Page 30 - A Practical Companion to Reservoir Stimulation

B. Rock Mechanics

EXAMPLE B-1
and Eq. 2-3 gives
Calculation of the Normal and Shear Stresses
at any Angle from the Principal Stresses 1
z = - (3000 - 4000) sin (2 x 20")
Assuming that the principal horizontal stress components at 2
depth are 3000 and 4000 psi, determine the normal and shear = -321 psi. (B-2)
stresses on a fault at 20" from the direction of the'minimum At the maximum sheer stress azh8 = 0, and thus from
horizontal stress. At what angle would the maximum shear Eq. 2-3 and remembering that z.~:,, = 0, then
stress be, what would be its value and what would be the
corresponding magnitude of the normal stress?

Solution (Ref. Section 2-2.1)
Figure B-l denotes the fault and the two principal stress
directions. In Eqs. 2-2 and 2-3, since or and CT, are the prin-
cipal stresses, then z,, = 0. (Definition of principal stress and, thus, 28= 90" or 8 = 45".
direction: shear stresses vanish.) Then, from Eq. 2-3 at 45",
As a result, Eq. 2-2 gives 1
z = - (3000 - 4000) sin 90" = -500 psi, (B-5)
CT = 4000 60s' 20" + 3000 sin2 20" 2
= 3883psi, (B- 1 ) and from Eq. 2-2 at 45",
CT = 4000 cos2 45" + 3000 sin2 45"

= 3500psi. (B-6)

B- 1