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TRANSPORTATION ENGINEERING I Tutorial II (IOE Pulchowk): Geometric Design of Highway — Complete Problem Solutions
TRANSPORTATION ENGINEERING I Tutorial II — Geometric Design of Highway. Comprehensive step-by-step solutions covering super-elevation design, transition curves, setback distances, summit and valley curves, momentum grades, and composite curve geometry. All problems solved with full formulas, substitutions, and final answers.
1
The design speed is 90 kmph and the maximum permissible super-elevation and coefficient of lateral friction are 0.07 and 0.15. A highway engineer is designing super-elevation for a curve of radius 280 m.
a) Find the equilibrium super-elevation
b) Design the super-elevation for the curve
c) If maximum super-elevation is not to be exceeded, calculate the maximum allowable speed
d) Find the ruling radius required to maintain 90 kmph
a) Find the equilibrium super-elevation
b) Design the super-elevation for the curve
c) If maximum super-elevation is not to be exceeded, calculate the maximum allowable speed
d) Find the ruling radius required to maintain 90 kmph
[marks]
Given:
• Design speed (V) = 90 kmph
• Maximum permissible super-elevation (emax) = 0.07
• Coefficient of lateral friction (fmax) = 0.15
• Radius of curve (R) = 280 m
• Design speed (V) = 90 kmph
• Maximum permissible super-elevation (emax) = 0.07
• Coefficient of lateral friction (fmax) = 0.15
• Radius of curve (R) = 280 m
a) Equilibrium Super-elevation
Equilibrium super-elevation (eeq) is the super-elevation required when lateral friction is zero (f = 0), so the pressure on both inner and outer wheels is equal.
Formula:
$$e_{eq} = \frac{V^2}{127R}$$
$$e_{eq} = \frac{90^2}{127 \times 280} = \frac{8100}{35560} = \mathbf{0.2278} \text{ (or 22.78\%)}$$
b) Design the Super-elevation for the Curve
As per IRC recommendations, the super-elevation should initially be designed for 75% of the design speed neglecting friction.
Step 1: Calculate e for 75% of design speed
$$e = \frac{(0.75V)^2}{127R} = \frac{V^2}{225R} = \frac{90^2}{225 \times 280} = \frac{8100}{63000} = 0.1286$$
Step 2: Compare with emax
Since calculated e (0.1286) > emax (0.07), we provide maximum allowable super-elevation.
Provided Super-elevation, e = 0.07
Step 3: Check for Lateral Friction (f)
Check if friction developed at full speed is within safe limit (0.15) when e = 0.07.
$$e + f = \frac{V^2}{127R} \implies 0.07 + f = \frac{90^2}{127 \times 280} = 0.2278$$
$$f = 0.2278 – 0.07 = 0.1578$$
Since calculated f (0.1578) > fmax (0.15), the design is unsafe for 90 kmph. Speed needs to be restricted.
c) Maximum Allowable Speed
Formula:
$$e_{max} + f_{max} = \frac{V_a^2}{127R}$$
$$0.07 + 0.15 = \frac{V_a^2}{127 \times 280}$$
$$0.22 = \frac{V_a^2}{35560}$$
$$V_a^2 = 0.22 \times 35560 = 7823.2$$
$$V_a = \sqrt{7823.2} = \mathbf{88.45 \text{ kmph}}$$
d) Ruling Radius Required for 90 kmph
Formula:
$$R_{ruling} = \frac{V^2}{127(e_{max} + f_{max})}$$
$$R_{ruling} = \frac{90^2}{127(0.07 + 0.15)} = \frac{8100}{127 \times 0.22} = \frac{8100}{27.94} = \mathbf{289.91 \text{ m}}$$
a) Equilibrium super-elevation = 0.2278 (22.78%)
b) Provided super-elevation = 0.07 (emax); f = 0.1578 > 0.15 → Unsafe for 90 kmph
c) Maximum allowable speed = 88.45 kmph
d) Ruling radius = 289.91 m
b) Provided super-elevation = 0.07 (emax); f = 0.1578 > 0.15 → Unsafe for 90 kmph
c) Maximum allowable speed = 88.45 kmph
d) Ruling radius = 289.91 m
2
A highway engineer wants to design a curve at an intersection angle of 130°. Due to restrictions the length of curve is fixed at 140 m.
a) Design the super-elevation for a design speed of 90 kmph on the straight section.
b) Find the restricted speed if required.
c) Calculate the ruling radius.
d) Determine the new length of curve using the ruling radius and chainage of PC and PT if chainage of PI is 1250 m.
a) Design the super-elevation for a design speed of 90 kmph on the straight section.
b) Find the restricted speed if required.
c) Calculate the ruling radius.
d) Determine the new length of curve using the ruling radius and chainage of PC and PT if chainage of PI is 1250 m.
[marks]
Given:
• Intersection Angle (I) = 130°
• Deflection Angle (Δ) = 180° − 130° = 50°
• Fixed length of curve (L) = 140 m
• Design Speed (V) = 90 kmph
• Chainage of PI = 1250 m
• emax = 0.07 (IRC standard), fmax = 0.15 (IRC standard)
• Intersection Angle (I) = 130°
• Deflection Angle (Δ) = 180° − 130° = 50°
• Fixed length of curve (L) = 140 m
• Design Speed (V) = 90 kmph
• Chainage of PI = 1250 m
• emax = 0.07 (IRC standard), fmax = 0.15 (IRC standard)
Preliminary: Find radius R from fixed curve length
$$L = \frac{\pi \cdot R \cdot \Delta}{180} \implies 140 = \frac{\pi \cdot R \cdot 50}{180} \implies R = \frac{140 \times 180}{50\pi} \approx \mathbf{160.43 \text{ m}}$$
a) Design Super-elevation for 90 kmph
Step 1: Calculate e for 75% design speed
$$e = \frac{V^2}{225R} = \frac{90^2}{225 \times 160.43} = \frac{8100}{36096.75} = 0.224$$
Since 0.224 > emax = 0.07 → Provide e = 0.07
Step 2: Check friction at full speed (V = 90 kmph)
$$0.07 + f = \frac{90^2}{127 \times 160.43} = \frac{8100}{20374.61} = 0.397$$
$$f = 0.397 – 0.07 = 0.327$$
Calculated f (0.327) > fmax (0.15) → Curve is unsafe for 90 kmph
b) Restricted Speed
$$e_{max} + f_{max} = \frac{V_a^2}{127R}$$
$$0.22 = \frac{V_a^2}{127 \times 160.43} = \frac{V_a^2}{20374.61}$$
$$V_a^2 = 0.22 \times 20374.61 = 4482.41$$
$$V_a = \sqrt{4482.41} \approx \mathbf{66.95 \text{ kmph}}$$
c) Ruling Radius
$$R_{ruling} = \frac{V^2}{127(e_{max} + f_{max})} = \frac{8100}{127 \times 0.22} = \frac{8100}{27.94} \approx \mathbf{289.91 \text{ m}}$$
d) New Curve Length, Chainages of PC and PT
1. New Length of Curve (L’) using R’ = 289.91 m:
$$L’ = \frac{\pi \cdot R’ \cdot \Delta}{180} = \frac{\pi \times 289.91 \times 50}{180} \approx \mathbf{252.99 \text{ m}}$$
2. Tangent Length (T):
$$T = R’ \tan\left(\frac{\Delta}{2}\right) = 289.91 \times \tan(25^\circ) = 289.91 \times 0.4663 \approx \mathbf{135.19 \text{ m}}$$
3. Chainages:
$$\text{Chainage of PC} = 1250 – 135.19 = \mathbf{1114.81 \text{ m}}$$
$$\text{Chainage of PT} = 1114.81 + 252.99 = \mathbf{1367.80 \text{ m}}$$
a) Provided e = 0.07; f = 0.327 > 0.15 → Unsafe for 90 kmph
b) Restricted speed = 66.95 kmph
c) Ruling radius = 289.91 m
d) New L’ = 252.99 m | Chainage of PC = 1114.81 m | Chainage of PT = 1367.80 m
b) Restricted speed = 66.95 kmph
c) Ruling radius = 289.91 m
d) New L’ = 252.99 m | Chainage of PC = 1114.81 m | Chainage of PT = 1367.80 m
3
The center-line elevation of a double-lane road is 410.2 m. The camber of pavement is 2.5%, cross-slope of shoulder is 5%, lane width is 3.5 m, and shoulder width is 1.5 m.
Find the elevation at the pavement edges, lane centers, and roadway edges for:
(i) Straight line camber
(ii) Parabolic camber
Find the elevation at the pavement edges, lane centers, and roadway edges for:
(i) Straight line camber
(ii) Parabolic camber
[marks]
Given:
• Centre-line elevation (Ecl) = 410.2 m
• Pavement camber (cp) = 2.5% (0.025)
• Shoulder cross-slope (cs) = 5% (0.05)
• Lane width (Wl) = 3.5 m → Distance from CL to lane center = 1.75 m
• Total pavement width (W) = 2 × 3.5 = 7.0 m
• Shoulder width (Ws) = 1.5 m
• Distance to roadway edge = 3.5 + 1.5 = 5.0 m
Note: Shoulders always have a straight-line cross-slope regardless of pavement type.
• Centre-line elevation (Ecl) = 410.2 m
• Pavement camber (cp) = 2.5% (0.025)
• Shoulder cross-slope (cs) = 5% (0.05)
• Lane width (Wl) = 3.5 m → Distance from CL to lane center = 1.75 m
• Total pavement width (W) = 2 × 3.5 = 7.0 m
• Shoulder width (Ws) = 1.5 m
• Distance to roadway edge = 3.5 + 1.5 = 5.0 m
Note: Shoulders always have a straight-line cross-slope regardless of pavement type.
(i) Straight Line Camber
For straight-line camber, elevation drops linearly from the centre-line outward.
Elevation at Lane Center (x = 1.75 m):
$$\text{Drop} = x \times c_p = 1.75 \times 0.025 = 0.04375 \text{ m}$$
$$\text{Elevation} = 410.2 – 0.04375 = \mathbf{410.156 \text{ m}}$$
Elevation at Pavement Edge (x = 3.5 m):
$$\text{Drop} = 3.5 \times 0.025 = 0.0875 \text{ m}$$
$$\text{Elevation} = 410.2 – 0.0875 = \mathbf{410.1125 \text{ m}}$$
Elevation at Roadway Edge (x = 5.0 m):
$$\text{Shoulder Drop} = W_s \times c_s = 1.5 \times 0.05 = 0.075 \text{ m}$$
$$\text{Elevation} = 410.1125 – 0.075 = \mathbf{410.0375 \text{ m}}$$
(ii) Parabolic Camber
For parabolic camber, the drop formula is:
$$y = \frac{2x^2}{nW} \quad \text{where } 1/n = 0.025 \Rightarrow n = 40, \; W = 7.0 \text{ m}$$
$$y = \frac{2x^2}{40 \times 7} = \frac{x^2}{140}$$
Elevation at Lane Center (x = 1.75 m):
$$y = \frac{1.75^2}{140} = \frac{3.0625}{140} = 0.021875 \text{ m}$$
$$\text{Elevation} = 410.2 – 0.021875 = \mathbf{410.178 \text{ m}}$$
Elevation at Pavement Edge (x = 3.5 m):
$$y = \frac{3.5^2}{140} = \frac{12.25}{140} = 0.0875 \text{ m}$$
$$\text{Elevation} = 410.2 – 0.0875 = \mathbf{410.1125 \text{ m}}$$
Note: Total drop at pavement edge for parabolic camber matches the straight-line camber by design.
Elevation at Roadway Edge:
Shoulder is always straight-line, same as case (i):
$$\text{Elevation} = 410.1125 – 0.075 = \mathbf{410.0375 \text{ m}}$$
Summary Table:
| Location | Straight Camber (m) | Parabolic Camber (m) |
|---|---|---|
| Centre-line | 410.200 | 410.200 |
| Lane Centre (x = 1.75 m) | 410.156 | 410.178 |
| Pavement Edge (x = 3.5 m) | 410.1125 | 410.1125 |
| Roadway Edge (x = 5.0 m) | 410.0375 | 410.0375 |
4
There is a horizontal highway curve of radius 400 m and length 250 m on the highway. Compute the setback distance required from the centerline of the pavement on the inner side of the curve so as to provide for:
a) Stopping sight distance of 85 m
b) Safe overtaking distance of 450 m
The distance between the centerline of road and inner lane is 1.9 m.
a) Stopping sight distance of 85 m
b) Safe overtaking distance of 450 m
The distance between the centerline of road and inner lane is 1.9 m.
[marks]
Given:
• Radius of horizontal curve (R) = 400 m
• Length of curve (L) = 250 m
• Distance from road CL to inner lane CL (d) = 1.9 m
• Radius of inner lane: R’ = R − d = 400 − 1.9 = 398.1 m
• Radius of horizontal curve (R) = 400 m
• Length of curve (L) = 250 m
• Distance from road CL to inner lane CL (d) = 1.9 m
• Radius of inner lane: R’ = R − d = 400 − 1.9 = 398.1 m
a) Setback for Stopping Sight Distance (SSD = 85 m)
Step 1: Compare L and S
L = 250 m, S = 85 m → L > S (sight line stays within curve)
Step 2: Subtended Angle (α/2)
$$\frac{\alpha}{2} = \frac{180 \times S}{2\pi \times R’} = \frac{180 \times 85}{2\pi \times 398.1} = \frac{15300}{2501.34} \approx 6.117^\circ$$
Step 3: Setback Distance (L > S formula)
$$m = R – R’\cos\left(\frac{\alpha}{2}\right) = 400 – 398.1 \times \cos(6.117^\circ)$$
$$m = 400 – 398.1 \times 0.9943 = 400 – 395.83 = \mathbf{4.17 \text{ m}}$$
b) Setback for Safe Overtaking Distance (OSD = 450 m)
Step 1: Compare L and S
L = 250 m, S = 450 m → L < S (sight line extends beyond curve)
Step 2: Subtended Angle for Curve Length
$$\frac{\alpha_c}{2} = \frac{180 \times L}{2\pi \times R’} = \frac{180 \times 250}{2\pi \times 398.1} = \frac{45000}{2501.34} \approx 17.99^\circ$$
Step 3: Setback Distance (L < S formula)
$$m = R – R’\cos\left(\frac{\alpha_c}{2}\right) + \frac{(S-L)}{2}\sin\left(\frac{\alpha_c}{2}\right)$$
$$m = 400 – 398.1\cos(17.99^\circ) + \frac{(450-250)}{2}\sin(17.99^\circ)$$
$$m = 400 – (398.1 \times 0.9511) + (100 \times 0.3088)$$
$$m = 400 – 378.63 + 30.88 = \mathbf{52.25 \text{ m}}$$
a) Setback distance for SSD (85 m) = 4.17 m
b) Setback distance for OSD (450 m) = 52.25 m
b) Setback distance for OSD (450 m) = 52.25 m
5
Calculate the minimum sight distance required to avoid a head-on collision of vehicles approaching from the opposite directions at 85 kmph. Use total perception reaction time of 2.5 seconds, coefficient of friction 0.35 and brake efficiency of 85%. The section of the road has a grade of 7%.
[marks]
Given:
• Speed (V) = 85 kmph = 85/3.6 = 23.61 m/s
• Reaction time (t) = 2.5 seconds
• Coefficient of friction (f) = 0.35
• Brake efficiency (η) = 85% = 0.85
• Grade (n) = 7% = 0.07
• Effective friction: feff = f × η = 0.35 × 0.85 = 0.2975
• Speed (V) = 85 kmph = 85/3.6 = 23.61 m/s
• Reaction time (t) = 2.5 seconds
• Coefficient of friction (f) = 0.35
• Brake efficiency (η) = 85% = 0.85
• Grade (n) = 7% = 0.07
• Effective friction: feff = f × η = 0.35 × 0.85 = 0.2975
Formula (SSD with grade):
$$SSD = vt + \frac{v^2}{2g(f_{eff} \pm n)}$$
a) SSD for Vehicle 1 (Ascending / Uphill)
Gravity assists braking → add grade (+n):
$$SSD_1 = (23.61 \times 2.5) + \frac{23.61^2}{2 \times 9.81 \times (0.2975 + 0.07)}$$
$$SSD_1 = 59.03 + \frac{557.43}{19.62 \times 0.3675} = 59.03 + \frac{557.43}{7.21} = 59.03 + 77.31 = \mathbf{136.34 \text{ m}}$$
b) SSD for Vehicle 2 (Descending / Downhill)
Gravity opposes braking → subtract grade (−n):
$$SSD_2 = (23.61 \times 2.5) + \frac{23.61^2}{2 \times 9.81 \times (0.2975 – 0.07)}$$
$$SSD_2 = 59.03 + \frac{557.43}{19.62 \times 0.2275} = 59.03 + \frac{557.43}{4.46} = 59.03 + 124.98 = \mathbf{184.01 \text{ m}}$$
Total Sight Distance:
$$\text{Total SSD} = SSD_1 + SSD_2 = 136.34 + 184.01 = \mathbf{320.35 \text{ m}}$$
Minimum sight distance to avoid head-on collision = 320.35 m
6
The speeds of overtaking and overtaken vehicles are 75 kmph and 60 kmph respectively on a two-way traffic road. If the acceleration of the overtaking vehicle is 2.4 kmph per second:
i) Calculate the safe overtaking sight distance.
ii) Determine the minimum length of the overtaking zone.
i) Calculate the safe overtaking sight distance.
ii) Determine the minimum length of the overtaking zone.
[marks]
Given:
• Speed of overtaking vehicle (V) = 75 kmph = 20.83 m/s
• Speed of overtaken vehicle (Vb) = 60 kmph = 16.67 m/s
• Acceleration (A) = 2.4 kmph/sec = 2.4 × (5/18) = 0.67 m/s²
• Road type: Two-way traffic
• Speed of overtaking vehicle (V) = 75 kmph = 20.83 m/s
• Speed of overtaken vehicle (Vb) = 60 kmph = 16.67 m/s
• Acceleration (A) = 2.4 kmph/sec = 2.4 × (5/18) = 0.67 m/s²
• Road type: Two-way traffic
i) Safe Overtaking Sight Distance (OSD = d₁ + d₂ + d₃)
Calculate d₁ (Reaction distance, t = 2.0 s IRC):
$$d_1 = v_b \times t = 16.67 \times 2.0 = \mathbf{33.34 \text{ m}}$$
Calculate d₂ (Overtaking distance):
$$s = 0.7v_b + 6 = 0.7 \times 16.67 + 6 = 11.67 + 6 = \mathbf{17.67 \text{ m}}$$
$$T = \sqrt{\frac{4s}{a}} = \sqrt{\frac{4 \times 17.67}{0.67}} = \sqrt{\frac{70.68}{0.67}} = \sqrt{105.49} = \mathbf{10.27 \text{ s}}$$
$$d_2 = (v_b \times T) + 2s = (16.67 \times 10.27) + (2 \times 17.67) = 171.20 + 35.34 = \mathbf{206.54 \text{ m}}$$
Calculate d₃ (Oncoming vehicle distance):
$$d_3 = v \times T = 20.83 \times 10.27 = \mathbf{213.92 \text{ m}}$$
Total OSD:
$$OSD = d_1 + d_2 + d_3 = 33.34 + 206.54 + 213.92 = \mathbf{453.80 \text{ m}}$$
ii) Minimum Length of Overtaking Zone
$$L_{min} = 3 \times OSD = 3 \times 453.80 = \mathbf{1361.40 \text{ m}}$$
Note: Desirable length = 5 × OSD = 2269.00 m
i) Safe Overtaking Sight Distance = 453.80 m
ii) Minimum Overtaking Zone Length = 1361.40 m
ii) Minimum Overtaking Zone Length = 1361.40 m
7
A two-lane highway with a ruling gradient of 6% has a compensated gradient of 4.25% at a horizontal curve section of length 80 m. The curve section has a sight obstruction 3 m from the edge of the carriageway. Determine the possible speed on the curve section based on stopping criterion. Assume coefficient of longitudinal friction of 0.35.
[marks]
Given:
• Ruling Gradient (G) = 6%
• Compensated Gradient (Gc) = 4.25%
• Length of Curve (L) = 80 m
• Obstruction from carriageway edge = 3 m
• Coefficient of longitudinal friction (f) = 0.35
• Reaction time (t) = 2.5 s (IRC standard)
• Two-lane road width (W) = 7.0 m
• Ruling Gradient (G) = 6%
• Compensated Gradient (Gc) = 4.25%
• Length of Curve (L) = 80 m
• Obstruction from carriageway edge = 3 m
• Coefficient of longitudinal friction (f) = 0.35
• Reaction time (t) = 2.5 s (IRC standard)
• Two-lane road width (W) = 7.0 m
Step 1: Find Radius R from Grade Compensation
$$\text{Grade Compensation} = G – G_c = 6.00\% – 4.25\% = 1.75\%$$
$$\frac{30+R}{R} = 1.75 \implies 30+R = 1.75R \implies 0.75R = 30 \implies R = \mathbf{40 \text{ m}}$$
Step 2: Calculate Setback Distance (m)
Distance from road CL to inner lane CL: d = 3.5/2 = 1.75 m
Radius of inner lane: R − d = 40 − 1.75 = 38.25 m
$$m = \frac{3.5}{2} + 3 = 1.75 + 3 = \mathbf{4.75 \text{ m}}$$
Step 3: Calculate Subtended Angle (α/2)
$$m = (R-d)\left[1 – \cos\left(\frac{\alpha}{2}\right)\right]$$
$$4.75 = 38.25 – 38.25\cos\left(\frac{\alpha}{2}\right)$$
$$\cos\left(\frac{\alpha}{2}\right) = \frac{33.5}{38.25} = 0.8758 \implies \frac{\alpha}{2} = \cos^{-1}(0.8758) = \mathbf{28.85^\circ}$$
Step 4: Available Sight Distance (S)
$$\frac{\alpha}{2} = \frac{90 \times S}{\pi(R-d)}$$
$$S = \frac{28.85 \times \pi \times 38.25}{90} = \mathbf{38.53 \text{ m}}$$
Check: 38.53 m < 80 m → assumption S < L is correct ✓
Step 5: Calculate Safe Speed (worst case: downhill at compensated gradient)
$$38.53 = 2.5v + \frac{v^2}{2 \times 9.81 \times (0.35 – 0.0425)}$$
$$38.53 = 2.5v + \frac{v^2}{19.62 \times 0.3075} = 2.5v + \frac{v^2}{6.033}$$
Multiply by 6.033:
$$v^2 + 15.08v – 232.45 = 0$$
$$v = \frac{-15.08 + \sqrt{(15.08)^2 + 4 \times 232.45}}{2} = \frac{-15.08 + \sqrt{227.4 + 929.8}}{2} = \frac{-15.08 + 34.02}{2} = \mathbf{9.47 \text{ m/s}}$$
$$V = 9.47 \times 3.6 = \mathbf{34.09 \text{ kmph}}$$
Possible safe speed on the curve section = 34.09 kmph ≈ 34 kmph
8
A transition curve is to be provided to connect a straight and a circular curve of radius 420 m on a two-lane road in plain terrain with moderate rainfall. Design the length of the transition curve assuming suitable data. The design speed is 75 kmph, and the rate of introduction of super-elevation is 1 in 180. Determine the elevations of the outer edge at two typical straight and circular sections if the center-line elevations are 318.2 m and 318.6 m respectively, with 3% parabolic camber.
[marks]
Given:
• Design speed (V) = 75 kmph → v = 75/3.6 = 20.83 m/s
• Radius (R) = 420 m
• Rate of super-elevation introduction (N) = 1 in 180
• Camber = 3% (0.03)
• CL elevation (straight) = 318.2 m; CL elevation (circular) = 318.6 m
• W = 7.0 m (two-lane), l = 6.0 m (standard wheelbase)
• Design speed (V) = 75 kmph → v = 75/3.6 = 20.83 m/s
• Radius (R) = 420 m
• Rate of super-elevation introduction (N) = 1 in 180
• Camber = 3% (0.03)
• CL elevation (straight) = 318.2 m; CL elevation (circular) = 318.6 m
• W = 7.0 m (two-lane), l = 6.0 m (standard wheelbase)
Design of Transition Curve Length (Ls)
Criterion A: Rate of Change of Centrifugal Acceleration (C)
$$C = \frac{80}{75+V} = \frac{80}{75+75} = \frac{80}{150} = 0.533 \text{ m/s}^3 \quad (\text{within IRC limits 0.5 to 0.8})$$
$$L_{s1} = \frac{v^3}{CR} = \frac{(20.83)^3}{0.533 \times 420} = \frac{9039.2}{223.86} = \mathbf{40.38 \text{ m}}$$
Criterion B: Rate of Introduction of Super-elevation
$$e = \frac{V^2}{225R} = \frac{75^2}{225 \times 420} = \frac{5625}{94500} = 0.0595 \text{ (5.95\%)}$$
Since 3% < 5.95% < 7% → Provide e = 0.0595
$$W_e = \frac{nl^2}{2R} + \frac{V}{9.5\sqrt{R}} = \frac{2 \times 36}{840} + \frac{75}{9.5 \times 20.49} = 0.086 + 0.385 = 0.471 \text{ m}$$
$$B = W + W_e = 7.0 + 0.471 = 7.471 \text{ m}$$
$$E = e \times \frac{B}{2} = 0.0595 \times 3.7355 = 0.222 \text{ m}$$
$$L_{s2} = E \times N = 0.222 \times 180 = \mathbf{39.96 \text{ m}}$$
Criterion C: IRC Empirical Formula (Plain/Rolling Terrain)
$$L_{s3} = \frac{2.7V^2}{R} = \frac{2.7 \times 75^2}{420} = \frac{15187.5}{420} = \mathbf{36.16 \text{ m}}$$
Design Length: Maximum of Ls1, Ls2, Ls3 = max(40.38, 39.96, 36.16) = 40.38 m
Adopted Ls = 41.0 m
Elevations of the Outer Edge
At the Straight Section (CL = 318.2 m):
Parabolic camber, no extra widening. Distance from CL to outer edge = W/2 = 3.5 m.
$$\text{Drop} = 3.5 \times 0.03 = 0.105 \text{ m}$$
$$\text{Elevation of Outer Edge} = 318.2 – 0.105 = \mathbf{318.095 \text{ m}}$$
At the Circular Section (CL = 318.6 m):
Super-elevation and extra widening fully applied. Outer edge raised above CL.
$$\text{Rise} = e \times \frac{B}{2} = 0.0595 \times 3.735 = 0.222 \text{ m}$$
$$\text{Elevation of Outer Edge} = 318.6 + 0.222 = \mathbf{318.822 \text{ m}}$$
Length of transition curve = 41.0 m (adopted)
Outer edge elevation at straight section = 318.095 m
Outer edge elevation at circular section = 318.822 m
Outer edge elevation at straight section = 318.095 m
Outer edge elevation at circular section = 318.822 m
9
The angle of intersection between two straights is 137.23°. The spiral angle for each transition curve is 8.35°. Calculate the length of transition curve, combined length of curve, and tangent length if the radius is 325 m.
