3 Spread Footing

8/13/2019 3 Spread Footing

1/17

Dr.: Youssef Gomaa Youssef

CE 402: Part AShallow Foundation Design

Lecture No. (3): Spread Footing

Fayoum UniversityFaculty of Engineering

Department of Civil Engineering


2/17


3/17

Design of Spread Footing

• Plain concrete footing (P.C.)

.* F La

P Area A B

q Eq.(1)

Assume thickness of P.C.:

t = (0.25 to 0.50)

Dim. of P.C. = A * B * tA

A1

t 1

t

P F.L

P G.S

q a

G.S

( ) ( ) A a B b Eq.(2) optional

Solve Eqs (1)&(2) to get A and B


4/17

Design of Spread Footing• Reinforced concrete footing (R.C.)

1 2 A A X

21( ) / 22 I n

A a M p

.

1 1*

G s

n

P p

A B

* cu

M d C

b F

1 covt d er

Steel cover=5.0 to 7.0cm

Dim. of R.C. = A1 * B 1 * t 1

A1

t 1

P G.S

p n

G.S

a

+

+ +

+

B.M.D

1 2 B B X

= 0.80 → 1.00 ∗


5/17


• Shear Stress:

1*2 s n

A aQ p d

0.75 cu suc

f q

* s

s su

Qq q

b d

Qs: shear force at critical sec. (II).

qs: shear stress.

qsu: ultimate shear strength.

If q s > q su , Increase d

Notes:

• No shear RFT in Footing .

t 1

A1

P G.S

(II)

G.S

a

(II)

(II) p n

d

+

-

S.F.D


6/17


• Punching Stress:

. * ( )*( ) p G s nQ P p a d b d

*2* ( ) ( ) p A d a d b d

p p

p

Qq

A

0.5 ( / ) / /cup cu c cu cq a b f f

If q p > q cup , Increase d

t 1

A1

P G.S

G.S

a

p n

d/2 a d/2

d/2

d/2

b


7/17


• Footing Reinforcement:

Which is required?

Top or bottom RFT

why?

* * I

s y

M A

f d j

Notes:

• Minimum number of bars per meter is five .

• Minimum diameter for main RFT is 12mm.

• Number of bars may be taken 5 to 8.

• Diameter of bars may be selected from 12 to 18mm.

t 1

G.S

a

+

+ +

+

B.M.D

G.S

A1

Main RFT5∅12/ ′

Main RFT5∅12/ ′


8/17


• Example(1):

Make a complete design for a footing supporting a 30cm X 60cm

column load of 230t at ground surface (G.S.). The foundation

level is 1.5 m below G.S. and the net allowable bearing capacity is1.75kg/cm2. Make the design considering the following two cases:

1- with plain concrete base

2- without Plain concrete base

a = 0.60m. B= 0.30m

p G.S = 230 t/m’

q a = 1.75kg/cm 2 = 17.5t/m 2 .

f cu = 250kg/cm 2 .

f y =3600kg/cm 2


9/17


• Plain concrete footing (P.C.)2. 230*1.15* 15.11

17.5 F L

a

P Area A B m

q

Assume thickness of P.C.: t = 0.30m

Dim. of P.C. = 4.05 * 3.75 * 0.30

A

A1

t 1

t

P F.L

P G.S

q a

G.SSolve Eqs (1)&(2) to get A and B

( ) ( ) A a B b

2

( 0.30) * 15.11 B B m

2 0.30* 15.11 0.0 B B

20.30 0.30 4*1*15.113.75

2*1 B m

3.75 0.30 4.05 A m

Eq.(1)2* 15.11 A B m

( ) 0.30 A B Eq.(2)0.30 A B


10/17

Design of Spread Footing• Reinforced concrete footing (R.C.)

1 4.05 2*0.30 3.45 A m

0.30 X m

21( ) / 22 I n

A a M p

2.

1 1

230*1.5031.75 /* 3.45*3.15

G sn

P p t m A B

532.23*105 56.77 60

* 100*250cu

M d C cm

b F

1 60 5 65t cm

Dim. of R.C. = 3.45 * 3.15* 65

A1

t 1

P G.S

p n

G.S

a

+

+ +

+

B.M.D

1 3.75 0.60 3.15 B m

2(3.45 0.60)/ 231.75 32.23 . / '

2 I M m t m


11/17


• Shear Stress:1*2 s n

A aQ p d

2500.75 0.75 9.68 /

1.50cu

suc

f q t m

3226.19*10 4.37 /

* 100*60 s

sQ

q t mb d

Qs: shear force at critical sec. (II).

qs: shear stress.

qsu: ultimate shear strength.

Notes:

• No shear RFT in Footing .

t 1

A1

P G.S

(II)

G.S

a

(II)

(II) p n

d

+

-

S.F.D

3.45 0.6031.75( 0.60) 26.19 / '

2 sQ t m

s suq q safe shear


12/17


• Punching Stress:

. * ( )*( ) p G s nQ P p a d b d

*2* ( ) ( ) p A d a d b d

3

24317.10*10 12.58 /2.52*10

p p p

Qq kg cm A

0.5 ( / ) / 250 /1.50 12.91cup cu cq b a f

t 1

A1

P G.S

G.S

a

p n

d/2 a d/2

d/2

d/2

b

230*1.50 31.75* (0.60 0.50)* (0.30 0.50) 317.1 pQ t

20.60* 2* (0.60 0.60) (0.30 0.60) 2.52 p A m

p cupq q safe punching


13/17


• Footing Reinforcement:

5232.23*10 21.50

* * 3600*60*0.695 I

s y

M A cm

f d j

As (short dir.)= 9 φ 18 mm/m’

As (long dir.)= 9 φ 18 mm/m’

0.65

G.S

0.60

+

+ +

+

B.M.D

G.S

3.45

(9 18/ ')m

(9 18/ ')m


14/17


• Problem(1):


column load of 150 t at ground surface (G.S.). The foundation

level is 1.5 m below G.S. and the net allowable bearing capacity is

1.50kg/cm2

. The column location with respect to neighbors isshown in Figure.

1.50m 0.50

0.25


15/17


• Problem(2):


column load of 120 t at ground surface (G.S.). The foundation

level is 1.5 m below G.S. Soil Profile under footing is shown in

the Figure. Allowable settlement is 2.0cm.

Cu=1.00kg/cm 2

mv= 0.10cm 2/kg

Cu=0.50kg/cm 2

mv =0.06cm 2/kg

3.00

5.00


16/17

Design of Spread Footing• Problem(3):

Prove that the unconfined compressive strength of saturated clay

can be considered as the allowable net bearing capacity of this

clay.

1 2 (ECP:202/3) Eq. (3-8)ult c c c f q q qq CN i D N i BN i

*5.14*1.30*1.0 6.68ult u c c c u uq C N i C C

F.S. from Table (3-3) ECP:202/3

6.682.60 2

. . 2.50ult u

a u u uq C

q C C q F S


17/17

Design of Spread Footing• Problem(4):

A large footing settles more than a small one resting on the same

ground profile and subjected to the same stresses. Give reasons

using neat sketches.

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