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Highway EngineeringDocument No TIU.FE.SGE.216

Validity Date 01-02-2021

Course Code SGE 409

Unit Faculty of Engineering / Surveying & Geomatics Dept. Page No 58

Department of Surveying and Geomatics Engineering Highway Engineering By Mohammed Qader Ali

Superelevation at horizontal curves (e): is the road's inclination due to rising the outer age

concerning the inner age to reduce overturning without appreciable reduction in speed or

safety and comfort.

a. Rotate pavement about the centerline (Suitable for cut and fill area)

b. Rotate about inner edge of pavement (Suitable for cut area)

c. Rotate about outside edge of pavement (suitable for fill area)



Course Code SGE 409





Course Code SGE 409



Tangent Run out Section Length: is the length of roadway needed to accomplish a change

in outside-lane cross slope from normal cross slope rate to zero. (The rotation about center

line)

Where;

Lt= (X): is the length of tangent run out (m).

ed : design super elevation rate (%).

enc: normal road cross slope rate (%).

Lr: length of run off (m)

Run off section length: is the length of roadway needed to accomplish a change in outside-

lane cross slope from 0 to full super elevation or vice versa. (The rotation about center line)

Where;

Lr: length of run off (m)

e: is full super elevation(%).

G: is relative gradient. from table e.g. it is (70,50 and 44 for design V= 40,80 and 100kph

respectively)

ually is 1 for two lane road and 1.5 for four lane road)



Course Code SGE 409



Super elevation transition length (T): is the length required to rotate the cross slope of a

highway from a normal crowned slope to a fully super elevated cross slope and vice versa.

T = Lt +Lr

How to Calculate Super elevation:

a. Using Superelevation Tables; by giving the (R) and (V);



Course Code SGE 409



b. Using Nomographs; (R) and (V) should be known;

c. Using Simple Curve Formula;

or

Where;

B: Breath of the Road R: the cure radius (m), V: the velocity (kph)

e: the supper elevation, g: Acceleration=9.8 mps2, and µ: coefficient of friction.



Course Code SGE 409



Minimum radius of circular curve: The curve radius is a function of velocity, super elevation

and coefficient of friction, so from the equation bellow we found that for a certain velocity we

can get a minimum radius when there were a maximum (e and fs).

Where;

Rmin: the minimum curve radius (m)

V: the design velocity (kph)

e: the supper elevation, emax=10% (SORB), =8% FOR NON SNOW ROAD AND = 16% FOR

SNOW ROAD

emin = camber (crown) rate= 1/48=2%, use emin if (-) result for safety only.

µ: coefficient of friction, usually = 0.15

Example 4.21: Calculate super elevation for a circular road having 300 meter and it is 8 meter

wide. Take 50 km/hr. as speed of vehicle. And take centerifugal ratio= 1/10.

Solution:

= = 0.53 m

Centerifugal ratio = 1/10

So super elevation limit = = 0.80m

so that our super elevation is with in limit, OK

Example 4.22: a two lane two side road have a design speed of (80)kph and its width is (7)m,

calculate (e) at a horizontal curve of (R=200)m and the different in elevation between the

inner and the outer ages?

Solution:

µ = 0.15,

therefore;

different in elevation between the inner and the outer ages = 0.1*7.0 = 0.7m= 70 cm.



Course Code SGE 409



Example 4.23: the same example (4.22) what's (e) if R=800m (flat curve)?

Solution: = - 0.087, (-) means use emin= 1/48=0.02= 2%

Different in elevation between the inner and the outer ages = 0.02*7.0 = 0.14m=

14 cm only.

Example 4.24: What is the minimum radius of curvature allowable for a roadway with a 100

km/h design speed, assuming that the maximum allowable superelevation rate is 0.12?

Compare this with the minimum curve radius of 490m (AASHTO). What is the actual

maximum super elevation rate allowable for a 100 km/h design speed, if the value of µ is the

maximum allowed is(0.12)?

Solution:

Therefore; for actual emax

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