A horizontal curve on a three-lane highway with 12-ft lanes has a PC located at station 10+00 and PT at station 13+70. The central angle is 25 degrees, the superelevation is 4%, and 40 ft of clearance is available between an obstruction and the centerline. Determine the maximum safe design speed to the nearest 5 mph.

Structural Analysis
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ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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A horizontal curve on a three-lane highway with 12-ft lanes has a PC located at station
10+00 and PT at station 13+70. The central angle is 25 degrees, the superelevation is 4%,
and 40 ft of clearance is available between an obstruction and the centerline. Determine the
maximum safe design speed to the nearest 5 mph.
Example prob below:
A horizontal curve on a two-lane highway with 12-ft lanes has a PC located at station
1125+35 and PT at station 1129+00. The central angle is 35 degrees, the superelevation is
8%, and 28 ft of clearance is available between an obstruction and the centerline. Determine
the maximum safe design speed to the nearest 5 mph.
Transcribed Image Text:A horizontal curve on a three-lane highway with 12-ft lanes has a PC located at station 10+00 and PT at station 13+70. The central angle is 25 degrees, the superelevation is 4%, and 40 ft of clearance is available between an obstruction and the centerline. Determine the maximum safe design speed to the nearest 5 mph. Example prob below: A horizontal curve on a two-lane highway with 12-ft lanes has a PC located at station 1125+35 and PT at station 1129+00. The central angle is 35 degrees, the superelevation is 8%, and 28 ft of clearance is available between an obstruction and the centerline. Determine the maximum safe design speed to the nearest 5 mph.
Solution:
Ry
Given:
2 lane road, 12 ft lane
PC = 1125 +35; PT = 1129 +00
max=0.08; 4= 35°
Distance between obstruction and centerline 28
Find:
V, Design Speed
R
L = PT-PC = (1129 +00)-(1125 +35) = 365 ft
365
L ==RAR=100-105
= = 597.5 ft
R₂=R-2=597.5-2=591.5 ft
M₁ = 28-
28-12-22 ft
Start with sight distance:
M₁ = R₂ {1-
- cos
L
90SSD
RR,
{}}
Try V-45 mph
From Table 3.1 → SSD - 360 ft
90x360
M.(req.)-591.5 (1-cos (55) - 27.2 ft> 22 → Does not satisfy sight distance
Try V-40 mph
From Table 3.1 SSD -305 ft
M,(req.) - 591.5 {1-cos(x)}-19.5 ft <22 Satisfies sight distance
Next, check to see if V-40 mph satisfies centripetal acceleration requirement:
V-40 mph; e-0.08
From Table 3.5
Remin=445ft
Since actual R₁ = 591.5> 587 ft V-40 mph is ok
Transcribed Image Text:Solution: Ry Given: 2 lane road, 12 ft lane PC = 1125 +35; PT = 1129 +00 max=0.08; 4= 35° Distance between obstruction and centerline 28 Find: V, Design Speed R L = PT-PC = (1129 +00)-(1125 +35) = 365 ft 365 L ==RAR=100-105 = = 597.5 ft R₂=R-2=597.5-2=591.5 ft M₁ = 28- 28-12-22 ft Start with sight distance: M₁ = R₂ {1- - cos L 90SSD RR, {}} Try V-45 mph From Table 3.1 → SSD - 360 ft 90x360 M.(req.)-591.5 (1-cos (55) - 27.2 ft> 22 → Does not satisfy sight distance Try V-40 mph From Table 3.1 SSD -305 ft M,(req.) - 591.5 {1-cos(x)}-19.5 ft <22 Satisfies sight distance Next, check to see if V-40 mph satisfies centripetal acceleration requirement: V-40 mph; e-0.08 From Table 3.5 Remin=445ft Since actual R₁ = 591.5> 587 ft V-40 mph is ok
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