Matrix Cracking in Ceramic and Polymer Matrix Laminated Composites
In-plane stress-strain relation of laminate
where
Stress-strain relation for each ply
with 1-axis parallel to fibers.
With ν m = ν f ≡ ν
Plane strain tensile modulus
The above assumes bonded fibers.
Reference: Xia, Carr, Hutchinson, 1993, Acta Mater 41, 2365-2376.
Matrix Cracking, continued
For specified Poisson ratios (ν f = ν m = 0.2) :
⎛ Ef ⎞
Gss E0
=
,c⎟
f
⎜
σ 2t
E
⎝ m ⎠
Finite element calculations (Xia, et al.) give
Gss E0
σ 2t
Ef
Isolated tunnel crack—plane strain assumed
for crack analysis.
Em
Representative of ceramic composites
but not epoxy/glass or epoxy/graphite composites
overall (average applied stress)= σ
tensile stress in the 90o layer (prior to cracking): σ 0 =
= 1, 2,3, 4,5
2ET
σ
EL + ET
Steady-state energy release rate for tunneling crack:
For uniform, isotropic material :
where δ ( x ) is the open displacement computed
from the plane strain solution. If there is initial
residual stress σ R then it must be added to σ 0 .
Gss E π
=
σ 2t 2
Matrix Cracking, continued—Limit of very stiff fibers
The previous method can be used for the case of very stiff fibers compared to matrix—see Xia, et al. We can also get
Insight from the results of Ho and Suo (???) for one isotropic layer sandwiched between two isotropic layers of different modulus.
#1
Gss E2
σ 02h
σ0
(h = 2t )
E1 − E2
, σ 0 = stress in layer 2
E1 + E2
E
For 1 >> 1, α → 1 and
E2
α=
Gss E2
≅ 0.5
2
σ0 h
For
G E
E2
>> 1, α → −1 and ss 2 2 becomes large!
E1
σ0 h
Matrix Cracking, continued—some representative numbers
For an epoxy matrix of thickness, h = 0.5mm, E = 5GPa, Γ IC = 50 Jm 2
Gss = 0.5
σ 02h
E2
= 0.5E2ε 0 2 h
Gss = Γ IC ⇒ ε 0 ≅ 0.02 & σ 0 ≅ 100 MPa
For an ceramic matrix of thickness, h = 0.5mm, E = 200GPa, Γ IC = 5 Jm 2
Gss =≅ 1
σ 02h
E2
= E2ε 0 2 h
Gss = Γ IC ⇒ ε 0 ≅ 0.2 ×10−3 & σ 0 ≅ 40MPa
Matrix Cracking, continued—periodic cracks
Gss E0
σ 2t
Isolated crack limit
2t / 2 L
The same technique can be used to compute Gss
assuming all the cracks tunnel simultaneously.
One makes use of a unit cell with periodic boundary conditions.
( E0 )cracked
E0
Matrix Cracking, continued—Sequential periodic cracks
s --periodic unit cell
s/2
s
The Gss for a second set of cracks bisecting an existing set with spacing s
( Gss )sequential = ( SEcracks ( s) − SEcracks ( s / 2) ) s
= ⎡⎣ 2 ( SEno cracks − SEcracks ( s / 2) ) s / 2 − ( SEno cracks − SEcracks ( s ) ) s ⎤⎦
= 2Gss ( s / 2) − Gss ( s )
With Gss E0 /(σ 2t ) = f (t / L)
for cracks with spacing 2L (see previous slide),
A new set of cracks tunneling between
a set of previous cracks.
s ≡ 4 L = spacing between existing cracks
s / 2 ≡ 2 L = spacing between new tunneling cracks
Recall that for a single set of cracks
with spacing s: Gss ( s ) =
( SEno cracks − SEcracks (s) ) s
( Gss )sequential E0 /(σ 2t ) = 2 f (2t / L) − f (t / L)
for cracks bisecting existing cracks with spacing 4 L.
Matrix Cracking, continued—crack spacing versus stress
With Gss E0 /(σ 2t ) = f (t / L) for cracks with spacing 2L (see previous slide),
( Gss )sequential E0 /(σ 2t ) = 2 f (2t / L) − f (t / L)
for cracks bisecting existing cracks with spacing 4 L.
Gss E0
σ 2t
⎛t⎞
= f⎜ ⎟
⎝L⎠
σ
t
E0 Γ IC
t/L
2t / 2 L
Relation between stress and crack spacing
(Gss ) sequential = Γ IC
⇒
Γ IC E0
= 2 f (2t / L) − f (t / L)
σ 2t
Note that we assume a new set of cracks is nucleated
and bisects the previous set of cracks—thus it is the
sequential G that is relevant.
Note, t/L increases roughly linearly with stress once
the crack spacing exceeds about twice the layer
thickness. The spacing does not correlate with
stress for large spacing—this behavior is
dominated by initial flaw statistics which is not
considered here.
