Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

trivial question? 1

Status
Not open for further replies.

gio1

Automotive
Jun 28, 2003
83

Hello

I have meshed a coil spring (with hexaedral elements) and loaded it in compression (prescribed displacement of one end, with the other fixed)

Looking at the results (VM) I am somewhat suprised to observe that the inner side of the coil is more stressed than the outside.

Is the coil spring not just a torsion spring? If so the stress in any section should be perfectly simmetrycal about its centre (neutral axis) right?

Am I missing something or is the solver giving wrong results?

Gio1
 
Replies continue below

Recommended for you

As a compression spring is compressed, the OD gets a little larger (the wire tends to unwind). Unless you are getting really wild numbers, I would say that the FEA is giving you a good picture of what is going on.
 

The difference I see between inner and outer side is not huge, but visible.

How does the increase in OD affect the stress? Would each section still not be subject to pure torsion?
 


No need to get strained rb1957! I'm going to give you a star for a very valuable post!
 
The difference is well described in any basic spring design textbook. Look for a description of the "curvature correction factor". Essentially, the fact that the spring is a helix (or that any segment of the spring is a curved bar) means that some bending stresses are formed, which cause the peak stresses at the inside of the coil to be higher than at the o.d., the difference increases as the ratio of spring diameter to wire diameter decreases. A formula for the correction factor to the torsion stress is

K = (4c-1)/(4c-4) + 0.615/c
where c = coil dia. / wire dia. = D/d

The torsion stress is then given by

Tau = 8PDK/(pi d3)

P = applied compressive load.

The above is an approximate solution, and generally valid when c>3.

Another effect is that the pitch of the coil changes as the spring deflects, which as cowski points out, means that the o.d. of the coil increases. More complex theory pertains.

See Wahl, Mechanical Springs, Spring Mfgr's. Institute and McGraw-Hill, chapter 5 and chapter 19. Or look for other texts at the SMI website.

 
Is this a geometrically linear model?

I know I should check before opening my big mouth, but, is it actually valid to neglect bending entirely in a coil spring?

Cheers

Greg Locock

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

Thanks for the formulas - I did not know this effect

The model is geometrically non-linear and includes self-contact between the coils - it's part of an assembly which I am solving with Abaqus/Explicit.

The FE model has a spring rate about 30% lower than that from the spring spec. sheet - which I find strange.
Even increasing artificially the Young's modulus by the amount needed to compensate for the slight loss of cross section area (due to the modelling with hexas the Xsection is a 8 side polygon) does not match the theoretical rate (from spring formulas).
 
Greg, is that question related to my post or the OP? The short answer is no, you can't ignore bending stresses, thus the correction term.

The correction factor given is to correct the shear stress derived from a pure torsion analysis of a coil spring. The correction itself attempts to account for bending effects (shift in the shear center of the wire cross section). the equation given is an approximation (linearization?) of the "real" solution, thus the c>3 limit (Wahl states that it should provide results within 2-3% of more complex calculations for c>3). The same text quoted above gives two Taylor-series expansions of the correction factors, from two additional sources.

The "corrected shear stress" is generally accepted to be a valid design criteria for springs, but perhaps this is as much because of the large volumes of tabulated/graphical data for yielding and fatigue limits of springs plotted versus the corrected shear stress.
 
gio1 -

I'm not sure I follow how you meshed the wire - how many elements are in the cross section - just eight (8)? Plot the shear stress across the wire - does it reach a maximum at the o.d.? Finally, try: do a hand-calc of the torsional constant for your meshed model section versus the real round wire, and adjust the model (make the "diameter" of the meshed wire larger) to give equivalent torsional stiffness.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor