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

Step 1: Set up Terzaghi's bearing capacity equation for a strip footing in cohesionless soil.
For a strip footing on cohesionless soil ($c=0$), Terzaghi's ultimate bearing capacity equation reduces to \(q_u = \gamma D_f N_q + 0.5\,\gamma B N_\gamma\), where \(D_f\) is the depth of embedment, \(B\) is the footing width, and \(N_q\), \(N_\gamma\) are bearing capacity factors that depend only on the soil friction angle.

Step 2: Write the equation at both given depths.
At \(D_f = 1\) m: \(q_{u1} = 18(1)N_q + 0.5(18)(1)N_\gamma = 18N_q + 9N_\gamma = 532.80\) kPa
At \(D_f = 2\) m: \(q_{u2} = 18(2)N_q + 0.5(18)(1)N_\gamma = 36N_q + 9N_\gamma = 864\) kPa
The width \(B = 1\) m and unit weight \(\gamma = 18\) kN/m³ are the same in both cases, so the \(N_\gamma\) term is identical in both equations.

Step 3: Eliminate the \(N_\gamma\) term.
Subtract the first equation from the second:
\[ (36N_q + 9N_\gamma) - (18N_q + 9N_\gamma) = 864 - 532.80 \]
\[ 18N_q = 331.20 \]

Step 4: Solve for \(N_q\).
\[ N_q = \frac{331.20}{18} = 18.4 \]

Step 5: Check the result.
Substituting back, \(18(18.4) + 9N_\gamma = 532.80 \Rightarrow 331.2 + 9N_\gamma = 532.80 \Rightarrow N_\gamma = 22.4\). Checking at 2 m depth: \(36(18.4) + 9(22.4) = 662.4 + 201.6 = 864\) kPa, which matches the given value, so the answer is consistent.

Final Answer:
\[ \boxed{N_q = 18.4} \]