Introduction
You've probably heard that the J-Integral is
path independent.
In fact, path independence is often the first thing mentioned when introducing
the J-Integral, even to the exclusion of its true importance as a measure of
energy release.
So what does path independence mean? It means that the equation will give the
same, correct value of energy release regardless of the path taken around
the crack tip; no matter how close or far away from it the path is.
The only requirement is that the path must enclose the crack tip.
However, we will see that the J-Integral is not always path independent after all.
The one exception occurs whenever significant body forces are present. They can come
in any form, including gravitational, magnetic, and even inertial forces due to
accelerations.
J-Integral Equation
Recall the J-Integral equation for a crack growing along the x-axis is
\[
J = \oint w \, n_x \, d\Gamma -
\oint {\bf T} \cdot {\partial {\bf u} \over \partial x} \, d\Gamma
\]
where \(d\Gamma\) is the differential increment along the path.
The x-component, \(n_x\), of the unit normal in the first integral,
and the partial derivative with respect to x, \(\partial()/\partial x\),
in the second integral are directly related to the fact that the
crack is growing along the x-axis.
Overview
Proof that the J-Integral is path independent is rather long.
While not overly complex, it does require basic knowledge of several topics.
The first is Tensor Notation. Information on this can be found on
these two pages:
Tensor Notation (Basics) and
Tensor Notation (Advanced).
The second topic consists of Traction Vectors and their relationship to stress.
Information on traction vectors is here:
Traction Vectors.
Finally, the divergence theorem will also be required. Information on the
divergence theorem is here:
Divergence Theorem.
The proof consists of two parts. The first is proof that the equation always
gives zero when integrated around a complete loop. (Note the standard integration
path is not a complete loop.) The second part uses the result of the first to
ultimately prove the path independence of the J-Integral.
Proof of Path Independence
Step 1 of 2
The first step in the proof is to show that integration of the equation around
a complete, closed loop equals zero, always (though only when body forces
are absent). It will be necessary to develop the proof using tensor notation
because vector notation is not sufficiently precise, at least in my opinion.
Start with the J-Integral equation with both terms contained within one integral.
\[
J = \oint \left( w \, n_x - {\bf T} \cdot {\partial {\bf u} \over \partial x} \right) d\Gamma
\]
Expressing it in tensor notation gives
\[
J = \oint \left( w \, n_1 - T_i \, {\partial u_i \over \partial x_1} \right) d\Gamma
\]
The next step is to insert the relationship between traction vectors
and stress ( \(T_i = \sigma_{ij} n_j\) ) into the equation.
\[
J = \oint \left( w \, n_1 - \sigma_{ij} \, n_j \, {\partial u_i \over \partial x_1} \right) d\Gamma
\]
Next, use the Kronecker Delta as the substitution operator.
Replace \(n_1\) with \(\delta_{1j} \, n_j\).
\[
J = \oint \left( w \, \delta_{1j} \, n_j - \sigma_{ij} \, n_j \, {\partial u_i \over \partial x_1} \right) d\Gamma
\]
And factor \(n_j\) from both terms.
\[
J = \oint \left( w \, \delta_{1j} - \sigma_{ij} \, {\partial u_i \over \partial x_1} \right) n_j \, d\Gamma
\]
The J-Integral equation is now in a form that permits application
of the Divergence Theorem. Specifically, the \(n_j \, d\Gamma\)
term makes this possible. Doing so gives
\[
J = \int_A {\partial \over \partial x_j} \left( w \, \delta_{1j} - \sigma_{ij} \, {\partial u_i \over \partial x_1} \right) dA
\]
Divergence Theorem
It is very important to be aware of the ramifications of the previous step
involving the Divergence Theorem. The first is to recognize that the
integral equality
\[
\oint (...) \, n_j \, d\Gamma = \int_A {\partial \over \partial x_j} (...) \, dA
\]
is true only if the path integral is a complete, closed loop.
This will be very important a little later.
The second important point is to recognize that \(dA\) is not the surface
area of a 3-D volume, but rather, the area of the 2-D region surrounded
by the path integral.
Integration Over a Closed Loop
As trivial as it may sound, it is nevertheless important to review the
simple concept of a
closed loop. And it is important to
first recognize that the integration path in the sketch at the top of this
page is NOT a closed loop! (Though it is a proper integration path for
evaluating the J-Integral.) This is because it is not continuous - the
head and tail of the path do not connect. And they cannot connect as long
as the crack is present and separates them. One cannot leap-frog the
crack because this constitutes jumping out of the material!
