Some background: I posted this question on the shear rigidity of a fluid cylinder in a sloppier form earlier.
Thanks to some helpful feedback from Chet Miller I went back and reviewed tensors before reframing the question.
I can rephrase my question better now, but my original doubt remains unresolved.
Before I proceed, please note that I am not asking why fluids do not have shear rigidity. My last question was pooled into another question to that effect.
My questions are specific to a paper1 which addresses the stresses on a thick, hollow, liquid cylinder (page 392, equations 4,5,6) under an axial load (Figure 1).
I do not understand their argument for arriving at the conclusion that the circumferential and longitudinal/axial normal stresses in such a cylinder are equal. First of all they do not prove that
the normal stresses ($\sigma_r,_r$, $\sigma_z,_z$,$\sigma_\phi,_\phi$) are actually also the principal stresses.
They appear to be asserting that the normal stresses shown are the principal stresses and that therefore it follows that the shear stresses on the planes of the cylindrical coordinate system are zero. Which leads them to the conclusion that the circumferential and axial/longitudinal normal stresses are equal. A conclusion that would not hold for a solid cylindrical object with a finite shear rigidity then.
Maybe I am just misreading that paper segment. Do correct me if that is the case.
This is the page:
In summary then, why are the radial, circumferential and axial directions (i.e. the primary directions of the cylindrical coordinate system) the principal directions for a fluid cylinder under an axial load? Secondly, why are the circumferential and axial principal stresses equal? I have not come across a simple proof.
Edit: The earlier draft of this post was less concise. I found drafting this question a useful exercise and usually any feedback I get here is useful. I will try to work out the principal stresses for a generic cylinder under an axial load and then update this post.
References
1 Mechanical Equilibrium of Thick, Hollow, Liquid Membrane Cylinders. Waugh and Hochmuth. 1987.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1330003/pdf/biophysj00162-0035.pdf
Updated References with two new additions on Oct 6, 2023
2 The Mechanics of Axially Symmetric Liposomes. Pamplona and Calladine. 1993.
https://pubmed.ncbi.nlm.nih.gov/8326721/
3 Mechanics of tether formation in liposomes. Calladine and Greenwood. 2002.
https://pubmed.ncbi.nlm.nih.gov/12405601/
4 Mechanics and Thermodynamics of Biomembranes. Evans and Skalak. 1980.
Following on my dialogue with Chet below, I am updating my post with a schematic of the type of biophysical experiment I am modeling.
It is on this type of membrane:
https://en.m.wikipedia.org/wiki/Lipid_bilayer
I have come up with a data analysis model for an experiment like the one shown below and want to ensure that the basic physics behind my analysis is sound. The formation of membrane cylinders from bilayers subjected to pulling forces is a highly non trivial process. I am merely digging through some old papers in the area addressing already formed cylinders and reviewing the basic physics to be certain there are no glaring errors in my analytic model.
Of the cartoons below, the first one depicts the initiation of the formation of such a membrane cylinder and the second one a formed cylinder. Since it is a complicated problem I tried to trim it down to a simple question to keep it within the scope of the forum. I am not sure I succeeded.
Looking at your question guidelines, I suspect this question is indeed off-topic here. I will probably follow Chet's advice and try physicsforums instead.
Edited Update: I updated this post with my initial understanding of the model Chet very graciously provided below. However, it is still a work in progress.
Okay I have updated the post with how I figure Chet modeled it now. I had two of the stresses flipped ($\sigma_r$ and $\sigma_z$ and $\sigma_z$ is now $\sigma_n$).
I updated Chet's differential force equation with the third stress that is normal to membrane as shown. I am writing out the third equation (the first two being the resolution of the three stresses parallel to the meridian's tangent and normal to it) -that is the moment balance equation and brings in the membrane bending stiffness. Finally I have to set the boundary conditions and solve for the three stresses.
Not sure that diagram/the equation is correct. I always have a hard time visualizing these catenoids that have two different radii of curvature.
