Chapter 13, Problem 13.1P

To determine

Find the percentage of load carried by the shaft.

Answer to Problem 13.1P

The percentage of load carried by the shaft is $\underset{\_}{0.62\%}$.

Explanation of Solution

Given information:

The diameter of the base $\left(D_b\right)$ is $5\text{ft}$.

The total length (L) is 27.0 ft.

Unit weight of loose sand $\left(\gamma \right)$ is $110{\text{lb/ft}}^{\text{3}}$.

Unit weight of dense sand $\left(\gamma \right)$ is $115{\text{lb/ft}}^{\text{3}}$.

The diameter of the shaft $\left(D_s\right)$ is $2.5\text{ft}$.

The length of the shaft $\left(L_1\right)$ is $24.0\text{ft}$.

The value of factor of safety (FOS) is 4.

Consider $\delta \text{'}=0.6\varphi \text{'}$.

Calculation:

Find the area of the base $\left(A_p\right)$ as follows:

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

Substitute $5\text{ft}$ for $D_b$.

$\begin{array}{c}{A}_p=\frac{\pi }{4}\times {\left(5\text{ft}\right)}^2\\ =19.64{\text{ft}}^2\end{array}$

From given Figure, find the effective stress of soil.

$\begin{array}{c}q\text{'}=24\text{ft}\times 110{\text{lb/ft}}^{\text{3}}+3\text{ft}\times 115{\text{lb/ft}}^{\text{3}}\\ =2,985{\text{lb/ft}}^{\text{2}}\end{array}$

Find the ratio $\frac{L}{D_b}$ as follows:

Substitute 27 ft for $L$ and 5 ft for $D_b$.

$\begin{array}{c}\frac{L}{D_b}=\frac{27\text{ft}}{5\text{ft}}\\ =5.4\end{array}$

Refer Table 13.3. “Berezantzev et al.’s value of $\omega$ ” in the textbook.

The value of $\omega$ is 0.83 for the soil friction angle of $38°$.

Find the bearing capacity factor $\left(N_q^{*}\right)$ using the relation:

$N_q^{*}=0.21e^{0.17\varphi \text{'}}$

Substitute $38°$ for $\varphi \text{'}$.

$\begin{array}{c}{N}_q^{*}=0.21e^{0.17\times 38°}\\ =134.2\end{array}$

Find the net ultimate load carrying capacity at the base $\left(Q_{p\left(\text{net}\right)}\right)$ using the relation:

$Q_{p\left(\text{net}\right)}=A_{p}q\text{'}\left(\omega N_q^{*}-1\right)$

Substitute $19.64{\text{ft}}^2$ for $A_p$, $2,985{\text{lb/ft}}^{\text{2}}$ for $q\text{'}$, 0.83 for $\omega$, and 134.2 for $N_q^{*}$.

$\begin{array}{c}{Q}_{p\left(\text{net}\right)}=19.64{\text{ft}}^2\times 2,985{\text{lb/ft}}^{\text{2}}\left(0.83\times 134.2-1\right)\\ =6,471.4\times {10}^3\text{lb}\\ =6,471.4\text{kip}\end{array}$

Find the shaft load $\left(Q_s\right)$ as follows:

Find the perimeter of the shaft $\left(p\right)$ as follows:

$p=\pi D_s$

Substitute 2.5 ft for $D_s$.

$\begin{array}{c}p=\pi \times 2.5\text{ft}\\ =7.85\text{ft}\end{array}$

Find the value of $\left(K_o\right)$ using the relation:

$K_o=1-\sin \varphi \text{'}$

Substitute $31°$ for $\varphi \text{'}$.

$\begin{array}{c}{K}_o=1-\sin 31\\ =0.485\end{array}$

Find the value of $\delta \text{'}$ using the relation:

$\delta \text{'}=0.6\varphi \text{'}$

Substitute $31°$ for $\varphi \text{'}$.

$\begin{array}{c}\delta \text{'}=0.6\left(31°\right)\\ =18.6°\end{array}$

Find the critical depth $\left(L\text{'}\right)$ using the relation:

$L\text{'}=15D_s$

Substitute 2.5 ft for $D_s$.

$\begin{array}{c}L\text{'}=15\times 2.5\text{ft}\\ =37.5\text{ft}\end{array}$

The value of critical depth 37.5 ft is more than the length $L_1=24\text{ft}$.

Find the frictional resistance $\left(f\right)$ along the pile shaft using the relation:

$f=K{\sigma }_o^{\text{'}}\tan \delta \text{'}$

At depth $z=0$:

$f=0$

At depth $z=24\text{ft}$:

Substitute 0.485 for K, $24\text{ft}\times 110{\text{lb/ft}}^{\text{3}}$ for ${\sigma }_o^{\text{'}}$, and $18.6°$ for $\delta \text{'}$.

$\begin{array}{c}f=0.485\times 24\text{ft}\times 110{\text{lb/ft}}^{\text{3}}\times \tan 18.6°\\ =430.9{\text{lb/ft}}^{\text{2}}\end{array}$

Find the load carrying capacity of the shaft $\left(Q_s\right)$ using the above calculated values as follows:

$\begin{array}{c}{Q}_s=f_{\text{av}}pL\\ =\left(\frac{0+430.9{\text{lb/ft}}^{\text{2}}}{2}\right)\times 7.85\text{ft}\times \text{24}\text{ft }\\ =40.6\times {10}^3\text{lb}\\ =40.6\text{kip}\end{array}$

Find the ultimate load $\left(Q_u\right)$ using the relation:

$Q_u=Q_p+Q_s$

Substitute $6,471.4\text{kip}$ for $Q_p$ and $40.6\text{kip}$ for $Q_s$.

$\begin{array}{c}{Q}_u=6,471.4\text{kip}+40.6\text{kip}\\ =6,512\text{kip}\end{array}$

Find the allowable load $\left(Q_{\text{all}}\right)$ using the relation:

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

Substitute $6,512\text{kip}$ for $Q_u$ and 4 for FOS.

$\begin{array}{c}{Q}_{\text{all}}=\frac{6,512\text{kip}}{\text{4}}\\ =1,628\text{kip}\end{array}$

Find the percentage of load carried by the shaft:

$\%\text{of}\text{load}\text{carried}\text{by}\text{the}\text{shaf}t=\frac{Q_s}{Q_u}\times 100$

Substitute $6,512\text{kip}$ for $Q_u$ and $40.6\text{kip}$ for $Q_s$.

$\begin{array}{c}\%\text{of}\text{load}\text{carried}\text{by}\text{the}\text{shaf}t=\frac{40.6\text{kip}}{6,512\text{kip}}\times 100\\ =0.62\%\end{array}$

Therefore, the percentage of load carried by the shaft is $\underset{\_}{0.62\%}$.