Geometric Design of Highways

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Geometric Design
of Highway
Highway Engineering
Prepared by: Bhavya Jaiswal

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Content of the Presentation
• Sight distance at intersection
• Horizontal Alignment
• Super-Elevation
• Radius of curve
• Extra widening
• Transition curve
• Setback Distance
• Vertical Alignment
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Sight distance at intersection
• Intersection sight distance could be understand as length
of the cross road, that must be clearly visible for the driver
of crossing or turning vehicle to be able to decide or
complete operation without any conflict with other
vehicles, approaching the intersection from the cross road.
• At intersections, where the two or more roads meet it is
essential to provide visibility for drivers, who are
approaching the intersections from the either sides.
• Sight distance should be provided for drivers on each side of the intersection, to make it easy
for them to find and see each other as shown in the below figure.
• IRC recommends that at non-controlled intersections, sufficient visibility should be provided
such that sight distance on each road is at-least equal to SSD corresponding to the design
speed of the road.
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Sight distance at intersection
• IRC recommends that a minimum visibility distance of 15 m along the minor road and a
distance of 220, 180, 145 and 110 m along the major road corresponding to design
speeds of 100, 80, 65 and 60 kmph. respectively may be provided.
Why it is needed?
1. To enable the change of speed for a vehicle,
2. Enable stopping of the vehicle nearing the approach,
3. Enabling the stopped vehicles on the minor road to safely cross the main road at
intersection.
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Headlight Distance
• It comes in a consideration during the night time.
• It could be defined as a distance which is visible
to the driver during the night time driving under
the enlightenment of the vehicle head light.
• This sight distance is critical in the case of up-
gradients and at an ascending portion of the
valley curve.
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Horizontal Alignment
• When a vehicle moves on the horizontal curve, it is subjected to the centrifugal force
which is acting in outward direction of the road. This load can be counteracted upto a
certain level by frictional force between the Tyre and pavement surface.
• This outward centrifugal force causes discomfort to the passenger and increases the
possibility of accidents.
• In such cases, the horizontal alignment comes into picture.
• The components covered in the horizontal alignment are…
Super-Elevation
Radius of curve/ minimum ruling radius
Extra widening
Transition curve
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Super-Elevation
• The super elevation can be defined as the Rising
of the outer edge of the pavement with respect to
the inner edge throughout the length of curve to
counteract the centrifugal force and reduce the
tendency of a vehicle to overturn is called Super-
Elevation.
• Derivation of the formula of Super-Elevation:
𝑃𝑐𝑜𝑠𝜃 = 𝑊𝑠𝑖𝑛𝜃 + 𝐹𝐴 + 𝐹𝐵
𝑃𝑐𝑜𝑠𝜃 = 𝑊𝑠𝑖𝑛𝜃 + 𝐹𝑅𝐴 + 𝐹𝑅𝐵
𝑃𝑐𝑜𝑠𝜃 = 𝑊𝑠𝑖𝑛𝜃 + 𝐹(𝑊𝑐𝑜𝑠𝜃) + 𝐹(𝑃𝑠𝑖𝑛𝜃)
𝑃𝑐𝑜𝑠𝜃 − 𝐹(𝑃𝑠𝑖𝑛𝜃) = 𝐹(𝑊𝑐𝑜𝑠𝜃) + 𝑊𝑠𝑖𝑛𝜃
𝑃𝑐𝑜𝑠𝜃[1 − 𝐹𝑡𝑎𝑛𝜃] = 𝑊𝑐𝑜𝑠𝜃[𝐹 + 𝑡𝑎𝑛𝜃]
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Super-Elevation
Hence,
𝑃
𝑊
=
𝐹+𝑡𝑎𝑛𝜃
1−𝐹𝑡𝑎𝑛𝜃
We know that…𝑃 =
𝑚𝑉2
𝑅
=
𝑊𝑉2
𝑔𝑅
𝑃
𝑊
=
𝑉2
𝑔𝑅
If 𝐹𝑒 is neglected than…
𝑉2
𝑔𝑅
=
𝐹+𝑒
1−𝐹𝑒
So, 𝑒 + 𝑓 =
𝑉2
𝑔𝑅
If we put V in kmph and f as 0.15 than the super elevation will be as 𝒆 + 𝒇 =
𝑽𝟐
𝟏𝟐𝟕𝑹
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Super-Elevation
• Here,
e= Super elevation
V= Speed of the vehicle, kmph
R= Radius of curve, m
f= Lateral Friction, 0.15
𝒆 + 𝒇 =
𝑽𝟐
𝟏𝟐𝟕𝑹
According to IRC:
𝑒𝑚𝑎𝑥 = 4% urban Road
𝑒𝑚𝑎𝑥 = 7% Plain terrain
𝑒𝑚𝑎𝑥 = 10% Hilly terrain
𝑒𝑚𝑖𝑛 = as the camber value
• If the friction factor is neglected than
the e with a formula e =
𝑽𝟐
𝟏𝟐𝟕𝑹
is
known as the Equilibrium Super
elevation.
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Design of the Super-Elevation
• Design of the super elevation for the mix traffic is a complex problem because various
vehicle moves with various speed.
• Maximum value of the super elevation can be archived by neglecting the lateral friction
that is also for the vehicles which are moving with design speed
• So, to compromise and to catch the practical conditions, for calculation of super
elevation speed is taken as 75% of the design speed.
𝒆 + 𝒇 =
𝑽𝟐
𝟏𝟐𝟕𝑹
After neglecting f, e =
𝑽𝟐
𝟏𝟐𝟕𝑹
Now for the mix traffic, V= 0.75V than e =
𝟎.𝟕𝟓𝑽𝟐
𝟏𝟐𝟕𝑹
so the super elevation will be as e =
𝑽𝟐
𝟐𝟐𝟓𝑹
Step-1
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Super-Elevation
If in the above equation Step-1, if e>0.07 than perform Step-2 as…
Put e=0.07 and find f
e + f =
𝑽𝟐
𝟏𝟐𝟕𝑹
f =
𝑽𝟐
𝟏𝟐𝟕𝑹
− 𝟎. 𝟎𝟕
Here, if f<0.15 than OK
But if f>0.15 than go for Step-3 as…
Put e=0.07 and f=0.15 and find new speed Va= allowable speed
e =
𝑽𝒂𝟐
𝟏𝟐𝟕𝑹
Step-2
Step-3
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Numerical Super-Elevation
Q. Design a super elevation for plain and rolling terrain if R=500m and design
speed=100kmph.
Solution:
Step-1:
e =
𝑽𝟐
𝟐𝟐𝟓𝑹
e= 0.08>0.07
So taking e=0.07
Step-2:
𝒆 + 𝒇 =
𝑽𝟐
𝟏𝟐𝟕𝑹
𝒇 =
𝑽𝟐
𝟏𝟐𝟕𝑹
𝟎. 𝟎𝟕
f=0.07<0.15 OK. Than we can provide e as 0.07
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Minimum Ruling Radius
The minimum curve radius is a limiting value of curvature for a given design
speed. In the design of horizontal alignment, smaller than the calculated
boundary value of minimum curve radius cannot be used. Thus, the minimum
radius of curvature is a significant value in alignment design.
The minimum ruling radius can be obtained by the following equation.
𝒆 + 𝒇 =
𝑽𝟐
𝟏𝟐𝟕𝑹
𝟎. 𝟎𝟕 + 𝟎. 𝟏𝟓 =
𝑽𝟐
𝟏𝟐𝟕𝑹
R =
𝑽𝟐
𝟐𝟕.𝟗𝟒
and R is in meter.
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Extra Widening
• Additional width of carriage way which is required
at horizontal curve is known as extra widening.
• The rare wheel follow the inner earth on the curve
as compared to front wheel, this phenomena is
called Off-Tracking.
The following are the reasons for which extra
widening is essential.
1. To avoid off-tracking.
2. To avoid transverse skidding.
3. To increase visibility of curve.
4. To encounter phycological tendency while
overtaking.
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Extra Widening
We = W mechanical + W phycological
We =
𝑛𝑙2
2𝑅
+
𝑉
9.5√𝑅
Here,
V= speed, kmph
n= number of lane
R= radius of curve, meter
l= wheel base length, meter
𝑙2
2𝑅
= off tracking
As per IRC:
If R<50m, widening only in inner edge.
If 50m<R<300m, widening in both edges.
If R>300m, no widening is required.
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Grade Composition
• When there is a horizontal curve in addition to gradient than there will be
incremental loss of traction due to both curve and gradient.
• So t is necessary to compensate the grade in such cases of horizontal curve.
It can be found out by the following formula.
• Grade Compensation =
30+𝑅
𝑅
or
75
𝑅
whichever is minimum.
• In no case, Compensated Grade not less than 4%
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Numerical of Grade Composition
Q. find the grade compensation and compensated grade for a vertical gradient of 7% and
horizontal curve 50m.
• Solution:
𝐺𝐶 =
30 + 𝑅
𝑅
= 1.5
𝐺𝐶 =
75
𝑅
= 1.6
𝑇ℎ𝑒 𝑐𝑜𝑚𝑝𝑒𝑛𝑠𝑎𝑡𝑒𝑑 𝑔𝑟𝑎𝑑𝑒 = 7 − 1.5 = 5.5%
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Transition Curve
• When a vehicle travelling on straight road and if enters into a
horizontal curve suddenly, it will cause discomfort to the driver.
• To avoid this inconvenience, it is required to provide a
transition curve.
• It a curve having varying radius and provided between two
curves of different radius.
Objectives:
1. To increase centrifugal force between straight and circular
curve gently.
2. To avoid the sudden jerk.
3. To introduce super elevation and extra widening gradually.
4. To improve aesthetic appearance.
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Length of the Transition Curve
1. On the basis of Rate of change of centrifugal acceleration: L=
𝑉3
𝑅𝐶
Where,
V= m/sec
C = rate of change in acc =
80
75+𝑉
2. Empirical formula by IRC: L=2.7
𝑉2
𝑅
for plain terrain and L=
𝑉2
𝑅
for hilly terrain.
Where, V=kmph
3. On the bases of change in super elevation:
L =
𝑒𝑁
2
(𝑤 + 𝑤𝑒) for rotated about center
L = 𝑒𝑁(𝑤 + 𝑤𝑒) for rotated about inner edge.
Where, 1/N = rate of change in super elevation, We = extra widening
The length of the transition curve is taken as maximum of all of above 3 approaches.
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Types of the Transition Curve
• According to IRC, ideal shape of transition curve is
Spiral because rate of change of centrifugal
acceleration is constant in case of spiral curve.
• As we can see in the picture, there are three types
of spiral curve:
1) Cubic Parabola
2) Bernoulli's lemniscate
3) Spiral Curve.
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Numerical on the Transition Curve
Q. Calculate length of the transition curve if the design speed=65kmph, Radius of
curve=220m, number of lane=2, pavement width=7.5m, super elevation=0.07, length of
wheel base=6m.
Solution:
Widening, We=
𝑛𝑙2
2𝑅
+
𝑉
9.5√𝑅
= 0.16 + 0.46 = 0.62𝑚
Length of TC:
1. L=
𝑉3
𝑅𝐶
= 46.82𝑚
2. L=2.7
𝑉2
𝑅
= 51.85𝑚
3. L = 𝑒𝑁 𝑤 + 𝑤𝑒 = 85.26𝑚
The maximum among all is 85.26m than the length of TC will be 85.26m.
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Set-Back Distance
• Set back is clearance between centerline of the curve to
the obstruction on the inner side of curve which is
applied to provide the adequate sight distance.
• Case-1: L > SD
• 𝑀 = 𝑂𝐶 − 𝑂𝐷
• 𝑀 = 𝑅 − 𝑅𝑐𝑜𝑠
∝
2
•
∝
2
=
SD
2R
∗
180
π
If double lane road than 𝑀 = 𝑅 − 𝑅 − 𝑑 𝑐𝑜𝑠
∝
2
• And
∝
2
=
SD
2 R−d
∗
180
π
also 𝑑 =
𝑤+𝑤𝑒
4
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Set-Back Distance
From the center of inner edge: Ed
𝑀 = 𝑅 − 𝑑 − 𝑅 − 𝑑 𝑐𝑜𝑠
∝
2
From the edge of road: Fd
𝑀 = 𝑅 − 2𝑑 − 𝑅 − 𝑑 𝑐𝑜𝑠
∝
2
So, in general
𝑀 = 𝑅 − 𝑅 − 𝑛 − 1 𝑑 𝑐𝑜𝑠
∝
2
∝
2
=
SD
2 R − n − 1 d
∗
180
π
𝑑 =
𝑤 + 𝑤𝑒
2𝑛
Where n = number of lanes.
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Set-Back Distance
Case-2: L < SD
𝑀 = 𝑅 − 𝑅𝑐𝑜𝑠
∝
2
+
𝑆𝐷 − 𝐿
2
𝑠𝑖𝑛
∝
2
∝=
𝐿
2𝑅
×
180
𝜋
And change all the formula accordingly.
