glacier dynamics slides 2011 - glaciers group
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Glacier DynamicsGlaciers 617
Andy Aschwanden
Geophysical Institute
University of Alaska Fairbanks, USA
October 2011
1 / 81
“The tradition of glacier studies that we in-herit draws upon two great legacies of theeighteenth and nineteenth centuries: classicalphysics and romatic enthusiasm for Nature.. . . It is easy to undervalue the romantic con-tribution, but in glaciology, it would be a mis-take to do so”Garry Clark, J. Glaciol., 1987)
Course Material
lecture is based on lectures by Ed Waddington & Martin Truffer Martin’s “Continuum Mechanics” script book recommendation: Greve and Blatter (2009)
Greve, R. and H. Blatter (2009). Dynamics of Ice Sheets and Glaciers.Advances in Geophysical and Enviornmental Mechanics andMathematics. Springer Verlag.
3 / 81
Introduction to glacier flow
4 / 81
How does a glacier move?
In a nut shell: The ice can deform as a viscous fluid The ice can slide over its substrate Well-described by continuum mechanics
Measurements & Observations 5 / 81
Measurements & Observations
1779 Gravitation theory by de SaussureH. B. de Saussure observes sliding
“. . . the weight of the ice might be sufficient to urge it down the slopeof the valley, if the sliding motion were aided by the water flowing atthe bottom.”
1827-1836 Hugi BlockJ. Hugi observed that a boulder moved 1315 m downstream between 1827and 1836
we would interpret this as clear evidence of glacier flow but back then, some people argued that a boulder slides on the
glacier surface, the glacier itself is motionless
Measurements & Observations 6 / 81
Measurements & Observations
1840-1846 Dilatation theory by L. Agassiz
glacier ice contains innumerable fissures and capillary tubes during the day, these tubes absorb the water and during the night, the water freezes this distension exerts a force and propels the glacier in the direction of
least resistance
Measurements & Observations 7 / 81
Measurements & Observations
1864-1930 Viscous flow theory by J. Forbes
made his own observations on Mer deGlace, France
glacier flows fastest in the center opposes Agassi’s theory if the dilatation theory were true then flow would be greatest at sunset and near the glacier margins
Measurements & Observations 8 / 81
Measuring surface velocities
Measuring angles (with Theodolite) anddistances (with Electronic Distance Meteror EDM) from fixed stations on theglacier margins
GPS (global positioning system) feature tracking in repeated aerial
photography or satellite images
Measurements & Observations 9 / 81
Measuring surface velocities
credit: USGS, Joughin et al., 2010
interferometric synthetic apertureradar (InSAR)
count fringes from a stationarypoint on bedrock
Measurements & Observations 10 / 81
Measuring borehole tilt
with tiltmeter(inclinometer)
Measurements & Observations 11 / 81
Kinematic vs. Dynamics
Kinematic description of flowIn a steady state,
Flow required to transport away upstream accumulation Glacier has to adjust its shape to make this flow happen
⇒ Rheological properties don’t figure in kinematic description
Dynamic description of flow
ice is a material with certain rheological properties (e.g. viscosity) Flow is determined by stresses applied to it
⇒ Accumulation/ablation don’t figure in dynamic description
Measurements & Observations 12 / 81
“Glacier Flow” Chart
climate
meteorological process
massbalance
ice dynamics processesthis lecture
glaciergeometry
Measurements & Observations 13 / 81
Why is it so hard to predict the future of an icesheet?
It’s easy because
composed of a single, largely homogenous material viscous flow is governed by the Navier-Stokes equations (19th century
physics) move very slowly (turbulence, Coriolis force, and other inertial effects
can be ignored)
It’s so hard because the stress resisting ice flow at the base can vary by orders of
magnitude ocean interactions could trigger instabilities
Measurements & Observations 14 / 81
Forces
a force is any influence that causes an object to undergo a change inspeed, a change in direction, or a change in shape
forces exist inside continuous bodies such as a glacier these forces can cause a glacier to deform
Forces, Stress & Strain 15 / 81
What is stress?
“Stress” is the force per unit area acting on a material
σ =F
A
Inside a glacier, stresses are due to weight of the overlying ice (overburden pressure) shape of the glacier surface (pressure gradients)
Forces, Stress & Strain 16 / 81
Types of stress
As a force per unit area, stress has a direction
Force can be directed normal to thearea
Result is pressure if the force is thesame on all faces of a cube.
