1. Find: αNAB N STAB=7+82 B STAA=5+47 3.9 5.1 6.4 7.8 D=3 20’ 00” A O o o
Find the deflection angle N A B. [pause] In this problem, points A and B ---- A O

2. Find: αNAB N STAB=7+82 B STAA=5+47 3.9 5.1 6.4 7.8 D=3 20’ 00” A O o o
are positioned at the beginning, and end of a horizontal curve. The stationing ---- A O

3. Find: αNAB N STAB=7+82 B STAA=5+47 3.9 5.1 6.4 7.8 D=3 20’ 00” A O o o
for these two points is given, as well as the degree of curvature --- A O

4. Find: αNAB N STAB=7+82 B STAA=5+47 3.9 5.1 6.4 7.8 D=3 20’ 00” A O o o
for the horizontal curve. [pause] The problem asks to find the deflection angle ---- A O

5. Find: αNAB N STAB=7+82 B STAA=5+47 3.9 5.1 6.4 7.8 D=3 20’ 00” A O o o
N A B, which is the angle a station set on A would have to turn if facing N, to face Point B. If we define the --- A O

6. Find: αNAB N STAB=7+82 B STAA=5+47 3.9 5.1 6.4 7.8 D=3 20’ 00” I A O o
interior angle of the curve, as Angle I, then we know our deflection angle equals ---- A O

7. Find: αNAB αNAB= αNAB N I 2 B I A O STAB=7+82 STAA=5+47 D=3 20’ 00” o
One half of angle I. [pause] Angle I equals --- A O
STAB=7+82
STAA=5+47 o
D=3 20’ 00”

8. Find: αNAB αNAB= αNAB N I LAB I= 2 R B I A O STAB=7+82 STAA=5+47
the length of the curve, from point A to point B, --- A O
STAB=7+82
STAA=5+47 o
D=3 20’ 00”

9. Find: αNAB αNAB= αNAB curve length N I LAB I= 2 R radius B I A O
divided by the radius of the curve, R. [pause] The length of the curve is computed by ---- A O
STAB=7+82
STAA=5+47 o
D=3 20’ 00”

10. Find: αNAB αNAB= αNAB curve length N I LAB I= 2 R radius B
LAB=STAB-STAA I
subtracting the stationing of Point B from the stationing of Point A. A O
STAB=7+82
STAA=5+47 o
D=3 20’ 00”

11. Find: αNAB αNAB= αNAB curve length N I LAB I= 2 R radius B
LAB=STAB-STAA I
782 feet minus 547 feet equals A O
STAB=7+82
STAA=5+47 o
D=3 20’ 00”

12. Find: αNAB αNAB= αNAB curve length N I LAB I= 2 R radius B
LAB=STAB-STAA
LAB=235 [ft] I
feet. [pause] The radius of the curve --- A O
STAB=7+82
STAA=5+47 o
D=3 20’ 00”

13. Find: αNAB αNAB= αNAB curve length N I LAB I= 2 R radius B
LAB=STAB-STAA
LAB=235 [ft] I
can be determined using the the degree of curvature value, ---- A O
STAB=7+82
STAA=5+47 o
D=3 20’ 00”

14. Find: αNAB αNAB= αNAB curve length N I LAB I= 2 R radius B
LAB=STAB-STAA
LAB=235 [ft] I
D. The radius of the curve, in feet, equals ---- A O
STAB=7+82
STAA=5+47 o
D=3 20’ 00”

15. Find: αNAB αNAB= αNAB LAB=235 [ft] N I LAB I= 2 R radius Bo R= D
in DD
form I
5,729.6, degrees feet, divided by the degree of curvature, in decimal degree form. Converting the degree of curvature ---- A O
STAB=7+82
STAA=5+47 o
D=3 20’ 00”

16. Find: αNAB αNAB= αNAB LAB=235 [ft] N I LAB I= 2 R radius Bo R= D
in DD
form I
from degrees minutes seconds form to decimal degrees from involves adding the degrees --- A O o
D=3 20’ 00”
DMS form

17. Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius Bo R= D
in DD
form I
to, the minutes divided by 60, and, to the seconds divided by 60 squared. This makes the degree of curvature ---- A O o
D=3 20’ 00”
DMS form 20 00
D=3+ + 60
602

18. Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius Bo R= D
in DD
form I
equal to degrees. [pause] Plugging in this value --- A O o
D=3 20’ 00”
DMS form o
D=3.333 20 00
D=3+ +
DD form 60
602

19. Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius Bo R= D
in DD
form I
for the degree of curvature, the radius of the curve computes to --- A O o
D=3 20’ 00”
DMS form o
D=3.333 20 00
D=3+ +
DD form 60
602

20. Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius Bo R= D
in DD
form I
1,719 feet. [pause] Plugging this value in ---- A O o
D=3 20’ 00”
DMS form o
D=3.333 20 00
D=3+ +
DD form 60
602

21. Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius Bo R= D
in DD
form I
for the radius, the internal angle, I, equals --- A O o
D=3 20’ 00”
DMS form o
D=3.333 20 00
D=3+ +
DD form 60
602

22. Find: αNAB αNAB= αNAB + 0.1367 [rad] LAB=235 [ft] N I LAB I= 2 R
radius B
1,719 [ft]
αNAB
5,729.6 [ *ft] o R= D
in DD
form I
radians. [pause] Plugging this value in --- A O o
D=3 20’ 00”
DMS form o
D=3.333 20 00
D=3+ +
DD form 60
602

23. Find: αNAB αNAB= αNAB + 0.1367 [rad] LAB=235 [ft] N I LAB I= 2 R Bo R= D
in DD
form I
for the internal angle, the deflection angle N A B equals, --- A O o
D=3 20’ 00”
DMS form o
D=3.333 20 00
D=3+ +
DD form 60
602

24. Find: αNAB αNAB= αNAB + 0.1367 [rad] LAB=235 [ft] N I LAB I= 2 R Bo R= D
in DD
form I
radians. [pause] To convert radians to decimal degrees, --- A O o
D=3 20’ 00”
DMS form o
D=3.333 20 00
D=3+ +
DD form 60
602

25. Find: αNAB αNAB= αNAB π + 0.1367 [rad] LAB=235 [ft] N I LAB I= 2 R Bo R=
180 D * π
in DD
form I
we’ll multiply by 180 degrees per radian, and deflection angle N A B equals, --- A O o
D=3 20’ 00”
DMS form o
D=3.333 20 00
D=3+ +
DD form 60
602

26. Find: αNAB αNAB= αNAB π αNAB=3.917 + 0.1367 [rad] LAB=235 [ft] N I LABo R=
180 D * π
in DD
αNAB=3.917 o
form
3.917 degrees. [pause] A O o
D=3 20’ 00”
DMS form o
D=3.333 20 00
D=3+ +
DD form 60
602

27. Find: αNAB αNAB= αNAB π αNAB=3.917 + 0.1367 [rad] LAB=235 [ft] N I LABo R=
180 D * π
αNAB=3.917 o o
3.9
5.1
6.4
7.8
When reviewing the possible solutions, --- o A o
D=3 20’ 00”
DMS o o 20 00
D=3+ + 60
602

28. Find: αNAB αNAB= αNAB π αNAB=3.917 + 0.1367 [rad] LAB=235 [ft] N I LABo R=
180 D * π
αNAB=3.917 o o
3.9
5.1
6.4
7.8
the answer is A.
answerA o A o
D=3 20’ 00”
DMS o o 20 00
D=3+ + 60
602

29. radians.

30. ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4