# Experimental Investigation on Sandstone Permeability under Plastic Flow: Permeability Evolution Law with Stress Increment

Authors: Peng Gong; Xiaoyan Ni; Zhanqing Chen; Yu Wu; Jiangyu Wu
Journal: Geofluids
 (2019)

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## Abstract

Permeability is an important physical property of rock. In rock and rock-like materials, permeability after yielding is closely related to plastic flow behavior. Theoretical analysis and experimental investigations are effective and reliable ways of studying the evolution of permeability. In this paper, fluid flow tests of sandstone samples under 11 stress states were conducted using the MTS 816.02 rock mechanics testing system and a self-designed permeation system. A new plastic flow rule was proposed based on the Mohr-Coulomb yield criterion and a nonassociated flow rule. The applicability of the flow rule was verified by comparing the estimated and experimental values of the equivalent shear strain of the Mohr–Coulomb criterion. Two new coefficients were defined to reflect the influence of the volume deformation and shear deformation on the permeability. A permeability model for plastic flow was established, which can be used to calculate both single-step and multistep permeability estimates of the sandstone samples during plastic flow. The errors of both estimation methods were analyzed. The experimental results showed that the plastic multiplier for the sandstone samples was positive during unloading conditions and negative during loading conditions. The experimental value error of the single-step permeability estimation was less than 16.5%. The experimental value errors of the multistep permeability estimations varied between 15.6% and 16.7%, indicating that the iterative format of the multistep permeability estimation method was generally stable and highly precise. A comparison of the permeability influence coefficients indicated that the influence of the equivalent shear strain on the permeability was smaller than that of the volumetric strain.


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## Body

## 1. Introduction

Sandstone is the primary type of sedimentary rock in which rock engineering problems arise, such as water disasters in coal mining engineering. Water inrush through roof or floor sandstone aquifers has become a serious problem that occurs in some deep coal mines [1].Water inrush can happen in coal mines when roof or floor sandstone aquifers have well-connected major flow paths. Fluid flow research of sandstone roof or floor under mining-induced stress conditions is the fundamental study on water inrush. The permeability evolution of sandstone is the key to studying the fluid flow behavior under mining. In the process of coal mining, deformation on the sandstone roof and floor is dominated by plastic displacements. Therefore, investigating the permeability evolution of sandstone under plastic deformation is essential.Permeability, the ability to transmit fluid is one of the most important physical properties of sandstone [2–4]. The evolution of permeability is associated with certain rock characteristics, and numerous researchers have attempted to correlate permeability with these characteristics. Some studies noted that the permeability of rocks is closely related to their porosity characteristics [5–9].Kenyon [10] proposed a power-law relationship between permeability and porosity:
(1)k=aϕb,where ϕ is porosity, k is permeability, b is a power exponent, and a is the limit of permeability when porosity tends to zero, i.e., a=limkϕ⟶0.Pape et al. [11] identified a relationship between the porosity and permeability of Rotliegend sandstone from a large dataset of experimental test results:
(2)k=3.10×10−11ϕ+7.46×10−09ϕ2+1.91ϕ10.Currently, the permeability-porosity relationship defined by rational fractions is the most widely used [12]:
(3)k=k01−ϕ01−ϕ2ϕϕ03,where ϕ0 is the initial porosity, and k0 the initial permeability, i.e., k0=kϕ=ϕ0.Rock deformation can result in a permeability change. Some experimental studies were conducted by researchers to investigate the deformation and permeability characteristics of rocks and to try to establish a relationship between permeability and strain. Main et al. [13] presented a simple geometrical model of the permeability changes due to increasing inelastic axial strain based on laboratory experiments. The model quantitatively explained the entire permeability cycle observed in the sandstone samples. Wang and Park [14] investigated the permeability of sedimentary rocks during the complete strain-stress process in a triaxial compression test. From the experimental results, they characterized the permeability into five stages according to the axial strain.Many researchers have investigated the relationship between rock permeability and stress components [15–22].The empirical expression between permeability and the effective normal stress was obtained by Louis [23] from water-drilling test data at a dam site:
(4)k=k0exp−a2σ,where σ is the effective normal stress and a2 is the coefficient to be determined.Snow [24] conducted a set of seepage tests of parallel fractures and proposed the following relationship between permeability and stress:
(5)k=k0logσσ03,where σ0 is the initial stress.When rock is deforming plastically, the stress-loading path has a clear effect on the variation in permeability [25–28]. The primary factor influencing the variation in rock permeability under a complex loading path is not only porosity but also fracture size, density, and direction.Bear [29] and Witherspoon et al. [30] successively derived the basic equations describing flow through a fracture. The simplest model of fracture flow is the well-known cubic law [31]. Javadi et al. [32] proposed the cubic law relationship between fracture permeability and fracture characteristics.
(6)k=wd312Af,where w is the fracture width, d is the aperture of the idealized parallel smooth fracture, and Af is the cross-sectional area.Relative displacement between fracture surfaces can induce changes in the volumetric strain. Several studies were conducted to determine the relationship between permeability and volumetric strain [28, 33, 34]. The relationship between the permeability and volumetric strain of fractured rock was given as follows [35]:
(7)k=k0exp1Kp1+v1−vp−p0−2E31−vεV−εV0,where the subscript 0 indicates initial values, Kp is the pore volume modulus, p is the fluid pressure in the fracture network, E is Young’s modulus, v is Poisson’s ratio, and εV is the volumetric strain.However, a rock sample undergoing plastic deformation is subjected to not only volumetric deformation but also shear deformation. Shear strain has an important and significant effect on permeability. In this paper, we report the results of conventional triaxial compression tests and fluid flow tests on sandstone samples using an MTS 816.02 rock-mechanics test system with a self-designed permeation system. These tests were conducted to investigate the permeability evolution law under complex loading path. Our aim was to explore the influence of plastic flow on the variation in permeability, based on the Mohr-Coulomb yield criterion and a nonassociated flow rule. This study proposes a new permeability prediction model that correlates the volumetric strain and the shear strain. The model was verified by comparing the predicted and experimental values of permeability.


## 2. Theoretical Principles

The relationship between permeability increments and stress increments can be established based on the strain increment-stress increment relationship and the permeability increment-strain increment relationship. In the following text, we briefly introduce the theory of plastic flow in rocks and the permeability prediction model.


### 2.1. Elasto-Plastic Constitutive Model of Rocks under Triaxial Compression

Axial strainε1, radial strain ε2, and circumferential strain ε3 occur in a rock sample under axial stress σ1, intermediate stress σ2, and confining stress σ3. Due to axis symmetry, it holds that σ3=σ2, and ε3=ε2. In the elastic deformation, the stress and strain components of a rock sample follow the generalized Hooke’s law [36], i.e.,
(8)σ1=ΛεV+2Gε1,σ3=ΛεV+2Gε3,where εV is the volumetric strain, εV=ε1+2ε3, and Λ and G are the Lame coefficients Λ=Ev/1+v1−2v, and G=E/21+v, respectively.Shear yielding in rock obeys the Mohr-Coulomb criterion. If the compressive stress is positive, the yield condition can be expressed as equation (9) if σ1≤σ2≤σ3:
(9)σ1−σ3tan2α+2ctanα=0,where c is the cohesion of rock, α is the angle of shear failure, α=90∘+φ/2, and φ is the internal friction angle.In the event that plastic flow occurs in the rock sample after yielding, the flow paths can be represented by a directed line segment on the deviatoric plane, as shown in Figure1. In principal strain space, the flow paths can be categorized into nine cases, as shown in Figure 1(a): (1) flow from one point P10 on the yield surface to another point P11 on the yield surface (P10⟶P11), (2) flow from one point P10 on the yield surface to one point P12 outside the yield surface (P10⟶P12), (3) flow from one point P10 on the yield surface to one point P13 inside the yield surface (P10⟶P13), (4) flow from one point P20 outside the yield surface to one point P21 on the yield surface (P20⟶P21), (5) flow from one point P20 outside the yield surface to one point P22 outside the yield surface (P20⟶P22), (6) flow from one point P20 outside the yield surface to one point P23 inside the yield surface (P20⟶P23), (7) flow from one point P30 inside the yield surface to one point P31 on the yield surface (P30⟶P31), (8) flow from one point P30 inside the yield surface to one point P32 outside the yield surface (P30⟶P32), and (9) flow from one point P30 inside the yield surface to another point P33 inside the yield surface (P30⟶P33).Figure 1
Flow paths on the deviatoric plane. (a) Principal strain space and (b) principal stress space.
(a)
(b)In principal stress space, the stress state must be located on the yield surface. Therefore, there are only four cases possible, with respect to plastic flow, on the deviatoric plane in principal stress space, as shown in Figure1(b): (1) flow from one point P10 on the yield surface to another point P11 on the yield surface (P10⟶P11), (2) flow from one point P10 on the yield surface to one point P13 inside the yield surface (P10⟶P13), (3) flow from one point P30 inside the yield surface to one point P31 on the yield surface (P30⟶P31), and (4) flow from one point P30 inside the yield surface to another point P33 inside the yield surface (P30⟶P33). The dashed line in Figure 1(b) indicates impossible flow paths.Plastic potential is related to not only the stress states but also the loading history. For rock, the plastic flow rule is nonassociated, and the plastic potential function can be described as a Mohr-Coulomb relationship, wherec and φ are no longer constant but a function of the loading history. If the plastic internal variable χ is used to represent the loading history, the plastic potential function can be expressed as follows [37–40]:
(10)Fσ1,σ3,χ=σ1−σ3tan245∘+φχ2−2cχtan45∘+φχ2.Equation (10) is the plastic potential function with an isotropic hardening (softening) model, which is a group of potential functions that change in terms of cχ and φχ. χ is used to describe the plastic deformation history. The plastic volumetric strain εVp of rock can be selected as the internal variable:
(11)χ=εVp=ε1p+2ε3p,Whenχ=0, we have cχχ=0=c and φχχ=0=φ, which correspond to equation (9). The values of cχ and φχ can be obtained using a triaxial compression test.Then, the flow rule is as follows:(12)Δε1p=Δλs∂F∂σ1=Δλs,Δε3p=Δλs∂F∂σ3=−Δλstan245∘+φχ2,where Δε1p and Δε3p are the plastic strain increments and Δλs is the plastic multiplier.When the stress state induces flow on the yield surface, the plastic multiplierΔλs can be expressed as follows:
(13)Δλs=1ΛFσ1T,σ3T1−ν/ν−2tan2α−1−1/νtan2αtan2α,where
(14)σ1T=σ1+Λ+2GΔε1+2ΛΔε3,σ3T=σ3+2ΛΔε1+Λ+2GΔε3.Equation (13) shows the plastic flow rules for geotechnical materials and is commonly used in geotechnical engineering [41].When the stress state is inside the yield surface, the plastic multiplierΔλs can no longer be calculated by equation (13). As shown in Figure 2, if the flow path of stress state is from B to B′ (B⟶B′), we can divide B⟶B′ into B⟶B″ and B″⟶B′ due to small deformation. The point B″ is on the yield surface. Therefore, we propose the following plastic multiplier expression according to equation (13):
(15)Δλs=1ΛFσ1T,σ3T−Fσ1N,σ3N1−ν/ν−2tan245∘+φχ/2−1−1/νtan245∘+φχ/2tan245∘+φχ/2.Figure 2
Flow paths inside the yield surface.Equation (15) shows the modified plastic flow rules for rock proposed in this paper.


### 2.2. Permeability Model for Plastic Flow

Assuming small deformation, the rock sample porosityϕ is related to the rock volumetric strain as follows:
(16)ϕ=ϕ0+εV,where ϕ0 is the initial porosity. Porosity is commonly measured by the mercury injection test.The power-exponential relationship between the porosity and permeability of the rock samples in the post-peak region is given by [42]
(17)k=k0ϕϕ0mk,where mk is the nondimensional coefficient.Equation (18) can be written by taking the logarithm on both sides of equation (17):
(18)lnk=lnk0+mklnϕϕ0.Equation (18) can be differentiated with respect to porosity ϕ to obtain
(19)dkdϕ=mkϕk.The permeability increment is easily obtained from equation (19):
(20)Δk=mkϕkΔϕ,where Δϕ is the porosity increment.Equation (21) can be written by combining equations (16) and (19):
(21)Δk=mkϕkΔεV.Equations (16) and (21) indicate that, in theory, only the volumetric strain changes the porosity and permeability, whereas the shear strain does not. However, the experimental results presented in this paper show that the shear deformation changes the permeability. To characterize the influence of the shear deformation on permeability, an equivalent shear strain is given by the following equation, according to the Mohr-Coulomb criterion:
(22)γMC=21−2ν1−νsec2αε1+21−2ν2ν−tan2αε3.Considering the influence of shear deformation, we have modified the equation to calculate permeability:(23)Δk=kϕλ1kΔεV+λ2kΔγMC,where λ1k and λ2k reflect the influence of the volumetric strain and equivalent shear strain, respectively, on the permeability. These parameters are defined as the permeability influence coefficients and can be estimated using the Monte Carlo method.Equation (23) describes the changes in permeability due to plastic deformation and is known as the permeability model under plastic flow.


## 2.1. Elasto-Plastic Constitutive Model of Rocks under Triaxial Compression

Axial strainε1, radial strain ε2, and circumferential strain ε3 occur in a rock sample under axial stress σ1, intermediate stress σ2, and confining stress σ3. Due to axis symmetry, it holds that σ3=σ2, and ε3=ε2. In the elastic deformation, the stress and strain components of a rock sample follow the generalized Hooke’s law [36], i.e.,
(8)σ1=ΛεV+2Gε1,σ3=ΛεV+2Gε3,where εV is the volumetric strain, εV=ε1+2ε3, and Λ and G are the Lame coefficients Λ=Ev/1+v1−2v, and G=E/21+v, respectively.Shear yielding in rock obeys the Mohr-Coulomb criterion. If the compressive stress is positive, the yield condition can be expressed as equation (9) if σ1≤σ2≤σ3:
(9)σ1−σ3tan2α+2ctanα=0,where c is the cohesion of rock, α is the angle of shear failure, α=90∘+φ/2, and φ is the internal friction angle.In the event that plastic flow occurs in the rock sample after yielding, the flow paths can be represented by a directed line segment on the deviatoric plane, as shown in Figure1. In principal strain space, the flow paths can be categorized into nine cases, as shown in Figure 1(a): (1) flow from one point P10 on the yield surface to another point P11 on the yield surface (P10⟶P11), (2) flow from one point P10 on the yield surface to one point P12 outside the yield surface (P10⟶P12), (3) flow from one point P10 on the yield surface to one point P13 inside the yield surface (P10⟶P13), (4) flow from one point P20 outside the yield surface to one point P21 on the yield surface (P20⟶P21), (5) flow from one point P20 outside the yield surface to one point P22 outside the yield surface (P20⟶P22), (6) flow from one point P20 outside the yield surface to one point P23 inside the yield surface (P20⟶P23), (7) flow from one point P30 inside the yield surface to one point P31 on the yield surface (P30⟶P31), (8) flow from one point P30 inside the yield surface to one point P32 outside the yield surface (P30⟶P32), and (9) flow from one point P30 inside the yield surface to another point P33 inside the yield surface (P30⟶P33).Figure 1
Flow paths on the deviatoric plane. (a) Principal strain space and (b) principal stress space.
(a)
(b)In principal stress space, the stress state must be located on the yield surface. Therefore, there are only four cases possible, with respect to plastic flow, on the deviatoric plane in principal stress space, as shown in Figure1(b): (1) flow from one point P10 on the yield surface to another point P11 on the yield surface (P10⟶P11), (2) flow from one point P10 on the yield surface to one point P13 inside the yield surface (P10⟶P13), (3) flow from one point P30 inside the yield surface to one point P31 on the yield surface (P30⟶P31), and (4) flow from one point P30 inside the yield surface to another point P33 inside the yield surface (P30⟶P33). The dashed line in Figure 1(b) indicates impossible flow paths.Plastic potential is related to not only the stress states but also the loading history. For rock, the plastic flow rule is nonassociated, and the plastic potential function can be described as a Mohr-Coulomb relationship, wherec and φ are no longer constant but a function of the loading history. If the plastic internal variable χ is used to represent the loading history, the plastic potential function can be expressed as follows [37–40]:
(10)Fσ1,σ3,χ=σ1−σ3tan245∘+φχ2−2cχtan45∘+φχ2.Equation (10) is the plastic potential function with an isotropic hardening (softening) model, which is a group of potential functions that change in terms of cχ and φχ. χ is used to describe the plastic deformation history. The plastic volumetric strain εVp of rock can be selected as the internal variable:
(11)χ=εVp=ε1p+2ε3p,Whenχ=0, we have cχχ=0=c and φχχ=0=φ, which correspond to equation (9). The values of cχ and φχ can be obtained using a triaxial compression test.Then, the flow rule is as follows:(12)Δε1p=Δλs∂F∂σ1=Δλs,Δε3p=Δλs∂F∂σ3=−Δλstan245∘+φχ2,where Δε1p and Δε3p are the plastic strain increments and Δλs is the plastic multiplier.When the stress state induces flow on the yield surface, the plastic multiplierΔλs can be expressed as follows:
(13)Δλs=1ΛFσ1T,σ3T1−ν/ν−2tan2α−1−1/νtan2αtan2α,where
(14)σ1T=σ1+Λ+2GΔε1+2ΛΔε3,σ3T=σ3+2ΛΔε1+Λ+2GΔε3.Equation (13) shows the plastic flow rules for geotechnical materials and is commonly used in geotechnical engineering [41].When the stress state is inside the yield surface, the plastic multiplierΔλs can no longer be calculated by equation (13). As shown in Figure 2, if the flow path of stress state is from B to B′ (B⟶B′), we can divide B⟶B′ into B⟶B″ and B″⟶B′ due to small deformation. The point B″ is on the yield surface. Therefore, we propose the following plastic multiplier expression according to equation (13):
(15)Δλs=1ΛFσ1T,σ3T−Fσ1N,σ3N1−ν/ν−2tan245∘+φχ/2−1−1/νtan245∘+φχ/2tan245∘+φχ/2.Figure 2
Flow paths inside the yield surface.Equation (15) shows the modified plastic flow rules for rock proposed in this paper.


## 2.2. Permeability Model for Plastic Flow

Assuming small deformation, the rock sample porosityϕ is related to the rock volumetric strain as follows:
(16)ϕ=ϕ0+εV,where ϕ0 is the initial porosity. Porosity is commonly measured by the mercury injection test.The power-exponential relationship between the porosity and permeability of the rock samples in the post-peak region is given by [42]
(17)k=k0ϕϕ0mk,where mk is the nondimensional coefficient.Equation (18) can be written by taking the logarithm on both sides of equation (17):
(18)lnk=lnk0+mklnϕϕ0.Equation (18) can be differentiated with respect to porosity ϕ to obtain
(19)dkdϕ=mkϕk.The permeability increment is easily obtained from equation (19):
(20)Δk=mkϕkΔϕ,where Δϕ is the porosity increment.Equation (21) can be written by combining equations (16) and (19):
(21)Δk=mkϕkΔεV.Equations (16) and (21) indicate that, in theory, only the volumetric strain changes the porosity and permeability, whereas the shear strain does not. However, the experimental results presented in this paper show that the shear deformation changes the permeability. To characterize the influence of the shear deformation on permeability, an equivalent shear strain is given by the following equation, according to the Mohr-Coulomb criterion:
(22)γMC=21−2ν1−νsec2αε1+21−2ν2ν−tan2αε3.Considering the influence of shear deformation, we have modified the equation to calculate permeability:(23)Δk=kϕλ1kΔεV+λ2kΔγMC,where λ1k and λ2k reflect the influence of the volumetric strain and equivalent shear strain, respectively, on the permeability. These parameters are defined as the permeability influence coefficients and can be estimated using the Monte Carlo method.Equation (23) describes the changes in permeability due to plastic deformation and is known as the permeability model under plastic flow.


## 3. Experimental Samples, Principles, and Procedures

### 3.1. Experimental Samples

The standard cylindrical sandstone rock samples of 50 mm diameter and 100 mm length were cut from sandstone blocks that were collected from a coal mine in Shanxi Lu’an, China. To ensure the reliability of the experimental results, rock samples without obvious macroscopic cracks and joints were chosen. The two ends of the rock samples were polished to remove uneven surfaces [1]. Parts of sandstone samples are shown in Figure 3. The mineral composition of the sandstone is mainly quartz, feldspar, calcite, oxides, and mica, which is detected by the powder crystal X-ray diffraction testing machine. The initial porosity of the sandstone is 5.7%, and the order of magnitude of the initial permeability is 10-17 m2, which measured by a PoreMaster 33 mercury intrusion meter.Figure 3
Sandstone samples.The triaxial compression tests were first performed to determine the sandstone sample mechanical parameters by using the MTS815.02 mechanics testing machine. The experiments were conducted under the confining pressure of 6.0 MPa. The values of the failure load, peak strength, peak strain, and elasticity modulus are shown in Table1. The mean value and variable coefficient of peak strength are 129.9 MPa and 4.30%, respectively. The mean value and variable coefficient of peak strain are 1.27% and 3.27%, respectively. The mean value and variable coefficient of the elasticity modulus are 9.64 GPa and 5.86%, respectively. The values of variable coefficient show that the dispersion of the sandstone samples is very small.Table 1
Mechanical parameters of sandstone samples under the confining pressure of 6.0 MPa.
Serial no.
Dia. (mm)
Height (mm)
Confining pressure (MPa)
Failure load (kN)
Peak strength (MPa)
Peak strain (%)
Elasticity modulus (GPa)
1
49.8
99.7
6
246.1
125.4
1.24
9.21
2
50.1
100.2
6
244.3
124.5
1.23
9.08
3
49.7
100.1
6
267.5
136.3
1.33
10.35
4
49.6
99.4
6
250.8
127.8
1.25
9.45
5
50.1
99.7
6
265.7
135.4
1.29
10.13


### 3.2. Experimental Principles

#### 3.2.1. Triaxial Compression Test

In triaxial compression tests, the confining pressure remains constant. The sampling period in this study isτ=1s. The axial stress σ1i, the axial strain ε1i, and the circumferential strain ε3i of each rock sample are measured at the moment ti=i−1τ, where i=1,2,⋯,N.We assume that the axial stress reaches a peak valueσ1p at the moment tNpNp<N, i.e., σ1p=σ1Np. For the elastic deformation, the relationship between stress and strain is described as follows:
(24)σ1i=λεVi+2Gε1i,σ3i=λεVi+2Gε3i,where εVi is the volumetric strain of the rock samples at sampling time, and εVi=ε1i+2ε3i.The strain of the rock sample can be divided into two parts, elastic strain and plastic strain, in which the elastic strain is given as follows:(25)ε1ei=1Eσ1i−2νσ3i,i=Np+1,Np+2,⋯,N,ε3ei=1Eσ3i−2νσ1i.Therefore, the plastic strain equation can be written as follows:(26)ε1pi=ε1i−ε1ei,i=Np+1,Np+2,⋯,N,ε3pi=ε3i−ε3ei,where the plastic internal variable is χi=εVpi=ε1pi+2ε3pi.We conduct the triaxial compression tests atnc different confining pressures, where σi3i=1,2,⋯,nc. Then, nc plastic internal variable axial stress (χ‐σ1) curves can be obtained. To establish the relationship between cχ, φχ, and χ, we set nvp typical values of χjj=1,2,⋯,nvp and obtain the corresponding axial stress σij1, where i=1,2,⋯,nc,j=1,2,⋯,nvp, under σi3 and χj. Using equation (9), we can obtain the values of cjχ and φjχ under different χj j=1,2,⋯,nvp by using linear regression.


