Lecture 9 – Footings - Icivil-hu.com

Reinforced Concrete II Hashemite University

Dr. Hazim Dwairi 1

The Hashemite University

Department of Civil Engineering

Lecture Lecture 9 9 –– FootingsFootings

Dr Hazim DwairiDr Hazim Dwairi

Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University

Dr. Hazim DwairiDr. Hazim Dwairi

Footings DefinitionFootings Definition

•• Footings are structural members used to support Footings are structural members used to support columns and walls to transmit and distribute theircolumns and walls to transmit and distribute theircolumns and walls to transmit and distribute their columns and walls to transmit and distribute their loads to the soil in such a way that the load loads to the soil in such a way that the load bearing capacity of the soil is not exceeded, bearing capacity of the soil is not exceeded, excessive settlement, differential settlement, or excessive settlement, differential settlement, or rotation are prevented and adequate safety rotation are prevented and adequate safety against overturning or sliding is maintained.against overturning or sliding is maintained.

Reinforced Concrete IIReinforced Concrete II

g g gg g g•• Since the soil is generally weaker than concrete Since the soil is generally weaker than concrete

columns and walls, the contact area between the columns and walls, the contact area between the footing and the soil is much larger than between footing and the soil is much larger than between the supported members and footing.the supported members and footing.

Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University


Dr. Hazim Dwairi 2

Types of FootingsTypes of Footings

a)a) Strip or Wall FootingStrip or Wall Footingb)b) S d F ti (Si l I l t d)S d F ti (Si l I l t d)b)b) Spread Footing (Single or Isolated)Spread Footing (Single or Isolated)



c)c) Spread Footing (Stepped)Spread Footing (Stepped)d)d) S d F ti (T d)S d F ti (T d)d)d) Spread Footing (Tapered)Spread Footing (Tapered)



Dr. Hazim Dwairi 3


e)e) Pile CapPile Capf)f) C bi d F tiC bi d F tif)f) Combined FootingCombined Footing



g)g) Cantilever or Strap FootingCantilever or Strap Footing



Dr. Hazim Dwairi 4


h)h) Mat or Raft FootingMat or Raft Footing


Distribution of Soil PressureDistribution of Soil Pressure

•• When the column load When the column load PP is applied on the is applied on the ppppcentroidcentroid of the footing, a of the footing, a uniform pressure is uniform pressure is assumed to develop on assumed to develop on the soil surface below the soil surface below the footing area.the footing area.


•• However the actual distribution of the soil is not However the actual distribution of the soil is not uniform, but depends on may factors especially uniform, but depends on may factors especially the composition of the soil and degree of the composition of the soil and degree of flexibility of the footing.flexibility of the footing.


Dr. Hazim Dwairi 5

Distribution of Soil PressureDistribution of Soil Pressure


Soil pressure distribution in

cohesionless soil. (Clay)

Soil pressure distribution in cohesive

soil. (Sand)

Design ConsiderationsDesign Considerations

Footings must be designed to carry the column Footings must be designed to carry the column loads and transmit them to the soil safely whileloads and transmit them to the soil safely whileloads and transmit them to the soil safely while loads and transmit them to the soil safely while satisfying code limitations.satisfying code limitations.1.1. The area of the footing based on the allowable The area of the footing based on the allowable

bearing soil capacity bearing soil capacity 2.2. TwoTwo--way shear or punch out shear.way shear or punch out shear.33 OO hh


3.3. OneOne--way shear way shear 4.4. Bending moment and steel reinforcement Bending moment and steel reinforcement

requiredrequired



Dr. Hazim Dwairi 6

Design ConsiderationsDesign Considerations

Footings must be designed to carry the column Footings must be designed to carry the column loads and transmit them to the soil safely whileloads and transmit them to the soil safely whileloads and transmit them to the soil safely while loads and transmit them to the soil safely while satisfying code limitations.satisfying code limitations.1.1. Bearing capacity of columns at their baseBearing capacity of columns at their base2.2. Dowel requirementsDowel requirements3.3. Development length of barsDevelopment length of bars


4.4. Differential settlementDifferential settlement


Eccentrically Loaded FootingEccentrically Loaded Footing

Locate column within the Kern of footing to prevent developing tensiondeveloping tension stresses in the soil underneath the footing



