6 stresses in soil - handouts

7/29/2019 6 Stresses in Soil - Handouts

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The University of New South Wales

School of Civil & Environmental Engineering

1

Soil Mechanics - CVEN2201

By Arman Khoshghalb

Stresses in soil

Soil MechanicsStress in Soils

Stresses in Soil

Sources of stresses in soil:

Self weight of soil

Due to the total unit weight of the soil (gt)

Surcharges

Uniform surcharges

Due to relatively wide foundations, embankments, etc.

Non-uniform loads

Due to tall buildings, etc.

2

When soil is dry There is only one constituent (air phase

usually does not carry any load) similar to solid

mechanics

Stresses in Dry Soil

z

q

sv

sh

sv = gd z +q

sh = Ko sv

svsh

Variation of stresses with depth:

z

Ko = Coefficient of lateral earth pressure at rest3

Stresses in Layered Soil

sv

sh

sv = gd1z1+gd2z2+q

q

sh = Ko2 sv

z1

z2

gd1, Ko1

gd2, Ko2

svsh

z

4





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By Arman Khoshghalb

Stresses in soil

Example

sv

sh

1m

1m

Calculate the vertical and horizontal stresses at apoint 2m below the ground surface:

gd1=16 kN/m3

Ko1=0.6

gd2=20 kN/m3

Ko2=0.3

0

16

36

sv sh

(kPa)

0

9.6

10.8

4.8

zsvsh

5

Stresses in Saturated SoilEffective stress concept:

Total stress at any depth is due to the

weigh of whatever is lying above thatlevel.

The total stress can be divided in twoparts:

A portion is carried by water in th e

continuous void space. This portion is

called pore water pressure (or neutral

stress) and acts with equal intensity in all

around.

The rest of the total stress is carried be

the soil solids at their points of contact.

This portion is called effective stress.

The principle of effective stress:

zt

WT

6

Stresses in Saturated Soil

u

u

u

u

ss

s

s

s

Soil under pressure:

7

Effective stress conceptIs the effective stress really the

same as the grain-to-grain

contact stress?

Answer: NO!

So, what is the physical meaning

of effective stress?

Vertical equilibrium:

P1(v) to Pn(v): vertical components of thecontact forces

as: Cross sectional area occupied by

solid-to-solid contacts.

z

A wavy line is passing only

through the point of

contacts of the solid

particles.

P1 P2 P3 P4 P5

Cross sectional area = A

hw

8





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By Arman Khoshghalb

Stresses in soil

Effective stress concept

z

A wavy line is p assing only



particles.

P1 P2 P3 P4 P5


hw

The effective stressis n ot really the sameas the grain-to-grain contact stress!

Rather it is the sum of the vertical

components of the contact f orces

divided by the total area:

in practical

problems :

W

9


z

A wavy line is passing only



particles.

P1 P2 P3 P4 P5


hw

The effective stress cannot be measured, itcan only be calculated!

To calculate the effective stress, firstcalculate the total stress and pore water

pressure, and then find the effectivestress using the principle of effective

stress:

For the case shown in the figure:

The effective stress at any point in the soilis independent of the depth of water

above the submerged soils.

W

10


Water cannot transmit any shear stress

Soils fail in shear

The effective stresses in a

soil mass control the

engineering properties (e.g.

shear strength and volumechange behaviour) of the

mass.

11

A note on the effect ive stress The effective stress is not really the grain-to-grain contact stress in

granular soils. It’s physical meaning is even more complicated in

fine-grainedcohesive soils.

Question: Given the above fact, Is it OK to use it in engineering

practice?

Answer:

Although it is not exactly the grain-to-grain contact stress, it

correlateswith it.

Experimental evidence has shown that the principle of

effective stress is an excellent approximation to reality and it isextremely useful for understanding soil behaviour, interpreting

laboratory test results and making engineering design

calculations.

The bottom line:

The concept works and that is why we use it.

12





4


By Arman Khoshghalb

Stresses in soil


The principle of

effective stress is thesingle most important

concept in geotechnical

engineering.

13

Stresses in Saturated Soil Effective stress is a tensor quantity with different

components

Vertical effective stress (one we have talked about so far):

sv = sv - uEffectivestress

Totalstress

Pore water pressure =gw zw

or measured by piezometer

sh = Ko svCoefficient of earth pressure

A property of the soil skeleton

sh = sh + u

Horizontal effective stress at any point depends on vertical

EFFECTIVE stress and coefficient of lateral earth pressure:

14

Stresses in Saturated Soil Calculate the effective stresses at a point 4m

below the ground surface. WT is at the surface.

4m

sv

sh

gsat=18 kN/m3

Ko=0.4

sv = gsat z =18 4 =72 kPa

u = gw zw = 9.8 4 =39.2 kPa

sv = s - u = 72 - 39.2 = 32.8 kPa

sh= Ko sv = 0.4 32.8 =13.1 kPash= sh + u = 13.1+39.2 =52.3 kPa

15


4m

sv

sh

gsat=18 kN/m3

Ko=0.4s

t

52.3 72

32.8

13.1

u = 39.2 kPa

u = 39.2 kPa

Calculate the effective stresses at a point 4mbelow the ground surface. WT is at the surface.

The difference between a total stressMohr circle and an effective stressMohr circle is pore pressure, u.

16





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By Arman Khoshghalb

Stresses in soil


4m

sv

sh

gsat=18 kN/m3

Ko=0.4

sv

u

z

sv

Variation of stresses with depth:

17 Autumn 2012 Stresses inSoils 18

Stresses in Saturated Soil Ground water table:

Steady state elevation at which

the pore water pressure isequal to zero.

It is not really a table!

Below the ground water table

the soil is saturated.

Depending on the grain size of

the soil above the groundwater

table, the soil may be saturated

because of capillarity, or it may

be unsaturated.

WT

Autumn2012 Stresses inSoils 19

Stresses in Saturated Soil Capillary tension:

z

u+

u-

Water rises above the water

table due to tension between soil

particles and water (water

surface tension).

Water is “hanging” on the soil

particles pulling the grains

together suction (capillary

pressure) – negative pore water

pressure

increase in effectivestress.

WT

Autumn 2012 Stresses inSoils 20

Stresses in Saturated Soil The height of capillary rise depends on pore sizes and their

distributions in the soils.

Terzaghi suggestion (based on capillary tube analogy):

Height of capillaryrise (m)

Empirical coefficient~ 0.03

(0.01-0.05)

Void ratio

Effective grainsize (mm)

ℎ =

∙





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By Arman Khoshghalb

Stresses in soil

Autumn2012 Stresses inSoils 21

For simplicity, we assume that Sr is 100% in capillary zone

For sands and gravels, assumed zero if not known Dry

above the water table.

Good example of capillary effect: sand castles.


WTh

c

z

u+

z=0

Pore water pressure in

capillary zone (capillary

pressure) = gw . z (negative)

Soi l type Height of capi l lary r ise

Gravel ~ 0

Sand 0 – 3 m

Silt 2 – 12 m

Clay > 10 m

0.5m

0.5m

1.0m

2.0m

gdry=15 kN/m3

gt=17.2 kN/m3

gsat=16.2 kN/m3

gsat=18.4 kN/m3

WT

sv(kPa)

0

7.5

16.1

32.3

69.1

u(kPa)

0

0

0

9.8

29.4

sv(kPa)

0

7.5

16.1

22.5

39.7z

sv

u

sv

22

Example 2Variation of stresses

with depth?

Gravel

Effects of WT Fluctuation

Example 3: A thick saturated clay deposit has initially a groundwater level 1m below the surface. Due to groundwater

extraction from an underlying aquifer the regional

ground water level is lowered by 2m. Calculate the

change in the vertical effective stress at depth z>3m.The unit weight of the clay, g, is constant with depth.

Initial GWL Lowered GWL

sv g z g z

u gw ( z-1 ) gw ( z-3 )

sv z (g - gw ) + gw z (g - gw ) + 3gw

23

Calculate the change in the vertical effective stress

at a point 5m below a river bed when the water

level in the river rises from 1m above the bed to

10m above the bed, due to a flood. The saturated

unit weight of soil is 20kN/m3.

Example 4

24





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By Arman Khoshghalb

Stresses in soil

Example 5

25

A 10 m deep layer of coarse sand overlays

bedrock. The water table is located 2 m below

the surface. The sand has the following

properties:

e=0.7 and Ko=0.5.

Assuming that the soil particles have a specific

gravity Gs = 2.7, calculate the effective stress

at a depth 5 m below the surface.

Calculate the stresses on a plane making an

angle 30o with the horizontal.

Application of any load at ground level

increases the total stresses in the soil.

For settlement calculations, we need to know

the change of vertical stress (∆ or ∆) with

depth in the soil due to surface loads ().

Methods:

If one-dimensional loading:

If three-dimensional loading:

2:1 Method

Theory of elasticity

2012 Stresses inSoils 26

STRESS DISTRIBUTION

∆ =


One dimensional loading If the dimensions of the loaded area are significantly

greater that the thickness of the soil l ayer, then one-

dimensional loading is assumed

the stress increase at any depth is equal to the applied

stress at the surface:

This assumption is not valid near the edge of end of the

loading area.

∆ =


One dimensional loadingq per unit area

Dsz

z

Dsz = q





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By Arman Khoshghalb

Stresses in soil


• The width of the loaded area is equal or less than the

thickness of the soil layer three dimensional loading

2:1 method


• 2:1 method: the simplest method to find ∆

• It computes the average stress increase at each depth.

• It is popular because it is simple, quick and easy to use.

• Empirical approach based on the assumption that the area

over which the load acts increases in a systematic way

with depth with a 2 vertical : 1 horizontal slope.

• The same vertical force is spread over an increasingly

larger area.

STRESS DISTRIBUTION

2012 Stresses inSoils

2:1 method Assumption: the area over which the load acts increases with

depth with a 2v:1h slope.

For uniformly loaded rectangle:q per unit area

B

B+z

1

2z

∆ =

+ +

2:1 method Assumption: the area over which

the load acts increases with depth

with a 2v:1h slope.

For strip footing:

2012 Stresses inSoils

q per unit area

∞

B

B+z

1

2z

∆ =×1

+ ×1


• The width of the loaded area is equal or less than the

thickness of the soil layer three dimensional loading

2:1 method


Solutions based on the theory of elasticity:

As long as the added stresses are well below failure.

Around 25% to 30% error

Boussinesq (1885)

For homogeneous, elastic and isotropic medium

Westergaard (1938)

Mainly for layered soils

STRESS DISTRIBUTION





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By Arman Khoshghalb

Stresses in soil


Elastic Solutions Point load:

Q

Dsz

z

r

R∆ =

3

2 + =

∆ =

1

1+ 2

=

Boussinesq’s solution

Westergaard’s solution

Equations have also been derived for radial, tangential and shear stresses.

We only consider increase in vertical

stress here since it is the important one

for settlement calculations.

Elastic Solutions

Point load:

Elastic Solutions In practice we rarely have loads that can be accurately

modeled as point load Engineering loads act on areas.

Solutions for different loaded areas can be developed by

integration of a point load over the area.

For any type of loading areas, equations and charts have

been developed based on both Boussinesq’s theory and

Westergaard’s theory.

only the equations and charts based on the Boussinesq’s

theory are presented here, s imilar equations and charts based

on the Westergaard’s theory are also available and can be

found in most advanced textbooks on soil mechanics.


Line load Like a point load, rarely used!

Dsz

z

x

R

q per unit length

y

z x

∆ =2





10


By Arman Khoshghalb

Stresses in soil

q per unit area

Dsz

y

z

ab

37

Strip loading

a is always positive and in radians.

×

=

b is negative under loaded area andpositive otherwise.

q per unit area

Dsz

z

r

38

Uniformly loaded circle


Uniformly loaded circle

q per unit area

Dsz

z

B=mz

40

Uniformly loaded Rectangle

Dsz = q Ir Ir obtained from Fadum’s chart.

m & n are interchangeable.

Under the corner of the loaded area

=

=





11


By Arman Khoshghalb

Stresses in soil

1010.10.01

0.00

0.05

0.10

0.15

0.20

0.25

Ir

m

0.2

0.4

0.6

0.8

1.0

2.0

0.1

0.5

0.3

1.4

n

mz

z

Fadum’s Chart

41

Elastic Solutions

a

d c

b

o

Uniformly loaded Rectangle:

Stress at other point under the rectangle can beobtained by superposition of the effects.

Dsz = q (I1+ I2 + I3 + I4)

a

d c

b

1

2

3

4

1

2

3

4

Dsz = q (I1- I2 - I3 + I4)

o

42

Newmark’s Chart

Stress under arbitrary shape:

Influence value = 0.005

O Qz

43

Newmark’s Chart Stress under arbitrary shape:

Newmark’s chart (influence chart):

Procedure:

Draw the loaded area to the scale “z” shown on the

chart;

Overlay the loaded area on the chart so that the

point of interest is located at the origin of the chart;

Count the number of squares “N” located within theloaded area;

The increase in vertical stress at the point of interest

is calculated as:

Dsz= q N (influence factor)

q is the pressure applied on the loaded area.44





12


By Arman Khoshghalb

Stresses in soil


For strip footing is deeper in the soilStress varies in vertical as well as

horizontal directions

Stress isobar Elastic Solutions The increase in vertical stress, found from the equations and charts

provided, must be added to the existing in situ overburden effective

stress, since the elastic solutions consider the half-space to be

weightless.

The Boussinesq and Westergaard theories give different results

Which theory should you use?

Answer: it is not that important which one you use! Both theories

are based on assumptions which are far from reality.

Although the assumptions of the Westergaard theory probably are

closer to reality for a layered soil deposit, Boussinesq’s solution is

usually preferred in geotechnical engineering problems.

The 2:1 method is used as often in practice as the solutions from

the theory of elasticity.

Example 6

a

d c

b

1

2

3

4

o

2m

3m

2m3m

47

Calculate the increase in vertical stress at a depth of 2m below point o due to the rectangular loading

shown in the opposite figure, using:

1- Boussinesq theory

2- Influence chart

3- 2:1 method.

The applied pressure over

the rectangle is 100kPa.

Example 7

Calculate the increase in vertical stress at a

depth of 1.5m below point o due to the

rectangular loading shown. The applied

pressure over the rectangle is 100kPa.

1m

2m

1m10m

a

d c

b

o

48





13


By Arman Khoshghalb

Stresses in soil


Example 8 A strip footing 4m wide carries a uniform pressure of

250 kN/m2 on the surface of a deposit of sand. The

water table is at the surface. The saturated unit

weight of the sand is 20 kN/m3

. Determine theeffective vertical stress at a point 3m below thecentre of the footing before and after the application

of the pressure.

6 stresses in soil - handouts

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