Tube support elements

Each tube support element has two nodes. One node represents the axis of the tube, the other is the center of the hole in the support plate (or midway between a pair of parallel sides of an “egg-crate”).

Let ${\mathbf{a}}_2$ be a unit vector along the axis of the tube, and let ${\mathbf{a}}_3$ be a unit vector along the axis of the interface element (that is, perpendicular to the parallel sides of the support in the “unidirectional” interface that is used with egg-crate supports, and parallel to the line joining the two nodes of the element in the “cylindrical” interface that is used with circular holes). It is assumed that ${\mathbf{a}}_3$ is in the cross-section of the tube and, hence, orthogonal to ${\mathbf{a}}_2$. We define a third basis vector as

Let ${\mathbf{x}}^N$ be the current position of node N of the element at any point in time (here N is 1 or 2).

Relative displacements in the element are measured from the position when the tube and the hole in the support plate are exactly aligned; that is, when the nodes of the element are at the same location. They are defined as follows:

axial to the interface element,

axial to the tube,

and

tangential, in the plane of the tube's cross-section,

for the unidirectional case:

and, for the cylindrical case:

The basis vector—${\mathbf{a}}_2$, along the axis of the tube and of the hole in the support plate—is assumed to be fixed. In the unidirectional element the ${\mathbf{a}}_1$ and ${\mathbf{a}}_3$ vectors are also fixed. In the cylindrical interface ${\mathbf{a}}_3$ is parallel to the line joining the nodes of the element, so

where

Therefore,

Thus, for this element

and

For simplicity we replace the integral with the backward difference approximation

The element generates an axial force—$P_3$, parallel to ${\mathbf{a}}_3$—and two shear forces—$Q_1$, parallel to ${\mathbf{a}}_1,$ and $Q_2$, parallel to ${\mathbf{a}}_2$. In addition, because the nodes of the element are at the center of the tube and at the center of the hole in the support, while the interaction forces between the tube and its support are transmitted at the point of contact of the tube with the support, these forces also cause moments at the nodes of the element. It is assumed that the moments caused by $P_3$ and by $Q_2$ are not significant and can be neglected. The only moments considered are $Q_1{d}_1/2$ at node 1, the center of the tube, and $Q_1{d}_2/2$ at node 2, the center of the hole. Here $d_1$ is the outside diameter of the tube and $d_2$ is the diameter of the hole for the cylindrical interface or is the distance between the parallel support plates for the uniaxial interface (see Figure 5).

Figure 5. Contact forces in the cross-section.

The virtual work contribution of the element is, then,

where $\delta {\varphi }_2^N$ is the virtual rate of rotation about the ${\mathbf{a}}_2$-axis at node N.

From this expression the contribution of the element to the Jacobian (stiffness) matrix of the equilibrium equations is immediately available as

The “initial stress” terms,

are only nonzero for the cylindrical interface, for which

so that

and so

Also, for this element,

and so

This term is not symmetric.

The “initial stress” terms for the cylindrical interface are, therefore,

The other terms in the stiffness matrix are associated with changes in the forces in the element, $d{P}_3$, $d{Q}_1$, and $d{Q}_2$. We assume $P_3$ is made up of a spring force that is a function of $u_3$ and a dashpot force that is a function of ${\dot{u}}_3$:

so that

For the implicit integration operator used for nonlinear dynamic analysis in Abaqus/Standard,

where

where $\mathrm{\Delta }t$ is the time increment and $\gamma$ and $\beta$ are the parameters of the integration operator.

Thus,

The values of $d{Q}_1$ and $d{Q}_2$ come from the friction theory and are defined from $d{u}_1$, $d{u}_2$, and $d{u}_3$ by that theory (see Coulomb friction).

In summary,

and

so the stiffness contribution of the element is

This matrix is not symmetric if $Q_1$, $d{Q}_1$, or $d{Q}_2$ is nonzero. Without friction they are zero, and the terms in $K_{33}$ and $P_3$ are the only nonzero terms. With relatively small friction coefficients in dynamic applications the terms $K_{\alpha 3}$ and—if the tube diameter is not very large—$\left(d_1/2\right)K_{1\alpha }$, $\left(d_1/2\right)K_{13}$, $\left(d_2/2\right)K_{1\alpha },$ and $\left(d_2/2\right)K_{13}$ can be neglected and, thus, a symmetric approximation to the Jacobian matrix used without serious degradation of the convergence rate of the Newton solution of the equilibrium equations.