13 Earth Materials - University College London Earth Materials.pdf · where E is the Young’s Modulus and υis the Poisson’s ratio. Poisson’s ratio varies between 0.2 and 0.3

GNH7/GG09/GEOL4002 EARTHQUAKE SEISMOLOGY AND EARTHQUAKE HAZARD

Earth Materials

Lecture 13

Earth Materials


Constitutive equationsThese are relationships between forces and deformation in a continuum, which

define the material behaviour.

Hooke’s law of elasticity

Robert Hooke (1635-1703) was a virtuoso scientist contributing to geology, palaeontology, biology as well as mechanics

LengthExtensionE

AreaForce

×=

σn = E εn

where E is material constant, the Young’s Modulus

Units are force/area – N/m2 or Pa

Hooke’s law

klijklij C εσ =


Shear modulus and bulk modulus

Shear or rigidity modulus:

sSS G εµεσ ==Bulk modulus (1/compressibility):

vKP ε=−Can write the bulk modulus in terms of the Laméparameters λ, µ:

K = λ + 2µ/3

and write Hooke’s law as:

σ = (λ +2µ) ε

Young’s or stiffness modulus:

nn Eεσ =

Mt Shasta andesite


Young’s Modulus or stiffness modulusYoung’s Modulus or stiffness modulus:

nn Eεσ =

Interatomic distance

Interatomic force


Shear Modulus or rigidity modulusShear modulus or stiffness modulus:

ss Gεσ =

Interatomic distance

Interatomic force


Hooke’s Law

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:

ijkkijij µεελδσ 2+=

1212

1133221111

22)(

µεσµεεεελσ

=+++=For example: Normal stress

Shear stress

This can be deduced from substituting into the Taylor expansion for stress and differentiating.

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.


Hooke’s Law

In terms of principal stresses and principal strains:

ijkkijij µεελδσ 2+=

3333221133

2233221122

1133221111

2)(2)(2)(

µεεεελσµεεεελσµεεεελσ

+++=+++=+++=

Hooke’s Law:

Consider normal stresses and normal strains:

3213

3212

3211

)2()2(

)2(

εµλελελσελεµλελσελελεµλσ

+++=+++=+++=


Hooke’s Law

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.

Can write in inverse form:

3213

3212

3211

1

1

1

σσυσυε

συσσυε

συσυσε

EEE

EEE

EEE

+−−=

−+−=

−−=


Constitutive equation: uniaxial elastic deformationAll components of stress zero except σ11:

3333221133

2233221122

1133221111

2)(02)(0

2)(

µεεεελσµεεεελσ

µεεεελσ

+++==+++==

+++=

11113322

111111

)(2

)23(

νεεµλ

λεε

εεµλ

µλµσ

−=+

−==

=++

= E

where E is Young’s Modulus and ν is Poisson’s ratio.

The solution to this set of simultaneous equations is:

σ11

ε11

dσ11/dε11 = E

σ11

σ22 = 0

σ11

σ33 = 0


Constitutive equations: isotropic compressionNo shear or strain; all normal stresses

equal to –p; all normal strains equal to εv /3.

VV KP εεµλ =⎟⎠⎞

⎜⎝⎛ +=−

32

where K is the bulk modulus; hence K = λ + 2/3µ

σ11 = -p

σ22 = -p

σ33 = -p

σ22 = -p

σ11 = -p

σ33 = -p

P = - 1/3 (σ11 + σ22 + σ33 ) = - 1/3 σii

332211 εεεε ++=∆

=VV

v -p

εv

-dp/dεv = K


Typical ERubber 7 MPaNormally consolidated clays 0.2 ~ 4 GPaBoulder clay (oversolidated) 10 ~20 GPaConcrete 20 GPaSandstone 20 GPaGranite 50 GPaBasalt 60 GPaSteel 205 GPaDiamond 1,200 GPa

Young’s Modulus (initial tangent) of Materials


50 MPa5 MPaGranite

40 MPa4 MPaBasalt

40 MPa4 MPaConcrete

10 MPa1 MPaSandstone

1 MPa300 kPaSoil

2,000 MPa30 MPaRubber

3,000 MPa3,000 MPaSteel piano wire

100 / 3 MPa100 / 3 MPaSpruce along/across grain

Compressive strength - unconfined

Uniaxial tensile strength

“Strength” of Materials


FractureCalculate the stress which will just separate two adjacent layers of atoms x layers apart x

