Interpretacion de Formulas PDF
Interpretacion de Formulas PDF
Interpretacion de Formulas PDF
DEFORMATION AND
STRAIN
Multimedia Course on Continuum Mechanics
Overview
Introduction Lecture 1
Deformation Gradient Tensor
Material Deformation Gradient Tensor Lecture 2
Inverse (Spatial) Deformation Gradient Tensor
Lecture 3
Displacements Lecture 4
Displacement Gradient Tensors
Strain Tensors
Green-Lagrange or Material Strain Tensor Lecture 5
Euler-Almansi or Spatial Strain Tensor
Variation of Distances
Stretch Lecture 6
Unit elongation
Variation of Angles Lecture 7
2
Overview (cont’d)
Physical interpretation of the Strain Tensors Lecture 8
Material Strain Tensor, E
Spatial Strain Tensor, e Lecture 9
Polar Decomposition Lecture 10
Volume Variation Lecture 11
Area Variation Lecture 12
Volumetric Strain Lecture 13
Infinitesimal Strain
Infinitesimal Strain Theory
Strain Tensors
Stretch and Unit Elongation Lecture 14
Physical Interpretation of Infinitesimal Strains
Engineering Strains
Variation of Angles
3
Overview (cont’d)
Infinitesimal Strain (cont’d)
Polar Decomposition Lecture 15
Volumetric Strain
Strain Rate
Spatial Velocity Gradient Tensor Lecture 16
Strain Rate Tensor and Rotation Rate Tensor or Spin Tensor
Physical Interpretation of the Tensors
Material Derivatives Lecture 17
Other Coordinate Systems
Cylindrical Coordinates Lecture 18
Spherical Coordinates
4
2.1 Introduction
Ch.2. Deformation and Strain
5
Deformation
Deformation: transformation of a body from
a reference configuration to a current configuration.
6
2.2 Deformation Gradient Tensors
Ch.2. Deformation and Strain
7
Continuous Medium in Movement
Ω0: non-deformed (or reference) Ω or Ωt: deformed (or present)
configuration, at reference time t0. configuration, at present time t.
X : Position vector of a particle at x : Position vector of the same particle
reference time. at present time.
ϕ ( X,t )
t0 Q t
Reference or Ωt
dX Q’
non-deformed dx
Ω0 P P’
X x Present or
deformed
8
Fundamental Equation of Deformation
The Equations of Motion:
not
xi i ( X1, X 2 , X 3 , t )
ϕ= xi ( X 1 , X 2 , X 3 , t ) i ∈ {1, 2,3}
not
= ( X, t ) x ( X, t )
x ϕ=
Differentiating w.r.t. X :
∂xi ( X, t ) not
dxi = dX j Fij ( X, t ) dX j i, j ∈ {1, 2,3}
∂X j
∂x ( X, t ) not Fundamental equation
d=
x X F ( X, t ) ⋅ dX
⋅ d= of deformation
∂X
(material) deformation
gradient tensor
9
Material Deformation Gradient Tensor
F(X,t):
is a primary measure of deformation
characterizes the variation of relative placements in the neighbourhood of a
material point (particle).
=dx F ( X, t ) ⋅ dX
10
Inverse (spatial) Deformation Gradient
Tensor
The inverse Equations of Motion:
not
( x1 , x2 , x3 , t ) X i ( x1 , x2 , x3 , t )
X i ϕ= i
−1
i ∈ {1, 2,3}
not
= ( x, t ) X ( x, t )
X ϕ= −1
Differentiating w.r.t. x :
∂X i ( X, t ) not
=dX i = dx j Fij−1 ( x, t ) dx j i, j ∈ {1, 2,3}
∂x j
∂X ( x, t ) not
=
dX =⋅ dx F −1 ( x, t ) ⋅ dx
∂x
11
Inverse (spatial) Deformation Gradient
Tensor
The spatial (or inverse) deformation gradient tensor:
= dX F −1 ( x, t ) ⋅ dx
∂
F ( x, t ) ≡ X(x, t ) ⊗ ∇
−1 REMARK x
The spatial Nabla 1
−1 ∂X i ∂
=
Fij i, j ∈ {1, 2,3} operator is defined as: [∇] =
∂x j ∂ x2
∇≡ eˆ i ∂
∂X 1 ∂X 1 ∂X 1 ∂xi
= ∇T ∂x1 ∂x2 ∂x3 x3
X1
∂ ∂ ∂ ∂X 2 ∂X 2 ∂X 2
F = [ X ⊗ ∇ ] = X 2 ∂x
−1
=
∂x2 ∂x3 ∂x1 ∂x2 ∂x3
X 3 1
∂X 3 ∂X 3 ∂X 3
F-1(x,t): = X
∂x1 ∂x2 ∂x3
is a primary measure of deformation
characterizes the variation of relative placements in the neighbourhood of a
spatial point.
It is not the spatial description of the material deformation gradient tensor
12
Properties of the Deformation
Gradients
The spatial deformation gradient tensor is the inverse of the material
deformation gradient tensor:
∂xi ∂X k ∂xi
= = δ ij F ⋅ F −1 = F −1 ⋅ F = 1
∂X k ∂x j ∂x j
∂x
If there is no motion, =
x X and =
F −1
= F= 1 .
∂X
13
Example
Compute the deformation gradient and inverse deformation gradient tensors
for a motion equation with Cartesian components given by,
X + Y 2t
[ x] Y (1 + t )
=
Zet
1.
Using the results obtained, check that F ⋅ F −1 =
14
X + Y 2t
[ ] ( )
Example - Solution =x Y 1 +
Zet
t
The Cartesian components of the inverse motion equation will be given by,
y 2t
X= x − 2 yt
( + )
2
1 t 1 0
(1 + t )
y
[ X=] ϕ −1 ( x, t =
) =
f (=
( ) x, t ) 0 1 + t
Y F X ( x , =
t ), t 0
1+ t
0 et
Z = ze − t f ( x ,t )
0
15
Example - Solution
The Cartesian components of the inverse deformation gradient tensor are,
2 yt
1 − 0
( + )
2
1 t
1
F −1 ( x, t ) = 0 0
1+ t
0 0 e − t
1:
And it is verified that F ⋅ F −1 =
2 yt 2 yt 2 yt
1 − 0 1 − + 0
0
2 yt
(1+ t) (1 + t ) (1 + t )
2 2 2
1
(1 + t ) 1 0 0
1 1+ t 0 1 0=
F ⋅ F −=
1
0 1 + t 0 ⋅ 0 0= 0 0 = [1]
0 1+ t 1+ t
0 et 0 0 1
0 0
0 e−t 0 et e − t
16
2.3 Displacements
Ch.2. Deformation and Strain
17
Displacements
Displacement: relative position of a particle, in its current (deformed)
configuration at time t, with respect to its position in the initial
(undeformed) configuration.
Displacement field: displacement of all the particles in the continuous
medium.
Material description (Lagrangian form):
( X, t ) x ( X, t ) − X
U= t0 t
U i ( X, t ) = xi ( X, t ) − X i i ∈ {1, 2,3} Ω
P U=u P’
Spatial description (Eulerian form): Ω0 X
x
u ( x, t )= x − X ( x, t )
ui ( x, t ) =
xi − X i ( x, t ) i ∈ {1, 2,3}
18
Displacement Gradient Tensor
U= ( X, t ) x ( X, t ) − X Taking partial derivatives of U w.r.t. X :
∂U i ( X, t ) ∂xi ( X, t ) ∂X i
U i ( X , t ) = xi ( X , t ) − X i i ∈ {1, 2,3}
def
= − = Fij − δ ij = J ij
∂X j ∂X j ∂X
j
∂U i δij
J = = Fij − δ ij i, j ∈ {1, 2,3} Fij
∂X j
ij
Material Displacement
def
Gradient Tensor
(
J X , t ) = U ( X, t ) ⊗ ∇ = F − 1
Taking partial derivatives of u w.r.t. x :
u ( x, t )= x − X ( x, t ) ∂ui (x, t ) ∂xi ∂X i (x, t ) def
= − = δ ij − Fij − δ ij = jij
−1
ui ( x, t ) =
xi − X i ( x, t ) i ∈ {1, 2,3} ∂x j ∂x j
∂x j
δij Fij−1
∂ui
ij ∂x =
j = δ ij − Fij−1 i, j ∈ {1, 2,3} Spatial Displacement Gradient
j
Tensor
def
j ( x, t ) = u ( x, t ) ⊗ ∇ =1 − F
−1
20
Strain Tensors
F characterizes changes of relative placements during motion but is not
a suitable measure of deformation for engineering purposes:
It is not null when no changes of distances and angles take place, e.g.,
in rigid-body motions.
21
Strain Tensors
Consider
F ( X,t )
t0 Q t d=
x F ⋅ dX
Ω dxi = Fij dX j
dS dX dx Q’
ds
d=
X F -1 ⋅ dx
Ω0 P P’
X x dX i = Fij−1dx j
22
Strain Tensors
d=
X F -1 ⋅ dx d=
x F ⋅ dX
dX i = Fij−1dx j dxi = Fij dX j
=
dS dX ⋅ dX =
ds dx ⋅ dx
( ) = ⋅ = [ ] ⋅ [ ] = [ ⋅ ] ⋅ [ F ⋅ dX ] = dX ⋅ FT ⋅ F ⋅ dX
2 T T
ds dx dx dx dx F dX
( ds ) dx
= = = =
2 T
k dx k F ki dX F
i kj dX j dX F F
i ki kj dX j dX i ik Fkj dX j
F
REMARK
not
( dS ) = dX ⋅ dX =[ dX ] ⋅ [ dX ] = F ⋅ dx ⋅ F ⋅ dx = dx ⋅ F ⋅ F ⋅ dx
2 T −1 T −1 −T −1 The convention
not
T
(•) −1 =
(•) −T
( dS )2 dX
= = k dX k Fki−1 dxi=
Fkj−1 dx j dxi Fki−=
1 −1
Fkj dx j dxi Fik−T Fkj−1dx j is used.
23
Green-Lagrange Strain Tensor
( ds ) = dX ⋅ FT ⋅ F ⋅ dX ( dS =
) dX ⋅ dX
2 2
Subtracting:
( ds ) − ( dS ) = dX ⋅ FT ⋅ F ⋅ dX − dX ⋅ dX = dX ⋅ FT ⋅ F ⋅ dX − dX ⋅ 1 ⋅ dX = dX ⋅ ( FT ⋅ F − 1 ) ⋅ dX = 2 dX ⋅ E ⋅ dX
2 2
def
= 2E
( )
E is symmetrical:
(2
1 T
) 2 F ⋅ ( F ) − 1= 2 ( F ⋅ F − 1=) E
1 T T T T 1 T
T
E=
T
F ⋅ F − 1=
=Eij E ji i, j ∈ {1,2,3}
24
Euler-Almansi Strain Tensor
( ds )= dx ⋅ dx ( dS ) =dx ⋅ F −T ⋅ F −1 ⋅ dx
2 2
Subtracting:
( ds ) − ( dS ) = dx ⋅ dx − dx ⋅ F −T ⋅ F −1 ⋅ dx = dx ⋅ 1 ⋅ dx − dx ⋅ F −T ⋅ F −1 ⋅ dx =
2 2
def
= 2e
= dx ⋅ (1 − F ⋅ F ) ⋅ dx = 2 dx ⋅ e ⋅ dx
−T −1
def
= 2e
The Euler-Almansi or Spatial Strain Tensor is defined:
e ( x, t ) =
1
2
( 1 − F −T
⋅ F −1
)
e ( x, t ) =
ij
1
2
( δ ij − Fki−1 Fkj −1 ) i, j ∈ {1, 2,3}
e is symmetrical: e =
T 1
2
( −T −1 T 1 T
2
−1 T
( −T T 1
2
)
1 − F ⋅ F ) = 1 − ( F ) ⋅ ( F ) = (1 − F −T ⋅ F −1 ) = e
( ds ) − ( dS )= 2 dX ⋅ E ⋅ dX
= 2 dx ⋅ e ⋅ d x
2 2
26
Strain Tensors in terms of Displacements
Substituting F −1= 1 - j and F= J + 1 into
E=
2
( F ⋅ F − 1) and e= 1 (1 − F −T ⋅ F −1 ) :
1 T
2
1 1
E= (1 + J T ) ⋅ (1 + J ) − 1= J + J T + J T ⋅ J
2 2
Eij= 1 ∂U i ∂U j ∂U k ∂U k
+ + i, j ∈ {1, 2,3}
2 ∂X j ∂X i ∂X i ∂X j
1
1 − (1 − jT ) ⋅ (1 − j) = j + jT − jT ⋅ j
1
e= 2
2
∂u
eij= 1 ∂ui + j − ∂uk ∂uk i, j ∈ {1, 2,3}
2 ∂x j ∂xi ∂xi ∂x j
27
Example
For the movement in the previous example, obtain the strain tensors in the
material and spatial description.
X + Y 2t
[ x] Y (1 + t )
=
Zet
29
Example - Solution
The deformation gradient tensor and its inverse tensor have already been
obtained: 2 yt
1 − 0
( + )
2
1 t
1 2Yt 0
0 1 + t 0 1
F=
F −1 =
0 0
1+ t
0 0 e t
0 0 e − t
30
Example - Solution
The spatial strain tensor :
2 yt
2 yt 1−1 − 0
1 − (1 + t )
2
0
0 (1 + t )
2
1 0
1 2
2
1 1 2 yt =
e=
1
(1 − F−T ⋅ F −1 )= 2 yt
1 − −
1
0 ⋅ 0
1
0 = −
(1 + t ) 2
1− −
2 yt
+
(1 + t ) 2 1 + t
0
(1+ t) 1+ t 1+ t
2
2 2 2
e − t 0 0 e−t −t −t
0 0
0 0 1− e e
2 yt
0 − 0
(1 + t )
2
2
2 yt 1 2
1 2 yt
= − 1− − 0
2 (1 + t ) 2 (1 + t ) 2 1 + t
0 0 1 − e −2 t
31
X + Y 2t
[ x] Y (1 + t )
Example - Solution =
Zet
2 yt
0 − 0 −
2Yt
(1 + t )
2
0 (1 + t )
0
2
1 2 yt 2 yt 1 2 2Yt 1
2 2
e ( x, t ) =
1 2Yt
− 1− − 0 e ( X, t ) = − 1 − − 0
2 (1 + t )
(1 + t ) 1 + t 2 (1 + t ) (1 + t ) 1 + t
2 2
y Y (1 + t )
=
0 0 1− e −2 t
0 0 1 − e −2t
spatial description material description
Observe that E ( x, t ) ≠ e ( x, t ) and E ( X, t ) ≠ e ( X, t ).
32
2.5 Variations of Distances
Ch.2. Deformation and Strain
33
Stretch
The stretch ratio or stretch is defined as:
def P´Q´ ds
stretch = λ T= λ=
t = (0 < λ < ∞)
PQ dS
t0 Q t t
dS dX Ω Q’ REMARK
dx
ds The sub-indexes (●)T and
(●)t are often dropped. But
Ω0 P P’
X x one must bear in mind that
stretch and unit elongation
always have a particular
direction associated to them.
34
Unit Elongation
The extension or unit elongation is defined as:
def ∆ PQ ds − dS
unit elongation = ε=
T ε=
t =
PQ dS
t0 Q t t
dS dX Ω Q’ REMARK
dx
ds The sub-indexes (●)T and
(●)t are often dropped. But
Ω0 P P’
X x one must bear in mind that
stretch and unit elongation
always have a particular
direction associated to them.
35
Relation between Stretch and Unit
Elongation
The stretch and unit elongation for a same point and direction are
related through:
ds − dS ds
ε= = − 1= λ − 1 ( −1 < ε < ∞ )
dS dS
=λ
= (ε 0 ) =
If λ 1= ds dS : P and Q may have moved in time but have kept
the distance between them constant.
36
Stretch and Unit Elongation in terms of
the Strain Tensors
Considering:
( ds ) − ( dS )= 2 dX ⋅ E ⋅ dX
2 2
dX = T dS
( ds ) − ( dS )= 2 dx ⋅ e ⋅ dx dx = t ds
2 2
Then:
× 1
( dS ) 2 = λ2
2
ds
( ds ) − ( dS ) = 2 ( dS ) T ⋅ E ⋅ T − 1= 2 T ⋅ E ⋅ T
2 2 2
dS REMARK
2
dS E(X,t) and e(x,t)
( ds ) − ( dS ) = 2 ( ds ) t ⋅ e ⋅ t 1− = 2 t ⋅e ⋅ t
2 2 2
ds contain information
× 1
( λ)
2
( ds )2 = 1 regarding the stretch
and unit elongation for
1
λ= any direction in the
λ = 1+ 2 T ⋅ E ⋅ T 1− 2 t ⋅e ⋅ t differential neighbour-
ε = λ −1 = 1 + 2 T ⋅ E ⋅ T −1 1 hood of a point.
