Ch 7 design of rcc footing

Design of Shallow Foundation: Academic use for students of Bachelor of Science in Civil Engineering

1. 84

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
CHAPTER SEVEN
ANALYSIS & DESIGN OF RCC FOOTING

7.1 Introduction
Footings are structural members used to support columns/walls and to transmit and
distribute their loads to the soil in such a way that
• Load bearing capacity of the soil is not exceeded
• Excessive settlement, differential settlement or rotation are prevented and
• Adequate safety against overturning or sliding is maintained.

7.2 Types of footing

Mainly two types, as shown in Figure 7.1.
 Shallow Foundation: If base of the footing is within 3 meters or 10' below ground
level, it is shallow foundation. Examples- Isolated footing, combined footing, mat
foundation, raft foundation etc.

 Deep Foundation: If base of the footing is more than 3 meters or 10' below ground
level, it is deep foundation. Examples- pile, caisson.














Figure 7.1: Types of footing

2. 85

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
7.3 Types of Shallow foundation

1) Isolated/Single footing
They are used to support single columns, as
shown in right figure. This is one of the most
economical types of footings and is used when
columns are spaced at relatively long distances.
They are divided into two categories:
i. Square Footing
ii. Rectangular Footing



2) Combined footing
Such footings usually support two columns, or three columns not in a row, as shown in
Figure 7.2 below. Combined footings are used when two columns are so close that single
footings cannot be used or when one column is located at or near a property line.



Figure 7.2: Combined footing

3. 86

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
3) Cantilever or strap footing

It consists of two single footings
connected with a beam or a strap and
support two single columns, as shown in
right figure. This type replaces a
combined footing and is more
economical.





4) Continuous footing

They support a row of three or more
columns, as shown in right figure. They
have limited width and continue under
all columns.




5) Raft or mat foundation

It consists of one footing usually placed
under the entire building area, as shown
in right figure. They are used, when soil
bearing capacity is low, column loads
are heavy, single footings cannot be
used, piles are not used and differential
settlement must be reduced.

4. 87

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
(b)
Figure 7.3: (a) Assumed uniform soil pr. distribution (b) Cohesionless soil pr. distribution
(c) Cohesive soil pr. distribution
(c)
(a)
7.4 Soil pressure distribution under footing

When the column load is applied on the
centroid of the footing, a uniform pressure is
assumed to develop on the soil surface below
the footing area, as shown in Figure 7.3(a).
However the actual distribution of the soil is
not uniform, as shown in Figures 7.3(b) and
(c). It depends on many factors especially the
composition of the soil and degree of
flexibility of the footing.


7.5 Design Considerations of footing
 Minimum Footing thickness is 6.0".
 Concrete cover is 3.0".
 In footing, long direction bars are placed below the short direction bars.
Average effective depth in both directions, d = h – clear cover = h – 4.0".
Effective depth in short direction, ds = h - 4.5″.
Effective depth in long direction, dl = h - 3.5″.
 #5~#11 bars are usually used as main bars in footing.

5. 88

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
 Effective soil pressure (qeff) & net soil pressure (qn)
Based on Figure 7.4, qeff and qn will be determined as below.






















Weight of backfill soil, Ws = Unit wt. of soil x Depth of backfill soil =ss
hw
Weight of base concrete, Wc = Unit wt. of conc. x Depth of base conc. = cc
hw
Effective upward soil pressure, qeff = Allowable soil pressure - Wt. of backfill - Wt. of
base
= csa
WWq 
Area of footing = effeff q
LLDL
q
P 

Provided area of footing, Ag = Width of footing, B X Length of footing, L
Net soil pressure, qn gg
u
A
LLDL
A
P 162.1 

Figure 7.4: Loads acting on footing
Column width, c
Pedestal column, (c+3")
Ground level
Depth of
base, hc
Depth of
backfill, hs
Depth of
footing, Df
Wt. of
backfill
Ws
Wt. of
backfill
Ws
Wt. of base, Wc
Allowable soil pressure, qa
Regular column

6. 89

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
 Critical sections in Footing
As shown in Figure 7.5, there three critical sections for RCC footing:
i. Critical section for beam shear or one way shear which is checked
at a distance d from the faces of the pedestal column in both
directions.
ii. Critical section for punching shear or two way shear which is
checked at a distance d/2 from all faces of pedestal column.
















iii. Critical section for moment which is checked at faces of pedestal
column in both directions.

