Lecture Lecture 9 9 Footings Footings Dr. Hazim Dwairi Dr - - PDF document

Reinforced Concrete II Hashemite University

• Dr. Hazim Dwairi

1

The Hashem ite University Departm ent of Civil Engineering

Lecture Lecture 9 9 – – Footings Footings

Dr Hazim Dwairi Dr Hazim Dwairi

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
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• Dr. Hazim Dwairi
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Footings Definition Footings 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 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.

Reinforced Concrete II Reinforced Concrete II

g g g g 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 Dwairi
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Ty p es of Footings Ty p es of Footings

a) a) Strip or Wall Footing Strip or Wall Footing b) 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)

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Ty p es of Footings Ty p es of Footings

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)

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Ty p es of Footings Ty p es of Footings

e) e) Pile Cap Pile Cap f) f) C bi d F ti C bi d F ti f) f) Combined Footing Combined Footing

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Ty p es of Footings Ty p es of Footings

g) g) Cantilever or Strap Footing Cantilever or Strap Footing

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Ty p es of Footings Ty p es of Footings

h) h) Mat or Raft Footing Mat or Raft Footing

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Distribution of Soil Pressure Distribution of Soil Pressure

• When the column load

When the column load P is applied on the is applied on the pp pp centroid centroid of the footing, a

• f 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.

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• 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.

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Distribution of Soil Pressure Distribution of Soil Pressure

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Soil pressure distribution in cohesionless soil. (Clay) Soil pressure distribution in cohesive

• soil. (Sand)

Design Consid era tions Design Consid era tions

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 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. Two

Two-

• way shear or punch out shear.

way shear or punch out shear. 3 O h h

Reinforced Concrete II Reinforced Concrete II

3.

• 3. One

One-

• way shear

way shear 4.

• 4. Bending moment and steel reinforcement

Bending moment and steel reinforcement required required

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Design Consid era tions Design Consid era tions

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 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 base

Bearing capacity of columns at their base 2.

• 2. Dowel requirements

Dowel requirements 3.

• 3. Development length of bars

Development length of bars

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4.

• 4. Differential settlement

Differential settlement

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Eccentrica lly Loa d ed Footing Eccentrica lly Loa d ed Footing

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

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Gross a nd Net Bea ring Ca p a city Gross a nd Net Bea ring Ca p a city

0.6m

10.8kN/m2

0.6m

γs=18kN/m3 γc=25kN/m3 15.0kN/m2 25.8kN/m2 10 8kN/m2

footing s gross net

W W q q − − =

soil

• f

weight =

s

W

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10.8kN/m 15.0kN/m2 25.8kN/m2

footing

• f

weight =

footing

W

Size of Footing Size 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 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

• f

Area = q

Service Load

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3.0 to 2.5

• f

range in the safety

• f

Factor FS = = FS q q

ult all

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Design for Tw o Design for Tw o-

• w a y Shea r

w a y Shea r

• For Slabs and footings,

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

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

Design for Tw o Design for Tw o-

• w a y Shea r

w a y Shea r

The shear force The shear force Vu acts at a acts at a section that has a length section that has a length section that has a length section that has a length b0 = = 4 4( (c+d c+d) ) or

• r 2(c

(c1+d) + +d) + 2 2(c (c2+d) +d) where where d d is the effective depth is the effective depth the section is subjected to a the section is subjected to a vertical downward load vertical downward load Pu and and vertical upward pressure vertical upward pressure q

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vertical upward pressure vertical upward pressure qu.