ε = λ −1 = −1
1− 2 t ⋅e ⋅ t
37
2.6 Variation of Angles
Ch.2. Deformation and Strain
38
Variation of Angles
dx( ) = t ( ) ds ( )
1 1 1
(1) (1) (1)
dX = T dS
dx( ) = t ( ) ds (
2 2 2)
( 2) ( 2) ( 2) T(1)
dX =T dS t
t0
Q t(1)
t(2)
T(2)
Θ Ω R’ θ
dS(1) Q’
R ds(2) ds(1)
dS(2)
Ω0 P P’
X x
39
Variation of Angles
d x( 2) =
x(1) ⋅ d ds (1) ds ( 2) cos θ
dx(1)
T
dx ( 2 )
(1)
dx = F ⋅ dX (1) T(1) ⋅ (1 + 2E ) ⋅ T( 2)
cos θ =
( 2) ( 2) 1 + 2 T(1) ⋅ E ⋅ T(1) 1 + 2 T( 2 ) ⋅ E ⋅ T( 2 )
dx = F ⋅ dX
= F ⋅ dX (1) ⋅ F ⋅ dX (2) = dX (1) ⋅ ( FT ⋅ F ) ⋅ dX (2)
T
(1) (2)
dx ⋅ dx
2E+1
dX(1) = T(1) dS (1)
dX( ) = T( ) dS ( )
2 2 2
1 1
dx (1)=
⋅ dx (2) T ⋅ ( 2E + 1 ) ⋅ T = ds (1)ds (2) (1) (2) T(1) ⋅ ( 2E + 1) ⋅=
(1) (1) (2) ( 2)
dS
dS T(2) ds (1)ds (2) cos θ
ds(1) ds(2) λ λ
(1) (2)
λ λ
λ = 1+ 2 T ⋅ E ⋅ T
T(1) ⋅ (1 + 2E ) ⋅ T( 2)
(1) (1)
1+ 2 T ⋅ E ⋅ T 1 + 2 T( 2 ) ⋅ E ⋅ T( 2 )
40
Variation of Angles
41
Variation of Angles
X(1) =
d X( 2) dS (1) dS ( 2) cos Θ
⋅ d
dX(1)
T
dX( 2)
t (1) ⋅ (1 − 2e ) ⋅ t ( 2)
dX=(1)
F −1 ⋅ dx(1) cos Θ =
( 2) −1
dX= F ⋅ dx
( 2) 1 − 2 t (1) ⋅ e ⋅ t (1) 1 − 2 t ( 2) ⋅ e ⋅ t ( 2)
dX (1) ⋅ dX (2) = F −1 ⋅ dx (1) ⋅ F −1 ⋅ dx (2) = dx (1) ⋅ ( F −T ⋅ F −1 ) ⋅ dx (2)
T
1−2e
dx(1) = t (1) ds (1)
( 2) ( 2) ( 2)
dx = t ds
REMARK λ=
1
E(X,t) and e(x,t) contain information 1− 2 t ⋅e ⋅ t
regarding the variation in angles t (1) ⋅ (1 − 2e ) ⋅ t ( 2)
between segments in the differential
(1) (1)
neighbourhood of a point. 1− 2 t ⋅e ⋅ t 1 − 2 t ( 2) ⋅ e ⋅ t ( 2)
42
Example
Let us consider the motion of a continuum body such that the spatial description
of the Cartesian components of the spatial Almansi strain tensor is given by,
0 0 −tetz
[ e ( x, t ) ] = 0 0 0
tz
−te 0 t ( 2etz − et )
Compute at time t=0 (the reference time), the length of the curve that at time
t=2 is a straight line going from point a (0,0,0) to point b (1,1,1).
The length of the curve at time t=0 can be expressed as,
1
∫ λ ( x, t ) ds
B b
=L ∫=
A
dS
a
= ds
λ
43
Example - Solution
The inverse of the stretch, at the points belonging to the straight line going
from a(0,0,0) to b(1,1,1) along the unit vector in the direction of the straight
line, is given by,
1
λ ( x, t ) = → λ −1 ( x, t ) = 1 − 2t ⋅ e ( x, t ) ⋅ t
1 − 2t ⋅ e ( x, t ) ⋅ t
Where the unit vector is given by,
1
[t ] = [1 1 1]
T
3
Substituting the unit vector and spatial Almansi strain tensor into the expression
of the inverse of the stretching yields,
2 t
λ −1
( x, t=) 1 + te
3
44
Example - Solution
The inverse of the stretch, which is uniform and therefore does not depends on
the spatial coordinates, at time t=2 reads,
4
λ −1 ( x, 2=
) 1 + e2
3
Substituting the inverse of the stretch into the integral expression provides the
length at time t=0,
4 2 4 2 (1,1,1)
( )
b b
∫Γ ∫a λ ∫a 3 ∫
−1
L= dS = x , 2 ds = 1 + e ds =1 + e ds =+
3 4e 2
3
(0,0,0)
= 3
45
2.7 Physical Interpretation of E and e
Ch.2. Deformation and Strain
46
Physical Interpretation of E
Consider the components of the material strain tensor, E:
E XX E XY E XZ E11 E12 E13
=
E = E XY EYY EYZ E E22 E23
12
E XZ EYZ EZZ E13 E23 E33
Stretching of
1 dS λ= 1 + 2E11 the material in
1
T (1)
≡ 0 dX ≡ dS T (1)
=
0
0 0
the X-direction
47
Physical Interpretation of E
Similarly, the stretching of the material in the Y-direction and the Z-
direction:
λ1 = 1 + 2 E11 ε X = λ X − 1= 1 + 2 E XX − 1
λ2 = 1 + 2 E22 ε Y = λY − 1= 1 + 2 EYY − 1
λ3 = 1 + 2 E33 ε Z = λZ − 1= 1 + 2 EZZ − 1
48
Physical Interpretation of E
Consider the angle between a segment parallel to the X-axis and a
segment parallel to the Y-axis, the angle is:
T ( ) ⋅ ( 1 + 2E ) ⋅ T ( )
1 2
cos θ =
1 + 2 T(1) ⋅ E ⋅ T(1) 1 + 2 T( 2 ) ⋅ E ⋅ T( 2 )
reference 1 0 T(1) ⋅ T( 2) =
0
T ⋅ E ⋅ T(1) =
(1)
configuration T(1) = 0 T( 2) = 1 E11
0
0
T(1) ⋅ E ⋅ T( 2) =E12
( 2) ( 2)
T ⋅E ⋅T = E22
2 E12
cos θ =
1 + 2 E11 1 + 2 E 22
deformed
configuration
2 E XY π 2 E XY
θ ≡ θ xy =arccos = − arcsin
1 + 2 EXX 1 + 2 EYY 2 1 + 2 EXX 1 + 2 EYY
49
Physical Interpretation of E
π 2 E XY
θ ≡ θ xy = − arcsin
2 1 + 2 EXX 1 + 2 EYY
2 E XY
θ xy − Θ
∆Θ XY =
=− arcsin
1 + 2 EXX 1 + 2 EYY
XY
π
2
reference
configuration
deformed
configuration
50
Physical Interpretation of E
Similarly, the increment of the final angle w.r.t. its initial value for
couples of segments oriented in the direction of the coordinate axes:
2 E XY
∆Θ XY = − arcsin
1 + 2 EXX 1 + 2 EYY
2 E XZ
∆Θ XZ = − arcsin
1 + 2 EXX 1 + 2 EZZ
2 EYZ
∆ΘYZ = − arcsin
1 + 2 EYY 1 + 2 EZZ
reference λ1 dX= 1 + 2 E XX dX
configuration
λ2 dY= 1 + 2 EYY dY
λ3 dZ= 1 + 2 EZZ dZ
2 E XY
∆Θ XY = − arcsin
1 + 2 EXX 1 + 2 EYY
2 E XZ
∆Θ XZ = − arcsin
1 + 2 EXX 1 + 2 EZZ
2 EYZ
∆ΘYZ = − arcsin
1 + 2 EYY 1 + 2 EZZ
52
Physical Interpretation of e
Consider the components of the spatial strain tensor, e:
exx exy exz e11 e12 e13
e ≡ exy eyy eyz = e12 e22 e23
exz eyz ezz e13 e23 e33
1 ds 1 Stretching of
λ1 = the material in
t (1) ≡ 0 dx ≡ 0 1 − 2e11
0 0 the x-direction
53
Physical Interpretation of e
Similarly, the stretching of the material in the y-direction and the z-
direction: 1 1
λ1 = ⇒ ε x = λx − 1 = −1
1 − 2e11 1 − 2exx
1 1
λ2 = ⇒ ε y = λy − 1= −1
1 − 2e22 1 − 2eyy
1 1
λ3 = ⇒ ε z = λz − 1 = −1
1 − 2e33 1 − 2ezz
54
Physical Interpretation of e
Consider the angle between a segment parallel to the x-axis and a
segment parallel to the y-axis, the angle is:
t (1) ⋅ (1 − 2e ) ⋅ t ( 2)
cos Θ =
reference deformed 1 − 2 t (1) ⋅ e ⋅ t (1) 1 − 2 t ( 2) ⋅ e ⋅ t ( 2)
configuration configuration
1 0 t (1) ⋅ t ( 2) = 0
(1) (1)
t ⋅ e ⋅t = e11
t (1) = 0 t (2) = 1
0 0 t (1) ⋅ e ⋅ t ( 2) =e12
( 2)
t ⋅e ⋅t = ( 2)
e22
−2 e12
cos Θ =
1 − 2 e11 1 − 2 e 22
π 2exy
Θ ≡ Θ XY= + arcsin
2 1 − 2 exx 1 − 2 eyy
55
Physical Interpretation of e
π 2exy
Θ ≡ Θ XY= + arcsin
2 1 − 2 exx 1 − 2 eyy
The increment of the angle in the reference configuration w.r.t. its value
in the deformed one:
2exy
∆θ xy = θ xy − Θ XY = − arcsin
1 − 2 exx 1 − 2 eyy
π
2
reference deformed
configuration configuration
56
Physical Interpretation of e
Similarly, the increment of the angle in the reference configuration w.r.t.
its value in the deformed one for couples of segments oriented in the
direction of the coordinate axes:
π 2exy
∆θ xy = − Θ XY = − arcsin
2 1 − 2 exx 1 − 2 eyy
π 2exz
∆θ xz = − Θ XZ = − arcsin
2 1 − 2 exx 1 − 2 ezz
π 2eyz
∆θ yz = − ΘYZ = − arcsin
2 1 − 2 eyy 1 − 2 ezz
57
Physical Interpretation of e
In short, reference
configuration
deformed
configuration
dx
= 1 − 2exx dx
λ1
dy
= 1 − 2eyy dy
λ2
dz
= 1 − 2ezz dz
λ3
π 2exy
∆θ xy = − Θ XY = − arcsin
2 1 − 2 exx 1 − 2 eyy
π 2exz
∆θ xz = − Θ XZ = − arcsin
2 1 − 2 exx 1 − 2 ezz
π 2eyz
∆θ yz = − ΘYZ = − arcsin
2 1 − 2 eyy 1 − 2 ezz
58
2.8 Polar Decomposition
Ch.2. Deformation and Strain
59
Polar Decomposition
Polar Decomposition Theorem:
“For any non-singular 2nd order tensor F there exist two unique
positive-definite symmetrical 2nd order tensors U and V, and a unique
orthogonal 2nd order tensor Q such that: ”
not
left polar
=
U F ⋅F
T
decomposition
not
V = F ⋅ FT F = Q ⋅ U = V ⋅Q
Q= F ⋅ U −1 =V −1 ⋅ F right polar
decomposition
The decomposition is unique. REMARK
Q: Rotation tensor An orthogonal 2nd
order tensor verifies:
U: Right or material stretch tensor
QT ⋅ Q = Q ⋅ QT = 1
V: Left or spatial stretch tensor
60
Properties of an orthogonal tensor
An orthogonal tensor Q when multiplied (dot product) times
a vector rotates it (without changing its length): y= Q ⋅ x
y has the same norm as x:
y = y ⋅ y = [ y ] [ y ] = [Q ⋅ x ] ⋅ [Q ⋅ x ] = x ⋅ Q
⋅Q⋅x = x
2 T T T 2
when Q is applied on two vectors x(1) and x(2), with the same origin,
the original angle they form is maintained:
T T
(1) (2)
y y
y = Q⋅x
(1) (1)
y ⋅y
(1)
x ⋅ QT ⋅ Q ⋅ x (2)
(2)
(1)
x (1) ⋅ x (2)
= = = cos α
y (2)= Q ⋅ x (2) y (1) y (2) y (1) y (2) x (1) x (2)
61
Polar Decomposition of F
Consider the deformation gradient tensor, F:
F = Q ⋅ U = V ⋅Q
stretching
rotation
dx =F ⋅ dX =( V ⋅ Q ) ⋅ dX =V ⋅ ( Q ⋅ dX )
(not)
F (•) ≡ stretching rotation (•)
rotation
stretching
dx =F ⋅ dX =( Q ⋅ U ) ⋅ dX =Q ⋅ ( U ⋅ dX )
REMARK
For a rigid body motion:
not
F(•) ≡ rotation stretching (•) U= V= 1 and Q = F
62
2.9 Volume Variation
Ch.2. Deformation and Strain
63
Differential Volume Ratio
Consider the variation of a differential volume associated to a particle
P:
=
dV0 ( dX
(1) ( 2)
× dX (3)
)=
⋅ dX
dX 1(1) dX 2(1) dX 3(1)
deformed ( 2) ( 2) ( 2)
configuration = det
= dX 1 dX 2 dX 3 M
( 3) ( 3) ( 3)
dX 1 dX 2
dX 3
M
reference
configuration
=
dVt ( dx( ) × dx( ) )=
1 2
⋅ dx( )
3
= =
M ij dX (i )
j mij dx (ji )
64
Differential Volume Ratio
Consider now: dx (i ) =
F ⋅ dX ( i ) i ∈ {1, 2,3} → Fundamental eq. of deformation
(i )
dx j = Fjk ⋅ dX k(i ) i, j ∈ {1, 2,3}
= =
M ij dX (i )
j and mij dx (ji )
m=
ij dx (j=
i)
Fjk dX k(=
i)
Fjk M ik= M ik FkjT = M ⋅ FT
m
Then:
dVt = m = M ⋅ FT = M FT = F M = F dV0 dVt = F dV0
dV
0
65
2.10 Area Variation
Ch.2. Deformation and Strain
66
Surface Area Ratio
Consider the variation of a differential area associated to a particle P:
dA := dAN → material vector "differential of area" → dA = dA
da := dan → spatial vector "differential of area" → da = dA
Deformed (current)
configuration ( 3)
dV0 = dH dA = d
X
⋅
N dA =
Reference (initial) dH
configuration =dX(3) ⋅ N
dA =d A ⋅ d X ( 3)
dA
dVt = dh da = x( ) ⋅ n da =
3
d
dh
=dx ( 3 ) ⋅ n
da =d a ⋅ d x ( 3)
da
67
Surface Area Ratio
Consider now:
da ⋅ dx (
3)
dV=
t
dx (3)= F ⋅ dX( )
3
dVt = F dV0
dV= dA ⋅ dX( )
3
0
( 3) ( 3) ( 3)
dVt = F d
A ⋅
dX =
d a ⋅ F ⋅ dX ∀dX ⇒ F dA = da ⋅ F
F dV dV0 dVt
0
da = F ⋅ dA ⋅ F −1 =
da n F N ⋅ F −1dA da F N ⋅ F −1 dA
=
dA = N dA
da = n da
68
2.11 Volumetric Strain
Ch.2. Deformation and Strain
69
Volumetric Strain
Volumetric Strain:
def dV ( X, t ) − dV ( X, t0 ) not dVt − dV0
e ( X, t ) =
dV ( X, t ) dV0
dVt = F dV0
F dV0 − dV0
e=
dV0
=
e F −1 REMARK
The incompressibility condition (null
volumetric strain) takes the form
e = J −1 = 0 ⇒ J = F = 1
70
2.12 Infinitesimal Strain
Ch.2. Deformation and Strain
71
Infinitesimal Strain Theory
The infinitesimal strain theory (also called t0 t
small strain theory) is based on the Ω
simplifying hypotheses: P u P’
Ω0 X
x
Displacements are very small w.r.t. the
typical dimensions in the continuum
medium,
As a consequence, u << ( size of Ω0 ) and the reference and deformed configurations are
considered to be practically the same, as are the material
not
and spatial coordinates:
Ω ≅ Ω0 and x = X+u ≅ X ( X, t ) u ( X, t ) ≡ u ( x, t )
U=
xi = X i + ui ≅ X i not
U i ( X, t ) =ui ( X, t ) ≡ ui ( x, t ) i ∈ {1, 2,3}
72
Infinitesimal Strain Theory
The material and spatial coordinates coincide, x = X +
u ≅X
≈0
Even though it is considered that u cannot be neglected when calculating
other properties such as the infinitesimal strain tensor ε.
There is no difference between the material and spatial
differential operators:
symb
∂ ∂
∇ = ˆ
e = eˆ i =
∇
∂ ∂
i
X i x i
J (=
X, t ) U( X, t )= ⊗ ∇ u(x, t )=
⊗ ∇ j(x, t )
=ε J + J=T
j + j=
T
(u ⊗ ∇ + ∇ ⊗ =
u ) ∇ su
2 2 2 ε is a symmetrical tensor and its components
ε ij =1 ∂ui ∂u j are infinitesimal: | ε ij |<< 1, ∀i, j ∈ {1, 2,3}
+ i, j ∈ {1, 2,3}
2 ∂x j ∂xi
74
Stretch and Unit Elongation
Stretch in terms of the strain tensors:
1
λT = 1 + 2
T⋅⋅T
E λt =
x 1− 2 t ⋅e ⋅ t
For a segment parallel to the x-axis, the stretch and unit elongation are:
λ ≅ 1+ t ⋅ε ⋅t
ε11 ε12 ε13 1
reference t ⋅ ε ⋅ t [1 0 0] ⋅ ε12 ε 22 ε 23 =
= ⋅ 0 ε11
configuration ε13 ε 23 ε 33 0
The diagonal components of the infinitesimal strain tensor are the unit
elongations of the material when in the x, y and z-directions.
