Page 289 - Physical Principles of Sedimentary Basin Analysis

8.6 Hydrofracturing 271
σ 1

S 0

σ 3
−T 0

Figure 8.13. The largest principal stress (σ 1 ) as a function of the least principal stress (σ 3 ). The
σ 1 -line cuts the σ 3 -axis at C 0 , which is the uniaxial compressive strength. The lowest σ 3 -value is the
tension cutoff −T 0 .

(This expression for σ M is derived in Exercise 8.4.) A fracture criterion that distinguishes
between shear fracture and tension fracture is summarized as follows:

! !
2 2
1 + μ − μ 0 σ 1 − 1 + μ + μ 0 σ 3 = 2S 0 for σ 1 >σ M
0 0 (8.30)
σ 3 =−T 0 for σ 1 <σ M .
Exercise 8.4 Show that expression (8.29) follows from equations (8.27) and (8.28) when
σ 3 = 0.

Exercise 8.5
1 1
(a) Show that the mean stress p = (σ 1 + σ 3 ) and the shear stress q = (σ 1 − σ 3 ) at
2 2
fracture are related as
q = p sin φ + S 0 cot φ. (8.31)

(b) Show that equation (8.27) for the largest principal stress as a function of the least
principal stress can be written as
1 + sin φ cos φ

σ 1 = σ 3 + 2S 0 . (8.32)
1 − sin φ 1 − sin φ
Solution: Figure 8.11 shows a Mohr’s circle that touches the fracture envelope. The center
1 1
of the circle is at (σ 1 + σ 3 ) and its radius is (σ 1 − σ 3 ). For the right-angled triangle in
2 2
Figure 8.11 we have
1 1
(σ 1 − σ 3 ) = S 0 cot φ + (σ 1 + σ 3 ) sin φ. (8.33)
2 2
This is equation (8.31), and we get equation (8.32) with the help of a little algebra.

8.6 Hydrofracturing
Fluid pressure changes the stress state of a saturated porous rock, and we have already seen
(Section 3.15) that it is the effective stress that controls the deformations of the rock. The
normal stress is reduced by the fluid pressure, while the shear stress is unaffected by the