Enough room for working is also an important consideration, 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)
Verbal directions in this book are for 15,000 gallons (57 cubic meters). Photo examples utilize a smaller sized model. Height is seven feet (2.13 meters). The calculations in this chapter work the same in cubic feet or cubic meters. The metric versions of this manual use 60 cubic meters, with a curve into the roof at two meters and a diameter of six meters. The seven foot height is chosen because welded wire is commonly available in seven foot wide rolls which are 200 feet long. The curve into the roof using welded wire from seven foot wide rolls begins at about six feet eight inches, which is approximately two meters. These calculation steps are for any volume.
πr2(7) = 15,000 gal ÷ 7.48 gallons per cubic foot = 2005 cubic feet
r2 = 2005 ft3 ÷ 7 ft ÷ 3.14 = 91.24 square feet
r = radius = 9.55 feet
2r = diameter = d = 19.1 feet
Fifteen thousand gallons is used in this example because many ferrocement tanks have been built of this size and there have been no problems, even after twenty-five to thirty years. Tanks of this age in the 50,000 to 100,00 gallon class have likewise shown no problems, Fifty thousand gallons is somewhat more difficult to build and 100,000 gallons is the beginning of a heavier construction project size. Ten foot (3 meter) height becomes economically logical as size approaches 50,000 gallons.
πd = 3.14 x 19.1 feet x 12 inches per foot = circumference = 719.8 inches.
Pressure per square inch (psi) = Depth in feet times .4335 pounds per square inch per foot of depth.
7 x .4335 = 3.03 psi
This means that there is 3.03 pounds of outward pressure on a 1 inch square at the bottom of the tank wall. Since the wall is 719.8 inches around, the total outward force on the bottom inch of wall is 3.03 x 719.8 = 2,180 pounds.
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. Common reinforcing steel is 50,000 pounds per square inch tensile strength and the welded wire is 90,000 pounds per square inch.
There are five horizontal wires and two reinforcing bars in the lowest foot of this fifteen thousand gallon tank. Ignore the welded wire bent to come up and out of the floor because it emerges at various angles. Standard welded wire is ten gauge wire on six inch squares. Ten gauge wire is 9/64 inch diameter (.14 inch) (radius = .07 inch).
πr2 = .015 square inches of steel times five wires = .077 square inches. Multiply this by 90,000 pounds per square inch = 6,930 pounds of tensile strength in the bottom foot of wall. Divide by 12 to compute the welded wire strength in an average inch of wall. 6930 ÷ 12 = 577.5 pounds of horizontal welded wire tensile strength per average vertical inch of wall.
The same calculation is done for two horizontal wraps of #4 bar (1/2 inch). πr2 multiplied by 2 multiplied by 50,000 pounds of tensile strength per square inch = 19,625 pounds of reinforcing bar tensile strength in the bottom foot of wall. Divide by 12 to find the average strength in an inch of wall. 19625 ÷ 12 = 1635.
The total wall steel strength is 1635 pounds + 577 pounds = 2,212 pounds of tensile strength. There is an additional #4 bar in the floor-to-wall key which brings the steel strength figure to 3030 pounds.
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 2180 pounds, 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; 3030 pounds at C and 3030 pounds at D. Total wall steel strength the water must break is thus 6060 pounds. Steel tensile strength divided by water force is 6060 ÷ 2180 = 2.78; the wall steel is 2.78 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 three 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 8000 psi compression strength. If a structural wall is three inches thick, points C and D would each add 24000 pounds to the 6060 pounds of steel strength. Arrow B must push with a force greater than 54,060 pounds to crush a one inch wide arc of ferrocement, at points C and D.
Wall area = 2πr(height) = 2π(9.55)(7) = 420 ft2
Roof: The roof steel extends down the wall and the roof is also an arc, add 25%.
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) = 570 ft2. Multiply this figure by the factors discussed previously. 570(1.2)(floor) + 570(1.25)(roof) = 1401 ≈ 1400 ft2 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. 420 ft2 multiplied by two = 840 ft2; multiply by 1.1 for wire overlaps and waste = 925 ft2.
The total for welded wire is 1400 ft2 for roof and floor plus 840 ft2 for the wall = 2325 ft2 of welded wire. The price of welded wire per ft2 multiplied by 2325 ft2 = 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 15 to 18 inches. Twenty feet is used further on in this book as a standard length. The method used to calculate reinforcing steel is to visualize a square with sides equal to the standard length of reinforcing steel. In this example it is a twenty foot square with an area of 400 square feet
Seventeen bars creates a spacing of 15 inches across 20 feet. This equals thirty four bars total. Divide 34 bars by 400 ft2 = 0.085 reinforcing steel bars per ft2. Add ten percent for waste and overlaps and there are 1.15 bars per ft2.
286 ft2 (roof) + 286 ft2 (floor) + 400 ft2 (wall) = 992 ft2 (total).
0.093 bars/ft2 multiplied by 992 ft2 = 92 bars of reinforcing steel at a 15 inch spacing.
This calculation at an 18 inch space between bars is 20 feet times 12 inches/foot divided by 18 inches, plus one bar = 14.33 bars. multiply this by two for the total bars = 28.66. Divide by 400 ft2 = .071 bars/ft2. Add ten percent = .079 bars/ft2. Multiply by the total area (992 ft2) and the reinforcing bars required equals 78.
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. 286(1.25) (roof) + 420(1.1) (wall) = 820 ft2.
Concrete is best estimated at 3 inch 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. 286 ft2 (floor area) multiplied by 0.25 ft (thickness) multiplied by 1.2 = 85.8 ft3 ÷ 273/cubic yard ≈ 3.2 yards.
Roof and wall is (286 ft2 + 420 ft2)(0.25 ft)(1.05) = 185.3 ft3 ÷ 273/cubic yard ≈ 7 yards.
Color pigments, extra cement water seal product, and glue (if the outside is to be colored).