geotech iv 4 permeability

7/30/2019 Geotech IV 4 Permeability

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Geotechnical EngineeringLecture

Civil Engineering Term IV

Prof. Kazunori Tabe, Ph.D.

Civil Engineering Section

School of Engineering and TechnologySharda University

Source: http://starb.on.coocan.jp/
http://starb.on.coocan.jp/gallery/p127/index0.htmlhttp://starb.cool.ne.jp/daily/daily2.html


2/85

Group Representatives will

give you Lecture Note! G1

Abhishek Singh (ID#007)->[email protected]

Asheesh Pandey(ID#033)->[email protected]

G2

Kshitish Jaiswal (ID#062)->[email protected]

http://in.groups.yahoo.com/group/sharda_geotech/

G3 Spalzes Dolma(ID#143)->[email protected]

G4

Sumit Kudesia (ID#145)->[email protected]
http://in.groups.yahoo.com/group/sharda_geotech/http://in.groups.yahoo.com/group/sharda_geotech/


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LMS will give you Lecture

Note!

Search: kazunori tabe

Click lecture note 1, 2, and 3


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Geotechnical Engineering-SyllabusLecture Contents

1. Syllabus and Introduction (4 hrs)

2. Soil Classification (4hrs)

3. Phase Diagram (4 hrs)4. Permeability of Soils (4 hrs)

5. Effective Stress (4 hrs)

6. Seepage and Flow Nets (4 hrs)

7. Soil Compaction (4hrs)

8. Consolidation and Settlements (4hrs)

9. Shear Strength (4hrs)

10. Slope Stability (4hrs)


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Geotechnical EngineeringTerm Periods and coverage

To Mid Term Exam1. 10th Jan,-14th Jan.

2. 17th Jan. 21st Jan.

3. 24th Jan. 28th Jan.

4. 31st

Jan.

4th

Feb.5. 7th Feb. 11st Feb.

6. 14th Feb. 18th Feb.

7. 21st Feb. 25th Feb.

8. 28th Feb. 4th Mar.

9. 9th Mar. 16th Mar.

Expected coverage:

Introduction (4 hrs),

Soil Classification(4hrs),

Phase Diagram (4 hrs),

Permeability of Soils (4 hrs),

Effective Stress (4 hrs),

Seepage and Flow Nets (4 hrs)

To End Term Exam

1. 21st Mar.-25th Mar.

2. 28th Mar. 1st Apr.

3. 4th Apr. 8th Apr.

4. 11st

Apr. 15th

Apr.5. 18th Apr. 22nd Apr.

6. 25th Apr. 29th Apr.

7. 2nd May 6th May

8. 9th May 16th May

Expected coverage:

Soil Compaction (4 hrs),

Consolidation and Settlement (4hrs),

Shear Strength and Triaxial Tests (4hrs)

Slope Stability (4 hrs)


6/85

What we learned in Geotech

lecture Clay

Atterberg Limit Tests

Soil Classification System

Phase Diagram More practical?

Sand

Sieve Analysis

Soil Classification System

Phase Diagram

More practical?

Soils as constructionmaterials

Clay

Water doesnt movethrough clay very

well.

Sand

Water moves throughsand quite well.


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4. Permeability of Soils (1/3)

http://starb.cool.ne.jp/daily/daily0.html


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Geotechnical Engineering -Syllabus

4. Permeability of Soils Learning Objectives;

Gain an understanding of Head and itsrole in fluid flow

Knowledge of Darcy's Law and its usage An understanding of hydraulic

conductivity, its measurement, andtypical values for different soils

An ability to perform flow calculations insoils

An ability to compute equivalenthydraulic conductivities for layeredsystems

1. Introduction (4 hrs)

2. Soil Classification (4hrs)

3. Phase Diagram (4 hrs)

4. Permeability of Soils (4

hrs)5. Effective Stress (4 hrs)

