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Transportation FE Civil Depth
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Table of Contents 1. Surveying
......................................................................................................................................
2
1.1 Bearings and Azimuths
..........................................................................................................
21.2 Stationing
...............................................................................................................................
2
2. Driver Performance and Behavior
................................................................................................
32.1 Information Processing and Perception
..................................................................................
32.2 Driver Expectancy
..................................................................................................................
32.3 Perception-Reaction Time
......................................................................................................
32.4 Stopping Sight Distance
.........................................................................................................
4
3. Horizontal Curves
.........................................................................................................................
63.1 Circular Curves
......................................................................................................................
63.2 Superelevation
........................................................................................................................
83.3 Stopping Sight Distance on Horizontal Curve Section
.......................................................... 9
4. Vertical Curves
...........................................................................................................................
104.1 Vertical Curve Elevations
....................................................................................................
104.2 Vertical Curve Design
..........................................................................................................
114.3 Intersection Sight Distance
...................................................................................................
12
5. Speed Characteristics
..................................................................................................................
145.1 Time Mean Speed
.................................................................................................................
145.2 Space Mean Speed
...............................................................................................................
145.3 Spot Speed Data Analysis
....................................................................................................
155.4 Speed, Flow and Density Relationships
...............................................................................
175.5 Speed, Distance, and Time Relationships
............................................................................
17
6. Signalized Intersections
..............................................................................................................
186.1 Change Interval
....................................................................................................................
186.2 Clearance Interval
................................................................................................................
19
7. Traffic Safety
..............................................................................................................................
208. Pavement Design
........................................................................................................................
219. Additional Problems for Self Study
............................................................................................
22 NOTE
CERM Civil Engineering Reference Manual for PE Exam, 11th
Edition by Michael Lindeburg
SRHB Supplied-Reference Handbook, 8th Edition, 2nd Revision by
NCEES
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1. Surveying
1.1 Bearings and Azimuths See definitions on CERM page 77-11
under section 23. PROBLEM 1 - Convert the following bearings to
azimuths from north:
(a) N 740 24 01 W (b) S 850 13 16 W (c) N 840 28 13 E (d) S 080
19 19 E
SOLUTION 1
(a) 2850 35 59 (b) 2650 13 16 (c) 840 28 13 (d) 1710 40 41
1.2 Stationing Stationing concept is used along horizontal
alignments for referencing purpose 1 station = 100 feet How do you
represent stationing?
o Specific location is represented as Sta 10+00 o Distance is
represented as 10.00 sta
PROBLEM 2 What is the station at Point B?
SOLUTION 2 Station at Point A = Sta 22+45 Station at Point B =
(Sta 22+45) + 1028 = 2245+1028 =3273
= Sta 32+73
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PROBLEM 3 The survey has identified the beginning and ending
points along an urban arterial highway that require new sidewalks.
The first station is located at 5+88 and the second station is
located at 10+05. What is the difference in length between stations
in feet and in stations? SOLUTION 3 Difference in length = (10+05)
(5+88) = 1005 588= 417 Difference in stations = 4.17 sta
2. Driver Performance and Behavior
2.1 Information Processing and Perception The time required to
respond successfully to any driving situation, such as an
emergency, involves four stages:
perception (detection and identification) decision reaction
response of the vehicle.
2.2 Driver Expectancy A fundamental component of driver
information processing and perception. Drivers operate with a set
of expectancies, e.g.:
freeway exits will be on the right side of the road (or the left
in Britain, Australia, etc.);
advance warning will be given of hazards in the roadway; other
drivers will obey traffic regulations, etc.
2.3 Perception-Reaction Time A significant variable in the
successful processing and use of information is the speed with
which this is done. Perception-Reaction Time (PRT) is a human
factor often cited by traffic engineers concerned with safety. PRT
is the interval between the appearance of some object or condition
in the drivers field of view and the initiation of a response such
as braking or changing course. Note that PRT involves the
initiation of a response (e.g. pressing the brake), not the
completion of the vehicle maneuver (stopping).
