FCE 545: TRANSPORTATION ENGINEERING IIIA WORKED TUTORIAL QUESTIIONS

1 TRANSPORT PLANNING

Example 1.1

Figure E1 shows a scaled road network for a medium sized town in Western Kenya.

Figure E1

Build a least-cost journey tree on node F using Moore’s algorithm. (Assume all centroid links are identical in cost, at 5 units)

Solution:

Assume all centroid links are identical in cost, at 5 units, and ignore in tree building process - therefore starting tree at node 8. Determine tree by moving forward one link at a time:

Link Cost Link combination

Total cost

Least cost to Node

8 - 1 5 18-2 5 21 -2 10 8-1-2 151 -4 10 8-1-4 15 41 - 5 20 8-1-5 25 51 - 6 15 8-1-6 20 62-4 10 8-2-4 152 - 3 20 8-2-3 25 32-9 30 8-2-9 35 9 (But to D via 3.)4-3 15 8-1-4-3 304-5 10 8-1-4-5 255 - 3 10 8-1-5-3 355-7 10 8-1-5-7 35 76-5 20 8-1-6-5 406 - 7 30 8-1-6-7 503-9 10 8-2-3-9 353 - 7 15 8-2-3-7 409 - 3 10 8-2-9-3 45

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2 JUNCTION DESIGN AND CONTROL

Example 2.1:

Figure E2.1 shows a local intersection. The obstruction is located 14 m from the centreline of the right lane of a local road and 20 m from the centreline of the right lane of an intersecting road.

Figure E2.1 Minimum Sight Triangle at a No-Control or Yield-Control Intersection Cases A and C

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Table 7.7 Suggested lengths and adjustments of Sight-Triangle Leg Case A – No Traffic

If the maximum speed limit on the intersecting road is 56 km/h, what should the speed limit on the local road be such that the minimum sight distance is provided to allow the drivers of approaching vehicles to avoid imminent collision by adjusting their speeds? Approach grades are 2%.

Solution:

Step 1: Determine the distance on the local road at which the driver first sees traffic on the intersecting road.

Speed limit on intersecting road = 56 km/h Distance required on intersecting road (da) = 50 m (from Table 7.7) Calculate the distance available on local road by using the equation

db=adada−b

= 20∗5050−14

=28m

Step 2: Determine the maximum speed allowable on the local road.

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The maximum speed allowable on local road is 32 km/h (from Table 7.7).

No correction is required for the approach grade as it is less than 3%.

Example 2.2

a) Figure E2.2 shows peak-hour volumes for a major intersection on an expressway.

Figure E2.2

Pedestrian volume at the intersection is negligible and a peak-hour factor of 0.95 is applicable.

The saturation flows for an intended four-phase system are shown below

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The influence of heavy vehicles and turning movements and all other factors that affect the saturation flow have already been considered. Amber (yellow) time is three seconds

Determine suitable signal timing for the intersection using the Webster method

Solution

Determine equivalent hourly flows by dividing the peak-hour volumes by the PHF (e.g. for left-turn lane group of phase A. equivalent hourly flow 222/0.95 = 234). See Figure E2.2.1 for all equivalent hourly flows.

Figure E2.2.1 Equivalent In in rlt flow

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Phase (ϕ) Critical lane Volume(veh/h)

A 488B 338C 115D 371

∑1312

Compute the total lost time. Since there is not an all-red phase - hat is. R = 0 - and there are four phases,

L = ∑l = 4 X 3.5 = 14 sec (assuming lost time per phase is 3.5 sec)

Phase A Phase B Phase C Phase DLane Group 1 2 1 2 1 2 1 2qij 234 976 676 135 26 194 371 322Sj 1615 3700 3700 1615 1615 3700 1615 3700Qij/Sj 0.145 0.264 0.183 0.084 0.016 0.052 0.230 0.087Yi 0.264 0.183 0.052 0.230

Determine Yi and ∑Yi .

∑Yi = (0.264 + 0.183 + 0.052 + 0.230) = 0.729

Determine the optimum cycle length using the equation:

Co=1.5L+5

1−∑i=1

ϕ

Y i

= (1.5x 14 )+51−0.729

=95.9 sec

Use 100 seconds as cycle lengths are usually multiples of 5 or 10 seconds.

