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. achange of shape) can be sustained only if a system of forces (loads) is applied to thebounding surfaces, so setting up internally a distribution of stress (i.e. force per unitarea). The defining characteristic of an elastic material is that when these loads areremoved the body returns exactly to its original shape. The original state in which noloads are applied is often called the natural state, or reference configuration.

Almost all engineering materials possess the property of elasticity provided the ex-ternal loads are not too large. If the loads are increased beyond a certain limit, differentfor each material, then the material will fail, either by fracture or by flow. In neithercase does the material return to its original shape when the loads are removed and wesay that the elastic limit has been exceeded. A solid material that fails by fracture issaid 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 andcontinuously distributed over the material body, so that the smallest part of the bodypossesses the same physical properties as the whole body. The theory of the mechanicalbehaviour of such materials is called continuum mechanics. We consider thereforeonly the macroscopic (large scale) behaviour of materials, which is adequate for mostengineering purposes, and ignore the microscopic (small scale) behaviour.

We further assume that the deformation is small so that the elastic material is lin-early elastic. For almost all engineering materials the linear theory of elasticity holdsif 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 forcescausing it.

A material body B occupies a region B0 of space at time t = 0, its reference config-uration, and a material particle has position vector X, with Cartesian componentsXi, i = 1, 2, 3, with respect to the orthonormal vectors {e1, e2, e3}:

X =3∑

i=1

Xi ei.

1

Figure 1: Reference configuration Current configuration

The body undergoes a motion, or deformation, during which the material particle atX when t = 0 is moved to

x = x(X, t) (2.1)

at time t and then has position vector

x =3∑

i=1

xi ei,

with Cartesian components xi, i = 1, 2, 3. Using the Einstein summation conven-tion, by which twice-occurring subscripts are summed over (from 1 to 3), we can write

X = Xi ei, x = xi ei.

We writex = 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. Thedisplacement 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 smallbut 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 placeof (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

v =∂x

∂t

∣

∣

∣

∣

∣

X

≈ ∂u

∂t

∣

∣

∣

∣

∣

x

,

and the particle acceleration is similarly approximated by

a =∂2x

∂t2

∣

∣

∣

∣

∣

X

≈ ∂2u

∂t2

∣

∣

∣

∣

∣

x

.

The deformation enters into the linear theory of elasticity only through the linearstrain tensor, e, which has components

eij =1

2

(

∂ui

∂xj

+∂uj

∂xi

)

(2.4)

(i, j = 1, 2, 3). For example,

e11 =1

2

(

∂u1

∂x1

+∂u1

∂x1

)

=∂u1

∂x1

and e12 =1

2

(

∂u1

∂x2

+∂u2

∂x1

)

.

A tensor is a linear map from one vector space to another; but all such maps canbe 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):

eji =1

2

(

∂uj

∂xi

+∂ui

∂xj

)

=1

2

(

∂ui

∂xj

+∂uj

∂xi

)

= eij .

The matrix (eij) of strain components is therefore real and symmetric and so, by thetheory of linear algebra, has 3 real eigenvalues with corresponding unit eigenvectorswhich are mutually orthogonal. The eigenvalues are known as the principal strainsand 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 = e1x1, u2 = e2x2, u3 = e3x3, (2.5)

independent of time t, in which ei are the constant (small) strains. Note that ei > 0corresponds to a stretch, and ei < 0 to a contraction.

From (2.5) we see that the strain components are given by

e11 =∂u1

∂x1

= e1, e22 =∂u2

∂x2

= e2, e33 =∂u3

∂x3

= e3,

with all the other components vanishing: e12 = e21 = e23 = e32 = e31 = e13 = 0. Thus

(eij) =

e1 0 00 e2 00 0 e3

3

Figure 2: Triaxial stretch

and we see that the principal strains are given by e1, e2, e3. Also, the principal axes ofstrain are aligned with the coordinate axes of Figure 2 and have components

100

,

010

,

001

,

respectively.After the triaxial stretch the new volume is

(1 + e1)(1 + e2)(1 + e3) = 1 + e1 + e2 + e3 + e1e2 + e2e3 + e3e1 + e1e2e3.

