It may be demonstrated by theoretical mechanics, that if a load is jointly supported by two kinds of material with dissimilar elasticities, the proportion of the loading borne by each will be in a ratio depending on their relative areas and moduli of elasticity. The formula for this may be developed as follows:

C = Total unit-compression upon concrete and steel in pounds per square inch = Total load divided by the combined area of the concrete and the steel; c = Unit-compression in the concrete, in pounds per square inch; s = Unit-compression in the steel, in pounds per square inch; p = Ratio of area of steel to total area of column;

Es/ Ec = Ratio of the moudli of elasticity; r =

εs = Deformation per unit of length in the steel;

ε = Deformation per unit of length in the concrete: As = Area of steel; Ac = Area of concrete.

The total compressive force in the concrete = Ac X c; and that in the steel = As X s.

The sum of these compressions = the total compression; and therefore,

C (Ac + As) = Ac c + As s.

The actual linear compression of the concrete equals that of the steel; therefore,

C/ Es = S / Es

From this equation, since r = Es/ Ec, we may write the equation rc = s

Solving the above equation for C, we obtain:

C = Ac c + As s / A+A

Substituting the value of s = rc, we have:

C = c (Ac + Asr)/ Ac+ As) = c ( As + Ac - As + As r) / Ac+ As

If p = the ratio of cross-section of steel to the total cross-section of the column, we have:

P = As /Ac + As

Substituting this value of As / Ac + Asin the above equation, we may


C = c (1 - p + pr). Solving this equation for p, we obtain: p = C-c / c (r -1) ..(41)

Example 1. A column is designed to carry a load of 160,000 pounds. If the column is made 18 inches square, and the load per square inch to be carried by the concrete is limited to 400 pounds, what must be the ratio of the steel, and how much steel would be required?

Answer. A column 18 inches square has an area of 324 square inches. Dividing 160,000 by 324, we have 494 pounds per square inch as the total unit-compression upon the concrete and the steel, which is C in the above formula. Assume that the concrete is 1:3:5 concrete, and that the ratio of the moduli of elasticity (r) is therefore 12. Substituting these values in Equation 41, we have: p = 494 - 400 = .0214

40 ( 12-1)

Multiplying this ratio by the total area of the column, 324 square inches, we have 6.93 square inches of steel required in the column. This would very nearly be provided by four bars 1 1/4 inches square. Four round bars 1 1/2 inches in diameter would give an excess in area.

Either solution would be amply safe under the circumstances, provided the column was properly reinforced with bands.

Example 2. A column 16 inches square is subjected to a load of 115,000 pounds, and is reinforced by four 7/8-inch square bars besides the bands. What is the actual compressive stress in the concrete per square inch?

Answer. Dividing the total stress (115,000) by the area (256), we have the combined unit-stress C = 449 pounds per square inch. By inverting one of the equations above, we can write: c = C/1-p+rp c =

In the above case, the four 7/8-inch bars have an area of 3.06 square inches; and therefore, p = 3.06/256 = .012; r = 12.

Substituting these values in the above equation, we may write:

c = 449 = 449 = 397 pounds per square inch.

1 - .012 + (.012 x 12) 1.132

The net area of the concrete in the above problem is 252.94 square inches. Multiplying this by 397, we have the total load carried by the concrete, which is 100,117 pounds. Subtracting this from 115,000 pounds, the total load, we have 14,883 pounds as the compressive stress carried by the steel. Dividing this by 3.06, the area of the steel, we have 4,864 pounds as the unit compressive stress in the steel. This is practically twelve times the unit-compression in the concrete, which is an illustration of the fact that if the compression is shared by the two materials in the ratio of their moduli of elasticity, the unit-stresses in the materials will be in the same ratio. This unit-stress in the steel is about one-third of the working stress which may properly be placed on the steel. It shows that we cannot economically use the steel in order to reduce the area of the concrete, and that the chief object in using steel in the columns is in order to protect the columns against buckling, and also to increase their strength by the use of bands.

It sometimes happens that in a building designed to be structurally of reinforced concrete, the column loads in the columns of the lower story may be so very great that concrete columns of sufficient size would take up more space than it is desirable to spare for such a purpose. For example, it might be required to support a load of 320,000 pounds on a column 18 inches square. If the concrete (1:3:5) is limited to a compressive stress of 400 pounds per square inch, we may solve for the area of steel required, precisely as was done in example 1. We should find that the required percentage of steel was 13.4 per cent, and that the required area of the steel was therefore 43.3 square inches. But such an area of steel could carry the entire load of 320,000 pounds without the aid of the concrete, and would have a compressive unit-stress of only 7,400 pounds. In such a case, it would be more economical to design a steel column to carry the entire load, and then to surround the column with sufficient concrete to fireproof it thoroughly. Since the stress in the steel and the concrete are divided in proportion to their relative moduli of elasticity, which is usually about 10 or 12, we cannot develop a working stress of, say, 15,000 pounds per square inch in the steel, without at the same time developing a compressive stress of 1,200 to 1,500 pounds in the concrete, which is objectionably high as a working stress.