Chapter 2, Problem 2.1CTP

Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples for laboratory testing from a recycled aggregate processing plant in Palm Beach County, Florida. Three samples, denoted by Soil A, Soil B, and Soil C, were collected from three locations of the aggregate stockpile, and sieve analyses were conducted (see Figure 2.35).

(a)

(b)

Figure 2.35 (a) Soil-aggregate stockpile; (b) sieve analysis (Courtesy of Khaled Sobhan, Florida Atlantic University, Boca Raton, Florida)

a. Determine the coefficient of uniformity and the coefficient of gradation for Soils A, B, and C.

b. Which one is coarser: Soil A or Soil C? Justify your answer.

c. Although the soils are obtained from the same stockpile, why are the curves so different? (Hint: Comment on particle segregation and representative field sampling.)

d. Determine the percentages of gravel, sand and fines according to Unified Soil Classification System.

To determine

Calculate the coefficient of uniformity $\left(C_u\right)$ and coefficient of gradation $\left(C_c\right)$ for soils A, B, and C.

Answer to Problem 2.1CTP

The uniformity coefficient of soil A is $\underset{\_}{26.42}$.

The coefficient of gradation of soil A is $\underset{\_}{3.64}$.

The uniformity coefficient of soil B is $\underset{\_}{35}$.

The coefficient of gradation of soil B is $\underset{\_}{3.35}$.

The uniformity coefficient of soil C is $\underset{\_}{46.67}$.

The coefficient of gradation of soil C is $\underset{\_}{1.15}$.

Explanation of Solution

Sketch the grain size distribution curve for soils A, B, and C as shown in Figure 1.

Refer to Figure 1.

For soil A:

The diameter of the particle corresponding to $60\%$ finer $\left(D_{60}\right)$ is $14\text{mm}$.

The diameter of the particle corresponding to $30\%$ finer $\left(D_{30}\right)$ is $5.2\text{mm}$.

The diameter of the particle corresponding to $10\%$ finer $\left(D_{10}\right)$ is $0.53\text{mm}$.

For soil B:

The diameter of the particle corresponding to $60\%$ finer $\left(D_{60}\right)$ is $8.1\text{mm}$.

The diameter of the particle corresponding to $30\%$ finer $\left(D_{30}\right)$ is $2.5\text{mm}$.

The diameter of the particle corresponding to $10\%$ finer $\left(D_{10}\right)$ is $0.23\text{mm}$.

For soil C:

The diameter of the particle corresponding to $60\%$ finer $\left(D_{60}\right)$ is $7\text{mm}$.

The diameter of the particle corresponding to $30\%$ finer $\left(D_{30}\right)$ is $1.1\text{mm}$.

The diameter of the particle corresponding to $10\%$ finer $\left(D_{10}\right)$ is $0.15\text{mm}$.

Calculate the uniformity coefficient $\left(C_u\right)$ using the relation.

$C_u=\frac{D_{60}}{D_{10}}$ (1)

For soil A:

Substitute $14\text{mm}$ for $D_{60}$ and $0.53\text{mm}$ for $D_{10}$ in Equation (1).

$\begin{array}{c}{C}_u=\frac{14}{0.53}\\ =26.42\end{array}$

Hence, the uniformity coefficient for soil A is $\underset{\_}{26.42}$.

For soil B:

Substitute $8.1\text{mm}$ for $D_{60}$ and $0.23\text{mm}$ for $D_{10}$ in Equation (1).

$\begin{array}{c}{C}_u=\frac{8.1}{0.23}\\ =35\end{array}$

Hence, the uniformity coefficient for soil B is $\underset{\_}{35}$.

For soil C:

Substitute $7\text{mm}$ for $D_{60}$ and $0.15\text{mm}$ for $D_{10}$ in Equation (1).

$\begin{array}{c}{C}_u=\frac{7}{0.15}\\ =46.67\end{array}$

Hence, the uniformity coefficient for soil C is $\underset{\_}{46.67}$.

Calculate the coefficient of gradation $\left(C_c\right)$ using the relation.

$C_c=\frac{D_{30}^2}{D_{60}\times D_{10}}$ (2)

For soil A:

Substitute $14\text{mm}$ for $D_{60}$, $5.2\text{mm}$ for $D_{30}$, and $0.53\text{mm}$ for $D_{10}$ in Equation (2).

$\begin{array}{c}{C}_c=\frac{{5.2}^2}{14\times 0.53}\\ =\frac{27.04}{5.512}\\ =3.64\end{array}$

Hence, the coefficient of gradation for soil A is $\underset{\_}{3.64}$.

For soil B:

Substitute $8.1\text{mm}$ for $D_{60}$, $2.5\text{mm}$ for $D_{30}$, and $0.23\text{mm}$ for $D_{10}$ in Equation (2).

$\begin{array}{c}{C}_c=\frac{{2.5}^2}{8.1\times 0.23}\\ =\frac{6.25}{1.863}\\ =3.35\end{array}$

Hence, the coefficient of gradation for soil B is $\underset{\_}{3.35}$.

For soil C:

Substitute $7\text{mm}$ for $D_{60}$, $1.1\text{mm}$ for $D_{30}$, and $0.15\text{mm}$ for $D_{10}$ in Equation (2).

$\begin{array}{c}{C}_c=\frac{{1.1}^2}{7\times 0.15}\\ =\frac{1.21}{1.05}\\ =1.15\end{array}$

Therefore, the coefficient of gradation for soil C is $\underset{\_}{1.15}$.

To determine

State which of the soil is coarser from soil A and C.

Answer to Problem 2.1CTP

Soil A is coarser than soil C.

Explanation of Solution

Refer to part (a).

