During plate load test, the settlement of a 35 cm plate is found to be 2 cm in a cohesive soil, then the settlement of square footing of 85 cm side under same loading conditions is ________.

Concept:

Plate Load Test:

It is a field test to determine the ultimate bearing capacity of soil and the portable settlement under a given loading.

Bearing Capacity Calculation for Clayey Soils

\(Ultimate\: bearing\: capacity\:[q_u(f)] = Ultimate\: load\: for\: plate\:[q_u(p)]\)

Bearing Capacity Calculation for Sandy Soils

\(Ultimate\: bearing\: capacity\:[q_u(f)] = {Width\: of\: pit (Bf) \over Size\: of\: Plate (Bp)}\times q_u(p)\)

Settlement of plate in clayey soil:

\(​\dfrac{S_P}{S_F} = \dfrac{B_P}{B_F}\)

Settlement of plate in sandy soil:

\(\dfrac{S_P}{S_F} = \left(\dfrac{B_P(B_F + 0.3)}{B_F(B_P + 0.3)}\right)^2\)

Where

Sf = settlement of foundation

Sp = settlement of plate

Bf = width of footing/foundation

Bp = width of plate

Calculation:

Given data,

Width of plate(Bp) = 35 cm, S_P = 2 cm

Width of footing(Bf) = 85 cm

Settlement of footing(S_F) in clayey soil:

\(​\dfrac{S_P}{S_F} = \dfrac{B_P}{B_F}\)

\(​\dfrac{2}{S_F} = \dfrac{35}{85}\)

S_F = 4.85 cm