study guide for an exam (geotechnical)

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7SettlementImmediateConsolidation1.ppt
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Geotechnical Engineering

7

Immediate settlement

&

Consolidation settlement

Contents:

Compression of soils

Immediate settlement

Consolidation settlement

3.1 One dimensional consolidation test

3.2 Interpretation of results

Secondary consolidation & creep

Tolerable differential settlement

1. COMPRESSION OF SOIL

Immediate settlement or elastic deformation

  • occurs immediately & is recoverable
  • only a small amount of settlement
  • assessed by Young’s modulus of soil

Consolidation or primary settlement

  • decrease in volume of voids
  • only recoverable to an extent - reduction in overburden pressure
  • for foundation design

Plastic deformation or secondary consolidation

  • lateral flow of soil particles
  • not recoverable
  • difficult to determine

Creep

  • collapse of fibrous matter
  • organic soils e.g. peat
  • difficult to determine
  • Of these four types of compression, the greatest contribution to the settlement of a foundation is generally consolidation or primary settlement.
  • Immediate settlement is generally comparatively small and can usually be assessed using Young’s modulus of compressibility for the soil, values of which are published.
  • Secondary consolidation and creep are difficult to evaluate and are not considered during this course.

How to calculate settlement

  • Cohesive soils use oedometer consolidation for both immediate and consolidation settlement – see section 3.0 in the notes
  • Granular soils difficulties in obtaining a undisturbed sample
  • Oedometer test neglects the influence of lateral deformation – occurs in sands
  • Therefore the settlement of granular soils cannot be predicted in the laboratory - Penetration Test

Penetration Test

  • Plate loading test - pushing a plate into the ground

220.unknown

Standard Penetration Test

  • SPT pushing rob into ground
  • Metal rob (20mm to 80mm)
  • dropping a 63.kg hammer
  • through 760mm
  • number of blows (N) for a penetration of 300mm
  • from ‘N’ settlement of foundations constructed on granular soils can be calculated
  • A number of methods of analysis - the most suitable two are: [1] D’Appolonia et al (1970) [2] Parry (1971)

2. IMMEDIATE SETTLEMENT

Find values of I and Eu

To find I use influence charts (Janbu, Bejerrum & Kjaernsli: 1956) I=m0 m1

Eu difficult to determine in laboratory tests – derived from Poisson's ratio together with the results of consolidation testing, typical values are:

Soil type

Eu (MN/m2)

Soft clay

2 - 5

Firm clay

4 - 8

Stiff clay

7 - 20

Sandy clay

30 - 40

Silty clay

7 - 20

Loose sand

10 - 25

Dense sand

50 - 90

Dense sandy clay

100 - 200

222.unknown

Example 1 – Calculate the immediate settlement beneath the centre of the 2 m square spread foundation

D/B =

L/B =

H1/B =

1/2 = 0.5

2/2 = 1

2.5/2 = 1.25

D/B =

L/B =

H2/B =

0.5

0.85

=10.6 mm

Eu=10MN/m2

Eu=70 MN/m2

B

D

L

P = 125 kN/m2

Dense SAND

Very loose to loose SAND

ROCK

1 m

2.5 m

3 m

m0 = 0.85

}

m1 = 0.5

}

H2

H1

H3=H2-H1(dense sand)

227.unknown

228.unknown

Eu=10MN/m2

Eu=70MN/m2

+

-

=

P = 125 kN/m2

Dense SAND

Very loose to loose SAND

ROCK

1 m

2.5 m

3 m

Eu = 10 MN/m2

Eu = 70 MN/m2

Eu = 70 MN/m2

H2

H1

H3=H2-H1(dense sand)

Eu= 70 MN/m2

D/B =

L/B =

H2/B =

1/2 = 0.5

2/2 = 1

5.5/2 = 2.75

H2

H1(dense sand)

= 0.3 mm

Total immediate settlement = 10.6 + 0.3 = 10.9 mm

m1 = 0.6

}

m0 = 0.85

}

B

D

L

P = 125 kN/m2

Dense SAND

Very loose to loose SAND

ROCK

1 m

2.5 m

3 m

m0 = 0.85

}

m1 = 0.5

}

H2

H1

H3=H2-H1(dense sand)

Example 1 - Calculate the immediate settlement beneath the centre of the 2 m square spread foundation

D/B =

L/B =

H1/B =

1/2 = 0.5

2/2 = 1

2.5/2 = 1.25

0.5

0.85

=10.6 mm

Eu=10MN/m2

Eu=70MN/m2

Given no depth (H) ‘infinite thickness’:

Example 2: A flexible foundation 6 m x 2 m, uniformly loaded 2500 kN, infinite thickness of saturated clay, E = 30 MN/m2, find immediate settlement beneath a corner.

