Brief introduction to continuum mechanics

Lecture notes to the course MEC-2410 Mechanics of materials - spring 2012

Reijo Kouhia

March 7th, 2012This is a very preliminary versio and contains many typos. I am very pleased to have anote of any error and any kind of idea on improving the text is wellcome. My e-mail

address is reijo.kouhia@tut.fi.

Chapter 1Introduction

1.1 The general structure of continuum mechanicsIn principle the general structure of equations in continuum mechanics is threefold. First,there is a balance equation (or balance equations), stating the equilibrium or force balanceof the system considerer. These equations relate e.g. the stress with external forces.Secondly, the stress is related to some kinematical quantity, like strain, by the constitutiveequations. Thirdly, the strain is related to displacements by the kinematical equations.

Balance equations are denoted as B = f , where B is the equilibrium operator,usually a system of differential operators. In the constitutive equations = C the oper-ator C can be either an algebraic or differential operator. Finally the geometrical relationi.e. the kinematical equations are denoted as = Bu. These three equations form thesystem to be solved in continuum mechanics, and it is illustrated in figure 1.1. The equi-librium operator B is the adjoint operator of the kinematical operator B. Therefore thereare only two independent quantities in the system.

Example - axially loaded bar. The equilibrium equation in terms of the axial force Nis

dNdx

= f, (1.1)

where f is the distributed load, [force/length], in the direction of the bars axis. Thus, theequilibrium operator B is

B = ddx

. (1.2)

The axial force is related to the strain via the elastic constitutive equation (containing thecross-section area as a geometric quantity)

N = EA. (1.3)

1

2 CHAPTER 1. INTRODUCTION

u

fB? = f

= C B?CBu = f

= Bu

6

-

Figure 1.1: The general structure of equations in mechanics.

In this case the constitutive operator C is purely algebraic constant C = EA. The kine-matical relation is

=du

dx, (1.4)

thus the kinematic operatorB =

d

dx, (1.5)

for which B is clearly the adjoint. The equilibrum equation expressed in terms of axialdisplacement is

BCBu = ddx

(EA

du

dx

)= f. (1.6)

Example - thin beam bending. The equilibrium equation in terms of the bending mo-ment M is

d2M

dx2= f, (1.7)

where f is the distributed transverse load [force/length]. Thus, the equilibrium operatorB is

B = d2

dx2. (1.8)

The bending moment is related to the curvature via the elastic constitutive equation (con-taining the inertiaa of the cross-section as a geometric quantity)

M = EI. (1.9)Again, the constitutive operator C is purely algebraic constant C = EI . The kinematicalrelation is

= d2v

dx2, (1.10)

1.1. THE GENERAL STRUCTURE OF CONTINUUM MECHANICS 3

thus the kinematic operator

B = d2

dx2, (1.11)

for whichB is clearly the adjoint, and also in this case B = B. The equilibrum equationexpressed in terms of axial displacement is

BCBu = d2

dx2

(EI

d2v

dx2

)= f. (1.12)

Example - linear 3-D elasticity. The equilibrium, constitutive and kinematical equa-tions are

div = b, (1.13) = C, (1.14) = sym gradu , (1.15)

where is the symmetric stress tensor, is the material density, b is the body force perunit mass, u is the displacement vector and C is the elasticity tensor. Thus the operatorsB, B and C are

B = div, (1.16)B = sym grad (1.17)C = C . (1.18)

The formal adjoint of the B = div operator is the gradient operator.

4 CHAPTER 1. INTRODUCTION

1.2 Tensor analysis

1.3 Nomenclature

Strain and stress

e, eij = deviatoric strain tensors, sij = deviatoric stress tensor

s1, s2, s3 = principal values of the deviatoric stress = shear strain

oct = octahedral shear strainij = strain tensoroct = octahedral strainv = volymetric strain

1, 2, 3 = princial strains = normal stress

, ij = stress tensorm = mean stressoct = octahedral stress

1, 2, 3 = principal stresses = shear stressm = mean shearing stressoct = octahedral shear stress

Invariants

I1(A) = trA = Aii = the first invariant of tensor AI2(A) =

12[tr(A2) (trA)2] = second invariant

I3 = detA = third invariantJ2(s) =

12tr s2 = second invariant of the deviatoric tensor s

J3 = det s = third invariant of a deviatoric tensor, , = the Heigh-Westergaard stress coordinates

= hydrostatic length = the length of the stress radius on the deviatoric plane = the Lode angle on the deviatoric plane

1.4. ON THE REFERENCES 5

Material parametersE = Youngs modulusG = shear modulusGf = fracture energyK = bulk modulusk = shear strengthm = f c = mf

t

, = Drucker-Prager murtoehdon parametreja = Poissons ratio = internal friction angle of the Mohr-Coulomb criterion

1.4 On the referencesThis lecture notes is mostly based on the following excellent books:

1. L.E. Malvern: Introduction to the Mechanics of a Continuous Medium. Beautifullywritten treatease on the topic.

2. G.A. Holzapfel: Nonlinear Solid Mechanics, A Continuum Approach for Engineers.A modern treatment of some basic material in Malverns book. Contains usefullmaterial for understanding nonlinear finite element methods.

3. J. Lemaitre, J.-L. Chaboche: Mechanics of Solid Materials.4. N.S. Ottosen, M. Ristinmaa: Mechanics of Constitutive Modelling.

6 CHAPTER 1. INTRODUCTION

Chapter 2Stress

2.1 Stress tensor and the theorem of CauchyLets consider a body B in a 3-dimensional space occupying a volume domain , seefigure 2.1. If the body B is divided into two parts by a surface S and the parts separatedfrom each other. The force acting on a small surface S is denoted by f . A tractionvector t is defined as

t = limS0

f

S=

df

dS. (2.1)

The traction vector depends on the position x and also on the normal direction n of thesurface, i.e.

t = t(x ,n), (2.2)a relationship, which is called as the postulate of Cauchy.1

In the rectangular cartesian coordinate system, the traction vectors action in threeperpendicular planes, parallel to the coordinate axes are denoted as t1, t2 and t3, seefigure 2.2. The components of the traction vectors are shown in the figure and expressedin terms of the unit vectors parallel to the coordinate axes e i the traction vectors are

t1 = 11e1 + 12e2 + 13e3, (2.3)t2 = 21e1 + 22e2 + 23e3, (2.4)t3 = 31e1 + 32e2 + 33e3. (2.5)

To obtain the expression of the traction vector in terms of the components ij , let usconsider a tetrahedra where the three faces are parallel to the coordinate planes and theremaining one is oriented in an arbitrary direction, see figure 2.3. In each of the faces,the average tranction is denoted as ti , where i = 1, 2, 3, n, and the area of the triangleA1A2A3 is denoted as S and S1,S2,S3 are the areas of triangles OA2A3, OA3A1and OA1A2, respectively. The body force acting on the tetrahedra is bV , where thevolume element V = 1

3hS, and h is the distance ON .

1Sometimes traction vector t is also called as a stress vector. However, in this lecture notes this namingis not used since the stress has a tensor character.

7

8 CHAPTER 2. STRESS

Figure 2.1: A continuum body and the traction vector.

Figure 2.2: Traction vectors in three perpendicular directions.

2.1. STRESS TENSOR AND THE THEOREM OF CAUCHY 9

Figure 2.3: Cauchys tetrahedra. Figure from [3, page 75].

10 CHAPTER 2. STRESS

Equilibrium equation for the tetrahedra is

tnS +13bhS t1S1 t2S2 + t3S3 = 0, (2.6)

which can be written as

S(tn +13bh n1t1 n2t2 + n3t3) = 0. (2.7)

Now, letting h 0 we get ti t i and

tn =3

i=1

n1t i = nit i

= n1(11e1 + 12e2 + 13e3) + n2(21e1 + 22e2 + 23e3)

+ n3(31e1 + 32e2 + 13e3), (2.8)

or

tn =

n111 + n221 + n331n212 + n222 + n332

n313 + n223 + n332

= nT = Tn . (2.9)

Notice the transpose in the stress tensor in the last expression. The stress tensori ,expressed in rectangular cartesian coordinate system is

=

11 12 1321 22 2331 32 33

=

xx xy xzyx yy yzzx zy zz

=

x xy xzyx y yzzx zy z

. (2.10)

The form of the right hand side is know as von Krmn notation and the -symbol in itdescribes the normal component of the stress and the shear stresses, respectively. Suchnotation is common in engineering literature.

The equation (2.9) is called the Cauchy stress theorem and it can be written as

t(x ,n) = [(x )]Tn , (2.11)

expressing the dependent quantities explicitely. It says that the traction vector dependslinearly on the normal vector n .

2.2 Coordinate transformationIf the stress tensor (or any other tensor) is known in a rectangular Cartesian coordinatesystem (x1, x2, x3) with unit base vectors e1, e2, e3 and we would like to know its compo-nents in other recangular Cartesian coordinate system (x1, x2, x3) with unit base vectors

2.3. PRINCIPAL STRESSES AND -AXES 11

e 1, e

2, e

3 a coordinate transformation tensor is needed. Lets write the stress tensor inthe xi-coordinate system as

= 11e1e1 + 12e1e2 + 13e1e3 + 21e2e1 + 22e2e2 + 23e2e3

+ 31e3e1 + 32e3e2 + 33e3e3. (2.12)This kind of representation is called the dyadic form, and many times the base vector parte iej is written either as e i e j , or in matrix notation e ieTj . It underlines the fact that atensor contains not only the components but also the base in which it is expressed. Usingthe Einsteins summation convention it is briefly written as

= ije iej =

ije

ie

j, (2.13)indicating the fact that the tensor is the same irrespectively in which coordinate system itis expressed.

Now, taking a scalar product by parts with the vector e k from the left and with e pfrom the right, we obtain

ij e

ke i ki

e je

p jp

= ij e

ke

i ki

e je

p jp

(2.14)

thuskp = kipjij or in matrix notation [] = [][][]T , (2.15)

where the compnents of the transformation matrix are ij = e iej . Notice that is thetransformation from xi-system to xi-cordinate system.

2.3 Principal stresses and -axesThe pricipal values of the stress tensor are obtained from the linear eigenvalue problem

(ij ij)nj , (2.16)where the vector ni defines the normal of the plane where the principal stress acts. Thehomogeneous system (2.16) has solution only if the coefficient matrix is singular, thus thedeterminat of it has to vanish, ad we obtain the characteristic equation

3 + I12 + I2 + I3 = 0. (2.17)The coefficients Ii, i = 1, . . . , 3 are

I1 = tr = ii = 11 + 22 + 33, (2.18)I2 =

12[tr(2) (tr)2] = 1

2(ijji I21 ), (2.19)

I3 = det(ij). (2.20)

12 CHAPTER 2. STRESS

Solution of the characteristic equation gives the principal values of the stress tensor, i.e.principal stresses 1, 2 ja 3, which are often numbered as: 1 2 3.

The coefficients I1, I2 and I3 are independent of the chosen coordinate system, thusthey are called invariants.2 Notice, that the principal stresses are also independent of thechosen coordinate system. Invariants have a central role in the development of constitutiveequations, as we will see in the subsequent chapters.

If the cordinate axes are chosen to coincide to the principal directions n i (2.16), thestress tensor will be diagonal

= [ij ] =

1 0 00 2 0

0 0 3

. (2.21)

The invariants I1, . . . , I3 have the following forms expressed in terms of the principalstresses

I1 = 1 + 2 + 3, (2.22)I2 = 12 23 31, (2.23)I3 = 123. (2.24)

2.4 Deviatoric stress tensorThe stress tensor can be additively decomposed into a deviatoric part, describing a pureshear state and an isoropic part describing hydrostatic pressure.

ij = sij + mij , (2.25)where m = 13I1 =

13kk is the mean stress and sij the deviatoric stress tensor, for which

the notation is also used in the literature. From the decomposition (2.25) it is observedthat the trace of the deviatoric stress tensor will vanish

trs = 0. (2.26)The principal valuse s of the deviatoric stress tensor s can be solved from

|sij sij | = 0, (2.27)giving the characteristic equation

s3 + J1s2 + J2s+ J3 = 0, (2.28)2The invariants appearing in the characteristic equation are usually called as principal invariants. No-

tice, that in this note the second invariant is often defined as of opposite sign. However, we would like todefine the principal invariants of the tensor and its deviator in a similar way.

2.5. OCTAHEDRAL PLANE AND STRESSES 13

where J1, . . . , J3 are the invariants of the deviatori stress tensor. They can be expressedas

J1 = trs = sii = sx + sy + sz = 0, (2.29)J2 =

12[tr(s2) (trs)2] = 1

2tr(s2) = 1

2sijsji (2.30)

= 16[(x y)2 + (y z)2 + (z x)2] + 2xy + 2yz + 2zx (2.31)

= 16[(1 2)2 + (2 3)2 + (3 1)2], (2.32)

J3 = det s . (2.33)The deviatoric stress tensor is obtained from the stress tensor by substracting the

isotropic part, thus the principal directions of the deviatoric stress tensor coincide to theprincipal directions of the stress tensor itself. Also the principal values of the deviatoricstress tensor are related to those of the stress tensor as

s1s2s3

=

123

mmm

. (2.34)

The deviatoric invariants exressed in terms of the principal values areJ2 =

12(s21 + s

22 + s

23), (2.35)

J3 =13(s31 + s

32 + s

33) = s1s2s3. (2.36)

2.5 Octahedral plane and stressesOctahedral plane is a plane whose normal makes equal angles with each of the princialaxes of stress. In the principal stress space the normal to the octahedral plane takes theform

n = [n1, n2, n3]T =

13[1, 1, 1]T (2.37)

The normal stress on the octahedral plane is thusoct = ijninj = 1n

21 + 2n

22 + 3n

23 =

13(1 + 2 + 3) = m (2.38)

and for the shear stress on the octahedral plane, the following equation is obtained 2oct = titi 2oct = ijiknjnk (ijninj)2. (2.39)

Expressed in terms of principal stresses, the octahedral shear stress is 2oct =

13(21 +

22 +

23) 19(1 + 2 + 3)2 (2.40)

= 19[(1 2)2 + (2 3)2 + (3 1)2]. (2.41)

By using the second invariant of the deviatoric stress tensor, the octahedral shear stresshas the form

oct =

23J2. (2.42)

14 CHAPTER 2. STRESS

2.6 Principal shear stressesIt is easy to see with the help of Mohrs circles that the maximun shear stress is one-halfof the largest difference between any two of the principal stresses and occurs in a planewhose unit normal makes an angle of 45 with each of the corresponding principal axes.The quantities

1 =12|2 3|, 2 = 12 |1 3|, 3 = 12 |1 2| (2.43)

are called as principal shear stresses and

max = max(1, 2, 3) (2.44)

or

max =12|1 3|, (2.45)

if the convention 1 2 3 is used.

2.7 Geometrical illustration of stress state and invariantsThe six-dimensional stress space is difficult to elucidate, therefore the principal stressspace is more convenient for illustration purposes. Lets consider a three-dimensionaleuclidean space where the coordinate axes are formed from the principal stresses 1, 2and 3, see figure 2.4.

Considering the stress point P (1, 2, 3), this the vector OP can be thought to rep-resent the stress. The hydrostatic axis is defined through relations 1 = 2 = 3, and itmakes equal angle to each of the principal stress axes, and thus the unit vector parallel tothe hydrostatic axis is

n =13[1, 1, 1]T . (2.46)

Since the deviatoric stress tensor vanishes along the hydrostatic axis, the plane perpen-dicular to it is called the deviatoric plane. The special deviatoric plane going through theorigin, i.e.

1 + 2 + 3 = 0, (2.47)is called the pi-plane, A stress state on the pi-plane is a pure shear stress state.

The vector OP can be divided into a component parallel to the hydrostatic axis ONand a component lying on the deviatoric plane NP , which are thus perpendicular to eachother.

The length of the hydrostatic part ON is

= | ~ON | = ~OP n = 13I1 =

3m =

3oct, (2.48)

2.7. GEOMETRICAL ILLUSTRATION OF STRESS STATE AND INVARIANTS 15

2

3

1

O

hydrostatic axis

nb

N

b

P (1, 2, 3)

Figure 2.4: Princial stress space.

and its component representation has the form

~ON =

mmm

= 1

3I1

11

1

. (2.49)

Respectively, the component NP on the devatoric plane is

~NP =

123

mmm

=

s1s2s3

. (2.50)

The length ~NP squared is

2 = | ~NP |2 = s21 + s22 + s23 = 2J2 = 3 2oct = 5 2m. (2.51)The invariants I1 and J2 have thus clear geometrical and physical interpetation. The

cubic deviatoric invariant J3 is ralated to the angle , defined on the deviatoric plane asan angle between the the projected 1-axis and the vector ~NP , see figure 2.5. The vectore1 is a unit vector in the direction of the projected 1-axis, and has the form

e1 =1

6

211

. (2.52)

The angle can then be determined by using the dot product of vectors ~NP and e1 as~NP e1 = cos , (2.53)

16 CHAPTER 2. STRESS

1

2 3

e1

N

P

Figure 2.5: Deviatoric plane. The projections of the principal stress axes are shown withdashed line.

which gives

cos =1

23J2

(2s1 s2 s3) = 32

s13J2

=21 2 3

23J2

. (2.54)

From the trigonometric identty it is obtained

cos 3 = 4 cos3 3 cos (2.55)

andcos 3 =

33

2

J3

J3/22

=

2J3 3oct

. (2.56)

A stress space described by the coordinates , ja is called the Heigh-Westergaardstress space.

