Linear Elasticity

Home , Elasticity (physics), 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 ex- ternal 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 lin- early 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 occupies a region B0 of space at time t = 0, its reference conﬁg- uration, and a materialB particle has position vector X, with Cartesian components Xi, i =1, 2, 3, with respect to the orthonormal vectors e1, e2, e3 : { } 3

X = Xi ei. i=1 X

1 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 = xi ei, i=1 X 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 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 × ∂ui hij = . ∂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 ∂x ∂u v = , ∂t ≈ ∂t X x and the particle acceleration is similarly approximated by

∂2x ∂2u a = 2 2 . ∂t ≈ ∂t X x

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

1 ∂ui ∂uj eij = + (2.4) 2 ∂xj ∂xi ! (i, j =1, 2, 3). For example,

1 ∂u1 ∂u1 ∂u1 1 ∂u1 ∂u2 e11 = + = and e12 = + . 2 ∂x1 ∂x1 ! ∂x1 2 ∂x2 ∂x1 ! 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:

T e = e, eji = eij, i.e. e12 = e21, e23 = e32, e31 = e13. We see this from (2.4):

1 ∂uj ∂ui 1 ∂ui ∂uj eji = + = + = eij. 2 ∂xi ∂xj ! 2 ∂xj ∂xi !

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 = e1x1, u2 = e2x2, u3 = e3x3, (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

∂u1 ∂u2 ∂u3 e11 = = e1, e22 = = e2, e33 = = e3, ∂x1 ∂x2 ∂x3 with all the other components vanishing: e12 = e21 = e23 = e32 = e31 = e13 = 0. Thus

e1 0 0 (eij)=  0 e2 0  0 0 e3     3 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 + e1e2 + e2e3 + e3e1 + e1e2e3.

Since the ei are small, the relative change in volume, known as the dilatation, is change in volume ∆= e1 + e2 + e3 = . 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 −1 ρ0 = , ρ = ρ = ρ0(1+∆) . 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

∂u1 Then = γ, with all other components of deformation gradient vanishing. So ∂x2

1 ∂u1 ∂u2 1 γ e12 = + = (γ +0)= = e21, 2 ∂x2 ∂x1 ! 2 2 with all other strain components vanishing: 1 0 2 γ 0 1 (eij)=  2 γ 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 (σij)=  σ21 σ22 σ23  . (3.1) 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 = n1e1 + n2e2 + n3e3 = niei, the traction vector is given by

t1 σ11 σ12 σ13 n1 σ t(n)= n, ti = σijnj,  t2  =  σ21 σ22 σ23   n2  , (3.2) t3 σ31 σ32 σ33 n3             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 = tini = σijnjni = σijninj, (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) tnn − with magnitude 2 2 1 2 τn = t(n) tnn = t(n) t , (3.4) | − | {| | − n} called the shear stress.

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 com- 3 4 ponents ( 5 , 0, 5 ), (b) ﬁnd the normal and shear stresses at P on the given plane.

Solution (a) The traction components are 3 1 2 3 5 3 t(n)=  2 4 6   0  =  6  3 6 1 4 13    5   5        (b) From (3.3) the normal stress is 3 3 4 97 tn = n t(n)= , 0,  6  = . · 5 5 · 13 25  5    From (3.4) the shear stress is

1 2 2 2 2 2 1 2 2 13 97 √22, 941 2 τn = t(n) tnn = t(n) tn = 3 +6 + = 6.059. | − | {| | − } ( 5 − 25 ) 25 ≈

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 the coordinate axes. The cuboid has volume δv = δx1δx2δx3. We calculate all the forces ∂σ22 acting on the element in the x2-direction. Unbalanced normal stress δx2 acts on an ∂x2 ∂σ22 area δx1δx3, giving a normal force on the element in the x2-direction of δx2 δx1δx3 ∂x2 × ∂σ22 ∂σ23 ∂σ21 = δv. Unbalanced shear stresses δx3 and δx1 act on areas δx1δx2 and ∂x2 ∂x3 ∂x1 ∂σ23 ∂σ21 δx2δx3, 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 ρb2δv 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:

2 ∂σ21 ∂σ22 ∂σ23 ∂ u2 δv + δv + δv + ρb2δv = ρδv 2 . ∂x1 ∂x2 ∂x3 ∂t On dividing by δv we obtain the equation of motion

2 ∂σ21 ∂σ22 ∂σ23 ∂ u2 + + + 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:

2 ∂σ11 ∂σ12 ∂σ13 ∂ u1 + + + ρb1 = ρ 2 , ∂x1 ∂x2 ∂x3 ∂t

9 2 ∂σ21 ∂σ22 ∂σ23 ∂ u2 + + + ρb2 = ρ 2 , (3.5) ∂x1 ∂x2 ∂x3 ∂t 2 ∂σ31 ∂σ32 ∂σ33 ∂ u3 + + + ρb3 = ρ 2 . ∂x1 ∂x2 ∂x3 ∂t These equations are conveniently written in suﬃx notation as

2 ∂σij ∂ ui + ρbi = ρ 2 (3.6) ∂xj ∂t 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:

T σ = 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 with corresponding principal axes of stress e1, e2, e3, along the coordinate axes. The traction on a plane with normal components (n1, n2, n3) is σ1 0 0 n1 σ1n1 t(n)=  0 σ2 0   n2  =  σ2n2  . 0 0 σ3 n3 σ3n3             The normal stress is therefore, from (3.3),

2 2 2 tn = σ1n1 + σ2n2 + σ3n3 (3.9) and the squared shear stress can, from (3.4), be shown to be given by

2 2 2 2 2 2 2 2 2 2 τ =(σ1 σ2) n1n2 +(σ1 σ3) n1n3 +(σ2 σ3) n2n3. (3.10) n − − − If the stress takes the form

σ = pI, σij = pδij − − in which δij denotes the Kronecker delta

1 i = j, δ = ij 0 i = j, ( 6 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, − σ11 σ σ12 σ13 n1 0 − σ21 σ22 σ σ23 n2 = 0 .  −      σ31 σ32 σ33 σ n3 0        −      There are non-trivial solutions (i.e. n = 0) if and only if the determinant of coeﬃcients vanishes: 6 σ11 σ σ12 σ13 − σ21 σ22 σ σ23 =0. − σ31 σ32 σ33 σ

−

In the present example this condition becomes 1 σ 2 3 − 2 4 σ 6 =0. − 3 61 σ

−

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 3 n1 0  2 4 6   n2  =  0  . 3 6 1 n3 0             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 = λeppδij +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 Lam´emoduli and must be measured experimentally for each material. In the absence of body force the equations of equilibrium (3.7) are

∂σ ij =0. (4.2) ∂xj

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 ∂u i = 0 α 0 ∂x   j ! 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 1 σpp p 2 K := 3 , or K = − = λ + µ (4.4) epp ∆ 3 is the bulk modulus of the material.

13 (b) Simple shear The displacements are

u1 = γx2, u2 =0, u3 =0, so that 1 0 2 γ 0 1 (eij)=  2 γ 0 0  and epp =0. 0 0 0     So from (4.1),

Figure 11: Simple shear

0 γµ 0 σij =2µeij =  γµ 0 0  . 0 0 0   1   Therefore, σ12 = γµ, e12 = 2 γ are the only non-zero stresses and strains. The quantity σ12 µ = 2e12 is known as the shear modulus of the material.

