· x aaaa bbb ccc aaa bbb ccc tensor analysis in cartesian coordinates tensor characteristics...

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tensor analysis

in cartesian coordinates

• literature

• motivation

• definition of tensors

basis vectors order of tensors (zero to four) identity tensors

• products of tensors

product with scalar dyadic product contraction of tensors

− single contraction− double contraction

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors

tensor analysis in cartesian coordinatesÜ 1

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tensor analysis

in cartesian coordinates

• tensor characteristics

symmetric and skew symmetric parts deviatoric and volumetric parts invariantes eigenvalues and eigenvectors coordinate transformation

• differentiation of tensors

derivations gradient & divergence total differential, variation & increment

• GAUSS integral theorem

• calculation rules

tensor analysis in cartesian coordinatesÜ 2

rotation e′′′i = Q3 · ei

e3 = e′′′3

e′′′1

e′′′2

e′′i = Q2 · e′′′i = Q2 ·Q3 · ei

e′′1

e′′2e′′3

e′i= Q1 ·e′′i = Q1 ·Q2 ·Q3 · ei

e′3α

• background: rotation of basis vectore′3 = Q1 ·Q2 · e3 about e1 and e2. coloredrepresentation of third component of e′3 asfunction of rotation angles α ∈ [0, 360o] andβ ∈ [0, 360o]. top: 3d rotation of basis vectorswith γ = 30o, β = −30o and α = 30o

• foreground: illustration of dyadic product oftensors

information about title picture ’tensor analysis’Ü 3

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• general tenor analysis

BASAR&KRÄTZIG (1985)

BASAR&WEICHERT (2000)

BETTEN (1987)

DE BOER (1982), DE BOER (2000) (Appendix B)

FLÜGGE (1972)

IBEN (1995)

KLINGBEIL (1989)

LIPPMANN (1996)

SCHADE (1997)

TROSTEL (1993)

WRIGGERS (2001)

• tensor analysis in cartesian coordinates

HOLZAPFEL (2000): Nonlinear Solid Mechanics. AContinuum Approach for Engineering. John Wiley & Sons

BONET&WOOD (1997): Nonlinear Continuum Mechanics forFinite Element Analysis. Cambridge University Press

tensor analysis - sourcesÜ 4

components matrices tensors

linear strain measureε11 = u1,1

ε22 = u2,2

ε33 = u3,3

ε12 =1

2[u1,2+u2,1]

ε23 =1

2[u2,3+u3,2]

ε13 =1

2[u1,3+u3,1]

ε = Dεuwith

DTε =

0 0∂

ε=∇sym

2[ui,j + uj,i] ei⊗ej

balance of momentum in linear continuum mechanics

ρ u1 = σ11,1 + σ12,2 + σ13,3 + ρ b1

ρ u2 = σ21,1 + σ22,2 + σ23,3 + ρ b2

ρ u3 = σ31,1 + σ32,2 + σ33,3 + ρ b3

ρ u = Dσσ + ρ b

0 0∂

ρ u= divσ + ρ b

= [σij,j + ρ bi] ei

NEUMANN boundary conditions in linear continuum mechanics

σ11 n1 + σ12 n2 + σ13 n3 = t?1

σ12 n1 + σ22 n2 + σ23 n3 = t?2

σ13 n1 + σ23 n2 + σ33 n3 = t?3

Dt σ = t?

n1 0 0 n2 0 n3

0 n2 0 n1 n3 0

0 0 n3 0 n2 n1

σ · n = t?

σij nj ei = t?i ei

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.1 motivation

motivation - tensor format continuum mechanicsÜ 5

first order tensor

-6qa e

6qae ′′1

e ′′2

-6qa e1

3a a′1

a′2 -

6qa e′1

-6qa e

• tensor is independent on coordinate system

a = ai ei = a′i e′i = a′′i e

′′i

• components of tensors depend on coordinate system

ai = a · ei a′i = a · e′i

• tensor components can be transformed betweencoordinate systems

a = a′i e′i = Q · [ai ei]

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.2 definition

definition - properties of tensorsÜ 6

right hand rule

ϕi, Mi

right thumb rulewithout interest in mechanics

e1rb6-

e1rb6-tpde2

6-tde3

• three dimensional - cartesian basis, right-handed (dextral) and orthonormal system

base vectors e1, e2 and e3 are perpendicular basis vectors have length one ‖ei‖ = 1 for i ∈ [1, 3]

generating a right-handed system positive rotations ϕi or moments Mi using right thumb rule

• two dimensional - cartesian basis, right-handed (dextral) and orthonormal system

using e1-e2 plane or e1-e3 plane third base vector defines positive rotations ϕi or moments Mi

cartesian coordinate systemsÜ 7

right hand rule

ϕi, Mi

right thumb rulewithout interest in mechanics

e1rb6-

e1rb6-tpde2

6-tde3

• three dimensional - cartesian basis, right-handed (dextral) and orthonormal system

base vectors e1, e2 and e3 are perpendicular basis vectors have length one ‖ei‖ = 1 for i ∈ [1, 3]

generating a right-handed system positive rotations ϕi or moments Mi using right thumb rule

• two dimensional - cartesian basis, right-handed (dextral) and orthonormal system

using e1-e2 plane or e1-e3 plane third base vector defines positive rotations ϕi or moments Mi

cartesian coordinate systemsÜ 7

spatialcoordinate

system

• thumb: e1

• forefinger: e2

• middle finger: e3

• spatial coordinate system

base vectors

e1, e2, e3

• planar coordinate systems

e1-e2-plane, base vectors

e1, e2

e2, e3

e1, e3

spatial and planar coordinate systemsÜ 8

spatialcoordinate

system

• thumb: e1

• forefinger: e2

base vectors

e1, e2, e3

e1, e2

e2, e3

e1, e3

spatialcoordinate

system

• thumb: e1

• forefinger: e2

base vectors

e1, e2, e3

e1, e2

e2, e3

e1, e3

spatialcoordinate

system

• thumb: e1

• forefinger: e2

base vectors

e1, e2, e3

e1, e2

e2, e3

e1, e3

cartesian base vectors andright hand rule

cartesian basis

representation of vectors

cartesian basis - right-handed (dextral) and orthonormal system

, e2 =

, e3 =

basis vectors ei, indices i, j ∈ 1, 2, 3

‖ei‖ = 1, ei · ej =

0 for i 6= j

1 for i = j

representation of tensors (first order tensor = vector)

a =3∑i=1

ai ei = a1e1+a2e2+a3e3 = ai ei = aj ej

• EINSTEIN summation conventionaddition over repeated indices• summation indices (twice available, dummy)

can be replaced, e.g. i→ j

tensor components

ai = a · ei

cartesian coordinates - basis vectorsÜ 9

cartesian base vectors andright hand rule

cartesian basis

representation of vectors

, e2 =

, e3 =

basis vectors ei, indices i, j ∈ 1, 2, 3

‖ei‖ = 1, ei · ej =

0 for i 6= j

1 for i = j

representation of tensors (first order tensor = vector)

