Geotechnical Engineering

DHARM

150 GEOTECHNICAL ENGINEERING

Neutral stress at 15 m below the bottom of the lake:

u = 21 × 9.81 kN/m^3 = 206.0 kN/m^2

Effective stress at 15 m below the bottom of the lake:

σ′ = 323.9 – 206.0 = 117.9 kN/m^2

Also, σ = 15 × γ′ = 15 × (γsat – γw)

= 15(17.67 – 9.81)

= 117.9 kN/m^2

The pressure diagrams are shown in Fig. 5.26.

Example 5.4. A uniform soil deposit has a void ratio 0.6 and specific gravity of 2.65. The
natural ground water is at 2.5 m below natural ground level. Due to capillary moisture, the
average degree of saturation above ground water table is 50%. Determine the neutral pres-
sure, total pressure and effective pressure at a depth of 6 m. Draw a neat sketch.

(S.V.U.—B. Tech., (Part-time)—April, 1982)

The conditions are shown in Fig. 5.27:

45.22

115115

UncertainUncertain Uncertain

24.53 24.53

++

45.22

34.34 80.66

All pressures

in kN/m^2

(a) Total pressure (b) Neutral pressure

(c) Effective pressure

S = 50% due to

capillary moisture

Saturated soil

2.52.5 mm

3.53.5 mm

Fig. 5.27 Soil profile (Example 5.4) Fig. 5.28 Pressure diagrams (Example 5.4)

Void ratio, e = 0.6

Specific gravity G = 2.65

γsat = Ge

e w

+

+

F

HG

I

KJ

=

+

1 +

265 060

1060

.γ (.. )

(.)

× 9.81 kN/m^3

= 19.93 kN/m^3

γ at 50% saturation

= GSe

e w

+

+

F

HG

I

KJ

=

×+

1 +

265 05 060

1060

. (... )
(.)

γ × 9.81 kN/m^3 = 18.09 kN/m^3.

Total pressure, σ at 6 m depth = 2.5 × 18.09 + 3.5 × 19.93

= 115 kN/m^2

Neutral pressure, u at 6 m depth = 3.5 × 9.81 = 34.34 kN/m^2

Effective pressure, σ at 6 m depth = (σ – u)

= 115.00 – 34.34 = 80.66 kN/m^2

The pressure diagrams are shown in Fig. 5.28.