Circumferential tangent in the cylinder

[decentfellow]

Homework Statement

A cylindrical vessel of height $H$ and radius $R$ contains liquid of density $\rho$. Determine the circumferential tension at a height $h$; also determine its maximum and minimum values.

This is the scan of the original question(solved), I couldn't understand what was done here so I tried to solve it by myself and it has frustrated me so much that I am about to cry.

Homework Equations

The Attempt at a Solution

So my line of thought was that what would have been the case when there was no liquid filled inside it, would there still be any kind of circumferential tension. So, what I want to ask here is that is there any kind of manufacturing stress present in the cylinder due to its curved shape. This has me stuck for quite some time.

Also, I tried attempting the question as follows:-

Consider a differential ring of height $dh$ situated a depth of $h$ as shown below:-

Now due to pressure of the liquid the force exerted on the walls of the differential element by water is given by

$$F_{\text{liq.}}=P(dh)(Rd\theta) \qquad \& \qquad P=\rho gh$$

Now, here is the part where I am having a doubt as to how to determine the tension that acts on the differential cylindrical element due to the expansion that occurs due to the pressure exerted by the liquid. I was confused b/w two cases:-

1. If there was already tension along the circumference due to the shape of the cylinder before it was filled with liquid. If so is the case then, does the expansion due to the liquid pressure add to it.
2. If there was no tension when there was no pressure due to the liquid(i.e. there was vacuum inside the cylinder), then after adding the liquid the cylinder expands due to the pressure exerted by the liquid, hence tension acts to not let the vessel expand more.

I don't know how to proceed with case-1, because I don't know how does the tension act in a cylinder which has no internal pressure due to any kind of fluid.

But with case-2, the forces that start acting on the circumference are as shown in the diagram below:-

After all the changes that can occur in the dimensions of the cylinder has occurred there will be static equilibrium, hence we get:-

$$2T\sin\left(\frac{d\theta}{2}\right)=P(dh)(Rd\theta)\\
\implies \require{cancel} T\cancel{d\theta}=P(dh)R\cancel{d\theta}\implies T=\rho Rg\;h\;dh$$

So, it seems that the circumferential tension depends on the differential height. That seems weird.

Also, doesn't the cylinder should have some thickness or else when it expands it will break immediately. Do correct me if my reasoning is wrong.

 

[Nidum]

Cut the cylinder with a vertical plane through the cylinder axis .

What are the forces acting on one of the resulting half cylinders ?

 

[Chestermiller]

What is the exact wording of the question?

 

[decentfellow]

What is the exact wording of the question?

I have included a scan of the original question and its solution (which I couldn't make head or tails of) in my original post.

 

[Nidum]

I don't think that whoever thought up that question had much understanding of the relevant subject material themselves .

We are not even told whether the cylinder is open or closed ended . Open ended is a fictional problem with a relatively easy solution . Closed ended is a real problem with a more difficult solution .

The maths shown seems to apply to the fictional open ended problem .

 

[Chestermiller]

I agree with Nidum. There is something very wrong with the problem statement and the solution that was presented.

 

[decentfellow]

I agree with Nidum. There is something very wrong with the problem statement and the solution that was presented.

I do know that the problem statement is quite atrocious but the thought and time that I had devoted to solve it has made me think that there does exist a nice relation between the tension and height if we assume that it tries to address the open ended cylinder problem. And also I think the author tries to find the tension averaged over the axial length.

In your opinion what would have been a more clear problem statement to make this question salvageable.

 

[haruspex]

Closed ended is a real problem with a more difficult solution .

Can't we get around that by supposing the base to be some elastic membrane resting on a hard frictionless surface? That should mean the base exerts no restraining tension on the cylinder.
What I'm less sure about is how the tangential strain at height h affects the tangential stress at height h+dh. But glossing over that...
The text is a little sloppy. First, the diagram shows "dh" as the thickness. One of the dashed lines is in the wrong place.
Next, the "Fh" term on the left is wrong. At first I thought it was supposed to be F(h), but now I think it should be Fdh. The author fixed that by making a compensating error, turning the dh on the right into h.
@decentfellow , your own mistake is related. Your T on the left should be TdhΔR.

 

[decentfellow]

Can't we get around that by supposing the base to be some elastic membrane resting on a hard frictionless surface? That should mean the base exerts no restraining tension on the cylinder.
What I'm less sure about is how the tangential strain at height h affects the tangential stress at height h+dh. But glossing over that...
The text is a little sloppy. First, the diagram shows "dh" as the thickness. One of the dashed lines is in the wrong place.
Next, the "Fh" term on the left is wrong. At first I thought it was supposed to be F(h), but now I think it should be Fdh. The author fixed that by making a compensating error, turning the dh on the right into h.
@decentfellow , your own mistake is related. Your T on the left should be TdhΔR.

So that I am not confused can you tell me what is $dh$ and $\Delta R$. And is there anyway I can change the title it was meant to be "Circumferential Tension(not tangent) in the cylinder."

 

[haruspex]

So that I am not confused can you tell me what is $dh$ and $\Delta R$. And is there anyway I can change the title it was meant to be "Circumferential Tension(not tangent) in the cylinder."

In Fig 3E.17, there are three diagrams. The one at the bottom shows an arc of the cylinder wall. The length is dl.
dh is shown, but inconsistently. The arrows are vertical, indicating it is the height (vertical thickness) of the element. Since "dh" implies a small increase in h, that makes sense. But the two dashed lines show it as the thickness of the cylinder wall, which is wrong.
Since the cylinder radius is R, ΔR stands for the wall thickness.
Tension in the wall is like a negative pressure, so it is a force per unit area. If the tension is T then the force on the element is TΔRdh.

 

[decentfellow]

In Fig 3E.17, there are three diagrams. The one at the bottom shows an arc of the cylinder wall. The length is dl.
dh is shown, but inconsistently. The arrows are vertical, indicating it is the height (vertical thickness) of the element. Since "dh" implies a small increase in h, that makes sense. But the two dashed lines show it as the thickness of the cylinder wall, which is wrong.
Since the cylinder radius is R, ΔR stands for the wall thickness.
Tension in the wall is like a negative pressure, so it is a force per unit area. If the tension is T then the force on the element is TΔRdh.

I drew a diagram which I think accurately tells the dimensions that you stated do tell if its correct or not.

For reference let's label it Fig.1

Also, I don't get how the tension is like a negative pressure is it due to the fact that the each layer along the thickness applies force on the one below, I can't seem to visualize how the force acts along the thickness.

 

[haruspex]

I drew a diagram which I think accurately tells the dimensions that you stated do tell if its correct or not.

https://www.physicsforums.com/attachments/111294

For reference let's label it Fig.1

Also, I don't get how the tension is like a negative pressure is it due to the fact that the each layer along the thickness applies force on the one below, I can't seem to visualize how the force acts along the thickness.

Yes, that diagram is better. I used ΔR, not dR, because the d notation is reserved for infinitesimal changes, which will tend to zero in the limit. The thickness of the wall is presumed small, but not infinitesimal.
The use of the word tension in the question is inaccurate. What they mean is stress, specifically hoop stress. Tension is a force, whereas hoop stress is a force per unit area (just like pressure).

 

[decentfellow]

Yes, that diagram is better. I used ΔR, not dR, because the d notation is reserved for infinitesimal changes, which will tend to zero in the limit. The thickness of the wall is presumed small, but not infinitesimal.
The use of the word tension in the question is inaccurate. What they mean is stress, specifically hoop stress. Tension is a force, whereas hoop stress is a force per unit area (just like pressure).

But the book's final answer is a force per unit length. After being confused about what the book wants to say I searched the web and found this http://www.sciencedirect.com/science/article/pii/0026286274900338. In that abstract it defines the circumferential tension as " the circumferential tension is defined as the arithmetic average of tensile forces per unit axial length acting perpendicularly to the section of tube walls cut by a plane passing through the tube axis." (I never opened that link as I thought that it was related to biology, but guess what biology includes mechanics).

In that case the circumferential tension becomes
$$T_{\text{circum}}=\dfrac{\displaystyle\int{\rho Rg\;h\;dh}}{\displaystyle\int{dh}}$$

But after writing this I was confused again whether the average was to be taken till the depth we at which we want to measure the circumferential tension or over the whole of the length of the cylinder. Well, in both the cases the answer is different than the answer given in the book.

 

[haruspex]

the circumferential tension is defined as the arithmetic average of tensile forces per unit axial length acting perpendicularly to the section of tube walls cut by a plane passing through the tube axis

Ok, let's go with that. So if the stress is σ, the circumferential tension (with that definition) is T, and the whole tensile force on the element in the diagram is F, we have:
$F=Tdh=\sigma \Delta Rdh$

 

[malemdk]

The problem has been solved in wrong way as noted by others

 

[haruspex]

The problem has been solved in wrong way as noted by others

Would you mind explaining more?
As I noted in post #8 there are errors in the algebra, and I have a doubt concerning how strain at one height affects stress at meighbouring heights, but so far no one has posted a definitive reason to say the method is wrong.

 

[malemdk]

Since the cylindrical wall experiences pressure gradient, the stress at the bottom is more than at the top
You will find solution in for this problem mechanics of solids,
Thin wall and thick wall cylindrical and spherical vessels
(Lame's equations)

 

[haruspex]

The solution approach in the text takes into account that the stress varies along the length of the cylinder. What it does not allow for is variation in stress through the radial thickness of the wall, i,e, this is the thin wall approximation. Lamé's equation is needed for the thick walled case. See e.g. https://en.m.wikipedia.org/wiki/Cylinder_stress#Hoop_stress.

 

[decentfellow]

Ok, let's go with that. So if the stress is σ, the circumferential tension (with that definition) is T, and the whole tensile force on the element in the diagram is F, we have:
$F=Tdh=\sigma \Delta Rdh$

If we go with that definition then we have,
$$2T\;\sin\left(\dfrac{d\theta}{2}\right)\;dh=P\;dh\;R\;d\theta=\rho\;ghRd\theta\;dh$$
The above equation results due to the equilibrium that is attained when the element has expanded the maximum amount it can. Hence, we get
$$\require{cancel}\cancel{2}T\;\dfrac{\cancel{d\theta}}{\cancel{2}}\;\cancel{dh}=\rho\;ghR\cancel{d\theta}\;\cancel{dh}\implies T=\rho\;ghR$$
Viola here is the answer.

So, to conclude I think we can restate the question as this:-

A cylindrical vessel of height $H$, radius $R$ and thickness $\Delta R$ contains a liquid of density $\rho$. Determine the circumferential tension at aheight $h$, if the circumferential tension is defined as the arithmetic average of tensile forces per unit axial length acting perpendicularly to the section of tube walls cut by a plane passing through the tube axis; Also determine it maximum and minimum values.

Is the Problem Statement somewhat correct now?

 

[haruspex]

If we go with that definition then we have,
$$2T\;\sin\left(\dfrac{d\theta}{2}\right)\;dh=P\;dh\;R\;d\theta=\rho\;ghRd\theta\;dh$$
The above equation results due to the equilibrium that is attained when the element has expanded the maximum amount it can. Hence, we get
$$\require{cancel}\cancel{2}T\;\dfrac{\cancel{d\theta}}{\cancel{2}}\;\cancel{dh}=\rho\;ghR\cancel{d\theta}\;\cancel{dh}\implies T=\rho\;ghR$$
Viola here is the answer.

So, to conclude I think we can restate the question as this:-

A cylindrical vessel of height $H$, radius $R$ and thickness $\Delta R$ contains a liquid of density $\rho$. Determine the circumferential tension at aheight $h$, if the circumferential tension is defined as the arithmetic average of tensile forces per unit axial length acting perpendicularly to the section of tube walls cut by a plane passing through the tube axis; Also determine it maximum and minimum values.

Is the Problem Statement somewhat correct now?

