Principle of Superposition
For a linear elastic system, the individual components of stress,
strain, displacement are additive. For example, two normal stress
in x direction caused by two different external load can be added to
get total stress
x 
P M
 y
A I
On simillar lines stress intensity
factors can also added
K total  K I (a)  K I (b)
P
P
a
P
P
= Ya a  Y b a
= Ya a  0
M
M
M
K total  K I (a)  0
M
M
P
a
M
P
Principle of Superposition
For example consider a plate having semicircular crack
subjected to an internal pressure
1.12 a
KI 
 g( )

3  a


,
2
8 8c
2
1/4
 2
a
2 
g( )   sin   2 cos  
c


For Semi circular crack

  , g( )  1
2
K I (total)  K I (b)  K I (c)
2
 1.12 p a  0

2.24
=
p a

2
a
2c
Crack Tip Plasticity
LEFM assumes a sharp crack tip, inducing infinite stress at the
crack tip. But in real materials, the crack tip radius is finite and
hence the crack tip stresses are also finite. In addition, inelastic
deformations due to plasticity in metals, crazing in polymers leads
to further relaxation of stresses.
For metals with yielding, LEFM solutions are not accurate.
A small region around the crack tip yields leading to a small
plastic zone around it.
For moderately yielding metals, LEFM
solutions can be used with simple
correction.
For extensively yielding metals,
alternative fracture parameters like
CTOD, J-integral are to be used taking
into account material non-linearity.
Small Scale Plasticity
Irwin’s Approach
Normal stress yy based on elastic analysis is given by
yy 
KI
q 
q
3q 
cos   1  sin   sin   
2r
 2
 2   2 
On the crack plane q = 0
 yy 
KI
2 r
As a first approximation yielding
occurs when yy = ys
2
KI
1  KI 
a  
 ys 
, ry 

 , or ry  

2   ys 
2   ys 
2ry
2
ry = first order estimate of plastic zone size. This is approximate
estimate, because of the fact that yy is based on elastic analysis
Irwin’s Approach
When yielding occurs, stresses redistribute in order to satisfy equilibrium conditions.
The cross hatched regions represents forces active in the elastic analysis that cannot be
carried in elasto-plastic analysis, because of the reason that the stresses cannot exceed
ys.To redistribute this excessive force, the plastic zone size must increase. This is
possible if the material immediately ahead of plastic zone carries more stress.
Irwin proposed that plasticity makes the crack behave as if it were
larger than actual physical size.
Let the effective crack size be aeff ,
such that aeff = a+d,
where d is the correction.
or
To permit redistribution of stresses , the areas A and B must be the same
2
 KI 

  plastic constraint factor
  ys 
effective crack size = a eff
1  KI 
 a  2ry = a+ 

   ys 
2
Effective stress in tensity factor K I  Y a eff
A long rectangular plate has a width of 100 mm, thickness
of 5 mm and an axial load of 50 kN. If the plate is made of
titanium Ti-6AL-4V,(KIC=115 MPa-m1/2, ys=p10 MPa)
150KN
what is the factor of safety against crack growth for a
crack of length a = 20 mm.
K I  Y  aeff
2
4

a
a 
  a  


Y   sec     1  0.025    0.06   
b
b 
  2b  

trial 1. a eff = a = 0.01m
1/ 2
1/ 2
X 0.2  

Y=  sec 

2

 
 1.02448
2
4
 1  0.025  0.2   0.06  0.2  
150 X 1000

 300 MPa
5 X 100
2a=20mm
2b=100mm
K I  1.02448 X 300 X 0.01
 54.525MPa  m1/ 2
Factor of Safety without plastic correction =
K IC
115

 2.11
K I 54.525
trial 2.
2
2
1  KI 
1  54.525 
2 ry = 
  
  0.00114275
   ys 
  910 
effective crack size = a eff  a  2 ry = 0.01  0.00114275  0.1114275
Y  1.3007
Effective SIF K I  Y  aeff  1.0307 X 300 X X 0.01114275  57.85
trial 3.
2 ry  0.001286
a eff  a  2ry = 0.011286
Effective stres s in tensity factor K I  58.25
Factor of Safety with plastic correction =
K IC
115

