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and so σ(σ2 ’ b2 ’ a2 ) = 0, with solutions σ = 0, ±(b2 + a2 )1/2 .
Thus the difference between the minimum and maximum principal stresses is
2(b2 + a2 )1/2 , i.e., [(‚ψ/‚x1 )2 + (‚ψ/‚x2 )2 ]1/2 .
For the choice ψ = (x2 ’ a2 )(x2 ’ a2 ),
1 2

‚ψ
= 2x2 (x2 ’ a2 ),
σ13 = 1
‚x2
‚ψ
= ’2x1 (x2 ’ a2 ),
σ23 =’ 2
‚x1
and so σ13 vanishes when x1 = ±a, and σ23 vanishes when x2 = ±a. Hence the
lateral faces of the cylinder are free from traction.
The couple on the end is given by

‚ψ ‚ψ
’ x1 + x2 dS
‚x1 ‚x2
end
a a
32a6
dx2 [2x2 (x2 ’ a2 ) + 2x2 (x2 ’ a2 )] =
=’ .•
dx1 12 21
9
a a




3.4 Virtual work rate principle
We can establish a useful set of integral relations for the interaction of the stress
¬eld with a test ¬eld v— , which reduce to the conservation of energy when v— is the
actual velocity ¬eld.
Consider material in a bounded region of volume V and surface S, subject to
surface tractions „ and body forces g, with a local acceleration ¬eld f.
The entire body must satisfy the equations of linear and angular motion

„ dS + g dm = f dm, (3.4.1)
S V V


x — „ dS + x — g dm = x — f dm. (3.4.2)
S V V
52 The Stress-Field Concept

A stress ¬eld σij is said to be compatible with these constraints if it possesses
the properties we have just established:
σij is symmetric,

σij = ρ(fj ’ gj) in V,
(3.4.3)
‚xi
niσij = „j on S,
ni[σij]+ = 0 on any jump surface J.


For a piecewise continuous and piecewise continuously differentiable stress ¬eld
σij and a ¬eld v— which is continuous and piecewise continuously differentiable:
‚—
„ · v— dS + g · v— dm = f · v— dm.
σij v dV + (3.4.4)
‚xi j
S V V V

Usually we would take v— to be a velocity or an in¬nitesimal displacement. When v
is velocity then as we shall see later (3.4.4) corresponds to the balance of externally
applied and internal work rates.
PROOF

„.v— dS = niσijv— dS
j
S S

(σijv— )dV + ni[σij]+v— dJ
= ’j
j
V ‚xi J
‚ ‚—
v—
= σij + σij v dV
j
‚xi j
‚xi
V
‚—
vj dV + v— ρ(fj ’ gj) dV,
= σij j
‚xi
V V

using the conditions (3.4.3). •


Converse principle
If σij is a symmetric tensor ¬eld such that (3.4.4) holds for arbitrary v— , then σij is
compatible with the constraints and conforms to (3.1.2) and (3.1.3).
Now

niσijv— dS = (σijv— ) dV + ni[σij]+v— dJ (3.4.5)
’j
j j
‚xi
S V J

and subtracting this from (3.4.4) we ¬nally obtain

(„j ’ niσij)v— dS + ni[σij]+v— dJ + σij ’ ρ(fj ’ gj) v— dV(3.4.6)
’j
j j
‚xi
S J V

and, since v— is arbitrary, each integrand must vanish. We are therefore able to
j
recover (3.4.3) and hence (3.1.2) and (3.1.3).
3.5 Stress from a Lagrangian viewpoint 53

3.5 Stress from a Lagrangian viewpoint
The stress tensor σij enables us to calculate traction vectors that act on an element
of area in the current (deformed) con¬guration. However, in some circumstances
we would like to be able to relate the stress back to the reference (undeformed
con¬guration).
From (2.1.11) we can relate an element of area in the deformed state dS to the
equivalent element in the reference state by
dS = det F F’T dΣ,
Σ (3.5.1)
which we can rewrite in terms of the normal vectors to the surface elements as
nj dS = J F’1νp dΣ. (3.5.2)
pj

The components of the traction vector „ acting on dS can therefore be expressed
as:
„i dS = σijnj dS = J F’1σijνp dΣ = σPKνp dΣ. (3.5.3)
pi
pj

The ¬rst Piola“Kirchhoff stress tensor
σPK = J F’1σij, σPK = J F’1σ, (3.5.4)
pi pj

is not symmetric, but from the symmetry of σ we ¬nd that [σPK]T FT = σPKF. The
¬rst Piola“Kirchhoff stress tensor is a two-point tensor since it links the current and
reference states. A second Piola“Kirchhoff tensor can be de¬ned that acts solely in
the reference state
σSK = J F’1σF’T = σPKF’T ; σPK = σSKFT . (3.5.5)
This second, symmetric tensor σSK plays an important role in the development
of elastic properties in terms of a Lagrangian strain-energy (see section 4.3). The
stress tensor σ in the deformed state can be recovered from
J σ = FσSKFT . (3.5.6)
4
Constitutive Relations




The balance laws for linear and angular momentum (3.1.2), (3.1.3) do not make
any reference to the properties of the body. The full speci¬cation of the mechanical
and thermal properties of a material requires further information as to the relation
between stress and strain, and thus to the relation between the geometry of
deformation and the resulting stress ¬eld. This information is provided by a
constitutive equation which gives a mathematical representation of the functional
dependence between stress and strain, designed to provide agreement with the
observed behaviour. The behaviour of real materials is both diverse and complex,
and so most constitutive equations are designed to capture the most important
features of the mechanical behaviour of a material in a particular situation. The
mathematical formulation may well be an approximation to the macroscopic
response induced by a speci¬c set of atomic scale processes such as, e.g., the
movement of dislocations in plastic deformation.
Mathematical and physical models of constitutive relations are most effective
when dealing with a single material, since it is dif¬cult to capture the full behaviour
of a composite with a single relation. However, although geological materials are
composed of an aggregate of minerals, many aspects of the behaviour of rocks can
be captured through fairly simple representations with temperature and pressure
dependence.


4.1 Constitutive relation requirements
A constitutive relation takes the form
stress = functional(deformation) (4.1.1)
and the possible behaviour is restricted by a set of postulates which are designed to
reconcile the mathematical and physical behaviour.

(a) Determinism:
The stress at time t depends only on the history of the deformation up to t
and not on future values.

54
4.1 Constitutive relation requirements 55

(b) Local Action:
The stress at ξ at some time t depends only on the deformation in the
neighbourhood of ξ
(c) Material Objectivity:
Constitutive equations are form invariant with respect to rigid motions of
the spatial frame of reference.
(d) Material Invariance (Symmetry)
Constitutive equations are form invariant with respect to a symmetry group.
For ¬‚uids the symmetry group is the full unimodular group. For solids the
symmetry group is restricted to a group of orthogonal transformations of
the material coordinates.