 Dimension of the Footing

For economic purposes, the footing base is tapered at
both sides as shown in right figure.
Figure 7.5: Critical sections for footing
2/3h
h h


15
ʹʹ





h
Main bars of the footing
Column bars

7. 90

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
7.6 Design Problem (Isolated Footing)
A reinforced concrete footing base of a low rise residential building is to be constructed 5′
below ground level. The allowable soil pressure obtained from soil test report is all
q = 4.5
ksf. The footing supports a 12″x14″ RCC column which is reinforced by 4#7 bars.

Including self-weight, footing has to carry the following loads:
Total un-factored dead load, DL = 307.56 k
Total un-factored live load, LL = 51.2 k

The material properties are as below:
fy = 60,000 psi
f’c= 4,000 psi
Unit weight of soil, s
w = 100 pcf
Unit weight of concrete, c
w = 150 pcf

Answer the followings:
a) Determine the size and area of the footing
b) Check effective depth, ‘d’
c) Calculate reinforcement, ‘Ast’ in both directions
d) Check for hook requirement of the main bars in both directions
e) Check for dowel bar and its hook requirement


Square footing

Given data:
Depth of base below G.L. = 5′-0″
Unit weight of soil, s
w = 100 pcf
Allowable soil pressure, all
q = 4.5 ksf
fy = 60,000 psi
f’c= 4,000 psi
Size of the column =12″x14″

Step-1: Load calculation

Total un-factored load, P = DL+LL = 307.56 + 51.2 = 358.8 k
Total factored load, u
P = 1.2DL+1.6LL = 1.2*307.11 + 1.6*51.2 = 450.5 k

8. 91

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Step-2 Determination of the footing size

Let the depth of footing base, h = 22″=1.83 ft.
Provided avg. effective depth in both directions, d = h – clear cover = 22.0 – 4.0 = 18.0″
Weight of soil, s
W = 66.316
12
22
5*100hw
s






 lb/ft
2

Weight of base concrete, c
W = 50.27483.1*150hw
c
 lb/ft
2

Effective upward soil pressure, qeff = csa
WWq  = 4500-316.66-274.50
= 3908.84 lb/ft
2
= 3.909 k/ft
2

Area of footing = 7.91
909.3
8.358


effq
LLDL ft
2

Size of footing = 10′x 10′
Provided area of footing, Ag = 100 ft
2

Net soil pressure, ksf
A
P
q
g
u
n 5.4
100
5.450



Step-3: Check for‘d’

Check by punching (two-way) shear

Punching shear is checked at a distance d/2 from all faces of column as shown in figure.
The pedestal column size = 15″x17″ "136)1817(2)1815(2)(2)(2
210
 dcdcb
"00.351817
"00.331815
2
1


dc
dc

Calculating the punching shear force, u
V
=   dcdcqP
nu

21 = 450.5-4.5*





12
35
*
12
0.33 = 414.4 k

For preventing punching failure,
Concrete shear strength = punching shear force uc Vdbfor 
0'4,

or, "1.16
136*000,4*4*75.0
1000*4.414
'4
0

bf
V
d
c
u

The required"00.81"1.16d , so ok.

9. 92

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
x
x
Check by beam (one-way) shear
Beam shear is checked at a distance d from any face of the column along the longer side of
footing as shown in figure.
The beam shear force, Vu = qn*b* x




For preventing one-way shear failure,
Concrete shear strength = beam shear force




The required, "0.81"4.11d , so ok.

Step-4: Calculation of the main steel (both directions)

Moment will be taken at the face of the column for calculating main steel as shown in
figure.

# Calculation of u
M







= 435.6 k-ft


# Calculation of
0026.0
18)1210(490.085.0
126.4352
11
60
485.0
85.0
2
11
85.0
22'
'
















xxxxx
xxx
bdf
M
f
f
c
u
y
c


k
d
cL
b qVor
nu
4.129
12
0.18
12*2
15
2
10
*10*5.4
22
1














 ''4.11
)12*10(*000,4*75.0*2
1000*4.129
,
4.129'2


dor
bd forVV
cuc
 10*
2
40.4
*40.4*5.4
*
2
**







 b
x
xqM
nu ft
cL
x 40.4
2
12
15
2
10
22
1



























10. 93

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
l
L
c
Cover
# Required reinforcement in both directions
(1) 62.50.18*)12*10(*0026.0 bdA
s
 in
2


# The minimum amount of steel for shrinkage and flexural are
(2)75.422*)12*10(*0018.00018.0  bhA
s in
2

(3) 2.7
000,60
0.18*)12*10(*200200

y
s
f
bd
A in
2

From (1~3), selected 2.7
s
A in
2

# Use #6 bar and the number of bars needed, 174.16
44.0
2.7

b
s
A
A
n nos.
# Spacing between bars,  
''0.7"13.7
117
3*212*10
1
cov*2







n
erL
S c/c
Use 17#6 @ 7.0″ c/c both ways.


