Chapter 11, Problem 11.1CTP

(a)

To determine

Calculate the total consolidation settlement under the action of fill load.

Answer to Problem 11.1CTP

The total consolidation settlement $\left(S_C\right)$ is $\underset{\_}{0.188\text{m}}$.

Explanation of Solution

Given information:

The thickness of fill material $\left(t\right)$ is $1.75\text{m}$.

The compacted unit weight of fill material $\left({\gamma }_{\text{fill}}\right)$ is $20.1\text{kN}/{\text{m}}^3$.

The length of the foundation $\left(L\right)$ is $8\text{m}$.

The breadth of the foundation $\left(B\right)$ is $8\text{m}$.

The depth of fill $\left(D_f\right)$ is $1.5\text{m}$.

The height of the layer silty sand $\left(H_1\right)$ is $3\text{m}$.

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

The height of the peat layer $\left(H_3\right)$ is $1.8\text{m}$.

The dry unit weight of sand $\left({\gamma }_d\right)$ is $17\text{kN}/{\text{m}}^3$.

The saturated unit weight of sand ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$ is $19.2\text{kN}/{\text{m}}^3$.

The saturated unit weight of clay ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$ is $18.8\text{kN}/{\text{m}}^3$.

The saturated unit weight of peat ${\left({\gamma }_{\text{sat}}\right)}_{\text{peat}}$ is $15\text{kN}/{\text{m}}^3$.

The time $\left(t\right)$ is $8\text{months}$.

The properties of clay and organic layers are given in the Table.

Calculation:

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

Calculate the distributed load $\left(q\right)$ as shown below.

$q=t{\gamma }_{\text{fill}}$

Substitute $1.75\text{m}$ for t and $20.1\text{kN}/{\text{m}}^{\text{3}}$ for ${\gamma }_{\text{fill}}$.

$\begin{array}{c}q=1.75\times 20.1\\ =35.02\text{kN}/{\text{m}}^2\end{array}$

Calculate the increase in vertical stress $\left(\Delta {\sigma }_z\right)$ below the center of a rectangular area using the relation as follows.

$\Delta {\sigma }_z=q{I}_4$ (1)

  Here, $I_4$ is the influence factor and as a function of $m_1$ and $n_1$.

For clay layer:

For the depth $\left(z\right)$ of $3\text{m}$:

Calculate the width $\left(b\right)$ as shown below.

$b=\frac{B}{2}$

Substitute $8\text{m}$ for B.

$\begin{array}{c}b=\frac{8}{2}\\ =4\text{m}\end{array}$

Calculate the ratio $\left(m_1\right)$ as shown below.

$m_1=\frac{L}{B}$

Substitute $8\text{m}$ for L and $8\text{m}$ for B.

$\begin{array}{c}{m}_1=\frac{8}{8}\\ =1\end{array}$

Calculate the ratio $\left(n_1\right)$ as shown below.

$n_1=\frac{z}{b}$

Substitute $4\text{m}$ for b and $3\text{m}$ for z.

$\begin{array}{c}{n}_1=\frac{3}{4}\\ =0.75\end{array}$

Similarly calculate the remaining values and tabulate as in Table 1.

Refer Table 10.11 “Variation of $I_4$ with $m_1$ and $n_1$” in the Text Book.

Take the value of $I_4$ as $0.892$, for the values $m_1$ of $1$ and $n_1$ of $0.6$.

Take the value of $I_4$ as $0.800$, for the values $m_1$ of $1$ and $n_1$ of $0.8$.

Calculate the value of $I_4$ for the values $m_1$ of $1$ and $n_1$ of $0.75$ by interpolation as shown below.

$\begin{array}{l}\frac{0.8-0.75}{0.8-0.6}=\frac{0.8-I_4}{0.8-0.892}\\ 0.25\times \left(-0.092\right)=0.8-I_4\\ I_4=0.823\end{array}$

Similarly calculate the remaining values and tabulate as in Table 1.

Calculate the increase in vertical stress $\left(\Delta {\sigma }_z\right)$ as shown below.

Substitute $35.02\text{kN}/{\text{m}}^2$ for $q$ and $0.823$ for $I_4$ in Equation (1).

$\begin{array}{c}\Delta {\sigma }_z=35.02\times 0.823\\ =28.82\text{kN}/{\text{m}}^2\end{array}$

Similarly calculate the increase in vertical stress values and tabulate as in Table 1.

Show the increase in vertical stress for each depth below the center of the loaded area as in Table 1.

