I know that there are a lot of factors to determine that exact number for load capacity of a I-beam, but is there a simple answer to this?

How tall, wide, thick will the I beam need to be to handle 3,000lbs or less and be supported on the ends only, 50 feet apart?

I looked at some formulas but it got confusing (chair)

Thanks

The formulas get confusing because they tend to calculate deflection and stress of the beam. To use the formulas, you have to determine what you consider "failure" of the beam. If it plastically deforms under the 3000lbs load? Can it deflect 1" in the center?

Do you mean actual load capacity, or are you concerned with deflection also? For most CNC applications, you are not concerned with load capacity because deflection will become excessive long before reaching maximum load capacity. Either way, you can download the free program BeamBoy to calculate stress and deflection. http://www.geocities.com/richgetze/

I am not really sure, but I believe the maximum bending stress for structural steel is around 20ksi. For a 50ft span, BeamBoy says it would take at least an S10x35 beam to stay under 20ksi with a 3000lb load, plus the distributed weight of the beam. The deflection with that load would be over 4 inches. At 35 lbs/ft that beam would weigh nearly a ton. What are you building?

Sounds like he wants to put a beam across his shop and put a trolley on it to move machinery around.

...How tall, wide, thick will the I beam need to be to handle 3,000lbs or less and be supported on the ends only, 50 feet apart?
Thanks

Approach the problem from the opposite direction. Find some standard I-beam profiles and the I value, that is the Moment of Inertia, for that profile then plug that into the formula along with your length and calculate the maximum deflection at the center. If the deflection is too much go to a bigger beam; bigger I and calculate again.

I can tell you 3,000 lbs in the center of a beam spanning 50 feet will require a hefty beam unless you can accept several inches of deflection.

2muchdtuff, you got it. It can bend and deflect all it wants as long as it says near the roof and not on my head.

So a 12" I beam will do the trick to be on the safe side.

When I get the winch to the shop need to see if I will have enought hight for the trolly, winch, i-beam.

What about two smaller I-beams side by side?

Years ago I had an engineer figure an I beam for my garage. He spec'd a 6" by 12" beam for a 24' span. With your 50' span I might want to go a little larger. As for the idea of side by side, I wouldn't. I'd weld the one on top of each other, this will give you less deflection.

Is it possible to support the center of the beam down from the roof or do you have an open steel truss girder arrangement holding up the roof deck.

I would think that at 50 feet, you are in the range of having buckling problems sideways as well as deflection problems. 2much is right, you should try to support it in the middle. Since the trolley will just ride on the lower flange, that leaves the upper flange free for support purposes.

And this is one application where going to an engineer that does these kind of things would pay off.

Here is a thought. Rather than mounting a huge beam, can you make some A-Frame supports that travel as well? I haven't seen this done, but if you can position a travelling A-Frame either side of your hoist, you can ensure smooth travel over a 50 foot difference and only slightly more inconvenience (due to positioning the supports).

Best,

BW

I know that there are a lot of factors to determine that exact number for load capacity of a I-beam, but is there a simple answer to this?

How tall, wide, thick will the I beam need to be to handle 3,000lbs or less and be supported on the ends only, 50 feet apart?

I looked at some formulas but it got confusing (chair)

Thanks

Dear DennisCNC,

You say you want to handle 3000lbs, and the beams will be, I assume, "simply supported". Do you want to support 3000lbs at the middle of the span, or will it be distributed along its length. The loading conditions are completey different, and may call for different bits of steel. If you can give more information, I'll get out my text books.


Best wishes,

Martin

^^ I measured the shop and it is 45' not 50' so thats a plus. I will use it move granite slabs around, from the cart to the saw and so on. Maybe my 3K lbs estimate is a little high. The slabs weight is 1,200lbs. max plus the winch, trolly, vacuum lifter, I think that will be another ~700lbs. So I think a better estimate would be 2,200 lbs. max load. This load will mostly be in the middle.
I can make the trolly support wheels farther apart to spread the load. If I happen to need to move something heavier i'll just use the forklift. With the over head crain one person can move the material around.

^^ I measured the shop and it is 45' not 50' so thats a plus. I will use it move granite slabs around, from the cart to the saw and so on. Maybe my 3K lbs estimate is a little high. The slabs weight is 1,200lbs. max plus the winch, trolly, vacuum lifter, I think that will be another ~700lbs. So I think a better estimate would be 2,200 lbs. max load. This load will mostly be in the middle.
I can make the trolly support wheels farther apart to spread the load. If I happen to need to move something heavier i'll just use the forklift. With the over head crain one person can move the material around.

Dear Dennis,

unterhaus is right on the button with with comments on buckling.

