Bolt preload vs failure in shear 1

[Chiwahua]

We have a case where we have a bolt that fails in shear.

Am I wrong to think that:

the higher the tension in the bolt, the less it takes to be able to shear the bolt?!

For example:

Bolt 1 is torqued at 600 lbs.ft
Bolt 2 is torqued at 200 lbs.ft

Let's pretend the friction due to the clamping force doesn't exist, it would takes less force to shear the bolt 1, right?

Thanks a lot for your help!

 

[rb1957]

In the analysis world, the shear loadpath between the plates is the bolt. bolt allowable is tension-shear interaction. So higher tension = less allowable shear.

In the real world there are two shear loadpaths ... the bolt and friction between the plates. then practically higher preload means more shear carried by friction !?

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

 

[drawoh]

Chiwahua,

Let's not ignore friction. If the bolts are clamped down hard enough, plates are held by friction, and there is no shear load on the bolt. If the bolt has come loose or is deliberately assembled loosely, there is no friction, and the joint is held entirely by shear. There is a transition where there is some friction clamping and some shear, but for most joints, this is a very tiny range of nut movement. Would you really try to design for this state? This is mostly a failure condition.

Your bolt torques have no meaning unless you tell us the size of your bolts and what they are clamping.

--
JHG

 

[SandCounter]

If a bolt is used properly it induces enough friction that the bolt is only in tension. Your observation is a good example of why bolts should be properly torqued.

I used to count sand. Now I don't count at all.

 

[EdStainless]

And if they are not torqued enough you might load them in bending also, then the whole thing is short.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, Plymouth Tube

 

[Tmoose]

" a bolt that fails in shear."

One bolt in a pattern of 6 holding a single component?
One bolt holding a single component, like serving as the axle for a pulley or wheel.

What is the evidence it is a shear failure ?
Does your bolt's fracture surface look like any of these?

http://people.exeter.ac.uk/TWDavies/solid mechanics/lecture 3/Image47.gif

http://www.onallcylinders.com/2014/03/13/diagnose-common-fastener-failures/

http://www.subaruforester.org/vbull...er-bolt-failure-oem-gates-others-img_1655.jpg

https://encrypted-tbn0.gstatic.com/...PE6-9y4rbVwFfT2JQgoZySGFjZyneVfMgiJiZfnAebiOw

Please post twenty seven eight-by-ten color glossy photographs with circles and arrows and a paragraph on the back of each one explaining what each one was to be used as evidence.

 

[Chiwahua]

Thanks for your answers so far.

At first my post was more about the theory that the higher the tension in the bolt, the less it would take to shear it.

But as I understand you want to know more about the reason why I ask the question.

So here is an image.

It's a load-lifting mechanism (cylinder is black and the mechanism is yellow) attached to a chassis (grey) and the pivot is orange and you see the locking mechanism that is a 5/8-11 Grade 8 bolt that is running through the "locking ring" welded onto the chassis. The shaft is 2"OD, the locking ring is 3" OD.

At the installation, the torque on the bolt is not controlled (no torque wrench). The bolt is installed with an impact wrench. From my research, the impact could give a torque up to 600lbs.ft. Not that I have reasons to believe it was torque this high but I just don't know how each mechanic "insists" on the impact.. The suggested torque for this fasteners is somewhere between 170-220 (lub-dry).

We've had failures in service that looks like double shear I believe:

So, obviously we will need to specify the torque, and since I suspect over-torquing being a part of the problem, I would tend to keep the torque minimal to allow more room to withstand higher shear but I know that doing this, I'm removing the friction from my side but from what FEA have shown me is that a 200 lbs.ft would not squeeze the shaft enough that the bolt would not see any shear.

Also, If you have good shaft-locking mechanisms you want to share, I'll be glad to consider.

Thanks again!

 

[drawoh]

Chiwahua,

Okay. That has no resemblance to anything I was conceiving, above. Regardless of how your bolt is tightened, you have a full shear load on it. If you tighten the bolt hard, you have combined stresses, and a rather complicated analysis. You need Von Mises stress (Tresca stress?), not a fastener manual.

Can you drill the bolt and use a castle nut and cotter pin?

--
JHG

 

[rb1957]

I don't understand the pictures relative to the drawing. The drawing shows a hinge pin joining a bunch of pieces together with a cross bolt at the end (to prevent travel of the hinge pin). The pictures look to have a plain slug in lieu of the bolt?

That said, what's being tightened by the impactor ? From the drwgs I assume the cross bolt ? So if the failure is these bolts are being Way over-tightened (nothing being clamped up, just a nut being tightened on shaft) and then there's some lateral load applied in operation, shearing the cross bolt. In this case it is a classical tension-shear interaction, Rt^2+Rs^2 = 1. My suggested solution is to torque the nut with a wrench; if you want to control the torque, then use a torque wrench. Or else add a spring element under the nut, so there is some compression available and some visible sign about how tight the nut is.

