soil settlement

8/10/2019 soil settlement

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Geosynthetics and

Reinforced Soil Structures

using geosynthetics

Professor of Civil Engineering

, ,

e-mail: [email protected]



Some Problems and Solutions in

soft clay soils

Low bearing capacityarge se emen s

Lateral flow of soils/slip circle failure

SOLUTIONS

Replace the soil- ,

Stone columns

Piles and reinforced concrete slab

Geocell mattress

Piles with geosynthetic reinforced platform (piled embankment)



-Sand drains

-

Vacuum assisted pre-consolidation



Schematic of

Sand Drain Principle

surcharge fill

t cT

vv

columns

Accelerated drainage achieved by reducing drainage path



Soil embankment placed as surcharge to drive the

consolidation process



-

Corrugated plastic core for

L

t

Geotextile filter



Connection arrangements

for wick drain installation











Rig for installing the PVDs



anu acture o s



General view after installation of PVD’s at a site



Continuous flow path even after shear

movement

No clogging due to geotextile filter

Re n orcement act on ue to t e tens estrength of the PVD



–Square pattern Triangular pattern

d

s

d

D

s

=

s

. .



Hansbo’s equation (Hansbo, 1979)

Time for consolidation,

2

2

2

3 18 4

ln D d d D D

t lnc

Neglecting the small quantity (d/D)2

2 10 75 D Dt ln . ln 1h U

, h –

d – Equivalent diameter of the PVD

– ameter o t e n uence area

U – The avereage degree of consolidation



PVD size = 100mm 5 mm

Consolidation to be achieved = 80%

= 2

Given

E uivalent diameter o circular drain havin same circum erence

2 100 5 d

66 84 0 0668 . mm . m



B Hansbo s e uation

Design of PVD …

2 11 0 75 D DTime, ear ln . ln

2

. .

0 0668 . n .

.

D LHS2 10.59

4 53.46

3.5 39.33.85 48.96 RHS

Diameter of the influence area, D = 3.85 m



Design of PVD …

Time for Consolidation Vs Spacing of drains

1000

d a y s )

d a t i o n ( =

U=80%

U=70%

500

r c o n s

o l =

U=50%

250

T

i m e f

0

0 1 2 3 4 5 6

Wick drain spacing, D (m)



Spacing of PVDs

Design of PVD …

1. By rectangular pattern = 3.85/1.128 = 3.4 m

. . . .

Triangular pattern is preferred as spacing is greater and overlappingo areas s ess.



-

Vacuum App cat on



















Increase in effective Stress due to

vacuum applicationhorizontal vacuum

pipelines

sand layer .

- .

Let dry and saturated unit weights of both soils be 17 and 20 kN/m3.



Stresses in sand layer before the application of vacuum

Total stress, v = Pa + dry* z = 100 + 18*z kPaPore pressure u = Pa = 100 kPa

Effective stress = -u = 18*z kPa

Stresses in sand layer after the application of vacuum After the a lication of vacuum, water level will raise from 3.0 m to 1.0 m below the

ground level.

Total stress, v = Pa + dry* 1 + sub*(z-1) = 100 + 18*1 + (z-1)*(20) kPa = 98+20*z kPa

= - * = - *,

Effective stress, v = v-u = 98+20*z – 10z+10 = 108 + 10*z

Increase in effective stress in sand la er increase in effective stress within sand layer due to vacuum =

= 108+10*z – 18*z = 108 – 8*z

i.e. change in effective stress is 100 kPa at 1 m depth

and 84 kPa at 3 m depth.

Stresses in clay layer (below water table) before



Stresses in clay layer (below water table) beforethe vacuum a lication

Total stress, v = Pa + 3*18+20*z, (z is measured from top of claylayer)

Pore pressure, u = Pa + 10*z

Effective stress, v = v-u =54+10*zStresses in clay layer after vacuum applicationTotal stress, v = Pa + 18+ 2*20+20*z = Pa + 58 + 20*z

Pore pressure, u = 20+10*zEffective stress, v = v-u = Pa+38+10*z=138+10*z

increase in effective stress in clay layer,

= 138+10*z-54-10*z=84 kPa

This increase in effective stress is constant withdepth!!!!!



depth

• No inherent increase in shear stress. Lateral roundmovements are compressive rather than expansive.

• Vacuum consolidation creates more uniform surfacese emen s.

• No surcharge fill is necessary to drive the system.

layer above the ground level, this layer acts like a semi-rigid

mat and hence construction equipment can be moved on thesite without waiting for consolidation to take place.

• Can get rid of secondary compressions which is not possible- .



a oa em an men

Soft

clay

es nc ne

Piles

Firm stratum



Schematic view of piled embankment concept .. Purpose of the

horizontal geogrid layer is to transfer the embankment load to

e p es an o equa ze e se emen s



CFA (Continuous Flight Auger) piles















Arching action (Terzaghi, 1943)

Soil layer-geosynthetic systems overlying

cav t es an vo s rou et a .

