Question

A horizontal curve with superelevation e = 0.05 (note, $e=tan\beta$ ) has a posted speed limit of 60 km/h. For this speed limit, only those vehicles with Y/X ratio(in figure) greater than 2 and travelling at speeds greater than 1.2 times the speed limit, will overturn. However, with recent increase in truck traffic the speed limit needs to be modified such that only those vehicles with Y/X ratio greater than 3 and travelling at more than 1.2 times the speed limit will overturn. What should be the new speed limit?(in km/hr round off to integer) (with respect to the figure, assume that the vehicle is travelling into the plane of the paper).

• A. 50

Answer 1

The correct option is A 50
Given,



$e=\tan \beta =0.05$
$V=60km/hr$
$=16.67m/s$

For no overturning,

$(P\cos \beta -W\sin \beta )y\le (W\cos \beta +P\sin \beta )x$
$Py\cos \beta -Wy\sin \beta \le Wx\cos \beta +Px\sin \beta x$
$Py\cos \beta -Px\sin \beta \le Wx\cos \beta +Wy\sin \beta$
$P(y\cos \beta -x\sin \beta )y\le W(x\cos \beta +y\sin \beta )$
$\frac{P}{W}\le \frac{x\cos \beta +y\sin \beta }{y\cos \beta -x\sin \beta }$

$\frac{V^2}{gR}=\frac{1+\frac{y}{x}tan\beta }{\frac{y}{x}-tan\beta }....\left(i\right)$

$\text{For}\frac{y}{x}<2;V=1.2v$

$=1.2\times 16.67$
$=20m/s$

From (i)
$\Rightarrow \frac{{20}^2}{9.81R}=\frac{1+2\times 0.05}{2-0.05}$

$R=72.28m$

Also given,

$\frac{y}{x}>3;V=1.2v$

Again from (i)

$\Rightarrow \frac{(1.2v{)}^2}{9.81\times 72.28}=\frac{1+3\times 0.05}{3-0.05}$
$v=13.85m/s$
$=49.86km/hr\simeq 50km/hr$