Discussion of the Griffith fracture criterion in piezoelectricity

Transcrição

Page 1
Workshop: Direct and Inverse Problems in Piezoelectricity
Linz, October 06-07, 2005
Discussion of the Griffith fracture
criterion in piezoelectricity
Anna-Margarete Sändig
[email protected]
Universität Stuttgart, Institut für Angewandte Analysis und Numerische Simulation
Outline
1. Introduction
2. The Griffith fracture criterion
3. 3D-Field equations
4. 2D-Field equations
5. Griffith formula
6. J-integral
7. Stress intensity factors
8. Conclusion
Page 2
Introduction
Page 3
The fracture behaviour in piezoelectric materials under mechanical and electrical
loading is not fully understood. In particular, it is not clear, whether the electric
field impedes or enhances crack propagation. Different fracture criterions for linear
models are discussed in the literature:
•
Total energy release rate. (Pak 90)
•
Mechanical strain energy release rate. (Park/Sun 95)
•
Stress intensity factors. (Suo/Kuo/Barnett/Willis 92)
•
Local energy release rate. (Gao et al. 97)
Introduction
Page 3
The fracture behaviour in piezoelectric materials under mechanical and electrical
loading is not fully understood. In particular, it is not clear, whether the electric
field impedes or enhances crack propagation. Different fracture criterions for linear
models are discussed in the literature:
•
Total energy release rate. (Pak 90)
•
Mechanical strain energy release rate. (Park/Sun 95)
•
Stress intensity factors. (Suo/Kuo/Barnett/Willis 92)
•
Local energy release rate. (Gao et al. 97)
We will discuss the Griffith energy criterion which is based on the total energy
release rate, in particular:
•
What is the total energy?
•
Which formulas for the energy release rate are available?
•
How are the Maz’ya-Plamenevski coefficient formulas related to the stress
intensity factors?
Griffith fracture criterion
The Griffith criterion is based on energy balance: The body is in an equilibrium
state, if the total energy is minimal, that means, the crack grows if there is a
neighboured configuration with smaller energy as the actual configuration.
g
δ
f
Page 4
Griffith fracture criterion
Page 4
The Griffith criterion is based on energy balance: The body is in an equilibrium
state, if the total energy is minimal, that means, the crack grows if there is a
neighboured configuration with smaller energy as the actual configuration.
g
δ
f
Total energy: Π = Û + D̂ = Ê − F + D̂
Û - potential energy, D̂- dissipative energy,
Ê- elasto-electrical energy (electric enthalpy),
R
R
1
Ê = 2 AB(U ) · B(U ) dy − Ω M B(U ) · (−∇ϕ),
Ω
R
R
F- external energy, F = f U dy +
gU da,
Ω
Field equations:
−B > ABU = f in Ω,
U = 0 on Γ,
ABU = (σ, D)> = 0 on Γ± ,
g = 0.
M B(U ) = D
ΓN
U = (u1 , u2 , u3 , ϕ)>
3D Field equations
Hooke’s law in 3D:
P3
(C)
γ
−
ijkl Ckl
k,l=1
k=1 ekij (E)k ,
P
P
(D)i = 3j=1 (ε)ij (E)j + 3j,k=1 eijk (γ)jk
(σ)ij =
P3
Voigt’s notation:
!
!
γ
σ
=A
= AB
D
E
Page 5
u
ϕ
!
,
3D Field equations
Voigt’s notation:
!
!
γ
σ
=A
= AB
D
E
Hooke’s law in 3D:
P3
(C)
γ
−
ijkl Ckl
k,l=1
k=1 ekij (E)k ,
P
P
(D)i = 3j=1 (ε)ij (E)j + 3j,k=1 eijk (γ)jk
(σ)ij =
Page 5
P3
transversal isotrop, poling in y3 -direction
A =
0
∂1
B
Div = @ 0
0
C
e
0
∂2
0
−e
ε
0
0
∂3
>
!
