DHARMLATERAL EARTH PRESSURE AND STABILITY OF RETAINING WALLS 527
γ = 18 kN/m^3 H = 10 m. φ = 30° δ = 20° β = 20° α = 90°
ψ = α – δ = 90° – 20° = 70°
φ + δ = 30° + 20° = 50°C
A b= 20°(+)
= 50°fd
10 mWall
60°BEf= 30°d cb10° Df= 70°
Repture lineGfad=( – )
= 70°aFig. 13.62 Coulomb’s analytical approach (Ex. 13.17)(^) AB = H = 10 m a =
10
70
60 9 216
sin
.sin.
°
°= m
d =
10
sin 70 °
. sin 50° (from ∆ABE) = 8.152 m
b =10
10110
sin.sin
°°(from ∆ABD) = 54.115 mc = bd=×8152 54 115.. m = 21.003 mx =ab
()bc..
+.=9 216 54 115×
75118 m = 6.639 mPa =1
21
2γψx^22 .sin =× × 18 ( .6 639) sin 70 °kN / m. run= 373 kN/m. run
Example 13.18: A retaining wall, 3.6 m high, supports a dry cohesionless backfill with a plane
ground surface sloping upwards at a surcharge angle of 10° 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 = 30°. Assuming wall friction angle of 12°, determine the total
active thrust by Rebhann’s method.
H = 3.6 m φ = 30° δ = 12° β = 10° α = 81° γ = 18.9 kN/m^3ψ = α – δ = 69° Pa =1
21
2γψx^22 sin =× ×18 9 2 40.. ×sin 69 °= 51 kN/m. run