Calculating Internal Pressure in Glass Tubes and Spherical Flasks

[Saitama]

If T is the tensile force per unit length, and the width of the shell is Δr, what is the tensile stress σ in the shell? What is the tensile stress in terms of P, Δr, and r? You need to determine how this tensile stress compares to the ultimate tensile stress σ_m.

I am not sure but the tensile force acts on both sides of the sphere (the inner surface and the outer surface), right?

If Δr is thickness, then the area of the thickness of the spherical cap I selected is $2\pi R\sin\theta Δr$. The tensile stress is 2T divided by this area, am I right?

 

[Chestermiller]

I am not sure but the tensile force acts on both sides of the sphere (the inner surface and the outer surface), right?

No. It acts within the shell metal. It is perpendicular to the radius at each point.

If Δr is thickness, then the area of the thickness of the spherical cap I selected is $2\pi R\sin\theta Δr$.

No. The force per unit length T you calculated is oriented tangent to the shell cap you selected. To get the stress, all you do is divide T by Δr. That will, of course, give you units of stress. You see, this is why I suggested you split the shell into hemispheres. Please try doing the problem using hemispheres. It is more clearcut that way.

The tensile stress is 2T divided by this area, am I right?

No. It's only T divided by Δr.

Chet

 

[voko]

If Δr is thickness, then the area of the thickness of the spherical cap I selected is $2\pi R\sin\theta Δr$. The tensile stress is 2T divided by this area, am I right?

Your $T$ is already force per length. If you divide that by area, you will get force per volume. Is that dimensionally correct?

 

[Saitama]

No. The force per unit length T you calculated is oriented tangent to the shell cap you selected. To get the stress, all you do is divide T by Δr. That will, of course, give you units of stress. You see, this is why I suggested you split the shell into hemispheres. Please try doing the problem using hemispheres. It is more clearcut that way.

Okay, I do it for the hemispheres. :)

By force balance, I get $P\pi R^2=T(2\pi R) \Rightarrow 2T/R=P$

Also, $\sigma_m=T/\Delta r$. From these two equations, we have $P=2\sigma_m \Delta r/R$ which is the correct answer. Have I done it correctly? What about my spherical caps? They too give the same equation as hemispheres.

Your $T$ is already force per length. If you divide that by area, you will get force per volume. Is that dimensionally correct?

Yes, very silly of me, thank you voko!

 

[Chestermiller]

Okay, I do it for the hemispheres. :)

By force balance, I get $P\pi R^2=T(2\pi R) \Rightarrow 2T/R=P$

Also, $\sigma_m=T/\Delta r$. From these two equations, we have $P=2\sigma_m \Delta r/R$ which is the correct answer. Have I done it correctly?

Yes. Nice job.
I would have proceeded a little differently by first calculating the stress σ for any arbitrary value of the pressure P:
σ=\frac{Pr}{2Δr}
To find the pressure at which the shell fails, I would then set the stress equal to the critical stress:
σ=\frac{Pr}{2Δr}=σ_m
This is the sequence of steps typically used in the design of more complicated mechanical structures to assure that the structure does not fail under load. Of course, for this simple example, it didn't matter.

What about my spherical caps? They too give the same equation as hemispheres.

Yes, as they should. But doing it with the hemispheres is much easier to see and to implement.

 

[haruspex]

What about my spherical caps? They too give the same equation as hemispheres.

Which tells you that all failure modes in which one fragment is a cap are equally possible. There is no worst case amongst them. It would be interesting to go on to analyse other shapes.

 

[Chestermiller]

Which tells you that all failure modes in which one fragment is a cap are equally possible. There is no worst case amongst them. It would be interesting to go on to analyse other shapes.

In solid mechanics, the conditions for failure are usually quantified in terms of the three principal stresses. In a typical failure criterion, failure will occur if some specific symmetric function of the three principal stresses exceeds some critical value. For example, the failure criterion is often expressed in terms of the sum of the squares of the differences between the principal stresses.

In this particular problem, the state of stress is transversely isotropic, with the principal stresses equal to σ, σ, and 0, and this state of stress applies uniformly throughout the spherical shell. The principal stress in the radial direction is 0, and the principal stresses in the latitudinal and longitudinal directions within the shell are both equal to σ everywhere in the shell. This means that failure can begin anywhere in the shell (as might be expected from the symmetry). In an actual spherical shell, failure will occur at the location in the fabrication that is statistically weakest.

 

[AlephZero]

Sure it's easier, but isn't it making an assumption as to which is the critical case? I agree it's intuitively likely, but it needs to be justified.

For a sphere, it is obvious by symmetry that the state of stress is independent of the latitude and longitude of the position on the sphere. So for an isotropic material, the failure is equally likely to occur anywhere.

So, you can cut the sphere up any way you like, to make the math easy. As Chestermiller said, cutting it into two hemispheres is about as easy as it gets.

FWIW I've seen job applicants attempting similar problems in a technical interview situation, and they often start by taking an arbitrary sized piece of shell like Voko suggested, and then tie themselves in knots with the trigonometry.

 

[AlephZero]

The principal stress in the radial direction is 0, and the principal stresses in the latitudinal and longitudinal directions within the shell are both equal to σ everywhere in the shell.

That is only an approximation. From the boundary conditions on the inside and outside surface, the radial stress component at the surface must equal the pressure. So the radial stress varies through the thickness and is of order of magnitude P.

But as the OP has found, the other stress components are of order of magnitude PR/t. Since R/t is likely to be much greater than 1, the radial stress can be ignored.

 

[Chestermiller]

That is only an approximation. From the boundary conditions on the inside and outside surface, the radial stress component at the surface must equal the pressure. So the radial stress varies through the thickness and is of order of magnitude P.

But as the OP has found, the other stress components are of order of magnitude PR/t. Since R/t is likely to be much greater than 1, the radial stress can be ignored.

Thanks AlephZero. Actually, I was aware of that, but I wanted to keep things simple. This is a so-called Strength of Materials thin wall approximation that is widely used when the wall thickness is small compared to the radius. (Actually the radial stress component at the surface is equal to minus the pressure, if you are using the usual solid mechanics convention in which tensile stresses are regarded as positive).

 

[haruspex]

For a sphere, it is obvious by symmetry that the state of stress is independent of the latitude and longitude of the position on the sphere. So for an isotropic material, the failure is equally likely to occur anywhere.

Being equally likely to occur anywhere is not the same as all fragment shapes/sizes being equally likely.

Edit:
Not sure if this is what you had in mind, but it's just dawned on me that the discussion of fragment shape can be avoided.
A break has to start somewhere, so just consider a short straight fracture. This will effectively be a short arc of a great circle, length Δs, say. Since the force transverse to the great circle is Pπr², the force across the short arc is PrΔs/2. Whether the fracture continues around the great circle or a cap or whatever is irrelevant.
If that's what you and Chet have been trying to tell me, I apologise for being slow to get it.

 

[voko]

FWIW I've seen job applicants attempting similar problems in a technical interview situation, and they often start by taking an arbitrary sized piece of shell like Voko suggested, and then tie themselves in knots with the trigonometry.

In this particular case, you can't avoid trigonometry in the hemispherical case unless you somehow know that the total force on the hemisphere is equal to the total force on a disk of the same radius, which is "obvious" only when you have some experience. If you don't know that, you still have to go through trigonometry and integration. It is indeed easier to evaluate the total tensile force at the edge of a hemisphere, but most people have problems with setting up the integral of pressure, so the simplification is marginal.

 

[Chestermiller]

In this particular case, you can't avoid trigonometry in the hemispherical case unless you somehow know that the total force on the hemisphere is equal to the total force on a disk of the same radius, which is "obvious" only when you have some experience. If you don't know that, you still have to go through trigonometry and integration. It is indeed easier to evaluate the total tensile force at the edge of a hemisphere, but most people have problems with setting up the integral of pressure, so the simplification is marginal.

You don't have to integrate the pressure over the hemisphere if you include the gas in the hemisphere as part of the free body.
Chet

 

[voko]

You don't have to integrate the pressure over the hemisphere if you include the gas in the hemisphere as part of the free body.

I am not debating that you cannot go without integrating. Yes you can. In fact, I do not even know how Pranav got the force due to pressure.

My point is that dealing with a spherical cap (and a hemisphere is a special case which I did not rule out) is quite straight-forward and requires no further arguments as to its validity. Ditto for pressure integral over the spherical cap. Any other shortcuts are not necessarily obvious and may not even seem correct.

On the other hand, a review of these shortcuts is fully appropriate now, after the result was obtained.

 

[Saitama]

Nice discussion everyone but its quite above my level. :shy:

In fact, I do not even know how Pranav got the force due to pressure.

Umm...I think its common to use the projection area or should I show the integrals? :)

I was thinking upon the tube case. As in the case of sphere, I selected a hemisphere, for this case I select a half cylinder. Let the tension force per unit length be T. Also, let R be the radius of tube and l be the length of tube. Then force due to gas pressure P is $P(2Rl)$. Balancing with the tensile force, I get: $T(2l)=P(2Rl) \Rightarrow P=T/R$. At critical case, $\sigma_m=T/\Delta r$. From the two equations, $P=\sigma_m \Delta r/R$.

For the circular disk, the force on it due to gas pressure is $P(\pi R^2)$. Balancing this with the tension force, I get $P(\pi R^2)=T(2\pi R) \Rightarrow P=2T/R$. Substituting $T=\sigma_m \Delta r$ gives $P=2\sigma_m \Delta r/R$. This is definitely greater than $\sigma_m \Delta r/R$ and hence, should be the answer but its not.

I would have proceeded a little differently by first calculating the stress σ for any arbitrary value of the pressure P:
σ=\frac{Pr}{2Δr}

How do you get this formula? It looks that it is the rearranged form of the equation I wrote but it seems like this is a common formula because of the way you state it, or am I missing something obvious?

 

[voko]

This is definitely greater than $\sigma_m \Delta r/R$ and hence, should be the answer but its not.

So you got two pressures that the tube can support at the same stress: higher at the end and lower in the middle. So what is the pressure that the tube can support as a whole?

I think its common to use the projection area or should I show the integrals? :)

Not really, but it is good that you know this.

 

[Saitama]

So you got two pressures that the tube can support at the same stress: higher at the end and lower in the middle. So what is the pressure that the tube can support as a whole?

I feel like an idiot now.

Thanks a lot voko!