A couple of books on elasticity, e.g. Gould and Soutas-Little, in the section on 2d elasticity in cylindrical coordinates, mention the case of 'quasi-axisymmetric' problems. These are problems where the stress and strain retain and axial symmetry (i.e., no dependence on the angular coordinate $\theta$), whereas the displacements $u_r$ and $u_\theta$ may carry $\theta$ dependence.

In particular, on writing the expression for $u_\theta$: $$ u_\theta = A*r\theta + B*\cos\theta + C*\sin\theta + D*r $$ with A,B,C,D constants, both mention that the last three terms rigid body motions, and hence do not contribute to any strain.

This statement per se I understand. Rigid body motions cannot give rise to strains or stresses. However, while I can imagine that the $\cos\theta$ and $\sin\theta$ represent rotations, I do not see how the $r$ term cannot give rise to strains! If $u_\theta$ is (say) zero at the inner radius and finite at the outer radius, it must surely give rise to strains! What is going on? What am I misunderstanding?

Addition to answer below --

Please see my comment to the answer for an explanation of the $r\theta$ term as well.


1 Answer 1


Consider a (small) rigid rotation $\alpha$ about the axis of the cylindrical coordinate system.

The displacement of a point at radius $r$ from the axis is $$\begin{align} u_r &= 0 \\ u_\theta &= \alpha r. \end{align}$$

That's where your $D \star r$ term comes from.

The $\cos\theta$ and $\sin\theta$ terms come from rigid body translations in the $\theta = 0$ and $\theta = \pi/2$ directions, not from rotations. For example, consider a (small) rigid translation $a$ in the $\theta =0$ direction. Describing this is cylindrical coordinates gives $$\begin{align} u_r &= a\cos\theta \\ u_\theta &= a\sin\theta. \end{align}$$

A rigid translation in the $\theta = \pi/2$ direction is similar, but with the $\cos$ and $\sin$ terms interchanged.

But the $A * r\theta$ term looks wrong. It doesn't make any sense, because it doesn't even represent a continuous displacement field. The values of $u_\theta$ when $\theta = 0$ and $\theta = 2\pi$ must be the same if the deformed shape of the structure "fits together" properly, but they are not!

  • $\begingroup$ Nice! For the $r\theta$ term, the word you are looking for is multi-valued. Indeed, such a term is inadmissible for an annulus, for which therefore $A=0$. However, it is admissible for sectors of an annulus where there is no such problem of multi-valuedness. Indeed, one of the textbook cases of this term is something called a 'rotational dislocation', which involves an annulus formed as open and separated by an angle $\alpha$, and then joined together. The BC is then $u_{\theta}(r,2\pi) =r\alpha$, which necessitates the $r\theta$ term to satisfy. $\endgroup$
    – ap21
    Sep 10, 2018 at 10:33

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