Chapter 17, Problem 17.17CTP

(a)

To determine

Find the settlement in sands in 10 years using the strain influence factor method.

Answer to Problem 17.17CTP

The settlement in sands in 10 years using the strain influence factor method is $\underset{\_}{7.55\text{mm}}$.

Explanation of Solution

Given information:

The width (B) of the continuous foundation is 2.0 m.

The cone penetration resistance ${\left(q_c\right)}_a$ for sand above clay layer is $10\text{MN}/{\text{m}}^{\text{2}}$.

The cone penetration resistance ${\left(q_c\right)}_b$ for sand below clay layer is $8\text{MN}/{\text{m}}^{\text{2}}$.

The depth of footing $\left(d_f\right)$ is 1.0 m.

The thickness $\left(t\right)$ of clay layer is 2 m.

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

The saturated unit weight $\left({\gamma }_{\text{sat}}\right)$ of the sand is $19{\text{kN/m}}^3$.

The stress $\left(\overline{q}\right)$ at the level of foundation is $125{\text{kN/m}}^2$.

The elasticity of sand layer is $E_s=3.5q_c$.

Calculation:

Find the applied stress (q) using the equation:

$q=\gamma D_f$

Substitute $17.5\text{kN}/{\text{m}}^{\text{3}}$ for $\gamma$ and 1.0 m for $D_f$.

$\begin{array}{c}q=17.5\times 1.0\\ =17.5\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the effective stress at the depth of $z_1$ before the construction of the foundation $\left[{q^{\prime }}_{z\left(1\right)}\right]$ using the equation:

${q^{\prime }}_{z\left(1\right)}=\gamma D$

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

$\begin{array}{c}{q^{\prime }}_{z\left(1\right)}=17.5\times 3\\ =52.5\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the maximum value of strain influence factor $I_{z\left(\text{m}\right)}$ using the relation:

$I_{z\left(\text{m}\right)}=0.5+0.1\sqrt{\frac{\overline{q}-q}{q_z{{}^{\prime }}_{(1)}}}$

Substitute $125\text{kN}/{\text{m}}^{\text{2}}$ for $\overline{q}$, $17.5\text{kN}/{\text{m}}^{\text{2}}$ for $q$, and $52.5\text{kN}/{\text{m}}^{\text{2}}$ for ${q^{\prime }}_{z\left(1\right)}$.

$\begin{array}{c}{I}_{z\left(\text{m}\right)}=0.5+0.1\sqrt{\frac{125-17.5}{52.5}}\\ =0.64\end{array}$

Find the correction factor $\left(C_1\right)$ without considering creep for the depth of foundation embedment using the equation:

$C_1=1-0.5\left(\frac{q}{\overline{q}-q}\right)$

Substitute $17.5\text{kN}/{\text{m}}^{\text{2}}$ for $q$ and $125\text{kN}/{\text{m}}^{\text{2}}$ for $\overline{q}$.

$\begin{array}{c}{C}_1=1-0.5\left(\frac{17.5}{125-17.5}\right)\\ =0.92\end{array}$

Find the correction factor $\left(C_2\right)$ to account for a creep in soil using the equation:

$C_2=1+0.2\log \left(\frac{t}{0.1}\right)$

Substitute 10 years for t.

$\begin{array}{c}{C}_2=1+0.2\log \left(\frac{10}{0.1}\right)\\ =1.4\end{array}$

Find the elasticity of the sand layer ${\left(E_s\right)}_a$ above the clay using the equation:

${\left(E_s\right)}_a=3.5\times {\left(q_c\right)}_a$

Substitute $10\text{MN}/{\text{m}}^{\text{2}}$ for ${\left(q_c\right)}_a$.

$\begin{array}{c}{E}_s=3.5\times 10\\ =35\text{MN}/{\text{m}}^{\text{2}}\end{array}$

Find the elasticity of the sand layer ${\left(E_s\right)}_b$ below the clay using the equation:

${\left(E_s\right)}_b=3.5\times {\left(q_c\right)}_b$

Substitute $8\text{MN}/{\text{m}}^{\text{2}}$ for ${\left(q_c\right)}_b$.

