Page 290 - Physical Principles of Sedimentary Basin Analysis

272 Rheology: fracture and flow
τ τ

φ p f φ p f
σ σ
σ = σ 1
3
σ
−T 0 σ 3 1 −T 0

(a) (b)
Figure 8.14. (a) A (positive) fluid pressure moves the Mohr’s circle to the right. (b) An isotropic stress
state can be moved leftwards until tensional fracture.

fluid pressure. Stress states represented by a Mohr’s circle are therefore shifted towards
the left by the value of the fluid pressure, because both the least and the largest principal
stress are normal stresses. Figure 8.14a shows how a stress state becomes translated to
the left. An increasing fluid pressure can therefore cause failure. The fluid pressure can
increase until the Mohr’s circle touches the fracture envelope, as shown in Figure 8.14a,
which is the condition for fracture. Fracturing caused by the fluid pressure is hydrofractur-
ing, and it leads to enough pathways for the fluid to prevent further increase in the fluid
pressure. Figure 8.14b also shows how increasing fluid pressure leads to tensile fracture for
an isotropic stress state. The effect of the fluid pressure is the same for sliding (Amontons’
law) as well as for fracturing.
The principal stresses σ 1 and σ 3 in the Coulomb fracture criterion (8.30) are therefore
replaced by the effective principal stress

σ = σ 1 − p f and σ = σ 3 − p f (8.34)
3
1
and the fracture criterion (8.30) becomes

! !
2 2
1 + μ − μ 0 σ − 1 + μ + μ 0 σ = 2S 0 for σ >σ M
0 1 0 3 1 (8.35)

σ =−T 0 for σ <σ M .
3 1
This fracture criterion can also be rewritten as an expression for the fluid pressure necessary
for fracturing. Some more notation is now introduced to simplify this task. We let a = (1+
2 1/2
2 1/2
μ ) −μ 0 , b = (1+μ ) +μ 0 and f = σ 3 /σ 1 . The latter coefficient express the amount
0 0
of stress anisotropy. The coefficient f is less than 1 (since σ 1 is the largest principal stress),
2
2 1/2
but sufficiently large for bf −a > 0, or alternatively f > a/b = ((1+μ ) −μ 0 ) .The
0
condition (8.35) for shear fracture is then
a (σ 1 − p f ) = b (σ 3 − p f ) + 2S 0 (8.36)