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

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

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

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

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

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

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”

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”

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”

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”

Find: αNAB αNAB= αNAB curve length N I LAB I= 2 R radius B LAB=STAB-STAA I 782 feet minus 547 feet equals 235 --- A O STAB=7+82 STAA=5+47 o D=3 20’ 00”

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”

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”

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”

Find: αNAB αNAB= αNAB LAB=235 [ft] N I LAB I= 2 R radius B o 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”

Find: αNAB αNAB= αNAB LAB=235 [ft] N I LAB I= 2 R radius B o 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

Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius B o 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

Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius B o R= D in DD form I equal to 3.333 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

Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius B o 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

Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius B o 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

Find: αNAB αNAB= αNAB + LAB=235 [ft] N I LAB I= 2 R radius B o 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

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 0.1367 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

Find: αNAB αNAB= αNAB + 0.1367 [rad] LAB=235 [ft] N I LAB I= 2 R B o 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

Find: αNAB αNAB= αNAB + 0.1367 [rad] LAB=235 [ft] N I LAB I= 2 R B o R= D in DD form I 0.06835 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

Find: αNAB αNAB= αNAB π + 0.1367 [rad] LAB=235 [ft] N I LAB I= 2 R B o 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

Find: αNAB αNAB= αNAB π αNAB=3.917 + 0.1367 [rad] LAB=235 [ft] N I LAB o 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

Find: αNAB αNAB= αNAB π αNAB=3.917 + 0.1367 [rad] LAB=235 [ft] N I LAB o 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

Find: αNAB αNAB= αNAB π αNAB=3.917 + 0.1367 [rad] LAB=235 [ft] N I LAB o 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

0.06835 radians.

? 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