[marks]
Given:
• Angle of Intersection (I) = 137.23°
• Deflection Angle (Δ) = 180° − 137.23° = 42.77°
• Spiral angle (θs) = 8.35°
• Radius (R) = 325 m
• Angle of Intersection (I) = 137.23°
• Deflection Angle (Δ) = 180° − 137.23° = 42.77°
• Spiral angle (θs) = 8.35°
• Radius (R) = 325 m
1. Length of Transition Curve (Ls)
$$\theta_s \text{ (deg)} = \frac{90 \times L_s}{\pi R} \implies L_s = \frac{\theta_s \times \pi \times R}{90}$$
$$L_s = \frac{8.35 \times \pi \times 325}{90} = \frac{8527.38}{90} = \mathbf{94.75 \text{ m}}$$
2. Combined Length of Curve (L)
$$\Delta_c = \Delta – 2\theta_s = 42.77^\circ – 2(8.35^\circ) = 42.77^\circ – 16.70^\circ = 26.07^\circ$$
$$L_c = \frac{\pi R \Delta_c}{180} = \frac{\pi \times 325 \times 26.07}{180} = \mathbf{147.85 \text{ m}}$$
$$L = L_c + 2L_s = 147.85 + 2(94.75) = 147.85 + 189.50 = \mathbf{337.35 \text{ m}}$$
3. Tangent Length (Ts)
$$p = \frac{L_s^2}{24R} = \frac{(94.75)^2}{24 \times 325} = \frac{8977.56}{7800} = \mathbf{1.15 \text{ m}}$$
$$k = \frac{L_s}{2} – \frac{L_s^3}{240R^2} = \frac{94.75}{2} – \frac{(94.75)^3}{240 \times (325)^2} = 47.375 – 0.033 = \mathbf{47.34 \text{ m}}$$
$$T_s = (R+p)\tan\left(\frac{\Delta}{2}\right) + k = (325+1.15)\tan(21.385^\circ) + 47.34$$
$$T_s = 326.15 \times 0.3916 + 47.34 = 127.72 + 47.34 = \mathbf{175.06 \text{ m}}$$
Length of Transition Curve (Ls) = 94.75 m
Combined Length of Curve (L) = 337.35 m
Tangent Length (Ts) = 175.06 m
Combined Length of Curve (L) = 337.35 m
Tangent Length (Ts) = 175.06 m
10
A vehicle moving on a horizontal curve at a design speed of 50 kmph develops a centrifugal ratio of 1/6. The deflection angle of the curve is 55°. Calculate:
a) Radius of the circular curve
b) Length of transition curve (rate of change of centrifugal acceleration criteria)
c) Total length of the composite curve
a) Radius of the circular curve
b) Length of transition curve (rate of change of centrifugal acceleration criteria)
c) Total length of the composite curve
[marks]
Given:
• Design speed (V) = 50 kmph → v = 50 × (5/18) = 13.89 m/s
• Centrifugal ratio (P/W) = 1/6
• Deflection angle (Δ) = 55°
• Design speed (V) = 50 kmph → v = 50 × (5/18) = 13.89 m/s
• Centrifugal ratio (P/W) = 1/6
• Deflection angle (Δ) = 55°
a) Radius of the Circular Curve
$$\frac{P}{W} = \frac{v^2}{gR} \implies \frac{1}{6} = \frac{(13.89)^2}{9.81 \times R} = \frac{192.93}{9.81R}$$
$$9.81R = 6 \times 192.93 = 1157.58 \implies R = \frac{1157.58}{9.81} = \mathbf{117.99 \text{ m}}$$
b) Length of Transition Curve
$$C = \frac{80}{75+V} = \frac{80}{75+50} = \frac{80}{125} = 0.64 \text{ m/s}^3 \quad (\text{within IRC 0.5–0.8})$$
$$L_s = \frac{v^3}{CR} = \frac{(13.89)^3}{0.64 \times 117.99} = \frac{2680.0}{75.51} = \mathbf{35.49 \text{ m}}$$
c) Total Length of the Composite Curve
Step 1: Spiral angle θs
$$\theta_s = \frac{90 \times L_s}{\pi R} = \frac{90 \times 35.49}{\pi \times 117.99} = \frac{3194.1}{370.67} = \mathbf{8.62^\circ}$$
Step 2: Central angle for circular curve
$$\Delta_c = \Delta – 2\theta_s = 55^\circ – 2(8.62^\circ) = 55^\circ – 17.24^\circ = \mathbf{37.76^\circ}$$
Step 3: Length of circular curve
$$L_c = \frac{\pi R \Delta_c}{180} = \frac{\pi \times 117.99 \times 37.76}{180} = \mathbf{77.75 \text{ m}}$$
Step 4: Total length
$$L = L_c + 2L_s = 77.75 + 2(35.49) = 77.75 + 70.98 = \mathbf{148.73 \text{ m}}$$
a) Radius = 117.99 m
b) Length of transition curve = 35.49 m
c) Total length of composite curve = 148.73 m
b) Length of transition curve = 35.49 m
c) Total length of composite curve = 148.73 m
11
A −2% gradient meets a +1.5% gradient at a chainage of 2000 m and at the reduced level of 600 m. If the design speed of the road is 80 kmph, determine the RL and chainage of the tangent points and lowest point on the curve. Assume height of headlight is 0.75 m, frictional coefficient f = 0.4, and α = 1°.
[marks]
Given:
• First gradient (n1) = −2% = −0.02; Second gradient (n2) = +1.5% = +0.015
• Deviation angle N = |−0.02 − 0.015| = 0.035
• Design Speed = 80 kmph → v = 22.22 m/s
• Chainage of V = 2000 m, RL of V = 600 m
• h1 = 0.75 m, α = 1°, f = 0.4, t = 2.5 s (IRC)
• First gradient (n1) = −2% = −0.02; Second gradient (n2) = +1.5% = +0.015
• Deviation angle N = |−0.02 − 0.015| = 0.035
• Design Speed = 80 kmph → v = 22.22 m/s
• Chainage of V = 2000 m, RL of V = 600 m
• h1 = 0.75 m, α = 1°, f = 0.4, t = 2.5 s (IRC)
Step 1: Stopping Sight Distance (SSD)
$$SSD = vt + \frac{v^2}{2gf} = (22.22 \times 2.5) + \frac{(22.22)^2}{2 \times 9.81 \times 0.4} = 55.55 + 62.91 = \mathbf{118.46 \text{ m}}$$
Step 2: Length of Valley Curve
Criterion A — Headlight Sight Distance (Assume L > SSD):
$$L_A = \frac{N \times SSD^2}{2(h_1 + SSD \cdot \tan\alpha)} = \frac{0.035 \times (118.46)^2}{2(0.75 + 118.46\tan 1^\circ)} = \frac{491.15}{2(0.75+2.068)} = \frac{491.15}{5.636} = 87.14 \text{ m}$$
87.14 m < SSD (118.46 m) → Assumption L > SSD is invalid
Assume L < SSD:
$$L_A = 2 \cdot SSD – \frac{2(h_1 + SSD\cdot\tan\alpha)}{N} = 2(118.46) – \frac{5.636}{0.035} = 236.92 – 161.03 = \mathbf{75.89 \text{ m}}$$
75.89 m < 118.46 m → valid ✓
Criterion B — Passenger Comfort (C = 0.6 m/s³):
$$L_B = 2\sqrt{\frac{Nv^3}{C}} = 2\sqrt{\frac{0.035 \times (22.22)^3}{0.6}} = 2\sqrt{639.95} = 2 \times 25.3 = \mathbf{50.60 \text{ m}}$$
Design L = max(75.89, 50.60) = 75.89 m → Adopted L = 76.0 m
Step 3: Chainage of Tangent Points
$$T_1 = 2000 – \frac{76}{2} = 2000 – 38 = \mathbf{1962 \text{ m}}$$
$$T_2 = 2000 + 38 = \mathbf{2038 \text{ m}}$$
Step 4: RL of Tangent Points
$$RL_{T1} = 600 – (-0.02 \times 38) = 600 + 0.76 = \mathbf{600.76 \text{ m}}$$
$$RL_{T2} = 600 + (0.015 \times 38) = 600 + 0.57 = \mathbf{600.57 \text{ m}}$$
Step 5: Lowest Point on Curve
$$x = L \times \frac{n_1}{N} = 76 \times \frac{0.02}{0.035} = \mathbf{43.43 \text{ m from } T_1}$$
$$\text{Chainage of Lowest Point} = 1962 + 43.43 = \mathbf{2005.43 \text{ m}}$$
$$RL_{lowest} = 600.76 – (0.02 \times 43.43) + \frac{0.035 \times (43.43)^2}{2 \times 76}$$
$$= 600.76 – 0.8686 + 0.4343 = \mathbf{600.326 \text{ m}}$$
Design length of valley curve = 76.0 m
Chainage T₁ = 1962 m, RL T₁ = 600.76 m
Chainage T₂ = 2038 m, RL T₂ = 600.57 m
Chainage of Lowest Point = 2005.43 m, RL = 600.326 m
Chainage T₁ = 1962 m, RL T₁ = 600.76 m
Chainage T₂ = 2038 m, RL T₂ = 600.57 m
Chainage of Lowest Point = 2005.43 m, RL = 600.326 m
12
Two approaching cars were caught in a head-on collision at a vertical summit curve connecting a 5% ascending gradient with a 4.5% descending gradient. The posted speed on this road is 80 kmph. The highway engineer found that the highest point of the vertical curve lies at a distance of 150 m from the beginning of the vertical curve. Check if the crash is due to a fault in the posted speed sign. (Driver eye height = 1.1 m, reaction time = 2.5 sec, f = 0.35). What should be the posted speed if there is a fault?
[marks]
Given:
• n1 = +5% = +0.05 (ascending); n2 = −4.5% = −0.045 (descending)
• Deviation angle N = n1 − n2 = 0.05 + 0.045 = 0.095
• Distance to highest point (x) = 150 m
• Posted speed = 80 kmph → v = 22.22 m/s
• h₁ = h₂ = 1.1 m, t = 2.5 s, f = 0.35
• n1 = +5% = +0.05 (ascending); n2 = −4.5% = −0.045 (descending)
• Deviation angle N = n1 − n2 = 0.05 + 0.045 = 0.095
• Distance to highest point (x) = 150 m
• Posted speed = 80 kmph → v = 22.22 m/s
• h₁ = h₂ = 1.1 m, t = 2.5 s, f = 0.35
Step 1: Find Curve Length (L)
$$x = \frac{n_1 \times L}{N} \implies 150 = \frac{0.05 \times L}{0.095} \implies L = \frac{150 \times 0.095}{0.05} = \mathbf{285 \text{ m}}$$
Step 2: SSD for 80 kmph
$$SSD = (22.22 \times 2.5) + \frac{(22.22)^2}{2 \times 9.81 \times 0.35} = 55.55 + 71.90 = \mathbf{127.45 \text{ m}}$$
$$\text{Required Distance} = 2 \times SSD = \mathbf{254.90 \text{ m}}$$
Step 3: Available Mutual Sight Distance (S)
For summit curve with h₁ = h₂ = 1.1 m (assuming S < L):
$$L = \frac{N \times S^2}{2(\sqrt{h_1}+\sqrt{h_2})^2}$$
$$285 = \frac{0.095 \times S^2}{2(\sqrt{1.1}+\sqrt{1.1})^2} = \frac{0.095 \times S^2}{2 \times (2 \times 1.0488)^2} = \frac{0.095 \times S^2}{8.80}$$
$$S^2 = \frac{285 \times 8.80}{0.095} = 26400 \implies S = \mathbf{162.48 \text{ m}}$$
Check: 162.48 m < 285 m → S < L valid ✓
Step 4: Conclusion
Available sight distance (162.48 m) < Required distance (254.90 m) → YES, crash is due to fault in the posted speed sign.
Step 5: Find Correct Posted Speed
For safety: 2 × SSD ≤ 162.48 m → SSDsafe ≤ 81.24 m
$$2.5v + \frac{v^2}{6.867} = 81.24 \implies v^2 + 17.167v – 557.87 = 0$$
$$v = \frac{-17.167 + \sqrt{(17.167)^2 + 4 \times 557.87}}{2} = \frac{-17.167 + 50.26}{2} = \mathbf{16.54 \text{ m/s}}$$
$$V = 16.54 \times 3.6 = \mathbf{59.54 \text{ kmph}} \approx 60 \text{ kmph}$$
Available sight distance = 162.48 m vs Required = 254.90 m
YES — crash is due to fault in posted speed limit
Correct posted speed limit should be 60 kmph
YES — crash is due to fault in posted speed limit
Correct posted speed limit should be 60 kmph
13
Design the length of valley curve with a descending grade of 1/35 and ascending grade of 1/45. The design speed is 80 kmph. Determine the RL of beginning, lowest and end point of curve if the RL of PVI is 212.36 m so as to fulfill both comfort condition and headlight sight distance for night visibility. Also determine the apex distance and mid-ordinate of the curve. Assume f = 0.35, C = 60 cm/s³.