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Numerical on Set-Back Distance
Q. A horizontal circular curve with a centerline radius of 160m is provided on two lane two way
straight highway section. Width of two lane is 7m, reaction time 2.4sec, design speed is 80kmph,
f=0.355. calculate the set back distance from center line of inner edge and length of curve is
300m.
Solution:
𝑆 = 𝑉𝑡 +
𝑉2
2𝑔𝑓
= 125𝑚 so L > SD.
From the center of inner edge,
𝑑 =
𝑤 + 𝑤𝑒
4
= 1.97𝑚
∝
2
=
SD
2 R − d
∗
180
π
= 22.66
𝑀 = 𝑅 − 𝑑 − 𝑅 − 𝑑 𝑐𝑜𝑠
∝
2
= 12.19𝑚
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Vertical curve
• Vertical curves are used in highway and street
vertical alignment to provide a gradual change
between two adjacent grade lines.
• There are two types of vertical curve in the
highway which are as below:
1. Summit Curve
2. Valley Curve
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1. Summit Curve
• It is a convex upward side and concave downward
side.
• The best shape of the summit curve is square
parabola but the ideal shape of the curve is
circular.
• Square parabola gives the good riding qualities
and uniform rate of change in grade.
• Length of summit curve depends on sight
distances.
• There is no problem of discomfort in summit curve
because gravity forces act downward and
centrifugal force acts upward.
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Length of the Summit Curve
𝐿 =
𝑁𝑆2
2𝐻 + 2ℎ
2
H = height of driver from pavement surface
H = height of object from pavement surface
H = 1.2m and h = 0.15m for SSD
H = 1.2m and h = 1.2m for OSD
 Case-1, L > Sight Distance
𝐿 =
𝑁𝑆2
4.4
for SSD
𝐿 =
𝑁𝑆2
9.6
for OSD
 Case-2, L < Sight Distance
𝐿 = 25 −
4.4
𝑁
for SSD
𝐿 = 25 −
9.6
𝑁
for OSD
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Numerical of the Summit Curve
Q. Find the length of the vertical curve when an ascending gradient of 1 in 50 meets a
descending gradient of 1 in 80 and the SSD is 180m.
Solution:
𝑁 = 𝑛1 + 𝑛2
𝑁 =
1
50
+
1
80
𝑁 = 0.0325
Here, SSD = 180m
𝐿 =
𝑁𝑆2
4.4
= 239.3~240𝑚
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2. Valley Curve
• Valley curve is concave upward and convex
downward.
• Valley curve is generally mid of two transition
curve of equal length without having circular
curve in between.
• Cubic parabola is the best shape for the valley
curves to gradually introduce the centrifugal
force.
• in case of valley curve, centrifugal force and
gravity force both are acting downward that
makes valley curves uncomfortable.
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2. Valley Curve
The design of the valley curve is done as per the following approaches:
Case-1: Based on comfort, 𝐿𝑉 = 2√
𝑁𝑉3
𝐶
Where,
V = m/sec
C = 0.6m/𝑆2
Case-2: Based on Headlight sight Distance
If L > HSD, 𝐿𝑉 =
𝑁𝑆2
1.5+0.035𝑆
If L < HSD, 𝐿𝑉 = 2𝑆 −
1.5+0.035𝑆
𝑁
Where, S = sight distance
Sight distance is not a problem for valley curve because in night time other vehicle with
headlight can be seen from a considerable distance.
Length of valley curve is taken as the maximum among both 31

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Numerical on Valley Curve
Q. A vertical curve is designed when a descending gradient of 1 in 25 meets with
ascending gradient of 1 in 30. Design the length of the Valley curve if the speed is
80kmph.
Solution:
𝑁 = 𝑛1 + 𝑛2
𝑁 = −(
1
25
+
1
30
)
𝑁 = −0.0733
Approach-1:
𝐿𝑉 = 2
𝑁𝑉3
𝐶
= 73.21𝑚
Approach-2:
𝐿𝑉 =
𝑁𝑆2
1.5 + 0.035𝑆
= 198.18𝑚
𝑆 = 𝑉𝑡 +
𝑉2
2𝑔𝑓
= 126.65𝑚
Length of valley curve is taken as the maximum among both so it will be 198.18m 32

Geometric Design of Highways

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