Aluminum Weldment Design Help 4

[Andre3]

I have recently moved from a mechanical design role into engineering at my small R&D company figured I would try reaching out to you folks for some help and direction as I do not have other colleges to consult with. I am looking to design a weldment between two 6061-T6 components supporting a very strong solenoid magnet. This had been a bolted connection but the bolts failed after the system was overloaded and now we would like to weld them together for a stronger, permanent connection. My thought was to chamfer the parts to make for a groove weld to hold them together but I am mostly unfamiliar with weldment design.

From what I have seen it looks like to be conservative I can expect the welded area to behave like 6061-O which is dramatically weaker. There is an axial magnetic force of about 250 kN downward and an outward radial force that is still being evaluated, for now I am treating it as 0.5 MPa acting on the inside wall of the support. Where should I start with evaluating the stress experienced at the weld and how it will behave? I have attached some snips of a simplified model for discussion.

 

[Andre3]

I don't recall claiming the bolts failed one at time other than them possibly rapidly unzipping and I don't see how the magnets being in separate chambers changes the problem I proposed at all. There is a lot going on with the system and I have been trying to keep from overloading with unnecessary information. This is my first post in the forum and I am new in the field, just doing my best and appreciate the help.

 

[3DDave]

The unreliable part is failing to provide: an accurate free-body diagram. The fact that the failure was explosive in nature. That the lid was in the load path. The total load capacity of the bolts. These should all be in the first post.

It's fun like a who-done-it mystery to pull these facts out one at a time.

Since there is room for the magnet to move there is a possibility have the bolts loose/ use precision spacers between the lid and the housing and to leave a 2 or 3 mm gap and try again, this time with a suitable spacer under the assembly to prevent more damage if bolts fail. If the bolts don't fail then the estimate of CTE of the magnet is likely wrong.

If the bolts do fail you probably need to find a different material for the bolts or to increase their number - it looks like there is enough room to triple the count.

 

[desertfox]

Hi Andre3
Thanks for the information, well we got there in the end, the bolts had to be in the load path for them to fail and I read all your posts very carefully, so eventually we came to the conclusion we agree on. You weren’t involved in the beginning of the design of this device, so I can appreciate that you inherited it and it’s hard after a failure to just become involved, under those circumstances I think you did your best to give as much information as you could. This is your first post to boot, so don’t be put off posting further problems,there are a good bunch of guys here that will help you.
Now the next problem is how to fix it, have you considered putting bolts, dowels at 90 degrees to the current bolts, this would mean putting them in shear but the bolt preload wouldn’t be affected by the contraction..

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[hydtools]

Make the cup part with an OD flange. Drop the cup into a hole in a supporting structure. Take the magnet load through the flange. The bolted-on cover then carries no load.

Ted

 

[Andre3]

Here are some notes on the joint :
[URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1609286770/tips/Bolted_Joint_Notes_adveoh.pdf[/url]
resources:

https://roymech.org/Useful_Ta bles/Screws/Preloading.html

https://mechanicalc.com/reference/bolted-joint-analysis

I made a guess at the input torque of 17 Nm after tightening one of the bolts down several times with an hex key and comparing it with a torque wrench. 17 Nm seems pretty reasonable to me, it was assembled by someone about my size so I think that is at least close. The bolts were not lubricated so I used a friction factor of .3, I think it was probably lower than this but that would help account for some of the local yielding of the aluminum reducing the preload.

With an estimated separation load of 5.2 kN and error bars from 107 N to 10.3 kN, it seems very likely that the joint had separated when it failed at 130 kN total attractive force or 4.64 kN/ bolt.

 

[3DDave]

According to this:

https://www.specialmetals.com/assets/smc/documents/inconel_alloy_718.pdf

the strength goes up as the temp goes down. Now that we know the area is for M6X1 screws, and the tensile area is 20.1mm^2 per screw and we know the number of bolts is 28, for a total section area of about 560 mm^2.

Inconel rod has a room temp yield strength of about 160 ksi or 1100 N/mm^2; so it should resist (1100*560)N or about 600kN. It should be higher at the low temperature.

So that leads to a potential problem of stress concentration - like single point sharp root cut threads.

Did you put your area as mechanical because that's your engineering degree or because that's the job you have now?

 

[3DDave]

Another thought occurs to me - the magnetic force won't be constant if the coils fail to keep the separation.

If the initial failure was in a few of the short helicoils then some of the remaining short helicoils would quickly follow. Once that starts it's a race between the helicoils and the bolts on that side. Once enough of the short-helicoil side fasteners fail it will move towards the long-helicoil side, causing the force on both to go up and failing the remaining bolts on both sides.

