The previous calculations have been made as if the percentage of the steel might be varied almost indefinitely. While there is considerable freedom of choice, there are limitations beyond which it is useless to pass; and there is always a most economical percentage, depending on the conditions. We have already determined that:

εc/ εs = k/1 - k.

But εc = c / Ec (1 - 1/2q); (see Equation 10), and

εs = s / Es therefore,

εc/ εs = c Es / sEc (1 - 1/2q) = cr / s (1-1/2q) = k / 1-k

Solving for K, we have: k =cr / cr + s (1 -1/2q)

Using as before the value of q = 2/3, the equation becomes k = cr / cr + .667s. Using the same value of q in Equation 15, and solving for p, we have: pr = 7 k2 / 18 (1 -k) Substituting the above value of k in this equation, we have, after considerable reduction:

P = 7/12 C/S CR / (cr +.667s). ..(18)

The above equation shows that we cannot select the percentage of steel at random, since it evidently depends on the selected stresses for the steel and concrete, and also on the ratio of their moduli. For example, consider a high-grade concrete (1:2:4) whose modulus of elasticity is considered to be 2,900,000, and which has a limiting compressive stress of 2,700 pounds (c'), which we may consider in conjunction with the limiting stress of 55,000 pounds in the steel. The values of c, s, and r are therefore 2,700, 55,000, and 10 respectively. Substituting these values in Equation 18, we compute p = .012.

Example. What percentage of steel would be required for ordinary stone concrete, with r = 15, c =2,000, and s = 55,000? Ans. 0.95 per cent.