tensor de almansi

Continuum mechanicsI. Kinematics in cartesian coordinates

Ales Janka

office Math [email protected]

http://perso.unifr.ch/ales.janka/mechanics

September 22, 2010, Universite de Fribourg

Ales Janka I. Kinematics

Kinematics: description of position and deformation

initial configuration x = x iei

x i . . . material coordinates

deformed config. y = y iei

y i . . . spatial coordinates

displacement u = y − x

Two possibilities:

Lagrange description:u(x) = y(x)− x

Euler description:u(y) = y − x(y)

e3

e2e1

x

y

dx

y

dy

x

u

u+dux+dxy+dy

Initial config.

Deformed config.


Material vs. spatial coordinates: deformation of a 1D rod

Let y1 = y1(x1, t) = [(x1)2 − 1] · t + x1.

Inversely, x1 = x1(y1, t) = 12t

(√1 + 4t(2t + y1)− 1

)


1. Lagrange description

x = x iei and y = y iei .

Choice: y = y(x)

Deformation gradient:

F ij (x) =

∂y i

∂x j= y i

,j

dx = dx iei

dy = dy iei =∂y i

∂x jdx jei

dy = dx j F ij (x) gi = dx j gj

e3

e1

e3

e2e1

x

y

dx

y(x)

dy

y(x+dx)

g2

g3

e2g1

^g

2

^g1

^g3

=

==

Initial config.

Deformed config.


1. Lagrange description: how to measure deformation?

The ”edge” dx in initial configuration is deformed to dy

Convenient measure of deformation:

dy2−dx2 = (dx i gi )·(dx j gj)−(dx igi )·(dx jgj) = dx i dx j (gij−gij)

where gij = gi · gj and gij = gi · gj

are metric tensors (matrices)

Green strain tensor:

εij(x) =1

2(gij − gij)

εij(x) =1

2

(F k

i (x) F `j (x) gk` − gij

)


1. Lagrange descript.: Green strain tensor in displacement

u(x) = y(x)− x

u = ui gi

dy = dx + du

du =∂ui

∂x jdx j gi

dy =(dx i + ui

,j dx j)

gi

e3

e2e1

x

y

y(x)

y(x+dx)

dy

(x)u

(x)u du

u(x+dx)dx

x

x+dx

Initial config.

Deformed config.


1. Lagrange descript.: Green strain tensor in displacement

(dy)2 =

(dx i + ui

,j dx j

)(dxk + uk

,` dx`

)gik

(dy)2−(dx)2 = ui,j dx j dxk gik + uk

,` dx` dx i gik + ui,j uk

,` dx j dx` gik

(dy)2−(dx)2 =(ui ,j + uj ,i + uk,i u

k,j

)dx i dx j

Green strain tensor in displacements:

εij =1

2

(ui ,j + uj ,i + uk,i u

k,j

)Green strain tensor in displacements in cartesian coordinates:

εij =1

2

(∂ui

∂xj+∂uj

∂xi+∂uk

∂xi

∂uk

∂xj

)Ales Janka I. Kinematics

1. Lagrange description: example 1 – rigid body motion

x

u

a

y

e3

e2

e1

Deformed config.

Initial config.

α

y1

y2

y3

=

cosα − sinα 0sinα cosα 0

0 0 1

·x1

x2

x3

+

a1

a2

a3



x

u

a

y

e3

e2

e1

Deformed config.

Initial config.

α

u1

u2

u3

=

cosα− 1 − sinα 0sinα cosα− 1 0

0 0 0

·x1

x2

x3

+

a1

a2

a3

ε11 =1

2

(u1,1 + u1,1 +

3∑k=1

uk,1uk,1

)

= cosα− 1 +(cosα− 1)2 + sin2 α

2= 0



x

u

a

y

e3

e2

e1

Deformed config.

Initial config.

α

u1

u2

u3

=

cosα− 1 − sinα 0sinα cosα− 1 0

0 0 0

·x1

x2

x3

+

a1

a2

a3

ε12 =1

2

(u1,2 + u2,1 +

3∑k=1

uk,1uk,2

)

=− sinα + sinα

2+− sinα (cosα− 1) + sinα (cosα− 1)

2= 0


1. Lagrange description: example 2

−5 −4 −3 −2 −1 0 1 2 3 4 5

−2

−1

0

1

Initial configuration is bent into the deformed configuration

Principal strain of Green strain tensor (Lagrange formulation)?



