Ge i-module4-rajesh sir

Geotechnical Engineering Geotechnical Engineering IIGeotechnical Engineering Geotechnical Engineering -- II

Stress Distribution,Stress Distribution,Slope StabilitySlope StabilitySlope StabilitySlope Stability

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 IV

Stress in soil:

Module IV

Stress in soil: Boussinesque's and Westergaard's equations for vertical loads-Pressure due to point loads and uniformly distributed loads –

Assumptions and limitations – Assumptions and limitations

Pressure bulb –

Newmark charts and their use –

Line loads and strip loads p

Stability of slopes:

Stability of finite slope-Stability of finite slope-Stability of infinite slope-Stability Number-Method of slices- The Swedish circle methodMethod of slices The Swedish circle methodThe friction circle method


Stress in soilStress in soil


Introduction

• When stresses in soil mass are relatively small, the soil may be assumed to behave elastically

• Based on the elastic theory, stresses at various points in the soil mass can be calculated

• Fi di t t i th il i i t t i ti ti th ttl t i • Finding out stresses in the soil is important in estimating the settlement in soil masses

• Many formulae available, most accepted among them are:y p g

• Boussinesq’s formula

• Westergaard’s formula


Boussinesq’s Formula for Point Loadsq

QPOINT LOAD, Q

Q

z

SEMI INFINITE MEDIUM

rP

zσ

Point load acting on surface of semi−infinite mass


Point load acting on surface of semi infinite mass

Assumptions:• Soil mass is elastic, isotropic, homogeneous and semi-infinite• The soil is weightless, compared to the applied loads• The load is a point load acting on the surface p g

Vertical stress at a point P due to a point load Q is:

( )( )5 22 2

3 12 1

zQz r z

σπ

=+

Q

( )( )1 r z+

z(J. Boussinesq, 1885)

rσ

P

2z BQ Iσ =

zσ2z Bz

BI Boussinesq stress coefficient(Boussinesq influence factor)


(Boussinesq influence factor)

Westergaard’s Formula for Point LoadsWestergaard s Formula for Point Loads

Vertical stress at a point P due to a point load Q is:

( )3 22

1z

Qσ =( )( )3 22 21 2

z z r zπ + Q

Westergaard (1938)

z

Westergaard (1938)

Qrσ

P2z W

Q Iz

σ =

zσ

WI Westergaard stress coefficient


Comparison of Boussinesq’s and Westergaard’s Formulae

Comparison of values of IW and IB


• The value of IW is less than IB by 33%, at r/z = 0

Variation of stress with depth under a unit point load


• Boussinesq’s equation assumes isotropic condition; Westergaard’s equation assumes stratified soil condition

H W d’ i i l fi ld di i i di • Hence Westergaard’s equation is closer to field conditions in sedimentary deposits

• However engineers prefer Boussinesq’s equation to compute stresses since


• However, engineers prefer Boussinesq s equation to compute stresses, since settlements computed from these stresses give conservative values

Problem 1. A concentrated load of 200 kN acts at the ground surface. Find thei l l h i f h l d i di l di f 5 dvertical stress along the axis of the load at a point at a radial distance of 5m and

a depth of 10m by a) Boussinesq’s and b) Westergaard’s Formulae. Neglect thedepth of foundation.

( )( )5 22 2 2

3 12 1

z BQ QIz z r z

σπ

= =+

a) Boussinesq’s Formula( )( )1 r z+

3 1I 3 10 2733

( )( )5 222 1BI

r zπ=

+ ( )( )5 222 1 5 10π=

+0.2733=

22kN

kN/m200 0.55

10z BIσ∴ = × =10


1Q QIσ = =) d’ l( )( )3 22 2 21 2

z WIz z r zσ

π= =

+a) Westergaard’s Formula

1 1

( )( )3 22

1

1 2WI

r zπ=

+ ( )( )3 22

1

1 2 r zπ=

+0.1733=

2kNkN/m

200 0 346Iσ∴ = =2 kN/m0.34610z WIσ∴


Line Loads LINE LOAD, q

ye oads q

x

R

SEMI INFINITE MEDIUM Pr

• Boussinesq’s equation provides a basis for finding stress at any point P in an z

elastic semi-infinite mass

• By applying this principle, the stress at any point P due to a line load of infinite extent in an elastic semi-infinite mass can be calculated

• State of stress - plain strain condition


• Strain in y-direction parallel to the line load is assumed to be zero

Consider an elementary length dy along the line load of intensity q/unit length.y g y g y q/ g

The load q.dy can be taken as a concentrated (point) load

V ti l t t i t P d t thi i t l d iΔ

( )

Vertical stress at a point P due to this point load is:zσΔ

( )( )( )5 22 2

3 12 1

z

qdyz r z

σπ

Δ =+ ( )( )

2 2 2r x y= +

( )( ) 5 22 2 2

3 12z

qdyz x y

σπ

Δ =⎛ ⎞+

z

( )21

x yz

⎛ ⎞++⎜ ⎟

⎜ ⎟⎝ ⎠

r


2 2 2 2 2 2R r z x y z= + = + +

( ) 3

5 2

3 qdy zσΔ =( )( )5 22 2 22z

z x yσ

πΔ

+ +

( ) 33 d+∞ +∞ 3 d∞( )( )( )

3

5 22 2 2

32z z

qdy z

z x yσ σ

π

+∞ +∞

−∞ −∞

= Δ =+ +

∫ ∫ ( )3

5 22 2 20

322

z qdy

z x yπ

∞

=+ +

∫( )

x and z are constants for a given point P. Hence, y is the only variable.

( )22

2 1z

qσ∴ = Vertical stress at a point P( )( )21z x zπ + due to a line load of intensity q/unit length

z Zq Iz

σ =ZI Influence factor

0 637 / 0


z =0.637, at x/z = 0

Strip Loadsp

Required to find the vertical stress at a point P on a vertical plane passing


Required to find the vertical stress at a point P on a vertical plane passing through the centre of the strip.

Vertical stress at a point P due to an elementary line load of intensity qdx is:

( )2

2 1qdxσΔ =

22 Bq dxσ ∫I t ti th idth f l d

( )( )221z z x z

σπ

Δ+

( )( )222 1

zBz x z

σπ −

=+

∫Integrating over the width of load,

2B

( )( )2

220

4

1

Bq dxz x zπ

=+

∫

tanxz

β=

( )Let 2secdx z β∴ =

( )2 22

222

4 sec 4 coszq qd d

θ θβσ β β β= =∫ ∫ ( )sinq θ θ+=


( )220 01 tan

z π πβ+∫ ∫ ( )

π

Uniform load on a circular areaU o oad o a c cu a a ea

UNIFORM LOAD ON CIRCULAR AREA, q/UNIT AREA

SEMI INFINITE MEDIUM


Required to find the stress at a point P under the load, on the vertical axis through the centroid of the loaded area


through the centroid of the loaded area.

Vertical stress at a point P due to a point load of intensity qdA is:

( )

( )5 22 2

3 12z

qdAσΔ =

( )( )

3

5 22 2

32qdA zπ

=( )( )5 22 22 1

z z r zπ + ( )2 22 r zπ +

h f d

( ) ( )3 33 3q z q z∑

Integrating over the ring of radius r,

( )( )

( )( )5 2 5 22 2 2 2

3 3 22 2z

q z q zdA rdrr z r z

σ ππ π

Δ = =+ +

∑

( )3

5 22 23z

zqrdrr z

σΔ =+( )

Integrating the above from r=0 to r=a gives the stress due to the entire load over the circular area


over the circular area

2 2 2 d d2 2 2r z n+ =Let r dr n dn=

0r n z= ⇒ =2 2r a n a z= ⇒ = +

2 2

353

a z

z zndnqzσ σ

+

= Δ =∫ ∫2 2

343

a z dnqz+

= ∫5z zA z

qn∫ ∫

3 21⎧ ⎫⎡ ⎤⎪ ⎪

4z

qn∫

( )211

1q

a z

⎧ ⎫⎡ ⎤⎪ ⎪= − ⎢ ⎥⎨ ⎬+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

Bq I=

I Boussinesq influence factor BI Boussinesq influence factor

for circular load

• The above formula can be used for finding stresses under ring loads also


• The above formula can be used for finding stresses under ring loads also.

Problem 2. A ring footing of external diameter 8m and internal diameter 4md h f 2 b l h d f I i l d f i irests at a depth of 2m below the ground surface. It carries a load of intensity

150 kN/m2. Find the vertical stress at a depths of 2m, 4m and 8m along theaxis of the footing below the footing base.

( )

3 2

211z qσ

⎧ ⎫⎡ ⎤⎪ ⎪= − ⎢ ⎥⎨ ⎬For circular load, Bq I=( )21z qa z

⎢ ⎥⎨ ⎬+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

B

For a ring load, ( )z zExt zInt BExt BIntq I Iσ σ σ= − = −

3 211I

⎡ ⎤⎢ ⎥

For z = 2m (below foundation) and a = 4m

( )211 4 2BExtI = − ⎢ ⎥+⎢ ⎥⎣ ⎦

(external radius),

0 911


0.911=

3 211I

⎡ ⎤⎢ ⎥F z 2 d a 2 (i t l di ) 0 646

( )211 2 2BIntI = − ⎢ ⎥+⎢ ⎥⎣ ⎦

For z = 2m and a = 2m (internal radius), 0.646=

Hence the stress at a depth of 2m below the load (foundation) along its axis,

( )I I

2kN/m39 7=

( )z BExt BIntq I Iσ = −

( )150 0 911 0 646σ∴ × kN/m39.7=( )150 0.911 0.646zσ∴ = × −

Similarly the stresses at depths of 4m and 8m can be determined as2 2kN/m and kN/m respectively.54 30 ,


Pressure distribution diagramsPressure distribution diagrams

( )5 22 2

3 12z

Qσ =

• Stress varies along depth as well as horizontal distance

( )( )2 22 1z r zπ +

• Vertical pressure on a given horizontal plane is same in all directions at equal radial distance from the axis of loading

• A contour connecting all points below the ground surface having equal vertical pressure: Isobar

• Isobar is a spatial surface in the form of a bulb• Isobar is a spatial surface in the form of a bulb

• The zone in a loaded soil mass bounded by an isobar of a given vertical pressure intensity is called a pressure bulb

• Vertical pressure at every point on the surface of pressure bulb is same



Isobars under a uniformly distributed circular area

Significant depth• Depth of pressure bulb corresponding to 20% of foundation contact pressure

• i e depth of the pressure isobar that represents a pressure of 0 2q where q is

g p

• i.e., depth of the pressure isobar that represents a pressure of 0.2q, where q is the vertical pressure applied

• Significant depth is approximately equal to 1.5 times the width of


g p pp y qsquare or circular footings

thca

nt d

ept

Sign

ific

• Terzaghi suggested (1936) that this depth is significantly responsible g gg ( ) p g y pfor the settlement of structures


h l l d d f l d l l b• When several loaded footings are placed closely, isobars merge into one large isobar, with a large significant depth


Problem 3. A single concentrated load 1000 kN acts at the ground surface.C i b f i l f i i 40 kN/ 2 UConstruct an isobar for a vertical pressure of intensity 40 kN/m2. UseBoussisesq formula.

3 1Q

Boussinesq’s Formula

( )( )5 22 2

3 12 1

zQz r z

σπ

=+ 1mz =? mr =

3 1000 140 × 2mz =

( )( )5 22 240

2 1z r zπ=

+3mz =

2 5

2

3 1000 1r z ×⎛ ⎞= −⎜ ⎟⎝ ⎠

3mz =

3.45mz =


22 40zπ⎜ ⎟×⎝ ⎠

Contact pressureContact pressure• Pressure at the contact surface between base of foundation and the

d l lunderlying soil mass

• Contact pressure distribution depends on:

• Flexural rigidity of footing

• Elastic properties of the sub grade soil

• If the footing is flexible, contact pressure is uniform, irrespective of the type of soil

• If the footing is perfectly rigid pressure distribution depends on whether the • If the footing is perfectly rigid, pressure distribution depends on whether the soil is cohesive or cohesionless



Vertical stress beneath loaded areas of irregular shape

Newmark’s Influence ChartsNewmark (1942)

