MAXIMU SHEAR OF THE PIN
MAXIMU SHEAR OF THE PIN
(OP)
Hi Guys,
How do I calculate a maximum shear strength of of the pin?
I know what the diameter of the pin is and also what yield strength of the material is.
I am surprised that there is nothing online regarding this. Could tell me the formula?
How do I calculate a maximum shear strength of of the pin?
I know what the diameter of the pin is and also what yield strength of the material is.
I am surprised that there is nothing online regarding this. Could tell me the formula?





RE: MAXIMU SHEAR OF THE PIN
RE: MAXIMU SHEAR OF THE PIN
be aware that your fastener could be limited by bearing (if the panels it's joining are thin).
RE: MAXIMU SHEAR OF THE PIN
Let's say that the pin has 80mm in diameter and it's max yield is 355 N/mm2
That means each mm2 of the calcualted area is able to take 355N before it shears that means an area of 5026mm2 (80mm dia pin) will take 5026x355= 1784 x 103 KN
RE: MAXIMU SHEAR OF THE PIN
I know that the pin will never shear but how will I calculate it?
RE: MAXIMU SHEAR OF THE PIN
P = stress*area
shank area sounds appropriate.
"yield stress" ... are you talking tension yield or shear yield ?
if you have tension yield then shear yield, for steel, is something like 0.57*fty
RE: MAXIMU SHEAR OF THE PIN
RE: MAXIMU SHEAR OF THE PIN
Tobalcane
"If you avoid failure, you also avoid success."
“Luck is where preparation meets opportunity”
"People get promoted when they provide value and when they build great relationships"
RE: MAXIMU SHEAR OF THE PIN
You say you “need a result in newtons,” but what you really need is a local mentor who can help you with these kinds of problems. This is pretty elementary engineering stuff and when you don’t know if you are using a tensile yield stress or a shear yield stress, you shouldn’t be doing this kind of problem if it could hurt anyone other than just you. You should invest in a good Strength of Materials text book and study it, if you are going to do this kind of work.
What is your engineering background? Please tell us.
RE: MAXIMU SHEAR OF THE PIN
I agree with dhengr too. This is sort of like a case of "If you've gotta ask how much it costs, you can't afford it." The best advice might be "If you've gotta ask how to do this, you don't need to be doing it."
RE: MAXIMU SHEAR OF THE PIN
i,e 0.4 x 355 N/mm2 = 142 N/mm2
RE: MAXIMU SHEAR OF THE PIN
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RE: MAXIMU SHEAR OF THE PIN
Where does the 40% comes from?
As I believe that would be rather low, and the 57% of yield strength mentioned by rb1957 seems more appropriate...
RE: MAXIMU SHEAR OF THE PIN
http://arch.umd.edu/Tech/Tech_II/Lectures/Allowabl...
A36 steel, Fy =36 ksi.... Allowable shear Stress is 14.5 ksi
36 ksi x .4 = 14.5 ksi
RE: MAXIMU SHEAR OF THE PIN
Shear stress = shearing force / shear area
Ted
RE: MAXIMU SHEAR OF THE PIN
RE: MAXIMU SHEAR OF THE PIN
Some use 0.500 yield strength, I prefer Maximum Energy Distortion Theory, 0.577.
Regards,
Cockroach
RE: MAXIMU SHEAR OF THE PIN
Ted
RE: MAXIMU SHEAR OF THE PIN
First of all, check if it is single shear or double shear construction on the pin, and calculate the shear stress accordingly. This is classroom practice.
Secondly, use 40% of the yield as the allowable as some folks have said or per AISC, and you will never go wrong.
The first week I got hired more than 20 years ago, the senior engineer told me to use 40% of the yield. And since then, never changed, never anything failed.
RE: MAXIMU SHEAR OF THE PIN
Second, my experiences have been positive with the Maximum Shear Theory using 0.577. Twenty-nine years of professional practice, I have volumes of tested mechanical design assemblies used in oilfield equipment based on this principle. The latest, two weeks ago was on eight brass shear screws rated at 3300 lbf/SW. The piston wetted hydraulic area gave a differential rating theoretical of 692 psi/scw, we stroked the assembly in the shop to measure an average of 685 psi/scw. Fairly close using 0.577 yield strenght as shear.
You can use forty percent, some calculations are fifty. But these are used in unwanted shear applications, so your factor of safety are higher in these cases. What happens when you intend for shear to activate a motion? Forty percent would be way too conservative.
Know what I mean? Application dependency.
Regards,
Cockroach
RE: MAXIMU SHEAR OF THE PIN
maximum shear stress = (1/3)^(-2) * maximum tensile yield = 1/1.732 * maximum tensile yield = 0.577 * maximum tensile yield.
Ted