Chapter 19, Problem 19.11CTP

(a)

To determine

Find the net allowable point bearing capacity of the drilled shaft ${\left(Q_p\right)}_{\text{all}}$.

Answer to Problem 19.11CTP

The net allowable point bearing capacity of the drilled shaft is $\underset{\_}{12,588.22\text{kN}}$.

Explanation of Solution

Given information:

The diameter of shaft $D_s$ is 1.5 m.

The base diameter $D_b$ is 0.75 m.

The length of the drilled shaft in clay layer 1 $L_1$ is 10 m.

The length of the drilled shaft in clay layer 2 $L_2$ is 0.75 m.

The total length of the drilled shaft L is 10.75 m

The unit weight of clay layer 1 ${\gamma }_1$ is $17{\text{kN/m}}^{\text{3}}$.

The unit weight of clay layer 2 ${\gamma }_2$ is $19{\text{kN/m}}^{\text{3}}$.

The soil friction angle ${\varphi }^{\prime }$ is $32°$.

The soil friction angle of clay layer 2 ${{\varphi }^{\prime }}_2$ is $40°$.

The factor of safety FOS is 4.

Calculation:

Find the area of the drilled shat in the clay layer 1 $A_p$ using the formula:

$A_p=\frac{\pi }{4}{D}_s^2$

Substitute 1.50 m for $D_s$.

$\begin{array}{c}{A}_p=\frac{\pi }{4}\times {\left(1.50\text{m}\right)}^2\\ =1.77{\text{m}}^{\text{2}}\end{array}$

Calculate the value of ratio $\left(L\text{/}{D}_b\right)$.

Substitute 10.75 m for L and 1.50 m for $D_s$.

$\begin{array}{c}\frac{L}{D_b}=\frac{10.75\text{m}}{1.5\text{m}}\\ =7.2\end{array}$

Calculate the vertical effective stress $q^{\prime }$ at the level of the bottom of the drilled shaft.

$q^{\prime }={\gamma }_1{L}_1+{\gamma }_2{L}_2$

Substitute $17{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, 10 m for $L_1$, $19{\text{kN/m}}^{\text{3}}$ for ${\gamma }_2$, and 0.75 m for $L_2$.

$\begin{array}{c}{q}^{\prime }=\left(17\frac{\text{kN}}{{\text{m}}^{\text{3}}}\times 10\text{m}\right)+\left(19\frac{\text{kN}}{{\text{m}}^{\text{3}}}\times 0.75\text{m}\right)\\ =184.3\frac{\text{kN}}{{\text{m}}^{\text{2}}}\end{array}$

Calculate the value of bearing capacity factor $N_q^{\ast }$ using the formula:

$N_q^{\ast }=0.21e^{0.17{{\varphi }^{\prime }}_2}$

Substitute $40°$ for ${{\varphi }^{\prime }}_2$.

$\begin{array}{c}{N}_q^{\ast }=0.21\times e^{0.17\times 40°}\\ =0.21\times e^{6.8}\\ =188.5\end{array}$

Calculate the net load-carrying capacity at the base of the drilled shaft $Q_{p\left(\text{net}\right)}$ using the formula:

$Q_{p\left(\text{net}\right)}=A_p{q}^{\prime }\left(\omega N_q^{\ast }-1\right)$

Get the value of correction factor, $\omega$, from Figure (19.8), “Variation of $\omega$ with ${\varphi }^{\prime }$ and $L\text{/}{D}_b$” corresponding to the soil friction angle value of $40°$ and ratio $L\text{/}{D}_b$ value of 7.2.

Take the value of correction factor as 0.82.

Substitute $1.77{\text{m}}^{\text{2}}$ for $A_p$, $184.3{\text{kN/m}}^2$ for $q^{\prime }$, 0.82 for $\omega$, and 188.5 for $N_q^{\ast }$.

