Chapter 9, Problem 9.19CTP

To determine

Plot the consolidation settlement versus time for the clay layer.

Explanation of Solution

Given information:

The thickness of the clay layer $\left(H\right)$ is $6\text{m}$.

The time $\left(t\right)$ is $12\text{months}$.

The final consolidation settlement $\left(S_p\right)$ is $220\text{mm}$.

The coefficient of consolidation of the clay $\left(c_v\right)$ is $3{\text{m}}^2/\text{year}$.

Calculation:

Consider the unit weight of water $\left({\gamma }_w\right)$ is $9.81\text{kN}/{\text{m}}^3$.

For the time $\left(t\right)$ of $3\text{months}$:

Calculate the maximum drainage path $\left(H_{\text{dr}}\right)$ for two way drainage using the relation.

$H_{\text{dr}}=\frac{H}{2}$

Substitute $6\text{m}$ for H.

$\begin{array}{c}{H}_{\text{dr}}=\frac{6}{2}\\ =3\text{m}\end{array}$

Calculate the time factor $\left(T_v\right)$ using the relation.

$T_v=\left(\frac{c_{v}t}{H_{dr}^2}\right)$ (1)

Substitute $3{\text{m}}^{\text{2}}/\text{year}$ for $c_v$, $3\text{months}$ for t, and $3\text{m}$ for $H_{\text{dr}}$ in Equation (1).

$\begin{array}{c}{T}_v=\frac{3{\text{m}}^2/\text{year}\times 3\text{months}\times \frac{\text{1 year}}{12\text{months}}}{{\left(3\text{m}\right)}^2}\\ =\frac{0.75}{9}\\ =0.083\end{array}$

Refer Figure 9.33 “$U-T_v$ relationships for ramp loading and instantaneous loading” in the Text Book.

Take the value of average degree of consolidation $\left(U\right)$ as $21\%$ for the value time factor $\left(T_v\right)$ of $0.083$.

Calculate the final consolidation settlement $\left(S_{p\left(\text{f}\right)}\right)$ with $25\%$ of the applied load using the relation.

$S_{p\left(\text{f}\right)}=\text{Percentage}\text{of}\text{load}\text{applied}\times S_p$ (2)

Substitute $25\%$ for percentage of load applied and $220\text{mm}$ for $S_p$ in Equation (2).

$\begin{array}{c}{S}_{p\left(\text{f}\right)}=\frac{25}{100}\times 220\\ =55\text{mm}\end{array}$

Calculate the consolidation settlement for $3\text{months}$ using the relation.

$\text{Consolidation}\text{settlement}=U\times S_{p\left(\text{f}\right)}$ (3)

Substitute $21\%$ for U and $55\text{mm}$ for $S_{p\left(\text{f}\right)}$ in Equation (3).

$\begin{array}{c}\text{Consolidation}\text{settlement}=\frac{21}{100}\times 55\\ =11.6\text{mm}\end{array}$

For the time $\left(t\right)$ of $6\text{months}$:

Calculate the time factor $\left(T_v\right)$ as shown below.

Substitute $3.0{\text{m}}^2/\text{year}$ for $c_v$, $6\text{months}$ for t, and $3\text{m}$ for $H_{\text{dr}}$ in Equation (1).

$\begin{array}{c}{T}_v=\frac{3{\text{m}}^2/\text{year}\times 6\text{months}\times \frac{\text{1 year}}{12\text{months}}}{{\left(3\text{m}\right)}^2}\\ =\frac{1.5}{9}\\ =0.167\end{array}$

Refer Figure 9.33 “$U-T_v$ relationships for ramp loading and instantaneous loading” in the Text Book.

Take the value of average degree of consolidation $\left(U\right)$ as $31\%$ for the value time factor $\left(T_v\right)$ of $0.167$.

Calculate the final consolidation settlement $\left(S_{p\left(\text{f}\right)}\right)$ with $50\%$ of the applied load as shown below.

Substitute $50\%$ for percentage of load applied and $220\text{mm}$ for $S_p$ in Equation (2).

$\begin{array}{c}{S}_{p\left(\text{f}\right)}=\frac{50}{100}\times 220\\ =110\text{mm}\end{array}$

Calculate the consolidation settlement for $6\text{months}$ using the relation.

Substitute $31\%$ for U and $110\text{mm}$ for $S_p$ in Equation (3).

