POST-TENSIONED CONCRETE COLUMN SUPPORTED SLAB DESIGN (FLAT PLATE SYSTEM)

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Ιόλη Αλεξανδρίδης
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1 POST-TNSIOND CONCRT COLUMN SUPPORTD SLAB DSIGN (FLAT PLAT SYSTM) DSIGND BY Mr. JAMALUDDIN CHALRMTHAI STRUCTURAL NGINR

2 TWO-WAY COLUMN-SUPPORTD POST-TNSIOND SLAB DSIGN CRITRIA MATRIAL CONDITIONS Concrete: fc' = 350 ksc β1 = 0.80 c = ksc Mild Steel: fy = 4000 ksc ( SD40 GRAD ) fu = 5600 ksc s = ksc Prestressing Steel: fpy = ksc (1860 GRAD) fpu = ksc 0.94fpy = ksc 0.80fpu = ksc Total Approximate Losses = 25 % ffective Losses = 12.5 % R = 0.88 fe = ksc fj = ksc p = ksc Due to the condition of symmetrically prestressed, the effective loss is equal to a half of total approximate loss. CRITRIA : 1 FROM 1 LOAD CONDITIONS DL : LOAD FACTOR 1.4 Concrete = 2400 kg/m 3 Other Super Imposed DL = 50 kg/m 2 LL : LOAD FACTOR 1.7 Super Imposed LL = 400 kg/m 2 L1 P L A N FLOOR TO FLOOR HIGHT FLOOR TO FLOOR HIGHT QUIVALNT FRAM SCTION

3 POST-TNSIOND FLAT PLAT DSIGN SPRADSHT PROJCT SLAB DSIGN FOR STIMAT SLAB COD FP6000 FP6000X6000 : 1 FROM 4 DIMNSION ANALYSIS Column Size (c 1 xc 2 ) = 0.60 x 0.60 m 2 Floor to Floor Height = 4.00 m L 1 = 6.00 m L 2 = 6.00 m ts min = L 2 /45 = cm Apply ts = 18 cm Concrete to Strand covering = 3 cm Concrete to Steel covering = 1.5 cm dp = 15 cm ds = 16.5 cm e = 6 cm y = 12 cm LOAD ANALYSIS Total Super Imposed DL = Total Super Imposed LL = Total Super Imposed DL+LL = Total Super Imposed Factored Load = 0.18x = 482 kg/m 2 = 400 kg/m = 882 kg/m 2 1.4x x400 = kg/m 2 PC-STRANDS ANALYSIS quivalent Balancing Loads = Total SDL = 482 kg/m 2 Balanced Distributed Loads = 482 x 6 = 2892 kg/m P e = Wb L 2 / 8y = 2892x6^2 /(8x0.12)= kg P j = P e /R = /0.875 = kg f pe = = ksc FROM SVN-WIRS STRAND Ø cm. A bp = cm 2 /strands Approximated nos. of Required Strands = /(14880x1.4)= 6 nos. APPLYING : SVN-WIRS STRAND Ø cm. x 9 nos. P e = 9x1.4x14880x0.875= kg COLUMN'S STIFFNSS ANALYSIS Kc = 4 c Ig /(L c - 2 t s )= (4x(282495x100^2)x(0.6x0.6^3/12))/(4-2x0.18) = kg-m ΣKc = 2xKc = 2x = kg-m CHK. bw+2hw = 60+2x18 = 96 cm bw+6hf = 60+6x18= 108 cm x1 = 18 cm y1 = 96 cm C = Σ(1-0.63x/y)x 3 y/3 = (1-0.63x18/96)x18^3x96/3 = cm 4 ΣKt = Σ9c C/[L 2 (1-c 2 /L 2 ) 3 ] = 9x(282495x100^2)x(164579/100^4)/[6x(1-0.6/6)^3] = kg-m From: Kec = (1/ΣKc+1/ΣKt) -1 FOR XTRIOR SPANS : Kec = (1/ / )^(-1) = kg-m FOR INTRIOR SPANS : Kec = [1/ /(2x )]^(-1) = kg-m SLAB'S STIFFNSS ANALYSIS Kes = Ks = 4 c Ig /(L 1 - c 1 /2) = [4x(282495x100^2)x(6x(18/100)^3/12)]/(6-0.6/2) = kg-m MOMNT DISTRIBUTION FACTOR ANALYSIS From: DFs = Kes /(Kec+Kes) DF exterior = /( ) = DF interior = /( x ) = PATTRN LOAD ANALYSIS βa = W SLL /W SDL = 400/482 = 0.83 > 0.75 α c = ΣKc/ΣKs = /(2x ) = 5.8 α 1 = cb I b / cs I s = 0/[cx(6x0.18^3/12)] = 0 L 2 /L 1 = 6/6 = 1.00 α min = Minimum value of α c = 0.70 < 5.8 NOT NCSSARY TO DTRMIN FFCTS FROM PATTRN OF LOADS FLAT-PLAT DSIGN SPRADSHTS / ACI