0
∂3
∂2
=
c12 c13 0
0
0
0
0
−e31
c12 c11 c13 0
0
0
0
0
−e31
c
c
c
0
0
0
0
0
−e33
B 13 13 33
0
0
−e15
0
B 0 0 0 c44 0
B 0 0 0 0 c44 “ 0 ” −e15 0
0
B
12
0
0
0
B 0 0 0 0 0 c11 −c
2
0 c11
@
∂3
0
∂1
0
0
0
0
0 e15
e15 0
e31 e31 e33 0
0
1
∂2
C
∂1 A ,
0
0
0
B =
0
0
0
ε11
0
0
Div
0
>
0
ε11
0
0
0
ε33
!
0
.
−∇
3D field equations: −B > A B U = f .
1
C
C
C
C,
C
A
u
ϕ
!
,
2D Field equations
Page 6
Reduction to 2D-Problems: U = U (x1 , x3 ) =⇒
Decoupling of the 3D-problem into a plane strain-state problem for (u 1 , u3 , ϕ) and
anti-plane strain-state problem for u2 .
Hookes law for the in-plane state (γ21 = γ22 = γ23 = 0, E2 = 0, U = (u1 , u3 , ϕ)> ):
1 0
σ11
c11
Bσ33 C Bc13
B C B
B C B
Bσ13 C = B 0
B C B
@ D1 A @ 0
e31
D3
0
c13
c33
0
0
e33
0
0
c44
e15
0
0
0
−e15
ε11
0
10
−e31
∂1
B
−e33 C
CB 0
CB
0 C B ∂3
CB
0 A@ 0
ε33
0
0
∂3
∂1
0
0
1
0
0 1
C
0 C u1
CB C
0 C @u3 A = ABU.
C
−∂1 A ϕ
−∂3
2D Field equations
Page 6
Reduction to 2D-Problems: U = U (x1 , x3 ) =⇒
Decoupling of the 3D-problem into a plane strain-state problem for (u 1 , u3 , ϕ) and
anti-plane strain-state problem for u2 .
Hookes law for the in-plane state (γ21 = γ22 = γ23 = 0, E2 = 0, U = (u1 , u3 , ϕ)> ):
1 0
σ11
c11
Bσ33 C Bc13
B C B
B C B
Bσ13 C = B 0
B C B
@ D1 A @ 0
e31
D3
0
c13
c33
0
0
e33
0
0
c44
e15
0
0
0
−e15
ε11
0
10
−e31
∂1
B
−e33 C
CB 0
CB
0 C B ∂3
CB
0 A@ 0
ε33
0
0
∂3
∂1
0
0
1
0
0 1
C
0 C u1
CB C
0 C @u3 A = ABU.
C
−∂1 A ϕ
−∂3
Hooke’s law for the out-of-plane strain state (γ11 = γ33 = γ13 = 0, E1 = E3 = 0):
σ23
σ21
!
=
c44
0
0
c11 −c12
2
!
∇u2 = A2 B2 u2 .
2D Field equations: −B > ABU = f, U = (u1 , u3 , ϕ), −B2T A2 B2 u2 = f2 .
2D Total energy
Page 7
Elasto-electrical energy (electric enthalpy):
Ê =
1
2
R
Ω
ABU · BU dy −
Exterior energy:
F =
Total energy:
R
Ω
R
f U dy +
Ω
R
M BU · (−∇ϕ)dy +
Ω
1
2
R
Ω
A2 B2 u2 · B2 u2 dy
f2 u2 dy
Π = Ê − F + D̂
2D Total energy
Page 7
Elasto-electrical energy (electric enthalpy):
Ê =
1
2
R
Ω
ABU · BU dy −
Exterior energy:
F =
Total energy:
R
Ω
R
f U dy +
Π = Ê − F + D̂
Ω
R
M BU · (−∇ϕ)dy +
Ω
1
2
R
Ω
A2 B2 u2 · B2 u2 dy
f2 u2 dy
PSfrag replacements
supp Θ y3
Straight crack propagation in y1 -direction,
Ωδ = domain with crack of the length l + δ,
Ω= domain with crack of the length l,
Πδ =total energy in Ωδ .
