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: [email protected]
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
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