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TRANSPORTATION ENGINEERING I Tutorial I (IOE Pulchowk): Complete Problem Solutions
TRANSPORTATION ENGINEERING I Tutorial I
with this comprehensive tutorial covering essential calculations for sight distance, overtaking maneuvers, skid resistance analysis, and accident investigation. Step-by-step solutions with formulas and explanations.
1. Calculate the minimum sight distance required to avoid a head-on collision of vehicles approaching from 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 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
• 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
Concept:
For a single-lane road with two-way traffic, the minimum sight distance to avoid a head-on collision is the sum of the Stopping Sight Distances (SSD) of both vehicles. One travels uphill (+7%), the other downhill (-7%).
Formula:
$$SSD = vt + \frac{v^2}{2g(\eta f \pm 0.01n)}$$
Step 1: SSD for Uphill Vehicle (SSDup)
$$SSD_{up} = (23.61 \times 2.5) + \frac{23.61^2}{2 \times 9.81 \times (0.85 \times 0.35 + 0.07)}$$
$$SSD_{up} = 59.03 + \frac{557.43}{19.62 \times (0.2975 + 0.07)}$$
$$SSD_{up} = 59.03 + \frac{557.43}{19.62 \times 0.3675}$$
$$SSD_{up} = 59.03 + \frac{557.43}{7.21} = 59.03 + 77.31 = 136.34 \text{ m}$$
Step 2: SSD for Downhill Vehicle (SSDdown)
$$SSD_{down} = (23.61 \times 2.5) + \frac{23.61^2}{2 \times 9.81 \times (0.85 \times 0.35 – 0.07)}$$
$$SSD_{down} = 59.03 + \frac{557.43}{19.62 \times (0.2975 – 0.07)}$$
$$SSD_{down} = 59.03 + \frac{557.43}{19.62 \times 0.2275}$$
$$SSD_{down} = 59.03 + \frac{557.43}{4.46} = 59.03 + 124.98 = 184.01 \text{ m}$$
Step 3: Total Sight Distance
$$\text{Total Distance} = SSD_{up} + SSD_{down}$$
$$\text{Total Distance} = 136.34 + 184.01 = 320.35 \text{ m}$$
The minimum sight distance required to avoid head-on collision is 320.35 meters.
2. 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:
a) Calculate the safe overtaking sight distance.
b) Determine the minimum length of the overtaking zone.
a) Calculate the safe overtaking sight distance.
b) Determine the minimum length of the overtaking zone.
[ marks]
Given:
• Overtaking vehicle speed (VA) = 75 kmph = 20.83 m/s
• Overtaken vehicle speed (VB) = 60 kmph = 16.67 m/s
• Acceleration (a) = 2.4 kmph/sec = 0.667 m/s²
• Design speed of oncoming vehicle (VC) = VA = 75 kmph = 20.83 m/s
• Overtaking vehicle speed (VA) = 75 kmph = 20.83 m/s
• Overtaken vehicle speed (VB) = 60 kmph = 16.67 m/s
• Acceleration (a) = 2.4 kmph/sec = 0.667 m/s²
• Design speed of oncoming vehicle (VC) = VA = 75 kmph = 20.83 m/s
a) Safe Overtaking Sight Distance (OSD)
OSD = d₁ + d₂ + d₃
Component d₁ (Reaction Distance):
Reaction time for overtaking typically t = 2.0 seconds
$$d_1 = V_B \times t = 16.67 \times 2 = 33.34 \text{ m}$$
Component d₂ (Overtaking Distance):
Spacing: $s = 0.2V_B + 6 = (0.2 \times 60) + 6 = 18 \text{ m}$
Overtaking time: $T = \sqrt{\frac{4s}{a}} = \sqrt{\frac{4 \times 18}{0.667}} = \sqrt{107.95} = 10.39 \text{ s}$
$$d_2 = V_B T + 2s = (16.67 \times 10.39) + (2 \times 18)$$
$$d_2 = 173.20 + 36 = 209.20 \text{ m}$$
Component d₃ (Oncoming Vehicle Distance):
$$d_3 = V_C \times T = 20.83 \times 10.39 = 216.42 \text{ m}$$
Total OSD:
$$OSD = d_1 + d_2 + d_3 = 33.34 + 209.20 + 216.42 = 458.96 \text{ m}$$
b) Minimum Length of Overtaking Zone
$$\text{Min Length} = 3 \times OSD = 3 \times 458.96 = 1376.88 \text{ m}$$
Note: Desirable length is typically 5 × OSD
a) Safe Overtaking Sight Distance = 458.96 m
b) Minimum Overtaking Zone Length = 1376.88 m
b) Minimum Overtaking Zone Length = 1376.88 m
3. In a road test for measuring skid resistance using skid resistance equipment, the timer indicated 4.25 seconds of brake application and braking distance indicated by color spray was measured to be 32.3 m before vehicle was brought to stop. What is the average skid resistance of the surface?
