”x
Figure 2.4. Transformation of a triad of vector elements from the reference to the current
state.
The volume is given by the scalar triple product
dV = [dx, δx, ”x],
= [F dξ, F δξ, F ”ξ],
ξ ξ ξ
= det F[dξ, δξ, ”ξ] = det F dV0 = J dV0,
ξξξ (2.1.6)
i.e., as would be expected the scaling is by the Jacobian of the transformation. The
condition J = det F = 0 prevents physically infeasible operations such as kinking,
tearing or inversion during the deformation.
If we think in terms of a ¬xed volume, (2.1.6) speci¬es the density after
2.1 Geometry of deformation 25
transformation. Since a mass element dm = ρ dV = ρ0 dV0, the density in the
current state is related to that in the reference state by
ρ(x) = J’1ρ0(ξ) = (det F)’1ρ0(ξ).
ξ ξ (2.1.7)
2.1.4 Deformation of an element of area
We can use the result for the deformation of a volume element to derive the
equivalent result for an element of area by considering the transformation of a skew
cylinder.
dx
Figure 2.5. Deformation of a skew
cylinder from the reference to the
dξ
current state.
dΣ
dS
Σ ξ
Consider a skew cylinder with base dΣ, generators dξ (Figure 2.5). Under the
deformation described by F the volume will transform to
dV = dx · dS = dxT dS. (2.1.8)
Thus,
dV = dxT dS = J dV0 = det F dξT dΣ,
ξΣ (2.1.9)
but dξ = F’1dx, and so
ξ
dxT dS = dxT det F(F’1)T dΣ.
Σ (2.1.10)
This relation must be true for any dx and so for any area we require
dS = det F F’T dΣ.
Σ (2.1.11)
Note: F’T ≡ (FT )’1 ≡ (F’1)T .
2.1.5 Homogeneous deformation
Since only the ratios of differentials occur in (2.1.2), such relations at any point P
are formally the same as those representing homogeneous deformation relative to P
x = Fξ (2.1.12)
with F independent of ξ. This mapping may be regarded as between the positions
of either a particle (material point) or a ¬bre (material line segment). The
deformation gradient F is uniquely determined by the mappings of any three
noncoplanar ¬bres and this provides a convenient way to ¬nd F experimentally.
Some simple deformations:
26 Description of Deformation
(a) Dilatation: with a simple rescaling of the coordinates
xi = »ξi, i = 1, 2, 3 (2.1.13)
the deformation gradient is simply F = »I, where I is the unit matrix, and the
volume change J = »3.
(b) Extension with lateral contraction or expansion: Suppose that
x1 = »1ξ1, x2 = »⊥ ξ2, x3 = »⊥ ξ3, (2.1.14)
then the deformation gradient F is given by
⎛ ⎞
»1 0 0
F = ⎝ 0 »⊥ 0 ⎠ , with J = »1»2 . (2.1.15)
⊥
0 0 »⊥
If »1 > 1 the deformation is a uniform extension in the 1direction, whereas
if »1 < 1 we have a uniform contraction. »⊥ measures the lateral contraction
(»⊥ < 1), or extension (»⊥ > 1), in the 2“3 plane. For an incompressible
material there can be no volume change, and so J = »1»2 = 1 with the result
⊥
that an extension in the 1direction must be accompanied by a lateral contraction
in the 2“3 plane.
(c) General extension: when the extensions along the three coordinate axes are
different so that
xi = »iξi, i = 1, 2, 3 (2.1.16)
the deformation gradient F = diag{»1, »2, »3}, with J = »1»2»3. If »1 = »’1
2
and »3 = 1 there will be no change of volume, areas are preserved in planes
normal to the x3 direction, and the deformation is a pure shear.
EXAMPLE: DEFORMATION OF CRYSTALS
The unit cell of a crystal can be represented through a vector triad ±i , where the vectors
are not required to be orthogonal.
Introduce the reciprocal triad βj such that ±T βj = ±i .βj = δij . Show that ±p βT = I
p
i
where I is the unit matrix.
Under homogeneous deformation the triad ±i is transformed into the triad ai ; show that
the deformation gradient tensor F can be represented as F = ap βT .
p
The reciprocal triad can be found by considering vector products,
e.g., β1 = (a2 — a3 )/(a1 .a2 — a3 ).
These vectors have the required property that ±T βj = ±i .βj = δij .
i
T T
Then (±p βp )±q = ±p (βp ±q ) = ±p δpq = ±q and, since a similar relation holds for
any linear combination of the ±i , we require ±p βT = I.
p
Under deformation ±i ’ ai , so ai = F±i and hence ap βT = F±p βT = F. •
p p
2.2 Strain 27
Consider a crystal lattice whose elementary cell is a unit cube. A latticeplane is de¬ned
to have intercepts ν’1 (i = 1, 2, 3) on the cube edges from node 0, where νi are positive
i
or negative integers. Suppose the crystal is deformed so that each cell becomes a paral
lelepiped with edges ai , and let a reciprocal lattice be de¬ned with cell edges bj such that
ai .bj = δij . Show that, under arbitrary deformation, the plane remains perpendicular to an
embedded direction in the reciprocal lattice, while the distance of the plane from 0 varies
as νj bj ’1 .
We have initially a cubic basis ±1 , ±2 , ±3 with reciprocal triad β1 , β2 , β3 .
The lattice plane with intercepts ν’1 has the equation
i
(1 ’ l ’ m) l m
ξ= ±1 + ±2 + ±3 ,
ν1 ν2 ν3
and the normal satisfying
1 1 1 1
ν· ±1 ’ ±2 =ν· ±1 ’ ±3 =0
ν1 ν2 ν1 ν3
i.e., ξ · ν = 1, is
ν = ν1 β1 + ν2 β2 + ν3 β3 .
After homogeneous deformation the orthogonal triad ±i is mapped to a vector triad ai
that will represent a parallepiped. Using an embedded basis, the equation of the plane
will transform to
1’l’m l m
x= a1 + a2 + a3 ,
ν1 ν2 ν3
and the perpendicular to the embedded directions x · n = 1 is given by
n = ν1 b1 + ν2 b2 + ν3 b3 ,
where bi is the reciprocal triad to ai .
The distance to the plane varies as 1/n = 1/νj bj . •
The reciprocal lattice is important in crystallography and controls the Xray diffraction
patterns from crystals.
2.2 Strain
The mechanical effects of deformation can be separated into a part which affects
the length of ¬bres and a rotation component. The rotation can be regarded as
occurring either before or after the stretch, and this leads to different algebraic
representations.
The differential deformation in the neighbourhood of ξ is de¬ned by dx = F dξ,ξ
and we now explore how this affects the length and orientation of a line segment
(¬bre).
2.2.1 Stretch
For a ¬bre in the direction ν,
dξ = dξν,
ξ ξν (2.2.1)
28 Description of Deformation
we introduce the stretch
dx ¬nal length
»(ν) = = . (2.2.2)
dξ
ξ initial length
The length of the line segment in the deformed state
dx2 = dxT dx = dξT (FT F)dξ.
ξ ξ (2.2.3)
Consider also the transformation of two vectors such that
dx = F dξ,
ξ dy = F d·. (2.2.4)
The scalar product in the current state is
dx · dy = dxT dy = dξT (FT F)d·;
ξ (2.2.5)
the ¬nal angle between the vectors in the deformed state is dx · dy/(dxdy).
Thus FT F is the metric for the deformed state relative to the reference state. Lengths
and angles are unchanged if and only if F is proper orthogonal (i.e. a rotation).
The stretch for the direction ν is to be found from
dξT (FT F)dx
dx2 ξ
= νT (FT F)ν.
2
(»(ν)) = = ν (2.2.6)
2 2
dξ
ξ dξ
ξ
A corresponding form can be found in terms of the deformed con¬guration. The
length of the line segment in the reference state is
dξ2 = dξT dξ = dxT (F’1)T F’1 dx
ξ ξξ
= dxT (FFT )’1 dx. (2.2.7)
Thus, if dx lies in the direction n,
(»(ν))2 = [nT (FFT )’1n]’1. (2.2.8)
2.2.2 Principal ¬bres and principal stretches
If we consider the eigenvectors dξr of the local metric tensor FT F we have
ξ
FT F dξr = »2 dξr,
ξ rξ (2.2.9)
and the eigenvalues »2 satisfy det FT F ’ »2I = 0, where I is the unit tensor.
r r
TT
In the reference state dξ F F dξ = const is an ellipsoid (the Lagrangian
ξ ξ
ellipsoid) with mutually orthogonal principal axes ±dξr. Under deformation this
ξ
ellipsoid transforms to a sphere (see Figure 2.6) “ for a principal ¬bre from (2.2.3):
dxr2 = dξT (FT F)dξr = »2dξr2,
ξr ξ rξ (2.2.10)
and thus the »r are the principal stretches. The triad of principal ¬bres dξr is termed
ξ
the Lagrangian triad (or material, referential triad).
If we premultiply the eigenrelation (2.2.9) by F we obtain
FFT F dξr = »2F dξr,
ξ ξ (2.2.11)
r
2.2 Strain 29
and so
FFT dxr = »2 dxr, (2.2.12)
r
so that the principal ¬bres are also ¬nally orthogonal. The dxr form a triad (the
Eulerian or spatial triad) of the principal axes of an ellipsoid dxT (FFT )’1 dxr =
r
const which results from the deformation of a sphere in the reference state (Figure
2.6).
Reference Deformed
ξ2 x2
x1
ξ1
ξ2 x2
x1
ξ1
Figure 2.6. Relation of Lagrangian (material) and Eulerian (spatial) triads.
The Lagrangian triad (2.2.9) and the Eulerian triad (2.2.12) generally have
different orientations as illustrated in Figure 2.6.
2.2.3 The decomposition theorem
We de¬ne the stretch tensor U to be coaxial with the Lagrangian triad of F, that is to
have the same principal axes, and to have eigenvalues »r. The array of background
components U can be constructed by the spectral formula
^ ^T ^
U= »rξrξr , ξr = 1. (2.2.13)
r
Considered as a mapping, U acts as a pure strain: it stretches the Lagrangian triad
of F just like F itself, but maintains the orientation of the triad. Hence F can differ
from U only by a ¬nal rotation
F = RU. (2.2.14)
30 Description of Deformation
In the context of matrix algebra this is an example of a polar decomposition.
Algebraically: since FT F is coaxial with U and has eigenvalues »2, FT F =
r
2 T 2
U . When F is given, the equation F F = U can be solved uniquely for a
positive de¬nite symmetric U. Then R de¬ned by (2.2.14) is automatically proper
orthogonal
FT F = UT RT RU = URT RU = U2, (2.2.15)
and so the rotation matrix R must satisfy
RT R = I, (2.2.16)
and det R > 0 since det F > 0 and det U > 0.
There is also a similar decomposition with ¬rst a rotation of the principal ¬bres,
and then a stretching:
F = RU = (RURT )R = VR. (2.2.17)
V = RURT is the array of background components of the pure strain that is
coaxial with the Eulerian triad of F and has principal values »r. V 2 = FFT is
the combination that we need in comparing the deformed to the reference state.
Note: The deformation gradient F can be speci¬ed independently of coordinate
frame by using the triads and principal stretches
T
^^ x^
F= xrξr , ^r, ξr = 1, (2.2.18)
r
which may be derived from the spectral form for U.
2.2.4 Pure rotation
A rotation by an angle θ about an axis n transforms a vector ξ to
x = ξ + sin θ(n — ξ) + (1 ’ cos θ)n — (n — ξ). (2.2.19)
We now introduce the skew matrix N(≡ n — ) associated with n, Nik = ijknj,
where the alternating symbol ijk takes the value 1 if {i, j, k} is an even permutation
of {1,2,3}, the value ’1 if {i, j, k} is an odd permutation of {1,2,3}, and the value
zero otherwise.
The rotation R can thus be represented as
R = I + N sin θ + N2(1 ’ cos θ). (2.2.20)
Now, N2 = nnT ’ 1, Nn = 0 and so N3 = ’N, N4 = ’N2. This enables us to
write R in the form
θ2r+1 θ2r
r
+ N2 r
R = exp(Nθ) = I + N (’1) (’1) ,
(2r + 1)! (2r)!
r r
θr
Nr
=I+ . (2.2.21)
r!
r
2.2 Strain 31
Once R is given, the angle θ may be computed from
tr R = 1 + 2 cos θ, (2.2.22)
and then N and n can be found from
N sin θ = 1 (R ’ RT ). (2.2.23)
2
EXAMPLE: DEFORMATION GRADIENT AND STRETCH MATRIX
The deformation gradient F is speci¬ed by
7 ’5 2
1
4 4 ’1 .
F=
9 ’5 7 2
Show that its Lagrangian triad is coaxial with (0, 0, 1), (1, 1, 0), (1, ’1, 0) and ¬nd the
principal stretches. Hence construct U and U’1 and by constructing R = FU’1 , verify
that it is a rotation of π/3 about (1, 1, 1). Finally show that the Eulerian triad is coaxial
with (2, ’1, 2), (1, 4, 1), (1, 0, ’1).
The stretch matrix is given by
90 ’54 0
1
U2 = FT F = ’54 90 0 ,
81 0 09
The Lagrangian triad is determined by
10ξ1 ’ 6ξ2 = 9»2 ξ1
’6ξ1 + 10ξ2 = 9»2 ξ2
ξ3 = 9»2 ξ3
and is thus coaxial with (1, 1, 0), (1, ’1, 0), (0, 0, 1)
with principal stretches 2 , 4 , 1 .
333
We can construct the stretch matrix U and its inverse U’1 from
^(1) ^(1)T + »2ξ(2)ξ(2)T + »3ξ(3)ξ(3)T ,
^^ ^^
U = »1ξ ξ
and
(1) (1)T (2) (2)T (3) (3)T
^ ^ ^ ^ ^ ^
U’1 = »’1ξ + »’1ξ + »’1ξ
ξ ξ ξ ,
1 2 3
so that
⎛ ⎞ ⎛3
⎞ 1
3 ’1 0 08 8
1
U = ⎝’1 3 0⎠ , U = 3⎝ 0⎠ .
1 3
3 8 8
0 01 0 01
The rotation
2 ’1 2
1
’1
2 2 ’1 .
R = FU =
3 ’1 2 2
32 Description of Deformation
Now trR = 2 = 1 + 2 cos θ and so cos θ = 1 , hence θ = π/3. Further
2
1 1
R1 1,
=
1 1
so (1, 1, 1) is the rotation axis.
To obtain the Eulerian triad we operate on the Lagrangian triad with R,
110 13 2
1
1 ’1 0 4 0 ’1
R =
3
001 1 ’3 2
and is coaxial with (1, 4, 1), (1, 0, ’1), (2, ’1, 2). •
2.2.5 Tensor measures of strain
We have seen that the change in shape of any cell can be calculated when we know
which three ¬bres in the initial con¬guration are principal, and by how much each
is stretched. Moreover, this is the minimal information for this purpose.
In this sense a suitable strain measure can be found by looking at the change in
ξ
the length of dξ, using (2.2.3) and (2.2.7)
dx2 ’ dξ2 = dξT FT F dξ ’ dξT dξ = 2 dξT E dξ,
ξ ξ ξ ξξ ξ ξ (2.2.24)
= dxT dx ’ dxT (FFT )’1 dx = 2 dxT e dx. (2.2.25)
Here we have introduced the Green strain,
E = 1 (FT F ’ I) = 1 (U2 ’ I), (2.2.26)
2 2
and the Cauchy strain,
e = 1 (I ’ (FFT )’1) = 1 (I ’ V ’2), (2.2.27)
2 2
associated with some speci¬ed background frame.
The components E of the Green strain generate a tensor whose axes de¬ne the
reference directions of the Lagrangian principal ¬bres and whose principal values
are
12
2 (»r ’ 1) r = 1, 2, 3. (2.2.28)
This strain measure is objective (i.e. unaffected by any ¬nal rotation) and it vanishes
when, and only when, the shape is unchanged.
The normalising factor 1 is included so that when the strain is in¬nitesimal,
2
the principal values of the measure are just the fractional changes in length of the
principal ¬bres; since, if »r = 1 + ,
+ »)2 ’ 1] ≈ ,
1
2 [(1 (2.2.29)
when is a ¬rstorder quantity.
We have comparable properties for the Cauchy strain e in terms of the Eulerian
2.2 Strain 33
triad, where the principal axes lie along the Eulerian principal ¬bres and the
principal values are
’ »’2)
1
2 (1 r = 1, 2, 3. (2.2.30)
r
Once again for in¬nitesimal strain the principal values reduce to the fractional
changes of length of the principal ¬bres.
