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7/31/2019 Eng. Khalid Footing-Design (1)
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Footing Design
7/31/2019 Eng. Khalid Footing-Design (1)
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Types of Footing
Wall footings are used to
support structural walls that
carry loads for other floorsor to support nonstructural
walls.
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Isolated or single footings
are used to support single
columns. This is one of themost economical types of
footings and is used when
columns are spaced at
relatively long distances.
Types of Footing
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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
column is located at or near
a property line.
Types of Footing
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Cantilever or strap footings
consist of two single
footings connected with abeam or a strap and support
two single columns. This
type replaces a combined
footing and is more
economical.
Types of Footing
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Continuous footings
support a row of three or
more columns. They havelimited width and continue
under all columns.
Types of Footing
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Rafted or mat foundation
consists of one footing
usually placed under the
entire building area. Theyare 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.
Types ofFooting
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Pile caps are thick slabs
used to tie a group of piles
together to support andtransmit column loads to the
piles.
Types of Footing
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Distribution of Soil Pressure
When the column loadPis 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 mayfactors especially the composition of
the soil and degree of flexibility of
the footing.
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Distribution of Soil Pressure
Soil pressure distribution incohesionless soil.
Soil pressure distribution in cohesivesoil.
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Eccentrically loaded footings
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Eccentrically loaded footings Example
Isolated FootingD.L = 900 kN
L.L = 450 kN
Ms = 150kN.m
Mu=200kN.mqall=200kpa
M
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Area required approximated
4
125.25.3
2
2
3
3
)(
2
)(
9.875.8
5.25.3
75.610200
10)450900(
200/20
3
mI
mA
mmuse
mq
PA
kPamtq
netall
sg
netall
kPaI
CM
A
P
kPaI
CM
A
P
P
Me
ss
ss
L
7.1249.8
150
75.8
1350
7.1839.8
150
75.8
1350
583.0111.01350
150
2
5.3
25.3
65.36
Check stress
183.7
124.7
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Ultimate pressure under footing
kPaI
CM
A
P
kPaI
CM
A
P
mkNM
kNP
uu
uu
u
u
1669.8
200
75.8
1800
2459.8
200
75.8
1800
.200
1800)450(6.19002.1
2
5.3
25.3
245
245
166
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Check Punching Shear
17985.25.3
5.24641000/37205303
2575.0
3
'
:equationsuitablecolumn thesquareFor3720)400530(4
4
)245166245166(kNV
kNdbf
V
isVcmb
U
o
c
C
C
o
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Check Beam Shear
5.567
5.2*53.05.1*2
245223
facecolumnfromat
13.8281000/25005306
2575.0
CU
U
U
C
VV
kN
V
dV
kNV
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directionlong/1675.131346100053000254.0
0.002542500*530*250.859.0
657.2*102-1-1
420
25*85.0
305d,2500b
.2.657)1(75.63)75.0(25.791)1()75.0(
75.63)5.2)(5.1)(211245(
25.791)5.2)(5.1(211
facecolumnfromdat
22
2
6
21
21
2
1
musecmmmA
mm
mkNPPM
kNP
kNP
M
S
U
U
Bending moment design
Long direction245
245
166
3.5m
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directionshort1.1010601000530002.0
0.0023500*530*250.859.0
719.25*102-1-1
420
25*85.0
305d,3500b
.25.7195.05.312
166245
facecolumnfromdat
22
2
6
cmmmA
mm
mkNM
M
S
U
U
Bending moment design
Short direction
245
245
166
2.5m
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0.832.42
12
1.42.5
3.5
Central band of short direction = 0.83 As = 0.83 (10.1)=8.6cm2
Central band ratio =
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14/m7 142142
16/m7
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Footing Design
Part II
Combined footing
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Example 1
Design a combined footing As shown
kPamtq netall 200/202
)( 225 mmNfc
2420 mmNfy
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Dimension calculation
The base dimension to get uniform distributed load
x2=6.2mx1=0.2m
800 kN 1200 kN
x
2000kN
A
800(0.2)+1200(6.2)=2000(x)
x = 3.8m
2x =7.6 m
800 kN 1200 kN
Try thickness
=80cm
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kPaPa
A
Pq
mq
PA
kNPP
kPamtq
uu
netall
sg
su
netall
190101908.1*6.7
102600
8.1*6.71010200
102000
2600)2000(3.1)(3.1
,200/20
33
2
3
3
)(
2
)(
Area required
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Check for punching Sheard = 730 mm
oK8.875190*765.0*13.1)3.1(800
60271000/22607301225
226073030275.0
12'2
3.20621000/22607303
2575.0
3
'
22601130)765(2
VkNV
kNdbf
b
dV
kNdbf
V
mmb
cU
o
csC
o
c
C
o
A
1.13m
0.765
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oK4.1317190*13.1*13.1)3.1(1200
5.133221000/452073012
25
4520
73040275.0
12
'2
4.41241000/45207303
2575.0
3
'
4520)400730(4
VkNV
kNdbf
b
dV
kNdbf
V
mmb
cU
o
csC
o
c
C
o
B
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Draw S.F.D & B.M.D
Stress under footing
= 190 *1.8 = 342 kN/m
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34.762facecolumnfromat.
25.8211000/180073062575.0
CU
U
C
VV
kNdVMax
kNV
Check for beam shear
b = 1800mm, d = 730mm
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Top/2095.28284710007300039.0
0.00391800*730*250.859.0
1366*102-1-1
420
25*85.0
307d,8001b
.1366
22
2
6
musecmmmA
mmmm
mkNMve
S
Bending moment Long direction
Bottom/1674.14144010008000018.0
0.00071800*730*250.859.0
246.7*102-1-1
420
25*85.0
307d,8001b
.7.246
22
min
min2
6
musecmmmA
mmmm
mkNMve
S
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Bending moment Short direction
/147116.11017658000018.0
765*730*250.859.0
141.6*102-1-1
420
25*85.0
307d,765b
6.1412
4.08.1
2
765.0
)765.0*8.1(
1040
22
min
min2
6
2
musecmmmA
mmmm
M
S
Under Column A
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/147116.11017658000018.0
1130*730*250.859.0
212.33*102-1-1
420
25*85.0
307d,1130b
33.2122
4.08.1
2
13.1
)13.1*8.1(
1560
22
min
min2
6
2
musecmmmA
mmmm
M
S
Under Column B
Shrinkage Reinforcement in short direction
/147116.11017658000018.0
22
min musecmmmAS
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Footing Design
Part IIICombined footing, strip footing, & Mat foundation
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Example 2
Design a combined footing As shown
kPamtq netall 180/182
)( 225 mmNfc
2420 mmNfy
Di i l l i
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Dimension calculation
The base dimension to get uniform distributed load
x1=0.2mx2=4.2
m
1200 kN 750 kN
x
1950kN
A
750(4.2)+1200(0.2)=1950 (x)
x = 1.75m
3
2
21
21 L
BB
BBx
Area required
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mB
mB
B
BLBBBBx
BB
BB
BB
LBB
m
q
PA
kPamtq
netall
sg
netall
4
1
29.045.175.1
335.4
55
32
5
5.22
8.1035.4
2
8.102
8.10
10180
101950
,200/20
1
2
2
2
21
21
21
21
21
21
2
3
3
)(
2
)(
Area required
kPaPa
A
Pq uu 23510235
8.10
1019503.1 33
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Check for punching Shearh= 750mm
d = 732 mm
oK6.1376235*733.0*065.1)3.1(1200
52221000/25906651225
259066530275.0
12'2
4.21601000/25906653
2575.0
3
'
25901065)732(2
VkNV
kNdbfbdV
kNdbf
V
mmb
cU
o
csC
o
c
C
o
A
B1=4m
B2=1m
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B
oK5.896235*633.0*965.0)3.1(800
52731000/2231665
12
25
2231
66530275.0
12
'2
5.18541000/22316653
2575.0
3
'
2231965)633(2
VkNV
kNdbf
b
dV
kNdbf
V
mmb
cU
o
csC
o
c
C
o
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Draw S.F.D & B.M.D
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mkN
wlM
wave
.11748
65.3705
8
705)235940(235
22
max
32
Mmax
68270.0975
1092)70.0(1560
Empirical S.F.D & B.M.D
m
Convert trapezoidal load to rectangle
Clear distance between column
B in moment design = ave. width = 2.5m
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668critical)mostthe(faceBcolumnfromat.
6961000/17006656
2575.0
17007.15.1)(21
0.150.815at
)(21
35.4965.0
CU
U
C
Lx
VV
kNdVMax
kNV
mmm
x
yb
Check for beam shear
d = 665mm
Y=1.5m
1m
4m
x
b
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Top/16012019951000665003.0
0.0032600*665*250.859.0
1260*102-1-1
420
25*85.0
307d.1260
22
2
6
musecmmmA
mmmkNMve
S
Bending moment Long direction
Bottom/1495.13135010007500018.0 22min musecmmmAS
260060.25.1)(2135.425.2 mb
Top
Bottom
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Bending moment Short direction
2001333325733665005.0
0.005733*665*250.859.0
583.57*102-1-1
420
25*85.0
665d
6.5832
4.075.3
2
733.0
)733.0*75.3(
1560
22
2
6
2
usecmmmA
mm
M
S
Under Column A
mmmb 35005.35.1)(21' 35.4 62.3
mmmmb 375075.32
45.3
mmmb 144044.15.1)(21'35.4633.0
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1466.86.8546337500018.0
633*665*250.859.0
84.6*102-1-1
420
25*85.0
665d,633b
6.842
3.022.1
2
633.0
)633.0*22.1(
975
22
min
min2
6
2
usecmmmA
mmmm
M
S
Under Column B
Shrinkage Reinforcement in short direction
/1495.13135075010000018.0 22min musecmmmAS
mmmmb 122022.12
144.1
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Reinforcement details
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Example 3 (Strip footing)Design a combined footing As shown
kPamtq netall 200/202
)( 225 mmNfc
2420 mmNfy
Dimension calculation
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Dimension calculation
The base dimension to get uniform distributed load
Assume
L1=0.6 x1=5.2m
800 kN 1280 kN
x
3040kN
A
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
960 kN
x2=10.7mL2
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mmmq
PA
kPamtq
netall
s
g
netall
8.13.119.1610180
103040
,180/18
2
3
3
)(
2
)(
kPaPa
A
Pq uu 19510195
8.13.11
1030403.1 33
Check for punching Shear
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p gh = 700 mm
d=630mm
oK1.1457195*03.1)3.1(1280
65841000/412063012
25
4120
63040
275.012
'
2
5.32441000/41206303
2575.0
3
'
4120))400630(4
2 VkNV
kNdb
f
b
d
V
kNdbf
V
mmb
cU
o
cs
C
o
c
C
o
B
Example
You can check other columns
Draw S.F.D & B.M.D Stress under footing
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= 195 *1.8 = 351 kN/m
Check for beam shear
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3.706)1009(7.0facecolumnfromat.
75.7081000/18006306
2575.0
CU
U
C
VV
kNdVMax
kNV
Check for beam shear
b = 1800mm, d = 630mm
L di i
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Top/2296.33336210006300053.0
0.00531800*630*250.859.0
1365*102-1-1
420
25*85.0
307d,8001b
.1366
22
2
6
musecmmmA
mmmm
mkNMve
S
Bending moment Long direction
Bottom/1486.12126010007000018.0
1800*730*250.859.0
81*102
-1-1420
25
*85.0
307d,8001b
.7.246
22
min
min2
6
musecmmmA
mmmm
mkNMve
S
Design Short direction as example 1 (lecture 11)
R i f d il
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Reinforcement details
Mat Foundation
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Mat Foundation
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Check for punching Shear
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Modified load
General Example, Ref. 2
G l i f t d t il
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General reinforcement details