Boehrnsen & Zahid - Constitutive Material Modes.

7/27/2019 Boehrnsen & Zahid - Constitutive Material Modes.

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Lecture Notes onConstitutive Material Models

Dr.Ing. Jens-Uwe Bohrnsenand

Muhammad Zahidand

Reza Kebriaei

Institute of Applied MechanicsSpielmannstr. 11

38106 Braunschweig

www.infam.tu-braunschweig.de

May 2010


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Contents

1 Introduction and mathematical preliminaries 1

1.1 Vectors and matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Indicial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Rules for matrices and vectors . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.4 Coordinate transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.5 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.6 Scalar, vector and tensor fields . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.7 Divergence theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.8 Summary of chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.10 Summary of chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.11 E xercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Foundation of Solid Mechanics 26

2.1 Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.1.1 Components of Stress . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.1.2 Stress on a normal plane . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.3 Principal stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.1.4 Stress invariants and special stress tensors . . . . . . . . . . . . . . 28

2.2 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.2.1 Physical principles . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

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2.3 Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.3.1 Position vector and displacement vector . . . . . . . . . . . . . . . 31

2.3.2 Strain tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.3.3 Linear theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.3.4 Properties of the strain tensor . . . . . . . . . . . . . . . . . . . . . 35

2.3.5 Compatibility equations for linear strain . . . . . . . . . . . . . . . 37

2.4 Material behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.4.1 Uniaxial behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.4.2 Generalized Hookes law . . . . . . . . . . . . . . . . . . . . . . . . 39

2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.5.1 Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.5.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.5.3 Deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

2.5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2.5.5 Material behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.5.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

3 Foundation of Constitutive Material Models 54

3.1 The Stress Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.2 Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3.3 Linear hyper-elasticity - Anisotropy . . . . . . . . . . . . . . . . . . . . . . 59

3.4 Viscoelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.4.1 Rheological Models . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.4.2 Maxwell Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.5 Plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.5.1 Phenomenological aspects . . . . . . . . . . . . . . . . . . . . . . . 63

3.5.2 Basic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.5.3 Classical theory of plasticity . . . . . . . . . . . . . . . . . . . . . . 67

3.6 Hardening, softening and failure . . . . . . . . . . . . . . . . . . . . . . . . 70

3.6.1 Frequently used failure and yield criteria . . . . . . . . . . . . . . . 71

3.6.2 Consistency condition for strain hardening materials . . . . . . . . 76

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4 Linear Elasticity and Failure Criteria for Concrete 79

4.1 Constitutive Modelling of Concrete . . . . . . . . . . . . . . . . . . . . . . 79

4.2 Mechanical Behaviour of Concrete . . . . . . . . . . . . . . . . . . . . . . . 80

4.2.1 Linear Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

4.2.2 Nonlinear Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . 80

4.2.3 Linear Perfectly Plastic . . . . . . . . . . . . . . . . . . . . . . . . . 81

4.3 Failure Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.3.1 OneParameter Model . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.3.2 TwoParameter Model . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.4 Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

4.4.1 Reinforcing Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.4.2 Interaction of Concrete and Reinforcement . . . . . . . . . . . . . . 89

5 Soil and Geomechanics 955.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.1.1 Effective Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.1.2 Total Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

5.2 Mechanical Behaviour of Soils . . . . . . . . . . . . . . . . . . . . . . . . . 97

5.2.1 The Nature of Soils and other Porous Media . . . . . . . . . . . . . 98

5.2.2 Common Soil Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

5.2.3 Stress Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.3 Failure Criteria of Soils and Bulk Solids . . . . . . . . . . . . . . . . . . . . 111

5.4 Material Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

5.4.1 Rate Dependent Model for cohesionless Soil and Bulk Solids . . . . 112

5.4.2 Theory of Porous Media, TPM . . . . . . . . . . . . . . . . . . . . 115

III


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6 Theory of Metal Plasticity 125

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

6.2 Uniaxial Plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

6.3 Yield Criteria and Hardening of Metals . . . . . . . . . . . . . . . . . . . . 125

6.3.1 Tresca Maximum Shear Stress Criterion . . . . . . . . . . . . . . . 126

6.3.2 Von Mises Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

6.3.3 Hardening Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

6.3.4 Isotropic Hardening . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

6.3.5 Kinematic Hardening . . . . . . . . . . . . . . . . . . . . . . . . . . 1296.3.6 Generalised Hardening . . . . . . . . . . . . . . . . . . . . . . . . . 131

6.4 Ductility and Fracture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

6.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

6.4.2 Ductile Fracture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

6.4.3 Brittle fracture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

6.4.4 Impact energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

6.5 Constitutive Model for the Plasticity of Metals . . . . . . . . . . . . . . . . 139

6.5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

6.5.2 Mechanisms on the Microscale . . . . . . . . . . . . . . . . . . . . . 140

6.5.3 Simulation of the Development of Dislocation Structures . . . . . . 141

6.5.4 Stochastic Constitutive Model . . . . . . . . . . . . . . . . . . . . . 143

6.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

7 Fluid aspect in Constitutive Material Modeling 1507.1 Fluid Flows and their Significance . . . . . . . . . . . . . . . . . . . . . . . 150

7.2 Solids and Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

7.3 Basic Equations of Fluid Mechanics . . . . . . . . . . . . . . . . . . . . . . 155

7.3.1 Mass Conservation (Continuity Equation) . . . . . . . . . . . . . . 155

7.3.2 Newtons Second Law . . . . . . . . . . . . . . . . . . . . . . . . 156

7.3.3 The NavierStokes Equations . . . . . . . . . . . . . . . . . . . 156

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7.3.4 Mechanical Energy Equation . . . . . . . . . . . . . . . . . . . . . . 161

7.3.5 Thermal Energy Equation . . . . . . . . . . . . . . . . . . . . . . . 161

7.4 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

7.4.1 Turbine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

7.4.2 Theory of Operation . . . . . . . . . . . . . . . . . . . . . . . . . . 163

7.5 FluidStructure Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . 169

7.5.1 Basic Idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

V


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1 Introduction and mathematical

preliminaries

1.1 Vectors and matrices

A vector is a directed line segment. In a cartesian coordinate system it looks as itis depicted in figure 1.1,

z

y

x

P

a

ez

ey

exax

ay

az P

a

x3

x2

x1

e3

e2

e1a1

a2

a3

Figure 1.1: Vector in a cartesian coordinate system. [29]

e.g., it can mean the location of a point P or a force. So a vector connects direction

and norm of a quantity. For representation in a coordinate system unit basis vectors

ex, ey and ez are used with |ex| = |ey| = |ez| = 1.| | denotes the norm, i. e., the length.Now the vector a is

a = axex + ayey + azez (1.1)

1


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2 CHAPTER 1. INTRODUCTION AND MATHEMATICAL PRELIMINARIES

with the coordinates (ax, ay, az)

= values/length in the direction of the basis vec-

tors/coordinate direction.

In continuum mechanics, it is more common to denote the axis with e1, e2 and e3

a = a1e1 + a2e2 + a3e3 (1.2)

Following are the different representations of a vector:

a =

a1

a2

a3

= (a1, a2, a3) (1.3)

with the length/norm (Euclidian norm)

|a| =

a21 + a22 + a

23 . (1.4)

A matrix is a collection of several numbers

A = A11 A12 A13 . . . A1n

A21 A22 A23 . . . A2n.... . .

...

Am1 Am2 Am3 . . . Amn

(1.5)

with n columns and m rows, i.e., an (m n) matrix. In the following mostlyquadratic(square) matrices, n m, are used.A vector is a one column matrix.

The graphical representation, as for a vector, is not possible. However, a physical

interpretation is often given, then tensors are introduced.

Special cases:

Zero vector or matrix: all elements are zero, e.g., a =

000

and A =

0 0 00 0 00 0 0

Symmetric matrix A = AT with AT is the transposed matrix, i.e., all elements

at the same place above and below the main diagonal are identical, e.g., A =

1 5 45 2 64 6 3


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1.2. INDICIAL NOTATION 3

1.2 Indicial Notation

Indicial notation is a convenient notation in mechanics for vectors and matrices/tensors.Letter indices as subscripts are appended to the generic letter representing the tensor

quantity of interest. Using a coordinate system with (e1, e2, e3) the components of

a vector a are ai (eq. 1.7) and of a matrix A are Aij with i = 1, 2, . . . , m and j =

1, 2, . . . , n (eq. 1.6). When an index appears twice in a term, that index is understood

to take on all the values of its range, and the resulting terms summed. In this so-called

Einstein summation, repeated indices are often referred to as dummy indices, since their

replacement by any other letter not appearing as a free index does not change the meaning

of the term in which they occur. In ordinary physical space, the range of the indices is

1, 2, 3.

Aii =m

i=1

Aii = A11 + A22 + A33 + . . . + Amm (1.6)

and

aibi = a1b1 + a2b2 + . . . + ambm. (1.7)

However, it is not summed up in an addition or subtraction symbol, i.e., ifai +bi or ai bi.

