Lec 12 - Design of Rectangular footing.PPT

lecture

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

2. DesignofRectangularFooting
Useinrestrictedlocations
Forreinforcementintheshorterdirection,aportionoftotal
reinforcementγ
sA
sisprovidedinthebandwidth.
Inshortdirection,larger%ageofreinforcementinband
width
Rest of reinforcement should meet at least temperature
reinforcement requirement.side short
side Long

1
2
direction short in orcementinfre Total
width band in orcementinfRe



γ
s =

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. f
c’=4000 psi , f
y=40,000 psi, q
a= 4000 lb/ft
2
. Assume
the base of the footing is 5ft below grade.
Solution
Assume 24” thick footing d = 19.5” 2
e
ft/lb3400
100
12
36
150
2
24
4000q


D + L = 185 + 150 = 335 k
Area Required2
ft53.98
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 ft
2
Solution2
u ft/k 25.5
98
)150)(7.1()185)(4.1(
q 


One-WayShear
b = 7’ = 84”562.4
12
5.19
52.6
52.6
2
5.12
5.141




V
u= (7) (4.625) (5.25) = 169.97 k
1.2 1.6
4.71
4.71152.5

5. .K.Oso28.18
844000285.0
1000x97.169
bf2
V
d
c
u




 V
u= (1.125)(14)(5.25)
= 82.69 k
Solution
One-WayShear
b = 14’ = 168”512.1
12
5.19
57.2
2
5.5
5.17



 .K.Oso85.4
1684000285.0
100069.82
bf2
V
d
c
u






152.5
0.75
19.13
4.71
74.2
74.2
4.65
0.75

6. Two-WayShear
b
o= (4)(18 + 19.5) = 150”k23.463
5.375.191825.5
12
5.37
714V
2
u














SteelinLong-Direction
Solution
=37.5/12
=3.125
-3.125
2( ) 4.71
415.6.K.Oso63.14
1504000485.0
100023.463
bf2
V
d
c
u






415.6
0.75
14.6ft.k 77.717
2
25.6
)25.5)(7)(25.6(M
u 





 psi 63.299
(19.5)840.9
1200077.717
bd
M
RquiredRe
22
u
n






4.71 643.95
643.95
268.80

7. 
)in13A(9#13Use
in696.32284002.0.minA
in87.12
5.1984007854.0bdA
7854.00
000,40
63.299765.112
11
765.11
1
f
mR2
11
m
1
765.11
400085.0
000,40
f85.0
f
m
2
s
2
s
2
s
y
n
c
y










 












 Solution
Dev.Length
Required for # 9 bars l
d= 25”
Available = 6.25 x 12 = 75” –3” = 72” O.K
268.80
0.00709
0.00709
11.62

8. LongitudinalSteelinShortDirection
765.11mNowpsi58
2.1912149.0
12000922.277
bd
M
RquiredRe
ft.k 922.277
2
75.2
)25.5)(14)(75.2(M
22
u
n
u














Solution005.0
000,40
200
f
200
in79.4
5.19121400146.0bd00146.0A
00146.0
000,40
58765.112
11
765.11
1
f
mR2
11
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. LongitudinalSteelinShortDirection
Solution
)in96.18A(bars8#24Use
in48.18121214005.0.minA
2
s
2
s

 3
2
12
2

1
2
orcementinfre Total
width band in orcementinfRe




 bars81624stRe
widthbandinbars16
3
2
24Use



10. Dev.Length
For # 8 bar l
d= 20”
Available (2.75)(12)-3=30” O.K 
k12.771
1000
1
18400085.07.0
A)f85.0(P
2
gcn









P
u= (1.4)(185) + (1.7)(150) = 514 k
Since øP
n> P
utherefore only four bars will be used as
dowels
A
s(min)= (0.005)(18)
2
= 1.62 in
2
Solution
Bearing Strength
0.65
716.04
1.2 1.6 462

11. Use 4 # 6 (A
s= 1.77 in
2
)
Required Dev. Length = 9”
Available is more than 9” O.K
Solution