Geotechnical Engineering

DHARM

196 GEOTECHNICAL ENGINEERING

Referring to Fig. 6.27, the effective stress at the top of sand stratum goes on getting
reduced as the excavation proceeds due to relief of stress, the neutral pressure in sand remain-
ing constant.

The excavation would fail when the effective stress reached zero value at the top of
sand.

Effective stress at the top of sand stratum,

σ = z. γsat – h.γw

If this is zero, hγw = z. γsat

or h = z
w

.γ

γ

sat (12.5 8) 18.64

9.81

=

−× = 8.55 m

Therefore, the water would have risen to a height of 8.55 m above the stratum of sand
into the drill hole before excavation under the influence of neutral pressure.

Example 6.5: Water flows at the rate of 0.09 ml/s in an upward direction through a sand
sample with a coefficient of permeability of 2.7 × 10–2 mm/s. The thickness of the sample is 120
mm and the area of cross-section is 5400 mm^2. Taking the saturated unit weight of the sand as
18.9 kN/m^3 , determine the effective pressure at the middle and bottom of the sample.

Here, q = 0.09 ml/s = 90 mm^3 /s, k = 2.7 × 10–2 mm/s

A = 5400 mm^2

i = q/kA =

90

2.7 10××−^25400

= 0.6173

γ′ = γsat – γw = (18.90 – 9.81) kN/m^3 = 9.09 kN/m^3

= 9.09 × 10–6 N/mm^3

For the bottom of the sample, z = 120 mm

σ = γ′z – izγw,

for downward flow, considering the effect of seepage pressure.

∴ (^) σ = (9.09 × 10–6 × 120 – 0.6173 × 120 × 9.81 × 10–6) N/mm^2
= 120 × 10–6 (9.09 – 0.6173 × 9.81) = 0.364 × 10–3 N/mm^2
= 364 N/m^2
For the middle of the sample, z = 60 mm
σ = γ′z – iz^ γw
= (9.09 × 10–4 × 60 – 0.6173 × 60 × 9.81 × 10–6) N/mm^2
= 0.182 × 10–3 N/mm^2 = 182 N/m^2.
Example 6.6: A deposit of cohesionless soil with a permeability of 3 × 10–2 cm/s has a depth of
10 m with an impervious ledge below. A sheet pile wall is driven into this deposit to a depth of
7.5 m. The wall extends above the surface of the soil and a 2.5 m depth of water acts on one
side. Sketch the flow net and determine the seepage quantity per metre length of the wall.
(S.V.U.— B.E. (R.R)—Nov., 1973)
The flow net is shown.
Number of flow channels, nf = 4
Number of equipotential drops, nd = 14