Contact Stress - Can we add it up?

[mecheng10]

I have calculated the contact stress for a sphere in a socket using formulas provided in Roark's formulas. But my problem is that the sphere is actually in contact with two sockets - upper and lower sides, so there are actually two contact areas.

Can I just multiply the contact stress I calculated for area 1 by 2 to get the final?

Experts - Can anyone please verify this?

Thanks

 

[Studiot]

Can I just multiply the contact stress I calculated for area 1 by 2 to get the final?

No, Contact stress is just that.

The local stress of the contact zone.

The stress in the rest (bulk) of the contacting bodies is obtained in the usual fashion by St Venant's theorem.

So you have two local areas with contact stresses on your ball.

If you wish to know whether the higher stresses due to contact are significant you will need to do some fracture mechanics analysis.

 

[john.phillip]

Also, look for Hertzian Stress, Tribology and Wear to get a better view about this interesting subject.

What is the required factor of safety for the contact stress you found?

Usually, the allowable stress is 4 for linear contact and 5.5 for point contact, but this varies a lot depending on the applicable code, so you might get your contact stress and divide it by 4, if suitable, before comparing with your allowable stress.

Another aspect is the subsurface shear stress induced by the contact force, as the mechanical failure is usually closely related to this shear stress rather directly by the contact stress. You might be required to obtain this shear stress and compare with the related allowable stress.

 

[mecheng10]

Thanks for the replies. Wont I be comparing the contact stress with the compressive strength of the material as the contact stress induced is due the compressive force? But you def bring up an imp subject of tribology..I will surely check that..

 

[Studiot]

Wont I be comparing the contact stress with the compressive strength of the material as the contact stress induced is due the compressive force?

Your original query has been completely answered. You clearly wish to discuss further - whcih is good.

Hertzian stresses (as set out in Roark) were mentioned. This is because Hertz first calculated analytical solutions for disks, balls and cylinders (1881).

No material can fail in compression. The failure always occurs where some internal stress field converts the compression to tension/shear. So the failure is always tensile or shear.

Your ball squashes out sideways due to a tensile field developed over most of its diameter and eventually fails about this diameter if homogeneous.
This, of course, assumes the material is ductile enough to do this, which is why I suggested looking at the fracture toughness of the material.

This is the basis for the so called Brazilian Crushing Test on cylinders, which fail by diametral splitting asunder.