PE Exam Review - Surveying
Demonstration Problem Solutions
I.
Demonstration Problem Solutions........................................................................... 2
1. Circular Curves – Part A ................................................................................... 2
2. Circular Curves – Part B ................................................................................... 9
3. Vertical Curves – Part A.................................................................................. 18
4. Vertical Curves – Part B.................................................................................. 25
5. Spiral Curves................................................................................................... 32
6. Grade Separation............................................................................................ 41
7. Earthwork ........................................................................................................ 49
Solved Problems – Surveying
2
Demo Problem Solutions
Module 1 Part a - Circular Curves
Situation
A circular curve is required to make the transition between two tangent sections in
a highway alignment. Use the data and sketch below to solve the following six
requirements.
• Design Speed
= 60 mph
• Side Friction Factor
= 0.15
• Maximum Superelevation Rate = 0.08
Requirements
ai. Based on the design speed, what is the minimum curve radius permitted?
aii. What is the maximum degree of curvature?
aiii. Using a degree of curvature of 05°30’00”, what is the station of the P.C.?
aiv. Using a degree of curvature of 05°30’00”, what is the station of the P.T.?
av. If a theodolite is set up at the P.C. of the curve with a 0°00’00” backsight on
the P.I. what is the required deflection angle to set a stake at the station 63+00 on
the curve?
avi. What is the bearing of the forward tangent?
© 2003 SmartPros Ltd.
Solved Problems – Surveying
Requirement 1ai
Steps
1. V (speed)
2. e (Superelevation rate)
3. f (Side friction factor)
= 60 mph
= 0.08
= 0.15
4. Rmin (Minimum curve radius) =
=
V2
(AASHTO green book:
15 • (e + f )
A Policy on Geometric
Design of Highways and
Streets, 1994, p 151).
60 2
( ft.)
15 • (0.08 + 0.15 )
= 1,043.48 ft.
© 2003 SmartPros Ltd.
3
Solved Problems – Surveying
Requirement 1aii
Steps
1. Rmin (Minimum curve radius)
2. Dmax (Maximum degree of curvature)
= 1,043.48 (from 1AB)
5,729.578
=
R min
=
5,729.578
1,043.48
= 5.490836
= 05°29’27”
© 2003 SmartPros Ltd.
4
Solved Problems – Surveying
Requirement 1aiii
Steps
1. Dmax (Maximum degree of curvature)
2. # (Deflection angle)
3. Station of P.I. (Point of intersection)
4. Rmin (Minimum curve radius)
= 5.5°
= 24.75°
= 62+48.56
5,729.578
=
Dmax
5,729.578
5.5
= 1,041.74 ft.
=
5. T (Length of subtangent)
6. Station of P.C. (Point of curvature)
∆ 
= Rmin • tan  
2
= 1,041.74 • tan (12.375)
= 228.56 ft.
= Station of P.I.– T
= 62+48.56 – 2+28.56
= 60+20.00
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 1aiv
Steps
1. Dmax (Maximum degree of curvature)
2. # (Deflection angle)
3. Station of P.C. (Point of curvature)
4. L (Length of circular arc)
= 5.5°
= 24.75°
= 60+20.00 (from 1AC)
∆
=
• 100
Dmax
24.75
• 100
5.5
= 450 ft.
= Station of P.C + L
= 60+20.00 + 4+50.00
= 64+70.00
=
5. Station of P.T.
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 1av
Steps
1.
2.
3.
4.
Station at given point
Station of P.C.
# (Deflection angle between tangents)
arc (Segment of arc length)
5. L (Length of circular arc)
6. Deflection Angle
= 63+00.00
= 60+20.00 (from 1AC)
= 24.75°
= Station at given point
– Station of P.C.
= 63+00.00 – 60+20.00
= 2+80.00
= 450 ft. (from 1AD)
arc ∆
=
•
L
2
280 24.75
•
=
450
2
= 07°42’00”
© 2003 SmartPros Ltd.
7
Solved Problems – Surveying
8
Requirement 1avi
Figure 1avi
N
78°26’30”
24°45’00”
α=S?
E
Back Tangent
N 78°26’30” E
Forward Tangent
S
Steps
1. Angle of back tangent
2. # (Deflection angle
between tangents)
3. α (Bearing of forward tangent
in Figure 1AF)
= 78°26’30”
= 24°45’00”
= 180° – Angle of back tangent – #
= 180 – 78°26’30” – 24°45’00
= S 76°48’30” E
© 2003 SmartPros Ltd.
Solved Problems – Surveying
9
Module 1 Part b - Circular Curves
Situation
The present highway layout (Curve 1) is too close to a historic building. The
proposed alignment (Curve 2) will shift the forward tangent 50’ west and parallel to
the original tangent. Using the data and sketch provided, solve the following seven
requirements.
• Design Speed is 55 miles per hour
• The P.C. of both curves is at the same station.
• Curve 1 has a degree curvature of 03°30’00”
• The railroad track is 80 feet south of and parallel to the back tangent
Requirements
bi. What is the subtangent length for Curve 1?
bii. What is the station of the P.C.?
biii. What is the degree of curvature for Curve 2?
biv. What is the station of the P.T. for Curve 2?
bv. Using a degree of curvature of 04°15’00”, what is the horizontal sight distance
for Curve 2?
bvi. Using a 55 mile per hour design speed, what is the maximum horizontal sight
distance (rounded for design) for Curve 2?
Bvii. Using a degree of curvature of 04015’00”, what is the station at the
intersection of the railroad track and Curve 2?
© 2003 SmartPros Ltd.
Solved Problems – Surveying
Requirement 1bi
Steps
1. D (Degree of curvature)
2. R (Curve radius)
3. ∆ (Deflection angle between tangents)
4. T (Length of subtangent)
= 03°30’00”
= 3.5°
5,729.578
=
D
5,729.578
=
3.5
= 1,637.02 ft,
= 32°36’00”
= 32.6°
∆ 
= R • tan  
2
= 1,637.02 • tan (16.3)
= 478.70 ft.
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 1bii
Steps
1. Station of P.I. (Point of intersection)
2. T (Length of subtangent)
3. Station of P.C. (Point of curvature)
= 75+98.70
= 478.70 ft. (from 1BA)
= Station of P.I. – T
= 75+98.70 – 4+78.70
= 71+20
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
12
Requirement 1biii
Figure 1BC-1
c
Back Tangent
32.6°
32.6°
50ft.
Forward Tangent
Steps
1. ∆ (Deflection angle between tangents)
2. c
= 32.6°
50
=
(Figure 1BC-1)
sin (∆ )
=
3. T1 (Length of subtangent of Curve 1)
4. T2 (Length of subtangent of Curve 2)
5. R2 (Radius of Curve 2)
6. D2 (Degree of curvature for Curve 2)
50
sin (32.6)
= 92.80 ft.
= 478.70 ft. (from 1BA)
= T1 – c
= 478.70 – 92.80
= 385.90 ft.
T2
=
∆ 
tan 
2
385.90
=
tan(16.3 )
= 1,319.68 ft.
5,729.578
=
R2
=
5,729.578
1,319.68
= 4.34°
= 04°20’30”
© 2003 SmartPros Ltd.
Solved Problems – Surveying
Requirement 1biv
Steps
1. ∆ (Deflection angle between tangents)
2. D2 (Degree of curvature for Curve 2)
3. L2 (Length of Curve 2)
= 32.6°
= 4.34° (from 1BC)
∆
=
• 100
D2
32.6
• 100
4.341667
= 750.86 ft.
=
4. Station of P.C.2 (Point of curvature
for Curve 2)
5. Station of P.T.2 (Point of tangency
for Curve 2)
= 71+20.00 (From 1BB)
= Station of P.C.2 + L2
= 71+20.00 + 7+50.86
= 78+70.86
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 1bv
Steps
1. D2 (Degree of curvature for Curve 2) = 4.25°
2. R2 (Radius of Curve 2)
=
5,729.578
D2
5,729.578
4.25
= 1,348.14 ft.
=
3. Sight distance for Curve 2(S2)
• Assume S2 < L2
¾ M2 (distance from the centerline
of roadway to obstruction)
¾
S2 (Sight distance)
=25 ft.
=
8 • R 2 • M2 (if S2 < L2)
=
8 • 1,348.14 • 25
= 519.26 ft.
¾
∆ (Deflection angle
between tangents)
D2 (Degree of curvature)
¾
L2 (Length of Curve)
¾
= 32.6°
= 4.25°
∆
=
• 100
D2
32.6
• 100
4.25
= 767.05 ft.
519.26 < 767.06, so assumption and S formula correct.
S2
= 519.26 ft.
=
•
•
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 1bvi
Table 1BF-1
Stopping Sight Distance
Design
Assumed Speed
Speed (mph)
(mph)
Computed (ft)
Rounded for
Design (ft)
20
20 - 20
106.7 - 106.7
125 - 125
25
24 - 25
138.5 - 146.5
150 - 150
30
28 - 30
177.3 - 195.7
200 - 200
35
32 - 35
217.7 - 248.4
225 - 250
40
36 - 40
267.0 - 313.3
275 - 325
45
40 - 45
318.7 - 382.7
325 - 400
50
44 - 50
376.4 - 461.1
400 - 475
55
48 - 55
432.0 - 537.8
450 - 550
60
52 - 60
501.5 - 633.8
525 - 650
65
55 - 65
549.4 - 724.0
550 - 725
70
58 - 70
613.1 - 840.0
625 - 850
From A Policy on Geometric Design of Highways and Streets, Copyright 1994
by the American Association of State Highway and Transportation Officials,
Washington, D.C. Used by permission.
Steps
1. Maximum stopping sight distance
= 550 ft. (Table 1BF-1)
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 1bvii
Figure 1BG-1
P.C.
Back Tangent
80 ft.
Curve 2
A
Railroad
Track
α
Radius
Center of
Circle
Steps
1. R2 (Radius of Curve 2)
2. A (Length from center to railroad track)
3. α (Angle at center of Curve 2)
= 1,348.14 ft.
= R2 – 80
= 1,348.14 – 80
= 1,268.14
 A 
 (Figure 1BG-1)
= arccos
 R2 
 1,268.14 
= arccos

