Perhitungan Penulangan Pile Cap

C. FOOT FOOTIN ING G FLEXURAL REINFORCEMENT ( Top & Bottom Reinforcement ) Uniform Load ¨ Equipment Testing Weight E(T) : 32,678 kN q E(T) = (E(T)/A)*b = (32678/(0. (32678/(0.25*3. 25*3.14*34 14*34.93^2 .93^2))*1 ))*1 = 34.101 kN/m

¨ Foundation Footing

= Hfooting x b x Unit weight of reinforced concrete = 0.5*1*23.536 = 11.768 kN/m

q total = 34.101+11.768 45.87 kN/m Span considered as pinned at both end ac ore

omen

2

xq oa x = . x = 1.4*0.125* 1.4*0.125*45.86 45.869*(4. 9*(4.3^2) 3^2) = 148.42 kN.m

Max factored moment = Mu footing effective thickness = d

,w ere assume

148.4 h - d'

= = =

=

.

m

kN.m

0.5-0.1

,assumed concrete cover = 0.10 m 0.40 m = 400 mm =

2

Ru

= Mu / (f x Width x d ) = 148.421/(0 148.421/(0.9*1* .9*1*0.4^2 0.4^2)/100 )/1000 0 1.03 MPa =

m

= fy / ( 0.85 0.85 * fc' ) = 395/(0 395/(0.85 .85*21 *21)) = .

required ρrequired

= (1/m) x (1 - (1 - 2m Ru/fy) Ru/fy) ) (1/22.129)*(1-(1-2*22.129*1.031/(395))^0.5) = 0.269%

ρbalanced

=

ρmax max

= 0.75 0.75 ρb   alanced = 1.74%

ρmin

= =

ρused

=

0.5

 β i * 0.85 0.8 . 5 fc' *

600 fy 600 + fy = 0.85*(0.85 0.85*(0.85*21/3 *21/395)*( 95)*(600/( 600/(600+39 600+395)) 5)) = 2.32%

0.35% (1.4/fy if 1.33 x r required > 1.4/fy) (0.18% if 1.33 x r required < 1.4/fy)

0.358% < 1.74%……...OK

As req

=

ρ*  b * h 2

= 1788.44906

Diameter of reinf. Bars As  No.of bars

spacing

mm

= = =

25 mm 490.87 mm2 As req / As 3.65 ≈

= = =

Width / no. of bars 0.27 m ≈ 273.97 mm

4

250 mm

Use D25 - 250 for top & bottom flexural reinforcement

Maximum Pile Reaction

=

1057

kN

-->

5.7.3.1 One Way Slab Action Shear  Allowable bending shear for slab without reinforcement : = ' taking b x d = Ac then: vc=φVc / Ac

.

.

.

.

-

= 0.75 x √fc' /6 = 0.75 x 21^0.5 /6 = 0.57 MPa

Footing is designed to resist pile reaction for one way slab action, at distance d from outer skin of pile

Span considered as pinned at both end q total = 34.101+11.768 45.87 kN/m shear force at d = 1.4 x (0.5*q*L-(q*(0.5*dpile+d)) = 1.4* (0.5*45.87*4.3-(45.87*((0.5*0.45)+0.5))) =

shear area

91.51 kN

,where assumed L=

4.30 m

2

A=   0.5*1 = 0.5 m Vu= 91.51/(0.5*1000) = . a Maximum ultimate shear stress = Vu1/Ac max = < 0.57 Mpa 0.18 Mpa ( No shear reinforcement needed )

(ok)

3.7.3.2 Punching Shear  Footing is designed to resist punching shear at d/2 from outer skin of pile

Dp d/2

Pile Pseudocritical

Punching shear area illustration Pmax =

1057.00

kN

(for single pile)

Since there 3 type of pile applied, that is corner, edge and interior pile, the most critical type, corner pile, will govern

Vc =

k x fc'^0.5 x bo x d

(Ref.3 sect 11.12.2.1 Eq. 11-37)

while k shall be the smallest of: 1. (2 + 4/βc)/12

=

(2+4/1)/12 =

0.5

2. ((αs d/bo) + 2)/12

=

((40*0.4/2.67)+2)/12 =

0.67

3. 4/12

= k used

where : Vc =  bc = αs = d=  bo =

=

0.33

0.33

shear strength provide by concrete ratio of long side to short side of pile or load (1 for circle area) constant, (20 for corner, 30 for edge, 40 for interior pile) Effective depth of footing Length of punching shear critical area = π (Dp + d) * as/40

=

1 40 0.4 m 2.67 m

Vup max

=

1057.00

kN

Allowable punching shear for slab without reinforcement : Vc

= k √fc' bo d (Ref.3 sect 11.12.2.1 Eq. 11-37) = 0.333*(21^0.5)*2670.354*400/1000 = 1,632 kN

φVc

= 0.75 x 1632 = 1,223.7

Vup max < f Vc ( No shear reinforcement needed )

kN

1057 < 1223.71..OK

. .

Maximum crack width on tension area shall be calculated based on Gergery-Lutz expression as follow: w=1.11*10  x

-6

β x fs x 3√( dc x A)

where w =  b = fs =

crack width (mm) height factor (1.2 for beam, 1.35 for slab) = reinforcement calculated stress at service load (MPa),

dc A

= =

(Ref.1 Sect. 8.8)

shall be 0.6fy for maximum crack thickness of concrete cover mm

= =  bar area (effective tension area divided by number of bar) = =

so: =

-6

1.11x10 . x

x fs x

3

dcxA

= 1.11e-6 x 1.35 x 237 x (100mm x 50000mm^2)^(1/3) 0.0607 mm < 0.33 mm ……….. OK  

1.35 237.0 MPa 0.10 . m (2*0.1*1)/(4) 2

0.05 m