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О”Оҫ
О”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, Оҙx, О”x],
= [F dОҫ, F ОҙОҫ, F О”Оҫ],
Оҫ Оҫ Оҫ
= det F[dОҫ, ОҙОҫ, О”Оҫ] = det F dV0 = J dV0,
ОҫОҫОҫ (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 п¬Ғxed volume, (2.1.6) speciп¬Ғes the density after
2.1 Geometry of deformation 25

transformation. Since a mass element dm = ПҒ dV = ПҒ0 dV0, the density in the
current state is related to that in the reference state by
ПҒ(x) = JвҲ’1ПҒ0(Оҫ) = (det F)вҲ’1ПҒ0(Оҫ).
Оҫ Оҫ (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
dОҫ
current state.
dОЈ
dS

ОЈ Оҫ
Consider a skew cylinder with base dОЈ, generators dОҫ (Figure 2.5). Under the
deformation described by F the volume will transform to
dV = dx В· dS = dxT dS. (2.1.8)
Thus,
dV = dxT dS = J dV0 = det F dОҫT dОЈ,
ОҫОЈ (2.1.9)
but dОҫ = FвҲ’1dx, and so
Оҫ
dxT dS = dxT det F(FвҲ’1)T dОЈ.
ОЈ (2.1.10)
This relation must be true for any dx and so for any area we require
dS = det F FвҲ’T dОЈ.
ОЈ (2.1.11)
Note: FвҲ’T вүЎ (FT )вҲ’1 вүЎ (FвҲ’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 = FОҫ (2.1.12)
with F independent of Оҫ. This mapping may be regarded as between the positions
of either a particle (material point) or a п¬Ғbre (material line segment). The
deformation gradient F is uniquely determined by the mappings of any three
non-coplanar п¬Ғbres and this provides a convenient way to п¬Ғnd F experimentally.
Some simple deformations:
26 Description of Deformation

(a) Dilatation: with a simple rescaling of the coordinates

xi = О»Оҫi, i = 1, 2, 3 (2.1.13)

the deformation gradient is simply F = О»I, where I is the unit matrix, and the
volume change J = О»3.
(b) Extension with lateral contraction or expansion: Suppose that

x1 = О»1Оҫ1, x2 = О»вҠҘ Оҫ2, x3 = О»вҠҘ Оҫ3, (2.1.14)

then the deformation gradient F is given by
вҺӣ вҺһ
О»1 0 0
F = вҺқ 0 О»вҠҘ 0 вҺ  , with J = О»1О»2 . (2.1.15)
вҠҘ
0 0 О»вҠҘ

If О»1 > 1 the deformation is a uniform extension in the 1-direction, whereas
if О»1 < 1 we have a uniform contraction. О»вҠҘ measures the lateral contraction
(О»вҠҘ < 1), or extension (О»вҠҘ > 1), in the 2вҖ“3 plane. For an incompressible
material there can be no volume change, and so J = О»1О»2 = 1 with the result
вҠҘ
that an extension in the 1-direction must be accompanied by a lateral contraction
in the 2вҖ“3 plane.
(c) General extension: when the extensions along the three coordinate axes are
different so that

xi = О»iОҫi, i = 1, 2, 3 (2.1.16)

the deformation gradient F = diag{О»1, О»2, О»3}, with J = О»1О»2О»3. If О»1 = О»вҲ’1
2
and О»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 О±i , where the vectors
are not required to be orthogonal.
Introduce the reciprocal triad ОІj such that О±T ОІj = О±i .ОІj = Оҙij . Show that О±p ОІT = I
p
i
where I is the unit matrix.
Under homogeneous deformation the triad О±i is transformed into the triad ai ; show that
the deformation gradient tensor F can be represented as F = ap ОІT .
p

The reciprocal triad can be found by considering vector products,
e.g., ОІ1 = (a2 Г— a3 )/(a1 .a2 Г— a3 ).
These vectors have the required property that О±T ОІj = О±i .ОІj = Оҙij .
i
T T
Then (О±p ОІp )О±q = О±p (ОІp О±q ) = О±p Оҙpq = О±q and, since a similar relation holds for
any linear combination of the О±i , we require О±p ОІT = I.
p
Under deformation О±i вҶ’ ai , so ai = FО±i and hence ap ОІT = FО±p ОІT = F. вҖў
p p
2.2 Strain 27

Consider a crystal lattice whose elementary cell is a unit cube. A lattice-plane is deп¬Ғned
to have intercepts ОҪвҲ’1 (i = 1, 2, 3) on the cube edges from node 0, where ОҪi 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 deп¬Ғned with cell edges bj such that
ai .bj = Оҙij . 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 |ОҪj bj |вҲ’1 .
We have initially a cubic basis О±1 , О±2 , О±3 with reciprocal triad ОІ1 , ОІ2 , ОІ3 .
The lattice plane with intercepts ОҪвҲ’1 has the equation
i

(1 вҲ’ l вҲ’ m) l m
Оҫ= О±1 + О±2 + О±3 ,
ОҪ1 ОҪ2 ОҪ3
and the normal satisfying
1 1 1 1
ОҪВ· О±1 вҲ’ О±2 =ОҪВ· О±1 вҲ’ О±3 =0
ОҪ1 ОҪ2 ОҪ1 ОҪ3
i.e., Оҫ В· ОҪ = 1, is
ОҪ = ОҪ1 ОІ1 + ОҪ2 ОІ2 + ОҪ3 ОІ3 .
After homogeneous deformation the orthogonal triad О±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
1вҲ’lвҲ’m l m
x= a1 + a2 + a3 ,
ОҪ1 ОҪ2 ОҪ3
and the perpendicular to the embedded directions x В· n = 1 is given by
n = ОҪ1 b1 + ОҪ2 b2 + ОҪ3 b3 ,
where bi is the reciprocal triad to ai .
The distance to the plane varies as 1/|n| = 1/|ОҪj bj |. вҖў
The reciprocal lattice is important in crystallography and controls the X-ray diffraction
patterns from crystals.

