Solid Mechanics: Linear Elasticity
N. H. Scott
(School of Mathematics, University of East Anglia,
Norwich, NR4 7TJ. United Kingdom)
1
Introduction
In a solid material (e.g. steel, rubber, wood, crystals, etc.) a deformation (i.e. a
change of shape) can be sustained only if a system of forces (loads) is applied to the
bounding surfaces, so setting up internally a distribution of stress (i.e. force per unit
area). The defining characteristic of an elastic material is that when these loads are
removed the body returns exactly to its original shape. The original state in which no
loads are applied is often called the natural state, or reference configuration.
Almost all engineering materials possess the property of elasticity provided the external loads are not too large. If the loads are increased beyond a certain limit, different
for each material, then the material will fail, either by fracture or by flow. In neither
case does the material return to its original shape when the loads are removed and we
say that the elastic limit has been exceeded. A solid material that fails by fracture is
said to be brittle and one that fails by flow is said to be plastic.
We do not consider atomic structure. We assume that matter is homogeneous and
continuously distributed over the material body, so that the smallest part of the body
possesses the same physical properties as the whole body. The theory of the mechanical
behaviour of such materials is called continuum mechanics. We consider therefore
only the macroscopic (large scale) behaviour of materials, which is adequate for most
engineering purposes, and ignore the microscopic (small scale) behaviour.
We further assume that the deformation is small so that the elastic material is linearly elastic. For almost all engineering materials the linear theory of elasticity holds
if the applied loads are small enough.
This unit discusses only the linear theory of elasticity.
2
The kinematics of deformation: strain
Kinematics is the study of motion (and deformation) without regard for the forces
causing it.
A material body B occupies a region B0 of space at time t = 0, its reference configuration, and a material particle has position vector X, with Cartesian components
Xi , i = 1, 2, 3, with respect to the orthonormal vectors {e1 , e2 , e3 }:
X=
3
X
i=1
1
Xi ei .
Figure 1: Reference configuration
Current configuration
The body undergoes a motion, or deformation, during which the material particle at
X when t = 0 is moved to
x = x(X, t)
(2.1)
at time t and then has position vector
x=
3
X
xi ei ,
i=1
with Cartesian components xi , i = 1, 2, 3. Using the Einstein summation convention, by which twice-occurring subscripts are summed over (from 1 to 3), we can write
X = Xi ei ,
x = xi ei .
We write
x = X + u(X, t),
xi = Xi + ui (X, t), i = 1, 2, 3,
(2.2)
so that u(X, t) is the particle displacement with components ui, i = 1, 2, 3. The
displacement gradient, h, is the 3 × 3 matrix with components
hij =
∂ui
.
∂Xj
In the linear theory of elasticity we assume not only that the components ui are small
but also that each of the derivatives hij is small. This has an important consequence:
since x − X = u is small we can afford to replace coordinates X by x and write in place
of (2.2)
x = X + u(x, t), xi = Xi + ui (x, t), i = 1, 2, 3,
(2.3)
and work in future only with the coordinates xi and the particle displacements ui (x, t).
2
The particle velocity is
∂u ∂x v=
≈
,
∂t X
∂t x
and the particle acceleration is similarly approximated by
∂ 2 x ∂ 2 u a=
≈
.
∂t2 X
∂t2 x
The deformation enters into the linear theory of elasticity only through the linear
strain tensor, e, which has components
∂ui ∂uj
+
∂xj
∂xi
1
eij =
2
!
(2.4)
(i, j = 1, 2, 3). For example,
e11
1
=
2
∂u1 ∂u1
+
∂x1 ∂x1
!
∂u1
=
∂x1
and e12
!
∂u1 ∂u2
.
+
∂x2 ∂x1
1
=
2
A tensor is a linear map from one vector space to another; but all such maps can
be represented by matrices. So, for tensor, think matrix.
The strain tensor e is symmetric:
eT = e,
eji = eij ,
i.e. e12 = e21 , e23 = e32 , e31 = e13 . We see this from (2.4):
1
eji =
2
∂uj
∂ui
+
∂xi
∂xj
!
1
=
2
∂ui ∂uj
+
∂xj
∂xi
!
= eij .
The matrix (eij ) of strain components is therefore real and symmetric and so, by the
theory of linear algebra, has 3 real eigenvalues with corresponding unit eigenvectors
which are mutually orthogonal. The eigenvalues are known as the principal strains
and the unit eigenvectors are the principal axes of strain.
Triaxial stretch
Consider a unit cube aligned with the coordinate axes and subjected to the deformation
u1 = e1 x1 ,
u2 = e2 x2 ,
u3 = e3 x3 ,
(2.5)
independent of time t, in which ei are the constant (small) strains. Note that ei > 0
corresponds to a stretch, and ei < 0 to a contraction.
From (2.5) we see that the strain components are given by
e11 =
∂u1
= e1 ,
∂x1
e22 =
∂u2
= e2 ,
∂x2
e33 =
∂u3
= e3 ,
∂x3
with all the other components vanishing: e12 = e21 = e23 = e32 = e31 = e13 = 0. Thus

e1

(eij ) =  0
0
3
0
e2
0

0
0 

e3
Figure 2: Triaxial stretch
and we see that the principal strains are given by e1 , e2 , e3 . Also, the principal axes of
strain are aligned with the coordinate axes of Figure 2 and have components