−5 −4 −3 −2 −1 0 1 2 3 4 5

−2

−1

0

1

−5 −4 −3 −2 −1 0 1 2 3 4 5

−2

−1

0

1

Wrong - this is Almansi strain (Euler formulation)!Ales Janka I. Kinematics


−5 −4 −3 −2 −1 0 1 2 3 4 5

−2

−1

0

1

−5 −4 −3 −2 −1 0 1 2 3 4 5

−2

−1

0

1

This is the correct Green strain (Lagrange formulation)Ales Janka I. Kinematics


−5 −4 −3 −2 −1 0 1 2 3 4 5

−2

−1

0

1

−5 −4 −3 −2 −1 0 1 2 3 4 5

−2

−1

0

1

This is the correct Green strain (Lagrange formulation)Ales Janka I. Kinematics

2. Euler description

x = x iei and y = y iei .

Choice: x = x(y)

Deformation gradient inverse:

F−1 ij =

∂x i

∂y j= x i

,j

dy = dy iei

dx = dx iei =∂x i

∂y jdy j ei

dx = dx j F−1 ij gi = dx j gj

e3

e2e1

x

y

dx

dy

(y)

y+dy

y

e1g1=

g2

e2

=

g1

~

g2

~

g3

~x(y+dy)

(y)x

e3g3 =

Initial config.

Deformed config.


2. Euler description: Almansi strain tensor

the deformed ”edge” dy corresponds to the undeformed dx

Difference of their (lengths)2:

dy2−dx2 = (dy igi )·(dy jgj)−(dy i gi )·(dy j gj) = dy i dy j (gij−gij)

where gij = gi · gj and gij = gi · gj

are metric tensors (matrices)

Almansi strain tensor:

Eij(y) =1

2(gij − gij)

Eij(y) =1

2

(gij − F−1 k

i F−1 `j gk`

)


2. Euler description: Almansi strain tensor in displacement

u(y) = y − x(y)

u = ui gi

dx = dy − du

du =∂ui

∂y jdy j gi

dx =(dy i − ui

,j dy j)

gi

e3

e2e1

x

y

dy

(y)u

(y)u du

u(y+dy)dx

ydx

(y)

y+dy(y+dy)x

(y)x

Initial config.

Deformed config.


2. Euler description: Almansi strain tensor in displacement

(dx)2 =

(dy i − ui

,j dy j

)(dyk − uk

,` dy `

)gik

(dy)2−(dx)2 = ui,j dy j dyk gik + uk

,` dy ` dy i gik − ui,j uk

,` dy j dy ` gik

(dy)2−(dx)2 =(ui ,j + uj ,i − uk,i u

k,j

)dy i dy j

Almansi strain tensor in displacements:

Eij =1

2

(ui ,j + uj ,i − uk,i u

k,j

)Almansi strain tensor in displacements in cartesian coordinates:

Eij =1

2

(∂ui

∂yj+∂uj

∂yi− ∂uk

∂yi

∂uk

∂yj

)Ales Janka I. Kinematics

3. Green and Almansi strain tensors: mutual relation

Green strain tensor:

εij(x) =1

2(gij − gij) =

1

2

(F k

i F `j gk` − gij

)εij =

1

2

(∂ui

∂xj+∂uj

∂xi+∂uk

∂xi

∂uk

∂xj

)Almansi strain tensor

Eij(y) =1

2(gij − gij) =

1

2

(gij − F−1 k

i F−1 `j gk`

)Eij =

1

2

(∂ui

∂yj+∂uj

∂yi− ∂uk

∂yi

∂uk

∂yj

)


3. Green and Almansi strain tensors: mutual relation

Relation between F ij = ∂y i

∂x j and F−1 jk = ∂x j

∂yk :

F ij · F−1 j

k =3∑

j=1

∂y i

∂x j

∂x j

∂yk=∂y i

∂yk= δik

by chain rule for the derivatives.

