Singly reinforced RCC beam: neutral axis and limiting moment
How to find the neutral axis depth, why IS 456 caps it, and how to design the steel for a given moment.
A singly reinforced beam has tension steel only. Concrete above the neutral axis carries compression, and the steel below carries tension.
Finding the neutral axis
Equate compression and tension:
0.36 fck b xu = 0.87 fy Ast
xu = 0.87 fy Ast0.36 fck b
Why xu is limited
If xu is too large, concrete crushes before steel yields: a sudden brittle failure. IS 456 caps xu so that steel yields first and the beam gives warning by cracking and sagging.
| Steel | xu,max / d | Mu,lim |
|---|---|---|
| Fe 250 | 0.53 | 0.149 fck b d² |
| Fe 415 | 0.48 | 0.138 fck b d² |
| Fe 500 | 0.46 | 0.133 fck b d² |
A section with xu below xu,max is under-reinforced, which is what we want.
Moment of resistance
Mu = 0.87 fy Ast d ( 1 − Ast fyb d fck )
Worked example
Design a beam 230 mm wide, effective depth 450 mm, for Mu = 120 kN·m. M20 concrete, Fe 500 steel.
Mu,lim = 0.133 × 20 × 230 × 450² = 123.9 kN·m
Since 120 kN·m is below 123.9 kN·m, a singly reinforced section works. Solving the moment equation for Ast:
Ast = 0.5 fckfy ( 1 − √(1 − 4.6 Mufck b d²) ) b d
Ast = 0.5 × 20500 ( 1 − √(1 − 4.6 × 120 × 10⁶20 × 230 × 450²) ) × 230 × 450 = 749 mm²
Provide 2 bars of 20 mm and 1 bar of 16 mm (829 mm²).
Checks you should not skip
- Minimum steel: Ast/(b d) ≥ 0.85/fy.
- Maximum steel: 4% of b D.
- Shear, development length, and deflection.
If Mu is more than Mu,lim, increase the depth or design a doubly reinforced section.
Quick check
For Fe 415 steel, the limiting moment of resistance of a singly reinforced rectangular section is: