Roark's Circular Ring Formulas

[JStephen]

See what you all think about this-
In Roark & Young's Formulas for Stress and Strain, they give stress and deflection for circular rings. In the 5th Edition, that's Table 17, page 220. In the discussion prior to the table, they say, "By superposition, these formulas can be combined so as to cover almost any condition of loading and support likely to occur."

So suppose you have a circular ring with uniform outward radial load, and apply any number of equally spaced radial loads to it. Deflection for the uniform load can be handily calculated, and is simply a uniform growth of the ring. Deflection due to the equally spaced point loads is from Load Case 7. So far, so good. Load situation is shown in the attached sketch, Figure A, and predicted deflection in Figure B.

The catch is that the uniform outward radial load will tend to hold the shape round, but this is not reflected in the formulas for Load Case 7. IE, the ring with the uniform outward load should be much stiffer than predicted by Load Case 7.

Note that the uniform outward load could be replaced by a large number of outward point loads equally spaced and produce similar results.

Two questions, then:
Where exactly is the Roark formulation getting off track? It would seem they are making additional assumptions not stated.
How DO you calculate the deflection in this case?

 

[fegenbush]

It seems that the point loads (4 in your attached sketch) would cause the cylindrical shape to act similar to a number of fixed-fixed uniformly loaded curved beams. The presence of the point loads disrupts the geometric stability of the circular shape. Without the point loads, the stress (and deflection) are primarily membrane. With the point loads, one must also consider bending stress in addition to the membrane stress. Thus, the deflection would increase.

When in doubt, break out Timoshenko and your favorite differential equation book.

 

[rb1957]

1) i don't think Roark is "off track" ... you superimpose the different loadings

2) superimpose the deflections from the different loads

Quando Omni Flunkus Moritati

 

[JStephen]

rb1957, take a look at the deflected shape shown in my Figure B. If you apply uniform internal pressure to that deflected hoop, will it change shape and become rounder?

Fegenbush, the formulas given are primarily based on bending already.

 

[rb1957]

ok, i don't think Roark includes pressure loads (i guess the engineers at Peenemünde could solve this easily, not being redundant).

i guess it's either ...
1) react the pressure with hoop stress and uniform radial displacement (small deflection theory)
2) (large deflection theory) apply the pressure to the deformed shape and get a (presumably) non-uniform axial load in the ring

Quando Omni Flunkus Moritati

 

[dhengr]

Jstephen:
I think this depends a lot on the magnitudes of the loads and the dia. and thickness of your 24-30" dia pipe, or 40' dia. tank. It might be better to use load case #1, and apply it twice. It might be instructive to compare the results of load case #1, applied twice, to load case #7 and see if you get the same general results. I would calc. and tabulate the stresses and deflections at 15̊ intervals around the circle, for each of my three load cases, then add them algebraically (superposition). As you said, the internal pressure problem is pretty straight forward. Then do case #1 with two loads at N & S; if you rotate this 90̊, you’ll have the E & W results too. I’d have to reread that section of Roark to comment further. It has been a long time since I’ve used that section. I do have the 5th Ed. so we would be looking at the same equations and pages, if we go further. I don’t see any calcs. or results in your attachment, so I’m not real sure what you think the problem is. Your Fig. B shows the right shape, but probably much too exaggerated. The ring will not really get stiffer because of the internal pressure. It will be uniformly stressed and expand in dia., E or EI or EI & t/d don’t change significantly, but I guess the internal pressure does offer some resistance to some forms of buckling.

 

[JStephen]

Load case 7 is for N loads uniformly spaced, so it can be applied directly.

Instead of internal pressure, you could apply the same load case with, say, 300 uniform loads, and get similar results. So it's not just a pressure vs point load issue.

A 3-D analogy of this would be a car tire, where you in fact depend on the internal pressure to increase the stiffness to any desired extent. You could calculate deflections for an uninflated tire subject to forces (such as the ground), calculate deflections due to internal pressure, and superimpose those, but the result wouldn't really be meaningful.

Where I'm running into this is calculating point deflections by Case 7 and wondering if that calculation is really meaningful at all with internal pressure.