If the top flange of the I beam is not restrained (ie prevented from moving sideways), and you have a clear span of 45 foot, the permissible bending stress for the beam will be far less that the nominal value for the steel. In order to get round this problem, you have to go for a larger section.

I have not done steel calculations for a while, but I thought I'd have a go.

My conclusion is that if the ends of the beam are not built in to something substantial, and if the top flange is not restrained, the major factor is keeping the bending stress to acceptable limits. If you have a beam of steel which has a basic permissible bending stress of about 23000 lbs per square inch, by the time you make allowances for the span and the lack of restraint, the actual bending stress that the beam can handle is down to about 6100 lbs per square inch under these conditions.

If my calculations are right (and it is possible they are not) you might consider this beam....

24" deep
9" wide

top and bottom flange thickness 11/16"
web thickness 7/16"
beam self weight 225 lbs per yard

The steel should have a basic permissible bending stress of at least 23000 lbs per square inch.

The total deflection at mid span (including that due to the self-weight of the beam) will only be a little over 1/4". The bending and shear stress are within limits but I have not checked for web crushing and buckling.

Finally, please don't go ahead and use this beam without getting a structural engineer to OK it. I can't afford the lawyers fees!

Best wishes

Martin

From post #12 martinw
24" deep
9" wide
3375 lbs

I think that qualifies as a 'hefty beam', even more hefty than I anticipated when I made the comment above.

The correspondence between Martin's specs and those from post #7 2muchstuff "He (an engineer) spec'd a 6" by 12" beam for a 24' span" is quite good; roughly four times the beam area, i.e. roughly 8 times I, for twice the span.

So now all that is needed is to spec out a beam capable of supporting 3375lbs so the beam capable of supporting 2200 lb can be installed.

From post #12 martinw
24" deep
9" wide
3375 lbs

I think that qualifies as a 'hefty beam', even more hefty than I anticipated when I made the comment above.

The correspondence between Martin's specs and those from post #7 2muchstuff "He (an engineer) spec'd a 6" by 12" beam for a 24' span" is quite good; roughly four times the beam area, i.e. roughly 8 times I, for twice the span.

So now all that is needed is to spec out a beam capable of supporting 3375lbs so the beam capable of supporting 2200 lb can be installed.



Dear Geof,

A span of 45 feet is an entirely different animal from one of 24 feet.

On long spans, and 45 feet is, for a simply supported, unrestrained steel beam, at about the limit of design, there is a danger that the compression forces in the top flange of an I beam reaches a level where it is acting as a column in compression and may buckle. Steel design takes this into account and reduces the allowable bending stress quite severely. It results in a deep, and wide section on long spans to reduce buckling of an unrestrained top flange of the I beam. The choice of "I" has , on long spans , little to do with deflection. All that matters is that you have a sufficient "Z", and an an l/ry ratio to keep the allowable bending stress within limits inter alia to prevent an un-restrained top flange from buckling.

IMHO, wear a crash hat and take out a lot of life insurance.

Best wishes

Martin

you also need to consider the whole weigth of the beam itself pretty closely thats a big distributed load...I can get out my mechanics book and pull the formulas but I don't think its really feasible. let me toss it in cosmos though with a standard-ish large I beam with the load at the center

UPDATE 7-24-06, pointed out by bma137, I used ultimate strength rather than yield strength, The numbers will be fixed

Oh the bad memories from Mechanics of Materials class..... I will take a look in my book, no gaurentees that I want to complete this;)

You want at least a factor of safety of 2, so lets say 5kips.(5000lbs) Theres also tons of different styles of beams. I wish I knew what was the most common...

wow this is hard to remember.

I'm going to calculate this for a wide flange shape (W shape)


From my notes in class:
Design should be Economical.
Choose beam with the smallest weight/unit length
1. Choose material, find σ ultimate for the material
2. Find σ allowable = σult/FS (FS= Factor of safety)
3. Draw shear and bending moment diagram.
4. Find Max Moment and location
5. Use eq σmax = Max moment/S
6. Find minimum S from Apendix C of my book, Properties of rolled - steel shapes.

The shear and bending moment diagram might not be necessary, it might confuse some of you but basicly you find the max moment, which in this case for the max stress would be in the center.

I'm going to use structural steel: ASTM-A36: σult=56KSI (56000 PSI)SHOULD BE YIELD STRENGTH OF 36KSI
Modulus of elasticity is 29*10^6 PSI(I dont remember if I need this yet. still gotta remember the symbol for it)

Getting to the simply supported beam, 2 fixed supports.

