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1
6. Elastic-Plastic Fracture Mechanics
Introduction
Applies when non-linear deformation is confined to a small region surrounding the crack tip
LEFM: Linear elastic fracture mechanics
Elastic-Plastic fracture mechanics (EPFM) :
Effects of the plastic zone negligible, linear asymptotic mechanical field (see eqs 4.36, 4.40).
Generalization to materials with a non-negligible plastic zone size: elastic-plastic materials
crack
Plastic zone
Elastic Fracture Contained yielding Full yielding Diffuse dissipation
D
L
a
, ,L a D B
B
L D a L D a
LEFM, KIC or GIC fracture criterion
EPFM, JC fracture criterion
Catastrophic failure, large deformations
2
Surface of the specimen Midsection Halfway between surface/midsection
Plastic zones (light regions) in a steel cracked plate (B 5.0 mm):
Slip bands at 45°Plane stress dominant
Net section stress 0.9 yield stress in both cases.
Plastic zones (dark regions) in a steel cracked plate (B 5.9 mm):
3
6.2 CTOD as yield criterion.
6.1 Models for small scale yielding : - Estimation of the plastic zone size using the von Mises yield criterion.
- Irwin’s approach (plastic correction).
- Dugdale’s model or the strip yield model.
Outline
6.5 Applications for some geometries (mode I loading).
6.3 The J contour integral as yield criterion.
6.4 Elastoplastic asymptotic field (HRR theory).
4
Von Mises equation:
6.1 Models for small scale yielding :
1 22 22
1 2 1 3 2 31
2e
e is the effective stress and i (i=1,2,3) are the principal normal stresses.
Recall the mode I asymptotic stresses in Cartesian components, i.e.
3cos 1 sin sin ...
2 2 22
3cos 1 sin sin ...
2 2 22
3cos sin cos ...
2 2 22
Ixx
Iyy
Ixy
K
r
K
r
K
r
yy
xy
xx
x
y
Oθ
r
LEFM analysis prediction:
KI : mode I stress intensity factor (SIF)
Estimation of the plastic zone size
5
We have the relationships (1) ,
1 222
1,2 2 2xx yy xx yy
xy
Thus, for the (mode I) asymptotic stress field:
1,2 cos 1 sin2 22
IK
r
rrr
x
y
Oθ
r
and in their polar form:Expressions are given in (4.36).
1 222
2 2rr rr
r
31 2
0 plane stress
plane strain
and
, ,rr r
3
0 plane stress
2cos plane strain
22IK
r
6
Substituting into the expression of e for plane stress
1 22 2 212cos sin cos (1 sin ) cos (1 sin )
2 2 2 2 2 22 2I
eK
r
Similarly, for plane strain
1 2
2 21 31 2 1 cos sin
22 2I
eK
r
(see expression of Y2
p 6.7 )
1 22 2 21
6cos sin 2cos2 2 22 2
IK
r
1 221 3
1 cos sin22 2
IK
r
1 2
2 2cos 4 1 3cos2 22
IK
r
1 22 2 2 2 21
4cos sin 2cos 2cos sin2 2 2 2 22 2
IK
r
1 22cos 4 3cos
2 22IK
r
7
Yielding occurs when: e Y Y is the uniaxial yield strength
Using the previous expressions (2) of e and solving for r,
22 2
22 2 2
1cos 4 3cos
2 2 2
1cos 4 1 3cos
2 2 2
I
Yp
I
Y
K
rK
plane stress
plane strain
Plot of the crack-tip plastic zone shapes (mode I):
2
1
4
p
I
Y
r
K
Plane strain
Plane stress (= 0.0)
increasing 0.1, 0.2, 0.3, 0.4, 0.5
8
1) 1D approximation (3) L corresponding to 0pr Remarks:
221 2
2I
Y
KL
Thus, in plane strain:
2) Significant difference in the size and shape of mode I plastic zones.
For a cracked specimen with finite thickness B, effects of the boundaries:
in plane stress:2
1
2I
Y
KL
- Essentially plane strain in the in the central region.
Triaxial state of stress near the crack tip:
- Pure plane stress state only at the free surface.
B
Evolution of the plastic zone shape through the thickness:
9
4) Solutions for rp not strictly correct, because they are based on a purely elastic:
3) Similar approach to obtain mode II and III plastic zones:
Alternatively, Irwin plasticity correction using an effective crack length …
Stress equilibrium not respected.
10
The Irwin approach Mode I loading of a elastic-perfectly plastic material:
02
Iyy
K
r
Plane stress assumed (1)
crackx
r1
Y
r2
(1)
(2)
(2)
To equilibrate the two stresses distributions (cross-hatched region)
r2 ?
Elastic:
Plastic correction 2,yy Y r r
r1 : Intersection between the elastic distribution and the horizontal line yy Y
12I
YK
r
1
20 2
rI
Y YK
r dxx
yy
21 2
11222
I IY
Y
K Kr r
2 1r r
2
11
2I
Y
Kr
11
Redistribution of stress due to plastic deformation:
Plastic zone length (plane stress):
Irwin’s model = simplified model for the extent of the plastic zone: - Focus only on the extent of the plastic zone along the crack axis, not on its shape.
