I was wondering, what beam offers greater ridigity (or has less deflection,) let's say a 4"x4" H beam or a 5" I (& about 3" wide) beam? The span is 17 feet if that matters.

- posted
16 years ago

- posted
16 years ago

I was wondering, what beam offers greater ridigity (or has less deflection,) let's say a 4"x4" H beam or a 5" I (& about 3" wide) beam? The span is 17 feet if that matters.

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- posted
16 years ago

You can download Beamboy.

I did the calcs with beamboy, just to see if it still worked. :) Used it quite a bit for a support project.

Your 4x4 H is proly a W 4x13 (13 is lbs/ft), and will deflect .261" under a

500 # point load at the center. Your 5x3 I beam is proly a S 5 x 10, and will deflect .24"; A 5 x 14.75 will deflect .194"The first number always gives more stiffness than the second two, ergo the lighter 5x10 is slightly stiffer, and the nearly-similar 5x14.75 is twice as stiff as the 4x13.. I'm surprised, tho, that the deflections were that small. Must be thinking of wood. :)

These are common sizes, iirc, in steel suppliers. About 50c/lb, here in N'Yawk.

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- posted
16 years ago

Look in Machinery's Handbook.

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- posted
16 years ago

If you go the MH route, I think you just want to compare X-X moments of inertia, to compare stiffnesses. But it won't give you deflections--you'd have to know how to calculate them. Presumably MH will give you that formula as well.

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- posted
16 years ago

Mr. P.V.'d,

Thanks for mentioning beamboy -- that'll come in handy for an upcoming project. FYI, when I googled for it, the first hit was URL:

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which lists a few other freeware progs (which I've yet to look at in detail.)
Thanks again

-- Larry

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- posted
16 years ago

No problem. Someone here a while ago had mentioned it, and I too was grateful. :)

Had a bit of a problem downloading it, iir. Think I had to try a couple of places. But then I'm a pyooter moron.

Don't know if I dropped the beamboy author a note of thanks. I think it's a good idea to let authors of freeware know when their stuff has proved helpful. Good karma.

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- posted
16 years ago

URL:

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To get reasonably accurate results it is necessary to describe the beam section with a minimum of: Beam height, flange width, web thickness, average (midpoint) flange thickness, ideally using a vernier caliper to take the dimensions while watching for burrs or excessive paint! For HSS -Hollow Structural Steel- section, the outside dimensions and the wall thickness are the governing factors.

Stating that a beam measures say 8" high x 8" wide, as determined by tape measure, is insufficient because there are at least 3 beams of differing section modulus that fall into this size range, and rolling mill tolerances don't make it any easier.

Occasionally even with all the listed dimensions given it is ambiguous which beam one is looking at. When I am faced with this dilemma I simply pick the next smaller size in the table and use its section modulus. This occasion arises when I am asked to certify the load capacity of an existing lift beam or mezzanine floor, for example.

Technical short-hand would describe an I-beam as W8x10, or 6x6x1/4 HSS, for example. The I-beam short hand is of the form of beam height x weight/foot of length. To really confuse the issue, consider an I- beam described as W12x190, superficially a 12" high I-beam weighing

190 lbs per foot of length. Looking in the beam tables you would find that this beam is really 14.38" high! So care is required in measuring beams and the use of tables.With the beam information as described above one can look in the tables of standard structural steel sections (I-beams, angles, channels, HSS, TEEs) and locate the section modulus (dimension in cubic inches), and the second moment of area, sometimes called moment of inertia, symbol 'I', for the applicable section.

With this section modulus and moment of inertia one can calculate the bending stress and beam deflection, given the loading condition, the beam span, and beam end support conditions. These calculations are fairly straight forward; the difficulty is sometimes in determining the bending moment in the beam when the loading is uneven or weirdly distributed.

A caveat: The allowable load (bending moment, actually) for a given beam also depends on the (laterally!) unsupported span of that beam. Consider a beam supporting a floor or a roof. This beam is continuously supported along its length and as a consequence can support a heavier load than say, an identical beam used as a runway for a monorail hoist. This load capacity difference is greater than one would get from the difference due to point loading (hoist) and floor/roof loading which is evenly distributed for design purposes.

Deflection of a steel beam is independent of the steel's strength (at least within the meaning of the concept of 'beam'). The allowable bending stress, however, is definitely a function of the steel's yield strength. 'A36' denotes a structural steel (cheapest grade) with a yield strength of 36,000 lbs. Structural steels are available with yield strengths of up to 100,000 lbs or so. Because floors are generally designed for deflection which is independent of steel strength, A36 is typically used in this application.

Columns are a whole 'nuther issue.

Trust this thoroughly fogs the issue! :-))

Wolfgang

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- posted
16 years ago

URL:

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Just a few things more:

When designing you need to apply an appropriate load factor to the various loads if they are dynamic, for example a reciprocating compressor on a mezzanine floor, or a powered bridge/trolley hoist. These factored loads get plugged into the design formulas.

And then you have to apply the right factor of safety FOS.

For building structures: 1.67 FOS For material hoists: 5 FOS For people hoists: 10 FOS and a whole bunch of other stuff for public use. Material lifts (fork lift truck) minimum 3 FOS but usually much more to allow for wear and tear and impact loading.

Hoisting (running) cables 8 FOS, often much more to allow for wear and tear.

Hence the art of engineering.

Wolfgang

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- posted
16 years ago

Beamboy has a table of standard beams to choose from. Whachoo see is whachoo get.

I translated for the OP his "beam size" to what he would find in the table/steel supply houses. With the help of a Ryerson structural catalog. :)

You stated deflection is independent of steel (yield) strength. I was about to compare a steel beam to an aluminum beam to dispute this, but then realized the density diffs would not allow the same geometry AND the same lbs/foot.

So, comparing steel w/ steel, you are saying that a really soft shitty HR steel beam would deflect the same amount as a super-tough alloy beam of the same dimension, with say, a point load dead center of the span? I guess you did, with A36 vs. A100!!

Counter-intuitive, but very inneresting, if true. How do you explain this? Don't bending moments directly impact on a property like yield strength, basically stretching??

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- posted
16 years ago

I was playing around with Beamboy, an nice freeware program.

For the "E" value , or modulus of elasticity, I use 29,000,000 which is the standard value for steel regardless of the alloy composition. Different alloys will not make much difference in this part of the calculation from what I've read.

On the X-X or Y-Y value for the moment of inertia, not quite sure which one to use or what that denotes.

At any rate I was hoping to provide some preliminary calulations to impress an archtitect, but there seems to be set beam sizes in the span tables R502.2.4 of the International Building Code, of which I don't have a copy and I'm not buying it, so I'll leave it to him.

Tony

"Proctologically Violated©®" wrote in message news:8qOli.1102$ snipped-for-privacy@newsfe12.lga...

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- posted
16 years ago

You not only have to come up with a beam size but you want to use a standard size so the price is less per pound. Some sizes cost more per pound than others. Hope he has that book.

John

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- posted
16 years ago

PV,

"=2E..if true..."??????!!!!!! :-))

I think what you are forgetting is that the allowable bending stress is a function of the yield strength of the steel. If you're fishing in the dark as it were (a new design), you would calculate the bending moment first, then check that the allowable bending moment has not been exceeded for that laterally unsupported span, then the bending stress, then the deflection. (Some people do the deflection first, on the assumption that if it is acceptable then the others will be OK. With much experience in this particular area one can exercise this judgement, especially if an employer pays for the errors and omissions insurance. :-))

If the deflection is too large pick a stiffer beam (larger section modulus) and try again. Same if the allowable bending moment was exceeded for that span. Published beam selection tables make this a straight forward task. The fine points of getting the load into the section, such as a large capacity lift beam for example, is where the art of engineering, and experience, come in.

Let's say you have 2 identical beams, same span, same section modulus, etc. etc., one A36, the other Q100.

The A36 beam can carry a load (point load if you wish) with the bending stress =3D 36,000/5 =3D 7,200 PSI. (let's ignore the fine point of shape factor here, which would permit a slightly higher stress for a safety factor of 5).

The Q100 beam could carry a load until the bending stress =3D 100,000/5 =3D 20,000 PSI. Of course the deflection with the Q100 beam stressed to

20,000 pSI would be greater than the A36 beam stressed to 7,200 PSI. (5 is the factor of safety for a material hoist).But, with both beams stressed to 7,200 PSI or less the deflections would be the same.

