MAXIMU SHEAR OF THE PIN 7

[barvas11]

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?

 

[SnTMan]

p/a

 

[rb1957]

"yield strength" ... tensile or shear ?

be aware that your fastener could be limited by bearing (if the panels it's joining are thin).

 

[barvas11]

should it not be A * Yield?
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

 

[barvas11]

It is a max shear. The pin is in the bearing and pivots 6t block.
I know that the pin will never shear but how will I calculate it?

 

[rb1957]

presumably it's bearing (and shear across the plain shank.

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

 

[barvas11]

Thank you.

 

[Twoballcane]

If there is a moment, this will not be pure sheer. You should be calculating principal stress or Von Mises stress and then compare these to yield strength. Now, if it is more than yield but less than ultimate means you’re in a plastic region and will have to make a decision if that is acceptable.

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"

 

[dhengr]

Barvas11:
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.

 

[tr1ntx]

Also, if the pin is the shaft through the bearing, the cross-sectional area will be total area, i.e., on both sides of the bearing.

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."

 

[som1973]

The Allowable shear stress is 40% of yield strength of the material of the pin.

i,e 0.4 x 355 N/mm2 = 142 N/mm2

 

[Walterke]

Can you make a sketch of how exactly your pin is loaded? Just to be sure it's pure shear and you don't need to take anything else into account.

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[kingnero]

The Allowable shear stress is 40% of yield strength of the material 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...

 

[som1973]

see following link (Refer 6th slide)

http://arch.umd.edu/Tech/Tech_II/Lectures/Allowable Stress & Factor of Safety.pdf

A36 steel, Fy =36 ksi.... Allowable shear Stress is 14.5 ksi
36 ksi x .4 = 14.5 ksi

 

[hydtools]

Maximum shear theory: maximum allowable shear stress = 0.5 * tensile yield point stress. Spots' Design of Machine Elements

Shear stress = shearing force / shear area

Ted

 

[chicopee]

Shear stresses presented by som1973 and hydtools are correctly calculated depending if you follow AISC code or follow classroom theoretical calculation via Mohr's circle. I would also input a S.o F. if I wanted to use a working S.S. value.

 

[Cockroach]

Shear is usually taken at 0.577 times the yield value of the material. For example, 3/8 - 24 UNF-2A brass set screw used for a shear pin in oilfield related product is rated at 3300 lbf/screw.

Some use 0.500 yield strength, I prefer Maximum Energy Distortion Theory, 0.577.

Regards,
Cockroach

 

[hydtools]

Choose a theory and a factor of safety to determine maximum allowed working stress. Maximum shear stress / FS = maximum allowed working stress.

Ted

 

[jtseng123]

Seems some misconcept for pin shear stress that raise so many issues for the maximum shear theory, vom Mises stress, mohr circle etc.
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.

 

[Cockroach]

JTSEng, we are taking shear. Von Mises pertains to principle states of stress, that means shear is zero. You would need to use Mohr's Sphere to rotate that shear out in the three dimensional case. So no misconception, but a fundamental error in understanding.

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