Subject (CSE570): Soil Behaviour and Geotechnical Engineering by Professor Jianhua YIN (ZS909), CEE, PolyU Lecture 1: Laboratory Tests Tests and Real Stress-Strain Behavi Beh aviour our of Soils Soils Lecture 2: Simple and Commonly Used Constitutive Models Lecture 3: 1-D Elastic Vi Visco-Plastic sco-Plastic Model and Consolidation Settlement Calculation
1
Abstracts:
This pa This part rt of th thee su subj bjec ectt (C (CSE SE57 570) 0) co cons nsis ists ts of th thre reee le lect ctur ures es.. Le Lect ctur uree 1 wi will ll co cove verr co conv nven enti tion onal al an and d ad adva vanc nced ed la lab b te test st ap appa para ratu tuse ses, s, ty typi pica call test te st da data ta,, in inte terp rpre reta tati tio on, th thee st stre ress ss-s -str trai ain n-s -str tren eng gth beh ehav avio iorr in 1D 1D,, 2D and 3D stress states. Lecture 2 will present simple and commonly used constitutive models. The framework of elastic plastic models is explained. After this, commonly used MohrCoulomb elastic-plastic model and critical state models are presented. The advantages and limitations of these models are explai ain ned ed.. Lec ectture 3 prese sen nts a 1-D Elas asttic Visco co--Plas asttic (E (EV VP) model and a simplified Hypothesis B method for consolidation settlement calculation of clayey soils. The main equations and validation of this new method are presented first. The final part incl in clud udes es wo work rked ed ex exam ampl ples es of cal calcu culat latio ion n of co cons nsol olid idat atio ion n set settl tlem emen entt of a si sing ngle le or tw two o la lay yer ered ed so soil ils. s.
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1
References: Essential References: [1] Mitchell, James K., “Fundamentals of Soil Behaviour”, Second Edition, John Wiley & Sons, Inc. (1993). [2] David M Potts and Lidija Zdravkovic, “Finite element element analysis in geotechnical engineering – theory”, Thomas Telford Telford Publishing Ltd, U.K. (ISBN: 0 7277 2753 2), (1999). [3] Papers by JH YIN (see relevant lectures) lectures) [4] Lecture notes of Prof JH Yin and Dr. ZY Yin Other References: [1] Geo-Slope (2004), Program manuals and software: SLOPE/W and SIGMA/W. [2] Plaxis 2D [3] David Muir Wood, “Soil Behaviour Behaviour and Critical State State Soil Mechanics”, Mechanics”, Cambridge University Press, (1990). [4] Britto, A.M. and Gunn, M.J., “Critical State Soil Mechanics via Finite Elements”, ELLIS HORWOOD LTD., (1990). [5] Chen, W.F W.F.. and Mizuno, E., “Nonlinear Analysis in Soil Mechanics”, Elsevier (1990). 3
Lecture Lectu re 1: Laboratory Laboratory Tests and Real Real Soil Behavior of Soils 1.1 1.2 1.3 1.4 1.5 1.6
Introduction Oedometer Test Test and 1D Behavior of Soils Direct Shear Tests Tests and 2D Behavior of Soils Triaxia riaxiall Tests Tests and 2D Behavior Beha vior of Soils True Triaxial Triaxial Tests Tests and 3D 3 D Behavior Behavio r of Soils Hollow Cylinder Triaxial Triaxial Tests Tests and 3D Behavio Behaviorr of Soils 1.7 Concluding Remarks
4
2
1.1 Introduction • Soil behavior is neither linear nor elastic • Soil failure will occur • Real behavior behavior – from 1-D compres compression sion tests, triaxial tests, to other advanced tests
5
1.2 Oedometer Tests and 1D Behaviour of Soils
• •
One degr degree ee of of freed freedom om (one (one ind indepe epende ndent nt var variab iable) le) Oedo Oe dome mete terr – si simu mula lati ting ng on one-d e-dim imens ensio iona nall (1D (1D)) compressions (or 1D straining), straining), example, a large area of reclamation on a horizontally uniform soil under uniform un iform pressure • Differ ereences fr fro om 1D stressing • Oed Oedome ometer ter test test:: elemen elementt test test or a sma small ll phys physica icall model model?? Element test: uniform strain and stress A small physical model: initial-boundary value problem – nonuniform strain and stress Oedometer test: both ok; an element test when ue=0 z 0; x y z
z 0; x y 0 z
0; 0 ' z
' x
' y
Only 1 strain z
z'
z' 0; x' y' 0 '
Only 1 stress z
6
3
The Hong Kong Polytechnic University
Reclamation of Hung Hom Bay completed in 1994 7
Oedometer test (1D straining or laterally confined consolidation test) Pre-loading fill Sand fill Water Table
Marine Deposits
Bedrock or soil
Confined or 1-D Straining Consolidation (or Oedometer) Condition: • Soil layers are horizontal and uniform • Loadi Loading ng is uniform (extensi (extensive ve UDL) • Deformation & water flow are in vertical only 8
4
Impermeable at time=0+ Water table
Uniform surcharge q=10 kPa Ai z A sat
z A
i' z A ' ,
us ue
( ' sat w )
f ' i' u ei i' q z A ' q
Impermeable bedr bedrock ock
Static porewater pressure:
us z A w
Initial excess porewater pressure:
uei q 10kPa
9
Consolidation analogy: q 10kPa
Valve closed
us z A w us z A w
uei q
10kPa
Valve opened q 10kPa
Valve Opened - final q 10kPa
us z A w ue ue (t )
us z A w ue 0
Ai z A sat i' z A '
A z A sat q
A z A sat q
A z A sat q
' i' z A '
' A u
' A u
A (us u e )
A (u s u e )
z A ' q ue
z A ' q
10
5
Effective stresses principle and equation:
Normal stress: stress: Shear stress: no change
Effective stresses control both deformation and shear resistance (or shear strength) since effective stresses reflect soil particle interaction. Why? q 10kPa Valve closed
us z A w uei q
10kPa
Normal stress is increased by 10 kPa. Any vertical deformation? deformation? No! Any shear resistance increased? No!
11
Terzaghi (1883-1963): Father of Soil Soil Mechanics 12
6
Peck in Taiwan 1998 13
14
7
Oedometers in Soil Mechanics Laboratory
15
v
F A
Ho
H1
Method: Water content measured at end of test= w1 Void ratio measured measured at at end of test= e1=w1Gs (assume S r =100%) Soil specimen thickness change H=H 1-H 0<0 V 0; H 0; But , 0; 0 since compression as possitive "" Void ratio at start of test= e0=e1-e or e1=e0+e v
V V e1 e0 V 1 V 0 V v1 V v 0 V 0 V e V HA H v z V V v V s 0 1 e0 1 e0 V 0 V 0 H 0 A H 0 1 V 0 v
v
s
s
s
V
v
V 0
v
s
e H e 1 e0 H H , , e (1 e0 ) (1 e1 e) H 0 H 0 1 e0 H 0 H H 0 H H H H H (1 e1 ) )e (1 e1 ) e (1 e1 ) e e (1 H 0 H 0 H 0 H 0 H 0 H
z
H H 0
z
is volume strain; is vertical strain;
z v
16
8
Compressibility characteristics: non-linear and elastic-plastic e V H v z ; e e0 e e0 v (1 e0 ) V 0 H 0 1 e0 e, V , H 0 Over-consolidated Over-consoli dated range Normally consolidated range (virgin compression range) i
Loading
i
Slope (1 e0 )mv i+1
i+1 Slope C e
Un-loading Re-loading
17
Behavior under 1-dimensional compression
NCL
1 independent variable in 1-D straining test Compression of a clay in 1-D straining test
18
9
1. The coefficient of volume compressibility m (mv) v v ,i 1 v ,i ' ; v mv (important for settlement) ' i'1 i' e ei e ei 1 e ei 1 e eo e ; o ; v ,i 1 o ; v ,i 1 v ,i i v 1 e0 1 e0 1 e0 1 e0 1 e0 mv
mv
v 1 ei ei 1 ' ' 1 e0 i 1 i'
v
mv
if i 0
e0 e1 ' 1 e0 1 0' 1
H i H i 1 H 0
v 1 H i H i 1 ' ' ' H 0 i 1 i
if i 0
1 H 0 H 1
' ' H 0 1 0
The mv varies with the vertical stress (not (not constant) constant) !
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2. The compression index (C c) in normally consolidated range and expansion (C e) (or unloading/reloading, swelling C s)
The compression index (C c) in normally consolidated range: ' ' ' ' ei ei 1 C c (log i 1 log i ) C c log( i 1 / i )
C c
ei ei 1 ' i 1
log(
e0 e1
if i 0
' i
/ )
log( 1' / 0' )
The expansion index ( C e) in un/re-loading or over-consolidated range: ei ei 1 C e (log i'1 log i' ) C e log( i'1 / i' )
C e
ei ei 1 ' i 1
log(
if i 1
' i
/ )
e1 e2 ' 2
Both C c and C e are constant !
' 1
log( / )
3. Relation between C c and mv
ei ei 1
mv
C c log( i'1 / i' )
e0 e C c log(
'
/ 0' )
e
C c '
ln10 '
1 e C c 1 0.434C c 1 v 1 e0 ' ' 1 e0 ' ln101 e0 '
20
10
VCL: Virgin Consolidation Line NCL: Normal Consolidation Consolidation Line e e0 C c log(
v' vo'
)
v' z' ; vo' 1kPa
C s=C e=C r
e ei C s log(
v' vi'
)
Idealized 1-D compression behavior
21
How to calculate settlements? Compression: for each layer H j (thickness), if mv and ’ are constant with depth z, then (loading): scj v H j mv H j '
Compression: for normally consolidated clay, using C c (loading): scj v H j
C c
1 e0
log
2' 1'
H j
2' 1' '
Heave/Swelling: for normally consolidated clay, using C e (un-loading): scj v H j
C e
1 e0
log
2' 1'
H j
Total compression S of multiple n-layers with H j: S
j n
j 1
scj
j n
j 1
vj H j
j n
j 1
mvj j' H j
or S
j n
C cj
1 e j 1
0 j
log
2' j 1' j
H j
22
11
Pre-consolidation pressure – how to to determine determine it? from laboratory oedometer test (by Dr. Arthur Arthur Casagrande) ' p' vp c' vc'
vi' v' 0
Initial effective stress vi' v' 0 OCR
' vp
v' 0
1/2
Select a point with maximum curvature D
1/2
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Vertical strain vs. log(time) for ratio C4 (100% HKMD and 0% sand), (a) loading, (b) unloading, and (c) reloading (Yi (Yin n 1999) L o g ( t i m e ) (m ( m i n ) (C (C 4 ) 0 .1
1
10
100
1000
1000 0
1 00 000
0 10kPa
2 4
25kPa
6 ) % (
n i a r t S l a c i t r e V
50kPa
8 1 0 1 2
100kPa
1 4 1 6
200kPa
1 8 2 0 2 2 2 4 2 6
400kPa
Creep: compression under a constant effective stress L o g ( t i m e ) (m (m i n ) ( C 4 )
Yin, J.-H. (1999). Properties and behaviour of Hong Kong marine deposits with different clay contents. Canadian Geotechnical Geotechnical Journal, Journal, Vol.36, Vol.36, No.6, pp.1085-1095 pp.1085-1095 24
12
L o g ( t i m e ) (m ( m i n ) (C (C 4 ) 0 .1 ) % (
n i a r t S l a c i t r e V
1
1 0
1 00
1000
10000
10000 0
2 2 50kPa
22.5 100kPa
2 3 23.5 200kPa
2 4
Unloading
24.5
L o g ( t i m e ) (m ( m i n ) (C (C 4 ) 0 .1
1
10
1 00
1000
10000
10000 0
21 75kPa 100kPa
) % (
n i a r t S l a c i t r e V
23
Reloading
200kPa 400kPa
25
Creep: compression under a constant effective stress
27
29
800kPa
31
25
Vertical strain vs. log(stress) with loading, unloading and reloading for mixing ratio C4 (Y (Yin in 1999) V e r ti t i c a l S t r e s s ( k P a ) (C (C 4 ) 10
100
1000
0
5
C 10 ) % (
n i a r t S l a c i t r e V
v v 0 C c log( e e0 C c log(
15 v v 0
C c
1 e 0
v' vo'
v' ' vo
); v v 0
log(
vo'
)
' vo
Creep: compression under a constant effective stress
)
C c
1 e 0
Creep 32 days
C
r r = 0 . 0 2 6 6 ,
C r
30
v'
C c
20
25
