Dear all,

We have to apply an imposed displacement say in +y (vertical) direction to a node (say node 10) where it is located a restraint in another direction (say X-direction).

Obviously it is not allowed to apply displacements where restraints is placed, so we would like to use cnode.
So we constraint the node (node 10) in y direction referring to a cnode (say cnode 11); then we apply a displacement to node 11 defining a new displacement vector say D1.
Load case we have to analyze are:
L1: W+T1+P1+D1 (OPE)
L2: W+P1 (SUS)
L3: L1-L2 (EXP)

It looks to me that adding D1 displacement at cnode 11, stiffness matrix of whole system is different compared with system without D1 displacement.
In fact load case L2(SUS) (where D1 displacement vector is not active) appears with node 10 constrained in y and not free in that direction.
We obtain the same think applying displacement directly to main node 10 (without restraint in x direction).

Is my guess correct?

If it is so, how can I simulate the effect of an imposed motion without impacting on global stiffnes matrix of the system? Note that the imposed motion is not due to a constraint settlement but to the effect (thermal expansion) of another (not-modelled) part of the piping system.

Can anyone help or comments?

Thanks

The rule is - If you specify a displacement and this displacement set is NOT included in the load case, the program will consider the piping fully restrained along that displacement vector.

I guess one way that you can have a pipe position specified in one load case and free in another is to use the snubber (SNB). The snubber can be turned on and off in the load case editor. Turn it on (with the associated displacement) for one case and turn it off for another.

Thank you Dave,

So the procedure that you suggest is the following (considering my previous example):

1. Constraint node 10 using restraint type YSNB
2. Associate to that restraint cnode 11
3. Apply displacement in +y direction to node 11 and define displacement vector D1
4. Let snubber active in load case L1(W+T1+P1+D1) and not-active in load case L2(W+P1)

In this way node 10 is displaced in the right location in load case L1 and completely free in load case L2.

Is that correct?

Thank you very much

Best regards

Dario Pioli

Yes.
Check your results...