Shaft Shear Stress

[Ian Punk]

Hi All

First time poster, please be gentle. I'd appreciate your collective thoughts on the following..............

In the past whenever I have considered simple shear (only) in a solid shaft I have basically used the formula as it is here:

http://image.slidesharecdn.com/e-sc...ring-science-lesson-7-5-638.jpg?cb=1352152858

When I consider the shear stress in a spline (root) I use the same formulae but include the fatigue life facto (Kf) and application factor (Ka) as in machinery's.

When I consider the material capability I would usually consider UTS x 0.577 as the point where yielding in shear would occur.

So to my question:

The material in question is a case carburised steel shaft...lets assume EN39B circa 50mm case depth 1mm.

Data Sheet example :

http://www.interlloy.com.au/our-products/case-hardening-steels/en39b/

So for my calculation as above I would use a UTS of 1300N/mm2

Max allowable shear stress in core yield = 1300 x 0.577 = 750 Mpa

Now....what about the 1mm case depth?

If I was to consider the UTS of the material based on surface hardness I would be looking at a UTS in excess of 2000Mpa

Max allowable shear stress in (1mm case depth section) yield = 2000 x 0.577 = 1154 Mpa

My assumption is to stick with my tried and tested calculation ignoring the effects of case hardening but my question is would the shaft be capable of taking the higher shear stress value?

Cheers!

 

[metengr]

No. Your design basis is correct in using the lower strength core material. The function of case hardening is for wear resistance and to prevent crushing from local contact or Hertzian stresses.

 

[EdStainless]

The most that you can count on the case for (other than wear resistance) is a slight reduction in the severity of stress concentration and fatigue factors. Continue using the core properties.

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P.E. Metallurgy, Plymouth Tube