1_kinematics-handout2

8/17/2019 1_kinematics-handout2

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Continuum mechanics

I. Kinematics in cartesian coordinates

Aleš Janka

office Math 0.107

[email protected]

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

September 22, 2010, Université de Fribourg

Aleš Janka I. Kinematics

Kinematics: description of position and deformation

initial configuration x = x i ei x i . . . material coordinates

deformed config. y = y i ei

y

i

. . . spatial coordinatesdisplacement 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+dx y+dy

Initial config.

Deformed config.



2/14

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

Let y 1 = y 1(x 1, t ) = [(x 1)2 − 1] · t + x 1.

Inversely, x 1 = x 1(y 1, t ) = 12t

1 + 4t (2t + y 1) − 1


1. Lagrange description

x = x i ei and y = y i ei .

Choice: y = y(x)

Deformation gradient:

F i j (x) = ∂ y i

∂ x j = y i , j

d x = dx i ei

d y = dy i ei = ∂ y i

∂ x j dx j ei

d y = dx j F i j

(x) gi = dx j ĝ j

e3

e1

e3

e2e1

x

y

dx

y(x)

dy

y(x+dx)

g2

g3

e2g

1

^g2

^g1

^g3

=

==

Initial config.

Deformed config.



3/14

1. Lagrange description: how to measure deformation?

The ”edge” d x in initial configuration is deformed to d y

Convenient measure of deformation:

d y2−d x2 = (dx i ĝi )·(dx j ĝ j )−(dx i gi )·(dx j g j ) = dx i dx j (ĝ ij −g ij )

where g ij = gi · g j and ĝ ij = ĝi · ĝ j are metric tensors (matrices)

Green strain tensor:

εij (x) = 1

2 (ĝ ij − g ij )

εij (x) = 1

2

F k i (x)F

j (x) g k − g ij


1. Lagrange descript.: Green strain tensor in displacement

u(x) = y(x) − x

u = u i

gi d y = d x + d u

d u = ∂ u i

∂ x j dx j gi

d y =dx i + u i , 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.



4/14

1. Lagrange descript.: Green strain tensor in displacement

(d y)2 =

dx i + u i , j dx

j

dx k + u k , dx

g ik

(d y)2−(d x)2 = u i , j dx j dx k g ik + u

k , dx

dx i g ik + u i , j u

k , dx

j dx g ik

(d y)2−(d x)2 =u i , j + u j ,i + u k ,i u

k , j

dx i dx j

Green strain tensor in displacements:

εij = 1

2

u i , j + u j ,i + u k ,i u

k , j

Green strain tensor in displacements in cartesian coordinates:

εij = 1

2

∂ u i ∂ x j

+ ∂ u j ∂ x i

+ ∂ u k ∂ x i

∂ u k ∂ x j


1. Lagrange description: example 1 – rigid body motion

x

u

a

y

e3

e2

e1

Deformed config.

Initial config.

α

y 1y 2y 3

=

cos α − sinα 0sin α cos α 0

0 0 1

·

x 1x 2x 3

+

a1a2a3



5/14


x

u

a

y

e3

e2

e1

Deformed config.

Initial config.

α

u 1u 2u 3

=

cos α− 1 − sinα 0sinα cos α− 1 0

0 0 0

·

x 1x 2x 3

+

a1a2a3

ε11 = 1

2

u 1,1 + u 1,1 +

3k =1

u k ,1u k ,1

= cos α− 1 + (cos α− 1)2 + sin2 α

2 = 0



x

u

a

y

e3

e2

e1

Deformed config.

Initial config.

α

u 1u 2u 3

=

cos α− 1 − sinα 0sinα cos α− 1 0

0 0 0

·

x 1x 2x 3

+

a1a2a3

ε12 =

1

2u 1,2 + u 2,1 +

3k =1

u k ,1u k ,2

= − sinα + sin α

2 +

− sinα (cos α− 1) + sin α (cos α− 1)

2 = 0



6/14

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)!Aleš Janka I. Kinematics


7/14


−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)Aleš 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)Aleš Janka I. Kinematics


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2. Euler description

x = x i ei and y = y i ei .