The uniformity coefficient of soil A is $26.42$.

The uniformity coefficient of soil C is $46.67$.

The percent of soil finer than $1\text{mm}$ for soil A is $20\%$.

The percent of soil finer than $1\text{mm}$ for soil C is $47\%$.

Hence, a higher percentage of soil C is finer than soil A.

Hence, soil A is coarser than soil C.

To determine

Explain the reason for curve different of soil A, B and C if it is obtained from same stockpile.

Explanation of Solution

The particle-size distribution curve shows the range of particle sizes present in a soil and the type of distribution of various-size particles.

Refer to Figure 1.

Particle separation of coarser and finer particles may take place in aggregate stockpiles. This makes representative sampling difficult.

Therefore, the particle-size distribution curve is different for soils A, B, and C.

To determine

Calculate the percentages of gravel, sand, and fines according to the Unified Soil Classification System.

Answer to Problem 2.1CTP

The percentage of gravel for soil A is $\underset{\_}{71\%}$.

The percentage of sand for soil A is $\underset{\_}{28\%}$.

The percentage of fines for soil A is $\underset{\_}{1\%}$.

The percentage of gravel for soil B is $\underset{\_}{55\%}$.

The percentage of sand for soil B is $\underset{\_}{43\%}$.

The percentage of fines for soil B is $\underset{\_}{2\%}$.

The percentage of gravel for soil C is $\underset{\_}{47\%}$.

The percentage of sand for soil C is $\underset{\_}{50\%}$.

The percentage of fines for soil C is $\underset{\_}{3\%}$.

Explanation of Solution

Refer to Figure 1.

For soil A.

The percent passing through $4.75\text{mm}$ sieve is $29\%$.

The percent passing through $0.075\text{mm}$ sieve is $1\%$.

Calculate the percentage of gravel as shown below.

$\begin{array}{c}\text{Percent}\text{of}\text{gravel}=100-\text{percent passing through 4}\text{.75}\text{mm}\text{sieve}\\ =100-29\\ =71\%\end{array}$

Hence, the percentage of gravel is $\underset{\_}{71\%}$.

Calculate the percentage of sand as shown below.

$\begin{array}{c}\text{Percent}\text{of}\text{sand}=\left(\begin{array}{l}\text{percent passing }\\ \text{through 4}\text{.75}\text{mm}\text{sieve}\end{array}\right)-\left(\begin{array}{l}\text{percent passing }\\ \text{through 0}\text{.075}\text{mm}\text{sieve}\end{array}\right)\\ =29-1\\ =28\%\end{array}$

Hence, the percentage of sand is $\underset{\_}{28\%}$.

Calculate the percentage of fines as shown below.

$\begin{array}{c}\text{Percent}\text{of}\text{fines}=\text{percent passing through 0}\text{.075}\text{mm}\text{sieve}-0\\ =1-0\\ =1\%\end{array}$

Hence, the percentage of fines is $\underset{\_}{1\%}$.

Refer to Figure 1.

For soil B.

The percent passing through $4.75\text{mm}$ sieve is $45\%$.

The percent passing through $0.075\text{mm}$ sieve is $2\%$.

Calculate the percentage of gravel as shown below.

$\begin{array}{c}\text{Percent}\text{of}\text{gravel}=100-\text{percent passing through 4}\text{.75}\text{mm}\text{sieve}\\ =100-45\\ =55\%\end{array}$

Hence, the percentage of gravel is $\underset{\_}{55\%}$.

Calculate the percentage of sand as shown below.

$\begin{array}{c}\text{Percent}\text{of}\text{sand}=\left(\begin{array}{l}\text{percent passing }\\ \text{through 4}\text{.75}\text{mm}\text{sieve}\end{array}\right)-\left(\begin{array}{l}\text{percent passing }\\ \text{through 0}\text{.075}\text{mm}\text{sieve}\end{array}\right)\\ =45-2\\ =43\%\end{array}$

Hence, the percentage of sand is $\underset{\_}{43\%}$.

Calculate the percentage of fines as shown below.

$\begin{array}{c}\text{Percent}\text{of}\text{fines}=\text{percent passing through 0}\text{.075}\text{mm}\text{sieve}-0\\ =2-0\\ =2\%\end{array}$

Hence, the percentage of fines is $\underset{\_}{2\%}$.

Refer to Figure 1.

For soil C.

The percent passing through $4.75\text{mm}$ sieve is $53\%$.

The percent passing through $0.075\text{mm}$ sieve is $3\%$.

Calculate the percentage of gravel as shown below.

$\begin{array}{c}\text{Percent}\text{of}\text{gravel}=100-\text{percent passing through 4}\text{.75}\text{mm}\text{sieve}\\ =100-53\\ =47\%\end{array}$

Hence, the percentage of gravel is $\underset{\_}{47\%}$.

Calculate the percentage of sand as shown below.

$\begin{array}{c}\text{Percent}\text{of}\text{sand}=\left(\begin{array}{l}\text{percent passing }\\ \text{through 4}\text{.75}\text{mm}\text{sieve}\end{array}\right)-\left(\begin{array}{l}\text{percent passing }\\ \text{through 0}\text{.075}\text{mm}\text{sieve}\end{array}\right)\\ =53-3\\ =50\%\end{array}$

Hence, the percentage of sand is $\underset{\_}{50\%}$.

Calculate the percentage of fines as shown below.

$\begin{array}{c}\text{Percent}\text{of}\text{fines}=\text{percent passing through 0}\text{.075}\text{mm}\text{sieve}-0\\ =3-0\\ =3\%\end{array}$

Hence, the percentage of fines is $\underset{\_}{3\%}$.