# Viscous stress in axisymmetric cylindrical coordinates

I'd like to write the viscous stress for incompressible Newtonian fluids in axisymmetric cylindrical coordinates $(r,z)$ with velocity component $(u,w)$ (for axisymmetric $v_\theta=0$ and $\partial_\theta()=0$). I get confused with the component of $\tau_{\theta\theta}=2\mu\frac{u}{r}$ because I don't know whether or not it should be included in the tensor. If I included $\tau_{\theta\theta}$, I got a 3*3 matrix, if not a 2*2 matrix. Please see

$\tau_{2*2}= \begin{pmatrix} 2\mu u_r & \mu (w_r+u_z) \\ \mu (w_r+u_z) & 2\mu w_z \end{pmatrix}, \quad \mbox{or} \quad \tau_{3*3}= \begin{pmatrix} 2\mu u_r & 0 &\mu (w_r+u_z) \\ 0 & \color{red}{2\mu\frac{u}{r}} &0 \\ \mu (w_r+u_z) & 0 & 2\mu w_z \end{pmatrix}$

I need the stress tensor to formulate an axisymmetric problem in cylindrical coordinates. Any one can explain which one is correct and why? Thanks in advance.

The 3x3 is correct. If you are going to do a force balance in the radial direction, the hoop stress must be included.

• @ Chester Miller Thanks a lot. But I still don't understand why one mush include the hoop stress (direct in circumferential direction) in the radial force balance... Do you mean a stress balance in the circumferential direction instead, in which $\tau_{\theta\theta}$ should be included. Btw, yes, I do need to do a stress balance in the radial direction. – jsxs Mar 14 '17 at 16:16
• Do the stress balance in the radial direction and you will see why. It's a curvature effect. – Chet Miller Mar 14 '17 at 16:42

Old post, but anyway. As Chet points out, the 3x3 is correct. Nonetheless, I believe that as the problem is axisymmetric, the contribution of the circumferential stress of every element belonging to a ring cancels out when integrated in the full ring (theta = 2.pi.r), so it doesn't need to be included. One can see an example of the momentum equation in the radial direction in the book "An introduction to Computational Fluid Dynamics" by Versteeg and Malalasekera, 2nd ed., p. 372.