TechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.PROJECTREINFORCED CONCRETEFLOOR DESIGN

PROJECT ADVISOR STUDENT: PALAGHIA IONUT-NICOLAE MARINELA BARBUTAANUL:IIIGRUPA: 3311

2009-2010TechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.1. Preliminary design of the structural elements1.1. Distance between beamsN=4Pn=6960N/m2L=11,6ml=5,6ma=2,9m =l/a=1.93 21.2 Preliminary design of the slab (one way reinforced)hp=a35=2,95*10^335=82,857 90mmOn choose beb=300mm bb=200mm1.3 Clear spans lc1=a-beb2-bb2=2,9-0,300/2-0,200/2= 2,65mlc2= lc3=a-bb2-bb2=2,9-0,200/2-0,200/2= 2,70mTechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A. 1.4 Loads (according to STAS 10101/OA-77 [3] and 10101/1-78 [4])A. Unfactored valuePermanent load-from own-weight of slab (hp=90mm)gnslab=0,9*25000=2250N/m2-from plaster weight (hplaster=30mm) gnplaster=0,03*21000=630N/m2-from mosaic weight (hmosaic=30mm) gnmosaic= 1000N/m2Total gn=3880N/m2Temporary load-service load given in project data:pnu=6960N/m2total load:gn+ pnu=qn=10840N/m2Permanent load from own-weight of slab: 2250*1,3=2925N/m2 from plaster weight: 630*1,3=820N/m2 from mosaic weight:1000*1,3=1300N/m2

Temporary load :pu=6960*1.3=9048N/m2Total load:q=14093N/m21.5 Statical ComputationTechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.Fig. 1.3M8=M1=111 q lc12=9339,8Nm MB=MH=-114 q lc12=-7338,4Nm M2=M3=M3=-116 q lc22=-6421,1Nm MC=MD=-116 q lc22=-6421,1Nmd=hp-cnom- /2=90-20-5/2=67,5mmcnom=cmin+c=10+10=20mm =M1bd2fcd=0.214 St,eff=200-2*30= 140 mmSupport B Left:The design shear force is:VB,redleft=VBleft-qd=173288,61-45965,15*460=152144,64 NThe capable shear force of the member without specifically shear force reinforcement is computed with relation:VRd,c=[CRd,c*1/*(1001*fck)1/3+k1*cp)*b*dTechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.Where cp = 0 axial force is neglectedCRd,c=0.18/c=0,18/1,5=0,121=1.0 for ordinary concretek=1+200d=1+200460=1.434 VRd, c min= 24.564 kNVB, redleft=173288,61 N> VRd,c=39.404 ctg =2.5VRd, max=cw*b*z*1*fcd*1ctg +tg =1*200*405*0.54*13.33*12.5+12.5=201.05*103 NIt can adopt for ctg =1.75The distance among stirrups for a diameter of 6mm and z=0.9d=0.9*450=405 is given by:As,w=2*28.3=56.6 mm2fywk=255 N/mm2 for OB37 and 6mm fywd=fyk1.15=255/1.15=221.7 N/mm2s=Asw*z*fywd*ctg VBredleft=(56.67*405*221,7*1.75)/( 173288,61)=51 mms=50 mmctg eff=VBredleft*sAsw*z*fywd=1.70VRd, max=cw*b*z*1*fcd*1ctg +tg =1*200*405*0.54*13.33*11.7+11.7=254.80*103 NVRd, max>VBredleftw,eff=Asws*b=56,650*200=0,00566 >min lb,rdq=1270 mmlbd=1*2*3*4*5* lb,rdq>lb, min1=1 for straight bars2=0.873=1.0=45=1.0lbd=1105 mmlbd, min=381 mm Where coefficients are :1 =1for straight bars( ) 90 . 01818 3015 . 0 1 / 15 . 0 12

,_

dcWhere cd =min(a/2, c1, c)=min(54,30)a=200-230-218=104 mm3= 4=1.015 without transversal pressuremm lbd1287 1430 1 1 1 90 . 0 1 bd bbrdqbdl mm lmmmmmm ll < ;' 429100180 18 10 10429 1430 3 . 0 3 . 0maxmin , min ,TechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.TWO WAY EDGE SUPPORTED SLABS2.1. Data and specificationsA concrete floor type two-way edge supported slab is considered as a part of a building. The sizes of the slab are (9.0m x 22.5 m) with two spans (L) of 4.5 m and five longitudinal spans (T) of 4.5 m. The floor is realized of concrete C20/25, the steel type PC 52. The finish is realized of mosaic (30 mm depth) on a layer of cement plaster of 30 mm depth.The live load is considered: pk = 4000 N/m2.Fire resistance 60 minutes R60.Exposure class to environmental conditions XC1Concrete C20/25.Steel type PC52.For concrete C20/25 the design strength in compression:33 . 135 . 1201 cckcc cdff N/mm2Tensile strength is obtained with the relation:15 . 15 . 1105 . 0 cctkct ctdffN/mm22 . 2 ctmf N/mm2Steel reinforcement type PC52 For mm 28 345 ykfN/mm2

