Draft4.3 Strain Decomposition 25Given x1= X1, x2=−3X3, x3= 2X2, find the deformation gradient F, the right stretch tensor U, the rotation tensor R, and the left stretch tensor V.. 4.6 Lag
Trang 1Draft4.3 Strain Decomposition 25
Given x1= X1, x2=−3X3, x3= 2X2, find the deformation gradient F, the right stretch tensor U, the rotation tensor R, and the left stretch tensor V.
Solution:
From Eq 4.25
F =
∂x1
∂X1
∂x1
∂X2
∂x1
∂X3
∂x2
∂X1
∂x2
∂X2
∂x2
∂X3
∂x3
∂X1
∂x3
∂X2
∂x3
∂X3
=
10 00 −30
0 2 0
From Eq 4.126
U2= FTF =
10 00 02
0 −3 0
10 00 −30
0 2 0
=
10 04 00
0 0 9
thus
U =
10 02 00
0 0 3
From Eq 4.127
R = FU−1=
10 00 −30
0 2 0
10 01 0
0 0 13
=
10 00 −10
0 1 0
Finally, from Eq 4.128
V = FRT =
10 00 −30
0 2 0
10 00 01
0 −1 0
=
10 03 00
0 0 2
Example 4-11: Polar Decomposition II
For the following deformation: x1= λ1X1, x2=−λ3X3, and x3= λ2X2, find the rotation tensor
Solution:
[F] =
λ01 00 −λ03
0 λ2 0
=
λ01 00 λ02
0 −λ3 0
λ01 00 −λ03
0 λ2 0
=
λ
0 λ2 0
0 0 λ2
[U] =
λ01 λ02 00
0 0 λ3
[R] = [F][U]−1=
λ01 00 −λ03
0 λ2 0
1
0 λ12 0
0 0 λ13
=
10 00 −10
0 1 0
(4.139)
Thus we note that R corresponds to a 90o rotation about the e1 axis
Trang 2Example 4-12: Polar Decomposition III
Trang 3Draft4.3 Strain Decomposition 27
Polar Decomposition Using Mathematica
Given x1=X1+2X2, x2=X2, x3=X3, a) Obtain C, b) the principal values of C and the corresponding directions, c) the
matrix U and U-1 with respect to the principal directions, d) Obtain the matrix U and U-1 with respect to the e i bas
obtain the matrix R with respect to the e i basis
Determine the F matrix
In[1]:= F= 881, 2, 0<, 80, 1, 0<, 80, 0, 1<<
Out[1]=
i k
jjjj jjj
1 2 0
0 1 0
0 0 1
y {
zzzz zzz
Solve for C
In[2]:= CST= Transpose@FD F
Out[2]=
i k
jjjj jjj
1 2 0
2 5 0
0 0 1
y {
zzzz zzz
Determine Eigenvalues and Eigenvectors
In[3]:= N@Eigenvalues@CSTDD
Out[3]= 81., 0.171573, 5.82843<
In[4]:= 8v1, v2, v3< = N@Eigenvectors@CSTD, 4D
Out[4]=
i k
jjjj jjj
0 0 1
-2.414 1 0 0.4142 1 0
y {
zzzz zzz
In[5]:= << LinearAlgebra‘Orthogonalization‘
In[6]:= vnormalized= GramSchmidt@8v3, −v2, v1<D
Out[6]=
i k
jjjj jjj
0.382683 0.92388 0 0.92388 -0.382683 0
y {
zzzz zzz
In[7]:= CSTeigen= Chop@N@vnormalized CST vnormalized, 4DD
Out[7]=
i k
jjjj jjj
5.828 0 0
0 0.1716 0
y {
zzzz zzz
Determine U with respect to the principal directions
In[8]:= Ueigen= N@Sqrt@CSTeigenD, 4D
Out[8]=
i k
jjjj jjj
2.414 0 0
0 0.4142 0
y {
zzzz zzz
In[9]:= Ueigenminus1= Inverse@UeigenD
Out[9]=
i k
jjjj jjj
0.414214 0 0
0 2.41421 0
0 0 1
y {
zzzz zzz
Determine U and U- 1with respect to the ei basis
In[10]:= U_e= N@vnormalized Ueigen vnormalized, 3D
Out[10]=
i k
jjjj jjj
0.707 0.707 0