Aijbj =Ai1b1 + Ai2b2 + . . . + Aikbk (1.8) free dummy

Further notation:

3

i=1 ai = a1 a2 a3 (1.9)

aixj

= ai,j with ai,i =a1x1

+a2x2

+ . . . (1.10)

orAijxj

=Ai1x1

+Ai2x2

+ . . . = Aij,j (1.11)

This is sometimes called comma convention!


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1.3 Rules for matrices and vectors

Addition and subtractionA B = C Cij = Aij Bij (1.12)

component by component, vector similar.

Multiplication

Vector with vector

Scalar (inner) product:

c = a b = aibi (1.13) Cross (outer) product:

c = a b =

e1 e2 e3

a1 a2 a3

b1 b2 b3

=a2b3 a3b2a3b1 a1b3

a1b2 a2b1

(1.14)Cross product is not commutative.

Using indical notation

ci = ijk ajbk (1.15)

with permutations symbol / alternating tensor

ijk =

1 i, j, k even permutation (e.g. 231)

1 i, j, k odd permutation (e.g. 321)0 i, j, k no permutation, i.e.

two or more indices have the same value

. (1.16)

Dyadic product:C = a b (1.17)

Matrix with matrix Inner product:

C = AB (1.18)

Cik = AijBjk (1.19)

Inner product of two matrices can be done with Falk scheme (fig. 1.2(a)). To

get one component Cij of C, you have to do a scalar product of two vectors ai


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1.3. RULES FOR MATRICES AND VECTORS 5

and bj, which are marked in figure 1.2 with a dotted line. It is also valid for

the special case of onecolumn matrix (vector) (fig. 1.2(b))

c = Ab ci = Aijbj . (1.20)

Bij

Aij Cij

(a) Product of matrix with matrix

bj

Aij ci

(b) Product of matrix with vector

Figure 1.2: Falk scheme. [29]

Remarks on special matrices:

Permutation symbol (see 1.16)

ijk =

1

2 (i j)(j k)(k i) (1.21) Kronecker delta

ij =

1 ifi = j0 ifi = j (1.22)so

ij

0 00 00 0

for i, j = 1, 2, 3 (1.23)

ijai = aj ijDjk = Dik (1.24)

Product of two unit vectorsei ej = ij (orthogonal basis) (1.25)

Decomposition of a matrixAij =

1

2(Aij + Aji)

Symmetric, Hydrostatic

+1

2(Aij Aji)

anti-symmetric/skrew symmetric, Shear

(1.26)


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1.4 Coordinate transformation

Assumption:

2 coordinate systems in one origin rotated against each other (fig. 1.3).

x1

x1

x2

x2

x3x3

Figure 1.3: Initial (x1, x2, x3) and rotated (x1, x

2, x

3) axes of transformed coordinate sys-

tem. [29]

The coordinates can be transformed

x1 = 11x1 + 12x2 + 13x3 = 1jxj (1.27)

x2 = 2jxj (1.28)

x3 = 3jxj (1.29)

xi = ijxj (1.30)

with the constant (only constant for cartesian system) coefficients

ij = cos(xi, xj ) direction cosine =

xj

x i= cos(ei, ej) = e

i

ej. (1.31)

In matrix notation we have

x = Rrotation matrix

x. (1.32)

Rij = xi,j (1.33)

So the primed coordinates can be expressed as a function of the unprimed ones

xi = xi(xi) x

= x(x). (1.34)


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1.5. TENSORS 7

If J = |R| does not vanish this transformation possesses a unique inverse

xi = xi(xi) x = x(x). (1.35)

J is called the Jacobian of the transformation.

1.5 Tensors

Definition:

Let E be an Euclidian vector space. A linear mapping:

4A : a A(a) (1.36)

is a second order tensor. If we speak of a tensor we tacitly mean a second order tensor.

If this is not the case it is explicitly said.

A mapping A is called linear if it is compatible with the two linear structures, i.e.

A(a + b) = A(a) + A(b) a, b A(a) = a , a (1.37)

The set of all tensors on E will be denoted by . The product of two tensors is defined as

the composition of two linear mappings:

A(B(a)) = (AB)(a) A, B , a (1.38)

From this follows the statement that the product does not commute, i.e. we have to

consider: AB = BA.The following rules hold:

comutative A+B= B+ A A,B assosiative (A)B = (A)B = (AB) , A, B distributive A(B + C) = AB + AC A, B, C


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distributive (A + B)C = AB + AC A, B, C zero element A0 = 0B = 0 A neutral elementof composition AI = IA = A A

To simplify notation, it will be assumed in the following that all tensors are element of

, all vectors element of E and all scalars element of .

Further definitions

Due to fact that complicated expressions include different levels of brackets we represent

the linear mapping b = A(a) from now on by means of square brackets: b = A[a].

Transpose of a tensor

Associated with an arbitrary tensor A there is a unique tensor AT , called the transpose

of A, such that:

5a.(AT[b]) = b.(AT[a]) (1.39)

We now replace A by BTand exchange a and b:

5b.((BT)T[a]) = a.(BT[b]) (1.40)

Comparing with above it is immediately understood that:

5(BT)T = B (1.41)

with composition and comparison of above mapping we can find:

5C = AB CT = BTAT (1.42)

Symmetric and skew-symmetric tensors

A tensor A is said to be symmetric if

5A = AT (1.43)

holds. It is called skew-symmetric if it has the property


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1.5. TENSORS 9

5A = AT (1.44)

The identity

5A =1

2(A + AT) +

1

2(A AT) (1.45)

shows that each tensor can be split uniquely into symmetric (Asym) and skew-symmetric

(Askw) parts.

Spherical tensors

A tensor A is said to be spherical if

5A = I = 0 (1.46)

is proportional to the identity tensor I.

Inverse of a tensor

If the linear mapping A is invertible (the inverse of the mapping is denoted by (A)1 we

can state

5AA1 = A1A = I (1.47)

The effect of a composition of two inverse mappings can be investigated as:

5(AB)1 = B1A1 (1.48)

Dyadic product

To an ordered pair of vectors (a; b) there corresponds a tensor, denoted by ab and called

the dyadic product or tensor product of a and b which is defined through its action onan arbitrary vector c by

5(a b)[c] = (b c)a (1.49)

The notion tensor product can be easily confused with the notion product of two tensors

so that we will here prefer the term dyadic product. with respekting to the main definition

we will find that:

5(a b)(c d) = (b c)a d (1.50)


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and so it yields:

5(c d)(a b) = (d a)c b == (b c)a d = (a b)(c d) (1.51)

It is now more easily understandable that the product of two tensors, i.e. the compositions

of two linear mappings, is not commutative. In the form of index notation we can show

the tensor A as:

5A = Aijei ej (1.52)

Product of two tensor

The product of two tensors C = AB is represented in index notation as:

5C = Aijei ej = Bklek el = AijBkljk ei el = AijBjl Cil

ei el (1.53)

Thus we arrive in matrix notation at the relation:

5C = AB (1.54)

In index notation we will show the zero tensor and identity tensor as:

50 = 0ijei ej 0ij = 0 (zerotensor) (1.55)

5I = Iijei ej (identitytesor) (1.56)

Trace of a Tensor

The trace of a dyadic product is defined as the scalar product of the two vectors:

5tr(a b) = a.b (1.57)

If we represent a tensor as usual by means of nine dyadic products we obtain the following:

5trA = tr(Aijei ej) = Aijtr(ei ej) = Aijij = Aii = A11 + A22 + A33 (1.58)


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1.6. SCALAR, VECTOR AND TENSOR FIELDS 11

Example

In many material models one differentiates between the so-called deviatoric and the vol-

umetric material response. Accordingly the stress tensor T is split into deviatoric Tdev

and volumetric Tvol parts. This can be done in the following way:

5T = (T 13

(trT)I) Tdev

+1

3(trT)I Tvol

(1.59)

The volumetric part Tvol is a spherical tensor. If the deviatoric part of the stress tensor

vanishes (which is for instance the case for many fluids) we have a hydrostatic stress state.

Tvol is often represented in terms of the pressure p:

5Tvol = pI (1.60)

Thus the trace operation is e.g. needed to carry out the above described split. The trace

of the deviatoric part vanishes. To show that we first determine the trace of the identity

tensor:

5trI = tr(ei ej) = 3 (1.61)

The trace of Tdev is computed with:

5Tdev = tr(T 13

(trT)I) = trT 13

(trT) trI

3= 0 (1.62)

1.6 Scalar, vector and tensor fields

A tensor field assigns a tensor T(x, t) to every pair (x, t) where the position vector x

varies over a particular region of space and t varies over a particular interval of time. The

tensor field is said to be continuous (or differentiable) if the components of T(x, t) are

continuous (or differentiable) functions of x and t. If the tensor T does not depend on

time the tensor field is said to be steady (T(x)).


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1. Scalar field: = (xi, t) = (x, t)

2. Vector field: vi = vi(xi, t) v = v(x, t)

3. Tensor field: Tij = Tij(xi, t) T = T(x, t)

Introduction of the differential operator : It is a vector called del or NablaOperator,defined by

= ei xi

and 2 = Laplacian operator

= = xi

xi

. (1.63)

A few differential operators on vectors or scalar:

grad = = ,iei (result: vector) (1.64)div v = v = vi,i (result: scalar) (1.65)

curl v = v = ijk vk,j (result: vector) (1.66)

Similar rules are available for tensors/vectors.

Gradient of a tensor field

The gradient of the tensor field T(x) is computed with:

5gradT(x) =Tijxk

ei ej ek = T(x)x

(1.67)

The gradient gradT is a third order tensor. As such its basis is represented by the dyadic

product of three vectors. We also introduce the abbreviation

5gradT = Tij,kei

ej

ek (1.68)

Divergence of a tensor field

To determine the divergence of a tensor field we have the definition:

5divT(x) = Tji,j (x)ei (1.69)

Example

The divergence of a tensor occurs for example in the balance of linear momentum


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1.6. SCALAR, VECTOR AND TENSOR FIELDS 13

Figure 1.4: Infinitesimal Volume Element

5divT(x) + f= 0 (1.70)

where T denotes the stress tensor and f a volume force. The derivation of the latter

equation can be carried out as follows. Consider the infinitesimal volume element plotted

in Figure 1.4.

The equilibrium conditions in x1, x2 and x3-direction read:

T11x1

dx1dx2dx3 +T21x2

dx1dx2dx3 +T31x3

dx1dx2dx3 + f1dx1dx2dx3 = 0 (1.71)

T12x

1

dx1dx2dx3 +T22x

2

dx1dx2dx3 +T32x

3

dx1dx2dx3 + f2dx1dx2dx3 = 0 (1.72)

T13x1

dx1dx2dx3 +T23x2

dx1dx2dx3 +T33x3

dx1dx2dx3 + f3dx1dx2dx3 = 0 (1.73)

The latter three equations can be also expressed as:

5Tji,j + fi = 0 (1.74)

Multiplying each equation with ei and summing them up yields the relation 1.67.


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1.7 Divergence theorem

For a domain V with boundary A the following integral transformation holds for a first-order tensor g

V

divgdV =

V

gdV =A

n gdA (1.75)V

gi,idV =

A

gi nidA (1.76)

and for a second-order tensor V

ji,j dV =

A

jinjdA (1.77)V

divdV =

V

dV =A

ndA. (1.78)

Here, n = niei denotes the outward normal vector to the boundary A.

1.8 Summary of chapter 1

Vectors

a =

a1a2a3

= a1 e1 + a2 e2 + a3 e3 = a110

0

+ a201

0

+ a300

1

Magnitude of a:

|a| =

a21 + a22 + a

23 is the length of a

Vector addition:

a1

a2

a3

+

b1

b2

b3

=

a1 + b1

a2 + b2

a3 + b3


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1.8. SUMMARY OF CHAPTER 1 15

Multiplication with a scalar:

c a1a2a3

= c a1c a2c a3

Scalar (inner, dot) product:

a b = |a||b| cos = a1 b1 + a2 b2 + a3 b3

Vector (outer, cross) product:

a b =

e1 e3 e3

a1 a2 a3

b1 b2 b3

= e1a2 a3b2 b3

e2a1 a3b1 b3+ e3a1 a2b1 b2

=a2b3 a3b2a3b1 a1b3

a1b2 a2b1

Rules for the vector product:

a

b =

(b

a)

(c a) b = a (c b) = c(a b)(a + b) c = a c + b ca (b c) = (a c) b (a b) c

Matrices

A = A11 A12 A13 ... A1n

A21 A22 A23 ... A2n......

......

Am1 Am2 Am3 ... Amn

= Aik

Multiplication of a matrix with a scalar:

c A = A c = c Aik e.g.: c

A11 A12

A21 A22

=

c A11 c A12c A21 c A22


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Addition of two matrices:

A + B = B + A = (Aik) + (Bik) = (Aik + Bik)e.g.:

A11 A12

A21 A22

+

B11 B12

B21 B22

=

A11 + B11 A12 + B12

A21 + B21 A22 + B22

Rules for addition of matrices:

(A + B) + C = A + (B + C) = A + B + C

Multiplication of two matrices:

Cik = Ai1B1k + Ai2B2k + ... + AilBlk =l

j=1

AijBjk i = 1, ..., m k = 1,...,n

e.g.:

B11 B12

B21 B22

A11 A12

A21 A22

A11B11 + A12B12 A11B12 + A12B22

A21B11 + A22B21 A21B12 + A22B22

Rules for multiplication of two matrices:A(BC) = (AB)C = ABC

AB = BA

Distributive law:

(A + B) C = A C + B C

Differential operators for vector analysis

Gradient of a scalar field f(x,y,z)

grad f(x1, x2, x3) =

f(x1,x2,x3)

x1f(x1,x2,x3)

x2f(x1,x2,x3)

x3

Derivative into a certain direction:

f

a(x1, x2, x3) =

a

|a| grad f(x1, x2, x3)


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Divergence of a vector field

div v(X(x1, x2, x3), Y(x1, x2, x3), Z(x1, x2, x3)) =

X

x1 +

Y

x2 +

Z

x3

Curl of a vector field

curlv(X(x1, x2, x3), Y(x1, x2, x3), Z(x1, x2, x3)) =

Zx2

Yx3

Xx3

Zx1

Yx1

Xx2

Nabla (del) Operator

= x1x2

x3

f(x1, x2, x3) =

f

x1f

x2f

x3

= grad f(x1, x2, x3)v(X(x1, x2, x3), Y(x1, x2, x3), Z(x1, x2, x3)) = Xx1 + Yx2 + Zx3 = div v

v = e1 e2 e3

x1

x2

x3

X Y Z

= curlvLaplacian operator

u = = div grad u = 2u

x21+

2u

x22+

2u

x23

Indical Notation Summation convention

A subscript appearing twice is summed from 1 to 3.

e.g.:

aibi =3

i=1

aibi

= a1b1 + a2b2 + a3b3

Djj = D11 + D22 + D33


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Comma-subscript convention

The partial derivative with respect to the variable xi is represented by the so-calledcomma-subscript convention e.g.:

xi= ,i = grad

vixi

= vi,i = divv

vixj

= vi,j

2vi

xjxk= vi,jk

1.9 Exercises

1. given: scalar field

f(x1, x2, x3) = 3x1 + x1ex2 + x1x2e

x3

(a)

gradf(x1, x2, x3) =?

(b)

gradf(3, 1, 0) =?


f(x1, x2, x3) = x21 +

3

2x22

Find the derivative of f in point/position vector

528

in the direction of a

304

.

3. given: vector field

V =

x1 + x22ex1x3 + sin x2x1x2x3

(a)

divV(X(x1, x2, x3), Y(x1, x2, x3), Z(x1, x2, x3)) =?

(b)

divV(1, , 2) =?


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V = x1 + x2

ex1+x2

+ x3x3 + sin x1

(a)

curlV(x1, x2, x3) =?

(b)

curlV(0, 8, 1) =?

5. Expand and, if possible, simplify the expression Dijxixj for

(a) Dij = Dji

(b) Dij = Dji.

6. Determine the component f2 for the given vector expressions

(a) fi = ci,jbj cj,ibj(b) fi = Bijf

j

7. Ifr2 = xixi and f(r) is an arbitrary function of r, show that

(a) (f(r)) = f(r)xr(b) 2(f(r)) = f(r) + 2f(r)

r,

where primes denote derivatives with respect to r.

1.10 Summary of chapter 1

Vectors

a =

a1a2a3

= a1 e1 + a2 e2 + a3 e3 = a110

0

+ a201

0

+ a300

1

Magnitude of a:

|a| =

a21 + a

22 + a

23 is the length of a


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Vector addition:

a1a2a3

+b1b2b3

= a1 + b1a2 + b2a3 + b3

Multiplication with a scalar:

c

a1

a2

a3

=

c a1c a2c

a3

Scalar (inner, dot) product:

a b = |a||b| cos = a1 b1 + a2 b2 + a3 b3

Vector (outer, cross) product:

a b = e1 e3 e3

a1 a2 a3

b1 b2 b3

= e1a2 a3b2 b3 e2a1 a3b1 b3+ e3a1 a2b1 b2 = a2b3

a3b2

a3b1 a1b3a1b2 a2b1

Rules for the vector product:

a b = (b a)(c a) b = a (c b) = c(a b)

(a + b) c = a c + b c

a (b c) = (a c) b (a b) c

Matrices

A =

A11 A12 A13 ... A1n

A21 A22 A23 ... A2n...

......

...

Am1 Am2 Am3 ... Amn

= Aik


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Multiplication of a matrix with a scalar:

c A = A c = c Aik e.g.: c A11 A12A21 A22

= c A11 c A12c A21 c A22

Addition of two matrices:

A + B = B + A = (Aik) + (Bik) = (Aik + Bik)

e.g.:

A11 A12A21 A22

+B11 B12B21 B22

= A11 + B11 A12 + B12A21 + B21 A22 + B22

Rules for addition of matrices:

(A + B) + C = A + (B + C) = A + B + C

Multiplication of two matrices:

Cik = Ai1B1k + Ai2B2k + ... + AilBlk =l

j=1

AijBjk i = 1, ..., m k = 1,...,n

e.g.:

B11 B12

B21 B22

A11 A12

A21 A22

A11B11 + A12B12 A11B12 + A12B22

A21B11 + A22B21 A21B12 + A22B22

Rules for multiplication of two matrices:

A(BC) = (AB)C = ABC

AB = BA

Distributive law:

(A + B) C = A C + B C


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Differential operators for vector analysis

Gradient of a scalar field f(x,y,z)

grad f(x1, x2, x3) =

f(x1,x2,x3)

x1f(x1,x2,x3)

x2f(x1,x2,x3)

x3

Derivative into a certain direction:

f

a(x1, x2, x3) =

a

|a

|

grad f(x1, x2, x3)

Divergence of a vector field

div v(X(x1, x2, x3), Y(x1, x2, x3), Z(x1, x2, x3)) =X

x1+

Y

x2+

Z

x3

Curl of a vector field

curlv(X(x1, x2, x3), Y(x1, x2, x3), Z(x1, x2, x3)) = Zx2

Yx3

Xx3

Zx1

Yx1

Xx2

Nabla (del) Operator

=

x1

x2

x3

f(x1, x2, x3) =

fx1f

x2f

x3

= grad f(x1, x2, x3)v(X(x1, x2, x3), Y(x1, x2, x3), Z(x1, x2, x3)) = Xx1 + Yx2 + Zx3 = div v

v =

e1 e2 e3

x1

x2

x3

X Y Z

= curlv


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1.11. EXERCISES 23

Laplacian operator

u = = div grad u =2u

x21+

2u

x22+

2u

x23

Indical Notation Summation convention

A subscript appearing twice is summed from 1 to 3.

e.g.:

aibi =3

i=1 aibi= a1b1 + a2b2 + a3b3

Djj = D11 + D22 + D33

Comma-subscript convention

The partial derivative with respect to the variable xi is represented by the so-called

comma-subscript convention e.g.:

xi= ,i = grad

vixi

= vi,i = divv

vixj

= vi,j

2vixjxk

= vi,jk

1.11 Exercises


f(x1, x2, x3) = 3x1 + x1ex2 + x1x2e

x3

(a)

gradf(x1, x2, x3) =?


32/183


(b)

gradf(3, 1, 0) =?


f(x1, x2, x3) = x21 +

3

2x22

Find the derivative of f in point/position vector

528

in the direction of a

304

.


V =

x1 + x

22

ex1x3 + sin x2

x1x2x3

(a)

divV(X(x1, x2, x3), Y(x1, x2, x3), Z(x1, x2, x3)) =?

(b)

divV(1, , 2) =?


V = x1 + x2

ex1+x2 + x3

x3 + sin x1(a)

curlV(x1, x2, x3) =?

(b)

curlV(0, 8, 1) =?

5. Expand and, if possible, simplify the expression Dijxixj for

(a) Dij = Dji

(b) Dij = Dji.

6. Determine the component f2 for the given vector expressions

(a) fi = ci,jbj cj,ibj(b) fi = Bijf

j

7. Ifr2 = xixi and f(r) is an arbitrary function of r, show that


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1.11. EXERCISES 25

(a) (f(r)) = f(r)xr

(b)

2(f(r)) = f(r) + 2f

(r)r

,

where primes denote derivatives with respect to r.


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2 Foundation of Solid Mechanics

2.1 Stress

2.1.1 Components of Stress

Assumption:

Cartesian coordinate system with unit vectors ei infinitesimal rectangular parallelepiped;

ti are not parallel to ei whereas the surfaces are perpendicular to the ei, respectively (fig.

2.1). So, all ei represents here the normal ni of the surfaces.

x1

x2

x3

e1

e2e3

t1

t2

t3

1112

13

Figure 2.1: Tractions ti and their components ij on the rectangular parallelepiped sur-

faces of an infinitesimal body. [29]

Each traction is separated in components in each coordinate direction

ti = i1e1 + i2e2 + i3e3 (2.1)

ti = ijej . (2.2)

26


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2.1. STRESS 27

With these coefficients ij a stress tensor can be defined

= 11 12 1321 22 2331 32 33

= ij , (2.3)with the following signconvention:

1. The first subscript i refers to the normal ei which denotes the face on which ti acts.

2. The second subscript j corresponds to the direction ej in which the stress acts.

3. ii (no summation) are positive (negative) if they produce tension (compression).

They are called normal components or normal stress

ij (i = j) are positive if coordinate direction xj and normal ei are both positiveor negative. If both differ in sign, ij (i = j) is negative. They are called shearcomponents or shear stress.

2.1.2 Stress on a normal plane

Interest is in the normal and tangential component of tn (fig. 2.2).

dAns

tnn

Figure 2.2: Normal and tangential component of tn. [29]

Normalvector: n = niei

Tangentialvector: s = siei (two possibilities in 3-D)

Normal component of stress tensor with respect to plane dAn:

nn = tn n = ijniej nkek= ijninkjk = ijnjni

(2.4)

Tangential component:

ns = tn s = ijniej skek = ijnisj (2.5)


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28 CHAPTER 2. FOUNDATION OF SOLID MECHANICS

2.1.3 Principal stress

2.1.3.1 Maximum normal stress

Question: Is there a plane in any body at any particular point where no shear stress

exists?

Answer: Yes

For such a plane the stress tensor must have the form

= (1) 0 0

0 (2) 00 0 (3)

(2.6)with three independent directions n(k) where the three principal stress components act.

Each plane given by these principal axes n(k) is called principal plane. So, it can be

defined a stress vector acting on each of these planes

t = (k)n (2.7)

where the tangential stress vector vanishes.

2.1.4 Stress invariants and special stress tensors

In general, the stress tensor at a distinct point differ for different coordinate systems.

However, there are three values, combinations of ij, which are the same in every coordi-

nate system. These are called stress invariants. They can be found as under

|ij (k)ij| = ((k))3 I1((k))2 + I2(k) I3 != 0 (2.8)

with

I1 = ii = tr (2.9)

I2 =1

2(iijj ijij) (2.10)

I3 = |ij| = det (2.11)


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2.1. STRESS 29

and represented in principal stresses

I1 = (1) + (2) + (3) (2.12)

I2 = ((1)(2) + (2)(3) + (3)(1)) (2.13)

I3 = (1)(2)(3), (2.14)

the first, second, and third stress invariant is given. The invariance is obvious because

all indices are dummy indices and, therefore, the values are scalars independent of the

coordinate system.

The special case of a stress tensor, e.g., pressure in a fluid,

= 01 0 00 1 00 0 1

ij = 0ij (2.15)is called hydrostatic stress state. If one assumes 0 =

ii3

= m of a general stress state,

where m is the mean normal stress state, the deviatoric stress state can be defined

S = mI =

11 m 12 1312 22 m 2313 23 33 m

. (2.16)In indical notation (I = ij: itendity-matrix (3x3)):

sij = ij ij kk3

(2.17)

where kk/3 are the components of the hydrostatic stress tensor and sij the components

of the deviatoric stress tensor.

The principal directions of the deviatoric stress tensor S are the same as those of the

stress tensor because the hydrostatic stress tensor has no principal direction, i.e., any

direction is a principal plane. The first two invariants of the deviatoric stress tensor are

J1 = sii = (11 m) + (22 m) + (33 m) = 0 (2.18)J2 = 1

2sijsij =

1

6[((1) (2))2 + ((2) (3))2 + ((3) (1))2], (2.19)

where the latter is often used in plasticity.

Remark: The elements on the main diagonal of the deviatoric stress tensor are mostly

not zero, contrary to the trace of deviatoric stress S, which is per definition equal zero

tr(S) = 0.


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2.2 Equilibrium

2.2.1 Physical principles

Consider an arbitrary body V with boundary A (surface) (fig. 2.3).

x1

x2

x3

fr

t

P

Figure 2.3: Body V under loading f with traction t acting normal to the boundary of the

body. [29]

In a 3-d body the following 2 axioms are given:

1. The principle of linear momentum is

V

fdV +A

t dA =V

d2

dt2u dV (2.20)

with displacement vector u and density .

2. The principle of angular momentum (moment of momentum)V

(r f) dV +A

(r t) dA =V

(r u) dV (2.21)

Considering the position vector r to point P(x)

r = xjej (2.22)

and further

r f = ijk xjfkei (2.23)r t = ijk xjtkei (2.24)


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2.3. DEFORMATION 31

The two principles, (2.20) and (2.21), are in indical notation

V

fidV + A

jinjdA = V

uidV note, that ( ) = d2dt2 () (2.25)V

ijk xj fkdV +

A

ijk xj lknldA =

V

ijk xjukdV, (2.26)

where the Cauchy theorem has been used. In the static case, the inertia terms on the

right hand side, vanish.

2.3 Deformation

2.3.1 Position vector and displacement vector

Consider the undeformed and the deformed configuration of an elastic body at time t = 0

and t = t, respectively (fig. 2.4).

x1X1

x2X2

x3X3

x

P(x) u p(X)

X

undeformeddeformed

t = 0 t = t

Figure 2.4: Deformation of an elastic body. [29]

It is convenient to designate two sets of Cartesian coordinates x and X, called material

(initial) coordinates and spatial (final) coordinates, respectively, that denote the unde-

formed and deformed position of the body. Now, the location of a point can be given in

material coordinates (Lagrangian description)

P = P(x, t) (2.27)

and in spatial coordinates (Eulerian description)

p = p(X, t). (2.28)


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Mostly, in solid mechanics the material coordinates and in fluid mechanics the spatial

coordinates are used. In general, every point is given in both

X = X(x, t) (2.29)

or

x = x(X, t) (2.30)

where the mapping from one system to the other is given if the Jacobian

J = Xix

j = |Xi,j| (2.31)exists.

So, a distance differential is

dXi =Xixj

dxj (2.32)

where ( ) denotes a fixed but free distance. From figure 3.1 it is obvious to define the

displacement vector by

u = X

x ui = Xi

xi. (2.33)

Remark: The Lagrangian or material formulation describes the movement of a particle,

where the Eulerian or spatial formulation describes the particle moving at a location.

2.3.2 Strain tensor

Consider two neighboring points p(X) and q(X) or P(x) and Q(x) (fig. 2.5) in both

configurations (undeformed/deformed)

x1X1

x2X2

x3X3

ds

Q(x + dx)

P(x)

u + du q(X + dX)

p(X)u

dS

Figure 2.5: Deformation of two neighboring points of a body. [29]


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2.3. DEFORMATION 33

which are separated by differential distances ds and dS, respectively. The squared length

of them is given by

|ds|2 = dxidxi (2.34)|dS|2 = dXidXi. (2.35)

With the Jacobian of the mapping from one coordinate representation to the other these

distances can be expressed by

|ds|2 = dxidxi = xiXj

xiXk

dXjdXk (2.36)

|dS

|2 = dX

idX

i=

Xi

xj

Xi

xkdx

jdx

k. (2.37)

To define the strain we want to express the relative change of the distance between the

point P and Q in the undeformed and deformed body. From figure 2.5 it is obvious that

ds + u + du dS u = 0 du = dS ds.

(2.38)

Taking the squared distances in material coordinates yield to

|dS|2

|ds|2

= Xi,jXi,kdxjdxk dxidxi= (Xi,jXi,k jk )

=2Ljk

dxjdxk (2.39)

with the Green or Lagrangian strain tensor Ljk , or in spatial coordinates

|dS|2 |ds|2 = dXidXi xiXj

xiXk

dXjdXk

= (jk

xi,jxi,k)

=2Ejk

dXjdXk(2.40)

with the Euler or Almansi strain tensor Ejk . Taking into account that

uixk

=Xixk

xixk

= Xi,k ik Xi,k = ui,k + ik (2.41)

orui

Xk=

XiXk

xiXk

= ik xi,k xi,k = ik ui,k (2.42)


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the Green strain tensor is

L

jk =

1

2[(ui,j + ij)(ui,k + ik) jk ]=

1

2[ui,jui,k + ui,jik + ijui,k + jk jk ]

=1

2[uk,j + uj,k + ui,jui,k]

with ui,j :=uiXj

(2.43)

and the Almansi tensor is

Ejk = 12[jk (ij ui,j)(ik ui,k)]

=1

2[uk,j + uj,k ui,jui,k]

with ui,j :=uixj

.

(2.44)

2.3.3 Linear theory

If small displacement gradients are assumed, i.e.

ui,juk,l ui,j (2.45)

the strain tensors of both configurations are equal.

ij = Lij =

Eij =

1

2(ui,j + uj,i) (2.46)

ij is called linear or infinitesimal strain tensor. This is equivalent to the assumption of

small unit extensions 2

, yielding

2(e) = 2eT EL e = 2eT EE e . (2.47)

With both assumptions the linear theory is established and no distinction between the

configurations respective coordinate system is necessary. The components on the main

diagonal are called normal strain and all other are the shear strains. The shear strains

here

ij =1

2(ui,j + uj,i) =

1

2ij (2.48)


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2.3. DEFORMATION 35

are equal to onehalf of the familiar engineering shear strains ij. However, only with

the definitions above the strain tensor

=

11 12 1312 22 2313 23 33

(2.49)has tensor properties. By the definition of the strains the symmetry of the strain tensor

is obvious.

2.3.4 Properties of the strain tensor

2.3.4.1 Principal strain

Besides the general tensor properties (transformation rules) the strain tensor has as the

stress tensor principal axes. The principal strains (k) are determined from the character-

istic equation

|ij (k)ij| = 0 k = 1, 2, 3 (2.50)

analogous to the stress. The three eigenvalues (k) are the principal strains. The corre-

sponding eigenvectors designate the direction associated with each of the principal strains

given by

(ij (k)ij)n(k)i = 0 (2.51)

These directions n(k) for each principal strain (k) are mutually perpendicular and, for

isotropic elastic materials, coincide with the direction of the principal stresses.

2.3.4.2 Volume and shape changes

It is sometimes convenient to separate the components of strain into those that cause

changes in the volume and those that cause changes in the shape of a differential element.

Consider a volume element V (a b c) oriented with the principal directions (fig. 2.6),then the principal strains are

(1) =a

a(2) =

b

b(3) =

c

c(2.52)

under the assumption of volume change in all three directions.


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a

bc

32

1

Figure 2.6: Volume V oriented with the principal directions. [29]

The volume change can be calculated by

V + V = (a + a)(b + b)(c + c)

= abc

1 +

a

a+

b

b+

c

c

+ O(2)

= V + ((1) + (2) + (3))V + O(2).

(2.53)

With the assumptions of small changes , finally,

V

V= (1) + (2) + (3) = ii (2.54)

and is called Dilation. Obviously, from the calculation this is a simple volume change

without any shear. It is valid for any coordinate system. The Dilation is also the first

invariant of the strain tensor, and also equal to the divergence of the displacement vector:

u = ui,i = ii (2.55)

Analogous to the stress tensor, the strain tensor can be divided in a hydrostatic part

M =

M 0 0

0 M 0

0 0 M

M =ii3

(2.56)

and a deviatoric part

D =

11 M 12 1312 22 M 2313 23 33 M

. (2.57)The mean normal strain M corresponds to a state of equal elongation in all directions

for an element at a given point. The element would remain similar to the original shape


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2.3. DEFORMATION 37

but changes volume. The deviatoric strain characterizes a change in shape of an element

with no change in volume. This can be seen by calculating the Dilation ofD:

trD = (11 M) + (22 M) + (33 M) = 0 (2.58)

2.3.5 Compatibility equations for linear strain

If the strain components ij are given explicitly as functions of the coordinates, the six

independent equations (symmetry of)

ij =1

2

(ui,j + uj,i)

are six equations to determine the three displacement components ui. The system is

overdetermined and will not, in general, possess a solution for an arbitrary choice of the

strain components ij. Therefore, if the displacement components ui are singlevalued and

continuous, some conditions must be imposed upon the strain components. The necessary

and sufficient conditions for such a displacement field are expressed by the equations

ij,km + km,ij ik,jm jm,ik = 0. (2.59)

These are 81 equations in all but only six are distinct

1.211x22

+222x21

= 2212

x1x2

2.222x23

+233x22

= 2223

x2x3

3.233x21

+211x23

= 2231

x3x1

4.

x1 23x1 + 31x2 + 12x3 = 211

x2x3

5.

x2

23x1

31x2

+12x3

=

222x3x1

6.

x3

23x1

+31x2

12x3

=

233x1x2

or x E = 0. (2.60)

The six equations written in symbolic form appear as

E = 0 (2.61)


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Even though we have the compatibility equations, the formulation is still incomplete in

that there is no connection between the equilibrium equations (three equations in six

unknowns ij), and the kinematic equations (six equations in nine unknowns ij and ui).We will seek the connection between equilibrium and kinematic equations in the laws of

physics governing material behavior, considered in the next chapter.

Remark on 2D:

For plane strain parallel to the x1 x2 plane, the six equations reduce to the singleequation

11,22 + 22,11 = 212,12 (2.62)

or symbolic E = 0. (2.63)

For plane stress parallel to the x1 x2 plane, the same condition as in case of plain strainis used, however, this is only an approximative assumption.

