Question 1.25: Derive the differential equation for equilibrium in polar co...
Derive the differential equation for equilibrium in polar coordinates.
Figure E. 1.34 shows an element under radial stress \sigma_r and tangential stress \sigma_\theta , respectively. The shear stress \tau{_\theta} _r equal to \tau _r {_\theta} derived earlier in Eq. (1.17). Considering equilibrium of forces in radial direction,
\tau_{xy}=\tau_{yx} (1.17d)
-\sigma_r r d \theta+\left(\sigma_r+\frac{\partial \sigma_r}{\partial r} d r\right)(r+d r) d \theta-\sigma_\theta d r
\sin \frac{d \theta}{2}-\left(\sigma_\theta+\frac{\partial \sigma_\theta}{\partial \theta} d \theta\right) d r \sin \frac{d \theta}{2}-\tau_{r \theta} d r
\cos \frac{d \theta}{2}+\left(\tau_{r \theta}+\frac{\partial \tau_{r \theta}}{\partial \theta} d \theta\right) d r \cos \frac{d \theta}{2}+F_r r d r d \theta = 0
Neglecting higher order differential terms, and taking,
\sin \frac{d \theta}{2} \approx \frac{d \theta}{2} \text { and } \cos \frac{d \theta}{2} \approx 1,
\frac{\partial \sigma_r}{\partial r}+\frac{1}{r} \frac{\partial \tau_{r \theta}}{\partial \theta}+\frac{\sigma_r-\sigma_\theta}{r}+F_r=0 (i)
Now consider equilibrium of forces in tangential direction,
-\tau_{r \theta} r d \theta+\left(\tau_{r \theta}+\frac{\partial \tau_{r \theta}}{\partial r} d r\right)(r+d r) d \theta-\sigma_\theta d r \cos \frac{d \theta}{2}+\left(\sigma_\theta+\frac{\partial \sigma_\theta}{\partial \theta} d \theta\right) d r \cos \frac{d \theta}{2}+
\tau_{r \theta} d r \sin \frac{d \theta}{2}+\left(\tau_{r \theta}+\frac{\partial \tau_{r \theta}}{\partial \theta} d \theta\right) d r \sin \frac{d \theta}{2}+
F_\theta r d r d \theta=0
which reduces to,
\frac{1}{r} \frac{\partial \sigma_\theta}{\partial \theta}+\frac{\partial \tau_{r \theta}}{\partial r}+2 \frac{\tau_{r \theta}}{r}+F_\theta=0 (ii)