I = influence factor

E = Soil modulus

u = 0.5 Poisson’s ratio

q = load / Area

q = 2500 kN / 6m x 2m =208 kN/m2

I = L/B = 6/2 = 3 therefore from table I = 0.89

= 0.00926 m = 9.26 mm

6 m

2 m

No depth

237.unknown

Poisson’s ratio

3. CONSOLIDATION SETTLEMENT

Consolidation in a fully saturated soil

  • apply a load
  • increase in PWP
  • pressure increase is gradually taken by the soil particles

Rate of load transfer of load from water to soil depends on permeability of soil

238.unknown

soil

Effective stress

soil

Total stress

Consolidation settlement

Load or pressure

When a saturated soil is loaded (F)

The increase in load causes an increase in total stress (s)

Which initially results in an equal increase in pore pressure (uw)

As pore water is squeezed out from the soil the increase in load transfers from the pore water to the solid particles (the spring)

Eventually all of the load transfers to effective stress (s’)

At all the time the relationship s= s’ + u is maintained

The Piston & Spring Analogy

F

F

σ’

uw

Transfer of stress involves movement of pore water therefore settlement – the rate of transfer depends on permeability of the soil

High Permeability soils (e.g. sands & gravels) settle quickly & increase in effective stress occurs over a short time period

Low Permeability soils (e.g. silts & clays) take a long time to settle or consolidate (years) thus the increase in effective stress is slow

W

valve open

Consolidation of the soil

Effective stress (spring)

kN/m2

Time (permeability of soil)

Water

pressure

kN/m2

Pressure in water

Pressure in spring (soil)

  • Consolidation in partially saturated soils will be almost immediate, as it is air that is being expelled. However, when an increase in load is applied to a fully saturated soil the resulting increase in pressure is first taken by the pore water and is then gradually transferred to the soil particles over a period of time. As has been discussed previously, the rate at which this transfer of load occurs will depend upon the permeability of the soil being consolidated.
  • In high permeability soils, such as sand and gravel, the transfer of pressure from the pore water to the soil skeleton will occur very quickly, thus consolidation will be rapid. Unfortunately, it is not possible to measure such movements in the laboratory as it is difficult, if not impossible, to obtain an undisturbed sample of sand or gravel. Moreover, the speed at which the pore water pressures dissipate would not permit valid measurements to be taken. Consequently, settlement of granular soil is generally evaluated using the results of insitu tests which are undertaken during the site investigation.
  • In clay and silt the transfer of pressure from the pore water to the soil skeleton will be slow as the permeability of these soils is low. Moreover, the cohesive nature of the soil would allow specimens to be prepared for laboratory testing. Consequently, it is the consolidation testing of cohesive soils which will form the subject of the work which follows

3. ONE DIMENSIONAL CONSOLIDATION TEST

The Oedometer Test (BS 1377: Part 5:1990), assesses the two important features related to consolidation:

The magnitude of consolidation settlement

The rate at which this settlement takes place

  • 75 mm dia. 20 mm thick
  • cored from a lager sample using a cutting ring - undisturbed & 1-D
  • thickness compressed
  • X-area is unaltered
  • top & bottom of sample capped with porous plates
  • immersed in a water bath (fully saturated)
  • loaded vertically
  • vertical settlement recorded

consolidation

Application of load

  • Load sample at 5 increments
  • each increment maintained until sample is fully saturated (24hrs)
  • each load is double the pervious one
  • initial pressure = effective overburden pressure
  • load applied via lever arm 10:1 ratio
  • after finial increment of load & 24 hrs consolidation all the loads are removed
  • sample swells for 24 hrs
  • moisture content taken

239.unknown

Recording the settlement (Oedometer test)

Magnitude of settlement

  • amount of vertical settlement at the end of each period of loading (24hrs) & final figure after swelling