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  , epp = α 2β. 0− 0 β −    −  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 :

αλ αµ(3λ +2µ) β = , T = . 2(λ + µ) λ + µ

We deﬁne the Young’s modulus E by

σ11 T µ(3λ +2µ) E := , so E = = (4.5) e11 α λ + µ and Poisson’s ratio ν by

e22 β λ ν := − , so ν = = , (4.6) e11 α 2(λ + µ) 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 Lam´econstants 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

9Kµ 3K 2µ E = , ν = − (4.8) 3K + µ 2(3K + µ)

15 and so inequalities (4.7) imply that 1 E > 0, 1 <ν< . (4.9) − 2 To obtain (4.9)2, write (4.8)2 as K 3 2 µ − ν = (4.10) K 2 3 +1 µ ! K 1 K and observe that ν 1 as 0, and ν as . −→ − µ −→ −→ 2 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 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 = λeppδij +2µeij. Taking the trace of both sides gives

σpp = (3λ +2µ)epp. (4.11) λ Thus σ = σ δ +2µe and so ij 3λ +2µ kk ij ij 1 λ eij = σpp δij + σij (4.12) 2µ (−3λ +2µ ) which is the strain-stress law, inverse to (4.1). The deviatoric stress and deviatoric strain are deﬁned by

′ 1 ′ 1 σ = σij σpp δij, e = eij epp δij, (4.13) ij − 3 ij − 3 respectively, and have the property that their traces vanish: ′ ′ σpp =0, epp =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, σij =2µeij, (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: 1 σpp K = 3 . 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 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 p 2 K = − = λ + µ. 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

e22 β 1 ν = − = = , e11 α 2 so that 1 ν = (4.18) 2 for all incompressible materials. We can see this another way — from (4.6) λ ν = 2(λ + µ) and for incompressibility λ , so that ν 1 , as at (4.18). −→ ∞ −→ 2 17 5 Conservation of energy: the strain energy function

The equations of motion (3.6) are

2 ∂σij ∂ ui + ρbi = ρ 2 . (5.1) ∂xj ∂t

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

∂σij ∂vi vi + ρbivi = ρ vi. ∂xj ∂t Thus ∂ ∂vi 1 ∂ (σijvi) σij + ρbivi = ρ (vivi). (5.2) ∂xj − ∂xj 2 ∂t Now ∂v ∂ ∂u ∂ ∂u ∂e ∂ω i = i = i = ij + ij , ∂xj ∂xj ∂t ∂t ∂xj ∂t ∂t where the strain tensor e and rotation ω are deﬁned by

1 ∂ui ∂uj 1 ∂ui ∂uj eij = + , ωij = . 2 ∂xj ∂xi ! 2 ∂xj − ∂xi ! Then ∂vi ∂eij ∂ωij σij = σij + , ∂xj ∂t ∂t ! ω T ∂ But ω is skew-symmetric, i.e. ω = −ω or ωji = ωij, so also is skew-symmetric − ∂t ∂ω and it follows that σ ij = 0. ij ∂t Then (5.2) may be written

∂ ∂eij 1 ∂ 2 (σijvi)+ ρbivi = σij + ρ (v ). ∂xj ∂t 2 ∂t

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

∂eij 1 ∂ 2 σijvinj da + ρb v dv = σij dv + ρ (v ) dv. Z∂Rt ZRt · ZRt ∂t ZRt 2 ∂t But t(n)= σn and ρ is assumed constant to this order:

∂eij ∂ 1 2 t(n) v da + ρb v dv = σij dv + ρ v dv. (5.3) Z∂Rt · ZRt · ZRt ∂t ∂t ZRt 2 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 The stress power per unit volume is deﬁned by ∂e P := σ ij . (5.4) ij ∂t 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 ∂e = σ ij . ∂t ij ∂t Now by the chain rule ∂W ∂eij ∂eij = σij ∂eij ∂t ∂t so that ∂W ∂eij σij =0. ∂eij − ! ∂t ∂e But at each position x and corresponding value of e, ij may be selected arbitrarily, ∂t ∂e so since the brackets do not depend on ij , we may conclude: ∂t ∂W σij = . (5.5) ∂eij