R 2= 0 . 9 9 9
e 0 e
1 e 0
); e0 e C c log( v'
c = 0 . 1 8 2 ,
C r
1 e 0
R 2= 0 . 8 4 1
Unloading and reloading Creep 18 days
35
26
13
1-D compression under isotropic stressing
z' x' y'
M e a n e f fe fe c t i ve v e s t r e s s p ' m ( k P a ) 100 1000 p 'm o
10
10000
0 C c -line C c 1 e 0
5 ) 10 % ( m v
Tes t: 1 day Un/reloading Loading
p m' ( z' x' y' ) / 3
15
n i a r 20 t s e m 2 5 u l o V 3 0
' p vm vm0 C c log( 'm ) pmo
C r C r - l1i n ee0
'
pm
vm vmi C r log( ' ) pmi
35
40 Yin, J.-H. and Zhu, J.-G. (1999). Measured and predicted time-dependent stress-strain behavior of Hong Kong marine deposits. Canadian Geotechnical Journal, Vol.36, No.4, pp.760-766
27
Time (min) 0 .1
10
1000
100000
8
) % ( 1 0 n i a r t 12 s e m1 4 u l o V1 6
Creep
“Primary” consolidation
18
Creep before and after EOP
End-of-“Primary” (EOP) consolidation
60 ) a 4 0 P k ( u 2 0
ue 0
0 0 .1
10
1000
100000
Time (min) 28
14
Old equation : e e EOP C e log t 0; log
t t EOP
t t EOP
; e ; right ?
New equation by Yin and Graham : e e0 C e log t 0;
e e0 ;
t o t t o definite;
right !
29
C r line
C c l i n e
Strain Stra in ra rate te ef effe fect cts: s: Do co cons nsta tant nt-r -rat atee-of of st stra rain in te test stss (a (aft fter er Yin an and d Gr Grah aham am 19 1989 89). ). The larger the strain rate, the higher the stress for the same strain value (the higher the prepre-conso consolidati lidation on press pressure) ure) Stresss rela Stres relaxation xation:: Ke Keep ep th thee st stra rain in co cons nsta tant nt,, th thee ef effe fect ctiv ivee st stre ress ss de decr crea ease sess wi with th ti time me Stress Str ess rat ratee ef effe fects cts:: Do co cons nsta tant nt-r -rat atee-of of st stre ress ss te test sts. s. Th Thee la larrge gerr th thee st stre ress ss,, th thee hig highe her r thee st th stre ress ss fo forr th thee sa same me st stra rain in va value lue.. 30
15
1.3 Direct Shear Tests Tests and 2D Behavior of Soils
• • • •
Two degr degrees ees of of freedo freedom m (two (two inde indepen penden dentt varia variable bles) s) Direct Dir ect shea shearr tests tests:: to get she shear ar streng strength th param paramete eters rs only only Simula Sim ulate te twotwo-dim dimensi ensiona onall (2D) (2D) fail failures ures and beh behavi avior or Limi Li mita tati tion onss of of dir direc ectt shea shearr tes tests ts?? (i) (ii) (iii) (iv) (v) (vi) (vii)
No contr control ol of drain drainag age: e: drain drained ed or und undrai raine ned? d? No measurement measurement of pore water pressure pressure so so that the effective effective stress stress may may not be known. Ok for sandy sandy soils soils since draine drained d condition condition is is kept. Data are used used for obtaining obtaining streng strength th parameter parameterss only. only. Deformation parameters (Y (Young’s oung’s modulus) cannot obtained. “Quick” shear or “slow” shear shear on clayey soils is questionab questionable. le. Many soil lab test reports reports present friction friction angle and cohesion cohesion from “quick” “quick” shear or “slow” shear tests. Be careful to use these strength parameters (effective (effective or total stress parameters?)
