SEILDEL – HABERFIELD’s [1] work was based on the analysis of the regular triangular asperities

and assumed that the asperities were rigid. They used normal stiffness shear stress (CNS) to measure

the joint dilatation rate and extended the LADANYI – ARCHAMBAULT’s [2] approach. The aim of

this research is to investigate the dependence of the constant normal load (CNL) on the rate of the

dilatation. Three points were chosen on the bilinear failure envelope: the first in the first linear part,

the second in transition stress, and the third in the second part of the bilinear failure envelope. 12

regular triangular cementmortar specimens were used to carry out this research. These cementmortar

specimens were investigated by TISA – KOVARI [3] before so all the material and shearing constants

were well-known. The other purpose of this research was to observe the influence of the normal stress

on the dilatation–displacement curves

Vásárhelyi, Balázs

English

Repository of the Academy's Library

PERIODICA POLYTECHNICA SER. CIV. ENG. VOL. 43, NO. 2, PP. 179–186 (1999)SHEAR FAILURE IN ROCK USING DIFFERENT CONSTANTNORMAL LOADBalázs VÁSÁRHELYIDepartment of Engineering GeologyTechnical University of BudapestH–1521 Budapest, HungaryReceived: July 1, 1998AbstractSEILDEL – HABERFIELD’s [1] work was based on the analysis of the regular triangular asperitiesand assumed that the asperities were rigid. They used normal stiffness shear stress (CNS) to measurethe joint dilatation rate and extended the LADANYI – ARCHAMBAULT’s [2] approach. The aim ofthis research is to investigate the dependence of the constant normal load (CNL) on the rate of thedilatation. Three points were chosen on the bilinear failure envelope: the first in the first linear part,the second in transition stress, and the third in the second part of the bilinear failure envelope. 12regular triangular cement mortar specimens were used to carry out this research. These cement mortarspecimens were investigated by TISA – KOVARI [3] before so all the material and shearing constantswere well-known. The other purpose of this research was to observe the influence of the normal stresson the dilatation–displacement curves.Keywords: direct shear test, constant normal load, shear–dilatation and shear–displacement, rockjoints failure envelope.1. IntroductionGenerally rock is not homogeneous and isotropic due to the presence of disconti-nuities. Therefore for practical applications a distinction between the terms ‘rockmaterial’ and ‘rock mass’ has to be made. Rock material refers to the continu-ous body found between joints, faults, etc. and consists of one or more lithologicunits which form a structure framework and may or may not contain joints, faults,etc. The term ‘discontinuity’ refers to a separation surface, i.e. a break in thespatial continuity of a material. In the case of rock mass discontinuities can bepresent in the form of joints, faults, cavities, stratification surfaces and shistositysurfaces. The prediction of the shear strength developed in rock joint undergoingshear displacements is a subject which has brought increased attention in recentyears.Stability analysis in jointed rock is often governed by the stability of a criticaljoint. In such cases, it is assumed that no interaction exists between the rock and thejoint; the joint can therefore be considered isolated. The mechanical characteristicsof the joint plane are necessary for assessing instability. The direct shear test is asuitable experiment to determine these parameters. Apart from a few exceptions, thenormal force has been generally held constant in shear tests reported in the literature.180 B. VÁSÁRHELYIThis experimental boundary condition corresponds in situ to a rock block slide atthe surface (Fig. 1). The block is able to slide freely and the normal force acting onthe joint remains constant during the slide.Fig. 1. Comparing the rock joint strength envelopeWhen non-planar surfaces are in sliding contact, they tend to dilate if theboundary conditions permit such deformation. The amount of dilatation dependson the geometry of the contacting surfaces, normal stiffness of the joint, deformabil-ity of the surrounding media. Dilatation causes enhanced shear strength due to theincreased normal stress. Proper modelling of the shear behaviour of rock disconti-nuities must take into account the conditions imposed by the rock mass rigidity. Inthis situation, the normal stress path should be properly understood for the accurateprediction of shear strength. Such behaviour is more realistically represented inthe controlled normal stiffness shear tests (CNS) rather than in the conventionalconstant normal load (CNL) tests.2. Theoretical BackgroundPATTON [4] performed a series of constant load stress direct shear tests with regularteeth inclination, i , at varying normal stresses. From these tests he established abilinear failure envelope – failure from an