Ah, I see. Assuming that the 3/8" hub had two 9-tooth sprockets chained to two 60-tooth sprockets on the above shaft attached to the arm, then there would be 1.5" of key to transfer the force (since both sprockets would be driven by the axle). However, I did indeed forget to consider the torque on the 48-tooth sprocket that drives that axle (with the chain running downward to a 22-tooth sprocket on the motor). Since a .375" AndyMark hub (of the type to which this sprocket would be mounted) has a width of 1" and its key would also have to take ~30 ft-lbs. of torque, that key would be the limiting factor.
1 * (3/32) = 3/32 = .09375
1920 / .09375 = 20,480 psi on the key of the sprocket driving the shaft, as compared to 13,653.3 psi on each of the keys of the two 9-tooth sprockets being driven by the shaft (if my math in my last post is correct). So 20,480 psi is the highest required by the design (plus some safety factors to compensate for acceleration and any downward impacts on the arm). Is that reasonable? What sort of material would it have to be made out of?
Just based on what I could find, shear strength is approximately 0.75 * ultimate tensile strength for alloy steel (according to
this site), and McMaster-Carr has key stock made of alloy 8630 steel with a tensile strength of 112,000 psi. 0.75 * 112,000 psi is 84,000 psi, which seems to be quite safely in excess of the 20,480 psi required at the key with the smallest cross section (the one on the 48-tooth sprocket that will be driving the shaft).
However, the
AndyMark hub that I found for .375" shafts is constructed from 6061 aluminum, which, according to this site, has a tensile strength of somewhere between 18 and 45 ksi (depending on the specific type of 6061, like -O, -T4, or -T6). According to the site from which I obtained tensile strength to shear strength conversion factors for the keys, aluminum alloys generally have a shear strength of about 0.65 times their tensile strength, so the shear strength of the aluminum in the hub should be somewhere roughly between 11.7 ksi and 29.25 ksi. However, I am not sure how to find the cross section in this case (or whether this is even an application that that formula is meant for), and hence cannot determine whether the hub could theoretically take 30 ft-lbs. I have made a request to AndyMark.biz for the maximum allowable torque on the hub, though, so hopefully I will know soon.
Thanks for all of your help!