INDUSTRIAL TRAINING BBR(INDIA) PVT LTD MCE,HASSAN KIRAN.H.L,4MC07CCS08 DESIGN DETAILS OF EXISTING TWO WAY Data: Clear short span = 4.50 m Clear long span = 6.30 m Grade ofconcrete = M25 = 25.00 Grade ofSteel = fy415 = 415.00 Support condtons ! Select end con #dthofthesupport $ = 0.23 m Sla" %oundary Condton = smply supported & ' & y f c( )*mm 2 f y )*mm 2
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DesignDESIGN DETAILS OF EXISTING TWO WAY SLABREFERENCESData: Clear short span Lx=4.50mClear long spanLy=6.30mGrade of concrete="1"0fck=25.00N/mm2Grade of Steel=fy415fy=415.00N/mm2Support conditions 9Select end condition case no. from the table26 Width of the support w=0.23mSlab Boundary Condition =simply supported
Ratio (Clear Long span/Clear short span) Ly/Lx = 6.30/4.50=1.40
The slab is to be designed for Two way
Slab thickness:Design of RC structures by N.Krishna Rajusimply supporteddepth=span/28Third edition (As per IS:456-2000),Pg-106continuousdepth=span/32CBS publisherscantileverdepth=span/7IS:456=2000,clause( 23.2.1),Pg-37Effective depthd=165.00mm5Ceiling for depth
Overall depth(D):Minimum clear cover for slab=15.00mmClear Cover Provided=20.00mmDia of the bar fe415=12.00mmOverall Depth=Effective depth + Clear cover + 0.50 x Dia of the barD=165.00+20.00+6.00D=195.00mm
Effective span(Leff):The least value of :1)Clear span+Effictive depth = 6.30+0.17=6.47m2)Centre to centre of supports = 6.30+0.23=6.53m
Therefore, Effective spanLeff=6.47m
Loads: Density of concrete=25KN/m3Unit width of the slab=1000mmSelf weigth of slab,DL=b x D x Density of concreteDL=1.00x0.20x25.00DL=4.88KN/m2Super imposed load LL=3.00KN/m2Floor finish FL=1.00KN/m2Total load,w=DL+LL+FLw =4.88+3.00+1.00w=8.88KN/m2Ultimate load,Wu= Load factor x Total loadWu=1.50x8.88Wu=13.31KN/m21.5Ulitimate load factor
Ultimate Design moments and shear force:Moment coefficientsIS:456=2000(Table-26),Pg-91x=0.085x: short directiony=0.056y: long directionUltimate momentsMux=x x Wu x L2Mu=0.09 x 13.31 x 6.47^2Mux=47.30KN-mMuy=y x Wu x L2Mu=0.06 x 13.31 x 6.47^2Muy=31.16KN-mUltimate shear force,Vu=0.5 x Wu x LVu=13.31 x 6.47/2Vu=43.03KN
Check for depth:for fe415,d'=(Mu /(0.138 x fck x b))1/2)d'=((47.30 x 10^6) / (0.138 x 25.00 x 1000.00))^(1/2)Requried depthd'=117.08mmd>d',hence,Effective depth is Safe
Reinforcement:a)Along shorter spanMu=0.36 x fck x b x d2 x (x/d) x (1-0.416 x (x/d))IS:456=2000,ANNEX-G,clause(38.1),Pg-96Neutral axis depth,(x/d)=1.2-Sqrt(1.4-(6.6 x Mux/(fck x b x d2)))x/d=1.2-Sqrt(1.4-(6.6 x 47.30 x 10^6/(25.00 x 1000.00 x 165.00^2)))x/d=0.21(x/d)
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