τv is the nominal shear stress in the beam = Vu / (b × d). τc is the design shear strength of concrete (IS 456 Table 19) depending on percentage steel and concrete grade. If τv τc, design stirrups to carry the excess (Vus = Vu − τc × b × d). If τv > τc,max (Cl. 40.2.3), section fails — increase size.

Beam Design Calculator — Singly Reinforced RCC per IS 456:2000

Mark	Dia	Shape	A	B	L (mm)	No.	Wt (kg)
B1	16φ		5000	240	5540	3	26.23
B2	12φ		6504	-	6504	2	11.55
S1	8φ		170	390	1280	13	6.57
S2	8φ		170	390	1280	20	10.10
TOTAL	38	54.45 kg

How beam design works — IS 456:2000 step by step

Beams transfer load from slabs to columns. In a typical 3-storey residential building, 30-60% of the total RCC volume is in beams, so getting beam design right has outsized economic impact. The calculator above runs the singly-reinforced limit-state design per IS 456 and flags when doubly reinforced or a deeper section is required.

You enter span, width, depth, factored loading (either direct UDL or explicit Mu + Vu), and — from the Design Context — concrete grade, steel grade, and cover. The output is the tension steel area Ast, main bar selection (number of bars × diameter), stirrup size and spacing, Mu,lim check, and shear stress check against IS 456 Cl. 40.

IS 456 clauses applied by the calculator:

Effective depth d = D − cover − bar diameter/2 (Cl. 26.4).
Limiting moment Mu,lim = 0.36 × fck × b × xu,max × (d − 0.42 × xu,max) (Annex G-1.1). xu,max per Cl. 38.1 depends on fy.
If Mu > Mu,lim → section fails singly reinforced, need doubly reinforced or deeper section.
Tension steel area Ast from Mu = 0.87 × fy × Ast × (d − Ast × fy / (0.36 × fck × b)). Solved iteratively.
Number of bars: ceil(Ast / area of selected bar diameter). Typical 16 mm, 20 mm, 25 mm for main beams.
Shear stress τv = Vu / (b × d). Compare against τc (Cl. 40.2, Table 19 — depends on % steel and fck) and τc,max (Cl. 40.2.3).
If τv ≤ τc → minimum stirrups per Cl. 26.5.1.6: Asv/b·Sv ≥ 0.4/(0.87·fy).
If τv > τc → design stirrups to carry Vus = Vu − τc × b × d per Cl. 40.4. Max spacing 0.75d or 300 mm (Cl. 26.5.1.5).
Minimum tension steel 0.2% (Cl. 26.5.1.1); maximum 4%.

Worked example — 5 m simply-supported beam

Example inputs

Span = 5000 mm, b = 230 mm, D = 450 mm, factored UDL = 30 kN/m, M25 concrete, Fe 500 steel, 25 mm cover. Output: effective depth d = 450 − 25 − 12.5/2 ≈ 418 mm. Factored moment Mu = 30 × 5² / 8 = 93.75 kN·m. Mu,lim for this section = 0.36 × 25 × 230 × (0.46 × 418) × (418 − 0.42 × 0.46 × 418) ≈ 122 kN·m > Mu ✓ singly reinforced OK. Ast ≈ 620 mm² → 2 × 20 mm + 1 × 16 mm = 829 mm² provided. Shear Vu = 30 × 5 / 2 = 75 kN. τv = 75000 / (230 × 418) = 0.78 MPa; τc for M25 at 0.86% steel ≈ 0.62 MPa. τv > τc → design stirrups: 8 mm @ 180 mm c/c (2-legged). Utilization 77%. Section sufficient; consider 300 mm depth for economy if span is shorter.

Common beam-design mistakes

Under-sizing depth. For spans > 5 m, beam depth below span/14 usually fails. Rule of thumb: depth ≈ span/12 to span/15 for typical residential beams.
Forgetting maximum stirrup spacing. Per Cl. 26.5.1.5, stirrup spacing cannot exceed 0.75d or 300 mm even if τv is low. Calculator enforces this.
Missing compression steel when Mu > Mu,lim. Calculator flags 'Doubly (increase depth)' — either increase D or add compression steel per Annex G-1.2.
Not providing minimum tension steel (0.2% of b·d) even at low moments. Anchoring + ductility requirement, not strength.
Treating continuous beams as simply supported. A 5 m continuous beam has mid-span Mu ≈ Wu·L²/12 (not /8) and support Mu ≈ Wu·L²/12 with opposite sign — support moment often governs steel size.
Using Fe 500 in zones III-V without toggling IS 13920. Seismic beams require Fe 500D; enable the toggle in Design Context.

Beam design FAQs

What is Mu,lim and why does the calculator flag 'doubly reinforced'?

Mu,lim is the limiting moment of resistance for a singly reinforced beam per IS 456 Annex G-1.1: Mu,lim = 0.36 × fck × b × xu,max × (d − 0.42 × xu,max). If factored moment Mu exceeds Mu,lim, singly reinforced fails and you need compression steel (doubly reinforced) OR a deeper beam.

How is stirrup spacing calculated?

Per IS 456 Cl. 40.4: Sv = 0.87 × fy × Asv × d / Vus, where Vus = Vu − τc × b × d. Maximum spacing limited to 0.75d or 300 mm (Cl. 26.5.1.5). Minimum shear reinforcement per Cl. 26.5.1.6. Calculator applies all three checks and reports the governing spacing.

Which beam width and depth should I start with?

Typical: width 200-230 mm residential, 230-300 mm commercial, 300-400 mm industrial. Overall depth 300-450 mm for span 3-5 m; 450-600 mm for 5-7 m; 600-900 mm for > 7 m. Calculator flags 'fails' if section can't carry the load; increase depth before width.

What is τv vs τc?

τv is the nominal shear stress = Vu / (b × d). τc is the design shear strength of concrete (IS 456 Table 19) depending on % steel and concrete grade. If τv <= τc, minimum stirrups only. If τv > τc, design stirrups for the excess. If τv > τc,max (Cl. 40.2.3), section fails — increase size.

Does the calculator handle continuous beams?

Currently simply supported only — you enter the factored maximum moment at mid-span. For continuous beams, compute support and mid-span moments using IS 456 Table 12 coefficients or analysis software, then enter the governing Mu manually. Continuous beam direct support is on the Tier 1 roadmap.

Should I use Fe 500 or Fe 500D?

Gravity-only beam in Zone II: Fe 500 is adequate and cheaper. Beam in Zone III+ or part of an SMRF (special moment resisting frame): Fe 500D is mandatory per IS 13920:2016 Cl. 5.2. Enable the 'IS 13920 seismic detailing' toggle in the Design Context — it auto-upgrades.