 1,348.14 
4. D2 (Degree of curvature for Curve 2)
5. L2 (Length of Curve 2
from P.C. to Railroad track)
= 19.83748°
= 19°50’15”
= 4.25°
=
α
• 100
D2
19.83748
• 100
4.25
= 466.76 ft.
=
6. Station of P.C.2 (Point of curvature
for Curve 2)
7. Station where Curve 2
= 71+20.00 (From 1BB)
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
intersects Railroad track
= Station of P.C.2 + L2
= 71+20.00 + 4+66.76
= 75+86.76
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
18
Module 2 Part a - Vertical Curves
Situation
A parabolic curve is used to make the vertical transition between the two tangent
sections in a highway alignment for an equal tangent crest curve. Using the design
data and the information in the sketch below, solve the following requirements.
• Design Speed
= 60 m.p.h.
• Elevation on top of 36” pipe = 137.16 ft.
Requirements
ai. Using the design speed and stopping sight distance (rounded for design), what
is the minimum length of the vertical curve required to make the grade transition?
aii. What is the elevation on the vertical curve at station 37+25.50?
aiii. What is the elevation of the high point on the curve?
aiv. If six feet of cover is required over the pipe at station 34+00, what is the
maximum length of vertical curve that can be used?
av. What is the actual passing sight distance for the vertical curve?
© 2003 SmartPros Ltd.
Solved Problems – Surveying
Requirement 2ai
Table 2ai
Stopping Sight Distance over a Crest Vertical Curve
Design Speed
(mph)
Assumed
Speed for
Condition
(mph)
Stopping
Sight
Distance
Rounded for
Design (ft.)
Rate of Vertical Curvature, K
(length [ft] per percent of A)
Computed
Rounded for
Design
20
20 - 20
125 - 125
8.6 - 8.6
10 - 10
25
24 - 25
150 - 150
14.4 - 16.1
20 - 20
30
28 - 30
200 - 200
23.7 - 28.8
30 - 30
35
32 - 35
225 - 250
35.7 - 46.4
40 - 50
40
36 - 40
275 - 325
53.6 - 73.9
60 - 80
45
40 - 45
325 - 400
76.4 - 110.2
80 - 120
50
44 - 50
400 - 475
106.6 - 160.0
110 - 160
55
48 - 55
450 - 550
140.4 - 217.6
150 - 220
60
52 - 60
525 - 650
189.2 - 302.2
190 - 310
65
55 - 65
550 - 725
227.1 - 394.3
230 - 400
70
58 - 70
625 - 850
282.8 - 530.9
290 - 540
From A Policy on Geometric Design of Highways and Streets, Copyright 1994
by the American Association of State Highway and Transportation Officials,
Washington, D.C. Used by permission.
Steps
1. K (Rate of vertical curvature)
Table 2AA-1)
2. g1 (Gradient 1)
3. g2 (Gradient 2)
4. A (Total change in grade of the curve)
5. L (Length of curve)
= 190 (minimum value in
= +3.5%
= -2.5%
= g1 – g2
= 3.5% – (–2.5%)
= 6.0%
=K•A
= 190 • 6
= 1,140 ft.
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 2aii
Steps
1. L
(Length of vertical curve) = 1,200 ft.
= 12+00.00
2. Station at PVI (Point of
vertical intersection)
= 38+50.00
3. Station of PVC
(Point vertical curvature)
4. ElevPVI (Elevation PVI)
5. g1 (Gradient 1)
6. ElevPVC (Elevation PVC)
L
2
= 38+50.00 – 6+00.00
= 32+50.00
= 160.25
= +3.5%
= Station of PVI –
L