asperity sliding and asperity shearingmode. The equations for the two portions of the failure envelope are given below:τ = σn tan(φµ + i), (1)τ = c j + σn tan φµ, (2)SHEAR FAILURE 181wherec j = cohesion;φµ = basic friction angle;i = regular teeth inclination.LADANYI – ARCHAMBAULT [2] provided an extension to Paton’s model toaccount for the sliding and shearing mechanisms found in natural rock joints. Theyconsidered the shear resistance of joints comprising regular triangular asperitieswith asperity angles of i . They suggested that the joint strength of natural jointscould be determined from the applied normal stress, the basic angle of rock andthe joint dilatation rate. They used the energy principles and rewrote the Patton’sequation in the asperity sliding mode asτ = σn tan(φµ + ν˙), (3)whereφµ = basic friction angle;ν˙ = the rate of dilatation at failure.SEIDEL – HABERFIELD [1] analyzed using extensions of LADANYI – AR-CHAMBAULT’s [2] approach. They suggested that the shear resistance of multi-asperity profiles in degrading rocks with varying asperity angles should be com-puted by combining the individual shear resistance of each individual asperity faceas follows:τ = σn tan[tan i + tan φµ1− tan φµ], (4)whereφµ = basic friction angle;ν˙ = the rate of dilatation at failure,i = regular teeth inclination.According to (4) the effective friction angle is given:φdeg rade = tan i + tan φµ1− tan φµ . (5)Even though the bilinear model is easier to use, the rock joint strength is bettercharacterized by a curve than by a bilinear envelope. JAEGER’S [5] equation hasbeen proposed for this conceptually simple case, and the contributions to jointstrength from different factors can be separated easily. Jaeger’s basic equation isshown with Patton’s equation in Fig. 2 and is given by:τ = c j[1− ebσn/qu]+ σn tan φµ, (6)wherec j = cohesion;b = empirical constant;σn = normal stress;qu = uniaxial compressive strength;φµ = basic friction angle.182 B. VÁSÁRHELYIFig. 2. Schematical view of the shear apparatus used in combination with a loading machine3. Test EquipmentThe CNL (constant normal load) equipment was designed and built in the RockMechanics Laboratory of the Swiss Federal Institute of Technology, Zurich [3]. Totransmit force to the specimen it was cast in epoxy resin in the usual manner. Theshear box fits exactly between two massive L-shaped steel pieces (Fig. 3). Thenormal force (N ) and shear force (S) were applied through one of these L-shapedsteel pieces. The other L-shaped steel piece provided the reactions, which were thelower loading plates of the servo-controlled press. This was achieved by resting theL-shaped steel piece against the front side of the frame as well as on a cylindricalbearing box with rests on the lower loading plate.Fig. 3. A typical stress–strain and dilatation–displacement curveThe shear apparatus is designed in such a way that both parts of the testspecimen can be displaced and rotated with respect to each other without constraints.This is made possible by load transfer through the cross roll and the spherical jointarrangement. The center of the sphere is aligned with the center of the joint surfaceat the start of the test.SHEAR FAILURE 183A fairly long hydraulic jack was chosen in order to minimize the componentof the jack force due to its rotation under the action of the shear deformations.The hinged load cell placed between the sack and the specimen was selected onaccount of its high accuracy, simple calibration and its direct electrical signal. Themeasuring system consists of load cells (S and N ) and 6 displacement transducers.The measurement of the relative displacement of the joint surfaces in the directionsof N and S is carried out directly at the joint itself, on both sides of test specimen.The maximum normal stress and shear stress which can be applied to a joint surfaceof the size of 24×12 cm are 20 N/mm2 and 100 N/mm2, respectively.4. Material DescriptionThe interpretation of the results of shear tests on real rocks is usually complicatedby sample variability. To overcome this difficulty the laboratory specimens weremade from an artificial material so that the shape, size and internal strength ofthe irregularities or ‘teeth’ on the surface of the test specimen could be evaluatedseparately.Cement mortar was selected as test material as it has rock-like properties.Table 1 presents the physical properties of the cement mortar. Basic shearingconstants were determined in the previous researches of TISA – KOVARI [3].Table 1. Physico-mechanical properties of the cement mortarPhysico-mechanical properties Index Unit ValueModulus of elasticity E GPa 17.5Poisson’s ratio ν – 0.1Maximum stress σ MPa 60Tensile stress σt MPa 6.5Basic friction angle φµ degree 33.64Asperity sliding angle α degree 59Cohesion c j N/mm2 2.13Transition stress σT N/mm2 2.1Density γ g/cm3 2.2All specimens were 150 mm wide and 140 mm long with four regular teeth wherethe distance