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).

enter image description here 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: enter image description here 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.


1 Mechanical Equilibrium of Thick, Hollow, Liquid Membrane Cylinders. Waugh and Hochmuth. 1987.


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.


4 Mechanics and Thermodynamics of Biomembranes. Evans and Skalak. 1980.

Following 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. enter image description here

enter image description here

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.

Flat Membrane

enter image description here

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 the 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.

enter image description here enter image description here

  • $\begingroup$ The format of this forum makes it very inconvenient to have a back-and-forth discussion of this type of analysis, particularly if equations are involved. Please submit your question to PhysicsForums.com, and I will be pleased to help you. $\endgroup$ Commented Sep 27, 2023 at 11:46
  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Sep 29, 2023 at 2:38
  • $\begingroup$ Small comment on notation: I would write tensor components without commas, $\sigma_{rr}~~ \sigma_{r\phi}~~ \sigma_{\phi \phi}$, etc. I'm surprised the quoted source uses commas, but subscript commas almost always mean derivatives: $ \sigma_{r,\phi} = \partial \sigma_r/\partial \phi$ $\endgroup$
    – RC_23
    Commented May 31 at 3:23
  • $\begingroup$ Thanks RC_23. I will correct that. My sources don’t use that notation. Thanks again for the correction. I had forgotten that notation and it would not work here at all. That was my dirty fix when tussling with MathJax. I will fix those. $\endgroup$
    – Nf23kdr
    Commented May 31 at 7:22

2 Answers 2


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: $$\frac{[r_0\sigma_E(\lambda_{r},\lambda_{\theta}) \hat{s}]}{\partial r_0}+\frac{[r_0\sigma_E(\lambda_{\theta},\lambda_{r}) \hat{\theta}]}{r_0d\theta}=0$$

  • $\begingroup$ See next installment. $\endgroup$ Commented Sep 28, 2023 at 23:50
  • $\begingroup$ I drew it on paper and am confused on the geometry in a couple of places and drawing a figure to add to the main post for you to confirm $\endgroup$
    – Nf23kdr
    Commented Sep 29, 2023 at 11:07
  • $\begingroup$ OK, I've only taken this a little further by introducing the "engineering stress function;" I want to give you a chance to figure out how to solve the equations and apply the boundary conditions. $\endgroup$ Commented Sep 29, 2023 at 11:45
  • $\begingroup$ Thank you for helping me with this Chet..it is really great. I will go over the whole thing -You are right-I would like to figure out how to proceed with what you have given me so far. (And though it is well in the future I will send you the paper once it is written up and gratefully acknowledge both you and PSE.) $\endgroup$
    – Nf23kdr
    Commented Sep 29, 2023 at 11:58
  • $\begingroup$ How are you visualizing $t$ Chet? $\endgroup$
    – Nf23kdr
    Commented Oct 4, 2023 at 9:37

(Please note this post is still a work in progress - I would like to make it about as useful as I can. But it is a bit unreadable because..among other things, MathJax and I are not friends yet ;-/. I will fix it in a week.)

I posted this question here almost exactly eight months ago on Sept 27, 2023. Thanks to some very helpful input from Chet Miller and after going over Evans and Skalak and some other material (see full reference list) again, I have the following solution to my problem. My understanding of the basic physics involved has improved in that time.

My original question can be boiled down to this: why are the axes corresponding to the cylindrical coordinate system also the principal axes (aka the axes of zero stress and maximal direct stress) for the specific problem I am interested in?

Further, I found the statements in the paper referred to in the OP (Waugh and Hochmuth, 1987) that the principal axial $σ_z$ and circumferential/hoop $σ_\phi$ stresses were equal confusing. I still have to revisit that, but at least I have a rudimentary proof indicating you can justify stating that $r,z,\phi$ (the coordinate axes for the cylindrical stress tensor) are the principal axes here.

As Chet recommended, I applied the differential force balance method. The following solution is adapted from Evans and Skalak Pages (52-64). I worked through it and reworked it a bit so it makes sense to me (assisted by the reply Chet posted above). Most of the primary references used are listed below.

Please note that this post belabors straightforward physics and would be too rudimentary for most posters here. But it is written so as to be legible to fellow neophytes ;-/ (to quote a post of Chet’s on traction and Wikipedia).

