# Failure Probability Model considering the Effect of Intermediate Principal Stress on Rock Strength

Authors: Yonglai Zheng; Shuxin Deng
Journal: Mathematical Problems in Engineering
 (2015)

---

## Abstract

A failure probability model is developed to describe the effect of the intermediate principal stress on rock strength. Each shear plane in rock samples is considered as a micro-unit. The strengths of these micro-units are assumed to match Weibull distribution. The macro strength of rock sample is a synthetic consideration of all directions’ probabilities. New model reproduces the typical phenomenon of intermediate principal stress effect that occurs in some true triaxial experiments. Based on the new model, a strength criterion is proposed and it can be regarded as a modified Mohr-Coulomb criterion with a uniformity coefficient. New strength criterion can quantitatively reflect the intermediate principal stress effect on rock strength and matches previously published experimental results better than common strength criteria.


---

## Body

## 1. Introduction

The stress state in three-dimensional space is defined by three mutually perpendicular stress components (σ
1, σ
2, σ
3) and it is an important subject in rock mechanics to study the rock failure behavior under complex stress conditions. The traditional considerations suggest that shear failure will occur when the shear stress along some plane in the sample is too large. The extreme values of shear stress in a material are related only to the largest and smallest principal stress (commonly denoted as σ
1 and σ
3). Based on such considerations, the influence of intermediate principal stress (σ
2) on the experimental results is not taken into account.However three-dimensional unequal stress states are very common in engineering practice and it is very important to predict the rock strength with the effect of intermediate principal stress. Many researchers [1–6] have conducted a large number of true triaxial tests on different rock types such as Dunham dolomite, Solnhofen limestone, and granite, to investigate the behavior under triaxial stress conditions. They found that the strength first increases and then reduces with the increase of σ
2. At the same time, researchers have developed many numerical models or used commercial software to study the failure processes of rocks under polyaxial stress conditions [7–11]. In these numerical tests, it was clear that the intermediate principal stress has an effect on rock strength and the mechanism of such an effect is discussed. The phenomenon of intermediate principal stress effect seems to be related to the heterogeneity of materials [10, 11]. Without material heterogeneity, local tensile stress cannot be generated in an overall compressive stress environment so that there is no crack initiation and propagation [7]. The intermediate principal stress confines the rock in such a way that fractures can be easier to develop in the direction parallel to σ
1 and σ
2. Fjær and Ruistuen’s numerical results [11] indicate that effect of σ
2 is related to stress symmetry, rather than the stress level. When σ
2 was close to σ
3 or σ
1, there were more possible directions for the failure planes with higher stress symmetry. The directions for the failure plane for which the theoretical failure criterion is first fulfilled may not coincide with the directions preferred by the rock’s heterogeneity. The rock will be weaker where the stress state is more symmetric and there are several equivalent directions of the failure plane to choose. These considerations are supported by Chang and Haimson’s observations [1]. They found that stress induced micro cracks in amphibolites were randomly oriented when σ
2
=
σ
3, while micro cracks became more aligned with the direction of σ
2 when σ
2
>
σ
3. Similar phenomenon was also observed by Cai [7] and Pan et al. [10].According to Fjær and Ruistuen’s considerations [11], σ
2 affects the failure probabilities in different directions under three-dimensional unequal stress. Which direction the failure plane eventually takes is determined by the heterogeneities of the rock. The heterogeneity is always incorporated by assuming the micro-units’ properties complying with a certain distribution and here Weibull distribution [12] is selected. Weibull distribution is introduced to explain the statistical size effect and later justified theoretically on the basis of some reasonable hypotheses about the statistical distribution and the role of microscopic flaws or micro cracks [13]. Unlike previous Weibull analysis [14–17] with volume consideration in this paper each potential shear failure plane is regarded as a micro-unit and failure probabilities of all directions are calculated. Thus, a new failure probability model considering the failure probabilities in different directions is proposed. By combining the new model with Mohr-Coulomb criterion, a new strength criterion is developed, which can quantitatively describe the effect of intermediate principal stress on rock strength.