$\begin{array}{c}{\left(E_s\right)}_b=3.5\times 8\\ =28\text{MN}/{\text{m}}^{\text{2}}\end{array}$

For continuous foundation $\left(L/B\ge 10\right)$, the value of strain influence factor at base of the foundation is 0.2.

Sketch the strain influence diagram as shown in Figure 1.

Refer Figure 1.

Find the strain influence factor $I_{z\left(4\text{m}\right)}$ from ground level using similar triangle method:

$\begin{array}{c}{I}_{z\left(4\text{m}\right)}=\frac{0.64}{6}\times 4\\ =0.427\end{array}$

Find the term ${\left(I_z\Delta z\right)}_{\text{sand}1}$ for sand layer 1 using Figure 1:

$\begin{array}{c}{\left(I_z\Delta z\right)}_{\text{sand}1}=\text{Area}\text{of}\text{trapezoidal}\text{ABFG}\\ \text{=}\frac{\left(0.2+0.64\right)}{2}\left(2-0\right)\\ =0.84\text{m}\end{array}$

Find the term ${\left(I_z\Delta z\right)}_{\text{sand}2}$ for sand layer 2 using Figure 2:

$\begin{array}{c}{\left(I_z\Delta z\right)}_{\text{sand}2}=\text{Area}\text{of}\text{triangle}\text{CDE}\\ \text{=}\frac{1}{2}\left(0.427\right)\left(8-4\right)\\ =0.854\text{m}\end{array}$

Find the term $\left(\Sigma \frac{I_z\Delta z}{E_s}\right)$ using the equation:

$\Sigma \frac{I_z\Delta z}{E_s}=\frac{{\left(I_z\Delta z\right)}_{\text{sand}1}}{{\left(E_s\right)}_a}+\frac{{\left(I_z\Delta z\right)}_{\text{sand}2}}{{\left(E_s\right)}_b}$

Substitute 0.84 m for ${\left(I_z\Delta z\right)}_{\text{sand}1}$, $35\text{MN}/{\text{m}}^{\text{2}}$ for ${\left(E_s\right)}_a$, 0.854 m for ${\left(I_z\Delta z\right)}_{\text{sand}2}$, and $28\text{MN}/{\text{m}}^{\text{2}}$ for ${\left(E_s\right)}_b$.

$\begin{array}{c}\Sigma \frac{I_z\Delta z}{E_s}=\frac{0.84}{35}+\frac{0.853}{28}\\ =0.0545\frac{\text{m}}{\frac{\text{MN}}{{\text{m}}^{\text{2}}}}\times \frac{1\frac{\text{MN}}{{\text{m}}^{\text{2}}}}{1\text{MPa}}\\ =0.0545\text{m}/\text{MPa}\end{array}$

Find the settlement in sand ${\left(S_e\right)}_s$ using the equation:

${\left(S_e\right)}_s=C_1{C}_2\left(\overline{q}-q\right)\left(\Sigma \frac{I_z\Delta z}{E_s}\right)$

Substitute 0.92 for $C_1$, 1.4 for $C_2$, $125\text{kN}/{\text{m}}^{\text{2}}$ for $\overline{q}$, $17.5\text{kN}/{\text{m}}^{\text{2}}$ for $q$, and $0.0545\text{m}/\text{MPa}$ for $\left(\Sigma \frac{I_z\Delta z}{E_s}\right)$.

$\begin{array}{c}{\left(S_e\right)}_s=0.92\times 1.4\left(125-17.5\right)\left(0.0545\frac{\text{m}}{\frac{\text{MN}}{{\text{m}}^{\text{2}}}}\times \frac{1\frac{\text{MN}}{{\text{m}}^{\text{2}}}}{{10}^3\frac{\text{kN}}{{\text{m}}^{\text{2}}}}\right)\\ =0.00755\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ =\text{7}\text{.55}\text{mm}\end{array}$

Therefore, the settlement in sands in 10 years using the strain influence factor method is $\underset{\_}{7.55\text{mm}}$.

