Highway design raport(final) (group 2)

Highway Design Senior Project 2010

Part- I

Section-1: Introduction

There is a growing universal demand for well prepared professionals in all disciplines. In
addition, increased pressure has consequently been placed in educational institution to prepare
the required number of qualified professional to fulfill society’s need. It is imperative that there
is a large need in the industry for engineers with training and experience, and the academic
should move successfully to fill the need. This is especially true for in the situation of Ethiopia
where there is a lack of well trained and experienced urban engineer’s.

Therefore, the integration of academic program and exposing students to more practical project
results in well-seasoned and, well-educated professionals.

Thus, this high way design project is intended to equip the students with practical design
reinforcing what they have attained theoretically in the class.

It is already known that, for rapid economic, industrial and cultural growth of any country, a
good system of transportation is very essential. One of the transportation systems that are
economical for developing countries like Ethiopia is road. A well – designed road network plays
an important role in transporting people and other industrial products to any direction with in
short time. Roads, to satisfy their intended purpose, must be constructed to be safe, easy,
economical, environmentally friend and must full fill the needs of inhabitants. Being safe, the
number of accidents that can occur will be minimized. Easiness decreases operation cost,
pollution and even time cost. Economical roads assure their feasibility according to their plans
and initiate further construction of roads. Schemes that do not satisfy the needs of localities may
not get the maximum utilization of the surplus man power that is really to exist in the rural
community and also its economical value may also decrease. Therefore, from this project it is
expected to understand and to get acquainted with the above facts by going through on the
following design aspects.

1.1 General Background

1

ECSC, IUDS, Urban Engineering Department (UE)


This high way design project is taken from the Hargele - Afder – Bare - Yet road project, which
is located in the Eastern part of the country in Somali National Regional State, Afder
Administrative Zone, Afder and Bare Woredas. The project is intended to facilitate the existing
and for the expected traffic load in the future, because the town is developing.

From this road we have given a stretch of 3 km emanating from station 12+500 to 15+500 for
this project to do geometric and pavement design in general.

1.2 Objectives

This final year design project on high-way has the following major objectives:-

 To expose the prospective graduates to a detail and organized design on road projects;
 To implement the knowledge that the prospective graduates have learned theoretically in
classes;
 To ensure a good carrier development;

1.3 Brief Description of The Project Area

The Hargele - Afder – Bare - Yet road project, is located in the Eastern part of the country in
Somali National Regional State, Afder Administrative Zone, Afder and Bare Woredas. The
project starts at Hargele (5º13’N and 42º 11’E) and pass through Hargele, Afder, Bare, town and
ends at Yet. The project length is estimated to be 142.4km. The Location map together with the
topographic map of the project area is shown below.

Fig. 1.1 Project Location Map

2



Location of the Project Road

Fig. 1.3.3 Digitized Proposed Project Alternative Alignments

3



Climate:

One of the environmental factors that affect performance of pavements structures is climate.
Hence, climate data of the project area mainly rainfall intensity, in terms of mean monthly and
mean annual and, temperature are required. According to the map shown on National Atlas of
Ethiopian Atlas, the project area is located in the region of the lowest annual rainfall. The mean
annual rainfall in this region is 300mm per year. The rainfall of the project area is characterized
by the following rainfall distribution:

 April, May and October  The wettest Months
 And in the remaining months  The driest months.

Topography:

The terrain of the project area through which the road alignment traverses is rolling in substantial
section of the project which is intercepted by mountainous terrain in some sections.

Potential of the area:

In the project area limited crop production, livestock and livestock products are available in the
area of influence of the road project even though the area is under attention to reverse food
deficit. There is an initiative to change the area that the potential resources of oil mining and salt
production may attract private investors and governmental agencies.

1.4 Scope of the project

The scope of the project goes as far as designing the geometry and pavement of a given road
section, with its appropriate drainage structures.

4



Section-2: Geometric design

2.1 Geometric design Control and Criteria

2.1.1 Terrain classification

2.1.1.1 Contour generation

The surveying data x, Y and Z coordinate taken from the road corridor using Hand Held GPS are
converted to a contour using GIS software.

2.1.1.2 Selection of center line

The center line of the road is delineated on the given road corridor using the contour elevations
by considering to have minimum earth work along the corridor.

2.1.1.3 Transverse terrain property
In order to know the type of the terrain along the selected center line or corridor, we took
horizontal distance perpendicular to the center line and vertical elevation measurements across
the road. Each measurement is taken longitudinally along the rod at 20m interval to get better
terrain classification. The values obtained are summarized in index table 2-1.
Slop= (vertical elevation / horizontal elevation)*100

Therefore, we generalize the following terrains classification along the road corridor:

STATION

From To TERRAIN AVG. SLOPE
CLASSIFICATION (%)

12+ 500 12+ 760 Rolling 23.14

12 + 760 13+ 080 Mountainous 26.63

13 + 080 13+ 520 Rolling 18.75

13 + 520 13+ 820 Mountainous 32.234

13 + 820 15 +500 Rolling 16.87

Table 2-2 Terrain Classification

5



Fig 2-1 Generated contour.

2.1.2 Design traffic volume

6



2.1.2.1 Traffic data analysis

In order to design the road, traffic data analysis is very important. Therefore, the secondary data
of traffic analysis we get from the project site comprises traffic volume before design, during
implementation and up to the design life time of the road. As the secondary data shows the
project life is 15 year. The traffic volume data and the design life time are expressed in the
following table.
T&
Year Car 4 WD S/ Bus L/ Bus S/ Truck M/ Truck L/ Truck TOTAL
T

2008 0 4 6 2 12 4 2 14 44

2009 0 5 7 2 13 5 3 16 51

2010 0 5 7 2 14 5 3 16 52

2011 0 6 8 3 14 5 3 17 56

2012 0 6 8 3 15 5 3 18 58

2013 0 15 16 6 31 20 28 34 149

2014 0 16 17 7 34 21 30 37 160

2015 0 19 19 8 36 22 32 39 174

2016 0 19 21 8 38 25 35 41 184

2017 0 19 21 9 40 26 36 44 193

2018 0 20 22 9 43 28 38 46 205

2019 0 21 25 11 44 31 42 49 221

2020 0 22 26 11 47 32 44 52 232

2021 0 22 26 12 49 34 46 53 241

2022 0 22 29 12 52 35 48 56 253

2023 0 25 30 13 55 36 51 59 267

2024 0 25 32 13 57 39 54 60 279

2025 0 26 33 14 60 40 57 64 292

2026 0 27 34 14 62 43 60 67 307

2027 0 28 37 16 66 44 63 70 323

Table 2-3 Traffic data analysis

From the above data,

o Traffic volume when the road open =149 veh/day
o Traffic volume at the end of the project life =323 veh/day
7



2.1.3 Road functional classification

Some of the factors which affect road design control and criteria are functional classification of
the road. In Ethiopian case, we have five functional classes based on AADT and importance of
the road.

Since, AADT of the project lies between 200-1000, and the road expected to serve centers of
provisional importance, the road could be main access road (class II).

2.2 Geometric Design Standard

Based on the traffic data obtained from the above table we decide the project design standard to
be (DS4).

Because:-

a) Even if the AADT at the opening of the road (2013) is 149 veh/day it will be greater than
200 veh/ day after five year and it is 323 veh/day at the end of design life (15 years). So it
fulfills the requirements of DS4. Since the recommended traffic volume for DS4 is 200-
1000 veh/day.(ERA)
b) The second reason is that since the area is an oil mining area, we expect the road will
accommodate the expected traffic volume during the design life time.
c) Based on the above reason, we decide the road to be DS4, to get full knowledge from the
whole project since the project is for academic purpose.
Therefore, we took the entire design element based on DS4. Refer the above information from
ERA manual Table 2.1.

From Design Standards vs. Road Classification and AADT table of ERA for DS4,

AADT=200 – 1000 vehicle/day

Surface type = paved

Carriageway = 6.7m

Shoulder width =1.5m for rolling

= 0.5m for mountainous
8



Design speed = 70km/hr for rolling

= 60km/hr =for mountainous

2.2.1 Horizontal Alignment
Based on our proposal of the center line of the road, we have tangents and curves. The curves are
curve1, curve2, curve3, curve4, curve5, and curve6.

Based on our terrain classification, the curves fall in to different terrain classification that leads
us to determine the radius and different elements of each curve.

Curve Terrain type

Curve 1 Rolling

Curve 2 Rolling

Curve 3 Rolling

Curve 4 Error! Not a
valid link.

Curve 5 Rolling

Curve 6 Rolling

Table 2-4 Horizontal curves and their terrain classification

Since our road is DS4, the minimum radius of each curve based on the terrain is:-

Minimum horizontal radius = 175m for rolling

= 125m for mountainous

Refer the following table for the rest of the design elements of DS4 (ERA standards)

Design Element Unit Flat Rolling Mountainous Escarpment Urban/Peri- Urban

Design Speed km/h 85 70 60 50 50

Min. Stopping Sight Distance m 155 110 85 55 55

Min. Passing Sight Distance m 340 275 225 175 175

% Passing Opportunity % 25 25 15 0 20

Min. Horizontal Curve Radius m 270 175 125 85 85

9



Transition Curves Required Yes Yes No No No

Max. Gradient (desirable) % 4 5 7 7 7

Max. Gradient (absolute) % 6 7 9 9 9

Minimum Gradient % 0.5 0.5 0.5 0.5 0.5

Maximum Super elevation % 8 8 8 8 4

Crest Vertical Curve k 60 31 18 10 10

Sag Vertical Curve k 36 25 18 12 12

Normal Cross fall % 2.5 2.5 2.5 2.5 2.5

Shoulder Cross fall % 4 4 4 4 4

Right of Way m 50 50 50 50 50

Table 2-5: Table 2-6 of ERA Geometric Design Parameters for Design Standard DS4 (Paved)

2.2.1.1 Horizontal curve elements

Curve-1 Design computation

a) Terrain type = Rolling

b) Deflection angle Δ = 390 (by measurement)

c) Point of intersection P.I=12+717.4m

d) Calculation of radius of the curve

Vd 2
Rmin =
127(e + f )

Where, Rmin=minimum radius

Vd=70km/hr…………….ERA, table 2.6

ed= 8% (max design super elevation rate, ERA, table 2.6)

f=0.14 (ERA. Table 8.1 for ed=8%)

70 2
Then, Rmin = =175.3m
127(0.08 + 0.14)

10



The calculated Rmin has no significant change from the recommended in ERA manual standard
(i.e., 175m), in addition to this, in order to minimize cut and fill, we use R min=175m from the
standard.

Therefore, radius of curve=Rc=175m

e) Tangent (T1)

 ∆
T1 = R * tan 
2

 39 
T1 = 175 * tan   = 61.97 m
 2 

f) Point of curvature (PC)

P.C1= P.I1 - T1

=12+717.4 – 0+061.97

=12+655.43m

g) Length of the curve (L)

 2Π 
L1 = ∆ * R *  
 360 

 2Π 
L1 = 390 *175 *   = 119.12m
 360 

h) Point of tangency (P.T)

P.T1= P.C1+L1

=12+655.43+119.12

=12+774.55m

i) External distance (E)

11



 ∆ 
E1 = R * sec  − 1
 2 

  39  
E1 = 175 * sec  −1 = 10.65m
  2  

j) Middle ordinate (M)

  ∆ 
M 1 = R * 1 − cos 
  2 

  39 
M 1 = 175 * 1 − cos  = 10.04m
  2 

k) Chord (Chord from P.C to P.T)

∆
C1 = 2 R sin  
2

 39 
C1 = 2 *175 * sin   = 116.83m
 2 

Fig.2.2 elements 0f curve-1
12





b) Deflection angle Δ = 330 (by measurement)



Vd 2
Rmin =
127(e + f )





70 2
127(0.08 + 0.14)

(i.e., 175m), in addition to this to prevent overlaps with curve 3, we use Rmin=175m from the
standard.


e) Tangent (T1)

Rmin = 175m

 ∆
T2 = R * tan  
2

 33 
T2 = 175 * tan   = 51.84m
 2 


13



P.C2= P.I2 – T2

=13+150.43– 0+051.84

=13+098.59m


 2Π 
L2 = ∆ * R *  
 360 

 2Π 
L2 = 330 *175 *   = 100.79m
 360 


P.T2= P.C2+L2

=13+98.59+100.79

=13+199.38m


 ∆ 
E2 = R * sec  − 1
 2 

  33  
E2 = 175 * sec  −1 = 7.52m
  2  


  ∆ 
M 2 = R * 1 − cos 
  2 

  33 
M 2 = 175 * 1 − cos  = 7.21m
  2 


14



 ∆
C2 = 2 R sin  
2

 33 
C2 = 2 *175 * sin   = 99.41m
 2 

Fig 2.3 elements of curve-2



b) Deflection angle Δ = 59.620 (by measurement)



Vd 2
Rmin =
127(e + f )




15



70 2
127(0.08 + 0.14)

(i.e., 175m), in addition to this to prevent overlaps with curve 2, we use R min=175m from the
standard.


e) Tangent (T3)

Rmin = 175m

∆
T3 = R * tan  
2

 59.62 
T3 = 175 * tan   = 100.26m
 2 


P.C3= P.I3 - T3

=13+363.64– 0+100.26

=13+263.38m


 2Π 
L3 = ∆ * R *  
 360 

 2Π 
L3 = 59.62 0 *175 *   = 182m
 360 


16



P.T3= P.C3+L3

=13+263.38+182m

=13+445.38m


 ∆ 
E3 = R * sec  − 1
 2 

  59.62  
E3 = 175 * sec  −1 = 26.69m
  2  


  ∆ 
M 3 = R * 1 − cos 
  2 

  59.62 
M 3 = 175 * 1 − cos  = 23.17 m
  2 


∆
C3 = 2 R sin  
2

 59.62 
C3 = 2 *175 * sin   = 173.99m
 2 





17



Vd 2
Rmin =
127(e + f )





70 2
Then, Rmin = =175.37 m
127(0.08 + 0.14)

(i.e., 175m), so we use Rmin=175m from the standard.

