Let FR and MR be the load and moment applied to the reference node. The statically admissible force distribution Fn among the coupling nodes satisfies
∑nFn=FR∑nxn×Fn=MR+xR×FR,
where xR and xn are the positions of the reference and coupling nodes, respectively. For an arbitrary number of coupling nodes there is no unique solution to Equation 1.
The force distribution adopted in Abaqus has the property that the linearized motion of the reference node is compatible with the coupling node group motion in an average sense. This compatibility can be described by considering the momentum of a moving coupling node group in a case where weight factors are considered as masses. In this case the reference node motion is identical to that of a point on a rigid body occupying the position of the reference node, where the center of mass of the rigid body is the center of mass of the coupling nodes and the rigid body moves with the same linear and angular momentum as the coupling node group. Since the element mass is distributed this way, the dynamic behavior of the element also has this property.
Fndef=ˆwn(FR+(T-1⋅ˆMR)×rn),
where
ˆMR=MR+rR×FR,rn=xn-ˉx,ˉx=∑nwnxn∑nwn=∑nˆwnxn,
and the coupling node arrangement inertia tensor is
T=∑nˆwn[(rn⋅rn)I-(rnrn)],
where I is the second-order identity tensor. This force distribution is recognized to be equivalent to the classic bolt-pattern force distribution when the weight factors are interpreted as bolt cross-section areas.