Step-5: Check for hook requirement for the main bars

The development length d
l for #6 bars of the footing required for proper bonding, "6.35
000,4*20
8
6
*000,60
'20

c
by
d
f
df
l


The available length beyond column face as shown in figure.






All of the main bars should be provided without hooks in both directions.


Step-6: Check for dowel bar and its hook requirement

Minimum area of dowels required = 0.005x Pedestal column area = 0.005*15*17 = 1.28
in
2

Extend 4#7 column bars into the footing acting as dowel and provided area = 4x0.60 =
2.40 in
2
>1.28 in
2
so ok.
 
.,"6.35"5.49
2
15
3
2
12*10
2
cov
2
okl
c
er
L
l
d



11. 94

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
l
h
The development length of the dowels #7 bar in compression is
1. "6.16
000,4
000,60*
8
7
*02.0
'
02.0

c
yb
d
f
fd
l
2. "75.15000,60*
8
7
*0003.00003.0 
ybd
fdl
3. "8
d
l
"17"6.16
d
l

As shown in figure, available length
= diabarsmainerh *2cov
= 22-3-[]
8
6
8
6

= 17.5" > "17
d
l , so ok.

No hook is required at the free ends of the column bars but will be bend at 90° and extend
up to a length required for binding of column bars with footing bars.

A detail of the reinforcement arrangement is shown in Figure 7.6.




h = 22ʹʹ h = 22ʹʹ
15ʹʹ
L = 10ʹ
17#6 @7ʹʹ both ways
h = 22ʹʹ
4#7 column bars
Figure 7.6: Details of reinforcement arrangement in square RCC footing

12. 95

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Rectangular footing

Given data:
Depth of base below G.L. = 5′-0″
Unit weight of soil, s
w = 100 pcf
Allowable soil pressure, all
q = 4.5 ksf
fy = 60,000 psi
f’c= 4,000 psi
Size of the column =12″x14″

Step-1: Load calculation

Total un-factored load, P = DL+LL = 307.56 + 51.2 = 358.8 k
Total factored load, u
P = 1.2DL+1.6LL = 1.2*307.11 + 1.6*51.2 = 450.5 k

Step-2 Determination of the footing size

Let the depth of footing base, h = 22″=1.83 ft.
Average effective depth in both directions, d = h – clear cover = 22.0 – 4.0 = 18.0″
Provided effective depth in short direction, ds = h - 4.5 = 22.0 – 4.5 = 17.5″
Provided effective depth in long direction, dl = h - 3.5 = 22.0 – 3.5 = 18.5″

Weight of soil, s
W = 66.316
12
22
5*100hw
s






 lb/ft
2

Weight of base concrete, c
W = 50.27483.1*150hw
c
 lb/ft
2

Effective upward soil pressure, qeff = csa
WWq  = 4500-316.66-274.50
= 3908.84 lb/ft
2
= 3.909 k/ft
2

Area of footing = 7.91
909.3
8.358


effq
LLDL ft
2

Size of footing = 8.0′x 11.5′
Provided area of footing, Ag = 92 ft
2

Net soil pressure, ksf
A
P
q
g
u
n 9.4
92
5.450


13. 96

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
x
Step-3: Check for‘d’

Check by punching (two-way) shear
Punching shear is checked at a distance d/2 from all faces of column as shown in figure.
The pedestal column size = 15″x17″ "136)1817(2)1815(2)(2)(2
210
 dcdcb
"00.351817
"00.331815
2
1


dc
dc


Calculating the punching shear force, u
V
=   dcdcqP
nu

21 = 450.5-4.9*





12
35
*
12
0.33 = 411.2 k
For preventing punching failure,
concrete shear strength = punching shear force uc Vdbfor 
0'4,

or, "94.15
136*000,4*4*75.0
1000*2.411
'4
0

bf
V
d
c
u

The required"50.81"94.15d , so ok

Check by beam (one-way) shear
Beam shear is checked at a distance d from any face of the column along the longer side of
footing as shown in figure.

