MINE 3514—Rock Mechanics Lab

Comparison of Strength and Failure Criterion Between Limestone and Granite

By: Daniel Delgado, Michael Devlin, John Herrin, and Sam Mast

November 10, 2009

Abstract

The purpose of this lab report is to find and compare the failure criterions and strength

characteristics of the Clark and Wilma rock samples. This was done experimentally using three

different tests; the uniaxial compressive test, the triaxial compressive test, and the indirect

shear test. The findings of this report are that Clark has a higher initial shear strength and

internal angle of friction while intact and jointed. This means Clark is stronger than Wilma in

either intact or jointed rock masses. This makes Clark the superior rock for structural

applications.

Summary of Results

Uniaxial Compression TestMethod Wilma ClarkCo (psi) 10040 32220

Failure Criterion |τ|= 2588 + σn*tan(37.33) |τ|= 4017 + σn*tan(51)

Triaxial Compression TestMethod Wilma Clarke

|τ| = 178.3 + σn*tan(59.3) |τ| = 1799 + σn*tan(64.1)p,q |τ| = 310 + σn*tan(69.6) |τ| = 2326 + σn*tan(73.5)

|τ| = 292 + σn*tan(60.6) |τ| = 2339.1 + σn*tan(59.0)

Indirect Shear TestMethod Wilma Clarke

σn, τ

σ1,σ3

Table of Contents

List of Tables…………………………………………………………………………………...2 List of Figures…………………………………………………………………………………..3

Introduction.…………………………………………………………………………………….4

Sample Preparation……………………………………………………………………………..6

Uniaxial Compressive Test……………………………………………………………………..7Theory………………………………………………………………….……………….7Experimental Apparatus and Procedures…………………………….…………………8Tabulation of Data…………………………………………………….………………..9Discussion of Results…………………………………………………………………..11

Triaxial Compressive Test…………………………………………………………………….14Theory………………………………………………………………………………...14Experimental Apparatus and Procedures……………………………………………. 16Tabulation of Data…………………………………………………………………….17Discussion of Results………………………………………………………………….21

Indirect Shear……….....……………………………………………………………………....24Theory………………………………………………………………………………....24Experimental Apparatus and Procedures……………………………………………....25Tabulation of Data……………………………………………………………………..26Discussion of Results…………………………………………………………………..29

Conclusion and Recommendations…………………………………………………………….20

References ……………………………………………………………………………………..32

Appendix………………………………………………………………………………………A-1Sample Calculations……….…………………………………………………………...A-1Raw Data …………….………………………………………….……………………..A-3

List of Tables

Table A: Results for Wilma from uniaxial compressive strength test……………………….….10

Table B: Results for Clark for uniaxial compressive strength test………………………….......10

Table C: Results for both Wilma and Clark rocks for compressive strength test……………....11

Table D: Results for Wilma rock from triaxial test………………….……………………….....19

Table E: Results for Clark rock from triaxial test.....…………………………………………....20

Table F: Results for indirect shear test for Clark rock………...…………………..…………….26

Table G: Results for indirect shear test for Wilma rock…....…………………………………...27

Table H: Mohr-Coulomb failure criterion determined by indirect shear test..…………………29

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List of Figures

Figure 1: Representative drawing of samples for indirect shear test………..……………………………6

Figure 2: Graph showing peak load versus displacement……………………………………..……….....9

Figure 3: Graph of stress versus strain curve with moduli…………………………..................................12

Figure 4: MTS with triaxial chamber prepared for testing…………………………..................................16

Figure 5: Failed sample with failure plane outlined……………………………………………………....17

Figure 6: P-Q diagram for Wilma samples………………………………………………………………..21

Figure 7: P-Q diagram for Clark samples………………………………………………………………....21

Figure 8: Graph of tau versus sigma normal for Wilma…………………………………………..………21

Figure 9: Graph of shear versus normal stress for the Clark rock from triaxial test……………...………22

Figure 10: Graph of principal stresses sigma 1 versus sigma 3 for Wilma from triaxial test…………….22

Figure 11: Graph of principal stresses sigma 1 versus sigma 3 for Clark from triaxial test……………...22

Figure 12: Graph of shear stress versus normal stress for Clark for indirect shear test…………………..28

Figure 13: Graph of shear stress versus normal stress for Wilma for indirect shear test ………………...28

3

Introduction

One of the most important properties of rocks is the compressive strength. Almost all

stresses on rock masses are compressive stresses. For most methods of comminution the

compressive strength is what determines the amount of energy required to break the rocks.

Most in situ stresses act in compression on pillars and openings as well so the compressive

strength will determine the factor of safety for pillars.

The triaxial compressive test and the indirect shear test can both be used to determine

failure criterion for a rock mass. The Mohr-Coulomb failure criterion relates the normal stress

to the shear strength of the rock. Three different methods for calculating the Mohr-Coulomb

failure criterion are going to be used and compared to each other. A p-q diagram, the principle

stresses, and a graph of the normal stress versus shear on the failure plane will each be used to

determine the failure criterion.