between the teeth was kept constant at 20 mm. The teeth were 5 mmhigh, 15 mm thick at the bottom and the inclination angle of the teeth was 26.57◦.184 B. VÁSÁRHELYI5. Results and DiscussionUsing cement mortar joints, 12 tests were carried out with 3 different normal stressvalues: 1.05; 2.10 and 3.15 N/mm2. These values were chosen because the aimof this research was to investigate the influence of the normal load in three points:in the first linear part, in the transition stress and in the second part of the bilinearfailure envelope. The measured maximum and residual stresses were in the knownlines. The measured results with the variances are summarized in Table 2.Table 2. Measured results with variances in parenthesesNormal Peak shear Res. shearforce σn stress τp stress τr tan−1(τp/σn)tan−1 (τr/σn)N/mm2 N/mm2 N/mm21.05 1.71 (0.19) 0.72 (0.10) 58.45 (2.88) 34.43 (3.14)2.1 2.99 (0.28) 1.29 (0.07) 54.91 (2.55) 31.71 (1.46)3.15 3.19 (0.27) 2.03 (0.09) 45.36 (2.31) 32.74 (1.19)Table 3 compares the effective friction angles calculated by different methodsaccording to Eq. (4) [1], Eq. (3) [2] (the variance in the bracket). The effectivefriction angle was 60.21◦ according to Patton’s Eq. (1).Table 3. The measured and calculated results with the different theoriesNormal force σn Measured Eq. (5) Eq. (3)N/mm2 dilatation angle ν˙ φdegrade φµ + ν˙1.05 24.76 (1.19) 59.07 (0.82) 58.40 (1.19)2.10 22.42 (1.18) 58.11 (0.57) 56.06 (1.18)3.15 12.48 (2.44) 53.83 (0.95) 46.12 (2.44)Comparing the results of the CNS tests, in the case of Eq. (3) all three pointsare closer to the original than the others. This Eq. (3) is correct where the otherequations fail. This means that this is a more general equation for the measureddilatation–displacement curve and it should be valid until the ‘teeth’ or irregularitiesare shorn off or till the deformation of the teeth can be measured. This is possiblythe point where the Jaeger’s joint failure envelope curve (Eq. (5)) ‘touches’ thePatton’s linear curve.HABERFIELD – JOHNSTONE [6] idealized the shear behaviour of concrete–rock joints under CNS conditions. The other aim was to study the character ofthe dilatation–displacement curve under CNL conditions. The rate of dilatation isnot the maximum at peak shear response (because of the sliding) and continues toSHEAR FAILURE 185increase at a decreased rate until the minimum shear strength is reached. Fig. 4shows a typical shear stress vs. displacement and dilatation vs. displacementrelation of a typical specimen. All the dilatation–displacement curves can be dividedinto three parts: in the first part (I) during the increasing strain-stress the angle ofthe teeth is constant. This angle is not changing in the second part (II) where thestress is decreasing. In the third part (III) during the decreasing shear-stress thechanging of the angle is successive and it becomes linear in the residual part of theshear-stress. The maximum dilatation is, of course, decreasing with the increasednormal load. In this case the average maximum dilatation was 1.10 mm at the1.05 N/mm2 normal load, 0.67 mm at 2.1 N/mm2 and 0.37 mm at the 3.15 N/mm2applied normal load.Fig. 4.6. ConclusionThe results show that the measured dilatation angle (ν˙) decreases with the increasednormal force and it is always present. However, Eq. (3) is also correct for the caseswhen the Patton’s and the Seidel and Haberfield equations fail. This means that this186 B. VÁSÁRHELYIis a more general equation and it should be valid until the ‘teeth’ (or irregularities)are not shorn off. This point is not at the transition stress, but in the meeting pointof the Jaeger’s curve and the bilinear curve. The measured dilatation–displacementcurves show that after the peak stress the rate of dilatation does not change for along time.AcknowledgementsThe author acknowledges the help of both Prof. K. Kovari and A. Tisa for theopportunity to carry out this research in the Rock Mechanics Laboratory of the SwissFederal Institute of Technology and thanks for the grant OTKA F02556.References[1] SEIDEL, J. P. – HABERFIELD, C. M. (1995): The Application of Energy Principles to theDetermination of the Sliding Resistance of Rock Joints. Rock Mech. Rock Engng., Vol. 28/4,pp. 211–226.[2] LADANYI, B. – ARCHAMBAULT, G. (1970): Simulation of Shear Behavior of Jointed RockMass. Proc. 11th Symp. on Rock Mechanics: Theory and Practice, AIME, New York, pp. 105–125.[3] TISA, A. – KOVARI, K. (1984): Continuous Failure State Direct Shear Tests. Rock Mech. RockEngng., Vol. 17/2, pp. 83–95.[4] PATTON, F. D. (1966): Multiple Modes of Shear Failure in Rock. 1st Congress of Int. Soc. RockMech., Lisbon, Vol. 1, pp. 509–513.[5] JAEGER, J. C. (1971): Friction of Rocks and Stability of Rock Slopes. Geotechnique, Vol. 21/2,pp. 97–134.[6] HABERFIELD, C. M. – JOHNSTONE, I. W. (1994): A Mechanistically-based Model for RoughRock Joints. Int. J. Rock Mech. Min. Sci. & Geomech. Abstr. Vol. 31/4, pp. 279–292.

