Chapter 9, Problem 9.20CTP

(a)

To determine

Calculate the consolidation settlement inone year.

Answer to Problem 9.20CTP

The consolidation settlement in one year $\left(S_p\right)$ is $\underset{\_}{206.25\text{mm}}$.

Explanation of Solution

Given information:

The thickness of sand layer $\left(H_1\right)$ is $2\text{m}$.

The thickness of clay layer $\left(H_2\right)$ is $4\text{m}$.

The moisture content $\left(w\right)$ is $35\%$.

The specific gravity of soil solids $\left(G_s\right)$ is $2.70$.

The compression index $\left(C_c\right)$ is $0.65$.

The coefficient of consolidation $\left(c_v\right)$ is $5{\text{m}}^2/\text{year}$.

The unit weight of sand is $\left(\gamma \right)$$17.5\text{kN}/{\text{m}}^3$.

The unit weight of fill $\left({\gamma }_{\text{fill}}\right)$ is $20\text{kN}/{\text{m}}^3$.

The depth of the compacted fill $\left(h\right)$ is $2\text{m}$.

Calculation:

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

Calculate the initial void ratio $\left(e_0\right)$ of the clay using the relation.

$e_0=w{G}_s$

Substitute $35\%$ for wand $2.7$ for $G_s$.

$\begin{array}{c}{e}_0=\frac{35}{100}\times 2.7\\ =0.945\end{array}$

Calculate the saturated unit weight $\left({\gamma }_{\text{sat}}\right)$ of the clay using the relation.

${\gamma }_{\text{sat}}=\left(\frac{G_s+e_0}{1+e_0}\right)\times {\gamma }_w$

Substitute $2.70$ for $G_s$, $0.945$ for $e_0$, and $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$.

$\begin{array}{c}{\gamma }_{\text{sat}}=\left(\frac{2.70+0.945}{1+0.945}\right)\times 9.81\\ =\frac{3.645}{1.945}\times 9.81\\ =18.4\text{kN}/{\text{m}}^{\text{3}}\end{array}$

At the middle of the clay layer:

Calculate the effective overburden pressure $\left({\sigma }_o{}^{\prime }\right)$ using the relation.

${\sigma }_o{}^{\prime }=\gamma H_1+\left({\gamma }_{\text{sat}}-{\gamma }_w\right)\frac{H_2}{2}$

Substitute $17.5\text{kN}/{\text{m}}^3$ for $\gamma$, $2\text{m}$ for $H_1$,$4\text{m}$ for $H_2$,$18.4\text{kN}/{\text{m}}^3$ for ${\gamma }_{\text{sat}}$, and $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$.

$\begin{array}{c}{\sigma }_o{}^{\prime }=17.5\times 2+\left(18.4-9.81\right)\times \frac{4}{2}\\ =35+17.18\\ =52.2{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the increase in vertical pressure $\left(\Delta {\sigma }^{\prime }\right)$ using the relation.

$\Delta {\sigma }^{\prime }={\gamma }_{\text{fill}}\times h$

Substitute $20\text{kN}/{\text{m}}^3$ for ${\gamma }_{\text{fill}}$ and $2\text{m}$ for h.

$\begin{array}{c}\Delta {\sigma }^{\prime }=20\times 2\\ =40\text{kN}/{\text{m}}^2\end{array}$

Calculate the final primary consolidation settlement $\left(S_p\right)$ using the relation.

$S_p=\frac{C_{c}H}{1+e_0}\log \left(\frac{{\sigma }_o{}^{\prime }+\Delta {\sigma }^{\prime }}{{\sigma }_o{}^{\prime }}\right)$

Substitute $0.65$ for $C_c$, $4\text{m}$ for H, $0.945$ for $e_0$, $52.2\text{kN}/{\text{m}}^2$ for ${\sigma }_o{}^{\prime }$, and $40\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$.

$\begin{array}{c}{S}_p=\frac{0.65\times 4}{1+0.945}\log \left(\frac{52.2+40}{52.2}\right)\\ =1.337\times 0.247\\ =0.330\text{m}\end{array}$

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

$T_v=\frac{c_{v}t}{H_{\text{dr}}^2}$

Substitute $5{\text{m}}^{\text{2}}/\text{year}$ for $c_v$, $1\text{year}$ for t, and $4\text{m}$ for $H_{\text{dr}}$.

$\begin{array}{c}{T}_v=\frac{5\times 1}{4^2}\\ =\frac{5}{16}\\ =0.3125\end{array}$

Refer Table 9.3 “Variation of Time Factor with Degree of consolidation” in the Text Book.

Take the value of Uas $62\%$ for the value $T_v$ of $0.307$.

Take the value of Uas $63\%$ for the value $T_v$ of $0.318$.

Calculate the value of U for the value $T_v$ of $0.3125$ using interpolation as shown below.

$\begin{array}{l}\frac{63-U}{63-62}=\frac{0.318-0.3125}{0.318-0.307}\\ 63-U=0.5\\ U=62.5\%\end{array}$

Calculate the final consolidation settlement $S_{p\left(\text{f}\right)}$ with $62.5\%$ of the load using the relation.

