ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo
Stress Intensity Factor
We have modeled a body by using the linear elastic theory. We have modeled a crack in the body by a flat plane, and the front of the crack by a straight line. Within this idealized model, the field around the front of the crack is singular. The singular field is clearly an artifact of the idealized model, but
Irwin and others made the singular field a centerpiece of fracture mechanics.
The mathematics of this singular field had been known long before Irwin entered the field. We will focus on the mathematics in this lecture, and will describe Irwin’s way of using the singular field in the following lecture.
In previous lectures, we have described seminal ideas of Inglis
( http://imechanica.org/node/7457 ) and of Griffith
( http://imechanica.org/node/7470 ). We have also described the reinterpretation of these ideas due to Irwin and Orowan ( http://imechanica.org/node/7507 ). Our descriptions have centered on two quantities: energy release rate and fracture energy. This energy-based approach leads us into applications of fracture mechanics ( http://imechanica.org/node/7531 ). Before showing you more applications, I’d like to tell you about a basic concept, possibly also due to Irwin, concerned with the modes of fracture.
Modes of fracture. Depending on the symmetry of the field around the tip of a crack, fracture may be classified into three modes.
•
Mode I: tensile mode, or opening mode.
•
Mode II: in-plane shear mode, or sliding mode.
•
Mode III: anti-plane shear mode, or tearing mode.
The modes describe the local condition around a point on the front. For example, consider a penny-shaped crack in an infinite body. The front of the crack is a circle. When a load pulls the body in the direction perpendicular to the plane of the crack, every point along the front of the crack is under the mode I condition.
When a load shears the body in the direction parallel to the plane of the crack, every point along the front of the crack is under a mixed mode II and mode III condition. Only a few special points on the front are under either a pure mode II condition, or a pure mode III condition.
The concept of the modes appeals to our intuition, but this concept is absent in the energy-based approach. Energy release rate by itself does not differentiate the modes of fracture. Energy release rate characterizes the amplitude of the applied load, but not the mode of the applied load. To describe the mode of the load will require us to talk about the field near the tip of the crack.
Energy release rate for a crack in a linear elastic material .
Consider a crack of length 2 a in a sheet subject to applied stress
appl
. The body is made of a homogeneous, isotropic, linearly elastic material, with Young’s modulus E and Poisson’s ratio ν . Linearity of the boundary-value problem and
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo dimensional consideration dictate that the energy release rate should take the form
The dimensionless number
G
= β
( σ appl
E
)
2 a
.
must be determined by solving the boundary-value problem.
Linear elastic field around the tip of a crack . Focus on the field near one tip of the crack. We model the tip of the crack as a point, and the two faces of the crack as straight lines. Let ( r ,
) be the polar coordinates centered at the tip of the crack. The two faces of the crack coincide with the lines
−
= +
and
=
.
A material particle in the body is at a distance r from the tip of the crack.
The material particle is said to be near the tip of the crack if r is much smaller than the length characteristic of the boundary-value problem; for example, r << a , where a is the length of the crack.
We want to determine the fields near the tip of the crack, including the stress
ij
( ) , strain
ij r ,
and displacement u i r ,
. To this end, we regard the body as infinite, and the crack as semi-infinite. The faces of the crack are traction-free, and the load is applied remotely from the tip of the crack. We represent the applied load by a single parameter: the energy release rate G . In doing so we change our perspective somewhat. In the past, we regard G as part of the solution of a boundary-value problem. We now regard G as the applied load in a boundary-value problem.
This mathematical model differs from a real crack in a real material in many aspects, but for the time being we stick to the model itself.
r
= +
= −
The near-tip stress field is square-root singular . I find a way to reach the square-root singularity without solving the boundary-value problem; rather, we invoke two elementary considerations: the boundary-value problem is linear, and the geometry of the problem is scale-free. You can skip this section if you do not like this line of reasoning.
The energy release rate G is the loading parameter of the boundary-value problem. As G increases, the fields of stress, strain and displacement increase.
In the linear theory of elasticity, the stress, strain, and displacement are linearly
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo proportional to the applied stress, but G is quadratic in the applied stress. Thus, the stress field scales with G as
ij
( ) ∝ G .
The infinite body and the semi-infinite crack provide no length scales— that is, the geometry of the boundary-value problem is scale-free. The ratio G / E has the dimension of length. The three quantities G , E and r form a single dimensionless group:
G
Er
.
