Shear

[Lutfi]

Okay, now I am going back to strength of materials days!

Let’s say that we have a 3/4 inch diameter rod. (Fy = 36 ksi, and Fu 58 ksi). AISC allows shear strength to be 0.4Fy.

My questions are:
1. What is the safety factor that the AISC uses for the allowable shear?

2. For A36 to shear, from my early college days and Mohr’s circle, steel will shear before it reaches its yield strength. Am I correct?

I would appreciate any updates on this.

Lutfi

 

[desertfox]

Hi Lufti

If you draw a Mohr stress circle based on the tensile yield stress the maximum shear stress value at 45degrees to the tensile yield stress would equal half the yield stress value, so if the yield stress was 36ksi then the shear stress maximum would be 18ksi.Now if we consider the factor of safety for shear which is the first part of your question
then

Factor of Safety= yield shear stress /allowable shear
stress

= 18ksi/(0.4*18ksi)

= 2.5

If we are talking about a ductile failure here and using the
"Maximum Shear Stress Theory" then this failure occurs for
the combined stress condition when the maximum shear stress equals the critical shear stress(yield or ultimate shear stress)produced in a element subjected to simple tension.
Therefore for this failure criteria, to reach the maximum shear stress, the material would have to have a tensile load
sufficient to produce the yield tensile stress.

 

[haynewp]

If this is an A36 threaded rod, the allowable shear stress is controlled by 0.17Fu for threads included in the shear plane and 0.22Fu for threads excluded.