Anti Backlash Nuts: Calculating Force & Backlash

[Mech_LS24]

TL;DR Summary

Forces with FBD diagram. I want to know how to calculate the backlash if the a force applied on the spindle is known

Hi,

On the internet I have seen "anti backlash nuts" Anti backlash nut example (see image), I searched on the forum but couldn't find any related topics. But it has made me interested how this should be considered as there is (in my opinion) another force namely, the Force occurred due to the spring (Fs). I made a FBD of the backlash nut connected with a spindle and a flange? How can be calculated how strong this spring should be if a force of 1000 N is applied to the spindle which should lead to a minimum backlash? How to determine the backlash after calculating the force on the spring?

 

[Dullard]

'Backlash' is the result of the force on the screw (1000N, in your example) changing polarity. Without the anti-backlash arrangement, the screw will (independent of rotation) move an amount determined by the 'slop' in the fit between the screw and nut. The anti-backlash arrangement keeps the net force on the screw from changing.

When the force is 'down' (as shown in your diagram), the screw rides against the lower edge of the threads in the fixed nut. The spring-nut isn't doing anything useful. When the force changes to 'up,' the spring force serves to keep the screw against the lower edge of the threads. I don't think that the amount of spring force may be correlated to the amount of backlash - the spring force is greater than the applied force, or it isn't. The down side of anti-backlash nuts is running friction (due the the 'extra' force required) and associated wear.

 

[jrmichler]

To build on what @Dullard said:

Anti backlash nuts are normally only used with ball screws. It's easier to understand them by sketching them with plain threads (easier to sketch). The sketch should show axial clearance between the nut and the screw, with the clearance on the correct side of each nut.

Make two such sketches, one with the force on the screw less than the spring force, and one with the screw force greater than the spring force. Then make two more sketches with the screw force in the opposite direction, for a total for four sketches. You should then fully understand how these nuts work, and how to calculate the forces.

 

[Mech_LS24]

Thanks.

I have drawn the sketches, see attached and below. I think sketch 3 or 4 isn't correct, maybe in sketch 3 the force from the spring creates a gap in the opposite direction, but doubting about that because this upper nut is attached to a 'fixed' part...

As well, I tried the calculation (see below). I assumed a wave spring with an force of 20 N, applied on both nuts. I looks like the spring doens't have any meaning in the calculation, because it rules out each other. This corresponds what @Dullard said: "The anti-backlash arrangement keeps the net force on the screw from changing". Is this the case for backlash nuts with lead screws?

 

[jrmichler]

Figure 1 is correct. The fixed nut threads are carrying the load, and the floating nut is pushed away by the spring.

Figure 2 is wrong. The fixed nut threads are correctly shown as carrying the load. The floating nut is pushed away by the spring. Since the fixed nut is carrying the load, the only load on the floating nut is from the spring.

Figure 3 is wrong. The spring force is greater than the load force, so the floating nut is carrying the load, and the screw is pushed down against the fixed nut by the spring. The screw has not shifted within the fixed nut, so no backlash.

Figure 4 is wrong. The screw force is greater than the spring force, so the screw is correctly shown shifted upwards in the fixed nut clearance. The spring pushes the floating nut down.

Don't worry about calculations until the four diagrams are correct. And good job on the clearly drawn diagrams.

EDIT: A simpler way to look at it is as follows. The screw has two forces on it, the external force and the force from the spring through the floating nut.

When the external force is downward, the spring force adds to it. The screw is seated downward on the fixed nut regardless of the magnitude of the spring force.

When the external force is upward, the spring force on the screw through the floating nut is still downward. The highest force controls the position of the screw in the fixed nut.

In all four cases, the only force on the floating nut is the spring force, therefore the floating nut never changes position relative to the screw.

 

[Mech_LS24]

@jrmichler thank you for the clear explanation!

I have edited the sketches and I quite confident I do understand the four situations now. See my sketches below:

For the calculation, I still feels a bit strange to me. Let's take situation 3, the force of the spring (Fs) has to be larger as the axial load force (F). Fs acts on the 'floating' nut downwards, F acts in the fixed nut which forces the screw downwards in the fixed nut due to larger Fs force. This means the should be no backlash, this still keeps me confused that the forces of the spring should always be higher as the axial load force (F)...

