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faltersainse [42]

4 years ago

10

A new interstate highway is being built with a design speed of 120 km/h. For one of the horizontal curves, the radius (measured

to the innermost vehicle path) is tentatively planned as 300 m. The limiting value for coefficient of side friction at 120 km/h is 0.09. What rate of superelevation is required for this curve at the 120 km/h design speed

1 answer:

nikklg [1K]4 years ago

4 0

Answer:

28.79%

Explanation:

Given

Design Speed, V = 120km/h = 33.33m/s

Radius, R = 300m

Side Friction, Fs = 0.09

Gravitational Constant = 9.8m/s²

Using the following formula, we'll solve the required rate of superelevation.

e + Fs = V²/gR where e = rate

e = V²/gR - Fs

e = (33.33)²/(9.8 * 300) - 0.09

e = 0.287853367346938

e = 28.79%

Hence, the required rate of superelevation for the curve is calculated as 28.79%

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zloy xaker [14]

Answer:

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Explanation:

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<em>Now, force normal to the inclined plane:</em>

$F_N=m.g.cos\theta$

$F_N=80\times 9.8\times cos25^{\circ}$

$F_N=710.54\,N$

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$f=\mu.F_N$

$f=0.15\times 710.54$

$f=106.58\,N$

<em>The component of weight along the inclined plane:</em>

$W_l=m.g.sin\theta$

$W_l=80\times 9.8\times sin25^{\circ}$

$W_l=331.33\,N$

<em>Now the total force required along the inclination to move at the top of hill:</em>

$F=f+W_l$

$F=106.58+331.33$

$F=437.91\,N$

<em>Hence the work done:</em>

$W=F.l$

$W=437.91\times 220$

$W=96340.80\,J$

<em>Now power:</em>

$P=\frac{W}{t}$

$P=\frac{96340.80}{138}$

$P=698.12\,W$

<u>So, power required for 30 such bodies:</u>

$P=30\times 698.12$

$P=20943.65\,W$

$P=\frac{20943.65}{745.7}$

$P=28.085\,hp$

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