Beam Design

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 f_r ≈ 0.62√f'c (ACI, MPa units). The corresponding moment is the cracking moment M_cr.
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 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.

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.

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.

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

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.

ACI 318-25 §22.2 Stress Block & Reinforcement

ACI 318-25 §9.6.1.2 Minimum & Maximum Reinforcement

ACI 318-25 Material Limits

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

ACI 318-25 §25.2 Minimum Bar Spacing

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.

ACI 318-25 §22.2 Design Summary

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

1. Compute R_n = M_u / (φ · b · d²)
2. Solve ρ = (0.85f'c / f_y) · [1 − √(1 − 2R_n / 0.85f'c)]
3. Check ρ_min ≤ ρ ≤ ρ_max (= 0.75 ρ_bal)
4. A_s = ρ · b · d
5. Check ε_t ≥ 0.004; confirm φ = 0.90
6. 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.

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)

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:

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

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:

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.

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.

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.

TS 500:2000 §7 Flexural Design

TS 500 uses the equivalent rectangular stress block — the same conceptual model as ACI 318 — but applies Turkish material factors (γ_c = 1.5, γ_s = 1.15) and its own ductility limit for the neutral axis depth.

TS 500:2000 §3.3 Material Factors

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

Example — ACI 318-25 Flexural Design (Singly Reinforced Rectangular Beam)

Given: Rectangular section, b = 300 mm, h = 550 mm, effective depth d = 500 mm (cover = 40 mm + Ø10 stirrup + Ø22/2). Concrete f'c = 28 MPa, steel f_y = 420 MPa. Factored moment: M_u = 250 kN·m.

1. Stress block factor β₁ — §22.2.2.4.3
   f'c = 28 MPa → β₁ = 0.85 (no reduction needed; limit applies for f'c > 28 MPa)
2. Design resistance coefficient R_n
   R_n = M_u/(φ·b·d²) = 250×10⁶/(0.90×300×500²) = 3.704 MPa
3. Required reinforcement ratio ρ
   ρ = (0.85f'c/f_y)·[1−√(1−2R_n/(0.85f'c))]
   = (0.85×28/420)·[1−√(1−2×3.704/23.80)]
   = 0.05667·[1−√0.6888] = 0.05667×0.1700 = 0.00963
4. Reinforcement limits — §9.6.1.2
   ρ_min = max(0.25√28/420, 1.4/420) = max(0.00315, 0.00333) = 0.00333
   ρ_bal = (0.85×0.85×28/420)·(0.003/0.00510) = 0.04817×0.5882 = 0.02833
   ρ_max = 0.75×0.02833 = 0.02125
   0.00333 ≤ ρ = 0.00963 ≤ 0.02125 ✓
5. Required steel area and bar selection
   A_s,req = ρ·b·d = 0.00963×300×500 = 1,445 mm²
   Select: 4Ø22 → A_s,prov = 4×380 = 1,520 mm²
6. Tension-controlled check — confirm φ = 0.90
   a = A_s·f_y/(0.85·f'c·b) = 1,520×420/(0.85×28×300) = 89.4 mm
   c = a/β₁ = 89.4/0.85 = 105.2 mm
   ε_t = (d/c − 1)·ε_cu = (500/105.2 − 1)×0.003 = 0.01126 ≫ 0.005
   → Tension-controlled failure ✓ → φ = 0.90 applies
7. Moment capacity verification
   φM_n = φ·A_s·f_y·(d − a/2)
   = 0.90×1,520×420×(500 − 44.7)/10⁶ = 261.5 kN·m ≥ M_u = 250 kN·m ✓

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)
• [8]
  TS 500:2000 — Requirements for Design and Construction of Reinforced Concrete Structures. Turkish Standards Institution, Ankara. §7 (Flexure, bond, and reinforcement detailing), §7.1.2 (A_st,min), §7.1.3 (A_st,max), §7.4 (Stress block), §7.5.4 (x_u/d ≤ 0.45 ductility limit).

