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.


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.


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.


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.


Continuous footingssupport a row of three or more columns. They have limited width and continue under all columns.under all columns.


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.


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


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<

=≈→


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

Documents

Footing Design