Acme Lead Screw - Resultant Nut Torque

[aulrich]

I am designing a lead screw assembly with a stationary, vertical screw and a moving nut in a nut carrier. I know how to calculate the torque required to raise and lower the load, but what I can't figure out is how to calculate the resultant torque on the nut (due to friction), ie the reaction torque that wants to spin the nut due to the friction in the threads.

This is important because the nut carrier is guided by some rollers which will have to support this resultant torque.

Anyone have ideas? I've done numerous web searches and no one mentions how to figure this resultant nut torque.

Regards,

Andri

 

[MikeHalloran]

Someday, the screw will get coated with dust/ rust/ crud/ wear particles/ who knows what, the nut will jam on it, and the torque applied to the nut will be limited only by whatever is turning it. Design for that.



Mike Halloran
Pembroke Pines, FL, USA

 

[aulrich]

Mike,

yes, assuming the resultant torque being equal to the torque input to the screw would certainly be a worst case scenario. However, I'd still like to know what the resultant nut torque would be during normal operation. This is because the guide rollers run on long, end supported aluminum extrusions which are going to bend some under load - and of course I'd like to know how much that bending deflection is.

Andri

 

[hydtools]

Maybe this will help:

http://nptel.iitm.ac.in/courses/Web...pur/Machine design1/pdf/module-6 lesson-1.pdf

The nut must resist the applied torque in order for the load to rise.

Ted

 

[aulrich]

Alright, I think I have something figured out. I know that the torque required to raise the load has in a sense two components: The torque to overcome friction in the threads, and the torque to actually raise the load.

I know that the total torque to raise the load is:

(Equation 1)

T_total=(F*d_m)/2 ((μ*secα+tanλ)/(1-μ*secα*tanλ))

where
F = nut load
d_m= mean screw diameter
μ = coeff. of friction
2α = thread angle
λ = lead angle

To calculate the torque to raise the load only (assuming no friction), I simply set μ=0 in the above equation:

T_raise=(F*d_m)/2 (tanλ/1)

(Equation 2)

Therefore, the torque to overcome friction only is the difference between equations 1 and 2.

Here is where I am a bit fuzzy: I assume that the resultant nut torque is then equal to the torque due to friction. This means that if the lead screw system was frictionless, there would not be any torque on the nut.

This kind of makes sense to me, but I believe Ted is saying that the resultant nut torque is actually equal to the total torque to raise the load.

Any comments on my theory?

Thanks,

Andri

 

[hydtools]

The nut must react to all torque input from the screw, less thrust collar or other bearing friction.

If the nut did not react to the screw thread input, friction or frictionless, the load would not move. Imagine that the nut is not attached to a non-rotating structure. The nut would turn with the screw and the load would not move.

For the frictionless case, the nut toque equals the screw torque.

For the friction case, the nut torque equals the screw torque.

Ted

 

[desertfox]

aulrich

I think your on the right track if you want to calculate the torque required just to overcome friction, so first calculate the total torque required to lift the load assumming a positive value for μ then set μ to zero then calculate torque again and the difference between them is the torque required just the overcon-me friction.

 

[Jboggs]

OK, I'm lost. And I've been designing mechanical systems for 35 years. Please check my facts as I understand them:
1. You have a stationary vertical screw.
2. You have a rotating nut supporting a moving payload.
3. That payload is guided vertically by some arrangement of rollers on rails.
3. A motor (mounted on the same frame as the payload) applies torque to the nut.
4. If that torque is enough to overcome the friction between the nut and the screw, the nut rotates and the payload rises.

If I understand your concern correctly, you know that applying that torque to the nut will create a resultant torque (force) on the frame members supporting the guide rollers, and you want to make sure they do not deflect. Is that right?

If so, the torque the frame members see will be the result of the total torque applied to the nut, whether that torque overcomes the nut/screw friction or not.

