Beam Design

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

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Clear Cover [m]

Material Properties

f'c — Concrete [MPa]

MPa

f_y — Rebar [MPa]

MPa

Compression Reinforcement (optional)

Include compression steel A'_s

Resistance Factor

φ — Flexure (tension-controlled)

—

ACI 318-25 §21.2.2: φ = 0.90 for tension-controlled sections (ε_t ≥ 0.005)

Factored Moment M_u [kN·m]

kN·m

📋

Enter values and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Nominal Cover [m]

Material Properties

f_ck — Concrete [MPa]

MPa

f_yk — Rebar [MPa]

MPa

Partial Safety Factors

γ_c — Concrete

Recommended: 1.50 (persistent)

γ_s — Steel

Recommended: 1.15

Compression Reinforcement (optional)

Include compression steel A'_s

Design Moment M_Ed [kN·m]

kN·m

📋

Enter values and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Clear Cover [m]

Main Bar Dia. [mm]

Material Properties

Concrete Grade (f_ck)

Steel Grade (f_y)

Factored Moment M_u [kN·m]

kN·m

📋

Enter values and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Clear Cover [m]

Material Properties

f'c — Concrete [MPa]

MPa

f_y — Rebar [MPa]

MPa

Resistance Factor

φ — Flexure

—

Tension Reinforcement

Bar Diameter [mm]

Number of Bars

bars

Compression Reinforcement (optional)

Bar Diameter [mm]

Number of Bars

bars

Second row of tension bars

Loading (optional — for D/C)

Applied Moment M_u [kN·m]

kN·m

📋

Enter values and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Nominal Cover [m]

Material Properties

f_ck — Concrete [MPa]

MPa

f_yk — Rebar [MPa]

MPa

Partial Safety Factors

γ_c — Concrete

γ_s — Steel

Tension Reinforcement

Bar Diameter [mm]

Number of Bars

bars

Compression Reinforcement (optional)

Bar Diameter [mm]

Number of Bars

bars

Second row of tension bars

Loading (optional — for D/C)

Applied Moment M_Ed [kN·m]

kN·m

📋

Enter values and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Clear Cover [m]

Material Properties

Concrete Grade (f_ck)

Steel Grade (f_y)

Tension Reinforcement

Bar Diameter [mm]

Number of Bars

bars

Loading (optional — for D/C)

Applied M_u [kN·m]

kN·m

📋

Enter values and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Clear Cover [m]

Material Properties

f'c [MPa]

MPa

f_y (stirrups) [MPa]

MPa

f_yl (long. bars) [MPa]

MPa

λ — Density factor

—

φ — Torsion/Shear

—

ACI 318-25: φ = 0.75 for shear/torsion

Design Actions

Shear V_u [kN]

Torsion T_u [kN·m]

kN·m

Transverse Reinforcement (Stirrups)

Stirrup Diameter [mm]

Stirrup Spacing [m]

Longitudinal Torsion Bars

Bar Diameter [mm]

Number of Bars

bars

📋

Enter values and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Nominal Cover [m]

Material Properties

f_ck [MPa]

MPa

f_yk [MPa]

MPa

γ_c — Concrete

γ_s — Steel

Design Actions

Shear V_Ed [kN]

Torsion T_Ed [kN·m]

kN·m

Transverse Reinforcement (Stirrups)

Stirrup Diameter [mm]

Stirrup Spacing [m]

Longitudinal Torsion Bars

Bar Diameter [mm]

Number of Bars

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.

📋

Enter values and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Clear Cover [m]

Main Bar Dia. [mm]

Material Properties

Concrete Grade (f_ck)

Steel Grade (f_y)

Design Actions

Shear V_u [kN]

Torsion T_u [kN·m]

kN·m

Bending Moment M_u [kN·m]

kN·m

Required for equivalent moment M_e per IS 456 Cl 41.3

Transverse Reinforcement (Stirrups)

Stirrup Dia. [mm]

Stirrup Spacing [m]

Longitudinal Torsion Bars

Bar Diameter [mm]

Number of Bars

bars

ℹ️

IS 456:2000 Cl 41 — Equivalent Shear/Moment Method: Combined torsion, bending, and shear are handled by designing for equivalent shear V_e = V_u + 1.6T_u/b and equivalent moments M_e1 (bottom), M_e2 (top). Stirrups are designed for the equivalent nominal shear stress τ_ve.

