Chapter 6, Problem 6.1P

For the following cases, determine the allowable gross vertical load-bearing capacity of the foundation. Use Terzaghi’s equation and assume general shear failure in soil. Use FS = 4.

Parameters for Problem 6.1

To determine

Find the allowable gross vertical load-bearing capacity of the continuous foundation.

Answer to Problem 6.1P

The allowable gross vertical load-bearing capacity of the continuous foundation is $\underset{\_}{5,195.45{\text{lb/ft}}^{\text{2}}}$.

Explanation of Solution

Given information;

The width (B) of the foundation is 3 ft.

The depth $\left(D_f\right)$ of the foundation is 3 ft.

The angle of friction ${\varphi }^{\prime }$ of soil is $28°$.

The cohesion $c^{\prime }$ of the soil is $400{\text{lb/ft}}^{\text{2}}$.

The specific weight $\gamma$ of soil is $110{\text{lb/ft}}^{\text{3}}$.

The factor of safety is (FS) is 4.0.

Calculation:

Refer Table (6.1), “Terzaghi’s bearing capacity factors” in the text book.

For ${\varphi }^{\prime }=28°$

Take the value of $N_C$ as 31.61.

Take the value of $N_q$ as 17.81.

Take the value of $N_{\gamma }$ as 13.70.

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

$\begin{array}{c}{q}_u=c^{\prime }{N}_C+q{N}_q+\frac{1}{2}\gamma B{N}_{\gamma }\\ =c^{\prime }{N}_C+\left(\gamma \times D_f\right)N_q+\frac{1}{2}\gamma B{N}_{\gamma }\\ =400\times 31.61+\left(110\times 3\right)17.81+\frac{1}{2}\times 110\times 3\times 13.70\\ =20,781.8{\text{lb/ft}}^{\text{2}}\end{array}$

Determine the allowable gross vertical load-bearing capacity $q_{\text{all}}$ using the formula;

$\begin{array}{c}{q}_{\text{all}}=\frac{q_u}{\text{FS}}\\ =\frac{20,781.8{\text{lb/ft}}^{\text{2}}}{4}\\ =5,195.45{\text{lb/ft}}^{\text{2}}\end{array}$

Hence, the allowable gross vertical load-bearing capacity $q_{\text{all}}$ of the continuous foundation is $\underset{\_}{5,195.45{\text{lb/ft}}^{\text{2}}}$.

To determine

Find the allowable gross vertical load-bearing capacity of the continuous foundation.

Answer to Problem 6.1P

The allowable gross vertical load-bearing capacity of the continuous foundation is $\underset{\_}{372.8{\text{kN/m}}^{\text{2}}}$.

Explanation of Solution

Given information;

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

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

The angle of friction ${\varphi }^{\prime }$ of soil is $35°$.

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

The specific weight $\gamma$ of soil is $17.8{\text{kN/m}}^{\text{3}}$.

The factor of safety is (FS) is 4.0.

Calculation:

Refer Table (6.1), “Terzaghi’s bearing capacity factors” in the text book.

For ${\varphi }^{\prime }=35°$

Take the value of $N_C$ as 57.75.

Take the value of $N_q$ as 41.44.

Take the value of $N_{\gamma }$ as 45.41.

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

$\begin{array}{c}{q}_u=c^{\prime }{N}_C+q{N}_q+\frac{1}{2}\gamma B{N}_{\gamma }\\ =c^{\prime }{N}_C+\left(\gamma \times D_f\right)N_q+\frac{1}{2}\gamma B{N}_{\gamma }\\ =0\times 57.75+\left(17.8\times 1.2\right)41.44+\frac{1}{2}\times 17.8\times 1.5\times 45.41\\ =1,491.38{\text{kN/m}}^{\text{2}}\end{array}$

Determine the allowable gross vertical load-bearing capacity $q_{\text{all}}$ using the formula;

$\begin{array}{c}{q}_{\text{all}}=\frac{q_u}{\text{FS}}\\ =\frac{1,491.38{\text{kN/m}}^{\text{2}}}{4}\\ =372.8{\text{kN/m}}^{\text{2}}\end{array}$

Hence, the allowable gross vertical load-bearing capacity $q_{\text{all}}$ of the foundation is $\underset{\_}{372.8{\text{kN/m}}^{\text{2}}}$.

To determine

Find the allowable gross vertical load-bearing capacity of the square foundation.

Answer to Problem 6.1P

The allowable gross vertical load-bearing capacity of the square foundation is $\underset{\_}{280.0{\text{kN/m}}^{\text{2}}}$.

Explanation of Solution

Given information;

The width (B) of the foundation is 3.0 m.

The depth $\left(D_f\right)$ of the foundation is 2.0 m.

The angle of friction ${\varphi }^{\prime }$ of soil is $30°$.

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

The specific weight $\gamma$ of soil is $16.5{\text{kN/m}}^{\text{3}}$.

The factor of safety is (FS) is 4.0.

Calculation:

Refer Table (6.1), “Terzaghi’s bearing capacity factors” in the text book.

For ${\varphi }^{\prime }=30°$

Take the value of $N_C$ as 37.16.

Take the value of $N_q$ as 22.46.

Take the value of $N_{\gamma }$ as 19.13.

Determine the ultimate load bearing capacity $q_u$ using the formula;

$\begin{array}{c}{q}_u=1.3c^{\prime }{N}_C+q{N}_q+0.4\gamma B{N}_{\gamma }\\ =1.3c^{\prime }{N}_C+\left(\gamma \times D_f\right)N_q+0.4\gamma B{N}_{\gamma }\\ =1.3\left(0\right)37.16+\left(16.5\times 2\right)22.46+0.4\times 16.5\times 3\times 19.13\\ =1,119.95{\text{kN/m}}^{\text{2}}\end{array}$

Determine the allowable gross vertical load-bearing capacity $q_{\text{all}}$ using the formula;

$\begin{array}{c}{q}_{\text{all}}=\frac{q_u}{\text{FS}}\\ =\frac{1,119.95{\text{kN/m}}^{\text{2}}}{4}\\ \approx 280.0{\text{kN/m}}^{\text{2}}\end{array}$

Hence, the allowable gross vertical load-bearing capacity $q_{\text{all}}$ of the square foundation is $\underset{\_}{280.0{\text{kN/m}}^{\text{2}}}$.