Chapter 12, Problem 12.1P

a.

To determine

Sketch the variation of pile load with respect to depth.

Explanation of Solution

Given information:

The variation of frictional resistance per unit area $f_{\left(z\right)}$ with respect to depth for first pile is given in the Figure.

The variation of pile load $Q_{\left(z\right)}$ with respect to depth for second pile is given in the Figure.

The applied load (Q) is 1,500 kN.

Length of the pile is 20 m.

Number of piles is 2.

Diameter of the pile (d) is 500 mm.

Calculation:

The pile load $Q_{\left(z\right)}$ decreases linearly with the soils 1 and 2 due to uniform $f_{\left(z\right)}$.

Find the perimeter of the pile (p) as follows:

$p=\pi \times d$

Substitute 500 mm for d.

$\begin{array}{c}p=\pi \times 500\text{mm}\\ =\pi \times 500\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\\ =1.571\text{m}\end{array}$

Find the variation of pile load $\left(d{Q}_{\left(z\right)}\right)$ using the relation:

$d{Q}_{\left(z\right)}=-p{f}_{\left(z\right)}dz$ (1)

Here, $dz$ is the depth.

Within soil 1:

From given Figure, the value of $f_{\left(z\right)}$ is $40{\text{kN/m}}^2$ for soil 1.

Substitute 1.571 m for p, $40{\text{kN/m}}^2$ for $f_{\left(z\right)}$, and 10 m for $dz$ in Equation (1).

$\begin{array}{c}d{Q}_{\left(z\right)}=-1.571\text{m}\times 40{\text{kN/m}}^2\times 10\text{m}\\ =-\text{628}\text{.4}\text{kN}\end{array}$

Find the pile load within soil 1:

$Q_{\left(z=10\text{m}\right)}=Q+d{Q}_{\left(z\right)}$

Substitute 1,500 kN for Q and $-\text{628}\text{.4}\text{kN}$ for $d{Q}_{\left(z\right)}$.

$\begin{array}{c}{Q}_{\left(z=10\text{m}\right)}=1,500\text{kN}-\text{628}\text{.4}\text{kN}\\ =\text{871}\text{.6}\text{kN}\end{array}$

Within soil 2:

From given Figure, the value of $f_{\left(z\right)}$ is $70{\text{kN/m}}^2$ for soil 2.

Substitute 1.571 m for p, $70{\text{kN/m}}^2$ for $f_{\left(z\right)}$, and 5 m for $dz$ in Equation (1).

$\begin{array}{c}d{Q}_{\left(z\right)}=-1.571\text{m}\times 70{\text{kN/m}}^2\times 5\text{m}\\ =-\text{549}\text{.9}\text{kN}\end{array}$

Find the pile load within soil 2:

$Q_{\left(z=15\text{m}\right)}=Q_{\left(z=10\text{m}\right)}+d{Q}_{\left(z\right)}$

Substitute $\text{871}\text{.6}\text{kN}$ for $Q_{\left(z=10\text{m}\right)}$ and $-\text{549}\text{.9}\text{kN}$ for $d{Q}_{\left(z\right)}$.

$\begin{array}{c}{Q}_{\left(z=15\text{m}\right)}=\text{871}\text{.6}\text{kN}-\text{549}\text{.9}\text{kN}\\ =\text{321}\text{.7}\text{kN}\end{array}$

Plot the variation of pile load $Q_z$ with respect to depth.

b.

To determine

Sketch the variation of frictional resistance per unit area with respect to depth.

Explanation of Solution

Calculation:

The pile load $Q_{\left(z\right)}$ varies linearly within the soils 1 and 2. Therefore, the value of $f_{\left(z\right)}$ is constant within the two soils.

Find the frictional resistance per unit area $\left(f_{\left(z\right)}\right)$ using the relation:

$f_{\left(z\right)}=-\frac{1}{p}\frac{d{Q}_{\left(z\right)}}{dz}$ (2)

Within soil 1:

From given Figure, the value of $d{Q}_{\left(z\right)}$ is $1,500\text{kN}-800\text{kN}$ for soil 1.

Substitute 1.571 m for p, $1,500\text{kN}-800\text{kN}$ for $d{Q}_{\left(z\right)}$, and 10 m for $dz$ in Equation (2).

$\begin{array}{c}{f}_{\left(z\right)}=\frac{1}{1.571\text{m}}\frac{1,500\text{kN}-800\text{kN}}{10\text{m}}\\ =44.6{\text{kN/m}}^2\end{array}$

Within soil 2:

From given Figure, the value of $d{Q}_{\left(z\right)}$ is $800\text{kN}-300\text{kN}$ for soil 1.

Substitute 1.571 m for p, $800\text{kN}-300\text{kN}$ for $d{Q}_{\left(z\right)}$, and 5 m for $dz$ in Equation (2).

$\begin{array}{c}{f}_{\left(z\right)}=\frac{1}{1.571\text{m}}\frac{800\text{kN}-300\text{kN}}{5\text{m}}\\ =63.7{\text{kN/m}}^2\end{array}$

Sketch the variation of pile load $Q_z$ with respect to depth and the variation of frictional resistance per unit area $\left(f_{\left(z\right)}\right)$ with respect to depth as shown in Figure 1.