$m_1=\frac{L}{B}$	$b=\frac{B}{2}$	Depth, $z\left(\text{m}\right)$	$n_1=\frac{z}{b}$	$I_4$	$\Delta {\sigma }_z\left(\text{kN}/{\text{m}}^2\right)$
1	4	3	0.75	0.823	28.82
1	4	5	1.25	0.599	20.98
1	4	7	1.75	0.403	14.11

Table 1

Refer to table 1.

Calculate the stress increase in the clay layer $\left(\Delta {\sigma }^{\prime }\right)$ using the relation.

$\Delta {\sigma }^{\prime }=\frac{\Delta {\sigma }_t{}^{\prime }+4\Delta {\sigma }_m{}^{\prime }+\Delta {\sigma }_b{}^{\prime }}{6}$ (2)

Here, $\Delta {\sigma }_t{}^{\prime }$ is the increase in effective stress at the top layer, $\Delta {\sigma }_m{}^{\prime }$ is the increase in effective stress at middle layer, and $\Delta {\sigma }_b{}^{\prime }$ is the increase in effective stress at the bottom layer.

Substitute $28.82\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }_t{}^{\prime }$, $20.98\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }_m{}^{\prime }$, and $14.11\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }_b{}^{\prime }$ in Equation (2).

$\begin{array}{c}\Delta {\sigma }^{\prime }=\frac{28.82+4\times 20.98+14.11}{6}\\ =\frac{126.85}{6}\\ =21.14\text{kN}/{\text{m}}^2\end{array}$

Calculate the average effective stress at the middle of the clay layer $\left({\sigma }_o{}^{\prime }\right)$ as shown below.

${\sigma }_o{}^{\prime }={\gamma }_d{D}_f+\left({\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}-{\gamma }_w\right)\left(H_1-D_f\right)+\left({\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}-{\gamma }_w\right)\times \frac{H_2}{2}$

Substitute $17\text{kN}/{\text{m}}^3$ for ${\gamma }_d$, $1.5\text{m}$ for $D_f$, $19.2\text{kN}/{\text{m}}^3$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$, $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$, $18.8\text{kN}/{\text{m}}^3$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$, $3\text{m}$ for $H_1$, and $4\text{m}$ for $H_2$.

$\begin{array}{c}{\sigma }_o{}^{\prime }=17\times 1.5+\left(19.2-9.81\right)\left(3-1.5\right)+\left(18.8-9.81\right)\times \frac{4}{2}\\ =25.5+14.085+17.98\\ =57.565\text{kN}/{\text{m}}^2\end{array}$

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

$S_C=\frac{C_{C}H}{1+e_0}\log \left(\frac{{\sigma }_o{}^{\prime }+\Delta {\sigma }^{\prime }}{{\sigma }_o{}^{\prime }}\right)$ (3)

Substitute $0.31$ for $C_C$, $4\text{m}$ for $H$, $1.08$ for $e_0$, $57.565\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_o{}^{\prime }$, and $21.14\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$ in Equation (3).

$\begin{array}{c}{S}_{C-\text{clay}}=\frac{0.31\times 4}{1+1.08}\log \left(\frac{57.565+21.14}{57.565}\right)\\ =0.59615\times 0.1358\\ =0.081\text{m}\end{array}$

For peat layer:

For the depth $\left(z\right)$ of $7\text{m}$:

Calculate the ratio $\left(n_1\right)$ as shown below.

$n_1=\frac{z}{b}$

Substitute $4\text{m}$ for b and $7\text{m}$ for z.

$\begin{array}{c}{n}_1=\frac{7}{4}\\ =1.75\end{array}$

Similarly calculate the remaining values and tabulate as in Table 2.

Refer Table 10.11 “Variation of $I_4$ with $m_1$ and $n_1$” in the Text Book.

Take the value of $I_4$ as $0.449$, for the values $m_1$ of $1$ and $n_1$ of $1.6$.

Take the value of $I_4$ as $0.388$, for the values $m_1$ of $1$ and $n_1$ of $1.8$.

Calculate the value of $I_4$ for the values $m_1$ of $1$ and $n_1$ of $1.75$ by interpolation as shown below.

$\begin{array}{l}\frac{1.8-1.75}{1.8-1.6}=\frac{0.388-I_4}{0.388-0.449}\\ 0.25\times \left(-0.061\right)=0.8-I_4\\ I_4=0.403\end{array}$

Similarly calculate the remaining values and tabulate as in Table 2.

Calculate the increase in vertical stress $\left(\Delta {\sigma }_z\right)$ as shown below.

Substitute $35.02\text{kN}/{\text{m}}^2$ for $q$ and $0.403$ for $I_4$ in Equation (1).