 

[desertfox]

Hi Andre3

Thanks for the notes however I’m not sure I agree with some of the figures you have shown, firstly the bolt preload at 17Nm ie 9.4KN would not cause the bolt head to become Embedded in the
Aluminium at room temperature based on a T6061 276 N/mm^2 yield stress.So what value of yield stress do you have for the Aluminium Alloy?

On cooling the bolt preload would fall to zero in my calculation, assuming the bolt is 36mm in length (approx) and given a preload of 9.4KN the bolt stretch at room temperature when assembled would be approximately o.o5mm which would relieve itself on cooling of the components and reduce the bolt preload to zero as stated earlier, however I can’t see the actual bolt length in your calculations could you please provide it along with the actual length of the helicoil? There appears to be at least three lengths of helicoil 1.5D,2D and 3D. I was trying to find the pullout load for M6 helicoils but obviously the length plays a part.

I suspect the bolts did fail in a domino fashion because the 250KN external load would not have broken the bolts had it been equally distributed however as you have stated in some area’s it appears to be the helicoils which have pulled out rather than the bolt failing so it would be interesting to know what the difference in length of the helicoils which failed compared to those which didn’t, assuming of course the helicoil threads were fully utilised.



“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Andre3]

desertfox
Radial pins might work, the total OD of the structure must remain fixed so the thickness of the flanges would have to be pretty small at about 9mm each. That may not be an issue though so I will look closer at that. I think I would use large diameter Aluminum pins and shrink fit them in place but I have to think more about it.

I have been using 270 N/mm^2 for 6061-T6 yield strength, the yielding I was considering was just surface effects. The top surface of the cover was fairly marred up after bolting and before the failure, I figured cutting the soft aluminum like that would also contribute to the preload loss, I believe I read that also but I am not sure where.

Thank you for pointing that out the bolt length, the grip length is not right for a joint with threads. Here

https://mechanicalc.com/reference/bolted-joint-analysis

they suggest using the length of the thru hole plus half of a bolt diameter for the grip length, so we would use 21mm for the grip length. factoring that in I get a separation load of 4.74 kN with error bars from 0 (totally loose bolts) to 9.69 kN. the bolts are 30mm long but I think it is the grip length that we use in these calculations.

The short helicoils were 1D and the long ones 1.5D. The 1D heilcoils should have failed around 15 kN and the long ones at around 25.5 kN. My understanding is that you want the strength of your threads to far exceed the bolt so even the 1.5D were marginal here.
[URL unfurl="true"]https://res.cloudinary.com/engineering-com/image/upload/v1609343008/tips/Tensile_Strength_Bulletin_Metric_LR_1_pcqv5g.pdf[/url]

Also to be clear, the failure happened around 130 kN (±10%), but the joint must be designed to operate at 250 kN continuously.

Dave
Yes the bolts should have been able to take over 600 kN, that is what is puzzling and why I am here. The attractive force increases at 9 kN/mm as the coils move toward each other and the total spring constant of the bolts when loose (AE/L) would be 6.35 MN/mm so I don't think there was a runaway with the attractive force. What has confused me from the start of the problem is how any bolt could fail before a short heilcoil and how could any of the short helicoils fail first if the force per joint was about 4.6 kN. That is why I am thinking now that the bolts on the top (with 1.5D helicoils) could have failed first from some symptom of the joint separating like shear and bending on the bolts with maybe stress concentrations, all made worse by the brittleness of the material. Once the top failed completely it gets closer and raises the total force to about 220 kN and causes the bottom to fail with half the bolts breaking for the same reason as the top and the remaining half rip out the short helicoils.

 

[desertfox]

Hi Andre3
Thanks for the information, I used a bolt length to get an idea of how much the bolt will stretch under a torque of 17Nm, I had to guess the length because I had nothing else to work with. I am not sure where you are getting the separation force from? Once that magnet is cooled to its working temperature the bolts are free of any tension. I will have a look at the information you gave and comeback later.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Andre3]

desertfox

To clarify, my claim is that with a room temperature preload of 9.4kN there would still be 3.4 kN of bolt tension after cooldown, I got the formulas from

https://roymech.org/Useful_Ta bles/Screws/Preloading.html

 

[desertfox]

Hi Andre3

There won’t be any preload left because the aluminium as contracted away from the bolt and the bolt stretch to get the preload you estimated is only 0.05mm, there is nothing to keep any tension in the bolt, what are the current depths of the parts being connected.

Ps the link you posted doesn’t work it just gives error.
“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Andre3]

desertfox

Sorry about the link here is a good one:

https://roymech.org/Useful_Tables/Screws/Preloading.html

The stiffness of joint factors into the thermal loading not just the relative contraction and the initial strain, I think I used the formulas correctly to get my conclusion but let me know what you think. The cover is 18mm thick, the total height of the bobbin is 97mm with a radial thickness of 18mm at the joint. The total OD of the structure is 470mm.