A combination of linearity and dimensional consideration dictates that the field of stress near the tip of the crack should take the form
ij
=
EG r f ij
( ) , where f ij
( ) are dimensionless functions of θ , and possibly also depend on
Poisson’s ratio ν . Thus, from these elementary considerations, we find out that the stress field near the tip of the crack is square-root singular in r .
Similar considerations show that the fields of strain and displacement near the tip of the crack take the following forms:
ij
( )
=
G
Er g ij
( ) , u i
( ) =
Gr
E h i
( ) .
These angular functions can be determined by solving the boundary-value problem, as described below.
Governing equations of linear elasticity . The state of a body is characterized by three fields: displacement u i
( x
1
, x
2
, x
3
) , strain
ij
( x
1
, x
2
, x
3
) , and stress
ij
( x
1
, x
2
, x
3
) . In the body, the three fields satisfy the governing equations:
•
Strain-displacement relations
ij
=
( u i , j
+ u j , i
) / 2
• Balance of forces
ij , j
= 0
• Stress-strain relations
ij
= C ijpq
pq
At every point on the surface of the body, in each direction, we prescribe either the displacement, or the traction
ij n j
. If these equations look vague to you, please review the notes on the elements of linear elasticity
( http://imechanica.org/node/205 ).
The singular field around the front of a crack . While a body is in a three-dimensional space, the singular field around the front of a crack is nearly two-dimensional. Let us clarify this reduction in dimension. In the three-
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo dimensional space, we model a body by a volume, a crack by a smooth surface, and the front of the crack by a smooth curve. For any point on the front, we use the point as the origin to set up local coordinates, with the axis x pointing in the direction of propagation of the front, y normal to the plane of the crack, and z tangent to the front.
Because the front is assumed to be a smooth curve, the field around the front is singular in x and y , but smooth in z . That is, for any component of the field, and f ( x , y , z ) , the derivative
f /
z is small compared to the derivatives
f /
x
f /
y . Consequently, we may drop all partial derivatives with respect to z in the governing equations. Consequently, the singular field around the front is locally characterized by a field of the form u ( x , y ) ( x , y ) ( x , y )
The field decouples into two types:
.
We further assume that the elastic behavior of the material is isotropic.
• Plane-strain deformation: u ( u x ,
( x ,
) y
=
) ≠
0 ,
0 , v ( v x ,
( x y
,
) y
=
)
0
≠ 0 , but w ( x , y ) = 0 .
•
Anti-plane deformation: y , but w ( x , y ) ≠ 0 .
The plane-strain deformation describes mode I and mode II cracks, and the antiplane deformation describes mode III cracks. In class we will mostly talk about mode I cracks.
Plane-strain conditions . To analyze this locally plane-strain field, we focus on an actual plane-strain problem:
•
The body is under the plane-strain conditions.
• The crack is a flat plane.
• The front of the crack is a straight line.
•
The field is elastic all the way to the front.
The fields reduced to
•
Displacement field:
•
Strain field: ε xx u ( x , y ) and
, ε yy
, ε xy v ( x , y ) . Note that w
=
0 .
are functions of x and y . Other strain components vanish.
•
Stress field: σ xx
, σ yy
, σ xy
, σ zz
are functions of x and y . Other stress components vanish.
Strain-displacement relations:
=
∂ u
,
∂ x
ε yy
=
∂ v
∂ y
, ε xy
=
1
2
⎛
⎜⎜
⎝
∂ u
∂ y
+
∂ v
∂ x
⎞
⎟⎟
⎠
.
Balance of forces:
ε xx
∂ σ
∂ x xx
+
∂ σ xy
∂ y
= 0 ,
∂ σ xy
∂ x
+
∂ σ yy
∂ y
= 0 .
Hooke’s law:
xx
=
1
E
(
xx
−
yy
−
zz
) ,
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo
yy zz
=
=
1
E
1
E xy
Under the plane strain conditions, zz
(
(
= zz
=
yy zz
−
−
xx xx
−
−
(
=
( 1 +
E
)
0 xy
,
, so that
xx
+
yy
) . zz
) yy
) ,
,
Substituting this relation to Hooke’s law, we obtain that
xx
=
1
−
E
2
⎛
⎝
xx
−
1
−
yy
⎞
⎠
,
yy
=
1
−
E
xy
2
=
⎛
⎝
yy
−
( 1 +
E
)
1
− xy
.
xx
⎞
⎠
,
The quantity
E
=
1 −
E
ν
2
, is known as the plane-strain modulus.
