How to determine crane drive torque

[jimsto]

I have a bridge crane endtruck drive which has failed. I am looking at the assembly and trying to determine how best to approximate the torque necessary to move the assembly. I was searching for a reference to correlate the crane load, wheel diameter and torque. Here is what I did...

Given:
Endtruck wheel diameter which has an integral gear: 5.25" diameter

Driving pinion diameter which is mounted on the ouput shaft of a worm gear reducer: 1.25" diameter

Total weight of load, bridge and endtruck: 17,250 pounds

Ramp up time: 6 seconds

Full speed: 90 feet per minute

Coefficient of rolling resistance out of Machinery's Handbook for iron on iron: 0.02

My Calcs:
Determine rolling resistance = (total wt. x Coefficient of rolling resistance) / radius of wheel = (17,250 x 0.02) / (5.25/2) = 131.429 lbs

Determine wheel speed = bridge speed / ((Pi x D))/12) = 90 / ((3.14 x 5.25)/12) = 65.481 RPM

Determine pinion speed = gear ratio x wheel speed = (5.25/1.25) x 65.481 = 275.02 RPM

Determine angular velocity = 275.02 Rev/Min x Min/60 sec x 2Pi Rad/Rev = 28.8 Rad/Sec

Determine angular acceleration = 28.8 Rad/Sec / 6 Sec = 4.8

Determine Torque = W/G x R^2 x angular acceleration = 131.429 lbs x (1.25/2)^2 x 4.8 = 426.429 in lbs

I'm pretty rusty with some of my mechincal engineering and would appreciate any help you can give me. Thanks.

 

[Neilmo]

Hi Jimsto,

I tackled the problem from a slightly different route and get this -

Rolling Force = W*f/r = 17250*0.02*2/5.25 = 131.4lb

Acceleration = v/t = 90/(60*6) = 0.25ft/sec^2

Acceleration Force = W*a/g = 17250*0.25/32.2 = 133.9lb

Total Force = 131.4 + 133.9 = 265.3lb

Power required = Force * Speed / 5250 = 265.3*90/5250 = 4.6HP

Personally, I would allow for a rolling resistance of slightly more than the `Machinery's Handbook` value of 0.02 to account for the initial static friction of the system.

Regards,
Neilmo

 

[Cranedoctor]

If a crane shaft has failed other things have to be checked.
1) Can the motor go from one speed to the other immediately on the controls.
2)Can the crane hit the stops
3)Can the crane be reveresed
4) Can cranes collide

Neilmo has the right idea and you can more or less work it back through the gearing

I havve formulae but may only add to confusion. If you use SEW drives they have a very nice computor disc with calcs on if you can prise one from them. Other gearbox/motor suppliers must have the same.

Cranedoctor

 

[Neilmo]

Jimsto,

One thing I forgot to add. The 4.6HP I calculated is the power required at the wheels. If you are coming to the wheels via a gearbox, you must allow for its efficiency and consequently this will `up` the actual power used.

Neilmo