Ge i-module2-rajesh sir

Geotechnical Engineering IGeotechnical Engineering - I

Soil water, Permeability,Shear Strength Shear Strength

Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur

Dept. of CE, GCE Kannur Dr.RajeshKN

1

Module II

Occurrence adsorbed and capillary water types effective stress total

Module IISoil water Occurrence - adsorbed and capillary water types - effective stress - totalstress - pore pressure - pressure diagrams

PermeabilityDefinition - Darcy's law - factors affecting permeability - laboratorydetermination stratified soils average permeability Seepagedetermination - stratified soils - average permeability. Seepage –downward and upward flow -quick sand

Shear Strength Definition - Mohr-Coulomb strength theory - Measurement of shearstrength – Types of Triaxial compression tests - measurement of pore

l d ff i U fi d C i

Shear Strength

pressure - total and effective stress – Unconfined Compression test - vaneshear tests –Direct shear test- strength parameters - choice of testconditions for field problems.


Soil water Modes of occurrence –Effective stress - total stress - pore pressure –p pPressure diagrams under various conditions –Flow of water through soil -


Modes of occurrence of soil water

Free water (Gravitational water)

• Water that is free to move through soil mass under gravity

Held water

• Structural water – chemically combined in the soil mineraly• Adsorbed water (Hygroscopic, contact moisture or surface bound

moisture) – held by adhesion• Capillary waterp y

Pore water

• Water that is free of strong soil attractive forces • Capillary water and free water are pore waters


Stress conditions in soilSt ess co d t o s so

Effective and neutral pressures

• At any plane of soil mass, the total pressure is the total load per unit area

• Pressure due to:• Self-weight• Over-burden

• Inter-granular pressure (effective pressure) and neutral pressure (pore )pressure)


I t l ( ff ti ) ′Inter-granular pressure (effective pressure)• Pressure transmitted from particle to particle through their contact points, through the soil mass above the plane

′σ

points, through the soil mass above the plane

• Effective in decreasing the voids ratio and in mobilising the shear strengthg

Neutral pressure (pore pressure) u w wu h= γ

• Pressure transmitted through pore fluid

• Equal to water load per unit area above the plane• Equal to water load per unit area above the plane

• No influence on voids ratio or shear strength

u′σ = σ +Hence, total pressure


Effective pressure under different conditions of soil-water system

1. Submerged soil mass1. Submerged soil mass

u′σ = σ − WaterEffective pressure

z1sat w w wz z h′σ = γ + γ − γAt A,

1z

wh( )1 1sat w wz z z z= γ + γ − + γ B

z( )sat wz= γ − γ

z ′= γzγAHere, effective pressure does not depend on 1z

Saturated soil


1 1 0w wu z z′σ = σ − = γ − γ =At B, Saturated soil

1z1

whB + =

z

A z ′γEff ti P

( )1 wz z+ γT t l

1sat wz zγ + γEffective pressure diagram

Pore pressure diagram

Total pressure diagram


2. Soil mass with surcharge

u′σ = σ −Effective pressure

1 sat wq z z z′σ = + γ + γ − γAt A, Surcharge, q

z

1 sat wq γ γ γ

1q z z ′= + γ + γMoist

1zB1 10q z q z′σ = + γ − = + γAt B,

Moist soil

z whSaturated soil

A


q

Moist l1z

q q

+soil1

B=

1z γ

Saturated soil

zwh

wzγP

A

Eff ti

( )1 sat wq z z+ γ + γ − γ

T t l

1 satq z z+ γ + γ


Effective pressure diagram

Total pressure diagram


3. Saturated soil with capillary fringe

u′σ = σ −( )1 sat wz z z′σ = + γ − γ 1 satz z′= γ + γAt A,

Effective pressure

( )1 sat wγ γ 1 satγ γCapillary

fringeAt x1,

z1x( ){ }1 1 1sat w

ux z x

′σ = σ −

′σ = γ − − − γ1z

Bx1( ){ }

( )1 1 1

1 1 1

sat w

sat wx z x

γ γ

′σ = γ + − γ

z whSaturated soil

′ ′

Capillarity induces downward pressure due to weight of water hanging below.

A( )1sat wx x z′σ = γ − − γAt x,

1 1 wx z′ ′∴σ = γ + γ


1 satz′σ = γAt B,

( )1sat w

1 wx z′= γ + γ

x xσ = γ = γz γ( )1 1 wz x− − γ 1 wz− γ

1z

1 sat satx xσ = γ = γ1 wz γ

x+ ( )1 wx z− γ

S t t d

B =x z′σ =′γ + γz Saturated

soilsatxσ = γ

1 wx zγ + γ

Pore pressure wzγ

A

T t l

( )1 satz z+ γ1 satz z′γ + γ

( )z z z′= + γ + γ diagram Total pressure diagram( )1 sat wz z z= + γ − γ

( )1 1 wz z z= + γ + γ

Effective pressure diagram


4. When flow takes place through soil

u′σ σ

a. Flow from top to bottom

Effective pressure uσ = σ −

( )1 wu z z h= + − γAt A,

Effective pressure

Pore pressure

z zσ = γ + γ 1z h

w

Total pressure

1sat wz zσ = γ + γB

S t t d

Flow

hEffective pressure

( )1 1sat w wz z z z h= γ + γ − + − γz Saturated

soilwh

A( )sat w wz h= γ − γ + γ

h′


wz h′= γ + γIncrease in effective pressure (downward seepage pressure)

1 wu z= γAt B, Pore pressure

1 wzσ = γ

1 wγp

Total pressure

1 1 0w wz z′σ = γ − γ =Effective pressure hiz

=

wxx hz

′ ′σ = γ + γAt x from B, wx ix′= γ + γ Hydraulic gradient

z

1z h

wxhz

γ

1

B

Flow

hx

x ′γz Saturated

soilwh

Dept. of CE, GCE Kannur Dr.RajeshKNPressure diagram

z ′γwhγ A

b. Flow from bottom to top

u′σ σ

p

Effective pressure uσ = σ −

( )1 wu z z h= + + γAt A,

Effective pressure

Pore pressure h

z zσ = γ + γ

Total pressure

1z h1sat wz zσ = γ + γ

Effective pressureB

S t t d

Flow

wh

( )1 1sat w wz z z z h= γ + γ − + + γz Saturated

soil

( )sat w wz h= γ − γ + γ

h′

A


wz h′= γ − γDecrease in effective pressure (upward seepage pressure)

1 wu z= γAt B, Pore pressure

1 wzσ = γ

1 wγp

Total pressure

1 1 0w wz z′σ = γ − γ =Effective pressure

wxx hz

′ ′σ = γ − γAt x from B, wx ix′= γ − γ h

1z

w whx′

z

B

Saturated

Flow

xwxx hz

′γ − γ

x z

A

Satu atedsoilw

xhz

γ

h′h


APressure diagram

wz h′γ − γwhγ

x ′γWhen submerged soil pressure balances seepage pressure ,ixγγWhen submerged soil pressure balances seepage pressure ,wixγ

0′σ =′γ

wx ix′γ = γ cw

i iγ⇒ = =

γCritical hydraulic gradient

When , cohesionless soils loose their shear strength, and a visible it ti (b ili lik h ) h > i k d

ci i=

1G

agitation (boiling-like phenomenon) happens => quick sand.

( )1Gγ 11cGie−

∴ =+

( )11

w Ge

γ −′γ =

+We have,

• To prevent quick sand during excavations below water table, pumping can be done before excavation, to lower water table.


• Quick sand doesn’t occur in clays, since cohesive forces prevent boiling.

• Problem 1: Water table is located at ad h f 3 b l d f idepth of 3m below ground surface in adeposit of sand of 11m thickness. Acapillary fringe is present above watert bl t d l l S t t d it

Water table3 m

Sa

table, upto ground level. Saturated unitweight of sand is 20kN/m3. Calculatethe effective pressure at depths 0, 3, 7 and11m from ground surface

4 m

and depo

11m from ground surface.4 m

osit

Depth Total pressure Porewater pressure

Effective pressure

0 0 –3x9.81= – 29.43 29.430 0 3x9.81 29.43 29.433 3x20=60 0 607 7x20=140 4x9.81=39.24 140 – 39.24=100.76


11 11x20=220 8x9.81=78.48 220 – 78.48=141.52

29.43−

1z3m

29.4329.43

z Saturated

603m

7m– = 60

soil

11m

140

220

39.24 100.76

11m

Total pressure di

220


78.48

Effective pressure

141.52

diagram diagram diagram


• Problem 2: For the given soil profile, compute the effective pressure at a depth of 12 mdepth of 12 m.

To find the total pressure:

L 1 0 t 3

n 0 4

Layer 1: 0 to 3 m.

1nen

=−

0.4 0.6671 0.4

= =−

Gγ1

wdGeγ

γ =+

332.65 9.81kN m 15.595kN m

1 0.667×

= =+


( )G e+ γ ( ) 32 65 0 667 9 81kN m+ ×( )1

wsat

G ee

+ γγ =

+( ) 32.65 0.667 9.81kN m

19.52 kN m1 0.667

+ ×= =

+

( )d r sat dSγ = γ + γ − γ ( ) 315.595 0.3 19.52 15.595 16.772 kN m= + × − =

L r 2 3m to 4mLayer 2: 3m to 4m.

wdGγ

γ =3

32.68 9.8kN m 16.432 kN m×= =

( ) 332.68 0.6 9.81kN m

20 111kN+ ×

1d eγ

+ 1 0.6+

( ) 320.111kN m1 0.6satγ = =+

( )S ( ) 316 432 0 8 20 111 16 432 19 375kN( )d r sat dSγ = γ + γ − γ ( ) 316.432 0.8 20.111 16.432 19.375kN m= + × − =

Layer 3: 4m to 6m


Layer 3: 4m to 6m. 320.111kN msatγ =

Layer 4: 6m to 9m.

( ) 332.1 3 9.8kN m

12.508kN m1 3sat

+ ×γ = =

+1 3+

Layer 5: 9m to 12m.

sate w G= 0.35 2.7 0.945= × =

( ) 332.7 0.945 9.81kN m

18.384 kN m1 0.945sat

+ ×γ = =

+

Total pressure at a depth of 12 m 3 16.772 1 19.375 2 20.111= × + × + ×23 12.508 3 18.384 202.59kN m+ × + × =

Effective pressure at a depth of 12 m 2202 59 8 9 8 124 11kN m= × =


Effective pressure at a depth of 12 m 202.59 8 9.8 124.11kN m= − × =

• Problem 3:

Bottom heave, when effective stress at top of gravel layer is p g yzero


( ) wG e+ γ ( )2.7 1.2 9.81+ × 3( )1

wsat e

γγ =

+( )

1 1.2=

+317.39 kN m=For the clay layer,

At the top of gravel strata, 12 17.39 15 9.81′σ = × − × 261.53kN m=

Let the bottom of excavation be h m above the top of gravel layer, so that the effective stress at top of gravel layer is zero.

0u′σ = σ − =

p g y

17.39 15 9.81u h∴σ = ⇒ = ×

8 5 mh∴ = 8.5 mh∴ =

12 8.5 3.5md∴ = − =


Permeability


Definition

Property of porous material that permits passage of water through interconnecting voids

• Material having continuous voids - permeable• Gravels - highly permeable; stiff clays - impermeable• Mostly in soils flow is laminar• Mostly in soils, flow is laminar


Head, gradient and potential, g p

When water flows through soil, the total head consists of:

1. Piezometric head or pressure head

2. Velocity head

3. Position head Negligible, for flow through soil

Rise of water in piezometric tube

3. Position head g g , g

With respect to any datum


A

( )h

A

H

wh

( )w ahB DATUM

H h

az zw

( )w bha

bz

b

b

Point Piezo head

Position head

Total head

b

head head head

a (hw)a za Hb (hw)b zb 0

H: Initial hydraulic head

h: Hydraulic head (potential)at any point h h z= ±


( w)b b

c hw z hw – z i=h/l: hydraulic gradientat any point wh h z= ±

Darcy's lawQuantity of water q flowing through a cross sectional area of soil mass under a hydraulic gradient can be expressed as:

Darcy s law

q kiA= (1)

Hydraulic gradient

ki

Coefficient of permeability Darcy, 1856

Total area of cross section

Hydraulic gradientiA

• Greater k, greater is the flow through soil

• Average discharge velocity

qv ki= = (2)v kiA

= =

Coefficient of permeability is the average discharge velocity under

(2)


p y g g yunit hydraulic gradient

Discharge and Seepage VelocityDischarge and Seepage Velocity

• In eqn. (1), is the Total area of cross sectionA

VOIDS SOLIDSA A A= +VOIDS SOLIDS

VOIDSA A<

Actual velocity sVOIDS

q qvA A

= >

Seepage velocity


VOIDSq vA v A= = s VOIDSq vA v A

1s

VOIDS VOIDS VOIDS

A v A Vv vA n A V n

⎧ ⎫= = = =⎨ ⎬

⎩ ⎭∵

VOIDS VOIDS VOIDS⎩ ⎭

k i (3)s pv k i= (3)Also,

Coefficient of percolation

1p psp

k i kv kkv ki k n n= = = ∴ =


v ki k n n

Coefficient of permeability influences:

• Water retaining capacity and stability of earth dams

• Capacity of pumping installations for the lowering of ground water p y p p g g gtable during excavations

• Rate of settlement of buildingsg

• etc.


Factors affecting permeability

1. Effect of size and shape of particles:

( ) 2 Permeability Grain size∝

k210.k C D=

A constant

k

C

Coefficient of permeability in cm/s

Effective size, in cm

(= 100, if D10 is in cm)C

10D

Allen Hazen 1892Allen Hazen, 1892


2. Effect of property of pore fluid

Coefficient of permeability is directly proportional to unit weight f fl id d i l ti l t it i it

wk γ∝

of fluid and inversely proportional to its viscosity

η

( )2. 0 1wCk eγ= C is a constant( )0.1k e= −

ηC

3. Effect of voids ratio3e

1eke

∝+

Taylor, 1948


4. Effect of structural arrangement of particles & stratificationg p

• Horizontal flow:• Horizontal flow:

is same for all stratai

1 1 2 2 3 3hQ k iZ k iz k iz k iz= = + + +

( )1 1 2 2 3 31

hk k iz k iz k izZ

∴ = + + +


Z

• Vertical flow:

From principle of continuity of flow

For the same area,

1 1 2 2Av A v= =

,

1 2v v= =

i.e., downward velocity is same for all stratav

1 1 2 2 3 3vk hv k z k z k zZ

= = = = =

Total loss of head, 1 2 3

1 1 2 2 3 3

h h h hz i z i z i

= + + += + + +


1 1 2 2 3 3z i z i z i+ + +

Z Z Z

( )vZ Zk v

hhv

= =1 2 3

vZk

z z zk k k

=⎛ ⎞

+ + +⎜ ⎟⎝ ⎠

That is,

1 2 3k k k⎝ ⎠

5. Effect of Degree of saturation: More entrapped air causesg ppless permeability

6. Effect of Adsorbed water: Reduces permeabilityp y


Thus, coefficient of permeability can be expressed as:

32. . .

1w

sek D Ce

γ=

η +(4)

D i th di t f h i l i hi h h th ti f sD is the diameter of spherical grain which has the same ratio of

volume to surface area collectively for the grains in a soil

is a shape constantC


Determination of permeability

Constant head testFalling head test

Laboratory methodsFalling head test

Pumping testsField methods p gBore hole tests

Indirect methodC t ti f i i di t ib tiComputation from grain size distribution


Constant head test:

Suitable for coarse grained soils


Suitable for coarse grained soils

C t t h d t i t f t b t t i il lConstant head permeameter consists of a tube to contain soil sample

Tube can be of any convenient dimension

The head h is kept a constant during the test

S il i t t d b f i th t tSoil is saturated before commencing the test

Test is performed by allowing water to flow through the soil sample

Measure the quantity of discharge Q in time t

Q k iA 1Q LkQq k iAt

= =Qkt h A

∴ =


Falling head test Suitable for fine grained soils

Soil sample is kept in a vertical cylindercylinder

A transparent stand pipe of cross-sectional area a is attached to the sectional area a is attached to the cylinder

Soil is saturated before commencing Soil is saturated before commencing the test

Test is performed by allowing water p y gto flow through the soil sample

Measure the elapsed time t2 – t1 for the Measure the elapsed time t2 t1 for the head to fall from initial h1 to final h2


• Let -dh be the change in head during a small time interval dt.

.dQ kiA dt=From Darcy’s Law,

khkiA dt dh Adt dh

.dQ a dh= −Also,

hi =∵. . .kiA dt a dh Adt a dhL

= − ⇒ = −

Integrating both sides,

iL

=∵

2 2t hAk dhdtL h

= −∫ ∫

g g ,

1 1t haL h∫ ∫

12.3 li aL hk1laL hk( )

110

2 1 2

lo. gi.e , kA t t h

=−( )

1

2 1 2

logekA t t h

=−


Constant head test was used to find coefficient of permeability of asand sample Diameter of sample = 10 cm length of sample = 20 cm

Problem 1:sand sample. Diameter of sample = 10 cm, length of sample = 20 cm,head of water = 35 cm, 110 cm3 of water was collected in 1 min 20 s,determine coefficient of permeability.

1Q Lkt h A

=Coefficient of permeability

3110Discharge, cmQ = 1min,20 80Time, s st = =g

20Length of sample, cmL = 35Constant water head, cmh =

1 110 20 1Q Lk 0 01009 8 648 d280 35 10

4

Qkt h A

∴ = = × × =⎛ ⎞π ×⎜ ⎟⎝ ⎠

0.01009 8.648cm s m day= =


Falling head test was used to find coefficient of permeability.Diameter of sample = 6 cm diameter of stand pipe = 2 cm initial

Problem 2:Diameter of sample = 6 cm, diameter of stand pipe = 2 cm, initialhead = 45 cm, final head after 2 min = 30 cm. Determine coefficientof permeability. 15Length of sample, cmL =

( )1

102 1 2

2.3 logaL hkA t t h

=−

Coefficient of permeability

222

4Area of stand pipe, cma π×

=

( )

42

264

Area of sample, cmA π×=

4

2 1 2 mint t− =1 245 , 30 cm cmh h= =

12.3 laL hk i


( )1

102 1 2

logkA t t h

∴ = =−

cm min 5.5 ?m day=

Problem 3:

4 cm has k3=7×10-4 cm/s. Assume that the flow is taking place perpendicular to the layers.


Zk =20

=1 2 3

1 2 3

vk z z zk k k

=⎛ ⎞

+ + +⎜ ⎟⎝ ⎠

4 4 48 8 4

2 10 5 10 7 10− − −

=⎛ ⎞+ + +⎜ ⎟× × ×⎝ ⎠

43.24 10 cm s−= ×

( )1

102.3 logaL hkA t t h

=−

Coefficient of permeability( )2 1 2A t t h

( ) 12 1 10

2.3 logaL ht tkA h

− =( ) 022

104 2

2.3 2cm 20cm 25log3 24 10 cm s 24cm 12

kA h

−

× ×=

× ×3.24 10 cm s 24cm 12× ×

3775.37s=


=1hr, 2min, 55.37s

• Assignment!

Explain field tests for determination of permeability (pumping Explain field tests for determination of permeability (pumping tests).


Shear Strength


Definition

• Ability to resist sliding along an internal surface within a soil mass: A very important property that determines the strength of a soil e y po ta t p ope ty t at dete es t e st e gt o a so mass

• Stability of foundations, slopes, embankments etc. depend on shear strength

Shear strength consists of:g

– Frictional resistance– Cohesion


50

Mohr-Coulomb failure theoryy• Material fails essentially by shear

• Shearing stress on the failure plane is a unique function of the normal stress acting on that plane

( )F

i i

( )s F σ=

51Friction

tanF N= φ

tanF NA A= φ

F NF N∝= μtans =σ φ (1)


tans =σ φ for purely granular soils(1)

• If cohesion also present,

for purely granular soils(1)

tans c= +σ φ (2) Coulomb Equation

experimentally determined i i l t

c ⎫⎬

E (2) th t i i d d t fc σ

empirical parametersφ ⎬⎭

At zero σ s c=

Eqn. (2) assumes that is independent ofc σ

At zero ,σ s c

That is, cohesion is the shear resistance at zero l


52

normal pressure

Plot between s and σ at failure (Coulomb envelope)


53

s s

φc

σφ

σCoulomb envelope for ideal (pure) friction material

Coulomb envelope for purely cohesive material


54

Mohr’s stress circle


55

σyτxyτxy

τyxσn

σxσx

τyx

τn

α

σ

τxy

yx

y xσ σ>σy

cos2 sin 22 2

y x y xn xy

σ σ σ σσ α τ α

− +⎛ ⎞= + +⎜ ⎟

⎝ ⎠⎝ ⎠

sin 2 cos22

y xn xy

σ στ α τ α

−= −


56

2n xy

20 sin 2 cos2 0

2y x

n xy

σ στ α τ α

−= ⇒ − =

2tan 2 xy

y x

τα

σ σ=

−

22

1 3 2 2y x y x

xy

σ σ σ σσ τ

+ −⎛ ⎞= ± +⎜ ⎟

⎝ ⎠

Principal stresses1,3 2 2 xy⎜ ⎟

⎝ ⎠stresses

2 22 2y x y xσ σ σ σ

σ τ τ+ −⎛ ⎞ ⎛ ⎞

+ +⎜ ⎟ ⎜ ⎟2 2y y

n n xyσ τ τ− + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

,02

y xσ σ+⎛ ⎞⎜ ⎟⎝ ⎠

22

2y x

xy

σ στ

−⎛ ⎞+⎜ ⎟

⎝ ⎠


57

⎝ ⎠ 2⎝ ⎠

( ),σ τ

22y xσ σ−⎛ ⎞

( ),y xyσ τ

τxyτ

2

2y x

xyτ⎛ ⎞+⎜ ⎟

⎝ ⎠

σyσx σσ σσ σ+

θ

τ

σσ3 σ1

2y xσ σ+

τxy

( )( ),x xyσ τ−1 2

tan 2xy

y x

τθ α

σ σ−= =

−


( )

σ σ

( ),σ τ

τ

1 3

2σ σ−

1 3 sin 22

σ σ α−

σσ3 σ1

2α

2

σσ3 σ1

1 3

2σ σ+ 1 3 cos2

2σ σ α−

σ1

σ3σ33

σ


59

σ1

1 3 1 3σ σ σ σ+ −1 3 1 3

1 3

cos22 2

sin 2

σ σ σ σσ α

σ στ α

+= +

−=

Stresses on any plane in terms of principal stresses

01 3−σ σ

sin 22

τ α=

01 3max , 45

2when= =

σ στ α

1 3

2on this pla e., n+

=σ σσ


60

τ

Coulomb envelope

sφ

failureτ

φ

c

σ 1σ3σ

c


61

φ′ ′ ′ Shearing strength based on effective stress( )

tantan

s cs c u

σ φσ φ

′ ′ ′= +

′ ′= + −(3)

tanu us c σ φ= + Shearing strength based on total stress(4)

uc Apparent cohesion based on total stress

Apparent angle of shearing resistance based on total stress

uφu Apparent cohesion based on total stress


62

′ ′ ′ ′1 3 1 3 cos2

2 2σ σ σ σσ α′ ′ ′ ′+ −′ = +

′ ′

Stresses on any plane in terms of principal stresses

(5)

1 3 sin 22

σ στ α′ ′−

=

p p

tans c σ φ′ ′ ′+ (3)

(6)

s is the shear strength

tans c σ φ= + (3)

τ is the shear stress on a planeg

s τ− is a minimum on the plane of failure

p

( ) 0d sd−

=τ

αdα

1 3 1 3 1 3cos2 tan sin 22 2 2

s c⎛ ⎞ ⎛ ⎞′ ′ ′ ′ ′ ′+ − −′ ′− = + + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟σ σ σ σ σ στ α φ α


63

2 2 2⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

φ

( )d s ⎛ ⎞ ⎛ ⎞′ ′ ′ ′− − −⎜ ⎟ ⎜ ⎟

τ σ σ σ σ( ) 1 3 1 30 tan 2sin 2 2cos2 02 2

d sd

⎛ ⎞ ⎛ ⎞′= ⇒ − − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

τ σ σ σ σφ α αα

( ) ( )( ) ( )1 3 1 3tan sin 2 cos2 0′ ′ ′ ′′⇒ − − − − =σ σ φ α σ σ α

tan sin 2 cos2′⇒ − =φ α α

t t 2 2 9 450′ ′+′

′+φφφtan cot 2 2 9 4502 f′ ′⇒ − = ⇒ = + ⇒ = ′=+φαα α φ αφ


Failure stress and maximum stress•Failure plane doesn’t carry maximum shear stress•Plane that has maximum shear stress is not the failure plane

Coulomb envelope

failureτmaxτ

sφ

c

σ 1σ3σ


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Measurement of shear strengthMeasurement of shear strength

• Direct shear test• Triaxial compression test• Unconfined compression test• Vane shear test• Vane shear test

Based on drainage conditions,

• Undrained test – no drainage allowed – no dissipation of pore pressure during the test

• Consolidated undrained test – drainage allowed only with initial normal stress

• Drained test - drainage allowed throughout the testDrained test - drainage allowed throughout the test


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Disadvantages of direct shear test

• Shear and normal stress distribution along failure surface is not if ti t th f il i t bili d if luniform entire strength of soil is not mobilised uniformly

• Plane of shear failure is pre-determined it may not be the weakest plane

• Effect of lateral restraint by the side walls of the shear boxy

• No control over drainage


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Direct shear test

N

F


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• Strain controlled test

• At failure, shear force F corresponding to the normal force N is noted

• Repeated for a number of identical samples

• From the set of values of σ and τ at failure, envelope is drawnFrom the set of values of σ and τ at failure, envelope is drawn

• Slope of the envelope represents angle of shearing resistance

τφφ

c


69σ

Triaxial compression test

Casagrande & Terzaghi – 1936


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σ1- σ3 Deviator stress

σ1

σ3

σ3σ3σ3 σ3 Cell pressure

σ3

σ1

σ1- σ3


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• The cylindrical soil specimen is subjected to an all round pressure (cell pressure,σ3) initially and then to a vertical pressure, σ1(cell pressure,σ3) initially and then to a vertical pressure, σ1.

• (σ1- σ3) is known as deviatoric stress

• During the test the deviatoric stress and vertical deformation of the sample are measured till failure

• Deviatoric stress and pore pressure corresponding to failure are noted

• Failure: at max value of stress or 20% axial strainFailure: at max value of stress or 20% axial strain

• A number of Mohr’s circles can be drawn from different sets of observations (σ1 and σ3)from which the failure envelope can be observations (σ1 and σ3)from which the failure envelope can be determined


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Coulomb envelope

ssφ

c

σ


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Stress state in soil specimen during triaxial compression

tanc′ ′ ′= +τ σ φ

τ

tanf c= +τ σ φ

FF

c’′φ 45

2fφα′

′ = +C A sin FC

KCφ′ =

σ′ 1′σ

3′σK O

KC


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But FC = Radius of Mohr’s circle M i h t ( )1 ′ ′= Maximum shear stress ( )1 3

12σ σ′ ′= −

( )1φ ′ ′′ ′

( )1 31

FC σ σ′ ′−

Also, KC=KO+OC ( )1 31cot2

c φ σ σ′ ′′ ′= + +

( )( )

1 3

1 3

2sin 1cot2

FCKC c

σ σφ

φ σ σ′∴ = =

′ ′′ ′ + +

21 3 tan 45 2 tan 45

2 2c

′ ′⎛ ⎞ ⎛ ⎞′ ′ ′= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

φ φσ σSimplifying,

21 3 tan 2 tanf fcσ σ α α′ ′ ′ ′ ′= +

2 2⎝ ⎠ ⎝ ⎠

1 3 2N c Nφ φσ σ′ ′ ′ ′ ′= + 2 2 0tan tan 45where, fNφφα′⎛ ⎞′ ′= = +⎜ ⎟

⎝ ⎠


1 3 φ φ 2, fφ ⎜ ⎟

⎝ ⎠

Stress state during triaxial compression …g p

I t f t t l I t f ff ti In terms of total stresses

In terms of effective stresses

21 3 tan 2 tanf fcσ σ α α′ ′ ′ ′ ′= +

21 3 tan 2 tanf u fcσ σ α α= +

2N N1 3 2N c Nφ φσ σ′ ′ ′ ′ ′= +

2 2 0tan tan 452fNφφα′⎛ ⎞′ ′= = +⎜ ⎟

⎝ ⎠

1 3 2 uN c Nφ φσ σ= +

2 2 0tan tan 452u

fNφφα ⎛ ⎞= = +⎜ ⎟

⎝ ⎠

(7)

2fφ ⎜ ⎟⎝ ⎠

(8)

2fφ ⎜ ⎟⎝ ⎠


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Advantages of triaxial test

Sh h ll h 3 d i di i b d i h • Shear strengths at all the 3 drainage conditions can be done with complete control

• Stress distribution on the failure plane is uniform

• State of stress at any stage of test is determinabley g


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Unconfined compression test

3 0σ =• A special case of triaxial test in which

11 3 2 tan 452

2 uuN c N cφ φ σσ φσ ⎛ ⎞=′ ′ ′ ′ ′= + ⇒ +⎜ ⎟

⎝ ⎠Hence,

Only one value of σ1 . Therefore, only one Mohr’s circle

So, this test can be applied only for saturated clays ( )0uφ =

, for saturated clays10 2u ucφ σ= ⇒ =

1

2 2u

uqc σ

∴ = = qu Unconfined compressive strength at failure


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sCoulomb envelope

failureτcu

1

2 2

failu e

uq σ= =

σ 1σ3 0σ =2 2


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Problem 1. An unconfined cylindrical specimen of clay fails under an axialf 240 kN/ 2 Th f il l i li d l f 550 hstress of 240 kN/m2. The failure plane was inclined at an angle of 550 to the

horizontal. Find the shear strength parameters c and ϕ for the soil.

2tan 2 tancσ σ α α= +1 3 tan 2 tanf u fcσ σ α α= +

3 0σ = 1 2 tanu fcσ α∴ =

0452fφα = + ( )02 45fφ α∴ = −

020=2 ( )f

11 2 tanu f uc c σσ α= ⇒ = 240

=284 kN m=

1 2 tanu f ufα 2 tan55


Problem 2. Two identical specimens of dry sand are tested in the triaxialapparatus with confining pressure of 150 N/mm2 and 250 N/mm2 respectivelyapparatus with confining pressure of 150 N/mm2 and 250 N/mm2 respectively.If the angle of friction for sand is 350 what are the values of axial stresses atfailure of the specimens?

21 3 tan 2 tanf u fcσ σ α α= +

For cohesionless dry sand, 0uc =

2 2 01 3 3tan tan 45

2fφσ σ α σ ⎛ ⎞∴ = = +⎜ ⎟

⎝ ⎠0

2 01

35150 tan 452

σ⎛ ⎞

= +⎜ ⎟⎝ ⎠

(1)2⎝ ⎠

2553.53 N mm=


02 0 35⎛ ⎞2 0

135250 tan 452

σ⎛ ⎞

= +⎜ ⎟⎝ ⎠

(2)

2922.543 N mm=


Problem 3. Two identical specimens of a soil were tested in the triaxialFi i f il d d i f 770 kN/ 2 h h llapparatus. First specimen failed at a deviator stress of 770 kN/m2 when the cell

pressure was 200 kN/m2. Second specimen failed at a deviator stress of 1370kN/m2 when the cell pressure was 400 kN/m2. Find c and ϕ for the soil.

21 3 tan 45 2 tan 45

2 2cφ φσ σ ⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠We have,

First specimen2

1 3 770 kN mσ σ− = 23 200 kN mσ =

21 970 kN mσ∴ =

2970 200 tan 45 2 tan 452 2

cφ φ⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(1)2 2⎝ ⎠ ⎝ ⎠

Second specimen21370 kN 2400 kN

21770 kN mσ∴ =21 3 1370 kN mσ σ− = 2

3 400 kN mσ = 1 1770 kN mσ∴ =

21770 400 tan 45 2 tan 45cφ φ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ (2)


1770 400 tan 45 2 tan 452 2

c= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(2)

From (1) and (2)From (1) and (2),

242.5 kN mc = 036.87φ =

Problem 4. The same soil as in the previous problem if tested in a direct shearapparatus estimate the shear stress at which sample will fail under a normalapparatus, estimate the shear stress at which sample will fail under a normalstress of 600 kN/m2.

tanf cτ σ φ= +

42 5 600 tan 36 87τ = +2492 5 kN m=42.5 600 tan 36.87fτ = + 492.5 kN m


Assignment: Vane shear testAssignment: Vane shear test


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SummarySummary

Soil water Occurrence - adsorbed and capillary water types - effective stress - totalstress - pore pressure - pressure diagrams

Soil water

PermeabilityDefinition - Darcy's law - factors affecting permeability - laboratoryDefinition Darcy s law factors affecting permeability laboratorydetermination - stratified soils - average permeability. Seepage –downward and upward flow -quick sand

Definition - Mohr-Coulomb strength theory - Measurement of shearstrength Types of Triaxial compression tests measurement of pore

Shear Strength

strength – Types of Triaxial compression tests - measurement of porepressure - total and effective stress – Unconfined Compression test - vaneshear tests –Direct shear test- strength parameters - choice of testconditions for field problems


86

conditions for field problems.