$\begin{array}{c}\text{Consolidation}\text{settlement}=\frac{31}{100}\times 110\\ =34.10\text{mm}\end{array}$

For the time $\left(t\right)$ of $12\text{months}$:

Calculate the time factor $\left(T_v\right)$ as shown below.

Substitute $3.0{\text{m}}^2/\text{year}$ for $c_v$, $12\text{months}$ for t, and $3\text{m}$ for $H_{\text{dr}}$ in Equation (1).

$\begin{array}{c}{T}_v=\frac{3{\text{m}}^2/\text{year}\times 12\text{months}\times \frac{\text{1 year}}{12\text{months}}}{{\left(3\text{m}\right)}^2}\\ =\frac{3}{9}\\ =0.333\end{array}$

Refer Figure 9.33 “$U-T_v$ relationships for ramp loading and instantaneous loading” in the Text Book.

Take the value of average degree of consolidation $\left(U\right)$ as $43\%$ for the value time factor $\left(T_v\right)$ of $0.333$.

Calculate the final consolidation settlement $\left(S_{p\left(\text{f}\right)}\right)$ with $50\%$ of the applied load as shown below.

Substitute $100\%$ for percentage of load applied and $220\text{mm}$ for $S_p$ in Equation (2).

$\begin{array}{c}{S}_{p\left(\text{f}\right)}=\frac{100}{100}\times 220\\ =220\text{mm}\end{array}$

Calculate the consolidation settlement for $12\text{months}$ using the relation.

Substitute $43\%$ for U and $220\text{mm}$ for $S_p$ in Equation (3).

$\begin{array}{c}\text{Consolidation}\text{settlement}=\frac{43}{100}\times 220\\ =94.6\text{mm}\end{array}$

For the time $\left(t\right)$ of $18\text{months}$ with $t_0$ of $12\text{months}$.

Calculate the time factor $T_v$ using the relation.

$T_v=\frac{c_v\left(t-t_0\right)}{H_{\text{dr}}^2}$ (4)

Substitute $3.0{\text{m}}^{\text{2}}/\text{year}$ for $c_v$, $\text{18}\text{months}$ for t,$12\text{months}$ for $t_0$,and $3\text{m}$ for $H_{\text{dr}}$ in Equation (4).

$\begin{array}{c}{T}_v=\frac{3{\text{m}}^2/\text{year}\times \left(12-6\right)\text{months}\times \frac{\text{1 year}}{12\text{months}}}{{\left(3\text{m}\right)}^2}\\ =\frac{1.5}{9}\\ =0.167\end{array}$

Refer Table 9.5“ $U-T_v$ Values for Non-Uniform Initial Pore Pressure Distributions” in the Text Book.

Take the value of $U_{t-t_0}$ as $0.3377$ for the value $T_v$ of $0.167$ for half-sine.

Calculate the average degree of consolidation $\left(U_t\right)$ using the relation.

$U_t=U_0+\left(1-U_0\right)U_{t-t_0}$

Substitute $43\%$ for $U_0$ and $0.3377$ for $U_{t-t_0}$.

$\begin{array}{c}{U}_t=\frac{43}{100}+\left(1-\frac{43}{100}\right)\times 0.3377\\ =0.43+0.1925\\ =0.6225\times 100\%\\ =62.25\%\end{array}$

Calculate the consolidation settlement for $18\text{months}$ using the relation.

Substitute $62.25\%$ for U and $220\text{mm}$ for $S_p$ in Equation (3).

$\begin{array}{c}\text{Consolidation}\text{settlement}=\frac{62.25}{100}\times 220\\ =136.95\text{mm}\end{array}$

For the time $\left(t\right)$ of $36\text{months}$ with $t_0$ of $12\text{months}$.

Calculate the time factor $\left(T_v\right)$ as shown below.

Substitute $3.0{\text{m}}^{\text{2}}/\text{year}$ for $c_v$, $36\text{months}$ for t, $12\text{months}$ for $t_0$, and $3\text{m}$ for $H_{\text{dr}}$ in Equation (4).

$\begin{array}{c}{T}_v=\frac{3{\text{m}}^2/\text{year}\times \left(36-12\right)\text{months}\times \frac{\text{1 year}}{12\text{months}}}{{\left(3\text{m}\right)}^2}\\ =\frac{6}{9}\\ =0.667\end{array}$

Refer Table 9.5 “ $U-T_v$ Values for Non-Uniform Initial Pore Pressure Distributions” in the Text Book.