A Mechanistically-based Model for Rough Rock Joints.

Continuous Failure State Direct Shear Tests. Rock Mech.

Friction of Rocks and Stability of Rock Slopes.

Multiple Modes of Shear Failure

Simulation of Shear Behavior of Jointed Rock Mass.

The Application of Energy Principles to the Determination of the Sliding Resistance of Rock Joints. Rock Mech.

Shear failure in rock using different constant normal load

\sc Seildel - Haberfield´s \cite1 work was based on the analysis of
the regular
triangular asperities and assumed that the asperities were rigid. They used
normal
stiffness shear stress (CNS) to measure the joint dilatation rate and
extended the \sc
Ladanyi - Archambault´s \cite2 approach. The aim of this research is to
investigate the
dependence of the constant normal load (CNL) on the rate of the dilatation.
Three points
were chosen on the bilinear failure envelope: the first in the first linear
part, the second in
transition stress, and the third in the second part of the bilinear failure
envelope. 12
regular triangular cement mortar specimens were used to carry out this
research. These
cement mortar specimens were investigated by \sc Tisa - Kovari \cite3
before so all the material and shearing constants were well-known. The
other
purpose of this research was to observe the influence of the normal stress
on the dilatation-displacement curves

Periodica Polytechnica (Budapest University of Technology and Economics)

