Highway Engineering

(Nandana) #1
The Design of Highway Intersections 137

Example 5.5 Contd


N

3.25m 3.25m Figure 5.30Diagram
of 2-lane approach.

The turning radius for all vehicles is 15 m. Zero gradient and one 30
second effective green period per 60 second traffic cycle can be assumed
(l=0.5). The traffic is assumed to be composed of 90% private cars and
10% heavy commercial vehicles.
Calculate the saturation flow for each of the two movements. Assume
two storage spaces exist within the junction.

Solution

For the nearside lane, the saturation flow can be calculated using
Equations 5.24 and 5.25:
S 1 =(S 0 - 140 dn)/(1 +1.5f/r) pcu/hr

where S 0 = 2080 - 42 dg ¥G+100(w -3.25)
The values of the relevant parameters are:
dn= 1
f =0.2
r = 15
G = 0
w =3.25

Therefore
S 0 = 2080 - (42dg ¥0) +100(3.25-3.25)
= 2080
and
S 1 =(2080 -140)/(1 +(1.5 ¥0.2/15))
=1902 pcu/hr

Contd
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