It doesn't hurt to use the axle's max torque load but if the engine/drivetrain is not capable of generating that load, then your over engineering.
Dave is right there. If your engine can't produce enough torque with your crawl ratio, then the axle will never see it.
I had a nice long chat with a friend of mine about this thread tonight. We were in the same classes in college. He mentioned that I should have included a ground force and it was at that point I realized that I had assumed a very worst case scenario. That image that I had attached did not include a ground force, because I had assumed that the tire was going to be in a clamped, or trapped, position. This means that ALL of the force goes into the links. I do realize that this could very well happen while rock crawling, but I don't think that you're going to get stuck and just mash on the gas till something gives. Because it won't be the links!
I didn't check the math because I left my calculator at work. So I'll assume your numbers are correct. I would say you are good on the 1250 lb force applied to the center of the link. You'll have a factor of safety of around 2. You don't compare this result to the buckling equation however. The buckling equation is only used for axial loading. This would be the forces applied to the links through the axle torque, the truck standing up on end, etc.
As for the load applied because of the axle torque, the forces applied to the links alone are going to be equal and opposite because of the torque moment applied through the axle. If you sum the forces in the x direction (horizontal) for the image I posted, the only two forces are the forces on the links. These are your "resolved" forces. Meaning the actual force in the links are broken down into their individual components. The actual forces in the links are going to depend on their angles (front to back and side to side) but they will be equal when broken down into their front to back horizontal force components.
The distances of a and b are the distances of the links from the centerline. It doesn't need to be broken up like this, since both equations are the same and both use (a+b), all you really need is the length, or distance between the links. One other thing to keep in mind, if your axle mounts are going to be swept forward from the axle (in the shape of a "C"), this is going to change (increase) the forces on the links because it will change the angle from tangential (angle "a") on the drawing.
The torque should be 66,000 in*lbs. (a+b) should be 10.5 inches. And the cosine 5 deg should be really close to 1. Your force on the links due to axle torque alone should be just over 6000 lbs. This is also assuming that the 5500 ft'lbs of max torque is what can be delivered to one side of the vehicle. Remember, you have two upper and two lower links. Your forces due to axle torque are small compared to the forces from landing on a rock in the middle of the span.
Which leads to my next question. Has anyone on here actually broken/bent a link from anything other than an impact or bad welds?
Bray D, would that be U of Iowa or U of Illinois?