Torque Power and Force

[DALOB]

I am trying to work out the torque required at a shaft so I can size a gearbox.
My shaft is connected to a sprocket of 260mm dia.
The force applied to the sprocket due to a chain is 340kN (this is the summation of all of the resistances)
The diameter of the shaft to which I will be attaching a gearbox is 125mm
The chain is running at 0.5m/s (it provide the require material throughput)
The calculated power required is 198kW (based on required F and velocity)

Now I know Torque = F*r. Can I state that the required torque is simply the torque from 340kN * 0.13m = 44kNm and therefore the tangential force at the shaft is 680kN

Thanks,

Jason

 

[chicopee]

We don't want to do your work as there are several other factors to consider so the is a NASA Publication, that's in the public domain, for reference :
1984
National Aeronaulrcs and Space Admrmstratqon
Scientific and Technical Information Branch
Design of Power- Transmitting Shafts

 

[DALOB]

I am sorry chicopee but I don't recall asking for you to do my work. A simple yes or no would have worked quite well.

My main reason for the question was I am trying to get my head around the concept of either varying force or torque based on radius. For instance, if I assume torque is constant then the force required will change (which I am assuming is correct) instead of the force being constant and then the resulting torque will change based on the radius. I may not be very clear in my question, but if my original question is correct then that will clear this up for me.

In my case, I have a known force at the sprocket which results in a torque at the applied radius. Now if my understanding is correct, this will be equal to a larger force required at the shaft diameter. Eg torque required at the shaft diameter is equal to torque produced at the sprocket.

BTW, thank you for the reference, I have downloaded it and while it contains a lot more info than I require right now, it will be useful for future calculations.

Jason

 

[GregLocock]

680 kN is wrong. Draw an FBD

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[DALOB]

Thanks Greg. There was a problem with the interface between the chair and the calculator...702kN

 

[dhengr]

DALOB:
Try thinking of it this way; you have a sprocket radius and a chain force, which, when multiplied gives you the torque. The torque doesn’t really change, except for frictional or other mechanical losses along the way, but the force required to produce the torque changes as a function of the radius at which the force is applied. And yes, there is still something in your chair which is malfunctioning, since (340kN * .13m)/(.125m/2) += 707.2kN.