3d and planar constitutive relations a school on … on mechanics/ppt... · 3d and planar...

3D and Planar Constitutive Relations

A School on Mechanics of Fibre Reinforced Polymer

Composites

Knowledge Incubation for TEQIP

Indian Institute of Technology Kanpur

PM Mohite

Department of Aerospace Engineering

Indian Institute of Technology Kanpur

22-25 January 2017

3D Constitutive Relations

• Generalized Hooke’s law of the proportionality of stress and strain:

Each of the six component of the stress at any point is a linear function

of the six components of strain at that point.

• Concept of initial state

• Loading under two situations:

- Isothermal and Reversible;

- Adiabatic and Reversible.

• Stress components are the partial differential coefficients of a function (W) of

the strain-components.

Generalized Hooke’s Law:

1

2ij

ij ji

W Wσ

ε ε

∂ ∂= +

∂ ∂

• Form of the Strain Energy Density Function (W):

Homogeneous quadratic function of the strain components.

• W is invariant

• and are tensors.

• W is taken to be zero when body is in the initial state in which are zero.

Then, constant is zero.


1constant

2ij ij ijkl ij klW C Cε ε ε= + +

ijC ijklC

ijε

• For unstrained and unstressed body, are zero.

This leads to

and

is a fourth order (stiffness) tensor/matrix.


ijC

, , , 1, 2,3ij ijkl klC i j k lσ ε= =

1

2ijkl ij klW C ε ε=

ijklC

( )4

3 81 independent constants!=

• Simply, you can view this as if you have a vector of 9 stress components which

is related to a vector of 9 strain components through a matrix of 9x9!


{ } [ ] { }9 1 9 19 9

Cσ ε× ××

=

[ ]

1111 1112 1113 1121 1122 1123 1131 1132 1133

1211 1212 1213 1221 1222 1223 1231 1232 1233

1311 1312 1313 1321 1322 1323 1331 1332 1333

2111 2112 2113 2121 2122 2123 2131 2132 2133

2211 2212 2213 22

C C C C C C C C C

C C C C C C C C C

C C C C C C C C C

C C C C C C C C C

C C C C C= 21 2222 2223 2231 2232 2233

2311 2312 2313 2321 2322 2323 2331 2332 2333

3111 3112 3113 3121 3122 3123 3131 3132 3133

3211 3212 3213 3221 3222 3223 3231 3232 3233

3311 3312 3313 3321 3322 3323 3331 33

C C C C C

C C C C C C C C C

C C C C C C C C C

C C C C C C C C C

C C C C C C C C 32 3333C

• Stress symmetry:

• six independent ways to express when i and j are taken together and still 9

ways to express k and l taken together.

Stress Tensor Symmetry:

jij iσ σ=

( )0 0 ijij kl jiklj kli C Cσ εσ −− ⇒ ==

ij ijkl klCσ ε= ji jikl klCσ ε=

ijkl jiklC C=

6 9 54 independent constants!× =

• Simply, you can view this as if you have a vector of 6 stress components which

is related to a vector of 9 strain components through a matrix of 6x9!

Stress Tensor Symmetry:

{ } [ ] { }6 1 9 16 9

Cσ ε× ××

=

[ ]

1111 1112 1113 1121 1122 1123 1131 1132 1133

2211 2212 2213 221 2222 2223 2231 2232 2233

3311 3312 3313 3321 3322 3323 3331 3332 3333

2311 2312 2313 2321 2322 2323 2331 2332 2333

1311 1312 1313 132

C C C C C C C C C

C C C C C C C C C

C C C C C C C C CC

C C C C C C C C C

C C C C

=

1 1322 1323 1331 1332 1333

1211 1212 1213 1221 1222 1223 1231 1232 1233

C C C C C

C C C C C C C C C

• Strain symmetry:

• six independent ways to express when i and j are taken together and 6 ways to

express k and l taken together.