But to make the curve smooth, we took R=236m, I.e. =RC=236m

e) Tangent (T4)

R = 236m

∆
T4 = R * tan 
2

 90.81 
T4 = 236 * tan   = 239m
 2 


P.C4= P.I4 – T4

=14+045.5– 0+239

=13+806.5m


 2Π 
L4 = ∆ * R *  
 360 

18



 2Π 
L4 = 90.810 * 236 *   = 374.m
 360 


P.T4= P.C4+L4

=13+806.5+374m

=14+180.5m


 ∆ 
E4 = R * sec  − 1
 2 

  90.81  
E4 = 236 * sec  −1 = 100.12m
  2  


  ∆ 
M 4 = R * 1 − cos 
  2 

  90.810 
M 4 = 236 * 1 − cos
 2  = 70.31m

  


 ∆
C4 = 2 R sin  
2

 90.810 
C 4 = 2 * 236 * sin 
 2  = 336.10 m

 



19






Vd 2
Rmin =
127(e + f )





70 2
127(0.08 + 0.14)

standard.


e) Tangent (T5)

Rmin = 175m

∆
T5 = R * tan 
2

 44.15 
T5 = 175 * tan   = 70.97 m
 2 


P.C5= P.I5 – T5

=14+756.69– 0+70.97m

20



=14+685.72m


 2Π 
L5 = ∆ * R *  
 360 

 2Π 
L5 = 44.150 *175 *   = 134.85m
 360 


P.T5= P.C5+L5

=14+685.72+134.85m

=14+820.57m


 ∆ 
E5 = R * sec  − 1
 2 

  44.150  
E5 = 175 * sec
  −1 = 13.84m

  2  


  ∆ 
M 5 = R * 1 − cos 
  2 

  44.15 
M 5 = 175 * 1 − cos  = 12.83m
  2 


∆
C5 = 2 R sin  
2

21



 44.150 
C5 = 2 *175 * sin 
 2  = 131.54m

 






Vd 2
Rmin =
127(e + f )





70 2
127(0.08 + 0.14)

standard.


e) Tangent (T6)

Rmin = 175m

 ∆
T6 = R * tan  
2

22



 32.48 
T6 = 175 * tan   = 50.97 m
 2 


P.C6= P.I6 – T6

=15+226.73m – 0+050.97

=15+175.76m


 2Π 
L6 = ∆ * R *  
 360 

 2Π 
L6 = 32.48 0 *175 *   = 99.20m
 360 


P.T6= P.C6+L6

=15+175.76m +99.20m

=15+274.96m


 ∆ 
E6 = R * sec  − 1
 2 

  32.480  
E6 = 175 * sec
  −1 = 7.27 m

  2  


  ∆ 
M 6 = R * 1 − cos 
  2 

23



  32.480 
M 6 = 175 * 1 − cos
 2  = 6.98m

  


∆
C6 = 2 R sin  
2

 32.48 0 
C6 = 2 *175 * sin 
 2  = 97.88m

 

2.2.1.2 Transition curve

When a vehicle traveling on a straight course enters a curve of finite radius, and suddenly
subjected to the centrifugal force which shock and sway. In order to avoid this it is customary to
provide a transition curve at the beginning of the circular curve having a radius equal to infinity
at the end of the straight and gradually reducing the radius to the radius of the circular curve
where the curve begins.

Mostly transition curves are introduced between:-

A/ between tangents and curves

B/ between two curves

Various forms of transition curves are suitable for high way transition, but the one most popular
and recommended for use is spiral.

Design of transition curve

Even if there are places to design transition curve, ERA design manual standard recommends
where and how to design this horizontal alignment design elements. Especially for Ethiopian
road, transition curves are a requirement for trunk and link road segments having a speed equal
to or greater than 80km/hr. (ERA)

But the characteristics of our project road segment is;-

Speed=60km/hr (for mountainous terrain)

24



Speed=70km/hr (for rolling terrain)

Terrain= mostly rolling and mountainous

Functional classification=Main access road.

Therefore, based on the ERA standard all curves in the project will not have transition curve. So,
it will be a simple curve with out transition curve.

2.2.1.3 Super elevation

Curve-1

When a vehicles moves in a circular path, it is forced radially by centrifugal force. The
centrifugal force is counter balanced by super elevation of the road way and/or the side friction
developed between the tire and the road surface. The centrifugal force is the result of design
speed, weight of car, friction, and gravitational acceleration having the following relation ship.

Wv 2
Fc =
gR

Where, Fc= centrifugal force

W=weight of the car

V=design speed

g= acceleration due to gravity

R= radius of the curve

So, super elevation rate is changing the road cross section from the normal road to elevate
towards the center of the curve. I.e., it counteracts a part of the centrifugal force, the remaining
part being resisted by the lateral friction.

Terms in super elevation:

 Tangent run out(Lt)
 Super elevation runoff(Lr)

25



Tangent run out (Lt)

It is the longitudinal length along the road designed to remove the adverse crown to a zero slope.
i.e., the outer edge of the road is raised from a normal cross slope to a zero slope which equal to
the grade level of the road (the level of the center line of the road).

Super elevation runoff length (Lr)

Super elevation run-off is a length of the road section from the point of removal of adverse
crown of the road to the full super elevated point on the curve.

Super elevation is equal to the length of transition curve when there is a transition curve. When
there is no transition curve i.e., when it is a simple curve,1/3 rd of the length is placed on the curve
and 2/3rd of the length is placed on the tangent part(ERA). Therefore, we follow the second
standard to design our super elevation since all the curves do not have transition curve.

