Problem 4
Question
Show that if a general state of stress is to be described in cylindrical coordinates, the requirement that \(\Sigma \mathbf{F}=0\) leads to the following three equations: Prob. \(4.4\) $$ \begin{aligned} &\frac{\partial \sigma_{r}}{\partial r}+\frac{1}{r} \frac{\partial \tau_{r \theta}}{\partial \theta}+\frac{\partial \tau_{z r}}{\partial z}+\frac{\sigma_{r}-\sigma_{\theta}}{r}=0 \\ &\frac{\partial \tau_{r \oplus}}{\partial r}+\frac{1}{r} \frac{\partial \sigma_{\theta}}{\partial \theta}+\frac{\partial \tau_{\theta z}}{\partial z}+2 \frac{\tau_{r \theta}}{r}=0 \\ &\frac{\partial \tau_{\pi}}{\partial r}+\frac{1}{r} \frac{\partial \tau_{\theta z}}{\partial \theta}+\frac{\partial \sigma_{z}}{\partial z}+\frac{\tau_{x c}}{r}=0 \end{aligned} $$
Step-by-Step Solution
VerifiedKey Concepts
Stress Tensor
- Radial stress, denoted by \(\sigma_r\), acting perpendicular to the surface of a cylindrical element.
- Circumferential or tangential stress, \(\sigma_\theta\), which acts along the circumference of the cylindrical surface.
- Axial stress, \(\sigma_z\), directed along the length of the cylinder.
- Shear stresses, \(\tau_{r\theta}, \tau_{\theta z}, \) and \(\tau_{zr}\), which denote forces parallel to the surface and between different planes.
Force Equilibrium
- Radial direction: Here, the sum of radial and shear forces clients to zero.
- Circumferential direction: Circumferential and tangential forces counterbalance for stability along the circular plane.
- Axial direction: Longitudinal forces, including shear components, align to maintain steadiness along the cylinder's length.