• To determine the vertical stress at any point under a uniformly loaded area of any shapeof any shape

• A chart consists of a number of concentric circles and radial lines

E h it l ti l t t th t f th di• Each area unit causes equal vertical stress at the centre of the diagram


5

Newmark’s Chart


Newmark’s Chart

Plan of a building foundation(drawn on tracing paper)(drawn on tracing paper)


Construction of Newmark’s Influence Chart

• Let a uniformly loaded circular area of radius r1 cm be divided into 20 sectors 20 sectors

• Let q is the load intensity, and σzis the vertical stress at a depth z

Op

below the centre of the area

• Each sector OA1B1 exerts a pressure of σz/20 at a depth zbelow the centre of the area

Construction of


Newmark’s Chart

Hence, from the equation for stress below a circular load,

( )

3 2

211

20 20 1z

fq i q

r zσ ⎧ ⎫⎡ ⎤⎪ ⎪= − =⎢ ⎥⎨ ⎬

+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭( )120 20 1 r z+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

3 21 1⎧ ⎫⎡ ⎤⎪ ⎪

( )21

1 1120 1fi

r z

⎡ ⎤⎪ ⎪= − ⎢ ⎥⎨ ⎬+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

Influence value

• Let us fix an arbitrary value of 0.005 for if

⎧ ⎫

( )

3 2

21

10.005 120 1qq

r z

⎧ ⎫⎡ ⎤⎪ ⎪∴ = − ⎢ ⎥⎨ ⎬+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭( )1⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

• For z = 5 cm (say), 1 1.35 cmr =


1

• Hence, if a circle is drawn with radius 1.35 cm, divided into 20 equal sectors, h t ill t f 0 005 t d th f 5 each sector will exert a pressure of 0.005 q at a depth of 5 cm.

• Let the radius of a second circle be r2 cm (to be determined).

• Let the annular space between two circles be divided into 20 equal parts (area units), like A1A2B2B1.

h t b h th t th ti l t th t d t h f th • r2 has to be such that the vertical pressure at the centre due to each of these area units is 0.005 q.

• The total pressure due to area units OA B and A A B B is 2x0 005 q• The total pressure due to area units OA1B1 and A1A2B2B1 is 2x0.005 q

• Therefore, vertical pressure due to OA2B2 is:

3 212 0 005 1qq

⎧ ⎫⎡ ⎤⎪ ⎪× = ⎢ ⎥⎨ ⎬ forcm 5 cm2r⇒( )2

2

2 0.005 120 1

qr z

× = − ⎢ ⎥⎨ ⎬+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

2 for cm, 5 cm2 zr =⇒ =


• Similarly a 3rd, 4th, 5th etc. circles can also be drawn

• For the 10th circle,

3 211 10 0 005q qq

⎧ ⎫⎡ ⎤⎪ ⎪− = × =⎢ ⎥⎨ ⎬ r⇒ = ∞( )2

10

1 10 0.00520 201

qr z

= × =⎢ ⎥⎨ ⎬+⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

10r⇒ = ∞

• Hence the Newmark’s Chart is drawn as in figure.


• In the Chart, each areaunit will exert an equalvertical stress of 0 005 q at a

5

vertical stress of 0.005.q at adepth of 5 cm at the centreof the diagram.

• Hence, N area units of theload will exert a totalpressure of 0.005.q.N at adepth of 5 cm at the centreof the diagram


How to Use Newmark’s Influence Chart

• A plan of the loaded area isdrawn on a tracing paper anddrawn on a tracing paper andplaced on the Chart, in such away that the point below whichstress is required coincides withstress is required coincides withthe centre of the Chart.

5


• The scale of the plan is such that the length AB on the Chart corresponds to the depth at which pressure is required corresponds to the depth at which pressure is required

• i.e., Let the distance AB is 5 cm (depth for which the Chart is drawn)

N if th t i i d t d th f 5 th th f d ti l • Now, if the stress is required at a depth of 5 m, then the foundation plan should be drawn to a scale of 5 cm = 5 m or 1 cm = 1 m

• Similarly, if the stress is required at a depth of 15 m, then the y, q p ,foundation plan should be drawn to a scale of 5 cm = 15 m or 1 cm=3 m

• The total number of area units (including fractions) covered by the plan of the l d d i t d Nloaded area is counted as N

• Hence, the stress exerted by the load at the point is 0.005.q.N

• Thus, Newmark’s chart can be used to determine the vertical stress at


any point under a uniformly loaded area of any shape

Stability of slopesStability of slopes


• Earth slopes can be:

• Natural slopes: those existing in nature, formed by natural causes

• Man-made slopes: Cuts, embankments, earth dams etc.

• Slopes are classified according to their extent as:

• Infinite slopes: Constant slope of infinite extent. e.g., mountains

• Finite slopes: Slopes of limited extent. e.g., cuts, embankments etc.

• Result of failure of a natural or man-made slope can be catastrophic

• Hence a study of slope stability is very importanty p y y p


• Causes of failure of slopes:

• Gravitational forceGravitational force

• Force due to seepage water

• Surface erosion due to flowing water• Surface erosion due to flowing water

• Sudden lowering of water adjacent to a slope

• Earthquake• Earthquake

•


Stability analysis of infinite slopes• Consider an infinite slope AB, inclined at angle i to the horizontal

• Failure of slope involves a sliding of soil mass along a plane parallel to

y y p

• Failure of slope involves a sliding of soil mass along a plane parallel to surface, at some depth

• Let CD is such a failure plane (critical surface) at depth z Bp ( ) p

• Consider a portion of soil mass PQRS of width b along inclined surface

i

P

Q

g

• Weight of this portion is

b

cosb iA

P

. . .cosW z b iγ=

• Vertical stress on CD is:zσz

D

. .coszW z ib

σ γ= = σR

S

i


τCi

• Components of vertical stress parallel to CD and perpendicular to CD are:

2cos cosz i z iσ σ γ= = sin cos .sinz i z i iτ σ γ= =

τ• The tangential component parallel to failure surface CD is the shear stress causing failure

fτ• This shear stress is resisted by the shear strength of the soil

τ fFττ

=• Hence factor of safety against sliding is:

• Shear strength includes cohesion and internal friction

tanf cτ σ φ= +

• Two cases are considered here: cohesionless soil and cohesive soil


Case 1: Cohesionless soilCase 1: Cohesionless soil

• OA is Coulomb’s failure envelope for cohesionless soil defined by the equation

tanfτ σ φ=equation


Failure condition for an infinite slope of cohesionless soil

• OB represents stresses on CD for varying depths• OB represents stresses on CD for varying depths2cos cosz i z iσ σ γ= =

sin cos .sinz i z i iτ σ γ= =sin tancos

z i ii

τ σσ σ= = A constant, for a given slope icosz iσ σ

tan iτ σ= Equation of curve OB

• For a given value of normal stress , failure happens when fτ τ>σ

• i.e., failure happens when i φ>

F t f f t i t lidi i i btan tanfF

τ σ φ φ• Factor of safety against sliding is given by:

tan tanfF

i iφ φ

τ σ= = =

• If the slope is less than , the slope is stableφ


• If the slope is larger than , the slope is unstable, for any depthφ

Case 2: Cohesive soilCase 2: Cohesive soil

• DA is Coulomb’s failure envelope for cohesive soil defined by the equation

tanf cτ σ φ= +

• If the slope is less than , the slope is stableφ

Failure condition for an infinite


slope of cohesive soil

• If the line OF representing stress on the failure plane meets the failure envelope at F

i φ>failure envelope at F

• At F, shear stress is equal to shear strength. i.e., failure happens at F.

• FF2 represents shear stress at some depth z

• C1C2 represents shear stress at some depth less than z. At this

• Hence, for any depth less than z, shear stress is less than shear strength and the slope is stable e g at C

depth, shear strength is CC2

strength, and the slope is stable. e.g., at C1

Failure condition for an infinite


slope of cohesive soil

• Hence if , slope is stable only for depths less than z, known as i φ>critical depth Hc.

• Factor of safety for any depth z less than H can be given as: • Factor of safety for any depth z less than Hc can be given as:

tanf cFτ σ φ+

= =τ τ

2cos cosi z iσ σ γ= =We have cos cosz i z iσ σ γ

sin cos .sinz i z i iτ σ γ= =

We have,

2cos tancos sin

f c z iFz i i

τ γ φτ γ

+∴ = =

cos .sinz i iτ γ


• At the critical depth H 1Fτ τ= ⇒ =• At the critical depth Hc , 1f Fτ τ= ⇒ =

2cos tani 1cc H iγ φ+ cH⇒ =i 1cos .sin

.e., c

cH i iγ φγ

= ( ) 2tan tan coscHi iγ φ

⇒ =−

• Thus, for given values of i and ϕ, Hc is proportional to cohesion c.

( ) 2tan tan cos nc

c i i SH

φγ

= − = Stability Number, a dimensionless quantity


Stability analysis of finite slopesStab ty a a ys s o te s opes

• Most commonly, failure occurs along curved surfaces

• Analysis based on the assumption that the curved surface is circular in shape• Analysis based on the assumption that the curved surface is circular in shape

• Some methods are:

• Swedish circle method (Method of Slices or Slip circle method)• Swedish circle method (Method of Slices or Slip circle method)

• Modified method of slices

Si lifi d Bi h ’ h d f li• Simplified Bishop’s method of slices

• Friction circle method

• Taylor’s stability numbers


• Types of failure of a finite slope:

• Slope failure: Slide surface intersects slope at or above toe; slope angle is high; soil close to the toe has a high strength

• Face failure: slide surface intersects slope above toe

• Toe failure: slide surface intersects at toe

• Base failure: Slide surface passes below the toe; Soil beneath the base is softer; slope angle is low

Face failure

Toe failurefailure failure


Base failure

The Swedish circle method

• Analysis of purely cohesive soil (ϕ=0 analysis)

Swedish Geotechnical Commission and W. Fellenius

• Analysis of soil possessing both cohesion and friction (c-ϕ analysis)

ϕ=0 analysis• Assume a number of slip circles and find the factor of safety of Assume a number of slip circles, and find the factor of safety of

each

• The slip circle with lowest factor of safety is taken as the critical slip circle


rLet AD be a trial slip circle (radius r)p ( )

r

Driving moment w.r.t O is resisted by the resisting moment

Driving moment .DM W x=

Weight of soil of the wedge ABDAWPerpendicular distance of centre of gravity of the wedge x


from the centre of rotation O

Resisting moment . .R uM c L r=g R u

Unit cohesionuc

C d l th f th li L2 rL π δ

Curved length of the slip arcL 360L=

. ..

R u

D

M c L rFM W x

= =Hence, factor of safety

u

m

cFc

=Alternatively,

where ..m

W xcL r

=

The shear resistance mobilised


Can be obtained by dividing the wedge into several slicesx

c-ϕ analysis• As in ϕ=0 analysis, assume a number of slip circles, and find the

factor of safety of each

Th li i l ith l t f t f f t i t k th iti l li • The slip circle with lowest factor of safety is taken as the critical slip circle

• Soil above the slip arc divided into several slices

• Weight of each slice, W, has a normal component N and a tangential component T

rg p

• Normal components will not cause driving moment


h lDriving moment on each slice

DM rTΔ = r

Resisting moment on each slice

( )( )tanRM r c L N φΔ = Δ +LΔ

( )tan tanr c cLM L N Nφφ ++ × ×Δ∑ ∑∑( )tan tanR

D

r cr

cLMFM

L NT

NTφφ ++ ×

=×

= =Δ∑

∑∑∑∑Factor of safety

( )tanF

c L N UT

φ′ ′=

+ −∑∑

Factor of safety in terms of effective values of cohesion

and friction


T∑and friction,

where U is the pore (neutral) pressure

Problem X:

( )tanF

c L N UT

φ′ ′+ × −= ∑

∑

Factor of safety w.r.t. shear strength and cohesion (in terms of effective values of

h d f ) T∑cohesion and friction),

( )20 27 900 216 tan18F

× + −1 694


( )450

F = 1.694=

Friction circle method

friction circle

r sinϕ

r

ϕ

• A small circle is drawn with O as centre and r sinϕ as radius. This circle is called the friction circle or ϕ-circle


called the friction circle or ϕ circle

friction circle

Fr sinϕ

D

r

EA

∆R

• Any line EF tangential to this circle cuts the failure arc AD at an obliquity ϕy g q y