$\begin{array}{c}\Delta {\sigma }_z=35.02\times 0.403\\ =14.11\text{kN}/{\text{m}}^2\end{array}$

Similarly calculate the increase in vertical stress values and tabulate as in Table 2.

Show the increase in vertical stress for each depth below the center of the loaded area as in Table 2.

$m_1=\frac{L}{B}$	$b=\frac{B}{2}$	Depth, $z\left(\text{m}\right)$	$n_1=\frac{z}{b}$	$I_4$	$\Delta {\sigma }_z\left(\text{kN}/{\text{m}}^2\right)$
1	4	7	1.75	0.403	14.11
1	4	7.9	1.975	0.342	11.98
1	4	8.8	2.2	0.302	10.58

Table 2

Refer to table 2.

Calculate the stress increase in the peat layer $\left(\Delta {\sigma }^{\prime }\right)$ using the relation.

Substitute $14.11\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }_t{}^{\prime }$, $11.98\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }_m{}^{\prime }$, and $10.58\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }_b{}^{\prime }$ in Equation (2).

$\begin{array}{c}\Delta {\sigma }^{\prime }=\frac{14.11+4\times 11.98+10.58}{6}\\ =\frac{72.61}{6}\\ =12.10\text{kN}/{\text{m}}^2\end{array}$

Calculate the average effective stress at the middle of the clay layer $\left({\sigma }_o{}^{\prime }\right)$ as shown below.

${\sigma }_o{}^{\prime }={\gamma }_d{D}_f+\left({\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}-{\gamma }_w\right)\left(H_1-D_f\right)+\left({\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}-{\gamma }_w\right)\times H_2+\left({\left({\gamma }_{\text{sat}}\right)}_{\text{peat}}-{\gamma }_w\right)\times \frac{H_3}{2}$

Substitute $17\text{kN}/{\text{m}}^3$ for ${\gamma }_d$, $1.5\text{m}$ for $D_f$, $19.2\text{kN}/{\text{m}}^3$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$, $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$, $18.8\text{kN}/{\text{m}}^3$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$, $15\text{kN}/{\text{m}}^3$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{peat}}$, $3\text{m}$ for $H_1$, $4\text{m}$ for $H_2$, and $1.8\text{m}$ for $H_3$.

$\begin{array}{c}{\sigma }_o{}^{\prime }=17\times 1.5+\left(19.2-9.81\right)\left(3-1.5\right)+\left(18.8-9.81\right)\times 4+\left(15-9.81\right)\times \frac{1.8}{2}\\ =25.5+14.085+35.96+4.671\\ =80.216\text{kN}/{\text{m}}^2\end{array}$

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

Substitute $7.2$ for $C_C$, $1.8\text{m}$ for $H$, $6.4$ for $e_0$, $80.216\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_o{}^{\prime }$, and $12.10\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$ in Equation (3).

$\begin{array}{c}{S}_{C-\text{Peat}}=\frac{7.2\times 1.8}{1+6.4}\log \left(\frac{80.216+12.10}{80.216}\right)\\ =1.7514\times 0.061\\ =0.107\text{m}\end{array}$

Calculate the total consolidation settlement under the action of fill load $\left(S_C\right)$ as shown below.

$S_C=S_{C-\text{clay}}+S_{C-\text{peat}}$

Substitute $0.081\text{m}$ for $S_{C-\text{clay}}$ and $0.107\text{m}$ for $S_{C-\text{peat}}$.

$\begin{array}{c}{S}_C=0.081+0.107\\ =0.188\end{array}$

Hence, the total consolidation settlement $\left(S_C\right)$ is $\underset{\_}{0.188\text{m}}$.

(b)

To determine

Calculate the time for $99\%$ primary consolidation for each layer.

Answer to Problem 11.1CTP

The time for $99\%$ primary consolidation for clay $\left(t_{\text{99-clay}}\right)$ is $\underset{\_}{138\text{days}}$.

The time for $99\%$ primary consolidation for peat $\left(t_{\text{99-peat}}\right)$ is $\underset{\_}{23\text{days}}$.

Explanation of Solution

Given information:

The thickness of fill material $\left(t\right)$ is $1.75\text{m}$.

The compacted unit weight of fill material $\left({\gamma }_{\text{fill}}\right)$ is $20.1\text{kN}/{\text{m}}^3$.

The length of the foundation $\left(L\right)$ is $8\text{m}$.

The breadth of the foundation $\left(B\right)$ is $8\text{m}$.

The depth of fill $\left(D_f\right)$ is $1.5\text{m}$.

The height of the layer silty sand $\left(H_1\right)$ is $3\text{m}$.