2.2 Strain
The mechanical effects of deformation can be separated into a part which affects
the length of п¬Ғbres 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 Оҫ is deп¬Ғned by dx = F dОҫ,Оҫ
and we now explore how this affects the length and orientation of a line segment
(п¬Ғbre).

2.2.1 Stretch
For a п¬Ғbre in the direction ОҪ,
dОҫ = |dОҫ|ОҪ,
Оҫ ОҫОҪ (2.2.1)
28 Description of Deformation

we introduce the stretch
|dx| п¬Ғnal length
О»(ОҪ) = = . (2.2.2)
|dОҫ|
Оҫ initial length
The length of the line segment in the deformed state
|dx|2 = dxT dx = dОҫT (FT F)dОҫ.
Оҫ Оҫ (2.2.3)
Consider also the transformation of two vectors such that
dx = F dОҫ,
Оҫ dy = F dО·. (2.2.4)
The scalar product in the current state is
dx В· dy = dxT dy = dОҫT (FT F)dО·;
Оҫ (2.2.5)
the п¬Ғnal angle between the vectors in the deformed state is dx В· dy/(|dx||dy|).
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 ОҪ is to be found from
dОҫT (FT F)dx
|dx|2 Оҫ
= ОҪT (FT F)ОҪ.
2
(О»(ОҪ)) = = ОҪ (2.2.6)
2 2
|dОҫ|
Оҫ |dОҫ|
Оҫ
A corresponding form can be found in terms of the deformed conп¬Ғguration. The
length of the line segment in the reference state is
|dОҫ|2 = dОҫT dОҫ = dxT (FвҲ’1)T FвҲ’1 dx
Оҫ ОҫОҫ
= dxT (FFT )вҲ’1 dx. (2.2.7)
Thus, if dx lies in the direction n,
(О»(ОҪ))2 = [nT (FFT )вҲ’1n]вҲ’1. (2.2.8)

2.2.2 Principal п¬Ғbres and principal stretches
If we consider the eigenvectors dОҫr of the local metric tensor FT F we have
Оҫ
FT F dОҫr = О»2 dОҫr,
Оҫ rОҫ (2.2.9)
and the eigenvalues О»2 satisfy det |FT F вҲ’ О»2I| = 0, where I is the unit tensor.
r r
TT
In the reference state dОҫ F F dОҫ = const is an ellipsoid (the Lagrangian
Оҫ Оҫ
ellipsoid) with mutually orthogonal principal axes В±dОҫr. Under deformation this
Оҫ
ellipsoid transforms to a sphere (see Figure 2.6) вҖ“ for a principal п¬Ғbre from (2.2.3):
|dxr|2 = dОҫT (FT F)dОҫr = О»2|dОҫr|2,
Оҫr Оҫ rОҫ (2.2.10)
and thus the О»r are the principal stretches. The triad of principal п¬Ғbres dОҫr is termed
Оҫ
If we premultiply the eigenrelation (2.2.9) by F we obtain
FFT F dОҫr = О»2F dОҫr,
Оҫ Оҫ (2.2.11)
r
2.2 Strain 29

and so
FFT dxr = О»2 dxr, (2.2.12)
r

so that the principal п¬Ғbres are also п¬Ғnally orthogonal. The dxr form a triad (the
Eulerian or spatial triad) of the principal axes of an ellipsoid dxT (FFT )вҲ’1 dxr =
r
const which results from the deformation of a sphere in the reference state (Figure
2.6).

Reference Deformed
Оҫ2 x2

x1
Оҫ1

Оҫ2 x2

x1
Оҫ1

Figure 2.6. Relation of Lagrangian (material) and Eulerian (spatial) triads.

different orientations as illustrated in Figure 2.6.

2.2.3 The decomposition theorem
We deп¬Ғne 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 О»r. The array of background
components U can be constructed by the spectral formula
^ ^T ^
U= О»rОҫrОҫr , |Оҫr| = 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 п¬Ғnal 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 О»2, FT F =
r
2 T 2
U . When F is given, the equation F F = U can be solved uniquely for a
positive deп¬Ғnite symmetric U. Then R deп¬Ғned 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 п¬Ғrst a rotation of the principal п¬Ғbres,
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 О»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 speciп¬Ғed independently of coordinate
frame by using the triads and principal stretches
T
^^ x^
F= xrОҫr , |^r|, |Оҫr| = 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 Оё about an axis n transforms a vector Оҫ to
x = Оҫ + sin Оё(n Г— Оҫ) + (1 вҲ’ cos Оё)n Г— (n Г— Оҫ). (2.2.19)
We now introduce the skew matrix N(вүЎ n Г— ) 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 вҲ’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 Оё + N2(1 вҲ’ cos Оё). (2.2.20)
Now, N2 = nnT вҲ’ 1, Nn = 0 and so N3 = вҲ’N, N4 = вҲ’N2. This enables us to
write R in the form
Оё2r+1 Оё2r
r
+ N2 r
R = exp(NОё) = I + N (вҲ’1) (вҲ’1) ,
(2r + 1)! (2r)!
r r
Оёr
Nr
=I+ . (2.2.21)
r!
r
2.2 Strain 31

