# Uniaxial Compression Mechanical Analysis of Cylindrical Rock Specimens Based on Space Axisymmetric Problem

Authors: Jianguang Li; Yanchun Wang; Chuanqi Su
Journal: Shock and Vibration
 (2021)

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## Abstract

The common failure forms in the uniaxial compression test of standard cylindrical rock specimens are symmetric cone failure and splitting failure. In the past, two kinds of failure forms were explained from the plane problems of elasticity theory. However, there was a lack of research based on space problems. In this paper, based on spatial axisymmetric elastic mechanics theory, three-dimensional stress function is constructed for the cylindrical rock specimen under uniform axial compression by adopting the semi-inverse method. According to the stress function, the stress components and the principal stress are deduced when considering the end effect. The failure form is speculated, which can well explain the two common types of damage in uniaxial compression. This work may enrich the basic research of rock mechanics.


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## Body

## 1. Introduction

As we all know, uniaxial compression experiment is the most common type of experiment to understand the mechanical properties of certain kinds of rocks. In various international standards, national standards, local standards, and industrial standards, there are clear test procedures, and the experimental results have become the classic textbook content. The two corresponding types of typical failure, i.e., split failure and symmetric cone failure, impressed people deeply. Traditionally, it is believed that the failure mechanism is related to the end effect of the compression head of the testing machine. But this explanation is not detailed and thorough enough from the view point of mathematical mechanics model. Some researchers tried to use the theory of elastic mechanics to derive the stress function. However, most of them are confined to the two-dimensional space, and most of the three-dimensional stress functions used to solve the rock-related problems are only based on fuzzy theoretical reasoning [1, 2], which lacks the applicability extension. In reference [1], the two-dimensional Airy stress function of cylindrical rock specimens under uniform load was given. In this paper, based on further derivation on the theory of spatial axisymmetry [3–5], the three-dimensional stress function and stress components of a standard cylindrical rock specimen under uniform axial load will be given on the basis of reference to predecessors' experience [6–12], and the destruction mechanism of uniaxial compression will be explained, which can well explain the two common types of failure in uniaxial compression. This work may enrich the basis research of rock mechanics.


## 2. Uniaxial Compression Mechanical Model of Cylindrical Rock Specimen

Figure1 shows the uniaxial compression mechanical model of a typical cylindrical specimen, with the heighth, the radius b, and the elastic modulus and Poisson’s ratio E and μ, respectively. The rock mass deforms in both axial and radial directions when the uniformly distributed load σv is applied in the axial direction (since the volume force is much smaller than σv, the volume force is isotropic and not considered). Some researchers have found that the force exerted on the surface of the specimen by the testing machine is nonuniform. But in practice, if the specimen meets the requirements of flatness and straightness, the assumption of axial uniform pressure is approximately true. Furthermore, by applying lubricant on the contact surface, the friction effect can be reduced and the uniformity of axial pressure can be further ensured. In the process of continuous deformation, the upper and lower surface of the rock specimen will produce friction binding force Fs, which is similar to the case in physics where the slider moves on a nonsmooth contact surface. In the limit case, when all the points at the same position on the two interfaces have relative motion, the value of Fs follows the friction law (take unit angle):(1)Fsmax=σνfb22,where Fsmax is the maximum value of the derived friction binding force and 30∘ is the friction coefficient of the contact surface.Figure 1
Uniaxial compression mechanics.As shown in Figure2, a space cylindrical coordinate system is established at the center of the specimen, with the z axis along the axial direction, the downward direction as the positive direction, and the r axis along the radial direction.Figure 2
Cylindrical coordinate system model of rock specimen.


