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torsion longitudinal beam reinforcement

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chekre

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May 8, 2013
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Hi guys,

i am dealing with big beam dimensions where the minimum torsion longitudinal reinforcement is coming too much and exceeding the Al required for torsion. (the beams width are ranging from 1.5 to 2m and depth around 1m).

according to ACI, torsion longitudinal reinforcement should be equally distributed to the 4 faces of the beam. This is also applied to the minimum longitudinal As.

I am wondering if this reinforcement(which is minimum) can't be calculated using benefit of the total reinforcement located at the bottom which is unstressed. for example, assume that the minimum As (minimum required for torsion) is 96cm2 and the bottom reinforcement as support (or d away from support) is 56cm2. cant i deduct (96-56)cm2 and divide the remaining As by 3 faces ??

Thank u
 
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i have a beam supporting from each side 2 different spans with different loading. So there is a considerable torsion at support. and at support, bottom reinforcement is not stressed since it will be in compression while the top bars will be in tension.

Torsion longi. Reinforcement required to resist the ultimat Tu is less than the minimum and the minimum is coming too much due to the beams dimensions.
 
Not sure what code you are using but ACI 318 says that the torsion reinforcement is additive to the required flexural reinforcement. This would apply at all sections. At each section considered either the top or bottom reinforcement wouldn't be stressed so it could be used to satisfy the torsion requirements.

The minimum reinforcement is required to accomplish aggregate interlock and a large beam would require a significant amount of reinforcement. It is like anything else, if the minimum is more than the required you have to use the minimum.
 
In the Canadian code (ACI may be similar), if the torsional moment exceeds 0.25 T[sub]cr[/sub] (cracking resistance of uncracked section) then torsional reinforcement must be provided. Otherwise, it can be neglected.

If torsional reinforcement is required, it must be distributed symmetrically around the section with a spacing not exceeding 300mm.

In your case, you need a total of 9600 mm[sup]2[/sup] or 9600/P per unit width of face where P is perimeter of section. If the unstressed reinforcement in the bottom exceeds the requirement for that face, you don't need to add more steel in that location, but you still need the specified amount in each of the other three faces.

BA
 
From the Commentary on ACI 318-05 11.6.3.8:

"The longitudinal reinforcement required for torsion is added at each section to the longitudinal reinforcement required for bending moment that acts at the same time as the torsion. The longitudinal reinforcement is then chosen for this sum, but should not be less than the amount required for the maximum bending moment at that section if this exceeds the moment acting at the same time as the torsion."

So essentially, torsion steel is additive to moment steel, for any moment that occurs at the same time as the torsion.

And from 11.6.5.3:

"Reinforced concrete beam specimens with less than 1 percent torsional reinforcement by volume have failed in pure torsion at torsional cracking. In the 1989 and prior codes, a relationship was presented that required about 1 percent torsional reinforcement in beams loaded in pure torsion and less in beams with combined shear and torsion, as a function of the ratio of shear stresses due to torsion and shear. Eq. 11-24 was simplified by assuming a single value of this reduction factor and results in a volumetric ratio of about 0.5 percent."

0.5% steel doesn't seem like a particularly excessive amount of steel, which is about what you're getting.

Brian C Potter, PE
Simple Supports - The history and practice of structural engineering.
ConstructionPic - Send annotated jobsite photos.
 
Like you said: it has to be evenly distributed……so counting on some help from some steel (just) at the bottom or top (intended for flexural/temperature) doesn’t seem appropriate. Me personally, I’ve typically just counted the 4 bars (at the 4 corners of the stirrups) for longitudinal reinforcement (I rectangular sections).
 
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