Spindle torque, belt driven

[aldumoul]

Wondering what the torque would be in a 2.5” dia. Spindle. Spindle speed=9,000 rpm. Spindle is driven by the motor, pulley ratio is 2.57:1. Shaft weighs 66 lbs.

Motor torque = 40.2 lb-ft
Motor hp = 15 hp
Motor speed = 3500 rpm
Motor shaft/rotor inertia = 0.474 LB-FT^2

I calculated the max allowable torque of the shaft, no bending stress. So looking to see if I’d be beyond that. Probably gave more info than needed, but want to make sure I’m not missing anything.

Thank you!

 

[MintJulep]

Well, it can't be more than the motor torque/gear ratio, can it?

Unless you also have a brake at the motor.

The maximum motor torque is likely different from what you listed, which is likely the full-load torque. Assuming that you have an AC induction motor.

https://www.engineeringtoolbox.com/electrical-motors-torques-d_651.html

 

[aldumoul]

You are correct used the full load torque. Breakdown torque is 103 lb-ft.

Sounds like I should’ve went with my original calc.. 2.57*motor torque of 103lb-ft

 

[EdStainless]

How about 103/2.57
Sped goes up, torque goes down.
You can't create HP out of thin air.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[BrianE22]

I get 15.6 (40.2/2.57).

 

[aldumoul]

Yes, 103/2.57. Realized my mistake after I submitted the post

Thanks!

 

[dvd]

HP = Torque (lb-ft) x RPM / 5252

 

[MintJulep]

So, relatively small torque for a relatively large shaft.

Stiffness matters.
Distance between supports, straightness and balance matter at 9,000 rpm.