Question

"A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed is 65 mi/h, and a maximum superelevation of 0.07 ft/ft is to be used. If the central angle of the curve is 38 degrees, design a curve for the highway by computing the radius and stationing of the PC and PT."

Answer 1

Answer:

Radius = 1565ft  ;  PC = 245 + 11.13  ; PT = 255 + 48.88

Explanation:

1. Accordingly to the law of mechanics;

Centrifugal factor =S+F = V²/15R

Where; S = Super-elevation slope = 0.07ft/ft

F= Slide friction factor

V= Design speed=65mi/h

R= Radius

From the graph (see attached), at design speed of 65mi/h, coefficient of slide friction factor, F =0.11

Applying the figures in the equation above;

0.07+0.11 = 65²/15R

R=281.67/0.18

R=1564.8

Approximately Radius = 1565ft

2. Stationing PC = Stationing PI - T

where T = Tangent distance

T= Rtan(Δ/2) where Δ = central angle = 38° & Stationing PI = 250 + 50

T=1565tan19°

T=538.87ft

Stationing PC = 250 + 50 - (5 + 38.87)

PC = 245 + 11.13

3. Stationing PT = Stationing PC + L

Where L = Length of the circular curve

L = π/180*(RΔ)

L=0.01745*1565*38

L=1037.75

Therefore;

Stationing PT= 245 + 11.13 + (10 +37.75)

PT = 255 + 48.88