314. Test for Overturning. Since the computed depth of the water is over 26 feet, we must calculate that the tank will he, say, 25 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 there fore 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-com pression 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 — 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 — = 45,337, 842 biquadratic inches. Calling c the unit-compression, we have, as the ultimate moment due to wind pressure: in which = 11S 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 shift ing 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 Finch bars were used for the vertical reinforcement, tl cy would need to be spaced only about 17 inches apart. This, 1- rvever, is on the basis that the neutral axis is at the center of the se ion, 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 require ments 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 sec tion. 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.