$\begin{array}{c}{Q}_{p\left(\text{net}\right)}=1.77{\text{m}}^{\text{2}}\times 184.3\frac{\text{kN}}{{\text{m}}^2}\times \left(0.82\times 188.5-1\right)\\ =50,096.22\text{kN}\end{array}$

Find the soil-pile friction-angle ${\delta }^{\prime }$ using the formula:

${\delta }^{\prime }=0.5{\varphi }^{\prime }$

Substitute $32°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{\delta }^{\prime }=0.6\times 31°\\ =18.6°\end{array}$

Calculate the earth pressure coefficient $K_0$ value using the formula:

$K_0=1-\sin {\varphi }^{\prime }$

Substitute $32°$ for ${\varphi }^{\prime }$.

$\begin{array}{c}{K}_0=1-\sin \left(32°\right)\\ =0.470\end{array}$

Calculate the perimeter p of the pile using the formula:

$p=\pi D_b$

Substitute 0.75 m for $D_b$.

$\begin{array}{c}p=\pi \times 0.75\text{m}\\ =2.36\text{m}\end{array}$

Find the critical depth of the drilled shaft $L^{\prime }$ using the formula:

$L^{\prime }=15D_b$

Substitute 0.75 m for $D_b$.

$\begin{array}{c}{L}^{\prime }=15\times 0.75\text{m}\\ =11.3\text{m}\end{array}$

The critical depth of the drilled shaft more than the length of the entire shaft, therefore, the frictional resistance along the pile shaft will increase over the entire length of the shaft.

Consider 0 ft from top of the pile.

$z=0\text{m}$

Calculate the magnitude of unit frictional resistance, $f_{z=0\text{m}}$, at a height of 0 ft from the top of the pile using the formula:

$\begin{array}{c}{f}_{z=0\text{m}}=K{{\sigma }^{\prime }}_o\tan {\delta }^{\prime }\\ =K\left(\gamma z\right)\tan {\delta }^{\prime }\end{array}$

Substitute 0 m for z.

$\begin{array}{c}{f}_{z=0\text{m}}=K\left(\gamma \times 0\right)\tan {\delta }^{\prime }\\ =0\frac{\text{kN}}{{\text{m}}^{\text{2}}}\end{array}$

The frictional resistance (skin friction), $Q_{s\left(z=0\text{m}\right)}$, at the top of the pile is zero as the unit frictional resistance, $f_{z=0\text{m}}$, at a height of 0 ft from the top of the pile is zero.

$Q_{s\left(z=0\text{m}\right)}=0\text{kN}$

Consider the pile to the depth of 10 m (critical depth of the pile) from the top of pile tip.

$\begin{array}{c}z=L_1\\ =10\text{m}\end{array}$

Calculate the magnitude of unit frictional resistance, $f_{z=10\text{m}}$, at a height of 10 m from the top of the pile using the formula:

$\begin{array}{c}{f}_{z=10\text{m}}=K{{\sigma }^{\prime }}_o\tan {\delta }^{\prime }\\ =K\left({\gamma }_{1}z\right)\tan {\delta }^{\prime }\end{array}$

Substitute 10 m for z, 0.470 for K, $17{\text{kN/m}}^{\text{3}}$ for ${\gamma }_1$, and $15°$ for ${\delta }^{\prime }$.

$\begin{array}{c}{f}_{z=10\text{m}}=0.470\times \left(17\frac{\text{kN}}{{\text{m}}^{\text{3}}}\times 10\text{m}\right)\times \tan \left(15°\right)\\ =21.41\frac{\text{kN}}{{\text{m}}^{\text{2}}}\end{array}$

Find the frictional resistance (skin friction), $Q_s$, at a depth of 10 m from the top of the pile tip the formula:

$\begin{array}{c}{Q}_s=p{L}_1{f}_{\text{av}}\\ =p{L}_1\left(\frac{f_{z=0\text{m}}+f_{z=10\text{m}}}{2}\right)\end{array}$

Substitute 2.36 m for p, 10 m for $L_1$, $21.41{\text{kN/m}}^{\text{2}}$ for $f_{z=10\text{m}}$, and $0{\text{kN/m}}^{\text{2}}$ for $f_{z=0\text{m}}$.

