Chapter 18, Problem 18.6P

To determine

Find the maximum allowable load on a driven pile.

Answer to Problem 18.6P

The maximum allowable load on a driven pile is $\underset{\_}{375.1\text{kN}}$.

Explanation of Solution

Given information:

The diameter of the driven pile D is 450 mm.

The length of the pile in the upper layer of the sand $L_1$ is 8 m.

The unit weight of the upper layer sand $\gamma$ is $17.0{\text{kN/m}}^{\text{3}}$.

The soil friction angle in the lower layer of sand ${{\varphi }^{\prime }}_1$ is $31°$.

The length of the pile in the lower layer of the sand $L_2$ is 10 m.

The saturated unit weight of the lower layer sand ${\gamma }_{\text{sat}}$ is $19.0{\text{kN/m}}^{\text{3}}$.

The soil friction angle in the lower layer of sand ${{\varphi }^{\prime }}_2$ is $33°$.

The soil-pile friction-angle ${\delta }^{\prime }$ is $0.65{\varphi }^{\prime }$.

The coefficient value K is $1.5K_o$.

The factor of safety $F_s$ is 3.

Calculation:

Draw the cross section of the pile as in Figure 1.

Calculate the earth pressure coefficient $K_0$ for the upper layer using the formula.

$K_{0,\text{UL}}=1-\sin {{\varphi }^{\prime }}_1$

Here, $K_{0,\text{UL}}$ is the value of $K_0$ for the upper layer.

Substitute $31°$ for ${{\varphi }^{\prime }}_1$.

$\begin{array}{c}{K}_{0,\text{UL}}=1-\sin \left(31°\right)\\ =0.485\end{array}$

Calculate the coefficient K for the upper layer using the formula.

$K_{\text{UL}}=1.5K_{0,\text{UL}}$

Here, $K_{\text{UL}}$ is the value of K for the upper layer.

Substitute 0.485 for $K_{0,\text{UL}}$.

$\begin{array}{c}{K}_{\text{UL}}=1.5\times 0.485\\ =0.727\end{array}$

Calculate the earth pressure coefficient $K_0$ for the lower layer using the formula.

$K_{0,\text{LL}}=1-\sin {{\varphi }^{\prime }}_2$

Here, $K_{0,\text{LL}}$ is the value of $K_0$ for the lower layer.

Substitute $33°$ for ${{\varphi }^{\prime }}_2$.

$\begin{array}{c}{K}_{0,\text{LL}}=1-\sin \left(33°\right)\\ =0.455\end{array}$

Calculate the coefficient K for the lower layer using the formula.

$K_{\text{LL}}=1.5K_{0,\text{LL}}$

Here, $K_{\text{LL}}$ is the value of K for the lower layer.

Substitute 0.455 for $K_{0,\text{LL}}$.

$\begin{array}{c}{K}_{\text{LL}}=1.5\times 0.455\\ =0.683\end{array}$

Calculate the area of pile $A_p$ using the formula.

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

Substitute 450 mm for D.

$\begin{array}{c}{A}_p=\frac{\pi }{4}\times {\left(450\text{mm}\right)}^2\\ =\frac{\pi }{4}\times {\left(450\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^2\\ =0.159{\text{m}}^{\text{2}}\end{array}$

Calculate the perimeter p of the pile using the formula.

$p=\pi D$

Substitute 450 mm for D.

$\begin{array}{c}p=\pi \times 450\text{mm}\\ =\pi \times \left(450\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)\\ =1.414\text{m}\end{array}$

Refer Figure (18.12), “Meyerhof’s bearing capacity factor, $N_q^{\ast }$”, corresponding to the soil friction angle” in the textbook.

Take the value of bearing capacity factor $N_q^{\ast }$, as 95 for ${\varphi }^{\prime }$ value of $33°$.

Calculate the load-carrying capacity $Q_p$ of the pile point using the formula.

$\begin{array}{c}{Q}_p=A_p{q}^{\prime }{N}_q^{\ast }\\ =A_p\left(\gamma L_1+L_2{\gamma }^{\prime }\right)N_q^{\ast }\\ =A_p\left[\gamma L_1+L_2\left({\gamma }_{\text{sat}}-{\gamma }_w\right)\right]N_q^{\ast }\end{array}$

Here, ${\gamma }_w$ is the unit weight of water and $q^{\prime }$ is the effective vertical stress at the level of the pile tip.

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

Substitute $0.159{\text{m}}^{\text{2}}$ for $A_p$, $17.0{\text{kN/m}}^{\text{3}}$ for $\gamma$, 8 m for $L_1$, 10 m for $L_2$, $19.0{\text{kN/m}}^{\text{3}}$ for ${\gamma }_{\text{sat}}$, $9.81{\text{kN/m}}^{\text{3}}$ for ${\gamma }_w$, and 95 for $N_q^{\ast }$.

