Limiting Moment of Resistance and xu,max/d for RCC Beams (IS 456) with Example

Short answer: in limit state design (IS 456:2000, Cl. 38.1), the neutral axis depth of a singly reinforced beam must not exceed x_u,max = 0.53d for Fe250, 0.48d for Fe415 and 0.46d for Fe500. At that depth the beam carries its limiting moment of resistance M_u,lim = 0.36 f_ck b x_u,max (d − 0.42 x_u,max), which works out to 0.149, 0.138 and 0.133 × f_ckbd² for the three steel grades. If the factored moment is bigger than this, you deepen the beam or design it as doubly reinforced — you never accept an over-reinforced section.

Where x_u,max comes from

At failure, IS 456 takes the concrete strain at the top fibre as 0.0035. For the steel to yield before the concrete crushes, the tension steel strain must reach at least 0.87fy/Es + 0.002 (E_s = 2×10⁵ N/mm²). Similar triangles in the strain diagram give:

xu,max/d = 0.0035 / (0.0055 + 0.87fy/Es) = 700 / (1100 + 0.87fy)

A beam with x_u smaller than x_u,max is under-reinforced: the steel yields first, the beam cracks and sags visibly, and people get a warning. A beam with x_u = x_u,max is balanced. A beam with x_u greater than x_u,max is over-reinforced: the concrete crushes suddenly before the steel yields, so the code does not allow it.

Limiting values for each steel grade

The limiting steel percentage comes from equating the concrete compression 0.36f_ckb x_u,max to the steel tension 0.87f_yA_st, so pt,lim = 41.38 (fck/fy)(xu,max/d). (The table uses the rounded x_u,max/d values; SP-16 uses the unrounded ratio, so its figures can differ by about 0.01%.) Notice that a higher steel grade gives a smaller x_u,max/d and a slightly smaller M_u,lim for the same section, because the steel needs more strain to yield.

The three formulas you actually use

Worked example

A simply supported beam is 230 mm wide with an effective depth of 410 mm. Concrete is M20 and steel is Fe500. The factored bending moment is 85 kN·m. Check the section and find the steel.

1. Limiting neutral axis depth: x_u,max = 0.46 × 410 = 188.6 mm.
2. Limiting moment: M_u,lim = 0.36 × 20 × 230 × 188.6 × (410 − 0.42 × 188.6) = 103.3 × 10⁶ N·mm = 103.3 kN·m (the quick formula 0.133f_ckbd² gives 102.8 kN·m).
3. Since 85 < 103.3 kN·m, a singly reinforced, under-reinforced section works.
4. M_u/bd² = 85 × 10⁶ / (230 × 410²) = 2.20 N/mm².
5. A_st = (0.5 × 20/500) [1 − √(1 − 4.6 × 2.20/20)] × 230 × 410 = about 560 mm².
6. Provide 3 bars of 16 mm: A_st = 603 mm² (p_t = 0.64%, below p_t,lim = 0.76% and above the minimum 0.85bd/f_y = 160 mm²).
7. Check: x_u = 0.87 × 500 × 603 / (0.36 × 20 × 230) = 158.4 mm < 188.6 mm, so the section is under-reinforced. M_R = 0.87 × 500 × 603 × (410 − 0.42 × 158.4) = 90.1 kN·m ≥ 85 kN·m. OK.

If the same beam had to carry 120 kN·m, it would exceed M_u,lim = 103.3 kN·m. Adding more tension steel would push x_u past 188.6 mm, so you would either increase the depth or add compression steel for the extra 16.7 kN·m (doubly reinforced design). With Fe415 instead of Fe500, M_u,lim for this section rises to 106.7 kN·m.

Common mistakes

• Using the overall depth D instead of the effective depth d (centre of tension steel to top fibre).
• Putting unfactored moments into limit state formulas — multiply service moments by 1.5 first.
• Adding bars “to be safe” until x_u exceeds x_u,max; extra steel beyond p_t,lim does not add usable strength in a singly reinforced beam.