Test Problem Summary
Problems are arranged by subject listed below. Not all solutions are
provided:
General
Minimum Radius Calc
Horizontal Curves
Horizontal Sight Distance
Superelevation
Earthwork
Stopping Distance
Vertical Curves
Vertical Sight Distance
Rational Method
Ditch Flow Problems
Culvert Problems
Aggregate Gradation
Equilivant Single Axel Loads
Asphalt Thickness Design
Temperature Steel Concrete Pavements
AASHTO Concrete Pavement Design
PCA Concrete Pavement Design
Traffic Flow Problems
Subject
General
1. The Interstate system of highways has approximately miles of
roadway?
45,000 miles
2. The Federal Highway Administration is a part of what cabinet
level Department?
Department of Transportation
3. According to the route designation system used in the USA, where
would US101 be located?
On the West Coast  either California or Washington
4. Use a short statement to define the following highway related term
or acronym.
A. Mass Diagram  cubic yards or cubic meters of cut/fill
versus station
B. Edge of Metal  edge of pavement
C. ISTEA  Intermodal Surface Transportation Efficiency Act
D. FHWA  Federal Highway Administration
E. Design Speed  maximum speed controlled by design
element
F. Clear Zone  edge of pavement to obstruction distance
5. What were the main new programs associated with the Federal Aid
Highway and Revenue Act of 1956?
A. The Interstate System
B. Highway Trust Fund
6. List the three major functional classifications for roads.
A. Arterial
B. Collector
C. Local
7. List four major factors to be considered when locating a highway
route.
A. Terrain
B. Soil
C. Economics
D. Safety
Also acceptable: historical preservation, political
8. ISTEA was landmark highway legislation passed in 1991. What
does the acronym stand for?
Intermodal Surface Transportation Efficiency Act
9. TEA-21 is the new highway legislation in 1997. What does this
acronym stand for?
Transportation Equity Act for the 21st Century
10.The major highway programs for funding purposes under ISTEA
and TEA-21 are?
1. Surface Transportation Program (STP)
2. National Highway System (NHS)
11.Sketch the relationship between mobility and access that is the
basis for the functional classification system. Label the axis and
add any notes that are appropriate.
12.Name one state traversed by the route with the designation
indicated.
A. I-5  CA, WA, OR
B. US-2  MI
13.Most road mileage in the United States is under the jurisdiction of
what level of government?
Local
14.List three highway related administrations that are part of the US
Department of Transportation
A. FHWA
B. NHTSA
C. FTA
15.Identify the following highway related organizations.
A. AASHTO  American Association of State Highway and
Transportation Officials
B. ITE  Institute of Transportation Engineers
C. ARTBA (for students with excellent memories)  American
Road and Transportation Builders Association
16.Name the principal source of funds for the nation’s highway
system (books words).
A. User tax
B. Property tax
C. Tolls
17.The Federal Aid Highway Act of 1921 established two major
principles related to future highway funding. They are:
A.
B.
18.TEA-21 is the new highway legislation. Where can I find
information about this important legislation?
19.Define qualitatively the relationship between mobility and access
for each of the three major classifications in the functional
classification system.
A. Arterial  higher mobility with a low degree of access
B. Collector  balance between mobility and access
C. Local  lower mobility with a high degree of access
20.Explain the basis for the Interstate route designation system.
(where are the high/low numbers and where are the even/odd)
The interstate routes are numbered from highest to lowest from
east to west and from north to south. Even-numbered interstate
routes run east-west while odd-numbered routes north-south.
21.What is the approximate (+/- 500000 mi) total road mileage in the
United States?
3.9 million miles
22.Write out the terms that identify the following acronyms or
symbol:
a.
b.
c.
d.
e.
CBR  California Bearing Ratio
AADT  Average Annual Daily Traffic
DHV  Design Hour Volume
Mr  Resilient Modulus
PSI  Change in Present Serviceability Index
Minimum Radius Calc
1. A horizontal curve at the site of an accident has a measured
superelevation of 3 %. The posted speed limit is 40 mph and the
radius of the curve is 1000 ft. Tests on the pavement at the site
indicate the transverse friction varies in a linear manner from 0.2 at 30
mph to 0.1 at 60 mph. Show by calculation the curve is
adequate/inadequate. (Would this be a good time to call my lawyer?)


V2
R

15( f  e) 


2


V
1000  

15(.3  .1V  .03) 


30


.1V
.1V
V 2  15,000 (.3 
 .03)  15,000 (.33 
)
30
30
V 2  4950  50V
V
2
 50V  4950  0
V 
V 
Vmax
 50 
50 2  4 * 1 * 4950
2 *1
 50  149.33
2
99.33

 49.67  50 MPH
2
Also acceptable
V2
R
15( f  e)
40 2
R
.1 * 40
15(.3 
 .03)
30
R40  542 ft  1000 ft
Safe for 40 MPH
2. An icy road has a side friction factor of 0.0, a 0.07 ft/ft maximum
superelevation and a radius of 1909 ft. What happens if the vehicle
speed is 20 mph. Show by calculating the developed forces.
tan   .07
  4
sin   .0699
Down slope forces  W * sin   .0699W
W V2
Up slope forces  Fc * cos   *
* cos 
g R
W
5280 2
1
Up slope forces  .998
* (20 *
) *
 .014W
32.2
3600
1909
.07W  .014W
Down slope forces  Up slope forces : vehicle sets to inside of curve
3. What is the minimum radius required on a horizontal curve such
that the centripetal and gravitational forces on a vehicle in the outside
lane are balanced by the frictional forces. Assume the normal crown
is 0.015 ft/ft, the side friction factor, f, is 0.024 and the design speed is
60 mph.
W * sin   f * N  W * sin  * f
f 
sin 
 tan   .015
cos 
not a function of R
W V2
 Wf
g R
5280 2
(60 *
)
V2
3600
R

g* f
32.2 * .024
R  10,020 ft
F
parallel to road
Fc cos   W sin   Fr  f ( N )
Fc cos   W sin   f ( Fc sin   W sin  )
Fc  W tan   f ( Fc tan   W )
W V2
W V2
 W tan   f (
tan   W )
g R
g R
V2
V2
f
tan   f  tan 
g*R
g*R
V2
(1  f tan  )  ( f  tan  )
g*R
V 2 1  f tan 
88 2 (1  .024 * 0.15)
(
)
g f  tan 
32.2 (.024  .015)
R  26721(1  .024 * .015)
R
R  26712 ft
R
V 2 (1  f tan  )
g
f e
88 2 (1  .024 * .015)
32.2 (.024  .015)
R  6167(1  .024 * .015)
R  6164 ft
R
Rmin  26712 ft
4. An icy road has a side friction factor of 0.0, a 7% maximum super
elevation and a radius of 600 m. Show by summing forces what
happens if the vehicle speed is reduced to 40 km/hr.
tan   .07
  4
sin   .0699
Down slope forces  W sin   .0699W
W V2
Up slope forces  Fc cos  
cos 
g R
W
1000 2 1
Up slope forces  .998
* (40 *
) *
 .021W
9.81
3600
600
.07W  .021W
Down slope forces  Up slope forces : vehicle sets to inside of curve
5. A vehicle travels around a horizontal curve with a radius of 500 ft
at 40 mph. If the side friction value is 0.15 and the road is flat (no
cross slope) show by force calculation that the vehicle will stay/leave
the roadway. Assume the vehicle weight is 2000 lb and the
centripetal acceleration is a = m*V2/R, where m is the mass, V is the
velocity and R is the radius of the curve.
2
WV
2000

g R
32.2
Fc  429.5lbs
Fc  ma 
5280 2
)
3600
500
(40 *
FR  f * W  .15 * 2000  300lbs
Fc  FR  vehicle will leave road
6. A vehicle on a tangent is traveling at 60 mph. If the cross slope of
the pavement is - 2 %, what is the factor of safety against sliding to
the side of the road.
FD  W sin 
FR  W cos  * f
FS F 
FR W cos  * f
f
.12



