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Solution

**Answer: D) 2500 PSF**

Refer to PE civil Reference handbook clause 3.4.2.1

Refer to Terzaghi bearing capacities factors

\(\text{get}\ for\ \phi=0\)

\(N_{c} = 5.7,\)

\(N_{q} = 1\),

\(N_{\gamma}\ = 0\)

General Terzaghi bearing capacity formula:

\[=CN_{c}S_{c} + \gamma D_{f} N_{q} + 0.5 \gamma B N_{\gamma} S_{\gamma}\]

\(\text {Shape factors from table where at}\ \phi =0\ \text {the following values should be considered.}\)

\[S_{c} = 1 + \frac{B_{f}}{5 L_{f}} \qquad For\ square\ foundation\ B_{f} = L_{f} \\\]

hence,

\(S_c = 1.2\\\)

\(S_{q} = 1.0 \\\)

\(S_{\gamma } = 1.0\)

\(C= 1100 PSF\)

\(\gamma = 120 PCF\)

\(D_{f} = Foundation\ level\ depth = 12 ft \\\)

\[q_{ult}=CN_{c}S_{c} + \gamma D_{f} N_{q} + 0.5 \gamma B N_{\gamma} S_{\gamma} \\\]

\[=1100 \times 5.7 \times 1.2 + 120 \times 12 \times 1 + 0.5 \times 120 \times 0 \times 1 = 8964\ PSF \\\]

\[q_{net} = q_{ult}– \gamma D_{f} = 8964 -120 \times 12 = 7524\ PSF \\\]

\[q_{net)all} = \frac{q_{net}}{FOS} = 7524 / 3.0 = 2508\ PSF \\\]