This is background physics work for an actual paper I am working on. My initial understanding of the physics behind this experiment was very rudimentary and I want to get the basics right this time.
Best Answer
As in your diagrams, let the membrane initially be horizontal at z = 0. Let $r_0$ be the radial location measured from the z axis in the initial flat membrane. Let the location of a material point in the deformed thin membrane be situated at $r=r(r_0,\theta)$ and $z=z(r_0,\theta)$.
Consider two closely neighboring material points in the initial undeformed membrane at ($r_0,\theta)$ and $(r_0+dr_0. \theta+d\theta)$. The length of the differential position vector joining these two material points initially is $\sqrt{(dr_0)^2+(r_0d\theta)^2}$. The differential position vector joining these same two material points in the deformed configuration is $$\hat{s}ds=\left(\hat{r}\frac{\partial r}{\partial r_0}+\hat{z}\frac{\partial z}{\partial r_0}\right)dr_0+\hat{\theta}\left(\frac{r}{r_0}\right)(r_0d\theta)$$where hatted quantities are unit vectors. The square of the length of this differential position vector is $$(ds)^2=\left[\left(\frac{\partial r}{\partial r_0}\right)^2+\left(\frac{\partial z}{\partial r_0}\right)^2\right](dr_0)^2+\left(\frac{r}{r_0}\right)^2(r_0d\theta)^2$$The ratio of the square of the deformed length to the square of the initial length is the square of the stretch ratio $\lambda$: $$\lambda^2=\lambda _r^2\cos^2{\alpha}+\lambda^2_{\theta}\sin^2{\alpha}$$with $$\lambda^2_r=\left[\left(\frac{\partial r}{\partial r_0}\right)^2+\left(\frac{\partial z}{\partial r_0}\right)^2\right]$$ $$\lambda^2_{\theta}=\left(\frac{r}{r_0}\right)^2$$ $$\cos{\alpha}=\frac{dr_0}{\sqrt{(dr_0)^2+(r_0d\theta)^2}}$$ and $$\sin{\alpha}=\frac{r_0d\theta}{\sqrt{(dr_0)^2+(r_0d\theta)^2}}$$
Now for a differential force balance on the deformed membrane. Consider the force balance on the "window" shaped element of the deformed membrane between $r_0$ and $r_0+dr_0$, and between $\theta$ and $\theta+d\theta$: $$\left(\sigma_r\lambda_{\theta}r_0d\theta t\hat{s}\right)_{r_0+dr_0}-\left(\sigma_r\lambda_{\theta}r_0d\theta t\hat{s}\right)_{r_0}+(\sigma_{\theta}\lambda_r dr_0 t\hat{\theta})_{\theta +d\theta}-(\sigma_{\theta}\lambda_r dr_0 t\hat{\theta})_{\theta}=0$$Dividing by $dr_0d\theta$ then gives: $$\frac{(\sigma_r\lambda_{\theta}r_0 t\hat{s})}{\partial r_0}+\frac{(\sigma_{\theta}\lambda_r t\hat{\theta})}{d\theta}=0$$Next, since the membrane material is incompressible, we have $$t=\frac{t_0}{\lambda_r \lambda_{\theta}}$$ Substituting this into the differential force balance then gives: $$\frac{[r_0(\sigma_r/\lambda_{r}) \hat{s}]}{\partial r_0}+\frac{[r_0(\sigma_{\theta}/\lambda_{\theta}) \hat{\theta}]}{r_0d\theta}=0$$
Define the engineering stress function $\sigma_E$ for a transversely isotropic membrane as follows: $$\sigma_E(\lambda_1,\lambda_2)=\frac{\sigma(\lambda_1,\lambda_2)}{\lambda_1}$$Then, our differential force balance becomes: $$\frac{[r_0\sigma_E(\lambda_{r},\lambda_{\theta}) \hat{s}]}{\partial r_0}+\frac{[r_0\sigma_E(\lambda_{\theta},\lambda_{r}) \hat{\theta}]}{r_0d\theta}=0$$