## 2. Modelling

### 2.1. Failure Probability Model with Volume Considerations

Assume that under the nominal stressR the density function of micro flaws per unit volume is n
(
R
). The amount of micro flaws in materials sample with a volume of V could be expressed as(1)
N
R
=
V
∫
0
R
n
x
d
x
.The strength of micro-units complies with Weibull distribution [12] and the probability density function is (2)
f
R
=
m
R
0
R
-
R
u
R
0
m
-
1
exp
⁡
-
R
-
R
u
R
0
m
,where · is Macaulay bracket; when x
≤
0, x
=
0; when x
>
0, x
=
x; R
u is stress threshold; R
0 is scalar parameter, R
0
>
0; m is shape parameter and it can be considered as the uniformity coefficient, m
>
1.Stress thresholdR
u is often set to zero. In this case, (2) has only two parameters (m and R
0). The failure probability of materials sample with one flaw can be obtained:(3)
F
R
=
∫
0
R
f
x
d
x
=
1
-
exp
⁡
-
R
R
0
m
.According to weakest link theory, the failure probability of materials sample with a volume ofV can be represented:(4)
F
V
R
=
1
-
1
-
F
R
N
R
.By inserting (1) and (3) into (4), we obtain(5)
F
V
R
=
1
-
exp
⁡
-
R
R
0
m
V
∫
0
R
n
x
d
x
=
1
-
exp
⁡
-
V
V
0
R
R
0
m
,where V
0
=
1
/
∫
0
R
n
(
x
)
d
x is considered as a reference volume.From (5), the average strength can be calculated:(6)
R
¯
=
∫
0
∞
s
f
s
d
s
=
∫
0
∞
s
d
F
s
=
∫
0
∞
1
-
F
V
s
d
s
.By inserting (5) into (6) and using variable substitution, (6) can be represented:(7)
R
¯
=
R
0
V
0
V
1
/
m
Γ
1
+
1
m
,where Γ
(
·
) is gamma function, Γ
(
x
)
=
∫
0
∞
t
x
-
1
exp
⁡
(
-
t
)
d
t.The variances
2 and variation coefficient ω can be obtained from (5) and (7):(8)
s
2
=
∫
R
u
∞
x
-
R
¯
2
f
x
d
x
=
∫
R
u
∞
x
-
R
¯
2
d
F
V
x
=
R
¯
2
Γ
1
+
2
/
m
Γ
2
1
+
1
/
m
-
1
,
(9)
ω
=
s
R
¯
=
Γ
1
+
2
/
m
Γ
2
1
+
1
/
m
-
1
.Considering the nonuniform force field, strength of each point inx
y
z coordinates can be expressed as R
f
(
x
,
y
,
z
). Here R is the maximum strength and f
(
x
,
y
,
z
) is a dimensionless coordinate function for each point. Thus (5) becomes(10)
F
V
R
=
1
-
exp
⁡
-
1
V
0
∫
R
f
x
,
y
,
z
R
f
x
,
y
,
z
R
0
m
d
V
.Accordingly, (6) becomes (11)
R
∗
¯
=
∫
0
∞
exp
⁡
-
g
R
V
0
d
R
,
(12)
g
R
=
∫
R
f
x
,
y
,
z
R
f
x
,
y
,
z
R
0
m
d
V
.Comparing the average strengthR
¯ in uniform force field and in nonuniform force field, from (6) and (11) we obtain(13)
R
∗
¯
=
R
¯
V
V
∗
1
/
m
,
(14)
V
∗
=
∫
V
f
x
,
y
,
z
m
d
V
.