📋

Enter values and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Cover to stirrup [m]

Material Properties

f'c [MPa]

MPa

f_yt Stirrup [MPa]

MPa

φ (shear)

—

λ (concrete)

—

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

A_sl (tension steel) [mm²]

mm²

ρ_w = A_sl/(b_w·d) · ACI 318-25 §22.5.5.1

Stirrup Layout

Stirrup Dia. [mm]

Legs n

Provided spacing s [m]

Applied Load

Factored Shear V_u [kN]

📊

Configure inputs and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Cover to stirrup [m]

Material Properties

f_ck [MPa]

MPa

f_ywk Stirrup [MPa]

MPa

γ_c

Recommended: 1.50

γ_s

Recommended: 1.15

Longitudinal Steel

A_sl (tension steel) [mm²]

mm²

ρ_l = A_sl/(b_w·d) · EN 1992-1-1 §6.2.2

Stirrup Layout

Stirrup Dia. [mm]

Legs n

Provided spacing s [m]

Applied Load

Design Shear V_Ed [kN]

📊

Configure inputs and click Calculate.

Input Parameters

kN-m · m

Units:

Section Geometry

Width b [m]

Total Depth h [m]

Cover to stirrup [m]

Material Properties

f_ck (MPa)

f_y Stirrup (MPa)

Longitudinal Steel

A_st (tension steel) [mm²]

mm²

Used to compute p_t = 100·A_st/(b·d) for τ_c (Table 19)

Stirrup Layout

Stirrup Dia. (mm)

Legs n

Provided spacing s_v [m]

Applied Load

Factored Shear V_u [kN]

📊

Configure inputs and click Calculate.

📚 Design Background & Code References

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 λ_s in §22.5.5.1; EC2 captures it via the k = 1 + √(200/d) term in V_Rd,c.

Design Approach

General Design Requirement

φV_n ≥ V_u (ACI) V_Rd ≥ V_Ed (EC2) V_u ≤ V_c + V_us (IS 456)

All three codes use the same additive model: concrete contribution + stirrup contribution ≥ factored shear demand. When V_u exceeds the concrete capacity alone, stirrups are required to carry the excess. When V_u 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 (V_u < φV_c), 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.

📚 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 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:

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 f_r ≈ 0.62√f'c (ACI, MPa units). The corresponding moment is the cracking moment M_cr.
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.
Ultimate limit state (ULS): The extreme compression fibre concrete reaches its crushing strain ε_cu. In a properly designed under-reinforced section, the tension steel has already yielded (ε_s ≫ ε_y), 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 C_c 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: C_c = 0.85 f'c · a · b
Tension resultant: T = A_s · f_y (at yield)
Moment capacity: φM_n = φ · A_s · f_y · (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 f_ck ≤ 50 MPa with analogous meaning.

Balanced Reinforcement Ratio ρ_bal

The balanced failure condition occurs when the extreme concrete fibre reaches ε_cu at the same instant that the outermost tension steel reaches its yield strain ε_y = f_y/E_s. 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)

ρ_bal = (0.85 β₁ f'c / f_y) · [ε_cu / (ε_cu + ε_y)]
ε_cu = 0.003 · ε_y = f_y / 200 000 (f_y in MPa)
Example (f'c = 28 MPa, f_y = 420 MPa): ρ_bal ≈ 0.0285

EC2 expresses the equivalent limit as x/d ≤ 0.45; IS 456 tabulates x_u,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.

Code	Ductility Requirement	Implication
ACI 318-25	Net tensile strain ε_t ≥ 0.004 at ULS (φ = 0.90 when ε_t ≥ 0.005)	ρ ≤ ≈ 0.75 ρ_bal
Eurocode 2	Neutral axis x/d ≤ 0.45 (for δ ≥ 0.7 redistribution)	K ≤ K' = 0.167
IS 456:2000	x_u/d ≤ x_u,max/d (by grade: Fe 415 → 0.48)	M_u ≤ M_u,lim

Key Assumptions at ULS

Tensile strength of concrete is neglected — only steel resists tension.
Concrete reaches its limiting compressive strain: ε_cu = 0.003 (ACI) or 0.0035 (EC2 for f_ck ≤ 50 MPa).
Stress distribution idealised as a rectangular (Whitney) stress block.
Steel has yielded: σ_s = f_y (ACI) or f_yd = f_yk/γ_s (EC2).
Section is singly-reinforced — compression steel neglected.

