POST-TENSIONED CONCRETE COLUMN SUPPORTED SLAB · PDF filepost-tensioned concrete column...

POST-TENSIONED CONCRETE COLUMN SUPPORTED SLAB DESIGN

DESIGNED BY

Mr. JAMALUDDIN CHALERMTHAI

STRUCTURAL ENGINEER

(FLAT PLATE SYSTEM)

CRITERIA : 1 FROM 1 TWO-WAY COLUMN-SUPPORTED POST-TENSIONED SLAB DESIGN CRITERIA

MATERIAL CONDITIONS

Concrete:

fc' = 350 ksc

β1 = 0.80

Ec = 282495 ksc

Mild Steel:

fy = 4000 ksc ( SD40 GRADE )

fu = 5600 ksc

Es = 2.04E+06 ksc

Prestressing Steel:

fpy = 17100 ksc (1860 GRADE)

fpu = 18600 ksc

0.94fpy = 16074 ksc

0.80fpu = 14880 ksc

Total Approximate Losses = 25 %

Effective Losses = 12.5 %

R = 0.88

fe = 14880 ksc

fj = 13020 ksc

Ep = 1.97E+06 ksc

LOAD CONDITIONS

DL : LOAD FACTOR 1.4

Concrete = 2400 kg/m3

Other Super Imposed DL = 50 kg/m2

LL : LOAD FACTOR 1.7

Super Imposed LL = 400 kg/m2

Due to the condition of symmetrically prestressed, the effective loss is equal to a half of total approximate loss.

Ln LnLn Ln

L2 L2L2

L1

L2 L2 L2L2

FLOOR TO FLOOR HEIGHT


L2

Ln LnLn Ln

Ln

Ln

Ln

Ln

P L A N

EQUIVALENT FRAME SECTION

L2 L2L2

L1

L2 L2 L2L2



L2

Ln LnLn Ln

Ln

Ln

Ln

Ln

P L A N

EQUIVALENT FRAME SECTION

FP6000X6000 : 1 FROM 4POST-TENSIONED FLAT PLATE DESIGN SPREADSHEETPROJECT SLAB DESIGN FOR ESTIMATE

SLAB CODE FP6000

DIMENSION ANALYSIS

Column Size (c1xc2) = 0.60 x 0.60 m2

Floor to Floor Height = 4.00 m

L1 = 6.00 m

L2 = 6.00 m

ts min = L2/45 = 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

0.18x2400+50 = 482 kg/m2

= 400 kg/m2

482+400 = 882 kg/m2

1.4x482+1.7x400 = 1354.8 kg/m2

PC-STRANDS ANALYSIS

Equivalent Balancing Loads = Total SDL = 482 kg/m2

Balanced Distributed Loads = 482 x 6 = 2892 kg/m

Pe = Wb·L2 / 8y = 2892x6^2 /(8x0.12)= 108450 kg

Pj = Pe/R = 108450/0.875 = 123943 kg

fpe = 14880 = 14880 ksc

FROM SEVEN-WIRES STRAND Ø 1.524 cm.

A = 1 40 2/ t d

Total Super Imposed DL =

Total Super Imposed LL =

Total Super Imposed DL+LL =

Total Super Imposed Factored Load =

Abp = 1.40 cm /strands

Approximated nos. of Required Strands = 123943/(14880x1.4)= 6 nos.

APPLYING : SEVEN-WIRES STRAND Ø 1.524 cm. x 9 nos.

Pe = 9x1.4x14880x0.875= 164052 kg

COLUMN'S STIFFNESS ANALYSIS

Kc = 4·Ec·Ig /(Lc - 2·ts)= 33526879 kg-m

ΣKc = 2xKc = 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)x3·y/3 = 164579 cm4

ΣKt = Σ9Ec·C/[L2(1-c2/L2)3] = 9566408 kg-m

From: Kec = (1/ΣKc+1/ΣKt)-1

FOR EXTERIOR SPANS :

Kec = 8371994 kg-m

FOR INTERIOR SPANS :

Kec = 14885465 kg-m

SLAB'S STIFFNESS ANALYSIS

Kes = Ks = 4·Ec·Ig /(L1 - c1/2) = 5780740 kg-m

MOMENT DISTRIBUTION FACTOR ANALYSIS

From: DFs = Kes /(Kec+Kes)

DF exterior = 0.408

DF interior = 0.219

PATTERN LOAD ANALYSIS

βa = WSLL/WSDL = 400/482 = 0.83 > 0.75

αc = ΣKc/ΣKs = 67053758/(2x5780740) = 5.8

α1 = Ecb Ib / Ecs Is = 0/[Ecx(6x0.18^3/12)] = 0

L2/L1 = 6/6 = 1.00

αmin = Minimum value of αc = 0.70 < 5.8

NOT NECESSARY TO DETERMINE EFFECTS FROM PATTERN OF LOADS

9x(282495x100^2)x(164579/100^4)/[6x(1-0.6/6)^3] =

(1-0.63x18/96)x18^3x96/3 =

[4x(282495x100^2)x(6x(18/100)^3/12)]/(6-0.6/2) =

(4x(282495x100^2)x(0.6x0.6^3/12))/(4-2x0.18) =

2x33526879 =

5780740/(8371994+5780740) =

5780740/(14885465+2x5780740) =

(1/67053758+1/9566408)^(-1) =

[1/67053758+1/(2x9566408)]^(-1) =

FLAT-PLATE DESIGN SPREADSHEETS / ACI 318-89

FP6000X6000 : 2 FROM 4FLEXURAL ANALYSIS

NET LOADED MOMENTS :

Wb = 8Pe·y/L2 = 4375 kg/m

WNET = WTSL - Wb = 917 kg/m

FIXED END MOMENT = 2751 kg-m

Moments Distribution

JOINT A E

MEMBER AB BA BC CB CD DC DE ED

DF 0.408 0.219 0.219 0.219 0.219 0.219 0.219 0.408

FEM -2751.00 2751.00 -2751.00 2751.00 -2751.00 2751.00 -2751.00 2751.00

BALANCE 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

BALANCE 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

BALANCE 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

BALANCE 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

BALANCE 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

CHECK ALLOWABLE STRESSES IN CONCRETE

Maximum Design Moments and Stresses :

M NEG = 3199 kg-m

M POS = 1694 kg-m

FROM : fc stresses = Pe/A ±My / Igs = 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.

NET BALANCED MOMENTS :

Wb = 4375 kg/m

FIXED END MOMENT = 13125 kg-m


JOINT A E

=

917x6^2/8-(1666+3199)/2 =

D

164052/(600x18)±319900x9/(600x18^3/12) =

4375x6^2/12 =

=

B C

D

8x164052x0.12/6^2 =

882x6-4375 =

917x6^2/12 =

B C

JOINT A E


DF 0.408 0.219 0.219 0.219 0.219 0.219 0.219 0.408

FEM 13125.00 -13125.00 13125.00 -13125.00 13125.00 -13125.00 13125.00 -13125.00

BALANCE -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

BALANCE 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

BALANCE -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

BALANCE 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

BALANCE -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 MOMENTS :

M PRIMARY = 9843 kg-m

JOINT A E


ΣM 0 9843 9843 9843 9843 9843 9843 0

SUMMATION MSECONDARY = M NET - M PRIMARY

NET MOMENTS 7947 15263 13724 12825 12825 13724 15263 7947

PRIMARY MOMENTS 0 9843 9843 9843 9843 9843 9843 0

SECONDARY MOMENTS 7947 5420 3881 2982 2982 3881 5420 7947

D

164052 x 0.06 =

B C

B C D


FP6000X6000 : 3 FROM 4

FACTORED MOMENTS :

WU = 8128.8 kg/m

FIXED END MOMENT = 24386.4 kg-m


JOINT A E


DF 0.408 0.219 0.219 0.219 0.219 0.219 0.219 0.408

FEM -24386.40 24386.40 -24386.40 24386.40 -24386.40 24386.40 -24386.40 24386.40

BALANCE 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

BALANCE 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

BALANCE 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

BALANCE 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

BALANCE 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

ULTIMATE MOMENTS :

JOINT A E


FACTORED MOMENT -14766 -28359 -25500 -23829 -23829 -25500 -28359 -14766

SECONDARY MOMENT 7947 5420 3881 2982 2982 3881 5420 7947

ULTIMATE MOMENTS : -6819 -22939 -21619 -20847 -20847 -21619 -22939 -6819

SUMMARY OF DESIGN MOMENTS

MFACTORED POS = 15017 kg-m

MU POS = 21701 kg-m

MU NEG = 22939 kg-m

REINFORCEMENT ANALYSIS

Mild Steel Reinforcements :

use D 12 mm

A = 1 13 2

8128.8x6^2/8-(14766+28359)/2 =

D

8128.8x6^2/12 =

B C D

B C

15017+(7947+5420)/2 =

=

1354.8 x 6 =

Ab = 1.13 cm

EFFECTIVE WIDTH OF REINFORCEMENTS = 3x18+60 = 114 cm

EFFECTIVE LENGTH OF REINFORCEMENTS = 600/3 = 200 cm

As min = 0.00075·ts·L2 = 8.1 cm2

16 cm

APPLY 10-D12@ 12.5 cm ( L= 200 cm )

SPAN-DEPTH RATIO = L2 / ts = 600/18 = 33 < 35

fse = Pe / Aps = 13020 ksc

p = Aps/bd = 0.0014

fps = fse + 700 + fc'/(300p) = 14553 ksc < O.K.

NOMINAL NEGATIVE FLEXURAL CAPACITY

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 POSITIVE FLEXURAL CAPACITY

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 SHEAR ANALYSIS

Outer Column

PROJECTED AREA = 6 x 3 = 18 m2

bo = c1+2c2+2dP = 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

PROJECTED AREA = 6 x 6 = 36 m2

bo = 2c1+2c2+4dP = 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.

a = (Aps·fps+As·fy)/(0.85fc'·b) =(12.6x14553+11.3x4000)/(0.85x350x600)=

a = Aps·fps/(0.85fc'·b) =(12.6x14553/(0.85x350x600)=

fse + 2000 = 16880 ksc

164052/(9x1.4) =

12.6/(600x15) =

13020+700+350/(300x0.0014) =

fpy = 17100 ksc

Maximum Bar Spacing = 1.13/8.1x114 =

0.00075x18x600 =


FP6000X6000 : 4 FROM 4

DEFLECTION CHECK

I panel = 600x18^3/12 = 291600 cm4

I col strip = I mid strip = 291600/2 = 145800 cm4

w net = = 917 kg-m

Δf = 5x9.17x600^4/(384x282495x291600)= 0.188 cm

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) =

ΔB = (3199-2877)x100x600/(8x14885465) = 0.162 cm

Δc = (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

Δallow = 600/240 = 1.667 cm O.K.

From Mpanel ; Mcol strip = 75%

Mmid strip = 25%

L2 = 6000 mm.

CO

LU

MN

ST

RIP

COLUMN STRIPMIDDLE STRIPCOLUMN STRIP

10-D12@125

7-SEVEN WIRE STRANDS

FP6000 ( T = 180 mm )2000 mm

N O T T O S C A L E

L1

= 6

000

mm

.




2000

mm

MID

DL

E S

TR

IPC

OL

UM

N S

TR

IP

10-D12@125

REBARS REINFORCEMENT DETAILN O T T O S C A L E


POST-TENSIONED CONCRETE COLUMN SUPPORTED SLAB · PDF filepost-tensioned concrete column...

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