2.4 Material behavior

2.4.1 Uniaxial behavior

Constitutive equations relate the strain to the stresses. The most elementary description

is Hookes law, which refers to a onedimensional extension test

11 = E11 (2.64)

where E is called the modulus of elasticity, or Youngs modulus.

Looking on an extension test with loading and unloading a different behavior is found

(fig. 2.7).

There is the linear area governed by Hookes law. In yielding occure which must be

governed by flow rules. is the unloading part where also in pressure yielding exist .

Finally, a new loading path with linear behavior starts. The region given by this curve is

known as hysteresis loop and is a measure of the energy dissipated through one loading

and unloading circle.

Nonlinear elastic theory is also possible. Then path is curved but in loading and

unloading the same path is given.


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2.4. MATERIAL BEHAVIOR 39

Figure 2.7: - diagram of an extension test. [29]

2.4.2 Generalized Hookes law

2.4.2.1 General anisotropic case

As a prerequisite to the postulation of a linear relationship between each component of

stress and strain, it is necessary to establish the existence of a strain energy density W

that is a homogeneous quadratic function of the strain components. The density function

should have coefficients such that W 0 in order to insure the stability of the body, withW(0) = 0 corresponding to a natural or zero energy state. For Hookes law it is

2W = Cijkmij km. (2.65)

The constitutive equation, i.e., the stressstrain relation, is a obtained by

ij =W

ij(2.66)

yielding the generalized Hookes law

ij = Cijkmkm. (2.67)

There, Cijkm is the fourthorder material tensor with 81 coefficients. These 81 coefficients

are reduced to 36 distinct elastic constants taking the symmetry of the stress and the strain

tensor into account. Introducing the notation

= (11 22 33 12 23 31)T (2.68)


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and

= (11 22 33 212 223 231)T (2.69)

Hookes law is

K = CKMM K, M = 1, 2, . . . , 6 (2.70)

and K and M represent the respective double indices:

1 = 11, 2 = 22, 3 = 33, 4 = 12, 5 = 23, 6 = 31.

From the strain energy density the symmetry of the materialtensor

Cijkm = Ckmij or CKM = CMK (2.71)

is obvious yielding only 21 distinct material constants in the general case. Such a material

is called anisotropic.

2.4.2.2 Planes of symmetry

Most engineering materials possess properties about one or more axes, i.e., these axes

can be reversed without changing the material. If, e.g., one plane of symmetry is the

x2 x3plane the x1axis can be reversed (fig. 2.8),

x1

x2

x3

(a) Original coordinate system

x1

x2

x3

(b) Onesymmetry plane

x2

x1

x3

(c) Twosymmetry planes

Figure 2.8: Coordinate systems for different kinds of symmetry. [29]

yielding a transformation

x =

1 0 00 1 0

0 0 1

x. (2.72)


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With the transformation property of tensors

ij = ikjl kl (2.73)

and

ij = ikjl kl (2.74)

it is

112233

122331

=

11

22

33

12

23

31

= C

112233

212

223231

= C

11

22

33

212

223

231

. (2.75)

The above can be rewritten

=

C11 C12 C13 C14 C15 C16C22 C23 C24 C25 C26

C33 C34 C35 C36C44 C45 C46

sym. C55 C56C66

(2.76)

but, since the constants do not change with the transformation, C14, C16, C24, C26, C34,

C36, C45, C56!

= 0 leaving 21 8 = 13 constants. Such a material is called monocline. Thecase of three symmetry planes yields an orthotropic material written explicitly as

=

C11 C12 C13 0 0 0

C22 C23 0 0 0

C33 0 0 0C44 0 0

sym. C55 0

C66

(2.77)

with only 9 constants. Further simplifications are achieved if directional independence,

i.e., axes can be interchanged, and rotational independence is given. This reduces the

numbers of constants to two, producing the familiar isotropic material. The number of

constants for various types of materials may be listed as follows:


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21 constants for general anisotropic materials;

9 constants for orthotropic materials;

2 constants for isotropic materials.

We now summarize the elastic constant stiffness coefficient matrices for a few selected

materials.

Orthotropic: 9 constants

C11 C12 C13 0 0 0

C22 C23 0 0 0

C33 0 0 0C44 0 0

sym. C55 0

C66

(2.78)

Isotropic: 2 constants

C11 C12 C12 0 0 0

C11 C12 0 0 0

C11 0 0 012

(C11 C12) 0 0sym. 1

2(C11 C12) 0

12

(C11 C12)

(2.79)

A number of effective modulus theories are available to reduce an inhomogeneous multi-

layered composite material to a single homogeneous anisotropic layer for wave propagation

and strength considerations.

2.4.2.3 Isotropic elastic constitutive law

Using the Lame constants , the stress strain relationship is

=

2 + 0 0 0

2 + 0 0 0

2 + 0 0 0

2 0 0

sym. 2 0

2

11

22

33

12

23

31

(2.80)


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or in indical notation using the stress and strain tensors

ij = 2ij + ijkk (2.81)

or vice versa

ij =ij2

ijkk2(2 + 3)

. (2.82)

Other choices of 2 constants are possible with

the shear modulus

= G =E

2(1 + ),

=E

(1 + )(1 2) , (2.83)

Youngs modulusE =

(2 + 3)

+ , (2.84)

Poissons ratio =

2( + ), (2.85)

bulk modulusK =

E

3(1 2) =3 + 2

3. (2.86)

From equation (2.83) it is obvious 1 < < 0.5 if remains finite. This is, however, trueonly in isotropic elastic materials. With the definition of Poissons ratio

= 2211

= 3311

(2.87)

a negative value produces a material which becomes thicker under tension. These mate-

rials can be produced in reality and are called auxetic materials.

The other limit = 0.5 (isochoric) can be discussed as: Taking the 1principal axes as

(1) = then both other are (2) = (3) = (see equation (2.87)). This yields thevolume change

V

V= ii = (1 2) (2.88)

Now, = 0.5 gives a vanishing volume change and the material is said to be incompress-

ible. Rubberlike materials exhibit this type of behavior.


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Finally, using the compression/bulk modulus K and the shear modulus G and further

the decomposition of the stress and strain tensor into deviatoric and hydrostatic part,

Hookes law is a given (eij are the components ofD)

kk = 3Kkk sij = 2Geij. (2.89)

2.5 Summary

2.5.1 Stress

Tractions

ti = ij ej

t1t2

t3

11

1213

21

22

2331

32

33

2.5.1.1 Stress Tensor

=

11 12 1321 22 2331 32 33

11, 22, 33 : normal components12, 13, 23: shear components

21, 31, 32

Stress at a point

ti = jinj

Transformation in another cartesian coordinate system

ij = ikjl kl = ikkllj

with direction cosine: ij = cos (xi, xj)


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2.5. SUMMARY 45

Stress in a normal plane

Normal component of stress tensor: nn = ijnjniTangential component of stress tensor: ns = ijnisj =

titi 2nn

Equilibrium

ij = ji = T

ij,j +fi = 0 (static case)

with boundary condition: ti = ijnj

Principal Stress

In the principal plane given by the principal axes n(k) no shear stress exists.

Stress tensor referring to principal stress directions:

=

(1) 0 00 (2) 00 0 (3)

with (1) (2) (3)

Determination of principal stresses (k) with:

|ij (k)ij| != 0 11 (k) 12 13

21 22 (k) 2331 32 33 (k)

!= 0with the Kronecker delta :

ij =

1 if i = j

0 if i = j


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Principal stress directions n(k):

(ij (k)ij) n(k)j = 0

(11 (k) n(k)1 + 12 n(k)2 + 13 n(k)3 = 021 n

(k)1 +(22 (k)) n(k)2 + 23 n(k)3 = 0

31 n(k)1 + 32 n

(k)2 + (33 (k)) n(k)3 = 0

Stress invariants

The first, second, and third stress invariant is independent of the coordinate system:

I1 = ii = tr = 11 + 22 + 33

I2 =1

2(iijj ijij)

= 1122 + 2233 + 3311 1212 2323 3131I3 = |ij| = det

Hydrostatic and deviatoric stress tensors

A stress tensor ij can be split into two component tensors, the hydrostatic stress tensor

M = M

1 0 0

0 1 0

0 0 1

Mij = Mij with M =kk

3

and the deviatoric tensor

S = MI =

11 M 12 1321 22 M 2331 32 33 M

ij = ijkk

3+ Sij.


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2.5. SUMMARY 47

2.5.2 Exercises

1. The state of stress at a point P in a structure is given by11 = 20000

22 = 1500033 = 3000

12 = 2000

23 = 2000

31 = 1000 .

(a) Compute the scalar components t1, t2 and t3 of the traction t on the planepassing through P whose outward normal vector n makes equal angles with

the coordinate axes x1, x2 and x3.

(b) Compute the normal and tangential components of stress on this plane.

