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ryldbl (Mechanical) (OP)
5 Jun 06 12:51
I am trying to design a shear pin that will connect two drive shafts.  I can calculate the shear stress that will be applied to the pin, however in order to design the pin to fail at a given stress level I have to employ a certain failure criterion.  Any suggestions on how to relate the yield/tensile strength of a given material to the shear stress at which the pin will actually separate?
CESSNA1 (Mechanical)
5 Jun 06 15:22
RYLDBL:  The ralationship between the tensile and shear moduli is

      E = 2*G*(1+v)

            E = Elastic modulus
            G = Shear modulus
            v = Poisson's ratio

so for common steels assuming E = 30,000,000 psi and v = .33 Then G = 11,278,196 psi

You should also use Mohr's Circle.

There are several failure theories that can be used such as the Von Mises and the Maximum Shear theory.  

Remember in shear pins the failure (breakage) occurs at the ultimate strength not at the yield strength.

Regards
Dave
rb1957 (Aerospace)
5 Jun 06 15:49
it is common, in designing strcutural fuzes, to incorporate a notch/groove to initiate the failur in a controlled manner.
CoryPad (Materials)
5 Jun 06 16:02
This question, in slightly different flavors, has been asked many times here at Eng-Tips, some even recently.  Elastic moduli won't help determine the fracture stress.  In fact, the only equation used is the approximation:

τ = 0.6 · σ

where

τ is the shear stress

and  

σ is the tensile stress

Testing is used when certainty is required.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

dvd (Mechanical)
5 Jun 06 17:49
Rexnord makes some pretty nice shear pin couplings. Voith makes a really cool coupling that works with hydraulic pressure and blow-out plugs. Consider buying a product that has already been proven.

Don't introduce sharp notches for failure because your pins will fatigue.

If you are making this product to sell to someone make sure to uniquely mark the pins so that you can tell if someone has duplicated your original pins.

Buy your material to a specification and test it.

Cockroach (Mechanical)
5 Jun 06 19:44
CoryPad is actually correct, although 0.577 rather than 0.6 is better recorded in the various engineering literatures under "Maximum Distortion Energy Theory".  Also correctly pointed out, it relates to the Von Mises-Hencky model.  Transverse loading (i.e. shear) is related to normal stress in this fashion, as pointed out by Cessna1

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

ornerynorsk (Industrial)
6 Jun 06 8:50
The shear pins I have used in the past do not have sharp notches, but rather a radiused groove to ensure failure at the correct point, so as not to bugger up the entire coupling and shaft arrangement.  If you want a sample in hand to work from, go down to your local hardware and ask for a shear pin from a garden tiller or snow blower.  
CoryPad (Materials)
6 Jun 06 10:09
Cockroach,

Von Mises is a YIELD criterion, not a fracture one.  Fracture is much more random, so three decimal precision is unwarranted in predicting shear fracture strength from tensile fracture strength.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

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