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Flow through soils in retaining walls
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EFFECTS OF UNBALANCED
HYDROSTATIC AND
SEEPAGE FORCES
ASSIGNMENT 2
EARTH RETAINING METHOD
RAMIREZ RYAN
()201383554
OUTLINE
I. REVIEW ON BERNOULLIS EQUATION
and FLOW NET
II. PROBLEM
III. SOLUTION
IV. FINAL ANSWER
1. Kinetic energy
datum
z
fluid particle
The energy of a fluid particle is made of:
2. Strain energy
3. Potential energy
- due to velocity
- due to pressure
- due to elevation (z) with respect to a datum
I. BERNOULLIS EQUATION
Total head =
datum
z
fluid particle
Expressing energy in unit of length:
Velocity head
+
Pressure head
+
Elevation head
I. BERNOULLIS EQUATION
Total head =
datum
z
fluid particle
For flow through soils, velocity (and thus
velocity head) is very small. Therefore,
Velocity head
+
Pressure head
+
Elevation head
0
Total head = Pressure head + Elevation head
I. BERNOULLIS EQUATION
h = h + h
If flow is from A to B, total head is
higher at A than at B.
water
AB
Energy is dissipated
in overcoming the
soil resistance and
hence is the head
loss.
I. BERNOULLIS EQUATION
I. BERNOULLIS EQUATION
concrete dam
impervious strata
soil
Stream/Flow line is simply the path of a water molecule.
datum
hL
hT = 0hT = hL
From upstream to downstream, total head steadily
decreases along the stream line.
I. SEEPAGE TERMINOLOGY
Equipotential line is simply a contour of constant total
head.
concrete dam
impervious strata
soil
datum
hL
hT = 0.8 hL
I. SEEPAGE TERMINOLOGY
hT = 0hT = hL
A network of selected stream lines and
equipotential lines.
concrete dam
impervious strata
soil
curvilinear square
90
I. FLOW NET
impervious strata
concrete dam
datum
X
z
hL
hT= hL hT = 0
Total head, hT = hL no. of drops from upstream x h
h
Elevation head, hE = -z
Pressure head, hP = Total head (hT) Elevation head (hE)
I. HEADS AT A POINT X
h =hL
Nd
hL, total head loss
h, head loss per element
Nd, no. of potential drop
Nf, no. of flow channel
Determine the net water pressure acting on the single row of sheet piles shown.
impervious strata
II. PROBLEM
1
34
5
6
2.50 m
4.00 m
4.70 m
4.80 m
El. 0 m
7
2
Single row of sheet piles
Downstream
Upstream
Datum
Nd = 12; Nf = 5; hL = 2.50 m
III. SOLUTION
SOLUTION TABLE
LOCATION/
POINT
ELEVATION
HEAD
hE (m)
TOTAL HEAD
hT (m)
PRESSURE
HEAD
hP (m)
WATER
PRESSURE
u (kPa)
1
2
3
4
5
6
7
III. SOLUTION
1. DETERMINE THE WATER PRESSURE ACTING AT THE BACK OF THE SHEET PILE.
1.1. Elevation head (hE) measured from datum
At point 1: hE = 0 m
At point 7: hE = -8.70 m
1.2. Total head (hTB)
At point 1:
At point 7:
LOCATION
ELEVATION
HEAD
hE (m)
TOTAL
HEAD
hTB (m)
SEEPAGE
PRESSURE
uB (kPa)
PRESSURE
HEAD
hPB (m)
WATER
PRESSURE
uB (kPa)
1 0.00 2.29
7 -8.70 1.25
hTB = 2.5 2.5
121 = .
hTB = 2.5 2.5
126 = .
III. SOLUTION
1. DETERMINE THE WATER PRESSURE ACTING AT THE BACK OF THE SHEET PILE
1.3. Seepage pressure (uB ) = WhTB
At point 1: uB = 9.8(2.29) = 22.44 kPa
At point 7: uB = 9.8(1.25) = 12.25 kPa
1.4. Pressure head (hPB) = hTB hE
At point 1: hPB = 2.29 0 = 2.29 kPa
At point 7: hPB = 1.25 (-8.7) = 9.95 kPa
LOCATION
ELEVATION
HEAD
hE (m)
TOTAL
HEAD
hTB (m)
SEEPAGE
PRESSURE
uB (kPa)
PRESSURE
HEAD
hPB (m)
WATER
PRESSURE
uB (kPa)
1 0.00 2.29 22.44 2.29
7 -8.70 1.25 12.25 9.95
III. SOLUTION
1. DETERMINE THE WATER PRESSURE ACTING AT THE BACK OF THE SHEET PILE
1.5. Water pressure (uB ) = WhPB
At point 1: uB = 9.8(2.29) = 22.44 kPa
At point 7: uB = 9.8(9.95) = 97.51 kPa
LOCATION
ELEVATION
HEAD
hE (m)
TOTAL
HEAD
hTB (m)
SEEPAGE
PRESSURE
uB (kPa)
PRESSURE
HEAD
hPB (m)
WATER
PRESSURE
uB (kPa)