Looks fine.
It is still possible that @malemdk or @Nidum will come back with a sound reason for saying the entire approach is fundamentally flawed. In the real world there would be vertical interaction, i.e. since the greatest stretching will be near the base, the tension at height h+dh will aid the band below it at height h. Maybe that is the objection.

 

[decentfellow]

In the real world there would be vertical interaction, i.e. since the greatest stretching will be near the base, the tension at height h+dh will aid the band below it at height h. Maybe that is the objection.

Vertical interaction?!? As in there would be a tension acting horizontally too? I don't understand what you imply when you say "the tension at height h+dh will aid the band below it at height h"

It is still possible that @malemdk or @Nidum will come back with a sound reason for saying the entire approach is fundamentally flawed.

Let them come and object...I will learn more XDXD.

 

[haruspex]

Vertical interaction?!? As in there would be a tension acting horizontally too? I don't understand what you imply when you say "the tension at height h+dh will aid the band below it at height h"

The approach used in the textbook treats the cylinder wall as though it is a stack of horizontal bands that can slide over each other. In practice, since the base layer is subject to greater stress than the layer above it, it wll expand more. This will lead to tension parallel to the cylinder axis, with the upper layers tending to pull in the lower layers, i.e. help them resist the stress.

 

[decentfellow]

The approach used in the textbook treats the cylinder wall as though it is a stack of horizontal bands that can slide over each other. In practice, since the base layer is subject to greater stress than the layer above it, it wll expand more. This will lead to tension parallel to the cylinder axis, with the upper layers tending to pull in the lower layers, i.e. help them resist the stress.

So there will also be shear stress acting on the differential layer due to the layers above pulling it towards the axis of the cylinder and those below are pulling it away from the axis of the cylinder.

 

[Nidum]

Primary objection was that the wording of the question was so unhelpful .

To analyse this problem properly it is necessary to consider :

(a) Force and deflection compatibility between adjacent hoop sections of the cylinder .

(b) The properties of the cylinder material .

(c) End conditions .

The analysis of this problem can be simplified if :

(d) The cylinder is thin walled .

(e) The pressure loading only changes gradually .

 

[decentfellow]

Primary objection was that the wording of the question was so unhelpful .

To analyse this problem properly it is necessary to consider :

(a) Force and deflection compatibility between adjacent hoop sections of the cylinder .

(b) The properties of the cylinder material .

(c) End conditions .

The analysis of this problem can be simplified if :

(d) The cylinder is thin walled .

(e) The pressure loading only changes gradually .

Okay let's consider that the cylinder is thin walled and has a thickness $\Delta R$ and its Young's Modulus is $Y$, what else would we be needing to make this question salvageable. I am still persistent to solve it(or a rather modified version of it), because I think I have got the gist(though its statement has been cruelly done to frustrate the solver) of the problem and would like to solve the correct version of it to test my understanding of the problem.

 

[malemdk]

Consider a unit length cylinder with inside diameter (d) loaded with pressure (P)
wall thickness (t) , hoop stress (S)
then
P x d = 2xSxt
S = Pxd/ 2 x t
= Px r/t
the stress experienced by the side wall

 

[haruspex]

So there will also be shear stress acting on the differential layer due to the layers above pulling it towards the axis of the cylinder and those below are pulling it away from the axis of the cylinder.

(a) Force and deflection compatibility between adjacent hoop sections of the cylinder .

The pressure loading only changes gradually .

I'm really no expert in this area, but my naive understanding is that the vertical interaction depends on the stiffness of the wall. If very stiff (despite being so thin!) then there is barely any deformation, so no vertical interaction.
If not so stiff, the lower part is distended more than the upper. This will lead not only to shear stress but to a shift in the circumferential stress, with some of it being transferred uowards.

 

[haruspex]

Consider a unit length cylinder with inside diameter (d) loaded with pressure (P)
wall thickness (t) , hoop stress (S)
then
P x d = 2xSxt
S = Pxd/ 2 x t
= Px r/t
the stress experienced by the side wall

I don't understand what this adds to the discussion. Using that simple approach, but a more detailed development, decentfellow already has that answer.