Geotechnical Engineering

DHARM

SEEPAGE AND FLOW NETS 195

With a factor of safety of 2.0 against piping,

Gradient, i =

ic

2

1 0725

2

=

.

= 0.53625

But i = h/L

L = h/i = 1.850/0.53625 m = 3.45 m

Available flow path = thickness of soil = 1.25 m.

∴ Depth of coarse sand required = 2.20 m.

Example 6.3: A glass container with pervious bottom containing fine sand in loose state (void
ratio = 0.8) is subjected to hydrostatic pressure from underneath until quick condition occurs
in the sand. If the specific gravity of sand particles = 2.65, area of cross-section of sand sample
= 10 cm^2 and height of sample = 10 cm, compute the head of water required to cause quicksand
condition and also the seepage force acting from below.

(S.V.U.—B.E.(R.R.)—Nov. 1974)

e = 0.8, G = 2.65

ic =

()

()

()

()

G

e

−

+

=

−

+

=

1

1

1

1

2.65

0.8

1.65

1.80 = 11/12 = 0.92

L = 10 cm. h = L. ic =

10 11

12

×

= 55/6 = 9.17 cm.

Seepage force per unit volume = i. γw

=

11

12

× 9.81 kN/m^3

Total seepage force =^11

12

× 9.81 ×

10 10

100 100 100

×

×× kN

≈ 0.0009 kN = 0.9 N.

Example 6.4: A large excavation was made in a stratum of stiff clay with a saturated unit
weight of 18.64 kN/m^3. When the depth of excavation reached 8 m, the excavation failed as a
mixture of sand and water rushed in. Subsequent borings indicated that the clay was under-
lain by a bed of sand with its top surface at a depth of 12.5 m. To what height would the water
have risen above the stratum of sand into a drill hole before the excavation was started?

12.5 m

8m

Excavation

h

Clay

Sand

z

Fig. 6.27 Excavation of clay underlain by sand (Example 6.4)