1. Find: STAC B C O A IAB 19+00 19+50 20+00 20+50 R STAA=15+50.00 IAB=60
Find the stationing at point C. [pause] In this problem, --- o
IAB=60
R=1,000 [ft]
O=200 [ft]

2. Find: STAC B C O A IAB 19+00 19+50 20+00 20+50 R STAA=15+50.00 IAB=60
a horizontal curve has an interior angle of 60 degrees and a radius of 1,000 feet. o
IAB=60
R=1,000 [ft]
O=200 [ft]

3. Find: STAC B C O A IAB 19+00 19+50 20+00 20+50 R STAA=15+50.00 IAB=60
Line C is parallel to Line B, and offset by 200 feet. o
IAB=60
R=1,000 [ft]
O=200 [ft]

4. Find: STAC B C O A IAB 19+00 19+50 20+00 20+50 R STAA=15+50.00 IAB=60
The stationing at Point A is given as [pause] The stationing at Point C can be determined by --- o
IAB=60
R=1,000 [ft]
O=200 [ft]

5. Find: STAC B C O STAC = STAA + LAC A IAB R STAA=15+50.00 IAB=60
adding Station A to the length of Arc AC. o
IAB=60
R=1,000 [ft]
O=200 [ft]

6. Find: STAC B C O STAC = STAA + LAC A IAB R STAA=15+50.00 IAB=60
Where Station A is given, and the length of Arc AC --- o
IAB=60
R=1,000 [ft]
O=200 [ft]

7. Find: STAC B C O STAC = STAA + LAC A IAB R LAC = R * IAC STAA=15+50.00
can be determined from the interior angle AC and the radius. o
IAB=60
R=1,000 [ft]
O=200 [ft]

8. Find: STAC B C O STAC = STAA + LAC A IAB R LAC = R * IAC STAA=15+50.00
From the problem statement, the radius of the horizontal curve is 1,000 feet, --- o
IAB=60
R=1,000 [ft]
O=200 [ft]

9. Find: STAC B C O STAC = STAA + LAC A IAB R LAC = R * IAC STAA=15+50.00
and interior angle AC can be determined using geometry. [pause] First we’ll divide ---
determine o
IAB=60
R=1,000 [ft]
using
geometry
O=200 [ft]

10. Find: STAC B C O ICB IAC STAC = STAA + LAC A IAB R LAC = R * IAC
Angle AB into Angles AC and CB. That way, Angle AC equals ---
determine o
IAB=60
R=1,000 [ft]
using
geometry
O=200 [ft]

11. Find: STAC B C O ICB IAC STAC = STAA + LAC A IAB R LAC = R * IAC
Angle AB minus Angle CB. The problem states Angle AB equals --- o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft]

12. Find: STAC B C O ICB IAC STAC = STAA + LAC A IAB R LAC = R * IAC
60 degrees. [pause] Next, we identify the center of the horizontal curve as --- o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft]

13. Find: STAC B C O IAC STAC = STAA + LAC A R P LAC = R * IAC
Point P, and the intersection of Line BP --- o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft]

14. Find: STAC B C O B’ IAC STAC = STAA + LAC A R P LAC = R * IAC
with Line C as Point B prime, then Triangle C P B prime --- o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft]

15. Find: STAC B C O B’ IAC STAC = STAA + LAC A R P LAC = R * IAC
is a right triangle, and Angle CB is equal to --- o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft]

16. Find: STAC B C O ICB’ B’ IAC STAC = STAA + LAC A R P LAC = R * IAC
Angle CB prime. [pause] From right triangle trigonometry, --- o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft]
ICB=ICB’

17. Find: STAC B C O ICB’ B’ IAC IAC=IAB-ICB A R ICB=ICB’ P B’P
STAA=
Angle CB prime is equal to the arc cosine of the length of B prime P, divided by the length of C P.
ICB’=cos-1 CP o
IAB=60
R=1,000 [ft]
O=200 [ft]

18. Find: STAC B C O ICB’ B’ IAC IAC=IAB-ICB A R ICB=ICB’ P B’P
STAA=
The length of C P equals the Radius, R, 1,000 feet. And the length of B prime P equals ---
ICB’=cos-1 CP o
IAB=60
R=1,000 [ft]
CP=R=1,000 [ft]
O=200 [ft]

19. Find: STAC B C O ICB’ B’ IAC IAC=IAB-ICB A R ICB=ICB’ P B’P
STAA=
the radius minus the offset between parallel Lines B and C, ---
ICB’=cos-1 CP o
IAB=60
R=1,000 [ft]
CP=R=1,000 [ft]
O=200 [ft]
B’P=R-O

20. Find: STAC B C O ICB’ B’ IAC IAC=IAB-ICB A R ICB=ICB’ P B’P
STAA=
or 1,000 feet minus 200 feet, which is 800 feet.
ICB’=cos-1 CP o
IAB=60
R=1,000 [ft]
CP=R=1,000 [ft]
O=200 [ft]
B’P=R-O=800 [ft]

21. Find: STAC B C O ICB’ B’ IAC IAC=IAB-ICB A R ICB=ICB’ P B’P
STAA=
Plugging these triangle side lengths into the arc cosine equation, ---
ICB’=cos-1 CP o
IAB=60
R=1,000 [ft]
CP=R=1,000 [ft]
O=200 [ft]
B’P=R-O=800 [ft]

22. Find: STAC B C O ICB’ B’ IAC IAC=IAB-ICB A R ICB=ICB’ P 36.870 B’P
STAA=
Angle CB prime equals degrees. Therefore, Angle CB also equals ---
ICB’=cos-1 CP o
IAB=60
R=1,000 [ft]
CP=R=1,000 [ft]
O=200 [ft]
B’P=R-O=800 [ft]

23. Find: STAC B C O ICB B’ IAC IAC=IAB-ICB A R ICB=ICB’ P 36.870 B’P
STAA=
degrees. [pause] Plugging in Angle AB and ---
ICB’=cos-1 CP o
IAB=60
R=1,000 [ft]
CP=R=1,000 [ft]
O=200 [ft]
B’P=R-O=800 [ft]

24. Find: STAC B C O 36.87 ICB B’ IAC IAC=IAB-ICB A R ICB=ICB’ P B’P
STAA=
Angle CB, we find Angle AC equals ----
ICB’=cos-1 CP o
IAB=60
R=1,000 [ft]
CP=R=1,000 [ft]
O=200 [ft]
B’P=R-O=800 [ft]

25. Find: STAC B C O 36.87 ICB B’ IAC IAC=IAB-ICB A 23.13 R ICB=ICB’ P B’P
STAA=
23.13 degrees, which converts to ----
ICB’=cos-1 CP o
IAB=60
R=1,000 [ft]
CP=R=1,000 [ft]
O=200 [ft]
B’P=R-O=800 [ft]

26. Find: STAC B C O ICB B’ IAC STAC = STAA + LAC A R P LAC = R * IAC
radians. [pause]
23.13
rad o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft] o
36.87

27. Find: STAC B C O ICB B’ IAC STAC = STAA + LAC A R P LAC = R * IAC
times the radius of the curve, 1,000 feet, equals ----
23.13
rad o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft] o
36.87

28. Find: STAC B C O ICB B’ IAC STAC = STAA + LAC A 403.7 [ft] R P
LAC = R * IAC o
STAA=
403.7 feet, the length of Arc AC. [pause] Adding together the length of Arc AC and the ----
23.13
rad o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft] o
36.87

29. Find: STAC B C O ICB B’ IAC STAC = STAA + LAC A 403.7 [ft] R P
LAC = R * IAC o
STAA=
stationing at Point A, the stationing at Point C equals ---
23.13
rad o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft] o
36.87

30. Find: STAC B C 19+53.70 ICB B’ IAC STAC = STAA + LAC A 403.7 [ft] R P
LAC = R * IAC o
STAA=
[pause]
23.13
rad o
IAC=IAB-ICB
IAB=60
R=1,000 [ft]
O=200 [ft] o
36.87

31. Find: STAC B C 19+53.70 ICB B’ IAC STAC = STAA + LAC A 403.7 [ft] R P
19+00
19+50
20+00
20+50
STAA=
When reviewing the possible solutions, ---- o
IAB=60
R=1,000 [ft]
O=200 [ft]

32. Find: STAC B C 19+53.70 ICB B’ IAC STAC = STAA + LAC A 403.7 [ft] R P
19+00
19+50
20+00
20+50
STAA=
the answer is B.
AnswerB o
IAB=60
R=1,000 [ft]
O=200 [ft]

33. ? 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