fracture mechanics, damage and fatigue linear … solution directly? sif: analytical methods...
TRANSCRIPT
University of Liège
Aerospace & Mechanical Engineering
Fracture Mechanics, Damage and Fatigue
Linear Elastic Fracture Mechanics – Stress Intensity Factor (SIF)
Fracture Mechanics – LEFM – SIF
Ludovic Noels
Computational & Multiscale Mechanics of Materials – CM3
http://www.ltas-cm3.ulg.ac.be/
Chemin des Chevreuils 1, B4000 Liège
Asymptotic solution of LEFM
• Summary
– Detailed expressions for the 3 modes
y
x
rq
Mode I Mode II Mode III
(opening) (sliding) (shearing)
y x
z
2016-2017 Fracture Mechanics – LEFM – SIF 2
• Computation of the SIFs– Analytical methods (for LEFM)
• Full field solution (see next slides)
– Limited to infinite planes
– SIFs obtained from asymptotic limit
• Superposition (see next slides)
– Of existing solutions
• Energetic approach (see previous lecture)
– Related to Griffith’s work
– Numerical
• Collocation method (see Annex 2)
• FEM
– Capture asymptotic solution
» Use singular (Barsoum, 1974) quadratic elements (central nodes moved to quarter edge)
– Energetic approach
– J integral
– Experimental
• Normalized experiments
• Strain Gauge Method
– Use of SIF handbook
LEFM: Computation of SIF
2016-2017 Fracture Mechanics – LEFM – SIF 3
• Reminder: method in linear elasticity
– Problem is governed by the bi-harmonic equation
– One solution of this equation has the form
where w(z) and W(z) are functions to be defined so
• The stress field
• The displacement field
satisfy the BCs
SIF: Analytical methods
Plane e
Plane s
2016-2017 Fracture Mechanics – LEFM – SIF 4
• Computation of SIF by analytical method (LEFM)
– Full field solution of a crack submitted to traction
• Infinite plane
• Mode I: , and ty (-x) = ty (x), ty (-y) = -ty (y)
– Westergaard approach for mode I
• As
and as for mode I, one should have
sxy = 0 for y = 0,
the solution is
• Indeed with this choice
SIF: Analytical methods
2a
x
y
ty
tx
2a
y
ty
x
2016-2017 Fracture Mechanics – LEFM – SIF 5
SIF: Analytical methods
• Computation of SIF by analytical method (LEFM) (2)
– Westergaard solution for mode I (2)
• The general stress/displacement fields read
• Using they become
2a
y
ty
x
Only W to be defined
2016-2017 Fracture Mechanics – LEFM – SIF 6
• Computation of SIF by analytical method (LEFM) (3)
– Full field solution of a crack submitted to traction
• Westergaard solution for mode I
• Solution W(x≥0) satisfying the BCs for a uniform traction (see next slides)
• NB: General solution (Sedov, 1972):
SIF: Analytical methods
2a
y
ty
x
2016-2017 Fracture Mechanics – LEFM – SIF 7
• Computation of SIF by analytical method (LEFM) (4)
– Check solution
• with
– Symmetry with respect to Ox and Oy
R (C) = R (C1) = I (C) = I (C1) = 0
SIF: Analytical methods
a
y
x
qr
z
n+
n-
2016-2017 Fracture Mechanics – LEFM – SIF 8
SIF: Analytical methods
• Computation of SIF by analytical method (LEFM) (5)
– Check solution (2)
• with
• On crack lips z = x ± i |e|, e → 0, |x| < a
Boundary condition on crack lips: t = n . s = 2 R (W’) = ± ty satisfied
2a
y
ty
x
a
y
x
qr
z
n+
n-
±
2016-2017 Fracture Mechanics – LEFM – SIF 9
SIF: Analytical methods
• Computation of SIF by analytical method (LEFM) (6)
– Asymptotic field on crack lips
• For z = a-r ± i |e|, e → 0
– Crack Opening Displacement (COD)
a
y
x
qr
z
n+
n-
2016-2017 Fracture Mechanics – LEFM – SIF 10
• Computation of SIF by analytical method (LEFM) (7)
– Asymptotic field ahead of crack
• For z = a+r ± i |e|, e → 0, r →0+
,
SIF: Analytical methods
a
y
x
qr
z
n+
n-
2016-2017 Fracture Mechanics – LEFM – SIF 11
• Computation of SIF by analytical method (LEFM) (8)
– Asymptotic stress field ahead of crack
• For z = a+r ± i |e|, e → 0, r →0+
,
SIF: Analytical methods
a
y
x
qr
z
n+
n-
2016-2017 Fracture Mechanics – LEFM – SIF 12
• Computation of SIF by analytical method (LEFM) (9)