FD
W sin 
tan  .02
FS F  6
7. A friction measuring trailer has a weight of 1000 lbs. If the
locked wheel force measured between the trailer and the towing
vehicle is 200 lbs at 40 mph, what is the friction factor between the
tires and the road surface?
8. A carnival rider drives a motorcycle up a circular vertical wall
and continues around the track until daylight. The radius of the track
is 20 m and the measured friction along the vertical wall is 0.1. How
fast does the driver have to go to maintain a perfect horizontal
position (km/h)?
9. A circus rider is attempting to ride around a track with a vertical
wall. If the side friction is 0.15 how fast must the rider go (mph) to
prevent sliding to the bottom of the wall? Radius of the track is 25 ft.
gR 32.2 * 25

 5366.7
f
.15
V  73.3 fps  49.95MPH
V2 
10. In the winter the side friction factor on a roadway may be
reduced to 0. In this case the only way to prevent sliding is to
increase/decrease the velocity of the vehicle. If the design speed of
the road way is 100 km/hr, emax is 10% and the radius of the curve is
the minimum determined by the conventional technique, what speed is
required to just keep a vehicle on the curve?
Horizontal Curve
1. The relationship between degree of curve and radius involves the
constant 5729.5780. Derive this constant from basic geometric
principles.
RD
 100
180
100 * 180
R
D
5729.5780
R
D
2. A horizontal curve has a central angle of 45o and a Point of
Intersection (PI) at Station 1+000.00 and radius of 1000.00 m. What
is the station of the Point of Tangency (PT)?
T  R tan

2
T  1000 * tan
L

45
 1000 * .41421  414.21
2
*R* 

180
180
PI  1  000.00
 414.21
PC  0  585.79
 785.40
PT  1  371.19
*1000 * 45  785.398m
3. A circular curve with a radius of 400 m and a long chord of 400
m. The PI Station is 1+000.000 find the following:
a. The PT Station
b. The External distance
c. A parabolic curve with the same long chord and tangent as in
problem 3 is used instead of the circular curve. What is the
external distance for this curve?
d. A horizontal curve has a deflection angle of 45o R with the PI
at Station 900.00 and a long chord of 500 m. What is the station of
the Point of Curvature (PC)?
4. A horizontal curve with a deflection angle of 30o has a radius of
500 m. At what distance (along the curve) from the PC is the
deflection angle for an observer at the PC equal to 15o ? What is the
PT Station if the PI station is 1+000.000?
Horizontal Sight Distance
1. A horizontal curve has a radius of 1000.00 ft. If the grade is flat
and the distance between the centerline and object is 30 ft, what is the
maximum stopping distance (assume no adjustment to center of the
inside lane). Remember the arc length is proportional to the
circumference of a circle.
30  R  R cos

2

30  1000(1  cos )
2

.030  1  cos
2

cos  1  .030  .97
2

 14.07
2
  28.13

S

360 2R
2 * 1000 * 28.13
S
 491 ft
360
2. The stopping distance (rounded for design) of a two lane road (12
ft lanes) is 450 ft. What is the required clear distance between the
drivers location and the edge of the forest (level terrain and no back
slope) if the radius to the centerline is 1000 ft?
Robj  1000 ft  6 ft  994 ft
 obj
S

2Robj 360
 obj 
S * 360 450 * 360

 25.94
2Robj
2 * 994
m  Robj (1  cos
 obj
)  994(1  cos
2
m  994(1  cos12.97)
m  25.4 ft
25.94
)
2
3. A two lane road with a 1000 ft horizontal curve (centerline), 12 ft
lanes and an interior angle (deflection angle) of 30 degrees defines the
line of sight for a vehicle on the curve. For the standard stopping
conditions (3.5 ft observer and 0.5 Ft object), what is the slope of the
line of sight (%)? Assume the grade is level.
slope 
elev
*100
LengthLC
LengthLC  2 Robs * sin
slope 

30
 2 * (1000  6) sin
 514.53 ft
2
2
3.5  0.5
*100  0.6%
514.53
4. For a horizontal curve on a two lane (3.6 m per lane) road with a
radius of 500 to the centerline, what is the distance of the line of sight
when an obstruction is located 10 m from the centerline of the road?
Assume the grade of the road is level

3.6
m  Robj (1  cos )  m  10 
 8.2
2
2


8.2
8.2  (500  1.8)(1  cos )  cos  1 
2
2
500  1.8

 10.41
2

L  2 Robj sin  2 * 498.2 sin( 10.41)
2
L  180m
5. Considering horizontal alignment, what slope is required to ensure
the line of sight around the curve is just tangent to the back slope
given the cross section and data provided below? (Needs Sketch)
6. The design speed of a road is 120 km/hr. Assuming adequate
stopping distance (high value) is provided, what is the required radius
to provide a 10 m clear zone from the edge of pavement to an
obstruction. Pavement width is 3.6 m and the road is a two lane
arterial route.
7. For a horizontal curve on a two lane (3.6 m per lane) road with
a radius of 500 m to the centerline, what is the stopping distance
available to a driver when an obstruction is located 10 m from the
centerline of the road? Assume the grade of the road is level.
8. A horizontal curve has a degree of curve of 3o and a design
speed of 60 mph. A hotel wants to erect a sign 20 ft from the edge of
a 12 ft pavement. Calculate the actual sight distance and the available
design speed for this condition. Should the hotel get permission to
erect the sign?
9. An object is located 30 ft to the inside of a 400 ft (CL)
horizontal curve. What is the maximum speed (calculated) that
provides adequate sight distance if the observer and object are directly
over the centerline.
R
Rm
400
400  30
* cos 1 (
)
cos 1 (
)
28.65
R
28.65
400
V2
V2
S  311.78  1.47 * V * t 
 1.47 * V * 2.5 
a
11.2
30(  G )
30(
 0)
g
32.2
S
311.78  3.675V  .096V 2
 3.67  3.67 2  4 * 0.96 * 311.78  3.67  11.54

2 * .096
.192
Vmax  41MPH
Superelevation
1. A discussions with my engineering colleagues, we determine
that slope of the outside edge of pavement on a superelevation
transition should be 0.5%. For a two lane road (12 ft lanes) and emax =
6%, what is the length of the superelevation runoff. Show on a sketch
the relationship between the centerline and the outside edge of
pavement. Assume the centerline is at a constant elevation of 100.00
ft.
.72
X
.72
X 
 144'
.005
.005 
2. Using the superelevation transition in Problem 4 and the
following curve data, what is the station of the beginning of the
superelevation runoff at the exit end of the curve?
PI Station = 10+10.00
Deflection angle = 45 o R
Degree of curve = 4.5 o
1/3 of superelevation runoff on the curve
R
5729.5780 5729.5780

 1273.24 ft
Dc
4.5
L
 * 100 45 * 100

 1000 ft
Dc
4.5

45
 1273.24 tan
 527.39 ft
2
2
PI  10  10.00
5  27.39
T  R tan
PC  4  82.61
10  00.00
PT  14  82.61
ST
1
 48.00  * 144
3
14  34.61
3. A horizontal curve has its PC at station 10+00.00. If the design
speed is 70 mph and the maximum superelevation is = 0.08 ft/ft, what
is the station of the beginning of the superelevation transition
assuming 60 percent of the transition is on the tangent. Assume a two
lane road, rotation about the center line and 12 ft lanes.
Table 14.13
super runoff  240 ft
.96
ft
 .004
240
ft
.24
X 
 60 ft
.004
L  240  60  300 ft
S
.6 L  .6 * 300  180 ft
BT  PC  1  80.00  10  00.00
1  80.00
BT  8  20.00
4. The super elevation runoff for a horizontal curve on a two-lane
road with a design speed of 65 mph and maximum superelevation of
0.08 ft/ft is rounded with a parabolic curve as shown below. What is
the outside edge of pavement elevation at the end of the transition?
Assume lane width is 12 ft and the centerline elevation is 100.00.
Table 14.13.......for e  .08 and V  65 MPH
Superelevation runoff  230 ft
12 * .08
G1 
 .00417 * 100  .417
230
G2  0
A  (G2  G1 )  (0  .417)  .417
(.417 * 50 2 )
O
 .052
(200 * 100)
elevation  100  .96  .052  100.9078 ft
5. Using the data from Problem 3, calculate the inside and outside
edge elevation at the PC of the curve if 70 % of the superelevation
runoff is on the tangent of the alignment. Assume the centerline
elevation is 100.00.
L1  230 * 70%  161 ft
 1  L * A  161 * .00417  .671
Outerelev  PCelev   1  100  .671  100.671 ft
L 2  30% * 230  69ft
 2  69 * .00417  .287
Innerelev  PCelev   2  100  .96  .287  99.328 ft
6. A superelevated road has a transition that is defined by some key
cross sections in relation to the pavement surface. For the case where
emax = 0.07 ft/ft, lane width is 12 ft, and the normal cross slope is
0.015 ft/ft, what is the left edge of pavement elevation for the
following points:
a. A normal cross section
99.82
b. A fully superelevated section
100.84
c. The end of the tangent runout/ beginning of super elevation
runoff.
100.00
Assume the grade is 0.0% and the centerline elevation is constant and
is 100.00.
a. elev  100  .015 * 12  99.82 ft
b. elev  100  .07 * 12  100.84 ft
c. elev  100
7. The outside edge of pavement on a superelevation runoff has a
slope of 0.44%, pavement width of 12 ft and a maximum
superelevation of 10 %( 0.1 ft/ft). What is the calculated length of the
runoff?
8. The point of curvature for a 1000 ft horizontal curve is at station
10+00.00. What is the station of the beginning of the tangent runout
and what is the superelevation (%) at the point of curvature. Assume
emax = 0.08, speed is 50 mph, lane width is 12 ft and 2/3 runoff on the
tangent.
Runoff  190 ft
e * width .08 *12
S  max

 .0050
Runoff
190
.24
Runout 
 47.5 ft
.0050
2
*190  127 ft
3
STA PC  10  00.00
 1  27.00
8  73.00
 0  47.50
STA BT  8  25.50
9. Use the sketch below and determine the inside edge, outside edge
and the centerline elevation for points A, B, C and E for the following
conditions:
A. 2 lane road
B. Lane width = 12 ft
C. Lane slope = -2 %
D. Rotation about outside edge
E. On the tangent before the transition the centerline elevation
= 100.0
F. emax = .1
Elevation
A
Inside Edge 99.76
Outside Edge 99.76
Centerline
100.00
B
99.52
99.76
99.76
C
99.28
99.76
99.52
E
97.36
97.76
98.56
10. If a superelevation transition is applied to a curve with a runoff
of 70 m, emax = 8% and a lane width of 3 m, what is the length of the
tangent runout? Assume the normal cross slope of the pavement is -2%
  width * emax  3m * .08  .24m

.24

 .00343
runoff
70
width * cross slope
3 * .02
TR 
* runoff 
* 70  17.5m

.24
S
11. A circular curve has 2 lanes (3.6 m each) and a maximum super
elevation of 10%. If the design speed is 110 km/h and 2/3 of the super
elevation runoff is on the tangent, what is the elevation of the inside
edge of pavement at the PC. Assume the elevation of the centerline is
100.00 and rotation is about the centerline
12. The outside edge and centerline profile for a typical two lane
pavement is shown below. Label the beginning of the superelevation
runoff, beginning of the superelevation runout and the relative
location of the point of curvature (PC).
13. A road has 8% maximum super elevation and a design speed of
70 mph. If the PC is located at station 5+00.00, what is the outside
edge of pavement elevation at station 4+00.00? Assume rotation
about the centerline, 2-lane pavement, 12 ft lanes and 60% of the
runoff on the tangent.
Runoff  205 Runout  51
.6 * 205  123 ft
PC  5  00.00
4  00.00
 1  23.00
3  77.00
FS  3  77.00
0  23.00
23
e% 
* 8  .90%
205
.90
elev  100 
* 12  100.11 ft
100
14. The maximum super elevation rate for a two-lane road is 0.08
ft/ft and design speed of 50 mph. What is the outside edge of
pavement elevation 50 ft from the end of the superelevation runoff.
Assume rotation about the centerline, a flat grade and elevation of
100.00 ft.
From Table 14.13  190 ft
x
e * width
140 * .08 * 12

x
 .707
190  50
190
190
elev  100  .71  100.71 ft
or
.96
slope 
 .0051
190
offset  .0051 * (190  50)  .71
elev  100  .71  100.71 ft
.24
Tangent runout 
 47 ft
.0051
TL  190  47  237 ft
Earthwork
1. From a mass diagram the cut at station 1+50.00 and 1+00.00 are
respectively 790 and 656 cubic yards. The area at each of the stations
is identical and defined be the cross section shown below. What is the
depth of the cut at the stations center line?
1
* 20 * x  20 x  30 x
2
A  A2 L
V  790  656  134  1
*
2
27
(30 x  30 x) 50 (60 x) * 25
134 
*

