1. 19
Reinforced Concrete Design
Design of Footings 1
Types of Footings
Bearing Pressure under Footing
Eccentrically Loaded Footing
Wall Footings
Column Footings
Mongkol JIRAVACHARADET
SURANAREE
UNIVERSITY OF TECHNOLOGY
INSTITUTE OF ENGINEERING
SCHOOL OF CIVIL ENGINEERING

2. Types of Footings
Wall
Property line

3. Isolated spread footing
Wall footing

4. Combined Footings
Property line
A
B
A
Rectangular, PA = PB
B
Rectangular, PA < PB
Property line
Property line
A
B
Rectangular, PA < PB
A
B
Strap or Cantilever

5. Pile cap
Piles
Weak soil
Bearing stratum
Mat Footing

6. Bearing pressure under footings
R
Axially Loaded Footings : Assume uniform pressure
Actual pressure is not uniform due to:
1) Footing flexibility
p, bearing pressure
2) Depth of footing below ground surface
3) Type of soil, e.g., clay or sand
R
Heave
R
Heave
Cohesionless soil
Cohesive soil

7. Eccentrically Loaded Footings
e
y
P
e
x
x
b
load
pmin =
y
P Mc
−
A
I
pmax =
P Mc
+
A
I
h
Tensile stress cannot be transmitted between soil and concrete.
For full compression, setting pmin = 0,
P Mc Pec
=
=
A
I
I
P
e=
I
Ac
emax = h/6
For rectangular footing of length h and width b,
I
bh 3 / 12 h
e=
=
=
Ac bh(h / 2) 6
h/3
h/3
h/3
Middle Third

8. Large eccentricity of load e > h/6
Centroid of soil pressure concurrent with applied load
e
P
a
R=
1
(3ab )pmax = P
2
where a = h/2 - e
R
3a
pmax
pmax =
2P
3ab

9. F
12.1
0.15 .
F F
F
1.8 . x1.2 .
F
F
80
0.40 .
e = 0.15
e
F
0.60 m
pmax =
Load
0.60 m
. < [h/6 = 0.30 .]
F
80
80 × 0.15 × 0.9
+
1.8 × 1.2 1.2 × 1.83 / 12
= 37.0 + 18.5 = 55.5 t/m2
0.90 m
pmin = 37.0 − 18.5 = 18.5 t/m2
0.90 m
e = 0.40
. > [h/6 = 0.30 .] F
a = 0.90 – 0.40 = 0.50
pmax =
F
.
2 × 80
= 88.9 t/m2
3 × 0.50 × 1.20

10. 12.1
.. .
. . 2522
( / . . .)
F
F
2
F
F
5
10
20
25
30
F
* F
F
F
100

11. F
F
12.2
1.5 .
γs = 2.0
γc = 2.4
/ . .
DL = 80 ton
LL = 40 ton
Grade
1.0 m
= (1.52)(0.5)(2.4) = 2.7
= (.32)(1.0)(2.4)
= 0.22
= (1.0)(1.52-0.32)(2.0) = 4.32
= 80+40
= 120
= 127.24
30x30 cm column
0.5 m
pgr =
127.24
= 56.55 t/m2
2
1.5
F
pn = pgr –
1.5
/ . .
F
F
= 56.55 – 1.5(2.0) = 53.55
/ . .

12. Wall Footings
Uniformly loaded wall
w
w
Wall
Bending deformation
Footing
1-m slice on which
design is based

13. Critical Section for Moment in Isolated Footings
b/2 b/2
Critical section
Critical section
s
b/4
Concrete column,
pedestal or wall
Masonry wall
s/2
Column with steel
base plate

14. Moment and Shear in Wall Footings
wu = 1.4wDL+1.7wLL
Required L = (wDL+wLL)/qa
qa = Allowable soil pressure, t/m2
b
Factored wall load = wu t/m
Factored soil pressure, qu = (wu )/L
d
2
d
qu
L
1 L−b
1
Mu = qu 
= qu (L − b )2

2  2 
8
L−b

Vu = qu 
−d
 2

Min t = 15 cm for footing on soil, 30 cm for footing on piles
Min As = (14 / fy ) (100 cm) d

15. EXAMPLE 12.3: Design of a Wall Footing to carries a dead load D of 12 t/m and a live
load L of 8 t/m. The max. soil pressure is 10 t/m2. f’c = 240 kg/cm2, fy = 4000 kg/cm2, and
γs = 2.0 t/m3.
D = 12 t/m
L = 8 t/m
Consider: 1-m strip
Assume footing t = 30 cm
25 cm
Net soil pressure:
Df = 1.50 m
5 cm
typical
8 cm
clear
pn = 10 - [0.3(2.4) + 1.2(2.0)]
= 6.88 t/m2
Req’d footing length:
L = (DL + LL) / pn = (12+8)/6.88
L
= 2.91 m
Ultimate soil pressure:
pu = (1.4 x 12 + 1.7 x 8) / 3.0 = 10.13 t/m
USE 3.0 m

16. Check Shear:
d = 22 cm
25 cm
115.5 cm
Vu = 10.13(1)(1.155) = 11.70
30 cm
φVc = 0.85(0.53)(100)(22)/1000
= 15.35
> Vu
10.13 t/m2
Flexural design:
25 cm
137.5 cm
Mu = 0.5 10.13 1.3752
= 9.58
-
Mu
9.58(105 )
Rn =
=
φ bd 2 0.9 × 100 × 222
= 21.98
10.13 t/m2
./ .2
OK

17. 0.85 f c′ 
2 Rn 
1 − 1 −
 = 0.0058 > [ ρ min = 0.0035]
ρ=

fy 
0.85 f c′ 

As = 0.0058(100)(22) = 12.82
F
.2/
DB16 @ 0.15 (As = 13.40
.2/
F
F
)
F
As = 0.0018(100)(30) = 5.4
F
OK
.2/
.2/
DB12 @ 0.20 (As = 5.65
25 cm
30 cm
3.00 m
)
DB12
DB12@0.20m
DB16@0.15m

18. Weight of footing ≈ 4-8 % of column load
Column Footings
Critical section for shear
2
d/2
d
1
Punching shear
2
Beam-shear short direction
3
Beam-shear long direction
1
d
3
Critical section for moment
2
1
2
1
Moment short direction
Moment long direction

19. Two-Way Action Shear (punching-shear)
On perimeter around column at distance d/2 from face of column
c1 + d
P
d/2
c2 + d
c2
b0
c1

20. Two-Way Action: cracking occur around column with periphery b0 at
distance d / 2 outside column. Vn is the smallest of

4 
Vn = Vc = 0.27  2 +
 fc′ b0 d
βc 

ACI Formula (11-35)
 αsd

Vn = Vc = 0.27 
+ 2  fc′ b0 d
 b0

ACI Formula (11-36)
Vn = Vc = 1.06 fc′ b0 d
ACI Formula (11-37)
where
b0 = perimeter of critical section at distance d /2 outside column
βc = ratio of long side to short side of column
αs = 40 for interior columns, 30 for edge columns and 20 for
corner columns

21. Distribution of Flexural Reinforcement
Footing Type
Square Footing
Rectangular Footing
s (typ.)
One-way
B
L
s (typ.)
AsL
As2
L
s (typ.)
Two-way
As1
L
B/2 B/2
L
As2
 2 
=
s1
 AsL
 β + 1
B As 2 = AsL − As1
2
AsB
L
β=
B
A

22. Transfer of Forces at Base of Column
For a supported column, bearing capacity is
φ Pnb = φ (0.85 fc′ A1 )
A1
where
A1 = loaded area (column area)
φ = 0.70
450
For a supporting footing,
2
1
A2 measured
on this plane
φ Pnb = φ (0.85 fc′ A1 )
A2
≤ 2 φ (0.85 fc′ A1 )
A1
where A2 = area of lower base of the largest
pyramid cone contained within footing having
side slope 1 vertical to 2 horizontal

23. EXAMPLE 12.4: Design of a Square Footing to support a 40 cm square column.
The column carries a dead load D of 40 ton and a live load L of 30 ton. The
allowable soil pressure 10 t/m2. f’c = 240 kg/cm2, fy = 4000 kg/cm2. Unit weight of
the soil above footing base = 2.0 t/m3.
D = 40 t
L = 30 t
(1) Determination of base area:
Assume footing depth = 40 cm
1.50 m
40 cm
Soil net pressure:
pn = 10 – [0.4(2.4) + 1.1(2.0)]
h
= 6.84 t/m2
Required area = (40+30)/6.84 = 10.23 m2
b
USE 3.2x3.2m square footing (10.24m2)
(2) Factored loads and soil reaction:
Pu = 1.4(40) + 1.7(30) = 107 tons
Ultimate pressure pu =
107
= 10.45 t/m2
10.24

24. Assume footing depth = 40 cm and effective depth d = 32 cm
Punching shear:
Vu = 10.45(3.22 – 0.722) = 101.6
40 cm
d/2=16 cm
bo = 4(72) = 288
.
φ Vc = 0.85(1.06) 240 (288)(32)/1000
72 cm
= 128.6
> Vu
OK
Beam shear:
40 cm
d=32 cm
Vu = 10.45(1.08)(3.2) = 36.12
φ Vc = 0.85(0.53) 240 (320)(32)/1000
= 71.47
108 cm
> Vu
OK

25. Flexural Design:
-
Mu = (0.5)(10.45)(3.2)(1.4)2 = 32.77
32.77(105 )
Rn =
= 11.11
0.9 × 320 × 322
./ .2
As = 0.0029(320)(32) = 29.70
ρ = 0.0029
.2
As,min = 0.0018(320)(40) = 23.04
.2 < As
OK
USE 15DB16# (As = 30.15 cm2)
40 cm
4DB25
DB16
Critical section
for moment
0.40 m
3.20 m
15DB16 #

26. Check development of reinforcement
Critical section for development is the same as that for moment (at face of column)
f
ld
αβγλ
= 0.28 y
db
fc′  c + K tr 


db 

Edge distance (bottom and side) = 8 cm
Center-to-center bar spacing = (320 - 2(8))/14 = 21.7 cm
8 cm (control)
c = minimum of
21.7 / 2 = 10.9 cm
Ktr = 0 (no transverse reinforcement)
c + K tr 8 + 0
=
= 5.0 > 2.5
db
1.6
USE 2.5

27. α = 1.0 (bottom bars)
β = 1.0 (uncoated reinforcement)
αβ = 1.0 < 1.7
γ = 0.8 (DB20 and smaller)
λ = 1.0 (normal weight concrete)
ld
4,000 1.0 × 1.0 × 0.8 × 1.0
= 0.28
= 23.1
db
2.5
240
ld = 23.1 x 1.6 = 37.0 cm > 30 cm
OK
Since ld = 37 cm < available embedment length (320/2 - 40/2 - 8 = 132 cm),
DB16 bars can be fully developed.

28. Transfer of Force at Base of Column
(1) Bearing strength of column
column bars
footing dowels
φPnb = φ (0.85f’c A1)
= 0.70(0.85x240x40x40)/1,000
32 cm
= 228.5 tons > 107 tons
OK
8 cm cover
(2) Bearing strength of footing
40 cm
Bearing strength of footing increased by factor
A2 A1 ≤ 2
where A2 is area of pyramid cone
having side slope 1 vertical to 2 horizontal
320 cm
A2
200 cm
A1
A2
=
A1
200 × 200
= 5 > 2, use 2
40 × 40

29. φPnb = 2φ (0.85f’c A1)
= 2(0.70)(0.85x240x40x40)/1,000
= 457 tons > 107 tons
OK
(3) Required dowel bars between column and footing:
Even though column and footing have enough bearing strength to transfer load,
area of reinforcement across interface ≥ 0.005(gross area of supported member)
As (min) = 0.005(40x40) = 8.0 cm2
Provide 4DB16 bars as dowels (As = 8.04 cm2)
(4) Development of dowel reinforcement in compression:
In column & footing:
For DB16 bars:
ld =
ld =
0.075d bfy
fc′
≥ 0.0043d bfy
0.075 × 1.6 × 4,000
240
= 31.0 cm
l d (min) = 0.0043 × 1.6 × 4,000 = 27.5 cm
(control)

30. Available length for development in footing
= footing thickness - cover - 2(footing bar dia.) - dowel bar dia.
= 40 - 8 - 2(1.6) - 1.6 = 27.2 cm ≈ 27.5 cm
OK
Therefore, the dowels can be fully developed in the footing.
Home work: Design a square spread footing with the following design conditions:
Service dead load = 150 ton
Service live load = 120 ton
Unit weight of soil = 2.0 ton/m3
Allowable soil pressure = 20 ton/m2
Column dimensions = 60 x 30 cm
P
Ground elev.
1.5 m