– Crack in an infinite plate under perpendicular loading
• Superposition
• Case 2: solved for ty = s∞
• Case 1:
&
&
SIF: Analytical methods
2ax
ys∞
s∞
s∞
s∞
x
y
2ax
y
s∞
s∞Case 1 Case 2
2016-2017 Fracture Mechanics – LEFM – SIF 13
• Computation of SIF by analytical method (LEFM) (10)
– Crack in an infinite plate under perpendicular loading
• Full field solution from W and case 1
• Asymptotic solution along y = 0, for x > a
• SIF:
SIF: Analytical methods
syy
x
Asymptotic syy
True syy
Zone of asymptotic
solution (in terms of
1/r1/2) dominance
Structural
responsePlasticity
∞
2ax
ys∞
s∞
0.1
1
10
100
0.001 0.01 0.1 1 10 100
r/a
syy
/ sasymptotic
full field
2016-2017 Fracture Mechanics – LEFM – SIF 14
• Computation of SIF by analytical method (LEFM) (11)
– Crack in an infinite plate under sliding
• Westergaard approach for mode II
– For mode II, syy = 0 for y = 0
– Crack subjected to a shearing tx
• Applying superposition principle and tx =t∞
SIF: Analytical methods
2ax
y t∞
t∞
2016-2017 Fracture Mechanics – LEFM – SIF 15
SIF: Analytical methods
• Computation of SIF by analytical method (LEFM) (12)
– Crack in an infinite plate under shearing
• Stress field (see Annex 1)
• Ahead of crack tip (y = 0 and x = a + r)
2ax
y t∞
t∞
x
y
rq
q1r1
r2
q2
2016-2017 Fracture Mechanics – LEFM – SIF 16
• Computation of SIF by analytical method (LEFM) (13)
– Here we have obtained the SIF using the full field solution so
– Why did we develop the asymptotic solution last time instead of using this full field solution directly?
SIF: Analytical methods
2016-2017 Fracture Mechanics – LEFM – SIF 17
• Solutions from SIF handbooks (see references)
– Obtained by using a variety of methods (analytical, numerical, …)
• Example: mode I for a crack in a finite plate (h/W > 3)
– General formula
– Numerical results based on Laurent expansion
» Isida, 1973
– Periodic crack approximation
» <5% error for a/W < 0.5
» Irwin, 1957
– Fit of Isida’s values
» <0.1% error
» Tada, 1973
SIF: Handbooks
2ax
ys
s
2Wh
0.8
0.85
0.9
0.95
1
0 0.5 1
a/W
(1-a
/W)^
1/2
f(a
/W)
2016-2017 Fracture Mechanics – LEFM – SIF 18
• Finite element model: extraction from stress and displacement fields
– Asymptotic crack tip stress and displacement fields determined by the SIF
&
– Stress correlation
• FEM computation
• Extract stress field s(r,q) for q=0 (or any other q)
• Extrapolate to get the SIFs
– Displacement correlation
• Idem but
with
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 19
• Finite element model: extraction from stress and displacement fields (2)
– Advantages of the method
• Simple
• Can be used with any FE software
• Only one post-processing to determine the 3 SIFs (if suitable loading)
• Can be used along a crack front in 3D
• When using displacement correlation, the field is primary solution
– Drawbacks
• The accuracy is strongly dependant on the mesh refinement
• The mesh refinement required depends on the element ability to capture the
singularity at crack tip (see next slide)
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 20
• Finite element model: Barsoum elements
– Previous method requires a fine mesh since a singularity in r½ is not naturally
captured by usual FE
– This singularity can be captured
• By enriching elements (as in XFEM for LEFM, see later)
• By using Barsoum (quarter-point) elements
– Quarter-point elements
• Quadratic elements with some mid-nodes located at a quarter of the edge
SIF: Numerical approaches
x
h
-1 1
1
-1x
h
0 1
1
x
y
3L/4L/4Mapping
x
y
3L/4L/4
Mapping
2016-2017 Fracture Mechanics – LEFM – SIF 21
• Finite element model: Barsoum elements (2)
– Quarter-point element: 1D example
• Shape functions
• Mapping
• Strain field
• So a strain field (and so for stress) in 1/r½ can be captured
SIF: Numerical approaches
x
-1 10
-0.2
0
0.2
0.4
0.6
0.8
1
-1 -0.5 0 0.5 1
