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Illustrative Problems Design of Square Footing Problem 9.1

This document provides an example problem for designing a square column footing to support a column with dead and live loads. It outlines the step-by-step process for determining the footing dimensions, reinforcement, and other design details. The solution involves iteratively selecting a trial footing depth, calculating the effective soil bearing capacity, sizing the footing, checking shear capacity, determining steel reinforcement area, and verifying column dowel requirements. The final designed footing has dimensions of 2.8m x 2.8m, a depth of 445mm to the top bars, and includes 13-25mm bars on each side with at least two column bars extended into the footing.

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Joshua De Leon
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2K views5 pages

Illustrative Problems Design of Square Footing Problem 9.1

This document provides an example problem for designing a square column footing to support a column with dead and live loads. It outlines the step-by-step process for determining the footing dimensions, reinforcement, and other design details. The solution involves iteratively selecting a trial footing depth, calculating the effective soil bearing capacity, sizing the footing, checking shear capacity, determining steel reinforcement area, and verifying column dowel requirements. The final designed footing has dimensions of 2.8m x 2.8m, a depth of 445mm to the top bars, and includes 13-25mm bars on each side with at least two column bars extended into the footing.

Uploaded by

Joshua De Leon
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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ILLUSTRATIVE PROBLEMS

DESIGN OF SQUARE FOOTING

Problem 9.1

A square column footing is to support a 400-mm square tied column that carries a dead load of 880 kN
and a live load of 710 kN, the column is reinforced with 8-25mm bars. The base of the footing is 1.50 m
below the natural grade where the allowable soil pressure is 235 kPa. The soil above the footing has a
weight of 15.6 kN/m3. Assuming fy = 27.5 MPa, fc = 27.5 MPa, and unit weight of concrete as 23.50
kN/m3, design the footing. Use 25 mm main bars.

Solution

Our first task in the design of footing is the determination of its depth. This requires several cycles of
trial and error procedure because its value affects the effective soil bearing capacity. There are several
rules of thumb used by designers for making initial thickness estimates, such as 20% of the footing width
plus 75mm. However, with the aid of computer (available at GERTC), this will become easier.

Initial estimate of footing depth:

A ftg= L2= (880+710)/235


L = 2.6 m
L = 2600 mm
Depth = 20% (2600) + 75 = 595 mm say 600 mm

Effective soil bearing capacity:


q c = q a−∑ γh
= 235-23.5 (0.6) – 15.6 (1.5 – 0.6)
q c = 206.86 kPa

Dimension of the footing:


Unfactored load
A ftg=
qc
880+710
=
206.86
A ftg = 7.69 m2 = L x L
L= 2.77 m say 2.8 m
Footing dimension = 2.8 m x 2.8 m
Factored load
q u=
Area of Footing
1.4 ( 880 ) +1.7(710)
¿ =311.1 kPa
2.8( 2.8)
q u=0.3111 MPa

Based on wide-beam shear:


V u=qu A shaded
= (0.3111)[2800(1200-d)]
V u=871.08 ( 1200−d ) N
1
V c = √ f ' c bw d
6
1
¿ √ 27.5(2800) d
6
V c =2447.2 d N

V u=ϕ v c
871.08(1200- d) = 0.85(2447.2d)
1200 – d = 2.388d
d= 354.2 mm

Based on the two-way or punching shear:


V u=qu A shaded
= (0.3111)[(2800)² - (400 + d)²]
= 0.3111 (2800² - 160000 – 800d - d²)
V u=0.3111(7,680,000−800 d−d2 )
1
V c= √ f ' c b o d
3
b o=4 ( 400+ d )
1
V c = √ 27.5 [ 4 ( 400+d ) ] d
3
V c =6.99( 400 d +d 2 )

[V ¿ ¿u=ϕ v c ]¿
0.3111(7,680,000 – 800d - d²) – 0.85[6.99(400d + d²)]
7,680,000 – 800d - d² - 7639d + 19d²
2
−8439 ± √ ( 8439 ) −4 ( 20 ) (−7,680,000)
d=
2(20)
d= 443.6 mm say 445 mm

Total depth of footing = 445 + 1.5(25) + 75


Total depth = 557.5 mm < 600 mm (OK)

Required Steel Area:


d= 445 mm
M u=( 311.1 )( 1.2 ) ( 2.8 )( 1.2/2 )
= 627.18 kN-m
M u=ϕ Ru b d 2

6 2
627.18 x 10 −0.9 R u ( 2800 ) ( 445 )
Ru=1.26 MPa

0.85 f ' c
ρ=
fy
1− 1−
[√2 Ru
0.85 f ' c ]
=
0.85(27.5)
275 [√ 1−
2(1.26)
0.85(27.5) ]
ρ = 0.00471

1.4 1.4
ρmin = =
f y 275
ρmin =0.005091

Use ρ = 0.005091
A s=ρbd
= 0.005091(2800)(445)
A s=6,343 mm 2

Number of 25-mm bars:


π
( 25 )2 N =6,343
4
N = 12.9 say 13
Development Length:
l db=0.02 Ab f y / √ f ' c
π
l db= 0.02 x ( 25 )2 (275)/ √ 27.5 = 515 mm
4
or l db=0.06 d b f y =0.06 ( 25 )( 275 ) = 412.5 mm
Furnished Ld - 1200 – 75 – 1125 mm > 515 mm (OK)

Verify if dowels or column bars extension are necessary:


Actual bearing strength = Pu=1.4 ( 880 ) +1.7 ( 710 )
Pu=2439 kN

Permissible bearing stress:


Φ 0.85 f c . A 1=0.7 ( 0.85 )( 27.5 ) ( 4002 )
= 2,618,000 N
Φ 0.85 f c . A 1=2,618 kN

But this may be multiplied by √ A 2 / A 1 ≤ 2


A1=0.4 x 0.4=0.16 m ²
A2=2.8 x 2.8=7.84 m²
√ A 2 / A 1=7 use 2
Permissible bearing stress = 2,618(2) = 5,236 kN > 2,439 kN (no need)

Minimum area of dowel or extension bar required by the Code:


Area = 0.005(400 x 400) = 800 mm²
At least two column bars (25-mm) must be extended into the footing.

Use 2.8 m x 2.8 m footing with an effective depth to top bars of 445 mm (total depth = 560 m), with 13-
25 mm bars on each side of the footing, and at least two column bars (25-mm as given) must be
extended into the footing.

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