ε xx ε xy ε xz
ε = ε xy ε yy ε yz
ε xz ε yz ε zz
77
Physical Interpretation of
Infinitesimal Strains
Consider the angle between a segment parallel to the X-axis and a
segment parallel to the Y-axis, the angle is Θ = π2 . XY
Applying: π 2E
θ ≡ θ xy = − arcsin XY
2 1 + 2 EXX 1 + 2 EYY
E XX = ε xx
E XY = ε xy
EYY = ε yy
π 2ε xy π π
reference
θ xy = − arcsin ≅ − arcsin 2ε xy = − 2ε xy
2 1 + 2ε xx 1 + 2ε yy 2 2
configuration ≈2ε xy
≈1 ≈1
REMARK
The Taylor linear series expansion of arcsin x yields
( x ) + ... = x + O ( x 2 )
d arcsin
arcsin ( x ) ≅ arcsin ( 0 ) +
dx x =0
78
Physical Interpretation of
Infinitesimal Strains
π
θ xy ≅ − 2ε xy
2
79
Physical Interpretation of
Infinitesimal Strains
In short,
reference deformed
configuration configuration
ε xx = ε x 1
ε xy =− ∆θ xy
2
ε yy = ε y 1
ε xz =− ∆θ xz
2
ε zz = ε z 1
ε yz =− ∆θ yz
2
80
Engineering Strains
Using an engineering notation, instead of the scientific notation, the
components of the infinitesimal strain tensor are
REMARK
Positive longitudinal strains indicate
increase in segment length.
cos(Θ + ∆θ ) =
T(1) ⋅ T(2) + 2T(1) ⋅ ε ⋅ T(2)
82
Variation of Angles
cos(Θ + ∆θ=) T(1) ⋅ T(2) + 2T(1) ⋅ ε ⋅ T(2)
T(1) and T(2) are unit vectors in the directions of the original segments,
therefore, T(1) ⋅ T(2)= T(1) T(2) cos Θ
= cos Θ
T(1) ≈ t (1)
Also, cos(Θ + ∆θ=) cos Θ ⋅ cos ∆θ − sinΘ ⋅ sin∆=
θ cos Θ − sinΘ ⋅ ∆θ T (2)
≈ t (2)
Θ ≈θ
≈1 ≈ ∆θ
2T(1) ⋅ ε ⋅ T(2) 2t (1) ⋅ ε ⋅ t (2)
θ cos Θ + 2T ⋅ ε ⋅ T
cos Θ − sinΘ ⋅ ∆= (1) (2)
∆θ =− =−
sinΘ sinθ
REMARK
The Taylor linear series expansion of sin x
and cos x yield
( x ) + ... = x + O ( x 2 )
d sin
sin ( x ) ≅ sin ( 0 ) +
dx x =0
( x ) + ... =1 + O ( x 2 )
d cos
cos ( x ) ≅ cos ( 0 ) +
dx x =0
83
Polar Decomposition
Polar decomposition in finite-strain problems:
left polar
not
decomposition
=
U FT ⋅ F
not
V = F ⋅ FT ⇒ F = Q ⋅ U = V ⋅Q
Q= F ⋅ U −1 =V −1 ⋅ F right polar
decomposition
REMARK
In Infinitesimal Strain Theory
x ≈ X , therefore,= ∂x
F ≈1
∂X
84
Polar Decomposition
In Infinitesimal Strain Theory:
U = FT F = (1 + J ) ⋅ ( 1 + J ) =
T
1 + J + JT + JT ⋅ J ≈ 1 + J + JT = 1 +
1
2
( J + JT )
<< J =x
=ε
U= 1 + ε
infinitesimal strain tensor
Similarly, REMARK
The Taylor linear series expansion of 1 + x
U = (1 + ε ) = 1 − ε = 1 − ( J + J T )
−1 1
and (1 + x ) yield
−1 −1
=x 2
dλ
x =1 + x + O ( x 2 )
1
λ ( x ) = 1 + x ≅ λ ( 0) +
=ε dx x =0 2
U −1 = 1 − ε λ ( x ) = (1 + x ) ≅ λ ( 0 ) +
−1 dλ
x =1 − x + O ( x 2 )
dx
infinitesimal x =0
strain tensor
85
Polar Decomposition
(1 + J ) ⋅ 1 − ( J + JT ) =+
1 J − ( J + J T ) − J ⋅ ( J + J T ) =+ ( J − JT )
1 1 1 1
F U −1 =
Q =⋅ 1
2 2 2 2
<< J = Ω
Q= 1 + Ω
The infinitesimal rotation tensor Ω is defined: REMARK
The antisymmetric or
def 1 1 def
Ω = (J − J T )= (u ⊗ ∇ − ∇ ⊗ u)= ∇ a u skew-symmetrical
2 2 gradient operator is
Ω 1 ∂ui ∂u j defined as:
= − << 1 i, j ∈ {1, 2,3}
ij 2 ∂x j ∂xi 1
0 Ω12 −Ω31
∇ a (•=
) [(•) ⊗ ∇ − ∇ ⊗ (•)]
2
The diagonal terms of Ω are zero: [ Ω ] = −Ω12 0 Ω 23
Ω31 −Ω 23 0
It can be expressed as
an infinitesimal rotation vector θ, ∂u3 ∂u2
∂x − ∂x
θ1 −Ω 23 2 3
REMARK
1 ∂u1 ∂u3 def 1
θ ≡ θ 2 = −Ω31 = − = ∇×u Ω is a skew-symmetric
θ −Ω 2 ∂x3 ∂x1 2 tensor and its components
3 12 ∂u2 ∂u1
− are infinitesimal.
∂x1 ∂x2
86
Polar Decomposition
From any skew-symmetric tensor Ω, it can be extracted a vector θ (axial
vector of Ω) exhibiting the following property:
Ω ⋅r =θ×r ∀r
As a consequence:
The resulting vector is orthogonal to r.
87
Proof of θ×r =
Ω⋅r ∀r
The result of the dot product of the infinitesimal rotation tensor, Ω, and a
generic vector, r, is exactly the same as the result of the cross product of
the infinitesimal rotation vector, θ, and this same vector.
0 Ω12 −Ω31 θ1 −Ω 23 r1
[Ω ] = −Ω12 0 Ω23 → θ ≡ θ 2 = −Ω31 ⇒ Ω ⋅ r =θ × r ∀r = r2
θ −Ω r
Ω31 −Ω 23 0 3 12 3
Proof:
eˆ1 eˆ 2 eˆ 3 eˆ1 eˆ 2 eˆ 3 Ω12 r2 − Ω31r3
θ × r = det θ1 θ 2 θ3 = det −Ω 23 −Ω12 =
not
−Ω31 −Ω12 r1 + Ω 23 r3
r1 r2 r3 Ω r −Ω r
r1 r2 r3 31 1 23 2
88
Polar Decomposition
Using:
J= F − 1
=ε
1
2
( J + JT ) 1 J =+
F =+ 1
1
2
( J + JT ) + ( J − JT )
1
2 F = 1+ε+Ω
Q= 1 + Ω =ε = Ω
REMARK
The infinitesimal rotation tensor
characterizes the rotation and, in the
small-strain context, maintains angles
89 and distances.
Volumetric Deformation
The volumetric strain:
=
e F −1
Considering: F= Q ⋅ U and U= 1 + ε
1 + ε xx ε xy ε xz
F = Q ⋅ U = Q U = U = 1 + ε = det ε xy 1 + ε yy ε yz =
ε xz ε yz 1 + ε zz
= 1 + ε xx + ε yy + ε zz + O ( ε 2 ) ≈ 1 + Tr ( ε )
= Tr ( ε )
e = Tr ( ε )
90
2.13 Strain Rate
Ch.2. Deformation and Strain
REMARK
We are no longer assuming an
infinitesimal strain framework
91
Spatial Velocity Gradient Tensor
Consider the relative velocity between two points in space at a given
(current) instant:
∂v
=
v P′ v= (x, t ) v ( x1 , x2 , x3 , t ) dv = ⋅ dx =⋅
l dx
∂
x
dv (x, t ) = v Q′ − v P′ = v ( x + dx, t ) − v ( x, t ) l
∂vi
=
dvi = dx j lij dx j
∂x j
lij i, j ∈ {1, 2,3}
( )= v ⊗ ∇
def ∂v x, t
l ( x , t )=
Spatial velocity ∂x
gradient tensor = lij ∂vi i, j ∈ {1, 2,3}
∂x j
92
Strain Rate and Rotation Rate (or Spin)
Tensors
The spatial velocity gradient tensor can be split into a symmetrical and
a skew-symmetrical tensor:
l= v ⊗ ∇
sym [ l ] + skew [ l ] =
l= :d + w
l ∂vi i, j ∈ 1, 2,3
=
ij ∂x { }
j
93
Physical Interpretation of d
The strain rate measures the rate of deformation of the square of the
differential length ds in the spatial configuration,
d d d d dx dx
( ds(t =
)) ( dx ⋅ d=x) ( dx ) ⋅ dx + dx ⋅ ( d=
x ) d ⋅ dx + dx ⋅ d = dv ⋅ dx + dx ⋅ dv
2
dt dt dt dt
dt
dt
dv = l ⋅ dx =v =v
1
= d (l + lT )
2
d
dt
( ds(t ) )
2
= (
d x ⋅ l ) ⋅ d x + d x ⋅ ( l ⋅ dx ) =
T
dx ⋅ lT + l ⋅ dx = 2dx ⋅ d ⋅ dx
dv
T dv = 2d
d
dt
( (ds (t )) 2 − (
dS ) 2 ) =
d
dt
( 2dX ⋅ E ( X, t ) ⋅ dX ) = 2dX ⋅
dE
dt
⋅ dX =
d
dt
( (ds (t )) 2 )
constant
2dX⋅E X,t ⋅dX notation
= E
94
Physical Interpretation of d
⋅ dX = dx ⋅ d ⋅ dx
dX ⋅ E
dx= F ⋅ dX
⋅ dX = dx ⋅ d ⋅ dx =
dX ⋅ E [
dx ] [ d ][ dx ] = [ F ⋅ dX ] [d ][ F ⋅ dX ] =
T
T
dX ⋅ ( FT ⋅ d ⋅ F) ⋅ dX
F⋅dX F⋅dX T T
dX F F dX
There is a direct relation between the material derivative of the material strain
tensor and the strain rate tensor but they are not the same.
and d will coincide when in the
E
reference configuration F |t =t0 = 1 .
REMARK
Given a 2 order tensor A,
nd
95
Physical Interpretation of w
To determine the (skew-symmetric) rotation rate (spin) tensor only three
different components are needed:
0 w12 w13
1 ∂vi ∂v j
w ij = − i, j ∈ {1, 2,3} [ w ] = − w12 0 w23
2 ∂x j ∂xi − w13 − w23 0
96
Physical Interpretation of w
It can be proven that the equality ω× r =w ⋅ r ∀r holds true.
Therefore:
ω is the angular velocity of a rotation movement.
ω x r = w · r is the rotation velocity of the point that
has r as its position vector w.r.t. the rotation centre.
dv =d ⋅ dx + w ⋅ dx
97
2.14 Material time Derivatives
Ch.2. Deformation and Strain
99
Deformation Gradient Tensor F
The material time derivative of the deformation gradient tensor,
∂xi ( X, t ) REMARK
=
Fij ( X, t ) i, j ∈ {1, 2,3} The equality of cross derivatives
∂X j
applies here: ∂ 2 (•) = ∂ 2 (•)
d ∂µi µ j ∂µ j µi
dt
dFij ∂ ∂xi ( X, t ) ∂ ∂xi ( X, t ) ∂Vi ( X, t ) ∂vi ( x ( X, t ) ) ∂x k
= = = = = lik Fkj
dt ∂t ∂X j ∂X j ∂t ∂X j ∂xk ∂X j
=Vi ( X, t ) =lik =Fkj
dF notation
dt = F= l ⋅ F
dF=ij
F= lik Fkj i, j ∈ {1, 2,3}
dt ij
100
Inverse Deformation Gradient Tensor F-1
The material time derivative of the inverse deformation gradient
tensor,
1
F ⋅ F −1 = REMARK
Do not mistake the material derivative
d of the inverse tensor for the inverse of
dt the material derivative of the tensor:
d ( F −1 )
( F ( X,t ) ) ≠ ( F
( X,t ) )
d −1 dF −1 d −1
(F ⋅ F ) = ⋅F + F⋅ = 0 −1
dt dt dt dt
d ( F −1 ) dF −1
⇒ F⋅ =− ⋅ F =−F ⋅ F −1
dt dt
Rearranging terms,
d ( F −1 )
−F −1 ⋅ F ⋅ F −1 =
= −F −1 ⋅ l ⋅ F ⋅ F −1 =
−F −1 ⋅ l
dt d ( F −1 )
= l⋅F =1 = −F −1 ⋅ l
dt
−1
dFij
dt = − F −1
ik lkj i, j ∈ {1, 2,3}
101
Strain Tensor E
The material time derivative of the material strain tensor has already
been derived for the physical interpretation of the deformation rate
tensor:
E= F ⋅d ⋅F
T
=
E
2
( F ⋅ F − 1)
1 T
F = l ⋅ F
d F=T
F T ⋅ lT
dt
dE 1 T
dt
= E=
2
( F ⋅ F + FT ⋅ F ) =
2
( F ⋅ l ⋅ F + FT ⋅ l ⋅ F ) = FT ⋅ ( l + lT ) ⋅ F = FT ⋅ d ⋅ F
1 T T 1
2
d
= FT ⋅ d ⋅ F
E
102
Strain Tensor e
The material time derivative of the spatial strain tensor,
e=
1
2
( 1 − F −T ⋅ F −1 )
F=−1
F −1 ⋅ l
d F −T= lT ⋅ F −T
dt
− ( F −T ⋅ F −1 + F −T ⋅ F −1 ) =( lT ⋅ F −T ⋅ F −1 + F −T ⋅ F −1 ⋅ l )
de 1 1
e =
=
dt 2 2
e = ( l ⋅ F ⋅ F + F −T ⋅ F −1 ⋅ l )
1 T −T −1
2
103
Volume differential dV
The material time derivative of the volume differential associated to a
given particle,
d F d F dFij −1 dFij −1
= = F F= F F F lik
d dt dFij dt
ji
dt
kj ji
( dV=) ( ∇ ⋅ v ) F dV0 lik Fkj ( F⋅F−1 ) δ
= ki
dt
= dV ki
∂v i dF
= F=
lii F = F ∇⋅v = F ∇ ⋅ v = (∇ ⋅ v ) F
d ∂xi
dt
( dV (x, t ) )= ∇ ⋅ v(x, t ) dV (x, t ) ∇⋅v
dt
104
Area differential vector da
The material time derivative of the area differential associated to a
given particle,
da ( x( X, t ), t ) = F ( X, t ) ⋅ dA( X) ⋅ F −1 ( X, t ) = F ⋅ dA ⋅ F −1
d
dt
dA ⋅ F −1 + F ⋅ dA ( F −1 )
d dF d
(t )
da=
dt dt dt
= F ∇⋅v = −F −1 ⋅ l
d
( da ) = ( ∇ ⋅ v ) F dA ⋅ F −1 − F dA ⋅ F −1 ⋅ l
dt
= da = da
d
( da ) = da ( ∇ ⋅ v ) − da ⋅ l = da ⋅ 1(∇ ⋅ v) − da ⋅ l = da ⋅ ( (∇ ⋅ v)1 − l )
dt da⋅1
105
2.15 Other Coordinate Systems
Ch.2. Deformation and Strain
106
Curvilinear Orthogonal Coord. System
A curvilinear coordinate system is defined by:
The coordinates, generically named {a, b, c}
Its vector basis, {eˆ a , eˆ b , eˆ c } , formed by unit vectors eˆ=
a eˆ=
b e=
ˆ c 1.
REMARK
A curvilinear orthogonal coordinate system can be seen as a mobile Cartesian
coordinate system { x′, y′, z ′} , associated to a curvilinear basis {eˆ a , eˆ b , eˆ c } .
108
Curvilinear Orthogonal Coord. System
A curvilinear orthogonal coordinate system can be seen as a mobile
Cartesian coordinate system {eˆ a , eˆ b , eˆ c } , associated to a curvilinear basis
{ x′, y′, z′} .
