Earth Materials

Earth Materials
Lecture 13
Earth Materials
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Hooke’s law of elasticity
Force
Extension
= E×
Area
Length
Hooke’s law
σn = E εn
where E is material constant, the
Young’s Modulus
Robert Hooke (1635-1703) was a virtuoso
scientist contributing to geology,
palaeontology, biology as well as mechanics
ß Constitutive equations
Units are force/area – N/m2 or Pa
σ ij = C ijkl ε kl
These are relationships between forces and deformation in a continuum, which
define the material behaviour.
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Shear modulus and bulk modulus
Young’s or stiffness modulus:
σ n = Eε n
Shear or rigidity modulus:
σ S = Gε S = µ ε s
Bulk modulus (1/compressibility):
− P = Kε v
Mt Shasta andesite
Can write the bulk modulus in terms of the Lamé
parameters λ, µ:
K = λ + 2µ/3
and write Hooke’s law as:
σ = (λ +2µ) ε
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Young’s Modulus or stiffness modulus
Young’s Modulus or stiffness modulus:
σ n = Eε n
Interatomic force
Interatomic distance
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Shear Modulus or rigidity modulus
Shear modulus or stiffness modulus:
σ s = Gε s
Interatomic force
Interatomic distance
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Hooke’s Law
σij and εkl are second-rank tensors so Cijkl is a fourth-rank tensor.
For a general, anisotropic material there are 21 independent elastic moduli.
In the isotropic case this tensor reduces to just two independent elastic
constants, λ and µ.
So the general form of Hooke’s Law reduces to:
σ ij = λδ ij ε kk + 2 µε ij
This can be deduced from substituting into the Taylor expansion
for stress and differentiating.
For example:
σ 11 = λ (ε 11 + ε 22 + ε 33 ) + 2 µε 11
σ 12 = 2 µε 12
Normal stress
Shear stress
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Hooke’s Law
Hooke’s Law:
σ ij = λδ ij ε kk + 2 µε ij
Consider normal stresses and normal strains:
σ 11 = λ (ε 11 + ε 22 + ε 33 ) + 2 µε 11
σ 22 = λ (ε 11 + ε 22 + ε 33 ) + 2 µε 22
σ 33 = λ (ε 11 + ε 22 + ε 33 ) + 2 µε 33
In terms of principal stresses and principal strains:
σ 1 = (λ + 2 µ )ε 1 + λ ε 2 + λ ε 3
σ 2 = λ ε 1 + (λ + 2 µ )ε 2 + λ ε 3
σ 3 = λ ε 1 + λ ε 2 + (λ + 2 µ )ε 3
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Hooke’s Law
Can write in inverse form:
υ
υ
1
σ1 − σ 2 − σ 3
E
E
E
υ
υ
1
ε2 = − σ1 + σ 2 − σ 3
E
E
E
υ
υ
1
ε3 = − σ1 − σ 2 + σ 3
E
E
E
ε1 =
where E is the Young’s Modulus and υ is the Poisson’s ratio.
Poisson’s ratio varies between 0.2 and 0.3 for rocks.
A principal stress component σi produces a strain σI /E in the
same direction and strains (-υ.σi / E) in orthogonal directions.
Elastic behaviour of an isotropic material can be characterized
either by specifying either λ and µ, or E and υ.
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Constitutive equation: uniaxial elastic deformation
All components of stress zero except σ11:
σ11
σ 11 = λ (ε 11 + ε 22 + ε 33 ) + 2 µε 11
σ 22 = 0 = λ (ε 11 + ε 22 + ε 33 ) + 2 µε 22
σ 33 = 0 = λ (ε 11 + ε 22 + ε 33 ) + 2 µε 33
dσ11/dε11 = E
ε11
The solution to this set of simultaneous equations is:
µ (3λ + 2 µ )
ε 11 = E ε 11
σ 11 =
λ+µ
λ
ε 22 = ε 33 = −
ε 11 = −νε 11
2(λ + µ )
σ33 = 0
σ11
where E is Young’s Modulus and ν is Poisson’s ratio.
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
σ22 = 0
σ11
Constitutive equations: isotropic compression
σ33 = -p
No shear or strain; all normal stresses
equal to –p; all normal strains
equal to εv /3.
