Tubing vs Round Pipe

The round would have less......here

Hello Smitty,

If you have the square tubing with the faces alternately vertical and horizontal, then the round pipe would sag less.

If you have the square pipe so it was installed with a corner at the top, another at the bottom, and 2 further corners on each side in the middle, horizontal axis passing through both the side corners, then the position is quite different.

I shall leave you to ponder why....

Sparky,

If the outside diameter of the round tube is equal to the outside dimension of the square tube, the square tube has greater stiffness and would sag less under load. However, square tubing has an outside dimension of 51 mm or exactly two inches whereas the round tubing has an outside diameter of 48 mm or about 1 7/8 inches.

The moment of inertia for the square tube is 0.319e6 mm⁴ or 0.766 in⁴ versus 0.157e6 mm⁴ or 0.377 in⁴ for the round tube. Thus the round tube would sag about twice as much as the square tube.

If you choose the next size up in round tubing, it has an outside diameter of 60 mm or about 2 3/8" and a moment of inertia of 0.397e6 mm⁴, so it would be stiffer than the square tube and would sag less under load (about 80%).

So far as the square tube is concerned, I believe the moment of inertia is identical whether its sides are vertical/horizontal or at 45 degrees. Thus it does not matter which way you mount it...the deflection would be the same.

As a cR4 member I hereby certify you "Caculation Engineer". Congrats..

(There is a thread on this).

......so, which is going to flex or sag more, the Round Pipe or the Square tubing?

Smitty,

I thought we were clear on that. The round pipe will sag 42% more than the square tubing.

I was going to look up the answers in the "book" when I saw ba/ael's answers. No need to do any looking up now, I can tell this guy knows his stuff.

Somebody who knows how to punch the "good answer" button should do it.

By the way, would one of you explain how a participant gets the "good answer" under his name.

To Close this discussion and Thank everyone involved..........

It is safe to say that the foloowing is true:

Given a 15' beam with 50. lb weights attached to each end, suspended from a ceiling ;

1. A 2" Round pipe will sag MORE than a 2x2 Tubing if both have approx. the SAME wall thickness
2. It makes NO DIFFERENCE if the 2x2 Tubing is positioned in its SQUARE form or in a DIAMOND "◊" form.

Correct?

Thank you all for your participation and answers.

I'm not convinced by the argument about the square section being the same strength on it's diagonal....

An I beam is obviously much stronger when the section is vertical, yet the cross section hasn't changed.

Maybe someone wants to elaborate...I dunno..I'm not a mech eng...in fact I'm just a cat

Nice answer thanks... (I have rated it as such)

Although..

'We have not mentioned anything so far about strength. We have been talking only about stiffness...'
Isn't this slightly nit picking? In this simple example they amount to much the same thing as I take it to be resistance to deflection....

Presumably 'strength' depends upon the application....hmmm that could be a thread in its own right?
Is a bow stave stronger before you taper it?
If you take it to the final deflection in it's un-tapered state it will break ...
Thus when we make it into a longbow do we it 'stronger' by removing material????

Cheers

Del

It is not nit picking. In structural analysis, strength and deformation are completely different animals. The strength of a steel beam is measured by its capacity at failure irrespective of deformation. The strength can be expressed by: Mp = Z.Fy where Mp is the plastic moment, Fy is the yield point of the steel and Z is the plastic modulus of the section being considered. The plastic modulus, Z carries units of length to the third power, whereas moment of inertia has units of length to the fourth power.

An example of a material with high stiffness and low strength is drywall eight feet high and, say 1/2" in thickness. As a beam, it does not deflect much but as soon as the load gets a little too much for it, it fails.

measured by its capacity at failure .

Cheers, this is a very interesting definition. I wasn't trying to criticise, more to understand.

'Nit picking' was the wrong choice of words. I still feel it's a subtle distinction when considering two tubular sections of the same material..the one which deforms more will also be the one which fails first?

Anyhow...
Back to my longbow/tapered beam question.(If you don't mind humouring me...or maybe I should post it as a new thread? Or is it trivially simple and I'm being thick?)

Say a steel beam (A) 36" long 2"x2" (width x depth) section fixed rigidly at one end is loaded at the tip.

A similar beam (B) tapered from 2x2 at the fixed end to 2 x 0.5 at the loaded end.

Which will fail first? B will obviously deform more than A but I haven't the maths or mech eng skill to quantify or answer the question?

