Plain Strainn

8/17/2019 Plain Strainn

1/73

Plain Strain

We will derive the transformation equations that relate the strains in

inclined directions to the strain in the reference directions.

State of plain strain - the only deformations are those in the xy plane,

i.e. it has only three strain components ε x , ε y and γ xy.

Plain stress is analogous to plane stress, but under ordinary conditions

they do not occur simultaneously, except when σ x = -σ y and when ν = 0

Strain components ε x , εy, and γxy in the xy plane (plane strain).


2/73

Comparison of plane stress and plane strain.


3/73

Transformation Equations for Plain Strain

Assume that the strain ε x , ε y and γ xy associated with the xy plane areknown. We need to determine the normal and shear strains (ε x1 and

γ x1y1) associated with the x 1 y1 axis. ε y1 can be obtained from the

equation of ε x1 by substituting θ + 90 for θ.


4/73

For an element of size δ x δ y

In the x direction, the strain ε x

produces an elongation ε x δ x .The diagonal increases in length by

ε x δ x cos θ.

In the y direction, the strain ε y produces an

elongation ε y δ y. The diagonal increases in

length by ε y δ y sin θ.

The shear strain γ xy produces a distortion.

The upper face moves γ xy δ y. This

deformation results in an increase of the

diagonal equal to: γ xy δ y cos θ


5/73

The total increaseΔd

of the diagonal is the sum of the preceding three

expressions, thus:

Δ

d =

ε

x

δ

x cos

θ

ε

y

δ

y sin

θ γ

xy

δ

y cos

θ

The normal strainε

x1

in thex

1

direction is equal to the increase in

length divided by the initial length δs of the diagonal.

ε

x1

= Δd / ds = ε

x

cos θ δx/δs ε

y

sin θ δy/δs γ

xy

cos θ δy/δs

Observing thatδx/δs = cos θ

andδy/δs = sin θ

θ θ γ

θ ε θ ε ε

θ θ γ θ ε θ ε ε

cossin22

sincos

cossinsincos

22

1

22

1

⎟ ⎠

⎞⎜⎝

⎛ ++=++=

XY Y X X

XY Y X X


6/73

γ

Shear Strain γ x1y1 associated with x 1 y1 axes.

This strain is equal to the decrease in angle between lines in the material

that were initially along the x 1 and y1 axes. Oa and Ob were the linesinitially along the x 1 and y1 axis respectively. The deformation caused

by the strains ε x , ε y and γ xy caused the Oa and Ob lines to rotate and

angle α and β from the x 1 and y1 axis respectively. The shear strain γ x1y1is the decrease in angle between the two lines that originally were at

right angles, therefore, γ x1y1 = α β.


7/73

The angle can be found from

the deformations produced by

the strains ε x , ε y and γ xy . Thestrains ε x and γ xy produce a cw-

rotation, while the strain ε y

produces a ccw-rotation.

Let us denote the angle of rotation produced by ε x , ε y and γ xy as α1 ,

α2 and α3 respectively. The angle α1 is equal to the distance ε x δ x

sinθ divided by the length δs of the diagonal:α1 = ε x sinθ dx/ds α2 = ε y cosθ dy/ds α3 = γ xy sinθ dy/ds

Observing that dx/ds = cos θ and dy/ds = sin θ. The resulting ccw-

rotation of the diagonal is

α

= - α1 + α2 - α3 = - ( ε x – ε y ) sinθ cosθ - γ xy sin

2θ


8/73

The rotation of line Ob which initially was at at 90o to the line Oa can

be found by substituting θ +90 for θ in the expression for α.

Because βis positive when clockwise. Thus

β = ( ε x – ε y ) sin( θ + 90) cos( θ + 90) + γ xy sin2( θ +90)

β = - ( ε x – ε y ) sinθ cosθ + γ xy cos2

θ

Adding α and β gives the shear strain γ x1y1

γ

x1y1 = + β = - 2(εx – εy) sinθ cosθ + γxy (cos2

θ - sin2

θ)To put the equation in a more useful form:

( )θ θ γ

θ θ ε θ θ ε

γ 2211sincos2cossincossin2 −++−=

XY Y X

Y X

θ θ γ

θ ε θ ε ε cossin22

sincos 221 ⎟ ⎠

⎞⎜⎝

⎛ ++= XY Y X X

θ θ γ θ ε θ ε ε cossin22

cossin 221 ⎟ ⎠ ⎞⎜

⎝ ⎛ −+= XY Y X Y


9/73

( )⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥

⎥

⎦

⎤

⎢⎢

⎢

⎣

⎡

−−−=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

2sincoscossincossincossin2cossin

cossin2sincos

2

22

22

22

11

1

1

XY

Y

X

Y X

Y

X

γ ε

ε

θ θ θ θ θ θ θ θ θ θ

θ θ θ θ

γ ε

ε

[ ]

[ ]

⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢⎢

⎢

⎣

⎡

×=

⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢⎢

⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

×=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

22

22

11

1

1

1

11

1

1

Y X

Y

X

XY

Y

X

XY

Y

X

Y X

Y

X

T

T

γ

ε

ε

γ

ε

ε

γ

ε

ε

γ

ε

ε [ ]

( )⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

θ θ θ θ θ θ

θ θ θ θ

θ θ θ θ

22

22

22

sincoscossincossin

cossin2cossin

cossin2sincos

T

[ ] Tensor Strain

zz yz xz

zy yy xy

zx yx xx

_

21

21

2

1

2

12

1

2

1

=

⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢⎢

⎣

⎡

=

ε γ γ

γ ε γ

γ γ ε

ε


10/73

Transformation Equations for Plain Strain

Using known trigonometric identities, the transformation equations

for plain strain becomes:

These equations are counterpart of the equations for plane stresswhere ε x1 , ε x , γ x1y1 and γ xy correspond to σ x1 , σ x , τ x1y1 and τ xyrespectively. There are also counterparts for principal stress and

Mohr’s circle.ε

x1 + ε y1 = ε x + ε y

( ) ( )

( )θ

γ θ

ε ε γ

θ γ

θ ε ε ε ε

ε

2cos2

2sin22

2sin

2

2cos

2211

1

XY Y X Y X

XY Y X Y X X

+−

−=

+−

++

=


11/73

Principal Strains

The angle for the principalstrains is

The value for the principalstrains are

Y X

XY P

ε ε

γ θ

−

=2tan

( )

( ) 222

22

1

222

222

⎟

⎠

⎞⎜

⎝

⎛ +⎟

⎠

⎞⎜

⎝

⎛ −−

+=

⎟

⎠

⎞⎜

⎝

⎛ +⎟

⎠

⎞⎜

⎝

⎛ −+

+=

XY Y X Y X

XY Y X Y X

γ ε ε ε ε ε

γ ε ε ε ε ε

Maximum Shear

The maximum shear

strains in the xy planeare associated with

axes at 45o to the

directions of the principal strains:

( )

( )22

or222

21

21

22

ε ε γ

ε ε γ γ ε ε γ

−

=

−=⎟ ⎠ ⎞⎜

⎝ ⎛ +⎟

⎠ ⎞⎜

⎝ ⎛ −+=

MAX

MAX XY Y X MAX


12/73

Mohr’s Circle

for Plane Strain


13/73

Example

An element of material in plane

strain undergoes the followingstrains: ε x =340x10

-6

ε y=110x10-6

γ

xy=180x10-6

Determine the following: (a)

the strains of an element

oriented at an angle θ = 30o

;(b) the principal strains and (c)

the maximum shear strains.

( ) ( )

( )θ

γ θ

ε ε γ

θ γ

θ ε ε ε ε

ε

2cos2

2sin22

2sin22cos22

11

1

XY Y X Y X

XY Y X Y X X

+−

−=

+−

++

=Solution


14/73

Then

εx1 = 225x10-6 + (115x10-6) cos 60o + (90x10-6) sin 60o = 360x10-6

½ γx1y1 = - (115x10-6) (sin 60o ) + ( 90x10-6)(cos 60o) = - 55x10-6

Therefore γx1y1 = - 110x10-6

The strain εy1 can be obtained from the equation εx1 + εy1 = εx+ εyεy1 = (340 + 110 -360)10-6 = 90x10-6

( )

( )22

2

22

1

222

222

⎟ ⎠ ⎞⎜

⎝ ⎛ +⎟

⎠ ⎞⎜

⎝ ⎛ −−+=

⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ −+

+=

XY Y X Y X

XY Y X Y X

γ ε ε ε ε ε

γ ε ε ε ε ε

(b) Principal Strains

The principal strains are readily

determine from the following equations:

ε1 = 370x10-6

ε2 = 80x10-6


15/73

(c) Maximum Shear Strain

The maximum shear strain is calculated from the equation:

½ γmax = SQR[((εx – εy)/2)2 + ( ½ γxy)2] or γmax = (ε1 – ε2 )Then γmax = 290x10-6

The normal strains of this element is εaver = ½ (εx + εy) = 225x10-6


16/73

For 3-D problems

[ ]

⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢

⎣

⎡

=

z zy zx

yz y yx

xz xy x

ε γ γ

γ ε γ

γ γ ε

ε

2

1

2

12

1

2

12

1

2

1Which is a symmetrical matrix. As in the

case of stresses:

0

21

21

2

1

2

12

1

2

1

0

0

0

2

1

2

121

21

2

1

2

1

=

−

−

−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎢

⎣

⎡

−

−

−

ε ε γ γ

γ ε ε γ

γ γ ε ε

ε ε γ γ

γ ε ε γ

γ γ ε ε

z zy zx

yz y yx

xz xy x

z zy zx

yz y yx

xz xy x

m

l

k

321 ε ε ε 〉〉


17/73


18/73

Dilatation (Volume strain)

Under pressure: the volume will change

p

p p

p

V-ΔV z y x

V

V ε ε ε ++=

Δ=Δ

222

3

222

2

1

32

2

1

3

2222222

222

0

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⋅−⎟

⎠

⎞⎜⎝

⎛ ⋅−⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⋅−⎟⎟

⎠

⎞⎜⎜⎝

⎛ ⋅⎟

⎠

⎞⎜⎝

⎛ ⋅⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ⋅+⋅⋅=

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ −⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −⋅+⋅+⋅=

++=

=−⋅+⋅−

xy

z xz

y

yz

x

yz xz xy

z y x

yz xz xy

z x z y y x

z y x

I

I

I

I I I

γ ε

γ ε

γ ε

γ γ γ ε ε ε

γ γ γ ε ε ε ε ε ε

ε ε ε

ε ε ε

S i D i εεε ++Δ


19/73

Strain Deviator

33

z y x ε ε ε ++=Δ

Mean strain

It produces a volume change (not a shape change)

[ ]

⎥

⎥⎥⎥⎥

⎥

⎦

⎤

⎢

⎢⎢⎢⎢

⎢

⎣

⎡

Δ−

Δ−

Δ−

=

3

1

2

1

2

12

1

3

1

2

12

1

2

1

3

1

z zy zx

yz y yx

xz xy x

D

ε γ γ

γ ε γ

γ γ ε

ε

[ ]

Strain Deviator Matrix

⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢⎢

⎣

⎡

Δ−

Δ−

Δ−

=

3100

03

10

003

1

3

2

1

ε

ε

ε

ε D


20/73

Application : Strain Gauge and Strain Rosette


21/73

(Hooke’s Law)When strains are small, most of materials are

linear elastic.Tensile:

σ Ε ε

Shear: τ= G γ

Poisson’s ratio

Nominal lateral strain(transverse strain) z

z z

l

l

0

Δ=

x

x x

l

l

0

Δ−=

Poisson’s ratio: z

x

straintensile

strainlateral

ε

ε

ν −=−=

Relationships between Stress and Strain


22/73

Relationships between Stress and Strain

For isotropic materials

Elastic Stress-Strain Relationships

0

0

3

2

11

=

==

σ

σ

ε σ E

Principal Stresses Principal Strains

E

E

E

13

12

11

σ ν ε

σ ν ε

σ ε

−=

−=

=Uniaxial

An isotropic material has a stress-strain relationships that areindependent of the orientation of the coordinate system at a point.