31
Shear strength (1) Shear failure (2) Shear strength tests
Shear sliding
When sliding Angle=?
N W
T
32
16
(1) Shear failure • It is is found found that that,, if at a poi point nt on on any pla plane ne with within in a soi soill mass, the shear stress becomes equal to the shear strength of the soil, then failure will occur at that point. • Coulomb first proposed a liner function between shear strength f and the normal total stress n: f c n tan c is the cohesion (line intercept) and is is the angle of shear resistance (slope angle of the line) The c and are also called total stress strength parameters.
The above equation is not GOOD, as Shear Resistance (friction) is provided by soil particles only. We shall use effective stresses. 33
(2) Shear strength tests The direct shear test
n
N A
l is the horizontal displacement h is the vertical displacement
T A
Soil Specimen
' u Direct shear box test: the most simple test for measuring shear strength of soils Problems: (i) pore pressure cannot be measured, (ii) the shear strain cannot be determined, (iii) stresses are not uniform. Applications: Measuring strength parameters para meters only, NOT only, NOT deformation parameters (e.g. no Young’s Young’s modulus modu lus E E and shear modulus G) 34
17
35
Four direct shear boxes in Soil Mechanics Laboratory
36
18
37
n'
f c ' n' tan tan '
c’ is a line intercept, may not be the true cohesion. True cohesion is due to cementation False cohesion is (a) due to line fitting produced intercept and (b) suction How to know?
n' What is the relationship of the failure criterion in terms of principal of principal stresses? stresses? 38
19
1.4 Triaxial Tests and 2D Behavior of Soils 1 3c 1 ( 1 F / A)
Conventional triaxial apparatus
3 3c Total stress path :
3 3c constant 1 3c 1
( 1 F / A)
39
l is the vertical displacement
1 F / A
Soil Specimen
1i 3
3 pcell
' u 40
20
Triaxial test: the most important and popular test for measuring shear strength and stress-strain curves of soils
Advantages: (i) pore pressure can be measured, (ii) the strains can be determined, (iii) stresses are uniform. Applications: Measuring strength parameters and deformation parameters (e.g. Young’s modulus mod ulus E and shear modulus G)
41
Pore pressure is measured electronic transducer.
42
21
Pore pressure is measured electronic transducer. 43
Two conventional triaxial apparatus in i n Soil Mechanics Laboratory
44
22
A stress-path triaxial apparatus in Soil Mechanics Laboratory
45
Triaxial test: the most important and popular test for measuring shear strength of soils l is the vertical displacement
1 F / A
1i 3
1 3 1 1 3 1
Soil Specimen
3 p
water pressure) (cell
46
23
Triaxi riaxial al test results results – how to define failure ?
3 150kPa 1i 3
1 3 1
3 100kPa
1 3 1 1' 1 u
