Roof & Compression Area Design for High Internal Pressure 17.5 -18 Kpa

Are you sure your units are correct? 18 kilopascal = 0.18 bar last time I looked.

Yeah it is correct 0.18 bar (18 Kpa)

Regards

ASP

So where does "very high pressure" come into it?

For this pressure.If you consider say tank of 40 Diameter:

1) compression area required as per Append.F Section F.5.1 would be very high.

2) Secondly if we consider rafter supported roof structure without columns.Then to design roof structure as per the requirement of Section 5.10.2.1 for loadings (then as per Append-R Section R.1 case (a) would be the critical one amongst all cases) would be very high.

Please correct if i am wrong

Regards

ASP

I haven't used these API specs for years, and can't remember most of what they say, but I don't see a problem designing it as a cone. If I remember right the specs give a minimum roof slope, I'll assume 15° from horizontal. In that case, and taking 40m dia. and roof thickness 6mm (ignoring corrosion allowance) I make max radial stress 116MPa, max circumferential stress 16MPa, resultant stress ~ 117MPa. This is for internal pressure only, ignoring any vacuum condition.

Where does the 14mm thickness for 23m dia. come from? Treating it as a flat roof needs much greater thickness.

Codey

Thanks for your reply,

14 thk i got treating as self supported dome roof with corossion allowance 1.6 mm.

Regards

ASP

Probably he means very high pressure compared with usual figure for the roof of an atmospheric storage tank. More typical would 3 - 5 kPa.

Codey

Dear ASP

• Can we design as cone -- Why NOT -GO AHEAD- but dished will have Techno/economic advantages

• designed for very high pressure say between 17.5 KPa to 18 KPa.-this is Low Pressure

• as it is to be designed for case a Dead load & internal pressure As per API 650 - Append-R Section : R.1 which would be critical among all cases from a to g. ---Design from Fundamentals of Stress and Strain --Not Anybody's Code
• What is your tank's diameter
• If you are in India- get SAIL Bokaro's Hi-Strength HRplates to halve dish mass.
  

Thanks for your reply,

Tanks are from Diameter 38 m to 42 m.

If we go for self supported dome roof (using only rafter not the column). Since Diameter is big then as per API 650 we can consider roof plate as 5 + corossion allowance in mm.Then this roof has to be stiffened by sections (say by rafter supports) as per the requirement of Section 5.10.2.1 for loadings (then as per Append-R Section R.1 case (a) would be the critical one amongst all cases).

Correct me if i am wrong in interpretation of Section R.1 case a)

Loading on rafters :-Dead Load (includes all dead load on roof say roof plates,nozzles,roof walkway platform) + internal pressure (18 Kpa)

Say roughly 18 Kpa (neglecting for time being dead loads). To design section of rafters for such load seems to be very high.

Please guide me .

Regards

ASP

38>42m dia -- average 40 This is big.

Your best bet is RCC thin Shell with rows of support column at diameters 30,24,16,8 m .

If the Roof is RCC Shell body could as well be RCC.

Make sure you use Super Plasticizers of latest PolyCarboxylate class.

You need Turnkey design Consultant for minimum Budget of Money and project time.

We can be.

mm

We have to go for Steel only.

POSSIBLE

Dished edges and optimum slopes to crown

And support Column/Beam system

mm

A few more comments.

What do you mean by Compression Area Design in original post?

Loading on rafters - Dead Load and internal pressure give forces in opposite directions.

For self-supporting dome (your #8) is this same as a dished end? I doubt whether you could get one 40m dia. I don't know whether the API codes include dished end design, but looking at BS5500 (pressure vessel code) it doesn't go down to the very low pressure/steel design stress ratio here. But for a typical dished end the thickness is only slightly greater than shell thickness (limited by longitudinal stress) which comes to something like 4mm (+ CA).

My comment in #5 re cone roof assumed self-supporting. If you add columns and rafters it alters things completely. It might not be easy to build a cone roof without (radial) rafters (maybe it's no problem, I don't know, I'm not an erector) but I don't think they're necessary to take the internal pressure. They would help resist external loads, but the cone alone should take the usual design vacuum condition, typically 25mm wg (0.25kPa)

Codey

Yeah you are correct i used high pressure with regard to atmospheric pressure not as per ASME STD pressure range.

Compression ring : i was referring to the compression area which is actually comprises of shell -Roof region (participating area).As per API 650 Append-F Section F.5.1 for such high pressure required compression area is very very high.I am finding it difficult to achieve this area.