Find the total effective green time:

Gte=C−L=100−14=86 sec.

Effective time for phase i is obtained from

Gei=Y i

Y 1+Y 2+…Y nGte=

Y i0.264+0.183+0.052+0.230

x86=Y i0.729

x 86

Yellow time λ = 3.0 sec; the actual green time Gai for each phase is obtained as follows:

Gai = Gei + li - 3.0

Actual green time for Phase A (GaA) = (0.264/0.729) x 86 + 3 5 - 3 0 = 32 secActual green time for Phase B (GaB) = (0.183/0.729) x 86 + 3,5 - 3.0 = 22 sec Actual green time for Phase C (GaC) = (0.052/0.729) x 86 + 3.5 - 3.0 = 7 secActual green time for Phase D (GaD) = (0.23/0.729) x 86 + 3.5 - 30 = 27 sec

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Example 2.3

A fixed time 2-phase signal is to be provided at an intersection having a North-South and an East-West road where only straight-ahead traffic is permitted. The design hour flows from the various arms and the saturation flows for these arms are given below:

North South East WestDesign hour flow (pcu/hr) 800 400 750 600Saturation flow (pcu/hr) 2400 2000 3000 3000

i) Calculate the optimum cycle time and green times for the minimum overall delay. The minimum inter-green time can be taken as 4 seconds lost time per phase due to starting delay, 2 seconds and the amber as 2 seconds.

ii) Sketch the timing diagram for each phase.

Solution:

N S E Wq 800 400 750 600s 2400 2000 3000 3000y =q/s 0.33 0.20 0.25 0.20y (max) values 0.33 0.25

The timing diagram is shown below

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3 ECONOMIC ANALYSIS

Example 3.1

An existing highway is 20 Km long connecting two small cities. It is proposed to improve the alignment by constructing a highway 15 Km long costing $500,000 per kilometre. Maintenance costs are likely to be $10,000 per kilometre per year. Land acquisition costs run ($75,000) $60,000 per kilometre. It is proposed to abandon the old road and sell the land for $10,000 per kilometre. Money can be borrowed at 8% per annum. It has been estimated that passenger vehicles travel at 35 kph at a cost of 20 cents per kilometre with a car occupancy of 1.5 persons per car. What should be the traffic demand for this new road to make this project feasible if the cost of the time of the car's occupants is assessed at $10 per hour? Give a brief discussion of your result.

Solution:

1. Basic costs:Construction 15 kilometres at $500,000 = $7,500,000 Land acquisition 15 kilometres at $75,000 60,000 = $ 900,000 Sub-total = $8,400,000

Deduct sale of land (old alignment) 20 x 10,000 = 200,000 Total $8,200,000

2. For simplicity, assume the new road will last indefinitely. Therefore, the investment need not be repaid, only the interest.

Annual cost of initial investment: 8% of $8,200,000 = $656,000

3. Net annual income: Assume that N vehicles per year use the road. 5 kilometres saved at 20 cents per kilometre. 5 x N x (20/100) = $N

Time saved per vehicle/trip:- 5 mi at 35 kph = 5/35 = 0.143 hr- Car occupancy = 1.5 persons/car - Value of time = $10/hr

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Therefore, Cost saving in time at $10/hr = 0.143 x 1.5 x N x 10 = 2.14N Maintenance savings = 5 km x $10,000 = $50,000 Total savings = $50,000 + $N + 2.14N = 50,000 + 3.14N

4. Benefit-cost ratio:

In order that the project could be justified, the B/C ratio must be equal to or greater than 1. Because the project is assumed to have an infinite life, B/C = Annual benefits / Annual Costs Or 50,000 + 3.14N = 656,000 N = 192,994/yr or about 528 veh/day

The project, therefore, would be justified if about 528 vehicles used the new road on a daily basis.

5. Discussion:

The following points should be noted regarding this problem.i) The social costs of land use have not been considered. For example, the

land on either side of the old road would decrease in value, whereas the land adjacent to the new route would increase in value. The net result would raise the value of N.

ii) When one works out the B/C ratio, it is generally advisable to use total benefits and total costs. In this problem, B/C was assumed to be 1 and is therefore permissible.

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