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)
Loading
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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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
📋
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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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Enter values and click Calculate.

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

Theory↑ Top

Shear Behaviour of Reinforced Concrete Beams

An uncracked reinforced concrete beam resists shear through a combination of concrete tensile and compressive stresses acting on inclined planes. Once diagonal (inclined) cracks form, the internal force path changes fundamentally — and the beam transitions from beam action to a more complex arch-and-truss mechanism.

Mechanisms of Shear Transfer

  • Compression zone contribution: The uncracked concrete above the neutral axis carries significant shear directly.
  • Aggregate interlock: The rough crack faces interlock and transfer shear across diagonal cracks; this is sensitive to crack width.
  • Dowel action: The longitudinal tension bars crossing a crack resist shear by bending — typically a minor contribution but not negligible.
  • Stirrup contribution: Vertical stirrups cross diagonal cracks and carry the remaining shear demand after the concrete mechanisms are exhausted.

Truss Analogy

The classical design model treats the cracked beam as a parallel-chord truss: the longitudinal steel acts as the tension chord, the concrete compression zone as the compression chord, inclined concrete struts carry compression, and stirrups act as vertical tension members. This model, due to Ritter (1899) and Mörsch (1902), underpins all three codes.

Eurocode 2 extends this to the variable-angle strut model — the strut angle θ is chosen by the designer between 21.8° and 45°, trading a shallower angle (fewer stirrups, more horizontal force in the chord) against a steeper one (more stirrups, less chord force).

Size Effect

Shear capacity per unit area decreases as beam depth increases — a phenomenon not captured by simple √f'c scaling. ACI 318-19 onward addresses this through the size effect factor λs in §22.5.5.1; EC2 captures it via the k = 1 + √(200/d) term in VRd,c.

Design Approach

General Design Requirement
φVn ≥ Vu   (ACI)    VRd ≥ VEd   (EC2)    Vu ≤ Vc + Vus   (IS 456)
All three codes use the same additive model: concrete contribution + stirrup contribution ≥ factored shear demand. When Vu exceeds the concrete capacity alone, stirrups are required to carry the excess. When Vu exceeds the maximum strut capacity, the section must be enlarged.

Why Minimum Stirrups Are Required

Even when the applied shear is below the concrete's capacity (Vu < φVc), all three codes mandate a minimum quantity of shear reinforcement in beams. This is because:

  • Diagonal cracking can occur suddenly; minimum stirrups prevent brittle failure after cracking.
  • Shear demand is often uncertain — construction loads, load redistribution, and dynamic effects can produce higher-than-designed shear.
  • Stirrups also confine the core concrete and hold the main bars in place against dowel splitting.
ACI 318-25↑ Top

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)
Eurocode 2↑ Top

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↑ Top

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
References↑ Top

References

  • [1]
    ACI 318-25 — Building Code Requirements for Structural Concrete. American Concrete Institute, 2025. §22.5 (Shear strength), §9.7.6.2 (Maximum stirrup spacing), §26.6.3 (Minimum shear reinforcement).
  • [2]
    EN 1992-1-1:2004 — Eurocode 2: Design of Concrete Structures. CEN, Brussels. §6.2 (Shear), §9.2.2 (Shear reinforcement in beams). Variable-angle strut inclination method.
  • [3]
    IS 456:2000 — Plain and Reinforced Concrete — Code of Practice, 4th Rev. Bureau of Indian Standards. Cl. 40 (Shear), Table 19 (Design shear strength τc), Table 20 (Maximum τc,max), Cl. 26.5.1 (Stirrup spacing limits).
  • [4]
    Ritter, W. — Die Bauweise Hennebique. Schweizerische Bauzeitung, Vol. 33, 1899. (Original truss analogy model for shear in RC beams.)
  • [5]
    Collins, M.P. & Mitchell, D. — Prestressed Concrete Structures. Prentice-Hall, 1991. Chapter 7 (Compression Field Theory — basis of EC2 variable-angle truss method).
  • [6]
    Wight, J.K. & MacGregor, J.G. — Reinforced Concrete: Mechanics and Design, 7th Ed. Pearson, 2016. Chapter 6 (Shear in Beams).
  • [7]
    Kani, G.N.J. — The Riddle of Shear Failure and Its Solution. ACI Journal, Vol. 61, No. 4, 1964. (Landmark study on size effect and shear span ratio influence.)
Disclaimer: For educational and preliminary design only. Verify all results with a licensed structural engineer.

