Lecture Notes - Linear Elasticity Theory

Ruhr-University Bochum

Faculty of Civil and Environmental Engineering

Institute of Mechanics

An Introduction to

Linear Continuum Mechanics

Klaus Hackl

Mehdi Goodarzi

2010

ii

Contents

1 Fundamentals 1

1.1 Stress and equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.1.1 Stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.2 Balance equations: dierential form . . . . . . . . . . . . . . . . . . 5

1.2 Strain and compatibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2.1 Compatibility conditions . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 The Field Equations of Linear Continuum Mechanics 17

2.1 Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Hooke's law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2.1 Hooke's law for isotropic materials . . . . . . . . . . . . . . . . . . . 192.2.2 Alternative formulations of Hooke's law for isotropic materials . . . 202.2.3 Plane-strain and plane-stress problems . . . . . . . . . . . . . . . . . 22

2.3 Navier and Beltrami-Michell equations . . . . . . . . . . . . . . . . . . . . . 23

3 Displacement Functions 27

3.1 Scalar and vector potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 Galerkin vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.2.1 Love's strain function . . . . . . . . . . . . . . . . . . . . . . . . . . 333.3 Displacement functions of Papkovich-Neuber . . . . . . . . . . . . . . . . . 35

iii

iv CONTENTS

Chapter 1

Fundamentals

The subject matter of mechanics is the study of motion, in how a physical object changesposition with time and why. Here we shall conne ourselves within principles of classicalmechanics. Studying how a body moves is called kinematics which is basically a geometri-cal investigation of motion dealing with fundamental concepts like space, time and frame of

reference. In a second step Newton's laws of motion are introduced based on other funda-mental concepts such as inertial frame of reference,mass and force to deal with the questionof why things move, and this is called kinetics. The alternative strategy is formulation ofmechanics based on conservation laws of energy, momentum and angular momentum whichturns out to have broader range of validity than Newton's formulation.

Scientic hypotheses try to explain physical observations based on a set of assumptionsand more often than not, neglecting certain eects. One of the most universal aspects ofobservations is the scale, that is the range of values of quantities during an observation.For instance, having a model proposed by an observation over the ordinary length scalesof laboratory, we can not draw the conclusion that the same model will hold over cosmiclength scales, or on sub-atomic length scales unless it is somehow tested or proved.

Considering length scales, mechanical model of a moving body can belong to one of thefollowing in order of sophistication

Particle is a model of a body of negligible dimensions so that it can be treated geomet-rically as a point, which means all the body mass is concentrated at its center ofmass. For example, in study of planetary motion dimensions of planets are negligi-ble compared to orbital distances therefore they are often treated as point masses.Kinematics of a particle deals with translation only. In this model, forces are free

vectors which means they can move freely without their mechanical eect being al-tered. Model equations for particle motion consist of balance of energy and linearmomentum.

Rigid body is an object of nite dimensions and negligible change in shape during motion.That is where the distance between every two points on the body remains constant.A rigid body is assumed to have a continuous mass distribution although in realitymatter is quantized at small scales. Kinematics of rigid body motion is expressedin terms of translation and rotation. Forces are sliding vectors viz they can slide on

1

2 CHAPTER 1. FUNDAMENTALS

their line of action without their mechanical eects being altered. Model equationsare balance of energy, linear momentum and angular momentum.

Deformable body is a mechanical object of continuous mass distribution whose relativedistance of points can change. This is ocially the starting point of continuum me-chanics although a rigid body has a continuous mass distribution too. Continuummechanics deals with the length scales large enough to neglect all the molecular ef-fects and, at the same time, small enough to observe the shape changes in the body.Together with translations and rotations we have deformation in kinematics studyof deformable bodies. As the point of action of forces is important here, forces arexed vectors. Balance equations of linear and angular momentum and energy are notsucient to model deformation. In addition we will need the so called constitutive

equations to have a closed model.

In the following section we work out the concept of forces in continuum media.

1.1 Stress and equilibrium

In the view of causality, forces are the cause for change in momentum and also for defor-mation. According Newton's third law, forces exist only in couples, to wit the presence ofany force is justiable only in the presence of its source. In this regard, we distinguish twotypes of forces

Field forces which are applied from a distance through a eld such as electromagnetic orgravitational eld.

Contact forces which are applied by direct contact of objects such as friction. Notethat these are also manifestation of eld forces that act at inter-molecular distances,therefore observed as direct contact at large scales.

Distribution of eld forces is usually expressed as force per unit volume or per unit mass,and distribution of contact forces is usually expressed as force per unit area. The exceptionis the so called concentrated force which is an attempt to model the situation when a niteamount of force is applied over a very small region of area or volume. In this specic casethe force distribution will not be nite-valued anymore. Therefore concentrated forces aretreated as individual force vectors rather than distributions, and presented by Dirac's delta

f(x)

t∂Ω

Ω

Fig. 1.1: Solid body.

function which will be introduced later on. Let Ω be a body atequilibrium with a nite and continuous distribution of solidmass m throughout its simply connected volume V , and bebounded by a smooth surface ∂Ω of area A. This is our typ-ical picture of a body under consideration in continuum solidmechanics. You will see potato-like illustrations (Fig. 1.1)!Assume a distribution of eld forces presented by a volumet-ric force eld f(x) and contact force (also called traction)distribution over ∂Ω denoted by t. Then equilibrium condi-

1.1. STRESS AND EQUILIBRIUM 3

tions for the body read∑F = 0 :

∫Ωf dV +

∫∂Ωt dA = 0 (1.1)∑

M = 0 :

∫Ωx× f dV +

∫∂Ωx× t dA = 0 (1.2)

This provides enough ground for analysis of the overall body motion which is a combinationof translation and rotation, as if the body was rigid. However, deformation is a localphenomenon which depends on the state of forces at each body point.

1.1.1 Stress tensor

In order to analyze the forces at each body point, we shall write the equilibrium equationsfor an innitesimal volume element at the body point. For an innitesimal element at

dX2

dX3

dX1

f

t1

t2

t3

(a)X1

X2

X3

n

−e1

−e2

−e3(b)

Fig. 1.2: a) Innitesimal volume element, and b) inintesimal tetrahedron.

point x with volume dV the volumetric force is given by dF V = f(x) dV . Also, there aretractions at element faces (Fig. 1.2.a), because an area element within a body of mass isan interface that experiences contact forces due to adhesion and/or cohesion. This tractionforce t at each point depends on the location x of area element dA, and on its directiondenoted by unit normal vector n

t = t(x,n) . (1.3)

This is Cauchy's assumption which is substantial in continuum mechanics, although itmight seem obvious. The functionality of t(x,n) is further specied by Cauchy's theoremthrough which the concept of stress tensor is introduced.

Cauchy's theorem

Assume that we want to write the equation of motion for an innitesimal tetrahedron (Fig.1.2.b) whose edges coincide with the three Cartesian axes. The four triangular faces haveunit normal vectors

−e1 , −e2 , −e3 , n .

We identify face triangles of the tetrahedron by their normal vectors i.e. 1, 2, 3 and n. Noticethat projection of the triangle n over coordinate planes are triangles 1, 2, 3. Therefore if


the center point of triangle n has the position

x = x1e1 + x2e2 + x3e3 , (1.4)

then the center points of the other three triangles are projections of x onto coordinateplanes, given by

xc1 = x2e2 + x3e3 , xc2 = x1e1 + x3e3 , x

c3 = x1e1 + x2e2 . (1.5)

Furthermore, if the triangle n has the area dA then the triangles 1, 2, 3 (which are itsprojections) have the areas

dAi = ei · n dA = nidA :

dA1 = n1 dAdA2 = n2 dAdA3 = n3 dA

(1.6)

Also note that the tetrahedron has the mass ρdV and acceleration a. Finally, the equationof motion is written as

ρa dV = f dV + t(x,n) dA+ t(xc1,−e1) dA1 + t(xc2,−e2) dA2 + t(xc3,−e3) dA3 . (1.7)

Substitution of dAi's according equation (1.6) gives

ρa dV = f dV + t(x,n) dA+ t(xc1,−e1)n1dA+ t(xc2,−e2)n2dA+ t(xc3,−e3)n3dA (1.8)

and dividing both sides by dA

ρadV

dA= f

dV

dA+ t(x,n) + t(xc1,−e1)n1 + t(xc2,−e2)n2 + t(xc3,−e3)n3 . (1.9)

However the fraction dV/dA tends to zero as the volume shrinks. This is because dV isproportional to L3 and dA is proportional to L2, where L is the characteristic length of theelement, hence

L→ 0 :dV

dL∝ L3

L2= L→ 0 .

