Help determining correct bolt size & number of bolts 2

[RohitGogna]

Hello,

I am trying to calculate the Moment and stress acting on the bolts connecting directly to the pan (please see attached images) to determine the bolt size and # of bolts to use. We are currently seeing the bolts fail way more often then they should be.

The method I have been using now is saying both ends of the pan are fixed and assuming the worst case scenario. Using the equation M = (w*L^2)/12 to determine the moment acting on the brackets.

Where w is the distributed load from the pan = 4720LB/29.145ft = 161.91 lb/ft
Where L is the entire length of the pan = 29.145 ft
Where M is the moment acting on the bolts.

From the equation I get that the moment acting on the bolts is 11460.94 lb-ft which is equal to 137531.37 lb-in.

Then using the equation sigma = (32*M)/((pi)*d^3) to determine the stress seen on 1 bolt (then divide stresses by proof strength of material to determine # of bolts required)

Where M is the moment we just calculated in the above equation (137531.37 lb-in)
Where d is the minor diameter of a 3/4" bolt which is 0.6255"
Where sigma is the stress seen on a single bolt.

From the equation I get that sigma is equal to 5,724,267.96 psi which to me does not seem to be reasonable and make much sense.

Is there something that I am missing? Am I going about solving this problem all wrong? Can someone please provide some insight as to how to approach this problem to determine the correct bolt & # of bolts to use?

Please let me know if you require any additional information in order to help me determine the proper solution to this problem.

 

[RohitGogna]

I have also attached a drawing showing the bolt location connecting to the pan

 

[chicopee]

First your brackets should have stiffners to minimize the brackets from flexing which creates a prying action on the horizontal bolts and vertical ones. You have not stated how the horizontal bolts are failing. Shear prevails on the horizontal bolts but under prying action they will also be under some tension. The vertical bolts will be under tension.

 

[RohitGogna]

I have attached a picture of how the bolts are failing. It seems as though both bolts are failing in shear, however the bolts on the left hand side end up breaking first and we believe that they bang around a bit before the bolts on the right side finally break which is why they have a rounded edge. Can someone please take a look at the pictures to confirm, or give their thoughts/opinion?

There are 2 sets of bolts in the picture from 2 different failures. One set of bolts is using Grade 5 Steel the other 316L stainless steel.

In this situation the bolts would be in single shear. therefore the equation to calculate the shear stress in the bolt is tau=F/A

Where tau is the Shear Stress in a single bolt (divide value by 2 since we have 2 bolts here)
Where F is the applied force on the bolts. In this case it would be half the weight of the pan. So it would be 2360lb
Where A is the cross sectional area of the bolt using the minor diameter of a 3/4" bolt which is 0.6255".

tau = (2360lb) / [(pi)*(0.6255"/2)^2]

This give me a Shear Stress of 7675.00psi. The shear allowable in grade 5 and 316L stainless steel can be assumed to be approx. 60% the tensile design load.
For a Grade 5 material we have a tensile load allowable of 120,000 psi which after you multiply by 0.6 gives you a shear allowable of 72,000 psi
For a 316L Stainless Steel we have a tensile load allowable of 75,000 psi which after you multiply by 0.6 gives you a shear allowable of 45,000 psi

These are both much larger than the calculated shear stress above. Is there something I may be missing/overlooking?

Thank you for taking the time to respond to this post.

 

[Screwman1]

Those horizontal bolts look like a fatigue failure to me. Shear allowable would be for a dead load case. A cyclic service load will create a fatigue situation.
If it was my joint, I would switch from Grade 5 to Grade 8 bolts and then torque them up to 75% of yield to freeze the joint and prevent lateral movement. Most fatigue problems can be solved with more clamp load.

 

[BrianE22]

It seems like shear-wise your bolts are o.k. (if they remain tight). Bending of the bracket would be a tough calculation given the ribs you have. Can you do an FEA of the bracket?

Does the pan lengthen or shorten from temperature changes?

 

[RohitGogna]

Unfortunately we do not have a software to perform an FEA on the bracket or bolts.