Result is normal stress if forces aredifferent on different faces
Force can be directed parallel to thearea
Result is shear stress
Forces, Stress & Strain 17 / 81
Magnitudes of stress & forces
Again, stress is force / (unit area) frictionless table water bottle in free fall stretching a rubber band
Forces, Stress & Strain 18 / 81
Pressure in a glacier
mass m = ρV
ρ = ice density ≈ 900 kg m−3
V = Volume = Area × depth = A · zSo pressure p at depth z is
p =m · g
A=
ρ ·A · z · gA
= ρgz
How deep do we have to drill into a glacier before the ice pressure is 1atmosphere?
Forces, Stress & Strain 19 / 81
Depth for 1 atm pressure?
z =p
ρ · g =?
So pressure rises by 1 atm for every x meters of depth in a glacier Does ice deform in response to this pressure?
Forces, Stress & Strain 20 / 81
Shear stress τ
Total stress t from ice column:
t = ρVg = ρ g h
How much of this weight will contribute to shear deformation? shear stress τ = ρ g h sin α
normal stress σ = ρ g h cos α
Forces, Stress & Strain 21 / 81
Shear stress in a glacier
h = 130 m α = 5
τ = ρ g h sin α
Shear stress at the glacier base, τb, is ≈ 1 bar, which is a typical value forbasal shear stress under a glacier
Forces, Stress & Strain 22 / 81
Are glacier thickness and slope related?
Suppose a glacier becomes thicker or steeper due to mass imbalance: it flows faster it quickly reduces thickness h or slope α, until τb ≈ 1 bar again
Can we then estimate glacier thickness (z = h) from its slope if we knowτb ≈ 1 bar?
τ = ρ g z sin α ⇒ h ∼ τbρ g sin α
How are shear stress and shear strain rate (velocity) related?
Forces, Stress & Strain 23 / 81
Strain rate γ
shear strain γ = 1
2
∆x∆z
= 1
2γ
shear strain rate γ = 1/2∆x/∆z
∆t= 1/2
∆u∆z
Forces, Stress & Strain 24 / 81
Material Science
Rheology
is the study of the flow of complex liquids or the deformation of softsolids.
different materials respond differently to applied stresses We are not going through this in great detail.
⇒ For further information, consult the handout or take Martin Truffers “Ice Physics 614” class
We need to define a relationship between shear stress and shear strain rate, τ = f (γ)
Material Science 25 / 81
Example: Shear Experiment
apply constant shear stress τ
τ is the applied shear stressγ is the shear angleγ is the shear rate
initial elastic deformation primary creep: shear rate γ decreases with time secondary creep: shear rate remains constant acceleration phase tertiary creep: constant shear rate (but higher)
Material Science 26 / 81
Example: Shear Experiment
τ is the applied shear stressγ is the shear angleγ is the shear rate
This suggests that the shear rate is a function of the applied shear stress, temperature
T , and pressure p
γ = γ (τ, T , p)
Material Science 27 / 81
Example: Shear Experiment
τ is the applied shear stressγ is the shear angleγ is the shear rate
The dynamic viscosity η relates shear stress and shear rate
τ = η (T , p, |γ|) γ
Material Science 28 / 81
Constitutive behavior of ice
γ = 2 A(T , p) τn−1τ
A(T , p) = A0e−Q/(RT ) Arrhenius law
A is called rate factor (ice softness) Q and R are the activation energy and the gas constant, respectively n is the exponent of the flow law, usually taken as n = 3
This is known as Glen’s flow law or Glen-Steinemann flow law but similarto Norton’s flow law for the flow of steel at high temperatures
Material Science 29 / 81
Temperature dependence
With some emperically-derived values for Q we get:
Temperature T [ C]
Rat
e fa
ctor
A [s
Pa
]
ice at 0 C is about 1000× softer than ice at -50 C
Material Science 30 / 81
Thin-film flow
∂u∂z = 2 A(T , p) τn−1τ ⇒
uh − ub = h
02 A (ρ g sin α z)n
dz =
We assume ub = 0, i.e. glacier is frozen to the bed
Material Science 31 / 81
Basal sliding
Despite decades of intensive research, basal sliding remains one of thegrand unsolved problems. We know:
sliding is a function of basal water pressure and bed roughness if ice is underlain by till, experiments suggest a plastic or
nearly-plastic behavior
Beyond the scope of this lecture
Material Science 32 / 81
Recap
so far, we have only considered shear stresses while they are often the most important ones in glacier flow, there are other stresses (e.g. normal stresses) which can be important
too we have assumed ice surface and bed surface are parallel let’s step back and look at the bigger picture yep, we will need basic calculus and some continuum mechanics
Material Science 33 / 81
Field equations for ice flow
Material Science 34 / 81
Recap
so far, we have only considered shear stresses while they are often the most important ones in glacier flow, there are other stresses (e.g. normal stresses) which can be important
too we have assumed ice surface and bed surface are parallel let’s step back and look at the bigger picture yep, we will need basic calculus and some continuum mechanics
Material Science 35 / 81
Continuum Mechanics
Continuum mechanics is the application of classical mechanics tocontinuous media. So
What is classical mechanics? What are continous media
Material Science 36 / 81
Classical MechanicsIn classical mechanics, we deal with two type of equations
conservation (balance) equations (fundamental to physics) material equations (less fundamental)
Balance Laws balance of mass balance of linear momentum balance of angular momentum balance of energy
Material Laws / Constitutive Equations / Closures
often obtained from experiments and theoretical considerations (suchas material objectivity)
Material Science 37 / 81
Classical Mechanics
Densities classical mechnics has concept of point mass determine velocity (and acceleration) by looking at changes (and rate
of changes) in position ⇒ Kinematics how do forces affect a mass? ⇒ Dynamics
Volume Ω has mass m, then
mass m =Ω ρ dv , ρ mass density
linear momentum mv =Ω ρv dv , ρv linear momentum density
internal energy U =Ω ρu dv , ρu energy density
(1)
Material Science 38 / 81
General Balance Laws
Imagine a volume Ω enclosed by a boundary ∂Ω
G(t) =
Ω
g(x , t) dv (2)
S(t) =
Ω
s(x , t) dv (3)
G , g
S, s
physical quantity and densitysupply and density
Material Science 39 / 81
General Balance Laws
G can only change if if there is a supply S within Ω
if there is a flux F of G through the boundary ∂Ω
F (t) =
∂Ω
(gv + φ(x , t)) ·n da (4)
gvφ(x , t)
transport of g with velocity v through ∂Ωflux density (defined later)
Material Science 40 / 81
General Balance Laws
General balance law is:
dG
dt= S − F (5)
d
dt
Ω
g(t) dv =
Ω
s(x , t) dv −
∂Ω
(gv + φ) ·n da (6)
Material Science 41 / 81
General Balance Laws
Assume that G = G(t) thus integral and differential can be exchanged Use Gauss’ Theorem:
∂Ω
(gv + φ) ·n da =
Ω
∇ · (gv + φ) dv
We can then write
Ω
∂g
∂tdv =
Ω
s(x , t) dv −
Ω
∇ · (gv + φ) dv (7)
Material Science 42 / 81
General Balance Laws
We made no assumption about the shape and size of Ω We can make Ω infinitely small
∂g
∂t= s(t) − ∇ · (gv + φ) (8)
ordg
dt=
∂g
∂t+ v ·∇g = s(t) − ∇ · (v + φ) (9)
Material Science 43 / 81
Mass Balance
the mass of a material volume cannot change, dM/dt = 0g = ρφ = 0s = 0
mass densityflux of g through the boundarysupply of g
from the above we get∂ρ
∂t+ ∇ · (ρv) = 0 (10)
⇒ known as the continuity equationWe assume ice is incompressible (density does not change), then
∇ · v = 0 (11)
Mass Balance 44 / 81
Momentum Balance
Newton’s Second Law total rate of change of momentum P equals sum of all forces F
dPdt
= F =
∂ω
tn da
internal
+
ω
f dv
external
(12)