### 2.2. Failure Probability Model with Direction Considerations

Shear failure is a basic failure mode of rocks under triaxial compression. Following Fjær and Ruistuen’s considerations [11], σ
2 affects the failure probabilities in different directions under three-dimensional unequal stress. In order to describe the intermediate principal stress effect, direction considerations instead of volume considerations are made to express failure probabilities in different directions. The shear planes in rock samples are considered as potential failure planes. In order to calculate the probability of each direction, each potential shear failure plane is regarded as a micro-unit. The effect of the intermediate principal stress can quantitatively be estimated by calculating the failure probabilities for all the shear planes and combining these into the total probability for failure.Figure1 shows a normal vector (ON) of a shear plane defined by the spherical coordinates α and φ, and the normal and shear stresses are given as(15)
σ
=
σ
1
cos
2
⁡
α
+
σ
2
sin
2
⁡
α
cos
2
⁡
φ
+
σ
3
sin
2
⁡
α
sin
2
⁡
φ
,
(16)
τ
=
σ
1
2
cos
2
⁡
α
+
σ
2
2
sin
2
⁡
α
cos
2
⁡
φ
+
σ
3
2
sin
2
⁡
α
sin
2
⁡
φ
-
σ
2
.Figure 1
Normal vector of a shear plane.The ranges ofα and φ are 0
≤
α
≤
π
/
2 and 0
≤
φ
≤
π
/
2. For each shear plane, it is assumed that shear failure will occur when the shear stress along some surface in the sample is too large. The maximum strength is(17)
R
m
a
x
=
σ
1
-
σ
3
2
.The nominal stress for each shear plane can be expressed asR
m
a
x
f
(
α
,
φ
) with a direction function f
(
α
,
φ
). Based on the assumption, failure occurs when(18)
R
m
a
x
f
α
,
φ
=
τ
.According to (16), (17), and (18), f
(
α
,
φ
) can be expressed as(19)
f
φ
,
α
=
2
sin
2
⁡
α
c
o
s
2
φ
1
-
b
2
+
sin
2
⁡
φ
-
sin
4
⁡
α
cos
2
⁡
φ
1
-
b
+
sin
2
⁡
φ
2
,where b is intermediate stress ratio, b
=
(
σ
2
-
σ
3
)
/
(
σ
1
-
σ
3
).For the direction consideration instead of volume consideration, thef
(
x
,
y
,
z
) in (10) could be replaced by f
(
α
,
φ
). Considering the special condition b
=
0 (σ
3
=
σ
2
<
σ
1, the conventional triaxial test), f
(
α
,
φ
) becomes(20)
f
α
,
φ
=
sin
⁡
2
α
.By inserting (20) into (14), the reference volume (denoted by V
0 for b
=
0) can be obtained as(21)
V
0
=
π
2
∫
0
π
/
2
sin
m
⁡
2
α
d
αwhen b
>
0, and according to (13), we obtain(22)
R
∗
¯
=
R
0
¯
V
0
V
∗
1
/
m
,where R
0
¯ is the mean strength when b
=
0, namely, the strength in conventional triaxial test. V
b can be calculated by (14)(23)
V
b
=
∫
0
π
/
2
∫
0
π
/
2
f
m
α
,
φ
d
α
d
φ
.In volume considerations, materials heterogeneity represents the different properties between different points. In direction considerations, materials heterogeneity represents the different properties between different shear planes. Equations (22) and (13) have similar form. However, meanings of “volume” are different. V
∗ and V in (13) are integrals of the dimensionless coordinate function f
(
x
,
y
,
z
) for volume considerations while V
∗ and V
0 in (22) are integrals of the direction function f
(
α
,
φ
) for direction considerations. Also, f
(
x
,
y
,
z
) in (10) is replaced by f
(
α
,
φ
) and the macro strength of materials sample is a synthetic consideration of all directions’ probabilities.The variances
2 and variation coefficient ω can be obtained from (8) and (9). Figure 2 shows the relationship curve between ω and m according to (9). From (19), (22), and (23) the relationship between R
b
/
R
0 and b can be obtained which is shown in Figure 3. The results shown in Figure 3 indicate that the intermediate principal stress significantly affects rock strength. The failure probability model in this paper is observed to reproduce the typical phenomenon of intermediate principal stress effect that occurs in some true triaxial experiments. The effect of intermediate principal stress is controlled by m. As m increases, the materials become more homogeneous and the effects of intermediate principal stress become less prominent and the variation coefficients of the results are lower (see Figure 2). When m
→
∞, the materials are absolutely homogeneous, there is no intermediate principal stress effect, and the curve of R
b
/
R
0
~
b is a horizontal line. The model developed in this paper becomes Mohr-Coulomb criterion, which does not include any effect of intermediate principal stress.Figure 2
Relationship betweenω and m.Figure 3
Relationship betweenR
b
/
R
0 and b for various m.