30015 . 1345 ydf N/mm2

TechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A. For mm 14

355 ykf N/mm2

30915 . 1355 ydfN/mm22.2. Slab computation2.2.1. Geometrical data of slabL= 7.3 mT=7.3Fig. 2.2.It must arrange the beams in such way that the ratio:5 . 121 ll3 . 73 . 7 = 1.00 < 1.5In this case the slab is reinforced on two directions.The depth of the slab must respect the following conditions:1825 . 0403 . 740 lhf mIf the variable action is big, the depth of the slab can be increased, so:hf = 150 mm2.2.2. Preliminary sizing of beamsFor beams, the minimum depth is:3 . 7201201min l h= 0.365 mFor establishing the width of beams, it must respect the condition:3 ... 2 bh, so 260 . 02 hb= 0.300 mOne chooses b = 300 mm.The sizes of the beam are : b=300 mm; h= 600 mmTechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.2.2.3. SpanFor establishing the spans the provisions are given in Appendix II.2 1a a l ln eff+ + fy y x xh hb tmmh ta a a a ,_

,_

;752150;2300min2;2min, 2 , 1 , 2 , 11 7150 / 7150 /7150 75 75 70007000230023007300, ,, ,, + + x eff y effy eff x effnx nyl lmm l lmm l lFor 1 the resistant reinforcement is on two directions2.2.5. LoadsTable 2.1.Type of loadsCharacteristic valueN/m2niDesign valuesN/m2A. Dead loads- slab weighthp b 1.0 1.0 =0.18250 25000 1.0 1.0 = = 4562.5 N/m21.1 5018.75- mosaic weight (hmosaic = 30 mm) 1000 N/m21.3 1300- plaster weight (hplaster = 30 mm)TechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.hplaster plaster 1.0 1.0 = 0.030 21000 1.01.0 = 630 N/m21.3 820TOTAL DEAD LOAD gn = 4562.5 + 1000 + 630 = 6192.5g = 5018.75 + 1300 + 820 = 7138.75B. Live loadpn = 2520 N/m2pn = 2520 1.3 p = 3276TOTAL LOAD qn = gn + pn = 6192.5 + 2520 = 8712.5q = g + p = 7138.75 + 3276 = 10414.752.2.6. Statical computation The computation is made in elastic domainDetermination of momentsMaximum negative edge moments are obtained when both panels adjacent to the particular edge carry full dead and live load.Forpositive momentsthere will be little, if any rotation at the continuous edges if dead load alone is acting, because the loads on both adjacent panels tend to produce opposite rotations which cancel, or nearly so. For this condition, the continuous edge can be regarded as fixed. On the other hand, the maximum live load moments are obtained when live load is placed only on the particular panel and not on any of adjacent panels. In this case, some rotation will occur at all continuous edges.If we consider a unit strip from the slab (Fig. 2), charged with g + p that can be replaced with q and q,2'pg q + 2"pq t TechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.Fig. 2The interior supports can be considered fixed and the external span is fixed at one edge and simple supported at the external edge.Under the action ofq, the rotations on the interior supports can be considered equal to zero.Under the action of loadq, the rotations are free at all supports; the stripcanbe considered as beam simple supported (Fig. 3).So, fortheloadqthepanelscanbeconsideredlikesinglepanelswithcorresponding edges (continuous or discontinuous) (Fig. 4).Clear spans and static schemes (elastic domain)Direction 1Fig.3aTechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.Fig. 4.a.Fig. 4.b.Direction 2Fig. 3.b.For the load q, the panels are simple supported at all four edges (Fig. 4.b).One uses the coefficients method, which makes use of tables of moment coefficients for a variety ofconditions. Thesecoefficientsarebasedonelasticanalysis, but alsoaccount forinelastic redistribution.In consequence, the design moment in either direction is smaller by an appropriate amount than the elastic maximum moment in that direction.2eff i xl q M 2eff i yl