Spring Force and Failure Point

After more searching I believe I have found the correct equations for determining stress and whether we will be in danger of permenant deformation of the spring. The equations pasted below show that Stress for .025" deflection will be 95663 psi. Hopefully, someone can do a sanity check on that. Given Yield Strength for spring steel of 159500 psi, I think we are okay, right?

Of course, in the below calcs I used the equations for a straight cantilevered beam, and ours is curved. Still looking for an answer on question #1.

Thanks,

Dave

A quick FEA analysis gives a stress of ~1300MPa in the curved section and ~1.7 mm deflection at the load point. This assumes fixed conditions in the spring support but in reality the bolt clamping never gives fixed condition so the deflection will be probably larger.

The numbers in the picture are difficult to see. It's .23" from undersurface of circuit board to the force vector, and the curve's radius is .10". Is that what you used?

We expect normal deflection to be about .025", max deflection limited to 0.35". During use, deflection will vary from .010" to .035".

Thank you for the FEA.

Dave

Yes I used 0.23 and the 0.1 radius.

Something doesn't add up. What force did you use? You found a deflection of 1.7mm which is 0.067", more than the OP stipulated.

1.8 lbf

1.8 lbf results in a stress of 124,200 psi or 856 MPa, not 1300 MPa.

Sorry for my mistake, I added the 0.23" to the 0.1" radius instead of including it inside the 0.23". The corrected FEA results are: Maximum stress (Von Mises) is 861 MPa. Maximum deflection at 0.23" is 0.59 mm.

Okay.

Okay. Here is the bottom line question:

Assume we constrain the spring movement at the point of force to between .25mm and .60mm when the device is being used. That should limit stress swing to between 350 MPa to 861 MPa, by my estimation. The material's yield strength is about 1100 Mpa. If usage demands that this swing be survived 5,000 times, does this spring have a chance of providing sustained reliable contact force (~ 1 lbf or more) over its life?

Thanks,

Dave

You can use these calculators http://www.fatiguecalculator.com/index.htm

I think you should limit the maximum stress to 60% of yield strength, i.e. 660 MPa. This should be adequate up to 20,000 cycles in the life of the spring. You could try higher stress levels on an experimental basis but I believe you have a high probability of tensile failure with 5,000 cycles at maximum stress of 80% yield strength.

Del,

As you have pointed out, a tapered member would be more efficient. The OP is using a strip of 0.2" x 0.01" spring steel. Tapering the width from 0.3 at the top of the curve down to 0.1 at the point of application of force may be a better choice.

AAARRRGGGHHH.
Who ever designed this spring has obviously never seen a tree or an longbow.

TAPER THE DAMNED THING! _{<slaps furry head with paw>}

Stop doing calculations, start doing experiments.
Del

Never designed a spring before, but I've seen a tree. By 'taper' do you mean that the bottom end should be narrower? Why?

Ah! Excellent. Thank you.

- Dave

1. It is not a very good approximation. You would overestimate the deflection. Assume the horizontal force is P. At the start of the curve, the moment is 0.13P. Within the curve, the moment at any point is (0.13 + r*sinΘ)P where Θ is the angle from a horizontal through the center to the point on arc. At the end of the arc, the moment is (0.13 + 0.1)P. Your method would give (0.13 + 0.157)P which is too high.

2. Under static load, the spring will not fail until the material reaches yield. For cyclic loading, fatigue would have to be considered.

3. You should specify Young's modulus, yield strength and ultimate strength.

Makes sense.

Thanks,

Dave

Sorry to react with a delay but I was on holidays.

You did well to compute first since the problem of many members of this "Club" is that they either do not master or they forgot how to compute. Trials are required but if the computations are made with a good model they only confirm results or allow an optimisation. The "cut and trial" method highly praised by -unfortunately- too many engineers with many years experience is costly and obsolete.

I got my degree more than 40 years ago but I always tried to keep abreast with development of methods and use simulation since more than 30 years so that I know what I write about.

I was confronted with a similar problem. The difficulty is to estimate the shock generated force. In some situations it can go as high as 5000 x weight of part. The pre-load has to be higher than the maximal inertial force. At this pre-load the stress level in the steel should be less 60% of the elastic limit to be with a safe reserve within the low-cycle fatigue life. It has as well the advantage of a high safety coefficient for static loads due to assembly and manufacturing tolerances.

The highest stress occurs at the end of the radius part and is not very depending on the radius value. The radius has only an effect on the maximal deflection. As it was already mentioned the Young modulus is directly involved in the deflection magnitude. A stainless steel has a lower module thus a higher deformation but also a lower yield limit so that it could be interesting to compare the designs using a plated steel plate and a stainless plate. The taper form increases the deflection and could reduce the sensitivity to tolerances but in any case the preload MUST be maintained in order to not loose the contact during a shock.

In my application I was obliged to consider a preload in exccess of 3000 x batteriy weight to maintain contact.

Hope it will help, if you need more input than use the private message channel since details are not of inetrest to every body.