An RCC beam having 200 mm wide and 400 mm effective depth is reinforced with Fe415 grade steel of area 1000 mm² in tension. Assume M20 grade concrete, what is approximate value of the depth of neutral axis for the equilibrium of forces at the limit state of collapse?
- A200 mm
- B250 mmCorrect
- C300 mm
- D350 mm
Solution
Concept
At collapse, concrete compression 0.36 fck b xu equals steel tension 0.87 fy Ast.
Formula
xu = 0.87 fy Ast / (0.36 fck b)
Step-by-step solution
1. Tension force T = 0.87 × 415 × 1000 = 361,050 N. 2. Compression force C = 0.36 × 20 × 200 × xu = 1440 xu. 3. Put C = T: 1440 xu = 361,050. 4. xu = 250.7 mm ≈ 250 mm.
Explanation
For equilibrium, total compression in concrete equals total tension in steel. Solving 0.36 fck b xu = 0.87 fy Ast gives xu ≈ 250 mm. Option B is correct.
Why the other options are wrong
A: 200 mm is too small for equilibrium. C and D: 300 mm and 350 mm give more compression than the steel tension.
Shortcut
xu = 361 × 1000 / 1440 ≈ 250 mm.
Common mistake
Using 0.45 fck instead of 0.36 fck, or forgetting the 0.87 factor on fy.
Correct answer: B) 250 mm