q M i, i tabulated moment coefficientsq uniform loadleff, length of span in both directions.TechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.A system of beams supports the slab; some panels, such as (4), have two discontinuous exterior edges, while the other edges are continuous with their neighbors.Panel (5) has one edge discontinuous and three continuous edges, the interior panel (6) has all edges continuous,and soon.Ata continuous edge in a slab, moments are negative; also the magnitude of the positive moments depends on the conditions of continuity at all four edges.The maximum and minimum moments at midspan on the two directions can be determined as the sum of the moments given by q and q that act on the single panels, in function of the ratio:ll2 2327675 . 71382' + + pg q= 8776.75 N/m216382" t t pq N/m2The coefficients and for determining the bending momentsMx,Myand on supports are given in Tables ...0 . 13 . 73 . 712 llA. Moments for panelsPanel type (1) 1 = 0.03646Panel type (4)4 = 0.02692Panel type (5)5 = 0.02268 5 = 0.01977 (A) Positive momentsPanel type (4)2124" 'eff eff y xl q l q M M + = = 0,02692 8776.75 7.152 +0.03646 1638 7.152 = 15131.8 NmPanel type (5)2125" 'eff eff xl q l q M + == 0,02268 8776.75 7.152 t0.03646 1638 7.152 = 13229.38 NmTechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.2125" 'eff eff yl q l q M t == 0.01977 8776.75 7.152 t0.03646 1638 7.152 = 5817.47 Nm(B) Moment on supportsFor determining bending moments on supports for continuous slabs supported on perimeter under uniform loads it can use the following relations (according to STAS ) (Fig. 4a)2 2515 . 7 75 . 10414 5 . 0101101 eff al q M = 26621.40Nm2 2515 . 7 75 . 10414 5 . 0121121 eff bl q M = 22184.5 Nm2 2515 . 7 75 . 10414 5 . 08181 eff cl q M = 33276.75 Nm( ) ( )2 2615 . 7 75 . 10414 5 . 0 11011101 eff dl q M = 26621.4 NmSteel designThe slab is reinforced on both directions, so it must determine the effective depth on two directions:2 /2 /1 11y s x s f yx s nom f xh dc h d ;' mmmmmmcs1010 5 1510maxminmm c c ctot nom20 10 10min + + TechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.mm dmm dyx115 2 / 10 10 20 150125 2 / 10 20 150 Fromthefireresistanceconditions, Table6.6-Kiss theminimumdistancetothe centroid of resistant steel is amin=10 mm. For a diameter of 10 mm the effective distance to the centroid is:mm a mm c as eff nom eff10 25 2 / 10 20 2 /min , > + + The necessary steel areas are: Span of panel 4:kNm M My x15.1318 245 . 0 242 . 033 . 13 125 30010 1318 . 15262 xcd xxxf d bM m mmffd b Aydcdx x x s/ 34 . 39630933 . 13125 300 245 . 02, One chooses constructive steel: 46 + 410 / m (As = 412 mm2/m)291 . 0 286 . 033 . 13 115 30010 1318 . 15262 ycd yyyf d bM m mmffd b Aydcdy y y s/ 09 . 43330933 . 13115 300 291 . 02, One chooses constructive steel: 412 / m (As = 452 mm2/m) Span of panel 5226 . 0 211 . 033 . 13 125 30010 229 . 1326 x x m mm Ax s/ 6 . 36530933 . 13125 300 226 . 02, One chooses constructive steel: 56 +58 / m (As = 393 mm2/m)TechnicalUniversityGh. AsachifromIasiFACULTYOFCIVILENGINEERINGANDBUILDINGSERVICESC.C.I.A.272 . 0 250 . 033 . 13 115 30010 229 . 1326 y y m mm Ay s/ 8 . 40430933 . 13115 300 272 . 02, One chooses constructive steel: 36 + 312 / m (As = 424 mm2/m) On supportsmm d dy x125 034 . 0 127 . 033 . 13 125 100010 621 . 26262 ycd xad af d bM m mm Aa s/ 34 . 18330933 . 13125 300 034 . 02, One chooses: 76 / m (As = 198 mm2/m)044 . 0 106 . 033 . 13 125 1010 184 . 222 36 y b m mm Ab s/ 2 . 23730933 . 13125 10 044 . 02 3, One chooses constructive steel: 36 + 38 / m (As = 236 mm2/m)067 . 0 159 . 033 . 13 125 1010 276 . 332 36 y c m mm Ac s/ 2 . 36130933 . 13125 10 067 . 02 3, One chooses: 38 + 310 / m (As = 386 mm2/m)