PERIODICA POLYTECHNICA SER. CIV. ENG. VOL. 43, NO. 2, PP. 179–186(1999)SHEAR FAILURE IN ROCK USING DIFFERENT CONSTANTNORMAL LOADBalázs VÁSÁRHELYIDepartment of Engineering GeologyTechnical University of BudapestH–1521 Budapest, HungaryReceived: July 1, 1998AbstractSEILDEL – HABERFIELD’s [1] work was based on the analysis of the regular triangular asperitiesand assumed that the asperities were rigid. They used normal stiffness shear stress (CNS) to measurethe joint dilatation rate and extended the LADANYI – ARCHAMBAULT ’s [2] approach. The aim ofthis research is to investigate the dependence of the constant normal load (CNL) on the rate of thedilatation. Three points were chosen on the bilinear failure envelope: the first in the first linear part,the second in transition stress, and the third in the second part of the bilinear failure envelope. 12regular triangular cement mortar specimens were used to carry out this research. These cement mortarspecimens were investigated by TISA – KOVARI [3] before so all the material and shearing constantswere well-known. The other purpose of this research was to observe the influence of the normal stresson the dilatation–displacement curves.Keywords: direct shear test, constant normal load, shear–dilatation and shear–displacement, rockjoints failure envelope.1. IntroductionGenerally rock is not homogeneous and isotropic due to the presence of disconti-nuities. Therefore for practical applications a distinction between the terms ‘rockmaterial’ and ‘rock mass’ has to be made. Rock material refers to the continu-ous body found between joints, faults, etc. and consists of one or more lithologicunits which form a structure framework and may or may not contain joints, faults,etc. The term ‘discontinuity’ refers to a separation surface, i.e. a break in thespatial continuity of a material. In the case of rock mass discontinuities can bepresent in the form of joints, faults, cavities, stratification surfaces and shistositysurfaces. The prediction of the shear strength developed in rock joint undergoingshear displacements is a subject which has brought increased attention in recentyears.Stability analysis in jointed rock is often governed by the stability of a criticaljoint. In such cases, it is assumed that no interaction exists between the rock and thejoint; the joint can therefore be considered isolated. The mechanical characteristicsof the joint plane are necessary for assessing instability. The direct shear test is asuitable experiment to determine these parameters. Apart from a few exceptions, thenormal force has been generally held constant in shear tests reported in the literature.180 B. VÁSÁRHELYIThis experimental boundary condition correspondsin itu to a rock block slide atthe surface (Fig. 1). The block is able to slide freely and the normal force acting onthe joint remains constant during the slide.Fig. 1. Comparing the rock joint strength envelopeWhen non-planar surfaces are in sliding contact, they tend to dilate if theboundary conditions permit such deformation. The amount of dilatation dependson the geometry of the contacting surfaces, normal stiffness of the joint, deformabil-ity of the surrounding media. Dilatation causes enhanced shear strength due to theincreased normal stress. Proper modelling of the shear behaviour of rock disconti-nuities must take into account the conditions imposed by the rock mass rigidity. Inthis situation, the normal stress path should be properly understood for the accurateprediction of shear strength. Such behaviour is more realistically represented inthe controlled normal stiffness shear tests (CNS) rather than in the conventionalconstant normal load (CNL) tests.2. Theoretical BackgroundPATTON [4] performed a series of constant load stress direct shear tests with regularteeth inclination,i , at varying normal stresses. From these tests he established abilinear failure envelope – failure from an asperity sliding and asperity shearingmode. The equations for the two portions of the failure envelope are given below:τ = σn tan(φµ + i), (1)τ = c j + σn tanφµ, (2)SHEAR FAILURE 181wherec j = cohesion;φµ = basic friction angle;i = regular teeth inclination.LADANYI – ARCHAMBAULT [2] provided an extension to Paton’s model toaccount for the sliding and shearing mechanisms found in natural rock joints. Theyconsidered the shear resistance of joints comprising regular triangular asperitieswith asperity angles ofi . They suggested that the joint strength of natural jointscould be determined from the applied normal stress, the basic angle of rock andthe joint dilatation rate. They used the energy principles and rewrote the Patton’sequation in