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inverted L-Beam effective width

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handex

Structural
Jul 1, 2010
56
Hey Guys,
might be a stupid question, but are the formulas for calculating the effective flange width of an inverted L beam the same as for a standard L beam? All the examples I have seen are when the flange is in compression. Basically I have an upturn at a slab edge and want to know how much slab steel can be used for Ieff and for ultimate calcs.

Cheers
 
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If the upturned leg is in compression, isn't the "slab" on the bottom, in tension, and thus not part of the Icr at all?

Not sure I understand your question.
 
yeah the slab is in tension, but I am trying to figure out how many bars I can use for the transformed section.
 
How wide is the slab? Usually with "L" beams like that you can engage all of the tension bars.

 
I don't know about your code, but often the bar diameter has to be 50% or greater than the largest bar before it can deemed useful.
 
Also, you can only use the area of the bars inside the shear ligs for shear calculations.
 
ACI has provisions for T beams with the flange in tension. That seems appropriate to use here - you just have a flange on one side.
 
I would think that the effective flange thickness for and inverted L-beam (upstand beam) would be the same as for an L-beam (downturn beam).

I use 4 times the slab thickness but your code will give an empirical equation for what width can be used.

And L-beam, be it upstand or downturned will need to be designed for positive and negative moment in general concrete design.
 
handex,
As I think you are in Australia, there is a provision in AS3600, Paragraph 8.1.10.2, which requires that "tensile reinforcement shall be well distributed...". What "well distributed" means is the question. It is advantageous for crack control and for placement of the concrete to place some of the bars outside the web. Have a read of what Warner et al say about it. My 1998 edition has a relevant section "Cross-sectional details" on pages 809 and 810.

SEIT is correct about the ACI provisions. Section 10.6 requires part of the tensile reinforcement to be distributed outside the web, within 1/10 of the span for T-beams. I suppose for an L-beam, that would mean 1/20 of the span.

The width of flange used for compression and the distribution of tensile reinforcement into the slab are really separate issues.
 
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