stress and strain - eth z

Chapter

4Stress and strain

In this chapter we introduce the concepts of stress and strain which are crucial forthe understanding of glacier flow. Any calculation of deformation rates and flowvelocities involves stresses and a flow law which relates them to the strain rates. Itis also very useful to take now a look at Appendix C “Vectors and Tensors”. Thenotation introduced there will be used below.

4.1 ForceThere are two different kinds of forces: body forces and surface forces.

Body forces act on each volume of mass, independent on the surrounding material.The gravity force is the body force that causes glaciers to flow. It exerts on eachvolume of ice a force that is proportional to the mass within that volume. Otherexamples of body forces are inertial forces such as the centrifugal force.

Surface forces arise from the action of one body on another across the surface ofcontact between them. A typical example is the force exerted from the glacier to itsbase, and vice versa. Moreover, across any internal surface of arbitrary orientationthat divides a block of material into two, one side of the block applies a surface forceon the other side.

A force F (typeset in bold face) is a vector quantity and may be divided into itscomponents along perpendicular directions. Vectors and tensors are explained inAppendix C.

x

z

F

Force

x

z

Fx

Fz

Force components

35

Chapter 4 Stress and strain

4.2 StressA force that acts on a surface is called a stress (or pressure when it is compressive).The intensity of the force depends on the area of the surface over which the forceis distributed. It is called a traction and is commonly represented in terms of itscomponents perpendicular and parallel to a surface.

Traction Traction components

In order to satisfy the requirement of mechanical equilibrium, any surface must havea pair of equal and opposite tractions acting on opposite sides of the surface. Thispair of tractions defines the surface stress vector Σ which is defined by the totalforce exerted on the surface divided by the surface area

Σ =F

A.

Stress and traction are measured in units of force per unit area[F

A

]=

N

m2= Pa, and the derived units 105 Pa = 1 bar = 0.1 MPa.

Surface stress

Σ(top)

Σ(bot)

Surface stress components

σ(top)s

σ(bot)s

σ(top)n

σ(bot)n

It is convenient to resolve the surface stress into components, one perpendicular tothe surface and two others parallel to the surface at right angles. These componentsare called normal stress and shear stress and are denoted by σn and σs1 resp. σs2 .

The condition of mechanical equilibrium implies F(top) + F(bot) = 0, and therefore

F(top)

A+

F(bot)

A= 0 ,

Σ(top) + Σ(bot) = 0. (4.1)

36

Physics of Glaciers HS 2020

Equation (4.1) asserts that the tractions on top and bottom of the surface are equaland opposite. The same must be true for the components of the surface stress

σ(top)n = −σ(bot)

n and σ(top)s = −σ(bot)

s . (4.2)

A pair of normal stresses that point towards each other is called a compressivestress (negative sign), a pair pointing away from each other is called a extensivestress (positive sign).

Stress equilibrium

For ease of presentation we consider two dimensions only (analogous relations holdin three dimensions). The stresses acting on a small volume of material exert forcesand moments that must balance for a mechanical equilibrium.

σ(top)zz

σ(top)zx

σ(bot)zz

σ(bot)zx

σ(lft)xx

σ(lft)xz σ

(rt)xx

σ(rt)xz

Σ(top)z

Σ(bot)z

Σ(lft)x

Σ(rt)x

dx

dz

From the balance of normal and shear tractions (Eq. 4.2) we obtain the relations

σ(rt)xx = −σ(lft)

xx σ(top)zz = −σ(bot)

zz

σ(rt)xz = −σ(lft)

xz σ(top)zx = −σ(bot)

zx . (4.3)

We also require that all moments with respect to the body center are balanced(otherwise the body would rotate). This involves only the shear components, sincethe moments of all normal components are zero (since they point to the center ofrotation). Denoting the areas of the surfaces Ax and Az, and the lengths of theblock 2dx and 2dz, and taking the moment anti-clockwise, we obtain

σ(top)zx Az︸︷︷︸Σ

(top)zx

dz + σ(bot)zx Az︸︷︷︸Σ

(bot)zx

dz − σ(lft)xz Ax︸︷︷︸Σ

(lft)xz

dx− σ(rt)xz Ax︸︷︷︸Σ

(rt)xz

dx!

= 0 . (4.4)

37


Using Ax = Az, dx = dz and Equation (4.3) we obtain 2σzx − 2σxz!

= 0, whichis equivalent to σzx = σxz. The stress state induced by [Σx,Σz] (four numbers) istherefore fully described by the three components

σxx, σxz = σzx, and σzz. (4.5)

An analogous relation holds in three dimensions where the stress components are

σxx, σyy, σzz

σxy = σyx, σxz = σzx, σyz = σzy. (4.6)

Stress tensor

The stress state at one pointx in a material is fully described by a stress tensor.This is fully analogous to, for example, temperature or body force:

Quantity Components mathematical objectTemperature 1 scalar (one number)Gravity force in 2D 2 vector (two numbers)Gravity force in 3D 3 vector (three numbers)Stress in 2D 2*2 = 4 tensor (four numbers)Stress in 3D 3*3 = 9 tensor (nine numbers)

The number of components of a tensor is just the number of force components (2 in2D, 3 in 3D) times the number of components to describe a plane (the face normalvector; again 2 in 2D, 3 in 3D).