Design computation

A/ computation of super elevation run-off

Super elevation runoff length can be obtained from table 8.5 (ERA) using radius (Rc) and super
elevation rate (e), or it can be computed from the following formula. (AASHTO)

Lr =
( wn1 ) ed ( b )
w
G

Where,

Lr=minimum super elevation run-off (m)

G=maximum relative gradient (percent)

n1=number of lanes rotated

Bw=adjustment factor for number of lane rotated

w=width of one traffic lane (in our case, w/2)

ed=design super elevation rate, percent

26



Then, n1=1, since the number of lane rotated is =1(AASHTO, exhibit 3-31)

bw=1, for one lane rotated(AASHTO, exhibit 3-31)

G=0.55%, for Vd=70km/hr (AASHTO, exhibit 3-31)

Design speed(Km/h)(Vd) Maximum relative Equivalent maximum relative
gradient(%)(G) slope (%)

20 0.80 1:125

30 0.75 1:133

40 0.70 1:143

50 0.65 1:150

60 0.60 1:167

70 0.55 1:182

80 0.50 1:200

90 0.45 1:213

100 0.40 1:227

110 0.35 1:244

120 0.30 1:263

130 0.25 1:286

Table2-6 (Exhibit 3-27 Maximum relative gradients of AASHTO)

 6 .7 
 *1 * 0.08
Therefore,  2 
Lr = (1) = 48.87m
0.55

But ERA recommends Lr=52m for ed=8% and Rc=175m. Thus, take Lr=52m

B/ computation of tangent run out (Lt)

Tangent run-out can be computed using the following equation. (AASHTO)

eNC
Lt = * ( Lr )
ed

Where,
27



Lt =minimum length of tangent run-out

eNC=normal cross slope rate, percent

ed =design super elevation, percent

Lr=super elevation runoff length

0.025
Then, Lt = * ( 52 ) = 16.25m
0.08

C/ Location of super elevation run-off (Lr)

Since there is no transition curve (spiral) between the tangent and the curve in the project, 2/3 rd
of the super elevation length is placed on the tangent and 1/3rd of the length is placed on the
curve part.

1
i.e., * 52 =17.33m (on the curve)
3

2
* 52 = 34.67 m (On the tangent)
3

Then,

The beginning of the super elevation runoff length is:-

=P.C-34.67m

=12+655.43-0+034.67

=12+620.76m

The end of the super elevation runoff length is:-

=P.C+17.33m

=12+655.43+0+017.33m

=12+672.76m

28



D/ location of tangent run-out length

Beginning=beginning of Lr minus Lt

=12+620.76-16.25m

=12+604.51m

End=12+620.76m

E/ station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%

It is a length(R) where total crown removal is attained.

So, R=2*Lt

=2*16.25

=32.50m,

Then, the station is,

Beginning= station of beginning of adverse crown removal

=12+604.51m

End=station of beginning of adverse crown removal plus +R

=12+604.51+32.50m

=12+637.01m

On the same process we can do the super elevation at the exit of the curve.

We know that the length of curve 1=119.12m

Then the part of the curve to be full super elevated is

=119.12-2*(1/3*Lr)

=119.12-2*(1/3*52)

=84.46m
29



F/ Then, the station of end of full super elevation is

=12+672.76+84.46m

=12+757.22m

G/ station of end of super elevation runoff is

=12+757.22+52m

=12+809.22m

H/ station of recovering adverse crown is

=12+809.22+16.25m

=12+825.47

Attainment of full super elevation:-

From three methods attaining full super elevation we use the method in which rotating the
surface of the road about the center line of the carriageway, gradually lowering the inner edge
and raising the upper edge, keeping the center line constant.

Illustration:

30



Fig.2-4 Attainment of super elevation

Based on the above super elevation attainment, the results are shown on the following figure.

31



Fig.2-5 Super elevation at entrance and exit for curve 1



Lr =
( wn1 ) ed ( b )
w
G

n1=1, since the number of lane rotated is =1(AASHTO, exhibit 3-31)


G=0.55%, (AASHTO, exhibit 3-31)

 6.7 
 *1 * 0.08
Lr = * (1) = 48.78m
0.55



32




eNC
Lt = * ( Lr )
ed

0.025
Then, Lt = * ( 52 ) = 16.25m
0.08


curve part.

1
i.e., * 52 =17.33m (on the curve)
3

2
3

Then,


=P.C-34.67m

=13+98.59-0+034.67

=13+63.92m


=P.C+17.33m

=13+98.59+0+017.33m

=13+115.92m


33




=13+63.92 -16.25m

=13+47.67m

End=13+63.92m



So, R=2*Lt

=2*16.25

=32.50m,

Then, the station is


=13+047.67m


=13+47.67m +32.50m

=13+080.17m


We know that the length of curve-2=100.79m


=100.79-2*(1/3*Lr)

=100.79-2*(1/3*52)

=66.12m

34



=end of Lr+L

=13+115.92 +66.12m

=13+182.04m

G/ station of end of super elevation runoff is

=13+182.04 +52m

=13+234.04m

H/ station of recovering adverse crown are:

=13+234.04+16.25m

=13+250.29m




Lr =
( wn1 ) ed ( b )
w
G



G=0.55%, (AASHTO, exhibit 3-31)

 6.7 
 *1 * 0.08
Lr = * (1) = 48.78m
0.55



35




eNC
Lt = * ( Lr )
ed

0.025
Lt = * ( 52 ) = 16.25m
0.08


curve part.

1
i.e., * 52 =17.33m (on the curve)
3

2
3

Then,


=P.C-34.67m

=13+263.38 -0+034.67 m

=13+228.71m


=P.C+17.33m

=13+263.38 +0+017.33m

=13+280.71m



36



=13+228.71-16.25m

=13+212.46m

End=13+228.71m



So, R=2*Lt

=2*16.25

=32.50m,



=13+212.46m


=13+212.46+32.50m

=13+244.96m


We know that the length of curve 3=182m


=182-2*(1/3*Lr)

=182-2*(1/3*52)

=147.33m


=13+280.71m +147.33m
37



=13+428.04m

G/ station of end of super elevation runoff is:

=13+428.04 +52m

=13+480.04m

H/ station of recovering adverse crown is:

=13+480.04+16.25m

=13+496.29m




Lr =
( wn1 ) ed ( b )
w
G



G=.55%, for Vd=70km/hr, (AASHTO, exhibit 3-31)

 6.7 
 *1 * 0.08
Lr = * (1) = 48.7m
0.55

But from ERA for ed=8% and v=70m/sec, by interpolation Lr=49.12m for Rc=236m. Thus, take
Lr=49.12m



38



eNC
Lt = * ( Lr )
ed

0.025
Lt = * ( 49.12 ) = 15.35m
0.08


curve part.

