The extreme durability of reinforced-concrete tanks, and their immunity from deterioration by rust, which so quickly destroys steel tanks, have resulted in the construction of a large and increasing number of tanks in reinforced concrete. Such tanks must be designed to withstand the bursting pressure of the water. If they are very high compared with their diameter, it is even possible that failure might result from excessive wind pressure.

The method of designing one of these tanks may best be considered from an example. Suppose that it is required to design a reinforced-concrete tank with a capacity of 50,000 gallons, which shall have an inside diameter of 18 feet. At 7.48 gallons per cubic foot, a capacity of 50,000 gallons will require 6,684 cubic feet. If the inside diameter of the tank is to be 18 feet, then the 18-foot circle will contain an area of 254.5 square feet. The depth of the water in the tank will therefore be 26.26 feet. The lowest foot of the tank will therefore be subjected to a bursting pressure due to 25.76 vertical feet of water. Since the water pressure per square foot increases 62 V pounds for each foot of depth, we shall have a total pressure of 1,610 pounds per square foot on the lowest foot of the tank. Since the diameter is 18 feet, the bursting pressure it must resist on each side is one-half of 18 X 1,610 = 1/2 X 28,980 = 14,490 pounds. If we allow a working stress of 15,000 pounds per square inch, this will require .966 square inch of metal in the lower foot. Since the bursting pressure is strictly proportional to the depth of the water, we need only divide this number proportionally to the depth to obtain the bursting pressure at other depths. For example, the ring one foot high, at one-half the depth of the tank, should have .483 square inch of metal; and that at one-third of the depth, should have .322 square inch of metal. The actual bars required for the lowest foot may be figured as follows: .9GG square inch per foot equals .0805 square inch per inch; 3/4-inch square bars, having an area .5625 square inch, will furnish the required strength when spaced 7 inches apart. At one-half the height, the required metal per linear inch of height is half of the above, or .040. This could be provided by using 3/4-inch bars spaced 14 inches apart; but this is not so good a distribution of metal as to use 5/8-inch square bars having an area of .39 square inch, and to space the bars nearly 10 inches apart. It would give a still better distribution of metal, to use 1/2-inch bars spaced 6 inches apart at this point, although the 1/2inch bars are a little more expensive per pound, and, if they are spaced very closely, will add slightly to the cost of placing the steel. The size and spacing of bars for other points in the height can be similarly determined. BRIDGE OF REINFORCED CONCRETE, AT PLAYA DEL REY, NEAR LOS ANGELES, CAL. Extreme Length, 205 feet 8 inches; Span, 146 feet; Width, 19 feet; Spring, 18 feet; Height above water, 20 feet.

A circle 18 feet in diameter has a circumference of somewhat over 56 feet. Assuming as a preliminary figure that the tank is to be 10 inches thick at the bottom, the mean diameter of the base ring would be 18.83 feet, which would give a circumference of over 59 feet. Allowing a lap of 3 feet on the bars, this would require that the bars should be about 62 feet long. Although it is possible to have bars rolled of this length, they are very difficult to handle, and require to be transported on the railroads on two flat cars. It is therefore preferable to use bars of slightly more than half this length, and to make two joints in each band.

The bands which are used for ordinary wooden tanks are usually fastened at the ends by screw-bolts. Some such method is necessary for the bands of concrete tanks, provided the bands are made of plain bars. Deformed bars have a great advantage in such work, owing to the fact that, if the bars are overlapped from 18 inches to 3 feet, according to their size, and are then wired together, it will require a greater force than the strength of the bar to pull the joints apart after they are once thoroughly incased in the concrete and the concrete has hardened.

## 314. Test For Overturning

Since the computed depth of the water is over 26 feet, we must calculate that the tank will be, say, 28 feet high. Its outer diameter will be approximately 20 feet. The total area exposed to the surface of the wind, will be 560 square feet.

We may assume that the wind has an average pressure of 50 pounds per square foot; but owing to the circular form of the tank, we shall assume that its effective pressure is only one-half of this; and therefore we may figure that the total overturning pressure of the wind equals 560 X 25 = 14,000 pounds. If this is considered to be applied at a point 14 feet above the ground, we have an overturning moment of 196,000 foot-pounds, or 2,352,000 inch-pounds.