1


 0 ,
0


0


 1 ,
0


0


 0 ,
1
respectively.
After the triaxial stretch the new volume is
(1 + e1 )(1 + e2 )(1 + e3 ) = 1 + e1 + e2 + e3 + e1 e2 + e2 e3 + e3 e1 + e1 e2 e3 .
Since the ei are small, the relative change in volume, known as the dilatation, is
∆ = e1 + e2 + e3 =
change in volume
.
original volume
But in this case e1 = e11 , e2 = e22 , e3 = e33 , so
∆ = e11 + e22 + e33 = tr e = ekk ,
(2.6)
where k is summed over. In fact, for any deformation, with e not necessarily diagonal,
the dilatation is given by (2.6). To see this we observe that any strain tensor e can be
diagonalized (because symmetric) by a suitable rotation of coordinates and then in the
new coordinates the deformation is necessarily of the form (2.5). So the dilatation is
given by (2.6). But tr e = ekk is an invariant quantity, unchanged by any rotation of
coordinates. Therefore (2.6) represents the dilatation for general e.
4
Conservation of mass
The mass m of the cube and the cuboid in Figure 2 are the same. Let ρ0 and ρ be the
initial and final densities:
m
m
⇒ ρ = ρ0 (1 + ∆)−1 .
ρ0 = , ρ =
1
1+∆
So, by the binomial theorem, with ∆ small, the conservation of mass reads
ρ = ρ0 (1 − ∆).
(2.7)
If the volume increases (∆ > 0), the density decreases (ρ < ρ0 );
if the volume decreases (∆ < 0), the density increases (ρ > ρ0 ).
Simple shear
Consider the same unit cube aligned with the coordinate axes but subjected to the shear
deformation
u1 = γx2 , u2 = 0, u3 = 0,
(2.8)
with γ a positive constant.
Figure 3: Simple shear
Then
∂u1
= γ, with all other components of deformation gradient vanishing. So
∂x2
e12
1
=
2
∂u1 ∂u2
+
∂x2 ∂x1
!
1
γ
= (γ + 0) = = e21 ,
2
2
with all other strain components vanishing:

0
 1
(eij ) =  2 γ
0
1
γ
2
0
0

0
0 
.
0
For general strain tensor e the diagonal components e11 , e22 , e33 are termed the
normal strains, whereas the off-diagonal elements e12 = e21 , e23 = e32 , e31 = e13 , are
termed the shear strains.
5
3
The theory of stress: equations of motion
The forces acting on a deformed body are of two kinds.
1. Body force, b, measured per unit mass, acts on volume elements, e.g. gravity,
inertial forces.
2. Contact force, t(n), measured per unit area.
Figure 4: Contact forces
Take an arbitrary surface element da with unit normal vector n. The material on
the side of da into which n points exerts a force t(n)da, across the surface element da,
on the material on the other side of da. The vector t(n) is the traction vector.
Example 1: Uniaxial tension A force per unit (deformed) area T is applied to a
Figure 5: Uniaxial tension
cuboid in the x1 -direction causing extension in the x1 -direction and (usually) lateral
contraction. Now
t(e1 ) = T e1 , but t(e2 ) = 0,
so t(n) depends on n, even for fixed x and t.
Figure 6: Simple shear
Example 2: Simple shear In this case t(e2 ) = T e1 , so t(n) is not necessarily parallel
to n.
6
Consider a cuboid of deformed material with sides δx1 , δx2 , δx3 parallel to the coordinate axes. The tractions on each face are as shown.
Figure 7: The stress components


σ11


The traction vector t(e1 ) on the face with normal e1 has components  σ21 ,
σ31




σ12


the traction vector t(e2 ) on the face with normal e2 has components  σ22 ,
σ32
σ13


the traction vector t(e3 ) on the face with normal e3 has components  σ23 .
σ33
These column vectors are put together to form the 3 × 3 matrix of components of the
stress tensor, σ:


σ11 σ12 σ13


(3.1)
(σij ) =  σ21 σ22 σ23  .
σ31 σ32 σ33
The diagonal elements σ11 , σ22 , σ33 are termed normal stresses and the off-diagonal
elements are termed shear stresses.
It can be shown that for arbitrary unit normal vector n = n1 e1 + n2 e2 + n3 e3 = ni ei ,
the traction vector is given by
t(n) = σn,
ti = σij nj ,





n1
σ11 σ12 σ13
t1





 t2  =  σ21 σ22 σ23   n2  ,
n3
σ31 σ32 σ33
t3
(3.2)
so the traction components ti are a linear combination of the unit surface normal components ni .
7
The normal component of traction is
tn = t(n) · n = ti ni = σij nj ni = σij ni nj ,
(3.3)
from (3.2). The normal stress tn is tensile when positive and compressive when
negative.
Figure 8: Normal and shear components of traction
The tangential component of traction is
t(n) − tn n
with magnitude
1
called the shear stress.
τn = |t(n) − tn n| = {|t(n)|2 − t2n } 2 ,
(3.4)
Example 3 For the stress components at a point P


1 2 3


 2 4 6 
3 6 1
(a) find the traction components at P on a plane whose outward unit normal has components ( 35 , 0, 54 ),
(b) find the normal and shear stresses at P on the given plane.
Solution (a) The traction components are
3
3
1 2 3


 5 

t(n) =  2 4 6   0  =  6 
13
4
3 6 1
5
5





(b) From (3.3) the normal stress is


3
3
97
4 

tn = n · t(n) =
· 6 = .
, 0,
5
5
25
13
5
From (3.4) the shear stress is
(
13
τn = |t(n)−tn n| = {|t(n)|2 −t2n } = 32 + 62 +
5
1
2
8
2
97
−
25
2 ) 12
=
√
22, 941
≈ 6.059.
25
The equations of motion
The balance of linear momentum
Figure 9: Balance of forces in the x2 -direction
Consider an elementary cuboid of material with sides δx1 , δx2 , δx3 parallel to the
coordinate axes. The cuboid has volume δv = δx1 δx2 δx3 . We calculate all the forces
∂σ22
δx2 acts on an
acting on the element in the x2 -direction. Unbalanced normal stress
∂x2
∂σ22
area δx1 δx3 , giving a normal force on the element in the x2 -direction of
δx2 ×δx1 δx3
∂x2
∂σ23
∂σ21
∂σ22
δv. Unbalanced shear stresses
δx3 and
δx1 act on areas δx1 δx2 and
=
∂x2
∂x3
∂x1
∂σ23
∂σ21
δx2 δx3 , respectively, giving shear forces in the x2 -direction of
δv and
δv ,
∂x3
∂x1
respectively. The mass of the element is ρδv and so the component of body force (per unit
mass) b2 in the x2 -direction contributes a force ρb2 δv on the element in that direction.
By Newton’s second law the total of these forces can be equated to the rate of change
of linear momentum in the x2 -direction:
∂σ22
∂σ23
∂ 2 u2
∂σ21
δv +
δv +
δv + ρb2 δv = ρδv 2 .
∂x1
∂x2
∂x3
∂t
On dividing by δv we obtain the equation of motion
∂ 2 u2
∂σ21 ∂σ22 ∂σ23
+
+
+ ρb2 = ρ 2 .
∂x1
∂x2
∂x3
∂t
By similar arguments in the x1 - and x3 -directions we deduce a complete set of three
equations of motion:
∂σ11 ∂σ12 ∂σ13
∂ 2 u1
+
+
+ ρb1 = ρ 2 ,
∂x1
∂x2
∂x3
∂t
9
∂σ21 ∂σ22 ∂σ23
∂ 2 u2
+
+
+ ρb2 = ρ 2 ,
∂x1
∂x2
∂x3
∂t
(3.5)
∂ 2 u3
∂σ31 ∂σ32 ∂σ33
+
+
+ ρb3 = ρ 2 .
∂x1
∂x2
∂x3
∂t
These equations are conveniently written in suffix notation as
∂σij
∂ 2 ui
+ ρbi = ρ 2
∂xj
∂t
(3.6)
for each i = 1, 2, 3, and summing over j.
In the static case, in which no quantities vary with time t then (3.6) reduce to the
equilibrium equations
∂σij
+ ρbi = 0.
(3.7)
∂xj
The balance of angular momentum
Figure 10: Shear stresses
Taking moments about the origin O shows that
σ21 = σ12 .
Similarly, we obtain
σ31 = σ13 ,
σ23 = σ32 .
So the balance of angular momentum reduces to the
symmetry of the stress tensor:
σ T = σ,
σji = σij .
(3.8)
(Recall also the symmetry of the strain tensor e).
Properties of the stress tensor, σ
It follows from this symmetry that σ has three real eigenvalues with corresponding real
unit eigenvectors. These three eigenvalues are termed the principal stresses and the
corresponding unit eigenvectors are termed the principal axes of stress. They are
mutually orthogonal.
10
If the stress has diagonal form


σ1 0 0


 0 σ2 0 
0 0 σ3
then the principal stresses are σ1 , σ2 , σ3
e1 , e2 , e3 , along the coordinate axes. The
(n1 , n2 , n3 ) is

σ1 0 0

t(n) =  0 σ2 0
0 0 σ3
with corresponding principal axes of stress
traction on a plane with normal components
The normal stress is therefore, from (3.3),




σ1 n1
n1




  n2  =  σ2 n2  .
σ3 n3
n3
tn = σ1 n21 + σ2 n22 + σ3 n23
(3.9)
and the squared shear stress can, from (3.4), be shown to be given by
τn2 = (σ1 − σ2 )2 n21 n22 + (σ1 − σ3 )2 n21 n23 + (σ2 − σ3 )2 n22 n23 .
(3.10)
If the stress takes the form
σ = −pI,
σij = −pδij
in which δij denotes the Kronecker delta
δij =
(
1 i = j,
0 i=
6 j,
in other words the components of the unit tensor I, it is said to be spherical (or
hydrostatic) and p is the pressure. For a surface segment with unit normal n
t(n) = σn = −pIn = −pn,
ti = −pni ,
so that the normal stress and shear stress are, from (3.3) and (3.4),
tn = (−pn) · n = −pn · n = −p and τn = 0.
If the stress is spherical, therefore, the traction is purely normal on every surface element,
so that there is no shear stress on any surface element.
Example 4 For the matrix of stress components of Example 3, namely,


1 2 3

(σij ) =  2 4 6 
,
3 6 1
find the principal stresses and the corresponding principal axes of stress.
11
Solution Let σ denote the eigenvalues and n the eigenvectors of the stress tensor σ.
Then, by definition, σn = σn, so that (σ − σI)n = 0, or, in components,





0
n1
σ11 − σ
σ12
σ13





n
σ
σ
−
σ
σ
 2  =  0 .

21
22
23
0
n3
σ31
σ32
σ33 − σ
There are non-trivial solutions (i.e. n 6= 0) if and only
vanishes:
σ −σ
σ12
σ13
11
σ21
σ22 − σ
σ23
σ31
σ32
σ33 − σ
In the present example this condition becomes
Expand by first row:
1−σ
2
3
2
4−σ
6
3
6
1−σ
if the determinant of coefficients
= 0.
= 0.
(1 − σ){(4 − σ)(1 − σ) − 36} − 2{2(1 − σ) − 18} + 3{12 − 3(4 − σ)} = 0.
Simplify the curly brackets:
(1 − σ)(σ 2 − 5σ − 32) − 2(−2σ − 16) + 9σ = 0.
Multiply out and collect terms:
−σ 3 + 6σ 2 + 40σ = 0.
Change overall sign and observe that σ = 0 is a root:
σ(σ 2 − 6σ − 40) = 0,
and factorise: σ(σ − 10)(σ + 4) = 0.
Therefore the principal stresses are
σ1 = 0,
σ2 = 10,
σ3 = −4.
We must now find the corresponding unit eigenvectors.
For σ = σ1 = 0 we must find components ni such that





0
n1
1 2 3





 2 4 6   n2  =  0  .
0
n3
3 6 1
2nd equation same as 1st. 1st and 3rd give n3 = 0, and then n1 = 2, n2 = −1 are
1
possible solutions. The unit vector n1 therefore has components 5− 2 (2, −1, 0).
1
For σ = σ2 = 10 we find the unit vector n2 has components 70− 2 (3, 6, 5).
1
For σ = σ3 = −4 we find the unit vector n3 has components 14− 2 (1, 2, −3).
12
4
Linear isotropic elasticity: constitutive equations
and homogeneous deformations
The constitutive equation of isotropic linear elasticity (the stress-strain law) is
σij = λepp δij + 2µeij .
(4.1)
known as the generalized Hooke’s law, with eij defined by (2.4). The material constants (or elastic moduli) λ and µ are known as the Lamé moduli and must be
measured experimentally for each material.
In the absence of body force the equations of equilibrium (3.7) are
∂σij
= 0.
∂xj
(4.2)
Homogeneous deformations
A deformation is said to be homogeneous if the strain tensor e is independent of x, i.e.
spatially uniform. From (4.1) the stress tensor σ is also uniform in space. Thus (4.2)
is satisfied trivially. Therefore, any homogeneous deformation satisfies the equations of
equilibrium and so is possible in an elastic material, in the absence of body force.
We examine various examples of homogeneous deformation.
(a) Uniform dilatation
An isotropic elastic sphere of radius a centred on the origin undergoes the uniform
dilatation
u1 = αx1 , u2 = αx2 , u3 = αx3 , (or u = αx),
(4.3)
where α is a dimensionless constant; α > 0 for expansion and α < 0 for contraction.
Now


!
α 0 0
∂ui


= 0 α 0 
∂xj
0 0 α
and so
eij = αδij .
The dilatation is ∆ = epp = 3α, and from (4.1),
σij = 3αλδij + 2µαδij = α(3λ + 2µ)δij
so
σ11 = σ22 = σ33 = −p, where p = −α(3λ + 2µ), with all shear stresses vanishing.
We have
p = −K∆
where
K :=
1
σ
3 pp
epp
,
or K =
is the bulk modulus of the material.
13
−p
2
=λ+ µ
∆
3
(4.4)
(b) Simple shear
The displacements are
u1 = γx2 ,
so that
So from (4.1),

u2 = 0,
0
 1
(eij ) =  2 γ
0
1
γ
2

0
0 

0
0
0
u3 = 0,
and epp = 0.
Figure 11: Simple shear


0 γµ 0


σij = 2µeij =  γµ 0 0  .
0 0 0
Therefore, σ12 = γµ, e12 = 21 γ
The quantity
are the only non-zero stresses and strains.
σ12
2e12
is known as the shear modulus of the material.
µ=
(c) Simple extension (uniaxial tension)
Suppose that a cylindrical rod lies with its generators parallel to the x1 -direction and
suffers a uniform tension T at each end:


T 0 0


(σij ) =  0 0 0  .
0 0 0
We assume a displacement
u1 = αx1 ,
u2 = −βx2 ,
u3 = −βx3 ,
where α > 0 corresponds to extension of the rod and β > 0 to lateral contraction. So


α 0
0


(eij ) =  0 −β 0  ,
0 0 −β
14
epp = α − 2β.
Figure 12: Uniaxial tension
From (4.1)
σ11 = T = λ(α − 2β) + 2µα,
σ22 = σ33 = 0 = λ(α − 2β) − 2µβ.
From the second equation we get β and then from the first we get T :
β=
αλ
,
2(λ + µ)
T =
αµ(3λ + 2µ)
.
λ+µ
We define the Young’s modulus E by
E :=
σ11
,
e11
so E =
T
µ(3λ + 2µ)
=
α
λ+µ
(4.5)
β
λ
=
,
α
2(λ + µ)
(4.6)
and Poisson’s ratio ν by
ν :=
−e22
,
e11
so ν =
both in terms of λ and µ. E is the tension per unit axial extension and ν is the transverse
contraction per unit axial extension.
Elastic constants
We have introduced elastic constants K, E, ν in addition to the Lamé constants λ, µ
and have provided clear mechanical interpretations of K, E, ν and µ. Any three of
λ, µ, K, E, ν can be expressed in terms of the other two by manipulating (4.4), (4.5)
and (4.6).
We shall suppose, on physical grounds, that
K > 0,
µ>0:
(4.7)
K > 0 implies that a compressive pressure produces a decrease in volume
µ > 0 implies a shear stress produces shear strain in the same direction,
not in the opposite direction.
By manipulating (4.4)–(4.6) we obtain
E=
9Kµ
,
3K + µ
ν=
15
3K − 2µ
2(3K + µ)
(4.8)
and so inequalities (4.7) imply that
1
−1 < ν < .
2
E > 0,
(4.9)
To obtain (4.9)2 , write (4.8)2 as
K
−2
µ
!
ν=
K
2 3 +1
µ
3
(4.10)
K
1
K
−→ 0, and ν −→ as
−→ ∞.
µ
2
µ
The inequalities (4.7) imply that
and observe that ν −→ −1 as
(a) pressure produces a decrease in volume in dilatation (K > 0)
(b) the shear is in the same direction as the shear stress
in simple shear (µ > 0)
(c) axial tension results in axial elongation in simple extension (E > 0)
Thus inequalities (4.7) ensure physically reasonable response of the material in the three
homogeneous deformations discussed above. However, (4.9)2 permits the possibility of
axial extension being accompanied by lateral expansion (ν < 0). But no known isotropic
1
material responds to simple extension in this way. Therefore in practice 0 < ν < .
2
The strain-stress law. Deviatoric tensors
The stress-strain law (4.1) is repeated here for convenience:
σij = λepp δij + 2µeij .
Taking the trace of both sides gives
σpp = (3λ + 2µ)epp .
Thus
σij =
λ
σkk δij + 2µeij
3λ + 2µ
(4.11)
and so
(
λ
1
−
σpp δij + σij
eij =
2µ
3λ + 2µ
)
(4.12)
which is the strain-stress law, inverse to (4.1).
The deviatoric stress and deviatoric strain are defined by
1
1
e′ij = eij − epp δij ,
σij′ = σij − σpp δij ,
3
3
respectively, and have the property that their traces vanish:
′
σpp
= 0,
e′pp = 0.
(4.13)
(4.14)
On substituting for σij and eij from (4.13) into (4.1) we find that the stress-strain law
(4.1) may be written
σpp = 3Kepp ,
σij′ = 2µe′ij ,
(4.15)
involving the bulk modulus K and the shear modulus µ.
16
From (4.15)1 we may recover our previous definition (4.4) of bulk modulus:
K=
1
σ
3 pp
epp
.
Equation (4.15)2 makes clear the proportionality of shear stresses and strains, connected
by the shear modulus µ.
Incompressibility
An incompressible material is one whose volume cannot be changed, though it can be
changed in shape, i.e. distorted.
In an incompressible material, therefore, every deformation is such that the dilatation
vanishes:
∆ = epp = 0.
The uniform dilatation (4.3), i.e. u = αx, cannot occur in an incompressible material
as epp = 0 implies α = 0.
For arbitrary hydrostatic pressure p there is no change of volume and so the stressstrain law (4.1) is replaced by
σij = −pδij + 2µeij
(4.16)
in which p(x, t) is an arbitrary pressure, not dependent on the strains eij .
The bulk modulus K is defined at (4.4) by
2
−p
= λ + µ.
K=
epp
3
In the limit of incompressibility epp −→ 0 and so
K −→ ∞,
λ −→ ∞
(4.17)
and µ is unaltered.
For the simple shear discussed previously, epp = 0 and so simple shear is possible in
any incompressible material.
For the uniaxial tension considered before:
u1 = αx1 ,
u2 = −βx2 ,
u3 = −βx3
to be possible in an incompressible material the dilatation must vanish, so that
β
1
epp = α − 2β = 0 ⇒
= .
α
2
But Poisson’s ratio is given by (4.6) to be
ν=
β
1
−e22
= = ,
e11
α
2
so that
1
2
for all incompressible materials. We can see this another way — from (4.6)
ν=
ν=
λ
2(λ + µ)
and for incompressibility λ −→ ∞, so that ν −→ 21 , as at (4.18).
17
(4.18)
5
Conservation of energy: the strain energy function
The equations of motion (3.6) are
∂ 2 ui
∂σij
+ ρbi = ρ 2 .
∂xj
∂t
(5.1)
Multiply by the velocity vi = ∂ui /∂t and sum over i:
∂σij
∂vi
vi + ρbi vi = ρ
vi .
∂xj
∂t
Thus
∂vi
1 ∂
∂
(σij vi ) − σij
+ ρbi vi = ρ (vi vi ).
∂xj
∂xj
2 ∂t
Now
(5.2)
∂vi
∂ ∂ui
∂ ∂ui
∂eij ∂ωij
=
=
=
+
,
∂xj
∂xj ∂t
∂t ∂xj
∂t
∂t
where the strain tensor e and rotation ω are defined by
1
eij =
2
∂ui ∂uj
+
∂xj
∂xi
Then
!
1
ωij =
2
,
∂vi
= σij
σij
∂xj
∂eij ∂ωij
+
∂t
∂t
∂ui
∂uj
−
∂xj
∂xi
!
!
.
,
∂ω
But ω is skew-symmetric, i.e. ω T = −ω or ωji = −ωij , so
also is skew-symmetric
∂t
∂ωij
and it follows that σij
= 0.
∂t
Then (5.2) may be written
∂eij 1 ∂ 2
∂
(σij vi ) + ρbi vi = σij
+ ρ (v ).
∂xj
∂t
2 ∂t
Integrate over an arbitrary sub-region Rt of the body Bt and use the divergence theorem:
Z
∂Rt
σij vi nj da +
Z
Rt
ρb · v dv =
Z
Rt
σij
Z
∂eij
1 ∂ 2
dv +
ρ (v ) dv.
∂t
∂t
Rt 2
But t(n) = σn and ρ is assumed constant to this order:
∂
∂eij
dv +
t(n) · v da +
ρb · v dv =
σij
∂t
∂t
∂Rt
Rt
Rt
Z
Z
Z
The terms of this equation are interpreted as follows:
1st term: rate of working of surface tractions
2nd term: rate of working of body forces
3rd term: stress power
4th term: rate of change of kinetic energy
18
Z
Rt
1 2
ρ v dv.
2
(5.3)
The stress power per unit volume is defined by
P := σij
∂eij
.
∂t
(5.4)
Suppose this is wholely derived as the rate of change of a single function W (e), the
strain energy, measured per unit volume, which depends on the strain e:
∂W
∂eij
= σij
.
∂t
∂t
Now by the chain rule
so that
∂W ∂eij
∂eij
= σij
∂eij ∂t
∂t
∂W
− σij
∂eij
!
∂eij
= 0.
∂t
∂eij
may be selected arbitrarily,
But at each position x and corresponding value of e,
∂t
∂eij
so since the brackets do not depend on
, we may conclude:
∂t
σij =
∂W
.
∂eij
(5.5)
Thus W (e) is a potential function for the stress σ. It is often termed the potential
energy. These ideas carry over into nonlinear elasticity.
The strain energy in isotropic linear elasticity
From (4.1) we see that the stress components σij are a linear combination of the strain
components eij . So from (5.5), W (e) must be a homogeneous quadratic in the strain
components.
From Euler’s result on the partial differentiation of a homogeneous function of degree n:
M
X
∂φ
n
xi
= nφ]
[Namely, if φ(λxi ) = λ φ(xi ) then
∂xi
i=1
∂W
eij = 2W
∂eij
so from (5.5)
1
(5.6)
W = σij eij .
2
Using the generalized Hooke’s law (4.1) allows the strain energy to be expressed entirely
in terms of the strain components eij :
1
W = λ(epp )2 + µeij eij .
2
19
(5.7)
Written out in full
1
W = λ(e11 + e22 + e33 )2 + µ{e211 + e222 + e233 + 2e212 + 2e223 + 2e231 }.
2
(5.8)
On physical grounds we expect the quadratic form (5.7) to be positive definite:
W (e) ≥ 0,
with W (e) = 0 only if e = 0.
(5.9)
Any straining of the body (e 6= 0) requires work to be done on the body (W > 0):
W < 0 for some e 6= 0 would imply that starting from an unstressed state of rest work
could spontaneously be done on the surroundings.
We now seek necessary and sufficient conditions on λ and µ for W (e) given by (5.8)
to be positive definite. From (4.4) the bulk modulus is K = λ + 23 µ and elimination of
λ from (5.8) in favour of K gives, after much manipulation,
o
1
1 n
W = K(e11 + e22 + e33 )2 + µ (e11 − e22 )2 + (e22 − e33 )2 + (e33 − e11 )2
2
3
o
n
+ 2µ e212 + e223 + e231 .
(5.10)
For the uniform dilatation e11 = e22 = e33 = α, eij = 0 for i 6= j, (5.10) becomes
W =
9K 2
α ,
2
so that W > 0 for α 6= 0 requires K > 0.
For the simple shear e12 = e21 = γ/2, all other eij vanishing, (5.10) becomes
W =
µ 2
γ ,
2
so that W > 0 for γ 6= 0 requires µ > 0. Therefore, necessary conditions for the
positive definiteness of W are
K > 0,
µ > 0.
(5.11)
These are also sufficient as they ensure that, from (5.10), W is expressed as a sum of
squares with positive coefficients such that W (e) = 0 requires e = 0.
But conditions (5.11) are precisely those adopted before, on different physical grounds,
see (4.7) and the following text.
At (4.13)2 we introduced the strain deviator
1
e′ij = eij − epp δij
3
e′pp = 0,
(5.12)
in terms of which it may be shown that the strain energy W , taken in the form (5.7),
may be written
1
W = K(epp )2 + µe′ij e′ij .
(5.13)
2
Conditions (5.11) may also be deduced from this form of W .
20
6
The torsion problem
We consider an isotropic elastic cylinder of length l and arbitrary cross-section, placed
with its generators parallel to the x3 -axis and with the origin 0 in one end face, as shown
in Figure 12. The cylinder is twisted about 0x3 by an amount α per unit length by the
application of shear tractions to its end faces, the curved surface remaining stress free.
The total angle of twist, αl, is sufficiently small for its square to be neglected.
Figure 12: Cylinder under torsion
Figure 13: The cross-section S
Figure 13 gives a plan view of the typical right cross-section S indicated in Figure 12.
If P is a typical point of S and 0′ the point of intersection of 0x3 with S, the deformation
rotates the line 0′ P anticlockwise through an angle αx3 , as shown. The components
of the displacement of P in the x1 - and x2 -directions are therefore, with r = 0′ P ,
x1 = r cos θ and x2 = r sin θ,
u1 = r cos(θ + αx3 ) − r cos θ
= r cos θ cos αx3 − r sin θ sin αx3 − r cos θ
2
= −r sin θ (αx3 ) + O(αx3 ) = −αx2 x3 ,
u2 = r sin(θ + αx3 ) − r sin θ
= r cos θ (αx3 ) + O(αx3 )2
= r sin θ cos αx3 + r cos θ sin αx3 − r sin θ
= αx1 x3 .







(6.1)






It follows from (6.1) and (2.4) that e11 = e22 = 0. We assume that e33 = 0 also,
the torsional deformation consisting entirely of shear with no axial extension. Thus
∂u3 /∂x3 = 0, implying that u3 is a function of x1 and x2 only:
u3 = αφ(x1 , x2 ).
The function φ, specifying the axial deviation of S from its initially plane form, is
called the warping function. The displacements of the simple torsion deformation are
21
therefore
u1 = −αx2 x3 ,
u2 = αx1 x3 ,
u3 = αφ(x1 , x2 ),
(6.2)
with strain components
(eij ) =
α
2










0
0
0
0
∂φ
∂φ
− x2
+ x1
∂x1
∂x2

∂φ
− x2 
∂x1



∂φ
+ x1 

∂x2

0
satisfying epp = 0, so that from (4.1) the stress components are

(σij ) =




αµ 




0
0
0
0
∂φ
∂φ
− x2
+ x1
∂x1
∂x2



∂φ
− x2 
∂x1



∂φ
+ x1 
.
∂x2

0
The equilibrium equations (3.7) in the absence of body force are
(6.3)


∂σij
= 0.
∂xj
(6.4)
The i = 1 and i = 2 equations are satisfied trivially and the i = 3 equation gives
∂
∂x1
!
∂φ
∂
− x2 +
∂x1
∂x2
!
∂φ
∂2φ ∂2φ
+ x1 =
+
= ∇2 φ = 0,
∂x2
∂x21 ∂x22
(6.5)
the warping function thus being a harmonic function.
The outward unit normal n to C, the boundary of S, has components (n1 , n2 , 0),
see Figure 13. The components of the stress vector acting on the curved surface of the
cylinder are, from (3.2) and (6.3),
t(n) = σn or





0
n1
0
0 σ31





0
0 σ32   n2  = 
.
 0
σ31 n1 + σ32 n2
0
σ31 σ32 0
Since this surface is stress free,
σ31 n1 + σ32 n2 = 0 on C
(6.6)
and substituting for the stress components from (6.3) gives
∂φ
∂φ
n1 +
n2 = x2 n1 − x1 n2
∂x1
∂x2
or
on C,
∂φ
= x2 n1 − x1 n2 on C,
(6.7)
∂n
∂φ/∂n being the directional derivative of φ along the outward normal to C. The warping
function φ thus satisfies the Neumann problem consisting of the plane Laplace equation
(6.5) in S together with the boundary condition (6.7) on C.
22
The twisting torque
The outward unit normal to the end face x3 = l has components (0, 0, 1). The traction
vector acting on this face therefore has components σi3 and the components of the
resultant force F acting on the face are,
Fi =
ZZ
S
σi3 da.
x3 =l
From (6.3), σ13 and σ23 are independent of x3 and σ33 = 0 so
F1 =
ZZ
S
σ13 da,
F2 =
ZZ
S
σ23 da,
F3 = 0.
From the i = 3 equilibrium equation (6.4) and the stress (6.3), and its symmetry
∂σ13 ∂σ23
∂σ31 ∂σ32
+
=
+
= 0.
∂x1
∂x2
∂x1
∂x2
It follows that
and
∂σ13 ∂σ23
∂(x1 σ13 ) ∂(x1 σ23 )
+
= x1
+
∂x1
∂x2
∂x1
∂x2
!
+ σ13 = σ13 ,
∂(x2 σ13 ) ∂(x2 σ23 )
∂σ13 ∂σ23
+
= x2
+
∂x1
∂x2
∂x1
∂x2
!
+ σ23 = σ23 ,
F1 =
ZZ (
∂(x1 σ13 ) ∂(x1 σ23 )
+
da =
∂x1
∂x2
F2 =
ZZ (
∂(x2 σ13 ) ∂(x2 σ23 )
da =
+
∂x1
∂x2
S
S
)
I
x1 (σ31 n1 + σ32 n2 ) ds = 0,
)
I
x2 (σ31 n1 + σ32 n2 ) ds = 0,
C
C
using the two-dimensional divergence theorem and the boundary condition (6.6). Thus
F = 0,
and, in exactly the same way, the resultant force on the end face x3 = 0 is zero.
The torque per unit area about 0 acting on the face x3 = l is
e1 e2 e3
l
x × t(e3 ) = x1 x2
σ13 σ23 σ33
,
and so, remembering that σ33 = 0, the resultant torque M acting on the face x3 = l has
components
M1 =
M2 =
M3 =
RR
S
RR
S
RR
S
= µα
(x2 σ33 − lσ32 )|x3 =l da
(lσ31 − x1 σ33 )|x3 =l da
= −lF2 = 0,
=
lF1
= 0,
(x1 σ32 − x2 σ31 )|x3 =l da
ZZ (
S
x1
!
∂φ
∂φ
+ x1 − x2
− x2
∂x2
∂x1
23
!)
da = αD,
where
D=µ
ZZ
S
x21
+
x22
!
∂φ
∂φ
da.
− x2
+ x1
∂x2
∂x1
(6.8)
Thus M = αDe3 , where e3 is the unit vector in the direction of x3 increasing, and the
resultant torque on the end face x3 = 0 is similarly −αDe3 .
The cylinder is therefore maintained in equilibrium by equal and opposite twisting
torques of magnitude αD acting about 0x3 on the end faces. The quantity D, defined
by (6.8), is called the torsional rigidity of S. D/µ has physical dimension (length)4
and depends only on the form of the cross-section S.
The Prandtl stress function
Since φ is a plane harmonic function, there exists an analytic function f of the complex
variable z = x1 + ix2 such that
φ(x1 , x2 ) = ℜe f (z).
Let
ψ(x1 , x2 ) = ℑm f (z),
so that
f (x1 + ix2 ) = φ(x1 , x2 ) + iψ(x1 , x2 ).
Then φ and ψ are conjugate harmonic functions satisfying the Cauchy-Riemann equations
∂φ
∂ψ
∂φ
∂ψ
=
,
=−
.
(6.9)
∂x1
∂x2
∂x2
∂x1
The unit vector t tangential to C in the anticlockwise sense has components
t1 = −n2 ,
t2 = n1 ,
t3 = 0,
(6.10)
see Figure 13. Using (6.9) and (6.10) we can rewrite (6.7) as follows:
∂φ
∂φ
n1 +
n2 = x2 n1 − x1 n2 on C,
∂x1
∂x
! 2
∂ψ
∂ψ
t2 + −
(−t1 ) = x2 t2 − x1 (−t1 ) on C,
∂x2
∂x1
∂ψ
∂ψ
t1 +
t2 = x1 t1 + x2 t2 on C.
∂x1
∂x2
Now by definition
∂ψ
∂ψ
∂ψ
t1 +
t2 = t · ∇ψ =
,
∂x1
∂x2
∂s
where s is arc length along C in the positive (anti-clockwise) sense. So (6.7) finally
reduces to
1 ∂ 2
∂ψ
=
(x + x22 ) on C.
∂s
2 ∂s 1
The Neumann problem (6.5), (6.7) is thus equivalent to the Dirichlet problem
∇2 ψ = 0 in S,
ψ =
1
(x21
2


+ x22 ) + constant on C. 
24
(6.11)
Simpler results are obtained by introducing the Prandtl stress function
1
Ψ = ψ − (x21 + x22 ).
2
(6.12)
From (6.11), Ψ satisfies the Poisson equation
∇2 Ψ = −2 in S,
(6.13)
and the boundary condition
Ψ = Ψ0
on C,
(6.14)
where Ψ0 is a constant.
In terms of Ψ, the non-zero shear stresses are given from (6.3), (6.9) and (6.12) by
!
!
∂φ
∂ψ
= µα
+ x1
= −µα
− x1
∂x2
∂x1
!
!
∂φ
∂ψ
= µα
− x2
= µα
− x2
∂x1
∂x2
σ23
σ31
∂Ψ
= −µα
∂x1
∂Ψ
= µα
,
∂x2
and the torsional rigidity from (6.8), (6.9) and (6.12) by
D=µ
ZZ
S
x21
+
x22
!






