Question

The ultimate bearing capacity of a 1 m wide strip footing is 532.80 kPa, when it is embedded at 1 m depth in dry cohesionless soil. The soil has a unit weight of 18 kN/m³. The ultimate bearing capacity is 864 kPa when the depth of embedment becomes 2 m.

Neglecting the effect of the depth factor, the bearing capacity factor \(N_q\) is (rounded off to one decimal place).

Hint:

Only the surcharge term \(\gamma D_f N_q\) changes with depth of embedment; subtract the two given bearing capacity equations to isolate and solve for \(N_q\).

Answer 1

Correct Answer: 18.4

This problem gives the ultimate bearing capacity of the same strip footing at two different depths of embedment, and asks us to back-calculate the bearing capacity factor $N_q$ using only that data, without needing the friction angle directly.

For a strip footing on a purely frictional (cohesionless) soil, the general bearing capacity expression has only two active terms since the cohesion term vanishes:

\[ q_u = \gamma D_f N_q + 0.5\gamma B N_\gamma \]

Notice that the surcharge term $\gamma D_f N_q$ is the only term that changes when $D_f$ changes from 1 m to 2 m; the self weight term $0.5\gamma B N_\gamma$ stays fixed because $B$, $\gamma$ and $N_\gamma$ do not depend on $D_f$. So the increase in $q_u$ going from $D_f = 1$ m to $D_f = 2$ m is caused purely by the increase in the surcharge term.

Increase in depth: $\Delta D_f = 2 - 1 = 1$ m. Increase in bearing capacity: $\Delta q_u = 864 - 532.80 = 331.20$ kPa.

Since the surcharge term is $\gamma D_f N_q$, its rate of change with depth is $\gamma N_q$ (kPa per metre of embedment), so:

\[ \gamma N_q \times \Delta D_f = \Delta q_u \]\[ 18 \times N_q \times 1 = 331.20 \]\[ N_q = \frac{331.20}{18} = 18.4 \]

Let's summarize:

• Only the surcharge term $\gamma D_f N_q$ depends on the depth of embedment; the self weight term stays constant between the two cases.
• The change in ultimate bearing capacity per metre of extra depth directly gives $\gamma N_q$, from which $N_q$ follows.
• Check: with $N_q = 18.4$, the self weight term works out to $532.80 - 18(18.4) = 201.6$ kPa, giving $N_\gamma = 22.4$, which reproduces both given bearing capacity values correctly.

So the bearing capacity factor is $N_q = 18.4$.