2
27
27
x  2.41 ft
A1  A2 
2. Use the profile shown below. Sketch the mass diagram and
determine the overhaul cost. Assume there is no free haul, balanced
earthwork, maximum fill of 1000 m3 at sta 0+500.00.and the cost of
overhaul is $10/m3sta. Also, what is the direction of the overhaul.
3. Sketch a mass diagram (label the axis) for a road project that
starts in a fill section (-) goes into a cut section (+) and ends with a net
waste of material.
4. Given the following earthwork data sketch the approximate
profile of the road and determine the waste/borrow for the job.
Station
0
100
200
300
400
500
600
700
800
900
1000
Cut(+)/Fill(-)
0
150
300
250
100
0
-150
-300
-100
-50
100
5. Given the cross section area at station 1+000.00 is 150 m2 (fill)
and 25 m2 (cut) and is 50 m2 (fill) and 100 m2 (cut) at station
1+020.00. What is the net volume (+ or -) of earthwork between the
stations if cut is + and fill is -.
e. Sketch a mass diagram (label the axis) for a road project that starts
in a fill section (-m3), goes into a cut section (+) and ends with a
net borrow of material. Show and label at least one balance line
somewhere on the diagram.
f. A mass diagram is shown below:
Find the following values
A. The first station where the section is in cut.
B. The over haul (free haul = 500 m). Use the approximate
method to find the centers of gravity.
Mass Diagram
600
500
1000
Cubic meters
400
300
200
100
0
-100 0
-200
-300
Mass Diagram
0
500
1000
1500
1500
2000
250
Distance, meters
7. The coordinates of a free haul line on a mass diagram are
(2+00.00, +600.00) and (12+00.00, +600.00). From these points the
mass diagram intersects the abscissa at 0+00.00 and 14+00.00.
Calculate the amount of overhaul in station cyds. Use the mid
ordinate method to determine the overhaul distance
8. Use the attached Mass Diagram and calculate the overhaul
(cuyds-ft). The free haul is 800 ft. Show and label the overhaul
ordinate and the borrow/waste (correctly labeled).
9. At what distance from the beginning does the actual profile go
from cut to fill?
Stopping Distance
1. A driver’s eye height is 2 ft above the pavement. What percent
reduction in stopping distance (over normal) will this condition
cause? Assume the following:
a. Object height = .5 ft
b. Crest vertical curve grade difference = 7%
c. Design speed = 70 mph
d. S < L
e. Normal stopping distance is based on the conservative value
for the design speed.
AS 2
7 * 850 2
where S  850 so L N 
 3805 ft
1329
1329
Ax 2
Offset 
200 L
2.0 * 200 * 3805
x1 
 466.3 ft
7
LN 
.5 * 200 * 3805
 233.1 ft
7
x1  x 2  699.4 ft
x2 
%reduction 
S  ( x1  x 2 )
850  699.4
* 100 
* 100  17.7%
850
850
Vertical Curves
1. Given the following measurements taken in the field find the
length of vertical curve and the station of the high point of the curve.
G1 = 5%
G2 = -2%
PC Station 10+00.00 Elev = 100.00
Station 11+00.00 Elev = 103.00
Ax 2
 7 * 100 2
L
 175 ft
200 L
200 * 2
dy 2 Ax G1


0
dx 200 L 100
G 200 L
5 200 * 175
x 1 *

*
 125 ft
100 2 A
100
2 * 7
G
Ax 2
5
 7 * 125 2
y  x * 1  100 ft 
 125 *
 100 
 103.12 ft
100
200 L
100
200 *175
STA  10  00.00
1  25.00
STA  11  25.00
Offset  2 ft 
2. The line of sight from an observer to an object along a crest
vertical curve is tangent at the high point of the curve. If the length
of the curve is 1000 ft and the observer’s eye height is 3.0 ft above
the pavement, what is the Station of the observer if the Station of the
PIVC is 10+00.00.
 G1 L
4 1000
* 
*
 666.67 ft 600
100 A 100
6
STA  5  00.00
6  66.67
STA  11  66.67
High pt. 
Ax 2
6x 2

3
200 L
200 *1000
600 * 1000
x2 
 100000
6
x  316.23 ft
STAobs  11  66.67
11  66.67
Offset 
STAobs
 3  16.23 or  3  16.23
 8  50.34 or 14  83.00
3. Using the data from Problem 3, calculate the inside and outside
edge elevation at the PC of the curve if 70 % of the superelevation
runoff is on the tangent of the alignment. Assume the centerline
elevation is 100.00.
4. A line of sight is tangent to a parabolic crest vertical curve at Sta
9+00.00. The observer is on the curve at Sta 6+00.00, what is the
height of the observer’s eye for the given conditions. Assume G1 =
2%, G2 = -3%, the PI Sta is 10+00.00 and the length of curve is
1000 ft. If all else fails remember y = ax2 + bx and solve for the
offset relationship.
h
Ax 2
(3  2) * 300 2

 2.25 ft
200 L
200 *1000
5. The general equation for a parabolic vertical curve is y =
ax2+bx+c. Assume the origin of the coordinate system is at the
PIVC, G1 = -5%, G2 = +3%, and the length = 1000 ft. Determine
the specific equation for this vertical curve.
6. The equation for a parabolic vertical curve is
y = (Ax^2)/200L + (G1/100)X
If the observer is at the point of curvature for the vertical curve
(VCPC=STA 10+78.61), what is the station at the point of tangency
of the line of sight to the curve if the observer is at the VCPC, L =
1000 ft, G1=3%, G2=-2% and the observer eye height is 3.5 ft above
the pavement.
A  G2  G1  (2  3)  5
Ax 2
offset 
x
200 L
STA  10  78.61
offset * 200 * L
3.5 * 200 *1000

 374.16 ft
A
5
3  74.16
STA  14  52.77
G
3
y  offset  1 * x  3.5 
* 374.16  7.72 ft
X
100
7. The equation
y
Ax 2 G1 x

 100
200 L 100
is for a parabolic curve with the origin at the PVC with x = 0.00
and y = 100.00. If the eye level of an observer is 2 m above the
pavement and the observer is at the PVC, what is the distance to the
point of tangency of the observer’s line of sight with the pavement?
G1 = 5%, G2 = -3% and the length of vertical curve is 200 m.
8. The equation for a parabolic vertical curve (beginning at the
PCVC) is: y 
Ax 2
G
 1 x
200 L 100
If the height of the observer is 2.0 m and the height of the object is
1.0 m, what is the available stopping distance over the crest vertical
curve if the vehicle and the object are on the curve?
9. For a design speed of 60 mph, what decrease/increase in length
of crest vertical curve occurs when the object height is lowered from
2.0 ft to 0.5 ft. Assume S<L and A= 6%.
10. Catch basins are located at the low point of a 600.0 sag vertical
curve with G1 =- 5% and G2=+2%. If the PI is at station 10+00.00
and 100.00 elevation, what is the station and elevation of the catch
basins
11. An observer is 6 ft (eye height) and is looking up station from
the PC. He is just able to see the head of another person 5 ft 6 in.
tall. What is the required length of vertical curve? Assume they are
both on the curve, at the same elevation, G1 = 3% and G2 = -5%
 8 x1
1200 L
 x1 
200 L
8
x  x1  x 2
2
6
x2 
1100 L
8
Ax 2 G1 x

0
200 L 100
8(12.25 L  11.73 L ) 2
3
0

(12.25 L  11.73 L )
200 L
100
8
3
0
(23.97 L ) 2 
(23.97 L )
200 L
100
0  22.99  .7191 L  .7191 L  22.99
L  1022.11 ft
y
12. A pipe 6 ft in diameter passes over a highway sag vertical
curve with a clearance of 16 ft. What is the lowest elevation and
station of the bottom of pipe? Assume the length of the vertical
curve is 400 ft, G1 = -3, G2 = 4%, PC Station is 5+00.00 and
elevation is 100.00.
G1
3
* 400 
* 400  171.43 ft
A
7
STA  5  00.00
Low pt 
1  71.43
STA  6  71.43
G
Ax 2
7 * 171.43 2
3
 1 * x  16  100 

* 171.43  16
200 L 100
200 * 400 100
elev  100  2.57  5.14  16  113.43 ft
elev  100 
13. A sag vertical curve has a –4% grade followed by a 5% grade.
The station at the PIVC is 1+000.000 with an elevation of 100.000.
What is the station and elevation of the low point on the curve?
14. I’m standing at the VCPC of a 120 m crest vertical curve. How
tall do I have to be to see the VCPT? G1 = +2% and G2 = -1%
Vertical Sight Distance
1. The eye height for a truck is 5.0 ft and for a car is 3.5 ft. If an
object on a crest vertical curve is 0.5 ft high, how much additional
stopping distance does the truck have over the car assuming the
design speed is 60 mph, length of vertical curve is 2000 ft, G1 = +4%
and G2 = -2%.
Offset Truck
AX 2 t
6 X 2t