The components of a vector and a tensor magnitude in the curvilinear
orthogonal basis will correspond to those in the given Cartesian local
system:
va v x′ Taa Tab Tac Tx′x′ Tx′y′ Tx′z′
v ≡ v b ≡ v y′ T ≡ Tba Tbb Tbc ≡ Ty′x′ Ty′y′ Ty′z′
Tca Tcc Tz′x′ Tz′z′
v c v z′ Tcb Tz′y′
109
Cylindrical Coordinate System
← z− coordinate line
← θ − coordinate line x = r cos θ
x( r , θ , z ) ≡ y = r sin θ
← r− coordinate line z = z
∂eˆ r ∂eˆθ
= eˆθ = −eˆ r
∂θ ∂θ
dV = r dθ dr dz
110
Cylindrical Coordinate System
Nabla operator ∂
∂r
∂ 1 ∂ ∂ 1 ∂
∇= eˆ r + eˆθ + eˆ z ⇒ ∇≡
∂r r ∂θ ∂z r ∂θ x = r cos θ
∂ x( r , θ , z ) ≡ y = r sin θ
∂z z = z
Displacement vector
ur
u= u r eˆ r + uθ eˆθ + u z eˆ z ⇒ u= uθ
u z
Velocity vector vr
v= v r eˆ r + vθ eˆθ + v z eˆ z ⇒ u= vθ
v z
111
Cylindrical Coordinate System
Infinitesimal strain tensor
ε x′x′ ε x′y′ ε x′z′ ε rr ε rθ ε rz
ε=
1
2
{
[u ⊗ ∇ ] + [u ⊗ ∇ ]
T
}
≡ ε x′y′ ε y′y′ ε y′z′ =
ε
rθ εθθ εθ z
ε x′z′ ε y′z′ ε z′z′ ε rz εθ z ε zz
∂u 1 1 ∂u r ∂uθ uθ
ε rr = r ε= + −
2 r ∂θ
rθ
∂r ∂r r
1 ∂uθ u r 1 ∂u r ∂u z
εθθ
= + =ε rz +
r ∂θ r 2 ∂ z ∂r
∂u 1 ∂uθ 1 ∂u z
ε zz = z εθ z
= +
∂z
2 ∂z r ∂θ
x = r cos θ
x( r , θ , z ) ≡ y = r sin θ
z = z
112
Cylindrical Coordinate System
Strain rate tensor
d x′x′ d x′y′ d x′z′ d rr d rθ d rz
d=
1
2
{
[v ⊗ ∇] + [v ⊗ ∇]
T
}
≡ d x′y′ d y′y′
d y′z′ =
d
rθ dθθ dθ z
d x′z′ d y′z′ d z′z′ d rz dθ z d zz
∂v 1 1 ∂v r ∂vθ vθ
d rr = r d= + −
2 r ∂θ
rθ
∂r ∂r r
1 ∂vθ v r 1 ∂v r ∂v z
=
dθθ + =
d rz +
r ∂θ r 2 ∂ z ∂r
∂v 1 ∂vθ 1 ∂v z
d zz = z = +
∂z dθ z
2 ∂z r ∂θ
x = r cos θ
x( r , θ , z ) ≡ y = r sin θ
z = z
113
Spherical Coordinate System
x = r sin θ cos φ
=x x ( r ,θ , ϕ ) =
≡ y r sin θ sin φ r− coordinate line →
z = r cos θ
dV = r 2 sin θ dr dθ dφ
114
Spherical Coordinate System
Nabla operator ∂
∂ r
∂ 1 ∂ ∂ 1 ∂
eˆ φ ⇒ ∇ ≡ x = r sin θ cos φ
1
∇= eˆ r + eˆθ +
∂r r ∂θ r sin θ ∂φ r ∂θ = x x ( r , θ , φ ) ≡=
y r sin θ sin φ
z = r cos θ
1 ∂
r sin θ ∂φ
Velocity vector vr
v= v r eˆ r + vθ eˆθ + vφ eˆ φ ⇒ u= vθ
vφ
115
Spherical Coordinate System
Infinitesimal strain tensor
ε x′x′ ε x′y′ ε x′z′ ε rr ε rθ ε rφ
ε=
1
2
{[u ⊗ ∇ ] + [u ⊗ ∇ ]
T
}
≡ ε x′y′ ε y′y′ ε y′z′ =
εθ r εθθ εθφ
ε x′z′ ε y′z′ ε z′z′ ε rφ εθφ ε φφ
∂u r
ε rr =
∂r
1 ∂uθ u r
εθθ
= + x = r sin θ cos φ
r ∂θ r
1 ∂uφ uθ u =x x ( r , θ , φ ) ≡=
y r sin θ sin φ
ε ϕϕ= + cotgφ + r z = r cos θ
r sin θ ∂φ r r
1 1 ∂u r ∂uθ uθ
ε= + −
2 r ∂θ
rθ
∂r r
1 1 ∂u r ∂uφ uφ
=ε rφ + −
2 r sin θ ∂φ ∂r r
1 1 ∂uθ 1 ∂uφ uφ
=εθφ + − φ
2 r sin θ ∂φ r ∂θ
cotg
r
116
Spherical Coordinate System
Deformation rate tensor
d x′x′ d x′y′ d x′z′ d rr d rθ d rφ
d=
1
2
{[v ⊗ ∇] + [v ⊗ ∇]
T
}
≡ d x′y′ d y′y′
d y′z′ =
d rθ dθθ dθφ
d x′z′ d y′z′ d z′z′ d rφ dθφ dφφ
∂v r
d rr =
∂r x = r sin θ cos φ
1 ∂vθ v r =x x ( r , θ , φ ) ≡=
y r sin θ sin φ
=
dθθ + z = r cos θ
r ∂θ r
1 ∂vφ vθ v
dφφ = + cotgϕ + r
r sin θ ∂φ r r
1 1 ∂v r ∂vθ vθ
d= + −
2 r ∂θ
rθ
∂r r
1 1 ∂v r ∂vφ vφ
= + −
2 r sin θ ∂φ
d rφ
∂r r
1 1 ∂vθ 1 ∂vφ vφ
= + − φ
2 r sin θ ∂φ r ∂θ
dθφ cotg
r
117
Chapter 2
Strain
rs
ee
s gin
2.1 Introduction
t d le En
r
ba
ge ro or
eS m
ci
f
Definition 2.1. In the broader context, the concept of deformation no
ra
C d P cs
longer refers to the study of the absolute motion of the particles as
b
a
i
seen in Chapter 1, but to the study of the relative motion, with respect
an an n
le
liv or ec
M
.A
d
uu
reference configuration Ω0 occupies the point in space P in the present config-
er
tin
.O
relative positions with respect to this particle in the reference and present times
given by dX and dx, respectively. The equation of motion is given by
C
©
not
x = ϕ (X,t) = x (X,t)
not . (2.1)
xi = ϕi (X1 , X2 , X3 ,t) = xi (X1 , X2 , X3 ,t) i ∈ {1, 2, 3}
41
42 C HAPTER 2. S TRAIN
rs
Figure 2.1: Continuous medium in motion.
ee
s gin
Equation (2.2) defines the material deformation gradient tensor F (X,t) 1 .
t d le En
⎧
⎪ not
⎨F = x⊗∇
r
ba
Material deformation
ge ro or
∂ xi
eS m
(2.3)
gradient tensor ⎪
⎩ Fi j = i, j ∈ {1, 2, 3}
ci
∂ Xj f
ra
C d P cs
b
a
i
an an n
⎡ ⎤
∂ x1 ∂ x1 ∂ x1
le
liv or ec
⎡ ⎤ ⎢ ⎥
x ⎢ ∂X ∂ X2 ∂ X3 ⎥
⎢ 1⎥ ∂
M
⎢ 1 ⎥
.A
∂ ∂ ⎢ ∂ x2 ∂ x2 ∂ x2 ⎥
[F] = x ⊗ ∇ = ⎢ ⎥
⎣ x2 ⎦ ∂ X1 , ∂ X2 , ∂ X3 = ⎢ ⎥ . (2.4)
⎢ ∂ X1 ∂ X3 ⎥
m
∂ X2
⎢ ⎥
d
x3 ⎣ ∂x ∂ x3 ⎦
uu
T ∂ x3
e
3
X Th
∇
er
∂ ∂ X2 ∂ X3
tin
[x] X1
on
.O
C
1 Here, the symbolic form of the material Nabla operator, ∇ ≡ ∂ êi /∂ Xi , applied to the
not
expression of the open or tensor product, [a ⊗ b]i j = [a b]i j = ai b j , is considered.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Deformation Gradient Tensor 43
rs
⎪ ∂ Xi (2.6)
⎩ dXi = dx j = Fi−1
j dx j i, j ∈ {1, 2, 3} .
ee
∂xj
s gin
The tensor defined in (2.6) is named spatial deformation gradient tensor or in-
verse (material) deformation gradient tensor and is characterized by2
t d le En
⎧
r
ba
⎪
ge ro or
not
⎨ F−1 = X ⊗ ∇
eS m
ci
Spatial deformation
∂ Xi f (2.7)
ra
gradient tensor ⎪ −1
⎩ Fi j = i, j ∈ {1, 2, 3}
C d P cs
∂xj
b
a
i
an an n
y ha
le
liv or ec
(2.6) and (2.7) as F−1 , is in effect the inverse of the (material) defor-
.A
∂ xi ∂ Xk ∂ xi not
uu
e
= = δi j =⇒ F · F−1 = 1 ,
X Th
∂ Xk ∂ x j ∂xj
er
tin
Fik F −1
on
.O
kj
C
∂ Xi ∂ xk ∂ Xi not
F−1 · F = 1 .
©
= = δi j =⇒
∂ xk ∂ X j ∂ Xj
Fik−1 Fk j
2 Here, the symbolic form of the spatial Nabla operator, ∇ ≡ ∂ êi /∂ xi , is considered. Note
the difference in notation between this spatial operator ∇ and the material Nabla ∇.
3 The two-index operator Delta Kronecker δ is defined as δ = 1 if i = j and δ = 0 if
ij ij ij
i = j. The second-order unit tensor 1 is given by [1]i j = δi j .
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
44 C HAPTER 2. S TRAIN
[X] ∂ x1 ∂ x2 ∂ x3
rs
ee
Example 2.1 – At a given time, the motion of a continuous medium is defined
s gin
by ⎧
⎨ x1 = X1 − AX3
⎪
t d le En
x2 = X2 − AX3 .
⎪
r
⎩
ba
x3 = −AX1 + AX2 + X3
ge ro or
eS m
ci
f
Obtain the material deformation gradient tensor F (X,t) at this time. By
ra
C d P cs
b
a
means of the inverse equation of motion, obtain the spatial deformation gra-
i
dient tensor F−1 (x). Using the results obtained, verify that F · F−1 = 1.
an an n
y ha
le
Solution
liv or ec
M
.A
X1 − AX3
d
T ⎢
⎥ ∂
uu
∂ ∂
e
not
F = x ⊗ ∇ ≡ [x] ∇ = ⎣ ⎢ ⎥
X2 − AX3 ⎦ ∂ X1 , ∂ X2 , ∂ X3
X Th
er
tin
−AX1 + AX2 + X3
on
.O
⎡ ⎤
−A
C
1 0
not ⎢ ⎥
©
F≡⎣ 0 1 −A ⎦ .
−A A 1
The inverse equation of motion is obtained directly from the algebraic inver-
sion of the equation of motion,
⎡ ⎤
X1 = 1 + A2 x1 − A2 x2 + Ax3
not ⎢ ⎥
X (x,t) ≡ ⎢ ⎥
⎣ X2 = A x1 + 1 − A x2 + Ax3 ⎦ .
2 2
X3 = Ax1 − Ax2 + x3
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Displacements 45
rs
A −A 1
ee
Finally, it is verified that
s gin
⎡ ⎤⎡ ⎤ ⎡ ⎤
0 −A 1 + A2 −A2
t d le En
1 A 1 0 0
not ⎢ ⎥⎢ ⎥ ⎢ ⎥ not
F · F−1 ≡ ⎣ 0 1 −A ⎦ ⎣ A2 1 − A2 A⎦ = ⎣0 1 0⎦ ≡ 1 .
r
ba
ge ro or
eS m
−A A 1 A −A 1 0 0 1
ci
f
ra
C d P cs
b
a
i
an an n
y ha
2.3 Displacements
le
liv or ec
M
.A
particle.
e
X Th
er
tin
on
.O
joins the points in space P (initial position) and P (position at the present time t)
©
of the particle (see Figure 2.2). The displacement of all the particles in the con-
tinuous medium defines a displacement vector field which, as all properties of
the continuous medium, can be described in material form U (X,t) or in spatial
form u (x,t) as follows.
U (X,t) = x (X,t) − X
(2.9)
Ui (X,t) = xi (X,t) − Xi i ∈ {1, 2, 3}
u (x,t) = x − X (x,t)
(2.10)
ui (x,t) = xi − Xi (x,t) i ∈ {1, 2, 3}
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
46 C HAPTER 2. S TRAIN
rs
ee
s gin
Figure 2.2: Displacement of a particle.
t d le En
r
2.3.1 Material and Spatial Displacement Gradient Tensors
ba
ge ro or
eS m
ci
Differentiation with respect to the material coordinates of the displacement vec-
f
ra
C d P cs
tor Ui defined in (2.9) results in
b
a
i
∂Ui ∂ xi ∂ Xi
an an n
de f
= − = Fi j − δi j = Ji j , (2.11)
y ha
∂ Xj ∂ Xj ∂ Xj
le
liv or ec
Fi j δi j
M
.A
⎧
uu
⎪
e
de f
⎨ J (X,t) = U (X,t) ⊗ ∇ = F − 1
X Th
er
Material displacement
tin
.O
C
⎧
⎪
©
⎨ U = J · dX
⎪ ∂Ui (2.13)
⎩ dUi = dX j = Ji j dX j i, j ∈ {1, 2, 3}
∂ Xj
Similarly, differentiation with respect to the spatial coordinates of the expres-
sion of ui given in (2.10) yields
∂ ui ∂ xi ∂ Xi de f
= − = δi j − Fi−1
j = ji j , (2.14)
∂xj ∂xj ∂xj
δi j Fi−1
j
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
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Strain Tensors 47
⎧
⎪
⎨ u = j · dx
⎪ ∂ ui (2.16)
⎩ dui = dx j = ji j dx j i, j ∈ {1, 2, 3}
rs
∂xj
ee
s gin
2.4 Strain Tensors
t d le En
Consider now a particle of the continuous medium that occupies the point in
r
space P in the material configuration, and another particle Q√in its differen-
ba
ge ro or
eS m
dS = dX · dX) from
ci
tial neighborhood separated a segment dX (with length
√
f
ra
the previous paticle, being dx (with length ds = dx · dx) its counterpart in
C d P cs
b
a
the present configuration (see Figure 2.3). Both differential vectors are related
i
an an n
through the deformation gradient tensor F (X,t) by means of equations (2.2) and
y ha
(2.6),
le
⎧
liv or ec
⎨ dx = F · dX and dX = F−1 · dx ,
M
.A
(2.17)
⎩ dxi = Fi j dX j and dXi = F −1 dx j i, j ∈ {1, 2, 3} .
m
ij
d
uu
e
Then,
X Th
⎧
er
tin
⎨ (ds)2 = dx · dx not
≡ [dx]T [dx] = [F · dX]T [F · dX] ≡ dX · FT · F · dX
not
on
.O
(2.18)
⎩ (ds)2 = dxk dxk = Fki dXi Fk j dX j = dXi Fki Fk j dX j = dXi F T Fk j dX j
ik
C
or, alternatively4 ,
⎧ −1 T −1
⎪
⎪ (dS) 2
= dX · dX
not
≡ [dX] T
[dX] = F · dx F · dx =
⎪
⎪
⎪
⎨ ≡ dx · F−T · F−1 · dx ,
not
(2.19)
⎪
⎪ (dS)2 = dXk dXk = Fki−1 dxi Fk−1 −1 −1
⎪
⎪ j dx j = dxi Fki Fk j dx j =
⎪
⎩
= dxi Fik−T Fk−1
j dx j .
T not
4 The convention (•)−1 = (•)−T is used.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
48 C HAPTER 2. S TRAIN
rs
ee
s gin
Figure 2.3: Differential segments in a continuous medium.
t d le En
r
2.4.1 Material Strain Tensor (Green-Lagrange Strain Tensor)
ba
ge ro or
eS m
ci
Subtracting expressions (2.18) and (2.19) results in
f
ra
C d P cs
b
a
(ds)2 − (dS)2 = dX · FT · F · dX − dX · dX =
i
an an n
= dX · FT · F · dX − dX · 1 · dX =
y ha
le
= dX · FT · F − 1 · dX = 2 dX · E · dX , (2.20)
liv or ec
M
.A
de f
= 2E
m
tensor as follows.
e
X Th
er
⎧
tin
⎪
Material ⎨ E (X,t) = 1 FT · F − 1
on
.O
2 (2.21)
(Green-Lagrange)
⎪ 1
strain tensor ⎩ Ei j (X,t) = Fki Fk j − δi j i, j ∈ {1, 2, 3}
C
2
©
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Strain Tensors 49
rs
which implicitly defines the spatial strain tensor or Almansi strain tensor as
ee
follows.
⎧
s gin
⎪
Spatial ⎨ e (x,t) = 1 1 − F−T · F−1
t d le En
(Almansi) 2 (2.23)
⎪
strain tensor ⎩ ei j (x,t) =
1
δi j − Fki−1 Fk−1 i, j ∈ {1, 2, 3}
r
2 j
ba
ge ro or
eS m
ci
f
ra
C d P cs
b
a
i
Remark 2.4. The spatial strain tensor e is symmetric. Proof is ob-
an an n
⎧
⎪
T = 1 1 − F−T · F−1 T = 1 1T − F−1 T · F−T T =
le
⎪
liv or ec
⎪
⎪ e
⎨ 2 2
M
.A
1 −T −1
⎪
⎪ = 1−F ·F =e,
⎪ 2
m
⎪
⎩e = e
d
ij ji i, j ∈ {1, 2, 3} .
uu
e
X Th
er
tin
on
.O
Example 2.2 – Obtain the material and spatial strain tensors for the motion
C
in Example 2.1.
