ESTIMATING CONCRETE.
In estimating the amount of concrete in a given piece of work and the quantities of materials required, the unit of measurement is usually the cubic yard (27 cubic feet). The following examples will explain best the method of determining the quantities required:
Example 1.—A wall 9 inches thick, 12 feet high, and 30 feet long has a door opening 3 feet wide and 6 feet high, also a footing 18 inches wide and 9 inches deep. The concrete is to be mixed in the proportions of 1:2:4.
The volume of the footing is found by multiplying together the dimensions expressed in feet, thus, 11/2 ×3/4 × 30 = 333/4 cubic feet. Similarly, the volume in the wall is3/4 × 12 × 30, less the door opening3/4 × 3 × 6 = 2561/2 cubic feet.
The total volume in footing and wall is 2901/4 cubic feet = 103/4 cubic yards.
To find the quantity of cement, sand, and gravel, multiply the amounts for 1 cubic yard, indicated in line 7 of Table 1, by 103/4, and it will be found that 65 sacks of cement, 4.83 cubic yards of sand, and 9.56 cubic yards of gravel are necessary to build the wall.
Example 2.—A pavement 27 feet long, 4 feet wide, and 6 inches thick has a 5-inch base mixed in the proportions of 1:3:5 and a 1-inch surface mixed in the proportions of 1:2.
The volume in the base is 27 × 4 ×5/12 = 45 cubic feet = 12/3 cubic yards.
The volume in the top is 27 × 4 ×1/12 = 9 cubic feet =1/3 cubic yard.
Multiplying the quantities in line 9 of Table 1 by 12/3 and those in line 2 by it is found that the base requires 7.74 sacks cement; 0.86 cubic yard sand; 1.43 cubic yards gravel; and the top requires 4.49 sacks cement; 0.33 cubic yard sand.
Example 3.[1]—A tank 9 feet inside diameter has walls 6 inches thick and 4 feet high (above the floor). The floor is 6 inches thick, the concrete is to be 1:2:4.
[1] A practical rule in finding the area of a circle is to multiply one-half the diameter (radius) by itself and the product by22/7. In finding the volume in the wall of a circular structure, such as a silo or tank, the area of the circle formed by the inside circumference is deducted from the area of the circle formed by the outside circumference and the remainder is multiplied by the height.
The volume in the floor is10/2 ×10/2 ×22/7 ×1/2 = 392/7 Cubic feet.
The area of the larger circle is 5 × 5 ×22/7 = 784/7 cubic feet.
The area of the smaller circle is 41/2 × 41/2 ×22/7 = 634/7 cubic feet.
The area of the wall, therefore, is 15 cubic feet and the volume is 15 × 4 = 60 cubic feet.
The total volume in the structure is 992/7 cubic feet or 32/3 cubic yards. Multiplying the quantities in line 7 of Table 1 by 32/3, it is found that the following material is needed: 22.14 sacks of cement; 1.65 cubic yards of sand; 3.27 cubic yards of gravel.
FORMS.
Forms are required to hold the concrete in place until it has attained sufficient strength to sustain itself and the initial loads to which it may be subjected. Concrete is plastic and will assume the shape of the form, thus any imperfection or impression on the face of the forms will be reproduced.
Wood is commonly used for forms, as it can be easily worked into different shapes, though various other materials sometimes are better adapted to special conditions. Cast iron, for instance, is suitable for casting small objects that are to be reproduced in quantities, such as concrete block or tile; plaster of Paris, glue, or moist sand are employed for casting ornaments or to produce a fine, smooth surface; sheet metal is suitable when the forms can be used repeatedly or for such circular structures as silos. When the sides of an excavation are not likely to cave in the earth may serve as a form.