1.Lect-1 Effective stress(geotechnical engineering).ppt

Effective stresses (Geotechnical engineering)

1. 1
CE-313 (2 Credit Hours)
Geotechnical Engineering-II
Introduction to Course and
Concept of Effective Stress
Instructor:
Prof-Dr Irshad Ahmad
Fall 2020
Department of Civil Engineering
University of Engineering and Technology, Peshawar

2. In this course following chapters will be discussed;
Chapter 1 Effective Stress Concept
Chapter 2 Consolidation
Chapter 3 Shear Strength of Soil
Chapter 4 Slope Stability
Chapter 5 Lateral Earth Pressure
2
Introduction to course

3. 3
Concept of Effective Stresses

4. 4
Consolidation

5. 5
Slopes

6. 6
Slope Failure

7. 7
Slope Failure

8. 8
Retaining Walls

9. 9
Retaining Wall Failure

10. 10
Ch-1
Effective
Stresses
Ch-3 Shear strength
Parameters of Soil
Ch-4 slope
stabililty
Ch-5 Lateral
earth
pressure
Ch-2
Consolidation
Sequence of Chapters

11. Course Title:GeotechnicalEngineering-II
Course Code:CE–313
Course Duration:OneSemester
Credit Units:02CreditHrs.
Level:5
th
Semester,3
rd
Year
Medium of Instruction:English
Prerequisites:GeotechnicalEngineering-I
Equivalent Courses:NotApplicable
11
Geotechnical Engineering-II
CE-313
(course specification form)
Part I: Course Information

12. 12
CourseAims
Afterthiscoursethestudentswill
1.Understandtheshearandconsolidationbehaviorofsoilsanddeterminationof
relevantsoilparametersusinglaboratorytests.
2.Carryoutslopestabilityanalysis
3.Findlateralearthpressures
CourseIntendedLearningOutcomes(CLOs)
1.Defineeffectivestress,Shearstrengthofsoil,soilconsolidationandlateral
earthpressureparameters,anddifferentslopefailures.
2.ExplainResponseofeffectivesoiltochangesintotalstress,differentshear
failuretheories,laboratoryandfieldtestsforshearstrengthconsolidation
parametersdetermination,consolidationprocess,lateralearthpressure
theories,andslopestabilitymethods.
3.Applytheconsolidationparameterstoestimateamountandtimerateof
settlement,lateralearthpressuretheoriestofindlateralearthpressures,and
shearstrengthparameterstofieldconditions
Evaluate the stability of slopes
Course Aims, Course Intended Learning Outcomes (CLOs)

13. 13
Courselearningoutcomeswillbeachievedthroughacombinationofthefollowing
teachingstrategies.
Quizzes
In-classactivities
Videopresentations
Readingassignments
Classroomdiscussions
Homeworkassignments
Mid-termmajorexamination
Finalcomprehensiveexamination
Solvedexamplesintheclassroom
Conductingtutorialsintheclassroom
Teaching and Learning Activities (TLAs)

14. 14
WEEKTOPIC CLOsQuiz/Assignm
ent
1 Principle of effective stresses, Effective Vertical Stress due to self
weight of the soil, Response of Effective Stresses to a change in total
stress, Spring Analogy
1,2
2 1D Consolidation, Oedometer Test, e-curve, e-log() curve, OCR,
OCC, and NCC, Compression index, recompression index,
Determination of Preconolidation Pressure by Casagrande Method
1,2Assignment-1
3 Schmertmann Procedure to obtain in-situ e-log() curve, coefficient
of volume compressibility, Consolidation settlement calculations for
NCC and OCC
1,2
4 Solved Examples 3 Quiz-1
5 Degree of Consolidation, Average degree of consolidation, Terzaghi’s
theory of 1D consolidation
12,3
6 Solution of consolidation equation, Isochrones, Determination of Cv
by Log-time method (due to Casagrande) and Root time method (due
to Tylor)
1,2,3
7 Solved Examples 3 Assignment-2
8 Importance, Shear strength of soil, Coulomb’s Theory, Modified
Coulomb’s Theory, Mohr-Coulomb Failure Criterion, Tresca
Failure Criterion
1,2 Quiz-2
9 MID TERM EXAM
Weekly schedule