 5.0
200 L 200 x 2000
X t  577
ot 
Offset Car
2
6X c
AX 2 c

 3.5
200 L 200 x 2000
X c  483
oc 
Difference  577  483  94
Note: The offset for the object is the same and the difference is
determined from the difference of the offset between the truck and
the car.
2. A crest vertical curve must pass through elevation 95.00 at
station 9+00. If the PIVC is at station 10+00.00 with elevation
100.00, what is the required length and the resulting design speed for
this vertical curve (h1=3.5 and h2=0.5). Assume that G1 = +2% and
G2 = -4%.
Elev at 9  00  100  2  98
Offset  98  95  3
L
6(  100)
2
AX
3
 2
200 L
200 L
L  746
AS 2
L
 746
1329
S  stopping _ disance  407
From Table 16.5
V  45 to 50mph
3. A truck driver’s eye height is 6 ft above the pavement. Based on
the usual criteria for stopping sight distance on a crest vertical curve,
what is the percent increase in sight distance for the truck driver
compared to the design driver?
2
AX 1
6
200 L
1200 L
X LT 
A
%increse 
ST

Sc
X Lc 
X 1T  X 2T
X 1c  X 2 c
700 L
A
X2 
.5 * 200 L
A
1200 L
100 L

A
A

700 L
100 L

A
A
L
( 1200  100 )
34.64  10 44.64
A


 1.22
26.46  10 36.46
L
( 700  100 )
A
4. The stopping sight distance over a crest vertical curve is given as
L
AS 2
100( 2h1  2h2 ) 2
What are the variables h1 and h2?
h1=height of observer
h2=height of object
5. A headlight for a truck is located 1000 mm above the PCVC. If
the light beam spreads upward at an angle of 2o above the horizontal,
what is the sight distance provided by this headlight system?
Assume G1=0% and G2=10% and the length of the vertical curve is
200 m.
6. The Conventional equation for a sag vertical curve is based on
the height of the headlights and a 1o cone of illumination by the
lights. The resulting equation is
L
AS 2
S<L
(120  3.5S )
Where S is the stopping distance and L is the length of vertical curve.
If the headlights are not aimed correctly and the cone of illumination
is 0o, what is the required length of vertical curve? Assume the
design speed is 80 km/hr.
7. On a crest vertical curve at the PCVC + 50 ft the offset from a
vertical tangent to the curve is 0.2 ft. If the grade difference is 8 %
what is the design speed for the vertical curve based on stopping
sight distance?
Ax 2
8 * 50 2
 .2 
200 L
200 L
L  500 ft
offset 
3.5 * 200 * 500
.5 * 200 * 500

 209.17  79.06  288.23 ft
8
8
V2
288.23  2.5V 
2 * 32.2 * .34
V  56.6 fps  39 MPH  40 MPH
S  x1  x 2 
Rational Method
1. Use the attached intensity curve and fill in the table given
below. Assume data is for Houghton.
CB
No.
Area % Sum
(Ac) Imp Area
(Ac)
1
2
.3
2
1
.4
3
1
.6
___
_
___
_
___
_
Com
b%
Imp
____
____
____
Time Flo Pipe Inc
Of
w Q size Time.
Conc (cfs (in) (min)
entrat
)
ion
(min)
15
___ 12
1.0
_
____ ___ 121.5
_
____ ___
_
2. Given two drainage areas that drain to a common point A as
shown below. With the data for each area given in the table below,
what is the maximum runoff at point A if the Rational Method is
used?
Property
Area (acres)
Distance to farthest point in area (ft)
Coefficient
Slope of ground
Area 1
5
1000
.5
2%
Area 2
2
500
.3
1%
Surface type
grass
bare ground
short
3. Given:
Area = 1 mi2
Streets = 10 %
Row crops (poor) = 40 %
Wood (poor) = 50 %
Time of concentration = 1 hr
Precipitation = 6 in
What is the difference between the runoff for antecedent moisture
condition I and III? Use the TR55 Method
4. Find the time of concentration associated with a drainage basin
with three components as listed below:
A. Short grass, length = 300 ft, slope = 2%.
B. Open channel, length = 2500 ft, n = .08, slope = .6%,
hydraulic radius = 6 ft.
C. 36 in storm sewer, flowing full, length = 1000 ft, n = .012,
slope = 1 %.
A. Figure15.4 V  1 fps
L 300 ft
Tc  
 300 sec  5 min
ft
V
1
sec
2
2
1
1
1.49
1.49 3
3
2
B. V 
*R *S 
* 6 * .006 2  4.76 fps
n
.08
2500
Tc 
 525 sec  8.75 min
4.76
 * 32
C. R 
 .75
4 * 3
2
1
1.49
V 
* .75 3 * .01 2  10.25 fps
.012
1000
Tc 
 98 sec  1.62 min
10.25
5. A watershed has the following characteristics:
No.
Area
Land Use
Soil Type
1
2
3
4
0.15 mi2
0.20 mi2
0.50 mi2
0.10 mi2
Meadow (good)
Residential (1/2 ac)
Forestland (fair)
Streets (paved)
A
C
B
D
AMC
I
II
I
III
Find the runoff if the rainfall is 5 in. Use the TR 55 Method
6. A typical section for a road is shown below. If the summit of the
drainage basin is 500 ft (horizontal) with a slope of 2 % above the
outlet, what is the time of concentration of the flow for this drainage
area? Assume the flow is only perpendicular and parallel to the
centerline of the road
From Figure17.4
ft
ft
V paved w / 4% slope  4.0
sec
sec
ft
ft
Vshort grass1 on 4  3.5
Vshort grass1 on 2  5
sec
sec
12 10 20 500
Tc1 



 4.3  2.5  5.7  227.3  239.77 sec
2.8 4.0 3.5 2.2
30 500
Tc 2 

 6  227.3  233.3 sec
5 2.2
Tc max  239.77 sec  .067 hr
V paved w / 2% slope  2.8
7. For the section described in problem 6, what is the runoff
associated with 1/2 of the cross section if the return period is 25 years
and the rainfall intensity can be determined using Fig 17.2 Assume
the time of concentration is 10 minutes and the average runoff
coefficient is 0.4 (Remember 1 acre = 43560 ft2).
Area 
Tc 
72 * 500
 .83 Acres
43560
10
 0.17
60
from Figure17.2 I  6.5
in
hr
Q  .4 * 6.5 * .83  2.1cfs
8. Circle the correct observation related to the TR 55 Method.
If the runoff, h, is large, the CN is high/low and the maximum
retention, S will be large/small.
9. A rural drainage area is:
 20% asphalt pavement
 30% turf meadow (good)
 50% cultivated fields (straight rows, poor)
Assuming that the soil type is A and the AMC is III, what is the
relationship between the runoff coefficient for the Rational Method
and the Curve Number for the TR 55 Method
10. A drainage basin is 30 % asphalt pavement with a runoff
coefficient of 0.9 and 70 % cultivated fields with a coefficient of 0.3.
What is the average runoff coefficient that would be used in the
equation Q=CIA?
11. A 10 acre drainage area has 10% asphalt pavement, 40% turf
meadows and 50% cultivated fields. If the rainfall intensity is 3 in/hr
how deep should a V shaped ditch (1 on 2 side slopes, n =0.1 and
slope = 2%) be to provide for adequate drainage for the area? Use
coefficients that will produce the largest runoff.
12. How large an area would the standard ditch section shown
below drain if the drainage area is 50% forested, 30% steep grassed
slopes and 20% concrete. Assume the storm duration is 30 min., the
storm return period is 10 years and Fig 11. 2 is valid in the design
area.
13. A 170 acre rural drainage area consists of four different
watershed areas as follows:
i. Steep grass-covered area = 40 %
C = .5-.7
ii. Cultivated area = 25 %
C = .2-.4
iii. Forested area = 30 %
C = .1-.3
iv. Turf meadow = 5 %
C = .1-.4
Using the rational formula determine the runoff rate for a storm of
100 yr frequency. Assume the rainfall intensity curve is Fig 17-2.
The following drainage basin characteristics apply.
a. overland flow length = 0.5 mile
b. average slope = 3 %
c. fallow or minimum tillage cultivated
C avg  .4 * .7  .35 * 4  .3 * .3  .05 * 4  .49
length
Tc 