## 3. Stress Function

### 3.1. Stress Function Form Analysis

According to the basic theory of elastic mechanics, the stress function of space axisymmetric problem is a function that is only related toz and ρ, but unrelated to θ. Therefore, the stress function is expressed as ϕρ,z, and the terms can be expressed in the form of exponential function, logarithmic function, exponential function form, and so on. But according to the definition domain and its physical meaning, the exponential function with positive integer is selected as the term of the stress function. Suppose the stress function has the form as(2)ϕρ,z=∑ρmznm,n∈N.The stress functionϕρ,z must be a biharmonic function and satisfy the compatible equation ∇4φ=0. The term of the stress function is expressed as(3)Fρ,z=ρmznm,n∈N,and(4)∇4Fρ,z=mm−1m−2m−3ρm−4zn+2mm−1m−2ρm−4zn−mm−1ρm−4zn+mrm−4zn+2mnm−1n−1ρm−2zn−2+2mnn−1ρm−2zn−2+nn−1n−2n−3ρmzn−4=m2m−22ρm−4zn+2m2nn−12ρm−2zn−2+nn−1n−2n−3ρmzn−4.Formula (4) shows that terms of the same degree can be substituted into the compatible equation to the same power, while terms of different degrees can be substituted into the compatible equation to different powers. Therefore, if the stress function ϕ satisfies the compatible equation, the combination of terms of the same degree must satisfy the compatible equation. When m (the exponent of ρ) is even and Fρ,z of the same power is substituted into the compatible equation, the results are linearly dependent. When the exponent m of ρ is odd and Fρ,z of the same power is substituted into the compatible equation, the results are linearly independent and the stress function cannot be constructed to satisfy the compatible equation. So, we know that the exponent of ρ has to be even.From the terms of the stress functionFρ,z, the terms of the stress component can be expressed as(5)σzF=∂∂z2−μ∇2Fρ,z−∂2Fρ,z∂z2,=2−μm2nρm−2zn−1+nn−1n−2ρmzn−3−nn−1n−2ρmzn−3,=2−μm2nρm−2zn−1+1−μnn−1n−2ρmzn−3,σρF=∂∂zμ∇2Fρ,z−∂2Fρ,z∂ρ2,=μm2nρm−2zn−1+nn−1n−2ρmzn−3−mnn−1ρm−2zn−1,=mnμm−m+1ρm−2zn−1+μnn−1n−2ρmzn−3,σθF=∂∂zμ∇2Fρ,z−1ρ∂Fρ,z∂r,=μm2nρm−2zn−1+nn−1n−2ρmz−3−mnρm−2zn−1,=mnμm−1ρm−2zn−1+μnn−1n−2ρmzn−3,τzρF=τρzF=∂∂ρ1−μ∇2Fρ,z−∂2Fρ,z∂z2,=1−μm2m−2ρm−3zn+mnn−1ρm−1zn−2−mnn−1ρm−1zn−2,=1−μm2m−2ρm−3zn−μmnn−1n−2ρm−1zn−2.According to thez-axis symmetry of the specimen, it can be known that σz must be z-axis symmetric, that is, σz is an even function of z. n is odd by