Displacement representations:
We introduce the displacement u between the deformed and reference
con¬gurations
u = x ’ ξ, (2.2.31)
and its gradients,
‚u ‚u
= F ’ I, F=I+ , (2.2.32)
‚ξ
ξ ‚ξ
ξ
‚u ‚u
= I ’ F’1, F’1 = I ’ . (2.2.33)
‚x ‚x
In terms of these displacement gradients the Green strain tensor is given by
T
‚u ‚u
T
2E = FF ’ I = I + I+ ’ I,
‚ξ
ξ ‚ξ
ξ
T T
‚u ‚u ‚u ‚u
= + + , (2.2.34)
‚ξ
ξ ‚ξ
ξ ‚ξ
ξ ‚ξ
ξ
and the Cauchy strain by
T
‚u ‚u
’T ’1
2e = I ’ F F =I’ I’ I’ ,
‚x ‚x
T T
‚u ‚u ‚u ‚u
= + ’ . (2.2.35)
‚x ‚x ‚x ‚x
The cartesian components of the strain tensors are therefore given by
i) Green strain tensor E:
‚ui ‚uj ‚uk ‚uk
1
Eij = + + , (2.2.36)
2 ‚ξj ‚ξi ‚ξi ‚ξj
and
ii) Cauchy strain tensor e:
‚ui ‚uj ‚uk ‚uk
1
eij = + ’ . (2.2.37)
2 ‚xj ‚xi ‚xi ‚xj
When the displacements and displacement gradients are suf¬ciently small, the
distinction between the two smallstrain de¬nitions is usually ignored, since the
secondorder terms are then unimportant.
34 Description of Deformation
2.3 Plane deformation
Introduce a unit vector n, and then in a deformation with displacement
u = x ’ ξ = γm(n · ξ) = γm(nTξ), (2.3.1)
particles are not displaced in the basal plane nTξ = 0 irrespective of the direction
of the unit vector m. The deformation gradient
F = I + γmnT , (2.3.2)
and the amount of deformation scales with the height of ξ above the basal plane
(n.ξ) (Figure 2.7).
ξ
n x
γ m(n . ξ )
Figure 2.7. Representation of plane
m
deformation in the direction m from
ξ
a basal plane with normal n.
Shear
An equivoluminal plane deformation is called a shear, the principal stretches are
then such that
»3 = 1 with »1»2 = 1. (2.3.3)
If mT n = 0 the translation components in (2.3.1) are parallel to the basal plane
and volume is conserved. This is simple shear
F = I + γmnT , mT n = 0. (2.3.4)
γ is called the amount of shear and a plane nTξ = const a shearing plane.
The simple shear process has the effect of rotating the principal strain axes, as
illustrated in Figure 2.8, and sets of initially perpendicular lines will be sheared and
end up at an oblique angle.
In contrast for the pure shear situation in which there is simultaneous lengthening
and perpendicular shortening, such that »1»2 = 1, the principal axes of strain
remain ¬xed (Figure 2.8). Any set of perpendicular lines inclined to the principal
axes will suffer shear strain and be no longer orthogonal after deformation.
In each case the twodimensional area and threedimensional volume will remain
unchanged in the homogeneous shear.
2.3 Plane deformation 35
Simple Shear
Pure Shear
Figure 2.8. Illustration of the action of a simple shear and a pure shear process. In each
homogeneous deformation the area of a body is preserved. The principal strain axes shown
by chain dotted lines rotate for simple shear, but remain ¬xed for pure shear. The solid ref
erence lines are shear strained in each case and are no longer orthogonal after deformation.
EXAMPLE: A SHEAR PROCESS
Consider a type of shearing
x1 = sξ1 + tξ2 /s, x2 = ξ2 /s, x3 = ξ3 , (2.3.5)
with s ≥ 1, t ≥ 0 (s ≡ 1 is a path of simple shear). Show that any resultant simple shear
can be obtained by a path of this type in (s, t) space such that the same ¬bres are principal
at every stage.
We have a homogeneous deformation gradient
s t/s
F= , det F = 1.
0 1/s
We require the principal axes of
s2 t
FT F =
t (t + 1)/s2
2
to be ¬xed. For a matrix of the form
ac
,
cb
the inclinations of the principal axes depend only on (a ’ b)/c.
Because we require the ¬nal state to be a simple shear the trajectory in (s, t) space has
36 Description of Deformation
to pass through s = 1, t = γ and s = 1, t = 0 to include the undeformed state. To keep
the inclination of the principal axes ¬xed we must have
t2 + 1
1 12
’ s2 = γ + 1 ’ 1 = γ,
2
t s γ
and so the path is given by
s4 + γs2 t ’ t2 = 1, 0 ¤ t ¤ γ,
as illustrated in Figure 2.9.
s2
Figure 2.9. Trajectory in (s, t) space
that retains the same principal ¬bres.
t
γ
2.4 Motion
We can follow the motion of a particle by tracking its position relative to its
reference position ξ (a Lagrangian viewpoint), or alternatively we may follow
the particle directly in an Eulerian viewpoint. In a similar way we can represent
a ¬eld qL(ξ, t) in terms of the Lagrangian frame so that qL follows the particle
ξ
initially at ξ. Or, alternatively, we can work in terms of the current con¬guration
of the medium with an Eulerian description qE(x, t) in terms of a ¬xed position
x. Because both representations link the value of q for the particle labelled by ξ
which is at x at time t:
qE(x(ξ, t), t) = qL(ξ, t).
ξ ξ (2.4.1)
When we follow a given particle ξ (a Lagrangian viewpoint) then the time rate
of change of interest is
‚
, (2.4.2)
‚t ξ
but, using the chain rule, we ¬nd that the ˜material time derivative™
D ‚ ‚ ‚ ‚
ψ≡ ψ= ψ+ xi ψ, (2.4.3)
Dt ‚t ‚t ‚t ‚xi
ξ ξ
x
in terms of an Eulerian decomposition.
