Civ E 315: Transportation Engineering I Winter 2011 Midterm

Civ E 315: Transportation Engineering I
Winter 2011
Midterm Questions and Solutions
Solution 1:
Estimate the 24-h volume for Wednesday using HEF
380 ∗ 31.36 440 ∗ 27.41 420 ∗ 20.18
3
10817.6
10818
Adjust the 24-h volume for Wednesday
10818 ∗ 6.96
7
Adjust for the month of June
10757 ∗ 1.018
AADT = 10951
10756.18
10757
10950.62
10951
Solution 2:
1. PIEV
Perception: see any object on road. Such as traffic control devises, markings, other road
users.
Identification: Identify the object as a stimulus.
Emotion: decide what action to take in response to the stimulus.
Volition: execution of the action decided on during the emotion process.
2. Sight Distance
The length of the roadway that a driver at the design speed can see ahead at any time
Types: Stopping; Decision; Passing
3. Spacing and Headway:
Spacing: distance between successive vehicles in a traffic stream as measured from
bumper to bumper
Headway: time between successive vehicles as they pass a point on a roadway.
4.
No.
5. Data resources:
Pedestrian counts
Pedestrian delay for at least 30 minutes in AM and PM
Condition Diagram---geometric condition or others
Collision Diagram---accident
Delay for AM and PM
Gaps: Number and distribution
85th speed on controlled approaches
85th speed on uncontrolled approaches
6. Shockwave
Any discontinuity in a state variable, whether it is due to changes in geometry or flow
conditions or abnormalities can potentially set up shock waves that travel through the
traffic stream.
We define a particular region of flow as being a section of freeway in which the speed,
density and flow are constant everywhere in that section.
Two adjacent sections in which any state variables are different from the values in the
other section are separated by a shockwave.
Solution 3:
65
1482
15
15 0.08 0.11
For the posted speed limit of 65 mpd, required curve radius =1482 ft.
Existing radius =750.
Thus, it is a hazardous location
Recommended actions:
(1) Increase radius to 1482 ft
(2) Reduce speed limit. We can calculate the appropriate speed limit by trial and error
15
Assuming u=40 mph, using the table
0.15
15 ∗ 750 0.08 0.15
50.9
0.14
Assuming u=50 mph, using table,
15 ∗ 750 0.08 0.14
49.7
Thus, for existing curve radius, safe speed limit = 50 mph
Solution 4:
From Table 15.5, k = 193
L = kA = 193 (2 – (-1)) = 579 ft
Station of BVC = (535+24.25) – (579 ft)/2 = 532+34.75
Station of EVC = (535+24.25) + (579 ft)/2 = 538+13.75
Elevation of BVC = 300 – (0.02)(579/2) = 294.21 ft
Elevation at any station on the leading tangent can be found in a similar manner.
The elevation on the curve can be found by subtracting the elevation on the leading tangent
by the offset, which can be found using Equation 15.15,
200
Using this procedure, the following table, which tabulates the elevation at 100 ft stations on
the curve, can be generated.
Station
Dist. from
Tangent Elev.
Offset
Curve Elev.
BVC
532+34.75
0
294.21
0
294.21
533+00
65.25
295.52
0.11
295.41
534+00
165.25
297.52
0.71
296.81
535+00
265.25
299.52
1.82
297.70
536+00
365.25
301.52
3.46
298.06
537+00
465.25
303.52
5.61
297.91
538+00
565.25
305.52
8.28
297.24
538+13.75
579.00
305.79
8.69
297.10
The distance from the BVC to the high point can be found as:
xhigh = LG1 / (G1 – G2) = (579)(2)/(2-(-1)) = 386 ft
The station of the high point is (532+34.75) + (386 ft) = 536+20.75
The difference between the elevation of the BVC and the elevation of the high point can be
found as:
yhigh = LG12 / (200(G1 – G2)) = (579)(2)2 / (200)(2-(-1)) = 3.86 ft
Therefore, the elevation of the high point is 294.21 + 3.86 = 298.07 ft
Solution 5:
(a)
150
30
0,
When
∞
When k=0,
For k=1,
150
148
/
(b)
150 30
150
30 (c)
From part (a), Jam density = 148 veh/mi/ln
Standard Greenberg model is:
Comparing the given equation with the standard form,
Optimum speed,
Optimum density,
Maximum flow,
30
54veh
/ /
54 ∗ 30 1620
/
/

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