Beam Design

Flexural design, flexural capacity, and torsion design for rectangular RC beams.

Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
MPa
MPa
Compression Reinforcement (optional)
Resistance Factor
—
ACI 318-25 §21.2.2: φ = 0.90 for tension-controlled sections (εt ≥ 0.005)
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kN·m
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Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
MPa
MPa
Partial Safety Factors
Recommended: 1.50 (persistent)
Recommended: 1.15
Compression Reinforcement (optional)
Loading
kN·m
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Input Parameters

kN-m · m

Section Geometry

m
m
m
mm
Material Properties
Loading
kN·m
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Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
MPa
MPa
Resistance Factor
—
Tension Reinforcement
mm
bars
Compression Reinforcement (optional)
mm
bars
Loading (optional — for D/C)
kN·m
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Enter values and click Calculate.

Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
MPa
MPa
Partial Safety Factors
Tension Reinforcement
mm
bars
Compression Reinforcement (optional)
mm
bars
Loading (optional — for D/C)
kN·m
📋
Enter values and click Calculate.

Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
Tension Reinforcement
mm
bars
Loading (optional — for D/C)
kN·m
📋
Enter values and click Calculate.

Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
MPa
MPa
MPa
—
—
ACI 318-25: φ = 0.75 for shear/torsion
Design Actions
kN
kN·m
Transverse Reinforcement (Stirrups)
mm
m
Longitudinal Torsion Bars
mm
bars
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Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
MPa
MPa
Design Actions
kN
kN·m
Transverse Reinforcement (Stirrups)
mm
m
Longitudinal Torsion Bars
mm
bars
ℹ️
Equilibrium vs Compatibility Torsion (EC2 §6.3.1): If this is equilibrium torsion (statically determinate — cannot be redistributed), full design is required. If this is compatibility torsion (statically indeterminate — can be redistributed), EC2 §6.3.1 permits neglecting torsion in ULS design provided cracking is acceptable and minimum reinforcement is provided. Do not neglect equilibrium torsion.
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Input Parameters

kN-m · m

Section Geometry

m
m
m
mm
Material Properties
Design Actions
kN
kN·m
kN·m
Required for equivalent moment Me per IS 456 Cl 41.3
Transverse Reinforcement (Stirrups)
mm
m
Longitudinal Torsion Bars
mm
bars
ℹ️
IS 456:2000 Cl 41 — Equivalent Shear/Moment Method: Combined torsion, bending, and shear are handled by designing for equivalent shear Ve = Vu + 1.6Tu/b and equivalent moments Me1 (bottom), Me2 (top). Stirrups are designed for the equivalent nominal shear stress τve.
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Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
MPa
MPa
—
—
φ = 0.75 (ACI 318-25 §21.2.1). λ = 1.0 NW · 0.85 sand-LW · 0.75 all-LW (§19.2.4).
Longitudinal Steel
mm²
ρw = Asl/(bw·d) · ACI 318-25 §22.5.5.1
Stirrup Layout
mm
m
Applied Load
kN
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Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
MPa
MPa
Recommended: 1.50
Recommended: 1.15
Longitudinal Steel
mm²
ρl = Asl/(bw·d) · EN 1992-1-1 §6.2.2
Stirrup Layout
mm
m
Applied Load
kN
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Configure inputs and click Calculate.

Input Parameters

kN-m · m

Section Geometry

m
m
m
Material Properties
Longitudinal Steel
mm²
Used to compute pt = 100·Ast/(b·d) for τc (Table 19)
Stirrup Layout
mm
m
Applied Load
kN
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Configure inputs and click Calculate.

📚 Design Background & Code References

ACI 318-25 §22.5 Shear Strength

The nominal shear strength of a section is Vn = Vc + Vs. The design requirement is φVn ≥ Vu, where φ = 0.75 for shear.

