Let's break down the problem step-by-step: --- # Step 1: Understand the given data - **Design force (Fcr):** 5,700 psi - **Design strength (F'c):** 24.1 MPa (which is 350 psi) - **Standard deviation of compressive strength (σ):** 2.4 MPa (which is 350 psi) - **Sample size (n):** more than 30 samples (assumed large enough for normal distribution approximation) - **Only air entrainer is allowed** --- # Step 2: Convert all units to the same system The design strength and standard deviation are given in MPa; convert them to psi: \[ 1 \text{ MPa} = 145.038 \text{ psi} \] - \( F'c = 24.1 \text{ MPa} \times 145.038 \approx 350 \text{ psi} \) - \( \sigma = 2.4 \text{ MPa} \times 145.038 \approx 348.1 \text{ psi} \) *(which is close to the given 350 psi, so we can use 350 psi for simplicity)* --- # Step 3: Find the strength reduction factor \( \phi \) The strength requirement (design strength \( F_{cr} \)) is related to the characteristic strength \( f_c \) (mean strength), the standard deviation, and the safety factor. Since the problem involves probabilistic considerations, the W/C ratio (water/cement ratio) impacts the mean strength \( f_c \), which is related to the standard deviation. --- # Step 4: Determine the relationship between W/C ratio and strength In concrete design, the strength \( f_c \) is typically related to the W/C ratio via empirical or code-based tables. The goal is to select a W/C ratio that achieves the required strength with sufficient safety. --- # Step 5: Use statistical basis for W/C ratio Given that the strength distribution is normal, the **proof load** corresponds to a certain percentile of the strength distribution. For a large sample, the probability of failure is minimized by selecting a W/C ratio that ensures the mean strength \( \mu \) exceeds the required strength \( F_{cr} \) by a certain number of standard deviations. The basic relation: \[ F_{cr} = \mu - k \sigma \] where \( k \) is the number of standard deviations corresponding to the desired confidence level. --- # Step 6: Find the appropriate \( k \) value The **W/C ratio** is typically controlled to achieve a certain mean strength \( \mu \). The W/C ratio for the concrete mix can be estimated from standard tables in the ACI workbook, correlating W/C ratio with mean strength. Given the data, the problem asks for the best **W/C ratio** based on the safety margin, which is related to the standard deviation. --- # Step 7: Calculate the safety margin Because only air entrainer is allowed, and considering the typical safety factor, the **W/C ratio** is related to the probability of failure, which is tied to the standard normal distribution: \[ \text{Probability} = P(Z \leq -k) \] Given the options, typical values of \(k\) relate to: | Probability | \(k\) (approximate) | |--------------|---------------------| | .34 | .41 | | .29 | .55 | | .45 | .12 | | .54 | .10 | But more precisely, the standard deviation and safety factor suggest a value around **.29** or **.34**. --- # **Final step: Choose the W/C ratio** Based on typical empirical data, a W/C ratio that corresponds to a strength safety margin near .29 (which indicates an acceptable safety margin for concrete strength) is selected. **Therefore, the best W/C ratio is approximately:** # **Answer: .29** --- # **Final Answer: .29**