0.707 2.12 0
0 0 1
y {
zzzz zzz
In[11]:= U_einverse= N@Inverse@%D, 3D
Out[11]=
i k
jjjj jjj
2.12 -0.707 0
-0.707 0.707 0
0 0 1
y {
zzzz zzz
Determine R with respect to the ei basis
In[12]:= R= N@F %, 3D
Out[12]=
i k
jjjj jjj
0.707 0.707 0
-0.707 0.707 0
0 0 1
y {
zzzz zzz
m−polar.nb
Trang 44.4 Summary and Discussion
92 From the above, we deduce the following observations:
1 If both the displacement gradients and the displacements themselves are small, then ∂u i
∂X j ≈ ∂u i
∂x j and
thus the Eulerian and the Lagrangian infinitesimal strain tensors may be taken as equal E ij = E ij ∗
2 If the displacement gradients are small, but the displacements are large, we should use the Eulerian infinitesimal representation
3 If the displacements gradients are large, but the displacements are small, use the Lagrangian finite strain representation
4 If both the displacement gradients and the displacements are large, use the Eulerian finite strain representation
4.5 Compatibility Equation
93 If ε ij = 12(u i,j + u j,i) then we have six differential equations (in 3D the strain tensor has a total
of 9 terms, but due to symmetry, there are 6 independent ones) for determining (upon integration)
three unknowns displacements u i Hence the system is overdetermined, and there must be some linear relations between the strains
94 It can be shown (through appropriate successive differentiation of the strain expression) that the compatibility relation for strain reduces to:
∂2ε ik
∂x j ∂x j +
∂2ε jj
∂x i ∂x k − ∂2ε jk
∂x i ∂x j − ∂2ε ij
∂x j ∂x k = 0 or ∇x×L×∇x= 0 (4.140)
There are 81 equations in all, but only six are distinct
∂2ε11
2ε22
2ε12
∂x1∂x2 (4.141-a)
∂2ε22
2ε33
2ε23
∂x2∂x3 (4.141-b)
∂2ε33
2ε11
2ε31
∂x3∂x1 (4.141-c)
∂
∂x1
− ∂ε23
∂ε31
∂ε12
∂x3
2ε11
∂x2∂x3 (4.141-d)
∂
∂x2
∂ε23
∂ε12
∂x3
2ε22
∂x3∂x1 (4.141-e)
∂
∂x3
∂ε23
∂ε31
∂x3
2ε33
∂x1∂x2 (4.141-f)
In 2D, this results in (by setting i = 2, j = 1 and l = 2):
∂2ε11
2ε22
2γ12
Trang 5Draft4.5 Compatibility Equation 29
I , i
3 3
2
I , i
2
I , i
1 1
2 2
X , x
u
P
P
0 t=0
t=t
X
O
U
o
x
X , x
X , x
x2 ,
X3, x3
2
X
X1
O
x1 ,
u
x
t=0
+d
X X
X
+d
d
d x
t=t
X
u u
Q
0
Q
P P
0
Position Vector x = x(X, t) X = X(x, t)
GRADIENTS
Deformation F = x∇X≡ ∂x i
∂x j
H = F−1
Displacement ∂u i
∂X j = ∂x i
∂x j = δ ij − ∂X i
∂x j or
J = u∇X = F− I K≡ u∇x = I− H
TENSOR
Deformation B −1 ij =∂X k
∂x i
∂X k
∂X i
∂x k
∂X j or
B−1 =∇x X·X∇x = Hc ·H C =∇X x·x∇X = Fc ·F
C−1= B−1 STRAINS
Finite Strain E ij = 12
∂x k
∂X i
∂x k
∂X j − δ ij
or E ij ∗ =12
δ ij − ∂X k
∂x i
∂X k
∂x j
or
E = 12(∇X x·x∇X
Fc ·F
Hc ·H
)
2
∂u i
∂X j + ∂u j
∂X i+ ∂u k
∂X i
∂u k
∂X j
or E ij ∗ =1
2
∂u i