Concrete Contribution — §22.5.5.1 (SI)
Vc = 0.66 λ ρw1/3 √f'c · bw · d   [N, MPa, mm]
ρw = Asl / (bw · d)   (longitudinal tension reinforcement ratio)
Stirrup Contribution — §22.5.8.5
Vs = Av · fyt · d / s   (vertical stirrups)
Vs,max = 0.66 √f'c · bw · d   (§22.5.1.2 — if exceeded, increase section)
Maximum Spacing — §9.7.6.2.2
If Vs ≤ 0.33√f'c · bw · d:   smax = min(d/2, 600 mm)
If Vs > 0.33√f'c · bw · d:   smax = min(d/4, 300 mm)
Minimum Shear Reinforcement — §26.6.3.1
Av/s ≥ max(0.062√f'c · bw/fyt,   0.35 · bw/fyt)

EN 1992-1-1 §6.2 Shear Resistance

Members without shear reinforcement rely on concrete tensile resistance (VRd,c). When VEd > VRd,c, vertical stirrups must carry the full design shear via the variable strut inclination method.

Concrete Resistance — §6.2.2
VRd,c = [CRd,c · k · (100ρlfck)1/3] · bw · d
CRd,c = 0.18/γc  ·   k = 1 + √(200/d) ≤ 2.0  ·   ρl = Asl/(bw·d) ≤ 0.02
VRd,c,min = 0.035 · k3/2 · √fck · bw · d
Stirrup Resistance — §6.2.3 (θ = 45°)
VRd,s = (Asw/s) · z · fywd   z = 0.9d, fywd = fywk/γs
VRd,max = 0.5 · bw · z · ν · fcd   (ν = 0.6(1−fck/250), fcd = fck/γc)
Maximum Spacing — §9.2.2
sl,max = 0.75 · d   (vertical stirrups, θ = 45°)
ρw,min = 0.08 √fck / fywk

IS 456:2000 Cl. 40 Shear Design

The nominal shear stress τv is checked against the design shear strength τc (Table 19) and maximum shear stress τc,max (Table 20). Stirrups carry the shear in excess of τc.

Nominal Shear Stress — Cl. 40.1
τv = Vu / (b · d)   [N/mm²]
Design Shear Strength — Table 19
τc depends on pt = 100·Ast/(b·d) and fck
τc,max from Table 20: M20 → 2.8, M25 → 3.1, M30 → 3.5, M35 → 3.7, M40+ → 4.0 N/mm²
Stirrup Contribution — Cl. 40.4
Vus = 0.87·fy·Asv·d / sv
Required Asv/sv = (Vu − τc·b·d) / (0.87·fy·d)
Minimum Shear Reinforcement — Cl. 26.5.1.6
Asv / (b·sv) ≥ 0.4 / (0.87·fy)
Max spacing: sv,max = min(0.75d, 300 mm) — Cl. 26.5.1.5

📚 Design Background & Code References

Flexural Behaviour of Reinforced Concrete Beams

A reinforced concrete beam resists bending through an internal force couple: compression carried by concrete in the compression zone, and tension carried by the steel reinforcement. At the ultimate limit state (ULS), the concrete is assumed to have reached its limiting compressive strain while the tension steel has yielded.

Both ACI 318 and Eurocode 2 are based on the Bernoulli-Euler hypothesis — plane sections remain plane after bending — with full strain compatibility between concrete and steel.

Key Assumptions at ULS

  • Tensile strength of concrete is neglected — only steel resists tension.
  • Concrete reaches its limiting compressive strain: εcu = 0.003 (ACI) or 0.0035 (EC2 for fck ≤ 50 MPa).
  • Stress distribution idealised as a rectangular (Whitney) stress block.
  • Steel has yielded: σs = fy (ACI) or fyd = fyk/γs (EC2).
  • Section is singly-reinforced — compression steel neglected.

Effective Depth

Effective Depth
d = h − cclear − dbar/2
h = total depth · c = clear cover to main bar face · dbar = main bar diameter (assumed 20 mm / 0.75 in). Stirrup thickness is conservatively included in the clear cover input.

