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DoITPoMS Teaching & Learning Packages Brittle Fracture The stresses ahead of the crack tip
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The stresses ahead of the crack tip

Consider the stresses acting across the plane in which the crack is growing and ahead of the crack tip. This analysis assumes that the normal stresses on the crack faces are zero and gives the stress ahead of the crack tip as

$$\sigma = {K \over {{{\left( {2\pi r} \right)}^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0em} 2}}}}}$$

$$\eqalign{ & = \int\limits_{\delta {\rm{c}}}^0 {{K \over {{{\left( {2\pi r} \right)}^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0em} 2}}}}}} {\rm{ 4 }}{K \over E}{\left( {{{\delta c - r} \over r}} \right)^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0em} 2}}}{\rm{d}}r \cr & = {2 \over \pi }{{{K^2}} \over E}\int\limits_{\delta {\rm{c}}}^0 {{{\left( {{{\delta c - r} \over r}} \right)}^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0em} 2}}}{\rm{d}}r} \cr} $$

We can integrate this by substituting for r using the expression

r = δc sin2ω

which gives dr as

dr = 2 δc sinω cosω dω

The limits of integration now become

$$\eqalign{ & r = 0,{\rm{ }}\omega = 0 \cr & r = \delta c,{\rm{ }}\omega = {\pi \over {\rm{2}}} \cr} $$

Integrating,

$${\int\limits_{\delta {\rm{c}}}^0 {\left( {{{\delta c - r} \over r}} \right)} ^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0em} 2}}} = \int\limits_{{\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0em} 2}}^0 {2{\rm{ }}\left( {{{\cos \omega } \over {\sin \omega }}} \right)} {\rm{ }}\sin \omega \cos \omega {\rm{ }}\delta c{\rm{ d}}\omega $$
$$\eqalign{ & = \int\limits_{{\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0em} 2}}^0 {\left( {1 + \cos 2\omega } \right){\rm{ }}} \delta c{\rm{ d}}\omega \cr & = \left[ {\omega + {1 \over 2}\sin 2\omega } \right]_{{\pi \mathord{\left/ {\vphantom {\pi 2}} \right. \kern-0em} 2}}^0\delta c \cr & = - {\pi \over 2}{\rm{ }}\delta c \cr} $$

The mechanical energy released, δUM, when the crack grows by δc is therefore

$$\displaylines{ \delta {U_{\rm{M}}} = - {2 \over \pi }{{{K^2}} \over E}{\pi \over 2}{\rm{ }}\delta c \cr = - {{{K^2}} \over E}{\rm{ }}\delta c \cr} $$

In the limit this gives

$$ - {{{\rm{d}}{U_{\rm{M}}}} \over {{\rm{d}}c}} = {{{K^2}} \over E}$$

and as

$$G = - {{{\rm{d}}{U_{\rm{M}}}} \over {{\rm{d}}c}}$$

we have

$$G = {{{K^2}} \over E}$$