Nature of Torsion in Reinforced Concrete

Torsion arises when an applied load acts eccentrically relative to the shear centre of a member, or when the structural system requires a beam to act as a support for perpendicular members. Unlike flexure and shear, torsion twists the cross-section about its own axis, producing a complex three-dimensional state of stress.

Two fundamentally different types of torsion arise in structural frames, and they must be treated differently in design:

• Equilibrium (Primary) Torsion: The torsional moment is required for static equilibrium — it cannot be redistributed without collapse of the structure. A cantilevering spandrel beam loaded off-axis is a classic example. Full torsional design is mandatory.
• Compatibility (Secondary) Torsion: The torsional moment arises from compatibility of deformations in a statically indeterminate structure. After cracking, the torsional stiffness drops sharply and the moment redistributes to adjacent members. EC2 §6.3.1 and ACI commentary permit neglecting compatibility torsion provided the structure can accommodate cracking and the member is detailed with minimum reinforcement.

Thin-Walled Tube (Bredt) Analogy

Solid rectangular sections under torsion are analysed using the thin-walled closed-section analogy. Once torsional cracks form, the uncracked core becomes ineffective and the section behaves as a thin-walled tube. A constant shear flow q circulates around the perimeter of the enclosed area A_o:

Space Truss Model

The design model for torsional reinforcement is the space truss analogy: diagonal concrete compression struts form on the faces of the section, resisted by transverse tension in closed stirrups and longitudinal tension in corner bars. This truss must be closed — open stirrups cannot carry torsion. The strut angle θ (measured from the member axis) may be taken as 45° (ACI, IS 456) or optimised between 21.8° and 45° (EC2 variable-angle method).

Combined Torsion and Shear

Torsion and shear co-exist in most real beams. Both produce shear stresses on the section walls, but their directions differ: on two opposite faces they add, and on the other two they subtract. Design codes therefore require checking the combined demand against the section's maximum capacity, typically through a circular or linear interaction equation. The reinforcement contributions add directly — stirrups from both shear and torsion sum.

ACI 318-25 §22.7 Torsional Strength

ACI 318-25 uses the thin-walled tube space-truss model at θ = 45°. The design shear flow area A_o = 0.85·A_oh, where A_oh is the area enclosed by the centreline of the outermost closed transverse reinforcement.

ACI 318-25 §9.7.5 Detailing Rules

• Stirrups must be closed (hooks at both ends engaging longitudinal bars)
• Maximum stirrup spacing: P_h/8 ≤ 300 mm
• Longitudinal bars: ≥ 1 bar per corner, spacing ≤ 300 mm around perimeter
• Longitudinal bars must have f_y ≥ 420 MPa; d_b ≥ s_stirrup/24 (§9.7.5.3)

EN 1992-1-1 §6.3 Torsional Resistance

EC2 models solid sections as thin-walled closed sections. No concrete contribution to torsional resistance is assumed — once cracked, all torsion is carried entirely by reinforcement (unlike shear, where V_Rd,c exists).

EN 1992-1-1 §9.2.3 Torsion Detailing

• Stirrups must be fully closed loops with 135° hooks or welded
• At least one longitudinal bar at each corner of the stirrup
• Stirrup spacing ≤ u_k/8 and ≤ min(b, h, 350 mm)
• Longitudinal bar spacing ≤ 350 mm

IS 456:2000 Cl. 41 Equivalent Shear-Moment Method

IS 456 does not use the space-truss or thin-walled tube model. Instead, torsion is converted into equivalent shear and equivalent bending moment, which are then added to the actual shear and bending moment for design. This simplified method is applicable to beams of compact cross-section under moderate combined loading.

Comparison of the Three Approaches

TS 500:2000 §9 Torsion Design

TS 500 §9 uses the thin-walled space-truss (Bredt) analogy — the same physical model as ACI 318 and EC2, but with a fixed strut angle of 45°. The section is replaced by an equivalent thin-walled tube, and shear flow is balanced by stirrups and longitudinal bars.

Example — ACI 318-25 Torsion Design (Rectangular Beam)

Given: Rectangular section b = 350 mm, h = 600 mm. Concrete f'c = 28 MPa, steel f_y = f_yt = 420 MPa. Clear cover = 40 mm to stirrup face, stirrup Ø10. Factored torsion: T_u = 60 kN·m. (Shear design handled separately; focus on torsion here.)