1
i.e., * 49.12 =16.37 m (on the curve)
3

2
* 49.12 = 32.75m (On the tangent)
3

Then,


=P.C-32.75m

=13+806.5-0+032.75

=13+773.75m


=P.C+16.37

=13+806.5+0+016.37m

=13+822.87m



39



=13+773.75 -15.35m

=13+758.4m

End=13+839.25m



So, R=2*Lt

=2*15.35

=30.7m



=13+823.39m


=13+823.39m +30.70m

=13+854.10m




=374-2*(1/3*Lr)

=374-2*(1/3*49.12)

=341.25m

40



=13+822.87+341.25m m

=14+164.12m


=14+164.12m +49.12m

=14+213.24m


=14+213.24 +15.35m

=14+228.59m




Lr =
( wn1 ) ed ( b )
w
G




 6.7 
 *1 * 0.08
Lr = * (1) = 48.78m
0.55




41



eNC
Lt = * ( Lr )
ed

0.025
Lt = * ( 52 ) = 16.25m
0.08


curve part.

1
i.e., * 52 =17.33m (on the curve)
3

2
3

Then,


=P.C-34.67m

=14+685.72m -0+034.67m

=14+651.05m


=P.C+17.33m

=14+685.72+0+017.33m

=14+703.05m



42



=14+651.05-16.25m

=14+634.80m

End=14+651.05m

E/ Station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%


So, R=2*Lt

=2*16.25

=32.50m,

Then, the station is;

Beginning=station of beginning of adverse crown removal

=14+634.80m


=14+634.80m +32.50m

=14+667.30m




=134.35-2*(1/3*Lr)

=134.35-2*(1/3*52)

=99.68m


=14+703.05m +99.68m
43



=14+802.73m

G/ station of end of super elevation runoff are:

=14+802.73m +52m

=14+854.73m


=14+854.73m +16.25m

=14+870.98m


A/ computation of super elevation run-off:


Lr =
( wn1 ) ed ( b )
w
G




 6.7 
 *1 * 0.08
Lr = * (1) = 48.78m
0.55


44





eNC
Lt = * ( Lr )
ed

0.025
Lt = * ( 52 ) = 16.25m
0.08


curve part.

1
i.e., * 52 =17.33m (on the curve)
3

2
3

Then,


=P.C-34.67m

=15+175.76m -0+034.67m

=15+141.10m


=P.C+17.33m

=15+175.76m +0+017.33m

=15+193.10m

45





=15+141.10m -16.25m

=15+123.85m

End=15+123.85m



So, R=2*Lt

=2*16.25

=32.50m,



=15+123.85m

End=station of beginning of adverse crown removal plus + R

=15+123.85m +32.50m

=15+156.35m




=99.20-2*(1/3*Lr)

=99.20-2*(1/3*52)

=64.53m
46




=15+193.10+64.53m

=15+257.63m


=15+257.63m +52m

=15+309.63m


=15+309.63m +16.25m

=15+325.88m

Super elevation overlaps:

The end of tangent run out (super elevation runoff length) for curve 2 and the beginning of
tangent run out (super elevation runoff length) of curve 3 overlaps with an amount of:

Over lap= (13+250.29)-(13+212.46)

=42.83m

Therefore, this overlap length has to distribute on the curve part of each curve according to the
following.

Half of the overlap distance has to be added to the part of the curve. I.e. if the overlap length is d,
the part of super elevation on the curve will be

=1/3rd (Lr) +d/2

=17.33+42.83/2m

=38.475m

But this length has to be 40% of length of the corresponding curve.

47



Check:

Lc of curve 2=100.79m

Then, 40%*100.79=40.32>38.745m…………….OK!

Lc of curve 3=182m,

Then, 0.4*182=72.8>38.475m………………………OK!

Re-adjustment for super elevation stations.

Curve-2

1. The beginning of the super elevation runoff length is:-

=P.C-(34.67-21.415) m

=13+98.59-(0+013.25)

=13+085.34m

2. The end of the super elevation runoff length is:-

=P.C+17.33m

=13+98.59+ (0+017.33+21.415) m

=13+137.34m

3. Location of tangent run-out length


=13+085.34m -16.25m

=13+069.09m

End=13+085.34m

4. Station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%

48



So, R=2*Lt

=2*16.25

=32.50m,



=13+069.09m


=13+069.09m +32.50m

=13+101.59m


We know that the length of curve-2=100.79m


=100.79-2*(1/3*Lr+21.415)

=100.79-2*(1/3*52+21.415)

=23.29m

5. Then, the station of end of full super elevation is

=end of Lr+23.29

=13+137.34m +23.29m

=13+160.63m

6. Station of end of super elevation runoff is

=13+160.63+52m

=13+212.63m
49



7. Station of recovering adverse crown is:

=13+212.63m +16.25m

=13+228.88m

Curve-3

1. The beginning of the super elevation runoff length is:-

=P.C-(34.67-21.415) m

=13+263.38 – (0+013.25) m

=13+250.13m

2. The end of the super elevation runoff length is:-

=P.C+ (17.33+21.415) m

=13+263.38 + (0+38.75) m

=13+302.13m

3. Location of tangent run-out length


=13+250.13m -16.25m

=13+233.88m

End=13+250.13m

4. Station where outer and inner edge of the road will have the same normal cross fall i.e., 2.5%


So, R=2*Lt

=2*16.25

=32.50m,
50





=13+250.13m


=13+250.13m +32.50m

=13+282.63m




=182-2*(1/3*Lr+d/2)

=182-2*((1/3*52) +42.83/2)

=104.50m

5. Then, the station of end of full super elevation is

=13+302.13m +104.50

=13+406.63m

6. Station of end of super elevation runoff is:

=13+406.63m + 52m

=13+458.63m

7/ station of recovering adverse crown is:

=13+458.63m +16.25m

=13+474.88m

51



Fig 2-6 profile, section and station of super elevation, tangent run out for all curves

52



STATIONS

CURVE NUMBER
A B C D E F G H

Curve 1 12+604.51 12+620.76 12+637.01 12+672.76 12+757.22 12+792.97 12+809.22 12+825.47

Curve 2 13+069.09 13+085.34 13+101.59 13+137.34 13+160.63 13+196.38 13+212.63 13+228.88

Curve 3 13+233.88 13+250.13 13+282.63 13+302.13 13+406.63 13+442.38 13+458.63 13+474.88

Curve 4 13+756.4 13+773.75 13+789.10 13+822.87 14+164.12 14+197.89 14+213.24 14+228.59

Curve 5 14+634.80 14+651.05 14+667.30 14+703.05 14+802.73 14+838.48 14+854.73 14+870.98

Curve 6 15+123.85 15+141.10 15+156.35 15+193.10 15+257.63 15+293.38 15+309.63 15+325.88

Table 2-7 stations of super elevation, tangent run out for all curves.

