CEE 3604: Introduction to Transportation Engineerings Spring 2021
Exam 1 (Take Home): Open Notes. Internet Use is Allowed.
Solution
Instructor: TraniProblem 1 : Answer with True or False(20%)
Question Answer
Subway guideway capacity increases with train cruise speed True if train cruises below the optimal speed for maximum capacity. Otherwise False. For the same speed, the sight stopping distances of a truck and a car are different FALSE
Runway capacity increases if aircraft fly faster on the final approach segment TRUE Increased dwell times at a station reduce the station capacity TRUE Automated People Movers at airports have guideway lengths greater than 6 miles FALSE Greenberg proposed the first traffic flow model in the US FALSE
Time-mean speed is always higher than space-mean speed Both
Problem 2 (25%) - Show Your Work and Calculations.
a) Calculate the sight stopping distance for a highway with a design speed of 115 km/hr and a local grade section of -2.65%. Use the AASHTO equivalent coefficient of friction in the calculation.
The friction coefficient is 0.28 at 115 km/hr (71 mph or 31.9 m/s)) The table shows the equivalent friction coefficients provided by AASHTO.
The reaction distance is (2.5 s * 31.9 m/s) = 80 meters Braking distance considering the -2.65% grade is 206 meters
Formula to estimate braking distance given an equivalent friction factor (f) and Grade G.
SSD = 80 + 206 = 286 meters
b) Compare the answer obtained in part (a) with the standard deceleration design value of 3.4 m/s-s. Comment on the difference between your solutions.
Using the AASTHO deceleration rate of 3.4 m/s-s the value of SSD is 243 meters.
The braking distance is reduced using 3.4 m/s-s because the deceleration rate is higher than (9.81 m/s-s * 0.28).
Problem 3 (25%)
A four-axle locomotive is used by Norfolk Southern for general cargo operations. Each locomotive develops 6500 HP of power with an efficiency factor of 0.8. At low speeds, the maximum tractive force developed by the locomotive is limited (flat line) to 3.8e5 Newtons to avoid wheel slippage between the locomotive wheels and the steel track. The train set is made up of two locomotives with a weight of 95 metric tons per locomotive and 45 flat cars with trailers behind (65 metric tons each).
a) Find the maximum speed of the train set on a guideway with 0.65% grade (uphill). Identify the key parameters of the problem:
K = 0.16; % Drag coefficient (dim) flat cars n = 4; % Number of axles
W = 95; % Total locomotive weight on rails (tons) w = W/n; % Weight per axle (tons)
noCars = 60; % Number of cars
wPerCar = 65; % Weight per cargo car (tons) grade = 0.65; % Grade in percent
g = 9.81; % Gravity constant (m/s-s) noLocomotives = 2; % Locomotives
Use the American Railway Engineering Association (AREA) equation to calculate the basic resistance in lbs/ton.
Calculate the total train resistance for various speeds and plot. remember that for low speeds, the maximum tractive force is limited to 3.8 e5 Newtons to avoid wheel slippage.
Figure 1. Train Resistance and Tractive Force. Grade = 0.65% and 45 Cars + 2 Locomotives. From the diagram, the maximum speed is 91 km/hr.
b) Plot the train resistance (in lb/ton) across the range of usable speeds (say 50-80 km/hr) for a guideway with 0.65% grade.
See Figure 2.
c) Could two locomotives and 60 flat cars could travel at the same grade at 60 km/hr? Explain and show your calculations.
Problem 4 (30%)
A vertical curve is proposed for a divided highway with a design speed of 105 km/hr. The curve has a -2.3% grade followed by a 2.9% grade. Due to design constraints, the elevation and station of the VPT point are 305 and 2+450 meters, respectively.
a) Estimate the curve length needed to satisfy the sight stopping distance criteria. Use the desired SSD values. Use the AASHTO equivalent coefficient of friction for the design speed in the calculation. Use the 2004 AASHTO height standards. Show all your calculations.
The value of the Reaction Distance is 73 in meters
The value of the Braking Distance is 153 in meters ( equivalent friction coefficient is f= 0.285 interpolated from values provided by AASHTO’s table - see Problem 1)
The value of SSD is 226 in meters (S =226 meters) Sketch the vertical curve to visualize the problem.
The vertical curve is a sag. A negative slope (-2.3%) is followed by a positive slope (2.9%).
The equations for sag vertical curve are:
The complete solution for stations and elevations of the VPI and VPC points are shown below.
c) Find the equation of the parabola to estimate the elevations on the vertical.
y = (0.029 - (-0.023)) / (2 * 292) * x^2 +-0.023 * x + 304.12
y = 8.9041e-05 * x^2 -0.023 * x + 304.12
Make a plot to check if the answer is correct.
e) Find the offset of the vertical curve at station 2+250 meters. The equation to estimate the offset is:
At station 2+250 meters the distance from VPC station is (2250 - 2158) = 92 meters. Hence the offset is:
Y = 8.9041e-05 * (92)^2 = 0.7536 meters
Y = 0.7536 meters