Chapter 18, Problem 18.25P

To determine

Find the consolidation settlement of the group.

Answer to Problem 18.25P

The consolidation settlement of the group is $\underset{\_}{225\text{mm}}$.

Explanation of Solution

Given information:

The total load $Q_g$ on the group pile is 2,500 kN.

The length $L_{p1}$ of pile in layer 1 is 3.0 m.

The length $L_{p2}$ of pile in layer 2 is 3.0 m.

The length $L_{p3}$ of pile in layer 3 is 8 0 m.

The unit weight ${\gamma }_1$ of the sand (layer 1) is $15{\text{kN/m}}^{\text{3}}$.

The saturated unit weight ${\gamma }_{\text{sat}\left(\text{2}\right)}$ of the sand (layer 2) is $17{\text{kN/m}}^{\text{3}}$.

The initial void ratio $e_0$ of the sand (layer 2) is 0.80.

The coefficient of consolidation $C_c$ of the sand (layer 2) is 0.24.

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

The initial void ratio $e_0$ of the normally consolidated clay (layer 3) is 0.85.

The coefficient of consolidation $C_c$ of the normally consolidated clay (layer 3) is 0.30.

The saturated unit weight ${\gamma }_{\text{sat}\left(4\right)}$ of the normally consolidated clay (layer 4) is $18{\text{kN/m}}^{\text{3}}$.

The initial void ratio $e_0$ of the normally consolidated clay (layer 4) is 1.0.

The coefficient of consolidation $C_c$ of the normally consolidated clay (layer 4) is 0.35.

The saturated unit weight ${\gamma }_{\text{sat}\left(5\right)}$ of the normally consolidated clay (layer 5) is $20{\text{kN/m}}^{\text{3}}$.

The initial void ratio $e_0$ of the normally consolidated clay (layer 5) is 0.7.

The coefficient of consolidation $C_c$ of the normally consolidated clay (layer 5) is 0.26.

Calculation:

Show the pressure diagram as in Figure 1.

Determine the total length L of pile using the formula.

$L=L_{p1}+L_{p2}+L_{p3}$

Substitute 3 m for $L_{p1}$, 3.0 m for $L_{p2}$, and 8.0 m for $L_{p3}$.

$\begin{array}{c}L=3.0\text{m}+3.0\text{m}+8.0\text{m}\\ =14\text{m}\end{array}$

The load starts transmitting to the soil at the pile depth of $\left(\frac{2L}{3}\right)$.

Determine the depth at which the load on the group pile $Q_g$, is transmitted to soil.

$\text{Depth}=\frac{2}{3}L$

Here, L is the depth of normally consolidated clay layer 1.

Substitute 10 m for L.

$\begin{array}{c}\text{Depth}=\frac{2}{3}\times 10\text{m}\\ =6.667\text{m}\end{array}$

Thus, the settlement starts in NCC layer 1.

Determine the stress ${{\sigma }^{\prime }}_{O\left(1\right)}$ distribution in normally consolidated clay (NCC) layer 1 using the formula.

${{\sigma }^{\prime }}_{O\left(1\right)}=\left({\gamma }_1{L}_1\right)+\left[\left({\gamma }_{\text{sat}\left(\text{2}\right)}-{\gamma }_w\right)L_2\right]+\left[\left({\gamma }_{\text{sat}\left(\text{3}\right)}-{\gamma }_w\right)L_3\right]$

Here, $L_1$ is the length of the pile in sand layer 1, ${\gamma }_w$ is the unit weight of the water, $L_2$ is the length of the pile in sand layer 2, and $L_3$ is the length of the pile in sand layer 3.

Take the unit weight of water as $9.81{\text{kN/m}}^{\text{3}}$.

Substitute $15{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, 3.0 m for $L_1$, $17{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}\left(\text{2}\right)}$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, 3.0 m for $L_2$, $19{\text{lb/ft}}^{\text{3}}$ for ${\gamma }_{\text{sat}\left(\text{3}\right)}$, and 6.667 m for $L_3$.

$\begin{array}{c}{{\sigma }^{\prime }}_{O\left(1\right)}=15\times 3+\left(17-9.81\right)\times 3.0+\left(19-9.81\right)\left(6.667\right)\\ =127.84{\text{kN/m}}^{\text{2}}\end{array}$

Determine the stress ${{\sigma }^{\prime }}_{O\left(2\right)}$ distribution in normally consolidated clay (NCC) layer 2 using the formula.

${{\sigma }^{\prime }}_{O\left(2\right)}={{\sigma }^{\prime }}_{O\left(1\right)}+\left[\left({\gamma }_{\text{sat}\left(\text{3}\right)}-{\gamma }_w\right)L_4\right]+\left[\left({\gamma }_{\text{sat}\left(\text{4}\right)}-{\gamma }_w\right)\left(\frac{L_4}{2}\right)\right]$

Here, $L_4$ is the length of the pile in NCC layer 4.

Substitute $127.84{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{O\left(1\right)}$, 3.333 m for $L_4$, $19{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}\left(\text{4}\right)}$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and $18{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}\left(\text{3}\right)}$.