 1.975
K I 58.25
Strip Yield Method (Proposed by Dugdale and Barrenblatt)
They assumed a narrow long slender plastic zone ahead of crack tip
in a non-hardening (ideal plastic) material in plane stress for a
through crack in a infinite plate.
The strip yield plastic zone assumes a crack length of 2a + 2r
Where r is length of plastic zone.
Approximate elasto-plastic behavior is obtained by superimposing two
elastic solutions (a) a through crack under remote tension (b) a
through crack under crack closure stress.
Concept: stress at the crack tip is no more a singular and it is a finite
value (ys), hence stress singularity term is zero. I.e. the length of the
plastic zone is such that the stress intensity factor due to remote
tension cancels with crack closure.
Strip Yield Method
SIF due to crack closure stress can be estimated by considering a normal
force P applied to the crack at a distance x from the center line of crack
SIF for two crack tips are given by
x
P ax
K I (a) 
a a  x
P ax
K I ( a ) 
a a  x
Crack closure force = P = - ys dx
For the problem considered
SIF is obtained by replacing P by
- ys dx and a by a+r, and adding the two
- ys
P
2a
 a+r  x
a+r  x 
K I (closure) 

 
dx
a+r  x 
  a+r  a  a+r  x
a+r
P
Strip Yield Method
a+r
1  a 
cos 


 a+r 
K I ( closure)  -2 ys
(A)
SIF due to remote tension
K I (tension )     a+r 
(B)
(A) = (B)
  
a
 cos 

a+r
2

 ys 
Expanding right hand side using Taylor's series
2
4
a
1   
1   
 1 
  
  
a+r
2!  2 ys  4!  2 ys 
Considering first two term of the Taylor's series
2
  a   KI 
r=
 
 ...... length of the plastic zone
2
8 ys
8   ys 
2
2
Comparison of plastic corrections with LEFM
LEFM solutions are linear
Irwin and strip yield predictions deviate from LEFM theory at
stresses greater than 0.5 ys
Two plasticty predictions deviate at 0.85 ys
Plastic Zone Shape
In the earlier calculation plastic zone size was calculated for q = 0
(along the crack plane), the plastic zone shape will be quite
different when all angles are considered. Plasticity are based on
von Mises and Tresca’s yield criteria. For mode I problem stress
field are obtained using Westergaard’ stress function., then
principal stresses are given by
1 , 2 
 xx   yy
and hence
2
1/ 2
  xx   yy 

2
 
  xy 

2



2
Using the principal stresses in von Mises yield criteria
Tresca:
 max   ys / 2
1  2
1  3
or
2
2
1  2
1

  max 
2
2
 max 
: for PSN
 max
: for PSS
(a) Von Mises yield criterion (b) Tresca yield criterion
Plane stress plastic zone sizes are larger than plane strain
plastic zone size.
Tresca plastic zones are larger than von Mises plastic zones.
Plastic zone shapes for sliding mode and tearing modes
Plane strain or plane stress
In general, the conditions ahead of a crack tip are neither plane stress
nor plane strain. There are limiting cases where a two dimensional
assumptions are valid, or at least provides a good approximation.
The nature of the plastic zone that is formed ahead of a crack tip
plays a very important role in the determination of the type of failure
that occurs. Since the plastic region is larger in PSS than in PSN,
plane stress failure will, in general, be ductile, while, on the other
hand, plane strain fracture will be brittle, even in a material that is
generally ductile. This phenomenon explains the peculiar thickness
effect, observed in all fracture experiments, that thin samples exhibit
a higher value of fracture toughness than thicker samples made of the
same material and operating at the same temperature. From this it can
be surmised that the plane stress fracture toughness is related
to both metallurgical parameters and specimen geometry while the
plane strain fracture toughness depends more on metallurgical factors
than on the others.
Due to presence of crack
tip, stress in a direction to
normal to crack plane yy
will be large near the crack
tip. This stress would in turn
tries to contract in x and z
direction. But the material
surrounding it will
constraint it, inducing
stresses in x and z direction,
there by a triaxial state of
stress exists near the crack
tip. This leads to plane
strain condition at interior.
At the plate surface zz is
zero and ezz is maximum.
This leads to plane stress
condition at exterior.
The state of stress is also dependent on size of plate thickness.
If the plastic zone size is small compared to the plate thickness,
plane strain condition exists.
If the plastic zone size is larger than the plate thickness, plane
stress condition prevails.
As the loading is increased, plastic zone size also increases
leading to plane stress conditions.
Effect of plate thickness on fracture toughness
Limits of LEFM
As per ASTM standard LEFM is applicable for components of size
 KI 
a  2.5 
  
 ys 
2
 KI 
t or B  2.5 
  
 ys 
2
 KI 
W or b  2.5 
  
 ys 
2
As per ASTM standard fracture toughness testing can be done on
Specimens of size
 K Ic 
a, B,W  2.5 
  
 ys 
2
Concept of Isoparametric Elements
Y

(-1,1)

(1,1)
(x3, y3)
(x7, y7)
(1,1)

(x4 , y 4)
(x 6, y6)
(-1,1)