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2

) ( d c q P V

u u u

+ − =

For Square column: For Rectangular column:

) )( (

2 1

d c d c q P V

u u u

+ + − =

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Design of One Design of One-

• w a y Shea r

w a y Shea r

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

bd f V

c c '

17 . =

b L

) (

1

d c L b V

L-Direction

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b c1 c2

) 2 2 (

1

d b q V

u u

− − =

Ld f V

c c '

17 . = ) 2 2 (

2

d c b L q V

u u

− − =

b-Direction d

Flexura l Strength a nd Footing Flexura l Strength a nd Footing reinforcem ent reinforcem ent

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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Flexura l Strength a nd Footing Flexura l Strength a nd Footing reinforcem ent reinforcem ent

• M

Mu = = 1 1/ /8 8 x x q qu

u x (L

x (L – – c) c)2

• Minimum area of steel

Minimum area of steel

2

y u s

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = a d f M A φ

Reinforced Concrete II Reinforced Concrete II

– Grade Grade-

• 40

40 As,min

s,min =

= 0 0. .002 002bh bh – Grade Grade-

• 60

60 As,min

s,min =

= 0 0. .0018 0018bh bh – Maximum spacing, Maximum spacing, S, is the , is the smallest of smallest of (3 3h or h or 450 450mm) mm)

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Flexura l Strength a nd Footing Flexura l Strength a nd Footing reinforcem ent reinforcem ent

• The reinforcement must be

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

• f

side short footing

• f

side long = β

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1 2 direction short in ent reinforcem Total width band in ent Reinforcem + = β

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Bea ring Ca p a city of Colum n a t Bea ring Ca p a city of Colum n a t Ba se Ba se

• 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

• 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. as: (ACI sec. 10 10. .17 17) )

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( ) ( )

bearing for 0.65 70 . 1 85 .

1 ' 1 2 1 '

= ≤ = φ φ φ A f A A A f N

c c

Bea ring Ca p a city of Colum n a t Bea ring Ca p a city of Colum n a t Ba se Ba se

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Dow els Reinforcem ent Dow els Reinforcem ent

• A minimum steel ratio

A minimum steel ratio ρ = = 0 0. .005 005 of the column

• f the column

section section as compared to as compared to ρ = 0 01 01 as minimum as minimum section section as compared to as compared to ρ = = 0.01 01 as minimum as minimum reinforcement for the column itself. The number reinforcement for the column itself. The number

• f dowel bars needed is four which may be
• f 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

Reinforced Concrete II Reinforced Concrete II

, g , g

• 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 the column by more than 4 4. .0 0 mm. mm.

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Dow els Reinforcem ent Dow els Reinforcem ent

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

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'

24 .

c b y db

f d f l =

development length for #7 bars

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Exa m p le Exa m p le 1 1 – – W a ll Footing W a ll 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 wall mm thick concrete wall The load on the wall mm thick concrete wall. The load on the wall mm thick concrete wall. The load on the wall consist of consist of 230 230 kN kN/m dead load (including the self /m dead load (including the self-

• weight of wall) and a

weight of wall) and a 146 146 kN kN/m live load. The base /m live load. The base

• f the footing is
• f the footing is 1200

1200 mm below final grade. mm below final grade. f fc’ = ’ = 21 21 MPa MPa, , f fy

y =

= 420 420 MPa MPa, the gross allowable soil , the gross allowable soil pressure = pressure = 240 240 kN kN/m /m2

2 , and the soil density is

, and the soil density is 18 18

Reinforced Concrete II Reinforced Concrete II

p , y , y kN kN/m /m3.

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Exa m p le Exa m p le 1 1 – – W a ll Footing W a ll Footing

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

2

/ 5 . 12 ) 5 . )( 25 ( 500 400 25 . 1 thickness wall 1.5) ~ (1 footing

• f

depth assume m kN h W mm

c Footing

= = = = × = × = γ

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( )

2 2

/ 9 . 214 6 . 12 5 . 12 240 / 6 . 12 ) 5 . 2 . 1 )( 18 ( m kN q m kN h W

net all s s soil

= − − = = − = = γ

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Exa m p le Exa m p le 1 1 – – W a ll Footing W a ll Footing

: load service total Calculate LL DL P

service

+ =

( )

80 . 1 75 . 1 376 : strip) m

• 1

(consider width footing Estimate 376 146 230 m b USE m P b

service service

= ⇒ = = = = + =

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( )

