1. Find: STAB I1=90 C 2,500 [ft] 2,000 [ft] B D A curve 1 R1=R2 F curve 2o
I1=90 C
2,500 [ft]
2,000 [ft] B D A
curve 1
R1=R2 F
curve 2 E
Find the stationing at point B. [pause] In this problem, ---
STAA=5+00 o
I2=45
15+86
16+58
C) 17+31
D) 18+23

2. Find: STAB I1=90 C 2,500 [ft] 2,000 [ft] B D A curve 1 R1=R2 F curve 2o
I1=90 C
2,500 [ft]
2,000 [ft] B D A
curve 1
R1=R2 F
curve 2 E
Curve 1, and Curve 2 are ---
STAA=5+00 o
I2=45
15+86
16+58
C) 17+31
D) 18+23

3. Find: STAB I1=90 C 2,500 [ft] 2,000 [ft] B D A curve 1 R1=R2 F curve 2o
I1=90 C
2,500 [ft]
2,000 [ft] B D A
curve 1
R1=R2 F
curve 2 E
connected at point D to create ---
STAA=5+00 o
I2=45
15+86
16+58
C) 17+31
D) 18+23

4. Find: STAB I1=90 C 2,500 [ft] 2,000 [ft] B D A curve 1 R1=R2 F curve 2o
I1=90 C
2,500 [ft]
2,000 [ft] B D A
curve 1
R1=R2 F
curve 2 E
a reverse curve, where the radius of the two curves are equal.
STAA=5+00 o
I2=45
15+86
16+58
C) 17+31
D) 18+23

5. Find: STAB I1=90 C 2,500 [ft] 2,000 [ft] B D A curve 1 R1=R2 F curve 2o
I1=90 C
2,500 [ft]
2,000 [ft] B D A
curve 1
R1=R2 F
curve 2 E
The interior angles of the two curves are provided, ----
STAA=5+00 o
I2=45
15+86
16+58
C) 17+31
D) 18+23

6. Find: STAB I1=90 C 2,500 [ft] 2,000 [ft] B D A curve 1 R1=R2 F curve 2o
I1=90 C
2,500 [ft]
2,000 [ft] B D A
curve 1
R1=R2 F
curve 2 E
as well as the lengths of segments AC and CE. The stationing at point A equals ----
STAA=5+00 o
I2=45
15+86
16+58
C) 17+31
D) 18+23

7. Find: STAB I1=90 C 2,500 [ft] 2,000 [ft] B D A curve 1 R1=R2 F curve 2o
I1=90 C
2,500 [ft]
2,000 [ft] B D A
curve 1
R1=R2 F
curve 2 E
To find the stationing at point B, ---
STAA=5+00 o
I2=45
15+86
16+58
C) 17+31
D) 18+23

8. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2 I1=90 Bo B
I1=90 D o
I2=45 A F
STAB=STAA+LAB E
we’ll add the the length of segment AB to the stationing at point A. The problem statement provides the --- I2

9. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2 I1=90 Bo B
I1=90 D o
I2=45 A F
STAB=STAA+LAB E
the length of segment AB equals ---- I2

10. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2 I1=90 Bo B
I1=90 D o
I2=45 A F
STAB=STAA+LAB E
the length of segment AC minus the length of segment BC. Length segment AC is given as --- I2
LAB=LAC-LBC

11. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2 I1=90 Bo B
I1=90 D o
I2=45 A F
STAB=STAA+LAB E
2,500 feet, and the length of BC is equal to the --- I2
LAB=LAC-LBC

12. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2 I1=90 Bo B
I1=90 D o
I2=45 A F
STAB=STAA+LAB E
tangent distance of curve 1, T1, which equals the radius of curve 1 times --- I2
LAB=LAC-LBC
T1=R1*tan(0.5*I1)

13. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2 I1=90 Bo B
I1=90 D o
I2=45 A F
STAB=STAA+LAB E
the tangent of one half times the interior angle of curve 1. The problem states the interior angle of --- I2
LAB=LAC-LBC
T1=R1*tan(0.5*I1)

14. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2 I1=90 Bo B
I1=90 D o
I2=45 A F
STAB=STAA+LAB E
curve 1 equals 90 degrees, and the radius of curve 1 equals ---- I2
LAB=LAC-LBC
T1=R1*tan(0.5*I1)

15. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2 I1=90 Bo B
I1=90 D o
I2=45 A F
STAB=STAA+LAB E
the radius of curve 2. For now, we’ll designate both radius 1 and radius 2 as --- I2
LAB=LAC-LBC
T1=R1*tan(0.5*I1)

16. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2=R I1=90o B
I1=90 D o
I2=45 A F
STAB=STAA+LAB E
R. [pause] Now our equation for T1 simplifies to ---- I2
LAB=LAC-LBC
T1=R1*tan(0.5*I1)

17. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2=R Bo B
I1=90 D o
I2=45 A o
T1=R*tan(0.5*90 ) F E
R times the tangent of 0.5 times 90 degrees, which simplifies to --- I2
T1=R1*tan(0.5*I1)

18. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2=R Bo B
I1=90 D o
I2=45 A o
T1=R*tan(0.5*90 ) F E
R. [pause] From the problem statement, we notice the length of segment ---
T1=R I2
T1=R1*tan(0.5*I1)

19. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] I1 C STAA=5+00 R1=R2=R Bo B
I1=90 D o
I2=45 A o
T1=R*tan(0.5*90 ) F E
CE equals 2,000 feet, which is equivalent to the sum of the ---
T1=R I2
T1=R1*tan(0.5*I1)

20. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2 o
T1=R*tan(0.5*90 ) F E
tangent distances T1 and T2. T2 equals R2 times ----
T1=R I2
T1=R1*tan(0.5*I1)

21. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2 o
T1=R*tan(0.5*90 ) F E
the tangent of one half the interior angle of curve 2. After substituting in the appropriate values for ----
T1=R I2
T2=R2*tan(0.5*I2)

22. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2 o
T1=R*tan(0.5*90 ) F E
R2 and I2, the tangent distance for curve 2 equals ---
T1=R I2
T2=R2*tan(0.5*I2)

23. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2 o
T1=R*tan(0.5*90 ) F E
0.414 times the radius. From the diagram, we know the length of CE equals ---
T1=R I2
T2=R2*tan(0.5*I2)
T2=0.414*R

24. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2 o
T1=R*tan(0.5*90 ) F E
T1 plus T2. After substituting in T1, T2 and the length of segment CE, ---
T1=R
T2=R2*tan(0.5*I2)
LCE=T1+T2
T2=0.414*R

25. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2 o
T1=R*tan(0.5*90 ) F E
the radii of both curves, R1 and R2 equal, ---
T1=R
T2=R2*tan(0.5*I2)
LCE=T1+T2
T2=0.414*R

26. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2 o
T1=R*tan(0.5*90 ) F E
1,414 feet. [pause] Therefore, the tangent distance of curve 1 equals ---
T1=R
T2=R2*tan(0.5*I2)
LCE=T1+T2
T2=0.414*R
2,000 [ft]=R+0.414*R
R=1,414 [ft]

27. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2 o
T1=R*tan(0.5*90 ) F E
1,414 feet. Now we can determine the length of segment AB by ---
T1=R
T1=1,414 [ft]
LCE=T1+T2
2,000 [ft]=R+0.414*R
R=1,414 [ft]

28. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2
T1=1,414 [ft] F E
subtracting this value of 1,414 feet from ----
LCE=T1+T2
2,000 [ft]=R+0.414*R
R=1,414 [ft]

29. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2
T1=1,414 [ft] F E
the length of segment AC, which was given as ----
LAB=LAC-LBC
LCE=T1+T2
2,000 [ft]=R+0.414*R
R=1,414 [ft]

30. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2
T1=1,414 [ft] F E
2,500 feet. This makes the length of segment AB equal to, ---
LAB=LAC-LBC
LCE=T1+T2
2,000 [ft]=R+0.414*R
R=1,414 [ft]

31. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2
T1=1,414 [ft] F E
1,086 feet. The stationing at point B equals ---
LAB=LAC-LBC
LCE=T1+T2
LAB=1,086 [ft]
2,000 [ft]=R+0.414*R
R=1,414 [ft]

32. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2
T1=1,414 [ft] F E
the stationing at point A, plus the length of segment AB. After substituting in ---
LAB=LAC-LBC
STAB=STAA+LAB
LAB=1,086 [ft]

33. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2
T1=1,414 [ft] F E
these two variables, the stationing at point B equals ----
LAB=LAC-LBC
STAB=STAA+LAB
LAB=1,086 [ft]

34. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2
T1=1,414 [ft] F E
1,586 feet. [pause]
LAB=LAC-LBC
STAB=STAA+LAB
LAB=1,086 [ft]
STAB=15+86

35. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2
15+86
16+58
C) 17+31
D) 18+23 F E
When reviewing the possible solutions, ----
STAB=STAA+LAB
STAB=15+86

36. Find: STAB LCE=2,000 [ft] LAC=2,500 [ft] C STAA=5+00 T1 R1=R2=R LCE Bo B
I1=90 o D
I2=45 A T2
15+86
16+58
C) 17+31
D) 18+23 F E
the answer is A
STAB=STAA+LAB
STAB=15+86
AnswerA

37. ? 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