When modelling verical dummy using the offset method. Why is it on the example shown on the application guide, there is a DX and DY on the dummy spreadsheet ? (DX being equal in magnitude but opposite in direction to the offset.)
Please explain what does this offsetting do ?

Thanks

CAESAR II uses "3D Beam Elements", which are essentially "infinitely thin sticks". CAESAR II only knows where the centerlines are - there is no sense of volume or surface.

In the example you reference, the element's delta dimensions (DX and DY) are necessary to go from the base node (205) to the bend mid point node (84). However, node (84) is on the centerline, which is not really the correct attachment location. The computation of the dimensional correction to get from the centerline to the actual attachment location is shown in the figure on the next page (in the Applications Guide). These corrections are defined in the input using the "Offset" dialog.

Thanks Mr. Ay, Just a follow up, is it OK for the dummy to appear inclined on the plot window (due to DX=4.390 in the example shown on the Application Guide) ?

Yes.

Thanks

Dear Mr.Richard,

I digged out the old query and want the confirmation again from the highly inspired person (Mr.Richard).

As per Caesar-II Aplication Guide, i modeled the trunnion/dummy support as per offset method. In the plot the dummy support is shown as tilted with reference to the (X,Y & Z coordinates) Refer attachment. Is it mean, the input as per offset method is correct even if the plot shows as a tilted dummy.

Since it is been long ago, the query was rised, is it still the plot for offset dummy was not updated after several revision of Caesar-II.

Thankyou.

The Piping Input Processor doesn't know where the bends nodes really are. This isn't determined until the model is sent through Error Checking, and the true coordinates are generated in the "analysis files". This is why you're seeing that tilt.

There is nothing to update or change here, unless we completely replace the "modeling methodology" we use (in the Piping Input Processor) for bends.

Dear Mr.Richard,

So you confirm that even the plot show as a tilted arrangement, if we input the offset option as per Caesar-II application guide, the dummy support is said to be modeled as per offset method.

With your confirmation let me model the same for our project.

Yes, follow the Application's Guide.

Dear Richard Ay,

I have followed the procedure for modeling trunnion on an elbow using the offset method as described in the Applications Guide.

As compared to the one modeled by "sillyman" (post #35611), the tilt of the trunnion on my model is somewhat going outward while the other inward.

please confirm if what i've done is correct..and if not, where i went wrong..

Good day everyone!

I was intrigued by this post of ammangz..And so i tried modeling the trunnion by offset method and came up to the same appearance of the trunnion (i mean the direction of the tilt)..

anybody who can confirm this?

Yes, he has modeled it correctly. The tilt won't be there for the analysis, only for the input. The offset will remove the tilt and allow the dummy leg to be aligned with the centerline of the pipe coming down.

Hello Stressers
I have one additional question to this type of support like dummy leg.
If I gave on the second end of support an anchor.
What is the difference if I connected this anchor to CNode or if I stay it without CNode.
Because in my case I have situation that if I give a CNode the system is correct, if not I have exceeded allowable stress !!!

Why ??

This depends on what you did with the CNODE. If you didn't do anything with the CNODE (like fix it with displacements), then that anchor isn't doing anything and you have essentially a free end.

Perhaps you should open a Support Request on Smart Support and send in the model so Tech Support can look at it?

Thank You Richard for Your answer.
So as I good understand when I have an anchor point in the end of dummy leg without Cnode - is fixed, but when I add a CNode like on the pictures below I have free end though I have assumed anchor ?



Regards
___________________________________

Adrian

From the above images:

1) The node you are trying to restrain is 10000.

2) If you add a restraint (anchor) at 10000, you have fixed node 10000 in global space.

3) If you add a restraint (anchor) at 10000, but define a CNODE (say 10001), then you have anchored node 10000 to 10001. So, what did you do with 10001? If the answer is nothing, then 10001 is free and therefore so is 10000. You must use 10001 as another node in your piping system, or apply displacements to 10001.

Thank You Richard for good explanation.
One more question.
I received information that on this bend I have a anchor point.
Additional question.
I anchored a node 10000 so the and a element from node 10000 to 240 is rigid as on the picture this is correct thinking or something should be changed ???
Because of course in anchor point displacements are 0 but in point 240 I have some displacement so it that means that rigid element can move on the end connected to bend ??
________________________
Adrian

If node 10000 is really anchored then you must model that node as an anchor. I question whether 10000-240 is really a rigid element - what is this? If this is a pipe or even a piece of structural steel it isn't rigid. Yes node 240 can have some movement (hence my question concerning the rigid element).

Richard is a Piece of steel structure like I-beam (double T-beam) but I make it as a piece of pipe with thicker wall because it is easier for me.
I think that the stiffness will be very similar to the beam from simmilar material.
But it is okey for me becasue You solved my problem

Dear Intergraph Experts
I have a question becasue in standard in Polish to use the trunnion on the elbow we should at a first step weld a pad to elbow.
So this is simple to do in Caesar II becasue we can add a fitting thickness.
But when we had a welded a piece of pipe we have something like a "Tee"
and in my opinion we should ad a SIF in this place !!
Is this is correct thinking ??
If Yes what kind of SIF we should implement ??


Thanks for Help

Adrian,

You can calculate these SIFs using FE/Pipe. Generally, there are 4 SIFs in this case at present time. In-plane/out-plane for the elbow, and in-plane/out-plane for the trunnion.

SIFs will vary wildly dependent on geometry.

But what can I do if I don't have this software ???

Alternatively, you could use the "SIF @ Bend" tool under the "Analysis" tool. However, it has limitations on ratios of sizes and thicknesses, and I find that most of the time it generates extremely conservative SIFs compared to FEA, and only sometimes it generates slightly non-conservative SIFs.

In addition, you could ignore the SIFs and model it within FEA software (there are free versions out there), but I can't attest to their ease of use or accuracy of solver.

Thanks for Your Help Michael.
It was very usefull for me.
Regards

Richard,

When using the offset method to model a stanchion under a bend, why are restraints that are applied to the bottom of the stanchion shown as remote from the node of application?

Thanks

Oops, forgot to dd the offsets.

The input processor only registers the tangent intersection point of the bend, not the "near", "mid" and "far" points. Your plotted images will be displayed at only "From" and "To" nodes.