This report is divided into three separate tests. The theory behind each test, the

procedure, results, and a discussion of results will be included for each test. The results are

displayed in tabular format. Significant results and recommendations are included in the

conclusions section.

This report shows a comparison between two rock types, a limestone (Clark) and a

granitic rock (Wilma). The purpose of these experiments is to determine several failure

criterion for the rock masses. Three different tests are performed to determine the failure

criterion, the Uniaxial compressive test, the triaxial compressive test, and the indirect shear

test. The triaxial test and indirect shear test are used to determine failure criterion for the

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rocks and the uniaxial compressive test is used to determine compressive strength of the rocks

and the elastic modulus for the rocks.

5

Sample Preperation

Samples for this lab used in the triaxial compression test and the uniaxial compression

test were four inch long samples prepared in the

same way as the four inch samples from the

previous lab. Samples for the indirect shear were cut

at a β of 58 degrees. One sample with a natural

foliation of β = 57 degrees was also included. The

sample was then taped together so the cut planes

were in contact.

6

Figure 1: Representative drawing of samples for indirect shear test

Uniaxial Compressive Test

Theory

Arguably the most important property of rocks for mining engineers is compressive

strength. Compressive strength is very important in determining the factor of safety of pillars in

a mine. The compressive strength also plays an important role in the comminution and blasting

of the rock. The compressive strength is determined by finding the peak stress on the sample

using the equation:

C0=PA

where P is the peak load on the sample, A is the cross sectional area of the sample, and C0 is

the compressive strength of the sample.

The strain of the samples can be determined using the formula:

ε= ΔLL

where ΔL was the deflection and L was the length of the sample. This can be used to determine

the elastic moduli of the samples later.

The Moduli for rock can be determined several ways. The elastic modulus can be

determined using the equation:

E=σε

where σ is the change in stress on the sample and ε is the change in strain of the sample while

the rock deforms elastically. Es, the Secant Modulus, can be found from a graph of stress versus

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strain over the entirety of the stress strain curve. It is calculated by determining the slope of a

line from the start of the graph of stress versus strain to the fracture point.

The angle of internal friction, Ф, is related to the angle of failure, θ, and is found using

the formula:

Ф=45+ θ2

The cohesion of the core sample, Si, is related to the compressive stress at failure and

the angle of failure by the equation:

C0=2∗Si∗tan (θ )

where C0 is the compressive strength of the sample.

Experimental Apparatus and Procedure

To perform the uniaxial compressive strength test, an MTS (Material Test System) was

used. An MTS consists of a hydraulic piston that can apply a maximum load of 1,000,000

pounds. For the uniaxial compressive test, the MTS was set to have a maximum load of

200,000 pounds of force. The piston is controlled by a computer which allows the tester to

monitor the load on the sample and the amount of deformation the sample is experiencing.

The samples were placed lengthwise beneath the piston with a metal disk above and below the

sample to provide uniform contact.

8

The test was performed using a displacement control of 0.00012 inches per second. The

piston was gradually loaded until the sample failed. The piston was reset and the shield and

sample removed. The peak load was then recorded for the sample. After the sample failed, the

pieces of the sample intact enough for analysis were then used to determine an angle of failure

by taking a protractor and determining the angle between the failure plane and the bottom of

the sample. This process was then repeated for

all samples. Some samples had no discernable

angle of failure because of the way in which the

sample broke so no angle was recorded.

During each loading, the computer

produced a graph of the load on the sample

versus the displacement of the piston, as shown

in Figure 1, which was later used to determine

the elastic and secant moduli for the sample.

Tabulation of Data

Ten samples of both the Wilma and Clark rocks were tested. The results of the tests are

in the tables below. The samples were all roughly four inches in length with diameters of about

1.87 inches. The first table includes the data for the Wilma rocks, the second table includes the

data for the Clark rocks, and the third table includes data for both the Wilma and Clark rocks.

9

Figure 2: Graph showing peak load versus displacement

Table A: Results for Wilma from uniaxial compressive strength test

Specimen Diameter (inch) P (lbs) Ɵ Area (inch^2) Ɵ in RadiansW-4-20 1.873 3.922 24669 60 2.755 8953 0.04809W-5-6 1.871 4.008 47733 68 2.749 17361 0.04799

W-4-22 1.871 3.926 31857 61 2.749 11587 0.04799W-4-23 1.884 3.990 36705 68 2.788 13167 0.04866W-4-7 1.872 3.997 21738 61 2.752 7898 0.04804W-4-4 1.867 3.961 25000 62 2.738 9132 0.04778W-4-5 1.872 4.057 12361 71 2.752 4491 0.04804W-4-6 1.869 3.969 19000 -------- 2.744 6925 --------

W-4-17 1.871 3.930 20121 63 2.749 7318 0.04799W-4-9 1.873 4.061 37355 59 2.755 13558 0.04809

Average 1.872 3.982 27653.9 63.6667 2.753 10039 0.04807Std. Dev. 0.0042673177 0.047834 10066.42 0.0125761693 3641.909732

Length (inch)

Compressive Strength (Psi)

Table B: Results for Clark from compressive strength test

Specimen P (lbs) Ɵ Ɵ in RadiansC-4-18 1.858 3.962 77599 -------- 2.711 28620 --------C-4-13 1.873 3.930 99322 -------- 2.755 36048 --------C-4-23 1.870 3.901 58390 -------- 2.746 21260 --------C-4-4 1.887 3.977 65480 68 2.797 23414 0.04881C-4-9 1.873 3.971 151471 -------- 2.755 54975 --------C-4-5 1.870 3.853 163656 -------- 2.746 59588 --------

C-4-22 1.870 3.938 59360 73 2.746 21613 0.04793C-4-15 1.862 3.948 84810 -------- 2.723 31146 --------C-4-3 1.868 4.017 87690 -------- 2.741 31997 --------C-4-2 1.867 3.938 37059 -------- 2.738 13537 --------

Average 1.870 3.944 88483.7 2.746 32220 0.048372Std. Dev. 0.0073 0.0423 38465 0.02137 13982

Diameter (inch)

Length (inch)

Area (inch^2)

Compressive Strength (Psi)

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Table C: Results for both Wilma and Clark rock for compressive strength test

Specimen Strain SiW-4-20 0.02363 0.006024 2.40E+06 2585 1.047 1.51E+06W-5-6 0.02233 0.005572 4271214 3507 1.187 3.10E+06

W-4-22 0.03359 0.008556 2631050 3211 1.065 1.31E+06W-4-23 0.03531 0.008849 2730652 2660 1.187 1.47E+06W-4-7 0.03756 0.009396 1962393 2189 1.065 9.64E+05W-4-4 0.04728 0.011936 1479130 2428 1.082 7.40E+05W-4-5 0.02628 0.006478 1188023 773 1.239 5.25E+05W-4-6 0.02317 0.005837 1750078 -------- -------- 1.17E+06

W-4-17 0.02434 0.006194 1713800 1864 1.100 1.14E+06W-4-9 0.02387 0.005878 3577961 4073 1.030 2.31E+06

Average 0.02974 0.00747 2370626 -------- -------- 1.42E+06Std. Dev. 0.00791 0.00201 918955 2602 1.056 724521C-4-18 0.02614 0.006598 5.61E+06 -------- -------- 4.33E+06C-4-13 0.02567 0.006531 6522442 -------- -------- 5.41E+06C-4-23 0.01656 0.004246 3513877 -------- -------- 3.02E+06C-4-4 0.03130 0.00787 5535788 4730 1.187 2.97E+06C-4-9 0.04126 0.010389 6052689 -------- -------- 5.29E+06C-4-5 0.05089 0.013208 5410970 -------- -------- 4.52E+06

C-4-22 0.08436 0.021422 2325990 3304 1.274 9.93E+05C-4-15 0.03947 0.009996 3849768 -------- -------- 3.06E+06C-4-3 0.02711 0.00675 5672388 -------- -------- 4.60E+06C-4-2 0.02749 0.006981 3931942 -------- -------- 1.91E+06

Average 0.03702 0.009399 4842172 -------- -------- 3.61E+06Std. Dev. 0.01829 0.004672 1275724 1825 0.551 1385204

Deflection (inch)

Elastic Modulus

Ɵ in Radians

Secant Modulus

Discussion of Results

The compressive strength of Clark was 32220 psi while the compressive strength of the

Wilma was only 10039 psi. The Clark also had a much larger elastic modulus, 3.84*10^6 psi,

than the Wilma, 1.44 *10^6 psi. From previous experiments, two other methods were used to

determine uniaxial compressive strength. First, a Schmidt hardness test was used to determine

compressive strength. A compressive strength of 13500 psi was determined for Clark and

12500 psi for Wilma. A point load test was then performed and compressive strengths of

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23677 psi for Wilma and 26258 psi for Clark. The uniaxial compressive test is going to be the

most accurate determination of the compressive strength because it tests the compressive

strength directly. Figure 2 shows how the secant and elastic moduli were determined

graphically on the stress versus strain curve.

The Clark rock is stronger in

compression than the Wilma rock

is. The higher compressive strength

of the Clark means that it would

perform better than the Wilma rock

given the same stresses. The Clark

rock also had a much larger elastic

modulus than the Wilma rock. The

Clark will deform much less than

the Wilma under the same stresses, which is better for a mine’s stability.

