Amount of kinetic energy converted into work to impulse shear four double-shear design shear pin?

[GerdB]

I am looking to confirm my thoughts on the impulse energy applied to a double-shear shear pin configuration, and how it compares to the total kinetic energy of a system.

Background: (assume all motions/forces remain in the same plane) A component of a piece of heavy equipment being transported at 15 MPH ( 22 ft/sec, or 264 inch/sec ) is secured in place by four shear pin in double-shear loading at four points on the frame of the heavy equipment machine. The pins are assumed to have a diameter of 1" (UTS and Yield not known at this time - leave as variable) , this component piece is 'stored' in place on the heavy frame of the heavy equipment machine which is loaded and rigidly attached/chained on a tractor trailer. The mass of the component is 232 slug, and the total mass of the system in motion is 3,510 slug. The Kinetic Energy calculated as 1/2 * M * v^2 (m in slug) of Component is ~149,500 ft lbs_force while entire 'system' is ~850,000 ft lbs_force. As a result of an impulse load condition acting upon the component alone, the component became immediately dislodged at once, the four pins sheared, and the remaining 'system' maintained its constant velocity of 15 MPH on level smooth terrain same plane of motion.

When the pin shear occurred, only a brief 'sound' , described as a bang was heard a small 'bump' was felt , and there was no perceivable change in the forward velocity of the system.

SOOO

the duration over which the shear of the pins - which were the only things holding component to remainder of system) occurred is estimated as the diameter of the pins, 1 inch, so 1.0 inch / 264 inch / second = ~0.00378787... second since velocity was 'unchanged' as forward momentum was ~850,000 ft pounds_force. Assume all four shear simultaneously.

And if I remember for pure shear from Von Mises yield criteria, shear strength is 1 / SQRooT ( 3 ) = 0.577 of yield strength or using Tresca it is 1/2 or 0.5 of yield strength, and for double lap shear loading Shear Stress = 2 x F / π x d^2 where d is pin diameter

so the force required is Pi * pin_diameter^2 * shear_strength . and because I have four pins and substituting back ultimate strength

Energy = (4 * Pi * Pin_Diameter^2 * ultimate strength * 0.577) acting for 0.00378787 second. And units are (inch * lb_force/inch^2)

and Tresca it is Energy = 4 * Pi * Pin_Diameter * shear_strength * 0.5 / 0.00378787 second


and work is Force * Distance

work is energy / time

so for ease of calculation lets use Von Mesis and yield_strength as 50,000 pound_force / inch squared and because we have 4 pins we get

4 * (3.14159 * (1.00^2) inch * 50,000 pound_force * 0.577/2) = ~181,270 inch pound_force or ~15,106 foot pound_force

and the total kinetic energy of the moving system prior to pins shearing has total Kinetic Energy of ~850,000 foot pound_force

and the shearing of the pins represents 15,106/850,000 or ~ 2.9x10^-4 % of the total Kinetic Energy being converted to work to shear the four pins ---> no perceivable velocity change observed and a bang being heard seem reasonable???


Thanks for responses. the pins are like hitch pins, put in to secure when transporting - removed for use - there is no indication of fatigue related failure, as these are more often than not left and replaced with new.

point of changing area is excellent - loaded area changes throughout shear from a full circle to nothing - hence the load/work would change.

impact testing uses a square sample - any test or results known for a round pin? -

and pins are in two telescoping box beam , one at each end, rails that alighn thanks again

Appreciate comments - thank you

 

[Sparweb]

You are closing in on the principle of fracture toughness. Try working it out in terms of "Work of Fracture" instead.
I believe the result should be much less than the value you've worked out above....and by an order of magnitude.
One flaw with using simple shear force is that the cross-section changes as the shear failure progresses.
I don't think the energy required to shear a pin is much dependent on the speed of the impact, either.

Back of the envelope for me gives about 500 foot-pounds of energy, per 1" diameter steel pin.

https://www.youtube.com/watch?v=tpGhqQvftAo

STF

 

[3DDave]

The velocity of the vehicle and object didn't change so the related energy is immaterial. I suspect a long period of fatigue crack growth followed by shearing of the remaining material.

Most of what I understand of the specifics of fracture mechanics involves staying away from it or make to test samples for the exact configuration of interest.

 

[racookpe1978]

But, would all four pins yield if a fatigue crack were the cause of failure in one pin?

The load would (try to) rotate and skid if the first yielded, then the 2nd and 3rd. No realistic way of simultaneous fatigue failure in four pins at the same time. If it were restrained as by rails on both sides, then failure of one pin could cause (later failure) of the remaining three, but there would still be evidence of twisting, wouldn't there?

 

[BUGGAR]

There are studies of race car impacts where individual pieces of the vehicle are sheared off, absorbing bits of energy as they go. You can find similar to what you are doing.

 

[rb1957]

why assume the presence of a crack ? why not assume that the impact load exceeded the static strength of the pins ? Simple number to calc.

agreed, with a sudden failure a crack is likely (I'd've thought that there would normally be some ductility in the failure), and should show up on the fracture surface. Then you can figure out the residual strength.

How much variability is there in the four attachments ? equal gapping (unlikely) or is one pin a tighter fit than the others (likely). If one of the four attmts fails "early" what happens to the remainder ? (I'd've thought that there would be some evident damage to the attmts as the load shunts.

another day in paradise, or is paradise one day closer ?

 

[3DDave]

I see the OP has changed the description of the problem. At this point, without pictures, there is no sensible description.

Someone is mistaken about how the damage happened.

The mention of velocity still has nothing to do with the shear failure.