DESIGN OF COLUMN FOOTING 1
DESIGN CRITERIA:
f’c = 20.7 Mpa (specified compression strength of concrete)
fy = 276 Mpa (specified yield strength of steel)
qu = 160 Kpa (Allowable Soil Bearing Pressure)
3
ϒsoil = 18 kN/m (Unit weight of soil)
3
ϒc = 23.54 kn/m (Unit weight of concrete)
hs = 2.0 m (Assumed height of soil above the footing to the NGL)
t = 450 mm (Assumed depth of footing)
CC = 75 mm (Minimum cover required by the code when exposed to
earth)
Size of column = 300 mm x 300 mm
Note: Depth of excavation of footing depends, until it reaches the stratum or hard part of
soil.
TOTAL LOADS CARRIED BY THE FOOTING:
FOR DEAD LOAD, PDL
PDL = Reaction from beam (T+L) + wt. of column + 10% of total weight of
column.
PDL = 179.78 kN+ 115.886 kN + (0.30m) (0.30m) (23.54kN/m3) (10.2) + 10 %
( 0.30m)
3
(0.30m) (23.54 kN/m ) (10.2m)
PDL = 319.437 kN
FOR LIVE LOAD, PLL
PLL = Reaction from beam (T+L)
PLL = 72.158 kN + 53.735 kN
PLL = 125.893 kN
PTOTAL = PDL + PLL
PTOTAL = 319.437 kN + 125.893 kN
PTOTAL = 445.330 kN
SOLVE FR THE EFFECTIVE SOIL BEARING CAPACITY (qe):
qe = qa - ϒsoilhs - ϒchc
qe = 160 kN/m2 – 18 kN/m3(2.0m) – 23.54 kN/m3(0.45m)
qe = 113.407 kN/m2
SOLVE FOR THE DIMENSION OF THE FOOTING:
Ptotal 445.330 kN
A= = = 3.927 mm2
qe 113.407 kN
√ 3.927 mm2 ;B = 1.982 m say 2m
Therefore, the dimension of the footing D = 2 m x 2 m (Area = 4m2 )
TOTAL FACTORED LOAD, (PU):
PU = 1.2 PDL + 1.6 PLL
PU = 1.2 (319.437 kN) + 1.6(125.893 kN)
PU = 584.753 kN
SOLVE FOR ULTIMATE SOIL BEARING CAPACITY:
P
qu ¿ u
Area
584.753 kN
qu ¿
4 m2
qu = 146.188 kN /m2
CHECKING SHEARING RESISTANCE OF THE FOOTING:
ONE WAY SHEAR:
d = hc – CC – 1/2φ bar
d = 450mm – 75 mm – ½(20mm)
d = 365 mm
CRITICAL SECTION FOR BEAM SHEAR (V U) IS AT DISTANCE d FROM THE
FACE OF THE COLUMN:
VU = qu(Areabeam shear)
VU = 146.188 kN /m2 (0.485m) (2m)
VU = 141.803 kN
ILLUSTRATIONS:
ONE WAY SHEAR AND PUNCHING SHEAR:
CRITICAL MOMENT:
SOLVING FOR NOMINAL SHEAR STRESS (Vn):
Vu
Vn = : where φ = 0.75 (NSCP 2010)
ϕbd
141.803(103 )N
Vn =
(0.75)(2000 mm)(365 mm)
Vn = 0.259 Mpa
Must be less than Vc =
√ f ' c = √ 20.7 Mpa = 0.758 Mpa allowable shear strength of
6 6
concrete for one way shear Vn < Vc , therefore SAFE.
TWO WAY OR PUNCHING SHEAR:
d = hc – CC – 1/2φ bar
d = 450mm – 75 mm – ½(20mm)
d = 365 mm
CRITICAL SECTION FOR PUNCHING SHEAR (V U) IS AT DISTANCE d/2
FROM THE FACE OF THE COLUMN:
VU = Pu - qu (Areapunching shear)
VU = 584.753 kN – 146.188 kN/m2 (0.665m)2
VU = 520.105 kN
SOLVING FOR NOMINAL SHEAR STRESS (Vn):
Vu
Vn = : where φ = 0.75 (NSCP 2010)
ϕ bo d
bo = 4(a+d) = 4(300mm+365mm) = 2660 mm
520.105(103 )N
Vn =
(0.75)(2660 mm)(365 mm)
Vn = 0.714 Mpa
Must be less than Vc =
√ f ' c = √ 20.7 Mpa = 0.758 Mpa allowable shear strength of
6 6
concrete for one way shear Vn < Vc , therefore SAFE.
SOLVING FOR THE REINFORCEMENTS:
CRITICAL FOR BENDING IS AT THE FACE OF THE COLUMN WHERE
MAXIMUM MOMENT OCCURS:
M = qu(Area)(x)
M = 146.188 kN/m2(2m)(0.85m)(0.85m/2)
M = 105.621 kN-m
SOLVING FOR THE REQUIRED PERCENTAGE OF STEEL (Rn):
Mu
Rn = ; where ϕ = 0.90 (NSCP 2010)
ø b d2
105.621 ( 106 ) N −mm
Rn = = 0.440 Mpa
(0.90)(2000 mm)¿ ¿
0.85 f ' c 2R n
ρ=
fy √
[1- 1−
0.85(20.7 Mpa)
0.85 f ' c
]
2(0.440 Mpa)
ρ=
(276 Mpa)
ρ = 0.00162
√
[1- 1−
0.85(20.7 Mpa)
¿]
1.4 1.4
ρmin = =
fy 276 Mpa
ρmin = 0.00507 but, ρmin> ρ< ρmax
therefore, use ρ = 0.00507
SOLVE FOR AREA OF STEEL:
As = ρbd
As = 0.00507(2000mm)(365mm)
As = 3702.899 mm2
SOLVE FOR NUMBER OF BARS:
Using 20 mm ϕ bar (Abar = 314.16 mm2)
As 3702.899mm 2
N= ¯ = = 11.787 pcs say 12 pcs.
A ¿ ¿ (314.16 mm2)
SOLVING FOR THE SPACING:
Abar 314.16 mm2
S= (1850 mm) = (2000mm)
As 3702.899mm 2
S = 169.683 mm say 150 mm
Therefore, use 12– 20 mm φ bar spaced @150 mm bothways.
Check for development length:
9 fy ∝ βγλ
Ld
= ' c + Ktr
d b 10 √ f ( )
db
Where: ktr = 0
C = 75mm
c+ Ktr 75 mm+0 c+ Ktr
= = 3.75 > 2.5 use = 2.5
db 20 mm db
Ld 9(276 Mpa)(1.3)(1.0)(1.0)(1.0)
= = 28.39 diameters
db 10 √ 20.7 Mpa (2.5)
3702.899mm 2
Ld = (28.19)( )(20mm) = 553.777 mm (required)
3769.92mm 2
Development length furnished:
Ld = 2000mm – 150mm – 75mm
Ld = 1775 mm> 553.777mm (OK)