The compressive strength of the Clark rock as determined by the uniaxial compressive

strength test was higher than either one of the other compressive strength calculations. This is

good because it means that if either one of the other strength calculations were used to design

a mine then the mine likely wouldn’t have any failures because both other estimates were

more conservative. The Wilma rock on the other hand had a lower compressive strength

determined through the uniaxial compressive test than either one of the other methods. This is

bad because if either one of the other tests were used for a mine plan as the compressive

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0.00E+00 5.00E-03 1.00E-020

2000

4000

6000

8000

10000

Strain versus Stress for Sample

C-4-2

Stress vs StrainSecant Modulus

Strain

Stre

ss (p

si)

Figure 3: Graph of stress versus strain curve with moduli

strength, then, as the factor of safety approaches one, failure will occur because the

compressive strength is drastically lower than the strength determined through either other

methods.

A t-test was performed on the data sets to determine if they were part of two separate

data sets with a degree of significance of the .05 level. The null hypothesis for all of the tests

was that the two data sets were part of the same set. The required t-score to reject the null

hypothesis using a two-tailed test was 2.2622. The t-scores for the peak load, compressive

strength, elastic modulus, and secant modulus were 4.838, 4.855, 4.971, and 4.423

respectively. The remaining t-scores were not large enough to reject the null hypothesis. The

four scores listed above are large enough to reject the null hypothesis meaning that the two

sets of samples came from different data sets at a level of significance of .05.

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Triaxial Compressive Test

Theory

The Mohr-Coulomb failure criterion of a rock is one of the most important modeling

tools that a mining engineer has at their disposal. The failure criterion allow the engineer to

predict the shear strength of the rock based on how the rocks are stressed. The Mohr-Coulomb

failure criterion of a rock is:

τ=S i+σ n∗tan (φ)

where τ is the critical shear stress, Si is the cohesion of the rock, σ n is the stress normal to the

plane, and φ is the angle of internal friction of the rock.

There are several different methods that can be used to determine the Mohr-Coulomb

failure criterion of the rock samples. First, a p-q diagram can be analyzed. The values for p are

graphed horizontally and the values for q are graphed vertically. The slope of the line of best fit

determined from the p-q diagram can be used to determine the angle of internal friction for the

rocks. The y-intercept of the line of best fit from the p-q diagram can be used to determine the

cohesion for the rocks. P and q are determined by the equations:

q=σ1−σ3

2p=

σ 1+σ3

2

where σ 3 is the minor principle stress, and σ 1 is the major principle stress. The cohesion of the

rock is determined using the equation:

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si=d /cos (φ)

where d is the y-intercept for the p-q diagram. The angle of internal friction is determined

using the equation:

φ=arcsin (tan (ϑ ) )

where tan (ϑ ) is the slope of the line of best fit from the p-q diagram. Another method used to

determine Mohr-Coulomb failure criterion of a rock is by graphing the principle stresses acting

on the sample. The equation:

φ=arcsin ( tan (ω )−1tan (ω)+1 )

is used to determine the angle of internal friction based on the graph of sigma1 v. sigma3

where tan (ω) is the slope of the line of best fit of the graph. The equation:

Si=C0(1−sin (φ ))

2cos (φ)

is used to determine the initial cohesion, Si, in the failure criterion where C0 is the y-intercept

of the graph of the principle stresses and φ is the angle of internal friction determined using the

equation above. Another way to find the Mohr-Coulomb failure criterion is by plotting the

shear stress acting on the failure plane against the normal stress to the failure plane. The y-

intercept of the graph is the cohesion, Si, and the slope is equal to tan (φ ). The stress normal to

the plane is determined using the equation:

σ n=12∗(σ1+σ 3 )+1

2∗(σ1−σ3 )∗cos (2ϑ )

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The shear stress acting on the plane is determined using the equation:

τ=12∗(σ1−σ3 )∗sin (2θ)

where τ is the shear stress acting on the plane, σ 1is the major principle stress, σ 3 is the minor

principle stress, and θ is the angle of failure of the sample.

Experimental Apparatus and Procedure

The triaxial compression test is performed using a Material Test System, or MTS, a

Franklin-Hoek triaxial chamber, and a hydraulic pump. A triaxial chamber, as seen in Figure 3,

consists of the metal

chamber, a valve connected

to a reservoir of hydraulic

fluid, and a rubber membrane

separating the hydraulic fluid

from the rock samples.

First, samples were

placed in the chamber. The

hydraulic pump was then

connected to the compression

chamber. The pump was

connected to an accumulator, a container filled with a mixture of a gas and a fluid under

pressure. The accumulator allows the MTS to compress the sample, which forces the sample

16

Figure 4: MTS with triaxial chamber prepared for testing

to expand diametrically, without increasing the confining pressure on the sample. The seals

were placed on the chamber and the chamber was placed in the MTS. The pump was used to

put a confining pressure, which varied from trial to trial, on the samples.

As loading began, the confining pressure was

maintained at a minimum point until the samples

failed. The confining pressure was closely watched

because it increased rapidly as the samples stopped

deforming elastically. When the samples failed, the

confining pressure and the peak load were recorded.

