Shear Stress vs Bending 1

[jdkuhndog]

When analyzing a clevis type connection...where the pin will be in double shear...at what point (distance between clevis supports) do you need to consider the bending stress in the pin?

For example...I have a typical application with a 3/4" pin seeing a load of 5.5 kips...the clevis supports are on a 1.34" center to center spacing (approx 1" inside to inside)

Area = 0.44 in^2
Sx = 0.041 in^3

Shear Stress = 1/2 x (5.5) / 0.44 = 6.25 ksi

but if I would analyze the bending here it would be...

M = PL/4 = 5.5*1.34/4 = 1.84 in*kips (if analyzed as simply supported)
M = PL/8 = 5.5*1.34/8 = 0.92 in*kips (if analyzed as fixed)

Bending Stress = M/S = M/0.041 = 44.8 ksi or 22 ksi

In either analysis(simply supported or fixed), the bending stress starts to get pretty high.

I usually analyze the bending stress, but is it necessary?


Thanks !

 

[liberty]

Have you considered combining the bending and shear stresses to determine the effective, or "von Mises" stress? You may have a case where instead of simple stress, you have biaxial or triaxial stresses which are more complicated to quantify. This could be wrong, but if your problem is a case of biaxial stress try this:

SIGMA' = (SIGMAa^2-SIGMAa*SIGMAb+SIGMAb^2)^(1/2)

where SIGMA' = von Mises stress
SIGMAa = principal bending stress
SIGMAb = principal shear stress

For better understanding of your problem, or if you discover you have a triaxial stress case, I recommend reviewing distortion-energy theory and other failure theories. See Chapter 6 of ME Design by Shigley/Mischke 5th edition 1989.

 

[jdkuhndog]

Thanks liberty.

Yes, I usually do combine stresses.

The reason I am asking about the bending in this case, is that I have seen calculations that don't even consider bending in this case. They assume that the clevis is a pure shear connection.

 

[liberty]

OK think about this: What minimum criteria does the clevis pin have to meet without failing? I ask because maybe the customer only needs the pin to hold together, without fracturing. What if it does yield in bending a little - does it matter to him, as long as he can easily remove/replace a bent clevis pin when necessary? I'll bet that's why bending was not considered in other examples you've found -- they only cared about meeting the ultimate shear strength requirement.

Sometimes we engineers design to requirements that we make up in our own heads, which the customer could care less about and would rather not pay for. I've done it lots of times...

 

[jdkuhndog]

No. The stresses imposed on components must meet a Standard... for bending ultimate/5 for transv. shear stress ultimate/(5 * sqrt(3)) for torsional ultimate/(5* sqrt(3))...for combined bending and torsional by Von Mises..allowable = ultimate/5.

so, I am aware what I need to check...was just curious if bending is really imposed in a clevis type arrangement. I assumed it was fixed in its support...I can even reduce the stress more by assuming a distributed load instead of a point load at the center.

Thanks.

 

[liberty]

Distributing the clevis pin support loading does sound more realistic - also distribute the loading on the pin between the supports if applicable. See what the shear and moment diagrams look like then. Good luck!

 

[GregLocock]

The simple approach is to use a 4 point loads on the pin, and check the resulting bending stresses. The 4 points are the inner edges of the clevis, and the outer edges of the bearing inside the clevis


V V
------------------------
^ ^

If your pin is 'too' long then of course the bending stresses can be significant.

Cheers

Greg Locock

 

[magicme]

this pin geometry (L/D = 1.33) really is not a beam in bending, and the textbook beam equations are derived for large L/D ( I think > 20) , so I have always treated this type of geometry as a shear connection, and customers have approved that approach.
Also .....I would not combine bending and shear stresses as max bending occurs at the outer fibers and shear is zero there; max shear occurs at the neutral axis and bending is zero there.

daveleo

 

[SuperStress]

The four point bending method suggested by Greg Locock is slightly unconservative because as the lug and clevis ears begin to yield in bearing, the effective distance used to compute the moment increases.

There are a variety of aerospace textbooks that deal with this topic. "Analysis and Design of Flight Vehicle Structures" by Bruhn is the "bible" and contains a conservative approach for computation of the bending in the pin.

The Bruhn approach is to compute the bending distance as tlug/4 + tclevis/2 + gap. (Where "gap" is the physical gap on each side between the lug and the clevis).

There are many less conservative methods found in company manuals and other texts.

As for when you need to check the bending, it's pretty much a function of how the various elements of the joint are designed. (e/D, thicknesses, gaps, pin diameter, hollow or solid pin, etc.)

Bending should definitely be checked whenever large tolerances allow significant fit-up gaps or when lug or clevis materials are soft relative to the hardness of the bolt. It's really not purely a function of distance between the clevis ears.

SuperStress

 

[chili]

As good engineers, we shouldn't overlook any of these issues. Check bending and shear....the unity check will normally take care of unusual geometry issues. Be sure to look at your lugs as well and don't forget bearing stress.

AISC is a good reference that has acceptable criteria for all of these cases.

Happy New Year!

 

[chicopee]

I would consider bending if you have a point load or a contact load along a partiallength of the exposed pin and combine your average and the bending stresses vectorially. If loading is along the entire exposed length of the pin then I would not consider the effect of bending and just determine the average shear stress. For a factor of safety use five which is the value prevalently used in the Crosby catalogue.