F i D
Footing
Design
i
Types of Footing
Wall footings are used to
support structural walls that
carry loads for other floors
or to support nonstructural
walls.
Types of Footing
Isolated or single footings
are used to support single
columns. This is one of the
most economical types of
footings and is used when
columns are spaced at
relatively long distances.
Types of Footing
Combined footings usually
support two columns, or
three columns not in a row.
Combined footings are used
when two columns are so
close that single footings
cannot be used or when one
col mn is located at or near
column
a property line.
Types of Footing
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.
economical
Types of Footing
Continuous footings
support a row of three or
more columns. They have
limited width and continue
under all columns.
Types of Footing
Rafted or mat foundation
consists of one footing
usually placed under the
entire building area. They
are used, when soil bearing
capacity
it is
i low,
l
column
l
loads are heavy single
footings cannot be used,
piles are not used and
differential settlement must
be reduced.
Types of Footing
Pile caps are thick slabs
used to tie a group of piles
together to support and
transmit column loads to the
piles.
Distribution of Soil Pressure
When the column load P is applied
on the centroid of the footing, a
uniform
if
pressure is
i assumedd to
develop on the soil surface below the
footing area.
area However the actual
distribution of the soil is not
uniform,
f
but depends
p
on mayy
factors especially the composition of
the soil and degree of flexibility of
the footing.
Distribution of Soil Pressure
Soil pressure distribution in
cohesionless soil.
Soil pressure distribution in cohesive
soil.
Eccentrically loaded footings
Eccentrically loaded footings Example
Isolated Footing
D.L = 900 kN
L.L = 450 kN
Ms = 150kN.m
Mu=200kN.m
qall =200kpa
M
Area required approximated
qall ( net )  20t / m 2  200kPa
Ag 
Ps
qall ( net )
(900  450) 103
2


6
.
75
m
200 103
use 3.5m  2.5m 
A  8.75m
2
I  3.5122.5  8.9m 4
Check stress
3
M 150
e

 0.111  L6  36.5  0.583
P 1350
Ps M s C 1350 150  32.5



 183.7 kPa
A
I
8.75
8 .9
Ps M s C 1350 150  32.5



 124.7 kPa
A
I
8.75
8.9
124.7
183 7
183.7
Ultimate p
pressure under footingg
Pu  1.2900   1.6(450)  1800kN
M u  200kN .m
Pu M u C 1800 200  32.5



 245kPa
A
I
8.75
8 .9
Pu M u C 1800 200  32.5



 166kPa
8 .9
A
I
8.75
245
166
245
Check Punchingg Shear
bo  4(530  400)  3720cm
F square column
For
l
the
th suitable
it bl VC equation
ti is
i :
fc '
25
 VC  
 530  3720 / 1000  2464.5kN
bo d  0.75 
3
3
VU 
(166  245 166  245 )
4
3.5  2.5  1798kN
Check Beam Shear
 VC  0.75 
25
 530  2500 / 1000  828.13kN
6
VU at d from column face 
 223  245 
VU  
 * 1.5  0.53* 2.5
2


 567.5kN
VU  VC
3.5m
Bendingg moment design
g
Long direction
M U at d from column face 
245
166
P1  211(1.5)(2.5)  791.25kN
P2  12 (245  211)(1.5)(2.5)  63.75kN
245
M U  P1 (0.75)  P2 (1)  791.25(0.75)  63.75(1)  657.2kN .m
b  2500, d  530mm

25 
2 106 * 657.2
657 2
  0.85 *
1 - 1   0.00254
2
0.90.8525 * 530 * 2500 
420 
AS  0.00254  530 1000  1346mm 2  13.5cm 2 use 716 / m long
g direction
Bendingg moment design
g
2 5m
2.5m
Short direction
M U at d from column face 
 245  166
1 3.50.5  719.25kN .m
MU  
2


b  3500, d  530mm
245

25 
2 106 * 719.25
  0.85 *
1 - 1   0.002
2
420 
0.90.85
0 8525 * 530 * 3500 
AS  0.002  530 1000  1060mm 2  10.1cm 2
245
166
short direction
Central band ratio =
3.5
 1.4
2.5
2
2

 0.83
  1 2.4

Central band of short direction = 0.83 As = 0.83 (10.1)=8.6cm2
214
716/m
714/m
214
Footing Design
Part II
C bi d ffooting
Combined
ti
Example 1
Design a combined footing As shown
qall ( net )  20t / m 2  200kPa
f c  25 N mm 2
f y  420 N mm 2
Dimension calculation
Th base
The
b
dimension
di
i to get uniform
if
di
distributed
ib d lload
d
2000kN
800 kN
1200 kN
A
x1=0.2m
x2=6.2m
x
800(0.2)+1200(6.2)=2000(x)
x = 3.8m
800 kN
1200 kN
Try thickness
=80cm
2x =7.6 m
Area required
qall ( net )  20t / m 2  200kPa,
Pu  1.3( Ps )  1.3(2000)  2600kN
Ag 
Ps
qall ( net )
2000 103
2

10
m
 7.6 *1.8

3
200 10
Pu 2600103
qu 

 190 103 Pa  190kPa
A
7.6 *1.8
Check for punching Shear
d = 730 mm
1.13m
A
0.765
bo  2(765)  1130  2260mm
fc '
25
bo d  0.75 
 730  2260 / 1000  2062.3kN
3
3
 d f '
30  730 
25


 VC    2  s  c bo d  0.75 2 

 730  2260 / 1000  6027 kN

b  12
2260  12


 VC  
VU  800(1.3)  1.13 * 0.765 *190  875.8kN  Vc
oK
K
B
bo  4(730  400)  4520mm
fc '
25
 VC  
 730  4520 / 1000  4124.4kN
bo d  0.75 
3
3
 s d  fc '
40  730 
25


 VC    2 
bo d  0.75 2 
 730  4520 / 1000  13322.5kN


b
12
4520
12




VU  1200(1.3)  1.13 *1.13 *190  1317.4kN  Vc
oK
Draw S.F.D & B.M.D
Stress under footing
f
g
= 190 *1.8 = 342 kN/m
Check for beam shear
b = 1800mm, d = 730mm
25
 VC  0.75 
 730  1800 / 1000  821.25kN
6
Max.  VU at d from column face  762.34kN
VU  VC
Bending moment Long
direction
 ve M  1366kN .m
b  1800mm, d  730mm

25 
2 106 *1366
  0.85 *
  0.0039
1 - 1 2
420 
0.90.8525 * 730 *1800 
AS  0.0039  730 1000  2847mm 2  28.5cm 2 use 9 20 / m Top
 ve M  246.7kN .m
b  1800mm, d  730mm

25 
2 106 * 246.7
  0.85 *
  0.0007   min
1 - 1 2
420 
0.90.8525 * 730 *1800 
AS min  0.0018  800 1000  1440mm 2  14.4cm 2 use 716 / m Bottom
B tt
Bending moment Short
direction
Under Column A
1040
0.765  1.8  0.4 

M

  141.6
(1.8 * 0.765)
2 
2

b  765mm, d  730mm
2

25 
2 106 *141.6
  0.85 *
   min
1 - 1 2
420 
0.90.8525 * 730 * 765 
AS min  0.0018  800  765  1101.6mm 2  11cm 2 use 714 / m
Under Column B
1560
1.13  1.8  0.4 
M


  212.33
(1.8 *1.13)
2 
2

b  1130mm, d  730mm
2

25 
2 106 * 212.33
  0.85 *
1 - 1    min
2
0.90.8525 * 730 *1130 
420 
AS min  0.0018  800  765  1101.6mm 2  11cm 2 use 714 / m
Shrinkage Reinforcement in short direction
AS min  0.0018  800  765  1101.6mm 2  11cm 2 use 714 / m
Footing
g Design
g
Part III
Combined footing, strip footing, & Mat foundation
Example 2
Design a combined footing As shown
qall ( net )  18 t / m 2  180kPa
f c  25 N mm 2
f y  420 N mm 2
Dimension calculation
The base dimension to get uniform distributed load
1200 kN
1950kN
750 kN
750(4.2)+1200(0.2)
750(4
2)+1200(0 2)=1950
1950 (x)
x = 1.75m
A
x1=0.2m
x2=4.2
m
x
 B1  2 B2  L

x  
 B1  B2  3
Area required
qall ( net )  20t / m 2  200kPa,
Ag 
Ps
qall ( net )
1950 103
2


10
.
8
m
180 103
 B1  B2 

 L  10.8
 2 
 B1  B2 

4.35  10.8
 2 
 B1  B2 

  2.5
 2 
B1  B2  5
 B  2 B2  L  5  B2  4.35
  
x   1

B

B
3
5

 3
2 
 1
1.75  1.45  0.29 B2
B2  1m
B1  4 m
Pu 1.31950  103
qu 

 235  103 Pa  235kPa
A
10.8
Check for punching Shear
h= 750mm
h
750
d = 732 mm
A
bo  2(732)  1065  2590mm
B2 1
B2=1m
B1=4m
fc '
25
bo d  0.75 
 665  2590 / 1000  2160.4kN
3
3
 d f '
30  665 
25


 VC    2  s  c bo d  0.75 2 

 665  2590 / 1000  5222kN

b  12
2590  12


 VC  
VU  1200(1.3)  1.065 * 0.733 * 235  1376.6kN  Vc
oK
K
B
bo  2(633)  965  2231 mm
fc '
25
 VC  
bo d  0.75 
 665  2231 / 1000  1854.5kN
3
3
 sd  fc '
30  665 
25


 VC    2 
bo d  0.75 2 
 665  2231 / 1000  5273kN


b
12
2231
12




VU  800(1.3)  0.965 * 0.633 * 235  896.5kN  Vc
oK
Draw S
S.F.D
FD & B
B.M.D
MD
Empirical S.F.D & B.M.D
975  0.70  682
m
Convert trapezoidal load to rectangle
wave  235  23 (940  235)  705
 M max 
wl
7053.65

 1174 kN .m
8
8
2
2
Clear distance between column
B in moment design = ave. width = 2.5m
1560  (0.70)  1092
Mmax
Check for beam shear
d = 665mm
b  1  2( Lx )  y
at x  0.815  0.15
1  2( 04.965
.35 )  1.5  1.7 m  1700 mm
25
 VC  0.75 
 665  1700 / 1000  696kN
6
Max.  VU at d from column B face ( the most critical)  668kN
VU  VC
b
Y=1.5m
x
4m
1m
Bending moment Long
Top
 ve M  1260kN .m
d  730mm
direction
b  1  2( 24..25
35 )  1.5  2.60m  2600

2 106 *1260
25 
  0.85 *
1 - 1   0.003
2
420 
0.90.8525 * 665 * 2600 
AS  0.003  665 1000  1995mm 2  20cm 2 use 1016 / m Top
Bottom
AS min  0.0018  750 1000  1350mm 2  13.5cm 2 use 914 / m Bottom
Bending moment Short
direction
Under Column A
b'  1  2( 34..62
35 )  1.5  3.5m  3500mm
b'  1  2( 04.633
.35 )  1.5  1.44m  1440mm
b
3.5  4
m  3.75m  3750mm
2
1560
0.733  3.75  0.4 

M

  583.6
(3.75 * 0.733)
2 
2

d  665mm
2

25 
2 106 * 583.57
  0.85 *
1 - 1   0.005
2
420 
0.90.8525 * 665 * 733 
AS  0.005  665  733  3325mm 2  33cm 2 use 10 20
Under Column B
b
1.44  1
m  1.22m  1220mm
2
975
0.633  1.22  0.3 

M

  84.6
(1.22 * 0.633)
2 
2

b  633mm, d  665mm
2

25 
2 106 * 84.6
  0.85 *
1 - 1    min
2
420 
0.90.8525 * 665 * 633 
AS min  0.0018  750  633  854.6mm 2  8.6cm 2 use 614
Shrinkage Reinforcement in short direction
AS min  0.0018 1000  750  1350mm 2  13.5cm 2 use 914 / m
Reinforcement details
Example 3 (Strip footing)
Design a combined footing As shown
qall ( net )  20t / m 2  200kPa
f c  25 N mm 2
f y  420 N mm 2
Dimension calculation
The base dimension to get uniform distributed load
800 kN
1280 kN
3040kN
960 kN
A
Assume
L1=0.6
x1=5.2m
x
x2=10 7m
x2=10.7m
800(0.6)+1280(5.1)+960(10.6)=
800(0
6)+1280(5 1)+960(10 6)
3040 (x)
x = 5.65m,
2(x)=11.3m
L2=11.3 - (10.6)=0.7
L2
qall ( net )  188t / m 2  180
80kkPaa,
Ag 
Ps
qall ( net )
3040 103
2

16
.
9
m
 11.3m 1.8m

3
180 10
Pu 1.33040 103
qu 

 195 103 Pa  195kPa
A
11.3 1.8
Check for punching Shear
h = 700 mm
d=630mm
Example
B
bo  4(630  400))  4120mm
fc '
25
 VC  
bo d  0.75 
 630  4120 / 1000  3244.5kN
3
3
 s d  fc '
40  630 
25


 VC    2 
bo d  0.75 2 
 630  4120 / 1000  6584kN


b
12
4120
12




VU  1280(1.3)  1.032 *195  1457.1kN  Vc
You can check other columns
oK
Draw S.F.D & B.M.D
Stress under footing
= 195 *1.8 = 351 kN/m
Check for beam shear
b = 1800mm, d = 630mm
25
 VC  0.75 
 630  1800 / 1000  708.75kN
6
Max.  VU at d from column face  0.7(1009)  706.3kN
VU  VC
Bending moment Long
direction
 ve M  1366kN .m
b  1800mm, d  730mm

25 
2 106 *1365
  0.85 *
1 - 1   0.0053
2
420 
0.90.8525 * 630 *1800 
AS  0.0053  630 1000  3362mm 2  33.6cm 2 use 9 22 / m Top
 ve M  246.7 kN .m
b  1800mm, d  730mm

25 
2 106 * 81
  0.85 *
1 - 1    mini
2
420 
0.90.8525 * 730 *1800 
AS min  0.0018  700 1000  1260mm 2  12.6cm 2 use 814 / m Bottom
Design Short
direction as example 1 (lecture 11)
Reinforcement details
Mat Foundation
The image part with relationship ID rId5 was not found in the file.
The image part with relationship ID rId6 was not found in the file.
Check for punching Shear
Rf 2
G
General
l Example,
E
l Ref.
Modified load
General reinforcement details