Geometric Design Of Highway And Planning Practice Questions

Given: $R = 200 , \text{m}$ (radius of curve) $CE = 92 , \text{m}$ (distance from C to E) From geometry: $CE = R \cdot \tan\left(\frac{\theta}{2}\right)$ Substitute values: $\tan\left(\frac{\theta}{2}\right) = \frac{92}{200} = 0.46$ $\frac{\theta}{2} = \tan^{-1}(0.46) \approx 24.8^\circ$ $\theta = 2 \times 24.8^\circ = \boxed{49.6^\circ}$ Final Answer: $\theta = 49.6^\circ$

As per IRC, extra widening on a horizontal curve is given by the formula: $W_e = W_m + W_{ph}$ Where: $W_m = \text{mechanical widening} = \frac{n l^2}{2R}$ $W_{ph} = \text{psychological widening} = \frac{v}{9.5 \sqrt{R}}$ Given: $n = 2,\ l = 7\ \text{m},\ R = 300\ \text{m},\ v = 80\ \text{km/h}$ Step 1: Mechanical Widening $W_m = \frac{2 \times 7^2}{2 \times 300} = \frac{98}{600} = 0.1633\ \text{m}$ Step 2: Psychological Widening $\sqrt{R} = \sqrt{300} \approx 17.32$ $W_{ph} = \frac{80}{9.5 \times 17.32} = \frac{80}{164.54} \approx 0.4863\ \text{m}$ Step 3: Total Extra Widening $W_e = 0.1633 + 0.4863 = \boxed{0.65\ \text{m}}$

Step 1: Algebraic difference of gradients $A = |+3 - (-5)| = 8$ Step 2: Conversion of speed $V = \frac{80 \times 1000}{3600} = 22.22 , \text{m/s}$ Step 3: Reaction distance $S_r = V \times t = 22.22 \times 2 = 44.44 , \text{m}$ Step 4: Braking distance $S_b = \frac{V^2}{2 g f} = \frac{(22.22)^2}{2 \times 9.81 \times 0.36} = 69.94 , \text{m}$ Step 5: Stopping Sight Distance (SSD) $S = S_r + S_b = 44.44 + 69.94 = 114.38 , \text{m}$ Step 6: Length of summit curve (SSD criterion) For summit curve, when $L \ge S$ $L = \frac{A S^2}{2(\sqrt{h_1} + \sqrt{h_2})^2}$ $\sqrt{h_1} = \sqrt{2.30} = 1.517, \quad \sqrt{h_2} = \sqrt{0.35} = 0.592$ $(\sqrt{h_1} + \sqrt{h_2})^2 = (2.109)^2 = 4.448$ $L = \frac{0.08 \times (114.38)^2}{2 \times 4.448} = \frac{1046.6}{8.896} = 117.70 , \text{m}$ Check condition: $L = 117.70 , \text{m} \ge S = 114.38 , \text{m}$ Hence, the condition is satisfied.

Given, $V=60 \mathrm{~km} / \mathrm{hr}, t_{R}=2.5 \mathrm{sec}$ Braking efficiency, $\eta=60 \%$ Coefficient of friction, $\mu=0.7 \%$ Down grace, $s=2 \%$ Stopping sight distance = Lag distance + Braking distance

Currently no explanation available

Given, $\left|G_{1}\right| \lt \left|G_{2}\right|$ and $\left|G_{1}\right| \neq\left|G_{2}\right| \neq 0$ Statement 1: $\left|G_{1}\right| \lt \left|G_{2}\right|$ This results in a vertical sag curve. Statement 2: $\left|G_{1}\right| \lt \left|G_{2}\right|$ This results in a vertical crest curve. Statement 3: $\left|G_{1}\right| \lt \left|G_{2}\right|$ This results in a vertical crest curve.

Radius of circular curve $(\mathrm{R})=300 \mathrm{~m}$ $\mathrm{V}_{\mathrm{d}}=15 \mathrm{~m} / \mathrm{s}$ Allowable jerk $(\mathrm{C})=0.75 \mathrm{~m} / \mathrm{sec}^{3}$

$\therefore \quad$ Minimum length of transition curve $=15 \mathrm{~m}$ .

SSD $=$ Braking distance + lag distance where, Braking distance $=$ Braking distance required to come to stop Lag distance $=$ distance travelled during the perception reaction time.

As per IRC $86: 2018$ Clause 8.6 Widening of Carriageway on Curves: At sharp horizontal curves, it is necessary to widen the carriageway to provide for safe passage of vehicles. The widening required has two components: (i) mechanical widening to components for the extra width occupied by a vehicle on the curve due to tracking of the rare wheels, and (ii) psychological widening to permit easy crossing of vehicles since vehicles in a lane tend to wander more on a curve than on a straight reach. On two-lane or wider roads, it is necessary that both the above components should be fully catered for so that the lateral clearance between vehicles of curves is maintained equal to the clearance available on straight. Position of single-lane roads however is somewhat different. Since during crossing maneuvers outer wheels of vehicles have in any case to use the shoulder whether on the straight or on the curve. It is, therefore sufficient on single lane roads if only the mechanical component of widening is taken into account. Based on the above considerations, the extra width of carriageway to be provided at horizontal curves on single and two-lane roads is given in Table For multi-lane roads, the pavement widening may be calculated by adding half the widening for two-lane roads to each lane. Table : Extra width of Pavement at Horizontal Curves

Given that, $n_1=+1$ and $n_2=-2$ $n=n_1-n_2=3$ SSD = 200 m and $h_1= 1.2 m$ and $h_2= 0.15 m$ as given $n_1$ up gradient, and $n_2$ - down gradient. So curve is summit curve. Assume L > SSD

Correction for elevation = It should increase at a rate of 7% per 300 m rise in elevation from MSL. Given that Basic runway length as MSL =1500m Elevation =300m Correction $=\frac{7}{100}\times \frac{300}{300}\times 1500=105m$ The actual length of runway= 1500+150=1605m

$e+f=\frac{V^2}{127R}$ $R=\frac{V^2}{127(e+f)}$

The ideal vertical curve is a $2^{\circ}$ parabola.

Given : f = 0.36; v = 60 km; g = 9.8 $m/s^2;\; t_R=2.5s$

As per IRC : 86-1983 Desirable minimum width of median in urban roads = 5 m