Rectangular Pressure vessel calculations vs. documented testing? 1

[sensiblesteve]

I'm currently trying to validate criteria that had been given, without approval, to a customer by a salesman who is no longer in our employ.
In short, I have a rectangular tank (cube), 2" radius corners all around.
Qualifing testing was at 65 kPa (9.4 psig) & 200kPa (29 psig) for 10 min. Results: no permenent deformation and lo leakage at 65 kPa, the unit was deformed but still no leakage at 200 kPa.
In trying to rate this unit for pressure I can't get any method to produce the rating that I witnessed and should be able to varify.
I tried from [P/V Handbook by Eugene Magyesy], [P/V Design Manual by Dennis Moss], [ASME VIII App 13] and [Rorks stress and strain]-flat plate condition(1g) uniform load plus uniform tension on all edges, which gave me the best results, but still not anything close to what I know to be true.
Can I use the confirmed hydralic pressure test ( 9.4 psig) data (x) .667 and use that as a rating.
Would this be deemed as acceptable.

I would be gratefull for any guidance on this matter as I am sure that most of you out there have a greater understaning of these matters than I. Steve

 

[rb1957]

0.667 looks suspiciously like the ultimate factor, 1.5. pressure vessels (a/c fuselages) have a higher ultimate factor.

another point would be how strong was the material of your test specimen compared with the spec minimums ?

i might use the 200kPa as an ultimate load, derate it for material over-strength and derate for limit (i'd use 0.50 as a factor, but that may be just an a/c thing)

 

[sensiblesteve]

Yes, a 1.5 factor is what I felt would be good.
Maybe I should clarify, This is not a pressure vessel per say.
The customer wants to put a 3.5 psig. vent on this unit as a replacement for the "open 180° return" which we installed [vapour pressure in adverse conditions], the oder is the issue with his people. We need some form of exceptable data to do this.

Shell; .135 ASTM A240-304 = tensile 72,300 psi [505 MPa]

 

[dhengr]

Steve:

That is really a crazy shape for a pressure vessel, whatever the length and side dimensions are, only a salesman would come up with that and then ask the engineers to put real numbers on it. All so that he could earn his commission and make engineering look inept. The 2" radius corners btwn. the sides and at the corners btwn. sides and the ends is where it is yielding, and you know that happened prior to the 29psig loading. Nothing else will happen, nothing worse, as long as you keep the pressure excursions within those you tested around, and you don’t allow some greater external or internal pressure.

You are going to put a 3.5psig vent (pressure relief valve?) on this; and you tested it to 9.4psig with no permanent deformation or leaking; then you tested it to 29psig with no leakage, but some permanent deformation, but no failure. You might step the testing up from 9.4, in increments, to see what pressure starts to cause permanent deformation, and where. Otherwise, poor fabrication could cause stress problems or leakage at any pressure. If you want to be safe, question how many 3.5psig vents you should put on the vessel in case the first on fails.

I’d rate it at (9.4/3.5) = 2.7 without deformation or leakage, or ( whatever/3.5) at yield and no leakage; and (29/3.5) = 8.3 with some permanent deformation, but no leakage or failure. The exact stresses and their location is really immaterial isn’t it, since the actual yield and ultimate stresses of the material can vary above their min. spec. values. Then tell them to pray for long operating life on their relief valves, and no crazy pressure excursions, because you don’t have any control over those later two items, and they would be the conditions which could really cause failure.

 

[prex]

I understand that you had no permanent deformation even at 29 psig, so can't see why you can't obtain satisfaction from coded procedures (but you should also go to ASME VIII Div.1 App.13).
ASME VIII Div.1 UG-101 has procedures for determining the maximum allowable working pressure of vessels by test. What you could be interested in is the Displacement Measurement Test Procedure, where you basically use displacement comparators to build a displacement/pressure curve and determine at which pressure yield starts.
Of course you didn't that, and also you have an austenitic steel that is excluded from that procedure, because it doesn't display a definite yield point.
However you could infer that the pressure of 9.4 psig is certainly below the hydrostatic test pressure coincident with the proportional limit of the weakest element of the component part tested. IMO you cannot extend this conclusion to the test pressure of 29 psi, even though no apparent permanent deformation was present, because yielding probably occurred in the corners.
ASME procedure, when you cannot determine the actual yield strength of the material under test (but you could do so if you have samples of the same material), requires the maximum allowable working pressure to be determined as P=0.4H , where H is the hydrostatic test pressure coincident ....
This procedure would give you P=0.4*9.4=3.76 psig.
If this can be OK for you...

prex

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[sensiblesteve]

prex,

Thank you, the ASME procedure that you referred to conserning P=0.4*H, where could I find this so I can review it with the customer for acceptance? The 3.76 psig. would be perfect if this can be referenced or considered standard protocol in a documented form.
I need to have something to back up the statement.