Therefore equation (1.9) yields

t(x,n) = −t(xc1,−e1)n1 − t(xc2,−e2)n2 − t(xc3,−e3)n3 . (1.10)

When the tetrahedron element shrinks, the center points of faces tend to each other

L→ 0 : xc1 → xc2 → xc3 → x

and equation (1.10) gives

t(x,n) = −t(x,−e1)n1 − t(x,−e2)n2 − t(x,−e3)n3 . (1.11)


Now writing components gives

t1(x,n) = −t1(x,−e1)n1 − t1(x,−e2)n2 − t1(x,−e3)n3 (1.12)



and in matrix form t1(x,n)t2(x,n)t3(x,n)

=

−t1(x,−e1) −t1(x,−e2) −t1(x,−e3)−t2(x,−e1) −t2(x,−e2) −t2(x,−e3)−t3(x,−e1) −t3(x,−e2) −t3(x,−e3)

n1

n2

n3

. (1.15)

The 3× 3 matrix in the latter equation is a second order tensor which is only a function ofx. It is called stress tensor and is usually denoted by σ. This equation can be written inthe more compact form

t(x,n) = σ(x) · n , (1.16)

which is called Cauchy's formula. Let's have a closerlook at the matrix form in equation (1.15). It is easyto show that −ti(x,−ej) = ti(x, ej). Then the com-ponent at the ith row and the jth column is

σij = ti(x, ej)

which is the ith component of the traction force appliedto the surface normal to the jth axis (Fig. 1.3). There-fore the second index j in the stress tensor σij showsthe surface to which the traction is applied, and therst index i shows the direction of the traction compo-nent.

X1

X2

X3

σ11

σ21

σ31

σ12

σ22

σ32

σ13σ23

σ33

Fig. 1.3: Stress components.

The importance of equation (1.16) is that the functionality of traction on x and nis separated as the stress tensor σ is a function of x only at the right-hand-side. If youremember, when introducing tensors, we mentioned that second order tensors are linearmappings from vectors to vectors. Cauchy's formula is a physical example of this, wherestress tensor maps directions denoted by n to tractions t exerted on an area element normalto n.

1.1.2 Balance equations: dierential form

As mentioned before we need the local state of forces in order to analyze deformation. Theintegrals in equations (1.1) and (1.2) are written over the whole body Ω, while they areequally true for any sub-domain of the body, say Ω ⊂ Ω which includes the body point x∫

Ωf dV +

∫∂Ωt dA = 0 (1.17)∫

Ωx× f dV +

∫∂Ωx× t dA = 0 (1.18)


Let's focus at the rst equation. If we substitute Cauchy's formula into (1.17) it yields∫Ωf dV +

∫∂Ωσ · n dA = 0 . (1.19)

Note that ∂Ω is a closed surface, therefore the second integral according Gauss theoremgives ∫

∂Ωσ · n dA =

∫Ωσ · ∇ dV (1.20)

Therefore equation (1.19) becomes∫Ωf dV +

∫Ωσ · ∇ dA =

∫Ω

(f + σ · ∇) dV = 0 . (1.21)

This integral vanishes for any arbitrary sub-body Ω only if the integrand equals zero

σ · ∇+ f = 0 (1.22)

which is the dierential equation for balance of forces.For the second balance equation (1.18), again we substitute Cauchy's formula to obtain∫

Ωx× f dV +

∫∂Ωx× (σ · n) dA = 0 (1.23)

however it holds that

x× (σ · n) = (x× σ) · n (1.24)

therefore ∫Ωx× f dV +

∫∂Ω

(x× σ) · n dA = 0 (1.25)

and once again Gauss theorem gives∫∂Ω

(x× σ) · n dA =

∫Ω

(x× σ) · ∇ dV (1.26)

which turns equation (1.25) into∫Ωx× f dV +

∫Ω

(x× σ) · ∇ dV =

∫Ω

(x× f + (x× σ) · ∇) dV = 0 . (1.27)

There is an identity that can be easily veried

(x× σ) · ∇ = ε : σ + x× (σ · ∇) (1.28)

where ε is permutation tensor. Putting this into equation (1.27) gives∫Ωx× (f + σ · ∇) + ε : σ dV = 0 (1.29)


however the term in parentheses vanishes due to (1.22), then∫Ωε : σ dV = 0 (1.30)

which holds for every sub-body Ω, therefore it must hold that

ε : σ = 0 . (1.31)

which can be true only if σ is a symmetric tensor

σ = σT . (1.32)

which is the local condition for balance of moments.

Exercise 1. Show that ε : σ = 0 results in σ = σT .

Exercise 2. Write equilibrium equations and Cauchy's formula in components.

Exercise 3. Consider a body with some uid exposed tohydrostatic pressure (Fig. 1.4).

σ = −p(x3) I =

−p(x3) 0 0

0 −p(x3) 00 0 −p(x3)

with body force: f = −ρge3. Solve equilibrium equationsfor p(x3) with boundary condition p(0) = p0.

X1 X2

X3

Fig. 1.4: Body lled with uid.

Example 1.1. Write equilibrium equations in cylindrical coordinates.

Solution. The equilibrium equation is a vector equation or a set of three scalar equationsgiven by (1.22) as

σ · ∇+ f = 0 .

The divergence operator in cylindrical coordinates is given by

σ · ∇ =

(1

r

∂

∂r(rσrr) +

1

r

∂σrϕ∂ϕ

+∂σrz∂z

)er +(

1

r

∂

∂r(rσϕr) +

1

r

∂σϕϕ∂ϕ

+∂σϕz∂z

)eϕ +(

1

r

∂

∂r(rσzr) +

1

r

∂σzϕ∂ϕ

+∂σzz∂z

)ez

and decomposition of body forces gives

f = frer + fϕeϕ + fzez .


Substitution into equilibrium equation results in

1

r

∂

∂r(rσrr) +

1

r

∂σrϕ∂ϕ

+∂σrz∂z

+ fr = 0 (1.33)

1

r

∂

∂r(rσϕr) +

1

r

∂σϕϕ∂ϕ

+∂σϕz∂z

+ fϕ = 0 (1.34)

1

r

∂

∂r(rσzr) +

1

r

∂σzϕ∂ϕ

+∂σzz∂z

+ fz = 0 . (1.35)

Exercise 4. Using the transformation from Cartesian to cylindrical coordinates and thenabla operator in Cartesian coordinates, show that in cylindrical coordinates the nablaoperator has the following form

∇ ≡ er∂r +1

reϕ∂ϕ + ez∂z .

Exercise 5. Calculate σrϕ(r) for a tube loaded by torsion(Fig. 1.5). There are no body forces and the boundary con-dition σrϕ(r1) = t1 is given. Note that

σ = σrϕ (ereϕ + eϕer) .

r0

r1

Fig. 1.5: Tube under torsion.

1.2 Strain and compatibility

As pointed out before, the motion of a continuum medium is a combination of translation,rotation and deformation. Deformation is the change in shape. Since the shape of an objectis characterized by the relative position of its points, in order to analyze deformation, wefocus on the change of innitesimal line elements that connect the neighboring points ofthe body. Let us assume a typical image of a continuum body (Fig. 1.6) at initial/unde-formed conguration whose points are denoted by the position vector X, and then in itsmoved/deformed conguration whose points are denoted by the position vector x. Thebasic assumption is that x is a one-to-one function ofX. This guarantees that the materialpoints are neither created, nor are they annihilated. So, the starting point is a descriptionof motion by

x = x(X, t) . (1.36)

It is also customary to dene displacement eld as

u = u(X, t) = x(X, t)−X . (1.37)

1.2. STRAIN AND COMPATIBILITY 9

u

xX

dXdx

Fig. 1.6: Undeformed and deformed congurations of a continuum body.

Excluding dynamic cases from our discussion, the time dependence is dropped. Moreover,we assume that the displacement eld is innitesimal

‖u‖ 1 . (1.38)

Now imagine two neighboring material points in the undeformed conguration con-nected by the dierential line element dX, and in the deformation conguration connectedby dx. We would like to study the relation dX 7−→ dx. According denition of displace-ment eld we have

dx = dX + u(X + dX)− u(X) . (1.39)

Then substitution of Taylor's expansion as

u(X + dX) = u(X) + u∇ · dX + · · · (1.40)

will result in

dx = dX + u∇ · dX = x∇ · dX . (1.41)

Here u∇ is the displacement gradient and x∇ the deformation gradient which are relatedclearly by

x∇ = I + u∇ . (1.42)

The displacement gradient is a second order tensor and can be decomposed into symmetric

and antisymmetric parts designated by

ε =1

2(u∇+∇u) (1.43)

ω =1

2(u∇−∇u) (1.44)

the symmetric part ε is called strain tensor and the antisymmetric part ω is the rota-

tion tensor. As their names suggest the symmetric part ε reects deformation and theantisymmetric part ω carries information about rotation only. It is immediately clear that

ε = εT , ω = −ωT . (1.45)


To stay within linear continuum theory the fundamental assumption is

‖ε‖ 1 , ‖ω‖ 1 . (1.46)

Every antisymmetric tensor has three independent elements only, and therefore can bespecied by a vector. So the rotation tensor ω can be expressed by the so called rotation

vector

w = (wi) =

(1

2εjikωjk

)= −1

2ε : ω . (1.47)

Exercise 6. Show that ωij = εikjwk.

By expanding the equations (1.47) and (1.44), one can show that

w =1

2∇× u . (1.48)

Physical meaning of the decomposition

Putting the decomposition u∇ = ε+ ω into (1.41) we obtain

dx = dX + u∇ · dX = dX + ε · dX + ω · dX . (1.49)

We show that dxr = ω · dX is the rotational part of deformation and dxs = ε · dX is thedeformational part of deformation. Let us focus at dxr rst

w

dv

dvr

ϕ

Fig. 1.7: Rotational part ofdisplacement.

[ω · dX]i = ωijdXj = εikjwkdXj (1.50)

which givesdxr = w × dX . (1.51)

This, according assumption (1.46), means that dxr is obtainedby a small rotation of dX around w. Note that dxr is normalto dX, and since ‖w‖ 1 then the ratio ‖dxr‖ / ‖dX‖ whichis equal to the angle of rotation (Fig. 1.7) is also small. Inshort form notation one can state

‖w‖ 1

ϕ ≈ ‖dxr‖ / ‖dX‖ = ‖w‖

⇒ ϕ 1 .

Now let us have a look at dxs. Deformation is expressed in terms of change in lengthsand angles. The length of the line element in the deformed conguration is obtained by

dx · dx = (dX +w × dX + ε · dX) · (dX +w × dX + ε · dX)

= dX · dX + 2dX · (w × dX) + 2dX · ε · dX + · · · (1.52)

where higher order terms, denoted by · · · , are neglected due to assumption (1.46). Also weknow that

dX · (w × dX) = 0


because dX is orthogonal to (w × dX). Therefore

dx2 = dX2 + 2dX · ε · dX (1.53)

and thendx

dX=

[1 + 2

dX · ε · dXdX2

]1/2

≈ 1 +dX · ε · dX

dX2, (1.54)

where Taylor's expansion is used because ‖ε‖ 1. The change of length is nally obtainedby

∆l = dx− dX =1

dXdX · ε · dX . (1.55)

Remember that this result is based on the assumption (1.46) of small strain and rotation.To study the changes in angle, assume two adjacent innitesimal vectors normal to

each other in the undeformed conguration denoted by dX1 and dX2, and their deformedcounterparts denoted by dx1 and dx2 respectively. Inner multiplication of the two vectorswill give us information on the angle between them

dx1 · dx2 = (dX1 +w × dX1 + ε · dX1) · (dX2 +w × dX2 + ε · dX2) (1.56)

= dX1 · dX2 + dX1 · (w × dX2) + (w × dX1) · dX2 + 2dX1 · ε · dX2 + · · ·

where the higher order terms are neglected. The rst term vanishes due to the assumptionof orthogonality dX1 ·dX2 = 0, and the sum of second and third terms is also zero because

dX1 · (w × dX2) = − (w × dX1) · dX2 .

Thereforedx1 · dx2 ≈ 2dX1 · ε · dX2 . (1.57)

On the other hand

dx1 · dx2 = dx1 dx2 cosϕ ≈ dX1 dX2 cosϕ = dX1 dX2 sin(π/2 − ϕ)

= dX1 dX2 sin(∆ϕ) ≈ dX1 dX2 ∆ϕ (1.58)

where π/2 is the angle in the undeformed conguration. Finally comparing (1.58) and (1.57)yields

∆ϕ = 2dX1 · ε · dX2

dX1 dX2. (1.59)

Again, note that this formula is obtained under the assumption (1.46). As can be seenfrom equations (1.55) and (1.59), deformation depends on the strain tensor ε only. There-fore our objective of interpreting the additive decomposition of displacement gradient intosymmetric (strain) and antisymmetric (rotation) parts is fullled.

Exercise 7. For the given displacement gradient

u∇ = 10−3

1 −2 03 0 00 1 1


1. calculate ε,ω,w.

2. For the two orthogonal vectors a =

101

mm and b =

−111

mm calculate the change

of their lengths and the angle between them.

Remark 1. Strain is dened as the symmetric part of displacement gradient. Sometimesthis is indicated by the short notation

ε = ∇su (1.60)

called symmetric gradient of displacement.

1.2.1 Compatibility conditions

If the equation (1.43) is expanded as

ε11 =∂u1

∂X1, ε12 = ε21 =

1

2

(∂u1

∂X2+∂u2

∂X1

),

ε22 =∂u2

∂X2, ε23 = ε32 =

1

2

(∂u2

∂X3+∂u3

∂X2

), (1.61)

ε33 =∂u3

∂X3, ε31 = ε13 =

1

2

(∂u3

∂X1+∂u1

∂X3

).

we see that the six independent components of the strain tensor are obtained from the threecomponents of the displacement vector, assuming that u is dierentiable. Now suppose thatthe strain components are given and the displacement components are to be determinedfrom these equations. In this case since we have six equations and three unknowns thereis not necessarily a unique acceptable solution for displacement eld. In other words theset of equations (1.61) poses a restriction on the strain eld, i.e. we can not choose straincomponents arbitrarily and they are somehow connected.

Denition 2. A compatible strain eld is the one for which a single-valued and continuousdisplacement eld can be found by solving the equation ε = ∇su.

Our objective then will be to nd the condition(s) under which a given strain eldis compatible. One would naturally think of three additional equations to connect straincomponents to each other. Applying the following identities from tensor calculus

∇×∇u = 0 and u∇×∇ = 0 . (1.62)

applying the curl operator from the left-hand-side to equation (1.43) we obtain

∇× ε = ∇×(

1

2(∇u+ u∇)

)=

1

2(

∇×∇u+∇× u∇) =1

2∇× u∇ (1.63)


and then from the right-hand-side

∇× ε×∇ =

(1

2∇× u∇

)×∇ =

1

2∇×

(u∇×∇) (1.64)

which nally leaves us with∇× ε×∇ = 0 , (1.65)

the so called compatibility condition. This condition is obtained from the denition ofstrain, therefore it is a necessary condition for compatibility, viz if the displacement eldis acceptable then the above condition should be fullled. In a next step we show that thiscondition is also sucient, i.e. if the condition is fullled then the strain eld is compatible,that is to say, a physically meaningful displacement eld exists such that ε = ∇su.

Since the expression ∇ × ε × ∇ is to vanish for a compatible strain eld, one woulddene its value as a measure of incompatibility. Therefore, the incompatibility of the strainis dened as

Inc(ε) ≡ ∇× ε×∇ , (1.66)

which makes sense as the compatibility condition can be restated as

Inc(ε) = 0 . (1.67)

Exercise 8. Prove that the alternative form of compatibility condition

tr(I ×∇× ε×∇× I) = 0

can also be expressed as

∆ε+∇tr(ε)∇ = ∇∇ · ε+ ε · ∇∇ . (1.68)

Now we intend to show that the condition (1.65) is sucient for the strain eld to becompatible in a simply connected domain. For this we assume that the strain eld ε isgiven in a simply connected body Ω, and it fullls (1.65), and then it has to be proven thata continuous single-valued displacement u can be obtained from (1.43) if the displacementis given at one point over the body.

X0

X

Ω

Fig. 1.8: Integration path.

