Hello there,

I have a doubt and I don't know if I will be clear in my query.

When we consider some types of events, such as earthquake and ship vessel inertial acceleration, as uniform loads in CAESAR II, we have several ways to obtain these loads by combination methods.

I will give 3 examples. All of then are considering the uniform load (U1) on horizontal direction.

1-The uniform load is calculated in a direct way.
L1 – W + P1 + T1 (OPE)
L2 – W + P1
L3 – U1

2-Uniform Load calculated from an operation condition combined with weight, pressure and temperature. Suggest by CAESAR II User Guide.
L1 – W + P1 + T1 + U1 (OPE)
L2 – W + P1 + T1 (OPE)
L3 – W + P1 (SUS)
L4 – L1 – L2 = U1 (SUS)

3-Uniform load calculated from a sustained load case combination with weight, pressure.
L1 – W + P1 + T1 (OPE)
L2 – W + P1 + U1 (OPE or SUS)
L3 – W + P1 (SUS)
L4 – L2 – L3 = U1 (SUS)

In the first example, the friction effect is ignored during the uniform load calculation. In the second example the same “problem” occurs, apparently due to the thermal presence. In the third example, the uniform load is calculated taking into consideration the friction effect.
This issue is clearly noted on a pipe running in the pipe rack and the uniform load is present on axial direction.

In my opinion, the friction effect is a restriction for the uniform load imposed movement and should be considered in uniform loads calculation. Without friction, we suggest that the pipe would slide freely or all calculated load would be imposed on the limit stop restraint.

If we apply a uniform load that is 20% of the vertical load, in a system that has friction (a maximum load of 30% of vertical load in the opposite direction) and no axial restraint, the displacements in axial direction, due to this uniform load only, should be 0 mm, right? As the expected uniform load is lower than the friction load, the pipe would not slide.

A test made in a system running on a pipe rack, with the 3 cases, has shown that in case 1 a 524 meters displacement was found. In case 2, the displacement calculated was 15 mm. And in case 3, the displacement was 0 mm.

I am really confused. I always thought that the case 2 was the correct one. But all results that I expected the reach in case 2, I am getting in case 3.

I wonder if CAESAR II considers that he friction effect has been overcome by the thermal displacement and the uniform load acts freely. Being this case, the worst possible case.

Regards

Marcelo Neves

Your example 2 is the correct one. If the pipe has lifted off due to themal loads then there will be no friction due to the uniform load as it is not in contact with the resting support. Example 3 does not consider the effect of thermal strain on your system and leads to incorrect results. The same can be said for Example 1. If the pipe is in contact with the restraint then you will see friction forces due to all the loads on the pipe (W, T, U).

Thank you Loren,

I understand your point. What I am not understanding is why, if we consider double acting vertical restraint in the entire system and our uniform load is lower than frintion, we still have displacement due to uniform load only on example 2.

Regards,

Marcelo Neves

There is a discussion on this exact topic in the Technical Reference Manual. Specifically, friction is approximated as a horizontal restraint with a defined (in the Configuration) stiffness. So you can still have movment, even if the load is less than mu*NormalForce.

Richard,

According to results from combination method suggested by CAESAR II Guide, a Line running along the Pipe Rack, subjected to a uniform load in axial direction, it would have all uniform load resisted by the transversal guide. I don’t know what I am missing, but in my understanding, part of this axial uniform load would be resisted by the friction and part would be resisted by the transversal guide. See attached file.
I am afraid that with these results, all axial uniform load being applied directly on the transversal guide, I am over sizing the Pipe Rack Anchorage Beam.

Thanks in advance,

Marcelo Neves

Friction resists movement. How can there be any movement at your two friction points if you have an axial stop at your guide?

Also, what is yoru normal load at these two friction points. Isthere deadweight? Friction load is mu times the normal load - no normal load, no friction.