Why does the 3 m deep WT have no effect?

It is deeper than B below the base

2076 Chaitra · 7 marks

Footing size in sand, 150 tons

Short answer

For sand, q_u = γ D_f N_q + 0.4 γ B N_γ. With FS=3, solving for B by trial gives B=1.7 m, providing a safe load of 1668 kN. Provide a 1.7 m × 1.7 m square footing.

Determine the size of the footing resting over sand to carry a column load of 150 tons. The bottom of the footing is 1.5 m below the ground level and water table is located at 3 m below the footing. Take unit weight of sand is $20\ \text{kN/m}^3$ . Assume $N_c = 55.2$ , $N_q = 39.51$ and $N_\gamma = 40.13$ .

Given: column load $= 150$ t $= 150 \times 9.81 = 1471.5$ kN, $D_f = 1.5$ m, $\gamma = 20$ kN/m³, sand ( $c = 0$ ), $N_q = 39.51$ , $N_\gamma = 40.13$ (and $N_c = 55.2$ , not needed). The water table is 3 m below the footing, which is more than $B$ , so it has no effect.

Assumptions: square footing, FS = 3 (not given), Terzaghi's equation, general shear, weight of footing included in the load.

\begin{aligned} q_u &= \gamma D_f N_q + 0.4\gamma B N_\gamma \\ &= 20 \times 1.5 \times 39.51 + 0.4 \times 20 \times 40.13\,B \\ &= 1185.3 + 321.0\,B\ \text{kN/m}^2 \end{aligned}

Safe load: $\dfrac{q_u}{3}B^2 = 1471.5$ , giving (by trial) $B = 1.61$ m.

$B$ (m)	$q_u$ (kN/m²)	Safe load $q_uB^2/3$ (kN)
1.6	1699.0	1449.8
1.7	1731.1	1667.6
1.8	1763.2	1904.2

Answer: provide a square footing of 1.7 m × 1.7 m (safe load 1668 kN > 1471.5 kN).

✓ Answer checked by Claude Sonnet 5.5

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