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inna [77]
4 years ago
11

A minor road intersects a major 4-lane divided road with a design speed of 55mph and a median width of 8 feet. The intersection

is controlled with a stop sign on the minor road. If the design vehicle is a passenger car, determine the minimum sight distance required on the major road that will allow a stopped vehicle on the minor road to safely turn left if the approach grade on the minor road is 3%.

Engineering
1 answer:
Vedmedyk [2.9K]4 years ago
8 0

Answer:

stopping sight diatnce is =158 meters

Explanation:

The Design Speed =55 MPH(Miles per Hour)

1 miles per hour =0.447 meter pe second

then 55MPH=55*0.447=24.58 meter per second

Consider recation time =2.5 second

Consider Co efficinet friction is 0.35

As per the Stopping sight distance based on the Breaking distance+ Lag distance

SSD=Vt+(V^2/(2gf))

SSD=24.58+(24.58^2/(2*9.81*0.35))

The total stopping sight distance is =150 meters

if we consider the Approaching garde on the minor road is 3%

SSD=Vt+(V^2/(2g(f-n/100))

SSD=24.58+(24.58^2/(2*9.81*(0.35-3/100))

then stopping sight diatnce is =158 meters

See attachment for workings

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Answer:

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Explanation:

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ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

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ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

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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 = 123.5 x 10⁻⁶ m   ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

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

ΔL = F(13.55 x 10⁻⁹ m/N)   --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

<u>F = 9.11 x 10³ N = 9.11 KN</u>

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3 years ago
The influent BOD5 to a primary settling tank is 345 mg/L. The average flow rate is 0.050 m3/s. If the BOD5 removal efficiency is
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The kilograms of BOD5 that are removed in the primary settling tank each day is  447.12 kg.

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