ACI 318-25 Strength Design Method

ACI 318-25 uses Load and Resistance Factor Design (LRFD). The design requirement is φMn ≥ Mu, where φ is the strength reduction factor and Mn is the nominal moment capacity.

Strength Reduction Factor φ — ACI 318-25 Table 21.2.2
φ = 0.90    tension-controlled (εt ≥ 0.005)
φ = 0.65 → 0.90    transition zone (0.002 < εt < 0.005)
φ = 0.65    compression-controlled (εt ≤ 0.002)
For beams under pure flexure, φ = 0.90 applies when ρ ≤ 0.75ρbal.

ACI 318-25 §22.2 Stress Block & Reinforcement

Stress Block Factor β₁
β₁ = 0.85   (f'c ≤ 28 MPa)    β₁ ≥ 0.65 (minimum)
β₁ = 0.85 − 0.05·(f'c − 28)/7   (f'c > 28 MPa)
Required Steel Area
Rn = Mu/(φ·b·d²)
ρ = (0.85f'c/fy)·[1−√(1−2Rn/0.85f'c)]
As = ρ·b·d

ACI 318-25 §9.6.1.2 Minimum & Maximum Reinforcement

Reinforcement Limits
ρmin = max(0.25√f'c/fy , 1.4/fy)   [MPa]
ρmax = 0.75·ρbal   (ensures εt ≥ 0.005)

ACI 318-25 Material Limits

ParameterMinimumMaximumReference
Concrete f'c17 MPa (2500 psi)70 MPa (10000 psi)*§19.2.1
Rebar fy280 MPa (40 ksi)550 MPa (80 ksi)§20.2.2
Strength factor φ0.650.90Table 21.2.2

* Higher f'c permitted with special provisions per §19.2.1.3

EN 1992-1-1 Partial Factor Method

Design Material Strengths
fcd = αcc·fck/γc    [αcc=1.0 recommended; γc=1.5]
fyd = fyk/γs              [γs=1.15]
γc and γs are Nationally Defined Parameters (NDPs) and may vary by country.

EN 1992-1-1 §6.1 Flexural Design

Design Procedure
K = MEd/(b·d²·fcd)
K' = 0.167   [limiting K, δ=1.0, no redistribution]
z = d·[0.5+√(0.25−K/1.134)] ≤ 0.95d
As = MEd/(fyd·z)

EN 1992-1-1 Material Limits

ParameterMinimumMaximumReference
Concrete fck12 MPa (C12/15)90 MPa (C90/105)Table 3.1
Rebar fyk400 MPa600 MPa§3.2.2
Partial factor γc1.0 (accidental)1.5 (persistent)Table 2.1N
Partial factor γs1.0 (accidental)1.15 (persistent)Table 2.1N

IS 456:2000 Code Overview

IS 456:2000 is the Indian Standard for Plain and Reinforced Concrete — Code of Practice. It covers the design of concrete structures for flexure, shear, torsion, and serviceability. Limit state design (Cl. 18) is the primary design method, targeting both the Limit State of Collapse and the Limit State of Serviceability.

IS 456 Cl. 38 Flexural Design (Limit State)

The limiting neutral axis depth ratio xu,lim/d depends on the steel grade (Fe 250 → 0.53, Fe 415 → 0.48, Fe 500 → 0.46, Fe 550 → 0.44). The design moment capacity is:

Singly Reinforced Beam — Cl. 38.1
Mu = 0.87 fy Ast d [1 − (Ast fy) / (b d fck)]
Mu,lim = 0.36 fck b xu,lim (d − 0.42 xu,lim)
Minimum / Maximum Steel — Cl. 26.5.1.1
Ast,min = 0.85 b d / fy
Ast,max = 0.04 b D   (D = total depth)

IS 456 Cl. 40 Shear Design

Nominal shear stress τv = Vu/(b·d) is compared with design shear strength τc (Table 19, function of pt and fck) and maximum τc,max (Table 20). Vertical stirrups carry excess shear.

