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Beam with cantenery action

Beam with cantenery action

Beam with cantenery action

(OP)
We have a project where we are checking a long steel tube that is on top of a steel plate wall - essentially the tube is the wall cap that goes around the perimeter (4 walls) of a steel tank.

The tank is filled with water and imparts lateral pressures on the tank's four walls. This creates an outward uniform force on the tube on all four sides.
The tank is about 10 ft. wide x 36 ft. long.

The tubes will naturally bend outward and have axial tension in them due to the orthogonal walls with their outward uniform load.

The question is - we need to determine the stress in the long 36 ft. tube due to axial, bending and perhaps any axial effect from what appears to maybe be a cantenary effect. How to do this?

Roark's has essentially a case where a beam is rigidly fixed at the ends with a uniform load on it.
It is labeled as Table 8.10, item 3 (Beams restrained against horizontal displacement at the ends - ends pinned to rigid supports, uniformly distributed transverse load on entire span.

However, the end walls of this tank are not perfectly rigid and will bow inward a bit with the cantenary tension on the long tube, right? So this will have the effect of reducing the axial tension in the tube a bit. Roark's does NOT have an item with axial spring supports at the ends.

We've modeled this in RISA 3D and get results but something tells me that the RISA results may not include non-linear effects of the cantenary action in this tube due to variations of the end supports with regard to axial loads.

I seem to understand that RISA does PDelta analysis but that only includes PDelta effects perpendicular to the member and does not include axial PDelta effects.

Any thoughts on this? I've inserted essentially the beam problem in a sketch.

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RE: Beam with cantenery action

Thats tricky. I'm not too familiar with RISA, but I suspect that you won't be able to treat the member as a beam (moment and shear) and a cable (tension due to cantenary action) at the same time. This is probably true for most analysis programs.

Some other thoughts - if you develop enough deflection to get true cantenary action, you're probably exceeding the "small deflections" assumptions in basic beam theory that you would use to get your moments and shear. Also, if you get enough deflection for cantenary action, are you going to have serviceability issues?

Are the tubes fixed to each other at the corners? That will add some moment at the ends of the beam in your sketch.

I think I would approach this problem by looking at it is a bending member only and check deflections. If the deflection was excessive and exceed the "small deflections" assumption, then I would treat it as essentially a cable and ignore the beam effects. Obviously the true answer lies somewhere in between those two methods, but that is beyond my analytical capabilities.

RE: Beam with cantenery action

1) use of "cantenery" is not correct. Catenary means deflections under self weight (only).

2) why go to the double cantilever beam ? why not simply supported as a (presumably) conservative model for a beam with a UDL ?

3) what would be the UDL ? presumably the loading is an enforced displacement (I can't see this rod reacting much of the applied load). Maybe Roark has this deflection in his plates section. His reference for plates is work done by Moody as part of the Corp of Engineers ... google "moody rectangular plates" (seriously).

another day in paradise, or is paradise one day closer ?

RE: Beam with cantenery action

Hmm. One possibility for a hand calculation:

1) Disregard wall below the HSS acting as a shear wall of sorts.

2) Calculate deflection in the HSS based on pure bending and uniform load.

3) Switch the beam model to a pair of rigid bars connected in the middle with a pin joint.

4) Use the tension that you know exists at the ends, and the deflection from #2, to calc a faux point load at mids-span that keeps the two bar model in equilibrium under the imposed displacement.

5) Switch back to the beam model, apply the faux load from #4 reversed, and calc some deflection the other way.

6) Go back to step #3.

Hopefully, a few iterations through steps three to six will close with a reasonable amount of effort. If that happens, you should have an equilibrium condition that you can design to and is probably about as accurate as anything else.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

RE: Beam with cantenery action

(OP)
rb1957 - cantenary is deflections under gravity....agree.
The tube is simply deflecting under a "similar" uniform load - whether it is gravity due to self weight or lateral load due to constant water pressure the tube "doesn't know".
So it is a similar problem - a long skinny, relatively flexible tube under constant load yes?
Or does the gravity load have a partial axial component when the "cable" droops?

