Lecture 9 Lecture 9 –– Footings Footings RC2...Example 1 Example 1 ––Wall Footing Wall...

Prestressed Concrete Hashemite University

Dr. Hazim Dwairi 1

The Hashemite University

Department of Civil Engineering

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

Lecture 9 Lecture 9 –– FootingsFootings

Dr. Hazim DwairiDr. Hazim Dwairi

Reinforced Concrete IIReinforced Concrete II

Footings DefinitionFootings Definition

•• Footings are structural members used to support Footings are structural members used to support columns 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.

•• 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) Spread Footing (Single or Isolated)Spread Footing (Single or Isolated)




c)c) Spread Footing (Stepped)Spread Footing (Stepped)d)d) Spread Footing (Tapered)Spread Footing (Tapered)



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e)e) Pile CapPile Capf)f) Combined FootingCombined Footing




g)g) Cantilever or Strap FootingCantilever or Strap Footing



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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 centroidcentroid 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 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.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 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 bars4.4. Differential settlementDifferential settlement



Eccentrically Loaded FootingEccentrically Loaded Footing


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


Dr. Hazim Dwairi 7


Gross and Net Bearing CapacityGross and Net Bearing Capacity


0.6m

0.6m

γγγγs=18kN/m3

γγγγc=25kN/m3

10.810.810.810.8kN/m2

15.015.015.015.0kN/m2

25.825.825.825.8kN/m2

10.810.810.810.8kN/m2

15.015.015.015.0kN/m2

25.825.825.825.8kN/m2

footingsgrossnet WWqq −−=

footing ofweight

soil ofweight

==

footing

s

W

W


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 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 =

3.0 to2.5 of

range in thesafety ofFactor FS=

=FS

q q ult

all

Service Load


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:


Where:bo = perimeter of critical sectionββββ = ratio of long side of column to short side < 2ααααs = 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 length section that has a length bb00 = 4(= 4(c+dc+d) ) or or 2(c2(c11+d) + 2(c+d) + 2(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 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 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

c1

c2)22

( 1 dcL

bqV uu −−=

LdfV cc'17.0=

)22

( 2 dcb

LqV uu −−=

L-Direction

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.



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•• MMuu = 1/8 x = 1/8 x qquu x (L x (L –– c)c)22

•• Minimum area of steelMinimum area of steel–– GradeGrade--40 40 AAs,mins,min = 0.002bh= 0.002bh–– GradeGrade--60 60 AAs,mins,min = 0.0018bh= 0.0018bh–– Maximum spacing, Maximum spacing, SS, is the , is the

smallest of smallest of (3h or 450mm)(3h or 450mm)Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University

2y

us

−=

adf

MA

φ



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


1

2

directionshort in ent reinforcem Total

widthbandin ent Reinforcem

+=

β

footing of sideshort

footing of side long=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 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. 10.17)as: (ACI sec. 10.17)


( ) ( )bearingfor 0.65

70.185.0 1'

1

21

'

=

≤=

φ

φφ AfA

AAfN 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 ρρρρρρρρ = 0.005 of the column = 0.005 of the column sectionsection as compared to as compared to ρρρρρρρρ = 0.01 as minimum = 0.01 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 reinforcement. 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 4.0 mm.the column by more than 4.0 mm.



Dowels ReinforcementDowels Reinforcement


'

24.0

c

bydb

f

dfl =

Compression development length for #7 bars

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


Dr. Hazim Dwairi 13


Example 1 Example 1 –– Wall FootingWall Footing

Design a plain concrete footing to support a 400 Design a plain concrete footing to support a 400 mm thick concrete wall. The load on the wall mm thick concrete wall. The load on the wall consist of 230 consist of 230 kNkN/m dead load (including the self/m dead load (including the self--weight of wall) and a 146 weight of wall) and a 146 kNkN/m live load. The base /m live load. The base of the footing is 1200 mm below final grade. of the footing is 1200 mm below final grade. ffcc’ = ’ = 21 21 MPaMPa, , ffyy = 420 = 420 MPaMPa, the gross allowable soil , the gross allowable soil pressure = 240 pressure = 240 kNkN/m/m2 2 , and the soil density is 18 , and the soil density is 18 kNkN/m/m33..




