Footing Design
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Transcript of Footing Design
F i D iFooting Design
Types of FootingTypes of Footing
Wall footings are used to support structural walls that carry loads for other floors or to support nonstructural walls.walls.
Types of FootingTypes of Footing
Isolated or single footingsare used to support single columns. This is one of the most economical types of footings and is used whenfootings and is used when columns are spaced at relatively long distances.
Types of FootingTypes of Footing
Combined footings usually support two columns, or three columns not in a row. Combined footings are used when two columns are sowhen two columns are so close that single footings cannot be used or when one col mn is located at or nearcolumn is located at or near a property line.
Types of FootingTypes of Footing
Cantilever or strap footingsconsist of two single footings connected with a beam or a strap and support two single columns. Thistwo single columns. This type replaces a combined footing and is more economicaleconomical.
Types of FootingTypes of Footing
Continuous footingssupport a row of three or more columns. They have limited width and continue under all columns.under all columns.
Types of FootingTypes of Footing
Rafted or mat foundationconsists of one footing usually placed under theusually placed under the entire building area. They are used, when soil bearing
it i l lcapacity is low, column loads are heavy single footings cannot be used, piles are not used and differential settlement must be reduced.be reduced.
Types of FootingTypes of Footing
Pile caps are thick slabs used to tie a group of piles together to support and transmit column loads to the piles.piles.
Distribution of Soil PressureDistribution of Soil Pressure
When the column load P is applied on the centroid of the footing, a
if i duniform pressure is assumed to develop on the soil surface below the footing area However the actualfooting area. However the actual distribution of the soil is not uniform, but depends on may f p yfactors especially the composition of the soil and degree of flexibility of the footing.
Distribution of Soil PressureDistribution of Soil Pressure
Soil pressure distribution in Soil pressure distribution in cohesiveSoil pressure distribution in cohesionless soil.
Soil pressure distribution in cohesive soil.
Eccentrically loaded footingsEccentrically loaded footings
Eccentrically loaded footings ExampleEccentrically loaded footings ExampleIsolated Footing M
D.L = 900 kNL.L = 450 kNMs = 150kN.mMu=200kN.mqall =200kpa
Area required approximated
23
2)(
75610)450900(
200/20
mPA
kPamtq
s
netall
=×+
==
==
( )
3)(
5.25.3
75.610200
mmuse
mq
Anetall
g
×
=×
==
183 7
412
5.25.3
2
9.8
75.83 mI
mA
==
=×
183.7124.7
Me L 58301110150 5.3 ==<===
Check stress
kPaICM
AP
Pe
ss 7.18398
150758
1350
583.0111.01350
25.3
66
=×
+=+
==<===
kPaICM
AP
IAss 7.124
9.8150
75.81350
9.875.825.3
=×
−=−
Ultimate pressure under footingp g( )
mkNMkNP
u
u
.2001800)450(6.19002.1
==+=
CMP
kPaICM
AP uu
2001800
2459.8
20075.8
1800
53
25.3
≈×
+=+
kPaICM
AP uu 166
9.8200
75.81800 2
5.3
≈×
−=−
245166
245
Check Punching Shearg
[ ]tiit bll thF
3720)400530(4iV
cmbo =+=φ
5.24641000/372053032575.0
3'
:equation suitablecolumn the squareFor
kNdbf
V
isV
oc
C
C
=×××==φφ
φ
( ) 17985.25.3
33
4)245166245166( kNVU =×≈ +++
Check Beam Shear
13.8281000/250053062575.0 C kNVφ =×××=
245223
facecolumn from at U dV
⎞⎛ +
⇒
( )
5.567
5.2*53.05.1*2
245223U
kN
V
=
−⎟⎠⎞
⎜⎝⎛ +
=
CU VV φ<
Bending moment design3.5m
facecolumn from dat MU ⇒
g g
Long direction245
166
2657)1(7563)750(25791)1()750(75.63)5.2)(5.1)(211245(
25.791)5.2)(5.1(211
21
2
1
kNPPMkNP
kNPU
=−=
== 245
657 2*10225
305d ,2500b.2.657)1(75.63)75.0(25.791)1()75.0(
6
21
mmmkNPPMU
⎤⎡ ×
===+=+=
( ) direction long /167 5.131346100053000254.0
0.00254 2500*530*250.859.0
657.2102-1-1 42025*85.0
22
2
musecmmmAS φ
ρ
≈=××=
=⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×=
gS φ
Bending moment design 2 5m
facecolumnfromdatM ⇒
g g
Short direction
245
2.5m
( ) ( ) .25.7195.05.312
166245facecolumn from dat
mkNM
M
U
U
=⎟⎠⎞
⎜⎝⎛ ×
+=
⇒ 245
245
166
( ) 0.002 3500*530*250 8590
719.25*102-1-1 42025*85.0
305d ,3500b
2
6
mm
=⎥⎥⎤
⎢⎢⎡ ×
=
==
ρ ( ) direction short 1.1010601000530002.0
3500*530*250.859.042022
2
cmmmAS ≈=××=
⎥⎦⎢⎣ρ
Central band ratio =Central band ratio =
1.42.53.5
==β
0.832.42
12
==+β
Central band of short direction = 0.83 As = 0.83 (10.1)=8.6cm2
14/m7φ 142φ 142φ
16/m7φφ
Footing DesignFooting Design Part II
C bi d f tiCombined footing
Example 1Example 1Design a combined footing As shown
kPamtq netall 200/20 2)( == 225 mmNfc =′
2420 mmNf y =
Dimension calculationTh b di i if di ib d l dThe base dimension to get uniform distributed load
800 kN 1200 kN
2000kN
1200 kN
A
x2=6.2mx1=0.2m
x
A
x 800(0.2)+1200(6.2)=2000(x)x = 3.8m
800 kN 1200 kN
Try thickness
2x =7.6 m
Try thickness =80cm
Area required
kNPP
kPamtq netall
2600)2000(31)(31
,200/20 2)( ==
Area required
mq
PA
kNPP
sg
su
8.1*6.71010200102000
2600)2000(3.1)(3.1
23
3
≈=××
==
===
( ) kPaPaAPq
q
uu
netall
1901019081*67102600
10200
33
)(
=×=×
==
×
A 8.1*6.7
Check for punching Sheard = 730 mm
A
1.13m
0.765
22601130)765(2 mmbo =+=
A0.765
2573030'
3.20621000/226073032575.0
3'
fd
kNdbf
V oc
C
α
φφ
⎞⎛ ×⎞⎛
=×××==
K8875190*7650*131)31(800
60271000/22607301225
226073030275.0
122
VkNV
kNdbf
bdV o
csC
φ
αφφ =×××⎠⎞
⎜⎝⎛ ×
+=⎠⎞
⎜⎝⎛ +=
oK 8.875190*765.0*13.1)3.1(800 VkNV cU φ<=−=
B
[ ]25'
4520)400730(4
f
mmbo =+=
B
5.133221000/45207301225
452073040275.0
12'
2
4.41241000/452073032575.0
3
kNdbf
bdV
kNdbf
V
ocs
C
oc
C
αφφ
φφ
=×××⎠⎞
⎜⎝⎛ ×
+=⎠⎞
⎜⎝⎛ +=
=×××==
oK 4.1317190*13.1*13.1)3.1(1200
12452012
V kNV
b
cU
oC
φ
φφ
<=−=
⎠⎜⎝⎠
⎜⎝
Draw S.F.D & B.M.D
Stress under footing f g= 190 *1.8 = 342 kN/m
Check for beam shearb = 1800mm, d = 730mm
25.8211000/180073062575.0 C kNVφ =×××=
34.762facecolumn from at .6
CU
U
VV
kNdVMax
φ<
=→
Bending moment Long direction
307d ,8001b.1366
mmmmmkNMve
===−
( ) 0.0039 1800*730*250.859.0
1366*102-1-1 42025*85.0 2
6
ρ =⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×=
Top /209 5.28284710007300039.0 22 musecmmmAS φ==××=
307d ,8001b.7.246
mmmmmkNMve
⎤⎡
===+
( )B tt/1674141440100080000180
0.0007 1800*730*250.859.0
246.7*102-1-1 42025*85.0
22
min2
6
A φ
ρρ <=⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×=
Bottom /167 4.14144010008000018.0 22min musecmmmAS φ==××=
Bending moment Short direction
2⎞⎛
Under Column A
307d ,765b
6.1412
4.08.12765.0
)765.0*8.1(1040
2
mmmm
M
==
=⎟⎠⎞
⎜⎝⎛ −
×=
( ) 765*730*250.859.0
141.6*102-1-1 42025*85.0
,
min2
6
ρρ <⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×=
/147 116.11017658000018.0 22min musecmmmAS φ==××=
Under Column B
33.2122
4.08.1213.1
)13.1*8.1(1560
2
M =⎟⎠⎞
⎜⎝⎛ −
×=
Under Column B
212.33*1021125*850
307d ,1130b22)13.18.1(
6
mmmm
ρρ <⎥⎤
⎢⎡ ×
=
==⎠⎝
( ) /147 116.11017658000018.0
1130*730*250.859.0
-1-1 420
85.0
22min
min2
musecmmmAS φ
ρρ
==××=
<⎥⎥⎦⎢
⎢⎣
=
Shrinkage Reinforcement in short direction
/147 116.11017658000018.0 22min musecmmmAS φ==××=minS φ
Footing Design g gPart III
Combined footing, strip footing, & Mat foundationCombined footing, strip footing, & Mat foundation
Example 2Example 2Design a combined footing As shown
kPamtq netall 180/ 18 2)( == 225 mmNfc =′
2420 mmNf y =
Dimension calculationThe base dimension to get uniform distributed load
1200 kN 750 kN
1950kN
750(4 2)+1200(0 2)=1950 (x)
x1=0.2mx2=4.2
A
750(4.2)+1200(0.2) 1950 (x)x = 1.75m
mx
32 21 LBBBBx ⎟
⎠
⎞⎜⎜⎝
⎛++
=321 BB ⎠
⎜⎝ +
kPamtq netall ,200/20 2)( ==
Area required
BB
mq
PAnetall
sg 8.10
10180101950 2
3
3
)(
⎞⎛
=××
==
BB
LBB
810354
8.102
21
21
⎞⎜⎛ +
=⎟⎠⎞
⎜⎝⎛ +
BB 5.22
8.1035.42
21
21
=⎠⎞
⎜⎝⎛ +
=⎠⎞
⎜⎝⎛
BLBB
BB
35.452
52
221
21
⎞⎜⎛ +⎞
⎜⎛ +
=+⎠⎝
B
BLBBBBx
29.045.175.1335.4
55
32
2
2
21
21
+=⎠⎞
⎜⎝⎛ +
=⎟⎠
⎞⎜⎜⎝
⎛++
=
mBmB
41
1
2
== ( ) kPaPa
APq u
u 235102358.10
1019503.1 33
=×=×
==
Check for punching Shearh 750h= 750mmd = 732 mm
A B2 1
25901065)732(2 mmbo =+=
A
B1=4m
B2=1m
2566530'
4.21601000/259066532575.0
3'
fd
kNdbf
V oc
C
α
φφ
⎞⎛ ×⎞⎛
=×××==
K61376235*7330*0651)31(1200
52221000/25906651225
259066530275.0
122
VkNV
kNdbf
bdV o
csC
φ
αφφ =×××⎠⎞
⎜⎝⎛ ×
+=⎠⎞
⎜⎝⎛ +=
oK 6.1376235*733.0*065.1)3.1(1200 VkNV cU φ<=−=
BB
2231965)633(2 mmbo =+=
52731000/223166525665302750'
2
5.18541000/223166532575.0
3'
kNdbfdV
kNdbf
V
cs
oc
C
αφφ
φφ
⎞⎜⎛ ×
+⎞⎜⎛ +
=×××==
oK5896235*6330*9650)31(800
52731000/2231665122231
275.012
2
VkNV
kNdbf
bV o
csC
φ
φφ
<=−=
=×××⎠⎞
⎜⎝⎛ +=
⎠⎞
⎜⎝⎛ +=
oK 5.896235633.0965.0)3.1(800 VkNV cU φ<==
Draw S F D & B M DDraw S.F.D & B.M.D
Empirical S.F.D & B.M.D
68270.0975 =×
m
1092)70.0(1560 =×Convert trapezoidal load to rectangle
( ) mkNwlM
wave
. 11748
65.37058
705)235940(23522
max
32
===−
=−+= Mmax
88
Clear distance between columnB in moment design = ave. width = 2.5m
Check for beam sheard = 665mm
0.150.815at )(21 L
x
xyb+=
×+=d = 665mm
6961000/170066525750
17007.15.1)(21 35.4965.0
kNV
mmm
φ =×××=
==×+
668critical)most the( face Bcolumn from at .
6961000/17006656
75.0
U
C
VV
kNdVMax
kNV
φ
φ
<
=→
=×××=
CU VV φ<
Y=1.5m b
1mx
4m
Bending moment Long direction
307d.1260
mmmkNMve =−
260060.25.1)(21 35.425.2 ==×+= mb
Top
( ) 0.003 2600*665*250.859.0
1260*102-1-1 42025*85.0
307d
2
6
mm
ρ =⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×=
=
( ) Top /1601 2019951000665003.0 22 musecmmmAS φ==××=
⎥⎦⎢⎣
Bottom
Bottom /149 5.13135010007500018.0 22min musecmmmAS φ==××=
Bending moment Short direction
Under Column A
mmmb 35005.35.1)(21' 35.462.3 ==×+=
mmmb 144044.15.1)(21' 35.4633.0 ==×+=
2⎞⎛
mmmmb 375075.32
45.3==
+=
665d
6.5832
4.075.32733.0
)733.0*75.3(1560
2
mm
M
=
=⎟⎠⎞
⎜⎝⎛ −
×=
( ) 0.005 733*665*250.859.0
583.57*102-1-1 42025*85.0 2
6
ρ =⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×=
2001 333325733665005.0 22 φusecmmmAS ==××=
Under Column B
mmmmb 122022.12
144.1==
+=
665d633b
6.842
3.022.12633.0
)633.0*22.1(975
2
mmmm
M
==
=⎟⎠⎞
⎜⎝⎛ −
×=
( ) 633*665*250.859.0
84.6*102-1-1 42025*85.0
665d ,633b
min2
6
ρρ
mmmm
<⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×=
146 6.86.8546337500018.0 22min φusecmmmAS ==××=
⎦⎣
Shrinkage Reinforcement in short directionShrinkage Reinforcement in short direction
/149 5.13135075010000018.0 22min musecmmmAS φ==××=
Reinforcement details
Example 3 (Strip footing)Example 3 (Strip footing)Design a combined footing As shown
kPamtq netall 200/20 2)( == 225 mmNfc =′
2420 mmNf y =
Dimension calculationThe base dimension to get uniform distributed load
800 kN 1280 kN
3040kN
960 kN
AssumeL1=0.6 x1=5.2m
A
x2=10 7m L2
x
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)=11.3mL2=11.3 - (10.6)=0.7
kPamtq netall ,180/18 2)( ==
mmmq
PA
k amtq
netall
sg
netall
8.13.119.1610180103040
,80/8
23
3
)(
)(
×≈=××
==
( ) kPaPaAPq u
u 195101958.13.111030403.1 3
3
=×=××
==
Check for punching Shearh = 700 mmd=630mmd=630mm
B
Example
25'
4120))400630(4
f
mmbo =+=
B
65841000/41206302563040275.0'
2
5.32441000/412063032575.0
3
kNdbfdV
kNdbf
V
ocs
C
oc
C
αφφ
φφ
=×××⎠⎞
⎜⎝⎛ ×
+=⎠⎞
⎜⎝⎛ +=
=×××==
oK 1.1457195*03.1)3.1(1280
12412012
2 V kNV
b
cU
oC
φ
φφ
<=−=
⎠⎜⎝⎠
⎜⎝
You can check other columns
Draw S.F.D & B.M.D Stress under footing = 195 *1.8 = 351 kN/m
Check for beam shearb = 1800mm, d = 630mm
75.7081000/180063062575.0 C kNVφ =×××=
3.706)1009(7.0facecolumn from at .6
CU
U
VV
kNdVMax
φ<
=≈→
Bending moment Long direction
307d ,8001b.1366
mmmmmkNMve
===−
( ) 0.0053 1800*630*250.859.0
1365*102-1-1 42025*85.0 2
6
ρ =⎥⎥⎦
⎤
⎢⎢⎣
⎡ ×=
Top /229 6.33336210006300053.0 22 musecmmmAS φ==××=
.7.246 mkNMve =+
81*102-1-125*850
307d ,8001b
i
6
mmmm
ρρ <⎥⎤
⎢⎡ ×
=
==
( ) Bottom /148 6.12126010007000018.0
1800*730*250.859.0
11 420
85.0
22min
min2
musecmmmAS φ
ρρ
==××=
<⎥⎥⎦⎢
⎢⎣
=
Design Short direction as example 1 (lecture 11)
Reinforcement details
Mat Foundation
Check for punching Shear
G l E l R f 2General Example, Ref. 2
Modified load
General reinforcement details