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RE: Beam with cantenery action

(OP)
KootK - the wall below is a deck-like wall that would flex like an accordion with axial strain changes in the tube so definitely did ignore any possible shear wall effect.

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RE: Beam with cantenery action

Back to the structural analysis:

Quote (CANPRO)

I suspect that you won't be able to treat the member as a beam (moment and shear) and a cable (tension due to cantenary action) at the same time. This is probably true for most analysis programs.

Any (half-decent) finite element analysis program that allows a beam element to resist an axial force and also purports to do a geometrically non-linear analysis should be able to capture this behaviour. The continuous updating of the element stiffness as the beam deflects is central to the idea of geometrically non-linear analysis and this should result in your structure transitioning from resisting loads through bending action to cable action.

Perhaps you haven't subdivided your mesh enough to capture this deformation? Also be careful with load increments, it may not work to apply all the load at once, better convergence will be achieved if you slowly apply the load in small increments.

Here's a quick example in Strand7, in which a slender beam with 20 elements is loaded with a UDL:




Axial force diagram (red) and bending moment diagram (green) with relatively small loading.


Load here is 1000 times higher than the first diagram - bending moment has only increased by a factor of 10 and is now represents a more constant distribution along the length of the beam; primary resistance of load is through axial force.


Load displacement diagram - the stiffening of the structure resulting from cable (membrane) action is apparent.

RE: Beam with cantenery action

That's an impressive bit of sport engineering HOTL.

Quote (JAE)

KootK - the wall below is a deck-like wall that would flex like an accordion with axial strain changes in the tube so definitely did ignore any possible shear wall effect

That's what I like to hear. Keeps things simple. In that case, I definitely like the method that I proposed (bias of course). Might be a good hand check if nothing else.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

RE: Beam with cantenery action

I get it, but it "simply" a beam with a distributed load, or more likely a distributed enforced displacement (since I don't see it really reacting much of the load ... yes, yes, the enforced displacement means it'll react some load ...). And this is why I first questioned the double cantilever assumption ... I see the corners opening up, as the rectangle ovalises under the water load; I don't see the reflex curve required for a double cantilever. I think you can find the max deflection of the top edge of the side of the tank, then assume a sin wave deformation, and figure out the internal moments and shears ...

another day in paradise, or is paradise one day closer ?

RE: Beam with cantenery action

I did a quick analysis with a 36-foot square frame of tubes with outward uniform load (just made up a load), with three different tube sizes: 3x3x1/8, 6x6x3/8, 8x8x1/2. Running it through a linear, PD with small disp, and PD with large disp. The linear and PD small disp gave me basically the same answers. PD with large disp gave me slightly larger axial force for larger tubes and significantly larger axial force (about 20% larger) for the small tube.

RE: Beam with cantenery action

On terminology:
It's catenary, not cantenary.
I disagree that the term only applies to chains hanging under gravity loads. That's where the name came from, but the catenary effect applies to any system where significant tension is generated by transverse loads on an element with restrained ends.

On the analysis:
I agree with handofthelion. The p-delta effect is essentially the same for tensile and compressive loads (except in tension it tends to reduce the deflection, rather than increase it).

It seems to me that the easiest way to model it would be to do a 3D analysis of 1 corner with symmetry boundary conditions. Include the steel plate as well as the top tubes, and apply the hydrostatic pressure.

Doug Jenkins
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RE: Beam with cantenery action

I think the loading of the tube is driven by the deflection of the side wall. Of course in deflecting it'll react a small amount of the load. I think it'd be reasonable to apply the max deflection of the wall due to the pressure. If you want to refine it, you could extract the internal shear reactions and determine the equivalent amount of water, ie an equivalent depth, and remove this from the deflection calc, and iterate.