σyield=36KSI
Max load = 3000lbs(FS goes with sigma σ )
FS=2
Mmax= 67500 ft-lbs at 22.5ft
σall = σult/FS = 36KSI/2 = 18ksi

S=Mmax/σall = 67500/1800lbs(must have units correct) = 37.5in^3

So going to the table we have these options:

PREVIOUS NUMBERS WERE INCORRECT
W12x30 Area = 8.79"^2 Flange width = 6.520 Thickness = .440 Web = .26 S=38.6

W16X26 Area = 7.68"^2 Flange width = 5.50 Thickness = .345 Web = .25 S=38.4

The second would be the ideal choice (W16x26) with it being the cheaper one.

There are a few extras that I did not list because your looking for the smallest number after the W# W is the width, so W12 is 12" wide(big flat to big flat) and the 30 kind of like the weight/ unit length,

So W16X26 would be the best choice.


Someone might want to double check these, its kind of early in the morning:)
-been doublechecked 7-24-06

Hope this helps more than just one of you! I included a sample problem with a shear and bending moment diagram, with a few loads and different support.

-edit- be careful, this is not engineering advice and I will not be liable of it.

Images have the wrong ones underlined because of reasons listed above.

Jon

http://www.efunda.com/math/areas/IbeamIndex.cfm - See if this will meet your needs.

Enjoy,

alright can't find the mechanics book but that looks right JF. Because of the vertical loading I would definitely aim more towards a narrow tall beam. Obviously I'm not giving out official engineering advice over the internet and all the legal speak but... Cosmos says factor of safety of around 4 with a W12x22 beam loaded with a 5k lb load at the center, the weigth of the beam and restrained at the end. I would be more worried about the realities of your building supporting all this safely.

avsfan, you kinda reminded me that I needed a disclaimer on that. Its more for a general idea of what you should do, and yes, the building and everythign else needs to be able to take the load.

Jon

JFettig, I noticed that you use the ultimate stress of A36 structural steel to find your allowable stress. I'm an ME undergrad as well and I know that you should never use the ultimate stress to design anything like that. You need to use the yield strength of the steel to find the allowable stress, and for A36 I believe it is a minimum of 36ksi.

Like I had said before, never use the Ultimate Stress to design unless you are designing it to fail.

P.S. that text book looks like the exact one we use.

I realize this sounds like cheating, but is it practical to buy a gantry? Wallace Crane (http://www.wallacecranes.com/gantries.htm) is one company that make portable gantries. I am very cautious about designing something to lift and carry heavy items around or over people. If you have a defined straight path you carry the load, you can put channel or track on the floor to guide the gantry.

Advantages: The beam can be short (size it to the load instead of the room). Someone else is liable for the design of beam and its capacity. You do not need to modify the building.

Disadvantages: You have to pay for it, but I do not know how that compares to modifying your building.

well I used some I-Beam on my house (it was cheaper the lvl-beams at the time)

BMA137, thanks for the correction, I bet your right.

Its possible that I screwd up and didn't remember what one to use, but its even more likely that my idiot teacher that has never had a job in the field, just got his degree and turned stupid taught us, I'm not just saying that, anyone who has had him would say that, my boss at work had him and says that.

I'll edit my post saying how its not correct and correct it.

Jon

You guys are over thinking this quite abit!

just look at a commercially available trolly system that can lift what you want and COPY IT! (barring any patent infringment of course)

If the company has been in business for very long they have all of the math worked out for the load that is advertised! If they didn't they would have been litigated out of existence!

Not to mention if you want swinging stability you will probably need 4 trolly rollers spaced out so now you math gets more complicated buy the fact that you have multiple contact points spaced away from center. if you don't do the 4 trollies you will probably be wearing out components very quickly from the swinging force.

And who ever said a factor of 2 for building should maybe go to maybe a factor of 5 instead. As all you have to do to blow a factor of 2 is to bounce the load and you now got a big bow in your beam.

^^ I like your thinking (are you Chinese?)

Thank you all for input of this!!
I'm renting the building that i'm in right now, I'll see how things are going to go before starting to put together a moster like that.


Dennis

^^ I like your thinking (are you Chinese?)

dude! :boxing:


It is legal to use even patented ideas for personal use!

I would never condone selling something that is a patented product (even if I disagreed with the patent issued)

copyrights however I have other opinions on!
(nuts)

[QUOTE=DennisCNC(are you Chinese?) [/QUOTE]
He's not Chiness and I don't think He drinks there beer ether.

^^ I like your thinking (are you Chinese?)

Thank you all for input of this!!
I'm renting the building that i'm in right now, I'll see how things are going to go before starting to put together a moster like that.