$\begin{array}{c}{Q}_s=2.36\text{m}\times 10\text{m}\times \left(\frac{21.41\frac{\text{kN}}{{\text{m}}^{\text{2}}}+0\frac{\text{kN}}{{\text{m}}^{\text{2}}}}{2}\right)\\ =252.64\text{kN}\end{array}$

Calculate the net allowable point bearing capacity of the drilled shaft using the formula:

${\left(Q_p\right)}_{\text{all}}=\frac{Q_s+Q_p}{\text{FOS}}$

Substitute 252.64 kN for $Q_s$, 50,096.22 kN for $Q_p$, and 4.0 for FOS.

$\begin{array}{c}{\left(Q_p\right)}_{\text{all}}=\frac{252.64\text{kN}+50,096.22\text{kN}}{\text{4}}\\ =\frac{50,352.86\text{kN}}{4}\\ =12,588.22\text{kN}\end{array}$

Thus, the net allowable point bearing capacity of the drilled shaft is $\underset{\_}{12,588.22\text{kN}}$.

(b)

To determine

Find the allowable load carrying capacity of the pile $Q_{\text{all}}$.

Answer to Problem 19.11CTP

The allowable load carrying capacity of the pile is $\underset{\_}{16,263.4\text{kN}}$.

Explanation of Solution

Calculation:

Find the ratio of allowable settlement to the diameter of the base using the relation:

$\begin{array}{c}\frac{\text{Allowable}\text{settlement}}{D_b}=\frac{15}{1,500}\times 100\\ =1.0\%\end{array}$

Refer Table (19.6), “Normalized Base loading transfer with settlement for cohesionless soils (based on average curve)” in the text book.

Based on the ratio of allowable settlement to the diameter of the base, take the value of end bearing $\frac{\text{End}\text{bearing}}{q_p{A}_p}$ as 0.32.

Find the allowable load $Q_{p\left(\text{all}\right)}$ carrying capacity of the pile using the formula:

$\frac{Q_{p\left(\text{all}\right)}}{Q_{p\left(\text{net}\right)}}=0.32$

Substitute 50,096.22 kN for $Q_{p\left(\text{net}\right)}$.

$\begin{array}{l}\frac{Q_{p\left(\text{all}\right)}}{50,096.22}=0.32\\ Q_{p\left(\text{all}\right)}=0.32\times 50,096.32\\ Q_{p\left(\text{all}\right)}=16,031\text{kN}\end{array}$

Find the ratio of allowable settlement to the diameter of the shaft base using the relation:

$\begin{array}{c}\frac{\text{Allowable}\text{settlement}}{D_s}=\frac{15}{750}\times 100\\ =2.0\%\end{array}$

Refer Table (19.5), “Normalized side load transfer with settlement for cohesionless soils (based on average curve)” in the text book.

Based on the ratio of allowable settlement to the diameter of the shaft base, take the value of $\frac{\text{Side}\text{load}\text{transfer}}{{\displaystyle \sum f_{i}p\Delta L_i}}$ as 0.92.

Find the allowable load $Q_{s\left(\text{all}\right)}$ carrying capacity of the shaft using the formula:

$\frac{Q_{s\left(\text{all}\right)}}{Q_s}=0.92$

Substitute 252.64 kN for $Q_s$.

$\begin{array}{l}\frac{Q_{s\left(\text{all}\right)}}{252.64}=0.92\\ Q_{s\left(\text{all}\right)}=0.92\times 252.64\\ Q_{s\left(\text{all}\right)}=232.4\text{kN}\end{array}$

Find the allowable load $Q_{\text{all}}$ carrying capacity of the pile using the formula:

$Q_{\text{all}}=Q_{p\left(\text{all}\right)}+Q_{s\left(\text{all}\right)}$

Substitute 16,031 kN for $Q_{p\left(\text{all}\right)}$ and 232.4 kN for $Q_{s\left(\text{all}\right)}$.

$\begin{array}{c}{Q}_{\text{all}}=16,031+232.4\\ =16,263.4\text{kN}\end{array}$

Thus, the allowable load carrying capacity of the pile is $\underset{\_}{16,263.4\text{kN}}$.