Loading on rafter: As said dead load & internal pressure will act in opposite direction.So net resultant pressure will be upward due to high internal pressure.To sustain such pressure section required would be of very big section size.If we donot weld rafter section to roof (Tank roof without column) then i think no internal pressure will come in to picture of rafter section design.We should only consider in this case dead load and live load only.Next regarding roof thickness we can consider min. thickness as per code with corrossion allowance without designing roof plate thickness.Please correct if i am wrong any where technically conceptwise.

But if we go for column with rafters (Ofcourse for cone roof as for dome we cant use column support) then no need to design roof thickness (can consider min. thk with corossion) as well as for participating area.We can consider top curb angle with min. angle section as per the STD at shell to roof junction.Regarding rafter design same as the above consideration of no weld of roof and rafter as mentioned in above paragraph.

Requsts you to please guide,correct & suggest.

Regards

ASP

Not sure about your 2^nd para. For cone my calculation for radial stress f_r is

f_r = P*D/(4*t*sin θ) where t = thickness and θ is angle from horizontal. Circumferential stress f_c = P*D*sin θ/(2*t)

Where does the high area required for compression come from?

Codey

As per API 650 -Append-F Sect:F.5.1

Total compression area required at the roof-to-shell junction :

A, (mm2)= 200 x D x D x (Pi - 0.08 x th) / Fy x tan(theta)

Where, D = Tank Diameter (m)

Pi = Design internal pressure(Kpa)

th = Nominal roof thickness (mm)

Fy = Lowest min. yield strength (Mpa)

thetha = Angle between the roof & hrozontal plane at the roof-to-shell junction (Degree)

Regards

ASP

I agree with #16.

Not having API 650 to hand, I don't understand the formula.

Why does it say compression area required? For internal pressure the roof-to-shell junction is in tension.

What is the 200?

What is the area A being calculated? Is it the total steel area round the tank? If so, the steel thickness th = A/(pi*D), do you agree? In that case, an iterative solution is needed to find th.

What value are you using for Fy? Using a reasonable 150MPa and just putting figures in, ignoring dimension problem below, gives A = 140000mm². That comes to th = 1.1mm, which is too small, and obviously impractical.

In the term (Pi - 0.08 x th) Pi and th have different dimensions. To get an answer I multiplied th by kPa/mm. This gives 0.08 x th = 0.445kPa, almost negligible vs 18kPa (and ignoring it gives a safe (higher) figure for A).

I assume API know what they're doing, but to me a more likely formula, ignoring (Pi - 0.08 x th) is A = Pi*pi*D²/(4*Fy*tan θ). This gives A = 560000mm² and th = 4.5mm. Leaving (Pi - 0.08 x th) in and treating it as above, an iterative solution gives A = 550000mm² and th = 4.4mm.

From my formula for f_r in #15, converting gives A = Pi*pi*D²/(4*Fy*sin θ) (on above assumption about meaning of A). This gives A = 580000mm² and th = 4.6mm.

A thought - is API 650 in imperial units, using foot for D and psi for Fy? In that case need a factor 4/pi*12² = 183, which somebody might have rounded to 200.

I don't think we've got to the bottom of this yet!

Codey

I think I must be misinterpreting what API 650 means by A.

I've checked that the formula in SI units gives the same area as in US Customary units, if D is in ft and Fy is in psi (which the spec you posted doesn't say). Still gives A = 140000mm² as my #18.

So A must be some other area from what I assumed. I still don't see where the 200 factor comes from, and in the US units formula I can't read the bit in brackets (Pi - ??). There's still the dimension query. The 0.08 must have dimensions to make it right, not just a number, but it's hard to know what without knowing what the formula is calculating. Tho as I said, it doesn't make a deal of difference to the result.

I think you need to study the codes in detail and work out what they mean.

I think I found a free download of at least one of the specs, but I didn't bother as it was going to take 10 minutes and take up disc space. I don't use it often enough to make it worthwhile.

Codey

You need to Upload a set of drawings to explain your basic concepts on CONE ANGLE, Column,Beam,Positive /negative pressures inside this 40m dia tank.

Codey and others can THEN help you reach your targeted optimum design.

By the way, what will be contained in this tank?And what ambient temp. range?

I THINK YOU NEED TO READ APPENDIX F-DESIGN OF TANKS FOR SMALL INTERNAL PRESSURES FROM API 650 THEN YOU FLOW THE RULES IN THIS APPENDIX IN UR DESIGN