hernma
Structural
- Apr 30, 2003
- 17
Statement: The value given in section 11.12.3.2 of ACI 318-99: Vn = 6(f´c) ½ bo d is too small, gives problem in design and should be increased, even reducing Vc. I will appreciate your opinions.
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Shear in flat slabs
- Thread starter hernma
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This provision is also in the candian code. I beleive that this is the limit for crushing of the concrete, regardless of how much steel is added. You can try adding a capital around your columns, or adding a thickened slab under equipment bases to increase your shear resistance locally.
The value could be increased but it needs incorporation of aggregate interlock effect in increasing shear carrying capacity. But if even if this value is increased, the flat slab shear carrying capacity might indicate lesser thickness of slab to be used, but i doubt if it will tie with meeting other effect such as flexure check.
- Thread starter
- #4
hernma
Structural
- Apr 30, 2003
- 17
ARH and benishak, thank you for your comments.
ARH: when you mention that the concrete crushes, I agree with you, because it means that you have a flexure mechanism of failure due to the extension of the upper steel, and this is what I have observed. I think it is also due to the torsion of the “beam” perpendicular to the direction being analysed. In other words the value of the unbalanced moment Mv transmitted by shear is less than the specified by ACI: vu = Vu/Ac + Mv/(J/c).
Normally the thickness t of the slab is taken around Lmx/20, so the problem is not to augment it, the aim goes to get more flexible provisions from ACI. Besides, changing the dimensions of elements is not always easy, because it can alter the architectural project.
benishak: maybe the interlock effect you are mentioning, is proportioned by the stirrups, which can never be absent, even in “unimportant” slabs. It is inconvenient to use thin slabs (say less than 20 cm), and I have to admit it is a problem to fight.
Mv becomes a real problem when the earthquake is considered. However ... we have suffered some strong earthquakes, maybe not the strongest in the world, but a good (¡...!) ones and I have never observed a punched flat slab. I can add that the behaviour of this buildings have been quite good (besides, I always arm the walls with a small quantity of horizontal steel).
I propose to my students the following formula:
Ductility = stirrups + compression steel.
¡Flat slabs have a tremendous ductility!. When you work with them, after a while, you can feel it. So, the problem is that your structure may become too flexible  use shear walls, and maintain the deflections < h/500. ( h is the height of the floor store).
In my exposition, I may refer to an author or article. My intention is far from criticize; it has to be understood as an example of a typical situation which is affecting our point of view. Another thing: if my English is not clear ...please guess ...
Vn = 6(f´c)½ bod as given in section 11.12.3.2 of ACI 318-02, is for stirrups, while it is taken Vn = 7(f´c)½ bod when studs or shear heads are used.
The difference may come from “Effectiveness of Shear Reinforcement in Slabs” by Ghaly and Hammill, Concrete International, January 1992.
They suppose that “ A short shear reinforcing bar ...can be ineffective because the cracks can pass above and below the bar without intersecting it.”
They also affirm : “Installing flexural reinforcement through row of stirrups in two orthogonal directions is extremely difficult.”
About the first clause: make the stirrups larger!, and besides, a specification can not be based assuming that a mistake is going to be committed.
About the second clause, I have been designing and constructing flat slabs for over 30 years, and it is not difficult al all. At intersections, combine closed with single leg stirrups.
It is frequent the use of blocks 40x20 (40x40cm) in plant to lighten weight, so 10 cm ribs go @ 50 cm, centre to centre.
Normally, we use a minimum of 4 bars continuously top and bottom (all) in a 2t to 3t wide “beam” along the axis of columns, with double stirrups Ø 10 mm @ 10 cm in a length of 3t from faces of columns and a single stirrup (say @ 20 cm) in the central part of the beam.
OK ...I don´t want to tire you ... more to come.
ARH: when you mention that the concrete crushes, I agree with you, because it means that you have a flexure mechanism of failure due to the extension of the upper steel, and this is what I have observed. I think it is also due to the torsion of the “beam” perpendicular to the direction being analysed. In other words the value of the unbalanced moment Mv transmitted by shear is less than the specified by ACI: vu = Vu/Ac + Mv/(J/c).
