Determing torque so that I may determine the diameter of my axle

[blinddog22]

A side view looking at either of the left/right beams. At eye level.

http://imgur.com/a/MTemv

 

[haruspex]

No, the 4 beams bolted together form the rectangular frame of a trailer. Something negligible will be placed on top of it to form a flat surface. Then a 1000lb load will be placed on it. The load has 4 legs that will all perfectly apply the loads to the 4 centers of the beams. The 4 squares with the X's through them signify the centers of the beams, which is also where the loads will be applied.

Ok, but something must be supporting the frame from below. Where does that act?

 

[blinddog22]

The axle from earlier will be. Looking at the picture, http://imgur.com/a/fBvpE , it will be located on the dashed line, bisecting the two longer plates on the bottom of the assembly.

 

[haruspex]

The axle from earlier will be. Looking at the picture, http://imgur.com/a/fBvpE , it will be located on the dashed line, bisecting the two longer plates on the bottom of the assembly.

Then you can simplify the picture. You have 250 down and 500 up at each of those points, so the net is 250 up.

 

[blinddog22]

So the two points over the axle will be 250 up and the two points on opposite sides of the axle will be 250 down.

This is what I officially have for the moment. http://imgur.com/a/NUGdm
For the right/left beam that is supporting the downward 250 lb load. I took the moment about point O. I included both bolts on the right side using the summation of moments I solved for the F the bolts are taking. This resulted in the 950 lb force. Near the bottom I added that force per bolt on acting downward. 2 bolts on either end of the beam.
I feel like this is very very wrong.

 

[haruspex]

In that case there is no torque.

Thought about this a bit more...
There are two sources of rolling resistance: frictional torque from the axle and deformation of the tyre/road. Both are proportional to the load, but we are given only the one coefficient.
What I wrote previously applies to the deformation component. It does not have a torque about the axle. Instead, it is a bit lke going uphill. The force from the road surface is not vertical, instead angling back against the direction of motion. The point on the ground it comes from is (on average) in front of the axle, so that the force line passes through the axle. So although it has no torque about the axle it contributes to the bending moment. Use Pythagoras to combine it with the vertical component you already have.
The frictional torque component will produce torque in the axle.

 

[haruspex]

Then you can simplify the picture. You have 250 down and 500 up at each of those points, so the net is 250 up.

Thinking about that some more...
We need to consider how the total 1000lb load is spread onto the four legs. If that part of the structure is more rigid than the rectangular frame with the bolts then almost the whole 1000lb will be borne by the parts of the frame above the axles. The midpoints of the other two sides will just flex down a little.

 

[malemdk]

Since the axle experiences both tensile and torsional stresses use principle stress equation to find the axle diameter