(you would still have to check the bending moment against the allowable for that span. For hoist beam design there may be a lesser value for the bending stress in the compression flange, than the normal beam design would permit).

Where deflection is not a consideration one could use a much lighter beam of Q100 steel for the same load.

The deflection formula for the above scenario with a simply supported beam is: delta X max. =3D (Pl******3 / 48 EI). Notice that the steel's yield strength does not enter into this calculation! P =3D point load at centre of span, l =3D span, E =3D modulus of elasticity =3D 29.5 x 10******6 for all carbon steels and does not vary very much with alloy steels. Many designers simply use 30 x 10******6. I =3D section's moment of inertia.

To go back to your comparison with the aluminum beam:

2 identically sized beams, identical loads and span, bending moment not to exceed the lower of the two permissible, what would be the relative deflections?E =3D 10 x 10******6 for aluminum. Consequently the aluminum beam would deflect 3x as much as the steel beam.

Clear as mud? :-))

Wolfgang

P=2ES.: If you know of a design program that reliably calculates the section modulus (inches******3) and the moment of inertia (inches******4) from a section's cad drawing, I'd really appreciate letting me know. I often run into cobbled-together-sections to analyze, and I'm tired of doing these tedious calcs by hand. W.

- Vote on answer
- posted
16 years ago

Then there is the method of: Sheeit, Jethro, it's not bending ***too much***, should be OK.... :)

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- posted
16 years ago

Use X-X. Beamboy uses a default E, I would stick with that.

Get a Ryerson or Thypin stock list book, if you can. Proly on line. Yer local steel distributor might have an in-house one, or sheets. Basic, useful dimensional stuff, with web thicknesses, flange widths, etc.

If you find such a stock book, esp. w/ laminated pages, guard it with yer life! :) Hide it.... no foolin...

- Vote on answer
- posted
16 years ago

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wFor a steel beam I usally do the strength calc first on the assmuption that the beam is stiff enough. On a wood beam I do the deflection calc first since that is normally the issue. But one must ALWAYS do both, and then do the shear calc if it is the least bit non standard.

As for steel beam data, this is a nice data base

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Note on the need to do the shear calc: About 15 years ago a contractor was rebuilding a 100' high concrete arch over the Mississippi river. They put in a temp beam bridge under it to support the arch during the project. 80' in the air, 192' long. Since it was TEMPORARY, they did the strength and deflection calc, skipped the shear calc. 400 tons of fresh concrete dropped the whole works into the river. The foreman got all the topside workers off, wound up riding in down. He survived but one worker down below was killed. 2 PE's lost their licenses.
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- posted
16 years ago

Most professional CAD packages have section property functions included. As for 'reliably,' a few test cases with known answers would be advisable, both to check accuracy and to verify the user's understanding of nomenclature/sign/units conventions being followed. In limited experience with several CAD programs I have encountered errors of both types. Before CAD software became available to me, I used to use a handy little BASIC program from a softcover book of engineering application programs for the Apple II+ computer :-)

David Merrill

P.S.: If you know of a design program that reliably calculates the section modulus (inches******3) and the moment of inertia (inches******4) from a section's cad drawing, I'd really appreciate letting me know. I often run into cobbled-together-sections to analyze, and I'm tired of doing these tedious calcs by hand. W.

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- posted
16 years ago

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Roy,

Good point on the shear stress!

Often I do this mentally from the beam reaction load and web section area because generally this stress is quite small; with shortish beams I definitely calculate it. I very seldom work with wooden structure, they are hard to weld. :-))

Wolfgang

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- posted
16 years ago

You just have to use the right rods for wood beams: Heat melt glue! ;-)

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- posted
16 years ago

I do mix and match with both steel and wood beams in old houses. Things like opening up a load bearing wall. 12x22 'W' shapes and similar.

For this sort of work it's interesting that a steel beam and a engineered wood beam have roughly the same cross section in terms of space taken up. ie a pair of 12" LVL beams are roughly equal to a 12"x22 steel section in strength, both take up about 12"x4" of building space. The steel is stiffer if that is the limiting factor but more work to attach joists and such to the sides. Easist way to attach to the steel is powder fasteners but 1-1/2" of wood and 1/4" of steel takes a bigger load than my tools will handle.

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- posted
16 years ago

I thought of 'welding' it with WELDWOOD.

Wolfgang

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