ĐỒ án ĐƯỜNG 2 Đại Học Xây Dựng

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NATIONAL UNIVERSITY OF CIVIL ENGINEERING

HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

CHAPTER I: CALCULATION OF ROADBED SLOPE STABILITY

I. Some method for slope stablity
1. Mechanical roadbed slope stability.

Soil block will slide

1.1.Calculation method:

i
Sliding surface

i

Ti

Pi

Pi.cos

Considering the equilibrium condition of a slide i on its slide surface we have:

Sliding forces : T = Q .sin
i
i
i

Holding force, obstructing the soil sliding: Ni=Q .Cos .tg  + c di
i
i
i
cos i

In which :
 Q i - Weight of slice ( Qi  di * hi *  i *1m )
  , c,  : density, cohesion of soil, angle of internal friction.
Equilibrium conditions:
Ti  Ni or tg i  tgi 

c
 hi .cos 2 i

To put it simply Professor Maslov added safety factor K:
tg i 

1
c
(tg 
).
K
 hi

1
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BRIDGES AND ROAD FACULTY

HIGHWAY AND TRAFFIC DEPARTMENT
ROADBED AND PAVEMENT DESIGN PROJECT

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 Ti  Pi sin  i (sliding force).
 Normal force Ni  Pc
i os i (causing friction N i tg i with cohesion cili is the
holding force).

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

2
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HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

Factor of Safety:
i n

K=

 M igiu
i 1
i n

n

=

�( Nitgi  cili )

 M ilat
i 1

1

n

Z
(Ti  Wi i )
�
R
1

n



�( Pcos tg
i

i

1
n

i

 cili )

�( Pi sin i  Wi
1

Zi
)
R

.

2.2. Comment:
-Advantage:
 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) .

3.1. Calculation method.
- Same as Fellenius Method except that it includes normal forces E i+1 and Ei-1 along sides of
slices.
- Factor of Safety :
n
Ptg
( i i  ci li )mi
�
cos i
1

K  n1
with mi  (1  tgi .tg i ) 1
Z
K
�( Pi sin  i  Wi i )
1

R

3.2. Comment.
-Advantage:
 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.
3
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NATIONAL UNIVERSITY OF CIVIL ENGINEERING

HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

4.Conclusion.
- In the methods mentioned above:
� Selection classical method Fellenius used to calculate.
II. Detailed calculation Classical method Fellennius.
1.Input data
- Check stabilize of soft ground
- Geological data:

 ( g / cm )
1.95
3

Embankment
c (kg/cm2)  (˚)
1.2
29

 ( g / cm )
1.65
3

Subgrade
c (kg/cm2)  (˚)
0.96
17

Ground
Embankment
10m

2.Checking method.
2.1- Determining the monolithic stability factor Kođ.
- The monolithic stability factor is determined follow formula:
 K1 * K * K * K * K * K

2
3
4
5
PP
Kođ
In which:
o K1 - the reliability of the data on the mechanical characteristics of the soil (such as
cohesion c and friction angle) K1=1.0 – 1.1 � choose K1=1.1.
o K2 - factor consider the significance of roadbed , depending on road level; K 2=1.0 –
1.03 � K2=1.02.
o K3 - factor consider extent of causing damage to the national economy if road is
undermined disrupting traffic; K3=1.0 – 1.2 � K3=1.15.
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 K 5=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.

4
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STUDENT : SU DUY LINH - 59CDE - 588859

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NATIONAL UNIVERSITY OF CIVIL ENGINEERING

HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

2.2- Using classic methods to find Kmin.

- Finding Kmin. 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 n

K=

 M igiu
i 1
i n

M

ilat

i 1

n

=

�( Nitgi  cili )
1

n

Z
(Ti  Wi i )
�
R
1

n



�( Pcos tg
i

1
n

i

i

 ci li )

Z
( Pi sin  i  Wi i )
�
R
1

.

