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GEOSYNTHETICS AND REINFORCED SOIL STRUCTURESREINFORCED SOIL STRUCTURES

Design and Construction of Pavements Using Geosynthetics-IUsing Geosynthetics I

P f K R j lProf K. RajagopalDepartment of Civil EngineeringIIT Madras, Chennai 600 036e-mail: gopalkr@iitm.ac.in

ROAD PAVEMENTSROAD PAVEMENTSRoad surface should be smooth and unyieldingy g

to enable free movement of vehicles.Should provide good support in all seasons.p g ppThe thickness of pavement should be sufficient

to reduce the tyre pressure on road surface toto reduce the tyre pressure on road surface toallowable bearing pressure of subgrade.– Flexible – consists of multiple layers which haveFlexible consists of multiple layers which have

very low or zero flexural strength– Rigid – consists of rigid layer which has goodRigid consists of rigid layer which has good

flexural strength and stiffness

Load Transfer in Flexible PavementsLoad Transfer in Flexible Pavements

l ibl fl h d f i f• Flexible pavements reflect the deformation of lower layers to upper layers

• Load transfer to lower layers is through grain to grain contact stressesg

• Good interlocking with larger & angular particles helps in spreading loads to larger areashelps in spreading loads to larger areas

• Binder also helps in load spreading ability• Top layer is strongest and lower layers made of inferior materials

Reflection of sub-surface undulations in upper layers due to lack of flexural strength

Flexible PavementFlexible Pavement

Bituminous wearing courseBituminous wearing course

Base course

Sub-base courseT

Subgrade soilSubgrade soilThickness of the pavement (T) depends on the strength of subgrade soil traffic loading design lifestrength of subgrade soil, traffic loading, design life, properties of layers

WBM material

WBMWBM

GSBGSB

WBM & GSB Layers in a flexible pavement

Typical Granular sub-base (GSB) layer in a flexible pavement

CBR Test Apparatus

INTERPRETATION OF CBR VALUE

CBR= (Pt/Ps)*100%

US Army Corps of EngineersUS Army Corps of Engineers

1751

pCBR

Ptc

175.1

AP

pc75.1

CBR

t= thickness of pavement (cm)P = wheel load (kg)CBR = California Bearing Ratio (%)CBR = California Bearing Ratio (%)pc = tyre pressure (kg/cm2)A = contact area (cm2)A contact area (cm )

f d f d h dU. S. Army Corps of Engineers Modified CBR Design Method

1 2

1.8)087.0log1275.0(

A

CBRPCt

where  t = the design thickness (inches)C = the traffic in terms of coveragesC  the traffic in terms of coveragesP = the equivalent single‐wheel load (in pounds), andA = the tire contact area (in square inches).( q )

• US ARMY CORPS of Engineers based on extensive• US ARMY CORPS of Engineers, based on extensive testing…..

63.034.228301.271log98.470log24.119'o C

rPNh

Where, 

uC

ho’ = Base course thickness under trafficN  = Number of traffic passes of an equivalent single wheel load, P (kN)

d i d h h kPcu = undrained shear strength, kPa

Recommended Pavement layer thicknesses (IRC 37 2001) t d d l l d 8160 k(IRC 37-2001) – standard axle load=8160 kg

80 kNSubgrade CBR = 5%

Cumulative Total Pavement compositiontraffic (msa)

thickness (mm)

BC (mm) DBM (mm) Granular base and sub base

10 660 40 70

Base = 250 mm

20 690 40 10030 710 40 120 250 mm

Sub-base 50 730 40 140100 750 50 150

= 300 mm150 770 50 170

Recommended Pavement layer thicknesses (IRC 37-2001) – standard axle load=8160 kg

Subgrade CBR = 10%Cumulative Total Pavement compositiontraffic (msa)

thickness (mm)

BC (mm) DBM (mm)

Granular base and sub base

10 540 40 50

Base = 250 mm

20 565 40 7530 580 40 90 250 mm

Sub-base 50 600 40 110100 630 50 130

= 200 mm150 650 50 150

Rigid PavementsRigid Pavements

• Consists of rigid reinforced concrete or pre‐stressed concrete which acts as a wearing gcourse and distribute the loads over wide area

Rigid RCC or PCC layer

b d ilbase/sub-base course

PCC layer

subgrade soil

• Load distribution due to flexural rigidity ofLoad distribution due to flexural rigidity of the pavement layers

Use of GeosyntheticsUse of Geosynthetics

h i h l i l f llGeosynthetics help in several ways as follows:o Reinforcement – helps in reducing subgrade stresses and prevents cracking of pavement due to swelling of foundation soil

o Separator – prevents mixing up of layerso Filter layer – prevents piping phenomenono Drainage layer – provides for safe disposal of water

o Asphalt reinforcement – helps in preventing the reflection cracks

Separation function of geosyntheticsSeparation function of geosynthetics

Loss of aggregate into soft ground

Stable aggregate layer due to separation

Lateral Restraint Function of geosynthetic

Lateral restraint by base friction and interlocking

Haliburton et al. (1981)

Improvement of Subgrade bearing capacity

I i b i it b l d di dIncrease in bearing capacity by load spreading and confinement of soil

Haliburton et al. (1981)

Membrane Support MechanismMembrane Support Mechanism

H lib t t l (1981)Haliburton et al. (1981)

Dual wheels and Equivalent Single Wheel Load (ESWL)

Heavy trucks may have dual wheels

d

s

d/2d/2

Axle load, P = 4 Ac pcP l l d

Area of two wheelsP = axle load, Ac = contact area of a single tyre, p = tyre inflation pressure

22 ABL cppc = tyre inflation pressure2

pp c

ec

G il R i f d U d R dGeotextile‐Reinforced Unpaved Road DesignDesign

Giroud and Noiray (1981)

AssumptionsAssumptions

Aggregate layer: CBR > 80Subgrade: soft clay (=0) soilSubgrade: soft clay ( 0) soilCuu = 30 CBR (kPa)No. of vehicle passes less than 10,000Standard axle load = 80 kN ( 8155 kg)Standard axle load = 80 kN ( 8155 kg)Failure is by plastic shear in the subgrade

ASSUMPTIONSASSUMPTIONS

S b d i f d l i d i d• Subgrade is soft saturated clay in undrained condition 

• Base course has CBR of at least 80• Rectangular contact area of tyresg y• Pyramidal stress distribution• Constant stress distribution angle in reinforced• Constant stress distribution angle in reinforced and unreinforced cases

• Bearing capacity increased form elastic to• Bearing capacity increased form elastic to ultimate with inclusion of geotextile

Assumptions• The angle of internal friction of the aggregate layer is large

enough to rule out shear failure within the aggregate.

• The interface friction between the aggregate layer and the geotextile is large enough to rule out sliding of base coursegeotextile is large enough to rule out sliding of base course and the geotextile.

• Analysis investigates only the bearing capacity of the foundation soil with and without geotextile reinforcement.

• The aggregate layer provides a pyramidal distribution of P/2.

• The wheels are assumed to travel along the same track along the road so that every cross section of the road receives thethe road so that every cross-section of the road receives the same load at the same location resulting in same deformation. Therefore, 2D analysis is valid

Assumptions (contd )Assumptions (contd..)

Th t d li l t l h i b d il• The study applies only to purely cohesive subgrade soil under fully saturated condition (undrained case)

• The placement of the geotextile on soft subgrade soil assumes a general shear failure, which otherwise g(unreinforced) would have been a local failure. (original paper says elastic limit).This leads that the Bearing capacity factor changes from to ( +2) Experiments bycapacity factor changes from to ( +2). Experiments by Barenberg and Bender (1978) showed that without fabric the bearing capacity was 3.3c but with fabric it is 6.0c.

• The geotextile provides restraint or confinement if placed at the interface which leads to improved load distributionat the interface which leads to improved load distribution capacity.

Stress distribution in unreinforced and reinforced sections

Thickness of aggregate layer without reinforcement h Thickness of aggregatereinforcement = ho Thickness of aggregate

layer with reinforcement = h

C I U i f dCase I: Unreinforced case= stress on soil subgrade interfaceg

hP

2 (1)ooooo

o hhLhB

p

)tan2)(tan2(

2

(2)ouo hCPP itelasticlim

Equating (1) and (2)Equating (1) and (2)…….h0

P(3)

)tan2)(tan2(2

u hLhB

PC

)tan2)(tan2( oooo hLhB

Using Eq. 3,  ho can be evaluated…..

Unreinforced caseUnreinforced case• Thickness of unreinforced sectionThickness of unreinforced section 

Pc

0 02 ( 2 tan )( 2 tan )2o o

c c

cP Ph hp p

Case II: With reinforcementp = stress on soil subgrade interfacep = stress on soil subgrade interface

= (4)PghP

2 ghLhB

)tan2)(tan2(

Pg = membrane support (vertical component)component)

(5)hCpp u )2(sheargeneral (5)

Equating (4) and (5)

pp u )(sheargeneral

Equating (4) and (5)……..

P(6)Pg

hLhB

PCu

)tan2)(tan2(2)2(

))((

KPg

2

21

aa

g

2 s

K‐ secant modulus strain developed strain developed

2a 2a2a’

tt

s sr

Evaluation pEvaluation pgCase (i)‐ chord length aa

bb parabolaoflengthcurve(i)               1

aabb 2

, parabolaoflengthcurvebb

(ii)aa

ars

2, lengthchordaa

aa loadbelowsettlements

Case (ii) aa

b (iii) 1ab

22ra(iv)22 32

2aaaa

ras

Case (iii) aa

b(v)

1ab

(vi)rs (vi)2

Design procedureDesign procedure

S 1 C l l h h hi k i h• Step1: Calculate h0, the aggregate thickness without geotextile, (static)S 2 C l l h hi k f h l i h• Step 2: Calculate h, thickness of the aggregate layer, with geotextile (static)S 3 C l l d i i hi k h h h• Step 3: Calculate reduction is aggregate thickness, h=ho‐h

• Step 4: Calculate h’0, the  aggregate thickness without il h ffi i k igeotextile when traffic is taken into account

• Step 5: Calculate the thickness of aggregate required,      h’ h ’ hh’=ho’‐h

Webster and Alford (1978)Webster and Alford (1978)

630log19.0 Nh s

o 63.0CBR

h o

ho = thickness of aggregate layer (m)Ns = no. of passes of standard axle load of 80 kNRut depth = 75 mm

Generalisation to other loadsGeneralisation to other loads

953 95.3

PN

PN is

PN si PN si

For rut depths other than 75 mmFor rut depths other than 75 mm,Replace logNs by [logNs – 2.34 (r 0.075)]

• US ARMY CORPS of Engineers based on extensive• US ARMY CORPS of Engineers, based on extensive testing…..

63.034.228301.271log98.470log24.119'o C

rPNh

Where, 

uC

ho’ = Base course thickness under trafficN  = Number of traffic passes of an equivalent single wheel load, P (kN)

d i d h h kPcu = undrained shear strength, kPa

P = 80 kNP 480 kPPc = 480 kPar = 300 mm