My instinct as a bow maker says that A will fail first...thus B has been made stronger by removing material.... this just amuses me as it would appear to be a paradox.

Regards

Del

Another thing to consider for possible incorporation into your longbow thread:

Suppose we make a 1" dia carbon fiber braid tube (like a Chinese finger trap). This is commonly done to make carbon fiber tubes. We take this braid, place it over a mandrel, and saturate it with epoxy resin, and then let it cure. (In comparison to the carbon fibers, the epoxy has negligible strength.) The resulting tube would be considered strong and stiff, particularly for its weight.

We glue this tube into a hole in the wall, and suspend weights from its end. It bends, and eventually fails: if the tube is 4' long, it might fail at a deflection of about 1', and with perhaps 100# at the end.

We make a second tube, but impregnate this one with RTV silicone. After cure, it is stiff enough to support its own weight fairly well when glued into the hole in the wall. We load it, and find that it doesn't fail with a snap crackle pop, but sags, and ends up supporting 3,000# hanging straight down from the hole in the wall.

Which tube is stronger, and should we tell Boeing to stop using epoxy?

Let's go to your bow first and maybe later come back to the "subtle distinction".

The prismatic section (A) will fail at the point of maximum moment when the section is fully yielded. The applied moment is P.y where P is the force and y is the distance from the force. The maximum moment occurs at the base where y = 36" and this is where we would expect failure to occur. Moment at base = 36P.

The plastic modulus Z is (2)³ /4 = 2 in^3. Assuming a yield stress of 36,000 psi (not very high nowadays) the plastic moment is 2 x 36000 = 72000"#.

Equating the maximum applied moment to the plastic moment, we would expect failure to occur when 36P = 72000"#, i.e. P = 2000 pounds.

Beam (B) may be tapered in such a way that at no point will the plastic moment be exceeded. However if you taper it as you have suggested, you will introduce a more critical stress situation away from the fixed base.

For example, a point 18" from the base will also be 18" from the applied load. The applied moment will be 18P, the thickness 1.25" and the plastic modulus 2.0*1.25²/4 ⁼ 0.78125 in^3. The plastic moment at that section is 28,125"#. Equating this to the applied moment, we find that failure occurs at that point under a load of only 1562#. There may be other more critical points than the midpoint.

So 1/2" is not a good choice for end thickness if you wish to create a bow of uniform strength throughout.

Excellent thanks...hmm, I should have quoted a more realistic bow profile... (although you do need some thickness at the end to attach the string .

It certainly highlights a glaring gap in my knowledge...something I shall have to explore further if I revist making bows from metal bar at some time.
I have made bows from high tensile aluminium alloy bar, these have been of even thickness with a simple tapered profile... if it is of interest we could discus it further...

Maybe my seat of the pants engineering is rather more luck than sound judgement (don't tell the others )

Thanks again (Ken also)

Del

I have never designed a longbow in my life, so I don't claim to be an expert. So far, I have been considering the load to be applied at the end of the cantilever at right angles to the axis of the member. In a real longbow, this would not be a good approximation as the string pulls at an angle quite different from that assumption. The amount of flex in the bow produces a sizable deflection. The axial component of the string tension contributes to bending in the bow which adds to the moment P.y.

It is quite straightforward to calculate the profile required for the horizontal load case. Simply equate the applied moment (multiplied by a suitable load factor) to the moment capacity of the section.

Try: Φ*P*y = Zy*Fy = b*x²*Fy/4 where x is the thickness required at distance y from the point of attachment of string. The factor Φ should likely be taken as 2.0 or 3.0 (sort of a safety factor). The force P is one half the force you expect to place at the midpoint of the bow.

Then x = (4Φ*P*y/(b*Fy))^1/2

You can solve for x at any distance y which should result in simultaneous failure at every point in the longbow under normal force P but neglects axial forces and eccentric moments or so called P-Δ effects. Your calculated value of x = 0 at the end must be rejected because (a) you will need a minimum cross section to resist shear force (which is constant for the entire length of bow) and (b) you will need sufficient section for attachment of the string.

I am sure that longbow manufacturers would scoff at the oversimplification of this analysis, but I think it touches on a few of the items you may wish to consider.

Wow..Thanks for that...