2
11
2 I
Y
Kr
- Equilibrium condition along the y-axis not respected.
yy
real crack x
Y
fictitious crack
Stress Intensity Factor corresponding to the effective crack of length aeff =a+r1
1 1,I effK a r K a r
2
11
23
I
Y
Kr
2eff
yyK
X
X
In plane strain, increasing of Y. : Irwin suggested in place of Y 3 Y
(effective SIF)
12
Application: Through-crack in an infinite plate
Effective crack length 2 (a+ry)
eff yK a r
21
2Y
effeff
KK a
Solving, closed-form solution:2
11
2Y
effa
K
(Irwin, plane stress)2
1
2I
yY
Kr
with
2a
ry ry
aeff
13
More generally, an iterative process is used to obtain the effective SIF:
eff eff effK Y a a
Convergence after a few iterations…
Initial:
IK Y a a 2
01
2I
Y
Ka a
m
1 plane stress
3 plane strainm
Yes
No
1i i
Algorithm:
( )I
ieff i iK K Y a a
Application:
Through-crack in an infinite plate (plane stress):
∞= 2 MPa, Y = 50 MPa, a = 0.1 m KI = 1.1209982Keff= 1.1214469 4 iterations
Y: dimensionless function depending on the geometry.
Do i = 1, imax :
( )1 1I
ii iK Y a a
2( )1
2I
i
iY
Ka a
m
( ) ( 1)I I
i iK K ?
14
Dugdale / Barenblatt yield strip model
2a cc
Long, slender plastic zone from both crack tips.
Perfect plasticity (non-hardening material), plane stress
Assumptions:
Plastic zone extent
Elastic-plastic behavior modeled by superimposing two elastic solutions:
application of the principle of superposition (see chap 5)
Crack length: 2(a+c)
Very thin plates, with elastic- perfect plastic behavior
Remote tension + closure stresses at the crack
15
2ac c
Principle:
Stresses should be finite in the yield zone:
• No stress singularity (i.e. terms in ) at the crack tip1 r
• Length c such that the SIF from the remote tension and closure stress cancel one another
SIF from the remote tension:
,IK a c a c
YY
16
SIF from the closure stress Y
Closure force at a point x in strip-yield zone:
YQ dx a x a c
Total SIF at A:
IB YQ ( a c ) x
K ,a c( a c ) x( a c )
YY
a c
xAB
I A YQ ( a c ) x
K ,a c( a c ) x( a c )
Recall first,
crack tip A
crack tip B
(see eqs 5.3)
a a c
Y YIA Y
( a c ) a
( a c ) x ( a c ) xK ,a c dx dx
( a c ) x ( a c ) x( a c ) ( a c )
(see equation 5.5)
By changing the variable x = -u, the first integral becomes,
1a
Y
( a c )
( a c ) u( ) du
( a c ) u( a c )
a cY
a
( a c ) udu
( a c ) u( a c )
17
a c
Y YIA Y
a
( a c ) x ( a c ) xK ,a c dx
( a c ) x ( a c ) x( a c ) ( a c )
a cY
a
( a c ) x ( a c ) xdx
( a c ) x ( a c ) x( a c )
2 22
a c
Ya
a c dx
( a c ) x
Thus, KIA can written
The same expression is obtained for KIB (at point B):
We denote KI for KIA or KIB thereafter.
(see eq. 6.15)
a a c
Y YIB Y
( a c ) a
( a c ) x ( a c ) xK ,a c dx dx
( a c ) x ( a c ) x( a c ) ( a c )
1a
Y
( a c )
( a c ) u( ) du
( a c ) u( a c )
a cY
a
( a c ) udu
( a c ) u( a c )
18
Recall that,1
2 2
a c
a
dx acos
a c( a c ) x
12I Y Ya c a
K ,a c cosa c
The SIF of the remote stress must balance with the one due to the closure stress, i.e.
0I I YK ,a c K ,a c
12 cosYa c a
a ca c
Thus, cos
2 Y
a
a c
By Taylor series expansion of cosines,
2 4 61 1 1
1 ...2! 2 4! 2 6! 2Y Y Y
a
a c
Keeping only the first two terms, solving for c22 2
2 88I
YY
a Kc
and /8 = 0.392
Irwin and Dugdale approaches predict similar plastic zone sizes.
19
eff effK a
Estimation of the effective stress intensity factor with the strip yield model:
cos2eff Y
a
a
-By setting effa a c
andcos
2
eff
Y
aK
tends to overestimate Keff because the actual aeff less than a+c
- Burdekin and Stone derived a more realistic estimation:
Thus,
1/ 2
2
8lnsec
2eff YY
K a
sec2 Y
a
(see Anderson, third ed., p65)