 Each sliding center gives Kmin value. Made with various sliding center will find
the coefficient Kmin.
 After that define Kmin required in min (the value of each center have found Kmin).
 Value Kmin newfound corresponding position which has the most dangerous
sliding 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 can
park around the roadbed distributed over 1m lengths . The load is equivalent to a
embankment whose height is determined by the formula hx hx 

n.G
.

 .B.l

5
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NATIONAL UNIVERSITY OF CIVIL ENGINEERING
BRIDGES AND ROAD FACULTY

HIGHWAY AND TRAFFIC DEPARTMENT
ROADBED AND PAVEMENT DESIGN PROJECT

 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.
3
o  - Soil density of embankment �   1.7 T/m .
o l- vehicle distribution range in the longitudinal direction m. � l = 4.2
m
o B- horizontal distribution width of vehicles, according to figures
� B = 6.9m.
� hx  0.51 m.
 Detailed calculations are shown in appendix.
III. Conclusion.
Result:

Safety factor K

B/2(O1)

B/4(O2)

B/8(O3)

2.572

3.38

3.603

The calculation process with the center O1, O2, O3 determining Safety factor
Kmin= 2.350.
Thus Kmin=2.350 > Kođ=1.2 � soft ground stability.

6
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STUDENT : SU DUY LINH - 59CDE - 588859


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NATIONAL UNIVERSITY OF CIVIL ENGINEERING
BRIDGES AND ROAD FACULTY

HIGHWAY AND TRAFFIC DEPARTMENT
ROADBED AND PAVEMENT DESIGN PROJECT

7

LECTURER : NGUYEN DUC NGHIEM
STUDENT : SU DUY LINH - 59CDE - 588859


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NATIONAL UNIVERSITY OF CIVIL ENGINEERING
BRIDGES AND ROAD FACULTY

HIGHWAY AND TRAFFIC DEPARTMENT
ROADBED AND PAVEMENT DESIGN PROJECT

8
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STUDENT : SU DUY LINH - 59CDE - 588859


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NATIONAL UNIVERSITY OF CIVIL ENGINEERING
BRIDGES AND ROAD FACULTY

CHAPTER II:

HIGHWAY AND TRAFFIC DEPARTMENT
ROADBED AND PAVEMENT DESIGN PROJECT

FLEXIBLE PAVEMENT DESIGN

Pavement is directly component subjected to the regular destruction of

vehicles and elements of the natural environment , it affects directly the quality of the
operation and exploitation of roads as well as construction costs. So pavement design
to ensure driving smoothly, economic, satisfies the technical requirements shall
require meticulous and precise. Therefore pavement must:
 There is enough strength to resist the shear stress deformation, bending and
stability of strength.
 Pavement must be flat to make vehicles running smoothly.
 Pavement must enough roughness to enhance the sticking coefficient between
the 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
Pressure on the pavement
P (kN)
p (Mpa)
100
0.6

120
0.6

Wheel streak diameter
D (cm)
33
36
9

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NATIONAL UNIVERSITY OF CIVIL ENGINEERING

HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

- Calculation time of pavement structure is taken by overhaul time of the most
sustainable surface layer, with class A1 layer 1 asphalt mediating particles T overhaul = 15
years, 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:
Nt = N0*(1+q)t .
Nt
� N0 
= 584 (veh/day)
(1  q)t
Table 1: Table of traffic volume over the time
Year
Ntt(veh/day)

0
257

5
414

10
668

15
1075

Passenger car

64

104

167

269

Light truck

64

104

167

269

Medium truck

77

124

200

323

26

41

67

108

26

41

67

108

Heavy truck with 2
axles
Heavy truck with 2
axles

2 Converted to calculate flow 100 kN
Traffic flow of different vehicles have converted to the standard axle loading
through cross-section of road at the end of the exploitation by the formula:

N



=  C1.C 2 .ni  Pi 
i 1
 Ptt 
k

tk

4. 4

(axles/day)

(3-1)
10

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NATIONAL UNIVERSITY OF CIVIL ENGINEERING

HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

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 :
Table 2
Axle loading Pi
(kN)
Front