### 2.3. Strength Criterion considering Intermediate Principal Stress Effect

For the shear failure of rock materials, computed strengthR can be expressed by the shear stress. When b
=
0, R
0 can be calculated by Mohr-Coulomb criterion:(24)
R
0
=
σ
10
-
σ
3
2
=
σ
3
sin
⁡
φ
+
c
cos
⁡
φ
1
-
sin
⁡
φ
,where φ is inner friction angel, c is cohesion, and σ
10 is axial stress when b
=
0 (σ
2
=
σ
3).From (24), (22) becomes(25)
σ
1
-
σ
3
2
=
σ
3
sin
⁡
φ
+
c
cos
⁡
φ
1
-
sin
⁡
φ
V
0
V
b
1
/
m
.Thus, a new strength criterion considering intermediate principal stress effect and heterogeneity of materials is obtained:(26)
τ
13
=
f
1
σ
3
·
f
2
μ
σ
,where(27)
τ
13
=
σ
1
-
σ
3
2
,
f
1
σ
3
=
σ
3
sin
⁡
φ
+
c
cos
⁡
φ
1
-
sin
⁡
φ
,
f
2
μ
σ
=
2
π
∫
0
π
/
2
sin
m
⁡
2
α
d
α
∫
0
2
π
∫
0
π
/
2
f
3
m
φ
,
α
,
μ
σ
d
α
d
φ
1
/
m
,
f
3
φ
,
α
,
μ
σ
=
2
sin
2
⁡
α
1
-
μ
σ
2
4
cos
2
⁡
φ
+
sin
2
⁡
φ
-
sin
4
⁡
α
1
-
μ
σ
2
cos
2
⁡
φ
+
sin
2
⁡
φ
2
,
μ
σ
=
2
σ
2
-
σ
1
-
σ
3
σ
1
-
σ
3
.μ
σ is Lode parameter, which is used to distinguish between the different shear stress states in three dimensions (3D), ranging from axisymmetric tension to biaxial tension with axisymmetric compression and passing through in-plane shear. The relationship between μ
σ and b is b
=
(
1
+
μ
σ
)
/
2. When m
→
∞, f
2
(
μ
σ
)
=
1, strength is independent of μ
σ or b and (26) is equivalent to Mohr-Coulomb criterion. Thus, (26) could be regarded as a modified Mohr-Coulomb criterion that can reflect the effect of intermediate principal stress.


## 2.1. Failure Probability Model with Volume Considerations

Assume that under the nominal stressR the density function of micro flaws per unit volume is n
(
R
). The amount of micro flaws in materials sample with a volume of V could be expressed as(1)
N
R
=
V
∫
0
R
n
x
d
x
.The strength of micro-units complies with Weibull distribution [12] and the probability density function is (2)
f
R
=
m
R
0
R
-
R
u
R
0
m
-
1
exp
⁡
-
R
-
R
u
R
0
m
,where · is Macaulay bracket; when x
≤
0, x
=
0; when x
>
0, x
=
x; R
u is stress threshold; R
0 is scalar parameter, R
0
>
0; m is shape parameter and it can be considered as the uniformity coefficient, m
>
1.Stress thresholdR
u is often set to zero. In this case, (2) has only two parameters (m and R
0). The failure probability of materials sample with one flaw can be obtained:(3)
F
R
=
∫
0
R
f
x
d
x
=
1
-
exp
⁡
-
R
R
0
m
.According to weakest link theory, the failure probability of materials sample with a volume ofV can be represented:(4)
F
V
R
=
1
-
1
-
F
R
N
R
.By inserting (1) and (3) into (4), we obtain(5)
F
V
R
=
1
-
exp
⁡
-
R
R
0
m
V
∫
0
R
n
x
d
x
=
1
-
exp
⁡
-
V
V
0
R
R
0
m
,where V
0
=
1
/
∫
0
R
n
(
x
)
d
x is considered as a reference volume.From (5), the average strength can be calculated:(6)
R
¯
=
∫
0
∞
s
f
s
d
s
=
∫
0
∞
s
d
F
s
=
∫
0
∞
1
-
F
V
s
d
s
.By inserting (5) into (6) and using variable substitution, (6) can be represented:(7)
R
¯
=
R
0
V
0
V
1
/
m
Γ
1
+
1
m
,where Γ
(
·
) is gamma function, Γ
(
x
)
=
∫
0
∞
t
x
-
1
exp
⁡
(
-
t
)
d
t.The variances
2 and variation coefficient ω can be obtained from (5) and (7):(8)
s
2
=
∫
R
u
∞
x
-
R
¯
2
f
x
d
x
=
∫
R
u
∞
x
-
R
¯
2
d
F
V
x
=
R
¯
2
Γ
1
+
2
/
m
Γ
2
1
+
1
/
m
-
1
,
(9)
ω
=
s
R
¯
=
Γ
1
+
2
/
m
Γ
2
1
+
1
/
m
-
1
.Considering the nonuniform force field, strength of each point inx
y
z coordinates can be expressed as R
f
(
x
,
y
,
z
). Here R is the maximum strength and f
(
x
,
y
,
z
) is a dimensionless coordinate function for each point. Thus (5) becomes(10)
F
V
R
=
1
-
exp
⁡
-
1
V
0
∫
R
f
x
,
y
,
z
R
f
x
,
y
,
z
R
0
m
d
V
.Accordingly, (6) becomes (11)
R
∗
¯
=
∫
0
∞
exp
⁡
-
g
R
V
0
d
R
,
(12)
g
R
=
∫
R
f
x
,
y
,
z
R
f
x
,
y
,
z
R
0
m
d
V
.Comparing the average strengthR
¯ in uniform force field and in nonuniform force field, from (6) and (11) we obtain(13)
R
∗
¯
=
R
¯
V
V
∗
1
/
m
,
(14)
V
∗
=
∫
V
f
x
,
y
,
z
m
d
V
.