©
Solution
The material strain tensor is ⎛⎡ ⎤⎡ ⎤ ⎡ ⎤⎞
1 0 −A 1 0 −A 1 0 0
1 T not 1
E (X,t) = F · F − 1 ≡ ⎝⎣ 0 1 A ⎦⎣ 0 1 −A ⎦−⎣ 0 1 0 ⎦⎠ =
2 2
−A −A 1 −A A 1 001
⎡ 2 ⎤
A −A2 −2A
1
= ⎣ −A2 A2 0 ⎦
2
−2A 0 2A2
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
50 C HAPTER 2. S TRAIN
rs
2A3
2
−2A − 2A −2A
ee
3 2A3 2
s gin
Observe that E = e.
t d le En
r
ba
ge ro or
eS m
ci
f
ra
C d P cs
Remark 2.5. The material strain tensor E and the spatial strain ten-
b
a
sor e are different tensors. They are not the material and spatial de-
i
an an n
le
(ds)2 − (dS)2 = 2dX · E · dX = 2dx · e · dx ,
liv or ec
M
.A
d
uu
er
tin
.O
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Variation of Distances: Stretch and Unit Elongation 51
1 1
E= 1 + JT · (1 + J) − 1 = J + JT + JT · J
2 2
(2.24)
1 ∂Ui ∂U j ∂Uk ∂Uk
Ei j = + + i, j ∈ {1, 2, 3}
2 ∂ X j ∂ Xi ∂ Xi ∂ X j
rs
ee
1 1
s gin
e= 1 − 1 − jT · (1 − j) = j + jT − jT · j
2 2
t d le En
(2.25)
1 ∂ ui ∂ u j ∂ uk ∂ uk
ei j = + − i, j ∈ {1, 2, 3}
r
2 ∂ x j ∂ xi ∂ xi ∂ x j
ba
ge ro or
eS m
ci
f
ra
C d P cs
b
a
i
an an n
.A
sponding positions in the present configuration are given by the points in space
P and Q such that the distance between the two particles in the reference con-
m
figuration, dS, is transformed into ds at the present time. The vectors T and t are
uu
e
er
tin
on
.O
is the length of the deformed differential segment P Q per unit of
length of the original differential segment PQ.
de f PQ ds
Stretch = λT = λt = = (0 < λ < ∞) . (2.26)
PQ dS
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
52 C HAPTER 2. S TRAIN
rs
ee
s gin
Figure 2.4: Differential segments and unit vectors in a continuous medium.
t d le En
r
ba
ge ro or
eS m
ci
f
Definition 2.4. The unit elongation, elongation ratio or extension of
ra
C d P cs
a material point P (or a spatial point P ) in the material direction T
b
a
i
5
(or spatial direction t ) is the increment of length of the deformed
an an n
y ha
.A
m
Δ PQ ds − dS
er
de f
tin
.O
C
Equations (2.26) and (2.27) allow immediately relating the values of the unit
©
elongation and the stretch for a same point and direction as follows.
ds − dS ds
ε= = −1 = λ − 1 (⇒ −1 < ε < ∞) (2.28)
dS dS
λ
5 Often, the subindices (•)T and (•)t will be dropped when referring to stretches or unit
elongations. However, one must bear in mind that both stretches and unit elongations are
always associated with a particular direction.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Variation of Distances: Stretch and Unit Elongation 53
rs
P and Q has shortened with the deformation of the medium.
ee
s gin
t d le En
2.5.1 Stretches, Unit Elongations and Strain Tensors
r
ba
ge ro or
Consider equations (2.21) and (2.22) as well as the geometric expressions
eS m
ci
dX = T dS and dx = t ds (see Figure 2.4). Then,
f
ra
C d P cs
⎧
b
a
⎪
⎪ (ds)2 − (dS)2 = 2 dX · E · dX = 2 (dS)2 T · E · T
i
⎪
an an n
⎪
⎨
y ha
dS T dS T (2.29)
le
⎪
⎪ (ds)2 − (dS)2 = 2 dx · e · dx = 2 (ds)2 t · e · t
liv or ec
⎪
⎪
⎩
M
.A
ds t ds t
m
√
X Th
er
tin
ds 2 λ = 1 + 2T · E · T
−1 = λ −1 = 2 T·E·T ⇒
2
√ (2.30)
dS ε = λ − 1 = 1 + 2T · E · T − 1
on
.O
C
λ
©
1
λ=√
dS 2 1 2
1 − 2t · e · t
1− = 1− = 2 t·e·t ⇒ 1 (2.31)
ds λ ε = λ −1 = √ −1
1 − 2t · e · t
1/λ
These equations allow calculating the unit elongation and stretch for a given
direction (in material description, T, or in spatial description, t ).
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
54 C HAPTER 2. S TRAIN
Remark 2.7. The material and spatial strain tensors, E (X,t) and
e (x,t), contain information on the stretches (and unit elongations)
for any direction in a differential neighborhood of a given particle,
as evidenced by (2.30) and (2.31).
rs
⎡ ⎤
ee
0 0 −tetz
not ⎢ ⎥
s gin
e (x,t) ≡ ⎣ 0 0 0 ⎦.
−tetz t (2etz − et )
t d le En
0
r
Calculate the length, at time t = 0, of the segment that at time t = 2 is recti-
ba
ge ro or
eS m
linear and joins points a ≡ (0, 0, 0) and b ≡ (1, 1, 1).
ci
f
ra
C d P cs
b
a
Solution
i
an an n
linear and the positions of its extremes A and B (see figure below) are not
M
.A
known. To determine its length, (2.31) is applied for a unit vector in the di-
rection of the spatial configuration t,
m
1 ds 1
uu
λ=√ = =⇒ dS = ds .
e
λ
X Th
1 − 2 t · e · t dS
er
tin
on
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Variation of Angles 55
√
To obtain the stretch in the direction t ≡ [1, 1, 1]T / 3, the expression t · e · t
not
is computed first as
⎡ ⎤⎡ ⎤
0 0 −tetz 1
not 1 ⎢ ⎥⎢ ⎥ 1 1 t
t·e·t ≡ √ [1, 1, 1] ⎣ 0 0 0 ⎦ ⎣ 1 ⎦ √ = − te .
3 3 3
−tetz 0 t (2etz − et ) 1
rs
1 " 1 3
λ=! =⇒ λ" =! =√ .
ee
1 + 23 tet t=2
1+ 3e
4 2 3 + 4e2
s gin
The length at time t = 0 of the segment AB is
t d le En
# B # b # b
1 1 1 1√
r
lAB = dS = ds = ds = lab =
ba
3
ge ro or
λ λ λ λ
eS m
A a a
ci
f
ra
lab
C d P cs
b
a
i
and replacing the expression obtained above for the stretch at time t = 2
an an n
y ha
finally results in $
le
lAB = 3 + 4e2 .
liv or ec
M
.A
m
d
uu
er
tin
Consider a particle P and two additional particles Q and R, belonging to the dif-
ferential neighborhood of P in the material configuration (see Figure 2.5), and
on
.O
the same particles occupying the spatial positions P , Q and R . The relationship
C
between the angles that form the corresponding differential segments in the ref-
©
and using the definitions of the unit vectors T(1) , T(2) , t(1) and t(2) that establish
the corresponding directions in Figure 2.5,
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
56 C HAPTER 2. S TRAIN
rs
ee
s gin
t d le En
r
ba
ge ro or
eS m
Figure 2.5: Angles between particles in a continuous medium.
ci
f
ra
C d P cs
⎧ ⎧
b
a
⎨ dX(1) = dS(1) T(1) ⎨ dx(1) = ds(1) t(1) ,
i
an an n
=⇒ (2.33)
y ha
⎧
⎧
m
⎪
⎪ 1
d
λ (1)
e
=⇒ (2.34)
X Th
⎪
tin
⎩ dS(2) = 1
ds(2) .
λ (2)
on
.O
Expanding now the scalar product6 of the vectors dx(1) and dx(2) ,
C
T
= F · dX(1) F · dX(2) ≡ dX(1) · FT · F · dX(2) = dX(1) · (2E + 1) · dX(2)
not
1 1
= dS(1) T(1) · (2E + 1) · T(2) dS(2) = (1) ds(1) T(1) · (2E + 1) · T(2) (2) ds(2) =
λ λ
(1) (2) 1 1 (1) (2)
= ds ds T · (2E + 1) · T ,
λ (1) λ (2)
(2.35)
6 The scalar product of two vectors a and b is defined in terms of the angle between them, θ ,
as a · b = |a| · |b| cos θ .
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Physical Interpretation of the Strain Tensors 57
rs
In an analogous way, operating on the reference configuration, the angle Θ
ee
between the differential segments dX(1) and dX(2) (in terms of t(1) , t(2) and e )
s gin
is obtained,
t d le En
t(1) · (1 − 2e) · t(2)
cosΘ = $ $ .
r
(2.38)
ba
ge ro or
1 − 2t(1) · e · t(1) 1 − 2t(2) · e · t(2)
eS m
ci
f
ra
C d P cs
b
a
i
Remark 2.8. Similarly to the discussion in Remark 2.7, the material
an an n
and spatial strain tensors, E (X,t) and e (x,t), also contain informa-
y ha
process. These facts will be the basis for providing a physical inter-
.A
d
uu
e
X Th
er
tin
.O
C
Consider a segment PQ, oriented parallel to the X1 -axis in the reference config-
uration (see Figure 2.6). Before the deformation takes place, PQ has a known
length dS = dX.
The length of P Q is sought. To this aim, consider the material strain tensor
E given by its components,
⎡ ⎤ ⎡ ⎤
EXX EXY EXZ E11 E12 E13
not ⎢ ⎥ ⎢ ⎥
E ≡ ⎣ EXY EYY EY Z ⎦ = ⎣ E12 E22 E23 ⎦ . (2.39)
EXZ EY Z EZZ E13 E23 E33
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
58 C HAPTER 2. S TRAIN
rs
ee
Figure 2.6: Differential segment in the reference configuration.
s gin
t d le En
Consequently,
r
ba
⎡ ⎤⎡ ⎤
ge ro or
eS m
E11 E12 E13 1
ci
⎣ f⎦ ⎣ 0 ⎦ = E11 .
ra
not T
T · E · T ≡ [T] [E] [T] = [1, 0, 0] E12 E22 E23 (2.40)
C d P cs
b
a
E13 E23 E33 0
i
an an n
y ha
.A
√ √ √
d
√ √ √
X Th
er
(2.41)
√ √ √
λ3 = 1 + 2E33 = 1 + 2EZZ ⇒ εZ = λZ − 1 = 1 + 2EZZ − 1
on
.O
C
Remark 2.9. The components EXX , EYY and EZZ (or E11 , E22 and
E33 ) of the main diagonal of tensor E (denoted longitudinal strains)
contain the information on stretch and unit elongations of the dif-
ferential segments that were initially (in the reference configuration)
oriented in the directions X, Y and Z, respectively.
• If EXX = 0 ⇒ εX = 0 : No unit elongation in direction X.
• If EYY = 0 ⇒ εY = 0 : No unit elongation in direction Y .
• If EZZ = 0 ⇒ εZ = 0 : No unit elongation in direction Z.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Physical Interpretation of the Strain Tensors 59
rs
ee
Figure 2.7: Angles between differential segments in the reference and present configu-
s gin
rations.
t d le En
r
Consider now the angle between segments PQ (parallel to the X1 -axis) and PR
ba
ge ro or
eS m
(parallel to the X2 -axis), where Q and R are two particles in the differential neigh-
ci
f
ra
borhood of P in the material configuration and P , Q and R are the respective
C d P cs
b
positions in the spatial configuration (see Figure 2.7). If the angle (Θ = π/2)
a
i
between the segments in the reference configuration is known, the angle θ in
an an n
y ha
the present configuration can be determined using (2.37) and taking into ac-
count their orthogonality ( T(1) · T(2) = 0 ) and the equalities T(1) · E · T(1) = E11 ,
le
liv or ec
.A
cos θ = $ $
d
uu
(2.42)
X Th
er
tin
2E12
=√ √ ,
1 + 2E11 1 + 2E22
on
.O
π 2EXY
θ ≡ θxy = − arcsin √ √ . (2.43)
2 1 + 2EXX 1 + 2EYY
The increment of the final angle with respect to its initial value results in
2EXY
ΔΘXY = θxy − ΘXY = − arcsin √ √ . (2.44)
1 + 2EXX 1 + 2EYY
π/2
Analogous results are obtained starting from pairs of segments that are ori-
ented in different combinations of the coordinate axes, resulting in
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
60 C HAPTER 2. S TRAIN
2EXY
ΔΘXY = − arcsin √ √
1 + 2EXX 1 + 2EYY
2EXZ
ΔΘXZ = − arcsin √ √ . (2.45)
1 + 2EXX 1 + 2EZZ
2EY Z
ΔΘY Z = − arcsin √ √
1 + 2EYY 1 + 2EZZ
rs
ee
Remark 2.10. The components EXY , EXZ and EY Z (or E12 , E13 and
s gin
E23 ) of the tensor E (denoted angular strains) contain the informa-
tion on variation of the angles between the differential segments that
t d le En
were initially (in the reference configuration) oriented in the direc-
tions X, Y and Z, respectively.
r
ba
ge ro or
eS m
• If EXY = 0 : The deformation does not produce a variation in the
ci
f
angle between the two segments initially oriented in the direc-
ra
C d P cs
tions X and Y .
b
a
i
• If EXZ = 0 : The deformation does not produce a variation in the
an an n
y ha
.A
d
uu
e
X Th
er
tin
.O
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Physical Interpretation of the Strain Tensors 61
rs
ee
s gin
Figure 2.8: Physical interpretation of the material strain tensor.
t d le En
r
ented in the direction of the coordinate axes in the present configuration,
ba
ge ro or
eS m
ci
f
ra
C d P cs
1 1 1
b
a
λ1 = √ =√ ⇒ εx = √ −1
i
1 − 2e11 1 − 2exx 1 − 2exx
an an n
y ha
1 1 1
λ2 = √ =$ ⇒ εy = $ −1 ,
le
(2.47)
liv or ec
.A
1 1 1
λ3 = √ =√ ⇒ εz = √ −1
m
er
while the components outside the main diagonal (angular strains) contain infor-
tin
mation on the variation of the angles between the differential segments oriented
on
.O
π 2exy
Δ θxy = −ΘXY = − arcsin √ $
2 1 − 2exx 1 − 2eyy
π 2exz .
Δ θxz = −ΘXZ = − arcsin √ √ (2.48)
2 1 − 2exx 1 − 2ezz
π 2eyz
Δ θyz = −ΘY Z = − arcsin $ √
2 1 − 2eyy 1 − 2ezz
Figure 2.9 summarizes the physical interpretation of the components of the spa-
tial strain tensor.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
62 C HAPTER 2. S TRAIN
rs
Figure 2.9: Physical interpretation of the spatial strain tensor.
ee
s gin
2.8 Polar Decomposition
t d le En
The polar decomposition theorem of tensor analysis establishes that, given a
r
ba
second-order tensor F such that |F| > 0, there exist an orthogonal tensor Q 7 and
ge ro or
eS m
ci
two symmetric tensors U and V such that8
f
ra
C d P cs
not √
⎫
b
a
U = FT · F ⎪
⎪
i
⎬
an an n
not √
y ha
V = F·F T =⇒ F = Q · U = V · Q . (2.49)
⎪
⎪
le
⎭
liv or ec
Q = F · U−1 = V−1 · F
M
.A
This decomposition is unique for each tensor F and is denominated left polar
m
Considering now the deformation gradient tensor and the fundamental re-
er
tin
.O
stretching
C
rotation
©
dx = F · dX = (V · Q) · dX = V · (Q · dX) (2.50)
not
F (•) ≡ stretching ◦ rotation (•)
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Polar Decomposition 63
rotation
stretching
dx = F · dX = (Q · U) · dX = Q · (U · dX) (2.51)
not
F (•) ≡ rotation ◦ stretching (•)
rs
rotation tensor and the mapping y = Q · x is denominated rotation.
ee
A rotation has the following properties:
s gin
• When applied on any vector x, the result is another vector
y = Q · x with the same modulus,
t d le En
y
2 = y·y ≡ [y]T ·[y] = [Q · x]T ·[Q · x] ≡ x·QT · Q ·x = x·x =
x
2 .
not not
r
ba
ge ro or
eS m
1
ci
f
ra
• The result of multiplying (mapping) the orthogonal tensor Q to
C d P cs
b
a
two vectors x(1) and x(2) with the same origin and that form an
i
an an n
.A
d
uu
er
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
64 C HAPTER 2. S TRAIN
rs
ee
s gin
t d le En
Figure 2.10: Polar decomposition.
r
ba
ge ro or
eS m
ci
f
ra
C d P cs
b
a
i
an an n
.A
by U = V = 1 and Q = F.
uu
e
X Th
er
tin
on
.O
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Volume Variation 65
tion are dx(1) = F · dX(1) , dx(2) = F · dX(2) and dx(3) = F · dX(3) . Obviously, the
relations
dx(i) = F · dX(i)
(i) (i) (2.52)
dx j = Fjk dXk i, j, k ∈ {1, 2, 3}
are satisfied. Then, the volumes10 associated with a particle in both configura-
tions can be written as
⎡ (1) (1) (1)
⎤
dX1 dX2 dX3
⎢ (2) ⎥
dV0 = dX(1) × dX(2) · dX(3) = det ⎢ (2) (2) ⎥
⎣ dX1 dX2 dX3 ⎦ = |M| ,
rs
ee
(3) (3) (3)
dX1 dX2 dX3
s gin
[M]
t d le En
⎡ (1) (1) (1)
⎤
r
dx1 dx2 dx3
ba
ge ro or
⎢ (2) ⎥
eS m
dVt = dx(1) × dx(2) · dx(3) = det ⎢ (2) (2) ⎥
⎣ dx1 dx2 dx3 ⎦ = |m| ,
ci
(2.53)
f
ra
C d P cs
(3) (3) (3)
b
a
dx1 dx2 dx3
i
an an n
[m]
y ha
le
liv or ec
(i) (i)
where Mi j = dX j and mi j = dx j . Considering these expressions,
M
.A
(i) (i)
mi j = dx j = Fjk dXk = Fjk dMik = dMik FkTj =⇒ m = M · FT (2.54)
m
d
uu
e
er
tin
.O
dV0 =⇒ dVt = |F|t dV0
⎪
C
⎪
dVt = dV (x (X,t) ,t) = |F (X,t)| dV0 (X, 0) = |F|t dV0 ⎭
©
(2.55)
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
66 C HAPTER 2. S TRAIN
rs
ee
s gin
Figure 2.11: Variation of a volume differential element.
t d le En
r
2.10 Area Variation
ba
ge ro or
eS m
ci
f
Consider an area differential dA associated with a particle P in the reference
ra
C d P cs
configuration and its variation along time. To define this area differential, con-
b
a
i
sider two particles Q and R in the differential neighborhood of P, whose relative
an an n
positions with respect to this particle are dX(1) and dX(2) , respectively (see Fig-
y ha
ure 2.12). Consider also an arbitrary auxiliary particle S whose relative position
le
liv or ec
.A
differential area, dA, is defined. The module of vector dA is dA and its direction
is the same as that of the unit normal vector in the material configuration N.
m
e
space P and will have an area differential da associated with it which, in turn,
X Th
normal vector in the spatial configuration. Consider also the positions of the
other particles Q , R and S and their relative position vectors dx(1) , dx(2) and
on
.O
dx(3) .
