SEEPAGE FORCES
Consider a random element of a flow net:
Each side has the same length, b
A
b
B
b
θ
D
C
the direction of flow is inclined at an angle of θ to the horizontal
lines AB and DC define the elemental flow channel
lines AD and BC are equipotentials, with a drop in head of ∆h when water
seeps from AD to BC
SEEPAGE FORCES
Geometrically:
A
D
B
θ
θ
θ
θ
C
θ
Fourdifference
The
congruent right
in elevation
angle triangles
between are
A and
formed
D is the
B
from
same
vertical
as
and horizontal
between
B and lines
D
C andprojected
is equal toinwards
bcosθ.
bsinθ. from the four corners of
the flow net element
Each has an angle θ as shown
SEEPAGE FORCES
The pore pressure distributions acting on each side
of the element are shown below:
If the
The
change
pore in
water
pore water
pressure between
at point
from
point
A
point
is
A uto
B,
A
pointpoint
and
due
is due
to and
a only
loss to
in
u B= is
wC
(h-z),
totalelevation
the
head -∆h
drop,
and bcosθ,
the
elevation
drop,
bsinθ
the change
in pore
water
pressure between point A
and point D is due only to
the elevation drop, bcosθ,
uD = uA + w bcosθ
uB = uA + w(bsinθ-∆h)
bsinθ
-∆h
bcosθ
uC = uAB + wwbcosθ
(bsinθ-∆h)
(bsinθ+bcosθ-∆h)
or
+ wbcosθ or
bcosθ
SEEPAGE FORCES
The
The
The
pore
net
equivalent
pressure
boundary
point
distribution
water
loadforce
(net
acting
boundary
acting
onon
AD
ABwater
BCwill
is:be
2cosθ
cancelled
cancelled
by
byDC
that
that
acting
on
BC,
DC
leaving:
leaving:
force)
b acting
x w(bsinθ-∆h)
on
is:acting
orb xwb2on
sinθ
bcosθ
∆h
or

b
b
w
ww
uD-uA = uC-uB = w bcosθ
w(bsinθ-∆h)
uB-uA = uC-uD = w(bsinθ-∆h)
uD = uA + w bcosθ
w bcosθ
uB = uA + w(bsinθ-∆h)
uC = uA + w(bsinθ+bcosθ-∆h)
w b2cosθ
SEEPAGE FORCES
What would the boundary water forces be if
seepage stopped? (i.e., the static case)
∆h would be 0, and the
The forces
only difference
on DC andbetween
BC wouldthe
be
2
static
and
and
seepage
wb2sinθcases
respectively,
is the force
w b cosθ
orthogonal
vectors
with a resultant
∆h
the seepage
force, J of
wb called
2 , (acting vertically)
If
the
average hydraulic gradient, i
w b
across the element is:
h
i
b
Δh
Then: J  Δhγwb 
γwb2  iγwb2
b
If b2 x 1 m is the volume of the
element, V then the seepage pressure,
j is defined as the seepage force per
unit volume:
j = i
w
w b2cosθ
SEEPAGE FORCES
the
total
weight
of
element
=gravitational
sat
b2stress
=conditions
vector
abseepage
How
If
In
Therefore,
the
will
effective
extremes:
The
seepage
let’s
concern
consider
stress
affect
ifthe
is
the
with
is
the
seepage
all
reduced
the
the
effective
support
direction
too
much
isby
downward,
at
and
upward
any
ofpoint
the
seepage,
the
soil.
in
forces
the
effective
soil
then
acting
mass?
theon
Boundary
water
force
on
wb2loads.
cosθ
vector
soil will
stress
the
soilwill
lose
element
beits
increased
ability
à la a vector
toorCD
support
if =upward
diagram.
the=First,
effective
thebdSEEPAGE
stress willcase:
be decreased
Boundary water force on BC = wb2sinθ-∆hwb = vector de
Resultant boundary water force = vector be
Resultant body force = vector ae = Effective Stress, σ’
w b2cosθ
SEEPAGE CASE
SEEPAGE FORCES
Nowtotal
consider
theofSTATIC
case:= satb2 = vector ab
the
weight
the element
Boundary water force on CD = wb2cosθ = vector bd
Boundary water force on BC = wb2sinθ = vector dc
Resultant boundary water force = wb2 = vector bc
Resultant body force =  ’b2 vector ac = Effective Stress, σ’
w b2cosθ
STATIC CASE
SEEPAGE FORCES
This brings up an alternative solution to the seepage case:
Effective weight of the element =  ’b2 = vector ac
Seepage force = ∆hwb = vector ce
Resultant body force vector ac = Effective Stress, σ’
To summarize, the resultant body force (effective stress)
can be obtained by considering:
B)
skeleton,
A) the equilibrium of the soil
whole
soil mass,
add the effective
weightweight
of theof
soil
mass
total saturated
the
soil (ac)
mass (ab)
OR
to the seepage
(be) water force (ce)
resultantforce
boundary
to find effective stress (ae)
SEEPAGE CASE (reprise)