(b) Mohr's diagram for unconfined compression lest
FIGURE 6Consider a soil prism in a state of triaxial stress. Let Q denote a point in this prism and
P a plane through Q. Let c = unit cohesive strength of soil; a = normal stress at Q on
plane P; Cr 1 and a 3 = maximum and minimum normal stress at Q 9 respectively; r = shear-
ing stress at Q, on plane P\ 6 = angle between P and the plane on which Cr 1 occurs; <£ = an-
gle of internal friction of the soil.
For an explanation of Mohr's circle of stress, refer to an earlier calculation procedure;
then refer to Fig. 6a. The shearing stress ED on plane P may be resolved into the cohesive
stress EG and the frictional stress GD. Therefore, r = c + a tan a. The maximum value of
a associated with point Q is found by drawing the tangent FH.
Assume that failure impends at Q. Two conclusions may be drawn: The angle between
FH and the base line OAB equals ^, and the angle between the plane of impending rup-
ture and the plane on which Cr 1 occurs equals one-half angle BCH. (A soil mass that is on
the verge of failure is said to be in limit equilibrium.)
In an unconfined compression test, the specimen is subjected to a vertical load without
being restrained horizontally. Therefore, (T 1 occurs on a horizontal plane.
Constructing Mohr's circle in Fig. 66, apply these values: (T 1 = 2040 lb/ft
2
(97 7 kPa)-
(T 3 = O; angle BCH= 2(56°) = 112°.
(a) Mohr's diagram for triaxial-stress condition