#### 3.2.2. Fluid Flow Test

In this study, we used a transient method to measure the permeability of a sandstone sample [1]. The principle of the method is shown in Figure 4.Figure 4
Principle of the transient method.The volume of storage tank 1 (ST1) is equal to the volume of storage tank 2 (ST2), and the pressures of the fluids in these tanks arep1 and p2, respectively. The sandstone samples are standard cylindrical specimens of height H and cross-sectional area S. At the start of the test, the pressure at the ends of a sandstone sample are p10 and p20 (p10>p20). During the process of the permeability tests, the fluid from ST1 flows into ST2 through the rock sample. The pressure p1 of the fluid in ST1 continuously decreases, while the pressure p2 of the fluid in tank 2 continuously increases. The pressure gradient p2−p1/H gradually decreases until the two storage tank pressures reach equilibrium, as shown in Figure 5.Figure 5
Variation curve of the storage tank pressure.If the mass flow from ST1 into the rock sample isq and the rock sample is presaturated, then the mass flow from the rock sample into ST2 is also q. The flow velocity within the rock sample is υ=q/ρS, where ρ is the density of the fluid; due to the compressibility of the fluid, the following relationship can be obtained:
(27)1cf=ρdp1dρ,where cf is the compressibility coefficient of the fluid.Sincedρ=−qdt/B and q=ρSυ, we arrive at the following expression:
(28)dp1dt=−SυcfV,where V is the volume of the storage tank.Similarly, we can write a similar expression for the other storage tank:(29)dp2dt=SυcfV,From equations (28) and (29), the expression for the flow velocity of the rock sample can be written as follows:
(30)υ=cfV2Sdp1−p2dt.According to Darcy’s law, equation (30) is rewritten as follows:
(31)υ=−kμp2−p1H,where k is the permeability of the rock specimen and μ is the dynamic viscosity of the fluid.Equation (32) can be derived by combining equations (30) and (31):
(32)dp1−p2p1−p2=−2SkcfVHμdt.In the experiments, the sampling frequency isτ, the total number of measurements is n, and tf=nτ can be calculated at the end of the sampling period. Equation (32) is integrated to obtain an equation in terms of the respective pressure values of the two storage tanks, p1f and p2f:
(33)k=cfVHμ2tfSlnp10−p20p1f−p2f.Deformation occurs when the rock sample is under considerable stress. The permeability variation is sensitive to the deformation of the rock samples. We testN stress states of the rock sample, i.e., Aiσ1i,σ2i,σ3i, corresponding to N permeability values, i.e., ki, i=1,2,⋯,N. AM is the peak stress state, and A1,A2,⋯,AM−1 are the elastic deformation states, while AM+1,AM+2,⋯,AN are the plastic flow states.The Lame coefficients,Λ and G, for determining the elastic constitutive relations can be obtained by using the least-squares method, as described below.The first step is to apply the generalized Hooke’s law to define the relationship between the stress and strain components ofA1,A2,⋯,AM−1:
(34)σ1i=ΛεVi+2Gε1i,i=1,2,⋯,M−1,σ3i=ΛεVi+2Gε3i.The second step is to construct a function based on the residual errors of equation (34):
(35)Π=∑i=14ΛεVi+2Gε1i−σ1i2+∑i=14ΛεVi+2Gε3i−σ3i2.The third step is to define a system of linear equations related toΛ and G, which is based on the functional extremum conditions ∂Π/∂Λ=0 and ∂Π/∂G=0:
(36)∑i=14ΛεViεVi+2Gε1iεVi−σ1iεVi+∑i=14ΛεViεVi+2Gε3iεVi−σ3iεVi=0,∑i=14ΛεViε1i+2Gε1iε1i−σ1iε1i+∑i=14ΛεViε3i+2Gε3iε3i−σ3iε3i=0.The fourth step is to introduce the sign convention:(37)C11C12C21C22=∑i=142εViεVi∑i=142εViε1i+ε3i∑i=14εViε1i+ε3i∑i=142ε1iε1i+ε3iε3i,B1B2=∑i=14εViσ1i+σ3i∑i=14σ1iε1i+σ3iε3i.The system of linear equations (36) of Λ and G are then transformed into a matrix:
(38)C11C12C21C22ΛG=B1B2.The last step is to obtain the optimal estimation ofΛ and G by resolving the linear equation for the two unknowns.The rock sample will yield at the pointAM where the principal stresses satisfy the Mohr-Coulomb criterion:
(39)σ1M−σ3Mtan2α+2ctanα=0.From pointAM to point AN, the rock sample is subjected to N−M flow steps. Step 1 includes the flow from AM to AM+1, written as AM⟶AM+1, and step 2 through steps N−M are written as AM+1⟶AM+2, AM+2⟶AM+3,……, and AN−1⟶AN, respectively.According to equation (12), the plastic strain increment in the flow step Ai⟶Ai+1 (i=M,M+1,⋯,N−1) is calculated as follows:
(40)Δε1pi=Δλsi,i=M+1,M+2,⋯,N,Δε3pi=−Δλsitan245∘+φχ2.The elastic strain increment is calculated as follows:(41)Δε1ei=12GΔσ1i−Λ3Λ+2GΔσ1i+2Δσ3i,i=M+1,M+2,⋯,N,Δε3ei=12GΔσ3i−Λ3Λ+2GΔσ1i+2Δσ3i.Therefore, the strain components of the rock sample under the stress stateAi+1(i=M,M+1,⋯,N−1) are as follows:
(42)ε1i=ε1i−1+Δε1ei+Δε1pi,i=M+1,M+2,⋯,N,ε3i=ε3i−1+Δε3ei+Δε3pi.Because stress stateAM is on the yield surface, the plastic multiplier in the flow step AM⟶AM+1 is calculated according to equation (13):
(43)ΔλsM+1=1ΛFσ1TM+1,σ3TM+11−ν/ν−2tan2α−1−1/νtan2αtan2α,where
(44)σ1TM+1=σ1M+Λ+2GΔε1M+1+2Λε3M+1,σ3TM+1=σ3M+2ΛΔε1M+1+Λ+2GΔε3M+1.Since the pointsAM+1,AM+2⋯⋯AN−1 are not on the yield surface, the plastic multipliers in the flow steps AM+1⟶AM+2,⋯⋯, and AN−1⟶AN are calculated according to equation (15):
(45)Δλsi=1ΛFσ1Ti,σ3Ti−Fσ1i,σ3i1−ν/ν−2tan245∘+φχ/2−1−1/νtan245∘+φχ/2tan245∘+φχ/2,i=M+2,M+3,⋯,N.Equations (43) and (45) constitute the computational format of the plastic multiplier.According to equations (40), (41), and (42), the computational format of the volumetric strain of the rock sample can be defined as follows:
(46)εVi=εVi−1+σmi−σmi−1K+1−2tan245∘+φχ2Δλsi,i=M+1,M+2,⋯,N,where σmi=σ1i+2σ3i/3 is the average principal stress.The experimental values of permeability are recorded askii=1,2,⋯,N. The values of permeability after yielding can be calculated according to equations (17) and (23), and the iterative format is as follows:
(47)kM=k0ϕMϕ0mk,k~i=ki−1+ki−1ϕi−1λ1kΔεVi+λ2kΔγMCi,i=M+1,M+2,⋯,N,where k~i is the estimated value of permeability in the stress state Ai, ΔεVi=εVi−εVi−1 is the volumetric strain increment during plastic flow Ai⟶Ai+1i=M,M+1,⋯,N−1, and ΔγMCi=γMCi−γMCi−1 is the equivalent shear strain increment during plastic flow Ai⟶Ai+1i=M,M+1,⋯,N−1.


### 3.3. Experimental System

The experimental system includes an axial loading system, a confining pressure loop, a permeation loop, and a permeameter. The main functions performed by the system are as follows: (1) applying the axial loading to the rock sample, (2) applying the confining pressure to the rock sample, and (3) applying the pore pressure (seepage pressure) to the upper and lower ends of the rock sample. The rock sample was enclosed in the permeameter (Figure6), and the MTS 816.02 rock mechanics testing system and TestStar II were used to apply an axial load to the permeameter, as shown in Figure 7. The permeameter is equipped with three nozzles: A, B, and C. Nozzle A is connected to the confining pressure loop, and nozzles B and C are connected to the permeation loop.Figure 6
Permeameter: (a) schematic diagram and (b) photograph. (1) Base plate, (2) lower steel plate, (3) porous disc, (4) cylindrical barrel, (5) upper steel plate-concave surface, (6) upper steel plate-convex surface, (7) bleeder screw for the main body, (8) bleeder screw for the needle plug, (9-10) O-type rubber sealing ring, (11) piston, (12) cover plate, (13) bolt, (14) nut, (15) plain washer, (16) spring washer, (17) screw, (18) nut, (19) axis pin, and (20-23) O-type rubber sealing rings.
(a)
(b)Figure 7
Connection between the permeameter and the MTS 816.02 rock mechanics testing system.The permeameter is composed of a base plate (#1), lower steel plate (#2), porous disc (#3), cylindrical barrel (#4), upper steel plate (#5, #6), bleeder screw (#7, #8), piston (#1), cover plate (#12), and other components, as shown in Figure6. There is a cavity between the rock sample and the cylindrical barrel, and the confining pressure system injects hydraulic oil into this cavity through nozzle A to apply the confining pressure (radial pressure) around the rock sample, as shown schematically in Figure 8. Nozzles B and C of the permeameter are connected to ST1 and ST2 in the permeation loop, respectively, and apply the pore pressure at both ends of the rock sample, as shown in Figure 9. The fluid in ST1 flows into the rock sample from nozzle B through the base plate, the lower steel plate, and the porous disc and flows into ST2 from nozzle C through another porous disc, the upper steel plate, and the piston. The cylindrical barrel is connected to the base plate and the cover plate by means of threaded fasteners (#13, #14, #15, and #16 in Figure 6). O-type rubber sealing rings are used to create seals between the cylindrical barrel and the cover plate, between the cylindrical barrel and the base plate, and between the cover plate and the piston (#20, #21, #22, and #23 in Figure 6). The upper steel plates (#5 and #6 in Figure 6) use a spherical contact structure to uniformly distribute the axial force. The air in the confining pressure cavity is evacuated via the bleeder screw (#7 and #8 in Figure 6).Figure 8
Schematic of the confining pressure system.Figure 9
Schematic of the permeation loop.In the confining pressure system (see Figure8), the valve spool of the directional control valve (DCV) is an M-type function valve. Before the fluid flow test, the spool is set to the left-hand position, and the hydraulic oil flows into the cavity via the plunger pump (PP) through both the DCV valve and the combined valve (a one-way valve and a throttle valve). At this point of the experiment, the one-way valve is not operational; the flow rate is adjusted by turning the hand wheel of the throttle valve (TV), and the confining pressure is adjusted by turning the hand wheel of the relief valve (RV). In the fluid flow test process, the confining pressure remains constant. After the test is completed, with the RV in the neutral position, the hydraulic oil flows directly to the oil tank (OT) through the PP. When the rock sample is removed, the cut-off valve (COV) is opened, and the RV will be placed in the right-hand position. At this point, the one-way valve in the combined valve is operational. The hydraulic oil in the cavity then flows back to the OT through the one-way valve and the COV.In the permeameter loop, the volume of the storage tanks is an index of the permeability measurement ranges. Before the fluid flow test, the pressure values of ST1 and ST2 are set top10 and p20. The process to set the initial storage tank pressures is as follows: turn on COV1, COV2, COV3, and COV4; start the electric test pump so that the pressure of the entire loop stabilizes to p20; turn off COV1, COV3, and COV4; pressurize the pipeline with the manual pump (MP); finally, stop the MP and turn off COV2 when the pressure of ST1 is stabilized at p10. During the test, COV1 and COV3 are turned on at the same time, and the pressure in the two tanks is recorded. After the test, COV1, COV2, COV3, and COV4 are turned on at the same time.


### 3.4. Experimental Procedure

Before the fluid flow test, the rock specimens were saturated with water. The test was divided into 11 steps, corresponding to the 11 stress statesAiσ1i,σ2i,σ3i and 11 strain states Aiε1i,ε2i,ε3i, (σ2i=σ3i, and ε2i=ε3i, i=1,2,⋯,11). The displacement controlling method is used in the test with constant strain rate of 2×10−5s−1. Each step of the fluid flow test had a sampling time of 1000 s, with a sampling rate of 1 Hz. The flow chart of the procedure for the fluid flow test is shown in Figure 10.Figure 10
Flow chart of the fluid flow test with triaxial compression.