2. Determine the body forces for which the following stress field describes a state of

equilibrium in the static case:

11 = 2x21 3x22 5x3

22 = 2x2

2 + 733 = 4x1 + x2 + 3x3 512 = x3 + 4x1x2 613 = 3x1 + 2x2 + 123 = 0

3. The state of stress at a point is given with respect to the Cartesian axes x1, x2 and

x3 by

ij = 2 2 02 2 00 0 2

.Determine the stress tensor ij for the rotated axes x

1, x

2 and x

3 related to the

unprimed axes by the transformation tensor

ik =

01

21

21

212

12

12

12

12

.


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4. In a continuum, the stress field is given by the tensor

ij = x

2

1x

2(1

x2

2)x

10

(1 x22)x1 (x3

23x2)

30

0 0 2x23

.Determine the principal stress values at the point P(a, 0, 2

a) and the correspond-

ing principal directions.

5. Evaluate the invariants of the stress tensors ij and ij, given in example 3 of chapter

2.

6. Decompose the stress tensor

ij =

3 10 010 0 300 30 27

into its hydrostatic and deviator parts and determine the principal deviator stresses!

7. Determine the principal stress values for

(a)

ij =

0 1 11 0 11 1 0

and

(b)

ij =

2 1 1

1 2 1

1 1 2

and show that both have the same principal directions.

2.5.3 Deformations

Linear (infinitesimal) strain tensor :

Lij = Eij = ij =

1

2(ui,j + uj,i)


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2.5. SUMMARY 49

=

u1,112

(u1,2 + u2,1)12

(u1,3 + u3,1)12

(u1,2 + u2,1) u2,212

(u2,3 + u3,2)

12 (u1,3 + u3,1) 12 (u2,3 + u3,2) u3,3

=

1112

1212

1312

21 2212

23

12 31 12 32 33

Principal strain values (k):

|ij (k)ij| != 0

Principal strain directions n(k):

(ij

(k)ij

) n(k)

j= 0

Hydrostatic and deviatoric strain tensors

A stress tensor ij can be split into two component tensors, the hydrostatic stain tensor

M = M

1 0 0

0 1 0

0 0 1

Mij = Mij with M =kk3

and the deviatoric strain tensor

(D) = MI =

11 M 12 1321 22 M 2331 32 33 M

Compatibility:

lm,ln + ln,lm

mn,ll = ll,mn

11,22 + 22,11 = 212,12 = 12,12

22,33 + 33,22 = 223,23 = 23,23

33,11 + 11,33 = 231,31 = 31,31

12,13 + 13,12 23,11 = 11,2323,21 + 21,23 31,22 = 22,3131,32 + 32,31 12,33 = 33,12


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2.5.4 Exercises

1. The displacement field of a continuum body is given byX1 = x1

X2 = x2 + Ax3

X3 = x3 + Ax2

where A is a constant. Determine the displacement vector components in both the

material and spatial form.

2. A continuum body undergoes the displacement

u =3x2 4x32x1 x3

4x2 x1

.Determine the displaced position of the vector joining particles A(1, 0, 3) and

B(3, 6, 6).

3. A displacement field is given by u1 = 3x1x22, u2 = 2x3x1 and u3 = x

23 x1x2. De-

termine the strain tensor ij and check whether or not the compatibility conditions

are satisfied.

4. A rectangular loaded plate is clamped along the x1- and x2-axis (see fig. 2.9). On

the basis of measurements, the approaches 11 = a(x21x2 + x

32); 22 = bx1x

22 are

suggested.x2, u2

x1, u1

Figure 2.9: Rectangular plate. [29]

(a) Check for compatibility!

(b) Find the displacement field and

(c) compute shear strain 12.


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2.5. SUMMARY 51

2.5.5 Material behavior

Generalized Hookes Law

ij = Cijkm km K = CKM M

with K, M = 1, 2, ..., 6 and K, M represent the respective double indices:

1=11, 2=22, 3=33, 4=12, 4=23, 6=31

11

22

33

12

23

31

=

C11 C12 C13 C14 C15 C16

C12 C22 C23 C24 C25 C26

C13 C23 C33 C34 C35 C36

C14 C24 C34 C44 C45 C46

C15 C25 C35 C45 C55 C56

C16 C26 C36 C46 C56 C66

11

22

33

12

23

31

Orthotropic material

11

22

3312

23

31

=

C11 C12 C13 0 0 0

C12 C22 C23 0 0 0

C13 C23 C33 0 0 00 0 0 C44 0 0

0 0 0 0 C55 0

0 0 0 0 0 C66

11

22

3312

23

31

Isotropic material

11

22

3312

23

31

=

2 + 0 0 0

2 + 0 0 0

2 + 0 0 00 0 0 2 0 0

0 0 0 0 2 0

0 0 0 0 0 2

11

22

3312

23

31


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Relation between Lame constants , and engineering constants:

= G = E2(1 + )

E =(2 + 3)

+

=E

(1 + )(1 2) =

2( + )

K =E

3(1 2)=

3 + 2

3

Thermal strains:

ij(T) = (T T0)ijij = 2ij + kkij ij(3 + 2)(T T0)

2.5.6 Exercises

1. Determine the constitutive relations governing the material behavior of a point hav-

ing the properties described below. Would the material be classified as anisotropic,

orthotropic or isotropic?

(a) state of stress:

11 = 10.8; 22 = 3.4; 33 = 3.0; 12 = 13 = 23 = 0

corresponding strain components:

11 = 10 104; 22 = 2 104; 33 = 2 104; 12 = 23 = 31 = 0(b) state of stress:

11 = 10; 22 = 2; 33 = 2; 12 = 23 = 31 = 0

corresponding strains:

11 = 10 104; 22 = 33 = 12 = 23 = 31 = 0(c) state of stress:

When subjected to a shearing stress 12, 13 or 23 of 10, the material develops

no strain except the corresponding shearing strain, with tensor component 12,

13 or 23, of 20 104.


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2.5. SUMMARY 53

2. A linear elastic, isotropic cuboid is loaded by a homogeneous temperature change.

Determine the stresses and strains of the cuboid, if

(a) expansion in x1 and x2-direction is prevented totally and if there is no preven-

tion in x3-direction.

(b) only in x1-direction, the expansion is prevented totally.

3. For steel E = 30 106 and G = 12 106. The components of strain at a point withinthis material are given by

= 0.004 0.001 0

0.001 0.006 0.0040 0.004 0.001

.Compute the corresponding components of the stress tensor ij.


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3 Foundation of Constitutive

Material Models

3.1 The Stress Plane

An octahedral plane is defined as a plane where the normal to the plane makes equal

angles to the three principal strain directions. Eight such planes exist and one example

is shown in figure 3.1 where the axes 1, 2 and 3 refer to the principal strain directions.

11

22

33

Aquisektrix

qp

Figure 3.1: Deviatoric or Octahedral Plane (plane): Hydrostatic Stress (pressure) p

and Deviatoric Stress q. [5]

For the normal to the octahedral plane shown in the figure, we have n = 13

111

In the coordinate system collinear with the principal strain directions, the strain tensor

54


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3.2. DILATION 55

takes the form =

1 0 0

0 2 0

0 0 3

The vector q is defined by q = n. It then follows from figure 3.2 that the normal strain

0 and tensorial shear strain 0/2 on the octahedral plane are given by

0 = nTq;

02

=

qTq 20,

Figure 3.2: The vector q = n and its components after direction m and n. [25]

where 0 is called the octahedral normal strain and 0 is called the octahedral shear strain.

The deviatoric or octhaedral plane shows the possible shear stresses at a specific hydro-

static pressure level. A typical shape of a yield function for i.e. soil can look like figure

3.4

3.2 Dilation

A specific quality of most bulk solids but also other materials as some kinds of artificial

ones show the effect of dilation. Dilation is characterised by a increase of volumetric

extension during deforming with shear strain in case of soil and bulk solids or while

stretching in case of auxetic materials.


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56 CHAPTER 3. FOUNDATION OF CONSTITUTIVE MATERIAL MODELS

Figure 3.3: Example: Single Surface Yield Function for Soil, 3D view of stress space.

Figure 3.4: Characteristic Single Surface Yield Function in deviatoric plane for Soil.[25]

A state of uniform Dilation occurs, if the strain tensor is given by

ij = bij,

where b is an arbitrary scalar. The deviatoric strain tensor eij becomes zero and the

strain state corresponds to a uniform Dilation, i.e. a volume change, where the extension-


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3.2. DILATION 57

Figure 3.5: (left) Uniform dilation; (mid) Auxetic material behaviour, auxetic hexagon;

(right) Dilating package of grains while shearing

Figure 3.6: Uniaxial strain. [25]

or contraction-in any direction is the same and equal to the parameter b, cf. Fig.3.5.

Uniaxial strain occurs if the displacement vector ui is given by

[ui] =

u1(xi, t)00

which implies that

11 = ui/xi

and all other strain components being zero, Fig. 3.6.