2.4 Motion 37
We de¬ne the velocity ¬eld corresponding to a particle initially at ξ as
‚ D
v= x(ξ, t) ≡
ξ x. (2.4.4)
‚t Dt
ξ
The acceleration ¬eld is then
‚ D
f= v(ξ, t) ≡
ξ v, (2.4.5)
‚t Dt
ξ
and in the Eulerian viewpoint
‚ ‚
f= v(x, t) + v · v, (2.4.6)
‚t ‚x
x
where we have written ‚/‚x for the gradient operator in the Eulerian con¬guration
(for ¬‚uids this is the most common viewpoint to use).
In general, then the mobile derivative following the particles is
D ‚ ‚
≡ + v· . (2.4.7)
Dt ‚t ‚x
x
If we consider a line element dx = F dξ, its material derivative is
ξ
D ‚L ‚
dx = dξ ·
ξ v= dx · v, (2.4.8)
Dt ‚ξ
ξ ‚x
a result that can also be derived from geometrical considerations.
EXAMPLE: TIMEDEPENDENT DEFORMATION
A body undergoes the deformation speci¬ed by
x1 = ξ1 (1 + a2 t2 ), x3 = ξ3 .
x2 = ξ2 ,
Find the displacement and velocity in both the material and the spatial description.
The displacement u = x ’ ξ in the material description is
u1 = ξ1 a2 t2 , u2 = 0, u3 = 0,
and in the spatial description
u1 = x1 a2 t2 /(1 + a2 t2 ), u2 = 0, u3 = 0.
The velocity, ‚u/‚t)ξ in the material description is
v1 = 2ξ1 a2 t, v2 = 0, v3 = 0,
and in the spatial description is
v1 = 2x1 a2 t/(1 + a2 t2 ), v2 = 0, v3 = 0.
38 Description of Deformation
2.5 The continuity equation
Consider a volume V0 in the undeformed state and monitor its state during
deformation. The conservation of mass requires that the integral
M0 = ρ0(ξ) dV0
ξ (2.5.1)
V0
should remain invariant. In terms of the deformed state, the mass
M0 = ρ(x, t) dV = ρ(x, t) J(t) dV0, (2.5.2)
V V0
where we have used (2.1.6). If we now equate the two expressions for M0 we have
the expression for the conservation of mass in the Lagrangian representation
[ρ0(ξ) ’ ρ(x, t)J(t)] dV0 = 0.
ξ (2.5.3)
V0
This relation must be true for any volume V0 and so the integrand must vanish,
which gives an alternative derivation of the density relation deduced above in
section 2.1.3.
Now the density in the undeformed state is a constant whatever the deformation,
and so
‚ ‚ D
ρ0(ξ) =
ξ ρ(x, t)J(t) = ρ(x, t)J(t) = 0. (2.5.4)
‚t ‚t Dt
ξ ξ
The material time derivative of density can therefore be found from
D D
ρ(x, t) + ρ(x, t)J’1(t) J(t) = 0. (2.5.5)
Dt Dt
From the properties of the determinant of the deformation gradient
D ‚ ‚
J(t) = J(t) = J(t) · v(x, t). (2.5.6)
Dt ‚t ‚x
ξ
in terms of the velocity ¬eld v(x, t). The material time derivative of the density is
thus determined by
D ‚
ρ(x, t) + ρ(x, t) · v(x, t) = 0. (2.5.7)
Dt ‚x
When we make use of the representation (2.4.7) for the material time derivative
we can write the Eulerian form of the equation of mass conservation in the form
‚ ‚
ρ(x, t) + · [ρ(x, t)v(x, t)] = 0, (2.5.8)
‚t ‚x
x
which is known as the continuity equation. The relation (2.5.8) can also be deduced
by considering a ¬xed volume in space and then allowing for mass ¬‚ux across the
boundary.
2.A Properties of the determinant of the deformation gradient 39
Material derivative of a volume integral
Consider some physical quantity Ψ (scalar, vector or tensor) associated with the
continuum so that ψ is the quantity per unit mass. The material rate of change of Ψ
for a ¬xed material volume V with surface S and outward normal n must be equated
^
to the rate of change of Ψ for the particles instantaneously within V and the net ¬‚ux
of Ψ into V across the surface S so that
‚
d
Ψ dV = Ψ dV + Ψv · n dS.^ (2.5.9)
V ‚t
dt V S
If we now represent Ψ as ψρ we have
‚
d
ψρ dV = (ψρ) dV + ψρv · n dS.
^ (2.5.10)
‚t
dt V V S
We apply the divergence theorem to the surface integral in (2.5.10), and then
‚ ‚
d
ψρ dV = (ψρ) + (ψρv) dV
‚t ‚x
dt V V
‚ ‚ ‚ ‚
= ψ ρ+ (ρv) + ρ ψ+v· ψ) dV. (2.5.11)
‚t ‚x ‚t ‚x
V
From the continuity equation (2.5.8) the ¬rst contribution to the volume integral
on the righthand side of (2.5.11) vanishes, and the second can be recognised as
ρ Dψ/Dt. Hence
D
d
ρψ dV = ρ ψ dV; (2.5.12)
Dt
dt V V
for ψ = 1 we recover the intergal form of the continuity condition (2.5.8).
Appendix: Properties of the determinant of the deformation gradient
In terms of the deformation gradient F = ‚x/‚ξ, Fij = ‚xi /‚ξj , the determinant
ξ
1
J = det F = 6 ijk pqr Fpi Fqj Frk (2A.1)
‚xp ‚xq ‚xr
1
= . (2A.2)
6 ijk pqr ‚ξ ‚ξj ‚ξk
i
Now, from the orthogonality relation
‚xi ‚ξj
= δik , (2A.3)
‚ξj ‚xk
we can apply Cramer™s rule for the solution of linear equations to obtain
‚ξj 1 ‚xp ‚xq
= , (2A.4)
jmp kqr
‚xk 2J ‚ξi ‚ξj
and thus we can recognise that
‚J ‚J ‚ξj
= =J . (2A.5)
‚Fij ‚(‚xi /‚ξj ) ‚xi
40 Description of Deformation
We can therefore establish the important relations
‚ 1 ‚xj
= 0, (2A.6)
‚xk J ‚ξk
and
‚ ‚ξj ‚ ‚xi ‚‚ ‚
J=J . =J xi = J .v , (2A.7)
‚t ‚xi ‚t ξ ‚ξj ‚xi ‚t ‚x
ξ ξ
which is used in (2.5.6).
