The question we looked at was a newspaper building (Q4 2005) and we all pretty much had at least one scheme that comprised the following layout. Roof to level 5, slab spans 6.25mx6.15m, either flat slabs or down stand beams Transfer structure at level 5, and slabs spanning 13.3mx 12.5m. Looking at the upper floors first, the floor span is relatively small (<7m) you can use span/28 (or 30) to give you a slab depth of 200mm or so. Now from my personal experience this is a little bit close to the mark for simply supported slabs for this span however continuous design is fine. If you were to have a flat slab, span/28 is about correct for this span of slab, although you will have more reinforcement (I wouldn’t worry too much about this unless the question specifically mentions access to materials) Remember though that for concrete, span/depth ratios is related to the type of slab and the relative span. Applying span/28 for your lower slabs will give you too shallow a depth! If you wanted to look at post tensioned, you can normally save 25mm of slab depth over the flat slab depth (plus obviously the rebar). You could try for a greater depth saving with post tensioned, however you need to be careful with creep and deflection. OK, now we have our slab depth we can work out what load is coming down our columns very quickly i.e. 24 (or 25 if using eurocodes!) x0.225=5.4kN/m2 Add in the live load and we have 8.4kN/m2 SLS or 12.36kN/m2 ULS. So a typical column load above level 5 = 12.36x6.35x6.15x7 floors = 3379kN ULS Now for column sizing, let’s assume 1% reinforcement and limit our compressive stress to 0.45fcu. This limiting stress value is a good starting point, regardless of the rebar, as it allows for notional bending without having to work it out, and self weight of the column So let’s use 50N/mm2 concrete (the loadings are going to be large) therefore our column needs to be 3379x1000/ (0.45x50)=150169mm2 or 387x387 round up to 400 x400mm So in about 1.5 minutes you have half the building designed! Now for the transfer deck/beam at level 5, we are transferring one column load across 12.5m, but we are also picking up other transfer beams at the same time (remember those column centres!) So let’s work out the shear on the worst case transfer beam to work out the size of the beam.
I am going to use span/10 to give me an initial depth of 1.25m, and our loading is 3379+ 3379/2 + 3379/2 + 12.36x12.5x13.3= 8813kN. Now we are going to limit our shear stress on the beam to 2N/mm2. This number is not that conservative, as if any of you have design concrete for shear, your applied shear stress is close to this number to allow you to design reasonable shear links. So our beam width needs to be 8813x1000 / (2x2x1250)= 1763mm Therefore our transfer beam is 1250dpx1800 wide. You may have realised at this stage we haven’t worried too much about self weight of the column and beams, the columns are not going to have a big influence (86kN from roof to level 5), but these transfer beams are quite a bit of weight each (675kN each). And for the next part of the scheme design we are looking at the long spanning slabs at levels 4 down to upper basement. Given that these are quite large, it is appropriate to use some form of voided/ribbed/ coffered slab, as it keeps the weight down and also save on materials. As these are not used as frequently anymore, it would be a good idea to at least have some typical sizes of ribbed slabs in your preparation folder, even have anticipated spans so that you have a bit more confidence writing the answer down if you have never designed one before. If you haven’t designed any concrete by hand before, at least do a few beams and slabs by hand and keep it in your preparation folder. One thing I didn’t go into detail on is ribbed slab designs, which are governed by the deflection of the beams that support them. For the ribs it is normal to limit the shear to 0.6N/mm2 and say use 150mm wide ribs at 600mm centres with 150mm structural top to give you a rough size. You can use any size you want in preliminary sizing and tweak it in part 2 if you wish. I did say we can use a span/15 or so (I was looking to get the slab to be 900mm deep as it “felt” like the correct depth) However the imposed loading of 15kN/m2 is a little bit outside of the design charts of the “Economic concrete Frame Elements” book by the Concrete Centre which I would always have looked at for some really quick sizing. So having looked at the justification , I am going to split up the slab span with beams just because the loading is so high. Sometimes going down a route of putting numbers to part 1 schemes and being unable to justify them, can be somewhat frustrating. However there is nothing wrong
with going back and adding extra beams here or there to make it work, that’s what tippex and erasers are for! Just make sure you are still complying with the clients brief So suddenly our slab is conveniently the same span as the other floors, the beams we can limit the shear stress to 2N/mm2 again, So total load on beam =12.36x12.5x13.3=2055kN Limiting stress to 2N/mm2 size of beam is 1250mmdpx850w Now we have the remainder of the floors sized we can size the columns in the centre Total load on column =3379x4 +675x2 +(2055x2x6)= 39526kN Therefore size of column (based on 0.45fcu)= 1325x1325mm Round up to 1500x1500mm for blast resistance (if you want to) This leave us with the retaining walls which are approximately 6m high, use height/10, which gives you 600mm thick walls. The last gravity element to scheme are the foundation sizes, and our sls load is 39526/1.5= 26351kN The rock has an allowable GBP=6000kN/m2 therefore size of pad=2x2m Last thing to scheme is lateral stability, assuming we are using shear walls; we just need to consider the shear wall as vertical cantilevers. Our two lower lifts are 4x14m long, plus another 2 lifts 4x9m. This means we will have at least 8no. 4m long shear walls for the lower portion of the building. So using H/15 for a cantilever wall, Our actual H/depth is 46.4/8=H/5.8 Therefore we don’t need any extra shear walls to help us out. So in about 5-10 minutes we have justified one of our schemes,
