CIVL3420: Transport Systems Engineering School of Civil Engineering

Semester 2, 2012 – Assignment 5 The University of Queensland

Assignment 5: Geometric Design Due: 5pm Mon 22 Oct 2012

To be submitted on BlackBoard via Turnitin Solutions

1. A level rail crossing with a two-way two-lane highway in a rural area is considered for

construction of a grade separated crossing. The road is overpassing with the

pavement surface 8 meters above the rail tracks and a design speed of 90 km/h. The

bridge should have a level segment of at least 50 meters to satisfy the rail clearance

requirements. What is the minimum length of the existing TWTL highway which must

be reconstructed to create the new vertical profile? Assume the initial highway grade

is zero.

[25 Marks]

Because of symmetry, only the minimum length on one side of the bridge can be

computed. The minimum length is when a sag curve is designed back to back with a

crest curve. Given the design speed:

39;38 == cs KK (sag and crest tables with regards to SSD)

Since the initial grade and final grades are zero, curve offsets are equal to change in

elevation:

55.48200200

;

8200200

22

=→=+→==

=+

AAKAK

AKLAKL

ALAL

cs

ccss

Cs

Therefore the total length of the two curves is:

mAKAKLL cscs 0.351=+=+

The minimum length of the reconstruction is:

m752502351 =+×

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2. A crest vertical curve has an initial grade of 4% and a design speed of 80 km/h. If the

curve is designed with an equal tangent, the highest point stationing is 3+012.5, and

PVT stationing is 3+200, determine the difference in the elevation of PVC and the

highest point on the curve. Assume the curve is designed with a minimum length.

[25 Marks]

The location of the highest point is

mkGxhl 1044261 =×==

Stationing of PVC=3+012.5-104=2+908.5, therefore,

mL 5.291)5.9082()2003( =+−+=

The change in grade can be calculated from:

0721.021.11 2 −=→== GK

LA

Using the equal tangent formula, the elevation of the highest point is determined:

myy

ycGbLGGacbxaxy

PVCx

PVC

hl08.2

,,2/)(, 112

2

=−

==−=++=

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3. A horizontal curve with Point of Intersection (PI) stationing at 2+200m is being

designed for a four-lane road in a mountainous terrain. The lane width is 3 m. The

deflection (central) angle (Δ) is 40 degrees and the curve tangent distance is 180.8m.

What is the stationing of the Point of Curvature (PC) and Point of Tangent (PT)? Also,

if the curve has a superelevation of 0.08 m/m, determine the safe vehicle speed on

this curve. The coefficient of side friction can be assumed 0.08.

[25 Marks]

Stationing PI =2+200 Stationing PC=2200-180.8=2+019.2

mRL

mTRRT

8.346180

7.496)2/tan(/)2/tan(

=∆=

=∆=→∆=π

Stationing PT=2019.2+346.8=2+365.99

Since the road has 2 lanes on each direction, the actual radius on the inner lane is:

hkmefRVef

VR

mRR

vv

v

/0.100)(127)(127

2.4922/332

=+×=→+

=

=−−=

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4. A horizontal curve for a two-way two-lane highway is being designed with a central

angle of 50 degrees and a design speed of 100 km/h. Assuming the superelevation is

0.05 m/m and lane width is 3.4 m,

a. determine the curve length and degree of curvature,

b. determine the clearance distance from the inside edge of the pavement to

ensure adequate SSD is provided.

[25 Marks]

mef

VRv 2.463

)05.012.0(127

100

)(127

22

=+

=+

=

Alternatively, you can use the table and interpolate between 435 and 490m.

mRR v 9.4642/4.3 =+=

Therefore,

deg76.35.30180

7.405180

=×

=

=∆=

RD

mRL

π

π

Part b: First we need SSD (assuming the design speed is a car)

mGf

VVRSSD T 4.170

)039.0(254

100

6.3

1005.2

)01.0(2546.3

22

=+

+×

=±

+=

Alternatively, you can use the table which results in SSD=185m.

mR

SSDRM

v

vs 81.7)(90

cos1 =

−=

π

Clearance from the edge of the pavement is:

7.81-3.4/2=6.11m