Geotechnical Engineering

DHARM

524 GEOTECHNICAL ENGINEERING

H = 9 m

γ = 18 kN/m^3

Pa = 180 kN/m. run

Initially,

Pa =

1

2

γHK^2

. a

∴ 180 =

1

2

×× ×18 9^2 Ka

Ka =

2 180

18 9^2

×

×

= 0.247

Wall

x

4.5 m

4.5 m

hm Backfill

Cinders = 9 kN/mg^3

x = 2.223(h + 4.5)

z

Granular backfill

g= 18 kN/m^3

(a) (b)
Fig. 13.60 Retaining wall with backfill partly of cinders (Ex. 13.14)
Let the increase in the height of wall be h m.
The depth of cinders backfill will be (h + 4.5) m and bottom 4.5 m is granular backfill
with Ka = 0.247. Since the friction angles for cinders is taken to be the same as that for the
granular soil, Ka for cinders is also 0.247, but γ for cinders is 9 kN/m^3.

The intensity of pressure at (h + 4.5)m depth = 0.247 × 9 (h + 4.5) kN/m^2

= 2.223 (h + 4.5) kN/m^2

Intensity of pressure at the base = 0.247 [9 (h + 4.5) + 18 × 4.5] kN/m^2

= 2.223 (h + 4.5) + 20 kN/m^2

Total thrust Pa′ = 1.112 (h + 4.5)^2 + 2.223 × 4.5 (h + 4.5) +

1

2 × 4.5 × 20

Equating this to the initial value Pa, or 180 kN, the following equation is obtained:

1.112 h^2 + 20h – 67.5 = 0

Solving, h = 2.90 m

Thus, the height of the wall may be increased by 2.90 m without increasing the thrust.

Example 13.15: A gravity retaining wall retains 12 m of a backfill. γ = 18 kN/m^3 , φ = 30° with
a uniform horizontal backfill. Assuming the wall interface to be vertical, determine the magni-
tude of active and passive earth pressure. Assume the angle of wall friction to be 20°. Deter-
mine the point of action also. (S.V.U.—Four-year B.Tech.—Dec., 1982)