Fayoum University
Faculty of Engineering
Department of Civil Engineering
CE 402: Part A
Shallow Foundation Design
Lecture No. (6): Eccentric Footing
Dr.: Youssef Gomaa Youssef
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
P  PD.L  PL.L
e
A
M
(P  Wf )
e < A/6
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.
CE 402: Foundation Engineering Design
The kern
To avoid zero pressure, the
eccentricity must stay within
the kern.
A/3
A
CE 402: Foundation Engineering Design
Eccentric Loads or Moments Cases
A
e <A/6
e =A/6
e >A/6
Case (a)
Case (b)
Case (c)
CE 402: Foundation Engineering Design
Eccentric Loads or Moments
For case A and B (not case C)
qmin
P
6e

(1  )
A* B
A
qmax
P
6e

(1  )
A* B
A
For case c
C
qmax 
2* P
3* C * B
3C  0.75 A
qmax  1.20qa
CE 402: Foundation Engineering Design
e >A/6
Case (c)
Footing Subjected to Double Moment
P
eA
eB
q
(1  6  6 )
A* B
A
B
For contact pressure to remain (+) ve everywhere,
6eA 6eB

 1.0
A
B
CE 402: Foundation Engineering Design
Design of Eccentric Footing
• Plain concrete footing (P.C.)
M
1.50 PG.S
Area  A * B 
qa
Assume
t1
M F .L  M  H * D f
Check
qmax  qa 
H
G.S
A  Area
PF .L  1.15 * PG.s
PG.S
PF.L
Df
t
e
M F .L
PF .L
A
e
6
PF .L
6e
(1  )
A* B
A
qmin
qmax
A1
A
B
Assume thickness of P.C.:
t = (0.25 to 0.50)
CE 402: Foundation Engineering Design
Dim. of P.C. = A * B * t
Design of Spread Footing
• Reinforced concrete footing (R.C.)
X  (0.8  1.0)t
A1  A  2X
e1 
f1, 2
B1  B  2X
a
M1
PG .S
PG.S
G.S
P
6e
 G.S (1  1 )
A1 * B2
A1
A a 2
A a 2 2
M I  f3 ( 1
) / 2  ( f1  f 3 )( 1
) *
2
2
3
f f
B b 2
M II  ( 1 2 ) * ( 1
)
2
2
M
d C
b * Fcu
t1  d  cover
CE 402: Foundation Engineering Design
t1
f2
f1
f3
A1
Dim. of R.C. = A1 * B1* t1
Design of Spread Footing
• Shear Stress:
Qs ( II ) 
PG.S
( f1  f 4 ) A1  a
*(
 d)
2
2
a
G.S
(II)
Qs ( III )
( f  f ) B b
 1 2 *( 1
 d)
2
2
qs 
t1
Qs
 q su
b *d
f2
f1
q su  0.75
f cu
d
f4
(II)
A1
c
If qs > qsu , Increase d
Notes:
•No shear RFT in Footing.
CE 402: Foundation Engineering Design
Qs: shear force at critical sec. (II).
qs: shear stress.
qsu: ultimate shear strength.
Design of Spread Footing
PG.S
• Punching Stress:
a
G.S
Q p  1.50 PG.S 
( f1  f 2 )
[(a  d ) * (b  d )]
2
t1
A p  d *2*(a  d )  (b  d )
f2
f1
qp 
Qp
d/2
Ap
qcup  0.5  (a / b ) f cu /  c  f cu /  c
If qp > qcup , Increase d
CE 402: Foundation Engineering Design
A1
d/2
b
d/2
a
d/2
Design of Spread Footing
• Footing Reinforcement:
MI
As1 
fy *d * j
a
G.S
As 2 
M II
fy *d * j
Main RFT
As2
t1
Main RFT
As1
A1
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.
CE 402: Foundation Engineering Design
Design of Eccentric Footing
• 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
fcu = 250kg/cm2.
pG.S = 120t
M=20m.t H=10t
fy =3600kg/cm2
qa = 0.80kg/cm2 = 8t/m2.
CE 402: Foundation Engineering Design
Design of Eccentric Footing
• Plain concrete footing (P.C.)
M
1.50 * PG.S 1.50 *120
Area  A * B 

 22.5m 2
qa
8
Assume
PG.S
H=10
G.S
A  Area  4.75  5.00m
t1
M F .L  20  10 * 2.00  40m.t
t
PF .L  1.15 * PG.s  1.15 *120  138t
e
M F .L 40

 0.290
PF .L 138
qmax 
2.00
PF.L
e
qmin
A 5
  0.833
6 6
138
6 * 0.290
(1 
)8
5* B
5.00
Assume thickness of P.C.:
t = 0.30
CE 402: Foundation Engineering Design
qmax
A1
A
B  4.65m
Dim. of P.C. = 5.00* 4.65 * 0.30
Design of Eccentric Footing
• Reinforced concrete footing (R.C.)
X  t  0.30m
A1  A  2 X  5.00  2 * 0.30  4.40m
0.60
B1  B  2 X  4.65  2 * 0.30  4.05m
e1 
M 1 20  10 *1.70

 0.308m
PG.S
120
f1, 2 
1.50 *120
6 * 0.308
(1 
)  14.34  5.85
4.40 * 4.05
4.40
PG.S
G.S
1.90
4.4  0.60 2
4.4  0.6 2 2
M I  10.67(
) / 2  (14.34  10.67)(
) *  47.36
2
2
3
14.34  5.85
4.05  0.30 2
M II  (
)*(
)  35.33
2
2
d 5
5
47.36 *10
100 * 250
 68.8  70cm
t1  70  5  75cm
CE 402: Foundation Engineering Design
t1
F2=5.85
F1=14.34
F3=10.67
A1=4.40
Dim. of R.C. = 4.40 * 4.05* 0.75
Design of Spread Footing
• Shear Stress:
f 4  5.85 
Qs ( II ) 
PG.S
(11.84  5.85)
* (1.90  0.60  0.6)  11.84
4.40
0.60
G.S
(14.34  11.84)
* (1.90  0.70)  15.71
2
(II)
1.90
d
t1
Qs ( III ) 
qs 
(14.34  5.85) 4.05  0.3
*(
 0.70)  11.86
2
2
15.71*1000
 2.24kg / cm 2  9.68
100 * 70
f2=5.85
f4=11.84
f1=14.34
(II)
A1=4.40
If qs > qsu , Increase d
Notes:
•No shear RFT in Footing.
CE 402: Foundation Engineering Design
Qs: shear force at critical sec. (II).
qs: shear stress.
qsu: ultimate shear strength.
Design of Spread Footing
PG.S
• Punching Stress:
a
G.S
Q p  1.50 *120 
(14.34  5.85)
[(0.60  0.70) * (0.30  0.70)]  166.88
2
t1
Ap  0.70 * 2 *[(0.60  0.70)  (0.30  .70)]  3.22
f2
f1
qp 
Qp
Ap
qp 
166.88 *1000
 5.18
3.22 *104
qcup  0.5  (a / b ) f cu /  c  f cu /  c
If qp > qcup , Increase d
CE 402: Foundation Engineering Design
A1
d/2
d/2
b
d/2
a
d/2
Design of Spread Footing
• Footing Reinforcement:
0.60
47.36 *105
As1 
 22.75
3600 * 70 * 0.826
G.S
As 2 
5
35.33 *10
 16.97
3600 * 70 * 0.826
Main RFT
618/m’
Main RFT
818/m’
0.75
4.40
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.
CE 402: Foundation Engineering Design
Crane Footing
Maximum stress on soil should be less than allowable bearing capacity
f max  qa
The ratio between maximum and minimum stresses should be less
than four
f max
4
f min
A  10e
CE 402: Foundation Engineering Design
f max (1  6e / A)

4
f min (1  6e / A)
Uniform Stress
below Footing Subjected to Moment
Uniform stress required that the eccentricity at foundation level equal
zero
e  0.0
A
a
c e
2
2
1.15PG.S
A* B 
qa
e
M F .L
 0.0
PG.s.
e
a/2
PG.S
M
G.S
A
PF.L
A
B
CE 402: Foundation Engineering Design
c
qa