Lecture 2 - Stresses in soils · (Stresses in soils) ENB371: Geotechnical Engineering 2 Chaminda...

Subsurface Stresses(Stresses in soils)

Subsurface Stresses(Stresses in soils)

ENB371: Geotechnical Engineering 2ENB371: Geotechnical Engineering 2Chaminda GallageChaminda Gallage

Contents…Contents…� Introduction: Subsurface stresses caused by surface

loading

� Subsurface stresses caused by the soil mass (Geostatic Stresses) – ENB272

• Vertical Stresses• Horizontal Stresses• Horizontal Stresses

� Vertical stress increase within a soil mass caused by vertical surface loading (foundation loads, wheel loads, ..etc)

• Infinitely loaded area • Point load (concentrated load)• Circular loaded area• Strip load • Rectangular loaded area

Introduction -1Introduction -1

WWWW

At a point within a soil mass, stresses will be developed as aresult of the soil lying above the point (Geostatic stress) a ndby any structural or other loading imposed onto that soilmass

The subsurface stress at a point isaffected by ground water table if it

Ground level

A A

hUnit weight

= γγγγ

Total stress= σσσσ= γγγγh Total stress= σσσσ= γγγγh+∆σ∆σ∆σ∆σ

affected by ground water table if itextends to an elevation above thepoint

Geostatic stresses at a point in soilGeostatic stresses at a point in soilGeostatic stresses at a point in soilGeostatic stresses at a point in soil

KKoo ≈≈≈≈≈≈≈≈ 1 1 –– sinsin φφφφφφφφKo = Coefficeint of earth

pressure at rest, φφφφ=friction angle

Introduction -2Introduction -2

Soil unit weight = Soil unit weight = γγγγγγγγss

zz σσσσσσσσvv = z * = z * γγγγγγγγss

σσσσσσσσ’’hh = K= K oo * * σσσσσσσσ’’ vv

Ko=σσσσ’h/σσσσ’v

Introduction -3Introduction -3Subsurface stresses in road pavements and airport runwaysare increased by wheel load on the surface

Introduction -4Introduction -4Subsurface stresses in soils are increased byfoundation loads

Burj Dubai (<700m)Dubai, U. A. E

Introduction - 5Introduction - 5Embankments and landfills cause to increasesubsoil stresses

Subsoil stress increase -1Subsoil stress increase -1

It is required to estimate the stress increase in thesoil due to the applied loads on the surface. Theestimated subsoil stress increase is used:

- to estimate settlement of foundation

- to check the bearing capacity of the foundation- to check the bearing capacity of the foundation

The Tower of Pisa in Italy, a widely-recognized foundation failure -settlement

Bearing capacity failure of a raft foundation

Subsoil stress increase -2Subsoil stress increase -2The calculation of subsurface stress increase due t o the following surface loading conditions will be discus sed.

• Infinitely loaded area • Point load (concentrated load)• Circular loaded area

Though the surface loading is caused to increase bo th vertical and horizontal stresses in soils, only the vertical str ess increase is discussed as it is the major stress component for most practi cal design problems

• Circular loaded area• Strip loading • Rectangular loaded area

Infinitely loaded area -1Infinitely loaded area -1The surface loading area is much larger than the depth of apoint where vertical stress increment ( ∆σ)∆σ)∆σ)∆σ) is calculated (e.g.land fills, preloading by soil deposition)

Infinitely loaded area -2Infinitely loaded area -2The vertical subsurface stress increment at any depth belowthe infinitely loaded area is considered to be the same as thesurface stress due to external load (filling material)

Ground Surface

h γ γhq =

σ z

q=∆=∆=∆ σσσ 21

1z

2z

1σ

2σ

1z

2z

11σσ ∆+

22σσ ∆+

Finitely loaded area Finitely loaded area If the surface loading area is finite (point, circular, stri p,rectangular, square), the vertical stress increment in the sub-soil decreases with increase in the depth and the distanceform the surface loading area.

Methods have been developed to estimate the vertical stressincrement in sub-soil considering the shape of the surfaceloading area.

Subsurface stress increment due to a point load -1Subsurface stress increment due to a point load -1

• For homogeneous (properties are the same everywhere ), and isotropic (properties are the same in all direct ion) soils

• Boussinesq (1885) (French mathematician) developed t he following equation to calculate vertical stress inc rement following equation to calculate vertical stress inc rement (∆σ∆σ∆σ∆σ) in soils due to point load on the surface:

2/5

22

1

1

2

3

+=∆

z

rz

Q

πσ 2/5

2)/(1

1

2

3

+=

zrI p π

Influence factor, I p

Then,pI

z

Q2

=∆σ Ip can be obtained numerically or using table where I p is given in terms of r/z

Influence factor, I P, for vertical stress increment due to point load

Influence factor, I P, for vertical stress increment due to point load

Subsurface stress increment due to a point load -2Subsurface stress increment due to a point load -2

Example -1 (1)Example -1 (1)In road pavement design, the standard vehicle axel is define das an axel with two single wheels as shown below.

P

L

For a particular vehicle group, the standard axel load (P) isgiven as 80 kN and the distance between two wheels (L) is1.8 m. What is the vertical stress increment in the sub-gradeat 4 m depth directly under a wheel if this axel is running.(consider wheel load as a point load)

L

Find: ∆σ∆σ∆σ∆σ

∆σ ∆σ ∆σ ∆σ = ∆σ∆σ∆σ∆σΑΑΑΑ + ∆σ∆σ∆σ∆σB

40kN

A

∆σ∆σ∆σ∆σ

x

y

Z=4m

r = 1.8 m

pIz

Q2

=∆σ

13132/52/5

Example -1 (2)Example -1 (2)40kN

B

z 301.0)4/8.1(1

1

2

3

)/(1

1

2

3)(

22=

+=

+=

ππ zrI Bp

Therefore,

kPaIz

QI

z

QBp

BAp

A 945.1301.04

40477.0

4

40)()(

2222=×+×=+=∆σ

477.0)4/0(1

1

2

3

)/(1

1

2

3)(

2/5

2

2/5

2=

+=

+=

ππ zrI Ap

Subsurface stress increment due to circular loaded area -1


2/522

0

12

)(3)(

+

=∆

z

rz

drdrqd

π

θσ

Considering force on the small element (black)And applying Boussinesq equation, the stressIncrease at A caused by this load



By integration of Boussinesq solution over the By integration of Boussinesq solution over the whole circular area, the vertical whole circular area, the vertical stress stress increment at depth z under the centre (at A)increment at depth z under the centre (at A)

cIq

z

Bq 02/32

0

21

11 =

+

−=∆σ The influence factor Ic can be calculated numerically or can be obtained from a table

2/5

22

1

1

2

3

+=∆

z

rz

Qz π

σ



Similar integration can be performed to obtain the vertical stress increase at A’, located a distance r from the centre of the loaded area at a depth z.This table can be used to calculate Ic corresponding to A’.

Subsurface stress increment due to strip loading -1Subsurface stress increment due to strip loading -1

Vertical stress increment due to strip area carrying uniform pressure

Vertical stress increment due to strip area carrying uniform pressure

• Soil is homogeneous, isotropic

αβ• The stress increment at point X ( ∆σ∆σ∆σ∆σz)

due to a uniform pressure q on a zσ∆

due to a uniform pressure q on a strip area of width B and infinite length is given by in terms of αααα and ββββ

{ })2cos(sin βαααπ

σ ++=∆ q

Vertical stress increment due to strip area carrying increasing pressure

Vertical stress increment due to strip area carrying increasing pressure

• Soil is homogeneous, isotropic

• The stress increment at point X ( ∆σ∆σ∆σ∆σz) due to increasing pressure from zero to q on a strip area of width B and

q

B

2R

X

to q on a strip area of width B and infinite length is given by in terms of αααα and β β β β the lengths R1 and R2

−=∆ βα

πσ 2sin

2

1

B

Xqz

z

αβ

zσ∆

X

1R2R

A strip footing 2 m wide carries a uniform pressure of 250 kPa on the surface of a deposit of sand. Water table is at the surface and the saturated unit weight of sand is 20 kN/m 3. Determine the effective vertical stress at a point 3 m below the centre of the footing before and after the appl ication of the surface pressure

A strip footing 2 m wide carries a uniform pressure of 250 kPa on the surface of a deposit of sand. Water table is at the surface and the saturated unit weight of sand is 20 kN/m 3. Determine the effective vertical stress at a point 3 m below the centre of the footing before and after the appl ication of the surface pressure

Example - 2

Before loadingBefore loading

Total vertical stress at 3 m depth, σσσσv

kPazsatv 60320 =×== γσ

Pore water pressure at 3 m depth, u

kPazu w 4.2938.9 =×== γEffective vertical stress at 3 m depth, σσσσ’v

kPauvv 6.304.2960' =−=−= σσ

Strip area carrying uniform pressure -Example

After loading

αβ

B = 2 m, q = 250 kPa, z = 3 m,

At the centre of the strip, β = β = β = β = −−−−α/2α/2α/2α/2Therefore, cos( α+2β)=1α+2β)=1α+2β)=1α+2β)=1

3

1

2tan =

α

zσ∆

{ })2cos(sin βαααπ

σ ++=∆ q

600.0sin

643.0'52363

1tan2 01

=

==

= −

α

α radians

The increase in stress at 3 m depth due to applied load

( ) kPaq

z 0.99)600.0643.0(250

sin =+=+=∆π

ααπ

σ

Hence, total effective vertical stress = 30.6+99.0 = 129.6 kPa

Subsurface stress increment due to rectangularly/ squarely loaded area -1

Subsurface stress increment due to rectangularly/ squarely loaded area -1

Vertical stress increment due to uniformly loaded rectangular area -1


[ ] 2/5222

30

2

)(3)(

zyx

zdydxqd

++=∆

πσ

Considering force on the small element (black)and applying Boussinesq equation, the stressIncrease at A caused by this load

2/5

22

1

1

2

3

+=∆

z

rz

Qz π

σ

By integration of By integration of BoussinesqBoussinesq solution solution over complete area, the vertical stress over complete area, the vertical stress increment at depth z under the centre increment at depth z under the centre (at A)(at A)

[ ] r

L

y

B

x

Iqzyx

zdydxq02/5222

30

0 0 2

)(3 =++

=∆ ∫ ∫= = π

σ

BL ≥

• The stresses increase under a cornercorner of a uniformly loaded a uniformly loaded flexibleflexiblerectangular area:rectangular area:

Influence factor for rectangular footing, Influence factor for rectangular footing, IIrr

rIq0=∆σ



Influence factor for rectangular footing, Influence factor for rectangular footing, IIrr

•• DefineDefine m = B/zm = B/z and and n = L/zn = L/z

•• Solution by numerically or tablesSolution by numerically or tables

IIrr = = 1 1 2mn(m2mn(m 22+n+n22+1)+1)1/21/2 . m. m 22+n+n22+2+2mm 22+n+n22--mm 22nn22+1+144ππππππππ mm 22+n+n22+1+1

+ tan+ tan --112mn(m2mn(m 22+n+n22+1)+1)1/21/2

mm 22+n+n22--mm 22nn22+1+1

Influence factor, Ir, for vertical stress increment due to uniformly loaded rectangular area -1


Rectangular areaRectangular area

The vertical stress increment at a depth z below po int O,

)( 43210 IIIIqz +++=∆σ

Example -3Example -3Determine the increase in stress at A and A’ below a rectangular area

)( 43210 IIIIq +++=∆σAt the centre of rectangular or square area,

4321 IIII ===

Then,)(4 0 Iq=∆σ

Then,)(4 0 Iq=∆σ

Then, 0

037.0,3.05

5.1

084.0,5.03

5.1

5.1

''

'' =====

=====

==

AA

AA

AA

AA

Iz

LorBnm

Iz

LorBnm

mLB

kPa

kPa

A

A

8.14037.0*100*4

6.33084.0*100*4

' ==∆==∆

σσ

Then, 0

Example – 4 (1)Example – 4 (1)A rectangular foundation 6 X 3 m carries a uniform pressure of 300 kPa near the surface of a soil mass. Determine the vert ical stress at a depth of 3 m below point A on the centre line 1.5 m outside the long edge of the foundation.

Hint: use the principle of superposition

Example - 4 (2)Example - 4 (2)Using the principle of superposition the problem is dealt as shown below

For the two rectangles (1) carrying a positive pres sure of 300 kPa,

B = 3.0 m, L = 4.5 m, m = B/z = 3.0/3.0 =1.0, n = L/z = 4.5/3.0 = 1.5 Therefore, I r = 0.193 (from the table)

For the two rectangles (2) carrying a negative pres sure of 300 kPa,B = 1.5 m, L = 3.0 m, m = B/z = 1.5/3.0 =0.5, n = L/z = 3.0/3.0 = 1.0 Therefore, I r = 0.120 (from the table)

kPaIqIq rrAz 44)120.03002()193.03002()(2)(2)( 2010 =××−××=+=∆σHence, vertical stress increase at A ( a depth of 3 m),

Influence chart for vertical stress increase - 1Influence chart for vertical stress increase - 1•Newmark (1942) constructed influence chart , based on the Boussinesq solution to determine the vertical stress increase at any point below an area of any shape carrying uniform pressure.

•Chart consists of influence areas which has a influence value of 0.005 per unit pressure• The loaded area is drawn on tracing paper to a scale such that the length of paper to a scale such that the length of the scale line on the chart is equal to the depth z•Position the loaded area on the chart such that the point at which the vertical stress required is at the centre of the chart.•The count the number of influence areas covered by the scale drawing, N

Then, vertical stress increase at z∆σ∆σ∆σ∆σz = 0.005 qo N

Influence chart for vertical stress increase -2Influence chart for vertical stress increase -2

•Use of Newmark’s chart to calculate stress increment at A, ( at a depth of 3.0 m)

•The scale line represents 3 m

Z = 3.0 mq0 = 300 kPa

Then, vertical stress increase at 3 m below point A∆σ∆σ∆σ∆σz = 0.005 qo N = 0.005 X 300 X 30 = 45 kPa

•The scale line represents 3 m•The scale to which the rectangular loading area must be drawn•The area is positioned on the chart such that A is at the centre•The number of influence areas covered by the rectangle, N = 30 (approximately)

Approximate method to determine the stress subsurface stress increment (60 0 approximation)

Approximate method to determine the stress subsurface stress increment (60 0 approximation)

According to this method, the increase in stress at depth z is

))((0

zLzB

LBq

++××=∆σ

Example – 5 (1)Example – 5 (1)

Compare the stress increase occurring 2m below the centre of a 3m by 3m square foundation imposing a bearing pressure of 145 kN/m2 when:

3 m

PLAN VIEW

kN/m2 when:

(i) The Boussinesq stress distribution is assumed

(ii) The 60 0 approximation is assumed

2/145 mkN

SECTIONAL VIEW

Example – 5 (2)Example – 5 (2)(i) The Boussinesq stress distribution

is assumed

L =3 m

PLAN VIEW

L’ =1.5 m

1

43

2)( 43210 IIIIq +++=∆σ

When the centre is considered,

4321 IIII ===

Then, )(4 Iq=∆σ2

/145 mkN

SECTIONAL VIEW

Then, )(4 0 Iq=∆σ

1367.0,75.02

5.1''

5.1''

=====

==

Iz

LorBnm

mLB

20 /145 mkNq =

kPam 3.791367.0*145*42 ==∆σ

Example – 5 (2)Example – 5 (2)

mLB 0.3==

mz 2=

At 2m depth,

20 /145 mkNq =

(ii) The 60 0 approximation is assumed

z

B

L

(B + z)

L + z

20 /145 mkNq =

2/145 mkN

SECTIONAL VIEW

B

σ∆

))((0

zLzB

LBq

++××=∆σ

2/2.5255

9145

)23)(23(

33145mkN=

××=

++××=∆σ

Lecture 2 - Stresses in soils · (Stresses in soils) ENB371: Geotechnical Engineering 2 Chaminda...

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Transcript of Lecture 2 - Stresses in soils · (Stresses in soils) ENB371: Geotechnical Engineering 2 Chaminda...