= Elev PVI −  g1 • 
2

= 160.25 − (3.5 • 6 )
7. P
8. Xp (Horizontal distance
from PVC to P)
9. g2 (Gradient 2)
10. R
= 139.25
= Point at 37+25.50
= 37+25.50 – 32+50.00
= 4+75.50
= –2.5%
g2 − g1
L
- 2.5 − 3.5
=
12
= –0.5
(Rate of change of grade) =
11. Yp (Elevation of P)
= YPVC
 R • XP 2 

+ (g1 • XP ) + 

2


 - 0.5 • 4.7550 2 

= 139.25 + (3.5 • 4.7550 ) + 
2


= 150.24
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 2aiii
Steps
1. P
2. L (Length of vertical curve)
3. g1 (Gradient 1)
4. g2 (Gradient 2)
5. Xp (Horizontal distance
from PVC to P)
= Highest point on curve.
= 1,200 ft.
= 12+00.00
= +3.5%
= –2.5%
=
g1 • L
g1 − g2
=
3.5 • 12
3.5 − ( −2.5)
= 7.0 stations
6. Station of PVC
(Point vertical curvature)
7. Station of P
8. ElevPVC (Elevation PVC)
9. R (Rate of change of grade)
10. Yp (Elevation of P)
= 32+50.00 (from 2AB)
= Station of PVC + Xp
= 32+50.00 + 7+00.00
= 39+50+00
= 139.25 (from 2AA)
= –0.5 (from 2AA)
 R • X2 