Plane strain or plane deformation occurs if the displacement vector ui is given by

[ui] =

u1(x1, x2, t)

u2(x1, x2, t)

0


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which implies

[ij] = 11 12 021 22 0

0 0 0

This strain state occurs often in practise when a long prismatic or cylindrical body is

loaded by forces which are perpendicular to the longitudinal elements and which do not

vary along the length. In this case, it can be assumed that all cross sections are in the

same state and if, moreover, the body is restricted from moving in the length direction, a

state of plane strain exists. An example is an internally pressurised tube with end sections

confined between smooth and rigid walls, Fig.3.7.

Figure 3.7: An example of plane strain. Pressurised tube with end sections confined

between smooth and rigid walls. [25]

So-called generalised plane strain or generalised plane deformation occurs if

[ui] = u1(x1, x2, t)u2(x1, x2, t)

u3(x1, x2, t)

which leads to

[ij] =

11 12 1321 22 2331 32 0


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3.3. LINEAR HYPER-ELASTICITY - ANISOTROPY 59

Figure 3.8: Simple shear, no volume change. [25]

3.3 Linear hyper-elasticity - Anisotropy

A material is anisotropic, if it behaves differently when loaded in the same manner in

different directions. As an illustration, consider the piece of wood shown in Fig. 3.9.

In Figure 3.9(a), we have uniaxial tension along the x1 axis and we may express therelation between 11 and 11 as 11 = Ea11, where Ea is some experimentally determined

stiffness parameter. Likewise, in Figure 3.9(b), we also have uniaxial tension along the

x1 axis and the relation between 11 and 11 can now be given as 11 = Eb11, where Ebis some experimentally determined stiffness parameter. Comparison of Figure 3.9(a) and

3.9(b) clearly shows that Ea = Eb. We are thereby led to the following general conclusion:Material anisotropy means that the constitutive relation takes different forms depending

on the Cartesian coordinate system we use.

Let us assume that the constitutive law between stresses and strains is linear and let us

investigate the general properties of this relation for a hyper-elastic material. The most

general linear relation must be of the form

ij = Dijklkl; Dijkl = Dijkl(xi), (3.1)

where Dijkl is the elastic stiffness tensor. That Dijkl indeed is a tensor of fourth order

follows from the quotient theorem and the fact that ij and ij are second order tensors.

The formulation (3.1) covers both anisotropic and isotropic elastic materials.


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Figure 3.9: Example of anisotropy. Piece of wood loaded by the same uniaxial tension in

different directions. [25]

3.4 Viscoelasticity

In classic elasticity there is no time delay between the application of a force and the defor-

mation that it causes. For many materials, however, there is additional time-dependent

deformation that is recoverable. This is called viscoelastic or anelastic deformation.

When a load is applied to a material, there is an instantaneous elastic response, but the

deformation also increases with time. This viscoelasticity should not be confused with

creep, which is time-dependent plastic deformation. Anelastic strains in metals and ce-

ramics are usually so small that they are ignored. In many polymers, however, viscoelastic

strains can be very significant. Anelasticity is responsible for the damping of vibrations.

A high damping capacity is desirable where vibrations might interfere with the precision

of instruments or machinery and for controlling unwanted noise. A low damping capac-

ity is desirable in materials used for frequency standards, in bells, and in many musical

instruments. Viscoelastic strains are often undesirable. They cause sagging of wooden

beams, denting of vinyl flooring by heavy furniture, and loss of dimensional stability in

gauging equipment. The energy associated with damping is released as heat, which often

causes an unwanted temperature increase. Study of damping peaks and how they are


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3.4. VISCOELASTICITY 61

affected by processing has been useful in identifying mechanisms.

3.4.1 Rheological Models

Anelastic behaviour can be modeled mathematically with structures constructed from

idealised elements representing elastic and viscous behaviour, as shown in Figure 3.10. A

spring models a perfectly elastic solid. The behaviour is described by

e = Fe/Ke, (3.2)

where e is the change of length of the spring, Fe is the force on the spring, and Ke is thespring constant.

Spring Dashpot

Figure 3.10: Rheological elements.

A dashpot models a perfectly viscous material. Its behaviour is described by

v = dv/dt = Fv/Kv, (3.3)

where v is the change in length of the dashpot, Fv is the force on it, and Kv is the

dashpot constant.

3.4.2 Maxwell Model

The Maxwell model consists of a spring and dashpot in series, as shown in Figure 3.11.

Here and in the following, and F, without subscripts, will refer to the overall elongation

and the external force. Consider how this model behaves in two simple experiments.


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v

Figure 3.11: Spring and dashpot in series (Maxwell model).

Figure 3.12: Strain relaxation predicted by the series model. The strain increases linearly

with time.


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3.5. PLASTICITY 63

First, let there be a sudden application of a force, F, at time t = 0, with the force being

maintained constant 3.12. The immediate response from the spring is e = F/Ke.

This is followed by a time-dependent response from the dashpot, v = Ft/Kv. The overall

response will be

= e + v = F/Ke + ft/Kv, (3.4)

so the strain rate will be constant. The viscous strain will not be recovered on unloading.

Now consider a second experiment. Assume that the material is forced to undergo a

sudden elongation, e, at time t = 0 and that this elongation is maintained for a period of

time, as sketched in 3.13. Initially the elongation must be accommodated entirely by thespring (e = e), so that the force initially jumps to a level Fo = Kee. This force

Figure 3.13: Stress relaxation predicted by the series model. The stress decays to zero.

causes the dashpot to operate, gradually increasing the strain v. The force in the

spring, F = Kee = ( v)Ke, equals the force in the dashpot, F = Kvdv/dt,( v)1dv = (Ke/Kv)dt. Integrating, ln[( v)/e] = (Ke/kv)t. Substituting( v) = F/Ke, Kee = Fo, and ln(F/Fo) = t/(Kv/Ke). Now defining a relaxation time, = Kv/Ke,

F = Foexp(t/). (3.5)

3.5 Plasticity

3.5.1 Phenomenological aspects

The uniaxial behaviour of materials shows that irreversible strain develops in a way which

depends on the type of material. In the case of metals such as mild steel, the observed


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behaviour in tension is schematised in 3.14(left), where it can be seen that the response

is elastic and linear until a point A is reached, from which plastic or irreversible strain

upon unloading appears. If the specimen is subjected to an increasing strain, the stressdoes not change until point E. Along the plateau ABDE, the material behaviour is known

as perfectly plastic. If the specimen is unloaded, both loading and unloading follow the

same path, without irreversible deformation. The level of stress at which plastic strains

appears does not change, and the material does not harden. Once a certain level of strain

has been reached (E), the stress again increases. If we unload at some point F and then

reload again, the material is able to resist a higher load until new plastic strains develop

(hardening). Finally, a maximum load is reached from which the stress decreases until

the material fails. In the case of soft soils such as saturated clays, the stress-strain curve

is different, as plastic strain are present from very early stages of the test

Figure 3.14: (left) Behaviour of mild steel; (right) Behaviour of soft clay. [32]

Finally, some geomaterials such as concrete present degradation due to damage caused by

the loading process to the structure of the material (3.15). Loading and unloading shows

clearly how the apparent elastic modulus of the material degrades as the test progresses.

Full understanding of this behaviour needs to take into account this process of degradation,

and theories such as Damage Mechanics provide a suitable framework. However, plastic

models can be developed to reproduce the observed behaviour with an acceptable degree

of accuracy.


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3.5. PLASTICITY 65

Figure 3.15: Behaviour of materials with damage. [32]

3.5.2 Basic theory

If the response of the material does not depend on the velocity at which the stress varies

the relationship between the increments of stress and strain can be written as d = (d)

where , is a function of the increment of the stress tensor d and variables describing

the state (or history) of the material. This is a general relation embracing most non-

linear, rate-independent constitutive laws.

An inverse form is

d = (d) (3.6)

As the material response does not depend on time,

d = (d), (3.7)

where +, is a positive scalar (Darve 1990). Consequently, is a homogeneousfunction of degree 1, which can be written

=

(d): d (3.8)

from which the increments of stress and strain are related by


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d = C : d d = D : d (3.9)

where

C =

(d)

is a fourth-order tensor, homogeneous, of degree zero in d. Before continuing, some basic

properties of C will be described.

We will consider a uniaxial loading-unloading-reloading test schematised in Figure 3.16,

where the constitutive tensor C is a scalar, the inverse of the slope at the point considered.As can be seen, the slope depends on the stress level, being smaller at higher stresses.

However, if we compare the slopes at points A1, A2andA3, they are not the same, and C

depends on past history (stresses, strains, modification of material microstructure, etc.)

Taking a closer look at point C, it can be seen that, for a given point, different slopes

are obtained in loading and unloading, which implies a dependence on the direction

of stress increment. This dependence is only on the direction, as C is a homogeneous

f

Boehrnsen & Zahid - Constitutive Material Modes.

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