The beam shear force, Vu = qn*b* x




For preventing one-way shear failure,
Concrete shear strength = beam shear force




The required, "50.81"24.15d , so ok. k
d
cL
bqVor
83.138
12
0.18
12*2
17
2
5.11
*8*9.4
22

1
nu














 ''24.15
)12*8(*000,4*75.0*2
1000*3.138
,
3.138'2


dor
bd forVV
cuc


14. 97

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
x
Step-4: Calculation of the main steel

A. Long direction reinforcement

Moment will be taken at the face of the column for calculating main steel as shown in
figure.

# Calculation of Mu 9.4
2
12
17
2
5.11
22


























cL
x
ft


= 470.6 k-ft


# Calculation of  0033.0
5.18)128(490.085.0
126.4702
11
60
485.0
85.0
2
11
85.0
22'
'
















xxxxx
xxx
bdf
M
f
f
c
u
y
c




# Required reinforcement in long direction,
(1) 82.55.18*)12*8(*0033.0 bdA
s
 in
2


# The minimum amount of steel for shrinkage and flexural are
(2) 80.322*)12*8(*0018.00018.0  bhA
s in
2

(3)92.5
000,60
5.18*)12*8(*200200

y
s
f
bd
A in
2

From (1~3), selected 92.5
s
A in
2

# Use #7 bar and the number of bars needed, 1087.9
60.0
92.5

b
s
A
A
n nos.
# Spacing between bars, 
"10
9
90
110
3*212*8
1
cov*2







n
erb
S c/c
Use 10#7@ 10.0″ c/c in long direction.


8*
2
9.4
*9.4*9.4
*
2
**







 b
x
xqM
nu

15. 98

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
l
L
c
Cover
Check for hook requirement:
# The development length d
l for #7 bars of the footing in long direction, "2.33
000,4*25
875.0*000,60
'25

c
by
d
f
df
l


The available length beyond column face as shown in figure.






So ok and no hook is required.

B. Short direction reinforcement
Moment will be taken at the face of the column for calculating main steel as shown in
figure.

# Calculation of Mu 38.3
2
12
15
2
8
22


























cb
y
ft 5.11*
2
38.3
*38.3*9.4
*
2
**







 L
y
yqM
nu

= 321.9 k-ft


# Calculation of  0017.0
5.17)125.11(490.085.0
129.3212
11
60
485.0
85.0
2
11
85.0
22'
'
















xxxxx
xxx
bdf
M
f
f
c
u
y
c



# Required reinforcement in short direction,
(1)15.45.17*)12*5.11(*0017.0 bdA
s
 in
2

 
"2.33"5.57
2
17
3
2
12*5.11
2
cov
2



dl
c
er
L
l

y

16. 99

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
# The minimum amount of steel for shrinkage and flexural are
(2) 5.522*)12*5.11(*0018.00018.0  bhA
s in
2

(3)05.8
000,60
5.17*)12*5.11(*200200

y
s
f
bd
A in
2

From (1~3), selected 05.8
s
A in
2

# Use #6 bar, compute the number of bars needed
193.18
44.0
05.8

b
s
A
A
n nos.

Arrangement of steel:
Maximum number of the total 19 nos. reinforcing bars will be placed in band width.

*Band width size
It is taken as equal to the footing width as shown in figure.

Bandwidth = width of footing
So, b = 8′

*Steel in band width



Where,
144.1
2
19
widthbandin Steel



Or, no. of required reinforcing bars in band width = 15.57  16 nos.

*Spacing for Band width
S = "0.64.6
116
12*8
1-n
widthBand




*Remaining length at each side of band width
= ½ (L- band width – 2*concrete cover) = ½ (11.5'-8'-2*3) = 1.5'
Excess reinforcing bars = 19-16 = 3 nos.
Place 2#6 at the mid of 1.5' i.e. 9 c/c at both sides of the band width.
44.1
8
5.11
footing of sideshort
footing of side long
 1
2


 steelTotal
widthband in Steel

17. 100

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
l
h
Check for hook requirement:
The development length d
l for #6 bars of the footing in short direction for proper bonding "6.35
000,4*20
75.0*000,60
'20

c
by
d
f
df
l

The available length beyond column face as shown in figure. 
.,"6.35"5.37
2
15
3
2
12*8
2
cov
2
oksol
c
er
L
l
d





All of the main bars should be provided without hooks in short direction.