📚 Design Background & Code References

Theory↑ Top

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 at every load level.

Three Stages of Beam Behaviour

A reinforced concrete beam passes through three behavioural stages as the applied moment increases from zero to failure:

  1. Uncracked elastic stage: The full concrete section — above and below the neutral axis — resists tension and compression. Both materials behave elastically. This stage ends when the extreme tension fibre stress reaches the modulus of rupture fr ≈ 0.62√f'c (ACI, MPa units). The corresponding moment is the cracking moment Mcr.
  2. Cracked elastic stage: Flexural cracks form at the tension face. Below the neutral axis, only the transformed steel area resists tension. Above the neutral axis, concrete carries compression elastically. This regime governs serviceability limit states — deflection and crack width checks.
  3. Ultimate limit state (ULS): The extreme compression fibre concrete reaches its crushing strain εcu. In a properly designed under-reinforced section, the tension steel has already yielded (εs ≫ εy), providing large deflections and visible cracking as warning before failure. This is the regime that controls strength design.

Whitney Equivalent Rectangular Stress Block

The actual parabolic-rectangular concrete stress distribution in the compression zone is replaced by an idealised rectangular stress block for calculation convenience. The block is calibrated so that its resultant force Cc equals that of the real distribution, and acts at the same centroid location.

ACI 318-25 Rectangular Block — §22.2.2
Block depth:   a = β₁ · c
Block stress:   0.85 f'c
Compression resultant:   Cc = 0.85 f'c · a · b
Tension resultant:   T = As · fy (at yield)
Moment capacity:   φMn = φ · As · fy · (d − a/2)
β₁ accounts for the shape of the actual distribution; it equals 0.85 for f'c ≤ 28 MPa and decreases linearly to a minimum of 0.65 for higher-strength concrete. EC2 uses λ = 0.8 and η = 1.0 for fck ≤ 50 MPa with analogous meaning.

Balanced Reinforcement Ratio ρbal

The balanced failure condition occurs when the extreme concrete fibre reaches εcu at the same instant that the outermost tension steel reaches its yield strain εy = fy/Es. At balanced failure, both concrete crushing and steel yielding occur simultaneously — a brittle failure with no ductile warning. All codes prohibit design at or above the balanced steel ratio.

Balanced Steel Ratio — ACI (SI)
ρbal = (0.85 β₁ f'c / fy) · [εcu / (εcu + εy)]
εcu = 0.003 ·   εy = fy / 200 000 (fy in MPa)
Example (f'c = 28 MPa, fy = 420 MPa): ρbal ≈ 0.0285
EC2 expresses the equivalent limit as x/d ≤ 0.45; IS 456 tabulates xu,max/d by steel grade. All three restrict the section to the under-reinforced side of balanced failure.

Ductility and the Tension-Controlled Requirement

Codes mandate ductile, under-reinforced failure to ensure large deflections and wide cracks occur before collapse — giving occupants warning and allowing load redistribution in indeterminate structures.

CodeDuctility RequirementImplication
ACI 318-25Net tensile strain εt ≥ 0.004 at ULS (φ = 0.90 when εt ≥ 0.005)ρ ≤ ≈ 0.75 ρbal
Eurocode 2Neutral axis x/d ≤ 0.45 (for δ ≥ 0.7 redistribution)K ≤ K' = 0.167
IS 456:2000xu/d ≤ xu,max/d (by grade: Fe 415 → 0.48)Mu ≤ Mu,lim

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↑ Top

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

ACI 318-25 §25.2 Minimum Bar Spacing

Clear Spacing Between Bars — §25.2.1
sclear ≥ max(db, 25 mm, 4/3 · aggregate size)
For bars in multiple layers: vertical clear spacing ≥ 25 mm
Adequate clear spacing allows concrete to flow around bars during placement and consolidation. Insufficient spacing leads to honeycombing and reduced bond strength.