$\begin{array}{c}{{\sigma }^{\prime }}_{O\left(2\right)}=127.84+\left(19-9.81\right)\times 3.333+\left(18-9.81\right)\left(\frac{4.0}{2}\right)\\ =127.84+47\\ =174.84{\text{kN/m}}^{\text{2}}\end{array}$

Determine the stress ${{\sigma }^{\prime }}_{O\left(3\right)}$ distribution in normally consolidated clay (NCC) layer 3 using the formula.

${{\sigma }^{\prime }}_{O\left(3\right)}={{\sigma }^{\prime }}_{O\left(2\right)}+\left[\left({\gamma }_{\text{sat}\left(\text{3}\right)}-{\gamma }_w\right)\frac{L_4}{2}\right]+\left[\left({\gamma }_{\text{sat}\left(\text{5}\right)}-{\gamma }_w\right)\left(\frac{L_5}{2}\right)\right]$

Here, $L_5$ is the length of the pile in NCC layer 4.

Substitute $174.85{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{O\left(2\right)}$, 4.0 m for $L_4$, $18{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}\left(\text{5}\right)}$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, $20{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}\left(4\right)}$, and 2.5 m for $L_5$.

$\begin{array}{c}{{\sigma }^{\prime }}_{O\left(3\right)}=174.85+\left(18-9.81\right)\times \frac{4.0}{2}+\left(20-9.81\right)\frac{2.5}{2}\\ =203.96{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the stress $\Delta {{\sigma }^{\prime }}_{\left(1\right)}$ increase at the middle of NCC Layer 1 using the formula.

$\Delta {{\sigma }^{\prime }}_{\left(1\right)}=\frac{Q_g}{{\left(B_g+L_4\right)}^2}$

Here, $Q_g$ is the total load on the group piles and $B_g$ is the width of the plan of pile group.

Substitute 2,500 kN for $Q_g$, 3.0 m for $B_g$, and 3.333 m for $L_g$.

$\begin{array}{c}\Delta {{\sigma }^{\prime }}_{\left(1\right)}=\frac{2,500}{{\left(3.0+3.333\right)}^2}\\ =62.33{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the stress $\Delta {{\sigma }^{\prime }}_{\left(2\right)}$ increase at the middle of NCC Layer 2 using the formula.

$\Delta {{\sigma }^{\prime }}_{\left(2\right)}=\frac{Q_g}{{\left(B_g+\left(L_3+\frac{L_4}{2}\right)\right)}^2}$

Substitute 2,500 kN for $Q_g$, 3.0 m for $B_g$, 6.667 for $L_3$, and 4.0 m for $L_4$.

$\begin{array}{c}\Delta {{\sigma }^{\prime }}_{\left(2\right)}=\frac{2,500}{{\left(3.0+\left(6.667+\frac{4.0}{2}\right)\right)}^2}\\ =18.36{\text{kN/m}}^{\text{2}}\end{array}$

Calculate the stress $\Delta {{\sigma }^{\prime }}_{\left(5\right)}$ increase at the middle of NCC Layer 3 using the formula.

$\Delta {{\sigma }^{\prime }}_{\left(3\right)}=\frac{Q_g}{{\left(B_g+L_3+L_4+\frac{L_5}{2}\right)}^2}$

Substitute 2,500 kN for $Q_g$, 3.0 m for $B_g$, 6.667 for $L_3$, 4.0 m for $L_4$, and 2.5 m for $L_5$.

$\begin{array}{c}\Delta {{\sigma }^{\prime }}_{\left(3\right)}=\frac{2,500}{{\left(3.0+6.667+4.0+\frac{2.5}{2}\right)}^2}\\ \frac{2,500}{{\left(3.0+11.92\right)}^2}\\ =11.23{\text{kN/m}}^{\text{2}}\end{array}$

Determine the value of $\left({{\sigma }^{\prime }}_{O\left(1\right)}+\Delta {{\sigma }^{\prime }}_{\left(1\right)}\right)$.

Substitute $127.84{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{O\left(1\right)}$ and $62.33{\text{kN/m}}^{\text{2}}$ for $\Delta {{\sigma }^{\prime }}_{\left(1\right)}$.

$\begin{array}{c}{{\sigma }^{\prime }}_{O\left(1\right)}+\Delta {{\sigma }^{\prime }}_{\left(1\right)}=127.84+62.33\\ =190.17{\text{kN/m}}^{\text{2}}\end{array}$

Determine the value of $\left({{\sigma }^{\prime }}_{O\left(2\right)}+\Delta {{\sigma }^{\prime }}_{\left(2\right)}\right)$.

Substitute $174.84{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{O\left(2\right)}$ and $18.36{\text{kN/m}}^{\text{2}}$ for $\Delta {{\sigma }^{\prime }}_{\left(2\right)}$.

$\begin{array}{c}{{\sigma }^{\prime }}_{O\left(2\right)}+\Delta {{\sigma }^{\prime }}_{\left(2\right)}=174.84+18.36\\ =193.20{\text{kN/m}}^{\text{2}}\end{array}$

Determine the value of $\left({{\sigma }^{\prime }}_{O\left(3\right)}+\Delta {{\sigma }^{\prime }}_{\left(3\right)}\right)$.