Chapter 3Kinematical relations

3.1 Motion of a continuum bodyMotion of a continuum body B embedded in a three-dimensional Euclidean space andoccupying a domain will be studied. Consider a point P which has an initial positionX at time t = 0. At time t > 0 the body occcupies another configuration and the motionof the particle P is described by mapping

x = (X , t), or in index notation xi = i(Xk, t). (3.1)

The motion is assumed to be invertibe and sufficiently many time differentiable. Thedisplacement vector is defined as

u = x X . (3.2)

3.2 Deformation gradientThe most important measure of deformation in non-linear continuum mechanics is thedeformation gradient, which will be introduced next. Consider a material curve at theinitial configuration, a position of a point on this curve is given as X = (), where denotes a parametrization, see figure 3.1. Notice that the material curve does not dependon time. During the motion, the material curve deforms into curve

x = (, t) = ((), t). (3.3)

The tangent vectors of the material and deformed curves are denoted as dX and dx ,respectively, and they are defined as

dX = ()d, (3.4)dx = (, t)d =

X()d = F dX , (3.5)

17

18 CHAPTER 3. KINEMATICAL RELATIONS

Figure 3.1: Deformation of a material curve, figure from [1, page 70].

since on the deformed curve x = (, t) = ((), t). The quantity F is called thedeformation gradient and it describes the motion in the neighbourhod of a point. It isdefined as

F =

X, or in indicial notation Fij =

iXj

. (3.6)

The deformation gradient reduces into identity tensor I if there is no motion, or themotion is a rigid translataion. However, rigid rotation will give a deformation gradientnot equal to the identity.

3.3 Definition of strain tensorsLets investigate the change of length of a line element. Denoting the length of a lineelement in the deformed configuration as ds and as dS in the initial configuration, thus

1

2[(ds)2 (dS)2] = 1

2(dx dx dX dX ) = 1

2(F dX F dX dX dX )

=1

2dX (F T F I )dX = dX E dX , (3.7)

where the tensorE =

1

2(F T F I ) (3.8)

is called the Green-Lagrange strain tensor.Lets express the Green-Lagrange strain in terms of displacement vector u . It is first

observed that the deformation gradient takes the form

F =

X= I +

u

X, (3.9)

3.3. DEFINITION OF STRAIN TENSORS 19

where the tensor u/X is called the displacement gradient. Thus the Green-Lagrangestrain tensor takes the form

E =1

2

[(I +

u

X

)T (I +

u

X

) I

]

=1

2

[u

X+

(u

X

)T+

(u

X

)T (u

X

)], (3.10)

or in index notationEij =

1

2

(uiXj

+ujXi

+ukXi

ukXj

). (3.11)

If the elements of the displacement gradient are small in comparison to unity, i.e.

uiXj

1, (3.12)

then the quadratic terms can be neglected and the infinitesimal strain tensor can be definedas the symmetric part of the displacement gradient

ij =1

2

(uiXj

+ujXi

) Eij . (3.13)

Lets define a stretch vector in the direction of a unit vector n0 as

= F n0, (3.14)

and the length of the stretch vector = || is called the stretch ratio or simply the stretch.The square of the stretch ratio is

2 = = n0FTF n0 = n0C n0, (3.15)

where the tensor C = F T F is called the right Cauchy-Green strain tensor. The atributeright comes from the fact that the deformation gradient operates on the right hand side.The right Cauchy-Green strain tensor is symmetric and positive definite tensor, i.e. C =C T and n C n > 0, n 6= 0.

For values 0 < < 1 a line element is compressed and elongated for values > 1.The deformation gradient can also be decomposed multiplicatively as

F = R U = V R, (3.16)

where R is an othogonal tensor (RT R = R RT = I ) describing the rotation of amaterial element and U and V are symmetric positive definite tensors describing the

20 CHAPTER 3. KINEMATICAL RELATIONS

deformation. The decomposition (3.16) is also called the polar decomposition. The tensorU is called as the right stretch tensor and V the left stretch tensor.

The square of the stretch can be expressed as

2 == = n0UTRT R U n0 = n0U

TU n0 = n0U

2n0. (3.17)

Other strain measures can be defined as

E (m) =1

m(Um I ). (3.18)

For m = 2 we obtain the Green-Lagrange strain tensor which have already been dis-cussed. With m = 0 we obtain the Hencky or logarithmic strain tensor

E (0) = lnU . (3.19)The logarithmic strain1 has a special position in non-linear continum mechanics, espe-cially in formulating constitutive equations, since it can be additively decomposed intovolymetric and isochoric parts similarly as the small strain tensor .

For m = 1 we obtainE (1) = U I , (3.20)

which is called the Biot strain tensor. If the defrmation is rotation free, i.e. R = I , theBiot strain tensor coincides the small strain tensor . It is much used in dimensionallyreduced continuum models, like beams, plates and shells.

3.4 Geometric intepretation of the strain componentsLets investigate the extension = 1 of a line element, for instance in a directionn0 = (1, 0, 0)

T, thus

(1) =C11, E11 =

12(C11 1) C11 = 1 + 2E11

=

1 + 2E11 =

1 + 2E11 1 (3.21)Secondly, lets compute the angle change of two unit vectors N 1 and N 2. In the

deformed configuration they are n1 = F N 1 and n1 = F N 2 and the angle betweenthem can be determined from

cos 12 =n1n2|n1||n2| =

N 1C N 2N 1C N 1

N 2C N 2

. (3.22)

1The logarithmic strain is sometimes called also as the true strain. Such naming is not used in this text, allproperly defined strain measures are applicable, since the definition of strain is a geometrical construction.Naturally, the choice of strain measure dictates the choise of the stress. However, deeper discussion on thistopic is beyong the present lecture notes.

3.5. DEFINITION OF INFINITESIMAL STRAIN 21

If we choose the directions N 1 and N 2 as

N 1 =

10

0

, N 2 =

01

0

, (3.23)

thencos 12 =

C12C11C22

=C12

(1)(2)=

2E12(1 + 2E11)(1 + 2E22)

. (3.24)

Using the trigonometric identity

sin(12pi 12) = cos 12 (3.25)

and if E11, E22 1 then12pi 12 2E12. (3.26)

Thus, the component E12 is approximately one half of the angle change of the two direc-tion vectors.

3.5 Definition of infinitesimal strainLets investigate the motion of two neghbouring points, which are denoted as P Q in theundeformed configuration. After deformation these points occupy the positions markedby p and q. Displacement of the point Q relative to P is defined as

du = uQ uP . (3.27)Length of the vector ~PQ is denoted as dS, thus

duidS

=uixj

dxjdS

, (3.28)

where the Jacobian matrix J = u/x can be divided additively into a symmetric andan antisymmetric part as

J = +, (3.29)where the symmetric part is the infinitesimal strain tensor

=

11 12 1321 22 2331 32 33

=

xx xy xzyx yy yzzx zy zz

=

x 12xy 12xz1

2yx y

12yz

12zx

12zy z

, (3.30)

and the antisymmetric part is the infinitesimal rotation tensor. Written in displacementcomponents the strain tensor has the expression

ij =12(ui,j + uj,i). (3.31)

It should be emphasised that the rotation matrix near the point P describes the rigid bodyrotation ony if the elements ij are small.

22 CHAPTER 3. KINEMATICAL RELATIONS

3.5.1 Principal strainsThe principal strains are obtained from the linear eigenvalue problem

(ij ij)nj = 0, (3.32)where the vector ni defines the normal direction of the principal strain plane. Thus, thecharacteristic polynomial has the form

3 + I12 + I2+ I3 = 0, (3.33)where the strain invariants Ii, i = 1, . . . , 3 are

I1 = tr = ii = 11 + 22 + 33, (3.34)I2 =

12[tr(2) (tr)2] = 1

2(ijji I21 ), (3.35)

I3 = det(ij). (3.36)If the coorinate axes are chosen to coincide with the axes of principal strains, the strain

tensor will be symmetric

= [ij ] =

1 0 00 2 0

0 0 3

. (3.37)

The invariants I1, . . . , I3 expressed in terms of the principal strains 1, . . . , 3 have theforms

I1 = 1 + 2 + 3, (3.38)I2 = 12 23 31, (3.39)I3 = 123. (3.40)

3.5.2 Deviatoric strainAs in the case of the stress tensor, the infinitesimal strain tensor can be additively decom-posed into a deviatoric part and an isotropic part as

ij = eij +13kkij, (3.41)

where the deviatoric strain tensor is denoted as e . In the literature also the notation isused. By definition the deviatoric strain tensor is traceless

tre = 0. (3.42)The eigenvalues of the deviatoric strain ei can be solved from the equation

|eij eij | = 0, (3.43)

3.5. DEFINITION OF INFINITESIMAL STRAIN 23

and the characteristitic equation is

e3 + J1e2 + J2e+ J3 = 0, (3.44)

where the invariants J1, . . . , J3 have expressions

J1 = tre = eii = ex + ey + ez = 0, (3.45)J2 =

12[tr(e2) (tre)2] = 1

2tr(e2) = 1

2eijeji (3.46)

= 16[(1 2)2 + (2 3)2 + (3 1)2], (3.47)

J3 = det e = tr(e3) = e1e2e3. (3.48)

For small strains the first invariant I1 = x + y + z v describes the relativevolume change.