Γ+
Γ−
δ
Crack perpendicular to the
polarisation in y3 direction
y1
Energy release rate
Griffith criterion:
Page 8
dΠδ
|δ=0 = 0
dδ
Assume D̂δ = 2δγ̂, γ̂ = γ̂(σc , Dc ), is known, then we have to calculate the
Energy Release Rate
ERR =
d(Eδ −Fδ )
|δ=0
dδ
= limδ→0
(Eδ −Fδ )−(E0 −F0 )
δ
Energy release rate
Griffith criterion:
Page 8
dΠδ
|δ=0 = 0
dδ
Assume D̂δ = 2δγ̂, γ̂ = γ̂(σc , Dc ), is known, then we have to calculate the
Energy Release Rate
ERR =
d(Eδ −Fδ )
|δ=0
dδ
= limδ→0
(Eδ −Fδ )−(E0 −F0 )
δ
Derivation of formulas for the ERR
Assume: Ωδ (x-coordinates) and Ω (y-coordinates) are connected by a mapping:
!
!
!
y1
x1
Θ(x)
=
−δ
,
0
y2
x2
where Θ ∈ C0∞ (Ωδ ), suppΘ ⊂ a neighborhood of the crack tip, Θ ≡ 1 near the
crack tip. Note, that these mappings are diffeomorphisms for 0 < δ < δ0 , since
∂(y1 ,y2 )
= 1 − δ∂1 Θ > 0.
(∂x1 ,x2 )
The Griffith formula
Matrix-vector notation for the energy
R 1
Ê = Ω 2 Cγ · γ − eγ · E − 12 εE · Edy,
e=
0
e31
0
e33
Page 9
e15
0
!
,
γ = Div > u
= elastic e. + piezo e. - electric e.
Matrix-vector notation for the energy
R 1
Ê = Ω 2 Cγ · γ − eγ · E − 12 εE · Edy,
e=
0
e31
0
e33
Page 9
e15
0
!
,
γ = Div > u
= elastic e. + piezo e. - electric e.
Theorem (Griffith formula)
ERR =
R
0
∂1 Θ
0
C
1
u
+
Cγ · γ∂1 θ) dy
∂
∂ 3 ΘC
1
2
A
B
B 0
(−Cγ
·
Ω
@
∂3 Θ
+
R
Ω (εE
· ∇ΘE1 − 12 εE · E∂1 Θ) dy
0
−
−
R
elastic part
∂1 Θ
∂1 Θ
B
+ Ω eγE1 · ∇Θ + e B
@ 0
∂3 Θ
R
1
0
R
Ω A2 ∇u2 ∂1 u2 · ∇Θdy +
Ω
R
electric part
1
C
∂ 3 ΘC
A ∂1 u · E − eγ · E∂1 Θ dy
∂1 Θ
coupled part
U ∂1 (f Θ)dy
exterior forces
1
Ω 2 A2 ∇u2 · ∇u2 ∂1 Θdy − Ω
R
u2 ∂1 (f2 Θ)dy
antiplane part
Proof:
•
•
•
Ideas from Khludnev/Sokolowski 99/00.
R
R
Transformation of the energy integrals Ω · · · dx into Ω · · · dy.
δ
convergence: kUδ − U0 kH 1 (Ω) → 0, kfδ − f0 kL2 (Ω) → 0.
Page 10
Proof:
•
•
•
Ideas from Khludnev/Sokolowski 99/00.
R
R
Transformation of the energy integrals Ω · · · dx into Ω · · · dy.
δ
convergence: kUδ − U0 kH 1 (Ω) → 0, kfδ − f0 kL2 (Ω) → 0.
Remarks:
Formula is independent of the choice of Θ.
Integration on supp Θ only!
Page 10
The J-integral
Page 11
Theorem ( J-integral)
R
R
R
R
ERR = Γ W n1 ds − Γ N σ · ∂1 uds + Γ DE1 · nds− Γ U · f n1 ds
+
R
1
A ∇u2
Γ 2 2
· ∇u2 n1 ds −
R
A ∇u2 ∂1 u2 · nds −
Γ 2
provided f ≡ const near the crack tip.
