How to Calculate Permissible Shear Stress? 1

[Fluid_Monkey]

Hello All,
I wanted to know how to work out the Permissible Shear Stress of Mild Steel having Yield Stress (YS) of 230 MPa and Ultimate Tensile Stress (UTS) of 410 MPa. Do you simply do 0.577xYS or do I need to consider a factory of Safety over and above this? 0.577 is 1/√3 and taken from Von Mises Criteria. Some websites seem to suggest that you need to take 0.577xUTS and then divide this by a reasonable Factor of Safety. Any help would be very much appreciated. I have attached the mechanical properties table for the material in question which is a low carbon mild steel.

 

[desertfox]

Hi Fluid Monkey

I usually take 0.577 times the yield stress and then put a safety factor of say 2 on that, it’s conservative but I can sleep at night.

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

 

[Jboggs]

The application must be considered. Above any code requirements, can failure result in human injury? If so, increase the safety factor significantly. Codes and rules of thumb are very important, but the bottom line eventually comes down to the judgment and personal responsibility of the designer.

 

[EdStainless]

please do not double post.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

 

[EnginerdNate]

If you start with yield stress, do the 1/sqrt(3) factor and then add a factor of safety on that you'll generally be pretty safe. You always need some factor of safety, with very few exceptions. Factor of safety will be dependent on your application. As an example, we are required to hold F.S. = 2.0 as a minimum on all tooling and other equipment on the production floor at my job, but flight hardware on aircraft would typically be designed to F.S. = 1.5. Unmanned spacecraft and rocket engine parts might be designed to F.S. = 1.15 or 1.1 because the application is so weight critical. On the other end of the spectrum, a bridge might be designed to F.S. = 5.0 because weight is much less critical and a catastrophic failure would have high potential for massive loss of life (and the service life is measured in decades not years or months or cycles). Factor of safety is determined by application.

All of that said, in critical applications it's better to use a reference backed by test data to get your allowables. Find a copy of MIL-HDBK-5 (The 'H' revision is pretty easy to find online) and look at the properties it has listed for 1025 steel and compare to your calculation. Most mild steels have very similar yield/ultimate properties for purposes of analysis. (A lot of structural steel specs allow for several different specific alloys as long as the yield/ultimate stress requirements are met).

Note that failure in shear is highly dependent on how purely shear the load condition is. Failure in pinned/bolted joints for example is highly dependent on how tightly the joint fits up, loose joints tend to bend the pin well before shear failure happens, whereas a tight fitting joint (both in the fitment between the bolted parts and in the fitment between those parts and the pin/bolt) is more likely to result in a more classical shear failure.

 

[soletto]

just use the mises calculation to recalculate equivalent stress from tau and then use a safety factor of 1,5

so allowable stress = Yield /(sqrt(3)*1,5) ≥ tau calculated

thats how i would do....

Maybe instead of 1,5 take 2 but it also can then be overconservative...