Hence:εk` = F i

k · Eij · F j`


3. Green and Almansi strain tensors: matrix form

Deformation gradient matrix:

F =

∂y1

∂x1∂y1

∂x2∂y1

∂x3

∂y2

∂x1∂y2

∂x2∂y2

∂x3

∂y3

∂x1∂y3

∂x2∂y3

∂x3

=[F i

j

]and F−1 =

[F−1 i

j

]

Green and Almansi matrix for cartesian coords (gi = ei , gij = δij):

[εij ] =1

2

(FT F− I

)and [Eij ] =

1

2

(I− F−T F−1

)Mutual relations:

[εij ] = FT · [Eij ] · F and [Eij ] = F−T · [εij ] · F−1


3. Green and Almansi strain tensors: small deformations

If ui ,j � 1 then uk,i · uk,j is negligible:

εij =1

2

(∂ui

∂x j+∂uj

∂x i+∂uk

∂x i

∂uk

∂x j

)≈ 1

2

(∂ui

∂x j+∂uj

∂x i

)= eij

We can replace Green strain εij by the Cauchy strain eij

Advantage: Cauchy strain eij(u) is linear in u

Eij =1

2

(∂ui

∂y j+∂uj

∂y i− ∂uk

∂y i

∂uk

∂y j

)≈ 1

2

(∂ui

∂y j+∂uj

∂y i

)=

1

2

(∂ui

∂xk

∂xk

∂y j+∂uj

∂yk

∂xk

∂y i

)≈ 1

2

(∂ui

∂x j+∂uj

∂x i

)= eij

because xk = yk − uk and ui ,k · uk,` is negligible.NB: Green and Almansi simplify to the same Cauchy strain!


4. Physical meaning of Cauchy strain in cartesian coords

dx dy

yx

e1

e2

e3

xyInitial config. Deformed config.

Special choices of the deformation mode:Let dx = dx1 e1 and dy = dy1 e1. Then:

dy2 − dx2 = (dy1)2 − (dx1)2 = 2 e11 (dx1)2

Hence (for small deformations dx1 + dy1 ≈ 2 dx1):

e11 =(dy1)2 − (dx1)2

2 (dx1)2=

(dy1 − dx1)(dy1 + dx1)

2 (dx1)2≈ dy1

dx1− 1

Meaning of ekk : relative elongation along ek



yx

e1

e2

e3

yx

dx2

dx1 dy1

dy2

Initial config. Deformed config.

θ

Special choices of the deformation mode:Let dx = dx1 + dx2 and dy = dy1 + dy2, dxk = dxk ek :

dy2−dx2 = (dy1)2+2 dy1 ·dy2+(dy2)2−(dx1)2−2 dx1 ·dx2−(dx2)2

= 2[e11 (dx1)2 + 2 e12 dx1 dx2 + e22 (dx2)2

]Hence (for small deformations ekk � 1 and θ is small):

e12 =θ

2

dy1

dx1

dy2

dx2≈ θ

2(1 + e11) (1 + e22)

≈ θ

2(1 + e11 + e22 + e11 e22) ≈ θ

2

Meaning of e12: half of shear angle θ in the plane (0, e1, e2)Ales Janka I. Kinematics


yx

e1

e2

e3

yx

dx2

dx1 dy1

dy2


θ

Special choices of the deformation mode:Let dx = dx1 + dx2 and dy = dy1 + dy2, dxk = dxk ek :

dy1 · dy2 = 2 e12 dx1 dx2

= |dy1| · |dy2| cos(π/2− θ) ≈ dy1 dy2 sin θ ≈ dy1 dy2 · θHence (for small deformations ekk � 1 and θ is small):

e12 =θ

2

dy1

dx1

dy2

dx2≈ θ

2(1 + e11) (1 + e22)

≈ θ

2(1 + e11 + e22 + e11 e22) ≈ θ

2

Meaning of e12: half of shear angle θ in the plane (0, e1, e2)Ales Janka I. Kinematics


e1

e2

e3

yxdx3

dy3dy1

dy2dx2dx1


Volume before (dV ) and after (dV ) deformation (small deformations):

dV = dx1 dx2 dx3 dV ≈ dy1 dy2 dy3

Relative change of volume (for small deformations ekk � 1):

dV − dV

dV=

dV

dV− 1 ≈ dy1

dx1

dy2

dx2

dy3

dx3− 1

= (1 + e11)(1 + e22)(1 + e33)− 1 ≈ e11 + e22 + e33

Meaning of trace of eij : relative change of volume


5. How to transform areas dS0 → dS?

Nanson’s relation:we know how to transform vectors:

dy i =∂y i

∂x jdx j

we know how to transform volumes:

dV = J·dV0 with J = det

[∂y i

∂x j

]

dx

e2

(x)ye3

e1

x

0dV

dy

dS

dS0

Initial config.

Deformed config.

dV

Idea: complete areas to volumes (for any dx, ie. any dy):

dSi dy i = dV = J · dV0 = J · dS0j dx j = J · dS0j∂x j

∂y idy i


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tensor de almansi