Although it is not strictly accurate to consider the moment of inertia of this circular section of the tank as it would be done if it were a strictly homogeneous material, since the neutral axis, instead of being at the center of the section, will be nearer to the compression side of the section, our simplest method of making such a calculation is to assume that the simple theory applies, and then to use a generous factor of safety. The effect of shifting the neutral axis from the center toward the compression side, will be to increase the unit-compression on the concrete, and reduce the unit-tension in the steel; but, as will be seen, it is generally necessary to make the concrete so thick that its unit compressive stress is at a very safe figure, while the reduction of the unit-tension in the steel is merely on the side of safety.

Applying the usual theory, we have, for the moment of inertia of a ring section, .049 (d14 - d4). Let us assume as a preliminary figure, that the wall of the tank is 10 inches thick at the bottom. Its outside diameter is therefore 18 feet + twice 10 inches, or 236 inches. The moment of inertia I = .049 (2364 - 2164) = 45,337,-842 biquadratic inches. Calling c the unit-compression, we have, as the ultimate moment due to wind pressure:

M = cI = c x 45,337,482 = 2,352,000 inch-pounds,

1/2 d1 1/2 d1 in which 1/2 d1 = 118 inches.

Solving the above equation for c, we have c equals a fraction less than 6 pounds per square inch. This pressure is so utterly insignificant, that, even if we double or treble it to allow for the shifting of the neutral axis from the center, and also double or treble the allowance made for wind pressure, although the pressure chosen is usually considered ample, we shall still find that there is practically no danger that the tank will fail owing to a crushing of the concrete due to wind pressure.

The above method of computation has its value in estimating the amount of steel required for vertical reinforcement. On the basis of 6 pounds per square inch, a sector with an average width of 1 inch and a diametral thickness of 10 inches would sustain a compression of about 60 pounds. Since we have been figuring working stresses, we shall figure a working tension of, say, 16,000 pounds per square inch in the steel. This tension would therefore require 60 =

16,000

.0037 square inch of metal per inch of width Even if 1/4-inch bars were used for the vertical reinforcement, they would need to be spaced only about 17 inches apart. This, however, is on the basis that the neutral axis is at the center of the setnon, which is known to be inaccurate.

A theoretical demonstration of the position of the neutral axis for such a section, is so exceedingly complicated that it will not be considered here. The theoretical amount of steel required is always less than that computed by the above approximate method; but the necessity for preventing cracks, which would cause leakage, would demand more vertical reinforcement than would be required by wind pressure alone.

## 315. Practical Details Of The Above Design

It was assumed as an approximate figure, that the thickness of the concrete side wall at the base of the tank should be 10 inches. The calculations have shown that, so far as wind pressure is concerned, such a thickness is very much greater than is required for this purpose; but it will not do to reduce the thickness in accordance with the apparent requirements for wind pressure. Although the thickness at the bottom might be reduced below 10 inches, it probably would not be wise to do so. It may, however, be tapered slightly towards the top, so that at the top the thickness will not be greater than 6 inches, or perhaps even 5 inches. The vertical bars in the lower part of the side wall must be bent so as to run into the base slab of the tank. This will bind the side wall to the bottom. The necessity for reinforcement in the bottom of the tank depends very largely upon the nature of the foundation, and also to some extent on the necessity for providing against temperature cracks, as has been discussed in a previous section. Even if the tank is placed on a firm and absolutely unyielding foundation, some reinforcement should be used in the bottom, in order to prevent cracks which might produce leakage. These bars should run from a point near the center, and be bent upward at least 2 or 3 feet into the vertical wall. Sometimes a gridiron of bars running in both directions is used for this purpose. This method is really preferable to the radial method. The methods of making tanks water-tight have already been discussed. TOP AND SIDE VIEW OF SPRING CANYON FLUME ON INTERSTATE CANAL, NORTH PLATTE IRRIGATION PROJECT, NEBRASKA.