Once R is given, the angle Оё may be computed from

tr R = 1 + 2 cos Оё, (2.2.22)

and then N and n can be found from

N sin Оё = 1 (R вҲ’ RT ). (2.2.23)
2

EXAMPLE: DEFORMATION GRADIENT AND STRETCH MATRIX
The deformation gradient F is speciп¬Ғed by
7 вҲ’5 2
1
4 4 вҲ’1 .
F=
9 вҲ’5 7 2

Show that its Lagrangian triad is coaxial with (0, 0, 1), (1, 1, 0), (1, вҲ’1, 0) and п¬Ғnd the
principal stretches. Hence construct U and UвҲ’1 and by constructing R = FUвҲ’1 , verify
that it is a rotation of ПҖ/3 about (1, 1, 1). Finally show that the Eulerian triad is coaxial
with (2, вҲ’1, 2), (1, 4, 1), (1, 0, вҲ’1).
The stretch matrix is given by
90 вҲ’54 0
1
U2 = FT F = вҲ’54 90 0 ,
81 0 09
The Lagrangian triad is determined by
10Оҫ1 вҲ’ 6Оҫ2 = 9О»2 Оҫ1
вҲ’6Оҫ1 + 10Оҫ2 = 9О»2 Оҫ2
Оҫ3 = 9О»2 Оҫ3

and is thus coaxial with (1, 1, 0), (1, вҲ’1, 0), (0, 0, 1)
with principal stretches 2 , 4 , 1 .
333
We can construct the stretch matrix U and its inverse UвҲ’1 from
^(1) ^(1)T + О»2Оҫ(2)Оҫ(2)T + О»3Оҫ(3)Оҫ(3)T ,
^^ ^^
U = О»1Оҫ Оҫ
and
(1) (1)T (2) (2)T (3) (3)T
^ ^ ^ ^ ^ ^
UвҲ’1 = О»вҲ’1Оҫ + О»вҲ’1Оҫ + О»вҲ’1Оҫ
Оҫ Оҫ Оҫ ,
1 2 3

so that
вҺӣ вҺһ вҺӣ3
вҺһ 1
3 вҲ’1 0 08 8
1
U = вҺқвҲ’1 3 0вҺ  , U = 3вҺқ 0вҺ  .
1 3
3 8 8
0 01 0 01
The rotation
2 вҲ’1 2
1
вҲ’1
2 2 вҲ’1 .
R = FU =
3 вҲ’1 2 2
32 Description of Deformation

Now trR = 2 = 1 + 2 cos Оё and so cos Оё = 1 , hence Оё = ПҖ/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 вҲ’1 0 4 0 вҲ’1
R =
3
001 1 вҲ’3 2
and is coaxial with (1, 4, 1), (1, 0, вҲ’1), (2, вҲ’1, 2). вҖў

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 п¬Ғbres in the initial conп¬Ғguration 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
Оҫ
the length of dОҫ, using (2.2.3) and (2.2.7)
|dx|2 вҲ’ |dОҫ|2 = dОҫT FT F dОҫ вҲ’ dОҫT dОҫ = 2 dОҫT E dОҫ,
Оҫ Оҫ Оҫ ОҫОҫ Оҫ Оҫ (2.2.24)
= dxT dx вҲ’ dxT (FFT )вҲ’1 dx = 2 dxT e dx. (2.2.25)
Here we have introduced the Green strain,
E = 1 (FT F вҲ’ I) = 1 (U2 вҲ’ I), (2.2.26)
2 2

and the Cauchy strain,
e = 1 (I вҲ’ (FFT )вҲ’1) = 1 (I вҲ’ V вҲ’2), (2.2.27)
2 2

associated with some speciп¬Ғed background frame.
The components E of the Green strain generate a tensor whose axes deп¬Ғne the
reference directions of the Lagrangian principal п¬Ғbres and whose principal values
are
12
2 (О»r вҲ’ 1) r = 1, 2, 3. (2.2.28)
This strain measure is objective (i.e. unaffected by any п¬Ғnal rotation) and it vanishes
when, and only when, the shape is unchanged.
The normalising factor 1 is included so that when the strain is inп¬Ғnitesimal,
2
the principal values of the measure are just the fractional changes in length of the
principal п¬Ғbres; since, if О»r = 1 + ,
+ О»)2 вҲ’ 1] вүҲ ,
1
2 [(1 (2.2.29)
when is a п¬Ғrst-order 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 п¬Ғbres and the
principal values are
вҲ’ О»вҲ’2)
1
2 (1 r = 1, 2, 3. (2.2.30)
r

Once again for inп¬Ғnitesimal strain the principal values reduce to the fractional
changes of length of the principal п¬Ғbres.