a =3∑i=1

ai ei = a1e1+a2e2+a3e3 = ai ei = aj ej

• EINSTEIN summation conventionaddition over repeated indices• summation indices (twice available, dummy)

can be replaced, e.g. i→ j

tensor components

ai = a · ei

basis vectors

products aiei

a1e1 =

a2e2 =

a3e3 =

EINSTEIN summation

, e2 =

, e3 =

representation of first order tensors

a =3∑i=1

ai ei = a1e1+a2e2+a3e3 = ai ei

using basis vectors and tensor components

a = a1

basis vectors

products aiei

a1e1 =

a2e2 =

a3e3 =

EINSTEIN summation

, e2 =

, e3 =

a =3∑i=1

a = a1

basis vectors

products aiei

a1e1 =

a2e2 =

a3e3 =

EINSTEIN summation

, e2 =

, e3 =

a =3∑i=1

a = a1

illustration tensor order

3. order

2. order

1. order

0. order

, second and fourth

order tensors

= ai ei

A11 A12 A13

A21 A22 A23

A31 A32 A33

= Aij ei ⊗ ej

A = = Aijkl ei ⊗ ej ⊗ ek ⊗ el

e.g. displacement-, stress- and constitutive tensors

= ui ei

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

= σij ei ⊗ ej

C = = Cijkl ei ⊗ ej ⊗ ek ⊗ el

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.3 characterization

3. order

2. order

1. order

0. order

first, second

and fourth

order tensors

= ai ei

A11 A12 A13

A21 A22 A23

A31 A32 A33

= Aij ei ⊗ ej

= ui ei

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

= σij ei ⊗ ej

3. order

2. order

1. order

0. order

first, second and fourth order tensors

= ai ei

A11 A12 A13

A21 A22 A23

A31 A32 A33

= Aij ei ⊗ ej

= ui ei

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

= σij ei ⊗ ej

3. order

2. order

1. order

0. order

zero,first, second and fourth order tensors

= ai ei

A11 A12 A13

A21 A22 A23

A31 A32 A33

= Aij ei ⊗ ej

= ui ei

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

= σij ei ⊗ ej

3. order

2. order

1. order

0. order

zero,first, second and fourth order tensors

= ai ei

A11 A12 A13

A21 A22 A23

A31 A32 A33

= Aij ei ⊗ ej

= ui ei

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

= σij ei ⊗ ej

c = a+ b

c = a+ b+ d

addition of tensors• only tensors of same order can be added• first order tensors

