Superelevation Application

7/21/2019 Superelevation Application

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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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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


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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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


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


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


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


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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For a right hand curve Looking at the left hand

side (LHS) first we see



Looking now at the

right hand side (RHS)

we see


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Combining both LHS and

RHS we get


Case where Superelevation Equals

Crossfall


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Example 1 Transition Curve


Diagram for Curve 1


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Diagram for Curves 2 & 3


Example 2 Plan Curve


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Calculating Level at Outer Edge of

Road


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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


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


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


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


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


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


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

Superelevation Application

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