[Travis Covington]

Originally Posted by IKE

Let's throw some numbers at this assume that while being pushedmost of the wieght is transferred to one wheel (100 pounds) and it is slipping give a coefficient of friction at 1.2(ish) and you have 120 pounds. for a 6 inch wheel this is 360 inch-pounds of torque in the system (120 pounds x 3 inch radius). If it is a 2 inch sprocket at the wheel, the chain force is then 1080 pounds. This gets added onto the tension that you put into the system. You can then run the calculation for the bending stress in a round beam constrained on both ends and then loaded in the middle (make sure you use the root diameter of the bolt if it is all thread).

We use two tensioners on each dead axle so that they are essentially in shear. Not only do they share the lods, but then you can directly connect them to the axle.

It should be noted that the 2 bearing block screws provide a significant clamping force on the side rail. The friction between the bearing blocks and the side rail adds a ton of resistance and limits the amount of bending force the tensioning screw sees while driving. The clamping force alone will hold the bearing block in place better than you might imagine. The more serious concern is a high speed impact on a wheel, which will provide significantly more stress on that screw than anything our little motors can provide. Even still, you'll likely be able to replace a broken/bent screw if it does fail.

For reference, we used this style of tensioning in 2005 and 2006 and had no issues while using a #10 screw.

However, more importantly (in my eyes, and as Cory mentioned), the lack of axial alignment between the inner and outer bearings/blocks IS an issue and should get more attention than the load this screw sees from drivetrain forces.