How would I go about calculating the permanent lateral deflection of a simply supported steel beam? The problem I have entails applying a central load and then measuring the permanent lateral deflection of the beam. Any clues would be much appreciated!

All you have to do (in theory) is take it past the point of yield, it will not spring back to where it started, but there is in reality springback. This is not simple to calculate and you have not provided sufficient information to do so. Consider two long steel members one has an I shape and the other rotated 90Â° to be an H section. Given the same force, they will not bend or yield the same depending on the dimensions of the cross section. Sounds like a homework problem given the lack of information.

It's not a homework question, I was just looking for some general guidance to point me in the right direction. I am a mechanical engineer but I've only just come into a related profession since graduating in 2008 and my memory is rather rusty - especially on yield theory as it wasn't really considered on my course. I will give you some more information - I hope you can help. I am designing a ladder to meet a European standard of manufacture. This requires the ladder to undergo some basic mechanical tests. One of these is a 'residual deflection' test. The ladder is set up horizontally between two trestles with roller supports and receives a 1100N load distributed over 100mm at the centre of the ladder. The load is sustained for 1 minute and the material is left to recover for 1 minute. The total residual deflection of the ladder must be less than 0.1% of the length. I am trying to get the most efficient design to pass the test and I need to predict how section designs will perform theoretically before ordering sample material to test. Please let me know what further information you might need. Any help would be appreciated!

Hi, Bit confused here, initially you asked for lateral bending now it appears you are looking at out-of-plane bending? BS EN ISO 14122 specifies both need to be carried out. It referes to EN 131-2 which I don't have so I can't see the specifics. Suffice to say the actual deflections will depend on the actual material used rather than the nominal 'book' values. I would be looking to design to meet the working elastic criteria then assuming common ductile materials and connections I would expect the ladder to meet the post yield check.

Sorry, there is a lateral test but the one i'm concerned with is as described. Working to the elastic criteria of the steel I'm using (with a yield strength of ~250Mpa), I would not get an cost-effective design. I need to understand how I can predict the amount of plastic deformation and the resulting residual curvature of the beam. It is not a standard profile, rather a rectangular section with semi-circles for the short edges rather than straight edges.

Your problem deals with, as you probably already realize, design in the plastic range rather than the elastic range. As mechanical engineers we generally stay in the elastic range and that's where we are comfortable. But in civil engineering plastic analysis is more common. Unfortunately most plastic analysis is intended to find strength rather than deflections so your problem is still a challenge. The theory in general: As the simply supported beam is loaded in the elastic range the stress in the outermost fibers increase to the yield point. Beyond the yield point the beam obviously does not immediately fail, but a plastic hinge begins to form as yielding begins at the extreme fibers and - if the load continues to increase - moves toward the beam's centroid. If the plastic hinge fully forms then the beam is at its load limit; any further increase in load and it will fail. (note: this discussion assumes that the wall thickness is sufficient to avoid local buckling and lateral bracing is sufficient to avoid lateral bucking). This load limit is given by the plastic moment: Mp = Fy*Zx where Fy = yield stress of the material and Zx = the plastic section modulus (Zx is not the more familiar elastic section modulus). So to avoid failure during the test yet be in the plastic range, the section needs to be designed so that the test load will produce a moment in the beam that is between the moment at elastic yield point (My = Fy*Sx where Sx = the elastic section modulus) and the plastic load limit Mp. So you will need to calculate both the Zx and Sx for your section to find this desired range and assure your section is designed such that the test load will put you in this range. Up to this point, the calculations are just to assure the beam is efficient yet does not fail in the test. To estimate the permanent deflection is tricky, but here is what I'd suggest: first, find the deflection at the elastic limit (yield moment) My using typical beam formulas. Then find the deflection at the actual moment expected during the test using the same formulas (this of course assumes the deflection will continue in the elastic range which is not completely accurate but should be conservative). Find the difference. This difference should be a reasonable estimate of the amount of deflection that occurred in the plastic range. The elastic part of the deflection should (mostly) recover; the plastic portion will not recover. Look around online for how to make the calculations of elastic and plastic section moduli for your custom section. If you have no luck, let me know and I can help.

You are not really designing for plastic deformation. You are designing to prevent it. johnstjohn has very good points. Most structural/mechanical design is to minimize deflection (elastic entirely, not plastic) - which is to say far from where yield occurs. If you already know the dimensions and material of existing and successful products, then you have little need for calculating plastic deformation. A thorough survey of existing and competitive products is a first order need. Similar to what he says, the exact material properties are probably an important second order effect, after general design is met. If you are optimizing the design (cheapening it more likely for a commodity type product) then you can expect to experience the plastic deformation you must avoid for some portion of your product. In this case where you must always meet these standards, you do not want to even approach the minimum deformation else you lose your rating and certification. Calculating the deformation is probably just going to get in the way of finishing your task. And even if you suss it out mathematically, you have to understand that due in part to material properties, thermal and work history, etc. You need to correlate the calculation to reality by testing and statistical means, in order to have well based faith in the calculation. It is very possible to overthink this sort of thing. The basic equations are for homogeneous materials with perfectly regular dimensions and properties. They are good for getting you in the ballpark (but you must apply factors of safety - because they are basically just wrong the closer you look at them), but if you have to get the most value from the manufacturing and safety as well, you can't come close to reality without testing. You need a design that is not going to deform and meets your cos targets. Then you should be concerned with manufacturing induced stress risers, finish, supplier quality, manufacturing control, etc. to keep an optimized product successful. Empirical stuff.

Probably worth also adding that the design of ladders is governed by human factors and design code guidance to achieve a safe useable product. These considerations alone are likely to prevent 'cheapening' the product such that plastic design becomes an issue, assuming common or garden materials!