Coulomb Active Earth Pressure

K_{a} = \frac{sin ^{2} ( θ + ϕ )}{sin ^{2} θ sin ( θ - δ ) [ 1 + \frac{s i n ( ϕ + δ ) s i n ( ϕ - β )}{s i n ( θ - δ ) s i n ( θ + β )} ] ^{2}}

Coulomb includes wall friction δ and wall backfill slope/back face angle. Rankine is the smooth vertical special case. Pa=½ Ka γ H² (plus cohesion terms if c≠0).

$δ$: Wall–soil friction
$θ$: Angle of back face to horizontal
$β$: Backfill slope

Step-by-step solved example

Smooth vertical Rankine limit of Coulomb: φ=30°, β=0, δ=0, θ=90°. Ka?

Coulomb wedge with wall friction δ vs Rankine smooth.

1. Rankine
Ka=(1−sinφ)/(1+sinφ)=(0.5)/(1.5)=0.333.
2. Pa
For H=6 m, γ=18: Pa=½×0.333×18×36=108 kN/m, at H/3.

Answer: Ka = 0.33 (Rankine special case)

10 practice questions

0/10 correct

1.Rankine Ka=(1−sinφ)/(1+sinφ) equals

2.Wall friction δ generally

3.Kp (passive) is about

4.A sloping backfill β>0

5.φ=0, Rankine Ka=

6.Cohesion c reduces Pa by

7.At-rest Ko≈1−sinφ is used when

8.Failure wedge in Coulomb is

9.Pa acts at

10.Surcharge q adds