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Helpful Member!  mikerascally (Mechanical) (OP)
19 May 06 17:34
How do you calculate for the Ultimate Shear Strength of a Material?

I have a SS416 heat treated at 800 (From Machinery Handbook) and now has:

Yeild Strength: 150000psi
Ultimate Tensile Strength: 195000psi
Hardness: C41, 390Bhn

One book says:
Ult. Shear Strength = 0.57 x Material Yield Strength

Machinery's Handbook says:
Ult.Shear Strength = 0.50 x Ult. Tensile Strength

They are both close in terms of results but which one if at all?!

desertfox (Mechanical)
19 May 06 17:47
Hi mikerascally

I would use the 0.5xUlt.Tensile strength


Helpful Member!  TVP (Materials)
20 May 06 13:40
Ultimate shear strength is commonly estimated to be 0.6*UTS.  The 0.57*TYS is probably taken from the von Mises/distortion energy/octahedral shear stress criterion, and it should be stated as shear yield strength = 0.577*tensile yield strength.
Helpful Member!  RPstress (Aerospace)
23 May 06 14:01
0.5 times tensile is the safe way to do it, as advocated in the "Introduction to Aircraft Stress Analysis" course at Cranfield, UK.

For round bar in shear (as used in the ESDU MMDH) anything up to 0.7 times may be achieved.

Examining MIL-HDBK-5, which has a mixture of shear test methods, with quite a lot of punch shear through sheet, 0.6 times is sort-of-usual.

Helpful Member!  Cockroach (Mechanical)
24 May 06 11:48
TVP is correct.

The Maximum Distortion Energy Theorem shows that shear stress is approximately 0.577 that of normal yield strength.  I've found this to be highly accurate comparing computed to laboratory results.

Kenneth J Hueston, PEng
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

PapaCormes (Mechanical)
1 Jun 06 10:58
What kind of Factor of Safety are you using on your Ult. Shear Strength
diamondjim (Mechanical)
2 Jun 06 11:45
I think part of the answer lies in how you
want to use this.  Do you want the part
to shear as in a shear pin or do you want it
not to shear.

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