The salient features of this problem are: 1) Thinness in one dimension; 2) Axisymmetry; 3) the fluid nature of a biological membrane and 4) incompressibility of the membrane. (I should note here that I don’t end up actually using either the incompressibility or the fluidity of the membrane in this proof.)

My initial confusion resulted from failure to comprehend why some of the components of the stress tensor vanish.

So, I started off with the most elementary concepts (1,2). Figure 1

An infinitesimal planar membrane element with no curvature (Figure 1), should have 9 different stress tensor components: 3 pairs of direct stresses balancing each other out and 6 pairs of shear stresses balancing each other out for force and moment balance. The shear stresses reduce to 3 equivalent components to ensure consistency with moment balance over the element. This yields the symmetry of the stress tensor in rectangular coordinates.

However, as this problem involves a cylindrical membrane tether, the cylindrical coordinate system $r,z,\phi$ is a better fit than the rectangular coordinate system $x,y,z$.

(Direct stresses are denoted throughout by $σ_i$$_j$ and shear stresses by $τ_i$$_j$. Here $j$ indicates the direction of the area normal of the face under consideration and $i$ indicates the direction of the force. This is not universally true as far as conventions go and it is very confusing when flipped, but generally this seems to be the most common way and so I am using it throughout.)

enter image description here

As you can see, force and moment balance (Figure 2), again ensure the symmetry of the cylindrical stress tensor. Next, I moved from the generic case to this specific one (Figure 3) where it starts to get a little more complicated. enter image description here enter image description here (Please note that in equations that follow later, due to the thinness of the membrane in $z$, a subset of the direct and shear forces here are treated as direct/shear force stress resultants, i.e. the integral of the stress over the thickness is used for convenience (apparently under the assumption that it doesn’t change as significantly over the membrane thickness $t≪rd\phi;t≪ds$ as it does over the other dimensions. This applies to only 6 of the 9 stress tensor components (ie the ones applicable to the thin faces): $T_\phi$$_\phi=σ_\phi$$_\phi t; T_m$$_m=σ_m$$_m t; T_\phi$$_m=τ_\phi$$_m t; T_m$$_\phi=τ_m$$_\phi t; Q_n$$_\phi=τ_n$$_\phi t; Q_n$$_m=τ_n$$_m t$.

I won’t address moment balance much in this post overall and right now I will just leave it at this: the stress tensor’s symmetry here applies only to one pair of shear stresses - the pair on the two faces of comparable dimensions. There, similar to the derivations above, it turns out that: $τ_\phi$$_m=τ_m$$_\phi$.

However, (and this was a source of some confusion to me) it appears that: $τ_m$$_n≠τ_n$$_m$. And similarly $τ_n$$_\phi≠τ_\phi$$_n$. I am still struggling with a couple of concepts connected with moments in the literature I reviewed for this. I will update this part of the post later.

But for now we are ready to write the differential force balance equations for the membrane. Before I do that, I wanted to spell out some basic differential calculus I stumbled over a few times.

In order to correctly estimate the components of these differential force balance equations, one has to understand how total derivatives work when the Chain Rule is applied to partial differentiation (3). The complete derivative of a function $f$ of $x,y$, where $y$ itself is a function of $x$ can be given in terms of partial derivatives as follows:


If y is a function of x alone, then $(df(x,y(x)))/dx=∂f/∂x+∂f/∂y.dy/dx$ (These two equations are taken directly from Wikipedia).

If the stress over a face is $σ$ (or $τ$ if a shear force is under consideration rather than a normal force) and the area of the face is $A$ and both are functions of system’s axes $x,y$, this can be applied to differential force analysis to compute the differential force as follows: $dF_f=A_f (∂σ/∂x) dx+σ (∂A_f/∂x) dx$

I cannot find that in Thomas’s Calculus or my other references at the moment, but maybe it is generally true that: $df(x,y,z)=(∂f(x,y,z)/∂x)dx=∂f(x,y,z)/∂y)dy=(∂f(x,y,z)/∂(xy))d(xy)=(∂f(x,y,z)/∂(xy))ydx=(∂f(x,y,z)/∂(xy))xdy$

As $r$ and $\phi$ typically appear here as the compound term $r\phi$ representing the arc length in the meridional direction, I got confused a couple of times when applying the Chain Rule and partial differentiation to write the differential force equations. I was initially computing it wrong, but I think: $df(r,\phi,s)≠(∂f(r,\phi,s)/∂r) dr+(∂f(r,\phi,s))/(∂\phi) d\phi$

Rather: $df(r,\phi,s)=(∂f(r,\phi,s)/∂(r\phi))d(r\phi)=(∂f(r,\phi,s)/∂r) dr=((∂f(r,\phi,s))/(∂\phi)) d\phi$

(Please let me know if I made an error there. I sometimes get confused when I use partial or total derivatives which is why I am spelling that out, but at least that checks out with the easier kinds of examples such as the differentiation of simple functions such as: $ye^x $ or $ycos⁡x$ wrt $xy$. And it works here, but then there are many terms that conveniently vanish here as well.).