combining with formula (5). At this point, there is Fρ,z=ρmzn(m, n∈N, m is even, and n is odd). Removing the independent stress term: z, which can be set(6)y0=A0z3+B0r2z,y1=A1z4+B1r2z2+C1r4y2=A2z5+B2r2z3+C2r4z,y3=A3z6+B3r2z4+C3r4z2+D3r6,y4=A4z7+B4r2z5+C4r4z3+D4r6z,⋮yk=Akz2k+3+Bkr2z2k+1+Ckr4z2k−1+⋅⋅⋅,k∈N,there are(7)ϕρ,z=y0+y2+y4+y6+⋅⋅⋅ynn∈Nandnis even.Substitutingy0,y2,y4,…,yn into the compatible equation and combining formula (4), we have(8)∇4y0=0,∇4y2=120A2+48B2+64C2z,∇4y4=840A4+160B4+64C4z3+120B4+120C4+480D4ρ2z,⋮∇4yn=⋯.It can be known from the above thaty0,y2,y4,…,yn is the biharmonic function. That is,(9)∇4y0=0,∇4y2=0,∇4y4=0,∇4y6=0⋅⋅⋅∇4yn=0.Let(10)σz0=∂∂z2−μ∇2y0−∂2y0∂z2,σz2=∂∂z2−μ∇2y2−∂2y2∂z2,σz4=∂∂z2−μ∇2y4−∂2y4∂z2,⋮σzn=∂∂z2−μ∇2yn−∂2yn∂z2.Substitutingy0,y2,y4,…,yn into formula (10), we can get(11)σz0=6A01−μ+4B02−μ,σz2=60A21−μ+12B22−μz2+6B21−μ+16C22−μρ2,σz4=210A41−μ+20B42−μz4+60B41−μ+48C42−μρ2z2,+6C41−μ+36D42−μρ4,⋮σzn=⋅⋅⋅.In Figure1, since FS is the derived friction binding force, the direction is outward along the radius and the z direction of the sandwich contact surface is only subject to uniform distribution stress σv′ actually, and the axial stress σz can be regarded as a function of z in the specimen, that is, σz is independent of ρ. According to formulae (8), (9), and (11), the conditions are right.(12a)Whenk=2,120A2+48B2+64C2=0,6B21−μ+16C22−μ=0,(12b)whenk=4,840A4+160B4+64C=0,120B4+120C4+480D4=0,60B41−μ+48C22−μ=0,6C41−μ+36D42−μ=0,⋮whenk=⋯.Solution to formula (12a) is obtained as follows:(13a)A2=83−μ151−μC2,(13b)B2=−82−μ31−μC2.Solution to formula (12b) shows that the equation system can only be established if and only if all of A4, B4, C4, and D4 are equal to 0. It can be seen from the analysis that each increase of n by 2 will increase a coefficient in the functionyn, and two equations will be added to the corresponding equation system and generate excessive constraints in formulae (12a) and (12b). The equation system can only be established if and only if all the unknowns contained are 0, that is, when the conditions of n>2cannot be satisfied. So, the stress function can be expressed as(14)ϕρ,z=y0+y2=A0z3+B0ρ2z+A2z5+B2ρ2z3+C2ρ4z.The stress components can be expressed as(15a)σz=6A01−μ+4B02−μ+60A21−μ+12B22−μz2,(15b)σρ=6A0μ+4B0μ−2B0+60A2μ+6B22μ−1z2+6B2μ+4C24μ−3ρ2,(15c)σφ=6A0μ+4B0μ−2B0+60A2μ+6B22μ−1z2+6B2μ+4C24μ−1ρ2,(15d)σz=6A01−μ+4B02−μ+60A21−μ+12B22−μz2.


### 3.2. Boundary Conditions

For the boundary conditions of rock mass specimens, the main boundary conditions should be exactly satisfied firstly, and the secondary boundary conditions should be satisfied as far as possible. The tension of normal stress is defined as “+,” and the pressure is defined as “−” The sign for shear stress is defined as follows: when the specimen is clockwise rotation, the shear stress is defined as “+.” For the counterclockwise rotation, the shear stress is defined as “−.” Next, the boundary conditions are substituted into the parameters in formulae (14) and (15a)–(15d).(1)
Thez-direction stress of the upper and lower end faces is −σv′:(16a)σzz=−h/2=−σv,(16b)σzz=h/2=−σv.Formulae (16a) and (16b) can be obtained.(17)6A01−μ+4B02−μ+60A21−μ+12B22−μ×h24=−σv.(2)
Resultant force of shear stress is zero of lateral free surface (unit angle):∫−h/2h/2τzρρ=bdz=0, natural meet.(3)
Resultant force of the shear stress of the upper and lower boundary, respectively, isFS and −FS (taking unit angle).Upper boundary:(18)∫0bτρzz=−h/2ρdρ=FS.Reduction to draw:(19)233B2μ−8C21−μhb3=FS.Lower boundary:(20)∫0bτρzz=h/2ρdρ=−FS.This simplifies to the same (3)–(17), (19), (21), and (23).Formulae (3)–((15a)–(15d)) and formulae (3)–(17), (19), (21), and (23) can be calculated:(21)A2=−3−μ10hb3FS,B2=2−μ2hb3FS,C2=−31−μ16hb3FS.(4)
The resultant of normal stress on the free side is zero:(22)∫−h/2h/2σρρ=bdz=0,which reduces to(23)6A0μ+4B0μ−2B0h+5A2μ+B2μ−12h3+6B2μ+4C24μ−3b2h=0.(5)
The resultant moment of the free side is zero:∫−h/2h/2σρρ=bzdz=0, natural meet.According to formulae (17) and (21), the following equation can be obtained:(24)6A01−μ+4B02−μ+3h2b3FS=−σv.According to formulae (21) and (23), the following equation can be obtained:(25)6A0μ+4B0μ−2B0−2+μh4b3FS+33+μ4hbFS=0.From formulae (24) and (25), we can get(26)A0=2−μ2+μh12b3μ+1FS−3+μ2−μ4hbμ+1FS+2μ−1h4b3μ+1FS+2μ−1σv′6μ+1,B0=33+μ1−μ8hbμ+1FS−2+μ1−μh8b3μ+1FS−3μh4b3μ+1FS−μσv′2μ+1.Thus, all undetermined coefficients are determined. The stress function and each stress component are determined accordingly, as shown in formulae (27) and (28), respectively.(27)ϕz,ρ=2−μ2+μh24b3μFS−3+μ2−μ8hbμFS+h4b3FS+σv′6−13+2−μ3μz3+33+μ1−μ16hbμFS−2+μ1−μh16b3μFS−3h8b3FS−σv′4+12−1−μ2μρ2z−3−μFS10hb3z5+2−μFS2hb3ρ2z3−31−μFS16hb3ρ4z.(28)σz=6FShb3z2−3FSh2b3−σv′,σρ=3μ+3FS4hb3ρ2−3μ+2FShb3z2+2+μFSh4b3−33+μFS4hb,σθ=33μ+1FS4hb3ρ2−3μ+2FShb3z2+2+μFSh4b3−33+μFS4hb,τzρ=τρz=−6FShb3ρz.


## 3.1. Stress Function Form Analysis

According to the basic theory of elastic mechanics, the stress function of space axisymmetric problem is a function that is only related toz and ρ, but unrelated to θ. Therefore, the stress function is expressed as ϕρ,z, and the terms can be expressed in the form of exponential function, logarithmic function, exponential function form, and so on. But according to the definition domain and its physical meaning, the exponential function with positive integer is selected as the term of the stress function. Suppose the stress function has the form as(2)ϕρ,z=∑ρmznm,n∈N.The stress functionϕρ,z must be a biharmonic function and satisfy the compatible equation ∇4φ=0. The term of the stress function is expressed as(3)Fρ,z=ρmznm,n∈N,and(4)∇4Fρ,z=mm−1m−2m−3ρm−4zn+2mm−1m−2ρm−4zn−mm−1ρm−4zn+mrm−4zn+2mnm−1n−1ρm−2zn−2+2mnn−1ρm−2zn−2+nn−1n−2n−3ρmzn−4=m2m−22ρm−4zn+2m2nn−12ρm−2zn−2+nn−1n−2n−3ρmzn−4.Formula (4) shows that terms of the same degree can be substituted into the compatible equation to the same power, while terms of different degrees can be substituted into the compatible equation to different powers. Therefore, if the stress function ϕ satisfies the compatible equation, the combination of terms of the same degree must satisfy the compatible equation. When m (the exponent of ρ) is even and Fρ,z of the same power is substituted into the compatible equation, the results are linearly dependent. When the exponent m of ρ is odd and Fρ,z of the same power is substituted into the compatible equation, the results are linearly independent and the stress function cannot be constructed to satisfy the compatible equation. So, we know that the exponent of ρ has to be even.From the terms of the stress functionFρ,z, the terms of the stress component can be expressed as(5)σzF=∂∂z2−μ∇2Fρ,z−∂2Fρ,z∂z2,=2−μm2nρm−2zn−1+nn−1n−2ρmzn−3−nn−1n−2ρmzn−3,=2−μm2nρm−2zn−1+1−μnn−1n−2ρmzn−3,σρF=∂∂zμ∇2Fρ,z−∂2Fρ,z∂ρ2,=μm2nρm−2zn−1+nn−1n−2ρmzn−3−mnn−1ρm−2zn−1,=mnμm−m+1ρm−2zn−1+μnn−1n−2ρmzn−3,σθF=∂∂zμ∇2Fρ,z−1ρ∂Fρ,z∂r,=μm2nρm−2zn−1+nn−1n−2ρmz−3−mnρm−2zn−1,=mnμm−1ρm−2zn−1+μnn−1n−2ρmzn−3,τzρF=τρzF=∂∂ρ1−μ∇2Fρ,z−∂2Fρ,z∂z2,=1−μm2m−2ρm−3zn+mnn−1ρm−1zn−2−mnn−1ρm−1zn−2,=1−μm2m−2ρm−3zn−μmnn−1n−2ρm−1zn−2.According to thez-axis symmetry of the specimen, it can be known that σz must be z-axis symmetric, that is, σz is an even function of z. n is odd by combining with formula (5). At this point, there is Fρ,z=ρmzn(m, n∈N, m is even, and n is odd). Removing the independent stress term: z, which can be set(6)y0=A0z3+B0r2z,y1=A1z4+B1r2z2+C1r4y2=A2z5+B2r2z3+C2r4z,y3=A3z6+B3r2z4+C3r4z2+D3r6,y4=A4z7+B4r2z5+C4r4z3+D4r6z,⋮yk=Akz2k+3+Bkr2z2k+1+Ckr4z2k−1+⋅⋅⋅,k∈N,there are(7)ϕρ,z=y0+y2+y4+y6+⋅⋅⋅ynn∈Nandnis even.Substitutingy0,y2,y4,…,yn into the compatible equation and combining formula (4), we have(8)∇4y0=0,∇4y2=120A2+48B2+64C2z,∇4y4=840A4+160B4+64C4z3+120B4+120C4+480D4ρ2z,⋮∇4yn=⋯.It can be known from the above thaty0,y2,y4,…,yn is the biharmonic function. That is,(9)∇4y0=0,∇4y2=0,∇4y4=0,∇4y6=0⋅⋅⋅∇4yn=0.Let(10)σz0=∂∂z2−μ∇2y0−∂2y0∂z2,σz2=∂∂z2−μ∇2y2−∂2y2∂z2,σz4=∂∂z2−μ∇2y4−∂2y4∂z2,⋮σzn=∂∂z2−μ∇2yn−∂2yn∂z2.Substitutingy0,y2,y4,…,yn into formula (10), we can get(11)σz0=6A01−μ+4B02−μ,σz2=60A21−μ+12B22−μz2+6B21−μ+16C22−μρ2,σz4=210A41−μ+20B42−μz4+60B41−μ+48C42−μρ2z2,+6C41−μ+36D42−μρ4,⋮σzn=⋅⋅⋅.In Figure1, since FS is the derived friction binding force, the direction is outward along the radius and the z direction of the sandwich contact surface is only subject to uniform distribution stress σv′ actually, and the axial stress σz can be regarded as a function of z in the specimen, that is, σz is independent of ρ. According to formulae (8), (9), and (11), the conditions are right.(12a)Whenk=2,120A2+48B2+64C2=0,6B21−μ+16C22−μ=0,(12b)whenk=4,840A4+160B4+64C=0,120B4+120C4+480D4=0,60B41−μ+48C22−μ=0,6C41−μ+36D42−μ=0,⋮whenk=⋯.Solution to formula (12a) is obtained as follows:(13a)A2=83−μ151−μC2,(13b)B2=−82−μ31−μC2.Solution to formula (12b) shows that the equation system can only be established if and only if all of A4, B4, C4, and D4 are equal to 0. It can be seen from the analysis that each increase of n by 2 will increase a coefficient in the functionyn, and two equations will be added to the corresponding equation system and generate excessive constraints in formulae (12a) and (12b). The equation system can only be established if and only if all the unknowns contained are 0, that is, when the conditions of n>2cannot be satisfied. So, the stress function can be expressed as(14)ϕρ,z=y0+y2=A0z3+B0ρ2z+A2z5+B2ρ2z3+C2ρ4z.The stress components can be expressed as(15a)σz=6A01−μ+4B02−μ+60A21−μ+12B22−μz2,(15b)σρ=6A0μ+4B0μ−2B0+60A2μ+6B22μ−1z2+6B2μ+4C24μ−3ρ2,(15c)σφ=6A0μ+4B0μ−2B0+60A2μ+6B22μ−1z2+6B2μ+4C24μ−1ρ2,(15d)σz=6A01−μ+4B02−μ+60A21−μ+12B22−μz2.