P and F are vector quantities identify the internal stress vector tn = Tn, see p. 31 T is the Cauchy stress tensor external forces f are, e.g. gravity or the Coriolis force (forces are a
source of momentum)
Momentum Balance 45 / 81
The Cauchy Stress Tensor
We’ve already introduced shear stress, but forces can be applied ondifferent faces and in different directions
T =
txx txy txztyx tyy tyztzx tzx tzz
We identify τ = txz
For example tez =
txztyztzz
is the stress vector of a cut along the xy-plane
Momentum Balance 46 / 81
Momentum Balance
∂ (ρv)∂t
= −∇ · (ρv ⊗ v) + ∇ ·T + f (13)
g = ρvφ = −Ts = f
momentum densityflux of g through the boundarysupply of g
By using the mass balance, we can write
d (ρv)dt
= ∇ ·T + f (14)
What’s new? shear stress τ replaced with Cauchy stress tensor T
Momentum Balance 47 / 81
Momentum Balance
For glacier ice, we assume ice is incompressible acceleration term d(ρv)
dt can be neglected supply of momentum is through gravity only, f = ρg
The momentum balance for glacier flow is
∇ ·T + ρg = 0 (15)
Momentum Balance 48 / 81
Angular Momentum Balance
no, I don’t want to do the whole derivation here if you’re interested (or bored), go through Equations 3.77 – 3.83 in
Greve and Blatter (2009) here, we just use the result the stress tensor T is symmetric
T = TT (16)
Angular Momentum Balance 49 / 81
Energy Balance
only the total energy (sum of kinetic and internal energy) is aconserved quantity
balance of kinetic energy is not an independent statement but a mereconsequence of the momentum balance
here, we formulate the balance of specific inner energy, u:
ρdu
dt= −∇ ·q + tr (T ·D) + ρr (17)
g = u
φ = qp = tr (T ·D)s = ρr
D = 1/2vT + v
specific internal energyheat fluxdissipation powerspecific radiation powerstrain rate or velocity gradient tensor
Energy Balance 50 / 81
Balance Equations
mass dρdt = −ρ∇ · v (1)
momentum ρdvdt = ∇ ·T + f (3)
internal energy ρdudt = −∇ ·q + tr (T ·D) + ρr (1)
left-hand sideρ (1)v (3)u (1)
right-hand sideT (6)q (3)
so we have 5 equations for 14 unknown fields the system is highly under-determined
⇒ closure relations required
Energy Balance 51 / 81
Closure Relations
balance equations are universally valid closure relations describe the specific behavior of a material closure relations are often called constitutive equations
Energy Balance 52 / 81
Material Science
Rheology
is the study of the flow of complex liquids or the deformation of softsolids.
We are not going through this in great detail.⇒ For further information, consult the handout or take Martin Truffers “Ice Physics 614” class
We need to define a constitutive equation for the heat flux q and for the internal energy u
a relationship between stress and strain, T = f (D)
Constitutive Equations 53 / 81
Generalization
Assume secondary creep incompressibility of ice ⇒ uniform pressure does not lead to
deformation T = −pIisotropic
+ T
deviatoric
with the pressure p = −1/3trT
The only non-straightforward step from
τ = η (T , p, |γ|) γ (18)
toT = η (T , p, | • |)D (19)
is the generalization of |γ|
Constitutive Equations 54 / 81
Generalization
T = η (T , p, | • |)D (20)
is a relation between two tensors (i.e. T and D) material objectivity tells us that such a relationship must be
independent of the chosen vector basis a second rank tensor has 3 scalar invariants
IT = trT (21)IIT = 1/2
trT2 − tr
T2
(22)
IIIT = det T (23)
trT = 0 for incompressible media σe =
√IIT is the effective stress
role of IIIT is not clear, but lab experiments indicate that the thirdinvariant is not important
Constitutive Equations 55 / 81
Generalization
Numerous lab experiments and field measurements suggest
1η (T , p, σe)
= 2A (T , p) f (σe) (24)
whereA(T , p) = A0e
−Q/(RT ) Arrhenius lawf (σe) = σn−1
e power law
Constitutive Equations 56 / 81
Generalization
What’s new?
