# Why is the partial derivative of stress in the circumferential direction equal to zero in axial symmetric bodies?

I've come across this phenomenon several times and never found any real reason for this.

When I consider a cylindrical object like a rod, I need to calculate several things. One of those is the stress tensor in cylindrical coordinates represented by $r$, $\theta$, and $z$ directions.

Now, in many books and in my lecture notes there's always this one line that says:

$$\frac{\partial\bullet}{\partial\theta}=0$$

I know that when compressing a steel rod (assuming there is little to no friction at the contact points) the steel rod should get compressed in the $z$-direction and expand in the radial direction. Why is there no stress in the circumferential direction?

I've looked in many places and I hope that I didn't accidentally miss the other SE question with respect to this specific topic.

• I'm going to change the title and wording to reflect the limited case where the partial wrt theta is zero. The stresses may be nonzero, but unchanging. – Phil Sweet Jan 9 '18 at 18:29

1) axisymmetry requires symmetry of the entire problem. The body and the loading and the material properties. For the bending case the boundary condition is not axi-symmetric and so you can not use an axisymmetry assumption to simplify the governing equations.

2) zero circumferential stress does not follow from axisymmetry. It happens to be so for this simple problem. You have zero transverse stress in a square bar in axial load too.

As a trivial example consider a cylinder under uniform pressure P over the entire surface. All normal stress components are equal to P, so clearly sigma_theta is not zero.

• I have proposed an edit to the question that may affect your answer – Phil Sweet Jan 9 '18 at 18:42
• @PhilSweet for what its worth I don't have rep to see the suggested edit. – agentp Jan 9 '18 at 19:09
• @agentp Thank you very much. Your answer was precise and simple. I realized how the plain tension test and the cylindrical tension test can have a one-dimensional loading, but a multidimensional strain. – fruitiest Punch Jan 10 '18 at 4:20

This assumption always requires justification, and when this justification is not given by your lecturer, you should ask for it if it is not clear to you. For example when investigating bending, derivatives in $\theta$ direction will not generally be zero.

The most common way to justify this assumption is to assume it in the first place, calculate a solution, and then check if the solution satisfies the full set of differential equations you had in the first place (without any simplifications). If it does, then you have found a valid solution. If it does not, then the assumption is nonsense. Sometimes the final check is not performed, probably because the authors think this is trivial.

The specific case of a uniaxially compressed rod, especially when neglecting friction at the contact surfaces, is one of the cases where this works.

• I have proposed an edit to the question that may affect your answer. – Phil Sweet Jan 9 '18 at 18:41