Given,
q = 36 KN/m²
P_a=K_a ?H+ K_a q= 3×18×6 + 3×36
P_p= 432 kN/m².

In Rankine?s ory, additional assumption of vertical and lateral stresses are conjugate is made. Being conjugate, both vertical and lateral pressures have same angle of obliquity ?.

If backfill is having its top surface inclined to horizontal at an angle ?, n passive earth pressure is given by,
\(p_p=?zcos? \frac{cos?+\sqrt{cos^2 ?-cos^2 ?}}{cos?-\sqrt{cos^2 ?-cos^2 ?}},\)
where, ?= surcharge angle
?=unit weight of soil
?= angle of internal friction.

For different value of ?, coefficient of active earth pressure is different. Since, lateral pressure intensity is due to:

• lateral pressure due to moist weight of backfill
• lateral pressure due to saturated weight of backfill
• lateral pressure due to water

Given,
q= 36 kN/m³
?= 18 kN/m³
\(Z_e = \frac{q}{?} = \frac{36}{18}=2m.\)

Given,
?= 18 kN/m³
H=6m
K_a=\(\frac{1}{3}\)
K_p=\(\frac{1}{K_a}=3 \)
Therefore P_p = K_p?H =3×18×6=324 KN/m³.

K0 can be calculated by using following equation as computed by Jacky in 1944:
K0 = 1 ? sin ?.

The Rankine?s active earth pressure (assuming wall to be smooth) can be found out by method of trial wedges.

The ratio K is conjugate ratio or Rankine?s lateral pressure ratio, which is given by,
\(K=\frac{cos?-\sqrt{cos^2 ?-cos^2 ?}}{cos?+\sqrt{cos^2 ?-cos^2 ?}}, \)
where, ?=surcharge angle
?=angle of internal friction.

Given,
?=450
K₀=1-sin?=1-sin45=0.293.