

I am trying to determine the size of IBeam needed to support a given weight at the center of a known span.
I.e., if I am building a bridge hoist for a trolley and chain fall, and the span is 9 feet between supports, what size beam do I need to support 3,000 lbs.? I have looked in Machinery's Handbook, but the info there is more than a bit confusing to me.
Can someone point me to any information regarding beam sizing/load for modern IBeams?

wilbilt, a number of configurations/shapes might suit your application. Do you already have a beam selected (or onhand material to weld up one)?
If so, you could provide us with the dimensions and material type  someone can most likely give you an estimate of the suitability for your needs.

I don't have a beam or other materials yet. I am hoping to be able to determine which size/material I should be looking for.
A local general engineering contractor has a pile of galvanized IBeams from highway signs, but they may be much larger than what I need.

A local general engineering contractor has a pile of galvanized IBeams from highway signs, but they may be much larger than what I need.
For your application, bigger is better (especially if the price is right). Can you obtain the dimensions of this material (and its composition, if possible) ?
Not trying to finesse your question about how to select the right structural shape but there are a few things to consider in the process, particularly where safety is concerned. This is one of those situations when hasty or incorrect information could lead to catastrophe.
Generally, few people would be comfortable providing design information longdistance but if you have a length of material, we might give you a conditional indication of whether the material is suitable for the application that you describe.

I have a gantry frame crane with a 7" 15.3 lb (per foot)I beam with supports nine feet apart. I have had 35004000 lbs hanging on it. This is a home built affair, been in service over 15 years.
John

I have a table here. Lessee. A 9 foot span with a concentrated load of 3000 lb in the center. Hmm.
My table shows a 5" x 3" 10.0 lb per ft American Standard I beam will support a 3800 lb concentrated at a 10 ft span given a 24,000 psi max fiber stress resulting in a 0.4" deflection. I would go to a 6" section unless you had some form of lateral stiffening to keep the upper flange from buckling.

Be VERY careful !
It's not a good idea to design from a set of generic tabular data without applying some thought to the problem ...
Rolling loads (the trolley), for example, can induce up to twice the static stress on the beam. And how about the effects of acceleration/deceleration? (Would you want to be underneath the load if the chain hoist slipped six inches and then caught?)
For safety, the beam should be rated for at LEAST ten times the static load requirement, IMO.
This isn't a simple problem and simple answers can be sloppy ones. If a quick answer is required, find a commercial unit with similar (or greater) load capacity and copy the beam design.
Copying an established design provides reasonable assurance that the problem has been considered, analyzed and that the appropriate safety margin has been included.
In fact, the company lawyers have probably looked over the engineering calculations

The Americian Institute for Steel Construction publishes tables of load capacity of various sections. Perhaps that was the one Forrestt was quoting. This is the standard reference. Maybe your Public Library has one in their reference room. This book is full of footnotes, all of which are important. The most important is the laterally unbraced length. A beam has to be braced against lateral movement if it is longer than a certain length. If you do not do it, the beam can collapse under less than design stress, suddenly, without warning. NOT GOOD. In some cases, bigger beams with the wrong shape can have a lateraly unsupported length shorter than smaller beams with the right shape.
Because of this, I agree with the advice to copy an existing commercial design.
Thermo1

In fact, the company lawyers have probably looked over the engineering calculations
randyc, I understand your concerns, and they are duly noted. I am generally one to err on the side of extreme safety.
Using the 7" x 3.660" 15.3 lb/ft beam as an example, I applied my dull mind to the formula in Machinery's Handbook and came up with 6,000 lbs. maximum fiber stress at the center of the span (assuming I used the right equations).
There are terms used in the book such as "moment of inertia", which I assume has nothing to do with the inertia of a "drop and stop" with 3,000 lbs. on the hook.
Thanks for the replies. I will investigate the locally available material for suitability. A 10X safety factor might require a beam that significantly reduces headroom in the shop, though...

Call up a big building materials supply house they can give you the specs on IBeams. But always figure 20% over your need for the work load. Then the question is how are you going to suspend this hunk of steel. What are your supports made out of?
6 Inch I Beams would work, with diagonal support at 70 degrees with strength equal to the 6 inch I beam and reinforce your uprights with 6 inch I Beams also. But I can not recommend this and take no liability in this matter, I asked my brother who is an ME and an NDT III.
Jerry

I am planning to build the supports from the same material as the beam with diagonal bracing.
The supports will have footplates and be encased in 30" deep concrete footings (no frost here).
3,000lb loads would be an extreme, with "normal" loads in the 6501000lb range.
I understand dynamic forces, safety factors and "the weakest link". I also understand liability.
Whatever I end up with, it will be overbuilt, but I still won't get under it.