A material is said to be homogenous if the material properties are

the same at all points in the body

Uniaxial Stresses


23/73

[ ]

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

xy

zx

yz

z

y

x

τ

τ

τ

σ σ

σ

σ

[ ]

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

=

xy

zx

yz

z

y

x

γ

γ

γ ε

ε

ε

ε

[ ]

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

0

0

0

0

0

xσ

σ

x x E ε σ =

E E E

x z

x y

x x

σ ν ε

σ ν ε

σ ε −=−==

[ ]

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

=

0

0

0

z

y

x

ε

ε

ε

ε ⎪

⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪

⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎢

⎣

⎡

−−

−−

−−

=

⎪

⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪

⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

0

0

0

0

0

100000

01

0000

001

000

0001

0001

0001

0

0

0

x

z

y

x

G

G

G

E E E

E E E

E E E

σ

υ υ

υ υ

υ υ

ε

ε

ε

Uniaxial Stresses



24/73

( )

( )

0

1

1

3

2

122

2

211

=

++

=+

+=

σ

ν

νε ε σ

ν

νε ε σ

E

E


E E

E E

21

3

122

211

σ

ν

σ

ν ε

σ ν

σ ε

σ ν

σ ε

−−=

−=

−= Biaxial

( ) ( )

( ) ( )

( ) ( )2

2133

2312

2

2

3211

21

1

211

21

1

ν ν

ε ε ν ν ε σ

ν ν ε ε ν ν ε σ

ν ν

ε ε ν ν ε σ

−−

++−=

−− ++−=

−−++−

=

E

E

E


E E E

E E E

E E E

213

3

3122

3211

σ

ν

σ

ν

σ

ε

σ ν σ ν σ ε

σ ν

σ ν

σ ε

−−=

−−=

−−=

Triaxial

1 ννTriaxial Stresses


25/73

[ ]

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

xy

zx

yz

z

y

x

τ

τ

τ

σ σ

σ

σ

[ ]

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

=

xy

zx

yz

z

y

x

γ

γ

γ ε

ε

ε

ε

1

1

1

z y x z

z y x y

z y x x

E E E

E E E

E E E

σ σ ν

σ ν

ε

σ ν

σ σ ν

ε

σ ν

σ ν

σ ε

+−−=

−+−=

−−=

[ ]

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

=

0

0

0

z

y

x

ε

ε

ε

ε

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎥⎥⎥⎥

⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎢⎢⎢⎢

⎢⎢

⎣

⎡

−−

−−

−−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

0

00

100000

01

0000

001000

0001

000

1

0001

0

00

z

y

x

z

y

x

G

G

G

E E E

E E E

E E E

σ

σ

σ

υ υ

υ υ

υ υ

ε

ε

ε

[ ]

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

=

0

0

0

z

y

x

σ

σ

σ

σ

Triaxial Stresses


26/73

For an isotropic material, the principal axes for stress and the

principal axes for strain coincide.

Y X

XY

ε ε

γ θ ε −

=2tan( )( )

G

E

xy

xy

y x y x

τ γ

σ σ ν ε ε

=

−−=− 1

1

( )( ) ( ) ( ) σ ε

θ σ σ

τ

ν

σ σ ν

τ

ε ε

γ θ 2tan

2

12

1

12tan =

−⋅

−=

−−

=−

= y x

xy

y x

xy

Y X

XY

G

E

E

G


27/73

Plane Stress ( )

( )

G

E

E

xy

xy

x y y

y x x

τ γ

νσ σ ε

νσ σ ε

=

−=

−=

1

1 ( )

0

0

==

+−=

zx

yz

y x z E

γ

γ

σ σ ν

ε

Plane Strain

( )( ) ( )[ ]

( )( ) ( )[ ] xy xy

x y y

y x x

G

E

E

γ τ

νε ε ν ν ν

σ

νε ε ν ν ν

σ

=

+−−+

=

+−−+

=

1211

1211 ( )( ) ( )

0

0

211

=

=

+−+=

xz

yz

y x z E

τ

τ

ε ε ν ν

ν σ

Tensors and Elasticity


28/73

Tensors

and

Elasticity Common misconception Cubic materials are isotropic, i.e. they have

the same

properties

in

every

direction.

Many

properties

are

isotropic

in cubic crystal, but elasticity, electrostriction and magnetostriction

are anisotropic even in cubic crystals.

Example turbine blade

(single crystal). Ni

based with

cuboidal

Ni3Ti intermetallic. It

shows variation in the

elastic constant with

the directions in the

material.

Some polycrystalline materials


29/73

Some polycrystalline materials

develop preferred orientations

during

processing.

They

will

show

a

degree of anisotropy that is

dependent on the degree of

preferred orientation or texture.

Tensor: A specific

type

of

matrix

representation

that

can

relate

the

directionality of either a material property (property tensors –

conductivity, elasticity) or a condition/state (condition tensors – stress,

strain).

