01 Stress and deformation tensor - tu-freiberg.de

Editor: Prof. Dr.-Ing. habil. Heinz Konietzky Layout: Gunther Lüttschwager TU Bergakademie Freiberg, Institut für Geotechnik, Gustav-Zeuner-Straße 1, 09599 Freiberg • [email protected]

Stress and deformation tensor Author: Prof. Dr.-Ing. habil. Heinz Konietzky (TU Bergakademie Freiberg, Geotechnical Institute)

1. Introduction .......................................................................................................... 2

2. Tensors ................................................................................................................... 3

2.1 Introduction ....................................................................................................... 3

2.2 Pseudotensors .................................................................................................. 5

2.3 Special tensors ................................................................................................. 6

2.4 Typical tensor operations .................................................................................. 7

2.5 Tensor analysis: simple examples .................................................................... 8

3. Stress tensor ..................................................................................................... 10

4. Deformation tensor ............................................................................................ 24

5. Compatibility condition ....................................................................................... 31

6. Equilibrium conditions ........................................................................................ 33

Stress and deformation tensor Only for private and internal use! Updated: 27. July 2021

page 2 of 32

1. Introduction Geomechanical calculations have to consider the following 3 fundamental relations:

• Equilibrium conditions

• Compatibility conditions

• Constitutive laws The coupling between the stresses and deformations is performed by the constitutive laws (material laws) as indicated by Fig. 1. In order to describe these relations effectively, the theory of tensors was developed at the end of the 19th century. During the 20th century, the use of tensors has extended beyond continuum mechanics and now includes among others the fields of special and general relativity, quantum mechanics, fluid mechanics and electromagnetism. In the context of geomechanics, we will use second-order tensors to describe stresses and deformations and fourth-order tensors to describe the constitu-tive laws. The scheme in Fig. 1 illustrates the interaction of the individual components, which are explained in more detail within the next chapters.

inner + outer Forces FI, FA

Displacements ui

Equilibrium conditions

Compatibility conditions

Stresses σij

Constitutive laws Deformations εij

Fig.1.1: Geomechanical calculation scheme


page 3 of 32

2. Tensors 2.1 Introduction

Let’s examine the known vector product in 3 . The vector product of two vectors pro-

duces a third vector 3, = × ∈z w x z 2.1

Understood as a function that maps ( )→x z x , the vector product is linear so that

( ) ( )

( ) ( ) ( ),α α× = ×

× + = × + ×

w x w xw x y w x w y

2.2

We will call such a linear function a tensor, in this specific case a second-order tensor. Any linear function in 3

can be described through a multiplication with a matrix, so that we can write

3 3, ×= × = ∈z w x Wx W 2.3

In the concrete case of the vector product, the matrix which describes the tensor takes the following form

3 2

3 1

2 1

00

0

w ww ww w

− = − −

W 2.4

Another example of a tensor is the rotation of a vector:

The rotation of a vector in 2 is a function which maps ( )→x y x . This function is once

again linear

( ) ( )( ) ( ) ( )m m=

+ = +

y x y xy w x y w y x

2.5

This function is therefore again called a (second-order) tensor and the rotation tensor can be described by means of matrix multiplication.


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( ) =y x Yx 2.6

With the rotation matrix

cos sinsin cos

α αα α

− =

Y 2.7

These two examples motivate the following definition: A multilinear function (i.e. a function which is linear in all its arguments) that acts on a vector and generates another vector is called a second-order tensor. Because vectors themselves can be used to represent lin-ear functions, they can similarly be understood as tensors of a lower order, with our ten-sors of second order acting on these lower-order tensors. This leads to the following in-ductive, though highly abstract definition of tensors:

Tensors of the order n r s= + are the multilinear functions between the two tensor spaces of the order r and s.

In the two examples given above, we discussed that second-order tensors can be de-scribed by matrices. Similarly, tensors of lower and higher order can be described by the generalization of matrices in different dimensions. This leads to the representation of ten-sors up to the fourth order in 3

as shown in Tab. 1. In addition to this index notation, different types of tensors can be can be expressed by means of dashes above the symbols or parenthesis: a scalar = zeroth-order tensor a or {a} vector = first-order tensor a or [a] 3 3× matrix = second-order tensor …

Because tensors are linear functions between vector spaces, every tensor can be ex-pressed through components with respect to a basis of the vector spaces. Let’s now ex-amine what happens when we change the basis of the vector space on which the tensor operates.

Let’s assume that ( )1,..., ne e′ ′ ′=e and ( )1,..., ne e=e are (ordered) bases of the n-dimen-sional vector space V. Every vector, including every basis vector can be described as a linear combination of the basis vectors.

