# I-Beam deflection/failure

• posted

This seems like it should be a fairly simple question but I'm no structural engineer so have no idea how to go about it. I have a W10x39 beam (I think it's grade A36) that is 10 ft long. The top of the beam is supported by overhead attachments 1 ft from each end leaving an 8 ft span in the middle. A weight is connected to the underside of the beam at its midpoint. Theoretically, what is the maximum weight that this size beam can support before it fails?

• posted

The only info I get on W10x39 beams is from here:

It gives the beam depth and width. Can you provide the flange and web thicknesses also? It's easy to calculate if you can give me that...

Dave

• posted

Dimensions of W10x39 beam:

Depth of Section = 9.92 in Flange Width = 7.985 in Flange Thickness = 0.530 in Web Thickness = 0.315 in

Thanks

• posted

If it were that easy, engineers would only need 2 weeks of school, and there would be no need for PEs. High school students could do it.

---------- Lets see what you have and briefly what it means to the final number-

First, you are not asking what the fixture you describe can support, or what the beam can support safely, rather you are asking what the beam itself can theoretically support before failure, clean and immediately after beam fabrication, at room temperature, assuming it has no fabrication defects out of the ordinary?

Since you want to take it to failure, all those things forgiven by safety factors and structural handbooks go out the window and now must be considered

a) It depends on the material strength. You are unsure of the material itself.

b) It depends on the connections' affect(s) on the theoretical shape. The connections here are unspecified - pin or fixed, welded and how and what kind and by what level of expertise, bolted and where and to what level, able to change to constrained or not. The geometry and connection method of the "bottom" especially has a great deal to do with the level at which the tensioned flange fails, and so does the freedom of the upper connections.

c) It depends on the ratios of stress types and the failure mode, both brittle/ductile (from axes) and tension/compression . Once the basics of the stress axes are determined so you can determine if the failure will be ductile or brittle (which changes the failure level) and any derates included for connection stress and material loss determined, you will need to determine the capacity of the unknown material in tension and in compression so the corresponding strength value may be used. They are different.

d) It depends on which criterion you need as failure. This is a big one. Fatigue, proof, yield, necking failure, or complete parting? Or shock.

e) It depends on the rate of application of the load (and I don't mean dynamic, which can be backed in to derate the static load ). Slow, fast so there are dynamics, or shock with energy concentrations?

f) It depends on the symmetry of loading on the long axis. Stable load, directly under the web, deflection on the flange limited, probably negligible. Swinging, moving, offset connection reduces capacity.

And a couple other things.

however, if you want to know where a theoretically-perfect simply-supported A36-spec-steel symmetrically-and-slowly-loaded beam theoretically yields (yield as defined by the A36 spec) for intellectual curiosity, that is easier.

I'll check the AISC cookbook for that value later today

• posted

That's a lot to consider. Thank you for the detailed response. For my purposes though, I'd just be interested in knowing the answer to this part------->" however, if you want to know where a theoretically-perfect simply-supported A36-spec-steel symmetrically-and-slowly-loaded beam theoretically yields (yield as defined by the A36 spec) for intellectual curiosity, that is easier"

• posted

Moment of inertial=210in^4

Looks like for every 1000 lbs you add, it'll increase the max stress in the beam by 1030psi. So if you're ultimate strength is 50ksi (which is pretty low for most steels), you can load it with:

50,000/1030*1,000lbs=48,500lbs

For different ultimate strengths, just change the 50,000 part.

Anyone feel free to double-check my math.

Dave

• posted

I'm afraid the calculation of ultimate strength failure is a bit more complex than the calculation of yield failure. See AISC formulae, and the definition of ultimate strength re: cross section remaining in the necked-down test sample

• posted

i'm from the metric side of the globle, but i want to add just this thought so you want to calculate how much weight you have to hang on before the deformation of the beam will make it horizontally (projected) 20% shorter ? that is a serious deformation, are you sure the attachments will hold ?

1) the formula's for force calculation with elastic materials are made for little displacements (neglecting second order derivatives) 2) the forces on the loadingspoint will not be pointed vertical (this is not really a problem, but you mind need it) 3) the beam (i don't know the W 10x39 but i guess it's a large one) will be deformed plastical (using after formula's) and you might want to look the strain-displament curve of the material

so it all depends on how accurate you want you answer to be

• posted

I don't think that's what the poster is trying to do. The situation (as I see it) is this.

There is a ten-foot beam.

One foot in from each end, there is a support attachment on the top, 'hanging' the beam from a ceiling, or other support.

therefore, there are eight feet 'free' in the middle (ten feet total, minus one foot off each end 'outside' of the supports).

kinda like this? (fixed width font)

• * _____ | __________________________________________ | ______

horizontal line is the beam, vertical are the supports.

I suppose you could look at it as an eight-foot beam with the appropriate support loads and replaced moments, but it won't ever really be an eight foot beam, unless I'm badly misunderstanding this myself. I have no idea what kind of metal would give you that kind of plastic deformation, either- is there anything?

regards,

k wallace

• posted

Building structural folks don't typically operate to yield, they comply with structural codes that allow a stress multiplier between the allowable load, and loads which cause permanent (yield) deflection

Aero people also place the maximal load factored by a given multiplier at the yield point.

41500 lbs is the allowable central point load between 8 ft pinned centers for this structural beam, by code, where bending is more critical than shear. I guess that a load of 60k would drive it into yield... That is not the same thing as failure, though.