Tensor of zero‐rank: scalar quantity (density, temperature).

Tensor of first‐rank: vector quantity (force, electric field, flux of atoms).

Tensor of

second

‐rank: relates

two

vector

quantities

(flux

of

atoms

with

concentration gradient).

Tensor third‐rank: relates vector with a second rank tensor (electric


30/73

Tensor third‐rank: relates vector with a second rank tensor (electric

field with strain in a piezoelectric material)

Tensor Fourth

‐rank: relates

two

second

rank

tensors

(relates

strain

and stress – Elasticity)

The key

to

understanding

property

or

condition

tensors

is

to

recognize

that tensors can be specified with reference to some coordinate system

which is usually defined in 3‐D space by orthogonal axes that obey a

right‐hand

rule.

Rotation Matrix and Euler Angles:Rotation Matrix and Euler Angles: Several schemes can be used to

produce a rotation matrix. The three Euler angles are given as three

counterclockwise rotations:

(a)A rotation about a z‐axis, defined as φ1 (b)A rotation about the new x‐axis, defined as Φ

(c)A rotation

about

the

second

z‐position, defined

as

φ2

The rotation matrix a is given by the matrix multiplication of the


31/73

g y p

rotation matrices of each individual rotations:

[ ]

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ΦΦ−Φ

ΦΦ+−Φ−−ΦΦ+Φ−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−×

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

ΦΦ−

ΦΦ×

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=⋅⋅= Φ

coscossinsinsin

cossincoscoscossinsincossincoscossinsinsinsincoscossincossinsincoscoscos

100

0cossin0sincos

cossin0

sincos0001

100

0cossin0sincos

11

221122112

221122112

11

11

22

22

12

φ φ

φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ

φ φ φ φ

φ φ φ φ

φ φ

a

aaaa

S hi P j i


32/73

Stereographic

Projection

Crystallographic directions, plane normals and

planes can be all represented in the stereographic

projection.

Locating a pole in a stereographic projection: 132001

132

⎤⎡


33/73

Locating a pole in a stereographic projection:

o

o

o

5.74141

132100cos

7.36141

132010cos

6.57141

132001cos

1-

1-

1-

=⎥⎦

⎤

⎢⎣

⎡ ⋅

=⎥⎦

⎤⎢⎣

⎡ ⋅

=⎥⎦

⎤⎢⎣

⎡ ⋅

Find the angle of the pole with the three

axes:

ExampleTh l ti hi b t i t ti d li d t i i t t


34/73

The relationship between orientation and applied stress is important

in describing the mechanical performance of many crystalline metals

and composites.

The relationship between applied stress and crystal direction is

essential in interpreting the microscopic deformation mechanisms

operating in

deforming

crystals.

[ ]

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡

−

=

100

030

002

T Consider a property‐tensor or a condition tensor T in the original {x y z} axes given by:

Find the tensor T’ , for the rotation

shown

below

from

the

initial

[100],

[010], [001] axes of a cubic crystal

o451 =φ [ ] ⎥⎥⎤

⎢⎢⎡

ΦΦ+−Φ−−ΦΦ+Φ−

= cossincoscoscossinsincossincoscossinsinsinsincoscossincossinsincoscoscos

221122112

221122112

a φ φ φ φ φ φ φ φ φ

φ φ φ φ φ φ φ φ φ


35/73

o

o

0

7.54

2 =

=Φ

φ

[ ]

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+−−−

+−

=

⎥⎥⎦⎢

⎢⎣ ΦΦ−Φ

7.54cos45cos7.54sin45sin7.54sin7.54sin45cos7.54cos45sin7.54cos

045sin45cos

][

7.54cos45cos7.54sin45sin7.54sin

0cos7.54sin0cos45cos7.54cos45sin0sin0cos45sin7.54cos45cos0sin

0sin7.54sin0sin45cos7.54cos45sin0cos0sin45sin7.54cos45cos0cos

][

coscossinsinsin 11

221122112

a

a

φ φ

⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢

⎣

⎡

−

−=

3

1

3

1

3

16

2

6

1

6

1

02

1

2

1

][a

[ ] [ ]T

aT aT ××='

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−

=

⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢

⎣

⎡

−

−

×

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

×

⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢

⎣

⎡

−

−=

33.165.1408.0

65.1167.0289.0

408.0289.05.2

3

1

6

2

0

3

1

6

1

2

13

1

6

1

2

1

100

030

002

3

1

3

1

3

16

2

6

1

6

1

02

1

2

1

]'[T

z y x σ νσ

νσ

ε =Isotropic Materials


36/73

E E E

E E E

E E E

z y x z

z y x y

x

σ σ ν

σ ν ε

σ ν

σ σ ν ε

ν ν ε

+−−=

−+−=

−−=

1

1

1

xy xy

zx zx

yz yz

G

G

G

τ γ

τ γ

τ γ

=

=

=

Isotropic Materials

xy zx yz z y x x S S S S S S τ τ τ σ σ σ ε 161514131211 +++++=


37/73

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

⎥⎥

⎥⎥⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢⎢

⎢⎢

⎣

⎡

=

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

S S S S S S

S S S S S S

S S S S S S S S S S S S

S S S S S S

S S S S S S

τ

τ

τ σ

σ

σ

γ

γ

γ ε

ε

ε

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

xy zx yz z y x y S S S S S S τ τ τ σ σ σ ε 262524232221 +++++=

xy zx yz z y x z S S S S S S τ τ τ σ σ σ ε 363534333231 +++++=

xy zx yz z y x yz S S S S S S τ τ τ σ σ σ γ 464544434241 +++++=

xy zx yz z y x zx S S S S S S τ τ τ σ σ σ γ 565554535251 +++++= xy zx yz z y x xy S S S S S S τ τ τ σ σ σ γ 666564636261 +++++=

[ ] [ ][ ]σ ε S =

S is the

compliance matrix

Isotropic Materials


38/73

An isotropic material has stress-strain relationships that are

independent of the orientation of the coordinate system at a point.The isotropic material requires only two independent material

constants, namely the Elastic Modulus and the Poisson’s Ratio.

