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= Energy Release Rate, G= The energy release rate $$G$$ is the rate at which energy is lost as a material undergoes fracture. This is an energy-per-unit-area, or in two-dimensional problems, an energy-per-unit-length. The energy release rate is mathematically understood as the decrement in total potential energy scaled by the increment in fracture surface area. Various energy balances can be constructed relating the energy released during fracture to the energy of the resulting new surface, as well as other dissipative processes such as plasticity and heat generation. The energy release rate is central to the field of fracture mechanics when solving problems and estimating material properties related to fracture and fatigue.

Definition


The energy release rate $$G$$ is defined as the instantaneous loss of total potential energy $$\Pi$$ per unit crack growth area $$s$$,



G \equiv -\frac{\partial \Pi}{\partial s} , $$

where the total potential energy is written in terms of the total strain energy $$\Omega$$, surface traction $$\mathbf{t}$$, displacement $$\mathbf{u}$$, and body force $$\mathbf{b}$$ by



\Pi = \Omega - \left\{ \int_{\mathcal{S}_t}\mathbf{t}\cdot\mathbf{u}\,dS + \int_\mathcal{V}\mathbf{b}\cdot\mathbf{u}\,dV \right\} . $$

The first integral is over the surface $$S_t$$ of the material, and the second over its volume $$V$$.

The figure on the right shows the plot of an external force $$P$$ vs. the load-point displacement $$q$$, in which the area under the curve is the strain energy. The white area between the curve and the $$P$$-axis is referred to as the complementary energy. In the case of a linearly-elastic material, $$P(q)$$ is a straight line and the strain energy equals the complementary energy.

Prescribed Displacement
In the case of prescribed displacement, the strain energy can be expressed in terms of the specified displacement and the crack surface $$\Omega(q,s)$$, and the change in this strain energy is only affected by the change in fracture surface area: $$ \delta \Omega =(\partial \Omega / \partial s) \delta s $$. Correspondingly, the energy release rate in this case is expressed as



G = -\left.\frac{\partial \Omega}{\partial s}\right|_q . $$

Here is where one can accurately refer to $$G$$ as the strain energy release rate.

Prescribed Loads
When the load is prescribed instead of the displacement, the strain energy needs to be modified as $$\Omega(q(P,s),s)$$. The energy release rate is then computed as



G = -\left.\frac{\partial}{\partial s}\right|_P \left(\Omega - Pq\right) . $$

Relation to Stress Intensity Factors
The energy release rate is directly related to the stress intensity factor associated with a given two-dimensional loading mode (Mode-I, Mode-II, or Mode-II) when the crack grows straight ahead. This is applicable to cracks under plane stress, plane strain, and antiplane shear.

For Mode-I, the energy release rate $$G$$ rate is related to the Mode-I stress intensity factor $$K_{I}$$ for a linearly-elastic material by



G = \frac{K^2_I}{E'}, $$

where $$E'$$ is related to Young's modulus $$E$$ and Poisson's ratio $$\nu$$ depending on whether the material is under plane stress or plane strain:



E' = \begin{cases} E,& \mathrm{plane~stress},\\ \\   \dfrac{E}{1-\nu^2},&\mathrm{plane~strain}. \end{cases} $$

For Mode-II, the energy release rate is similarly written as



G = \frac{K^2_{II}}{E'}. $$

For Mode-III (antiplane shear), the energy release rate now is a function of the shear modulus $$\mu$$,



G = \frac{K^2_{III}}{2\mu}. $$

For an arbitrary combination of all loading modes, these linear elastic solutions may be superposed as



G = \frac{K^2_{I}}{E'} + \frac{K^2_{II}}{E'} + \frac{K^2_{III}}{2\mu}. $$

These relations can be used to calculate the fracture toughness of the material $$K_{IC}$$, the minimum stress intensity factor required to initiate crack growth, in an experiment where the energy release rate, loading conditions, material geometry, and material properties are known.

Calculating G
There are a variety of methods available for calculating the energy release rate given material properties, specimen geometry, and loading conditions. Some are dependent on certain criteria being satisfied, such as the material being entirely elastic or even linearly-elastic, and/or that the crack must grow straight ahead. The only method presented that works arbitrarily is that using the total potential energy. If two methods are both applicable, they should yield identical energy release rates.

Total Potential Energy
The only method to calculate $$G$$ for arbitrary conditions is to calculate the total potential energy and differentiate it with respect to the crack surface area. This is typically done by: all in terms of the crack surface area.
 * calculating the stress field resulting from the loading,
 * calculating the strain energy in the material resulting from the stress field,
 * calculating the work done by the external loads,

Compliance Method
If the material is linearly-elastic, the computation of its energy release rate can be much simplified. In this case, the Load vs. Load-point Displacement curve is linear with a positive slope, and the displacement per unit force applied is defined as the compliance, $$C$$
 * $$ C = \frac{q}{P}. $$

The corresponding strain energy $$\Omega$$ (area under the curve) is equal to

\Omega = \frac{1}{2}Pq = \frac{1}{2} \frac{q^2}{C} = \frac{1}{2} P^2 C. $$ Using the compliance method, one can show that the energy release rate for both cases of prescribed load and displacement come out to be

G = \frac{1}{2} P^2 \frac{\partial C}{\partial s}. $$

Multiple Specimen Methods for Nonlinear Materials
In the case of prescribed displacement, holding the crack length fixed, the energy release rate can be computed by



G = -\int^q_0 \frac{\partial P}{\partial s}\,dq, $$

while in the case of prescribed load,



G = \int^P_0 \frac{\partial q}{\partial s}\,dP. $$ As one can see that, in both cases, the energy release rate $$G$$ times the change in surface $$ds$$ returns the area between curves, which indicates the energy dissipated for the new surface area as illustrated in the right figure

G ds = -ds \int^q_0 \frac{\partial P}{\partial s}\,dq = ds \int^P_0 \frac{\partial q}{\partial s}\,dP. $$

Crack Closure Integral
Since the energy release rate is defined as the negative derivative of the total potential energy with respect to crack surface growth, the energy release rate may be written as the difference between the potential energy before and after the crack grows. After some careful derivation, this leads one to the crack closure integral



G = \lim_{\Delta s\to 0} -\frac{1}{\Delta s}\int_{\Delta s} \frac{1}{2}\,t_i^0\left(\Delta u_i^+ - \Delta u_i^-\right)\,dS, $$

where $$\Delta s$$ is the new fracture surface area, $$t_i^0$$ are the components of the traction released on the top fracture surface as the crack grows, $$\Delta u_i^+ - \Delta u_i^-$$ are the components of the crack opening displacement (the difference in displacement increments between the top and bottom crack surfaces), and the integral is over the surface of the material $$S$$.

The crack closure integral is valid only for elastic materials, but is still valid for cracks that grow in any direction. Nevertheless, for a two-dimensional crack that does indeed grow straight ahead, the crack closure integral simplifies to



G = \lim_{\Delta a\to 0} \frac{1}{\Delta a}\int_0^{\Delta a} \sigma_{i2}(x_1,0)u_i(\Delta a-x_1,\pi)\,dx_1, $$

where $$\Delta a$$ is the new crack length, and the displacement components are written as a function of the polar coordinates $$r=\Delta a-x_1$$ and $$\theta=\pi$$.

J-Integral
In certain situations, the energy release rate $$G$$ can be calculated using the J-integral, i.e. $$G=J$$, using


 * $$J = \int_\Gamma \left(Wn_1 - t_i\,\frac{\partial u_i}{\partial x_1}\right)\,d\Gamma,$$

where $$W$$ is the elastic strain energy density, $$n_1$$is the $$x_1$$component of the unit vector normal to $$\Gamma$$, the curve used for the line integral, $$t_i$$are the components of the traction vector $$\mathbf{t}=\boldsymbol{\sigma}\cdot\mathbf{n}$$, where $$\boldsymbol{\sigma}$$ is the stress tensor, and $$u_i$$are the components of the displacement vector.

This integral is zero over a simple closed path and is path independent, allowing any simple path starting and ending on the crack faces to be used to calculate $$J$$. In order to equate the energy release rate to the J-Integral, $$G=J$$, the following conditions must be met:


 * the crack must be growing straight ahead, and
 * the deformation near the crack (enclosed by $$\Gamma$$) must be elastic (not plastic).

The J-integral may be calculated with these conditions violated, but then $$G\neq J$$. When they are not violated, one can then relate the energy release rate and the J-integral to the elastic moduli and the stress intensity factors using


 * $$G = J = \frac{K_I^2}{E'} + \frac{K_{II}^2}{E'} + \frac{K_{III}^2}{2\mu}.$$