FIGURE 15
In Fig. 15, O is the center of an assumed arc of failure AC. Let W= weight of soil mass
above arc AC; R = resultant of all normal and frictional forces existing along arc AC; C =
resultant cohesive force developed; L 0 = length of arc AC; L 0 = length of chord AC. The
soil above the arc is in equilibrium under the forces W, R 9 and C. Since W is known in
magnitude and direction, the magnitude of C may be readily found if the directions of H
and C are determined.
Locate the first trial position of O by setting a = 25° and /3 = 35°.
- Draw the arc AC and the radius OM bisecting this arc
- Establish rectangular coordinate axes at O, making OM the
y axis - Obtain the needed basic data
Scale the drawing or make the necessary calculations. Thus, r = 78.8 ft (24.02 m); L 0 =
154.6 ft (47.12 m); L 0 = 131.0 ft (39.93 m); area above arc - 4050 ft^2 (376.2 m^2 ); W =
4050(120) - 486,000 Ib (2,161,728 N); horizontal distance from A to centroid of area =
66.7 ft (20.33 m). - Draw the vector representing W
Since the soil is homogeneous, this vector passes through the centroid of the area. - State the equation for C; locate its action line
Thus,
C = Cx = CL, (27)
4>- circle