Chapter 5, Problem 5.1P

To determine

Find the gross allowable load carried by the foundation.

Answer to Problem 5.1P

The gross allowable load carried by the foundation is $\underset{\_}{9,209\text{kN}}$.

Explanation of Solution

Given information;

The width of the foundation (B) is 1.5 m.

The length of the foundation (L) is 2.5 m.

The depth of the foundation $\left(D_f\right)$ is 1.2 m.

The depth of base (H) is 0.9 m.

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

The cohesion of the soil $c^{\prime }$ is 0.

The unit weight of the soil $\left(\gamma \right)$ is $17{\text{kN/m}}^{\text{3}}$.

The factor of safety is (FS) is 3.0.

Calculation:

Determine the ratio of $\left(\frac{H}{B}\right)$.

$\begin{array}{c}\left(\frac{H}{B}\right)=\frac{0.9}{1.5}\\ =0.6\end{array}$

Refer Figure (5.4), “Mandel and Salencon’s bearing capacity factor $N_q^{\ast }$ ” in the text book.

Take $N_q^{\ast }$ value as 380 for ${\varphi }^{\prime }=40°$ and $\left(\frac{H}{B}\right)$ of 0.6.

Refer Figure (5.5), “Mandel and Salencon’s bearing capacity factor $N_{\gamma }^{\ast }$ ” in the text book.

Take $N_{\gamma }^{\ast }$ value as 200 for ${\varphi }^{\prime }=40°$ and $\left(\frac{H}{B}\right)$ of 0.6.

Refer Figure (5.6), “Mandel and Salencon’s bearing capacity factor $N_{\gamma }^{\ast }$ ” in the text book.

Take $m_1$ value as 0.46 and $m_2$ value as 0.52 for ${\varphi }^{\prime }=40°$ and $\left(\frac{H}{B}\right)$ of 0.6.

Determine the shape factor $F_{qs}^{\ast }$ using the formula;

$\begin{array}{c}{F}_{qs}^{\ast }=1-m_1\left(\frac{B}{L}\right)\\ =1-0.46\left(\frac{1.5}{2.5}\right)\\ =0.724\end{array}$

Determine the shape factor $F_{\gamma s}^{\ast }$ using the formula;

$\begin{array}{c}{F}_{\gamma s}^{\ast }=1-m_2\left(\frac{B}{L}\right)\\ =1-0.52\times \frac{1.5}{2.5}\\ =0.688\end{array}$

Determine the ultimate load bearing capacity $q_u$ of the rectangular foundation using the formula;

$\begin{array}{c}{q}_u=q{N}_q^{\ast }{F}_{qs}^{\ast }+\frac{1}{2}\gamma B{N}_{\gamma }^{\ast }{F}_{\gamma s}^{\ast }\\ =\left(\gamma \times D_f\right)N_q^{\ast }{F}_{qs}^{\ast }+\frac{1}{2}\gamma B{N}_{\gamma }^{\ast }{F}_{\gamma s}^{\ast }\\ =\left[\begin{array}{l}\left(17\times 1.2\right)\left(380\right)\left(0.724\right)\\ +\frac{1}{2}\left(17\right)1.5\left(200\right)\left(0.688\right)\end{array}\right]\\ =7,366.8{\text{kN/m}}^{\text{2}}\end{array}$

Determine the gross allowable load carried by the foundation .

$\begin{array}{c}{Q}_{\text{all}}=\frac{q_u\times B\times L}{\text{FS}}\\ =\frac{7,366.85\times 1.5\times 2.5}{3}\\ =9,209\text{kN}\end{array}$

Therefore, the gross allowable load carried by the foundation is $\underset{\_}{9,209\text{kN}}$.