With that said, we can now write the equations (Figure 4). enter image description here

As far as I can tell the thinness of the membrane and axisymmetry impact how differential force analysis ends up being applied to this problem. First, I shall examine the force balances in the two dimensions which the membrane is not thin in.

Now comparing and contrasting the elements contributing to circumferential and meridional force balance gives us:

  1. For both axes, a differential normal force balance term. These terms are respectively:

$dF_\phi$$_\phi=(∂(σ_\phi$$_\phi tds cos⁡\alpha)/∂(r\phi)) d(r\phi); dF_m$$_m=(∂ \sigma_m$$_m (trd\phi)))/∂s ds$

  1. For both axes, a differential shear force balance term: $dF_\phi$$_m=(∂(τ_\phi$$_m trd\phi)/∂s ds; dF_m$$_\phi=(∂τ_m$$_\phi) cos⁡(\alpha) (tds))/∂(r\phi) d(r\phi)$

  2. Next, due to the trapezoidal shape of the element when unfolded or projected on to the tangent plane, the shear force $τ_m$$_\phi$ has a component affecting the circumferential force balance and the normal force $σ_\phi$$_\phi$ adds a term to the meridional force balance equation: $F_m$$_\phi,circ = τ_m$$_\phi sin⁡\alpha(tds) +((τ_m$$_\phi)+(∂τ_m$$_\phi/∂(r\phi) d(r\phi)) sin⁡\alpha (tds)$

$F_\phi$$_\phi,meri=-(σ_\phi$$_\phi) cos⁡(π/2-\alpha)+((\sigma_\phi$$_\phi)+(∂σ_\phi$$_\phi/∂(r\phi)) d(r\phi)) cos⁡(π/2-α) )(tds)$

(These are not differential elements but resolved components of these forces along those directions. I do not get why $F_\phi,_\phi,meri$ is negative rather than positive, but that is how it is in both Evans and Skalak and Schumacher et. al, so I am leaving it that way for now. Anyway it doesn’t impact my final analysis here).

  1. In contrast to how the differential analysis proceeds for the other shear forces, here:

$dF_m$$_n≠(∂(τ_m$$_n dsrd\phi)/∂t_n ) dt_n;dF_\phi$$_n≠ (∂(τ_\phi$$_n dsrd\phi))/(∂t_n ) dt_n$ Instead, two total surface shear tractions along $∅,m$ are respectively given by: $τ_\phi$$_n (rd\phi)ds$ and $τ_m$$_n (rd\phi)ds$. I attribute this treatment to the thinness of the membrane along $r,n$. This analysis presumably applies whether it is a membrane element one layer of which is the innermost or outermost surface or an element both sides of which are intra-membrane surfaces. Finally, we have two terms that arise due to the curvatures of the membrane which results in the two transverse shears having terms that when resolved along the circumferential and meridional directions give respectively (Figure 4): $2τ_n$$_\phi (tds) (rd\phi)/(2R_\phi )$; $2τ_n$$_m (trd\phi)ds/(2R_m )$ (While they don’t spell it out, some of the terms above such as $dF_m$$_m$ or $dF_\phi$$_\phi$ should strictly speaking be multiplied by $cos⁡\alpha_m , cos\alpha_\phi$, but since $cos⁡ \alpha_m ≈ cos⁡ \alpha_\phi≈cos⁡\alpha≈1$, those factors are dropped). Here: $R_\phi=r/sin⁡ψ ; R_m= 1/(∂ψ⁄∂s)$

(I don’t off-hand know the source or the derivation for these radii of curvature, but they are used in all the papers on membrane tethers I consulted).

