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Strain Analysis
Ramadas Chennamsetti
Summary
Deformation gradient
Green-Lagrange strain
Almansi strain
Rotation vector
Strain transformation
Principal strains
Strain decomposition
Strain compatibility
2
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Introduction
The effect of forces applied to a body – Newton’s
second law – Stress analysis
Applied forces => deformations
Concerned with study of deformations – geometric
problem – no material properties
Strains => kinematics of material deformation
Change in relative position of any two points in a
continuous body => deformed or strained
Distance remains constant – rigid body motion
Rigid body => Translations & Rotations
Continuum mechanics approach
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Introduction
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Deformation – consists of relative motion
between particles and rigid body motion
A”
A’
y
x
Change in position
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A
Introduction
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Particle P occupies
(X1, X2, X3)
Final position
Initial position
Body ‘B’
x2
P’
X2
P
u(X1, X2, t)
X(X1, X2)
x(X1, X2)
O
X1
x1
Body undergoes
macroscopic
changes within
the body
‘Deformation’
Co-ordinate system
refers to material
particles in the body
=> Material coordinate system
Material point P occupies P’ in final config
Deformation csys x1x2x3 => Spatial co-ordinate system5
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Description of motion
Displacement
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Displacement in a body
Mapping from reference csys
to deformed csys - function
Application of force
Initial state of body
Final state of body
A’
u(X1, X2, t)
A
X(X1, X2)
x(X1, X2)
O
X1
OA = X ( X 1 , X 2 , X 3 )
~
~
OA' = x( X 1 , X 2 , X 3 )
~
~
'
OARamadas
= OA+ AA
' => x = X + U
Chennamsetti
~
~
~
~
~
~
AA' = U ( X 1 , X 2 , X 3 )
~
~
6
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X2
Mapping
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Material co-ordinate system – spatial coordinate system
'
χ :P→P
=> x = χ X
~
()
~
Existence of inverse mapping
Substitute spatial co-ordinates =>
−1
Material co-ordinates – which
material occupies the spatial
~
~
7
position
X = χ ( x)
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Substitute material co-ordinates of the particle in
mapping function => Final position of the particle
Mapping
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Variable – velocity, density etc vector /
scalar
( )
φ = φ X ,t
~
Expressed in material csys
( )
Mapping : x = χ X
()
=> φ = φ (X , t ) = φ (χ (x ), t )
=> φ = φ (x , t )
~
~
X = χ −1 x
~
−1
~
~
~
Expressed in spatial co-ordinate system
Ramadas Chennamsetti
8
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~
Displacement
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In indices notation
x = X + U => xi = X i + ui ( X 1 , X 2 , X 3 )
~
~
This relates the current deformed geometry with
initial undeformed geometry of a point
Displacement field => movement of each point in
initial configuration to final configuration
U => rigid body motion and deformation in the
body
Deformation – important
9
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~
Normal strain
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Body subjected to loads –
AB => A’B’
AB // A’B’ & AB = A’B’
⇒Rigid body translation
AB // A’B’ & AB = A’B’
A
B
B’
⇒Rigid body rotation
AB // A’B’ & AB ≠ A’B’
⇒Deformation / normal
Strain
10
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A’
Shear strain
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Shear strain => distortion
C
A’
A
θo B
θ
B’
No change in angle
=> shear strain = 0
Change in angle –
measure of shear
strain / distortion
11
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C’
Change in angle
between two lines
=> shear strain
Displacement
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General motion of a solid –
y
y
x
Undeformed state
x
Rigid body rotation
x
Horizontal extension
Rigid body motion
y
y
x
Vertical extension
x
Shear deformation
Ramadas
Chennamsetti
doesn’t effect strain
field – no effect on
stresses
12
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y
Strain
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(
final length ) − (initial length )
E=
2
~
(initial length )
~
2
2
13
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Strain – measure of displacement of a body
Various measures of strain - definition of strain is
not unique – equally valid definition of strain
Engineering normal strain – change in
length/original length
Green-Lagrange strain
Green – Lagrange strain
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Green – Lagrange strain
Q' ( x1 + dx1 , x2 + dx2 , x3 + dx3 )
P' (x1 , x2 , x3 )
X2, x2
II
I
X3, x3
P( X 1 , X 2 , X 3 )
X1, x1
Q( X 1 + dX 1 , X 2 + dX 2 , X 3 + dX 3 )
Xi – undeformed configuration, xi – deformed configuration
14
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o
Green – Lagrange strain
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PQ = dS, P’Q’ = ds
PQ 2 = dS . dS = dSi ei .dS j e j = dSi dS j ei . e j
~
~
~
~
~
~
=> PQ 2 = dSi dS jδ ij = dSi dSi = dSi2 = dS 2
P ' Q'2 = dsi2 = ds 2
Similarly,
2. dX . E . dX = ds − dS = dx . dx − dX . dX
2
~
~
~
~
~
dx = dX + dU
~
~
T
2
~
~
~
dX = dX 1 e1 + dX 2 e2 + dX 3 e3
~
~
~
dx = dx1 e1 + dx2 e2 + dx3 e3
Ramadas
~
~
Chennamsetti
~
~
T
~
~
15
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Green – Lagrange strain (E) is defined as,
Green – Lagrange strain
R&DE (Engineers), DRDO
dU – infinitesimal displacement
function of (X1, X2, X3)
dU = dU ( X 1 , X 2 , X 3 )
~
~
it is a vector => dU = du1 e1 + du2 e2 + du3 e3
~
dui = dui ( X 1 , X 2 , X 3 )
~
~
Use chain rule
∂ui
∂ui
∂ui
dui =
dX 1 +
dX 2 +
dX 3 = ∇ui • dX
~
∂X 1
∂X 2
∂X 3
16
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~
Green – Lagrange strain
R&DE (Engineers), DRDO
Writing in vector form
(
)
dU = dui ei = ∇ui • dX ei
~
~
~
 ∂ui

