1. No category

MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
Sample Questions & Worked Out Examples
For
Min 04027
ROCK MECHANICS
B Tech. (Second Year)
Mining Engineering
ROCK MECHANICS
MIN-04027
B Tech., Second Year
Mining Engineering
2
CHAPTER 1
SOIL AND ROCK MECHANICS
1.* Explain the standard test procedure of the apparent porosity test in laboratory. (10
marks)
2.* Write the short notes on the following items. (10 marks each)
(i) The mechanical properties of rock material
(ii) The basic elements of the strength of materials
3.* In a series of apparent porosity tests of granite specimens, the following test
results are recorded as shown in table (3). Determine the mean value, standard
deviation and the coefficient of variation of the results. (10 marks)
Table (3) Results of Porosity Test
Sample
Diameter,
No.
cm
1
DHY-33
6.07
2
DHY-34
6.07
3
DHY-35
6.07
4
DHY-36
6.07
5
DHY-38
4.75
6
DHY-39
4.75
7
DHY-41
4.75
8
DHY-44
4.75
9
DHY-48
4.75
10
DHY-49
4.75
Length,
cm
7.44
7.25
9.14
6.43
7.22
7.27
7.33
7.24
8.68
7.36
Wo
gm
543.0
532.0
675.0
472.0
330.0
332.0
333.0
330.0
395.0
332.0
Wm
gm
547.0
535.0
680.0
475.0
333.0
335.0
335.0
332.0
397.0
335.0
Remark: Apparent porosity, %, Wo = weight of oven-dried specimens
Wm = weight of saturated specimens
Answer: The mean value = 1.8163 %, Standard deviation = 0.3920 %,
The coefficient of variation = 0.2158
4.* Explain the standard test procedures and formula to determine the apparent
specific gravity, ASG, the natural moisture capacity Wn, the total moisture
capacity, Wt, the yield of water, Y, the coefficient of water saturation, Kws of rock.
(20 marks)
3
5. ** The laboratory test results of granite rock are recorded as shown in table (5).
Determine the mean value, standard deviation and the coefficient of variation of
the results for the apparent specific gravity, the natural moisture capacity Wn, the
total moisture capacity, Wt, the yield of water, Y, the coefficient of water
saturation, Kws of rock. (5 marks each)
Table (5) Results of laboratory test
Sample
No.
1
2
3
4
Diameter,
cm
4.740
4.755
4.755
4.755
Length,
cm
7.430
7.285
7.240
7.095
Volume,
cm3
131.1098
129.3661
128.5670
125.9921
Wo
gm
328.0
330.0
330.0
325.0
W1
gm
330.0
332.0
333.0
328.0
Ww
gm
333.0
335.0
338.0
331.0
Remark: Apparent porosity, %, Wo = weight of oven-dried specimens
Ww = weight of saturated specimens, W1 = weight of air dried specimens
6. ** In a series of apparent porosity tests of granite specimens, the following test
results are recorded as shown in table (6). Determine the mean value, standard
deviation and the coefficient of variation of the results. (20 marks)
Table (6) Results of laboratory test
Sample No.
1
2
3
4
5
6
7
8
9
10
DHY- 4
DHY- 6
DHY- 6
DHY-10
DHY-12
DHY-16
DHY-19
DHY-22
DHY-26
DHY-27
Diameter,
cm
4.73
4.73
4.73
4.50
4.50
4.50
4.50
4.73
4.73
4.75
Length,
cm
5.81
7.615
7.615
5.38
4.81
4.90
4.15
7.61
9.51
9.53
Wo
gm
261.40
346.20
343.10
218.90
196.40
200.40
151.00
343.90
437.00
436.00
Wm
gm
263.20
348.10
344.60
220.20
197.40
201.30
151.80
345.90
440.00
440.0
Rock Type
Granite
Granite
Granite
Granite
Granite
Granite
Granite
Granite
Granite
Granite
Answer: The mean value = 1.5670 %, Standard deviation = 0.3597 %
The coefficient of variation = 0.2296
7. *** Explain the characteristics of loosened rock and its application in mining
practice. (10 marks)
4
CHAPTER 2
THEORY OF ELASTICITY
1.* Basic assumptions of elastic properties and Hooke's law for idealized elastic
properties. (10 marks)
2.* Write the short notes on the following mechanical properties of rock within the
elastic range. (10 marks each)
(i) Young Modulus, (E)
(ii) Poisson's Ratio (υ)
3.* Write the short note on the followings. (10 marks each)
(i) The characteristic of the acoustic properties
(ii) Determination of the elastic properties of rock and rock mass based on the
acoustic properties
4. ** In the interpretation of laboratory test results, discuss, with necessary sketch,
how to calculate the Young’s modulus of rock specimens, which depends on the
variation of loading condition. (10 marks)
5. ** Discuss the nature of the acoustic properties of rock mass and its application in
mining practice. (10 marks)
6. *** Discuss briefly how to apply the mechanical properties of rock to obtain the
rational design or be technically successful in miming and rock excavation works.
(10 marks)
5
CHAPTER 3
STRENGTH OF ROCKS
1.* Explain the factors to be considered (or the influencing factors on rock properties)
in sample preparation for precise rock testing. (10 marks)
or
Explain, briefly, the factors, which influence the strength and deformation of intact
rock, to be considered in specimen preparation. (10 marks)
(or)
Explain the significant factors in specimen preparation to obtain the reliable
laboratory test results. (10 marks)
2. * Explain the precaution and procedure of the specimen preparation for precise
laboratory rock testing. (10 marks)
3.* Draw the Mohr’s diagram or Mohr’s circle to represent the values of shear
stresses and normal stress by using graphical method for the given major principal
stress, σ1 and minor principal stress, σ3. Give an numerical example. (10 marks)
4.* From the following results of a series of triaxial compression test on Granite rock
specimens, draw the Mohr's circle and measure the effective shear stress, τeff and
normal stress, σN. Compare these results with calculated values.
Specimen
No.
1
2
3
4
Diameter,
cm
3.05
3.05
3.05
3.05
Area,
cm2
7.3062
7.3062
7.3062
7.3062
Max.
Load, KN
90.0
100.0
150.0
170.0
σ1
MPa
147.8203
169.7196
186.1441
202.5686
5.* Write the short notes on the followings. (10 marks each)
(i) Basic concept of maximum shear stress theory
(ii) Basic concept of effective shear stress theory
(iii) Basic concept of Hoek and Brown failure theory
σ3
MPa
2.00
4.00
6.00
8.00
6
6.* Discuss the importance or application of the mechanical properties of rock and
rock mass in consideration for excavation design calculation. (10 marks)
7.* In a series of Point Load Strength Index tests of granite specimens, the following
test results are recorded as shown in table (20). Determine the mean value,
standard deviation and the coefficient of variation of the results. (20 marks)
Table (20) Results of Point Load Strength Test
1
2
3
4
5
6
7
8
9
Sample
No.
DHY-42
DHY-43
DHY-44
DHY-45
DHY-46
DHY-47
DHY-49
DHY-51
DHY-52
Diameter,
cm
4.750
4.755
4.760
4.760
4.765
4.760
4.750
4.750
4.750
Length,
cm
7.220
7.210
7.330
7.320
7.120
7.015
7.245
7.440
7.140
Max. Load,
KN
14.3
14.2
13.2
11.7
13.0
13.0
13.2
13.0
13.0
Rock Type
Granite
-
Answer: The mean value = 5.8272 MPa, Standard deviation = 0.3416 MPa
The coefficient of variation = 0.0586
8.* In a series of uniaxial compressive strength tests of granite specimens, the
following test results are recorded as shown in table (21). Determine the mean
value, standard deviation and the coefficient of variation of the results.
Table (21) Results of Uniaxial Compressive Strength Test
Specimen
Diameter,
Max. Load,
Remarks
No.
cm
KN
1
DHY-40
4.755
224.0
Medium grained size
2
DHY-41
4.760
148.0
Medium grained size
3
DHY-42
4.755
90.0
Medium grained size
4
DHY-43
4.755
95.0
Medium grained size
5
DHY-44
4.760
116.0
Coarse grained size
6
DHY-45
4.760
148.0
Medium grained size
7
DHY-46
4.760
114.0
Medium grained size
8
DHY-47
4.765
110.0
Coarse grained size
9
DHY-48
4.760
160.0
Medium grained size
10
DHY-49
4.755
70.0
Coarse grained size
Remark: The ratio of length and diameter is between 2.0 and 2.5,
Answer: The mean value = 71.6921 MPa, Standard deviation = 24.8882 MPa
The coefficient of variation = 0.3472
7
9. * In a series of triaxial compression test on granite, the following test results are
obtained as shown in table (22). (20 marks)
Table (22) Results of triaxial compression strength test
Max.
σ1
Specimen Diameter,
Area,
Load, KN
MPa
No.
cm
cm2
1
3.04
7.2583
90.0
123.9954
2
3.04
7.2583
104.0
143.2835
3
3.04
7.2583
120.0
165.3272
4
3.04
7.2583
124.0
170.8381
σ3
MPa
2.00
4.00
6.00
8.00
Uniaxial compressive strength of intact rock = 106.4176 MPa
(1) Determine the Hoek and Brown strength parameters, ‘m’ and ‘s’.
(2) Calculate the value of σ1 at the failure if σ3 = 10.5 MPa using Hoek and
Brown yield criterion.
10. * In a series of Brazilian tensile tests of granite specimens, the following test
results are recorded as shown in table (23). Determine the mean value, standard
deviation and the coefficient of variation of the results. (20 marks)
Table (23) Results of Brazilian Tensile Strength Test
Specimen
Diameter, Length,
Max. Load,
No.
cm
cm
KN
1
DHY-3
4.725
7.82
27.00
2
DHY-4
4.73
5.81
20.0
3
DHY-5
4.70
7.615
27.00
4
DHY-6
4.70
7.615
30.00
5
DHY-7
4.70
7.60
22.00
6
DHY-8
4.70
7.30
20.00
7
DHY-9
4.50
10.16
40.00
8
DHY-10
4.50
5.38
17.00
9
DHY-11
4.50
10.19
22.00
10
DHY-16
4.50
7.750
25.00
11
DHY-17
4.50
7.68
15.00
12
DHY-18
4.50
7.62
20.00
13
DHY-19
4.50
7.845
22.00
14
DHY-20
4.73
7.61
27.00
15
DHY-21
4.725
7.70
35.00
Rock Type
Granite
-
Answer: Avg. = 4.3987 MPa, SD = 4.3987 MPa, COD = 0.2055
11. ** Explain the procedure of the irregular lump test in laboratory rock testing and
correlation between the results of irregular lump test and uniaxial compressive
strength test. (10 marks)
8
12. *** In a series of triaxial compression test, the following test results are recorded
as shown in table (25). (20 marks)
Table (25) Results of Triaxial compression test
Specimen
No.
A
B
C
D
Diameter,
cm
3.77
3.77
3.77
3.77
Area,
cm2
11.1628
11.1628
11.1628
11.1628
Max.
Load, KN
90.0
108.0
132.0
144.0
σ1
MPa
80.6250
96.7500
118.2501
129.0001
σ3
MPa
3.00
6.00
9.00
12.00
(i) Determine uniaxial compressive strength, σc, coefficient of angle of internal
friction, tanφ and cohesion of intact rock.
(ii) Calculate the value of σ1 at the failure if σ3 = 12.5 MPa using Mohr-Coulomb
criterion.
(iii) Draw the Mohr's envelop curve
CHAPTER 4
PLASTIC AND RHEOLOGICAL PROPERTICS OF ROCKS
1. * Explain the creep law or rheological properties of rock. (10 marks)
2. * Discuss, briefly, the followings:
(i) The importance of the prediction of creep behavior of rock in mining practice.
(ii) Two elastic constants of rocks (Young Modulus and Poisson's Ratio)
3. ** Discuss the importance of the prediction of the creep behavior in mining
practice. (10 marks)
4. *** Explain the application of rheological properties of rocks in mining practice.
(10 marks)
9
CHAPTER 5
APPLICATION OF TECHNICAL INDICES OF ROCKS
1. * Write the short note on the following technical indices of rock in mining practice.
(10 marks each)
(i) Hardness of rock (static and dynamic hardness) and its applications in
mining practice
(ii) Abrasiveness and its applications in mining practice
2. ** Write the short note on the following technical indices of rock in mining
practice. (10 marks each)
(i) Crushability of Rock
(ii) Drillability of Rock
(iii) Blastability of Rock
3. ** Write the short note on the following technical indices of rock in mining
practice. (10 marks each)
(i) Brittleness of rock,
(ii) Toughness of rock
CHAPTER 6
CLASSIFICATION OF ROCKS AND ROCK MASS
1. * Write the short note on the following topics. (5 marks each)
(i)
Intact rock
(ii) Moderately jointed rock
(ii) Stratified rock
(iv) Blocky and seamy rock
2. * Write the short note on the following topics. (5 marks each)
(i)
Crushed or chemically intact rock
(ii) Swelling rock
(iii) Squeezing rock
(iv) Rock Quality Designation (RQD)
10
3. * Explain the basic consideration factors for the determination of the tunneling
quality index (Q-system). (20 marks)
4. * Explain the basic concept of CSIR (Bieniawski) rock mass classification for the
determination of the rock mass quality. (20 marks)
5. * Discuss briefly the influence of clay seams and fault gouge on the rock mass
strength. (10 marks)
6. * Discuss briefly the influence of the characteristics of discontinuity on the strength
of rock mass. (10 marks)
7. ** Discuss as much as you know the followings. (10 marks each)
(i) Quantification of the geological parameters, which influence the
engineering properties of rock mass.
(ii) Influence of the characteristics of discontinuity on the strength of rock
mass.
CHAPTER 7
Analytical Solution
1. *Discuss the basic assumption on stress distribution around underground opening.
2. *Describe the general types of stress field and explain this stress field.
3. **Briefly, discuss the procedure for designing or evaluating the stability of single opening in
competent rock.
4. **Discuss, briefly, time dependent deformation of rock.
5. ***A circular tunnel of 15 ft diameter is to be excavated through a section of totally crushed
rock (Assume as cohesionless ground). Tunnel will be driven at a depth of 300 ft overburden.
The properties of rock mass around the tunnel are recorded as follow:
Unit weight of rock mass = 150 pcf
Friction angle
= 37°
Calculate the maximum and minimum support pressure when the clastic zone extends to 15 ft
from the center of tunnel.
6. ***A vertical mine opening of 4.0 m diameter (final diameter) is to be connect from level 11
level 10 through the jointed rock mass as shown in following figure. The result of laboratory
data and the following information are recorded as follow.
Ground surface
Level 10
2m
4m
Z=300.0
30.0m
Level 11
Given
- Uniaxial compressive strength of intact rock = 30.50 MPa
-
Angle of internal friction of rock mass = 35 degree
-
Hoek and Brown strength parameter ‘s ‘ = 0.0025
-
Unit weight of rock mass = 0.0265 MN/m3
-
Swell factor = 1.1
-
Radius of plastic zone = 4.0 m
Calculate the required support pressure at the beginning point of level 11
CHAPTER 8
Stability of Slope
1. **Write down the short on the following slope failure.
(i) Plane failure
(ii) Wedge failure
(iii) Circular failure
2. ***The cross of a non-working slope in a strip mining operation is shown in the following
figure. The mechanical properties of upper layer are c= 15 KN/m2, φ =20° and γ =17 KN/m3.
The lower layer has c = 25 KN/m2, φ =15° and γ =19 KN/m3.
Assume that both materials are homogeneous and slope is located permanently above the
water table. Determine the factor of safety for the slip circle shown.
9m
r =30 m
5
1.5
2
20 m
4
3
15m
Upper layer
2
1
Lower layer
3. *** The dimension of slope and the mechanical properties of each rock layer are shown in the
following figure.
8.7m
90°
r =30 m
3
2
Upper layer
2
1
30.2 m
1
4
12.5m
20 m
Lower layer
Upper layer
Cohesion = 25.0 KN/m2
φ =29° γ =20.0 KN/m3.
Lower layer
Cohesion = 55.0 KN/m2
φ =20.5° γ =22.0 KN/m3
Assume that both materials are homogeneous and slope is permanently situated above the
ground water table. Determine the factor of safety for the slip circle shown in above figure.
(Note: Tension crack is allowed.)
4. ** *A proposed cutting is to have the dimensions shown in the following figure.
5m
8m
3m
6m
9m
The soil has the following properties: φ = 15°, c 13.5 KN/m2, γ = 19.3 KN/m3.
Determine the factor of safety against slipping for the slip circle shown. (i) ignoring tension
crack and (ii) allowing for a tension crack. Answer (i) 1.7 (ii) 1.6
5.*** Investigate the stability of the embankment shown in the following figure.
4.75 m
9.15 m
7.64 m
9.15 m
The embankment consist of two soils both with bulk densities of 19.3KN/m3; the upper
soil has c = 7.2 KN/m2 and φ = 30°, whilst the lower soil has c =32.5 KN/m2 and φ = 0°. Analyze
the slip circle shown (ignore tension crack). Answer FS =1.2
6.*** Using Bishop’s, determine the factor of safety for the following slopes.
(i) Pore ratio =0.5,c′ = 5.37 KN/m2, φ= 40° ,γ= 14.4 KN/m3 , H = 15.2 m, slope =3:1
(ii) Pore ratio =0.3,c′ = 7.20 KN/m2, φ= 39° ,γ= 12.8 KN/m3 , H = 76.4m, slope =2:1
(i) Pore ratio =0.5,c′ = 20.0 KN/m2, φ= 25° ,γ= 17.7 KN/m3 , H = 25 m, angle of slope =3:1
Assume that rainfall is intense enough to completely saturate the block, then the saturated
zone extends completely to the surface.
7.*** The cross section of overall pit slope is shown in following figure. The information
obtained from field investigation is given as follows. Assume that materials are homogeneous
and slope is located permanently above ground water table. Upper layer: c= 25 KN/m2, φ=13°,
γ=20 KN/m3. Lower layer: 27KN/m3, φ=17°, γ=20 KN/m3.Determine the factor of safety for the
slip circle shown in figure.