3 50kPa
3' 3 u
u is zero for drained test,
1' 3'
3 50kPa
( 1 u ) ( 3 u )
3 100kPa 3 150kPa
1 3 q q is called deviator stress or the principal stress difference. 47
Failure plane
'
f
1 2
( 1' 3' ) sin 2 ; 'f
1 2
( 1' 3' )
1 2
( 1' 3' ) cos 2
is the angle between major principal major principal plane and the failure plane
2 90 ; 45 o
'
o
' 2
48
24
Consider vertical and horizontal force equilibrium of the triangular area of the specimen: sin 2 2 sin con 1 tan
Failure plane
2 tan 1 tan 2
2
cos 2
1 tan 2
h d
2
1' d f ' (d / cos ) cos f ( d / cos ) sin 0 ' ' 3 ( d tan ) f (d / cos ) sin f (d / cos ) cos 0 1' d f ' d f d tan 0 ; ' ' 3 ( d tan ) f d tan f d 0
(vertical F v 0) (horizontal F h 0)
f ' f tan 1' ; ' ' f tan f 3 tan
f ' f tan 1' ' ' f f / tan 3
(1) (2)
(1) ( 2) : f tan f / tan 1' 3' 1 3 2 tan '
f ( 1 3 ) /(tan 1 / tan ) '
'
'
1 tan 2
2
1 3 '
'
2
1 3 '
sin 2
'
2
sin( 2 )
(3)
49
Consider vertical and horizontal force equilibrium of the triangular area of the specimen: sin 2 2 sin con cos 2
From (3) : f
1 tan 2
Failure plane
2 tan 1 tan 2
1 tan 2
1' 3' 2
h
sin( 2 )
d
1 3 '
'
2
2 ' 1
' 3
2
f '
From (1) and (3) :
f 1 f tan 1 ( 1 3 ) '
'
'
1 3 '
'
2 ' 1
' 3
2
1 3 '
'
2
(1
'
'
2 tan 2 1 tan 2
1 tan 2 1 3 '
'
2
2
1 3 '
tan 2
)
1' 3'
1' 3'
'
2
2
1 3 '
[
cos( 2 ) '
2
( 1' 3' ) 2 tan 2 2
1 tan 2
]
' ' 2 2 1 3 1 tan 2 tan ( ) 2 2 1 tan
1 tan 3' 1' 3' ' 3' 1' 3' ( ) cos 2 1 cos( 2 ) 2 2 1 tan 2 2 2 2 ' 1
' 3
2
' 1
The above is proof of Mohr circle : f c ' f ' tan '
50
25
1 ( 1' 2
'
3' )
'
c cot cot
1 ( 1' 2
3' ) ' ' ' ' c cot 12 ( 1 3 )
From Fig.4.2:
1 ( 1' 3' ) 2 ' ' ' c cot 12 ( 1
sin '
3' )
( 1' 3' ) ( 1' 3' ) sin ' 2c ' cos ' tan ( 45
'
) 2c tan(45 ) 2 2 This is Mohr-Coulomb failure criterion ! 1'
'
or
3'
2
o
'
o
51
( 1' 3' ) ( 1' 3' ) sin ' 2c ' cos ' 1' (1 sin ' ) 3' (1 sin ' ) 2c ' cos ' 1' 3' 1'
3'
1'
3'
1'
3'
tan
1'
2
(1 sin ) '
(1 sin ) '
2c '
'
cos
(1 sin ' ) 2 cos 2 ' (1 sin ' )2
2c
cos(90o ' ) sin '
cos '
'
(1 sin ' )
(1 sin ' )
[1 cos(90o ' )]2
3'
cos '
'
sin 2 (90o ' )
2c
cos2 ' sin 2 1 sin(90o ' ) cos '
(1 sin ' )
(1 sin ' )(1 sin ' )
2c
'
sin(90o ' ) 1 cos(90o ' )
sin 1 cos tan (45 2
o
' 2
) 2c tan( 45 '
o
' 2
)
This is Mohr-Coulomb failure criterion ! 52
26
( 1' 3' ) ( 1' 3' ) sin ' 2c ' cos ' 1' (1 sin ' ) 3' (1 sin ' ) 2c ' cos ' 1'
3'
1'
3'
1'
3'
1'
3'
(1 sin ' ) (1 sin ' ) (1 sin ' ) (1 sin ) '
(1 sin ' ) (1 sin ' ) (1 sin ' ) (1 sin ' )
2c
'
2c
'
2c
'
2c
'
cos ' (1 sin ' ) 1 sin 2 ' (1 sin ' ) (1 sin ' )(1 sin ' ) (1 sin ' ) 2 1 sin ' 1 sin '
This is Mohr-Coulomb failure criterion – another form !
53
1 ( 1' 2
3' ) 12 [( 1 u ) ( 3 u )]
12 ( 1 3 ) 1 ( 1' 2
3' ) 12 [( 1 u ) ( 3 u )]
12 ( 1 3 ) u
1 ( 1' 2
3' ) a ' 12 ( 1' 3' ) tan '
( 1' 3' ) ( 1' 3' ) tan ' 2a '
This is Mohr-Coulomb failure criterion – another form !
