dr.: youssef gomaa youssef
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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.
CE 402: Foundation Engineering Design
The kern To avoid zero pressure, the
eccentricity must stay within
the kern.
A
A/3
CE 402: Foundation Engineering Design
Eccentric Loads or Moments Cases
A
e <A/6
Case (a)
e =A/6
Case (b)
e >A/6
Case (c)
CE 402: Foundation Engineering Design
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
CE 402: Foundation Engineering Design
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
CE 402: Foundation Engineering Design
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
CE 402: Foundation Engineering Design
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
CE 402: Foundation Engineering Design
Design of Spread Footing
• 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
CE 402: Foundation Engineering Design
Design of Spread Footing
• Punching Stress:
*2* ( ) ( )pA d a d b d
p
p
p
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
CE 402: Foundation Engineering Design
Design of Spread Footing
• 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
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
pG.S = 120t M=20m.t H=10t
qa = 0.80kg/cm2 = 8t/m2.
fcu = 250kg/cm2.
fy =3600kg/cm2
CE 402: Foundation Engineering Design
Design of Eccentric Footing
• 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
CE 402: Foundation Engineering Design
Design of Eccentric Footing
• 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
CE 402: Foundation Engineering Design
Design of Spread Footing
• 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
CE 402: Foundation Engineering Design
Design of Spread Footing
• Punching Stress:
p
p
p
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
CE 402: Foundation Engineering Design
Design of Spread Footing
• 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’
CE 402: Foundation Engineering Design
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
CE 402: Foundation Engineering Design
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
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