Substitute $203.96{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{O\left(3\right)}$ and $11.23{\text{kN/m}}^{\text{2}}$ for $\Delta {{\sigma }^{\prime }}_{\left(3\right)}$.

$\begin{array}{c}{{\sigma }^{\prime }}_{O\left(3\right)}+\Delta {{\sigma }^{\prime }}_{\left(3\right)}=203.96+11.23\\ =215.19{\text{kN/m}}^{\text{2}}\end{array}$

Determine the consolidated settlement $\Delta S_{c\left(1\right)}$ of NCC layer 1 using the formula.

$\Delta S_{c\left(1\right)}=\left[\frac{C_c{L}_1}{1+e_0}\right]\log \left(\frac{{{\sigma }^{\prime }}_{O\left(1\right)}+\Delta {{\sigma }^{\prime }}_{\left(1\right)}}{{{\sigma }^{\prime }}_{O\left(1\right)}}\right)$

Here, $e_0$ is the initial void ratio in NCC layer 1.

Substitute 0.30 for $C_c$, 0.85 for $e_0$, 6.667 m for $L_1$, $190.17{\text{kN/m}}^{\text{2}}$ for $\left({{\sigma }^{\prime }}_{O\left(1\right)}+\Delta {{\sigma }^{\prime }}_{\left(1\right)}\right)$, and $127.84{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{O\left(1\right)}$.

$\begin{array}{c}\Delta S_{c\left(1\right)}=\left[\frac{0.30\times 6.667}{1+0.85}\right]\times \log \left(\frac{190.17}{127.84}\right)\\ =1.0811\times \log \left(1.48756\right)\\ =0.186\text{m}\end{array}$

Determine the consolidated settlement $\Delta S_{c\left(2\right)}$ of NCC layer 2 using the formula.

$\Delta S_{c\left(2\right)}=\left[\frac{C_c{L}_2}{1+e_0}\right]\log \left(\frac{{{\sigma }^{\prime }}_{O\left(2\right)}+\Delta {{\sigma }^{\prime }}_{\left(2\right)}}{{{\sigma }^{\prime }}_{O\left(2\right)}}\right)$

Here, $e_0$ is the initial void ratio in NCC layer 2.

Substitute 0.35 for $C_c$, 1 for $e_0$, 4.0 m for $L_2$, $193.20{\text{kN/m}}^{\text{2}}$ for $\left({{\sigma }^{\prime }}_{O\left(2\right)}+\Delta {{\sigma }^{\prime }}_{\left(2\right)}\right)$, and $174.84{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{O\left(2\right)}$.

$\begin{array}{c}\Delta S_{c\left(2\right)}=\left[\frac{0.35\times 4.0}{1+1}\right]\times \log \left(\frac{193.20}{174.84}\right)\\ =0.7\times \log \left(1.105\right)\\ =0.03\text{m}\end{array}$

Determine the consolidated settlement $\Delta S_{c\left(3\right)}$ of NCC layer 3 using the formula.

$\Delta S_{c\left(3\right)}=\left[\frac{C_c{L}_3}{1+e}\right]\log \left(\frac{{{\sigma }^{\prime }}_{O\left(3\right)}+\Delta {{\sigma }^{\prime }}_{\left(3\right)}}{{{\sigma }^{\prime }}_{O\left(3\right)}}\right)$

Here, $e_0$ is the initial void ratio in NCC layer 3.

Substitute 0.26 for $C_c$, 0.7 for $e_0$, 2.5 m for $L_4$, $215.19{\text{kN/m}}^{\text{2}}$ for $\left({{\sigma }^{\prime }}_{O\left(3\right)}+\Delta {{\sigma }^{\prime }}_{\left(3\right)}\right)$, and $203.96{\text{kN/m}}^{\text{2}}$ for ${{\sigma }^{\prime }}_{O\left(3\right)}$.

$\begin{array}{c}\Delta S_{c\left(3\right)}=\left[\frac{0.26\times 2.5}{1+0.7}\right]\times \log \left(\frac{215.19}{203.96}\right)\\ =0.382\times \log \left(1.055\right)\\ =0.0089\text{m}\end{array}$

Determine the total $\Delta S_c$ consolidated settlement of the group pile using the formula.

$\Delta S_c=\Delta S_{c\left(1\right)}+\Delta S_{c\left(2\right)}+\Delta S_{c\left(3\right)}$

Substitute 0.186 m for $\Delta S_{c\left(1\right)}$, 0.03 m for $\Delta S_{c\left(2\right)}$, and 0.0089 m for $\Delta S_{c\left(3\right)}$.

$\begin{array}{c}\Delta S_c=0.186+0.03+0.0089\\ =0.2249\text{m}\times \left(\frac{\text{1,000}\text{mm}}{\text{1}\text{m}}\right)\\ =224.9\text{mm}\\ \approx 225\text{mm}\end{array}$

Therefore, the consolidated settlement of the group pile is $\underset{\_}{225\text{mm}}$.