Table of Contents
Nonzero nodal displacements and rotations
Linear problems
Nonzero components of nodal displacement and rotation (a non-homogeneous boundary condition) must be prescribed in the first load case and apply to all load cases.
In the name.i2 file, a vector is first defined using a NV set:
NV set number T 1 V $u$ $v$ ; 2 components NV set number T 1 V $u$ $v$ $w$ ; 3 components NV set number T 1 V $u$ $v$ $w$ $\alpha$ $\beta$ ; 5 components NV set number T 1 V $u$ $v$ $w$ $\varphi_x$ $\varphi_y$ $\varphi_z$ ; 6 components
depending on the number of components at the node, or:
NV set number T 1 C component number list V prescribed values
and the set thus defined is then assigned to the selected nodes:
AS 1 /$\dots$ /N set number N node number list /$\dots$
The first set form is used to prescribe all degrees of freedom at a node. The vector can only be assigned to nodes that have the same or a greater number of degrees of freedom than the number of prescribed values. Any missing values are treated as zero.
The second set form is used when it is necessary to prescribe the values of only some of the degrees of freedom at a node, while the others are to remain free. The vector can only be assigned to nodes that have all their components listed after the key letter C. The component (degree of freedom) numbers follow from the table:
| Number | Component | Meaning |
|---|---|---|
| 1 | $u$ | displacement in the direction of the $x$ axis $\text{[m]}$ |
| 2 | $v$ | displacement in the direction of the $y$ axis $\text{[m]}$ |
| 3 | $w$ | displacement in the direction of the $z$ axis $\text{[m]}$ |
| 4 | $\alpha$ | the rotation angle of the semi-loof element edge $\text{[rad]}$, see the Reference Manual |
| 4 | $\varphi_x$ | the rotation angle about the $x$ axis $\text{[rad]}$, see the Reference Manual |
| 5 | $\beta$ | the rotation angle of the semi-loof element edge $\text{[rad]}$, see the Reference Manual |
| 5 | $\varphi_y$ | the rotation angle about the $y$ axis $\text{[rad]}$, see the Reference Manual |
| 6 | $\varphi_z$ | the rotation angle about the $z$ axis $\text{[rad]}$, see the Reference Manual |
Example
Let us consider two NV sets with the identification numbers 1 and 2:
NV 1 T 1 V 0 0.001 0 NV 2 T 1 C 6 V 1e-3
The first set describes a displacement of $1~\text{mm}$ in the direction of the $y$ axis (the other DOFs are fixed).
The second set prescribes the rotation angle $\varphi_z=10^{-3}~\text{rad}$ (the other DOFs are free).
The set NV 1 can be assigned to any node, the set NV 2 only to a node of a beam element.
The assignment might, for example, take the form
… /N 1 N 23 24 109/N 2 N 35/ …
in which case the displacement vector of nodes 23, 24, and 109 will be fixed as $[0;0.001;0]$, or $[0;0.001;0;0;0]$, or $[0;0.001;0;0;0;0]$, depending on the type of node. Node 35 must belong to a beam element, and only the rotation $\varphi_z$ will be prescribed at it; the other DOFs will remain unknown.
Nonlinear problems
In nonlinear problems, the PMD system allows only a single way of enforcing a nonzero displacement (displacement-controlled loading). That is the penalty function method. Let us consider the part of the mesh shown in the figure, where a displacement $\mathbf{u}_0$ is prescribed at a node.
First, a spring with stiffness $k_n$ is assigned to the chosen node, in the direction given by the vector $\mathbf{u}_0$. The stiffness $k_n$ should be approximately 6 orders of magnitude greater than the local stiffness of the body (which can at least be estimated by order of magnitude). Next, a nodal force $\mathbf{F}_0=k_n\mathbf{u}_0$ is assigned, which enforces the displacement $\mathbf{u}_0$ on the spring. Since the spring stiffness is substantially greater than the stiffness of the body, the actual displacement can be expected to be close to $\mathbf{u}_0$ as well.
All necessary springs are defined in batch AS 1, and the corresponding nodal forces in batches AS 2 and higher. If any load case contains a discrete force $\mathbf{F}$ at a node with a very stiff spring, this means that in that load case, the displacement $\mathbf{u}=\mathbf{F}/k_n$ will be enforced. If the prescription of such a force is missing in some load case, the displacement of the node will be approximately zero.