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How can I calculate the stresses in corners where different geometries meet?

This image is a quick example of the positions i am referring to.

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Additional information

In class we studied the membrane model for thin walled pressure vessels, and we used that to calculate $\sigma_m$ and $\sigma_t$ in a random location along the cylindrical and conical profile.

We didn't mention of a way to calculate stresses in proximity of changes in profile, and this annoys me a bit as I have no knowledge of the material behaviour there, but due to the abrupt change in geometry I guess there will be stress concentrations there.

First I attempted to model the transition zone as a corner with a small fillet, as to remove the discontinuity, and analyze that, but I got quickly lost in math as the volume of revolution was not so easy to calculate.

Then I assumed the pressure constant along the fillet, as it's small compared to the whole vessel, but I got constant stresses back, so I guess that's not the right answer.

I checked Peterson for stress concentrations and found nothing, same for Roark's. The only place where I found something is on Strength of materials by Timoshenko, but he explains the answer using a formula from plates in bending, a topic which I have yet to cover, so I am not comfortable with plate theory yet.

So the question is:

How can I calculate the stresses in corners where different geometries meet?

Eventually, are there simpler paths other than plate theory or am I forced to learn that before understanding how to calculate these stresses?

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1 Answer 1

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Junction between cylindrical and conical shell

Simple conservative solution

You can use longitudinal membrane forces and put them into equilibrium in vertical direction. Cylindrical shell has the whole longitudinal force $N_{\vartheta V}$ in the vertical direction, but in case of conical shell, you have to split the longitudinal force $N_{\vartheta K}$ into vertical and radial components.

enter image description here

As you can see, $N_{\vartheta K}\cdot \cos(\vartheta)$ has to have the same magnitude as $N_{\vartheta V}$. From membrane theory perspective, the radial force $N_R = N_{\vartheta K}\cdot \sin(\vartheta)$ is a force with which the conical shell is trying to deform edge of the cylindrical shell and membrane shell has no local radial stiffness to resist this. But you can design a ring stiffener just for this purpose.

In case of internal pressure $P$ and using mean cylindrical shell diameter $D_m$, the $N_{\vartheta V}$ force is: $$N_{\vartheta V} = \frac{P\cdot D_m}{4}$$

It is also true, that: $$N_{\vartheta V} = N_{\vartheta K}\cdot \cos(\vartheta) = N_R\cdot \frac{\cos(\vartheta)}{\sin(\vartheta)}$$

Therefore: $$N_R = N_{\vartheta V}\cdot \tan(\vartheta) = \frac{P\cdot D_m}{4}\cdot \tan(\vartheta)$$

A ring stiffener with cross-section $A$ can be welded to the junction. In case of internal pressure, it will be compressed and the compressive stress $\sigma_c$ can be calculated from the radial force equilibrium (this is similar to calculating hoop stress in cylindrical shell):

$$2\cdot \sigma_c \cdot A = N_R\cdot D_m$$

So the compressive stress will be:

$$\sigma_c = N_R\cdot \frac{D_m}{2A} = \frac{P\cdot D_m^2\cdot \tan(\vartheta)}{8A}$$

If you put a transition zone defined by radius there, it is more complicated, but this zone will also have to be able to resist the radial force. And as you decrease the radius, the force does not change much, but the size of the zone does, so the membrane stress there rises up to infinity for zero radius.

Reality

In reality, the shell junction does have some resistance against the radial deformation, so you can take this into account. Derivation is quite complex, but important thing is, that the length of the shell you can include into the stiffening area is calculated using formula:

$$L = k\cdot \sqrt{e_n\cdot D_m}$$

where $e_n$ is nominal thickness and $k$ is a factor, 1.4 in this case I think. So the required additional stiffening area is less than calculated in the previous solution and it can take a form of local thickness increase instead of ring stiffener. Transition zone defined by radius also helps.

This is used for example in vertical cylindrical storage tank design (API 620), where you have a junction between the conical roof and the cylindrical shell. Pressure vessels could also be based on this, but more advanced techniques are used in some standards.

Point on the axis

Regarding the point on the axis, the stresses there would be basically 0. The vessel would probably not end with a sharp spike like this.

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  • $\begingroup$ Thank you for your answer, I like this method as it ties to what I already know. I want to clarify some things: - Does the weld on the ring need to be air/watertight, so the pressure won't act on the other side of the ring as well, or is it only a matter of the bending force being transmitted through welds? -Can I model the reinforcing ring as a curved beam as well, by maybe taking half a circle and loading it with a distributed load representing the radial force? -Where can I find more about the formula that gives you the length to include in the stiffening area (the last formula)? $\endgroup$ Jun 16, 2023 at 9:12
  • $\begingroup$ The ring is used only for low pressure applications and it would be welded from the outside on already welded junction, so there is no reason for the stiffener weld to be air/water tight. There should not be much bending force transmitted through the welds, although the junction deformation also includes a little bit of rotation and if you axisymmetrically rotate the ring stiffener, the internal stress response is very similar to bending of straight beam. You can model the ring as you propose, although the analysis should lead just to axisymmetric deformation. $\endgroup$ Jun 17, 2023 at 19:07
  • $\begingroup$ The last formula may be the most common in many engineering standards, although I have never seen its origin story. I think you can see it if you look at formulae for axisymmetric bending of steel cylindrical shells. Effects of local loads on the shell fall down very quickly as you go further away from the load. $\endgroup$ Jun 17, 2023 at 19:15

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