Ex39-1 A road 10m wide is to deflect through an angle of 65° with the centre line radius 350 m, the chainage of the intersection point being 1006 m. A Transition curve is to be used at each end of the circular curve of such a length that the rate of gain of radial acceleration is 0.4 m3/sec, when the speed is 60 kmph. Find out
(a) Length of the transition curve
(b) Chainage of all junction points
(c) offset at x = L/4, L/2, 3L/4 and L
Solution :
Figure Example 39.1
Given, deflection angle D = 65°
Radius of the circular curve, R = 350 m
Road width, B = 10 m
Chainage of I = 1006 m
Speed of vehicle, V = 60 kmph = 16.67 m sec-1
Radial acceleration, a = 0.4 m3/sec
Length of the transition curve L =
= 33.07 m
= 0.13 m
Length of the tangent, IT = (R + S) tan
+
= (350 + 0.13) tan
+
= 239.59 m
Chainage of tangent T1 = Chainage of I - IT
= 1006.00 - 239.59 = 766.41 m
Length of the composite curve,
Chainage of tangent T2 = Chainage of T1 + l
= 766.41 + 430.13 = 1196.54 m
Chainage of junction point, A = Chainage of T1+ L
= 766.41 + 33.07 = 799.48 m
Chainage of junction point, A = Chainage of T2- L
= 1196.54 - 33.07 = 1163.47 m
Offset length from tangent at a distance, X along transition curve is given by