It is only at this stage that we meet any distinction between ¬‚uids and solids. The
¬rst principle (a) allows for material with memory as in viscoelastic behaviour, e.g.,
plastics, the Earth™s upper mantle.


4.1.1 Simple materials
A important class of material behaviour is described by simple materials for which
the stress depends on the local deformation through a functional in time:

σ(ξ, t) = f(Ft, ξ, t),
ξ (4.1.2)
where Ft denotes the history up to time t of the deformation gradient F(ξ, t) and f
ξ
is a tensor functional in time alone.
In this case the principles (a) and (b) are satis¬ed automatically. In order to meet
the requirement of material objectivity (c) we consider the effect of a rigid rotation.
We rotate the spatial frame by QT , so that

F— = QF.
x— = Qx, (4.1.3)
The stress transforms as a second-rank tensor under this rotation:

σ— = QσQT . (4.1.4)
After rotation we must have the same functional form for the dependence of stress
on deformation as in (4.1.2) i.e.,

σ— (t) = f(F—t, ξ, t), (4.1.5)
and thus

Q f(Ft) QT = f([QFt]), for all Q. (4.1.6)
The condition for invariance under rotation (4.1.6) thus imposes a constraint on the
possible form of the functional f.
56 Constitutive Relations

We can proceed further by making the polar decomposition F = RU. Since
(4.1.6) is an identity in Q, choose Q = RT . Then
RT f(Ft, ξ, t)R = f([RT F]t, ξ, t) = f(Ut, ξ, t), (4.1.7)
and so, from the orthogonality of R, we have
f(Ft, ξ, t) = R f(Ut, ξ, t)RT , (4.1.8)
which satis¬es (a), (b) and (c) with
σ(ξ, t) = R f(Ut, ξ, t) RT .
ξ (4.1.9)
The stress therefore depends on the history of deformation only through the stretch
U or equivalently on any strain measure such as
E = 1 (FT F ’ I) = 1 (U2 ’ I). (4.1.10)
2 2

If we write
f(Ut) = U g(Et) UT , (4.1.11)
in terms of a new functional g, then the stress tensor
σ = RU g(Et) UT RT = F g(Et) FT . (4.1.12)
We may therefore express the stress behaviour for the simple material in the form
σ(ξ, t) = F g(Et, ξ, t) FT .
ξ (4.1.13)
in terms of a functional of strain and time.


4.1.2 Material symmetry
When we use the decomposition F = RU, we represent the form of the deformation
in terms of ¬rst stretching an element and then rotating it. If the material possesses
some intrinsic symmetry, rotations within the symmetry group of the material
applied before the deformation RU will leave the resulting stress the same.
Suppose the material has symmetry with respect to the orthogonal transformation
P, such that PPT = I, then states FP and F will be indistinguishable. We therefore
require that the stress representation
σ = f(F) = f(FP), (4.1.14)
for all F and any P in the symmetry group of the material. If the body is isotropic
then (4.1.14) holds for any orthogonal matrix P.
With the representation of the stress (4.1.13) as a function of strain and time, the
symmetry property (4.1.14) gives
Fg(E)FT = FPg(PT EP)PT FT , (4.1.15)
4.2 Energy balance 57


since E = 1 (FT F ’ I). As a result we ¬nd that the functional g needs to have the
2
property,
PT g(E)P = g(PT EP), (4.1.16)
as an expression of the required symmetries.


4.1.3 Functional dependence
We can describe different classes of material by the functional dependence
appearing in the constitutive relation. For viscoelastic materials we need the full
dependence on the history of deformation, but for many materials such effects are
not important.
We can characterise a ¬‚uid by the assumption that the deviatoric part of the stress
depends on the current strain rate ™ = ‚v/‚x, i.e.,
σ = ’pI + f( ™ , ρ), (4.1.17)
where I is the unit tensor and p is the pressure. If the dependence on ™ is a linear
function, we have a Newtonian viscous ¬‚uid.
For solid materials, the deformation itself is a more natural variable than strain
rate. Thus elastic behaviour can be characterised by
stress = function(deformation), (4.1.18)
i.e. σ = f(F), (4.1.19)
and so from (4.1.9) σ = R f(U) RT .
This approach, Cauchy elasticity, may be used with material symmetry
arguments to ¬nd the form of the function f. However, a slightly more restrictive
development based on the idea of a work function (due to Green) seems to be in
close accord with experimental viewpoints, and we will adopt this approach in our
subsequent development.


4.2 Energy balance
We cannot separate the mechanical aspects of deformation from thermal and
other changes and have therefore to introduce the concepts of continuum
thermodynamics. In many applications thermal effects are small and are often
neglected, but we should be aware how they may enter into the consideration of
the deformation of a medium.
As in section 3.4 we consider a volume V with surface S (and outward normal
n) subject to surface tractions „ and body forces g. The work rate of the body and
^
surface forces for a velocity ¬eld v is then
d
„ · v dS +
Y= g · v dm. (4.2.1)
dt S V
58 Constitutive Relations

We can recast the right-hand side of (4.2.1) in terms of the acceleration ¬eld f and
the stress tensor σij by using the virtual work-rate principle (3.4.4) with v— set equal
to the actual velocity v,

d
Y= σij vj dV + f · v dm. (4.2.2)
‚xi
dt V V

· v dm as the rate of change of kinetic energy
We can recognise Vf
D 12
d d
12 12
f · v dm = = 2 ρv dV = ρ ( v )dV,
2 v dm (4.2.3)
Dt 2
dt dt
V V V V
where we have employed (2.5.12) to recast the material derivative of the volume
integral.
The rate of increase of thermal energy for a material volume is the resultant of
the heat production by internal sources, h per unit volume, and the heat conducted
across the surface S:
d
H= h dV ’ q · n dS.
^ (4.2.4)
dt V S
q is the heat ¬‚ux vector which can be related to the temperature gradient by
Fourier™s law of heat conduction

q = ’k T, (4.2.5)
‚x
where k is the thermal conductivity of the material and T is the temperature. We
can recast the expression for the rate of change of the thermal energy in terms of a
volume integral by using the divergence theorem
‚ ‚
d
H= h+ ·k T dV. (4.2.6)
‚x ‚x
dt V
Conservation of energy requires that the gain in the internal and kinetic energy
balance the sum of the work rate by external forces and the rate of increase of
thermal energy. In terms of the internal energy density U per unit mass, we obtain
d d
ρ(U + 1 v2)dV = (H + Y). (4.2.7)
2
dt dt
V
From (2.5.12) we can rewrite the left-hand side of (4.2.7) in terms of an integral
over the material derivative of the speci¬c energy in the continuum,
D
d dH dY
ρ(U + 1 v2)dV = (U + 1 v2)dV =
ρ + . (4.2.8)
2 2
Dt
dt dt dt
V V
Now from (4.2.2) and (4.2.3) we can express the work rate of the various forces as
‚ D 12
dY
= σij vj dV + ρ ( v )dV. (4.2.9)
Dt 2
‚xi
dt V V
With some rearrangement of (4.2.8) using (4.2.9) and the expression (4.2.6) for
4.2 Energy balance 59

the rate of change of thermal energy, we can now ¬nd an expression for the volume
integral of the material rate of change of the speci¬c internal energy,