Thinking about the corners ... I'd use a single pin joint and remove any fixity. But there's an interesting question ... what is the angle between the sides doing ? how to analyze ??

Of course the best (most accurate) answer is the FE the heck out of it.

another day in paradise, or is paradise one day closer ?

RE: Beam with cantenery action

Quote (rb1957)

I think the loading of the tube is driven by the deflection of the side wall.
Agree.

The problem sounds like the mirror image of the top wale ring of a rectangular cofferdam. In this case, pressure in the tank pushing out, versus pressure on the outside of a cofferdam pushing in. I'm sure its more complicated than that, but believe rb1957's approach is effective.

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RE: Beam with cantenery action

If the members have zero capacity to resist bending moment, they act like cables. None is a catenary. They all deflect in a parabolic shape because the load is uniform.

The tension at midpoint of the long cables is w(36)2/8d1 = 162w/d1 where d1 is the deflection at midspan. Deflection d2 of each half endwall results from the force 162w/d1 - 5w acting on it. Deflections d1 and d2 must be geometrically consistent with the change in length of the member.

BA

RE: Beam with cantenery action

I think a part of this problem has been made too complex. I think determining the tension in the tubes is fairly straight forward - it can only be what is supplied by the adjacent wall. If the uniform load on the tube is W, then the tension in the 36' long tube must be 1/2*W*10'. Now for analyzing the 36' tube, you have a uniform load, and axial loads applied at the end. If you assume initially that the horizontal deflection is zero, the first iteration is basic statics. If the tubes are fixed together at the corners this step might get a bit more tricky.

The first iteration will give you a deflected shape and moment/shear/tension along the length of the tube. Using the deflected shape, and any change in length due to axial loads, you can determine the horizontal displacements and run a second iteration.

handofthelion - that looks like a great tool. I could have used that in the past. In RISA I just modeled a W4x13 spanning hundreds of feet with a huge load on it and it returned zero axial. That's exactly what I expected and the results I've gotten from similar programs in the past. Granted, I rarely turn to these programs in my day to day work and I'm sure I only use a fraction of their capabilities, so maybe I can do this in RISA. I'm not familiar with strand7, is it comparable in capabilities to RISA and similar programs?

RE: Beam with cantenery action

I’m not sure RISA accounts for an axial tension p-delta working to straighten a bending member but it does p-little delta for compresion. RISA doesn't take into account FE plate p-delta according to the manual.

I would use the tank generator module and make the walls plates that are plane stress to get the catenary type tension forces from the walls onto the tubes. Then design the tubes as normal bending and axial members without any catenary action of the tubes themselves. There may be some straightening effect from tension on the tubes if RISA doesn’t do p-delta for tension.

RE: Beam with cantenery action

And I just tried modeling that in RISA and got enormous wall deflections assuming a 10 ft deep tank with 0.5" walls so maybe that is not a good approach...

RE: Beam with cantenery action

@CANPRO, I've not used RISA before but a quick look at their documentation - looks like they're doing a pseudo-non-linear analysis by using an algorithm to account only for a reduction of flexural stiffness due to P-Delta effects. Strand7 seems to be more of a general finite element program whereas RISA seems to be more of a design program(?). Strand7 therefore implements a generalised geometrically non-linear analysis that is not limited to certain effects.

my structural engineering blog

RE: Beam with cantenery action

The tension will cause a P-(little delta) moment that will reduce the overall moment. I would start with the B1 factor from the AISC B1-B2 approach.

RE: Beam with cantenery action

JAE, if you are really interested in a beam with axial springs, why not use Roark's formulae for beams with simultaneous axial and transverse loading?
To do so, you need to iterate on the axial load by calculating the shortening of the beam due to the transverse deflection. This calculation is however not in the Roark, but you can use this sheet from xcalcs.com
However I don't understand why you don't treat it as an edge beam to a flat plate?

prex
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RE: Beam with cantenery action

thank you for the info handofthelion. I learned how to use Abaqus years ago, which sounds like more of a comparable program to what you used.

