Study guide
3.3 Fracture Analysis & Failure Locations
ACI Concrete Strength Testing Technician Exam
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1 Chapter 1: Introduction & Exam Overview
1.1 About the ACI Strength Testing Exam 1.2 Study Preparation & Performance Exam Tips
2 Chapter 2: ASTM C39 — Compressive Strength of Cylindrical Concrete Specimens
2.1 Scope, Significance, & Specimen Preparation 2.2 Testing Machine Requirements & Verification 2.3 Compressive Loading Rates & Controls 2.4 Specimen End Tolerances & Dimensional Measurements 2.5 Cylinder Fracture Classifications & Calculations 2.6 Load Indicators, Density Determination & Report Contents
3 Chapter 3: ASTM C78 — Flexural Strength of Concrete (Third-Point Loading)
3.1 Testing Apparatus & Specimen Setup 3.2 Loading Rates & Modulus of Rupture Calculations 3.3 Fracture Analysis & Failure Locations 3.4 Machine Requirements, Specimen Conformance & Reporting
4 Chapter 4: ASTM C617 — Capping Cylindrical Concrete Specimens with Sulfur Mortar
4.1 Capping Equipment & Material Specifications 4.2 Capping Procedures & Planeness Verification 4.3 Cap Thickness Limitations 4.4 Sulfur Safety & Hardening Time Rules
5 Chapter 5: ASTM C1231 — Practice for Use of Unbonded Caps
5.1 Neoprene Pad Specifications & Durometer Selection 5.2 Retainer Rings & Equipment Dimensions 5.3 Pad Inspection & Reuse Limits
6 Chapter 6: General Laboratory Practice & Quality Assurance (ASTM C1077 / ACI 318)
6.1 Laboratory Accreditation & Personnel 6.2 Machine Verification & Documentation
3.3 Fracture Analysis & Failure Locations
Key Takeaways Explain This Topic AI
After failure, the technician must take three width and three depth measurements at the fracture cross-section to calculate the average dimensions (b and d) to the nearest 0.05 in.
If the fracture initiates outside the middle third of the span by 5% or less of the span length, the modulus of rupture is calculated using the modified formula R = 3Pa / (b * d^2).
If the fracture initiates outside the middle third by more than 5% of the span length (greater than 0.9 inches for an 18-inch span), the test results must be discarded.
Flexural failures always initiate on the bottom (tension) face of the beam, and the point of initiation is measured along the centerline of this face from the nearest support.
Tests breaking deep in the shear span are rejected because the stress state is complex, and failures there typically indicate a specimen defect or machine misalignment.
Last updated: July 2026
- 3.3 Fracture Analysis & Failure Locations
- Post-Test Inspection & Specimen Measurements
Once the concrete beam has ruptured, the technician must perform a detailed inspection of the specimen. The primary objective is to measure the dimensions at the fracture plane and locate the exact point where the fracture initiated.
Dimensional Measurements: The technician must take three measurements of the width and three measurements of the depth at the fracture cross-section. The three measurements for each dimension are taken at the edges and center of the cross-section. These measurements are averaged to the nearest 0.05 inches (1 mm) to determine the average width ( b b b ) and average depth ( d d d ) used in the modulus of rupture calculation. Using nominal dimensions (like exactly 6.00 inches) instead of the actual measured dimensions is a major violation of ASTM C78 and will invalidate the test.
Locating the Fracture Initiation: The technician must inspect the tension surface of the beam—which is the bottom surface during the loading test—to identify the point where the crack initiated. Because concrete is weak in tension, flexural failure always begins as a tensile crack on the bottom surface and propagates upward toward the top (compression) surface. The location of the fracture is measured along the centerline of the tension face, determining its distance from the nearest support block.
The Three Failure Location Cases
Depending on where the fracture initiates on the tension face relative to the support span, the test is classified into one of three cases. The support span ( L L L ) is divided into three equal segments: two outer-third segments and a middle-third segment. For a standard 18-inch span, the middle-third segment lies between 6.0 and 12.0 inches from either support.
Case 1: Fracture within the Middle Third
If the fracture initiates on the tension face within the middle-third of the span length, the bending moment is constant and at its maximum, while the shear stress is zero. This is the ideal failure mode.