c = a+ b = ai ei + bi ei = [ai + bi] ei =

a1 + b1

a2 + b2

a3 + b3

• second order tensors

A+B = [Aij +Bij ] ei ⊗ ej

A11 +B11 A12 +B12 A13 +B13

A21 +B21 A22 +B22 A23 +B23

A31 +B31 A32 +B32 A33 +B33

• calculation rules associative law

A+ [B +C] = [A+B] +C

commutative law

A+B = B +A

identical and inverse elements

A+ 0 = A, A+ [−A] = 0

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.4 addition

addition of tensorsÜ 12

KRONECKER symbol

dyadic product

1⊗10

• KRONECKER symbol

δij = ei · ej =

0 for i 6= j

1 for i = j

• identity tensors

second order identity tensor 1

fourth order identity tensor I′

dyadic product 1⊗ 1

symmetric fourth order identity tensor I

1 = δij ei ⊗ ej

I′ = δikδjl ei ⊗ ej ⊗ ek ⊗ el

1⊗ 1 = δij δkl ei ⊗ ej ⊗ ek ⊗ el

2[δikδjl + δilδjk] ei ⊗ ej ⊗ ek ⊗ el

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.5 fundamental tensors

identity tensors 1,1⊗1,I ′ and IÜ 13

KRONECKER symbol

dyadic product

1⊗10

• KRONECKER symbol

δij = ei · ej =

0 for i 6= j

1 for i = j

• identity tensors

second order identity tensor 1

fourth order identity tensor I′

dyadic product 1⊗ 1

symmetric fourth order identity tensor I

1 = δij ei ⊗ ej

I′ = δikδjl ei ⊗ ej ⊗ ek ⊗ el

identity tensors 1,1⊗1,I ′ and IÜ 13

repetition: KRONECKER symbol & fourth order identity tensors

δij = ei · ej =

0 for i 6= j

1 for i = j

1⊗ 1 = δij δkl ei ⊗ ej ⊗ ek ⊗ elI =

fourth order constitutive tensor C

C = 2µ I + λ 1⊗ 1 =[µ [δilδjk + δikδjl] + λ δijδkl

]ei ⊗ ej ⊗ ek ⊗ el

= Cijkl ei ⊗ ej ⊗ ek ⊗ el

developement of components C1111, C1122, C1112 and C1212

C1111 = µ [δ11δ11 + δ11δ11] + λ δ11δ11

= µ [1 · 1 + 1 · 1] + λ 1 · 1 = 2 µ+ λ

C1122 = µ [δ12δ12 + δ12δ12] + λ δ11δ22

= µ [0 · 0 + 0 · 0] + λ 1 · 1 = λ

C1112 = µ [δ12δ11 + δ11δ12] + λ δ11δ12

= µ [0 · 1 + 1 · 0] + λ 1 · 0 = 0

C1212 = µ [δ12δ21 + δ11δ22] + λ δ12δ12

= µ [0 · 0 + 1 · 1] + λ 0 · 0 = µ

identity tensors (examples)Ü 14

δij = ei · ej =

0 for i 6= j

1 for i = j

C1111 = µ [δ11δ11 + δ11δ11] + λ δ11δ11

= µ [1 · 1 + 1 · 1] + λ 1 · 1 = 2 µ+ λ

C1122 = µ [δ12δ12 + δ12δ12] + λ δ11δ22

= µ [0 · 0 + 0 · 0] + λ 1 · 1 = λ

C1112 = µ [δ12δ11 + δ11δ12] + λ δ11δ12

= µ [0 · 1 + 1 · 0] + λ 1 · 0 = 0

C1212 = µ [δ12δ21 + δ11δ22] + λ δ12δ12

= µ [0 · 0 + 1 · 1] + λ 0 · 0 = µ

δij = ei · ej =

0 for i 6= j

1 for i = j

C1111 = µ [δ11δ11 + δ11δ11] + λ δ11δ11

= µ [1 · 1 + 1 · 1] + λ 1 · 1 = 2 µ+ λ

C1122 = µ [δ12δ12 + δ12δ12] + λ δ11δ22

= µ [0 · 0 + 0 · 0] + λ 1 · 1 = λ

C1112 = µ [δ12δ11 + δ11δ12] + λ δ11δ12

= µ [0 · 1 + 1 · 0] + λ 1 · 0 = 0

C1212 = µ [δ12δ21 + δ11δ22] + λ δ12δ12

= µ [0 · 0 + 1 · 1] + λ 0 · 0 = µ

δij = ei · ej =

0 for i 6= j

1 for i = j

C1111 = µ [δ11δ11 + δ11δ11] + λ δ11δ11

= µ [1 · 1 + 1 · 1] + λ 1 · 1 = 2 µ+ λ

C1122 = µ [δ12δ12 + δ12δ12] + λ δ11δ22

= µ [0 · 0 + 0 · 0] + λ 1 · 1 = λ

C1112 = µ [δ12δ11 + δ11δ12] + λ δ11δ12

= µ [0 · 1 + 1 · 0] + λ 1 · 0 = 0

C1212 = µ [δ12δ21 + δ11δ22] + λ δ12δ12

= µ [0 · 0 + 1 · 1] + λ 0 · 0 = µ

δij = ei · ej =

0 for i 6= j

1 for i = j

C1111 = µ [δ11δ11 + δ11δ11] + λ δ11δ11

= µ [1 · 1 + 1 · 1] + λ 1 · 1 = 2 µ+ λ

C1122 = µ [δ12δ12 + δ12δ12] + λ δ11δ22

= µ [0 · 0 + 0 · 0] + λ 1 · 1 = λ

C1112 = µ [δ12δ11 + δ11δ12] + λ δ11δ12

= µ [0 · 1 + 1 · 0] + λ 1 · 0 = 0

C1212 = µ [δ12δ21 + δ11δ22] + λ δ12δ12

= µ [0 · 0 + 1 · 1] + λ 0 · 0 = µ

δij = ei · ej =

0 for i 6= j

1 for i = j

C1111 = µ [δ11δ11 + δ11δ11] + λ δ11δ11

= µ [1 · 1 + 1 · 1] + λ 1 · 1 = 2 µ+ λ

C1122 = µ [δ12δ12 + δ12δ12] + λ δ11δ22

= µ [0 · 0 + 0 · 0] + λ 1 · 1 = λ

C1112 = µ [δ12δ11 + δ11δ12] + λ δ11δ12

= µ [0 · 1 + 1 · 0] + λ 1 · 0 = 0

C1212 = µ [δ12δ21 + δ11δ22] + λ δ12δ12

= µ [0 · 0 + 1 · 1] + λ 0 · 0 = µ

δij = ei · ej =

0 for i 6= j

1 for i = j

C1111 = µ [δ11δ11 + δ11δ11] + λ δ11δ11

= µ [1 · 1 + 1 · 1] + λ 1 · 1 = 2 µ+ λ

C1122 = µ [δ12δ12 + δ12δ12] + λ δ11δ22

= µ [0 · 0 + 0 · 0] + λ 1 · 1 = λ

C1112 = µ [δ12δ11 + δ11δ12] + λ δ11δ12

= µ [0 · 1 + 1 · 0] + λ 1 · 0 = 0

C1212 = µ [δ12δ21 + δ11δ22] + λ δ12δ12

= µ [0 · 0 + 1 · 1] + λ 0 · 0 = µ

δij = ei · ej =

0 for i 6= j

1 for i = j

C1111 = µ [δ11δ11 + δ11δ11] + λ δ11δ11

= µ [1 · 1 + 1 · 1] + λ 1 · 1 = 2 µ+ λ

C1122 = µ [δ12δ12 + δ12δ12] + λ δ11δ22

= µ [0 · 0 + 0 · 0] + λ 1 · 1 = λ

C1112 = µ [δ12δ11 + δ11δ12] + λ δ11δ12

= µ [0 · 1 + 1 · 0] + λ 1 · 0 = 0

C1212 = µ [δ12δ21 + δ11δ22] + λ δ12δ12

= µ [0 · 0 + 1 · 1] + λ 0 · 0 = µ

δij = ei · ej =

0 for i 6= j

1 for i = j

C1111 = µ [δ11δ11 + δ11δ11] + λ δ11δ11

= µ [1 · 1 + 1 · 1] + λ 1 · 1 = 2 µ+ λ

C1122 = µ [δ12δ12 + δ12δ12] + λ δ11δ22

= µ [0 · 0 + 0 · 0] + λ 1 · 1 = λ

C1112 = µ [δ12δ11 + δ11δ12] + λ δ11δ12

= µ [0 · 1 + 1 · 0] + λ 1 · 0 = 0

C1212 = µ [δ12δ21 + δ11δ22] + λ δ12δ12

= µ [0 · 0 + 1 · 1] + λ 0 · 0 = µ

first order tensors

α < 1 < β

characterization of products

• multiplication by a scalar

• dyadic product

• contraction

simple contraction double contraction

multiplication by a scalar

• first order tensors

α a = α ai ei = [α ai] ei =

• second order tensors

α A = α Aij ei ⊗ ej =

αA11 αA12 αA13

αA21 αA22 αA23

αA31 αA32 αA33

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.6 products • e1.6.1 multiplication by scalars

products of tensorsÜ 15

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geometrical interpretationC = a⊗ b

a⊗ b = aibj ei ⊗ ej

C = A⊗ b

A⊗ b = Aijbk ei ⊗ ej ⊗ ek

dyadic products

• increasing order• symbol ⊗two first order tensors

a⊗ b = [ai ei]⊗ [ej bj ] = ai bj ei ⊗ ej

higher order tensors

C = Cijk ei ⊗ ej ⊗ ek =A ⊗ b

= Aij bk ei ⊗ ej ⊗ ekC = Cijkl ei⊗ej⊗ek⊗el =A ⊗ B

= Aij Bkl ei ⊗ ej ⊗ ek ⊗ el

e.g. consitutive tensor

C = 2µ I+ λ 1⊗ 1

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.6 products • e1.6.2 dyadic product

