Non-rect beam strenght

I want to make a beam that will be triangular, rather than rectangular, and hollow.
Does anyone know a source or have the knowledge to work out how strong
such a beam would be. It sort of feels as if it would be prop. to the sin of the agles of the sides, or whatever, but I am not sure, and it probably gets more complex than that.
Any advice appreciated. ***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
d
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
The second step is to plug it into....... f sub b = M c / eye

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sat, 31 Jul 2004 07:06:32 -0700, "larsen-tools"
......and in reply I say!:
remove ns from my header address to reply via email I did ask for "any advice"
Appreciated.

***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Advice......... You are in over your head.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sat, 31 Jul 2004 17:27:19 -0700, "larsen-tools"
......and in reply I say!:
remove ns from my header address to reply via email

Which is how we all learned to swim.... ***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
snipped-for-privacy@dodo.com.au says...

"Properties of Sections" is what you want, here's the first two Google hits...
http://www.dow.com/styron/design/guide/properties.htm
http://ourworld.compuserve.com/homepages/MJVanVoorhis/T101 . htm
Strength will be proportional to Z (section modulus), stiffness proportional to I (moment of inertia). Use superposition (subtract the properties of the "missing" material from that of a solid prism) to account for your hollow setion. If you want to do this analytically, the subject you're looking for is "Strength of Materials". That's also the name of the classic college text by Timoshenko.
This is all you need to compare the strength or stiffness of various sections. If you want the actual numbers you of course need to account for beam material, span, type of support(s), and load distribution.
Ned Simmons
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
In article < snipped-for-privacy@news.suscom-

One more thing, these properties depend on the orientation of the beam relative to the loading, for example, whether the load is acting parallel or perpendicular to a side of your triangle.
p.s.s. "Roark's Formulas for Stress and Strain" is the Cliff Notes version for all this stuff, but probably not very useful on its own unless you already have a pretty good understanding of the subject.
Ned Simmons
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
vaguely proposed a theory ......and in reply I say!:
remove ns from my header address to reply via email
Yes. Thanks. That will be awkward, and I probably have to allow for worst case. It's for a loader rake bucket. So the direction of roces could be anywhere.
Maybe I should just go to round, and work that way? Problem is I am using an alloy for lightweight strength. You can't get tube, and making it would cost a fortune.

***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Old Nick wrote:

What alloy? Are you thinking of welding it up out of flat stock? If so, consider square. Square tube is a very efficient shape for bending strength and also has excellent resistance to twisting. If you give me dimensions, I can run the beam calc on my software. Otherwise, if you give position of support(s) and load(s) with amount of load and material strength, I could give you dimensions for an appropriate square tube.
Ted
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sun, 01 Aug 2004 00:43:22 GMT, Ted Edwards
......and in reply I say!:
remove ns from my header address to reply via email
It's Bisalloy. Pre tempered etc.
I can do square / rect no problems. IT was the triangular that was a problem. I want it mainly to allow better sheeding of soil, when dealing with rock or logs etc. Maybe I should just shake the oil off and go square! <G>

***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Old Nick wrote:

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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Depending on the scale of the thing........ why not use a rectangular shape or CRS flat bar on edge w/ an equal leg angle as a " ^ " top?.... no big deal.

Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Mon, 2 Aug 2004 13:12:40 -0700, "larsen-tools"
......and in reply I say!:
remove ns from my header address to reply via email

Ted and...ummm larsen <G>
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.

***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
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.

solids
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Mon, 2 Aug 2004 19:47:07 -0700, "larsen-tools"
......and in reply I say!:
remove ns from my header address to reply via email
No. I didn't think you were _just_ being a smartarse! <G>
I know you are right. But I'm willing to have a go. This stuff does get done... It may take some time.

***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Mon, 02 Aug 2004 20:37:09 GMT, Ted Edwards
......and in reply I say!:
remove ns from my header address to reply via email

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.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Old Nick wrote:

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 2. (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)

radius of 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.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Wed, 04 Aug 2004 17:25:15 GMT, Ted Edwards
......and in reply I say!:
remove ns from my header address to reply via email
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! <G>
***************************************************** It's not the milk and honey we hate. It's having it rammed down our throats.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Old Nick wrote:

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
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.