(x 8, y8)
(x5, y5)
(-1,-1)
(1,-1)
(x2 , y2 )
(x1, y1)
(-1,-1)
(1,-1)
X
 Elements are defined in local coordinate system (,) with straight
well defined edges.
Elements defined in local coordinates have advantage that numerical
integration limits vary –1 to +1
Elements from local coordinated when mapped on to global cartesian
coordinates, it can be distorted to a new curvilinear set as shown in the
figure.
 In Isop. Formulation, with distorted elements, numerical integration
are still carried with limits of local coorinates i.e. –1 to +1
In the conventional FE formulations final stiffness
matrix is given by [K] =  [B]T[D][B] dv
 N i

0 

 x

N i 

where  B  0
y 


 N i N i 
 y x 


and dv = dxdy
If an appropriate transformation is obtained for [B] and dv from
(x,y) to (,) we have equation for [K] set in (,) system where
the integration can be done with in the limits –1 to +1
In order to achieve such a transformation ( [B] and dv) two sets of
nodes ate defined for each element.
One set of nodes (marked as )are used to
interpolate coordinates of a point with in the
element using nodal coordinates. For example
If N1, N2, N3…… are the interpolation
function used to interpolate shape variation,
such that
X = N1 X1+ N2 X2 + N3 X3 +………
Y = N1 Y1+ N2 Y2 + N3 Y3 +………
For an eight node quadratic element
N1 = - ¼ .(111)
N5 = - ½ .(121
The interpolation functions N1, N2, N3……
are generally called here as ‘Shape function’
as they define the shape/distortions
4
7
3
8
X (x,y)
6
1
5
2
Nodes to define shape (coordinate
interpolation with in the element)
Another set of nodes (marked as )are used to
interpolate displacements of a point with in the
element using nodal displacements. For
example if N1, N2, N3…… are the interpolation
function used to interpolate shape variation,
such that
u = N1 u1+ N2 u2 + N3 u3 +………
v = N1 v1+ N2 v2 + N3 v3 +………
For a four node bilinear element
N1 = - ¼ .(11
N2 = - ½ .(11
The interpolation functions N1, N2, N3…… are
generally called here as ‘Displacement
function’ as they define the displacement
variation with in the element.
4
7
3
X (u,v)
8
1
6
5
2
Nodes to define displacement
(displacement interpolation
with in the element)
4
If number of nodes used to define shape (identified by
and Ni ) and the number of nodes used to define the
displacement variation (identified by and Ni ) are
equal, say for example 8 nodes for shape and 8 nodes
for displacement, then such elements are called
ISOPARAMETRIC ELEMENTS.
7
3
8
1
6
5
2
Isoparametric element
4
7
3
If number of nodes used to define shape are more than
6
the number of nodes used to define the displacement 8
variation say for example 8 nodes for shape and 4
nodes for displacement, then such elements are called Superparametric
1
5
element2
SUPERPARAMETRIC ELEMENTS.
4
If number of nodes used to define shape are less than
the number of nodes used to define the displacement
variation say for example 4 nodes for shape and 8
nodes for displacement, then such elements are called
SUBPARAMETRIC ELEMENTS.
7
3
8
1
6
5
2
Subparametric element
N i N i x N i y



x  y 
N i N i x N i y



x  y 
 N i   x
    


 N i   x
    
y   N i 
   x 



N
y   i 
   y 
 N i 
 N i 
  
 x 

   J   N 
 N i 
 i
 y 
  
(A)
N i 
 N i
   x i   yi 

where J   
N i 
 N i x
yi


i
 
 
J  is called Jacobian matrix
For any curved element formulations (iso, super or sub) [J] is
obtained using shape functions (defined for shapes)
Re writing eqn. (A)
 N i 
 N i 


 x 
1   
 N    J  

 i
 N i 
 y 
  
 N i

0 

 x


N i 
note  B  0
y 


 N i N i 
 y
x 

 N i 
  
note that 
 is obtained by displacement functions
 N i 
  
 N i 
 x 
The terms of strain-displacement matrix [B] are given by 


N
 i
 y 
of the prvious equation and dv  dxdy  det J d d 
Crack tip singular elements
4
7
6
Collapse
side
1,8,4
8
1
5
Crack tip Move mid-side
Nodes at 41 points
3
7
7
6
5
1
5
2
1,8,4
6
Crack tip
Crack tip
Crack tip
3
6
3
3
8
7
4
Use of conventional elements near the crack
tip, even with very fine mesh near would not
simulate the stress field conditions of the
crack tip (1/ r ) . Barsoum showed that by
moving mid-side nodes of a 8 node
isoparametric element to quarter points
induced 1/ r singularity and improved the
performance of the analysis enormously.
2
5
Move mid-side
Nodes at 41 points
2
2
In the FEM formulations strain vector
 N i

