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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, which is an energy-per-unit-area. 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

Plot of Load vs. Load-Displacement

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 is equal to 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

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

If the material is linearly-elastic, then $\Omega =Pq/2$ and one may instead write

$G=\left.{\frac {\partial \Omega }{\partial s}}\right|_{P}.$

G in Two-Dimensional Cases

In the cases of two-dimensional problems, the change in crack growth area is simply the change in crack length times the thickness of the specimen. Namely, $\partial s=B\partial a$. Therefore, the equation for computing $G$ can be modified for the 2D case:

• Prescribed Displacement: $G=-\left.{\frac {1}{B}}{\frac {\partial \Omega }{\partial a}}\right|_{q}.$
• Prescribed Load: $G=-\left.{\frac {1}{B}}{\frac {\partial }{\partial a}}\right|_{P}\left(\Omega -Pq\right).$
• Prescribed Load, Linear Elastic: $G=\left.{\frac {1}{B}}{\frac {\partial \Omega }{\partial a}}\right|_{P}.$

One can refer to the example calculations embedded in the next section for further information. Sometimes the strain energy is written using $U=\Omega /B$, an energy-per-unit thickness. This gives

• Prescribed Displacement: $G=-\left.{\frac {\partial U}{\partial a}}\right|_{q}.$
• Prescribed Load: $G=-\left.{\frac {\partial }{\partial a}}\right|_{P}\left(U-{\frac {Pq}{B}}\right).$
• Prescribed Load, Linear Elastic: $G=\left.{\frac {\partial U}{\partial a}}\right|_{P}.$

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

$G={\frac {K_{I}^{2}}{E'}},$

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

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

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

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

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

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

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:

• 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,

all in terms of the crack surface area.

Example calculation of $G$ using the total potential energy

Double cantilever beam (DCB) specimen under tensile load. This problem is two-dimensional and has a fixed load, so with $s=aB$, $G=-{\frac {1}{B}}\left.{\frac {\partial }{\partial a}}\right|_{P}(\Omega -Pq).$ Since the material is linearly-elastic, $\Omega =Pq/2$ and thus $G={\frac {1}{B}}{\frac {\partial \Omega }{\partial a}}.$ The stresses in the DCB are due to the bending stresses in each cantilever beam, $\sigma _{11}(x,y)=-{\frac {M(x)y}{I}}=-{\frac {12Pxy}{Bh^{3}}},$ where $B$ is the length into the page. Using $W=\sigma _{11}^{2}/2E$, one now has the strain energy $\Omega =\int _{V}W\,dV=2B\int _{-h/2}^{h/2}\int _{0}^{a}{\frac {1}{2E}}\left({\frac {12Pxy}{Bh^{3}}}\right)^{2}\,dx\,dy,$ where the factor of 2 out front is due to there being 2 cantilever beams. Solving, $\Omega ={\frac {P^{2}}{B^{3}}}{\frac {4a^{3}}{Eh^{3}}},$ then taking a derivative with respect to $a$ and dividing by $B$, $G={\frac {P^{2}}{B^{2}}}{\frac {12a^{2}}{Eh^{3}}}.$

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

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

Example calculation of $G$ using the compliance method

A Double Cantilever Beam Specimen with Thickness B (not shown) Consider a double cantilever beam (DCB) specimen as shown in the right figure. The displacement of a single beam is $\nu _{max}={\frac {Pa^{3}}{3EI}},\quad {\text{with}}\quad I={\frac {Bh^{3}}{12}}.$ The resulting load-point displacement $q$ is therefore $2\nu _{max}$. Substitute into the equation for compliance and simplify: $C={\frac {q}{P}}={\frac {8a^{3}}{EBh^{3}}}.$ Now, ${\partial C}/{\partial s}$ is computed as ${\frac {\partial C}{\partial s}}={\frac {\partial C}{\partial a}}{\frac {\partial a}{\partial s}}=\left({\frac {24a^{2}}{EBh^{3}}}\right)\left({\frac {1}{B}}\right).$ Finally, the energy release rate of this DCB specimen can be expressed as $G(P,a)={\frac {1}{2}}P^{2}{\frac {\partial C}{\partial s}}={\frac {P^{2}}{B^{2}}}{\frac {12a^{2}}{Eh^{3}}}.$ Note that alternatively, the energy release rate can be expressed in terms of $q$ and $a$: $G(q,a)={\frac {3}{16}}{\frac {Eh^{3}q^{2}}{a^{4}}},$ indicating that $G$ decreases with the crack length $a$ for the case of fixed displacement, and vice versa for the case of fixed load.

Multiple Specimen Methods for Nonlinear Materials

Graphical Illustration of G under Fixed Displacement and Fixed Load Conditions

In the case of prescribed displacement, holding the crack length fixed, the energy release rate can be computed by

$G=-\int _{0}^{q}{\frac {\partial P}{\partial s}}\,dq,$

while in the case of prescribed load,

$G=\int _{0}^{P}{\frac {\partial q}{\partial s}}\,dP.$

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

$Gds=-ds\int _{0}^{q}{\frac {\partial P}{\partial s}}\,dq=ds\int _{0}^{P}{\frac {\partial q}{\partial s}}\,dP.$

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 $\theta =\pi$.

Example calculation of $G$ using the crack closure integral

Double cantilever beam (DCB) specimen under tensile load. Consider the crack in the DCB specimen shown in the figure. The nonzero stress and displacement components are given by as $\sigma _{22}(x_{1},0)={\frac {K_{I}}{\sqrt {2\pi x_{1}}}},\qquad u_{22}(\Delta a-x_{1},\pi )={\frac {8K_{I}}{E}}{\sqrt {\frac {\Delta a_{1}-x_{1}}{2\pi }}}.$ The crack closure integral for this linearly-elastic material, assuming the crack grows straight ahead, is $G={\frac {2K_{I}^{2}}{\pi E}}\,\lim _{\Delta a\to 0}{\frac {1}{\Delta a}}\int _{0}^{\Delta a}{\sqrt {\frac {\Delta a_{1}-x_{1}}{x_{1}}}}\,dx_{1}.$ Consider rescaling the integral using $\xi =x_{1}/\Delta a$ for $G={\frac {2K_{I}^{2}}{\pi E}}\int _{0}^{1}{\sqrt {\frac {1-\xi }{\xi }}}\,d\xi ,$ where one computes the simpler integral to be $\int _{0}^{1}{\sqrt {\frac {1-\xi }{\xi }}}\,d\xi ={\frac {\pi }{2}},$ leaving the energy release rate as $G={\frac {K_{I}^{2}}{E}},$ the expected relation. In this case it is not straightforward to obtain $K_{I}$ directly from the loading and geometry of the problem, but since the crack grows straight ahead and the material is linearly-elastic, the energy release rate here should be the same as the energy release rate calculated using the other methods. This allows one to indirectly retrieve the stress intensity factor for this problem as $K_{I}={\frac {P}{B}}{\frac {2a}{h}}{\sqrt {\frac {3}{h}}}.$

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

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

Example calculation of $G$ using the J-Integral

Double cantilever beam (DCB) specimen under tensile load. J-integral path for the DCB specimen under tensile load. Consider the double cantilever beam specimen shown in the figure, where the crack centered in the beam of height $2h$ has a length of $a$, and a load $P$ is applied to open the crack. Assume that the material is linearly-elastic and that the crack grows straight forward. Consider a rectangular path shown in the second figure: start on the top crack face, (1) go up to the top at $h$, (2) go to the right past the crack tip, (3) go down to the bottom at $-2h$, (4) go along the bottom to the left, and (5) go back up to the bottom crack face. The J-Integral is zero along many parts of this path. The material is effectively unloaded behind the crack, so both the strain energy density and traction are zero along (1) and (2), and hence the J-Integral. Along (2) and (4) one has $n_{1}=0$ as well as $\mathbf {t} =\mathbf {0}$ (no traction on the free surface), so the J-Integral is zero on (2) and (4) as well. This leaves only (3); assuming one is far enough from the crack on (3), the traction term is zero since $u_{1}=0$ and $\sigma _{12}=0$ far from the crack, leaving $J=\int _{\Gamma _{3}}Wn_{1}\,d\Gamma _{3}.$ $n_{1}=1$ along (3), and $W$ is due to bending stress for a cantilever beam $\sigma _{11}(y)=-{\frac {My}{I}}=-{\frac {12Pay}{Bh^{3}}},$ where $B$ is the length into the page. Using $W=\sigma _{11}^{2}/2E$, one now has $J=2\int _{-h/2}^{h/2}{\frac {1}{2E}}\left({\frac {12Pay}{Bh^{3}}}\right)^{2}\,dy,$ where the factor of 2 out front is due to there being 2 cantilever beams. Solving, $G=J={\frac {P^{2}}{B^{2}}}{\frac {12a^{2}}{Eh^{3}}}.$

See also

• Fracture mechanics
• Stress intensity factor
• Fracture toughness
• J-Integral

External links

• Fracture Mechanics Notes by Prof. Alan T. Zehnder (from Cornell University)
• Nonlinear Fracture Mechanics Notes by Prof. John Hutchinson (from Harvard University)
• Griffith's Strain Energy Release Rate on www.fracturemechanics.org

Category:Fracture mechanics Category:Solid mechanics Category:Mechanics