Properly modeling Restraint Stiffness

Posted by: joeseagle

Properly modeling Restraint Stiffness - 11/01/19 10:54 AM

Probably a simple question here! Can someone walk me through the proper inputs on a restraint?

Example support criterea: A pipe sitting on top of a very tall T-pole. There's no guides or axial stops. The friction of the pipe shoe is the only horizontal loads acting on the T-pole.

In order to more accurately determine loads at the support, I want to include the stiffness of the T-pole, which I can get from my Structural Engineer. At this support location in my model:
-What are the exact restraint types (ie +y, x, z, etc)
-Which do you apply friction to if any
-Which do you apply stiffness values of the T-pole to?

Thanks!
Posted by: anubis512

Re: Properly modeling Restraint Stiffness - 11/04/19 06:46 AM

If the support is a simple un-guided pipe shoe, you have only +y.

You would apply friction at this restraint (typ 0.3 for metal/metal or maybe 0.1 for teflon or whatever your situation may be).

Stiffness is an iterative process. You'd likely start w/ the default Caesar assumption that it's rigid, calculate the force, pass that to structural and go back and forth until the structural is happy with the load and displacement. Like you said, with only friction the horizontal displacement is going to be very low unless there's wind/seismic.
Posted by: Michael_Fletcher

Re: Properly modeling Restraint Stiffness - 11/04/19 11:03 AM

Case 1:
Consider a T-pole where Kaxial = infinity lb/in. (Axial in this regard is vertical for the t-pole.)
It also has Klateral = 0 lb/in, because it's infinitely long and infinitely thin.
It's steel on steel contact.

In Caesar, this would be a +Y WITHOUT friction. This is because any friction load that exists between the pipe and the support just moves the T-pole without any effort.

Also, we don't input a stiffness for the +Y, because 1e12 lb/in (default value) was selected for software stability.

Case 2:
Kaxial = infinity lb/in
Klateral = infinity lb/in
Steel/steel

This is also a +Y, but with friction factor, f. (Generally 0.3, but different folks like different values, so I won't tell you what friction factor to use.)

This is very much the idealized scenario that almost everyone uses.

Case 3:
Kaxial = not 0, not infinity, but impactful lb/in
Klateral = infinity lb/in
Steel/Steel

+Y with calculated vertical stiffness and friction factor f. If adjacent supports are left at default stiffness, the vertical load will be lessened at "our" t-pole by a factor based on rigidity of our pipe, which will affect stresses accordingly.

Case 4:
Kaxial = infinity lb/in
Klateral = not 0, not infinity, but impactful lb/in
Steel/steel

The friction between the pipe and support is large enough to cause the structure to bend before the pipe slides.

Option 1: +Y with friction factor f. Overestimates the lateral loads on t-pole and how much the pipe slides.
Option 2: +Y with reduced friction factor f. Probably closer to reality, but not easily solved for.
Option 3: +Y without friction, X2/Z2 with release load F and spring load L. Probably closest to reality you can possibly get, but extremely not easily solved for.
Posted by: joeseagle

Re: Properly modeling Restraint Stiffness - 11/04/19 03:30 PM

Michael, thank you much for the detailed explanation!

I think case 4 is exactly what I'm trying to model. I fully agree option #3 gives most accurate representation. You say: "The friction between the pipe and support is large enough to cause the structure to bend before the pipe slides". I'm trying to understand what the stiffness represents in K1 vs K2 if the piping isn't sliding. Wouldn't I just want to use regular X/Z with stiffness values (based on the physical properties of my T-pole provided by the structural engineer) in each direction?

The result would look like:

Option 3: +Y without friction, X/Z with stiffness values of the T-pole in both directions provided by my structural engineer.
Posted by: Michael_Fletcher

Re: Properly modeling Restraint Stiffness - 11/05/19 10:40 AM

I think you meant option 4 (I'm going to assume so). Option 4 is fine so long as your static friction is guaranteed to never allow the pipe to slide. In which case, you'd probably just want to put actual physical X/Z supports there to guarantee the analysis matches reality.

There's also options 5 and 6, which should've been mentioned previously, and I shouldn't have assumed the intent was to omit it.

Option 5: Model the T-pole in the structural modeler, and pull it in.
This is fairly laborious and less intuitive than "normal" CAESAR operation. I'll note the limitation that CAESAR only considers one moment of inertia for a beam, so it'll rotate the beam along its axis to whatever CAESAR prefers, and you'll have to accept it.

Option 6: Approximate the T-pole as a piping element whose moment of inertia matches that of the beam for the axis line of action in question. I'd sooner do this than option 5, though.
Posted by: Faizal K

Re: Properly modeling Restraint Stiffness - 11/05/19 12:01 PM

I think modeling the steel is quite easy if this is just one tall column with a short beam at the top. Sounds like all you need for the support point is to model one element column.

Michael,
I'm not sure what you meant by CAESAR only considering one moment of inertia for a beam. To my knowledge, one of the advantages of using the structural modeler is that you can use two area moments of inertia for each element. And you can also rotate it to the proper orientation.
Posted by: Michael_Fletcher

Re: Properly modeling Restraint Stiffness - 11/06/19 09:33 AM

If the stress analyst is modeling (and sizing) steel, the whole CSA -> Piping Input Editor -> Piping Output Editor -> CSA loop is tedious if you have complicated structure.

Regarding point 2, either I misremembered my CAESAR training with COADE, or that's an update that I never knew was made. If so, then disregard my earlier statement.