chapter4

A force F typeset in bold face is a vector quantity and may be divided into its components along perpendicular directions.. It is called a traction and is commonly represented in terms o

Stress and strain

In this chapter we introduce the concepts of stress and strain which are crucial for

the understanding of glacier flow Any calculation of deformation rates and flow

velocities involves stresses and a flow law which relates them to the strain rates It

is also very useful to take now a look at Appendix C “Vectors and Tensors” The

notation introduced there will be used below

4.1 Force

There are two different kinds of forces: body forces and surface forces

Body forces act on each volume of mass, independent on the surrounding material Body forces

The gravity force is the body force that causes glaciers to flow It exerts on each

volume of ice a force that is proportional to the mass within that volume Other

examples of body forces are inertia forces such as the centrifugal force

Surface forces arise from the action of one body on another across the surface of Surface forces

contact between them A typical example is the force exerted from the glacier to its

base, and vice versa Moreover, across any internal surface of arbitrary orientation

that divides a block of material into two, one side of the block applies a surface force

on the other side

A force F (typeset in bold face) is a vector quantity and may be divided into its

components along perpendicular directions Vectors and tensors are explained in

Appendix B

x

z

F Force

x

z

Fx

Fz

Force components

4.2 Stress

A 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 force

is distributed It is called a traction and is commonly represented in terms of its

traction

components perpendicular and parallel to a surface

Traction Traction components

In order to satisfy the requirement of mechanical equilibrium, any surface must have

a pair of equal and opposite tractions acting on opposite sides of the surface This pair of tractions defines the surface stress vector Σ which is defined by the total

surface stress

force exerted on the surface divided by the surface area

A. Stress and traction are measured in units of force per unit area

 F A



m2 = Pa, and the derived units 105Pa = 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 to the surface and two others parallel to the surface at right angles These components are called normal stress and shear stress and are denoted by σnand σs1 resp σs2

normal stress

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

F(top)

F(bot)

A = 0 ,

Σ(top)+ Σ(bot)= 0 (4.1)

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

σ(top)n = −σ(bot)n and σs(top) = −σs(bot) (4.2)

A pair of normal stresses that point towards each other is called a compressive stress (negative sign), a pair pointing away from each other is called a extensive stress (positive sign)

Stress equilibrium

For ease of presentation we consider two dimensions only (analogous relations hold

in three dimensions) The stresses acting on a small volume of material exert forces and moments that must balance for a mechanical equilibrium

σ(top) zz

σ(top) zx

σ(bot) zz

σ(bot) zx

σ(lft)

(lft)

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

σxx(rt)= −σxx(lft) σzz(top) = −σ(bot)zz

σxz(rt)= −σxz(lft) σzx(top) = −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, since the moments of all normal components are zero Denoting the surface areas Ax and

Az, and taking the moment anti-clockwise, we obtain

σ(top)zx Azdz + σzx(bot)Azdz − σ(lft)xz Axdx − σxz(rt)Axdx= 0 ! (4.4)

Using Ax = Az, dx = dz and Equation (4.3) we obtain 2σzx − 2σxz = 0, which!

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

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

σxx, σyy, σzz

σxy = σyx, σxz = σzx, σyz = σzy (4.6)

Stress tensor

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

σ = [σij] =Σx

Σz



=σxx σxz

σxz σzz



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

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

σ = [σij] =





Σx

Σy

Σz



=





σxx σxy σxz

σxy σyy σyz

σxz σyz σzz



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

For the stress vector Σ and the unit normal ˆn on an arbitrary surface the following important relation holds

If the stress tensor is known in one coordinate system K, it can be calculated in any other system K0 The transformation formula is the same as in Equation (C.25)

σij0 = αipαjqσpq or σ0 = RσRT (4.11)

where R = [α] is an arbitrary rotation

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

Example The components of the stress tensor are

σ = [σij] =





2 −1 1





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 ˆn

Solution: The unit normal on the surface is

ˆ

n =





∂F

∂x 1

∂F

∂x 2

∂F

∂x 3





= √1 2





1 1 0





and the traction on the surface is

Σ(ˆn) = σ ˆn =





2 −1 1





1

√ 2





1 1 0



= √1

2





3 1 4





The angle θ is

cos θ = Σ(ˆn) · ˆn

|Σ(ˆn)| =

1

√ 2

4

√

Stress invariants

From the examples in section C.2 we know how to calculate quantities that are

independent of the orientation of the coordinate axes For a second order tensor in

three dimensions three invariants can be constructed The first is

Iσ = 1

3σii =

1

3trσ =

1

3(σxx+ σyy+ σzz), (4.12)

For incompressible materials like glacier ice, the isotropic mean stress does not

contribute to deformation It is therefore useful to characterize the stress state by

the stress deviator The deviatoric stress tensor is that part of the stress tensor deviator

which is extra from the isotropic stress state

σij(d) := σij − σmδij = σij− 1

The second invariant of the deviatoric stress tensor is defined by second invariant

