lecture

1. Reinforced Concrete Design-II
Lecture 12: Design Column Footing-
rectangular footing
Semester – Fall 2020
Dr.Tahir Mehmood
Reference
Design of Concrete Structures 14th Ed. by Nilson, Darwin and Dolan.

2. Design of Rectangular Footing
 Use in restricted locations
 For reinforcement in the shorter direction, a portion of total
reinforcement γsAs is provided in the band width.
 In short direction, larger % age of reinforcement in band
width
 Rest of reinforcement should meet at least temperature
reinforcement requirement.
side
short
side
Long
1
2
direction
short
in
orcement
inf
re
Total
width
band
in
orcement
inf
Re





γs =

3. Problem
 Design a rectangular footing for an 18” square column with
a dead load of 185 k and a live load of 150 k. Make the
length of the long side equal to twice the width of the short
side. fc’=4000 psi , fy=40,000 psi, qa = 4000 lb/ft2. Assume
the base of the footing is 5ft below grade.
Solution
 Assume 24” thick footing d = 19.5”
2
e
ft
/
lb
3400
100
12
36
150
2
24
4000
q






 D + L = 185 + 150 = 335 k
Area Required 2
ft
53
.
98
4
.
3
335



4. b=4 L=24.6
b=5 L=19.7
b=6 L=16.4
b=7 L=14.08
 So Use 7’ x 14’ = 98 ft2
Solution
2
u ft
/
k
25
.
5
98
)
150
)(
7
.
1
(
)
185
)(
4
.
1
(
q 


One-Way Shear
b = 7’ = 84”
5
62
.
4
12
5
.
19
5
2
.
6
5
2
.
6
2
5
.
12
5
.
1
4
1











 Vu = (7) (4.625) (5.25) = 169.97 k
1.2 1.6 4.71
4.71 152.5

5. .
K
.
O
so
2
8
.
18
84
4000
2
85
.
0
1000
x
97
.
169
b
f
2
V
d
c
u










Vu = (1.125)(14)(5.25)
= 82.69 k
Solution
One-Way Shear
b = 14’ = 168”
5
12
.
1
12
5
.
19
5
7
.
2
2
5
.
5
5
.
1
7










.
K
.
O
so
8
5
.
4
168
4000
2
85
.
0
1000
69
.
82
b
f
2
V
d
c
u











152.5
0.75
19.13
4.71
74.2
74.2
4.65
0.75

6. Two-Way Shear
bo = (4)(18 + 19.5) = 150”
k
23
.
463
5
.
37
5
.
19
18
25
.
5
12
5
.
37
7
14
V
2
u




















Steel in Long-Direction
Solution
=37.5/12
=3.125
-3.125 2
( ) 4.71
415.6
.
K
.
O
so
6
3
.
14
150
4000
4
85
.
0
1000
23
.
463
b
f
2
V
d
c
u











415.6
0.75
14.6
ft
.
k
77
.
717
2
25
.
6
)
25
.
5
)(
7
)(
25
.
6
(
Mu 







psi
63
.
299
(19.5)
84
0.9
12000
77
.
717
bd
M
R
quired
Re 2
2
u
n







4.71 643.95
643.95
268.80

7.  
)
in
13
A
(
9
#
13
Use
in
696
.
3
22
84
002
.
0
.
min
A
in
87
.
12
5
.
19
84
007854
.
0
bd
A
7854
.
00
000
,
40
63
.
299
765
.
11
2
1
1
765
.
11
1
f
mR
2
1
1
m
1
765
.
11
4000
85
.
0
000
,
40
f
85
.
0
f
m
2
s
2
s
2
s
y
n
c
y

















 



















Solution
Dev. Length
Required for # 9 bars ld = 25”
Available = 6.25 x 12 = 75” – 3” = 72” O.K
268.80
0.00709
0.00709
11.62

8. Longitudinal Steel in Short Direction
 
765
.
11
m
Now
psi
58
2
.
19
12
14
9
.
0
12000
922
.
277
bd
M
R
quired
Re
ft
.
k
922
.
277
2
75
.
2
)
25
.
5
)(
14
)(
75
.
2
(
M
2
2
u
n
u

















Solution
005
.
0
000
,
40
200
f
200
in
79
.
4
5
.
19
12
14
00146
.
0
bd
00146
.
0
A
00146
.
0
000
,
40
58
765
.
11
2
1
1
765
.
11
1
f
mR
2
1
1
m
1
y
min
2
s
y
n
















 














4.71
249.33
249.33
52
52
0.00131
0.00131 0.00131
4.30

9. Longitudinal Steel in Short Direction
Solution
 
)
in
96
.
18
A
(
bars
8
#
24
Use
in
48
.
18
12
12
14
005
.
0
.
min
A
2
s
2
s






3
2
1
2
2
1
2
orcement
inf
re
Total
width
band
in
orcement
inf
Re






bars
8
16
24
st
Re
width
band
in
bars
16
3
2
24
Use





10. Dev. Length
 For # 8 bar ld = 20”
 Available (2.75)(12)-3=30” O.K
   
k
12
.
771
1000
1
18
4000
85
.
0
7
.
0
A
)
f
85
.
0
(
P
2
g
c
n













 Pu = (1.4)(185) + (1.7)(150) = 514 k
 Since øPn > Pu therefore only four bars will be used as
dowels
 As(min) = (0.005)(18)2 = 1.62 in2
Solution
Bearing Strength
0.65
716.04
1.2 1.6 462

11.  Use 4 # 6 (As = 1.77 in2)
 Required Dev. Length = 9”
 Available is more than 9” O.K
Solution