53



2.2.1.4 Curve widening

Widening on a curve is giving extra width on a road curves. This is because:-

 It has been found that the drivers on curves have difficulty in steering their
vehicles to outer edge of road as they are able to on the straight because the rear
wheels do not follow precisely the same path as the front wheels when the
vehicles negotiates a horizontal curve or makes a turn.
 Also there is psychological tendency to drive at greater clearance, when passing
vehicle on curved than on straights. Hence, there is dire necessity for widening
the carriage way on curves.
 On curves the vehicles occupy a greater width because the rear wheels track
inside the front wheels.
Analysis of extra widening on horizontal curves

When vehicles negotiate a curve, the rear wheel generally do not follow the same track as
that of the front wheels. It has been observed that except at very high speed, the rear axle
of a motor vehicles remains in line with the radius of the curve. Since the body of the
vehicle is rigid, therefore, the front wheel will twist themselves at one angle to their axle,
such that vertical plane passing through each wheel is perpendicular to the radius of the
curve in order to trace the path on the curve. This is known as ‘off tracking’.

To determine width (W) it is necessary to select an appropriate design vehicle. The
design vehicle should usually be a truck because the off tracking is much greater for
trucks than for passenger car. (AASHTO) There fore, widening on horizontal curves
depend on:

 The length and width of the vehicle
 Radius of curvature

54



Fig 2-7 widening of pavements on horizontal curves

Let;

L= length of wheel base of vehicle in m.

b=width of the road in m,

w=extra width in m,

R1=radius of the outer rear wheel in m,

R2= radius of the outer front wheel in m,

n=number of lanes

Rc= radius of curvature

The formula obtained from the above geometries for extra widening for more than one
lane (mechanical widening) is:-

n * L2
mechanical..widening = wm =
2 * Rc

The extra widening needed for psychological reasons mentioned above is assumed as:-

55



v
psycho log icalwidening = w p =
10 Rc

There fore, total widening w will be:-

n * L2 v
w= +
2 * Rc 10 Rc

Widening attainment on curves

The following rules apply for attaining widening on both ends of the curve. (AASHTO)

A. widening should be done gradually and has to be realized on the inside edge of un-
spiraled curve (on simple curve) pavements.

B. In the case of a circular curve with transition curves, widening may be applied on the
inside edge or divide equally on either side of the center line.

C. On highway curves without transition curves widening should preferably be attained
along the length of super elevation runoff. A smooth fitting alignment would result from
attaining widening on-one half to two-third along the tangent and the remaining along the
curve.

D. Widening is not necessary for large radius greater than 250m.

Curve-1, 2, 3, 5, and 6 Design computations

Design data: Rc = 175m, n=2

L= take 6m (for the design vehicle usually a truck, corresponding to AASHTO, Single
unit (SU))

V=70m/sec

n * L2 v
w= +
2 * Rc 10 Rc

2 * 62 70
w= + = 0.73m
2 *175 10 175
56



For all curves having a radius between 120 to 250m ERA recommends a minimum of
widening width equal to 0.6m. But we recommend the calculated value 0.73m. So, all the
curves will have the corresponding value unless they are no less than the recommended
value by ERA. Therefore, this widening will be introduced at the inner edge of the
curves. Because all the curves are un spiraled curves.

Fig2-8.widening of pavement on curves

WIDENING STARTING STARTING LAST PT OF END POINT REMARK
WIDTH(M) POINT OF POINT OF FULL OF
WIDENING FULL WIDENING WIDENING
WIDENING

0.73 12+620.76 12+672.76 12+757.22 12+809.22 12+620.76

Table 2-8 widening stations for curve 1


Design data: Rc=236m, N=2, L= take 6m, V=70m/se

57



n * L2 v
w= +
2 * Rc 10 Rc

2 * 62 70
w= + = 0.61m
2 * 236 10 236

CUR WIDENI STARTING STARTING LAST PT OF END POINT
VE NG POINT OF POINT OF FULL OF
NO. WIDTH( WIDENING FULL WIDENING WIDENING
M) WIDENING
C1 0.73 12+620.76 12+672.76 12+757.22 12+809.22

C2 0.73 13+085.34 13+137.34 13+160.63 13+212.63

C2 0.73 13+250.13 13+302.13 13+406.63 13+458.63

C3 0.73 13+839.25 13+822.87 14+164.12 14+213.24

C4 0.61 14+651.05 14+703.05 14+802.73 14+854.73

C5 0.73 15+141.10 15+193.10 15+257.63 15+309.63

C6 0.73 12+620.76 12+672.76 12+757.22 12+809.22

Table2-9 Widening length and stations for all curves.

2.2.1.4 Site distance

Another element of horizontal alignment is the site distance across the inside of the
curves. Sight distance is the distance visible to the driver of a passenger car or the
roadway ahead that is visible to the driver. For highway safety, the designer must provide
sight distances of sufficient length that drivers can control the operation of their vehicles.
They must be able to avoid striking an unexpected object on the traveled way.

Where there are site obstruction( such as walls, cut slops, buildings and longitudinal
barriers) on the inside of curves or the in side of the median lane on divided highways, a
design may need adjustment in the normal high way cross section or change in the
alignment if removal of the obstruction is impractical to provide adequate site distance.
Because of the many variables in alignment, in cross section and in the number, type and
58



location of potential obstructions, specific study is usually need for each individual curve.
With site distance for the design speed as a control, the designer should check the actual
conditions on each curve and make the appropriate adjustment to provide adequate
distance.

Two-lane rural highways should generally provide such passing sight distance at frequent
intervals and for substantial portions of their length.

Stopping site distance

Stopping sight distance is the distance required by a driver of a vehicle traveling at a
given speed to bring his vehicle to a stop after an object on the road way becomes visible.
The minimum stopping sight distance is determined from the following formula, which
takes into account both the driver reaction time and the distance required to stop the
vehicle. The formula is:

d= (0.278) (t) (v) +v2/ 254f

Where:

d = distance (meter)

t = driver reaction time, generally taken to be 2.5 seconds

V = initial speed (km/h)

F = coefficient of friction between tires and roadway (see Table 7-1)

OR the stopping site distance is given in ERA manual in the following table.

Design Speed Coefficient Stopping Sight Passing Sight Reduced Passing
Sight Distance
(km/h) of Friction (f) Distance (m) Distance (m) for design (m)
from formulae

20 0.42 20 160 50

30 0.40 30 217 75
59



40 0.38 45 285 125

50 0.35 55 345 175

60 0.33 85 407 225

70 0.31 110 482 275

85 0.30 155 573 340

100 0.29 205 670 375

120 0.28 285 792 425

Table 2-10: Sight Distances

The coefficient of friction values shown in Table 2-10 have been determined from test
using the lowest results of the friction tests. The values shown in the third column of the
above table for minimum stopping sight distance are rounded from the above formula.
For the general use in the design of horizontal curve, the sight line is a chord of the curve,
and the stopping site distance is measured along the center line of the inside lane around
the curve.

