How do you design a T-beam? 3

There is a considerable amount of literature available about the design of T-Beams. I am not sure what kind of help you are looking for.

In positive moment regions, T-Beams provide a wide compression block with a high center of gravity. This increases effective depth and reduces the amount of positive reinforcement required.

In negative moment regions, however, a T-Beam behaves pretty much the same as a rectangular section.

BA

BA, thank your quickly reply.
I do as same as you said. But I want to know how arrange the top reinforcements (negative moment)
Option 1: All arrange in the top of rectangle
Option 2: parts arrange in the top of rectangle and the other parts in the two side of T flange?
I think the negative moment is resisted by the rectangle with two side of flange together, consequently I think the option 2 is right.
Do you think so?

Option 2 sounds good,
Don't want to be putting all the reo in the rectangle, because then you have trouble getting the vibrator down for the bottom steel.

Don't know your code that you are designing too, so don't know if there is any rules for effective width of the T beam, bar spacing ect. So how best to distribute these bars will have to be in accordance with your code in regards to the flange.




When in doubt, just take the next small step.

In addition, before I send you up the path, remember to keep enough longitudinal reo in the rectangle for the shear stirps as a minimum.

When in doubt, just take the next small step.

Rowingengineer,
Thanks your reply.
I am designing according to ACI318, but I don't find how percent reinforcements shall be arranged in the two side flange?
I know the range of the flange from the ACI318 clause 8.10

You may even arrange it uniformly on the top width of the head, provided that the shearing force at the interface between web and flange protrusions stays in permissible range. It is easy to calculate, (assuming a negative moment bending situation) count the numbers of bars in the protrusion to one side, multiply for area and standing stress in the bars; do that in 2 sections 1 ft apart, near support; the difference between the two is the shearing force in such foot of an arm of concrete head; for the stress, divide by the area in such ft of the interface, i.e. 1 ft x thickness at root of arm, and compare the standing shearing stress with the factored or nonfactored stress permitted for the case. If too much, you can't put that number of bars in the protrusion; or more exactly said, you must reinforce in shear the shear interface to meet the code provisions; a shear-friction calculation may be permissible for the case but since the section will be for the factored loads likely cracked in tension make sure whatever the code says about compressive struts or limitations on the allowable shear force due to extenuation of the compressive capacity of the concrete for the situation is met. It is better to assume a reasonably low permissible stress in shear at the interface, say, 1.5 MPa under factored forces, or what the code allows, and if not met, place more main rebar in the web projection, and less outside.

All what above mainly meaning that the amounts can be quite precisely calculated; you only need a section analyzer that gives you the status of the section, in your case being of interest the stresses in the steel. I did some worksheets on Mathcad that give the status of T-beam sections under positive and negative bending action, and should still be available at mathsoft's site in the Collaboratory section for Mathcad 2000.

Two things to consider

1. For serviceability design, use the effective flange in the calculation for I, this will stiffen up the section for deflections.

2. For ultimate design, only use the solid rectangular section for the ultimate design, this will be critical for negative moment where you require steel in the top face, I would not distribute the tensile steel over the entire flange width of your Tee setction.

Tension steel should ideally be well distributed across the full effective width. When you have a lot of bars in the flanges you should also ensure the slab has sufficient longitudinal shear capacity to develop those bars.

There is a clause in ACI that requires you to distribute some tension steel in the flange and not just on the rectangular portion.

Take a look at clause 10.6.6 and R10.6.6 for crack control.

When in doubt, just take the next small step.

Yup, that's the one.

Just because i'm aussie i will also give the AS3600 referances, Cl 8.6.1 (B) and CL 8.1.8.2

When in doubt, just take the next small step.

I try to avoid using Tee beams for negative bending. When you add a good amount of reinforcment into the flange(even just for crack control, or shrinkage or such) then you can end up "over-reinforcing" the beam based on the relative small width of the compression stem. This then leads to potential non-ductile beam failures....

That being said section 10.6.6 of ACI gives some specific requirements for reinforcing the flange. though I'm not sure that really helps us deal with the issue above very much.

As far as spreading the reinforcement around the stem vs. just in the Tee. I will try to put as little reinforcent in the flange as possible.... Whatever is based on Temp/shrinkage or crack control. If more reinforcement is required then I think it would be preferable to put it in the tension side of the stem. But, it would also probably be costlier as well.

I agree with Josh. There is nothing which says that a rectangular beam with attached slab has to be considered as a tee beam. The top bars need to be within the stem width or close thereto. If there are too many bars to fit into the stirrups, either design with two layers or move a couple outside the stirrups. Distributing over the full width (as the codes suggest), in my opinion, moves the bars too far from where they are required. Now as to crack control, the "flange" will crack due to differential restraint shrinkage, and those cracks need to be controlled for serviceability reasons by added shrinkage reinforcement.

thanks everybody
my question is the attached figure, maybe you dont fully understand me.

Why are you providing stirrups in the flange area? Your maximum shear core is going to be through the rectangular web of your beam, this is the only location where stirrups would be required.

I agree with KBVT - limit the stirrups to the rectangular section.

Mike McCann
MMC Engineering

Option 1 has 2 layers of 6 bars for a total of 12 bars. Option 2 has 5 + 6 + 5 = 16 bars and a greater effective depth, so the options are not equivalent.

In Option 2, you could use 3 + 6 + 3 = 12 bars which would be permissible if the force in the bars in the slab can be transferred to the web by shear in the slab. Slab bars should be fairly close to the web of the beam, say within L/20 where L is the beam span.

I don't think you should ever place more than 50% of the top bars outside the beam web. For an L beam, this would be 25%.


BA

BA,sorry i dont draw clearly.
the number of the reinforcements is only signal and not real.
i think 20% reinforcements can be distributed in the flanges, and 75% in the rectangle.
right?