JAE, I mainly asked handofthelion this because I suspected that RISA isn't capturing any catenary effects at all. I've been down this road before with a different problem but looking for the same effect using RISA and similar programs. I think thats probably why you're not feeling like RISA is giving you the true answer.

In your first post you stated "So this will have the effect of reducing the axial tension in the tube a bit". I don't think the axial tension (or at least the component perpendicular to the adjacent wall) can change. On the 36' tube, the tension should be 1/2*W*10', and the 10' tube should have a tension of 1/2*W*36'. Where W is the uniform reaction at the top of the tank wall.

RE: Beam with cantenery action

(OP)
Thanks for all your input.

Yes, we were concerned with the fact that RISA doesn't do PDelta with the axial component of the member.

The tank is only about 5.5 ft. tall and 8 ft. x 36 ft. in plan. This is an actual tank that was built and filled with water - then began failing with excessive outward bowing.
We needed to verify the behavior and understand the progress through loading, yielding in some areas, and initially I was concerned about the axial component that was "missing" from the RISA model.

It turns out that the stresses are way above yield such that we could properly discount the axial component of the member as the end walls would essentially bow inward (i.e. the end springs I show in the sketch above have small "k" values).

The tank sides are a corrugated deck-like system so in plane they are just accordions as the welding is on the outer faces of the decking and allows the tapered deck sides to move relative to the tube.

So the discussion above is all helpful and valid for the theoretical treatment but for this particular project - the numbers overwhelm any subtlety in the second order axial numbers.

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RE: Beam with cantenery action

If the water is 5' deep, w = 260#/'.
M = 260(36)2/8 = 42,120'#

If the member cannot resist the above moment, it could try to act as a cable. Then, if Δ1 is L/36 = 1', the tension would be 42,120#. Depending on Fy, the member would need an area in the order of 2.0in2 to resist axial tension. Apart from having a low "k" value, the end springs may not be strong enough to resist such a large horizontal force.

BA

RE: Beam with cantenery action

"excessive outward bowing" ... at about mid depth at the centre of the long side ?

sorry, but can't see the short sides bowing inward ?

how'd the corners behave ? with sandwich side side-walls I'd expect a joint on both faces, or only one ??

I'm not sure how effective this tie-rod around the top edge would be in helping the walls react pressure. I'd've added (vertical) beams on the long sides.

I guess you could run the tank without the tie-rod, and with. If RISA is missing some axial load you could estimate this by the change in length. I wonder what happens to the tie-rod in the corners ??

another day in paradise, or is paradise one day closer ?

RE: Beam with cantenery action

(OP)
The outward bowing was about 12 to 16 inches per the owner.
They then took out the water and the permanent outward bowing is now 1 1/2" or so.

The end walls appeared straight but they were only 8 ft. long.

BAretired - not sure we would generate that much axial load as the end walls would simply bow inward (not much stiffness there) and the large axial loads just wouldn't be possible....thus my end springs in the drawing I posted above. Infinite rigidity (a pin support) creates way more load than a spring condition with minimal stiffness.

Analysis suggested that the tubes on top of the 36 ft. long wall yielded first at the corners, then secondarily at midspan.

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RE: Beam with cantenery action

thx. don't really get the end support issue. I'd consider the tie-rod as that, a tension member, rather than as a beam with axial restraint and lateral load.

the corner of the tie-rods will always give you trouble, as the tension load tries to straighten out the corners (and ovalise the rectangle).

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RE: Beam with cantenery action

any cross-bracing in the middle?

Dik

RE: Beam with cantenery action

Quote (dik)

any cross-bracing in the middle?

How about three straight tension ties 8' long, dividing the side members into four nine foot spans?

BA

RE: Beam with cantenery action

(OP)
No cross bracing at all.
Can't really do tension ties as this is a swimming pool believe it or not.
Originally a large shipping container cut in half longitudinally and then reinforced to serve as an open tank-pool.

