Calculating laminate thickness for high surface pressure

General Composites Discussions Engineering Talk
#1

Hi,
Being a practical person, I haven’t learned how to calculate laminate thickness for specific strength. I now need to figure out what the theoretical layup of a cylinder should be, which needs to support 7.5 bar water pressure. Here are the specifications:

ID: 40mm
inner length: 400mm
Outer dimensions not fixed
Fiber: T700 + Inner layer of Kevlar
Matrix: recommendations for marine use?

I’m not looking for just an answer. I’m hoping someone can either walk me through or point me in the right direction.:smiley:

Thanks in advance.

#2

Well I’ll point you in the right direction. The primary stress that you are going to need to calculate is the hoop stress. The longitudinal stress will be half that of the hoop stress. So if you use a woven fabric on 0/90 and design for the hoop stress the longitudinal stress will be taken care of by a factor of 2.

The internet has many examples for calculating the hoop stress in a pressure vessel.

You’re going to find out that it’s not going to take very much material. :wink:

Let us know if you need more help.

#3

I agree with the above. Do a Wikipedia search for “Cylinder Stresses” and it should give you all the information and formulas you need. I would use a safety factor of at least 3 (i.e. design the cylinder for 22.5 bar) when you calculate the laminate thickness you need.

The laminate thickness required will probably be much smaller than the thickness required to ensure that the cylinder is not damaged when it is handled, dropped and or bumped.

I would stay away from PE and stick with VE or Epoxy resin since you are using carbon/aramid. Just remember for a marine application that aramid is hygroscopic (absorbs water).

#4

Thanks for the feedback. I’ve spent the last 2 days trying to get my head around how to apply hoop stress here but all the info I find is for vessels with internal pressure. The fibers would the be in tension and the walls very thin. Mine needs to withstand external pressure (7.65 kg/sqcm) so the fiber are in compression.

I punched the number into Laminator for a layup of 0/90/45/-45/90/0. It says I’m Ok according to Tsai-Wu falure criteria but I have my doubts because of the limitations of the program. It was written for flat plates and XY loads. I have to trick it by setting these to 0 and applying only shear loads.

At the moment I have the feeling I’m going in the completely wrong direction.

#5

The formula for calculating internal pressure is the same as for calculating external pressure. The only difference, as you stated, is that with internal pressure the walls undergo tensile stress. With external pressure the walls undergo compression stress. This does complicate things because thin walled structures will buckle before they reach their compression yield strength. I would think that the flexural modulus of the wall would be part of the calculation.

Hopefully, Infusible can shine some more light of the subject. :smiley:

#6

Again I agree with wyowindworks, the equations give the correct results for external or internal pressure just that the fibers will be in compression for an external pressure.

Just think of it this way, if you look at a very small segment of the part, say .001 inchess X .001 inchess this is effectively a flat plate. Lets say that the y direction runs the length of the vessel, the x direction in the width direction, and z is normal to the surface. The compression stress in the x-direction is twice the compression stress in the y-direction as given by the equations we mentioned above. Does this make sense?

Like mentioned above, buckilng will probably be the limiting factor if you make the laminate very thin not to mention if would be very easy to damage. You can do some research online for buckling of curved composite plates but I would just use a large safety factor on you design pressure and overbuild the part.

Remember the KISS principle: “Keep it simple, stupid!”

#7

If a pressurized tube has to be made with thick walls, there’s nothing like filament winding. But in this case just rolling UD prepreg is good enough.
I don’t see the use of kevlar in this case.

#8

First off, thanks for the help. Since I didn’t study engineering, I’m trying to learn the theory and maths on my own. I thought this would be a simple enough real project to try. I already know what the layup would be from a practical stand point. I wanted to see what it looked like on paper. I’m missing some basic mechanics theory so I’m not quite there yet.

The vessel will hold Li-Ion batteries so I chose kevlar to line it for three reasons. 1) it’s non-conductive, 2) it’s very resilient to rubbing and 3) Li-Ion batteries can explode so the kevlar would limit the amount shrapnel flying about in a failure state. I realize that I could get similar effects using glass instead but kevlar would be an additional marketing point.

As soon as I finish I will post the results here but it will take a few more days.