Roller Chain Tensile Strength vs. Horse Power

[dberg]

Hello, I am working on modifying a hammer mill to be driven by roller chain.
I want to power this with a 150 HP tractor and want to make sure that the chain drive is not the "weak link" in this.
I am thinking that I will use a ~5" dia. sprocket to drive another 5" sprocket and use a double #60 roller chain that has a tensile strength of
17000 lbs. The speed will be 540 RPM.

The question is, will this idea hold up to the 150 HP that could potentially be put on this?

Thanks

[PinkertonD]

Horsepower is not going to be the issue, Torque is. You would need to consider the load on the driven sprocket, also consider both shaft diameters. That is probably going to be 2" diameter alone, so a 5" sprocket would have to be machined from solid.

There is a lot more involved than just the tensile limit of the chain. If you give us the details we can provide more appropriate advise.

[zeke]

Looks like you have a factor of safety of about 2. Considering startup impact and fatigue I would use at least 3.5 for a factor of safety on the chain and shaft.

[PinkertonD]

Just thinking, but is this mill currently belt-driven? The constant shock-loading of the drive train might be an issue. Belts would soften that out with slip. Chains may have a shorter life. You are talking almost 1500-ft-lbs at max load and speed. At 5" diameter that's gonna start getting into big numbers. Sometimes things are done for a reason.

[dberg]

Thanks for the help.
Regarding the load on this......it is unknown to me, but will basically be limited by the HP of the tractor. So, the rate of the grain fed into hammer mill will be regulated to allow the tractor to maintain 540 RPM on the PTO.
I'm simply trying to make sure that the drive chain for this is of sufficient size/strength. My info shows a double #60 to be 17000 lbs tensile strength, and if I go up to a double #80 it will be 29000 lbs.

Zeke, if I understand correctly, the safety factor with a double 60 = 2, and you suggest 3.5, then that would basically put it at a requiremetn of 29000 lbs.??
Am I understanding this correctly?

Thanks

[PinkertonD]

Regarding the load on this......it is unknown to me, but will basically be limited by the HP of the tractor.

But that's what I am trying to point out, that there is 1460-ft-lbs force available for the tractor to apply. Using James Watt's (JW) calculation for torque, HP x 5252(a Constant JW derived from 1 theoretical HP) / RPM = ft-lbs. So, 150hp x 5252 / 540rpm = 1460-ft-lbs of torque.

Going down in sprocket size will increase the mechanical advantage the tractor has to turn the smaller sprocket. Try to think of it this way. The tractor has a 12" lever to apply it's 1460-lbs turning force to the shaft, but the sprocket has a 2.5" (5" diameter) lever to pull the chain. That means the mechanical advantage the tractor has is 4.8:1 (12 / 2.5). That in turn means if you load up the mill to the 150hp of the tractor, the chain would need to pull 7008-lbs on the chain.

If the sprocket was 12" radius then the max load the tractor could apply to the chain would be 12/12 = 1 and 1 x 1460 = 1460-lbs.

If the sprocket was 24" radius then the max load the tractor could apply (remember the tractor only has a 12" lever to apply the force) to the chain would be 12/24 = 0.5 and 0.5 x 1460 = 730-lbs at the chain.

If the sprocket was 6" radius then the max load the tractor could apply to the chain would be 12/6 = 2 and 2 x 1460 = 2920-lbs.

I am just trying to indicate who has what mechanical advantage for sizing the chain. The actual sizing is more complicated taking into consideration, tooth load, teeth in engagement, stretch etc. This also ignores maybe 15-percent in mechanical losses and no safety factor yet built in.