As a related concern, am I also hearing a question about how to calculate that friction? Do you have access to Machinery's Handbook? It's in there. We don't know what thread form you're using, and that is a significant factor. We also don't know the materials involved, the fits, the cleanliness, ...

As a side comment, I'm curious why you are trying to do such precise calculations when we know there are unknown factors in the real world. You know how much torque is required to lift the load, right? If you have a halfway decent lead screw design, torque from friction should be some fraction of that. My advice would be to double the calculated torque and move on, unless there is something I don't know here.

 

[MintJulep]

Question : how much torque is required on the head of a bolt to prevent it from rotating while tightening the nut?

Answer : all of it.

 

[desertfox]

aulrich
In your last post your assuming that the resultant nut torque is all due to friction but it isn't, your resultant nut torque is made up of:-

resultant nut torque= torque req (for load)+ torque req {fric}
So if you want the load to lift but only apply sufficient torque to lift the load and ignore friction; the load won't be moving;now if you apply torque to lift the load+torque to overcome friction then; the load starts to rise.

http://www.roymech.co.uk/Useful_Tables/Cams_Springs/Power_Screws_1.html

If you draw a freebody diagram of the stationary screw and place the forces acting on it via the nut, then you will have a vertical load W (weight of load} acting through the centre of the screw, plus a tangential force acting about the centre axis of the screw (torque} due purely to friction which is proportional to the weight of the actual load W.
See the third diagram down in the link supplied above.

 

[aulrich]

Thanks to all who have responded.

First, I'd like to address Jboggs' comments. You are correct in your facts 1,2,3 and 5. However, number four should be:

4. A motor applies a torque to one end of the lead screw.

I am interested in knowing the resultant torque on the nut. That torque on the nut will cause a force on the support rails and I want to make sure they do not deflect excessively.

I am not really interested in calculating the torque due to friction (though I have the formulas to do so).

Based on other comments I have received, the consensus seems to be that the resultant torque that the nut exerts on the support rails is equal to the input torque (from the motor) required to raise the payload.

See attached figure.

 

[desertfox]

aulrich

The only torque transferred to the stationary lead screw is the friction torque, the rest is the vertical load through the screw through trying to lift it

 

[desertfox]

The diagram you've posted suggests the leadscrew isn't stationary but is rotating and the earlier posts suggest that the torque is only applied to the nut which now conflicts with you're last post.
My last post was based on the nut rotating only and not the screw.

 

[hydtools]

desertfox, it's all relative. It just depends where the observer is, on the nut or on the screw. Your approach would still be valid.

aulrich, you know everything you need to know.

Ted

 

[desertfox]

Hi hydtools

thankyou for that, I was thinking it changed things but it doesn't

 

[aulrich]

Desertfox,

sorry for the confusion. I realize now that in my first post I noted that the lead screw was stationary - what I should have said is that it is rotating, but axially stationary. The nut, of course, is constrained from rotating, but moves along the screw axially to raise a load.

Ted, so is your previous comment still valid:

The nut must react to all torque input from the screw, less thrust collar or other bearing friction.

If the nut did not react to the screw thread input, friction or frictionless, the load would not move. Imagine that the nut is not attached to a non-rotating structure. The nut would turn with the screw and the load would not move.

For the frictionless case, the nut toque equals the screw torque.

For the friction case, the nut torque equals the screw torque.

Andri

 

[hydtools]

Andri,
Yes

Ted

 

[Jboggs]

So, now that you have revised your original statement about the "stationary" lead screw, the whole thing makes a lot more sense.
And now I see that your bottom line question is this: "how much torque will be transferred to the nut through friction?"
I refer you back to my original comments above. Do you have access to Machinery's Handbook? It can help you calculate the friction associated with different thread forms, which in turn can tell you how much torque will be required to produce useful work. There are MANY other factors to consider also, such as lubrication, environmental contamination, wear, alignment, ...
My recommendation is to build the guide frame work to be stiff enough to absorb all of the torque applied to the screw.