Let's solve each part step-by-step: --- ## **a. Determine the percent finer than (i.e. % passing) each sieve** ### **Step 1: Prepare the Data Table** | Sieve Size (mm) | Mass Retained (g) | |-----------------|-------------------| | 4.75 | . | | 3.35 | 30. | | 2.00 | 48.7 | | .850 | 127.3 | | .425 | 96.8 | | .250 | 76.6 | | .150 | 55.2 | | .075 | 43.4 | | Pan | 22. | ### **Step 2: Calculate Cumulative Mass Retained** Add each mass from the top down: | Sieve Size (mm) | Mass Retained (g) | Cumulative Mass Retained (g) | |-----------------|-------------------|------------------------------| | 4.75 | . | . | | 3.35 | 30. | 30. | | 2.00 | 48.7 | 78.7 | | .850 | 127.3 | 206. | | .425 | 96.8 | 302.8 | | .250 | 76.6 | 379.4 | | .150 | 55.2 | 434.6 | | .075 | 43.4 | 478. | | Pan | 22. | 500. | ### **Step 3: Calculate Percent Passing Each Sieve** Total mass = 500. g \[ \text{Percent passing} = \left( \frac{\text{Total mass} - \text{Cumulative mass retained}}{\text{Total mass}} \right) \times 100 \] | Sieve Size (mm) | Cumulative Mass Retained (g) | Percent Passing (%) | |-----------------|------------------------------|---------------------| | 4.75 | . | 100. | | 3.35 | 30. | 94. | | 2.00 | 78.7 | 84.26 | | .850 | 206. | 58.8 | | .425 | 302.8 | 39.44 | | .250 | 379.4 | 24.12 | | .150 | 434.6 | 13.08 | | .075 | 478. | 4.4 | | Pan | 500. | | ### **Step 4: Plot Grain Size Distribution Curve** - **X-axis**: Sieve Size (log scale, mm) - **Y-axis**: Percent Passing (%) - **Plot**: Points from table above, connect smoothly. --- ## **b. Determine \( D_{10}, D_{30}, D_{60} \) from the Curve** - \( D_{10} \): Size at 10% passing - \( D_{30} \): Size at 30% passing - \( D_{60} \): Size at 60% passing **Interpolate between points using the data above**: #### **Between .150 mm (13.08%) and .250 mm (24.12%)** \[ \text{For } D_{10}: \] \[ D_{10} = .150 + \left(\frac{10 - 13.08}{24.12 - 13.08}\right) \times (.250 - .150) \] \[ = .150 + \left(\frac{-3.08}{11.04}\right) \times .10 \] \[ = .150 - .0279 = .122~\text{mm} \] #### **For \( D_{30} \): Between .250 mm (24.12%) and .425 mm (39.44%)** \[ D_{30} = .250 + \left(\frac{30 - 24.12}{39.44 - 24.12}\right) \times (.425 - .250) \] \[ = .250 + \left(\frac{5.88}{15.32}\right) \times .175 \] \[ = .250 + .0672 = .317~\text{mm} \] #### **For \( D_{60} \): Between .850 mm (58.8%) and 2.00 mm (84.26%)** \[ D_{60} = .850 + \left(\frac{60 - 58.8}{84.26 - 58.8}\right) \times (2.00 - .850) \] \[ = .850 + \left(\frac{1.2}{25.46}\right) \times 1.15 \] \[ = .850 + .0542 = .904~\text{mm} \] --- ## **c. Calculate the Uniformity Coefficient, \( C_u \)** \[ C_u = \frac{D_{60}}{D_{10}} \] \[ C_u = \frac{.904}{.122} = 7.41 \] --- ## **d. Calculate the Coefficient of Curvature, \( C_c \)** \[ C_c = \frac{(D_{30})^2}{D_{10} \times D_{60}} \] \[ C_c = \frac{(.317)^2}{.122 \times .904} \] \[ = \frac{.100}{.110} = .91 \] --- ## **Final Answers Summary** | Parameter | Value | |-----------------------------|------------| | \( D_{10} \) | .122 mm | | \( D_{30} \) | .317 mm | | \( D_{60} \) | .904 mm | | Uniformity Coefficient \( C_u \) | 7.41 | | Coefficient of Curvature \( C_c \) | .91 | --- ### **Grain Size Distribution Curve** *(Plot as per data above: X-axis, log sieve size; Y-axis, percent passing)* --- Let me know if you need the plot or further explanation!