## 3.1. Experimental Samples

The standard cylindrical sandstone rock samples of 50 mm diameter and 100 mm length were cut from sandstone blocks that were collected from a coal mine in Shanxi Lu’an, China. To ensure the reliability of the experimental results, rock samples without obvious macroscopic cracks and joints were chosen. The two ends of the rock samples were polished to remove uneven surfaces [1]. Parts of sandstone samples are shown in Figure 3. The mineral composition of the sandstone is mainly quartz, feldspar, calcite, oxides, and mica, which is detected by the powder crystal X-ray diffraction testing machine. The initial porosity of the sandstone is 5.7%, and the order of magnitude of the initial permeability is 10-17 m2, which measured by a PoreMaster 33 mercury intrusion meter.Figure 3
Sandstone samples.The triaxial compression tests were first performed to determine the sandstone sample mechanical parameters by using the MTS815.02 mechanics testing machine. The experiments were conducted under the confining pressure of 6.0 MPa. The values of the failure load, peak strength, peak strain, and elasticity modulus are shown in Table1. The mean value and variable coefficient of peak strength are 129.9 MPa and 4.30%, respectively. The mean value and variable coefficient of peak strain are 1.27% and 3.27%, respectively. The mean value and variable coefficient of the elasticity modulus are 9.64 GPa and 5.86%, respectively. The values of variable coefficient show that the dispersion of the sandstone samples is very small.Table 1
Mechanical parameters of sandstone samples under the confining pressure of 6.0 MPa.
Serial no.
Dia. (mm)
Height (mm)
Confining pressure (MPa)
Failure load (kN)
Peak strength (MPa)
Peak strain (%)
Elasticity modulus (GPa)
1
49.8
99.7
6
246.1
125.4
1.24
9.21
2
50.1
100.2
6
244.3
124.5
1.23
9.08
3
49.7
100.1
6
267.5
136.3
1.33
10.35
4
49.6
99.4
6
250.8
127.8
1.25
9.45
5
50.1
99.7
6
265.7
135.4
1.29
10.13


## 3.2. Experimental Principles

### 3.2.1. Triaxial Compression Test

In triaxial compression tests, the confining pressure remains constant. The sampling period in this study isτ=1s. The axial stress σ1i, the axial strain ε1i, and the circumferential strain ε3i of each rock sample are measured at the moment ti=i−1τ, where i=1,2,⋯,N.We assume that the axial stress reaches a peak valueσ1p at the moment tNpNp<N, i.e., σ1p=σ1Np. For the elastic deformation, the relationship between stress and strain is described as follows:
(24)σ1i=λεVi+2Gε1i,σ3i=λεVi+2Gε3i,where εVi is the volumetric strain of the rock samples at sampling time, and εVi=ε1i+2ε3i.The strain of the rock sample can be divided into two parts, elastic strain and plastic strain, in which the elastic strain is given as follows:(25)ε1ei=1Eσ1i−2νσ3i,i=Np+1,Np+2,⋯,N,ε3ei=1Eσ3i−2νσ1i.Therefore, the plastic strain equation can be written as follows:(26)ε1pi=ε1i−ε1ei,i=Np+1,Np+2,⋯,N,ε3pi=ε3i−ε3ei,where the plastic internal variable is χi=εVpi=ε1pi+2ε3pi.We conduct the triaxial compression tests atnc different confining pressures, where σi3i=1,2,⋯,nc. Then, nc plastic internal variable axial stress (χ‐σ1) curves can be obtained. To establish the relationship between cχ, φχ, and χ, we set nvp typical values of χjj=1,2,⋯,nvp and obtain the corresponding axial stress σij1, where i=1,2,⋯,nc,j=1,2,⋯,nvp, under σi3 and χj. Using equation (9), we can obtain the values of cjχ and φjχ under different χj j=1,2,⋯,nvp by using linear regression.


### 3.2.2. Fluid Flow Test

In this study, we used a transient method to measure the permeability of a sandstone sample [1]. The principle of the method is shown in Figure 4.Figure 4
Principle of the transient method.The volume of storage tank 1 (ST1) is equal to the volume of storage tank 2 (ST2), and the pressures of the fluids in these tanks arep1 and p2, respectively. The sandstone samples are standard cylindrical specimens of height H and cross-sectional area S. At the start of the test, the pressure at the ends of a sandstone sample are p10 and p20 (p10>p20). During the process of the permeability tests, the fluid from ST1 flows into ST2 through the rock sample. The pressure p1 of the fluid in ST1 continuously decreases, while the pressure p2 of the fluid in tank 2 continuously increases. The pressure gradient p2−p1/H gradually decreases until the two storage tank pressures reach equilibrium, as shown in Figure 5.Figure 5
Variation curve of the storage tank pressure.If the mass flow from ST1 into the rock sample isq and the rock sample is presaturated, then the mass flow from the rock sample into ST2 is also q. The flow velocity within the rock sample is υ=q/ρS, where ρ is the density of the fluid; due to the compressibility of the fluid, the following relationship can be obtained:
(27)1cf=ρdp1dρ,where cf is the compressibility coefficient of the fluid.Sincedρ=−qdt/B and q=ρSυ, we arrive at the following expression:
(28)dp1dt=−SυcfV,where V is the volume of the storage tank.Similarly, we can write a similar expression for the other storage tank:(29)dp2dt=SυcfV,From equations (28) and (29), the expression for the flow velocity of the rock sample can be written as follows:
(30)υ=cfV2Sdp1−p2dt.According to Darcy’s law, equation (30) is rewritten as follows:
(31)υ=−kμp2−p1H,where k is the permeability of the rock specimen and μ is the dynamic viscosity of the fluid.Equation (32) can be derived by combining equations (30) and (31):
(32)dp1−p2p1−p2=−2SkcfVHμdt.In the experiments, the sampling frequency isτ, the total number of measurements is n, and tf=nτ can be calculated at the end of the sampling period. Equation (32) is integrated to obtain an equation in terms of the respective pressure values of the two storage tanks, p1f and p2f:
(33)k=cfVHμ2tfSlnp10−p20p1f−p2f.Deformation occurs when the rock sample is under considerable stress. The permeability variation is sensitive to the deformation of the rock samples. We testN stress states of the rock sample, i.e., Aiσ1i,σ2i,σ3i, corresponding to N permeability values, i.e., ki, i=1,2,⋯,N. AM is the peak stress state, and A1,A2,⋯,AM−1 are the elastic deformation states, while AM+1,AM+2,⋯,AN are the plastic flow states.The Lame coefficients,Λ and G, for determining the elastic constitutive relations can be obtained by using the least-squares method, as described below.The first step is to apply the generalized Hooke’s law to define the relationship between the stress and strain components ofA1,A2,⋯,AM−1:
(34)σ1i=ΛεVi+2Gε1i,i=1,2,⋯,M−1,σ3i=ΛεVi+2Gε3i.The second step is to construct a function based on the residual errors of equation (34):
(35)Π=∑i=14ΛεVi+2Gε1i−σ1i2+∑i=14ΛεVi+2Gε3i−σ3i2.The third step is to define a system of linear equations related toΛ and G, which is based on the functional extremum conditions ∂Π/∂Λ=0 and ∂Π/∂G=0:
(36)∑i=14ΛεViεVi+2Gε1iεVi−σ1iεVi+∑i=14ΛεViεVi+2Gε3iεVi−σ3iεVi=0,∑i=14ΛεViε1i+2Gε1iε1i−σ1iε1i+∑i=14ΛεViε3i+2Gε3iε3i−σ3iε3i=0.The fourth step is to introduce the sign convention:(37)C11C12C21C22=∑i=142εViεVi∑i=142εViε1i+ε3i∑i=14εViε1i+ε3i∑i=142ε1iε1i+ε3iε3i,B1B2=∑i=14εViσ1i+σ3i∑i=14σ1iε1i+σ3iε3i.The system of linear equations (36) of Λ and G are then transformed into a matrix:
(38)C11C12C21C22ΛG=B1B2.The last step is to obtain the optimal estimation ofΛ and G by resolving the linear equation for the two unknowns.The rock sample will yield at the pointAM where the principal stresses satisfy the Mohr-Coulomb criterion:
(39)σ1M−σ3Mtan2α+2ctanα=0.From pointAM to point AN, the rock sample is subjected to N−M flow steps. Step 1 includes the flow from AM to AM+1, written as AM⟶AM+1, and step 2 through steps N−M are written as AM+1⟶AM+2, AM+2⟶AM+3,……, and AN−1⟶AN, respectively.According to equation (12), the plastic strain increment in the flow step Ai⟶Ai+1 (i=M,M+1,⋯,N−1) is calculated as follows:
(40)Δε1pi=Δλsi,i=M+1,M+2,⋯,N,Δε3pi=−Δλsitan245∘+φχ2.The elastic strain increment is calculated as follows:(41)Δε1ei=12GΔσ1i−Λ3Λ+2GΔσ1i+2Δσ3i,i=M+1,M+2,⋯,N,Δε3ei=12GΔσ3i−Λ3Λ+2GΔσ1i+2Δσ3i.Therefore, the strain components of the rock sample under the stress stateAi+1(i=M,M+1,⋯,N−1) are as follows:
(42)ε1i=ε1i−1+Δε1ei+Δε1pi,i=M+1,M+2,⋯,N,ε3i=ε3i−1+Δε3ei+Δε3pi.Because stress stateAM is on the yield surface, the plastic multiplier in the flow step AM⟶AM+1 is calculated according to equation (13):
(43)ΔλsM+1=1ΛFσ1TM+1,σ3TM+11−ν/ν−2tan2α−1−1/νtan2αtan2α,where
(44)σ1TM+1=σ1M+Λ+2GΔε1M+1+2Λε3M+1,σ3TM+1=σ3M+2ΛΔε1M+1+Λ+2GΔε3M+1.Since the pointsAM+1,AM+2⋯⋯AN−1 are not on the yield surface, the plastic multipliers in the flow steps AM+1⟶AM+2,⋯⋯, and AN−1⟶AN are calculated according to equation (15):
(45)Δλsi=1ΛFσ1Ti,σ3Ti−Fσ1i,σ3i1−ν/ν−2tan245∘+φχ/2−1−1/νtan245∘+φχ/2tan245∘+φχ/2,i=M+2,M+3,⋯,N.Equations (43) and (45) constitute the computational format of the plastic multiplier.According to equations (40), (41), and (42), the computational format of the volumetric strain of the rock sample can be defined as follows:
(46)εVi=εVi−1+σmi−σmi−1K+1−2tan245∘+φχ2Δλsi,i=M+1,M+2,⋯,N,where σmi=σ1i+2σ3i/3 is the average principal stress.The experimental values of permeability are recorded askii=1,2,⋯,N. The values of permeability after yielding can be calculated according to equations (17) and (23), and the iterative format is as follows:
(47)kM=k0ϕMϕ0mk,k~i=ki−1+ki−1ϕi−1λ1kΔεVi+λ2kΔγMCi,i=M+1,M+2,⋯,N,where k~i is the estimated value of permeability in the stress state Ai, ΔεVi=εVi−εVi−1 is the volumetric strain increment during plastic flow Ai⟶Ai+1i=M,M+1,⋯,N−1, and ΔγMCi=γMCi−γMCi−1 is the equivalent shear strain increment during plastic flow Ai⟶Ai+1i=M,M+1,⋯,N−1.


## 3.2.1. Triaxial Compression Test

In triaxial compression tests, the confining pressure remains constant. The sampling period in this study isτ=1s. The axial stress σ1i, the axial strain ε1i, and the circumferential strain ε3i of each rock sample are measured at the moment ti=i−1τ, where i=1,2,⋯,N.We assume that the axial stress reaches a peak valueσ1p at the moment tNpNp<N, i.e., σ1p=σ1Np. For the elastic deformation, the relationship between stress and strain is described as follows:
(24)σ1i=λεVi+2Gε1i,σ3i=λεVi+2Gε3i,where εVi is the volumetric strain of the rock samples at sampling time, and εVi=ε1i+2ε3i.The strain of the rock sample can be divided into two parts, elastic strain and plastic strain, in which the elastic strain is given as follows:(25)ε1ei=1Eσ1i−2νσ3i,i=Np+1,Np+2,⋯,N,ε3ei=1Eσ3i−2νσ1i.Therefore, the plastic strain equation can be written as follows:(26)ε1pi=ε1i−ε1ei,i=Np+1,Np+2,⋯,N,ε3pi=ε3i−ε3ei,where the plastic internal variable is χi=εVpi=ε1pi+2ε3pi.We conduct the triaxial compression tests atnc different confining pressures, where σi3i=1,2,⋯,nc. Then, nc plastic internal variable axial stress (χ‐σ1) curves can be obtained. To establish the relationship between cχ, φχ, and χ, we set nvp typical values of χjj=1,2,⋯,nvp and obtain the corresponding axial stress σij1, where i=1,2,⋯,nc,j=1,2,⋯,nvp, under σi3 and χj. Using equation (9), we can obtain the values of cjχ and φjχ under different χj j=1,2,⋯,nvp by using linear regression.


