Plain Strainn
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Transcript of Plain Strainn
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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).
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Comparison of plane stress and plane strain.
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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.
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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
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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
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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.
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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
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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
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( )⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−−−=
⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢
⎣
⎡
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
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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
+−
−=
+−
++
=
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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
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Mohr’s Circle
for Plane Strain
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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
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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
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(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
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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 〉〉
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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
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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
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Application : Strain Gauge and Strain Rosette
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(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
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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
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[ ]
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
=
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
Principal Stresses Principal Strains
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( )
( )
0
1
1
3
2
122
2
211
=
++
=+
+=
σ
ν
νε ε σ
ν
νε ε σ
E
E
Principal Stresses Principal Strains
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
Principal Stresses Principal Strains
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
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[ ]
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
=
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
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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
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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
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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
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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
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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
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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
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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
⎤⎡
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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
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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
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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
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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 +++++=
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⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
⎥⎥
⎥⎥⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢⎢⎢
⎢⎢
⎣
⎡
=
⎪⎪
⎪⎪
⎭
⎪⎪
⎪⎪
⎬
⎫
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
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
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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
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( )( )( ) ( )( ) ( )( )
( )( )( )
( )( ) ( )( )
( )( ) ( )( )
( )
( )( )
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−+
−
−+−+
−+−+−
−+
−+−+−+
=
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
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
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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
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
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⎪⎪⎪
⎪
⎭
⎪⎪⎪
⎪
⎬
⎫
⎪⎪⎪
⎪
⎩
⎪⎪⎪
⎪
⎨
⎧
⎥⎥⎥
⎥⎥⎥⎥
⎥
⎦
⎤
⎢⎢⎢
⎢⎢⎢⎢
⎢
⎣
⎡
=
⎪⎪⎪
⎪
⎭
⎪⎪⎪
⎪
⎬
⎫
⎪⎪⎪
⎪
⎩
⎪⎪⎪
⎪
⎨
⎧
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
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 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
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[ ] [ ][ ]
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎢⎢⎢⎢
⎣
⎡
=
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧=
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
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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
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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.
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Crystal Systems
-
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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
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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
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TRICLINIC (21)
MONOCLINIC (13)
ORTHORHOMBIC (9)
CUBIC (3)
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(7) TETRAGONAL (6)
HEXAGONAL (5) ISOTROPIC (2)
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(7) TRIGONAL (6)
Orthotropic Materials
L t id t i l ith d l f l ti t
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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
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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
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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
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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
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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
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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
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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
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⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
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
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
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
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⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎥
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎢⎢⎢⎢
⎣
⎡
−
−
=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
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;
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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
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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
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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
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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 ,
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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
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⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
−
−
⎥
⎥⎥
⎦
⎤
⎢
⎢⎢
⎣
⎡=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
−
−
−
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
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( )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
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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⎥⎤
⎢⎡
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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
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[ ] 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
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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
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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
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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
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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
=
=++⎟ ⎠
⎞⎜⎝
⎛ −−−−=