Take the value of $U_{t-t_0}$ as $0.7725$ for the value $T_v$ of $0.600$ for half-sine.

Take the value of $U_{t-t_0}$ as $0.8222$ for the value $T_v$ of $0.700$ for half-sine.

Calculate the value of $U_{t-t_0}$ for the value $T_v$ of $0.667$ by interpolation as shown below.

$\begin{array}{l}\frac{0.700-0.667}{0.700-0.600}=\frac{0.8222-U_{t-t_0}}{0.8222-0.7725}\\ 0.33\times 0.0497=0.8222-U_{t-t_0}\\ 0.0164=0.8222-U_{t-t_0}\\ U_{t-t_0}=0.806\end{array}$

Calculate the average degree of consolidation $\left(U_t\right)$ using the relation.

$U_t=U_0+\left(1-U_0\right)U_{t-t_0}$

Substitute $43\%$ for $U_0$ and $0.806$ for $U_{t-t_0}$.

$\begin{array}{c}{U}_t=\frac{43}{100}+\left(1-\frac{43}{100}\right)0.806\\ =0.43+0.4594\\ =0.8894\times 100\%\\ =88.9\%\end{array}$

Calculate the consolidation settlement for $36\text{months}$ using the relation.

Substitute $88.9\%$ for U and $220\text{mm}$ for $S_p$ in Equation (3).

$\begin{array}{c}\text{Consolidation}\text{settlement}=\frac{88.9}{100}\times 220\\ =195.58\text{mm}\end{array}$

For the time $\left(t\right)$ of $60\text{months}$ with $t_0$ of $12\text{months}$.

Calculate the time factor $\left(T_v\right)$ as shown below.

Substitute $3.0{\text{m}}^{\text{2}}/\text{year}$ for $c_v$, $60\text{months}$ for t, $12\text{months}$ for $t_0$, and $3\text{m}$ for $H_{\text{dr}}$ in Equation (4).

$\begin{array}{c}{T}_v=\frac{3{\text{m}}^2/\text{year}\times \left(60-12\right)\text{months}\times \frac{\text{1 year}}{12\text{months}}}{{\left(3\text{m}\right)}^2}\\ =\frac{12}{9.0}\\ =1.33\end{array}$

Refer Table 9.5 “ $U-T_v$ Values for Non-Uniform Initial Pore Pressure Distributions” in the Text Book.

Take the value of $U_{t-t_0}$ as $0.9152$ for the value $T_v$ of $1$ for half-sine.

Take the value of $U_{t-t_0}$ as $0.9753$ for the value $T_v$ of $1.5$ for half-sine.

Calculate the value of $U_{t-t_0}$ for the value $T_v$ of $1.33$ by interpolation as shown below.

$\begin{array}{l}\frac{1.5-1.33}{1.5-1}=\frac{0.9753-U_{t-t_0}}{0.9753-0.9152}\\ 0.34\times 0.0601=0.9753-U_{t-t_0}\\ 0.0204=0.9753-U_{t-t_0}\\ U_{t-t_0}=0.955\end{array}$

Calculate the average degree of consolidation $\left(U_t\right)$ using the relation.

$U_t=U_0+\left(1-U_0\right)U_{t-t_0}$

Substitute $43\%$ for $U_0$ and 0.955for $U_{t-t_0}$.

$\begin{array}{c}{U}_t=\frac{43}{100}+\left(1-\frac{43}{100}\right)0.955\\ =0.43+0.5444\\ =0.9744\times 100\%\\ =97.4\%\end{array}$

Calculate the consolidation settlement for $60\text{months}$ using the relation.

Substitute $97.4\%$ for U and $220\text{mm}$ for $S_p$ in Equation (3).

$\begin{array}{c}\text{Consolidation}\text{settlement}=\frac{97.4}{100}\times 220\\ =214.3\text{mm}\end{array}$

Show the values of time and consolidation settlement as in Table 1.

Time (months)	Consolidation settlement $\left(\text{mm}\right)$
3	11.60
6	34.10
12	94.60
18	136.95
36	195.58
60	214.30

Table 1

Refer to Table 1.

Sketch the plot between consolidation settlement and time as in Figure 1.