• Conversely, any vector representing reaction ∆R (resultant friction) at an obliquity ϕ to an element of the failure arc AD will be tangential to the f i ti i l


friction circle

• Let cm is the mobilised unit cohesion for equilibrium

• Let a is the position of mobilised cohesive force , assumed acting parallel to AD

mc Lr

( ) ( )m mc L a c L r=L L

rLaL

∴ =L

L is length of the chord ADPosition of mobilised cohesive force c L

L

cohesive force mc L

• Vector mc L is drawn parallel to AD, at a distance a from O


• Vector W is the weight of the wedgeg g

mc L• is the vector of mobilised cohesive force

a

W R

c

b mc L

• Vector R is the total frictional resistance, drawn with its line of actionpassing through the intersection of the above 2 vectors, and tangential to thefriction circle

• From the force triangle abc, knowing W, the magnitude of , and thus can be determined

mc Lc


thus , can be determined. mc

• The Factor of safety F with respect to cohesive strength can be The Factor of safety Fs with respect to cohesive strength can be determined as:

scFc

=mc

• A number of slip circles are taken, and factor of safety of each is found p yin the manner described above.

• The slip circle with lowest factor of safety is taken as the critical slip circlecircle


Taylor’s Stability Numberay o s Stab ty Nu be• Taylor proposed the concept of using an abstract number (Taylor’s Stability Number) for finding factor of safety of slopesStability Number) for finding factor of safety of slopes

• He studied a large number of slopes with various slope angles and angles of internal friction…

• …and came up with an expression for Taylor’s Stability Number as:

mn

c

c cSF H Hγ γ

= = , assuming full mobilisation of internal friction

• Knowing slope angle and angle of internal friction, Taylor’s Stability Number can be found from the tables and charts presented by him.

• Knowing Taylor’s Stability Number, factor of safety of slope can be found as: cF


cn

FS Hγ

=

Problem X: An excavation is made with a vertical face in a clay soil which hasc = 50 kN/m2 bulk unit weight = 18 kN/m3 Determine the maximum depthcu = 50 kN/m2, bulk unit weight = 18 kN/m3. Determine the maximum depthof excavation so that the excavation is stable. The following are given for ϕ=0

Slope angle 600 750 900Slope angle 600 750 900

Stability number 0.191 0.219 0.261

For the excavation, slope angle = 900 because it has a vertical face

Hence stability number Sn is 0.261, from table.

250 kN/mc

n

cHSγ

∴ = 3

50 18

kN/mkN/m 0.261

=×

m10.64 =


Problem X: Calculate the factor of safety with respect to cohesion of a clayslope laid 1 in 1 to a height of 8 m if the angle of internal friction ϕ = 200, c = 30p g g ϕkN/m2, bulk unit weight = 19 kN/m3. What will be the critical height of slopein this soil? The following are given for ϕ = 200

Slope angle 300 450

Stability number 0.0625 0.062

For the given slope, slope angle ( )1 01tan 451−= =( )1

Hence Stability Number Sn is 0.062, from table.

F t f f t ith c

n

cFS Hγ

=2

3

30 0.062

kN/mkN/m19 8 m

=× ×

3.18=Factor of safety with

respect to cohesion

ccH

S γ=Critical height

2

3

30 0 062 19

kN/mkN/m

=×

m25.5=


nS γ 0.062 19 kN/m×

SummarySummary

Stress in soil: Stress in soil: Boussinesque's and Westergaard's equations for vertical loads-Pressure due to point loads and uniformly distributed loads –

Line loads and strip loads Line loads and strip loads

Assumptions and limitations –

Pressure bulb –

Newmark charts and their use – Newmark charts and their use

Stability of slopes:

Stability of finite slope-Stability of finite slope-stability of infinite slope-Stability Number-Method of slices- The Swedish circle methodMethod of slices The Swedish circle methodThe friction circle method


69

Success happens when chance meets a prepared mind.
