Chapter 11, Problem 11.21P

(a)

To determine

Calculate the average stress increase in the clay layer due to the applied load.

Answer to Problem 11.21P

The average stress increase in the clay layer $\left(\Delta {\sigma }^{\prime }\right)$ is $\underset{\_}{18.63\text{kN}/{\text{m}}^2}$.

Explanation of Solution

Given information:

The foundation load $\left(P\right)$ is $478\text{kN}$.

The length of the foundation $\left(L\right)$ is $2.5\text{m}$.

The breadth of the foundation $\left(B\right)$ is $2.5\text{m}$.

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

The height of the clay layer 1 $\left(H_1\right)$ is $3.25\text{m}$.

The height of the clay layer 2 $\left(H_2\right)$ is $3.5\text{m}$.

The dry unit weight of sand $\left({\gamma }_d\right)$ is $16\text{kN}/{\text{m}}^3$.

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

The liquid limit $\left(LL\right)$ is $37$.

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.

$q=\frac{P}{BL}$

Substitute $478\text{kN}$ for P, $2.5\text{m}$ for B, and $2.5\text{m}$ for L.

$\begin{array}{c}q=\frac{478}{2.5\times 2.5}\\ =\frac{478}{6.25}\\ =76.48\text{kN}/{\text{m}}^2\end{array}$

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 the depth $\left(z\right)$ of $1.75\text{m}$:

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

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

Substitute $2.5\text{m}$ for B.

$\begin{array}{c}b=\frac{2.5}{2}\\ =1.25\text{m}\end{array}$

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

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

Substitute $2.5\text{m}$ for L and $2.5\text{m}$ for B.

$\begin{array}{c}{m}_1=\frac{2.5}{2.5}\\ =1\end{array}$

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

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

Substitute $1.25\text{m}$ for b and $1.75\text{m}$ for z.

$\begin{array}{c}{n}_1=\frac{1.75}{1.25}\\ =1.4\end{array}$

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 $0.522$, for the values $m_1$ of $1$ and $n_1$ of $1.4$.

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 $76.48\text{kN}/{\text{m}}^2$ for $q$ and $0.522$ for $I_4$ in Equation (1).

$\begin{array}{c}\Delta {\sigma }_z=76.48\times 0.522\\ =39.92\text{kN}/{\text{m}}^2\end{array}$

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	1.25	1.75	1.4	0.522	39.92
1	1.25	3.5	2.8	0.210	16.06
1	1.25	5.25	4.2	0.10	7.65

Table 1

Refer to table 1.

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

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

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 $39.92\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }_t{}^{\prime }$, $16.06\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }_m{}^{\prime }$, and $7.65\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }_b{}^{\prime }$.

$\begin{array}{c}\Delta {\sigma }^{\prime }=\frac{39.92+4\times 16.06+7.65}{6}\\ =\frac{111.81}{6}\\ =18.63\text{kN}/{\text{m}}^2\end{array}$

Hence, the average stress increase in the clay layer $\left(\Delta {\sigma }^{\prime }\right)$ is $\underset{\_}{18.63\text{kN}/{\text{m}}^2}$.

(b)

To determine

Calculate the primary consolidation settlement.

Answer to Problem 11.21P

The primary consolidation settlement $\left(S_C\right)$ is $\underset{\_}{184\text{mm}}$.

Explanation of Solution

Given information:

The foundation load $\left(P\right)$ is $478\text{kN}$.

The length of the foundation $\left(L\right)$ is $2.5\text{m}$.

The breadth of the foundation $\left(B\right)$ is $2.5\text{m}$.

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

The height of the clay layer 1 $\left(H_1\right)$ is $3.25\text{m}$.

The height of the clay layer 2 $\left(H_2\right)$ is $3.5\text{m}$.

The dry unit weight of sand $\left({\gamma }_d\right)$ is $16\text{kN}/{\text{m}}^3$.

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

The liquid limit $\left(LL\right)$ is $37$.

The moisture content $\left(w\right)$ is $19\%$.

The specific gravity of soil solids $\left(G_s\right)$ is $2.71$.

The preconsolidation pressure $\left({\sigma }_c{}^{\prime }\right)$ is $65\text{kN}/{\text{m}}^2$.

The swell index $\left(C_S\right)$ is $\frac{1}{5}{C}_C$.

Calculation:

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

The stress at the middle of the clay layer $\left(\Delta {\sigma }^{\prime }=q\right)$ is $76.48\text{kN}/{\text{m}}^2$.

Calculate the compression index $\left(C_C\right)$ using the relation.

$C_C=0.009\left(LL-10\right)$

Substitute $37$ for LL.

