Question

1) A crest vertical curve of a highway segment is designed to connect upgrades of 5 percent while the downgrade on the other side is 3 percent. What’s the minimum length of the curve for adequate stopping sight distance if the stopping sight distance is 400 ft for the design speed is 45 mph and drives’ eye height is 3.5 ft and the object height 2.0 ft?

A. 430
B. 493
C. 593
D. 530

2) What is the minimum radius in feet for a horizontal curve with a superelevation of 0.04? Use AASHTO’s design value (0.12) for the coefficient of side friction, assuming a target design speed of 60 mph.

A. 1210
B. 1510
C. 1310
D. 1410

Answer 1

Sol 0 Given that n, = 5%(upward gradient M2= 3% (down word gradient) (+= 57.) (-M2= 13.3) =) General equation for the longth of a Crest Vertical Curve in terms of Algebraic difference in grades: when Stopping Sight distance (s) is Jess than Length of curre (L) : NS L = Curve, (Ft) loo (J2h + √2h2 whert L = Length of vertical S= Stopping Sight distance N = algebric difference in grade eye ha= height of object above Roadway above Roadway (f) hi height of Surface (ft)

No In-ng -15-(-3) Z Hi= 3.5 ft (given) he aft (given) Stopping sight distance = 400 ft (givens So Length of clara (L355D) assume La 8 x(40097 100 (12835 + 2x2) 593.06 FH 9 = So Longth of a Caest Vertical Clyve is 593.06 ft Hence option c is the correct Answfa

Sol.2 given that Super clevation Ce) = 0,04 Cafficient of side friction = 0.12 target design speed to mph V= Gomiles per hour V=60 x 1.60934 km ph ) V= 96-5606 kmphil [imph = 1.60934 kmph) 2 < e + f = 127 R V= Speed (kmph) R = Radius [0.04 +0.123 = (96.5606) 187 XR R = 458.856 meter [imeter = 3.78084 feet) R$ 458.856 x 3.28084 RE 1505.43 fract But Code Says in Range of lo.. [ Ramin 1510 Ft ft Told So Ans B is right answer