## 2.2. Failure Probability Model with Direction Considerations

Shear failure is a basic failure mode of rocks under triaxial compression. Following Fjær and Ruistuen’s considerations [11], σ
2 affects the failure probabilities in different directions under three-dimensional unequal stress. In order to describe the intermediate principal stress effect, direction considerations instead of volume considerations are made to express failure probabilities in different directions. The shear planes in rock samples are considered as potential failure planes. In order to calculate the probability of each direction, each potential shear failure plane is regarded as a micro-unit. The effect of the intermediate principal stress can quantitatively be estimated by calculating the failure probabilities for all the shear planes and combining these into the total probability for failure.Figure1 shows a normal vector (ON) of a shear plane defined by the spherical coordinates α and φ, and the normal and shear stresses are given as(15)
σ
=
σ
1
cos
2
⁡
α
+
σ
2
sin
2
⁡
α
cos
2
⁡
φ
+
σ
3
sin
2
⁡
α
sin
2
⁡
φ
,
(16)
τ
=
σ
1
2
cos
2
⁡
α
+
σ
2
2
sin
2
⁡
α
cos
2
⁡
φ
+
σ
3
2
sin
2
⁡
α
sin
2
⁡
φ
-
σ
2
.Figure 1
Normal vector of a shear plane.The ranges ofα and φ are 0
≤
α
≤
π
/
2 and 0
≤
φ
≤
π
/
2. For each shear plane, it is assumed that shear failure will occur when the shear stress along some surface in the sample is too large. The maximum strength is(17)
R
m
a
x
=
σ
1
-
σ
3
2
.The nominal stress for each shear plane can be expressed asR
m
a
x
f
(
α
,
φ
) with a direction function f
(
α
,
φ
). Based on the assumption, failure occurs when(18)
R
m
a
x
f
α
,
φ
=
τ
.According to (16), (17), and (18), f
(
α
,
φ
) can be expressed as(19)
f
φ
,
α
=
2
sin
2
⁡
α
c
o
s
2
φ
1
-
b
2
+
sin
2
⁡
φ
-
sin
4
⁡
α
cos
2
⁡
φ
1
-
b
+
sin
2
⁡
φ
2
,where b is intermediate stress ratio, b
=
(
σ
2
-
σ
3
)
/
(
σ
1
-
σ
3
).For the direction consideration instead of volume consideration, thef
(
x
,
y
,
z
) in (10) could be replaced by f
(
α
,
φ
). Considering the special condition b
=
0 (σ
3
=
σ
2
<
σ
1, the conventional triaxial test), f
(
α
,
φ
) becomes(20)
f
α
,
φ
=
sin
⁡
2
α
.By inserting (20) into (14), the reference volume (denoted by V
0 for b
=
0) can be obtained as(21)
V
0
=
π
2
∫
0
π
/
2
sin
m
⁡
2
α
d
αwhen b
>
0, and according to (13), we obtain(22)
R
∗
¯
=
R
0
¯
V
0
V
∗
1
/
m
,where R
0
¯ is the mean strength when b
=
0, namely, the strength in conventional triaxial test. V
b can be calculated by (14)(23)
V
b
=
∫
0
π
/
2
∫
0
π
/
2
f
m
α
,
φ
d
α
d
φ
.In volume considerations, materials heterogeneity represents the different properties between different points. In direction considerations, materials heterogeneity represents the different properties between different shear planes. Equations (22) and (13) have similar form. However, meanings of “volume” are different. V
∗ and V in (13) are integrals of the dimensionless coordinate function f
(
x
,
y
,
z
) for volume considerations while V
∗ and V
0 in (22) are integrals of the direction function f
(
α
,
φ
) for direction considerations. Also, f
(
x
,
y
,
z
) in (10) is replaced by f
(
α
,
φ
) and the macro strength of materials sample is a synthetic consideration of all directions’ probabilities.The variances
2 and variation coefficient ω can be obtained from (8) and (9). Figure 2 shows the relationship curve between ω and m according to (9). From (19), (22), and (23) the relationship between R
b
/
R
0 and b can be obtained which is shown in Figure 3. The results shown in Figure 3 indicate that the intermediate principal stress significantly affects rock strength. The failure probability model in this paper is observed to reproduce the typical phenomenon of intermediate principal stress effect that occurs in some true triaxial experiments. The effect of intermediate principal stress is controlled by m. As m increases, the materials become more homogeneous and the effects of intermediate principal stress become less prominent and the variation coefficients of the results are lower (see Figure 2). When m
→
∞, the materials are absolutely homogeneous, there is no intermediate principal stress effect, and the curve of R
b
/
R
0
~
b is a horizontal line. The model developed in this paper becomes Mohr-Coulomb criterion, which does not include any effect of intermediate principal stress.Figure 2
Relationship betweenω and m.Figure 3
Relationship betweenR
b
/
R
0 and b for various m.