After the rock failed, the samples were removed from

the chamber, and, if possible, an angle of failure was

recorded by measuring the angle between one of the

two ends of the sample and the failure plane. If there was no clear failure plane, no angle was

recorded. The failure plane for a sample can be seen in Figure 4.

Tabulation of Data

Ten samples of both the Wilma and Clark rocks were tested. The results of the tests are

in the tables below. The samples were all roughly four inches in length with diameters of about

1.87 inches. The first table includes the data for the Wilma rocks, the second table includes the

data for the Clark rocks, the third table includes data for the Wilma rocks, and the fourth table

includes data for the Clark rocks.

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Figure 5: Failed sample with failure plane outlined

Table D: Sigma 1 and sigma 3 results for the Wilma rock for the triaxial test

Specimen D (in) P (lbs) comments

w-4-15 1.872 600 48482 17615 67

w-15-6-148.2 1.875 800 53179 19260 56

w-4-25 1.872 800 30297 11008 65w-4-12 1.872 1500 50824 18466 60w-4-14 1.884 1400 52143 18704 62

w-4-24 1.874 1000 25608 9284 60w-5-6-142.6 1.874 1000 60613 21975 68w-5-6-157.7 1.874 1500 113886 41290 70

w-4-1 1.874 840 30928 11213 60w-4-16 1.874 1200 56414 20453 68average 1.8745 1064 52237.4 18927 63.6

sigma3 (psi)

sigma1 (psi)

theta (degrees)

failed along foliation

failed along foliation

failed along foliation

Specimen p (psi) q (psi) phi (degrees) tau (psi)w-4-15 9107 8507 3198 44 6120

10030 9230 6572 - -w-4-25 5904 5104 2623 - -w-4-12 9983 8483 5741 30 7346w-4-14 10052 8652 5214 34 7173w-4-24 5142 4142 3071 - -

11488 10488 3943 46 7285

21395 19895 6155 50 12788w-4-1 6027 5187 3433 30 4492

w-4-16 10827 9627 3902 46 6687

sigma n (psi)

w-15-6-148.2

w-5-6-142.6

w-5-6-157.7

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Table E: Results for Clark rock from triaxial test

Specimen D (in) P (lbs) comment

c-4-12 1.872 550 95721 34778 -

c-4-11 1.873 900 65540 23787 77c-4-17 1.871 1000 58693 21348 66c-4-1 1.868 1250 74514 27189 65

c-4-14 1.87 600 75338 27431 -

c-4-24 1.869 1100 74514 27160 66c-4-16 1.872 1200 57427 20865 66c-4-6 1.874 1500 132417 48008 -

c-4-10 1.874 2300 139835 50698 68.5c-4-19 1.873 800 91590 33242 -

average 1.8716 1120 86558.9 31450 68.083333

sigma3 (psi)

sigma1 (psi)

theta (degrees)

failed on discontinuity

failed on discontinuity

specimen p (psi) q (psi) phi (degrees) tau (psi)c-4-12 17664 17114 -------- -------- --------c-4-11 12344 11444 2058 -------- --------c-4-17 11174 10174 4366 42 7561c-4-1 14220 12970 5883 40 9935

c-4-14 14015 13415 -------- -------- --------c-4-24 14130 13030 5411 42 --------c-4-16 11032 9832 4453 42 7307c-4-6 24754 23254 -------- -------- --------

c-4-10 26499 24199 8801 47 16504c-4-19 17021 16221 -------- -------- --------

average 16285 15165 31450 42.6 12097

sigma n (psi)

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Discussion of results

A p-q diagram, along with several other methods, was used to determine the Mohr-

Coulomb failure criterion for the rock samples.

The Mohr-Coulomb failure criterion for the

Wilma samples based on the p-q diagram, as

seen in figure 5, is:

τ=310+σ n∗tan (69.6)

where τ is the shear strength of an intact rock, and

σ n is the stress normal to the plane of failure. The

failure criterion for the Clark rocks from the p-q

diagram, as seen in figure 6, is:

τ=2326+σn∗tan(73.5)

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10000 15000 20000 25000 300000

50001000015000200002500030000

f(x) = 0.937876241578035 x − 108.299266606957R² = 0.994340833805251

p-q Diagram (Wilma)

p (sigma1+sigma3)/2)

q ((s

igm

a1-s

igm

a3)/

2)

0 5000 100001500020000250000

5000

10000

15000

20000

25000

f(x) = 0.959494710855087 x − 659.133839553948R² = 0.996500111086318

p-q diagram (Clark)

p ((sigma1+sigma3)/2)

q ((s

igm

a1-s

igm

a3)/

2)

30003500

40004500

50005500

60006500

0

5000

10000

15000

f(x) = 1.68238861536524 x − 178.376655337885R² = 0.590892558160062

Tau v. Sigma Normal for Wilma

Sigma Normal (psi)

Tau

(psi)