Yeh, obviously a real longbow made of Yew doesn't bear analysis,(being a 'natural composite') but for a crossbow made of flat bar that is pretty usefull... I might have to do some 'armchair' design for fun. The bow in question is below...you can see the taper if fairly gentle..it will be fun to see how the 'ideal' taper compares with the empirical/practical one.

Once again thanks for your input on this niggling question which has nagged at me for the last 40yrs or so

Del,

At the risk of confusing you even further, I was under the impression that you were tapering the thickness of the bow. I see from your photo that you are tapering the width, not the depth. Silly me...what do I know about bow making?

Let's call the width 'b' and the thickness 'd'. In the previous example, Beam (B) had a dimension b x d. I thought the depth, d varied from 2" at the middle to 1/2" at the end when, in fact, your photograph shows that the width, b is the dimension which tapers. Sorry about that.

Now the plastic modulus, Z = b.d²/4. The previous analysis was okay for Beam (A) but not for Beam (B). At mid-length of the bow, i.e. 18", the moment is still 18P but width b = 1.25" and the plastic modulus is 1.25 in^3. Equating the applied moment to the flexural strength, we get 18P = Z.Fy = 1.25 x 36,000. The value of P at 18" required to fail the bar is 2,500 pounds.

So your method of tapering the beam is okay after all. In fact, it is clear that the strength is critical at the middle of the bow where the section is maximum. If bending strength was the only item to be considered, you could taper the width b from 2" down to 0, but for reasons previously dicussed, you would want to keep a minimum dimension of 1/2".

So your gut feeling was perfectly good. However, it is not correct to say that you have made the bow stronger by tapering it. The critical section will still fail when the lateral force reaches 2,000 pounds. Sorry for the confusion.

Now, if you want to taper both b and d, well, that is another story.

You were not confused!
You interpreted my original Q correctly...I just asked the wrong question. .
Mind it does prove even a cat will get it right by purre chance some of the time.
Of course a real longbow tapers in both directions at once and has BIG deflections.
You can see the sap wood/heart wood line on the bow..sap wood good in tension, heart wood good in compression...beautiful timber to play with. They say a longbow at full draw is 7/8th broken . I dare not draw it much further than that shown...

Del

Isn't this slightly nit picking? In this simple example they amount to much the same thing as I take it to be resistance to deflection....

Gosh golly no!! It's not nit picking at all. Stiffness and strength are very different things. (Calling it nit picking is a little like saying that voltage and current are really pretty much the same. That's a bit of an exaggeration, but not too far off.)

All materials are elastic (rubbery) to lesser or greater extent. Some, like rubber, will stretch a great deal and return to their original length (in other words, they have not yielded). Some polyurethanes will stretch to 600% of the original length. Carbon fibers are at the other extreme, with high modulus (stiff) fibers stretching only about 1% before they snap.

Saltine crackers are very stiff (on edge, one might support 1 pound without any noticeable deflection) but not very strong, as compared to a rubber band. A rubber band might deflect significantly or under a one pound load, but will still be intact (but very long) with a five pound load.

Usually, the term "stiffness" is used to denote a property of a particular structure, and modulus (Young's modulus) is used as a material property (in either tension or compression) -- although this rule is anything but hard and fast: people talk about stiff materials all the time, but I've never heard anyone talk about the modulus of a structure.

The longbow question would make a good thread. In fact, when you remove material you are making the long bow less stiff as a structure: you'd get more drawstring pull (force) from a 6' 2x4, but the deflection for that pull would be small. If you turned the 2x4 so you are pushing against its narrow side, the deflection would be even smaller: if you rigged the string loosely, so it made an angle of 30 degrees or so with the ends when the slack was taken out, I'd guess that you could pull several hundred pounds worth before you deflected the 2x4 more than an inch or two.

I'm rating this a good answer, too. I think that if you have an answer rated "good" by two people it shows up on your good answer tally. If that is true, then you will have one official good answer out of 27 posts, making your ratio at least an order of magnitude better than mine.

Correct! But if you want to compare efficiency of sections, that is another matter. Clearly, the weight of the round section is in the order of 3/4 of the weight of the square tube. You could increase the diameter of the round section and arrive at a stiffer beam.

By the same token, you could use HSS64x64x4.8 (metric) which is HSS2.5x2.5x3/16 in Imperial measure. The weight would be 4% larger but the moment of inertia would increase by a factor of 1.86 which means the sag would decrease by almost half.

