PERMISSIBLE SHEAR STRESS OF MILD STEEL

I've seen both used (57.7% of YS and UTS).
For safety pins that work in shear, of high-strength pins/bolts (that are likely to show a brittle fracture after what likely will be a shock load), 57% of UTS is common.
If you suspect a ductile behaviour or overloading, 57.7% of yield is the way to go (and is also more conservative).

FoS for shear is the same as for tensile. Most likely depends on the application.

What kind of service? Is there any fatigue? Any risk of injury when it fails?
Given that in shear you expect risk starting near 58% UTS you need to work back from there.
I have worked in applications where the SF has been everything from 2x to 8x, with 3.5-4x being the most common.

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P.E. Metallurgy, consulting work welcomed

Dear EdStainless,
Thank you for your response. However, my main question is whether permissible shear stress is 0.577xYS/FoS or 0.577xUTS/FoS.
It is for a Hydraulic Application where we use FoS of 2.5.
If it is 0.577xYS, then do we need to divide it by a FoS to get the permissible shear stress?
FoS = Factor of Safety
YS = Yield Stress of Material
UTS = Ultimate Tensile Stress of Material

Doesn't my reply answers both of your questions?

- YS or UTS: depends on the application. Hydraulic doesn't mean anything useful. Be more specific, especially about the expected failure mode.

- FoS: if you use 2.5 for everything else, then that's what I would use for this situation as well. But be aware that there are FoS for loads, for materials, for dangerous situations, and in general a whole lot of fudge factors. For low carbon/low alloy steels in european codes, factor of safety is generally between 1.0 and 1.2, because the material behaviour is well known and understood. Doesn't mean you can't apply a FoS on the loads, but that's a different thing.

Dear kingnero,

It is for the design of a clevis eye that can fail in both tensile or shear on different planes 'AA' and 'BB' as shown in attached image. Assume it is a pulling type load and the part is in tension. Therefore, it can undergo tensile failure along section 'BB' or shear failure along section 'AA'.

What I am particularly interested in is whether my understanding of Von Mises criteria is correct whereby :
Permissible Shear Stress = 0.577 x Yield Stress / Factor of Safety

People are sometimes interpreting this as :
Permissible Shear Stress = 0.577 x UTS / Factor of Safety

You are suggesting that the decision whether to use YS or UTS to work out the permissible shear stress is dependent on the application? As per my understanding, both of the above cannot be correct and the latter would result in a higher permissible value for the same FoS and therefore might be misleading.

Yes, that is exactly what I am suggesting.
BTW, your shear planes won't be one in the center of the part but two at (a little bit less than) D (= diameter of the pin) apart
and it won't be true shear but shear + bending. I'd use 58% of yield for that. Especially since it's mild steel, brittle fracture is unlikely (unless again working temp. Is far below zero Celsius)

Yes, you are right. In this case, only the part to the right hand side of the hole on section AA is subject to shear stresses in pulling mode.

Have a look at these two links, I would go with .577 times yield divided with a FoS


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