Stress and Strain Tensor Symmetry:

jij iε ε=

( )0 0 ijij kl ijlki klj C Cσ εσ −− ⇒ ==

ij ijkl klCσ ε= ij ijlk lkCσ ε=

ijkl ijlkC C=

6 6 36 independent constants!× =

• Or simply, you can view this as if you have a vector of 6 stress components

which is related to a vector of 6 strain components through a matrix of 6x6!


{ } [ ] { }6 1 6 16 6

Cσ ε× ××

=

[ ]

1111 1122 1133 1123 1113 1112

2211 2222 2233 2223 2213 2212

3311 3322 3333 3323 3313 3312

2311 2322 2333 2323 2312 2312

1311 1322 1333 1323 1313 1312

1211 1222 1233 1223 1212 1212

C C C C C C

C C C C C C

C C C C C CC

C C C C C C

C C C C C C

C C C C C C

=

• In other words,


1111 1122 1133 1123 1113 111211

2211 2222 2233 2223 2213 221222

3311 3322 3333 3323 3313 331233

2311 2322 2333 2323 2312 231223

1311 1322 1333 1323 1313 131213

1211 12212

C C C C C C

C C C C C C

C C C C C C

C C C C C C

C C C C C C

C C

σ

σ

σ

σ

σ

σ

=

11

22

33

23

13

2 1233 1223 1212 1212 12C C C C

ε

ε

ε

ε

ε

ε

• Using Voigt notation - a way to represent a symmetric tensor by reducing its

order

• For stress components:

• Strain Components:

Voigt Notation:

11 12 13

22 23

33

σ σ σ

σ σ

σ

1

2

3

4

6 5

{ } { }11 22 33 23 13 12 1 2 3 4 5 6σ σ σ σ σ σ σ σ σ σ σ σ=

{ } { }11 22 33 23 13 12 1 2 3 4 5 62 2 2ε ε ε ε ε ε ε ε ε ε ε ε=

• Instead of writing C as a fourth order tensor, written as a second order tensor

and stress and strains tensors are written as vectors !

Stress and Strain Tensor Symmetries:

1 11 12 13 14 15 16 1

2 21 22 23 24 25 26 2

3 31 32 33 34 35 36 3

4 41 42 43 44 45 46 4

5 51 51 53 54 55 56 5

6 61 62 63 64 65 66 6

C C C C C C

C C C C C C

C C C C C C

C C C C C C

C C C C C C

C C C C C C

σ ε

σ ε

σ ε

σ ε

σ ε

σ ε

=

• Existence of W :

Hyperelastic materials

Invariant

Positive Definite

Existence of W:

( ) 0ij ji i jC C ε ε− =

1

2ij i jW C ε ε=

1

2ij j iW C ε ε=

ij jiC C=

•

Existence of W:

1 11 12 13 14 15 16 1

2 22 23 24 25 26 2

3 33 34 35 36 3

4 44 45 46 4

5 55 56 5

6 66 6

C C C C C C

C C C C C

C C C C

C C C

C C

C

σ ε

σ ε

σ ε

σ ε

σ ε

σ ε

=

21 independent constants!

• Stress symmetry

Minor Symmetries

• Strain symmetry

• Existence of W:

Major Symmetry

Symmetries:

ijkl jiklC C=

ijkl jiklC C=

ijkl klijC C=

• Transformations

• Prime denotes the transformed coordinates.

• aij denotes the components of a transformation matrix

Transformations:

• Further reduction in constants obtained by material symmetry

• Symmetry Definition: Any geometrical figure which can be brought to

coincidence with itself, by an operation which changes the position of any of its

points, is said to possess “symmetry”.

• Rotation and Reflection

Material Symmetry:

• Quadratic in strain components:

• Note that the strains used are engineering strains.

