capacity check - shear links 2

[oneintheeye]

Forgive me here but this is relevent to a British standard clause and I realise most here are US.

BS8110 (Concrete) states that all tension bars in a beam must be within 150mm of a vertical leg. It also gives other min dims. I don't know if you have similar clauses in US.

My question is this;

I am looking at an existing stucture for some increased loading, the check calls for links for the shear. The area is fine, however the link arrangment does not meet the detailing clause above. I am guessing that maybe in an older code version this was acceptable or the original design didn't need links so they put in the minimum area (albeit not to BS8110 detailing rules, but they could in theory be omitted totally).

To my mind the links are required to halt any shear crack propogating down to the main bars and reducing the capacity of the section be breaking the bond between the main steel and the concrete, Leading to failure. Hence the spacing only 150mm from the tension bar. Therefore in my assessment the non confoming detail will mean the beam in idadequate in shear due to the detail even though the area is above that required.

Any views?

 

[oneintheeye]

I will try and answer all points;

BAretired - the existing design loads require minumum links only but do not meet code in the arrangment.

The beam is 1000mm deep.

This is definatly a beam rather than a slab I would say. there are no alternative load distrubution paths. It is a beam sat on piles.

There is no other load path for what the client wants to do.

The new load requires design links and not minumum only.

there is a difficulty in doing any strengthening works due to ownership/liability issues.

think I'm stuck!!!

 

[ishvaaag]

Following with the strut and tie scheme to the stirrups at the sides ... it is not to forget that apart from checking the tie the compressive struts are also to be checked. Furthermore, the aduced reason for closeness (longitudinally) of stirrups is "confinement of the struts (in the analogy of Mörsch)". So we need to devise a feasible strut mechanism able to stand the loads. For loads atop (invert the position if otherwise), one of this can be a fan (an inverted triangle) taking loads from atop, then dividing in 2 struts, one towards each lateral leg of a stirrup. And the "tributary" load to be taken as the load passed in the truss analogy. In section projection the strut can be to some extent somewhat wide, since at bottom has a net of receiving strong longitudinal bars, and atop the node fan-struts. Contrarily looking from the sides the critical strut can't be taken as wide since loading in legs of stirrups too far apart.Say, 15 cm, 20 cm, 25 cm perhaps for the side width of the strut. And then you need to check for the strength of this tridimensionally confined strut. This will provide some extra strength but better not be too optimistic, it is supposed we are trying to check a thing under strut and tie provisions in the code. If the struts and the tie then work, the structure may work at the limit state and if the values you have used in the check are code-derived you have a handle to the code problem as well.

 

[oneintheeye]

just an update; i have got in front of me the handbook to the code. I will repeat the comment for the relevent clause below; Sorry for the lengthy post.

The logic behind the limit on the lateral spacing is less clear but experimental evidence suggests a reason why such a limit is valuable. One of the reasons of stirrup is to inhibit dowel failure of the tension steel....Clearly the effectiveness of the stirrup in achieving this will reduce with increasing distance of the vertical leg from the bar considered. Clearly, if a bar is placed further than 150mm from a stirrup leg, it can still be used to provide flexural strength but should be ignored in assessing Vc (for non UK this is concrete shear strength based on longitudinal reinforcement). The situation in slabs can be used to extend this interperation further. In slabs, stirrups are not required until V > Vc. It seems logical to argue from this that the requirement relating to the spacing of stirrup legs is to ensure that Vc can be maintained in circumatances where V > Vc. It therefore seems reasonable to conclude that the limitations on lateral spacing may be ignored where V < Vc.

 

[oneintheeye]

now my own comment on my post above;

the code for beams requires links if 0.5Vc < V < Vc + 04 for minumum links and designed if greater value.

Slabs it is simply V > Vc.

So to me the above interpretation doesn't add up for the following reasons;

1)Beams and slabs are treated differently the last sentance seems to go against the code equations above
2) If we have to ignore longitudinal bars in beams that are not within 150mm of stirrup how can we use these bars in slabs? Surely if beams and slabs are comaprable you would have no longitudinal steel contributing to shear strength since most slabs are detailed without shear reinforcement.