So what does a closed loop integration path look like? It looks like
the sketch here. This 4-segment path is indeed a continuous closed loop.
Granted, it looks different from the J-Integral path shown above, but
that is OK. It will be resolved shortly. And it should be clear
that the J-Integral path is in fact a subset of the closed path
here. This will prove to be critical to the overall proof of path
independence.
The next step is logical. It is to apply the \(\partial / \partial x_j\) derivative.
Doing so gives
\[
J = \int_A \left( {\partial w \over \partial x_j} \, \delta_{1j} -
{\partial \sigma_{ij} \over \partial x_j} \, {\partial u_i \over \partial x_1} -
\sigma_{ij} \, {\partial \over \partial x_j} \left( {\partial u_i \over \partial x_1} \right) \right) dA
\]
Note the product rule of differentiation, applied to the second term, has produced an additional
one, making three total. Each of the three terms will now be addressed. Beginning with
the first, apply the substitution property of the Kronecker Delta as follows.
\[
{\partial w \over \partial x_j} \, \delta_{1j} = {\partial w \over \partial x_1}
\]
This gives the following form for the J-Integral.
\[
J = \int_A \left( {\partial w \over \partial x_1} -
{\partial \sigma_{ij} \over \partial x_j} \, {\partial u_i \over \partial x_1} -
\sigma_{ij} \, {\partial \over \partial x_j} \left( {\partial u_i \over \partial x_1} \right) \right) dA
\]
Now focus on the \(\partial \sigma_{ij} / \partial x_j\) derivative in the second term.
It is the same derivative that appears in the equilibrium differential equation, which is
the differential form of \(F = m a\).
See
equilibrium.html for more information on the equilibrium equation.
\[
{\partial \sigma_{ij} \over \partial x_j} + f_i = \rho \, a_i
\]
At this point, the second key step in the proof arrives. It is to
impose that no body forces are present in the problem, \(f_i = 0\). These also
include inertial forces too, \(\rho \, a_i = 0\). (Technically, they can be
present, as long as they are negligible compared to \(\sigma_{ij,j}\).)
This leaves
\[
{\partial \sigma_{ij} \over \partial x_j} = 0
\]
and for this reason, the entire 2nd term in the J-Integral equation goes away!
\[
J = \int_A \left( {\partial w \over \partial x_1} -
\sigma_{ij} \, {\partial \over \partial x_j} \left( {\partial u_i \over \partial x_1} \right) \right) dA
\]
Finally, focus on the derivative portion of the last term.
The \({\partial \over \partial x_j} \left( {\partial u_i \over \partial x_1} \right) \) expression
can be written several different ways.
\[
{\partial \over \partial x_j} \left( {\partial u_i \over \partial x_1} \right) =
{\partial^2 u_i \over \partial x_j \partial x_1} =
{\partial u_{i,j} \over \partial x_1}
\]
Inserting the last form into the J-Integral equation gives
\[
J = \int_A \left( {\partial w \over \partial x_1} -
\sigma_{ij} \, {\partial u_{i,j} \over \partial x_1} \right) dA
\]
Focus again on the last term: \(\sigma_{ij} \, {\partial u_{i,j} \over \partial x_1}\).
This term is equal to \(\sigma_{ij} \, {\partial \epsilon_{ij} \over \partial x_1}\).
The proof of this equality follows below.
Equality Proof
Start with the \(\sigma_{ij} \, {\partial \epsilon_{ij} \over \partial x_1}\) term
and insert the following relationship
\[
\epsilon_{ij} = {1 \over 2} ( u_{i,j} + u_{j,i} )
\]
to obtain
\[
\sigma_{ij} \, {\partial \epsilon_{ij} \over \partial x_1} =
{1 \over 2} \left( \sigma_{ij} \, {\partial u_{i,j} \over \partial x_1} +
\sigma_{ij} \, {\partial u_{j,i} \over \partial x_1} \right)
\]
But since \(\sigma_{ij}\) is symmetric, the last two terms are equal. Therefore
\[
\sigma_{ij} \, {\partial \epsilon_{ij} \over \partial x_1} =
{1 \over 2} \left( \sigma_{ij} \, {\partial u_{i,j} \over \partial x_1} +
\sigma_{ij} \, {\partial u_{j,i} \over \partial x_1} \right) =
\sigma_{ij} \, {\partial u_{i,j} \over \partial x_1}
\]
So \( {\partial u_{i,j} / \partial x_1} \) can be replaced with
\( {\partial \epsilon_{ij} / \partial x_1} \), though only because
both are multiplied by \(\sigma_{ij}\).