6. Seepage and Flow Nets (4hrs)

7. Soil Compaction (4hrs)

8. Consolidation andSettlements (4hrs)

9. Shear Strength (4hrs)10. Slope Stability (4hrs)

Source: meted.ucar.edu


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Darcys Law (1856)

q=k*i*A

q=Flow rate

k=hydraulic conductivity orcoefficient of permeability

i=hydraulic gradient (Dh/L)

A=area of cross-section of thesoil

Soil Specimen, k

Dh

LCross-sectional area, A


10/85

Typical Values of Hydraulic

Conductivity of Saturated SoilsSoil type K (cm/sec)

Clean

gravel

100-1.0

Coarse

sand

1.0-0.01

(Water moves through sand quite

well!!)

Fine

sand

0.01-0.001

Silty

sand

0.001-0.00001

Clay


11/85

Clay (assumed impervious)

EL+17.68 m

GWT(EL+22.86 m)GWT(EL+22.86 m)

EL+18.29 m

GWT(EL+17.68 m)

Slurry wall (k=0.000001cm/sec)

0.91 m

q q

Plan View

365.76 m

48.7

7m

A slurry wall cutoff system is proposed for a deep building excavation asshown below. Compute a steady state pumping quantity for the construction.

Construction Dewatering

(Question)


12/85


(Solution)

WallSlurryofAreaSurfaceWetA

GradientHydraulici

WallSlurryoftyCondactiviHydraulick

where

iAkq

Slurry

Slurry

:

A slurry wall cutoff system is proposed for a deep buildingexcavation as shown below. Compute a steady state pumpingquantity for the construction.


EL+17.68 m


EL+18.29 mGWT(EL+17.68 m)


0.91 m

q q

Plan View

365.76 m

48.7

7m


13/85


(Solution) A slurry wall cutoff system is proposed for a deep building

excavation as shown below. Compute a steady state pumpingquantity for the construction.


EL+17.68 m


EL+18.29 mGWT(EL+17.68 m)


0.91 m

q q

Plan View

365.76 m

48.7

7m

22

2

386.0

000,10

000,1

sec246060

sec000001.0

sec000001.0

m

lpd

m

cm

cm

liter

day

cmcmk

ka

slurry

Slurry


14/85


(Solution)

91.0

18.518.568.1786.22

91.0

)(

D

D

L

hi

hmdiffernceheadhydraulic

Lmwallslurryofthickness

GradientHydraulicib



EL+17.68 m


EL+18.29 mGWT(EL+17.68 m)


0.91 m

q q

Plan View

365.76 m

48.7

7m


15/85


(Solution)

2295,4

68.1786.22277.48276.365

)(

m

A

areasurfacewetc



EL+17.68 m


EL+18.29 mGWT(EL+17.68 m)


0.91 m

q q

Plan View

365.76 m

48.7

7m


16/85


(Solution)

lpd

kiAq

000,21

)295,4()91.0

18.5()86.0(



EL+17.68 m


EL+18.29 mGWT(EL+17.68 m)


0.91 m

q q

Plan View

365.76 m

48.7

7m


17/85

Core text page

P. 137-148.


18/85

Permeability

P.137.

The permeability of a soil is a soil propertywhich described quantitatively, the ease withwhich water flows through that soil.


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Darcys Law (1856)

P.137-140.

Darcy discovered that

water travels in soilvoids with a velocity,v = (k)(i), where k ishydraulic conductivityof soil and i is

hydraulic gradient,which is equaled to i=h/L.