Dr. Hazim Dwairi 7

Gross and Net Bearing CapacityGross and Net Bearing Capacity

0.6m 10.8kN/m2

0.6m

γs=18kN/m3

γc=25kN/m3

15.0kN/m2

25.8kN/m2

10 8kN/m2

footingsgrossnet WWqq −−=soil ofweight =sW


10.8kN/m

15.0kN/m2

25.8kN/m2

footing ofweight =footingW

Size of FootingSize of Footing

•• The area of footing can be determined from the The area of footing can be determined from the actual external loads such that the allowable soilactual external loads such that the allowable soilactual external loads such that the allowable soil actual external loads such that the allowable soil pressure is not exceeded.pressure is not exceeded.

( )pressure soil allowable Gross

selfweight including load Total footing of Area =

q

Service Load


3.0 to2.5 of range in thesafety ofFactor FS =

=FSq q ult

all


Dr. Hazim Dwairi 8

Design for TwoDesign for Two--way Shearway Shear

•• For Slabs and footings, For Slabs and footings, VVcc is the smallest of a, b is the smallest of a, b and c:and c:and c:and c:


Where:bo = perimeter of critical sectionβ = ratio of long side of column to short side < 2αs = 40 for interior columns, 30 for edge columns and 20 for corner columns.

Design for TwoDesign for Two--way Shearway Shear

The shear force The shear force VVuu acts at a acts at a section that has a lengthsection that has a lengthsection that has a length section that has a length bb00 = = 44((c+dc+d) ) or or 22(c(c11+d) + +d) + 22(c(c22+d) +d) where where d d is the effective depthis the effective depththe section is subjected to a the section is subjected to a vertical downward load vertical downward load PPuu and and vertical upward pressurevertical upward pressure qq


vertical upward pressurevertical upward pressure qquu..


2)( dcqPV uuu +−=For Square column:

For Rectangular column:))(( 21 dcdcqPV uuu ++−=


Dr. Hazim Dwairi 9

Design of OneDesign of One--way Shearway Shear

For footings with bending For footings with bending action in one direction theaction in one direction theaction in one direction the action in one direction the critical section is located a critical section is located a distance distance dd from face of from face of column. column.

bdfV cc'17.0=

b

L

)( 1 dcLbVL-Direction


b

c1

c2)22

( 1 dbqV uu −−=

LdfV cc'17.0=

)22

( 2 dcbLqV uu −−=b-Direction

d

Flexural Strength and Footing Flexural Strength and Footing reinforcement reinforcement

The bending moment in The bending moment in each direction of the footing each direction of the footing must be checked and the must be checked and the appropriate reinforcement appropriate reinforcement must be provided.must be provided.



Dr. Hazim Dwairi 10


•• MMuu = = 11//8 8 x x qquu x (L x (L –– c)c)22

•• Minimum area of steelMinimum area of steel

2y

us

⎟⎠⎞

⎜⎝⎛ −

=adf

MAφ


–– GradeGrade--40 40 AAs,mins,min = = 00..002002bhbh–– GradeGrade--60 60 AAs,mins,min = = 00..00180018bhbh–– Maximum spacing, Maximum spacing, SS, is the , is the

smallest of smallest of ((33h or h or 450450mm)mm)Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University


•• The reinforcement must be The reinforcement must be distributed across the entiredistributed across the entiredistributed across the entire distributed across the entire width of the footing.width of the footing.

footing of sideshort footing of side long

=β


12

directionshort in ent reinforcem Total widthbandin ent Reinforcem

+=β


Dr. Hazim Dwairi 11

Bearing Capacity of Column at Bearing Capacity of Column at BaseBase

•• The column applies a concentrated load on The column applies a concentrated load on footing This load is transmitted by bearingfooting This load is transmitted by bearingfooting. This load is transmitted by bearing footing. This load is transmitted by bearing stresses in the concrete and the stresses in the stresses in the concrete and the stresses in the dowels crossing the joint.dowels crossing the joint.

