How is this related to Calculating Shaft Torque?

In summary, PF, it seems that you are designing a cart that needs to move a load of 40,000 lbs horizontally for a distance of 70 ft. You have determined that the cart will need to move at a rate of 1 ft/s and the diameter of the cart wheels is 15 inches. You have also considered the coefficient of friction between steel-on-steel to be 0.8 and the diameter of the shaft to be 2 3/16 inches. After calculating the Full Load Torque of a 2.5 HP motor, you have determined that the driving force at each wheel will be 1375.77 lbs. You have also explored the possibility of using a 3 HP or 5
  • #1
vwmeche94
10
0
PF,

I need to design a cart that will move (horizontally/linearly) a load of ~40,000 lbs including its own weight (about 6500 lbs) at a total distance of 70 ft on a set of rails.

Known data:
1. Cart will need to move at a distance of 1 ft/s
2. Diameter of the cart wheels (4) is 15 in.
3. Coefficient of friction, COF (steel-on-steel) = 0.8
4. The shaft diameter (4140 steel) is 2 3/16"
5. Electric motor of 2.5 HP

I have calculated the diameter to yield 15.27 rpm and FLT (Full Load Torque) of a 2.5 HP motor to be 859.85 lb-ft using that rpm.

Based on an equation shared with me from a Machine Design Book,

W (weight) x mu (COF) / r (radius) = 40,000 lb x 0.8 ÷ 7.5 in = 4266 lb/in - load at each wheel.

What I need to determine is the shaft torque under load and the type of gearing (gear box) I should use to accelerate to and maintain a speed of 1 ft/s with a 40,000 lb load. I also need to determine if the 2.5 HP motor is adequate enough to handle repeated moving of the load.

Thank you in advance,
 
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  • #2
At this point, I've got the following:

Full Load Torque, T=5252 x 2.5 HP / 15.27 rpm = 859.85 lb-ft

This calculated torque creates a driving force of [859.85 lb-ft / (7.5/12) =] 1375.77 lbs of force at the wheels (real-wheel driven when cart fully loaded). If I select a 3 HP electric motor, the driving force will be 1650.92 lbs at the wheels.

Load at each wheel was figured up to be 4,266 lb (40,000lb x 0.8 ÷ 7.5 in)

So now my questions with a gearbox ratio of 118:1 chain drive between the gearbox and wheel is 1:1, are

  • Is 1651 lbs of force (using 3 HP motor) at the rear two cart wheels more than enough to move the cart and give me 60 ft/min travel rate?

  • Would 1375 lbs (using 2.5 HP motor) be enough to move the cart at the rate of 60 ft/min?

  • Would 1100 lbs (using 2 HP motor) be enough to move the cart at the rate of 60 ft/min?

Any advice, insight, or opinion would be greatly appreciated.
 
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  • #3
See attached sketch which represents total load of 40,000 lbs, including the parts and cart itself.
 

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  • #4
All the basic principles are on http://hpwizard.com/car-performance.html" .

You are interested in http://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html" (and not friction coefficient).

Drag shouldn't be a factor at such a low speed, but your acceleration requirement might be important (the time taken to go from 0 to 60 ft/min). The power needed for accelerating (especially such a large weight) is usually very large compared to constant speed travel.
 
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  • #5
Thank you jack action. The force required to get that size of a load moving is a concern for me; I am still working on this and have decided to increase the size of the electric motor from 3 to 5 HP (1750 RPM, 15.4-14.2/7.1 AMPS) as well as the shaft diameter. While reading through my Machinery Handbook, I stumbled across the shaft equation

D=4.6 * 4th sqrt (P/N), where P is power in kw (assumed HP) and N the RPM's. Plugging in the value of 5 HP and getting a resultant diameter of 3.23" got me to thinking seriously about the shaft snapping on me while trying to move a full load of 40,000 lbs.

Any additional insight appreciated.
 
  • #6
I'm no expert on shaft design but, if I understood correctly what you wrote, it seems you are using an equation to find the drive shaft diameter (although the ones I've seen are elevated to 3rd power not the 4th; Was that a typo on your part or are you using another equation?)

But, what you are designing is not a drive shaft, it's an axle. The difference is that you have bending moments due to the weight supported by the axle in addition of the torsion due to the power transmission. In your case, I'm sure it will make a big difference.

I think this type of equation is more relevant:

[PLAIN]http://www.roymech.co.uk/images24/shaft_17.gif

More details on http://www.roymech.co.uk/Useful_Tables/Drive/Shaft_design.html" .

Again, I have not study your case in depth.
 
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  • #7
The worst case steel-on-steel rolling resistance coefficient is about 0.0025, or 0.25%. Thus the force to push a 40,000 lb cart on a level straight track is 0.0025 x 40,000 lbs = 100 lbs. To push it at 1 foot per second will require 100 foot-pounds per second, or 1/5 HP.

Let's do this again in mks. The power required is force times velocity. 18,200 kg x 9.81 m/sec2 x 0.0025 x 0.3 meters per second = 134 watts (about 1/5 HP).

The required axle torque is

T = 134 watts x 60/(2 π 15.3 RPM)= 84 Newton-meters.

The kinetic energy of the moving cart at full speed is ½mv2 = ½ x 18,200 kg x (0.3m/sec)2 = 820 joules. To accelerate to full speed in 5 seconds will require about another 164 watts, or about another 1/4 HP.

Your biggest problem will be accelerating and decelerating, so an induction motor is not a good choice. Induction motors should not be used at other than their design RPM. You should use a dc traction motor, possibly brushless.

Bob S
 
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  • #8
jack action said:
I'm no expert on shaft design but, if I understood correctly what you wrote, it seems you are using an equation to find the drive shaft diameter (although the ones I've seen are elevated to 3rd power not the 4th; Was that a typo on your part or are you using another equation?)

But, what you are designing is not a drive shaft, it's an axle. The difference is that you have bending moments due to the weight supported by the axle in addition of the torsion due to the power transmission. In your case, I'm sure it will make a big difference.

I think this type of equation is more relevant:

[PLAIN]http://www.roymech.co.uk/images24/shaft_17.gif

More details on http://www.roymech.co.uk/Useful_Tables/Drive/Shaft_design.html" .

Again, I have not study your case in depth.

Yes, I am using an equation found on pp. 277 of the 25th editiong Machinery's Handbook which I thought was appropriate.
 
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  • #9
Bob S said:
The worst case steel-on-steel rolling resistance coefficient is about 0.0025, or 0.25%. Thus the force to push a 40,000 lb cart on a level straight track is 0.0025 x 40,000 lbs = 100 lbs. To push it at 1 foot per second will require 100 foot-pounds per second, or 1/5 HP.

Let's do this again in mks. The power required is force times velocity. 18,200 kg x 9.81 m/sec2 x 0.0025 x 0.3 meters per second = 134 watts (about 1/5 HP).

The required axle torque is

T = 134 watts x 60/(2 π 15.3 RPM)= 84 Newton-meters.

The kinetic energy of the moving cart at full speed is ½mv2 = ½ x 18,200 kg x (0.3m/sec)2 = 820 joules. To accelerate to full speed in 5 seconds will require about another 164 watts, or about another 1/4 HP.

Your biggest problem will be accelerating and decelerating, so an induction motor is not a good choice. Induction motors should not be used at other than their design RPM. You should use a dc traction motor, possibly brushless.

Bob S

Bob, your explanation more closely correlates to what I seeing in the field wrt 2 - 3 HP motors moving heavy loads; after reading your post, I researched a little more to see where the Cf I've been using is much to large and none applicable in this case since it is a rolling friction and not static (sliding) friction I'm dealing with.
 

Related to How is this related to Calculating Shaft Torque?

1. What is shaft torque and why is it important?

Shaft torque is the twisting force applied to a rotating shaft. It is important because it helps determine the power and efficiency of a machine or system.

2. How do you calculate shaft torque?

Shaft torque can be calculated by multiplying the force applied to the shaft by the distance from the center of rotation to the point where the force is applied.

3. What units are used to measure shaft torque?

Shaft torque is typically measured in units of newton-meters (Nm) or pound-feet (lb-ft).

4. What factors affect the amount of shaft torque required?

The amount of shaft torque required depends on factors such as the speed of rotation, the weight of the load being moved, and the efficiency of the system.

5. How can I optimize shaft torque for my machine or system?

To optimize shaft torque, you can adjust the gear ratio, increase the power input, or reduce friction and resistance within the system. It is also important to regularly maintain and lubricate the shaft to ensure smooth operation.

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