6-Eccecntric Footing.pdf shajasjakssjssjwjs

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
M
e
P W


. .
D L L L
P 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
(
*
min
A
e
B
A
P
q 
 )
6
1
(
*
max
A
e
B
A
P
q 

B
C
P
q
*
*
3
*
2
max 
For case c
e >A/6
Case (c)
C
A
C 75
.
0
3 
a
q
q 20
.
1
max 

Footing Subjected to Double Moment
For contact pressure to remain (+) ve everywhere,
)
6
6
1
(
* B
e
A
e
B
A
P
q B
A



0
.
1
6
6


B
e
A
e B
A

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
S
G
q
P
B
A
Area .
50
.
1
* 

Area
A 
Assume
f
L
F D
H
M
M *
. 

L
F
L
F
P
M
e
.
.

s
G
L
F P
P .
. *
15
.
1

6
A
e 
Check A
A1
t1
t
PF.L
PG.S
qmax
G.S
M H
Df
qmin
)
6
1
(
*
.
max
A
e
B
A
P
q
q L
F
a 

 B

Design of Spread Footing
• Reinforced concrete footing (R.C.)
1 2
A A X
 
* cu
M
d C
b F
 1 cov
t d er
 
Dim. of R.C. = A1 * B1* t1
1 2
B B X
 
t
X )
0
.
1
8
.
0
( 

)
6
1
(
* 1
1
2
1
.
2
,
1
A
e
B
A
P
f S
G


S
G
P
M
e
.
1
1 
3
2
*
)
2
)(
(
2
/
)
2
( 2
1
3
1
2
1
3
a
A
f
f
a
A
f
MI





2
1
2
1
)
2
(
*
)
2
(
b
B
f
f
MII



A1
t1
PG.S
f1
G.S
a
f2
f3

• Shear Stress:
0.75 cu
su
c
f
q


*
s
s su
Q
q 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
)
( 1
4
1
)
( d
a
A
f
f
Q II
s 



)
2
(
*
2
)
( 1
2
1
)
( d
b
B
f
f
Q III
s 



t1
PG.S
(II)
G.S
a
(II)
(II)
d
A1
f1
f2
f4

• Punching Stress:
 
*2* ( ) ( )
p
A d a d b d
   
p
p
p
Q
q
A

 
0.5 ( / ) / /
cup cu c cu c
q 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 2
1
. d
b
d
a
f
f
P
Q S
G
p 





• 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
j
d
f
M
A
y
II
s
*
*
2 
j
d
f
M
A
y
I
s
*
*
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
.
22
8
120
*
50
.
1
*
50
.
1
* m
q
P
B
A
Area
a
S
G




m
Area
A 00
.
5
75
.
4 


Assume
t
m
M L
F .
40
00
.
2
*
10
20
. 


290
.
0
138
40
.
.



L
F
L
F
P
M
e
t
P
P s
G
L
F 138
120
*
15
.
1
*
15
.
1 .
. 


833
.
0
6
5
6



A
e A
A1
t1
t
PF.L
PG.S
qmax
G.S
M H=10
2.00
qmin
8
)
00
.
5
290
.
0
*
6
1
(
*
5
138
max 


B
q m
B 65
.
4


• Reinforced concrete footing (R.C.)
Dim. of R.C. = 4.40 * 4.05* 0.75
m
t
X 30
.
0


85
.
5
34
.
14
)
40
.
4
308
.
0
*
6
1
(
05
.
4
*
40
.
4
120
*
50
.
1
2
,
1 



f
m
P
M
e
S
G
308
.
0
120
70
.
1
*
10
20
.
1
1 



36
.
47
3
2
*
)
2
6
.
0
4
.
4
)(
67
.
10
34
.
14
(
2
/
)
2
60
.
0
4
.
4
(
67
.
10 2
2






I
M
33
.
35
)
2
30
.
0
05
.
4
(
*
)
2
85
.
5
34
.
14
( 2




II
M
m
X
A
A 40
.
4
30
.
0
*
2
00
.
5
2
1 




m
X
B
B 05
.
4
30
.
0
*
2
65
.
4
2
1 




cm
d 70
8
.
68
250
*
100
10
*
36
.
47
5
5



cm
t 75
5
70
1 


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
.
0
90
.
1
(
*
2
)
84
.
11
34
.
14
(
)
( 



II
s
Q
86
.
11
)
70
.
0
2
3
.
0
05
.
4
(
*
2
)
85
.
5
34
.
14
(
)
( 




III
s
Q
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
.
0
60
.
0
90
.
1
(
*
40
.
4
)
85
.
5
84
.
11
(
85
.
5
4 





f
68
.
9
/
24
.
2
70
*
100
1000
*
71
.
15 2


 cm
kg
qs

• Punching Stress:
p
p
p
Q
q
A

 
0.5 ( / ) / /
cup cu c cu c
q 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
.
0
30
.
0
(
*
)
70
.
0
60
.
0
[(
2
)
85
.
5
34
.
14
(
120
*
50
.
1 





p
Q
22
.
3
)]
70
.
30
.
0
(
)
70
.
0
60
.
0
[(
*
2
*
70
.
0 




p
A
18
.
5
10
*
22
.
3
1000
*
88
.
166
4


p
q

• 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
.
16
826
.
0
*
70
*
3600
10
*
33
.
35 5
2 

s
A
75
.
22
826
.
0
*
70
*
3600
10
*
36
.
47 5
1 

s
A
Main RFT
618/m’
Main RFT
818/m’

Crane Footing
Maximum stress on soil should be less than allowable bearing capacity
a
q
f 
max
The ratio between maximum and minimum stresses should be less
than four
4
)
/
6
1
(
)
/
6
1
(
min
max




A
e
A
e
f
f
4
min
max

f
f
e
A 10


Uniform Stress
below Footing Subjected to Moment
Uniform stress required that the eccentricity at foundation level equal
zero
0
.
0

e 0
.
0
.
.
.


s
G
L
F
P
M
e
A
PF.L
PG.S
qa
G.S
M
e c
a/2
e
a
c
A



2
2
A
a
S
G
q
P
B
A .
15
.
1
*  B

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