Benefits of the geosynthetic



Benefits of the geosynthetic

reinforcements over the piles

Allows the piles to be spaced at greater spacing,

,

Prevents the soil yielding near the edges of the

piles,

Prevents the lateral spread of piles at the

ex rem es

Eliminates the necessit for rakin of the iles.

Geosynthetic reinforced piled



Geosynthetic reinforced piled

embankment

Gabion block to counteractEmbankment

Geosynthetic

Lateral Thrust

reinforcement

GL

H

Pile

caps

a

Piles



1. Estimation of the degree of arching in the

fill.

2. Calculation of the tension in the

geosynthetic rein orcement layer .

the extremities

For successful design of piled



For successful design of piled

embankments. e area ra o o e a op e or p e

caps.2. The lateral thrust on the piles at the

extremities due to lateral soil movement.

3. Relating the settlements at the pile head

analytical solution for this)

.the embankment.



- A quantity used for the comparison of various methods

=

Overburden pressure due to the embankment fill

pS H

After Russell and Pierpoint (1997)



q

z

reinforcement

Width of

long void,

(s-a)

Stress reduction ratio, 2 2

2 2 4

3 1

aHK tan

s a D

s aS e



2 2

3 2

2 8

c D

. s pS s a

H

s a H

Stress reduction ratio

1

12 6

T

r

s a

T a Tensile force in the

geosynthetic per unit width,

pc=stress on pile cap

aC p cc Cc=arching coefficient

v=Hv

Design of Geosynthetic reinforced



Design of Geosynthetic reinforced

Piled Embankment - Example

Puverised flyash filled

embankment

9 m = 14kN/m3

Pile caps

(1.1 m s uare)

Soft clay

(Without piles

settlement = 700 mm)

4 m

Embankment Details



•

• Tensile safety factor = 3.0

• ea extens on at a ure =

Geotextiles LongitudinalStren th kN/m

TransverseStren th kN/m



A - . .

Assuming2

Geosynthetic

R G

b = 0.2 0.7= 0.14 m

21

b tan

b TT

11 03.

a

2G

a R sin

1

2T

R bGWeight of the fill , W

G .

52 08W . kN m

T

Circular arc method



Circular arc method…..

Considering the reaction force as

0 15 18.9 kN m BW . h

,

251.5 kN/mT R W W

With two layers of geotextile, B laid in cross direction

The total strength = 1050 kN/m

The strain Gin the geotextile, 12 2 91050

. % . %

G a .

Circular arc method



Circular arc method…..

As εG < the predicted

Try with b = 0.19 m

= 14.93º

R G = 5.63 m

W = 38.08 kN/m

TT = 108 kN/m

G, .

From the geometry, εG = 1.2%

As these two are compatible the tension in the geosynthetic

T = 108 kN/m.

ε

G = 1.2 %



Tension in the geosynthetic

2

21 1

T aT WT WB a

2 21 16 4 16b a b b

2G e aa a

1 69 0 12cc

. hC . B

Loading coefficient

Catenary Deformation analysis….




1D Arching: Pressure ratio = CcBc/h

2D Arching Pressure ratio = (CcBc/h)2





Loading Coefficient, 1 69 0 12 13.71c. hC .

B

Pressure ratio – (1D) = CcBc/h = 1.676

Pressure ratio – (2D) = (1.676)2 = 2.809

In any 4 square piles,

Pile area = 1.21 m2

Total area = 16 m2

= 2 .

Total load = 16149 = 2016 kN

= = . . Load on soil = 2016-428 = 1588 kN = 107.4 kN/m2




y y

WT = 107.4 kN/m

WB = 0.15 h = 18.9 kN/m

s per e equa ons s own ear er

TT = 309.8 kN/m

From load-extension data εG = (309.8/1050)12 = 3.5 %‘ ’ = G , G .

As the two values are in close agreement further iteration is not

necessary.



-

• To minimise the differential settlement at the top of the embankment

0 7 H . s a

• In the present case, H = 0.7(4 – 1.1) = 2.03 m < 9 m

• The Arching coefficient (considering end bearing pile)

1 95 0 18. H Cc .a

•

= 15.77

2 2215 77 1 1cC a . .

9c v H



-

• or . s-a , e s r u e oa carr e y egeosynthetic reinforcement

2 2

2 2

1 4c

. s s a p

WT s a

= 176.85 kN/m

v

• ens on n t e re n orcement or stra n

1T W s a

• Tension due to lateral thrust,

.2 6r a

0 5 170 1 kN/m L

T . Ka H .

• Total tension = 656.3 kN/m



By Circular arc method

T = m; εG = . ; T = . m

B Catenar deformation method

TT = 310 kN/m; εG = 3.4 %; WT = 107.4 kN/m

y me o

T = 656.3 kN/m; ε = 5 %; W = 176.85 kN/m



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