Soil Specimen, k

Dh

LCross-sectional area, A


20/85

Measurement of permeabilityof Soils (Laboratory Tests)

P. 140-142.

Constant Head Permeability Test

Falling Head Permeability Test


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Constant Head Test (p.140-141)

Soil Specimen, k

Dh

L

Cross-sectional area, A

Constant Level

)(

)(

)(sec

(sec),)(arg

)/(

:

)24.6(sec/

2

3

cmspecimensoilofLengthL

cmheadindiffernceh

cmsampleofareationalcrossA

ttimeincollectedcmedischQ

scmtypermeabilioftcoefficienk

where

cmAht

QL

iA

qk

kiAq

Good for gravel and sand sample


22/85

Falling Head Test (p.141-142)

2

110

21

2

110

12

2

1

12

log)(

303.2

log)(

303.2

ln)(

)25.6(

h

h

tAhh

VwLk

orhh

AttaLk

orh

h

ttA

aLk

see

WhereVw= volume of water flow through the specimenL= length of specimenh1= beginning head differenceh2= ending head differencet = test duration=t2-t1a=area of buretteA= area of specimen

Good for fine sand or silty sand

Soil Specimen, k

h1

L

Cross-sectional area, A

h2

Cross sectional area, a


23/85

Range of HydraulicConductivity for Various Soils

Soil Types Hydraulic Conductivity (cm/sec)

Medium to coarse gravel Greater than 10-1

Coarse to fine sand 10-1 to 10-3

Fine sand, silty sand 10-3 to 10-5

Silt, clayey silt, silty clay 10-4 to 10-6

Clays 10-7 or less

Das, B.M. (1998) Principles of Geotechnical Engineering 4th edition, PWSPublishing OCmpany, MA


24/85

Thank you.Source: http://starb.on.coocan.jp/
http://starb.on.coocan.jp/daily/daily0.html


25/85

Geotechnical EngineeringLectureCivil Engineering Term IV


Civil Engineering Section

School of Engineering and TechnologySharda University



26/85

Announcement

Geotech lab will start on Monday, 7th, February.

Place: Surveying Lab

Make 5-student Team. (8 teams /each Group) Each Team choose one topic (water content,

permeability, etc) regarding Geotechnical Engineering orSoil Mechanics.

Make 10 minute presentation for Each team will perform a 10 min. presentation on

Monday, 7th February.

Each Team should bring laptop for the presentation

then.


27/85




28/85

Const./Falling Head TestCalculation 1

Flow is taking place under a head, h, through a soil specimen of length, L.What will be the change in the discharge if the head is doubled, and the length ofthe specimen reduced by one half? If the area is triple and length is triple?


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Determine the flow rate:q1=k(h/L)Aq2=k(2h/0.5L)A=4k(h/L)A



30/85


Determine the flow rate:q1=k(h/L)Aq2=k(2h/0.5L)A=4k(h/L)A

Determine the flow rate:

q1=k(h/L)Aq2=k(h/3L)3A=k(h/L)A=q1



31/85

A constant head permeability test was performed on a specimen of fine sand.The following values were provided:

Length of the specimen=40cmDiameter of the specimen=20cmHead difference=30cmVolume of water collected in 4 min.=250cm3

What is the hydraulic conductivity?



32/85

Const./Falling head TestQuestion 2




Given: L=40cm, D=20cm, Dh=30cm, q=(250/4) cm3/min


33/85




Given: L=40cm, D=20cm, Dh=30cm, q=(250/4) cm3/minCalculate flow rate:i=Dh/L=0.75, A=pD2/4=314.2cm2q=kiA, k=q/(iA), q=1.04 cm3/seck=q/(iA)=0.0044 cm/sec=4.4*10(-3) cm/s



34/85

30cm

20cm

Three permeability tests are performed onthe soil and the test result are as follows:

Is this the constant head test or falling headtest?What is the hydraulic conductivity of thesoil?

Test 1 Test 2 Test 3

Beginning Head

Difference (cm)

20 25 21

Ending Head Difference

(cm)

11 17 12

Test Duration (sec) 18 16 18

Volume of water collected

(cm3)

15 18 16



35/85

30cm

20cm

Given: D=20cm, L=30cm



36/85

30cm

20cm


Determine type of permeability test->It is a falling head test.