This is a structural engineering problem, not mechanical. Allowable loads from the Steel Manual (this is the bible for this stuff and surely available in your library) incorporates safety factors suitable for buildings. I agree that a crane that will hoist estimated loads, experience impact and rolling loads, and whose failure will cause injury or property damage, should be more conservatively designed, especially as you are only buying one of these, not a whole skyscraper full, and you don't have redundant members. But there's no point going overboard or there's no end to it. If you hurt yourself placing an oversized member (or have to skimp on some other piece of safety equipment because you spent too much on this), you haven't made a good deal.
Steel beams are rolled in three basic shapes that you might find used: wide flange columns, wide flange beams, and Ibeams (American Standard beams), an obsolete shape but still provided in small sizes (what you want). Each size (height) comes in different weights per foot, and the strength of "a 6" beam" can vary widely.
You may also need to consider deflection; a very strong long (in proportion to its height) beam might still sag enough to start your hoist rolling toward the center from a quarter point.
The calculations are not daunting and you can learn how to do them for a simple shape like this, or a portal (the beam plus its columns) in any structural design textbook. Or perhaps you can swap some machining service with a civil or structural engineer in your area for help with this design.

Find a copy of Kent's handbook. It has simple formulabased span info. Point and distributed loads, etc, etc, etc.
You might need an older one, I haven't seen a new one.
If you might have to meet a building code, the code already has the requirements spelled out for ordinary situations, no need for much calculation.

My Brother is a ME and NDT III that does testing on bridges, factory buildings and other industrial structures.
I am the Civil Engineer in the family, but I don't do that stuff any more, I do my metal art and leave the techy stuff to others.
Jerry

Machinery's Handbook is confusing because it's just a reference, not a howto  it assumes the reader has studied structural engineering and just needs to look up the formulas and constants. But calculating deflection of a point load on a simple beam is extremely easy. There are only three things you need from the book: 1) the modulus of elasticity (known by the symbol "E") of the material, 2) the moment of intertia of the beam shape (symbol "I"), and 3) the formula to calculate the deflection for your configuration.
All of these are in the chapter "Strength of Materials". Look at the section with tables for "Steel WideFlange Sections". Wideflange is the standard construction shape that most people colloquially call "I beams". Ibeams are actually a different and mostly obsolete shape but still available, as someone else pointed out. Wideflange beams are known by a designation like "W6x25", "W14x136", etc. The first number is the height and the second number is the weight per foot. There are typically around a dozen different weights for each height. The weight, and structural properties, vary because the width and the thickness of the flanges varies for each size within a given height. The WideFlange table in the Handbook lists the full dimensions of each size, along with the moment of inertia for each shape.
Moment of Inertia (I) is too complicated to explain and not necessary to understand beyond knowing that it is a number that reflects how rigid a given cross section is. For example, a solid round is more rigid than a tube of the same diameter, and has a different I (the specific value which of course varies, depending on diameter and wall thickness) For your purposes, you just need to find out the I value for the beams you might choose to use. These are in the Handbook tables.
Modulus of elasticity (E) is a very simple concept. It is merely a measure of how much a material stretches under load. Every linear elastic material has a constant E, and you just need to know what it is and plug it into the formulas. (A linear elastic material is one that will strech a given amount per given load. In other words if you applied n pounds of tension to a wire and it stretched x inches, and you apply 2n pounds and it stretches 2x, it is linear.) Steel is linear elastic, and its E value is 2.9E7. It varies slightly depending on alloy, but not enough to be relevant here.
Then there are tables titled "Stresses and Deflections in Beams", which has a series of drawings showing beam and load configurations. For example, a simple beam with a single load in the middle; a simple beam with a single load at some point that is not the middle; a beam with a uniforming distributed load; etc. The pictures are largely selfexplanatory. Foe each configuration, the table lists formulas to calculate stress and deflection at any given point along the length of the beam, as well as handy shortcut formulas for common things, like maximum deflection at the center of a beam.
So, with I and E (which describe the material and the shape of the beam), plus your parameters of length and load, you have all the ingredients to do the calcs. Note that for a simple point load at the center of a beam, the maximum deflection is simply the formula:
W * length^3 / 48 * E * I
So if I pick a W6x12, its moment of inertia (I value) is 21.7. Plugging your parameters in, gives us this:
3000 * 108^3 / 48 * 2.9E7 * 21.7 = 0.125"
3,000 pounds on the center of a 108" long W6x12 will cause it to sag 1/8" in the middle.
So that's how you interpret what you saw in Machinery's Handbook.
Here's the formula applied to the example Forrest cited from some handbook:
I of S5x10 = 12.3 (Ibeams are designated "Snxn")
3800 * 120^3 / 48 * 2.9E7 * 12.3 = 0.384"
His book said it would sag .4", so we're basically deadon here.
In closing, I would note that your question was what size beam is "needed to support", but the need must be defined. Probably you mean "need not to break and fall on your head." But note that no material starts out rigid and then just breaks. First, metals flex under load and return to prior shape when the load is removed; this is the elastic range. Second, when the load is high enough, the material flexes under load but does not return to prior shape, ie it bends. Third, when the load is really high, the material may break.
3,000 pounds is not a lot of weight relative to WideFlange beams, and even small ones will not bend, let alone break under a static load (3,000 pounds dropping 50' and hitting a W4x13 would be a different story of course.) So in many applications, deflection is the design constraint, not failure. For example, the Uniform Building Code specifies minimum building floor joist sizes not to a safety standard, but to a comfort standard  people don't like to feel the floor under them flex, even if it is perfectly safe.
So in summary, something from an S5x10 to a S7x15.3 (like John has) would be in order. A middle ground is a S6x12 which is twice as strong as a S5x10, and about 60% as strong as the 7 incher. None of these pose a failure risk in a relatively static load, home shop environment  they will just vary in the amount of flex.