 

[rb1957]

have you plotted increasing numbers of smaller point loads, to see if the deflection trends to the pressure calc;
and the internal moments should trend to zero ?

i'd go back to more basic theory, our favourite irish stressman (tim o'shenko), as Roark is only talking about maximum reactions/deflections, i think you need the general solution for a pair of opposite forces, and superimpose and it should tend toward the hoop stress solution. The rings as analyzed consider axial load, shear and in-plane moment; under internal pressure (= hoop stress) the shear and moments are zero, and only axial load = hoop is reacting the applied pressure.

another way to look at the problem, is to apply pairs of loads P. the first pair stretches that the loads and compresses (?) at the 90deg points. now if you apply P at the 90 deg points at four points the deflection will be the large +ve stretch due to the load, and a -ve due to the other load, ie the deflction is less than the single pair, which is inline with your intuition that pressure loading should be stiffener.

Quando Omni Flunkus Moritati

 

[prex]

As already stated above, with small deflection theory the superposition of effects is of course applicable, and it is entirely valid for this case.
And also you get with this the stiffening effect due to pressure: if you only apply the radial loads, you get, say, a displacement d under the load. Now you apply also the pressure, but the displacement is no more d, might be zero or even outwards. If you want to have the same displacement as before, you need to increase the radial loads, so you could say that the ring is now stiffer due to the internal pressure.
The example of superposition with tires doesn't work because deflections are generally not small there, and, more importantly, in plane stiffness is unrelated (and much higher) to bending stiffness in a tire because of the metal reinforcing mesh: so the internal pressure allows for a shell effect to develop, that wouldn't intervene otherwise. In a similar reasoning with the rubber tube alone, you get back to the same conclusion as for steel, provided you stay with small deflections.

prex
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[JStephen]

Consider a linear situation that seems to me to have similarities.

See the attached sketch. Suppose first off, you have a completely flexible cable under initial tension T, and you apply a force P at mid-span. (This is different from the cable problem posted a day or two ago, in which the cable had zero force initially. This is the guitar-string version.)
The cable deflects an amount δ.
Equating forces, we get 2Tsinθ = P
For small angles/deflections, take sinθ = tanθ = δ/(L/2)
Then get 2T δ/(L/2) = P, or P/δ = 4T/L
IE, in this situation, there is a stiffness k of 4T/L even for small deflections of the cable. (Note that increased cable tension due to deflection has been neglected here.)

Now, suppose instead this is just a slender beam. From standard beam deflection equations, P/δ = 48EI/L^3.

But, suppose this is a slender beam with high axial loading and a lateral load. How do we combine the loading? The effects of the axial tension alone are just a stretching in the axial direction. The effects of the bending alone are just the 48EI/L^3 deflection. But, if that 4T/L stiffness is comparable to the 48EI/L^3 stiffness, the deflection result will be off, possibly way off, even for small deflections.

 

[rb1957]

sorry but wtf ? a cable with transverse load is nothing like a ring ??

Quando Omni Flunkus Moritati

 

[prex]

A cable or a beam under tension are two examples of structures that do not have a linear solution, hence superposition does not work there.
A ring instead has linear solutions, because it's not straight. The curvature allows for in plane strain to work against bending and indeed there is in plane strain in the bending solution (unlike in a beam).

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[JStephen]

Look at it from another angle. See attached sketches.

For uniform pressure, you have the situation shown in Figure A. Summing forces in the Y direction should equal zero in this case, the tension in the ring times the sine of that angle will match the Y-component of the pressure force.

For a concentrated load, you have the situation in B. Summing forces in the Y direction should equal zero in this case as well, the shear and tension in the ring times the sine/cosine of the respective angles will match the applied force. It’s not obvious, but I believe the derivation assumes that the original angles are used in this case.

Figure C shows the deflected shape with the concentrated load. The tension and shear forces are of the same magnitude as the concentrated load. This being the case, the sum of the loads will still approximately equal zero if the deflection angles are “small”.