O
R= 51 m
90°
2.5
R= 51 m
6
2
15 m upper layer
5
30 m
4
2
3
Lower layer
17.25 m
************************************************************************
NOTE
* Must Know Questions
** Should Know Questions
*** Could Know Questions
11
QUESTIONS AND ANSWERS
No (1) In a series of apparent porosity tests of granite specimens, the following test
results are recorded as shown in table (1). Determine the mean value, standard
deviation, and the coefficient of variation of the results.
Solution:
The apparent porosity can be calculated by the following equation and
recorded as shown in following table (1).
P=
Wm − Wo
× 100
dwV
Where, P = apparent porosity, %, dw = density of water, V = Volume of specimen
Wo = weight of oven-dried specimens
Wm = weight of saturated specimens
Table (1) Results of Porosity Test
Sample
No.
DHY-33
DHY-34
DHY-35
DHY-36
DHY-38
DHY-39
DHY-41
DHY-44
DHY-48
DHY-49
Sample
No.
DHY-35-1
DHY-35-2
DHY-36-1
DHY-36-2
DHY-38
DHY-39
DHY-41
DHY-44
DHY-48
DHY-49
Dia.,
cm
6.07
6.07
6.07
6.07
4.75
4.75
4.75
4.75
4.75
4.75
Length,
cm
7.44
7.25
9.14
6.43
7.22
7.27
7.33
7.24
8.68
7.36
Wo
gm
543.0
532.0
675.0
472.0
330.0
332.0
333.0
330.0
395.0
332.0
Wm
gm
547.0
535.0
680.0
475.0
333.0
335.0
335.0
332.0
397.0
335.0
Porosity
%
1.85789
1.42993
1.89041
1.61229
2.34481
2.32868
1.53975
1.55889
1.30027
2.30020
Results: The mean value = 1.8163 %
Standard deviation = 0.3920 %
The coefficient of variation = 0.2158
-------------------------------------------------------------------------------
No (2) Basic assumptions of elastic properties and Hooke's law for idealized elastic
properties. (10 marks)
12
Answer:
When any body is subjected an external force, the body exhibits an opposite
reaction or internal forces develop in it, tending to restore its original shape. The
surface density of the force developed in each element of the body is known as stress.
Stress is a vector quantity depending both on the internal properties of the rock, and
on the shape of the external forces. Under the action of external forces the rock may
undergo changes in its linear dimension, volume or shape, all of which are known as
strains (or deformations).
Before relationships between stress and strain can be developed, the elastic
properties of rocks must be specified. The different branches of the science of
mechanics are based on certain idealized elastic properties of material. The classical
theory of elasticity is restricted to solid materials processing the following idealized
elastic properties.
(1) Linear between stress and strain - If a body is subjected to a stress, then the
strain in the direction of the stress is directly proportional to the applied
stress as shown in figure (2). That is Hooke’s law applies.
σ
E
σ=Eε
ε
Figure (2) Elastic properties of rock within elastic limit
(2) Homogeneity – The material of a body is uniformly distributed throughout
the volume of the body and the elastic properties of the material are the same
at all points in the body.
(3) Isotropy – The elastic properties of the material are the same in all directions.
(4) Perfectly elastic – Upon removal of deforming forces, the size and shape of a
body return precisely to their original state.
No actual materials satisfy exactly all of these requirements. However, the
deviations from the assumed idealized conditions for many materials are so slight that
results predicted on the basis of the theory of elasticity are verified by experiment.
13
Most structural materials and many rocks posses characteristics and elastic properties,
which permit the theory of elasticity to be used in practice.
------------------------------------------------------------------No (3) In the interpretation of laboratory test results, discuss, with necessary sketch,
how to calculate the Young’s modulus of rock specimens, which depends on the
variation of loading condition. (10 marks)
Answer:
Young’s Modulus
For perfectly elastic material, the axial stress-strain relation obtained from
conventional uniaxial test would be a straight line with constant slope, E as shown in
figure (1):
σ =Eε
---------
(a)
Where, σ = longitudinal stress (compressive or tensile)
ε = strain
σ, Stress
E = Young Modulus
E
ε, Strain
Figure 1 stress-strain diagram
In actual condition, the stress-strain curve for most rock is not linear; instead,
the slope of the curve varies with the stress level (or loading condition). Furthermore,
slope varies with loading rate, and it differs according to whether load is applied or
removed. In this circumstances it is not strictly proper to use term Young’s modulus
from the results of laboratory test, but for the practical purposes it is convenient to
regard the slope of the curve, either at a specific point or average over a certain
section as Young’s modulus (see figure 2).
14
σ, Stress
Q
R
M
P
ε, Strain
Figure (2) Definition of modulii for non-linear stress-strain diagram
Where, M = initial modulus, P = secant modulus
R = cord modulus,
Q = tangent modulus
When the term modulus is used without qualification, it is usually taken to be
the tangent modulus at 50% of the ultimate crushing strength of rock. All modulii
have dimensions expressing stressed, N/m3 (in the M.K.S system).
----------------------------------------------------------------------------------No (4) Discuss the nature of the acoustic properties of rock mass and its application in
mining practice. (10 marks)
Answer:
Acoustic Properties
The nature of the propagation of elastic waves in rock is defined by:
(i) The velocity of propagation of elastic waves,
(ii) Acoustic absorption coefficients and
(iii) Wave impedance.
Rock masses are characterized by coefficients of reflection and indices of
refraction of elastic waves.
(i) The velocity of propagation of elastic waves
The velocity of propagation of elastic waves in an infinite absolutely elastic
and isotropic medium depends on the mechanical properties of rock mass such as
Young modulus and Poisson’s ratio. There are three types of elastic waves:
longitudinal waves, transverse waves and surface waves. The following relation of
propagation velocity of longitudinal waves, transverse waves and
(Love waves) in a rock mass is:
surface waves
15
vp > vs > vL
(ii) Acoustic absorption coefficients
The propagation of elastic waves is accompanied with a gradual drop of
intensity the further they spread from their source. In most cases, two factors are
responsible for this loss of intensity:
(1) absorption of a part of the elastic energy by the rock and its transformation
into heat, owing to friction between the vibration particles; and
(2) dissipation of the sound energy in various directions owing to
inhomogeneities in the rock (pore s, fractures, disseminations, etc.).