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

The height of the peat layer $\left(H_3\right)$ is $1.8\text{m}$.

The dry unit weight of sand $\left({\gamma }_d\right)$ is $17\text{kN}/{\text{m}}^3$.

The saturated unit weight of sand ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$ is $19.2\text{kN}/{\text{m}}^3$.

The saturated unit weight of clay ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$ is $18.8\text{kN}/{\text{m}}^3$.

The saturated unit weight of peat ${\left({\gamma }_{\text{sat}}\right)}_{\text{peat}}$ is $15\text{kN}/{\text{m}}^3$.

The time $\left(t\right)$ is $8\text{months}$.

The properties of clay and organic layers are given in the Table.

Calculation:

The degree of consolidation $\left(U\right)$ is $99\%$.

The clay layer is permeable and having less void ratio compared to peat layer. Hence, double drainage condition is assumed for the clay layer.

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

Refer Table 11.7 “Variation of $T_v$ with U” in the Text Book.

Take the value of $T_v$ as $1.781$, for the value U of $99\%$.

Calculate the length of maximum drainage path $\left(H_{\text{dr}}\right)$ for clay layer using the relation.

$H_{\text{dr}}=\frac{H_2}{2}$

Substitute $4\text{m}$ for $H_2$.

$\begin{array}{c}{H}_{\text{dr}}=\frac{4}{2}\\ =2\text{m}\end{array}$

Calculate the time for $99\%$ consolidation $\left(t_{\text{99-clay}}\right)$ using the relation.

$T_v=\frac{c_{v}t}{H_{\text{dr}}^2}$ (4)

Substitute $1.781$ for $T_v$, $0.006{\text{cm}}^{\text{2}}/\sec$ for $c_v$, and $2\text{m}$ for $H_{\text{dr}}$ in Equation (4).

$\begin{array}{l}1.781=\frac{0.006{\text{cm}}^{\text{2}}/\sec \times t_{\text{99-clay}}}{{\left(2\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)}^2}\\ 0.006t_{\text{99-clay}}=71,240\\ t_{\text{99-clay}}=11,873,333.333\text{sec}\times \frac{1\text{min}}{60\text{sec}}\times \frac{1\text{hr}}{60\text{mins}}\times \frac{1\text{day}}{24\text{hrs}}\\ t_{\text{99-clay}}=138\text{day}s\end{array}$

Hence, the time for $99\%$ primary consolidation for clay $\left(t_{\text{99-clay}}\right)$ is $\underset{\_}{138\text{days}}$.

The peat layer is low permeable and having high void ratio compared to clay layer. Hence, single drainage condition is assumed for the peat layer.

Calculate the length of maximum drainage path $\left(H_{\text{dr}}\right)$ for clay layer using the relation.

$H_{\text{dr}}=H_3$

Substitute $1.8\text{m}$ for $H_3$.

$H_{\text{dr}}=1.8\text{m}$

Calculate the time for $99\%$ consolidation $\left(t_{\text{99-peat}}\right)$ using the relation.

Substitute $1.781$ for $T_v$, $0.029{\text{cm}}^{\text{2}}/\sec$ for $c_v$, and $1.8\text{m}$ for $H_{\text{dr}}$ in Equation (4).

$\begin{array}{l}1.781=\frac{0.029{\text{cm}}^{\text{2}}/\sec \times t_{\text{99-peat}}}{{\left(1.8\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)}^2}\\ 0.029t_{\text{99-peat}}=57,704.4\\ t_{\text{99-peat}}=1,989,806.897\text{sec}\times \frac{1\text{min}}{60\text{sec}}\times \frac{1\text{hr}}{60\text{mins}}\times \frac{1\text{day}}{24\text{hrs}}\\ t_{\text{99-peat}}=23\text{day}s\end{array}$

Hence, the time for $99\%$ primary consolidation for peat $\left(t_{\text{99-peat}}\right)$ is $\underset{\_}{23\text{days}}$.

(c)

To determine

Calculate the secondary compression in each layer up to end of $18$ months.

Answer to Problem 11.1CTP

The secondary compression for clay $\left(S_{s-\text{clay}}\right)$ is $\underset{\_}{0.055\text{m}}$.

The secondary compression for peat $\left(S_{s-\text{peat}}\right)$ is $\underset{\_}{0.096\text{m}}$.

Explanation of Solution

Given information:

The thickness of fill material $\left(t\right)$ is $1.75\text{m}$.

The compacted unit weight of fill material $\left({\gamma }_{\text{fill}}\right)$ is $20.1\text{kN}/{\text{m}}^3$.

