Transcript FLOW NETS - Mohawk College

EXAMPLE 3.3 (Craig)
First,flow
The
direction
net forofseepage
under a sheet
seepage
is leftpile
to right.
wall is
shown
in Figure
3.8 a),
The total
head driving
the saturated
seepage
is 8 m.unit weight
of
thenumber
soil being
Then,
the20
3.
kN/m
equipotentials:
Determine the values of
effective stress at A and
B.
12
11
0
+
+
10
9
8
sat = 20 kN/m3
7
6
5
4
1
3
2
EXAMPLE 3.3 (Craig)
3A is uatwater
The
First
The
4m
volume
resultant
pore
forofPoint
water
water
of
Apressure
AD
(effective
11m
D exerts
x
on1m
the
stress,
x=another
column
1m
=σ
11A’)at
m
boundary
acting
Therefore,
uA body
=column
wabove
(hforce
) acting
=is
9.8(5.5+7)
122.5kPa
A A is:
A-z
Apoint
3 x 11
force
equal
tocolumn
9.8 x
=122.5kN
39.2
kN
consider
the
elevation
head
A,
soil
AD,
11
m high
and
long and
Therefore,column
220kN
The
boundary
 the
+ 39.2kN
water
AD
at4of
-force
weighs
on
20kN/m
the
= bottom
136.7kN
1m
m231m
==122.5kN
220kN
 1m
wide
the total head at A:
1 m2
D
39.2 kN
11.00 m
zA σA’ 7m
= 136.7 kN
12
11
0
10
A
220 kN
8.2
hA  8.00 
 5.46  5.5m
122.5kN
12
8.2
9
8
sat = 20 kN/m3
+
+
7
6
5
4
1
3
2
EXAMPLE 3.3 (Craig)
3 column
3
We
The
could
resultant
have
body
solved
(effective
’ using
the
stress,
Effective
σA’) acting
Weight
at11
of
Amis:
The
’
= 20-9.8
seepage
= 10.2
force
kN/m
Jforce
=3 iand
VAAD
the
, VAD
volume
= 11
mof
AD is
A for
wσ
column
AD
and
the
seepage
∆h
from
Aweight
is 8.0
=force,
2.5AD
over
ax
distance
of kN
11 m:
The
112.2kN
effective
D +to
24.5kN
 of
=– 5.5
136.7kN
column
m,Jis
A 10.2
11 = 112.2

m
Ji A 2.5
= 0.227x9.8x11=24.5kN

 0.227 m/m
1 m2
11 m
D
12
11.00 m
11
0
10
A
σA’ = 136.7 kN
112.2 kN
24.5 kN
8.2
9
8
sat = 20 kN/m3
+
+
7
6
5
4
1
3
2
EXAMPLE 3.3 (Craig)
3B is u atwater
Now
The
pore
1m
volume
resultant
forof
Point
water
water
body
B
column
above
acting
(effective
6 Cmexerts
xon1 the
mstress,
column
1 m =σ6
at
boundary
macting
B is:
Therefore,
uof
pressure
(hforce
) =is
9.8(1.6+7)
=xanother
84.28
kPa
B’)
B
B=
w
B-zBC
Bpoint
3kN
3==84.3
force
equal
tocolumn
9.8
=84.3
9.8on
kN
consider
the
elevation
head
B,
soil
BC,
6=kN/m
m
high
long
and
The
Therefore,column
120
kN
boundary
 the
+ 9.8
water
kN x
BC
at1of
-force
weighs
kN
20
the
45.5
bottom
xand
61m
m21m
120kN
kN
1m
wide
the total head at B:
C
1 m2
9.8 kN
12
11
0
6.00 m
zB  7m
σB’ = 45.5 kN
B
2.4
hB  8.00 
 1.6m
12
120 kN
84.3 kN
+
+
10
9
8
sat = 20 kN/m3
7
6
5
4
1
2.4
3
2
EXAMPLE 3.3 (Craig)
3
The
Now
resultant
finding= σ10.2
’ using
Effective
σB’)
column
acting
Bmis:
’ = 20-9.8
seepage
force
kN/m
Jforce
=3iand
VBCthe
, VBCvolume
= Weight
6stress,
m3 of of
column
BC BC
isat6and
Bbody
Bthe
w(effective
the
seepage
force,
JB–of
∆h
Cweight
iskN
1.6
= 1.6
m,
a distance
ofkN
6 m:
Thefrom
61.2
effective
kN B-to
15.7
= 0.0
45.5
column
kN BC
 over
is 10.2
x 6 = 61.2

m
JiA =1.6
0.266x9.8x6=15.7
kN 
 0.26 m/m
6m
1 m2
12
C
11
0
6.00 m
B
σA’ = 45.5 kN
61.2 kN
15.7 kN
+
+
10
9
8
sat = 20 kN/m3
7
6
5
4
1
2.4
3
2