The octahedral strains are defined similarly as for the stress

oct =13I1 =

13v, (3.49)

2oct =83J2. (3.50)

For the first sight the equation (3.50) might look strange as copared to the expression ofthe octahedral stress, but we have to remember that xy = 2xy, etc.

24 CHAPTER 3. KINEMATICAL RELATIONS

Chapter 4Constitutive models

4.1 IntroductionConstitutive equations describe the response of a material to applied loads. In continuummechanics, distinction between fluids and solids can be characterized in this stage. It isimportant to notice that the balance equations and the kinematical relations described inthe previous sections are equally valid both fr fluids and solids. In this lecture notes onlyphenomenological models will be introduced, which roughly means that mathematicalexpressions are fitted to experimental data. Phenomenological models are not capableto relate the actual physical mechanisms of deformation to the underlying mcroscopicphysical structure of the material.

The constitutive equations should obey the thermodynamic principles, (i) the conser-vation of energy and (ii) the dissipation inequality, i.e. the nonnegativity of the entropyrate.

Excellent texts for materials modelling are [2, 4].

4.2 Elastic constitutive modelsElasticity means that the response of a material is independent of the load history. Themost general form of elasticity is called as Cauchy-elasticity and it essentially means thatthere exists one-to-one relation between stress and strain

ij = fij(kl), or ij = gij(kl). (4.1)

The tensor valued tensor functions fij and gij are called as response functions. For non-linear Cauchy-elastic models, the loading-unloading process may yield hysteresis, whichis incompatible with the notion of elasticity, where the response should be reversible.However, in this lecture notes the Couchy-elasticity is not treated.

Another form of elasticity, where the constitutive equations are expressed in rate-form

ij = fij(kl, mn) (4.2)

25

26 CHAPTER 4. CONSTITUTIVE MODELS

is called hypo-elastic. If the material is incrementally linear, it can be written in the form

ij = Cijkl(mn)kl. (4.3)

The most rigorous form of elasticity is called as hyper-elasticity, and the constitutiveequations of a hyper-elastic model can be derived from a potential, i.e. the strain energyfunction W = W (ij) as

ij =W

ij. (4.4)

4.2.1 Isotropic elasticityA material which behaviour is independent of the direction in which the response is mea-sured is called isotropic. Therefore also the strain energy density should be an isotropictensor valued scalar function

W = W () = W () = W (T ) = W (I1, I2, I3), (4.5)

where I1, I2 and I3 are the principal invariants of the strain tensor and is the transfor-mation tensor from the x -coordinate system to the x -system, i.e. x = x . Alternativelythe strain energy density function W can be written as

W = W (I1, J2, J3), or W = W (I1, I2, I3), (4.6)

where J2 and J3 are the invariants of the deviatoric strain tensor and I2, I3 the genericinvariants defined as

I2 =12tr(2), I3 =

12tr(3). (4.7)

Equations (4.5) and (4.6) are spacial forms of representation theorems, for which an al-ternative form can be written as: the most general form of an isotropic elastic materialmodel can be written as

= a0I + a1+ a22, (4.8)

where the coefficients a0, a1 and a2 can be non-linear functions of the strain invariants.Alternatively (4.8) can be formulated as

= b0I + b1 + b22, (4.9)

where b0, b1 and b2 can be non-linear functions of stress invariants.From (4.8) and (4.9) it can be easily seen that the principal directions of the strain-

and stress tensors coincide.

4.2. ELASTIC CONSTITUTIVE MODELS 27

4.2.2 Transversally isotropic elasticityA material is called transversally isotropic if the behaviour of it is isotropic in a plane anddifferent in the direction of the normal of that isotropy plane. The strain energy densityfunction can now be written as

W = W (,M ) = W (T ,MT ), (4.10)where M = mmT is called the structural tensor and the unit vector m defines the normalof the isotropy plane.

