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

[amp_mcq option1=”14 cm2″ option2=”15 cm2″ option3=”16 cm2″ option4=”17 cm2″ correct=”option1″]

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.