Chapter 14, Problem 14.1P

State whether the following are true or false.

1. a. The higher the friction angle, the higher the value of K₀.
2. b. K₀ is greater for normally consolidated clays than overconsolidated clays.
3. c. Active earth pressure coefficient is greater than the passive one.
4. d. The larger the cohesion, the larger is the depth of the tensile cracks in clays in active state.
5. e. Lateral earth pressures increase linearly with depth.

To determine

Whether the statement “The higher the friction angle, the higher the value of $K_0$” is true or false.

Answer to Problem 14.1P

The given statement is $\underset{\_}{\text{False}}$.

Explanation of Solution

Write the equation of coefficient of earth pressure at rest $\left(K_0\right)$.

$K_o=1-\sin {\varphi }^{\prime }$ (1)

Here, ${\varphi }^{\prime }$ is drained friction angle.

Find the coefficient of earth pressure at rest when ${\varphi }^{\prime }$ is $30°$.

Substitute $30°$ for ${\varphi }^{\prime }$ in Equation (1).

$\begin{array}{c}{K}_o=1-\sin \left(30°\right)\\ =0.5\end{array}$

Find the coefficient of earth pressure at rest when ${\varphi }^{\prime }$ is $60°$.

Substitute $60°$ for ${\varphi }^{\prime }$ in Equation (1).

$\begin{array}{c}{K}_o=1-\sin \left(60°\right)\\ =0.134\end{array}$

Tabulate the values of coefficient of earth pressure at rest with two different drained friction angles of $30°$ and $60°$ as shown in Table 1.

Drained friction angle, ${\varphi }^{\prime }$	Coefficient of earth pressure at rest, $K_o$
$30°$	0.5
$60°$	0.134

Table 1

Refer Table 1.

The coefficient earth pressure at rest $\left(K_0\right)$ decreases with increase of drained friction angle $\left({\varphi }^{\prime }\right)$.

Therefore, the given statement is $\underset{\_}{\text{False}}$.

To determine

Whether the statement “$K_0$ is greater for normally consolidated clays than over consolidated

clays” is true or false.

Answer to Problem 14.1P

The given statement is $\underset{\_}{\text{False}}$.

Explanation of Solution

The value of over consolidated ratio (OCR) is a whole number.

Write the relationship between earth pressure at rest for over consolidated clay $\left[K_{o\left(\text{Over}\text{consolidated}\right)}\right]$ and normally consolidated clay $\left[K_{o\left(\text{Normally}\text{consolidated}\right)}\right]$.

$K_{o\left(\text{Over}\text{consolidated}\right)}=K_{o\left(\text{Normally}\text{consolidated}\right)}\sqrt{OCR}$

The multiplication of square root of over consolidated ratio (OCR) with earth pressure at rest for normal consolidated clay gives the earth pressure at rest for over consolidated clay. Hence, the value earth pressure at rest for over consolidated clay is greater than the earth pressure at rest for normal consolidated clay.

Therefore, the given statement is $\underset{\_}{\text{False}}$.

To determine

Whether the statement “Active earth pressure coefficient is greater than the passive one” is true or false.

Answer to Problem 14.1P

The given statement is $\underset{\_}{\text{False}}$.

Explanation of Solution

Write the equation of active earth pressure coefficient $\left(K_a\right)$.

$K_a={\tan }^2\left(45-\frac{{\varphi }^{\prime }}{2}\right)$ (2)

Here, ${\varphi }^{\prime }$ is the friction angle.

Find the active earth pressure coefficient $\left(K_a\right)$ when ${\varphi }^{\prime }$ is $30°$.

Substitute $30°$ for ${\varphi }^{\prime }$ in Equation (2).

$\begin{array}{c}{K}_a={\tan }^2\left(45-\frac{30°}{2}\right)\\ =0.333\end{array}$

Write the equation of passive earth pressure coefficient $\left(K_p\right)$.

$K_p={\tan }^2\left(45+\frac{{\varphi }^{\prime }}{2}\right)$ (3)

Substitute $30°$ for ${\varphi }^{\prime }$ in Equation (3).

$\begin{array}{c}{K}_p={\tan }^2\left(45+\frac{30}{2}\right)\\ =3\end{array}$

Tabulate the values of active and passive earth pressure coefficients with corresponding friction angle of $30°$ as shown in Table 1.

Friction angle, ${\varphi }^{\prime }$	Active earth pressure coefficient, $K_a$	Passive earth pressure coefficient, $K_p$
$30°$	0.333	3

Refer Table 1.

The active earth pressure coefficient $\left(K_a\right)$ is lesser than the passive earth pressure coefficient $\left(K_p\right)$ for same friction angle $\left({\varphi }^{\prime }\right)$ of $30°$.

Therefore, the given statement is $\underset{\_}{\text{False}}$.

To determine

Whether the statement “The larger the cohesion, the larger is the depth of the tensile cracks in

clays in active state” is true or false.

Answer to Problem 14.1P

The given statement is $\underset{\_}{\text{True}}$.

Explanation of Solution

Write the equation of depth of tensile crack $\left(z_o\right)$.

$z_o=\frac{2c^{\prime }}{\gamma \sqrt{K_a}}$ (4)

Here, $c^{\prime }$ is the cohesion, $\gamma$ is the unit weight of the soil, and $K_a$ is the active earth pressure coefficient.

From Equation (4), cohesion $\left(c^{\prime }\right)$ is directly proportional to the depth of tensile crack $\left(z_o\right)$. Hence, the larger value of cohesion provides larger value of depth of the tensile cracks in clays in active state

Therefore, the given statement is $\underset{\_}{\text{True}}$.

To determine

Whether the statement “Lateral earth pressures increase linearly with depth” is true or false.

Answer to Problem 14.1P

The given statement is $\underset{\_}{\text{True}}$.

Explanation of Solution

Write the equation of lateral earth pressure $\left({{\sigma }^{\prime }}_h\right)$.

${{\sigma }^{\prime }}_h=K_o\gamma z$

Here, $K_o$ is the earth pressure coefficient at rest, $\gamma$ is the unit weight of the soil, and z is the depth.

The lateral earth pressure equation is a linear equation and the lateral earth pressure $\left({{\sigma }^{\prime }}_h\right)$ is directly proportional to depth (z). Hence, the increase in depth increases the lateral earth pressure linearly.

Therefore, the given statement is $\underset{\_}{\text{True}}$.