## 2.3. Strength Criterion considering Intermediate Principal Stress Effect

For the shear failure of rock materials, computed strengthR can be expressed by the shear stress. When b
=
0, R
0 can be calculated by Mohr-Coulomb criterion:(24)
R
0
=
σ
10
-
σ
3
2
=
σ
3
sin
⁡
φ
+
c
cos
⁡
φ
1
-
sin
⁡
φ
,where φ is inner friction angel, c is cohesion, and σ
10 is axial stress when b
=
0 (σ
2
=
σ
3).From (24), (22) becomes(25)
σ
1
-
σ
3
2
=
σ
3
sin
⁡
φ
+
c
cos
⁡
φ
1
-
sin
⁡
φ
V
0
V
b
1
/
m
.Thus, a new strength criterion considering intermediate principal stress effect and heterogeneity of materials is obtained:(26)
τ
13
=
f
1
σ
3
·
f
2
μ
σ
,where(27)
τ
13
=
σ
1
-
σ
3
2
,
f
1
σ
3
=
σ
3
sin
⁡
φ
+
c
cos
⁡
φ
1
-
sin
⁡
φ
,
f
2
μ
σ
=
2
π
∫
0
π
/
2
sin
m
⁡
2
α
d
α
∫
0
2
π
∫
0
π
/
2
f
3
m
φ
,
α
,
μ
σ
d
α
d
φ
1
/
m
,
f
3
φ
,
α
,
μ
σ
=
2
sin
2
⁡
α
1
-
μ
σ
2
4
cos
2
⁡
φ
+
sin
2
⁡
φ
-
sin
4
⁡
α
1
-
μ
σ
2
cos
2
⁡
φ
+
sin
2
⁡
φ
2
,
μ
σ
=
2
σ
2
-
σ
1
-
σ
3
σ
1
-
σ
3
.μ
σ is Lode parameter, which is used to distinguish between the different shear stress states in three dimensions (3D), ranging from axisymmetric tension to biaxial tension with axisymmetric compression and passing through in-plane shear. The relationship between μ
σ and b is b
=
(
1
+
μ
σ
)
/
2. When m
→
∞, f
2
(
μ
σ
)
=
1, strength is independent of μ
σ or b and (26) is equivalent to Mohr-Coulomb criterion. Thus, (26) could be regarded as a modified Mohr-Coulomb criterion that can reflect the effect of intermediate principal stress.