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CCCCCCCCC

geometrical interpretationC = a⊗ b

a⊗ b = aibj ei ⊗ ej

C = A⊗ b

A⊗ b = Aijbk ei ⊗ ej ⊗ ek

dyadic products

• increasing order• symbol ⊗two first order tensors

a⊗ b = [ai ei]⊗ [ej bj ] = ai bj ei ⊗ ej

higher order tensors

C = Cijk ei ⊗ ej ⊗ ek =A ⊗ b

= Aij bk ei ⊗ ej ⊗ ekC = Cijkl ei⊗ej⊗ek⊗el =A ⊗ B

= Aij Bkl ei ⊗ ej ⊗ ek ⊗ el

e.g. consitutive tensor

C = 2µ I+ λ 1⊗ 1

A11A11A11A11A11A11A11A11A11

A11e1⊗e1

+ · · · =

A12A12A12A12A12A12A12A12A12

A12e1⊗e2

A11A11A11A11A11A11A11A11A11

A21A21A21A21A21A21A21A21A21

A31A31A31A31A31A31A31A31A31A12A12A12A12A12A12A12A12A12

A22A22A22A22A22A22A22A22A22

A32A32A32A32A32A32A32A32A32

A13A13A13A13A13A13A13A13A13

A23A23A23A23A23A23A23A23A23

A33A33A33A33A33A33A33A33A33

Aijei⊗ej

, e2 =

, e3 =

representation of second order tensors

A =3∑i=1

3∑j=1

Aij ei⊗ej = Aij ei⊗ej

= A11 e1⊗e1 + A12 e1⊗e2 + A13 e1⊗e3

+ A21 e2⊗e1 + A22 e2⊗e2 + A23 e2⊗e3

+ A31 e3⊗e1 + A32 e3⊗e2 + A33 e3⊗e3

A = A11

+ · · ·

A11 A12 A13

A21 A22 A23

A31 A32 A33

= Aij ei⊗ej

dyadic product of base vectorsÜ 17

A111A111A111A111A111A111A111A111A111

A111e1⊗e1⊗e1

0A231A231A231A231A231A231A231A231A231

A231e2⊗e3⊗e1

Aijkei⊗ej⊗ek

+ · · ·+

+ · · · =

, e2 =

, e3 =

representation of third order tensors

A =3∑i=1

3∑j=1

3∑k=1

Aijk ei⊗ej⊗ek

= Aijk ei⊗ej⊗ek= A111 e1⊗e1⊗e1 +A112 e1⊗e1⊗e2 + · · ·

representation of fourth order tensors

A =3∑i=1

3∑j=1

3∑k=1

3∑l=1

Aijkl ei⊗ej⊗ek⊗el

= Aijkl ei⊗ej⊗ek⊗el= A1111 e1⊗e1⊗e1⊗e1

+ A1112 e1⊗e1⊗e1⊗e2 + · · ·

dyadic product of base vectorsÜ 18

e.g. dyadic product of basis tensors ei

e1⊗e1 =

[1 0 0

e2⊗e3 =

[0 0 1

detailed calculation of a dyadic product

a⊗ b = [ai ei]⊗ [ej bj ] = aibj ei⊗ej= a1bj e1⊗ej + a2bj e2⊗ej + a3bj e3⊗ej= a1b1 e1⊗e1 + a1b2 e1⊗e2 + a1b3 e1⊗e3 + a2b1 e2⊗e1 + · · ·+ a3b3 e3⊗e3

= a1b1

+ a1b2

+ a1b3

+ a2b1

+ · · ·+ a3b3

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

C11 C12 C13

C21 C22 C23

C31 C32 C33

= Cij ei⊗ej

Cij = aibj

dyadic product ⊗ (example)Ü 19

e1⊗e1 =

[1 0 0

e2⊗e3 =

[0 0 1

a⊗ b = [ai ei]⊗ [ej bj ] = aibj ei⊗ej

= a1bj e1⊗ej + a2bj e2⊗ej + a3bj e3⊗ej= a1b1 e1⊗e1 + a1b2 e1⊗e2 + a1b3 e1⊗e3 + a2b1 e2⊗e1 + · · ·+ a3b3 e3⊗e3

= a1b1

+ a1b2

+ a1b3

+ a2b1

+ · · ·+ a3b3

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

C11 C12 C13

C21 C22 C23

C31 C32 C33

= Cij ei⊗ej

Cij = aibj

e1⊗e1 =

[1 0 0

e2⊗e3 =

[0 0 1

a⊗ b = [ai ei]⊗ [ej bj ] = aibj ei⊗ej= a1bj e1⊗ej + a2bj e2⊗ej + a3bj e3⊗ej

= a1b1 e1⊗e1 + a1b2 e1⊗e2 + a1b3 e1⊗e3 + a2b1 e2⊗e1 + · · ·+ a3b3 e3⊗e3

= a1b1

+ a1b2

+ a1b3

+ a2b1

+ · · ·+ a3b3

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

C11 C12 C13

C21 C22 C23

C31 C32 C33

= Cij ei⊗ej

Cij = aibj

e1⊗e1 =

[1 0 0

e2⊗e3 =

[0 0 1

= a1b1

+ a1b2

+ a1b3

+ a2b1

+ · · ·+ a3b3

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

C11 C12 C13

C21 C22 C23

C31 C32 C33

= Cij ei⊗ej

Cij = aibj

e1⊗e1 =

[1 0 0

e2⊗e3 =

[0 0 1

= a1b1

+ a1b2

+ a1b3

+ a2b1

+ · · ·+ a3b3

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

C11 C12 C13

C21 C22 C23

C31 C32 C33

= Cij ei⊗ej

Cij = aibj

e1⊗e1 =

[1 0 0

e2⊗e3 =

[0 0 1

= a1b1

+ a1b2

+ a1b3

+ a2b1

+ · · ·+ a3b3

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

C11 C12 C13

C21 C22 C23

C31 C32 C33

= Cij ei⊗ej

Cij = aibj

e1⊗e1 =

[1 0 0

e2⊗e3 =

[0 0 1

= a1b1

+ a1b2

+ a1b3

+ a2b1

+ · · ·+ a3b3

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

C11 C12 C13

C21 C22 C23

C31 C32 C33

= Cij ei⊗ej

Cij = aibj

e1⊗e1 =

[1 0 0

e2⊗e3 =

[0 0 1

= a1b1

+ a1b2

+ a1b3

+ a2b1

+ · · ·+ a3b3

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

C11 C12 C13

C21 C22 C23

C31 C32 C33

= Cij ei⊗ej

Cij = aibj

first order tensors

simple contraction

ϑ ·a · b = |a| |b| cosϑ

·aaaaaaaaa

ccccccccc

bbbbbbbbb

AAAAAAAAA

ccccccccc

bbbbbbbbb

contractive products→ reduction of order

• simple contraction, symbol ·• double contraction, symbol :

simple contraction

a ·b =aiei ·ejbj =ai δijbj =aibi

A ·b =Aijei⊗ej ·ekbk =Aijei δjkbk=Aijbj ei

A ·B=Aijei⊗ej ·ek⊗elBkl= =AijBjl ei⊗el

double contraction

A :B=Aij [ei⊗ej ] : [ek⊗el]Bkl=Aij δikδjlBkl

=Aij Bij

requirements of present course

c = = a · b = ai bi

c = =A :B = Aij Bij

c = ci ei =A · b = Aij bj ei

C = Cij ei ⊗ ej =A ·B = Aik Bkj ei ⊗ ejC = Cij ei ⊗ ej = A :B = Aijkl Bkl ei ⊗ ej