Step-5: Check for dowel bar and its hook requirement

Minimum area of dowels required = 0.005x Pedestal column area = 0.005*15*17 = 1.28
in
2

Extend 4#7 column bars into the footing acting as dowel and provided area = 4x0.60 =
2.40 in
2
>1.28 in
2
so ok.

The development length of the dowels #7 bar in compression is
1. "6.16
000,4
000,60*
8
7
*02.0
'
02.0

c
yb
d
f
fd
l
2. "75.15000,60*
8
7
*0003.00003.0 
ybd
fdl
3. "8
d
l
"17"6.16
d
l
As shown in figure, available length = diabarsmainerh *2cov
= 22-3-[]
8
7
8
6

= 17.4" > "17
d
l , so ok.

No hook is required at the free ends of the column bars but will be bend at 90° and extend
up to a length required for binding of column bars with footing bars.
l
L
c
Cover

18. 101

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
A detail of the reinforcement arrangement is shown in Figure 7.7.



























16 # 6 @ 6ʹʹ
h = 22ʹʹ
L = 11.5ʹ
2#6 @ 9ʹʹ
h = 22ʹʹ
15ʹʹ
h = 22ʹʹ
Figure 7.7: Details of reinforcement arrangement in rectangular RCC footing
10 #7 @10.0" Band width = 8ʹ

4 # 7 column bars

19. 102

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
7.7 Design Problem (Combined Footing)

Design a combined footing to support two square columns, as shown in figure below. The
exterior pedestal column (I) has a section 16"x16", which carries DL of 180 k and a LL of
120 k. The interior pedestal column (II) has a section of 20"x20", which carries a DL of
250 k and a LL of 140 k. The base of the footing is 5' below final grade and allowable soil
pressure is 5 k/ft
2
. Use f’c = 4 ksi and fy = 60 ksi. The external column is located 2' from
the property line.







Solution

Step-1: Load calculation
Un-factored load, Column I: P = DL+LL = 180 + 120 = 300 k
Column II: P = DL+LL = 250 + 140 = 390 k
Total un-factored loads for both columns = 300+390 = 690 k

Factored load, Column I: Pu = 1.2DL+1.6LL= 1.2*180+1.6*120 = 408 k
Column II: Pu = 1.2DL+1.6LL= 1.2*250+1.6*140 = 524 k
Total factored loads for both columns = 408+524 = 932 k

Step-2: Determination of the centroid of the footing









I II
a = 4'
x
Col. II Col. I
16' 2'
P = 300 k
Pu= 408 k
P = 390 k
Pu= 524 k Resultant R

20. 103

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Consider the resultant of the all un-factored loads acts at distance ‘x’ from the centerline
of ‘Column I’.




Extend the footing up to the property line of column I,
half footing length = 9' + 2' = 11'
So the length of the footing = 2 x 11 = 22'.
The length, a = 22-(2+16) = 4'.

Step-3: Determination of the footing size

Let the depth of footing base, h = 36″=3'.
Average effective depth in both directions, d = h – clear cover = 36.0 – 4.0 = 32.0″
Provided effective depth in short direction, ds = h - 4.5 = 36.0 – 4.5 = 32.5″
Provided effective depth in long direction, dl = h - 3.5 = 36.0 – 3.5 = 33.5″

Weight of soil, s
W = 200
12
36
5*100 





hw
s lb/ft
2

Weight of base concrete, c
W = 4503*150hw
c lb/ft
2

Effective upward soil pressure, qeff = csa
WWq  = 5,000-200-450
= 4,350 lb/ft
2
= 4.35 k/ft
2

Area of footing = 62.158
35.4
690


effq
LLDL ft
2

Length of the footing, L = 22'
Hence, width of the footing, B = '50.721.7
22
62.158

Size of footing = 7.5′x 22.0′
Provided area of footing, Ag = 165 ft
2

Net soil pressure, ksf
A
P
q
g
u
n 65.5
165
932
  
ft. 9 Use ft.
300 x ft 390 x 16ft
F
Fx
x
i
ii






04.9
390300
0

21. 104

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Step-4: SFD and BMD


1.33' 1.33' 14.5' 1.67' 3.17'
SFD
BMD
20"x20" 16"x16"
4'
Col. II Col. I
16' 2'
Pu = 408 k Pu = 524 k
qu= 5.65 ksf = 5.65x7.5= 42.38 k/ft
-323.38 k
319.38 k
56.37 k
84.62 k
-295.13 k
-134 k
-169.3 k
354.7 k
212.17 k-ft
338.6 k-ft
-1149.1 k-ft
84.62 k-ft
37.5 k-ft

22. 105

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Step-5: Check for‘d’

Check by punching (two-way) shear
Punching shear will be checked at a distance d/2 from all faces of the loaded column II as
shown in figure.