2

/ 1 . 283 80 . 1 146 6 . 1 230 2 . 1 80 . 1 75 . 1 9 . 214 m kN q m b USE m q b

u net all

= × + × = ⇒

Exa m p le Exa m p le 1 1 – – W a ll Footing W a ll Footing

2) 2) Check one Check one-

• way shear

way shear ! . 241 5 . 412 1000 21 17 . 75 . 40 . 81 . 1 4125 . 2 40 . 80 . 1 1 . 283 @ 5 . 412 5 . 12 75 500 Estimate OKAY V V kN V kN m d V mm d

c u

= × × × × = = × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = = − − = φ φ

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! OKAY V V

c u

⇒ < φ

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Exa m p le Exa m p le 1 1 – – W a ll Footing W a ll Footing

3) 3) Design flexural reinforcement Design flexural reinforcement A A a d f A M m kN M

s y s n u

420 5 . 412 420 9 . 10 4 . 69 ) 2 / ( . 40 . 69 2 40 . 80 . 1 2 1 . 283

6 2

φ φ ⎟ ⎞ ⎜ ⎛ × − × × × = × − = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × =

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mm m USE mm A mm A A

s s

200 @ 14 5 OR / 14 5 5 . 742 5 . 412 1000 0018 . ; 451 1000 21 85 . 5 . 412 420 9 . 10 4 . 69

2 min 2

φ φ = × × = = ⇒ ⎟ ⎠ ⎜ ⎝ × × × × × ×

Exa m p le Exa m p le 1 1 – – W a ll Footing W a ll Footing

4) 4) Check development length Check development length d f 14 420 9 9 USE mm l mm l mm f d f l

d c b y d

HOOK 90 700 2 400 1800 1155 1155 21 10 14 420 9 10 9

'

⇒ = − = > = = × × × = =

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mm mm f d f l USE

c b y dh

700 308 21 14 420 24 . 24 . HOOK 90

'

< = × × = = ⇒

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Exa m p le Exa m p le 1 1 – – W a ll Footing W a ll Footing

700 mm 500 mm 168 mm

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1800 mm

φ14@200mm φ14@200mm

Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

Design a square footing to support a Design a square footing to support a 450 450 mm mm-

• square tied interior column reinforced with

square tied interior column reinforced with 8 8φ φ25 25 square tied interior column reinforced with square tied interior column reinforced with 8 8φ φ25 25

• bars. The column carries an
• bars. The column carries an unfactored

unfactored axial axial dead load of dead load of 1000 1000 kN kN 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 kN kN/m /m2

• 2. Use

. Use f fc’ = ’ = 28 28 MPa MPa and and f fy = = 420 420 MPa MPa. .

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c y

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Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

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

2 2

P S il All bl (N t) Eff ti / 8 . 10 18 ) 6 . 2 . 1 ( / 15 25 6 . 600mm h depth footing Assume m kN W m kN W

soil footing

= × − = = × = =

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( )

2

/ 2 . 214 8 . 10 15 240 : Pressure Soil Allowable (Net) Effective m kN q

net all

= − − =

Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

Load Service LL DL P + = =

2

87 . 8 214.2 1900 footing

• f

Area 2640 ) 900 ( 6 . 1 ) 1000 ( 2 . 1 6 . 1 2 . 1 Load Factored 1900 900 1000 m kN LL DL P kN

u

= = = + = + = = = + =

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2

/ 3 . 293 3 3 2640 . 3 . 3 214.2 m kN q m m USE

u

= × = ×

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Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

2) 2) Check thickness for two Check thickness for two-

• way shear

way shear ⎪ ⎪ ⎪ ⎧ ⎟ ⎞ ⎜ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = + = = × − − = d d b f d b f mm b Perimeter mm d Average

• c
• c
• '

'

51 . 2 1 17 . 3750 ) 5 . 487 450 ( 4 5 . 487 25 5 . 1 75 600 α β

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⎪ ⎪ ⎪ ⎩ ⎪ ⎨ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = d b f d b f d b f b d

• f

mallest V

• c
• c
• c
• s

c ' ' '

33 . 59 . 2 083 . s α

Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

( )

3 . 2382 9375 . 2 . 293 2640 2 . 2394 5 . 487 3750 28 33 . 75 .