Stirrup Capacity — Cl. 40.4 (b)
Vus = 0.87 fy Asv d / sv
sv,max = min(0.75d, 300 mm) — Cl. 26.5.1.5

IS 456 Cl. 41 Torsion Design

IS 456 converts torsion to equivalent shear (Ve) and equivalent bending moment (Mt) using the equivalent moment method. Both are checked against the section capacity.

Equivalent Shear — Cl. 41.3.1
Ve = Vu + 1.6 Tu / b
τve = Ve / (b · d) ≤ τc,max
Equivalent Moment — Cl. 41.4.2
Mt = Tu (1 + D/b) / 1.7
Design for Mu1 = Mu + Mt (tension side) and Mu2 = Mt − Mu (if positive, compression side)

IS 456 Table 2 Concrete Grades

Standard grades: M15, M20, M25, M30, M35, M40, M45, M50, M55. The number denotes the characteristic compressive strength fck (cylinder) in MPa at 28 days. M25 (25 MPa) is the minimum grade for reinforced concrete in moderate exposure.

IS 456 Table 1 Steel Grades

Fe 250 (mild steel, fy = 250 MPa), Fe 415 (HYSD, most common), Fe 500, Fe 550. Design yield stress = 0.87 fy in the limit state method.

Design Limits Comparison — ACI 318-25 vs. Eurocode 2 vs. IS 456:2000

ParameterACI 318-25Eurocode 2IS 456:2000
Concrete strain at ULSεcu = 0.003εcu = 0.0035εcu = 0.0035
Resistance / partial factorφ = 0.90 (flexure)γc=1.5 · γs=1.15γc=1.5 · γs=1.15 (implicit)
Stress block depth factorβ₁ = 0.85 → 0.65λ = 0.8 (fck ≤ 50 MPa)0.36·fck·b·xu (parabolic-rect.)
Stress block intensity0.85·f'cη·fcd, η=1.00.36·fck (equiv. rect. block)
Minimum reinforcementmax(0.25√f'c/fy, 1.4/fy)·b·dmax(0.26fctm/fyk, 0.0013)·b·d0.85·b·d / fy (Cl. 26.5.1.1a)
Maximum reinforcementρ ≤ 0.75·ρbalK ≤ K' = 0.1670.04·b·D (Cl. 26.5.1.1b)
Neutral axis limit (xu/d)Not directly (εt ≥ 0.004)x/d ≤ 0.45Fe415: 0.48 · Fe500: 0.46 · Fe250: 0.53
Maximum lever armNot explicitly limitedz ≤ 0.95dNot explicitly limited

References

  • [1]
    ACI 318-25 — Building Code Requirements for Structural Concrete and Commentary. American Concrete Institute, 2025. §9.6, §21.2.2, §22.2.
  • [2]
    EN 1992-1-1:2004 — Eurocode 2: Design of Concrete Structures. CEN, Brussels. §6.1, §9.2.1.
  • [3]
    IS 456:2000 — Plain and Reinforced Concrete — Code of Practice, 4th Rev. Bureau of Indian Standards (BIS), New Delhi. Cl. 38 (Flexure), Cl. 40 (Shear), Cl. 41 (Torsion).
  • [4]
    MacGregor, J.G. & Wight, J.K. — Reinforced Concrete: Mechanics and Design, 7th Ed. Pearson, 2016.
  • [5]
    Mosley, Bungey & Hulse — Reinforced Concrete Design to Eurocode 2, 7th Ed. Palgrave Macmillan, 2012.
  • [6]
    Whitney, C.S. — Plastic Theory of Reinforced Concrete Design. Trans. ASCE, Vol. 107, 1942.
  • [7]
    Pillai, S.U. & Menon, D. — Reinforced Concrete Design, 3rd Ed. Tata McGraw-Hill, 2009. (IS 456-based reference text)
Disclaimer: For educational and preliminary design only. All results must be verified by a licensed structural engineer. Always refer to the current edition of the applicable code.
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