Fail Safe Pin Shear 1

[mxrethoret]

I am a recent graduate that has been assigned with the task of determining the maximum vertical force a fail safe pin can hold. At the point of failure, the load will be fixed directly down the center of the pin. The pin is supported on either side by circular guides. Correct me if I'm wrong, but I believe that this is a beam problem with fixed ends. So... how do I find this force? Can I assume, like others on this forum have said, that the maximum shear force is equal to 0.6*U.T.S., then multiply this result with the area to find the maximum vertical force? The material is SS321 and the pin is 1" in diameter.

Thanks in advance for your help.

 

[israelkk]

It depends on the length of the pin and the width of the circular guides. If they are too wide there will be bending in addition to shear. It also depends on the parts tolerances and it will be affected by cyclic application of the load that will widen tand distort the circular guides with time.

 

[desertfox]

Hi mxrethoret

Can you provide a sketch so we can see the pin length and other related dimensions.

desertfox

 

[dhengr]

MX:

Provide a sketch, you can see it, we can’t. You say a “1" dia. pin and supported on either side by circular guides,” but how long is the pin? And, you say that “At the point of failure, the load will be fixed directly down the center of the pin.” Does this mean the shear area will be (1" dia.)(pin length) and are the circular guides prevented from moving apart at the shear plane, to truly hold the shear pin captive, and shear it? i.e. the shearing load is applied parallel to the length of the pin, on the two guide halves containing the semi-circular guides.

Or are you shearing across the same pin area/plane but with the load applied 90° to the previous example; that is along the length of the pin and on its round surface at 12 & 6 o’clock? The semi-circular guides will not support this pin loading properly, without the two halves being forced apart.

Or are you shearing the pin through its circular area, a shear area of .785 sq.inches?

 

[mxrethoret]

Hey Guys,

Attached is a sketch of the problem. It can be assumed that the load is applied at the midpoint of the pin.

Dhengr... the sketch should answer the majority of your questions. The only thing holding the pin on the shear plane is a pnuematic mechanism that pushes the pin into place. Can you clarify why this is important to consider in this calculation? Is it because a horizontal constraint would reduce bending deflection and increase stress concentrations at the end?

Israelkk.. there is no cyclic loading.

 

[hydromarine]

You have given the clearance size of .125 between the pin and hole on each end is this correct? seems a lot of clearance

 

[mxrethoret]

Hydro,

This was not my design. I was simply asked to analyze the properties of the pre-existing design.

 

[rb1957]

it could be a dbly cantilevered beam, depending on the nuts on the ends, bearing against the supporting plates.

how does the difference in internal moment (between SS and dbly fixed) affect your design ? this is a teaching problem ... don't expect us to give you the answers (what'll you learn in that case ?)

this is meant to be your analysis of a simple problem. if you want the ends to be fixed, "make it so" ... what things would need to be defined in order to make the ends fixed ? what if the fixity relaxes under load ?? what is the critical mode of failure ?? (there's more than one)

research "structural fuzes" ... things designed to break at a predetermined load in a controlled manner.

 

[rb1957]

impact loading ??

 

[mxrethoret]

Thank you for the questions rb. I have come up with a number that seems to be independent of whether or not it is cantilevered at the end.

sigma = M*c/I ==> F(x=L/2) = sigmamax*I*(L/2)/c = 2,509 lb

I am a little concerned that the boundary conditions are not coming into play here as my intuition is telling me that they are important. If it was cantilevered, I would anticipate higher stress concentrations at the walls. In particular, there should be a reaction moment.. and this would mean an increase in overall moment. So... a cantilevered beam would be less capable of supporting the load. Is my train of thinking on the correct path here?

I will continue to research impact loading as well. Thanks again.

 

[chicopee]

first calculate the deflection as a simply supported beam to determine if the end clearances bind. If they dont than you are OK as a simply supported beam and use your formula. If binding occurs, then I would assume the pin has moments and rotation on each end and solve it that way but not with the ends fixed.

 

[dhengr]

MX:
“the load will be fixed directly down the center of the pin” has a number of potential different meanings; down the pin lengthwise, at its center, or across the round shape of the pin, at its center span, for example. Draw every possible alternative that your wording might allow/suggest, and you might see why we could be unsure of what you really meant. Again, you are looking at it, but we can’t see it from here, so you must be very careful that your verbiage only allows one word picture to be drawn, out of many. There isn’t anything wrong with a free hand sketch, but proportions are important, so reasonable scale and dimensioning is important, and that almost always shows things better than our verbiage.

By your sketch, I would say, it appears that you mean ‘the pin spans btwn. two .5" thk. support plates, 1.5" clear distance apart (btwn. them); and is loaded at its mid span by a third .5" thk. plate, and is shearing the pin through its round area (.785 sq.in.), in double shear.’ With the proportions your system has, I would say you have a simple beam with a span length of about 2", that gives you the worst bending stress on the pin; and you did say fail safe, so be conservative. The support plates aren’t thick enough to fix the pin ends, and the .125" dia. clearance certainly won’t fix them. In its simplest form the shear stress is Tau = Fmax./(2 x .785sq.in.). Then you might combine these two stresses and compare then to ultimate strength. You might determine that yielding is failure. And, in any case you will have some factor of safety or margin of safety against yield or ultimate. That’s called engineering judgement and is dependent upon the specifics of the problem. Finally, the specs. on the materials involved will give you some min. yield and ultimate tensile strength, and some chemistry, % elongation and the like, which start to indicate toughness, hardness, etc., but the actual yield for a given pin will most likely be higher than the min. value.

Furthermore, you will get considerable yielding in bearing (a Hertz stress problem) btwn. the pin and the plates before you shear the pin. And, any imperfections in the pin can drastically influence the actual failure number and mode. Impact loading and fatigue loading are also very important. This is actually a fairly complex little problem, and what you are looking for is a safe and educated force value that your system can withstand and still allow you to sleep at night, not hurt anyone, or seriously damage any equipment. You are not looking for an exact force value.

 

[rb1957]

sorry MX, but you can't get the same result independent of end fixity ... the max bending moment in a dbly cantilevered beam is about 1/2 that of a SS beam. your M (=PL/4) as pointed out above is for a SS beam. you're also only looking at elastic stresses ... what about plasticity ? (since you're concerned about breaking the pin in two)

like i said, there is more than one failure mode.