Effective Depth

d = h − c_clear − d_bar/2

h = total depth · c = clear cover to main bar face · d_bar = 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 φM_n ≥ M_u, where φ is the strength reduction factor and M_n is the nominal moment capacity.

Strength Reduction Factor φ — ACI 318-25 Table 21.2.2

φ = 0.90    tension-controlled (ε_t ≥ 0.005)
φ = 0.65 → 0.90    transition zone (0.002 < ε_t < 0.005)
φ = 0.65    compression-controlled (ε_t ≤ 0.002)

For beams under pure flexure, φ = 0.90 applies when ρ ≤ 0.75ρ_bal.

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

R_n = M_u/(φ·b·d²)
ρ = (0.85f'c/f_y)·[1−√(1−2R_n/0.85f'c)]
A_s = ρ·b·d

ACI 318-25 §9.6.1.2 Minimum & Maximum Reinforcement

Reinforcement Limits

ρ_min = max(0.25√f'c/f_y , 1.4/f_y) [MPa]
ρ_max = 0.75·ρ_bal (ensures ε_t ≥ 0.005)

ACI 318-25 Material Limits

Parameter	Minimum	Maximum	Reference
Concrete f'c	17 MPa (2500 psi)	70 MPa (10000 psi)*	§19.2.1
Rebar f_y	280 MPa (40 ksi)	550 MPa (80 ksi)	§20.2.2
Strength factor φ	0.65	0.90	Table 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

s_clear ≥ max(d_b, 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/f_s) − 2.5·c_c [mm]
s ≤ 300 · (280/f_s) (interior exposure limit)
f_s = 2/3 · f_y (approximate service bar stress, MPa)
c_c = clear cover to flexural tension reinforcement

For f_y = 420 MPa with c_c = 40 mm: s_max ≈ 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:

Compute R_n = M_u / (φ · b · d²)
Solve ρ = (0.85f'c / f_y) · [1 − √(1 − 2R_n / 0.85f'c)]
Check ρ_min ≤ ρ ≤ ρ_max (= 0.75 ρ_bal)
A_s = ρ · b · d
Check ε_t ≥ 0.004; confirm φ = 0.90
Verify bar spacing ≥ s_clear,min and ≤ s_crack,max

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 f_cd and f_yd that are used throughout all capacity calculations.

Design Material Strengths

f_cd = α_cc·f_ck/γ_c [α_cc=1.0 recommended; γ_c=1.5]
f_yd = f_yk/γ_s [γ_s=1.15]

γ_c and γ_s are Nationally Defined Parameters (NDPs) and may be modified by National Annexes. The recommended values are γ_c = 1.5 (persistent/transient) and γ_s = 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 = M_Ed/(b·d²·f_cd)
K' = 0.167 [limiting K for δ=1.0, no redistribution]
z = d·[0.5+√(0.25−K/1.134)] ≤ 0.95d (lever arm)
A_s = M_Ed/(f_yd·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 f_ck ≤ 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)

A_s,min = max(0.26·f_ctm/f_yk, 0.0013) · b·d
f_ctm = 0.30·f_ck^2/3 (f_ck ≤ 50 MPa)
Example (C25/30, f_yk=500): f_ctm=2.56 MPa → ρ_min=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 M_cr ≈ M_n.

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:

System	Lightly loaded (ρ = 0.5%)	Normally loaded (ρ = 1.5%)
Simply supported beam/slab	20	14
End span of continuous	26	18
Interior span of continuous	30	20
Cantilever	8	6

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

EN 1992-1-1 Material Limits

Parameter	Minimum	Maximum	Reference
Concrete f_ck	12 MPa (C12/15)	90 MPa (C90/105)	Table 3.1
Rebar f_yk	400 MPa	600 MPa	§3.2.2
Partial factor γ_c	1.0 (accidental)	1.5 (persistent)	Table 2.1N
Partial factor γ_s	1.0 (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 x_u,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

M_u = 0.87 f_y A_st d [1 − (A_st f_y) / (b d f_ck)]
M_u,lim = 0.36 f_ck b x_u,lim (d − 0.42 x_u,lim)

Minimum / Maximum Steel — Cl. 26.5.1.1

A_st,min = 0.85 b d / f_y
A_st,max = 0.04 b D (D = total depth)

IS 456 Cl. 40 Shear Design

Nominal shear stress τ_v = V_u/(b·d) is compared with design shear strength τ_c (Table 19, function of p_t and f_ck) and maximum τ_c,max (Table 20). Vertical stirrups carry excess shear.