= YPVC + (g1 • X ) + 
 2 
 - 0.5 • 7.0 2 

= 139.25 + (3.5 • 7.0 ) + 
2


= 151.50
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 2aiv
Steps
1. ElevTopPipe
(Elevation at top pipe)
2. D (Depth of cover)
3. Elevcurve
(Elevation of curve)
4. L
5. P
6. Station at P
7. Station at PVI
(Point vertical intersection)
8. ElevPVI
(Elevation of PVI)
9. XP
10. g1 (Gradient 1)
11. g2 (Gradient 2)
12. R (Rate of change
of grade)
= 137.16 ft.
= 6 ft.
= ElevTopPipe + D
= 137.16 + 6
= 143.16 ft.
= Length of new vertical curve.
= Point at top of new curve.
= 34+00.00
= 38+50.00
= 160.25 ft.
= Horizontal distance from PVC to P.
L
– (Station at PVI – Station at P)
=
2
L
=
– (38+50.00 – 34+00.00)
2
L
=
– (4.5 stations)
2
= +3.5%
= –2.5%
g2 − g1
L
- 2.5 − 3.5
=
L
−6
=
L
=
 R • XP 2 

= YPVC + (g1 • XP ) + 

2


New PVC is unknown, so express in terms of PVI, g1, and L
13. Yp (Elevation of P)
 R • XP 2 

L


=  Elev PVI − g1 •   + (g1 • XP ) + 

2 
2




© 2003 SmartPros Ltd.
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Solved Problems – Surveying
XP unknown, so express in terms of L

=  Elev PVI


= 160.25

2


 R •  L − 4.5 
2
 
L 

L
 


− g1 •   +  g1 •  − 4.5  + 

2
2
2

 

 






L 

L
− 3.5 •   +  3.5 •  − 4.5
2 

2

14. Simplifying equation for YP
L2 – (19.79 • L) + 81
=0
L
(b
)
6
2
L
 
•  − 4.5 
2
 

L •2



−4•a•c
2•a
= 14 stations or 5.8 stations
= 1,400 ft. (max length as requested)
Solving for L with X =
−b±

 
  + 
 


2
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
24
Requirement 2av
Steps
1. Sight distance for Curve (S)
• Assume S < L
¾ L (Curve length)
¾ A (Total change
in grade of the curve)
¾
S
= 1,200 ft.
= 6.0% (from 2AA)
= 55.61•
L
(if S < L; AASHTO Green
A
Book)
1,200
6
= 786.44 ft.
796.44 < 1,200, so formula is correct.
S
= 786 ft. (rounded down)
= 55.61•
•
•
© 2003 SmartPros Ltd.
Solved Problems – Surveying
25
Module 2 Part b - Vertical Curves
Situation
A parabolic vertical sag curve is used to make the transition between two equal
tangent sections in a highway alignment. Using the design data and the
information on the sketch below, solve the following requirements:
• Design Speed
= 50 m.p.h.
• Elevation on underside of bridge structure = 147.36 ft.
• Bridge located at station 74+65.50
Requirements
B. At what station on the vertical would a pair of catch basins or drainage inlet
structures be required?
C. What is the curve elevation at the lowest point on the vertical curve?
D. If the river flowing under the highway floods to elevation 133.0, what is the
minimum station to which the flooding would extend?
E. If the river flowing under the highway floods to elevation 133.0, what is the
maximum station to which the flooding would extend?
F. What is the bridge clearance at station 74+65.50?
G. Using the design speed and stopping sight distance (rounded for design), what
is the maximum length of the vertical curve required to make the grade
transition?
© 2003 SmartPros Ltd.
Solved Problems – Surveying
Requirement 2bi
Steps
1. Station of PVI
(Point of vertical intersection)
2. L (Length of curve)
3. Station of PVC
(Point of vertical curvature)
4.
5.
6.
7.
g1 (Gradient 1)
g2 (Gradient 2)
P
XP (Horizontal distance
from PVC to P)
= 76+35.00
= 8 stations.
L
= Station of PVI −  
2
= 76+35.00 – 4
= 72+35.00
= –2.2%
= +1.8%
= Point at lowest point of curve.
=
g1 • L
g1 − g2
− 2.2 • 8
- 2.2 − 1.8
= 4.4 stations
= Station of PVC + XP
= 72+35.00 + 4+40.00
= 76+75.00
=
8. Station at P
© 2003 SmartPros Ltd.
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Solved Problems – Surveying
Requirement 2bii
Steps
1. g1 (Gradient 1)
2. g2 (Gradient 2)
3. L (Length of curve)
4. R (Rate of change in grade)
= –2.2%
= +1.8%
= 8 stations.
g − g1
= 2
L
1.8 − (− 2.2)
=
8
= 0.5
5. ElevPVI (Elevation of Point
of vertical intersection)
6. ElevPVC (Elevation of Point
= 126.80 ft.
of vertical curvature)
7. P
8. XP (Horizontal distance
from PVC to P)
9. YP (Vertical distance
from PVC to P)

L  
= Elev PVI −  g1 •   
2 

= 126.80 – (–2.2 • 4)
= 135.60 ft.
= Point at lowest point of curve.
= 4.4 stations
 R • X 2P 

