# Ultimate shear strength

Can anyone here provide the ultimate shear strengths of copper (both
hard and annealed), yellow brass, and A36 or 1018 steel? Thanks.
Bert
The rule of thumb is shear strength = ~1/3 of the tensile strength. There are some exceptions...like carbon thread, glass fiber, etc. iirc, 1018C runs about 30 ksi, so shear strength is ~10ksi.
StaticsJason
Did you mean to say 2/3? Everything I've read indicates that shear strength is typically 60 to 90% of tensile strength. But I'm looking for something a little more precise for the materials in question. Surely that information must be available somewhere, but I haven't found it on the web or in my limited home library.
Bert
"Statics" wrote:
I've made cursory investigations (ie, talking to everyone I know who knows more about this stuff than me...) and have yet to come across a theory that explains how one would calculate one from the other, and how they are related physically.
Jim
================================================== please reply to: JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com ==================================================
Get thee to an elementary engineering text. Mechanics of Materials by EP Popov will do nicely.
Look for "failure theories" Maximum shear stress theory and Maximum distortion energy theory.
Erich
Easy name to remember, I will search it out next week. Thank you!
Jim
================================================== please reply to: JRR(zero) at yktvmv (dot) vnet (dot) ibm (dot) com ==================================================
Co> >
You are welcome.
Erich
Or "Introduction to Mechanics of Solids" by same author (if indeed it is not the same book). I still have my school copy, purchased for \$14.95!
Thanks, Tom. For future reference, may I ask where you found this data?
Bert
I wonder if I recall this right... Ultimate Shear Strength (USS) is related to Ultimate Tensile Strength (UTS) by Poisson's Ratio. Poisson's Ratio is simply the ratio of the original cross-sectional area of the tensile sample divided by the area of the failed surface (normal to the pull axis) on the tensile sample. So, if we have a one inch diameter tensile sample of mild steel and pull it to failure, we should have seen a tensile strength maximum of about 55,000 psi.
Let's see ... The original sample had about 0.784 sq. in of area, the failed section had an area of about 0.470 sq. in. UTS from the test was about 55,000 psi, so:
0.470 ------ * 50,000 = about 30,000 psi for USS. 0.785
Look up the UTS in a table, multiply it times Poisson's ratio to get USS, or do the ratio of failed sample area (often given in tables) to original sample area and multiply times the UTS.
Cliff - Did I remember it right??
Jeff Thompson
Al wrote:
In my haste, I gave you figures for yield strength rather than ultimate. Are you designing shearing dies or parts that need to sustain shear loading?
If designing parts to survive shear loading keep in mind that in ductile materials, deformation and/or necking (in the case of tensile loading) can increase the stress in the part without significant increase in load when in the failure region of the stress - strain curve.
Designing a ductile part assuming that it will stay the same shape (and load bearing cross section) all the way up to the ultimate stress can be dangerous.
If you are intentionally cutting something, ultimate stress is your figure; if you are designing a part for safe use, yield stress is what you should design with.
"Mechanics of Materials" by Gere & Timoshenko is a good reference, though the other book(s) mentioned in the thread would likely do fine.
hth, StaticsJason
PS: A36 is so named because its tensile yield strength is 36ksi.

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