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.6 products • e1.6.3 contraction

contraction of tensors · and :Ü 20

first order tensors

simple contraction

ϑ ·a · b = |a| |b| cosϑ

·aaaaaaaaa

ccccccccc

bbbbbbbbb

AAAAAAAAA

ccccccccc

bbbbbbbbb

simple contraction

double contraction

=Aij Bij

c = = a · b = ai bi

c = =A :B = Aij Bij

first order tensors

simple contraction

ϑ ·a · b = |a| |b| cosϑ

·aaaaaaaaa

ccccccccc

bbbbbbbbb

AAAAAAAAA

ccccccccc

bbbbbbbbb

simple contraction

double contraction

=Aij Bij

c = = a · b = ai bi

c = =A :B = Aij Bij

first order tensors

simple contraction

ϑ ·a · b = |a| |b| cosϑ

·aaaaaaaaa

ccccccccc

bbbbbbbbb

AAAAAAAAA

ccccccccc

bbbbbbbbb

simple contraction

double contraction

=Aij Bij

c = = a · b = ai bi

c = =A :B = Aij Bij

strain vector ε and strain tensor ε

ε =[ε11 ε22 ε33 2ε12 2ε23 2ε13

]T ε =

ε11 ε12 ε13

ε22 ε23

ε33sym

@@@@@R

stress vector σ and stress tensor σ

σ =[σ11 σ22 σ33 σ12 σ23 σ13

]T σ =

σ11 σ12 σ13

σ22 σ23

σ33sym

@@@@@R

specific strain energy

2W = σ : ε = σij εij

= σ1j ε1j + σ2j ε2j + σ3j ε3j

= σ11 ε11 + σ12 ε12 + σ13 ε13 + σ21 ε21 + σ22 ε22 + σ23 ε23

+ σ31 ε31 + σ32 ε32 + σ33 ε33

= σ11 ε11 + σ22 ε22 + σ33 ε33

+ [σ12 ε12 + σ21 ε21] + [σ23 ε23 + σ32 ε32] + [σ13 ε13 + σ31 ε31]

= σ11 ε11 + σ22 ε22 + σ33 ε33 + 2 σ12 ε12 + 2 σ23 ε23 + 2 σ13 ε13

= σ · ε

examples of productsÜ 21

scalar dyadic single doubleproduct product contraction contraction

associative law

α[βA] = [αβ]A αa⊗b = a⊗[αb] αA·b = A·[αb] αA : B = A : [αB]

distributive

[α+β]A = αA+βA

α[A+B] = αA+αB

A⊗[b+c] = A⊗b+A⊗c

[A+B]⊗c = A⊗c+B⊗c

A·[b+c] = A·b+A·c[A+B]·c = A·c+B ·c

A : [B+C] = A : B+A : C

[A+B] : C = A : C+B : C

commutative law

αA = Aα

identical element

1A = A 1·a = a I : A = A

inverse element

1A + [−1]A = 0

zero element

0⊗a = 0 0·a = 0 O : A = 0

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.6 products • e1.6.4 calculation laws

products of tensorsÜ 22

symmetric & skew symmetric

• transposition of a tensor

A = Aij ei ⊗ ej

A = Aijkl ei⊗ej⊗ek⊗el

AT = Aji ei⊗ej

AT = Ailjk ei⊗ej⊗ek⊗el

• symmetric and skew symmetric tensors

A =Asym +Askw =1

[A−AT

]A = Asym + Askw =

[A + AT

[A−AT

• transposition of a dyadic product

[a⊗ b]T = b⊗ a

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.7 tensor characteristics • e1.7.1 symmetric tensors

transposition of a tensorÜ 23

• additive decomposition of a second order tensor in volumentric and deviatoric parts

A = Avol +Adev

• volumetric tensor (three dimensional and two dimensional)

Avol =1

3[A : 1] 1 Avol =

2[A : 1] 1

• deviatoric tensor (three dimensional and two dimensional)

Adev = A−1

3[A : 1] 1 Adev = A−

2[A : 1] 1

• calculations rules

tr[Avol] = tr[A] tr[Adev] = 0

• general formulation of deviatoric tensor in ND spatial dimensions

Avol =1

ND[A : 1] 1

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.7 tensor characteristics • e1.7.2 deviator

volumetric deviatoric decompositionÜ 24

invariants of second order tensors

• invariants are scalar valued quantities of a tensor which are independent on the coordinatesystem• invariants of three and two dimensional second order tensors

IA = tr[A]

IIA =1

[tr2[A]− tr[A ·A]

]IIIA = det[A] = |A|

IA = tr[A]

IIA = det[A] = |A|

• trace of a tensor

tr[A] = A : 1 = Aii = A11 +A22 +A33

• determinant of a tensor

three dimensional

det[A] = |A| = A11A22A33 +A21A32A13 +A31A12A23

− A11A23A32 −A22A31A13 −A33A12A21

two dimensional

det[A] = |A| = A11A22 −A21A12

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.7 tensor characteristics • e1.7.3 invariants

invariants of tensorsÜ 25

calculation rules for invariants of tensors

• trace of a tensor

tr[1] = 3

tr[AT ] = tr[A]

tr[A ·B] = tr[B ·A]

tr[αA+ βB] = αtr[A] + βtr[B]

tr[A ·BT ] = tr[A ·B]

tr[A] = tr[A · 1] = A : 1

• determinant of a tensor

det[1] = 1

det[AT ] = det[A]

det[A ·B] = det[A] det[B]

det[αA] = α3 det[A]

det[a⊗ b] = 0

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.7 tensor characteristics • e1.7.3 invariants

invariants of tensorsÜ 26

• second order tensors with i, j ∈ [1, 3]

A = Aij ei ⊗ ej

• eigenvalue problem of arbitrary second order tensors

[A− λ 1] ·Φ = 0

• eigenvalues (non-trivial solutions)

det [A− λ 1] = 0

• characteristic polynomial, three dimensional i, j ∈ [1, 3], invariants of tensors A

λ3 − IA λ2 + IIA λ− IIIA = 0 λ1|2|3 = . . .

• characteristic polynomial, two dimensional i, j ∈ [1, 2], invariants of tensors A

λ2 − IA λ+ IIA = 0 λ1|2 =IA

√I2A

4− IIA

• yields ND (spatial dimension) eigenvalues λi• related eigenvectors Φi are solutions of linear systems of equations

[A− λi 1] ·Φi = 0

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.7 tensor characteristics • e1.7.4 eigenvalues

eigenvalue problem of tensorsÜ 27

• general quadratic characteristic polynomial

λ2 + 2p λ+ q = 0

• two solutions

λ1|2 = −p±√D

• discriminant

D = p2 − q

• characterization of solution properties

D > 0: two real solutions D = 0: one double real solution D < 0: two conjugate complex solutions

zero values of quadratic polynomialsÜ 28

• general cubic characteristic polynomial

λ3 + aλ2 + bλ+ c = 0

• substitution λ = λ+ a/3, discriminant D

λ3 + 3pλ+ 2q = 0, 2q =2a3

27−ab

3+ c, 3p = b−

3, D = p3 + q2

• characterization of solution properties D > 0: one real and two conjugate complex solutions D = 0, q 6= 0: two real solutions, one is a double solution D = 0, q = 0: tripple real solution D < 0: three different real solutions

