Stress at c'sink holes of enclosed box with internal pressure 1

[patrickv]

I have an issue I'm not quite sure how to approach. A customer is requesting that we prove that our battery pack chassis will not burst or deform if additional gases are introduced inside the enclosed box (the batteries could vent).

The chassis is assumed to be a sealed .05" thick 5052 AL enclosure with dims shown (attached). There are four countersunk screws attached to standoffs securing a PC board inside. The volume of air in the box is approximately 205.8 mL. If 2400 mL is introduced, how would I find the pressure exerted on the inside walls of the box and, more importantly how do I illustrate that the thinnest portion of the wall (at the countersinks) won't give. Pressure inside and outside the box during installation is 11.777 psi. Thank you.

 

[btrueblood]

Assuming the battery evolves gas slowly, the process is essentially isotherm (no temperature change of the gas). The final pressure is given by the ratio of gas masses, or roughly 11.8 psi * (2400+206)/206 = 149 psi. Roughly, because the gas evolved by the battery is likely hydrogen, not O2/N2 mix, so you'd have to use the partial pressure law (Charles' law?).

If the pressure rises quickly (fast enough that no heat transfer to the walls occurs), then the process is more nearly adiabatic, and the compression will follow more closely

Pf = Pinit * (Vf/Vinit)^gamma

Again assuming air (which it won't be), this would give a pressure ratio of about 35 or so, with a final pressure nearer 410 psi.

How do you show the countersink holes won't give? Fastest way would be to fill it with water and pump to 410 psi, or whatever your calculation says (including all real gas effects, temperature variations, etc.).

...and after you sweep up the pieces, look for a cheap way to vent the batteries.

 

[rb1957]

if the camber is really at 11.777psi during manufacture, either you have a very special low pressue assembly factory or you're at a pretty high altitude ... wonderering why it isn't 14.7psi ??

let's take btrueblood's 149psi, over 16sq.in = 2400 lbs. i'd be more worried about the support of the cover ... it looks like the reaction to the pressure is carried by the stepped flange around the perimeter ... the screws attach to a PCB, and the PCB attaches to the rest-of-the-world ?? if so the PCB is going to be the weak link. if the stand-offs carry on directly to the base of the bax (and thence to the rest-of-the-world) then you might be putting 600 lbs tension on a -4 screw ... that sounds a little too much to me.

 

[Cockroach]

I agree with RB1957, doing a generalized finite element analysis (SolidWorks 2009) based on your model specifications give a peak stress of 22,686 psi at the counter sink holes.

Several assumptions here. I went with 5052-H32 aluminum alloy with yield of 28,280 psi. Confirming the previous computation that introduction of 2400 mL of air, the internal pressure would be 149.2 psi. The board can is assumed to be held rigid along the bottom edge of the geometry, opposite the 4-40 UNC c'sink screws defined as the upper face. You fail to mention fillets internal and external to the can, I went with sharp which obviously generates very high stress razors in the corners of the walls. I expect this to be the problem, noting the mathematics associated with flat plate deflection for boundary clamped condition, rectangular - square. I also placed the c'sink holes 1/2 inch off the walls of the can, position not being specified in your earlier post.

Hand waving type argument, the holes are almost at material yield, factor of safety less than 1.246. I've attached the FEA as a MS Word 2003 document.

Hope this helps somewhat.

Kenneth J Hueston, PEng
Principal
Sturni-Hueston Engineering Inc
Edmonton, Alberta Canada

 

[gwolf2]

A .05" enclosure under 149 psi will briefly look like a balloon and then explode. Should be fun fitting all the bits back together with tape though.

 

[aroundhere]

We had a hydraulic tank (1/8" thick 5052-h32 that our guy applied 120psi shop air to. No matter what, I coudlnt hake him understand that f=p*a and that he was putting thousands of pounds of force on the walls. He learned when he came in the next morning and had 30 gallons of hydraulic oil on the floor from cracked welds (gorgeous welds too)

Sorry, that doenst help you but I just felt I needed to tell that story...carry on.