Figure D shows the deflected shape with both concentrated load and uniform pressure. For the forces to approximately sum to zero in this case, the product of the deflection angles and hoop tension must be “small”. Even if the deflection angle is “small”, if the hoop tension is “large”, forces will not balance. Since the deflection angle is from one load case and the hoop tension from another load case, they are not connected, and you can have results that are not at equilibrium even with actual deflections that are “small”. IE, the derivation doesn’t assume that deflections are small, it assumes that the product of deflections and hoop tension are small, but this is not one of the stated limitations. To bring figure D back into equilibrium, it is necessary to reduce the deflection angles- IE, the ring must deflect less with internal pressure present- even with small deflections.

 

[rb1957]

basic structural analysis assumes small deflections, that deflections don't change the structure; in your case this means applying the pressure to the undeflected ring.

large deflection theory says deflections change the structure. in your case this is applying the pressure to the deflected ring.

there are also applications (like your cable red-herring) that are non-linear and so superposition doesn't apply.

Quando Omni Flunkus Moritati

 

[paddingtongreen]

I disagree with: "The catch is that the uniform outward radial load will tend to hold the shape round" What drives this statement?

From first principles: As long as we neglect the local effects of the load application, the uniform/closely spaced loads do not cause a bending moment in the ring, just pure tension. There is no reason to think they will hold the ring in shape after other loads are applied.

The four load case will cause a continuously variable moment around the ring, tension on the inside at the load points and on the outside between load points. It will also cause a varying axial compression, equal to half an applied load at the load point and √2 times an applied load midway between. As long as the stresses remain in the elastic range and the thing doesn't buckle, you can superimpose.

The ratio of the loads has a great effect. If each single load placed at the four quadrant points approaches the value of the continuous force on one quarter of the ring, the axials almost disappear.


Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[SAIL3]

Internal pos press does stiffen the ring and increases it's stability. For the majority of cases, stability only becomes an issue when there are compressive forces present. In the OP's case, one can consider the membrane stresses from internal pressure as diminishing or, if high enough, canceling out the compressive component that is present from the moment due to the conc loads.All well and good, but, the questions raised in this post has left me with some doubt on the validy of combining the results of these loads or load cases.
For example: assume that the 4 conc loads are applied first and causes the ring to deflect and as a result is not circular anymore. Then start applying internal presure to this deflected shape. Can I still use the standard theory to obtain the membrane stresses in the ring even though it is not circular anymore?.As was pointed out that it may be only valid for small deflections. For some reason(maybe ignorance) this does not concern me as the final state of the combined stresses will always be more stable than the initial 4-point load alone and all I have to worry about is the max combined stress. How I got to that final state is extremely complicated theoretically.

 

[paddingtongreen]

The pressure works in the same way as a sail fills when in the wind, but the material itself is not stiffened.

I think that it does not matter which order the loads are applied.
If you take Fig. B, (the deformed shape shows an exaggerated picture of the elastic deformation) the pressure on the four "flats" is not radial, it is perpendicular to the surface, it will deflect the flats in the center, rounding them; it will push the flats apart, thus straightening the corners. Both the corners and the flats move towards the original position until the superposition is complete.



Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin

 

[JStephen]

" it will deflect the flats in the center, rounding them; it will push the flats apart, thus straightening the corners. "
Note that the principle of superposition says that adding uniform pressure will not change that deflected shape- the whole thing will just get larger, per superposition.

 

[prex]

Again: Roark's formulae and the principle of superposition, cited from Roark in the original post, are generally valid for linear elasticity, that is for small deflections (with other possible limitations related to geometry, constraints and material).
As many ways of reasoning above are clearly with large deflections, then is no use questioning the Roark, where this discussion started from, and of course a non linear large deflections approach must be used.
But now, which is the question we are discussing about?

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[paddingtongreen]

"Note that the principle of superposition says that adding uniform pressure will not change that deflected shape- the whole thing will just get larger, per superposition."

I agree, I was responding to SAIL3 who worried about the effect of the point loads being applied first. I had the exaggerated drawings in mind and that did make it seem as though I left the realm of small deflections..

Michael.
"Science adjusts its views based on what's observed. Faith is the denial of observation so that belief can be preserved." ~ Tim Minchin