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Effects of Loading Rate and Pore Pressure on
Compressive Strength of Rocks
S. Khamrat
K. Fuenkajorn
Geomechanics Research Unit
Institute of Engineering
Suranaree University of Technology
The 11th International Conference on Mining, Materials and Petroleum Engineering
The 7th International Conference on Earth Resources Technology
ASEAN Forum on Clean Coal Technology
November 11-13, 2013, Chiang Mai
2
Outline
Background and Rationale
Objectives
Rock Specimens and Preparation
Laboratory Testing
Test Results
Conclusions and Discussions
Background and Rationale
Ground water
Dry zone
Saturated zone
Dead weight
water
(bridge)
Before
Ground water
Dry zone
Saturated zone
Dead weight
water
After
3
4
Background and Rationale…
Srinakarin Dam http://www.thai-tour.com
Bhumipol Dam http://roggerroll.wordpress.com
Background and Rationale…
Masuda (2001) studies the effects of water
on rock strength in granite and andesite.
The failure strength decreased linearly as
the logarithm of the strain rate decreased.
5
Background and Rationale…
Cobanoglu and Celik (2012) determine the
uniaxial compressive strength tested in the
dry and saturated conditions.
The average saturated to dry strength ratios
of travertines is 0.922.
6
Background and Rationale…
Vasarhelyi (2003) determined the unconfined
compressive strength of British sandstones.
Statistically the saturated UCS is 75.6% of
the dry (UCSsat = 0.759UCSdry), while the
saturated tangent and secant moduli are
76.1 and 79.0% of the dry samples
respectively.
7
Background and Rationale…
Li et al. (2012) study the influence of water
content and anisotropy on the strength and
deformability of sedimentary rocks.
The influence of water are reflected as a
reduction of Young's modulus and increase
of Poisson's ratio.
8
9
Objectives
Determine the effects of pore pressure on the compressive strength and elastic properties of granite, marl and marble.
Determine stress rate and confining pressure effect on the rock compressive strength.
Assess the predictive capability of three-dimensional failure criteria that can be applied in the design and stability analysis of rock embankments and foundations under dry and saturated conditions.
10
Rock Specimens and Preparation
Granite Marl Marble
11
Rock Specimens and Preparation…
The specimens are submerged under water in a pressure vacuum chamber.
Water Content
The water content (W) in rock can be calculated by:
W = (Ww / Ws) 100 (1)
where Ww = mass of water in rock
Ws = dry mass of rock
Every two hours
Laboratory Testing
12
Granite Marl
Marble
Wave = 0.141 0.03
Laboratory Testing…
13
Wave = 0.141 0.03
Wave = 2.705 0.62
Wave = 0.093 0.03
Laboratory Testing…
A polyaxial load frame (Fuenkajorn & Kenkhunthod, 2010) 14
Cantilever beam
Laboratory Testing…
Applied loading rate varies from 0.001, 0.01, 0.1,
1 and 10 MPa/s
Applied confining pressure (3) varies from 0, 3,
7, 12 MPa.
15
Laboratory Testing…
16
Dry condition
1
33
1
33
(Neoprene sheet)
Saturated condition
1
33
(Perforated Neoprene sheet)
1
33
Laboratory Testing…
Hydraulic Cylindrical 17
Laboratory Testing…
18
Laboratory Testing…
19 Lateral stain
Axial stain
Laboratory Testing…
Axial stresses (1) is increased until failure
occurs.
The axial strain, lateral strain, and time are
monitored.
20
Dry Saturated
3 =
0 MPa
3 MPa
7 MPa
12 MPa
1 = 1 MPa/s.
SaturatedDry
1 = 0.001 MPa/s.
21
Shear
failure mode
Extension
failure mode
Rock Samples after Testing
Test Results
Coulomb criterion
- Shear strength,
- Cohesion, c
- Internal friction angle,
Elastic parameters
- Elastic modulus, E
- Poisson’s ratio,
Strength of rock Maximum compressive strength
Strain energy density criterion
- Distortional strain energy, Wd
- Mean strain energy, Wm
Test Data
Data Analysis
Test Results…
23
1
,f (
MP
a)
0.0001 0.001 0.01 0.1 1 10
Granite
0
50
100
150
200
250
300
3
7
12
3 = 0 MPa
Dry
1
,f (
MP
a)
0.0001 0.001 0.01 0.1 1 10
Marl
0
50
100
150
3
7
12
3 = 0 MPa
Dry
¶1/¶t (MPa/s)¶1/¶t (MPa/s)
1
,f (
MP
a)
0.0001 0.001 0.01 0.1 1 10
Marble
0
50
100
150
¶1/¶t (MPa/s)
3
7
12
3 = 0 MPa
Dry
Saturated
1
,f (
MP
a)
0.0001 0.001 0.01 0.1 1 10
Granite
0
50
100
150
200
250
300
3
7
12
3 = 0 MPa
Dry Saturated
1
,f (
MP
a)
0.0001 0.001 0.01 0.1 1 10
Marl
0
50
100
150
3
7
12
3 = 0 MPa
Dry
¶1/¶t (MPa/s)¶1/¶t (MPa/s)
1
,f (
MP
a)
0.0001 0.001 0.01 0.1 1 10
Marble
0
50
100
150
¶1/¶t (MPa/s)
3
7
12
3 = 0 MPa
Dry Saturated
Test Results…
Coulomb Criterion
The shear strength () can be represented by
this equation
= c + n tan (2)
where n = the normal stress,
c = the cohesion,
= friction angle.