Stirrup Capacity — Cl. 40.4 (b)

V_us = 0.87 f_y A_sv d / s_v
s_v,max = min(0.75d, 300 mm) — Cl. 26.5.1.5

IS 456 Cl. 41 Torsion Design

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

Equivalent Shear — Cl. 41.3.1

V_e = V_u + 1.6 T_u / b
τ_ve = V_e / (b · d) ≤ τ_c,max

Equivalent Moment — Cl. 41.4.2

M_t = T_u (1 + D/b) / 1.7
Design for M_u1 = M_u + M_t (tension side) and M_u2 = M_t − M_u (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 f_ck (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

Parameter	ACI 318-25	Eurocode 2	IS 456:2000
Concrete strain at ULS	ε_cu = 0.003	ε_cu = 0.0035	ε_cu = 0.0035
Resistance / partial factor	φ = 0.90 (flexure)	γ_c=1.5 · γ_s=1.15	γ_c=1.5 · γ_s=1.15 (implicit)
Stress block depth factor	β₁ = 0.85 → 0.65	λ = 0.8 (f_ck ≤ 50 MPa)	0.36·f_ck·b·x_u (parabolic-rect.)
Stress block intensity	0.85·f'c	η·f_cd, η=1.0	0.36·f_ck (equiv. rect. block)
Minimum reinforcement	max(0.25√f'c/f_y, 1.4/f_y)·b·d	max(0.26f_ctm/f_yk, 0.0013)·b·d	0.85·b·d / f_y (Cl. 26.5.1.1a)
Maximum reinforcement	ρ ≤ 0.75·ρ_bal	K ≤ K' = 0.167	0.04·b·D (Cl. 26.5.1.1b)
Neutral axis limit (x_u/d)	Not directly (ε_t ≥ 0.004)	x/d ≤ 0.45	Fe415: 0.48 · Fe500: 0.46 · Fe250: 0.53
Maximum lever arm	Not explicitly limited	z ≤ 0.95d	Not explicitly limited

Beam Design

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📚 Design Background & Code References

Shear Behaviour of Reinforced Concrete Beams

Mechanisms of Shear Transfer

Truss Analogy

Size Effect

Design Approach

Why Minimum Stirrups Are Required

ACI 318-25 §22.5 Shear Strength

EN 1992-1-1 §6.2 Shear Resistance

IS 456:2000 Cl. 40 Shear Design

References

📚 Design Background & Code References

Flexural Behaviour of Reinforced Concrete Beams

Three Stages of Beam Behaviour

Whitney Equivalent Rectangular Stress Block

Balanced Reinforcement Ratio ρbal

Ductility and the Tension-Controlled Requirement

Key Assumptions at ULS

Effective Depth

ACI 318-25 Strength Design Method

ACI 318-25 §22.2 Stress Block & Reinforcement

ACI 318-25 §9.6.1.2 Minimum & Maximum Reinforcement

ACI 318-25 Material Limits

ACI 318-25 §25.2 Minimum Bar Spacing

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

ACI 318-25 §22.2 Design Summary

EN 1992-1-1 Partial Factor Method

EN 1992-1-1 §6.1 Flexural Design

EN 1992-1-1 §9.2.1.1 Minimum Reinforcement

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

EN 1992-1-1 Material Limits

IS 456:2000 Code Overview

IS 456 Cl. 38 Flexural Design (Limit State)

IS 456 Cl. 40 Shear Design

IS 456 Cl. 41 Torsion Design

IS 456 Table 2 Concrete Grades

IS 456 Table 1 Steel Grades

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

References

Related Calculators & Guides

Balanced Reinforcement Ratio ρ_bal