Competition between crack penetration and deflection at an interface
σ
A little background: Consider the feasibility of tunnel cracking in a homogeneous isotropic material.
Gsides = π
2a
G front =
θ
r
Fig. B1
Fig. B2
Fig. A
Fig. B3
λ is given by:
2( β − α ) α + β 2
+
cos πλ = (1 − λ )
1+ β
1− β 2
(Zak and Williams, J. Appl. Mech. 30, 142-143, 1963)
See Next Slide for λ
2
Ref: He and Hutch, IJSS 25, 1053-1067,1989
& He, Evans, Hutch, IJSS 31, 3443-34551994
σ 2a
Cracks will not tunnel
in a homogeneous, isotropic
material because it is more
favorable for them to spread
2
π σ a along their sides.
E
2 E
If the crack is in a layer and
the sides are along an interface,
there are three possibilities:
(i) the crack is arrested along the sides
(ii) the crack penetrates the interface
(iii) the crack deflects into the interface
We first consider the stress field near the crack tip
in Fig. A for the tip at the interface. The stress field
is symmetric about the crack assuming the loading
is symmetric. The stresses have the form:
σ ij =
k
fij (θ )
(2π r )λ
where the dimensionless functions fij (θ ) depend
on the Dundurs' mismatch parameters, α and β ,
as does λ --see to the left.
Note that k has units of stress ilength λ = Paimλ !!
Competition between crack penetration and deflection at an interface, continued
k
fij (θ )
(2π r )λ
σ ij =
λ
k has units Paimλ
Consider a short crack of length a in B1.
λ = 1/ 2, isotropic
β =0
α=
The stress intensity factor K of this crack depends
linearly on k. Dimensional arguments require:
E1 − E1
E1 + E2
K1 = c1ka − λ +1/ 2
G penetration = (c1k )2 a1− 2 λ / E1
where c1 is a dimensionless function of α and β .
Note! G → ∞ as a → 0 if λ > 1/ 2 (α <0).
Now consider a short crack of length a in B2 & B3.
B3
B1
Gdeflection
G penetration
B2
⎡⎣ d + d 2 ⎤⎦ E
1
=
2
c1
E*
2
1
tan Ψ deflection ≡
The stress intensity factors K1 & K2 of this crack depends
linearly on k. For each case, dimensional arguments require:
2
K2 d2
=
K1 d1
Note that the above ratios are independent of load!
⇐
K1 = d1ka − λ +1/ 2 , K 2 = d 2 ka − λ +1/ 2 ,
Gdeflection = ⎡⎣(d1k )2 + (d 2 k ) 2 ⎤⎦ a1− 2 λ / E *
1 1⎛ 1
1 ⎞
for β =0 and * = ⎜ + ⎟
E
2 ⎝ E1 E2 ⎠
where d1 and d 2 are dimensionless functions of α (β = 0).
Competition between crack penetration and deflection at an interface, continued
Crack Advance Criteria:
Penetration ⇒ G penetration = (Γ IC ) material1
Deflection ⇒ Gdeflection = (ΓC (ψ ))interface
Assuming roughly the same flaw size, a,
for both the interface and the penetrating crack,
(ΓC (ψ ))interface Gdeflection
<
⇒ Deflection wins
(Γ IC ) material1
G penetration
(ΓC (ψ ))interface Gdeflection
>
⇒ Penetration wins
G penetration
(Γ IC ) material1
Gdeflection
See plot. Of course the load must be sufficient such that
Gdeflection ≥ (ΓC (ψ ))interface or G penetration ≥ (Γ IC ) material1
β =0
G penetration
When the elastic mismatch is small,
Gdeflection
1
≅
G penetration 4
Deflection wins
Penetration wins
(ΓC (ψ ))interface
(Γ IC ) material1
Competition between crack advance in interface and kinking out of interface
Reference: He & Hutchinson, J. Appl. Mech. 1989, 270-278.
Crack in the interface:
Stress intensity factors & energy release for crack on interface:
K1 & K 2 , tanψ 0 =
K2
1
1 1⎛ 1
1 ⎞
, G0 = * ( K12 + K 2 2 ) ,
= ⎜ + ⎟
*
K1
E
E
2 ⎝ E1 E2 ⎠
Crack kinking out of the interface:
Stress intensity factors & energy release for kinked crack ( β = 0):
K
1
K I & K II , tanψ = II , Gkink =
K I 2 + K II 2 )
(
KI
E2
Assume ψ 0 > 0 so kinked crack propagates into material #2.
K I & K II are linear functions of K1 & K 2 . Dimensional analysis implies:
K I = a11 K1 + a12 K 2 , K II = a21 K1 + a22 K 2
where the aij (ω , α ) depend on ω and the first Dundurs' parameter α, but independent of a.
Brief sketch of solution procedures to determine KI & KII based on integral equation methods.