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


( ) 2

2

2

/9.2146.125.12240

/6.12)5.02.1)(18(

/5.12)5.0)(25(

50040025.1

thicknesswall1.5)~(1 footing ofdepth assume

mkNq

mkNhW

mkNhW

mm

netall

sssoil

cFooting

=−−=

=−==

====×=

×=

γ

γ


Dr. Hazim Dwairi 14




( )2/1.283

80.1

1466.12302.1

80.1 75.19.214

376

:strip) m-1(consider width footing Estimate

376

146230

:load service totalCalculate

mkNq

mbUSEmq

Pb

LLDLP

u

netall

service

service

=×+×=

=⇒===

=+=+=



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

OKAYVV

kNV

kNmdV

mmd

cu

c

u

⇒<=××××=

=×

−−=

=−−=

φφ


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3)3) Design flexural reinforcementDesign flexural reinforcement


mmmUSE

mmAmmA

AA

adfAM

mkNM

s

ss

ysn

u

200@145 OR /145

5.7425.41210000018.0;451

10002185.0

4205.4124209.0104.69

)2/(

.40.692

40.080.1

2

1.283

2min

2

6

2

φφ

φφ

=××==⇒

×××−×××=×

−=

=

−×=



4)4) Check development lengthCheck development length


mmmmf

dfl

USE

mmlmml

mmf

dfl

c

bydh

d

c

byd

70030821

1442024.024.0

HOOK 90

7002

40018001155

11552110

144209

10

9

'

0

'

<=××==

⇒

=−=>=

=×

××==


Dr. Hazim Dwairi 16




1800 mm

700 mm

500 mm168 mm

φφφφ14@200mm φφφφ14@200mm


Example 2 Example 2 –– Single FootingSingle Footing

Design a square footing to support a 450 mmDesign a square footing to support a 450 mm--square tied interior column reinforced with 8square tied interior column reinforced with 8φφ25 25 bars. The column carries an bars. The column carries an unfactoredunfactored axial axial dead load of 1000 dead load of 1000 kNkN and an axial live load of and an axial live load of 900 900 kN.kN. The base of the footing is 1200 mm The base of the footing is 1200 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..



Dr. Hazim Dwairi 17



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


( ) 2

2

2

/2.2148.1015240

:Pressure Soil Allowable (Net) Effective

/8.1018)6.02.1(

/15256.0

600mmhdepth footing Assume

mkNq

mkNW

mkNW

netall

soil

footing

=−−=

=×−=

=×=

=




2

2

/3.29333

2640

0.30.3

87.8214.2

1900footing of Area

2640)900(6.1)1000(2.1

6.12.1 Load Factored

19009001000

Load Service

mkNq

mmUSE

m

kN

LLDLP

kN

LLDLP

u

u

=×

=

×

==

=+=+==

=+=+==


Dr. Hazim Dwairi 18



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


=

+

=

+

=

=+==×−−=

dbf

dbfdbfb

d

dbfdbf

ofmallestV

mmbPerimeter

mmdAverage

oc

ococo

s

ococ

c

o

'

''

''

33.0

59.02083.0

51.02

117.0

s

3750)5.487450(4

5.487255.175600

αβ



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


( )OKAYshear way -Two

3.23829375.02.2932640

2.23945.48737502833.075.02

cu

u

c

VV

kNV

kNV

φ

φ

<=×−=

=××××=

OKAYshear way -One

7.9865.48730002817.075.0

9.6923)4875.02

45.0

2

3(3.293

cu

c

u

VV

kNV

kNV

φφ

<=××××=

=×−−×=


Dr. Hazim Dwairi 19



4)4) Design the flexural reinforcementDesign the flexural reinforcement


mmsmmUSE

mmA

mmA

AA

mkNM

s,

s

ss

u

5.237113

7523000 ; 63702513

324060030000018.0

6325

30002885.02

4205.4874209.0109.1121

.9.112132

45.00.33.293

:column of edge at themoment ultimate Calculate

2

2min

2

6

=−

×−==

=××=

=⇒

××××−×××=×

=×

−×=

φ



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


( )

OKAYkNNkNP

kNN

NkNN

kNN

mA

mA

u ⇒=<==⇒

>=×××=

=×××==++=

=×=

62812640

281,6

865,19203.0

123.8)203.0(2865.085.0

281,6203.02865.07.1

123.82.145.02.1

203.045.045.0

:Stress BearingCheck

12

1

222

21


Dr. Hazim Dwairi 20




yb

c

ybdc

y

u

fdf

fdl

mmUSE

mm

.;f

N-P

044.024.0

dowels oflength t Developmen

1256204

1013)450450(005.0dowels of Area

650 required dowels of Area

:PN caseIn

'

2

u

>=

==×>

==∴

>

φ

φφ




OKAY 120047628

4202524.0

:Hook 90 Provide

17861200752

4503000

17862810

254209

10

9

:25ent reinforcem footingfor length t Developmen

46242025044.047628

4202524.0 25

37042020044.038128

4202024.0 20

o

'

mmmml

mmmmlAvailable

mmf

dfl

mmmmlfor

mmmmlfor

dh

c

byd

dc

dc

<=××=

<=−−=

=×

××==

=××>=××=→

=××>=××=→

φ

φ

φ


Dr. Hazim Dwairi 21




3000 mm

600 mm

476 mm

476 mm

381 mm

13φφφφ254φφφφ20

8φφφφ25

450 mm


Example 3 Example 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 section columns. The exterior column (I) has a section 400 x 400 mm, which carries DL of 800 400 x 400 mm, which carries DL of 800 kNkN and a and a LL of 530 LL of 530 kN.kN. The interior column (II) has a section The interior column (II) has a section of 500 x 500mm, which carries a DL of 1100 of 500 x 500mm, which carries a DL of 1100 kNkNand a LL of 620 and a LL of 620 kN.kN. The base of the footing is The base of the footing is 1500 mm below final grade and allowable soil 1500 mm below final grade and allowable soil pressure is 240 pressure is 240 kNkN/m/m22. Use . Use ffcc’ = 28 ’ = 28 MPaMPa and and ffyy = = 420 420 MPaMPa. The distance between column is 5.0 m . The distance between column is 5.0 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

5000 mm2820 mm 2820 mm

mmlUSE

mml

6500

54002002005000min

==++=

3250 mm3250 mm




( )

2

2

2

/2545.25.6

4120

5.2 3.25.6

81.14

81.14206

3050footing of Area

/2065.018125240

1000 footing ofdepth Assume

412011506.119002.1

305011501900

mkNq

mBUSEmB

m

mkNq

mmh

kNP

kNP

u

netall

u

service

=×

=

=⇒==

==

=×−×−===×+×=

=+=


Dr. Hazim Dwairi 23




6500 mm

2500 mm

500400

2820

3250

230

3250

8204550

1080

1300 1400

1400



Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite UniversityOK! 504090056002833.075.0

5600)900500(4

1814

)9.05.0(254)6206.111002.1(

:IIcolumn for shear way -Two

OK! 428490047602833.075.0

4760)1300210802(

1451

)3.108.1(254)5306.18002.1(

:Icolumn for shear way -Two

90025751000 Average

2

uc

o

u

uc

o

u

VkNV

mmb

kN

V

VkNV

mmb

kN

V

mmd

>=××××=

=+×==

+×−×+×=

>=××××=

=×+×==

××−×+×=

=−−=

φ

φ


Dr. Hazim Dwairi 24




633.8 kN/m2

272.53

1808 kN 2312 kN

1535.47

1633.53

678.47

58.6 362.8

2.42 m

1799.3

Vu, kN

Mu, kN.m

16.8 103



Dr. Hazim DwairiDr. Hazim Dwairi The Hashemite UniversityThe Hashemite University223111700100065000018.0

22124500100025000018.0

5702 2215 USE

5406

25002885.0

4209004209.0101800

.1800

!!

225025009002817.075.0

7.904)9.025.0(8.63353.1633@

:shearway -One

2min,

2min,

2

2

6

φ

φφ

φ

⇒=××=

⇒=××=

=⇒

=⇒

×××−×××=×

=

>=××××=

=+×−=

+

mmA

mmA

mmA

mmA

AA

mkNM

OKAY

VkNV

kNdV

s

s

s

s

ss

veu

uc

u


Dr. Hazim Dwairi 25




15φφφφ22

12φφφφ2231φφφφ22

230 400 4550 500 820



•• 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


Lecture 9 Lecture 9 –– Footings Footings RC2...Example 1 Example 1 ––Wall Footing Wall...

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Transcript of Lecture 9 Lecture 9 –– Footings Footings RC2...Example 1 Example 1 ––Wall Footing Wall...