The length of the foundation $\left(L\right)$ is $8\text{m}$.

The breadth of the foundation $\left(B\right)$ is $8\text{m}$.

The depth of fill $\left(D_f\right)$ is $1.5\text{m}$.

The height of the layer silty sand $\left(H_1\right)$ is $3\text{m}$.

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

The height of the peat layer $\left(H_3\right)$ is $1.8\text{m}$.

The dry unit weight of sand $\left({\gamma }_d\right)$ is $17\text{kN}/{\text{m}}^3$.

The saturated unit weight of sand ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$ is $19.2\text{kN}/{\text{m}}^3$.

The saturated unit weight of clay ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$ is $18.8\text{kN}/{\text{m}}^3$.

The saturated unit weight of peat ${\left({\gamma }_{\text{sat}}\right)}_{\text{peat}}$ is $15\text{kN}/{\text{m}}^3$.

The time $\left(t\right)$ is $8\text{months}$.

The properties of clay and organic layers are given in the Table.

Calculation:

Refer to part (b).

The time for $99\%$ primary consolidation for clay $\left(t_{\text{99-clay}}\right)$ is $138\text{days}$.

The time for $99\%$ primary consolidation for peat $\left(t_{\text{99-peat}}\right)$ is $23\text{days}$.

For clay:

Calculate the primary void ratio $\left(\Delta e_{\text{primary}}\right)$ as shown below.

$\Delta e_{\text{primary}}=C_C\log \left(\frac{{\sigma }_o{}^{\prime }+\Delta {\sigma }^{\prime }}{{\sigma }_o{}^{\prime }}\right)$ (5)

Substitute $0.31$ for $C_C$, $57.565\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_o{}^{\prime }$, and $21.14\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$ in Equation (5).

$\begin{array}{c}\Delta e_{\text{primary}}=0.31\log \left(\frac{57.565+21.14}{57.565}\right)\\ =0.31\times 0.1358\\ =0.042\end{array}$

Calculate the void ratio at the end of primary consolidation $\left(e_p\right)$ as shown below.

$e_p=e_0-\Delta e_{\text{primary}}$ (6)

Substitute $1.08$ for $e_0$ and $0.042$ for $\Delta e_{\text{primary}}$ in Equation (6).

$\begin{array}{c}{e}_p=1.08-0.042\\ =1.038\end{array}$

Calculate the magnitude of secondary compression index $\left(C_{\alpha }{}^{\prime }\right)$ as shown below.

$C_{\alpha }{}^{\prime }=\frac{C_{\alpha }}{1+e_p}$ (7)

Here, $C_{\alpha }$ is the secondary compression index.

Substitute $0.048$ for $C_{\alpha }$ and $1.038$ for $e_p$.

$\begin{array}{c}{C}_{\alpha }{}^{\prime }=\frac{0.048}{1+1.038}\\ =0.0235\end{array}$

Calculate the secondary compression $\left(S_s\right)$ as shown below.

$S_s=C_{\alpha }{}^{\prime }H\log \left(\frac{t_2}{t_1}\right)$ (8)

Substitute $0.0235$ for $C_{\alpha }{}^{\prime }$, $4\text{m}$ for H, $18\text{months}$ for $t_2$, and $138\text{days}$ for $t_1$ in Equation (8).

$\begin{array}{c}{S}_{s-\text{clay}}=0.0235\times \left(4\text{m}\right)\log \left(\frac{18\text{months}\times \frac{30\text{days}}{1\text{month}}}{138\text{days}}\right)\\ =0.094\times 0.5925\\ =0.055\text{m}\end{array}$

Hence, the secondary compression for clay $\left(S_{s-\text{clay}}\right)$ is $\underset{\_}{0.055\text{m}}$.

For peat:

Calculate the primary void ratio $\left(\Delta e_{\text{primary}}\right)$ as shown below.

Substitute $7.2$ for $C_C$, $80.216\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_o{}^{\prime }$, and $12.10\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$ in Equation (5).

$\begin{array}{c}\Delta e_{\text{primary}}=7.2\log \left(\frac{80.216+12.1}{80.216}\right)\\ =7.2\times 0.061\\ =0.44\end{array}$

Calculate the void ratio at the end of primary consolidation $\left(e_p\right)$ as shown below.

Substitute $6.4$ for $e_0$ and $0.44$ for $\Delta e_{\text{primary}}$ in Equation (6).

$\begin{array}{c}{e}_p=6.4-0.44\\ =5.96\end{array}$

Calculate the magnitude of secondary compression index $\left(C_{\alpha }{}^{\prime }\right)$ as shown below.

