CE 310: Lec 6
Note: In this ppt, figures/text are adopted from various sources;
this is only for class explanation purpose.
Course content
Introduction to transportation systems engineering:
• Introduction to transportation Engineering; and Definition of
transportation engineering
• Issues with transportation system
Highways:
• Geometric design (Alignment planning and design, design
controls and criteria, sight distance, horizontal alignment, vertical
alignment, Geometric design at intersection);
• Highway materials testing and characterization;
• Pavement analysis and design (Flexible and Rigid);
• Highway construction, maintenance, management and
rehabilitation.
Horizontal Alignment
• Often changes in the direction are necessitated in
highway alignment due to various reasons such as
topographic consideration, obligatory points.
• Geometric design elements pertaining to horizontal
alignment of highway should consider safe and
comfortable movement of vehicles.
• Sudden change In direction with sharp curves or
reverse curves, should be avoided.
• Improper design of horizontal alignment of roads
would result in higher accident rate and increase in
vehicle operating cost.
Design of Horizontal Alignment
Horizontal Alignment
Design elements to be considered for setting horizontal
alignment are:
• Super-elevation
• Widening of pavement on curves
• Radius of circular curve
• Set-back distance
• Type and length of transition curve
Centrifugal Force on Curves
• Curve in plan provides change in direction to centre line of road
• Forces acting on vehicle while traversing the curve
➢ Centrifugal force (acting outwards through CG of vehicle)
➢ Transverse Frictional Resistance (opp. to Centrifugal force)
• Equation of Centrifugal Force
➢ P: Centrifugal force, kg
➢ W: Vehicle Weight, kg
➢ R: Radius of circular curve, m
➢ v: Vehicle Speed, m/s
➢ g: Acceleration due to gravity = 9.81 m/s2
• Effects due to centrifugal force:
(1) Overturning
(2) Lateral skidding
1. Overturning Effect
Centrifugal force tends the vehicle to overturn about the
outer wheel B on horizontal curve without superelevation.
h= height of C.G of vehicle above the road surface
b= width of the wheel base
•Overturning Moment due to Centrifugal Force
=Ph
•Restoring Moment due to Vehicle Weight
= W b/2
•Equilibrium condition
P h = W b/2
or P/W = b/2h
•Danger of Overturning
P/W = v2/g R ≥ b/2h
P/W is called centrifugal ratio or impact factor
• Increase in height (overloading)
• R is short
• Speed is high
2. Transverse Skidding Effect
Centrifugal force also has tendency to push the vehicle outwards in the
transverse direction.
Vehicle will skid if centrifugal force exceeds maximum transverse friction force or
transverse skid resistance counteracting the centrifugal force
•Transverse skid resistance due to friction
FA + FB = f (RA + RB) = f W
where, f = lateral friction coefficient between
tire & pavement (usually 0.15)
•Equilibrium condition
P=fW
•Danger of Skidding
when Centrifugal Ratio
P/W = v2/g R ≥ f
• Speed is high
• R is short
• Friction is less (wet surface, old tire)
Summary of Overturning and Skid
– and Geometry
• Thus for Horizontal Curve with no superelevation
– If
– If
P/W ≥ b/2h
P/W ≥ f
then
then
overturning occurs
skidding occurs
• To counteract Centrifugal Force and to reduce tendency to
overturn or skid
Outer edge of pavement is raised with respect to inner edge
• This inclination is called Superelevation (or Cant or Banking)
Do not get confused with
camber or cross slope (The
purpose of camber was to
drain water off the pavement
surface (Typically 1.7 to 2.5 %
for flexible and concrete
pavements)
Superelevation
Superelevation
• Rate of
Superelevation
• But as θ is very
small
Analysis of Superelevation
• Analysis of Superelevation
For Equilibrium
At limiting equilibrium
i.e. when FA and FB
reach max values of
f RA and f RB resp,
and
Thus, FA+FB = f (RA+RB)
Superelevation (contd.)
• Analysis of Superelevation
Dividing by Wcosθ,
f = 0.15, and tanθ seldom exceeds 0.07 or 1/15
Thus ftanθ = 0.01 and 1-ftanθ ≈ 1
Therefore,
Superelevation (contd.)
• Maximum superelevation
Maximum limit of superelevation 7% or
(0.07) in plane and rolling terrain.
Superelevation (contd.)
• Minimum superelevation
Superelevation Design: IRC Method
• Step 1
– Superelevation for 75% design speed is calculated, neglecting
friction (assumption, 75% speed will be taken care by e, rest by
friction (f)). (Designing it for high speed, v may cause problem
for slow moving vehicle and vice versa).
➢ If e < 0.07, then the value obtained is provided,
➢ If e ≥ 0.07, then consider e = 0.07 and proceed to Step 2
• Step 2
– Find f at e = 0.07 and full design speed as,
Superelevation Design: IRC Method
• Step 2 (Contd.)
– If f < 0.15, then e = 0.07 is safe for design speed
If f ≥ 0.15, then calculate restricted speed as in Step 3
• Step 3: Calculate safe allowable speed
i.e.
Thus, safe allowable speed
– Speed is limited to lower of design and allowable speeds
Superelevation
• Very flat curve large radius, P is small, so normal camber
may be retained. However, it will provide negative superelevation (2 lane 2 way highway)
• In this situation, check provided “f”, (Usually cumulative
effect of P and negative e, may not be significant
• Minimum Super-elevation: From drainage consideration,
minimum cross slope to drain off water, if e < camber, then
provide camber slope
Attaining Superelevation
• Attainment of Superelevation can be split into 2 parts
– Elimination of crown of the cambered section
– Rotation of pavement to attain full superelevation
• Elimination of crown can be done by two methods
Outer Edge rotated about
the crown
Drainage may issue at
some point, where it
becomes horizontal or <
camber
Crown shifted outwards
(Diagonal crown method)
High negative super-elevation or sharp
camber, dangerous.
Attaining Superelevation (contd.)
• Rotation of pavement to attain full superelevation
– When the crown of the camber is eliminated, the
superelevation at this section is equal to the camber
– But if design superelevation to be provided is more than
camber, then pavement section has to be rotated further
– This can be done by two methods
• By rotating section
about centre line
• By rotating section
about inner edge
Attaining Superelevation (contd.)
22
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Source: http://user.gs.rmit.edu.au/rod/files/publications/Horizontal%20Curves.pdf
Superelevation
Class participation-3
1. The radius of a horizontal curve is 100 m and design speed is 50Kmph. Consider
coefficient of lateral friction as 0.15 .
Find e if full lateral friction is assumed to develop, and
Find f needed when no superelevation is provided
Solution:
V=50kmph
R=100m
f=0.15
e + f = V2/(127*R) ➔ e = 0.047
Rate of superelevation for full f is 0.047 or 1 in 21.2
When e = 0,
f = V2/(127*R) ➔ f = 0.19
Class participation-3
2) Design superelevation rate (using IRC method) for a horizontal
curve of radius 500m and speed 100kmph
Solution:
Step 1: Superelevation for 75% of design speed is,
e = V2/(225*R) ➔ e = 0.089 > 0.07 ➔ e = 0.07
Step 2: Checking ‘f’ developed for full speed,
f = (V2/(127*R)) – 0.07 ➔ f = 0.087 < 0.15 ➔ Design safe
Provide superelevation of 0.07
“Sentiment is a chemical defect
on the losing side”- Sir Arthur
Conan Doyle