Module 3: Geometric design of highways

Lecture 15 Horizontal alignment II

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Solution

Assumptions

The ruling design speed for NH passing through a rolling terrain is 80 kmph. The coefficient of lateral friction $f$=0.15. The maximum permissible super elevation $e$=0.07.

Case: Radius = 450m

Step 1

Find $e$ for 75 percent of design speed, neglecting $f$, i.e $e_1=\frac{(0.75v)^2}{gR}$. $v=\frac{V}{3.6} = \frac{80}{3.6}=22.22 m/sec$ $e_1=\frac{(0.75\times22.22)^2}{{9.81\times 450}}=0.0629$

Step 2

$e_1 \le 0.07$. Hence the design is sufficient.

Answer: Design superelevation: 0.06.

Case: Radius = 150m

Step 1

Find $e$ for 75 percent of design speed, neglecting $f$, i.e $e_1=\frac{(0.75v)^2}{gR}$. $v=\frac{V}{3.6} = \frac{80}{3.6}=22.22 m/sec$ $e_1=\frac{(0.75\times22.22)^2}{{9.81\times 150}}=0.188$ Max.$e$ to be provided = $0.07$

Step 3

Find $f_1$ for the design speed and max $e$, i.e $f_1=\frac{v^2}{gR}-e=\frac{22.22^2}{9.81\times150}-0.07 =0.265$.

Step 4

Find the allowable speed $v_a$ for the maximum $e=0.07$ and $f=0.15$, $v_a=\sqrt{0.22gR}=\sqrt{0.22\times 9.81\times 150}= 17.99 m/sec=17.99\times 3.6 =64 kmph$