Transcript PTYS 554 Evolution of Planetary Surfaces Tectonics I PYTS 554 – Tectonics I  Tectonics I        Vocabulary of stress and strain Elastic, ductile and viscous deformation Mohr’s.

PTYS 554
Evolution of Planetary Surfaces
Tectonics I
PYTS 554 – Tectonics I

Tectonics I







Vocabulary of stress and strain
Elastic, ductile and viscous deformation
Mohr’s circle and yield stresses
Failure, friction and faults
Brittle to ductile transition
Anderson theory and fault types around the solar system
Tectonics II




Generating tectonic stresses on planets
Slope failure and landslides
Viscoelastic behavior and the Maxwell time
Non-brittle deformation, folds and boudinage etc…
2
PYTS 554 – Tectonics I

Compositional vs. mechanical terms

Crust, mantle, core are compositionally different
 Earth has two types of crust

Lithosphere, Asthenosphere, Mesosphere, Outer Core and Inner Core are
mechanically different
 Earth’s lithosphere is divided into plates…
3
PYTS 554 – Tectonics I

How is the lithosphere defined?

Behaves elastically over geologic time

Warm rocks flow viscously
 Most of the mantle flows over geologic time

Cold rocks behave elastically
 Crust and upper mantle

Rocks start to flow at half their melting
temperature


Melosh, 2011
Thermal conductivity of rock is ~3.3 W/m/K
At what depth is T=Tm/2
4
PYTS 554 – Tectonics I

5
Relative movement of blocks of crustal material
Moon & Mercury –
Earth –
Wrinkle Ridges
Pretty much everything
Europa – Extension and strike-slip
Mars –
Extension and compression
Enceladus - Extension
PYTS 554 – Tectonics I

6
The same thing that supports topography allows tectonics to occur



Materials have strength
Consider a cylindrical mountain, width w and height h
How long would strength-less topography last?
w
æ w ö2
F = r g h pç ÷
è2ø
1 é4 æ wö
F = ê pç ÷
2 êë 3 è 2 ø
h
3
Weight of the mountain
ù ·
úr v
úû
F=ma for material in
the hemisphere
æ 3g ö
v = ç ÷h
èwø
·
v
æ w ö2 · 1
æ w ö2
p ç ÷ h = 4p ç ÷ v
è2ø
è2ø
2
·
Conserve volume
1 ··
h = 2v or v = h
2
Solution for h: h = h0 e
-t
t
where t =
w
6g
i.e. mountains 10km across would collapse in ~13s
·
PYTS 554 – Tectonics I

7
Response of materials to stress (σ) – elastic deformation
L
ΔL
L
ΔL
Linear (normal) strain (ε) = ΔL/L
Shear Strain (ε) = ΔL/L
s s = G es
s l = E el
G is shear modulus (rigidity)
E is Young’s modulus
Volumetric strain = ΔV/V
dP = K
dV
V
K is the bulk modulus
PYTS 554 – Tectonics I

8
Stress is a 2nd order tensor

Combining this quantity with a vector describing the orientation of a plane
gives the traction (a vector) acting on that plane
si j
i describes the orientation of a plane of interest
j describes the component of the traction on that plane
These components are arranged in a 3x3 matrix
æ s
ç 11 s 12 s 13
s i j = ç s 21 s 2 2 s 23
ç
ç s 31 s 32 s 33
è
ö
÷
÷
÷
÷
ø
T = s i j ei
s 11 s 22 s 33
s 13 s 23 s 12
Are normal stresses, causing normal strain
(Pressure is - 1 (s 11 + s 22 + s 33 ))
3
Are shear stresses, causing shear strain
We’re only interested in deformation, not rigid body rotation so:
s ji = s ij
PYTS 554 – Tectonics I

The components of the tensor depend on the coordinate system used…
ss
s N2
=
s N1
s N2

9
s N1
Shear stresses in one
coordinate system
can appear as normal
stresses in another
There is at least one special coordinate system where the components of the
stress tensor are only non-zero on the diagonal i.e. there are NO shear stresses
on planes perpendicular to these coordinate axes
æ s
0 0
ç 1
si j = ç 0 s2 0
çç
è 0 0 s3
ö
÷
÷
÷÷
ø
Where:
s1 ³ s 2 ³ s 3
These are principle stresses that act parallel
to the principle axes
The tractions on these planes have only one
component – the normal component
Pressure again: - 1 3 (s 1 + s 2 + s 3 )
PYTS 554 – Tectonics I

Principle stresses (s 1 s 2 s 3 ) produce strains in those directions


Principle strains – all longitudinal
(e1 e2 e3 )
Stretching a material in one direction usually means it wants to contract
in orthogonal directions



10
Quantified with Poisson’s ratio
This property of real materials means shear stain is always present
Extensional strain of σ1/E in one direction implies orthogonal
compression of –ν σ1/E


Where ν is Poisson’s ratio
Range 0.0-0.5
Linear strain (ε) = ΔL/L
s l = E el
E is Young’s modulus
ΔL
or
L
Where λ is the Lamé
parameter
G is the shear modulus
PYTS 554 – Tectonics I

