Steel Built-Up Beam 1

[JrStructuralEng]

I am adding 127x127x4.8 HSS to the bottom of an existing W360x33 in order to obtain increased moment capacity inreased moment of inertia.

Are there any considerations I need to take into account when doing this from a design perspective? The Beam is laterally braced. I am assuming I use the larger 'Sy' when determining the bending capacity of the beam.

Any help would be great! Thanks.

(Also is there any relatively easy way of determining minimum weld lengths along connection.)

This is probably a lame quesiton for some of you, but I have never done a built up section before.

 

[Lion06]

I don't know why you would use Sy. Why not Sx, unless you are concerned about weak axis bending (but then you would probably be welding a different section in a different location).
Either way, you will have to calc out your stresses and superimpose them. Unless you are jacking the existing beam to relieve the dead load stress you must calc the dead load stress on the existing beam (unreinforced section), then calc the "additional" load stress (can be super imposed DL and LL or just LL, depends on your situation), and superimpose that onto your DL stress on the unreinforced section. You will need to ensure that neither allowable stress is exceeded (either the top for the existing strength or at the bottom for the strength of the HSS - technically, you should check at the bottom of the existing shape to ensure you are not overstressed there, but you shouldn't be).
As far as the weld, if you are jacking to relieve stress, use VQ/It to calc the required weld strength. If you aren't jacking, use the same formula, but strip out the shear from DL. You can be conservative and leave in the DL shear regardless of your approach. This will likely result in a stitch weld. You also need to provide a continuous weld at each end of the reinforcement to ensure it can develop the strength required. I would use Fy*Ag of the reinforcement and size the weld at the end to develop that strength to ensure the entire reinforcement section can yield before the weld fails.

 

[csd72]

longitudinal shear v =VQ/It

 

[JrStructuralEng]

Yes I meant Sx.
You kind of lost me with the superimposed "additional" load line.

Can I simplify determining the capacity by using:

Mr = Phi*Sx*Fy

If so, would I use the larger Sx with for the new section? Since that would produce the lower yield?

Does anyone know any good example documentation or StructEIT, can you give me your home address so I can come over for a few lectures. I think I slept through this class in school.

 

[Lion06]

By "additional" load, I just meant any load imposed after the section is reinforced. That may not be limited to LL only as some additional DL could be imposed after the section is reinforced.

The equation you listed should use Zx (the plastic section modulus) and I don't think it's appropriate to use that since your steel will have different yield strengths - The WF is either 36ksi of 50 ksi yield (most likely), while the square HSS is 46 ksi yield (most likely). You could do that, but it's probably not worth the effort. Superimposing the working stresses is probably the easiest way to go.

I do have a few examples from projects that I've worked on. I might scan one in if I can dig it up.

 

[JrStructuralEng]

If could scan one, that would be great. I'm using limit states design but the approach should be similar.

 

[hokie66]

The SMALLER section modulus gives you the highest stress, but both top and bottom need to be checked against the stress for that material.

Quite frankly, and no insult intended, your questions indicate a lack of basic knowledge, and you should seek assistance from a more experience engineer in your company.

 

[HgTX]

I had sophomore-level homework problems along these lines. Take another look at your strength of materials book.

And then talk to an experienced engineer.

Hg

Eng-Tips policies: faq731-376

 

[Lion06]

Here is an example calc from a project I worked on.
Let me know if you have any specific questions.

 

[JrStructuralEng]

Thanks HgTX ...very encouraging.

 

[JrStructuralEng]

Thanks for the post StructuralEIT.

I follow most of your post, makes sense. Still my confusion lies in the first part of the problem

Snew,top > Mpos,max - Mdlmax / [18 - (63(12)/88.4)]

Where are you getting the 18ksi?

I believe what your doing is taking the "remaining" moment once the deal load is removed over your additional stress to find the new elastic section requirement. Is that correct?

 

[Lion06]

The 18ksi is the allowable bending stress for the original section (It is old and has an Fy=30ksi, so 0.6Fy=18ksi)

 

[JrStructuralEng]

So when your detemining Fbot = 7.82ksi < 30ksi (on fourth page) wouldn't you actually have two different allowable bending stresses? because a portion of the section is assumed to be 18ksi, and the new is 30ksi?

 

[Lion06]

And that is why I am checking fbottom against 30ksi because that is the allowable bending stress of the reinforcement (0.6*50ksi = 30ksi).
ftop is checked against 18ksi.

 

[JrStructuralEng]

Yes, I understand that. What I mean is if your neutral axis in the "old" section (See how yours is at 11.95"). Then the 'S-bott' actually encompasses two different material types. In your case it wasn't a concern...

but in certain cases, like perhaps if you were to use a flat plate and had much higher stress, you would need to check the stress at the old section, as well as the new for F-bott. y/n?

 

[Lion06]

You can prove this to yourself, but if you are only adding reinforcement to the bottom, the stress at the bottom of the existing section will ALWAYS be less than the stress at the top of the existing section. Here is why.
Before the reinforcement is added, the stresses at the top and bottom of the existing section are equal.
For the stresses on the composite section, use Fb=My/I. Since you are adding area to the bottom of the section, the ENA moves down from its original location. Therefore ytop will always be greater than ybottom of existing section, therefore the additional stress on the existing section will always be higher at the top. Again, this assumes reinforcement is added to the bottom only.

 

[JrStructuralEng]

makes good sense, thanks for taking the time to explain it out. I appreciate that you are willing to help me out regardless of my experience level with built-up sections.