The horizontal site line offset needed for clear site areas that satisfy stopping site distance
can be derived from the geometry for the several dimension explained in the following
figure.

60



Fig 2-9 Site distance for horizontal curves

Relevant formulae are as follows:

∆
Siteline( S ) = 2 R sin
2

 ∆
Middle..ordinate(d ) = R1 − cos 
 2

Where ∆ = Deflection angle

R=radius (from the center line of the inner lane)

Design computation

Using the above formulas the stopping site distance(d), the line of site(S) and middle
ordinate(M) of each horizontal curves can be calculated from the data’s of each curve
organized in the following table below.

driver
deflection Radius speed(V) reaction Coefficient of
curve no time
angle(D) (R),m km/hr friction(f)
(t) in sec.
Curve 1. 39 173.325 70 2.5 0.31

Curve 2. 33 173.325 70 2.5 0.31

Curve 3. 59.62 173.325 70 2.5 0.31

Curve 4. 90.81 234.325 70 2.5 0.31

Curve 5. 44.15 173.325 70 2.5 0.31

Curve 6. 32.48 173.325 70 2.5 0.31

Table 2-11 different data about each curve

∆
Siteline( S ) = 2 R sin
2

61



 ∆
Middle..ordinate( d ) = R1 − cos 
 2

v2
Stoppingsitedist..(d ) = 0.278vt +
254 f

Curve Site line (S) Middle Stopping site distance(m)
in m. ordinate (M)
in m. Calculated Recommended by
distance in m ERA

curve 1 115.714 9.94 510.55 110

curve 2 98.454 7.14 510.55 110

curve 3 172.329 22.93 510.55 110

curve 4 333.72 69.81 510.55 110

curve 5 130.278 12.76 510.55 110

curve 6 96.945 6.92 510.55 110

Table2-12 Site distance elements

62



Fig 2-10 stopping site distance of curve 1

Passing site distance

Passing sight distance is the minimum sight distance on two-way single roadway roads
that must be available to enable the driver of one vehicle to pass another vehicle safely
without interfering with the speed of an oncoming vehicle traveling at the design speed.
Within the sight area the terrain should be the same level or a level lower than the
roadway. Otherwise, for horizontal curves, it may be necessary to remove obstructions
and widen cuttings on the insides of curves to obtain the required sight distance. The
passing sight distance is generally determined by a formula with four components, as
follows:

d1 = initial maneuver distance, including a time for perception and reaction

d2 = distance during which passing vehicle is in the opposing lane

d3 = clearance distance between vehicles at the end of the maneuver

d4 = distance traversed by the opposing vehicle

The formulae for these components are as indicated below:
63



d1 = 0.278 t1 (v – m + at1/2)

Where,

t1 = time of initial maneuver, s

a = average acceleration, km/h/s

v = average speed of passing vehicle, km/h

m = difference in speed of passed vehicle and passing vehicle, km/h

d2 = 0.278 vt2

Where,

t2 = time passing vehicle occupies left lane, sec.

v = average speed of passing vehicle, km/h

d3 = safe clearance distance between vehicles at the end of the maneuver, is dependent on
ambient speeds as per Table 7-2 of ERA standard:

Table 7-2: Clearance Distance (d3) vs. Ambient Speeds

Speed Group (km/h)

Speed group(km/hr) 50-65 66-80 81-100 101-120

D3(m) 30 55 80 100

d4 = distance traversed by the opposing vehicle, which is approximately equal to 2/3 rd of
d2 whereby the passing vehicle is entering the left lane, estimated at:

d4 = 2d2/3

The minimum Passing Sight Distance (PSD) for design is therefore:

PSD = d1+ d2 + d3 + d4

64



Even if it is calculated using the above formula ERA recommends passing site distance,
so we use the value given by ERA design manual.

Sample calculation

Curve 1

Data:

Design speed=70km/hr=v of passing vehicle

Assume the following values

T1=3.5 sec, T2=3sec, a=1.0m/sec2

V of passing vehicle=70km/hr

V of passed vehicle=65km/hr

i.e., m=70-65=5km/hr

Then,

d1= 0.278 t1 (v – m + at1/2)

d1 = 0.278 *3.5* (70 – 5 + (1*3)/2) =64.71m

d2= 0.278 vt2= 0.278 *70*3 =58.38m

d3=55m, for design speed group=66km/hr-80km/hr

d4= 2d2/3 = (2*58.38)/3 =38.92m

Therefore, total passing site distance is,

PSD=d1+d2+d3+d4 = Error! Not a valid link.Error! Not a valid link.Error! Not a valid
link.Error! Not a valid link. =218.95m

65



Fig 2-11 Components of passing maneuver used in passing site distance.

2.2.2 Design of vertical alignment

The two major aspects of vertical alignment are vertical curvature, which is governed by
sight distance criteria, and gradient, which is related to vehicle performance and level of
service. The purpose of vertical alignment design is to determine the elevation of selected
points along the roadway, to ensure proper drainage, safety, and ride comfort. So it is
important to use different series of grades and to create a smooth transition between these
grades parabolic curves are used. The vertical alignment includes:

 Joining the grades with smooth curve.
 Location of appropriate gradients.
2.2.2.1 Design consideration

2.2.2.1.1 Gradient and grade controls

Changes of grade from plus to minus should be placed in cuts, and changes from a minus
grade to a plus grade should be placed in fills.Highway should be designed to encourage
uniform operation throughout the stretch.In the analysis of grades and grade control, one
of the most important considerations is the effect of grades on the operating of the motor

66



vehicle.Determination of grades for vertical alignment the following are taken in to
consideration for;

1. The maximum limit of grades.

 Visibility related to sight distance.
 Stopping sight distance.
 Passing sight distance.
 Rider and passenger comfort.
 Cost of vehicle operation
 General appearance
 Cut and fill (earth work)

2. The minimum limit of grades.

 Drainage purpose
In this project the two cases are taken in to account as recommended by ERA 2001.

2.2.2.1.2 Vertical curves

A vertical curve provides a smooth transition between two tangent grades. There are two
types of vertical curves. Crest vertical curves and sag vertical curves.

 When a vertical curve connects a positive grade with a negative grade, it is
referred to as a crest curve.
 When a vertical curve connects a negative grade with a positive grade, it is
termed as a sag curve.
In this project crest and sage curves are applied to create a smooth transition between
these grades.

Length of vertical curves

Crest curves:

67



For crest curves, the most important consideration in determining the length of the curve
is the sight distance requirement.

