Geotechnical Engineering

DHARM

SHEARING STRENGTH OF SOILS 303

Hence σ 1 = 200 + 200 = 400 kN/m^2 for σ 3 = 200 kN/m^2.
Example 8.8: The stresses acting on the plane of maximum shearing stress through a given
point in sand are as follows: total normal stress = 250 kN/m^2 ; pore-water pressure = 88.5 kN/
m^2 ; shearing stress = 85 kN/m^2. Failure is occurring in the region surrounding the point. De-
termine the major and minor principal effective stresses, the normal effective stress and the
shearing stress on the plane of failure and the friction angle of the sand. Define clearly the
terms ‘plane of maximum shearing stress’ and ‘plane of failure’ in relation to the Mohr’s rup-
ture diagram. (S.V.U.—B.E., (R.R.)—Nov., 1973)

85

Shear stress, kN/mO

2

f¢= 31°45

76.5 116 246.5 s

Normal stress, kN/m^2

t

C

2 q¢cr= 121°45

161.5

B

D

F

f

E

Mohr’s circle

of effective stress

q¢cr= 60°52

A

Fig. 8.51 Mohr’s circle of effective stresses (Ex. 8.8)

Total normal stress = 250 kN/m^2

Pore water pressure = 88.5 kN/m^2

Effective normal stress on the plane of maximum shear = (250 – 88.5) = 161.5 kN/m^2

Max. Shear stress = 85 kN/m^2

Graphical solution:

The normal stress on the plane of maximum shear stress is plotted as OC to a suitable

scale; CD is plotted perpendicular to σ-axis as the maximum shear stress. With C as centre

and CD as radius, the Mohr’s circle is established. A tangent drawn to the circle from the

origin O establishes the strength envelope. The foot of the perpendicular E from the point of

tangency F is located. The principal effective stresses and the stresses on the plane of failure

are scaled-off. The angle of internal friction is measured with a protractor. (Fig. 8.51.)

The results are: Major effective principal stress = (OB) = 246.5 kN/m^2

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

Angle of internal friction, φ (angle FOB) = 31 ° 45 ′

Normal effective stress on plane of failure (OE) = 116 kN/m^2

Shearing stress on the plane of failure (EF) = 72 kN/m^2

Analytical solution:

σσ 13

2

F +

HG

I

KJ = Normal stress on the plane of maximum shear = 161.5 ...(1)

σσ 13

2

F −

HG

I

KJ = Maximum shear stress = 85 ...(2)