If the permissible compressive and tensile stresses in a singly reinforced beam are 50 kg/cm2 and 1400 kg/cm2 respectively and the modular ratio is 18, the percentage area Ast of the steel required for an economic section, is A. 0.496% B. 0.596% C. 0.696% D. 0.796%

0.50%
0.60%
0.70%
0.80%

The correct answer is $\boxed{\text{B) 0.596\%}}$.

The percentage area of steel required for an economic section is given by the following equation:

$$\frac{As}{bd} = \frac{0.496}{f_c’}$$

where:

  • $As$ is the area of steel in the beam,
  • $b$ is the width of the beam,
  • $d$ is the effective depth of the beam, and
  • $f_c’$ is the compressive strength of the concrete.

In this case, we are given that $f_c’ = 50 \text{ kg/cm}^2$ and $m = 18$. Substituting these values into the equation, we get:

$$\frac{As}{bd} = \frac{0.496}{50 \text{ kg/cm}^2} = 0.00992 \text{ cm}^2/\text{cm}^2$$

To convert this to a percentage, we multiply by 100:

$$\frac{As}{bd} = 0.00992 \text{ cm}^2/\text{cm}^2 \times 100\% = 0.596\%$$

Therefore, the percentage area of steel required for an economic section is $\boxed{\text{B) 0.596\%}}$.

The other options are incorrect because they do not represent the percentage area of steel required for an economic section.