Velocity
.5mi * 5280
.8
ft
sec
ft
mi  3300 sec* 1  .92hr
3600
I  3.1
Q  C avg * I * A  .49 * 3.1 * 170  258cfs
Ditch Flow Problems
1. For the conditions given below find the incremental loss in flow
capacity after the ditch is partially filled with silt. Assume the
slope=2% and n=0.01.
2. A ditch has an original shape as shown below. To increase the
flow the cross section is changed by excavating material to increase
the base width to 6 ft. What is the flow for the improved cross
section? Assume the slope is 1 %, n = .01, fore slope is on 4 and the
back slope is 1 on 2. Height of the water remains constant
3. A V ditch with bottom 5 ft below the centerline grade must
provide 3 ft of freeboard above the high water mark. What area can
the ditch drain if the 10 yr storm intensity is 5 in/ hr and the other
details are shown below?
4. An area is 40% steep grass, 25% cultivated, 30% forested and 5
% turf meadow. For a 5 acre plot, 10 year storm and 15 minutes
time of concentration, what size ditch section is required if it is semi
circular flowing full and has n = 0.012 and Slope = 1 %?
5. The ditch for a low volume road is 4 ft below center line grade.
If the water level cannot be within 2 ft of the centerline grade, what
is the maximum drainage area that can be serviced by the ditch
section shown below?
6. For Q = 100 cfs, Slope = 2%, and n = 0.012, what is the
difference in depth of flow for the open channels shown in Fig 17.6
versus Fig 17.7?
7. For the ditch shown below, what is the maximum drainage area
that can be served if the flow velocity is 4 fps? Assume that Cavg =
0.5 and the rainfall intensity is 5 in/hr.
8. A V ditch with side slopes of 1 on 2 has a discharge of 100 cfs
and a maximum depth of flow of 3 ft. What material should be used
to line the ditch? Assume the slope of the ditch is 2 %.
A
1
* base * width  .5 * 12 * 3  19 ft 2
2
WP  2 6 2  3 2  13.4 ft
R
A
18

 1.34
WP 13.4
2
1
1.49
n
* 18 * 1.34 3 * .02 2  .046
100
From Table17.9
light brush on banks .035 - .05
Bermuda grass mowed to 2 in. .05 - .035
"
" length 4 to 6 in.
.06 - .04
Dense growth of weeds .035 - .05
Some light brush .035 - .05
Bottom of gravel, cobbles, and few boulders
.04 - .05
Also acceptable
High grass, mature field crops, cleared land with or without sprouts
9. What is the maximum flow that can occur in the gutter shown in
the sketch below without over topping the curb? Assume the
roadway grade is 2%.
10. A rectangular shaped ditch shown below if flowing full. If the
contributing area is 20 Hectares and it is 50% forest, 30% cultivated
fields and 20% pavement, what is the occurrence period associated
with this condition? Use an appropriate runoff coefficient that will
produce the maximum runoff. Assume a time of concentration of 15
min, slope of the ditch is 2% and the Manning coefficient is 0.01.
11. A V ditch with a depth of 2 ft and 1 on 1 side slopes drains a
roadway with a 2 % slope. What is the maximum area this ditch can
drain if the average runoff coefficient is 0.5 and the 10 year rainfall
intensity is 2 in/hr. The Manning’s coefficient is 0.012 for the
vegetation in the ditch.
12. Area A (steep grassy slope) is 2.5 acres and has a time of
concentration of 30 min. Area B (forested) is 5.0 acres and has a
time of concentration of 1 hr. Both areas contribute to MH # 1. The
outlet at MH#1 is a 15 in pipe. What is the maximum velocity of
flow in the pipe (assume flowing full). Use a 25 yr storm frequency
and Fig 17.2.
13. A typical MDOT ditch has 1 on 4 side slopes and a 6 ft bottom.
The maximum depth is 3 ft. What is the maximum flow if n =0.01
and the roadway slope is 2 %.
14. Maximum runoff for a V ditch (1 on 2 backslope and 1 on 4
foreslope) is 100 cfs. If n = 0.1, slope is 2 % and clearance from
centerline elevation to maximum water level is 3 ft. What is the
required depth of the ditch measured from the centerline
elevation15. A ditch section is shown below. Assume the
contributing area is 20 hectares and is 50% forested, 30% turf
meadows, 10% gravel shoulders and 10% asphalt pavement. What
is the expected design period for the section that would allow water
to rise to the edge of pavement (EOP). The rainfall duration is 1 hr
and the region is typical of Fig 11.2.
15. A ditch section is shown below. Assume the contributing area
is 20 hectares and is 50% forrested, 30% turf meadow, 10% gravel
shoulders and 10% asphalt pavement, what is the expected design
period for a section that would allow water to rise to the edge of the
pavement (EOP) . The rainfall duration is 1 in / hr and the region is
typical of Fig. 11.2.
Culvert Problems
1. A culvert (entrance type 1) is needed to move the water
transported in the ditch from prob. 2 under the road. Assume the
runoff is 200 cfs and the headwater elevation/diameter ratio is 1.
What size pipe (ft) is need and what is the invert elevation (ft)
assuming the water elevation at the inlet end is 100.00 ft.
2. Find the invert elevation at points A and B assuming the water
elevation for inlet control is equal to the water elevation for outlet
control. Assume Q = 200 cfs, 5 ft diameter concrete pipe, n = .012
,ke = .5 tailwater elevation = 2 ft.
3. A 4 ft diameter concrete pipe ( square edge with headwall) has a
slope of 1 %, length = 250 ft, n = .012, ke = 0.2, entrance invert of
130.00 and tailwater of 8 ft. If 3 ft of freeboard is required above
the maximum water level, what is the plan grade elevation at the
station of the culvert? Assume the flow is 200 cfs.
4. A 4 ft diameter concrete pipe (n = 0.012) with square edge and
headwall has its entrance invert at 100.00. If the flow is 100 cfs ,
what is the minimum elevation of the roadway centerline. Assume
the vertical distance between the centerline and water level is 4 ft,
tailwater is 8 ft, slope of the pipe is 2 %, entrance coefficient is 0.2
and the length of the pipe is 200 ft
5. A 24 in concrete pipe with square edge and headwall is 450 ft
long and has a fall of 3 ft in its length. It will flow full and the outlet
will be submerged by 2 ft above the crown of the pipe, what flow
will the pipe carry when the inlet is submerged by 3 ft of water
above the crown of the pipe and the slope of the pipe is adjusted so
inlet control equals outlet control. n = 0.012, ke = 0.5
6. A 48 in pipe passes below an embankment and must provide 3 ft
of clearance between the high water elevation and the plan grade.
The flow through the pipe is 200 cfs and the length is 150 ft. Find
the slope of the pipe when the calculated water levels from inlet
control and outlet control are equal.
7. A culvert is placed with its entrance invert at an elevation of
100.00. If a 150 ft culvert pipe is placed at a slope of 1 %, what is
the expected maximum water elevation if the tailwater is 10 ft, Q =
300 cfs, n = 0.012 and the culvert is six ft in diameter with a k e value
of 0.2. This is a circular concrete pipe with grooved end projecting
Inlet
Q
AD
1
2