D ‚ ‚ ‚
ρ U dV = h+ ·k T + σij vj dV. (4.2.10)
Dt ‚x ‚x ‚xi
V V

Since the volume is arbitrary we may equate the integrands on the two sides of this
equation and so the local form of the principle of conservation of energy is

D ‚ ‚ ‚
ρ U=h+ ·k T + σij vj. (4.2.11)
Dt ‚x ‚x ‚xi



Neglect of thermal effects
The internal heat generation h from, e.g., radioactive materials in the Earth occurs
suf¬ciently slowly that it is not important except for very long-term deformation.
The heat transport contribution to the rate of change of internal energy,

‚ ‚
·k T , (4.2.12)
‚x ‚x

can be neglected in discussions of deformation in two simplifying circumstances:

(i) Static Problems: if the time scale over which the deformation is occurring
is suf¬ciently long, heat ¬‚ow will occur to equalise temperature and
conditions will be isothermal. Such is the case in static deformation in
the laboratory or in the long-term deformation of Earth materials.
(ii) Dynamic Problems: if the time scale of deformation is short enough, the
thermal state has no time to adjust to the disturbance and the effect
is conditions comparable to thermal isolation. This is an adiabatic or
isentropic state, appropriate to most elastic wave propagation, but not, e.g.,
for rubber-like materials. For seismic waves in the Earth with periods of
around a second and wavelengths of a few kilometres, the thermal time
constant is many thousands of years and so the adiabatic assumption is
assured. However for small specimens of the same material around 5 mm
in size, the thermal time scale is much shorter (about 20 seconds) and the
adiabatic assumption requires high-frequency waves (1 kHz or greater).

The form of the internal energy function will differ in the two cases, and so the
mechanical response to deformation will be different. For intermediate time scales,
the roles of the deformation and thermal behaviour are closely linked and need to
be treated with continuum thermodynamics.
60 Constitutive Relations

4.3 Elastic materials
In those cases where thermal effects can be neglected, we can equate the rate of
change of the energy of deformation (the stress power) to the deformation term in
(4.2.2):
D ‚
W dV0 = σij vj dV, (4.3.1)
V0 Dt ‚xi
V
where W is called the work density or strain energy.
We can recast the left-hand side of (4.3.1) in terms of the current volume V by
using the relation of dV to dV0 from (2.1.6). We ¬nd
D ‚
(det F)’1 W dV = σij vj dV, (4.3.2)
Dt ‚xi
V V
but once again the volume V is arbitrary and so
D ‚
(det F)’1 W = σij vj. (4.3.3)
Dt ‚xi
We can ¬nd an alternative expression for the mobile derivative DW/Dt by
employing the chain rule in terms of the deformation gradient;
‚W D ‚xj
D ‚W D
W= Fjk = , (4.3.4)
Dt ‚Fjk Dt ‚Fjk Dt ‚ξk
‚vj ‚W ‚xi ‚vj
‚W
= = . (4.3.5)
‚Fjk ‚ξk ‚Fjk ‚ξk ‚xi
With this representation for DW/Dt inserted in (4.3.3) we have
‚vj ‚vj
‚W
(det F)’1 Fik = σij , (4.3.6)
‚Fjk ‚xi ‚xi
and, since this is true for an arbitrary velocity ¬eld,
‚W
det F σij = Fik . (4.3.7)
‚Fjk
We already know that the stress σ depends only on the stretch U (or strain E), we
therefore recast (4.3.7) in terms of the Green strain E. The elements of the Green
strain are
2Ekl = FjkFjl ’ δkl, (4.3.8)
and we can express the increment in the strain energy as,
‚W 1 ‚W ‚W
dW = (Fjk dFjl + Fjl dFjk) = Fjl dFjk = dFjk. (4.3.9)
‚Ekl 2 ‚Ekl ‚Fjk
The representation (4.3.7) for the stress tensor can therefore be written in the
form,
‚W
det F σij = FikFjl , (4.3.10)
‚Ekl
4.4 Isotropic elastic material 61

or, symbolically, as,
‚W T ‚W
U RT .
det F σ = F F =R U (4.3.11)
‚E ‚E
We can now identify (4.3.11) as having precisely the form (4.1.13) with the
functional g(E) identi¬ed with the tensor derivative ‚W/‚E. Further, from (3.5.6),
we can recognise g(E) = ‚W/‚E as the second Piola“Kirchhoff tensor in the
reference state.
From the symmetry properties (4.1.16) we can deduce
‚W T ‚W
(P EP) = PT P. (4.3.12)
‚E ‚E
for all orthogonal P in the symmetry group of the elastic medium.


Material symmetry
The strain energy function W is required to be invariant under the symmetry group
of the material {P}, so that
W(F) ≡ W(FP), (4.3.13)
and since a rigid post-rotation makes no difference
W(PT FP) = W(F) = W(U), (4.3.14)
and thus W depends only on the stretch matrix U.


4.4 Isotropic elastic material
If the material properties are unaffected by any rotation, the solid is said to be
isotropic. In these circumstances there are some simpli¬cations in the form of the
relations between stress and strain.


4.4.1 Effect of rotation
For all rotations Q, from (4.3.14),
W(QFQT ) = W(F) = W(U) = W(QT UQ). (4.4.1)
If we choose Q as the rotation from background axes to the Lagrangian triad, then
QT FQ = diag{»1, »2, »3} (4.4.2)
in some ordering of the principal stretches.
Thus for any isotropic solid, W(F) is a symmetric function of the principal
stretches, since it is unchanged by a rotation of π/2 about the x3 principal axis
which takes the indices 1’2 etc.
62 Constitutive Relations

4.4.2 Coaxiality of the Cauchy stress tensor and the Eulerian triad
From the symmetry property of stress (4.1.16) and the stress formula (4.3.11) we
have the relation (4.3.12)
‚W T ‚W
(Q EQ) = QT Q, (4.4.3)
‚E ‚E
which now must hold for all rotations Q.
A consequence of isotropy is that ‚W/‚E and E are coaxial. The tensor E
is symmetric with principal axes along the Lagrangian triad and principal values
12
2 (»r ’ 1), r = 1, 2, 3. In this principal basis consider a rotation P of π about the
3-axis: which takes 1’ ’1, and 2’ ’2, then PT EP = E, and so from (4.3.13)
‚W T ‚W
(P EP) = PT P. (4.4.4)
‚E ‚E
We write g(E) = ‚W/‚E, and thus g = PT gP. Since P is orthogonal Pg = gP.
The effect of the rotation by π about the 3-axis is then that the elements g13, g23
would change sign, but this is not allowed by the symmetry operation and hence
g13 = g23 = 0. (4.4.5)
Similarly we may show that all off-diagonal terms vanish, i.e., with respect to the
Lagrangian triad
‚W
g(E) = = diag{g1, g2, g3}, (4.4.6)
‚E
and so ‚W/‚E and E are coaxial. This is in fact a general theorem for an isotropic
tensor function W.
Now from the stress relation (4.3.11) we have
‚W
U RT ,
det F σ = R U (4.4.7)
‚E
and so
‚W
RT σ R = U U (det F)’1. (4.4.8)
‚E
With the Lagrangian triad as basis the right-hand side of (4.4.8) is diagonal. The
action of the rotation R is to take the Lagrangian into the Eulerian triad and so, for
an isotropic solid, the principal axes of σ lie along the Eulerian triad. Thus the
Cauchy stress tensor and the Eulerian triad are coaxial in the case of isotropy.


4.4.3 Principal stresses
For a pure strain deformation, Fij = Uij, and referred to the Eulerian triad Fij =
diag{»1, »2, »3}. From (4.3.7) the principal stress σr is given by
‚W
det F σr = »1»2»3 σr = »r (no sum). (4.4.9)
‚»r
4.4 Isotropic elastic material 63

Suppose the work function is given as W(±, β, γ) with ± = log(»1»2»3), β =
»1 + »2 + »3, γ = 1 (»2 + »2 + »2). Then the principal stress
21 2 3

1 ‚W ‚W ‚W 2
σr = + »r + », (4.4.10)
‚γ r
»1»2»3 ‚± ‚β
and since σ is coaxial with the Eulerian triad, »r are the principal values of RURT ,
since F = RU = (RURT )R. Further det F = »1»2»3 so we can equate principal
values,
‚W ‚W ‚W 2
σr = (det F)’1R U RT .
I+ U+ (4.4.11)
‚± ‚β ‚γ r
Transferring to general axes we have
‚W ‚W ‚W 2 T
σ = (det F)’1R I+ U+ UR (4.4.12)
‚± ‚β ‚γ
‚W ‚W ‚W
= (det F)’1 (FFT )1/2 + (FFT ) ,
I+ (4.4.13)
‚± ‚β ‚γ
which, as we would expect, depends on FFT .


4.4.4 Some isotropic work functions
The simplest type of work function is W = φ(»1»2»3) that depends only on change
of volume, for which from (4.4.9) each of the principal stresses σr = φ (»1»2»3),
and so the stress tensor is purely hydrostatic.
A second type of work function is provided by W = ψ(»1) + ψ(»2) + ψ(»3)
for which the principal stresses are σr = »r ψ (»r )/(»1»2»3). However this second
type of work function has the defect that it predicts no change in lateral dimensions
in a tension or compression test
σ1 = 0, σ2 = σ3 = 0, ’ »1 = »2 = 1. (4.4.14)
To remedy this problem we combine the two forms for the work function to give
W = φ(»1»2»3) + ψ(»1) + ψ(»2) + ψ(»3), (4.4.15)
for which the principal stresses are
σr = φ (»1»2»3) + »r ψ (»r )/(»1»2»3). (4.4.16)
Such a representation for the work function is suitable for small deformations in
most materials. For large elastic deformation it provides an adequate description of
the static deformation of vulcanised or synthetic rubbers (i.e. random assemblages
or cross-linked, long-chain molecules) for dimensional changes of a factor of 10 or
so. Large elastic deformations are also feasible in rocks at high temperatures.
Near the reference state we can consider incremental deformation by setting
64 Constitutive Relations

»r = 1 + er , where er is a ¬rst-order quantity and can denote any normalised
measure of strain. Then the principal stresses are given by
σr = [φ (1) ’ ψ (1)](e1 + e2 + e3)
+ [ψ (1) + ψ (1)]er + [φ (1) + ψ (1)] + O(e2). (4.4.17)
r

If the ground state is stress-free, so that φ (1) + ψ (1) = 0, then the expressions
for the principal stress simplify somewhat:
σr = [φ (1) + φ (1)](e1 + e2 + e3) + [ψ (1) + ψ (1)]er + O(e2). (4.4.18)
r

We may identify the elastic moduli in the ground state by comparison with a general
linearised development (see Section 5.2.1).
Note that we may also make an expansion »r = »r0 + er about a distorted state
but then the moduli for incremental deformation need not be isotropic (see Section
6.4).


4.5 Fluids
We have so far considered elastic solids for which the stress depends on strain. We
can characterise a ¬‚uid by a constitutive equation in which stress depends on the
rate of strain. A suitable constitutive relation is thus
σij = sij(‚vp/‚xq, ρ, T ), (4.5.1)
in terms of the velocity gradient ‚vp/‚xq, density ρ and temperature T .
We introduce the rate of deformation tensor
‚vi ‚vj
Dij = 12 ‚x + ‚x , (4.5.2)
j i

which represents the symmetric part of the velocity gradient in the Eulerian
con¬guration. The corresponding anti-symmetric part
‚vi ‚vj
1
= ’ , (4.5.3)
ij 2 ‚xj ‚xi
is called the spin or vorticity tensor.
We will now impose the requirement on the ¬‚uid that the stress σ is independent
of any superimposed rigid body rotation. We rewrite (4.5.1) in the form
σ = s(D, , ρ, T ) (4.5.4)
to separate the dependence on the symmetric and anti-symmetric parts of the
velocity gradient. Suppose we have a deformation process
x = x(ξ, t),
ξ v = v(x, t), (4.5.5)
and superpose a time-dependent rigid rotation so that in the new ¬‚ow
x = Q(t)x(ξ, t),
ξ
¯ (4.5.6)
4.5 Fluids 65

where Q(t) is an orthogonal rotation tensor. The associated velocity
D ™ ™
v= x = Qx + Q™ = Qx + Qv,
¯ ¯ x (4.5.7)
Dt
and the velocity gradient
‚¯ ‚¯ ‚x ‚v
v v ™ QT .
= = Q+Q (4.5.8)
‚¯ ‚x ‚¯ ‚x
x x
The rate of strain tensor for the new ¬‚ow is given by
‚¯ ‚¯T
v v ™T
¯ ™
= 1 (QQT + QQ ) + QDQT ,
1
D= + (4.5.9)
2 2
‚¯ ‚¯
x x
but, since QQT = I, the ¬rst term vanishes and thus
¯
D = QDQT . (4.5.10)
The spin tensor for the new ¬‚ow
‚¯ ‚¯T
v v
¯= 1

