pavement (turapan) - universiti sains malaysia 366/pavementslides.pdf · reg 366 hbp usm 2007...

REG 366 HBP USM 2007

PAVEMENT(TURAPAN)

MATERIALS, DESIGN AND CONSTRUCTION

FOUNDATION THICKNESS


MATERIALS

• SOILS• AGGREGATES• CEMENTS• CONCRETE• ASPHALT • LIGHTWEIGHT MATERIALS


Engineering Definition• Soil is considered to be

any loose sedimentary deposit, such as gravel, sand, silt, clay or a mixture of these materials. They are readily separated into relatively small pieces in which it may contain air, water or organic material (from vegetation).

• Rock is the naturally occurring material composed of mineral particles so firmly bonded together that great effort is required to separate the particle.


Main Soil Group

Granular Soils/Coarse-grained soils(Frictional)

Sand and Gravel

Fine-grained soils(Cohesive)

Silt and Clay

Organic soils Peat, organic clays, organic silt.


GRANULAR SOILS

• Excellent foundation materials for supporting structures and roads

• The best embankment materials• The best backfill material for retaining

walls• Might settle under vibratory load or blasts• Dewatering can be difficult due to high

permeability


Cohesive soils• Very often posses low shear strength• Plastic and compressible• Loses part of shear strength upon wet• Loses part of shear strength upon disturbance• Shrink upon drying and expand upon wetting• Poor material for backfill• Practically impervious• Clay slopes are prone to landslides• Difficult to compact (silt)


Organic Soils

• Any soil containing a sufficient amount of organic matter to influence its engineering properties is called an organic soil.


Organic matter

• Decayed vegetative and/or animal matter in various state of decomposition

OBJECTIONABLE BECAUSE……

1. REDUCES LOAD CARRYING CAPACITY

2. INCREASE COMPRESSIBILITY

3. FREQUENTLY CONTAIN TOXIC GASES THAT ARE RELEASED DURING EXCAVATION PROCESS.


Soil Analytical Representation• Soil is made up of various

sized particle packed together. Spaces between particles are known as voids.

• The voids usually contain a mixture of water and another part air or other gas. Or it can be completely water or air.

Air

Water

Solid

Void


Collapsible soils

• Soil deposit that experience significant decrease in volume when exposed to water.


Liquefaction• Liquefaction can occur

when saturated cohesionless sand deposit exist in relatively loose condition.

• When subjected to vibration or shock wave, the soil grain will quickly move into denser arrangement

• However water prevented the particle to particle contact causing shear strength lost.


Expansive clay• Clays containing

montmorillonite expand in volume when in contact with water.

• The expansive force created by clay undergoing an increase in water content is capable of lifting heavy structure and imposing excessive lateral pressure.

• Swell forces can go beyond 500kPa


Laterites• Residual soil formed from

weathering of igneous rock under conditions of high temperature and high rainfall.

• Usually reddish in color but not always.

• Agriculturally – not suitable because of low nutrients.

• Acceptable construction material for road bases and embankment or foundation.


Lightweight Geo-material


Road Applications

• Reduce loading consequently settlement and avoiding potential sliding and soil treatment requirements

• Minimizing earthwork operations on hill slopes and reducing load on ground and retaining systems.

TireShred Based Geocomposite

Soil Cover

Soft Soil Area


Hilly Slopes


RW and Bridge Abutment• As compressible layers to

induce active conditions for retaining wall, at the same time reduce lateral and vertical pressure.

• Control differential settlement, eliminate sudden bump and maintenance, reduce lateral and vertical vressure, thus reduce foundation requirements

Retaining Wall

Granular Backfill or Geocomposite


Bridge Abutment



Full Wall Pressure Reduction 5.00 5.50 6.00 6.50 7.00 7.50 8.00 8.50 9.00 9.50 10.00 10.50

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

Effective normal stressesExtreme effective normal stress -25.10 kN/m2

5.00 5.50 6.00 6.50 7.00 7.50 8.00 8.50 9.00 9.50 10.00 10.50 11.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

4.00

Effective normal stressesExtreme effective normal stress -11.07 kN/m2

11 kPaMax

25 kPaMax


ROAD FOUNDATIONThe Purpose of the foundation is to transfer the loading from the road to the soil or sub-grade.

The foundation for the purpose of ROAD CONSTRUCTION is defined as the sub-base and any associated strengthening materials

used.


Road Foundation

• If the road is new then the first step is the assessment of the sub-grade on which the road is to be built. If this is not carried out correctly then the information on which the designs are based may be incorrect.

• If the road is a not a new road then the first step is the assessment of the existing foundations.


Sub-grade Strength• The strength of the sub-grade (soil below the

pavement) is assessed using a test known as the California Bearing Ratio test. This was developed in California in the 1930's and makes no attempt to determine any of the standard soil properties such as density. It is merely a value and it is integral to the process of road design. Nearly all design charts for the road foundations are based on the CBR value for the sub-grade.


CBR VALUES

• To determine the CBR for a soil the designer has two options, they can either use the California Bearing Ratio (CBR) test if equipment is available or they can use the table below to estimate the CBR.


Type of soil

Plasticity index (%)

CBR (%) Depth of formation below water

table More than 600mm 600mm or less

Heavy clay 70 2 1*

60 2 1.5* 50 2.5 2 40 3 2

Silty clay 30 5 3 Sandy clay 20 6 4

10 7 5 Silt - 2 1* Sand (poorly graded) non-plastic 20 10 Sand (well graded) non-plastic 40 15 Well graded sandy gravel non-plastic 60 20

Table 1 Estimation of CBR values


CBR Test• The test involves the equipment shown below. The

plunger is then seated into the soil using a force of 50N for an expected CBR below 30% or 250N for greater than 30%. The plunger is then penetrated into the soil at a constant rate of 1mm/min and the forces recorded at penetration intervals of 0.25mm. The total penetration should not exceed 7.5mm. These results are then compared to a standard curve for a value of 100% CBR. The forces on the standard curve are 13.2kN at 2.5mm penetration and 20.0kN at 5.0mm penetration. The CBR is then a simple ratio of the corresponding values and where a difference between the value at 2.5mm and 5mm occurs, the higher value is taken. Annular weights are sometimes used to represent a surcharge.


Test Setup


Compaction

Densification of soil by removal of air using mechanical energy


Compaction vs. ConsolidationCONSOLIDATION:

REDUCTION OF SOIL VOLUME UNDER STRESS/LOADING. STRESS OR LOADING CAUSES

1) DEFORMATION OF SOIL PARTICLE

2) RELOCATION OF SOIL PARTICLE

3) EXPULSION OF WATER OR AIR ( IN MOST PRACTICAL CASES, ITS WATER)


Effect of moisture content• Water added act as

softening agent• Soil slip onto each

other and into a more dense packed position

• Beyond certain m%, dry unit weight is reduced because the moist/water takes up spaces of soil !!!


Optimum Moisture Content

Water

Moisture content, m

Unit Weight

γ=γd


Proctor compaction test

MOULD

Extension

HAMMER

25 BLOWS


Standard & Modified Proctor

Original Ilustration by Prof. Bengt B. Broms inFoundation Engineering

http://www.geoforum.com/misc/broms.asp




Test Concept

• Soil is mixed with varying amount of water then compacted.

• Unit wt =

W = weight of compacted soil in the moldVm = volume of mold

• Moisture content w% is determined in lab.

mVW

=γ


Plotting dry density against w%

• Dry density is calculated as:

• Plot the various γd vs. w%

100%1 wd

+=

γγ

w%

γd


S=100 and 80 curve• S=100% is the ZAV

curve – zero air void• ZAV plot can be

calculated using:

• Both ZAV and S80 can also be calculated using:

s

wzav

Gw 1+

=γγ

%1001×⎟⎟⎠

⎞⎜⎜⎝

⎛−=

sd

w

GSw

γγ

w%

γd

ZavS80


Specs. For Field Compaction

• Typical requirements: 90-95% MDD

• Specs. normally in term of relative compaction Cr.

( ) %100max

×=labd

dRC

γγ

Type of works Minimum CR

Building platform or road

90%

Upper 150mm of sub-grade below roadway

95%

Dam 100%


Site Procedure

• Soil is normally compacted in layers of 200-300mm.

• Constant check of field density should be carried out to ensure compliance.

• Rate of construction in embankment work should be control to prevent build up of pore pressure

• Heavier roller gives better compaction.


Site Procedure (cont’d)• For embankment work,

the best procedure is to compact a trial area and measure the dry density.

• When relative compaction is satisfactory, the number of roller passes is used for the actual embankment.


How to determine field density?

• Field density tests– Sand Cone Test/Sand Replacement Method– Rubber balloon test– Nuclear density test– Water ring test– Drive cylinder test


Sand Replacement Method

• ASTM D1556 or BS 1377

100%1 w

WW field

dry

+=

Compacted fill

Test hole filled with std. sand

VWdry

dry =γ


Example 1: Proctor compaction test1 2 3 4 5

Compacted Soil + mold 3762 3921 4034 4091 4040

Mass of tray 20.11 21.24 19.81 20.30 20.99

Mass of tray + wet soil 240.85 227.03 263.45 267.01 240.29

Mass of tray + dry soil 231.32 212.65 241.14 238.81 209.33

Mass of mold = 2031 g, Volume of mold = 9.44 E-4 m3

1) Compute γd and w% for each data point and plot results

2) Calculate S80 and S100 using Gs = 2.69

3) Determine MDD and wo


Calculate moisture content for each data points:

%100% ×−

−=

traydry

drywet

MMMM

w

%10011.2032.23132.23185.240

1 ×−−

=w

= 4.5 %

Moisture content


Dry density

mVW

=γ

3/22.17045.0199.17

100%1

mkNwd =+

=+

=γγ

Remember W = mass x gravity

Gravity = 9.81 m/s2

W1 = (3.762-2.031) x 9.81

= 16.981 kg.m/s2 or N

= 0.016981 kN

γ = 0.016981 / 9.44E-4

= 17.99 kN/m3


Summary of data points

1 2 3 4 5

γ (kN/m3) 17.99 19.64 20.81 21.41 20.88

w 4.5% 7.5% 10.1% 12.9% 16.4%

γd (kN/m3) 17.22 18.27 18.90 18.96 17.94


S80 and S100

s

wzav

Gw 1+

=γγ %1001

×⎟⎟⎠

⎞⎜⎜⎝

⎛−=

sd

w

GSw

γγ

Use γd = 16, 18 and 20 (i.e. within data points range)

%3.19%10069.21

1681.9%8080 =×⎥⎦

⎤⎢⎣⎡ −=w


S80 and S100 data points

γd S80 S100

16 19.3% 24.1%

18 13.9% 17.3%

20 9.5% 11.9%


Chart and Report

15

16

17

18

19

20

21

0 5 10 15 20 25 30

Moisture content (%)

Dry

Den

sity

(kN

/m3)

curves80s100

OMC

MDD


Example 2: Sand Replacement test

• Determination of Dry Density of soil on site

• BS 1377• Specs.

Requirement of Compaction ratio is 90% MDD of ex.#1

• Initial mass of sand: 5.912 kg• Remaining sand after

pouring: 2.378 kg• Mass of soil from hole: 2.383

kg• Moisture content w = 7.0 %• Density of sand: 1490 kg/m3• Volume of cone: 1.114E-3 m3• Calculate γd and Compaction

ratio.