3
The StressField Concept
The stress tensor provides a description of the force distribution through the
continuum and is most readily treated in the deformed state (an Eulerian treatment).
We suppress the details of complex interatomic forces and make the simplifying
assumption that the effect of all the forces acting on a given surface may be
represented through a single vector ¬eld de¬ned over the surface.
3.1 Traction and stress
Consider any point P of a continuum and any in¬nitesimal plane element δS with P
as centroid. Let x be the position vector of P, and n the unit normal to δS (in some
consistent sense).
δS
n
Figure 3.1. The traction „ acting on an
P „ element with normal n.
Then the mechanical in¬‚uence exerted across δS by the material on the side to
which n points is postulated to be represented by a force
„(n, x)δS (3.1.1)
through P. The negative of this force is exerted across δS on this material. „ is
called the traction vector and has the dimensions of force per unit area.
Note: a mutual couple could also be introduced across δS, but the need for it has
never been indisputably established in any particular situation.
41
42 The StressField Concept
The properties of traction can be determined from a hypothesis due substantially
to Euler: the vector ¬elds of force and massacceleration in a continuum have equal
linear and rotational resultants over any part of the continuum.
Consider a volume V, bounded by a surface S with outward normal n, and dm a
generic mass element in V, and set
f “ local acceleration,
g “ local body force per unit mass (e.g., gravitational or ¬ctitious),
μ “ local body moment per unit mass (e.g., electrostatic or magnetic).
Then we can write the equations of conservation of linear and angular
momentum in the form
„(x) dS + g(x) dm = f(x) dm, (3.1.2)
S V V
x — „(x) dS + x — g(x) dm + μ(x) dm = x — f(x) dm. (3.1.3)
S V V V
Tensor character of stress
At a point P consider a plane element perpendicular to the background axis xi. Let
σi denote the vector traction exerted over this element by the material situated on
its positive side (Figure 3.2).
δS
σi
Figure 3.2. Traction σi acting on an
xi
P element perpendicular to the xi axis.
The components σij of σi constitute the stressmatrix. σ11, σ22, σ33 are called the
normal components and the others shear components.
The stress matrix fully speci¬es the mechanical state at P, in the sense that it
determines the traction „ for any n. Consider an in¬nitesimal tetrahedron at P with
three faces parallel to the coordinate planes and the fourth with outward normal n
and area 2 (Figure 3.3).
We now apply the equation (3.1.2) for the linear force resultant to this volume.
Suppose that n falls in the ¬rst octant; the facial areas are then
(n1, n2, n3, 1) 2. (3.1.4)
The surface integral over the tetrahedron is thus
2
+ O( 3) =
„ dS = („ ’ niσi) (f ’ g) dm. (3.1.5)
S V
3.1 Traction and stress 43
3
n
Figure 3.3. Resultant forces on an
elementary tetrahedron
2
1
The same value is obtained when n falls in any other octant, or when n is
perpendicular to one axis (in which case the tetrahedron is replaced by a triangular
prism).
Now the mass integrals over the small volume are of order O( 3) because we
can take f, g constant within the volume, and so, as ’ 0,
„ dS ’ 0, (3.1.6)
S
Thus
„(n) = niσi, i.e. „j(n) = niσij. (3.1.7)
which is known as Cauchy™s relation.
We note that (3.1.7) also implies that σij transforms like a secondrank tensor.
Consider a new set of rectangular axes indicated by stars and set lpi to be the
direction cosine between the pth starred axis and the ith old axis (Figure 3.4).
Figure 3.4. Rotation of coordinate
*
xq
axes
^
lqj = cos qj
^
qj
xj
Then
σ— = σijlpilqj, (3.1.8)
pq
since
„— = lqj„j, σi = lpiσ— , lipljp = δij.
as (3.1.9)
q p
44 The StressField Concept
Alternatively, in terms of the rotation R which carries the unstarred axes into the
starred ones
Rip = lpi, (3.1.10)
and
σ— = RT σR. (3.1.11)
3.2 Local equations of linear motion
We rewrite (3.1.2) in component form
„j(x) dS = σij(x)ni(x) dS = [fj(x) ’ gj(x)] dm, (3.2.1)
S S V
and then use the tensor divergence theorem
‚ ‚ dm
σij(x)ni(x) dS = σij(x) dV = σij(x) , (3.2.2)
‚xi ‚xi ρ(x)
S V V
to represent the effects of the surface tractions though a volume integral. Here
ρ(x) is the local density and the stress tensor σij is assumed to be continuous and
differentiable within V.
From (3.2.1) and (3.2.2) we have the relation
‚ dm
σij(x) = [fj(x) ’ gj(x)] dm, (3.2.3)
‚xi ρ(x)
V V
and since the region V is arbitrary
‚
σij(x) + ρgj(x) = ρfj(x). (3.2.4)
‚xi
We see that it is the stress gradient, in conjunction with the body force, that
determines the local acceleration.
If ‚σij/‚xi = 0, the stress ¬eld is said to be selfequilibrated
Note: the local equations (3.1.7) [„j = σijni] and (3.2.4) imply the global equation
(3.1.2).