(b)

To determine

Find the elastic settlement in the clay assuming undrained conditions.

Answer to Problem 17.17CTP

The elastic settlement in the clay assuming undrained conditions is $\underset{\_}{2.9\text{mm}}$.

Explanation of Solution

Given information:

The elasticity of soil is $E_s=20{\text{MN/m}}^2$

Calculation:

Consider two homogenous clay layers below the base of the foundation. One is the clay layer up to 4 m depth below foundation and another one is clay layer up to 2 m below foundation.

Sketch the cross section of clay layer 1 as shown in Figure 2.

Here, the depth of clay layer (H) is 4 m.

Find the ratio of depth of footing to breadth of footing $\left(\frac{D_f}{B}\right)$.

Substitute 1 m for $D_f$ and 2 m for B.

$\begin{array}{c}\left(\frac{D_f}{B}\right)=\frac{1}{2}\\ =0.5\end{array}$

Find the ratio of length of footing to breadth of footing $\left(\frac{L}{B}\right)$.

Substitute $\infty$ for L and 2 m for B.

$\begin{array}{c}\left(\frac{L}{B}\right)=\frac{\infty }{2}\\ =\infty \end{array}$

Find the ratio of depth of clay layer to breadth of footing $\left(\frac{H}{B}\right)$.

Substitute 4 m for H and 2 m for B.

$\begin{array}{c}\left(\frac{H}{B}\right)=\frac{4}{2}\\ =2\end{array}$

Find the net applied pressure $\left(q_0\right)$ using the equation:

$q_0=\overline{q}-q$

Substitute $125\text{kN}/{\text{m}}^{\text{2}}$ for $\overline{q}$, $17.5\text{kN}/{\text{m}}^{\text{2}}$ for $q$.

$\begin{array}{c}{q}_0=125-17.5\\ =107.5\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the factor $\left(A_1\right)$.

Refer Table 17.1, “Variation of $A_1$ with $D_f/B$” in the textbook.

Take the value of $A_1$ with corresponding $D_f/B$ of 0.5 as 0.975.

Determine the factor $\left(A_2\right)$.

Refer Table 17.2, “Variation of $A_2$ with $L/B$ and $H/B$” in the text book.

Take the value of $A_2$ with corresponding $L/B$ of $\infty$ and is $H/B$ of 2 as 0.64.

Find the elastic settlement of clay layer 1 $\left(S_{e1}\right)$ using the equation:

$S_{e1}=A_1{A}_2\frac{q_{0}B}{E_s}$

Substitute 0.975 for $A_1$, 0.64 for $A_2$, $107.5\text{kN}/{\text{m}}^{\text{2}}$ for $q_0$, 2 m for B, and $20\text{MN}/{\text{m}}^{\text{2}}$ for $E_s$.

$\begin{array}{c}{S}_{e1}=0.975\times 0.64\times \frac{107.5\times 2}{\left(20{\text{MN/m}}^2\times \frac{{10}^3{\text{kN/m}}^2}{1{\text{MN/m}}^2}\right)}\\ =0.0067\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ =\text{6}\text{.7}\text{mm}\end{array}$

Sketch the cross section of clay layer 2 as shown in Figure 3.

Here, the depth of clay layer (H) is 2 m.

Find the ratio of depth of footing to breadth of footing $\left(\frac{D_f}{B}\right)$.

Substitute 1 m for $D_f$ and 2 m for B.

$\begin{array}{c}\left(\frac{D_f}{B}\right)=\frac{1}{2}\\ =0.5\end{array}$

Find the ratio of length of footing to breadth of footing $\left(\frac{L}{B}\right)$.

Substitute $\infty$ for L and 2 m for B.

