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