We denote the point where the displacement is given byX0 and the given value by u0. From (1.41) we have

du = ε · dX + ω · dX (1.69)

whose integration gives

u(X) = u(X0) +

∫ X

X0

ε(X)· dX +

∫ X

X0

ω(X)· dX (1.70)

where the integral is taken over any continuous path C connecting X0 and X. Using indexnotation the last term can be calculated as∫ X

X0

ω · dX =

∫ X

X0

ω · I · dX =

∫ X

X0

ω ·[(X −X

)∇]· dX (1.71)

=

∫ X

X0

ωij ∂k(Xj −Xj

)dXk = ωij

(Xj −Xj

) ∣∣∣XX0

−∫ X

X0

∂kωij(Xj −Xj

)dXk


here integration by parts is used. Then we have

ωij(Xj −Xj

) ∣∣∣XX0

= ω(X) · (X −X)− ω(X0) · (X0 −X) = ω(X0) · (X0 −X) (1.72)

and using the result in exercise 6

∂kωij(Xj −Xj

)= ∂kεiljwl

(Xj −Xj

)= ∇w ×

(X −X

)= ∇

(1

2∇ × u

)×(X −X

)= −1

2

(∇u× ∇

)×(X −X

)= −

(ε× ∇

)×(X −X

)which nally gives

u(X) = u(X0) + ω(X0) · (X −X0) +

∫ X

X0

dX ·[ε+

(ε× ∇

)×(X −X

)](1.73)

the so called Michell-Cesaro formula. Now we have to show that the integral term inthe above formula is path-independent i.e. u(X) is a single-valued function. However, weremember that path-independence of a line integral is fullled if and only if the integralvanishes over every closed path, viz we should prove that∮

CdX ·

[ε+

(ε× ∇

)×(X −X

)]= 0 . (1.74)

Using Stokes' theorem and the compatibility condition (1.65) we have∮CdX ·

[ε+

(ε× ∇

)×(X −X

)]=

∫AdAn · ∇ ×

[ε+

(ε× ∇

)×(X −X

)](1.75)

=

∫AdAn · (∇× ε×∇)×

(X −X

)= 0 . (1.76)

Proof of the second step is left as exercise 10.

Exercise 9. Let the tensor A be given as

A = ax21e1e1 + bx1x2e2e2 .

For which values of a and b holds A×∇ = 0? Calculate

u(X3) =

∫ X3

0A · ds

for the two paths indicated in Fig. 1.9.

O

x2

x1

X2 X3

X1

Fig. 1.9: Paths of integra-tion.

Exercise 10. Prove the following identity

∇ ×[ε+

(ε× ∇

)× (x− x)

]=(∇ × ε× ∇

)× (x− x) . (1.77)


Example 1.2. Let us have a look at the classical solution tothe cantilever beam problem, considering its compatibility.The solution to the cantilever beam problem shown in thegure is of the form

ε = −κ(x) y exex . (1.78)

where κ(x) is the curvature at each cross section. The beam isbuilt-in at its left end x = 0, where u(0) = 0 and ω(0) = 0.Check if ε is compatible. If yes, nd the displacement eldcomponents ux(x, y) and uy(x, y).

x0 x

yεxx

Fig. 1.10: Cantilever beam.

Solution. We start with the compatibility condition, applying the curl operator from theright-hand-side to the strain

ε×∇ = (−κ(x) yexex)× (∂xex + ∂yey + ∂zez) = −κ(x) exez (1.79)

and then from the left

∇× ε×∇ = (∂xex + ∂yey + ∂zez)× (−κ(x) exez) = 0 . X (1.80)

Now with the compatibility condition being fullled, we want to calculate the displacementeld from Michell-Cesaro formula. Applying the boundary conditions at x = 0 we obtain

u(x) =

∫ x

0dx ·

[ε+

(ε× ∇

)× (x− x)

](1.81)

in which(ε× ∇

)× (x− x) = (−κ(x) exez)× ((x− x) ex + (y − y) ey + (z − z) ez)

= −κ(x)(

(x− x) exey − (y − y) exex). (1.82)

Substitution into (1.81) yields

u(x) =

∫ x

0[exdx+ eydy] ·

[κ(x)

(− yexex − (x− x) exey + (y − y) exex

)]=

∫ x

0dx κ(x) [−yex − (x− x) ey] . (1.83)

Therefore

ux(x, y) = −y∫ x

0κ(x) dx (1.84)

and

uy(x, y) = −∫ x

0κ(x) (x− x) dx . (1.85)

Then it is also clear thatu′′y = κ(x) and ux = −y u′y . (1.86)


Exercise 11. For the given strain tensor and boundary conditions

ε = x22e1e1 + x1x2 (e1e2 + e2e1)

u(0) = 0 , ω(0) = ω0 (e1e2 + e2e1)

calculate the displacement eld u(x).

Chapter 2

The Field Equations of Linear

Continuum Mechanics

Until now we have developed the basic kinematical and dynamical notions concerning ma-terial bodies. As mentioned before, balance laws of linear momentum, angular momentumand energy are not sucient to model mechanical behavior of deformable bodies, and weadditionally need the so called constitutive laws. Constitutive laws try to express materialdeformation versus loading, based on specic thermo-mechanical assumptions. Dependingon the type of constitutive formulation one can classify materials to e.g. elastic, plastic,viscoelastic etc.

In this chapter we briey address constitutive laws and governing equations of linearelasticity. Derivation of these laws and the respective thermodynamics is postponed to thesecond part of these lectures in nonlinear continuum mechanics where hyper-elasticity isstudied.

2.1 Elasticity

So far we have established two separate sets of equations, namely equilibrium equations(1.22) and kinematic equations of strain-displacement (1.61), and now we want to studystress-strain relations. As will be seen, this completes the set of equations, i.e. there willbe as many independent equations as unknowns.

We shall conne our study within the so called elastic range, that is the range of strainfor which the material behavior remains elastic. A body of material is called elastic if ateach body point the strain is a one-to-one function of stress at that point regardless of thehistory of loading. In formal notation

σ(x)←→ ε(x) . (2.1)

This ensures that stress and strain are both single-valued functions of each other whichin turn means that the material follows the same stress-strain curve during loading andunloading, and shows no hysteresis eect. This is more rigorously presented within theframework of hyper-elasticity later on.

17

18CHAPTER 2. THE FIELD EQUATIONS OF LINEAR CONTINUUM MECHANICS

2.2 Hooke's law

Experimental data for most hard solids shows a linear relation between load and strainwithin a specic interval during loading and unloading. In the most general representationthis linearity is of the form of generalized Hooke's law as

σ(x) = C(x) : ε(x) (2.2)

which says that each of stress components at any point of a body is a linear function of thestrain components at that point. Here the so called stiness tensor C is introduced whichis a fourth-order tensor. This linearity could be also represented as

ε(x) = S(x) : σ(x) (2.3)

where S is called compliance tensor which is a fourth-order tensor satisfying the relation

S : C = C : S = I . (2.4)

and I is the fourth-order identity tensor

I = δikδjl eiejekel . (2.5)

The functionality of position x is emphasized in equation (2.2), however we usuallywrite

σ = C : ε . (2.6)

Specially for a homogeneous∗ body when C does not depend on x, we have

σ(x) = C : ε(x) .

To further clarify the above equations we can write them in index notation as

σij = Cijkl εkl . (2.7)

The stiness tensor, being of fourth-order, has 3×3×3×3 = 81 components. However,these components are not independent. Looking into equation (2.2), since stress and straintensors are both symmetriceach having six independent componentsthe stiness tensorcan only have 6×6 = 36 independent components. Because the most general linear relationbetween the six stress components and the six strain components can take the form of

σ11

σ22

σ33

σ23

σ31

σ12

=

K11 K12 K13 K14 K15 K16

K21 · · ·

K31. . .

...

K61 · · · K66

ε11

ε22

ε33

2ε23

2ε31

2ε12

(2.8)

which is called contracted notation. Here the original stress and strain tensors are recastin a vectorial form. The factor 2 in front of shear strains has technical reasons beyond ourscope, however it can be omitted in principle.

∗A homogeneous body is the one whose physical properties are the same for all of its points.

2.2. HOOKE'S LAW 19

Remark 3. Note that the symmetry of stress and strain tensors is fullled only if

Cijkl = Cjikl = Cijlk (2.9)

which is called minor symmtery of the stiness tensor.

We will see later on that the existence of strain energy potential requires the fourthorder tensor C to satisfy

Cijkl = Cklij (2.10)

the so called major symmetry, which in turn requires the symmetry of the stiness matrixK in equation (2.8). Therefore the matrix K has only 21 independent components.

The above arguments which prove the reduction of number of independent stinesscomponents from 81 to 21 can be summarized as follows

Summary 4. Since the stiness tensor has to fulll major and minor symmetries

Cijkl = Cklij = Cjikl = Cijlk (2.11)

it has only 21 independent components for a linear elastic material.

2.2.1 Hooke's law for isotropic materials

As yet, we have simplied the general Hooke's law based on symmetry properties (2.11).Now we consider material symmetries based on which we can further reduce the numberof required stiness components.