300

2
1
2
 4.3  4 submerged
*6 *6
4
From Table17.10
c  .0317 y  .69
HW
Q 2
300
 c(
)  y  .5S  .0317(
) 2  .69  .5 * .01  1.29
1
1
D
 2 2
AD 2
*6 *6
4
HW  6 *1.29  7.74
ELi  100  HW  100  7.74  107.74 ft
Outlet
29n 2 L V 2
29 * .012 2 *150 300 2
1
H  (1  k e  1.33 )( )  (1  .2 
(
) *(
)
 2
 2
2g
2 * 32.2
R
*6
*6
(4
) 4
6
H L  2.73 ft
EL  100  .01 *150  TW  H L  98.50  10  2.73  111.23 ft
ELmax  111.23 ft
8. A 5 ft diameter 100 ft long concrete pipe with square edge and
headwall has its inlet invert at 100.00 elevation. If the slope of the
pipe is 1% and the tail water is 7 ft, what is the required centerline
elevation if a minimum of 5 ft is required between the maximum
water elevation and the centerline elevation? Assume Q=200 cfs
and ke=0.2.
9. A 1500 mm concrete pipe with n = 0.012 and square edge with
wing walls entrance conditions has a discharge of 10 cms. If the
entrance invert elevation is 100.00, the pipe is 100 m with a slope of
½%, tail water elevation of 2.0 m and Ke 0.5, what is the maximum
water elevation for these conditions? Is this inlet control or outlet
control?
10. A concrete culvert is subject to inlet control and has a flow of
10 cms. The maximum water level at the upstream end is 100 m and
the entrance invert must be placed at an elevation of 96.00 m. What
size pipe must be used to meet these conditions? Assume entrance
condition (1).
11. For the condition given in Prob 3, what size CMP pipe is
needed at the bottom of the drainage area assuming maximum
HW/D = 1.1, inlet control and a projecting end.
12. A 36 in culvert is under Outlet control. The maximum water
elevation is 110. Find the exit invert elevation associated with outlet
control if the pipe has Q = 50 cfs, n=0.012, ke =.5, the tail water is
3.0 ft and length of 150 ft.
13. Determine whether a 6 ft 6 in reinforced concrete pipe with
ke= 0.5, grooved end with projecting and carrying a flow of 200 ft3/s
will operate under inlet or outlet control condition. Assume:
a. Design headwater elevation = 105 ft
b. Tail water depth = 6 ft
c. Length of culvert = 200 ft
d. Slope of culvert = 1.5%
e. n= .012
INLET CONTROL
HW
 .85  From Figure17.17
D
HW  .85 * 78  58.5in  4.87 ft
InvertUS  105  5.52  99.48 ft
1.5
* 200  99.48 ft
100
OUTLET CONTROL
Invert DS  99.48 
H L  (1  k e 
29n 2 L V 2
29(0.12) 2 * 200
200 2
1
)

(
1

0
.
5

)(
) *(
)
1.33
4
2
2
2g
2 * 32.2
R
 3.25 3  3.25
(
)
2 * 3.25
H L  1.09
ho  TW or
d c  3.75
dc  D
which ever is greater
2
from Figure17.24
d c  D 3.75  6.5