"208)3220(4)(4
0
 dcb
"00.523220 dc


Calculating the punching shear force, u
V
= dcdcqP
nu
 = 524- 5.65*





12
52
*
12
52 = 417.91 k
For preventing punching failure, concrete shear strength = punching shear force uc Vdbfor 
0'4,

or, "96.11
208*000,4*4*75.0
1000*91.417
'4
0

bf
V
d
c
u

The required d = 11.96" < Provided d = 33.5", so ok

Check by beam (one-way) shear
Beam shear will be checked at a distance d from left face of the column II along the longer
side of footing.

From SFD, it is found that maximum shear force at the face of Column II is 319.38 k.
The beam shear force @ d distance from that face,

 k k dqVV
u
37.206
12
32
*38.4238.319
max




d
22'
7.5'
c+d=20+32=52"
d/2

23. 106

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
For preventing one-way shear failure, concrete shear strength = beam shear force




The required d = 24.2" < Provided d = 33.5", so ok

Step-6: Calculation of the main steel


A. Long direction reinforcement

 Top -ve steel

Moment will be taken from BMD.
# Maximum –ve moment, u
M = 1149.1 k-ft.
# Calculation of 0026.0
5.33)125.7(490.085.0
121.11492
11
60
485.0
85.0
2
11
85.0
22'
'
















xxxxx
xxx
bdf
M
f
f
c
u
y
c



# Required reinforcement
(1) 80.75.33*)12*5.7(*0026.0 bdA
s
 in
2


# The minimum amount of steel for shrinkage and flexural are
(2)83.536*)12*5.7(*0018.00018.0  bhA
s in
2

(3) 05.10
000,60
5.33*)12*5.7(*200200

y
s
f
bd
A in
2

From (1~3), selected 05.10
s
A in
2

# Use #8 bar and the number of bars needed, 1372.12
79.0
05.10

b
s
A
A
n nos.
# Spacing between bars,  
"0.7
113
3*212*5.7
1
cov*2







n
erL
S c/c
Use 13#8 @ 7.0″ c/c at top.

 Bottom +ve steel

Moment will be taken from BMD.
# Maximum +ve moment, u
M = 338.6 k-ft. ''2.24
)12*5.7(*000,4*75.0*2
1000*37.206
,
37.206'2


dor
bd forVV
cuc


24. 107

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
# Calculation of 0008.0
5.33)125.7(490.085.0
126.3382
11
60
485.0
85.0
2
11
85.0
22'
'
















xxxxx
xxx
bdf
M
f
f
c
u
y
c



# Required reinforcement
(1) 23.25.33*)12*5.7(*0008.0 bdA
s
 in
2


# The minimum amount of steel for shrinkage and flexural are
(2)83.536*)12*5.7(*0018.00018.0  bhA
s in
2

(3) 05.10
000,60
5.33*)12*5.7(*200200

y
s
f
bd
A in
2

From (1~3), selected 05.10
s
A in
2

# Use #8 bar and the number of bars needed, 1372.12
79.0
05.10

b
s
A
A
n nos.
# Spacing between bars,  
"0.7
113
3*212*5.7
1
cov*2







n
erL
S c/c
Use 13#8 @ 7.0″ c/c at bottom.


B. Short direction reinforcement









'08.3
212
"16
2
'5.7
1

X
Y
'92.2
212
"20
2
'5.7
2

X
Y

7.5'
Y1 Y2
22'
7.5'
2.0+3.75 = 5.75'

25. 108

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Bottom steel for Col. I
# ftk
Y
YqM
nu  1.154
2
08.3
'*75.5*08.3*65.5
2
*)75.5**(
1
1
# Calculation of 0005.0
5.32)1275.5(490.085.0
121.1542
11
60
485.0
85.0
2
11
85.0
22'
'
















xxxxx
xxx
bdf
M
f
f
c
u
y
c



# Required reinforcement
(1) 06.15.32*)12*75.5(*0005.0 bdA
s
 in
2


# The minimum amount of steel for shrinkage and flexural are
(2)47.436*)12*75.5(*0018.00018.0  bhA
s in
2

(3) 45.7
000,60
5.32*)12*75.5(*200200

y
s
f
bd
A in
2

From (1~3), selected 45.7
s
A in
2

# Use #7 bar and the number of bars needed, 1345.12
60.0
45.7

b
s
A
A
n nos.
# Spacing between bars,  
"5.5
113
312*75.5
1
cov75.5







n
er
S c/c
Use 13#7 @ 5.5″ c/c under column I.

