Dear All ,

I still consider a newbie in pipe stress analysis, i faced some problem in my stress analysis this morning .

1. it is a displacement problem .What i wish to know is does the displacement affect the calculated stress result . currently what i am working is a ballast system , exactly a straight long pipe in the ship. The loadcase what i concern in the stress analysis is as below
L1 W+T1+P1 (OPE)
L2 W+T1+P1+D1 (OPE)
L3 W+T1+P1-D1 (OPE)

The reason i include the displacement is i am considering the sagging and hogging effect in the ship eventually the pipe will delfect up and down. What my concern is the result for this 3 case have similar in term of calculated stress result , even the loading also similar after i generated the output. I wish to the what main reason cause this things happen , suppose it should be different in the result .

So , I wonder i am making some mistake during input or it is really have same some result.

Regard Glenn

You may try to create 2 more load case which are
L4 D1
L5 -D1

from the 2 load cases, you can see how it react and to check whether both of it have the same result or not. And compare the deflection with the OPE case.

if your L4 and L5 are very small, then it might have similar calculated result.

Besides, check the Comb Method whether it is correct or not. did you select ABS?

ps: as richard mention, it reminds me something. Normally, the Displacement of sagging and hogging will be input as Cnode, so that it will not be fix.

Also be aware that when you specify a displacement at a node point you are declaring a "boundary condition" (in other words a restraint). Once this DOF (degree of freedom) is restrained, it is restrained for all load cases, even if you don't include the "D1" vector in the load case.

As an example, consider this:

Node 150, dy=+10mm
L1 W+P1+T1 (OPE)
L2 W+P1+T1+D1 (OPE)

In case 2 node 150 will be displaced up 10mm, the other 5 degrees of freedom will be free to do anything necessary. In case 1, node 150 will be fixed in the "DY" direction and have a resulting displacement of zero. This is because the "DY" direction has been fixed by your displacement specification, and in the real world you can't run out and change your boundary conditions between load cases.

Make sure you don't have the "tail wagging the dog here". Don't specify a boundary condition (either a restraint or a displacement) unless you really have the pipe attached to something. If you need to model "real world displacements" at an arbitrary point in the piping system, that is not attached to something, then you should consider applying a concentrated force. Using a force allows the node to move as necessary in all load cases, applying a displacement fixes that node at either zero or the specified displacement.

Ok . thanks alot . i do apply the Cnode in displacement , i not sure how to deal with the ABS ,SRSS,Algebraic . Can you briefly explain to me ? in the L1,L2,L3 , i empty the combination .What is your advise should i put int the combination method?

At this stage I would suggest you use the CAESAR II default recommendations. You won't need the ABS, SRSS, etc until you start playing with multi-directional occasional load components.