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.