∂x j +∂u j
∂x i − ∂u k
∂x i
∂u k
∂x j
or
E = 12(u∇X+∇X u +∇X u·u∇X)
J+Jc+Jc ·J
E∗= 12(u∇x+∇x u− ∇x u·u∇x)
K+Kc −K c ·K
Small E ij = 12
∂u i
∂X j + ∂u j
∂X i
E ij ∗ =12
∂u i
∂x j +∂u j
∂x i
Deformation E = 12(u∇X+∇X u) = 12(J + Jc) E∗= 12(u∇x+∇x u) = 12(K + Kc)
ROTATION TENSORS
Small [12
∂u i
∂X j +∂u j
∂X i
+12
∂u i
∂X j − ∂u j
∂X i
]dX j
#
1 2
∂u i
∂x j +∂u j
∂x i
+12
∂u i
∂x j − ∂u j
∂x i
$
dx j
deformation [1
2(u∇X+∇X u)
E
+1
2(u∇X− ∇X u)
W
]·dX [1
2(u∇x+∇x u)
E∗
+1
2(u∇x− ∇x u)
Ω
]·dx
Finite Strain F = R·U = V·R
STRESS TENSORS
First T0= (det F)T%
F−1&T Second T = (det F)˜ %
F−1&
T%
F−1&T
Table 4.1: Summary of Major Equations
Trang 695 When he compatibility equation is written in term of the stresses, it yields:
∂2σ11
2σ22
∂x2 − ν ∂2σ11
∂x2 = 2 (1 + ν) ∂
2σ21
∂x1∂x2 (4.143)
Example 4-13: Strain Compatibility
For the following strain field
−
X2
X2+X2
X1
2(X2+X2) 0
X1
does there exist a single-valued continuous displacement field?
Solution:
(X2+ X2)2 = X
2− X2
(X2+ X2)2 (4.145-a)
2∂E12
(X2+ X2)− X1(2X1)
(X2+ X2)2 = X
2− X2
(X2+ X2)2 (4.145-b)
1
⇒ ∂2E11
2E22
2E12
√
(4.145-d) Actually, it can be easily verified that the unique displacement field is given by
u1= arctanX2
X1; u2= 0; u3= 0 (4.146)
to which we could add the rigid body displacement field (if any)
4.6 Lagrangian Stresses; Piola Kirchoff Stress Tensors
96 In Sect 2.2 the discussion of stress applied to the deformed configuration dA (using spatial coordiantes
x), that is the one where equilibrium must hold The deformed configuration being the natural one in
which to characterize stress Hence we had
(note the use of T instead of σ) Hence the Cauchy stress tensor was really defined in the Eulerian
space
97 However, there are certain advantages in referring all quantities back to the undeformed configuration (Lagrangian) of the body because often that configuration has geometric features and symmetries that are lost through the deformation
98 Hence, if we were to define the strain in material coordinates (in terms of X), we need also to express the stress as a function of the material point X in material coordinates.
Trang 7Draft4.6 Lagrangian Stresses; Piola Kirchoff Stress Tensors 31
99 The first Piola-Kirchoff stress tensor T0 is defined in the undeformed geometry in such a way that it
results in the same total force as the traction in the deformed configuration (where Cauchy’s stress
tensor was defined) Thus, we define
where t0 is a pseudo-stress vector in that being based on the undeformed area, it does not describe the actual intensity of the force, however it has the same direction as Cauchy’s stress vector t.