Normally the thickness t of the slab is taken around Lmx/20, so the problem is not to augment it, the aim goes to get more flexible provisions from ACI. Besides, changing the dimensions of elements is not always easy, because it can alter the architectural project.
benishak: maybe the interlock effect you are mentioning, is proportioned by the stirrups, which can never be absent, even in “unimportant” slabs. It is inconvenient to use thin slabs (say less than 20 cm), and I have to admit it is a problem to fight.
Mv becomes a real problem when the earthquake is considered. However ... we have suffered some strong earthquakes, maybe not the strongest in the world, but a good (¡...!) ones and I have never observed a punched flat slab. I can add that the behaviour of this buildings have been quite good (besides, I always arm the walls with a small quantity of horizontal steel).
I propose to my students the following formula:
Ductility = stirrups + compression steel.
¡Flat slabs have a tremendous ductility!. When you work with them, after a while, you can feel it. So, the problem is that your structure may become too flexible  use shear walls, and maintain the deflections < h/500. ( h is the height of the floor store).
In my exposition, I may refer to an author or article. My intention is far from criticize; it has to be understood as an example of a typical situation which is affecting our point of view. Another thing: if my English is not clear ...please guess ...
Vn = 6(f´c)½ bod as given in section 11.12.3.2 of ACI 318-02, is for stirrups, while it is taken Vn = 7(f´c)½ bod when studs or shear heads are used.
The difference may come from “Effectiveness of Shear Reinforcement in Slabs” by Ghaly and Hammill, Concrete International, January 1992.
They suppose that “ A short shear reinforcing bar ...can be ineffective because the cracks can pass above and below the bar without intersecting it.”
They also affirm : “Installing flexural reinforcement through row of stirrups in two orthogonal directions is extremely difficult.”
About the first clause: make the stirrups larger!, and besides, a specification can not be based assuming that a mistake is going to be committed.
About the second clause, I have been designing and constructing flat slabs for over 30 years, and it is not difficult al all. At intersections, combine closed with single leg stirrups.
It is frequent the use of blocks 40x20 (40x40cm) in plant to lighten weight, so 10 cm ribs go @ 50 cm, centre to centre.
Normally, we use a minimum of 4 bars continuously top and bottom (all) in a 2t to 3t wide “beam” along the axis of columns, with double stirrups Ø 10 mm @ 10 cm in a length of 3t from faces of columns and a single stirrup (say @ 20 cm) in the central part of the beam.
OK ...I don´t want to tire you ... more to come.
- Thread starter
- #5
hernma
Structural
- Apr 30, 2003
- 17
As a second point, I mention the modelling followed by ACI when the shear is checked as a beam in one direction, assuming the critical section extends across the entire width.
According to 11.12, the flat slab works in exactly the same form as does a footing, taking the full width as contributing.
While it is logical in a cantilevered foundation, it is not in a flat slab.
Again, taking as typical the Example 18.3 exposed in “PCA Notes on ACI 318-99”, Chapter 18, “Shear in Slabs” wich consists of a flat slab with panel sizes 21 ft x 21 ft, slab thickness 7.5 and d = 6 inches.
It is expressed, pg.18-23: “Wide-beam action rarely control the shear strength of two-way slab systems”,... as a consecuence of using the full width, 21 ft.
This assumption is not even concordant with flexure specifications, which stablish stringent conditions (but at least sound ones) regarding the concentration of flexural steel near the column and between the column strip.
Now, the flexure is ductile , the shear, by nature is non ductile , how can the shear, specially due to earthquake be taken for such a length, 21 ft ?!
Besides, notice that for purposes of controlling deflections, it is practically compulsory the use of shear walls, where the punching shear is not problem (and very unlike to happen!), even with vc = 2 (f´c)^1/2, psi.
