Enough area for working is also important, especially on the uphill side. Make the site large enough so dirt and rocks don’t fall into the steel armature. Contamination entangled in the structure is a problem to avoid during construction. The area made up of excavated fill is a good place for the access road to terminate and to store materials. If this is a large tank and the excavated material is a mountain of dirt poised to cause damage below during a flood year, then it should be placed on a cut bench cut of its own and be compacted for stability and safety.
πr2h = volume (where π = 3.14, r = radius, and h = height)
The following example is for a tank of sixty cubic meters; height is 2.13 meters.
πr2(2.13) = 60 cubic meters (sixty thousand liters)
r2 = 60 cubic meters ÷ (2.13 x 3.14) = 8.971 m2
r = radius = 3 meters
2r = diameter = d = 6 meters
Convert the depth into pressure, measured in grams per square centimeter and calculate the circumference in centimeters.
πd = 3.14 x 6 meters = circumference = 1884 centimeters.
The pressure on a square centimeter (kg/cm2) = the depth of 2.13 meters = 0.213 kilograms per square centimeter.
This means that there is 0.213 kilograms of outward pressure on a one centimeter square at the bottom of the tank wall. Since the wall is 1884 centimeters around, the total outward force on the bottom centimeter of wall is 0.213 x 1884 = 401 kilograms.
The next step is to determine the strength of the wall as it resists this outward pressure. The concrete plaster is only considered as waterproofing for the steel in this calculation. All the strength is assumed to be in the steel. Add up the horizontal strands of welded wire and the bars which encircle the tank. Count the welded wire and the reinforcing bars separately since they are different strengths of steel. Reinforcing steel is 3515 kilograms of tensile strength per square centimeter and the welded wire is 6328kg/cm2.
There are five horizontal wires and two reinforcing bars in the bottom thirty centimeters of this sixty cubic meter tank. Ignore the welded wire bent to come up and out of the floor until further along the discussion. Standard welded wire is ten gauge wire on 7.5 centimeter squares. Ten gauge wire is 0.356 cm diameter.
πr2 = 0.1 square centimeters of steel times five wires = 0.5 square centimeters. Multiply this by 6328 kilograms per square centimeter = 3164 kilograms of tensile strength in the bottom 30 centimeters of wall. Divide by 30 to compute the welded wire strength in an average centimeter of wall. 3164 ÷ 30 = 105 kilograms of horizontal welded wire tensile strength per average vertical centimeter of wall. The same calculation is done for two horizontal wraps of #4 bar (1.27 centimeters).
πr2 multiplied by 2 multiplied by 3515 kilograms of tensile strength per square centimeter = 7030 kilograms of tensile strength in the reinforcing bar, in the bottom 30 cm of wall. Divide by 30 to find the average strength in a centimeter of wall. 7030 ÷ 30 = 234.
The total wall steel strength is 234 + 105 kilograms = 339 kilograms of tensile strength in the steel. There is an additional #4 bar in the floor-to-wall key which brings the steel strength figure to 456 kilograms.
The final step in comparing steel tensile strength to water force is to draw a circle and quarter it as pictured below.
Imagine all the water force as concentrated in one direction along arrow B. The small circle at A is an anchor. Arrow B pulls with a force of 401 kilograms, which is the total outward water force on the bottom centimeter of wall (calculated above).
Imagine next that the tank wall is infinitely strong except where the line CD cuts the tank in half. At points C and D the wall is the tensile strength of the steel calculations; 456 kilograms at C and 456 kilograms at D. Total wall steel strength the water must break is thus 912 kilograms. Steel tensile strength divided by water force is 912 ÷ 401 = 2.3; the wall steel is 2.3 times stronger than the water force.
Note 1: The welded wire coming out of the floor adds enough to bring the steel strength figure to almost 2.5 times stronger than water force, assuming that all the wires are at 45 degrees.
Note 2: An impression of just how strong ferrocement is for structures other than tanks is gained by reversing arrow B; push instead of pull. Well cured ferrocement easily has 550 kilograms of compression strength per square centimeter. If a structural wall is eight centimeters thick, points C and D would add 8800 kilograms to the 912 kilograms of steel strength. Arrow B must push with a force greater than 9700 kilograms to crush a one centimeter wide arc of ferrocement, at points C and D.
Wall area = 2πr(height) = 2π(3)(2) = 37.5 m2
Roof: The roof steel extends down the wall and the roof is also an arc.
Floor: To estimate floor steel add ten percent for waste and ten percent for the steel which extends beyond the circumference line before bending it to vertical position.
The result is (1.2)πr2 = floor area calculation for steel. Add a little more for roof arc and use (1.25)πr2 = roof area calculation for roof steel.
Floor or roof area multiplied by 2 (two layers of welded wire) = 56.5 m2. Multiply this figure by the factors discussed previously. 56.5(1.2)(floor) + 56.5(1.25)(roof) = 138.4 ≈ 138m2 of welded wire in the roof and the floor.
Conclude the welded wire computation by adding the wall.
There are two layers of welded wire in the wall. 37.5m2 multiplied by two = 75m2; add 10 m2 for wire overlaps and waste = 85 m2.
The total for welded wire is 138m2 for roof and floor plus 85m2 for the wall = 223 m2 of welded wire. The price of welded wire per m2 multiplied by 223 m2 = total cost of welded wire.
Calculation of reinforcing bars depends upon the spacing chosen between the bars and the length of a standard bar. Chapter two uses the grid space of 30 to 45 centimeters. Six meters is used further on in this book as a standard length. The method used to calculate reinforcing steel is to visualize a square with equal to the standard length of reinforcing steel. In this example it is a six meter square with an area of 36m2.
Nineteen bars creates a spacing of 33.33 centimeters across six meters. This equals thirty eight bars total. Divide 38 bars by 36 m2 = 1.05 reinforcing steel bars per m2. Add ten percent for waste and overlaps and there are 1.15 bars per m2.
28.26m2 (roof) + 28.26m2 (floor) + 37.5m2 (wall) = 94m2 (total).
1.15 bars/m2 multiplied by 94m2 = 108 bars of reinforcing steel at a 33.33 centimeter spacing.
This calculation at a 45 centimeter space between bars is 6 m divided by 45 cm, plus one bar = 14.33 bars. multiply this by two for the total bars = 28.66. Divide by 36m2 = .79 bars/m2. Add ten percent = .9 bars/m2. Multiply by the total area (94m2) and the reinforcing bars required equals 85.
Multiply the price of one reinforcing steel bar by the number of bars to compute the total cost of reinforcing steel bars.
Expanded metal for the inside of the roof and wall is wall plus roof areas multiplied by their use factors. 28.26(1.25) (roof) + 37.5(1.1) (wall) = 76.5 m2.
Concrete is best estimated at 7.75 centimeter thickness multiplied by the total area plus approximately five percent for waste. The floor is estimated separately and done first.
A small volume factor (0.2) for the joint between wall and floor is added to the floor estimate. 28.26 m2 (floor area) multiplied by 0.0775 m (thickness) multiplied by 1.2 = 2.6 m3.
Roof and wall is (28.26 m2 + 37.5 m2)(0.0775)(1.05) = 5.35 m3.
Color pigments, extra cement water seal product, and glue (if the outside is to be colored).