it is required to connect two straight directions making an angle of 6. 60° at t

Question

it is required to connect two straight directions making an angle of 6. 60° at the point of intersection (PI) by a horizontal circular curve PC assing through a point A as shown in the diagram. If the stations of points Pl and A are 4+615.41 m and 4+552.31 m, respectively, and the deflection angle at A is 10°, calculate the following: (Total-20 marks) 8-10 P1. (4+615A) 4+552 (a) Radius of the curve and hence the degree of the curve in me tric units. (6 marks) (b) Stations of the beginning and end of curve, PC and PT, respectively. (4 marks) Chord lengths c1 and c2 joining PC to A and A to PT, respectively. (6 marks) (c) Deflection angles for PT and a point along the curve at station 4+600.00m. (4 marks) (d) R-

Explanation / Answer

given that degree of curve(D) = 600

now radious of curve = 1146/D

R = 1146/ 60 = 19.10 m

stations of PC and PT

length of tangent T = R* tan(60/2)

= 19.10*tan(30)

= 11.03 m

but chainage of PI given that 4+615.41 = 92.308 m

now chainage of PC = 92.308 - 11.03 = 81.278 m

length of tangent L = R*D * pi/1800 = 19.10* 600 * pi/ 1800 = 20.00 m

now chainge of PT = L + chainage of PC = 20.00 + 81.278 = 91.278 m

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