Rear

Light truck

18

56

1

2-wheels

64

Medium truck
Heavy truck 2
axles

Heavy truck 3
axles

25.8

69.6

1

2-wheels

77

48.2

100.0

1

2-wheels

26

45.4

90.0

2

2-wheels

Type of vehicle

No of wheels
in rear grove

Rear axle
distance
(m)

No of
vehicle
ni
veh/day

No
of
rear

<3.0

26

Table 3 : Table of equivalent standard axle in years
In 0th year
Type of vehicle
Light truck
Medium
truck
Heavy truck

with 2 axles

Front
Rear
Front
Rear
Front
Rear
Front

Pi
(kN)
18
56
25.8
69.6
48.2
100.0
45.4

C1
1
1
1
1
1
1
1

C2

C1.C2.ni.()4,4

ni
6.4
1
6.4
1
6.4
1.0
6.4

64
64
77
77
26
26
26

*
10
2
25
16
63
12
11

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HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

Heavy truck
with 3 axles

Rear

90.0

ROADBED AND PAVEMENT DESIGN PROJECT

2.2

1.0

26

87

 Ntk = 217( axles / day)
In 5th year

Type of vehicle
Light truck
Medium
truck
Heavy truck
with 2 axles
Heavy truck
with 3 axles

Front
Rear
Front
Rear
Front
Rear
Front
Rear

Pi
(kN)
18
56
25.8
69.6
48.2
100.0
45.4
90.0

C1

C2

1
1
1
1
1
1
1
2.2

C1.C2.ni.()4,4

ni
6.4
1
6.4
1
6.4
1.0
6.4
1.0

168
168
168
168
84
84

84
84

*
13
3
34
22
84
17
116

 Ntk = 289 (axles /day )
In 10th year
Type of vehicle
Light truck
Medium
truck
Heavy truck
with 2 axles
Heavy truck
with 3 axles

Front
Rear
Front
Rear
Front
Rear
Front

Rear

Pi
(kN)
18
56
25.8
69.6
48.2
100.0
45.4
90.0

C1

C2

1
1
1
1
1
1
1
2.2

C1.C2.ni.()4,4

ni
6.4

1
6.4
1
6.4
1.0
6.4
1.0

224
224
224
224
112
112
112
112

*
17
4
45
29
112
22
155

 Ntk = 385 (axles /day )
In 15th year
Type of vehicle
Light truck

Front
Rear
Front

Pi
(kN)
18
56
25.8

C1
1
1
1

C2

C1.C2.ni.()4,4

ni
6.4
1
6.4

300
300
300

*

23
5
12

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NATIONAL UNIVERSITY OF CIVIL ENGINEERING
BRIDGES AND ROAD FACULTY

Medium
truck
Heavy truck
with 2 axles
Heavy truck
with 3 axles

Rear
Front
Rear
Front
Rear

HIGHWAY AND TRAFFIC DEPARTMENT
ROADBED AND PAVEMENT DESIGN PROJECT

69.6

1

1

300

61

48.2
100.0
45.4
90.0

1
1
1
2.2

6.4
1.0
6.4
1.0

150
150
150
150

39

150
30
208

 Ntk = 515 (axles /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 following
formula (3-1)
• f1 =0.55 :Loading distribution factor of the two-lane road
Table 4 : Table of calculate axle loading in years.
Year
Ntk (axles /day)
Ntt (axles /lane.day)

0
217
119

5
289
159

10
385
212

15
515
283

+ Determine the number of accumulated standard axle:
(1  q)15  1
N 15 365 = 1.605 10 6 (axle)
Naccumulated =
q (1  q )14

IV. Selecting grade of pavement.
+ With Ntt0 =119, Ntt5 =159, Ntt10 = 212 , Ntt15 = 283 (axles/day.lane) .
 Lookup table 3-4 and table 3-5( 22TCN211-06). We have selected pavement
grade and requirements of elastic modulus table are shown in the following
table:
Table 5 : Table of elastic modulus
No of axles
E yctt =max{Eyc,Emin}
Eyc
Emin
Class of
Calculated
Type axle/day.la
road
year
Mpa
Mpa
Mpa
ne
IV

0
A2
124.5
100
124.5
119
13
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HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

IV

5

A2

159

129.7

100

129.7

IV

10

A1

212

160.7

130

160.7

IV

15

A1

283

165

130

165

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

Loam

0.95

Relative
humidity
a=

W
Wnh

E
(Mpa)

Cohesion
C(Mpa)

Angle of
friction 
(degree)

42

0.096

17

0.60

+ Define the parameters for calculation of materials used pavement structure layer:
-With design speed of route 60(Km/h) , based on the organizational capacity of
contractors and construction materials along with local sources (hilly, and near major
rivers should take advantage of the raw materials gravel and rubble) so we can take
advantage of materials to make pavement and then we get the parameters calculated as
Table 7: The characteristics of pavement structure materials.

No

1
2
3

Materials

E (Mpa)
Rebound Shear

BendRku
C

deflection stress
tensile
(Mpa) (Mpa) (degree)
standard condition stress
t = 30oC t = 60oC condition

AC wearing course
9.5mm
AC binder course
12.5mm
Gravel-cement treated

420

300

1800

2.8

350

250

1600

2.0

600

600

600

0.8
14

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HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

aggregate
4 Graded aggregate type I

300

300

300

5 Graded aggregate type II
Natural aggregate
6
(gravels)
7
Ground: Loams

250

250

250

200

200

200

42

0.032

24

V. Method of pavement construction

1. The principles
- Design of the pavement on the principles of the general design of roadbed and
pavement, pavement must be tight and thermal stability.
- Have to take advantage of local materials, using the experience of road
construction and exploitation of local conditions.
- Pavement structure must conform to the mechanical construction and road
maintenance.
- The pavement must be sufficient strength, stability, good ebrasion resistant under
the effect of vehicle load and climate.
- The material in the structure must have strength decreases from top to bottom
matching stress distribution state to reduce costs.
- Structures do not have too many layers of complication for technological lines
construction.

2. Gathering Investment (15years)
2.1. Basic of selection:
Gathering Investment is the plan needs a large amount of initial capital to
construct a road reaching the standard with a 15 years life cycle. Road has V tt=
60(km/h) and by request of investors and lenders which we use A1 pavement using
asphalt concrete with life cycle is 15 years.
With road level III recommend using realibility factor 0.9. Thus strength
dv
coefficient K cd =1.10 ( for 2 alternatives ).

The minimum general elastic modulus:
tt

E chmin = K cddv . E yc =1.10 165=181.5 (Mpa)

So using Ntt15= 283 (axle/day.lane) , Ech =181.5 (Mpa) for calculation.

15
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HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

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
Ech2
Diagram:

E ch
E Ch1

E1 , h1

E Ch2

E2 , h2

E Ch3

E3 , h3
E4

,

E Ch4

h4
Subgrade E0 =

42

MPa

Surface course
`
1

Materials
AC wearing course

2

AC binder course

Ech=181.5 Mpa

hi (cm)
5

E(Mpa)
420

6

350

h1 5

 0.152

D 33
  Looking up mathematic diagram Kogan:
E ch 181.5

0.432

E1
420
Ech1
=0.43  Ech1 =195.72 Mpa
E1
h2
6

 0.182


D 33
  Looking up mathematic diagram Kogan:
E ch1 195.72

0.56 

E2
350

16
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HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

E ch2
= 0.598Ech2 = 209.3 Mpa
E2

So that Ech2 = 209.3 Mpa
2.2. Basement structure and select the basement alternatives

Basement has to ensure the requirements for strength, simplicity of construction
technology, concentrated local materials, lower costs, suitable for pavement and
surface level
2.2.1. Propose 2 alternatives.
+Alternative 1:
Layer

Material

Ech2 =158.2 Mpa

hi (cm)

Ei (Mpa)

3

Graded aggregate type I

300

4

Graded aggregate type II

250

Subgrade

E0 =42(Mpa)

Table calculate the thickness of the material layers of alternative 1
Table 8
Opti
ons

h3
(cm)

Ech2/E3

h3/D

Ech3/E3

Ech3
(Mpa)