Although taller buildings have been done , I feel that 20 stories high is a reasonable limit to use flat slabs. Drop panels have not been used.
The shear, as a beam, near or in the vicinity of shear walls is high, and we could be in the case of coupling beams, Section 21.7.7 of ACI 318.
Due to this situation it is often necessary to add a tall beam between a shear wall and the next column.
Again, can a beam 21 ft wide and d = 6 in be effective against this force ? ¡...impossible...!
The effective width of flat slabs (.20 to .25(L+L´)) to be used in the analysis of beam rigidities, which is not of capital importance, has deserved a lot of thoughts and research.
As a contrast, the use of the total width (very sensitive aspect) as effective against shear has been adopted without enough sustain.
I have always designed taking as maximum the column strip as effective width, and the stirrups I get frequently dominates over those for punching purposes.
In summary:
a) vn = 6 (f´c)^ ½ , psi, is a too small limit when considering Mv.
I suggest, taking in account the ductility of flat slabs ""reinforced in the form as explained above"", to increment it at least to vn = 8 (f´c)^ ½ ,psi, even taking vc= 0 (in beams, it is allowed vn = 10 (f´c)^ ½).
b) The tributary width have to be reduced, the modelling is not correct and , very clearly, it does not cover the frequent case when shear walls are used.
As a final comment, pure whafle slabs without stirrups must be proscribed in seismic countries. They can be combined with the already mentioned wide beams.
I will appreciate your comments and experiences in this field.
According to 11.12, the flat slab works in exactly the same form as does a footing, taking the full width as contributing.
While it is logical in a cantilevered foundation, it is not in a flat slab.
Again, taking as typical the Example 18.3 exposed in “PCA Notes on ACI 318-99”, Chapter 18, “Shear in Slabs” wich consists of a flat slab with panel sizes 21 ft x 21 ft, slab thickness 7.5 and d = 6 inches.
It is expressed, pg.18-23: “Wide-beam action rarely control the shear strength of two-way slab systems”,... as a consecuence of using the full width, 21 ft.
This assumption is not even concordant with flexure specifications, which stablish stringent conditions (but at least sound ones) regarding the concentration of flexural steel near the column and between the column strip.
Now, the flexure is ductile , the shear, by nature is non ductile , how can the shear, specially due to earthquake be taken for such a length, 21 ft ?!
Besides, notice that for purposes of controlling deflections, it is practically compulsory the use of shear walls, where the punching shear is not problem (and very unlike to happen!), even with vc = 2 (f´c)^1/2, psi.
Although taller buildings have been done , I feel that 20 stories high is a reasonable limit to use flat slabs. Drop panels have not been used.
The shear, as a beam, near or in the vicinity of shear walls is high, and we could be in the case of coupling beams, Section 21.7.7 of ACI 318.
Due to this situation it is often necessary to add a tall beam between a shear wall and the next column.
Again, can a beam 21 ft wide and d = 6 in be effective against this force ? ¡...impossible...!
The effective width of flat slabs (.20 to .25(L+L´)) to be used in the analysis of beam rigidities, which is not of capital importance, has deserved a lot of thoughts and research.
As a contrast, the use of the total width (very sensitive aspect) as effective against shear has been adopted without enough sustain.
I have always designed taking as maximum the column strip as effective width, and the stirrups I get frequently dominates over those for punching purposes.
In summary:
a) vn = 6 (f´c)^ ½ , psi, is a too small limit when considering Mv.
I suggest, taking in account the ductility of flat slabs ""reinforced in the form as explained above"", to increment it at least to vn = 8 (f´c)^ ½ ,psi, even taking vc= 0 (in beams, it is allowed vn = 10 (f´c)^ ½).
b) The tributary width have to be reduced, the modelling is not correct and , very clearly, it does not cover the frequent case when shear walls are used.
As a final comment, pure whafle slabs without stirrups must be proscribed in seismic countries. They can be combined with the already mentioned wide beams.
I will appreciate your comments and experiences in this field.
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