Non-rect beam strenght

How about square but rotated 45 degrees to give a diamond shape? That would shed the soil and allow you to use a standard structural shape.

Do you have Machinery's Hanbook? If so, formulae for square in usual and rotated are given on pg 189 in my 24ed. If you _really_ want triangular, formulae for solid is on pg 198 (24ed). If the wall thickness is uniform, you may calculate hollow by calculating two solids and subtracting.

If you don't have MH, post or e-mail me and I'll copy the formulae out for you.

Ted

On Mon, 2 Aug 2004 13:12:40 -0700, "larsen-tools" vaguely proposed a theory ......and in reply I say!:

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Ted and...ummm larsen

Yes. All of these things had occurred to me.....seriously.

It may be a bigger deal than it seems to make a compound thingy. I am willing to allow much extra to make up for (a) the total unknown of the forces involved (b) my own ignorance. But the slimmer these things are the better. So if I used a rect section, and placed a trianlg eon top, I would still be interested in the added strength of the triangular section as it adds significant height to the tine.

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On Mon, 02 Aug 2004 20:37:09 GMT, Ted Edwards vaguely proposed a theory ......and in reply I say!:

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Yes. I had thought of this. At first sight, I felt that it would also strengthen the thing sideways. But I am now beginning to doubt that. Although it gives a deeper profile to a sideways force, it is a weak profile.

They cost Aud$250 over here...sigh! I see you guys get it for $85. Bum!

The best I could hope for was about Aud$190. Although I see there is CD version. That would cut postage hugely. Books and postage do not go together.

Ok. Ted. Thanks. I do not have MH. I would appreciate the formulae. I have found (was pointed to) a couple of sources on the Web, but it's always interesting to see if there's another derivation (lets my slow old brain maybe use it in another cointext :-< ), or to check what I find.

***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.

Nick, I have no idea what you are going to make but it sounds like "eye-ball engineering" is the way to go...... since the loading conditions are unknown, I think. If you are working w/ built-up steel beams, presuming they act monolithically, my first two replies are correct.

When sizing standard rolled steel beams the formula becomes...... f sub b = M / S ...... f sub b = fiber bending stress, 24,000 psi for A-36 steel M = bending moment ft-lbs S = the section modulus .... from tables of standard shapes, in^3 , ...... S = eye / c eye = capital "I" which looks like little L in this font c = distance from neutral axis to extreme surface.

You know f sub b, you calculate M and then pick a section w/ an adequate S. That's for gradual, not impact, loading. Then you check some other things such as shear & size the connections. We don't talk about "the strength of a beam" we select beams with an adequate S to handle the loading conditions.

The formulas/ information that I gave work for beams in bridges, buildings or bird houses........ and you thought I was just being a smart-ass.

Another idea, but not sure if it will give you the o/s shape you need...

Instead of welding the legs of the triangle directly to each other, place a small/medium round or square rod at each corner of the triangle and weld the legs to those. Two advantages: it increases the area at the o/s corners, and it cuts down the sharp corner which is point of highest stress. In other words, it decreases "c" and increases "I", in Mc/I. The final o/s profile is a triangle w/ rounded corners. But perhaps you need sharp corners for the soil?

A simple 60 degree V fixture would help keep things in position during tacking/welding.

David

On Sat, 31 Jul 2004 16:19:15 +0800, Old Nick vaguely proposed a theory ......and in reply I say!:

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One of these days I am going to spill my tittles corectly.

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On Sat, 31 Jul 2004 16:19:15 +0800, Old Nick vaguely proposed a theory ......and in reply I say!:

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Ok. Here is what I am looking at.

I am not sure why this happened, but I was looking at a formula for a triangular section, and the formula included a divide by 36. I thought to myself, AH! Aell, then! A rect section would be divide by twelve........and it was!

Unfortunately this sort of thing happens so rarely to me that it was probably luck :-<

However, I am looking at starting with a rectangle at the tip (to give "dig"), basing the sizes on that alone, then gradually "lifting" a triangle on the top as I move back toward the root.

I am not sure yet whether I will design to include the triangle part for strength, using the idea that it's about 1/3 as good as a rectangle of the same height, or whether to simply be happy that the part that sheds soil will also add a bit of strength.

To be honest, I reckon the tines will hold, based on my previous tests of this alloy. The interface between them and the back of the bucket is where the greatest forces (and the possible weaknesses) will be. I reckon this will be a trial and _error_ part of the design.

So thanks all for bouncing it around.

Now to work out the best way to get the alloy cut....BEEN THERE DONE THAT HERE ALREADY THANKS , and had a lot of useful suggestions.

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On Mon, 2 Aug 2004 19:47:07 -0700, "larsen-tools" vaguely proposed a theory ......and in reply I say!:

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No. I didn't think you were _just_ being a smartarse!

I know you are right. But I'm willing to have a go. This stuff does get done... It may take some time.

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Not really at least not greatly so. Interestingly, the moment of inertia (I) and the radius of gyration (r) are identical for the square tube whether oriented with sides vertical/horizontal or rotated 45 degrees. However the section modulus (Z) for the rotated is less by a factor of square root

1. (This is because the distance to the extreme fiber is greater.) Note these statements apply to vertical or horizontal loads. Since you stated that the load could be applied in _any_ direction, you should really consider the worst (weakest) direction. The best strength/weight in such a case would probably be round.

In the following formulae, a is the length of the outside side and b is the length of the inside side except for round tube where a is OD and b is ID. I is moment of inertia, c is distance from neutral axis to extreme fibre and A is Area of material in section.

shape I c A _ square |_| 0.0833(a^4 - b^4) 0.5 a a^2 - b^2

square /\ \/ " 0.707 a "

Equilateral Triangle 0.0541(a^4 - b^4) 0.577 a 0.433 (a^2 - b^2)

round tube 0.0491(a^4 - b^4) 0.500 a 0.7854 (a^2 - b^2)

gyration from r = Sqrt(I/A) and the shock load factor from Q = I/(c^2).

I will do an example of each. SInce you are concerned with weight, I will compare the sections of equal section area and thus weight if the same material (steel) is used for all. Note that you want I as large as possible for stiffness, S as large as possible for strength and Q as large as possible for best shock resistance. The reference section will be 6" diameter,

0.25" wall round tube. The others will also be 0.25" wall with a chosen to give the same section area, A, of 4.516 in^2.

Shape a(in.) I(in^4) c(in) S(in^3) Q(in^2) r(in) tube 6 18.699 3 6.233 2.078 2.035 square 4.766 15.398 2.383 6.461 2.711 1.846 diamond 4.766 15.398 3.370 4.569 1.356 1.846 triangle 6.454 13.715 3.726 3.681 0.988 1.743

So, I would offer the opinion that, for equal weight, the round tube is the stiffest, the square is the strongest and the most shock resistant but not much better than the round tube and the triangle looses on all counts. Given your desire to avoid a flat top surface, it seems to me that a round tube is the best bet. Note that, at least here in Canada, standard pipe is not as strong an alloy as structural tubing.

Hope this helps.

Disclaimer: I am not a licensed structural engineer. Use at your own risk.

On Wed, 04 Aug 2004 17:25:15 GMT, Ted Edwards vaguely proposed a theory ......and in reply I say!:

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Thanks for that Ted. Round tube is out of the question, because of material constraints, but I realise it's the "safest".

Sure does.

I will demand a full refund of your fees upon failure of the structure!

***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.

What are the material constraints? How long is this thing? Is this to be built from flat stock you already have? Noting that square is better than triangular and round better yet, I would conjecture that pentagonal, hexagonal, ... would approach the performance of round. Hmmm. Pentagon could even have a soil shedding peak at the top. :-)

Ted

On Thu, 05 Aug 2004 17:40:41 GMT, Ted Edwards vaguely proposed a theory ......and in reply I say!:

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OK. Some of this was buried in all the posts, but in bits, I admit.

I will share my (admittedly fractured and generalised) thoughts. I am not arguing with you, but laying out the facts as I know them.

I do have a sheet of this (expensive) stuff sitting out there, bought before I knew all of this :=

Pentagon would have one fewer weld and would have better earth shedding with a point up.

Do you have TIG? 309SS filler would probably be good.

Do you have a plasma cutter? You could make a jig to cut out the 15x5 pieces. Cut out some 6 or 7 pentagons of sizes suitable for a form and drill a hole in the middle. Thread them on a piece of rebar and tack them in place. Fasten your cut and dressed pieces to the form with hose clamps, tack the tine between the clamps and tap the form out. Remove the hose clamps and weld 'er up.

Piece of cake - for the guy who doesn't have to do it. :-)

Ted

On Sat, 07 Aug 2004 03:29:53 GMT, Ted Edwards vaguely proposed a theory ......and in reply I say!:

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Sorry. That's what I meant.

Hmmm..pentastuff....mark of the devil and all that.....bummer.

But nobody said it had to be _accurate_. This job will simply consign me to somewhere very hot _for a long time_, and not enternity, right?

No. TIG is a fascinating but remote chimera

Chime´ra n. 1. (Myth.) A monster represented as vomiting flames, and as having the head of a lion, the body of a goat, and the tail of a dragon. 2. A vain, foolish, or incongruous fancy, or creature of the imagination; as, the chimera of an author.

.......hey! not far off in both counts!

for me. But there are reccommended MIG fillers, and careful pre-warming.

Actually, I wonder about TIG and this stuff. TIG has a high, slow heat input like flame, right? Not good.

No. _AND_ "WE" just had a fight in which the phrase "and look at all the toys and shit you have about the place" came up.

Let me let this slide for a while.

Seriously. I have considered a plasma cutter. But they are so _expensive_ (double USD) for anything that is really worth buying. A lot of my work consists of "railway iron" and other unspecified thicknesses.....we are talking kilobucks here!

Yep.

Hang on....

Sorry...........HOSE CLAMPS! HAH!

But what about torsional distortion? This stuff is truly amazing. I would see more like "tack weld the pentagon forms to a large basis....done well! Weld up the top ones that you can do. Allow to cool. Grind tacks. Turn over. Repeat with the rest."

I am rather proud of my tack welding for this sort of work. This stuff has made me start again........PING!

FWIW...Suddenly I like you!

***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.

Much faster than flame, not as fast as MIG but unless you are very good with MIG, there's the risk of good looking weak welds. I do know that the homebuilt aircraft folks don't like MIG for welding 4130.

Ted