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RE: Beam with cantenery action

how about tension ties around the outside ? like running a cable around the damn'd thing.

if you want to make it look respectable ... cables on each side, with eyes at both ends, pins joining the cables together, turnbuckles in the middle of each cable to tension them. Maybe at 2/3 depth and 1/3 ?

another day in paradise, or is paradise one day closer ?

RE: Beam with cantenery action

(OP)
I think the cheapest would be to slap up an additional perimeter set of tubes to supplement the undersized existing tubes.
That way the interior pool area is not interrupted and the work is all on the outside.

Thanks for the suggestions.

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RE: Beam with cantenery action

I agree with JAE on the cheapest fix.

BA

RE: Beam with cantenery action

Quote (rb1957)

how about tension ties around the outside ? like running a cable around the damn'd thing.

if you want to make it look respectable ... cables on each side, with eyes at both ends, pins joining the cables together, turnbuckles in the middle of each cable to tension them. Maybe at 2/3 depth and 1/3 ?

That will not work unless the end walls can resist the tension in the long cables. And even if they could, the long cables would deflect substantially under load.

BA

RE: Beam with cantenery action

Quote (BAretired)

If the member cannot resist the above moment, it could try to act as a cable. Then, if Δ1 is L/36 = 1', the tension would be 42,120#. Depending on Fy, the member would need an area in the order of 2.0in2 to resist axial tension. Apart from having a low "k" value, the end springs may not be strong enough to resist such a large horizontal force.

Quote (JAE)

BAretired - not sure we would generate that much axial load as the end walls would simply bow inward (not much stiffness there) and the large axial loads just wouldn't be possible....thus my end springs in the drawing I posted above. Infinite rigidity (a pin support) creates way more load than a spring condition with minimal stiffness.

I know this has been more or less solved on JAE's end. But just on the issue of the tension in the 36' long tube; I don't think the spring stiffness or deflection has anything to do with it. If the water is 5' deep and the reaction at the top wall is 260 lbs/ft, then the reaction at the end of the 10' tube is 1/2x10'x260lbs/ft = 1300lbs. Maybe I'm over simplifying this, but I don't see how the tension in the 36' tube can be any more or less than the force created by the water acting on the 10' segment of wall. With that said, the tension in 36' long tube appears to be insignificant compared to the bending.

RE: Beam with cantenery action

I agree. My solution involved constant tension for the same reason.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

RE: Beam with cantenery action

CANPRO, I agree that, if the members act as a beam in bending, the reaction from the 10' and 36' long members would be 1300# and 4680# respectively.

I did not know the size of the members, but it was clear from the OP that the 36' long members were too small to act as a beam. As an approximation, I assumed that they had zero bending strength and carried the load as a cable. In order to do that, the corner posts had to resist an axial force in the long members. The magnitude of the axial force at midspan is the bending moment 'M' of a 36' beam divided by its sag. In this case, sag is measured horizontally.

M is 260*36*36/8 = 42,120'#. If the sag of the member at midspan is assumed to be one foot, the axial force at midspan is 42,120#. If the sag is 2', the axial force is only 21,060#. The horizontal reaction going into the corner post would be the axial force minus 1300#.

KootK, I'm not quite sure with whom you are agreeing but the tension in the 36' long member, while nearly constant, is not constant. It is slightly greater at the ends than at midspan because of the slope.



BA

RE: Beam with cantenery action

the corner post could be a reinforcement added to the side walls.

As the cables on each side make a ring, could the reaction from the long sides be reacted by the cables on the short sides, and vis versa; naturally increasing the tension in each.

I thought you could analyze the side walls as plates, without the cable supports, and this'd tell you the maximum deflection, which'd tell you the change in length and so the tension in the cable. Clearly adding this to the structure relieves the side walls somewhat.

another day in paradise, or is paradise one day closer ?