Alternatively, you could opt for a rectangular section with height larger than width and improve the stiffness with less weight.

Any of these shapes could be used. The only criteria are the appearance and the permissible deflection. Your choice!

Thank you Del,

It seems both Sparky and Ba/el have opinions on the diamond shape vs square shape.

I am way out of my league on this so lets wait and see who clears this up.

At least i know to use Tubing instead of round pipe.

Cheers.

Smitty

Hello Smitty,

If you carefully read the second part of my reply:

http://cr4.globalspec.com/comment/160821/Re-tubing-vs-round-pipe

You may notice the careful wording, which no replier yet seems to have understood.

".....the position is quite different."

Of course the position is quite different, because the square tubing has been rotated 45 degrees from its horizontal earlier position

The first part of my reply was designed to encourage others to think about the problem....

You may notice the careful wording, which no replier yet seems to have understood.

Perhaps it is not so much that repliers failed to understand, but that they did not want to seem overly critical. The first part of your original post is not correct: in other words, the square tube would sag less -- not the other way around. If you turn the square tube 45 degrees (so it appears diamond shaped from the end) its stiffness remains the same. What you gain in height you loose by placing more material closer to the node. Ba/ael addressed this above, I believe. If the tube is rectangular (but not square) in cross section, then rotating it (relative to the load direction) can make a large difference in stiffness.

Sparky,

I did understand what you meant. And, you're absolutely right. You get a "good'n" for it.

Oops. Quick edit. Or else I'm arguing with myself. The saving grace of being an OF is we can be wacky and people just chalk it up to senility.

My agreement with Sparky was on the "position" comment, not the conclusion that the round sagged less.

Thank you again Sparky,

i do not have an analytical mind nor am I am overly gifted with any 'engineering" smarts......so I can not figure out these formulas and such.

I just need to know before we build this spreader bar system, that square tubing will NOT sag as much as Round.

If someone has a better answer as to which way to turn the tubing then I'll go that direction too.

I will look up the machinists handbook that was discussed at the end of this string and see what they say.

Thank you.

smitty

I just need to know before we build this spreader bar system, that square tubing will NOT sag as much as Round.

That is correct if the outer dimension and thickness are equal.

If someone has a better answer as to which way to turn the tubing then I'll go that direction too.

Turn the tubing any direction you wish. It doesn't matter.

I will look up the machinists handbook that was discussed at the end of this string and see what they say.

Look at the column marked 'I' for moment of inertia. That is a direct measure of stiffness. My choice would be HSS2.5X2.5X1/8". It is only 72% of the weight of the HSS2x2x1/4 and is 38% stiffer.

This structure will behave like a seesaw (teeter totter). When hanging the 50# weights, be sure to support the first until the second is in place.

Good luck and Happy New Year!

I just need to know before we build this spreader bar system, ...

I wonder... maybe this is really a spreader bar and not a beam? If you hung the two weights from cables that were run to the center point, and used the tube to spread the cables so that the weights were 15' apart, then the structure is entirely different, and a much lighter tube could be used, because it would be in compression rather than under a bending load. In that case, the tube would be called a spreader -- especially by sailors.

Also, be aware that if the two weights are not the same, (and the tube is used as a beam hung from its middle, i.e., not as a spreader) the tube will tip a long way from horizontal: You might think that if one weight is 50# and the other is 50.5# that the tilt will only be slight. Depending upon how the tube is supported in the middle, a slight imbalance could make the bar closer to vertical than horizontal.

Also be aware that if the tube is used as a beam, that the tension load on the upper side and compression load on the lower will be fairly high: on the order of 4000#. You'd want to make sure you do not drill into the upper or lower sides of the beam.

Finally, if people might be underneath one of these weights, you might want to check out the final structure with a structural engineer. Better safe than sorry.

Ken makes a good point in that the OP did talk about a "spreader" and that will be different in analysis. Although a 15 foot spreader with only a 100 lb load seems peculiar. And why the requirement for rigidity in a spreader? Normally you would be more concerned about buckling. And, did I miss the material? I assume steel, but???

Smitty,

Talk to us. It's New Year's Day and some of us have time to kill between football games (no, Del, not soccer - this is the colonies) and we can probably over-analyze this thing to death. Give us some more info. When you add the self-weight of the beam/spreader, the stress (as Ken calculated) is getting a little high for me to stand under it unless I find my steel-toe shoes (I don't bother with hard hats anymore - I keep my remaining neuron in a beaker at the National Aviary).