 ∂ui

dU =  e j
• dX k ek  ei = 
dX k e j • ek  ei
 ~ ∂x
 ~  ∂x
~
~
~
~
~
j
j




 ∂ui

∂ui


dX k δ jk ei =
dX j ei = dui ei
=> dU =
 ∂X
 ~ ∂X
~
~
~
j
j


∂ui
dX j
=> dui =
∂X j
17
Ramadas Chennamsetti
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~
Green – Lagrange strain
R&DE (Engineers), DRDO
The new position is,
use
dxi = dX i + dui
∂ui
dui =
dX j
∂X j

∂ui
∂ui

=> dxi = δ ij dX j +
dX j = dX j δ ij +

∂X j
∂X j



=> dx = dX •  I + ∇ U 
~
~
~ 
~
~


Ramadas Chennamsetti




18
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∂ui
dxi = dX i +
dX j
∂X j
Green – Lagrange strain
R&DE (Engineers), DRDO
Green-Lagrange strain is defined as,
2. dX . E . dX = dx . dxT − dX . dX
T
~
~
~
~
~
~
~
T
~
T


 


T
T
2. dX . E . dX =  dX . I + ∇U  . dX . I + ∇U   − dX . dX
 ~ ~
~
~
~
~   ~  ~
~ 
~
~
~
~
~

 





 
 T
T
T



2. dX . E . dX = dX . I + ∇U  .  I + ∇U . dX  − dX . dX
 ~ ~
 ~ ~
~
~
~
~   ~
~  ~
~
~
  ~




 T
T
T
T 

2. dX . E . dX = dX . I + ∇U + ∇U + ∇U .∇U . dX − dX . dX
~
~
~
~ ~
~
~
~
~  ~
~
~
~
~


1
E = ∇U + ∇U T + ∇U .∇U T
~
~
~
~
2 ~
~
T
(
Ramadas Chennamsetti
)
19
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T
Green – Lagrange strain
R&DE (Engineers), DRDO
Green-Lagrange Strain – Normal strains
εX
2
2
2

∂u 1  ∂u   ∂v   ∂w  
=
+ 
 +
 +
 
∂ X 2  ∂ X   ∂X   ∂ X  
2
2
2
∂ w 1  ∂ u   ∂ v   ∂ w  
+ 
εZ =
 +
 +
 
∂Z 2  ∂Z   ∂Z   ∂Z  
20
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2
2
2
∂ v 1  ∂ u   ∂ v   ∂ w  
εY =
+ 
 +
 +
 
∂Y 2  ∂Y   ∂Y   ∂Y  
Green – Lagrange strain
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Green-Lagrange Strain – Shear strains
∂u ∂v
γ XY = +
∂Y ∂X
 ∂u ∂u ∂v ∂v ∂w ∂w
+
+
+

∂
X
∂
Y
∂
X
∂
Y
∂
X
∂
Y


∂u ∂w  ∂u ∂u ∂v ∂v ∂w ∂w
γ XZ = + + 
+
+
∂Z ∂X  ∂X ∂Z ∂X ∂Z ∂X ∂Z 
21
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∂w ∂v  ∂u ∂u ∂v ∂v ∂w ∂w
+
+
γ YZ = + + 
∂Y ∂Z  ∂Z ∂Y ∂Z ∂Y ∂Z ∂Y 
Green – Lagrange strain
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Green-Lagrange strain captures finite deflections,
strains and rotations – non-linear terms
Infinitesimal strains, rotations => linear terms –
non-linear terms – much smaller than linear terms
Small strains known as engineering strains
Engineering strains – denoted by => εij
22
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∂ui
<< 1
∂x j
Engineering strain
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C’
D
u+
∂u
dx
∂x
A’(u,v)
A
∂v
v + dy − v
∂v
∂y
εy =
=
dy
∂y
C
dy
y
∂u
u + dx − u
∂u
∂
x
εx =
=
dx
∂x
dx
B’
u+
∂u
dx
∂x
B
x
23
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∂v
v + dy
∂y D’
Engineering strain
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
∂u
∂v 
 u +
dy, v + dy 
∂y
∂y 


∂v  D’
 u, v + dy 
∂y 

C’
D”
D
C
dy
θ1
∂u


u
+
dx
,
v


B’
∂x


A’(u,v)
A
dx
B
x
24
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∂u
∂v 

 u + dx, v + dx 
∂x
∂x 
B” 
θ2
y
C”
Engineering strain
R&DE (Engineers), DRDO
Shear strain – Change in the angle
∂v
v + dx − v
∂v
∂
x
tan θ1 =
=
dx
∂x
∂v
θ1 ≈
∂x
γ xy
∂u
u + dy − u
∂u
∂y
tan θ 2 =
=
dy
∂y
∂u
θ2 ≈
∂y
∂u ∂v
= θ1 + θ 2 ≈
+
∂y ∂x
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Initial angle => 900
Final angle => (900 - θ1- θ2)
Change in angle => θ = θ1+ θ2
Engineering strain
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The six strain components –
Normal strains –
∂v
εy =
∂y
∂u
εx =
∂x
∂w
εz =
∂z
Shear strains –
Indices notation -
∂w ∂v
γ yz = +
∂y ∂z
1
ε ij = (ui , j + u j ,i )
2
∂w ∂u
γ zx = +
∂x ∂z
Infinitesimal strains => Green-Lagrange
strain = Engineering Strain
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∂u ∂v
γ xy = +
∂y ∂x
Engineering strain
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Rigid body rotations –
OP
'
~
~
~
OP = x i + y j
y
~
U
P’(y, -x)
= OP + U
OP
P(x, y)
~
~
'
~
= − y i+ x j
~
~
Linear strains –
x
εxx= εyy = -1, γxy = 0
Non-linear strains –
Rotated by
900
εxx= εyy = 0, γxy = 0
Linear strains can’t capture large rotations use nonlinear strains
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O
Engineering strain
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Assuming engineering – linear strains
εx

