Stresses on a hinge pin 2

[tnteng]

Can anyone help on this problem?

A hinge pin is loaded such that the the entire length of the pin is bearing against the hinge padeyes. There is an upper padeye and a lower padeye on one side of the hinge and only one middle padeye on the other side of the hinge. Double shear of the pin will be considered. Would bending of the pin need to be considered and combined with the shear stress as a combined loading?

If bending is considered, what is the typical loading pattern used in this analysis?

Thanks in advance,

 

[israelkk]

From my experience after seeing many broken hinge pins, double shear is an optimistic approach, although it is the usual way it is presenred in the engineering books and where the reaction forces are taken at the middle of the padeye.

However, this is only true when the hinge is new, in time with wear and deformations the bearing holes become larger and then the reactions move to the farmost sides of the bearing padeye. Unless the pin is wery thick (more friction moments loses in the hinge) the pin deflects. Therefore, as the distance between the reaction forces is now larger than the original distance (when the hinge was new), the bending forces become very dominant.

Therefore, a good design should make the hinge as narrow as possible, meaning, shorter pin and as a result less bending moments. The designer should assume that the reactions may move to the edges of the pin and calculate the bending stresses in order to be on the safe side.

Few years ago I had a chance to invesigate and solve such a problem after many occasion of pin break before time. The hinge width was ~20mm wide and the pin was 8mm diameter. My calculations showed that taking into consideration that the reaction moved to the end of the pin, the pin should be brake at the middle, and this was the case.

The new design was 8 mm wide hinge with 5 mm dia pin and this never break.

 

[dooron]

The pin will usually bend before it shears. The lugs should also be checked for tearout etc. Bruhn "analysis and design of flight vehicle structures" sec.D1.14 has some guidelines on where the load is. Apint load at the centre of the beam with a beam span equal to the outer extremes of the lugs is probably a good start for bending stress check.

 

[cb68]

This is very interesting. I have always assumed that it is a double shear problem. When considering the bending stresses do I also need to combine these stresses with that of the shear or is it a case of when the pin is new it is a double shear problem and when the pin is worn then it is a pure bending problem.

Excuse my ignorance if this is trivial but if I need to combine these 2 stresses how would I do this?

 

[zvivik]

in order to check how far is the part from failure in case of biaxial or triaxial stress ,various criterions were developed.these are called failure theories.
in my field (lifting appliances) codes define the equivalent stress as:sigma=sqrt(sigmax^2+sigmay^2-sigmax*sigmay+3*tau^2).this should be <=sigma allowable.
hope this answers your question
Zvi

 

[cb68]

Thanks for the response. I'm therefore assuming that your equation will simplify in this case to sigma=sqrt(sigmax^2+3*tau^2). Ie sigmay=0. This is the case with bending and shear however in this example would I need to use bending and shear combined or are they mutually exclusive events. Ie one when pin is new and one when pin is worn.

 

[israelkk]

When you assume a worn joint ideally the full load is in the middle of the pin and the reactions are at the ends of the pin (worst case). Therefore, it is a symetrical case and you can analyse it as a beam with 1/2 pin length fixed at one end (was the middle of the pin) and loaded at the end by 1/2 of the full load on the joint.

When you design the joint it has to withstand both the shear and the bending stresses simultaniously because at the fixed side the bean fills the shear stress as the result of the load at the end and the tensile stress from the bending caused by the load a the end.

 

[cb68]

israelkk,
So if I'm understanding correctly following on from zvivik we get

sigmax=(My)/I where M=1/2 Pin Length * 1/2 Load
tau=F/A where F=1/2 Load

Then use failure criterian (Is this Von Mises??) as presented by zvivik.

Have I understood you correctly??

 

[chili]

The bending moment, worst case, would be M=PL/4. Where
L is the maximum length between the &quot;simple&quot; supports
P is the maximum load in the center.
You can calculate your section modulus as Z=d^3/6. (All units in inches of course.)
Your bending stress (using AISC terms) is fb=M/Z. Y
our allowable stress for this case is Fb=.75*Fy.

For the shear, fv = P/A and the allowable Fv=.4Fy

One way I have done it is via the unity check where Fv/fv + Fb/fb =< 1.0

At least that is what us old farts have always done.....Chili

 

[chili]

sorry....i screwed up the unity check, but you know what I meant....Chili

 

[israelkk]

cb68:

Yes, you are correct.