Tiêu đề MSTE 38 Solutions.pdf ✅

38 Transportation Engineering: Road Analysis and Design Solutions SITUATION 1. A 20 km stretch of a highway had the following reported accidents: Year Fatal Injury Property Damage ADT 1991 4 42 110 1000 1992 2 54 210 1200 1993 5 60 182 1250 1994 7 74 240 1300 1995 6 94 175 1350 TOTAL 24 324 917 6100 ▣ 1. Find the severity ratio. [SOLUTION] SR = 24 + 324 24 + 324 + 917 = 0.275 ▣ 2. Determine the rate of injury accidents. [SOLUTION] Ra = (24 + 324) × 106 6100 × (20 × 0.62 mi) = 1257.70 Note: The total ADT in 5 years will be used instead of Average ADT x 5 years since each year has different ADTs. ▣ 3. Compute the rate of total accidents. [SOLUTION] R = (24 + 324 + 917) × 106 6100(20 × 0.62) = 4571.795 ▣ 4. Twelve vehicles are observed in a 400 m section of the extension of SCTEX. Average time of headway is 4 seconds. Determine the space mean speed. [SOLUTION] q = kμs
Traffic density: k = 12 veh 0.4 km = 30 veh/km Traffic volume: q = 1 veh 4 3600 hr = 900 veh/hr μs = 900 veh hr 30 veh km μs = 30 kph ▣ 5. In a certain portion of highway, the recorded Peak Hour Factor (PHF) during rush hour is 0.90. The highest 5-minute volume is 250 vehicles and the space mean speed is 90 kph. Find the volume of traffic. [SOLUTION] 0.9 = V ( 60 min 5 min ) (250 veh) V = 2700 veh SITUATION 2. A car having a weight of 40 kN is moving at a certain speed around the curve. Assuming no lateral pressure between the tire and pavement and considering a centrifugal ratio of 0.30. ▣ 6. Compute the force that will tend to pull the car away from the center of the curve. [SOLUTION] That force is the centrifugal force. CF = Wv 2 gR From the given, v 2 gR = 0.3 CF = (40 kN)(0.3) CF = 12 kN
▣ 7. If the degree of curve is 4° determine the maximum speed that the car could move around the curve. [SOLUTION] R = 3600 4π = 286.479 m 0.3 = v 2 gR 0.3 = v 2 (9.81)(286.479) v = 29.036 m s v = 104.53 kph ▣ 8. Compute the angle of embankment to be provided for this speed if the skid resistance is 0.12. [SOLUTION] tan(φ + θ) = v 2 gR tan(tan−1 0.12 + θ) = 0.3 θ = 9.86° ▣ 9. Compute the thickness of a rigid pavement with a wheel load capacity of 54 kN, if the allowable tensile stress of concrete is 1.6 MPa. Neglect the effects of dowels. [SOLUTION] t = √ 3W f t = √ 3(54 000N) 1.6 MPa = 318.20 mm ▣ 10. A 53.5 kN wheel load has a maximum tire pressure of 0.62 MPa. This pressure is to be uniformly distributed over the area of tire contact on the roadway. Assuming the subgrade pressure is not to exceed 0.14 MPa, determine the required thickness of flexible pavement structure, according to the principle of the cone pressure distribution. [SOLUTION] σ = F A 0.62 MPa = 53 500 N πR2 → R = 165.7319 mm