I've been thinking about this tonight and I was thinking to maybe compute 3000lbs on each end of a 2ft tube dropped at 1ft onto a point. That seems like an extreme condition and the variables could be backed off to find approximate failure points from there. Having said that, I'm not a materials engineer so, my methodology is likely to be naive or just plain wrong.
That sounds close, usually with basic beam bending for this type of scenario you calculate the force it will see (in this case directed towards the middle as worst case scenario) and treat the ends as 100% statically supported. After this is completed, another interesting thing to look at in the system is the strength of the bolts they are attached to:
Some links and link skids come with 9/16" Grade 8 bolts. Grade 8 minimum yield strength = 130,000 psi (this is the force to bend, not break) Shear strength is approx 60% tensile strength: 130k x .60 = 78,000 psi. Tensile stress area for 9/16 bolt= .182" squared. Multiply this area by 2 because of two bolts, and by 2 again because they are in double shear= .728 in sq of area.
Force (lbs) = P (psi) x A (sq in)
F= 78,000 x .728
F= 56,800 lbs to bend the bolts that mount the lower links given an evenly distributed force.
Divide this # by two to see what a single bolt can do if the load was on one end.
Now the factory metric hardware should be close to this # too.
There is yall's basic statics lesson for today. Part two is the strength of the lower links...