The pan is in the same environment (slightly corrosive due to all the chemicals and water) the entire time and it does not really change in temperature (might change by 1 or 2 degrees but nothing substantial).

So I do not think it we need to worry about the pan shrinking or expanding due to temperature changes.

 

[SwinnyGG]

The second part of your calculation (where you're getting 5.5 million psi) doesn't make sense, because your choice of equation is incorrect. It looks to me like you're using some form of the equation for stress due to torsion, or something.

But, are the bolts being loaded directly in torsion due to the live load? No.

Are the bolts being loaded in bending due to the applied moment? No.

Because of your bracket design, your bolt is loaded in two ways- shear from the direct gravitational load, and tension from prying action on the bolts.

Shear stress is relatively easy. You have four bolts, two at each end. Assuming that your 'pan' is level and is actually evenly loaded, your shear force is 1/4 of the gravitational load.

4720 lb / 4 =1180 lb per bolt.

Shear stress = F/A = 1180 lb / .302 in^2 = 3907 psi shear stress per bolt

If you assume that the bracket doesn't deflect, than you would stop here, and you'd be scratching your head about these failed bolts. The problem is that this quick analysis neglects prying action- which, judging by your drawings, is huge in this case.

As the bracket is loaded, the vertical bolts cause the bracket to want to act like a fixed cantilever. As that cantilever deflects down, it also pulls away from the 'pan', as the end of the cantilever wants to follow an arc. This puts the horizontal part of the bracket in tension, which in turn puts the vertical part of the bracket in bending. The only thing resisting this bending moment is a couple between the other end of the flange, which is under compression against the side of the 'pan', and the tension force in the bolt.

The equation then gets complicated, because to arrive at true bolt force you have to balance the deflection due the gravity and the prying moment- these two forces resist each other, so calculating either one in a vacuum gives a bad result.

By some rough calculation and estimation of how those forces will balance (which I'm not putting here because this post is already really long), the bracket will deflect downward in the realm of .040", which means that the vertical part of the flange, at its upper end, will deflect away from the wall of the pan something like .010". That means the bending moment in the bracket is around 12,000 in lb (3,000 lbf x 4 in effective length) per bolt.

The moment arm that the bolt has available to resist this moment is very very small- its the distance between the bolt centerline and edge of the bolt head.

12,000 in lb / 1.125" (width of .75" bolt head) / 2 (because we are using bolt centerline) = 21,300 lb (!)

21,300 lb / .302 in^2 = 70,500 psi (!)

The proof strength of a grade 5 bolt is only 85,000 psi.

That static stress in each bolt is about 75,000 psi, which is about 90% of the proof stress. This leaves very, very little margin for load dynamics, vibration, or other effects; this also means that the bolts will become vulnerable to failure due to very minor corrosion.

If I were you, my very first move would be to change to grade 8 bolts. Proof stress for grade 8 is about 40% higher, which will give you a lot more margin.

If the material of the 'pan' is not strong enough to handle the preload required for a grade 8 bolt to fully develop, than the next best option would be to re-design the mounting brackets to avoid prying action, since that is the dominant stress component that is causing failure.

 

[RohitGogna]

That was an amazing explanation jgKRI. Thank you so much for explaining this to me.

If you don't mind me asking how did you get the 3000lb x 4" effective length of the both to achieve the moment in the bracket of 12,000 lb-in? I am failing to understand where the 3000lb comes from. I am assuming that you assumed the 4" effective length for the bolt? and that this should actually be the length of the bolt engaged with the threads?

And if its not to much trouble can you please post the information about how you balanced the forces? I would like to learn and understand exactly what you did behind the scenes as well so in the future if a similar problem arises I would be able to solve it on my own.

Thank you again. This was extremely helpful.

 

[SwinnyGG]

3,000 lb is a rough estimate of tensile force in the horizontal part of the bracket.

4 in. is the length of the bracket (taken from your drawing) between the upper surface and the centerline of the horizontal bolts.

'effective length' as I have used it is a property of the bracket, not the bolts.

 

[BrianE22]

You could reduce the bending moment (and force on the bolts) by making the brackets less stiff in bending. Your pan would deflect more though.