Groups of two of the previous parameters describe the elastic response of a
homogenous isotropic solid

Conversions between parameters are straightforward
11
PYTS 554 – Tectonics I

Typical numbers (Turcotte & Schubert)
12
PYTS 554 – Tectonics I

13
Materials fail under too much stress


Elastic response up to the yield stress
Brittle or ductile failure after that
Strain hardening
Special case of
plastic flow
Strain Softening
Brittle failure

Ductile (distributed) failure
Material usually fails because of shear stresses first


Wait! I thought there were no shear stresses when using principle axis…
How big is the shear stress?
PYTS 554 – Tectonics I

14
How much shear stress is there?



Depends on orientation relative to the principle stresses
In two dimensions…
æs1 - s 3 ö
ss =ç
÷ sin 2q
Normal and shear stresses form
è 2 ø
a Mohr circle
s1
ss
æs1 + s 3 ö æs1 - s 3 ö
÷+ç
÷ cos2q
è 2 ø è 2 ø
sN =ç
æs -s 3 ö
radius = ç 1
÷
è 2 ø
s3
q
2q
sN
æs1 + s 3 ö
ç
÷
è 2 ø
Maximum shear stress:
On a plane orientated at 45° to the principle axis
Depends on difference in max/min principle stresses
Unaffected (mostly) by the intermediate principle stress
s2
PYTS 554 – Tectonics I

Consider differential stress
1
 Failure when:
(s 1 - s 3 ) ³ s y
2


15
Failure when:
1 é(s 1 - s 3 ) 2 + (s 1 - s 2 ) 2 + (s 2 - s 3 ) 2 ù ³ s y
û
2ë
Increase confining pressure


(Tresca criterion)
Increases yield stress
Promotes ductile failure

(Von Mises criterion)
Increase temperature


Decrease yield stress
Promotes ductile failure
PYTS 554 – Tectonics I

Low confining pressure


Weaker rock with brittle faulting
High confining pressure (+ high temperatures)

Stronger rock with ductile deformation
16
PYTS 554 – Tectonics I
17

What sets this yield strength?

Mineral crystals are strong, but rocks are packed with microfractures

Crack are long and thin

Approximated as ellipses
a >> b
Effective stress concentrators

Larger cracks are easier to grow


σ
b
a
σ
æ
aö
s Tip = s o ç1+ 2 ÷
è
bø
PYTS 554 – Tectonics I

18
Failure envelopes

When shear stress exceeds a critical value then failure occurs
Critical shear stress increases with increasing pressure
Rocks have finite strength even with no confining pressure

Coulomb failure envelope


s s = Yo + fF p
 Yo is rock cohesion (20-50 MPa)
 fF is the coefficient of internal friction (~0.6)
What about fractured rock?
Cohesion = 0
Tensile strength =0
Byerlee’s Law:
ìï 0.85s
s N £ 200MPa
N
ss = í
ïî 50 + 0.6 s N s N ³ 200MPa
Melosh, 2011
PYTS 554 – Tectonics I
Why do faults stick and slip?

Basically because the coefficients of static and dynamic friction are different

Stick-slip faults store energy to release as Earthquakes

Shear-strain increases with time as:

Stress on the fault is:
e s ( t ) = uot b
s = s fd + 2Ges
 G is the shear modulus
 σfd (dynamic friction) left over from previous break
 Fault can handle stresses up to σfs before it breaks (Static friction)
t=
b
s fs - s fd )
(
2Guo

Breaks after time:

Fault locks when stress falls to σfd (dynamic friction)
If σfd < σfs then you get stick-slip behavior

19
PYTS 554 – Tectonics I

20
Brittle to ductile transition

Confining pressure increases
with Depth (rocks get stronger)

Temperature increases with
depth and promotes rock flow

Upper 100m – Griffith cracks

P~0.1-1 Kbars, z < 8-15km,
shear fractures

P~10 kbar, z < 30-40km
distributed deformation
(ductile)

This transition sets the depth
of faults
Melosh, 2011
Golembek
PYTS 554 – Tectonics I

Back to Mohr circles…

Coulomb failure criterion is a straight line




Intercept is cohesive strength
Slope = angle of internal friction
Tan(slope) = fs
In geologic settings



21
Coefficient of internal friction ~0.6
Angle of internal friction ~30°
Angle of intersection gives fault orientation
(180 - 2q ) + f + 90
= 180
q = 45 + f 2

So θ is ~60°

θ is the angle between the fault plane and the minimum principle stress,
s3
PYTS 554 – Tectonics I

Anderson theory of faulting



All faults explained with shear stresses
No shear stresses on a free surface means
that one principle stress axis is perpendicular
to it.
Three principle stresses







σ1 > σ2 > σ3
σ1 bisects the acute angle (2 x 30°)
σ2 parallel to both shear plains
σ3 bisects the obtuse angle (2 x 60°)
So there are only three possibilities
One of these principle stresses is the one
that is perpendicular to the free surface.
Note all the forces here are compressive….
Only their strengths differ
σ2
22
PYTS 554 – Tectonics I

Before we talk about faults….

