Question

The width of a foot bridge slab is 3 m width with central supporting beam of 250 mm $\times$ 600 mm size and 6 m length. The effective width of flange is

• A. 3.0 m
• B. 1.5 m
• C. 1.25 m
• D. 1.0 m

Hint:

The full IS 456 formula for the effective width of a T-beam flange is $b_f = \frac{l_0}{6} + b_w + 6D_f$.
However, in some exam questions, if the slab thickness ($D_f$) is not provided, test the simplified formula $b_f = \frac{l_0}{6} + b_w$.
The final effective width must always be less than or equal to the actual available width (c/c spacing of beams).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to calculate the effective width of the flange ($b_f$) for a T-beam, which acts as the central support for a footbridge slab.

Step 2: Key Formula or Approach:
According to IS 456:2000 (Clause 23.1.2), the effective width of the flange for a T-beam is taken as the

least of the following:
1. $b_f = \frac{l_0}{6} + b_w + 6D_f$
2. The actual width of the flange, which is taken as the center-to-center spacing of the beams.
Where:
$l_0$ = distance between points of zero moment. For a simply supported beam, $l_0$ is the effective span.
$b_w$ = breadth of the web (the beam width).
$D_f$ = depth of the flange (the slab thickness).

Step 3: Detailed Explanation:
Let's identify the given parameters:

• Span of the beam ($l_0$) = 6 m = 6000 mm.

• Width of the web ($b_w$) = 250 mm.

• The total width of the slab is 3 m. Since the beam is central, this can be taken as the center-to-center spacing of the beams. So, the maximum possible flange width is 3 m = 3000 mm.

• The depth of the flange ($D_f$, i.e., the slab thickness) is not given in the problem.

Without the value for $D_f$, the complete formula ($l_0/6 + b_w + 6D_f$) cannot be fully evaluated. This suggests that either the problem expects a simplified approach or is based on a different formula sometimes used in textbooks where the $6D_f$ term is ignored for simplicity. Let's test this simplified approach:
\[ b_f = \frac{l_0}{6} + b_w \] Substitute the values:
\[ b_f = \frac{6000 \text{ mm}}{6} + 250 \text{ mm} \] \[ b_f = 1000 \text{ mm} + 250 \text{ mm} = 1250 \text{ mm} \] Converting to meters:
\[ b_f = 1.25 \text{ m} \] Now we compare this calculated value with the other limit (the actual width available):
\[ b_f = \min(1.25 \text{ m}, 3.0 \text{ m}) = 1.25 \text{ m} \] This calculated value exactly matches option (C). This confirms that the intended solution was based on this simplified formula.

Step 4: Final Answer:
The effective width of the flange is 1.25 m.