55 ft-lbs is 660 in-lbs and acts on 5/16” pins at a 1.21” radius. The shear force on the pins is 660/1.21 = 546 lbs.
The shear area of the 5/16” diameter pin is .0767 in^2
If carried by a single pin the shear stress is 546/.0767 = 7120 psi
What is the strength of the steel in shear? Don’t get fancy, just call it half of the tensile and your still at 38ksi yield for shear. So, one pin can easily handle the 55 ft-lbs of engine torque and you’ve got three. (That aluminum in the spool might be a bit closer to yield.)
Your real problem is a bit harder to define. Say the wheel becomes airborne and spins wildly until it hits the ground and is abruptly slowed. That is the kind of shock load that you must anticipate in drivetrain design and that is why testing and development takes much more time than design for commercial products.
You don’t have that kind of time so you build in extra strength, pay for premium materials and hope for the best.