C
The volumes dV0 and dVt of the corresponding parallelepipeds can be calcu-
©
lated as
dV0 = dH dA = dX(3) · N dA = dX(3) · N dA = dA · dX(3)
dH dA (2.56)
(3)
dVt = dh da = dx · n da = dx · n da = da · dx(3)
(3)
dh da
and, taking into account that dx(3) = F · dX(3) , as well as the expression for
change in volume (2.55), results in
da · F · dX(3) = da · dx(3) = dVt = |F| dV0 = |F| dA · dX(3) ∀dX(3) . (2.57)
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
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Infinitesimal Strain 67
rs
ee
Figure 2.12: Variation of an area differential.
s gin
t d le En
Comparing the first and last terms12 in (2.57) and considering that the relative
r
position of particle S can take any value (as can, therefore, vector dX(3) ), finally
ba
ge ro or
eS m
ci
yields
f
ra
da · F = |F| dA =⇒ da = |F| dA · F−1 .
C d P cs
(2.58)
b
a
i
an an n
To obtain the relation between the two area differential scalars, dA and da,
y ha
( (
M
.A
er
tin
.O
12 Here, the following tensor algebra theorem is taken into account: given two vectors a and
b, if the relation a · x = b · x is satisfied for all values of x, then a = b.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
68 C HAPTER 2. S TRAIN
rs
ee
s gin
Figure 2.13: Infinitesimal strain in the continuous medium.
t d le En
r
ba
ge ro or
eS m
In accordance with the first hypothesis, the reference configuration Ω0 and
ci
f
the present configuration Ωt are very close together and are considered to be
ra
C d P cs
b
indistinguishable from one another. Consequently, the material and spatial co-
a
i
ordinates coincide and discriminating between material and spatial descriptions
an an n
y ha
.A
⎩ xi = Xi + ui ∼
= Xi ⎩ Ui (X,t) not
= ui (X,t) ≡ ui (x,t) i ∈ {1, 2, 3}
m
(2.60)
d
uu
" "
er
tin
" ∂ ui "
" " ∀ i, j ∈ {1, 2, 3} .
"∂xj "
1 (2.61)
on
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Infinitesimal Strain 69
⎧
⎪
⎪ 1
T + JT · J ∼ 1 J + JT ,
⎪
⎪ E = J + J =
⎪
⎨ 2 2
1 ∂ ui ∂ u j ∂ uk ∂ uk ∼ 1 ∂ ui ∂ u j (2.63)
⎪
⎪ Ei j = + + = + ,
⎪
⎪ 2 ∂ x j ∂ xi ∂ xi ∂ x j 2 ∂ x j ∂ xi
⎪
⎩
1
where the infinitesimal character of the second-order term (∂ uk ∂ uk /∂ x j ∂ xi ) has
been taken into account. Operating in a similar manner with the spatial strain
tensor,
⎧
rs
⎪
⎪ 1
T − jT · j ∼ 1 j + jT = 1 J + JT ,
⎪
ee
⎪
⎪ e = j + j =
⎨ 2
2 2
s gin
1 ∂ ui ∂ u j ∂ uk ∂ uk ∼ 1 ∂ ui ∂ u j (2.64)
⎪
⎪ e ij = + − = + .
⎪
⎪ ∂ ∂ ∂ ∂ ∂ ∂
t d le En
⎪ 2 x j x i x i x j 2 x j xi
⎩
1
r
ba
ge ro or
eS m
ci
Equations (2.63) and (2.64) allow defining the infinitesimal strain tensor (or
f
ra
small strain tensor) ε as13
C d P cs
b
a
⎧
i
an an n
⎪
⎪ 1 not
y ha
⎪
⎨ε = J + JT = ∇s u
le
Infinitesimal 2
liv or ec
(2.65)
strain tensor ⎪
⎪ 1 ∂ ui ∂ u j
⎪ ε
⎩ ij = + i, j ∈ {1, 2, 3}
M
.A
2 ∂ x j ∂ xi
m
d
uu
e
X Th
er
tin
.O
strain tensor.
E (x,t) = e (x,t) = ε (x,t)
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
70 C HAPTER 2. S TRAIN
Example 2.4 – Determine under which conditions the motion in Example 2.1
constitutes an infinitesimal strain case and obtain the infinitesimal strain ten-
rs
sor for this case. Compare it with the result obtained from the spatial and
ee
material strain tensors in Example 2.2 taking into account the infinitesimal
strain hypotheses.
s gin
t d le En
Solution
r
The equation of motion is given by
ba
ge ro or
eS m
⎧
ci
⎨ x1 = X1 − AX3
⎪ f
ra
C d P cs
b
a
x2 = X2 − AX3 ,
⎪
i
⎩
an an n
x3 = −AX1 + AX2 + X3
y ha
le
liv or ec
⎡ ⎤
.A
U1 = −AX3
not ⎢ ⎥
m
U (X,t) = x − X ≡ ⎣ U2 = −AX3 ⎦.
d
uu
e
U3 = −AX1 + AX2
X Th
er
tin
.O
finitesimal (A
1). Now, to obtain the infinitesimal strain tensor, first the
displacement gradient tensor J (X,t) = j (x,t) must be computed,
C
⎡ ⎤ ⎡ ⎤
−AX3 0 0 −A
not ⎢ ⎥ ∂ ∂ ∂ ⎢ ⎥
J = U⊗∇ ≡ ⎢ ⎣ −AX3 ⎥
⎦ ∂ X1 , ∂ X2 , ∂ X3 = ⎣ 0 0 −A ⎦ .
−AX1 + AX2 −A A 0
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Infinitesimal Strain 71
The material and spatial strain tensors obtained in Example 2.2 are, respec-
tively,
⎡ 2 ⎤
A −A2 −2A
not 1 ⎢ ⎥
E (X,t) ≡ ⎣ −A2 A2 0 ⎦ and
2
−2A 0 2A2
⎡ ⎤
−3A2 − 2A4 A2 + 2A4 −2A − 2A3
not 1 ⎢ ⎥
e (X,t) ≡ ⎣ A2 + 2A4 A2 − 2A4 2A3 ⎦.
2
−2A − 2A 3 3 −2A 2
rs
2A
ee
Neglecting
4 the second-order
and higher-order infinitesimal terms
A
A3
A2
A results in
s gin
⎡ ⎤ ⎡ ⎤
t d le En
0 0 −A 0 0 −A
not ⎢ ⎥ not ⎢ ⎥
0 −A ⎦ =⇒ E = e = ε ,
r
E≡⎣ 0 0 −A ⎦ and e≡⎣ 0
ba
ge ro or
eS m
−A A −A A
ci
0 0
f
ra
C d P cs
b
a
which is in accordance with Remark 2.14.
i
an an n
y ha
Considering
the general expression
√ (2.30) of the unit elongation in the direction
M
.A
yields
uu
e
√
X Th
er
1 + 2t · ε · t ∼
tin
λt = = 1+t·ε ·t
(2.66)
on
x
.O
εt = λt − 1 = t · ε · t
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
72 C HAPTER 2. S TRAIN
rs
ee
Consider a differential segment PQ oriented in the reference configuration
s gin
parallel to the coordinate axis x1 ≡ x. The stretch λx and the unit elongation εx
in this direction are, according to (2.66) with t = [1, 0, 0]T ,
t d le En
λx = 1 + t · ε · t = 1 + εxx =⇒ εx = λx − 1 = εxx .
r
(2.68)
ba
ge ro or
eS m
ci
This allows assigning to the component εxx ≡ ε11 the physical meaning of unit
f
ra
elongation εx in the direction of the coordinate axis x1 ≡ x. A similar interpre-
C d P cs
b
a
tation is deduced for the other components in the main diagonal of the tensor
i
an an n
Given now the components outside the main diagonal of ε , consider the dif-
M
.A
the coordinate directions x and y, respectively. Then, these two segments form
d
an angle Θxy = π/2 in this configuration. Applying (2.43), the increment in the
uu
e
er
tin
π εxy ∼
Δ θxy = θxy − = −2 arcsin $ $ = −2 arcsin εxy = −2εxy ,
on
.O
2 1 + 2εxx 1 + 2εyy
εxy
C
1 1
©
(2.70)
where the infinitesimal character of εxx , εyy and εxy has been taken into account.
Consequently, εxy can be interpreted from (2.70) as minus the semi-increment,
produced by the strain, of the angle between the two differential segments ini-
tially oriented parallel to the coordinate directions x and y. A similar interpre-
tation is deduced for the other components εxz and εyz ,
1 1 1
εxy = − Δ θxy ; εxz = − Δ θxz ; εyz = − Δ θyz . (2.71)
2 2 2
15 The Taylor series expansion of arcsin x around x = 0 is arcsin x = x + O x2 .
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
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Infinitesimal Strain 73
engineering notation
⎡
scientific notation
⎤
⎡ ⎤ ⎡ ⎤
ε11 ε12 ε13 εxx εxy εxz εx 2 γxy 2 γxz
1 1
⎢ ⎥ ⎢ ⎥ ⎢ ⎥
ε22 ε23 ⎦ ≡ ⎣ εxy εyy εyz ⎦ ≡ ⎢ ⎥
not
ε ≡ ⎣ ε12
rs
⎣2
1
γxy ε y
1
2 γyz ⎦ (2.72)
ee
ε13 ε23 ε33 εxz εyz εzz
2 γxz 2 γyz εz
1 1
s gin
t d le En
r
Remark 2.17. The components in the main diagonal of the strain ten-
ba
ge ro or
eS m
sor (named longitudinal strains) are denoted by ε(•) and coincide
ci
f
ra
with the unit elongations in the directions
of the coordinate axes.
C d P cs
Positive values of longitudinal strains ε(•) > 0 correspond to an
b
a
i
increase in length of the corresponding differential segments in the
an an n
y ha
reference configuration.
le
liv or ec
M
.A
m
er
.O
de f
T
ε ∈ R6 ε = εx , εy , εx , γxy , γxz , γyz
(2.73)
longitudinal angular
strains strains
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
74 C HAPTER 2. S TRAIN
rs
1
1
ee
where T(1) and T(2) are the unit vectors
s gin
( ( (in the( directions of PQ and PR and,
(1) (2) ( (1) ( ( (2) (
therefore, the relation T · T = (T ( (T ( cosΘ = cosΘ is fulfilled. Con-
t d le En
sidering the infinitesimal character of the components of ε and Δ θ , the follow-
ing holds true16 .
r
ba
ge ro or
eS m
ci
f
cos θ = cos (Θ + Δ θ ) = cosΘ · cos Δ θ − sinΘ · sin Δ θ =
ra
C d P cs
b
a
≈1 ≈ Δθ
i
an an n
y ha
= cosΘ
le
liv or ec
.A
≈1 ≈1
d
uu
er
tin
.O
Δθ = − =− , (2.76)
©
sinΘ sin θ
where the infinitesimal character of the strain has been taken into account and,
thus, it follows that T(1) ≈ t(1) , T(2) ≈ t(2) and Θ ≈ θ .
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Infinitesimal Strain 75
rs
ee
Figure 2.15: Variation of the angle between two differential segments in infinitesimal
strain.
s gin
t d le En
ing (2.12) and the infinitesimal character of the components of the tensor J
r
(see (2.61)), tensor U in (2.49) can be written as17
ba
ge ro or
eS m
$ $ ⎫
ci
U = FT · F = (1 + JT ) · (1 + J) = f ⎪
ra
⎪
C d P cs
⎪
⎬
b
a
$ $
i
1
= 1 + J + J + J · J ≈ 1 + J + J = 1 + J + J ⎪ =⇒ U = 1 + ε .
an an n
T T T T
⎪
y ha
⎪
2
⎭
J ε
le
liv or ec
(2.77)
M
.A
1
uu
U−1 = (1 + ε )−1 = 1 − ε = 1 −
e
J + JT . (2.78)
X Th
2
er
tin
.O
⎫
1 ⎪
C
−1
Q = F · U = (1 + J) · 1 − J + J T = ⎪
⎪
⎪
⎪
©
2 ⎬
1 1 1 =⇒ Q = 1 + Ω .
= 1 + J − J + JT − J · J + JT = 1 + J − JT ⎪ ⎪
⎪
2 2 2 ⎪ ⎪
⎭
J Ω
(2.79)
√ √
17 The Taylor series expansions of tensor 1 + x around x = 0 is 1 + x = 1 + x/2 + O x2 .
18 The Taylor series expansions of tensor (1 + x)−1 around x = 0 is (1 + x)−1 = 1 − x +
O x2 .
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
76 C HAPTER 2. S TRAIN
rs
⎧
ee
⎨ Ω T = 1 J − JT T = 1 JT − J = −Ω Ω
s gin
2 2 .
⎩ Ω = −Ω i, j ∈ {1, 2, 3}
ji ij
t d le En
Consequently, the terms in the main diagonal of Ω are zero, and its
r
ba
ge ro or
eS m
matrix of components has the structure
ci
⎡ f ⎤
ra
C d P cs
0 Ω12 −Ω31
b
a
Ω] = ⎣ −Ω12 Ω23 ⎦ .
i
[Ω 0
an an n
y ha
Ω31 −Ω23 0
le
liv or ec
M
.A
⎡ ⎤
on
.O
Infinitesimal ∂ u 3 ∂ u2
rotation vector: ⎡ ⎤ ⎡ ⎤ ⎢ − ⎥
C
θ1 −Ω23 ⎢ ∂ x2 ∂ x3 ⎥
⎢ ⎥ 1
©
not ⎢ ⎥ ⎢ ⎥ 1⎢ ⎥ de f
θ ≡ ⎣ θ2 ⎦ = ⎣ −Ω31 ⎦ = ⎢ ∂ u1 − ∂ u3 ⎥ = ∇ × u . (2.81)
2 ⎢ ∂ x3 ∂ x1 ⎥ 2
θ3 −Ω12 ⎢ ⎥
⎣ ∂u ∂ u1 ⎦
2
−
∂ x1 ∂ x2
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Infinitesimal Strain 77
rs
Figure 2.16) coincide. Indeed,
ee
⎡ ⎤⎡ ⎤ ⎡ ⎤
s gin
0 Ω12 −Ω31 r1 Ω12 r2 − Ω31 r3
not ⎢ ⎥⎢ ⎥ ⎢ ⎥
Ω · r ≡ ⎣ −Ω12 0 Ω23 ⎦ ⎣ r2 ⎦ = ⎣ −Ω12 r1 + Ω23 r3 ⎦ ,
t d le En
Ω31 −Ω23 0 r3 Ω31 r1 − Ω23 r2
r
ba
ge ro or
⎡ ⎤ ⎡ ⎤
eS m
ci
ê1 ê2 ê3 ê1 ê2 ê3
f
ra
not ⎢ ⎥ ⎢ ⎥
C d P cs
θ × r ≡ ⎣ θ1 θ2 θ3 ⎦ = ⎣ −Ω23 −Ω31 −Ω12 ⎦ =
b
a
i
an an n
r1 r2 r3 r1 r2 r3
⎡ ⎤
y ha
Ω12 r2 − Ω31 r3
le
⎢ ⎥
liv or ec
= ⎣ −Ω12 r1 + Ω23 r3 ⎦ .