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PRT depends on the situation. Response time is generally
quickest when there is one specific response to be made to a single
stimulus (brake lights of vehicle ahead). In the case of choice
reaction time, in which there is more than one stimulus and/or more
than one possible response (e.g. toll plaza), reaction time
increases as a function of the number of possibilities. A driver
may, for example, have to decide whether to steer or brake, or
both, to avoid a pedestrian. The PRT used for design standards by
AASHTO includes 1.5 sec for perception and decision, 1.0 sec for
making a response, for a total of 2.5 sec, which is generally
considered adequate for all but the most complex driving
situations. PROBLEM 4 Which of the following factors affect driver
performance and behavior?
A. cell phone use B. fatigue C. traffic D. drugs and alcohol E.
young / old age F. law enforcement G. All of the above
SOLUTION 4 The correct answer is G. All of the above affect
driver performance and behavior.
2.4 Stopping Sight Distance See Chapter 3 in A Policy on
Geometric Design of Highways and Streets 2004 by AASHTO. This book
is popularly known as Green Book Stopping Sight Distance is the sum
of two distances: (1) the distance traversed by the vehicle from
the instant the driver sights an object necessitating a stop to the
instant the brakes are applied; and (2) the distance needed to stop
the vehicle from the instant the brake application begins. These
are referred to as brake reaction distance and braking distance,
respectively. The AASHTO GB provides the following equations for
calculating braking distance and SSD, with and without the effect
of grades.
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The following equation includes terms for both the brake
reaction distance and braking distance (Use equation 78.43(b) on
page 78-9 in CERM. Also see the same equation on page 162 of
SRHB):
First part of the equation represents brake reaction distance
and the second part represent braking distance PROBLEM 5 A motorist
is traveling on a level grade at 50 mph. A tree has fallen across
the road and forces the motorist to stop. Assuming a 2.5 sec PRT
and 11.2 ft/sec2 deceleration rate, determine the brake reaction
distance and braking distance in feet.
A. 147 and 154 B. 165 and 194
C. 184 and 240 D. 165 and 290
SOLUTION 5
The correct answer is C, 184 and 240
2VSSD = 1.47V t +
3032.2
mphmph a G
Brake reaction distance = 1.47Vt = 1.47(50)(2.5) = 184'
where:
SSD = Stopping Sight Distance, ft; V = design speed, mph; t =
breaking reaction time, 2.5 sec
2
a = deceleration rate, 11.2 ft/sec
G = percent of grade divided by 100, it is in decimal
2 2V 50Braking distance = = = 240' 11.230 30 0
32.2 32.2
mph
a G
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3. Horizontal Curves
3.1 Circular Curves Horizontal circular curve is a circular arc
between two straight lines known as
tangents.
See equations 78.1 to 78.12 and Figure 78.1 on pages 78-2 of
CERM. Also see page 164 of SRHB.
PROBLEM 6 For the following circular curves having radius R,
what is their degree of curve by Arc definition and Chord
definition? SOLUTION 6
(a) Roadway curve with 500.00 ft 05729.578' 11 27'33"
500aD
ft
(b) Roadway curve with 1500.00 ft 05729.578' 3 49'10.99"
1500a ftD
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PROBLEM 7 Compute T, L, E, HSO, R, and stations of the PC and PT
for the circular curve described below: Highway curve with R =
750.000 ft, I = 180 30, and PI Sta 123+24.80 SOLUTION 7
0 0 0
0
0
0 0
5729.578' 7 38'22"; I = 18 +30' 60' = 18.5 ; 750
I 18.5T = R tan = 750 tan = 122.145 ft2 2
18.5L = 2 R = 2 750 = 242.164 ft360 360
aD ft
I
0 0
0
I I 18.5 18.5E = R tan tan = 750 tan tan 2 4 2 4
= 750 0.1629 0.0809 = 9.881 ftI 18.5HSO = R 1 - cos =750 1-cos =
750(0.013) = 9.72 2
53ftPC Sta. = PI Sta. - T = 12324.800 - 122.145 = 122+02.655PT
Sta. = PC Sta. + L = 12202.655 + 242.164 = 124+44.819
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PROBLEM 8 Compute interior (or intersection) angle for the
following circular curve: SOLUTION 8
0 0 0 0180 60 60 60I
3.2 Superelevation Used at horizontal curves Use equations
78.37(b) on page 78-7 of CERM. Also see on page 163 of
SRHB. PROBLEM 9 What is the minimum radius, Rmin, that can be
used on a horizontal curve with a 70 mph design speed, a maximum
superelevation, emax = 0.08, and a side friction factor, f = 0.10?
