shear in flat slabs

huangyhg

shear in flat slabs
statement: the value given in section  11.12.3.2  of   aci 318-99:   vn = 6(f麓c) 陆 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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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.
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  鈥渂eam鈥?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 鈥渦nimportant鈥?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 (隆...!) 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.
隆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麓c)陆 bod  as given in section 11.12.3.2 of aci 318-02, is  for  stirrups, while it is taken vn = 7(f麓c)陆 bod when studs or shear heads are used.
the difference may come from 鈥淓ffectiveness of shear reinforcement in slabs鈥?by ghaly and hammill, concrete international, january 1992.
they suppose that 鈥?a short shear reinforcing bar ...can be ineffective because the cracks can pass above and below the bar without intersecting it.鈥?br>
they also affirm : 鈥淚nstalling flexural reinforcement through row of stirrups in two orthogonal directions is extremely difficult.鈥?br>
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 鈥渂eam鈥?along the axis of columns, with double stirrups 脴 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麓t want to tire you ... more to come.
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  鈥淧ca notes on aci 318-99鈥? chapter 18, 鈥淪hear in slabs鈥?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:  鈥淲ide-beam action rarely control the shear strength of two-way slab systems鈥?... 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麓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 ? 隆...impossible...!
the effective width  of flat slabs (.20 to .25(l+l麓)) 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麓c)^ 陆 , 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麓c)^ 陆 ,psi, even taking  vc= 0 (in beams, it is allowed  vn = 10 (f麓c)^ 陆).
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.