37/85

30cm

20cm



Determine the area: A=pD^2/4=314 cm2



38/85

30cm

20cm




2

110

21

log)(

303.2

h

h

tAhh

VwLk



39/85

30cm

20cm




sec

005.012

21log

))314)(18)(1221((

)30(16303.2

sec005.0

17

25log

))314)(16)(1725((

)30(18303.2

sec005.0

11

20log

))314)(18)(1120((

)30(15303.2

log)(

303.2

103

102

101

2

110

21

cmk

cmk

cmk

h

h

tAhh

VwLk

test

test

test



40/85

30cm

20cm





sec

105005.0

3

005.0005.0005.0

sec005.0

12

21log

))314)(18)(1221((

)30(16303.2

sec005.0

1725log

))314)(16)(1725(()30(18303.2

sec005.0

11

20log

))314)(18)(1120((

)30(15303.2

log)(

303.2

3

103

102

101

2

110

21

cmk

cmk

cmk

cmk

h

h

tAhh

VwLk

average

test

test

test


41/85


A drainage pipe is clogged with sandwhose hydraulic conductivity is found to be9.0 cm/s. The average difference inheadwater and trailwater elevation is 1.2 mand it has been observed that there is aflow of 330 cm3/s through the pipe. If thepipe is 6m long and has a cross-sectionarea of 20 cm2, what length of the pipe isfilled with sand?

1.2m

6m


42/85


Given: k=9.0 cm/s, Dh=1.2m, q= 330cm3/s , A= 20 cm2, L=6m

1.2m

6m


43/85



Length of clogged pipe:

1.2m

6m


44/85




q=kiA=k(h/L(clogged))A,k=q(L(clogged))/(hA),L(clogged)=kDhA/q=(9)(120)(20)/330=65.5 cm

1.2m

6m


45/85




q=kiA=k(h/L(clogged))A,k=q(L(clogged))/(hA),L(clogged)=kDhA/q=(9)(120)(20)/330=65.5 cm

Percentage pf pipe that is clogged:%clogged=L(clogged)/L*100=65.5/600*100=10%

1.2m

6m


46/85

Permeability of Stratified Soils

P.147

When a subsoil in the field consists of a

number of strata which have differentpermeabilities, calculation of seepagethrough such a stratified deposit

requires the computation of averagevalues of k applicable for the wholedeposit.


47/85

Permeability of Stratified Soils

P.147

There are two such average values: the

average horizontal coefficient ofpermeability, k(h) when the flow isparallel to the strata and the vertical

coefficient of permeability, k(v), whenthe flow is normal to the strata.


48/85

Equivalent Hydraulic Conductivity(horizontal case p.147)


49/85

Equivalent Hydraulic Conductivity(horizontal case p.147)


50/85

Equivalent Hydraulic Conductivity(vertical case p.147-148)


51/85

Equivalent Hydraulic Conductivity(vertical case p.147-148)


52/85



53/85

Geotechnical EngineeringLectureCivil Engineering Term IV


Civil Engineering SectionSchool of Engineering and Technology

Sharda University



54/85

Announcement

Geotech lab will start on Monday, 7th, February.

Place: Surveying Lab

Make 5-student Team. (8 teams /each Group)

Each Team choose one topic (water content,permeability, etc) regarding Geotechnical Engineering orSoil Mechanics.

Make 10 minute presentation for

Each team will perform a 10 min. presentation onMonday, 7th February.

Each Team should bring laptop for the presentation

then.