[ε ] = ε xy
ε xz

ε xy
εy
ε yz

εx

ε xz  
1

ε yz  =  γ xy
2

ε y  1
γ
 2 xz
εij – Tensorial strain
γij – Engineering shear strain
Ramadas Chennamsetti
1
γ xy
2
εy
1
γ yz
2
1 
γ xz 
2
1 
γ yz 
2 
ε y 

1
ε ij = γ ij
2
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Strain tensor
Displacement gradient
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Displacement gradient tensor – second order
tensor
∂ui
H ij =
=> H = ∇ U
~
~
∂x j
~
H ij = ui , j
 ∂u1

 ∂x1
∂u 2

=
 ∂x1
 ∂u
 3
 ∂x1
∂u1
∂x2
∂u2
∂x2
∂u3
∂x2
Ramadas Chennamsetti
∂u1 

∂x3 
∂u2 
∂x3 
∂u3 
∂x3 
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In matrix form
Displacement gradient
R&DE (Engineers), DRDO
Square matrix – sum of symmetric and skew
symmetric matrices
1
1
T
[M ] = M + M + M − M T
2
2
[ M ] = [ S ] + [ A]
(
)
(
)
)
(
1
1
T
T
[ S ] = M + M , [ A] = M − M
2
2
[ S ]T = [ S ], [ A ]T = −[ A ]
)
[S] – Symmetric tensor, [A] – Skew-symmetric
tensor – Main diagonal terms => 0
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(
Displacement gradient
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Decomposing displacement gradient
∂u1
∂x2
∂u 2
∂x2
∂u3
∂x2
∂u1
∂u1  

∂x1
∂x3  
∂u2   1  ∂u1 ∂u2 

=  
+

∂x3
2  ∂x2 ∂x1 

∂u3   1  ∂u ∂u 
 1 + 3 

∂x3   2 ∂x ∂x
1 
  3
Zeros in main diagonal

0


 1  ∂u1 ∂u 2 

−
 
 2  ∂x2 ∂x1 
 1  ∂u1 ∂u3 

−
 
 2  ∂x3 ∂x1 
1  ∂u1 ∂u2 


+
2  ∂x2 ∂x1 
∂u 2
∂x2
1  ∂u2 ∂u3 


+
2  ∂x3 ∂x2 
1  ∂u1 ∂u2 


−
2  ∂x2 ∂x1 
0
1  ∂u 2 ∂u3 


−
2  ∂x3 ∂x2 
Ramadas Chennamsetti
1  ∂u1 ∂u3  

 
+
2  ∂x3 ∂x1  
1  ∂u 2 ∂u3 

 +
+
2  ∂x3 ∂x2 

∂u3

∂x3

1  ∂u1 ∂u3  

 
−
2  ∂x3 ∂x1  
1  ∂u 2 ∂u3 


−
2  ∂x3 ∂x2 

0


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ui , j
 ∂u1

 ∂x1
∂u
= 2
 ∂x1
 ∂u
 3
 ∂x1
Displacement gradient
R&DE (Engineers), DRDO
Displacement gradient is sum of infinitesimal
strain tensor and rotation tensor
H =ε+Ω
~
~
~
~
~
~
1
1
ui , j = (ui , j + u j ,i ) + (ui , j − u j ,i )
2
2
1
=> ε ij = (ui , j + u j ,i ) => Symmetric tensor
2
1
Ω ij = (ui , j − u j ,i ) => Skew - symmetric tensor
2
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ui , j = ε ij + Ω ij
Displacement gradient
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If displacement gradient is symmetric
=> ui,j = uj,i
Rotation tensor becomes zero – only strain tensor
exists
Displacement gradient = strain tensor
ui , j
1
= (ui , j + u j ,i )
2
33
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u i , j = u j ,i
Rotation vector
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Rotation vector is defined as
1
2
ω = ∇ ×U
~
~
1 ∂
1 ∂u j
×u j ej =
ω = ei
ei × e j
~
2 ~ ∂xi
2 ∂xi ~ ~
~
1  ∂u1 ∂u3 

−
Say, k = 3, i, j = 1, 2 => ω 3 = 
2  ∂x3 ∂x1 
ω represents components of rigid body rotations
34
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1 ∂u j
1 ∂u j
ω=
ε ijk ek => ω k =
ε ijk
~
2 ∂xi
2 ∂xi
~
Rotation vector
R&DE (Engineers), DRDO
Relation between rotation vector and tensor
1 ∂u j
eijk
ωk =
2 ∂xi
Displacement gradient =>
∂u j
∂xi
= ε ji + Ω ji
eijk – skew symmetric matrix in ‘i’, ‘j’
1
1
ω k = − Ω ji e jik => elmkω k = − Ω ji e jik elmk
2
2
Ramadas Chennamsetti
35
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1
1
ωk = (ε ji + Ω ji ) eijk = − (ε ji e jik + Ω ji e jik )
2
2
Rotation vector
R&DE (Engineers), DRDO
Multiplication of two permutations
1
1
elmkωk = − Ω ji (δ jlδ im − δ jmδ il ) = − (Ωlm − Ωml )
2
2
− elmkωk = Ωlm
36
Ramadas Chennamsetti
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1
1
ωk = − Ω jie jik => elmkωk = − Ω ji e jik elmk
2
2
e jik elmk = δ jlδ im − δ jmδ il
Rotation vector
R&DE (Engineers), DRDO
From shear strain
 ∂u
∂v 
∂v
∂u
γ xy =  +  = 0 => = − = constant = c
∂x
∂y
 ∂y ∂x 
∂v
∂u
= c,
= −c
∂x
∂y
∆y
Integrating these two equations
-∂u/∂y
u = uo − cy, v = vo + cx
∂v/∂x
∆x
For constant rotation
1  ∂v ∂u 
1
ω z =  −  => (c + c ) => c = ω z
2  ∂x ∂y 
2
Ramadas Chennamsetti
37
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=>
Rotation vector
R&DE (Engineers), DRDO
The displacements are
u = uo − ω z y, v = vo + ω z x
∂u ∂v
γ xy = + = 0
∂y ∂x
∂w ∂u
∂w ∂v
γ yz = + = 0 γ zx = + = 0
∂x ∂z
∂y ∂z
Integrate these equations – integration constants
38
Ramadas Chennamsetti
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Extending the same procedure to 3D rigid body
rotations – all shear strains => 0
Rotation vector
R&DE (Engineers), DRDO
Integration constants – rigid body translations and
rotations
u = uo − ω z y + ω y z
v = vo − ω x z + ω z x
Rigid body translations and rotations don’t
contribute to strains and stresses – may be dropped
Basic assumption – small deformation theory
Small deformation theory – Principle of
superposition
39
Ramadas Chennamsetti
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w = wo − ω y x + ω x y
Transformation of strain
R&DE (Engineers), DRDO
[ε ] = [T ][ε ][T ]
T
'
[T] – Transformation matrix => direction cosines
40
Ramadas Chennamsetti
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Strain – second order tensor
Transformation rules of any second order tensor –
applicable – ex. Stress tensor
'
σ
ij = lim l jnσ mn
Transformation of stress tensor
Transformation of strain tensor
ε ij' = liml jnε mn
In matrix notation
Principal strains
R&DE (Engineers), DRDO
Principal strains – similar to principal stresses –
maximum strains acting on three mutually
perpendicular planes, where shear stresses => 0
ε y = lε xy + mε yy + nε yz = mε
ε z = lε xz + mε yz + nε zz = nε
Writing in indices notation
(ε l
ij j
− δ ij εl j ) = 0
For non-trivial solution
determinant = 0
41
Ramadas Chennamsetti
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ε x = lε xx + mε xy + nε xz = lε
Principal strains
Determinant
(ε
− δ ij ε ) l j = 0
ij
=> ε ij − δ ij ε = 0
ε − I1ε + I 2ε − I 3 = 0
3
R&DE (Engineers), DRDO
Expand this
2
Ii - Invariant
εx
I2 =
ε xy
ε xy ε x
+
ε y ε xz
εx
I3 =
symm
ε xz ε y
+
ε z ε yz
ε xy
εy
ε yz
εz
ε xz
ε yz
εz
Ramadas Chennamsetti
First invariant I1
known as cubical
dilatation
42
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I1 = ε x + ε y + ε z
Strain decomposition
Hydrostatic part represents volume deformation –
isotropic tensor – same in all direction
Volumetric strains – accounts for change in size or
volume only
Deviatoric strain tensor – accounts for change in
shape – distortion of material
Physically εd => deviation of strain from 1/3rd of
pure cubical dilatation
Tr[εd] = 0 => Trace of deviatoric strain tensor is
zero
43
Ramadas Chennamsetti
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R&DE (Engineers), DRDO
Strain compatibility
R&DE (Engineers), DRDO
Two dimensional stress analysis – three
strain components => εxx, εyy and γxy
Writing in terms of displacements
ε yy
∂v
=
∂y
∂ 2ε xx
∂ 3u
∂ 2  ∂u 
 
