Chapter 2, Problem 2.15CTP

In concrete work, Fuller and Thompson (1907) suggested that a dense packing of grains can be achieved if the percent finer (p) and grain size (D) are related by the following equation, where n is a constant varying in the range of 0.3-0.6. D_max is the size of the largest grain within the soil.

$p={\left(\frac{D}{D_{\max }}\right)}^n\times 100$

This equation is sometimes used in roadwork for selecting the aggregates.

1. a. For n = 0.5, show that the soil is well graded.
2. b. If n = 0.5 and D_max = 19.0 mm, find the percentages of gravel, sand, and fines within the soil. Use the Unified Soil Classification System.

To determine

Show that the soil is well graded for constant as 0.5.

Answer to Problem 2.15CTP

The soil is well graded for $n=0.5$.

Explanation of Solution

Given information:

The range of the constant $\left(n\right)$ is $0.3-0.6$.

The constant value $\left(n\right)$ is $0.5$.

Calculation:

Show the equation of percent finer as follows.

$p={\left(\frac{D}{D_{\max }}\right)}^n\times 100$ (1)

Here, D is the grain size and $D_{\max }$ is the size of the largest grain within the soil.

For $10\%$ finer:

Substitute $30\%$ for p and $0.5$ for n in Equation (1).

$10={\left(\frac{D_{10}}{D_{\max }}\right)}^{0.5}\times 100$ (2)

For $30\%$ finer:

Substitute $30\%$ for p and $0.5$ for n in Equation (1).

$30={\left(\frac{D_{30}}{D_{\max }}\right)}^{0.5}\times 100$ (3)

For $60\%$ finer:

Substitute $60\%$ for p and $0.5$ for n in Equation (1).

$60={\left(\frac{D_{60}}{D_{\max }}\right)}^{0.5}\times 100$ (4)

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

Divide Equation (4) by Equation (2).

$\begin{array}{l}\frac{60}{10}=\frac{{\left(\frac{D_{60}}{D_{\max }}\right)}^{0.5}\times 100}{{\left(\frac{D_{10}}{D_{\max }}\right)}^{0.5}\times 100}\\ 6={\left(\frac{D_{60}}{D_{10}}\right)}^{0.5}\\ \frac{D_{60}}{D_{10}}=36\\ C_u=36\end{array}$

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

Apply $\frac{{\left(\text{Equation}\left(\text{3}\right)\right)}^2}{\text{Equation}\left(2\right)\times \text{Equation}\left(4\right)}$.

$\begin{array}{l}\frac{{30}^2}{60\times 10}=\frac{{\left({\left(\frac{D_{30}}{D_{\max }}\right)}^{0.5}\times 100\right)}^2}{{\left(\frac{D_{60}}{D_{\max }}\right)}^{0.5}\times 100\times {\left(\frac{D_{10}}{D_{\max }}\right)}^{0.5}\times 100}\\ \frac{9}{6}=\frac{D_{30}}{D_{60}^{0.5}\times D_{10}^{0.5}}\\ 1.5=\frac{D_{30}}{D_{60}^{0.5}\times D_{10}^{0.5}}\\ \frac{D_{30}^2}{D_{60}\times D_{10}}=2.25\end{array}$

$C_c=2.25$

The value of uniformity coefficient is greater than 6 $\left(C_u>6\right)$ and the coefficient of gradation $\left(C_c=2.25\right)$ lies between 1 and 3. Hence, the soil is termed as well graded soil.

Hence, the soil is well graded.

To determine

Calculate the percentages of gravels, sands, and fines in the soils based on the Unified Soil Classification System.

Answer to Problem 2.15CTP

The soil contains $\underset{\_}{\text{43}\text{.7\%}\text{of}\text{gravel,}50\text{\%}\text{of}\text{sand,}\text{and}6.3\text{\%}\text{of}\text{fines}}$.

Explanation of Solution

Given information:

The range of the constant $\left(n\right)$ is $0.3-0.6$.

The constant value $\left(n\right)$ is $0.5$.

The size of the largest grain within the soil $\left(D_{\max }\right)$ is $19\text{mm}$.

Calculation:

For percentage of fines the grain size $\left(D\right)$ is $0.075\text{mm}$.

For percentage of sand and fines the grain size $\left(D\right)$ is $4.75\text{mm}$.

Calculate the percentage of fines in the soil.

Substitute $0.075\text{mm}$ for D, $19\text{mm}$ for $D_{\max }$, and $0.5$ for n in Equation (1).

$\begin{array}{c}p={\left(\frac{0.075}{19}\right)}^{0.5}\times 100\\ =0.0628\times 100\\ =6.3\%\end{array}$

Hence, the percentage of fines is $\underset{\_}{6.3\%}$.

Calculate the percentage of sand and fines in the soil.

Substitute $4.75\text{mm}$ for D, $19\text{mm}$ for $D_{\max }$, and $0.5$ for n in Equation (1).

$\begin{array}{c}p={\left(\frac{4.75}{19}\right)}^{0.5}\times 100\\ =0.5\times 100\\ =50\%\end{array}$

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

Calculate the percentage of gravel in the soil.

$\begin{array}{c}\text{Gravel}=100-\%\text{of}\text{sand}\text{and}\text{fines}-\%\text{of}\text{fines}\\ =100-50-6.3\\ =43.7\%\end{array}$

Therefore, the percentage of gravel is $\underset{\_}{43.7\%}$.