 

[jrmichler]

NOW you have it. To better understand the calculations, do a free body diagram (FBD) of the screw. There is a force F, a spring force Fs (applied through the floating nut), and the difference between those two forces is the force on the fixed nut. Remember that the forces must sum to zero, so the force on the fixed nut is the difference between the external applied force and the spring force through the floating nut.

Do this for all four cases. By then, it should all make sense and you should not be confused.

 

[Mech_LS24]

I made the FBD for the 1st situation (I try to finish the other this night). EDIT: added situation 2 & 3. EDIT: added situation 4

As well I added the Sum of Forces in the Y-direction. Looks like the Fs doesn't affect the axial load because the Fr handles this force. Right?

 

[jrmichler]

Check your arrows. Fspring is always pushing the floating nut away from the fixed nut (downward in your diagrams).

Force F is given, Force Fs is also given, so Force Fr, the resultant force on the fixed nut, is calculated from sum of forces equals zero. Solve that equation for Fr (Fr = ...).

 

[Mech_LS24]

Check your arrows. Fspring is always pushing the floating nut away from the fixed nut (downward in your diagrams).

Of course, should have known that. When the F < Fs, the spring 'pushes' the floating nut downwards. I don't know what I was thinking there last night.

Now I take the following input:

• F = 225 N
• Fs = 375 N

For situation 1:
Fr = F+Fs
Fr = 225 + 375 = 600 N. There is no Backlash if the load F reverse. Because Fs is larger as F. Sketch 1 & 3 confirms this, as the red gaps (backlash) stays in the same area.

I think I can conclude that if zero-backlash is demanded, the force from the spring should always be higher as the load (F) applied on the screw. Correct?

On the other hand, a force from the spring which is higher as the axial load force results in a very high friction coefficient?

Or is the pressure between the screw and nut the limiting factor here? Due to the high amounts of forces applied on the fixed nut.

UPDATED SKETCHES:

 

[jrmichler]

I think I can conclude that if zero-backlash is demanded, the force from the spring should always be higher as than the load (F) applied on the screw. Correct?

Fixed that for you.

On the other hand, a force from the spring which is higher as the axial load force results in a very high friction coefficient?

No. The friction coefficient stays the same. The torque to turn the screw increases.

 

[Mech_LS24]

Thanks.

This lead to a restriction for this backlash nut right? Assume I want to apply a load of 750 N (75 kg) which isn't that much in my opinion. A spring with a minimum of force of 750 N is needed, this is a heavy spring!

Solve that equation for Fr (Fr = ...).

But @jrmichler , why you are referring calculating the Fr load? I think the engineer should be more interested in the Force needed for the spring or the minimum load (F) which the spindle can handle?

Or is the pressure between the screw and nut the limiting factor here? Due to the high amounts of forces applied on the fixed nut.

I think this should be a new topic? Brings me to the next interesting step :)

 

[Mech_LS24]

This lead to a restriction for this backlash nut right? Assume I want to apply a load of 750 N (75 kg) which isn't that much in my opinion. A spring with a minimum of force of 750 N is needed, this is a heavy spring!

But @jrmichler , why you are referring calculating the Fr load? I think the engineer should be more interested in the Force needed for the spring or the minimum load (F) which the spindle can handle?

I can assume both above quotes are correct?

 

[Mech_LS24]

@jrmichler , which examples, documentation or search hits (e.g. Google) would you suggest for calculating the pressure between the flanges of the threads? I found the term 'screw stresses', is this the case?

Thanks in advance.

 

[jrmichler]

There is more to this than finding a simple equation. Your best approach is to get a book on machine design. Any book that is used for a junior level mechanical engineering course in machine design will work. Here are links to two such books:

https://www.amazon.com/dp/9813151005/?tag=pfamazon01-20

https://www.amazon.com/dp/1119723604/?tag=pfamazon01-20