  0 


ui 
1
e  J   B    ,
N i 

where  B  0
vi 
 


J = jacobian matrix
 N i N i 
,  parametric coordinates of a point
   


on an element
is given by
The strain vector can be singular if either J  or  B is sigular
-1
 (x,y)
=0
J  can be sigular when det J =
 (,)
consider the quadratic element as shown
-1
let us evaluate the boundary nodes
between 1 and 2 in x direction
4
7
3
6
8
1
5
2
Crack tip Move mid-side
Nodes at 41 points
1
1
N1   (1  ), N1  (1  ),
2
2
x  N1 x1  N 2 x2  N 5 x5
N1  (1  2 )
1
1
x   (1  ) x1  (1  ) x2  (1  2 ) x5
2
2
x1  0, x2  L, x5  L / 4
y
L
L/4
1
2 L
x  (1  ) L  (1   )
2
4
x
solving   1  2
L
x L
Jacobian J  
 (1  ) 
 2
2
5
1
x
L
as x  0; det J  0......leading to singularity
x
Let us evaluate the strain
1
1
u  N1u1  N 2 u2  N 5u5   (1  )u1  (1  )u2  (1  2 )u5
2
2
x
using   1  2
L
1
x 
x
1
x  x 
u    1  2
2  2
 u1   1  2
2
 u2
2
L 
L
2
L  L 
 x x
 4
  u5
 L L
u
1 3
4
1
1
4
4
 2
ex 
 
  u1   
  u2  
  u5
x
2  xL L 
2
xL L 
 xL L 
1
1


......leading to 1/ r singularity
x
r
Virtual Crack Extension Method to Evaluate SIF
For a two-dimensional cracked
body in mode I, the total potential
in terms of FE solutions is given
by
U = ½ {u}T[K]{u}- {u}T[F]
The strain energy release rate
Is defined as
U
G=a
 u
= a
T
crack tip
FE mesh
1 T  K
T  F
 K u  F   2 u a u  u a
The first term in above expression is zero (equilibrium condition).
In the absence of traction on crack face the third term is also zero.
Hence
U K I2
1 T  K
G=
  u
u
a E 
2
a
where E   E for plane stress
E
=
for plane strain
2
1- 
For a given FE mesh for a body with
crack length a, and to extend the crack
length by Da it is not necessary to alter
entire mesh. This can be achieved by
moving a few elements near the crack
tip and keeping rest intact. If N
number of elements that are effected
then
1 T  N  K 
G =  u  
 u
2
 i 1 a 
The strain energy release rate
is proportional to the
derivative of the stiff ness
matrix with respect to crack
length.
a
Da
Stress Approach to Evaluate SIF
yy 
KI
q 
q
3q 
cos   1  sin   sin   
2r
 2
 2   2 
On the crack plane q = 0
KI
 yy 
2 r
Re arranging
r1 r2 r3 r4
K I   yy 2r
y1
y2
y3
At various points r1 , r2 ...measure  y1 ,  y2 ..
Hence a set of K I (r1 ), K I (r2 )...
can be calculated.
Plot these values againist r/a.
Extend the best fit line to meet
abcissa to get the required K I for
the given structural conponent
and loading.
r2
r1
KI
x
x
x
x
x
x
r3
x
r/a
y4
r4
Displacement Approach to Evaluate SIF
2(1  )
r
q   1

2  q 
v
KI
sin   
 cos   
E
2  2   2
 2 
where
v = displacements in y directions
3  
  (3  4) for PSS and   
 for PSN
 1  
At some given angle q=q1
K I (r) 
v1
2
  f  q1  v(r)
r
r2
q= q1
r1
At various points r1 , r2 ...measure v1 , v 2 ..
v3
v2
r
Hence a set of K I (r1 ), K I (r2 )...can be
calculated. Plot these values againist r/a.
Extend the best fit line to meet abcissa to
get the required K I for the given structural
KI
x
x
x
x
x
x
x
conponent and loading.
r/a
r3
v4
r4
Strain Energy Release Rate to Evaluate SIF
1
2
a
1
2
a + Da
For the given structural component and loading conduct FE analysis
for the initial crack length 'a' and find the total strain energy in the system U(a).
Now physically shift the crack tip by one element length by suitablly altering the
boundary condition, and find the new strain energy for the extended crack, U(a+Da).
Now SE release rate is given by
DU
U(a  Da)  U(a)
=Da
Da
using the above computed calue of G, SIF is evaluated using
G
K I = GE for PSS and K I =
GE
for PSN cases.
2
1-