A tensor can be represented by a matrix, i.e. a square array of numbers, representingits components.

The stress components in Equation (4.5) form a two-dimensional tensor

σ = [σij] =

[Σx

Σz

]=

(σxx σxzσxz σzz

). (4.7)

Since σxz = σzx, the stress tensor is symmetric, that is

σ = [σij] = [σji] = σT . (4.8)

Using similar arguments in the three-dimensional case, it is possible to show thatthe stress tensor is also symmetric and has the general form

σ = [σij] =

Σx

Σy

Σz

=

σxx σxy σxzσxy σyy σyzσxz σyz σzz

. (4.9)

Notice that only six components are independent, since balance of moments (Eq. 4.6)leads to σxy = σyx, σyz = σzy and σxz = σzx.

38


For the stress vector Σ and the unit normal n on an arbitrary surface the followingimportant relation holds (normal matrix multiplication in shorthand notation)

Σn = σ · n = σiknk. (4.10)

If the stress tensor is known in one coordinate system K, it can be calculated in anyother system K ′. The transformation formula is the same as in Equation (C.25)

σ′ij = αipαjq σpq or σ′ = RσRT . (4.11)

where R = [α] is an arbitrary rotation.

The above transformation explains, why the shear stress components change theirvalue by moving from a vertically aligned to a tilted coordinate system.

Example The components of the stress tensor are

σ = [σij] =

1 2 32 −1 13 1 0

Find the traction on a plane defined by

F (x) = x1 + x2 − 1 = 0.

Also determine the angle θ between the stress vector Σ and the surface normal n.

Solution: The unit normal on the surface is

n =

∣∣∣∣∣∣∣∣∣∣∣∣ ∂F

∂x1∂F∂x2∂F∂x3

∣∣∣∣∣∣∣∣∣∣∣∣ =

1√2

110

and the traction on the surface is

Σn = σn =

1 2 32 −1 13 1 0

1√2

110

=1√2

314

.

The angle θ is

cos θ =Σn · n|Σn|

=1√2

4√26

⇒ θ = 56◦.

39


Stress invariants

From the examples in section C.2 we know how to calculate quantities that areindependent of the orientation of the coordinate axes. For a second order tensor inthree dimensions three invariants can be constructed. The first is

Iσ =1

3σii =

1

3trσ =

1

3(σxx + σyy + σzz) = σm, (4.12)

and is also called the mean stress σm.

For incompressible materials like glacier ice, the isotropic mean stress does notcontribute to deformation. It is therefore useful to characterize the stress state bythe stress deviator. The deviatoric stress tensor is that part of the stress tensorwhich is extra from the isotropic stress state

σ(d)ij := σij − σmδij = σij −

1

3σiiδij . (4.13)

The second invariant of the deviatoric stress tensor is defined by

(IIσ(d))2 =1

2σ(d)ij σ

(d)ij =

1

2(σ(d))2

=1

2

((σ(d)

xx )2 + (σ(d)yy )2 + (σ(d)

zz )2 + 2(σ(d)xy )2 + 2(σ(d)

xz )2 + 2(σ(d)yz )2

). (4.14)

It is also called the octahedral stress or the effective shear stress, and is often denotedby τ or σe. It will be important for the formulation of the ice flow law.

The third invariant IIIσ(d) is the determinant of the deviatoric stress tensor

IIIσ(d) = det(σ(d)ij ) =

1

3σ(d)ij σ

(d)jk σ

(d)ki . (4.15)

It is seldom used in glaciology.

Principal stresses

A face Fn with unit normal n is free of shear forces, if the stress vector Σn is parallelto n. In this case the vectors Σn and n differ only by a numerical factor so that wecan write

Σn = σ · n = λn . (4.16)

The proportionality constant λ is an eigenvalue and the vector n an eigenvector ofthe tensor σ. An eigenvector of the stress tensor always fulfills equation (4.16). Ittherefore follows that an eigenvector of σ defines the orientation of a face withoutshear stresses. Furthermore, the eigenvalue is the normal stress on this face.

40


A short reminder of some properties of symmetric tensors:

• All eigenvalues are real numbers.

• Two eigenvectors that belong to different eigenvalues are perpendicular to eachother.

• There exists at least one coordinate system in which the representation of thetensor has nonzero values only on the main diagonal. All off-diagonal entriesare zero.

For the Cauchy stress tensor, a coordinate system can always be found in which thetensor is purely diagonal. The three eigenvectors of σ, designated with s(1), s(2) ands(3), are perpendicular to each other and define a orthogonal coordinate system. Inthis coordinate system σ has the form

σ =

λ1 0 00 λ2 00 0 λ3

. (4.17)

Since σ · s(i) = λis(i) (no summation convention!), the eigenvalues λ1, λ2 and λ3 are

the normal stresses. No tangential stresses act on the faces with unit normal s(i).The eigenvalues λi are called principal stress and the eigenvectors s(i) principalaxes.

The eigenvalues can be found by solving the problem

σ · s = λs

which, written in components, reads

σijnjsj − λsi = 0

or (σij − λδij)sj = 0 .

The trivial solution is sj = 0. The requirement for a non-trivial solution is

det(σij − λδij) = 0 .

This equation leads to a polynomial of third order in λ, which can be written as

λ3 − I1λ2 + I2λ− I3 = 0,

where use has been made of the following invariants of the stress tensor

I1 := trσ = σii ,

I2 :=1

2(σiiσjj − σijσij) ,

I3 := det(σ) .

Note: these invariants are different from the ones used before, but can be combinedto yield the same forms.

41


4.3 DeformationA rigid body motion (translation, rotation) induces no change of the body shape.The strain of a body is the change in size and shape that the body has experiencedduring deformation. The strain is homogeneous if the changes in size and shapeare proportionately identical for each small part of the body and for the body as awhole. The strain is inhomogeneous if the changes in size and shape of small partsof the body are different from place to place: straight lines become curved, planesbecome curved surfaces, and parallel planes and lines do not remain parallel afterdeformation.

Linear strain

The stretch sn of a material line segment is defined as the ratio of the deformedlength lf to its undeformed length lo

sn :=lflo. (4.18)

The extension en of a material line segment is the ratio of change in length ∆l toits initial length lo

en :=lf − lolo

=∆l

lo= sn − 1. (4.19)

(Note the sign convention: a positive extension is lengthening, a negative extensionis shortening the body.) The above definition gives the average extension after alength change. Going to very small extension increments, one defines the strain ε

ε :=dl

l, (4.20)

that is, the ratio of the infinitesimal current extension increment dl with respectto the current length l. To obtain the finite strain of the extension from lo tolf we have to integrate Equation (4.20) with respect to l (the reference length l isincreasing with increasing extension)

ε :=

∫ lf

lo

1

ldl = ln

(lflo

)= ln(sn) . (4.21)

For obvious reasons ε is also called logarithmic strain.

42


Strain

We now consider the deformation of an arbitrary body by studying the relative dis-placement of three neighboring points P , P ′, P ′′ in the body. If they are transformedto the points Q, Q′, Q′′ in the deformed configuration, the change in area and anglesof the triangle is completely determined if we know the change in length of the sides.

a3, x3

a1, x1

a2, x2

b

bb

PP ′

P ′′

b

bb

Q

Q′

Q′′

(a1, a2, a3)(x1, x2, x3)

Consider an infinitesimal line element connecting the point P (a1, a2, a3) to a neigh-boring point P ′(a1 + da1, a2 + da2, a3 + da3). The square of the length dso of PP ′ inthe original configuration is given by

ds2o = da21 + da22 + da23 = dai dai .

When P and P ′ are deformed to the points Q(x1, x2, x3) and Q′(x1 + dx1, x2 +dx2, x3 + dx3), respectively, the square of the length ds of the new element QQ′ is

ds2 = dx21 + dx22 + dx23 = dxi dxi .

We may express the transformation from the a coordinate system into the x coor-dinate system and its inverse by the expressions

xi = xi(a1, a2, a3) and ai = ai(x1, x2, x3) . (4.22)

Therefore, using the Kronecker delta, we can write (with an arbitrary but convenientchoice of index labels)

ds2o = δkl dak dal = δkl∂ak∂xi

∂al∂xj

dxi dxj ,

ds2 = δij dxi dxj = δij∂xi∂ak

∂xj∂al

dak dal . (4.23)

43


The difference between the squares of the length elements may be written as

ds2 − ds2o =

(δij

∂xi∂ak

∂xj∂al− δkl

)dak dal , (4.24)

or as

ds2 − ds2o =

(δij − δkl

∂ak∂xi

∂al∂xj

)dxi dxj . (4.25)

We define the strain tensor in two variants

Green - St. Venant Ekl =1

2

(δij

∂xi∂aj

∂xk∂al− δkl

), (4.26)

Cauchy eij =1

2

(δij − δkl

∂ak∂xi

∂al∂xj

), (4.27)

so that (remember that index names are arbitrary)

ds2 − ds2o = 2Eij dai daj , (4.28)ds2 − ds2o = 2eij dxi dxj . (4.29)

The Green strain tensor Eij is the strain with reference to the original, undeformedstate and is often referred to as Lagrangian. We will mainly use the Cauchy straintensor which is defined with respect to the momentaneous configuration. It is oftenreferred to as Eulerian.

Eij and eij are tensors in the coordinate systems {ai} and {xi}, respectively. Obvi-ously both are symmetric

Eij = Eji , eij = eji . (4.30)

An immediate consequence of Equations (4.28) and (4.29) is that ds2 − ds2o = 0implies Eij = eij = 0 and vice versa. Therefore, a deformation in which the lengthof every line element remains unchanged is a rigid-body motion (translation or ro-tation).

Strain components

If we introduce the displacement vector u with the components

up = xp − ap

then we can write

∂xp∂ai

=∂up∂ai

+ δpi,∂ap∂xi

= δpi −∂up∂xi

,

44


and the strain tensors reduce to the simpler form

Eij =1

2

[δpq

(∂up∂ai

+ δpi

)(∂uq∂aj

+ δqj

)− δij

]=

1

2

[∂uj∂ai

+∂ui∂aj

+∂uq∂ai

∂up∂aj

]and

eij =1

2

[δij − δpq

(−∂up∂xi

+ δpi

)(−∂uq∂xj

+ δqj

)]=

1

2

[∂uj∂xi

+∂ui∂xj− ∂uq∂xi

∂up∂xj

]

We now write out the components for e (the expressions for E are completely anal-ogous), and use the more conventional variable names x, y, z instead of x1, x2, x3,and u, v, w instead of u1, u2, u3 (notice that we use here u, v, w to designate dis-placements). This leads to nine terms of the general form

exx =∂u

∂x− 1

2

[(∂u

∂x

)2

+

(∂v

∂x

)2

+

(∂w

∂x

)2],

exy =1

2

[∂u

∂y+∂v

∂x−(∂u

∂x

∂u

∂y+∂v

∂x

∂v

∂y+∂w

∂x

∂w

∂y

)]. (4.31)

If the components of displacement ui are such that their first derivatives are verysmall and the squares and products of the derivatives of ui are negligible, then eijreduces to Cauchy’s infinitesimal strain tensor

εij =1

2

(∂ui∂xj

+∂uj∂xi

). (4.32)

In unabridged notation it reads

εxx =∂u

∂x, εxy =

1

2

(∂u

∂y+∂v

∂x

)= εyx ,

εyy =∂v

∂y, εxz =

1

2

(∂u

∂z+∂w

∂x

)= εzx , (4.33)

εzz =∂w

∂z, εyz =

1

2

(∂v

∂z+∂w

∂y

)= εzy .

In the case of infinitesimal displacement, the distinction between the Lagrangian andEulerian tensor disappears, since it is unimportant whether the derivatives of thedisplacements are calculated at the position of a point before or after deformation.Four common cases are shown in Figure 4.1.

45


x

zu u+

∂u

∂xdx

Case 1:∂u

∂x> 0, w = 0

x

zu u+

∂u

∂xdx

Case 2:∂u

∂x< 0, w = 0

x

z

Slope to vertical = ∂u

∂z

Slope = ∂w

∂x

Case 3:∂u

∂z> 0,

∂w

∂x> 0

x

z

Case 4:∂u

∂z> 0,

∂u

∂x=

∂w

∂x= 0

Figure 4.1: Different strain states: uniaxial extension (case 1), uniaxial compression(case 2), shear (case 3) and simple shear (case 4).

Rotation

Consider the infinitesimal displacement field ui(x1, x2, x3). We then can form thecartesian tensor

ωij =1

2

(∂uj∂xi− ∂ui∂xj

), (4.34)

which is antisymmetric, i.e.ωij = −ωji. (4.35)

Therefore the rotation tensor ωij has only three independent components – ω12, ω23

and ω31 – because ω11 = ω22 = ω33 = 0.

We can therefore write any relative movement of two points as the sum of a rotationand a deformation. Consider a point P with coordinates xi and a point P ′ in theneighborhood with coordinates xi+dxi. The relative displacement of P ′ with respect

46


to P isdui =

∂ui∂xj

dxj . (4.36)

This can be rewritten as

dui =1

2

(∂ui∂xj

+∂uj∂xi

)dxj +

1

2

(∂ui∂xj− ∂uj∂xi

)dxj = (εij + ωij) dxj . (4.37)

Strain rate

For the study of glacier flow, we are concerned with the velocity field v(x, y, z),which describes the velocity of every particle of the body. At every point (x, y, z),the velocity field is expressed by the components (from now on u, v and w are usedto denote the components of the velocity vector)

u(x, y, z), v(x, y, z), w(x, y, z),

or by vi(x1, x2, x3) in index notation.

Note: we reuse the letter u, v, w to designate velocity components instead ofdisplacement components. Since the velocity is just the change in time of the in-finitesimal displacement, the equations from section (4.3) apply unaltered. Insteadof the infinitesimal strain tensor, we now look at the strain rate tensor

εij :=1

2

(∂vi∂xj

+∂vj∂xi

). (4.38)

The only change with respect to Equation (4.32) is the dot (denoting the rate ofdeformation), the use of velocity vi instead of displacement ui. Remember that thedot is part of the symbol used to designate “strain rate” and does not indicate atime derivative. Notice that other authors (e.g. K. Hutter; Greve and Blatter) usethe symbol Dij. In the current literature mostly our notation is used.

47

Bibliography

Cuffey, K. and Paterson, W. (2010). The Physics of Glaciers. Elsevier, Burlington,MA, USA. ISBN 978-0-12-369461-4.

Duval, P. (1977). The role of water content on the creep rate of polycristalline ice.In Isotopes and impurities in snow and ice, pages 29–33. International Associationof Hydrological Sciences. Publication No. 118.

Hutter, K. (1983). Theoretical glaciology; material science of ice and the mechanicsof glaciers and ice sheets. D. Reidel Publishing Company/Tokyo, Terra ScientificPublishing Company.

Paterson, W. S. B. (1999). The Physics of Glaciers. Butterworth-Heinemann, thirdedition.

Smith, G. D. and Morland, L. W. (1981). Viscous relations for the steady creep ofpolycrystalline ice. Cold Regions Science and Technology, 5:141–150.

0

Appendix

AList of Symbols

Latin lettersSymbol Description Units

A softness parameter, a constant in Glen’s flow law MPa−3 a−1

b specific mass balance rate kg m−2 a−1

bi specific volumetric mass balance rate m a−1

B(T ) temperature dependence of viscosityC specific heat capacity J kg−1 K−1

E Young’s modulus of elasticity MPag acceleration due to gravity m s−2

g vertical gradient of balance rate ∂bi/∂z a−1

h vertical coordinate, depth below surface mH ice thickness mk heat conductivity W m−1 K−1

n exponent in Glen’s flow lawp pressure MPaP heat production WQ heat flux W m−2

q ice flux, water flux m3 s−1

t time (seconds, years) s, aT temperature ◦Cubal balance velocity m a−1

u, v, w components of the velocity vector v m a−1

v velocity vector, v = (u, v, w) m a−1

w.eq. water equivalentx, y, z space coordinates mx position vector, x = (x, y, z) mz vertical coordinate, pointing upwards mzb bedrock elevation mzELA equilibrium line altitude mzs surface elevation m

1

Appendix A List of Symbols

Greek lettersSymbol Description Units

α surface slope tanα = dzsdx

◦

β bed slope tan β = dzbdx

◦

ε strain rate tensor with components εij a−1

η shear viscosity MPa · aγ Clausius-Clapeyron constant [ ∼ 0.074 K MPa−1 ] K MPa−1

κ thermal diffusivityν elastic Poisson ratioρi density of ice [ 900− 917 kg m−3 ] kg m−3

ρw density of water kg m−3

σe effective uniaxial stress [σe := (32σ(d)ij σ

(d)ij )

12 =√

3τ ] MPaσm mean stress [σm := 1

3σkk] MPa

σ stress tensor with components σij MPa

σ(d) stress deviator tensor [σ(d)ij := σij − 1

3σkkδij = σij − σmδij] MPa

τ effective shear stress [τ := (12σ(d)ij σ

(d)ij )

12 = 1√

3σe] MPa

A 2

Appendix

BUseful quantities

Quantity Symbol Value Unit

Mechanical properties

Density of water (0◦C) ρw 999.84 kg m−3

Density of bubble free ice (0◦C) ρi 917 kg m−3

Young modulus of ice E 8.7 · 109 PaShear modulus of ice µ 3.8 · 109 PaPoisson ratio of ice ν 0.31Creep activation energy (<−10◦C) Q 78 kJ mol−1

Thermal properties

Specific heat capacity of water Cw 4182 J K−1 kg−1

Specific heat capacity of ice Ci 2093 J K−1 kg−1

Thermal conductivity of ice (at 0◦C) k 2.1 W m−1 K−1

Thermal diffusivity of ice (at −1◦C) κ 1.09 · 10−6 m2 s−1

Latent heat of fusion (ice/water) L 333.5 kJ kg−1

Depression of melting point (Clausius-Clapeyron constant)- pure ice and air-free water γp 0.074 K MPa−1

- pure ice and air-saturated water γa 0.098 K MPa−1

Constants

Gravity acceleration g 9.81 m s−2

Triple point temperature Ttp 273.16 KTriple point pressure ptp 611.73 PaGas constant R 8.31 J mol−1 K−1

Avogadro number NA 6.023 · 1023

Boltzmann constant kb 1.3807 · 10−23 J K−1

Stefan-Boltzmann constant σsb 5.67 · 10−8 W m−2 K−4

Solar constant (radiation) Qsolar 1368 W m−2

1

Appendix B Useful quantities

Flow law parameter

T (◦C) A ( s−1 Pa−3) A ( a−1 MPa−3) AP ( s−1 kPa−3) AP ( a−1 MPa−3)

0 2.4 · 10−24 75.7 (6.8 · 10−15) (215)-2 1.7 · 10−24 53.6-5 9.3 · 10−25 29.3 (1.6 · 10−15) (50.5)-10 3.5 · 10−25 11.0 (4.9 · 10−16) (15.5)-15 2.1 · 10−25 6.62 (2.9 · 10−16) (9.2)-20 1.2 · 10−25 3.78 (1.7 · 10−16) (5.4)-30 3.7 · 10−26 1.17 (5.1 · 10−17) (1.6)-40 1.0 · 10−26 0.315 (1.4 · 10−17) (0.44)-50 2.6 · 10−27 0.082 (3.6 · 10−18) (0.11)

Table B.1: Flow law parameter A recommended by Cuffey and Paterson (2010), andthe older values AP recommended by Paterson (1999).

It is common to assume that the flow law parameter A can be split into a con-stant rate factor at a reference temperature A0 and a parameter absorbing thetemperature dependence B(T ) (e.g. Hutter, 1983; Paterson, 1999). At tempera-tures below −10◦C the rate factor is of Arrhenius type with an activation energy ofabout 60 kJ mol−1 (Paterson, 1994). A double exponential fit derived by Smith andMorland (1981, eq. 21) is often used

B(T ) = 0.9316 exp(0.32769T ) + 0.0686 exp(0.07205T ) , T ≥ −7.65◦C, (B.1)B(T ) = 0.7242 exp(0.59784T ) + 0.3438 exp(0.14747T ) , T < −7.65◦C, (B.2)

where T is the Celsius temperature. This parameterization is almost identical tothe values given in Table B1 (Paterson, 1999, p. 97).

The rate factor A in Glen’s flow law is also affected by the percentage of water µwithin the ice (Duval, 1977; Paterson, 1999)

A(µ) = (3.2 + 5.8µ) · 10−15 kPa−3 s−1

= (101 + 183µ) MPa−3 a−1 , (B.3)

B.1.

B 1

Appendix

CVectors and tensors

C.1 VectorsA vector x can be represented in a (not necessarily orthogonal) coordinate system

x = x1e1 + x2e2 + x3e3

= xpep (C.1)

where ei are unit length basis vectors. They form a base for a cartesian (orthogonal)coordinate system if all of the following conditions are fulfilled

ei · ej = δij . (C.2)

The symbol δij is the Kronecker symbol and is defined by

δij :=

{1 if i = j,

0 if i 6= j.

In Equation (C.1) we have used the summation convention: we sum over allindices that appear twice.

We now consider another orthogonal coordinate system K ′ that is rotated withrespect to the original coordinate system K, but has the same origin. The new basevectors e′i also fulfill the condition

e′i · e′j = δij . (C.3)

The vector x can then be written in the new base as

x = x′1e′1 + x′2e

′2 + x′3e

′3 (C.4)

= x′pe′p .

The vector x stays the same, its representation is different in both coordinate systems(the components xi and x′i are different).

3

Appendix C Vectors and tensors

e1

e2

e3 K

e1

e2

e3 = e′

3 K, K’

e′

1

e′

2

θ

θ

Figure C.1: The original coordinate system K (solid lines) and the rotated systemK ′ (dashed lines)

Rotation matrix

We now derive the connection between the representations of the vector x in bothcoordinate systems. For this we first define the direction cosine

αij := e′i · ej = cos(e′i, ej)︸︷︷︸direction cosine

(C.5)

The quantity αij is the scalar product of the unit vectors e′i and ej and therefore alsothe cosine of the angle between the vectors e′i and ej (remember a · b = ab cos θ).

With help of the summation convention (Eq. C.1) we write

x = xpep, (C.6)

an make the scalar product with the unit vector ei

x · ei = xpep · ei= xpδpi

= xi (C.7)

and therefore

xi = x · ei= x′pe

′p · ei (Equation C.4)

= x′pαpi . (Equation C.5) (C.8)

Therefore we have shown that the representations of the vector x in both coordinatesystems K and K ′ are linked by

xi = αpix′p . (C.9)

C 4


It can also be shown that the inverse transformation is given by

x′i = αipxp . (C.10)

We now derive the same rules for the direction cosines αij. First we write

x′i = xpαip (Equation C.10)= αjpx

′jαip (Equation C.9)

= αjpαipx′j .

and use x′i = δijx′j so that we obtain δijx′j = αjpαipx

′j. This is true for all values of

x′j so that we arrive atαipαjp = δij . (C.11)

Similarly it can be shown thatαpiαpj = δij . (C.12)

It is convenient to write the direction cosines as a matrix

[αij] =

α11 α12 α13

α21 α22 α23

α31 α32 α33

(C.13)

Equations (C.11) and (C.12) can then be written more compactly as

αipαjp = δij or [αij][αij]T = 1 ,

αpiαpj = δij or [αij]T [αij] = 1 , (C.14)

and Equation (C.9) is written as

x = [αij]Tx′ . (C.15)

The matrix [αij] is called the rotation matrix. As Equation (C.14) shows, it hasthe important property that the transpose of the rotation matrix is identical to itsinverse

[αij]T = [αij]

−1 . (C.16)

The inverse matrix [αij]−1 is defined through

[αij][αij]−1 = [αij]

−1[αij] = 1 . (C.17)

Example We assume that the coordinate system K ′ is rotated by the angle θ withrespect to the coordinate system K (Figure C.1). The components of the rotationmatrix can be obtained by calculating the scalar products ei · e′j:

· e1 e2 e3

e′1 cos θ sin θ 0e′2 − sin θ cos θ 0e′3 0 0 1

C 5


For the calculation of the components α21 and α12 in the table we have made use ofthe relations

cos(θ − π/2) = sin(θ)

cos(θ + π/2) = − sin(θ)

For example

α21 = cos(e′2, e1) = cos(−π/2− θ)= − sin(θ) .

Therefore the rotation matrix is

[αij] =

cos θ sin θ 0− sin θ cos θ 0

0 0 1

(C.18)

Next we take a point P which position vector has the coordinates (2,1,3) in thecoordinate system K

P = 2e1 + e2 + 3e3.

We want to calculate the coordinates of P in the system K ′. With the relation(Eq. C.10)

x′i = αipxp

we obtain

x′1 = α11x1 + α12x2 + α13x3

= cos θ · 2 + sin θ · 1 + 0 · 3x′2 = α21x1 + α22x2 + α23x3

= − sin θ · 2 + cos θ · 1 + 0 · 3x′3 = α31x1 + α32x2 + α33x3

= 0 · 2 + 0 · 1 + 1 · 3 ,

and arrive at

P = (2 cos θ + sin θ)e′1 + (−2 sin θ + cos θ)e′2 + 3e′3.

C 6


Comma notation

We consider the scalar field f = f(x) (e.g. temperature field), the vector fieldfk = fk(x) (e.g. velocity field) and the tensor field fpq = fpq(x) (e.g. stress field). Weintroduce the compact comma notation f,i for the derivative of f with respect tospatial direction xi

f,i :=∂f

∂xii = 1, 2, 3.

Examples of the comma notation

(I) (xkfk),i = fi + xkfk,i

(II) (xkfk),ij = fi,j + fj,i + xkfk,ij

Proofs:

(I) (xkfk),i = xk,ifk + xkfk,i

= δkifk + xkfk,i

= fi + xkfk,i

(II) (xkfk),ij = (fi + xkfk,i),j

= fi,j + xk,jfk,i + xkfk,ij

= fi,j + fj,i + xkfk,ij

C 7


C.2 Tensors

First order tensor

A vector a can be written in two different bases ek and e′k as

a = apep anda = a′pe

′p. (C.19)

Further valid expressions are

a′i = αipap andai = αpia

′p (C.20)

where αij are the components of a rotation matrix. We now define a (cartesian)tensor of order 1 as a quantity, which is represented by three real numbers thatunder the change from the {xi}-system to the {x′i}-system are transformed as

a′i = αipap. (C.21)

Second order tensor

The above definition is extendable. We consider two vectors a and b. Analogous toEquation (C.20) we can write

b′i = αipbp andbi = αpib

′p. (C.22)

We now form the product aibj and look at its transformation behavior

a′ib′j = (αipap)(αjqbq)

= αipαjqapbq andaibj = (αpia

′p)(αqjb

′q)

= αpiαqja′pb′q (C.23)

These equations yield the relation between aibj and a′ib′j. We now write the productaibj as a new quantity

cij := aibj

and alsoc′ij := a′ib

′j .

C 8


The new quantity [aibj] = [cij] can be written as 3 × 3 matrix, the tensor productof the vectors a and b

[aibj] =

a1b1 a1b2 a1b3a2b1 a2b2 a2b3a3b1 a3b2 a3b3

= a⊗ b︸︷︷︸

tensor product

With the above definitions, Equations (C.23) can be written

cij = aibj = αpiαqja′pb′q = αpiαqjc

′pq and

c′ij = a′ib′j = αipαjqapbq = αipαjqcpq . (C.24)

A tensor of order 2 is defined by this transformation rule, i.e. a tensor A of order 2with the components [A]ij = Aij always transforms like

A′ij = αipαjqApq . (C.25)

Tensor of order n

Definition: Given the 3n numbers a′i1,i2,··· ,in that transform as

a′i1,i2,··· ,in = αi1,j1αi2,j2 · · ·αin,jnaj1,j2,··· ,jn (C.26)

under the change from the cartesian coordinate system xi to x′i. These numbers arecalled cartesian tensor of order n

A tensor is defined by its transformation properties. To test whether a given quantityis a tensor, the components have to transform according to Equation (C.26).

Example: We take some scalar field φ = φ(x), which could be for example thetemperature field. We define the quantity ai := φ,i (the temperature gradient) inany coordinate system K. Are the three numbers (a1, a2, a3) the components of atensor?