The representation theorem of a transversally isotropic solid says that the strain en-ergy density function can depend on five invariants

W = W (I1, I2, I3, I4, I5), (4.11)where the invarinats Ii are

I1 = tr , I2 =1

2tr(2), I3 =

1

3tr(3), I4 = tr(M ), I5 = tr(

2M ). (4.12)

The invariants I4 and I5 can also be written as

I4 = tr(M ) = mTm , I5 = tr(

2M ) = mT2m . (4.13)The constitutive equation is thus

=W

=W

I1I +

W

I2+

W

I32 +

W

I4M +

W

I5(M +M ). (4.14)

If we restrict to linear elasticity, the cefficients W/Ii has to satisfy

W

I1= 1I1 + I4, (4.15)

W

I2= 2, (4.16)

W

I3= 0, (4.17)

W

I4= 3I1 + 4I4, (4.18)

W

I1= 5, (4.19)

since all the terms in (4.14) have to be linear in . Due to the identity2W

IiIj=

2W

IjIi, (4.20)

28 CHAPTER 4. CONSTITUTIVE MODELS

we have now

I4

(W

I1

)=

I1

(W

I4

)thus = 3. (4.21)

Transversally isotropic linear solid has thus five material coefficients, and the constitutiveequation can be written as

= (1 tr +3 tr(M ))I+2+(3 tr +4 tr(M ))M +5(M +M ). (4.22)

4.2.3 Orthotropic materialA material is called orthotropic if it has three perpendicular symmetry planes. Lets denotethe unit vectors normal to the symetry planes as m1,m2 and m3. Due to the orthogonalitym im j = ij . The structural tensors associated with these direction vectors are M i =mmT , and they satisfy

M 1 +M 2 +M 3 = I , (4.23)due to the orthogonality. Thus, only two structural tensors are necessary to describe thebehaviour of an orthotropic material

W = W (,M 1,M2 ) = W (T ,M 1

T ,M 2T ). (4.24)

The representation theorem of an orthotropic solid says that the strain energy densityfunction can depend on seven invariants

W = W (tr , 12tr(2), 1

3tr(3), tr(M 1), tr(M 2), tr(

2M 1), tr(2M 2)). (4.25)

It can be written in a form, where all the structural tensors M i are symmetricaly present.Notice that

M 1 + M 2M 3 = (M 1 +M 2 +M 3) = , (4.26)M 1+M 2+M 3 = (M 1 +M 2 +M 3) = , (4.27)

thus summing by parts gives

= 12(M 1 +M 1) +

12(M 2 +M 2) +

12(M 3 +M 3), (4.28)

andtr = tr(M 1) + tr(M 2) + tr(M 3). (4.29)

In a similar way it can be deduced

tr (2) = tr(2M 1) + tr(2M 2) + tr(

2M 3). (4.30)In other words, the invariants tr , tr(M 1) and tr(M 2) can be replaced by the invari-ants I1 = tr(M 1), I2 = tr(M 2) and I3 = tr(M 3). In a similar way the invariants

4.2. ELASTIC CONSTITUTIVE MODELS 29

tr(2), tr(2M 1) and tr(2M 2) can be replaced by the invariants I4 = tr(2M 1), I5 =tr(2M 2) and I6 = tr(2M 3). If we now denote the cubic invariant I7 = 13 tr(

3), thestrain energy density function for an orthotropic material can be written as a function ofthese seven invariants as

W = W (I1, . . . , I7), (4.31)and the constitutive equation has the form

=W

=

7i=1

W

Ii

Ii

=W

I1M 1 +

W

I2M 2 +

W

I3M 3 +

W

I4(M 1 +M 1)

+W

I5(M 2 +M 2) +

W

I6(M 3 +M 3) +

W

I72. (4.32)

If we now restrict to a linear model, the coefficients W/Ii has to satisfy the follow-ing conditions

W

I1= 1I1 + 1I2 + 2I3, (4.33)

W

I2= 2I1 + 3I2 + 3I3, (4.34)

W

I3= 4I1 + 5I2 + 6I3, (4.35)

W

I4= 7, (4.36)

W

I5= 8, (4.37)

W

I6= 9, (4.38)

W

I7= 0. (4.39)

Due to the identity of the second derivatives (4.20), we have

I2

(W

I1

)=

I1

(W

I2

) 1 = 2, (4.40)

I3

(W

I1

)=

I1

(W

I3

) 2 = 4, (4.41)

I3

(W

I2

)=

I2

(W

I3

) 3 = 5. (4.42)

30 CHAPTER 4. CONSTITUTIVE MODELS

Bibliography

[1] G.A. Holzapfel. Nonlinear Solid Mechanics, A Continuum Approach for Engineers. JohnWiley & Sons, 2000.

[2] J. Lemaitre and J.-L. Chaboche. Mechanics of Solid Materials. Cambridge University Press,1990.

[3] L.E. Malvern. Introduction to the Mechanics of a Continuous Medium. Prentice Hall, Engle-wood Cliffs, New Jersey, 1969.

[4] N.S. Ottosen and M. Ristinmaa. The Mechanics of Constitutive Modeling. Elsevier, 2005.

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