x
N
N
N
1
2
3
x
0 LL/4
2016-2017 Fracture Mechanics – LEFM – SIF 22
• Finite element model: Barsoum elements (3)
– Quarter-point element: 2D example
• Fine mesh at crack tip: mid-nodes of first ring are moved to quarter points
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 23
• Finite element model: global energy & compliance
– For cracks growing straight ahead (if only one mode is involved)
• G=J, and in linear elasticity
• Prescribed loading: (in linear elasticity)
• Prescribed displacement:
– Perform 2 computations
• With the same loading (displacement) but with two different crack lengths
• So, per unit thickness:
• Could be rewritten using compliance C in terms of the generalized loading Q
(linear elasticity)
– Advantages• Simple
• Can be used with any FE software and requires only post-processing
• Less mesh sensitive than correlation methods as a global variable is used
– Drawbacks• 2 computations needed
• Only one mode (so one SIF) can be considered at a time
• In 3D, SIF variation along crack front cannot be determined
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 24
• Finite element model: crack closure integral
– Cracks growing straight ahead
– For such cracks, in linear elasticity
SIF: Numerical approaches
y
x
a Da
r
q
syy = 0,
uy ≠ 0
syy ≠ 0,
uy = 0
y
x
a Da
r q
syy = 0,
uy ≠ 0
syy = 0,
uy ≠ 0
2016-2017 Fracture Mechanics – LEFM – SIF 25
• Finite element model: crack closure integral (2)
– This integral can be computed using FE software
• Nodal release
– First computation: nodes j on both crack lips are constrained together
– Second computation: constrain on node j is released
• SIFs can be computed from
– Reaction forces of first computation &
– Displacement jumps of second computation
– , &
SIF: Numerical approaches
j-1 j j+1j-2
a Da
Du
j-1 j j+1j-2
a Da
F
-F
2016-2017 Fracture Mechanics – LEFM – SIF 26
• Finite element model: crack closure integral (3)
– Advantage
• Requires only post processing
– Drawbacks
• If the crack is not a line of symmetry, the node displacements have to be
constrained (using Lagrange multipliers e.g.)
• Two computations are needed (can be improved using modified nodal release)
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 27
• Finite element model: direct computation of J-integral
– For linear elasticity and for any contour G embedding a straight crack
&
– Assuming symmetry & a structured meshed
• dl = dy on G1, dl = -dx on G1,dl = -dy on G3
• As values are computed on Gauss points,
ideally, the integrals will add Gauss points’
contributions
SIF: Numerical approaches
G1
G2
G3
x
y
2016-2017 Fracture Mechanics – LEFM – SIF 28
• Finite element model: J-integral by domain integration
– For linear elasticity and for any contour G embedding a straight crack
&
– Let us define a contour C=G1+G-+G+-G
• Interior of this contour is region D
• Define q(x, y) such that
– q = 1 on G
– q = 0 on G1
• As crack is stress free
SIF: Numerical approaches
x
y
G
G1
G+
G-
-
2016-2017 Fracture Mechanics – LEFM – SIF 29
• Finite element model: J-integral by domain integration (2)
– Computation of
• C is closed divergence theorem
• But & =0
• q is discretized using the same
shape functions than the elements
– This integral is valid for any annular region around the crack tip
• As long as the crack lips are straight
SIF: Numerical approaches
x
y
C
D
q=1
q=0
2016-2017 Fracture Mechanics – LEFM – SIF 30
• Finite element model: J-integral
– Advantages
• Accurate (especially using domain integration)
• Does not require Barsoum elements
– Drawbacks
• Only one mode at the time
• Either modifying the FE code in order to have easy post processing or
• Post processing difficult when using a standard FE software
– As this method is really accurate and computationally efficient, can it be
modified in order to extract the SIF of each mode?
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 31
• Finite element model: J-integral (2)
– How to extract the different SIFs?
• Compute J for the solution u of the considered problem J
• Compute J for another field uaux to be specialized Jaux
• Compute J for the sum of u & uaux to be specialized Js
• Relation between the different Js?
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 32
• Finite element model: J-integral (3)
– How to extract the different SIFs (2)?
• Relation between the different Js can be deduced from
–
–
• So is rewritten
– The right term is called the interaction integral
» What is its expression in terms of the SIFs ?
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 33
• Finite element model: J-integral (4)
– How to extract the different SIFs (3)?
• It has been found that
• With &
• Direct substitution leads to
–
–
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 34
• Finite element model: J-integral (5)
– How to extract the different SIFs (4)?
• As
– With
• And as these two last relations are symmetric in K and Kaux
– By analogy with
– One can feel that
• This is obtained explicitly after substitution of f and g by their closed forms
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 35
• Finite element model: J-integral (6)
– The SIFs are deduced from the so-called interaction integral
•
• Indeed, if uaux is chosen such that only Kiaux ≠ 0, Ki is obtained directly
– This method
• Is very accurate compared to correlation methods
• But it requires
– More computations
– Extensive modifications of the FE code
SIF: Numerical approaches
2016-2017 Fracture Mechanics – LEFM – SIF 36
• Strain gauge method
– Asymptotic solution for mode I
• with l = -½, 0, ½, …
• The asymptotic solution in r-½ cannot be matched accurately using a gauge
• So the solution considered is limited to the 3rd order
SIF: experimental methods
2016-2017 Fracture Mechanics – LEFM – SIF 37
• Strain gauge method (2)
– Position of the strain gauge
• Located at (r, q ) of the crack tip
• With an orientation a
– One** can show that in the referential O’x’y’
with
• The gauge is located at (r, q*, a* ) such that
&
– Mounting of the gauge is critical
SIF: experimental methods
y
x
rq
y’
x’
a
**Dally JW and Sanford RJ (1988), Strain gage methods for measuring the opening mode stress intensity
factor KI , Experimental Mechanics 27, 381–388.
2016-2017 Fracture Mechanics – LEFM – SIF 38
• Strain gauge method (3)
– Strain gauges are now replaced by Digital Image Correlation (DIC)
• Markers on the surface are tracked optically
• Displacements and strains can be computed
– These strains can be used to deduce the SIFs
SIF: experimental methods
2016-2017 Fracture Mechanics – LEFM – SIF 39
• Flawed cylinder
– A piston is used to increase inner pressure
• From 0 to 55 MPa
– Cylinder made of
• Peaked-aged aluminum alloy
– 7075-T651
• Yield sp0 = 550 Mpa
• Toughness KIC = 30 MPa m1/2
– Malfunction
• Cylinder burst
• Post failure analyses
– Initial elliptical flaw at inner wall
» 4.5 mm long
» 1.45 mm deep
» Normal to hoop stress
– Origin of burst?
Exercise 1
L=
20
cm
t = 1 cm
Din = 9 cm
2c = 4.5 mm
a = 1.45 mm
sqq
sqq
2016-2017 Fracture Mechanics – LEFM – SIF 40
References
• Lecture notes
– Lecture Notes on Fracture Mechanics, Alan T. Zehnder, Cornell University,
Ithaca, http://hdl.handle.net/1813/3075
• Other references
– « on-line »
• Fracture Mechanics, Piet Schreurs, TUe,
http://www.mate.tue.nl/~piet/edu/frm/sht/bmsht.html
– Book
• Fracture Mechanics: Fundamentals and applications, D. T. Anderson. CRC press,
1991.
• Fatigue of Materials, S. Suresh, Cambridge press, 1998.