Displacement representations:
We introduce the displacement u between the deformed and reference
conп¬Ғgurations
u = x вҲ’ Оҫ, (2.2.31)
вҲ‚u вҲ‚u
= F вҲ’ I, F=I+ , (2.2.32)
вҲ‚Оҫ
Оҫ вҲ‚Оҫ
Оҫ
вҲ‚u вҲ‚u
= I вҲ’ FвҲ’1, FвҲ’1 = I вҲ’ . (2.2.33)
вҲ‚x вҲ‚x
In terms of these displacement gradients the Green strain tensor is given by
T
вҲ‚u вҲ‚u
T
2E = FF вҲ’ I = I + I+ вҲ’ I,
вҲ‚Оҫ
Оҫ вҲ‚Оҫ
Оҫ
T T
вҲ‚u вҲ‚u вҲ‚u вҲ‚u
= + + , (2.2.34)
вҲ‚Оҫ
Оҫ вҲ‚Оҫ
Оҫ вҲ‚Оҫ
Оҫ вҲ‚Оҫ
Оҫ
and the Cauchy strain by
T
вҲ‚u вҲ‚u
вҲ’T вҲ’1
2e = I вҲ’ F F =IвҲ’ IвҲ’ IвҲ’ ,
вҲ‚x вҲ‚x
T T
вҲ‚u вҲ‚u вҲ‚u вҲ‚u
= + вҲ’ . (2.2.35)
вҲ‚x вҲ‚x вҲ‚x вҲ‚x
The cartesian components of the strain tensors are therefore given by
i) Green strain tensor E:
вҲ‚ui вҲ‚uj вҲ‚uk вҲ‚uk
1
Eij = + + , (2.2.36)
2 вҲ‚Оҫj вҲ‚Оҫi вҲ‚Оҫi вҲ‚Оҫj
and
ii) Cauchy strain tensor e:
вҲ‚ui вҲ‚uj вҲ‚uk вҲ‚uk
1
eij = + вҲ’ . (2.2.37)
2 вҲ‚xj вҲ‚xi вҲ‚xi вҲ‚xj
When the displacements and displacement gradients are sufп¬Ғciently small, the
distinction between the two small-strain deп¬Ғnitions is usually ignored, since the
second-order 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 вҲ’ Оҫ = Оіm(n В· Оҫ) = Оіm(nTОҫ), (2.3.1)
particles are not displaced in the basal plane nTОҫ = 0 irrespective of the direction
of the unit vector m. The deformation gradient
F = I + ОіmnT , (2.3.2)
and the amount of deformation scales with the height of Оҫ above the basal plane
(n.Оҫ) (Figure 2.7).
Оҫ

n x

Оі m(n . Оҫ )
Figure 2.7. Representation of plane
m
deformation in the direction m from
Оҫ
a basal plane with normal n.

Shear
An equivoluminal plane deformation is called a shear, the principal stretches are
then such that
О»3 = 1 with О»1О»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 + ОіmnT , mT n = 0. (2.3.4)
Оі is called the amount of shear and a plane nTОҫ = 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 О»1О»2 = 1, the principal axes of strain
remain п¬Ғxed (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 two-dimensional area and three-dimensional 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 п¬Ғxed 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 = sОҫ1 + tОҫ2 /s, x2 = Оҫ2 /s, x3 = Оҫ3 , (2.3.5)
with s вүҘ 1, t вүҘ 0 (s вүЎ 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 п¬Ғbres 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 п¬Ғxed. For a matrix of the form
ac
,
cb
the inclinations of the principal axes depend only on (a вҲ’ b)/c.
Because we require the п¬Ғnal state to be a simple shear the trajectory in (s, t) space has
36 Description of Deformation

to pass through s = 1, t = Оі and s = 1, t = 0 to include the undeformed state. To keep
the inclination of the principal axes п¬Ғxed we must have
t2 + 1
1 12
вҲ’ s2 = Оі + 1 вҲ’ 1 = Оі,
2
t s Оі
and so the path is given by
s4 + Оіs2 t вҲ’ t2 = 1, 0 вү¤ t вү¤ Оі,
as illustrated in Figure 2.9.

s2

Figure 2.9. Trajectory in (s, t) space
that retains the same principal п¬Ғbres.

t
Оі

2.4 Motion
We can follow the motion of a particle by tracking its position relative to its
reference position Оҫ (a Lagrangian viewpoint), or alternatively we may follow
the particle directly in an Eulerian viewpoint. In a similar way we can represent
a п¬Ғeld qL(Оҫ, t) in terms of the Lagrangian frame so that qL follows the particle
Оҫ
initially at Оҫ. Or, alternatively, we can work in terms of the current conп¬Ғguration
of the medium with an Eulerian description qE(x, t) in terms of a п¬Ғxed position
x. Because both representations link the value of q for the particle labelled by Оҫ
which is at x at time t:
qE(x(Оҫ, t), t) = qL(Оҫ, t).
Оҫ Оҫ (2.4.1)
When we follow a given particle Оҫ (a Lagrangian viewpoint) then the time rate
of change of interest is
вҲ‚
, (2.4.2)
вҲ‚t Оҫ

but, using the chain rule, we п¬Ғnd that the вҖ˜material time derivativeвҖ™
D вҲ‚ вҲ‚ вҲ‚ вҲ‚
ПҲвүЎ ПҲ= ПҲ+ xi ПҲ, (2.4.3)
Dt вҲ‚t вҲ‚t вҲ‚t вҲ‚xi
Оҫ Оҫ
x

in terms of an Eulerian decomposition.
2.4 Motion 37

We deп¬Ғne the velocity п¬Ғeld corresponding to a particle initially at Оҫ as
вҲ‚ D
v= x(Оҫ, t) вүЎ
Оҫ x. (2.4.4)
вҲ‚t Dt
Оҫ

The acceleration п¬Ғeld is then
вҲ‚ D
f= v(Оҫ, t) вүЎ
Оҫ v, (2.4.5)
вҲ‚t Dt
Оҫ

and in the Eulerian viewpoint
вҲ‚ вҲ‚
f= v(x, t) + v В· v, (2.4.6)
вҲ‚t вҲ‚x
x

where we have written вҲ‚/вҲ‚x for the gradient operator in the Eulerian conп¬Ғguration
(for п¬‚uids this is the most common viewpoint to use).
In general, then the mobile derivative following the particles is
D вҲ‚ вҲ‚
вүЎ + vВ· . (2.4.7)
Dt вҲ‚t вҲ‚x
x

If we consider a line element dx = F dОҫ, its material derivative is
Оҫ
D вҲ‚L вҲ‚
dx = dОҫ В·
Оҫ v= dx В· v, (2.4.8)
Dt вҲ‚Оҫ
Оҫ вҲ‚x
a result that can also be derived from geometrical considerations.