## 3. Model Validation and Verification

### 3.1. Comparison with Experimental Results

Comparisons between the failure probability model as described in this paper and results of the experimental results [5] are plotted in Figure 4. Fjær and Ruistuen [11] have developed a numerical method to calculate the intermediate principal stress effect on rock strength. Their results are also plotted in Figure 4 for comparison. The strength parameters can be obtained by fitting the strength envelopes. Three strength parameters are as follows: c
=
31
M
P
a, φ
=
27.37
°, and m
=
2.5. From Figure 4, both model predictions in this paper and Fjær and Ruistuen’s numerical results [11] are observed to be in good agreement with the results from the experimental results [5]. Failure probability model in this paper matches the experimental results [5] better than Fjær and Ruistuen’s numerical results [11], especially when σ
2 is low.Figure 4
Relationships betweenσ
1 and σ
2 for constant σ
3.


### 3.2. Comparison with Common Strength Criteria

Common strength criteria are Mohr-Coulomb criterion, Drucker-Prager criterion, 3D Griffith criterion, and twin shear stress criterion. Mathematical forms of these strength criteria are as follows.Mohr-Coulomb criterion is as follows:(28)
σ
1
-
σ
3
2
=
σ
1
+
σ
3
2
sin
⁡
φ
+
c
cos
⁡
φ
.Drucker-Prager criterion [18] is as follows:(29)
J
2
-
α
I
1
-
k
=
0
,where α and k are material constants, I
1
=
σ
1
+
σ
2
+
σ
3 is the first stress invariant, and J
2
=
[
σ
1
-
σ
2
2
+
σ
2
-
σ
3
2
+
σ
3
-
σ
1
2
]
/
6 is the second deviatoric stress invariant.3D Griffith criterion [19] is as follows:(30)
σ
1
-
σ
2
2
+
σ
2
-
σ
3
2
+
σ
3
-
σ
1
2
=
24
T
0
σ
1
+
σ
2
+
σ
3
,where T
0 is a material constant.Twin shear stress criterion [20] is as follows:(31)
σ
1
-
1
2
σ
2
+
σ
3
=
σ
t
,
w
h
e
n
σ
2
≤
1
2
σ
1
+
σ
3
,
1
2
σ
1
+
σ
2
-
σ
3
=
σ
t
,
w
h
e
n
σ
2
≥
1
2
σ
1
+
σ
3
,where σ
t is a material constant.A comparison of the criteria listed above and the strength criterion developed in this paper is shown in Figure5 for σ
3
=
15
M
P
a. The material constants for each criterion have been chosen so that all predict the same strength when σ
2
=
σ
3
=
15
M
P
a. The material constants are as follows: c
=
31
M
P
a, φ
=
27.37
°, m
=
2.5, α
=
0.209, k
=
38.51
M
P
a, T
0
=
8.05
M
P
a, and σ
t
=
137.58
M
P
a. Figure 5 shows that Mohr-Coulomb criterion does not depend on the intermediate principal stress, contrary to the experimental observations. Drucker-Prager criterion and 3D Griffith criterion overestimate the effect of σ
2. Both twin shear stress criterion and the strength criterion developed in this paper can well describe the intermediate principal stress effect on rock strength. However, twin shear stress criterion cannot reflect the observation [19] that the strength of a sample under extension loading (σ
2
=
σ
1) is greater than that under compression loading (σ
2
=
σ
3). On the other hand, the failure surface for twin shear stress criterion is a set of planes, forming sharp corners at the intersections. As such, the failure surface is not differentiable at the corners, which sometimes causes problems in numerical calculations involving the criterion. New strength criterion developed in this paper could overcome these problems.Figure 5
Comparison of common criteria and the strength criterion developed in this paper.


## 3.1. Comparison with Experimental Results

Comparisons between the failure probability model as described in this paper and results of the experimental results [5] are plotted in Figure 4. Fjær and Ruistuen [11] have developed a numerical method to calculate the intermediate principal stress effect on rock strength. Their results are also plotted in Figure 4 for comparison. The strength parameters can be obtained by fitting the strength envelopes. Three strength parameters are as follows: c
=
31
M
P
a, φ
=
27.37
°, and m
=
2.5. From Figure 4, both model predictions in this paper and Fjær and Ruistuen’s numerical results [11] are observed to be in good agreement with the results from the experimental results [5]. Failure probability model in this paper matches the experimental results [5] better than Fjær and Ruistuen’s numerical results [11], especially when σ
2 is low.Figure 4
Relationships betweenσ
1 and σ
2 for constant σ
3.


