finding equilibrium state of stress at a point in an elliptical pressurized vessel

iqjump123

1. The problem statement, all variables and given/known data

derive the equilibrium state of the stress at a point in a circular pressurized vessel, membrane equation, in terms of σ_a and σ_b and internal pressure b. Figure is picture of a thin walled pressurized elliptical vessel (like a sphere vessel except with horizontal diameter(a) not equal to vertical diameter (b)).

2. Relevant equations

For a sphere, the equilibrium stress is denoted as pr/2t, with r being the radius and t being the thickness.

3. The attempt at a solution

There are plenty of solutions available for the stress involving a spherical vessel, and for a cylindrical vessel, but I haven't seen anything for the elliptical one. Any help will be appreciated!!

nvn

iqjump123: Is this a school assignment or test question?

From a top view, your pressure vessel looks like a circle, correct? And from a side view, your pressure vessel looks like an ellipse with its major (large) diameter (a) horizontal, and its minor (small) diameter (b) vertical, correct?

iqjump123

nvn, thanks for your prompt reply, and happy new year. :)

This is actually a preparation question for an exam I am taking.

your description is almost correct- the pressure vessel from the side view, it looks like an ellipse with major diameter being vertical, minor diameter being horizontal. So like an upright egg shape, so to say.

Thanks!

nvn

iqjump123: Typically, a and b are the ellipse semi-axes. Are you required to use a and b as the ellipse diameters? If not, you might want to change a and b to the ellipse semi-axes. The vertical semi-axis is b.

The y axis is vertical; the x axis is horizontal. The origin of the xy coordinate system is at the ellipsoid centerpoint. The xy plane creates a vertical section cut passing through the ellipsoid centerpoint. You only need to consider the xy plane (a 2-D object) for this derivation, not a 3-D object. Therefore, the 2-D object on the xy plane is an ellipse, with the xy coordinate system origin at the ellipse center. This ellipse is a meridian of the ellipsoid.

(1) Expressions for the principal radii of curvature r1 and r2 will be required. Therefore, first derive r1 and r2.

We cannot give you the relevant equations, nor do the homework for you. We can only check your math. But I will give you this small hint, just to get you started on item 1. Hint 1: Write the equation of the above-mentioned meridional ellipse (in terms of a, b, x, and y), and solve for y, because you will need the equation of the ellipse, y, as a function of x, to use in the derivation of r1 and r2.

iqjump123

Thanks nvn. To clarify-I never did and do not intend to just get answers nor have anybody do the work for me on this forum. I just want to get the help I need to learn. After all, the reason I am solving these problems is to get myself ready so I myself can solve them in the exam.

Back to topic.
It is a little confusing how this can be viewed as a 2-d problem- the expression we will get for the principle radius, I presume will be involving the z axis (in the page, out of the page). Would it be that after we get the principle radius, we won't have to worry about the z axis?

Anyways, I went ahead with your hint and derived the equation of an ellipse point with a and b as the semi axis made the equation to be

x^2/a^2+y^2/b^2=1.
solving this for y will give the respective coordinates for any x(which i didn't do here but is trivial).

so then for r1 involving the minor axes (a-a), the principle radius r1 will be just a?
then for r2 involving the major axes (b-b), assuming that the angle is known,

r2=[√{(1-x^2/a^2)b^2}-x)]/sin(2), where 2 indicates phi, the angle.

I included the image of the figure diagram given in the problem.

any further help will be appreciated.

Attached Thumbnails

 

nvn

iqjump123: The given problem can be viewed as a 2-D problem, because it is axisymmetric (identical) about the y axis. Therefore, you only need to solve a 2-D problem, involving one meridional ellipse (on the xy plane), as a function x or angle phi. You do not need angle theta.

The principal radii of curvature r1 and r2 are not the ellipse semi-axes. Parameters r1 and r2 are the principal radii of curvature of the ellipse at any angle phi (for r2) and delta-phi (for r1). Principal radius r1 is the ellipse radius of curvature (in the meridian plane) at any point. Principal radius r2 generates the shell surface in the direction perpendicular to the direction of the tangent to the meridian curve.

If you are not familiar with principal radii of curvature r1 and r2, study membrane stress on any surface of revolution (or shell of revolution), to learn what r1 and r2 are. And this will explain r1 and r2 better than I can. Afterwards, derive r1 and r2.

iqjump123

Yes it feels I will need more background information in order for me to solve this problem. thanks for the help.

If you can recommend in any of the textbooks, articles to help me understand this, it will be helpful.

nvn

iqjump123: Another way to say it is, principal radius r2 is the radius of curvature that generates a circle in the xz plane, at any x coordinate. Principal radius r1 is the radius of curvature of the meridional ellipse (in the xy plane), at any x coordinate. First, derive r1 and r2, as a function of x.

nvn

iqjump123: By the way, you do not need to hit the Quote button each time you reply. Typically, just hit the "New Reply" button at the bottom of the thread page, when you reply.
Yes, very good. Now, solve this for y.
No, good try, but this is wrong. (I do not understand how you derived this, but regardless, it is incorrect.)

Because you have not made too much progress yet, I will go ahead and give you a big hint for deriving r2, just to get you started. Deriving r1 is more difficult than deriving r2; therefore, you can derive r1 later, as a separate task. First, derive r2, since it is slightly easier. Hint 2: See Fig. 1 in the attached file. Now, see if you can derive r2, as a function of x. Show your work, if you want us to check your math.

Attached Thumbnails

 

iqjump123

hello nvn, I appreciate the help, as always.

I have one question about the figure. I noticed that there is an arbitrary value of h, and the point c. I presume that is the line that is intersecting at the axes. Can I assume that any of these values (h, location of pt c, angle phi) is known to me? Without either of these information available, I am lost on how to obtain the r value.

My earlier calculation was based on if our pivot point was on the origin, instead of if it was in an arbitrary point, such as c, shown in your figure. I used the equation I derived for the ellipse to come up with the radius. However, it seems that we can't assume that the principle radius intersects with the origin at all times, and therefore this method won't work.
I included a pdf attachment of the math I went through to obtain the answer as an attachment.

Thanks again.

Attached Files

My approach_prob2.pdf (103.3 KB, 4 views)

nvn

iqjump123: Variables h, phi, and point C are not known. They are variables, not constants. They are functions of x. You must write expressions for these unknowns, as a function of x.

Notice, in Fig. 1 in post 9, principal radius of curvature r2 always starts on the y axis, always ends at the ellipse, and is always normal (perpendicular) to the ellipse. Radius r2 generates circles in horizontal planes (because this is a surface of revolution).

Your radius r in your attached file in post 10 is not a radius of curvature. You can see, it is not normal to the ellipse.

Derive the length of radius r2, as a function of x, using Fig. 1 in post 9. Hint 3: Use trigonometry. Hint 4: Notice the pink line in Fig. 1. What does it tell you, and how can you use that to derive the length of r2, as a function of x? Show your work, if you want us to check your math.