Dennis

Interesting Thread

JFettig, understood, I know exactly what you mean with your teachers.

nope.

I dont drink chineese beer but if you got something I should try, give it up man!

:D
:cheers:

[QUOTE=DennisCNC(are you Chinese?)
He's not Chiness and I don't think He drinks there beer ether.[/QUOTE]

Off topic but what does Chinese Beer look and taste like? Schafers? Schmits? Dog Urine? :rolleyes:

Miljnor could not possibly be Chinese, he writes english very well and I don't have a hard time understanding him.

I am a Mutt but I don't believe chinees is in my pedigree. ;)

as far as beers go. I don't drink alot of common beers as most tast like dog urine, but as far back as I can remember chinees are right up there with the common beers.

there are you happy lakeside? you are evil! changing yet another post into a beer post! :D

and I always get the blame! damn! ;)

:cheers:

You undergrads and recent grads are making me feel old!

Especially in a rented building, I still recommend buying a portable gantry. If you change locations, you disassemble the gantry, put the parts in a truck and take it to the new building.

If you use a short beam, you will have more usable headroom (more headroom is always better). Between the beam, the trolley and the hoist, headroom disappears quickly.

As a different approach, is it practical to get a cart of some sort under the load? It changes the problems a bit, but the part of the building that matters is the floor. It is something to ponder.

Deflection might be significant for you. When you lift the load and the beam deflects, the load will try to drift toward the center of the beam. Depending on the load and the deflection, the force could be large. You also have to overcome that force when you move the load away from the center of the beam.

In my experience, if a tool or piece of equipment is available off-the-shelf, it is usually less expensive in the long run to buy it than to build it in-house. Let the experts design and build it.

In this era of plentiful lawsuits, do not assume any liability needlessly (that is my PE license talking). Yes, you are still liable for selecting the proper equipment, assembling it properly, maintaining it properly and operating it properly, but you can avoid the problems designing the equipment, buying the proper materials (it that beam really the grade you think it is?) and fabricating it. You can also improve the safety of the equipment (and reduce your liability) by paying someone to inspect your lifting equipment, by locking it when no one is supposed to be using it (we had someone break the gearbox in a 5-ton hoist by misusing the hoist afterhours) and by teaching everyone who might use the hoist on how to use it correctly.

Oh, my goodness! I sound like the dreaded Safety Inspector! It has been that kind of a week already. In short, be careful and do not let your computations get in the way of your common sense.

Bob

who would have though that a post on I-beam selection could get this out of hand?

aint life grand. :D

Yes I should not talk about Beer
So to all you kiddies out there a mind like a good beer is a shame to waste.
I'm Irish and Polish so I have dumb luck

I'm building a bridge for a 5 ton load (tractor and load of firewood)
It is a 20 foot span, and I've used 3 10inch I-beams (scrounged up for free)
I think it is overdesigned, but it would be interesting to know for sure.

I don't care about the amount of deflection (an inch or two is fine).
What is the deflection? Is it safe for a 5 ton load? Is it safe for a 10 ton load? Assume the load is at the mid-point.

This is probably a very simple calculation for those who have done it before.

Thanks
Pete

Dear Pete,

Presumably you thinking of arranging the beams "plank-wise"? Can you give any more information about the dimensions and properties of your beams?

Best wishes,

Martin

My engineering (???? lots of big questions marks because I am not an engineer) is done by seat-of-the-pants and gut feeling. And both of them are saying do not drive your five ton tractor, let alone your ten ton tractor over your three 10" I beam bridge at least until after you have written your will, assigned an executor and got all your affairs in order.

For bridges a good overload factor is maybe 5x so you should build something that can support a static load of 25 or 50 tons. That way when your 5 or 10 ton load bumps and bounces over it you do not overload it with the shock loads.

Find a reference website that will give you tables for the value of I for standard I beams and another site that gives you the formula for calculating the midpoint deflection of a simply supported beam and do the calculations. If your bridge is going to deflect more than about 1" with the 25 ton load I am not going to drive your tractor over it.

And if you don't know what I is or what a simply supported beam is don't build any bridges until you do.

dude! :boxing:


It is legal to use even patented ideas for personal use!

I would never condone selling something that is a patented product (even if I disagreed with the patent issued)

copyrights however I have other opinions on!
(nuts)

What makes you think that you can legally "use patented ideas for personal use"? (And by the way, one can patent machines, one can patent methods, one cannot patent ideas.)

I'm not a lawyer, nor do a play one on TV; but I have interacted with patent lawyers.

Ken

Dear Pete,

Let's look at your bridge from another angle.