To answer this question we inquire the transformation properties of ai. Followingthe definition of ai, which is valid for all coordinate systems, we have

a′i =∂φ

∂x′i,

and with the chain rule

a′i =∂φ

∂x′i=

∂φ

∂xk

∂xk∂x′i

= φ,k∂xk∂x′i

.

C 9


Equation shows that∂xk∂x′i

= αik ,

such that

a′i = αikφ,k = αikak .

Comparison with Equation (C.26) shows that ai is indeed a tensor of first order.

Example: We now figure out how the second order tensor A looks in theK ′-system,when it has the following form in the K-system

A =

0 γ 0γ 0 00 0 1

The K ′-system is rotated with respect to the K-system about the e3 axis by anangle of θ = π/4.

Since A is a second order tensor, we have

a′ij = αikαjlakl with

[αij] =

cos θ sin θ 0− sin θ cos θ 0

0 0 1

,

with θ = π/4. Using the transformation formula we obtain

a′11 = α1kα1lakl

= α11(α11a11 + α12a12 + α13a13)

+ α12(α11a21 + α12a22 + α13a23)

+ α13(α11a31 + α12a32 + α13a33)

= cos θ(0 cos θ + γ sin θ + 0)

+ sin θ(γ cos θ + 0 sin θ + 0)

+ 0 (. . . )

= 2γ cos θ sin θ = 2γ1√2

1√2

= γ.

Therefore we have

a′11 = γ.

C 10


Similarly for the next component we get

a′12 = α1kα2lakl

= α11(α21a11 + α22a12 + α23a13)

+ α12(α21a21 + α22a22 + α23a23)

+ α13(α21a31 + α22a32 + α23a33)

= cos θ(−0 sin θ + γ cos θ + 0)

+ sin θ(−γ sin θ + 0 cos θ + 0)

+ 0 (. . . )

= γ(cos2 θ − sin2 θ)

= γ(cos2 θ − sin2 θ)|(θ=π/4) = 0 ,

and thereforea′12 = γ.

With this we arrive at the representation of tensor A in the K ′ system

A =

γ 0 00 −γ 00 0 1

.

C 11


Invariants

A quantity which is independent of the orientation of the coordinate system is calledan invariant. This is best explained with some examples.

Example: a and b are vectors with components ai and bj. Show that the scalarproduct c = a · b is invariant.

To show this we have to prove that the scalar product is independent of the orien-tation of the coordinate system, i.e. that c′ = a′ib

′i and c = aibi are the same

c′ = a′ib′i

= αipapαiqbq

= αipαiqapbq

= δpqapbq

= apbp

= aibi

= c

Example: cij are the components of a second order tensor. Show that the trace ciiis an invariant.

c′ii = αipαiqcpq

= δpqcpq

= cpp = cii

Example: Show that cikcki is an invariant.

c′ikc′ki = αiqαkpcqpαkrαiscrs

= αiqαisαkpαkrcqpcrs

= δqsδprcqpcrs

= cqpcpq = cikcki

C 12


The permutation symbol εijkThe permutation symbol εijk has the value zero if at least two of the indices arethe same. Otherwise the symbol has the value +1 or -1, depending on whether theorder of the indices is cyclic or anticyclic. Thus we have ε123 = ε231 = ε312 = 1 (cyclicindices), ε132 = ε213 = ε321 = −1 (anticyclic indices), and εiij = εijj = εjii = 0 (nosummation over repeated indices!) for i and j = 1, 2, 3.

εijk :=

+1, i, j, k in cyclic order−1, i, j, k in anticyclic order0, two or more indices have the same value

The permutation symbol is also known as the Levi-Civita ε symbol. This symbol ismainly used to write the vector product in index notation. One valid expression is

εijk = ei · (ej × ek).

A useful relation between the Kronecker symbol and the permutation symbol is theδ − ε relation

εijkεkpq = δipδjq − δjpδiq. (C.27)

Isotropic tensors

An isotropic tensor is a tensor with the same entries in each coordinate system.For each isotropic tensor of order n this relation holds

A′ijk... = Aijk....

The unit tensor δij is an example of an isotropic tensor, since

δ′ij = αipαjqδpq (Eq. C.25)= αipαjp

= δij (Eq. C.11) ,

and therefore δ′ij = δij.

C 13


We note some relations without proof, that will be useful in further sections.

• Each isotropic second order tensor A with components [A]ij can be written inthe form

[A]ij = α δij

where α is a scalar.

• Each isotropic third order tensor A with components [A]ijk can be written inthe form

[A]ijk = α εijk

where α is a scalar.

• Each isotropic fourth order tensor A with components [A]ijkl can be writtenin the form

[A]ijkl = α δijδkl + β δikδjl + γ δilδjk

where α, β and γ are a scalar quantities.

C 14

stress and strain - eth z

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