• CARDANO solution formula

λ1 = u+ + u−, λ2 = ρ+u+ + ρ−u−, λ3 = ρ−u+ + ρ+u−

u± =3√−q ±

√D, ρ± :=

2[−1± i

for one real discriminat D ≥ 0 u+ and u− are uniquely determined in general both complex real third roots u± have to be determined such that u+u− = −p is

fulfilled

• back substitution λi = a/3− λi

zero values of cubic polynomialsÜ 29

tdq e1

ϕ3 cosϕ3

sinϕ3

cosϕ3

cosϕ3 = e1 · e′1

rotation of tensor basis

• outline

orthogonal tensors

2d transformation of tensor basis

3d transformation of tensor basis

transformation of first order tensor components

transformation of second order tensor components

• schedule

two dimensional explanation and illustration

three dimensional extension

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.8 coordinate transformation

transformation of tensorsÜ 30

e3 ϕ2

ϕ2ϕ3

base vectors

e1, e2, e3

rotations

e1-e2-plane

ϕ =[ϕ3

] e2-e3-plane

ϕ =[ϕ1

] e1-e3-plane

ϕ =[ϕ2

rotations in 3d and 2d coordinate systemsÜ 31

e3 ϕ2

ϕ2ϕ3

base vectors

e1, e2, e3

rotations

e1-e2-plane

ϕ =[ϕ3

e2-e3-plane

ϕ =[ϕ1

] e1-e3-plane

ϕ =[ϕ2

e3 ϕ2

ϕ2ϕ3

base vectors

e1, e2, e3

rotations

e1-e2-plane

ϕ =[ϕ3

] e2-e3-plane

ϕ =[ϕ1

e1-e3-plane

ϕ =[ϕ2

e3 ϕ2

ϕ2ϕ3

base vectors

e1, e2, e3

rotations

e1-e2-plane

ϕ =[ϕ3

] e2-e3-plane

ϕ =[ϕ1

] e1-e3-plane

ϕ =[ϕ2

elementary rotation about e1

e′2e′3

ϕ1 ϕ1 rb

e′2e′3

e3 = e′3

rotation tensor Q1

0 cosϕ1 − sinϕ1

0 sinϕ1 cosϕ1

transformation rule

e′i = Q1 · ei

rotation tensor Q2

cosϕ2 0 sinϕ2

− sinϕ2 0 cosϕ2

transformation rule

e′i = Q2 · ei

rotation tensor Q3

cosϕ3 − sinϕ3 0

sinϕ3 cosϕ3 0

transformation rule

e′i = Q3 · ei

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.8 coordinate transformation • e1.8.3 3d rotation tensor

elementary rotations about e1, e2, e3Ü 36

e′2e′3

ϕ1 ϕ1 rb

e′2e′3

e3 = e′3

e3 = e′′′3

e′′′1

e′′′2

ϕ3 = 30o

e′′i = Q2 · e′′′i = Q2 ·Q3 · ei

e′′1e′′2

e′′3

ϕ2 = −30o

e′i= Q1 ·e′′i = Q1 ·Q2 ·Q3 · ei

e′3ϕ1

ϕ1 = 30o

cumulative rotation about e1, e2, e3Ü 38

e′2e′3

ϕ1 ϕ1 rb

e′2e′3

e3 = e′3

e3 = e′′′3

e′′′1

e′′′2

ϕ3 = 30o

e′′i = Q2 · e′′′i = Q2 ·Q3 · ei

e′′1e′′2

e′′3

ϕ2 = −30o

e′i= Q1 ·e′′i = Q1 ·Q2 ·Q3 · ei

e′3ϕ1

ϕ1 = 30o

e′2e′3

ϕ1 ϕ1 rb

e′2e′3

e3 = e′3

e3 = e′′′3

e′′′1

e′′′2

ϕ3 = 30o

e′′i = Q2 · e′′′i = Q2 ·Q3 · ei

e′′1e′′2

e′′3

ϕ2 = −30o

e′i= Q1 ·e′′i = Q1 ·Q2 ·Q3 · ei

e′3ϕ1

ϕ1 = 30o

e′2e′3

ϕ1 ϕ1 rb

e′2e′3

e3 = e′3

e3 = e′′′3

e′′′1

e′′′2

ϕ3 = 30o

e′′i = Q2 · e′′′i = Q2 ·Q3 · ei

e′′1e′′2

e′′3

ϕ2 = −30o

e′i= Q1 ·e′′i = Q1 ·Q2 ·Q3 · ei

e′3ϕ1

ϕ1 = 30o

e′2e′3

ϕ1 ϕ1 rb

e′2e′3

e3 = e′3

e′2e′3

ϕ1 ϕ1

ϕ1 = 30o

e′′i = Q2 · e′′′i = Q2 ·Q1 · ei

e′′1

e′′2

e′′3

ϕ2ϕ2

ϕ2 = −30o

e′i= Q3 ·e′′i = Q3 ·Q2 ·Q1 ·ei

ϕ3 = 30o

e′2e′3

ϕ1 ϕ1 rb

e′2e′3

e3 = e′3

e′2e′3

ϕ1 ϕ1

ϕ1 = 30o

e′′i = Q2 · e′′′i = Q2 ·Q1 · ei

e′′1

e′′2

e′′3

ϕ2ϕ2

ϕ2 = −30o

e′i= Q3 ·e′′i = Q3 ·Q2 ·Q1 ·ei

ϕ3 = 30o

e′2e′3

ϕ1 ϕ1 rb

e′2e′3

e3 = e′3

e′2e′3

ϕ1 ϕ1

ϕ1 = 30o

e′′i = Q2 · e′′′i = Q2 ·Q1 · ei

e′′1

e′′2

e′′3

ϕ2ϕ2

ϕ2 = −30o

e′i= Q3 ·e′′i = Q3 ·Q2 ·Q1 ·ei

ϕ3 = 30o

rotation sequence Q3, Q2, Q1

resulting rotation

e′i = Q1 ·Q2 ·Q3 · ei

resulting rotation

e′i = Q3 ·Q2 ·Q1 · ei

• rotation sequence Q3, Q2,Q1

e′i = Q1 ·Q2 ·Q3 · ei

e′i = Q3 ·Q2 ·Q1 · ei

• rotation sequence can notchanged (rotation is notcommutative)