Example 2: Solution

3

33

2

2

3333

33

)(

/36.1707.01

58.181

/58.1810258.110337.2

10337.21000

181.9383.2

10258.1)10114.1()10372.2(

10372.21490

534.3383.2

534.3378.2912.5

mkNw

mkNVW

kNgMW

mVVV

mV

M

M

d

hole

soil

soilsoil

coneholeconehole

holecone

wetsoil

holeconesand

=+

=+

=

=××

==

×=⎟⎠⎞

⎜⎝⎛××=×=

×=×−×=−=

×==

=

=−=

−

−

−

−−−+

−+

+

γγ

γ


Example 2:cont’d

• Relative compaction What if CR < 90%?

– Ripping, mixing and re-compacting

– Add water( )

OK

Clabd

dR

%36.91%1000.1936.17

%100max

=×=

×=γγ


Relative density• In place density

compared to its laboratory max and minimum density

• Max – compacted in laboratory

• Min – loosely filled into a test container

• Then compacted and loose density including void ration can be calculated

%100%minmax

max ×⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

eeee

D fieldR

⎥⎥⎦

⎤

⎢⎢⎣

⎡×⎥⎥⎦

⎤

⎢⎢⎣

⎡

−

−=

)(

(max)

(min)(max)

(min))(%fieldd

d

dd

dfielddRD

γγ

γγγγ

or


Typical values

Relative density (%)

Unit Weight (kN/m3)

Loose <35 <14

Medium dense 35-65 14-17

Dense 65-85 17-20

Very dense >85 >20


Pavement Structure

Wearing Course

Binder Course

Base Course

Sub Base Course

SUBGRADE


SUBGRADE

The uppermost part of soil, natural or imported, supporting the load transmitted from the overlying layers


Sub Base

• Serve as an aid to disperse load from the base course before transmitting to sub-grade. (some design omit this part)

Quary dust Sand


Base

• Specified material built up to certain thickness. This layer plays an important role in the support and dispersion of traffic loads

Crusher run


Binder

• Resisting shear + supporting and dispersing loads

Wearing

• Topmost layer. Direct contact with traffic load. Must resist abrasion and prevent skidding


Example:road pavement


Example: pavement details


Vertical alignment


Thickness Design

• CBR and the percentage of heavy vehicles are the main factors in thickness design of the pavement


Traffic EstimationAverage Daily Traffic (ADT) – both ways

Get the percentage of heavy commercial vehicles - Pc

Get annual traffic growth – r

Then:

Calculate Vo (initial annual traffic for one direction)

)100/(3655.0 co PADTV ×××=


Total number of commercial vehicles, Vc

( )[ ]rrVV

xo

c11 −+

=Vc = total number of commercial vehicles for x yrs

Vo = Initial yearly commercial traffic

r = rate of annual traffic growth


Check total traffic volume

( )xx rVV += 11

Vx = volume of daily traffic after x yrs in one direction

V1 = initial daily traffic in one direction

X = design period (yrs)


Maximum hourly traffic flow, c

c = I x R x Tc= maximum one way hourly capacity

I= ideal hourly capacity as in Table 3.2

R=roadway factor in Table 3.3

T=traffic reduction factor table 3.4

C = 10 x c

C = 24 hrs one way traffic capacity


Estimate equivalent factor,e

Percentage of heavy vehicles

0-15% 16-50% 51-100%

e Local1.2

Trunk2.0 3.0 3.7


Equivalent Standard Axle, ESA

ESA = Vc x e

BASED ON CBR AND ESA,

DETERMINE EQUIVALENT THICKNESS TA FROM CHART


Composite thicknesses

TA = a1D1 + a2D2 + …….anDn

a1, a2 = structural coefficient from table 3.5

D1, D2, Dn = thickness of each layer

pavement (turapan) - universiti sains malaysia 366/pavementslides.pdf · reg 366 hbp usm 2007...

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