## 3.2. Boundary Conditions

For the boundary conditions of rock mass specimens, the main boundary conditions should be exactly satisfied firstly, and the secondary boundary conditions should be satisfied as far as possible. The tension of normal stress is defined as “+,” and the pressure is defined as “−” The sign for shear stress is defined as follows: when the specimen is clockwise rotation, the shear stress is defined as “+.” For the counterclockwise rotation, the shear stress is defined as “−.” Next, the boundary conditions are substituted into the parameters in formulae (14) and (15a)–(15d).(1)
Thez-direction stress of the upper and lower end faces is −σv′:(16a)σzz=−h/2=−σv,(16b)σzz=h/2=−σv.Formulae (16a) and (16b) can be obtained.(17)6A01−μ+4B02−μ+60A21−μ+12B22−μ×h24=−σv.(2)
Resultant force of shear stress is zero of lateral free surface (unit angle):∫−h/2h/2τzρρ=bdz=0, natural meet.(3)
Resultant force of the shear stress of the upper and lower boundary, respectively, isFS and −FS (taking unit angle).Upper boundary:(18)∫0bτρzz=−h/2ρdρ=FS.Reduction to draw:(19)233B2μ−8C21−μhb3=FS.Lower boundary:(20)∫0bτρzz=h/2ρdρ=−FS.This simplifies to the same (3)–(17), (19), (21), and (23).Formulae (3)–((15a)–(15d)) and formulae (3)–(17), (19), (21), and (23) can be calculated:(21)A2=−3−μ10hb3FS,B2=2−μ2hb3FS,C2=−31−μ16hb3FS.(4)
The resultant of normal stress on the free side is zero:(22)∫−h/2h/2σρρ=bdz=0,which reduces to(23)6A0μ+4B0μ−2B0h+5A2μ+B2μ−12h3+6B2μ+4C24μ−3b2h=0.(5)
The resultant moment of the free side is zero:∫−h/2h/2σρρ=bzdz=0, natural meet.According to formulae (17) and (21), the following equation can be obtained:(24)6A01−μ+4B02−μ+3h2b3FS=−σv.According to formulae (21) and (23), the following equation can be obtained:(25)6A0μ+4B0μ−2B0−2+μh4b3FS+33+μ4hbFS=0.From formulae (24) and (25), we can get(26)A0=2−μ2+μh12b3μ+1FS−3+μ2−μ4hbμ+1FS+2μ−1h4b3μ+1FS+2μ−1σv′6μ+1,B0=33+μ1−μ8hbμ+1FS−2+μ1−μh8b3μ+1FS−3μh4b3μ+1FS−μσv′2μ+1.Thus, all undetermined coefficients are determined. The stress function and each stress component are determined accordingly, as shown in formulae (27) and (28), respectively.(27)ϕz,ρ=2−μ2+μh24b3μFS−3+μ2−μ8hbμFS+h4b3FS+σv′6−13+2−μ3μz3+33+μ1−μ16hbμFS−2+μ1−μh16b3μFS−3h8b3FS−σv′4+12−1−μ2μρ2z−3−μFS10hb3z5+2−μFS2hb3ρ2z3−31−μFS16hb3ρ4z.(28)σz=6FShb3z2−3FSh2b3−σv′,σρ=3μ+3FS4hb3ρ2−3μ+2FShb3z2+2+μFSh4b3−33+μFS4hb,σθ=33μ+1FS4hb3ρ2−3μ+2FShb3z2+2+μFSh4b3−33+μFS4hb,τzρ=τρz=−6FShb3ρz.


## 4. Uniaxial Compression Stress Distribution of Cylindrical Rock Specimens

In order to directly reflect the results, the assignment calculation is adopted based on experience and standard. If the standard cylinder specimen is taken and the height-diameter ratio is equal to2:1, then H=4b, that is, the full height of the specimen H=100mm, the radius b=25mm, the modulus of elasticity E=800MPa, uniformly distributed axial load σv=2MPa, Poisson’s ratioμ=0.28, friction coefficient between specimen and end face f=0.21, and maximum frictional force Ff=σvfb2/2. All these can be substituted into the stress component expression (28), which can be further simplified to(29)σz=0.000504z2−3.26,σρ=0.00020664ρ2−0.00057456z2+0.34965,σφ=0.00011592ρ2−0.00057456z2+0.34965,τzρ=−0.000504ρz.Becauseτρφ=τφρ=τzφ=τφz=0, σφ can be thought of as a principal stress, and the stresses on the cross sections orthogonal to the principal plane are independent of the principal stress σφ. Therefore, the three-dimensional problem can be transformed into a plane problem, combined with the calculation formula of plane principal stress:(30)σ1σ3=σz+σρ2±σz−σρ22+τzρ2,and if specific values are substituted, Table 1 can be obtained.Table 1
Stress distribution details of sample internal key points (unit of stress: MPa).
ρzf=0.21,μ=0.28,σv′=2Mpa0b/2b−h/2σz−2−2−2σρ1.0868−1.05450.9576σφ/σ2−1.0868−1.06861.0143τzρ00.3150.63σ1−1.0868−0.9592−0.5284σ3−2−2.0953−2.4276−h/4σz−2.945−2.945−2.945σρ−0.00950.02280.1197σφ/σ2−0.00950.008660.063τzρ00.15750.315σ1−0.00950.03110.1517σ3−2.945−2.9533−2.9770σz−3.26−3.26−3.26σρ0.34970.38190.4788σφ/σ20.34970.36780.4221τzρ000σ10.34970.38190.4788σ3−3.26−3.26−3.26h/4σz−2.945−2.945−2.945σρ−0.00950.02280.1197σφ/σ2−0.00950.008660.063τzρ0−0.1575−0.315σ1−0.00950.03110.1517σ3−2.945−2.9533−2.977h/2σz−2−2−2σρ−1.0868−1.0545−0.9576σφ/σ2−1.0868−1.0686−1.0143τzρ0−0.315−0.63σ1−1.0868−0.9592−0.5284σ3−2−2.0953−2.4276The following conclusions can be drawn from the analysis of formulae (28) and (29):(1)
The stress component at different positions of rock specimen is related to the uniform load applied, the friction coefficient between the specimen and the applied force, and the height-diameter ratio of the specimen. Among them,σρand σφare also related to Poisson's ratio of the specimen. The greater the applied load, the greater the stress component.(2)
The normal stress of rock specimen has a quadratic function relation with the height and diameter of the specimen, and the shear stress has a linear function distribution with the height and diameter of the specimen.By comparing Table1 and above, more accurate conclusions can be drawn:(1)
All normal stress components and principal stress are distributed symmetrically around the surfaceρoφ.(2)
Normal stressσz is compression stress, which is the smallest at both ends of the specimen and the largest in the middle of the specimen, whose value is not related to ρ.(3)
The existence of friction force is like a hoop applied to the specimen. The normal stress presents a compression distribution at both ends fromh/4 to h/2 of the specimen. The closer the same height is to the center, the greater the absolute value of the stress will be. Just as stress accumulates from all sides towards the center, stress is in the tensile state from the middle of h/4 to h/4, and the outward tensile stress at the same height increases.(4)
The absolute value of normal stress is relatively large fromh/2 to ±h/4 at both ends of the specimen, while the stress at ±h/4 is small and almost negligible. But it gets bigger in the opposite direction at h=0.(5)
The tangential stress is distributed in mirror image onρoφ plane, whose absolute value increases with the increase of the absolute value of h andρ. It indicates that the closer the specimen is to the surface, the greater the shear stress will be. The shear stress is zero at z=0 or ρ=0. In summary, due to the existence of friction force, the stress value of the specimen is distributed centrally at both ends, which leads to complicated stress situation at both ends. The direction of stress will also change along the direction of height and radial direction.(6)
By analyzing the maximum principal stress, it can be found that the absolute value of maximum principal stress increases from both ends of the rock mass to the middle, which reaches the maximum value at the middle height of the rock mass, and is equal toσz in value.(7)
The distribution of minimum principal stressσφis similar to the above. It indicates that the direction of principal stress of rock mass is consistent with the direction of coordinate axes at h=0. The place of h=0 and ρ=b is the place where rock mass is most prone to failure.According to Mohr criterion, bidirectional compressive strength > unidirectional compressive strength > bidirectional tensile strength. Therefore, considering the friction, the specimen in the zones of① and ② forms a tight hoop effect, not easy to damage, while the specimen in the ⑤ zone damages first (the radius of the stress circle is the maximum, bidirectional tensile), and the damage extends to the zones of ③ and ④ and forms the conical failure pattern (namely, the conjugate shear failure surface under Mohr–Coulomb criterion), as shown in Figure 3. This is consistent with the common failure patterns of uniaxial compression specimens.Figure 3
Stress distribution of uniaxial compression specimen considering friction.FS=0 when the friction effect at the end is not considered. Formula (28) can be further reduced to(31)σZ=−σv′,σρ=σθ=τzρ=τρz=0.Obviously, each point is a unidirectional stress state, andσz is the main stress. In the transverse direction, due to the Poisson effect, the lateral strain is generated. The strain is the same at the same ρ position. When the strain reaches the maximum of elongation line strain, it will crack along the parallel direction of the main stress. Considering the inevitable nonuniformity of the material, the cracks may be multiple and finally present the splitting failure pattern. In this way, logical and reasonable explanations are given for the causes of the two kinds of failure modes commonly seen in uniaxial compression experiments of rock specimens (see Figure 4).Figure 4
Two common failure modes of rock under uniaxial compression.The compressive strength is always obtained byσc=F/A in standard rock compression experiments, provided that each point inside the specimen is in a unidirectional stress state. In fact, the points in the specimen are not in a unidirectional stress state when the friction effect of the end is considered. Obviously, the unidirectional stress state can only be approximately considered when the specimen is prepared in full compliance with national standards and the end face is treated with special lubrication during loading.



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*Source: 8813115-2021-03-30.xml*