σ

σ

ε

strain energy / m2 = ½ stress x strain x vol

Ue = ½ σn εn x

σ

εHooke’s law: εn = σn / E

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:

sn UE

x 22

2

≈σ

or xEU

xEU ss

n ≈≈ 2σ

For steel: Us = 1 J/m; E = 200 GPa;

x = 2 x 10-10 m⇒ σmax = 30 GPa ≈ E / 10


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


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


Engineering behaviour of soils• Soils are granular materials – their behaviour is quite different to crystalline rock

Uniaxial deformation

Shear deformation

• Properties are highly dependent on water content

• The curvature of the stress-strain is largest near the origin

• Deformation is strongly non-linear

• The constitutive relation for shear deformation, found from hundreds of experiments is:

rs

rss G

εεεεσ

+= 0

εr is the reference strain


Constitutive equation for soilsSoils are fractal materialsThere 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 :

d

s

s

s

s cdd

⎟⎟⎠

⎞⎜⎜⎝

⎛=

εσ

εσ

constitutive equation for soils


Constitutive equation for soilsd

s

s

s

s cdd

⎟⎟⎠

⎞⎜⎜⎝

⎛=

εσ

εσ

From fractals:

Integrating and setting d = 2:rs

rss G

εεεεσ

+= 0

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


Stress-strain curve of a soil as compared with that of a crystalline rock – note different definition of rigidity

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

Liquefaction of soils: phase transitionThis aspect of soil behaviour is completely different from crystalline rock


Constitutive equation: viscous flowIncompressible viscous fluidsFor viscous fluids the deviatoric stress is proportional to strain-rate:

where η is the shear viscosityijij

•

= '' 2 εησ

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.

ε

σ1/2η


Experimental techniques to study friction

Shear box

Rotary shearTriaxial test

Direct shear


Experimental resultsAt low normal stresses (σN < 200 MPa)

Linear friction law observed: σS = µ σNA significant amount of variation between rock types: µcan vary between 0.2 and 2.0 but most commonly between 0.5 – 0.9Average for all data given by: σS = 0.85 σN

At higher normal stresses (σN > 200 MPa)

Very little variation between wide range of rock types (with some notable exceptions – esp. clay minerals which can have unusually low µ

But friction does not obey Amonton’s Law (i.e. straight line through origin) but Coulomb’s Law

Best fit to all data given by:

σS = 50 + 0.6 σN


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

σN

φµ= tan φ

φ is the “angle of friction”slope µ


Field observationsWe are concerned with friction related to earthquakes, i.e., friction on faultsFaults 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


Summary: Byerlee’s Friction Laws

All data may be fitted by two straight lines:σN < 200 MPa σS = 0.85 σN

σN > 200 MPa σS = 50 + 0.6 σN

These are largely independent of rock typeIndependent of roughness of contacting surfacesIndependent of rock strength or hardnessIndependent of sliding velocityIndependent of temperature (up to 400oC)


Experimental results of triaxial deformation tests

σ3 σ3

σ1 σ1 σ1

σ1 σ1 σ1 σ1

σ1

ConfiningPressure PC

Differential Stress (σ1 - σ3) Total

AxialStress σ1

PCHydrostaticPC applied inall directionsprior to thedifferentialloading.

PC PC = σ2 = σ3

Modes of brittle fracture in a triaxial system


Bottom steelFv520 piston

Pressure Vessel

Fibrous alumina insulation

Bottom plug

Bottom waveguide

Top wave-guide

Pore fluid inlet

Rock Specimen

Load Cell

Alumina coil support

Alumina Disc

Top steelFv520 piston

Top pyrophillite enclosing disc

Bottomenclosing pyrophillite

block

Insulating filler

To AE transducer

Fluid outlet fitting Thermocouple feedthrough

Pressure fittings

Actuator applying axial load


Experimental resultsSchematic 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.


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

µi = tan φi

φi is the “angle of internal friction”

σS

σN

φi

slope µ i

Tensile fracture

(σ2 = -σT)

Shear fracture

CσT – tensile strength

Documents

13 Earth Materials - University College London Earth Materials.pdf · where E is the Young’s Modulus and υis the Poisson’s ratio. Poisson’s ratio varies between 0.2 and 0.3