A Very Stable Challenge

With linear structural analysis, a number is a number. However, reality starts to creep in when physics dictates that you need to account for nonlinear effects. The three member structure shown below is a perfect example–Example1.1.tcl, one of the most OG OpenSees examples.

Have you ever paid attention to the member sizes though? If so, you know it’s a good thing this model is linear. Assuming square cross-sections, the slenderness ratio, L/r, for each member is 131, 186, and 210.

Plus, the elastic modulus assigned to each member is E=3000 (not 30,000). Assuming base units of kip and inch, the material is one order of magnitude less stiff than steel, about as stiff as low strength concrete. The elastic buckling forces assuming simple end conditions (K=1) are 17.1, 4.28, and 3.35 kip.

Nonetheless, in the OG example script, the members are linear-elastic truss elements and the model is numerically stable. So OpenSees (or any analysis software) can find a solution for any load factor, $\lambda$.

When $\lambda$=1, the applied loads are 100 kip and 50 kip and linear analysis gives member axial forces 43.9 kip (T), 57.5 kip (C), and 55.3 kip (C). For the members in compression, the axial force is an order of magnitude higher than the elastic buckling forces. If this structure was real, it could not handle $\lambda$=1.

Now here’s the modeling challenge: accounting for geometric nonlinearity, what is the load factor, $\lambda$, when the horizontal displacement of the top joint reaches 0.25 inch? Keep the reference loads in the proportion shown in the figure above.

I recommend you use displacement control and switch to frame elements. Assume square cross-sections when computing I. I don’t think you need to worry about material nonlinearity. Don’t consider out of plane buckling.

I look forward to your responses. I will share anonymized results in another post after the deadline.

RESULTS