Formula: Use the standard equation: R = P ⋅ L b ⋅ d 2 R = \frac{P \cdot L}{b \cdot d^2} R = b ⋅ d 2 P ⋅ L
Example: On an 18-inch span, a fracture initiating at 7.5 inches from the nearest support is within the middle third (6 to 12 inches). The technician calculates R R R using the standard formula.
Case 2: Fracture Outside the Middle Third by ≤ 5 % \le 5\% ≤ 5% of Span
If the fracture initiates on the tension face outside the middle third of the span length by not more than 5% of the span length, the stress distribution is no longer uniform, and shear forces are present. However, the ASTM C78 standard allows the calculation of the modulus of rupture by incorporating the distance from the fracture line to the nearest support.
5% Span Limit Calculation:
For a standard 18-inch span: 18 ⋅ 0.05 = 0.9 inches 18 \cdot 0.05 = 0.9 \text{ inches} 18 ⋅ 0.05 = 0.9 inches . The boundaries of the middle third are at 6.0 inches and 12.0 inches. Therefore, a failure can occur between 5.1 and 6.0 inches, or between 12.0 and 12.9 inches from a support.
For a 12-inch span: 12 ⋅ 0.05 = 0.6 inches 12 \cdot 0.05 = 0.6 \text{ inches} 12 ⋅ 0.05 = 0.6 inches . The middle third boundaries are at 4.0 and 8.0 inches. A failure can occur between 3.4 and 4.0 inches, or between 8.0 and 8.6 inches.
Formula: Use the modified equation: R = 3 ⋅ P ⋅ a b ⋅ d 2 R = \frac{3 \cdot P \cdot a}{b \cdot d^2} R = b ⋅ d 2 3 ⋅ P ⋅ a
Where a a a is the average distance between the line of fracture and the nearest support, measured on the tension surface of the beam. Note that a a a will always be less than L / 3 L/3 L /3 in this case.
Example Calculation: A concrete beam is tested on an 18-inch span. The average width is 6.00 in., and the average depth is 6.00 in. The beam fails at a load of 4,800 lbf, and the fracture initiates on the bottom surface at a distance of 5.5 inches from the nearest support.
Determine if the test is valid:
The middle-third boundary is at 6.0 inches.
The fracture is at 5.5 inches, which is 6.0 − 5.5 = 0.5 inches 6.0 - 5.5 = 0.5 \text{ inches} 6.0 − 5.5 = 0.5 inches outside the middle third.
The 5% limit for an 18-inch span is 0.9 inches. Since 0.5 in. ≤ \le ≤ 0.9 in., the test is valid under Case 2.
Apply the modified formula: R = 3 ⋅ P ⋅ a b ⋅ d 2 = 3 ⋅ 4 , 800 ⋅ 5.5 6.00 ⋅ 6.00 2 = 79 , 200 216.00 = 366.67 psi R = \frac{3 \cdot P \cdot a}{b \cdot d^2} = \frac{3 \cdot 4,800 \cdot 5.5}{6.00 \cdot 6.00^2} = \frac{79,200}{216.00} = 366.67 \text{ psi} R = b ⋅ d 2 3 ⋅ P ⋅ a = 6.00 ⋅ 6.0 0 2 3 ⋅ 4 , 800 ⋅ 5.5 = 216.00 79 , 200 = 366.67 psi
Round to the nearest 5 psi: 365 psi .
Case 3: Fracture Outside the Middle Third by > 5 % > 5\% > 5% of Span
If the fracture initiates on the tension surface outside the middle third of the span length by more than 5% of the span length, the test results must be discarded (rejected) .
Example Scenario: On an 18-inch span, a fracture initiates 4.8 inches from the nearest support on the bottom face.
The distance from the middle third boundary is 6.0 − 4.8 = 1.2 inches 6.0 - 4.8 = 1.2 \text{ inches} 6.0 − 4.8 = 1.2 inches .
Since 1.2 inches is greater than the 5% span limit of 0.9 inches, the technician must discard the test results and report that the specimen broke outside the acceptable zone.
Why are these tests discarded? Near the support blocks, the concrete is subjected to high shear stresses, local compression from the reaction forces, and complex multi-axial stress states. If a specimen fails this close to the support, it indicates that the failure was not caused by pure bending tension. Instead, it was likely triggered by a localized defect (such as a large aggregate pocket, void, or crack) or severe eccentric loading caused by equipment misalignment. Simple beam theory cannot accurately calculate the modulus of rupture under these conditions, and any calculated strength would be highly inaccurate and misleading.