1

Soil Mechanics (ECIV 3451)

Discussion of Chapter 6: Permeability

General Review:

Hydraulic gradient (i) = height loss per unit length = L

h∆.

Darcy law:

Velocity (V) = k i

q= VA = K i A

V :دي ��� ا��� ����ن ا ��ء �� ���� ا ���� ا �������� . �� ا

q : ��"و� ،���� �ة 'ل �% ��� ا) �ل ����ن ا ��ء �� ا ���� و��دة ��+ ("��* ��Rate of seepage.

Seepage velocity (Vs) = V / n.

Vs :���� ����ن ا ��ء �� ا �0ا/�ت �.% (-.-�ت ا ����.

K : ����� ).Hydraulic conductivity( أو ) Coefficient of permeability(����3 ا �02ذ��

Constant head test:

Aht

QLK =

K: Hydraulic conductivity.

Q: Quantity of water that flows

through the soil specimen.

L: Height of soil specimen.

A: Cross section of soil specimen.

h: Constant head between inlet

and outlet.

t: Required time to collect quantity

Q of water.

2

Falling Head test:

=

2

1log303.2h

h

At

aLk

Equivalent Hydraulic conductivity:

1- Equivalent Hydraulic conductivity for layered soil with horizontal flow:

ر ا ��ء �% '-�� إ � أ�8ى( -� ;. �ة ، و) ��� ��-' 3< ر ا ��ء -�.(

AnikAikAikAikAik

qqqqq

nneqeq

neq

......

...

333222111

321

++++=

++++=

a: Cross section of stand pipe.

A: Cross section of soil specimen.

L: Length of soil specimen.

h1: difference in height of water in stand

pipe at time t1.

h2: difference in height of water in stand

pipe at time t2.

t: Time interval during falling of water head

from h1 to h2(i.e. t=t2-t1)

K η

γ k w=

3

2- Equivalent Hydraulic conductivity for layered soil with Vertical flow:

AnikAikAikAikAik

qqqqq

nneqeq

neq

......

...

333222111

321

=====

=====

4

Problem 1) 6.5 Text bookProblem 1) 6.5 Text bookProblem 1) 6.5 Text bookProblem 1) 6.5 Text book

• For constant head test, the following data were given:

• L = 400 mm.

• A = 135 cm2.

• h = 450 mm.

• Water collected in 3 minutes = 640 cm3.

• Void ration of sand = 0.54

Determine :

a. Hydraulic conductivity (k cm/sec).

b. Seepage velocity.

Solution

a. K=??

sec/023.060345135

40640cmk

Aht

QLk =

××××=⇒=

b. Vs=??

sec/075.035.0

026.0

35.054.01

54.0

1,

sec/026.0135

180/640/

cmV

ne

en

n

VV

cmA

tQ

A

qV

s

s

==

=+

=→+

==

====

Problem Problem Problem Problem 2222) 6.) 6.) 6.) 6.6666 Text bookText bookText bookText book

For variable head permeability test, the following data were given:

• Length of soil specimen = 20 in.

• Area of soil specimen = 2.5 in2.

• Area of stand pipe = 0.15 in2.

• Head difference at t=0.00 is 30 in.

• Head difference at t=8 min is 16in.

Determine:

a. Hydraulic conductivity ( k in/min).

b. What is the head difference at time t=6 min.

Solution

a. K=??

( ) min/0943.016

30log

085.2

2015.0303.2log303.2

2

1 inkkh

h

At

aLk =→

−××=⇒

=

b. At t2=6min, h2=??

( )

inheh

hh

h

hh

h

At

aLk

72.18/30

47.0)/30ln(2047.010ln

)/30ln(2047.0

30log

30log

065.2

2015.0303.20943.0log303.2

247.0

2

22

2

22

1

=⇒=

=→=→=

∴

−××=⇒

=

5

Problem 3)Problem 3)Problem 3)Problem 3)

An inclined premeameter tube is filled with three layers of soil of different permeability as

shown below. Find the head at points (A – B – C – D) with respect to given datum, for the

following cases:

a. Assume k1 = k2 = k3.

b. Assume 3k1 = k2 = 2k3.

Solution

a. k1 = k2 = k3

� آ<=�2 =����3 �� �� @�8ى ، و�� ��-' %� 3��2.� : C)D=vertical flow أن '-��ت ا ���� ���-� �A.B أن ا ��ء

321

321

321

333222111

321

iiii

kkkkk

AAAA

AikAikAikAik

qqqq

eq

eq

eqeq

eq

===∴

====⇒

===⇒

===

===

mh

i

A

eq

2028

5

1648

51020

=

=++

−−=

No loss of head as A is first point

cmhLihh

cmhLihh

cmhLihh

DcD

cBc

BAB

151628

585.17

85.17428

557.18

57.18828

520

33

22

11

=×

−=⇒−=

=×

−=⇒−=

=×

−=⇒−=

6

b. 3k1 = k2 = 2k3 → k2 = 3k1 , k3 = 1.5 k1.

