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

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

TWO-WAY COLUMN-SUPPORTD POST-TNSIOND SLAB DSIGN CRITRIA MATRIAL CONDITIONS Concrete: fc' = 350 ksc β1 = 0.80 c = 282495 ksc Mild Steel: fy = 4000 ksc ( SD40 GRAD ) fu = 5600 ksc s = 2.04+06 ksc Prestressing Steel: fpy = 17100 ksc (1860 GRAD) fpu = 18600 ksc 0.94fpy = 16074 ksc 0.80fpu = 14880 ksc Total Approximate Losses = 25 % ffective Losses = 12.5 % R = 0.88 fe = 14880 ksc fj = 13020 ksc p = 1.97+06 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

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 = 13.33 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 = 0.18x2400+50 = 482 kg/m 2 = 400 kg/m 2 482+400 = 882 kg/m 2 1.4x482+1.7x400 = 1354.8 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)= 108450 kg P j = P e /R = 108450/0.875 = 123943 kg f pe = 14880 = 14880 ksc FROM SVN-WIRS STRAND Ø 1.524 cm. A bp = 140 1.40 cm 2 /strands Approximated nos. of Required Strands = 123943/(14880x1.4)= 6 nos. APPLYING : SVN-WIRS STRAND Ø 1.524 cm. x 9 nos. P e = 9x1.4x14880x0.875= 164052 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) = 33526879 kg-m ΣKc = 2xKc = 2x33526879 = 67053758 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 = 164579 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] = 9566408 kg-m From: Kec = (1/ΣKc+1/ΣKt) -1 FOR XTRIOR SPANS : Kec = (1/67053758+1/9566408)^(-1) = 8371994 kg-m FOR INTRIOR SPANS : Kec = [1/67053758+1/(2x9566408)]^(-1) = 14885465 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) = 5780740 kg-m MOMNT DISTRIBUTION FACTOR ANALYSIS From: DFs = Kes /(Kec+Kes) DF exterior = 5780740/(8371994+5780740) = 0.408 DF interior = 5780740/(14885465+2x5780740) = 0.219 PATTRN LOAD ANALYSIS βa = W SLL /W SDL = 400/482 = 0.83 > 0.75 α c = ΣKc/ΣKs = 67053758/(2x5780740) = 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 318-89

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 = 882x6-4375 = 917 kg/m FIXD ND MOMNT = 917x6^2/12 = 2751 kg-m Moments Distribution FP6000X6000 : 2 FROM 4 JOINT A B C D DF 0.408 0.219 0.219 0.219 0.219 0.219 0.219 0.408 FM -2751.00 2751.00-2751.00 2751.00-2751.00 2751.00-2751.00 2751.00 BALANC 1122.41 0.00 0.00 0.00 0.00 0.00 0.00-1122.41 C.O. 0.00 561.20 0.00 0.00 0.00 0.00-561.20 0.00 BALANC 0.00-122.90-122.90 0.00 0.00 122.90 122.90 0.00 C.O. -61.45 0.00 0.00-61.45 61.45 0.00 0.00 61.45 BALANC 25.07 0.00 0.00 0.00 0.00 0.00 0.00-25.07 C.O. 0.00 12.54 0.00 0.00 0.00 0.00-12.54 0.00 BALANC 0.00-2.75-2.75 0.00 0.00 2.75 2.75 0.00 C.O. -1.37 0.00 0.00-1.37 1.37 0.00 0.00 1.37 BALANC 0.56 0.00 0.00 0.00 0.00 0.00 0.00-0.56 ΣM -1666 3199-2877 2688-2688 2877-3199 1666 CHCK ALLOWABL STRSSS IN CONCRT Maximum Design Moments and Stresses : M NG = = 3199 kg-m M POS = 917x6^2/8-(1666+3199)/2 = 1694 kg-m FROM : fc stresses = Pe/A ±My / Igs = 164052/(600x18)±319900x9/(600x18^3/12) = 15.19±9.87 fcc = 25.06 ksc fct = 5.32 ksc Allowable Stresses : fcc allow = 0.3x350 = 105 ksc > fcc : O.K. fct allow = -1.6 (350) = -29.93 ksc < fct : O.K. NT BALANCD MOMNTS : Wb = = 4375 kg/m FIXD ND MOMNT = 4375x6^2/12 = 13125 kg-m Moments Distribution JOINT A B C D DF 0.408 0.219 0.219 0.219 0.219 0.219 0.219 0.408 FM 13125.00-13125.00 13125.00-13125.00 13125.00-13125.00 13125.00-13125.00 BALANC -5355.00 0.00 0.00 0.00 0.00 0.00 0.00 5355.00 C.O. 0.00-2677.50 0.00 0.00 0.00 0.00 2677.50 0.00 BALANC 0.00 586.37 586.37 0.00 0.00-586.37-586.37 0.00 C.O. 293.19 0.00 0.00 293.19-293.19 0.00 0.00-293.19 BALANC -119.62 0.00 0.00 0.00 0.00 0.00 0.00 119.62 C.O. 0.00-59.81 0.00 0.00 0.00 0.00 59.81 0.00 BALANC 0.00 13.10 13.10 0.00 0.00-13.10-13.10 0.00 C.O. 6.55 0.00 0.00 6.55-6.55 0.00 0.00-6.55 BALANC -2.67 0.00 0.00 0.00 0.00 0.00 0.00 2.67 ΣM 7947-15263 13724-12825 12825-13724 15263-7947 PRIMARY MOMNTS : M PRIMARY = 164052 x 0.06 = 9843 kg-m JOINT A B C D ΣM 0 9843 9843 9843 9843 9843 9843 0 SUMMATION M SCONDARY = M NT - M PRIMARY NT MOMNTS 7947 15263 13724 12825 12825 13724 15263 7947 PRIMARY MOMNTS 0 9843 9843 9843 9843 9843 9843 0 SCONDARY MOMNTS 7947 5420 3881 2982 2982 3881 5420 7947 FLAT-PLAT DSIGN SPRADSHTS / ACI 318-89

FP6000X6000 : 3 FROM 4 FACTORD MOMNTS : W U = 1354.8 x 6 = 8128.8 kg/m FIXD ND MOMNT = 8128.8x6^2/12 = 24386.4 kg-m Moments Distribution JOINT A B C D DF 0.408 0.219 0.219 0.219 0.219 0.219 0.219 0.408 FM -24386.40 24386.40-24386.40 24386.40-24386.40 24386.40-24386.40 24386.40 BALANC 9949.65 0.00 0.00 0.00 0.00 0.00 0.00-9949.65 C.O. 0.00 4974.83 0.00 0.00 0.00 0.00-4974.83 0.00 BALANC 0.00-1089.49-1089.49 0.00 0.00 1089.49 1089.49 0.00 C.O. -544.74 0.00 0.00-544.74 544.74 0.00 0.00 544.74 BALANC 222.26 0.00 0.00 0.00 0.00 0.00 0.00-222.26 C.O. 0.00 111.13 0.00 0.00 0.00 0.00-111.13 0.00 BALANC 0.00-24.34-24.34 0.00 0.00 24.34 24.34 0.00 C.O. -12.17 0.00 0.00-12.17 12.17 0.00 0.00 12.17 BALANC 4.96 0.00 0.00 0.00 0.00 0.00 0.00-4.96 ΣM -14766 28359-25500 23829-23829 25500-28359 14766 ULTIMAT MOMNTS : JOINT A B C D FACTORD MOMNT -14766-28359 -25500-23829 -23829-25500 -28359-14766 SCONDARY MOMNT 7947 5420 3881 2982 2982 3881 5420 7947 ULTIMAT MOMNTS : -6819-22939 -21619-20847 -20847-21619 -22939-6819 SUMMARY OF DSIGN MOMNTS M FACTORD POS = 8128.8x6^2/8-(14766+28359)/2 = 15017 kg-m M U POS = 15017+(7947+5420)/2 = 21701 kg-m M U NG = = 22939 kg-m RINFORCMNT ANALYSIS Mild Steel Reinforcements : use D 12 mm A b = 113 1.13 2 cm FFCTIV WIDTH OF RINFORCMNTS = 3x18+60 = 114 cm FFCTIV LNGTH OF RINFORCMNTS = 600/3 = 200 cm As min = 0.00075 t s L 2 = 0.00075x18x600 = 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 = 164052/(9x1.4) = 13020 ksc p = Aps/bd = 12.6/(600x15) = 0.0014 fpy = 17100 ksc fps = fse + 700 + fc'/(300 p) = 13020+700+350/(300x0.0014) = 14553 ksc < O.K. NOMINAL NGATIV FLXURAL CAPACITY fse + 2000 = 16880 ksc a = (Aps fps+as fy)/(0.85fc' b) =(12.6x14553+11.3x4000)/(0.85x350x600)= 1.28 cm ØMn=Ø[Aps fps(dp-a/2)+as fy(ds-a/2)]= 0.9x[12.6x145.53x(15-1.28/2)+10x1.13x40(16.5-1.28/2)]= 30150 kg-m > 22939 kg-m O.K. NOMINAL POSITIV FLXURAL CAPACITY a = Aps fps/(0.85fc' b) =(12.6x14553/(0.85x350x600)= 1.027 cm ØMn=ØAps fps(dp-a/2)= 0.9x12.6x145.53x(15-1.027/2)= 23907 kg-m > 21701 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 1354.8 = 24386.4 kg vup = 24386.4/(210x15)= 7.742 ksc Øvc = 1.06x0.85 (350) = 16.856 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 1354.8 = 48772.8 kg-m vup = 48772.8/(300x15) = 10.838 ksc Øvc = 1.06x0.85 (350) = 16.856 ksc O.K. FLAT-PLAT DSIGN SPRADSHTS / ACI 318-89

FP6000X6000 : 4 FROM 4 DFLCTION CHCK I panel = 600x18^3/12 = 291600 cm 4 I col strip = I mid strip = 291600/2 = 145800 cm 4 w net = = 917 kg-m Δf = 5x9.17x600^4/(384x282495x291600)= 0.188 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 ] = 0.282 cm Δf mid strip = Df [M mid / M panel x I panel / I mid ] = 0.094 cm Δ = M net L /(8Kec) = Δ B = (3199-2877)x100x600/(8x14885465) = 0.162 cm Δ c = (2688-2688)x100x600/(8x14885465) = 0 cm SUMMARY ΣΔ col strip = 0.282+0.162+0 = 0.444 cm ΣΔ mid strip = 0.094+0.162+0 = 0.256 cm Δ allow = 600/240 = 1.667 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@125 10-D12@125 RBARS RINFORCMNT DTAIL N O T T O S C A L FLAT-PLAT DSIGN SPRADSHTS / ACI 318-89