Geotechnical Engineering

DHARM

SHEARING STRENGTH OF SOILS 309

φ = 40°; θcr = 65° with the horizontal. (Fig. 8.57)

Analytical method:

σ 3 = 150 kN/m^2

(σ 1 – σ 3 ) = 535 kN/m^2

∴σ 1 = 685 kN/m^2

σ 1 = σ 3 Nφ, where Nφ = tan^2 (45° + φ/2)

∴ Nφ = σ 1 /σ 3 =

685

150

= 4.57

(45° + φ/2) = 64°55′

∴φ/2 = 19°55′

or φ = 39°50′

Hence, θcr = (45° + φ/2) = 64 ° 55 ′.

The graphical values compare very well with these results.

Example 8.15: The shearing resistance of a soil is determined by the equation s = c′ + σ′ tan φ′.
Two drained triaxial tests are performed on the material. In the first test the all-round pres-
sure is 200 kN/m^2 and failure occurs at an added axial stress of 600 kN/m^2. In the second test
all-round pressure is 350 kN/m^2 and failure occurs at an added axial stress of 1050 kN/m^2.
What values of c′ and φ′ correspond to these results? (S.V.U.—B.E., (R.R.)—Nov., 1973)
Graphical method:

f¢= 36°30

s

Normal stress, kN/m^2

t

600

400

200

0 200 400 600

Shear stress, kN/m

2

800 1000 1200 1400

f¢

II

I

Strength envelope

(effective stresses)

c=0¢

Fig. 8.58 Mohr’s circle for effective stress (Ex. 8.15)

Since the cell pressures (σ 3 ) and the added axial stresses (σ 1 – σ 3 ) are known, σ 1 -values

are also obtained by addition. The Mohr’s circles for the two tests are drawn. The common

tangent to the two circles is seen to pass very nearly through the origin and is sketched. The

inclination of this line, which is the strength envelope in terms of effective stresses, with the σ-

axis is the effective friction angle. The value of c′ is zero; and the value of φ′, as measured with

a protractor, is 36 ° 30 ′. (Fig. 8.58)

Analytical method:

Since the tests are drained tests, we may assume c′ = 0. On this basis, we may obtain Nφ.

From both tests, Nφ = σ 1 /σ 3 = 4