M
.A
Ω31 r1 − Ω23 r2
m
.O
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
78 C HAPTER 2. S TRAIN
rs
ee
Figure 2.16: Product of the infinitesimal rotation vector and tensor on a vector r.
s gin
t d le En
Consider now a differential segment dX in the neighborhood of a particle P
in the reference configuration (see Figure 2.17). In accordance with (2.82), the
r
ba
ge ro or
stretching transforms this vector into vector dx as follows.
eS m
ci
f
ra
stretching rotation
C d P cs
b
a
dx = F · dX = (1 + ε + Ω) · dX = ε · dX + (1 + Ω) · dX
i
an an n
(2.83)
y ha
le
F (•) ≡ stretching (•) + rotation (•)
liv or ec
M
.A
m
d
uu
.O
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Volumetric Strain 79
rs
ee
s gin
t d le En
2.12 Volumetric Strain
r
ba
ge ro or
eS m
ci
f
ra
C d P cs
Definition 2.6. The volumetric strain is the increment produced by
b
a
i
the deformation of the volume associated with a particle, per unit of
an an n
le
liv or ec
M
.A
dV (X, 0) dV0
X Th
er
tin
on
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
80 C HAPTER 2. S TRAIN
• Infinitesimal strain
Considering (2.49) and recalling that Q is an orthogonal tensor (|Q| = 1), yields
rs
⎡ ⎤
ee
1 + εxx εxy εxz
⎢ ⎥
s gin
|F| = |Q · U| = |Q| |U| = |U| = |1 + ε | = det ⎣ εxy 1 + εyy εyz ⎦ ,
εxz εyz 1 + εzz
t d le En
(2.86)
r
ba
where (2.77) has been considered. Taking into account that the components of ε
ge ro or
eS m
ci
are infinitesimal, and neglecting in the expression of its determinant the second-
f
ra
order and higher-order infinitesimal terms, results in
C d P cs
b
a
⎡ ⎤
i
an an n
⎢ ⎥
|F| = det ⎣ εxy 1 + εyy εyz ⎦ = 1 + εxx + εyy + εzz +O ε 2 ≈ 1 + Tr (εε ) .
le
liv or ec
.A
(2.87)
m
Then, introducing (2.87) into (2.85) yields, for the infinitesimal strain case
d
⎫
uu
e
er
tin
.O
dV0
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
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Strain Rate 81
Then, ⎧
⎪
⎪ ∂v
⎪
⎪
rs
⎨ dv = · dx = l · dx
∂x
ee
, (2.90)
⎪
⎪
⎪ ∂ vi
s gin
⎪
⎩ dvi = dx j = li j dx j i, j ∈ {1, 2, 3}
∂xj
t d le En
where the spatial velocity gradient tensor l (x,t) has been introduced.
r
ba
⎧
ge ro or
eS m
⎪ de f ∂ v (x,t)
ci
⎪ l (x,t) =
⎪
⎪ f
ra
⎪
⎨ ∂x
C d P cs
b
a
Spatial velocity
l = v⊗∇
i
(2.91)
gradient tensor ⎪
an an n
⎪
⎪ ∂ vi
y ha
⎪
⎪ i, j ∈ {1, 2, 3}
⎩ li j =
∂xj
le
liv or ec
M
.A
m
d
uu
e
X Th
er
tin
on
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
82 C HAPTER 2. S TRAIN
rs
ij i, j ∈ {1, 2, 3}
2 ∂ x j ∂ xi
ee
rate (2.93)
⎪
tensor ⎪
⎪
⎪
⎡ ⎤
s gin
⎪
⎪ d 11 d 12 d 13
⎪
⎪
⎪
⎪ [d] = ⎣ d12 d22 d23 ⎦
⎩
t d le En
d13 d23 d33
r
ba
ge ro or
eS m
ci
and w is an antisymmetric tensor denominated rotation rate tensor or spin ten-
f
ra
sor, whose expression is
C d P cs
b
a
i
⎧
an an n
⎪
⎪ 1 1
y ha
de f
⎪
⎪ w = skew (l
l ) = l − l T not
= (v ⊗ ∇ − ∇ ⊗ v) = ∇a v
⎪
⎪
le
⎪ 2 2
liv or ec
Rotation⎪
⎪
⎨ w = 1 ∂ vi − ∂ v j
⎪
i, j ∈ {1, 2, 3}
M
.A
rate ij
2 ∂ x j ∂ xi (2.94)
(spin) ⎪
⎪ ⎡ ⎤
tensor ⎪
m
⎪
⎪ 0 w12 −w31
d
⎪
⎪
uu
⎪
⎪ [w] = ⎣ −w12 0 w23 ⎦
e
⎪
⎩
X Th
er
w31 −w23 0
tin
on
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Strain Rate 83
rs
dt dt
ee
Replacing (2.2) into (2.97) results in22
s gin
.
dX · E · dX = dx · d · dx ≡ [dx]T [d] [dx] = [dX]T FT · d · F [dX]
not
t d le En
. .
=⇒ dX · FT · d · F − E · dX = 0 ∀ dX =⇒ FT · d · F − E = 0
r
ba
ge ro or
eS m
.
ci
E = FT · d · F .
f (2.98)
ra
C d P cs
b
a
i
an an n
y ha
.
the material strain tensor E (X,t), providing a physical interpreta-
M
.A
exactly the same. Both tensors will coincide in the following cases:
uu
"
e
= 1.
er
tin
t=t0
∂x
on
∂X
C
22 Here, the following tensor algebra theorem is used: given a second-order tensor A, if
x · A · x = 0 is verified for all vectors x = 0, then A ≡ 0.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
84 C HAPTER 2. S TRAIN
Figure 2.20: Differential segment between particles of the continuous medium along
rs
time.
ee
s gin
2.13.4 Physical Interpretation of the Rotation Rate Tensor
t d le En
Taking into account the antisymmetric character of w (which implies it can be
r
defined using only three different components), the vector
ba
ge ro or
eS m
⎡ ⎤
ci
∂ v 2 ∂ v3 f
ra
⎢−
C d P cs
− ⎥ ⎡ ⎤
b
a
⎢ ⎥
⎢ ∂ x3 ∂ x2 ⎥ −w23
i
an an n
not 1 ⎢ ∂ v 3 ∂ v1 ⎥
1 1
ω = rot (v) = ∇ × v ≡ ⎢ ⎥=⎢ ⎥
y ha
⎢ − − ⎥ ⎣ −w31 ⎦ (2.99)
2 2 2⎢ ∂ x1 ∂ x3 ⎥
le
⎢ ⎥ −w12
liv or ec
⎣ ∂ v 1 ∂ v2 ⎦
− −
M
.A
∂ x2 ∂ x1
m
er
tin
ω ×r = w·r ∀r (2.100)
on
.O
rotation motion, and ω × r = w · r as the rotation velocity of the point that has r
©
as the position vector with respect to the rotation center (see Figure 2.21). Then,
considering (2.90) and (2.92),
dv = l · dx = (d + w) · dx = d · dx + w · dx , (2.101)
stretch rotation
velocity velocity
which allows describing the relative velocity dv of the particles in the neigh-
borhood of a given particle P (see Figure 2.22) as the sum of a relative stretch
23 Observe the similarity in the structure of tensors Ω and θ in Section 2.11.6 and of tensors
w and ω seen here.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
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Material Time Derivatives of Strain and Other Magnitude Tensors 85
rs
ee
Figure 2.21: Vorticity vector.
s gin
t d le En
r
ba
ge ro or
eS m
ci
f
ra
C d P cs
b
a
i
an an n
y ha
le
liv or ec
M
.A
d
uu
e
velocity (characterized by the strain rate tensor d) and a relative rotation velocity
X Th
ω ).
er
.O
Magnitude Tensors
©
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
86 C HAPTER 2. S TRAIN
dF not .
= F = l ·F
dt (2.102 (cont.))
dFi j .
= Fi j = lik Fk j i, j ∈ {1, 2, 3}
rs
dt
ee
s gin
where (2.91) has been taken into account for the velocity gradient tensor l . To
obtain the material time derivative of tensor F−1 , the time derivative of the iden-
tity F · F−1 = 1 is performed25 .
t d le En
r
d dF −1 d F−1
ba
ge ro or
−1 −1
F · F = 1 =⇒ F·F = ·F +F· =0
eS m
ci
−1 dt dt dt
f
ra
d F .
C d P cs
=⇒ = −F−1 · F · F−1 = −F−1 · l · F · F−1 = −F−1 · l =⇒
b
a
dt
i
an an n
l ·F 1
y ha
−1
le
liv or ec
d F
= −F−1 · l
M
dt
.A
(2.103)
dFi−1
= Fik−1 lk j
j
m
i, j ∈ {1, 2, 3}
d
dt
uu
e
X Th
er
tin
.O
1 T . 1 .T .
©
dE
E= F · F − 1 =⇒ =E= F · F + FT · F =
2 dt 2
1 T T 1
= F · l · F + FT · l · F = FT · l + l T · F = FT · d · F
2 2
.
=⇒ E = F · d · F .
T 2d (2.104)
25 The material time derivative of the inverse tensor d F−1 /dt must not be confused with
. −1
the inverse of the material derivative of the tensor: F . These two tensors are completely
different tensors.
26 Observe that the result is the same as the one obtained in (2.98) using an alternative pro-
cedure.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
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Material Time Derivatives of Strain and Other Magnitude Tensors 87
Using (2.23) and (2.103) for the spatial strain tensor e yields
1 −T −1
de . 1 d −T −1 d −1
e= 1−F ·F ⇒ =e=− F · F + F−T · F =
2 dt 2 dt dt
1 T −T −1
= l · F · F + F−T · F−1 · l
2
. 1 T −T −1
=⇒ e = l · F · F + F−T · F−1 · l . (2.105)
2
rs
2.14.3 Volume and Area Differentials
ee
The volume differential dV (X,t) associated with a certain particle P varies
s gin
along time (see Figure 2.23) and, in consequence, it makes sense to calculate
t d le En
its material derivative. Differentiating (2.55) for a volume differential results in
r
d d |F|
ba
ge ro or
dV (X,t) = |F (X,t)| dV0 (X) =⇒ dV (t) = dV0 .
eS m
(2.106)
ci
dt dt
f
ra
C d P cs
Therefore, the material derivative of the determinant of the deformation gradient
b
a
tensor |F| is27
i
an an n
y ha
dt dFi j dt dt
M
.A
lik Fk j [ F·F−1 ] =δ
ki ki
m
∂ vi d |F|
uu
∂ xi dt
X Th
er
tin
where (2.102) and (2.91) have been considered. Introducing (2.107) into (2.106)
on
.O
d
©
27 The derivative of the determinant of a tensor A with respect to the same tensor can be
written in compact notation as d |A|/dA = |A| · A−T or, in index notation, as d |A|/dAi j =
|A| · A−1
ji .
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
88 C HAPTER 2. S TRAIN
rs
Figure 2.23: Variation of the volume differential.
ee
s gin
pression, results in
t d le En
d d d |F| d −1
(da) = |F| · dA · F−1 = dA · F−1 + |F| · dA =
r
F
ba
ge ro or
dt dt dt
dt
eS m
ci
|F| ∇ · v −F−1 · l
f
ra
C d P cs
b
a
= (∇ · v) |F| dA · F−1 − |F| dA · F−1 · l =⇒
i
an an n
y ha
da da
le
liv or ec
d
(da) = (∇ · v) da − da · l = da · ((∇ · v) 1 − l ) , (2.109)
M
.A
dt
m
er
tin
on
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Motion and Strains in Cylindrical and Spherical Coordinates 89
rs
Remark 2.23. An orthogonal curvilinear coordinate system (gener-
ee
ically referred to as {a, b, c}), is characterized by its physical unit
s gin
basis {êa , êb , êc } (
êa
=
êb
=
êc
= 1), whose components are
orthogonal to each other (êa · êb = êa · êc = êb · êc = 0), as is also the
t d le En
case in a Cartesian system. The fundamental difference is that the
r
orientation of the curvilinear basis changes at each point in space
ba
ge ro or
(êm ≡ êm (x) m ∈ {a, b, c}). Therefore, for the purposes here, an
eS m
ci
f
orthogonal curvilinear coordinate system can be considered as a mo-
ra
bile Cartesian coordinate system {x , y , z } associated with a curvi-
C d P cs
b
a
linear basis {êa , êb , êc } (see Figure 2.25).
i
an an n
y ha
le
liv or ec
M
.A
er
tin
⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
va vx Taa Tab Tac Tx x Tx y Tx z
on
.O
not ⎢ ⎥ ⎢ ⎥ not ⎢ ⎥ ⎢ ⎥
v ≡ ⎣ vb ⎦ ≡ ⎣ vy ⎦ T ≡ ⎣ Tba Tbb Tbc ⎦ ≡ ⎣ Ty x Ty y Ty z ⎦
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
90 C HAPTER 2. S TRAIN
rs
• Nabla operator, ∇
ee
T
s gin
∂ 1 ∂ ∂ ∂
not 1 ∂ ∂
∇ = êr + êθ + êz =⇒ ∇≡ , , (2.111)
∂r r ∂θ ∂z ∂r r ∂θ ∂z
t d le En
r
ba
ge ro or
⎡ ⎤
eS m
x = r cos θ
ci
f x (r, θ , z) ≡ ⎣ y = r sin θ ⎦
not
ra
C d P cs
b z=z
a
i
an an n
y ha
le
liv or ec
M
.A
m
d
uu
e
X Th
er
tin
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Motion and Strains in Cylindrical and Spherical Coordinates 91
u ≡ [ur , uθ , uz ]T
not
u = ur êr + uθ êθ + uz êz =⇒ (2.112)
v ≡ [vr , vθ , vz ]T
not
v = vr êr + vθ êθ + vz êz =⇒ (2.113)
rs
not ⎢ ⎥ ⎢ ⎥
ε= (u ⊗ ∇) + (u ⊗ ∇)T ≡ ⎣ εx y εy y εy z ⎦ = ⎣ εrθ εθ θ εθ z ⎦
ee
2
εx z εy z εz z εrz εθ z εzz
s gin
∂ ur 1 ∂ uθ ur ∂ uz
t d le En
εrr = εθ θ = + εzz =
∂r r ∂θ r ∂z
r
ba
ge ro or
1 1 ∂ ur ∂ uθ uθ ∂ ur ∂ uz
eS m
1
εrθ = + − εrz = +
ci
2 r ∂θ ∂r f ∂z ∂r
ra
r 2
C d P cs
b
a
1 ∂ u θ 1 ∂ uz
i
an an n
εθ z = + (2.114)
2 ∂z r ∂θ
y ha
le
liv or ec
.A
Figure (2.26).
m
⎡ ⎤ ⎡ ⎤
er
tin
.O
2
C
∂ vr 1 ∂ vθ vr ∂ vz
drr = dθ θ = + dzz =
∂r r ∂θ r ∂z
1 1 ∂ vr ∂ vθ vθ 1 ∂ v r ∂ vz
drθ = + − drz = +
2 r ∂θ ∂r r 2 ∂z ∂r
1 ∂ vθ 1 ∂ vz
dθ z = + (2.115)
2 ∂z r ∂θ
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
92 C HAPTER 2. S TRAIN
rs
• Nabla operator, ∇
ee
s gin
∂ 1 ∂ 1 ∂ ∂not 1 ∂ 1 ∂ T
∇ = êr + êθ + êφ =⇒ ∇≡ , ,
∂r r ∂θ r sin θ ∂ φ ∂ r r ∂ θ r sin θ ∂ φ
t d le En
(2.117)
r
ba
ge ro or
eS m
ci
• Displacement vector, u, and velocity vector, v f
ra
C d P cs
b
a
not T
i
u = ur êr + uθ êθ + uφ êφ =⇒ u ≡ ur , uθ , uφ
an an n
(2.118)
y ha
not T
le
v = vr êr + vθ êθ + vφ êφ =⇒ v ≡ vr , vθ , vφ (2.119)
liv or ec
M
.A
m
d
uu
e
X Th
er
tin
⎡ ⎤
x = r sin θ cos φ
on
.O
z = z cos θ
©
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Motion and Strains in Cylindrical and Spherical Coordinates 93
∂ ur 1 ∂ uθ ur
εrr = εθ θ = +
∂r r ∂θ r
1 ∂ uφ uθ ur
rs
εφ φ = + cot φ +
r sin θ ∂ φ r r
ee
1 1 ∂ ur ∂ uθ uθ 1 ∂ ur ∂ uφ uφ
s gin
1
εrθ = + − εrφ = + −
2 r ∂θ ∂r r 2 r sin θ ∂ φ ∂r r
t d le En
1 1 ∂ uθ 1 ∂ uφ uφ
r
εθ φ = + − cot φ (2.120)
ba
ge ro or
2 r sin θ ∂ φ r ∂θ r
eS m
ci
f
ra
The components of ε are presented on the corresponding differential element in
C d P cs
b
a
Figure 2.28.
i
an an n
y ha
le
• Strain rate tensor, d
liv or ec
⎡ ⎤ ⎡ ⎤
M
.A
2
uu
er
tin
∂ vr 1 ∂ vθ vr
drr = dθ θ = +
on
∂r r ∂θ
.O
r
1 ∂ vφ vθ
C
vr
dφ φ = + cot φ +
©
r sin θ ∂ φ r r
1 1 ∂ vr ∂ vθ vθ 1 1 ∂ vr ∂ vφ vφ
drθ = + − drφ = + −
2 r ∂θ ∂r r 2 r sin θ ∂ φ ∂r r
1 1 ∂ vθ 1 ∂ vφ vφ
dθ φ = + − cot φ (2.121)
2 r sin θ ∂ φ r ∂θ r
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
94 C HAPTER 2. S TRAIN
rs
ee
s gin
t d le En
r
ba
ge ro or
Figure 2.28: Differential element in spherical coordinates.
eS m
ci
f
ra
C d P cs
b
a
i
an an n
y ha
le
liv or ec
M
.A
m
d
uu
e
X Th
er
tin
on
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Problems and Exercises 95
P ROBLEMS
Problem 2.1 – A deformation that takes place in a continuous medium has the
following consequences on the triangle shown in the figure below:
1. The segment OA increases its initial length in (1 + p).
2. The angle AOB decreases in q radians its initial value.
rs
3. The area increases its initial value in (1 + r).
ee
4. p, q, r, s
1.
s gin
The deformation is uniform and the z-axis is one of the principal directions of
t d le En
the deformation gradient tensor, which is symmetric. In addition, the stretch in
this direction is known to be λz = 1 + s. Obtain the infinitesimal strain tensor.
r
ba
ge ro or
eS m
ci
f
ra
C d P cs
b
a
i
an an n
y ha
le
liv or ec
M
.A
m
d
uu
e
X Th
er
tin
on
.O
Solution
C
A uniform deformation implies that the deformation gradient tensor (F) does
©
not depend on the spatial variables. Consequently, the strain tensor (E) and the
stretches (λ ) do not depend on them either. Also, note that the problem is to be
solved under infinitesimal strain theory.