1

n

j ij ii

e a e=

′ = ∑ 2.8

This means that a change of basis is described through a series of coefficients 𝑎𝑎𝑖𝑖𝑖𝑖. If ijT are the components of the Tensor T with respect to the basis e, then, because of the linearity of tensors, we can obtain the components of T with respect to e’ through


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1 1

n n

kl kj li iji j

T a a T= =

′ =

∑ ∑ 2.9

with k, l = 1,2…n, or using the shorter Summation Convention kl kj li ijT a a T′ = 2.10

Tab. 1: Matrix and tensor definition (index notation)

symbol matrix type tensor order no. of values phys. example a scalar zeroth 1 density

ia vector first 3 displacement

ija 3 3× second 9 stress

ijka 3 3 3× × third 27 --

ijkla 3 3 3 3× × × fourth 81 stiffness matrix

Going forward, this summation will always be implied if an index appears twice in a mul-tiplicative term. It is worth noting that there are different ways to define tensors. Occa-sionally, the described transformation behavior of the describing matrices is used as an equivalent definition to the one we used. 2.2 Pseudotensors If tensors can be described through generalized matrices, one can ask the question why we bothered with our original definition, which is certainly less intuitive. In short, not eve-rything that can be described as a n-dimensional matrix behaves like a tensor. For exam-ple, let’s examine the permutation symbol, also called the Levi-Civita-symbol or ε-tensor. This symbol is defined by the sign of a permutation of the numbers 1, 2, …, n for an integer n. The permutation symbol can be defined in any dimension greater than one. In two dimensions, it is

( ) ( )( ) ( )

1 if , 1,2

1 if , 2,1 0 if

ij

i j

i ji j

ε

+ =

= − = =

2.11

and arranged into a 2 2× antisymmetric matrix:

0 11 0ijε

= −

. 2.12


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In three dimensions, it is,

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

1 if , , 1,2,3 or 2,3,1 or 3,1,2 (even permutations)

1 if , , 3,2,1 or 1,3,2 or 2,1,3 (uneven permutations) 0 if , ,

ijk

i j k

i j ki j i k j k

ε

+ =

= − = = = =

2.13

The ε-tensor is completely antisymmetric (skew-symmetric): ε = ε = ε = −ε = −ε = −ε =123 231 312 321 132 213 1, 2.14 all other elements are zero! Arranged into a 3 x 3 x 3 matrix:

, 2.15 while the permutation symbol has a representation as a generalized matrix, it does not follow the transformation rules of a tensor. Under certain orthogonal transformations, for example a reflection in an odd number of dimensions, it should be multiplied by -1 if it were a tensor. However, the permutation symbol does not change at all and is therefore not a proper tensor. 2.3 Special tensors Zero tensor: All elements of the so-called zero tensor are zero:

0 0 00 0 00 0 0

=

ija 2.16

Symmetric tensor: A tensor is symmetric if non-diagonal elements are paarewise identical, e.g.: ij jia a= , that means: 12 21 23 32 13 31a a , a a and a a .= = = Antisymmetric tensor: A tensor is antisymmetric if non-diagonal elements are paarewise identical ij abloute value, but with opposite sign, e.g.: ij jia a for i j= − ≠ , that means: 12 21 23 32 13 31a a , a a and a a= − = − = −


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2.4 Typical tensor operations

Transpose of a tensor (matrix): The transposed matrix is created by reflection over the main diagonal or with other words: by writing raws as columns and vice versa.

r r

r rs s rs

a a a a

a a a a

=

T11 1 11 1

1 1

2.17

e.g.: Tij jia a=

111 12 13 11 21 31

21 22 23 12 22 32

31 32 33 11 23 33

a a a a a aa a a a a aa a a a a a

− =

2.18

Inverse of a tensor (matrix): The product of a matrix and the corresponding invertible matrix results is the unit matrix (all diagonal elements = 1). e.g.: δ−⋅ =1

ij ij ija a and ( ) 11ij ija a

−− = 2.19 Addition and Substraction of a tensor (matrix): Only tensors of same rank can be added or subtracted. Sum or difference of two ten-sors of same rank is also a tensor of the same rank. e.g.: 1... 1... 1...i in i in i ina b s+ = 2.20

Tensor product (cross product: b x c):

i ijk j ka b cε= ⋅ ⋅ 2.21 Tensor product (dot product: b ∙ c): The tensor product is obtained by simply multiplying components of two tensors togehther, pair by pair, so that the result of the product of a tensor with rank n with a tensor of rank m is a tensor of rank m+n.

1.... 1... 1... 1... 1...m n m ni i j j j i i j ja b c= ⋅ 2.22 e.g.: i jk ijka b c⋅ = 2.23


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Determinant of a tensor (matrix):

1 2 3 11 22 33 21 32 13 31 12 23 11 32 23 12 21 33 13 22 31ij ijt i j ta a a a a a a a a a a a a a a a a a a a a aε= = + + − − − 2.24 Einstein’s summation convention: Summation over equal indices is peformed: e.g.: 11 22 33iia a a a= + + 2.25 Replacement rule: Change of indices (e.g. from k to i): e.g.: i ik ka aδ= 2.26 Contraction: Contraction occurs either when a pair of literal indices of the tensor are set equal to each other and summed over or if during the multiplication of two tensors of order n ≥ 2 one index of the left factor is equal to the right factor. In both cases the rank of the final tensor is reduced by two.

e.g.: ij j i

ijk jq ikq

a b c

a b c

=

= or iik ka b= or ij ijk Ka bδ = 2.27

Derivative (comma convention): The derivative with respect to another physical or geometrical quantity (coordinate, time etc.) is indicated by a comma:

e.g.: ,i

i jj

uux

∂=

∂ or

2

, 2i ttd xxdt

= 2.28

2.5 Tensor analysis: simple examples The following equations document the tensor handling with index notation.

11 22 33iia a a a= + +

ij ijkl kleσ ε= ⋅

i j ij i ja b a b cδ⋅ ⋅ = ⋅ =

i ij ja b c= ⋅


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ij ik kja b c= ⋅

Tij jia a=

TT

ij ija a =

ij ij ij ija b b a c⋅ = ⋅ =

1 1ij ij ij ija a a a I− −⋅ = ⋅ =

ij ji iia a bδ⋅ = =

ij kj ika a δ⋅ =

1 1 2 2 3 3ik ki k k k k k ka b a b a b a b⋅ = ⋅ + ⋅ + ⋅

1 1 2 2 3 3i ia b a b a b a b⋅ = ⋅ + ⋅ + ⋅

ijk i j ka b u v w= ⋅ ⋅ ⋅

ik ij kl jlc b b a= ⋅ ⋅

ijk jl ikla b c⋅ =

i j ija b c⋅ =

i i ic b a= −

ij ijk iik ka b cδ ⋅ = =

ik jk kl lm mib a a a a a= ⋅ ⋅ ⋅ ⋅

,i

i jj

aax

∂=

∂

2

ij i js dx dxδ∂ = ⋅ ⋅

6ijk ijkε ε⋅ =

2 2 1 3 3 1jk j ka b c b c b cε= ⋅ ⋅ = ⋅ − ⋅


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3. Stress tensor Load is generated by outer forces FA (area force) or inner force FI (volume forces) ac-cording to Fig. 3.1. For an arbitrary orientated cut a stress vector t is obtained, assuming that only forces and no moments are transferred. A denotes the area, where the force vector is considered.

0limA

FtA∆ →

= ∆ 3.1

The stress state can be defined in a cartesian coordinate system as illustrated in Fig. 3.2. Along the three faces of the cube three stress vectors t1, t2 and t3 can be obtained, whereby { }i1 i2 i3, ,σ σ σ represent the three stress components on the particular cube faces (Fig. 3.2). In detail the stress tensor can be described as follows:

[ ]11 12 13

1 2 3 21 22 23

31 32 33

, ,xx xy xz

Tij yx yy yz

zx zy zz

t t t = = =

σ σ σ σ τ τσ σ σ σ τ σ τ

σ σ σ τ τ σ 3.2

Fig. 3.1:Solid body with volume and area forces


page 11 of 32

Fig. 3.2: 3-dimentional stress components at a cube

The first index specifies the normal of the particular face under consideration, the second index the impact direction of the stress component. According to eq. 3.2 the stress tensor consists of 9 elements. However, assumed that the sum of the moments is zero, pairwise identical shear stresses are obtained. This feature is also called ‘Boltzmann-Axiom’ and explained in more detail in Fig. 3.3 for the 2-dimensional case (the extension to 3D is straightforward) and by eq. 3.3.

2 2

2 2

2 2

0 4 4

0 4 4

0 4 4

= = ⋅ ∆ ⋅ ∆ − ⋅ ∆ ⋅ ∆ ⇒ =

= = ⋅ ∆ ⋅ ∆ − ⋅ ∆ ⋅ ∆ ⇒ =

= = ⋅ ∆ ⋅ ∆ − ⋅ ∆ ⋅ ∆ ⇒ =

∑∑∑

xy xy yx xy yx

xz xz zx xz zx

yz yz zy yz zy

M l l l l

M l l l l

M l l l l

τ τ τ τ

τ τ τ τ

τ τ τ τ

3.3

From eq. 3.3 it follows, that the stress tensor is symmetric, that means:

ij ji=σ σ or T

σ = σ 3.4

Therefore, the number of stress values is reduced from 9 to 6 (three pairwise identical shear stresses meaning no rotations). The relationship between stress vector and stress tensor is obtained on the basis of the equilibrium conditions in direction of the coordinates xi (Fig. 3.4):

( )cos ,i in n x= 3.5

d di iA n A= 3.6

where ni is the unit normal vector.