Figure 6: P-Q diagram for Wilma samples

Figure 8:Graph of tau versus sigma normal for Wilma

Figure 7: P-Q diagram for Clark samples

Two other methods were used to determine the Mohr-Coulomb failure criterion. One that uses

the principle stresses and the shear stress, and another that uses the stress normal to the

failure plane and the shear stress on that

plane. The shear and normal stress were

plotted together, as seen in Figures 8 and 9,

and used to derive failure criterion for the

rocks. The failure criterion derived from the

normal stress and the shear stress for the

Wilma rock is:

τ=178.3+σ n tan (59.3)

The failure criterion determined from the normal

stress and shear stress for the Clark rock is:

τ=1799+σ n tan(64.1)

The principle stresses were plotted against each other, as seen in Figures 9 and 10, and the line

of best fit was used to determine failure criterion. The failure

criterion for the Wilma rock from the principle stresses is:

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400 600 800 1000 1200 1400 16000

1000020000300004000050000

f(x) = 15.6206292876623 x + 2306.43035793207R² = 0.309483539906549

Sigma 1 Versus Sigma 3 (Wilma)

sigma 3 (psi)

sigm

a 1

(psi)

4000 5000 6000 7000 8000 9000100000

5000

10000

15000

20000

f(x) = 2.06368185259237 x − 1799.22960078587R² = 0.99430569933914

Tau v. Sigma Normal for Clark

Sigma Normal (psi)

Tau

(psi)

Figure 9: Graph of shear versus normal stresses for the Clark rock from triaxial test

Figure 10: Graph of principle stresses sigma 1 versus sigma 3 for Wilma from triaxial test

Figure 11: Graph of principle stresses sigma 1 versus sigma 3 for Clark from triaxial test

τ=292+σ n tan (60.6)

The failure criterion for the Clark rock derived

from the principle stresses is:

τ=2339.1+σ n tan (59.0)

The Clark rock was typically weaker than the Wilma rock. The failure criterion for the

Clark would expect a higher maximum shear strength than that for the Wilma in every case

except for when the p-q diagram was used to determine the failure criterion.

Figures 7 and 8 show plots of sigma 3 versus sigma 1, which can be used to determine Mohr-

Coulomb failure criterion for the samples.

22

0 500 1000 1500 2000 25000

100002000030000400005000060000

f(x) = 13.0146240359381 x + 16874.1145299759R² = 0.399336382654405

Sigma 1 Versus Sigma 3(Clark)

Sigma 3 (psi)

Sigm

a 1

(psi)

Indirect Shear Test

Theory

Shear strength is a very important rock property that is used extensively in mine design.

Shear along with compressive strength determine what excavations may be made in a rock

mass. Shear strength in rock is determined by the imperfections such as joints or foliations

present within the rock. These imperfections are normally the weakest part of a rock mass and

control the overall strength of that mass.

The shear strength can be defined in a rock mass by the Mohr-Coulomb failure criterion.

This criterion is defined by two characteristic values initial shear strength (Si) and the angle of

internal friction (ϕs). When the Mohr-Coulomb failure criterion is used in a rock with

discontinuities the initial shear strength of the discontinuity (Ss) must be used. The initial

strength of a discontinuity plane is the maximum shear stress that can supported along the

discontinuity plane without failing when no normal stress is applied to the plane. The tangent

of the angle of internal friction acts as a friction coefficient for the discontinuity plane and thus

defines the amount of frictional stress present based upon the normal stress applied to that

plane. These values can be found by graphing the normal (σn) and shear (τ ) stresses which are

found using the following equations:

σn= σ1+σ3

2+σ1−σ3

2cos2 β

τ= σ1−σ3

2sin 2 β Si

23

Using the normal vs. shear stress graph and an added linear trend line the initial shear

strength (Ss) of the discontinuity can be found by locating the y-intercept. Next the slope of the

graph can be used to find the angle of internal friction(ϕs) with the following equation:

ϕs=tan−1(slope)

Once the angle of internal friction and the initial shear strength have been determined

they can be plugged into the following equation which defines the Mohr-Coulomb failure

criterion.

|τ|= Ss + σn tanϕs

Experimental Apparatus and Procedure

The indirect shear test uses the same experimental apparatus as the triaxial stress test.

However there are significant differences in the procedure. The sample was placed in the

Franklin-Hoek triaxial chamber with care so that the tape around the sample did not break.

The indirect shear test differs in its modulation of the confining pressure. A sample was

placed under a specific confining pressure and then stressed axially until failure. The samples

were run through the test several times at different confining pressures. The axial and

confining pressures at every failure point were recorded.

24

Tabulation of Data

Three samples of Clark and Wilma were tested. The results are recorded in the tables

below. The first table contains the data for the Clark samples and the second table contains the

results for the Wilma samples. The graphs necessary to finding the Mohr-Coulomb failure

criterion as well as the criterions themselves are also present.

