DHARM
LATERAL EARTH PRESSURE AND STABILITY OF RETAINING WALLS 529
Pa =
1
2
1
2
γψx^22 sin =× × ×19 3 sin 70 °
= 80.34 kN/m.
Example 13.20: A retaining wall 8 m high is battered at a positive angle of 15° and retains a
cohesionless backfill rising at 20° with the horizontal from the crest of the wall. φ = 30° and δ
= 20°. Bulk density of the fill is 17.1 kN/m^3. Determine the active thrust, its direction and point
of action on the wall by Rebhann’s method. (S.V.U.—B.E. (N.R.)—Sep., 1967)
H = 8 m α = 90° – 15° = 75° β = 20° φ = 30° δ = 20° ψ = α – δ = 55°
Since the φ-line does not meet the ground surface within the drawing, the special case
when β ≈ φ is applied, the construction being performed with an arbitrary location, D 1 for D.
AG is drawn parallel to A 1 G 1 , G being on the φ-line. Triangle CGL, the weight of which gives
Pa, is completed, with |CGL = ψ.
Pa =
1
2
1
2
γψx^22 .sin =× × 171. (. ) sin7 0 55 °= 343 kN/m run
The location of Pa is at (1/3) H or 2.67 m above the base and the direction is at δ or 20 °
with the horizontal.
a= 75°
20°
C
M
G
D 1
L
f-line
x = 7.0 m
5.8 m
y= 55°
AG || to A G
s 11
A
Rupture plane
8m
( + )=50°fd Pa
Wall
G 1
f= 30°
y= 55°
E 1
2.67 m
A
B
F 1
15°
Fig. 13.65 Special case of Rebhann’s construction when β ≈ φ (Ex. 13.20)
Example 13.21: A retaining wall 3.6 m high supports a dry cohesionless backfill with a plane
ground surface sloping upwards at a surcharge angle of 20° from the top of the wall. The back
of the wall is inclined to the vertical at a positive batter angle of 9°. The unit weight of the
backfill is 18.9 kN/m^3 and φ = 20°. Assuming a wall friction angle of 12°, determine the total
active thrust by Rebhann’s method.