Rock failure relationships

Understanding rock failure relationships is important because under reservoir pressure and stress conditions, production can induce rock failure, sometime with catastrophic effects. By applying strength criteria, within reservoir simulators we can predict when problems might occur.

Rock failure definition

Stress strain relationships in rocks examined the elastic behavior of rocks, which was largely reversible. Here we deal with permanent deformation. By rock failure, we mean the formation of faults and fracture planes, crushing, and relative motion of individual mineral grains and cements.

Failure can involve formation of discrete fracture zones and the more "ductile" or homogeneous deformation. The latter deformation is caused by a broad distribution of fracture zones or general grain crushing during compaction. We will not consider deformation caused by plastic strains of the mineral components, as is common in salt and in calcite at higher temperatures. In our analysis, several assumptions are made:

• Material is isotropic and homogeneous
• Stresses are applied uniformly
• Textural characteristics such as grain size and sorting have no influence
• Temperature and strain rate are ignored
• Intermediate stresses are presumed to play no role.

Each of these assumptions can be violated, and some have been demonstrated to have major influences on rock strength.

Coulumb-Navier failure

To begin with, a brief review of the standard Mohr failure criteria will be examined to introduce concepts and define terms, as well as to establish the basic mathematics behind the strength relationships.

Units of stress and strength are the same as pressure and are covered in Stress strain relationships in rocks. More detailed descriptions can be found in standard textbooks (i.e., Jaeger).

Mohr circles and linear failure envelope

Mohr circles and a linear failure envelope are the most common methods used to plot stresses and indicate strength limits. This technique predicts failure when stresses surpass both the intrinsic strength of a rock and internal friction.

The primary terms and characteristics are shown in Fig. 1.

• Normal stresses across any plane are plotted on the horizontal axis, and shear stresses are plotted on the vertical axis.
• Compressive stresses are defined as positive (as opposed to the mechanical engineering convention of tensional stresses being positive).
• For the hydrostatic case, all stresses are equal; this stress state is represented by a point on the horizontal axis.
• When stresses differ, the maximum principal stress, σ₁, and minimum stress, σ₃, are plotted on the horizontal axis and the possible shear stresses along any plane fall on a hemisphere connecting σ₁ and σ₃ (Fig. 2).

• 

  Fig. 1 – Schematic diagram defining stress orientations along and across a plane.

• 

  Fig. 2 – Generalized rock failure with a linear envelope.

The mean stress, σ_m, and radius of this circle, r, are simple sums and differences of the principal stresses.

....................(1)

....................(2)

The normal stress across any plane, σ_n, and the shear stress along the plane, τ, are functions of the principal stresses and the plane orientation.

....................(3)

....................(4)

....................(5)

....................(6)

where θ is the angle between the plane and the σ₃ direction.

From Eqs. 5 and 6, the maximum shear occurs along a plane oriented at (45°). However, because of friction, rocks do not fail along this plane. Instead, failure occurs along some rotated plane where friction is lower, yet shear stress is still high. This failure point (or plane) is shown in Fig. 1 as the nearly diagonal line. Fig. 1 also shows the associated normal and shear stresses. If numerous failure tests are made and plotted, an envelope is defined as in Fig. 2. In this case, friction is assumed to be a simple linear function of normal stress, and the resulting envelope is also linear. The slope of this envelope is α, and we define μ as the angle of internal friction

....................(7)

Within this framework, we can define several important properties of the rock as shown in Fig. 2: C₀ = Uniaxial or unconfined compressive strength (σ₃ = 0) C_u = Cohesive strength or the intercept of the envelope with σ_n = 0 C_t = Tensional strength

The failure envelope is then defined by the line

....................(8)

If the rock has already been broken, or a fracture already exists, then both C_u and C_t will be close to zero.

Several useful equations can be derived from the geometric relationships shown so far. From the equation for a circle,

....................(9)

At the intersection of the envelope and the circle, we must have

....................(10)

which leads to

....................(11)

Using the general solution to a second-order polynomial gives

....................(12)

Because we want only a point where the circle touches the envelope, the square root term must vanish.

....................(13)

After some algebraic manipulation, we find

....................(14)

and

....................(15)

Substitution of C_u (defined in Eq.14) into Eq. 12 gives an expression for normal stress.

....................(16)

If the envelope could be continued into the tensional region, the tensional strength could easily be obtained:

....................(17)

Under tension, the stresses are negative, although the tensional strength is a positive number. Thus, if rocks could fail according to a constant internal friction, we would have a simple way to relate the stresses involved and need only a couple of material constants, such as C_u and α.

Limitation of Mohr's approach

We are immediately faced with two problems when we try to apply Coulomb-Navier failure criteria:

1. Rocks do not generally have a linear failure envelope
2. Material properties controlling failure must be obtained either through logs or assumed behavior.

Fig. 3 shows the type of envelope commonly seen. In fact, we know that the slope must change as stresses are increased because rocks begin yielding and act more plastically. Fig. 4 shows the generalized behavior expected. At normal stresses above the brittle-ductile transition, failure can no longer be maintained on a single plane, but is distributed more homogeneously throughout the sample. We must develop different failure criteria, one that produced an appropriately curved envelope, and we expect it to have a strong porosity dependence (Fig. 5).

• 

  Fig. 3 – Measured failure strengths for the B-d #1 sandstone with straight-line segments and curved envelope fits to failure.

• 

  Fig. 4 – Curved rock failure envelope with schematic failure styles at different mean pressure ranges.

• 

  Fig. 5 – Failure mechanisms as a function of porosity for sandstones (after Scott).