(iii) Wave impedance
The wave impedance of rocks is a factor governing their capacity to reflect
and refract elastic waves. Reflection and refraction takes place either at the interface
between rocks of different acoustic properties, or when elastic waves pass into rock
from an external medium (voids).
When sound energy passes from a medium with low wave impedance to a
medium with high impedance, the bulk of the energy is reflected. Thus 99.8 % is
reflected when elastic waves pass from into water, and 85 % when they pass from
water into rock.
Application of Acoustic Properties in Excavation and Mining
Elastic waves can be induced in rocks by explosions, blows, mechanical
vibrators etc. Explosives are employed to produce seismic wave, the mechanical
method is used mainly to induce waves of infrasonic and audio frequencies.
Acoustic methods are widely used in mining to obtain information about rock
and rock masses. They are based on the relationship between the various acoustic
properties of rocks and their other physical properties (elastic constants), mineral
composition, structure(porosity), external condition (fracturing and disturbance, stress
state, moisture content, degree of freezing of the water in rocks etc.).
Young modulus can be calculated from the velocity of longitudinal waves:
v2γ
E=
g
The induced stresses within rock mass due to vibration effect due to explosion
can be calculated from the velocity of longitudinal and transverse waves:
Ev p
Ev s
, and σ =
σ=
Vs
Vp
Where, E = Young modulus, vp = velocity of longitudinal wave
16
vs= velocity of transverse wave, Vp = sonic velocity of longitudinal wave
Vs = sonic velocity of transverse wave
-------------------------------------------------------------------------
No (5) Explain the precaution and procedure of the specimen preparation for precise
laboratory rock testing. (10 marks)
Answer:
The precaution and procedure of the specimen preparation for Precise Tests
(1) Collection and storage
a. Test material is collected from rough block, dressed blocks, or drill cores.
b. Samples should not be collected from material, which has been modified by
blasting, weathering, and contamination of rough handling.
c. To represent each of the areas of interest 3 to 10 specimens should be
collected.
d. - Sample should be delivered to the laboratory as soon as possible for
subsequent specimen preparation and testing.
e. Sample should be moisture-proofed immediately after collection.
f. Sample should be transported and stored from excessive changes in
humidity and temperature.
(2) Avoid of contamination
a. Contamination of the external surface of finished surface of finished
samples
should be avoided.
b. If cutting oil or dirty water must be used, then the rock should be
thoroughly saturated with clean water before machining start.
(3) Sawing
(a) A precise cut off machine should be used.
(b) To avoid the problem of spalling and lip formation at the end of the cut the
core on both sides of the cut should be supported.
(4) Surface grinding
(5) Lapping
- Final smooth finished and ground samples.
(6) Sample measurement (Quality control)
17
(a) Final dimensions are normally measured with a micrometer and reported to
nearest 0.001 in.
(b) The impressions are made by sandwiching a sheet of carbon paper and a
sheet of white paper between the sample end and a smooth surface. The
upper end of the sample is given a light blow with a rubber or plastic
hammer, and an imprint is formed on the white paper.
------------------------------------------------------------------------No (6) In a series of triaxial compression test on granite, the following test results are
obtained as shown in table (22). (20 marks)
Table (22) Results of triaxial compression strength test
Specimen Diameter,
Area,
Max.
σ1
No.
cm
cm2
Load, KN
MPa
1
3.04
7.2583
90.0
123.9954
2
3.04
7.2583
104.0
143.2835
3
3.04
7.2583
120.0
165.3272
4
3.04
7.2583
124.0
170.8381
σ3
MPa
2.00
4.00
6.00
8.00
Uniaxial compressive strength of intact rock, σc = 106.4176 MPa
(1) Determine the Hoek and Brown strength parameters, ‘m’ and ‘s’.
(2) Calculate the value of σ1 at the failure if σ3 = 10.5 MPa using Hoek and
Brown yield criterion.
Answer:
(1)
Table (1)
Specimen
No.
σ3
σc
1
2
3
4
0.0188
0.0376
0.0564
0.0752
 σ1 − σ 3 


 σc 
1.3142
1.7131
2.2416
2.3415
2
2
 σ − σ3 
σ
 are plotted in the following
From the table, the values of 3 and  1
σc
σ
c


graph:
18
 σ1 − σ 3 


 σc 
2
2.75
2.5
y = 19.204x + 1
R2 = 0.945
2.25
2
1.75
1.5
1.25
1
0.75
0.5
0.25
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
σ3
0.08σ c
Graph (1) Calculation of Hoek and Brown strength parameters ‘m’ and ‘s’ values
From Graph (1) m = 19.204 and s = 1.0
(Answer)
(2) To calculate the value of σ1 at the failure if σ3 = 10.5 MPa, use the following
equation:
σ1 = σ3 + (mσcσ3 + s σ2c)0.5
σ1 = 10.5 + ((19.204×106.4176 ×10.5) + (1.0 × 106.4176 × 106.4176)0.5
= 191.5607 MPa
(Answer)
------------------------------------------------------No (7) In a series of triaxial compression test, the following test results are recorded
as shown in table (25). (20 marks)
Table (25) Results of Triaxial compression test
Specimen
No.
A
B
C
D
Diameter,
cm
3.77
3.77
3.77
3.77
Area,
cm2
11.1628
11.1628
11.1628
11.1628
Max.
Load, KN
90.0
108.0
132.0
144.0
σ1
MPa
80.6250
96.7500
118.2501
129.0001
σ3
MPa
3.00
6.00
9.00
12.00
(i) Determine uniaxial compressive strength, σc, coefficient of angle of
internal friction tanφ and cohesion of intact rock.
(ii) Calculate the value of σ1 at the failure if σ3 = 12.5 MPa using MohrCoulomb criterion.
19
(iii) Draw the Mohr's envelop curve
Solution:
(i) According to Mohr-Coulomb Friction Law, the stress conditions at failure is:
cosφ
1 + sin φ
σ1 = 2c
+ σ3
……………
(a)
1 - sinφ
1 − sin φ
cosφ
σ c = 2c
……………
(b)
1 - sinφ
Therefore, from equation (a), the incremental strength due to confining
pressure σ3, is:
σ1 = 2c
cosφ
1 + sin φ
+ σ3
1 - sinφ
1 − sin φ
∆ σ1 = ∆σ3 (1+sinφ)/(1-sinφ)
To calculate angle of internal friction, the incremental strength of rock at low
confining pressure will be used.