Fault geometry


Dip measures the steepness of the fault plane
Strike measures its orientation
23
PYTS 554 – Tectonics I

Largest principle (σ1) stress
perpendicular to surface

Typical dips at ~60°
24
PYTS 554 – Tectonics I
25
Extensional Tectonics

Crust gets pulled apart

Final landscape occupies more area than
initial

Can occur in settings of


Steeply dipping
Uplift (e.g. volcanic dome)
Edge of subsidence basins (e.g. collapsing
ice sheet)
Shallowly dipping
PYTS 554 – Tectonics I

Horst and Graben






Graben are down-dropped blocks of crust
Parallel sides
Fault planes typically dip at 60 degrees
Horst are the parallel blocks remaining
between grabens
Width of graben gives depth of fracturing
On Mars fault planes intersect at depths of
0.5-5km
26
PYTS 554 – Tectonics I
27
PYTS 554 – Tectonics I

In reality graben fields are complex…


Different episodes can produce different orientations
Old graben can be reactivated
Ceraunius Fossae - Mars
Lakshmi -Venus
28
PYTS 554 – Tectonics I

Smallest (σ3) principle stress
perpendicular to surface

Typical dips of 30°
29
PYTS 554 – Tectonics I
30
Compressional Tectonics

Crust gets pushed together

Final landscape occupies less area than initial

Can occur in settings of

Center of subsidence basins (e.g. lunar maria)

Overthrust – dip < 20 & large displacements

Blindthrust – fault has not yet broken the surface
Steeply dipping
Shallowly dipping
PYTS 554 – Tectonics I
31
Montesi and Zuber, 2003.
PYTS 554 – Tectonics I

Intermediate (σ2) principle stress
perpendicular to surface

Fault planes typically vertical
32
PYTS 554 – Tectonics I
Shear Tectonics

Strike Slip faults



Shear forces cause build up of strain
Displacement resisted by friction
Fault eventually breaks
Right-lateral
Left-lateral
(Dextral)
(Sinistral)

Vertical Strike-slip
faults = wrench faults

Oblique normal and
thrust faults have a
strike-slip component
Europa
33
PYTS 554 – Tectonics I

Tectonics I







Vocabulary of stress and strain
Elastic, ductile and viscous deformation
Mohr’s circle and yield stresses
Failure, friction and faults
Brittle to ductile transition
Anderson theory and fault types around the solar system
Tectonics II




Generating tectonic stresses on planets
Slope failure and landslides
Viscoelastic behavior and the Maxwell time
Non-brittle deformation, folds and boudinage etc…
34
PYTS 554 – Tectonics I

Random extras
35
PYTS 554 – Tectonics I

How to faults break?

Shear zone starts with formation of Riedel shears (R and R’)
Orientation controlled by angle of internal friction


36
Formation of P-shears


Mirror image of R shears
Links of R-shears to complete the shear zone
Revere St., San Francisco
(Hayward Fault)
PYTS 554 – Tectonics I

37
Wrinkle ridges




Surface expression of
blind thrust faults (or
eroded thrust faults)
Associated with
topographic steps
Upper sediments can be
folded without breaking
Fault spacing used to
constrain the brittle to
ductile transition on Mars
Montesi and Zuber, 2003.
PYTS 554 – Tectonics I

·
Rocks flow as well as flex
Stress is related to strain rate
Viscous deformation is irreversible



Where η is the
dynamic viscosity
s = he
·
Motion of lattice defects, requires activation energies
Viscous flow is highly temperature dependant

38
e µe
-Q
RT
w
Back to our mountain example
h
s = rgh
·
·
and e = h
Solution for h:
h = h0 e
w
-t
t
where t =
v
Works in reverse too…
In the case of post-glacial rebound
τ ~ 5000 years
w ~ 300km
Implies η ~ 1021 Pa s – pretty good
h
r gw
PYTS 554 – Tectonics I

How to quantify τfs

Sliding block experiments
Increase slope until slide occurs

Normal stress is:

Shear stress is:

Sliding starts when:


Experiments show:







 f   fs
 fs  f s N
Amonton’s law – the harder you press the fault
together the stronger it is
So fs=tan(Φ)
fs is about 0.85 for many geologic materials
In general:

mgcos(f )
A
mgsin(f )
tf =
A
sN =
t f s = Co + f ss N = Co + tan(f )s N
Coulomb behavior – linear increase in strength
with confining pressure
Co is the cohesion
Φ is the angle of internal friction
In loose granular stuff Φ is the angle of repose
(~35 degrees) and Co is 0.
39
PYTS 554 – Tectonics I

Effect of pore pressure



Reduces normal stress…
And cohesion term…
Material fails under lower stresses
s S = Co + (s N - Pf ) tan(f )

Pore pressure – interconnected full pores
Pf   g y



Density of water < rock
Max pore pressure is ~40% of overburden
Landslides on the Earth are commonly
triggered by changes in pore pressure
40