2 ‚¯ ‚¯
x x
™T
™ ™
= 1 (QQT ’ QQ ) + Q QT = Q(QT Q + )QT . (4.5.11)
2

If σ is the stress from the original deformation process, the stress associated with
the new ¬‚ow including the time-dependent rotation is
σ = QσQT ,
¯ (4.5.12)
¯
and in terms of D, ¯ we can express σ as
¯
¯
σ = s(D, ¯ , ρ, T ).
¯ (4.5.13)
Now, equating the representations for σ from (4.5.12) with (4.5.1) and (4.5.13),
¯
using (4.5.10), (4.5.11), we require

s(QDQT , Q(QT Q + )QT , ρ, T ) = Qs(D, , ρ, T )QT , (4.5.14)

for all rotations Q. Suppose now that Q = I, Q = 0, then

s(D, Q + , ρ, T ) = s(D, , ρ, T ), (4.5.15)
and so s must be independent of the spin tensor . We can therefore reduce
(4.5.14) to the form
s(QDQT , ρ, T ) = Qs(D, ρ, T )QT , (4.5.16)
for all rotations Q.
Following the arguments of Section 4.4.2, the isotropy property (4.5.16) requires
that σ = s(D, ρ, T ) and D are coaxial. The most general form for σ is therefore
σ = aI + bD + cD2. (4.5.17)
66 Constitutive Relations

where a, b, c are functions of ρ and T and the invariants of D: tr D, det D,
2 2
1
2 [tr D ’ (tr D) ], which are symmetric functions of the principal values d1, d2,
d3.
In the absence of motion D ≡ 0 and
σ = s(0, ρ, T ) = ’p(ρ, T )I, (4.5.18)
where p(ρ, T ) is the hydrostatic pressure.
The classical Newtonian viscous ¬‚uid is a special case of (4.5.17) in which the
stress is linear in the velocity gradient
σij = ’p(ρ, T )δij + bijklDkl. (4.5.19)
The bijkl are the components of a fourth-order isotropic tensor with i ” j, k ” l
symmetry so
bijkl = ζδijδkl + ·(δikδjl + δilδjk), (4.5.20)
and thus we can write
σij = [’p(ρ, T ) + ζ(ρ, T )]δij + 2·(ρ, T )Dij. (4.5.21)
For a dilatational ¬‚ow from the origin
v = dr, dij = dδij,
(4.5.22)
σij = [’p(ρ, T ) + (ζ + 2 ·)]δij.
3

and so ζ + 2 · is called the coef¬cient of bulk viscosity (and is often assumed to be
3
zero).
For a shear ¬‚ow
v1 = dx2, v2 = v3 = 0,
(4.5.23)
d12 = d21 = 1 d, all other dij = 0,
2
and
σ12 = σ21 = ·d, σ11 = σ22 = σ33 = ’p, σ23 = σ13 = 0, (4.5.24)
so that · is known as the coef¬cient of shear viscosity (more commonly just
viscosity).
The explicit form of the stress tensor when the bulk viscosity vanishes (a Stokes
¬‚uid) is
‚vi ‚vj ‚vk
’ 2·
σij = ’pδij + · + δij, (4.5.25)
3
‚xj ‚xi ‚xk
and this gives an excellent description of many ¬‚uids e.g. water, air.
The additional dependence on D2 in (4.5.17) is insuf¬cient to provide an
adequate description of the ¬‚ow of many composite materials, e.g., blood,
polymeric liquids which deviate from linear dependence on rate of strain, these
materials have a memory of deformation which can be described using viscoelastic
constitutive laws.
4.6 Viscoelasticity 67

4.6 Viscoelasticity
We can characterise a solid by a stress-state which depends on strain and a ¬‚uid
by a stress-state which depends on the rate of strain. The intermediate situation
in which stress depends on the strain and strain-rate via the history of deformation
covers a wide variety of behaviour described as viscoelastic.
We will con¬ne our attention to situations in which the strains are small so that
the relation between stress and strain is linear. Even this development of linear
viscoelasticity will allow the description of quite complex behaviour.
We assume that the stress tensor σ depends on the Cauchy strain tensor e by a
linear transformation
σ(x, t) = C{et}, (4.6.1)
where, as in Chapter 4, et denotes the history of strain up to time t. This linear
relation is assumed to be single-valued and to possess a unique inverse relation
e(x, t) = J{σt}, (4.6.2)
Because the relation is linear C, J represent tensor functions, e.g.,
σij(x, t) = Cijpq{et }, (4.6.3)
pq

with ij, pq symmetry associated with the symmetries of the stress and strain tensor.


Models of viscoelastic behaviour
One of the most useful models that can be built from the mechanical elements
introduced in Section 1.1.3 is the standard linear solid consisting of a spring in
series with a spring and dashpot combination.

m1
m2




·1

Figure 4.1. Mechanical model for a standard linear solid, built of springs and dashpot
elements.

If the strain in the parallel spring and dashpot is e1 then the stress is
σ = m1e1 + ·1e1 = m2e2,
™ (4.6.4)
since this must also balance the strain e2 in the second spring. The total strain is
e = e1 + e2, (4.6.5)
68 Constitutive Relations

and thus
m1m2e + ·1m2e = (m1 + m2)σ + ·1σ.
™ ™ (4.6.6)
We may rewrite this relation as
σ e
σ+ = m2 e + ,
™ ™ (4.6.7)
„R „C
and the characteristic time scales for relaxation („R) and creep („C) are given by
·1 ·1
„R = , „C = . (4.6.8)
m1 + m2 m1
This model gives a three parameter description (m2, „R, „C) of a viscoelastic
material and covers all the mechanical features of such solids. For very rapid
deformations the terms in σ and e are important and the material behaves like an
™ ™
elastic solid with dynamic modulus m2; for very slow deformation the behaviour
is again elastic with a static modulus „Rm2/„C.
A Burgers material provides a further generalisation of allowed behaviour
by including the possibility of long-term viscosity, with the introduction of a
further dashpot. The Burgers model thus consists of a Maxwell element and a
Kelvin“Voigt element in series.

m1
·2
m2



·1



Figure 4.2. Mechanical model for a Burgers solid.

The total strain = 1 + 2 is the combination of the strains in the Maxwell
element 1 and the Kelvin“Voigt element 2. The stress in the Kelvin“Voigt
element is
σ = m1( ’ 2) + ·1( ™ ’ ™ 2), (4.6.9)
and for the Maxwell element
σ σ

2= +. (4.6.10)
m2 ·2
Now, by eliminating 2 between (4.6.9) and (4.6.10), we can derive the constitutive
relation for the Burgers model in the form
m1m2 m1 + m2 m2 m1m2
m2 ¨ + ™ =σ+ + σ+ σ,
¨ ™ (4.6.11)
·2 ·1 ·2 ·1·2
4.7 Plasticity and ¬‚ow 69

which we can rewrite as
m1 1 1 1
m2 ¨ + ™ =σ+ + σ+ σ.
¨ ™ (4.6.12)
„M „R „M „C„M
where we have introduced the Maxwell time „M = ·2/m2, in addition to the
characteristic time scales for relaxation („R) and creep („C) introduced above.
Under constant stress σ0, the evolution of strain for the Burgers model takes the
form
σ0 σ0 t σ0
= + 1 ’ exp ’ + t. (4.6.13)
m2 m1 „C ·2
The strain relation (4.6.13) separates rather neatly into an initial elastic response
from the ¬rst spring element, continuing creep from the Kelvin“Voigt element and
viscous ¬‚ow from the ¬nal dashpot. The ¬‚exibility of the Burgers model means
that it can provide a useful representation of a wide range of behaviour.