EXAMPLE: TIME-DEPENDENT DEFORMATION
A body undergoes the deformation speciп¬Ғed by

x1 = Оҫ1 (1 + a2 t2 ), x3 = Оҫ3 .
x2 = Оҫ2 ,
Find the displacement and velocity in both the material and the spatial description.
The displacement u = x вҲ’ Оҫ in the material description is

u1 = Оҫ1 a2 t2 , u2 = 0, u3 = 0,
and in the spatial description

u1 = x1 a2 t2 /(1 + a2 t2 ), u2 = 0, u3 = 0.
The velocity, вҲ‚u/вҲ‚t)Оҫ in the material description is

v1 = 2Оҫ1 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 = ПҒ0(Оҫ) dV0
Оҫ (2.5.1)
V0

should remain invariant. In terms of the deformed state, the mass

M0 = ПҒ(x, t) dV = ПҒ(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

[ПҒ0(Оҫ) вҲ’ ПҒ(x, t)J(t)] dV0 = 0.
Оҫ (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
вҲ‚ вҲ‚ D
ПҒ0(Оҫ) =
Оҫ ПҒ(x, t)J(t) = ПҒ(x, t)J(t) = 0. (2.5.4)
вҲ‚t вҲ‚t Dt
Оҫ Оҫ

The material time derivative of density can therefore be found from
D D
ПҒ(x, t) + ПҒ(x, t)JвҲ’1(t) J(t) = 0. (2.5.5)
Dt Dt
From the properties of the determinant of the deformation gradient
D вҲ‚ вҲ‚
J(t) = J(t) = J(t) В· v(x, t). (2.5.6)
Dt вҲ‚t вҲ‚x
Оҫ

in terms of the velocity п¬Ғeld v(x, t). The material time derivative of the density is
thus determined by
D вҲ‚
ПҒ(x, t) + ПҒ(x, t) В· v(x, t) = 0. (2.5.7)
Dt вҲ‚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
вҲ‚ вҲ‚
ПҒ(x, t) + В· [ПҒ(x, t)v(x, t)] = 0, (2.5.8)
вҲ‚t вҲ‚x
x
which is known as the continuity equation. The relation (2.5.8) can also be deduced
by considering a п¬Ғxed volume in space and then allowing for mass п¬‚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 ОЁ (scalar, vector or tensor) associated with the
continuum so that ПҲ is the quantity per unit mass. The material rate of change of ОЁ
for a п¬Ғxed material volume V with surface S and outward normal n must be equated
^
to the rate of change of ОЁ for the particles instantaneously within V and the net п¬‚ux
of ОЁ into V across the surface S so that
вҲ‚
d
ОЁ dV = ОЁ dV + ОЁv В· n dS.^ (2.5.9)
V вҲ‚t
dt V S
If we now represent ОЁ as ПҲПҒ we have
вҲ‚
d
ПҲПҒ dV = (ПҲПҒ) dV + ПҲПҒv В· n dS.
^ (2.5.10)
вҲ‚t
dt V V S
We apply the divergence theorem to the surface integral in (2.5.10), and then
вҲ‚ вҲ‚
d
ПҲПҒ dV = (ПҲПҒ) + (ПҲПҒv) dV
вҲ‚t вҲ‚x
dt V V
вҲ‚ вҲ‚ вҲ‚ вҲ‚
= ПҲ ПҒ+ (ПҒv) + ПҒ ПҲ+vВ· ПҲ) dV. (2.5.11)
вҲ‚t вҲ‚x вҲ‚t вҲ‚x
V
From the continuity equation (2.5.8) the п¬Ғrst contribution to the volume integral
on the right-hand side of (2.5.11) vanishes, and the second can be recognised as
ПҒ DПҲ/Dt. Hence
D
d
ПҒПҲ dV = ПҒ ПҲ dV; (2.5.12)
Dt
dt V V
for ПҲ = 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 = вҲ‚x/вҲ‚Оҫ, Fij = вҲ‚xi /вҲ‚Оҫj , the determinant
Оҫ
1
J = det F = 6 ijk pqr Fpi Fqj Frk (2A.1)
вҲ‚xp вҲ‚xq вҲ‚xr
1
= . (2A.2)
6 ijk pqr вҲ‚Оҫ вҲ‚Оҫj вҲ‚Оҫk
i

Now, from the orthogonality relation
вҲ‚xi вҲ‚Оҫj
= Оҙik , (2A.3)
вҲ‚Оҫj вҲ‚xk
we can apply CramerвҖ™s rule for the solution of linear equations to obtain
вҲ‚Оҫj 1 вҲ‚xp вҲ‚xq
= , (2A.4)
jmp kqr
вҲ‚xk 2J вҲ‚Оҫi вҲ‚Оҫj
and thus we can recognise that
вҲ‚J вҲ‚J вҲ‚Оҫj
= =J . (2A.5)
вҲ‚Fij вҲ‚(вҲ‚xi /вҲ‚Оҫj ) вҲ‚xi
40 Description of Deformation

We can therefore establish the important relations
вҲ‚ 1 вҲ‚xj
= 0, (2A.6)
вҲ‚xk J вҲ‚Оҫk
and
вҲ‚ вҲ‚Оҫj вҲ‚ вҲ‚xi вҲ‚вҲ‚ вҲ‚
J=J . =J xi = J .v , (2A.7)
вҲ‚t вҲ‚xi вҲ‚t Оҫ вҲ‚Оҫj вҲ‚xi вҲ‚t вҲ‚x
Оҫ Оҫ

which is used in (2.5.6).
3
The Stress-Field 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 п¬Ғeld deп¬Ғned over the surface.

3.1 Traction and stress
Consider any point P of a continuum and any inп¬Ғnitesimal plane element ОҙS with P
as centroid. Let x be the position vector of P, and n the unit normal to ОҙS (in some
consistent sense).

ОҙS
n
Figure 3.1. The traction П„ acting on an
P П„ element with normal n.

Then the mechanical inп¬‚uence exerted across ОҙS by the material on the side to
which n points is postulated to be represented by a force

П„(n, x)ОҙS (3.1.1)

through P. The negative of this force is exerted across ОҙS on this material. П„ is
called the traction vector and has the dimensions of force per unit area.
Note: a mutual couple could also be introduced across ОҙS, but the need for it has
never been indisputably established in any particular situation.

41
42 The Stress-Field Concept

The properties of traction can be determined from a hypothesis due substantially
to Euler: the vector п¬Ғelds of force and mass-acceleration 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 вҖ“ local acceleration,
g вҖ“ local body force per unit mass (e.g., gravitational or п¬Ғctitious),
Ој вҖ“ 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

П„(x) dS + g(x) dm = f(x) dm, (3.1.2)
S V V

x Г— П„(x) dS + x Г— g(x) dm + Ој(x) dm = x Г— 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
Пғi denote the vector traction exerted over this element by the material situated on
its positive side (Figure 3.2).

ОҙS
Пғi
Figure 3.2. Traction Пғi acting on an
xi
P element perpendicular to the xi axis.

The components Пғij of Пғi constitute the stress-matrix. Пғ11, Пғ22, Пғ33 are called the
normal components and the others shear components.
The stress matrix fully speciп¬Ғes the mechanical state at P, in the sense that it
determines the traction П„ for any n. Consider an inп¬Ғnitesimal 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 п¬Ғrst 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) =
П„ dS = (П„ вҲ’ niПғi) (f вҲ’ 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 вҶ’ 0,

П„ dS вҶ’ 0, (3.1.6)
S
Thus
П„(n) = niПғi, i.e. П„j(n) = niПғij. (3.1.7)
which is known as CauchyвҖ™s relation.
We note that (3.1.7) also implies that Пғij transforms like a second-rank 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
ПғвҲ— = Пғijlpilqj, (3.1.8)
pq

since
П„вҲ— = lqjП„j, Пғi = lpiПғвҲ— , lipljp = Оҙij.
as (3.1.9)
q p
44 The Stress-Field Concept

Alternatively, in terms of the rotation R which carries the unstarred axes into the
starred ones
Rip = lpi, (3.1.10)
and
ПғвҲ— = RT ПғR. (3.1.11)

3.2 Local equations of linear motion
We rewrite (3.1.2) in component form

П„j(x) dS = Пғij(x)ni(x) dS = [fj(x) вҲ’ gj(x)] dm, (3.2.1)
S S V

and then use the tensor divergence theorem
вҲ‚ вҲ‚ dm
Пғij(x)ni(x) dS = Пғij(x) dV = Пғij(x) , (3.2.2)
вҲ‚xi вҲ‚xi ПҒ(x)
S V V

to represent the effects of the surface tractions though a volume integral. Here
ПҒ(x) is the local density and the stress tensor Пғij is assumed to be continuous and
differentiable within V.
From (3.2.1) and (3.2.2) we have the relation
вҲ‚ dm
Пғij(x) = [fj(x) вҲ’ gj(x)] dm, (3.2.3)
вҲ‚xi ПҒ(x)
V V

and since the region V is arbitrary
вҲ‚
Пғij(x) + ПҒgj(x) = ПҒfj(x). (3.2.4)
вҲ‚xi
We see that it is the stress gradient, in conjunction with the body force, that
determines the local acceleration.
If вҲ‚Пғij/вҲ‚xi = 0, the stress п¬Ғeld is said to be self-equilibrated
Note: the local equations (3.1.7) [П„j = Пғijni] 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
вҲ‚ dm
ijkxjП„k dS = (xjПғlk) , (3.2.5)
ijk
вҲ‚xl ПҒ
S V

=вҲ’ Ојi dm + ijkxj[fk вҲ’ gk] dm. (3.2.6)
V V
3.2 Local equations of linear motion 45

From the equation of motion (3.2.4)

вҲ‚ вҲ‚ dm
(xjПғlk) вҲ’ xj Пғlk =вҲ’ Ојi dm, (3.2.7)
ijk
вҲ‚xl вҲ‚xl ПҒ
V V

and so,

ijkОҙjlПғlk dm = ijkПғjk dm = вҲ’ПҒ Ојi dm. (3.2.8)
V V V

Since the region V is again arbitrary

ijkПғjk(x) = вҲ’Ојi(x), (3.2.9)

and we may recover (3.1.3) from these local equations.
We will subsequently suppose that body moments are absent (Ојi вүЎ 0) and then

Пғjk(x) = Пғkj(x), (3.2.10)

and the stress matrix is symmetric.