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σ

τ

yy

σxx

σyy

σxx

yx

τxy

τxy

τyx

∆l

Fig. 3.3: Equilibrium considerations for a volume element (2D, x-y-plane)

Fig. 3.4: Orientation of stress tensor and stress vector

Force equilibrium in 1-, 2- and 3-direction:

1 11 1 21 2 31 3

2 12 1 22 2 32 3

3 13 1 23 2 33 3

t dA dA dA dAt dA dA dA dAt dA dA dA dA

σ σ σσ σ σσ σ σ

= + +

= + +

= + +

3.7

Using (3.5) and (3.6) eq. 3.7 can be simplified as follows:

1 11 1 21 2 31 3

2 12 1 22 2 32 3

3 13 1 23 2 33 3

t n n nt n n nt n n n

= σ + σ + σ

= σ + σ + σ

= σ + σ + σ. 3.8


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Equation 3.8 can be rewritten in tensor form as follows:

i ji j ij j

T

t n n

n n

= σ = σ

= σ = σ. 3.9

Equation 3.9 documents the equality of pairwise shear stresses. The so defined second-order stress tensor is called ‘Cauchy stress tensor’ or ‘true’ stress tensor or ‘Euler stress tensor’. The Cauchy stress tensor σij relates the current force vector to the current (de-formed) area element.

i ji jdF dA= σ 3.10 Fi: current force vector Aj: current area element with d dj jA n A= Alternatively, the current force vector Fi can be related to the original area A° (that means before any deformation!). Such a stress tensor is called ‚Nominal stress tensor‘, ‘La-grange stress tensor’ or ‘First Piola-Kirchhoff tensor’ Tij: d di ji jF T A= 3.11

The stress tensor can be decomposed into normal and shear components (n: normal vector; m: tangential vector) as illustrated by Fig. 3.5:

i i i ij jn t n nσ σ= = ⋅ 3.12 or

i i i ij jm t m nτ = = σ 3.13

In detail, equations 3.12 and 3.13 can also be written as:

1 11 1 1 12 2 1 13 3

2 21 1 2 22 2 2 23 3

3 31 1 3 32 2 3 33 3

n n n n n nn n n n n nn n n n n n

σ σ σ σσ σ σσ σ σ

= + +

+ + +

+ + +

3.14

From equation 3.14 the following instances can be deduced:

100

n =

→ 11n =σ σ and


page 14 of 32

001

n =

→ 33nσ σ=

For the shear stress follows:

1 11 1 1 12 2 1 13 3

2 21 1 2 22 2 2 23 3

3 31 1 3 32 2 3 33 3

m n m n m nm n m n m nm n m n m n

τ σ σ σσ σ σσ σ σ

= + +

+ + +

+ + +

3.15

From equation 3.15 the following instances can be deduced:

100

n =

; 010

m =

→ 21n =τ σ

001

n =

; 010

m =

→ 23nτ σ=

n

m

t

.σ

τ

n

m

Fig. 3.5: Decomposition of stress vector t into normal and shear stress component

If n i ji jm nτ σ= , then: 100

n =

; 010

m =

→ 12nτ σ= .

Thereby, it always holds: ni ni = 1 and mi mi = 1 Now we consider specific directions, where only normal stresses σ exist, but no shear stress τ. For such a constellation it holds:


page 15 of 32

and = σ ⋅ = σ ⋅ δ ⋅i ij i i ij jt n t n , 3.16

where nj characterizes the principal stress directions. Equalization of both expressions in eq. 3.16 yields: ( ) or 0ij j ij j ij ij jn n nσ ⋅ = σ ⋅ δ ⋅ σ − δ ⋅ σ = 3.17

Equation 3.17 describes an eigenvalue problem with eigenvalues σ und nj. The non-trivial solution is obtained if the coefficient determinant of eq. 3.18 vanishes:

( )det 0ij ijσ σδ− = 3.18

or 11 12 13

12 22 23

13 23 33

0σ σ σ σ

σ σ σ σσ σ σ σ

−− =

− 3.19

The solution of equation 3.19 is a characteristic equation of third order:

3 21 2 3 0I I Iσ σ σ− + − = , 3.20

where the following holds:

1 11 22 33KK ij ijI σ σ σ σ σ δ= = + + = , 3.21

( ) 11 13 22 2311 122

31 33 32 3321 22

2 2 211 22 22 33 11 33 12 23 31

12 ii jj ij jiI

σ σ σ σσ σσ σ σ σ

σ σ σ σσ σ

σ σ σ σ σ σ τ τ τ

= − = + +

= + + − − −

, 3.22

( )3

2 2 211 22 33 11 23 22 13 33 12 12 23 31

1 1 3det3 2 2

2

ij ii jj KK ij jK Ki ij ji KKI σ σ σ σ σ σ σ σ σ σ

σ σ σ σ τ σ τ σ τ τ τ τ

= = + −

= − − − +

. 3.23

The values I1, I2, I3 are called ‘main invariants’ (I1: first main invariant, I2: second main invariant, I3: third main invariant) of the stress tensor, that means that they are independ-ent of the coordinate systems (independent of translations or rotations of the reference system). Besides these main invariants there are the so called ‘basic invariants’, which can be considered as a special subset of the main invariants. They are defined as follows:


page 16 of 32

1 1

22 1 2

33 1 1 2 3

1 12 21 13 3

kk

ij ji

ij jk ki

J I

J I I

J I I I I

= σ =

= σ σ = −

= σ σ σ = − +

3.24

Besides the cartesian representation it is also possible to find a formulation in form of the principal stresses:

1 1 2 3

2 1 2 2 3 1 3

3 1 2 3

III

= + +

= + +

=

σ σ σσ σ σ σ σ σσ σ σ

3.25

An interesting decomposition of the stress tensor is possible, if a mean normal stress is defined as follows:

( )0 11 22 331 13 3KKσ σ σ σ σ= = + + 3.26

σ0 is also called ‘hydrostatic stress state’ or ‘mean stress’ or ‘spherical stress’. Based on these definitions the stress tensor can be written as:

0ij ij ijsσ σ δ= + 3.27

In terms of matrix notation this means:

σ σ σ σ σ σ σ σσ σ σ σ σ σ σ σσ σ σ σ σ σ σ σ

σσ

σ

− = + − −

= +

11 12 13 0 11 0 12 13

21 22 23 0 21 22 0 23

31 32 33 0 31 32 33 0

0 11 12 13

0 21 22 23

0 31 32 33

0 00 00 0

0 00 00 0

s s ss s ss s s

3.28

where sij is referred as deviatoric stress part. For the spherical tensor as well as for the stress deviator invariants can be defined. The main invariants for the spherical tensor are given as follows:

I I I= = =2 31 0 2 0 3 0

332

σ σ σ 3.29

The corresponding basic invariants are:

2 31 0 2 0 3 0

332

J J Jσ σ σ= = = 3.30

For the deviatoric part the main invariants are:


page 17 of 32

( ) ( ) ( )

( )( )( ) ( )( ) ( )( )

( )

1 11 0 22 0 33 0

2

2 2 211 0 22 0 22 0 33 0 11 0 33 0 12 23 31

3

0

12

det

1 1 33 2 2

Dkk

Dii jj ij ji

Dij

ii jj kk ij jk ki ij ji kk

I s

I s s s s

I s

s s s s s s s s s

= = − + − + − =

= −

= − − + − − + − − − − −

=

= + −

σ σ σ σ σ σ

σ σ σ σ σ σ σ σ σ σ σ σ σ σ σ 3.31

The basic invariants for the deviatoric part are:

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

1

2 2 2 2 2 22 11 0 22 0 33 0 12 23 31

2 2 2 2 2 211 22 22 33 33 11 12 23 31

2 2 21 2 2 3 3 1

3 1 0 2 0 3 0

0

1 1 2 2 22 2

1616

13

Dkk

Dij ji

Dij jk ki

J s

J s s

J s s s

= =

= = σ − σ + σ − σ + σ − σ + σ + σ + σ

= σ − σ + σ − σ + σ − σ + σ + σ + σ

= σ − σ + σ − σ + σ − σ

= = σ − σ ⋅ σ − σ ⋅ σ − σ

3.32

Quite often stress components are defined, which are related to the octahedral plane. The octahedral plane is equally inclined to the principal stress directions (hydrostatic axis). The principal stresses act along the x1, x2 and x3 direction:

1

2

3

0 00 00 0

ij

σ σ = σ σ

3.33

The stress vector tj is defined by the three principal stress components σ1, σ2 and σ3. Regarding the normal on the octahedral plane the stress vector tj has the following carte-sian components:

Ni ij jt n= σ 1

3jn = 3.34


page 18 of 32

σα

x1

x2

x3

2

σ3

σ1

tj

nj

αα

σ1

σ2

σ3

1arc cos 54,73

α = ≈ °

[ ]1 2 3, ,= σ σ σjt

Fig. 3.6: Representation of octahedral stresses

The projection and summation of the components on the vektor nj (hydrostatic axis) pro-vides the octahedral normal stress:

( )31 21 2 3 0

1 133 3 3 3OCT

σσ σ σ = + + = σ + σ + σ = σ

3.35

The octahedral normal stress is equivalent to mean stress (Eq. 3.26). The subtraction of the octahedral normal stresses from the principal stresses leads to the deviatoric stresses:

1 1 0

2 2 0

3 3 0

sss

= σ − σ

= σ − σ

= σ − σ

3.36

These deviatoric stresses can also be referred to the octahedral plane and given as Car-tesian components:

31 21 2 33 3 3s s s ss st t t= = = 3.37

The addition of vectors leads to the octahedral shear stresses:


page 19 of 32

( ) ( ) ( )

( )

2 2 2

1 2 3

22 231 2

2 2 21 2 3

2

3 3 3132 13 3

OCT

Dij ij

t t t

ss s

s s s

J s s

= + +

= + +

= + +

= =

τ

3.38

Another very popular quantity is the so-called ‚von-Mises equivalent stress‘ σF. This stress value is based on a strength criterion, which relates the yield stress σF to the stress devi-ator:

220 3 D

FJ= − σ 3.39

This implies that:

( ) ( ) ( )2 2 22 1 2 2 3 1 3

3 132 2

DF ij ijJ s sσ σ σ σ σ σ σ= = = − + − + − 3.40

and

22 23 3OCT F F= =τ σ σ 3.41

Principal stresses and principal stress directions: The stress tensor as a symmetric linear operator has the characteristic, that it can be diagonalised. That means, there are three orientations (directions) perpendicular to each other in space, where the corresponding normal stresses reach extreme values (principal stresses or principal normal stresses) and the shear stresses vanish. In this case, only the trace of the tensors has non-vanishing values:

1

2

3

0 00 00 0

ij

σ σ = σ σ

3.42

The stress vectors on these specific surface areas coincide with the directions of the normal vectors of these surface areas. Therefore, the stress vectors have only one non-vanishing component. Thus, for the stress vector at the considered surface area it holds:

i j ijt n σ=


page 20 of 32

and

1 1 1 1

2 2 2 2

3 3 3 3

t n lt n mt n n

= σ = σ

= σ = σ

= σ = σ

3.44

The normal vector { }, ,in l m n= describes the principal normal stress directions. For the unit vector the following holds in general:

32 2 2 2

11i

in l m n

=

= + + =∑ 3.45

squaring equation 3.44 yields:

2 2 21 12 2 22 22 2 23 3

t l

t m

t n

=

=

=

σ

σ

σ

3.46

and

22 1

212

2 2222

2 323

tl

tm

tn

=σ

=σ

=σ

3.47

The addition of the eq. 3.47 under consideration of eq. 3.45 gives:

tt t

+ + =22 231 2

2 2 21 2 3

1σ σ σ

3.48

Eq. 3.48 describes an ellipsoid, that means the values σ1, σ2 and σ3 represent the half-axes of the ellipsoid (Fig. 3.7). The surface of the ellipsoid represents all possible stress vectors. If two principal stresses are equal, a spheroid is coming up. If all principal stresses are equal (isotropic stress state) a sphere is coming up. In geomechanics, especially in soil mechanics, descriptions on the basis of the deviatoric stress plane, see Fig. 3.8, are very common.


page 21 of 32

Fig. 3.7: Prinzipal stress ellipsoid

Fig. 3.8: Decomposition of the stress state into hydrostatic and deviatoric part, where the stress vector t

defines the stress point T

σ

1

3

2

σ

σ

t

h

s σ = σ = σHydrostatische Achse1 2 3

σ + σ + σ = const.Deviatorebene

1 2 3

T (σ , σ , σ )1 2 3

=

33arccos


page 22 of 32

Fig. 3.9: Illustration of Lode angle θ in the π-plane

( )1 2 3 1

2 2 21 2 3 2

3 33 3

2 D

h I

s s s s J

= σ + σ + σ =

= + + = 3.49

On the deviatoric plane it holds: ( )1 2 3 const.+ + =σ σ σ 3.50

The deviatoric plane through the coordinate system is also called π-plane (Fig. 3.9). It holds:

( )( )

332

2

332

2

3 3cos 3 and2

1 3 3arccos3 2 ( )

D

D

D

D

J

J

J

J

=

=

θ

θ

3.51

In geotechnical engineering the follwoing two modified invariants are often used: Roscoe invariants p und q as well as Lode angle θ. Thereby, it holds:

1

2

332

2

1 ,3

3 and

1 3 3arccos .3 2 ( )

D

D

D

p

q J

J

J

= Ι

=

θ =

3.52

σ'1

σ'2

σ'3

θT


page 23 of 32

For the conventional triaxial test the following expressions can be deduced:

1 3

1 3

21 3 1 2 3

1 ( 2 ) ,3

and1 arccos (3 6 ) 3 6 .3

p

q

s s s s s

= +

= −

= ⋅ =

σ σ

σ σ

θ

3.53

Fig. 3.10: Illustration of Lode angle in the principal stress space


page 24 of 32

4. Deformation tensor Deformations in terms of strain (length change and angle change) can be defined in quite different ways. This is illustrated for a 1-dimensionaler beam under elongation, where l = final length and l0 = initial length.

0

0

l ll−

ε = engineering (technical) formulation (Lagrange)

0l ll−

ε = engineering (technical) formulation (Euler) 2 2

020

l l12 l

−ε = quadratic formulation (Lagrange)

2 20

2

l l12 l

−ε = quadratic formulation (Euler)

0

llnl

ε = logarithmic formulation

All the above mentioned definitions have the following common characteristics: value of 0, if l = l0. for small deformations (small strain) all above given definitions deliver nearly the

same value. for large deformations (large strain), the above given definitions result in signifi-

cant different values.

Proof of approximate equality Deformationen for small strain:

(a) for quadratic approach:

( )( ) ( )( )2 20 0 0 00 0

2 2 20 0 0 0

l l l l 2l l ll l l l1 1 12 l 2 l 2 l l

+ − −− −ε = = = =

(b) for logarithmic approach:

Taylor-series: ( ) ( )nn 1

n 1

x 1ln(x) 1

n+∞

=

−= −∑

Based on series expansion (Taylor series) the logarithmic approach yields:

2 3 4

0 0 0 0 0

l l 1 l 1 l 1 lln 1 1 1 1 ...l l 2 l 3 l 4 l

= − − − + − − − +

2 3 4

0 0 0 0 0

0 0 0 0 0 0

l l l l l l l l l ll 1 1 1ln ...l l 2 l 3 l 4 l l

− − − − −= − + − + ≈


page 25 of 32

Example: stretching by 50%: engineering procedure: ε = 0.5 quadratic procedure: ε = 0.277 logarithmic procedure: ε = 0.405 stretching by 1%: engineering procedure: ε = 0.01000 quadratic procedure: ε = 0.00985 logarithmic procedure: ε = 0.00995 For the coordinates of a point at the initial and final deformed state the following inverse

relations exist: o

ji ix x x =

and o o

i i jx x x =

.

