1919

� Types of Footings

� Bearing Pressure under Footing

� Eccentrically Loaded Footing

�Wall Footings

� Column Footings

Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Design of Footings 1

Types of Footings

Wall

Property line

Isolated spread footing Wall footing

Combined Footings

A B

Rectangular, PA = PB

A B

Rectangular, PA < PB

Property line

A B

Rectangular, PA < PB

Property line

A B

Strap or Cantilever

Property line

Mat Footing

Pile cap

PilesWeak soil

Bearing stratum

Bearing pressure under footings

Axially Loaded Footings : Assume uniform pressure

Actual pressure is not uniform due to:

1) Footing flexibility

2) Depth of footing below ground surface

3) Type of soil, e.g., clay or sand

R

p, bearing pressure

R

Cohesionless soil

HeaveHeave

R

Cohesive soil

Eccentrically Loaded Footings

P Mc Pec

A I I= =

Ie

Ac=

For rectangular footing of length h and width b,

3 /12

( / 2) 6

I bh he

Ac bh h= = =

x x

y

y

e

h

load

b

eP

max

P Mcp

A I= +

min

P Mcp

A I= −

Tensile stress cannot be transmitted between soil and concrete.

For full compression, setting pmin = 0, P

emax = h/6

h/3 h/3 h/3

Middle Third

Large eccentricity of load e > h/6

Centroid of soil pressure concurrent with applied load

a

3a

eP

pmaxR

max

1(3 )

2R ab p P= = max

2

3

Pp

ab=

where a = h/2 - e

��������� 12.1 �������� �������� ����� 1.8 . x1.2 . ����������������� 80 ����������� ��!�0.15 . ���#��!$�%�&��'�(��� �&)*���+�'�&����*���������� ,���(+-���.��,���&/��������!�0!1�.&#��!$0�2���� 0.40 .

0.60 m

0.60 m

0.90 m 0.90 m

e

Load

����� 0�1 .&��� e = 0.15 . < [h/6 = 0.30 .]/� ���( ��2(!'�&4��5�!��&

max 3

2

80 80 0.15 0.9

1.8 1.2 1.2 1.8 /12

37.0 18.5 55.5 t/m

p× ×

= +× ×

= + =

2

min 37.0 18.5 18.5 t/mp = − =

0�1 .&��� e = 0.40 . > [h/6 = 0.30 .] �6�/ �����2(!'�&/� 0�%� !

a = 0.90 – 0.40 = 0.50 .

2

max

2 8088.9 t/m

3 0.50 1.20p

×= =

× ×

�������� 12.1 ������������� ���� �.�.�. ����������� �.�. 2522

�������������������(���/�.�. .)

*����������� ���������������������� ������������������������������� ��!�

�*�.2.���1.�*� '�2�0�7 �� 2

�*�����%�&��1.���!�2(� 5

�*�'�2���1.���!�!�� 10

��(���1.�*���� 20

�*��*���� 25

�*������1.�*����! 30

�*�.�,���� !�&4 2'��/9�) 100

��������� 12.2 ,���(+'�&������&� �'%�'�&���/��:*�� ����.&������/� 0�%� ! ������/���� 1.5 .

��2(!���������.&�*� γs = 2.0 ���/%�. . '%�,.����� γc = 2.4 ���/%�. .

0.5 m

1.0 m 30x30 cm column

Grade

DL = 80 ton

LL = 40 ton

����� �����������������&� ���������

�������������� = (1.52)(0.5)(2.4) = 2.7 ���

��������0/� = (.32)(1.0)(2.4) = 0.22 ���

���������*� = (1.0)(1.52-0.32)(2.0) = 4.32 ���

�������������� = 80+40 = 120 ���

���������( ���&� � = 127.24 ���

2

2

127.2456.55 t/m

1.5grp = ='�&����*����&� �

'�&����*�/��:*5�!%�0%!,(� '���2�&���(2�&��2(!���������.&�*�'%�,.�����,1.

pn = pgr – ���������*�%B� 1.5 0 �� = 56.55 – 1.5(2.0) = 53.55 ���/��. .

