RC Beam Design Step-by-Step: Flexure, Shear, and Deflection
A complete walkthrough of reinforced concrete rectangular beam design per ACI 318-25, covering preliminary sizing, flexural reinforcement, shear stirrup design, minimum steel requirements, and deflection control — illustrated with a fully worked numerical example.
2. Factored Loads and Design Moment
Using ASCE 7-22 / ACI 318-25 load combinations:
For continuous beams, use ACI moment coefficients (Table 6.5.2) or an elastic frame analysis. Moment redistribution of up to 20% is permitted at supports (ACI §6.6.5).
5. Shear Design — Stirrups
ACI 318-25 §22.5 (detailed method):
Rearranging for required stirrup spacing: s = Avfytd / Vs
| Zone | Condition | Max spacing |
|---|---|---|
| No stirrups required | Vu ≤ 0.5φVc | — |
| Min stirrups only | 0.5φVc < Vu ≤ φVc | d/2 ≤ 600 mm |
| Standard shear zone | Vu ≤ φ(Vc+4Vc) | d/2 ≤ 600 mm |
| High shear zone | Vs > 4Vc | d/4 ≤ 300 mm |
Minimum stirrup area: Av,min/s = max(0.062√f'c/fyt, 0.35/fyt) × bw
Critical section for shear: located at d from the face of support (not at the support itself). The shear force at d from the support governs stirrup design in most cases.
6. Deflection Control
If h ≥ hmin from Table 9.3.1.1, no deflection calculation is needed. Otherwise, compute:
Long-term deflection multiplier per ACI §24.2.4:
| Span Condition | Deflection Limit (ACI Table 24.2.2) |
|---|---|
| Immediate live load, flat roofs | L/180 |
| Immediate live load, floors | L/360 |
| Total (creep+shrinkage+LL), non-structural elements attached | L/480 |
| Total (creep+shrinkage+LL), no non-structural elements | L/240 |
7. Worked Example — 6 m Simply-Supported Beam
Given: L=6.0 m, Dead load D=20 kN/m (includes self-weight), Live load L=25 kN/m. f'c=28 MPa, fy=420 MPa. Try b=300 mm, h=550 mm (hmin=6000/16=375 mm ✓).
Step 1 — Factored loads:
wu = 1.2×20 + 1.6×25 = 24 + 40 = 64 kN/m
Mu = 64×6²/8 = 288 kN·m | Vu,face = 64×6/2 = 192 kN
Step 2 — Effective depth: d = 550 − 40 − 10 − 12 = 488 mm (Ø24 main bar)
Step 3 — Flexural design:
a = 488 − √(488² − 2×288×10⁶/(0.90×0.85×28×300)) = 488 − √(238,144 − 90,240) = 488 − √147,904 = 488 − 384.6 = 103.4 mm
As = 288×10⁶ / (0.90×420×(488−51.7)) = 288×10⁶ / (0.90×420×436.3) = 288×10⁶/164,890 = 1,746 mm²
Use 4Ø25 (As,prov=1,963 mm²) arranged in one row. Check spacing: (300−2×40−2×10−4×25)/(3) = 46.7 mm > 25 mm ✓
Step 4 — Minimum steel: As,min = 0.00333×300×488 = 488 mm² < 1,746 mm² ✓
Step 5 — Shear at d from support: Vu,d = 192 − 64×0.488 = 192 − 31.2 = 160.8 kN
ρw = 1963/(300×488) = 0.01342
Vc = [8×1.0×(0.01342)1/3×(28)1/3] × 300×488/6×10⁻³ = [8×0.2389×3.037]×24,400/6×10⁻³ = 5.809×4,067×10⁻³ = 128.4 kN
φVc = 0.75×128.4 = 96.3 kN < Vu,d=160.8 kN → stirrups required
Vs = (160.8−96.3)/0.75 = 86.0 kN
Use Ø10 double-leg stirrups (Av=2×78.5=157 mm²): s = 157×420×488/(86,000) = 374 mm < d/2=244 mm → use s=200 mm
Step 6 — Deflection: h=550 > hmin=375 mm → no calculation required per ACI Table 9.3.1.1 ✓