= Elev PVC + (g1 • XP ) + 
 2 
 0.5 • 4.4 2 

= 135.60 + (- 2.2 • 4.4 ) + 
2


= 130.76 ft.
© 2003 SmartPros Ltd.
27
Solved Problems – Surveying
Requirement 2biii
Steps
1. P
2. g1 (Gradient 1)
3. g2 (Gradient 2)
4. R (Rate of change in grade)
5. ElevPVC (Elevation of Point
of vertical curvature)
6. XP
7. YP (Vertical distance PVC to P)
= Point where elevation is 133 ft.
= –2.2%
= +1.8%
= 0.5 (from 2BB)
= 135.60 ft (from 2BB)
= Horizontal distance PVC to P
 R • XP 2 

= Elev PVC + (g1 • XP ) + 

2


 0.5 • XP 2 

= 135.60 + (- 2.2 • XP ) + 

2


133.0
8. Simplifying equation for XP
XP2 – (8.8 • X) + 10.4 = 0
Solving for XP with X =
XP
9. Station of PVC
(Point of vertical curvature)
10. Minimum station of flooding
−b±
(b
)
−4•a•c
2•a
= 1.4067 stations & 7.3933 stations
2
= 72+35.00 (from 2BA)
= Station of PVC + Minimum XP
= 72+35.00 + 1+40.67
= 73+75.67
© 2003 SmartPros Ltd.
28
Solved Problems – Surveying
Requirement 2biv
Steps
1. Station of PVC
(Point of vertical curvature)
2. XP (Maximum)
3. Maximum station of flooding
= 72+35.00 (from 2BA)
= 7.3933 stations (from 2BC)
= Station of PVC + Maximum XP
= 72+35.00 + 7+39.33
= 79+74.33
© 2003 SmartPros Ltd.
29
Solved Problems – Surveying
Requirement 2bv
Steps
1. P
2. Station of PVC
(Point of vertical curvature)
3. ElevPVC
4. g1 (Gradient 1)
5. R (Rate of change in grade)
6. XP
= Point on curve at Station 74+65.50.
= 72+35.00 (from 2BA)
= 135.60 ft (from 2BB)
= –2.2%
= 0.5 (from 2BB)
= Horizontal distance from PVC to P
= Station at P – Station at PVC
= 74+65.50 – 72+35.00
= 2+30.50
= 2.305 stations
7. YP
(Vertical distance PVC to P)
= Elev PVC
 R • XP 2 

+ (g1 • XP ) + 

2


 0.5 • 2.305 2 

= 135.60 + (- 2.2 • 2.305 ) + 
2


= 131.86 ft
8. ElevBridge
(Elevation bridge structure)
9. Bridge clearance
= 147.36 ft.
= ElevBridge – YP
= 147.36 – 131.86
= 15.50 ft.
© 2003 SmartPros Ltd.
30
Solved Problems – Surveying
Requirement 2bvi
Table 2bvi
Stopping Sight Distance on a Sag Vertical Curve
Design Speed
(mph)
Stopping
Sight
Distance
Rounded for
Design (ft.)
Assumed
Speed for
Condition
(mph)
Rate of Vertical Curvature, K
(length [ft] per percent of A)
Computed
Rounded for
Design
20
20 - 20
125 - 125
14.7 - 14.7
20 - 20
25
24 - 25
150 - 150
21.7 - 23.5
30 - 30
30
28 - 30
200 - 200
30.8 - 35.3
40 - 40
35
32 - 35
225 - 250
40.8 - 48.6
50 - 50
40
36 - 40
275 - 325
53.4 - 65.6
60 - 70
45
40 - 45
325 - 400
67.0 - 84.2
70 - 90
50
44 - 50
400 - 475
82.5 - 105.6
90 - 110
55
48 - 55
450 - 550
97.6 - 126.7
100 - 130
60
52 - 60
525 - 650
116.7 - 153.4
120 – 160
65
55 - 65
550 - 725
227.1 - 394.3
130 - 180
70
58 - 70
625 - 850
282.8 - 530.9
150 - 220
From A Policy on Geometric Design of Highways and Streets, Copyright 1994
by the American Association of State Highway and Transportation Officials,
Washington, D.C. Used by permission.
Steps
1.
2.
3.
4.
K (Rate of vertical curvature)
g1 (Gradient 1)
g2 (Gradient 2)
A (Total change
in grade of curve)
5. L (Length of curve)
= 110 (maximum value in Table 2BF-1)
= –2.2%
= +1.8%
= g2 – g1
= 1.8% – (–2.2%)
= 4.0%
=K•A
= 110 • 4
= 440 ft.
© 2003 SmartPros Ltd.
31
Solved Problems – Surveying
32
Module 3 - Spiral Curves
Situation
Two tangent sections in a horizontal highway alignment study intersect at an angle
of 40°20’00”. The change in direction will be accomplished using a circular curve
with a transition spiral at each end. Using the data and sketch below, solve the
following requirements:
• Design Speed
= 60 m.p.h.
• Station of the PI
= 46+12.32
• Degree of curvature
= 6°00’
Requirements
a. Based on the design speed, what is the minimum length of the spiral required?
b. Using a 400 foot long spiral curve, what is the value of the spiral angle (θs)?
c. What is the total tangent distance (from PI to TS)?
d. What is the station of the SC?
e. What is the length of the circular arc between the SC and CS?
f. What is the station of the ST?
g. What is the value of the total external distance (ES)?
h. A theodolite is set up at the TS with a 00°00’00” backsight on the PI. What is
the deflection angle to set a stake at station 43+00.00?
© 2003 SmartPros Ltd.
Solved Problems – Surveying
Requirement 3a
Steps
1. D (Degree of curvature)
2. V (Design speed)
3. C (Rate of increase of
centripetal acceleration)
= 6°
= 60 mph
4. RC (Curve radius)
= 2 (for comfort and safety)
5,729.578
=
D
= 954.93 ft.
5. L (Minimum length of spiral)
3.15 • V 3
=
(AASHTO Green book)
RC • C
3.15 • 60 3
954.93 • 2
= 356 ft (rounded down)
=
© 2003 SmartPros Ltd.
33
Solved Problems – Surveying
Requirement 3b
Steps
1. L S (Length from TS to SC)
2. DC (Degree of curvature)
3. θS (Spiral angle)
= 400 ft.
= 6°
L • DC
= S
200
400 • 6
=
200
= 12°
© 2003 SmartPros Ltd.
34
Solved Problems – Surveying
Requirement 3c
Steps
1. θS (Spiral angle)
2. L (Length of spiral)
3. p (Offset distance from
initial tangent to Point Curvature)
4. k (Abscissa of shifted PC referred
to Tangent to spiral point [TS])
5. RC (Curve radius)
6. ∆ (Angle between tangents)
7. DC (Degree of curvature)
8. TS (Total tangent distance)
= 12° (from 3A)
= 400 ft. (from 3B)
= 0.017426 • L (From Hickerson,
Route location
and design,
5th Edition,
McGraw-Hill.)
= 0.017426 • 400
= 6.97 ft.
= 0.499270 • L (From Hickerson)
= 0.499270 • 400
= 199.71
= 954.93 ft. (from 3A)
= 40°20’00”
= 6°
∆ 
= (R C + p ) • tan  + k
2
 40 o 20'00" 
 + 199.71
= (954.93 + 6.97 ) • tan
2