R
Γ
u2 f2 n1 ds,
The J-integral
Page 11
Theorem ( J-integral)
R
R
R
R
ERR = Γ W n1 ds − Γ N σ · ∂1 uds + Γ DE1 · nds− Γ U · f n1 ds
+
R
1
A ∇u2
Γ 2 2
· ∇u2 n1 ds −
R
A ∇u2 ∂1 u2 · nds −
Γ 2
provided f ≡ const near the crack tip.
σ = Cγ − e> E
n1
0
0
n3
Γ
u2 f2 n1 ds,
Γ
D = eγ + εE
N=
R
n3
n1
!
PSfrag replacements
Ω∗
W = 12 Cγ · γ − eγ · E − 12 εE · E
Proof: Partial integration
The J-integral
The Griffith criterion yields
d(Ê − F + D̂)δ
dΠδ
|δ=0 =
|δ=0 = 0
dδ
dδ
Assume, D̂δ = 2δγ̂. Therefore
dD̂δ
|
dδ δ=0
= 2γ̂(σc , Dc ).
The Griffith fracture criterion reads:
Page 12
The J-integral
The Griffith criterion yields
d(Ê − F + D̂)δ
dΠδ
|δ=0 =
|δ=0 = 0
dδ
dδ
Assume, D̂δ = 2δγ̂. Therefore
dD̂δ
|
dδ δ=0
= 2γ̂(σc , Dc ).
The Griffith fracture criterion reads:
The crack propagates, if
– ERR ≥ critical value = 2γ̂(σc , Dc ).
Page 12
Stress intensity factors
Page 13
From the general theory of elliptic boundary value problems follows:
0 1 0 1 0 1
u1
u1
0
Bu C B 0 C Bu C
B 2C B C B 2C
U =B C=B C+B C=
@u 3 A @ u 3 A @ 0 A
0
ϕ
ϕ
0
s11 (ω)
1
0
s21 (ω)
1
0
s41 (ω)
1
0
1
0
C
B 0 C
B 0 C
B 0 C
B 3
1 B
1 B
1 B
1 Bs2 (ω)C
C
C
C
= c1 r 2 B 1
C + Ureg ,
C + c2 r 2 B 2
C + c4 r 2 B 4
C + c3 r 2 B
@ s3 (ω) A
@ s3 (ω) A
@ s3 (ω) A
@ 0 A
0
ϕ1 (ω)
ϕ2 (ω)
ϕ4 (ω)
3
where Ureg = O(r 2 ).
Page 13
From the general theory of elliptic boundary value problems follows:
0 1 0 1 0 1
u1
u1
0
Bu C B 0 C Bu C
B 2C B C B 2C
U =B C=B C+B C=
@u 3 A @ u 3 A @ 0 A
0
ϕ
ϕ
0
s11 (ω)
1
0
s21 (ω)
1
0
s41 (ω)
1
0
1
0
C
B 0 C
B 0 C
B 0 C
B 3
1 B
1 B
1 B
1 Bs2 (ω)C
C
C
C
= c1 r 2 B 1
C + Ureg ,
C + c2 r 2 B 2
C + c4 r 2 B 4
C + c3 r 2 B
@ s3 (ω) A
@ s3 (ω) A
@ s3 (ω) A
@ 0 A
0
ϕ1 (ω)
ϕ2 (ω)
ϕ4 (ω)
3
where Ureg = O(r 2 ).
Inserting this decomposition of U into the J-integral we get:
ERR =
P
2
a
c
c
+
a
c
ij
i
j
3
3
i,j=1,2,4
Page 14
1
The coefficients aij , ci can be calculated, if the singular vectors r 2 S i (ω)
0 1
0 2
0 4
0
1
1
1
1
s1 (ω)
s1 (ω)
s1 (ω)
0
B 0 C
B 0 C
B 0 C
B 3
C
1 B
1 B
1 B
1 Bs2 (ω)C
C
C
C
r2 B 1
C
C, r2 B 2
C, r2 B 4
C, r2 B
@ s3 (ω) A
@ s3 (ω) A
@ s3 (ω) A
@ 0 A
0
ϕ1 (ω)
ϕ2 (ω)
ϕ4 (ω)
1
and the so called dual singular vectors r − 2 S −i (ω), i = 1, 2, 3, 4,
0 −1
0 −2
0 −4
1
1
1
s1 (ω)
s1 (ω)
s1 (ω)
B 0 C
B 0 C
B 0 C
1 B
1 B
B
C
C
C
−1
−
−
−1
r 2 B −1
C , r 2 B −2
C , r 2 B −4
C, r 2
@ s3 (ω) A
@ s3 (ω) A
@ s3 (ω) A
ϕ− 1(ω)
ϕ−2 (ω)
ϕ−4 (ω)
are known.