SOLUTION 9
Use equation 78.37(b) emax = 8%; V = 70 mph; fmax = 0.10
2
minmax max
VR =
15(e + )mph
f
2
min
70R = 1814.8015(0.08 + 0.10)
ft
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3.3 Stopping Sight Distance on Horizontal Curve Section
Obstructions along the inside of curves can limit the available
(chord) sight distance. A curve must be designed that will
simultaneously provide the required stopping sight distance while
maintaining a clearance from a roadside obstruction. See equation
78.45 and figure 78.9 on page 78-10 in CERM. Also see on page 163
of SRHB. PROBLEM 10 A four-lane undivided highway has a design
speed of 40 mph. The lanes are 12 ft wide. The centerline Degree of
Curvature, D is 10o 45 Determine the required clearance from the
center of the curves inside lane based on Stopping Sight Distance
criteria.
A) 22.41 ft B) 21.67 ft C) 305 ft D) 533 ft SOLUTION 10 Using
SSD equation (page 162 of SRHB), for V = 40 mph, S = 305 ft. D =
10o 45 is equivalent to R= 532.98 ft (using equation on page 164 of
SRHB) The centerline of the inside lane is offset 18 ft (12 ft + 6
ft) from the roadway centerline. Therefore R lane centerline =
532.98 ft 18 ft = 514.98 ft
Using equation 78.45 in CERM (Also see on page 163 of SRHB),
28.65 28.65 3051 cos 514.98 1 cos 22.41
514.98SHSO R ft
R
Answer A
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4. Vertical Curves Vertical curves are used to change the grade
of a highway. Most vertical curves take the shape of an
equal-tangent parabola. Such curves
are symmetrical about the vertex. Two types of vertical curves
Crest and Sag
See figure 78.10 and equations 78.46 to 78.49 on pages 78-11 of
CERM. Also see on page 165 of SRHB.
4.1 Vertical Curve Elevations PROBLEM 11 A +3.25% grade
intersects a -2.00% grade at Sta. 45+25 and elevation 695.42 ft. A
1000 ft vertical curve connects the two grades. Determine:
a) the station and elevation of turning point b) the elevations
along the curve at Sta. 45+00; Sta. 50+25
SOLUTION 11 a) Using equation 78.46 on page 78-11,
2 1 2.00 3.25 0.52510Sta
G GRL
Highest Point Location: x = -G1/R = 6.1905 Sta. (Eqn. 78.48)
Sta. = BVC Sta. + x = 4025 ft + 619.05 ft = 4644.05 ft = Sta
46+44.05 Using equation 78.47, Elev. = (R/2)x2+G1(x)+BVC Elev.
Elevation at BVC = 695.42-500(0.0325) = 679.17 ft Elev. =
(R/2)x2+G1(x)+BVC Elev. = 689.23 ft b)
Using equation 78.47, Elev. = (R/2)x2+G1(x)+BVC Elev., results
are tabulated
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At Sta. 45+00 xsta = 4.75 At Sta. 50+25 xsta = 10.00
Sta. (R/2) xsta 2 G1* xsta BVC Elev. Elev.
45+00 -5.92 15.44 679.17 ft 688.69 ft 50+25 -26.25 32.5 679.17
ft 685.42 ft
4.2 Vertical Curve Design Using AASHTO Guidelines Minimum
vertical curve length is computed Based on sight distance criteria
Use Table 78.4 in CERM (See page 163 of SRHB)
PROBLEM 12 You are designing a vertical curve on a two-lane
highway with G1=+3.50%; G2= -2.25%; PVI 85+00; PVI elevation =
457.59 feet; and a Design Speed of 65 mph. What is the most
appropriate length of the curve you should design? SOLUTION 12 From
Table 78.4 of CERM (Also see page 163 of SRHB), Using SSD equation,
for V=65 mph, S=645 ft
Since S = 645 feet < L = 1108.5 feet, our assumption is
Good.