Under the plane strain conditions, the elastic field is represented by 2 displacements, 3 strains, and 4 stresses. The 9 functions are governed by 9 field equations (3 strain-displacement relations, 3 stress-strain relations, 2 equations to balance forces, and 1 relation between
z
,
x
,
y
. If you have not studied plane-strain problems before, http://imechanica.org/node/319 . please review the notes
Airy’s function . We now have 9 equations for 9 functions. As usual we can eliminate some functions by combining equations. Many approaches of elimination have been devised. Here we will follow the approach due to Airy
(1863), who reduced the system of equations to one equation for one function.
Recall a theorem in calculus. If functions f ( x , y ) and g ( x , y ) satisfy the following relation
∂ f
∂ g
=
,
∂ x
∂ y then a function
( x , y ) exists, such that f =
∂
∂
y
, g =
∂
∂ x
.
According to this theorem, one equation of force balance,
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo implies that a function
∂
∂ x xx
= −
∂
xy
∂ y
( x , y ) exists, such that
,
xx
=
∂
∂ y
The other equation of force balance,
,
xy
= −
∂
∂ x
. implies that a function
∂
xy
= −
∂
yy
∂ x
∂ y
( x , y ) exists, such that
,
yy
=
∂
∂ x
,
xy
= −
∂
∂ y
.
In the above, we have expressed the shear stress
xy
in two ways.
Equating the two expressions, we obtain that
∂
∂
x
=
∂
∂
y
.
According to the theorem in calculus, this equation implies that a function
( x , y ) exists, such that
=
∂
∂ y
,
=
∂
∂ x
.
Summing up, we can express the three stresses in terms of one function:
σ xx
=
∂
2
φ
∂ y 2
, σ yy
=
∂
2
φ
∂ x 2
, σ xy
= −
∂
2
φ
∂ x
∂ y
.
The function
( x , y ) is known as Airy’s function. Stresses expressed by Airy’s function satisfy the equations of force balance.
Strains in terms of Airy’s function . Using Hooke’s law, we express the strains in terms of Airy’s function:
xx
yy
=
=
1
1
2
E
E
2
2
y 2
2
x 2
1
1
2
x 2
,
2
y 2
,
xy
= −
( 1
+
E
) ∂
2
∂ x
∂ y
.
The equation of compatibility . Eliminate the displacements from the strain-displacement relations, and we obtain that
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo
∂
∂
2
ε y xx
2
+
∂
∂
2
ε x yy
2
=
2
∂ ε xy
∂ x
∂ y
.
This equation is known as the equation of compatibility.
The compatibility equation can be expressed in terms of Airy’s function:
∂
4
φ
∂ x 4
+
2
∂
4
φ
∂ x 2
∂ y 2
+
∂
4
φ
∂ y 4
=
0 .
This equation is known as the bi-harmonic equation . According to Meleshko, this equation was first derived by Maxwell when asked by Stokes to review the paper by Airy. The plane-strain problem is governed by the bi-harmonic equation and the boundary conditions. Once
is solved, one can determine the stresses, strains, and displacements.
The biharmonic equation is often written as
⎛
⎜⎜
∂
2
∂ x 2
+
∂
2
⎞
⎟⎟
⎛
⎜⎜
∂
2
+
∂
2
⎞
⎟⎟
=
0 .
∂ y 2
∂ x 2
∂ y 2
Polar coordinates . When Airy’s function is expressed as a function of the polar coordinates,
( ) , the polar components of the stress are expressed as
σ rr
=
∂
2
φ r 2
∂ θ
2
+
The bi-harmonic equation is
∂ φ r ∂ r
, σ
θθ
=
∂
2
φ
∂ r 2
, σ r θ
= −
∂
∂ r
⎛
⎝
∂ φ r ∂ θ
⎞
⎠
.
⎛
⎜⎜
⎝ r 2
∂
∂
2
θ
2
+ r
∂
∂ r
+
∂
∂ r
2
2
⎞ ⎛
⎟⎟
⎠
⎜⎜
⎝ r 2
∂
∂
2
θ
2
+ r
∂
∂ r
+
∂
∂ r
2
2
⎟⎟
⎠
⎞
φ =
0 .
