Practice Question: Bearing Capacity of Square Footing

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Solution

The following are values of the bearing capacity factors for friction angle \(φ=12^°\)  ;

\(N_c=9.3\)

\(N_q=3\)

\(N_{\gamma }=1.7\)

For a soil with small angle of internal friction, cohesion is approximately
half the unconfined compression strength \(= 1,200\ lb/ft^2\).

Shape factors for square footing are:

\(S_c=1+\left(\frac{B_f}{L_f}\right)\left(\frac{N_q}{Nc}\right)=1.3\)

\(S_q=1+\left(\frac{B_f}{L_f}tan\varphi \right)=1.2\)

\(S_{\gamma }=1-0.4\left(\frac{B_f}{L_f}\right)=\ 0.6\)

 

For a square footing, bearing capacity equation yields:

 

\(q_{ult}\ =\ cN_cs_c\ +\ qN_qs_q+\ 0.5\gamma B_fN_{\gamma }s_{\gamma }\)

\(q_{ult}\ =\ 1200\times 9.3\times 1.3\ +(120\times 3)\times 3\times 1.2+\ 0.5\times 120\times 6\times 1.7\times 0.6\)

\(q_{ult}\ =\ 16,171\ {lb}/{ft^2}\)

\(q_{all}\ =\ 5,390\ {lb}/{ft^2}\)

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