## 3.2.2. Fluid Flow Test

In this study, we used a transient method to measure the permeability of a sandstone sample [1]. The principle of the method is shown in Figure 4.Figure 4
Principle of the transient method.The volume of storage tank 1 (ST1) is equal to the volume of storage tank 2 (ST2), and the pressures of the fluids in these tanks arep1 and p2, respectively. The sandstone samples are standard cylindrical specimens of height H and cross-sectional area S. At the start of the test, the pressure at the ends of a sandstone sample are p10 and p20 (p10>p20). During the process of the permeability tests, the fluid from ST1 flows into ST2 through the rock sample. The pressure p1 of the fluid in ST1 continuously decreases, while the pressure p2 of the fluid in tank 2 continuously increases. The pressure gradient p2−p1/H gradually decreases until the two storage tank pressures reach equilibrium, as shown in Figure 5.Figure 5
Variation curve of the storage tank pressure.If the mass flow from ST1 into the rock sample isq and the rock sample is presaturated, then the mass flow from the rock sample into ST2 is also q. The flow velocity within the rock sample is υ=q/ρS, where ρ is the density of the fluid; due to the compressibility of the fluid, the following relationship can be obtained:
(27)1cf=ρdp1dρ,where cf is the compressibility coefficient of the fluid.Sincedρ=−qdt/B and q=ρSυ, we arrive at the following expression:
(28)dp1dt=−SυcfV,where V is the volume of the storage tank.Similarly, we can write a similar expression for the other storage tank:(29)dp2dt=SυcfV,From equations (28) and (29), the expression for the flow velocity of the rock sample can be written as follows:
(30)υ=cfV2Sdp1−p2dt.According to Darcy’s law, equation (30) is rewritten as follows:
(31)υ=−kμp2−p1H,where k is the permeability of the rock specimen and μ is the dynamic viscosity of the fluid.Equation (32) can be derived by combining equations (30) and (31):
(32)dp1−p2p1−p2=−2SkcfVHμdt.In the experiments, the sampling frequency isτ, the total number of measurements is n, and tf=nτ can be calculated at the end of the sampling period. Equation (32) is integrated to obtain an equation in terms of the respective pressure values of the two storage tanks, p1f and p2f:
(33)k=cfVHμ2tfSlnp10−p20p1f−p2f.Deformation occurs when the rock sample is under considerable stress. The permeability variation is sensitive to the deformation of the rock samples. We testN stress states of the rock sample, i.e., Aiσ1i,σ2i,σ3i, corresponding to N permeability values, i.e., ki, i=1,2,⋯,N. AM is the peak stress state, and A1,A2,⋯,AM−1 are the elastic deformation states, while AM+1,AM+2,⋯,AN are the plastic flow states.The Lame coefficients,Λ and G, for determining the elastic constitutive relations can be obtained by using the least-squares method, as described below.The first step is to apply the generalized Hooke’s law to define the relationship between the stress and strain components ofA1,A2,⋯,AM−1:
(34)σ1i=ΛεVi+2Gε1i,i=1,2,⋯,M−1,σ3i=ΛεVi+2Gε3i.The second step is to construct a function based on the residual errors of equation (34):
(35)Π=∑i=14ΛεVi+2Gε1i−σ1i2+∑i=14ΛεVi+2Gε3i−σ3i2.The third step is to define a system of linear equations related toΛ and G, which is based on the functional extremum conditions ∂Π/∂Λ=0 and ∂Π/∂G=0:
(36)∑i=14ΛεViεVi+2Gε1iεVi−σ1iεVi+∑i=14ΛεViεVi+2Gε3iεVi−σ3iεVi=0,∑i=14ΛεViε1i+2Gε1iε1i−σ1iε1i+∑i=14ΛεViε3i+2Gε3iε3i−σ3iε3i=0.The fourth step is to introduce the sign convention:(37)C11C12C21C22=∑i=142εViεVi∑i=142εViε1i+ε3i∑i=14εViε1i+ε3i∑i=142ε1iε1i+ε3iε3i,B1B2=∑i=14εViσ1i+σ3i∑i=14σ1iε1i+σ3iε3i.The system of linear equations (36) of Λ and G are then transformed into a matrix:
(38)C11C12C21C22ΛG=B1B2.The last step is to obtain the optimal estimation ofΛ and G by resolving the linear equation for the two unknowns.The rock sample will yield at the pointAM where the principal stresses satisfy the Mohr-Coulomb criterion:
(39)σ1M−σ3Mtan2α+2ctanα=0.From pointAM to point AN, the rock sample is subjected to N−M flow steps. Step 1 includes the flow from AM to AM+1, written as AM⟶AM+1, and step 2 through steps N−M are written as AM+1⟶AM+2, AM+2⟶AM+3,……, and AN−1⟶AN, respectively.According to equation (12), the plastic strain increment in the flow step Ai⟶Ai+1 (i=M,M+1,⋯,N−1) is calculated as follows:
(40)Δε1pi=Δλsi,i=M+1,M+2,⋯,N,Δε3pi=−Δλsitan245∘+φχ2.The elastic strain increment is calculated as follows:(41)Δε1ei=12GΔσ1i−Λ3Λ+2GΔσ1i+2Δσ3i,i=M+1,M+2,⋯,N,Δε3ei=12GΔσ3i−Λ3Λ+2GΔσ1i+2Δσ3i.Therefore, the strain components of the rock sample under the stress stateAi+1(i=M,M+1,⋯,N−1) are as follows:
(42)ε1i=ε1i−1+Δε1ei+Δε1pi,i=M+1,M+2,⋯,N,ε3i=ε3i−1+Δε3ei+Δε3pi.Because stress stateAM is on the yield surface, the plastic multiplier in the flow step AM⟶AM+1 is calculated according to equation (13):
(43)ΔλsM+1=1ΛFσ1TM+1,σ3TM+11−ν/ν−2tan2α−1−1/νtan2αtan2α,where
(44)σ1TM+1=σ1M+Λ+2GΔε1M+1+2Λε3M+1,σ3TM+1=σ3M+2ΛΔε1M+1+Λ+2GΔε3M+1.Since the pointsAM+1,AM+2⋯⋯AN−1 are not on the yield surface, the plastic multipliers in the flow steps AM+1⟶AM+2,⋯⋯, and AN−1⟶AN are calculated according to equation (15):
(45)Δλsi=1ΛFσ1Ti,σ3Ti−Fσ1i,σ3i1−ν/ν−2tan245∘+φχ/2−1−1/νtan245∘+φχ/2tan245∘+φχ/2,i=M+2,M+3,⋯,N.Equations (43) and (45) constitute the computational format of the plastic multiplier.According to equations (40), (41), and (42), the computational format of the volumetric strain of the rock sample can be defined as follows:
(46)εVi=εVi−1+σmi−σmi−1K+1−2tan245∘+φχ2Δλsi,i=M+1,M+2,⋯,N,where σmi=σ1i+2σ3i/3 is the average principal stress.The experimental values of permeability are recorded askii=1,2,⋯,N. The values of permeability after yielding can be calculated according to equations (17) and (23), and the iterative format is as follows:
(47)kM=k0ϕMϕ0mk,k~i=ki−1+ki−1ϕi−1λ1kΔεVi+λ2kΔγMCi,i=M+1,M+2,⋯,N,where k~i is the estimated value of permeability in the stress state Ai, ΔεVi=εVi−εVi−1 is the volumetric strain increment during plastic flow Ai⟶Ai+1i=M,M+1,⋯,N−1, and ΔγMCi=γMCi−γMCi−1 is the equivalent shear strain increment during plastic flow Ai⟶Ai+1i=M,M+1,⋯,N−1.


## 3.3. Experimental System

The experimental system includes an axial loading system, a confining pressure loop, a permeation loop, and a permeameter. The main functions performed by the system are as follows: (1) applying the axial loading to the rock sample, (2) applying the confining pressure to the rock sample, and (3) applying the pore pressure (seepage pressure) to the upper and lower ends of the rock sample. The rock sample was enclosed in the permeameter (Figure6), and the MTS 816.02 rock mechanics testing system and TestStar II were used to apply an axial load to the permeameter, as shown in Figure 7. The permeameter is equipped with three nozzles: A, B, and C. Nozzle A is connected to the confining pressure loop, and nozzles B and C are connected to the permeation loop.Figure 6
Permeameter: (a) schematic diagram and (b) photograph. (1) Base plate, (2) lower steel plate, (3) porous disc, (4) cylindrical barrel, (5) upper steel plate-concave surface, (6) upper steel plate-convex surface, (7) bleeder screw for the main body, (8) bleeder screw for the needle plug, (9-10) O-type rubber sealing ring, (11) piston, (12) cover plate, (13) bolt, (14) nut, (15) plain washer, (16) spring washer, (17) screw, (18) nut, (19) axis pin, and (20-23) O-type rubber sealing rings.
(a)
(b)Figure 7
Connection between the permeameter and the MTS 816.02 rock mechanics testing system.The permeameter is composed of a base plate (#1), lower steel plate (#2), porous disc (#3), cylindrical barrel (#4), upper steel plate (#5, #6), bleeder screw (#7, #8), piston (#1), cover plate (#12), and other components, as shown in Figure6. There is a cavity between the rock sample and the cylindrical barrel, and the confining pressure system injects hydraulic oil into this cavity through nozzle A to apply the confining pressure (radial pressure) around the rock sample, as shown schematically in Figure 8. Nozzles B and C of the permeameter are connected to ST1 and ST2 in the permeation loop, respectively, and apply the pore pressure at both ends of the rock sample, as shown in Figure 9. The fluid in ST1 flows into the rock sample from nozzle B through the base plate, the lower steel plate, and the porous disc and flows into ST2 from nozzle C through another porous disc, the upper steel plate, and the piston. The cylindrical barrel is connected to the base plate and the cover plate by means of threaded fasteners (#13, #14, #15, and #16 in Figure 6). O-type rubber sealing rings are used to create seals between the cylindrical barrel and the cover plate, between the cylindrical barrel and the base plate, and between the cover plate and the piston (#20, #21, #22, and #23 in Figure 6). The upper steel plates (#5 and #6 in Figure 6) use a spherical contact structure to uniformly distribute the axial force. The air in the confining pressure cavity is evacuated via the bleeder screw (#7 and #8 in Figure 6).Figure 8
Schematic of the confining pressure system.Figure 9
Schematic of the permeation loop.In the confining pressure system (see Figure8), the valve spool of the directional control valve (DCV) is an M-type function valve. Before the fluid flow test, the spool is set to the left-hand position, and the hydraulic oil flows into the cavity via the plunger pump (PP) through both the DCV valve and the combined valve (a one-way valve and a throttle valve). At this point of the experiment, the one-way valve is not operational; the flow rate is adjusted by turning the hand wheel of the throttle valve (TV), and the confining pressure is adjusted by turning the hand wheel of the relief valve (RV). In the fluid flow test process, the confining pressure remains constant. After the test is completed, with the RV in the neutral position, the hydraulic oil flows directly to the oil tank (OT) through the PP. When the rock sample is removed, the cut-off valve (COV) is opened, and the RV will be placed in the right-hand position. At this point, the one-way valve in the combined valve is operational. The hydraulic oil in the cavity then flows back to the OT through the one-way valve and the COV.In the permeameter loop, the volume of the storage tanks is an index of the permeability measurement ranges. Before the fluid flow test, the pressure values of ST1 and ST2 are set top10 and p20. The process to set the initial storage tank pressures is as follows: turn on COV1, COV2, COV3, and COV4; start the electric test pump so that the pressure of the entire loop stabilizes to p20; turn off COV1, COV3, and COV4; pressurize the pipeline with the manual pump (MP); finally, stop the MP and turn off COV2 when the pressure of ST1 is stabilized at p10. During the test, COV1 and COV3 are turned on at the same time, and the pressure in the two tanks is recorded. After the test, COV1, COV2, COV3, and COV4 are turned on at the same time.


## 3.4. Experimental Procedure

Before the fluid flow test, the rock specimens were saturated with water. The test was divided into 11 steps, corresponding to the 11 stress statesAiσ1i,σ2i,σ3i and 11 strain states Aiε1i,ε2i,ε3i, (σ2i=σ3i, and ε2i=ε3i, i=1,2,⋯,11). The displacement controlling method is used in the test with constant strain rate of 2×10−5s−1. Each step of the fluid flow test had a sampling time of 1000 s, with a sampling rate of 1 Hz. The flow chart of the procedure for the fluid flow test is shown in Figure 10.Figure 10
Flow chart of the fluid flow test with triaxial compression.