(6.15)





!
ZZ
∂ψ
∂Ψ
∂Ψ
∂ψ
da = −µ
da.
− x2
+ x2
x1
− x1
∂x1
∂x2
∂x1
∂x2
S
(6.16)
Since
∂Ψ
∂(x1 Ψ) ∂(x2 Ψ)
∂Ψ
+ x2
=
+
− 2Ψ,
∂x1
∂x2
∂x1
∂x2
the application of the two-dimensional divergence theorem gives
x1
ZZ
S
!
∂Ψ
∂Ψ
x1
+ x2
da = −2
∂x1
∂x2
ZZ
= −2
ZZ
S
S
Ψ da + µ
I
Ψ(x1 n1 + x2 n2 )ds
Ψ da + µ
I
Ψ0 (x1 n1 + x2 n2 )ds,
C
C
where (6.14) has been used. Thus
D=2
ZZ
S
Ψ da − µ
I
C
Ψ0 (x1 n1 + x2 n2 )ds.
(6.17)
If S is simply connected, so that C is a single closed curve we can replace Ψ by Ψ−Ψ0 .
Equations (6.13), (6.15) and (6.16) are unchanged and (6.14) and (6.17) simplify to
Ψ = 0 on C,
D = 2µ
ZZ
25
S
Ψ da.
(6.18)
(6.19)
Figure 14: The elliptical cylinder
Examples
1. Torsion of an elliptical cylinder
Suppose that C is an ellipse and that 0x3 is the line of centres of the right cross-sections.
If a1 and a2 are the semi-axes of the ellipse in the x1 - and x2 -directions, the equation of
C is
x21 x22
+
= 1.
a21 a22
The function
!
x21 x22
Ψ(x1 , x2 ) = A 2 + 2 − 1
a1 a2
clearly satisfies the boundary condition (6.18), i.e. vanishes on C, and
!
∂2Ψ ∂2Ψ
2
2
∇ Ψ=
+
=A 2 + 2 .
2
2
∂x1
∂x2
a1 a2
2
The Poisson equation (6.13) therefore holds if
2
2
A 2+ 2
a1 a2
!
= −2
⇒
A=−
a21 a22
,
a21 + a22
and so the Prandtl stress function for the elliptical cylinder is
a2 a2
Ψ=− 21 22
a1 + a2
!
a21 a22 − a22 x21 − a21 x22
x21 x22
+
−
1
=
.
a21 a22
a21 + a22
From (6.12), i.e. ψ = Ψ + 21 (x21 + x22 ),
a21 a22 − a22 x21 − a21 x22 1 2
+ (x1 + x22 )
a21 + a22
2
1
=
(2a2 a2 − 2a22 x21 − 2a21 x22 + a21 x21 + a21 x22 + a22 x21 + a22 x22 )
2
2(a1 + a22 ) 1 2
o
n
1
2
2
2
2
2 2
(a
−
a
)(x
−
x
)
+
2a
a
=
1
2
1
2
1
2
2(a21 + a22 )
o
n
1
= ℑm
(a21 − a22 )iz 2 + 2ia21 a22
2
2
2(a1 + a2 )
ψ(x1 , x2 ) =
26
where z = x1 + ix2 and so z 2 = x21 − x22 + 2ix1 x2 and iz 2 = i(x21 − x22 ) − 2x1 x2 . The
warping function φ is therefore
n
o
1
2
2
2
2 2
(a
−
a
)iz
+
2ia
a
1
2
1
2
2(a21 + a22 )
φ(x1 , x2 ) = ℜe
= −
a21 − a22
x1 x2 .
a21 + a22
When a1 > a2 the warping pattern is as shown in Figure 15. The curves φ = constant
are rectangular hyperbolae.
Figure 15: The warping pattern for an elliptical cylinder
The torsional rigidity D is, from (6.19),
D = 2µ
2µa2 a2
Ψ da = 2 1 22
a1 + a2
S
ZZ
ZZ
S
!
x2 x2
1 − 21 − 22 dx1 dx2 .
a1
a2
Calculation of D. Because S is elliptical we transform the double integral using
modified polar coordinates
x1 = a1 r cos θ,
x2 = a2 r sin θ.
x21 x22
+
= r 2 the limits of integration are 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.
a21 a22
The Jacobian is
Since
∂(x1 , x2 )
=
∂(r, θ)
∂x1
∂r
∂x2
∂r
∂x1
∂θ
∂x2
∂θ
=
a1 cos θ
a2 sin θ
− a1 r sin θ a2 r cos θ
= a1 a2 r(cos2 θ + sin2 θ) = a1 a2 r,
so that dx1 dx2 7→ a1 a2 rdrdθ. Then the double integral is
ZZ
S
!
x2 x2
1 − 21 − 22 dx1 dx2 =
a1
a2
Z
2πZ 1
θ=0 r=0
a2 r 2 cos2 θ a22 r 2 sin2 θ
1− 1 2
−
a1 a2 rdrdθ
a1
a22
!
27
= a1 a2
Z
2πZ 1
θ=0 r=0
Therefore
1−r
2
rdrdθ = 2πa1 a2
"
r2 r4
−
2
4
#1
=
0
πa1 a2
.
2
2µa21 a22
πa1 a2
πµa31 a32
×
=
,
a21 + a22
2
a21 + a22
the torsional rigidity of an elliptical cylinder.
D=
Torsion of a circular cylinder In the case of a circular cylinder, a1 = a2 = a and
1 2
(a − x21 − x22 ),
2
φ(x1 , x2 ) = 0,
1
πµa4 .
D =
2
Because φ = 0, there is no warping when the circular cylinder is twisted about its central
axis.
Ψ(x1 , x2 ) =
2. Torsion of a circular tube
Figure 16: The circular tube
We consider the cross-section S bounded by the concentric circles C1 , C2 of radii a
and b (< a). The axis of torsion is once again the axis of symmetry.
The function
1
Ψ(x1 , x2 ) = (a2 − x21 − x22 ),
2
obtained above for the circular cylinder, satisfies the Poisson equation (6.13), vanishes
on C1 and is constant on C2 . It is therefore an acceptable Prandtl stress function for
the circular tube and, as before, φ = 0, so that the right cross-sections are unwarped.
Because the boundary of the tube consists of disjoint curves we cannot use the fomula
(6.19) for the torsional rigidity. Instead, we obtain from (6.16) the expression
D = −µ
ZZ S
−x21
−
x22
da = µ
1
= πµ(a4 − b4 )
2
for the torsional rigidity of a circular tube.
28
Z
b
a
1
r .2πrdr = 2πµ r 4
4
2
a
b