Solids have crystal structures at the atomic scale. These crystal structures are classiedbased on their symmetry properties. A detailed study of how these symmetries inuencethe stiness tensor goes beyond our agenda (see [4]), but regardless of crystallography atypical solid material is composed of microscopic grains with random distributions andorientations throughout the body. Although each grain has a specic crystal structure withits respective symmetries, the overall behavior of the material at the macro scale seemsisotropic due to this randomness.

Denition 5. An isotropic material is the one that has identical thermo-mechanical prop-erties in all directions at each material point.

This requires that the elastic constants† be invariant under an arbitrary rotation of coor-dinate system

Cijkl = C ′ijkl (2.12)

and consequently we come up with two substantial results.

Theorem 6. For an isotropic material, the principal axes of the strain tensor coincide with

the principal axes of the stress tensor.

†Components of stiness tensor are also called elastic constants.


Theorem 7. For an isotropic material, the stiness tensor in principal coordinates follows

the relationships

C1111 = C2222 = C3333 = C (2.13)

C1122 = C2233 = C3311 = C2211 = C1133 = C3322 = λ (2.14)

and the rest of components are zero.

So, there are only two independent stiness components for an isotropic material inprincipal coordinates. But this holds in any other coordinates as well because the rotationtransformation from principal coordinates to any other coordinates leaves us with a stinesstensor whose components are expressed in terms of the two components above. For anarbitrary coordinate system one can show the following constitutive equation

σ = λ tr(ε) I + 2µ ε (2.15)

where λ and µ are Lamé parameters. Where λ is the same as in (2.14), and µ is related tothe constant C in (2.13) by 2µ = C − λ. The above equation in index notation is given by

σij = λ εkk δij + 2µ εij . (2.16)

The inverse form of equation (2.15) is given by

ε =1

2µσ − λ

2µ (3λ+ 2µ)tr(σ) I . (2.17)

In view of Hooke's law as given by (2.2) and (2.3) stiness and compliance tensors for anisotropic material are given in terms of Lamé parameters as

C = λI ⊗ I + 2µI and S = − λ

2µ (3λ+ 2µ)I ⊗ I +

1

2µI , (2.18)

where I and I are the second-order and fourth-order identity tensors respectively.

2.2.2 Alternative formulations of Hooke's law for isotropic materials

We proceed with reformulations of Hooke's law which are suitable for dierent types ofproblems.

Volumetric-deviatoric decomposition

The strain tensor can be additively decomposed into the so called volumetric and deviatoric

parts

ε = vol(ε) + dev(ε) : vol(ε) =1

3tr(ε) I , dev(ε) = ε− 1

3tr(ε) I . (2.19)

Kinematically, the volumetric part reects dilatation which is the relative change of volume

dv − dVdV

= tr(ε) (2.20)

2.2. HOOKE'S LAW 21

and the deviatoric part reects distortion. Interesting is that if we decompose the stress inthe same fashion into deviatoric and volumetric parts as

σ = vol(σ) + dev(σ) : vol(σ) =1

3tr(σ) I , dev(σ) = σ − 1

3tr(σ) I (2.21)

then the Hooke's law is decomposed such that volumetric stress depends on volumetricstrain only, and deviatoric stress depends on deviatoric strain only by

σvol = (3λ+ 2µ) εvol , σdev = 2µεdev (2.22)

The so called bulk modulus or compression modulus of elasticity is introduce as

K = λ+2

3µ (2.23)

which based on (2.22) yieldsvol(σ) = 3K vol(ε) (2.24)

Remark 8. The deviatoric parts of stress and strain tensors are traceless i.e. tr(dev(•)) = 0.

Maybe the most important aspect of the above decomposition is that deviatoric andvolumetric parts are orthogonal to each other in the sense that

vol(σ) : dev(σ) = 0 , vol(ε) : dev(ε) = 0 . (2.25)

Then it follows thatσ : ε = vol(σ) : vol(ε) + dev(σ) : dev(ε) (2.26)

which means the elastic energy is decomposed into the volumetric elastic energy and thedeviatoric elastic energy.

Young's modulus, shear modulus and Poisson's ratio

Hooke's law for isotropic solids is often expressed in terms of a set of material constantswhich are more favorable in applied mechanics because they can be directly measured inexperiments. Those constants are Young's modulus

E =µ (2λ+ 2µ)

λ+ µ, (2.27)

Poisson's ratio

ν =λ

2 (λ+ µ), (2.28)

and usually renaming one of Lamé parameters to shear modulus

G = µ . (2.29)

Then Hooke's law is reformulated as

σ =Eν

(1 + ν) (1− 2ν)tr(σ) I +

E

1 + νε (2.30)

and its inverse as

ε =1 + ν

Eσ − ν

Etr(σ) I . (2.31)


Exercise 12. During the tension test of a prismatic bar with square cross section havingthe dimensions l = 500mm and a = 30mm, the changes in the dimensions of ∆l = 8mm and∆d = −0.2mm are obtained after application of a force F = 2000N (Fig. 2.1). Calculate the

F Fa a−∆a

l

l + ∆l

Fig. 2.1: Prismatic bar subjected to tension.

material constants: λ, µ,E, ν.

2.2.3 Plane-strain and plane-stress problems

Whenever a general theory is developed there are special cases that draw special attention,sometimes due to their broad range of application and other times for the impact theyhave in development of models. Two well know examples of such cases in elasticity areplane-strain and plane-stress problems.

Plane-strain

As the name suggests, plane-strain is a type of deformation in a material body when givena specic plane, all the o-plane components of strain are zero or negligibly small. Assumethat the plane of strain is normal to X3 axis in Cartesian coordinates. Then we may expressthe plane-strain condition by

εij = 0 for i = 3 or j = 3 . (2.32)

Then Hooke's law for an isotropic material becomes

σ11 = λ (ε11 + ε22) + 2µ ε11 (2.33)

σ22 = λ (ε11 + ε22) + 2µ ε22 (2.34)

σ12 = 2µ ε12 (2.35)

σ33 = λ (ε11 + ε22) (2.36)

σ13 = σ23 = 0 (2.37)

Plane-stress

Now the meaning of plane-stress should be immediately clear by analogy. Plane stress isa condition where for a given plane all the o-plane stress components are zero. If theplane of stress is normal to X3 axis in Cartesian coordinates then we state the plane-stresscondition as

σij = 0 for i = 3 or j = 3 . (2.38)

2.3. NAVIER AND BELTRAMI-MICHELL EQUATIONS 23

With introduction of a new constant

λ =2λµ

λ+ 2µ(2.39)

the Hooke's law for isotropic materials becomes

σ11 = λ (ε11 + ε22) + 2µ ε11 (2.40)

σ22 = λ (ε11 + ε22) + 2µ ε22 (2.41)

σ12 = 2µ ε12 . (2.42)

Also note that as a result of condition (2.38) we have

ε33 = − λ

λ+ 2µ(ε11 + ε22) (2.43)

ε13 = ε23 = 0 . (2.44)

Exercise 13. Prove the following relationships

ν =3K − E

6K, λ =

3K − 2µ

3, E =

λ (1 + ν) (1− 2ν)

ν.

2.3 Navier and Beltrami-Michell equations

Now we shall be able to write the governing equations of isotropic elasticity as Hooke's lawcompletes the set of equations. Let us restate all those equations together

ε =1

2(∇u+ u∇) εij =

1

2(∂iuj + ∂jui) (2.45)

∇ · σ + f = 0 ∂iσij + fj = 0 (2.46)

σ = λ tr(ε) I + 2µ ε σij = λ εkk δij + 2µ εij . (2.47)

In the rst step we would like to have an equation in terms of displacement eld only.For this we have to substitute ε from (2.45) into (2.47) which yields

σij = λ1

2(∂kuk + ∂kuk) δij + 2µ

1

2(∂iuj + ∂jui)

= λ∂kuk δij + µ (∂iuj + ∂jui) (2.48)

and then plug in this result into (2.46)

∂i (λ∂kuk δij + µ (∂iuj + ∂jui)) + fj = λ∂i∂kukδij + µ∂i∂iuj + µ∂i∂jui + fj

= λ∂j∂kuk + µ∂i∂iuj + µ∂j∂kuk + fj

= (λ+ µ) ∂j∂kuk + µ∂i∂iuj + fj = 0 . (2.49)

This equation can be written in operator notation as

(λ+ µ)∇∇ · u+ µ∆u+ f = 0 , (2.50)

which is the so called Navier equation. Since u is a vector the equation is basically a systemof three partial dierential equations.


Exercise 14. Prove the alternative form of the Navier equation as

(λ+ 2µ)∇∇ · u− µ∇×∇× u+ f = 0 .

