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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
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Superelevation Application
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Superelevation Development
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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Two Key Lengths
Superelevation Development Length (Le) is the
distance over which superelevation is
developed from the normal crossfall to full
superelevation
Transition Length (Lp) is the length of plan
transition spiral
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Location of Superelevation
Development on Transition Curves
Transition placed if Shift (S) > 0.25
Plan transition is placed halfway either side of
the Original Tangent Point (TP)
Superelevation begins along the straight prior tothe transition at the SS
Superelevation is developed such that at the
start of the transition curve (TS) the outer edge
of the pavement is level
At the start of the circular curve (SC) the
superelevation is fully developed
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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Location of Superelevation
Development on Circular Curves
If shift S is < 0.25m then no transition required.
Usually it is expected that the superelevation
will be fully developed once the straight starts
the curve.
However in practice designers place
approximately 70% of the development on the
straight prior to the curve and 30% on the curve
Usually determined by the rate of rotation of thepavement
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Superelevation Development Length
(Le)
May be determined by two methods
Rate of rotation of pavement
Relative Grade
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Rate of Rotation Method
Should not exceed 2.5% per second of travel withan absolute maximum of 3.5% for lower speeds
Le = (e1-e2)*V/0.09 for 2.5 %/sec (> 70 km/hr)
Or
Le = (e1-e2)*V/0.126 for 3.5 %/sec (< 70 km/hr)
Where
V = design speed in km/hre1 = normal crossfall at start
e2 = maximum superelevation
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Rate of Rotation Method - Example
Design Speed = 100km/hr, max e of 6%
Le = (e1-e2)*V/0.09 for 2.5 %/sec
Le = (-0.03-0.06)*100/0.09 = 100m (ignore sign)
Also, we can estimate Lp, by
Lp = Le 0.4 V
So
Lp = 100 0.4*100 = 60
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Relative Grade Method
Difference between grade at edge of carriageway
and grade for axis of rotation (centreline)
Le = W (e1-e2)*100/G
Where
W = lane width (nominally 3.5m)
G = relative grade in % from tablese1 = normal crossfall at start
e2 = maximum superelevation
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Relative Grade Method - Example
V= 100km/hr, W = 3.7m and e max = 6%
Le = W (e1-e2)*100/G
From tables G = 0.43
Le = 3.7*(-0.03-0.06)*100/0.43
= 3.7*0.09*100/0.43
= 77.4m (ignore sign)
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Tables
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Superelevation Diagram
Enables designers to show rate of change of
crossfall graphically
Crossfall plotted with respect to distance along
the road (chainage) Drawn to normal drafting standards
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Superelevation Diagram
For a right hand curve Looking at the left hand
side (LHS) first we see
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Superelevation Diagram
Looking now at the
right hand side (RHS)
we see
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Superelevation Diagram
Combining both LHS and
RHS we get
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Case where Superelevation Equals
Crossfall
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Example 1 Transition Curve
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Diagram for Curve 1
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Diagram for Curves 2 & 3
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Example 2 Plan Curve
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Superelevation Diagram
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Calculating Level at Outer Edge of
Road
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T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Calculating Level at Outer Edge of
Road
RL edge = RL centre +/- W.e
Where W = lane width
e = superelevation at the point
Eg RL edge = 100.00 + 3.5 * 0.05
= 100.00 +0.175= 100.175
e can be scaled off diagram
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Superelevation Chainage
Calculations on Simple Circular
Curve
1. Determine the radius of a simple curve (R)
given a design (operating) speed of 100 km/hr
and maximum superelevation of 5%. Round the
radius up to the nearest 10m.
2. If the curve is to be placed on a horizontal
alignment that has an intersection angle of 35o
and chainage of 2507.56 at the IP then
determine the chainages of SS1,TP1, TP2 and
SS2.
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Firstly Compute R
Rmin = V2/127 (emax + fmax)
For desirable minimum use desirable
maximum for side friction
So Rmin = 1002/127 (0.05max + 0.12max)
= 10000/ (127*0.17)
= 463.177m
Round up to 470m
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Compute Le
Le = (e1-e2).V/0.09 by rate of rotation
= (-0.03-0.05)* 100/0.09
= -0.08 *100/0.09 = 88.9m
Using standard rule of 70% of super elevation
development on the straight and 30% in the
curve, gives
70% of Le before TP = 0.7*88.9 =62.2m
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Now compute chainages
1. Tangent Length = R tan (I/2)
= 470 * tan (35/2)
= 148.190m
2. TP1 = 2507.56 148.190 = 2359.370
3. SS1 = TP1 - 0.7*88.9 = 2359.370 62.2 = 2297.17
4. Arc = R*I (in radians) = 470*35*PI/180 = 287.107
5. TP2 = TP1 + Arc = 2359.37 + 287.107 = 2646.477
6. SS2 = TP2 + 0.7*88.9 = 2646.477 + 62.2 =2708.677
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Superelevation Chainage
Calculations on Transitioned Curve
You have been asked to design a proposed minimum
horizontal curve on a rural road which has an
intersection angle of 35o, the chainage at the IP being
2507.56. The maximum crossfall for the curve is to be
5% and design speed of 100 km/hr.
Compute SS1, TS, SC, CS, ST and SS2
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Firstly Compute R
Rmin = V2/127 (emax + fmax)
For desirable minimum use desirable
maximum for side friction
So Rmin = 1002/127 (0.05max + 0.12max)
= 10000/ (127*0.17)
= 463.177m
Round up to 470m
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Next Compute Lp
Lp = (e1-e2).V/0.09 by rate of rotation
= (0.00-0.05)* 100/0.09
= 0.05 *100/0.09 = 55.5m
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Compute Shift to See if Transition
Required
The shift can be calculated by:
Shift = S =Lp2/24R
Shift = 55.52 /24*470
S = 0.273 therefore transition required
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Compute Le
Le = (e1-e2).V/0.09 by rate of rotation
= (-0.03-0.05)* 100/0.09
= -0.08 *100/0.09 = 88.9m
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Compute Tangent Length
Tangent Length = (R +S) tan (I/2)
= (470+0.273) tan (35/2)
= 148.277m
Then
The distance from the IP to the TS (start of transition)
= (R +S) tan (I/2) + Lp/2
= 148.277 +55.5/2
= 176.027m
Length of circular curve = arc = R*I (in radians) - Lp= (470* 35*pi/180) 55.5
= 287.607 55.5
= 231.607m
T h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n dT h e U n i v e r s i t y o f S o u t h e r n Q u e e n s l a n d
Compute ChainagesChainage of TS = IP Chainage (IP to TS)
= 2507.56 176.027
= 2331.533
Chainage of SS1 = TS (Le - Lp)
= 2331.533 (88.9-55.5)= 2298.133
Chainage of SC = TS + Lp
= 2331.533 + 55.5
= 2387.033
Chainage of CS = SC + Arc
= 2387.033 + 231.607
= 2618.64
Chainage of ST = CS + Lp
= 2618.64 + 55.5
= 2674.14
Chainage of SS2 = ST + (Le - Lp)
= 2674.14 + (88.9-55.5)
= 2707.54