So finally, we arrive at these two differential force balance equations for the circumferential and meridional directions:

$(∂T_\phi$$_\phi/∂\phi)+T_\phi$$_\phi/(d\phi) (∂(ds))/(∂\phi)+(∂τ_\phi$$_m)/∂s tr+(τ_\phi$$_m) tcos⁡ψ+τ_\phi$$_m/(d\phi) tr (∂(d\phi))/∂s+(2τ_m$$_\phi) t sin\alpha+(τ_\phi$$_n) r+(τ_n$$_\phi) t r/R_\phi = 0 $

Where $sin⁡\alpha=1/2 cos⁡ψ d\phi+1/2 r (d(d\phi))/ds$

$(∂T_m$$_m/∂s) r+T_m$$_m cos⁡ψ+T_m$$_m (r/d\phi) (∂d\phi)/∂s+(∂T_m$$_\phi/∂\phi)+T_m$$_\phi (∂(ds)/∂\phi)-2 (T_\phi$$_\phi/(d\phi)) sin⁡\alpha+τ_m$$_n (r) + (τ_n$$_m (tr))/R_m =0$

The final equation for force balance (in the normal direction) is simpler and has just six components: Due to the curvature of the membrane (Figure 4) again there are two components that affect force balance in the normal direction, i.e. the components resulting from the resolution of $\sigma_\phi$$_\phi$,$\sigma_m$$_m$ in the $n$ and $\phi,m$ directions respectively. The direct stress on the broad face again is treated differently from the treatment of $σ_\phi$$_\phi$,$σ_m$$_m$ in the previous force balance equations and instead of a differential component (much like the shear tractions $τ_m$$_n,τ_\phi$$_n$) there is a total force term which is the difference in pressure between the inner and outer sides, $∆P$ corresponding to direct stress $σ_n$$_n$. Again, please correct me if I am wrong, but I am attributing that to the thinness of the membrane. Finally, there are the differential force balance components arising due to the two transverse shears $Q_n$$_m,Q_n$$_\phi$. That gives us the equation for normal force balance:

$σ_n$$_n (dsrd\phi)-(σ_\phi$$_\phi)(tds)(rd\phi/R_\phi)-σ_m$$_m (trd\phi)(ds/R_m )+∂(τ_n$$_\phi) (tds)/∂(r\phi) d(r\phi)+∂(τ_n$$_m rd\phi)/∂s ds=0$

The next thing to do was inspect the components of the cylindrical strain tensor and figure out why so many of the terms vanish - indeed the entire circumferential force balance equation essentially vanishes!). Whilst materials I consulted stated that axisymmetry makes some components vanish (which is vaguely intuitive), even so I was lost without a formal derivation, which I finally found (4). I found the geometrical derivation a bit painful to follow spatially and simplified it a bit for my comprehension-decomposing each shear strain into two parts as shown. I am inserting the derivations in the images without text (Figure 5) because…I am a bit tired of wrestling with MathJax... enter image description here enter image description here enter image description here enter image description here Here, $u_r,u_z,u_\phi$ are the respective displacements in $r,z,\phi$. Important note: This is a different element and here $dr$ has a different meaning from the term $dr$ when applied to the earlier differential force analysis with the initial cylindrical infinitesimal element for instance. Here it corresponds to the membrane thickness $t$ from that analysis.

Anyway, to cut a long story short, for axisymmetric components as there is no net change in any components as a function of the angle about the axis of symmetry $\phi$, both terms containing derivatives with respect to this angle $(∂/(∂\phi)$ as well as the displacement, $u_\phi$ vanish leaving the following stresses only: $σ_\phi$$_\phi$,$σ_m$$_m,σ_n$$_n,\tau_r$$_z$. Now, my original question would be better phrased as follows: for an arbitrarily oriented cylindrical infinitesimal element (Figure), why are the principal axes aligned with the axes of the cylindrical coordinate system. In other words, why does $𝟁=π/2$, correspond to the orientation of the principal axes? To answer this, one needs to write the differential force balance equations for the system (whilst ensuring that moment balance criterion are met). Then it is possible to use Fermat’s Theorem (6) and differentiate throughout with respect to $𝟁$ and solve for the cases where all the normal stresses are at their maximum values and the shear stresses are correspondingly at their minimum value (i.e. zero). Now our system of force balance equations reduces to the following two equations: $(∂T_m$$_m)/∂s r+(T_m$$_m cos⁡ψ)+T_m$$_m r/(d\phi) (∂d\phi)/∂s+2 T_\phi$$_\phi/(d\phi) sin⁡\alpha+τ_m$$_n (r)+ (τ_n$$_m (tr))/R_m =0$