Therefore, L =1108.5 ft
2 2 22 1
Assume S < L; ( ) 5.75*645L= = = =1108.5 ft
2,158 2,158 2,158AS ABS G G S
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4.3 Intersection Sight Distance Reference AASHTO Geometric
Design of Highways and Streets pages 651-675
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Recommended dimensions of the sight triangles vary with the type
of traffic control used at an intersection. But, the equation to
calculate side length b which is called Intersection Sight Distance
(ISD) is same for all scenarios. The equation is: b = ISD =
1.47Vmajor tg Where Vmajor = Design speed in MPH on major street tg
= Time gap in seconds varies by vehicle type and movement type
(AASHTO has established guidelines to compute this value) PROBLEM
13 A passenger vehicle is stopped on a minor street at a stop sign
of stop controlled intersection. The design speed of the major road
is 60 mph. Stopped passenger vehicle is getting ready to make right
turn from minor street to major road. If this vehicle requires a
time gap of 6.5 to avoid a collision with the vehicle approaching
the intersection, at least how far the approaching vehicle has to
be from the intersection before the passenger car makes right turn?
SOLUTION 13 ISD = 1.47Vmajor tg = 1.47 * 60 * 6.5 = 573.3 ft
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5. Speed Characteristics
5.1 Time Mean Speed From spot speed studies which are measured
as vehicles pass a point on the
road It is the average (mean) of all vehicles passing a point on
a road over some
specified time period
5.2 Space Mean Speed From travel time runs for a given section
of the roadway It is the average (mean) of all vehicles occupying a
given section of a highway
over some specified time period PROBLEM 14 Six travel time runs
are made on a 1000 feet section and the data is presented in the
following table. Compute average speed
Vehicle Measured Time (sec) 1 18 2 20 3 22 4 19 5 20 6 20
SOLUTION 14 6 1000 50.4 fps
119sS
1
Where: speed of the vehicle, and number of vehicles included in
the measurement sample
n
ii
t
thi
SS
nS in
1
Where: length of the segment travel time of the vehicle to
traverse the section
ns
ii
thi
nLSt
Lt i L
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Example Graphs
5.3 Spot Speed Data Analysis Mean speed or Time mean speed
defined in the previous section 85th percentile speed speed at
which 85 percent of free-flowing vehicles are
traveling at or below. To establish speed zones nearest 5 mph
increment at or below the 85th percentile speed.
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PROBLEM 15 The following speed data was collected using a radar
gun on an arterial street:
Estimate the following: a) Mean speed b) 85th percentile
speed
SOLUTION 15 Expand the table as shown here:
a) Mean speed: 13820x = = 38.39 mph;
360
b) 85th percentile speed of vehicles: Step 1: Total #
observations x 0.85 = 360 x 0.85 = 306
Step 2: The 306th observation is in the 46-50 mph speed group.
Therefore the
85th percentile speed is 48 mph.
Speed Group (mph) Frequency
16 to 20 10 21 to 25 17 26 to 30 36 31 to 35 66 36 to 40 84 41
to 45 70 46 to 50 50 51 to 55 21 56 to 60 6
Speed Group Frequency Assumed Speed Total
Speed Cumulative Frequency
16 to 20 10 18 180 1021 to 25 17 23 391 2726 to 30 36 28 1008
6331 to 35 66 33 2178 12936 to 40 84 38 3192 21341 to 45 70 43 3010
28346 to 50 50 48 2400 33351 to 55 21 53 1113 35456 to 60 6 58 348
360
360 13820
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5.4 Speed, Flow and Density Relationships Speed (S) Space mean
speed (mph) Flow/Rate of Flow/Volume (v) number of vehicles per
hour per lane (vphpl) Density (D) Number of vehicles per mile per
lane (vpmpl)
vDS
Spacing Distance between common points (e.g. the front bumper)
on
successive vehicles (ft/veh) 5280 ft/mileSpacing in ft
D
Reference - Speed/flow/density relationships graphs from Page
73-7 of CERM (Also see Page 163 of SRHB)
5.5 Speed, Distance, and Time Relationships Distance = Speed *
Time PROBLEM 16 Refer to the figure at right. At t = 0, vehicle A,
traveling at a speed of 60 km/h, passes a road marker on a straight
section of a four-lane divided highway. Vehicle B, traveling with a
speed of 90 km/h, passes the marker 2 sec later. Find the time when
vehicle B overtakes vehicle A, and the corresponding distance from
the road marker. SOLUTION 16 The figure above shows a sketch of the
events. S is the distance from the marker where B overtakes A at
time t. From the figure, s = sA = sB
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Using s = vt , sA = vAt = 16.7t and sB = vB(t -2) = 25(t-2)
16.7t = 25 (t-2) t = 6.02sec s = sA = 16.7(6.02) = 101 m
6. Signalized Intersections Cycle length the time required for
one complete sequence of all signal indications Phase the
right-of-way (green), change (yellow), and clearance (all red)
intervals in a cycle that are assigned to an independent traffic
movement or combination of movements Green interval the
right-of-way interval during which the signal indication is green
Yellow Change interval the first interval following the green
interval or which the signal indication is yellow Clearance
interval an interval that follows a yellow change interval and
preceds the next conflicting green interval
6.1 Change Interval Also known as Yellow Interval
2Gg2a
v ty Where: y = length of yellow interval (sec) t = driver
perception/reaction time (1.0 sec generally used) v = velocity of
approaching vehicle (fps) a = deceleration rate (10 fps2 generally
used) G = acceleration due to gravity (32.2 fps2 generally used) g
= grade of approach (percent/100) Also see page 162 of SRHB for the
above equation.
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Transportation FE Civil Depth
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6.2 Clearance Interval Also known as Red Clearance Interval and
All Red
w Lr - If there is no pedestrian trafficv
P Lr - If there is pedestrian traffic or the crosswalk is
protected by ped. signalv
Where: r = length of red clearance interval (sec) w = width of
intersection; more precisely, the length of the vehicle path from
the departure stop line to the far side of farthest conflicting
traffic lane (ft) P = width of intersection; more precisely, the
length of the vehicle path from the departure stop line to the far
side of farthest conflicting ped crosswalk (ft) L = Length of
vehicle (20 ft generally used) v = speed of vehicle through
intersection (fps)
Also see Page 162 of SRHB for the above equation PROBLEM 17
Estimate change interval for an approach with a grade -2% and an
approach speed of 30 mph. SOLUTION 17
For -2% grade, 5280 ft/mile30 miles/hour*
3600 sec/hour1.02*10 2*32.2 (-2 0
)
y /1 0
= 3.4 sec
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7. Traffic Safety Crash rates are normally considered better
indicators of risk than crash frequencies.
Crash rates for intersections are normally expressed in terms of
crashes per million entering vehicles (MEV), using the following
equation:
V*T*365
10*A R6
int
Crash rates for roadway segments are normally expressed in terms
of crashes per 100 million vehicle-miles (100 MVM), using the
following equation:
L*V*T*365
10*A R8
sec Where: Rint = crash rate for the intersection
Rsec = crash rate for the roadway section A = number of reported
crashes T = time period of the analysis (years) V = annual average
daily traffic volumes (veh/day) L = length of the segment
(miles)
PROBLEM 18 An intersection has a total entering traffic volume
of 42,000 vehicles per day. During the past three years, there have
been a total of 35 reported intersection-related crashes. What is
the crash rate for this intersection? SOLUTION 18
6
int
35 10R 365 3 42,000
= 0.76 crashes per MEV PROBLEM 19 A five-mile long section of
two-lane road has an AADT of 8,000. There have been six crashes on
this section of road during the past two years. What is the crash
rate? SOLUTION 19
8
sec
6 10R 365 2 8,000 5
= 20.5 crashes per 100MVM
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8. Pavement Design Asphalt surface Using AASHTO Guidelines
o Estimate Design Traffic o Determine the Structural Number (SN)
for the given traffic, roadway, and
soil characteristics o Compute layer-thickness using the
following equation (see on page 166
of SRHB):
SN = a1D1 + a2D2 + +anDn Where SN = structural number of the
pavement ai = strength coefficient of the ith layer Di = thickness
of the ith layer in inches
PROBLEM 20 Determine the thickness of flexible pavement layer 2
given the following: FACTS: Total Structural Number required = 7.0
Total number of layers = 2 Layer 1 consists of asphalt concrete
with a strength coefficient of 0.46 Layer 1 thickness = 6 inches
Layer 2 consists of granular base with a strength coefficient of
0.15 SOLUTION 20 Using equation 75.29 in CERM, 1 1 2 2SN D a D
a
1 12
2
7.0 6 0.46 28.3"0.15
SN D aDa
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9. Additional Problems for Self Study PROBLEM 21 The local
survey crew measures a distance of 1,135 ft of new drainage pipe
that is required for the new road construction. The beginning point
for the drainage is located at station 8+77. What is the station
number of the ending point?
A) 12.20 Sta B) Sta 12+20 C) 20.12 Sta D) Sta 20+12 SOLUTION 21
End point station = (Sta 8+77) + 1,135 = 877 + 1,135 = 2012 = Sta
20+12 Answer D PROBLEM 22 A motorist is traveling down a 6% grade
at 65 mph and needs to stop because of a crash scene. Assuming a
2.0 sec PRT and 12.0 ft/sec2 deceleration rate, determine the total
SSD in feet.
A) 524 ft B) 642 ft C) 750 ft D) 869 ft
SOLUTION 22
2 2V 65SSD = 1.47Vt + = 1.47(65)2 + 1230 30 0.06
32.2 32.24225 = 191.1+ = 641.52' 9.380
a G
Answer B
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PROBLEM 23 Compute T, L, E, HSO, R, and stations of the PC and
PT for the circular curve described below. Highway curve with Da =
30 30, I = 240 45, and PI Sta. 30+00.00. SOLUTION 23
0 0a
0 0
0
0
0 0
5729.578' 5729.578' 5729.578'R = = = = 1637.022 ftD 3 30'
3.5
I = 24 + 45' 60' = 24.75 ;
I 24.75T = R tan = 1637.022 tan = 359.174 ft2 2
24.75L = 2 R = 2 1637.022 = 707.143 ft360 360
E = R t
I
0 0
0
I I 24.75 24.75an tan = 1637.022 tan tan 2 4 2 4
E = 1637.022 0.2194 0.10841 = 38.9368 ft.
I 24.75HSO = R 1 - cos =1637.022 1-cos = 16372 2
.022(0.0232) = 38.0348 ft
PC Sta. = PI Sta. - T = 3000.00 - 359.174 = 2640.826 ~
26+40.83PT Sta. = PC Sta. + L = 2640.826 + 707.143 = 3347.969 ~
33+47.97
PROBLEM 24 A -3.00% grade intersects a +1.25% grade at Sta.
85+80 and elevation 210.00 ft. A 350-ft vertical curve connects the
two grades. Determine the station and elevation of turning point.
SOLUTION 24 R = (1.25+3.00)/3.5= 1.214 Elev. at BVC =
210.00+175(0.03) = 215.25 ft Low Point Station and Elevation
Location: x = -G1/R = 2.4706 sta Sta. = BVC Sta. + x = 8405 ft +
247.06 ft= 8652.06 ft = Sta 86+52.06 Elev. = (R/2)x2+G1(x)+BVC
Elev. = 211.54 ft
ahmed youssef ([email protected])This copy is given to the
following student as part of School of PE course. Not allowed to
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Transportation FE Civil Depth
24
PROBLEM 25 At a large port facility near New York City
tractor-trailer trucks are lined-up in long queues as their drivers
wait to have their containers inspected and documents processed.
Assume the trucks have an average length of 73.5 ft and the average
space between the rear and front bumpers of successive vehicles is
10 feet. What is the best estimate for the jam density
(trucks/mile) in one lane of trucks? A) 50 B) 63 C) 77 D) 84
SOLUTION 25 By definition, jam density is the density at zero
speed. Based on the information provided the average spacing
between the front bumpers of successive vehicles is about 83.5
feet, the jam density can be estimated as follows:
5280 feet vehicleJam density = = 63 vehicles mile1 mile 83.5
feet
Answer B
ahmed youssef ([email protected])This copy is given to the
following student as part of School of PE course. Not allowed to
distribute to others.