Brian Whatcott Altus OK

• posted

I have no info off hand on what is this grade A36 steel. If it is ordinary structural steel that has a sharp yield point, the ultimate fully plastic load for an I beam is just a bit higher (perhaps 10%) than the load to cause yielding at the top and bottom flanges.

• posted

it all depends on the moment of inertia and the maximal elastic strain of the material if the OP would like to put it all in SI-units and maybe a little ascii-art for the W-profile (i'm more of a scolar type engineer) and a little more info on the supports i wouldn't mind giving it a go

• posted

"Guru" wrote in news:42fdd0c2\$ snipped-for-privacy@news.starhub.net.sg:

At which point a fully plastic hinge would develop and the load would start to be taken in tension. I'm a bit surprised by the 10% figure, frankly.

Cheers

Greg Locock

• posted

You are right. When at a certain load, the section with the highest bending moment has gone fully plastic, a plastic hinge will form at this location. A simply supported beam with two supports will at this load will collapse (deflection increasing rapidly) theoretically.

About the ratio of the fully plastic moment to the initial yielding moment, this is 1.5 for a rectangular section beam, 1.3 for a circular section beam, and about 1.1 for a I section beam. The initial yielding moment is the bending moment which causes the stress at the outer fibres of the beam to reach yield point.

• posted

Tension and bending in a pseudo curved beam after the center yields.

• posted

Yes, i.e. engineering stress vs. true stress. For most metals, there's not a lot of difference between true stress and engineering stress. It sounds like Sanh just wants a good ballpark, and isn't really concerned with ultra-high accuracy.

You really only need to worry about the difference when you're dealing with very ductile metals, where there's a lot of strain occuring beyond the linear elastic region of the material.

A second point is that yield failure isn't nessesarily simple... are you using Von Mises stresses, principle stresses, energy/volume criteria, etc...?

BTW Sanh, I'd probably throw in a safty factor of 3 before you load it up.

Dave

• posted

For A36, yield strength = 36 ksi min & ultimate strength = 59 ksi.

If can be, but doesn't have to be. Inelastic bending would show additional capability.

What you describes is a true stress. However published material allowables are based on engineering stress. That could be different if you did your own testing..

• posted

"Jeff Finlayson" wrote in message news: snipped-for-privacy@corp.supernews.com...

It always is - the only exception that I can think of after 35 years of engineering and testing is in theoretically inelastic materials - bricks or unreinforced concrete. Old engineers have the joke about the new engineer asking the old one for info on the ultimate strength of maraging steel - and the only reasonable response is "having trouble with the design'?

Only a fool or an inexperienced engineer would use ultimate strength as a criterion for ductile material bending failure. Period.

( 1) When one uses ultimate as a criterion, any fatigue considerations and forgiveness are automatically out - those factors and levels were not determined at ultimate. 2) Reusing a part that is designed for stress effectiveness is automatically out - for obvious reasons. 3) Assurance that the part maintains the parameters of deflection necessary to apply bending moment theory when the material is yielding is a bitch. 4) PEs cannot defend ultimate strength ductile designs in court against the PE on the other side who points out a) the technical problems with using ultimate that degrades the design, and b) the mere pennies saved and injuries caused by not using yield. 5) My engineering managers had better have a damn good reason why they decided to go through the extra expense of using all that extra engineering time to design to ultimate, and if they didn't know extra time was needed, they don't get into management. 6) Many of the criterion used in shape design relies on yield design in the design circles, not ultimate strength. 7) Unlike yield, ultimate is not, to the best of my knowledge, a statistically determined minimum value of the material specs. 8) The per-cent difference, and type of difference, between yield and ultimate varies with type, class, and processing of the particular material. 9) And the biggest reason ultimate isn't used in good designs involving ductile materials is basic engineering - less reliable and more equations needs more margin of safety factor for the same assurance- increasing back the amount of material needed Ultimate has more considerations and equations needed to assure the design and thus, except when the engineer's ego is bigger than his brain, the larger number of equations necessary to determine design response requires a larger safety factor than does yield in order to assure the same level of integrity of design. But those FEA programs have the same limits as the old equations derived from experiment, except wiht the equations you knew there were more equations in play and you got to use the first derivative of those equations to get an error, instead of being lulled into thinking that all computer output has the same reliability. The level of error of the output is now inside the program, with few of them giving error estimates. The more equation components, and the less used the equations are, the less reliable the output and the greater the safety factor required

In the end that means the engineer usually has wasted his time calculating, because he has to add extra margin in his safety factor for each time he added an equation component - and since the advantage of doing all that calculation and attendant added margin it requires presupposes a decent difference in the two strengths to be useful, and since a large difference generally means a greater scattering and thus less reliability in the ultimate material strength which requires additional safety factor margin, its a rare design indeed that benefits from using ultimate.

Unless the test department redesigns the design anyway. )

• posted

My point is simple linear analysis is conservative. Things don't have to be super complicated..

Well, ultimate strength is the main criteria in aerospace. But ultimate plastic analysis is used sparely..

• posted

Hmmm....0.1% yield is the usual aviation design criterion in a material with a soft knee like most aluminum alloys.

But there are composites with essentially no plastic range - and they are used with a generous safety factor from UTS I expect it is this class that Jeff has in mind?

Brian Whatcott Altus OK

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