55/85




56/85

Equivalent Hydraulic ConductivityCalculation 1

Given: h1+h2=5

cm

cmcm


57/85


Given: h1+h2=5

cm

cmcm

Given: h1+h2=5, A1=A2=A


58/85


Given: h1+h2=5

cm

cmcm


Calculate head at point A:

Q1=Q2K1(h1/L1)A=k2(h2/L2)A,k1(h1/6)=3k1(h2/3), h1=6h2


59/85


Given: h1+h2=5

cm

cmcm




6h1+h2=5, h2=5/7=0.71H1=6-h2=6-0.71=4.29


60/85


Given: h1+h2=5

cm

cmcm




6h1+h2=5, h2=5/7=0.71H1=6-h2=6-0.71=4.29

Head at point A is 0.71 cm


61/85

1.5m

2.1m

3m

1.5m

2.1m

3m

30cm

150cm

75cm

30cm

75cm

150cm

Three horizontal stratified uniformedsoils have hydraulic conductivity of 30cm/day, 75 cm/day, and 150 cm/day,

respectively. Find the average hydraulicconductivity of the soil layer inhorizontal and vertical directions.



62/85

H1=1.5m

H2=2.1m

H3=3m

H1=1.5m

H2=2.1m

H3=3m

30cm

150cm

75cm

30cm

75cm

150cm


Given: H1=1.5m, H2=2.1m, H3=3mk1=30cm/day, k2=75cm/day, k3=150cm/day


63/85

H1=1.5m

H2=2.1m

H3=3m

H1=1.5m

H2=2.1m

H3=3m

30cm

150cm

75cm

30cm

75cm

150cm



Average hydraulic conductivity in horizontal direction:For horizontal flow->

n

j

j

n

j

jj

eq

d

dk

k

1

1


64/85

H1=1.5m

H2=2.1m

H3=3m

H1=1.5m

H2=2.1m

H3=3m

30cm

150cm

75cm

30cm

75cm

150cm




day

m

HHH

HkHkHkk

d

dk

k

H

n

j

j

n

j

jj

eq

19931.25.1

31501.2755.130

321

332211

1

1

l d l d


65/85

H1=1.5m

H2=2.1m

H3=3m

H1=1.5m

H2=2.1m

H3=3m

30cm

150cm

75cm

30cm

75cm

150cm




Average hydraulic conductivity in vertical direction:

For vertical flow->

day

m

HHH

HkHkHkk

d

dk

k

H

n

j

j

n

j

jj

eq

19931.25.1

31501.2755.130

321

332211

1

1

i l d li C d i i


66/85

H1=1.5m

H2=2.1m

H3=3m

H1=1.5m

H2=2.1m

H3=3m

30cm

150cm

75cm

30cm

75cm

150cm





For vertical flow->

day

m

HHH

HkHkHkk

d

dk

k

H

n

j

j

n

j

jj

eq

19931.25.1

31501.2755.130

321

332211

1

1

n

j j

j

n

j

j

eq

k

d

d

k

1

1

E i l H d li C d i i


67/85

H1=1.5m

H2=2.1m

H3=3m

H1=1.5m

H2=2.1m

H3=3m

30cm

150cm

75cm

30cm

75cm

150cm





For vertical flow->

day

m

HHH

HkHkHkk

d

dk

k

H

n

j

j

n

j

jj

eq

19931.25.1

31501.2755.130

321

332211

1

1

day

cm

k

H

k

H

k

H

HHHk

k

d

d

k

H

n

j j

j

n

j

j

eq

3.67

150

3

75

1.2

30

5.1

31.25.1

3

3

2

2

1

1

321

1

1


68/85

Permeability of Soils Additional 1

A 4.5 m thick clay deposit isinterbedded with a couple of thinlayers of silt a quarter m thick each.

Determine the ratio of horizontaland vertical hydraulic conductivity ofclay deposit if the hydraulicconductivity of silt is 100 times thatof clay.What is the new ratio if one layer of

25 cm thick clay deposit is foundinstead of two layers? Assume thehydraulic conductivity remains thesame.