The Williams expansion . Within the linear elastic theory, the field in a body is determined by a boundary-value problem. The field depends on the boundary conditions, namely, the size and the shape of the crack and the body, as well as the magnitude and the distribution of the load. Some such boundaryvalue problems had been solved before Williams entered the field. Williams took a different approach. Instead of solving individual boundary-value problems, he focused on the singular field around the tip of the crack, in a zone so small that the crack can be assumed to be semi-infinite, and the boundary of the body infinitely far away. He discovered that the form of the singular field is universal, independent of the shape of the body and the crack.
Let ( r ,
) be the polar coordinates, centered at a particular point on the front of the crack. The crack propagates in the direction of the crack coincide with
= ±
= 0 , and the two faces
. The two faces of the crack are traction-free.
We solve the biharmonic equation using the method of separation of variables. Each term in the bi-harmonic equation has the same dimension in r .
For such an equi-dimensional equation, the solution is r to some power. Write the solution in the form
= r λ + 1 f ,
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo where the constant
and the function f ( ) are to be determined. Insert this form into the biharmonic equation, and we obtain an ordinary differential equation (ODE):
⎡
⎢
⎣ d d θ
2
2
+ ( λ − 1 ) 2
⎤
⎥
⎡
⎢
⎣ d d θ
2
2
+ ( λ + 1 ) 2
⎤
⎥
⎦ f ( ) = 0 .
⎦
This is an ODE with constant coefficients. The solution is of the form f ( )
= exp ( ) , where b is to be determined. Inserting this form into the ODE, we find that
( b 2
+ (
1 ) 2 ) ( b 2
+ (
+ 1 ) 2 )
= 0 .
This is a fourth-order algebraic equation for b . The four roots are b
(
1 ) i ,
(
1 ) i , where i
= f
( )
1
=
. The general solution to the ODE is
A cos ( λ +
1 ) θ +
B cos ( λ
1 ) θ +
C sin ( λ +
1 ) θ +
D sin ( λ
1 ) θ , where A , B , C and D are constants. For the mode I crack, the field is symmetric with respect to the x -axis, so that C
=
D
=
0 .
Summarizing, we find that the solution for a mode I crack is
( ) = r λ + 1 [ A cos (
+ 1 )
+ B cos (
1 )
] , where
, A and B are constants to be determined by using the boundary conditions.
The stresses are
σ rr
=
σ
θθ
− λ r
=
λ − 1 [ A ( λ +
1 ) ( λ +
1 ) θ +
B ( λ −
3 ) ( λ −
1 ) θ
( λ +
1 ) λ r λ −
1 [ A cos ( λ +
1 ) θ +
B cos ( λ
1 ) θ
]
]
At
σ r θ
=
( λ −
1 ) r λ − 1 [ A ( λ +
1 ) ( λ +
1 ) θ +
B
=
, both components of the traction vanish,
( λ
θθ
−
1
=
) (
r θ
λ
=
−
0
1 ) θ ] .
, namely,
⎡
⎢
⎣
(
(
λ
λ
2
+
1
−
)
1
λ
) cos sin
λπ
λπ (
( λ
λ
+
−
1
1 )
) λ
2 cos sin
λπ
λπ
⎤
⎥
⎦
⎡
⎢
⎣
A
B
⎤
⎥
⎦
=
⎡
⎢
⎣
0
0
⎤
⎥
⎦
This pair of linear algebraic equations for A and B form an eigenvalue problem.
To have a solution such that A and B are not both zero, the determinant must vanish, namely,
(
2
1 ) sin
cos
=
0 .
The solutions are
= ...
− 1 , −
1
2
, 0 ,
1
2
, 1 ,
3
2
,...
Consequently, the field of stress takes the form of an expansion:
ij
=
+ ∞
m
= −∞ a m r m / 2 f ij
( ) ( ) .
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo
The functions f ij
( ) ( ) are determined by the eigenvalue problem. The amplitudes a m
, however, are not determined by the eigenvalue problem, and should be determined by the full boundary-value problem.
The square-root singularity . Each term in the Williams expansion corresponds to a solution to the biharmonic equation. Which term should we choose? Note that
~ r λ − 1 ,
~ r λ − 1 ,
Recall the stress concentration for the ellipse. Require that the displacement to be bounded, so that
>
0 two requirements force us to choose u ~ r λ .
. Require the stress to be singular, so that
<
1 . The
= +
1
2
.