$\begin{array}{c}\left(\frac{L}{B}\right)=\frac{\infty }{2}\\ =\infty \end{array}$

Find the ratio of depth of clay layer to breadth of footing $\left(\frac{H}{B}\right)$.

Substitute 2 m for H and 2 m for B.

$\begin{array}{c}\left(\frac{H}{B}\right)=\frac{2}{2}\\ =1\end{array}$

Find the net applied pressure $\left(q_0\right)$ using the equation:

$q_0=\overline{q}-q$

Substitute $125\text{kN}/{\text{m}}^{\text{2}}$ for $\overline{q}$ and $17.5\text{kN}/{\text{m}}^{\text{2}}$ for $q$.

$\begin{array}{c}{q}_0=125-17.5\\ =107.5\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the factor $\left(A_1\right)$.

Refer Table 17.1, “Variation of $A_1$ with $D_f/B$” in the textbook.

Take the value of $A_1$ with corresponding $D_f/B$ of 0.5 is 0.975.

Determine the factor $\left(A_2\right)$.

Refer Table 17.2, “Variation of $A_2$ with $L/B$ and $H/B$” in the text book.

Take the value of $A_2$ with corresponding $L/B$ of $\infty$ and is $H/B$ of 1 is 0.36.

Find the elastic settlement of clay layer 2 $\left(S_{e2}\right)$ using the equation:

$S_{e2}=A_1{A}_2\frac{q_{0}B}{E_s}$

Substitute 0.975 for $A_1$, 0.36 for $A_2$, $107.5\text{kN}/{\text{m}}^{\text{2}}$ for $q_0$, 2 m for B, and $20\text{MN}/{\text{m}}^{\text{2}}$ for $E_s$.

$\begin{array}{c}{S}_{e2}=0.975\times 0.36\times \frac{107.5\times 2}{\left(20{\text{MN/m}}^2\times \frac{{10}^3{\text{kN/m}}^2}{1{\text{MN/m}}^2}\right)}\\ =0.0038\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ =3.8\text{mm}\end{array}$

Find the elastic settlement of the clay layer $\left(S_e\right)$ using the equation:

$S_e=S_{e1}-S_{e2}$

Substitute 6.7 mm for $S_{e1}$ and 3.8 mm for $S_{e2}$.

$\begin{array}{c}{S}_e=6.7-3.8\\ =2.9\text{mm}\end{array}$

Therefore, the elastic settlement in the clay assuming undrained conditions is $\underset{\_}{2.9\text{mm}}$.

(c)

To determine

Find the consolidation settlement in the clay.

Answer to Problem 17.17CTP

The consolidation settlement in the clay is $\underset{\_}{108.2\text{mm}}$.

Explanation of Solution

Calculation:

Find the increase in vertical stress at the top $\left(\Delta {\sigma }_{\text{t}}\right)$ using 2:1 method:

$\left(\Delta {\sigma }_{\text{t}}\right)=\frac{\overline{q}t}{t+h_t}$

Here, t is the thickness of the clay layer and $h_{\text{t}}$ is the depth of clay layer at the top.

Substitute $125\text{kN}/{\text{m}}^{\text{2}}$ for $\overline{q}$, 2 m for t, and 2 m for $h_t$.

$\begin{array}{c}\left(\Delta {\sigma }_{\text{t}}\right)=\frac{125\times 2}{2+2}\\ =62.5\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the increase in vertical stress at the middle $\left(\Delta {\sigma }_{\text{m}}\right)$ using 2:1 method:

$\left(\Delta {\sigma }_{\text{m}}\right)=\frac{\overline{q}t}{d+h_{\text{m}}}$

Here, $h_{\text{m}}$ is the depth of clay layer at the middle.

Substitute $125\text{kN}/{\text{m}}^{\text{2}}$ for $\overline{q}$, 2 m for t, and 3 m for $h_{\text{m}}$.