$\begin{array}{c}{C}_C=0.009\left(37-10\right)\\ =0.243\end{array}$

Calculate the swell index $\left(C_S\right)$ as shown below.

$C_S=\frac{1}{5}{C}_C$

Substitute $0.243$ for $C_C$.

$\begin{array}{c}{C}_S=\frac{1}{5}\times 0.243\\ =0.0486\end{array}$

Calculate the saturated unit weight of clay layer $\left({\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}\right)$ as shown below.

${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}=\frac{\left(1+w\right){\gamma }_w{G}_s}{1+w{G}_s}$

Substitute $19\%$ for w, $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$, and $2.71$ for $G_s$.

$\begin{array}{c}{\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}=\frac{\left(1+\frac{19}{100}\right)\times 9.81\times 2.71}{1+\frac{19}{100}\times 2.71}\\ =\frac{1.19\times 26.5851}{1.5149}\\ =20.88\text{kN}/{\text{m}}^3\end{array}$

Calculate the void ratio $\left(e_0\right)$ as shown below.

$e_0=w{G}_s$

Substitute $19\%$ for w and $2.71$ for $G_s$.

$\begin{array}{c}{e}_0=\frac{19}{100}\times 2.71\\ =0.515\end{array}$

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

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

Substitute $16\text{kN}/{\text{m}}^3$ for ${\gamma }_d$, $1.5\text{m}$ for $D_f$, $18.8\text{kN}/{\text{m}}^3$ for $\left({\gamma }_{\text{sat}}\right)$, $9.81\text{kN}/{\text{m}}^3$ for ${\gamma }_w$, $20.88\text{kN}/{\text{m}}^3$ for ${\left({\gamma }_{\text{sat}}\right)}_{\text{clay}}$, $3.25\text{m}$ for $H_1$, and $3.5\text{m}$ for $H_2$.

$\begin{array}{c}{\sigma }_o{}^{\prime }=16\times 1.5+\left(18.8-9.81\right)\left(3.25-1.5\right)+\left(20.88-9.81\right)\times \frac{3.5}{2}\\ =24+15.7325+19.3725\\ =59.1\text{kN}/{\text{m}}^2\end{array}$

The effective stress ${\sigma }_o{}^{\prime }=59.1\text{kN}/{\text{m}}^{{}^{\text{2}}}\le {\sigma }_c{}^{\prime }=65\text{kN}/{\text{m}}^{\text{2}}$. Hence, the clay is over consolidated.

Check for the condition ${\sigma }_o{}^{\prime }+\Delta {\sigma }^{\prime }>{\sigma }_c{}^{\prime }$ as shown below.

Substitute $59.1\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_o{}^{\prime }$, $76.48\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$, and $65\text{kN}/{\text{m}}^2$ for ${\sigma }_c{}^{\prime }$.

$\begin{array}{l}59.1+76.48>65\\ 135.58\text{kN}/{\text{m}}^2>65\text{kN}/{\text{m}}^2\end{array}$

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

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

Substitute $0.243$ for $C_C$, $0.0486$ for $C_S$, $3.5\text{m}$ for $H$, $0.515$ for $e_0$, $59.1\text{kN}/{\text{m}}^{\text{2}}$ for ${\sigma }_o{}^{\prime }$, $76.48\text{kN}/{\text{m}}^2$ for $\Delta {\sigma }^{\prime }$, and $65\text{kN}/{\text{m}}^2$ for ${\sigma }_c{}^{\prime }$.

$\begin{array}{c}{S}_C=\frac{0.0486\times 3.5}{1+0.515}\log \left(\frac{65}{59.1}\right)+\frac{0.243\times 3.5}{1+0.515}\log \left(\frac{59.1+76.48}{65}\right)\\ =0.1123\times 0.0413+0.5614\times 0.3193\\ =0.184\text{m}\times \frac{1,000\text{mm}}{1\text{m}}\\ =184\text{mm}\end{array}$

Therefore, the primary consolidation settlement $\left(S_C\right)$ is $\underset{\_}{184\text{mm}}$.

(c)

To determine

Calculate the degree of consolidation after 2 years.

Answer to Problem 11.21P

The degree of consolidation after 2 years $\left(U\right)$ is $\underset{\_}{25\%}$.

Explanation of Solution

Given information:

The settlement after 2 years $\left(S_{C\left(t\right)}\right)$ is $46\text{mm}$.

The foundation load $\left(P\right)$ is $478\text{kN}$.

The length of the foundation $\left(L\right)$ is $2.5\text{m}$.

The breadth of the foundation $\left(B\right)$ is $2.5\text{m}$.

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

The height of the clay layer 1 $\left(H_1\right)$ is $3.25\text{m}$.