 5.1 ft
2
2
ELho  ELo  ho  H L  99.48  6  1.09  103.57 ft
105  103.57 so inlet controls
14. A 100 m concrete pipe culvert with outlet flow control
conditions and 2 m3/s flow has a maximum water elevation of
100.00 m. The culvert is 1m in diameter and the tail water is 1 m.
What is the culvert’s invert elevation at the downstream end?
Assume n = 0.012 and ke = .5.
Aggregate Gradation
1. A base course is combined with a borrow material. If the gradation
curves show that 30% of the borrow and 70% of the base course pass
the # 4 sieve (4.5 mm) and the required spec limits at the # 4 are 70 –
40 % passing, what is the minimum proportion of borrow that will
produce a combined mix that just meets the midpoint of the specs at the
#4 sieve
2. An aggregate has 40% passing the Number 4 (4.84 mm) sieve and
100 % passing the 2 in sieve. If the specifications require the aggregate
to be graded according to the power rule + 10%. Does the aggregate
meet the specifications (show by calculation)?
The power law is:
Px  100(
SS x 0.5
)
SS Max
Where
SSx = sieve size at screen x
SSMax = sieve size maximum
3. Two aggregates are blended to meet specifications at the number
200 sieve. If the specifications require 10% passing the 200 sieve and
aggregate A has 15% passing the 200 and Aggregate B has 3 % passing
the 200 sieve, what is the proportion of aggregate A and B that will
provide a mix that has 10% passing the 200 sieve?
Equilivant Single Axel Loads
1. The traffic conditions at a site produces the following data
a. AADT = 10000
b. % Trucks=20%
c. Two lane facility
d. directional split = 50%
e. annual growth = 5%
f. Truck factor = 1.00 (based on truck traffic)
What are the expected ESAL after 20 years?
2. If the relationship between load (kips) and number of load
repetition to failure is:
W = 100 - 11 log 10(NR)
where:
W = the load in kips
and
NR = the number of load repetitions.
Based on the number of repetitions to failure, what would be the
single 18 kip load equivalence factor for a load of 10 kips.
10  100  11log 10 ( NR)
90
 152 *10 6 repetitions
11
18  100  11 log 10 ( NR)
log 10 ( NR) 
log 10 ( NR) 
F
82
 28.5 *10 6 repetitions
11
28.5
 .187
152
3. A coal truck has one single axle with 12 kips and five tandem
axles with 32 kips each and makes five trips a day to Baraga. How
many trips would be required by a passenger car with two axles with
2 kips on each axle to apply the same number of 18 kip single axle
loads as the truck?
4. In trying to assess the effect of different vehicles the following
data was obtained
Axle Load
Displacement
Number of repetitions to
produce crack
10 Kip
0.001 in
18 Kip
0.01 in
10000
1000
If the pavement wear is directly related to the product formed from
displacement and number of repetitions to first crack, what is the
appropriate load factor relating the 10 Kip load to the 18 Kip load.
5. A four-lane highway has an AADT of 100000 VPD with an
expected growth of 3% per year. If the truck traffic is 20 % of the
total with an average equivalent load factor of 0.4 and the passenger
cars are on average 4000 lbs, what is the number of ESAL in the
design lane if the expected life of the pavement is 25 yrs?
6. A traffic stream for a two lane road has an AADT of 10000
vehicles with an ESAL factor of 0.2. If the design period is 20 yr.
with a growth of 2% per year, what are the ESALs (18kip) if the
lane distribution is 50% in each direction and the growth factor is
GF 
((1  r ) n  1)
r
7. Calculate the number of equivalent single 18 Kip axle loads
(ESALs) for a two lane traffic stream with an AADT of 10000
vehicle per day (vpd), 3 % annual growth and a traffic factor of 2.0
ESAL/veh. The design period is 20 years.
8. Which vehicle shown below is the most efficient in terms of
ESALs?
40 K
40 K
20 K
20 K 10 K
40 K
10 K
9. A traffic stream has the following 2 axle vehicles
1000 veh/day @ 2000 lbs / axle
500 veh/day @ 10000 lbs/axle
What is the daily 2 axle ESAL18K factor for this traffic stream?
10. US 41 (other principal) has an AADT of 10000 vehicles per
day with 20% trucks. Of the trucks 5% are multi unit with 5 axles.
What are the ESAL per lane contributed by the 5 axle trucks for a
period of 1000 days?
11. The predicted traffic mix for a proposed urban four-lane
interstate is:
a. Passenger cars = 69 %
b. Single unit trucks
1. 2 - axle 4 tire = 20 %
2. 2 - axle 6 tire = 5%
3. 3 axle or more = 4%
c. Tractor semitrailer and combinations
1. 3-axle = 2 %
ESALs  AADT * % * 365 * f i * G f * f d
f d  .45
from Table 20.7
G f  29.78
ESALs2 axle 4tire
(1.04 20  1)
 29.78
.04
 3500 * .20 * 365 * .002 * 29.78 * .45  6848
from Table 20.6, also G f 
ESALs2 axle6tire  3500 * .05 * 365 * .17 * 29.78 * .45  145518
ESALs3axle  3500 * .04 * 365 * .61 * 29.78 * .45  417723
ESALs3axle  3500 * .02 * 365 * .98 * 29.78 * .45  335547
Total ESALs  335547  417723  145518  6848  .906 * 10 6 ESALs
The projected AADT during the first year is 3500 (both
directions) and the growth rate is 4 %. What is the ESAL (based
only on trucks) for a 20 yr design period?
12. The traffic stream on a 4 lane rural freeway is expected to
carry an AADT of 10000 during the first year of operation with an
expected growth of 5% per year over the 20 year design life. The
expected traffic is 60% passenger cars, 20% 2-axle 4-tire trucks,
10% 2-axle 6-tire trucks, 5% 3-axle trucks and 5% tractor-trailers.
What are the ESALs for the design life on the design lane?
Asphalt Thichkness Design
1. Using the AASHTO Design procedure and the following data
what is the thickness of the base material?
SN = 4.0 when the Mr = 10000 psi (sub grade)
SN = 3.0 when the Mr = 15000 psi (sub base)
SN = 2.0 when the Mr = 30000 psi (base)
2. From the AASHTO design procedure for asphalt pavements the
following information is obtained
SN = 2.0 (Based on Mr base aggregate)
SN = 3.0 (Based on Mr sub base aggregate)
SN = 4.0 (Based on Mr subgrade soil)
What is the thickness of the base assuming that no sub base is
used, m2 and m3 are = 0.8, a1 = 0.44, a2 = .14 and a3 = .11.
3. The subgrade soil, traffic and serviceability index for a site near
Houghton provides a SN of 4.0. If the calculated thickness of base
is 12 in and asphalt is 4 in, what is the resilient modulus of the
subbase? Assume SN = .4 D1+.2 D2+.1 D3m3, D PSI =2, R=90%,
So = .3, m3 =.8 and ESAL=5000000.
From AASHTO Design Chart
M r  7  8 *10 3 psi
Use Design Eqn
log 10 W18  Z R S o  9.36 log 10 ( SN  1)  0.20 
log 10 [PSI
0.40  [1094
(4.2  1.5)
]
 2.32 log 10 M r  8.07
( SN  1) 5.19
log 10 [ 2
]
(4.2  1.5)
log 10 (5000000 )  1.282 (.3)  9.36 log 10 (4  1)  0.20 
 2.32 log 10 M r  8.07
0.40  [1094
5.19 ]
(4  1)
6.699  .3846  6.542  0.20  .198  2.32 log 10 M r  8.07
9.0096  2.32 log 10 M r
M r  7646 psi
5. From the AASHTO design procedure for asphalt pavements the
following information is obtained
SN = 2.0 (Based on Mr base aggregate)
SN = 3.0 (Based on Mr sub base aggregate)
SN = 4.0 (Based on Mr subgrade soil)
What is the thickness of the base assuming that no sub base is
used, m2 and m3 are = 0.8, a1 = 0.44, a2 = .14 and a3 = .11.
Temperature Steel Concrete Pavements
1. The longitudinal steel requirement for a concrete pavement is
0.1 in2 per linear ft. The only steel available is a ¼ in diameter
wire. What is the required spacing if this wire is used?
2. The cross sectional area of transverse temperature steel for a
concrete slab 12 ft wide, 40 ft long (between joints) and 10 inches
thick is 0.125 in2/ft. What is the required length and spacing for a
0.375 in diameter tie bar? Assume concrete bond strength is = 400
psi and steel yield stress of the tie bar is 30000 psi.
Aact 
 2
4

 (3.75) 2
4
 .11045in 2
in 2
Areq 'd  .125
ft
A
.11045
Spacing  act 
 .88 ft  10.6in
Areq 'd
.125
t
2 Aact * f s
2 * .11045 * 30000

 14in
.375
2 * r * f b
2 *
* 400
2
3. The longitudinal steel requirement for a concrete pavement is
0.1 in2 per linear ft. The only steel available is a ¼ in diameter
wire. What is the required spacing if this wire is used?
AASHTO Concrete Pavement Design
1. List the three sources of the stresses that are usually considered
in the design of concrete pavements.
a. ___________________________________________
b. ___________________________________________
c. ____________________________________________
2. List two methods used to control cracks in concrete pavements
a.
_____________________________________________
g. ____________________________________________
3. Sketch a typical contraction joint in a reinforced jointed
concrete pavement with load transfer device (dowel bar).
PCA Concrete Pavement Design
1. The PCA method of concrete pavement design has a fatigue
analysis and a erosion analysis design. The allowable repetitions
for a 15 kip single axle load is 30000 and this is 20 % of the fatigue
life, what is the number of loads applied assuming the load factor is
1.2.
2. The Portland Cement Association rigid pavement design
method is based on the percent of fatigue and erosion resistance
used. Given the data below what is the percent erosion resistance
used by 70.0/day - 50 kip single axel loads? Given
Combined k = 400 pci
Modulus of rupture of concrete = 700 psi
Local Access Road
Trial thickness = 10 in
Doweled joints without concrete shoulders
Design life = 10 years
Growth = 0 %
70
days
* 365
*10 years  255500 * .5  127750
day
year
From Table 21.13 Erosion factor  2.49
# Load 
From Figure 21.16 Allowable Loads  400000
255500
%
 64%
400000
127750
%
 32%
400000
3. The PCA method of concrete pavement design has a fatigue
analysis and an erosion analysis design component. The allowable
repetitions for a 20 kip single axle load is 30000 and this is 20 % of
the fatigue life, what is the number of 20 kip loads applied
assuming the load factor is 1.2.
Traffic Flow Problems
1. Traffic on a road section follows the relationship
u  50  0.5k
where u is the speed and k is the concentration. What is the
spacing when the speed is 25 mph?
2. If q = ku and spacing = u*t where u is in fps and t = 2.5 sec,
what is the density (vehicles / mile) and volume (vehicles / hr) on
the road at u = 30 mph?
ft
veh
1
1
veh
d 
 48
ft
1mile
S
mi
110
*
veh 5280 ft
veh
veh
q  uk  30 MPH * 48
 1440
mi
hr
S  u * t  30 MPH * 2.5 sec  110
3. Traffic on a road section follows the uniform acceleration
relationship
a=3.4 m/s2
where a is the deceleration of a braking vehicle. What is the
distance required for a vehicle to go from 100 to 0 km/sec?