Geotechnical Engineering

DHARM

306 GEOTECHNICAL ENGINEERING

Example 8.11: In a triaxial shear test conducted on a soil sample having a cohesion of 12 kN/
m^2 and angle of shearing resistance of 36°, the cell pressure was 200 kN/m^2. Determine the
value of the deviator stress at failure. (S.V.U.—B.E., (R.R)—Nov., 1974)
The strength envelope is drawn through E on the τ-axis, OE being equal to C = 12 kN/m^2
to a convenient scale, at an angle φ = 36° with the σ-axis. The cell pressure, σ 3 = 200 kN/m^2 is
plotted as OA. With centre on the σ-axis, a circle is drawn to pass through A and be tangential
to the envelope, by trial and error. AC is scaled-off, C being the centre of the Mohr’s circle,
which is (σ 1 – σ 3 )/2. The deviator stress is double this value. In this case the result is
616 kN/m^2. (Fig. 8.54).

0

Shear stress, kN/m

2

f= 36°

s

Normal stress, kN/m^2

t

C

D

c= E

12 kN/m^2

Strength envelope

A

s 3 = 200 kN/m^2 (ss 13 – )/2 = 308 kN/m^2

Fig. 8.54 Mohr’s circle for triaxial test (Ex. 8.11)

Analytical solution:

c = 12 kN/m^2

φ = 36°

σ 3 = 200 kN/m^2

σ 1 = σ 3 Nφ + 2c Nφ

where Nφ = tan^2 (45° + φ/2).

Nφ = tan^2 (45° + 18°) = tan^2 63° = 3.8518

Nφ = tan 63° = 1.9626

∴ σ 1 = 200 × 3.8518 + 2 × 12 × 1.9626 = 817.5 kN/m^2

Deviator stress = σ 1 – σ 3 = (817.5 – 200) kN/m^2 = 617.5 kN/m^2

The result from the graphical solution agrees well with this value.

Example 8.12: A triaxial compression test on a cohesive sample cylindrical in shape yields the

following effective stresses:

Major principal stress ... 8 MN/m^2

Minor principal stress ... 2 MN/m^2

Angle of inclination of rupture plane is 60° to the horizontal. Present the above data, by

means of a Mohr’s circle of stress diagram. Find the cohesion and angle of internal friction.

(S.V.U.—Four-year, B.Tech.—June, 1982)