2016-2017 Fracture Mechanics – LEFM – SIF 41
• Stress field
– Consider thick cylinder with
• rin = 0.045 m & rout = 0.055 m
• Inner pressure p
Exercise 1: Solution
L=
20
cm
t = 1 cm
Din = 9 cm
2c = 4.5 mm
a = 1.45 mm
sqq
sqq
2016-2017 Fracture Mechanics – LEFM – SIF 42
• SIF
– The wall is not perforated
• Use SIF for semi-elliptical crack in large plate
• See SIF handbook
• Geometrical effect
• Plasticity correction
– SSY criterion:
– Not fully satisfied plasticity correction
• SIF with
– Rupture:
• For r = rin, p would be 100 MPa, this is the critical value
Exercise 1: Solution
2c = 4.5 mm
a = 1.45 mm
sqq
sqq
2016-2017 Fracture Mechanics – LEFM – SIF 43
• Rupture mode
– For r = rin, p would be 100 Mpa, this is the critical value
– This is out of the range of the piston activity
• So rupture should come from fatigue
• Cyclic loading
– p from 0 to 55 Mpa
• Hoop stress from 0 to
– SIF
• Assuming a/c remains constant during crack propagation
Exercise 1: Solution
2016-2017 Fracture Mechanics – LEFM – SIF 44
• Cyclic loading
– p from 0 to 55 Mpa
&
– Due to initial flaw
• Assuming curves are valid for R=0
• We are in Paris regime
crack propagation
– Rupture will happen for
• Number of cycles?
Exercise 1: Solution
2016-2017 Fracture Mechanics – LEFM – SIF 45
• Cyclic loading (2)
– As
• Assuming curves are valid for R=0
• We are in Paris regime
• !!!Life strongly depends on the maximum pressure reached during accidents
Exercise 1: Solution
2016-2017 Fracture Mechanics – LEFM – SIF 46
• Computation of SIF by analytical method (LEFM)
– Crack in an infinite plate under shearing
• Mode III:
uz is the imaginary part of a function z(z)
• This function has to be found to satisfy the BCs
• Stress field
– As
– One has
– And
Annex 1: Analytical methods
2ax
y t∞
t∞
2016-2017 Fracture Mechanics – LEFM – SIF 47
Annex 1: Analytical methods
• Computation of SIF by analytical method (LEFM) (2)
– Crack in an infinite plate under shearing (2)
• Solution of the problem?
–
– with
uz = I(z)
• Symmetry?
– uz(-y) = -uz(y) satisfied
• Far away field?
– For y→±∞: syz → ±t∞ satisfied
2ax
y t∞
t∞
2016-2017 Fracture Mechanics – LEFM – SIF 48
Annex 1: Analytical methods
• Computation of SIF by analytical method (LEFM) (3)
– Crack in an infinite plate under shearing (3)
• Solution of the problem?
–
– with
• Crack lips stress free?
satisfied
2ax
y t∞
t∞
x
y
rq
q1r1
r2
q2
2016-2017 Fracture Mechanics – LEFM – SIF 49
• Computation of SIF by Boundary Collocation Method (LEFM)
– Example: crack in a finite circular plate
• Considering mode III (can be done for I and II)
• Laurent series
with
• But for the crack lies in q=p, so we use
is satisfied
– Finite displacement n = 0, 1, …
Annex 2: Semi analytical methods
a=R
x
y
R
q
Tz = f(q)
2016-2017 Fracture Mechanics – LEFM – SIF 50
• Computation of SIF by Boundary Collocation Method (LEFM) (2)
– Example: crack in a finite circular plate (2)
• Unknowns an are obtained by defining m collocation points on the boundary
– At these points
for k=1, .., m, & assuming an expansion up to order p, p+1 ≤ m
Annex 2: Semi analytical methods
a=R
x
y
R
qk
Tz = f(qk)
• Example: R=1, & loading
– For accuracy m = 2p+2
– Collocation points only on upper
side (avoid trivial solution)
– Least squares resolution
2016-2017 Fracture Mechanics – LEFM – SIF 51
100
101
102
1.5
2
2.5
3
p
KIII
m=2(p+1)
p
KII
I