Stability of Rock Wedges and Excavation Surfaces - Rock Mechanics Based on an Anisotropic Jointed Rock Model - page 324

Q 1 and Q 2 can be determined from the balance of forces (11.87) considering the com-
ponents perpendicular and parallel to D2. Using the notation given in Fig. 11.29 the
balance of forces can be formulated by the following equations:

R N + Q 1N + Q 2N = 0, (11.90)

R T + Q 1T + Q 2T = 0, (11.91)

in which

R N = R ⋅ cos β D1 , (11.92)

R T = R ⋅ sin β D1 , (11.93)

Q 1T = Q 1N ⋅ tan φ D1,d , (11.94)

Q 2T = Q 2N ⋅ tan δ , (11.95)

where β D1 is the dip angle of D1. Considering the angular sum at triangles ACD and

AEF in Fig. 11.29 the angle δ can be expressed by β D1 , φ D,2 and the dip angle β D2 of D2:

δ = 90° + β D2 - β D1 - φ D2 . (11.96)

Inserting (11.92) to (11.95) into (11.90) and (11.91) and solving for Q 1N and Q 2N yields:

(11.97)

(11.98)

The absolute values of {Q 1 } and {Q 2 } can be calculated as follows:

(11.99)

(11.100)

where , Q 1N and Q 2N are given by (11.96) to (11.98).

Should the reaction force {Q 2 } be more steeply inclined than the resultant {R}, the closing
of the force polygon requires a tensile reaction force {Q 1 } that cannot be carried, so no bal-
ance of forces is possible (Wittke 1990). In such a case a resisting anchor force is required.
An investigation of the case where {R} intersects D2 leads to the result that only a ro-
tation about {r m } is possible.

δ