Past exam of the solid mechanics course of the University of Oxford / 2010 / 3 / c / Solution

For a differentiable crack path, the local quasistatic Griffith laws with constant positive toughness areThey express irreversibility, local stability, and energy balance during growth. A strictly subcritical release rate forces the crack to stop.

Fix a crack parameter $s$ and let $u(s,t)$ denote the equilibrium displacement with the prescribed time-dependent loads. The scalar equilibrium problem is linear and has a unique solution for its fixed Dirichlet data. Both body force and boundary displacement scale by $t$, soThe minimized potential energy is quadratic under this scaling: the strain term scales by $t^2$ and so does the load term $-f(t)u(s,t)$. Hence $E(s,t)=t^{2}E(s,1)$ andEquivalently the crack-tip coefficient scales by $t$ and its squared energy-release formula gives the same result.

Put $s_0=\ell (0)$. Finiteness and continuity of $G(s,1)$ make it bounded by some finite $M\ge 0$ on a neighborhood of $s_0$. Continuity of $\ell$ supplies a positive interval on which $\ell (t)$ stays in that neighborhood. Shorten it to $[0,t_0]$ so that $t_0^{2}M<G_c$. Then $G(\ell (t),t)\le t_0^{2}M<G_c$, and complementarity forces $\dot{\ell }(t)=0$ throughout. ThereforeNo differentiating an unknown crack path or assuming its length already constant is needed; its continuity first gives the local release-rate bound.