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