Geotechnical Engineering

DHARM

302 GEOTECHNICAL ENGINEERING

Plot the Mohr’s circles and hence determine the strength envelope and angle of internal

friction of the soil.

The data indicate that the tests are triaxial compression tests; the Mohr’s circles are

plotted with (σ 1 – σ 3 ) as diameter and the strength envelope is obtained as the common tan-

gent.

The angle of internal friction of found to be 30 °, by measurement with a protractor from

Fig. 8.49.

Example 8.7: A particular soil failed under a major principal stress of 300 kN/m^2 with a

corresponding minor principal stress of 100 kN/m^2. If, for the same soil, the minor principal

stress had been 200 kN/m^2 , determine what the major principal stress would have been if

(a) φ = 30° and (b) φ = 0°.

Graphical solution:

300

200

100

0

Shear stress, kN/m

2

Strength envelope = 30°

f

30°

200 300 600 s

Normal stress, kN/m^2

t

100 400 500 700

Strength envelope = 0°f

Fig. 8.50 Mohr’s circle and strength envelope (Ex. 8.7)

The Mohr circle of stress is drawn to which the strength envelope will be tangential; the

envelopes for φ = 0° and φ = 30° are drawn. Two stress circles, each starting at a minor princi-

pal stress value of 200 kN/m^2 , one tangential to φ = 0° envelope, and the other tangential to φ

= 30° envelope are drawn.

The corresponding major principal stresses are scaled off as 400 kN/m^2 and 600 kN/m^2.

Analytical solution:

(a) φ = 30° ;

σ 3 = 100 kN/m^2 , σ 1 = 300 kN/m^2

σ

σ

3

1

=

1

1

130

130

−

+

= −°

+°

sin

sin

sin

sin

φ

φ

= 1/3

The given stress circle will be tangential to the strength envelope with φ = 30°.
With σ 3 = 200 kN/m^2 , σ 1 = 3 × 200 = 600 kN/m^2 ,
if the circle is to be tangential to the strength envelope φ = 30° passing through the origin.
(b) φ = 0° ;
If the given stress circle has to be tangential to the strength envelope φ = 0°, the enve-
lope has to be drawn with c = τ = 100 kN/m^2. The deviator stress will then be 200 kN/m^2 ,
irrespective of the minor principal stress.