2072 Chaitra · 6 marks

Mat 8 m × 10 m, Skempton net safe pressure

Short answer

For mat 8×10 m, Df=5 m, cu=50 kPa, Df/B=0.625, Nc=6.53, qnu=326.7 kPa. With FS=3, net safe bearing capacity = 108.9 kPa (gross safe 203.9 kPa).

A mat foundation of size 8 m × 10 m is resting at a depth of 5 m. The foundation is resting on saturated cohesive soil having undrained cohesion of $50\ \text{kPa}$ . The soil has unit weight of $19\ \text{kN/m}^3$ . Find the net safe bearing capacity using Skempton's method.

Given: mat $B = 8$ m, $L = 10$ m, $D_f = 5$ m, saturated clay $c_u = 50$ kPa, $\gamma = 19$ kN/m³.

Skempton's method ( $\phi = 0$ ):

\begin{aligned} \frac{D_f}{B} &= \frac{5}{8} = 0.625 < 2.5 \\ N_{c(sq)} &= 6\left(1 + 0.2 \times 0.625\right) = 6.75 \\ N_{c(rect)} &= \left(0.84 + 0.16 \times \frac{8}{10}\right)(6.75) = 0.968 \times 6.75 = 6.53 \\ q_{nu} &= c_uN_c = 50 \times 6.534 = 326.7\ \text{kPa} \end{aligned}

With a factor of safety $F = 3$ (assumed, usual value):

q_{ns} = \frac{q_{nu}}{F} = \frac{326.7}{3} = 108.9\ \text{kPa}

Answer: net ultimate capacity $= 326.7$ kPa and net safe bearing capacity $\approx 108.9$ kPa (gross safe pressure $= 108.9 + 19 \times 5 = 203.9$ kPa).

✓ Answer checked by Claude Sonnet 5.5

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