$S_{p\left(\text{f}\right)}=U\times S_p$

Substitute $62.5\%$ for Uand $0.33\text{m}$ for final consolidation settlement.

$\begin{array}{c}{S}_{p\left(\text{f}\right)}=\frac{62.5}{100}\times 0.33\\ =0.20625\text{m}\times \frac{1,000\text{mm}}{1\text{m}}\\ =206.25\text{mm}\end{array}$

Therefore, the consolidation settlement in one year is $\underset{\_}{206.25\text{mm}}$.

(b)

To determine

Plot the in situ variation of pore water pressure and effective stress with depth for the clay layer.

Explanation of Solution

Given information:

The thickness of sand layer $\left(H_1\right)$ is $2\text{m}$.

The thickness of clay layer $\left(H_2\right)$ is $4\text{m}$.

The moisture content $\left(w\right)$ is $35\%$.

The specific gravity of soil solids $\left(G_s\right)$ is $2.70$.

The compression index $\left(C_c\right)$ is $0.65$.

The coefficient of consolidation $\left(c_v\right)$ is $5{\text{m}}^2/\text{year}$.

The unit weight of sand is $\left(\gamma \right)$$17.5\text{kN}/{\text{m}}^3$.

The unit weight of fill $\left({\gamma }_{\text{fill}}\right)$ is $20\text{kN}/{\text{m}}^3$.

The depth of the compacted fill $\left(h\right)$ is $2\text{m}$.

Calculation:

Calculate the total stress $\left(\sigma \right)$ at the top of the clay layer using the relation.

$\sigma ={\gamma }_{\text{sat}}\times h$

Substitute $17.5\text{kN}/{\text{m}}^3$ for $\gamma$ and $2\text{m}$ for h.

$\begin{array}{c}\sigma =17.5\times 2\\ =35\text{kN}/{\text{m}}^2\end{array}$

Calculate the pore water pressure $\left(u\right)$ at the top of the clay layer using the relation.

$u={\gamma }_w\times h$

Substitute $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$ and $0\text{m}$ for h.

$\begin{array}{c}u=9.81\times 0\\ =0\end{array}$

Calculate the total stress $\left({\sigma }^{\prime }\right)$ at the top of the clay layer using the relation.

${\sigma }^{\prime }=\sigma -u$

Substitute $35\text{kN}/{\text{m}}^2$ for $\sigma$ and $0$ for u.

$\begin{array}{c}{\sigma }^{\prime }=35-0\\ =35{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the total stress $\left(\sigma \right)$ at the bottom of the clay layer using the relation.

$\sigma =\gamma H_1+{\gamma }_{\text{sat}}{H}_2$

Substitute $17.5\text{kN}/{\text{m}}^{{}^3}$ for $\gamma$ and $2\text{m}$ for $H_1$, $18.4\text{kN}/{\text{m}}^3$ for ${\gamma }_{\text{sat}}$, and $4\text{m}$ for $H_2$.

$\begin{array}{c}\sigma =17.5\times 2.0+18.4\times 4.0\\ =108.6{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the pore water pressure $\left(u\right)$ at the bottom of the clay layer using the relation.

$u={\gamma }_{w}H$

Substitute $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$ and $4\text{m}$ for H.

$\begin{array}{c}u=9.81\times 4\\ =39.24\text{kN}/{\text{m}}^2\end{array}$

Calculate the total stress $\left({\sigma }^{\prime }\right)$ at the bottom of the clay layer using the relation.

${\sigma }^{\prime }=\sigma -u$

Substitute $108.6\text{kN}/{\text{m}}^2$ for $\sigma$ and $39.24\text{kN}/{\text{m}}^2$ for u.

$\begin{array}{c}{\sigma }^{\prime }=108.6-39.24\\ =69.36{\text{kN/m}}^{\text{2}}\end{array}$

Show the variation of pore water pressure and effective stress with depth for the clay layer as in Figure 1.

(c)

To determine

Plot the variation of pore water pressure and effective stress with depth after on year.

Explanation of Solution

Given information:

The thickness of sand layer $\left(H_1\right)$ is $2\text{m}$.

The thickness of clay layer $\left(H_2\right)$ is $4\text{m}$.

The moisture content $\left(w\right)$ is $35\%$.

The specific gravity of soil solids $\left(G_s\right)$ is $2.70$.

The compression index $\left(C_c\right)$ is $0.65$.

The coefficient of consolidation $\left(c_v\right)$ is $5{\text{m}}^2/\text{year}$.

The unit weight of sand is $\left(\gamma \right)$$17.5\text{kN}/{\text{m}}^3$.

The unit weight of fill $\left({\gamma }_{\text{fill}}\right)$ is $20\text{kN}/{\text{m}}^3$.

The depth of the compacted fill $\left(h\right)$ is $2\text{m}$.

Calculation:

Consider the degree of consolidation $U_z$ as $1$.