Wall Footings

1-m slice on which

design is based

Wall

Footing

wUniformly loaded wall

w

Bending deformation

Critical Section for Moment in Isolated Footings

Concrete column,

pedestal or wall

Critical section

Masonry wall

b/2 b/2

b/4

Column with steel

base plate

s

s/2

Critical section

b

d

wu = 1.4wDL+1.7wLL

d

L

qu

Factored wall load = wu t/m

Factored soil pressure, qu = (wu )/L

Required L = (wDL+wLL)/qa

qa = Allowable soil pressure, t/m2

Moment and Shear in Wall Footings

2

21 1( )

2 2 8u u u

L bM q q L b

− = = −

2u u

L bV q d

− = −

Min t = 15 cm for footing on soil, 30 cm for footing on piles

Min As = (14 / fy ) (100 cm) d

EXAMPLE 12.3: Design of a Wall Footing to carries a dead load D of 12 t/m and a live

load L of 8 t/m. The max. soil pressure is 10 t/m2. f’c = 240 kg/cm2, fy = 4000 kg/cm

2, and

γs = 2.0 t/m3.

D = 12 t/m

L = 8 t/m

25 cm

L

8 cm

clear

5 cm

typical

Df = 1.50 m

Consider: 1-m strip

Assume footing t = 30 cm

Net soil pressure:

pn = 10 - [0.3(2.4) + 1.2(2.0)]

= 6.88 t/m2

Req’d footing length:

L = (DL + LL) / pn = (12+8)/6.88

= 2.91 m USE 3.0 m

Ultimate soil pressure:

pu = (1.4 x 12 + 1.7 x 8) / 3.0 = 10.13 t/m

Check Shear:

115.5 cm

d = 22 cm

25 cm

10.13 t/m2

30 cm

Vu = 10.13(1)(1.155) = 11.70 ���

φVc = 0.85(0.53)(100)(22)/1000

= 15.35 ��� > Vu OK

Flexural design:

137.5 cm25 cm

10.13 t/m2

Mu = 0.5�10.13 � 1.3752

= 9.58 ���-0 ��5

2 2

9.58(10 )

0.9 100 22

un

MR

bdφ= =

× ×

= 21.98 ��./- .2

]0035.0[0058.085.0

211

85.0min =>=

′−−

′= ρρ

c

n

y

c

f

R

f

fOK

0%1.��6�0�%7� DB12 @ 0.20 (As = 5.65 - .2/0 ��)

25 cm

3.00 m

30 cm

DB12DB12@0.20m

DB16@0.15m

As = 0.0058(100)(22) = 12.82 - .2/0 ��

0%1.��6�0�%7� DB16 @ 0.15 (As = 13.40 - .2/0 ��)

�������"�#��� ��� �6�0)1 .����������'�����(

As = 0.0018(100)(30) = 5.4 - .2/0 ��

Column Footings

1

d/2 d

d

3

2

1

3

2

Punching shear

Beam-shear short direction

Beam-shear long direction

Critical section for shear

Weight of footing ≈ 4-8 % of column load

1

2

Critical section for moment

1

2

Moment short direction

Moment long direction

Two-Way Action Shear (punching-shear)

P

On perimeter around column at distance d/2 from face of column

c1 + d

c2c2 + d

c1

d/2

b0

0

40.27 2n c c

c

V V f b dβ

′= = +

Two-Way Action: cracking occur around column with periphery b0 at

distance d / 2 outside column. Vn is the smallest of

0

0

0.27 2sn c c

dV V f b d

b

α ′= = +

01.06n c cV V f b d′= =

ACI Formula (11-35)

ACI Formula (11-36)

ACI Formula (11-37)

where

b0 = perimeter of critical section at distance d /2 outside column

βc = ratio of long side to short side of column

αs = 40 for interior columns, 30 for edge columns and 20 for

corner columns

Distribution of Flexural Reinforcement

Footing Type

One-way

Two-way

Square Footing Rectangular Footing

L

B

s (typ.)

L

L

s (typ.)

L

B

B/2 B/2

As2As2 As1

AsL

AsB

s (typ.)

1

12

2

1

2

s sL

sL ss

A A

A AA

L

B

β

β

= +

−=

=

Transfer of Forces at Base of Column

For a supported column, bearing capacity is

where A1 = loaded area (column area)

φ = 0.70

1(0.85 )nb cP f Aφ φ ′=A1

450

2

1

A2 measured

on this plane

For a supporting footing,

21 1

1

(0.85 ) 2 (0.85 )nb c c

AP f A f A

Aφ φ φ′ ′= ≤

where A2 = area of lower base of the largest

pyramid cone contained within footing having

side slope 1 vertical to 2 horizontal

EXAMPLE 12.4: Design of a Square Footing to support a 40 cm square column.

The column carries a dead load D of 40 ton and a live load L of 30 ton. The

allowable soil pressure 10 t/m2. f’c = 240 kg/cm2, fy = 4000 kg/cm

2. Unit weight of

the soil above footing base = 2.0 t/m3.