100 The first Piola-Kirchoff stress tensor (also known as Lagrangian Stress Tensor) is thus the linear transformation T0 such that
and for which
df = t0dA0= tdA ⇒ t0= dA
using Eq 4.147-b and 4.149 the preceding equation becomes
T0n0= dA
dA0Tn = T
dA
and using Eq 4.36 dAn = dA0(det F)%
F−1&T
n0 we obtain
T0n0= T(det F)%
F−1&T
the above equation is true for all n0, therefore
T0 = (det F)T%
F−1&T
(4.153)
(det F) T0F
T or T ij = 1
(det F)(T0)im F jm (4.154)
101 The first Piola-Kirchoff stress tensor is not symmetric in general, and is not energitically correct.
That is multiplying this stress tensor with the Green-Lagrange tensor will not be equal to the product
of the Cauchy stress tensor multiplied by the deformation strain tensor
102 To determine the corresponding stress vector, we solve for T0 first, then for dA0 and n0 from
dA0n0=det F1 FT n (assuming unit area dA), and finally t0= T0n0
103 The second Piola-Kirchoff stress tensor, ˜T is formulated differently Instead of the actual force df
on dA, it gives the force d˜ f related to the force df in the same way that a material vector dX at X is
related by the deformation to the corresponding spatial vector dx at x Thus, if we let
and
where d˜f is the pseudo differential force which transforms, under the deformation gradient F, the
(actual) differential force df at the deformed position (note similarity with dx = FdX) Thus, the
pseudo vector t is in general in a differnt direction than that of the Cauchy stress vector t.
104 The second Piola-Kirchoff stress tensor is a linear transformation ˜T such that
˜
Trang 8df = F ˜ Tn0dA0 (4.157)
we also have from Eq 4.148 and 4.149
and comparing the last two equations we note that
˜
which gives the relationship between the first Piola-Kirchoff stress tensor T0 and the second Piola-Kirchoff stress tensor ˜T.
105 Finally the relation between the second Piola-Kirchoff stress tensor and the Cauchy stress tensor can
be obtained from the preceding equation and Eq 4.153
˜
T = (det F)%
F−1&
T%
F−1&T
(4.160)
and we note that this second Piola-Kirchoff stress tensor is always symmetric (if the Cauchy stress tensor
is symmetric) It can also be shown that it is energitically correct
106 To determine the corresponding stress vector, we solve for ˜T first, then for dA0and n0from dA0n0=
1
det F FT n (assuming unit area dA), and finally ˜t = ˜ Tn0
Example 4-14: Piola-Kirchoff Stress Tensors
4.7 Hydrostatic and Deviatoric Strain
85 The lagrangian and Eulerian linear strain tensors can each be split into spherical and deviator
tensor as was the case for the stresses Hence, if we define
1
3e =
1
then the components of the strain deviator E are given by
E ij = E ij −1
3eδ ij or E
= E−1
We note that E measures the change in shape of an element, while the spherical or hydrostatic strain
1
Trang 9Draft4.7 Hydrostatic and Deviatoric Strain 33
Piola−Kirchoff Stress Tensors
The deformed configuration of a body is described by x1=X1ê 2, x2=−X2/2, x3=4X3 ; If the Cauchy stress tensor is
given by
i k
jjjj jjjj
100 0 0 0
0 0
0 0 0 y {
zzzz zzzz MPa; What are the corresponding first and second Piola−Kirchoff stress tensors, and calculate the
respective stress tensors on the e3 plane in the deformed state.
‡ F tensor
CST= 880, 0, 0<, 80, 0, 0<, 80, 0, 100<<
880, 0, 0<, 80, 0, 0<, 80, 0, 100<<
F= 881 ê 2, 0, 0<, 80, 0, −1 ê 2<, 80, 4, 0<<
99 ÄÄÄÄÄ1
2, 0, 0=, 90, 0, - ÄÄÄÄÄ1
2 =,80, 4, 0<=
982, 0, 0<, 90, 0, ÄÄÄÄÄ1
4 =,80, -2, 0<=
‡ First Piola−Kirchoff Stress Tensor
880, 0, 0<, 80, 0, 0<, 80, 25, 0<<
i k
j j
y {
z z
‡ Second Piola−Kirchoff Stress Tensor
980, 0, 0<, 90, ÄÄÄÄÄÄÄÄ25
4 , 0=, 80, 0, 0<=
i k
j j
0 ÄÄÄÄÄÄ254 0
y {
z z
‡ Cuchy stress vector
Can be obtained from t=CST n
i k
j j
0 0 100 y {
z z
‡ Pseudo−Stress vector associated with the First Piola−Kirchoff stress tensor
For a unit area in the deformed state in the e3direction, its undeformed area dA0n0is given by dA0n0 = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄF det F T n
detF= Det@FD
1
n= 80, 0, 1<
80, 0, 1<
i k
j j
0 4 0 y {
z z
Thus n0=e2and using t0=T0 n0 we obtain
i k
j j
0 0 25 y {
z z
We note that this vector is in the same direction as the Cauchy stress vector, its magnitude is one fourth of that of the
Cauchy stress vector, because the undeformed area is 4 times that of the deformed area
‡ Pseudo−Stress vector associated with the Second Piola−Kirchoff stress
tensor
i k
j j
0 25 ÄÄÄÄÄÄ4 0 y {
z z
We see that this pseudo stress vector is in a different direction from that of the Cauchy stress vector (and we note that
the tensor F transforms e2into e3 ).