Other failure approaches

Numerous failure criteria have been proposed that are primarily empirically based. Table 1 shows some of the criteria proposed both for general purposes and for specific rock types or conditions. Observed failure envelopes are smooth forms so simple exponential or power-law functions can usually be found that fit the data well. The relations of Bienlawski and Hoek and Brown are most common. Much of the recent work in rock mechanics has been directed toward ascertaining the constants of these relationships in terms of easily measurable rock properties. Note that these relationships apply primarily to the brittle failure regime and cannot be used for grain crushing or pore collapse (as discussed in Compressive strength of rocks) or when substantial ductile or plastic deformation is involved. We will examine these proposed forms to interrelate terms and reduce unknowns to variables that can be derived from logs.

• 

  Table 1 - Strength laws, sandstones

Hoek-Brown equations

Hoek and Brown compiled extensive data on a variety of rock types and produced relationships that are simple and can be developed into forms amenable to well-log analysis. A primary feature of this failure criterion is a relation between the maximum and minimum stresses when both are normalized by the uniaxial compressive strength

....................(18)

This formulation was motivated by the systematic behavior seen in many tests as shown in Fig. 6. InEq. 18,mand sare material constants dependent on the overall quality of the rock mass, and mis also dependent on the rock type ('Table 2).

• 

  Fig. 6 – Failure envelope for sandstones after Hoek and Brown. Stresses are normalized by rock uniaxial compressive strength.

• 

  Table 2 - Hoek and Brown lithologic and quality coefficients

Note that we could derive the value for m from a mineralogic analysis. In our analysis, we will presume that the local rock mass of interest is intact, and thus

....................(19)

For applications that are in sandstones, numeric results can often use

....................(20)

Eq. 18 can be rewritten to give one principal stress in terms of the other:

....................(21)

....................(22)

Such normalized stress states were used to construct the curved envelope in Fig. 4.

The tensional strength, the stress at which an envelope would cross the horizontal axis, is found by equating σ₁ to σ₃ in Eq. 21 (note that C_t is defined as a positive number).

....................(23)

For sandstones, this results in , or 0.067 C₀. The 15 uniaxial tensional strength σ_τ* is slightly different and is defined as the value at which the maximum stress, σ₁, equals zero. From Eq. 21, we get

....................(24)

Other basic properties are not so simply derived.

We must produce from the stress relationships (Eqs. 18 or 21) an equation for a failure envelope that permits us to resolve the shear and normal stresses on a failure plane, its orientation, and an approximation of the internal friction, and simply predict regions of instability. The general envelope shapes seen in Figs. 4 and 6 suggest a form like that proposed by Murrell and Bienlawski:

....................(25)

where A, b, and n are material constants. Because the envelope intersects the horizontal axis when the normal stress equals the tensional strength,

....................(26)

When the normal stress is zero, the envelope intersects the vertical axis at the cohesion value C_u. From Eq. 25, this requires

....................(27)

Therefore, the general form for an envelope is

....................(28)

To derive the slope, α, at any point, we note that the envelope is only slowly varying over a small stress range and could be locally approximated by a line. If we use a pseudo-cohesion defined by Eq. 14 for the stress condition, σ_m, r we can subtract the same from a slightly different stress condition, σ_m’, r’. Solving for α gives

....................(29)

....................(30)

The Hoek-Brown stress criteria allow us to redefine the mean, σ_m, and differential, r, stresses

....................(31)

....................(32)

By substituting these relations into Eq. 30 for two stresses σ₃ and σ₃ + δ σ₃, expanding the result and allowing the stress difference, δσ₃, to approach zero (what a pain!), we find

....................(33)

As we found previously in Eq. 16, the normal stress is then,

....................(34)

The cohesion is the shear stress value when σ_n equals zero. This will occur for σ₃ somewhere between zero and −C_t. In other words, σ_n = 0 for

....................(35)

where β is a value around 0.5. We could substitute this term into Eqs. 33 and 34 and solve for β . However, this results in a rather complicated root to a third-order polynomial. Fortunately, by iteration, we can show that β is relatively constant at about 0.62 with little dependence on m. Using this value of β in Eq. 35 and substituting into the previous equations gives us our cohesion. For a sandstone with m = 15, we get

....................(36)

The definition of our curved envelope in Eq. 12 is not strictly compatible with the Hoek-Brown stress relations. However, we can get an estimate of the exponent, n, by using our tensile and cohesion strengths and some reasonable value of σ_n such as σ_n = C₀. From Fig. 2, we can see that τ is approximately 1.1 C₀ at this point. From Eq. 28, with m equal to 15,

....................(20)

This value falls within the range of 0.65 to 0.75 suggested by Yudhbir et al. Thus, from a presumed simple relation between σ₁ and σ₃, almost all the necessary parameters can be derived.

Nomenclature

C0	=	uniaxial or unconfined compressive strength, GPa or MPa
Ct	=	tensional strength, GPa or MPa
Cu	=	cohesive strength, GPa or MPa
m	=	Hoek-Brown strength coefficient
s	=	Hoek-Brown strength coefficient
r	=	radius of stress "circle," GPa or MPa
n	=	strength envelope exponent
σm	=	mean stress, GPa or MPa
σn	=	normal stress, GPa or MPa
σshear	=	shear stress components, GPa or MPa
α	=	aspect ratio
β	=	strength factor, numeric
σ1	=	stress in direction 1, GPa or MPa
σ2	=	stress in direction 2, GPa or MPa
σ3	=	stress in direction 3, GPa or MPa
θ	=	wave propagation angle to symmetry axis
τ	=	shear stress, GPa or MPa
A	=	strength material constant
b	=	strength envelope intercept, GPa or MPa

Noteworthy papers in OnePetro

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External links

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See also

Compressive strength of rocks

Pore fluid effects on rock mechanics

Subsurface stress and pore pressure

Stress strain relationships in rocks

PEH:Rock_Properties