3.2.1 Symmetry of the stress tensor
With the aid of the tensor divergence theorem the component form of the
conservation of angular momentum equation (3.1.3) can be written as
‚ dm
ijkxj„k dS = (xjσlk) , (3.2.5)
ijk
‚xl ρ
S V
=’ μi dm + ijkxj[fk ’ gk] dm. (3.2.6)
V V
3.2 Local equations of linear motion 45
From the equation of motion (3.2.4)
‚ ‚ dm
(xjσlk) ’ xj σlk =’ μi dm, (3.2.7)
ijk
‚xl ‚xl ρ
V V
and so,
ijkδjlσlk dm = ijkσjk dm = ’ρ μi dm. (3.2.8)
V V V
Since the region V is again arbitrary
ijkσjk(x) = ’μi(x), (3.2.9)
and we may recover (3.1.3) from these local equations.
We will subsequently suppose that body moments are absent (μi ≡ 0) and then
σjk(x) = σkj(x), (3.2.10)
and the stress matrix is symmetric.
EXAMPLE: STRESS FIELDS
Show that the stress ¬eld σij = pδij + ρvi vj is selfequilibrated, where p, ρ and v are the
pressure, density and velocity in a steady ¬‚ow of some inviscid compressible ¬‚uid.
The gradient of the stress tensor σ
‚σij ‚p ‚vi ‚vj ‚ρ
= + ρvj + ρvi + vi vj
‚xj ‚xj ‚xj ‚xj ‚xj
= [∇p + ρ(v · ∇)v]i + vi [∇ · (ρv)].
In steady ¬‚ow,
∇p + ρ(v · ∇)v = 0,
and from the continuity equation
‚ρ
+ ∇ · (ρv) = 0,
‚t
but since ¬‚ow is steady ‚ρ/‚t = 0. Thus
‚σij
= 0,
‚xj
and so the stress ¬eld is selfequilibrated.•
Show that the stress ¬eld σij = vi vj ’ 1 v2 δij exerts tractions on any closed surface which
2
are statically equivalent overall to body forces v(∇ · v) per unit volume, when v is any
irrotational ¬eld (i.e. ∇ — v = 0).
46 The StressField Concept
The gradient of the stress tensor σ
‚σij ‚vj ‚vi ‚vk
= vi + vj ’ vk
‚xj ‚xj ‚xj ‚xi
‚vj ‚vi ‚vj
= vi + vj ’
‚xj ‚xj ‚xi
= vj ∇ · v,
since ∇ — v = 0 and we can recognise the bracketed term as a component of ∇ — v. In
a static scenario, we have from (3.2.1) and (3.2.2)
‚σij
dV = „i (x)dS = ’ gi dV
‚xj
V S V
and the body force per unit volume g is v(∇ · v) •
3.2.2 Stress jumps (continuity conditions)
A discontinuity in material properties across an internal interface generally causes
a discontinuity in the stress tensor.
Such interfaces are present in simple laminates like plywood and in composite
solids of any kind. Well known composites are rocks, reinforced concrete,
glass¬bre resins (used in the hull of boats), cermets “ sintered mixtures of ceramic
inclusions in a metallic matrix (used in cutting tools), and ceramic ¬brereinforced
materials (used in turbine blades).
Major internal discontinuities in the Earth occur at the base of the crust, where
the Mohoroviˇ i´ discontinuity can in places be very sharp, and at the phase
cc
boundaries in the mantle transition zone. Important ¬‚uid“solid discontinuities are
at the core“mantle boundary between the silicate mantle and the metallic ¬‚uid in
the outer core and the transition from the ¬‚uid outer core to the solid inner core.
„+
n
+
’
Figure 3.5. Interface conditions on
traction
„’
µ
Possible jumps in the stress components at an interface are, however, restricted
by the inviolable continuity of the traction vector acting on the interface
[„j]+ = ni[σij]+ = 0, (3.2.11)
’ ’
3.2 Local equations of linear motion 47
where n is the interface normal. We may prove continuity of traction by applying
(3.1.2) to a vanishingly thin disc enclosing an interfacial area A (Figure 3.5),
„+ dA+ + „’ dA’ + O( ) = 0, (3.2.12)
A A
where the O( ) term arises from tractions on the edge, body forces and ¬nite
massaccelerations (we exclude idealised ˜shocks™ with a jump in velocity).
In the limit as ’ 0
„+(n) + „’(’n) = 0, i.e. [„(n)]+ = 0. (3.2.13)
’
Since (3.2.11) imposes only three constraints on six independent components
of σij, the stress tensor is not necessarily continuous at an interface. Consider a
local coordinate scheme with x1, x2 in the tangent plane and x3 normal: then „ in
(3.2.11) becomes σ3 and so
[σ31] = [σ32] = [σ33] = 0; (3.2.14)
consequently [σ13] = [σ23] = 0, but we cannot require [σ11], [σ22], [σ12] to be
zero, simply from (3.2.11).
3
Figure 3.6. Unconstrained elements
of the stress tensor from the condition
2
of continuity of traction at an inter
σ
face.
22
’σ21
σ12
σ11
+

1
In a similar way, at a body surface, such ˜interior™ tractions are not constrained
by the applied loads. In particular, even where the load is zero, the local stress
tensor does not have to vanish completely “ such a case is provided by a rod under
uniform tension.
EXAMPLE: REPRESENTATION OF STRESS JUMPS
Show that every statically admissable jump in the stress tensor is of the type
(δir ’ ni nr )(δjs ’ nj ns )prs
for some symmetric prs , where n is the unit normal to the jump surface.
Set the stress jump pij = (δik ’ni nk )(δjl ’nj nl )pkl . Consider the normal component
^
of the associated traction
pij nj = (δik ’ ni nk )(nl ’ n2 nl )pkl
^
= 0, since n = 1.