 Sight distance
— stopping and
— passing sight distance
Sag curves:

For sag curves, the criteria for determining the length of the curve are:

 vehicle headlight distance,
 rider comfort,
 drainage control and
 General appearance.
When the computed curve length for the above requirements is less than the minimum
curve length recommended by ERA2001, this recommended value is taken as curve
length.

Error! Not a valid link.Site distance (Both stopping and passing)

For Crest Vertical Curve
The stopping sight distance is the controlling factor in determining the length of a crest
vertical curve.
Minimum Length required for safe stopping calculated (from AASHTO)

When Sd ≥ Lvcmin

When Sd ≤ Lvcmin

The 100 in the above equations are to convert A from % into decimals.
68



Where Lvc min = Minimum length of vertical curve compute

Sd = Min. Stopping Sight Distance = 85 m for mountainous terrain.

Psd = Min. Passing Sight Distance = 225 m for mountainous terrain.

Sight distances should be checked during design, and adjustments made to meet the
minimum requirements. The following values should be used for the determination of
sight lines. Shown in the figures below:

Fig 2-12 Site distance for crust curve

ERA Manual recommends that:

h1= Driver's eye height = 1.07 meters

h2 = Object height for stopping sight distance = 0.15 meters

= Object height for passing sight distance: = 1.30 meters

For sag Vertical Curve
69



Figure below shows the driver’s sight limitation when approaching a sag vertical curve.
The problem is more obvious during the night time when the sight of the driver is
restricted by the area projected by the headlight beams of vehicle. Hence, the angle of the
beam from the horizontal plane is also important. This design control criteria is known as
headlight sight distance. The headlight height of h = 0.6 m and upward angle for the
headlight projection cone of β =1° is normally assumed. The governing equations are
(from AASHTO)

When Sd ≥ Lvcmin

When Sd ≤ Lvcmin

Fig 2-13 Site distance for sag curve

A driver may experience discomfort when passing a vertical curve. The effect of
discomfort is more obvious on a sag vertical curve than a crest vertical curve with the
same radius, because the gravitational and centripetal forces are in the opposite
directions. Some of the ride discomfort may be compensated by combination of vehicle
weight, suspension system and tire flexibility. The following equation has been

70



recommended by AASHTO as the minimum length of a vertical curve that will provide
satisfactory level of ride comfort.

Design standards from ERA manual:

Urban/Peri- Urban
Design Element Unit Flat

Mountainous

Escarpment
Rolling

Design Speed km/h 85 70 60 50 50

Min. Stopping Sight Distance m 155 110 85 55 55

Min. Passing Sight Distance m 340 275 225 175 175

% Passing Opportunity % 25 25 15 0 20

Max. Gradient (desirable) % 4 5 7 7 7

Max. Gradient (absolute) % 6 7 9 9 9

Minimum Gradient % 0.5 0.5 0.5 0.5 0.5

Crest Vertical Curve k 60 31 18 10 10

Sag Vertical Curve k 36 25 18 12 12

Table 2-13 Design Parameters for Design Standard DS4 (Paved)

Phasing: Even if we face phasing problem on vertical curve 1 with horizontal curve 3
and vertical curve 3 with horizontal curve 5, we took a corrective action by separating
them again vertical curve 2 and horizontal curve 4 corrected by making the ends of the
curves to end at a common station in the design process according to ERA.

2.2.2.2. Computation of gradients
1. Gradient of the first alignment (g1)
71



To calculate the first gradient;
Elevation of the first point = 1386 m
Elevation of the second point = 1395.4 m
Elevation difference = 1395.4-1386 = 9.4 m
Horizontal distance b/n the two points = (13+572)-(12+500) = 1072 m
Gradient (Slope) = elevation difference/horizontal distance
= (9.4/1072) = 0.0088
Gradient (Slope) g1 = 0.88 %
2. Gradient of the second alignment (g2)
To calculate the second gradient;

Elevation of the first point = 1395.4 m

Elevation of the second point = 1375 m

Elevation difference = 1375-1395.4 = -20.4 m

Horizontal distance b/n the two points = (14+000)-(13+572) = 428 m

Slope (gradient) = elevation difference/ horizontal distance

= -20.4/430 = -0.0477

Gradient (Slope) g2 = -4.77 %

3. Gradient of the third alignment (g3)

To calculate the third gradient



Elevation difference = 1377-1375 = 2m

Horizontal distance b/n the two points = (14+480)-(14+000) = 480m

72



Gradient (Slope) = elevation difference/ horizontal distance

= (2/480) = 0.0042

Gradient (Slope) g3 = 0.42 %

4. Gradient of the forth alignment (g4)

To calculate the forth gradient



Elevation difference = 1352-1377 = -25

Horizontal distance b/n the two points = (15+500)-(14+480) = 1020m

Slope (gradient) = elevation difference/ horizontal distance

= -25/1020 = -0.0245

Gradient (Slope) g4= -2.45%

Elevation station
Horizontal
second Second Slope
Grade First point point Elev. diff. First point point distance(m) (%)

g1 1386 1395.4 9.4 12500 13572 1072 0.88

g2 1395.4 1375 -20.4 13572 14000 428 -4.77

g3 1375 1377 2 14000 14480 480 0.42

g4 1377 1352 -25 14480 15500 1020 -2.45

Table 2-14: Summery of gradients of vertical alignment

2.2.2.5 Computation of vertical curve elements

There are three vertical curves in this project;

73



The first vertical curve is a crest curve connects a positive grade with a negative grade;
i.e. 0.88 % and -4.77 %.

The second curve is a sag curve connects a negative grade with a positive grade ;

i.e. -4.77 % and 0.42 %.

The third curve is a crest curve connects a positive grade with a negative grade;

i.e. 0.42 % and -2.45 %.

1. For Curve one (crest curve)
Station of PVI = 13+572

Elevation PVI = 1395.4 m

Gradient, g1 = 0.88 %

Gradient, g2 = -4.77 %

Grade Algebraic difference of grades (A)

A = g2-g1 =0.88 - (-4.77) = /5.64/ = 5.64 %

Computation of the curve length

a) Curve length required for minimum curvature, k

The value of K = 18 for DS4 from design standard, and Mountainous

Lvcmin = AK = 5.64*18 = 101.58 m

But to get smooth vertical curve to different safety purpose we increase LVC from
101.58 to 120 m

b) Length required for safe stopping

When Sd ≥ Lvcmin

74


Highway design raport(final) (group 2)

Highway design raport(final) (group 2)

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (20)

Similar to Highway design raport(final) (group 2)

Similar to Highway design raport(final) (group 2) (20)

Highway design raport(final) (group 2)