Static Load help with steel bals

[nickjk]

I am looking at a wedge application that places hardened spherical balls between wedge surfaces to allow a reduced wedge angle for greater mechanical advantage without locking problems caused by friction. Travel distance is very small, approx .090“.
From research, I have found the maximum contact stress applied to a hardened steel ball and race to be 609,000 psi. This contact stress will deform the ball or race approx. .0001 x the diameter of the ball. The above stated deformation is assumed safe for static applications.
I have been using Roark’s Formula for stress and strain Table 14.1 case 1a (Sphere on flat plate) for my calculations.
How does the contact Stress ?c relate to tensile strength when it is almost twice the value.
What is meant by ?t which is .133 ?c and is ?t what I use to evaluate tensile strength.
Calculations for a reference project show contact stress to be approx 877,346 psi with a .00028” deformation on a .079” diameter ball. Would this fail even if there is no or very little rotation.
Thanks,
Nickjk

 

[electricpete]

j

Calculations for a reference project show contact stress to be approx 877,346 psi with a .00028" deformation on a .079" diameter ball.

So the ratio of deformation/diameter is 0.079/0.0028 = 0.035 ?
That is 350 times the thumbrule of deformation/diameter = 0.0001

If we scale the 877kpsi down by a factor of 350, it would be something like 2.5 kpsi which is far below yield in tension, compression or shear, right? That seems consistent with the thrumbule.

Sorry if I have missed the point.

=====================================
(2B)+(2B)' ?

 

[nickjk]

electricpete,

The ratio of deformation/diameter is .00028/.079= .0035
35 times the thumbrule of deformation/diameter=.0001

scale 877kpsi down by factor of 35, it would be 25 kpsi.

I thought for hardened steel balls deformation/diameter=.0001 is achieved at approx 609,000 psi contact stress. For 440c stainless steel I thought we could assume approx 85% of that contact stress which would be 517,000 psi. Even 517,000 psi contact stress seems to be to high when compared to yield or tensile stress.

That is why I wanted to know what is the definition of Roarks ?t in the formula and is this the value I use to determine if the stress is safe. Note: ?t = .133 * 877kpsi = 116kpsi which appears to be with-in 2/3 the tensile strength.

I do not have enough experience in this area, I may be all wet.

Thank you for your response,

Nickjk

 

[nickjk]

Sorry,

I should be calling ?c compressive stress

Nickjk

 

[dinjin]

Nick,
From research, I have found the maximum contact stress applied to a hardened steel ball and race to be 609,000 psi. This contact stress will deform the ball or race approx. .0001 x the diameter of the ball.
.0001 x ball diameter yield .0000079 allowable deformation.
How are you getting .0028 with a 877000psi contact stress?
Your deformation is at least 35 times the recommended, why
would you think it would survive? The subsurface stresses
would be phenomenal. I would hope the allowable would have
a safety factor of 3 before catastrophic failure.

 

[israelkk]

nickjk

Because both the ball and the flat surface are compressed the deformation you are looking for is approximately half "y = relative motion of approach along the axis of loading", and not "a" = radius of circular contact area". The reason such a high compressive loads are allowed compared to the tensile stresses is that this a triaxial load case and not a one dimension tensile specimen load case. On page 692 Roark gives the crush load formula as tested by SKF too.

 

[nickjk]

dinjin,

The deformation is .00028" and not .0028". The value was calculated using Roark's formula to solve for y
Table 14.1 condition 1a using P= 20.07 lbs, E= 30,000,000 psi, Kd=D= .079".
The compression stress ?c was also calculated using Roark's with v= .293, Ce= .0000001, a= .0033"
If I understand correctly the deformation of .0001 x the ball diameter is the amount of deformation that can be tolerated without harming the application of a normal bearing. Please note, the reference project is not used as a normal bearing application. There is no rotation and linear motion is very small, .094". Deformation is greater than normal .00028" and can be seen on reference project race but this did not seem to harm its operation during the 500,000 test cycles.

Thank you for your response,

Nickjk

 

[nickjk]

israelkk,

The value .00028 is deformation "y". I do understand your thoughts of deflection being approx 1/2 that value.

From SKF empirical formula for crushing load, the crushing load for a .079" ball would be 878 lbs. The reference project is using 20 lbs. of load per ball.

From Roark's page 689 it states that max compressive stress calculated occurs at the center of the surface of contact and is not the maximum tensile stress, which occurs at the boundary of the contact area and is normal.

If I believe a deformation of .00028" or .00014" is acceptable, would it be wrong to use the Roark formula .133(?c)max for my design stress?

Thank you for your thoughts and help,

Nickjk

 

[israelkk]

You are working in a gray area where no bearing manufacturer can help you. Use your own judgement, reasoning and tests to see if you can use loads that distort the ball/roller beyond the 0.0001 ratio of ball/roller distortion compared to the ball/roller diameter. You can look at Harris book "Rolling bearing analysis" where you will find more about loading bearings beyond the static and dynamic values given in bearing catalogs. From your last two posts I see that you are in the right direction. However, note that to guaranty a definitive life cycle, use a safety factor for at least 4. If you can make it to last 10 times more than you need to guaranty it is best. Note that metal properties varies between different batches, heat treatments, environment, surface conditions, etc. Therefore, if you need to guaranty 500,000 operations without fail you have to show at least 2,000,000 operation and to be completely sure 5,000,000 operations.

 

[Tmoose]

How will this be lubricated? How many balls will be sharing the load? The support structures will be stout indeed, and the manufacturing gorgeous to get good load sharing. Unequally shared loads can murder the tall bearings.

The static load rating is the limit for occasional, VERY slow events, and as you found, based on a relatively small permanent deformation. I'd expect vibration analysis to likely detect that deformation. And I'd consider if it might hinder the free motion you're looking for.

The basic dynamic load rating is the loading that can be survived thru some basic number ( a million, perhaps) of load cycles without fatigue (starting subsurface) failure most of the time. The interaction of the contact faces (no lube, insufficient lube) can initiate a destructive failure from surface stress and distress long before subsurface fatigue ever sets in.

Precision bearings using excellent steel, superb machining and very good lubrication are often designed for a "contact stress" of less than 2000 MPa/290,000 psi in an attempt to achieve infinite life.

 

[nickjk]

israelkk,

I agree and thank you for your time and help.

Nickjk

 

[nickjk]

Tmoose,

This is not a typical ball bearing application. There is no rotation and travel is limited to approx .09 axial with the full load applied at the very end. I do not see how vibration would be an issue. Because of this I believe greater deformation can be tolerated.

Thank you for your time and suggestions.

Nickjk

 

[Tmoose]

Was the test rig identical to the actual part, in regards number of rolling elements? If your testing proved the deformation did not impair function that may be the end of it.

My reference to vibration was that deformations from static overload are mechanically measureable.

I'd say a 0.079' diameter ball rotates about 130 degrees for the center to advance .09 inch. If the 0.09 inch is the travel of a platen on top of the ball ~ 65 degrees of rotation is required. The contact ellipse would be tiny in comparison.

Surface distress or spalling will over time conspire with the shallow dent to form a pothole at the highly loaded end point.

http://www.vibanalysis.co.uk/vibcases/vibch02/vibch2p1.jpg.

When a ball parks there I suspect some added effort may be required to get things rolling again.