[marks]
Given:
• Time of braking (t) = 4.25 s
• Braking distance (L) = 32.3 m
• Final velocity (v) = 0 m/s
• Time of braking (t) = 4.25 s
• Braking distance (L) = 32.3 m
• Final velocity (v) = 0 m/s
Step 1: Find Initial Velocity (u)
Using average velocity formula:
$$S = \frac{u + v}{2} \times t$$
$$32.3 = \frac{u + 0}{2} \times 4.25$$
$$32.3 = u \times 2.125$$
$$u = \frac{32.3}{2.125} = 15.2 \text{ m/s}$$
Step 2: Find Deceleration (a)
$$v = u – at$$
$$0 = 15.2 – a(4.25)$$
$$a = \frac{15.2}{4.25} = 3.576 \text{ m/s}^2$$
Step 3: Find Skid Resistance (f)
$$f = \frac{a}{g} = \frac{3.576}{9.81} = 0.364$$
The average skid resistance of the surface is 0.364.
4. A vehicle hits a bridge abutment at a speed estimated by investigation as 20 kmph. Skid marks of 30 m on the pavement (f=0.35) followed by skid marks of 60 m on the gravel shoulder approaching the abutment (f=0.3). What was the initial speed of vehicle?
[ marks]
Given:
• Impact speed (v) = 20 kmph = 5.56 m/s
• Pavement: s₁ = 30 m, f₁ = 0.35
• Gravel: s₂ = 60 m, f₂ = 0.30
• Impact speed (v) = 20 kmph = 5.56 m/s
• Pavement: s₁ = 30 m, f₁ = 0.35
• Gravel: s₂ = 60 m, f₂ = 0.30
Method: Work-Energy Principle
Initial Kinetic Energy = Final Kinetic Energy + Work done against friction on both surfaces
$$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + (mgf_1 s_1) + (mgf_2 s_2)$$
Step 1: Simplify Equation
$$u^2 = v^2 + 2g(f_1 s_1 + f_2 s_2)$$
Step 2: Substitute Values
$$u^2 = (5.56)^2 + 2 \times 9.81 \times [(0.35 \times 30) + (0.30 \times 60)]$$
$$u^2 = 30.91 + 19.62 \times [10.5 + 18]$$
$$u^2 = 30.91 + 19.62 \times 28.5$$
$$u^2 = 30.91 + 559.17 = 590.08$$
Step 3: Calculate Initial Speed
$$u = \sqrt{590.08} = 24.29 \text{ m/s}$$
$$\text{Convert to kmph: } 24.29 \times 3.6 = 87.44 \text{ kmph}$$
The initial speed of the vehicle was 87.44 kmph.
5. Drivers must slow down from 100 kmph to 70 kmph to negotiate a severe curve on a rural highway. A warning sign for the curve is clearly visible for a distance of 30 m. How far in advance of the curve must the sign be posted to ensure vehicles can safely decelerate? Assume level grade, f = 0.30, perception reaction time of 2 seconds.
[marks]
Given:
• Initial speed (u) = 100 kmph = 27.78 m/s
• Final speed (v) = 70 kmph = 19.44 m/s
• Sign visibility = 30 m
• Perception reaction time (t) = 2.0 s
• Friction coefficient (f) = 0.30
• Level grade
• Initial speed (u) = 100 kmph = 27.78 m/s
• Final speed (v) = 70 kmph = 19.44 m/s
• Sign visibility = 30 m
• Perception reaction time (t) = 2.0 s
• Friction coefficient (f) = 0.30
• Level grade
Step 1: Reaction Distance (Dr)
$$D_r = u \times t = 27.78 \times 2 = 55.56 \text{ m}$$
Step 2: Braking Distance (Db)
$$D_b = \frac{u^2 – v^2}{2gf}$$
$$D_b = \frac{27.78^2 – 19.44^2}{2 \times 9.81 \times 0.30}$$
$$D_b = \frac{771.73 – 377.91}{5.886} = \frac{393.82}{5.886} = 66.91 \text{ m}$$
Step 3: Total Distance Required
$$D_{required} = D_r + D_b = 55.56 + 66.91 = 122.47 \text{ m}$$
Step 4: Sign Placement Calculation
Since the sign is visible 30 m before reaching it, the driver begins the process 30 m “earlier”:
$$\text{Distance from curve} = D_{required} – 30$$
$$\text{Distance from curve} = 122.47 – 30 = 92.47 \text{ m}$$
The warning sign must be posted at least 92.47 meters in advance of the curve.
6. The driver of a vehicle approaching a signalized intersection at a speed of 30 kmph applied brakes on seeing the signal changing from green to amber and the vehicle was brought to stop on the prescribed stop line during the amber time of 5 seconds. If the reaction time of the driver is assumed as 1.5 seconds, compute the average friction coefficient developed.
[marks]
Given:
• Approach speed (u) = 30 kmph = 8.33 m/s
• Final speed (v) = 0 m/s
• Amber time (Tamber) = 5 s
• Reaction time (tr) = 1.5 s
• Approach speed (u) = 30 kmph = 8.33 m/s
• Final speed (v) = 0 m/s
• Amber time (Tamber) = 5 s
• Reaction time (tr) = 1.5 s
Step 1: Braking Time Available
The total maneuver must be completed within the amber time:
$$\text{Braking Time } (t_b) = T_{amber} – t_r = 5.0 – 1.5 = 3.5 \text{ s}$$
Step 2: Calculate Required Deceleration (a)
$$v = u – a t_b$$
$$0 = 8.33 – a(3.5)$$
$$a = \frac{8.33}{3.5} = 2.38 \text{ m/s}^2$$
Step 3: Calculate Friction Coefficient (f)
$$f = \frac{a}{g} = \frac{2.38}{9.81} = 0.243$$
The average friction coefficient developed is 0.243.