Question

In: Civil Engineering

Given: - Level terrain - extended freeway segment - Urban area with 0.9 interchanges/km - DDHV...

Given:

- Level terrain - extended freeway segment

- Urban area with 0.9 interchanges/km

- DDHV 4050 vehicles/hour

- 10% trucks, no buses, no RVs

- PHF 0.95

- Full shoulder and 3.6m wide lanes

Determine:

- Number of lanes to operate at LOS=C

Solutions

Expert Solution

Assume Base free flow speed (BFFS) = 90 kmph

Lane width = 3.6 m

Reduction in speed corresponding to lane width, fLW = 0 kmph

Lateral Clearance = 1.8 m

Reduction in speed corresponding to lateral clearance, fLC kmph

Interchanges/Ramps = 0.9/km

Reduction in speed corresponding to Interchanges/ramps, fID = 8.1 kmph

No. of lanes in one direction = n

Reduction in speed corresponding to number of lanes, fN kmph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID= 90 – 0 - 8.1 - fLC – fN kmph

Peak Flow, V = 4050 veh/hr

Peak-hour factor = 0.95

Trucks = 10 %

Level Terrain

fHV = 1/ (1 + 0.10 (1.5-1)) = 1/1.05 = 0.952

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = 4050/ (0.95*n*0.952*1.0) = 4478.11/n ~ 4478/n veh/hr/ln

S = FFS

Density = Vp/S

Hence Level of Service is C

Density of LOS C should lie between 11 – 16 veh/km/ln

a) Assume 2 - lanes in each direction

FFS = 81.9 - 7.3 - 0 = 74.6 kmph

Flow rate = 4478/2 = 2239 veh/hr/ln

Density = 2239/74.6 = 30.03 veh/km/ln

Not possible

b) Assume 3 - lanes in each direction

FFS = 81.9 - 4.8 - 0 = 77.1 kmph

Flow rate = 4478/3 = 1493 veh/hr/ln

Density = 1493/77.1 = 19.36 veh/km/ln

Not possible

C) Assume 4 - lanes in each direction

FFS = 81.9 - 2.4 - 0 = 79.5 kmph

Flow rate = 4478/4 = 1120 veh/hr/ln

Density = 1120/79.5 = 14.08 veh/km/ln

Possible

Hence, Four- lanes in each direction should be considered to obtain LOS C

Total number of lanes = 8


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