A simply supported beam 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg/m including its self weight. If the lever arm factor is 0.85 and permissible tensile stress of steel is 1400 kg/cm2, the area of steel required, is A. 14 cm2 B. 15 cm2 C. 16 cm2 D. 17 cm2

The correct answer is A. 14 cm2.

The formula for calculating the area of steel required is:

$A = \frac{wL}{bd}f_y$

where:

• $A$ is the area of steel required (in cm2)
• $w$ is the uniformly distributed load (in kg/m)
• $l$ is the length of the beam (in m)
• $b$ is the effective depth of the beam (in cm)
• $d$ is the lever arm factor (in dimensionless)
• $f_y$ is the permissible tensile stress of steel (in kg/cm2)

In this case, we have:

• $w = 2400 \frac{kg}{m}$
• $l = 6 m$
• $b = 50 cm = 0.5 m$
• $d = 0.85$
• $f_y = 1400 \frac{kg}{cm^2}$

Substituting these values into the formula, we get:

$A = \frac{(2400 \frac{kg}{m})(6 m)}{(0.5 m)(0.85)(1400 \frac{kg}{cm^2})} = 14 cm^2$

Therefore, the area of steel required is 14 cm2.