$\begin{array}{c}{Q}_p=0.159\times \left\{\left(8\times 18.0\right)+\left[10\times \left(19-9.81\right)\right]\right\}\times 95\\ =15.105\times \left(144+91.9\right)\\ =3,563\text{kN}\end{array}$

Check the calculated value of load-carrying capacity of the pile point using Meyerhof’s equation.

$\begin{array}{c}{Q}_p=A_p{q}_l\\ =A_p\left(50N_q^{\ast }\tan {\varphi }^{\prime }\right)\end{array}$

Substitute $0.159{\text{m}}^{\text{2}}$ for $A_p$, 95 for $N_q^{\ast }$, and $33°$ for ${{\varphi }^{\prime }}_2$.

$\begin{array}{c}{Q}_p=0.159{\text{m}}^{\text{2}}\times \left(50\times 95\times \tan 33°\right)\\ =490.5\text{kN}\end{array}$

Use the lowest of the calculated value of load-carrying capacity of the pile point.

$\begin{array}{c}{Q}_p=\min \left\{490.5\text{kN,}\text{3563}\text{kN}\right\}\\ =490.5\text{kN}\end{array}$

Calculate the soil-pile friction-angle ${{\delta }^{\prime }}_{\text{UL}}$ for the upper layer of sand using the formula.

${{\delta }^{\prime }}_{\text{UL}}=0.65{\varphi }^{\prime }$

Substitute $31°$ for ${{\varphi }^{\prime }}_1$.

$\begin{array}{c}{{\delta }^{\prime }}_{\text{UL}}=0.65\times 31°\\ =20.2°\end{array}$

Calculate the soil-pile friction-angle ${{\delta }^{\prime }}_{\text{LL}}$ for the upper layer of sand using the formula.

${{\delta }^{\prime }}_{\text{UP}}=0.65{\varphi }^{\prime }$

Substitute $33°$ for ${{\varphi }^{\prime }}_2$.

$\begin{array}{c}{{\delta }^{\prime }}_{\text{UP}}=0.65\times 33°\\ =21.5°\end{array}$

Calculate the critical depth of the pile $L^{\prime }$ using the formula.

$L^{\prime }=15D$

Substitute 450 mm for D.

$\begin{array}{c}{L}^{\prime }=15\times 450\text{mm}\\ =15\times \left(450\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)\\ =6.75\text{m}\end{array}$

Calculate the unit frictional resistance at the upper layer of sand.

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.

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

Substitute 0 m for z.

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

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 6.75 m (critical depth of the pile) from the top of pile tip.

$\begin{array}{c}z=L^{\prime }\\ =6.75\text{m}\end{array}$

Calculate the magnitude of unit frictional resistance $f_{z=6.75\text{m}}$, at a height of 6.75 m from the top of the pile using the formula.

$\begin{array}{c}{f}_{z=6.75\text{m}}=K_{\text{UL}}{{\sigma }^{\prime }}_o\tan {{\delta }^{\prime }}_{\text{UL}}\\ =K_{\text{UL}}\left(\gamma z\right)\tan {{\delta }^{\prime }}_{\text{UL}}\end{array}$

Substitute 6.75 m for z, 0.727 for $K_{\text{UL}}$, $17{\text{kN/m}}^{\text{3}}$ for $\gamma$, and $20.2°$ for ${{\delta }^{\prime }}_{\text{UL}}$.

$\begin{array}{c}{f}_{z=6.75\text{m}}=0.727\times \left(17.0\times 6.75\right)\times \tan \left(20.2°\right)\\ =30.7{\text{kN/m}}^2\end{array}$

Calculate the magnitude of unit frictional resistance $f_{z=8\text{m}}$, at a height of 8 m from the top of the pile (end of upper layer of sand) using the formula.

$\begin{array}{c}{f}_{z=8\text{m}}=K_{\text{UL}}{{\sigma }^{\prime }}_o\tan {{\delta }^{\prime }}_{\text{UL}}\\ =K_{\text{UL}}\left(\gamma z\right)\tan {{\delta }^{\prime }}_{\text{UL}}\end{array}$

Substitute 6.75 m for z, 0.727 for $K_{\text{UL}}$, $17{\text{kN/m}}^{\text{3}}$ for $\gamma$, and $20.2°$ for ${{\delta }^{\prime }}_{\text{UL}}$.