D = 40 t

L = 30 t

40 cm

b

h1.50 m

Assume footing depth = 40 cm

Soil net pressure:

pn = 10 – [0.4(2.4) + 1.1(2.0)]

= 6.84 t/m2

Required area = (40+30)/6.84 = 10.23 m2

USE 3.2x3.2m square footing (10.24m2)

210710.45 t/m

10.24up = =Ultimate pressure

(1) Determination of base area:

(2) Factored loads and soil reaction:

Pu = 1.4(40) + 1.7(30) = 107 tons

40 cm

72 cm

d/2=16 cm

Punching shear:

Vu = 10.45(3.22 – 0.722) = 101.6 ���

bo = 4(72) = 288 - .

φ Vc = 0.85(1.06) (288)(32)/1000

= 128.6 ��� > Vu OK

240

Beam shear:

40 cm

108 cm

d=32 cmVu = 10.45(1.08)(3.2) = 36.12 ���

φ Vc = 0.85(0.53) (320)(32)/1000

= 71.47 ��� > Vu OK

240

Assume footing depth = 40 cm and effective depth d = 32 cm

Flexural Design:

3.20 m

0.40 m

15DB16 #

DB16 ����.�4DB2540 cm

Mu = (0.5)(10.45)(3.2)(1.4)2 = 32.77 ���-0 ��

Rn = = 11.11 ��./- .2 ρ = 0.0029

As = 0.0029(320)(32) = 29.70 - .2

As,min = 0.0018(320)(40) = 23.04 - .2 < As OK

USE 15DB16# (As= 30.15 cm2)

5

2

32.77(10 )

0.9 320 32× ×

Critical section

for moment

Check development of reinforcement

Critical section for development is the same as that for moment (at face of column)

0.28yd

b trc

b

fl

d c Kf

d

αβγλ=

′ +

Edge distance (bottom and side) = 8 cm

Center-to-center bar spacing = (320 - 2(8))/14 = 21.7 cm

c = minimum of8 cm (control)

21.7 / 2 = 10.9 cm

Ktr = 0 (no transverse reinforcement)

8 05.0 2.5 .

1.6

tr

b

c K

d

+ += = > USE 2 5

α = 1.0 (bottom bars)

β = 1.0 (uncoated reinforcement)

αβ = 1.0 < 1.7

γ = 0.8 (DB20 and smaller)

λ = 1.0 (normal weight concrete)

4,000 1.0 1.0 0.8 1.00.28 23.1

2.5240

d

b

l

d

× × ×= =

ld = 23.1 x 1.6 = 37.0 cm > 30 cm OK

Since ld = 37 cm < available embedment length (320/2 - 40/2 - 8 = 132 cm),

DB16 bars can be fully developed.

A1

40 cm

320cm

Transfer of Force at Base of Column

8 cm cover

32 cm

column bars

footing dowels

(1) Bearing strength of column

φPnb = φ (0.85f’c A1)

= 0.70(0.85x240x40x40)/1,000

= 228.5 tons > 107 tons OK

(2) Bearing strength of footing

200cmA2

Bearing strength of footing increased by factor

2 1 2A A ≤ where A2 is area of pyramid cone

having side slope 1 vertical to 2 horizontal

2

1

200 2005 2, use 2

40 40

A

A

×= = >

×

φPnb = 2φ (0.85f’c A1)

= 2(0.70)(0.85x240x40x40)/1,000

= 457 tons > 107 tons OK

(3) Required dowel bars between column and footing:

Even though column and footing have enough bearing strength to transfer load,

area of reinforcement across interface ≥ 0.005(gross area of supported member)

As (min) = 0.005(40x40) = 8.0 cm2

Provide 4DB16 bars as dowels (As = 8.04 cm2)

(4) Development of dowel reinforcement in compression:

In column & footing:0.075

0.0043b y

d b y

c

d fl d f

f= ≥

′

For DB16 bars:

(min)

0.075 1.6 4,00031.0 cm (control)

240

0.0043 1.6 4,000 27.5 cm

d

d

l

l

× ×= =

= × × =

Available length for development in footing

= footing thickness - cover - 2(footing bar dia.) - dowel bar dia.

= 40 - 8 - 2(1.6) - 1.6 = 27.2 cm ≈ 27.5 cm OK

Therefore, the dowels can be fully developed in the footing.

Home work: Design a square spread footing with the following design conditions:

1.5 m

Ground elev.

PService dead load = 150 ton

Service live load = 120 ton

Unit weight of soil = 2.0 ton/m3

Allowable soil pressure = 20 ton/m2

Column dimensions = 60 x 30 cm