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 de- gree n: M n ∂φ [Namely, if φ(λxi)= λ φ(xi) then xi = nφ] i=1 ∂xi X ∂W eij =2W ∂eij so from (5.5) 1 W = σ e . (5.6) 2 ij ij 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(e )2 + µe e . (5.7) 2 pp ij ij

19 Written out in full

1 2 2 2 2 2 2 2 W = λ(e11 + e22 + e33) + µ e11 + e22 + e33 +2e12 +2e23 +2e31 . (5.8) 2 { } 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 = 0) requires work to be done on the body (W > 0): 6 W < 0 for some e = 0 would imply that starting from an unstressed state of rest work could spontaneously6 be done on the surroundings. We now seek necessary and suﬃcient conditions on λ and µ for W (e) given by (5.8) 2 to be positive deﬁnite. From (4.4) the bulk modulus is K = λ + 3 µ and elimination of λ from (5.8) in favour of K gives, after much manipulation,

1 2 1 2 2 2 W = K(e11 + e22 + e33) + µ (e11 e22) +(e22 e33) +(e33 e11) 2 3 − − − n o 2 2 2 +2µ e12 + e23 + e31 . (5.10) n o For the uniform dilatation e11 = e22 = e33 = α, eij = 0 for i = j, (5.10) becomes 6 9K W = α2, 2 so that W > 0 for α = 0 requires K > 0. 6 For the simple shear e12 = e21 = γ/2, all other eij vanishing, (5.10) becomes µ W = γ2, 2 so that W > 0 for γ = 0 requires µ > 0. Therefore, necessary conditions for the positive definiteness of 6W 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 = eij epp δij e =0, (5.12) ij − 3 pp 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(e )2 + µe′ e′ . (5.13) 2 pp ij ij 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) = αx2x3,  − −  (6.1)  u2 = r sin(θ + αx3) r sin θ = r sin θ cos αx3 + r cos θ sin αx3 r sin θ  − 2 − = r cos θ (αx3)+ O(αx3) = αx1x3.   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 = αx2x3, u2 = αx1x3, u3 = αφ(x1, x2), (6.2) − with strain components ∂φ 0 0 x2  ∂x1 −  α ∂φ (e )=  0 0 + x1  ij   2  ∂x2     ∂φ ∂φ     x2 + x1 0   ∂x1 − ∂x2    satisfying epp = 0, so that from (4.1) the stress components are ∂φ 0 0 x2  ∂x1 −  ∂φ (σ )= αµ  0 0 + x1  . (6.3) ij    ∂x2     ∂φ ∂φ     x2 + x1 0   ∂x1 − ∂x2    The equilibrium equations (3.7) in the absence of body force are ∂σ ij =0. (6.4) ∂xj The i = 1 and i = 2 equations are satisﬁed trivially and the i = 3 equation gives 2 2 ∂ ∂φ ∂ ∂φ ∂ φ ∂ φ 2 x2 + + x1 = 2 + 2 = φ =0, (6.5) ∂x1 ∂x1 − ! ∂x2 ∂x2 ! ∂x1 ∂x2 ∇ 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),

0 0 σ31 n1 0 σ t(n)= n or  0 0 σ32   n2  =  0  . σ31 σ32 0 0 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 ∂φ n1 + n2 = x2n1 x1n2 on C, ∂x1 ∂x2 − or ∂φ = x2n1 x1n2 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 = σi3 da. Z ZS x3=l

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

F1 = σ13 da, F2 = σ23 da, F3 =0. Z ZS Z ZS From the i = 3 equilibrium equation (6.4) and the stress (6.3), and its symmetry

∂σ31 ∂σ32 ∂σ13 ∂σ23 + = + =0. ∂x1 ∂x2 ∂x1 ∂x2 It follows that

∂(x1σ13) ∂(x1σ23) ∂σ13 ∂σ23 + = x1 + + σ13 = σ13, ∂x1 ∂x2 ∂x1 ∂x2 !

∂(x2σ13) ∂(x2σ23) ∂σ13 ∂σ23 + = x2 + + σ23 = σ23, ∂x1 ∂x2 ∂x1 ∂x2 ! and ∂(x1σ13) ∂(x1σ23) F1 = + da = x1(σ31n1 + σ32n2) ds =0, Z ZS ( ∂x1 ∂x2 ) IC ∂(x2σ13) ∂(x2σ23) F2 = + da = x2(σ31n1 + σ32n2) ds =0, Z ZS ( ∂x1 ∂x2 ) IC 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 x t(e3)= x1 x2 l , × σ13 σ23 σ33

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

M1 = (x2σ33 lσ32) da = lF2 = 0, S − |x3=l − M2 = RR (lσ31 x1σ33) da = lF1 = 0, S − |x3=l M3 = RR (x1σ32 x2σ31) da S − |x3=l RR ∂φ ∂φ = µα x1 + x1 x2 x2 da = αD, Z ZS ( ∂x2 ! − ∂x1 − !) 23 where 2 2 ∂φ ∂φ D = µ x1 + x2 + x1 x2 da. (6.8) Z ZS ∂x2 − ∂x1 ! 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)= e f(z). ℜ Let ψ(x1, x2)= m f(z), ℑ 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 = x2n1 x1n2 on C, ∂x1 ∂x2 − ∂ψ ∂ψ t2 + ( t1) = x2t2 x1( t1) on C, ∂x2 −∂x1 ! − − − ∂ψ ∂ψ t1 + t2 = x1t1 + x2t2 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 ∂ψ 1 ∂ 2 2 = (x1 + x2) on C. ∂s 2 ∂s The Neumann problem (6.5), (6.7) is thus equivalent to the Dirichlet problem

2ψ = 0 in S, ∇ (6.11) 1 2 2  ψ = 2 (x1 + x2) + constant on C. 

24  Simpler results are obtained by introducing the Prandtl stress function

1 2 2 Ψ= ψ (x1 + x2). (6.12) − 2 From (6.11), Ψ satisﬁes the Poisson equation

2Ψ= 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 ∂Ψ σ23 = µα + x1 = µα x1 = µα ∂x2 ! − ∂x1 − ! − ∂x1  (6.15)  ∂φ ∂ψ ∂Ψ  σ31 = µα x2 = µα x2 = µα ,  ∂x1 − ! ∂x2 − ! ∂x2   and the torsional rigidity from (6.8), (6.9) and (6.12) by 

2 2 ∂ψ ∂ψ ∂Ψ ∂Ψ D = µ x1 + x2 x1 x2 da = µ x1 + x2 da. (6.16) Z ZS − ∂x1 − ∂x2 ! − Z ZS ∂x1 ∂x2 ! Since ∂Ψ ∂Ψ ∂(x1Ψ) ∂(x2Ψ) x1 + x2 = + 2Ψ, ∂x1 ∂x2 ∂x1 ∂x2 − the application of the two-dimensional divergence theorem gives

∂Ψ ∂Ψ x1 + x2 da = 2 Ψ da + µ Ψ(x1n1 + x2n2)ds Z ZS ∂x1 ∂x2 ! − Z ZS IC = 2 Ψ da + µ Ψ0(x1n1 + x2n2)ds, − Z ZS IC where (6.14) has been used. Thus

D =2 Ψ da µ Ψ0(x1n1 + x2n2)ds. (6.17) Z ZS − IC 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 = 0on C, (6.18) D = 2µ Ψ da. (6.19) Z ZS

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 of

C is 2 2 x1 x2 2 + 2 =1. a1 a2

The function 2 2 x1 x2 Ψ(x1, x2)= A 2 + 2 1 a1 a2 − ! clearly satisﬁes the boundary condition (6.18), i.e. vanishes on C, and 2 2 2 ∂ Ψ ∂ Ψ 2 2 Ψ= 2 + 2 = A 2 + 2 . ∇ ∂x1 ∂x2 a1 a2 ! The Poisson equation (6.13) therefore holds if 2 2 2 2 a1a2 A 2 + 2 = 2 A = 2 2 , a1 a2 ! − ⇒ − a1 + a2 and so the Prandtl stress function for the elliptical cylinder is 2 2 2 2 2 2 2 2 2 2 a1a2 x1 x2 a1a2 a2x1 a1x2 Ψ= 2 2 2 + 2 1 = − 2 2− . − a1 + a2 a1 a2 − ! a1 + a2 1 2 2 From (6.12), i.e. ψ =Ψ+ 2 (x1 + x2), 2 2 2 2 2 2 a1a2 a2x1 a1x2 1 2 2 ψ(x1, x2) = − 2 2− + (x1 + x2) a1 + a2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 = 2 2 (2a1a2 2a2x1 2a1x2 + a1x1 + a1x2 + a2x1 + a2x2) 2(a1 + a2) − − 1 2 2 2 2 2 2 = 2 2 (a1 a2)(x1 x2)+2a1a2 2(a1 + a2) − − n o 1 2 2 2 2 2 = m 2 2 (a1 a2)iz +2ia1a2 ℑ 2(a1 + a2) − n o 26 2 2 2 2 2 2 where z = x1 + ix2 and so z = x1 x2 +2ix1x2 and iz = i(x1 x2) 2x1x2. The warping function φ is therefore − − −

1 2 2 2 2 2 φ(x1, x2) = e 2 2 (a1 a2)iz +2ia1a2 ℜ 2(a1 + a2) − n o 2 2 a1 a2 = 2 − 2 x1x2. − a1 + a2

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),

2 2 2 2 2µa1a2 x1 x2 D =2µ Ψ da = 2 2 1 2 2 dx1dx2. Z ZS a1 + a2 Z ZS − a1 − a2 !

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

x1 = a1r cos θ, x2 = a2r sin θ. 2 2 x1 x2 2 Since 2 + 2 = r the limits of integration are 0 r 1, 0 θ 2π. a1 a2 ≤ ≤ ≤ ≤ The Jacobian is

∂x1 ∂x1 a1 cos θ a1r sin θ ∂(x1, x2) ∂r ∂θ − = = ∂(r, θ) ∂x2 ∂x2 a2 sin θ a2r cos θ

∂r ∂θ

2 2 = a1a2r(cos θ + sin θ)= a1a2r, so that dx1dx2 a1a2 rdrdθ. Then the double integral is 7→ 2 2 2π 1 2 2 2 2 2 2 x1 x2 a1r cos θ a2r sin θ 1 2 2 dx1dx2 = 1 2 2 a1a2 rdrdθ =0 =0 Z ZS − a1 − a2 ! Zθ Zr − a1 − a2 ! 27 1 2π 1 2 4 2 r r πa1a2 = a1a2 1 r rdrdθ =2πa1a2 = . θ=0 r=0 − " 2 − 4 # 2 Z Z 0 Therefore 2 2 3 3 2µa1a2 πa1a2 πµa1a2 D = 2 2 = 2 2 , a1 + a2 × 2 a1 + a2 the torsional rigidity of an elliptical cylinder.

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

1 2 2 2 Ψ(x1, x2) = (a x1 x2), 2 − − φ(x1, x2) = 0, 1 D = πµa4. 2 Because φ = 0, there is no warping when the circular cylinder is twisted about its central axis.

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 2 2 2 Ψ(x1, x2)= (a x1 x2), 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 a a 2 2 2 1 4 D = µ x1 x2 da = µ r .2πrdr =2πµ r − S − − b 4 b Z Z Z 1 = πµ(a4 b4) 2 − for the torsional rigidity of a circular tube.