1
0
Bs−3 (ω)C
C
B 2
C
B
@ 0 A
0
0
Page 14
1
The coefficients aij , ci can be calculated, if the singular vectors r 2 S i (ω)
0 1
0 2
0 4
0
1
1
1
1
s1 (ω)
s1 (ω)
s1 (ω)
0
B 0 C
B 0 C
B 0 C
B 3
C
1 B
1 B
1 B
1 Bs2 (ω)C
C
C
C
r2 B 1
C
C, r2 B 2
C, r2 B 4
C, r2 B
@ s3 (ω) A
@ s3 (ω) A
@ s3 (ω) A
@ 0 A
0
ϕ1 (ω)
ϕ2 (ω)
ϕ4 (ω)
1
and the so called dual singular vectors r − 2 S −i (ω), i = 1, 2, 3, 4,
0 −1
0 −2
0 −4
1
1
1
s1 (ω)
s1 (ω)
s1 (ω)
B 0 C
B 0 C
B 0 C
1 B
1 B
B
C
C
C
−1
−
−
−1
r 2 B −1
C , r 2 B −2
C , r 2 B −4
C, r 2
@ s3 (ω) A
@ s3 (ω) A
@ s3 (ω) A
ϕ− 1(ω)
ϕ−2 (ω)
ϕ−4 (ω)
are known.
1
0
Bs−3 (ω)C
C
B 2
C
B
@ 0 A
0
0
Stroh’s formalism is used by Lothe/Barnett(76), Deeg(80) to calculate the
singular terms. Explicit closed form ??
In preparation: Computation of the singular terms by numerical solution of a
quadratic eigenvalue problem (W.Geis)
Page 15
Calculation of the coefficients c1 , c2 , c3 , c4 by Maz’ya-Plamenevski coefficient
formulas (76):
ci = bi
R
f ·r
Ω
−1
2
S
−i
dy +
R
1
1
>
− 2 −i
>
− 2 −i
N
ABU
·
r
S
−
U
·
N
ABr
S ds
∂Ω
bi = scaling constants. Shortly
R
R
ci = bi Ω f ·S−i dy+ ∂Ω g·S−i −U ·N > ABS−i ds.
Page 15
Calculation of the coefficients c1 , c2 , c3 , c4 by Maz’ya-Plamenevski coefficient
formulas (76):
ci = bi
R
f ·r
Ω
−1
2
S
−i
dy +
R
1
1
>
− 2 −i
>
− 2 −i
N
ABU
·
r
S
−
U
·
N
ABr
S ds
∂Ω
bi = scaling constants. Shortly
R
R
ci = bi Ω f ·S−i dy+ ∂Ω g·S−i −U ·N > ABS−i ds.
We have assumed, that!the normal mechanical and electric stresses
σ
g = N > ABU = N >
= 0.
D
Using FEM and BEM the coefficients c1 can be computed by postprocessing at
least for linear elasticity. Maybe, it can be done for piezoelectric ceramics in a
similar way.
Note, the coefficients ci are fixed numbers, depending on the volume and surface
loads.
Page 16
We remind of the K-factors in linear
! isotropic elasticity (plane strain):
!
