TORQUE QUESTION

[LovinLife]

Hi Everyone,
Would like a little help please.
Have a pc. of 1/2-13 lubed all-thread running thru the middle of a compression spring. As I drive this screw it will bring a threaded block up the screw and compress the spring. When compressed the spring force is 75lbs. Travel will be 6 inches per min. or 78 r.p.m. on the screw. I'm coming up with 3in/lbs req'd. For some reason doesn't look right to me. Driving the screw with a 1.25 p.d. gear. Thanks to all who respond, you have a wonderful forum.

 

[Tmoose]

http://www.holo-krome.com/pdf/techbk30-33.pdf

1/2-13 = unrc (coarse)
1800 lb-inch gets about 15760 lb clamp force, so 75 lbs would require 75/15760 of 1800, or about 8.6 lb-in, depending on assumptions of lubrication, etc.

 

[GregLocock]

Does the spring start uncompressed, and is then compressed up to 75 lbf in 6 inches?



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[LovinLife]

Hi Tmoose,
Not sure this is the figure I'm looking for since my threaded block is not coming up against a stop after which the tension would be induced.
Hi Greg,
The 75lbs. is after the spring is compressed. It will start at about 16lbs. and finish at 75lbs.
Thanks

 

[byrdj]

not sure what you are asking?

is it the torque to compress the spring to the 75 pound force? this would just be a normal bolt torque calc with the very low clamping force.the torque as the spring begins to compress will be lower than final torq-ue at full compression.

or are you looking for the "power" (torque-time) to run the nut the 6".

 

[TheTick]

Lube or not, friction is present and will increase linearly with spring compression.

 

[eromlignod]

I'm assuming that you want to know the minimum torque required for the screw to compress the spring. The maximum load would be at full compression, or 75#.

The no-friction formula for axially-loaded screw torque is

T = F*p / 2*pi

where F is the load and p is the pitch or lead of the screw. In your case the lead is 1/13 = .077" and the load is 75 lbs, which gives you

T = 75 * .077 / 6.28 = .92 lb-in

which isn't much.

You can waste a lot of time getting technical with friction figures, but you can also just triple this figure and still get less than 3 lb-in of torque.

In my experience (21 years) of sizing motors for odd applications, my advice would be: "oversize, oversize, oversize". The increase in cost is minuscule in comparison to undersizing and having to replace everything and start over.

Don
Kansas City