2 u c

kN V kN V φ = × − = = × × × × = 3) 3) Check one Check one-

• way shear

way shear

( )

OKAY shear way

• Two

c u u

V V φ < 9 . 692 3 ) 4875 . 2 45 . 2 3 ( 3 . 293

u

kN V = × − − × =

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OKAY shear way

• One

7 . 986 5 . 487 3000 28 17 . 75 .

c u c

V V kN V φ φ < = × × × × =

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Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

4) 4) Design the flexural reinforcement Design the flexural reinforcement : column

• f

edge at the moment ultimate Calculate A A A m kN M

s s u

6325 3000 28 85 . 2 420 5 . 487 420 9 . 10 9 . 1121 . 9 . 1121 3 2 45 . . 3 3 . 293

2 6

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × × × − × × × = × = × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − × =

Reinforced Concrete II Reinforced Concrete II

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mm s mm USE mm A mm A

s, s

5 . 237 1 13 75 2 3000 ; 6370 25 13 3240 600 3000 0018 . 6325

2 2 min 2

= − × − = = = × × = = ⇒ φ

Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

5) 5) Design column Design column-

• footing connection

footing connection : Stress Bearing Check

( )

kN N m A m A = × × × = = + + = = × = 123 8 281 , 6 203 . 28 65 . 7 . 1 123 . 8 2 . 1 45 . 2 . 1 203 . 45 . 45 . : Stress Bearing Check

1 2 2 2 2 1

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OKAY kN N kN P kN N N kN N

u

⇒ = < = = ⇒ > = × × × = 6281 2640 281 , 6 865 , 19 203 . 123 . 8 ) 203 . ( 28 65 . 85 .

1 2

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Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

N-P 6 i d d l f A : P N case In

u

> φ

y u

mm USE mm . ; f N P dowels

• f

length t Developmen 1256 20 4 1013 ) 450 450 ( 005 . dowels

• f

Area 65 required dowels

• f

Area

2

= = × > = = ∴ φ φ φ

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y b c y b dc

f d f f d l 044 . 24 .

'

> =

Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

420 25 24 370 420 20 044 . 381 28 420 20 24 . 20 mm mm l for

dc

× × = × × > = × × = → φ 450 3000 1786 28 10 25 420 9 10 9 : 25 ent reinforcem footing for length t Developmen 462 420 25 044 . 476 28 420 25 24 . 25

'

mm f d f l mm mm l for

c b y d dc

− = × × × = = = × × > = × × = → φ φ

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OKAY 1200 476 28 420 25 24 . : Hook 90 Provide 1786 1200 75 2 450 3000

• mm

mm l mm mm l Available

dh

< = × × = < = − =

Reinforced Concrete II Hashemite University

• Dr. Hazim Dwairi

21

Exa m p le Exa m p le 2 2 – – Single Footing Single Footing

8φ25 450 mm 476 mm 381 mm 13φ25 4φ20

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University

3000 mm 600 mm 476 mm 381 mm

Exa m p le Exa m p le 3 3 – – Com bined Footing Com bined 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

• columns. 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 kN kN 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

• f
• f 500

500 x x 500 500mm, which carries a DL of mm, which carries a DL of 1100 1100 kN kN and 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

Reinforced Concrete II Reinforced Concrete II

g pressure is pressure is 240 240 kN kN/m /m2. Use . Use f fc’ = ’ = 28 28 MPa MPa and and f fy = = 420 420 MPa

• MPa. The distance between column is

. The distance between column is 5 5. .0 0 m m center to center (cc). center to center (cc).

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University

Reinforced Concrete II Hashemite University

• Dr. Hazim Dwairi

22

Exa m p le Exa m p le 3 3 – – Com bined Footing Com bined Footing

DL = 800 kN LL = 530 kN DL = 1100 kN LL = 620 kN DL = 1900 kN LL = 1150 kN LL = 530 kN LL = 620 kN 5000 mm 2820 mm 2820 mm 3250 mm 3250 mm

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University

mm l USE mm l 6500 5400 200 200 5000

min

= = + + =

Exa m p le Exa m p le 3 3 – – Com bined Footing Com bined Footing

4120 1150 6 . 1 1900 2 . 1 3050 1150 1900 kN P kN P

u service

= × + × = = + =

( )

2 2

81 14 81 . 14 206 3050 footing

• f

Area / 206 5 . 18 1 25 240 1000 footing

• f

depth Assume m m kN q mm h

net all u

= = = × − × − = =

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University

2

/ 254 5 . 2 5 . 6 4120 5 . 2 3 . 2 5 . 6 81 . 14 m kN q m B USE m B

u

= × = = ⇒ = =

Reinforced Concrete II Hashemite University

• Dr. Hazim Dwairi

23

Exa m p le Exa m p le 3 3 – – Com bined Footing Com bined Footing

6500 mm 2500 mm 2820 3250 3250 1080 1300 1400 1400

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University

500 400 230 820 4550

Exa m p le Exa m p le 3 3 – – Com bined Footing Com bined Footing

) 3 1 08 1 ( 254 ) 530 6 1 800 2 1 ( : I column for shear way

• Two

900 25 75 1000 Average V mm d × × × + × = = − − = : II column for shear way

• Two

OK! 4284 900 4760 28 33 . 75 . 4760 ) 1300 2 1080 2 ( 1451 ) 3 . 1 08 . 1 ( 254 ) 530 6 . 1 800 2 . 1 (

u c

• u

V kN V mm b kN V > = × × × × = = × + × = = × × − × + × = φ

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University

OK! 5040 900 5600 28 33 . 75 . 5600 ) 900 500 ( 4 1814 ) 9 . 5 . ( 254 ) 620 6 . 1 1100 2 . 1 (

2 u c

• u

V kN V mm b kN V > = × × × × = = + × = = + × − × + × = φ

Reinforced Concrete II Hashemite University

• Dr. Hazim Dwairi

24

Exa m p le Exa m p le 3 3 – – Com bined Footing Com bined Footing

1808 kN 2312 kN 633.8 kN/m2 272.53 1535.47 1633.53 678.47 2.42 m

Vu, kN

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University

535 58.6 362.8 1799.3

Mu, kN.m

16.8 103

Exa m p le Exa m p le 3 3 – – Com bined Footing Com bined Footing

2250 2500 900 28 17 75 7 . 904 ) 9 . 25 . ( 8 . 633 53 . 1633 @ : shear way

• One

φ > = × × × × = = + × − = V kN V kN d Vu 2500 28 85 . 420 900 420 9 . 10 1800 . 1800 ! ! 2250 2500 900 28 17 . 75 .

6

φ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × × − × × × = × = > = × × × × =

+

A A m kN M OKAY V kN V

s s ve u u c

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University

22 31 11700 1000 6500 0018 . 22 12 4500 1000 2500 0018 . 5702 22 15 USE 5406

2 min , 2 min , 2 2

φ φ φ ⇒ = × × = ⇒ = × × = = ⇒ = ⇒ mm A mm A mm A mm A

s s s s

Reinforced Concrete II Hashemite University

• Dr. Hazim Dwairi

25

Exa m p le Exa m p le 3 3 – – Com bined Footing Com bined Footing

230 400 4550 500 820 15φ22

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University

12φ22 31φ22

Exa m p le Exa m p le 3 3 – – Com bined Footing Com bined Footing

• Things still need to be checked:

Things still need to be checked:

√ Cases of loading Cases of loading √ Bearing stresses under columns Bearing stresses under columns √ Development length Development length √ Dowel bars if needed Dowel bars if needed

Reinforced Concrete II Reinforced Concrete II

• Dr. Hazim Dwairi
• Dr. Hazim Dwairi

The Hashemite University The Hashemite University