Chapter 11, Problem 11.1P

To determine

Draw the plot for ${\gamma }_d\left(\text{lb}/{\text{ft}}^{\text{3}}\right)$ versus the liquid limit and identify the zone where the soil is collapsing on saturation.

Explanation of Solution

Given information:

The specific gravity of the loessial soil is $G_s=2.72$.

The liquid limit of the soil is $LL=24$.

The dry density of the soil is ${\gamma }_d=92.2\text{lb}/{\text{ft}}^{\text{3}}$.

Calculation:

Refer to the Equation (11.5) in the textbook.

Consider the density of water is ${\gamma }_w=62.4\text{lb}/{\text{ft}}^{\text{3}}$.

Consider the liquid limit of 10.

Find the natural dry unit weight $\left({\gamma }_d\right)$ using the relation.

${\gamma }_d=\frac{G_s{\gamma }_w}{1+\frac{LL}{100}{G}_s}$

Substitute $G_s=2.72$, ${\gamma }_w=62.4\text{lb}/{\text{ft}}^{\text{3}}$, and $LL=10$.

$\begin{array}{c}{\gamma }_d=\frac{2.72\times 62.4}{1+\frac{10}{100}\times 2.72}\\ =133.434\text{lb}/{\text{ft}}^{\text{3}}\end{array}$

Tabulate the remaining calculated values as in Table 1.

LL	${\gamma }_d\left(\text{lb}/{\text{ft}}^{\text{3}}\right)$
10	133.434
15	120.545
20	109.927
25	101.029
30	93.4626
35	86.9508
40	81.2874

Table 1

Draw the graph between liquid limit versus the natural dry unit weight as in Figure 1.

Mark the liquid limit of 24 corresponding to the dry unit weight of $92.2\text{lb}/{\text{ft}}^{\text{3}}$. The point lies below the curve of the liquid limit.

Thus, collapse will occur.