## API 620

## API 620

(OP)

Hello everyone.

Im designing a couple of tanks with a cone bottom, I'm using API 620 for this. Im not sure If I got my numbers right. Can you help me?

This is the information for the tank

Wshell = 3345 lb

Wimpeller = 486 lb

Wliqshell = 52,712 lb

Wtop = 638 lb

Wcone = 445 lb

P at 20ft = 8.66PSI

Density = 1

Tank dims:

Shell height = 20ft

Shell diameter = 44in

Cone height = 10"

Cone diameter = 44in

Cone angle = 12.8 degrees

It has a skirt right where the cone touches the cylindrical shell, the skirt is supporting the cone.

I used Structural Analysis and Design of Process Equipment 3rd. Edition to do the calculations.

If I look at the point where the cylinder touches the cone above the cone I get these numbers

Sum Fy = 0

Wt + V - P = 0

57,181 + 88*pi*V - 8.66*pi*R^2 = 0

V = -16.31 lb/in, using V to get T1, i get that T1 = V = -16.31, the shell is under compression

T2 = PR = 8.66 * 44

T2 = 381 lb/in

Im using a t = 3/16"

My actual compressive stress = 16.31/0.1875 = 86.98 PSI

Actual tensile stress = 381/0.1875 = 2032 PSI

Using 1.8x10^6 (t/R) to get allowable compressive stress, i get Sca = 7,670 PSI

M = 86.98/7670 = 0.01

N = 2032/16000 = 0.12

N^2 + N*M + M^2 <1

0.0157 < 1 so using a 3/16" for the course (cylinder) should be ok. correct?

Now, if I look below the line under consideration, this is where I get confused, I don't know how to treat the skirt

these are my calculations

Since the V calculated before is connecting to the cone, V should be the same for the cone area due to continuity, correct?

if that is the case V = 16.31 lb/in

My T1 = V/sin12.8

T1 = 74.73 lb/in

T2 = PR/sin12.8

T2 = (6.66)(44)/0.22

T2 = 1,732 lb/in

Both forces are in tension.

t = T1/SE , t = T2/SE

T2 > T1 so I'm using T2 to calculate my thickness.

t = 1,732/16000 (0.85) = 0.127", I'm using a 1/4" material for the cone, so i should be ok, correct?

If I move my line halfway into the cone (6" down) and consider what's above my line.

I get the W of the initial calculations + Whalf cone + Win the cone

That gives me a new Wt = 57,826 lb

If I do my FBD I get the following

Sum Fy = 0

Wt + V - P = 0

Wt + 44*pi*V - 8.84*pi*R^2

V = -321.09 lb/in

using V to calculate T1 i get:

T1 = V/sin12.8

T1 = -1,459.5 lb/in

T2 = 855.7 lb/in

T1 is under compression and T2 is under tension.

I do the same procedure and i get the following

t = 1/4" for the cone

Sc = 1459.5/0.25 = 5,838 PSI

St = 855.7/0.25 = 3,423 PSI

Sca = 1.8x10^6 (t/R)

Sca = 20,454 PSI

N = 0.2139

M = 0.2854

0.19 < 1

Using a 1/4" for the cone should be ok?

am I missing something else?

I did not use the formulas on the standard because I could not figure out the direction of (Wt + F)/At on both cylinders and cones. Also, I was not sure which At I needed to use since Wshell and Wcone is acting only on the thickness and Wliq is acting on the area of the cyl/cone

Thank you for your help.

Im designing a couple of tanks with a cone bottom, I'm using API 620 for this. Im not sure If I got my numbers right. Can you help me?

This is the information for the tank

Wshell = 3345 lb

Wimpeller = 486 lb

Wliqshell = 52,712 lb

Wtop = 638 lb

Wcone = 445 lb

P at 20ft = 8.66PSI

Density = 1

Tank dims:

Shell height = 20ft

Shell diameter = 44in

Cone height = 10"

Cone diameter = 44in

Cone angle = 12.8 degrees

It has a skirt right where the cone touches the cylindrical shell, the skirt is supporting the cone.

I used Structural Analysis and Design of Process Equipment 3rd. Edition to do the calculations.

If I look at the point where the cylinder touches the cone above the cone I get these numbers

Sum Fy = 0

Wt + V - P = 0

57,181 + 88*pi*V - 8.66*pi*R^2 = 0

V = -16.31 lb/in, using V to get T1, i get that T1 = V = -16.31, the shell is under compression

T2 = PR = 8.66 * 44

T2 = 381 lb/in

Im using a t = 3/16"

My actual compressive stress = 16.31/0.1875 = 86.98 PSI

Actual tensile stress = 381/0.1875 = 2032 PSI

Using 1.8x10^6 (t/R) to get allowable compressive stress, i get Sca = 7,670 PSI

M = 86.98/7670 = 0.01

N = 2032/16000 = 0.12

N^2 + N*M + M^2 <1

0.0157 < 1 so using a 3/16" for the course (cylinder) should be ok. correct?