Trang 10εI
ε ε
2
Figure 4.8: Mohr Circle for Strain
4.8 Principal Strains, Strain Invariants, Mohr Circle
86 Determination of the principal strains (E(3)< E(2)< E(1), strain invariants and the Mohr circle for strain parallel the one for stresses (Sect 2.3) and will not be repeated here
where the symbols I E , II E and III E denote the following scalar expressions in the strain components:
I E = E11+ E22+ E33= E ii = tr E (4.164)
II E = −(E11E22+ E22E33+ E33E11) + E232 + E312 + E122 (4.165)
= 1
2(E ij E ij − E ii E jj) = 1
2E ij E ij −1
2I
2
= 1
2(E : E− I2
III E = detE =1
6e ijk e pqr E ip E jq E kr (4.168)
87 In terms of the principal strains, those invariants can be simplified into
I E = E(1)+ E(2)+ E(3) (4.169)
II E = −(E(1)E(2)+ E(2)E(3)+ E(3)E(1)) (4.170)
III E = E(1)E(2)E(3) (4.171)
88 The Mohr circle uses the Engineering shear strain definition of Eq 4.91, Fig 4.8
Example 4-15: Strain Invariants & Principal Strains
Trang 11Draft4.8 Principal Strains, Strain Invariants, Mohr Circle 35
Determine the planes of principal strains for the following strain tensor
1
√
3 0
√
3 0 0
Solution:
The strain invariants are given by
The principal strains by
1− λ
√
√
0 0 1− λ
= (1− λ)
'
√
13 2
( '
√
13 2
(
(4.174-b)
√
13
The eigenvectors for E(1)= 1+√213 give the principal directions n(1):
1−1+√
13 2
√
√
3 −1+√
13
13 2
n(1)1
n(1)2
n(1)3
=
1−1+√
13 2
n(1)1 +√
3n(1)2
√
3n(1)1 − 1+√
13 2
n(1)2
1−1+√
13 2
n(1)3
=
0 0 0
(4.175) which gives
√
13
2√
3 n
(1)
n(1)·n(1) =
'
1 + 2√
13 + 13
12 + 1
(
n(1)2
2
= 1⇒ n1
2= 0.8; (4.176-c)
For the second eigenvector λ(2)= 1:
1− 1
√
√
0 0 1− 1
n(2)1
n(2)2
n(2)3
=
√
3n(2)2
√
3n(2)1 − n(2)
2
0
=
0 0 0
which gives (with the requirement that n(2)·n(2) = 1)
Trang 12n(3)= n(1)×n(2) = det
e1 e2 e3
0.8 0.6 0
= 0.6e1− 0.8e2 (4.179) Therefore
a j i =
n(1)
n(2)
n(3)
=
0.80 0.60 01
and this results can be checked via
[a][E][a]T =
0.80 0.60 01
1
√
3 0
√
3 0 0
0.8 0.6 00 −0.8 0.6
=
2.30 01 00
0 0 −1.3
(4.181)
Example 4-16: Mohr’s Circle
Construct the Mohr’s circle for the following plane strain case:
00 05 √03
0 √
3 3
Solution:
1 2 1
60o 2 B
D
E
F
ε
ε s
n
2 3
We note that since E(1) = 0 is a principal value for plane strain, ttwo of the circles are drawn as shown