Chapter 9, Problem 9.1P

A 20 m long concrete pile is shown in Figure P12.2. Estimate the ultimate point load Q_p by

1. a. Meyerhof’s method
2. b. Vesic’s method
3. c. Coyle and Castello’s method

Use m = 600 in Eq. (12.28).

To determine

Find the ultimate point load using Meyerhof’s method.

Answer to Problem 9.1P

The ultimate point load using Meyerhof’s method is $\underset{\_}{5,001.3\text{kN}}$.

Explanation of Solution

Given information:

The length of the concrete pile is 20 m.

Area of the concrete pile $\left(A_p\right)$ is $460\text{mm}\times 460\text{mm}$.

Soil friction angle $\left(\varphi \text{'}\right)$ is $30°$.

Unit weight of sand $\left(\gamma \right)$ is $18.6{\text{kN/m}}^{\text{3}}$.

Depth $\left(D\right)$ is 20 m.

Calculation:

Find the bearing capacity $\left(q\text{'}\right)$ using the relation.

$q\text{'}=\gamma D$

Substitute 20 m for D and $18.6{\text{kN/m}}^{\text{3}}$ for $\gamma$.

$\begin{array}{c}q\text{'}=18.6{\text{kN/m}}^{\text{3}}\times 20\text{m}\\ \text{= }372{\text{kN/m}}^{\text{2}}\end{array}$

Refer Table 12.6. “Interpolated values of $N_q^{*}$ based on Meyerhof’s theory” in the textbook.

The value of $N_q^{*}$ is 525 for the friction angle of $30°$.

Take the atmospheric pressure $p_a$ as $100{\text{kN/m}}^2$.

Find the value of ultimate point load $\left(Q_p\right)$ using the relation:

$Q_p=A_{p}q\text{'}{N}_q^{*}\le A_p{q}_l$ (1)

Here, $q_l$ is the limiting point resistance.

Find the value of $A_{p}q\text{'}{N}_q^{*}$ as follows:

Substitute $460\text{mm}\times 460\text{mm}$ for $A_p$, $372{\text{kN/m}}^{\text{2}}$ for $q\text{'}$, and 525 for $N_q^{*}$.

$\begin{array}{c}{A}_{p}q\text{'}{N}_q^{*}=460\text{mm}\times 460\text{mm}\times 372{\text{kN/m}}^{\text{2}}\times 525\\ =0.460\text{m}\times 0.460\text{m}\times 372{\text{kN/m}}^{\text{2}}\times 525\\ =41,325.5\text{kN}\end{array}$

Find the value of $A_p{q}_l$ as follows:

Substitute $0.5p_a{N}_q^{*}\tan \varphi \text{'}$ for $q_l$.

$A_p{q}_l=A_p\times 0.5p_a{N}_q^{*}\tan \varphi {\text{'}}_2$

Substitute $460\text{mm}\times 460\text{mm}$ for $A_p$, $100{\text{kN/m}}^2$ for $p_a$, 525 for $N_q^{*}$, and $30°$ for ${\varphi }_2^{\text{'}}$.

$\begin{array}{c}{A}_p{q}_l=460\text{mm}\times 460\text{mm}\times 0.5\times 100{\text{kN/m}}^2\times 525\times \tan 42°\\ =0.460\text{m}\times 0.460\text{m}\times 0.5\times 100{\text{kN/m}}^2\times 525\times \tan 42°\\ =5,001.3\text{kN}\end{array}$

The ultimate load should satisfy,

$Q_p=A_{p}q\text{'}{N}_q^{*}\le A_p{q}_l$

Substitute $41,325.5\text{kN}$ for $A_{p}q\text{'}{N}_q^{*}$ and $5,001.3\text{kN}$ for $A_p{q}_l$.

$Q_p=41,325.5\text{kN}>5,001.3\text{kN}$

So take the value of $Q_p$ as 5,001.3 kN.

Therefore, the ultimate point load using Meyerhof’s method is $\underset{\_}{5,001.3\text{kN}}$.

To determine

Find the ultimate point load using Vesic’s method.

Answer to Problem 9.1P

The ultimate point load using Vesic’s method is $\underset{\_}{5,232.3\text{kN}}$.

Explanation of Solution

Calculation:

Find the modulus of elasticity $\left(E_s\right)$ of the relation:

$E_s=m{p}_a$

Substitute 600 for m and $100{\text{kN/m}}^2$ for $p_a$.

$\begin{array}{c}{E}_s=600\times 100{\text{kN/m}}^2\\ =60,0000{\text{kN/m}}^2\end{array}$

Find the Poisson’s ratio of the soil using the relation:

${\mu }_s=0.1+0.3\left(\frac{\varphi \text{'}-25}{20}\right)$

Substitute 42 for $\varphi \text{'}$.

