Chapter 11, Problem 11.6P

(a)

To determine

Find the expected primary consolidation settlement.

Answer to Problem 11.6P

The expected primary consolidation settlement $\left(S_p\right)$ is $\underset{\_}{\text{245}\text{.2}\text{mm}}$.

Explanation of Solution

Given information:

The compression index $C_c$ is 0.5.

The thickness of the clay layer (H) is 4 m.

The average permanent load on the clay layer $\Delta {\sigma }_{\left(p\right)}$ is $50\text{kN}/{\text{m}}^{\text{2}}$.

The average effective overburden pressure ${{\sigma }^{\prime }}_o$ is $80\text{kN}/{\text{m}}^{\text{2}}$.

The initial void ratio $e_0$ is 0.72.

Calculation:

Determine the maximum primary consolidation settlement caused by the structural load $\left(S_p\right)$:

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

Substitute 0.5 for $C_c$, 4 m for H, 0.72 for $e_0$, $80\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_o$, and $50\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_{\left(p\right)}$.

$\begin{array}{c}{S}_p=\frac{0.5\times 4}{1+0.72}\times \log \left[\frac{80+50}{80}\right]\\ =0.2452\text{m}\times \frac{\text{1,000}\text{mm}}{\text{1}\text{m}}\\ =245.2\text{mm}\end{array}$

Therefore, the expected primary consolidation settlement is $\underset{\_}{\text{245}\text{.2}\text{mm}}$.

(b)

To determine

Find the pre-compression time.

Answer to Problem 11.6P

The pre-compression time $\left(t_2\right)$ is $\underset{\_}{263\text{days}}$.

Explanation of Solution

Given information:

The average permanent load on the clay layer $\Delta {\sigma }_{\left(p\right)}$ is $50\text{kN}/{\text{m}}^{\text{2}}$.

The average effective overburden pressure ${{\sigma }^{\prime }}_o$ is $80\text{kN}/{\text{m}}^{\text{2}}$.

The surcharge load per unit area $\Delta {\sigma }_{\left(f\right)}$ is $60\text{kN}/{\text{m}}^{\text{2}}$.

The coefficient of consolidation $c_v$ is $2.5{\text{m}}^{\text{2}}/\text{yr}$.

The thickness of the clay layer (H) is 4 m.

Calculation:

Determine the ratio of $\left(\Delta {\sigma }_{\left(p\right)}/{{\sigma }^{\prime }}_o\right)$.

Substitute $50\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_{\left(p\right)}$ and $80\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_o$.

$\begin{array}{c}\frac{\Delta {\sigma }_{\left(p\right)}}{{{\sigma }^{\prime }}_o}=\frac{50}{80}\\ =0.625\end{array}$

Determine the value of $\left(\Delta {\sigma }_{\left(f\right)}/\Delta {\sigma }_{\left(p\right)}\right)$.

Substitute $60\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_{\left(f\right)}$ and $50\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_{\left(p\right)}$.

$\begin{array}{c}\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}=\frac{60}{50}\\ =1.2\end{array}$

Refer Figure 11.6, “Plot of U against $\Delta {\sigma }_{\left(f\right)}/\Delta {\sigma }_{\left(p\right)}$ for various values of $\Delta {\sigma }_{\left(p\right)}/{{\sigma }^{\prime }}_o$”, in the text book.

For $\frac{\Delta {\sigma }_{\left(p\right)}}{{{\sigma }^{\prime }}_o}=0.625$ and $\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}=1.2$,

The degree of consolidation is $U=57\%$

Refer Figure 11-8, “Plot of mid plane degree of consolidation against $T_v$” in the textbook.

For $U=57\%$,

The time factor is $T_v=0.45$

Determine the pre-compression time $t_2$ using the formula:

$t_2=\frac{T_v{H}_{\text{dr}}^2}{c_v}$

Here, $H_{\text{dr}}$ is the maximum drainage path ($H/2$ for two-way drainage and H for one-way drainage, and $t_2$ is the time duration of pre-compression.

Here, the drainage is two-way. Therefore, $H_{\text{dr}}=\frac{H}{2}$.

Substitute 0.45 for $T_v$, $\left(\frac{4}{2}\right)\text{m}$ for $H_{\text{dr}}$, and $2.5{\text{m}}^{\text{2}}/\text{yr}$ for $c_v$.

$\begin{array}{c}{t}_2=\frac{0.45\times {\left(\frac{\text{4}}{2}\right)}^2}{2.5\frac{{\text{m}}^{\text{2}}}{\text{yr}}}\\ =0.72\text{yr}\times \frac{\text{365}\text{days}}{\text{1}\text{yr}}\\ =263\text{days}\end{array}$

Thus, the pre-compression time is $\underset{\_}{263\text{days}}$.

(c)

To determine

Find the surcharge pressure that would eliminate the primary consolidation settlements beyond the preloading period.

Answer to Problem 11.6P

The surcharge pressure that would eliminate the primary consolidation settlements beyond the preloading period is $\left(\Delta {\sigma }_{\left(f\right)}\right)$ is $\underset{\_}{196\frac{\text{kN}}{{\text{m}}^{\text{2}}}}$.

Explanation of Solution

Given information:

The coefficient of consolidation $c_v$ is $2.5{\text{m}}^{\text{2}}/\text{yr}$.

The required time (t) is 6 months.

The thickness of the clay layer (H) is 4 m.

The average permanent load on the clay layer $\Delta {\sigma }_{\left(p\right)}$ is $50\text{kN}/{\text{m}}^{\text{2}}$.

The average effective overburden pressure ${{\sigma }^{\prime }}_o$ is $80\text{kN}/{\text{m}}^{\text{2}}$.

Calculation:

Determine the time factor using the formula:

$\begin{array}{c}{T}_v=\frac{t_2{c}_v}{H_{\text{dr}}^2}\\ =\frac{t_2{c}_v}{{\left(\frac{H}{2}\right)}^2}\\ =\frac{4t_2{c}_v}{H^2}\end{array}$

Substitute 6 months for $t_2$, 4 m for H, and $2.5{\text{m}}^{\text{2}}/\text{yr}$ for $c_v$.

$\begin{array}{c}{T}_v=\frac{4\times 6\text{months}\times 2.5\frac{{\text{m}}^{\text{2}}}{\text{yr}}}{{\left(4\text{m}\right)}^2}\\ =\frac{4\times \left(6\text{months}\times \frac{1\text{yr}}{12\text{months}}\right)\times 2.5\frac{{\text{m}}^{\text{2}}}{\text{yr}}}{{\left(4\text{m}\right)}^2}\\ =0.31\end{array}$

Refer Figure 11-8, “Plot of mid plane degree of consolidation against $T_v$” in the textbook.

For $T_v=0.31$,

The value is $U_{\left(z/H_{\text{dr}}=1\right)}=0.423$

Determine the value of $\left(\Delta {\sigma }_{\left(p\right)}/{{\sigma }^{\prime }}_o\right)$ ratio.

Substitute $50\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_{\left(p\right)}$ and $80\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_o$.

$\begin{array}{c}\frac{\Delta {\sigma }_{\left(p\right)}}{{{\sigma }^{\prime }}_o}=\frac{50\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{80\frac{\text{kN}}{{\text{m}}^{\text{2}}}}\\ =0.625\end{array}$

Determine the surcharge pressure using the formula:

$U=\frac{\log \left[1+\frac{\Delta {\sigma }_{\left(p\right)}}{{{\sigma }^{\prime }}_o}\right]}{\log \left\{1+\frac{\Delta {\sigma }_{\left(p\right)}}{{{\sigma }^{\prime }}_o}\left[1+\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}\right]\right\}}$

Substitute 0.625 for $\left(\Delta {\sigma }_{\left(p\right)}/{{\sigma }^{\prime }}_o\right)$, 0.41 for U, and $80\text{kN}/{\text{m}}^{\text{2}}$ $\Delta {\sigma }_{\left(p\right)}$.

$\begin{array}{c}\frac{\log \left[1+0.63\right]}{\log \left\{1+0.63\times \left[1+\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}\right]\right\}}=0.423\\ \frac{0.212}{\log \left\{1+0.63\times \left[1+\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}\right]\right\}}=0.423\\ log\left\{1+0.63\times \left[1+\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}\right]\right\}=\frac{0.212}{0.423}\\ \log \left\{1+0.63\times \left[1+\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}\right]\right\}=0.501\end{array}$

Simplify the Equation.

$\begin{array}{l}1+0.63\times \left[1+\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}\right]={10}^{0.501}\\ 0.63\times \left[1+\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}\right]=3.1735-1\\ 1+\frac{\Delta {\sigma }_{\left(f\right)}}{\Delta {\sigma }_{\left(p\right)}}=\frac{2.1735}{0.63}\\ \frac{\Delta {\sigma }_{\left(f\right)}}{80\frac{\text{kN}}{{\text{m}}^{\text{2}}}}=3.45-1\end{array}$

$\begin{array}{c}\Delta {\sigma }_{\left(f\right)}=2.45\times 80\frac{\text{kN}}{{\text{m}}^{\text{2}}}\\ \Delta {\sigma }_{\left(f\right)}=196\frac{\text{kN}}{{\text{m}}^{\text{2}}}\end{array}$

Thus, the surcharge pressure that would eliminate the primary consolidation settlements beyond the preloading period is $\underset{\_}{196\frac{\text{kN}}{{\text{m}}^{\text{2}}}}$.