321

3121111332211

1

1113

3

2

2

1

1

1312

321

333222111

321

5.134.1

5.134.1

4.1

5.1

16

3

48

1648

5.1,3

iiii

ikikikikikikikik

kk

kkk

k

k

H

k

H

k

H

Hk

kkkk

AAAA

AikAikAikAik

qqqq

eq

eqeqeq

eqeqeq

eqeq

eq

===

===⇒===∴

=⇒

+

+

++=→

+

+

=

==⇒====⇒

===

===

1. 25.028

54.14.11 =×== eqii

2. 12

1

28

5

3

4.1

3

4.12 =×== eqii

3. 6

1

28

5

5.1

4.1

5.1

4.13 =×== eqii

mhA 20=

cmhLihh

cmhLihh

cmhLihh

DcD

cBc

BAB

15166

167.17

67.17412

118

18825.020

33

22

11

=×

−=⇒−=

=×

−=⇒−=

=×−=⇒−=

���� ا����� �� � ������ ����ت ا���� آ�� ه �� ، :$� ا�!#ال ا�!�

إ�0 ���� وا/�ة و ذ�, 2، 1$��ة ا�'� ه� أن )' ل ا������&

��9د ����� ا���7ذ�� ا����$6 وه5ا 4& ��� ا���) ن ا��3ص :�$ اB$�� و ��A ا��أ<� @? <�=�� ا��!>�� ��4رة 4& ���

�(�Cا� ?>��� �� � ��& و)'�DE آ�� <� وآ�� ه�F.

7

Soil Mechanics (ECIV 3451)

Discussion of Chapter 7: Seepage

General notes:

- Flow net : series of interconnecting of flow lines and equipotential lines.

- Flow line: Line that connect between upstream and downstream .

- Equipotential line: Line that connect between points of equal head.

** Flow lines and equipotential lines intersect at right angle.

- Flow channel: The region surrounded by two consecutive flow lines .

** Flow channels segments should be square or has constant width to length ratio.

** Number of flow channels is called Nf.

- potential drop: the region surrounded by two consecutive equipotential lines.

** Number of potential drops is called Nd.

- Rate of seepage through all flow channels:

l

bn

HHH

nN

NkHq

d

f

=

−=

=

21

b: Side length of flow channel segment perpendicular to flow direction

l: Side length of flow channel segment parallel to flow direction.

i = H/Nd and nN

Hkq

d

=∆ (Rate of seepage through one flow channel)

8

9

10

11

��.Fx/b - �� �sheet �� (�ل آ�ن ال �>ن

pile ��� ا �=I ا�H�% ، أي �>; �� ه

���"� �+، و��� آ3 (�ل ��� ��x �J��.F +��

� ة (�� '�ف " LM�2� %�sheet pile ا

12

Problem 1) For the flow net shown in the figure below:

- Determine the rate of seepage under the weir (k = 1 x 10-3

cm/sec).

- Calculate the uplift force at the base of the weir per meter length.

Solution

A- Find q:

( ) mmN

NkHq

NN

mH

d

f

df

sec//42.214

45.110101

14,4

5.85.110

323 =××××==

===−=

−−

B- Find uplift force:

ة ا��$H ا� ب /!�ب Eط ا��� �K 0E4 ���( آ� ��4��� إ��9د �MN ا���ء �4 �������ة 0E4 ا�!�، و� ���

Equipotential lines

�4� إ��9د ا�� ة أن ��'�0 ا�MNO ��4رة 4& ���Dأ)�� <����� $� ا�� P�/و ، ?>��� �'� ��C ا����ط ا�� ���( �KQ أول و � .T�U ��'�ف ������ $�R!'( M ا���E� MNOء �4

B NO � و ا �� ���J�P ا" أول =��� ه� ا ���2 ا "��0 ��� �"�ر ا

C NO � و ا �� ���J�P ا" �8Q =��� ه� ا ���2 ا "��0 ��� ��.% ا

( )[ ]( )[ ] 2

2

/028.6281.911607.0310

/617.11581.92607.0310

607.014

5.110

mkNU

mkNU

N

H

C

B

d

=××−+=

=××−+=

=−=

Uplift force = ( ) mkN /4.328637028.62617.1152

1 =×+

A

B C

115.617

kN/m2

62.028

kN/m2

13

Problem 2) For the flow net shown in the figure below, determine the

rate of seepage per meter length.

Z

X

Solution

???

9,5

5.35.04

=⇒=

===−=

kN

NkHq

NN

mH

d

f

df

Tا �S=* ��.�2 ("�ب ،���Hو ا ��د '-��.% ��0��O.% �% ا ���� و =�Vا Hن �"�ر ا ��ء �� ا�U �ه.% ا �أW =�Vا ��Hا �� �Uا �� X��<� . zوا�U �� ا �أ�� �x ����3 ا �02ذ�� ا

( ) ( ) ( )[ ]( ) sec/1056.3

1016

5

102

510

sec/109510101651010210

1

8

88

82626

mk

mk

zeq

xeq

−

−−

−−−−−

×=

×+

×

=

×=×××+×××=

The equivalent soil seems to be as as (Anisotropic) soil, so the seepage rate is calculated as

follow:

( ) ( ) mmN

NHkkq

d

f

zeqxeq sec//101.19

55.31056.3109 3788 −−− ×=×××××==

sec/102 61 cmk −×=

sec/1016 62 cmk −×=