$\begin{array}{c}\left(\Delta {\sigma }_m\right)=\frac{125\times 2}{2+3}\\ =50\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the increase in vertical stress at the bottom $\left(\Delta {\sigma }_{\text{b}}\right)$ using 2:1 method:

$\left(\Delta {\sigma }_{\text{b}}\right)=\frac{\overline{q}t}{t+h_{\text{b}}}$

Here, $h_{\text{b}}$ is the depth of clay layer at the bottom.

Substitute $125\text{kN}/{\text{m}}^{\text{2}}$ for $\overline{q}$, 2 m for t, and 4 m for $h_{\text{b}}$.

$\begin{array}{c}\left(\Delta {\sigma }_b\right)=\frac{125\times 2}{2+4}\\ =41.67\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the average increase in vertical stress $\left(\Delta {{\sigma }^{\prime }}_{\text{av}}\right)$ using Simpson’s rule:

$\Delta {{\sigma }^{\prime }}_{\text{av}}=\frac{\Delta {\sigma }_{\text{t}}+4\Delta {\sigma }_{\text{m}}+\Delta {\sigma }_{\text{b}}}{6}$

Substitute $62.5\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_{\text{t}}$, $50\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_{\text{m}}$, and $41.67\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {\sigma }_{\text{b}}$.

$\begin{array}{c}\Delta {{\sigma }^{\prime }}_{\text{av}}=\frac{62.5+4\times 50+41.67}{6}\\ =50.7\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the initial effective over burden stress at the middle of the clay layer $\left({{\sigma }^{\prime }}_0\right)$ using the equation:

${{\sigma }^{\prime }}_0=D{\gamma }_{\text{dry}\left(\text{sand}\right)}+D_f\left({\gamma }_{\text{sub}}-{\gamma }_w\right)$

Substitute 3 m for D, $17.5\text{kN}/{\text{m}}^{\text{2}}$     for ${\gamma }_{\text{dry(sand)}}$,1 m for $D_f$, $19.5\text{kN}/{\text{m}}^{\text{3}}$ for ${\gamma }_{\text{clay}}$ and $9.81\text{kN}/{\text{m}}^{\text{3}}$ for ${\gamma }_w$.

$\begin{array}{c}{{\sigma }^{\prime }}_0=\left(3\times 17.5\right)+1\left(19.5-9.81\right)\\ =62.2\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Find the consolidation settlement in the clay $\left(S_p\right)$ using the equation:

$S_p=\frac{C_{c}H}{1+e_0}\log \left(\frac{{{\sigma }^{\prime }}_0+\Delta {{\sigma }^{\prime }}_{\text{av}}}{{{\sigma }^{\prime }}_0}\right)$

Substitute 0.4 for $C_c$, 2 m for H, 0.95 for $e_0$, $62.2\text{kN}/{\text{m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_0$, and $50.7\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta {{\sigma }^{\prime }}_{\text{av}}$.

$\begin{array}{c}{S}_p=\frac{0.4\times 2}{1+0.95}\log \left(\frac{62.2+50.7}{62.2}\right)\\ =0.106\text{m}\times \frac{{10}^3\text{mm}}{1\text{m}}\\ \text{=106}\text{mm}\end{array}$

Therefore, the consolidation settlement in the clay is $\underset{\_}{106\text{mm}}$.

(d)

To determine

Find the settlement of the footing in 10 years.

Answer to Problem 17.17CTP

The settlement of the footing in 10 years is $\underset{\_}{116.45\text{mm}}$.

Explanation of Solution

Calculation:

Find the settlement (total settlement) of the footing in 10 years ${\left(S_e\right)}_t$ using the equation:

${\left(S_e\right)}_t={\left(S_e\right)}_s+S_e+S_p$

Substitute 7.55 mm for ${\left(S_e\right)}_s$, 2.9 mm for $S_e$, and 106 mm for $S_p$.

$\begin{array}{c}{\left(S_e\right)}_t=7.55+2.9+106\\ =116.45\text{mm}\end{array}$

Therefore, the settlement of the footing in 10 years is $\underset{\_}{116.45\text{mm}}$.