Correct Answer - Option 2 : Both 1 and 2
Concept:
For a plate load test for sands.
\({q_f} = {q_p} \times \frac{{{B_f}}}{{{B_p}}}\)
Where,
qf = ult bearing capacity of footing
qp = ult. Bearing capacity of plate.
Bf, Bp = width of footing & plate respectively so
qf ∝ Bf
⇒ Ultimate bearing capacity of a footing on sand increases with increase in its width.
Settlement of footing (Sf) on sand increases with increase in width (Bf), can be explained as
\(\frac{{{S_{{f_1}}}}}{{{S_p}}} = {\left[ {\frac{{{B_{{f_1}}}\left( {{B_p} + 0.3} \right)}}{{{B_p}\left( {{B_{{f_1}}} + 0.3} \right)}}} \right]^2}\) (for 1st footing)
\(\frac{{{S_{{f_2}}}}}{{{S_p}}} = {\left[ {\frac{{{B_{{f_1}}}\left( {{B_p} + 0.3} \right)}}{{{B_p}\left( {{B_{{f_2}}} + 0.3} \right)}}} \right]^2}\) (for 2nd footing)
\(\therefore \frac{{{S_{{f_1}}}}}{{{S_{{f_2}}}}} = {\left[ {\frac{{{B_{{f_1}}}\left( {{B_{{f_2}}} + 0.3} \right)}}{{{B_{{f_2}}}\left( {{B_{{f_1}}} + 0.3} \right)}}} \right]^2} = {\left( {\frac{{1 + \frac{{0.3}}{{{B_{{f_2}}}}}\;}}{{1 + \frac{{0.3}}{{{B_{{f_1}}}}}}}} \right)^2}\)
If \({B_{{f_2}}} > {B_{{f_1}}}\;;\;\frac{{{S_{{f_1}}}}}{{{S_{{f_2}}}}} < 1\)
∴ \({S_{{f_2}}} > {S_{{f_1}}}\)
Hence settlement increases.