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. Body forcesThe 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 inertia forces such as the centrifugal force.

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

x

z

F

Force

x

z

Fx

Fz

Force components

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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 itstractioncomponents 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 which is defined by the totalsurface stressforce exerted on the surface divided by the surface area

=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

(top)

(bot)

Surface stress components

(top)s

(bot)s

(top)n

(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 n and s1 resp. s2 .normal stress

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

F(top)

A+

F(bot)

A= 0 ,

(top) + (bot) = 0. (4.1)

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Physics of Glaciers I HS 2013

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

(top)n = (bot)n and (top)s = (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.

(top)zz

(top)zx

(bot)zz

(bot)zx

(lft)xx

(lft)xz

(rt)xx

(rt)xz

(top)z

(bot)z

(lft)x

(rt)x

dx

dz

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

(rt)xx = (lft)xx (top)zz = (bot)zz(rt)xz = (lft)xz (top)zx = (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. Denoting the surface areas Axand Az, and taking the moment anti-clockwise, we obtain Kreuzprodukt fr dasMoment!!

(top)zx Azdz + (bot)zx Azdz (lft)xz Axdx (rt)xz Axdx

!= 0 . (4.4)

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Chapter 4 Stress and strain

Using Ax = Az, dx = dz and Equation (4.3) we obtain 2zx 2xz!

= 0, whichis equivalent to zx = xz. The stress state induced by [x,z] (four numbers) istherefore fully described by the three components

xx, xz = zx, and zz. (4.5)

An analogous relation holds in three dimensions where the stress tensor componentsare

xx, yy, zz

xy = yx, xz = zx, yz = zy. (4.6)

Stress tensor

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

= [ij] =

[xz

]=

(xx xzxz zz

). (4.7)

Since xz = zx, the stress tensor is symmetric, that is

= [ij] = [ji] = T . (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

= [ij] =

xyz

=xx xy xzxy yy yzxz yz zz

. (4.9)Notice that only six components are independent, since balance of moments (Eq. 4.6)leads to xy = yx, yz = zy and xz = zx.

For the stress vector and the unit normal n on an arbitrary surface the followingimportant relation holds

i(n) = n = iknk. (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)

ij = ipjq pq or = RRT . (4.11)

where R = [] 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.

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Physics of Glaciers I HS 2013

Example The components of the stress tensor are

= [ij] =

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 between the stress vector and the surface normal n.

Solution: The unit normal on the surface is

n =

Fx1F

x2Fx3

= 12

110

and the traction on the surface is

(n) = n =

1 2 32 1 13 1 0

12

110

= 12

314

.The angle is

cos =(n) n|(n)|

=12

426

= 56.

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

I =1

3ii =

1

3tr =

1

3(xx + yy + zz), (4.12)

and is also called the mean stress m. mean stress

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 tensor deviatorwhich is extra from the isotropic stress state

(d)ij := ij mij = ij

1

3iiij . (4.13)

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

37

Chapter 4 Stress and strain

(II(d))2 =

1

2

(d)ij

(d)ij =

1

2((d))2

=1

2

(((d)xx )

2 + ((d)yy )2 + ((d)zz )

2 + 2((d)xy )2 + 2((d)xz )

2 + 2((d)yz )2). (4.14)

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

The third invariant III(d) is the determinant of the deviatoric stress tensorthird invariant

III(d) = det((d)ij ) =

1

3

(d)ij

(d)jk

(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 (n) isparallel to n. In this case the vectors (n) and n differ only by a numerical factorso that we can write

(n) = n = n . (4.16)The proportionality constant is an eigenvalue and the vector n an eigenvector ofthe tensor . An eigenvector of the stress tensor always fulfills equation (4.16). Ittherefore follows that an eigenvector of defines the orientation of a face withoutshear 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 eachother.

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

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

=

1 0 00 2 00 0 3

. (4.17)Since s(i) = is(i) (no summation convention!), the eigenvalues 1, 2 and 3 arethe normal stresses. No tangential stresses act on the faces with unit normal s(i).

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Physics of Glaciers I HS 2013

The eigenvalues i are called principal stress and the eigenvectors s(i) principalprincipal stressaxes.

The eigenvalues can be found by solving the problem

s = s

which, written in components, reads

ijnjsj si = 0or (ij ij)sj = 0 .

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

det(ij ij) = 0 .

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

3 I12 + I2 I3 = 0,

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

I1 := tr = ii ,

I2 :=1

2(iijj ijij) ,

I3 := det() .

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

39

Chapter 4 Stress and strain

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 experiencedstrainduring 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 deformedstretchlength 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 toextensionits 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 strain

:=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 tofinite strainlf we have to integrate Equation (4.20) with respect to l (the reference length l isincreasing with increasing extension)

:=

lflo

1

ldl = ln

(lflo

)= ln(sn) . (4.21)

For obvious reasons is also called logarithmic strain.

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Physics of Glaciers I HS 2013

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 + da

22 + da

23 = 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 + dx

23 = 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 = kl dak dal = klakxi

alxj

dxi dxj ,

ds2 = ij dxi dxj = ijxiak

xjal

dak dal . (4.23)

41

Chapter 4 Stress and strain

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

ds2 ds2o =(ij

xiak

xjal kl

)dak dal , (4.24)

or as

ds2 ds2o =(ij kl

akxi

alxj

)dxi dxj . (4.25)

We define the strain tensor in two variantsstrain tensor

Green - St. Venant Ekl =1

2

(ij

xiaj

xkal kl

), (4.26)

Cauchy eij =1

2

(ij kl

akxi

alxj

), (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

xpai

=upai

+ pi,apxi

= pi upxi

,

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Physics of Glaciers I HS 2013

and the strain tensors reduce to the simpler form

Eij =1

2

[pq

(upai

+ pi

)(uqaj

+ qj

) ij

]=

1

2

[ujai

+uiaj

+uqai

upaj

]and

eij =1

2

[ij pq

(upxi

+ pi

)(uqxj

+ qj

)]=

1

2

[ujxi

+uixj uqxi

upxj

]

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 Cauchys infinitesimal strain tensor

ij =1

2

(uixj

+ujxi

). (4.32)

In unabridged notation it reads

xx =u

x, xy =

1

2

(u

y+v

x

)= yx ,

yy =v

y, xz =

1

2

(u

z+w

x

)= zx , (4.33)

zz =w

z, yz =

1

2

(v

z+w

y

)= zy .

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.

43

Chapter 4 Stress and strain

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 = uz

Slope = wx

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

ij =1

2

(ujxi uixj

), (4.34)

which is antisymmetric, i.e.ij = ji. (4.35)

Therefore the rotation tensor ij has only three independent components 12, 23and 31 because 11 = 22 = 33 = 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

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Physics of Glaciers I HS 2013

to P isdui =

uixj

dxj . (4.36)

This can be rewritten as

dui =1

2

(uixj

+ujxi

)dxj +

1

2

(uixj ujxi

)dxj = (ij + ij) 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 strain rate

ij :=1

2

(vixj

+vjxi

). (4.38)

The only change with respect to Eq. (4.32) is the dot, the use of velocity vi insteadof displacement ui. Remember that the dot is part of the symbol used to designatestrain rate and does not indicate a time derivative. Notice that other authors(e.g. K. Hutter) use the symbol Dij.

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