Plain cruciform (not flanged cruciform) member design using AISC 360

[AWDMIKE]

We have a situation where we are using a plain cruciform shaped beam in a simply supported situation. The cruciform is symmetric about the vertical axis but not about a horizontal axis (see attached image).

The beam cross section is nearly a inverted "T" shape, however, the bottom of the flange of the beam is not flush with the bottom of the web of the beam. My understanding is that this is called a "plain cruciform". I apologize if the terminology I am using is incorrect. I borrowed the terminology from:

http://www.newsteelconstruction.com/wp/wp-content/uploads/TechPaper/NSCApril06_Tech.pdf

What I am trying to determine is if I can use the provisions in AISC 360 F9 for this situation. Table F1.1 shows a cross section of F9 that is not exactly like my situation. That section would cover yielding, LTB, and FLB.

Any recommendations or guidance would be appreciated. Thanks very much.

 

[KootK]

If the proportions that you've sketched are accurate, I would just design the beam as a plate, ignoring the flange. That should be simple and sufficiently accurate. Alternately, treat it as a tee section ignoring the web below the flange.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[SlideRuleEra]

I agree with KootK about ignoring the flange and designing as a vertical plate. Also, treating the shape as a "shortended" tee is reasonable as long as you realize the tee is upside-down (as shown in the sketch, the (Edit) stems flanges are in tension during bending). This has a significant effect on the calcs since in the as-shown position the stems do not brace the compression part of the beam.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[JLNJ]

Agreeing with the gents above...if you look at the section modulus without the little horizontal plate, and then with it, you will see that adding the plate does little to nothing to reduce the compression stresses in the top of the stem.

 

[AWDMIKE]

Thanks to everyone who has responded thus far.

SlideRuleEra - Did you mean to say that the flanges (not stems) do not brace the compression part of the beam?

All - I need the additional stiffness provided to me by the flanges at the bottom, so I'm inclined to treat it as a t-shape. As mentioned, the flanges are too far from the compression stresses in the top of the stem to be of any real help, but that brings me back to my problem on which I should have mentioned in my first post.

Section F9.1 has criteria for (a) stems in tension and (b) stems in compression:
For tension .........Mp = FyZx <= 1.6 * My
For compression...Mp = FyZx <= 1.0 * My

Section F11.1 has criteria:

.................Mp = FyZ <= 1.6 * My

Am I missing something? If I meet the Lb*d/t^2 criteria in F11 and can use the equation above, my rectangular bar will have a much higher capacity than if I treat it like a tee-beam with the small flanges attached near the bottom (at least, if the stem is in compression)?

Thanks again.

 

[KootK]

Tees perform better than plates primarily because the flange braces that one edge of the web from participating in web buckling. Rationally, the flanges need to possss some minimum amount of strength, stiffness, and stability of their own in order to do that job. To my knowledge, AISC says nothing about what those minimums are. And that's probably because those minimums are pretty easily achieved for normally proportioned members. However, if your proportions were sketched accurately, there is doubt in my mind as to whether or not your flanges will be up to snuff for bracing the web properly.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[WinelandV]

You're neglecting the LTB for the plate only case.

--> Lb*d/(t^2) <= 0.08E/Fy for LTB to not control/occur.

Unless you're dealing with an incredibly short span (or have the beam braced at very short intervals) you're going to be looking at Equation (F11-3).

--> Mn = Fcr*Sx

where Fcr = (1.9*E*Cb)/(Lb*d/(t^2))

Also, don't forget the LTB check for the T-shape either. There's no quick "if this ratio is less than this ratio" check to see if LTB controls for the T shape, so you get to check for LTB regardless of your braced length.

 

[AWDMIKE]

More excellent responses! Thank you everyone.

My point was, for the situation where yielding would control (either a rectangular beam or tee beam), the capacities appear to be quite different. If the beam was proportioned such that LTB, FLB of tees, and local buckling of tee stems did not occur, I would have a limit on the nominal moment of My for a tee (stem in compression) and 1.6*My for a rectangular beam.

Is this odd to anyone? Am I missing something obvious?

 

[SlideRuleEra]

AWDMIKE - I did misspeak in a previous post, should have said "flange" instead of "stem". Sorry.

http://www.SlideRuleEra.net

http://www.VacuumTubeEra.net

 

[BAretired]

My point was, for the situation where yielding would control (either a rectangular beam or tee beam), the capacities appear to be quite different. If the beam was proportioned such that LTB, FLB of tees, and local buckling of tee stems did not occur, I would have a limit on the nominal moment of My for a tee (stem in compression) and 1.6*My for a rectangular beam.

Is this odd to anyone? Am I missing something obvious?

It seems odd to me. What do you mean be 'nominal moment' and why would the limit be 1.6*My for a rectangular beam?

BA

 

[structSU10]

I think some of this comes from the general proportions of the sections you are talking about. A WT stem is generally fairly thin. the Lb*d/t^2 < 0.08E/Fy is pretty tough to achieve (a 6" deep plate braced every 12" needs to be 1.25" thick), and you likely never get that condition for a WT stem.

The other interesting thing here is the commentary explicitly states to use Cb=1 for WT with stems in compression, citing testing that shows premature failure, but allow for the Cb factor to rectangular plates.

 

[AWDMIKE]

In AISC 360, Mn is the nominal flexural strength. I apologize if I created confusion by calling it a nominal moment.

Hi BAretired - Rectangular beams meeting the geometric criteria in AISC 360 Chapter F11.1 would use the nominal strength criteria shown in equation F11-1. I'll attach it for reference in case you don't have AISC 360 handy. The upper limit happens to be 1.6*My, but for a rectangular shape I'd imagine that Fy*Z would govern, where Z would just be 1.5 times the elastic section modulus for a rectangle.

Nominal is just the term they use for the strength before it's knocked down by the resistance factor (phi) or the safety factor (omega). I'm using AISC and their terminology.

 

[KootK]

One significant difference between a WT with it's stem in compression and a rectangular plate is that the plastic neutral axis on the WT would be typically be much further down on the section. That means that, in order to achieve Mp, the extreme compression fibers in the WT would need to undergo much more strain (and possibly strain hardening) than the extreme compression flange in the rectangular plate. And that implies a more critical stability condition with regard to stem local buckling.

I don't know that this is the answer but it may be related.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[BAretired]

I agree that, for a rectangular section Mp = Fy*Z or 1.5*My (where My = moment at first yield). I wasn't sure where the 1.6 factor came from but I guess it is part of the AISC specification.

BA