A singly reinforced rectangular beam is 300 mm wide with an effective depth of 500 mm. It uses M20 concrete and Fe415 steel, with 3 bars of 20 mm diameter as tension reinforcement. Using IS 456 limit state design, find the depth of the neutral axis, check whether the section is under-reinforced, and calculate the ultimate moment of resistance.
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x_u = 0.87 f_y A_st / (0.36 f_ck b). For Fe415, x_u,max = 0.48d.
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Worked answer.
Step 1: Steel area.
A_st = 3 × (π/4) × 20² = 3 × 314.16 ≈ 942.5 mm².
Step 2: Neutral axis depth. Equating compression in concrete (0.36 f_ck b x_u) to tension in steel (0.87 f_y A_st):
x_u = (0.87 × 415 × 942.5) / (0.36 × 20 × 300) = 340,281 / 2160 ≈ 157.5 mm.
Step 3: Limiting neutral axis depth. For Fe415, x_u,max / d = 0.48, so x_u,max = 0.48 × 500 = 240 mm.
Since x_u = 157.5 mm < 240 mm, the section is under-reinforced. The steel yields before the concrete crushes, giving a ductile failure with warning.
Step 4: Moment of resistance. For an under-reinforced section, taking the lever arm from the IS 456 stress block:
M_u = 0.87 f_y A_st (d − 0.42 x_u)
= 340,281 × (500 − 0.42 × 157.5)
= 340,281 × (500 − 66.2)
= 340,281 × 433.8 ≈ 147.6 × 10⁶ N·mm = 147.6 kN·m.
Check 1: Compression side. M_u = 0.36 f_ck b x_u (d − 0.42 x_u) = 2160 × 157.5 × 433.8 ≈ 147.6 kN·m. The same answer, as it must be.
Check 2: Limiting moment. M_u,lim = 0.138 f_ck b d² = 0.138 × 20 × 300 × 500² = 207 kN·m. Since 147.6 < 207, the result is consistent with an under-reinforced section.
IS 456 Annex G writes this as M_u = 0.87 f_y A_st d (1 − A_st f_y / (b d f_ck)), which gives about 148.0 kN·m, practically the same.
Interpretation. The beam can resist a factored moment of about 147.6 kN·m. With a load factor of 1.5, the safe service moment is about 147.6 / 1.5 ≈ 98.4 kN·m.
Common mistake. Using 0.36 f_ck b d (with d instead of x_u), or using 0.48 for every steel grade. The limiting ratio is 0.53 for Fe250, 0.48 for Fe415 and 0.46 for Fe500.
Likely follow-up. "If the required M_u were 250 kN·m, what would you do?" It exceeds M_u,lim, so either increase the depth or design a doubly reinforced section with compression steel.
Original UniCircle question. Last reviewed 28 Oct 2026.