Hello,

I have a question about the basic functionality of CAESAR II.

Let's say I have a very simple model, a single pipe with restraints on both nodes, 10 and 20. These restraints can either be just +Y (simply supported) or anchors (fixed). This pipe is going to sag due to gravity, and this can be verified by looking at the bending stresses at nodes 10 and 20. However, the deflection of this sag is not captured in the results since there is no internal node between 10 and 20. Is there any way to obtain the deflection information of this pipe other than inserting internal nodes between 10 and 20?

Thanks.

No.

Add a mid-point node. It will cost you nothing and provide to data you want.

In Joel P's model he described above, how can we tell what the actual max bending stress is between nodes 10 and 20?

Also, why does each node have different bending stresses on either side of it (assuming this was a bigger model)?

Usually, long pipe run should be broken while modeling in CAESAR II even though not required from modeling point of view. Otherwise it may miss high sagging and stress values at those points.
Need attension!

Why does the output give two different bending stress values at each node? I remember this being discussed in the training, but I forgot what the reason was.

If there's one pipe in and one pipe out, with no restraint, you have the same load on both pipes framing into the node. So you figure you should get the same stress?
Remember that the stress we calculate (or the Code calculates) is not a true stress but, let's say, a nominal stress.
These bending moments are intensified to reflect the reduced strength of the individual component - a bend is weaker than a girth butt weld, a tee is weaker than a girth butt weld. A stress intensification factor is used to include this weakness in the stress calculation. So, if the node in question joins a bend to a straight pipe, then you can get different stresses with the same set of loads.

Dave, thanks for your reply. Can you take a look at the data below?
NODE Axial Stress lb./sq.in. Bending Stress lb./sq.in. Torsion Stress lb./sq.in. Hoop Stress lb./sq.in. Max Stress Intensity lb./sq.in. SIF In Plane SIF Out Plane Code Stress lb./sq.in. Allowable Stress lb./sq.in. Ratio % Piping Code

240 4138.1 736.9 -7.4 9205.8 12415.4 1 1 4875 19350 25.2
248 4138.1 148.7 7.4 9205.8 12415.4 1 1 4286.8 19350 22.2

248 4138.1 190.9 -7.4 9205.8 12415.4 1.288 1.073 4329 19350 22.4
249 4106.7 61.3 -3.2 9205.8 12415.4 1.288 1.073 4167.9 19350 21.5

249 4106.7 61.3 3.2 9205.8 12415.4 1.288 1.073 4167.9 19350 21.5
250 4092.4 158.5 -12.6 9205.8 12415.4 1.288 1.073 4250.9 19350 22

250 4092.4 123.4 12.6 9205.8 12415.4 1 1 4215.8 19350 21.8
260 4085.7 136.9 -12.6 9205.8 12415.4 1 1 4222.6 19350 21.8


Sorry I wasn't sure how to format it to read easier. This model has two straight lengths of pipe coming into a bend (Node 250). I understand that the SIFs apply to the bend and not the straight lengths of pipe in the model.

If I consider a Free Body Diagram at each node, I would expect the bending stresses to be equal. If we look at Node 248, one side of it has a bending stress of 148.7psi and a SIF of 1. On the either side, the bending stress is 190.9psi and SIFs of 1.288 and 1.073.

Can you please explain where 148.7psi and 190.9psi come from?

Thanks!

Except for the "Code Stress" column, I would call those other columns textbook stresses. That is, they have the general form that you would expect. I, personally, do not have much interest in a "non-Code" stress.
It looks to me like the bending stress is an intensified bending stress. Your Code stress changes by 42 psi and so does your bending stress. I would say that bending stress = SQRT[(ii*Mi)^2+(io*Mo)^2]/Z

Dave,

Thanks again for your reply.

Is there a way to output what CAESAR calculates as it's Z value?

Thanks!

I'm assuming you're using B31.3.
The Tee SIF scratchpad in the input processor will show the effective section modulus (Ze) for reduced branches. If you specify a size-on-size tee, the (full size) branch section modulus will be what you want.
Be aware, though, that reduced branches have a different calculation for Ze, the effective section modulus