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Design of Square Footing

This document summarizes the design of a square footing with the following key details: 1. The footing is 2m x 2m in plan with a depth of 475mm to support a 1106.9kN load. 2. Reinforcement consists of 12 bars of 20mm diameter steel placed at 200mm centers both ways. 3. Additional details include extending at least two column bars into the footing to develop the full bearing capacity.

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raymark mendoza
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172 views3 pages

Design of Square Footing

This document summarizes the design of a square footing with the following key details: 1. The footing is 2m x 2m in plan with a depth of 475mm to support a 1106.9kN load. 2. Reinforcement consists of 12 bars of 20mm diameter steel placed at 200mm centers both ways. 3. Additional details include extending at least two column bars into the footing to develop the full bearing capacity.

Uploaded by

raymark mendoza
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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DESIGN OF SQUARE FOOTING

f’c = 20.7 Mpa, fy = 276 Mpa.


Use qall = 350 KPa
Unit weight of 10.05 kN / m3
C = 300 mm x 300 mm
Pu = 1106.9kN
1. Assume depth of footing (1 t0 2) time 5. Base on wide beam shear:
the width of column 0.30m

Tf = 2Tcol = 2(0.3) = 0.6m

2. Effective soil bearing capacity, Qe 200mm 0.30m

Qe = Qall –Ʃδh
Qe = 350 – 10.05 (1.5 – 0.6)-23.544(0.6)
d
Qe = 326.8286KPa 200mm-d

3. Dimension of footing Vu = QuAshaded


Af = Pu/Qe Vu = 326.52(2)(2-d)
Af = 1106.9kN /326.8286KPa Vu = 0.327(2000)(2000-d)
Af = 3.39m2 Vu = 654(2000-d)
A=L2
L=√3.39 = 1.84m≈ 2m Vc = 1/6√fc’bwd
Vc = 1/6√21(2000)(d)
4. DEPTH OF FOOTING Vc = 1527.53d
Qu = (factored load)/Ar
Qu =1106.9/3.39 = 326.52KPa Vu = øVc
654(2000-d) = 0.85(1527.53d)
d = 669.95 ≈ 675mm
6. Base on two way or punching shear Mu = øRubd2
Vu = QuAshaded Ru = 117.55x106/0.9(2000)(350)2
Ru = 0.54MPa

300+d
9.
300 + d 300mm

Ρ=
0.85𝑓𝑐’
𝑓𝑦
(1-√1 − 2Ru
0.85fc’
)
300mm

P=
0.85(20.7)
276
(1-√1 − 2(0.54)
0.85(20.7)
)
Vu = 0.327(20002-(300+d)2]
P = 0.002
Vu = 0.327(4,000,000-90000-600d-d2)

Pmin = 1.4/276 = 0.0051


Vc = 1/3√fc’bod
Vc = 1/3√20.7(300+d)(4)d Pmax = 0.75 [0.85(0.85)(20.7)(600)
276 (600+276)
]
Vc = 6.07(300d+d2)
Pmax = 0.028
0.0051>0.002<0.028
Vu = øVc
CONDITION:
0.327(4,000,000-90000-600d-d2)
Ƿmin< Ƿact < Ƿmax − − − −Use Ƿact
= 0.85[6.07(300d+d2)]
Ƿmin> Ƿact < Ƿmax − − − −Use Ƿmin
d = 349.29 mm ≈ 350mm
Ƿmin< Ƿact < Ƿmax − − − −Use Ƿmax
Use 0.0051
7. Total depth of footing, df
Df = 350+1.5(25)+75
10. As = Pbd
Df = 462.5 say 475mm
As = 0.0051 (2000)(350)
475 < 600 ok
As = 3570 mm2
Use 20mm ɸbar RSB
8. Required steel area: 3570
2 N= Π(202 )
= 11.36 say 12 - 20mm dia. Bars
Mu = QuB(x /2)
4
Mu = 326.52(2)(0.62/2)
Mu = 117.55 kN.m
11. Check for spacing: 13. DETAILINGS
S= [2000 – 75(2)-12(20)]/11
300mm
S = 146.36 mm>200mm
Use S=200mm

475mm
12. Check for development length, Ld 350mm
Ld = (2000/2)-300 -75 2m
Ld = 625mm ok!! 12-20mm dia. RSB@200mmO.C. Bothways

Verifying if dowels or column bars


extension is necessary:
Actual bearing strength = Pu = 1106.9kN
Permissible bearing stress:
Ø0.85fc’A1 = 0.7(0.85)(20.7)(300)2
A1 = Acolumn = 1108.485 𝑚𝑚2

But this may be multiplied by √A2/A1 <2


A1 = 0.3x0.3= 0.9 m2
A2 = 2 x 2= 4 m2
√A2/A1 = 4 m2/0.9 m2
√A2/A1 = 4.44 > 2 use 2

Permissible bearing stress = 1108.485 (2) =


2216.97 >1106.9
Min. area of dowel required by the code:
Area = 0.005(300x300) = 450mm2

At least 2 column bars(20mm) must be


extended into the footing.

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