•• Maximum bearing load on the concrete is given Maximum bearing load on the concrete is given as: (ACI sec. as: (ACI sec. 1010..1717))


( ) ( )bearingfor 0.65

70.185.0 1'

1

21

'

=

≤=

φ

φφ AfAAAfN cc

Bearing Capacity of Column at Bearing Capacity of Column at BaseBase



Dr. Hazim Dwairi 12

Dowels ReinforcementDowels Reinforcement

•• A minimum steel ratio A minimum steel ratio ρρ = = 00..005 005 of the column of the column sectionsection as compared toas compared to ρρ == 00 0101 as minimumas minimumsectionsection as compared to as compared to ρρ = = 00..01 01 as minimum as minimum reinforcement for the column itself. The number reinforcement for the column itself. The number of dowel bars needed is four which may be of dowel bars needed is four which may be placed at the four corners of the column. The placed at the four corners of the column. The dowel bars are usually extended into the footing, dowel bars are usually extended into the footing, bent at the ends, and tied to the main footing bent at the ends, and tied to the main footing


, g, greinforcement. The dowel diameter shall not reinforcement. The dowel diameter shall not exceed the diameter of the longitudinal bars in exceed the diameter of the longitudinal bars in the column by more than the column by more than 44..0 0 mm.mm.


Dowels ReinforcementDowels Reinforcement

Compression

Larger of (Compression development length for #8 bar OR compression Lap-splice for #7 bar)


'

24.0

c

bydb

f

dfl =

development length for #7 bars


Dr. Hazim Dwairi 13

Example Example 1 1 –– Wall FootingWall Footing

Design a plain concrete footing to support a Design a plain concrete footing to support a 400 400 mm thick concrete wall The load on the wallmm thick concrete wall The load on the wallmm thick concrete wall. The load on the wall mm thick concrete wall. The load on the wall consist of consist of 230 230 kNkN/m dead load (including the self/m dead load (including the self--weight of wall) and a weight of wall) and a 146 146 kNkN/m live load. The base /m live load. The base of the footing is of the footing is 1200 1200 mm below final grade. mm below final grade. ffcc’ = ’ = 21 21 MPaMPa, , ffyy = = 420 420 MPaMPa, the gross allowable soil , the gross allowable soil pressure = pressure = 240 240 kNkN/m/m2 2 , and the soil density is , and the soil density is 18 18


pp , y, ykNkN/m/m33..



1)1) Estimate the size of the footing and the Estimate the size of the footing and the factored net pressurefactored net pressurefactored net pressure.factored net pressure.

2/5.12)5.0)(25(50040025.1

thicknesswall1.5)~(1footing ofdepth assume

mkNhWmm

cFooting ===

=×=×=

γ


( ) 2

2

/9.2146.125.12240

/6.12)5.02.1)(18(

mkNq

mkNhW

netall

sssoil

=−−=

=−== γ


Dr. Hazim Dwairi 14


:loadservice totalCalculateLLDLPservice +=

( ) 80.175.1376:strip) m-1(consider width footing Estimate

376 146230

mbUSEmPb service

service

=⇒===

=+=


( )2/1.283

80.11466.12302.1

80.1 75.19.214

mkNq

mbUSEmq

b

u

netall

=×+×

=

⇒


2)2) Check oneCheck one--way shearway shear

!0.2415.41210002117.075.0

40.810.14125.02

40.080.11.283@

5.4125.1275500 Estimate

OKAYVVkNV

kNmdV

mmd

c

u

=××××=

=×⎟⎠⎞

⎜⎝⎛ −

−=

=−−=

φφ


!OKAYVV cu ⇒<φ


Dr. Hazim Dwairi 15


3)3) Design flexural reinforcementDesign flexural reinforcement

AA

adfAM

mkNM

s

ysn

u

4205.4124209.0104.69

)2/(

.40.692

40.080.12

1.283

6

2

φφ

⎟⎞

⎜⎛ ×

−×××=×

−=

=⎟⎠⎞

⎜⎝⎛ −

×=


mmmUSEmmAmmA

A

s

s

200@145 OR /145 5.7425.41210000018.0;451

10002185.05.4124209.0104.69

2min

2

φφ=××==⇒

⎟⎠

⎜⎝ ××

××××


4)4) Check development lengthCheck development lengthdf 1442099

USE

mmlmml

mmf

dfl

d

c

byd

HOOK90

7002

40018001155

11552110

14420910

9

0

'

