Geotechnical Engineering

DHARM

304 GEOTECHNICAL ENGINEERING

Solving (1) and (2), σ 1 = 246.5 kN/m^2 (Major principal effective stress)

σ 3 = 76.5 kN/m^2 (Minor principal effective stress)

sin φ = ()/

()/

σσ

σσ

13

13

2

2

−

+

=^85

161 5.

= 0.526

∴ Angle of internal friction φ = 31 ° 45 ′ nearly.

Normal stress on the failure plane

=

σσ 13

2

F +

HG

I

KJ –

σσ 13

2

F −

HG

I

KJ. sin φ

= 161.5 –

85 85

161 5

×

.

= 116.76 kN/m^2

Shear stress on the failure plane

=

σσ 13

2

F −

HG

I
KJ. cos φ
= 85 × cos 31°45′ = 72.27 kN/m^2
The answers from a graphical approach compare very well with those from the analyti-
cal approach. The planes of maximum shear, i.e., the planes on which the shearing stress is the
maximum, are inclined at 45° with the principal planes.

The failure plane i.e., the plane on which the resultant has max. obliquity, is inclined at
(45° + φ/2) or 61°52′ (Counterclockwise) with the major principal plane. These observations are
confirmed from the Mohr’s circle of stress.

Example 8.9: In an unconfined compression test, a sample of sandy clay 8 cm long and 4 cm in
diameter fails under a load of 120 N at 10% strain. Compute the shearing resistance taking
into account the effect of change in cross-section of the sample.

(S.V.U.—B.Tech. (Part-time)—May, 1983)

Size of specimen = 4 cm dia. × 8 cm long.

Initial area of cross-section = (π/4) × 4^2 = 4π cm^2.

Area of cross-section at failure =

A 0

() 1 −ε

=

4

1010

π

(.)−

= 4π/0.9 = 40 π/9 cm^2

Load at failure = 120 N.

Axial stress at failure =

120 9

40

×

π

N/cm^2

= 2.7/π N/cm^2

= 8.6 N/cm^2

Shear stress at failure =

1

2

× 8.6 = 4.3 N/cm^2

The corresponding Mohr’s circle is shown in Fig. 8.52.