Bottom steel for Col. II
# ftk
Y
YqM
nu
 7.180
2
92.2
*5.7*92.2*65.5
2
2
*)5.7**(
2
# Calculation of 0004.0
5.32)125.7(490.085.0
127.1802
11
60
485.0
85.0
2
11
85.0
22'
'
















xxxxx
xxx
bdf
M
f
f
c
u
y
c



# Required reinforcement,
(1) 24.15.32*)12*5.7(*0004.0 bdA
s
 in
2


# The minimum amount of steel for shrinkage and flexural are
(2)83.536*)12*5.7(*0018.00018.0  bhA
s in
2

(3) 75.9
000,60
5.32*)12*5.7(*200200

y
s
f
bd
A in
2

26. 109

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
From (1~3), selected 75.9
s
A in
2

# Use #7 bar and the number of bars needed, 1725.16
60.0
75.9

b
s
A
A
n nos.
# Spacing between bars, "5.5"63.5
117
12*5.7
1
12*5.7





n
S c/c
Use 17#7 @ 5.5″ c/c under column II.

Check for hook requirement:
The development length d
l for #7 bars of the footing in short direction for Column II "2.33
000,4*25
8
7
*000,60
'25

c
by
d
f
df
l


The available length beyond column face,  
.,"2.33"32
2
20
3
2
12*5.7
2
cov
2
oknotsol
c
er
b
l
d


So, provide 90° hooks at the end of all bars under Col. I and II in short direction.
A detail of the reinforcement arrangement is shown in Figure 7.8.
Figure 7.8: Reinforcement arrangement in Combined Footing
22'
36"
5.75' 7.5'
Col. II Col. I
13#8 @ 7.0" in long direction
17#7 @ 5.5" in
short direction
13#7 @ 5.5" in
short direction
T & S Steel

27. 110

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
7.8 Theoretical questions

1. What do you mean by footing? Mention different types of shallow foundation.
2. Distinguish between ‘shallow’ and ‘deep’ foundation.
3. Distinguish between ‘isolated’ and ‘combined’ footing.
4. What are the critical sections that should be checked in a RCC footing?

7.9 Exercise Problems

Exercise Problem-01: Design the reinforcement in short and long direction of a
combined footing as shown in Figure 2, given that fc
’
= 4 ksi, fy = 50 ksi, qall = 5 ksf. The
footing base is 5' below the grade level. The edge of column #1 is at the property line, and
the spacing between columns is 18' center-to-center. The length of the footing is 22' and
the total depth is 36".

Column #1 (18"x18") Column #2 (24"x24")
DL = 100 kips DL = 150 kips
LL = 181 kips LL = 214 kips

The required SFD and BMD are also given in figure for the references.

28. 111

CE 317: Design of Concrete Structures II, Md. Mahbub-ul-Alam, Asst. Prof, CEN, SUB
Exercise Problem-02: A square footing has to support 18" square tied interior column
reinforced with 8#9 bars. The column carries an unfactored axial dead load of 250 k and
an axial live load of 300 k. The base of the footing is 4.0' below final grade and allowable
soil pressure is 4.5 ksf. Use f’c = 4 ksi and fy = 60 ksi. The total depth of the footing is
22.0" and the average effective depth is 18.0".

Check the depth of the footing for punching and beam shear failure.

Exercise Problem-03: A Rectangular footing having size 18'x10', as shown figure below,
is to support a 16" square tied column reinforced with 4#8 bars. Soil pressure, qa = 4.0 ksf
and qn = 4.9 ksf. Use f’c = 4 ksi and fy = 60 ksi. The total depth of the footing is 24.0" and
the average effective depth is 20.0" which has been checked for one-way and two-way
shear.
18'



10'





Design the reinforcement in short direction. Also check for hook requirements for the main bars.

=============
16"