( 1' 3' ) ( 1' 3' ) sin ' 2c ' cos ' 2a ' 2c ' cos ' a ' c ' cos ' tan ' sin ' ' tan 1 (sin ' ) or c ' a ' / cos ' ; ' sin 1 (tan ' )
54
27
q ( 1' 3' )
( 1' 3' ) ( 1 3 ) p ' 13 ( 1 2 3 ) u
p u
p ' 13 ( 1' 2 3' )
This is Mohr-Coulom Mohr-Coulomb b failure criterion – another form !
a q a p ' tan ;
( 1' 3' ) a 13 ( 1' 2 3' ) tan
q 1' 3' , t 1' 3'
q t 2 1' , 1' (t q) / 2, t q 2 3' , 3' (t q ) / 2
q a 13 ( 1 2 3 ) tan a 13 (
t q
t q) tan a 13 ( 2 6q 6a (3t q) tan 6q q tan 6a 3t tan '
'
3t q
q(6 tan ) 6a 3t tan (below obtained ) :
( 1' 3' )
6a 6 tan
( 1' 3' )
3 tan
( 1' 3' ) 2c ' cos ' ( 1' 3' ) sin ' 3 tan 6 tan 6a 6 tan
sin ' tan
6 sin ' 3 sin
'
) tan
Compare (4) to (5) :
3 tan 3 M 6 tan 6 M 6a 3a 1 2c ' cos ' c' 6 tan 6 tan cos '
sin '
( 4)
6 tan
2
(5)
M tan 1 (
2c ' cos ' a c ' cos ' (6 tan ) / 3
6 sin ' 3 sin '
)
6 cos ' 3 sin '
55
Example 1.1
The results shown in the following first table were obtained at failure in a series of triaxial tests on specimens of a saturated clay initially 38mm in diameter by 76 mm long. Determine the values of the shear strength parameters with respect to (a) total stress and (b) effective stress. Discuss why UU tests show zero friction angle angle in terms of total stress. Solution:
The principal stress difference at failure in each test is obtained by dividing the axial load by the cross-sectional cross-sectional area area of the specimen specimen at failure failure (second table). The corrected corrected crosssectional area is calculated: A A0
1 v 1 a
The is no volume change during an undrained test on a saturated clay. The initial values are: l0 76 mm , A0 1135 mm 2 ; V 0 86 10 3 mm 3 The total stress parameters, representing the unconsolidated and undrained (UU) strength of the clay, are: 2 cu 85 kN / m ,
u
The effective stress parameters, representing the consolidated and drained (CD) strength of the clay, are: c ' 0, ' 27 o 56
28
Ao
Area correction The corrected area A is: is: v
V o V Lo Ao LA V o
A Ao
L
Lo
Lo Ao
Lo
A L
Lo
(1 v ) L Lo L o L
A Ao
Lo
Lo ( L o L) Lo
1 v
1
( Lo L) Lo
1 a v
1 a
V o V V o
V V o
L L L is the volume strain; a o Lo Lo a is axial strain; Compression strains and stresses are positive in Soil Mechanics. v
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3 Unconsolid ated Undrained (UU ) Consolidat ed Drained (CD )
Effective stress
ue 0
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29
From 3 UU tests :
From 3 CD tests :
c u 85kPa, u 0
' ' o c 0, 27
why u 0 ?
Answer: The three soil specimens must come from the same depth of the ground at which, the initial effective stress stress was the same. Undrained consolidation does not change the initial effective stress. In fact, c fact, cu increases with the initial effective stress:
c u (0.2 ~ 0.4) v'
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Example 1.2
Based on the data in Table 1, use the method of q-p’ plot to determine the values of the shear strength parameters parameters with respect to (a) total stress and (b) effective stress. Solution:
See tables and figure below Cell pressure 3 (kPa)
UU UU UU CD CD CD
test test test test test test
200 400 600 200 400 600
3 (kPa) UU UU UU CD CD CD
test test test test test test
Axial Axial deformation Volume change A xia l lo ad ad (N) (mm) (ml=cm^2) 222 9.83 0.00 215 10.06 0.00 226 10.28 0.00 403 10.81 6.60 848 12.26 8.20 1265 14.17 9.50
1=l/lo 200 400 600 200 400 600
v=v /Vo 0.1293 0.1324 0.1353 0.1422 0.1613 0.1864
A (mm^2) 0.0000 0.0000 0.0000 0.0766 0.0951 0.1102
1302.6 1307.1 1311.5 1220.9 1223.6 1240.4
q=1-3 p=(1+23)/3 (kPa) (kPa) (kPa) 170.4 370.4 256.8 164.5 564.5 454.8 172.3 772.3 657.4 330.1 530.1 310.0 693.0 1093.0 631.0 1019.8 1619.8 939.9
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30
1200 1000
y = 1.0952x - 5.7259 R² = 0.9996
800 ) a P k (
sin '
600
q
400
y = 0.0049x + 166.86 R² = 0.0564
200
c'
3 tan 3 M 6 tan 6 M 3a 1
6 tan cos '
0 0
100
200
300
400
500
600
700
800
900
1000
p (kPa)
1200 1000
y = 1.0874x R² = 0.9996
800 ) a P k (
600
q
400
y = 0x + 169.08
200 0 0
100
200
300
400
500
600
700
800
900
1000
p (kPa)
UU t es es ts ts (u=0)
UU t es es ts ts a= tan()= sin(')= = c=
166.86 0.0049 0.0024 0.14 83.36
169.08 kPa 0 0 0.00 degree 84.54 kPa
CD tests
CD test s (c '=0)
-5.7259 kPa 1.0952 0.46 27.60 degree 27.60 degree -2.73 kPa -2.73 kPa
0 kPa 1.087 0.46 27.41 degree 27.41 degree 0.00 kPa 0.00 kPa
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1.5 True Triaxial Tests and 3D Behavior of Soils Truly Triaxial Triaxial System (TTS): control of 3 independent parameters Test Types: (i) Change to any principle principle stress – varyin varying g b-value (ii) Plane strain tests Problems: Interference at the corners (a) non-uniform stresses (b) small compression
1
3
2
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Development of failure criteria and constitutive models Extension to/study in a general stress space: advanced lab facilities are needed. Useful to engineers: Mohr-Coulomb failure criterion gives more conservative strength – on the the safe safe side !
Failure surfaces in plane 63
63
Type 1: 6 sliding rigid plate
Rigid sliding plates
Cubical soil specimen inside
(a) A schematic view of a rigid boundary true triaxial loading frame with six sliding rigid plates plate s (Hambly 1969, Pearce 1972, Arrey and Wood 1988) 1 988) 64 64
32
Cambridgee type TTS - six sliding rigid plates Cambridg plates Photo courtesy of Professor Muir-Wood Muir-Wood
65
65
A special chamber filled with de-aired water (or oil) to apply confining pressure
Piston to apply vertical force with LVDT outside to measure vertical displacement
Load cell inside –one vertical and two horizontal
Sliding design
Piston to apply horizontal force with LVDT outside to measure horizontal displacement
Flexible tubing for water drainage/back water pressure
Sliding design
New sliding loading plates and setup
(both left and right sides)
A brick shaped soil specimen in sealed rubber membrane
Sliding design
Flexible tubing for water drainage/back water pressure
Sliding design
No Interference at the corners using sliding plates 66
66
33
New sliding loading plates and setup
No Interference at the corners using sliding plates
67
67
Yin’s Sliding Plates Inside
68
68
34
Chamber is filled with water and closed
Yin’s Sliding Plates Inside
69
69
70
70
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The whole system 71
71
HK CDG (Completely Decomposed Granite)
72
72
36
Effect of the magnitude of the intermediate principal stress on sand b
0 triaxial compression 1 3 1 triaxial extension
2 3
' 47.8 o
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1.6 Concluding Remarks • Re Real al soi soill beh behav avio iorr is is com compl plic icat ated ed • No mode models ls can can capt capture/ ure/repr reprodu oduce ce all all soi soill beha behavio viorr features • Sim Simple ple mod models els are devel develope oped d and and used. used. But the they y shall shall at least reproduce the soil behavior that is dominant in the problem under investigation,
for example: (a) for soil instability (slope failure), the soil failure shall be reproduced (b) for settlement analysis, proper stiffness shall be reproduced.
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