effects of unbalanced hydrostatic and seepage forces
DESCRIPTION
Flow through soils in retaining wallsTRANSCRIPT
-
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
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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
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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 = .
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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
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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
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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
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IV. FINAL ANSWER
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THANK YOU!
ASSIGNMENT 2
EARTH RETAINING METHOD
RAMIREZ RYAN
()201383554