The height of the clay layer 2 $\left(H_2\right)$ is $3.5\text{m}$.

Calculation:

Refer to part (b).

The primary consolidation settlement $\left(S_C\right)$ is $184\text{mm}$.

Calculate the degree of consolidation after 2 years $\left(U\right)$ as shown below.

$U=\frac{S_{C\left(t\right)}}{S_C}$

Substitute $46\text{mm}$ for $S_{C\left(t\right)}$ and $184\text{mm}$ for $S_C$.

$\begin{array}{c}U=\frac{46}{184}\\ =0.25\times 100\%\\ =25\%\end{array}$

Hence, the degree of consolidation $\left(U\right)$ is $\underset{\_}{25\%}$.

(d)

To determine

Calculate the coefficient of consolidation for the pressure range.

Answer to Problem 11.21P

The coefficient of consolidation of the clay $\left(c_v\right)$ is $\underset{\_}{0.075{\text{m}}^{\text{2}}/\text{year}}$.

Explanation of Solution

Given information:

The settlement after 2 years $\left(S_{C\left(t\right)}\right)$ is $46\text{mm}$.

The foundation load $\left(P\right)$ is $478\text{kN}$.

The length of the foundation $\left(L\right)$ is $2.5\text{m}$.

The breadth of the foundation $\left(B\right)$ is $2.5\text{m}$.

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

The height of the clay layer 1 $\left(H_1\right)$ is $3.25\text{m}$.

The height of the clay layer 2 $\left(H_2\right)$ is $3.5\text{m}$.

Calculation:

Refer to part (c).

The degree of consolidation $\left(U\right)$ is $25\%$.

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 $0.0491$, for the value U of $25\%$.

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

$H_{\text{dr}}=\frac{H_3}{2}$

Substitute $3.5\text{m}$ for H.

$\begin{array}{c}{H}_{\text{dr}}=\frac{3.5}{2}\\ =1.75\text{m}\end{array}$

Calculate the coefficient of consolidation $\left(c_v\right)$ using the relation.

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

Substitute $0.0491$ for $T_v$, $2\text{year}$ for $t$, and $1.75\text{m}$ for $H_{\text{dr}}$.

$\begin{array}{l}0.0491=\frac{c_v\times 2}{{1.75}^2}\\ 2c_v=0.1504\\ c_v=0.075{\text{m}}^{\text{2}}/\text{year}\end{array}$

Hence, the coefficient of consolidation of the clay $\left(c_v\right)$ is $\underset{\_}{0.075{\text{m}}^{\text{2}}/\text{year}}$.

(e)

To determine

Calculate the settlement in 3 years.

Answer to Problem 11.21P

The time settlement in 3 years $\left(S_{C\left(3\right)}\right)$ is $\underset{\_}{56\text{mm}}$.

Explanation of Solution

Given information:

The foundation load $\left(P\right)$ is $478\text{kN}$.

The length of the foundation $\left(L\right)$ is $2.5\text{m}$.

The breadth of the foundation $\left(B\right)$ is $2.5\text{m}$.

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

The height of the clay layer 1 $\left(H_1\right)$ is $3.25\text{m}$.

The height of the clay layer 2 $\left(H_2\right)$ is $3.5\text{m}$.

The time $\left(t\right)$ is $3\text{years}$.

Calculation:

Refer to part (b).

The primary consolidation settlement $\left(S_C\right)$ is $184\text{mm}$.

Refer to part (d).

The coefficient of consolidation of the clay $\left(c_v\right)$ is $0.075{\text{m}}^{\text{2}}/\text{year}$.

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

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

Substitute $3\text{years}$ for $t$, $0.075{\text{m}}^{\text{2}}/\text{year}$ for $c_v$, and $1.75\text{m}$ for $H_{\text{dr}}$.

$\begin{array}{l}{T}_v=\frac{0.075\times 3}{{1.75}^2}\\ =0.073\end{array}$

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

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

Take the value of $U$ as $\approx 30.5\%$, for the value $T_v$ of $0.073$.

Calculate the settlement in 3 years $\left(S_{C\left(3\right)}\right)$ as shown below.

$U=\frac{S_{C\left(3\right)}}{S_C}$

Substitute $30.5\%$ for U and $184\text{mm}$ for $S_C$.

$\begin{array}{l}\frac{30.5}{100}=\frac{S_{C\left(3\right)}}{184}\\ S_{C\left(3\right)}=56.12\text{mm}\\ S_{C\left(3\right)}\approx 56\text{mm}\end{array}$

Therefore, the time settlement in 3 years $\left(S_{C\left(3\right)}\right)$ is $\underset{\_}{56\text{mm}}$.