Lec13 Continuous Beams and One Way Slabs(3) Footings (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

1. 26-Apr-13
1
Footings
• Footings are structural members used to support columns and
walls and to transmit and distribute their loads to the soil.
• Load bearing capacity of the soil must not be exceeded;
excessive settlement, differential settlement, or rotation must
be prevented, and adequate safety against overturning or
sliding must be maintained.
• As soil strength is less than that of concrete, footings have
therefore dimensions larger than the supported members
(columns or walls).
• Types of footings:
• Wall footings are used to support structural walls that carry
loads for other floors or to support nonstructural walls.
• The footing has the same length as the wall but is wider.

2. 26-Apr-13
2
Isolated / single / spread footings support single columns.
This is one of the most economical types of footings and is used
when columns are spaced at relatively long distances
Combined footings usually support two or three columns in a row.
Combined footings are used when tow columns are so close that
single footings cannot be used or when one column is located at or
near a property line

3. 26-Apr-13
3
Cantilever or strap footings consist of two single footings connected
with a beam or a strap and support two single columns.
This type replaces a combined footing and is more economical
Continuous footings support a row of three or more columns.
They have limited width and continue under all columns.

4. 26-Apr-13
4
Rafted or mat foundation consists of one footing usually placed
under the entire building area. They are used, when soil bearing
capacity is low or column loads are heavy.
Pile caps are thick slabs used to tie a group of piles together to
support and transmit column loads to the piles.

5. 26-Apr-13
5
Distribution of Soil Pressure
When the column load P is applied on the centroid of the footing, a
uniform pressure is assumed to develop on the soil surface below
the footing area.
However the actual distribution of the soil is not uniform, but
depends on may factors especially the composition of the soil and
degree of flexibility of the footing.
Distribution of Soil Pressure

6. 26-Apr-13
6
Combined loading (axial force and bending moment)
For a square footing with width L, total soil pressure is given by:
e is the eccentricity:
To avoid tension in the soil the equivalent eccentric load must be
applied within the kern:







L
e
L
P
L
M
L
P
I
My
A
P
q
6
1
6
232
max
P
M
e 
6
L
e 
Spread footing under compression force only
A footing is usually subjected to column load, its self weight, the
weight of the soil above as well as a possible top surcharge such as
ground slab and its load.
This loading is uniform apart from column load P.

7. 26-Apr-13
7
Total soil pressure is:
The pressure causing bending in the footing is caused by the
concentrated force P only.
It is therefore convenient to compute the net soil pressure as:
 0qhh
A
P
q ss  
 
A
P
qhhqq ssn  0
Footing analysis and design
Footings must be designed to carry the column loads and transmit
them to the soil safely while satisfying code limitations.
Footing dimensions are based on allowable soil bearing capacity
using service loads:
RC design of footings uses ultimate loads:
LDLiveDead PPPPP 
LDu PPP 7.14.1 

8. 26-Apr-13
8
Analysis and design steps
• 1- Footing area (dimensions)
• 2- Check thickness for one and two way shear
• 3- RC design of footing
• In step 1, we use net allowable pressure and
service load P
• In steps 2 and 3, we use net ultimate pressure
Minimum footing area (dimensions)
Net soil pressure must not exceed the net allowable soil pressure:
The net allowable soil pressure is deduced from the allowable soil
pressure as:
Minimum footing area is therefore:
The adopted footing area must be equal or greater than
nan q
A
P
q 
na
LD
na q
PP
q
P
A

min
 0qhhqq ssana  
minA

9. 26-Apr-13
9
Strength requirements
• For strength requirements (shear and bending), net
ultimate soil pressure is used.
• The net ultimate pressure is obtained using ultimate
axial force and actual adopted footing area.
• L1 and L2 are the final adopted footing dimensions
21LL
P
A
P
q uu
nu 
Shear strength
• The spread footing supporting a single column acts like a two-
way slab panel (flat plate) for an interior column.
• Both one-way and two-way shear must be checked.
• We must check in both cases that: uc VV 

10. 26-Apr-13
10
Two-way shear
Assume a value for average steel depth d
Usually: d = h – 100 (mm)
Determine critical perimeter length b0
For rectangular columns of sides c1 and c2
b0 = 2(c1+d) +2(c2+d) = 2(c1 + c2 + 2d)
For square columns where one side = c
b0 = 4(c + d)
Ultimate two-way shear
• The ultimate shear is computed over the loaded area defined
as the total footing area minus the critical area:
• Rectangular column:
• Square column:
     dcdcLLqAAqV nucnuu  21210
  dcdcqPV nuuu  21
 2
dcqPV nuuu 

11. 26-Apr-13
11
Nominal two-way shear


























)3(
3
)2(
12
2
)1(
6
2
1
0
'
0
'
0
0
'
db
f
db
f
b
d
db
f
MinV
c
cs
c
c
c


side)short/side(longratioaspectcolumn:c
column)(interior40s
One-way shear
For footings with bending action in
one direction the critical section is
located a distance d from column face
Ultimate shear force at section m-m:






 d
cL
LqV
22
1
2nuu
dL
f
V c
c 2
'
6

Nominal one-way shear strength:

12. 26-Apr-13
12
Flexural Strength and Footing reinforcement
The footing is modeled as a double
cantilever beam subjected to uniform
load as shown.
The maximum ultimate moment is
located at the column face section n-n
RC design is performed using this
moment value.
 
8
)(
2
2
2
1
2
2
1
cL
Lq
cL
wM nuuu







 

Example 1: Square spread footing
Design a square spread footing supporting a square column with
section 450x450, transmitting a service dead axial load of 1780 kN
and a service live axial load of 1200 kN.
The buried footing supports a 150 mm thick top soil layer with unit
weight of 19 kN/m3 and a 150 mm thick ground slab subjected to a
loading of 4.5 kN/m2.

13. 26-Apr-13
13
• RC Material data:
• Soil allowable bearing capacity is:
•
• The service axial loads are:
• The net allowable soil pressure is:
• Soil unit weight is:
• The top surcharge q0 is composed of the ground slab weight
and its loading:
2
0 /1.85.4150.024 mkNxq 
MPafc 25'

3
/24 mkNc  MPafy 420
2
/300 mkNqa 
kNPD 0.1780 kNPL 0.1200
 0qhhqq sscana  
3
/19 mkNs 
Footing thickness
Footing thickness h is estimated as varying between one to
two times the dimension of the column section:
Thus:
We choose an initial thickness of 800 mm.
The net allowable soil pressure is then:
chc  20.1 900450  h
  2
/85.2691.8150.019800.0240.300 mkNqna 

14. 26-Apr-13
14
The required footing area must satisfy:
For a square footing, the length must satisfy:
We take L = 3.4 m.
Net ultimate soil pressure is:
Before performing RC design, we must check the footing
thickness with respect to shear.
2
053.11
85.269
12001780
m
q
PP
A
na
LD





mL 323.3
2
/04.392
56.11
0.4532
4.34.3
7.14.1
mkN
PP
A
P
q LDu
nu 



The average steel depth is estimated as:
Two way shear check
The footing with the column is similar to a flat plate with an
interior column.
The critical perimeter dimensions are:
The critical perimeter length is:
mmhhhd 700100800100257525cover 
mmdcppp 115070045021 
mmpb 46001150440 

15. 26-Apr-13
15
3.4 m
3.4 m1150
1150
3.4 m
3.4 m450
450
Ultimate shear is computed over the loaded area defined
as the total footing area minus the critical area:
   
  kN
ppLLqAAqV nucnuu
5.401315.115.14.34.304.392
21210


For two-way shear, concrete nominal shear strength is given
by the minimum of equations (1) to (3) :



























column)(Internal40)3(
3
0.1
450
450
)2(
12
2
4600700)1(
6
2
1
Min
0
'
0
'
0
00
'
s
c
c
cs
c
c
c
db
f
db
f
b
d
mmbmmddb
f
V




kNV
kNN
kNN
kNN
V cc 67.5366
67.536667.53666667004600
3
25
0.10850108500007004600
12
25
4600
70040
2
0.805080500007004600
6
25
1
2
1
Min 
















 









kNVkNV uc 5.40130.402567.536675.0 
Two-way shear is thus OK

16. 26-Apr-13
16
One-way shear check
3.4
3.4m
d
p
md
cL
p 775.07.0
2
45.0
2
4.3
22

The ultimate shear given by the
shaded loaded area is:
kN
pLqV nuu
0.1033
4.3775.004.392


For one-way shear, concrete nominal shear strength is:
kNNdL
f
V c
c 67.991667.99166667003400
6
25
6
2
'

kNVc 5.743767.991675.0 
It is much greater than Vu.
One-way shear is thus also OK
RC Design
wu = qnu x 3.4
3.4m
f
0.45
3.4
3.4m
f
The ultimate moment at the fixed end is:
mkN
cL
Lq
f
wM nuuu .0.1450
8
)45.04.3(
4.304.392
8
)(
2
222






17. 26-Apr-13
17
RC design of a rectangular section with dimensions:
b = 3.4 m = 3400 mm h = 800 mm
Steel depth: d = h – cover – 25 = 800 – 75 – 25 = 700 mm
We find the required steel ratio:
Required steel area is: As = 5610.6 mm 2
Minimum steel Asmin = 0.0018bh
= 0.0018 x 3400 x 800 = 4896.0 mm 2
We use eighteen 20-mm bars (5654.87 mm 2) in both
directions.
This bottom reinforcement must be distributed across the
entire footing width.
0023574.0
Dowel bars
At least four column corner bars (called dowel bars) must be
extended into the footing, bent at the ends, and tied to the
main footing reinforcement. The dowel diameter shall not
exceed the diameter of the longitudinal bars in the column
by more than 4 mm.

18. 26-Apr-13
18
Example 2: Wall (strip) footing
Design a strip footing supporting a 300-mm thick wall
transmitting a service dead axial load of 145 kN/m and a
service live axial load of 180 kN/m.
The buried footing supports a 1.5 m thick top soil layer with
unit weight of 18 kN/m3.
We analyze and design 1-m strip.
RC Material data:
Soil allowable bearing capacity is:
The service axial loads are:
The net allowable soil pressure is:
Soil unit weight is:
There is no top surcharge: q0 = 0
Footing thickness estimated as varying between one to one
and half times the dimension of the wall width.
We use: h = 400 mm
MPafc 25'

3
/24 mkNc  MPafy 420
2
/240 mkNqa 
kNPD 0.145 kNPL 0.180
 0qhhqq sscana  
3
/18 mkNs 
450300  h

19. 26-Apr-13
19
The net allowable soil pressure is then:
 2
/4.2035.1184.024240 mkNqna 
Minimum footing area (for 1-m strip):
Thus:
We use: L = 1.6 m
Net ultimate soil pressure is:
Before performing RC design, we must check the footing
thickness with respect to shear.
For wall footings, only one-way (beam) shear has to be
checked.
2
minmin 598.1
4.203
180145
1 m
q
PP
LA
na
LD




 mL 598.1min 
2
/125.318
0.16.1
1807.11454.1
mkN
A
P
q u
nu 



One-way shear check
The ultimate shear given by the
shaded loaded area is:
For one-way shear, concrete nominal shear strength is:
It is greater than Vu.
One-way shear is thus OK
1.6
1.0 m
d
p
md
wL
p 335.0315.0
2
3.0
6.1
22

kNpLqV nuu 6.1060.1335.0125.3182 
kNNdb
f
V w
c
c 5.2622625003151000
6
25
6
'

kNVc 875.1965.26275.0 

20. 26-Apr-13
20
RC Design
Footing modeled as double cantilever
Ultimate moment at fixed end is:
wu =qnu x 1.0
1.6 m
f
0.30
mkNx
f
wM uu .2.67
8
)3.06.1(
0.1125.318
2
22



RC design of a rectangular section with dimensions:
b = 1.0 m = 1000 mm h = 400 mm
Steel depth: d = 315 mm
We find: As = 574.6 mm2
Asmin = 0.0018bh = 0.0018 x 1000 x 400 = 720.0 mm2
We thus use minimum steel area
This requires four 16-mm bars (per meter), spacing = 250 mm
Provide shrinkage steel in longitudinal direction
Example 3: Restricted spread footing
Design a restricted spread footing supporting a square
column 450 x 450 mm, transmitting a service dead axial load
of 1100 kN and a service live axial load of 900 kN.
The buried footing supports a 1.2 m thick top soil layer with
unit weight of 18 kN/m3.
The footing width is restricted to 2.5 m
Material data:
MPafc 25'
 3
/24 mkNc  MPafy 420
2
/240 mkNqa 
Choose footing depth h = 700 mm
3
/18 mkNs 
Net allowable soil pressure is:  0qhhqq sscana  
  2
/6.2012.1187.024240 mkNqna 
There is no top surcharge: q0 = 0

21. 26-Apr-13
21
The required footing area must satisfy:
Footing width limited to 2.5 m
Minimum footing length is then:
We use:
Net ultimate soil pressure is:
Before performing RC design, we must check the footing
thickness with respect to shear.
2
29.9
6.201
9001100
m
q
PP
A
na
LD





mL 97.3
5.2
29.9
1 
2
/0.307
0.10
0.3070
5.20.4
7.14.1
mkN
x
PP
A
P
q LDu
nu 


mL 5.22 
mL 0.41 
The average steel depth is estimated as:
Two way shear check
The footing with the column is similar to a flat plate with an
interior column.
The critical perimeter dimensions are:
The critical perimeter length is:
mmhhhd 600100700100257525cover 
mmdcppp 105060045021 
mmpb 42001050440 

22. 26-Apr-13
22
Ultimate shear is computed over the loaded area defined
as the total footing area minus the critical area:
   
  kN
ppLLqAAqV nucnuu
53.273105.105.15.20.40.307
21210


For two-way shear, the concrete nominal shear strength is
given by the minimum of equations (1) to (3)



























column)(Internal40)3(
3
0.1
450
450
)2(
12
2
4200600)1(
6
2
1
Min
0
'
0
'
0
00
'
s
c
c
cs
c
c
c
db
f
db
f
b
d
mmbmmddb
f
V




kNV
kNN
kNN
kNN
V cc 0.4200
0.42000.42000006004200
3
25
0.8100810000006004200
12
25
4200
60040
2
0.630063000006004200
6
25
1
2
1
Min 
















 









kNVkNV uc 53.27310.31500.420075.0 
Two-way shear is thus OK

23. 26-Apr-13
23
One-way shear check
4.0
2.5m
d
p
md
cL
p 175.16.0
2
45.0
2
0.4
22
1

The ultimate shear given by the
shaded loaded area is:
kN
pLqV nuu
8.901
5.2175.10.3072


For one-way shear, concrete nominal shear strength is:
kNNdL
f
V c
c 0.125012500006002500
6
25
6
2
'

kNVc 5.9370.125075.0 
It is greater than Vu.
One-way shear is thus also OK
RC Design
Design must be performed in both directions
wu =qnu x 2.5
4.0m
f
0.45
4.0
2.5m
f
The ultimate moment at the fixed end is:
mkN
cL
Lq
f
wM nuuu .05.1209
8
)45.00.4(
5.20.307
8
)(
2
22
1
2
2





RC design in long direction

24. 26-Apr-13
24
RC design of a rectangular section with dimensions:
b = 2.5 m = 2500 mm h = 700 mm
Steel depth: d = h – cover – 25 = 700 – 75 – 25 = 600 mm
Required steel area is: As = 5532.6 mm 2
Minimum steel Asmin = 0.0018bh
= 0.0018 x 2500 x 700 = 3150.0 mm 2
We use eighteen 20-mm bars (5654.87 mm 2)
Corresponding spacing S = 139 mm = 2500/18
The ultimate moment at the fixed end is:
mkN
cL
Lq
f
wM nuuu .08.645
8
)45.05.2(
0.40.307
8
)(
2
22
2
1
2





RC design in short direction
wu =qnu x 4.0
2.5m
f
0.45
2.5
4.0m
f

25. 26-Apr-13
25
RC design of a rectangular section with dimensions:
b = 4.0 m = 4000 mm h = 700 mm
Steel depth: d = h – cover – 25 = 700 – 75 – 25 = 600 mm
Required steel area is: As = 2878.4 mm 2
Minimum steel Asmin = 0.0018bh
= 0.0018 x 4000 x 700 = 5040.0 mm 2
We use seventeen 20-mm bars (5340.7 mm 2)
Corresponding spacing S = 235 mm = 4000/17