2 ⎞
⎛
− P = ⎜ λ + µ ⎟ε V = Kε V
3 ⎠
⎝
σ22 = -p
σ11 = -p
∆V
εv =
= ε11 + ε 22 + ε 33
V
σ11 = -p
σ22 = -p
σ33 = -p
-p
P = - 1/3 (σ11 + σ22 + σ33 ) = - 1/3 σii
-dp/dεv = K
where K is the bulk modulus;
hence K = λ + 2/3µ
εv
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Young’s Modulus (initial tangent) of Materials
Rubber
Normally consolidated clays
Boulder clay (oversolidated)
Concrete
Sandstone
Granite
Basalt
Steel
Diamond
Typical E
7 MPa
0.2 ~ 4 GPa
10 ~20 GPa
20 GPa
20 GPa
50 GPa
60 GPa
205 GPa
1,200 GPa
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“Strength” of Materials
Uniaxial tensile
strength
Compressive strength
- unconfined
Soil
300 kPa
1 MPa
Sandstone
1 MPa
10 MPa
Concrete
4 MPa
40 MPa
Basalt
4 MPa
40 MPa
Granite
5 MPa
50 MPa
Rubber
30 MPa
2,000 MPa
Spruce along/across grain
100 / 3 MPa
100 / 3 MPa
Steel piano wire
3,000 MPa
3,000 MPa
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
σ
Fracture
ε
Calculate the stress which will just separate two
adjacent layers of atoms x layers apart
σ
strain energy / m2 = ½ stress x strain x vol
x
Ue = ½ σn εn x
ε
Hooke’s law: εn = σn / E
σ
Ue = σn2 x / 2E
If Us is the surface energy of the solid per square metre, then the total
surface energy of the solid per square metre would be 2Us per square metre
Suppose that at the theoretical strength the whole of the strain energy
between two layers of atoms is potentially convertible to surface energy:
σ n2 x
2E
≈ 2U s
or
Us E
Us E
σn ≈ 2
≈
x
x
For steel: Us = 1 J/m; E = 200 GPa;
⇒ σmax = 30 GPa ≈ E / 10
x = 2 x 10-10 m
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Griffith energy balance
Microcrack in lava
The reason why rocks don’t reach their theoretical strength is because they
contain cracks
Crack models are also used in modelling earthquake faulting
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Dislocations (line defects) in shear
The reason why rocks don’t reach their theoretical shear strength is because
they contain dislocations
Dislocation models are also used in modelling earthquake faulting
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Engineering behaviour of soils
• Soils are granular materials – their
behaviour is quite different to crystalline
rock
• Deformation is strongly non-linear
• The curvature of the stress-strain is largest
near the origin
• Properties are highly dependent on
water content
Uniaxial deformation
• The constitutive relation for shear
deformation, found from hundreds
of experiments is:
εs εr
σ s = G0
εs + εr
εr is the reference strain
Shear deformation
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Constitutive equation for soils
Soils are fractal materials
There is a lognormal distribution
of grain sizes (c.f. crack lengths
in rocks)
Suppose we subject a soil to a
simple shear strain. The shear
forces applied to each grain must
be lognormally distributed since
they are proportional to the grain
surfaces. So the shear modulus
and rigidity must be related by a
power law:
G = c µd
where d is the fractal dimension
of the grain size distribution
replacing G and µ by their
definitions in terms of shear stress
σs and shear strain εs :
⎛σs ⎞
dσ s
= c⎜⎜ ⎟⎟
dεs
⎝ εs ⎠
d
constitutive equation for soils
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Constitutive equation for soils
From fractals:
⎛σs ⎞
dσ s
= c⎜⎜ ⎟⎟
dεs
⎝ εs ⎠
Integrating and setting d = 2:
d
εs εr
σ s = G0
εs + εr
This is the same as the empirical constitutive equation!
This is a hyperbolic stress-strain relation (i.e., like a deformation stress-strain curve)
It may be interpreted as saying that the shear modulus G = dσ/dε of a soil decays
inversely as (1 + τ) where τ = εs / εr is the normalised strain
Note that the stress-strain behaviour of soils cannot be linearized at small
strain
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Liquefaction of soils: phase transition
This aspect of soil behaviour is
completely different from
crystalline rock
Soil liquefaction: Kobe port area
Motion on soft ground to strong
earthquake is fundamentally
different to small earthquakes
because sediments go through a
phase transition and liquefy
Stress-strain curve of a soil as
compared with that of a crystalline
rock – note different definition of
rigidity
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Constitutive equation: viscous flow
Incompressible viscous fluids
For viscous fluids the deviatoric stress
is proportional to strain-rate:
ε
•
σ ij' = 2η ε ' ij
where η is the shear viscosity
1/2η
σ
Viscosity is an internal property of a fluid that offers resistance to flow.