EXAMPLE: STRESS FIELDS
Show that the stress п¬Ғeld Пғij = pОҙij + ПҒvi vj is self-equilibrated, where p, ПҒ and v are the
pressure, density and velocity in a steady п¬‚ow of some inviscid compressible п¬‚uid.

The gradient of the stress tensor Пғ
вҲ‚Пғij вҲ‚p вҲ‚vi вҲ‚vj вҲ‚ПҒ
= + ПҒvj + ПҒvi + vi vj
вҲ‚xj вҲ‚xj вҲ‚xj вҲ‚xj вҲ‚xj
= [вҲҮp + ПҒ(v В· вҲҮ)v]i + vi [вҲҮ В· (ПҒv)].

вҲҮp + ПҒ(v В· вҲҮ)v = 0,

and from the continuity equation
вҲ‚ПҒ
+ вҲҮ В· (ПҒv) = 0,
вҲ‚t
but since п¬‚ow is steady вҲ‚ПҒ/вҲ‚t = 0. Thus
вҲ‚Пғij
= 0,
вҲ‚xj
and so the stress п¬Ғeld is self-equilibrated.вҖў

Show that the stress п¬Ғeld Пғij = vi vj вҲ’ 1 v2 Оҙij exerts tractions on any closed surface which
2
are statically equivalent overall to body forces v(вҲҮ В· v) per unit volume, when v is any
irrotational п¬Ғeld (i.e. вҲҮ Г— v = 0).
46 The Stress-Field Concept

The gradient of the stress tensor Пғ
вҲ‚Пғij вҲ‚vj вҲ‚vi вҲ‚vk
= vi + vj вҲ’ vk
вҲ‚xj вҲ‚xj вҲ‚xj вҲ‚xi
вҲ‚vj вҲ‚vi вҲ‚vj
= vi + vj вҲ’
вҲ‚xj вҲ‚xj вҲ‚xi
= vj вҲҮ В· v,
since вҲҮ Г— v = 0 and we can recognise the bracketed term as a component of вҲҮ Г— v. In
a static scenario, we have from (3.2.1) and (3.2.2)
вҲ‚Пғij
dV = П„i (x)dS = вҲ’ gi dV
вҲ‚xj
V S V

and the body force per unit volume g is v(вҲҮ В· v) вҖў

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-п¬Ғbre resins (used in the hull of boats), cermets вҖ“ sintered mixtures of ceramic
inclusions in a metallic matrix (used in cutting tools), and ceramic п¬Ғbre-reinforced
Major internal discontinuities in the Earth occur at the base of the crust, where
the MohoroviЛҮ iВҙ discontinuity can in places be very sharp, and at the phase
cc
boundaries in the mantle transition zone. Important п¬‚uidвҖ“solid discontinuities are
at the coreвҖ“mantle boundary between the silicate mantle and the metallic п¬‚uid in
the outer core and the transition from the п¬‚uid outer core to the solid inner core.

П„+
n

+
вҲ’
Figure 3.5. Interface conditions on
traction

П„вҲ’
Оµ

Possible jumps in the stress components at an interface are, however, restricted
by the inviolable continuity of the traction vector acting on the interface
[П„j]+ = ni[Пғij]+ = 0, (3.2.11)
вҲ’ вҲ’
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),

П„+ dA+ + П„вҲ’ dAвҲ’ + O( ) = 0, (3.2.12)
A A
where the O( ) term arises from tractions on the edge, body forces and п¬Ғnite
mass-accelerations (we exclude idealised вҖ˜shocksвҖ™ with a jump in velocity).
In the limit as вҶ’ 0
П„+(n) + П„вҲ’(вҲ’n) = 0, i.e. [П„(n)]+ = 0. (3.2.13)
вҲ’

Since (3.2.11) imposes only three constraints on six independent components
of Пғij, 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 П„ in
(3.2.11) becomes Пғ3 and so
[Пғ31] = [Пғ32] = [Пғ33] = 0; (3.2.14)
consequently [Пғ13] = [Пғ23] = 0, but we cannot require [Пғ11], [Пғ22], [Пғ12] 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-
Пғ
face.
22

вҲ’Пғ21
Пғ12
Пғ11
+
-
1

In a similar way, at a body surface, such вҖ˜interiorвҖ™ 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 вҖ“ 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
(Оҙir вҲ’ ni nr )(Оҙjs вҲ’ nj ns )prs
for some symmetric prs , where n is the unit normal to the jump surface.
Set the stress jump pij = (Оҙik вҲ’ni nk )(Оҙjl вҲ’nj nl )pkl . Consider the normal component
^
of the associated traction
pij nj = (Оҙik вҲ’ ni nk )(nl вҲ’ |n2 |nl )pkl
^
= 0, since |n| = 1.
48 The Stress-Field Concept

The traction in a direction m orthogonal to the normal n, with m В· n = 0, would be
pij mj = (Оҙik вҲ’ ni nk )(ml вҲ’ nj mj nl )pkl
^
= (Оҙik вҲ’ ni nk )ml pkl , as m В· n = 0
= mj pij вҲ’ (nk ml pkl )ni .
Thus the traction jump in the normal direction is zero, and in orthogonal directions is
non-zero as required for a statically admissable stress jump. вҖў

3.3 Principal basis for stress
The traction vector П„(n) over a plane element is purely normal when П„ = Пғn for
some Пғ. Thus since П„j = Пғijni, we have