1. Section geometry
   A_cp = 350×600 = 210,000 mm² · P_cp = 2(350+600) = 1,900 mm
   x₁ = 350 − 2(40+5) = 260 mm · y₁ = 600 − 2(40+5) = 510 mm
   A_oh = 260×510 = 132,600 mm² · P_h = 2(260+510) = 1,540 mm
   A_o = 0.85×132,600 = 112,710 mm²
2. Threshold torsion — §22.7.4
   T_th = 0.083×1.0×√28×(210,000²/1,900) / 10⁶ = 0.083×5.292×23,210,526/10⁶ = 10.19 kN·m
   φT_th = 0.75×10.19 = 7.64 kN·m
   T_u = 60 kN·m >> φT_th → Torsion design required
3. Required transverse reinforcement — §22.7.6.1
   A_t/s = T_u/(φ·2·A_o·f_yt·cot θ) = 60×10⁶/(0.75×2×112,710×420×1.0)
   = 60×10⁶/71,007,300 = 0.845 mm²/mm per stirrup leg
   For Ø10 closed stirrups: A_t = 78.5 mm² per leg
   Required s ≤ 78.5/0.845 = 92.9 mm → Use s = 90 mm
   A_t,prov/s = 78.5/90 = 0.872 mm²/mm > 0.845 ✓
   Check max spacing: P_h/8 = 1,540/8 = 192 mm → 90 mm < 192 mm ✓
4. Required longitudinal torsion steel — §22.7.6.1
   A_l = (A_t/s)·P_h·(f_yt/f_y)·cot²θ
   = 0.845×1,540×(420/420)×1.0² = 1,301 mm²
   Minimum A_l (§9.7.5.1): 0.42×5.292×210,000/420 − 0.845×1,540 = 1,112 − 1,301 < 0 → not governing
   A_l = 1,301 mm² governs
5. Distribute longitudinal bars
   Spacing ≤ 300 mm around perimeter; one bar per corner required
   Perimeter ≈ P_h = 1,540 mm → need ≥ 1,540/300 ≈ 6 bars
   Use 8Ø16 (8×201 = 1,608 mm² ≥ 1,301 mm²) evenly spaced at ~193 mm ✓

References

• [1]
  ACI 318-25 — Building Code Requirements for Structural Concrete. American Concrete Institute, 2025. §22.7 (Torsional strength), §9.7.5 (Longitudinal torsional reinforcement), §9.7.6.3 (Torsional reinforcement detailing).
• [2]
  EN 1992-1-1:2004 — Eurocode 2: Design of Concrete Structures. CEN, Brussels. §6.3 (Torsion), §9.2.3 (Torsion reinforcement detailing). Variable-angle truss method.
• [3]
  IS 456:2000 — Plain and Reinforced Concrete — Code of Practice, 4th Rev. Bureau of Indian Standards. Cl. 41 (Torsion — equivalent shear and bending moment method), Cl. 26.5.1.7 (Stirrup spacing for torsion).
• [4]
  Hsu, T.T.C. — Unified Theory of Reinforced Concrete. CRC Press, 1993. Chapter 4 (Torsion — softened truss model and space-truss analogy).
• [5]
  Collins, M.P. & Mitchell, D. — Prestressed Concrete Structures. Prentice-Hall, 1991. Chapter 6 (Combined flexure, shear and torsion).
• [6]
  MacGregor, J.G. & Wight, J.K. — Reinforced Concrete: Mechanics and Design, 7th Ed. Pearson, 2016. Chapter 7 (Torsion in Beams — derivation of ACI design equations).
• [7]
  TS 500:2000 — Requirements for Design and Construction of Reinforced Concrete Structures. Turkish Standards Institution, Ankara. §9 (Torsion — thin-walled tube analogy), §9.3.2 (Stirrup requirement), §9.3.3 (Longitudinal reinforcement), §9.3.4 (Section adequacy — 0.22f_cd limit).