Choice: x = x(y)

Deformation gradient inverse:

F −1 i

j = ∂ x i

∂ y j = x i , j

d y = dy i ei

d x = dx i ei = ∂ x i

∂ y j dy j ei

d x = dx j F −1 i j gi = dx

j g̃ j

e3

e2e1

x

y

dx

dy

(y)

y+dy

y

e1g1=

g2e2=

g1~

g2

~

g3~

x(y+dy)

(y)x

e3g3 =

Initial config.

Deformed config.


2. Euler description: Almansi strain tensor

the deformed ”edge” d y corresponds to the undeformed d x

Difference of their (lengths)2:

d y2−d x2 = (dy i gi )·(dy j g j )−(dy

i g̃i )·(dy j g̃ j ) = dy

i dy j (g ij −g̃ ij )

where g ij = gi · g j and g̃ ij = g̃i · g̃ j are metric tensors (matrices)

Almansi strain tensor:

E ij (y) = 1

2 (g ij − g̃ ij )

E ij (y) = 12

g ij − F

−1 k

i F −1

j g k



9/14

2. Euler description: Almansi strain tensor in displacement

u(y) = y − x(y)

u = u i gi

d x = d y − d u

d u = ∂ u i

∂ y j dy j gi

d x =

dy i − u i , 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

(d x)2 =

dy i − u i , j dy

j

dy k − u k , dy

g ik

(d y)2−(d x)2 = u i , j dy j dy k g ik + u

k , dy

dy i g ik − u i , j u

k , dy

j dy g ik

(d y)2−(d x)2 =u i , j + u j ,i − u k ,i u k , j

dy i dy j

Almansi strain tensor in displacements:

E ij = 1

2

u i , j + u j ,i − u k ,i u

k , j

Almansi strain tensor in displacements in cartesian coordinates:

E ij = 1

2

∂ u i ∂ y j

+ ∂ u j ∂ y i

− ∂ u k ∂ y i

∂ u k ∂ y j



10/14

3. Green and Almansi strain tensors: mutual relation

Green strain tensor:

εij (x) = 1

2

(ĝ ij − g ij ) = 1

2F k i F j g k − g ij

εij = 1

2

∂ u i ∂ x j

+ ∂ u j ∂ x i

+ ∂ u k ∂ x i

∂ u k ∂ x j

Almansi strain tensor

E ij (y) = 1

2 (g ij − g̃ ij ) =

1

2 g ij − F −1 k

i F −1

j g k E ij =

1

2

∂ u i ∂ y j

+ ∂ u j ∂ y i

− ∂ u k ∂ y i

∂ u k ∂ y j


3. Green and Almansi strain tensors: mutual relation

Relation between F i j = ∂ y i

∂ x j and F −1

j k =

∂ x j

∂ y k :

F i j · F

−1 j

k =

3 j =1

∂ y i

∂ x j ∂ x j

∂ y k = ∂ y i

∂ y k = δ i k

by chain rule for the derivatives.

Hence:εk = F

i k · E ij · F

j



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3. Green and Almansi strain tensors: matrix form

Deformation gradient matrix:

F =

∂ y 1

∂ x 1∂ y 1

∂ x 2∂ y 1

∂ x 3

∂ y 2

∂ x 1

∂ y 2

∂ x 2

∂ y 2

∂ x 3

∂ y 3

∂ x 1∂ y 3

∂ x 2∂ y 3

∂ x 3

= F i j and F−1 = F −1

i

j

Green and Almansi matrix for cartesian coords (gi = ei , g ij = δ ij ):

[εij ] = 1

2 FT F − I and [E ij ] =

1

2 I − F−T F−1

Mutual relations:

[εij ] = FT · [E ij ] · F and [E ij ] = F

−T · [εij ] · F−1


3. Green and Almansi strain tensors: small deformations

If u i , j 1 then u k ,i · u k , j is negligible:

εij = 1

2

∂ u i

∂ x j +

∂ u j

∂ x i +

∂ u k

∂ x i ∂ u k

∂ x j

≈

1

2

∂ u i

∂ x j +

∂ u j

∂ x i

= e ij

We can replace Green strain εij by the Cauchy strain e ij Advantage: Cauchy strain e ij (u) is linear in u

E ij = 1

2

∂ u i

∂ y j +

∂ u j

∂ y i −

∂ u k

∂ y i ∂ u k

∂ y j

≈

1

2

∂ u i

∂ y j +

∂ u j

∂ y i

= 1

2

∂ u i

∂ x k ∂ x k

∂ y j +

∂ u j

∂ y k ∂ x k

∂ y i

≈

1

2

∂ u i

∂ x j +

∂ u j

∂ x i

= e ij

because x k = y k − u k and u i ,k · u k , is negligible.NB: Green and Almansi simplify to the same Cauchy strain!



12/14

4. Physical meaning of Cauchy strain in cartesian coords

dx dy

yx

e1

e2

e3x

yInitial config. Deformed config.

Special choices of the deformation mode:Let d x = dx 1 e1 and d y = dy

1 e1. Then:

d y2 − d x2 = (dy 1)2 − (dx 1)2 = 2 e 11 (dx 1)2

Hence (for small deformations dx 1 + dy 1 ≈ 2 dx 1):

e 11 = (dy 1)2 − (dx 1)2

2 (dx 1)2 =

(dy 1 − dx 1)(dy 1 + dx 1)

2 (dx 1)2 ≈

dy 1

dx 1 − 1

Meaning of e kk : 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 d x = d x1 + d x2 and d y = d y1 + d y2, d xk = dx

k ek :

d y2−d x2 = (d y1)2 +2 d y1 ·d y2 +(d y2)

2−(d x1)2−2 d x1 ·d x2−(d x2)

2

= 2e 11 (dx

1)2 + 2 e 12 dx 1 dx 2 + e 22 (dx

2)2

Hence (for small deformations e kk 1 and θ is small):

e 12 =

θ

2

dy 1

dx 1

dy 2

dx 2 ≈

θ

2 (1 + e 11) (1 + e 22)

≈ θ

2 (1 + e 11 + e 22 + e 11 e 22) ≈

θ

2

Meaning of e 12: half of shear angle θ in the plane (0, e1, e2)Aleš Janka I. Kinematics


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yx

e1

e2

e3

yx

dx2

dx1 dy1

dy2


θ

Special choices of the deformation mode:Let d x = d x1 + d x2 and d y = d y1 + d y2, d xk = dx

k ek :

d y1 · d y2 = 2 e 12 dx 1 dx 2

= |d y1| · |d y2| cos(π/2 − θ) ≈ dy 1 dy 2 sin θ ≈ dy 1 dy 2 · θ

Hence (for small deformations e kk 1 and θ is small):

e 12 = θ

2

dy 1

dx 1dy 2

dx 2 ≈

θ

2 (1 + e 11) (1 + e 22)

≈ θ

2 (1 + e 11 + e 22 + e 11 e 22) ≈

θ

2

Meaning of e 12: half of shear angle θ in the plane (0, e1, e2)Aleš Janka I. Kinematics


e1

e2

e3

yxdx3

dy3dy1

dy2dx2dx1


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

dV = dx 1 dx 2 dx 3 d ̂V ≈ dy 1 dy 2 dy 3

Relative change of volume (for small deformations e kk 1):

d ̂V − dV

dV =

d ̂V

dV − 1 ≈

dy 1

dx 1

dy 2

dx 2

dy 3

dx 3 − 1

= (1 + e 11)(1 + e 22)(1 + e 33) − 1 ≈ e 11 + e 22 + e 33

Meaning of trace of e ij : relative change of volume



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5. How to transform areas d S0 → d S?

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

dy i = ∂ y i

∂ x j dx j

we know how to transform volumes:

dV = J ·dV 0 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 d x, ie. any d y):

dS i dy i = dV = J · dV 0 = J · dS 0 j dx

j = J · dS 0 j ∂ x j

∂ y i dy i


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