S. K. Konyukhov, Prof., STI.

Calculation of Metal Tanks According to Stiglitz

Large-capacity tanks are most often encountered in chemical engineering. We use them for storing a great many valuable chemical liquids and gases: kerosene, petroleum, fuel oil, gasoline, lubricating oils, vegetable oils, train oil, fats, ammonia water, alkalis, molasses, sulfuric acid, water, etc. The most favored type of tank is an open or closed cylinder with a flat or concave bottom. The axis of the cylinder is usually arranged vertically. Tanks with a flat bottom are encountered most often, probably because their manufacture is cheaper and, in weight, they are lighter.

A good example of the widespread use of tanks for public needs may be furnished by the water-tower tanks of municipal water-supply and railway stations, and by gas works.

Tanks are built both of small size, less than 100 cubic meters in capacity, and very substantial ones exceeding 15,000 cubic meters.

Whatever the principal dimensions of such tanks may be, the basis of their design must be the principle of the least expenditure of construction material for a given capacity.

Let us first consider tanks whose capacity does not exceed 700 cubic meters.

Let the cubic content of the side walls and bottom of such a tank be \(H\) cubic meters, and the capacity \(J\) cubic meters. It is necessary to design this tank so that

\[ \frac{H}{J} \]

attains the smallest value.

If the height of the tank is denoted by \(h\), the diameter of the base by \(D\), the thickness of the wall of the lateral surface by \(S_1\), and the thickness of the bottom by \(S_2\), then two fundamental geometrical equations may be written:

\[ H = \pi D h S_1 + \frac{\pi D^2}{4} S_2 \qquad \cdots\cdots (1) \]

\[ J = \frac{\pi D^2}{4} h \qquad \cdots\cdots (2) \]

From these equations, by eliminating \(h\), the following is easily obtained:

\[ H = \frac{4 J S_1}{D} + \frac{\pi D^2}{4} S_2 \qquad \cdots\cdots (3) \]

Let us examine the function \(H\), as dependent on \(D\), in order to find its minimum.

\[ \frac{dH}{dD} = -\frac{4 J S_1}{D^2} + \frac{\pi D S_2}{2} \qquad \cdots\cdots (4) \]

\[ D = 2 \sqrt[3]{\frac{J S_1}{\pi S_2}} \qquad \cdots\cdots (5) \]

Here it is assumed that the tank has neither angle irons, nor cover plates, nor lap strips, nor rivets, i.e., as though it were welded from a single piece. However, such large tanks cannot everywhere be made by welding,

whereas in the old way they rivet them. With a riveted joint the weight of the tank will increase by approximately 8%. This is what we shall keep in mind.

Example 1. Find the diameter and height of an open cylindrical iron tank with a capacity \(J = 700\) cubic meters. The tank is intended for water and has wall thickness \(S_1 = 6\) mm, and thickness of the flat bottom \(S_2 = 5\) mm. We may now write, using equation (5),

\[ D = 2 \sqrt[3]{\frac{700 \cdot 6}{\pi \cdot 5}} = 12.886 \simeq 12.9 \text{ meters}; \]

\[ h = \frac{4J}{\pi D^2} = \frac{4 \cdot 700}{3.14 \cdot 12.9^2} = 5.36 \text{ meters} \simeq 5.4 \text{ meters}. \]

If in equation (5) we replace \(J\) by \(\dfrac{\pi D^2 h}{4}\), then we easily obtain the following relation

\[ D/2 : h = S_1 : S_2 \ . . . . . . . . . . . \ (6) \]

Thus, for an open tank of cylindrical form with the most advantageous dimensions, the radius of the base is related to the height of the tank as the thickness of the side wall is related to the thickness of the bottom. If \(S_1 = S_2\), then \(D = 2h\). The tank should be squat, not tall.

If the base of the vessel or tank is a square with side \(a\), then the conclusion remains the same

\[ a/2 : h = S_2 : S_2 \ . . . . . . . . . . . \ (7) \]

Let the vessel now be closed and again of cylindrical form. We introduce the roof into the calculation, using the same equation (5). Since people not infrequently move about on the roof, it is usually made convex, with a rise from \(1/10\) to \(1/12\) of the base diameter. Let us denote the thickness of the iron for the roof by \(S_3\). Then the material required for it will be

\[ \left(\frac{D^2}{4} + f^2\right)\pi S_3 \ . . . . . . . . . \ (8), \]

where \(f\) is the rise of the vault.

If \(f = \dfrac{D}{10}\), then, per square meter of the tank’s base area, the thickness will be

\[ S_3' = 1.04 S_3 \ . . . . . . . . . . . \ (9) \]

If \(f = \dfrac{D}{12}\), then

\[ S_3' = 1.028 S_3 \ . . . . . . . . . . . \ (10) \]

At the bottom of the roof a tie is made, in the manner of Schwedler’s rafters. The weight of this tie depends on the magnitude of the tank diameter and ranges from 12 to 35 kilograms per square meter of base area. If one represents the weight of this rafter tie by the weight of a fictitious circle of diameter \(D\) and thickness \(S_4\), then in practice the weight of the tie is found as follows. Let, for example, the weight of the rafter truss under the tank roof be chosen as 16 kg/sq. meter of base area. Taking, in round numbers, the specific weight of iron as equal to 8, we obtain the thickness

\[ S_4 = \frac{16}{8} = 2 \text{ mm}. \]

Example. Calculate the diameter and height of a tank for storing kerosene with a capacity of \(J=600\) cubic meters. The thickness of the walls is \(S_1=5\) mm, of the bottom \(S_2=6\) mm, of the roof \(2.5\) mm \(=S_3\), and the weight of the rafters is 16 kg/square meter of base area.

Thus, we have

\[ S_1=5\ \text{mm};\quad S_2=6\ \text{mm},\quad S'_3=1.04S_3=1.04\times2.5=2.6 \]

and

\[ S_4=\frac{16}{8}=2\ \text{mm}. \]

Consequently,

\[ S_2+S'_3+S_4=6+2.6+2=10.6\ \text{mm}. \]

And then

\[ D=2\sqrt[3]{\frac{600\cdot5}{\pi\cdot10.6}}=2\sqrt[3]{90}=8.963=9\ \text{m}. \]

\[ h=\frac{9\cdot10.6}{2.5}=9.54\ \text{m}=9.5\ \text{m}. \]

To determine the thickness of the tank walls, let us cut out from the wall of the vessel an arc corresponding to the central angle \(d\varphi\). From Fig. 1 it is evident that on this strip, of height \(dh\), the forces \(dP\) and \(\sigma \cdot s \cdot dh\) will act, the direction of which is indicated by the arrows. Since these three forces are in equilibrium, they must form a closed triangle (Fig. 2). It is not difficult to see that

\[ dP=\gamma h\,\frac{D}{2}\,dh\,d\varphi \ldots\ldots\ldots (11) \]

On the other hand,

\[ dP=\sigma s\,dh\,d\varphi \ldots\ldots\ldots (12) \]

Hence,

\[ \sigma s=\gamma \frac{D}{2}h \ldots\ldots\ldots (13) \]

Here \(\sigma\) is the mean stress, \(\gamma\) is the density of the liquid, and \(h\) is the depth of the liquid.

From this equation we obtain

\[ S=\frac{D\gamma h}{2\sigma}\ldots\ldots\ldots (14) \]

If one takes into account that rivet holes weaken the sheets at the joints, it will be necessary to introduce a correction coefficient into the calculations.

Calling the pitch of the riveted joint \(t\), and the diameter of the rivet shank \(d\), we obtain the correction coefficient \(\varphi\) in the form of the ratio

\[ \varphi=\frac{t-d}{d}\ldots\ldots\ldots (15) \]

If the rivets are arranged in two rows and the joint is made with overlap, then \(\varphi=0.74\); with cover plates \(\varphi=0.80\), and with cover plates above and below \(\varphi=0.90\).