For specimen No. 1 and No.2,
∆σ1 = 96.7500 – 80.6250 = 16.1250
∆σ3 = 6.0 – 3.0 = 3.0
16.125 = 3.0 (1+sinφ)/(1-sinφ)
φ = 43° 20′ 9.32″
tan φ = 0.9435
(Answer)
By using Equation (a) for specimen No.1,
80.6250 = (2c cos43.3359/(1 – sin43.3359)) + (3(1+sin43.3359)/(1-sin43.3359))
c = 13.9104 MPa
(Answer)
By using Equation (b) for specimen No.1,
σc = 2×13.9104×cos43.3359/(1-sin43.3359)
= 64.4998 MPa
(Answer)
(ii) By using Equation (a)
σ1 = 64.4995 + 12.5((1+sin 43.3359)/(1-sin43.3359))
= 131.6869 MPa
20
τ, Shear stress
(iii) To draw the Mohr’ envelop,
Mohr's envelop
σND
σNC
cohesion
σNB
σNA
σN
σ3=0
σ3D
σc
σ1A
σ1B
σ1C
σ1D σN, Normal stress
Figure (1) Mohr's circles
Scale: 1 in = 30 MPa
-------------------------------------------------------
No (8) Write the short note on the following technical indices of rock in mining
practice. (10 marks each)
(i) Crushability of Rock
(ii) Drillability of Rock
(iii) Blastability of Rock
Answer:
(i) Crushability of Rock
In mining practice, there are indices of the breakability of rocks in certain
processes. Crushability is a generalized parameter of various mechanical properties of
rocks including elastic, plastic, and strength properties.
Crushability expresses the energy consumed in grinding a rock under a
dynamic load. Laboratory methods of determining crushability are usually based on
estimation of the power required to crush a definite volume of rock.
21
By one of these methods, crushability determined by dropping a load of 16 kg
onto a specimen from a height of 0.5 m. Crushability is established by the volume of
particles with a diameter under 7 mm obtained as a result of this blow.
Since the crushability (C) of rocks is essentially the reciprocal of toughness, it
may be determined by using the physical constant involved in calculations of
toughness.
C=
E
1
= pl , m3/J
T Eσ c
……..
Equation 5.3
It should be noted that the technological parameters determined by different
methods are not the same and should not therefore be converted to one another.
Crushability plays the same role in mining as strength. Both are generalized
parameters of the breakability of rocks, with the single difference that crushability is
an index based on the energy requirements of breaking rocks while strength is linked
only with the firmness of rocks.
(ii) Drillability of Rock
Drillability is the degree of resistance of a rock to breaking by a drilling tool.
It embraces such mechanical characteristics of rocks as their elastic properties,
strength and plasticity, and also technological indices like hardness, abrasiveness, etc.
Drillability is technological parameter indication the depth of hole drilled in
the rock being tested in one clear minute of drilling time in standard conditions.
Conversely it may give the amount of clear drilling time required to drill one linear
meter of hole in the same conditions.
Drillability depends on the design of the drilling equipment used and its
operating conditions. Therefore, in drawing up any technological scale of rocks based
on drillability one has strictly to observe the standard conditions, such as a definite
tool, strengthened by standard alloys, a definite diameter of blast hole and operating
conditions (pressure in the air piping, etc.).
Obviously, any scale so prepared will be applicable only to a definite drilling
tool. Consequently a scale of drillability is only adoptable for calculating the amount
of drilling equipment of a given type required, determining its productivity, and
standardizing the speed of drilling.
(iii) Blastability of Rock
Blastability characterizes the degree of resistance of rocks to explosives
required to crush 1 m3 of rock in a mass (powder factor (or) specific charging).
22
Blastability includes the elastic (dynamic Young’s modulus), Strength (ultimate
tensile and shear strengths), and plastic properties of the rock. It is utilized to establish
the consumption of explosive to work a deposit, to draw up standards for blasting
work, and to fix working norms for the blaster.
The quantitative indices of blast ability depend not only upon the properties of
rock but also upon the type of explosive used and the technique of placing (drilling
pattern) it in the rock; therefore all these conditions must also be strictly controlled
when drawing up any scale of rocks on the basis of blastability.
In Soviet mining practice rocks are grouped in 16 categories on the basis of
the unit consumption of ammonite No.9, which produces a crater of standard throw.
In most cases in mining, explosive charges of normal breaking are used, by
which only the rock of the volume of the crater is broken. The unit consumption of
explosives for such charges comes to a third of the unit expenditure for charges of
normal throw.
----------------------------------------------------------------------------------------No (9) Write the short note on the following topics. (5 marks each)
(i) Intact rock
(ii) Stratified rock
(iii) Moderately jointed rock
(iv) Blocky and seamy rock
Answer:
(i) Intact rock:
Intact rock contains neither joints nor hair cracks. Hence, if it breaks, it breaks
across the sound rock. On account of the injury to the rock due to blasting, spalls may
drop off the roof several hours or days after blasting. This is known as a spalling
condition. Hard, intact rock may also be encountered in the popping condition
involving the spontaneous and violent detachment of rock slabs from the sides or roof.
(ii) Stratified rock:
Stratified rock consists of individual strata with little or no resistance against
separation along the boundaries between strata. The strata may or may not be
weakened by transverse joints. In such rock the spalling condition is quite common.
(iii) Moderately jointed rock:
This rock contains joints and cracks, but the blocks between joints are locally
grown together or so intimately interlocked that vertical walls do not require lateral
23
support. In rocks of this type, both spalling and popping conditions may be
encountered.
(iv) Blocky and seamy rock:
This rock consists of chemically intact or almost intact rock fragments, which
are entirely separated from each other and imperfectly interlocked. In such rock,
vertical walls may require lateral support.