4 FLXURAL ANALYSIS NT LOADD MOMNTS : W b = 8P e y/l 2 = 8x164052x0.12/6^2 = 4375 kg/m W NT = W TSL - W b = 882x = 917 kg/m FIXD ND MOMNT = 917x6^2/12 = 2751 kg-m Moments Distribution FP6000X6000 : 2 FROM 4 JOINT A B C D DF FM BALANC C.O BALANC C.O BALANC C.O BALANC C.O BALANC ΣM CHCK ALLOWABL STRSSS IN CONCRT Maximum Design Moments and Stresses : M NG = = 3199 kg-m M POS = 917x6^2/8-( )/2 = 1694 kg-m FROM : fc stresses = Pe/A ±My / Igs = /(600x18)±319900x9/(600x18^3/12) = 15.19±9.87 fcc = ksc fct = 5.32 ksc Allowable Stresses : fcc allow = 0.3x350 = 105 ksc > fcc : O.K. fct allow = -1.6 (350) = ksc < fct : O.K. NT BALANCD MOMNTS : Wb = = 4375 kg/m FIXD ND MOMNT = 4375x6^2/12 = kg-m Moments Distribution JOINT A B C D DF FM BALANC C.O BALANC C.O BALANC C.O BALANC C.O BALANC ΣM PRIMARY MOMNTS : M PRIMARY = x 0.06 = 9843 kg-m JOINT A B C D ΣM SUMMATION M SCONDARY = M NT - M PRIMARY NT MOMNTS PRIMARY MOMNTS SCONDARY MOMNTS FLAT-PLAT DSIGN SPRADSHTS / ACI

5 FP6000X6000 : 3 FROM 4 FACTORD MOMNTS : W U = x 6 = kg/m FIXD ND MOMNT = x6^2/12 = kg-m Moments Distribution JOINT A B C D DF FM BALANC C.O BALANC C.O BALANC C.O BALANC C.O BALANC ΣM ULTIMAT MOMNTS : JOINT A B C D FACTORD MOMNT SCONDARY MOMNT ULTIMAT MOMNTS : SUMMARY OF DSIGN MOMNTS M FACTORD POS = x6^2/8-( )/2 = kg-m M U POS = ( )/2 = kg-m M U NG = = kg-m RINFORCMNT ANALYSIS Mild Steel Reinforcements : use D 12 mm A b = cm FFCTIV WIDTH OF RINFORCMNTS = 3x18+60 = 114 cm FFCTIV LNGTH OF RINFORCMNTS = 600/3 = 200 cm As min = t s L 2 = x18x600 = 8.1 cm 2 Maximum Bar Spacing = 1.13/8.1x114 = 16 cm APPLY 10-D12@ 12.5 cm ( L= 200 cm ) SPAN-DPTH RATIO = L 2 / t s = 600/18 = 33 < 35 fse = Pe / Aps = /(9x1.4) = ksc p = Aps/bd = 12.6/(600x15) = fpy = ksc fps = fse fc'/(300 p) = /(300x0.0014) = ksc < O.K. NOMINAL NGATIV FLXURAL CAPACITY fse = ksc a = (Aps fps+as fy)/(0.85fc' b) =(12.6x x4000)/(0.85x350x600)= 1.28 cm ØMn=Ø[Aps fps(dp-a/2)+as fy(ds-a/2)]= 0.9x[12.6x145.53x( /2)+10x1.13x40( /2)]= kg-m > kg-m O.K. NOMINAL POSITIV FLXURAL CAPACITY a = Aps fps/(0.85fc' b) =(12.6x14553/(0.85x350x600)= cm ØMn=ØAps fps(dp-a/2)= 0.9x12.6x145.53x( /2)= kg-m > kg-m O.K. PUNCHING SHAR ANALYSIS Outer Column PROJCTD ARA = 6 x 3 = 18 m 2 bo = c 1 +2c 2 +2d P = 60+2x60+2x15 = 210 cm Vup = 18 x = kg vup = /(210x15)= ksc Øvc = 1.06x0.85 (350) = ksc O.K. Inner Column PROJCTD ARA = 6 x 6 = 36 m 2 bo = 2c 1 +2c 2 +4d P = 2x60+2x60+4x15 = 300 cm Vup = 36 x = kg-m vup = /(300x15) = ksc Øvc = 1.06x0.85 (350) = ksc O.K. FLAT-PLAT DSIGN SPRADSHTS / ACI

6 FP6000X6000 : 4 FROM 4 DFLCTION CHCK I panel = 600x18^3/12 = cm 4 I col strip = I mid strip = /2 = cm 4 w net = = 917 kg-m Δf = 5x9.17x600^4/(384x282495x291600)= cm From M panel ; M col strip = 75% M mid strip = 25% Δf col strip = Df [M col / M panel x I panel / I col ] = cm Δf mid strip = Df [M mid / M panel x I panel / I mid ] = cm Δ = M net L /(8Kec) = Δ B = ( )x100x600/(8x ) = cm Δ c = ( )x100x600/(8x ) = 0 cm SUMMARY ΣΔ col strip = = cm ΣΔ mid strip = = cm Δ allow = 600/240 = cm O.K. = 6000 mm. MIDDL STRIP L1 = 6000 mm. 2-SVN WIR STRANDS 2-SVN WIR STRANDS MIDDL STRIP 7-SVN WIR STRANDS 7-SVN WIR STRANDS 2000 mm FP6000 ( T = 180 mm ) N O T T O S C A L 2000 mm 10-D12@ D12@125 RBARS RINFORCMNT DTAIL N O T T O S C A L FLAT-PLAT DSIGN SPRADSHTS / ACI

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