Another step would be to set up a system of dierential equations in terms of stresstensor. Stress tensor has six independent components. The equation (2.46) provides threepartial dierential equations which is not sucient. On the other hand equations (2.45)and (2.47) do not provide more equations in terms of stress. Therefore we need additionalequations. It turns out the additional equation is the alternative form of compatibilitycondition (1.68) that we repeat here

∆ε+∇tr(ε)∇ = ∇∇ · ε+ ε · ∇∇ . (2.51)

From equation (2.22) we have

tr(ε) =1− 2ν

Etr(σ) , (2.52)

and the inverse form of Hooke's law (2.31)

ε =1 + ν

Eσ − ν

Etr(σ) I . (2.53)

Now we put (2.53) and (2.52) into (2.51) to obtain

1 + ν

E∆σ − ν

E∆tr(σ) I +

1− 2ν

E∇tr(σ)∇ =

1 + ν

E(∇∇ · σ + σ · ∇∇)− ν

E(∇∇ · (tr(σ) I) + (tr(σ) I) · ∇∇) . (2.54)

For the second term on the left-hand-side based on equilibrium equation (2.46) we have

∆tr(σ) = ∂m∂mσkk = ∇ · σ · ∇ = −∇ · f . (2.55)

For the rst term on the right-hand-side again based on (2.46) we have

∇∇ · σ + σ · ∇∇ = − (∇f + f∇) , (2.56)

because ∇ · σ = σ · ∇ = −f . Considering the second term on the right-hand-side

(∇∇ · (tr(σ) I))ij = ∂i∂kσmmδkj = ∂i∂jσmm = (∇tr(σ)∇)ij . (2.57)

The same can be easily shown for (tr(σ) I) · ∇∇ therefore

∇∇ · (tr(σ) I) = (tr(σ) I) · ∇∇ = ∇ tr(σ)∇ . (2.58)

Substituting all these results into (2.54) yields

∆σ +1

1 + ν∇ tr(σ)∇+

ν

1− ν∇ · f I +∇f + f∇ = 0 , (ν 6= −1) (2.59)

the so called Beltrami-Michell formulation. Note that Beltrami-Michell equation is in factcompatibility equation in terms of stress which together with equilibrium equation ∇ ·σ+f = 0, establishes a complete system of partial dierential equations for the stress eld.

2.3. NAVIER AND BELTRAMI-MICHELL EQUATIONS 25

Exercise 15. Assume f = const., using Beltrami-Michell equation show that

∆∆σ = 0 and ∆∆ε = 0 .

Exercise 16. Assuming that

σ = σ(r) and f = f(r) ,

derive Beltrami-Michell equation in cylindrical coordinates.


Chapter 3

Displacement Functions

An elastostatics problem is basically a boundary value problem, with boundary conditionsof the type Dirichlet, Neumann or mixed. In chapter 2 we derived the governing equationsof linear isotropic elasticity in terms of displacement eld (Navier equation) and in termsof stress eld (Beltrami-Michell and equilibrium equations).

In this chapter we present some classical solutions to the displacement formulation in acoherent way. All the methods are meant to simplify the solution of the original boundaryvalue problem. We shall often assume that the body forces are not present which is nota serious restriction as in practical situations, nding the particular solution belonging tobody forces is not complicated∗. Throughout our derivations we frequently employ tensorialand vectorial identities.

Our point of departure is the Navier equation with zero body force

(λ+ µ)∇∇ · u+ µ∆u = 0 , (3.1)

with prescribed boundary conditions

u(x) = u∗ , x ∈ Γu (3.2)

σ(x) · n = t∗ , x ∈ Γσ . (3.3)

The idea is to substitute u with a combination of scalar or vector functions, called dis-placement functions, or their derivatives so that the governing equation in terms of thesefunctions takes the form of harmonic or biharmonic equations which are basically simplerto solve than the Navier equation. However, the boundary conditions expressed in termsof displacement functions become more complicated.

3.1 Scalar and vector potentials

The rst approach we are going to present is based on the so called Helmholtz theorem asfollows.

∗We remember that the solution to an inhomogeneous dierential equation is the sum of the generalsolution to the homogenized equation and the particular solution to the inhomogeneous equation.

27

28 CHAPTER 3. DISPLACEMENT FUNCTIONS

Theorem 9. Every nite vector eld which is uniform, continuous and vanishing at in-

nity may be decomposed into the sum of a curl-free (irrotational) and a divergence-free

(solenoidal) eld.

We know that a curl-free vector eld has the form of v = ∇φ, and a divergence-freevector eld has the form of w = ∇ × ψ. Therefore if the displacement eld u fullls theassumptions of Helmholtz theorem we can write

u = ∇φ+∇×ψ , (3.4)

where φ is a scalar and ψ is a vector eld. Now if u is substituted from (3.4) into (3.1) itgives

(λ+ µ)∇∇ · u+ µ∆u = (λ+ µ)∇∇ · [∇φ+∇×ψ] + µ∆ [∇φ+∇×ψ]

= (λ+ ν) [∇∇ · ∇φ+∇((((((∇ · (∇×ψ)] + µ∆∇φ+ µ∆∇×ψ

= (λ+ µ)∇∆φ+ µ∇∆φ+ µ∇×∆ψ = 0 , (3.5)

which nally gives(λ+ 2µ)∇∆φ+ µ∇×∆ψ = 0 . (3.6)

Any set of φ and ψ that satises this equation will also satisfy (3.1) when u = ∇φ+∇×ψ,and for any solution u to (3.1) there exits at least one set of φ and ψ that satises (3.6),which is not unique because u is expressed in terms of rst derivatives of φ and φ. Howeverthis is not important because we are interested in nding the solution in terms of u.

One particular solution of (3.6) is the solution to

∆φ = const. and ∆ψ = const. (3.7)

and as long as this solution is capable of satisfying the boundary conditions our objectiveis fullled. The two above equations are called Laplace's equations or harmonic equations.

Remark 10. The reader may wonder why we reduced the original problem to the particularcase of harmonic functions. The answer is: simplication. However it might be the casethat the problem is over-simplied, but we do not know until we try the harmonic solutionand see if the boundary conditions are fullled. The good news is that harmonic solutionsare rich enough to reect the boundary conditions in many engineering problems we face.

Example 3.1. Wave propagation. The dynamic case of wave propagation can be studiedwith Helmholtz decomposition. Instead of equilibrium equation we have the balance ofmomentum

∇ · σ + f = ρu , (3.8)

where ρ is density and u is acceleration. Consequently Navier equation becomes

(λ+ µ)∇∇ · u+ µ∆u+ f = ρu . (3.9)

After substitution of Helmholtz decomposition and derivations we get the particular caseof

(λ+ 2µ) ∆φ = ρφ , µ∆ψ = ρψ , (3.10)

3.1. SCALAR AND VECTOR POTENTIALS 29

which are called wave equations. The solution will be of the form

φ = φ(k · x− cpt) , ψ = ψ(k · x− cst) , (3.11)

with k being a unit vector called wave-number vector or propagation vector and cp and csare compression wave speed and shear wave speed. In order to obtain the wave speed wehave

φ = φ(kjxj − cpt) =⇒

∆φ = ∂i∂iφ = kikiφ

′′ = φ′′

φ = ∂t∂tφ = c2pφ′′ (3.12)

which by substitution into the wave equation for φ yields

(λ− 2µ) φ′′ = c2pρφ′′ , (3.13)

therefore

cp =

√λ+ 2µ

ρ. (3.14)

In the same way we can show that the shear wave speed is

cs =

√2µ

ρ. (3.15)

hr

z

p(r)

Fig. 3.1: Axially symmetric slab.

Example 3.2. A two-dimensionally innite slab is loaded at the top with a given axiallysymmetric distribution p(r) in cylindrical coordinates, and constrained with a frictionlesssurface at the bottom. We want to nd the solution in terms of Helmholtz functions.