 

[ishvaaag]

I have also some incongruence in design provided by one spanish structural design package, yet the cause may be (I think it is) EHE code itself. You have a slab on columns, check it, simply you deduce it is mandatory a thicker slab at support, whatever the rebar shear you add as per two way slabs design. Now, you include a cross of embedded beams (rebar shearhead) in the slab ... and complies without further thickening! It is clear then that the code must be more sympathetic to beam design than two way slabs for shear.

 

[apsix]

The Australian code, AS3600, is more consistent. (This may not help your problem but should help your understanding)

By AS3600, 0.5Vc doesn't apply for beams if D does not exceed the greater of 250mm and half the width of the web. The limit is Vc for those cases.

From the commentary to AS3600; "Concrete beams can possess onsiderable strength without shear reinforcement. However, this strength will be reduced if cracking occurs. Cracking resulting from restrained shrinkage and restrained thermal deformations have been responsible for a number of shear failures in members without shear reinforcement. Since shear failure can be quite sudden, the Standard adopts a conservative approach to the utilization of the strength of beams without shear reinforcement."

For some reason the same degree of conservatism wasn't necessary for slabs &shallow beams.

 

[oneintheeye]

can I have another opinion on this. How would you say the duration of loading would effect the shear failure? The load applied will be very short term. A one off crane lift.

 

[ishvaaag]

The process of loading through a cable of relatively small stiffness makes that the impact factor when applying the maximum load must be small, and even so more thant enough, it is my view, to overcome any loss of strength that would be required for long term applications. The assumed strength in the codes uses to contain around a 15% of diminished strength in compression for concrete for long term application of loads. So in general, being in shear a fragile failure, it is likely that if it survives short term would survive long term, with the crane load applied for some time, and normally not a very special other consideration of impact factor than as per recommendation of the crane operator needs to be accounted. If he says, here 35 tons reaction, put those there; most likely is a conservative estimate, both in value and since not truly long term, for shear.

 

[oneintheeye]

I'm sorry I am not understanding your reply. I am asking for opinions on the load duration for shear failure. Not whether there will be a long term problem even if it stays up during the crane lift.
The crane load given causes a problem as per the earlier posts.

 

[ishvaaag]

Essentially I am saying that you need not to use a particular impact factor on a sudden load, that ordinary calculations will cover for that once you use the calculated value of the reactions. Then as it is a temporary load, you may use the reduced safety factors allowed for works.

Conversely, I don't see a way of acknowledging a better behaviour for the concrete respect shear because the load lasts, say, 1 day or part of it. If the load brings the member to its (code) shear strength plus safety factor (the load safety factor consumed), you bring the structure at a situation that is not resisted every 95% of the times, or so goes the limit states theory.

So the safety factor on loading is what separates the beam of that degree of safety, to be safe 95% of the times. And this level of safety is sought to be attained anywhere reliably in the structure by following the normative clauses. So for longterm structures you keep 1.5 or 1.6 times the required strength to be a sound structure 95% of the times, and for temporary works, 1.25 or 1.35 times such required strength (less, but still well over the required strength to be safe 95% of the times).

Shear is a fragile failure and the mere average shear stress is a good indicator of its closeness to failure. For ordinary concretes of say 20 MPa compressive strength of the past decades, if you showed to have over 3.5 MPa in shear in a section, you were looking at rubble in a steel cage.

Nawy puts the limit of shear stress of plain concrete at about 20% of the compressive strength (without concomitant compression). Essentially, we do NOT want the plain concrete to crack in shear since a fragile condition, so we normally project our sections to limit values that forfeit before than anything the shear failure of the plain concrete within. We are when reinforcing in shear, in a situation akin to when reinforcing glass with some mesh, or some normal (quite deep) footing with a mesh. In all the three cases, it is expected that the plain material will crack before the steel does what it can; for the glass case most of the times just keep the pieces together; for the footing case, it is unlikely you will have more strength from the mesh rebar than of the bare concrete section itself but at significant amounts of rebar unusual till recently mandated at some places; and for the shear case, if you let to go the average stress in shear (without concomitant compression over the section) to 20% of the actual compressive strength, you will have a ruined member.