$σ_\phi$$_\phi (tds)(rd\phi/R_\phi)+σ_m$$_m (trd\phi)(ds/R_m )-∂(τ_n$$_m rd\phi)/∂s ds=0$ (The pressure difference across these membranes is typically zero.) Differentiating the meridional force balance equation with respect to $ψ$ results in some unwieldy and fairly intractable looking equations, so I turned to the normal force balance equation. The transverse shear is frequently relatively negligible here (and anyway with it included I again found the problem intractable). However, if the transverse shear is neglected (rationalizing it by noting that according to literature it is generally smaller) the proof looks more tractable. Next, the fact that the membrane adopts the shape of a catenary prior to tube formation can be exploited to ascribe attributes of the catenary to this differential element (5), i.e.: $(dψ)/ds=(cos⁡ψ )^2/α_c $
(where $\alpha_c$ is a parameter characterizing the catenary) Now the normal force balance equation reduces to: $σ_m$$_m r(cos⁡ψ)^2+\alpha_c σ_\phi$$_\phi sin⁡ψ=0$ Differentiating throughout wrt $ψ$ and substituting $∂r/∂ψ=(∂r/∂s)⁄(dψ/ds); ((∂σ_\phi$$_\phi))/∂ψ=(∂σ_m,_m)/∂ψ=0)$, gives $cos⁡ψ=0$ i.e. $ψ=π/2$ as at least one solution which was my original question. The values of $σ_m$ $σ_\phi$ can also be obtained (as functions of $\alpha_c$ anyway). I had a few more points to add about moments and a few other things. But for now, I am posting this for a start.

For now, while MathJax and I remain estranged, here are the main equations as images. I still have to check the whole post over and add a section on moments which I will do in a month or so: enter image description here enter image description here

Final Comments: My background is not in mechanical engineering (or applied physics for that matter) and that is reflected in the initial post. I was attempting to work backwards from primary literature (which as chemomechanics correctly pointed out is not intended to address fundamentals). Chet’s earlier suggestion that I get an elementary grasp of tensors followed by his other responses have helped a lot and I reviewed some of the foundational materials required to have at least a rudimentary grasp of the mechanics needed to understand this problem. It is still a little clunky and inelegant and I am still working out many related points (especially tied to the moments), but at least I don’t think it is flat out wrong. Any comments, corrections, admonishments etc. are welcome :o). I am updating this post to illustrate the background learning process I had to go through to be sure about some fundamental concepts tied to the research problem I am interested in. It is definitely harder to work non-linearly when doing research than in the conventional direction (basics first and more advanced research on that foundation), but it happens to me all the time in interdisciplinary work. Leaves one feeling a bit like an eternal neophyte, but oh well.. I would be curious about your comments if you are around Chet-I ended up using the catenary shape of the curve to justify my proof. I am curious as to what you had in mind.

You and others really perform a yeoman’s service on this site :). I do thank you for getting me to go over the physics. I was and am trying, but there is always so much of it ;-/ ….

I need to go over this post and check that all the equations and statements are correct as far as I know. And clean it up so it is less unreadable, but for now I posting this before MathJax and the figures and the post sort of crash.

Thanks again Chet..appreciate your feedback. It is why I typed this response up (such as it is ;-/)

Citations 1)Kelly, PA. Mechanics Lecture Notes: An introduction to Solid Mechanics. Available from http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/index.html

  1. https://archive.nptel.ac.in/content/storage2/courses/112106141/Pdfs/2_1.pdf

  2. https://en.wikipedia.org/wiki/Total_derivative

  3. https://ocw.mit.edu/courses/22-314j-structural-mechanics-in-nuclear-power-technology-fall-2006/db946c4cb4732459c6ea2d854836c3b0_m11a.pdf

  4. https://en.wikipedia.org/wiki/Catenary

  5. https://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx

The other main references are already cited in the previous posts. I will post this for now and clean it up later. MathJax has wiped me out for now. But please let me know if it doesn't conform to community standards or suffer from other issues. I didn’t end up addressing moment balance adequately for now. I will improve the post later-it is a bit of a hideous, unreadable mess. Please bear with me-I will clean it up when I get the time. There are still many conceptual issues with this post that need fixing as well.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.