Substitute $0.273$ for $C_{\alpha }$ and $5.96$ for $e_p$.

$\begin{array}{c}{C}_{\alpha }{}^{\prime }=\frac{0.273}{1+5.96}\\ =0.039\end{array}$

Calculate the secondary compression $\left(S_s\right)$ as shown below.

Substitute $0.039$ for $C_{\alpha }{}^{\prime }$, $1.8\text{m}$ for H, $18\text{months}$ for $t_2$, and $23\text{days}$ for $t_1$ in Equation (8).

$\begin{array}{c}{S}_{s-\text{clay}}=0.039\times \left(1.8\text{m}\right)\log \left(\frac{18\text{months}\times \frac{30\text{days}}{1\text{month}}}{23\text{days}}\right)\\ =0.0702\times 1.3707\\ =0.096\text{m}\end{array}$

Hence, the secondary compression for peat $\left(S_{s-\text{peat}}\right)$ is $\underset{\_}{0.096\text{m}}$.

(d)

To determine

Calculate the total settlement after 18 months.

Answer to Problem 11.1CTP

The total settlement after 18 months is $\underset{\_}{0.339\text{m}}$.

Explanation of Solution

Given information:

The thickness of fill material $\left(t\right)$ is $1.75\text{m}$.

The compacted unit weight of fill material $\left({\gamma }_{\text{fill}}\right)$ is $20.1\text{kN}/{\text{m}}^3$.

The length of the foundation $\left(L\right)$ is $8\text{m}$.

The breadth of the foundation $\left(B\right)$ is $8\text{m}$.

The depth of fill $\left(D_f\right)$ is $1.5\text{m}$.

The height of the layer silty sand $\left(H_1\right)$ is $3\text{m}$.

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

The height of the peat layer $\left(H_3\right)$ is $1.8\text{m}$.

The dry unit weight of sand $\left({\gamma }_d\right)$ is $17\text{kN}/{\text{m}}^3$.

The saturated unit weight of sand ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$ is $19.2\text{kN}/{\text{m}}^3$.

The saturated unit weight of clay ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$ is $18.8\text{kN}/{\text{m}}^3$.

The saturated unit weight of peat ${\left({\gamma }_{\text{sat}}\right)}_{\text{peat}}$ is $15\text{kN}/{\text{m}}^3$.

The time $\left(t\right)$ is $8\text{months}$.

The properties of clay and organic layers are given in the Table.

Calculation:

Refer to part (a).

The total consolidation settlement $\left(S_C\right)$ is $0.188$.

Refer to part (c).

The secondary compression for clay $\left(S_{s-\text{clay}}\right)$ is $0.055\text{m}$.

The secondary compression for peat $\left(S_{s-\text{peat}}\right)$ is $0.096\text{m}$.

Calculate the total settlement after 18 months as shown below.

$\text{Total}\text{settlement}=S_C+S_{s-\text{clay}}+S_{s-\text{peat}}$

Substitute $0.188\text{m}$ for $S_C$, $0.055\text{m}$ for $S_{s-\text{clay}}$, and $0.096\text{m}$ for $S_{s-\text{peat}}$.

$\begin{array}{c}\text{Total}\text{settlement}=0.188+0.055+0.096\\ =0.339\text{m}\end{array}$

Hence, the total settlement after 18 months is $\underset{\_}{0.339\text{m}}$.

(e)

To determine

Calculate the excess pore water pressure at point A two months after the application of the fill load.

Answer to Problem 11.1CTP

The excess pore water pressure at point A $\left(u_z\right)$ is $\underset{\_}{2.114\text{kN}/{\text{m}}^2}$.

Explanation of Solution

Given information:

The thickness of fill material $\left(t\right)$ is $1.75\text{m}$.

The compacted unit weight of fill material $\left({\gamma }_{\text{fill}}\right)$ is $20.1\text{kN}/{\text{m}}^3$.

The length of the foundation $\left(L\right)$ is $8\text{m}$.

The breadth of the foundation $\left(B\right)$ is $8\text{m}$.

The depth of fill $\left(D_f\right)$ is $1.5\text{m}$.

The height of the layer silty sand $\left(H_1\right)$ is $3\text{m}$.

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

The height of the peat layer $\left(H_3\right)$ is $1.8\text{m}$.

The dry unit weight of sand $\left({\gamma }_d\right)$ is $17\text{kN}/{\text{m}}^3$.

The saturated unit weight of sand ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$ is $19.2\text{kN}/{\text{m}}^3$.

The saturated unit weight of clay ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$ is $18.8\text{kN}/{\text{m}}^3$.

