Civil Engineering Reference
In-Depth Information
desired
T
/
B
curve, the value of
N
c
can be determined. Using Fig. 8.8, values of
N
c
for dif-
ferent
T
/
B
ratios are as follows:
T
/
B
N
c
Percent reduction in
N
c
0
0
100
0.25
0.7
87
0.50
1.3
76
1.00
2.5
55
1.50
3.8
31
5.5
0
Example Problem for Cohesive Surface Layer.
This example problem illustrates the use
of Eq. (8.6). Use the data from the example problem in Sec. 8.2.2.
Solution.
To calculate the factor of safety in terms of a bearing capacity failure for the
strip and spread footings, the following values are used:
P
50 kN/m for strip footing and 500 kN for spread footing
T
2.5 m i.e., total thickness of unliquefiable soil layer minus footing embed-
ment depth
3 m
0.5 m
2.5 m
c
1
s
u
50 kPa
50 kN/m
2
upper cohesive soil layer
c
2
0 kPa
0 kN/m
2
liquefied soil layer
B
1 m strip footing
B
L
2 m spread footing
N
c
5.5 for strip footing, using Fig. 8.8 with
T
/
B
2.5/1
2.5 and
c
2
/
c
1
0
N
c
3.2 for spread footing, using Fig. 8.8 with
T
/
B
2.5/2
1.25 and
c
2
/
c
1
0
Substituting the above values into Eqs. (8.6
a
) and (8.6
b
), gives
q
ult
cN
c
s
u
N
c
(50 kN/m
2
) (5.5)
275 kN/m
2
for strip footing
q
ult
s
u
N
c
(
1
0.3
B
L
)
1.3
s
u
N
c
(1.3) (50 kN/m
2
) (3.2)
208 kN/m
2
for
spread footing
The ultimate load is calculated as follows:
__
Q
ult
q
ult
B
(275 kN/m
2
) (1 m)
275 kN/m
for strip footing
Q
ult
q
ult
B
2
(208 kN/m
2
) (2 m)
2
832 kN
for spread footing
And finally the factor of safety is calculated as follows:
FS
Q
ult
P
275 kN/m
___
_________
50kN/m
5.5
for strip footing
FS
Q
ult
P
832 kN
___
_______
500 kN
1.7
for spread footing
These values are similar to the values calculated in Sec. 8.2.2 (i.e., FS
5.0 for the strip
footing and FS
2.0 for the spread footing).
