3.2 Loading Rates & Modulus of Rupture Calculations

3.2 Loading Rates & Modulus of Rupture Calculations

Load Application Rate Requirements

To obtain accurate and reproducible flexural strength measurements, the load must be applied continuously and without shock. The rate of loading must be controlled so that the increase in the extreme fiber tensile stress remains constant. Under ASTM C78, the load must be applied at a rate that increases the extreme fiber stress between 125 and 175 psi/min (0.86 to 1.21 MPa/min) until rupture occurs.

This stress rate is not programmed directly into most testing machines as a stress value; instead, the technician must calculate the equivalent load rate in pounds-force per minute (lbf/min) or Newtons per minute (N/min) based on the specimen's geometry. The formula used to determine the rate of load application ($P_{rate}$) is: $P_{rate} = \\frac{S_{rate} \\cdot b \\cdot d^2}{L}$

Where:

• $P_{rate}$ = rate of load application, lbf/min (or N/min)
• $S_{rate}$ = specified rate of increase in extreme fiber stress, psi/min (or MPa/min)
• $b$ = average width of the specimen, in. (or mm)
• $d$ = average depth of the specimen, in. (or mm)
• $L$ = span length, in. (or mm)

Calculating Loading Rates for Standard Specimens

Let us calculate the acceptable loading rate range for a standard 6 in. x 6 in. (150 mm x 150 mm) concrete beam tested on an 18 in. (450 mm) span.

1. Lower Limit (125 psi/min stress rate): $P_{rate} = \\frac{125 \\cdot 6 \\cdot 6^2}{18} = \\frac{125 \\cdot 6 \\cdot 36}{18} = 125 \\cdot 12 = 1,500 \\text{ lbf/min}$

2. Upper Limit (175 psi/min stress rate): $P_{rate} = \\frac{175 \\cdot 6 \\cdot 6^2}{18} = \\frac{175 \\cdot 6 \\cdot 36}{18} = 175 \\cdot 12 = 2,100 \\text{ lbf/min}$

Therefore, the technician must adjust the testing machine controls to apply the load at a constant rate between 1,500 and 2,100 lbf/min (approximately 6.7 to 9.3 kN/min). For other beam dimensions, such as a 4 in. x 4 in. beam on a 12 in. span, the calculation yields a lower limit of 667 lbf/min and an upper limit of 933 lbf/min.

The Physical Impact of Loading Rates

Controlling the loading rate is critical because concrete exhibits viscoelastic behavior. If the load is applied too rapidly (above 175 psi/min), the concrete specimen does not have sufficient time to distribute the internal stresses or initiate microcracking at the aggregate-paste interfaces. As a result, the beam will withstand a higher maximum load than it would under standard loading, leading to an overestimation of the flexural strength (potentially by 10% or more). Conversely, if the load is applied too slowly (below 125 psi/min), sustained-loading effects and progressive, slow microcracking can accumulate over the extended duration of the test, causing the beam to rupture at a lower maximum load, which underestimates the concrete's strength.

Modulus of Rupture (MOR) Calculation

When the beam ruptures, the maximum applied load ($P$) indicated by the testing machine is recorded. To calculate the modulus of rupture ($R$), the technician must first inspect the fracture surface to verify that the failure initiated on the tension (bottom) face within the middle third of the span.

For a fracture that initiates in the tension surface within the middle third of the span, the modulus of rupture is calculated using the standard formula: $R = \\frac{P \\cdot L}{b \\cdot d^2}$

Where:

• $R$ = modulus of rupture, psi (or MPa)
• $P$ = maximum applied load indicated by the testing machine, lbf (or N)
• $L$ = span length, in. (or mm)
• $b$ = average width of the specimen at the fracture, in. (or mm)
• $d$ = average depth of the specimen at the fracture, in. (or mm)

Note on Dimension Measurements: The width ($b$) and depth ($d$) must be determined by averaging three measurements taken across the fracture cross-section. Because the depth ($d$) is squared in the denominator, any small error in depth measurement will propagate as a squared error in the final strength value. For instance, a minor under-measurement of depth by 0.1 inches on a 6-inch beam will result in a 3.4% overestimation of the calculated modulus of rupture.

Worked Calculation Examples

Example 1: Standard Beam with Nominal Dimensions A concrete beam is tested on an 18.0-inch span. The maximum applied load is 5,400 lbf. The average width at the fracture is 6.00 inches, and the average depth is 6.00 inches. $R = \\frac{5,400 \\cdot 18.0}{6.00 \\cdot 6.00^2} = \\frac{97,200}{6.00 \\cdot 36.00} = \\frac{97,200}{216.00} = 450.0 \\text{ psi}$

Example 2: Specimen with Measured Irregular Dimensions A concrete beam is tested on an 18.0-inch span. The maximum applied load is 5,680 lbf. The average width ($b$) at the fracture is measured as 6.04 inches, and the average depth ($d$) is measured as 5.96 inches.

1. Calculate $d^2$: $5.96^2 = 35.5216 \text{ in}^2$
2. Calculate the denominator ($b \cdot d^2$): $6.04 \cdot 35.5216 = 214.550 \text{ in}^3$
3. Calculate $R$: $R = \\frac{5,680 \\cdot 18.0}{214.550} = \\frac{102,240}{214.550} = 476.53 \\text{ psi}$
4. Rounding Rule: ASTM C78 requires rounding the final modulus of rupture to the nearest 5 psi (or 0.05 MPa for metric calculations).
   • 476.53 psi rounded to the nearest 5 psi is 475 psi.

Example 3: Metric Calculation A beam is tested on a 450 mm span. The maximum load ($P$) is 25,400 N. The average width is 151 mm, and the average depth is 149 mm.

1. Calculate $d^2$: $149^2 = 22,201 \text{ mm}^2$
2. Calculate denominator: $151 \cdot 22,201 = 3,352,351 \text{ mm}^3$
3. Calculate $R$ in MPa (since $1 \text{ N/mm}^2 = 1 \text{ MPa}$): $R = \\frac{25,400 \\cdot 450}{3,352,351} = \\frac{11,430,000}{3,352,351} = 3.41 \\text{ MPa}$
4. Rounding to the nearest 0.05 MPa gives 3.40 MPa.