Refer to part (a).

The time factor $\left(T_v\right)$ is $0.3125$.

Calculate the effective stress $\left({\sigma }^{\prime }\right)$ at the top of the clay layer using the relation.

$\begin{array}{c}{\sigma }^{\prime }={\sigma }_0{}^{\prime }+\Delta {\sigma }^{\prime }\\ ={\gamma }_{\text{sat}}{H}_1+\Delta {\sigma }^{\prime }\end{array}$

Substitute $17.5\text{kN}/{\text{m}}^3$ for ${\gamma }_{\text{sat}}$,$2\text{m}$ for $H_1$, and $40\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$.

$\begin{array}{c}{\sigma }^{\prime }=17.5\times 2+40\\ =75\text{kN}/{\text{m}}^2\end{array}$

Calculate the pore water $\left(u\right)$ pressure at the top of the clay layer using the relation.

$u=u_0+\Delta u$

Substitute $0$ for u and $0$ for $\Delta u$.

$u=0$

Calculate the ratio of height of soil to the maximum drainage depth as shown below.

$\text{Ratio}=\frac{z}{H_{dr}}$

Substitute $2\text{m}$ for z and $4\text{m}$ for $H_{dr}$.

$\begin{array}{c}\frac{z}{H_{\text{dr}}}=\frac{2}{4}\\ =0.5\end{array}$

Refer Figure 9.20 “Variation of $U_z$ with $T_v$ and $\frac{z}{H_{dr}}$” in the Text Book.

Take the value of $U_z$ as $0.61$, for the values $T_v$ of $0.3125$ and $\frac{z}{H_{\text{dr}}}$ of $0.5$.

Calculate the effective stress $\left({\sigma }^{\prime }\right)$ at the middle of the clay layer using the relation.

${\sigma }^{\prime }={\sigma }_0{}^{\prime }+\Delta {\sigma }^{\prime }{U}_z$

Substitute $52.2\text{kN}/{\text{m}}^2$ for ${\sigma }_0{}^{\prime }$, $40\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$, and $0.61$ for $U_z$.

$\begin{array}{c}{\sigma }^{\prime }=52.2+40\times 0.61\\ =76.9\text{kN}/{\text{m}}^2\end{array}$

Calculate the pore water $\left(u\right)$ pressure at the middle of the clay layer using the relation.

$\begin{array}{c}u=u_0+\Delta u\\ ={\gamma }_w\frac{H_2}{2}+\Delta u\left(1-0.61\right)\end{array}$

Substitute $4\text{m}$ for $H_2$, $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$, and $40\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$.

$\begin{array}{c}u=2\times 9.81+40\left(1-0.61\right)\\ =35.2\text{kN}/{\text{m}}^2\end{array}$

At the bottom of the clay layer, after one year:

Calculate the ratio of height of soil to the maximum drainage depth;

$\text{Ratio}=\frac{z}{H_{dr}}$

Substitute $4\text{m}$ for z and $4\text{m}$ for $H_{\text{dr}}$.

$\begin{array}{c}\frac{z}{H_{\text{dr}}}=\frac{4}{4}\\ =1\end{array}$

Refer Figure 9.20 “Variation of $U_z$ with $T_v$ and $\frac{z}{H_{dr}}$” in the Text Book.

Take the value of $U_z$ as $0.42$, for the values $T_v$ of $0.3125$ and $\frac{z}{H_{\text{dr}}}$ of $1$.

Calculate the effective stress $\left({\sigma }^{\prime }\right)$ at the bottom of the clay layer using the relation.

$\begin{array}{c}{\sigma }^{\prime }={\sigma }_0{}^{\prime }+\Delta {\sigma }^{\prime }\\ =\gamma H_1+\left({\gamma }_{\text{sat}}-{\gamma }_w\right)H_2+\Delta {\sigma }^{\prime }{U}_z\end{array}$

Substitute $17.5\text{kN}/{\text{m}}^3$ for $\gamma$, $2\text{m}$ for $H_1$, $4\text{m}$ for $H_2$, $18.4\text{kN}/{\text{m}}^3$ for ${\gamma }_{\text{sat}}$, $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$, $40\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$, and $0.42$ for $U_z$.

$\begin{array}{c}{\sigma }^{\prime }=17.5\times 2+\left(18.4-9.81\right)\times 4+40\times 0.42\\ =35+34.36+16.8\\ =86.2\text{kN}/{\text{m}}^2\end{array}$

Calculate the pore water $\left(u\right)$ pressure at the bottom of the clay layer using the relation.

$\begin{array}{c}u=u_0+\Delta u\\ ={\gamma }_w{H}_2+\Delta u\left(1-0.42\right)\end{array}$

Substitute $4\text{m}$ for $H_2$, $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$, and $40\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$.

$\begin{array}{c}u=4\times 9.81+40\left(1-0.42\right)\\ =39.24+23.2\\ =62.4\text{kN}/{\text{m}}^2\end{array}$

Show the variation of the pore water pressure and the effective stress with depth as in Figure 1.