(Пғij вҲ’ ПғОҙij)ni = 0, (3.3.1)

The eigenvectors В±nr (r = 1, 2, 3) are called the principal axes of stress and the
eigenvalues Пғr given by the roots of det(Пғij вҲ’ ПғОҙij) = 0 are called the principal
stresses.
Since Пғij is symmetric, the principal stresses Пғr are real and the axes are orthogonal.
Each Пғr acts in opposite directions on opposite faces of a вҖ˜principalвҖ™ box. The
vectors nr can also be thought of geometrically as the principal axes of the stress

Пғijxixj = const. (3.3.2)

When the principal axes and stresses are known, stress components on
background axes are given by the spectral formula

Пғ r nr nT .
Пғ= (3.3.3)
r
r

Note: any quantity which is solely a function of the Пғr is an invariant of Пғij. The
cubic equation for the principal stresses Пғr can be written as

det(Пғij вҲ’ ПғОҙij) = вҲ’Пғ3 + IПғПғ2 вҲ’ IIПғПғ + IIIПғ = 0 (3.3.4)

in terms of the invariants
IПғ = Пғ1 + Пғ2 + Пғ3 = Пғkk
IIПғ = Пғ1Пғ2 + Пғ2Пғ3 + Пғ3Пғ1 = 1 [ПғiiПғjj вҲ’ ПғijПғji] (3.3.5)
2
IIIПғ = Пғ1Пғ2Пғ3 = det Пғ.
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

Пғ
Пғ n
3
1

Figure 3.7. Transformation
Пғ
2
from the principal basis for
stress to normal and shear
m
Пғ
2
tractions on a surface.
Пғ
1
Пғ
3

Пғ2

__ (Пғ +Пғ )
1
21 2

Figure 3.8. Conп¬Ғguration with
Пғ1 Пғ1
maximum shear stress.
1
__ (Пғ вҲ’ Пғ2 )
21

Пғ2

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 В· n = 0, |m| = 1,
(Figure 3.7), then we should use
П„(n) = (n1Пғ1, n2Пғ2, n3Пғ3), (3.3.6)
making use of the relation П„ = niПғi, with n = (n1, n2, n3) in a principal basis.
Then
n В· П„(n) = n2Пғ1 + n2Пғ2 + n2Пғ3, (3.3.7)
1 2 3

m В· П„(n) = m1n1Пғ1 + m2n2Пғ2 + m3n3Пғ3 = n В· П„(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, Пғ3, and the tractions on the
other faces have no components in that direction.
50 The Stress-Field Concept

П„
Пғ2 m.П„

n
Оё
2Оё
Пғ1
Пғ1
Пғ2 Пғ1 П„
0 n.П„
m

Пғ2

Figure 3.9. Stress-circle construction for normal and shear tractions.

^
Consider the normal in the 12 plane at an angle Оё to the Пғ1 axis, then
n = (cos Оё, sin Оё, 0)m = (sin Оё, вҲ’ cos Оё, 0) (3.3.9)
^
so that the normal and shear stress components in the 12 plane are
n В· П„ = cos2 Оё Пғ1 + sin2 ОёПғ2 = 1 (Пғ1 + Пғ2) + 1 (Пғ1 вҲ’ Пғ2) cos 2Оё
2 2
(3.3.10)
1
mВ·П„= 2 (Пғ1 вҲ’ Пғ2) sin 2Оё
As Оё is varied from 0 to ПҖ, the locus of m В· П„ versus n В· П„ is a circle: centre
1 1
2 (Пғ1 + Пғ2), radius 2 (Пғ1 вҲ’ Пғ2), where we have assumed that Пғ1 is greater than Пғ2
(Figure 3.9).
The shear stress is largest when Оё = ПҖ/4
(m В· П„)max = 1 (Пғ1 вҲ’ Пғ2) (3.3.11)
2

and then the normal tractions are equal to 1 (Пғ1 + Пғ2) 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 cross-section is the
square вҲ’a вү¤ x1 вү¤ a, вҲ’a вү¤ x2 вү¤ a is subjected to torsion by couples acting over its
ends x3 = 0 and x3 = L. The stress components are given by Пғ13 = вҲ‚ПҲ/вҲ‚x2 , Пғ23 =
вҲ’вҲ‚ПҲ/вҲ‚x1 , Пғ11 = Пғ12 = Пғ22 = Пғ33 = 0, where ПҲ = ПҲ(x1 , x2 ).
(a) Show that this stress tensor is self-equilibrated;
(b) show that the difference between the maximum and minimum stress components is
2[(вҲ‚ПҲ/вҲ‚x1 )2 + (вҲ‚ПҲ/вҲ‚x2 )2 ]1/2 , and п¬Ғnd the principal axis which corresponds to the zero
principal component;
(c) for the special case ПҲ = (x2 вҲ’ a2 )(x2 вҲ’ 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 п¬Ғeld is self-equilibrated because

вҲ‚2 ПҲ вҲ‚2 ПҲ
вҲ‚Пғ13 вҲ‚Пғ23
+ = вҲ’ =0
вҲ‚x1 вҲ‚x2 вҲ‚x2 вҲ‚x1 вҲ‚x1 вҲ‚x2
and all other components of the gradient of the stress tensor are zero.
Now set a = вҲ‚ПҲ/вҲ‚x2 , b = вҲ’вҲ‚ПҲ/вҲ‚x1 , then the principal stresses satisfy

Пғ0a
0 Пғ вҲ’b = 0,
det
a вҲ’b Пғ
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