Geotechnical Engineering

DHARM

SHEARING STRENGTH OF SOILS 311

Example 8.17: Given the following data from a consolidated undrained test with pore water

pressure measurement, determine the total and effective stress parameters:

σ 3 100 kN/m^2 200 kN/m^2

(σ 1 – σ 3 )f 156 kN/m^2 198 kN/m^2

uf 58 kN/m^2 138 kN/m^2

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

(a) Total stresses: (b) Effective stresses:

σ 3 100 kN/m^2 200 kN/m^2 σ 3 (100 – 58) = 42 kN/m^2 62 kN/m^2

σ 1 (100 + 156) = 256 kN/m^2 398 kN/m^2 σ 1 (256 – 58) = 198 kN/m^2 260 kN/m^2

These principal stresses are used to draw the corresponding Mohr’ circles (Fig. 8.60).

f¢= 38°

0 100 200 300 400

Shear stress, kN/m

2

200

100

c=36

c=16¢

Total stresses

Effective stresses

t

Normal stress, kN/m^2

f= 12°

Fig. 8.60 Effective stress and total stresses envelopes (Ex. 8.17)

Total stress parameters:

c = 36 kN/m^2 ; φ = 12°

Effective stress parameters:

c′ = 16 kN/m^2 ; φ′ = 38 °

The clay should have been overconsolidated, since c′ < c and φ′ > φ.

Example 8.18: A thin layer of silt exists at a depth of 18 m below the surface of the ground.
The soil above this level has an average dry density of 1.53 Mg/m^3 and an average water
content of 36%. The water table is almost at the surface. Tests on undisturbed samples of the
silt indicate the following values:
cu = 45 kN/m^2 ; φu = 18°; c′ = 35 kN/m^2 ; φ′ = 27°
Estimate the shearing resistance of the silt on a horizontal plane, (a) when the shear
stress builds up rapidly and (b) when the shear stress builds up very slowly.
Bulk unit weight, γ = γd (1 + w)
= 1.53 × 1.36 = 2.081 Mg/m^3
Submerged unit weight, γ′ = 2.081 – 1.0 = 1.081 Mg/m^3