2.094 — Finite Element Analysis of Solids and Fluids
Fall ‘08
Lecture 11 - Deformation, strain and stress tensors
Prof. K.J. Bathe
MIT OpenCourseWare
We stated that we use
�
�
t
t
τij δ teij d V =
tV
Reading:
Ch. 6
t
0 Sij
0V
δ 0t�ij d 0V = tR
(11.1)
The deformation gradient We use txi = 0xi + tui
⎤
⎡ ∂ tx
∂ tx
∂ tx
1
⎢
⎢
⎢
t
⎢
X
=
0
⎢
⎢
⎣
1
1
∂ 0x1
∂ 0x2
∂ 0x3
t
t
t
∂ x2
∂ 0x1
∂ x2
∂ 0x2
∂ x2
∂ 0x3
∂ tx3
∂ 0x1
∂ tx3
∂ 0x2
∂ tx3
∂ 0x3
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(11.2)
⎤
d tx1
d tx = ⎣ d tx2 ⎦
d tx3
⎡ 0
⎤
d x1
d 0x = ⎣ d 0x2 ⎦
d 0x3
⎡
(11.3)
(11.4)
Implies that
d tx = 0tX d 0x
(11.5)
(0tX is frequently denoted by 0tF or simply F , but we use F for
force vector)
We will also use the right Cauchy-Green deformation tensor
t
0C
= 0tX T 0tX
Some applications
45
(11.6)
MIT 2.094
11. Deformation, strain and stress tensors
The stretch of a fiber ( tλ):
� t �2
� t �2
d txT d tx
ds
λ = 0 T 0 = 0
d x d x
d s
(11.7)
The length of a fiber is
�
�1
d 0s = d 0xT d 0x 2
(11.8)
� 0 T t T��t
�
0
� t �2
d x 0X
0X d x
λ =
,
d 0s · d 0s
from (11.5)
(11.9)
Express
� �
d 0x = d 0s 0n
(11.10)
0
0
n = unit vector into direction of d x
� t �2 0 T t 0
⇒ λ = n 0 C n
∴ tλ =
(11.11)
(11.12)
� 0 T t 0 � 12
n 0C n
(11.13)
Also,
�
d tx̂
�T � t � � t � � t �
· d x = d ŝ d s cos tθ,
From (11.5),
�
ˆT
d 0x̂T 0tX
cos tθ =
��
t
0X
d 0x
d tsˆ d ts
0 0 T t 0
d ŝ n̂ 0 C n d 0s
=
d tsˆ · d ts
(a · b = �a��b� cos θ)
(11.14)
�
�
t ˆ
0X
≡ 0tX
�
(11.15)
(11.16)
0
∴ cos tθ =
n̂T 0tC 0n
(11.17)
tλ̂ tλ
Also,
0
t
ρ=
ρ
det 0tX
(see Ex. 6.5)
(11.18)
Example
Reading:
Ex. 6.6 in
the text
46
MIT 2.094
11. Deformation, strain and stress tensors
1
(1 + 0x1 )(1 + 0x2 )
4
h1 =
(11.19)
..
.
t
xi = 0xi + tui
=
4
�
hk txki ,
(11.20)
(i = 1, 2)
(11.21)
k=1
where txki are the nodal point coordinates at time t ( tx11 = 2, tx12 = 1.5)
Then we obtain
�
1
5 + 0x2
t
1
0X =
4 2 (1 + 0x2 )
At 0x1 = 0, 0x2 = 0,
�
�
1 5
�
t
=
0 X �0
4 12
xi = 0x2 =0
1 + 0x1
1
0
2 (9 + x1 )
1
�
(11.22)
�
(11.23)
9
2
The Green-Lagrange Strain
t
0�
=
� 1 �t
�
1 �t T t
0X 0X − I =
0C − I
2
2
∂
∂ txi
=
∂ 0xj
(11.24)
�0
�
xi + tui
∂ tui
=
δ
+
ij
∂ 0xj
∂ 0xj
We find that
�
1 �t
t
t
t
t
0�ij =
0ui,j + 0uj,i + 0uk,i 0uk,j ,
2
(11.25)
sum over k = 1, 2, 3
(11.26)
where
t
0ui,j
=
∂ tui
∂ 0xj
(11.27)
47
MIT 2.094
11. Deformation, strain and stress tensors
Polar decomposition of 0tX
t
0X
= 0tR 0tU
(11.28)
where 0tR is a rotation matrix, such that
t T t
0R 0R
=I
(11.29)
and 0tU is a symmetric matrix (stretch)
Ex. 6.9 textbook
�
t
0X
=
√
3
2
1
2
1
−
√2
3
2
��
4
3
0
0
�
(11.30)
3
2
Then,
� �2
= 0tX T 0tX = 0tU
�
1 �� t �2
t
−I
0� =
0U
2
t
0C
(11.31)
(11.32)
This shows, by an example, that the components of the Green-Lagrange strain are independent of a
rigid-body rotation.
48
MIT OpenCourseWare
http://ocw.mit.edu
2.094 Finite Element Analysis of Solids and Fluids II
Spring 2011
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.