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Islamic University of Gaza
Lectures notes: Foundation Engineering
Bearing Capacity of Shallow Foundations
· Ultimate bearing capacity (qu)
Shear failure أقل حمل مقسوم على المساحة يؤدي لحدوث
· Allowable bearing capacity (qall)
Shear failure حمل مقسوم على المساحة يمكن ان تتحمله التربة بدون حدوث
qall should be adequate to prevent excessive settlement and shear failure
· Types of shear failure:
1- General shear failure: For Dense sand.
2- Local shear failure: For Medium compaction soil.
3- Punching shear failure: For loose soil
Calculation of ultimate bearing capacity of shallow foundations without eccentricity:
1- Terzaghi’s theory:
Assumption for Terzaghi’s theory:
· The foundation is considered to be shallow if
(
)
B
D
f
£
, in recent studies the foundation is considered to be shallow if
(
)
4
/
£
B
D
f
. Other wise it is considered to be deep foundation.
· Foundation is considered to be strip if
(
)
00
.
0
/
®
L
B
.
· The soil from ground surface (سطح الأرض الطبيعية) to the bottom of the foundation (سطح التأسيس) is replaced by stress
f
D
q
g
=
.
For General shear failure:
Type of foundation
Ultimate bearing capacity qu
Strip Footing
g
g
BN
qN
cN
q
q
c
u
2
1
+
+
=
Square footing
g
g
BN
qN
cN
q
q
c
u
4
.
0
3
.
1
+
+
=
Circular footing
g
g
BN
qN
cN
q
q
c
u
3
.
0
3
.
1
+
+
=
c: Cohesive.
f
D
q
g
=
B: Foundation width (Diameter if circular).
g
N
N
N
q
c
,
,
: Bearing capacity factors given from table 3.1 P.158 as function of angle of friction .
For Local shear failure:
Type of foundation
Ultimate bearing capacity qu
Strip Footing
'
'
'
2
1
3
2
g
g
BN
qN
cN
q
q
c
u
+
+
=
Square footing
'
'
'
4
.
0
867
.
0
g
g
BN
qN
cN
q
q
c
u
+
+
=
Circular footing
'
'
'
3
.
0
867
.
0
g
g
BN
qN
cN
q
q
c
u
+
+
=
'
'
'
,
,
g
N
N
N
q
c
: Factors for bearing capacity given from table 3.2 P.160
Or from table 3.1 P158 but replace
'
F
by
F
:
÷
ø
ö
ç
è
æ
F
=
F
-
tan
3
2
tan
1
'
2- Meyerhof’s equations (General bearing capacity equation):
Terzagi equations neglect:
· Rectangular footings.
· Inclination of loads.
· Shear strength of soil above the foundation.
Meyerhof’s equation takes in consideration theses variables:
i
d
s
qi
qd
qs
q
ci
cd
cs
c
u
F
F
F
BN
F
F
F
qN
F
F
F
cN
q
g
g
g
g
g
5
.
0
+
+
=
g
N
N
N
q
c
,
,
: Table 3.4 P.168
P.169
3.5
Table
factors.
n
Inclinatio
,
,
factors.
Depth
,
,
factors.
Shape
,
,
Þ
ú
ú
ú
û
ù
ê
ê
ê
ë
é
Þ
Þ
Þ
i
qi
ci
d
qd
cd
s
qs
cs
F
F
F
F
F
F
F
F
F
g
g
g
3- Skempton’s equation for clay without inclination:
÷
ø
ö
ç
è
æ
+
÷
÷
ø
ö
ç
ç
è
æ
+
=
L
B
B
D
c
q
f
u
2
.
0
1
2
.
0
1
5
4- Vesic’s equation (Consider compressibility of soil):
c
d
s
qc
qd
qs
q
cc
cd
cs
c
u
F
F
F
BN
F
F
F
qN
F
F
F
cN
q
g
g
g
g
g
5
.
0
+
+
=
g
N
N
N
q
c
,
,
: Table 3.4 P.168
Þ
c
qc
cc
F
F
F
g
,
,
Soil Compressibility factors.
Effect of water table in bearing capacity equations:
Case I) Water table is located at depth D1 so that 0 ≤ D1 ≤ Df:
2
'
1
D
D
q
g
g
+
=
(الشق الثاني من المعادلة)
w
sat
g
g
g
g
-
=
=
'
(الحد الأخير من المعادلة)
Case II) Water table is located at depth d below the foundation so that 0 ≤ d ≤ B:
f
D
q
g
=
(الشق الثاني من المعادلة).
(
)
'
'
g
g
g
g
g
-
+
=
=
B
d
الحد الأخير من المعادلة) )
Case III) Water table is located at depth d below the foundation so that d > B:
No changes in equations.
Factor of safety:
Ultimate bearing capacity من خلال المعادلات السابقة حسبنا
و هذه القيمة تمثل الاجهاد الذي اذا أثر على التربة تنهار عنده، ولهذا لا يجب استخدامه عند التصميم بل نستخدم قيمة أقل منه وهذا من خلال قسمة هذه القيمة على معامل أمان.
(
)
(
)
(
)
load.
Ultimate
capacity
bearing
allowable
Net
capacity
bearing
allowable
Gross
capacity
bearing
ultimate
Net
capacity
bearing
ultimate
Gross
Gross
Q
q
q
q
q
q
q
u
net
all
all
u
net
u
u
Þ
Þ
Þ
Þ
-
=
Þ
(
)
(
)
FS
q
q
FS
q
q
FS
q
q
FS
q
q
all
u
net
u
net
all
u
all
-
=
-
=
=
Þ
=
Þ
FS = (3 – 4) for bearing capacity
Factor of safety with respect to shear:
Example 1)
Determine the size of square footings to carry gross allowable load (295 KN) given that:
00
.
0
.
35
.
/
15
.
18
00
.
1
3
3
=
=
F
=
=
=
C
m
KN
D
FS
f
o
g
Use Terzagi equations assuming general shear failure.
295KN
Df=1.00
B
C=0.00
F
=35
g
=18.15KN/m3
Solution
92cm
B
:
error
and
By trial
684
.
2
2814
.
2
41
.
45
15
.
18
4
.
0
44
.
41
1
15
.
18
0
885
41
.
45
,
44
.
41
,
75
.
57
35
At
4
.
0
3
.
1
:
footing
square
For
.
885
3
295
.
295
2
3
2
2
2
2
=
=
+
Þ
´
´
´
+
´
´
+
=
=
=
=
Þ
=
F
+
+
=
=
´
=
´
=
=
=
´
=
B
B
B
B
N
N
N
BN
qN
cN
q
B
B
FS
q
q
B
A
Q
q
Area
q
Q
q
c
q
c
u
all
u
all
all
all
all
g
g
g
Example 2)
Determine the net allowable load that foundation can carry (no inclination), Use Meyerhof equation given that:
.
/
50
.
25
.
/
10
00
.
2
4
2
3
m
KN
C
m
KN
D
FS
w
f
=
=
F
=
=
=
o
g
Solution
i
d
s
qi
qd
qs
q
ci
cd
cs
c
u
F
F
F
BN
F
F
F
qN
F
F
F
cN
q
g
g
g
g
g
5
.
0
+
+
=
88
.
10
,
66
.
10
,
72
.
20
25
At
=
=
=
Þ
=
F
g
N
N
N
q
c
The water table is at depth = 1m
.
/
4
.
9
10
4
.
19
.
/
2
.
26
1
)
10
4
.
19
(
1
8
.
16
3
'
2
m
KN
m
KN
q
=
-
=
=
=
´
-
+
´
=
g
g
· Shape factors:
(
)
733
.
0
3
2
4
.
0
1
4
.
0
1
311
.
1
25
tan
3
2
1
tan
1
343
.
1
72
.
20
66
.
10
3
2
1
1
=
÷
ø
ö
ç
è
æ
-
=
-
=
=
+
=
F
+
=
=
´
+
=
÷
÷
ø
ö
ç
ç
è
æ
+
=
L
B
F
L
B
F
N
N
L
B
F
s
qs
c
q
cs
g
· Depth factors:
(
)
(
)
1
313
.
1
2
2
25
sin
1
25
tan
2
1
4
.
1
2
2
4
.
0
1
1
2
/
2
/
2
=
=
´
-
+
=
=
÷
ø
ö
ç
è
æ
+
=
=
=
d
qd
cd
f
F
F
F
B
D
g
· Inclination factors:
Due to absence of inclined load, the inclination factor is 1 every where.
(
)
(
)
(
)
(
)
(
)
KN
A
q
Q
m
KN
FS
q
q
m
KN
q
q
q
m
KN
q
q
net
all
net
all
net
u
net
all
u
net
u
u
u
1
.
3716
6
35
.
619
/
35
.
619
4
4
.
2477
/
4
.
2477
2
.
26
6
.
2503
.
/
6
.
2503
1
1
733
.
0
88
.
10
2
4
.
9
5
.
0
1
313
.
1
311
.
1
66
.
10
2
.
26
1
4
.
1
343
.
1
72
.
20
50
2
2
2
=
´
=
´
=
=
=
=
=
-
=
-
=
=
´
´
´
´
´
´
+
´
´
´
´
+
´
´
´
´
=
HOMEWORK CH (3)
Use
3
3
/
4
.
62
/
10
ft
Ib
m
KN
w
=
=
g
Q.1) Find the ultimate bearing capacity of square footing ( 2.5m x 2.5m) which is placed 2.5m below the ground surface of soil having the shown properties:
2.5 m
2.5 m
g
d=18KN/m3
C=0.00
F
=30
h
The water table is located at distance (h) below the ground surface; if sat=19KN/m2
Find the ultimate bearing capacity using Terzagi equation for the following cases:
· h = 7m.
· h = 4m.
· h = 1m.
· h = 0m.
Comment on the results.
Q.2) For the square foundation shown in the
Figure, find the gross allowable load that
Foundation can carry for the following cases:
A- No water table is observed.
B- Water table at depth 0.5m below the
Bottom of the foundation.
Use FS = 3 with Meyerhof equation.
Q.3) for the shown square footing (2.5m x 2.5m) if the allowable load P=800 KN, use FS=6 to determine the allowable resisting moment (M). Use Terzagi equations (=35o)
HW …..3.4 , 3.5 , 3.8 , 3.11
elastal