-----------------------------------------------------------------No (10) Explain the basic concept of CSIR (Bieniawski) rock mass classification for
the determination of the rock mass quality. (20 marks)
Answer:
In classification of the quality of jointed rock mass behavior, Bieniawski
suggested that the rock mass should be:
1. divided into groups of similar behavior;
2. provided a good basic for understanding the characteristics of the rock mass;
3. facilitated the planning and the design of structures in rock by yielding
quantitative data and
4. Provided a common basic for effective communication among all persons
concerned with geomechanics problems.
Therefore, the adopted classification is:
1. simple and meaningful in terms; and
2. based on measurable parameters, which is determined quickly and cheaply
in the field.
In order to satisfy these requirements, Bieniawski proposed his Geomechanics
Classification, which should incorporate the following parameters:
1. Rock Quality Designation (RQD)
2. State of weathering,
3. Uniaxial compressive strength of intact rock,
4. Spacing of joints and beddings,
5. Strike and dip orientations,
6. Separation of joints,
7. Continuity of joints, and
8. Ground water inflow.
24
In the practical application of the original CSIR Geomechanics Classification,
Bieniawski modified his classification system by eliminating the state of weathering
as a separate parameter since its effect is accounted for by the uniaxial compressive
strength, and by including the separation and continuity of joints in a new parameter,
the condition of joints in a new parameter. In addition, the strike and dip orientations
of joints were removed from the list of basic parameters had been considered.
Therefore, Bieniawski has considered the five basic classification parameters.
1. Strength or intact rock material.
2. Rock Quality Designation.
3. Spacing of joints.
4. Condition of joints.
5. Ground water conditions.
----------------------------------------------------------------------------------------No (11) Discuss briefly the influence of clay seams and fault gouge on the rock mass
strength. (10 marks)
Answer:
In classification of the rock mass quality, it is not enough to determine the
physical condition of rock mass. Brekke and Howard point out the importance to
qualify discontinuity infillings, which have a significant influence upon the
engineering behavior of the rock mass containing these discontinuities. Although their
list does not constitute a rock mass classification, it is include in this discussion
because of the important engineering consequences, which can result from neglecting
these facts when designing and excavation.
Brekke and Howard’s comments on discontinuity infilling are as follows:
1. Joints, seams and sometimes ever-minor faults may be healed through
precipitation from solution of quartz or calcite. In this instance, the
discontinuity may be ‘welded’ together. Such discontinuities may, however,
have broken up again, forming new surface.
2. Clean discontinuities, i.e., without fillings or coatings. Many of the rough
joints or partings will have favorable character. Close to the surface, however,
it is imperative not to confuse clean discontinuities with ‘empty’
discontinuities where filling material has been leached and washed away due
to surface weathering.
25
3. Calcite fillings may, particularly when they are porous or flaky, dissolve
during the lifetime of the underground opening. Their contribution to the
strength of the rock mass will then, of course, disappear. This is a long-term
stability (and sometimes fluid flow) problem, which can easily be overlooked
during design and construction. Gypsum fillings may be having the same way.
4. Coating or filling of chlorite, talc and graphite give very slippery (i.e., low
strength) joint, seams or faults, in particular when wet.
5. Inactive clay material in seams and faults naturally represents a very weak
material that may squeeze or be washed out.
6. Swelling clay may cause serious problems through free swell and consequent
loss of strength, or through considerable swelling pressure when confined.
7. Material that has been altered to a more cohesionless material (sand-like) may
run or flow into the tunnel immediately following excavation.
In contrast to the comment by Merritt that joints containing clay fillings may
occur near the surface, Brekke and Selmerolsen report that clay fillings with a
very low degree of consolidation have been encountered at considerable depth.
Hence, the underground excavation designer can never afford to ignore the
danger, which can arise as a result of the pressure of these features.
-----------------------------------------------------------------------------------------
12.*** A vertical mine opening of 4.0 m diameter (final diameter) is to be connect from level
11 level 10 through the jointed rock mass as shown in following figure. The result of
laboratory data and the following information are recorded as follow. (Min. 07028 Chapter I)
Ground surface
Level 10
2m
4m
Z=300.0
30.0m
Level 11
Given
- Uniaxial compressive strength of intact rock = 30.50 MPa
-
Angle of internal friction of rock mass = 35 degree
-
Hoek and Brown strength parameter ‘s ‘ = 0.0025
-
Unit weight of rock mass = 0.0265 MN/m3
-
Swell factor = 1.1
-
Radius of plastic zone = 4.0 m
Calculate the required support pressure at the beginning point of level 11
Answer
σ cj = sσ ci
2
= 0.0025 × (30.50) 2 = 1.525MPa
σ v = γh = 0.0265 × 300 × 1.1 = 8.745MPa
σ v > σ cj ∴Plastic condition
1 − sinφ
= 0.25
1 − sinφ
σ h = i(σ v − σ cj )
i=
= 0.25(8.745 - 1.525) = 1.805 MPa
σt= 2 σh =2 × 1.805 =3.61 MPa
σt>σh (unstable condition)
1
2
σ rb = (σ t − σ cj )(1 − sin φ )
= 0.415 MPa
j=
1
i
a
2
Pi = σ rb ( ) j −1 = 0.415( ) 4−1 = 0.051MPa (Ans:)
b
4
13.*** A shaft of 10 ft diameter is to be sunk from ground surface to the depth of 500 ft through
the joint rock mass. As the results of laboratory testing and field testing, the following
information was recorded.(20marks) Chapter I (8th Semester)
- Uniaxial compressive of intact rock = 3500 psi
- Friction angle of rock mass, degree = 30
- CSIR(RMR) rating
= 72
- Unit weight of rock mass
= 165 pcf
Assume that the rock mass is homogenous and hydrostatic stress at which failure occur.
Determine the depth at which failure will occur and the required support pressure at the deepest
end when the clastic zone extends to 20 ft from the center of shaft.