• free choice of sequences ofelementary rotations aboutbase vectors ei• our decision

elementary rot. about e3

• 3d rotation tensor

Q = Q1 ·Q2 ·Q3

resulting rotation

e′i = Q1 ·Q2 ·Q3 · ei

resulting rotation

e′i = Q3 ·Q2 ·Q1 · ei

e′i = Q1 ·Q2 ·Q3 · ei

e′i = Q3 ·Q2 ·Q1 · ei

Q = Q1 ·Q2 ·Q3

resulting rotation

e′i = Q1 ·Q2 ·Q3 · ei

resulting rotation

e′i = Q3 ·Q2 ·Q1 · ei

e′i = Q1 ·Q2 ·Q3 · ei

e′i = Q3 ·Q2 ·Q1 · ei

Q = Q1 ·Q2 ·Q3

• rotation sequence Q3, Q2, Q1

first rotation Q3 with ϕ3 = 90o

second rotation Q2 with ϕ2 = 90o

third rotation Q1 with ϕ1 = 90o

thrid rotation Q3 with ϕ3 = 90o

• result is different to inverse rotation sequenceQ3, Q2, Q1

• rotation is not commutative

study of rotation sequenceÜ 41

third rotation Q1 with ϕ1 = 90o

thrid rotation Q3 with ϕ3 = 90o

• result is different to inverse rotation sequenceQ3, Q2, Q1

• rotation is not commutative

study of rotation sequenceÜ 41

• elementary rotation about basis vector e1, rotated tensor basis, rotated basis vectors

0 cosϕ1 − sinϕ1

0 sinϕ1 cosϕ1

e′1 =

, e′2 =

cosϕ1

sinϕ1

, e′3 =

− sinϕ1

cosϕ1

• elementary rotation about basis vectors e2 and e3

cosϕ2 0 sinϕ2

− sinϕ2 0 cosϕ2

cosϕ3 − sinϕ3 0

sinϕ3 cosϕ3 0

• rotated basis vectors after elementary rotations about e2 and e3

e′1 =

cosϕ2

− sinϕ2

, e′2 =

, e′3 =

sinϕ2

cosϕ2

e′1 =

cosϕ3

sinϕ3

, e′2 =

− sinϕ3

cosϕ3

, e′3 =

• three dimensional rotation tensor Q = Q1 ·Q2 ·Q3

cosϕ2 cosϕ3 − cosϕ2 sinϕ3 sinϕ2

cosϕ1 sinϕ3 + sinϕ1 sinϕ2 cosϕ3 cosϕ1 cosϕ3 − sinϕ1 sinϕ2 sinϕ3 − sinϕ1 cosϕ2

sinϕ1 sinϕ3 − cosϕ1 sinϕ2 cosϕ3 sinϕ1 cosϕ3 + cosϕ1 sinϕ2 sinϕ3 cosϕ1 cosϕ2

three-dimensional rotation tensorÜ 42

-tdq e1

ϕ3 cosϕ3

sinϕ3

cosϕ3

cosϕ3 = e1 · e′1

a′1 a

coordinate transformation

• tensors are independent on coordinate system

only tensor components are dependent oncoordinate system

transformation of tensor components is required

• representation of first order tensor in original and rotated basisvectors

a = aj ej = aj Qji e′i a = a′j e

′j = a′j Qij ei

• comparision of both representations

a′i = aj Qji = Qji aj ai = a′j Qij = Qij a′j

• all components of tensor in original and rotated coordinates([a])

[a]′ = QT · [a] [a] = Q · [a]′

• test (example on left hand side)

a′1 = a1 cosϕ3 + a2 sinϕ3 = a1Q11 + a2Q21

a′2 = −a1 sinϕ3 + a2 cosϕ3 = a1Q12 + a2Q22

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.8 coordinate transformation • e1.8.4 first order tensors

transformation of first order tensorsÜ 43

• transformation of basis vectors, tensor basis

e′j = Qij ei ej = Qji e′i

• representation of second order tensors in original and rotated basis vectorstransformation of basis vectors

A = Akl ek ⊗ el = Akl Qki Qlj e′i ⊗ e′j = A′ij e

′i ⊗ e′j

A = A′kl e′k ⊗ e

′l = A′kl Qik Qjl ei ⊗ ej = Aij ei ⊗ ej

• comparision of both representations

A′ij = Qki Akl Qlj Aij = Qik A′kl Qjl

• all components of tensor in original and rotated coordinates ([A])

[A]′ = QT · [A] ·Q [A] = Q · [A]′ ·QT

alternative derivation of second order tensor transformation relations• components of second order tensor A = Akl ek ⊗ el

ei ·A · ej = Akl [ei · ek]⊗ [el · ej ] = Akl δikδlj = Aij

• with respect to rotated coordinate system

Aij = ei ·A · ej = [Qik e′k] ·A · [e′l Qjl] = Qik A

′kl Qjl

A′ij = e′i ·A · e′j = [Qki ek] ·A · [el Qlj ] = Qki Akl Qlj

d.kuhl, wes.online, university of kasselÜÜlinear computational structural mechanics • e1 tensors • e1.8 coordinate transformation • e1.8.5 second order tensors

transformation of second order tensorsÜ 44

functions of one variable functions of two variablesde

=∂f(X

A=b ∫ a

∂f(X

1,∂f(X

V=b2 ∫ a2

b1 ∫ a1

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.9 derivatives

derivatives of tensorsÜ 45

derivatives of tensors with respect to tensor valued variables→ increase order of a tensor by order of tensor valued variable

∂a∂b

∂b1∂a

∂b2∂a

= ∂a∂bi

∂b3∂a2

∂b3∂a3

= ∂ai∂bj

ei ⊗ ej

∂B11

∂B12

∂B13∂a

∂B21

∂B22

∂B23∂a

∂B31

∂B32

∂B33

= ∂a∂Bij

ei ⊗ ej

∂B= =

∂Aij∂Bkl

ei ⊗ ej ⊗ ek ⊗ el

differentiation of tensorsÜ 46

∂a∂b

∂b1∂a

∂b2∂a

= ∂a∂bi

∂b3∂a2

∂b3∂a3

= ∂ai∂bj

ei ⊗ ej

∂B11

∂B12

∂B13∂a

∂B21

∂B22

∂B23∂a

∂B31

∂B32

∂B33

= ∂a∂Bij

ei ⊗ ej

∂B= =

∂Aij∂Bkl

∂a∂b

∂b1∂a

∂b2∂a

= ∂a∂bi

∂b3∂a2

∂b3∂a3

= ∂ai∂bj

ei ⊗ ej

∂B11

∂B12

∂B13∂a

∂B21

∂B22

∂B23∂a

∂B31

∂B32

∂B33

= ∂a∂Bij

ei ⊗ ej

∂B= =

∂Aij∂Bkl

∂a∂b

∂b1∂a

∂b2∂a

= ∂a∂bi

∂b3∂a2

∂b3∂a3

= ∂ai∂bj

ei ⊗ ej

∂B11

∂B12

∂B13∂a

∂B21

∂B22

∂B23∂a

∂B31

∂B32

∂B33

= ∂a∂Bij

ei ⊗ ej

∂B= =

∂Aij∂Bkl

∂a∂b

∂b1∂a

∂b2∂a

= ∂a∂bi

∂b3∂a2

∂b3∂a3

= ∂ai∂bj

ei ⊗ ej

∂B11

∂B12

∂B13∂a

∂B21

∂B22

∂B23∂a

∂B31

∂B32

∂B33

= ∂a∂Bij

ei ⊗ ej

∂B= =

∂Aij∂Bkl

gradient (∇) & symmetric part of gradient (∇sym)→ derivative of vector a with repect to position vector X