Viscosity is measured in units of Pa s (Pascal seconds), which is a unit of
pressure times a unit of time. This is a force applied to the fluid, acting for
some length of time. A marble (density 2800 kg/m3) and a steel ball bearing
(7800 kg/m3) will both measure the viscosity of a liquid with different
velocities. Water has a viscosity of 0.001 Pa s, a Pahoehoe lava flow 100 Pa
s, an a'a flow has a viscosity of 1000 Pa s. We can mentally imagine a sphere
dropping through them and how long it might take.
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Experimental techniques to study friction
Shear box
Triaxial test
Direct shear
Rotary shear
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Experimental results
ß
At low normal stresses (σN < 200 MPa)
a Linear friction law observed: σS = µ σN
a A significant amount of variation between rock types: µ
can vary between 0.2 and 2.0 but most commonly
between 0.5 – 0.9
a Average for all data given by: σS = 0.85 σN
ß At higher normal stresses (σN > 200 MPa)
a Very little variation between wide range of rock types (with
some notable exceptions – esp. clay minerals which can have
unusually low µ
a But friction does not obey Amonton’s Law (i.e. straight line
through origin) but Coulomb’s Law
a Best fit to all data given by:
a σS = 50 + 0.6 σN
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Simple failure criteria
(a) Friction – Amonton’s Law
1st: Friction is proportional normal load (N)
Hence: F = µ N
- µ is the coefficient of friction
2nd: Friction force (F) is independent of the areas in contact
So in terms of stresses: σS = µ σN = σN tanφ
May be simply represented on a Mohr diagram:
σS
µ
e
p
slo
φ
µ= tan φ
φ is the “angle of friction”
σN
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Field observations
ß
ß
ß
We are concerned with friction related to earthquakes, i.e.,
friction on faults
Faults are interfaces that have already fractured in previously
intact material and have subsequently been displaced in shear
(i.e., have slipped)
Hence they are not “mated” surfaces (unlike joints)
Joint
Fault
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Summary: Byerlee’s Friction Laws
ß
ß
ß
ß
ß
ß
All data may be fitted by two straight lines:
a σN < 200 MPa
σS = 0.85 σN
a σN > 200 MPa
σS = 50 + 0.6 σN
These are largely independent of rock type
Independent of roughness of contacting surfaces
Independent of rock strength or hardness
Independent of sliding velocity
Independent of temperature (up to 400oC)
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Experimental results of triaxial deformation tests
Differential Stress
(σ1 - σ3)
Confining
Pressure PC
Modes of brittle fracture in a triaxial system
Total
Axial
Stress
σ1
PC
Hydrostatic
PC applied in
all directions
prior to the
differential
loading.
PC
PC = σ2 = σ3
σ1
σ1
σ1
σ1
σ3
σ3
σ1
σ1
σ1
σ1
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Actuator
applying
axial load
Fluid outlet fitting
To AE transducer
Thermocouple
feedthrough
Top wave-guide
Pressure Vessel
Load Cell
Insulating filler
Top pyrophillite
enclosing disc
Top steel
Fv520 piston
Alumina coil
support
Alumina Disc
Rock Specimen
Pore fluid inlet
Fibrous alumina
insulation
Bottom steel
Fv520 piston
Bottom
enclosing pyrophillite
block
Bottom wave
guide
Pressure fittings
Bottom plug
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Experimental results
Schematic
stress-strain
curves for rock
deformation
over a range of
confining
pressure
Dependence of
differential
stress at shear
failure in
compression on
confining
pressure for a
wide range of
igneous rocks
Strength of Westerly granite as
a function of confining
pressure. Also shown is
frictional strength.
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD
Simple failure criteria
(b) Faulting – Coulomb’s Law
σS = C + µi σN = σN tanφi
C is a constant – the cohesion
µi is the coefficient of “internal” friction
Tensile fracture
Shear fracture
σS
(σ2 = -σT)
σT – tensile strength
C
µi
e
p
slo
µi = tan φi
φi
σN
φi is the “angle of internal friction”
GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD