Torsion Formula - Yield strength limit states design

[Woody1515]

Hi everyone,

For the typical torsion formula T = J*max shear stress/r, what is the maximum shear stress for a steel circular shaft? Is it just the yield strength? For limit states design here in Canada, is there a specific resistance factor that needs to be used beside the typical 0.9?

Thanks!

 

[Woody1515]

I’ve noticed S16 does not have any torsion formulas to follow.

 

[dik]

I generally try to avoid it and remove it by other means. If I can't, I try to use HSS sections. The AISC has several good publications on torsion, and I generally follow their methods.

So strange to see the singularity approaching while the entire planet is rapidly turning into a hellscape. -John Coates

-Dik

 

[Woody1515]

I can’t avoid it unfortunately. It seems odd how it is not covered in depth in the Canadian steel code. Is the maximum yield strength simply just the yield strength when dealing with torsion?

 

[desertfox]

Hi Woody1515

Normally in the field mechanical engineering the maximum shear stress would be the maximum shear stress allowable ie meaning the 0.9 * material yield stress * safety factor like 0.6.
Hope this helps

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[bugbus]

The stress limit would be some strength reduction factor (e.g. 0.9) times the shear yield strength (i.e. fy / sqrt(3)). Some sources use 0.6*fy instead.

Don't use the basic yield strength of the material directly.