Table F: Results for indirect shear test for Clark rocks

Specimen No. Diameter (in) Peak P (lbs) β (degree) τ (psi)C-A4-2 1.854 700 4585 1698 58 980.4 448.7

850 5763 2135 58 1211 577.31100 7899 2926 58 1613 820.61300 9667 3581 58 1940 10251500 11730 4346 58 2299 12791700 13470 4988 58 2623 14771900 15300 5668 58 2958 16932100 17010 6302 58 3280 18881000 4900 1815 58 1229 366.31200 9789 3626 58 1881 10901500 13400 4963 58 2473 15561800 16280 6030 58 2988 19012000 18290 6775 58 3341 2146

C-A2-B 1.877 1400 11390 4117 58 2163 12211600 12900 4663 58 2460 13771800 14440 5219 58 2760 15372000 16000 5782 58 3062 1700

C-3-0 1.858 1000 4915 1813 58 1228 365.31200 6215 2292 58 1507 490.91400 7346 2709 58 1768 588.41600 8362 3084 58 2017 6671800 9504 3505 58 2279 766.42000 10550 3890 58 2531 849.42200 11680 4307 58 2792 946.92400 12670 4674 58 3039 1022

σ3 (psi) σ1 (psi) σn (psi)

25

Table G: Results for Wilma rocks from indirect shear test

Specimen No. Diameter (in) Peak P (lbs) β (degree) τ (psi)W-3-0 1.875 2000 10940 3961 58 2551 881.1

2200 13610 4928 58 2966 12262400 16120 5837 58 3365 15452650 18720 6779 58 3809 1855

W-5-6 1.871 500 3476 1264 58 714.6 343.5700 4714 1715 58 984.9 455.9900 6135 2231 58 1274 598.3

1100 7456 2712 58 1553 724.41300 9102 3311 58 1865 903.51500 10580 3847 58 2159 10551700 12260 4458 58 2474 12391900 13770 5009 58 2773 13971000 5451 1983 58 1276 441.6800 4200 1528 58 1004 327

1000 7143 2598 58 1449 718.11200 8800 3201 58 1762 899.11400 10330 3757 58 2062 10591600 12100 4401 58 2387 12591800 14100 5128 58 2735 1496

W-8-8 1.871 600 3876 1410 57 840.2 369.9800 5576 2028 57 1164 561

1000 7296 2654 57 1491 755.41200 9272 3372 57 1844 992.31400 11000 3999 57 2171 11871600 12790 4653 57 2506 13941800 14890 5417 57 2873 16522000 16640 6050 57 3201 1850800 4981 1812 57 1100 462.1

1000 5900 2146 57 1340 523.41200 6990 2542 57 1598 613.21400 8476 3083 57 1899 768.71600 9985 3632 57 2203 9281800 11420 4153 57 2498 1075

σ3 (psi) σ1 (psi) σn (psi)

26

500 1000 1500 2000 2500 3000 35000

500

1000

1500

2000

2500

f(x) = 0.599134528891965 x − 262.788859966586R² = 0.695329814173208

Sigma n vs. τ for Clarke

Sigma n

τ

Figure 12: Graph of shear stress versus normal stress for Clark from indirect shear test

500 1000 1500 2000 2500 3000 3500 40000

200400600800

100012001400160018002000

f(x) = 0.523367977901942 x − 93.700009165644R² = 0.903099364365246

Sigma n vs. τ for Wilma

Sigma n

τ

Figure 13: Graph of shear stress versus normal stress for Wilma from indirect shear test

27

Table H: Mohr-Coulomb failure criterion determined by indirect shear test

Failure Criterion

Clarke

Wilma

|τ| = 262.8 + σn*tan(30.93)

|τ| = 93.7 + σn*tan(27.63)

Discussion of Results

The Mohr-Coulomb failure criterion of the discontinuity plane present in the Wilma

samples is |τ| = 93.7 + σn * tan(27.63). The criterion of the discontinuity plane in the Clark

samples is |τ| = 262.8 + σn * tan(30.93). When the criterions are compared two significant

comparisons can be made. The first is that Clark has greater initial shear strength. This means

that when no normal stress is applied to the discontinuity plane Clark rock will be

approximately 2.8 times stronger. The second is Clark has a greater angle of internal friction

than Wilma. This is significant because it means that the normal stress applied on the

discontinuity plane of Clark will have a larger affect on its shear strength than in Wilma. This

fact is further evidenced by the larger slope seen in Figure 11 as compared to Figure 12. This

information is useful to engineers planning excavations in highly jointed rock masses of Clark

and Wilma. The accuracy of these results is somewhat difficult to determine. This is due to the

relatively number of samples tested. Based on Figures 11 and 12, the results found for Wilma

are far more accurate than Clark as can be seen by the R² value of 0.9031 which is far closer to

1 which indicates a perfect fit of the trend line to the data than the R² value associated with the

Clark samples of 0.6953.

28

Conclusions and Recommendations

Based on the tests performed in the lab, five conclusions and comparisons can be

reached regarding the failure criterion as well as strength characteristics of Wilma and Clark.

