Ⅰ、Ⅱ、Ⅲ型裂纹应力

Ⅰ、Ⅱ、Ⅲ型裂纹应力及坐标变换公式对比表

类型	极坐标应力公式	直角坐标应力公式	坐标变换公式
Ⅰ型	$ \begin{cases} \sigma_{rr} = \frac{K_I}{\sqrt{2\pi r}}\cos\frac{\theta}{2}\left(1+\sin^2\frac{\theta}{2}\right) \ \sigma_{\theta\theta} = \frac{K_I}{\sqrt{2\pi r}}\cos^3\frac{\theta}{2} \ \sigma_{r\theta} = \frac{K_I}{\sqrt{2\pi r}}\sin\frac{\theta}{2}\cos^2\frac{\theta}{2} \end{cases} $	$ \begin{cases} \sigma_x = \frac{K_I}{\sqrt{2\pi r}}\cos\frac{\theta}{2}\left(1-\sin\frac{\theta}{2}\sin\frac{3\theta}{2}\right) \ \sigma_y = \frac{K_I}{\sqrt{2\pi r}}\cos\frac{\theta}{2}\left(1+\sin\frac{\theta}{2}\sin\frac{3\theta}{2}\right) \ \tau_{xy} = \frac{K_I}{\sqrt{2\pi r}}\cos\frac{\theta}{2}\sin\frac{\theta}{2}\cos\frac{3\theta}{2} \end{cases} $	\(\begin{cases} \sigma_x = \sigma_{rr}\cos^2\theta + \sigma_{\theta\theta}\sin^2\theta - 2\sigma_{r\theta}\sin\theta\cos\theta \\ \sigma_y = \sigma_{rr}\sin^2\theta + \sigma_{\theta\theta}\cos^2\theta + 2\sigma_{r\theta}\sin\theta\cos\theta \\ \tau_{xy} = (\sigma_{rr}-\sigma_{\theta\theta})\sin\theta\cos\theta + \sigma_{r\theta}(\cos^2\theta-\sin^2\theta) \end{cases}\)
Ⅱ型	$ \begin{cases} \sigma_{rr} = -\frac{K_{II}}{\sqrt{2\pi r}}\sin\frac{\theta}{2}\left(1-\cos^2\frac{\theta}{2}\right) \ \sigma_{\theta\theta} = -\frac{K_{II}}{\sqrt{2\pi r}}\sin\frac{\theta}{2}\cos^2\frac{\theta}{2} \ \sigma_{r\theta} = \frac{K_{II}}{\sqrt{2\pi r}}\cos\frac{\theta}{2}\left(1+\sin^2\frac{\theta}{2}\right) \end{cases} $	$ \begin{cases} \sigma_x = -\frac{K_{II}}{\sqrt{2\pi r}}\sin\frac{\theta}{2}\left(2+\cos\frac{\theta}{2}\cos\frac{3\theta}{2}\right) \ \sigma_y = \frac{K_{II}}{\sqrt{2\pi r}}\cos\frac{\theta}{2}\sin\frac{\theta}{2}\cos\frac{3\theta}{2} \ \tau_{xy} = \frac{K_{II}}{\sqrt{2\pi r}}\cos\frac{\theta}{2}\left(1-\sin\frac{\theta}{2}\sin\frac{3\theta}{2}\right) \end{cases} $	同Ⅰ型变换式
Ⅲ型	$ \begin{cases} \sigma_{r\theta}=0 \ \sigma_{\theta z} = -\frac{K_{III}}{\sqrt{2\pi r}}\sin\frac{\theta}{2} \ \sigma_{rz} = \frac{K_{III}}{\sqrt{2\pi r}}\cos\frac{\theta}{2} \end{cases} $	$ \begin{cases} \tau_{xz} = \frac{K_{III}}{\sqrt{2\pi r}}\cos\frac{\theta}{2}\left(1+\sin^2\frac{\theta}{2}\right) \ \tau_{yz} = -\frac{K_{III}}{\sqrt{2\pi r}}\sin\frac{\theta}{2}\cos^2\frac{\theta}{2} \end{cases} $	\(\begin{cases} \tau_{xz} = \sigma_{rz}\cos\theta - \sigma_{\theta z}\sin\theta \\ \tau_{yz} = \sigma_{rz}\sin\theta + \sigma_{\theta z}\cos\theta \end{cases}\)