Chapter 17, Problem 17.1P

To determine

Find the area ratio for each case and appropriate sample for the given soil characterization tests.

Answer to Problem 17.1P

The area ratio for tube 1 is $\underset{\_}{5.06\%}$.

The area ratio for tube 2 is $\underset{\_}{4.47\%}$.

The area ratio for tube 3 is $\underset{\_}{4.98\%}$.

Explanation of Solution

Given information:

The outside diameter $\left(D_o\right)$ for sample 1 is $50.8\text{mm}$.

The outside diameter $\left(D_o\right)$ for sample 2 is $76.2\text{mm}$.

The outside diameter $\left(D_o\right)$ for sample 3 is $127\text{mm}$.

The wall thickness (t) for sample 1 is $1.24\text{mm}$.

The wall thickness (t) for sample 2 is $1.65\text{mm}$.

The wall thickness (t) for sample 3 is $3.05\text{mm}$.

Calculation:

Calculate the inside diameter $\left(D_i\right)$ for tube 1 using the relation as follows:

$\left(D_i\right)=D_o-t$ (1)

Substitute $50.8\text{mm}$ for $\left(D_o\right)$ and $1.24\text{mm}$ for t in Equation (1).

$\begin{array}{c}\left(D_i\right)=50.8-1.24\\ =49.56\text{mm}\end{array}$

Find the area ratio for $\left(A_r\right)$ tube 1 using the Equation as follows:

$A_r\left(\%\right)=\frac{D_o^2-D_i^2}{D_i^2}\times 100$ (2)

Here, $D_i$ inside diameter of sample 1.

Substitute $50.8\text{mm}$ for $D_o$ and $49.56\text{mm}$ for $D_i$ in Equation (2).

$\begin{array}{c}{A}_r\left(\%\right)=\frac{{50.8}^2-{49.56}^2}{{49.56}^2}\times 100\\ =\frac{2,580.64-2,456.19}{2,456.19}\times 100\\ =\frac{124.44}{2,456.19}\times 100\\ =5.06\%\le 10\%\end{array}$

Thus, the area ratio for tube 1 is $\underset{\_}{5.06\%}$.

Calculate the inside diameter $\left(D_i\right)$ for tube 2 using the relation as follows:

Substitute $76.2\text{mm}$ for $\left(D_o\right)$ and $1.65\text{mm}$ for t in Equation (1).

$\begin{array}{c}\left(D_i\right)=76.2-1.65\\ =74.55\text{mm}\end{array}$

Find the area ratio for $\left(A_r\right)$ tube 2 using the Equation as follows:

Substitute $76.2\text{mm}$ for $D_o$ and $74.55\text{mm}$ for $D_i$ in Equation (2).

$\begin{array}{c}{A}_r\left(\%\right)=\frac{{76.2}^2-{74.55}^2}{{74.55}^2}\times 100\\ =\frac{5,806.44-5,557.7}{5,557.7}\times 100\\ =\frac{248.74}{5,557.7}\times 100\\ =4.47\%\le 10\%\end{array}$

Thus, the area ratio for tube 2 is $\underset{\_}{4.47\%}$.

Calculate the inside diameter $\left(D_i\right)$ for tube 3 using the relation as follows:

Substitute $127\text{mm}$ for $\left(D_o\right)$ and $3.05\text{mm}$ for t in Equation (1).

$\begin{array}{c}\left(D_i\right)=127-3.05\\ =123.95\text{mm}\end{array}$

Find the area ratio for $\left(A_r\right)$ tube 3 using the Equation as follows:

Substitute $127\text{mm}$ for $D_o$ and $123.95\text{mm}$ for $D_i$ in Equation (2).

$\begin{array}{c}{A}_r\left(\%\right)=\frac{{127}^2-{123.95}^2}{{123.95}^2}\times 100\\ =\frac{16,129-15,363.60}{15,363.60}\times 100\\ =4.98\%\le 10\%\end{array}$

Thus, the area ratio for tube 3 is $\underset{\_}{4.98\%}$.

A soil sample usually can be considered undisturbed if the area ratio is less than or equal to 10%.

• All the given three samples are considered as undisturbed sample, hence, it is suitable for consolidation and unconfined compression test.
• The Grain size distribution, atterberg limits are conducted only when the soil on remolded or disturbed soil, hence, sample disturbance is not a factor.

Thus, the all three samples are appropriate for all tests.