## 4. Experimental Results

### 4.1. Triaxial Compression Test Results

We conducted a series of conventional triaxial compression tests under confining pressures of 2 MPa, 4 MPa, 6 MPa, 8 MPa, and 10 MPa. The test results are presented in Table2.Table 2
Triaxial compressive strength of the sandstone samples.
No.
Serial no.
Dia. (mm)
Height (mm)
Confining pressure (MPa)
Failure load (kN)
Compressive strength (MPa)
1
R-TS1
50.1
99.9
2.0
174.07
88.7
2
R-TS2
50.3
99.3
171.52
87.4
3
R-TS3
49.9
100.1
177.21
90.3
4
R-TS4
49.6
99.6
4.0
233.93
119.2
5
R-TS5
49.5
99.4
230.40
117.4
6
R-TS6
50.1
100.3
236.48
120.5
7
R-TS7
49.7
99.3
6.0
268.67
136.9
8
R-TS8
49.2
100.4
262.39
133.7
9
R-TS9
50.1
99.2
256.30
130.6
10
R-TS10
49.7
99.6
8.0
281.03
143.2
11
R-TS11
50.2
100.2
288.68
147.1
12
R-TS12
49.6
99.4
283.58
144.5
13
R-TS13
50.4
100.2
10.0
299.48
152.6
14
R-TS14
50.3
99.8
315.37
160.7
15
R-TS15
50.1
100.4
304.97
155.4The plastic volumetric strain of rock samples R-TS1, R-TS4, R-TS8, R-TS12, and R-TS15 was calculated according to the method described in Section3.1. Then, we obtained 5 plastic internal variable versus axial stress curves, one for each of the confining pressures tested, as shown in Figure 11.Figure 11
Plastic internal variable versus axial stress curves under various confining pressures.We calculated the values ofcχ and φχ corresponding to the plastic volume strains of 0.0, 0.001, 0.002, 0.003, 0.004, and 0.005 by using equation (10), and the results are presented in Figure 12 and Table 3.Figure 12
The change inc and φ with χ.Table 3
The calculated results ofcχ and φχ with varying χ.
χ
cχ
φχ
0
11.2
54.3
0.001
12.2
51.1
0.002
8.66
53.3
0.003
5.53
54.3
0.004
2.98
54.9
0.005
1.29
54.3


### 4.2. Fluid Flow Test Results and Analysis

The fluid flow tests were carried out underN=11 stress states, specifically Aii=1,2,⋯,11. Each of the rock samples had a diameter of 49.6 mm and a height of 99.2 mm. Water was used as the flow liquid; the water had a mass density of ρ=1000kg/m3, dynamic viscosity of μ=1.002×10−3Pa⋅s at 20°C, and a compressibility coefficient of cf=0.458×10−9Pa−1.At the start of the fluid flow test, the pressures at the ends of the sandstone sample arep10=5MPa and p20=3MPa; therefore, the pore pressure difference is 2 MPa. In the pre-peak strength state, the confining pressure is set as 6 MPa. To investigate the influence of plastic flow on the permeability of the sandstone, the axial stress and confining pressure of the rock samples were both changed after the seventh step of the permeability test.The axial stress-strain curve of the sandstone sample is shown in Figure13. As seen in Figure 13, the axial stress reaches a peak value at point A5. At that point, the rock sample is considered to be deforming via shear yielding. Therefore, points A1,⋯,A4 describe the pre-peak strength state, and points A6,⋯,A11 describe the plastic flow state. The rock sample went through a 6-step plastic flow, represented by points A5−A11 in Figure 13. The values of the axial stress σ1, confining pressure σ3, axial strain ε1, and circumferential strain ε3 measured during the test are presented in Table 4. The sign convention of ε1, ε3, and εV is that ε1 and ε3 in compression are positive and εV in dilation is positive.Figure 13
Relationship obtained between the axial stress and strain of the sandstone sample.Table 4
Axial stress, confining pressure, axial strain, and circumferential strain.
State
σ1 (MPa)
σ3 (MPa)
ε1 (%)
ε3 (%)
State
A1
10.75
6
0.051
0.022
Pre-peak
A2
37.78
6
0.434
0.033
Pre-peak
A3
56.29
6
0.625
0.040
Pre-peak
A4
76.68
6
0.816
0.100
Pre-peak
A5
97.43
6
1.008
-0.189
Shear yielding
A6
32.92
6
1.372
-1.421
Plastic flow
A7
24.95
7.25
1.324
-1.353
Plastic flow
A8
28.19
5
1.419
-1.300
Plastic flow
A9
18.87
5.66
1.324
-1.307
Plastic flow
A10
23.51
5.9
1.372
-1.332
Plastic flow
A11
10.7
5.9
1.132
-1.123
Plastic flowForAii=1,⋯,4, in the elastic deformation state, the values of the Lame coefficients Λ and G of the sandstone sample were obtained according to the method described from equation (34) to equation (38), where
(48)C11C12C21C22=2.42×10−42.85×10−41.42×10−44.62×10−4andB1B2=1.74×1062.13×106.Then, the values of the Lame coefficients areΛ=2.75GPa and G=3.76GPa.A total of 6 plastic flow steps occurred after yielding of the rock sample,Ai⟶Ai+1 (i=5,⋯,10). The values of the stress increments (Δσ1 and Δσ3), strain increments (Δε1 and Δε3), plastic potential increment (ΔF=∂F/∂σ1Δσ1+∂F/∂σ3Δσ3), and plastic multiplier Δλsii=6,⋯11 in each flow step are presented in Table 5.Table 5
Stress increment and strain increment during plastic flow of the rock sample.
Step
Δσ1 (MPa)
Δσ3 (MPa)
Δε1 (%)
Δε3 (%)
Δλs
ΔF
State
A5⟶A6
-64.5
0
0.364
-1.23
6.79E−3
-64.50
Unloading
A6⟶A7
-7.97
1.25
-0.048
0.068
9.02E−4
-15.63
Unloading
A7⟶A8
3.24
-2.25
0.095
0.053
−4.55E−4
17.03
Loading
A8⟶A9
-9.32
0.66
-0.095
-0.007
1.01E−3
-13.37
Unloading
A9⟶A10
4.64
0.24
0.048
-0.025
−4.75E−4
3.17
Loading
A10⟶A11
-12.8
0
-0.24
0.209
1.35E−3
-12.80
UnloadingAs shown in Table5, the flow steps A5⟶A6, A6⟶A7, A8⟶A9, and A10⟶A11 show the unloading process, and A7⟶A8 and A9⟶A10 show the loading process. The plastic multiplier lies between −4.7×10−4 and 6.79×10−3, and it is positive during the unloading steps (ΔF<0) and negative during the loading steps.The volumetric strain, porosity, equivalent shear strain, and permeability of each stress state after the sample yieldedAii=6,7,⋯,11 are presented in Table 6.Table 6
Volumetric strain, porosity, equivalent shear strain, and permeability of each stress state.
State
εV (%)
ϕ (%)
γMC (%)
km−2
A6
1.47
7.50
6.03
9.28×10−17
A7
1.38
7.23
5.77
9.00×10−17
A8
1.18
6.88
5.83
8.37×10−17
A9
1.29
6.99
5.66
8.71×10−17
A10
1.29
7.20
5.81
8.72×10−17
A11
1.11
6.81
4.86
8.17×10−17As seen in Table7, the volumetric strain increments during the first unloading process A5⟶A6 and A8⟶A9 are positive, and the volumetric strain increments during the later unloading processes A6⟶A7 and A10⟶A11 are negative. In addition, the volumetric strain increment during the loading process A7⟶A8 is negative, and the volumetric strain increment during the loading process A9⟶A10 is zero. These changes in the sign of the volumetric strain increment indicate that there is no one-to-one relationship between the volumetric strain/variation and either loading or unloading. Similarly, there is also no one-to-one relationship between the equivalent shear strain variation and either loading or unloading.Table 7
Strain increment and permeability increment during plastic flow of the rock sample.
Step
ΔεV (%)
ΔγMC (%)
Δkm−2
State
A5⟶A6
2.10
3.66
5.38×10−17
Unloading
A6⟶A7
-0.27
-0.25
−2.7×10−18
Unloading
A7⟶A8
-0.20
0.05
−6.28×10−18
Loading
A8⟶A9
0.11
-0.16
3.38×10−18
Unloading
A9⟶A10
0.00
0.15
6.00×10−20
Loading
A10⟶A11
-0.18
-0.96
−5.49×10−18
UnloadingShear deformation occurred in the rock sample during the loading steps ofA9⟶A10, without volume deformation. A permeability increment magnitude on the order of 1×10−20 was observed at this step, and a permeability increment magnitude on the order of 1×10−17 to 1×10−18 was observed during the other steps. Therefore, the influence of the shear strain on permeability is less than that of the volumetric strain.The region of search forλ1k was 2,3, and the region of search for λ2k was 0.01,0.5. By using the Monte Carlo method, the optimal estimated values of the permeability influence coefficients were obtained as λ1k=2.55 and λ2k=0.102. We found that λ2k<λ1k, which proves that the influence of the shear strain on permeability is less than that of the volumetric strain.From the fluid flow test results and the calculated values ofλ1k and λ2k, it can be concluded that both shear strain and volumetric strain have an obvious effect on permeability. These two results both verify that the volumetric strain has a major effect on permeability.The estimated values of permeability under every stress state after shear yielding, calculated by using equation (47), are presented in Figure 14 and Table 8.Figure 14
Single-step estimation of the permeability after shear yielding.Table 8
Experimental values, estimated values, and relative errors of the permeability.
State
Permeability (m-2)
Experimental values
Single-step estimated values
Relative error (%)
A5
3.90×10−17
3.90×10−17
—
A6
9.28×10−17
7.75×10−17
16.5%
A7
9.00×10−17
9.03×10−17
0.28%
A8
8.37×10−17
8.34×10−17
0.41%
A9
8.71×10−17
8.73×10−17
0.26%
A10
8.72×10−17
8.70×10−17
0.22%
A11
8.17×10−17
8.27×10−17
1.20%Table8 shows that the error between the estimated values and the experimental values in the stress state A6 is 16.5%, which is because the axial stress decreases sharply as soon as the rock sample reaches the peak stress state. The errors in the stress states A7,⋯,A11 were between 0.22% and 1.20%. These small errors indicate that the iterative format for estimating permeability, represented by equation (47), is generally stable.


## 4.1. Triaxial Compression Test Results

We conducted a series of conventional triaxial compression tests under confining pressures of 2 MPa, 4 MPa, 6 MPa, 8 MPa, and 10 MPa. The test results are presented in Table2.Table 2
Triaxial compressive strength of the sandstone samples.
No.
Serial no.
Dia. (mm)
Height (mm)
Confining pressure (MPa)
Failure load (kN)
Compressive strength (MPa)
1
R-TS1
50.1
99.9
2.0
174.07
88.7
2
R-TS2
50.3
99.3
171.52
87.4
3
R-TS3
49.9
100.1
177.21
90.3
4
R-TS4
49.6
99.6
4.0
233.93
119.2
5
R-TS5
49.5
99.4
230.40
117.4
6
R-TS6
50.1
100.3
236.48
120.5
7
R-TS7
49.7
99.3
6.0
268.67
136.9
8
R-TS8
49.2
100.4
262.39
133.7
9
R-TS9
50.1
99.2
256.30
130.6
10
R-TS10
49.7
99.6
8.0
281.03
143.2
11
R-TS11
50.2
100.2
288.68
147.1
12
R-TS12
49.6
99.4
283.58
144.5
13
R-TS13
50.4
100.2
10.0
299.48
152.6
14
R-TS14
50.3
99.8
315.37
160.7
15
R-TS15
50.1
100.4
304.97
155.4The plastic volumetric strain of rock samples R-TS1, R-TS4, R-TS8, R-TS12, and R-TS15 was calculated according to the method described in Section3.1. Then, we obtained 5 plastic internal variable versus axial stress curves, one for each of the confining pressures tested, as shown in Figure 11.Figure 11
Plastic internal variable versus axial stress curves under various confining pressures.We calculated the values ofcχ and φχ corresponding to the plastic volume strains of 0.0, 0.001, 0.002, 0.003, 0.004, and 0.005 by using equation (10), and the results are presented in Figure 12 and Table 3.Figure 12
The change inc and φ with χ.Table 3
The calculated results ofcχ and φχ with varying χ.
χ
cχ
φχ
0
11.2
54.3
0.001
12.2
51.1
0.002
8.66
53.3
0.003
5.53
54.3
0.004
2.98
54.9
0.005
1.29
54.3


