Force in lead screw to prevent rotation

[MGZmechanical]

Designing a lead screw. Usually we have a very strong guide that prevents the screw from rotating. But now we have a case with a weak guide and we were thinking about the necessary force to prevent it from rotating.

My assumption is: we need to lift 1000 kg with a 24 mm screw --> Necesary torque (aprox) =25 Nm
The force to prevent rotation (if friction factor stem-nut = 0,15) aprox 0,15·1000 =150 kg (or 4 N·m aplied at stem)

Am I the right way or I am over or under stimating the neccesary force?
Best regrads

 

[drawoh]

ManoloGalarraga,

Are you talking about a thread with a lead angle that prevents rotation under load, or something else?

--
JHG

 

[MintJulep]

Necesary torque (aprox) =25 Nm

 

[GregLocock]

Yeah, put MJ's reasonable haton, do you really think you could stop a tonne from moving with the grip of two fingers?

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[gruntguru]

Your calculation for back-driving force should include the lead angle (or pitch and dia) of the screw thread. (If not a square thread section, it should also include the thread angle)

je suis charlie

 

[tbuelna]

Also consider that there is a difference between static (breakaway) and dynamic (sliding) frictions. If you are relying on friction in your lead screw thread contacts to maintain position of your load, it would be prudent to use a very conservative static friction coefficient in your calculations. If there is the possibility of any lubricant being present on the lead screw threads, a static friction coefficient of 0.15 is probably too optimistic.

 

[MGZmechanical]

We have a nut that turns and pulls a stem upside. The stem is according DIN 103 and has not back drive effect.
The problem is that when the stems goes up it also has a tendency to rotate (is a secondary effect I guess due to friction). As usually the load has a strong guide we don't care about it. But we have a case were we'll have to hold this tendency to rotation with a pin. And now we're trying to guess what pin to use.
Hope this helps.

 

[MintJulep]

Necesary torque (aprox) =25 Nm

 

[MGZmechanical]

Sorry, the real figures are not those. Those were the numbers I made at home to explain the problem. The real one is a motor with 1000 Nm and a vertical lift of 185 kN (18500 kg) with a 60 mm stem.

Now you can see my problem. If the residual torque is 15% I have 150 Nm which is a torque that needs some figures to be made.

Best regards,