\[
J = \int_A \left( {\partial w \over \partial x_1} -
\sigma_{ij} \, {\partial \epsilon_{ij} \over \partial x_1} \right) dA
\]
However, \(\sigma_{ij}\) is the partial derivative of strain energy density,
\(w\), with respect to strain, \(\epsilon_{ij}\). So replace
\(\sigma_{ij}\) with \( \partial w / \partial \epsilon_{ij}\).
\[
J = \int_A \left( {\partial w \over \partial x_1} -
{ \partial w \over \partial \epsilon_{ij} } \,
{\partial \epsilon_{ij} \over \partial x_1} \right) dA
\]
And the two \(\partial \epsilon_{ij}\) terms cancel-out, giving
\[
J = \int_A \left( {\partial w \over \partial x_1} -
{\partial w \over \partial x_1} \right) dA = 0
\]
This completes the proof that the J-Integral gives zero when integrated over a complete,
closed loop in the absence of body forces, including inertial forces. It is the first
of the two major steps needed to prove the J-Integral equation is path independent.
Although it is only one-half of the two major steps involved in proving the path
independence, it constitutes 90% of the total amount of work. So fortunately, we're
almost done.
Step 2 of 2
The second step involves evaluating the J-Integral over the complete
loop shown in the figure. The path is partitioned into the four
regions shown. The corresponding integral is
\[
J = \int_1 (...) \, d\Gamma +
\int_2 (...) \, d\Gamma +
\int_3 (...) \, d\Gamma +
\int_4 (...) \, d\Gamma = 0
\]
A key result of this partition is that the integral is identically
zero for regions
2 and
4. This is because both
\(n_x = 0\) and \(\sigma_{ij} = 0\) in these regions. This leaves
\[
J = \int_1 (...) \, d\Gamma +
\int_3 (...) \, d\Gamma = 0
\]
Therefore
\[
\int_1 (...) \, d\Gamma = - \int_3 (...) \, d\Gamma
\]
which is in fact a profound result. The first thing of note is not
actually profound, but rather, a simple fact. It is the fact that
\(\int_3 (...) \, d\Gamma\) is negative simply because it takes a clockwise
path around the crack tip. But otherwise, the two integrals are equal
even though they are evaluated along two
different paths,
1
and
3. Since
1 and
3 represent two different arbitrary
paths, the only conclusion to be drawn is that
any path will also evaluate
to the same value. Hence, the J-Integral is
path independent!
Review
There was so much math in the above proof that it's easy to lose track
of the the few steps that were truly critical versus the many steps that
were little more than algebraic manipulation. So let's review those
key steps. First, start with the J-Integral equation itself.
\[
J = \oint \left( w \, n_x \, -
{\bf T} \cdot {\partial {\bf u} \over \partial x} \right) d\Gamma
\]
Recall the equation computes the amount of energy released
(consumed, used, etc) per unit increase of crack surface area
as the crack propagates. Note the crack is assumed to propagate
along the x-axis, and that is why \(x\)
(as opposed to \(y\) or \(z\)) is present
on \(n_x\) and \(\partial/\partial x\).
We only encountered one limiting condition during the entire proof
of path independence. It is that the body forces, \(f_i\),
including inertial forces, \(\rho \, a_i\), must be negligible
compared to local stress gradients, \(\sigma_{ij,j}\).
Note the body and inertial forces need not be zero, only negligible.
This limitation arose when dealing with the equilibrium equation.
\[
{\partial \sigma_{ij} \over \partial x_j} + f_i = \rho \, a_i
\]
Neglecting the body forces and inertial forces leaves
\[
{\partial \sigma_{ij} \over \partial x_j} = 0
\]
which is critical to the proof.
That is in fact the only limiting condition required to ensure
path independence of the J-Integral.
References
- Rice, J.R., "A Path Independent Integral and the Approximate Analysis of Strain Concentration by Notches and Cracks." Journal of Applied Mechanics, Vol. 35, pp. 379-386, 1968.
- Griffith, A.A., "The Phenomena of Rupture and Flow in Solids," Philosophical Transactions, Series A, Vol. 221, pp. 163-198, 1920.
