Shear strength and 304 stainless steel 1

[teejer]

Does anyone know the relationship between shear strength and yield strength of 304 stainless steel or any type of stainless steel for that matter. My Machinists handbook does not list that value for stainless steel. I found on some other old forums on the internet that .577 x yield or .6 x yield as a rule of thumb works for shear, but I don't know the source of that info or how it is derived. I look forward to your replies.

 

[MartinSr00]

It depends on your assumed failure mode. There are various failure theories.

A popular on (I believe used by the ASME Code) is Tresca or maximum shear stress theory of failure. It assumes that the material is highly ductile, which 304 is. In such materials, the failure stress in pure shear is half the tensile stress failure. I don't think those theories deal with yield failures, so I'm unclear if the same ratios apply.

Other theories include maximum principal stress failure (usually brittle materials) and Von Mises (constant energy of distortion failure theory, I think).

The failure envelopes, though don't follow precise patterns. There is statistical scatter.

I think the reason you're finding scatter is that materials don't behave ideally, don't fail at one discrete stress, and may not be anisotropic (meaning their properties aren't alwasy the same in all three coordinate directions).

 

[metengr]

Actually, the 0.57 value of yield in shear is obtained from the von Mises yield criterion. You can Google for more information.

shear stress to cause yield = tensile yield/[√]3

 

[teejer]

Thanks for the explanations. The stainless application that I am looking into is a nut that I had some lab tests done on to determine the yield point for a connection. When I multiply the shear area for the internal thread by the theoretical shear yield stress (30,000 to 32,000 psi x .577), I come very close to the lab results for the yield point for the nut. I have just never had a lab result come so close to the theoretical calculation, but I wish I understood the .577 number better. I am going to have to read up on the Von Mises yield criterion, because I like to understand everything that I work with. It's been a while since I've had to go beyond simple stress calculations!