II. Kinematics in curvilinear coordinates Continuum mechanics

Continuum mechanics
II. Kinematics in curvilinear coordinates

Aleˇs Janka

office Math 0.107
[email protected]
http://perso.unifr.ch/ales.janka/mechanics

December 22, 2010, Universit´e de Fribourg

Aleˇs Janka II. Kinematics in curvilinear coordinates

1. Strain in cartesian coordinates (recapitulation)

Green strain tensor: Lagrange formulation

εij = 1 ∂ui + ∂uj + ∂uk ∂uk
2 ∂xj ∂xi ∂xi ∂xj

Cauchy strain tensor: linearized strain for small deformations

eij = 1 ∂ui + ∂uj
2 ∂xj ∂xi

Almansi strain tensor: Euler formulation

Eij = 1 ∂ui + ∂uj − ∂uk ∂uk
2 ∂yj ∂yi ∂yi ∂yj

Aleˇs Janka II. Kinematics in curvilinear coordinates

1. Strain in curvilinear coordinates

Green strain tensor: Lagrange formulation

εij = 1 ∇j ui + ∇i uj + ∇i uk · ∇j uk
2

Cauchy strain tensor: linearized strain for small deformations

eij = 1 (∇j ui + ∇i uj )
2

Aleˇs Janka II. Kinematics in curvilinear coordinates

2. Example of using curvilinear coordinates

A rotational cylinder is being deformed into a rotational
hyperboloid. Calculate the Cauchy strain tensor.

It’s advantageous to use the cylindrical coordinates: 

ξ1 cos ξ2 y(x) = ξξ11 cos ξ2 − ξ3 sin ξ2 
x = ξ1 sin ξ2  Z

ξ3 −→ sin ξ2 + ξ3 cos ξ2
Z

ξ3

Aleˇs Janka II. Kinematics in curvilinear coordinates

2. Example of using curvilinear coordinates

Advantages of using curvilinear coordinates:
Simpler analytical formulae for particular deformation modes
and particular geometries
Better intuitive understanding of deformation modes
Particularly useful for shells and membranes or anisotropic
materials

Remember the inflated baloon demonstration?

tr r0
σ r

p

Aleˇs Janka II. Kinematics in curvilinear coordinates

3. Cauchy strain in cylindrical coordinates

Cauchy strain in curvilinear coordinates:

eij = 1 (∇i uj + ∇j ui )
2

Covariant derivative:

∇j ui = ∂ui − Γij u
∂ξj

Cylindrical coordinates:

ξ1 cos ξ2 1 0 0

x(ξ1, ξ2, ξ3) =  ξ1 sin ξ2  , [gij ] = 0 ξ1 2 0
ξ3 0 0 1

Christoffel symbols of 2nd kind: for cylindrical coordinates

Γ122 = −ξ1 , Γ212 = Γ221 = 1 , Γij = 0 otherwise.
ξ1

Aleˇs Janka II. Kinematics in curvilinear coordinates

3. Cauchy strain in cylindrical coordinates

e11 = ∇1u1 = ∂u1
∂ξ1

»–
1 1 ∂u2 ∂u1
e12 = 2 (∇1u2 + ∇2u1) = 2 ∂ξ1 − Γ221u2 + ∂ξ2 − Γ122u2

»–
1 ∂u2 ∂u1 2
= 2 ∂ξ1 + ∂ξ2 − ξ1 u2

»–
1 1 ∂u3 ∂u1
e13 = 2 (∇1u3 + ∇3u1) = 2 ∂ξ1 + ∂ξ3

e22 = ∇2u2 = ∂u2 − Γ122 u1 = ∂u2 + ξ1 u1
∂ξ2 ∂ξ2

1 1 »–
2 2 ∂u3 ∂u2
e23 = (∇2u3 + ∇3u2) = ∂ξ2 + ∂ξ3

e33 = ∇3u3 = ∂u3
∂ξ3

Note that physical units of eij are quite inhomogeneous here!

Aleˇs Janka II. Kinematics in curvilinear coordinates

3. Cauchy strain in cylindrical coordinates

Non-homogeneity of physical units for eij and ui

Units of cylindrical coordinates: ξ1 in [m], ξ2 in [rad], ξ3 in [m].

Covariant basis: gi = ∂x :
∂ξi

 ξ2 −ξ1 sin ξ2  
cos 0

g1 = sin ξ2  , g2 =  ξ1 cos ξ2 , g3 = 0

0 01

in [1] in [m] in [1]

Contravariant basis:  sin ξ2 
 cos ξ2
− 1
ξ1

g1 = g1 , g2 = 1 , g3 = g3
ξ1

in [1] 0 in [1]

in [1/m]

Aleˇs Janka II. Kinematics in curvilinear coordinates

3. Cauchy strain in cylindrical coordinates

Non-homogeneity of physical units for eij and ui

Units for ui and ui : displacement u = ui gi = ui gi should be in [m]:

coordinate its unit coordinate its unit
[m]
u1 [1] u1 [m]
u2 [m] u2 [m2]
u3 u3
[m]

Hence, units for eij :

coordinate its unit coordinate its unit
[1] [m2]
e11 [m] e22
e12 [1] e23 [m]
e13 e33
[1]

Correction of unit inhomogeneity:

introduction of physical components e(ij) and u(i) by:
eij = √gii · gjj · e(ij) and ui = √gii · u(i)

Aleˇs Janka II. Kinematics in curvilinear coordinates

3. Cauchy strain in cylindrical coordinates

Transforming covariant components to physical components

For cylindrical coordinates:

e11 = e(11) e21 = ξ1 · e(21) e31 = e(31)

e12 = ξ1 · e(12) e22 = (ξ1)2 · e(22) e32 = ξ1 · e(32)
ξ1 · e(23)
e13 = e(13) e23 = e33 = e(33)

u1 = u(1) → ∂u1 = ∂ u(1)
∂ξj ∂ξj

u2 = ξ1 · u(2) → ∂u2 = u(2) + ξ1 ∂ u(2) , ∂u2 = ξ1 ∂ u(2)
∂ξ1 ∂ξ1 ∂ξ2 ∂ξ2

∂u2 = ξ1 ∂ u(2)
∂ξ3 ∂ξ3

u3 = u(3) → ∂u3 = ∂ u(3)
∂ξj ∂ξj

Physical components in cylindrical coordinates usually written

u(1) = ur , u(2) = uθ , u(3) = uz
e(23) = eθz
e(11) = err , e(12) = erθ , ...

Aleˇs Janka II. Kinematics in curvilinear coordinates

3. Cauchy strain in cylindrical coordinates

Transforming covariant components to physical components

err = ∂ur
∂r

1 „«
2 ∂uθ 1 ∂ur uθ
er θ = ∂r + r ∂θ − r

„«
1 ∂uz ∂ur
erz = 2 ∂r + ∂z

eθθ = 1 ∂uθ + ur
r ∂θ r

1 „«
2 1 ∂uz ∂uθ
eθz = r ∂θ + ∂z

ezz = ∂uz
∂z

Aleˇs Janka II. Kinematics in curvilinear coordinates

4. Back to cylinder → hyperboloid

A rotational cylinder is being deformed into a rotational
hyperboloid. Calculate the Cauchy strain tensor.

Use the cylindrical coordinates:
 −ξξ11
ξ1 cos ξ2 − ξ3 sin ξ2 cos ξ2   −ξ1 ξ3 sin ξ2 
Z ξ2   ξ1 Z cos ξ2
u = y−x = ξ1 sin
sin ξ2 + ξ3 cos ξ2 ξ3 = ξ3
Z Z

ξ3 0

Aleˇs Janka II. Kinematics in curvilinear coordinates

4. Back to cylinder → hyperboloid

 −ξ1 ξ3 sin ξ2   2  − 1  
 ξ1 Z cos ξ2 cos ξ1 sin ξ2 0
ξ ξ3  cos ξ2 +0
ξ3 Z 0 0
u = Z = 0  sin ξ2  +(ξ1)2 1
ξ1 1
0
0

g1 g2 g3

Hence

u1 = u3 = 0 , u2 = (ξ1)2 ξ3 and u(2) = uθ = ξ1 ξ3
Z Z

Resulting Cauchy strain:

err = erθ = erz = eθθ = ezz = 0 and eθz = r
2Z

ie. pure shear (ie. distortion of angles) in the (θ, z) tangent-plane.

Aleˇs Janka II. Kinematics in curvilinear coordinates

5. Yet another cylinder → hyperboloid example

But different from the previous one!

A rotational cylinder is being deformed into a rotational

hyperboloid in the following way (in cylindrical coordinates):

2
 1 cos ξ2 y(x) = ξξ11 1+ ξ3 2 cos ξξ22
ξ 1+ Z sin

x = ξ1 sin ξ2  −→ ξ3
Z
ξ3

ξ3

The resulting shape is the same, but the deformation tensor
is different! Why?

Aleˇs Janka II. Kinematics in curvilinear coordinates