Substituting this value to the algebraic equations, we find that B
=
3 A . In particular, we find the hoop stress is
θθ
( )
= −
B r cos 3
⎛
⎝
2
⎞
⎠
.
The above justifications for the choice of the square-root singularity are flimsy. The significance of this choice has to be understood later, when we see how this singularity is used in practice. For a discussion, see C.Y. Hui and Andy
Ruina, Why K? High order singularities and small scale yielding, International
Journal of Fracture 72, 97-120 (1995).
Stress intensity factor . Williams solved an eigenvalue problem because no load was specified: both the field equations and the boundary conditions are homogeneous. Like any other eigenvalue problem, this eigenvalue problem leaves the amplitude undetermined. In this case, all the field is determined up to the constant B . By convention, the constant B is written as
B = − K / 2 π . Consequently, the stress at a distance r directly ahead the crack tip is
θθ
( ) =
K
2
r
.
The legend has it that Irwin chose the letter K after J.A. Kies, one of his coworkers.
The field of stress around the tip of the crack is given by
σ rr
=
K
2 π r
θθ
= cos
⎛
⎜
⎝
θ
2
⎞
⎟
⎠
⎛
⎜⎜
⎝
1
K
2
r cos 3
+ sin
⎛
⎝
2
2
⎞
⎠
⎛
⎜
⎝
θ
2
⎞
⎟
⎠
⎞
⎟⎟
⎠
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r θ
=
K
2
r cos 2
⎛
⎝
2
⎞
⎠ sin
⎛
⎝
2
⎞
⎠
One can also determine the displacement components. In particular, the crack opening displacement a distance r behind the crack tip is
8 K r
=
,
E 2
where E
=
E / ( 1
ν
2 ) under the plane strain conditions, and plane stress conditions.
Notes
E
=
E under the
• K is called the stress intensity factor. Its magnitude is undetermined in the eigenvalue problem.
• K is the amplitude of the field around the tip of the crack.
•
The factor
• The r and conditions.
( ) − 1 / 2 is set by convention.
dependence are independent of the external boundary
• By modeling the crack front as a mathematical curve, the linear elastic theory does not account for any process of fracture.
Determine the stress intensity factor by solving a boundaryvalue problem. For a given boundary-value problem, K can be determined.
For example, consider the Griffith crack, a crack of length 2 a in an infinite plate, subject to remote stress
. The boundary-value problem was solved by Inglis
(1913). The field in the body is expressed in analytical terms. For example, the stress ahead of the crack is given by
yy
= x 2
x
2
, x
> a
− a
The distance of a point x ahead of the crack tip is given by r
= x
− a r , and we obtain that
yy
=
( r
+ a )
( r
+
2 a ) r
When the tip of the crack is approached, r « a , we have
. Replace x by
yy
=
a
2 r
.
This crack-tip field is obtained from the boundary-value problem, and it recovers the form determined by the eigenvalue problem, namely,
yy
=
K
2
A comparison of the two expressions gives
r
,
K
=
a .
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This is the stress intensity factor for the Griffith crack.
Alternatively, one can determine the stress intensity factor by using the displacement field. The boundary-value problem solved by Inglis also gives the opening displacement of the crack:
( )
=
4
E a 2
− x 2 , x
< a .
The distance of a point x behind the crack tip is given by r , and we obtain that r
= a
− x . Replace x by
( )
=
4
E
When the tip of the crack is approached,
( r r
«
+ a
2 a ) r .
, we have
( )
=
4
E
2 ar .
This crack-tip field is obtained from the boundary-value problem, and it recovers the form determined by the eigenvalue problem, namely,
=
8 K
E
A comparison of the two expressions gives r
2
.
K
=
a .
Handbooks for K . Stress intensity factor for a given cracked body is determined by solving a boundary-value problem. Many configurations of cracked bodies have been solved. The results are collected in handbooks (e.g., H.
Tada, P.C. Paris and G.R. Irwin, The Stress Analysis of Cracks Handbook, Del
Research, St. Louis, MO., 1995). In general, the stress intensity factor takes the form where
K
=
Y
a ,
is an applied stress, a is a length scale characterize the crack geometry, and Y is a dimensionless number. Two examples follow. is
For a crack at the edge of a semi-infinite plane, the stress intensity factor
K
=
1 .
1215
a .
For a penny-shaped crack in an infinite body, the stress intensity factor is
K
=
2
a .
For the two examples, can you explain why the first example has a larger stress intensity factor than the Griffith crack, and the second example has a smaller one? In any event, the difference in the stress intensity factors for the three cases is small. Thus, for a small crack in a large body, the stress intensity factor is determined by the size of the crack, but is insensitive to the shape of the crack.
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo
Linear superposition . Elasticity problem is linear. For a given body with a given crack, if a force P causes the stress intensity factor
Q causes the stress intensity factor K = and Q causes the stress intensity factor K =
P +
Q .
K =
P , and force
Q . The combined action of the forces P
Finite element method to determine K . For a complicated structure with a crack, once can determine the elastic field using the finite element method, and then extract from the field the stress intensity factor. A brute force method is that you use the finite element method to determine the displacement field, and then fit the crack opening to
=
8 K
E r
2
, with K as the fitting parameter. There are a number of more clever methods.
We’ll mention them later at suitable points.
Irwin’s G-K relation . Consider a crack in an elastic body subject to a load. The energy release rate G is the decrease in the elastic energy associated with the crack advancing by a unit area, while the load is rigidly held. The stress intensity factor K is the amplitude of the field around the tip of the crack. Irwin discovered the following relation between the stress intensity factor and the energy release rate:
G
=
K
E
2
(Mode I, plane strain).
Proof . Consider two bodies, 1 and 2. Both have the same configuration: a sheet of unit thickness, and a semi-infinite crack. The crack in Body 2 is longer than that in Body 1 by a length b . The displacement of the applied load is fixed, so that the applied force does no work when the crack extends. Let U and U be the strain energy stored in the two bodies, respectively. By definition, the energy release rate is
G
=
U
−
U b
.
Let K
the stress intensity factor of the crack in Body 1. The stress field ahead the crack in Body 1 is
σ =
K
2 π x
.
Let K
the stress intensity factor of the crack in Body 2. The opening displacement of the crack in Body 2 is
=
8 K
E b − x
2
.
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo
The difference in the energy in the two bodied is due to the work done by the closing traction:
U
−
U
=
1
2
0 b
dx .
This gives
G
=
2
1 b b
0
K
2 π x
8 K
E b
2
−
π x dx
=
2 K
π
K
E
1
0
1
− t t dt
=
K K
E
The change in the length of the crack b is small compared to the size of body or the total length of the crack a . As b / a → 0 , the stress intensity factors of the cracks in the two bodies approach each other, K
Irwin’s G-K relation
=
K
=
K . Thus, we reach
G =
K
E
2
.
For some problems, the energy release rate is easier to determine than the stress intensity factor. We have seen two examples: the double-cantilever beam, and the channel crack. Once the energy release rate is determined, one can obtain the stress intensity factor by using Irwin’s relation.
Mode II.
Directly ahead the crack tip
r θ
=
K
II
2
r
Behind the crack tip, the sliding displacement between the two crack faces is
II
=
8 K
E
II
2 r
.
The energy release rate relates to the stress intensity factor as
G =
K
E
2
II .
Mode III.
Directly ahead the crack tip
rz
=
K
2
III
r
Behind the crack tip, the tearing displacement between the two crack faces is
III
=
4 K
III r
2
,
=
2
E
( 1
+
)
.
The energy release rate relates to the stress intensity factor as
G =
K 2
III
2
.
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ES 247 Fracture Mechanics http://imechanica.org/node/7448 Zhigang Suo
Historical Notes
For a very interesting historical account involving Airy, Stokes and Maxwell, see
V.V. Meleshko, Selected topics in the history of the two-dimensional biharmonic problem, Applied Mechanics Review 56, 33-85 (2003).
The square-root singularity near the tip of a crack was known before Irwin entered the field. Here are two papers that clearly displayed the square-root singularity.
•
H.M. Westergaard, Bearing pressures and cracks. Journal of Applied
Mechanics 6, A49-A53 (1939). Westergaard was a professor at Harvard.
•
I.N. Sneddon, The distribution of stress in the neighborhood of a crack in an elastic solid. Proceedings of the Royal Society of London A 187, 229-
260 (1946).
M.L. Williams, On the stress distribution at the base of a stationary crack.
Journal of Applied Mechanics 24, 109-115 (1957).
G.R. Irwin, Analysis of stresses and strains near the end of a crack traversing a plate, Journal of Applied Mechanics 24, 361-364 (1957).
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