STRESS ANALYSIS ON A BOLT IN SINGLE SHEAR

[Anonymous]

Hi all,

Wondering if someone could help me. The problem is simply reconciling stress analysis results with hand calculations and I suspect I'm simply setting up constraints incorrectly or something.

A simple problem of an M6 bolt in single shear as in the image below. I am using the forumla in image to calculate the shear stress. The diameter (being an M6 bolt) is 6mm and hence radius is 3mm with a shear force of 1000 N being used. Using the forumla in the image below and the figures above, I calculate the shear stress in the bolt to be 35.4 MN/m^2 or 35.4 MPa if you prefer.

My problem is that when I simulate this problem in Inventor 2013 stress analysis the results (for any kind of stress analysis; Von Mises, 1st and 3rd principal, all the planes etc...) are way way off what the hand calculation is, ranging from 160 MPa to around 800 MPa.

Could somebody with the time to do so possibly talk me/post images of how to set this example up in Inventor correctly so that you get the same (or near enough) result.

Many thanks,

Sam

[JDMather]

Attach the ipt file here.

Your formula gives the area of a circle where the image shows the force across the face area of two cylindrical faces.

[JDMather]

Any progress on this?

Force over area of a circle or area of a cylindrical face?

[Anonymous]

Hi JD,

Still no luck really. I appreciate what you're saying in that in my forumla the force is acting over the cross section of a circle and doesn't take into account the thickness. I've used a different forumla: Shear stress = F/Dia.*(t1+t2) but still similarly disproportinate results. Besides all the text books I've used state the forumla I originally posted.

Here's a link to the IPT file if you wanted to have a look at how I've set it up (which may well be incorrect). It was slightly too big to attach directly.

http://dl.dropbox.com/u/35359353/Shear%20stress%20in%20bolt.ipt

Regards,

Sam

[JDMather]

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@Anonymous wrote:

Here's a link to the IPT file ...

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ipt files can be attached directly here.  I don't go to 3rd party sites.

[Anonymous]

It wouldn't let me attach it directly as it said the file size was too big. It's just a dropbox link, if you prefer I could email it to you. Or is there a way to shrink the file size?

[JDMather]

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@Anonymous wrote:

Or is there a way to shrink the file size?

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Find the red End of Part marker at the bottom of the browser.
Drag the red EOP to the top of the feature tree hiding all features.
Now save the part with the EOP in this rolled up state, the file should be much smaller.

If it is still too large (over 1.5M) then right click on the filename in Windows Explorer and select Send to Compressed (zipped) Folder.  Attach the resulting *.zip file here.

I am a bit curious how geometry as simple as that which you depict could be so large a file size, so later when I get to my "safe" test machine where I can download anything I will try your dropbox link.

[Anonymous]

Did what you said, also deleted the simulation from it and put it in a .zip folder.

[JDMather]

I think something is wrong with this file but I haven't figured out what, yet.  The file size is much larger than I would expect.

[karthur1]

The shear stress (4F/PI*D^2) is the formula for AVERAGE Shear stress on the cross section.  Your math is correct for this single shear problem (35 MPa).

The stress JD is talking about is the Nominal BEARING stress which equals P/A.  Where A= D*t.  In your case, D=6 and t=3.  So the nominal bearing stress would be 1000/(6*3) or 55.6 MPa.

I tried to setup the stress analysis on my own part also, but I couldnt get Inventor to give me the correct answer.  I suspect it is because FEA is looking at an infinitesimally small point that cause the stress to increase.

[Anonymous]

That clarifies things hugely!

The result you got there in your simulation is just what I was looking for. Although not 35MPa, 43MPa is close enough for me, as you say it's an infintesimally small point so the FEA result you got is the one I'm after.

2 Things...

What constrains did you use?

Also your result is the 'Stress in XZ'...maybe a stupid question but how did you know that was the right one to use?

[karthur1 · Accepted solution]

For the constraint, I used a fixed constraint on the band of material between the two planes.

In order to know which plane to look at for the stress. I have the Origin 3D Indicator turned on.  If you dont see it on your screen, you can turn it on in Tools>Application options, Display Panel. Check the "Show Origin 3D indicator" box.

[Anonymous]

Many thanks!

All sorted now.

One last question... Although I've managed to resolve this issue, and established that the shear stress (in the XZ plane in your case) is 43MPa, because the Von Mises result is much bigger, I assume that means that it will still somehow see that stress and therefore the 43MPa shear stress isn't what I should be worrying about.

After all, what I'm trying to find out is if this bolt will fail when subjected to a 1KN load in this configuration.

[kstate92]

I have no advice to offer, but this thread reminded me of what Dr. Jess Comer said in the SAE Bolted Joint seminar I attended: "Bolts make lousy pins."