= 552.98 ft
© 2003 SmartPros Ltd.
35
Solved Problems – Surveying
Requirement 3d
Steps
1. Station of PI (Point of intersection)
2. TS (Total tangent distance)
3. Station of TS (Tangent to spiral point)
4. L S (Length of spiral)
5. Station of SC (Spiral to curve)
= 46+12.32
= 552.98 ft. (from 3C)
= Station of PI – TS
= 46+12.32 – 5+52.98
= 40+59.34
= 400 ft. (from 3B)
= 4+00.00
= Station of TS + L S
= 40+59.34 + 4+00.00
= 44+59.34
© 2003 SmartPros Ltd.
36
Solved Problems – Surveying
Requirement 3e
Steps
1.
2.
3.
4.
∆ (Angle between tangents)
θS (Spiral angle)
DC (Degree of curvature)
∆C (Central angle of circular arc)
5. L C (Length of circular arc)
= 40°20’00”
= 12° (from 3A)
= 6°
= ∆ – (2 • θS)
= 40°20’00” – 24°00’00”
= 16°20’00”
∆ • 100
= C
DC
16.33 • 100
6
= 272.22 ft.
=
© 2003 SmartPros Ltd.
37
Solved Problems – Surveying
Requirement 3f
Steps
1.
2.
3.
4.
Station of SC (Spiral to curve)
L C (Length of circular arc)
L S (Length of spiral)
Station of CS (Curve to spiral)
5. Station of ST (Spiral to tangent)
= 44+59.34 (from 3D)
= 2+72.22 (from 3E)
= 4+00.00 (from 3B)
= Station of SC + L C
= 44+59.34 + 2+72.22
= 47+31.56
= Station of CS + L S
= 47+31.56 + 4+00.00
= 51+31.56
© 2003 SmartPros Ltd.
38
Solved Problems – Surveying
Requirement 3g
Steps
1. RC (Curve radius)
2. p (Offset distance from
initial tangent to Point Curvature)
3. ∆ (Angle between tangents)
4. ES (Total external distance)
= 954.93 ft. (from 3A)
= 6.97 ft. (from 3C)
= 40°20’00”