The saturated unit weight of peat ${\left({\gamma }_{\text{sat}}\right)}_{\text{peat}}$ is $15\text{kN}/{\text{m}}^3$.

The time $\left(t\right)$ is $8\text{months}$.

The properties of clay and organic layers are given in the Table.

The depth $\left(z\right)$ is $3.2\text{m}$.

Calculation:

Refer to part (a).

The pore water pressure $\left(u_o=\Delta {\sigma }^{\prime }\right)$ is $21.14\text{kN}/{\text{m}}^2$.

Calculate the length of maximum drainage path $\left(H_{\text{dr}}\right)$ using the relation.

$H_{\text{dr}}=\frac{H_2}{2}$

Substitute $4\text{m}$ for H.

$\begin{array}{c}{H}_{\text{dr}}=\frac{4}{2}\\ =2\text{m}\end{array}$

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

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

Substitute $0.006{\text{cm}}^2/\sec$ for $c_v$, $2\text{months}$ for $t$, and $2\text{m}$ for $H_{\text{dr}}$.

$\begin{array}{c}{T}_v=\frac{0.006{\text{cm}}^2/\sec \times 2\text{months}\times \frac{30\text{days}}{1\text{month}}\times \frac{24\text{hrs}}{1\text{day}}\times \frac{60\text{mins}}{1\text{hr}}\times \frac{60\text{sec}}{1\min }}{{\left(2\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)}^2}\\ =0.7776\end{array}$

Calculate the ratio $\frac{z}{H_{\text{dr}}}$ as shown below.

Substitute $3.2\text{m}$ for z and $2\text{m}$ for $H_{\text{dr}}$.

$\begin{array}{c}\frac{z}{H_{\text{dr}}}=\frac{3.2}{2}\\ =1.6\end{array}$

Calculate the degree of consolidation $\left(U\right)$ as shown below.

Refer Figure 11.29 “Variation of $U_z$ with $T_v$ and $\frac{z}{H_{\text{dr}}}$” in the Text Book.

Take the value of U as $\approx 0.9$, for the values $T_v$ of $0.7776$ and $\frac{z}{H_{\text{dr}}}$ of $1.6$.

Calculate the excess pore water pressure after 2 months $\left(u_z\right)$ as shown below.

$U=1-\frac{u_z}{u_o}$

Substitute $0.9$ for U and $21.14\text{kN}/{\text{m}}^2$ for $u_o$.

$\begin{array}{l}0.9=1-\frac{u_z}{21.14}\\ \frac{u_z}{21.14}=0.1\\ u_z=2.114\text{kN}/{\text{m}}^2\end{array}$

Hence, the excess pore water pressure at point A $\left(u_z\right)$ is $\underset{\_}{2.114\text{kN}/{\text{m}}^2}$.

(f)

To determine

Calculate the effective stress at point A two months after the application of the fill load.

Answer to Problem 11.1CTP

The effective stress at point A is $\underset{\_}{87.38\text{kN}/{\text{m}}^2}$.

Explanation of Solution

Given information:

The thickness of fill material $\left(t\right)$ is $1.75\text{m}$.

The compacted unit weight of fill material $\left({\gamma }_{\text{fill}}\right)$ is $20.1\text{kN}/{\text{m}}^3$.

The length of the foundation $\left(L\right)$ is $8\text{m}$.

The breadth of the foundation $\left(B\right)$ is $8\text{m}$.

The depth of fill $\left(D_f\right)$ is $1.5\text{m}$.

The height of the layer silty sand $\left(H_1\right)$ is $3\text{m}$.

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

The height of the peat layer $\left(H_3\right)$ is $1.8\text{m}$.

The dry unit weight of sand $\left({\gamma }_d\right)$ is $17\text{kN}/{\text{m}}^3$.

The saturated unit weight of sand ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$ is $19.2\text{kN}/{\text{m}}^3$.

The saturated unit weight of clay ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$ is $18.8\text{kN}/{\text{m}}^3$.

The saturated unit weight of peat ${\left({\gamma }_{\text{sat}}\right)}_{\text{peat}}$ is $15\text{kN}/{\text{m}}^3$.

The time $\left(t\right)$ is $8\text{months}$.

The properties of clay and organic layers are given in the Table.

The depth $\left(z\right)$ is $3.2\text{m}$.

Calculation:

Refer to part (a).

The pore water pressure $\left(u_o=\Delta {\sigma }^{\prime }\right)$ is $21.14\text{kN}/{\text{m}}^2$.

Refer to part (e)

The excess pore water pressure at point A $\left(u_z\right)$ is $2.114\text{kN}/{\text{m}}^2$.