RE: Beam with cantenery action

Quote (rb1957)

As the cables on each side make a ring, could the reaction from the long sides be reacted by the cables on the short sides, and vis versa; naturally increasing the tension in each.
Cables on all sides make a ring, but the reaction from the long sides is acting at 90o to the reaction from the short sides, so neither has any effect on the other.

BA

RE: Beam with cantenery action

@JAE: I modeled this in STAAD and it became more of a arching action thing than anything else. I did a beam with moments released and horizontal springs at the end. To initiate the arching action (i.e. axial load), I introduced a bit of imperfection (about 0.5") at mid span and did a P-Delta (the P-Delta didn't have that big of an impact as far as I can tell). The key factor in reducing moment in the beams and getting axial force was the stiffness of the horizontal spring: the stiffer it was the less moment developed and the more axial load there was. This was the case regardless of how much initial imperfection I introduced.

So I guess the question is: are your end supports really stiff enough develop the kind of axial forces you are concerned about?

(And sorry I was late to the party.....been on vacation for about the last 4 days.)

RE: Beam with cantenery action

"but the reaction from the long sides is acting at 90o to the reaction from the short sides" ...

the reaction from the long sides would be acting along the cable on the short sides, and vis versa; so couldn't the short side cable react the loads from the long side ?

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RE: Beam with cantenery action

oops, that's dumb ... getting cables and beams mixed up !? ... of course the reaction for a cable is axial and the reaction for a beam is lateral ...

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RE: Beam with cantenery action

JAE:
Why not sell the idea of a deck all the way around the top of this pool? That way you could hide the stiff horiz. beams and corner details which should have been there in the first place to make this crazy idea work. They thought they were being clever (and/or cheap) in the first place, and sooner or later they have to pay the piper if they want this to really work.

RE: Beam with cantenery action

I wonder if a complete container would work, ie half fill a container before you cut it in half ? The sides would be much stiffer, and a nice loadpath into the roof. I know it isn't practical but it does give you a design point.

another day in paradise, or is paradise one day closer ?

RE: Beam with cantenery action

Quote (BART)

KootK, I'm not quite sure with whom you are agreeing

I was agreeing with CANPRO. I've been under the impression that when one posts a comment without specifying a target recipient by name, that target recipient is assumed to be the poster of the previous comment. That or OP, I go both ways.

Quote (BART)

...but the tension in the 36' long member, while nearly constant, is not constant. It is slightly greater at the ends than at midspan because of the slope.

I suppose that it would have been more precise to say that the horizontal thrust in the member at the ends is constant.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

RE: Beam with cantenery action

Quote (KootK)

I was agreeing with CANPRO. I've been under the impression that when one posts a comment without specifying a target recipient by name, that target recipient is assumed to be the poster of the previous comment. That or OP, I go both ways.

Ordinarily, I would have been under the same impression but in this case, I didn't think you were agreeing with the entirety of CANPRO's comment.

Quote (KootK)

I suppose that it would have been more precise to say that the horizontal thrust in the member at the ends is constant.

Ordinarily, with a cable sagging in a vertical plane, that would have been more precise, but in this case, the member deflects in a horizontal plane, so thrust is always horizontal.

BA

RE: Beam with cantenery action

Sizing the tube not to fail is pretty straight forward and avoids the axial tension complication since the pipe stays relatively straight under load, as it should. For the following calcs assumed each 36' long tube has simple supports with maximum deflection at mid-span (an arbitrary) 2.40" (L/180). Choose what ever deflection is wanted.



Per BAretired, UDL on the tube is 260 lb/ft and tension is 1300 lb.

Calculate the required moment of inertia to meet the above criteria (36' long steel member, deflection of 2.40", 260 lb/ft loading).
I = 141 in4

For a 10" diameter, "standard weight" steel pipe:
I = 161 in4

Check 10" pipe bending stress... fb = 16.9 KSI

Check 10" pipe tensile stress... t = 109 PSI

Unless I'm missing something, done.

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