If I have a flat span of steel material, 3.0m x 150mm wide and 5mm thick, and apply an increasing downward force to the middle, at some point it will fail.

But if the profile of that piece of steel was T shaped instead of a flat span, it is much stronger.

Is the Safe Working Load (SWL) for different profiles calculatable by a specific formula?

Assuming so, otherwise how could you design things like cranes, does the same formula apply to different materials e.g. aluminium v steel and different thicknesses of the same material?

  • $\begingroup$ Related: engineering.stackexchange.com/questions/68/… $\endgroup$ – user16 Jul 18 '15 at 1:55
  • $\begingroup$ By T piece you mean a stiffening rib that improves SWL? $\endgroup$ – Narasimham Jul 20 '15 at 18:17
  • $\begingroup$ @Narasimham yep $\endgroup$ – Steve Jul 21 '15 at 7:12
  • $\begingroup$ So your question / query is about different profiles, different applications (cranes etc) different thicknesses and different materials (steel, aluminum) for given load to stay within SWL, Right? $\endgroup$ – Narasimham Jul 21 '15 at 17:02
  • $\begingroup$ I was just curious about formulae and whether they carry across to different materials $\endgroup$ – Steve Jul 21 '15 at 23:46

In answer to the question Is the "SWL" for different profiles calculatable by a specific formula? - Yes, although it’s not one formula, there are lots of different things to consider, all with sets of formulae depending on the exact situation. Calculating Safe Working Loads (SWL) (or, indeed, starting from a desired working load and calculating the required structural member) is a large part of what structural engineers do.

A beam can “fail” in many different ways. Most likely for your given case would be exceeding the plastic capacity of the beam in bending, such that a hinge forms in the middle of the beam, causing significant deformation which then pulls the beam off one of its supports. The plastic capacity of a metal beam is relatively easy to calculate, you can use the formula

s = M/Z

where s is the yield stress of the material, M is the applied bending moment, and Z is the plastic section modulus of your chosen profile.

s is material dependant, so changing the material will change this. You can get various grades of steel, with yields from 230 to 260 MPa. Aluminium can be supplied with yields from 15 to 20 MPa.

M depends on the applied loading. In your case it is due to a force F applied in the middle of a 3.0m span, which gives a maximum moment of 0.75F. (Calculating bending moments is a whole separate topic).

Z depends on your profile. Wikipedia gives some formulae for some basic sections. (Calculating section modulae is a big topic too).

This formula will give your theoretical maximum load, and not your SWL. The SWL also takes into account factors of safety, which are used to ensure that the applied load doesn't exceed the capacity. There are design standards which specify these factors (which standards apply depend on the situation).

To answer your headline question “How much strength does adding a T piece add?” - it all depends on the dimensions of the T piece. And, as alluded to by grfrazee, as increase the depth of the T you run the risk of the failure method changing to lateral torsional buckling.

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One can determine the section properties of a shape by using the parallel axis theorem, as is taught in an elementary mechanics of materials course. I have found the following website that will calculate the section properties of a T-shape when one inserts the proper input. Note that this website also lists the formulations behind the calculations.

Yours could possibly have flange buckling issues given the slenderness of the flanges, but it's hard to tell without running any calculations. The following snippet from the American Institute of Steel Construction (AISC) 14th Edition Steel Construction Manual specifies the limiting b/t ratio that separated nonslender and slender compression elements. This may not be applicable depending in your jurisdiction since you are in Australia. However, the Australian Steel Institute should have a similar provision codified in its code.

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The formulations for section properties of the shape are material-independent, so long as the material can be considered isotropic, homogenous, and elastic (steel and aluminum are, concrete is not).

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  • $\begingroup$ Are you sure this is in a buckling regime? Without doing the calculations, a 3.0m T-shaped section with with 150mm width and depth sounds like it would be pretty stiff. $\endgroup$ – regdoug Jul 18 '15 at 12:28
  • $\begingroup$ Whoops, I read it as 150cm width. I work in the US, and metric is a little hard for me to visualize sometimes. $\endgroup$ – grfrazee Jul 18 '15 at 21:05
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    $\begingroup$ It's worth mentioning that the relevant code (even in the US) may be different, since this is apparently a mechanical engineering question (by the tag), while the AISC code only applies to structures. The overall concept behind the codes should be the same, but the exact limits and safety factors may very well be different. $\endgroup$ – Wasabi Jul 20 '15 at 15:59
  • $\begingroup$ Good clarification. I saw the word "crane" and jumped to building structures by default. $\endgroup$ – grfrazee Jul 20 '15 at 16:22
  • $\begingroup$ I also understood his question as seeking general improvement considerations. $\endgroup$ – Narasimham Jul 21 '15 at 17:06

There are better ways of uniformly stiffening a plate e.g., by the Sandwich principle. A somewhat compromised version is the Isogrid, these are the distributed I-beams.

In both methods, effective bending rigidity is increased by making plate central parts hollow and so lighter in weight. Stresses are reduced.

Please google under these names. In place of 5 mm is a thick plate, you can use 1/8 inch plate and tack weld on isogrid cells or at least orthogonal ribs/grills of same 1/8 inch thickness, say 3 inch cell sizes. The webs are like bottle crates. Stiffness could go up by 4/5 times. The extra fabrication is worth it.

If alternate material choice is ok, even a 1-inch plywood sheet can be bonded with 1/16 thick steel sheets usind a good adhesive on either side for reduced deformation.

When improving designs, detail is important.

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  • $\begingroup$ This answer misses the point. The OP is not asking "what is the best way to improve my design?", but rather "is it possible to calculate the effect of different designs?" $\endgroup$ – AndyT Jul 21 '15 at 12:36

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