ĐỒ án ĐƯỜNG 2 Đại Học Xây Dựng

- Calculation time of pavement structure is taken by overhaul time of the mostsustainable surface layer, with class A1 layer 1 asphalt mediating particles Toverhaul = 15years, so the cal

CHAPTER I: CALCULATION OF ROADBED SLOPE STABILITY

I Some method for slope stablity

1 Mechanical roadbed slope stability.

i i

d c

1.2 Comment:

- Advantage:

 Simple calculation, easy to understand

-Disadvantage:

 Less accuracy and reliability

 Scope of application is small, rarely used in practice

2 Classical Method (Ordinary Method).

2.1 Calculation method.

Considering the plane problem , circular cylinder sliding mass divided into slices as shown,assuming the soil block will slide at a time

Thus each slices will subject to the effects of self weight Pi ; Pi divided into two parts:

 Tangential force at the sliding surface T i P isini (sliding force).

 Normal force N i Pc i osi (causing friction N tg i i with cohesion cili is theholding force)

 Considering the effect of the earthquake, each piece was subjected to a slidingforces Wi , the lever arm compared with O is Zi

 Results calculated with relative accuracy.

 Apply popular in practice

-Disadvantage:

 Relatively complex calculations

 Assuming no exact reality (the pieces do not interact with each other)

3 Bishop Method (Simplified)

K

Z P

 Method for the most accurate results of the three methods

 Assumptions and calculation close to reality

-Disadvantage:

 Methods of finding groping Kmin

 Calculations quite long

- In the methods mentioned above:

� Selection classical method Fellenius used to calculate.

II Detailed calculation Classical method Fellennius.

2.1- Determining the monolithic stability factor K ođ

- The monolithic stability factor is determined follow formula:

o K4 - factor consider degree of fit between the calculated diagram with hydrogeological

at road construction site K4=1.0 – 1.05�K4=1.05

o K5 - factor consider type of soil and its work in roadbed structure K5=1.0 – 1.05

� K5=1.05

o Kpp - factor consider degree of reliability of calculation methods �Kpp=1.01

� The monolithic stability factor: Kođ=1.1x1.02x1.15x1.05x1.05x1.01=1.4.

2.2- Using classic methods to find K min

- Finding K min based on finding the position sliding center of the most dangerous sliding

surface based on the experience locus sliding center

- Detailed calculations:

 Determine the the experience locus sliding center

 On experience locus sliding center , take a point , at that point draw a slip with different radii, each radii will give stability coefficient K according to the formula:

i ilat

n i

i igiu

M M

 After that define Kmin required in min (the value of each center have found Kmin)

 Value Kmin newfound corresponding position which has the most dangeroussliding surface

2.3 Assessment and conclusions:

-From Kmin compared with Kođ if:

 Kmin > Kođ � roadbed slope stability

 Kmin < Kođ � roadbed slopes unstable , should propose measures to strengthen

3 Detailed calculations.

 The calculation is done with 3 sliding center O1, O2, O3 with different radii

 In the calculation process, taking into account vehicle weight following II.4.3 [2]

 Loads are considered the load of maximum heavy vehicles at the same time canpark around the roadbed distributed over 1m lengths The load is equivalent to aembankment whose height is determined by the formula hx x ..

n G h

B l



 In which :

o G- The heaviest weight of a vehicle (T) �G=13T.

o n- The maximum number of vehicles put on the road, n=2

o  - Soil density of embankment � 1.7 T/m3

o l- vehicle distribution range in the longitudinal direction m � l = 4.2m

o B- horizontal distribution width of vehicles, according to figures

CHAPTER II: FLEXIBLE PAVEMENT DESIGN

Pavement is directly component subjected to the regular destruction ofvehicles and elements of the natural environment , it affects directly the quality of theoperation and exploitation of roads as well as construction costs So pavement design

to ensure driving smoothly, economic, satisfies the technical requirements shallrequire meticulous and precise Therefore pavement must:

 There is enough strength to resist the shear stress deformation, bending andstability of strength

 Pavement must be flat to make vehicles running smoothly

 Pavement must enough roughness to enhance the sticking coefficient betweenthe wheels and the pavement

 Minimize the amount of dust caused by the pavement, avoiding pollution,pavement must have good abrasion resistance

Suitable with the construction capability of local area

The design standards:

- Highway − Specifications for design TCVN 4054-2005 [1]

- Flexible pavement design standard 22TCN211-06 [2]

- Pavement Material Specifications [3]

- Highway design volume IV, [4];

I Loads and Times (22TCN 211 – 06) [2]

- Standard axle loading :

 a single axle, double-wheel grove

 P=100kN

 p=0.6Mpa

 D=33cm

Table 1: The characteristics of standard axle loading

Standard axle loading

- Calculation time of pavement structure is taken by overhaul time of the mostsustainable surface layer, with class A1 layer 1 asphalt mediating particles Toverhaul = 15years, so the calculation time of pavement structure is 15 years

II Traffic flow and vehicle components

Traffic flow crossing 2 points A27-B27 in the year 15th : 1500 veh/day

+ Vehicle component:

25% passenger car (Volga)

25% light truck (Gaz-51)

30% medium truck (Zil-150)

20% heavy truck

Traffic volume growth rate : q = 10%

III Converted to calculate traffic flow 100kN:

1 The formula for calculating the trafic flow:

2 Converted to calculate flow 100 kN

Traffic flow of different vehicles have converted to the standard axle loadingthrough cross-section of road at the end of the exploitation by the formula:

Ntk =

4 4

i

(axles/day) (3-1)

In which:

• Ntk : Number of equivalent standard axle loading (axles/day)

• ni: Number of real axle loading of vehicle i

•C1: Axle factor which is determined following formula (3-2):

C1=1 + 1.2(m-1) (3-2)

With the distance of axle grove <3m, when distance between 2 axle groves>3m, have

to convert separately for each axle

•C2: the factor that consider effect of the number of wheels in one grove: if 1 wheel

C2=6.4;with 2 wheels C2=1.0; with 4 wheels C2= 0.38

Thus equivalent traffic flow to standard loading axle over calculation years are shown in the following table :

No of wheels

in rear grove

Rear axle distance (m)

No of vehicle

n i veh/day

+ Determine the number of axles calculated on a lane

Two-lane road so traffic flow is calculated reality:

Ntt= Ntk f1 (axle/day) (3-3)

In which :

• Ntk : Number of equivalent standard axle loading (axles/day) is determined followingformula (3-1)

• f1 =0.55 :Loading distribution factor of the two-lane road

Table 4 : Table of calculate axle loading in years

+ Determine the number of accumulated standard axle:

Naccumulated = 365

)1(

1)1(

15 14

Table 5 : Table of elastic modulus

E yc Mpa

E min Mpa

tt yc

IV 5 A2 159 129.7 100 129.7

Value Eyc , Emin in accordance with Table 3-4 and Table 3- 5(22TCN211-06)

+ Defining the calculation parameters of the ground:

Based on the type of land (survey results), the ground is loam soil with

mechanical properties as the table below

Table 6 : Table of characteristic parameters of the ground

Type

of soil Compactness

Relative humidity a=

Wnh W

E (Mpa)

Cohesion C(Mpa)

Angle of friction 

(degree)

+ Define the parameters for calculation of materials used pavement structure layer: -With design speed of route 60(Km/h) , based on the organizational capacity ofcontractors and construction materials along with local sources (hilly, and near majorrivers should take advantage of the raw materials gravel and rubble) so we can takeadvantage of materials to make pavement and then we get the parameters calculated as

Table 7: The characteristics of pavement structure materials

E (Mpa)

R ku (Mpa)

C (Mpa)



(degree)

Rebound deflection standard

t = 30 o C

Shear stress condition

t = 60 o C

tensile stress condition

4 Graded aggregate type I 300 300 300

5 Graded aggregate type II 250 250 250

With road level III recommend using realibility factor 0.9 Thus strength coefficient dv

Surface layer of pavement was selected as follows

Layer 1: AC wearing course 9.5mm E1 =420 Mpa, h1 = 5 cm

Layer 2: AC binder course 12.5mm E2 = 350 Mpa, h2 = 6 cm

From these figures we determine a general elastic modulus of basement and subgrade

181.5

E

E

0.152 33

195.72

E

E

0.182 33

ch2 2

= 0.598Ech2 = 209.3 Mpa

So that E ch2 = 209.3 Mpa

2.2 Basement structure and select the basement alternatives

Basement has to ensure the requirements for strength, simplicity of constructiontechnology, concentrated local materials, lower costs, suitable for pavement andsurface level

2.2.1 Propose 2 alternatives

+Alternative 1:

h 4 (cm)

h 4 Select

catalog ter prices

AD.112

22

Gradedaggregate type

3 15950000 213165 1192262 17355427

AD.121

11

Gravel-cementtreatedaggregate 6% 100m

3 28987135 1585073 4081272 34653480

To compute the cost of the alternatives , we have divided into the erection

layer, the thickness of this layer suitable for rolling equipment

Often for the layer thickness h = 20-40cm is divided into 2 classes: Above layer 0.4h ; Under layer 0.6h

We have total prices of alternatives in the following table

Table of total prices of alternative 1

Prices (dong/100m 2 )

h 4 (cm)

Prices (dong/100m 2 )

E ch3 (Mpa) E ch3 /E 4 E 0 /E 4 H 4 /D

h 4 (cm)

h 4 select

Graded aggregate

(dong/100m2)

h 3 (cm)

Prices (dong/100m 2 )

h 4 (cm)

Prices (dong/100m2)

Conclusion: By comparing the price of the alternatives chose alternative 1with

h3 = 18 cm, h4 = 32 cm with total prices 8878007 đ /100m2

Pavement structure in gathering investment:

Eyc=165 (Mpa)

H1= 5cm AC wearing course E1 = 420 (MPa)

H2 = 6cm AC binder course E2 = 350 (MPa)

H3=18 cm : Graded-aggregate type I, E3 =300 (MPa)

H4 = 32cm: Graded-aggregate type II, E4 = 250 (MPa)

Subgrade E0= 42 (MPa)

2.2 Checking the pavement structures.

a Check the Rebound deflection standard:

- Convert 2 layers to 1 layer by formula (3-5):

3 3 / 1 1

1

1

So multi-layer structure is converted into 2-layer structure with above layer 60cm

thick with average elastic modulus : E tb dc  '

tb

E =1.2� 285.81 = 342.97(Mpa)

Using mathematic diagram Figure 3.1 determining general elastic modulus

85 , 1 33

E E

Conclusion : Pavement structure satisfy the rebound deflection standard

b Check the shear stress condition:

- Convert 2 layers to 1 layer by formula (3-5):

3 3 / 1 1

1

1

Tax : maximum sheer stress (MPa)

Tav : is the shear stress due to selfweight of material layers located on at the point(MPa)

85.133

61



cm H

Looking up the Diagram 3.4 [1] � Tav = - 0.0013(MPa)

 Detemine Ctt

Value Ctt is determined following formula : Ctt = C.K1.K2.K3

In which :

C : bonding strength of embankment or incohesive materials C= 0.032 (MPa)

K1: the factor that shows the decrease of slide sheer resistance of soil or incohesivematerials under moving and shaking load : K1=0.6;

K2: the factor that shows elements causing structural inhomogeneity Because of

Ntt =254.1 (axle/day.lane)< 1000 so that following Table 3.8 [1] �K2 = 0.8;

K3: the factor that shows the increase in slide sheer resistance of soil or incohesivematerials provided that its function in structure is different from samples,subgrade

= 0.0245 (MPa)

Conclusion : Pavement structure satisfy the shear stress condition

c Checking bended-tensile stress condition (15 o C) :

Checking condition:

ku tt

ku ku cd

R K

ku : Maximum bending tensile stresses arising in the bottom of monolithicmaterials layer under the effect of the load wheel.

+ For binder course layer :

H1 = 11 (cm) ; 1690 91

6 5

6 1600 5 1800

Value E’tb of 2 layers graded-aggregate type I and graded-aggregate type II

E’tb = 267.3 (MPa) (follow the result in Table 13) with the thickness of 2 layers H’ =

18 + 32 = 50 (cm)

This value must also consider the adjustment coefficient  f � �� �H D

� �

 with5151

1

42

5151 1 33

ku

 : bending tensile stress unit;

p : tire pressure of standard loading axle , p = 0.6 (MPa)

91.1690

333.033

h K h

(cm)

E’ tb (MPa)

42

697 1 33

1800

121.033

Calculation bending tensile strength of monolithic materials is determined byformula:

k1 : actor considering the strength decline due to material fatigue under the effect

of loads, k1 is taken according to the formula below:

1 0 , 22 6 0 , 22

) 10 065 1 (

11 11 11

11

ku 1 2

tt ku

R R k k = 2.8 x 0.524 x 1.0 = 1.468 (MPa) Coefficient K cd ku = 0.94 (Table 3.7[1]) for road level III with realiability 0.9

+ Check for AC binder course :

)(115.194.0

048.198

k

R ku dc

ku tt

+ Check for AC wearing course :

)(562.194.0

468.1995

k

R ku dc

ku tt

Conclusion : The asphalt concrete layer ensures bending-tensile stress conditions

3 Phasing of investment (2 stages):

- Phasing of investment is a part of an overall plan, must always be consideredphasing of investment matching the rule of the traffic growth as well as theappropriate for time invested

- Base on the structure of gathering of investment , it is studying the possibility ofinvesting diverging in two stages The early years, traffic volume is small constructtemporarily 2 basement layers, with surface treatment using asphalt layer.

 Pavement structure of phasing of investment plan:

3 3 / 1 1

'

1

1

Therefore : E’tb = 267.30 (MPa) Consider the adjustment coefficient  f � �� �H D

.

314

42

5151

a Check rebound deflection standard.

- Calculation trafic volume: Ntt= 5

tt

N =159 (axle/day,lane) with Eyc=129.7 (Mpa) Check condition: Ech > K cd dv.Eyc

We got: Ech = 155.55 (MPa) > 1.1 x 129.7 = 142.67 (MPa) � OK

Conclusion : Pavement structure ensures rebound deflection standard.

b Check shear stress condition.

We got : E’TB = 267.30 (MPa) Consider the adjustment coefficient  f � �� �H D

� �

 với

5151 1 33

88.314

5151.133

50



cm H

Looking up diagram figure 3.4 [1] � Tav = - 0.0011 (MPa)

+ Detemine Ctt

Value Ctt is determined following formula : Ctt = C.K1.K2.K3

In which :

C : bonding strength of embankment or incohesive materials C= 0.032 (MPa)

K1: the factor that shows the decrease of slide sheer resistance of soil or incohesivematerials under moving and shaking load ,K1=0.6;

K2: the factor that shows elements causing structural inhomogeneity Because of

Ntt =192.17 (axle/day.lane)< 1000 so that following Table 3.8 [1] �K2 = 0.8;

K3: the factor that shows the increase in slide sheer resistance of soil or incohesivematerials provided that its function in structure is different from samples,subgrade

General conclusion: Pavement structures of first 5-year ensures the bearing

capacity.

2 Stage II (Next 10 years) :

- The strength of pavement down 10% so elastic modulus Ech of pavementremaining: Ech1 = 0.9x155.55 = 140(MPa)

-To upgrade pavement A1 , we add AC wearing course 5cm and AC binder course7cm

Structure of new pavement: