The RBE3 is a powerful tool for distributing applied forces and mass in a finite element model (FEM). Unlike some of the other R-type elements, such as RBAR and RBE2, the RBE3 element does not add stiffness in the FEM. Forces and moments applied at a reference point (also known as the dependent node) are distributed to a set of independent nodes based on the RBE3 geometry and local weight factors. To evaluate RBE3 forces, consider the following figure:
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In this figure, a force $\vec{F} = \begin{bmatrix} 0 & 0 & 100 \end{bmatrix}^{T}$ and a moment $\vec{M} = \begin{bmatrix} 10 & 0 & 0 \end{bmatrix}^{T}$ are acting on the node 5 (dependent node), and we want to distribute this force-moment system to the four nodes of the CQUAD4 element. To do this in MSC Nastran, the three translational degrees-of-freedom (DOFs) at nodes 1 through 4 are set as independent, whereas all six degrees-of-freedom at node 5 are set as dependent (Side note: these dependent DOFs are stored in a special set known as m-set in MSC Nastran terminology, and are eliminated from the final system of equations).
Note that in this example, the weight factor for all independent DOFs is 1. This means that the force-moment system acting on the dependent node is distributed among the independent DOFs based on geometry alone. This holds true whenever all weights are equal.
To evaluate the forces acting on the independent nodes, consider a rigid body made up of discrete points which occupy the same positions in the space as the independent nodes and the masses of these discrete points are equal to the weight factors of the independent nodes. As a result of this one-to-one correspondence between the points on the rigid body and the independent nodes, the centre-of-mass, total mass, and moment of inertia tensor of the rigid body and the system of independent nodes are equal. Additionally, we assume that the rigid body and the system of independent nodes are under the action of the same force-moment system. As a result of this equivalence, the problem of evaluating the accelerations (and thus forces) of RBE3 independent nodes reduces to the problem of determining the rigid-body accelerations (and forces) of the discrete points on the hypothetical rigid-body. To determine the rigid-body accelerations of the discrete points, we translate the force-moment system acting at the dependent node to the centre-of-mass, and then determine the 'total linear accelerations' of the discrete points. The total linear acceleration will be the sum of translational acceleration and linear acceleration due to rotation.
The first step in the evaluation of the RBE3 forces is the determination of the weighted centre-of-mass (CM).
$$\vec{x}_{CM} = \frac{\sum w_{i} \vec{x}_{i}}{\sum w_{i}}$$
where, $\vec{x}_{CM}$ and $\vec{x}$ are the position vectors of the CM and independent nodes, respectively.
In the second step, the force-moment system acting at the reference node is translated to the weighted CM. This is achieved using:
$$\vec{F}_{CM} = \vec{F}$$
$$\vec{M}^{CM} = \vec{M} + \vec{r}_{Dependent}^{CM} \times \vec{F}$$
where, $\vec{r}_{Dependent}^{CM}$ is the relative position of the dependent node with respect to the weighted CM: $\vec{r}_{Dependent}^{CM} = \vec{x}_{Dependent} - \vec{x}_{CM}$.
The linear acceleration at the centre-of-mass (or any other point on the rigid-body) is simply $\frac{\vec{F}_{CM}}{\sum w_{i}}$. The corresponding force at the $i-$th independent node is then $w_{i} \frac{\vec{F}_{CM}}{\sum w_{i}}$.
The linear acceleration of any independent node due to the rotation of the rigid body is given by $\vec{\alpha}^{CM} \times \vec{r}_{i}^{CM}$, where, $\vec{\alpha}^{CM}$ is the rotational acceleration of the rigid-body about the centre-of-mass, and $\vec{r}_{i}^{CM}$ is the relative position of the $i-$th node with respect to the centre-of-mass: $\vec{r}_{i}^{CM} = \vec{x}_{i} - \vec{x}_{CM}$. The rotational acceleration of the rigid-body about the centre-of-mass is given by $\vec{\alpha}^{CM} = (\underline{I}^{CM})^{-1} \vec{M}^{CM}$, where, $\underline{I}^{CM}$ is the moment of inertia tensor of the rigid body about the centre-of-mass. For a system of discrete nodes, the moment of inertia tensor is given by:
$$ \underline{I} = \begin{bmatrix} \sum w_{i} \left( (y_{i}^{CM})^2 + (z_{i}^{CM})^2 \right) & -\sum w_{i} x_{i}^{CM} y_{i}^{CM} & -\sum w_{i} x_{i}^{CM} z_{i}^{CM} \\ -\sum w_{i} y_{i}^{CM} x_{i}^{CM} & \sum w_{i} \left( (x_{i}^{CM})^2+ (z_{i}^{CM})^2 \right) & -\sum w_{i} y_{i}^{CM} z_{i}^{CM} \\ -\sum w_{i} z_{i}^{CM} x_{i}^{CM} & -\sum w_{i} z_{i}^{CM} y_{i}^{CM} & \sum w_{i} \left( (x_{i}^{CM})^2 + (y_{i}^{CM})^2 \right) \end{bmatrix}$$
where, $[x_{i}^{CM}$, $y_{i}^{CM}$, and $z_{i}^{CM}]^{T}$ is the relative position of the $i-$th independent node with respect to the CM: $x_{i}^{CM} = x_{i} - x_{CM}$, $y_{i}^{CM} = y_{i} - y_{CM}$, and $z_{i}^{CM} = z_{i} - z_{CM}$.
So, the linear acceleration at the $i-$th node due to a moment acting the centre-of-mass is given by $(\underline{I}^{CM})^{-1} \vec{M}^{CM} \times \vec{r}_{i}^{CM}$. The resulting force on the $i-$th node will then be $w_i (\underline{I}^{CM})^{-1} \vec{M}^{CM} \times \vec{r}_{i}^{CM}$.
Finally, the total force acting on the $i-$th independent node due to a force-moment system acting at the dependent node is given by:
$$ \vec{F}_{i} = w_{i} \frac{\vec{F}_{CM}}{\sum w_{i}} + w_i (\underline{I}^{CM})^{-1} \vec{M}^{CM} \times \vec{r}_{i}^{CM} $$