C. FOOTINGFLEXURALREINFORCEMENT(Top&BottomReinforcement)Uniform LoadUniform Load

EquipmentTesting Weight E(T) : 32,678 kNq E(T) = (E(T)/A)*b

= (32678/(0.25*3.14*34.93^2))*1= 34.101 kN/m

FoundationFooting = Hfooting x b x Unit weight of reinforced concrete

* *= 0.5*1*23.536= 11.768 kN/m

q total = 34.101+11.76845.87 kN/m

Span considered as pinned at both endFactored Moment 1 = 1 4 x 1/8 x q total x L2 where assumed L= 4 30 mFactored Moment 1 = 1.4 x 1/8 x q total x L ,where assumed L= 4.30 m

= 1.4*0.125*45.869*(4.3^2)= 148.42 kN.m

Max factored moment = Mu = 148.4 kN.mfooting effective thickness = d = h - d' ,assumed concrete cover = 0.10 m

= 0.5-0.1 = 0.40 m= 400 mm

Ru = Mu / (f x Width x d2)= 148.421/(0.9*1*0.4^2)/1000= MPa

m = fy / ( 0.85 * fc' ) = 395/(0.85*21)=

1.03

22 13=

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

=

balanced = i * 0.85 fc' * 600

22.13

0.269%

balanced i 0.85 fc 600 fy 600 + fy

= 0.85*(0.85*21/395)*(600/(600+395))=

max = 0.75 balanced=

2.32%

1.74%

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

used = < 1.74%...OK0.358%

0.35%

As req = * b * h= 1788.44906 mm2

Diameter of reinf. Bars 25 mmAs = mm2No.of bars = As req / As

= 4490.87

3.65

spacing = Width / no. of bars= m= mm 250 mm

Use D25 - 250 for top & bottom flexural reinforcement

0.27 273.97

Maximum Pile Reaction = 1057 kN -->

5.7.3.1 One Way Slab Action ShearAllowable bending shear for slab without reinforcement :

Vc = fc' b d / 6 (Ref 3 Sect 11 3 Eq 11-3)Vc = fc b d / 6 (Ref.3 Sect. 11.3 Eq. 11-3)taking b x d = Ac then:

vc=Vc / 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 endp pq total = 34.101+11.768

45.87 kN/mshear force at d = 1.4 x (0.5*q*L-(q*(0.5*dpile+d)) ,where assumed L= 4.30 m

= 1.4* (0.5*45.87*4.3-(45.87*((0.5*0.45)+0.5)))= 91.51 kN

shear area

A= 0.5*1= 0.5 m2Vu= 91.51/(0.5*1000)

= 0 183 MPa= 0.183 MPaMaximum ultimate shear stress = Vu1/Ac max = 0.18 Mpa < 0.57 Mpa (ok)

( No shear reinforcement needed )

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

Dp

d/2

Pseudocritical

Pile

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/c)/12 = (2+4/1)/12 = 0.5

2. ((s d/bo) + 2)/12 = ((40*0.4/2.67)+2)/12 = 0.67

3. 4/12 = 0.33

kused = 0.33

where :Vc = shear strength provide by concretebc = ratio of long side to short side of pile or load (1 for circle area) 1s = constant, (20 for corner, 30 for edge, 40 for interior pile) 40d = Effective depth of footing = 0.4 m

bo = Length of punching shear critical area = (Dp + d) * as/40 = 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

Vc = 0.75 x 1632= 1,223.7 kN

Vup max < f Vc 1057 < 1223.71..OK( No shear reinforcement needed )

5 7 4 Maximum crack width on tension area5.7.4 Maximum crack width on tension areaMaximum crack width on tension area shall be calculated based on Gergery-Lutzexpression as follow:

w=1.11*10-6 x x fs x 3( dc x A) (Ref.1 Sect. 8.8)

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

shall be 0.6fy for maximum crack = MPadc = thickness of concrete cover (mm) = m

1.35

237.0 0.10 dc thickness of concrete cover (mm) m

A = bar area (effective tension area divided by number of bar) = (2*0.1*1)/(4)= m2

so:w = 1.11x10-6 x x fs x 3( dc x A)

0.05

0.10

w = 1.11x10 x x fs x ( dc x A)= 1.11e-6 x 1.35 x 237 x (100mm x 50000mm^2)^(1/3)

0.0607 mm < 0.33 mm .. OK