ω
ω
√
√
1
1
sin( 2 )(κ + 2 + cos ω)
cos( 2 )
KI r
KII r
1
2
2
2
√
√
r S = 2µ 2π (κ − cos ω)
S
=
,
r
2µ 2π
sin( ω2 )
cos( ω2 )(κ − 2 + cos ω
0
1
1
σ11
B 1 C
Bσ C
@ 22 A
1
σ12
=
KI cos( ω )
2
√
2πr
0
1
B
B1
@
1
3ω
− sin( ω
2 ) sin( 2 )C
3ω C ,
+ sin( ω
2 ) sin( 2 )A
3ω
sin( ω
2 ) cos( 2 )
0
1
2
σ11
B 2 C
Bσ C
@ 22 A
2
σ12
0
=
1
− sin( ω
)(2 + cos( ω
) cos( 3ω
2
2
2 ))C
KII B
ω
3ω
B
C
√
sin( ω
A
2 ) cos( 2 ) cos( 2 )
2πr @
ω
ω
3ω
cos( 2 )(1 − sin( 2 ) sin( 2 ))
Page 16
We remind of the K-factors in linear
! isotropic elasticity (plane strain):
!
ω
ω
√
√
1
1
sin( 2 )(κ + 2 + cos ω)
cos( 2 )
KI r
KII r
1
2
2
2
√
√
r S = 2µ 2π (κ − cos ω)
S
=
,
r
2µ 2π
sin( ω2 )
cos( ω2 )(κ − 2 + cos ω
0
1
1
σ11
B 1 C
Bσ C
@ 22 A
1
σ12
=
KI cos( ω )
2
√
2πr
0
1
B
B1
@
1
3ω
− sin( ω
2 ) sin( 2 )C
3ω C ,
+ sin( ω
2 ) sin( 2 )A
3ω
sin( ω
2 ) cos( 2 )
0
1
2
σ11
B 2 C
Bσ C
@ 22 A
2
σ12
0
=
√
√
1
KI = limr→0 2πr σ22 (ω = 0) = limr→0 2πr σ22 (ω = 0)
√
√
2
KII = limr→0 2πr σ12
(ω = 0) = limr→0 2πr σ12 (ω = 0)
Break down of this method in piezoelectricity ( Pak 92, Park/Sun 95)
or anisotropic elastic case:
√
√
lim lim 2πr σ.. (r, ω) 6= lim lim 2πr σ.. (r, ω)
r→0 ω→0
1
− sin( ω
)(2 + cos( ω
) cos( 3ω
2
2
2 ))C
KII B
ω
3ω
B
C
√
sin( ω
A
2 ) cos( 2 ) cos( 2 )
2πr @
ω
ω
3ω
cos( 2 )(1 − sin( 2 ) sin( 2 ))
ω→0 r→0
Page 17
These authors have investigated an infinite piezoelectric medium containing a
center crack with far-field loading. They observed that for ω = 0 the stress and
electrial fields are not coupled.
Citation Park/Sun 95: This leads us to conclude that stress intensity factor is not
suitable as fracture criterion for piecielectric materials.
They used the mechanical part of the energy release rate (crack closure integral)
instead.
Page 17
These authors have investigated an infinite piezoelectric medium containing a
center crack with far-field loading. They observed that for ω = 0 the stress and
electrial fields are not coupled.
Citation Park/Sun 95: This leads us to conclude that stress intensity factor is not
suitable as fracture criterion for piecielectric materials.
They used the mechanical part of the energy release rate (crack closure integral)
instead.
Mathematical point of view:
To overcome these difficulties use the correct formulas for the computation of the
stress intensity factors!
Conclusion
Page 18
•
Crack growth in piezoceramics can be enhanced or retarded by an electric
field, depending on its magnitude and direction. In particular, if the crack is
perpendicular to the poling direction, the resulting field equations for the
mechanical and electric fields are strongly coupled.
•
The Griffith criterion, based on the total energy balance, is discussed. The
energy release rate is given by a Griffith formula from which the J-integral
follows. The influence of volume forces (e.g. temperature) is regarded. Surface
forces (tractions, charges) can be considered too.
•
It is possible to express the total energy release rate by stress intensity factors
in a correct way.
•
Since the computation of the stress intensity factors demands the knowledge
of the dual singular terms, it seems to be more efficient to compute the
J-integral.
•
Numerical experiments should be done in future.

Discussion of the Griffith fracture criterion in piezoelectricity

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