= 2 =
2
∂y
∂y ∂x ∂y∂x  ∂y 
Add these two equations
∂u ∂v
γ xy = +
∂y ∂x
∂ 2ε yy
∂x 2
∂ 3v
∂ 2  ∂v 
= 2 =
 
∂x ∂y ∂y∂x  ∂x 
2
2
∂ 2ε xx ∂ ε yy ∂ γ xy One compatibility equation
+
=
2
2
44
∂y Ramadas
∂x Chennamsetti
∂x∂y
rd_mech@yahoo.co.in
ε xx
∂u
=
∂x
Strain compatibility
R&DE (Engineers), DRDO
In 3D => six strain compatibility equations
∂ ε xx ∂ ε yy ∂ γ xy
+
=
2
2
∂y
∂x
∂x∂y
(i = j = 1, k = l = 2)
∂ 2ε xx
∂  ∂γ yz ∂γ xz ∂γ xy 
 −
 = 2
+
+
∂x  ∂x
∂y
∂z 
∂z∂y
(i = j = 1, k = 2, l = 3)
∂ 2 ε yy
∂  ∂γ yz ∂γ xz ∂γ xy

−
+
∂y  ∂ x
∂y
∂z
2
2
∂ γ yz
∂ 2 ε zz
+
=
2
2
∂y∂z
∂z
∂y
(i = j = 2 , k = l = 3 )
2
∂ 2 ε xx ∂ 2 ε zz
∂ 2 γ xz
=
+
2
2
∂z
∂x
∂x∂z
(i = j = 3, k = l = 1)
∂ ε yy

 = 2
∂ x ∂z

(i = j = 2, k = 3, l = 1)
2
∂  ∂γ yz ∂γ xz ∂γ xy

+
−
∂z  ∂x
∂y
∂z

∂ 2ε zz
 = 2
∂ x∂ y

(i = j = 3, k = 1, l = 2 )
45
Ramadas Chennamsetti
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2
Strain compatibility
Six strain compatibility equations – no
displacement term appears
Displacements – integration of strain –
displacement relations
Six equations – three displacements => over
constraining – can’t get a single valued
displacements – ill-conditioned
For well conditioned system => Number of
equations = number of unknowns
Three more extra equations required
46
Ramadas Chennamsetti
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO
Strain compatibility
R&DE (Engineers), DRDO
There exist relations among six strain
components => Strain compatibility
equations – six equation
Only three of them are independent
Equation Strain-displacement relations => 6
Independent strain compatibility equations => 3
=> Number of unknowns = number of equations
47
Ramadas Chennamsetti
rd_mech@yahoo.co.in
Displacements => 3
Strain compatibility
Satisfy strain compatibility equations =>
single valued displacement in the domain
Conversely – all six strain compatibility
equations satisfied if displacement field is
single valued
Arbitrary selection of strains – fails to
satisfy strain compatibility equations
48
Ramadas Chennamsetti
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO
Strain compatibility
R&DE (Engineers), DRDO
Strains are calculated
from displacements
Compatible – no gaps
and overlaps
Integrated to get
displacements
Gaps & overlaps incompatibility
Ramadas Chennamsetti
49
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Arbitrary strain field not
satisfying compatibility
eqns.
Polar co-ordinates
R&DE (Engineers), DRDO
Use strain-displacement relations in cartesian
co-ordinates – convert to polar co-ordinates
Derive from polar co-ordinates
50
Ramadas Chennamsetti
rd_mech@yahoo.co.in
Some geometries – cylindrical – easy to
handle in cylindrical co-ordinates
Evaluate stress, strains, displacements etc in
that co-ordinate system – ex. Radial and
Hoop stresses
Two different ways –
Polar co-ordinates
R&DE (Engineers), DRDO
From cartesian csys – ex, ey – unit vectors in
cartesian csys - er, eθ - unit vectors in polar csys
z
ez
P
eθ
y
ey
eθ
er
er
P(x, y)
y
θ
x
r
θ
eθ
er
x
(
1
T
Cartesian strain => ε ij = ∇ U + ∇ U
~
2 ~
Ramadas Chennamsetti
)
ex
51
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r
Polar co-ordinates
R&DE (Engineers), DRDO
ex = er cos θ − eθ sin θ
~
~
~
e y = er sin θ + eθ cosθ
~
~
~
ez = ez
~
x = r cosθ
y = r sin θ
z=z
~
r =x +
2
2
 