Calculate the increase in effective stress $\left(\Delta u\right)$ after 2 months as shown below.

$\Delta u=u_o-u_z$

Substitute $21.14\text{kN}/{\text{m}}^2$ for $u_o$ and $2.114\text{kN}/{\text{m}}^2$ for $u_z$.

$\begin{array}{c}\Delta u=21.14-2.114\\ =19.026\text{kN}/{\text{m}}^2\end{array}$

Calculate the average effective stress at the point A $\left({\sigma }_{o-A}{}^{\prime }\right)$ as shown below.

${\sigma }_{o-A}{}^{\prime }={\gamma }_d{D}_f+\left({\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}-{\gamma }_w\right)\left(H_1-D_f\right)+\left({\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}-{\gamma }_w\right)\times z$

Substitute $17\text{kN}/{\text{m}}^3$ for ${\gamma }_d$, $1.5\text{m}$ for $D_f$, $19.2\text{kN}/{\text{m}}^3$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$, $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$, $18.8\text{kN}/{\text{m}}^3$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$, $3\text{m}$ for $H_1$, and $3.2\text{m}$ for z.

$\begin{array}{c}{\sigma }_{o-A}{}^{\prime }=17\times 1.5+\left(19.2-9.81\right)\left(3-1.5\right)+\left(18.8-9.81\right)\times 3.2\\ =25.5+14.085+28.768\\ =68.353\text{kN}/{\text{m}}^2\end{array}$

Calculate the final effective stress at point A as shown below.

$\text{Final}\text{effective}\text{stress}={\sigma }_{o-A}{}^{\prime }+\Delta u$

Substitute $68.353\text{kN}/{\text{m}}^2$ for ${\sigma }_{o-A}{}^{\prime }$ and $19.026\text{kN}/{\text{m}}^2$ for $\Delta u$.

$\begin{array}{c}\text{Final}\text{effective}\text{stress}=68.353+19.026\\ =87.38\text{kN}/{\text{m}}^2\end{array}$

Hence, the effective stress at point A is $\underset{\_}{87.38\text{kN}/{\text{m}}^2}$.

(g)

To determine

Calculate the piezometer reading at point A two months after the application of the fill load.

Answer to Problem 11.1CTP

The piezometer reading at point A is $\underset{\_}{48.22\text{kN}/{\text{m}}^2}$.

Explanation of Solution

Given information:

The thickness of fill material $\left(t\right)$ is $1.75\text{m}$.

The compacted unit weight of fill material $\left({\gamma }_{\text{fill}}\right)$ is $20.1\text{kN}/{\text{m}}^3$.

The length of the foundation $\left(L\right)$ is $8\text{m}$.

The breadth of the foundation $\left(B\right)$ is $8\text{m}$.

The depth of fill $\left(D_f\right)$ is $1.5\text{m}$.

The height of the layer silty sand $\left(H_1\right)$ is $3\text{m}$.

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

The height of the peat layer $\left(H_3\right)$ is $1.8\text{m}$.

The dry unit weight of sand $\left({\gamma }_d\right)$ is $17\text{kN}/{\text{m}}^3$.

The saturated unit weight of sand ${\left({\gamma }_{\text{sat}}\right)}_{\text{sand}}$ is $19.2\text{kN}/{\text{m}}^3$.

The saturated unit weight of clay ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$ is $18.8\text{kN}/{\text{m}}^3$.

The saturated unit weight of peat ${\left({\gamma }_{\text{sat}}\right)}_{\text{peat}}$ is $15\text{kN}/{\text{m}}^3$.

The time $\left(t\right)$ is $8\text{months}$.

The properties of clay and organic layers are given in the Table.

The depth $\left(z\right)$ is $3.2\text{m}$.

Calculation:

Refer to part (e)

The excess pore water pressure at point A $\left(u_z\right)$ is $2.114\text{kN}/{\text{m}}^2$.

The piezometer reading is the total pore water pressure.

Calculate the piezometer reading $\left(u_{\text{piezometer}}\right)$ as shown below.

$u_{\text{piezometer}}=\left(H_1-D_f+z\right){\gamma }_w+u_z$

Substitute $3\text{m}$ for $H_1$, $1.5\text{m}$ for $D_f$, $3.2\text{m}$ for z, $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$, and $2.114\text{kN}/{\text{m}}^2$ for $u_z$.

$\begin{array}{c}{u}_{\text{piezometer}}=\left(3-1.5+3.2\right)9.81+2.114\\ =48.221\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Hence, the piezometer reading at point A is $\underset{\_}{48.22\text{kN}/{\text{m}}^2}$.