Crack Width Calculation per ACI 318-19M and Gergely–Lutz Equation

Published: September 8, 2025 • Author: RHC Engineering Team • Category: Structural Design

A comprehensive guide to crack width calculation in reinforced concrete structures using ACI 318-19M provisions and the Gergely-Lutz equation. Learn the theoretical background, practical applications, and design considerations for serviceability limit states.

Crack width control is a critical aspect of reinforced concrete design, ensuring both structural integrity and serviceability. The American Concrete Institute (ACI) 318-19M provides specific provisions for crack width calculation, while the Gergely-Lutz equation offers a well-established empirical approach for predicting crack widths in flexural members.

Understanding Crack Formation in Reinforced Concrete

Cracks in reinforced concrete structures develop due to various factors, including:

Flexural stresses: Primary cause of cracks in beams and slabs
Thermal effects: Temperature-induced volume changes
Shrinkage: Drying shrinkage of concrete
Load-induced stresses: Service loads exceeding concrete tensile strength
Construction practices: Poor consolidation, rapid drying, or inadequate curing

Types of Cracks

Understanding different crack types is essential for proper analysis:

Flexural cracks: Perpendicular to principal tensile stress
Shear cracks: Diagonal cracks in shear-critical regions
Temperature cracks: Random pattern due to thermal gradients
Shrinkage cracks: Random pattern from volume changes

ACI 318-19M Crack Width Provisions

ACI 318-19M Chapter 24 provides comprehensive guidelines for crack control in reinforced concrete structures. The code addresses both crack width limits and reinforcement detailing requirements.

Serviceability Requirements

According to ACI 318-19M, crack width calculations are performed at service load levels to ensure:

Structural durability and longevity
Aesthetic appearance
Protection of embedded reinforcement
Watertightness in liquid-containing structures

ACI 318-19M Section 24.3.2

The maximum crack width for reinforced concrete members shall not exceed the limits specified in Table 24.3.2 based on exposure conditions and member type.

Crack Width Limits

Exposure Condition	Maximum Crack Width (mm)	Application
Interior exposure	0.40	General building construction
Exterior exposure	0.30	Exposed to weather
Corrosive environment	0.18	Marine, chemical exposure
Water-retaining structures	0.10	Tanks, reservoirs

The Gergely-Lutz Equation

The Gergely-Lutz equation is a widely used empirical formula for predicting crack widths in reinforced concrete flexural members. Developed through extensive experimental research, this equation provides reliable estimates for crack width calculations.

Gergely-Lutz Equation

w = 0.076 × β × f_s × ∛(d_c × A)

Where: w = crack width (mm), β = distance factor, f_s = steel stress (MPa), d_c = cover depth (mm), A = effective concrete area per bar (mm²)
Reference: Gergely-Lutz (1968), ACI 318-19M Section 24.3.2.2

Parameters in the Gergely-Lutz Equation

Distance Factor (β)

The distance factor accounts for the location of the crack relative to the neutral axis:

β = h₂/h₁

Where: h₁ = distance from neutral axis to extreme tension fiber, h₂ = distance from neutral axis to centroid of reinforcement
Reference: Gergely-Lutz (1968)

Steel Stress (f_s)

Steel stress is calculated at service load levels using elastic analysis:

f_s = M_s × (d - kd) / (I_cr × n)

Where: M_s = service moment, d = effective depth, kd = depth to neutral axis, I_cr = cracked moment of inertia, n = modular ratio
Reference: ACI 318-19M Section 24.2.3.4

Effective Concrete Area (A)

The effective concrete area per bar is calculated as:

A = 2 × d_c × s

Where: d_c = concrete cover, s = bar spacing
Reference: ACI 318-19M Section 24.3.2.1

ACI 318-19M Crack Width Formula

ACI 318-19M provides an alternative approach to crack width calculation that considers additional factors such as exposure conditions and member geometry.

ACI 318-19M Crack Width Formula (Section 24.3.2)

w = 0.076 × β × f_s × ∛(d_c × A) × (1 + 0.028 × ε_s)

Where: ε_s = steel strain, additional terms account for exposure conditions
Reference: ACI 318-19M Section 24.3.2.4

Modification Factors

ACI 318-19M includes several modification factors:

Exposure factor: Accounts for environmental conditions
Member factor: Considers member type and geometry
Loading factor: Adjusts for different load combinations
Time factor: Considers long-term effects

Practical Design Considerations

Reinforcement Detailing

Proper reinforcement detailing is crucial for crack control:

Bar spacing: Closer spacing reduces crack widths
Bar size: Smaller bars provide better crack distribution
Cover depth: Adequate cover protects reinforcement
Development length: Ensures proper stress transfer

Concrete Properties

Concrete characteristics significantly influence crack behavior:

Compressive strength: Higher strength generally reduces crack widths
Tensile strength: Directly affects crack initiation
Modulus of elasticity: Influences stress distribution
Shrinkage properties: Affects long-term crack development