We have replacedγ = η (T , p, |τ |) τ (25)
byD = η (T , p, |σeff |)T (26)
Constitutive Equations 57 / 81
Calorimetric Equation of State
Internal energy, u depends linearly on temperature, T ,
u = u0 + c (T − T0) (27)
u0 is a reference value at a reference temperature T0
c is the heat capacity (a measure how much heat is needed to raisethe temperature by a certain amount)
we obtaindu
dt= c
dT
dt(28)
Constitutive Equations 58 / 81
Heat Flux
Fourier’s law of heat conduction
Tj-1 Tj Tj+1
t
modified after Christophe Dang Ngoc Chan, wikicommons image
heat flows from higher to lowertemperature
this is the second law ofthermodynamics
q = −k∇T
the thermal conductivity k is ameasure for how fast the heatspreads
Constitutive Equations 59 / 81
Recap
Balance Equations
mass 0 = ∇ · v (1)momentum ρdv
dt = −∇ ·T + ∇p + f (3)temperature ρc
dTdt = −∇ ·q + tr (T ·D) + ρr (1)
Flow Law
Glen’s flow law T = 2 η D = A(T )−1/nσ1−ne D (6)
now we have 15 equations for 15 unknown fields the system is now well-determined
Recap 60 / 81
Large-Scale Dynamics
Ice sheets are glaciers too! we will focus on the ice sheet part and, for now, ignore ice shelves ice is assumed to be incompressible, ∇ · v = 0
Glaciers 61 / 81
The Incompressible Navier-Stokes Equation
Momentum Balance
ρdvdt
acceleration
= − ∇ ·T
deviatoric
+ ∇pisotropic
+ ρggravity
− 2ρΩ × v Coriolis force
(29)
This is the Navier-Stokes equation for incompressible flow
still rather complicated do we really need all these terms? Scale Analysis is a power- and useful concept to assess the relative
importance of terms
Scale analysis 62 / 81
Scale Analysis
Some typical values for an ice sheet
horizontal extend [L] = 1000 kmvertical extend [H] = 1 km
horizontal velocity [U] = 100 ma−1
vertical velocity [W ] = 0.1 ma−1
pressure [P] = ρg [H] = 10 MPatime-scale [T ] = [L]/[U] = 104 a1
The aspect ratio is defined as
=[H]
[L]=
[W ]
[L]= 10−3 for an ice sheet
The scaling argument for valley glaciers is almost the same
Scale analysis 63 / 81
Scale Analysis
Froude numberThe Froude number Fr is the ratio of acceleration and pressure gradient.In the horizontal we have
Fr =ρ[U]/[t]
[P]/[L]=
ρ[U]2/[L]
ρg [H]/[L]=
[U]2
g [H]≈ 10−15
and in the vertical
Fr =ρ[W ]/[t]
[P]/[L]=
ρ[W ]2/[L]
ρg [H]/[L]=
[U]2
g [H]≈ 10−21
⇒ The acceleration term is negligible
Scale analysis 64 / 81
Scale Analysis
Rossby numberThe Rossby number Ro is the ratio of acceleration and Coriolis force
Ro =ρ[U]/[t]
2ρΩ[U]=
ρ[U]2/[L]
2ρΩ[U]=
[U]
2Ω[L] ≈ 2 × 10−8
and thus the Coriolis to pressure gradient is
2ρΩ[U]
[P]/[L]=
Fr
Ro≈ 5 × 10−8
⇒ The Coriolis term is also negligible
Scale analysis 65 / 81
Stokes Equation
By neglecting both the acceleration term Coriolis term
the Navier-Stokes equation simplifies to
∇ ·η
∇v + ∇vT
− ∇p = −ρg (30)
gravitational force exerted on the ice is balanced by stress within theice
this is called the Stokes equation
Stokes Equation 66 / 81
Temperature Equation
Recall the equation for temperature
ρdT
dt= −∇ ·q + tr (T ·D) + ρr (31)
except for the uppermost few centimeters of ice,radiation is negligible
ρdT
dt= −∇ ·q + tr (T ·D) (32)
Energy Balance 67 / 81
Boundary Conditions
Balance Equations
mass ∇ · v = 0momentum ∇ ·
η
∇v + ∇vT
− ∇p = −ρg
temperature ρcdTdt − ∇ · (k∇T ) = tr (T ·D)
Boundary ConditionsTo solve this system of equations, we need boundary conditions:
mass (kinematic) momentum (dynamic) temperature (thermodynamic)
Boundary Conditions 68 / 81
Ice-Thickness Equation
∂H
∂t= (qin − qout) + as − ab
= −∇ ·Q + as − ab (33)
as and ab are in the vertical direction the volume flux Q is defined as
Q =
QxQy
=
hb vx dz
hb vy dz