The initial and final lengths of a segment parallel to the x-axis are related as
follows.
# A # A ⎫
OA f inal = λx dX = λx dX = λx OAinitial ⎬
O O =⇒ λx = 1 + p
⎭
OA f inal = (1 + p) OAinitial
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
96 C HAPTER 2. S TRAIN
Also, an initial right angle (the angle between the x- and y-axes) is related to its
corresponding final angle after the deformation through
⎫
π ⎪
⎪
initial angle = ⎬ q
2 =⇒ Δ Φ xy = −γxy = −2ε xy = −q =⇒ εxy = .
π ⎪
⎪ 2
final angle = + Δ Φ xy ⎭
2
In addition, F is symmetric and the z-axis is a principal direction, therefore
⎡ ⎤
∂ ux ∂ ux ∂ ux
rs
⎡ ⎤ ⎢1+ ∂x ∂y ∂z ⎥
⎢ ⎥
ee
F11 F12 0
not ⎢ ⎥ not not ⎢ ∂ uy ∂ uy ∂ uy ⎥
F ≡ ⎣ F12 F22 0 ⎦ ≡ 1 + J ≡ ⎢ ⎢ ⎥,
s gin
1+ ⎥
⎢ ∂ x ∂ y ∂ z ⎥
0 0 F33 ⎣ ∂ uz ∂ uz ∂ uz ⎦
t d le En
1+
∂x ∂y ∂z
r
ba
ge ro or
eS m
which reveals the nature of the components of the displacement vector,
ci
⎧ f
ra
C d P cs
⎪ ∂ ∂
bux (x, y) ,
a
⎪
⎪ u x u y
⎨ = = 0 =⇒
i
an an n
∂z ∂z uy (x, y) ,
y ha
⎪
⎪ ∂ u z ∂ uz
⎪
⎩ = = 0 =⇒ uz (z) .
le
liv or ec
∂x ∂y
M
.A
d
1 ∂ ux ∂ u z
uu
εxz = + = 0 =⇒ εxz = 0
e
2 ∂z ∂x
X Th
er
tin
1 ∂ ux ∂ u z
εxz = + = 0 =⇒ εxz = 0
on
.O
2 ∂z ∂x ⎫
∂ uz ⎬
C
εzz = = λz − 1
∂z =⇒ εzz = s
©
λ = 1+s ⎭
z
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Problems and Exercises 97
rs
not ⎢ ⎥ not ⎢ ⎥
dA0 ≡ ⎣ 0 ⎦ =⇒ dA0 · F−1 ≡ ⎢ ⎣
0 ⎥.
⎦
ee
1
dA0 dA0
s gin
1+s
t d le En
Then, taking into account that |F| = Tr (εε ) + 1, and neglecting second-order
terms results in
r
ba
⎫
ge ro or
eS m
dA = (1 + r) dA0 ⎬
ci
f
=⇒ εyy = r − p .
ra
1
C d P cs
dA = (1 + p + s + εyy ) dA0 ⎭
b
a
1+s
i
an an n
y ha
q
M
.A
⎢p 2
0⎥
not ⎢ q ⎥
ε ≡⎢ 0⎥⎥ .
m
⎢2 r− p
d
⎣ ⎦
uu
e
0 0 s
X Th
er
tin
on
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
98 C HAPTER 2. S TRAIN
Then,
a) Justify why the infinitesimal strain theory cannot be used here.
b) Determine the deformation gradient tensor, the possible values of p and the
displacement field in its material and spatial forms.
c) Draw the deformed solid.
Solution
rs
a) The angle AOC changes from 90◦ to 45◦ therefore, it is obvious that the
ee
deformation involved is not infinitesimal. In addition, under infinitesimal strain
theory Δ Φ
1 is satisfied and, in this problem, Δ Φ = π/4 ≈ 0.7854.
s gin
Observation: strains are dimensionless; in engineering, small strains are usually
t d le En
considered when these are of order 10−3 − 10−4 .
r
ba
ge ro or
eS m
b) The conditions in the statement of the problem must be imposed one by one:
ci
f
1. Considering that F (X,t) = F (t) and knowing that dx = F · dX, the latter
ra
C d P cs
b
can be integrated as
a
i
# # #
an an n
y ha
⎡ ⎤ ⎡ ⎤
F11 F12 F13 C1
M
.A
not ⎢ ⎥ not ⎢ ⎥
with F ≡ ⎣ F21 F22 F23 ⎦ and C ≡ ⎣ C2 ⎦ ,
m
er
.O
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
0 0 0
C
⎢ ⎥ ⎢ ⎥ not ⎢ ⎥
⎣ 0 ⎦ = [F] ⎣ 0 ⎦ + C =⇒ C ≡ ⎣ 0 ⎦
©
0 0 0
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Problems and Exercises 99
rs
ee
0 0 F33
s gin
2. The condition in the statement imposes that V f inal = pVinitial .
t d le En
Expression dV f = |F| dV0 allows to locally relate the differential volumes at
different instants of time. In this case, F is constant for each fixed t, thus, the
r
ba
ge ro or
expression can be integrated and the determinant of F can be moved outside the
eS m
ci
integral, # # #
f
ra
C d P cs
Vf = dV f = |F| dV0 = |F| dV0 = |F| V0 .
b
a
i
V V0 V0
an an n
le
liv or ec
3. The condition in the statement imposes that lAC, f inal = √p lAC, initial .
2
M
.A
⎡ ⎤⎡ ⎤ ⎡ ⎤
X Th
er
tin
1 0 F13 0 a F13
not ⎢ ⎥⎢ ⎥ ⎢ ⎥
xC = F · XC ≡ ⎣ 0 1 F23 ⎦ ⎣ 0 ⎦ = ⎣ a F23 ⎦ and
on
.O
0 0 F33 a ap
C
p p √
= √ lAC = √ 2 a = p a .
2 2
Therefore,
!
(F13 − 1)2 + F23
2 + p2 = p ⇒ (F − 1)2 + F 2 = 0 ⇒ F = 1; F = 0
13 23 13 23
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
100 C HAPTER 2. S TRAIN
4. The condition in the statement imposes that AOC f inal = 45◦ = π/4.
rs
Considering dX(1) ≡ [1, 0, 0] and dX(2) ≡ [0, 0, 1], the corresponding vectors
not not
ee
in the spatial configuration are computed as
s gin
⎡ ⎤⎡ ⎤ ⎡ ⎤
1 0 1 1 1
(1) not ⎢ ⎥⎢ ⎥ ⎢ ⎥
t d le En
(1)
dx = F · dX ≡ ⎣ 0 1 0 ⎦ ⎣ 0 ⎦ = ⎣ 0 ⎦ ,
r
0 0 p 0 0
ba
ge ro or
eS m
ci
⎡
⎤⎡ ⎤ ⎡ ⎤
f
ra
0 1 0 1
1
C d P cs
(2) not ⎢ ⎥⎢ ⎥ ⎢ ⎥
b
a
(2)
dx = F · dX ≡ ⎣ 0 1 0 ⎦ ⎣ 0 ⎦ = ⎣ 0 ⎦ .
i
an an n
y ha
0 0 p 1 p
le
liv or ec
Then, √
dx(1) · dx(2) 2
M
.A
◦
cos AOC f inal = cos 45 = "" (1) "" "" (2) "" =
dx dx 2
m
is imposed, with
uu
e
.O
such that √
C
1 2 1
$ = =√ =⇒ p = ±1 .
©
1 + p2 2 2
But |F| = p > 0, and, consequently, p = 1. Then, the deformation gradient tensor
is
⎡ ⎤
1 0 1
not ⎢ ⎥
F ≡ ⎣0 1 0⎦ .
0 0 1
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Problems and Exercises 101
which allows determining the displacement field in material and spatial descrip-
tions as
⎡ ⎤ ⎡ ⎤
rs
Z z
not ⎢ ⎥ not ⎢ ⎥
ee
U (X,t) = x − X ≡ ⎣ 0 ⎦ and u (x,t) ≡ ⎣ 0 ⎦ .
s gin
0 0
t d le En
c) The graphical representation of the deformed tetrahedron is:
r
ba
ge ro or
eS m
ci
f
ra
C d P cs
b
a
i
an an n
y ha
le
liv or ec
M
.A
m
d
uu
e
X Th
er
tin
on
.O
UY = UZ = 0 , ∀ X, Y, Z
"
UX "X=0 = 0 , ∀ X, Y
"
UX "X=L = δ
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
102 C HAPTER 2. S TRAIN
c) The possible values (positive and negative) that δ may take. Justify the an-
swer obtained.
d) The material and spatial strain tensors and the infinitesimal strain tensor.
e) Plot the curves EXX − δ /L, exx − δ /L and εx − δ /L for all possible values
of δ , indicating every significant value.
rs
ee
s gin
t d le En
r
ba
ge ro or
eS m
ci
f
ra
Solution
C d P cs
b
a
a) A uniform deformation implies that F (X,t) = F (t) , ∀t, X. The deformation
i
an an n
# #
M
.A
∂ U (X,t)
J= =⇒ dU = J dX =⇒ dU = J dX
∂X
m
# #
uu
e
=⇒ dU = J dX =⇒ U = J · X + C (t) .
X Th
er
tin
where C (t) is an integration constant. Then, the general expression of the mate-
on
.O
b) Using the previous result and applying the boundary conditions given in the
statement of the problem will yield the values of J and C.
Boundary conditions:
UY = UZ = 0 , ∀ X, Y, Z ⇒ Points only move in the X-direction.
"
UX "X=0 = 0 , ∀ Y, Z ⇒ The YZ plane at the origin is fixed.
"
UX "X=L = δ , ∀ Y, Z ⇒ This plane moves in a uniform manner
in the X-direction.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Problems and Exercises 103
If the result obtained in a) is written in component form, the equations and con-
clusions that can be reached will be understood better.
rs
UZ = 0 , ∀ X, Y, Z =⇒ J31 = J32 = J33 = C3 = 0
ee
s gin
From the second boundary condition:
"
t d le En
UX "X=0 = 0 , ∀ Y, Z =⇒ J12 = J13 = C1 = 0
r
ba
ge ro or
From the third boundary condition:
eS m
ci
" f
ra
δ
C d P cs
UX "X=L = δ , ∀ Y, Z =⇒ J11 L = δ ⇒ J11 =
b
a
L
i
an an n
y ha
Finally,
⎡ ⎤ ⎡ ⎤
⎡ ⎤ δ
le
δ
liv or ec
0 0 0 ⎢L ⎥X
⎢ L ⎥ not ⎢ ⎥ not ⎢ ⎥
M
.A
J≡⎢ ⎥
not
⎣0 0 0⎦; C ≡ ⎣0⎦ =⇒ U (X) = J · X + C ≡ ⎢ 0 ⎥ .
⎣ ⎦
m
0 0 0 0
d
0
uu
e
X Th
er
tin
c) In order to justify all the possible positive and negative values that δ may
on
.O
take, the condition |F| > 0 must be imposed. Therefore, the determinant of F
must be computed,
C
⎡ ⎤
δ
1+ 0 0
not ⎢ ⎥
F = 1+J ≡ ⎢
L ⎥ =⇒ |F| = 1 + δ > 0 =⇒ δ > −L .
⎣ 0 1 0 ⎦ L
0 0 1
d) To obtain the spatial and material strain tensors as well as the infinitesimal
strain tensor, their respective definitions must be taken into account.
1
Spatial strain tensor: e= 1 − F−T · F−1
2
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
104 C HAPTER 2. S TRAIN
1 T
Material strain tensor: E=
F ·F−1
2
1 T
Infinitesimal strain tensor: ε = J ·J
2
Applying these definitions using the values of F and J calculated in b) and c),
the corresponding expressions are obtained.
⎡ ⎤
exx 0 0 -
⎢ ⎥ δ 1 δ2 δ 2
rs
not
e ≡ ⎣ 0 0 0 ⎦ with exx = + 1 +
L 2 L2 L
ee
0 0 0
⎡ ⎤ ⎡ ⎤
s gin
δ
EXX 0 0 ⎢L 0 0 ⎥
not ⎢ ⎥
⎥ with EXX = δ + 1 δ
2
not ⎢ ⎥
t d le En
E≡⎢ ⎣ 0 0 0 ⎦ ; ε ≡ ⎢ 0 0 0 ⎥
L 2L 2 ⎣ ⎦
r
ba
ge ro or
0 0 0 0 0 0
eS m
ci
f
ra
C d P cs
b
a
i
an an n
le
liv or ec
Here,
M
.A
• EXX is a second-order
parabola that contains the
m
er
EXX = −1/2.
tin
.O
origin).
©
It can be concluded, then, that for small δ /L strains the three functions have a
very similar behavior and the same slope at the origin. That is, the same result
will be obtained with any of the definitions of strain tensor. However, outside
this domain (large or finite strains) the three curves are clearly different.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Problems and Exercises 105
E XERCISES
rs
a) The material description of v1 and the spatial description of v2 (consider
ee
t = 0 is the reference configuration).
s gin
b) The density distribution in both cases (consider ρ0 is the initial density).
c) The material and spatial descriptions of the displacement field as well as
t d le En
the material (Green-Lagrange) and spatial (Almansi) strain tensors for the
r
velocity field v1 .
ba
ge ro or
eS m
d) Repeat c) for configurations close to the reference configuration (t → 0).
ci
f
ra
C d P cs
e) Prove that the two strain tensors coincide for the conditions stated in d).
b
a
i
an an n
y ha
x = X +Y t , y=Y , z=Z.
M
.A
Obtain the length at time t = 2 of the segment of material line that at time t = 1
m
er
tin
on
.O
⎡ ⎤
©
0 tetX 0
⎢ ⎥
E≡⎢ 0 ⎥
not
⎦.
tX
⎣ te 0
0 0 tetY
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
106 C HAPTER 2. S TRAIN
x=X , y=Y , z = Z − Xt .
Calculate the angle formed at time t = 0 by the differential segments that at time
t = t are parallel to the x- and z-axes.
rs
2) It is antisymmetric with re-
ee
spect to plane Y = 0, that is, the
following is satisfied:
s gin
U (X,Y, Z) = −U (X, −Y, Z)
t d le En
∀ X,Y, Z
r
ba
ge ro or
eS m
ci
3) Under said displacement field,
f
ra
C d P cs
the volume of the element in the
b
a
figure does not change, its an-
i
an an n
.A
Determine:
d
uu
e
a) The most general expression of the given displacement field, such that condi-
X Th
er
tin
.O
deformation gradient tensor and the material strain tensor. Draw the de-
C
formed shape of the element in the figure, indicating the most significant
©
values.
c) The directions (defined by their unit vectors T) for which the deformation is
reduced to a stretch (there is no rotation).
NOTE: Finite strains must be considered (not infinitesimal ones).
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
Problems and Exercises 107
2.6 – The solid in the figure undergoes a uniform deformation such that points
A, B and C do not move. Assuming an infinitesimal strain framework,
a) Express the displacement field in terms of “generic” values of the stretches
and rotations.
b) Identify the null components of the strain tensor and express the rotation
vector in terms of the stretches.
rs
2) The volume becomes (1 + q)
ee
times its initial value.
s gin
3) The angle θ increases its value
in r (given in radians).
t d le En
Under these conditions, deter-
r
ba
mine:
ge ro or
eS m
ci
c) The strain tensor, the rotation
f
ra
C d P cs
vector and the displacement
b
a
i
field in terms of p, q and r.
an an n
y ha
.A
glected.
m
d
uu
2.7 – The solid in the figure undergoes a uniform deformation with the following
e
consequences:
X Th
er
tin
not move.
C
mains constant.
3) The angle θxy remains con-
stant.
4) The angle θyz increases in r
radians.
5) The segment AF becomes
(1 + p) times its initial length.
6) The area of the triangle
ABE becomes (1 + q) its ini-
tial value.
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961
108 C HAPTER 2. S TRAIN
Then,
a) Express the displacement field in terms of “generic” values of the stretches
and rotations.
b) Identify the null components of the strain tensor and express the rotation
vector in terms of the stretches.
c) Determine the strain tensor, the rotation vector and the displacement field in
terms of p, q and r.
NOTE: The values of p, q and r are small and its second-order infinitesimal
terms can be neglected.
rs
ee
2.8 – The sphere in the figure undergoes a uniform deformation (F = const.)
s gin
such that points A, B and C move to positions A , B and C , respectively. Point
O does not move. Determine:
t d le En
a) The deformation gradient tensor in terms of p and q.
r
ba
ge ro or
b) The equation of the deformed external surface of the sphere. Indicate which
eS m
ci
type of surface it is and draw it.
f
ra
C d P cs
c) The material and spatial strain tensors. Obtain the value of p in terms of q
b
a
i
when the material is assumed to be incompressible.
an an n
y ha
d) Repeat c) using infinitesimal strain theory. Prove that when p and q are small,
the results of c) and d) coincide.
le
liv or ec
M
.A
m
d
uu
e
X Th
er
tin
on
.O
C
X. Oliver and C. Agelet de Saracibar Continuum Mechanics for Engineers.Theory and Problems
doi:10.13140/RG.2.2.25821.20961