the asperity sliding mode asτ = σn tan(φµ + ν̇), (3)whereφµ = basic friction angle;ν̇ = the rate of dilatation at failure.SEIDEL – HABERFIELD [1] analyzed using extensions of LADANYI – AR-CHAMBAULT ’s [2] approach. They suggested that the shear resistance of multi-asperity profiles in degrading rocks with varying asperity angles should be com-puted by combining the individual shear resistance of each individual asperity faceas follows:τ = σn tan[tani + tanφµ1 − tanφµ], (4)whereφµ = basic friction angle;ν̇ = the rate of dilatation at failure,i = regular teeth inclination.According to (4) the effective friction angle is given:φdeg rade= tani + tanφµ1 − tanφµ . (5)Even though the bilinear model is easier to use, the rock joint strength is bettercharacterized by a curve than by a bilinear envelope. JAEGER’ S [5] equation hasbeen proposed for this conceptually simple case, and the contributions to jointstrength from different factors can be separated easily. Jaeger’s basic equation isshown with Patton’s equation inFig. 2 and is given by:τ = c j[1 − ebσn/qu] + σn tanφµ, (6)wherec j = cohesion;b = empirical constant;σn = normal stress;qu = uniaxial compressive strength;φµ = basic friction angle.182 B. VÁSÁRHELYIFig. 2. Schematical view of the shear apparatus used in combination with a loading machine3. Test EquipmentThe CNL (constant normal load) equipment was designed and built in the RockMechanics Laboratory of the Swiss Federal Institute of Technology, Zurich [3]. Totransmit force to the specimen it was cast in epoxy resin in the usual manner. Theshear box fits exactly between two massive L-shaped steel pieces (Fig. 3). Thenormal force (N ) and shear force (S) were applied through one of these L-shapedsteel pieces. The other L-shaped steel piece provided the reactions, which were thelower loading plates of the servo-controlled press. This was achieved by resting theL-shaped steel piece against the front side of the frame as well as on a cylindricalbearing box with rests on the lower loading plate.Fig. 3. A typical stress–strain and dilatation–displacement curveThe shear apparatus is designed in such a way that both parts of the testspecimen can be displaced and rotated with respect to each other without constraints.This is made possible by load transfer through the cross roll and the spherical jointarrangement. The center of the sphere is aligned with the center of the joint surfaceat the start of the test.SHEAR FAILURE 183A fairly long hydraulic jack was chosen in order to minimize the componentof the jack force due to its rotation under the action of the shear deformations.The hinged load cell placed between the sack and the specimen was selected onaccount of its high accuracy, simple calibration and its direct electrical signal. Themeasuring system consists of load cells (S andN ) and 6 displacement transducers.The measurement of the relative displacement of the joint surfaces in the directionsof N andS is carried out directly at the joint itself, on both sides of test specimen.The maximum normal stress and shear stress which can be applied to a joint surfaceof the size of 24×12 cm are 20 N/mm2 and 100 N/mm2, respectively.4. Material DescriptionThe interpretation of the results of shear tests on real rocks is usually complicatedby sample variability. To overcome this difficulty the laboratory specimens weremade from an artificial material so that the shape, size and internal strength ofthe irregularities or ‘teeth’ on the surface of the test specimen could be evaluatedseparately.Cement mortar was selected as test material as it has rock-like properties.Table 1 presents the physical properties of the cement mortar. Basic shearingconstants were determined in the previous researches of TISA – KOVARI [3].Table 1. Physico-mechanical properties of the cement mortarPhysico-mechanical propertiesIndex Unit ValueModulus of elasticity E GPa 17.5Poisson’s ratio ν – 0.1Maximum stress σ MPa 60Tensile stress σt MPa 6.5Basic friction angle φµ degree 33.64Asperity sliding angle α degree 59Cohesion c j N/mm2 2.13Transition stress σT N/mm2 2.1Density γ g/cm3 2.2All specimens were 150 mm wide and 140 mm long with four regular teeth wherethe distance between the teeth was kept constant at 20 mm. The teeth were 5 mmhigh, 15 mm thick at the bottom and the inclination angle of the teeth was 26.57◦.184 B. VÁSÁRHELYI5. Results and DiscussionUsing cement mortar joints, 12 tests were carried out with 3 different normal stressvalues: 1.05; 2.10 and 3.15 N/mm2. These values were chosen because the aimof this research was to investigate the influence of the normal load in three points:in the first linear part, in the transition stress