Form of W:

• One Plane of Material Symmetry: Monoclinic Materials

Material Symmetry:

• Transformation of axes:

• Transformation matrix:

• Transformation of strains:

Material Symmetry: One Plane of Material Symmetry

• Transformation of stresses:


• Transformation of stiffness:

• Comparison of stress components:


'ij ijC C=

14 15 0C C⇒ = =

Similarly,

13 independent constants


Second Approach: Invariance of W

W for

Hyper-

elastic

material


Second Approach: Invariance of W

• For W to be invariant the product terms

must vanish, that is,


1 4 1 5 2 4 2 5 3 4 3 5 4 6 5 6, , , , , , ,ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε ε

14 15 24 25 34 35 46 56, , , , , , , are zeroC C C C C C C C

• Two Orthogonal Planes of Material Symmetry: Orthotropic Materials

• Transformation of axes:

Transformation matrix:

Transformation of strains:

Material Symmetry: Two Orthogonal Planes of Material Symmetry

• Transformation of stresses:

• Transformation of stiffness:


'ij ijC C=

• Comparison of stresses:

• Similarly,


16 0C⇒ =

• Stiffness Tensor:

9 independent constants


• Second approach: Invariance of w

• W for monoclinic material:

• or form of W:


2 2 2 211 1 22 2 33 3 44 4

2 255 5 66 6 12 1 2 13 1 3

16 1 6 23 2 3 26 2 6 36 3 6

45 4 5

1

2

C C C C

C C C CW

C C C C

C

ε ε ε ε

ε ε ε ε ε ε

ε ε ε ε ε ε ε ε

ε ε

+ + + + + + + +

= + + + +

2 2 2 2 2 21 2 3 4 5 6

1 2 1 3 1 6

2 3 2 6 3 6 4 5

, , , , , ,

, , ,

, , ,

W W

ε ε ε ε ε ε

ε ε ε ε ε ε


=

• For W to be invariant under the strain transformations

• the product terms

must vanish, which is possible when

W for orthotropic material:


1 6 2 6 3 6 4 5, , ,ε ε ε ε ε ε ε ε

16 26 36 45, , , are zeroC C C C

2 2 2 2 2 211 1 22 2 33 3 44 4 55 5 66 6

12 1 2 13 1 3 23 2 3

1

2

C C C C C CW

C C C

ε ε ε ε ε ε

ε ε ε ε ε ε

+ + + + + +=

+ +

• When material has two orthogonal planes of symmetry then it also symmetric

about a plane which is mutually orthogonal earlier two planes!

•Such materials are called Orthotropic Materials.

• Select now the remaining plane x1-x3 as the third orthogonal plane of material

symmetry in addition to earlier two planes

• Follow the same procedure, either comparing the stresses or invariance of W

• There will be no change in the final stiffness tensor. Number of independent

constants will be still 9!


• When material has two orthogonal planes of symmetry then it also symmetric

about a plane which is mutually orthogonal earlier two planes!

• Alternately:

• Now select the plane x1-x3 as the second orthogonal plane of material

symmetry in addition to x1-x2 plane


• Isotropic behaviour of UD lamina in the cross-sectional plane (perpendicular

to fibre s’ length)


Isotropy in a Plane:

• Transformation of strains:



2 2 2 2 2 21 2 3 4 5 6

1 2 1 3 1 6

2 3 2 6 3 6 4 5

, , , , , ,

, , ,

, , ,

W W

ε ε ε ε ε ε

ε ε ε ε ε ε


=

W for monoclinic material: For invariance

1 6 2 6 3 6 4 5, , , 0ε ε ε ε ε ε ε ε =

16 26 36 45, , , are zeroC C C C

• Trigonometric identities for the strains:

• Form of W:


( ) ( )

( ) ( ) ( ) ( )

' '22 33 22 33

22 ' ' '22 33 23 22 33 23

2 222 ' '12 13 12 13

,

,

ε ε ε ε

ε ε ε ε ε ε

ε ε ε ε

+ = +

− = + −

+ = +

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

2 22

11 22 33 22 33 23 12 13

2 2 2' ' ' ' ' ' ' '11 22 33 22 33 23 12 13

, , ,

, , ,

W W

W W



= + − +

= + − +

• Strain energy density function for orthotropic material:

• rearranging


211 11 12 11 22 13 11 33

2 2 222 22 33 33 23 22 33 44 23

2 255 13 66 12

2 2

+ 2 4

+ 4 4

W C C C

C C C C

C C

ε ε ε ε ε

ε ε ε ε ε

ε ε

= + +

+ + +

+

( )211 11 11 12 22 13 33

2 255 13 66 12

2 2 222 22 33 33 23 22 33 44 23

2

+ 4 4 +

+ 2 4

W C C C

C C

C C C C

ε ε ε ε

ε ε

ε ε ε ε ε

= + +

+

+ + +

• In the second bracket, we take

• In the third bracket, we take

• Rearrange the last bracket with and unchanged


( )211 11 11 12 22 13 33

2 255 13 66 12

2 2 222 22 33 33 23 22 33 44 23

2

+ 4 4 +

+ 2 4

W C C C

C C

C C C C

ε ε ε ε

ε ε

ε ε ε ε ε

= + +

+

+ + +

12 13C C=

55 66C C=

( )

2 2 222 22 33 33 23 22 33 44 23

2 222 22 33 22 22 33 23 22 33 44 23

2 4

2 2 4

C C C C

C C C C

ε ε ε ε ε

ε ε ε ε ε ε ε

+ + +

= + − + +

22 33C C= 23C

• Rearrange the last bracket further as

• we need


( ) ( )

2 2 222 22 33 33 23 22 33 44 23

2 222 22 33 22 23 22 33 44 23

2 4

2 2

C C C C

C C C C

ε ε ε ε ε

ε ε ε ε ε

+ + +

= + − − −

22 2344

2

C CC

−=

• Stiffness tensor

• 5 independent constants

• Such materials are called Transversely isotropic materials.

• Define: where


11 12 12

22 23

22

22 23

66

66

0 0 0

0 0 0

0 0 0

0 02

0

ij

C C C

C C

C

C C C

Sym C

C

= −

Transverse Isotropy with an Additional Orthogonal Plane:

• Consider isotropy in x1-x2 plane as well

• strain

transformation


• Trigonometric identities:

• Form of W:

( ) ( )

( ) ( ) ( ) ( )

' '11 22 11 22

22 ' ' '11 22 12 11 22 12

2 22 2 ' '13 23 13 23

,

,

ε ε ε ε

ε ε ε ε ε ε

ε ε ε ε

+ = +

− = + −

+ = +

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

2 22

11 22 33 11 22 12 13 23

2 2 2' ' ' ' ' ' ' '11 22 33 11 22 12 13 23

, , ,

, , ,

W W

W W



= + − +

= + − +

• In the second bracket, we take

• In the third bracket, we take

• Rearrange the last bracket with and unchanged


12 23C C=

11 22C C= 12C

• Two independent constants !

• Define:

where

Isotropy:

11 12 12

11 12

11

11 12

11 12

11 12

0 0 0

0 0 0

0 0 0

0 02

02

2

ij

C C C

C C

C

C CC

C C

C CSym

− = − −

• Generalized Hooke’s Law: 81 independent constants

• Stress tensor symmetry: 54 independent constants

• Strain tensor symmetry: 36 independent constants

• Existence of W (Hyperelastic/Aelotropic): 21 independent constants

• Existence of one plane of material symmetry: 13 independent constants

• Existence of two/three mutually perpendicular planes of symmetry:

(Orthotropic Material) 9 independent constants

• One plane of isotropy: 5 independent constants

• Two/three/infinite planes of isotropy: 2 independent constants

3D Constitutive Relations: Quick Review

3D Constitutive Relations for

Orthotropic Materials

• Strain-stress Relations

• Normal stresses and

strains

Constitutive Relations for Orthotropic Materials:

• Shear stresses and strains

• Poisson’s ratio:

(no sum over i, j)

• In general,


ij jiν ν≠

• Determination of Engineering Constants:


• Matrix-vector form:

where,


Always work with compliance

tensor.

It is easy to remember.

• Important Relations:

• Reciprocal relation

where,


• Stiffness Relations:

where,


• For strain energy to be positive definite both Compliance and Stiffness

tensors must be positive definite.

• Strain energy to be positive definite the diagonal entries of the Compliance

tensor must be positive.

• Similarly, the diagonal entries of the Stiffness tensor must be positive

• and the determinant must also be positive

Constraints on Engineering Constants:


• Constraint on Poisson’s ratio:

From constraint on determinant:

• For transverse isotropic material:

with

we get, and

Finally, leads to the condition

For isotropic materials:


Constitutive Relations: Transformations

• 123 – Principal material directions

• xyz – global reference directions

Transformation matrix

for rotation about z-axis:

• Stress transformation:

For example,

that is,


Constitutive Relations: Stress Transformations

'ij ki lj kla aσ σ=

Constitutive Relations: Stress and Strain Transformations


where, m=cosθ and n=sinθ

• Stress transformation:

For example,

that is,

Now using


'ij ki lj kla aε ε=


• Strain transformation:


• Stress Transformation:

• Strain Transformation:


• From the first principles:

Writing in global coordinates

leads to

and

Constitutive Relations: Stiffness Transformations

• Results in monoclinic

behaviour!

Constitutive relations:


• From the first principles:

Writing in global coordinates

Or

and

Constitutive Relations: Compliance Transformations

• The transformed Stiffness and Compliance tensors are symmetric!

• From invariance of W one can show


[ ] [ ] [ ] [ ]1 1

1 2 2 1 and T T

T T T T− −

= =

• Coefficient of thermal expansion is different in 3 directions!

Constitutive Relations: Thermal Effects

• Thermal strain in principal material directions

where,

These strains will not produce stresses unless restricted!

Transforming strains into

global coordinates

We get,


That is,

where,

and


Total strains:

Mechanical strains:

Thus,

gives the thermo-elastic

constitutive equations as

Constitutive Relations: Thermo-Elastic Equations

Stresses in global direction:

where,

Constitutive Relations: Thermo-Elastic Equations

Hygral strains:

where,

are coefficient s of hygral expansion

Constitutive Relations: Hygral Effects

Constitutive Relations: Hygro-Thermal Effects

Total strains:

with,

The hygro-thermo-elastic

constitutive equation:

2D Constitutive Relations for

Orthotropic Materials

Constitutive relation in 3D:

Transverse stresses are zero:

leads to

Constitutive Relations: Planar Equations

0zz xz yzσ τ τ= = =

44 45

45 55

0

0

yz yz xz

xz xz yz

S S

S S

γ τ τ

γ τ τ

= + =

= + =

Transverse normal strain:

Constitutive Relations: Transverse Strain

13 23 36 0zz xx yy xyS S Sε σ σ τ= + + ≠

Constitutive Relations: In-Plane Stresses

Constitutive relation in 3D (Principal Directions):

Transverse stresses are zero:

leads to

Constitutive Relations: Planar Equations

Transverse normal strain:

that is,

Therefore, for planar case

Planar Relations: Principal Directions

33 13 11 23 22 36S S Sε σ σ= + + 12 0τ ≠

33 13 11 23 22 0S Sε σ σ= + ≠

Transverse normal strain from stiffness relations:

Transverse normal stress:

gives


Stresses in principal directions:

Putting


33 11 22 in terms of ,ε ε ε


And can be written in a form as

Qij and Cij are not same

Inverse f orm:

Sij are same as in 3D relations


Reduced Stiffness Matrix:

Stiffness

Compliance in terms

of engineering

constants



Strains in principal directions:

Planar Relations: Transformation of Stresses and Strains


Stresses in principal directions

Stresses in global directions:

Planar Relations: Hygro-thermo-elastic Relations

Thank You

3d and planar constitutive relations a school on … on mechanics/ppt... · 3d and planar...

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