15. 15
10 Direct Shear Test, Peak, Ultimate, and Residual Shear Strengths, Solved
Examples
2
11 Triaxial Compression Test (UU, CU, and CD), Solved Examples.1,2 Assignment-
3
12 Application of UU, CU, and CD Tests to field conditions, Unconfined
compression Test, Drained and Undrained Shear Strength, Shear Strength
of Sand, Liquefaction,
1,2,3
13 Behavior of NC and OC clays in drained and undrained test, Mohr’s
Circles for OC and NC clays under CU tests, Failure envelop for OC clays,
Types of Analysis for OC and NC clays, Vane Shear Test
2.3 Quiz-3
14 Active, Passive and At-Rest Earth Pressures, Lateral Strain Vs Earth
Pressure, Rankine’s Lateral Earth pressure theory
1.2
15 LEP for surcharge loads, stratified Soil, Drained & undrained Analysis,
Sloping soil surface, Solved Examples on Rankine’s Theory, Coulomb’s
Theory of Earth pressures
1,2,3Assignment-
4
16Types of Slope Failure, limit equilibrium method, Slope stability analysis
for u=0 soil, Taylor Stability Number, Solved Examples clays and clayey
silts
1,4
17 Method of Slices with Swedish and Bishop Routine Solutions, Plane
Translational Slip, Solved Examples
1,4 Quiz-4
18
Final Term Exam
Weekly schedule

16. 16
ATAs CLOs Weight % Remarks
Final-Term Exam 1-4 50% 2hrswrittenexams
Mid-Term Exam 1-3 25% 2hrswrittenexams
Home Assignments 1-4 10% Total4assignments
Quizzes 1-4 15% Total4in-classtests
Total (%)100 % Thefinalgradingisrelative
Assessment Tasks Activities (ATAs)
Indicative of likely activities and tasks designed to assess how well students achieve
the CLOs. Final details will be provided to students in their first week of attendance in
this course.
Assessment Tasks Activities (ATAs)

17. 17
Grade
Letter
Grade
Points
Grade Definitions
A
A-
4.0
3.67
Excellent
Strongevidenceoforiginalthinking;goodorganization,
capacitytoanalyze&synthesize;superiorgraspofsubject
matter;evidenceofextensiveknowledgebase.
B+
B
B-
3.33
3.00
2.67
Good
Evidenceofgraspofsubject,someevidenceofcriticalcapacity
andanalyticalability;reasonableunderstandingofissues;
evidenceoffamiliaritywithliterature.
C+
C
C-
2.33
2.00
1.67
Adequate
Studentwhoisprofitingfromtheuniversityexperience;
understandingofthesubject;abilitytodevelopsolutionsto
simpleproblemsinthematerial.
D+
D
1.33
1.00
Marginal
Sufficientfamiliaritywiththesubjectmatterstoenablethe
studenttoprogresswithoutrepeatingthecourse.
F 0.00Failure
Littleevidenceoffamiliaritywiththesubjectmatter;weakness
incriticalandanalyticalskills;limited,orirrelevantuseof
literature.
Grading of Students Achievements
The grading for this course is based on the Academic Regulations criterion of the
University.
Assessment Tasks Activities (ATAs)

18. 18
Syllabus
Chapter-01:EffectiveStress
Chapter-1 Effective Stress

Principle of effective stresses, Effective Vertical Stress due to self weigthof the soil,
Response of Effective Stresses to a change in total stress, Spring Analogy
Chapter-2 Consolidation
1D Consolidation, Oedometer Test, e-curve, e-log() curve, OCR, OCC, and NCC,
Compression index, recompression index, Determination of Preconolidation Pressure
by Casagrande Method
Schmertmann Procedure to obtain in-situ e-log() curve, coefficient of volume
compressibility, Consolidation settlement calculations for NCC and OCC
PartIII:Syllabus,RecommendedBooks,OnlineSources,CourseCoordinator

19. 19
Chapter-3 Time Rate of Consolidation Settlement
Degree of Consolidation, Average degree of consolidation, Terzaghi’s theory of 1D
consolidation
Solution of consolidation equation, Isochrones, Determination of Cvby Log-time
method (due to Casagrande) and Root time method (due to Tylor), solved examples
Chapter-4 Shear Strength of Soil
Importance, Shear strength of soil, Coulomb’s Theory, Modified Coulomb’s
Theory, Mohr-Coulomb Failure Criterion, Tresca Failure Criterion
Direct Shear Test, Peak, Ultimate, and Residual Shear Strengths, Solved
Examples
Triaxial Compression Test (UU, CU, and CD), Solved Examples.
Application of UU, CU, and CD Tests to field conditions, Unconfined compression
Test, Drained and Undrained Shear Strength, Shear Strength of Sand,
Liquefaction,
Behaviour of NC and OC clays in drained and undrained test, Mohr’s Circles for
OC and NC clays under CU tests, Failure envelop for OC clays, Types of Analysis
for OC and NC clays, Vane Shear Test
PartIII:Syllabus,RecommendedBooks,OnlineSources,CourseCoordinator

20. 20
Chapter-5 Lateral Earth Pressures
Active, Passive and At-Rest Earth Pressures, Lateral Strain Vs Earth Pressure,
Rankine’s Lateral Earth pressure theory
LEP for surcharge loads, stratified Soil, Drained & undrained Analysis, Sloping soil
surface, Solved Examples on Rankine’s Theory, Coulomb’s Theory of Earth
pressures
Chapter-5 Slope Stability
Types of Slope Failure, limit equilibrium method, Slope stability analysis for u=0
soil, Taylor Stability Number, Solved Examples clays and clayey silts
Method of Slices with Swedish and Bishop Routine Solutions, Solved Examples
Plane Translational Slip, Solved Examples
PartIII:Syllabus,RecommendedBooks,OnlineSources,CourseCoordinator

21. 21
Dr.IrshadAhmad
Professor
CivilEngineeringDepartment
UETPeshawar,KhyberPakhtunkhwa.
[email protected]
OfficeLocation:
CE:A209,
CivilEngineeringDepartment,
UETPeshawar.
Recommended Books and References
1.Soil Mechanics by R.F Craig
2.Soil Mechanics and Foundations by Muni Budhu
Online Resources
1.Watch online videos/animations for shear and consolidation tests.
Course Coordinator
PartIII:Syllabus,RecommendedBooks,OnlineSources,CourseCoordinator

22. 22
Chapter-1
CONCEPT OF EFFECTIVE STRESS

23. 23
CONTENTS
•Introduction
•Principle of effective stress
•Effective vertical stress due to self weight of soil
•Response of effective stress to a change in total
stresses
•Consolidation and its analogy
•Examples

24. 24
Introduction

25. 25
The Principle of Effective Stress

26. 26
Total
stresses
Effective
Stresses
Pore
water
Pressure
Fully saturated soil

27. 27
Effective Vertical Stress due to self weight of the soil
Saturated unit
weight of
soil
sat

28. A saturated layer of clay 4 m thick is overlain by sand 5m
deep, the water table being 3m below the surface. The
saturated unit weights of the clay and sand are 19 kN/m
3
and
20 kN/m
3
, respectively; above the water table the unit weight
of the sand is 17 kN/m
3
. Plot the values of total vertical stress
and effective vertical stress against depth. If sand to a height
of 1 m above the water table is saturated with capillary water,
how are the above stresses affected.
28
Example 3.1 (SM by R.F. Craig)
5 m
4 m
3 m
Sand
=17kN/m
3
clay

29. 29
5 m
4 m
3 m
Sand
=17kN/m
3
clay
3m
5m
0m
9m

sat
=19 kN/m
3

sat
=20 kN/m
3
- 
17*3=51
91+19*4=167
51+20*2=91
17*3=51
91-19.6=71.4
167-58.86=108.2
9.81*2=19.6
9.81*6=58.86
Total stresses
(kPa)
Effective
stresses (kPa)
Pore water
Pressure(kPa)

30. 30
5 m
4 m
3 m
Sand
=17kN/m
3
ꞌ=17kN/m
3
clay

sat
=19 kN/m
3
ꞌ(19-9.8)=9.19

sat
=20 kN/m
3
ꞌ(20-9.8)=10.2
(17-0)*3=51
51+10.2*2=71.4
71.4+9.2*4=108.2
Effective stresses (kPa)
ꞌ=ꞌz = (
sat-
w) z
Alternate Method
0m
3m
5m
9m

31. 31

32. The water table is at level at which pore water is at
atmospheric (u=0). Above the water table water is held
under negative pressure and even if the soil is saturated
above the water table, does not contribute to hydrostatic
pressure below the water table. The only effect of one
meter capillary rise is, therefore, to increase the total unit
weight of sand between 2 and 3 meter depth from 17 to 20
kN/m
3
,an increase of 3 kN/m
3
. Both total and effective
stresses below 3 meter depth are therefore increased by the
same amount 3x1 = 3.0 kN/m
2
, pore water pressures, pore
water pressures being unchanged.
32
5 m
4 m
3 msand
clay
1 m
When soil 1 m above water table is saturated by capillary rise
=17 kN/m
3

sat
=20 kN/m
3

sat
=19 kN/m
3

33. 33
Spring Analogy

34. 34
Spring Analogy
time
Total Force
Spring Force
Water Force
Force

35. 35
Spring Analogy
Springsoil skeleton
waterPore water in soil
Bore diameter permeability of soil
Rigid cylinder wallszero lateral
Strain in soils

