dr.: youssef gomaa youssef

Dr.: Youssef Gomaa Youssef

CE 402: Part A

Shallow Foundation Design

Lecture No. (6): Eccentric Footing

Fayoum University Faculty of Engineering

Department of Civil Engineering

Eccentric Footing

Eccentric footing: A spread or wall footing that also must

resist a moment in addition to the axial column load.

A

e < A/6

Eccentric Loads or Moments

( )f

MeP W

. .D L L LP P P A

e < A/6

CE 402: Foundation Engineering Design

Eccentric Footing

Combined axial and bending stresses increase the pressure on one

edge or corner of a footing. We assume again a linear distribution

based on a constant relationship to settling.

If the pressure combination is in tension, this effectively means the

contact is gone between soil and footing and the pressure is really

zero.

To avoid zero pressure, the eccentricity must stay within the kern. The

maximum pressure must not exceed the net allowable soil pressure.


The kern To avoid zero pressure, the

eccentricity must stay within

the kern.

A

A/3


Eccentric Loads or Moments Cases

A

e <A/6

Case (a)

e =A/6

Case (b)

e >A/6

Case (c)


For case A and B (not case C)

Eccentric Loads or Moments

)6

1(*

minA

e

BA

Pq )

61(

*max

A

e

BA

Pq

BC

Pq

**3

*2max

For case c

e >A/6

Case (c)

C

AC 75.03

aqq 20.1max


Footing Subjected to Double Moment

For contact pressure to remain (+) ve everywhere,

)661(* B

e

A

e

BA

Pq BA

0.166

B

e

A

e BA


Design of Eccentric Footing

• Plain concrete footing (P.C.)

Assume thickness of P.C.:

t = (0.25 to 0.50)

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

a

SG

q

PBAArea .50.1

*

AreaA Assume

fLF DHMM *.

LF

LF

P

Me

.

.sGLF PP .. *15.1

6

Ae Check A

A1

t1

t

PF.L

PG.S

qmax

G.S M H

Df

qmin

)6

1(*

.max

A

e

BA

Pqq LF

a B


Design of Spread Footing

• Reinforced concrete footing (R.C.)

1 2A A X

* cu

Md C

b F

1 covt d er Dim. of R.C. = A1 * B1* t1

1 2B B X

tX )0.18.0(

)6

1(* 1

1

21

.2,1

A

e

BA

Pf SG

SGP

Me

.

11

3

2*)

2)((2/)

2( 21

31

213

aAff

aAfM I

2121 )2

(*)2

(bBff

M II

A1

t1

PG.S

f1

G.S

a

f2

f3



• Shear Stress:

0.75 cusu

c

fq

*

ss su

Qq q

b d

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

qs: shear stress.

qsu: ultimate shear strength.

If qs > qsu , Increase d

Notes:

•No shear RFT in Footing.

)2

(*2

)( 141)( d

aAffQ IIs

)2

(*2

)( 121)( d

bBffQ IIIs

t1

PG.S

(II)

G.S

a

(II)

(II)

d

A1

f1

f2

f4



• Punching Stress:

*2* ( ) ( )pA d a d b d

p

p

p

Qq

A

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

If qp > qcup , Increase d

t1

A1

PG.S

G.S

a

d/2 a d/2

d/2

d/2

b

f1

f2

)](*)[(2

)(50.1 21

. dbdaff

PQ SGp



• Footing Reinforcement:

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.

t1

G.S

a

G.S

A1

jdf

MA

y

IIs

**2

jdf

MA

y

Is

**1

Main RFT As2

Main RFT As1



• Example(1):

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

column load of 120t at ground surface (G.S.), 20m.t moment

and 10t horizontal force at G.S. The foundation level is 2.00 m

below G.S. and the net allowable bearing capacity is 0.80kg/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

pG.S = 120t M=20m.t H=10t

qa = 0.80kg/cm2 = 8t/m2.

fcu = 250kg/cm2.

fy =3600kg/cm2



• Plain concrete footing (P.C.)

Assume thickness of P.C.:

t = 0.30

Dim. of P.C. = 5.00* 4.65 * 0.30

2. 5.228

120*50.1*50.1* m

q

PBAArea

a

SG

mAreaA 00.575.4 Assume

tmM LF .4000.2*1020.

290.0138

40

.

. LF

LF

P

Me

tPP sGLF 138120*15.1*15.1 ..

833.06

5

6

Ae A

A1

t1

t

PF.L

PG.S

qmax

G.S M H=10

2.00

qmin

8)00.5

290.0*61(

*5

138max

Bq mB 65.4



• Reinforced concrete footing (R.C.)

Dim. of R.C. = 4.40 * 4.05* 0.75

mtX 30.0

85.534.14)40.4

308.0*61(

05.4*40.4

120*50.12,1 f

mP

Me

SG

308.0120

70.1*1020

.

11

36.473

2*)

2

6.04.4)(67.1034.14(2/)

2

60.04.4(67.10 22

IM

33.35)2

30.005.4(*)

2

85.534.14( 2

IIM

mXAA 40.430.0*200.521

mXBB 05.430.0*265.421

cmd 708.68250*100

10*36.475

5

cmt 755701

t1

PG.S

F1=14.34

G.S

F2=5.85

F3=10.67

A1=4.40

1.90

0.60



• Shear Stress:

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

qs: shear stress.

qsu: ultimate shear strength.

If qs > qsu , Increase d

Notes:

•No shear RFT in Footing.

71.15)70.090.1(*2

)84.1134.14()(

IIsQ

86.11)70.02

3.005.4(*

2

)85.534.14()(

IIIsQ

t1

PG.S

(II)

G.S

0.60

(II)

(II)

d

A1=4.40

1.90

f2=5.85

f1=14.34 f4=11.84

84.11)6.060.090.1(*40.4

)85.584.11(85.54

f

68.9/24.270*100

1000*71.15 2 cmkgqs



• Punching Stress:

p

p

p

Qq

A

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

If qp > qcup , Increase d

t1

A1

PG.S

G.S

a

d/2 a d/2

d/2

d/2

b

f1

f2

88.166)]70.030.0(*)70.060.0[(2

)85.534.14(120*50.1

pQ

22.3)]70.30.0()70.060.0[(*2*70.0 pA

18.510*22.3

1000*88.1664

pq



• Footing Reinforcement:

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.

0.75

G.S

0.60

G.S

4.40

97.16826.0*70*3600

10*33.35 5

2 sA

75.22826.0*70*3600

10*36.47 5

1 sA

Main RFT

618/m’

Main RFT 818/m’


Crane Footing

Maximum stress on soil should be less than allowable bearing capacity

aqf max

The ratio between maximum and minimum stresses should be less

than four 4

)/61(

)/61(

min

max

Ae

Ae

f

f4

min

max f

f

eA 10


Uniform Stress below Footing Subjected to Moment

Uniform stress required that the eccentricity at foundation level equal

zero

0.0e 0.0..

. sG

LF

P

Me

A

PF.L

PG.S

qa

G.S M

e c

a/2

ea

cA

22

A

a

SG

q

PBA .15.1

* B

dr.: youssef gomaa youssef

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