Question

The minimum tension reinforcement required for a rectangular beam of width 250 mm and effective depth 400 mm using Fe 500 grade steel reinforcement is

• A. 120 mm$^2$
• B. 150 mm$^2$
• C. 170 mm$^2$
• D. 180 mm$^2$

Hint:

The formula for minimum tension steel, $\frac{A_{st}}{bd} = \frac{0.85}{f_y}$, is crucial to prevent brittle failure of a beam due to sudden cracking of concrete before the steel can engage.
Always remember this formula for beam design checks.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the minimum area of tension steel ($A_{st,min}$) required in a rectangular RC beam, as per IS 456.

Step 2: Key Formula or Approach:
According to IS 456:2000 (Clause 26.5.1.1a), the minimum area of tension reinforcement shall not be less than that given by the following equation:
\[ \frac{A_{st}}{bd} = \frac{0.85}{f_y} \] where:
$A_{st}$ = Minimum area of tension reinforcement
$b$ = Width of the beam
$d$ = Effective depth of the beam
$f_y$ = Characteristic strength of the reinforcement in N/mm$^2$

Step 3: Detailed Explanation:
We are given:
- Width ($b$) = 250 mm
- Effective depth ($d$) = 400 mm
- Steel grade is Fe 500, so the characteristic strength ($f_y$) = 500 N/mm$^2$.
Rearrange the formula to solve for $A_{st}$:
\[ A_{st,min} = \frac{0.85 \times b \times d}{f_y} \] Substitute the given values:
\[ A_{st,min} = \frac{0.85 \times 250 \text{ mm} \times 400 \text{ mm}}{500 \text{ N/mm}^2} \] \[ A_{st,min} = \frac{0.85 \times 100,000}{500} \] \[ A_{st,min} = \frac{85,000}{500} \] \[ A_{st,min} = \frac{850}{5} = 170 \text{ mm}^2 \]

Step 4: Final Answer:
The minimum tension reinforcement required is 170 mm$^2$.