∇a =∂a

∂Xjei⊗ej = ai,j ei⊗ej ∇syma =

2[ai,j+aj,i] ei ⊗ ej

transposition of gradient

∇Ta = [∇a]T = aj,i ei⊗ej

divergence (div) of a tensor→ reduction of order of a tensor

diva = tr[∇a] = ai,i =a1,1+a2,2+a3,3 divA =

A11,1+A12,2+A13,3

A21,1+A22,2+A23,3

A31,1+A32,2+A33,3

= Aij,jei

e.g. displacement gradient and stress tensor

∇u =

u1,1 u1,2 u1,3

u2,1 u2,2 u2,3

u3,1 u3,2 u3,3

= ui,j ei ⊗ ej divσ =

σ11,1+σ12,2+σ13,3

σ21,1+σ22,2+σ23,3

σ31,1+σ32,2+σ33,3

= σij,j ei

e.g. strain tensor (geometrically linear theory)

ε = ∇symu =1

[∇u+∇Tu

2[ui,j + uj,i] ei ⊗ ej = εij ei ⊗ ej

gradient and divergenceÜ 47

∇a =∂a

A11,1+A12,2+A13,3

A21,1+A22,2+A23,3

A31,1+A32,2+A33,3

= Aij,jei

∇u =

u1,1 u1,2 u1,3

u2,1 u2,2 u2,3

u3,1 u3,2 u3,3

σ11,1+σ12,2+σ13,3

σ21,1+σ22,2+σ23,3

σ31,1+σ32,2+σ33,3

= σij,j ei

ε = ∇symu =1

[∇u+∇Tu

∇a =∂a

A11,1+A12,2+A13,3

A21,1+A22,2+A23,3

A31,1+A32,2+A33,3

= Aij,jei

∇u =

u1,1 u1,2 u1,3

u2,1 u2,2 u2,3

u3,1 u3,2 u3,3

σ11,1+σ12,2+σ13,3

σ21,1+σ22,2+σ23,3

σ31,1+σ32,2+σ33,3

= σij,j ei

ε = ∇symu =1

[∇u+∇Tu

∇a =∂a

A11,1+A12,2+A13,3

A21,1+A22,2+A23,3

A31,1+A32,2+A33,3

= Aij,jei

∇u =

u1,1 u1,2 u1,3

u2,1 u2,2 u2,3

u3,1 u3,2 u3,3

σ11,1+σ12,2+σ13,3

σ21,1+σ22,2+σ23,3

σ31,1+σ32,2+σ33,3

= σij,j ei

ε = ∇symu =1

[∇u+∇Tu

f + df

∂f∂Xf

f + dfrb rb

functions of one variable

functions of two variable

• illustration of total differential df by means of functions of onevariable

df is the change of the function value f for incrementalchange of the variable dX

mathematical formulation using the derivative ∂f/∂X

df =∂f

∂XdX

• total differential df of a function of two variables

df is the change of the function value f for incrementalchanges of variables dX1 and dX2

mathematical formulation using the partial derivatives∂f/∂X1 and ∂f/∂X2

df =∂f

∂X1dX1 +

∂X2dX2

• total differential of function df of multiple variables

df is the change of the function value f for incrementalchanges of the variables

• total differential, variation & increment are formally identical,just the interpretation and the symbols (d, δ and ∆) are different

total differential, variation & increment of tensorsÜ 48

total differential, variation and increment of a tensor valued function, e.g. A(B, c)

dA(B, c) =∂A(B, c)

∂B: dB +

∂A(B, c)

∂c· dc

δA(B, c) =∂A(B, c)

∂B: δB +

∂A(B, c)

∂c· δc

∆A(B, c) =∂A(B, c)

∂B: ∆B +

∂A(B, c)

∂c·∆c

chain rule of total differential, variation and increment of a tensor valued function, e.g.

δA(B(C(d))) =∂A(B(C(d)))

∂B:∂B(C(d))

∂C:∂C(d)

∂d· δd

=∂A(B(C(d)))

∂B:∂B(C(d))

∂C: δC(d)

=∂A(B(C(d)))

∂B: δB(C(d))

example - increment of second PIOLA-KIRCHHOFF stress tensor (geometrically non-linear)

∆S(E(F (∇u))) =∂S

∂E:∂E

∂F:∂F

∂∇u: ∆∇u =

∂E:∂E

∂F:∆F =

∂E:∆E = C :∆E

product rule, example - increment of specific internal virtual work (geometrically non-linear)

δ[A : B] = δA : B +A : δB ∆[δE : S] = ∆δE : S + δE : ∆S

total differential, variation & increment of tensorsÜ 49

GAUSS integral theorem → transformation of volume integral over domain Ω of gradient into anintegral over surface of body Γ∫Ω

∇a dV =

a n dA∫Ω

∇a dV =

a⊗ n dA∫Ω

∇A dV =

A⊗ n dA

a,i dV =

a ni dA∫Ω

ai,j dV =

ai nj dA∫Ω

Aij,k dV =

Aij nk dA

multiplication by KRONECKER symbol δij → form, relevant for course∫Ω

diva dV =

a· n dA∫Ω

divA dV =

A· n dA

ai,i dV =

ai ni dA∫Ω

Aij,j dV =

Aij nj dA

e.g. development of weak form of balance of linear momentum - principle of virtual work∫Ω

div[δu · σ] dV =

δu · σ · n dA =

∫Γσ

δu · σ · n dA =

∫Γσ

δu · t dA

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.10 Gauß integral theorem

tensor analysis - GAUSS integral theorem Ü 50

double contraction of tensor A with a symmetric tensor B = BT

A : B = Asym : B Aij Bij =1

2[Aij +Aji] Bkl

both side double contraction of tensor C with Cijkl = Cjikl = Cjilk = Cijlk

A : C : B = Asym : C : Bsym AijCijklBkl=1

4[Aij+Aji]Cijkl[Bkl+Blk]

gradient and divergence of tensors - formulas

∇[a b] = a ∇b + b⊗∇a∇[a · b] =∇Ta · b +∇Tb · a∇[a B] = a ∇B +B ⊗∇a∇A : 1 = divA

div[a b] = a divb + b · ∇adiv[a⊗ b] = a divb +∇a · bdiv[a ·B] =∇a : B + a · divB

div[a B] = a divB +B · ∇a

chain rule of divergence div[a ·B] = ∇a : B + a · divB and GAUSS integral

theorem consitute divergence theorem∫Ω

a · divB dV =

div[a ·B] dV −∫Ω

∇a : B dV =

a ·B · n dA−∫Ω

∇a : B dV

d.kuhl, wes.online, university of kasselÜ Ülinear computational structural mechanics • e1 tensors • e1.11 tensor reformulations

tensor analysis - more formulas ...Ü 51

· x aaaa bbb ccc aaa bbb ccc tensor analysis in cartesian coordinates tensor characteristics...

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