Question

Problem 4. A section of a two-lane highway has 12-ft lanes and a design speed of 75 mi/h. The section contains a vertical curve and a horizontal curve. The vertical curve connects a-2.5% grade and a +1.5% grade. The PVT of this vertical curve is at station 36 + 50. The PC of the horizontal curve is located 294 ft before the PVC of the vertical curve. The horizontal curve has superelevation of 8% and central angle of 38 degrees. If the vertical and horizontal curves are designed to meet the minimum required lengths for the given design speed, calculate the stations of the PC and PT. (Hint for Problem 4: in this problem we have both (combined) vertical and horizontal curve on the same roadway section. This kind of design problem usually can be split into two 2- 1 dimensional problems: vertical alignment and horizontal alignment. They are solved separately, with consideration of design speed that is common for both vertical and horizontal alignment.) Design Control Tables:

Answer 1

4) Given,

Vertical curve and a horizontal curve are constructed on the same section of highway for same design speed.

Highway has two 12-ft lanes

Design Speed = 75 mph

Vertical Curve:

G1 = -2.5%

G2 = +1.5%

A = |G1 – G2| = 4%

Station of PVT = 36 + 50

K = 206 for V = 75 mph

Length of Vertical curve, L = KA = 206 * 4 = 824 ft

Station of PVC = Station of PVT – L = (36 + 50) – (8 + 24) = (28 + 26)

Horizontal Curve:

The PC of horizontal curve is located 294 ft before the PVC of vertical curve

Central Angle = 38 degrees

Super elevation = 8% = 0.08

f = 0.07

e + f = V²/gR

0.08 + 0.07 = 110²/(32.2 * R)

R = 2210.45 ft ~ 2210 ft

T = R tan(Δ/2) = 2210 * tan (38/2) = 760.96 ft

L = R Δ = 2210 * 38 * 3.14/180 = 1464.98 ft

Station of PC = Station of PVC – (2 + 94) = (28 + 26) – (2 + 94) = (25 + 32)

Station of PT = Station of PC + L = (25 + 32) + (14 + 64.98) = (39 + 96.98)