Now, if I look below the line under consideration, this is where I get confused, I don't know how to treat the skirt

these are my calculations

Since the V calculated before is connecting to the cone, V should be the same for the cone area due to continuity, correct?

if that is the case V = 16.31 lb/in

My T1 = V/sin12.8

T1 = 74.73 lb/in

T2 = PR/sin12.8

T2 = (6.66)(44)/0.22

T2 = 1,732 lb/in

Both forces are in tension.

t = T1/SE , t = T2/SE

T2 > T1 so I'm using T2 to calculate my thickness.

t = 1,732/16000 (0.85) = 0.127", I'm using a 1/4" material for the cone, so i should be ok, correct?

If I move my line halfway into the cone (6" down) and consider what's above my line.

I get the W of the initial calculations + Whalf cone + Win the cone

That gives me a new Wt = 57,826 lb

If I do my FBD I get the following

Sum Fy = 0

Wt + V - P = 0

Wt + 44*pi*V - 8.84*pi*R^2

V = -321.09 lb/in

using V to calculate T1 i get:

T1 = V/sin12.8

T1 = -1,459.5 lb/in

T2 = 855.7 lb/in

T1 is under compression and T2 is under tension.

I do the same procedure and i get the following

t = 1/4" for the cone

Sc = 1459.5/0.25 = 5,838 PSI

St = 855.7/0.25 = 3,423 PSI

Sca = 1.8x10^6 (t/R)

Sca = 20,454 PSI

N = 0.2139

M = 0.2854

0.19 < 1

Using a 1/4" for the cone should be ok?

am I missing something else?

I did not use the formulas on the standard because I could not figure out the direction of (Wt + F)/At on both cylinders and cones. Also, I was not sure which At I needed to use since Wshell and Wcone is acting only on the thickness and Wliq is acting on the area of the cyl/cone

Thank you for your help.

## RE: API 620

At the bottom of the shell, A is the inside shell area, P is hydrostatic pressure plus design pressure at that point, W is weight of shell, roof, shell contents, roof live/snow/equipment loads, etc. W acts opposite to the pressure at this point.

In the cone, A is the inside shell (or cone) area, P is hydrostatic pressure plus design pressure at that point, W is the weight of the cone and cone contents. W acts in the same direction as the pressure at this point.

Normally, in a skirt, T2 is zero and you just have compressive stress as P/A +/- MC/I, where the MC/I is due to wind or seismic overturn, P is the sum of all the weights above.

For a 42" diameter tank, a lot of your details will come up with minimum thickness, etc.

## RE: API 620

Thank you for taking the time to view my post and answer me.

I was struggling with the fact that we are subtracting a pressure value (when you are considering the cylinder), understanding that there is no negative pressure (unless you have a vacuum).

I'm also having a difficult time understanding why the book "Structural Analysis and Design of Process Equipment 3rd. Edition" is saying that the V force acting vertically along the cyl and cone is the same due to continuity.

Base on the definition you gave me (below the line, cone area) We have to consider the cone weight and liquid inside. when I do that, I get different values, see calculations below

Using continuity from "Structural Analysis and Design of Process Equipment 3rd. Edition".

V = 13.61 lb/in, then T1 = V/sin12.8 i get a T1 = 74.73 lb/in.

But when I used the formula from A620

T1 = (R/2Cos(90-12.8))(P + ((W+F)/A))

T1 = (44/2(0.22)(8.66 + (445 + 730)/6082)

T1 = 99.3(8.66 + 0.19)

T1 = 878.8 lb/in

It's not even close to the one above.

Im also confused with the statement about the angle.

Alpha = one-half the included apex angle of the conical or bottom, what does that mean?

I also have a question regarding the compressive ring design.

Do you have to design a compressive ring when the forces in your cone are both positive (tension)?

If I move my line just a couple of inches down and analyze above my line, I will have compressive stress but it won't be much, if I keep moving down, my compressive stress increases but I'm moving away from the intersection between the cyl and cone.

Thank you for your help

## RE: API 620

You need to design this using basic stress concentration formula and / or FEA.

Remember - More details = better answers

Also: If you get a response it's polite to respond to it.

## RE: API 620

I think you are talking about API 650, that one works with flat bottomed cylindrical tanks.

API 620 comes into play when you have cone tops or bottoms.

With the tank sizes I'm working on, if I had a flat bottomed tank, I think I would use AWWA D-100, because I'm dealing with water-liked liquids.

## RE: API 620

Immediately below the cone junction, the weights being supported by the skirt are the shell weight itself, the roof weight, any live or equipment loads on the roof, the cone weight, the cone contents weight, and the shell contents weight.

The T1 forces in the cone have a substantial vertical component, so the T1 forces in the shell/skirt just above and just below the cone will be a lot different.

You do need to check the compression ring area. The T1 force in the cone is positive, meaning it's pulling inwards on that ring, which introduces a compressive force in the compression ring region.

## RE: API 620

I understand the compressive ring now based on your response, that makes sense.

Now, if I look immediately below the cone junction, you said that I only have the weight of the liquid in the cone + the weight of the cone shell and they go in the same direction as pressure.

Based on that I do get a higher T1 than T1 on the cylinder shell.

T1(cone) = 878 lb/in

T1(Cyl) = -13.61 lb/in

But when you do the FBD immediately below the cone junction, Do I need to include the reaction force created by the Skirt supporting all the weight?

Thanks