Let br (η ) & bθ (η ) be components of an edge dislocation at z0 . The problem
noted in the figure where the dislocation interacts with a semi-infinite crack can
be solved in closed form. The tractions acting on the plane at angle ω
at a point z = te − iω are given by (see He & Hutch, 1989)
⎛⎛ 1 1
⎞
⎞
+ Hθθ (t ,η , ω ) ⎟ bθ (η ) + Hθ r (t ,η , ω )br (η ) ⎟
⎠
⎝ ⎝ 4π t − η
⎠
σ θθ (t ) = E ⎜ ⎜
⎛⎛ 1 1
⎞
⎞
+ H rθ (t ,η , ω ) ⎟ br (η ) + H rθ (t ,η , ω )bθ (η ) ⎟
⎠
⎝ ⎝ 4π t − η
⎠
σ rθ (t ) = E ⎜ ⎜
Competition between crack advance in interface and kinking out of interface: continued
The stress on the plane at z due to the applied intensity factors is (classic crack tip fields)
σ θθ (t ) =
K1
K2
K1
K2
f θθ(1) ( −ω ) +
f θθ(2) (−ω ), σ rθ (t ) =
f rθ(1) (−ω ) +
f rθ(2) (−ω ),
2π t
2π t
2π t
2π t
The integral equations for the distributions br (η ) & bθ (η ) are
a⎛⎛ 1
⎞
⎞
K
K2
1
+ Hθθ (t ,η , ω ) ⎟ bθ (η ) + Hθ r (t ,η , ω )br (η ) ⎟ dη = − 1 fθθ(1) (−ω ) −
E∫ ⎜⎜
fθθ(2) (−ω )
0
2π t
2π t
⎠
⎝ ⎝ 4π t − η
⎠
⎛⎛ 1 1
⎞
⎞
K
K2
f r(2)
+ H rθ (t ,η , ω ) ⎟ br (η ) + H rθ (t ,η , ω )bθ (η ) ⎟dη = − 1 f rθ(1) (−ω ) −
E∫⎜⎜
θ ( −ω )
4π t − η
2π t
2π t
⎠
⎠
0 ⎝⎝
a
These are called Cauchy-type integral equations. There are powerful numerial methods
for solving these equations (Erdogan and Gupta, 1972, Q. Appl. Math. 29, 525-534).
The desired stress intensity factors, K I and K II , and thus the coefficients, aij , are simply
related to the distribution of the dislocations as t → a.
Alternatively, finite element methods could be used to obtain the intensity factors and
coefficients. However, given the interest in all orientations ω , integral equation methods
are probably more efficient and somewhat more accurate.
Competition between crack advance in interface and kinking out of interface: continued
KINKING IN A HOMOGENEOUS MATERIAL UNDER MIXED MODE LOADING
E1 = E2 = E , ν 1 = ν 2 = ν ;
K1 & K 2 are prescribed.
Contending criteria for advance of the kinked crack in a material with isotropic
and homogenerous elastic and fracture properties.
A) ω is determined by K II = 0; advance requires K I = K IC
B) ω is determined by maximizing G; advance requires G = Γ IC
C ) ω is determined by maximizing σ θθ associated with K1 & K 2
Criterion C was set as a homework problem. It give a reasonable
ψ0
estimate of ω (compared to A or B) as long as ψ 0 < 45o
ψ 0 = tan −1 ( K 2 / K1 )
=ψ 0
α =0
β =0
ψ0
There is very little difference between A & B
Competition between crack advance in interface and kinking out of interface: continued
Bi-material case
ψ0
=ψ 0
=ψ 0
Kink angle associated with K2=0
ψ0
=ψ 0
Competition between crack advance in interface and kinking out of interface: continued
Bi-material case
Employ maximum G criterion B--(essentially indentical to A).
ψ0
Crack advance criterion for material 2 below interface: Gkink = Γ (2)
IC
β =0
Crack advance criterion for interface crack: G0 = ΓC(interface) (ψ 0 )
Consider curves where kinking and crack advance in
interface are equally likely :
Kinking into material 2
Γ (2)
Gkink
IC
= (interface)
G0
(ψ 0 )
ΓC
α=
Advance in interface
Γ
(2)
IC
/Γ
(interface)
C
(ψ 0 )
These curves are plotted in the figure to the left. That is, the solid
curves are ψ 0 vs. Gkink / G0 .
For a prescribed mode mix, ψ 0 :
(interface)
Kinking into material 2 will occur if Γ (2)
IC / Γ C
Curves corresponding to equally likely kinking and advance in
the interface for a wide range of elastic mismatch.
is to the left of the curve (for a given α ), while
Crack advance along with interface will occur if
For example, for α =0 and ψ 0 = 45o :
(interface)
is to the right of the curve.
Γ (2)
IC / Γ C
(interface)
Kinking will occur if Γ (2)
(45o ) < 1.6
IC / Γ C
(interface)
Interface advance will occur if Γ (2)
(45o ) > 1.6
IC / Γ C
The more compliant is material 2, the greater must be its toughness
to avoid kinking.