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LenaWriter [7]
3 years ago
12

A divided multilane highway in a recreational area has four lanes (two lanes in each direction) and is on rolling terrain. The h

ighway has 10-ft lanes with a 5-ft right-side shoulder and a 3-ft left-side shoulder. The posted speed is 55 mi/h. Previously, there were six access points per mile, but recent development has increased the number of access points to 12 per mile. Before development, the peak-hour factor was 0.95, and the directional hourly volume was 2300 vehicles with 15% heavy vehicles. After development, the peak-hour directional flow is 2700 vehicles with the same vehicle percentages and peak-hour factor.
1. What is the LOS before and after the development?
Engineering
1 answer:
sertanlavr [38]3 years ago
8 0

Answer:

Explanation:

Before: PT= 0.10, PB= 0.03 (given) ET = 2.5 ER = 2.0 (Table 6.5)

fHV= 0.847 (Eq. 6.5) PHF = 0.95, fp= 0.90, N=2, V = 2200 (given)

vp= V/[PHF⋅fHVTB⋅fp⋅N] = 1518.9 (Eq. 6.3)

BFFS = 50+5, BFFS =55 (given) fLW= 6.6

TLC=6+3=9 fLC= 0.65

fM= 0.0

fA= 1.0

FFS = BFFS −fLW−fLC–fM–fA= 46.75 (Eq. 6.7)

Use FFS=45 D= vp/S = 33.75pc/mi/ln Eq (6.6)

After: fA= 3.0

FFS = BFFS −fLW−fLC–fM–fA= 44.75 (Eq. 6.7)

Use FFS=45 Vnew= 2600 Vp= Va/[PHF⋅fHB⋅fp⋅N] = 1795 (Eq. 6.3) D= vp/S = 39.89pc/mi/ln

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Calculate the angle of banking on a bend of 100m radius so that vehicles can travel round the bend at 50km/hr without side thrus
saw5 [17]

Answer:

11.125°

Explanation:

Given:

Radius of bend, R = 100 m

Speed around the bend = 50 Km/hr = \frac{5}{18}\times50 = 13.89 m/s

Now,

We have the relation

\tan\theta=\frac{v^2}{gR}

where,

θ = angle of banking

g is the acceleration due to gravity

on substituting the respective values, we get

\tan\theta=\frac{13.89^2}{9.81\times100}

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3 0
3 years ago
An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% fil
den301095 [7]

Answer:

Explanation:

From the given information:

Addition of all the materials = 65+ 36+ 5 +6 = 112  which is higher than 100 percentage; SO we need to find;

The actual percentage of each material which can be determined as follows:

Percentage of the coarse aggregate will be = 65 × 112/100

= 72.80%

Percentage of the Fine aggregate will be = 36 × 112/100

= 40.32%

Percentage of the filler  will be = 5 × 112/100

= 5.6%

Percentage of the   asphalt binder will be = 6 × 112/100

= 6.72 %

So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:

Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)

Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)

Gt = 100/ (50.722)

Gt =1.972

Also;Given that the bulk density = 143.9 lb/ft³

LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft

Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43

=2.305

The percentage of air  void  = (Gt -Gm )× 100/ Gt

= (1.972 - 2.305) ×  100/ 1.972

= -16.89%

The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)

= 617.647/54.42

= 11.35%

Thus; the percentage voids in mineral aggregate =  -16.89% + 11.35%

the percentage voids in mineral aggregate = -5.45%

The percent voids filled with asphalt. = 100 × 11.35/-5.45

The percent voids filled with asphalt = - 208.26 %

7 0
3 years ago
Find the resultant of the force system on the body OABC as shown .find the points where the resultant will cut the X and Y axis?
MAXImum [283]

Explanation:

the resultant force =

\sqrt{}  {x}^{2}  +  {y}^{2}

5 0
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