## Cone bottom thickness

## Cone bottom thickness

(OP)

Hi,

Im new working with API620 and need some guidance.

Im designing several storage vertical cylindrical tanks base on API 620.

One of the tanks has the following data:

Radius = 72 in

Shell Height = 34ft.

Wshell = 12,500 lbs.

Wroof = 831 lbs.

Density = 62.4lbs.ft3

Cone bottom.

Cone Radius = 72in

Cone Height = 6in.

Opened to atmosphere.

If I'm analyzing at the bottom of the shell where it connects to the cone, this are my numbers:

T1 = (Rc/2)(P + ((W+f)/A)))

Since i dont have external forces, F = 0

I end up with the following:

Rc = 72

P = Pliq + Pair, since the tank is open to atmosphere, Pair = 0.

Pliq at that point is = 14.84 PSI

W = Wroof + WShell + Wliq)

W = 831 + 12,500 + 239,947.31....... Wliq = 62.4*Pi*6^2*34

A = 16,286.. Area = Pi*r^2 = Pi*(72)^2

All the forces acting up are positive on my diagram.

I end up with:

T1 = (72/2)(14.84 + (-(831+12,500+239,947)/16,286))

T1 = -25.63 lb/in

T2 = PR/Sin(Theta)

Theta = 4.76 degrees

T2 = (14.84 * 72)/0.0829

T2 = 12,888.78 lb/in

I believe my calculations are correct, but im not sure.

This is where I start having questions.

Since my T1 is under compression and T2 under tension I have to use Figure 5.1 to calculate the the computed compressive stress, scc, shall not exceed a value of the allowable compressive stress, sca, determined from Figure 5-1 by entering the computed value of N and the value of t/R associated with the compressive unit stress and reading the value of sc that corresponds to that point. The value of sc will be the limiting value of sca for the given conditions

I thinking of using a t = 0.25in.

t/R = 0.25/72

t/R = 0.0034

With that number i get a value of N close to 0.74

using that N, my Sta = 16,000 x 0.74

Sta = 11,840 PSI

Now, to calculate my Scc I used the following formula.

Scc = T2/t

Im using T2 to compare against my Sta because T2 is my highest number.

Scc = 12,888.78/0.25

It gives me close to 52,000 PSI, that is way bigger than 11,840, which mean it will fail.

I either change the thickness of my cone or increase my Theta.

Am i correct?

Thank you

Im new working with API620 and need some guidance.

Im designing several storage vertical cylindrical tanks base on API 620.

One of the tanks has the following data:

Radius = 72 in

Shell Height = 34ft.

Wshell = 12,500 lbs.

Wroof = 831 lbs.

Density = 62.4lbs.ft3

Cone bottom.

Cone Radius = 72in

Cone Height = 6in.

Opened to atmosphere.

If I'm analyzing at the bottom of the shell where it connects to the cone, this are my numbers:

T1 = (Rc/2)(P + ((W+f)/A)))

Since i dont have external forces, F = 0

I end up with the following:

Rc = 72

P = Pliq + Pair, since the tank is open to atmosphere, Pair = 0.

Pliq at that point is = 14.84 PSI

W = Wroof + WShell + Wliq)

W = 831 + 12,500 + 239,947.31....... Wliq = 62.4*Pi*6^2*34

A = 16,286.. Area = Pi*r^2 = Pi*(72)^2

All the forces acting up are positive on my diagram.

I end up with:

T1 = (72/2)(14.84 + (-(831+12,500+239,947)/16,286))

T1 = -25.63 lb/in

T2 = PR/Sin(Theta)

Theta = 4.76 degrees

T2 = (14.84 * 72)/0.0829

T2 = 12,888.78 lb/in

I believe my calculations are correct, but im not sure.

This is where I start having questions.

Since my T1 is under compression and T2 under tension I have to use Figure 5.1 to calculate the the computed compressive stress, scc, shall not exceed a value of the allowable compressive stress, sca, determined from Figure 5-1 by entering the computed value of N and the value of t/R associated with the compressive unit stress and reading the value of sc that corresponds to that point. The value of sc will be the limiting value of sca for the given conditions

I thinking of using a t = 0.25in.

t/R = 0.25/72

t/R = 0.0034

With that number i get a value of N close to 0.74

using that N, my Sta = 16,000 x 0.74

Sta = 11,840 PSI

Now, to calculate my Scc I used the following formula.

Scc = T2/t

Im using T2 to compare against my Sta because T2 is my highest number.

Scc = 12,888.78/0.25

It gives me close to 52,000 PSI, that is way bigger than 11,840, which mean it will fail.

I either change the thickness of my cone or increase my Theta.

Am i correct?

Thank you

## RE: Cone bottom thickness

In the cone, the supported weight is weight of cone plate and cone contents only (plus any piping hanging off of it), pressure acts in the same direction as the weight, and you should get T1 and T2 both positive, usually with T2 controlling cone thickness and T1 having major influence on compression ring design.

In the shell above the cone, you should get relatively low T1, either positive or negative depending on how much internal pressure relative to the weight, and T2 should be positive.

In the shell below the cone, you should get T2 as zero and T1 as simply weight divided by area, using allowable buckling stress in a cylinder.

Usually, it will help to make a suspended bottom steeper than 4.76 degrees.

## RE: Cone bottom thickness

I did the analysis for the shell, and I did get a low T1 = -25.63 Lb/in, T2 at that point was 1,068 lb/in.

T2 = P*Rc - 14.84 x 72

You are right on the cone, I was doing my FBD wrong, T1 and T2 are Positive.

T1 = 309.16 lb/in

T2 = 12,888 lb/in

Since both are positive, I use the following formula to calculate thickness:

t = (T2/(Sts*E))+ c

I get a t = 1 1/8"

That is a very thick plate.

Are my calculations right? (see attached picture).

Thank you for your help.