I wonder... maybe this is really a spreader bar and not a beam? If you hung the two weights from cables that were run to the center point, and used the tube to spread the cables so that the weights were 15' apart, then the structure is entirely different, and a much lighter tube could be used, because it would be in compression rather than under a bending load. In that case, the tube would be called a spreader -- especially by sailors.

I agree with Ken completely. if this is the case, the whole discussion so far has been based on the wrong premise. Smitty, we need clarification.

Which would have less sag?

The Square one.

MM

Thank you .

Square Tubing it is.

I am not concerned with strength as neither the 2" tubing or 2" round pipe will be under enormous strain. I just need rigidity.

Thank you all.

Smitty

Smitty,

Get a copy of Machinery's Handbook. The whole sordid story is there, although without any candid photos. And, it's even pretty easy to use. Just look up in the index for section properties.

TV,

Is it Machinery's Handbook or Machinist's Handbook?

Thank you .

Smitty

It's Machinery's Handbook and the edition doesn't matter for many things like this so you might find a used copy for less than $30. Another good book for learning about stuff like this is Simplified Engineering for Architects and Builders by Parker and Ambrose. Again, there should be a ton of cheap used ones out there, probably around $10. You can either go to Amazon or to ABE for used books.

Happy New Year.

For people interested in figuring out problems like the original question that was asked, I have a book published by a steel distributor which lists all types of beams and tubes---also gives the section modulus for each item calculated for various lengths and the deflection. This makes it really easy and fast to figure out the answers. I would think that most bigger distributors would have a similar book (or maybe online) for those who don't want to buy the latest version of the Machinist's Handbook.

Del,

Thanks for your trying to answer me about the good answers. I understood the "rate" button. I guess I wasn't very clear. For instance on Blink, right under the number of posts he's made, it shows "good answers 6". What I'm trying to understand is how the "good answers" are registered which show up there. What's the process? I would like to see ba/ael receive a "good answer" which shows up there...he deserves the credit.

Thanks again,

Randy

Hi Randy,

I rated ba/ael's answer as good too. The good answer tally that shows up under one's name are the number of answers which were rated to be good answers 2 or more times. Therefore, when two of us rated ba/ael's answer "good" it showed up under his name as 1 good answer. CR4 is saying, in effect, that for an answer to get credit as "good", at least two people have to consider it good.

Hello Again,

Sorry for the long pause......I had the makings of the flu and it was potent.

What i am doing is making an overkill contraption to slide under some light fixtures in our shop. The center of the 15' beam will be welded to yet another overkill bracket .

There will be little or no sway once we get it set up like we want.

I am picking up small parts to sand blast and paint. Nothing will weigh over a few pounds at most.

Not sure of how the final version will come out but I wanted maximum rigidity on the spreader beam. I was told by my welder that i could have used a smaller beam than this considering nothing will weigh more than 5-15 lbs....

I do everything overkill to avoid problems later.

The information you all gave is excellent and just what i needed. I just wanted rigidity and the 2x2 tubing will suffice. I am checking with the local steel works for what size walls they have and I may have to go slightly smaller or even thicker. However, this will work for us.

Thank you all for your concerns on safety. Trust me when i say that I have had it looked at by a ship fitter. he agrees this is way over kill and can handle the weights.

the 50 lbs on each end are the little winches and i missed their weight . I thought they were 20kg. They are less than 20 lbs each with cable...once again, overkill.

All is well.

Happy New Years

Just to throw a wrench into the works, what about 2x2x2 triangular tubing?

Hello geoformer,

Here is your wrench warning sign....

Where § is the modulus of elasticity, Ω is the electrical resistance per unit length, equals the cosmical radius of polygonality, and Ψ is the ratio of water-miscibility, we get the equation:

ζ√15/2.2.2 x ^Þ/_{50 =} ζΨ³/_{5 X 19√50} x §

Q.E.D.

I trust that answers your rather easy question.

Kind Regards....

Just to throw a wrench into the works, what about 2x2x2 triangular tubing?

Strong enough to carry the load, but about one fifth the stiffness of the square tube for an equivalent wall thickness. So deflection could be a problem. Anyway, who markets triangular tubes?

Mabe the Bermuda Triangle