ACI 318-25 §24.3 Crack Control — Maximum Bar Spacing

Even when flexural strength is adequate, bar spacing in the tension zone is limited to control crack widths under service loads. Wider crack widths accelerate corrosion and are visually alarming to occupants.

Maximum Spacing in Tension Zone — §24.3.2
s ≤ 380 · (280/fs) − 2.5·cc   [mm]
s ≤ 300 · (280/fs)   (interior exposure limit)
fs = 2/3 · fy   (approximate service bar stress, MPa)
cc = clear cover to flexural tension reinforcement
For fy = 420 MPa with cc = 40 mm: smax ≈ 213 mm. This is often the governing spacing limit in slabs and wide beams.

ACI 318-25 §22.2 Design Summary

The complete ACI flexural design procedure for a singly reinforced rectangular section:

  1. Compute Rn = Mu / (φ · b · d²)
  2. Solve ρ = (0.85f'c / fy) · [1 − √(1 − 2Rn / 0.85f'c)]
  3. Check ρmin ≤ ρ ≤ ρmax (= 0.75 ρbal)
  4. As = ρ · b · d
  5. Check εt ≥ 0.004; confirm φ = 0.90
  6. Verify bar spacing ≥ sclear,min and ≤ scrack,max
Eurocode 2↑ Top

EN 1992-1-1 Partial Factor Method

Eurocode 2 uses a partial factor method rather than a single capacity reduction factor φ. Safety is achieved by factoring the material strengths directly, rather than the resulting capacity. This produces design strengths fcd and fyd that are used throughout all capacity calculations.

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 be modified by National Annexes. The recommended values are γc = 1.5 (persistent/transient) and γs = 1.15. For accidental design situations both reduce to 1.0.

EN 1992-1-1 §6.1 Flexural Design

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

If K > K': compression steel required or section must be enlarged
The factor 1.134 = 2·0.567 where 0.567 = λ·η/(2) = 0.8·1.0/2·(1/0.8) — derived from the rectangular stress block parameters for fck ≤ 50 MPa. The lever arm formula is equivalent to solving the force equilibrium and moment equations directly.

EN 1992-1-1 §9.2.1.1 Minimum Reinforcement

Minimum As (§9.2.1.1)
As,min = max(0.26·fctm/fyk, 0.0013) · b·d
fctm = 0.30·fck2/3   (fck ≤ 50 MPa)
Example (C25/30, fyk=500): fctm=2.56 MPa → ρmin=0.0013
Minimum reinforcement ensures the section does not fail immediately upon cracking. The formula links the minimum steel to the tensile capacity of the concrete, ensuring the steel can carry the tension released at cracking without yielding immediately. At minimum steel, the cracking moment Mcr ≈ Mn.

EN 1992-1-1 Table 7.4N Span-to-Depth Ratios (Deflection)

EC2 allows deflection to be checked implicitly by limiting the span-to-effective-depth ratio l/d rather than computing deflections explicitly. The limiting ratios are:

SystemLightly loaded (ρ = 0.5%)Normally loaded (ρ = 1.5%)
Simply supported beam/slab2014
End span of continuous2618
Interior span of continuous3020
Cantilever86

Values from EC2 Table 7.4N for simply supported beams with fyk = 500 MPa. Multiply by 0.8 for flanged beams (beff/bw > 3) and by 1.5 for flat slabs.

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↑ Top

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.

Limits & Comparison↑ Top

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↑ Top

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.

Related Calculators & Guides

Column Design (PMM) Slab Design Foundation Design Crack Width Development Length Retaining Wall (ACI) ACI 318 Design Guide Eurocode 2 Design Guide IS 456 Design Guide TS 500 / TBDY Design Guide
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