25 cm

4.5m


69/85


25 cm

4.5m

Given:H(silt)=2*25cm=50cm, H(clay)=4.5m-50cm=4m,k(silt)=100k(clay)


70/85


25 cm

4.5m


Average hydraulic conductivity in horizontal direction:


71/85


25 cm

4.5m



clayH

clayclay

claysilt

clayclaysiltsilt

H

kk

kk

hh

HkHkk

1240050

40050100


72/85


25 cm

4.5m




clayH

clayclay

claysilt

clayclaysiltsilt

H

kk

kk

hh

HkHkk

1240050

40050100

clayV

clayclayclay

clay

silt

silt

claysilt

V

kk

k

cm

k

cm

cmcm

k

H

k

H

HHk

11.1

400

12

50

40050


73/85


25 cm

4.5m




Ratio of horizontal and vertical hydraulic conductivity:

clayH

clayclay

claysilt

clayclaysiltsilt

H

kk

kk

hh

HkHkk

1240050

40050100

clayV

clayclayclay

clay

silt

silt

claysilt

V

kk

k

cm

k

cm

cmcm

k

H

k

H

HHk

11.1

400

12

50

40050

8.1011.1

12

clay

clay

V

H

k

k

k

kRatio


74/85


25 cm

4.5m




Ratio of horizontal and vertical hydraulic conductivity:

New ratio of horizontal and vertical hydraulic conductivity:

Since the overall thickness of silt and clay layers arethe same, the ratio of horizontal and vertical hydraulicconductivity remains as 10.8

clayH

clayclay

claysilt

clayclaysiltsilt

H

kk

kk

hh

HkHkk

1240050

40050100

clayV

clayclayclay

clay

silt

silt

claysilt

V

kk

k

cm

k

cm

cmcm

k

H

k

H

HHk

11.1

400

12

50

40050

8.1011.1

12

clay

clay

V

H

k

k

k

kRatio


75/85


65 cm2


76/85


Given:g1=1833 kg/m3, d1=100cm, k1=10(^-6)cm/sg2=1600kg/m3, d2=75cm, k2=10(^-7)cm/sH1=100cm, H2=50cm

65 cm2


77/85



Calculate equivalent hydraulic conductivity, keq:

sec1005.2

10

75

10

100

75100 7

76

2

2

1

1

21 cm

k

d

k

d

ddkeq

65 cm2


78/85




Find hydraulic gradient:

sec1005.2

10

75

10

100

75100 7

76

2

2

1

1

21 cm

k

d

k

d

ddkeq

85.075100

50100

21

21

dd

HHi

65 cm2


79/85




Find hydraulic gradient:

Calculate flow rate:

sec1005.2

10

75

10

100

75100 7

76

2

2

1

1

21 cm

k

d

k

d

ddkeq

85.075100

50100

21

21

dd

HHi

s

cmiAkq eq

3

571013.16585.01005.2

65 cm2


80/85


18m

4.5m

6.5m

9m

6.5m

5.4m

34m

In the flow test shown adischarge of 27m3/hr of watertraveled in the sand layerbetween the two piezometers.What is the permeability if thewidth of the sand layer(perpendiculat to the paper) is

100 m?


81/85


82/85


18m

4.5m

6.5m

9m

6.5m

5.4m

34m

Given: q=27 cm3/hr, H(clay)=6.5m, H(sand)=9m,DH=5.4mD=34m, d=18m, Width=100m, Dh=4.5m

Determine the area:

Tan a =DH/D=5.4/34=0.1588-> a=9.02 degree.A=H(sand)*cos a * width=(9)(cos9.02)(100)=889 m2


83/85


18m

4.5m

6.5m

9m

6.5m

5.4m

34m


Determine the area:


Find hydraulic gradient, i:L=d/(cos(a))=18/cos 9.02=18.23mi=Dh/L=4.5/18.23=0.25


84/85


18m

4.5m

6.5m

9m

6.5m

5.4m

34m


Determine the area:


Find hydraulic gradient, i:L=d/(cos(a))=18/cos 9.02=18.23mi=Dh/L=4.5/18.23=0.25

Calculate hydraulic conductivity, k:

K=q/(iA)=27/(0.25*889)=0.12 cm/hr=2.02*10^(-3) cm/s


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geotech iv 4 permeability

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