 

[mfgenggear]

Quote "The overload would be 300 kN, so you propose that the bolts be preloaded so they will have 10.7 kN of tension when cold and ready to operate?
The bolts are M6x1.0 Inconel 718 age hardened.
The magnet is charged with DC current, I am not sure if that answers your question." Unquote

the fact is the bolts failed
so the conclusion the bolts have to be more robust such as minimum of M9 bolts or larger.
in the English version 3/8 or even better 1/2 inch, it is really very evident the bolts were inadequate.
Kindly advise if the aluminum cover distorted. from the moment.

 

[hydtools]

Make the cap in a conical shape to eliminate the "oil canning" effect of the flat plate shape to reduce or eliminate the bending load applied to the screws.

Specify rounded or radiused thread root on the screws to reduce the sharp notch effect in the screw thread.

How quickly is the magnetic field established? There could be a case of tensile impact on the screws.

Ted

 

[Andre3]

mfgenggear

I agree, I came here looking to move past the failure to come up with a solution which I thought may be welding, but I am glad we have gone down this path-I think that I am understanding the failure better. I have very little flexibility on dimensional changes to the structure, so I am stuck with smaller bolts. The radial thickness of the bobbin at the joint is 18 mm and limits the bolt diameter, especially considering the aluminum will need to be helicoiled. The covers are minimally distorted, they rock slightly on a surface plate but I cant say that it is any more than when we received them.

hydtools
Is this sort of what you had in mind?:

I like that, to the extent that I have enough clearance I will try to incorporate that profile.
I Definitely will use bolts with round roots in the future, especially with more brittle materials like the 718.
The field is ramped very slowly over the course of hours so I wouldn't expect any impact from the ramp. We did hear some pinging/clicking noises accompanied by voltage spikes as we got close to the failure load which indicates some type of coil motion. It could have been bolts starting to pop or settling of the other structural components not shown but that could have been some shock to the screws.

 

[desertfox]

Hi Andre3

You haven’t shared the joint stiffness calculations but try to imagine this, when you tighten the bolt to get the preload you quote it needs to stretch about 0.05mm so now you have the 9.4kN and the bolt is clamping the Aluminium alloy, the joint is now cooled and the aluminium alloy contracts by 97 * 0.00435= 0.421mm but the bolt contracts 30* 0.00238 = 0.0714mm. Now the cover alone is 18mm thick and that contracts during cooling by 18 * 0.00435 = 0.0783mm.
If the bolt is a spring and it is stretched by 0.05mm and then released it will return to its equilibrium position and it tension force returns to zero, given that the lid itself contracts more than the original bolt was stretched suggests to me that on cooling the bolts have zero preload. I will look into it again in case I have made a mistake, however if you are correct the design you have might work but you would have to increase the preload at room temperature to compensate the loss you have calculated and without embedding the screw head into the aluminium alloy, personally I don’t think that’s possible but let’s see the joint stiffness calculations you have and go from there.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[desertfox]

Andre3

The inconel 718 is not brittle at -250C according to literature I have seen.

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[mfgenggear]

Andree

let me try to under stand the issue.
excuse my terminology I am old school from the ranks.

even thou the assembly is in cryogen, the coils are at elevated temperature.
therefore it has thermal expansion radially and axially. if I am correct
then there is moment on the cap of the assembly. causing the bolts to fail.

therefore if the bolts size can not be changed how to eliminate the moment.
if the expansion of the magnets can be calculated. allow clearance between the
magnets and housing, and cap to allow for the expansion, is that possible.
all speculation if there a instantaneous super heat from the magnets causing a rapid change in
size, it would put instantaneous force on the housing in all directions.
thus exceeding the ultimate stress of the bolts. rapid failure.

 

[Andre3]

The formulas I used are shown in my first set of notes, I got them from

https://mechanicalc.com/reference/bolted-joint-analysis

. With a revised grip length of 21mm I get a bolt stiffness of 2.268e8 N/m and a grip stiffness of 4.575e8 N/m.

I will have to think more about your explanation on the thermal loading, I have trouble imaging how the the material below the joint factors into the loading. Using this estimation I would be thinking more like the cover and portion of the bobbin engaged with the bolt contacts 30*0.00435 = .1305mm and the bolt contacts 30*.00238 = 0.0714mm with a difference of .0591. Since the bolt stretched .030*9.4e3/(2.21e11*20.1e-6)= .0592mm we would still have .0001mm of stretch in the bolt.

*I made an error. When calculating the initial strain I entered E= 2.37e11 into the calculator (which is at 3 K) to get .0592mm. Either way gets the point across.*

I spoke to an engineer today with the company that manufactured the bolt and his qualitative take was the 718 is fairly brittle relative other metals at room temperature and that the brittleness would worsen at low temperature and could make them shock sensitive.

As a comparison here is the room temperature failure of 718 vs 316 SS :