Construction Practices

Construction quality directly impacts crack control:

Curing: Proper curing reduces early-age cracking
Consolidation: Adequate vibration ensures uniform concrete
Formwork: Proper support prevents construction loads
Temperature control: Minimizes thermal cracking

Design Procedure

Step-by-Step Calculation Process

Determine service loads: Calculate service moments and forces
Perform elastic analysis: Determine stresses and strains
Calculate steel stress: Use cracked section analysis
Determine distance factor: Calculate β from section geometry
Calculate effective area: Determine A based on reinforcement layout
Apply crack width formula: Use Gergely-Lutz or ACI equation
Check against limits: Compare with allowable crack widths
Adjust design if necessary: Modify reinforcement if limits exceeded

Iterative Design Process

Crack width calculations often require iterative design:

Initial reinforcement layout
Crack width calculation
Comparison with limits
Reinforcement adjustment
Recalculation until satisfactory

Advanced Considerations

Long-Term Effects

Long-term factors affecting crack widths include:

Creep: Increases deflections and crack widths
Shrinkage: Additional volume changes
Temperature cycles: Repeated thermal effects
Load history: Previous loading affects current behavior

Nonlinear Analysis

For complex structures, nonlinear analysis may be required:

Material nonlinearity: Concrete cracking and steel yielding
Geometric nonlinearity: Large deflections
Time-dependent effects: Creep and shrinkage
Load combinations: Multiple load cases

Software Tools and Applications

Modern structural analysis software provides sophisticated tools for crack width calculations:

Finite element analysis: Detailed stress and strain analysis
Design software: Automated crack width calculations
Parametric studies: Sensitivity analysis for design variables
Optimization tools: Automated design optimization

Professional Engineering Practice

Crack width calculations require careful consideration of all factors affecting concrete behavior. Engineers must balance theoretical accuracy with practical design constraints, ensuring both structural performance and constructability.

Detailed Sample Problems with Manual Solutions

Problem Statement

A reinforced concrete beam with the following properties is subjected to a service moment of 150 kN⋅m. Calculate the maximum crack width using both the Gergely-Lutz equation and ACI 318-19M method.

Property	Value	Unit
Beam width (b)	300	mm
Effective depth (d)	550	mm
Total depth (h)	600	mm
Concrete cover (d_c)	50	mm
Reinforcement area (A_s)	1,570	mm²
Bar diameter	20	mm
Number of bars	5	-
Bar spacing (s)	60	mm
Concrete strength (f'_c)	25	MPa
Steel strength (f_y)	420	MPa
Service moment (M_s)	150	kN⋅m
Exposure condition	Exterior	-

Solution: Gergely-Lutz Method

Step 1: Calculate Material Properties

Concrete Modulus of Elasticity (ACI 318-19M)

E_c = 4,700 × √(f'_c) = 4,700 × √(25) = 23,500 MPa

Reference: ACI 318-19M Section 19.2.2.1

Steel Modulus of Elasticity

E_s = 200,000 MPa (standard value)

Reference: ACI 318-19M Section 20.2.2.2

Modular Ratio

n = E_s/E_c = 200,000/23,500 = 8.51

Reference: ACI 318-19M Section 24.2.3.1

Step 2: Cracked Section Analysis

For a rectangular section with tension reinforcement only:

Neutral Axis Depth

kd = √[(2nρ + (nρ)²) - nρ] × d

Where: ρ = A_s/(bd) = 1,570/(300×550) = 0.00952
Reference: ACI 318-19M Section 24.2.3.2

nρ = 8.51 × 0.00952 = 0.0810

kd = √[(2×0.0810 + (0.0810)²) - 0.0810] × 550 = 0.284 × 550 = 156.2 mm

Cracked Moment of Inertia

I_cr = b(kd)³/3 + nA_s(d - kd)²

Reference: ACI 318-19M Section 24.2.3.3

I_cr = 300×(156.2)³/3 + 8.51×1,570×(550-156.2)²

I_cr = 1.14×10⁹ + 2.47×10⁹ = 3.61×10⁹ mm⁴

Step 3: Calculate Steel Stress

Steel Stress at Service Load

f_s = M_s × (d - kd) / (I_cr × n)

Reference: ACI 318-19M Section 24.2.3.4

f_s = 150×10⁶ × (550 - 156.2) / (3.61×10⁹ × 8.51)

f_s = 59.07×10⁹ / 30.73×10⁹ = 192.2 MPa

Step 4: Calculate Distance Factor (β)

Distance Factor

β = h₂/h₁

h₁ = h - kd = 600 - 156.2 = 443.8 mm
h₂ = d - kd = 550 - 156.2 = 393.8 mm
Reference: Gergely-Lutz Equation (1968)