1 0.00 2.29 22.44 2.29 22.44
7 -8.70 1.25 12.25 9.95 97.51
OR, Water pressure (uB ) = Wz + WhTB
At point 1: uB = 9.8(0) + 22.44 = 22.44 kPa
At point 7: uB = 9.8(8.70) + 12.25 = 97.51 kPa
III. SOLUTION
2. DETERMINE THE WATER PRESSURE ACTING IN FRONT OF THE SHEET PILE.
2.1. Elevation head (hE) measured from datum
At point 1: hE = 0 m
At point 7: hE = -8.70 m
2.2. Total head (hTF)
At point 1:
At point 7:
LOCATION
ELEVATION
HEAD
hE (m)
TOTAL
HEAD
hTF (m)
SEEPAGE
PRESSURE
uF (kPa)
PRESSURE
HEAD
hPB (m)
WATER
PRESSURE
uF (kPa)
1 0.00 0.00
7 -8.70 1.04
hTF = m
hTF = 2.5 2.5
127 = .
III. SOLUTION
2. DETERMINE THE WATER PRESSURE ACTING IN FRONT OF THE SHEET PILE
2.3. Seepage pressure (uB ) = WhTF
At point 1: uB = 9.8(0) = 0 kPa
At point 7: uB = 9.8(1.04) = 10.19 kPa
2.4. Pressure head (hPF) = hTF hE
At point 1: hP = 0 0 = 0 kPa
At point 7: hP = 1.04 (-8.7) = 9.74 kPa
LOCATION
ELEVATION
HEAD
hE (m)
TOTAL
HEAD
hTF (m)
SEEPAGE
PRESSURE
uB (kPa)
PRESSURE
HEAD
hPF (m)
WATER
PRESSURE
uF (kPa)
1 0.00 0.00 0.00 0.00
7 -8.70 1.04 10.19 9.74
III. SOLUTION
2. DETERMINE THE WATER PRESSURE ACTING IN FRONT OF THE SHEET PILE
2.5. Water pressure (uB ) = WhPF
At point 1: uB = 9.8(0) = 0 kPa
At point 7: uB = 9.8(9.74) = 95.45 kPa
LOCATION
ELEVATION
HEAD
hE (m)
TOTAL
HEAD
hT (m)
SEEPAGE
PRESSURE
uB (kPa)
PRESSURE
HEAD
hP (m)
WATER
PRESSURE
uF (kPa)
1 0.00 0.00 0.00 0.00 0.00
7 -8.70 1.04 10.19 9.74 95.45
3.1. Net water pressure (uNET) = uB uF
At point 1: uNET = 22.44 0 = 22.44 kPa
At point 7: uNET = 97.51 95.45 = 2.06 kPa
3. DETERMINE THE NET WATER PRESSURE ACTING ON THE SHEET PILE
IV. FINAL ANSWERLO
CA
TIO
N/
PO
INT
BACK OF THE SHEET PILE FRONT OF THE SHEET PILE
NET
WA
TER
PR
ESSU
RE
uN
ET(k
Pa
)
ELE
VA
TIO
N
HEA
D
hE(m
)
TOTA
L
HEA
D
hTB
(m)
SEEP
AG
E
PR
ESSU
RE
u
B(k
Pa
)
PR
ESSU
RE
HEA
D
hP
B(m
)
WA
TER
PR
ESSU
RE
uB
(kP
a)
TOTA
L
HEA
D
hTF
(m)
SEEP
AG
E
PR
ESSU
RE
u
F(k
Pa
)
PR
ESSU
RE
HEA
D
hP
F(m
)
WA
TER
PR
ESSU
RE
uF
(kP
a)
1 0.00 2.29 22.44 2.29 22.44 0.00 0.00 0.00 0 22.44
2 -2.70 2.08 20.38 4.78 46.84 0.00 0.00 2.7 26.46 20.38
3 -4.00 1.98 19.40 5.98 58.60 0.00 0.00 4 39.2 19.4
4 -5.50 1.83 17.93 7.33 71.83 0.21 2.06 5.71 55.96 15.87
5 -7.10 1.67 16.34 8.77 85.95 0.50 4.90 7.6 74.48 11.47
6 -8.30 1.46 14.31 9.76 95.65 0.84 8.23 9.14 89.57 6.08
7 -8.70 1.25 12.25 9.95 97.51 1.04 10.19 9.74 95.45 2.06
COMPLETE SOLUTION TABLE
IV. FINAL ANSWER
THANK YOU!
ASSIGNMENT 2
EARTH RETAINING METHOD
RAMIREZ RYAN
()201383554
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