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Ipatiy [6.2K]
3 years ago
8

A single lane highway has a horizontal curve. The curve has a super elevation of 4% and a design speed of 45 mph. The PC station

is 105+00 and the PI is at 108+75.
What is the station of the PT?
Engineering
1 answer:
andreyandreev [35.5K]3 years ago
6 0

Answer: 112 + 19.27

Explanation:

Super elevation is an inward transverse slope provided through out the length of the horizontal curve which ends up serving as a counteract to the centrifugal force and checks tendency of overturning. It changes from infinite radius to radius of a transition curve.

Super curve elevation (e) = 4%

4/100= 0.04

e= V^2/gR

Make R the subject of the formula.

egR= V^2

R= V^2/eg

V= 45mph

=45 × 0.44704m/s

=20.1168m/s

g (force due to gravity) =9.81

Therefore,

R= (20.1168)^2/9.81 × 0.04

= 1031.31m

Tangent Length( T) = PI - PC

Tangent Length= 10875 - 10500

=375m

T= R Tan(I/2)

375= 1031.31 × Tan(I/2)

I= 39.96

Also,

L= πRI/180

= 719.27m

Station PT= Stat PC+ L

10500 + 719.27

=11219.27

=112 + 19.27

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Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D
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Explanation:

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Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

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ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

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Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

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Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini
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<u>Explanation:</u>

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4 0
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