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 bediagonalized (because symmetric) by a suitable rotation of coordinates and then in thenew coordinates the deformation is necessarily of the form (2.5). So the dilatation isgiven by (2.6). But tr e = ekk is an invariant quantity, unchanged by any rotation ofcoordinates. 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 theinitial and final densities:

ρ0 =m

1, ρ =

m

1 + ∆⇒ ρ = ρ0(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 sheardeformation

u1 = γx2, u2 = 0, u3 = 0, (2.8)

with γ a positive constant.

Figure 3: Simple shear

Then∂u1

∂x2

= γ, with all other components of deformation gradient vanishing. So

e12 =1

2

(

∂u1

∂x2

+∂u2

∂x1

)

=1

2(γ + 0) =

γ

2= e21,

with all other strain components vanishing:

(eij) =

0 1

2γ 0

1

2γ 0 00 0 0

.

For general strain tensor e the diagonal components e11, e22, e33 are termed thenormal strains, whereas the off-diagonal elements e12 = e21, e23 = e32, e31 = e13, aretermed 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 onthe 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) lateralcontraction. Now

t(e1) = Te1, but t(e

2) = 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) = Te1, so t(n) is not necessarily parallel

to n.

6

Consider a cuboid of deformed material with sides δx1, δx2, δx3 parallel to the coor-dinate axes. The tractions on each face are as shown.

Figure 7: The stress components

The traction vector t(e1) on the face with normal e1 has components

σ11

σ21

σ31

,

the traction vector t(e2) on the face with normal e2 has components

σ12

σ22

σ32

,

the traction vector t(e3) on the face with normal e3 has components

σ13

σ23

σ33

.

These column vectors are put together to form the 3 × 3 matrix of components of thestress tensor, σ:

(σij) =

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

. (3.1)

The diagonal elements σ11, σ22, σ33 are termed normal stresses and the off-diagonalelements are termed shear stresses.

It can be shown that for arbitrary unit normal vector n = n1e1 +n2e2 +n3e3 = niei,the traction vector is given by

t(n) = σn, ti = σijnj ,

t1t2t3

=

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

n1

n2

n3

, (3.2)

so the traction components ti are a linear combination of the unit surface normal com-ponents ni.

7

The normal component of traction is

tn = t(n) · n = tini = σijnjni = σijninj , (3.3)

from (3.2). The normal stress tn is tensile when positive and compressive whennegative.

Figure 8: Normal and shear components of traction

The tangential component of traction is

t(n) − tnn

with magnitudeτn = |t(n) − tnn| = {|t(n)|2 − t2n}

1

2 , (3.4)

called the shear stress.

Example 3 For the stress components at a point P

1 2 32 4 63 6 1

(a) find the traction components at P on a plane whose outward unit normal has com-ponents (3

5, 0, 4

5),

(b) find the normal and shear stresses at P on the given plane.

Solution (a) The traction components are

t(n) =

1 2 32 4 63 6 1

3

5

04

5

=

3613

5

(b) From (3.3) the normal stress is

tn = n · t(n) =(

3

5, 0,

4

5

)

·

3613

5

=97

25.

From (3.4) the shear stress is

τn = |t(n)−tnn| = {|t(n)|2−t2n}1

2 =

{

32 + 62 +(

13

5

)2

−(

97

25

)2} 1

2

=

√22, 941

25≈ 6.059.

8

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 thecoordinate axes. The cuboid has volume δv = δx1δx2δx3. We calculate all the forces

acting on the element in the x2-direction. Unbalanced normal stress∂σ22

∂x2

δx2 acts on an

area δx1δx3, giving a normal force on the element in the x2-direction of∂σ22

∂x2

δx2×δx1δx3

=∂σ22

∂x2

δv. Unbalanced shear stresses∂σ23

∂x3

δx3 and∂σ21

∂x1

δx1 act on areas δx1δx2 and

δx2δx3, respectively, giving shear forces in the x2-direction of∂σ23

∂x3

δv and∂σ21

∂x1

δv ,

respectively. The mass of the element is ρδv and so the component of body force (per unitmass) b2 in the x2-direction contributes a force ρb2δv on the element in that direction.

By Newton’s second law the total of these forces can be equated to the rate of changeof linear momentum in the x2-direction:

∂σ21

∂x1

δv +∂σ22

∂x2

δv +∂σ23

∂x3

δv + ρb2δv = ρδv∂2u2

∂t2.

On dividing by δv we obtain the equation of motion

∂σ21

∂x1

+∂σ22

∂x2

+∂σ23

∂x3

+ ρb2 = ρ∂2u2

∂t2.

By similar arguments in the x1- and x3-directions we deduce a complete set of threeequations of motion:

∂σ11

∂x1

+∂σ12

∂x2

+∂σ13

∂x3

+ ρb1 = ρ∂2u1

∂t2,

9

∂σ21

∂x1

+∂σ22

∂x2

+∂σ23

∂x3

+ ρb2 = ρ∂2u2

∂t2, (3.5)

∂σ31

∂x1

+∂σ32

∂x2

+∂σ33

∂x3

+ ρb3 = ρ∂2u3

∂t2.

These equations are conveniently written in suffix notation as

∂σij

∂xj

+ ρbi = ρ∂2ui

∂t2(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

∂xj

+ ρbi = 0. (3.7)

The balance of angular momentum

Figure 10: Shear stresses

Taking moments about the origin O shows that

σ21 = σ12.

Similarly, we obtainσ31 = σ13, σ23 = σ32.

So the balance of angular momentum reduces to thesymmetry of the stress tensor:

σT = 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 realunit eigenvectors. These three eigenvalues are termed the principal stresses and thecorresponding unit eigenvectors are termed the principal axes of stress. They aremutually orthogonal.

10

If the stress has diagonal form

σ1 0 00 σ2 00 0 σ3

then the principal stresses are σ1, σ2, σ3 with corresponding principal axes of stresse1, e2, e3, along the coordinate axes. The traction on a plane with normal components(n1, n2, n3) is

t(n) =

σ1 0 00 σ2 00 0 σ3

n1

n2

n3

=

σ1n1

σ2n2

σ3n3

.

The normal stress is therefore, from (3.3),

tn = σ1n2

1+ σ2n

2

2+ σ3n

2

3(3.9)

and the squared shear stress can, from (3.4), be shown to be given by

τ 2

n = (σ1 − σ2)2n2

1n2

2+ (σ1 − σ3)

2n2

1n2

3+ (σ2 − σ3)

2n2

2n2

3. (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 (orhydrostatic) 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,

(σij) =

1 2 32 4 63 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,

σ11 − σ σ12 σ13

σ21 σ22 − σ σ23

σ31 σ32 σ33 − σ

n1

n2

n3

=

000

.

There are non-trivial solutions (i.e. n 6= 0) if and only if the determinant of coefficientsvanishes:

∣

∣

∣

∣

∣

∣

∣

σ11 − σ σ12 σ13

σ21 σ22 − σ σ23

σ31 σ32 σ33 − σ

∣

∣

∣

∣

∣

∣

∣

= 0.

In the present example this condition becomes

∣

∣

∣

∣

∣

∣

∣

1 − σ 2 32 4 − σ 63 6 1 − σ

∣

∣

∣

∣

∣

∣

∣

= 0.

Expand by first row:

(1 − σ){(4 − σ)(1 − σ) − 36} − 2{2(1 − σ) − 18} + 3{12 − 3(4 − σ)} = 0.

Simplify the curly brackets:

(1 − σ)(σ2 − 5σ − 32) − 2(−2σ − 16) + 9σ = 0.

Multiply out and collect terms:

−σ3 + 6σ2 + 40σ = 0.

Change overall sign and observe that σ = 0 is a root:

σ(σ2 − 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

1 2 32 4 63 6 1

n1

n2

n3

=

000

.

2nd equation same as 1st. 1st and 3rd give n3 = 0, and then n1 = 2, n2 = −1 arepossible solutions. The unit vector n1 therefore has components 5−

1

2 (2,−1, 0).

For σ = σ2 = 10 we find the unit vector n2 has components 70−1

2 (3, 6, 5).

For σ = σ3 = −4 we find the unit vector n3 has components 14−1

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 = λeppδij + 2µeij. (4.1)

known as the generalized Hooke’s law, with eij defined by (2.4). The material con-stants (or elastic moduli) λ and µ are known as the Lame moduli and must bemeasured experimentally for each material.

In the absence of body force the equations of equilibrium (3.7) are

∂σij

∂xj

= 0. (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 ofequilibrium 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 uniformdilatation

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

(

∂ui

∂xj

)

=

α 0 00 α 00 0 α

and soeij = 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

3σpp

epp

, or K =−p∆

= λ+2

3µ (4.4)

is the bulk modulus of the material.

13

(b) Simple shear

The displacements are

u1 = γx2, u2 = 0, u3 = 0,

so that

(eij) =

0 1

2γ 0

1

2γ 0 00 0 0

and epp = 0.

So from (4.1),

Figure 11: Simple shear

σij = 2µeij =

0 γµ 0γµ 0 00 0 0

.

Therefore, σ12 = γµ, e12 = 1

2γ are the only non-zero stresses and strains.

The quantity

µ =σ12

2e12

is known as the shear modulus of the material.

(c) Simple extension (uniaxial tension)

Suppose that a cylindrical rod lies with its generators parallel to the x1-direction andsuffers a uniform tension T at each end:

(σij) =

T 0 00 0 00 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

(eij) =

α 0 00 −β 00 0 −β

, epp = α− 2β.

14

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

α=µ(3λ+ 2µ)

λ+ µ(4.5)

and Poisson’s ratio ν by

ν :=−e22e11

, so ν =β

α=

λ

2(λ+ µ), (4.6)

both in terms of λ and µ. E is the tension per unit axial extension and ν is the transversecontraction per unit axial extension.

Elastic constants

We have introduced elastic constants K, E, ν in addition to the Lame constants λ, µand have provided clear mechanical interpretations of K, E, ν and µ. Any three ofλ, 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 volumeµ > 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 =

3K − 2µ

2(3K + µ)(4.8)

15

and so inequalities (4.7) imply that

E > 0, −1 < ν <1

2. (4.9)

To obtain (4.9)2, write (4.8)2 as

ν =3K

µ− 2

2

(

3K

µ+ 1

) (4.10)

and observe that ν −→ −1 asK

µ−→ 0, and ν −→ 1

2as

K

µ−→ ∞.

The inequalities (4.7) imply that

(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 threehomogeneous deformations discussed above. However, (4.9)2 permits the possibility ofaxial extension being accompanied by lateral expansion (ν < 0). But no known isotropic

material responds to simple extension in this way. Therefore in practice 0 < ν <1

2.

The strain-stress law. Deviatoric tensors

The stress-strain law (4.1) is repeated here for convenience:

σij = λeppδij + 2µeij.

Taking the trace of both sides gives

σpp = (3λ+ 2µ)epp. (4.11)

Thus σij =λ

3λ+ 2µσkk δij + 2µeij and so

eij =1

2µ

{

− λ

3λ+ 2µσpp δij + σij

}

(4.12)

which is the strain-stress law, inverse to (4.1).The deviatoric stress and deviatoric strain are defined by

σ′

ij = σij −1

3σpp δij, e′ij = eij −

1

3epp δij , (4.13)

respectively, and have the property that their traces vanish:

σ′

pp = 0, e′pp = 0. (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, σ′

ij = 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

3σpp

epp

.

Equation (4.15)2 makes clear the proportionality of shear stresses and strains, connectedby the shear modulus µ.

Incompressibility

An incompressible material is one whose volume cannot be changed, though it can bechanged in shape, i.e. distorted.

In an incompressible material, therefore, every deformation is such that the dilatationvanishes:

∆ = epp = 0.

The uniform dilatation (4.3), i.e. u = αx, cannot occur in an incompressible materialas epp = 0 implies α = 0.

For arbitrary hydrostatic pressure p there is no change of volume and so the stress-strain 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

K =−pepp

= λ+2

3µ.

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

epp = α− 2β = 0 ⇒ β

α=

1

2.

But Poisson’s ratio is given by (4.6) to be

ν =−e22e11

=β

α=

1

2,

so that

ν =1

2(4.18)

for all incompressible materials. We can see this another way — from (4.6)

ν =λ

2(λ+ µ)

and for incompressibility λ −→ ∞, so that ν −→ 1

2, as at (4.18).

17

5 Conservation of energy: the strain energy func-

tion

The equations of motion (3.6) are

∂σij

∂xj

+ ρbi = ρ∂2ui

∂t2. (5.1)

Multiply by the velocity vi = ∂ui/∂t and sum over i:

∂σij

∂xj

vi + ρbivi = ρ∂vi

∂tvi.

Thus∂

∂xj

(σijvi) − σij

∂vi

∂xj

+ ρbivi =1

2ρ∂

∂t(vivi). (5.2)

Now∂vi

∂xj

=∂

∂xj

∂ui

∂t=

∂

∂t

∂ui

∂xj

=∂eij

∂t+∂ωij

∂t,

where the strain tensor e and rotation ω are defined by

eij =1

2

(

∂ui

∂xj

+∂uj

∂xi

)

, ωij =1

2

(

∂ui

∂xj

− ∂uj

∂xi

)

.

Then

σij

∂vi

∂xj

= σij

(

∂eij

∂t+∂ωij

∂t

)

,

But ω is skew-symmetric, i.e. ωT = −ω or ωji = −ωij , so

∂ω

∂talso is skew-symmetric

and it follows that σij

∂ωij

∂t= 0.

Then (5.2) may be written

∂

∂xj

(σijvi) + ρbivi = σij

∂eij

∂t+

1

2ρ∂

∂t(v2).

Integrate over an arbitrary sub-region Rt of the body Bt and use the divergence theorem:

∫

∂Rt

σijvinj da+∫

Rt

ρb · v dv =∫

Rt

σij

∂eij

∂tdv +

∫

Rt

1

2ρ∂

∂t(v2) dv.

But t(n) = σn and ρ is assumed constant to this order:

∫

∂Rt

t(n) · v da+∫

Rt

ρb · v dv =∫

Rt

σij

∂eij

∂tdv +

∂

∂t

∫

Rt

1

2ρ v2 dv. (5.3)

The terms of this equation are interpreted as follows:1st term: rate of working of surface tractions2nd term: rate of working of body forces3rd term: stress power4th term: rate of change of kinetic energy

18

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), thestrain energy, measured per unit volume, which depends on the strain e:

∂W

∂t= σij

∂eij

∂t.

Now by the chain rule∂W

∂eij

∂eij

∂t= σij

∂eij

∂t

so that(

∂W

∂eij

− σij

)

∂eij

∂t= 0.

But at each position x and corresponding value of e,∂eij

∂tmay be selected arbitrarily,

so since the brackets do not depend on∂eij

∂t, we may conclude:

σij =∂W

∂eij

. (5.5)

Thus W (e) is a potential function for the stress σ. It is often termed the potentialenergy. 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 straincomponents eij. So from (5.5), W (e) must be a homogeneous quadratic in the straincomponents.

From Euler’s result on the partial differentiation of a homogeneous function of de-gree n:

[Namely, if φ(λxi) = λnφ(xi) thenM∑

i=1

xi

∂φ

∂xi

= nφ]

∂W

∂eij

eij = 2W

so from (5.5)

W =1

2σijeij . (5.6)

Using the generalized Hooke’s law (4.1) allows the strain energy to be expressed entirelyin terms of the strain components eij :

W =1

2λ(epp)

2 + µeijeij . (5.7)

19

Written out in full

W =1

2λ(e11 + e22 + e33)

2 + µ{e211

+ e222

+ e233

+ 2e212

+ 2e223

+ 2e231}. (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 workcould 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 + 2

3µ and elimination of

λ from (5.8) in favour of K gives, after much manipulation,

W =1

2K(e11 + e22 + e33)

2 +1

3µ{

(e11 − e22)2 + (e22 − e33)

2 + (e33 − e11)2}

+ 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

2α2,

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

2γ2,

so that W > 0 for γ 6= 0 requires µ > 0. Therefore, necessary conditions for thepositive 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 ofsquares 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

e′ij = eij −1

3epp δij 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

W =1

2K(epp)

2 + µe′ije′

ij. (5.13)

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, placedwith its generators parallel to the x3-axis and with the origin 0 in one end face, as shownin Figure 12. The cylinder is twisted about 0x3 by an amount α per unit length by theapplication 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 deformationrotates the line 0′P anticlockwise through an angle αx3, as shown. The componentsof 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 cosαx3 − r sin θ sinαx3 − r cos θ

= −r sin θ (αx3) +O(αx3)2 = −αx2x3,

u2 = r sin(θ + αx3) − r sin θ = r sin θ cosαx3 + r cos θ sinαx3 − r sin θ

= r cos θ (αx3) +O(αx3)2 = αx1x3.

(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, iscalled the warping function. The displacements of the simple torsion deformation are

21

thereforeu1 = −αx2x3, u2 = αx1x3, u3 = αφ(x1, x2), (6.2)

with strain components

(eij) =α

2

0 0∂φ

∂x1

− x2

0 0∂φ

∂x2

+ x1

∂φ

∂x1

− x2

∂φ

∂x2

+ x1 0

satisfying epp = 0, so that from (4.1) the stress components are

(σij) = αµ

0 0∂φ

∂x1

− x2

0 0∂φ

∂x2

+ x1

∂φ

∂x1

− x2

∂φ

∂x2

+ x1 0

. (6.3)

The equilibrium equations (3.7) in the absence of body force are

∂σij

∂xj

= 0. (6.4)

The i = 1 and i = 2 equations are satisfied trivially and the i = 3 equation gives

∂

∂x1

(

∂φ

∂x1

− x2

)

+∂

∂x2

(

∂φ

∂x2

+ x1

)

=∂2φ

∂x21

+∂2φ

∂x22

= ∇2φ = 0, (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 thecylinder are, from (3.2) and (6.3),

t(n) = σn or

0 0 σ31

0 0 σ32

σ31 σ32 0

n1

n2

0

=

00

σ31n1 + σ32n2

.

Since this surface is stress free,

σ31n1 + σ32n2 = 0 on C (6.6)

and substituting for the stress components from (6.3) gives

∂φ

∂x1

n1 +∂φ

∂x2

n2 = x2n1 − x1n2 on C,

or∂φ

∂n= x2n1 − x1n2 on C, (6.7)

∂φ/∂n being the directional derivative of φ along the outward normal to C. The warpingfunction 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 tractionvector acting on this face therefore has components σi3 and the components of theresultant force F acting on the face are,

Fi =∫ ∫

Sσi3

∣

∣

∣

∣

x3=l

da.

From (6.3), σ13 and σ23 are independent of x3 and σ33 = 0 so

F1 =∫ ∫

Sσ13 da, F2 =

∫ ∫

Sσ23 da, F3 = 0.

From the i = 3 equilibrium equation (6.4) and the stress (6.3), and its symmetry

∂σ31

∂x1

+∂σ32

∂x2

=∂σ13

∂x1

+∂σ23

∂x2

= 0.

It follows that

∂(x1σ13)

∂x1

+∂(x1σ23)

∂x2

= x1

(

∂σ13

∂x1

+∂σ23

∂x2

)

+ σ13 = σ13,

∂(x2σ13)

∂x1

+∂(x2σ23)

∂x2

= x2

(

∂σ13

∂x1

+∂σ23

∂x2

)

+ σ23 = σ23,

and

F1 =∫ ∫

S

{

∂(x1σ13)

∂x1

+∂(x1σ23)

∂x2

}

da =∮

Cx1(σ31n1 + σ32n2) ds = 0,

F2 =∫ ∫

S

{

∂(x2σ13)

∂x1

+∂(x2σ23)

∂x2

}

da =∮

Cx2(σ31n1 + σ32n2) ds = 0,

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

x × t(e3) =

∣

∣

∣

∣

∣

∣

∣

e1 e2 e3

x1 x2 lσ13 σ23 σ33

∣

∣

∣

∣

∣

∣

∣

,

and so, remembering that σ33 = 0, the resultant torque M acting on the face x3 = l hascomponents

M1 =∫∫

S (x2σ33 − lσ32)|x3=l da = −lF2 = 0,

M2 =∫∫

S (lσ31 − x1σ33)|x3=l da = lF1 = 0,

M3 =∫∫

S (x1σ32 − x2σ31)|x3=l da

= µα∫ ∫

S

{

x1

(

∂φ

∂x2

+ x1

)

− x2

(

∂φ

∂x1

− x2

)}

da = αD,

23

where

D = µ∫ ∫

S

(

x2

1+ x2

2+ x1

∂φ

∂x2

− x2

∂φ

∂x1

)

da. (6.8)

Thus M = αDe3, where e3 is the unit vector in the direction of x3 increasing, and theresultant torque on the end face x3 = 0 is similarly −αDe3.

The cylinder is therefore maintained in equilibrium by equal and opposite twistingtorques of magnitude αD acting about 0x3 on the end faces. The quantity D, definedby (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 complexvariable z = x1 + ix2 such that

φ(x1, x2) = ℜe f(z).

Letψ(x1, x2) = ℑm f(z),

so thatf(x1 + ix2) = φ(x1, x2) + iψ(x1, x2).

Then φ and ψ are conjugate harmonic functions satisfying the Cauchy-Riemann equa-tions

∂φ

∂x1

=∂ψ

∂x2

,∂φ

∂x2

= − ∂ψ

∂x1

. (6.9)

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

∂x1

n1 +∂φ

∂x2

n2 = x2n1 − x1n2 on C,

∂ψ

∂x2

t2 +

(

− ∂ψ

∂x1

)

(−t1) = x2t2 − x1(−t1) on C,

∂ψ

∂x1

t1 +∂ψ

∂x2

t2 = x1t1 + x2t2 on C.

Now by definition∂ψ

∂x1

t1 +∂ψ

∂x2

t2 = t · ∇ψ =∂ψ

∂s,

where s is arc length along C in the positive (anti-clockwise) sense. So (6.7) finallyreduces to

∂ψ

∂s=

1

2

∂

∂s(x2

1+ x2

2) on C.

The Neumann problem (6.5), (6.7) is thus equivalent to the Dirichlet problem

∇2ψ = 0 in S,

ψ = 1

2(x2

1+ x2

2) + constant on C.

(6.11)

24

Simpler results are obtained by introducing the Prandtl stress function

Ψ = ψ − 1

2(x2

1+ x2

2). (6.12)

From (6.11), Ψ satisfies the Poisson equation

∇2Ψ = −2 in S, (6.13)

and the boundary conditionΨ = 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

σ23 = µα

(

∂φ

∂x2

+ x1

)

= −µα(

∂ψ

∂x1

− x1

)

= −µα ∂Ψ∂x1

σ31 = µα

(

∂φ

∂x1

− x2

)

= µα

(

∂ψ

∂x2

− x2

)

= µα∂Ψ

∂x2

,

(6.15)

and the torsional rigidity from (6.8), (6.9) and (6.12) by

D = µ∫ ∫

S

(

x2

1+ x2

2− x1

∂ψ

∂x1

− x2

∂ψ

∂x2

)

da = −µ∫ ∫

S

(

x1

∂Ψ

∂x1

+ x2

∂Ψ

∂x2

)

da. (6.16)

Since

x1

∂Ψ

∂x1

+ x2

∂Ψ

∂x2

=∂(x1Ψ)

∂x1

+∂(x2Ψ)

∂x2

− 2Ψ,

the application of the two-dimensional divergence theorem gives

∫ ∫

S

(

x1

∂Ψ

∂x1

+ x2

∂Ψ

∂x2

)

da = −2∫ ∫

SΨ da+ µ

∮

CΨ(x1n1 + x2n2)ds

= −2∫ ∫

SΨ da+ µ

∮

CΨ0(x1n1 + x2n2)ds,

where (6.14) has been used. Thus

D = 2∫ ∫

SΨ da− µ

∮

CΨ0(x1n1 + x2n2)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, (6.18)

D = 2µ∫ ∫

SΨ da. (6.19)

25

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 ofC is

x2

1

a21

+x2

2

a22

= 1.

The function

Ψ(x1, x2) = A

(

x2

1

a21

+x2

2

a22

− 1

)

clearly satisfies the boundary condition (6.18), i.e. vanishes on C, and

∇2Ψ =∂2Ψ

∂x21

+∂2Ψ

∂x22

= A

(

2

a21

+2

a22

)

.

The Poisson equation (6.13) therefore holds if

A

(

2

a21

+2

a22

)

= −2 ⇒ A = − a2

1a2

2

a21 + a2

2

,

and so the Prandtl stress function for the elliptical cylinder is

Ψ = − a2

1a2

2

a21 + a2

2

(

x2

1

a21

+x2

2

a22

− 1

)

=a2

1a2

2− a2

2x2

1− a2

1x2

2

a21 + a2

2

.

From (6.12), i.e. ψ = Ψ + 1

2(x2

1+ x2

2),

ψ(x1, x2) =a2

1a2

2− a2

2x2

1− a2

1x2

2

a21 + a2

2

+1

2(x2

1+ x2

2)

=1

2(a21 + a2

2)(2a2

1a2

2− 2a2

2x2

1− 2a2

1x2

2+ a2

1x2

1+ a2

1x2

2+ a2

2x2

1+ a2

2x2

2)

=1

2(a21 + a2

2)

{

(a2

1− a2

2)(x2

1− x2

2) + 2a2

1a2

2

}

= ℑm1

2(a21 + a2

2)

{

(a2

1− a2

2)iz2 + 2ia2

1a2

2

}

26

where z = x1 + ix2 and so z2 = x2

1− x2

2+ 2ix1x2 and iz2 = i(x2

1− x2

2) − 2x1x2. The

warping function φ is therefore

φ(x1, x2) = ℜe1

2(a21 + a2

2)

{

(a2

1− a2

2)iz2 + 2ia2

1a2

2

}

= − a2

1− a2

2

a21 + a2

2

x1x2.

When a1 > a2 the warping pattern is as shown in Figure 15. The curves φ = constantare rectangular hyperbolae.

Figure 15: The warping pattern for an elliptical cylinder

The torsional rigidity D is, from (6.19),

D = 2µ∫ ∫

SΨ da =

2µa2

1a2

2

a21 + a2

2

∫ ∫

S

(

1 − x2

1

a21

− x2

2

a22

)

dx1dx2.

Calculation of D. Because S is elliptical we transform the double integral usingmodified polar coordinates

x1 = a1r cos θ, x2 = a2r sin θ.

Sincex2

1

a21

+x2

2

a22

= r2 the limits of integration are 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.

The Jacobian is

∂(x1, x2)

∂(r, θ)=

∣

∣

∣

∣

∣

∣

∣

∣

∣

∂x1

∂r

∂x1

∂θ

∂x2

∂r

∂x2

∂θ

∣

∣

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

a1 cos θ − a1r sin θ

a2 sin θ a2r cos θ

∣

∣

∣

∣

∣

∣

∣

= a1a2r(cos2 θ + sin2 θ) = a1a2r,

so that dx1dx2 7→ a1a2 rdrdθ. Then the double integral is

∫ ∫

S

(

1 − x2

1

a21

− x2

2

a22

)

dx1dx2 =∫

2π

θ=0

∫

1

r=0

(

1 − a2

1r2 cos2 θ

a21

− a2

2r2 sin2 θ

a22

)

a1a2 rdrdθ

27

= a1a2

∫

2π

θ=0

∫

1

r=0

(

1 − r2)

rdrdθ = 2πa1a2

[

r2

2− r4

4

]1

0

=πa1a2

2.

Therefore

D =2µa2

1a2

2

a21 + a2

2

× πa1a2

2=πµa3

1a3

2

a21 + a2

2

,

the torsional rigidity of an elliptical cylinder.

Torsion of a circular cylinder In the case of a circular cylinder, a1 = a2 = a and

Ψ(x1, x2) =1

2(a2 − x2

1− x2

2),

φ(x1, x2) = 0,

D =1

2πµa4.

Because φ = 0, there is no warping when the circular cylinder is twisted about its centralaxis.

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 aand b (< a). The axis of torsion is once again the axis of symmetry.

The function

Ψ(x1, x2) =1

2(a2 − x2

1− x2

2),

obtained above for the circular cylinder, satisfies the Poisson equation (6.13), vanisheson C1 and is constant on C2. It is therefore an acceptable Prandtl stress function forthe 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∫ ∫

S

(

−x2

1− x2

2

)

da = µ∫ a

br2.2πrdr = 2πµ

[

1

4r4

]a

b

=1

2πµ(a4 − b4)

for the torsional rigidity of a circular tube.

28