Solution. To understand the problem we have to express boundary conditions. At the topwe have a vertical force distribution therefore

z = h : σzz = −p(r) , σrz = 0 (3.16)

and at the bottom there is no vertical movement and tangential force (no friction!)

z = 0 : uz = 0 , σrz = 0 . (3.17)


Axial symmetry means that u, ε and σ are independent of ϕ. Back to equations (3.7) weassume the special case

ψ = 0 (3.18)

∆φ = 0 , (3.19)

and we have to see if the solution can fulll all the boundary conditions. If it does not thenwe will assume a more sophisticated case. Now we need to express the boundary conditionsin terms of φ. Starting from the above assumption

ε =1

2(∇u+ u∇) = ∇φ∇ , (3.20)

therefore based on Hooke's law (2.15)

σ = λ tr(ε) I + 2µ ε = λ∆φ I + 2µ∇φ∇ = 2µ∇φ∇ . (3.21)

Then in cylindrical coordinates

ur = ∂rφ , uz = ∂zφ (3.22)

σrr = 2µ∂2rφ , σϕϕ = 2µ

1

r∂rφ (3.23)

σzz = 2µ∂2zφ , σrz = 2µ∂r∂zφ , (3.24)

and the boundary conditions (3.16) and (3.17) in terms of φ are obtained as

z = h : 2µ∂2zφ = −p(r) , 2µ∂r∂zφ = 0 (3.25)

z = 0 : ∂zφ = 0 , 2µ∂r∂zφ = 0 (3.26)

The Laplace's equation (3.19) in cylindrical coordinates reads

∆φ = ∂2rφ+

1

r∂rφ+ ∂2

zφ = 0 . (3.27)

For separation of variables† we put

φ(r, z) = f(r) g(z) , (3.28)

which results in

f ′′g +1

rf ′g + fg′′ = 0 (3.29)

after factorizationf ′′ + 1/r f

f= −g

′′

g= ±k2 , (3.30)

where ±k2 reects the possibility of a positive or negative constant. Since the rst expres-sion in the above equation is a function of r only and the second is a function of z only,therefore the only possible way for them to be equal is to be constant.

†Separation of variables is a standard method to reduce PDE's to ODE's.

3.1. SCALAR AND VECTOR POTENTIALS 31

The negative choice −k2 leads to contradiction with boundary values. Therefore wetake the positive constant +k2 which yields two ordinary dierential equations

f ′′ +1

rf ′ − k2f = 0 , g′′ + k2g = 0 . (3.31)

The rst equation is the so called Bessel's equation of zeroth order which has the generalsolution

f(r) = AJ0(kr) +BH0(kr) , (3.32)

where J0 is Bessel function of zeroth order and H0 is Hankel function of zeroth order.The second equation has the well know real solution

g(z) = C sin(kz) +D cos(kz) . (3.33)

SinceH0(x)→∞ for x→ 0 (3.34)

in order to have a nite solution in the neighborhood of r = 0 it must hold that B = 0,then the general solution becomes

φ = J0(kr) (C sin(kz) +D cos(kz)) . (3.35)

Applying the boundary conditions gives

uz(0) = ∂zφ(0) = J0(kr) k C = 0 ⇒ C = 0 (3.36)

σrz(0) = 2µ∂r∂zφ(0) = 2µk J ′0(kr) k C = 0 X (3.37)

σrz(h) = sµk J ′0(kr) (−Dk sin(kh)) = 0 ⇒ sin(kh) = 0 ⇒ kn =nπ

h, n = 0, 1, . . .

(3.38)

As you can see there are innitely many possible values for k and therefore there areinnitely many solutions of the form

φn(r, z) = J0(knr) cos(knz) , (3.39)

which are called normal modes. The general solution is the linear combination of all possiblesolutions as

φ =∞∑n=0

an φn . (3.40)

This situation is a bit tricky because there are innitely many constants an's to determinebut there is only one remaining boundary condition to be fullled as

σzz(h) = 2µ∂2zφ = −2µ

∑n

kna2nJ0(knr) cos(knh) =

− 2µ∑n

(−1)n ank2nJ0(knr) = −p(r) . (3.41)


The key idea is the orthogonality property of Bessel functions as∫ ∞0

J0(kir) J0(kjr) r dr = 0 for i 6= j . (3.42)

Therefore if we multiply both sides of (3.1) with J0(kir) r and then integrate over [0,∞]we obtain

2µ(−1)iaik2i

∫ ∞0

(J0(kir))2 r dr =

∫ ∞0

p(r) J0(kir) r dr (3.43)

because all the terms except the ith term vanish due to (3.42), and ai is obtained as

ai =1

2µ(−1)ik2i

∫∞0 p(r) J0(kir) r dr∫∞0 (J0(kir))

2 r dr(3.44)

which nally yields

φ =

∞∑n=0

1

2µ(−1)ik2n

∫∞0 p(r) J0(knr) r dr∫∞0 (J0(knr))

2 r drJ0(knr) cos(knz) . (3.45)

Exercise 17. Using Helmholtz theorem calculate u, ε and σ for the given scalar and vectorpotentials

φ = k(x2

1 − x22

); ψ = l

x21 − x2

2

00

.

Exercise 18. A circular plate with radius R and thickness h is constrained at the top andbottom by frictionless surfaces (Fig. 3.2). A load distribution p(z) is applied on its verticalboundary. Using Helmholtz theorem derive solution for axially symmetric problem.

z

r

p (z)

R

h

Fig. 3.2: Circular plate under radial loading.

Hint: If the ansatz φ = J0(kr) [C sin(kz) +D cos(kz)] is used, the solution in the formφn = J0(knr) cos knz will be obtained. Then Fourier expansion in z should be used.

3.2 Galerkin vector

Helmholtz decomposition gives displacement eld in terms of rst derivatives of a scalarand a vector potential. It is also possible to express u in terms of second derivatives of avector eld. Let us introduce the Galerkin vector V which is related to u by

u = k∆V −∇∇ · V (3.46)

where k is a constant.

3.2. GALERKIN VECTOR 33

Theorem 11. For any vector eld u and constant k there exists a vector eld V that

satises the relation (3.46).

Therefore the relation (3.46) provides a completely general solution if V is determinedsomehow. If the Galerkin vector is plugged into Navier equation (2.50) it gives

(λ+ µ) k∇∇ ·∆V − (λ+ µ)∇∇ · ∇∇ · V + µk∆∆V − µ∆∇∇ · V + f = 0 (3.47)

and then

[(λ+ µ) k − (λ− 2µ)]∇∇ ·∆V + µk∆∆V + f = 0 , (3.48)

as we are allowed to chose the constant k, the best choice is the one that simplies theabove equation the most, which turns out to be

k =λ+ 2µ

λ+ µ= 2 (1− ν) , (3.49)

and this gives

2 (1− ν)µ∆∆V + f = 0 , (3.50)

which is a vectorial biharmonic equation. Biharmonic equations are well studied in analysisand nding the answer to the above equation is basically simpler than the original Navierequation. Then we obviously need the relation (3.46) to transform the answer in terms ofV to u.

Remark 12. Galerkin vector is related to Helmholtz potentials by

φ = −∇ · V , ∇×ψ = 2 (1− ν) ∆V . (3.51)

3.2.1 Love's strain function

There is a special case of Galerkin vector which can be used in axially symmetric problems.Assuming that the only non-vanishing component of Galerkin vector is the X3 componentwe have

V1 = V2 = 0 , V3 = L (3.52)

where L is Love's strain function. Then (3.50) simplies to

2 (1− ν)µ∆∆L+ f3 = 0 . (3.53)

Note that f1 = f2 = 0 which is the symmetry requirement. The relation (3.46) in terms ofLove's function reads

u1 = −∂1∂3L , u2 = −∂2∂3L , u3 = 2 (1− ν) ∆L− ∂23L . (3.54)

Example 3.3. In the case of spherical symmetry L is a function of r only

L = L(r) : r =√xixi = ‖x‖ (3.55)


and Laplacian in spherical coordinates is given by

∆ ≡ ∂2r +

2

r∂r . (3.56)

Therefore in the absence of body forces equation (3.53) becomes

∆∆L =

(∂2r +

2

r∂r

)(∂2r +

2

r∂r

)L = ∂4

rL+4

r∂3rL = 0 . (3.57)

The general solution to this equation has the form

L = C1r−1 + C2 + C3r + C4r

2 , (3.58)

with Ci's being integration constants to be determined by application of boundary condi-tions.

Kelvin's problem

f0e3

Fig. 3.3: Kelvin's problem.