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

G

G

G

E E E

E E E

E E E

τ

τ τ

σ

σ

σ

ν ν

ν ν

ν ν

γ

γ γ

ε

ε

ε

100000

01

0000

001

000

0001

0001

0001

( ) ⎤⎡ − E E E ν ν ν 000

1


39/73

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

( )( )( )

( )( ) ( )( )

( )( ) ( )( )

( )

( )( )

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−+

−

−+−+

−+−+−

−+

−+−+−+

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

G

G

G

E E E

E E E

γ

γ γ

ε

ε

ε

ν ν

ν

ν ν

ν

ν ν

ν ν ν

ν ν ν

ν ν ν

ν

ν ν ν ν ν ν

τ

τ τ

σ

σ

σ

00000

00000

00000

000

211

1

211211

000211211

1211

000211211211

[ ] [ ][ ]ε σ C =

C is the elastic or stiffness matrix

The isotropic material requires

only two independent materialconstants, namely the Elastic

Modulus and the Poisson’s

Ratio.

Anisotropic Materials


40/73

pUp to this point we have limited the study of the properties of materials

to isotropic materials. For the most general linearly elastic anisotropicmaterials, a particular component of stress is assumed to depend of all

six components of strain.

xy zx yz z y x x C C C C C C γ γ γ ε ε ε σ 161514131211 +++++=Where Cij are constants if the material is homogeneous

⎪⎪

⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎪⎪

⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

C C C C C C

C C C C C C

C C C C C C C C C C C C


γ

γ

γ

ε

ε ε

τ

τ

τ

σ

σ σ

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

Taking energy considerations the coefficients of this matrix are


41/73

⎪⎪⎪

⎪

⎭

⎪⎪⎪

⎪

⎬

⎫

⎪⎪⎪

⎪

⎩

⎪⎪⎪

⎪

⎨

⎧

⎥⎥⎥

⎥⎥⎥⎥

⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢

⎢

⎣

⎡

=

⎪⎪⎪

⎪

⎭

⎪⎪⎪

⎪

⎬

⎫

⎪⎪⎪

⎪

⎩

⎪⎪⎪

⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x


C C C C C C

C C C C C C

C C C C C C

C C C C C C

γ γ

γ

ε

ε

ε

τ τ

τ

σ

σ

σ

665646362616

565545352515

464544342414

363534332313

262524232212

161514131211

g gy

symmetric. Hence, instead of 36 independent constant, we have

21 independent constants

[ ]

⎥⎥⎥

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢⎢

⎣

⎡

=

665646362616

565545352515

464544342414

363534332313

262524232212

161514131211


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 is referred to as the elastic matrix

or stiffness matrix.

Hence, we can also write


42/73

[ ] [ ][ ]

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧=

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

S S S S S S

S S S S S S

S S S S S S

S S S S S S S S S S S S

S S S S S S

S

τ

τ

τ

σ σ

σ

γ

γ

γ

ε ε

ε

σ ε

665646362616

565545352515

464544342414

363534332313

262524232212

161514131211

The matrix S is referred to as the compliance matrix and the

elements of S are the compliances.

21 elastic constants are required to describe the most general


43/73

21 elastic constants are required to describe the most general

anisotropic material (fully anisotropic). This is in contrast to an

isotropic material for which there are only two independent elasticconstants (typically the Young Modulus and the Poisson’s ratio).

( )( )( ) ( )( )

( )( ) ( )[ ] )εε E

)εε E E

z y x x

z y x x

++−−+=

+−+

+−+

−=

(1211

(211211

1

ν ε ν ν ν

σ

ν ν ν ε

ν ν ν σ

Many materials of practical interest contain certain material


44/73

symmetries with respect to their elastic properties (elastic symmetries).

Other type of symmetries are possible optical, electrical and thermal properties.

Let us determine the structure of the elastic matrix for a material

with a single plane of elastic symmetry. Crystals whose crystalline

structure is monoclinic as examples of materials possessing a single

plane of elastic symmetry. Example Iron aluminide, gypsum, talc,

ice, selenium

Materials with one plane of symmetry are referred to as Monoclinic

materials.


45/73

Crystal Systems


46/73

y y

Crystallographers have shown that only

seven different types of unit cells arenecessary to create all point lattice

Cubic a= b = c ; α = β = γ = 90

Tetragonal a= b ≠ c ; α = β = γ = 90Rhombohedral a= b = c ; α = β = γ ≠ 90

Hexagonal a= b ≠ c ; α = β = 90, γ =120

Orthorhombica≠ b ≠ c ; α = β = γ = 90

Monoclinic a≠ b ≠ c ; α = γ = 90 ≠ β

Triclinic a≠ b ≠ c ; α ≠ γ ≠ β ≠ 90

Monoclinic Materials


47/73

Let us assume that the z-plane is the plane of elastic symmetry.

For such a material the elastic coefficients in the stress-strain law

must remain unchanged when subjected to a transformation that

represents a reflection in the symmetry plane.

For monoclinic materials (due to one plane of elastic symmetry) thenumber of independent elastic constants is reduced from 21 to 13.