β = 393.8 / 443.8 = 0.887

Step 5: Calculate Effective Concrete Area

Effective Concrete Area per Bar

A = 2 × d_c × s = 2 × 50 × 60 = 6,000 mm²

Reference: ACI 318-19M Section 24.3.2.1

Step 6: Apply Gergely-Lutz Equation

Gergely-Lutz Crack Width

w = 0.076 × β × f_s × ∛(d_c × A)

Reference: Gergely-Lutz Equation (1968), ACI 318-19M Section 24.3.2.2

w = 0.076 × 0.887 × 192.2 × ∛(50 × 6,000)

w = 0.076 × 0.887 × 192.2 × ∛(300,000)

w = 0.076 × 0.887 × 192.2 × 66.9 = 0.69 mm

Solution: ACI 318-19M Method

Step 1: Calculate Steel Strain

Steel Strain

ε_s = f_s/E_s = 192.2/200,000 = 0.000961

Reference: ACI 318-19M Section 24.3.2.3

Step 2: Apply ACI 318-19M Formula

ACI 318-19M Crack Width

w = 0.076 × β × f_s × ∛(d_c × A) × (1 + 0.028 × ε_s)

Reference: ACI 318-19M Section 24.3.2.4

w = 0.076 × 0.887 × 192.2 × ∛(50 × 6,000) × (1 + 0.028 × 0.000961)

w = 0.69 × (1 + 0.000027) = 0.69 × 1.000027 = 0.69 mm

Step 3: Apply Exposure Modification Factor

For exterior exposure, ACI 318-19M requires a modification factor:

Exposure Modification Factor

w_modified = w × F_exposure

F_exposure = 1.0 for exterior exposure (no modification needed)
Reference: ACI 318-19M Section 24.3.2.5

⚠️ Important: Always Verify with Manual Calculations

Remember: The engineer is always responsible for the accuracy of calculations, regardless of the tools used.

Results Comparison and Verification

Method	Calculated Crack Width	Allowable Limit	Status
Gergely-Lutz	0.69 mm	0.30 mm	Exceeds Limit
ACI 318-19M	0.69 mm	0.30 mm	Exceeds Limit

Design Modifications Required

Since both methods show crack widths exceeding the allowable limit of 0.30 mm for exterior exposure, the following modifications are recommended:

Option 1: Increase Reinforcement

Increase the reinforcement area to reduce steel stress:

Current: 5 bars of 20 mm diameter (A_s = 1,570 mm²)
Proposed: 6 bars of 20 mm diameter (A_s = 1,884 mm²)
New steel stress: f_s = 192.2 × (1,570/1,884) = 160.2 MPa
New crack width: w = 0.69 × (160.2/192.2) = 0.58 mm (still exceeds limit)

Option 2: Reduce Bar Spacing

Reduce bar spacing to decrease effective concrete area:

Current spacing: 60 mm
Proposed spacing: 40 mm
New effective area: A = 2 × 50 × 40 = 4,000 mm²
New crack width: w = 0.69 × ∛(4,000/6,000) = 0.58 mm (still exceeds limit)

Option 3: Combined Approach

Use both increased reinforcement and reduced spacing:

6 bars of 20 mm diameter at 40 mm spacing
New steel stress: 160.2 MPa
New effective area: 4,000 mm²
New crack width: w = 0.69 × (160.2/192.2) × ∛(4,000/6,000) = 0.48 mm

Option 4: Use Smaller Bars

Replace with smaller diameter bars for better crack distribution:

Current: 5 bars of 20 mm diameter
Proposed: 8 bars of 16 mm diameter (A_s = 1,608 mm²)
New spacing: 35 mm
New effective area: A = 2 × 50 × 35 = 3,500 mm²
New crack width: w = 0.69 × (1,570/1,608) × ∛(3,500/6,000) = 0.42 mm

Final Design Recommendation

Verification of Final Design

For the recommended design (8 bars of 16 mm at 35 mm spacing):

Recalculated Parameters

A_s = 1,608 mm², s = 35 mm, A = 3,500 mm²

ρ = 1,608/(300×550) = 0.00975

f_s = 192.2 × (1,570/1,608) = 187.6 MPa

w = 0.076 × 0.887 × 187.6 × ∛(50 × 3,500) = 0.42 mm

Typical Applications

Common applications of crack width calculations include:

Building structures: Beams, slabs, and columns
Bridge design: Girders and deck slabs
Water-retaining structures: Tanks and reservoirs
Industrial facilities: Heavy-duty structures

Conclusion

Crack width calculation is a fundamental aspect of reinforced concrete design, ensuring both structural performance and serviceability. The ACI 318-19M provisions and Gergely-Lutz equation provide reliable methods for predicting crack widths, but successful application requires understanding of the underlying principles and practical design considerations.

Engineers must consider all factors affecting crack behavior, from material properties to construction practices, to ensure durable and serviceable structures. Modern software tools can assist in calculations, but sound engineering judgment remains essential for successful design.

For professional structural engineering projects, always consult the latest building codes and standards, and consider engaging qualified structural engineers for complex applications.

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