Geotechnical Engineering

DHARM

516 GEOTECHNICAL ENGINEERING

Active pressure at base of wall = Ka. γH =

125

125

−°17 7 12

+°

sin ××

sin

.

= 86.2 kN/m^2

The distribution of pressure is triangular as shown in Fig. 13.53 (b).

Total active thrust per metre run of wall =

1

2

1

2

γHK^2 a=× ×12 86 2. = 517.2 kN

This acts at (1/3)H or 4 m above the base of the wall.
(b) Water table at 6 m from surface:
Active pressure at 6 m depth = 0.406 × 17.7 × 6 = 43.1 kN/m^2
Active pressure at the base of the wall = Ka(γ. 6 + γ ′. 6) + γw .6
= 0.406 (17.7 × 6 + 10 × 6) + 9.81 × 6 = 67.5 + 58.9 = 126.4 kN/m^2
(This is obtained by assuming γ above the water table to be 17.7 kN/m^2 and the sub-
merged unit weight γ ′, in the bottom 6 m zone, to be 10 kN/m^2.

The pressure distribution is shown in Fig. 13.53 (c).

Total active thrust per metre run = Area of the pressure distribution diagram

=^1

2

643164311

2

62441

2

××..+× + ××.+ ×× 6589.

= 129.3 + 258.6 + 73.2 + 176.7 = 637.8 kN

The height of its point of application above the base is obtained by taking moments.

z =

(.... )

.

129 3 8 258 6 3 73 2 2 176 7 2

637 8

×+ ×+ ×+ ×

= 3.62 m

Total thrust increase by 120.6 kN and the point of application gets lowered by 0.38 m.

Example 13.4: A wall, 5.4 m high, retains sand. In the loose state the sand has void ratio of 0.63
and φ = 27°, while in the dense state, the corresponding values of void ratio and φ are 0.36 and
45° respectively. Compare the ratio of active and passive earth pressure in the two cases,
assuming G = 2.64.

(a) Loose State:

G = 2.64 e = 0.63

γd =

G

e

. w
()

.

(.)

γ

1

264 1

+ 1063

=

×

+ = 16.2 kN/m

3

φ = 27°

Ka =

127

127

0 376

−°

+°

sin

sin

.; Kp =

127

127

2 663

+°

−°

=

sin

sin

.

Active pressure at depth H m = Ka.γ.H = 0.376 × 16.2 H = 6.09. H kN/m^2

Passive pressure at depth H m = Kp. γH = 2.663 × 16.2 H = 43.14 H kN/m^2

(b) Dense State:

G = 2.64 e = 0.36

γd =

264 10

1036

. 19 4
(.)

×.

+

= kN / m^3