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Slender Column Bolted Joints 2

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cb68

Mechanical
Jul 5, 2003
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I originally posted this on 'Mechanical engineering other topics'. I have been asked to now post on this forum instead. Apologies if anyone has already read.

I have a slender square aluminium (6082 T6) column that is 44mm square and 160mm long. A load of 2250N is taken through the centre of the column in the 160mm direction. I have clearance holes for M5 A2 Class 70 stainless steel studding in the 4 corners of the column drilled the full length of the 160mm. The centre of each hole is mounted 5.4mm in from each respective edge.

The column is mounted on a flat stainless steel 316L plate with four tapped holes of 12mm full thread depth into which the studding is screwed and held with loctite in position. The other end is stainless steel flat washer and A2 nut.

I have assumed:
1. I am torquing the nuts to 3Nm.
2. Each stud carries 1/4 of the load ie 562.5N
3. That the stiffness of the studding is greater than the member stiffness. In fact I have said that the stud carries the full load.

So

Using T=0.2Pd where T is tightening torque, P is tension induced by torque and d is nominal diameter I end up with a P of 3000N. Add the external load of 562.5N to 3000N = 3562.5N. The tensile stress area for an M5 is 14.2mm2. Therefore stress is 250MPa. Now A2 stainless has a Proof Stress of 450MPa giving me 56% of proof stress.

Now I have heard rumours that for a joint that is dissassembled you can typically run up to 75% of proof. Can anybody shed some light on this in particular with relation to A2 stainless?

Next I thought would the studding strip from the 316L plate. Now Juvinall - Fundamentals of Machine Component Design presents a formula F=pi.d(0.75t)(0.58Sy) which is the tensile force to strip a nut of thickness t & nominal diameter d. With an Sy of 240MPa for 316L I end up with an F of 19,700N which gives a big safety factor of 5.5.

As I am not an experienced stress engineer do these assumptions and calculations seem respectable? Is there any other failure mode I have missed (eg should I be checking the nut)?

Thanks
 
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These calculations are respectable, although they do not represent the utmost in accuracy. I recommend you obtain Guide to Design Criteria for Bolted and Riveted Joints available at:


You should check the nut by insuring that its material and property class are compatible with the screw. One other failure mode to check is allowable surface pressure on the column.

Good luck.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
Thanks for your input CoryPad

When you say they do not represent the utmost in accuracy do you mean they are conservative or that a more detailed analysis is reqd? I am happy with conservative in this application as it is a low volume design and so I cannot justify extended analysis.
 
SteveBraune

No you have not misread my post. In hindsight I used bad terminology. I was trying to get across that the bolts were running in the direction of the longest edge.

Also following on from CoryPad's advice I did some more research in attempt to learn a little more.

A couple of things. With the geometry I have I think I might struggle to work out the relative stiffnesses of both the column and the studding so I am tempted to stick with the assumption that the studding carries the full load. Does this seem reasonable?

In addition I looked into the F=pi.d(0.75t)(0.58Sy). Now from this formula if I'm correct the 0.58Sy is just distortion energy theory. Therefore according to this formula the Thread Shear Area is pi.d(0.75t) which for an M5 yields 11.8tmm2. Now I have found that ISO898-1 Mechanical properties of fasteners made of carbon steel and alloy steel - Part 1: Bolts, screws and studs lists the shear area as 0.5.pi.t(d-0.64952p) where p is the pitch. For an M5 this yields shear area of 7.0tmm2. This is far below that proposed by Juvinall.

Now taking this a little further and determining the thread length required to ensure the studding fails before the thread strips. The 316L has a yield 250MPa which gives a proof stress of approximately 216MPa with the studding having 450MPa. Now the studding will fail when it's tensile stress area is 14.2mm2. If we use distortion energy theory with Max Shear = 0.58 Max Tensile the ISO formula yields 14.2=(216/450)(7.0t)x0.58 which gives a t of 7.2mm and if I use Juvinall (which I used in my original post) we get 14.2=(216/450)(11.8t)x0.58 which gives a t of 4.3mm.

Following this methodology my Juvinall formula to ISO formula yields extremely different results or am I missing something here. Can anybody shed some light on what should be used?
 
cb68,

ISO 898-1 does not show any shear area information. Can you recheck your equation and source?

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
CoryPad,

My source was which references ISO 898-1 at the very top. The equations seem to be correct.

Thinking about this again I think I may now see my error. The Juvinall equation is the tensile force to strip a nut where the equation I have presented as ISO 898-1 appears to be for the screw. If my understanding is correct then these would be different with the screw shear area being less than the nut. Does this sound reasonable?

Can you confirm if the Juvinall formula is one that you are familiar with or if their is something more appropriate?

This has now gone well above my work remit but I'm now intrigued.

Thanks
 
Those equations are NOT from ISO 898-1. I have seen them elsewhere - they are correct. You are correct about screw versus nut stripping areas.


Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
 
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