(Min. 07028 Chapter I)
Answer
RMR − 100 

σ cj = σ ci exp(
18.75 

72 − 100 

= 5.459 psf
= σ ci exp(
18.75 

Assume the depth of the failure is equal hf
σ v = γh f = 165 x hf
For hydrostatic stress field
σv = σh = 165 x hf
σt = 2σh = 2 x165 x hf = 330 hf
For equilibrium condition
σt = σcj
330 hf = 5.459
hf = 0.0165 ft
σv = γh =165 x 500 =8250 psf
σv > σcj ( plastic condition)
1 − sinφ
= 0.333
1 − sinφ
σ h = i(σ v − σ cj ) = 27470.682 psf
i=
σt > σcj (unstable condition)
σt= 2 σh =2 × 27470.682 =54941.364 psf
1
2
σ rb = (σ t − σ cj )(1 − sin φ )
= 13733.976 psf
j=
1
=3
i
5
a
Pi = σ rb ( ) j−1 = 13733.976( ) 3−1 = 858.3735psf (Ans:)
20
b
14.*** A circular tunnel of 15 ft diameter is to be excavated through a section of totally crushed
rock (Assume as cohesionless ground). Tunnel will be driven at a depth of 300 ft overburden.
The properties of rock mass around the tunnel are recorded as follow: (Min. 07028 Chapter I)
Unit weight of rock mass = 150 pcf
Friction angle
= 37°
Calculate the maximum and minimum support pressure when the clastic zone extends to 15 ft
from the center of tunnel. (20 marks) Chapter I (8th Semester)
Answer
a = 7.5 ft b= 15.0 ft
γ = 150 pcf
φ = 37 degree
σ v = γh
= 150 x300 = 45000 psf
i = 0.25
σh = i σv = 0.25 x 45000 = 11250 psf
σth = 3σh - σv
σtv = 3σv - σh
= 3 x11250 – 45000
= 3 x 45000 – 11250
= - 11250 psf
= 123750 psf
1
2
σ rbv = (σ tv − σ cj )(1 − sin φ )
= 24637. 695 psf
1
2
σ rbh = (σ th − σ cj )(1 − sin φ )
= 2239.790 psf
For vertical support pressure
Piv =
PiAv + PiBv
PiAv = σrb (a /b)j-1 = 3074.711 psf
PiBv =
3 γa
a
(1 − ( ) j − 2 ) = 632.812 psf
2 j−2
b
Piv = 3712.523 psf
For horizontal support pressure
PiAh = σrb (a /b)j-1= 279.973 psf
PiBh = 0
Pih = 279.973 psf
Maximum support pressure Piv = 3712.523 psf
c = 0 (cohesion less)
Minimum support pressure Pih = 279.973 psf
15.*** The cross of a non-working slope in a strip mining operation is shown in the following
figure. The mechanical properties of upper layer are c= 15 KN/m2, φ =20° and γ =17 KN/m3.
The lower layer has c = 25 KN/m2, φ =15° and γ =19 KN/m3.
Assume that both materials are homogeneous and slope is located permanently above the
water table. (Min. 07028Chapter II)
9m
r =30 m
5
1.5
2
20 m
4
3
Upper layer
2
1
Answer
For upper layer
c = 15 KN/m2
φ =20°
γ =17 KN/m2
φ =15°
γ =19 KN/m2
For lower layer
c = 25 KN/m2
Weight = Area x density
Normal component, N = W cos α
Tangential component, T = W sin α
Lower layer
15m
Slice
Area m2
α(degree)
Weight,
(KN)
W Normal
Tangential
component
component
N (KN)
T (KN)
1
5.85
10
111.15
109.46
-19.30
2
59.85
10
1137.15
1119.87
197.46
3
134.60
29
2557.40
2236.75
1239.85
4
62.50
45
1062.50
751.30
751.30
5
25.00
64
425.00
186.31
381.98
ΣWT= 2591.24
for upper layer
Σ N = 751.3 + 186.31 = 937.61
Σ N tan φ = 937.61 x tan 20°
For lower layer
Σ N =109.46 + 1119.87 + 2236.45 = 3466.08
Σ N tan φ = 3466.08 x tan 15°
Σ N tan φ(total) = 937.61+3466.08 = 1269.99 KN
crθ = 15 x 30 x 36/180 x π + 25 x 30 x 56/180 x π =1015.77 KN
F=
crθ + ∑ N tan φ
∑W
T
= 0.895 (Ans:)
16.*** The dimension of slope and the mechanical properties of each rock layer are shown in the
following figure.
8.7m
90°
r =30 m
3
4
Upper layer
2
1
30.2 m
Upper layer
12.5m
20 m
Lower layer
Cohesion = 25.0 KN/m2
φ =29° γ =20.0 KN/m3.
Lower layer
Cohesion = 55.0 KN/m2
φ =20.5° γ =22.0 KN/m3
Assume that both materials are homogeneous and slope is permanently situated above the
ground water table. Determine the factor of safety for the slip circle shown in above figure.
(Note: Tension crack is allowed.) (Min. 07028 Chapter II)
Answer
For upper layer
c = 25 KN/m2
φ =29°
γ =20 KN/m2
φ =20.5°
γ =22 KN/m2
For lower layer
c = 55 KN/m2
Weight = Area x density
Normal component, N = W cos α
Tangential component, T = W sin α
Slice
Area m2
α(degree)
Weight,
(KN)
W Normal
Tangential
component
component
N (KN)
T (KN)
1
38.5
7
847.0
840.68
-103.22
2
101.10
13
2224.2
2167.19
500.33
3
122.5
35
2695.50
2208.02
1546.07
4
43.75
60
875.0
437.50
757.77
ΣWT= 2700.95
For upper layer
Σ N = 437.50
Σ N tan φ = 437.50 x tan 20°
for lower layer
Σ N =840.68 + 2167.19 + 2208.02 = 5125.89
Σ N tan φ = 5125.89 x tan 15°
Σ N tan φ(total) = 2192.65 KN
crθ = 25 x 30 x 14/180 x π + 55 x 30 x 76/180 x π =2371.90 KN
crθ + ∑ N tan φ
F=
∑WT
= 1.69 (Ans:)
If tension crack is allowed:
hc =
2c
φ
tan (45 + ) = 4.24 m
γ
2
L = rθ′
4.24 = rθ′
θ′ = 0.141
= 79.92°- 5°= 74.92°
crθ = 25 x 30 x 5/180 x π + 55 x 30 x 76/180 x π =2357.09 KN
crθ + ∑ N tan φ
F=
∑WT
= 1.64 (Ans:)
&&&&&&&&&&&&&&&&&&&&&&&&&&&