Question

A shallow strip footing of width 2 m is embedded at a depth of 1.5 m below the ground surface in a homogeneous pure clay with angle of internal friction equal to zero. The unit weight of soil is 20 kN/m³ and the undrained cohesion of soil is 20 kN/m². Due to a rise of the ground water table from far below the founding depth to the ground surface during the monsoon season, the magnitude of percentage change in the net ultimate bearing capacity of the footing as per Terzaghi's theory is ______ (rounded off to two decimal places).

Hint:

For \(\phi=0\) soils, Terzaghi's \(N_q=1\) makes the overburden term cancel exactly in the net bearing capacity, so \(q_u(\text{net})=cN_c\) regardless of the water table.

Answer 1

Correct Answer: 0

Step 1: Write down what changes when the water table rises.
Before the rise, the footing sits above the water table, so the overburden pressure above the base uses the given (moist) unit weight $\gamma=20$ kN/m$^3$: $q_{before} = \gamma D_f = 20\times1.5=30$ kPa.
After the rise, water fills the ground up to the surface, so this soil becomes buoyant. Taking the submerged unit weight as roughly $\gamma' \approx \gamma - \gamma_w \approx 20-9.81=10.19$ kN/m$^3$, $q_{after}=\gamma' D_f = 10.19\times1.5=15.29$ kPa.

Step 2: Compute the gross ultimate bearing capacity in both cases.
$q_u = cN_c+qN_q$, with $N_c=5.7$, $N_q=1$ for $\phi=0$ clay:
Before: $q_u(\text{gross,before}) = 20\times5.7+30\times1 = 114+30=144$ kPa.
After: $q_u(\text{gross,after}) = 20\times5.7+15.29\times1=114+15.29=129.29$ kPa.
So the gross capacity does drop when the water table rises, by about 14.7 kPa.

Step 3: Subtract the overburden to get the net capacity in both cases.
Net capacity removes exactly the surcharge $q$ that was already acting before construction:
Before: $q_u(\text{net,before}) = 144-30=114$ kPa.
After: $q_u(\text{net,after}) = 129.29-15.29=114$ kPa.

Step 4: Compare the two net values.
Both net values come out to exactly 114 kPa, since $q_u(net)=cN_c$ only whenever $N_q=1$, which is always true for $\phi=0$ soil. The rise in water table changes the gross capacity but leaves the net capacity, which is what the footing actually has to resist beyond the weight of soil it replaces, unchanged.

Step 5: State the percentage change.
\[ \%\text{change} = \frac{114-114}{114}\times100=0\% \]
\[ \boxed{0.00\%} \]