4.7 Plasticity and ¬‚ow
Although a linear dependence of stress on strain-rate provides a reasonable
description of many ¬‚uids, it is less representative of the slow deformation of solids.
At high temperature and low stresses, polycrystalline silicate minerals frequently
show a power-law dependence of strain rate on stress
™ = Aσn, with n > 1. (4.7.1)
A similar relation has also been used to describe glacier ¬‚ow. This simple
power-law dependence re¬‚ects the macroscopic effects of plastic deformation and
rearrangement within the polycrystalline materials.
The in¬‚uence of weak non-linearity will be most pronounced when comparing
deformation at different time scales, since the effective viscosity will depend on
strain rate and hence the frequency of deformation. For example, although the
long-term effects of glacial rebound following the melting of the major ice sheets
of the last glacial can be quite well represented using a Newtonian viscosity for the
Earth™s mantle, the effective viscosity deduced from modern high-precision studies
using satellite based geodesy is slightly different. The difference is suf¬cient to
suggest that weakly non-linear viscosity is present in the upper mantle.


EXAMPLE: NON-LINEAR VISCOSITY
The behaviour of certain viscous ¬‚uids can be modelled by the constitutive equation
σij = ’pδij + 2·(K2 )Dij
(n’1)/2
where K2 = 2Dij Dij , for rate-of-strain tensor Dij , and ·(K2 ) = kK2 , where k and
n are positive constants (n = 1 corresponds to a Newtonian ¬‚uid).
Such a power-law ¬‚uid undergoes a simple shearing ¬‚ow between two large parallel
plates a distance h apart, such that one plate is held ¬xed and the other moves with a
70 Constitutive Relations

constant speed U in its plane. Show that the shearing force per unit area on the plates is
k(U/h)n and that the apparent viscosity is k(U/h)n’1 as a function of the shear rate U/h.
Take cartesian axes such that the 1-axis is directed along the ¬‚ow, and the 3-axis across
¬‚ow. The ¬‚uid will be in stationary contact with the plates so that v1 (0) = 0, v1 (h) =
U. The linear velocity ¬eld v = (Uz/h, 0, 0) The rate-of-deformation tensor Dij will
only have non-zero components D13 = D31 = U/2h, so that K2 = 2Dij Dij =
U2 /h2 . Thus ·(K2 ) = k(U/h)n’1 and
σij = ’pδij + k(U/h)n ;
hence
σ13 = σ13 = k(U/h)n , σ11 = σ22 = σ33 = p.
The traction on the top plate is „j = σ3j and so
„1 = k(U/h)n , „2 = 0, „3 = ’p.
The magnitude of the shearing force per unit area is therefore k(U/h)n . The apparent
viscosity · such that
¯
σij = ’pδij + ·Dij ,
¯
is · = k(U/h)n’1 . •
¯
5
Linearised Elasticity and Viscoelasticity




The only known continua which are capable of ˜¬nite™ elastic strain and which have
also comparable resistance to shear and compression are the new strong ¬bres or
whiskers of ceramics based on carbon, boron etc.
In common metals the elastic range of strain is only in¬nitesimal and may be
treated by a linearised theory. In the Earth as well, small additional disturbances
can be treated by linearising about an existing stress state.
5.1 Linearisation of deformation
The assumption we shall make in subsequent work is that the strain is small, i.e.,
the difference U ’ I is ¬rst order. However, for bodies such as slender columns
or thin panels, the rotation can be large even though the strain is small, and in this
case we have to retain rotation terms R. In general, we can take R ’ I to also be
small, and work with a fully linearised theory.
In terms of particle displacement
u = x ’ ξ = Aξ,
ξ A = F ’ I, (5.1.1)
where A is the displacement gradient. To ¬rst order in A, the Green strain tensor
E = 1 (FT F ’ I) = 1 (A + AT ) + · · · , (5.1.2)
2 2

and also the incremental stretch
(U ’ I) = 1 (A + AT ) + · · · . (5.1.3)
2

Now, from (5.1.1) we can express the displacement gradient as
A = RU ’ I = (R ’ I) + (U ’ I) + · · · , (5.1.4)
so that we have a representation for the rotation R
(R ’ I) = 1 (A ’ AT ) + · · · , (5.1.5)
2

but for a ¬rst-order rotation we also have
(R ’ I) = θN = θ n — . (5.1.6)

71
72 Linearised Elasticity and Viscoelasticity

To ¬rst order the Green strain tensor elements are given by
‚ui ‚uj
1
Eij = + + ··· , (5.1.7)
2 ‚ξj ‚ξi
and the distinction between Eij and the Cauchy strain tensor eij is negligible. The
fractional dilatation
‚uk
Ekk = + · · · = div u + · · · . (5.1.8)
‚ξk
The rotation yields
θ n = 1 ∇ — u + · · · = 1 curl u + · · · . (5.1.9)
2 2

In this linearised theory we will use x rather than ξ to denote the reference position
of a particle.


5.2 The elastic constitutive relation
We make an expansion of the strain energy function about the reference state
2W(E) = a + bijEij + cijklEijEkl + O(E3), (5.2.1)
and since Eij is symmetric we can arrange i ” j and k ” l symmetry in the
coef¬cients and also ij ” kl symmetry.
bij = bji, cijkl = cjikl = cijlk = cklij. (5.2.2)
If the reference state is stress-free bij = 0. In the Earth, for example, there is a
large hydrostatic pressure ¬eld due to self-gravitation and then 1 bij = ’pδij.
2
Note: the fact that earthquakes occur means that there must also be non-hydrostatic
stresses representable by the second term in (5.2.1).
Now from the relation between the stress tensor and the strain derivative of the
strain energy, (4.3.11), we have
‚W T
det F σ = F F. (5.2.3)
‚E
Thus, in the absence of non-hydrostatic pre-stress, when E is ¬rst order but R is
unrestricted
‚W T
R + O(E2),
σ=R (5.2.4)
‚E
as noted above this semi-linearised relation is needed for slender bodies.
For full linearity A ’ I is ¬rst order, and
‚W
+ O(E2),
σ= (5.2.5)
‚E
and so the component representation of the stress tensor, using (5.1.1), becomes
σij = 1 bij + cijklEkl + O(E2). (5.2.6)
2
5.2 The elastic constitutive relation 73

For a stress-free ground state (or working in terms of incremental stresses) we
have the generalised form of Hooke™s Law
σij = cijklekl + O(e2), W = 1 σijeij + O(e3), (5.2.7)
2

where we have used the traditional notation for the strain, in view of the equivalence
of the Cauchy and Green strain de¬nitions for small deformation.
In this full anisotropic case the fourth order tensor of elastic moduli cijkl is
restricted by the i ” j, k ” l and ij ” kl symmetries to 21 independent
components. The presence of material symmetry further reduces the number of
independent moduli, e.g. for hexagonal symmetry there are ¬ve moduli.

5.2.1 Isotropic response
If the material is isotropic then cijkl must be an isotropic tensor subject to the
required symmetries in permutation of indices, and so
cijkl = »δijδkl + μ(δikδjl + δilδjk). (5.2.8)
Thus the stress“strain relation is
σij = »δijekk + 2μeij; (5.2.9)
» and μ are known as the Lam´ moduli.
e
The inverse relation to (5.2.9) is most conveniently expressed in terms of another
set of constants
Eeij = (1 + …)σij ’ …σkkδij, (5.2.10)
where E is Young™s modulus and … is Poisson™s ratio.

5.2.2 Nature of moduli
The interpretation of these moduli can be obtained by looking at simple
deformations.
(a) Consider a purely volumetric strain
eij = 0, i = j,
(5.2.11)
e11 = e22 = e33 = e,
then the stress components are
σij = 0, i = j, σ11 = 3κe, etc.
and (5.2.12)
κ = » + 2 μ is called the bulk modulus and we have a hydrostatic stress ¬eld.
3
(b) If we consider a simple shear deformation
x1 = ξ1 + γξ2, x2 = ξ2, x3 = ξ3, (5.2.13)
the only non-zero in¬nitesimal strain component is e12 = 1 γ and the 2
corresponding stress σ12 = μγ, with all other σij = 0. μ is called the shear or
rigidity modulus. In general, if the dilatation k ekk vanishes σij = 2μeij.
74 Linearised Elasticity and Viscoelasticity

(c) Consider uniaxial tension σ11 = σ1, σ22 = σ33 = 0, σij = 0, i = j, then
e11 = σ1/E,
e22 = ’…σ1/E, eij = 0, i = j, (5.2.14)
e33 = ’…σ1/E,
so … is the quotient of transverse contraction and longitudinal extension.
In terms of κ and μ the work function can be written as
W = 1 κe2 + μ(eij ’ 1 ekkδij)(eij ’ 1 eqqδij), (5.2.15)
kk
2 3 3
and this will be positive de¬nite if κ > 0, μ > 0.


5.2.3 Interrelations between moduli
The interrelations between the moduli may be obtained by comparing the
descriptions of different deformations
1 ’ 2…
σkk = 3κekk, ekk = σkk, (5.2.16)
E
and
»
2μeij = σij ’ σkkδij
3» + 2μ (5.2.17)
Eeij = (1 + …)σij ’ …σkkδij.
On combining the results of (5.2.16) and (5.2.17) we obtain
E E
= 2μ, = 3κ,
1+… 1 ’ 2…
(5.2.18)
3 1 1 3κ ’ 2μ
=+, 2… = .
E μ 3κ 3κ + μ
For positive-de¬nite W we note that κ > 0 with μ > 0 is equivalent to E > 0 with
’1 < … < 1 . Normally … is positive since there is transverse contraction under
2
compression.
The limiting case of incompressibility (or second-order volume change) is
obtained by letting κ ’ ∞ with μ ¬xed so that E ’ 3μ, and … ’ 1 .2
The notation we have used for the elastic moduli is common in elasticity and
seismology. However, in the engineering and materials physics literature the
notation conventions are somewhat different: the bulk modulus κ is designated
K and the shear modulus μ is represented by G.


5.2.4 An example of linearisation
Suppose we have a work function speci¬ed in terms of principal stretches as in
(4.4.16), so that
W = φ(»1»2»3) + ψ(»r ), (5.2.19)
r
5.2 The elastic constitutive relation 75

with
c cm
[(»1»2»3)’mn ’ 1],
φ= ψ(»r ) = [» ’ 1], (5.2.20)
mr
mn
for which the principal stress is

σr = c[»m ’ v’mn]/v with v = »1»2»3. (5.2.21)
r

Near the reference state

σr = mc[n(e1 + e2 + e3) + er ] + O(e2). (5.2.22)

More general W may be constructed by summing over various (c, m) pairs keeping
n ¬xed.
Comparing with the isotropic constitutive relation in the principal frame

σr = 2μer + »(e1 + e2 + e3), (5.2.23)

we can identify the shear modulus μ = 1 mc, and Poisson™s ratio … = n/(1 +
2
2n).
Under a uniaxial load σ1 with arbitrary magnitude

»2 = (»1»2)’n, since »2 = »3,
2
(5.2.24)
log »2 = ’… log »1.


5.2.5 Elastic constants
Typical values of material parameters at 300 K are listed in the following tables.
They relate to random aggregates of anisotropic single crystals (at least 1012 m’3 ).
Such aggregates are effectively isotropic and homogeneous at a macroscopic level
(the volume of a test specimen being of order 10’5 m3 . The values are means of the
best data available and have been ˜normalised™ so as to be theoretically consistent
(within the accuracy of the decimal places given).

Common metals
» μ[G] κ[K] E …
Aluminium 0.60 0.265 0.775 0.715 0.35
Copper 1.09 0.45 1.39 1.22 0.355
Nickel 1.39 0.76 1.90 2.01 0.32
Lead 0.38 0.074 0.43 0.21 0.42
Gold 1.48 0.28 1.67 0.795 0.42
Silver 0.82 0.27 1.00 0.755 0.37
Iron 1.18 0.83 1.73 2.145 0.29
Molybdenum 1.66 1.60 2.73 4.02 0.255
76 Linearised Elasticity and Viscoelasticity

Material properties in the Earth
» μ[G] κ[K] E …
2500 km deep 4.195 2.723 6.011 7.096 0.303
1000 km deep 2.180 1.820 3.420 4.630 0.272
20 km deep 0.429 0.323 0.645 0.830 0.285

Units for », μ[G], κ[K], E are 1011 N m’2 .


In metals, the elastic range of strain is terminated by plastic yielding, e.g., for Cu
and Al the tensile yield stress ≈ 17.5 — 106 N m’2 so that
eyield ≈ 1.4 — 10’4 for Cu,
(5.2.25)
’4
eyield ≈ 2.4 — 10 for Al.
Even for stronger constructional materials such as mild steel and aluminium alloys,
the yield strains are around 10’3 .
For Earth materials, the linear approximation works well for strains less than
’5 which is rarely exceeded except in the immediate vicinity of earthquakes and
10
underground nuclear explosions.


5.2.6 The uniqueness theorem
Locally the stress-¬eld in a body satis¬es (3.2.4)

σij + ρgj = ρfj, (5.2.26)
‚xi
where g is the body force and f is the acceleration.
In the linearised approximation, the acceleration f = ‚2u/‚t2 in terms of the
displacement u. The advected terms v · ∇ can be neglected. Consider a bounded
elastic region V with given boundary conditions on S:
(i) „ speci¬ed on part of the surface,
(ii) u speci¬ed,
(iii) normal component of u and tangential component of „ (or vice versa),
and a locally positive-de¬nite strain energy density W = 1 σijeij.
2
Suppose we have Cauchy initial conditions
‚u
u(x, 0) = u(x), (x, 0) = u (x), (5.2.27)
‚t
where u, u and the body force g are bounded in V. Let u, u be two distinct
solutions under these conditions and denote their difference by δu.
Then δu satis¬es
‚2 ‚ ‚
ρ 2 δui = δσij, δu = δu = 0 at t = 0, (5.2.28)
‚t ‚xj ‚t
5.2 The elastic constitutive relation 77

and δu · δ„ = 0 on the surface S.
Consider
t
‚2
‚ ‚
0= dV δui ρ 2 δui ’ δσij
dt
‚t ‚t ‚xj
0 V
2 t t
‚ ‚‚ ‚
1
= δui + dt dV δui δσij ’ dt dS njδσij δui
dm
2 ‚t ‚t ‚xj ‚t
V 0 V 0 S
t t
‚ ‚
= dV K(δu) + dt dV δeij δσij ’ dS δ„i δui.
dt (5.2.29)
‚t ‚t
V 0 V 0 S
The last term vanishes identically, the ¬rst term is the positive de¬nite kinetic
energy and the second is the total elastic energy associated with δu
1
W= 2 δeij δσij dV, (5.2.30)
V
which by hypothesis is positive de¬nite.
Thus ‚(δu)/‚t and δeij are zero everywhere in V and so δu is a static
displacement with δu = 0 at t = 0. This can only be achieved if
δu = 0 in V for all t > 0, (5.2.31)
i.e., the solution is unique.

Kirchhoff™s Theorem
Positive-de¬nite W at every point x suf¬ces for uniqueness.
For isotropic media this condition reduces to
κ>0 μ > 0,
with
(5.2.32)
’1 < … < 1 .
E>0
or with 2



EXAMPLE: STATIC ELASTIC FIELDS
Show that in an isotropic elastic body which does not include the origin and for which
σij = »δij ekk + 2μeij the two displacement ¬elds (a) u1 (x) = cx, (b) u2 (x) = dx/r3 ,
and c, d small constants, are self-equilibrating.
Consider a spherical shell of internal radius a and external radius b subject to uniform
external pressure p. Neglecting body forces, ¬nd the equilibrium displacement and stress
¬elds. Identify the principal stresses.
If the material fails as soon as the largest principal stress exceeds F and deformation can
be regarded as elastic up to failure, ¬nd the shell thickness h needed to sustain a pressure
p when h a.
For a self-equilibrated ¬eld, the stress gradient ‚σij /‚xj = 0.
(a) The displacement ui = cxi and the strain
‚ui ‚uj
1
eij = + = cδij
2 ‚xj ‚xi
78 Linearised Elasticity and Viscoelasticity

The associated stress tensor
σij = »δij 3c + 2μcδij = (3» + 2μ)cδij
is uniform, so the stress gradient vanishes and the ¬eld is self-equilibrated.
(b) The displacement ui = dxi /r3 and r‚r/‚xj = xj . The strain
δij xi xj
eij = d ’3 5
r3 r
and the associated stress
3 xk xk δij xi xj
σij = »δij d ’3 5 + 2μ d 3 ’ 3 5 ,
r3 r r r
the ¬rst term vanishes so we are left with just the shear components. The stress gradient
‚σij δij xj 3 3 15
= 2μd ’3 5 ’ 5 δij xj ’ 5 3xi + 7 xj xi xj
‚xj r r r r
μd
= 6 5 [’x ’ i ’ xi ’ 3xi + 5xi ] = 0,
r
and so the ¬eld is self-equilibrated.
Both u1 and u2 are radial and we may superpose them to give a radial ¬eld to match the
pressure loading.
(a) u1 gives a hydrostatic pressure (3» + 2μ)c.
(b) u2 in a locally radial frame has principal components (’2, 1, 1)2μd/r3 .
Set u = u1 + u2 , then the radial stress is
4μd
σrr = (3» + 2μ)c ’ .
r3
At r = a we require no radial traction at the free surface, and at r = b we must match
the external pressure p. Thus we require
4μd 4μd
0 = (3» + 2μ)c ’ , ’p = (3» + 2μ)c ’ 3 .
a3 b
so that c and d can be determined from
1 1
b3 p = (3» + 2μ)c(a3 ’ b3 ).
p = 4μd ’3,
b3 a
The displacement ¬eld is thus
b3 p 1 a3
1
u= 3 + x,
a ’ b3 3» + 2μ 4μ r3
with associated stress ¬eld
b3 p δij 3
δij + 1 a3
σij = 3 ’ 5 xi xj .
2
a ’ b3 3
r r
Consider a locally radial frame, e.g., pointing along (x, 0, 0) , then
b3 p a3 a3 a3
σxx , σyy , σzz =3 1 ’ 3 ,1 + 3,1 + 3 .
a ’ b3 r 2r 2r
The radial principal stress has functional form 1 ’ a3 /r3 and the hoop stresses in the
5.3 Integral representations 79

orthogonal directions behave as 1 + a3 /2r3 .
The largest principal stress is · = b3 p(1 + a3 /2r3 )/(a3 ’ b3 ). When h = b ’ a is
small, · ≈ a3 1 p/3ha2 = 1 pa/h.
3 2
Thus if · must be less than F to avoid failure, 1 pa/h ¤ F, and so h ≥ pa/2F to sustain
2
the pressure p. •




5.3 Integral representations
We can supplement the differential equations for a single elastic ¬eld with a set
of integral relations that connect the properties of two different ¬elds. We are
thereby able to use the behaviour of known solutions to infer information about
other situations without undertaking a full solution. We can also derive a useful
representation for the displacement ¬eld in terms of the Green™s tensor for the
medium. This relation will prove to be very useful in the description of sources
of elastic energy, e.g., faulting in an earthquake.
Consider the displacement ¬eld in an elastic body under different loadings
(i) u(x) with „ on S, g in V ,
(ii) u— (x) with „— on S, g— in V.
Then

ρ‚ttui = ρgi + ‚jσij, (5.3.1)

ρ‚ttu— = ρg— + ‚jσ— . (5.3.2)
i i ij


Take the scalar product of (5.3.1) with u— and of (5.3.2) with ui and then take their
i
difference. Now integrate throughout V and over time to get
t
dm (‚ttuiu— ’ ui‚ttu— )
dt i i

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