24
n
33
1
1
Test Results…
Cohesion
c = b ((1-sin) / (2-cos) (3)
Internal friction angle
= arc sin (m-1) / (m+1) (4)
c = χ· ln(∂σ1/∂t) + ψ (5)
= ω· ln(∂σ1/∂t) + ι (6)
The parameters χ, γ, ω, ι are empirical parameters.
Test Results…
Substituting equations (5) and (6) into (2)
Equation (2) can be rewritten as
= [ χ·ln(∂1/∂t) + ψ] + σn tan [ ω·ln(∂1/∂t) + ι] (7)
26
Test Results…
27
c (
MP
a)
0.001 0.01 0.1 1 10
Marl
cDry = 0.497ln(¶1/¶t)+11.58 MPa
0
5
10
15
0.0001
c (
MP
a)
0.0001 0.001 0.01 0.1 1 10
Marble
cDry = 0.147ln(¶1/¶t)+10.52 MPa
0
5
10
15
¶1/¶t (MPa/s)
c (
MP
a)
0.0001 0.001 0.01 0.1 1 10
Granite
cDry = 0.269ln(¶1/¶t)+9.98 MPa
0
5
10
15
c (
MP
a)
cSat = 0.479ln(¶1/¶t)+11.48 MPa
0.001 0.01 0.1 1 10
Marl
cDry = 0.497ln(¶1/¶t)+11.58 MPa
0
5
10
15
0.0001
c (
MP
a)
cSat = 0.170ln(¶1/¶t)+10.58 MPa
0.0001 0.001 0.01 0.1 1 10
Marble
cDry = 0.147ln(¶1/¶t)+10.52 MPa
0
5
10
15
¶1/¶t (MPa/s)
c (
MP
a)
cSat = 0.290ln(¶1/¶t)+9.55 MPa
0.0001 0.001 0.01 0.1 1 10
Granite
cDry = 0.269ln(¶1/¶t)+9.98 MPa
0
5
10
15
Test Results…
28 ¶1/¶t (MPa/s)
Dry = 0.431ln(¶1/¶t) + 61.4 degrees
0.0001 0.001 0.01 0.1 1 10
(
De
gre
es)
Marble
0
10
20
30
40
50
60
70
Dry = 0.431ln(¶1/¶t) + 61.4 degrees
0.0001 0.001 0.01 0.1 1 10
(
De
gre
es)
Granite
0
10
20
30
40
50
60
70
Dry = 0.263ln(¶1/¶t) + 43.8 degrees
0.0001 0.001 0.01 0.1 1 10
(
De
gre
es)
Marl
0
10
20
30
40
50
60
70
¶1/¶t (MPa/s)
Dry = 0.431ln(¶1/¶t) + 61.4 degrees
Sat = 0.364ln(¶1/¶t) + 59.6 degrees
0.0001 0.001 0.01 0.1 1 10
(
De
gre
es)
Marble
0
10
20
30
40
50
60
70
Dry = 0.431ln(¶1/¶t) + 61.4 degrees
Sat = 0.364ln(¶1/¶t) + 59.6 degrees
0.0001 0.001 0.01 0.1 1 10
(
De
gre
es)
Granite
0
10
20
30
40
50
60
70
Dry = 0.263ln(¶1/¶t) + 43.8 degrees
Sat = 0.166ln(¶1/¶t) + 42.8 degrees
0.0001 0.001 0.01 0.1 1 10
(
De
gre
es)
Marl
0
10
20
30
40
50
60
70
Test Results…
Elastic Parameters
The elastic modulus (E), Poisson’s ratio () can
be determined by:
G = (1/2)(oct/oct) (8)
= (1/3) [(3m /) 2G] (9)
E = 2G (1+) (10)
= /(2(+G)) (11)
29
1 1
2
3
2
3
Test Results…
The elastic parameters can be determined as
a function of the loading rate as:
E = κ (∂1/∂t)ξ (12)
= α ln (∂1/∂t)+β (13)
The parameters κ, ξ, α, β are empirical parameters.