48 The StressField Concept
The traction in a direction m orthogonal to the normal n, with m · n = 0, would be
pij mj = (δik ’ ni nk )(ml ’ nj mj nl )pkl
^
= (δik ’ ni nk )ml pkl , as m · n = 0
= mj pij ’ (nk ml pkl )ni .
Thus the traction jump in the normal direction is zero, and in orthogonal directions is
nonzero as required for a statically admissable stress jump. •
3.3 Principal basis for stress
The traction vector „(n) over a plane element is purely normal when „ = σn for
some σ. Thus since „j = σijni, we have
(σij ’ σδij)ni = 0, (3.3.1)
The eigenvectors ±nr (r = 1, 2, 3) are called the principal axes of stress and the
eigenvalues σr given by the roots of det(σij ’ σδij) = 0 are called the principal
stresses.
Since σij is symmetric, the principal stresses σr are real and the axes are orthogonal.
Each σr acts in opposite directions on opposite faces of a ˜principal™ box. The
vectors nr can also be thought of geometrically as the principal axes of the stress
quadric surface
σijxixj = const. (3.3.2)
When the principal axes and stresses are known, stress components on
background axes are given by the spectral formula
σ r nr nT .
σ= (3.3.3)
r
r
Note: any quantity which is solely a function of the σr is an invariant of σij. The
cubic equation for the principal stresses σr can be written as
det(σij ’ σδij) = ’σ3 + Iσσ2 ’ IIσσ + IIIσ = 0 (3.3.4)
in terms of the invariants
Iσ = σ1 + σ2 + σ3 = σkk
IIσ = σ1σ2 + σ2σ3 + σ3σ1 = 1 [σiiσjj ’ σijσji] (3.3.5)
2
IIIσ = σ1σ2σ3 = det σ.
On occasion one may need the stress tensor components with respect to a
particular set of axes, and it is often easier to use the general tensor transformation
rule (3.1.8) from the principal basis, rather than use the spectral form (3.3.3).
3.3 Principal basis for stress 49
σ
σ n
3
1
Figure 3.7. Transformation
σ
2
from the principal basis for
stress to normal and shear
m
σ
2
tractions on a surface.
σ
1
σ
3
σ2
__ (σ +σ )
1
21 2
Figure 3.8. Con¬guration with
σ1 σ1
maximum shear stress.
1
__ (σ ’ σ2 )
21
σ2
If, in particular, we seek the normal component of traction on an element with
normal n, and the shear component in a direction m with m · n = 0, m = 1,
(Figure 3.7), then we should use
„(n) = (n1σ1, n2σ2, n3σ3), (3.3.6)
making use of the relation „ = niσi, with n = (n1, n2, n3) in a principal basis.
Then
n · „(n) = n2σ1 + n2σ2 + n2σ3, (3.3.7)
1 2 3
m · „(n) = m1n1σ1 + m2n2σ2 + m3n3σ3 = n · „(m). (3.3.8)
From (3.3.8), any normal stress lies between the algebraically least and greatest
principal stresses.
Stress circle
Consider the tractions on a box with one pair of faces perpendicular to a principal
axis. Then the traction on this pair is purely normal, σ3, and the tractions on the
other faces have no components in that direction.
50 The StressField Concept
„
σ2 m.„
n
θ
2θ
σ1
σ1
σ2 σ1 „
0 n.„
m
σ2
Figure 3.9. Stresscircle construction for normal and shear tractions.
^
Consider the normal in the 12 plane at an angle θ to the σ1 axis, then
n = (cos θ, sin θ, 0)m = (sin θ, ’ cos θ, 0) (3.3.9)
^
so that the normal and shear stress components in the 12 plane are
n · „ = cos2 θ σ1 + sin2 θσ2 = 1 (σ1 + σ2) + 1 (σ1 ’ σ2) cos 2θ
2 2
(3.3.10)
1
m·„= 2 (σ1 ’ σ2) sin 2θ
As θ is varied from 0 to π, the locus of m · „ versus n · „ is a circle: centre
1 1
2 (σ1 + σ2), radius 2 (σ1 ’ σ2), where we have assumed that σ1 is greater than σ2
(Figure 3.9).
The shear stress is largest when θ = π/4
(m · „)max = 1 (σ1 ’ σ2) (3.3.11)
2
and then the normal tractions are equal to 1 (σ1 + σ2) on all four faces as illustrated
2
in Figure 3.8.
EXAMPLE: PRINCIPAL STRESSES
A cylinder whose axis is parallel to the x3 axis and whose normal crosssection is the
square ’a ¤ x1 ¤ a, ’a ¤ x2 ¤ a is subjected to torsion by couples acting over its
ends x3 = 0 and x3 = L. The stress components are given by σ13 = ‚ψ/‚x2 , σ23 =
’‚ψ/‚x1 , σ11 = σ12 = σ22 = σ33 = 0, where ψ = ψ(x1 , x2 ).
(a) Show that this stress tensor is selfequilibrated;
(b) show that the difference between the maximum and minimum stress components is
2[(‚ψ/‚x1 )2 + (‚ψ/‚x2 )2 ]1/2 , and ¬nd the principal axis which corresponds to the zero
principal component;
(c) for the special case ψ = (x2 ’ a2 )(x2 ’ a2 ) show that the lateral surfaces are free from
1 2
traction and that the couple acting on each end face is 32a6 /9.
3.4 Virtual work rate principle 51
The ¬eld is selfequilibrated because
‚2 ψ ‚2 ψ
‚σ13 ‚σ23
+ = ’ =0
‚x1 ‚x2 ‚x2 ‚x1 ‚x1 ‚x2
and all other components of the gradient of the stress tensor are zero.
Now set a = ‚ψ/‚x2 , b = ’‚ψ/‚x1 , then the principal stresses satisfy
σ0a
0 σ ’b = 0,
det
a ’b σ