$\begin{array}{c}{f}_{z=8\text{m}}=0.727\times \left(17.0\times 6.75\right)\times \tan \left(20.2°\right)\\ =30.7{\text{kN/m}}^2\end{array}$

Below the upper layer $\left(8\text{m}<z<15\text{m}\right)$, the unit frictioanl resistance remains constant $\left(30.7{\text{kN/m}}^{\text{2}}\right)$.

Calculate the frictional resistance (skin friction) $Q_{s\left(z=\text{0}\text{m}\text{to}\text{6}\text{.75}\text{m}\right)}$, at a depth of 6.75 m from the top of the pile tip using the formula.

$\begin{array}{c}{Q}_{s\left(z=\text{0}\text{m}\text{to}\text{6}\text{.75}\text{m}\right)}=p{L}^{\prime }{f}_{\text{av}}\\ =p{L}^{\prime }\left(\frac{f_{z=0\text{m}}+f_{z=6.75\text{m}}}{2}\right)\end{array}$

Substitute 1.414 m for p, 6.75 m for $L^{\prime }$, $30.7{\text{kN/m}}^{\text{2}}$ for $f_{z=6\text{m}}$, and $0{\text{kN/m}}^{\text{2}}$ for $f_{z=0\text{m}}$.

$\begin{array}{c}{Q}_{s\left(z=\text{0}\text{m}\text{to}\text{6}\text{.75}\text{m}\right)}=1.414\times 6.75\times \left(\frac{30.7+0}{2}\right)\\ =146.51\text{kN}\end{array}$

Calculate the frictional resistance (skin friction) $Q_{s\left(z=\text{6}\text{.75}\text{m}\text{to}\text{8}\text{m}\right)}$, from the critical depth of the pile to the end of upper layer of sand using the formula.

$Q_{s\left(z=\text{6}\text{.75}\text{m}\text{to}8\text{m}\right)}=p\left(L_1-L^{\prime }\right)f_{z=6.75\text{ft}}$

Substitute 1.414 m for p, 6.75 m for $L^{\prime }$, 15 m for $L_1$, and $30.7{\text{kN/m}}^{\text{2}}$ for $f_{z=6.75\text{m}}$.

$\begin{array}{c}{Q}_{s\left(z=\text{6}\text{m}\text{to}15\text{m}\right)}=1.414\text{m}\times \left(8\text{m}-6.75\text{m}\right)\times 30.7\frac{\text{kN}}{{\text{m}}^{\text{2}}}\\ =54.26\text{kN}\end{array}$

Calculate the frictional resistance (skin friction) $Q_{s\left(z=\text{8}\text{m}\text{to}\text{18}\text{m}\right)}$, in the lower layer of sand using the formula.

$Q_{s\left(z=\text{8}\text{m}\text{to}18\text{m}\right)}=p{L}_2{f}_{z=6.75\text{m}}$

Substitute 1.414 m for p, 10 m for $L_2$, and $30.7{\text{kN/m}}^{\text{2}}$ for $f_{z=6.75\text{m}}$.

$\begin{array}{c}{Q}_{s\left(z=\text{8}\text{m}\text{to}18\text{m}\right)}=1.414\times 10\times 30.7\\ =434\text{kN}\end{array}$

Calculate the ultimate load on the pile $Q_u$.

$Q_u=Q_{s\left(z=\text{0}\text{m}\text{to}\text{6}\text{.75}\text{m}\right)}+Q_{s\left(z=\text{6}\text{.75}\text{m}\text{to}\text{8}\text{m}\right)}+Q_{s\left(z=\text{8}\text{m}\text{to}\text{18}\text{m}\right)}+Q_p$

Substitute 146.51 kN for $Q_{s\left(z=\text{0}\text{m}\text{to}\text{6}\text{.75}\text{m}\right)}$, 54.26 kN for $Q_{s\left(z=\text{6}\text{.75}\text{m}\text{to}\text{8}\text{m}\right)}$, 434 kN for $Q_{s\left(z=\text{8}\text{m}\text{to}\text{18}\text{m}\right)}$, and 490.5 kN for $Q_p$.

$\begin{array}{c}{Q}_u=146.51+54.26+\text{434}+490.5\\ =1,125.27\text{kN}\end{array}$

Calculate the allowable load on the pile $Q_{\text{all}}$ using the formula.

$Q_{\text{all}}=\frac{Q_u}{F_s}$

Substitute 1,125.27 kN for $Q_u$ and 3 for $F_s$.

$\begin{array}{c}{Q}_{\text{all}}=\frac{\text{1,125}\text{.27}}{\text{3}}\\ =\text{375}\text{.1}\text{kN}\end{array}$

Therefore, the load carrying capacity of the pile is $\underset{\_}{\text{375}\text{.1}\text{kN}}$.