30
Test Results…
31
Granite
5
10
15
0.0001 0.001 0.01 0.1 1 10
EDry = 12.546(¶1/¶t)0.094
GPaE
(G
Pa
)
0
5
10
15
00.0001 0.001 0.01 0.1 1 10
EDry = 9.9409(¶1/¶t)0.0858
GPa
E (
GP
a)
Marl
Marble
5
10
15
00.0001 0.001 0.01 0.1 1 10
¶1/¶t (MPa/s)
E (
GP
a)
EDry = 9.11(¶1/¶t)0.076
GPa
Granite
5
10
15
0.0001 0.001 0.01 0.1 1 10
ESat = 10.110(¶1/¶t)0.079
GPa
EDry = 12.546(¶1/¶t)0.094
GPaE
(G
Pa
)
0
5
10
15
00.0001 0.001 0.01 0.1 1 10
ESat = 8.457(¶1/¶t)0.0755
GPa
EDry = 9.9409(¶1/¶t)0.0858
GPa
E (
GP
a)
Marl
Marble
5
10
15
00.0001 0.001 0.01 0.1 1 10
¶1/¶t (MPa/s)
ESat = 7.14(¶1/¶t)0.076
GPa
E (
GP
a)
EDry = 9.11(¶1/¶t)0.076
GPa
Test Results…
32
Granite
0.0001 0.001 0.01 0.1 1 10
Dry
Dry = -0.0008ln(¶1/¶t)+0.279
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Marl
0.0001 0.001 0.01 0.1 1 10
Dry
Dry = 0.0011ln(¶1/¶t)+0.297
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Marble
0.0001 0.001 0.01 0.1 1 10
¶1/¶t (MPa/s)
Dry
Dry = 0.001ln(¶1/¶t)+0.280
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Granite
0.0001 0.001 0.01 0.1 1 10
Saturated
Dry
Dry = -0.0008ln(¶1/¶t)+0.279
Sat = -0.0008ln(¶1/¶t)+0.282
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35 Saturated
Marl
0.0001 0.001 0.01 0.1 1 10
Dry
Dry = 0.0011ln(¶1/¶t)+0.297
Sat = 0.0001ln(¶1/¶t)+0.292
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Marble
0.0001 0.001 0.01 0.1 1 10
¶1/¶t (MPa/s)
Saturated
Dry
Dry = 0.001ln(¶1/¶t)+0.280
Sat = 0.003ln(¶1/¶t)+0.272
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
Strain Energy Density Criterion
Strain energy is the energy
stored by a system undergoing
deformation in 3D.
The strain energy density
principle is applied here to
describe the rock strength and
deformation under different
loading rates.
33
1 1
2
3
2
3
Test Results…
Test Results…
34
The distortional strain energy (Wd) at failure can
be calculated as follows (Jaeger et al., 2007).
(14)
The mean strain energy (Wm) at failure can can
be calculated as follows
(15)
Test Results…
The elastic parameters G and K can be
determined for each specimen using the following
relations:
(16)
(17)
where E = Elastic modulus
= Poisson’s ratio
35
Test Results…
The octahedral shear strength can be determined as:
oct = [(1/3)[(12)2 + (23)
2 + (31)2 ]]1/2 (18)
The mean stresses can be determined as:
m = (1/3)(1+(2 3)) (19)
36
Wd,Dry = 4.394Wm,Dry + 0.065
R² = 0.99
0 0.2 0.4 0.6 0.8
Wd (
MP
a)
Granite
1
2
3
4
0
Marble
Wm (MPa)
0 0.1 0.2
Wd,Dry = 2.218Wm,Dry + 0.066
R² = 0.99
Wd (
MP
a)
0
0.1
0.2
0.3
0.4
0.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0 0.1 0.2
Wd,Dry = 2.865Wm,Dry + 0.075
R² = 0.98
Wd (
MP
a)
Marl
37
Test Results…
Wm (MPa)
0 0.1 0.2
Wd.Sat = 2.018Wm,Sat + 0.087
R² = 0.95
Wd (
MP
a)
0
0.1
0.2
0.3
0.4
0.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0 0.1 0.2
Wd,Sat = 2.501Wm,Sat + 0.094
R² = 0.99
Wd (
MP
a)
Wd (
MP
a)
Wd,Sat = 3.923Wm,Sat + 0.106
R² = 0.99
0 0.2 0.4 0.6 0.8
1
2
3
4
0
Parameters Conditions
Dry Saturated
Compressive strength, 1
Elastic Modulus, E
Poisson’s ratio,
Cohesion, c
Friction angle,
Distortional strain energy, Wd
Mean strain energy, Wm
38
Summary properties of rock
Conclusions and Discussion
The compressive strength of dry specimens is higher
than that of the saturated specimens due to effect of
pore pressure.
The strength of rock under low loading rate is lower
than of high loading rate because under low loading
rate rocks respond to stresses by ductile behavior
not brittle behavior.
Conclusions and Discussion…
The strength of the saturated specimens under high
loading rates is reduced the trapped pore water
builds up the pore pressure.
On the other hand, under low loading rates thus the
pore water has sufficient time to seep out from the
specimens, the rock behavior is similar to dry
condition.
Conclusions and Discussion…
The elastic modulus of the dry specimens is higher
than that of the saturated specimens that agrees with
Li et al. 2012; Vasarhelyi, 2003.
The strength criterion can be used to predict the
strength and deformation of in-situ rocks under dry
and saturated conditions.
Acknowledgements
42
Funded by Suranaree University of Technology
and by the Higher Education Promotion and National Research University of Thailand
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