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

⎥⎥

⎥⎥⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢⎢

⎢⎢

⎣

⎡

=

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

C C C C

C C

C C C C C C

C C C C

C C C C

γ

γ

γ ε

ε

ε

τ

τ

τ σ

σ

σ

66362616

5545

4544

36332313

26232212

16131211

00

0000

000000

00

00

KEY TO

NOTATION


48/73

TRICLINIC (21)

MONOCLINIC (13)

ORTHORHOMBIC (9)

CUBIC (3)


49/73

(7) TETRAGONAL (6)

HEXAGONAL (5) ISOTROPIC (2)


50/73

(7) TRIGONAL (6)

Orthotropic Materials

L t id t i l ith d l f l ti t


51/73

Let us consider a material with a second plane of elastic symmetry.

The y-plane and the z-plane are the planes of elastic symmetry and are perpendicular to each other. Again, for such a material the elastic

coefficients in the stress-strain law must remain unchanged when

subjected to a transformation that represents a reflection in the

symmetry plane. For orthotropic materials (due to the two planes of

elastic symmetry) the number of independent elastic constants is

reduced from 21 to 9.

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

⎥⎥

⎥⎥⎥⎥

⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢⎢

⎢⎢

⎣

⎡

=

⎪⎪

⎪⎪

⎭

⎪⎪

⎪⎪

⎬

⎫

⎪⎪

⎪⎪

⎩

⎪⎪

⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

C

C

C

C C C

C C C

C C C

γ

γ

γ ε

ε

ε

τ

τ

τ σ

σ

σ

66

55

44

332313

232212

131211

00000

00000

00000

000

000

000

Materials possessing two perpendicular planes of elastic symmetry

must also possess a third mutually perpendicular plane of elastic


52/73

must also possess a third mutually perpendicular plane of elastic

symmetry. Materials having three mutually perpendicular planes of elastic

symmetry are referred to as orthotropic (orthogonally anisotropic)

materials.

Long Fiber Composite

Transversely Isotropic MaterialsM i l h i i i l


53/73

Materials that are isotropic in a plane.

Transversely isotropic materials require five independent materialconstants.

( )⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

C C

C

C

C C C

C C C

C C C

γ

γ γ

ε

ε

ε

τ

τ τ

σ

σ

σ

2

00000

00000

00000

000

000

000

1211

44

44

331313

131112

131211

Isotropic MaterialsThe isotropic material requires only two independent material


54/73

The isotropic material requires only two independent material

constants, namely the Elastic Modulus and the Poisson’s Ratio.

( )

( )( ) ⎪

⎪⎪

⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥

⎥⎥⎥⎥

⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎢⎢⎢⎢

⎢⎢

⎣

⎡

−−

−=

⎪⎪⎪

⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎪

⎩

⎪⎪⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

C C

C C

C C C C C

C C C

C C C

γ

γ

γ

ε ε

ε

τ

τ

τ

σ σ

σ

200000

02

0000

002

000

000

000

000

1211

1211

1211

111212

121112

121211

( )( )( ) ( )( )

( )( )

G E C C E

C E

C =+

=−

−+=

−+−

=ν ν ν

ν

ν ν

ν

122

211

211

1 12111211

Engineering Material Constants for Orthotropic Materials


55/73

The quantities appearing in the coefficient matrix can be writtenin terms of well understood engineering constants such as the

Young Modulus and the Poisson’s ratio.

For the x, y and z coordinate axes we can write:

z z z

y y y

x x x

E

E

E

ε σ

ε σ

ε σ

==

=

Where the Young Modulus in the x-, y- and z-

directions are not necessarily equal.

Any extension in the x-axis is accompanied by acontraction in the y- and z- axis. However, this

quantities are not necessarily equal in orthotropic

materials.Where

ν xy is the contraction in the y-direction

due to the stress in the x-direction

x xz z

x xy y

ε ν ε

ε ν ε

−=

−=

If all three stresses are applied

simultaneously then:

[ ] [ ][ ]

⎪⎫

⎪⎧

⎥⎤

⎢⎡

⎪⎫

⎪⎧

=

x x S S S

S

σ ε

σ ε

131211 000


56/73

simultaneously, then:

z

z

y

y

yz

x

x

xz y

z z

zy

y y

x x

xy

y

z

z

zx y

y

yx

x

x

x

E E E

E E E

E E E

σ σ ν

σ ν

ε

σ

ν

σ σ

ν

ε

σ ν σ ν σ ε

1

1

1

+−−=

−+−=

−−=

⎪⎪

⎪⎪

⎭

⎪⎪⎪

⎪

⎬

⎪⎪

⎪⎪

⎩

⎪⎪⎪

⎪

⎨

⎥⎥

⎥⎥⎥⎥⎥

⎥

⎦⎢⎢

⎢⎢⎢⎢⎢

⎢

⎣

=

⎪⎪

⎪⎪

⎭

⎪⎪⎪

⎪

⎬

⎪⎪

⎪⎪

⎩

⎪⎪⎪

⎪

⎨

xy

zx

yz

z

y

xy

zx

yz

z

y

S

S

S

S S S

S S S

τ

τ

τ

σ

σ

γ

γ

γ

ε

ε

66

55

44

332313

232212

131211

00000

00000

00000

000

000

Comparing with the compliance matrix

for orthotropic materials:

1

1

1

231312

332211

y

yz

z

zy

x

xz

z

zx

x

xy

y

yx

z y x

E E S

E E S

E E S

E S

E S

E S

ν ν ν ν ν ν −=−=−=−=−=−=

===

Whereν

xy is the contraction in the y-direction due to the stress inthe x-direction

Whereas with

isotropic materials

1

1

1 zx zx yz yz xy xy

GGG=== τ γ τ γ τ γ


57/73

p

the relationship

between shear

stress and shear

strain is the same in

any coordinate planes, for

orthotropic

materials theserelationships are not

the same.

1 1 1 665544 xy zx yz

zx yz xy

GS

GS

GS

GGG

===

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥

⎥⎥⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

z y

yz

x

xz

z

zy

y x

xy

z

zx

y

yx

x

xy

zx

yz

z

y

x

G

G

G

E E E

E E E

E E E

τ

τ

τ

σ

σ

σ

ν ν

ν ν

ν ν

γ

γ

γ

ε

ε

ε

100000

01

0000

001

000

0001

0001

0001

⎤⎡ ++− νννννννν1


58/73

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

Δ−

Δ+

Δ+

Δ

+

Δ−

Δ

+Δ

+

Δ

+

Δ

=

⎪⎪⎪⎪

⎭

⎪⎪⎪⎪

⎬

⎫

⎪⎪⎪⎪

⎩

⎪⎪⎪⎪

⎨

⎧

xy

zx

yz

z

y

x

xy

zx

yz

y x

yx xy

y x

yx xz yz

y x

yz xy xz

z x

xy zx zy

z x

xz zx

z x

zy xz xy

z y

zy yx zx

z y

yz zx yx

z y

zy yz

xy

zx

yz

z

y

x

G

G

G

E E E E E E

E E E E E E

E E E E E E

γ

γ

γ

ε

ε

ε

ν ν ν ν ν ν ν ν



τ

τ

τ

σ

σ

σ

00000

00000

00000

0001

0001

0001

z y x

zx yz xy xz zx zy yz yx xy

E E E

ν ν ν ν ν ν ν ν ν 21 −−−−=Δ

In 2-D


59/73

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

−

−

=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y x

xy

y

yx

x

xy

y

x

G

E E

E E

τ

σ

σ ν

ν

γ

ε

ε

100

01

01

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y

x

C

C C

C C

γ

ε

ε

τ

σ

σ

33

2212

1211

00

0

0

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥⎥⎥

⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎢⎢

⎣

⎡

−−

−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

yx xy

y

yx xy

y xy

yx xy

x yx

yx xy

x

xy

y

x

G

E E

E E

γ

ε

ε

ν ν ν ν ν

ν ν

ν

ν ν

τ

σ

σ

00

011

011

Neumann’s Principle

This is the most important concept in crystal physics It states;


60/73

This is the most important concept in crystal physics. It states;

……………... the

symmetry

of

any

physical

property

of

a

crystal

must include the symmetry elements of the point group of the crystal . This

means that measurements made in symmetry‐related directions will

give

the

same

property

coefficients.Example: NaCl belongs to the m3m group . The [100] and [010]

directions are equivalent.

Since these

directions

are

physically

the same, it should be expected that

measurements of permittivity,

elasticity or

any

other

physical

property

will be the same in these two

directions.

7 Crystal Systems


61/73

Crystal System External Minimum Symmetry Unit Cell Properties

Triclinic None a, b, c, al, be, ga,

Monoclinic One 2‐fold

axis,

||

to

b

(b

unique)

a,

b,

c,

90,

be,

90

Orthorhombic Three perpendicular 2‐folds a, b, c, 90, 90, 90

Tetragonal One 4‐fold axis, parallel c a, a, c, 90, 90, 90

Trigonal One 3‐fold axis a, a, c, 90, 90, 120

Hexagonal One 6‐fold

axis a,

a,

c,

90,

90,

120

Cubic Four 3‐folds along space diagonal a, a, ,a, 90, 90, 90

triclinictrigonal

hexagonal

cubic tetragonal

monoclinic orthorhombic

Anisotropy Factor

Cubic Symmetry


62/73

Cubic Symmetry

For cubic

crystals,

there

are

four

three

‐fold

symmetry

axes (along the body diagonals) such that:

665544

312312

332211

S S S

S S S S S S

==

====

There

is

a

reduction

of

the

nine

constants

for

orthotropic symmetry to three. An anisotropic factor

A, can be defined for cubic crystals

44

1211 )(2

S

S S A

−=

Using

the

direction

cosines

l,

m,

n for

a

particular

direction, one can determine the elastic properties

of a cubic single crystal in a particular direction by

the relationship:

222222

44121111

'

112

12

1nl nmml S S S S S

E hkl ++⎥⎦

⎤⎢⎣

⎡ −−−==

Isotropy

When the anisotropy factor is equal to one, there are just two


63/73

independent components, e.g. C11

and C12

. In this instance, the

rigidity or shear modulus G is given by: ( )

44

121144

1

2

1

S C C C G =−==

And λ is given by: 12C =

These

two

constants

are

known

as

the

Lame

constants

and

are

used

to

describe

all

the

elastic

constants

of

isotropic

materials

Poisson’s

ratio can de expressed in terms of Lame constants: ⎟

⎠

⎞⎜⎝

⎛ +=

+−=−=

λ

υ GC C

C

S

S

12

1

1211

12

11

12

The compressibility

( β )

or

bulk

modulus

(K)relate hydrostatic or mean stress to volume

strain3

21 G K mean +=

Δ== λ σ

β

⎟ ⎠ ⎞⎜

⎝ ⎛ +

⎟ ⎠

⎞⎜⎝

⎛ +

=λ

λ

G

GG

E 1

23

Example

An orthotropic material has the following properties E x=7,500ksi ,


64/73

Solution:

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

yx xy

y

yx xy

y xy

yx xy

x yx

yx xy

x

xy

y

x

G

E E

E E

γ

ε

ε

ν ν ν ν

ν

ν ν

ν

ν ν

τ

σ

σ

00

011

011

12 x

xy

y

yx

E E S

ν ν

−=−=

p g p p x

E y= 2,500ksi , G xy = 1,250ksi andν xy= 0.25. Determine the principal

stresses and strains at a point on a free surface where the following

strains were measured: ε x =-400μ ; ε y=600μ ; γ xy=-500μ . Consider

plane stress conditions

083.07500

250025.0 =

×=== y

x

xy

yx

x

xy

y

yx E

E E E

ν ν

ν ν


65/73

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−

⎥

⎥⎥

⎦

⎤

⎢

⎢⎢

⎣

⎡=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

−

−

−

6

6

6

10500

10600

10400

125000

02.25533.638

03.6387660

x

x

x

xy

y

x

τ

σ

σ

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

−

−

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

xy

y

x

xy

y x

xy

y

yx

x

xy

y

x

G

E E

E E

τ

σ

σ ν

ν

γ

ε

ε

100

01

01

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

−

−

=⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

psi psi

psi

xy

y

x

6256.1276

2681

τ σ

σ

psi psi

psi

Max 1.20754.2777

9.1372

2

1

=−=

=

τ σ

σ

μ γ μ ε

μ ε

1118459

659

2

1

=−=

=

Max

5.0600400

5002tan =

−−−=

−=

Y X

XY

ε ε

γ θ ε


66/73

( )316.0

6.12762681

)625(222tan =

−−−⋅=

−=

y x

xy

σ σ

τ θ σ

Different angles to obtain the principal stresses and the principal

strains.

Example

S ppose e start ith a state of strain (in strain) strain⎥⎤

⎢⎡

μ3020050

2050300


67/73

Suppose we start with a state of strain (in μ strain) strain−

⎥⎥⎦⎢⎢⎣

μ

1003020

3020050

Consider an orthotropic material where :

GPa

⎥⎥⎥⎥

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎢⎢⎢⎢

⎣

⎡

6.2700000

0100000

0045000

000754025

000405055

0002555103

We need to change the strain tensor for a strain vector

300⎥⎤

⎢⎡


68/73

610

10040

60

100200

1003020

3020050

2050300−×

⎥⎥⎥

⎥⎥⎥⎥⎥

⎦⎢⎢⎢

⎢⎢⎢⎢⎢

⎣

=−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ strainμ

[ ] GPa

⎥

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

6.2700000

0100000

0045000

000754025

000405055

0002555103

σ 6

10

100

40

60

100

200

300

−

×

⎥

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

×

⎤⎡⎥⎥⎥⎤

⎢⎢⎢⎡

4.076.24.4400023

500.30

400.44


69/73

[ ] MPa MPa⎥⎥⎥⎦⎢

⎢⎢⎣

=

⎥⎥

⎥⎥⎥⎥

⎦⎢⎢

⎢⎢⎢⎢

⎣

=0.2370.24.0

70.25.3076.2

760.2

400.0

700.2

000.23σ

Eigen‐values Eigen‐vector (cosines from x‐y‐z angles to the

principal axes)

MPa

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

9656.4400

08154.300

00119.22

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−−−

0418.03110.09495.0

1948.09296.03130.0

9800.01980.00217.0

Example

The orthotropic elastic constants for bovine (cow) femoral (leg) bone


70/73

has been

reported

from

measurements

using

ultrasound.

The

values

vary on the basis of the position around the bone and along its length.

The elastic constants can be determined using piezoelectric crystals to

propagate and

measure

the

speed

of

sound

in

the

material.

Two

types

of elastic constants can be determined. Propagation of dilatational

waves can be used to measure longitudinal stiffness (e.g. C11) and

propagation shear

waves

can

be

used

to

measure

the

shear

moduli

(e.g. C44)

wavestransverseof speed waveV

wavesal dilatationof speed waveV

density

V C V C

trans

dil

transdil

_ _ _ _

_ _ _ _

2442

11

=

=

=

==

ρ

ρ ρ

The approximately

reported

stiffness

values

are:

⎥⎤

⎢⎡ 0008.43.614

Find the Young Modulus


71/73

MPa

⎥⎥⎥

⎥⎥⎥⎥

⎥

⎦⎢⎢⎢

⎢⎢⎢⎢

⎢

⎣ 3.50000003.60000

007000

0002578.400074.183.6

Find the Young Modulus

along the

bone

length

(z

‐

direction)? And along the

radial direction (x and y

directions).

Find the Poisson’s ratios?

Convert the

stiffness

matrix

into

a compliance

matrix.

1

19.000000

016.00000

0014.0000

000046.0014.0009.0

000014.007.0026.0

000009.0026.0086.0

−

⎥⎥

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−−

MPa

11

373.00260 7.21046.0

11

12

21

2

21

33

3 =⇒−==== υ

υ

. E

υ

-GPaS

E


72/73

1044.00090 28.1407.0

11

3016.00260 6.11086.0

11

13

1

13

22

2

121

12

111

=⇒−====

=⇒−====

υ

υ

. E

υ

-GPaS

E

. E

υ

-GPaS E

Example

Determine the modulus of elasticity for iron

single crystals in the , and muvw

Cos

l wvu

uvwCos

==

=++

=222

))(010(

1))(100(

β

α


73/73

g y ,

directions.

3

1

3

1

3

1111

02

1

2

1110

001100nml Directions

nwvu

uvwCos

wvu

=++

=++

222

222

1

))(001(1

γ

β

60.88.20.8

10 44121113

−

−−

Fe

S S S GPa

( )

GPaE

E

125

0.8)0(2

6.88.20.820.8

1

100

100

=

=⎟ ⎠

⎞⎜⎝

⎛ −−−−=

( )

GPa E E

270

7.3)

9

1

9

1

9

1(

2

6.88.20.820.8

1

111

111

=

=++⎟

⎠

⎞⎜

⎝

⎛ −−−−=

( )

GPa

E

210

75.4)004

1(

2

6.88.20.820.8

1

110

110

=

=++⎟ ⎠

⎞⎜⎝

⎛ −−−−=

Plain Strainn

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Transcript of Plain Strainn