 

[JStephen]

It sounds like maybe the designs are based on plate bending, but you're getting enough deflection that the plate is acting as a membrane as well, which means the designs underestimate the strength considerably.

 

[prex]

The procedure is in ASME VIII Div.1 UG-101, as referenced above.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

http://www.levitans.com

: Air bearing pads

 

[racookpe1978]

I see the 2 inch "edge" radius, but not a corner radius - where the three edges meet. That would also be 2 inches, right?

What is the overall height, length, and depth and wall thickness? Seems that the centers of your "real world" vessel are somewhere between the simple approximation of a pressurized "rectangular box" (with 8 right-angle corners and flat sides) and the overly gross simplification of a sphere (with perfectly-rounded "corner radii" equal to the "sphere radii")

 

[Cockroach]

I have done this in the past, some years now.

Pulling my old notes, I considered the four walls of the channel as a flat plate of infinite length. The boundary of such plates where fixed, immovable. Then I applied the closed form solution set from Roark and evaluated the maximum deflection, centre of the plate; this also gave me the maximum moment at the edge of the plate. My reject criterion was material yield at Factor of Safety of 1.25.

From this I used Finite Element, which was Algor in that day. I would do the same today, but use SolidWorks with the Simulation package. My reject criterion as the test statistic was scientific error for stress at the boundary, that is, five percent.

hope this helps.

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

 

[nuche1973]

Interesting problem. Did you have a resolution? I have a few questions, since I am in a similar situation: 1. do you have stiffeners? 2. what material are you using? 3. Operating temperatures? 4. what yeild stress values are you using? 5.What are the oreintation of your stiffeners? My situation is similar in the fact that I work in non-code vessels and our stiffeners are parallel to the vertical edges. Thus, ASME VIII appedix 13 is not fully applicable. Also, the yield stress values used in ASME are at higher temperatures. Therefore, I mainly use Roarks formulas for stress an strain inconjunction with AISC flexure formulas. I have found that our vessels operate at a higher pressure than what I have calculated, especially when deformation is ignored.

 

[racookpe1978]

Why would you prefer stiffeners to be parallel to the vertical edges?

Seems that you'd need (?) a flat flange at the "top" to bolt on the lid (for lifting and installation as well) so the tank is reinforced up there - parallel to the ground.

The flat (?) bottom is going to act like a stiffener so the bottom is OK.

But wouldn't it be more efficient to add a flat bar (rib) parallel to the ground at about the midpoint? That would reduce outward movement of the middle of the 4 walls better, right?

 

[nuche1973]

racookpe1978: the answer can be summed up in the phrase "we've been doing it like this for years." and since I am the new guy, it will take a lot of proof from the field inorder to instigate change. The orientation of our stiffeners is a source of my dillema. However, when the people who pay the bills and call the shots do not understand the mathematics behind the design then we, or I, have to pick the battles. How do you tell a 70 year old man, who's your boss, that the design he has been using is incorrect? Especially when the vessels we are working with are non-code vessels? (under 15 psig)


There are days when I wake up feeling like the dumbest man on the planet, then there are days when I confirm it.

 

[rb1957]

"the design he has been using is incorrect" ... not optiminal certainly, but surely not incorrect as it works. equally certainly, if you tell the boss his design is "Wrong" and it's been working for the last X years i suspect he'll tell you where to go (and take your box of stuff from your desk with you). but if you tell him that there is a "better" design, that'll save him money, he might listen.

optiminal would, i suspect, take more time/money than it's worth to determine.

 

[dhengr]

Rb1957... Nicely put, and right on the money. Except I’m not sure we or he knows which is actually the optimal design, at this point.

Nuche1973... I’d want to know much more about the materials used, thicknesses and all of the dimensions and loads on the tank. If it’s a long tank the horiz. stiffeners may have pretty long and inefficient span lengths. Their terminations at the four vert. tank corners would be fairly complicated. Maybe spanning them from the bot. pl. to a top stiffener/bolting lip would be better. Maybe the stiffeners should be inside, not outside. I would want my primary welds at tank corners, sides to bot., sides to sides at the vert. corners and sides to the top flange, on the side (inside or outside) so that the joint moments &/or deformations did not put the root of the welds in tension. Clip the corners of the stiffeners so you don’t weld into the corner btwn. the bot. and the side, etc., this tri-axial stress condition and stress raiser is nasty. Same thing at the four corners t&b btwn. two sides and the bot. or top lip, you might want to reinforce that in some way. Look for ways to improve fabrication and material utilization. And, check the design both ways, or several ways and take the best one to the boss, with some proof of improvement or savings. That’s selling your ideas, and maybe scoring a few points. It’s pretty tough to argue with something that has worked successfully for years.