First the initial shear strength of Clark is greater than that of Wilma.

Second, internal angle of friction is larger in Clark than in Wilma.

Third, Clark has an elastic modulus of 3.84*10^6 psi which is far greater than the

elastic modulus of Wilma which is 1.44*10^6 psi.

Fourth, the initial shear strength of a discontinuity in Clark of 262.8 psi is 2.8

times bigger than the initial shear strength of a discontinuity in Wilma.

Fifth, the angle of internal friction of a discontinuity in Clark of 30.39 degrees is

larger than the angle of internal friction of a discontinuity found in Wilma of

27.63 degrees.

The initial shear strength of Clark as found by the triaxial stress test are 1799 psi for σ n

versus τ , 2326 psi for p-q diagram, and 2339 psi for σ 1 versus σ 3. These values are clearly larger

than the Wilma shear strengths found using the triaxial test of 178.3 psi for σ n versus τ , 310 psi

for p-q diagram, and 292 psi for σ 1 versus σ 3. This is significant because it means when no

normal stress is placed on the plane of failure Clark will be significantly stronger than Wilma.

Thus it is far less likely to fail in shear and a better host rock for almost any underground

application. The same is true of the uniaxial compressive strength of Clark 32220 psi when

compared to Wilma 10040 psi. Again the data demonstrates that Clark is the stronger of the

29

two rocks. This trend extends into jointed rock masses where the internal shear strength of

Clark is also larger than Wilma.

The internal angles of friction mirror the same trends as the internal shear strength.

They are consistently larger in Clark where the triaxial test yielded the following values: 64.1

degrees for σ n versus τ , 73.5 degrees for the p-q diagram, and 59 degrees for σ 1 versus σ 3.

These values are clearly larger than the internal angles of friction found during the triaxial test

for Wilma of: 59.3 degrees for σ n versus τ , 69.6 degrees for the p-q diagram, and 60.6 degrees

for σ 1 versus σ 3. As with the initial shear strength this trend is also present within jointed rock

masses of Clark and Wilma. Finally because the Elastic Modulus of Clark is significantly larger

than Wilma Clark will deform less under the same stresses. All this information when put

together leads to the conclusion that Clark is a stronger and therefore better host rock in

underground mine applications than Wilma.

30

References

Karfakis, Mario. Rock Mechanics Notes. Virginia Tech. 2009

Delgado et al. Comparison Between Wilma and Clark Stones. Virginia Tech. 2009

31

Appendix

Sample Calculations

The following represents a set of sample calculations for sample w-4-15, the p-q diagram for Clark from the triaxial test, sample w-4-20, the plot of the principle stresses for the Wilma rock,

Principle Stress

σ 1=PA

= 48482

(1.8722 )

2

∗π

=17615 psi

Angle of Internal Friction

φ=2θ−90=2∗67−90=44degrees

P

p=σ 1+σ3

2=17615+600

2=9107 psi

Q

q=σ1−σ3

2=17615−600

2=8507 psi

Normal Stress to the Failure Plane

σ n=12∗(σ1+σ 3 )+1

2∗(σ1−σ3 )∗cos (2θ )=1

2∗(17615+600 )+ 1

2∗(17615−600 )∗cos (2∗67 )=3198 psi

Cohesion (from normal and shear stresses)

Si=σ 1−σ3 tan (θ )2

2∗tan (θ )=

17615−600∗tan (67 )2

2∗tan (67 )=3032 psi

Cohesion (from principle stresses)

Si=σ 1−σ3∗( 1+sin (φ )

1−sin (φ ) )2∗( cos (φ )

1−sin (φ ) )=¿

Angle of Internal Friction (from p-q diagram)

A-1

φ=arcsin (tan (α ) )=arcsin ( .959 )=73.5degrees

Cohesion (from p-q diagram)

Si=a

cos (φ )= 695.7

cos (73.5 )=2326 psi

Strain

ε= δLL

=.02363.922

=.00602

Cohesion (from uniaxial compressive test)

Si=C0

2∗tan (θ )= 8953

2∗tan (60 )=2585 psi

Cross Sectional Area

A=( D2 )2

∗π=( 1.8732 )

2

∗3.141592=2.755 i n2

Elastic Modulus

E= δσδϵ

= 6046.382−3549.688

4.50∗10−3−3.46∗10−3=2.40∗106 psi

Secant Modulus

E sec=σϵ= 8953

.006024=1.51∗106 psi

Angle of Internal Friction (from principle stresses)

φ=arcsin ( tan (ω )−1tan (ω)+1 )=arcsin ( 15.62−1

15.62+1 )=60.6degrees

Cohesion (from principle stresses)

Si=C0(1−sin (φ ))

2cos (φ)=

2306∗(1−sin (60.6 ) )2∗cos (60.6 )

=292 psi

Raw Data

A-2

A-3

A-4

A-5

A-6

A-7

A-8

A-9

A-10

A-11