1
2
3
4

15
16
17
18

0.689
0.689

0.689
0.689

0.455
0.485
0.515
0.545

0.773
0.778
0.784
0.789

232
233
235
237

Ech3/E4 E0/E4 H4/D
0.928
0.934
0.941
0.947

0.168
0.168
0.168
0.168

h4

(cm)

h4
Select

1.121 36.99
1.063 35.06
1.003 33.1
0.963 31.78

37
36
34
32

Table unit price of construction materials based on unit prices of Thai Nguyen
province in 2013
Table 9

No

Unit prices

Parame-

Material

Worker

Machine

Unit
17

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NATIONAL UNIVERSITY OF CIVIL ENGINEERING

HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

catalog

ter

prices

AD.112
22

Graded
aggregate type
I (New road).

100m3

17005000

240494

1222676

18468170

AD.112
12

Graded
aggregate type
II(New road)

100m3

15950000

213165

1192262

17355427

AD.121
11

Gravel-cement
treated
aggregate 6%

100m3

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
Table 10

Options

1
2
3
4

Graded-aggregate
type I

Graded-aggregate
type II

h3
(cm)

Prices
(dong/100m2)

h4
(cm)

Prices
(dong/100m2)

15
16
17
18

2770226
2954907
3139589
3324271

37

36
34
32

6421508
6247954
5900845
5553736

Total prices
(dong/100m2)
9191734
9202861
9040434
8878007

* Propose the options 3 with h3 = 18 cm, h4 = 34 cm with total prices
8878007 đ/100m2 for comparision.

+ Alternative 2:
Layer
Material

Ech2 =158.2(Mpa)

hi (cm)

Ei (Mpa)
18

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BRIDGES AND ROAD FACULTY

3
4

Gravel-cement treated
aggregate(6%)
Graded aggregate type II
Subgrade

HIGHWAY AND TRAFFIC DEPARTMENT
ROADBED AND PAVEMENT DESIGN PROJECT

600
250
Eo =42 (Mpa)

Table calculate the thickness of the material layers of alternative 2
Table 11
Opti
on

h3

(cm)

Ech2/E3

Ech3/E4

E0/E4

H4/D

3

Ech3
(Mpa)

1

12

0.35

0.364

0.46

273.6

1.094

0.168

0.901 29.73

30

2

13

0.35

0.394

0.47

280.2

1.121

0.168

0.838 27.65

28

3

14

0.35

0.424

0.48

286.2

1.145

0.168

0.772 25.47

26

4

15

0.35

0.455

0.49

292.2

1.171

0.168

0.701 23.13

24

h3/D

Ech3/E

h4
(cm)

h4
select

Table: Total price of alternative 2
Table 12
Gravel-cement
Graded aggregate
treated
type II
Total prices
aggregate(6%)
Option
(dong/100m2)
h3
Prices
h4
Prices
(cm) (dong/100m2) (cm) (dong/100m2)

1
12
4158418
30
5206628
9365046
2
13
4504952
28
4859520
9364472
3
14
4851487
26
4512411
9363898
4
15
5198022
24
4165303
9363325
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)

19
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NATIONAL UNIVERSITY OF CIVIL ENGINEERING

HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

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):
1  k .t 1 / 3 
Etb  E1 

 1 k 

In which

k=

3

(3-5).

h2
E2
; t = E ; htb = h1 + h2
h1
1

Calculation result:

Table 13
Layer
Gradedaggregate
type II
Gradedaggregate
type I
AC binder
course
AC wearing
course

Ei(Mpa)

t=

E2
E1

250

420

K=

H2
H1

htb=h2+h1(cm)

Etb' (Mpa)

32

250

50

267.30

56

275.45

61

285.81

32
300
1.2
250

300
350

hi(cm)

350
1.309
267.30
420
1.525
275.45

18
6
5

18
0.5625
32
6
0.12
50
4
0.0714
56

H/D = 61/33=1.85 looking up Table 3.6 (Page 134 22TCN211-06) :
H�
f�
� ��  = 1.2
�D �
So multi-layer structure is converted into 2-layer structure with above layer 60cm
thick with average elastic modulus : Etbdc . Etb' =1.2 �285.81 = 342.97(Mpa).
Using mathematic diagram Figure 3.1 determining general elastic modulus.
20
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HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

H 61

 1,85

D 33
  Looking up mathematic diagram Kogan

E0
42

0.1225

Etbdc 342.97
�

Ech
= 0.53
Etbdc

� Ech = Etbdc x 0.53 = 342.97 x 0.53 = 182.66 (Mpa)
dv
Check condition Ech ≥ K cd
.Eyc
tt
Therefore Ech =182.66(Mpa) > Echmin = K cddv . E yc =1.10 165=181.5 (Mpa)

Conclusion : Pavement structure satisfy the rebound deflection standard
b. Check the shear stress condition:
- Convert 2 layers to 1 layer by formula (3-5):
1  k .t 1 / 3 
Etb  E1 

 1 k 
h2

3

(3-5).

E2

k = h ; t = E ; htb = h1 + h2
1
1
The result is shown in Table 13 . With Etbdc = 342.97 Mpa.
Ctt
Checking condition: Tax  Tav �K tr
cd
In which:
Tax : maximum sheer stress (MPa)
Tav : is the shear stress due to selfweight of material layers located on at the point
(MPa)
K trcd : the coefficient of sliding shear strength selected depending on reliability
.With road level III, two lanes by selected reliability 0.9 � K trcd = 0.94



Ctt : Calculation cohesion of roadbed or less adhesive material (MPa) in the state of
humidity, density calculations.
Determine maximum sheer stress Tax

21
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BRIDGES AND ROAD FACULTY

HIGHWAY AND TRAFFIC DEPARTMENT
ROADBED AND PAVEMENT DESIGN PROJECT

H 61

 1.85

D 33

T
Etbdc 342.97


8.166 => Looking up mathematic diagram 3.3 [1] � ax 0,0127
p
E0
42


 24 0


� Tax = p x 0.0127 = 0,6x 0.0127 = 0.0076 (MPa).



Detemine the shear stress due to selfweight of material layers T av.
H 61cm
  Looking up the Diagram 3.4 [1] � Tav = - 0.0013(MPa).
 24 0 



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 incohesive
materials 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 incohesive
materials provided that its function in structure is different from samples,subgrade
is loams � K3 = 1.5.
� Ctt = C.K1.K2.K3 = 0.032 x 0.6 x0.8 x1.5 = 0.023 (Mpa)
Ctt
0.023
So: Tax + Tav = 0.0082 – 0.0013 = 0,0069(MPa) < K tr =
= 0.0245 (MPa)
0.94
cd
Conclusion : Pavement structure satisfy the shear stress condition.
c. Checking bended-tensile stress condition (15oC) :
Checking condition:

 ku

Rttku
� ku
K cd

In which :
ku
K cd

: the coefficient of sliding shear strength selected depending on reliability

tr
.With road level III, two lanes by selected reliability 0.9 � K cd = 0.94;

R ttku : The design bending bearing strength of solid mateial;
22
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BRIDGES AND ROAD FACULTY

HIGHWAY AND TRAFFIC DEPARTMENT
ROADBED AND PAVEMENT DESIGN PROJECT

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

+ For binder course layer :

monolithic

1800 5  1600 6
1690.91 (Mpa)
56

H1 = 11 (cm) ; E1 

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).
�H �
This value must also consider the adjustment coefficient   f � � with
�D �
H 50
 1.5151
D 33

Looking up Table 3-6 [1] =>  = 1.178.
dc
'
So Etb   .Etb = 1.178 x 267.30 = 314.88(MPa).

H 50

 1.5151

Ech.m

D 33
 Looking up Kogan Diagram � dc = 0.494

E0
42
Etb

0.1334
dc

314.88
Etb

That: Ech.m = Etbdc x 0.494 = 314.88 x 0.494= 155.55 (MPa).
Determine  ku:

 ku   ku . p.kb

In which:
ku : Maximum bending tensile stresses arising in the bottom of monolithic
material layers under the effect of the load wheel;
kb : coefficient that consider stress distribution properties in the pavement under
the effect of the calculation load is double or single wheel. I use the term should
double wheels grove kb = 0.85;

 ku : bending tensile stress unit;
p : tire pressure of standard loading axle , p = 0.6 (MPa).
H 1 11

 0.333


D 33
  Looking up mathematic diagram 3.5 [1]
E1
1690.91

10.87 
Ech ,m 155.55


 ku = 1.93(MPa)
23
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NATIONAL UNIVERSITY OF CIVIL ENGINEERING

HIGHWAY AND TRAFFIC DEPARTMENT

BRIDGES AND ROAD FACULTY

ROADBED AND PAVEMENT DESIGN PROJECT

Thus :  ku   ku . p.kb = 1.93x0.6x0.85 = 0.984(MPa).
+ For AC wearing course:
H1 = 4 (cm) ; E1 = 1800 (MPa)
Value E’tb of 3 under layers is determine following Table :

Table 14
Ei
(Mpa)

Layer
Graded-aggregate
type II
Graded-aggregate
type I
AC binder course

t

E2
E1

250
300
1600

hi
(cm)

K

h2
h1

Htb
E’tb

(cm) (MPa)

32
300
1.2
250
1600
5.986
267.30

18
6

18
0.5625
32
6
0.120
50

32

250

50

267.30

56

343.68

H 56
�H �
 1.697
Consider the adjustment coefficient   f � �with
D 33
�D �

Looking up Table 3-6 [1] :  = 1.187.
Thus

Etbdc   .Etb' = 1.187 x 343.68 = 407.95 (MPa).
H 56

 1.697

Ech.m
D 33
 Looking up Kogan diagram � dc = 0.485

E0
42
Etb

0.11
dc
407.95
Etb


Thus : Ech.m = Etbdc x0.485 = 407.95 x 0.485 = 197.86 (MPa).
+ Determine  ku:  ku   ku . p.kb
H
5

 0.121

D 33
  Looking up diagram Figure 3.5 [1] :  ku = 1.95 (MPa)
E1
1800

9.1

E ch.m 197.86

Thus :  ku   ku . p.kb = 1.95 x0.6x0.85 = 0.995 (MPa).
+ Determine R ttku :
24
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BRIDGES AND ROAD FACULTY

HIGHWAY AND TRAFFIC DEPARTMENT

ROADBED AND PAVEMENT DESIGN PROJECT

Calculation bending tensile strength of monolithic materials is determined by
formula:

Rttku  Rku .k1.k2
In which:
Rku : Bending tensile strength limit at calculations temperatures in calculation
perfusions under the effect of load is applied 1 times;
k2 : factor considering the declining strength over time compared with the agent of
climate and weather. With asphalt concrete type I take k2 = 1.0;
k1 : actor considering the strength decline due to material fatigue under the effect
of loads, k1 is taken according to the formula below:
11 .11
11 .11
k1  0, 22 
=0.524
Ne
(1.065 10 6 ) 0, 22

Thus:
Calculation bending tensile strength of AC binder course layer:

Rttku  Rku .k1.k2 = 2.0 x 0.524 x 1.0 = 1.048 (MPa)
Calculation bending tensile strength of AC wearing course lay er:

Rttku  Rku .k1.k2 = 2.8 x 0.524 x 1.0 = 1.468 (MPa)
ku
Coefficient K cd
= 0.94 (Table 3.7[1]) for road level III with realiability 0.9

+ Check for AC binder course :
 ku

Rttku 1.048
0.98  ku 
1.115( Mpa) � OK
0.94
k dc

+ Check for AC wearing course :
 ku 0.995 

Rttku 1.468

1.562( Mpa) � OK
0.94
k dcku

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 considered
phasing of investment matching the rule of the traffic growth as well as the
appropriate for time invested.

25
LECTURER : NGUYEN DUC NGHIEM
STUDENT : SU DUY LINH - 59CDE - 588859