Equation (14) now takes the following form:

\[ S=\frac{D\gamma h}{2\varphi\sigma_z}\ldots\ldots\ldots (16) \]

where \(\sigma_z\) is the allowable stress per unit area. Let, for example, \(D=12\) m, \(h=8\) m, \(\varphi=0.74\), and \(\gamma=800\) kg (cu. m for kerosene), and \(\sigma_z=1000\) kg/cm\(^2\), or \(10^7\) kg/m\(^2\). Then

\[ S=\frac{12\cdot 8\cdot 800}{2\cdot 0.74\cdot 10^7} =0.0052\ \text{m}=5.2\ \text{mm}. \]

Very often equation (16) is written in the following form

\[ S=\frac{D\gamma h}{20\varphi\sigma_z}\ . . . . . . . . . . . .\ (17) \]

where \(S\) is obtained in millimeters, \(D\) and \(h\) are taken in meters, \(\gamma\) in kilograms (cu. m), and \(\sigma_z\) in kilograms per sq. centimeter.

To obtain a strong and tight closing seam, caulking is necessary. But sheets thinner than 5 mm are difficult to caulk. Therefore, for tanks with a capacity of up to 700 meters, iron 5 mm thick is taken for the bottom and for the walls of the tanks.

Equation (14) gives grounds for concluding that, for an ideal vessel, the thickness of the tank walls should vary according to the law of the equation of a straight line, since \(S\) is a linear function of \(h\). Equation (14) has no constant term, and, consequently, the straight line passes through the origin of coordinates. The limiting values for \(S\) will be, on the one hand, zero, and on the other, for points near the bottom,

\[ S=\frac{D\gamma h}{2\varphi\sigma}\ . . . . . . . . . . . . .\ (18) \]

or

\[ S=\frac{D\gamma h}{20\varphi\sigma_z}\ . . . . . . . . . . . .\ (19) \]

If we suppose that the weight of the tank roof with the rafter tie is uniformly distributed over the bottom, and denote the ideal bottom thickness under this condition by \(S_{11}\), then we obtain

\[ H=\frac{\pi D^2}{4}S_{11}+\pi D h\cdot\frac{D\gamma h}{2\cdot 20\varphi\sigma_z}\ . .\ (20) \]

or, likewise,

\[ H=\frac{J S_{11}}{h}+\frac{J h\gamma}{10\varphi\sigma_z}\ . . . . . . . .\ (21) \]

Here the 2 in the denominator is obtained from the fact that the mean wall thickness will be \(S/2\).

If \(H\) is analyzed to find the minimum, we obtain

\[ \frac{dH}{dh}=-\frac{J S_{11}}{h^2}+\frac{J\gamma}{10\varphi\sigma_z}\ . . . . .\ (22) \]

Whence it follows that

\[ h=\sqrt{\frac{10\varphi\sigma_z\cdot S_{11}}{\gamma}}\ . . . . . . . . .\ (23) \]

In practice, a tank with a wall thickness tapering to nothing is difficult to make, and therefore the wall thickness is made stepped, in courses with the course thickness increasing downward from 0.5 mm to 3.5 mm; while for the upper part one thickness \(S_o=(5—6)\) mm is taken over the height \(h_o\). To determine the height \(h_o\), we shall use the equation

\[ h_o=\frac{20\varphi\sigma_z S_o}{D\gamma}\ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ . \ (24) \]

Now it will be possible to proceed to the calculation of a tank whose walls will form steps.

According to Fig. 3, an area \(\dfrac{S_o h_o}{2}\) will be added to the upper part of the wall section. The volume of this additional ring will be

\[ R_o=\pi D\frac{S_o h_o}{2}=\frac{10\pi\varphi\sigma_z S_o^2}{\gamma}\ . \ . \ . \ (25) \]

Analyzing equation (25), it is not difficult to see that \(R_o\) depends neither on \(D\) nor on \(h_o\). This means that \(R_o\) will be the same for tanks of different diameters and different heights. Next it will be necessary to investigate the additional rings with heights \(\Delta h_1\), \(\Delta h_2,\ldots\) (Fig. 3), the sections of which will be

\[ \frac{\Delta S_1\,\Delta h_1}{2},\quad \frac{\Delta S_2\,\Delta h_2}{2},\quad . \ . \ . \ . \ . \ . \ . \]

Summing the volumes of these rings, we obtain

\[ R_1+R_2+R_3+=\frac{10\pi\varphi\sigma_z}{\gamma} \left(\Delta S_1^2+\Delta S_2^2+\right)\ . \ . \ (26) \]

This mass of iron, with variable heights and diameters, is nevertheless close to a constant quantity, and therefore plays no role in differentiation. Thus, in determining the diameter and height of the tank, one may use the previous equations.

Example. Calculate a tank for gasoline with a capacity of

\[ J=1500\ \text{cub. mtr.},\quad \text{if } \gamma=700\ \text{kg/mtr.}^3. \]

We take \(\varphi=0.74\) and \(\sigma_z=1000\ \text{kg/cm.}^2\).

Let the roof truss give a weight of \(27\ \text{kg/sq. mtr.}\) of base surface. Then

\[ S_4=\frac{27}{8}=\ . \ . \ . \ . \ . \ . \ . \ 3.4\ \text{mm.} \]

\[ \text{Roof } S_3=2.5\ \text{or } 1.04\cdot 2.5\ . \ . \ 2.6 \]

\[ \text{Bottom } S_2=\ . \ . \ . \ . \ . \ . \ . \ 7 \]

\[ \overline{\qquad\qquad\qquad} \]

\[ S_{11}=\ . \ . \ 13\ \text{mm.} \]

Thus, we obtain

\[ h=\sqrt{\frac{10\cdot0.74\cdot1000\cdot13}{700}}=11.7\ \text{mtr.} \]

\[ D=\sqrt{\frac{4\cdot1500}{\pi\cdot11.7}}=12.8\ \text{mtr.} \]

The thickness of the tank walls at the bottom will be

\[ S=\frac{12.8\cdot11.7\cdot700}{2\cdot0.74\cdot1000}=9\ \text{mm.} \]

Usually the side surface is divided into courses so that the height of each course is between 1 and 2 meters.

If we stop at a course height of 1.3 meters, then, beginning from the bottom, we obtain courses with wall thicknesses of 9 mm, 8 mm, 7 mm, 6 mm, and the last course with wall thickness of 5 mm. The height of this last course will be

\[ h = 11.7 - 1.3 \cdot 4 = 6.5 \text{ m}. \]

This picture is given in Fig. 4.

Example 2. Determine the principal dimensions for a tank intended for kerosene. The capacity of the tank is \(J = 2500\) cubic meters; \(\gamma = 960\) kg/m\(^3\). We take \(\varphi = 0.78\) and \(\sigma_z = 1200\) kg/cm\(^2\).

If the roof is taken with thickness \(S_3 = \ldots\ 3\) mm, then

\[ S'_3 = 1.04 \cdot 3 = \ldots\ 3.12 \text{ mm}. \]

bottom thickness \(S_2 = \ldots\ldots\ldots\ldots\ldots\ 8\) mm.

Rafters \(30\) kg/m\(^2\) \(\ldots\ldots\ldots\ldots\ldots\ 3.75\) mm.

\[ S_{11} = \ldots\ 14.87 \text{ mm}. \]

Consequently,

\[ h = \sqrt{\frac{10 \cdot 0.78 \cdot 1200 \cdot 14.87}{960}} = 12 \text{ m}. \]

\[ D = \sqrt{\frac{4 \cdot 2500}{\pi \cdot 12}} = 16.3 \text{ m}. \]

The wall thickness in the lower course is

\[ S = \frac{16.3 \cdot 12 \cdot 960}{2 \cdot 0.78 \cdot 1200} = 10 \text{ mm}. \]

Fig. 5 shows that the lower course has wall thickness 10 mm and height 1.8 m; the next course has thickness 8.5 mm and height 1.8 m; the third course has wall thickness 7 mm and height 1.8 m; finally, the next course has wall thickness 5.5 mm and height 6.6 m.

Example 3. A tank for gasoline with capacity \(J = 850\) cubic meters and \(\gamma = 900\) kg/m\(^3\).

We take \(\varphi = 0.77\) and \(\sigma_z = 1200\) kg/cm\(^2\).

A roof 2.5 mm thick gives \(1.04 \times 2.5 = \ldots\ 2.6\) mm.

Rafters \(16\) kg/m\(^2\) \(\ldots\ldots\ldots\ldots\ldots\ 2\) mm.

Bottom of 5 mm \(\ldots\ldots\ldots\ldots\ldots\ 5\) mm.

\[ S_{11} = \ldots\ 9.6 \text{ mm}. \]

Thus,

\[ h = \sqrt{\frac{10 \cdot 0.77 \cdot 1000 \cdot 9.6}{900}} = 9.1 \text{ m}. \]

\[ D = \sqrt{\frac{4 \cdot 900}{\pi \cdot 9.1}} = 6 \text{ m}. \]

Thickness of the lower course

\[ S = \frac{11.3 \cdot 9 \cdot 1 \cdot 900}{2 \cdot 0.77 \cdot 1000} = 6 \text{ mm}. \]

The thickness \(S\) proved to be small, and therefore it may be taken as uniform over the entire height of the tank; then, by equation (6),

\[ D_2 : h = S_1 : S_2 = 6 : 9.6 = 5.65 : 9.1 . \]

The basic equation for calculating the principal dimensions shows that \(J\) does not enter into it. Indeed, the dimensions of the tank depend on the thickness of the bottom, the roof, the rafters, and \(\varphi\).

If we pass to large tanks with a diameter of 15–35 m and a capacity \(J = 1500\text{–}12000\) cubic meters, then in them the weight of the rafters per square meter of axial surface is approximately constant. If for all large tanks one adopts the same thickness of the bottom and roof, then equation (23) indicates that all tanks should have the same height. Here an interesting phenomenon is observed: for tanks with capacities from 2000 cubic meters upward, the influence of the weight \(R_0\) becomes more and more attenuated as \(J\) increases, while the least unit of weight, in kilograms, falling on one cubic meter of liquid, decreases slowly with increasing \(J\), so that it may be taken as constant.

Thus it is equally advantageous to build one tank of 12,000 cubic meters capacity, or three tanks of 4000 cubic meters capacity each.

If all additional rings \(R\) are neglected, we obtain

\[ H = \frac{J S_{11}}{h} + \frac{J h \gamma}{10 \varphi \sigma_z}; \quad \frac{H}{J} = \mathrm{Const} \ldots \ldots \ldots \ldots \tag{27} \]

Let us solve the problem of the least weight of an iron tank with a capacity exceeding 1500 cubic meters.

Practice shows that the most advantageous height of large-capacity tanks varies within the limits between 10 and 13 meters, while the height of the courses is approximately equal to 1.5 meters.

If such a division into courses is carried out over the whole height, then 8 courses will be obtained (Fig. 6). The sum of the areas hatched horizontally and horizontally and vertically is \(12.5\%\) of the area \(\dfrac{hS}{2}\). About \(6.5\%\) goes to the angles. On the cover plates, lap overlap, and rivet heads there goes on average \(10\%\). Let us denote by \(R'_0\) the magnitude of the stepped area hatched vertically. Then we obtain:

\[ H = 1.1 \cdot 5.85 \left( \frac{J S_{11}}{h} + J h \frac{\gamma}{10 \varphi \sigma_z} \cdot \frac{100 + 12.5 + 6.5}{100} + R'_0 \right) \ldots \ldots \tag{28} \]

or

\[ H = 8.635 \left( 2.19 J h \frac{\gamma}{10 \varphi \sigma_z} + R'_0 \right) \ldots \ldots \ldots \ldots \ldots \tag{29} \]

But the quantity \(R'_0\) is approximately equal to \(R_0\), i.e.

\[ R'_0 = R_0 = \frac{10\pi \varphi \sigma_z}{\gamma} S_0^2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \tag{30} \]

Thus,

\[ H = 8.635 \left( 2.19 J h \frac{\gamma}{10 \varphi \sigma_z} + \frac{10\pi \varphi \sigma_z}{\gamma} S_0^2 \right) \ldots \ldots \ldots \tag{31} \]

or

\[ \frac{H}{J} = \frac{1.9 h}{k} + \frac{270 S_0^2 k}{J} \ldots \ldots \ldots \ldots \ldots \tag{32} \]

where

\[ k = \frac{\varphi \sigma_z}{\gamma}. \]

Example. How great should the weight be of a tank with a capacity of \(j = 3000\) cubic meters?

Take \(S_0 = 6\) mm. For the bottom, cover, and rafters, take \(S_{11} = 14.9\) mm. If \(\gamma = 1000\ \mathrm{kg}/\mathrm{m}^3\), \(\varphi = 0.75\), and \(\sigma = 1200\ \mathrm{kg}/\mathrm{cm}^2\), then

\[ K = \frac{\varphi \sigma_z}{\gamma} = \frac{0.75 \cdot 1200}{1000} = 0.9 \]

Now we have

\[ h = \sqrt{\frac{10 \varphi \sigma_z S_{11}}{\gamma}} = \sqrt{\frac{10 \cdot 0.75 \cdot 1200 \cdot 14.9}{1000}} = 11.6\ \mathrm{m}. \]

\[ \frac{H}{j} = \frac{1.9 \cdot 11.6}{0.9} + \frac{270 \cdot 36 \cdot 0.9}{3000} = 24.5 + 2.916 = 27.416\ \mathrm{kg}/\mathrm{m}^3. \]

Consequently, the tank weighs

\[ 27.416 \times 3000 = 82248\ \mathrm{kg} \simeq 82\ \text{tons}. \]

The change in the unit weight with increasing tank height may be seen in Fig. 7, where the ratios \(\frac{H}{j}\) in \(\mathrm{kg}/\mathrm{m}^3\) are given for a tank with a capacity of 6000 cubic meters.

In the construction of one tank of 6000 cubic meters, the following were taken:

\[ S_2 = 7\ \mathrm{mm},\quad S_3 = 2.5\ \mathrm{mm},\quad S_4 = 30\ \mathrm{kg}/\mathrm{m}^2,\quad S_0 = 5\ \mathrm{mm}. \]

Angles at the top were \(90 \times 90 \times 11\), and at the bottom \(120 \times 120 \times 13\). Further, it was taken that \(\varphi = 0.76\), \(\sigma_z = 1200\ \mathrm{kg}/\mathrm{cm}^2\), and \(\gamma = 1000\ \mathrm{kg}/\mathrm{m}^3\). The smallest unit weight according to diagram 7 is obtained as \(23.5\ \mathrm{kg}/\mathrm{m}^3\) at a tank height \(h = 11\) meters. But the water tank was built only 7 meters high. Thus, an excess amount of iron was used:

\[ 6000(25.7 - 23.5) = 13200\ \mathrm{kg}. \]

This amounts to about 13% of the total weight.

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

Fig. 7.

Calculation of metal tanks according to Stiglitz

To the article by Prof. S. N. Kononkov

Fig. 6.

Fig. 5.