derivation on p. 70 – 72 this is the central evolution equation in ice sheet dynamics
Boundary Conditions 69 / 81
Surface Dynamic Boundary Condition
Continuity of the stress vector
T ·n = Tatm ·n (34)
Tatm is the Cauchy stress tensor in the atmosphere atmospheric pressure and wind stress are small compared to stress in
the iceT ·n = 0 (35)
this is a stress-free condition
Boundary Conditions 70 / 81
Basal Dynamic Boundary Condition
T ·n = Tlith ·n (36)
Tlith is the Cauchy stress tensor in the lithosphere unfortunately, we have no information about the stress state of the
lithosphere this boundary condition is not very useful empirical relationship is required relate bed-parallel shear stress τb and bed-parallel basal velocity ub
ub = f (τb)
Boundary Conditions 71 / 81
Basal Dynamic Boundary Condition
we can distinguish two cases ice is frozen to the bed if temperature below pressure melting point Tm
ice is sliding if temperature at pressure melting point Tm
ub =
0, T < Tm Dirichlet conditionf (τb), T = Tm Robin condition
despite 50 years of work (and lots of progress), this remains one ofthe grand unsolved problems in glaciology
Boundary Conditions 72 / 81
Surface Thermodynamic Boundary Condition
the surface temperature Ts can be well approximated by the meanannual surface air temperature, as long as the latter is ≤ 0 C
we therefore simply prescribe the surface temperature (Dirichletcondition):
T = Ts
Boundary Conditions 73 / 81
Basal Thermodynamic Boundary Condition
Cold Basethe heat flux into the ice equals the geothermal heat flux q
⊥geo
q ·n = q⊥geo
Temperate Basethe ice is at the pressure melting point Tm:
T = Tm
Boundary Conditions 74 / 81
Thermomechanically-Coupled Glacier Flow
Balance Equations
mass ∇ · v = 0momentum ∇ ·
η
∇v + ∇vT
− ∇p = −ρg
temperature ρc dTdt − ∇ · (k∇T ) = tr (T ·D)
Boundary Conditions
mass∂h∂t + vx
∂h∂x + vy
∂h∂y − vz = as surface
∂b∂t + vx
∂b∂x + vy
∂b∂y − vz = ab base
momentum T · n = 0 surfaceub = 0, f (τb) base
temperature T = Ts surfaceT = Tm or q · n = q⊥
geo base
Boundary Conditions 75 / 81
Numerical Modeling
the presented Stokes problem is solvable on a valley glacier scale but still very hard on a whole ice sheet scale for ice sheets, the shallowness (small aspect ratio) can be further
exploited, this leads to a hierarchy of approximations Ed Bueler’s Karthaus lecture is a good starting point
Boundary Conditions 76 / 81
Basics and Notation We assume a fixed orthonormal basis ex , ey , ez Einstein summation convention says that double indices (here i) imply
summation
Nabla
∇ = ex∂
∂x+ ey
∂
∂y+ ez
∂
∂z= ei
∂
∂xi
Gradient the gradient of the scalar quantity g is thus
∇g = ex∂g
∂x+ ey
∂g
∂y+ ez
∂g
∂z= eig,i =
∂g/∂x
∂g/∂y
∂g/∂z
Math Stuff 77 / 81
Basics and Notation
Divergence
the divergence of the quantity g is
∇ · g =∂gx∂x
+∂gy∂y
+∂gz∂z
= gi ,i
Total Derivative or material derivative
dg
dt=
∂g
∂t+ v ·∇g
where v is the velocity
Math Stuff 78 / 81
Basics and Notation
Trace the trace of the n×n matrix A is the sum of the diagonal elements
trA = A11 + A22 + . . . + Ann = Aii
Math Stuff 79 / 81
Recall from Vector Calculus
Divergence Theorem
∂ω
φ ·n da =
ω
∇ ·φ dv (37)
The sum of all sources minus the sum of all sinks gives the net flowout of a region
Reynolds Transport Theorem
ddt
ω
φ dv =
ω
∂φ
∂tdv +
∂ω
φv ·n da (38)
What was already there plus what goes in minus what goes out iswhat is there
Math Stuff 80 / 81
The Cauchy Stress Tensor
T =
txx txy txztyx tyy tyztzx tzx tzz
For example
tez =
txztyztzz
is the stress vector of a cut along the xy-plane
Math Stuff 81 / 81
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