Force resulting from two opposing torsion springs

[ensoll]

Imagine you could put a coil torsion spring in place of each of your shoulders. One leg of each spring would act against your collarbone (fixed) and the other leg of each spring would act against your adjacent arms (moving). The springs would be trying to keep your hands together, effectively turning your arms into a clamp. Held in your hands would be a meter to read the force being felt by the two springs.

The springs need to apply 1.0 Lbf on the meter. Logically, you might assume that each spring should be applying 1/2 Lbf to the meter. If your arms were 2 ft long, you would need a spring torque of 1 ft-lb.

Unfortunately, this does not seem to be the case in reality. For some reason, in my tests, a much LOWER force is read on the meter, somewhere closer to 0.6 Lbf. Can anyone offer a reason for this? What should be done to correct it?

I suspect what is happening is that the two springs "combine coils" to effectively represent the output force of one spring with the total number of coils as the two springs combined. For example, if each of the two springs have 10 coils, the "effective" spring has 20 coils. As we know, when the coil count goes up, the strength of the spring goes down, with all other parameters being equal.

One last point: by removing one of the springs from the system, the other is not able to act on the meter. The collarbone is not a "fixed plane" but rather only fixed to the other spring arm (i.e. not attached to your spine)

Any thoughts?

 

[desertfox]

Hi ensoll

What angle are the moving legs of each spring relative to the clamp arm you may need to resolve this force into horizontal and vertical components, which means that not all the force generated at the end of the spring arm is going into clamping your component.


regards

desertfox

 

[ivymike]

re angle - wouldn't that have the opposite effect to the one observed (hypotenuse of a right triangle is longer than the legs)?

if the system is in static equilibrium, with each arm applying a force of 1/2 lbf along the direction of motion of the scale, with the respective arms acting in opposite directions, the scale should read 1/2lbf, shouldn't it? (replace either arm with "ground" and what does the scale read?)

now if you take your 1/2 lbf and account for the angularity mentioned above, maybe you can get to 0.6lbf...?

regarding the theory of "combining coils," if you have two springs of constant stiffness k (torsional or otherwise), and you have a fixed plate between them, and you apply the same magnitude opposing forces F (or torsion) to the free end of each spring, then each spring should deflect by x=F/k (for a total displacement, end-to-end, of 2x). If you took out the plate, the combined stiffness of the two springs would be 1/(1/k + 1/k), or 1/2k. The stiffness would be halved... now apply the force F to each end, and you get a displacement of 2x. Same answer.

 

[israelkk]

Can you provide more info about your spring actual dimensions, deflections and forces, material, number of coils, arm length where the force reaction apply, springs arrangement with respect to the shaft axis inside them, etc. A sketch will be helpful.

http://www.webspawner.com/users/israelkk/

 

[ivymike]

It sounds to me like he's decribing something along the lines of a big "clothes pin" or spring clamp. The OP has made the mistake of thinking that pushing against each side of a scale with 0.5lbf should make the scale read 1lbf (it should read 0.5lbf). In his experiment, he was expecting to see 1.0lbf (as above) and he saw 0.6lbf. I suggest that he should have been expecting to see 0.5lbf. Desertfox's angularity suggestion could be one thing contributing to the difference between an expected 0.5lbf and an observed 0.6lbf.

 

[ensoll]

I will work on a sketch and the actual spring configuration as soon as I can. This problem has me perplexed and I greatly appreciate all of your discussion.

Chris P

 

[ensoll]

I've worked up a sketch, but I'm not sure it's necessary. Today, I used two springs which were supposed to give 3.0 " lbs of torque at 90 degrees deflection. Working together, I expected to get a reading of about 1.7 Lbs f on a force reading device using strain guages. The arm length of the clamping system is 3.5" long per side.

3 "lbs / 3.5" gives ~.86 lbs force PER SPRING.

Strangely, with both springs working against each other, the force guage only read ~.8-.9 lbs f. Can anyone explain this phenomenon? It looks like the strength of the system is limited to the strength of one spring. I was expecting the force guage to indicate .86 lbs from one side and .86 lbs from the other, totalling ~1.7 lbs f on the meter.

 

[israelkk]

If you hold a plate between both your hands and push with both hands at a 10 lbf at each hand the force on the plate is 10lb on each side of the plate not 20. The plate is in equilibrium because 10 lb from the right is balanced by the 10 lb from the left but the total force is 10 lb.

If you place the plate on the floor and press from the top with 10 lb force the floor reacts from beneath with an equal and opposite force of 10 lb.

The second spring in your system replaces the floor.

Therefore, if you need ~1.7 lb then you have to find/design a spring with 6 lb-in assuming the arm is still 3.5" long.

http://www.webspawner.com/users/israelkk/

 

[ivymike]

maybe the third time will be the charm...

 

[ensoll]

I feel stupid. I can't believe I forgot about statics 101. What got me off on the wrong track was that I was getting .6-.7 Lbsf at one point with springs that were supposed to yield .5 each. Perhaps something was off with the spring deflection angle. If I had gotten .4-.5, I would have known right away. I thought I was getting friction and flexure losses. I became thoroughly confused when, as I increased the spring rate, the force only raised insignificantly.

~thanks and sorry to have wasted all of your mental beauty