## 3.2. Comparison with Common Strength Criteria

Common strength criteria are Mohr-Coulomb criterion, Drucker-Prager criterion, 3D Griffith criterion, and twin shear stress criterion. Mathematical forms of these strength criteria are as follows.Mohr-Coulomb criterion is as follows:(28)
σ
1
-
σ
3
2
=
σ
1
+
σ
3
2
sin
⁡
φ
+
c
cos
⁡
φ
.Drucker-Prager criterion [18] is as follows:(29)
J
2
-
α
I
1
-
k
=
0
,where α and k are material constants, I
1
=
σ
1
+
σ
2
+
σ
3 is the first stress invariant, and J
2
=
[
σ
1
-
σ
2
2
+
σ
2
-
σ
3
2
+
σ
3
-
σ
1
2
]
/
6 is the second deviatoric stress invariant.3D Griffith criterion [19] is as follows:(30)
σ
1
-
σ
2
2
+
σ
2
-
σ
3
2
+
σ
3
-
σ
1
2
=
24
T
0
σ
1
+
σ
2
+
σ
3
,where T
0 is a material constant.Twin shear stress criterion [20] is as follows:(31)
σ
1
-
1
2
σ
2
+
σ
3
=
σ
t
,
w
h
e
n
σ
2
≤
1
2
σ
1
+
σ
3
,
1
2
σ
1
+
σ
2
-
σ
3
=
σ
t
,
w
h
e
n
σ
2
≥
1
2
σ
1
+
σ
3
,where σ
t is a material constant.A comparison of the criteria listed above and the strength criterion developed in this paper is shown in Figure5 for σ
3
=
15
M
P
a. The material constants for each criterion have been chosen so that all predict the same strength when σ
2
=
σ
3
=
15
M
P
a. The material constants are as follows: c
=
31
M
P
a, φ
=
27.37
°, m
=
2.5, α
=
0.209, k
=
38.51
M
P
a, T
0
=
8.05
M
P
a, and σ
t
=
137.58
M
P
a. Figure 5 shows that Mohr-Coulomb criterion does not depend on the intermediate principal stress, contrary to the experimental observations. Drucker-Prager criterion and 3D Griffith criterion overestimate the effect of σ
2. Both twin shear stress criterion and the strength criterion developed in this paper can well describe the intermediate principal stress effect on rock strength. However, twin shear stress criterion cannot reflect the observation [19] that the strength of a sample under extension loading (σ
2
=
σ
1) is greater than that under compression loading (σ
2
=
σ
3). On the other hand, the failure surface for twin shear stress criterion is a set of planes, forming sharp corners at the intersections. As such, the failure surface is not differentiable at the corners, which sometimes causes problems in numerical calculations involving the criterion. New strength criterion developed in this paper could overcome these problems.Figure 5
Comparison of common criteria and the strength criterion developed in this paper.


## 4. Conclusions

This paper has developed a new failure probability model to predict the effect of the intermediate principal stress on rock strength. Each shear plane in rock samples is considered as a micro-unit. The strengths of these micro-units are assumed to match Weibull distribution. Direction considerations instead of the volume considerations are made to express failure probabilities in different directions. The macro strength of rock sample is a synthetic consideration of all directions’ probabilities. New failure probability model reproduces the typical phenomenon of intermediate principal stress effect that occurs in some true triaxial experiments. Material heterogeneity plays a major role for intermediate principal stress effect. The intermediate principal stress effect becomes more prominent for more heterogeneous rocks. When materials are absolutely homogeneous, there is no intermediate principal stress effect and the failure model becomes Mohr-Coulomb criterion.Based on the new model with direction considerations, computed strength is expressed by the shear stress and the computed strength whenσ
2
=
σ
3 is calculated by Mohr-Coulomb criterion. In this way a strength criterion is developed to quantitatively describe the effect of intermediate principal stress on rock strength. When uniformity coefficient m is infinitely large, the new criterion is equivalent to Mohr-Coulomb criterion. Therefore, the proposed strength criterion can be regarded as a modified Mohr-Coulomb criterion that can reflect the effect of intermediate principal stress.By comparing with the experimental results and common strength criteria, it is found that strength criterion developed in this paper can well describe the intermediate principal stress effect on rock strength. New strength criterion matches the experimental results better than common strength criteria. In addition, new strength can avoid the nondifferentiable problem in numerical calculations involving the criterion.



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*Source: 960973-2015-11-23.xml*