e 
e
− sinθ
0 r 
 ~x  cosθ
  
~

cosθ
0eθ 
ey  = sinθ
~   0
~

0
1
ez 
ez  
~
~ 
 
 
e
=
[
T
]
 xyz 
erθz 
2
 ~ 
 ~ 
y
y
Tanθ =
x
52
Ramadas Chennamsetti
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Relation between unit vectors –
Polar co-ordinates
R&DE (Engineers), DRDO
∂
∂ ∂r
=
∂x ∂r ∂x
∂
∂ ∂r
=
∂y ∂r ∂y
∂
∂
∇ = ex
+ ey
~ ∂x
~ ∂y
∂ ∂θ
∂ sin θ
+
= cos θ
−
∂θ ∂x
∂r
r
∂ ∂θ
∂ cos θ
+
= sin θ
+
∂θ ∂y
∂r
r
∂
∂θ
∂
∂θ
Substitute all these
∂
1 ∂
∂
∇ = er
+ eθ
+ ez
~ ∂r
~ ∂z
~ r ∂θ
Ramadas Chennamsetti
53
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In cartesian csys - gradient
Polar co-ordinates
R&DE (Engineers), DRDO
Displacement vector in polar csys –
U = u r er + uθ eθ + u z ez
~
~
~
~
Gradient of displacement vector –
 1  ∂ur

∂uθ 
1
1 ∂u z

 r  ∂θ − uθ  eθ er + r  ur + ∂θ  eθ eθ + r ∂θ eθ ez  +
~ ~ 
~ ~

~ ~
 
∂uθ
∂u z
 ∂ur

e
e
+
e
e
+
e
e
 ∂z ~z ~r ∂z ~z ~θ ∂z ~z ~z 


54
Ramadas Chennamsetti
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∂uθ
∂u z
 ∂ur

∇U = 
er er +
er eθ +
er ez  +
~
 ∂r ~ ~ ∂r ~ ~ ∂r ~ ~ 
Polar co-ordinates
R&DE (Engineers), DRDO
Strain =>
ε =
~
~
[
( ) ]
1
∇U + ∇U
~
~
2
T
Stain components in polar co-ordinates
Normal strains
∂ur
∂u
1
, ε θθ =  ur + θ
∂r
r
∂θ
∂u z

=
,
ε

zz
∂z

Shear strains
1  ∂uθ 1 ∂u z 
+
,
2  ∂z r ∂θ 
εθz = 
1  ∂ur ∂u z 
1  1 ∂ur ∂uθ uθ 
+
+
− 
, ε rθ = 
2  ∂z ∂r 
2  r ∂θ ∂r
r 
ε zr = 
55
Ramadas Chennamsetti
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ε rr =
56
Ramadas Chennamsetti
rd_mech@yahoo.co.in
R&DE (Engineers), DRDO