[marks]
Given:
• n1 = −1/35 = −0.02857 (descending); n2 = +1/45 = +0.02222 (ascending)
• N = |−0.02857 − 0.02222| = 0.05079
• Design speed = 80 kmph → v = 22.22 m/s
• RL of PVI = 212.36 m, f = 0.35, C = 0.6 m/s³
• h₁ = 0.75 m, α = 1°, t = 2.5 s (IRC)
• n1 = −1/35 = −0.02857 (descending); n2 = +1/45 = +0.02222 (ascending)
• N = |−0.02857 − 0.02222| = 0.05079
• Design speed = 80 kmph → v = 22.22 m/s
• RL of PVI = 212.36 m, f = 0.35, C = 0.6 m/s³
• h₁ = 0.75 m, α = 1°, t = 2.5 s (IRC)
Step 1: SSD
$$SSD = (22.22 \times 2.5) + \frac{(22.22)^2}{2 \times 9.81 \times 0.35} = 55.55 + 71.90 = \mathbf{127.45 \text{ m}}$$
Step 2: Length of Valley Curve
Criterion A — Comfort Condition:
$$L_c = 2\sqrt{\frac{Nv^3}{C}} = 2\sqrt{\frac{0.05079 \times (22.22)^3}{0.6}} = 2\sqrt{928.66} = 2 \times 30.47 = \mathbf{60.94 \text{ m}}$$
Criterion B — Headlight Sight Distance (Assume L > SSD):
$$L_h = \frac{N \times SSD^2}{1.5 + 0.035 \times SSD} = \frac{0.05079 \times (127.45)^2}{1.5 + 0.035 \times 127.45} = \frac{825.01}{5.96} = \mathbf{138.42 \text{ m}}$$
138.42 m > 127.45 m → assumption L > SSD valid ✓
Design L = max(60.94, 138.42) = 138.42 m → Adopted L = 139.0 m
Step 3: Reduced Levels (L/2 = 69.5 m from PVI)
$$RL_{BVC} = 212.36 + \left(\frac{1}{35} \times 69.5\right) = 212.36 + 1.986 = \mathbf{214.346 \text{ m}}$$
$$RL_{EVC} = 212.36 + \left(\frac{1}{45} \times 69.5\right) = 212.36 + 1.544 = \mathbf{213.904 \text{ m}}$$
Step 4: Lowest Point
$$x = L \times \frac{n_1}{N} = 139 \times \frac{0.02857}{0.05079} = 139 \times 0.5625 = \mathbf{78.19 \text{ m from BVC}}$$
$$RL_{lowest} = 214.346 – (0.02857 \times 78.19) + \frac{0.05079 \times (78.19)^2}{2 \times 139}$$
$$= 214.346 – 2.234 + 1.117 = \mathbf{213.229 \text{ m}}$$
Step 5: Apex Distance / Mid-Ordinate
$$e = \frac{N \times L}{8} = \frac{0.05079 \times 139}{8} = \mathbf{0.882 \text{ m}}$$
Length of valley curve = 139.0 m
RL of BVC = 214.346 m | RL of EVC = 213.904 m
Lowest point at x = 78.19 m from BVC, RL = 213.229 m
Apex Distance / Mid-Ordinate = 0.882 m
RL of BVC = 214.346 m | RL of EVC = 213.904 m
Lowest point at x = 78.19 m from BVC, RL = 213.229 m
Apex Distance / Mid-Ordinate = 0.882 m
14
A setback distance for a four-lane road was fixed as 25 m. The length of the horizontal curve was designed to be 150 m, with deflection angle of 80° between tangents. On the next section, a vertical curve was to be designed with 3% descending and 2% ascending gradient. Find:
a) The design speed on the horizontal curve section
b) The length of the vertical curve
A sewer pipe of 5 m diameter was to be placed at the lowest point with minimum fill of 1.2 m above it. If the pipe was at chainage of 1095 m and elevation 312.5 m, find the chainage and elevation of beginning and end of the vertical curve.
a) The design speed on the horizontal curve section
b) The length of the vertical curve
A sewer pipe of 5 m diameter was to be placed at the lowest point with minimum fill of 1.2 m above it. If the pipe was at chainage of 1095 m and elevation 312.5 m, find the chainage and elevation of beginning and end of the vertical curve.
[marks]
Given:
• Setback distance (m) = 25 m; Lc = 150 m; Δ = 80°
• 4-lane road (total width = 14 m)
• Vertical curve: n₁ = −3%, n₂ = +2% → N = 0.05
• Sewer pipe: diameter = 5 m, min fill = 1.2 m
• Chainage of lowest point = 1095 m, elevation = 312.5 m
• Setback distance (m) = 25 m; Lc = 150 m; Δ = 80°
• 4-lane road (total width = 14 m)
• Vertical curve: n₁ = −3%, n₂ = +2% → N = 0.05
• Sewer pipe: diameter = 5 m, min fill = 1.2 m
• Chainage of lowest point = 1095 m, elevation = 312.5 m
Part A: Design Speed on Horizontal Curve
Step 1: Radius R
$$L_c = \frac{\pi R \Delta}{180} \implies 150 = \frac{\pi R \times 80}{180} \implies R = \frac{150 \times 180}{80\pi} = \mathbf{107.43 \text{ m}}$$
Step 2: Inner Lane Geometry (4-lane, width = 14 m)
Inner half = 7 m. Inner lane CL from road CL: d = 7.0 − 1.75 = 5.25 m
$$R’ = R – d = 107.43 – 5.25 = 102.18 \text{ m}$$
Step 3: Available Sight Distance (assuming S < L)
$$m = R – R’\cos\left(\frac{\alpha}{2}\right) \implies 25 = 107.43 – 102.18\cos\left(\frac{\alpha}{2}\right)$$
$$\cos\left(\frac{\alpha}{2}\right) = \frac{82.43}{102.18} = 0.8067 \implies \frac{\alpha}{2} = 0.6322 \text{ rad}$$
$$S = 2 \times 0.6322 \times 102.18 = \mathbf{129.20 \text{ m}}$$
129.20 m < 150 m → valid ✓
Step 4: Design Speed from SSD
$$129.20 = 2.5v + \frac{v^2}{6.867} \implies v^2 + 17.167v – 887.19 = 0$$
$$v = \frac{-17.167 + \sqrt{(17.167)^2 + 4 \times 887.19}}{2} = 22.41 \text{ m/s} \implies V = 22.41 \times 3.6 = \mathbf{80.68 \approx 80 \text{ kmph}}$$
Part B: Length of Vertical Valley Curve
At V = 80 kmph: SSD = 127.47 m
$$L_v = \frac{N \times SSD^2}{1.5 + 0.035 \times SSD} = \frac{0.05 \times (127.47)^2}{1.5 + 0.035 \times 127.47} = \frac{812.43}{5.96} = \mathbf{136.28 \text{ m}}$$
Part C: Chainages and Elevations
Step 1: Road elevation at lowest point
$$E_{lowest} = 312.5 + 5.0 + 1.2 = \mathbf{318.70 \text{ m}}$$
Step 2: Distance from BVC to lowest point
$$x = L_v \times \frac{n_1}{N} = 136.28 \times \frac{0.03}{0.05} = \mathbf{81.77 \text{ m}}$$
Step 3: Chainages
$$\text{Chainage BVC} = 1095 – 81.77 = \mathbf{1013.23 \text{ m}}$$
$$\text{Chainage EVC} = 1013.23 + 136.28 = \mathbf{1149.51 \text{ m}}$$
Step 4: Elevation of BVC
$$318.70 = E_{BVC} – (0.03 \times 81.77) + \frac{0.05 \times (81.77)^2}{2 \times 136.28}$$
$$318.70 = E_{BVC} – 2.453 + 1.226 \implies E_{BVC} = \mathbf{319.93 \text{ m}}$$
Step 5: Elevation of EVC
$$E_{PVI} = 319.93 – 0.03 \times 68.14 = 317.88 \text{ m}$$
$$E_{EVC} = 317.88 + 0.02 \times 68.14 = \mathbf{319.24 \text{ m}}$$
a) Design speed = 80 kmph
b) Length of vertical curve = 136.28 m
Chainage BVC = 1013.23 m, Elevation BVC = 319.93 m
Chainage EVC = 1149.51 m, Elevation EVC = 319.24 m
b) Length of vertical curve = 136.28 m
Chainage BVC = 1013.23 m, Elevation BVC = 319.93 m
Chainage EVC = 1149.51 m, Elevation EVC = 319.24 m
15
A vertical curve of length 35 m is designed to connect a mild ascending gradient of 3% to a steep ascending gradient of 9.5% based on headlight sight distance condition (neglecting the effect of gradient).
i) Calculate the design speed of the highway section.
ii) Determine the maximum allowable length of the steeper gradient to design it as a momentum grade with an exit speed of 15 kmph. Assume headlight beam angle = 1°, height of headlight = 0.75 m, f = 0.35, coefficient of inertia = 0.85.
i) Calculate the design speed of the highway section.
ii) Determine the maximum allowable length of the steeper gradient to design it as a momentum grade with an exit speed of 15 kmph. Assume headlight beam angle = 1°, height of headlight = 0.75 m, f = 0.35, coefficient of inertia = 0.85.