∆  
= (R C + p ) •  sec   − 1 + p
2 

= (954.93 + 6.97 ) • (sec[20.167] − 1) + 6.97
= 69.79 ft.
© 2003 SmartPros Ltd.
39
Solved Problems – Surveying
Requirement 3h
Steps
1.
2.
3.
4.
5.
Station of Point P
Station of TS (Tangent to spiral point)
L S (Length of spiral)
θS (Spiral angle)
L (Length along spiral curve)
= 43+00.00
= 40+59.34 (from 3D)
= 4+00.00 (from 3B)
= 12° (from 3A)
= Station of P – Station of TS
= 43+00.00 – 40+59.34
= 2+40.66
2
6. θ (Deflection angle)
1  L 
= •   • θ S
3  LS 
2
1  240.66 
= •
 • 12
3  400 
= 1°26’52”
© 2003 SmartPros Ltd.
40
Solved Problems – Surveying
41
Module 4 - Grade Separation
Situation
As part of a design process for a new by-pass highway, a grade separation
structure (bridge) is required to carry traffic on an existing road (Smith Lane) over
the by-pass. Based on the design data and sketch below, solve the following
requirements:
• By-pass Grade
= –4.5% from West to East
• Smith Lane Grade
= Crest Vertical Curve
• PVI Station
= 63+50.00
• PVI Elevation
= 126.50
• Curve Length
= 600 ft.
= +3.2%
• g1
• g2
= –3.2%
• Intersection Equation: Bypass
= Station 107+35; Smith Lane = 62+10
• Minimum Bridge Clearance
= 16 ft.
• By-pass Transverse or Cross Slope = 1/4” per foot
• Smith Lane Cross Slope
= 3/8” per foot
© 2003 SmartPros Ltd.
Solved Problems – Surveying
42
Requirements
a. What is the centerline station on Smith Lane opposite Point X?
b. What is the centerline elevation on the vertical curve on Smith Lane at Station
61+70.90?
c. What is the elevation at X on Smith Lane bridge?
d. Using the minimum bridge clearance permitted, what is the maximum elevation
at X on the by-pass?
e. What is the maximum elevation on the centerline of the by-pass at station
106+00.00?
© 2003 SmartPros Ltd.
Solved Problems – Surveying
Requirement 4a
Figure 4a
Steps
1. θ (Angle between roads)
2. A (Angle DXE)
3. WB (Width by-pass lane)
4. WS (Width Smith Lane)
5. Station at C
6. CD
= 67°30’00”
= 180° – 90°– 67°30’00”
= 22°30’00”
= 30 ft.
= 16 ft.
= 62+10’00”
WB
=
cos(A )
=
7. DE
8. Center of Smith opposite X
30
cos(22.5 )
= 32.47 ft
= WS • tan(A)
= 16 • tan(22.5)
= 6.63 ft.
= Station at C – CD – DE
= 62+10’00” – 0+32.47 – 0+06.63
= 61+70.90
© 2003 SmartPros Ltd.
43
Solved Problems – Surveying
Requirement 4b
Steps
1. ElevPVI (Elevation of Point of
vertical intersection)
2. Station of PVI
3. g1 (Grade 1)
4. g2 (Grade 2)
5. L (Curve length)
6. R (Rate of change of grade)
7. P
8. ElevPVC (Elevation of Point of
vertical curvature)
= 126.50 ft.
= 63+50.00
= +3.2%
= –3.2%
= 600 ft.
g − g1
= 2
L
- 3.2 − 3.2
=
6
= –1.0667
= Point at Station 61+70.90

L  
= Elev PVI −  g1 •   
2 


6
= 126.50 −  3.2 •   
2

= 116.90 ft.
9. Station of PVC
L
= Station of PVI −  
2
6
= 63+50.00 –  
2
= 60+50.00
10. XP (Horizontal distance
from PVC to P)
11. YP (Elevation at P)
= Station of P – Station of PVC
= 61+70.90 – 60+50.00
= 1+20.90
 R • X 2P 

= ElevPVC + (g1 • R) + 
2


= 116.90 + (3.2 • –1.0667)
 - 1.0667 • 1.2090 2 

+ 
2


= 119.99 ft.
© 2003 SmartPros Ltd.
44
Solved Problems – Surveying
Requirement 4c
Steps
1.
2.
3.
4.
P
YP (Elevation at P)
θ (Cross slope)
W (Pavement width)
5. ElevX (Elevation of X)
= Point X on Smith Lane Bridge
= 119.99 ft. (from 4B)
= 3/8” per foot.
= 32 ft.
W 