and in the second part of the bilinearfailure envelope. The measured maximum and residual stresses were in the knownlines. The measured results with the variances are summarized inTable 2.Table 2. Measured results with variances in parenthesesNormal Peak shear Res. shearforceσn stressτp stressτr tan−1(τp/σn)tan−1 (τr/σn)N/mm2 N/mm2 N/mm21.05 1.71 (0.19) 0.72 (0.10) 58.45 (2.88) 34.43 (3.14)2.1 2.99 (0.28) 1.29 (0.07) 54.91 (2.55) 31.71 (1.46)3.15 3.19 (0.27) 2.03 (0.09) 45.36 (2.31) 32.74 (1.19)Table 3 compares the effective friction angles calculated by different methodsaccording toEq. (4) [1], Eq. (3) [2] (the variance in the bracket). The effectivefriction angle was 60.21◦ according to Patton’sEq. (1).Table 3. The measured and calculated results with the different theoriesNormal forceσn Measured Eq. (5) Eq. (3)N/mm2 dilatation anglėν φdegrade φµ + ν̇1.05 24.76 (1.19) 59.07 (0.82) 58.40 (1.19)2.10 22.42 (1.18) 58.11 (0.57) 56.06 (1.18)3.15 12.48 (2.44) 53.83 (0.95) 46.12 (2.44)Comparing the results of the CNS tests, in the case ofEq. (3) all three pointsare closer to the original than the others. ThisEq. (3) is correct where the otherequations fail. This means that this is a more general equation for the measureddilatation–displacement curve and it should be valid until the ‘teeth’ or irregularitiesare shorn off or till the deformation of the teeth can be measured. This is possiblythe point where the Jaeger’s joint failure envelope curve (Eq. 5)) ‘touches’ thePatton’s linear curve.HABERFIELD – JOHNSTONE[6] idealized the shear behaviour of concrete–rock joints under CNS conditions. The other aim was to study the character ofthe dilatation–displacement curve under CNL conditions. The rate of dilatation isnot the maximum at peak shear response (because of the sliding) and continues toSHEAR FAILURE 185increase at a decreased rate until the minimum shear strength is reached.Fig. 4shows a typical shear stress vs. displacement and dilatation vs. displacementrelationof a typical specimen. All the dilatation–displacement curves can be dividedinto three parts: in the first part (I) during the increasing strain-stress the angle ofthe teeth is constant. This angle is not changing in the second part (II) where thestress is decreasing. In the third part (III) during the decreasing shear-stress thechanging of the angle is successive and it becomes linear in the residual part of theshear-stress. The maximum dilatation is, of course, decreasing with the increasednormal load. In this case the average maximum dilatation was 1.10 mm at the1.05 N/mm2 normal load, 0.67 mm at 2.1 N/mm2 and 0.37 mm at the 3.15 N/mm2applied normal load.Fig. 4.6. ConclusionThe results show that the measured dilatation angle (ν̇) decreases with the increasednormal force and it is always present. However,Eq. (3) is also correct for the caseswhen the Patton’s and the Seidel and Haberfield equations fail. This means that this186 B. VÁSÁRHELYIis a more general equation and it should be valid until the ‘teeth’ (or irregularities)are not shorn off. This point is not at the transition stress, but in the meeting pointof the Jaeger’s curve and the bilinear curve. The measured dilatation–displacementcurves show that after the peak stress the rate of dilatation does not change for along time.AcknowledgementsThe author acknowledges the help of both Prof. K. Kovari and A. Tisa for theopportunity to carry out this research in the Rock Mechanics Laboratory of the SwissFederal Institute of Technology and thanks for the grant OTKA F02556.References[1] SEIDEL, J. P. – HABERFIELD, C. M. (1995): The Application of Energy Principles to theDetermination of the Sliding Resistance of Rock Joints.Rock Mech. Rock Engng., Vol. 28/4,pp. 211–226.[2] L ADANYI , B. – ARCHAMBAULT, G. (1970): Simulation of Shear Behavior of Jointed RockMass.Proc. 11th Symp. on Rock Mechanics: Theory and Practice, AIME, New York, pp. 105–125.[3] T ISA, A. – KOVARI, K. (1984): Continuous Failure State Direct Shear Tests.Rock Mech. RockEngng., Vol. 17/2, pp. 83–95.[4] PATTON, F. D. (1966): Multiple Modes of Shear Failure in Rock.1st Congress of Int. Soc. RockMech., Lisbon, Vol. 1, pp. 509–513.[5] JAEGER, J. C. (1971): Friction of Rocks and Stability of Rock Slopes.Geotechnique, Vol. 21/2,pp. 97–134.[6] HABERFIELD, C. M. – JOHNSTONE, I. W. (1994): A Mechanistically-based Model for RoughRock Joints.Int. J. Rock Mech. Min. Sci. & Geomech. Abstr. Vol. 31/4, pp. 279–292.

Shear Failure in Rock Using Different Constant Normal Load

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Shear failure in rock using different constant normal load

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