[marks]
Given:
• n1 = +3% = +0.03; n2 = +9.5% = +0.095
• N = |0.03 − 0.095| = 0.065
• L = 35 m (valley curve)
• h = 0.75 m, α = 1°, f = 0.35, Ci = 0.85
• Exit speed v₂ = 15 kmph = 4.167 m/s
• n1 = +3% = +0.03; n2 = +9.5% = +0.095
• N = |0.03 − 0.095| = 0.065
• L = 35 m (valley curve)
• h = 0.75 m, α = 1°, f = 0.35, Ci = 0.85
• Exit speed v₂ = 15 kmph = 4.167 m/s
Part I: Design Speed
Step 1: Find Sight Distance S (assuming L < S)
$$L = 2S – \frac{2h + 2S\tan\alpha}{N}$$
$$35 = 2S – \frac{2(0.75) + 2S\tan(1^\circ)}{0.065} = 2S – \frac{1.5 + 0.0349S}{0.065}$$
$$35 = 2S – 23.077 – 0.5369S$$
$$35 + 23.077 = 1.4631S$$
$$S = \frac{58.077}{1.4631} = \mathbf{39.70 \text{ m}}$$
Check: 35 m < 39.70 m → assumption L < S valid ✓
Step 2: Find Design Speed (SSD = S = 39.70 m, t = 2.5 s)
$$39.70 = 2.5v + \frac{v^2}{6.867} \implies v^2 + 17.1675v – 272.62 = 0$$
$$v = \frac{-17.1675 + \sqrt{(17.1675)^2 + 4 \times 272.62}}{2} = \frac{-17.1675 + 37.218}{2} = \mathbf{10.025 \text{ m/s}}$$
$$V = 10.025 \times 3.6 = \mathbf{36.09 \approx 36 \text{ kmph}}$$
Part II: Maximum Length of Momentum Grade
Step 1: Change in Kinetic Energy
v₁ = 10.025 m/s (entry), v₂ = 4.167 m/s (exit)
$$\Delta v^2 = v_1^2 – v_2^2 = (10.025)^2 – (4.167)^2 = 100.50 – 17.36 = 83.14 \text{ m}^2/\text{s}^2$$
$$\text{Loss of KE} = \frac{W}{2g}(1+C_i)\Delta v^2 = \frac{W \times 1.85}{19.62} \times 83.14 = 7.84W$$
Step 2: Excess Grade Resistance
$$\text{Excess Decelerating Force} = W(n_2 – n_1) = W(0.095 – 0.03) = 0.065W$$
Step 3: Momentum Grade Length
$$7.84W = 0.065W \times L_m \implies L_m = \frac{7.84}{0.065} = \mathbf{120.6 \text{ m}}$$
i) Design speed = 36.09 ≈ 36 kmph
ii) Maximum allowable length of momentum grade = 120.6 m
ii) Maximum allowable length of momentum grade = 120.6 m
16
A national highway of two lanes has a curve of 300 m radius to be set out to connect two straights in hilly terrain. The maximum speed of moving vehicles on this curve is restricted to 60 kmph. Transition curves are to be introduced at each end of the circular curve. Calculate:
a) Suitable length of transition curve
b) Necessary shift of the circular curve
c) Chainage at the beginning and end of the curve
Given: Angle of intersection = 110° 20′, Rate of change of centrifugal acceleration = 0.60 m/s³, Chainage at PI = 1020 m
a) Suitable length of transition curve
b) Necessary shift of the circular curve
c) Chainage at the beginning and end of the curve
Given: Angle of intersection = 110° 20′, Rate of change of centrifugal acceleration = 0.60 m/s³, Chainage at PI = 1020 m
[marks]
Given:
• Design speed (V) = 60 kmph → v = 60 × (5/18) = 16.67 m/s
• Radius (R) = 300 m
• Intersection Angle (I) = 110° 20′
• Deflection Angle (Δ) = 180° − 110° 20′ = 69° 40′ (69.667°)
• C = 0.60 m/s³
• Chainage of PI = 1020 m, Terrain: Hilly
• Design speed (V) = 60 kmph → v = 60 × (5/18) = 16.67 m/s
• Radius (R) = 300 m
• Intersection Angle (I) = 110° 20′
• Deflection Angle (Δ) = 180° − 110° 20′ = 69° 40′ (69.667°)
• C = 0.60 m/s³
• Chainage of PI = 1020 m, Terrain: Hilly
a) Suitable Length of Transition Curve (Ls)
Criterion 1: Rate of Change of Centrifugal Acceleration
$$L_{s1} = \frac{v^3}{CR} = \frac{(16.67)^3}{0.60 \times 300} = \frac{4632.4}{180} = \mathbf{25.74 \text{ m}} \approx 25.72 \text{ m (precise)}$$
Criterion 2: IRC Empirical Formula (Hilly Terrain)
$$L_{s2} = \frac{V^2}{R} = \frac{60^2}{300} = \frac{3600}{300} = \mathbf{12 \text{ m}}$$
Adopted Ls = 25.72 m (maximum of both criteria)
b) Shift of the Circular Curve (S)
$$S = \frac{L_s^2}{24R} = \frac{(25.72)^2}{24 \times 300} = \frac{661.52}{7200} = \mathbf{0.092 \text{ m}}$$
c) Chainages at Beginning and End of Curve
Step 1: Tangent Length (T)
$$T = (R+S)\tan\left(\frac{\Delta}{2}\right) + \frac{L_s}{2}$$
$$\frac{\Delta}{2} = 34^\circ 50’$$
$$T = (300+0.092)\tan(34^\circ 50′) + \frac{25.72}{2} = (300.092 \times 0.6958) + 12.86 = 208.80 + 12.86 = \mathbf{221.66 \text{ m}}$$
Step 2: Total Length of Curve System
$$L_{total} = \frac{\pi R \Delta}{180} + L_s = \frac{\pi \times 300 \times 69.667}{180} + 25.72 = 364.77 + 25.72 = \mathbf{390.49 \text{ m}}$$
Step 3: Chainages
$$\text{Chainage of PC (Start)} = 1020 – 221.66 = \mathbf{798.34 \text{ m}}$$
$$\text{Chainage of PT (End)} = 798.34 + 390.49 = \mathbf{1188.83 \text{ m}}$$
a) Length of transition curve = 25.72 m
b) Shift of circular curve = 0.092 m
c) Chainage at beginning (PC) = 798.34 m | Chainage at end (PT) = 1188.83 m
b) Shift of circular curve = 0.092 m
c) Chainage at beginning (PC) = 798.34 m | Chainage at end (PT) = 1188.83 m