## 4.2. Fluid Flow Test Results and Analysis

The fluid flow tests were carried out underN=11 stress states, specifically Aii=1,2,⋯,11. Each of the rock samples had a diameter of 49.6 mm and a height of 99.2 mm. Water was used as the flow liquid; the water had a mass density of ρ=1000kg/m3, dynamic viscosity of μ=1.002×10−3Pa⋅s at 20°C, and a compressibility coefficient of cf=0.458×10−9Pa−1.At the start of the fluid flow test, the pressures at the ends of the sandstone sample arep10=5MPa and p20=3MPa; therefore, the pore pressure difference is 2 MPa. In the pre-peak strength state, the confining pressure is set as 6 MPa. To investigate the influence of plastic flow on the permeability of the sandstone, the axial stress and confining pressure of the rock samples were both changed after the seventh step of the permeability test.The axial stress-strain curve of the sandstone sample is shown in Figure13. As seen in Figure 13, the axial stress reaches a peak value at point A5. At that point, the rock sample is considered to be deforming via shear yielding. Therefore, points A1,⋯,A4 describe the pre-peak strength state, and points A6,⋯,A11 describe the plastic flow state. The rock sample went through a 6-step plastic flow, represented by points A5−A11 in Figure 13. The values of the axial stress σ1, confining pressure σ3, axial strain ε1, and circumferential strain ε3 measured during the test are presented in Table 4. The sign convention of ε1, ε3, and εV is that ε1 and ε3 in compression are positive and εV in dilation is positive.Figure 13
Relationship obtained between the axial stress and strain of the sandstone sample.Table 4
Axial stress, confining pressure, axial strain, and circumferential strain.
State
σ1 (MPa)
σ3 (MPa)
ε1 (%)
ε3 (%)
State
A1
10.75
6
0.051
0.022
Pre-peak
A2
37.78
6
0.434
0.033
Pre-peak
A3
56.29
6
0.625
0.040
Pre-peak
A4
76.68
6
0.816
0.100
Pre-peak
A5
97.43
6
1.008
-0.189
Shear yielding
A6
32.92
6
1.372
-1.421
Plastic flow
A7
24.95
7.25
1.324
-1.353
Plastic flow
A8
28.19
5
1.419
-1.300
Plastic flow
A9
18.87
5.66
1.324
-1.307
Plastic flow
A10
23.51
5.9
1.372
-1.332
Plastic flow
A11
10.7
5.9
1.132
-1.123
Plastic flowForAii=1,⋯,4, in the elastic deformation state, the values of the Lame coefficients Λ and G of the sandstone sample were obtained according to the method described from equation (34) to equation (38), where
(48)C11C12C21C22=2.42×10−42.85×10−41.42×10−44.62×10−4andB1B2=1.74×1062.13×106.Then, the values of the Lame coefficients areΛ=2.75GPa and G=3.76GPa.A total of 6 plastic flow steps occurred after yielding of the rock sample,Ai⟶Ai+1 (i=5,⋯,10). The values of the stress increments (Δσ1 and Δσ3), strain increments (Δε1 and Δε3), plastic potential increment (ΔF=∂F/∂σ1Δσ1+∂F/∂σ3Δσ3), and plastic multiplier Δλsii=6,⋯11 in each flow step are presented in Table 5.Table 5
Stress increment and strain increment during plastic flow of the rock sample.
Step
Δσ1 (MPa)
Δσ3 (MPa)
Δε1 (%)
Δε3 (%)
Δλs
ΔF
State
A5⟶A6
-64.5
0
0.364
-1.23
6.79E−3
-64.50
Unloading
A6⟶A7
-7.97
1.25
-0.048
0.068
9.02E−4
-15.63
Unloading
A7⟶A8
3.24
-2.25
0.095
0.053
−4.55E−4
17.03
Loading
A8⟶A9
-9.32
0.66
-0.095
-0.007
1.01E−3
-13.37
Unloading
A9⟶A10
4.64
0.24
0.048
-0.025
−4.75E−4
3.17
Loading
A10⟶A11
-12.8
0
-0.24
0.209
1.35E−3
-12.80
UnloadingAs shown in Table5, the flow steps A5⟶A6, A6⟶A7, A8⟶A9, and A10⟶A11 show the unloading process, and A7⟶A8 and A9⟶A10 show the loading process. The plastic multiplier lies between −4.7×10−4 and 6.79×10−3, and it is positive during the unloading steps (ΔF<0) and negative during the loading steps.The volumetric strain, porosity, equivalent shear strain, and permeability of each stress state after the sample yieldedAii=6,7,⋯,11 are presented in Table 6.Table 6
Volumetric strain, porosity, equivalent shear strain, and permeability of each stress state.
State
εV (%)
ϕ (%)
γMC (%)
km−2
A6
1.47
7.50
6.03
9.28×10−17
A7
1.38
7.23
5.77
9.00×10−17
A8
1.18
6.88
5.83
8.37×10−17
A9
1.29
6.99
5.66
8.71×10−17
A10
1.29
7.20
5.81
8.72×10−17
A11
1.11
6.81
4.86
8.17×10−17As seen in Table7, the volumetric strain increments during the first unloading process A5⟶A6 and A8⟶A9 are positive, and the volumetric strain increments during the later unloading processes A6⟶A7 and A10⟶A11 are negative. In addition, the volumetric strain increment during the loading process A7⟶A8 is negative, and the volumetric strain increment during the loading process A9⟶A10 is zero. These changes in the sign of the volumetric strain increment indicate that there is no one-to-one relationship between the volumetric strain/variation and either loading or unloading. Similarly, there is also no one-to-one relationship between the equivalent shear strain variation and either loading or unloading.Table 7
Strain increment and permeability increment during plastic flow of the rock sample.
Step
ΔεV (%)
ΔγMC (%)
Δkm−2
State
A5⟶A6
2.10
3.66
5.38×10−17
Unloading
A6⟶A7
-0.27
-0.25
−2.7×10−18
Unloading
A7⟶A8
-0.20
0.05
−6.28×10−18
Loading
A8⟶A9
0.11
-0.16
3.38×10−18
Unloading
A9⟶A10
0.00
0.15
6.00×10−20
Loading
A10⟶A11
-0.18
-0.96
−5.49×10−18
UnloadingShear deformation occurred in the rock sample during the loading steps ofA9⟶A10, without volume deformation. A permeability increment magnitude on the order of 1×10−20 was observed at this step, and a permeability increment magnitude on the order of 1×10−17 to 1×10−18 was observed during the other steps. Therefore, the influence of the shear strain on permeability is less than that of the volumetric strain.The region of search forλ1k was 2,3, and the region of search for λ2k was 0.01,0.5. By using the Monte Carlo method, the optimal estimated values of the permeability influence coefficients were obtained as λ1k=2.55 and λ2k=0.102. We found that λ2k<λ1k, which proves that the influence of the shear strain on permeability is less than that of the volumetric strain.From the fluid flow test results and the calculated values ofλ1k and λ2k, it can be concluded that both shear strain and volumetric strain have an obvious effect on permeability. These two results both verify that the volumetric strain has a major effect on permeability.The estimated values of permeability under every stress state after shear yielding, calculated by using equation (47), are presented in Figure 14 and Table 8.Figure 14
Single-step estimation of the permeability after shear yielding.Table 8
Experimental values, estimated values, and relative errors of the permeability.
State
Permeability (m-2)
Experimental values
Single-step estimated values
Relative error (%)
A5
3.90×10−17
3.90×10−17
—
A6
9.28×10−17
7.75×10−17
16.5%
A7
9.00×10−17
9.03×10−17
0.28%
A8
8.37×10−17
8.34×10−17
0.41%
A9
8.71×10−17
8.73×10−17
0.26%
A10
8.72×10−17
8.70×10−17
0.22%
A11
8.17×10−17
8.27×10−17
1.20%Table8 shows that the error between the estimated values and the experimental values in the stress state A6 is 16.5%, which is because the axial stress decreases sharply as soon as the rock sample reaches the peak stress state. The errors in the stress states A7,⋯,A11 were between 0.22% and 1.20%. These small errors indicate that the iterative format for estimating permeability, represented by equation (47), is generally stable.


## 5. Discussion

In this study, we conducted fluid flow tests under 11 stress states. The permeability variation law for the sandstone samples after shear yielding was discussed. The strain increments after yielding included both the elastic strain increments and plastic strain increments. The former is directly given according to the generalized Hooke’s law, while the latter is given by the flow rule, and the total strain increments are as follows:(49)Δε1=12GΔσ1−Λ3Λ+2GΔσ1+2Δσ3+Δλs,Δε3=12GΔσ3−Λ3Λ+2GΔσ1+2Δσ3−Δλstan245∘+φχ2.According to equations (22) and (49), the volumetric strain increment ΔεV and the equivalent strain increment ΔγMC can be written as follows:
(50)ΔεV=12GΔσ1−Λ3Λ+2GΔσ1+2Δσ3+Δλs+1GΔσ3−Λ3Λ+2GΔσ1+2Δσ3−2Δλstan245∘+φχ2,(51)ΔγMC=21−2ν1−νsec245∘+φχ212GΔσ1i−Λ3Λ+2GΔσ1i+2Δσ3i+Δλs+21−2ν2ν−tan245∘+φχ212GΔσ3i−Λ3Λ+2GΔσ1i+2Δσ3i−Δλstan245∘+φχ2.The expression between the permeability increment and the stress increment in plastic flow can be obtained by substituting equations (50) and (51) into equation (23):
(52)Δk=kϕλ1k12GΔσ1−Λ3Λ+2GΔσ1+2Δσ3+Δλs+1GΔσ3−Λ3Λ+2GΔσ1+2Δσ3−2Δλstan245∘+φχ2+kϕλ2k21−2ν1−νsec245∘+φχ212GΔσ1−Λ3Λ+2GΔσ1+2Δσ3+Δλs+21−2ν2ν−tan245∘+φχ212GΔσ3i−Λ3Λ+2GΔσ1+2Δσ3−Δλstan245∘+φχ2.Based on equation (52), an iterative format is constructed to estimate the permeability increment in each step and the permeability at each stress-strain state, after shear yielding.The performances of the plastic flow rule and the permeability model are discussed below.


### 5.1. Plastic Flow Rule

Substituting equations (40) and (41) into equation (42), we obtain the following equation:
(53)ε1i=ε1i−1+12GΔσ1i−Λ3Λ+2GΔσ1i+2Δσ3i+λsiε3i=ε3i−1+12GΔσ3i−Λ3Λ+2GΔσ1i+2Δσ3i−λsitan245∘+φ2,where Δσ1i=σ1i−σ1i−1 and Δσ3i=σ3i−σ3i−1 for i=6,7,⋯,11.The flow rule represented by equation (53) is a mapping of R2⟶R2; however, geometrically comparing the estimated value of the strain component with the experimental value is difficult. Therefore, we introduce the equivalent shear stress that corresponds to the Mohr-Coulomb yield criterion:
(54)τMC=σ1−σ3tan245∘+φχ2.Before yielding, the equivalent shear stressτMC is linearly related to the equivalent shear strain γMC, i.e., τMC=GγMC. After yielding, τMC≤2ctanα and τMC≠GγMC.The equivalent shear stressτMCi and the equivalent shear strain γMCi, where i=6,7,⋯,11 are calculated for each stress state according to equations (54) and (22). The increment of equivalent shear strain is found to be ΔγMCi=γMCi−γMCi−1 by combining equations (22) and (54), where i=6,7,⋯,11, and the estimated value of the equivalent shear strain is γ~MCi=γ~MCi−1+ΔγMCi, where i=6,7,⋯,11; these values are also calculated for each stress state.Figure15 shows the estimated values and experimental values of the equivalent shear strain corresponding to the equivalent shear stress τMCi, where i=6,7,⋯,11. Figure 15 shows that the estimated equivalent shear strain values are approximately equal to the experimental values in each state after yielding. It can be argued that the flow rule estimation of the equivalent shear strain, as proposed in this paper, is highly accurate.Figure 15
Estimated result of the equivalent shear stress and shear strain.


### 5.2. Permeability Model

The permeability model represented by equation (47) estimates the permeability k~i at the Ai state, based on the experimental value of permeability ki−1 at the Ai−1 state, which is a single-step estimation. Although the single-step estimation is highly precise, it largely depends on the test. To facilitate engineering applications, we estimate the permeability k~i at the Ai state by means of the estimated permeability value k~i−1 at the Ai−1 state and propose the following multistep estimation format for permeability:
(55)kM=k0ϕMϕ0mk,k~M+1=kM+kMϕMλ1kΔεVM+1+λ2kΔγMCM+1,k~i=k~i−1+k~i−1ϕi−1λ1kΔεVi+λ2kΔγMCi,i=M+2,M+3,⋯,N.The values of permeability under the statesA6,A7,⋯,A11 estimated by equation (55) are presented in Table 9.Table 9
Multistep permeability estimation.
State
Permeability (m-2)
Experimental values
Multistep estimated values
Relative error (%)
A5
3.90×10−17
3.90×10−17
0.00%
A6
9.28×10−17
7.74×10−17
16.63%
A7
9.00×10−17
7.53×10−17
16.40%
A8
8.37×10−17
6.97×10−17
16.73%
A9
8.71×10−17
7.27×10−17
16.51%
A10
8.72×10−17
7.26×10−17
16.69%
A11
8.17×10−17
6.89×10−17
15.68%Table9 shows that the error between the estimated and experimental values of permeability is 16.6% at the stress state A6, while the errors obtained at the other stress states A7,⋯,A11 are between 15.6% and 16.7%. Therefore, the iterative format represented by equation (55) is also generally stable.Figure16 shows a comparison of the multiple-step estimated values and the single-step estimated values of permeability. As shown in Figure 16, the multistep estimation at the stress state A7 significantly differs from the single-step estimation (a difference of approximately 16.5%), and at later stress states, the error does not significantly increase.Figure 16
Comparison between the single-step estimation and multistep estimation.The permeability model proposed in this paper can be used to only estimate the permeability after shear yielding. The variation in permeability after yielding is mainly caused by the relative displacement of the shear yield surface of the rock sample. Figure17 shows the relative movement between the shear yield surfaces of a rock sample. The relative displacement Δun in the normal direction of the shear plane results in the volumetric strain increment ΔεVi, and the relative displacement Δut in the shear plane results in the equivalent shear strain increment ΔγMCi, where i=M+1,M+2,⋯,N.Figure 17
Relative displacement of the shear plane: (a) normal displacement and (b) tangential displacement.
(a)
(b)


### 5.3. Applicability of Permeability Model for Plastic Flow

We conducted permeability tests of sandstone samples under the complete stress-strain process in order to fully verify the applicability of the permeability model for plastic flow. The test results are presented in Table10. Iterative calculations are performed using equations (47) and (55) to obtain single-step estimated and multistep estimated values of permeability under every stress state after shear yielding, which are presented in Figure 18 and Table 11.Table 10
Values of axial stress, confining pressure, axial strain, circumferential strain, and permeability.
State
σ1 (MPa)
σ3 (MPa)
ε1 (%)
ε3 (%)
km−2
State
1
12.43
6.00
0.147
0.011
9.24×10−17
Pre-peak
2
44.22
6.00
0.468
0.041
8.20×10−17
Pre-peak
3
63.24
6.00
0.628
0.070
7.81×10−17
Pre-peak
4
80.57
6.00
0.789
0.112
7.53×10−17
Pre-peak
5
97.34
6.00
1.109
-0.195
6.97×10−17
Shear yielding
6
40.65
6.00
1.124
-0.698
1.10×10−16
Plastic flow
7
53.25
6.00
1.140
-0.758
1.15×10−16
Plastic flow
8
44.56
6.00
1.116
-0.793
1.19×10−16
Plastic flow
9
48.85
6.00
1.229
-0.800
1.14×10−16
Plastic flow
10
35.56
6.00
1.125
-0.842
1.23×10−16
Plastic flow
11
34.42
6.00
1.109
-0.845
1.25E−16
Plastic flowFigure 18
Comparison between experimental and estimated values under complete stress-strain process.Table 11
Estimated values and relative errors of the permeability under complete stress-strain process.
State
Permeability (m-2)
Experimental values
Single-step estimated values
Relative error (%)
Multistep estimated values
Relative error (%)
A5
6.79×10−17
6.79×10−17
—
6.79×10−17
—
A6
1.10×10−16
1.07×10−16
2.55%
1.07×10−16
4.21%
A7
1.15×10−16
1.14×10−16
0.61%
1.10×10−16
4.13%
A8
1.19×10−16
1.21×10−16
1.76%
1.14×10−16
3.88%
A9
1.14×10−16
1.06×10−16
7.60%
1.09×10−16
5.01%
A10
1.23×10−16
1.31×10−16
6.40%
1.18×10−16
3.96%
A11
1.25×10−16
1.26×10−16
1.04%
1.19×10−16
4.05%Table11 shows that the errors between the single-step estimated values and the experimental values in the stress state A7,⋯,A11 were from 0.61% to 7.60%. The errors between the multistep estimated values and the experimental values were from 3.88% to 5.01%. These errors indicate that the iterative format for estimating permeability, represented by equation (47) and (55), are generally stable.


## 5.1. Plastic Flow Rule

Substituting equations (40) and (41) into equation (42), we obtain the following equation:
(53)ε1i=ε1i−1+12GΔσ1i−Λ3Λ+2GΔσ1i+2Δσ3i+λsiε3i=ε3i−1+12GΔσ3i−Λ3Λ+2GΔσ1i+2Δσ3i−λsitan245∘+φ2,where Δσ1i=σ1i−σ1i−1 and Δσ3i=σ3i−σ3i−1 for i=6,7,⋯,11.The flow rule represented by equation (53) is a mapping of R2⟶R2; however, geometrically comparing the estimated value of the strain component with the experimental value is difficult. Therefore, we introduce the equivalent shear stress that corresponds to the Mohr-Coulomb yield criterion:
(54)τMC=σ1−σ3tan245∘+φχ2.Before yielding, the equivalent shear stressτMC is linearly related to the equivalent shear strain γMC, i.e., τMC=GγMC. After yielding, τMC≤2ctanα and τMC≠GγMC.The equivalent shear stressτMCi and the equivalent shear strain γMCi, where i=6,7,⋯,11 are calculated for each stress state according to equations (54) and (22). The increment of equivalent shear strain is found to be ΔγMCi=γMCi−γMCi−1 by combining equations (22) and (54), where i=6,7,⋯,11, and the estimated value of the equivalent shear strain is γ~MCi=γ~MCi−1+ΔγMCi, where i=6,7,⋯,11; these values are also calculated for each stress state.Figure15 shows the estimated values and experimental values of the equivalent shear strain corresponding to the equivalent shear stress τMCi, where i=6,7,⋯,11. Figure 15 shows that the estimated equivalent shear strain values are approximately equal to the experimental values in each state after yielding. It can be argued that the flow rule estimation of the equivalent shear strain, as proposed in this paper, is highly accurate.Figure 15
Estimated result of the equivalent shear stress and shear strain.


## 5.2. Permeability Model

The permeability model represented by equation (47) estimates the permeability k~i at the Ai state, based on the experimental value of permeability ki−1 at the Ai−1 state, which is a single-step estimation. Although the single-step estimation is highly precise, it largely depends on the test. To facilitate engineering applications, we estimate the permeability k~i at the Ai state by means of the estimated permeability value k~i−1 at the Ai−1 state and propose the following multistep estimation format for permeability:
(55)kM=k0ϕMϕ0mk,k~M+1=kM+kMϕMλ1kΔεVM+1+λ2kΔγMCM+1,k~i=k~i−1+k~i−1ϕi−1λ1kΔεVi+λ2kΔγMCi,i=M+2,M+3,⋯,N.The values of permeability under the statesA6,A7,⋯,A11 estimated by equation (55) are presented in Table 9.Table 9
Multistep permeability estimation.
State
Permeability (m-2)
Experimental values
Multistep estimated values
Relative error (%)
A5
3.90×10−17
3.90×10−17
0.00%
A6
9.28×10−17
7.74×10−17
16.63%
A7
9.00×10−17
7.53×10−17
16.40%
A8
8.37×10−17
6.97×10−17
16.73%
A9
8.71×10−17
7.27×10−17
16.51%
A10
8.72×10−17
7.26×10−17
16.69%
A11
8.17×10−17
6.89×10−17
15.68%Table9 shows that the error between the estimated and experimental values of permeability is 16.6% at the stress state A6, while the errors obtained at the other stress states A7,⋯,A11 are between 15.6% and 16.7%. Therefore, the iterative format represented by equation (55) is also generally stable.Figure16 shows a comparison of the multiple-step estimated values and the single-step estimated values of permeability. As shown in Figure 16, the multistep estimation at the stress state A7 significantly differs from the single-step estimation (a difference of approximately 16.5%), and at later stress states, the error does not significantly increase.Figure 16
Comparison between the single-step estimation and multistep estimation.The permeability model proposed in this paper can be used to only estimate the permeability after shear yielding. The variation in permeability after yielding is mainly caused by the relative displacement of the shear yield surface of the rock sample. Figure17 shows the relative movement between the shear yield surfaces of a rock sample. The relative displacement Δun in the normal direction of the shear plane results in the volumetric strain increment ΔεVi, and the relative displacement Δut in the shear plane results in the equivalent shear strain increment ΔγMCi, where i=M+1,M+2,⋯,N.Figure 17
Relative displacement of the shear plane: (a) normal displacement and (b) tangential displacement.
(a)
(b)


## 5.3. Applicability of Permeability Model for Plastic Flow

We conducted permeability tests of sandstone samples under the complete stress-strain process in order to fully verify the applicability of the permeability model for plastic flow. The test results are presented in Table10. Iterative calculations are performed using equations (47) and (55) to obtain single-step estimated and multistep estimated values of permeability under every stress state after shear yielding, which are presented in Figure 18 and Table 11.Table 10
Values of axial stress, confining pressure, axial strain, circumferential strain, and permeability.
State
σ1 (MPa)
σ3 (MPa)
ε1 (%)
ε3 (%)
km−2
State
1
12.43
6.00
0.147
0.011
9.24×10−17
Pre-peak
2
44.22
6.00
0.468
0.041
8.20×10−17
Pre-peak
3
63.24
6.00
0.628
0.070
7.81×10−17
Pre-peak
4
80.57
6.00
0.789
0.112
7.53×10−17
Pre-peak
5
97.34
6.00
1.109
-0.195
6.97×10−17
Shear yielding
6
40.65
6.00
1.124
-0.698
1.10×10−16
Plastic flow
7
53.25
6.00
1.140
-0.758
1.15×10−16
Plastic flow
8
44.56
6.00
1.116
-0.793
1.19×10−16
Plastic flow
9
48.85
6.00
1.229
-0.800
1.14×10−16
Plastic flow
10
35.56
6.00
1.125
-0.842
1.23×10−16
Plastic flow
11
34.42
6.00
1.109
-0.845
1.25E−16
Plastic flowFigure 18
Comparison between experimental and estimated values under complete stress-strain process.Table 11
Estimated values and relative errors of the permeability under complete stress-strain process.
State
Permeability (m-2)
Experimental values
Single-step estimated values
Relative error (%)
Multistep estimated values
Relative error (%)
A5
6.79×10−17
6.79×10−17
—
6.79×10−17
—
A6
1.10×10−16
1.07×10−16
2.55%
1.07×10−16
4.21%
A7
1.15×10−16
1.14×10−16
0.61%
1.10×10−16
4.13%
A8
1.19×10−16
1.21×10−16
1.76%
1.14×10−16
3.88%
A9
1.14×10−16
1.06×10−16
7.60%
1.09×10−16
5.01%
A10
1.23×10−16
1.31×10−16
6.40%
1.18×10−16
3.96%
A11
1.25×10−16
1.26×10−16
1.04%
1.19×10−16
4.05%Table11 shows that the errors between the single-step estimated values and the experimental values in the stress state A7,⋯,A11 were from 0.61% to 7.60%. The errors between the multistep estimated values and the experimental values were from 3.88% to 5.01%. These errors indicate that the iterative format for estimating permeability, represented by equation (47) and (55), are generally stable.


## 6. Conclusions

Conventional triaxial compression tests were conducted on sandstone samples. A fluid flow test was conducted using the MTS 816.02 rock mechanics testing system and a self-designed permeation system to obtain the values of permeability under different stress states. The proposed iterative formats of the plastic multiplier and permeability estimations were based on the permeability model. The main conclusions were as follows:(1)
A new plastic flow rule in the stress space was proposed based on the Mohr-Coulomb yield criterion and the isotropic hardening (softening) model. The plastic multiplier in each flow step after shear yielding was calculated, and the plastic multiplier was found to be positive under unloading conditions and negative under loading conditions
(2)
The equivalent shear strain increment in each flow step after shear yielding was calculated using the new plastic flow rule. The difference between estimated and experimental values of the equivalent shear strain increment was very small, which indicated that the new plastic flow rule is highly accurate
(3)
The permeability of the sandstone samples during the plastic state was estimated using the single-step and multistep estimation models. In the single-step estimation, the error between the estimated value and the experimental value of permeability was 16.5% at stress stateA6, and the error in the remaining states was less than 1.2%. In the multistep estimation, the error between the estimated value and the experimental value of permeability was 16.6% at stress state A6, and the error was maintained between 15.6% and 16.7% in stress states A7,⋯,A11. These errors indicate that the iterative format of the multiple-step permeability estimation method is generally stable. The multiple-step permeability estimation model required the experimental value of permeability only at state A5, not the experimental values of permeability under states A6,A7,⋯,A11. Therefore, the experimental value of permeability from state A5 was used to estimate the permeability at states A6,A7,⋯,A11. The multistep estimation method is therefore satisfactory for engineering applications and is highly precise
(4)
The influence coefficients of the volumetric strain increments and equivalent shear-strain increments of permeability wereλ1k=2.55 and λ2k=0.102, respectively. Since λ2k<λ1k, the influence of the shear strain on permeability was less than that of the volumetric strain
(5)
In the flow stepA9⟶A10, the volumetric strain increment was zero, and the magnitude of the increment of permeability was on the order of 1×10−20, which is significantly smaller than those of the other five flow steps that have nonzero volumetric strain increments
(6)
The relative errors between the estimated values and the experimental values of permeability under complete stress-strain process are from 0.61% to 7.60%. The errors show that the permeability model for plastic flow is applicable



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