The classical problem of Kelvin deals with a concentratedforce in an innite medium. This is an important case whosesolution is used to solve more complicated problems. Let usassume the concentrated force at the origin of coordinate sys-tem and directed towards X3 axis as

f = f0 δ(x) e3 (3.59)

where δ(x) is Dirac's delta function in 3-dimensional space.Due to the axial symmetry we use the Love's formulation ofNavier equation with body force

2 (1− ν)µ∆∆L = −f0 δ(x) . (3.60)

Integrating both sides within a sphere of radius R surrounding the origin yields

2 (1− ν)µ

∫r≤R

∆∆LdV = −f0

∫r≤R

δ(x) dV = −f0 , (3.61)

where we have used the fundamental property of Dirac's delta∫r≤R δ(x) dV = 1. On the

other hand we know that ∆∆L = ∇ · ∇∆L, and using Gauss theorem∫r≤R

∆∆LdV =

∫r≤R∇ · ∇∆LdV =

∫r=R

n · ∇∆LdS =

∫r=R

∂r∆LdS . (3.62)

The general solution of the form (3.58)

L = C1r−1 + C2 + C3r + C4r

2

is simplied toL = C1r

−1 + C2 (3.63)

3.3. DISPLACEMENT FUNCTIONS OF PAPKOVICH-NEUBER 35

because otherwise

x→∞ : L→∞ (3.64)

which is not acceptable as the solution has to stay nite. Then plugging (3.63) into thesurface integral in (3.62) gives∫

r=R∂r∆LdS = 4πR2

(−2C3

1

R2

)= −8πC3 (3.65)

then putting back into (3.61) results in

16π (1− ν)µC3 = f0 (3.66)

and nally the answer is

L =f0

16π (1− ν)µr . (3.67)

Exercise 19. For the Love strain function in (3.67) calculate u and σ.

3.3 Displacement functions of Papkovich-Neuber

So far we have managed to replace the Navier equation by the fourth-order biharmonicequation (3.50) and the third-order equation (3.6). Note that the harmonic solutions to(3.6) are special cases that satisfy the original third-order equation, not the general solutionto it. Now it would be nice if we could have an equivalent formulation to Navier equationwhich is of the second-order i.e. one order lower with respect to (3.6) and two orders lowerwith respect to (3.50).

Starting from Galerkin vector (3.46), if the vector eld A and the scalar eld b areintroduced such that

A =1

2∆V , b = ∇ · V (3.68)

then we can write displacement eld in terms of A and b

u = 4 (1− ν)A−∇b . (3.69)

Therefore

∆A =1

2∆∆V = 0 X , see (3.50) (3.70)

∆b = ∆∇ · V = ∇ ·∆V = 2∇ ·A . (3.71)

Exercise 20. For the Papkovich-Neuber solution u = 4 (1− ν)A−∇b show that

∆ (A ·X) = 2∇ ·A . (3.72)


Using the result in the recent exercise the general form of b is given by

b = A · x+ a , ∆a = 0 . (3.73)

Note that a can be any harmonic function in general. Then this all comes down to

u = 4 (1− ν)A−∇ [A · x+ a] (3.74)

∆A = 0 , ∆a = 0 . (3.75)

This is called Papkovich-Neuber solution which is equivalent to Galerkin's solution, andsince Galerkin's solution is equivalent to Navier equation, then the above formulation iscompletely general and equivalent to our original formulation of Navier.

Example 3.4. Problem of Leon. Find the displacement eld in a uni-axially loadedinnite medium with a spherical cavity (Fig. 3.4).

R

σ

Fig. 3.4: Leon's problem.

Solution. Laplace's equation in spherical coordinates (inde-pendent of ϕ due to axial symmetry of our problem), reads

∆f =∂2f

∂r2+

2

r

∂f

∂r+

cos θ

r2 sin θ

∂f

∂θ+

1

r2

∂2f

∂θ2= 0 (3.76)

and we have to solve (3.75)

∆A = 0 , ∆a = 0 . (3.77)

Because of axial symmetry around the axis of loading the onlynonzero component of A is Az, which means Ax = Ay = 0and we are left with

∆Az = 0 , ∆a = 0 . (3.78)

By separation of variables we seek solutions of the form

f(r, θ) = R(r) Θ(θ) (3.79)

that ends up in

1

R

d

dr

(r2dR

dr

)= − 1

sin θΘ

d

dθ

(sin θ

dΘ

dθ

)= l (l + 1) , l ∈ 0, 1, 2, . . . . (3.80)

The special form of the constant l (l + 1), for l being a non-negative integer, guarantees aconvergent solution of Legendre equation that appears here in terms of η = cos θ‡(

1− η2) d2Θ

dη2− 2η

dΘ

dη+ l (l + 1) Θ = 0 . (3.81)

‡To derive (3.80) into (3.81) we need to know from the chain rule that

d

dθ= − sin θ

d

dη⇒ d

dη= − 1

sin θ

d

dθ

.

3.3. DISPLACEMENT FUNCTIONS OF PAPKOVICH-NEUBER 37

The general solution is given by

f =

(Arl +B

1

rl+1

)Pl(cos θ) , (3.82)

where Pl(x) is Legendre polynomial of degree l. Then the general solution to (3.78) isobtained

a =∞∑l=0

(clr

l + dl1

rl+1

)Pl(cos θ) (3.83)

Az =∞∑l=0

(Clr

l +Dl1

rl+1

)Pl(cos θ) . (3.84)

The boundary conditions are given as

r →∞ : σzz =σ (3.85)

r = R : σrr = σrθ = 0 . (3.86)

In order to homogenize the boundary condition the following re-parameterization is intro-duced

σ = σ − σ ,

σ =

σezez . (3.87)

Then according transformation rules for second-order tensors

σrr =

σ cos2 θ ,

σθθ =

σ sin2 θ ,

σrθ = −

σ sin θ cos θ . (3.88)

And in terms of σ the boundary conditions are

r →∞ : σ → 0 ⇒ cl = 0 , Cl = 0 (3.89)

r = R : σrr = − σrr , σrθ = − σrθ . (3.90)

Considering the number of boundary conditions to be fullled we can start from a limitednumber of terms in (3.83) and (3.84), say l = 0, 1, 2 to see if the boundary conditions arefullled. If we need more terms we can add them later on, so let us assume

a = d01

r+ d1

1

r2cos θ + d2

1

r3

(3 cos2 θ − 1

)(3.91)

Az = D01

r+D1

1

r2cos θ +D2

1

r3

(3 cos2 θ − 1

). (3.92)

(3.93)

To express the boundary conditions in terms of a and Az we use the following result whoseproof is left as an exercise

ur = 4 (1− ν)Az cos θ − ∂r (Azr cos θ + a) (3.94)

uθ = −4 (1− ν)Az sin θ − 1

r∂θ (Azr cos θ + a) . (3.95)


We also need the following formulas

εrr = ∂rur , εθθ =1

r∂θuθ +

1

rur (3.96)

εrθ =1

2

(1

r∂θur + ∂ruθ −

1

ruθ

)(3.97)

εϕϕ =1

rur +

1

rcot θ uθ . (3.98)

It is left to the reader as an important exercise to apply the boundary conditions and obtainthe following results

d0 = − 1

4µ

6− 5ν

7− 5νR3 σ , D0 = 0 , (3.99a)

d1 = 0 , D1 =1

4µ

5

7− 5νR3 σ , (3.99b)

d2 = − 1

4µ

1

7− 5νR5 σ , D2 = 0 . (3.99c)

Exercise 21. Using Papkovich-Neuber solution

u = 4 (1− ν)A−∇b , b = A ·X + a

for the ansatz

a =∞∑n=0

(cnr

n + dn1

rn+1

)pn(cos θ)

A1 = A2 = 0

A3 =

∞∑n=0

(Cnr

n +Dn1

rn+1

)pn(cos θ)

with pn(cos θ) being Legendre polynomial, show that

ur = 4 (1− ν)A3 cos θ − ∂r (A3r cos θ + a)

uθ = −4 (1− ν)A3 sin θ − 1

r∂θ (A3r cos θ + a) .

Exercise 22. Using Papkovch-Neuber solution for a sphere with radius R and surfacedisplacement ur(R, θ) = 0 and uθ(R, θ) = u0 sin θ nd σ(r, θ).Hint: use ansatz of the previous assignment but only up to n = 1 for a and A3.

Bibliography

[1] A.I. Borisenko and I.E. Tarapov. Vector and Tensor Analysis with Applications. DoverPub. Inc, 1968.

[2] P.C. Chou and N.J. Pagano. Elasticity: Tensor, Dyadic, and Engineering Approaches.Dover Pub. Inc, 1992.

[3] G.A. Holzapfel. Nonlinear Solid Mechanics: A Continuum Approach for Engineering.Wiley, 2000.

[4] A.E. Love. Treatise on the Mathematical Theory of Elasticity. Dover Pub. Inc, 4th

edition, 1927.

[5] R.W. Ogden. Non-Linear Elastic Deformations. Dover Pub. Inc, 1997.

[6] M.R. Spiegel. Mathematical Handbook of Formulas and Tables. McGraw-Hill (Schaum'soutline), 1979.

[7] T.C.T. Ting. Anisotropic Elasticity: Theory and Applications. Oxford University Press,1996.

39

Date post:	28-Oct-2015
Category:	Documents
Upload:	zcapg17
View:	238 times
Download:	7 times

Lecture Notes - Linear Elasticity Theory

Documents