Do you understand where the lateral load is coming from ? Is it expected ? Or is it a load not intended by the original design. If so, then you should re-design the joint to account for this lateral load.



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

 

[drawoh]

Chiwahua,

Have you analyzed this thing at all? For a 5/8-11UNC Grade[ ]8 bolt, at 90% of yield stress and no lubrication, I am getting about 250lb.ft of torque maximum. That bolt looks long enough that you can control the tension by calculating turns of the nut. What is pulling your pivot out of the hole?

--
JHG

 

[Chiwahua]

So from your comments, having a minimal torque (let's say 50 lbs.ft) would kind of make sense? To have plenty of room for the shearing.

Here is a non-exploded view:

The pictures above showed a shaft half-removed (man-removed, and not a sideways load) and remainings of bolts. That looks to have double sheared, right?

rb1957, the equation Rt^2+Rs^2 = 1 puzzles me a bit.

Rt would stand for (tension stress ratio) and
Rs would stand for (shear stress ratio),

Rt^2+Rs^2 = 1 should be <1 instead, right?

drawoh, I know the bolt would not torque up to 600 lbs.ft It would break much before. I'm just saying the impact could torque up to 600, meaning that maybe the bolt was over-torqued but not knowing at what point.

 

[rb1957]

yes ... MS = 1/sqrt(Rt^2+Rs^2) -1

the parts (the small pin) in the pic just don't look like a bolt.

part of your problem is that you've got a "gorilla" with an impacter and he's torquing the cr@p out of the nut ... surprised you're not finding the bolt in two pieces (on installation). Find the picture about "who decides the torque on a bolt". As currently designed I think the joint is too rigid for the simple tools (and tool holders?) that you want/need to use. I thinkyou need to introduce some soft element (copper washer ?) that'll collapse under load.

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

 

[drawoh]

Chiwahua,

I repeat -- you need to do an analysis.

[ol 1]
[li]What is the shear force of your pivot?[/li]
[li]Should there be a shear force on your pivot? Do you understand your mechanism?[/li]
[li]How strong must your pivot retainer be to withstand your shear force? Perhaps your shear force is beyond the capacity of any 5/8 bolt![/li]
[/ol]

--
JHG

 

[hydtools]

I see no reason for there to be any shear load. Use a lynch pin instead of a bolt and nut.

Ted

 

[Chiwahua]

Drawoh:

yes, I have started an analysis on this. It's just, that in the mean time, I'm asked to specify a torque for this bolt before I got all the results and I obviously want to be on the safe side.

yes it makes sense that the pivot is forced to turn (and shear the bolt, the lifting mechanism rotates around the pivot and friction is present.

so far, the FEAs give me around ~11 000 lbs shearing force at each interface.

rb1957: what do you mean the "joint is too rigid"?

 

[gtaw]

You have a problem that is different that what you suspect. The keeper's, i.e., the bolt, purpose is to keep the pin from working out when the machine is in operation. The various components,i.e., the hydraulic cylinder should be rotating freely on the pin. If there is insufficient or no lubrication on the pin, the hydraulic cylinder is forcing the pin to rotate and shear both ends of the keeper (as seen in the photograph). I suspect the pin and the hydraulic cylinder are galled and roughened. Now, no amount of lubrication is going to help. The clevis end of the hydraulic cylinder needs to be repaired and a new pin with grease fittings installed. The pins do have a grease fitting right?

Best regards - Al

 

[gtaw]

Almost forgot to complement Tmoose for the photographs. Excellent my friend!

Best regards - Al

 

[Chiwahua]

gtaw: There is grease that should be applied through a grease fitting in the pivot pin and you can see the greasing holes in the pivot. We have no way of knowing if the grease is applied regularly at our customer's.

 

[drawoh]

Chiwahua,

I have another thought here. Is your bolt failing at 1.5[&times;]capacity, or at 15[&times;]capacity? Perhaps someone has unwittingly designed a shear pin, and it is preventing a more catastrophic failure. You need to thoroughly understand how your pivot works. This is not FEA. This is mechanics of machines and machine design.

--
JHG

 

[rb1957]

"so far, the FEAs give me around ~11 000 lbs shearing force at each interface." but this shear is carried by the hinge pin; this is not the shear that is causing the failure (which looks to be axial along the pin, yes?) or is this force acting along the hinge pin axis ?

rb1957: what do you mean the "joint is too rigid"? ... it looks to me has though you are putting a bolt into a hole in a solid shaft (as a cross bolt) then torquing the nut against the shaft. There is nothing to deflect under the tension from the torque, not flexibility in the joint, you're tightening the head of the bolt with the nut on the opposite side of the solid shaft. Without careful control of the torque it is easy to overload the bolt, it is difficult to know the torque you're applying.

as others have noted, there are many possible failure causes or contributors

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