What is the Nc for the rectangular mat (B = 18, L = 22, Df = 3) in the answer?

6.20 for the square, 6.02 after the rectangle shape factor

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2081 Baisakh · 6 marks
2073 Shrawan · 4 marks

Mat 18 m × 22 m, safe building weight

Short answer

Use Skempton: c_u=37.5 kN/m², Df/B=0.167, N_c(sq)=6.20, N_c(rect)=6.02, q_nu=225.7 kN/m², q_ns=75.2, gross safe pressure=135.2 kN/m². Safe total weight = 135.2 × 396 = 53 557 kN (≈53.6 MN).

A mat 18 m × 22 m in plan has its base 3 m below the surface of the deposit of clay with unit weight of $20\ \text{kN/m}^3$ . The unconfined compressive strength of clay is $75\ \text{kN/m}^2$ . The factor of safety against bearing capacity failure must be 3. Determine total weight of building plus the foundation the raft can safely support.

Given: mat $B = 18$ m, $L = 22$ m, $D_f = 3$ m, $\gamma = 20$ kN/m³, $q_u$ (unconfined) $= 75$ kN/m², so $c_u = 37.5$ kN/m², FS = 3.

Use Skempton's equation for clay ( $\phi = 0$ ).

Bearing capacity factor

\begin{aligned} \frac{D_f}{B} &= \frac{3}{18} = 0.167 < 2.5 \\ N_{c(sq)} &= 6\left(1 + 0.2 \times 0.167\right) = 6.20 \\ N_{c(rect)} &= \left(0.84 + 0.16\frac{B}{L}\right)N_{c(sq)} = (0.84 + 0.1309)(6.20) = 6.02 \end{aligned}

Safe pressure

\begin{aligned} q_{nu} &= c_uN_c = 37.5 \times 6.02 = 225.7\ \text{kN/m}^2 \\ q_{ns} &= \frac{225.7}{3} = 75.2\ \text{kN/m}^2 \\ q_s &= q_{ns} + \gamma D_f = 75.2 + 20 \times 3 = 135.2\ \text{kN/m}^2 \end{aligned}

Safe total weight

Q_{safe} = q_s \times A = 135.2 \times (18 \times 22) = 135.2 \times 396 = 53\,557\ \text{kN}

Answer: the raft can safely support a total weight (building plus foundation) of about 53 600 kN (53.6 MN).

✓ Answer checked by Claude Sonnet 5.5

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