$\begin{array}{c}{\mu }_s=0.1+0.3\left(\frac{42-25}{20}\right)\\ =0.355\end{array}$

Find the value of average volumetric strain in the plastic zone below the pile point $\left(\Delta \right)$ using the relation:

$\Delta =0.005\left(1-\frac{\varphi \text{'}-25}{20}\right)\frac{q\text{'}}{p_a}$

Substitute 42 for $\varphi \text{'}$, $372{\text{kN/m}}^{\text{2}}$ for $q\text{'}$, and $100{\text{kN/m}}^2$ for $p_a$.

$\begin{array}{c}\Delta =0.005\left(1-\frac{42-25}{20}\right)\frac{372{\text{kN/m}}^{\text{2}}}{100{\text{kN/m}}^2}\\ =0.00279\end{array}$

Find the value of rigidity index $\left(I_r\right)$ using the relation:

$I_r=\frac{E_s}{2\left(1+{\mu }_s\right)q\text{'}\tan \varphi \text{'}}$

Substitute $60,0000{\text{kN/m}}^2$ for $E_s$, $0.355$ for ${\mu }_s$, $372{\text{kN/m}}^{\text{2}}$ for $q\text{'}$, and 42 for $\varphi \text{'}$.

$\begin{array}{c}{I}_r=\frac{60,0000{\text{kN/m}}^2}{2\left(1+0.355\right)372{\text{kN/m}}^{\text{2}}\times \tan 42}\\ =66.1\end{array}$

Find the value of reduced rigidity index $\left(I_{rr}\right)$ using the relation:

$I_{rr}=\frac{I_r}{1+I_r\Delta }$

Substitute 66.1 for $I_r$ and 0.00279 for $\Delta$.

$\begin{array}{c}{I}_{rr}=\frac{66.1}{1+66.1\times 0.00279}\\ =55.8\end{array}$

Find the mean effective normal ground stress $\left({\overline{\sigma }}_o^{\text{'}}\right)$ at the level of the pile point using the relation:

${\overline{\sigma }}_o^{\text{'}}=\left(\frac{1+2\left(1-\sin \varphi \text{'}\right)}{3}\right)q\text{'}$

Substitute $372{\text{kN/m}}^{\text{2}}$ for $q\text{'}$ and 42 for $\varphi \text{'}$.

$\begin{array}{c}{\overline{\sigma }}_o^{\text{'}}=\left(\frac{1+2\left(1-\sin 42\right)}{3}\right)372{\text{kN/m}}^{\text{2}}\\ =206.06{\text{kN/m}}^{\text{2}}\end{array}$

Refer Table 12.8. “Bearing capacity factors $N_{\sigma }^{*}$ based on the theory of expansion of cavities” in the textbook.

The value of $N_{\sigma }^{*}$ is 120 for the $\varphi \text{'}$ of $42°$.

Find the value of ultimate point load $\left(Q_p\right)$ using the relation:

$Q_p=A_p{\overline{\sigma }}_o^{\text{'}}{N}_{\sigma }^{*}$

Substitute $460\text{mm}\times 460\text{mm}$ for $A_p$, $206.06{\text{kN/m}}^{\text{2}}$ for ${\overline{\sigma }}_o^{\text{'}}$, and 120 for $N_{\sigma }^{*}$.

$\begin{array}{c}{Q}_p=460\text{mm}\times 460\text{mm}\times 206.06{\text{kN/m}}^{\text{2}}\times 120\\ =0.460\text{m}\times 0.460\text{m}\times 206.06{\text{kN/m}}^{\text{2}}\times 120\\ =5,232.3\text{kN}\end{array}$

Therefore, the ultimate point load using Vesic’s method is $\underset{\_}{5,232.3\text{kN}}$.

To determine

Find the ultimate point load using Coyle and Castello’s method.

Answer to Problem 9.1P

The ultimate point load using Coyle and Castello’s is $\underset{\_}{6,454.6\text{kN}}$.

Explanation of Solution

Calculation:

Find the value of $\frac{L}{D}$ using the relation:

Substitute 20 m for L and 460 mm for D.

$\begin{array}{c}\frac{L}{D}=\frac{20\text{m}}{460\text{mm}}\\ =\frac{20\text{m}}{460\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}\\ =43.48\end{array}$

Refer Figure 12.20, “Variation of $N_q^{*}$ with $\frac{L}{D}$” in the textbook.

The value of $N_q^{*}$ is 82 for $\frac{L}{D}$ of 43.48 and 42 for $\varphi \text{'}$.

Find the value of ultimate point load $\left(Q_p\right)$ using the relation:

$Q_p=q\text{'}{N}_q^{*}{A}_p$

Substitute $460\text{mm}\times 460\text{mm}$ for $A_p$, $372{\text{kN/m}}^{\text{2}}$ for $q\text{'}$, and 82 for $N_q^{*}$.

$\begin{array}{c}{Q}_p=372{\text{kN/m}}^{\text{2}}\times 460\text{mm}\times 460\text{mm}\times 82\\ =372{\text{kN/m}}^{\text{2}}\times 0.460\text{m}\times 0.460\text{m}\times 82\\ =6,454.6\text{kN}\end{array}$

Therefore, the ultimate point load using Coyle and Castello’s method is $\underset{\_}{6,454.6\text{kN}}$.