Rate of settlement (& permeability)

  • reading of settlement against time, normally at 0.25, 0.5 1, 2, 4, 8, 15 & 30 mins then 1, 2, 4, 8 & 24 hrs after the application of each load

The two pieces of information normally required from this test are:

3.2 Interpretation of test results

Coefficient of compressibility ‘mv’

or

mv = volume change per increase in effective stress (m2/MN) inverse of pressure

P

H

soil

dH

P+dP

soil

Amount of
Consolidation Settlement

Total settlement = average settlement calculate for each layer

Use thinner layers because the stress will change significantly over a thick layer

mv

where: dH = consolidation settlement (mm)

H = thickness of layer (m)

mv = coefficient of compressibility (m2/MN)

sv = vertical stress at depth z (kN/m2)

Applied load (p)

H

sv

Compressible layer (e.g. clay)

Sand

Sand

sv1

mv1

H1

sv2

H2

mv2

sv3

mv3

H3

Applied load (p)

Clay

Example 3: Estimate the consolidation settlement of the layer of soft silty CLAY beneath the 2 m square foundation shown below

P = 150 kN/m2

soft CLAY mv=0.8 m2/MN

1 m

1.5 m

0.75 m

= 0.03348m => 33.48 mm

I = 0.093

=55.8 kN/m2

1m

1m

246.unknown

4. SECONDARY CONSOLIDATION & CREEP

No reliable method of estimating the magnitude of secondary consolidation or creep in soil

5. TOLERABLE DIFFERENTIAL SETTLEMENT

Structural safety

Appearance of the building

Limitations imposed by special equipment

251.unknown

252.unknown

6. Gross and Net Foundation Pressure

  • Gross (P) = total applied load divided by the area of foundation
  • Net is the increase in pressure at foundation level, therefore

Net (PN) = gross foundation pressure – effective overburden pressure (Po)

PN = P - Po

*

Example 4: calculate the gross and net foundation pressures for the 1 m x 2 m foundation shown below

Net = Gross - Po

Net = 200 – 38 = 162 kN/m2

1 m

1 m

ϒ = 18 kN/m3

ϒ‘ = 20 kN/m3

Total load including footing 400 kN

*

7. TIME SETTLEMENT:

Example 5: A layer of clay 5 m thick with a bulk unit weight (g) of 20 kN/m3 overlies a sand. The clay is in turn overlain by a 3 m thick layer of sand (g = 18.3 kN/m3). Consolidation tests reveal the clay to have a coefficient of volume change (mv) of 4 x l0-4m2/kN and a coefficient of consolidation (cv) of 5x10-8 m2/s. A structure is to be erected quickly, ultimately imposing a uniform stress of 150 kN/m2 on a raft foundation 8 m square, the base of which is located 2 m below the surface of the upper sand. Stress changes at the mid-height of the clay may be regarded as being representative of the whole layer. Construct a time settlement curve for the foundation due to the consolidation of the clay for a period of 3 ½ years.

Charts of Fadums Influence factors, and degree of consolidation versus time factor are shown in the figures below respectively.

CLAY

  • = 20 kN/m3

mv = 4x10-4 m2/kN

cv = 5x10-8 m2/s

2 m

8 m

150 kN/m2

SAND

g = 18.3 kN/m2

8 m

sv

255.unknown

256.unknown

Foundation stress causing settlement:

2

Consolidation of clay layer:

i) Vertical stress at centre of layer

2 m

8 m

150kN/m2

SAND

g = 18.3 kN/m3

8 m

sv

2.5 m

5 m

1 m

4 m

4 m

0.185

1.14

1.14

260.unknown

ii) Consolidation of clay layer

= 0.1678 m x1000 = 167.8 mm = 170 mm

ii) Time settlement curve

where: cv = 5x10-8 m2/s

t = time in days = variable

d = H/2 =2.5

H = 5 m

Clay

Sand

Sand

d

d

seconds to days

91.25 182.5 273.75 365 547.5 730 912.5 1095 1277.5

0.06 0.13 0.19 0.25 0.38 0.50 0.63 0.76 0.88

0.06

0.11

Case 1 = wide or extensive applied pressure compared to thickness of the soil layer, e.g. an embankment and lowering the of the water table.

Case 2 = applied pressure over a small area such as a foundation.

Case 3 = self-weight of soil forming an embankment

Strip or spread loads take case 2

0.11 0.25 0.35 0.46 0.59 0.70 0.79 0.84 0.88

From part ii) above consolidation settlement d = 170 mm

d x U = 170 x 0.11 = 19

19 42 60 78 100 119 134 143 150

270.unknown

Sheet1

Tabulate
Years 1/4 1/2 3/4 1 1 1/2 2 2 1/2 3 3 1/2
Days (t)
Tv
U from graph
d x u

Sheet2

Sheet3

Chart1

0.25
0.5
0.75
1
1.5
2
2.5
3
3.5
Years
Settlement
19
42
60
78
100
119
134
143
150

Sheet1

Tabulate
Years 1/4 1/2 3/4 1 1 1/2 2 2 1/2 3 3 1/2
Days
Tv
U from graph
d x u
0.25 0.50 0.75 1.00 1.50 2.00 2.50 3.00 3.50
19 42 60 78 100 119 134 143 150

Sheet1

0
0
0
0
0
0
0
0
0
Years
Settlement
0
0
0
0
0
0
0
0
0

Sheet2

Sheet3

Chart1

0.25
0.5
0.75
1
1.5
2
2.5
3
3.5
Years
Settlement
19
42
60
78
100
119
134
143
150

Sheet1

Tabulate
Years 1/4 1/2 3/4 1 1 1/2 2 2 1/2 3 3 1/2
Days
Tv
U from graph
d x u
0.25 0.50 0.75 1.00 1.50 2.00 2.50 3.00 3.50
19 42 60 78 100 119 134 143 150

Sheet1
Years
Settlement

Sheet2

Sheet3

u

i

E

pBI

=

d

10

5

.

0

85

.

0

2

125

´

´

´

=

i

d

70

5

.

0

85

.

0

2

125

70

6

.

0

85

.

0

2

125

´

´

´

-

´

´

´

=

i

d

(

)

I

E

qB

.

1

2

n

d

-

=

(

)

89

.

0

30000

5

.

0

1

2

208

2

´

-

´

=

d

δH = mv.δP.H

dH=m

v

.dP.H

mv = 1 . δH H δP

m

v

= 1 . dH

H dP

δH =mv.σ v.H

dH=m

v

.s

v

.H

δH =mv1.σ v1.H1 +mv2.σ v2.H2 +.....

dH=m

v1

.s

v1

.H

1

+m

v2

.s

v2

.H

2

+.....

1

0.53

1.875

L

m

z

===

1

0.53

1.875

B

n

z

===

2

40.093150/

z

IPkNm

s

==´´

H

dP

m

dH

v

.

.

=

3

0.81055.80.75

dH

-

=´´´

m

m

kN

kN

m

m

´

´

=

2

2

/

/

Gross = 400 kN 1m×2 m

= 200 kN /m2

Gross=

400kN

1m´2m

=200kN/m

2

Po = (1m×18kN /m3)+(1m×20kN /m3 )=38kN /m2

Po=(1m´18kN/m

3

)+(1m´20kN/m

3

)=38kN/m

2

23

218.3

150

kNk

P

N

m

mm

æö

=-´

ç÷

èø

∴P =113.4kN /m2

\P=113.4kN/m

2

4

4()1.14

3.5

vzz

L

IPwhereImn

Z

s

======

σ v =4(IzP)

=4 0.185×113.4( ) =83.92kN /m2

s

v

=4(I

z

P)

=40.185´113.4

()

=83.92kN/m

2

vv

mH

ds

=

∴δ =83.92 kN m2

×4×10−4 m2

kN ×5m

\d=83.92

kN

m

2

´4´10

-4

m

2

kN

´5m

(

)

(

)

(

)

8

2

2

2

510

2.

/

5

ms

t

T

ays

m

v

d

-

´´

=

2

v

ct

Tv

d

=

(

)

(

)

9

810

606024

da

t

T

ys

s

v

-

´´

\=´´´

(

)

4

6.91210

as

v

dy

Tt

-

\=´

Tabulate

Years1/41/23/411 1/222 1/233 1/2

Days (t)

Tv

U from graph

x u

0

20

40

60

80

100

120

140

160

0.000.501.001.502.002.503.003.504.00

Years

Settlement