 

[hokie66]

Duration of load is not a consideration in concrete design for shear. Beam shear does not in fact occur instantaneously, but no allowance is made for short term loading as is done with wood.

 

[apsix]

Perhaps a reduction in safety factor can be used, but a careful assessment & control of risks is required.

 

[rapt]

Handbook to British Standard by Palladian Publications suggests

1 The transverse spacing limit is basically to prevent dowel failure of the tension reinforcement
2 Any bar > 150mm from a stirrup leg should be ignored in the calculation of vc, but ccan still be included for flexure.
3 The spacing limit can be ignored if vu < vc.

 

[oneintheeye]

rapt, yes I am aware of this explanation. Folowing on would you then utilise the shear links (vertical bars) as resisting shear.

i.e. if you have your Vc based on the bars in 150mm of a leg, in my case the first two bars on the outside of the section. The Vu is greater than Vc can you include the links in the shear resistance? I only have vertical legs on the outside (side faces) of the beam. Probably explained better by my sketch attached in an earlier post.

 

[ishvaaag]

You can, but when in the shear calculation stating the longitudinal steel only count the first two bars, each side. Or at least, that is not in lack of logic or prohibited by the code. Furthermore, quite limited minimum longitudinal steel serves to ensure quite efficient behaviour of the stirrups.

But this can't make for the lack of compliance on transversal separation of vertical legs of shear reinforcement.

 

[oneintheeye]

yes I understand that interpretation of the code but I would then have Vc based on 4 bars (2 outer bars each side 150mm apart) in combination with 2 shear legs at these bars, so one vertical leg each side of the section. In this case The beam is 1500mm side so I have 1500 - 2 x 200 = 1100mm width of concrete in the middle with a) no shear links b) no effective longitudinal reinforcement from a shear resistance point of view.

 

[ishvaaag]

True. Can you define the section, what reinforcement above, what below, and shear links @ center ? this way we can make some assessment by other means, worksheets etc. I made a worksheet following a procedure by Vijaya Rangan that gave nicely (assume it works for your case) and adjusted minimum shear steel for a standing longitudinal steel (that defined at BS, tensile, above, etc) Also others following Collins etc that in order to get appraisal of the problem. On the other hand, I think for the proportions of your sketch and being a service level situation, the load most likely will arch towards existing (enough) stirrup steel at the sides...initially seems something more of pure code compliance than one actual problem at works ... as it used to be said, RC structures behave as we reinforce them. Here the designer has placed it seems enough shear steel but at faces, then structure will follow developing arcing action towards the stirrups and the only worrying things would be that in one strut and tie scheme compressive strengj exceed the limit and not enough tie be present. If the strut and tie scheme is solid, the structure will behave well.

 

[oneintheeye]

please see attached section. Appreciate all your input in this

 

[ishvaaag]

I attach a printout of your case as an entry in the as per Vijaya Rangan model article worksheet. I have assumed an actual compression capacity of 20 MPa, so characteristic compressive strength should be about 30 MPa in the actual beam. For this data, smeared shear (not an area where strut and tie would make a better model) the worksheet indicates that your section could sustain some aesthetic factored 30 m·tonne moment in concurrence with factored 120 tonne shear. The struts neither would crash nor show cracking. The entry for the strength of concrete is one assumed to exist in the beam, so you can go from the actual strength of the concrete in the beam (85% of it as a precaution against sustained loads that fail structures 1 day after application).

A repeat of the worksheet with the same moment and fc at 10 MPa gives 85 tonne shear concurrent with 30 m·tonne (both factored), and with fc entry at 30 MPa the 30 m·tonne factored moment can be concurrent with 155 tonne factored shear. By tonne understand metric ton. It seems all quite linear growth in bearable shear with strength. In all three fc cases, at separation 15 cm you are over twice closer than required as a minimum (as the sheet was made). In all cases shear cracking and strut compression remains safely apart from limit values, so a failure of the strut and tie scheme seems unlikely.

Please look the worksheet and look specially for some area of bad input (input goes in blue), gladly may approach more another copy to your case.

 

[ishvaaag]

The case for fc= 10 MPa wouldn't be covered, really, see the margins o applicability. But from 20 MPa would apply to the actual strength of almost any practical modern structural concrete since the eighties.