soil shear strength_lecture

7/30/2019 Soil Shear Strength_Lecture

1/80

PBWN302

CEM/WEE/STE

Dr.Asmaa

Moddather

SoilMechanicsandFoundations

FacultyofEngineering CairoUniversity


2/80


3/80

Soil fail in shear.

Shear strength of soil is the internal resistance that the

soil mass offer to resist failure/sliding along any plane

.

ear a ure w occur at po nts w ere s ear stresses exceeds soils shear stren th S .

Dr.Asmaa Moddather PBWN302 Fall2012


4/80

Shear stresses are generated into the soil mass due to

.

The engineer needs to know the nature of shearing

o bearing capacity of foundations.

o stability of slopes.

o lateral pressure on retaining walls.



5/80

FailuresurfaceStablemass

Bearing

capacity

of

Foundation



6/80

Failuresurface

Stablemass

ta ty o s opes



7/80

Directionofmovement

Failuresurface

Stablemass

lateral ressureonretainin walls

Failuresurface



8/80

1. oca a ure ta es p ace at po nts w ere s ear stresses() > shear strength (S).

2. When local failure occurs at sufficiently large numberof oints within the soil mass a eneral failure takesplace.

3. a ure a es e orm o s ng o a so oc over aFailure/Sliding/Slip surface within the soil mass.

To study shear failure at a point, we need to calculate:

1. Stresses (, ) on any plane through this point


2. Shear resistance S at this point


9/80

ons er ng a certa n po nt ns ethe soil mass and knowing the

y

norma an s ear s ressesacting on two planes at this point:xy

x

o What is the maximum and

minimum normal stresses

xx

xy

magnitude and direction ?

o What is the maximum shear

xy

stresses (magnitude anddirection)?

y

o What is the normal () and shear

stresses actin on an lane?



10/80

t s a grap ca met o to presentthe state of stress along any plane

pass ng roug any po n w n esoil mass. y

Need to define and signxy

x

o For :

xx

xy

Compression

+ve sign

Tension ve signxy

o For :

Rotation clockwiseve si n

y


Rotation anticlockwise +ve sign


11/80

y

ane

tressesonp anea: x,xy

Stressesonplaneb:( , x )

xy

xy

ane a

xx

xyMohrs circle

xy

y

a

o

yxy

bxy



12/80

Mohrs circle

x

yxy

xy

Plane a

xy

xy

a

y

ox

yxy

Every point on the circle represents the state of stress acting on a

plane passing through the soil element.


There are infinite number of planes passing through the element.


13/80

Plane b

maxc

xx

xyxy

Plane a

aMinor principal

plane (3, 0)Major principal

plane (1, 0)xy

xy

b

y

mind

Maximumandminimumnormalstresses:

o Majorprincipalplane(1,0)

3,

Maximumandminimumshearstresses:

o Planec

+ /2


o Planed((1+3)/2,min)


14/80

Plane b

xx

xyxy

Plane a

a xyxy

,

b

y

Pole

Define: Pole

parallel to any plane, the line will intersect the circle at a point

whose coordinates are and actin on this lane.



15/80

Plane b

xx

xyxy

Plane a

axy

xy

plane

by

Poleorizonta

plane

HowtofindthePole?

Needtoknowthefollowingaboutasingleplane:

o Stateofstress( and)


o Directionofplane


16/80

Plane b

maxc

xx

xyxy

Plane a

aMinor principal

plane (3, 0)Major principal

plane (1, 0)xy

xy

1

b

y

Pole

23

mind

Knowingthelocationofthepole,determinethedirectionof:

1, ,

o Minorprincipalplane(2,withhorizontal,clockwise)

o

Plane

with

maximum

shear

stress

( ,

with

horizontal,

clockwise)


o Planewithminimumshearstress(4,withhorizontal,anticlockwise)


17/80

estateo stresson2perpen cu ar

laneswithinasoilelementisshownin2 t/m2

theoppositefigure.DrawMohrscircle, 1.5 t/m2

anddetermine:

5 t/m25 t/m

.

whichtheyact. 2

ii.Maximum

shear

stress

and

the

Soil Element

.

iii.Thestateofstressonaplaneinclined


20o anticlockwisewithhorizontal.


18/80

22 t/m

2

2Plane b

, .

Stresses on plane b: (+2, +1.5) t/m25 t/m2

5 t/m2

Plane a

2 t/m2Note: horizontal

and

vertical

axes

MUST

be

drawn

withthesamescale 2

b Pole

o (t/m2)

a



19/80

t/m2

max

b3 Pole

o 1

3

2

(t/m2)

a

i.Principle stresses ( = 0):

1: Major principle stress = 5.62 t/m2, 1 = 68

o clockwise, with horizontal

3: Minor principle stress =1.37 t/m2, 2 = 158

o clockwise, with horizontal


Maximum shear stress: max = 2.12 t/m2, 3 = 23

o anticlockwise, with horizontal


20/80

t/m2

b20

Pole

(t/m2)

(1.39, 0.18)

(t/m2)

a

iii.

State of stress at plane inclined 20o anticlockwise with horizontal:

= 1.39 t/m

2


= 0.18 t/m2


21/80

o Frictionandinterlockingbetweensoilparticles Friction

componento Interparticleattractionforces(duetoelectrochemicaleffects)

Cohesioncomponent

function of the effective normal stress () acting on plane of

failure: ''' tancS +=

where, c and are the shear strength parameters

c: effective cohesion


: effective angle of shearing resistance


22/80

Mohr-Coulomb shear strength

failure envelope

c

'''

f tancS +==



23/80

Shear strength

failure envelope

(f, f)

c

Elastic equilibrium

ast c equ r um



24/80

,

between particles ():

v.loose

versus

v.

dense

o Gradation:

poor ygra e versuswe gra e

small large

o

Particle

shape:roundedversusangular

small large


o Particlesurface

roughness

(as

roughness

increases,increases)


25/80

orcoarsegra ne so :

o

ranges

rom

27o

to

45o

dry wet



26/80

Forfine

rained

soils,

S

de ends

on

friction

and

interlockin

o Stresshistor overconsolidation ratio :

betweenparticles()aswellascohesion(c):

asOCR

increases,

S

increases

,

floculated hashigherS

:

asdisturbanceincreases,Sdecreases

o Soi permea i ity water rainage :

Shearstrength

of

soil

loaded

under

drained

conditions(slowly)isdifferentfromthat

c

loadedunderundrained conditions(quickly)

Drainedshearstrengthversusundrained shear

stren th



27/80

Shear strength parameters for a particular soil are

determined by means of laboratory tests on specimens

sampled from insitu soil.

Great care is required in sampling, storage, and handling

o samp es pr or o es ng, espec a y n case o

undisturbed samples where it is necessary to preserve the

in

situ structure and water content of soil.



28/80

Shearboxapparatus.

ProvingringShearbox

Vertica isp acement ia gage

Motor,applieshorizontal


Weightstoapplynormalload


29/80



30/80

s a ev ce use to measure orce. t cons sts

of an elastic ring of known diameter with a

measuring device located in the center of the

r ng t at measures ts e ect on.

,

by which we convert the readings of the

measuring device (deflection) into force.

Force = proving ring reading x calibration

constant.



31/80

c e m a co s ear oxappara us:

oa ngp a e

Samplemaybecircularorsquare

~

2


Boxissplitinto2halves


32/80

1. ssem e eappara us samp e,porouss ones, erpaperneeded),shearboxparts)andfillshearboxwithwaterifneeded.

2. pp y

norma

orce

on

top

o

s ear

ox

1 .3. Shearforceisappliedtospecimenbymovingonehalfofthebox

relativetotheothertocausefailureinthesoilspecimen.Shearingforceismeasuredbyprovingringorloadcell.

Pictureofsampleafterfailure

4. ert ca v an or zonta h sp acementsaremon tore duringthetest.


5. Repeatsteps

2,

3,

and

4

for

different

normal

forces

(N2,N3).


33/80

1. Foreach

test,

calculate:

Normalstress()=normalforce/areaofsample

Shearstress()=shearforce/areaofsample

2. Plot:

Peak

ress,

Densesand

shearstrength

Densesandec

imen,

v

nsion

Shearst

f heightofsp

Expa

ion Sheardisplacement,h

sandf

==constantLoosesand

Chang

ein

Com

pres

Loosesandandnormallyconsolidatedclayresultinsimilar(h andvh

Sheardisplacement,h


Densesandandoverconsolidated clayresultinsimilar(h andvhrelationships)


34/80

3. or es 1,ge 1 an max1=f1.

Fortest2,get2 andmax2=f2.

, 3 max3= f3.

4. Plot:

Test (2)

Test (3)

(3, f3)

Test (1)

(1,f1)

, ote: or zonta an vert ca

axesMUST

be

drawn

with

the

samescale

c



35/80

1. enera y, rec s ear es s are con uc e on ry san , sa ura e san ,

and saturated clay.

2. For sands, soil has high permeability, water drains instantaneously

ur ng e es , u = 0 ur ng s ear ng, ere ore, = roug ou

the test.

3. For clays, to maintain = throughout the test (similar to sands), the

es s con uc e s ow y so excess pore wa er pressure u can ra n

during shearing.

4. Measured shear strength parameters: c, called drained/effective


s ear strengt parameters.


36/80

oreac es ,we now an f on a urep ane,w c s or zon a .

(,f) Pole

cMinor principal

plane (3, 0)Major principalplane (1, 0)



37/80

s ear ox test carr e out on a s ty c ay so , gave t e

followin results:

235.6176.6117.758.9Verticalload(N)

8..1..2Provin

rin

dial

au ereadin

mm

If the shear box area is 36 cm2 and the proving ring

,

and the angle of shearing resistance.



38/80

xamp examp e

235.6176.6117.758.9Verticalload(kg)

8.97.15.33.2rov ngr ng a gaugerea ng

(mm)

132.75105.7578.75

=3.2x15

Horizontalforce, H

(kg) =47.25

=6.544.913.27

.

=1.64 (kg/cm2)

3.692.942.19=47.25/36

=

(kg/cm2)


.


39/80

xamp examp e4.00

3.50

2.50

.

(kg/cm

2)

1.50

2.00

arStress,

1.00Sh

0.00

.

0.00 0. 0 1.00 1. 0 2.00 2. 0 .00 . 0 .00 . 0 .00 . 0 6.00 6. 0 .00

c

Normalstress, (kg/cm2)c = 0.60 kg/cm2


= 25o


40/80

1. Simple

2. Not expensive

Disadvantages:

. ,weakest plane (might overestimate strength).

. , ,

are adjusted such that u = 0..

specimen is not uniform (stress concentration at corners).



41/80

Provingring

Test

specimen

Dialgageforvertical

displacementTriaxialcell

Triaxialcell



42/80

Filterpaper

Rubbermembrane

OringsTop/bottom


Porousstones


43/80

Verticalload

Transparentcylinder

Orings

Confiningpressure



44/80



45/80

Air is released SamplePlacement

Open valve

Air release valve

Top drainage

Loading ram

Transparent cylinder

Porous stone

Specimen

Porous stone

Membrane

Open valve


All around

Pressure supply

o om ra nage


46/80

SamplePlacement:

1. Trim

sample

to

required

dimensions

(h/d

~ 2).

2. P acesamp eon ottom i terpaperan porousstone.

3. ace

op

er

paper,

porous

s one,

an

oa ng

cap.

. .

6. Lowerloadin bar usttorestonloadin ca .



47/80

oa ng ages:

o Sta eI

l in confinin ressure( )

Openvalve

Open/closedvalve


aroun

Pressuresupply


48/80

Applyingaxialload(N)

Deviatorstress

(d)

oa ng ages:

o Sta eII

l in confinin ressure( )

Openvalve

Open/closedvalve


aroun

Pressuresupply


49/80

o Stage1:

Apply

confining

pressure

(c)

= cell

pressure.o Sta e2:

Increaseaxialload(N)uptofailure.Axialload/areaofspecimen

iscalled

deviator

stress

.

c c+df

cc cc

+


Stage(1) Stage(2)


50/80

ra nage con ons ur ng s ages 1 an 2 are se accor ng orequired type of test.

c c + df

cc cc

Valve (open/closed) Valve (open/closed)

c

c + df

Valve opened

Valve closed

Valve opened

Valve closed


Consolidated, C

Unconsolidated, U

Drained, D

Undrained, U


51/80

c c +df

cc cc


Stage (1)

cStage (2)

c + df

Consolidated, C Unconsolidated, U Drained, D Undrained, U

Stage 1 Stage 2 Test Type

-

C U Consolidated-Undrained (CU)

U U Unconsolidated-Undrained UU


U D Unconsolidated-Drained (UD) ever one


52/80

c c df

cc cc

c

c+df

Note:

Stage 2

c 3

c

+df

=1Stage 1

Minor principal plane Major principal plane


Pole(c, 0) (c+df, 0)


53/80

c c + df

cc cc

Valve (open) Valve (open)

Stage (1)

cStage (2)

c + df

Consolidated, C Drained, D

= =

c c + df1 =

1 =

3 = 3

3 = 3


1 1 1 1


54/80

StressstrainRelationships:

s,d

OCC

OCCecimen,

V

nsion

e

viatorStre

NCC

df

v

olumeofs

Expa

sion Axial Strain, v

D

NCCChangein

Compres

Axial Strain, v

d v V v

relationships)


oosesan

an

norma y

conso a e

c ay

resu

n

s m ar

d

van

Vv

relationships)


55/80

Mohr circle at failure

dStage I

tage

1fdf3

1



56/80

1f(2) 3(2)1f(1)3(1)c=0



57/80

OCC NCC

pc

c1



58/80

c c + df

cc cc

Valve (open) Valve (closed)

Stage (1)

cStage (2)

c + df

Consolidated, C Undrained, U

= =3 c c

c c + df1 =

1 =u =

3 = 3

u = va ue measure uring test

3 = 3 - u


1 = 1 1 = 1 - u


59/80

StressstrainRelationships:

s,d OCC

OCCressure

,u

ve

e

viatorStre

NCC

df

p

orewater

Axial Strain, v

D

NCCChangein

+ve

Axial Strain, v

Dense sand and overconsolidated cla result in similar ( and u

relationships)


v v

relationships)


60/80

u = +ve =

1 = 1u decreases

Same diameter



(Total stresses, measured)

ective stresses, ca cu ate

'c=0


u u


61/80

= -c c, ,

3 = 3u increases

1 = 1u increases

o Ifc>pc, soil behaves as NCC, u = +ve

3 = 3u decreases

=



62/80

Overconsolidated Cla (OCC)



OCC

ect ve stresses, ca cu ate(Total stresses, measured)

2

c1

pc

u1 u2


3 4u=-veu=+ve


63/80

Drainage conditions during stages 1 and 2 are set according to

re uired t e of test.

c

c

+ df

cc cc


c

c + df

Valve opened

Draina e allowed

Valve closed

Draina e not allowed

Valve opened

Draina e allowed

Valve closed

Draina e not allowed


Consolidated, C Unconsolidated, U Drained, D Undrained, U


64/80

c c + df

cc cc

Valve (closed) Valve (closed)

Stage (1)

cStage (2)

c + df

Unconsolidated, U Undrained, U

= =

c c + df1 =

1 =d . . dc . . c

Measured u during shearing = uc + ud +ve (NCC), -ve (OCC)

Dr.Asmaa Moddather PBWN302 Fall20123 = 3 - u 1 = 1 - u


65/80

One effective stress circle to all

total stress circles calculatedSame diameter


Total stresses measured

u= 0

1

'3 1(1)3(1)

cu

3(2) 1(2)

u(1) (-ve)

u(2) (+ve)c =undrained shearstren th

Note: increasingc doesntresultinanyincreasein3


Sinceall

3

prior

to

shearing

is

the

same

in

all

samples,

the

strength

of

the

sampleswhenshearedwillbethesame.


66/80

Loadingframe

Provingring

(measuresaxialdisplacement)

Sample

SpecialcaseofUUtest

o c =0


o Verysimple

and

quick


67/80

amp e acemen :

. ~ .

. .

as drying will alter samples characteristics considerably.

3. Lower loading piston until it contacts specimen.

4. Begin the test, continue until load values decrease or until20 strain is reac e .



68/80

Ao A

Bulging

H

Ho H

ConstantV =

HAAH oo

testbeforeafter test

=

AAHA oooo

ooo

===

=

%

AA o=


)(1)HH(1H)(H

o

o 100


69/80

Readings:Load(F),verticaldisplacement(H)

Data

Reduction: a o

o where: Ho =initialsampleheight

o Stress: =F/Ac

c o a

Ao =initial

x

sectional

area

of

the

sample

DataPlotting:


o Plotstress

() versus

axial

strain

(a)


70/80

Stress,

qu

Strain, (%)

a,f

qu =unconfinedstrength


cu =undrained shear

strength

=

qu/2


71/80

qu

c

= 0 c = 0

u = 0

cu

qu0.0

u

3

1

=

= qu

0

qu =unconfinedstrength


cu =undrained shear

strength

=

qu/2

UnconsolidatedUnconsolidatedUndrainedUndrained (UU Test(UU Test

versusUnconfinedCompressionTestversusUnconfinedCompressionTest


72/80

pp

One effective stress circle to all

total stress circles (calculated)Same diameter

Mohr circle at failure(Total stresses, measured)

c = 3 = 0

u= 0

1 '3 1(1)3(1)

u

3(2) 1(2)

u(UC)u(UU1)

u(UU2)

=


u


73/80

ra ne

s ear

s reng

parame ers:

o c' OCC

o : NCC,sand

Theseparametersareobtainedfrom:

o CDTest

o Usedtodesignstructuresforlongtermcondition Drained.



74/80

Undrained shearstrengthparameters:

o cu, u=0 ,

eseparametersareo ta ne rom:

o

es


Undrained.

xamp examp e 11


75/80

The following results were obtained at failure in a series of

.

i. Determine the shear strength parameters of this soil.

ii. If a specimen is subjected to a confining pressure of 150 kPa

, .

on n ngpressure

(kN/m2)

ev a ors ress

(kN/m2)Test

1003501

210102


xamp examp e 11


76/80

1= c+df3= cc

2

df2

Test

4501001003501

6202102104102

' .

c'=119kN/m2

620c'

'=12o

450100 210

Note: sincec>0,thisclayis


OCC,and

pc >210kN/m2

xamp examp e 11


77/80

ii. If a specimen is subjected to a confining pressure of 150 kPa and a

deviator stress of 200 kPa would it fail? Comment.

350150

c

Elastic equilibrium


xamp examp e 22


78/80

Aclayspecimenfailedatunconfinedstrengthvalueof400kN/m2

when lacedinanunconfinedtesta aratuswhatwouldbethe

undrained shearstrength

for

the

tested

specimen?

u =0

Undrained Shearstrength:

0.0

400

u

cu =

qu/2

=

400/2

=

200

kN/m2


xamp examp e 33


79/80

,

(sampleheight=10.4cm,samplediameter=4.2cm),Findtheunconfined

strengthandtheundrained shearstrengthforthetestedsample.(Ao =13.85

cm2,Ho =10.4cm).

1112171613940Load(kg)

0.3500.3000.1110.0 00.0530.0340.0130.0

3.3652.8851.0670.7690.5100.327=0.013x100/10.40.0 =H/Ho (%)

=0.125

14.3414.2714.0013.9613.9213.9013.85A

(cm2

)0.125

13.85=A

=13.87

100

)100

(%)(1

=

Dr.Asmaa Moddather PBWN302 Fall20120.770.841.211.150.930.65

=0.290Stress(kg/cm2) 87.13

4.0=

xamp examp e 33


80/80

1.20

1.40

qu

1.00

(kg/cm

2)

0.60

.

ialStress,

0.20

.Ax

0.00

0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00

Axial Strain, a (%)

Unconfinedstrength:qu =1.21kg/cm2


Un raine S earstrengt :

cu =

qu/2

=

0.61

g/cm

2

soil shear strength_lecture

Documents