Compress Spring question 3

[byrdj]

I have search (googled) but can not find an answer.

For a coil wire compression spring, if one end is free to rotate, does it rotate as it is compressed and how to calculate rotation and resulting torsional force.

the concern is due to anti rotation guide damage on spring plate for large valve springs (ie >1" diameter wire)

Thanks

 

[MikeHalloran]

Yes, it's going to rotate, or try real hard to do so, as you have apparently found out the hard way.

Somebody will chime in with an equation pretty soon, I'm sure.



Mike Halloran
Pembroke Pines, FL, USA

 

[GregLocock]

Roughly the axial torsional stiffness of a coil spring is given by

(E*d^4)/(128*n*R)

so if you can work out how much the thing twists you can work out the Torque to untwist it.

I don't know how much a spring twists for a given axial load though .

Cheers

Greg Locock

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

 

[israelkk]

The angular deflection (unwinding) of the ends for compression spring due to axial loading is given in page 249 of Mechanical Springs, A. M. Wahl 2nd Edition 1963. Therefore, the maximum torque will be the torque to prevent it. For this you need to treat the spring as a torsion spring and calculate the torque that will create/negate this deflection. However, the axial force, coefficient of friction and spring diameter create a moment that may ne smaller than the maximum moment.

 

[byrdj]

Thanks Greg

That will at least allow me to quantify the as found forced angular offset prior to disassembly even though I still can not calculated the expected angular deflection with respect to axial.

 

[byrdj]

I have two springs
wire diameter d"= 1.8 & 1.3
active coils n= 7.4 & 9.8
mean raidus R"= 8.8 & 6.5
assuming Young's E=30E6

It looks like if my guided plate is twisted 1" on the spring OD, that will have a load of 34,000 pounds on guide roller. (from wear, I'm guessing actual is about 1/4" on the OD)

I still would like to be able to calculated how much I twist from free to installed, then installed to max compression.

 

[israelkk]

byrdj

The unwinding calculation is not a simple formula. It depends on the pitch angle, free length and deflection. You have to use the basic mathmatics that was used to develop the known spring deflection formula.

Anyway, to calculate the unwinding of your spring/s according to Wahl's formulations the spring free length and the deflection is needed. I used the 34000 lb to calculate the deflection for the 1.8" wire diameter spring (it is ~9") however, the free length is missing. Can you give more detailed data concerning each spring?

 

[byrdj]

Here is all the information from the bid/acceptance drawing

wire diameter d"= 1.8 & 1.3
active coils n= 7.4 & 9.8
mean raidus R"= 8.8 & 6.5
free lenght L"= 44.3 & 43.9
lenght installed L1"= 31 & 30.6
lenght max compressed L2"= 18.8 & 18.4
load installed F1= 4339 & 2548
load max F2= 8344 & 4900
gradiant k= 327 & 192
wind right & left


thus axial deflections would be 13.3 & 25.5

the 34,000 pound force I spoke about would be for an angular deflection of the guided plate of 6.5 degrees (1"on the spring OD). From survalance camera, I estimated wear in guide at 1/4" ( unaccesable enviroment while in operation)

If you would be so kind as to provide the expected twisted from Wahl's equations (I have not been able to locate copy), It would provide me with an insight and would be GREATLY appricated.

However, I would not be able to "present" the information due to the involvement of variuos industry oversight orginations with out me performing the calcs.

My intuiation is a angular preload at assembly, not something from design.

I have made an interesting observation in these latter years. while my job has been to set the controls, there is an every increasing need to venture into the other areas of engineering. Last week it was to determine the cooling water requirement for an hydrulic pumping unit for an abnormal operation condition

 

[CoryPad]

To purchase your own copy of Wahl, go to SMI:

http://www.smihq.org/public/pstorep.html#sftyposter

 

[israelkk]

byrdj

The calculated unwinding angles for the 8.8" / 6.5" radiuses respectively are:
For the first load point the unwinding angles are 1.729 deg (0.585" periperally) / 2.987 deg (0.678" periperally)
For the second load point the unwinding angles are 4.397 deg (1.489" periperally) / 7.544 deg (1.712" periperally).
The periperally travel is based on the mean diameter of the spring.

Therefore, the periperally travel is 1.712"-0.678" = 1.034" for the 1.3" spring and the periperally travel is 1.489"-0.585" = 0.904" for the 1.8" spring.

 

[byrdj]

THANKS,
I owe you big time,
Joe

The amounts were a little more than I expected. IF and WHEN I get a chance to perform actual measurement, I'll post. The Plans are to inspect and repair within the next 4 weeks. The OEM is purposing a beefer guide, even though this wear had not been observed in the first 40 years of operation

 

[volpegrigia]

I read the question late. In automotive suspensions (i.e. McPherson struts), designers tend (or at least did so it the 80's) to include thrust bearings to accommodate the spring rotation during compression/extension to improve the NVH of the vehicle itself. These bearing cartridges were sitting between the coil and the strut tower and allowed the spring rotation without any torsion load transmitted to the tower. One end of the srping can be pinned and the other one is free to rotate, thus no friction induced moment may results due to the spring motion while compressing.
Perhaps this type of solution will eliminate any possible issues you may face.