Is it a short piece of rough ground that needs to be a bit flatter for your tractor, or does is span a yawning 20 ft wide ravine that is 130 feet deep and has vertical walls??? If it is the latter, I personally would consider the consequences of structural failure a bit more seriously than if it was the former. (Mind you, a 5 ton load of metal and lumber toppling on to you even on a flat surface will probably have the same result).

End of party-pooper section....

OK, you say you have 3 12" deep I beams.

How wide are the top and bottom flanges, and how thick are they?

How thick is the web?

How far apart are the three beams, and what is the "deck" that forms the bridge?

Can you imagine a loading situation in which (say by driving the tractor in a wonky manner) only one of the three beams takes the load for a second or two.

Please do not get me wrong. I'm not trying to design your bridge, I'm just trying to find out if your beams are complete "no-hopers" or otherwise.

Best wishes,

Martin

Hi Martin,
These 3 beams are in standard "I" orientation, 10 inches high, 4 inches wide.
It is a 20 foot span, held at each end.
I pulled these out of my cousin's construction scrapyard, but they look like common steel extruction.
What I am interested in is the load capability per beam. For instance, what is the deflection for a two-ton midpoint load? If it is minimal, I know that my 3-beam bridge is safe for a 5-ton load with a one-ton wooden deck of 3x6 hemlock planks.

Thanks,
Pete

Dear Pete,

Presumably you thinking of arranging the beams "plank-wise"? Can you give any more information about the dimensions and properties of your beams?

Best wishes,

Martin

It's free. http://www.geocities.com/richgetze

I calculate a deflection of .31 inches at the center of the beam with a load of 4000 lbs. That's for an S10 x 25.4 beam. The beams in BeamBoy can be selected from a list of standard beams, or you can specify your own. S10 x 25.4 means that it is 10" deep and weighs 25.4 lbs per linear foot. That is the lightest 10" beam I saw listed.

Ken

....If it is minimal, I know that my 3-beam bridge is safe for a 5-ton load with a one-ton wooden deck of 3x6 hemlock planks.

Thanks,
Pete

No you do not.

When you drive anything over any bridge there is no way that the load is guaranteed to be distributed evenly. Your bridge has to be strong enough so that it can support your working load concentrated at any point on the bridge. In other words each beam needs to be able to support your working load individually then you may be able to conclude that it is safe.

you could also hang an extra trolly with threaded shaft hanging down. then tap a hole in plate welded to pipe. slide trolly where you want it and screw pipe down to extend to floor for extra support. when finished screw pipe up and slide it out of the way.
just my thoughts Dar

Dear Pete,

I guessed what the weight per foot was of your beam, and hand calculated that you would get a deflection at mid span of about 11mm (somewhere between 3/8" and 1/2". That is with a STATIC load of two tons at mid span.

OK, you may say, "I can live with that"...well err, no.

With a reasonably long and narrow beam, the top flange tends to buckle sideways under load, unless it is laterally constrained. In order to take this into account, the basic permissible bending stress for steel is reduced.

I did the bending stress calculation, and your beam will be asked to take twice the stress that it safely could .

Add to this the point raised by Geof about dynamic forces being considerably greater than the static one of two tons, and I would say you are in very dangerous territory indeed.

Sorry.

Best wishes,

Martin

Dear Dennis,

unterhaus is right on the button with with comments on buckling.

If the top flange of the I beam is not restrained (ie prevented from moving sideways), and you have a clear span of 45 foot, the permissible bending stress for the beam will be far less that the nominal value for the steel. In order to get round this problem, you have to go for a larger section.

I have not done steel calculations for a while, but I thought I'd have a go.

My conclusion is that if the ends of the beam are not built in to something substantial, and if the top flange is not restrained, the major factor is keeping the bending stress to acceptable limits. If you have a beam of steel which has a basic permissible bending stress of about 23000 lbs per square inch, by the time you make allowances for the span and the lack of restraint, the actual bending stress that the beam can handle is down to about 6100 lbs per square inch under these conditions.

If my calculations are right (and it is possible they are not) you might consider this beam....

24" deep
9" wide

top and bottom flange thickness 11/16"
web thickness 7/16"
beam self weight 225 lbs per yard

The steel should have a basic permissible bending stress of at least 23000 lbs per square inch.

The total deflection at mid span (including that due to the self-weight of the beam) will only be a little over 1/4". The bending and shear stress are within limits but I have not checked for web crushing and buckling.

Finally, please don't go ahead and use this beam without getting a structural engineer to OK it. I can't afford the lawyers fees!

Best wishes

Martin

Do factory inspectors look for engineering certification for cranes?
Fork lift is already approved.
If you cant afford the structural engineer, use the fork.
:nono: