Bolted joint resistances. 1

[dbecker]

Hello,

I am trying to calculate roughly the value for bolted copper busbar resistance.

I have viewed this website

http://www.copperinfo.co.uk/busbars/pub22-copper-for-busbars/sec7.htm#Joint

But the information is less consistent than reality.

I am seeing more like 5-7 micro-ohms resistance across a bolted connection. About half inch thick copper on both ends overlapping about 2 inches, with belleville washers.

Does that sound reasonable?

Thanks

 

[SparkyLarks]

5 to 7 micro ohms seames reaosonable.

I think that national gid say 5 ohms for a fixed connectin and 10 micro ohms for a flexible connection.

I did some testing last week and it was stainless steel lugs bolted to copper busbar, and saw 1.7 to 3.3 micro ohms.

Got one reading of 14micro ohms because the swong washer was used.

What size are the bolts, what torque is used?

 

[dbecker]

Thank you for this valuable information.

Do you have any documentation that says what the lug connection value should be? You were mentioning something about the national gid, is this a code I can follow? Do you have a link?

Regards

 

[desertfox]

Hi dbecker

What is your joint area? how many bolts? what torque are you using and finally which belleville washer are you using?
If you can provide more information like that above we may be able to help further.

desertfox

 

[dbecker]

Hello desertfox,

The contact area is 25.4 mm by 25.4 mm with two holes centered, each hole is 12.7 mm in diameter.

Hardware, two M12 bolts with one belleville washer on each side. I do not know which kind, which are most typically used in industry? We can assume that for now.

I do not have a torque value, assume a typical value of 15 ft-lbs? Does that sound legitimate?

Any kind of ballpark answer with the above information will greatly help me at this time.

Material is fresh copper on copper, no oxidation.

Anything between 3 to 7 micro-ohms is what I am looking for, I just need confirmation.

Thanks again,

- D

 

[desertfox]

Hi dbecker

Yes your figure sound reasonable to me, I calculate your pressure to be around 13N/mm^2

desertfox

 

[dbecker]

Thank you desertfox, that is much appreciated information. I thought I was going crazy for a second there.

 

[dbecker]

Hi desertfox, can you please walk me through the hand calculations you performed? I would like to know how you did that.

Thank you,


- D

 

[panelman]

15 ftlbs looks light for an M12, that's more like what I'd put on an M8.

M12 end up at 30 or 40 ftlbs

 

[dbecker]

Hello, we can assume 30-40 ft-lbs. I just needed a starting number for the calculations, can someone walk me through how the calculation was performed? I am in dire need of this and cannot find much information on the web.

The website i posted up in the beginning of the thread is giving me a result thats 100x what desertfox and I agreed on. Which is why I am curious how to do this calculation. Please help, thank you.

 

[desertfox]

Hi dbecker

Haven't got time now to go into great detail, however I got the 13N/mm^2 pressure by calculating the axial load from the bolts based on the first torque figures you gave, then dividing that by your joit area ie:- 50.8mm overlap X 25.4
face area.
Now what does look in question is the units on the graph because it gives micro ohms per mm^2, which implys that you multiply the calculated pressure figure by the value obtained off the graph, not sure thats correct.
I'll look at it again tomorrow when I'm home.

Regards

desertfox

 

[desertfox]

Hi dbecker

Okay I have sorted it I think, first calculate the clamping load of each bolt from the 15 ft-lbs using the equation:-

F= T/(0.2*d) where T= torque
d = bolt dia
0.2 = friction factor

F = 15*12/(0.2*0.5) = 1000lbf

now you have two bolts :- 2 * 1000lbf = 2000lbf

joint pressure = 2000/(2"*1") = 1000lbf/in^2

convert to N/mm^2 = 12.44 N/mm^2

Now using the graph on the link you provided go along the horizontal axis till you get to 12.5N/mm^2, then go up vertically till you intersect the curve, then read across on the vertical axis a value for Y, in your case I read approx 2500 micro ohms/mm^2, finally divide this figure by the cross sectional area of the joint:-

2500/(2"*1"*25.4)^2 = 1.93 micro ohms.