= YP − θ •

2 • 12 

 3 32 
= 119.99 −  •

 8 24 
= 119.49 ft.
© 2003 SmartPros Ltd.
45
Solved Problems – Surveying
Requirement 4d
Steps
1.
2.
3.
4.
5.
P
ElevX (Elevation of X)
CMin (Minimum Clearance)
DSlab (Depth of slab)
DBeams (Depth of beams)
= Point X on Smith Lane Bridge
= 119.49 ft. (from 4C)
= 16 ft.
= 12 inches.
= 36 inches.
D
+ DBeams
6. D (Depth of bridge structure)
= Slab
12
12 + 36
=
12
= 4 ft.
7. ElevMax (Maximum elevation of X) = ElevX – D – CMin
= 119.49 – 4 – 16
= 99.49 ft
© 2003 SmartPros Ltd.
46
Solved Problems – Surveying
Requirement 4e
Figure 4E-1
Steps
1.
2.
3.
4.
5.
6.
7.
8.
P
A (Intersection angle)
WS (Width Smith Lane)
WB (Width by-pass lane)
gBP (Grade by-pass lane)
Station at C
Station at IP (Intersection point)
ElevEP
(Elevation at edge pavement)
9. θ (Cross slope)
10. CF
= Point at Station 106+00.00
= 22°30’00”
= 16 ft.
= 30 ft.
= 4.5%
= 62+10’00”
= 107+35.00
= 99.49 ft. (from 4D)
= 1/4” per foot.
WS
=
cos(A )
=
11. FG
12. Station at G
13. ElevG (Elevation of G)
16
cos(22.5 )
= 17.32 ft
= WB • tan(A)
= 30 • tan(22.5)
= 12.43 ft.
= Station at IP – CF – FG
= 107+35.00” – 0+17.32 – 0+12.43
= 107+05.25
θ • WB
= ElevEP +
12
1 • 30
= 99.49 +
4 • 12
© 2003 SmartPros Ltd.
47
Solved Problems – Surveying
14. XP (Horizontal distance G to P)
15. YP (Elevation at P)
= 100.12 ft.
= Station at G – Station at P
= 107+05.25 – 106+00.00
= 1+05.25
= ElevG + gBP • XP
= 100.12 + (4.5 • 1.0525)
= 104.86 ft.
© 2003 SmartPros Ltd.
48
Solved Problems – Surveying
49
Module 5 - Earthwork
Situation
The centerline of a proposed highway has been established in the field. Cross
sections showing pre-construction conditions have been taken at every half station
(50 feet), and other selected stations where there is any significant change in
ground conditions. A template showing the final design of the highway has been
plotted on each of the cross sections. Based on the design information plotted on
each of the cross sections and using a planimeter, the areas of cut and fill have
been determined. Shown below is a listing of these areas for a 400-foot section on
the highway. Using the given design data, solve the following requirements.
Station
10+00
10+40
10+50
11+00
11+50
11+80
12+00
12+50
13+00
13+30
13+50
14+00
Earthwork Data
Cut Area (ft2) Fill Area (ft2)
120
490
106
620
85
1,180
163
1,375
640
920
1,002
240
1,365
110
1,260
90
1,139
120
710
180
505
360
240
369
Requirements
a. If the shrinkage of the soil is 12%, how much of a surplus exists, if any, when
using the average end area method to calculate the amount of material needed
to construct the highway subgrade.
b. What is the station of the balance point on the mass diagram? (Assume the
mass diagram ordinate at station 10+0 = 0).
c. For this requirement, assume a freehaul of 1,000 feet, the center of mass of
the excavated material is at station 22+85, and the center of mass of the
embankment is at station 41+25. Based on these assumptions and the
portion of a mass diagram shown with the requirement, what is the overhaul
in cubic yard stations?
© 2003 SmartPros Ltd.
Solved Problems – Surveying
50
Requirement 5a
Table 5a
Station
10+00
10+40
10+50
11+00
11+50
11+80
12+00
12+50
13+00
13+30
13+50
14+00
Total
Cut Area
(ft2)
120
106
85
163
640
1,002
1,365
1,260
1,139
710
505
240
Cut Volume
(yds3)
167
35
230
744
912
877
2,430
2,221
1,027
450
690
9,783
Fill Area
(ft2)
490
620
1,180
1,375
920
240
110
90
120
180
360
369
• Ai
• L
= Cut Area at Station i
= Length between stations
= 50 ft.
(A i + A i-1 ) • L (divide by 27 to convert from ft. to yds3)
• Cut Volumei =
2 • 27
• Bi
= Fill Area at Station i
(B + Bi-1 ) • L • 112 (add 12% for shrinkage)
• Fill Volumei = i
2 • 27 • 100
Steps
1. Surplus = Total Fill Volume – Total Cut Volume (Table 5A-1)
= 9,783 – 8,783
= 1,000 yds3
© 2003 SmartPros Ltd.
Fill Volume
(yds3)
921
373
2,650
2,380
722
145
207
218
187
224
756
8,783
Solved Problems – Surveying
Requirement 5b
Table 5B-1
Station
10+00
10+40
10+50
11+00
11+50
11+80
12+00
12+50
13+00
13+30
13+50
14+00
Cut Volume
(yds3)
167
35
230
744
912
877
2,430
2,221
1,027
450
690
Fill Volume
(yds3)
921
373
2,650
2,380
722
145
207
218
187
224
756
Mass Diagram
Ordinate (yds3)
0
-754
-1,092
-3,512
-5,148
-4,958
-4,226
-2,003
0
+840
+1,066
Steps
1. Mass Diagram Ordinatej =
2. Balance Point
Cut Volume i − Fill Volume i ) yds3
(
i=0
j
∑
= Station where Mass diagram ordinate
is zero.
= 13+00.00
© 2003 SmartPros Ltd.
51
Solved Problems – Surveying
Requirement 5c
Figure 5C-1
Steps
1. CMEmbankment
(Center of mass for embankment)
2. CMExcavation
(Center of mass for excavation)
3. LFreehaul (Length of freehaul)
4. LOverhaul (Length of overhaul)
5. HLOrdinate (Horizontal-line ordinate
in Figure 5C-1)
6. VOverhaul (Volume of overhaul)
= 41+25.00
= 22+85.00
= 10 stations
= CMEmbankment – CMExcavation – LFreehaul
= 41+25.00 – 22+85.00 – 10+00.00
= 8+40.00
= 2,260 yds3
= HLOrdinate • LOverhaul
= 2,260 • 8.4
= 18,984 cubic yard stations
© 2003 SmartPros Ltd.
52