You can do a good calculation using the formula S=Mc/I.
S= stress (psi)
M= bending moment (must be inchlbs)
c= distance from neutral axis to max stress (inches)
I= moment of inertia for the cross section (in^4)
I'm purposefully making this a little technical so nobody gets hurt by just plugging in some numbers.
The span is 9 feet, 108" and the max stress will occur at the middle, at 54". The bending moment applied is then 54 x 3000 = 162000 inlbs.
Since we are sizing for "c" and "I" simultaneously, we'd better plug in some values rather than trying to solve for them.
S, the allowable stress in the beam, definitely needs to be below 36ksi. But for lifting structures there MUST BE A SAFETY FACTOR of 2 to 5. Let's pick 4, so we want the S to be below 9ksi.
For a 6" Sshape x 12.5 lbs/ft (S is most common, the Wide Flange W or "H" beams are not as common), Ixx = 22.1 and c = 3. We get 22 ksi. Too much stress.
The right choice is a S10 x 25.4, Ixx = 124, and c = 5. S = 6.5 ksi. Anything smaller and you are making a foolish choice that will get somebody hurt.
So you want a 10" Ibeam (measured vertically parallel to the web). It weighs 25.4 lbs per foot of length, so the total beam is going to weigh 230lbs....better plan ahead for handling it while mounting or hanging...nothing worse than wrestling a monster piece of steel up in the sky and your arms getting tired. If you give out it's gonna nut you in the head....
BTW: Nothing personal to anybody but I would not trust my life to any calculations done based on beam deflection.
Good luck.
Matt

Go down to your local demolition or salvage yard and buy a 12 inch thick wall Ibeam for a hundred bucks and forget about it.Nine feet is nothing and bolt everything together so movement will not crack anything.This is if you have the height which you did not say anything about.I say if this collapses than sue me,I stand behind my over engineering.

If you look in the Grainger catalog there are several gantry cranes, in the material handling section. They have the specs in the catalog, for various sizes. The 400# gantry shows an 8" beam 4" flange, with a max span of 10'. These are supported with beam going straight down to the base. These are commercial use products. You could check McMasterCarr's web site also, they have gantrys too.
Diclaimer, I have no connection with either company and assume no risk in you using their design.
I agree with the others use as big a beam as you can get and put up, you may want the extra capacity later on.
Ben

Do you have room to add some welded on triangulation to the length of the IBeam?
That, combined with angled side supports may do it for you.

I am a mechanical engineer and have designed and used beams and channels to suport loads of 2000 pounds over about a span of 9 feet. It worked for me and corresponds with some of the replys, but I do not quarantee any advice I give here.
Safety factor was mentioned a few times but one was about 11/2 to one, no way would I use that low number. I also think 10 is an overkill. Machinery's handbook sugest 5 as a safety factor for structial steel if it has no unusual conditions like a moving load. I use that 5 number.
In the 12th edition of Mashinery's page 376 has a very simple formula that is simple when using standard American beams. It is for a concentrated load at the center ( worst case). It is s=Wl divided by 4Z. s is the stress at the beams center,W is the load in pounds. l is the span in inches. Z is the section modulus of a beam which is listed on pages 368 and 369. The Z very much simplifies the problem.
So using the formula s=Wl divided by 4Z and solving for the Z that fits the required beam you want. Here is my take Z= load times length in inches divided by 4 times stress you use.
So Z=3000 X 9 X 12 all divided by 4 X 6000 ( 6000 is the safety factor of 5 by dividing 30,000 psi yield strength of structal steel beams)
Z=13.5 . On page 369 a beam of 8" X 17.5 lb./foot at14.6 works. Or any of the other 8" will work.
If you support on pipe or similat you have to be very careful of buckling which is another problem.
Walt
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