⇒

=−

=>=

=×

××==


mmmmf

dfl

USE

c

bydh 700308

211442024.024.0

HOOK 90

'<=

××==

⇒


Dr. Hazim Dwairi 16


700 mm

500 mm168 mm


1800 mm

φ14@200mm φ14@200mm

Example Example 2 2 –– Single FootingSingle Footing

Design a square footing to support a Design a square footing to support a 450 450 mmmm--square tied interior column reinforced withsquare tied interior column reinforced with 88φφ2525square tied interior column reinforced with square tied interior column reinforced with 88φφ25 25 bars. The column carries an bars. The column carries an unfactoredunfactored axial axial dead load of dead load of 1000 1000 kNkN and an axial live load of and an axial live load of 900 900 kN.kN. The base of the footing is The base of the footing is 1200 1200 mm mm below final grade and allowable soil pressure is below final grade and allowable soil pressure is 240 240 kNkN/m/m22. Use . Use ffcc’ = ’ = 28 28 MPaMPa and and ffyy = = 420 420 MPaMPa..


cc yy



Dr. Hazim Dwairi 17


1)1) Estimate the footing size and the factored net Estimate the footing size and the factored net soil pressuresoil pressuresoil pressure.soil pressure.

2

2

PS ilAll bl(N t)Eff ti/8.1018)6.02.1(

/15256.0600mmhdepth footing Assume

mkNW

mkNW

soil

footing

=×−=

=×=

=


( ) 2/2.2148.1015240:PressureSoilAllowable(Net) Effective

mkNq netall =−−=


Load Service LLDLP +==

287.8214.21900footing of Area

2640)900(6.1)1000(2.1 6.12.1 Load Factored

19009001000

m

kNLLDLP

kN

u

==

=+=+==

=+=


2/3.29333

26400.30.3

214.2

mkNq

mmUSE

u =×

=

×


Dr. Hazim Dwairi 18


2)2) Check thickness for twoCheck thickness for two--way shearway shear

⎪⎪⎪⎧

⎟⎞

⎜⎛

=⎟⎟⎠

⎞⎜⎜⎝

⎛+

=+==×−−=

d

dbfdbf

mmbPerimetermmdAverage

ococ

o

'' 51.02117.0

3750)5.487450(4 5.487255.175600

αβ


⎪⎪⎪

⎩

⎪⎨ =⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

dbf

dbfdbfb

dofmallestV

oc

ococo

sc

'

''

33.0

59.02083.0 s α


( ) 3.23829375.02.2932640

2.23945.48737502833.075.02

u

c

kNV

kNVφ

=×−=

=××××=

3)3) Check oneCheck one--way shearway shear

( )OKAYshear way -Two cu

u

VV φ<

9.6923)4875.0245.0

23(3.293u kNV =×−−×=


OKAYshear way -One 7.9865.48730002817.075.0

cu

c

VVkNV

φφ

<=××××=


Dr. Hazim Dwairi 19


4)4) Design the flexural reinforcementDesign the flexural reinforcement:column of edge at themoment ultimate Calculate

A

AA

mkNM

ss

u

632530002885.02

4205.4874209.0109.1121

.9.112132

45.00.33.293

2

6 ⎟⎠⎞

⎜⎝⎛

××××

−×××=×

=×⎟⎠⎞

⎜⎝⎛ −

×=


mmsmmUSE

mmA

mmA

s,

s

5.237113

7523000 ; 63702513

324060030000018.0

6325

2

2min

2

=−×−

==

=××=

=⇒

φ


5)5) Design columnDesign column--footing connectionfooting connection:StressBearingCheck

( )kNN

mA

mA

=×××==++=

=×=

1238

281,6203.02865.07.1123.82.145.02.1

203.045.045.0:Stress BearingCheck

1

222

21


OKAYkNNkNPkNN

NkNN

u ⇒=<==⇒

>=×××=

62812640 281,6

865,19203.0123.8)203.0(2865.085.0 12


Dr. Hazim Dwairi 20


N-P 60i dd lfA

:PN caseIn u>

φy

u

mmUSEmm

.;f

N P

dowels oflength t Developmen1256204

1013)450450(005.0dowels of Area

650required dowels of Area

2=

=×>

==∴

φ

φφ


yb

c

ybdc fd

f

fdl 044.0

24.0'

>=


42025240

37042020044.038128

4202024.0 20 mmmmlfor dc

××

=××>=××

=→φ

4503000

17862810

25420910

9:25ent reinforcem footingfor length t Developmen

46242025044.047628

4202524.0 25

'mm

f

dfl

mmmmlfor

c

byd

dc

−

=×

××==

=××>=××

=→

φ

φ


OKAY 120047628

4202524.0:Hook 90 Provide

17861200752

4503000

o

mmmml

mmmmlAvailable

dh <=××

=

<=−=


Dr. Hazim Dwairi 21


8φ25

450 mm

476 mm

381 mm

13φ254φ20


3000 mm

600 mm

476 mm

381 mm

Example Example 3 3 –– Combined FootingCombined Footing

Design a rectangular footing to support two square Design a rectangular footing to support two square columns The exterior column (I) has a sectioncolumns The exterior column (I) has a sectioncolumns. The exterior column (I) has a section columns. The exterior column (I) has a section 400 400 x x 400 400 mm, which carries DL of mm, which carries DL of 800 800 kNkN and a and a LL of LL of 530 530 kN.kN. The interior column (II) has a section The interior column (II) has a section of of 500 500 x x 500500mm, which carries a DL of mm, which carries a DL of 1100 1100 kNkNand a LL of and a LL of 620 620 kN.kN. The base of the footing is The base of the footing is 1500 1500 mm below final grade and allowable soil mm below final grade and allowable soil


ggpressure is pressure is 240 240 kNkN/m/m22. Use . Use ffcc’ = ’ = 28 28 MPaMPa and and ffyy = = 420 420 MPaMPa. The distance between column is . The distance between column is 55..0 0 m m center to center (cc).center to center (cc).



Dr. Hazim Dwairi 22


DL = 800 kNLL = 530 kN

DL = 1100 kNLL = 620 kN

DL = 1900 kNLL = 1150 kN

LL = 530 kN LL = 620 kN

5000 mm2820 mm 2820 mm

3250 mm3250 mm


mmlUSEmml

6500 54002002005000min

==++=


412011506.119002.1305011501900

kNPkNP

u

service

=×+×==+=

( )2

2

8114

81.14206

3050footing of Area

/2065.0181252401000 footing ofdepth Assume

m

mkNqmmh

netall

u

==

=×−×−=

=


2/2545.25.6

4120

5.2 3.25.681.14

mkNq

mBUSEmB

u =×

=

=⇒==


Dr. Hazim Dwairi 23


6500 mm

2500 mm

2820

3250 3250

1080

1300 1400

1400


500400230 8204550


)31081(254)5306180021(:Icolumn for shear way -Two90025751000 Average

V

mmd

×××+×=

=−−=

:IIcolumn for shear way -TwoOK! 428490047602833.075.0

4760)1300210802(1451

)3.108.1(254)5306.18002.1(

uc

o

u

VkNV

mmbkN

V

>=××××=

=×+×==

××−×+×=

φ

Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite UniversityOK! 504090056002833.075.0

5600)900500(41814

)9.05.0(254)6206.111002.1( 2

uc

o

u

VkNV

mmbkN

V

>=××××=

=+×==

+×−×+×=

φ


Dr. Hazim Dwairi 24

Example Example 3 3 –– Combined FootingCombined Footing1808 kN 2312 kN

633.8 kN/m2

272.53

1535.47

1633.53

678.47

2.42 mVu, kN


535

58.6 362.8

1799.3

Mu, kN.m

16.8 103


2250250090028170750

7.904)9.025.0(8.63353.1633@:shearway -One

φ >=××××=

=+×−=

VkNV

kNdVu

25002885.04209004209.0101800

.1800!!

225025009002817.075.0

6

φ

⎟⎠⎞

⎜⎝⎛

×××

−×××=×

=

>=××××=

+

AA

mkNMOKAY

VkNV

ss

veu

uc

Reinforced Concrete IIReinforced Concrete IIDr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University223111700100065000018.0

22124500100025000018.0

5702 2215 USE

5406

2min,

2min,

2

2

φ

φ

φ

⇒=××=

⇒=××=

=⇒

=⇒

mmA

mmA

mmA

mmA

s

s

s

s


Dr. Hazim Dwairi 25


230 400 4550 500 820

15φ22


12φ2231φ22


•• Things still need to be checked:Things still need to be checked:√√√√ Cases of loadingCases of loading√√ Bearing stresses under columnsBearing stresses under columns√√ Development lengthDevelopment length√√ Dowel bars if neededDowel bars if needed


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