t & (\lambda_{2p \times 6n})^T \\ \lambda_{2p \times 6n} & 0 \end{bmatrix}\begin{bmatrix} \delta p \\ F_J \end{bmatrix} = \begin{bmatrix} g \\ 0 \end{bmatrix}$$

Where $\lambda$ is the Lagrange multiplier which for this two joint system is calculated as follows like this: $$\lambda=\begin{bmatrix} 0 & \cdots &0&1_{1,6i-5} & 0 &\cdots& 0 & -1_{1,6j-5} & 0 & \cdots &0\\ 0 & \cdots &0&1_{2,6i-4} & 0 &\cdots& 0 & -1_{2,6j-4} & 0 & \cdots &0 \\0 & \cdots &0&1_{3,6i-3} & 0 &\cdots& 0 & -1_{3,6j-3} & 0 & \cdots &0\\0 & \cdots &0&1_{4,6k-5} & 0 &\cdots& 0 & -1_{4,6m-5} & 0 & \cdots &0\\ 0 & \cdots &0&1_{5,6k-4} & 0 &\cdots& 0 & -1_{5,6m-4} & 0 & \cdots &0 \\0 & \cdots &0&1_{6,6k-3} & 0 &\cdots& 0 & -1_{6,6m-3} & 0 & \cdots &0\end{bmatrix}$$

When I run the code, the two bodies part company immediately. Any ideas where I'm going wrong? I'm just simulating a double pendulum with 10 elements in each body for now.

• If all you need are simple spherical joints, this can be done easily without Lagrange multipliers. It is just a matter of defining the same equations in the global matrices for the translational DOFs at each beam end at the joint. The rotational DOFs for the two beams at the joint are independent. – Bill Greene Aug 21 '14 at 12:34
• I agree with @BillGreene: unless you have to compute the internal forces $F_J$ (which could be useful for correctly engineering the joints), it is much more easy to have a single "node" at the joint with 9 degrees of freedom: 3 translational DOF and 6 rotational ones (i.e 3 rotational DOFs for each beam.) This will ensure that the beams are always correctly connected. – Stefano M Aug 21 '14 at 12:57
• Hi @Stefano M, The reason I'm doing it in this way is to keep the joint forces like you suggest. I've just realised I'd made a programming error and what I've written above actually works fine. – Peter Greaves Aug 21 '14 at 14:55