(IIσ(d))2 = 1

2σ

(d)

ij σij(d) = 1

2(σ

(d))2

= 1

2 (σ

(d)

xx)2+ (σ(d)yy)2+ (σzz(d))2+ 2(σxy(d))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 denoted

by 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

third invariant

IIIσ(d) = det(σ(d)ij ) = 1

3σ

(d)

ij σjk(d)σki(d) (4.15)

It is seldom used in glaciology

Principal stresses

A face Fn ˆ with unit normal ˆn is free of shear forces, if the stress vector Σ(ˆn) is parallel to ˆn In this case the vectors Σ(ˆn) and ˆn differ only by a numerical factor

so that we can write

The proportionality constant λ is an eigenvalue and the vector ˆn an eigenvector of the tensor σ An eigenvector of the stress tensor always fulfills equation (4.16) It therefore follows that an eigenvector of σ defines the orientation of a face without shear stresses Furthermore, the eigenvalue is the normal stress on this face

A short reminder of some properties of symmetric tensors:

• All eigenvalues are real numbers

• Two eigenvectors that belong to different eigenvalues are perpendicular to each other

• There exists at least one coordinate system in which the representation of the tensor has only nonzero values on the main diagonal

For the Cauchy stress tensor, a coordinate system can always be found in which the tensor is purely diagonal The three eigenvectors of σ, designated with s(1), s(2) and

s(3), are perpendicular to each other and define a orthogonal coordinate system In this coordinate system σ has the form

σ =







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) principal

principal stress

axes

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 combined

to yield the same forms

4.3 Deformation

A 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 experienced

strain

during deformation The strain is homogeneous if the changes in size and shape are proportionately identical for each small part of the body and for the body as a whole The strain is inhomogeneous if the changes in size and shape of small parts

of the body are different from place to place: straight lines become curved, planes become curved surfaces, and parallel planes and lines do not remain parallel after deformation

Linear strain

The stretch sn of a material line segment is defined as the ratio of the deformed

stretch

length lf to its undeformed length lo

sn:= lf

lo

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

extension

its initial length lo

en := lf − lo

lo =

∆l

(Note the sign convention: a positive extension is lengthening, a negative extension

is shortening the body.) The above definition gives the average extension after a length change Going to very small extension increments, one defines the strain ε

strain

ε := dl

that is, the ratio of the infinitesimal current extension increment dl with respect

to the current length l To obtain the finite strain of the extension from lo to

finite strain

lf we have to integrate Equation (4.20) with respect to l (the reference length l is increasing with increasing extension)

¯

ε :=

Z lf

l o

1

l dl = ln

 lf

lo



For obvious reasons ¯ε is also called logarithmic strain

We now consider the deformation of an arbitrary body by studying the relative dis-placement of three neighboring points P , P0, P00in the body If they are transformed

to the points Q, Q0, Q00in the deformed configuration, the change in area and angles

of the triangle is completely determined if we know the change in length of the sides

a3, x3

a1, x1

a2, x2

P

P ′

P ′′

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 P0(a1+ da1, a2+ da2, a3+ da3) The square of the length dso of P P0 in the original configuration is given by

ds2o = da21+ da22 + da23 = daidai

When P and P0 are deformed to the points Q(x1, x2, x3) and Q0(x1 + dx1, x2 +

dx2, x3+ dx3), respectively, the square of the length ds of the new element QQ0 is

ds2 = dx21+ dx22+ dx23 = dxidxi

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 convenient choice of index labels)

ds2o = δkldakdal= δkl∂ak

∂xi

∂al

∂xj dxidxj,

ds2 = δijdxidxj = δij∂xi

∂ak

∂xj

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

ds2 − ds2o =



δij ∂xi

∂ak

∂xj

∂al − δkl



or as

ds2 − ds2

o =



δij− δkl ∂ak

∂xi

∂al

∂xj



We define the strain tensor in two variants

strain tensor

2



δij ∂xi

∂aj

∂xk

∂al

− δkl



2



δij − δkl∂ak

∂xi

∂al

∂xj



so that (remember that index names are arbitrary)

ds2− ds2

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

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

An immediate consequence of Equations (4.28) and (4.29) is that ds2− ds2

o = 0 im-plies Eij = eij = 0 and vice versa Therefore, a deformation in which the length of every line element remains unchanged is a rigid-body motion (translation or rota-tion)

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 ,

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



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

εij = 1 2

 ∂ui

∂xj +

∂uj

∂xi



In unabridged notation it reads

εxx = ∂u

1 2

 ∂u

∂y +

∂v

∂x



= εyx,

εyy = ∂v

1 2

 ∂u

∂z +

∂w

∂x



εzz = ∂w

1 2

 ∂v

∂z +

∂w

∂y



= εzy

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

z

u u + ∂u∂xdx

Case 1: ∂u

∂x > 0, w = 0

x

z

u 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 the cartesian tensor

ωij = 1 2

 ∂uj

∂xi − ∂ui

∂xj



which is antisymmetric, i.e

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 rotation and a deformation Consider a point P with coordinates xi and a point P0 in the neighborhood with coordinates xi+dxi The relative displacement of P0with respect