Okay so I calculate 2 micro ohms for your joint, however a lot of factors can effect that figure ie: variation in bolt clamping load, surface finish of the copper, plating or tinning of the mating surface.
The tightening of the bolts is important, however you should not base it on the strength of the bolt material but on the stress in the copper joint due to bolt torque.
Your pressure on the copper looks about right to me but you should also calulate the stress in copper after the joint as expanded due to the normal service current generating a temperature rise.

desertfox

 

[dbecker]

Excellent desertfox, I will review this calculation and I may come back with a question. Thank you for taking the time with this it is extremely helpful for our project.

- Dan

 

[desertfox]

Hi dbecker

Your welcome its difficult to respond in the week as I am away from my main computor, however weekends I have good access.
Anyway this paper on bolted joints should help, see section 2.4 it analyses at busbar joint.

http://docs.google.com/viewer?a=v&q=cache:WtP95ocdxHoJ:

http://www.solonmfg.com/springs/pdfs

/DavetPaper.pdf+plain+washer+stiffness+in+bolt+joint+
analysis&hl=en&gl=uk&pid=bl&srcid=ADGEESip_PZWUsnbT-9G_fUtUf1zAMSIDfJcKn7LaYnA5xE1iwWA7Ms6Vzxa_
EVG6DjqkPbEbMjXnldawFuMj5giR2C8dL58Q1NJ4Am_MfQeZ3UKWH-mZpcrsOUGKnvuORY9f5x-9C9h&sig=AHIEtbQB1IPCSyo4zySPTKJ-MkKt7StKCQ

desertfox

 

[desertfox]

If the link doesn't work properly try this one its the same paper:-

http://www.solonmfg.com/springs/pdfs/DavetPaper.pdf

desertfox

 

[dbecker]

Hello desertfox,

I read through that document and there was no explicit way to compute resistance for a bolted joint. Although it was very helpful in understanding the belleville washer, it was more geared towards universal usage of them and only one segment devoted to electrical bus connections without any examples on contact resistance.


I will use your technique which you outlined using the website I first posted. That works fine for my case.

Thanks!

 

[dbecker]

desertfox,

When you say divide by the cross sectional area of the joint to obtain the 2 micro ohms, which cross sectional area do you mean? The face to face contact area? Or the area of the thickness of the entire joint (imagine a plane section through the joint).

Thanks,

D

 

[dbecker]

I think the section of the joint is what you meant. I perform the calculation and get a similar result.

desertfox would you happen to know what the resistance is for a large lug connection, such as the following:

http://www.specialized.net/Speciali...ession-One-Hole-40-AWG-PURPLE--Long-6182.aspx

After the crimp, what would the resistance typically be? This is for 646 kcmil cable.

 

[desertfox]

Hi dbecker

the area for the calculation should be the area of contact and from your earlier posts I believed your coppers overlapped by 2" and in the other plane 1".
The link I left wasn't to help you calculate resistance but to show how to calculate bolt and copper stresses due to a temperature rise and also to indicate that bolt clamping forces can vary a lot which in turn effects the electrical contact pressure and hence the joint resistance.
I think you can estimate the resistance of the lug by working out the bolt load due to tightening torque and hence the pressure by dividing in by its contact area.
I assume this is linked to your earlier posts in the Heat transfer Forum relating to some lug failures you experienced.

desertfox

 

[dbecker]

desertfox you are right this does pertain to an old post I made about lug temp failures.

This is what's confusing me though;

"Now using the graph on the link you provided go along the horizontal axis till you get to 12.5N/mm^2, then go up vertically till you intersect the curve, then read across on the vertical axis a value for Y, in your case I read approx 2500 micro ohms/mm^2, finally divide this figure by the cross sectional area of the joint:-

2500/(2"*1"*25.4)^2 = 1.93 micro ohms."

I dont understand why you squared the area term, if the overlap is 1X2 inches it shouldnt be squared? I am missing something.

I did the bolt calc and got the 12.5 N/mm2, that is right. I found the 2500 micro ohm/mm2 from the graph; I just need to understand which area I divide by, the contact face area or the cross sectional area of the joint (thickness of the joint).


Thanks again,

- D