Final Exam (Semester II) with Answer
CE 4350: Highway and Traffic Engineering
13/07/1433 H
Time allowed: 180 Minutes
Instructor : Dr. Khaled KHEDER
Student Name / IDExercises and Questions / Marks
Exercise n° 1 / 10
Exercise n° 2 / 10
Exercise n° 3 / 10
Exercise n° 4 / 5
Question n° 1 / 5
Total / 40
Exercise n° 1 :
A plus 3.0 percent grade intersects a minus 2.0 percent grade at station 3 + 20 and at an elevation of 320.40 ft. Given that a 180 ft length of curve is utilized.
Give your answer about the following :
● Draw a sketch of the vertical curve;
● Determine the station of the PVC;
● Determine the elevation of PVC;
● Determine the station of PVT;
● Determine the elevation of PVT;
● Calculate the elevation at every 25 ft station;
● Locate the station of the high point;
● Calculate the high point of the curve.
ANSWER Exercise n° 1 :
● The curve is a crest with :
● A the first we convert foot to meter : 1 ft corresponds to 0.3048 m so 180 ft corresponds to 54.864 m and then :
● The middle point of the curve is 0.343 m below the VPI at station 3 + 20. The VPC and the VPT are located at either side of the VPI at distances of :
● The high point is located at a distance :
● The following table illustrates the use of equations below to calculate the required curve elevations :
Point P (station) / x(m) / Tangent elevation / Offset y
(m) / Curve elevation
2 + 92.568 / 000.00 / 79.225 / 0.00 / 79.225
3 + 00 / 7.432 / 91.225 / -0.27 / 90.955
3 + 20 / 27.432 / 97.658 / -0.34 / 97.318
3 + 25.486 / 32.9184 / 103.658 / -0.67 / 102.988
3 + 30 / 37.432 / 105.883 / -4.25 / 101.633
3 + 47.432 / 54.864 / 111.883 / -8.36 / 103.523
Exercise n° 2 :
For a horizontal curve with an external angle of 100°, the design speed is 80 km/h, the corresponding value of ƒmax is 0.13, and the maximum design value for e is 0.10.
Give your answer about the following :
● Draw a sketch for the horizontal curve;
● Calculate the maximum degree of curve;
● Calculate the minimum radius of simple circular curve;
● Calculate the design value for e for a curve that has a radius of 245 m.
ANSWER Exercise n° 2 :
● By either equation 1 or equation 2 :
And using equation :
The external angle does not enter these calculations. For a radius of 245 m the following equation gives a designed superelevation (edes) :
Exercise n° 3 :
Referring to the Figure 3.1 below shown a horizontal alignment of a highway. Two straight lines intersecting at the PI can be connected by an infinite number of circular curves. Each of these curves may be defined by its radius R or by its degree of curve D.
Give your answer about the following :
● Prove that the length of tangent : T = R*tan(Δ/2);
● Prove that the middle ordinate distance : M = R*(1-cos(Δ/2));
● Prove that the external distance : E = R*(sec(Δ/2) -1);
● Prove that the long chord : LC = 2R*sin(Δ/2).
ANSWER Exercise n° 3 :
● Referring to the figure below we can prove the upper relationship as following :
Exercise n° 4 :
Determine a suitable cross section for a channel to carry an estimated runoff of 340 ft3/sec if the slope of the channel is 1 percent and Manning’s, n, is 0.015.
Give your answer using two different ways :
● Manning’s formula;
● Manning’s chart.
ANSWER Exercise n° 4 :
● Select a channel section and then use Manning’s formula to determine the flow depth required for the estimated runoff. Assume a rectangular channel 6 ft wide.
Flow depth = d;
Cross-sectional area = 6*d;
Wetted perimeter = 6 + 2*d;
Hydraulic radius R = 6*d/(6 + 2*d).
Using flow equation below :
According to the data above and after converting foot into meter, we have :
This equation is solved by trial and error to obtain :
Alternatively, the Manning’s chart shown in Figure 1 below can used since the width is 1.8 m. Enter the chart at Q=9.6 m3/s (after concerting the foot into the meter), move vertically to intersect the channel slope of 1% (0.01), and then read the normal depth of 1.2 m from the normal depth lines.
Figure 1 : Graphical solution of Manning’s equation for rectangular channel.
Question n° 1 :
Define the following concepts either using full names or using sketch graphics :
● PC :
● PT :
● Superelevation :
● Long profile :
● Cross section :
● VPC :
● VPI :
● VPT :
● AASHTO :
● Normal crown :
ANSWER Question n° 1 :
The definition of the following concepts according to the textbook is given as following :
● PC : Point of curvature for horizontal alignment.
● PT : Point of tangency for horizontal alignment.
● Superelevation : the cross section of the road when we have a curvilinear form presents an angle β with horizontal.
● Long profile : along the axe of the road we draw the line of elevation of the ground, the red project line in order to estimate the cut and fill quantities.
● Cross section : at each point of the long profile in the perpendicular direction we draw the line of elevation of the ground, the red project line in order to estimate the cut and fill quantities.
● VPC : Vertical Point of Curvature for vertical alignment.
● VPI : Vertical Point of Intersection for vertical alignment.
● VPT : Vertical Point of Tangency for vertical alignment.
● AASHTO : American Association of State Highway and Transportation Officials
● Normal crown : in cross section the road has a double slope starting from the axe and going to the sides in order to make a good drainage when it is raining.
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