36. 36
Response of Effective Stress to a change in total stress

37. 37
Response of Effective Stress to a change in total stress

38. 38
Response of Effective Stress to a change in total stress

39. z
39
Response of Effective Stress to a change in total stress
u
s=
wz = hydrostatic pwp
ꞌ
o=
o–u
s= effective stress

o=
satz = total stress

40. z
t=0 (undrained condition with ue=ui)
u= u
s+ ∆

v≠0

l0
∆Change in total stress
ue=u
i= Initial Excess pore water
pressure
u
i∆
ꞌꞌ
o

o+∆
ꞌ+u= ꞌ+(u
s+∆)
ꞌ= ꞌ
o
40
∆~ q

41. z
u= u
s+ u
e
41
@ t > 0 (Dissipation of Excess PWP)
u=u
s+u
e
u
eExcess pore water pressure at
t>0
u
e< u
i
ꞌ
o> ꞌ& ꞌꞌ
o+(∆-u
e)Drainage
∆s
ꞌ> ꞌ
o
∆
time
time
0
u
u
s
ꞌ
o
ꞌ
∆

o

o
t

42. z
u u
s
u
e= Excess pore water pressure at
t= t
f& u
e=0
ꞌꞌ
o+∆
42
@ t = tf (Drained Condition ue=0)
u

o=
satz
s
f
u
i
∆
u
s
0
time
time
ꞌ
o
ꞌ
t=t
f
∆

43. 43
Terminology Learnt
•Total stresses
•Effective stresses
•Static Pore water Pressure
•Excess Pore Water Pressure
•Dissipation of excess PWP
•Drainage of Pore water
•Fully Saturated soil
•Drained and Undrained condition
•One dimensional Consolidation

44. 44
Example 3.2
A 5 m depth of sand overlies a 6 m layer of clay, the
water table being at the surface; the permeability of the
clay is very low. The saturated unit weight of the sand is
19 kN/m
3
and that of clay is 20 kN/m
3
. A 4 m depth of
fill material of unit weight 20 kN/m
3
is placed on the
surface over an extensive area.
Determine the Effective vertical stress , Total Vertical
stress , pore water pressure (u) at the center of the clay
layer;
(a)Immediately of after the fill has been placed,
assuming this to take place rapidly and
(b)Many years after the fill has been placed
4 m Fill
20kN/m
3
5 m
6 m
sand
8 m
3 m

sat19kN/m
3
Clay

sat20kN/m
3

45. Change in Total Stress = ∆= 4 x 20 = 80 kPa
(a)Immediately After Construction
Total Vertical Stress

v= 
o+∆

v(19*5)+(3*20)+80= 235 kPa
Effective Vertical Stress
Since clay is of low permeability, all the ∆(change in
total stress) is taken by pore water as excess pore water
pressure. Hence no increase in the effective stress due
to ∆
ꞌ
v5*(9.2)+(3*10.2)=76.6 kPa
Static PWP (u
s)
u
s9.8*8=78.4 kPa
Excess PWP (u
e)
u
e
∆80 kPa
Total PWP (u)
u u
s
+∆78.4+80=158.4 kPa
4 m
5 m
6 m
8 m
3 m
Fill
20kN/m
3
Sand

sat19kN/m
3
ꞌ9.2 kN/m
3
Clay

sat20 kN/m
3
ꞌ10.2 kN/m
3
Solution

46. 46
Solution
(b) Long time after the fill has been placed
Total Vertical Stress

vꞌ+(u
s+ ∆)= 235 kPa
Effective Vertical Stress
After long time the excess PWP will dissipate and the change in stress (∆) will be
taken by the soil skeleton and hence an increase of 80 kPa in effective will occur.
ꞌ
v76.6+80 = 156.6 kPa
u
s
9.8*8 78.4 kPa
Static PWP (u
s)
u
s
9.8*8=78.4 kPa
Excess PWP (u
e)
u
e
= 0
Total PWP (u)
u= u
s
78.4 kPa

47. 47

48. 48

49. 49
Exercises

50. 50
Exercises