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

An antiplane displacement has the form $v(x_1,x_2,x_3)=(0,0,u(x_1,x_2))$, independent of the out-of-plane coordinate. Only $e_{13}=e_{31}=u_{x_1}/2$ and $e_{23}=e_{32}=u_{x_2}/2$ are nonzero, so its strain energy per unit out-of-plane thickness is $\mu |\nabla u{|}^2/2$.

The equilibrium minimizes total potential energyHere $f$ is the out-of-plane body-force component; the two crack faces have independent traces. For a variation $\eta$ with zero outer-boundary trace, stationarity givesIntegrating by parts first for interior test functions and then for variations free on either crack face yieldsThese are the weak equilibrium equation, prescribed displacement, and zero traction. The tip behavior is selected by finite energy; it is not an additional prescribed tip value.

Use $-\pi <{\theta }_s<\pi$, with the crack along the negative radial axis. In the coefficient convention used here, the leading crack-tip expansion isThe square-root term is harmonic and has zero angular derivative on both crack faces. Regular contributions may include a constant and bounded-gradient terms under smooth loading.

If $E(s)$ is the minimized elastic-plus-load potential, the energy release rate is $\boxed{G(s)=-dE(s)/ds}$. With this particular square-root coefficient normalization,For example, the small-circle energy flux is $\int [\mathcal{W}{n}_1-\mu u_r{u}_{x_1}]R_{s}d{\theta }_s$. The singular term gives $\mathcal{W}=\mu k_s^2/(8R_s)$, $u_r=k_s\sin ({\theta }_s/2)/(2\sqrt{R_s})$ and $u_{x_1}=-k_s\sin ({\theta }_s/2)/(2\sqrt{R_s})$. Integration over $(-\pi ,\pi )$ produces the stated factor. Other stress-intensity normalizations change that factor.

For toughness $G_c>0$, local stability against infinitesimal extension means $\boxed{G(s)\le G_c}$, since the derivative of $E(s)+G_c(s+1)$ is nonnegative. Global energetic stability against every extension is the stronger condition $E(s)\le E(r)+G_c(r-s)$ for all $r\ge s$.

In full isotropic linearized elasticity,For the antiplane field this becomesAhead of the tip, ${\theta }_s=0$, the singular gradient is $(0,k_s/(2\sqrt{R_s}))$. ThusThis is stronger than the printed remainder $o(R_s^{-1})$. If $k_s$ denotes a nonnegative intensity, it has the printed leading coefficient; for a signed singular coefficient its absolute value is essential because the left side is a norm.

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.