### Stress Intensity Factor $K_{I}$ is determined by geometry and loading conditions $ K_{I} = Q \sigma_{\infty}\sqrt{ \pi a } $ $Q=$ configuration factor, $Q=1$ for infinite plate $a=$ crack radius, $2a =$ crack width $\sigma_{\infty} =$ stress applied perpendicular to crack ### Critical Stress Intensity factor $K_{IC}$ is a function of material properties $ K_{IC} = \sigma_{c}\sqrt{ \frac{\pi a_{0}}{2} } $ $a_{0} =$ lattice spacing of atoms $\sigma_{c} =$ critical stress Crack propagates when $K_{I} = K_{IC}$ (if factor of safety, then $K_{I} = \frac{K_{IC}}{FoS}$) $K_{I}=K_{Ic}$ only if $r_{Ic} \ll \{a, (w-a), B, h\}$ ### Critical Stress $ \sigma_{c} = 2\sqrt{ \frac{a}{a_{0}} } \sigma^{\infty } $ $a_{0} =$ lattice spacing of atoms ### Local Stress $ t(r_{k},\theta) = \frac{K_{I}}{\sqrt{ 2\pi r }}[f(\theta)] \cdot \hat{\underline{n}} $ $ \sigma_{22}(r,\theta=0) = \frac{K_{I}}{\sqrt{ 2\pi r }} \leq \sigma_{y} $ $\sigma_{22} =$ stress component that tries to open the crack ### Small Scale Yielding Condition The plastic deformation zone radius is given by $ r_{Ip} = \frac{1}{2\pi}\left( \frac{K_{I}}{\sigma_{y}} \right)^{2} $ For linear elastic fracture mechanics to apply, specimen dimensions much be much greater than $r_{Ip}$: $ \{ a, w-a, h, B \} > 15r_{Ip} $ ![[Pasted image 20240519014614.png|300]] #### Superposition By linearity, we can superpose loading conditions and stress intensity factors: $ \sigma_{\infty} =\sigma_{\infty,1} + \sigma_{\infty,2} + \sigma_{\infty,3} + \dots + \sigma_{\infty, n} $ $ K_{I} = K_{I,1} + K_{I,2} + K_{I, 3} + \dots + K_{I,n} $ $ K_{Ic} = \sum_{i=1}^{n}K_{I,i} $