Strain energy release rate

The energy release rate G {displaystyle 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.The energy release rate G {displaystyle G}   is defined as the instantaneous loss of total potential energy Π {displaystyle Pi }   per unit crack growth area s {displaystyle s}  ,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.This problem is two-dimensional and has a fixed load, so with s = a B {displaystyle s=aB}  ,Consider a double cantilever beam (DCB) specimen as shown in the right figure. The displacement of a single beam isConsider the crack in the DCB specimen shown in the figure. The nonzero stress and displacement components are given by asConsider the double cantilever beam specimen shown in the figure, where the crack centered in the beam of height 2 h {displaystyle 2h}   has a length of a {displaystyle a}  , and a load P {displaystyle 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 {displaystyle h}  , (2) go to the right past the crack tip, (3) go down to the bottom at − 2 h {displaystyle -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 {displaystyle n_{1}=0}   as well as t = 0 {displaystyle 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 {displaystyle u_{1}=0}   and σ 12 = 0 {displaystyle sigma _{12}=0}   far from the crack, leavingA handful of methods exist for calculating G {displaystyle G}   with finite elements. Although a direct calculation of the J-integral is possible (using the strains and stresses outputted by FEA), approximate approaches for some type of crack growth exist and provide reasonable accuracy with straightforward calculations. This section will elaborate on some relatively simple methods for fracture analysis utilizing numerical simulations. G 1 MCCI = 1 2 Δ a F 2 j Δ u 2 j − 1 {displaystyle G_{1}^{ ext{MCCI}}={frac {1}{2Delta a}}F_{2}^{j}{Delta u_{2}^{j-1}}}   G 1 MCCI = 1 2 Δ a ( F 2 j Δ u 2 j − 2 + F 2 j + 1 Δ u 2 j − 1 ) {displaystyle G_{1}^{ ext{MCCI}}={frac {1}{2Delta a}}left(F_{2}^{j}{Delta u_{2}^{j-2}}+F_{2}^{j+1}{Delta u_{2}^{j-1}} ight)}  The J-intregral may be expressed over the full contour as follow: x ( ξ , η ) = ∑ i = 1 8 N i ( ξ , η ) x i {displaystyle x(xi ,eta )=sum _{i=1}^{8}N_{i}(xi ,eta )x_{i}}   y ( ξ , η ) = ∑ i = 1 8 N i ( ξ , η ) y i {displaystyle y(xi ,eta )=sum _{i=1}^{8}N_{i}(xi ,eta )y_{i}}   u ( ξ , η ) = ∑ i = 1 8 N i ( ξ , η ) u i {displaystyle u(xi ,eta )=sum _{i=1}^{8}N_{i}(xi ,eta )u_{i}}   v ( ξ , η ) = ∑ i = 1 8 N i ( ξ , η ) v i {displaystyle v(xi ,eta )=sum _{i=1}^{8}N_{i}(xi ,eta )v_{i}}   N 1 = − ( ξ − 1 ) ( η − 1 ) ( 1 + η + ξ ) 4 {displaystyle N_{1}={frac {-(xi -1)(eta -1)(1+eta +xi )}{4}}}   N 1 ( ξ , − 1 ) = − ξ ( 1 − ξ ) 2 {displaystyle N_{1}(xi ,-1)=-{frac {xi (1-xi )}{2}}}   γ x x = ∂ u ∂ x = ∑ i = 1 , 2 , 5 ∂ N i ∂ ξ ∂ ξ ∂ x u i {displaystyle gamma _{xx}={frac {partial u}{partial x}}=sum _{i=1,2,5}{frac {partial N_{i}}{partial xi }}{frac {partial xi }{partial x}}u_{i}}   x ( ξ ) = ξ ( 1 + ξ ) 2 L + ( 1 − ξ 2 ) L 4 {displaystyle x(xi )={frac {xi (1+xi )}{2}}L+(1-xi ^{2}){frac {L}{4}}}   ξ ( x ) = − 1 + 2 x L {displaystyle xi (x)=-1+2{sqrt {frac {x}{L}}}}   ∂ ξ ∂ x = 1 x L {displaystyle {frac {partial xi }{partial x}}={frac {1}{sqrt {xL}}}}   γ x x = 4 L ( u 2 2 − u 5 ) + 1 x L ( 2 u 5 − u 2 5 ) {displaystyle gamma _{xx}={frac {4}{L}}left({frac {u_{2}}{2}}-u_{5} ight)+{frac {1}{sqrt {xL}}}left(2u_{5}-{frac {u_{2}}{5}} ight)}   u = u 3 + x L [ 4 u 6 − 3 u 3 − u 1 ] + x L [ 2 u 1 + 2 u 3 − 4 u 6 ] {displaystyle u=u_{3}+{sqrt {frac {x}{L}}}left+{frac {x}{L}}left}   γ x x = ∂ u ∂ x = 1 x L [ − u 1 2 − 3 u 3 2 + 2 u 6 ] + 1 L [ 2 u 1 + 2 u 3 − 4 u 6 ] {displaystyle gamma _{xx}={frac {partial u}{partial x}}={frac {1}{sqrt {xL}}}left+{frac {1}{L}}left}