Which of the given option provides the effective width of an isolated T-beam of RCC as per IS 456-2000? Assume - bf = effective width of flange, l₀ = distance between points of zero moments in the beam, bw = breadth of the web, Df = thickness of flange and b = actual width of the flange.
- A[0.5 l₀ / ((l₀ /
- B+ 4)] + bw B. (l₀ / 12) + bw + 3 Df
- C(l₀ / 6) + bw + 6 Df
- D[l₀ / ((l₀ / b) + 4)] + bwCorrect
Solution
Concept
IS 456 gives a special formula for the effective flange width of an isolated T-beam, and it must not exceed the actual flange width b.
Formula
bf = [l₀ / ((l₀ / b) + 4)] + bw ≤ b
Step-by-step solution
1. IS 456:2000 gives separate effective flange width rules for monolithic and isolated beams. 2. Isolated T-beam: bf = [l₀ / ((l₀ / b) + 4)] + bw. 3. Isolated L-beam: bf = [0.5 l₀ / ((l₀ / b) + 4)] + bw. 4. Monolithic T-beam: bf = l₀/6 + bw + 6 Df. Monolithic L-beam: bf = l₀/12 + bw + 3 Df. 5. In all cases bf must not be more than the actual width b. 6. So option D is correct.
Explanation
For an isolated T-beam the flange is not part of a continuous slab, so IS 456 uses the formula bf = l₀/((l₀/b)+4) + bw. This value must not be more than the actual flange width b. Option D matches this formula.
Why the other options are wrong
A: 0.5 l₀/((l₀/b)+4) + bw is for an isolated L-beam. B: l₀/12 + b
Correct answer: D) [l₀ / ((l₀ / b) + 4)] + b_w