DHARM
298 GEOTECHNICAL ENGINEERING
tan φ = τ/σ = 4/10 = 0.4
φ = 21 ° 48 ′
θ = 45° + φ/2 = 45° +
21 48
2
° ′
= 55 ° 54 ′.
Graphical solution (Fig. 8.45):
The procedure is first to draw the σ-and τ-axes from an origin O and then, to a suitable scale,
set-off point D with coordinates (10,4), Joining O to D, the strength envelope is got. The Mohr
Circle should be tangential to OD to D. DC is drawn perpendicular to OD to cut OX in C, which
is the centre of the circle. With C as the centre and CD as radius, the circle is completed to cut
OX in A and B.
D
4 kN/m
2
C
2 = 111°45q¢
A B
s
Strength envelop
t e
O
10 kN/m^2
s 3 = 7.25 kN/m^2
s 3 = 15.9
sr= 10.8 kN/m
2
f= 22°
Fig. 8.45 Mohr’s circle (Ex. 8.1)
By scaling, the resultant stress = OD = 10.8 kN/m^2.
With protractor, φ = 22° and θ = 55°53′
We also observe than σ 3 = OA = 7.25 kN/m^2 and σ 1 = OB = 15.9 kN/m^2.
Example 8.2: Clean and dry sand samples were tested in a large shear box, 25 cm × 25 cm and
the following results were obtained :
Normal load (kN) 5 10 15
Peak shear load (kN) 5 10 15
Ultimate shear load (kN) 2.9 5.8 8.7
Determine the angle of shearing resistance of the sand in the dense and loose states.
The value of φ obtained from the peak stress represents the angle of shearing resistance
of the sand in its initial compacted state; that from the ultimate stress corresponds to the sand
when loosened by the shearing action.
The area of the shear box = 25 × 25 = 625 cm^2.
= 0.0625 m^2.
Normal stress in the first test = 5/0.0625 kN/m^2 = 80 kN/m^2