The definition of the deformation tensor can be made in two systems:

1. In relation to the undeformed initial system (= Lagrange approach), that means ui is a function of the initial coordinates

i i ju u x =

4.1

2. In relation to the deformed final system

(= Euler approach), that means ui is a function of the final coordinates. ~

i i ju u x =

4.2

x

x

x

2

1

3

x2

x3

x1

uiP

P

„Lagrange“


page 26 of 32

x

x

x

2

1

3

x2

x3x1

uiP

P

„Euler“

Fig. 4.1: Euler and Lagrange approaches in respect to deformations

The general definition of the deformation tensor (quadratic approach) reads as follows: L

K Kij

i j

x x

x x

∂ ∂=

∂ ∂

ε (Lagrange) 4.3

and

j

K

i

KE

ij xx

xx

∂∂

∂∂

=ε

(Euler) 4.4

With the help of the gradient tensors (= displacement gradients) i

j

u

x

∂

∂

and i

j

ux

∂∂

, respec-

tively, the deformation tensor can be defined as follows: „Lagrange“:

with andi ii i i i ij

jj

Li i

jK ij ij

j K

jK i ijK

j K j K

x ux x u xx x

u u

x xuu u u

x x x x

∂ ∂ = + = + ∂ ∂

∂ ∂ = + + ∂ ∂

∂∂ ∂ ∂= + + +

∂ ∂ ∂ ∂

δ

ε δ δ

δ

4.5


page 27 of 32

„Euler“:

( ) with andi ii i j ij

j j

Ei K i i

jK jKK j j K

uxx x u xx x

u u u ux x x u

∂∂= − = −

∂ ∂

∂ ∂ ∂ ∂= − − +

∂ ∂ ∂ ∂

δ

ε δ

4.6

For the Lagrangian approach the grid follows the deformations. For the Euler approach the material ‘flows’ through the stiff grid. Besides the displacement gradient and the deformation tensor, the deformation gradient Fij is of vitial importance:

L iij ij

j

xF Fx

∂= =

∂

or ( 1)jEij ij

i

xF F

x−∂

= =∂

4.7

The deformation gradient is a second-rank tensor. He projects the line element vector d is (initial configuration) to line element vector ds

(current configuration). Thereby, the same material points are considered (Fig. 4.3). The illustration of the fundamental distinc-tion between Euler and Lagrange approaches using numerical meshing is shown in Fig. 4.2.


page 28 of 32

a) Lagrange Same nodes, but different ‘geographic’ coordinates

A (2, 2)

B (2, 4)

B (2, 4)A (2, 2)

Original Deformed

b) Euler new nodes, but old ‘geographic’ coordinates

A (2, 2)

B (2, 4)

B (2, 2)A (2, 1)

Original Deformed

Fig. 4.1: Langange vs. Euler scheme

y

x

Bahnlinien d s

d s°

Fig. 4.3: Illustration of deformation gradient


page 29 of 32

It holds:

( 1)

d d and

d di ij j

i ij j

s F s

s F s−

= ⋅

= ⋅

4.8

From the engineering point of view the deformation gradient can be defined according to eq. 4.5 as:

12

12

G L

jk j K jK

j K i i

K j j K

u u u u

x x x x

ε ε δ = − ∂ ∂ ∂ ∂ = + + ∂ ∂ ∂ ∂

4.9

or according to eq. 4.6 as:

12

12

A E

jK jK jK

j K i i

K j j K

u u u uu x x x

ε δ ε = − ∂ ∂ ∂ ∂

= + − ∂ ∂ ∂ ∂

4.10

Expression 4.9 is called ‘Green deformation tensor’, the expression 4.10 is called ‘Al-mansi deformation tensor’. In the engineering praxis the Green deformation tensor is pre-ferred. Moreover, most often the quadratic term is neglected under the assumption, that

1i

j

u

x

∂

∂

. Thus, for small deformation, the distinction between Langrangian and Eulerian

approaches disappears and the simplified deformation tensor is given as:

12

jiij

j i

uu

x xε

∂∂ = + ∂ ∂

4.11

The deformation tensor according to equation 4.11 can be extended to include rotations:

( ) ( ), , , ,1 12 2ij i j j i j i i j

ij ij

Deformations Rotations

u u u u

e w

ε = + + −

= +

4.12


page 30 of 32

Fig.4.4: Illustration of rotation and deformation (2D)

It holds:

12 13 12 21

21 23 13 31

31 32 23 32

11 12 13 12 21

21 22 23 13 31

31 32 33 23 32

00 with

0and

with

ij

ij

w w w ww w w w w

w w w w

e e e e ee e e e e e

e e e e e

= − = = − = −

= = = =

4.13

Thus, the deformation tensor can be written as:

11 12 12 13 13

21 21 22 23 23

31 31 32 32 33

ij

e e w e we w e e we w e w e

ε+ +

= + + + +

4.14

with

( )12ij ij jie ε ε= + and ( )1 for

2ij ji ijw i jε ε= − ≠ .

eij is called deformation tensor, wij is called rotation tensor. It holds:

ij ije i j1 for 2

= ≠κ 4.15

Where κij are shear strain components and e11, e22 and e33 are direct strain components (elongations or shortenings).


page 31 of 32

The volumetric strain εv is given by the following expression:

11 22 33d

dv KKV

V∆

ε = = ε = ε + ε + ε 4.16

The mean direct strain (elongation or shortening) 0ε is given by:

01 13 3KK v= =ε ε ε 4.17

In most cases rotations are neglected and it holds:

11 12 13

21 22 23

31 32 33

ij

e e ee e ee e e

ε =

with 12 21

23 32

13 31

e ee ee e

===

4.18

In complete analogy to the stress tensor invariants can be defined also for the deformation tensor, e.g.:

= + +1 11 22 33I e e e ,

2 11 22 22 33 11 33I e e e e e e= + + and 4.19

3 11 22 33I e e e= .

5. Compatibility condition From expression 5.1 the strain components can be obtained in a unique manner. Other-wise, the displacements can not be obtained in a unique manner based on given strains only. The compatibility conditions (= conditions of integrability) are necessary additional requirements to deduce displacements on the basis of given strain components by inte-gration. The consideration of the compatibility conditions guarantees that strains lead to a ‘correct’ displacement field and the continuum is not disturbed. Starting point is the deformation tensor:

( ), ,12ij i j j iu uε = + 5.1

Second derivatives of equation 5.1 with corresponding index permutations give the fol-lowing four expressions:


page 32 of 32

( )

( )

( )

( )

, , ,

, , ,

, , ,

, , ,

12121212

ij kl i jkl j ikl

kl ij k lij l kij

ik jl i kjl k ijl

jl ik j lik l jik

u u

u u

u u

u u

ε

ε

ε

ε

= +

= +

= +

= +

5.2

Due to the fact that the sequence of differentation is arbitrary, through addition and sub-traction of the expressions 5.2 the following expression is obtained:

, , , , 0ij kl kl ij ik jl jl ikε ε ε ε+ − − = 5.3

From expression 5.3 the 6 compatibility conditions can be deduced under the condition

ij ji i j for = ≠ε ε as follows:

11, 22 22,11 12,12

22, 33 33, 22 23, 23

33,11 11, 33 13,13

11, 23 23,11 13, 21 12, 31

22, 31 31, 22 21, 32 23,12

33,12 12, 33 32,13 31, 23

2 02 02 0

000

ε ε ε

ε ε ε

ε ε ε

ε ε ε ε

ε ε ε ε

ε ε ε ε

+ − =

+ − =

+ − =

+ − − =

+ − − =

+ − − =

5.4

First equation in 5.4 can exemplary also be written as:

2 22

2 2 2yy xyxx

y x x yε εε ∂ ∂∂

+ =∂ ∂ ∂ ∂

5.5

Under plain strain conditions all strain components and derivations in respect to the third direction in space vanish, that means only equation 5.5 left over. Equation 5.5 indicates, that the second derivations of the strains and the second derivations of the angular dis-tortions have to be in due proportion.


page 33 of 32

6. Equilibrium conditions For any volume element inside a body the forces and moments have to be in equilibrium. Usually it is assumed, that the solid body does not rotate and therefore the sum of the moments is zero by default. According to Fig. 6.1 the following yields:

0 :

d d d d d d d d

d d d d d

d d d d d

x

yxxx x yx

x

zxyx zx

zx x

F

x y z y z y z xy

z x z x yz

x y F x y z

τσσ σ τ

ττ τ

τ

=

∂ ∂= + − + + ∂ ∂

∂ − + + ∂ − +

∑

6.1

0 :

d d d d d d d

d d d d d

d d d d d

y

y zyy y zy

xyxy xy

zy y

F

y x z x z z dx yy z

z y x y zx

x y F x y z

σ τσ σ τ

ττ τ

τ

=

∂ ∂ = + − + + ∂ ∂

∂ − + + ∂ − +

∑

6.2

0 :

d d d d d d d d

d d d d d

d d d d d

z

zyzz z zy

y

xzzy xz

xz z

F

z x y x y y x zz y

x z x y zx

y z F x y z

τσσ σ τ

ττ τ

τ

=

∂ ∂= + − + + ∂ ∂

∂ − + + ∂ − +

∑

6.3


page 34 of 32

Fig. 6.1: Force equilibrium at volume element (Fi: volume forces)

Eq. 6.1 to 6.3 can be simplified in the following way:

0

0

0

yxx zxx

xy y zyy

yzxz zz

Fx y z

Fx y z

Fx y z

∂∂ ∂+ + + =

∂ ∂ ∂∂ ∂ ∂

+ + + =∂ ∂ ∂

∂∂ ∂+ + + =

∂ ∂ ∂

τσ τ

τ σ τ

ττ σ

6.4

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