concrete Footing Design

**************************************DESIGN OF FOOTING************************* Carmel B. Sabado Prof. Geronides P. Ancog CE-162 10-Aug-09 I.Column details: 1.(type the letter corresponding to the shape of column) a A. Square B. Rectangular C. Circular 2.Dimensions of column: (put the values in the green colored spaces provided) a=b= 600 mm *length and width of the column are equal *not applicable *not applicable *not applicable

II. Information Needed 1.Material Properties: f'c= 27.5 Mpa fy= 414 Mpa 2. Load to be carried: 1580 KN PDL = PLL = 1200 KN 3. Allowable soil Pressure: 285 Kpa 4.Shape of Footing to be Designed: (type the letter corresponding to the shape of footing) A. Square B. Rectangle

length= width= B= 5. Diameter of Bar Reinforcement: rebars long φb= rebars short φb= 6. Assumed Footing weight:

3.3 4.4 m 2.4 m 2.4 m 25 30 mm 20 mm

b

*longer dimension of the footing *shorter dimension of the footing

*reinforcement in the longer dimension *reinforcement in the shorter dimension

7 %Qu

III. Design Computations: 1.Dependable Ultimate Soil Bearing:

1. 2 D L1.6 L L qa D LL L

=

391.21 kPa

Q u =1 . 2 DL 1 . 6 LL

=

3816 kPa

q u= 2.Required Footing Area:

10.44 m² Actual Footing Area 3. Net soil Pressure:

=

10.56 m²

*design footing area is okay

369.8 kPa

4. Critical Section of the Footing: 1. Footing depth as controlled by BEAM SHEAR:

1 B−a vc =  f c ' −d 6 2 B−a V u =q n L v =q n −d 2

Lv =



φv c =

Vu bd

; d=



=

0.87 Mpa

a= b=

600 mm 1000 mm

* consider a 1 meter strip

V u *with these working equations we can get the value of d: φv c b





B−a −d 2 d= φv c  b=1000  qn

in the short dimension trial d: computed d: 320 288.71 304.35 296.49 300.42 298.45 299.44 298.94 299.19 299.06

in the long dimensio trial d: 678 643.14 634.38 632.18

2. Footing Depth as controlled by Punching Shear

vc =

1  f c' 3

=

1.75 Mpa

Vu = qn[Af-(a+d)²] b = 4(a+d)

φv c =

Vu bd

; d=

[ 

qn A f − d=

Vu φv c b

 a2d  1000

] 2

x 1000

*with these working equations we can get the value of d:

 φv c  [ 4  ad  ] trial d:

computed d: 452.52 471.15 473.19 473.4 473.42 ***beam shear controls!

500 476.26 473.71 473.45 473.42

Reinforcements in the long Direction: Ll'= (L-a)/2= 1.9 MuL = qn(Ll')²/2= 667.48 Trial dl= 680 mm

5.Design as Singly Reinforced Rectangular Beam:

[

ρ min =

1.4  f c' , f y 4f y

]

Reinforcements in the short Direction: Ls' = (B-a)/2= 0.9 Mus = qn(Ls')²/2= 149.77 Trial ds=dl-((φbl/2)+(φbs/2)) 655

use b=1000mm 0

min

ρ max =. 75

ρ=

[ 

[

. 85 f c ' β 1 . 003 E s

]

f

y

. 003 E s  f

[ 

2R u 2ωR u . 85 f c ' 1 = 1− 1− 1− 1− fy .85 f c ' ω fy

y

]

]

=

=

Ru= ρ= As= Rebars= =

N=int([B or L]As/Ao+1)

0.02

in the long direction: 1.6 0 ok! 2731.61 mm2/mm 30 mm 10 bars

Check validity of assumed footing weight of : Total depth : h= d+100 = Wf = (1.40) B² h wc = Actual footing weight : Assumed footing weight : %Qu=

in the short direction: 0.39 0 okay! 618.84 mm2/mm 20 mm 9 bars

780 mm 272.14 Kn 267.12 Kn

Wc=

revision should be done!

23.6 Kn/m3

6. Short Rebars Distribution: Ast= As=Ast(2/(L/B)+1)= central strip= No. of bars at the outer strip=

2827.43 mm2 1995.84 mm2 7 2

-20 mm φbars -20 mm φbars

7 20mm φbars

2

780 mm

2.4 meters

0.8 meters

END OF THE PROGRAM!!!=) ******NOTHING FOLLOWS******

in the long dimension computed d: 608.27 625.63 629.99 631.08

mm

20mm φ bars

680 mm 100 mm

FOOTING REBARS:

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100 mm

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FOOTING DIMENSION:

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100

100 mm

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