Torque required to move a pipe on driven rollers 1

[jimjnr]

I have a problem with calculating the required torque to be applied to rollers to move a pipe horizonally.

The pipe sits on 5 rollers, three of which are to be driven by gear motors. The radius of contact between the centre of the rollers and the pipe is 93.7 mm, and the mass of the pipe is 3000 kg. The rotation of the rollers needs to be 41 rpm.

Can anyone help?

 

[zekeman]

The torque would depend on the acceleration you need to bring the speed up to the final velocity.
Let's say you want to come to speed in 1 minute and say the roller diameter is 5 cm
The final RPM is 41 so the final linear speed is
41*2pi*5/2=643cm/min=6.43 m/min

From this and the one minute time to get to speed the acceleration is

a=6.4/t=6.4/60sec=0.11m/sec^2 From Newton' Law

F=Ma
F=3000kg*0.11=330Newtons

Now you have to make sure the coefficient of friction is sufficient to supply this force.

let's say that 2 rollers are at 45 degrees then each will get 0.7 of the vertical weight so
0.7*3000*9.8=20,580 N
330/2/20,580=.008 fraction of the normal force the rollers need to transmit
As long as this is less than the coefficient of friction the motion postulated can be sustained.
Friction for these cases are usually in the range of 0.1 and 0.3 so you can increase the acceleration by factor of 10 but you would increase the drive power by that factor.

 

[jimjnr]

Thanks zekeman

So I take it from there it would be a simple case of Torque = F.r

I did get a bit lost when you were talking about 2 rollers at 45 degrees though.

If you are looking at the system side on, the five rollers are horizontal and the length of the pipe sits along the top of them.

 

[zekeman]

Do you mean that the 5 rollers are in the same plane and their axes are at right angles to the pipe axis.
If so, the equations of motion are still the same and
F=mu*N
for each driving wheel (assuming equal contact)
N=3000*9.8/5N=5800 N
And
The total normal force for the 3 drivers is

5800*3=17600
and the fraction of normal force required for the problem I posed
330/17600==.018

Again, as long as this value is less than the friction coefficient you're OK.

Is my picture correct?

And finally, yes F*r is the torque

 

[jimjnr]

"Do you mean that the 5 rollers are in the same plane and their axes are at right angles to the pipe axis."

Yes this is correct.

So this is what I have so far, and please correct me if I am wrong:

Known Variables:
M = 3000 kg
rotational Speed = 4.338 rads/sec
radius of contact = 0.0937 m
number of rollers = 5
number of driven rollers = 3
linear acceleration = 3 m/s^2
coefficient of friction(steel on polyurethane) = 0.6

therefore:
F = M * linear acceleration = 3000 * 3 = 9000 N

Torque = F * radius = 9000 * 0.0937 = 843.3 Nm

Torque per driven roller = 843.3 / 3 = 281.1 Nm

Power per motor = torque * rotational speed = 281.1 * 4.338 = 1.22 kW

from what I can gather the coefficient of friction would only be acting on the driven rolers therefore:

0.2 * 3000 * 9.81 = 5886 N per roller

9000 / 3 / 5886 = 0.51 fraction of the normal force (<0.6)

or would this only be affected by the force acting on each roller giving:

9000 / 5 / 5886 = 0.31

I then just need to check that I dont need to add on any torque to compensate for load friction in the bearings and moment of inertia of the rollers?

Thanks again

 

[zekeman]

"9000 / 3 / 5886 = 0.51 fraction of the normal force (<0.6)"

That one is correct, but the margin ,6 vs .51 is problematical. You may need to add force to the pipe or reduce the acceleration requirements, since you can't be assured that the drivers are taking 3/5 of the pipe weight as the normal force.

 

[jimjnr]

Thats a very good point. I will have to check the positioning of the pipe at start of the operation. If memory serves me correct, the pipe starts off with a longer section protruding further over the drive end.

If you imagine the rollers as 3 x driven then 2 x non driven, and the pipe will be travelling in the direction towards the driven rollers.

Thanks for all your help, it has been invaluable. How quickly the fundamentals start to slip your memory from uni days when you dont use them constantly.