Highways & Transportation (ECIV4333)

The Islamic University of Gaza Civil Engineering Department

1st Semester, NOV 2016 (A) Time Allowed 30 minutes

:Instructors

Dr. Yahya R. Sarraj T.A :Hesham A. Hanoona

MIDTERM EXAM

الرقم الجامعي اسم الطالب

Notes Mark Question No.

1

2

total

Question 1:

Traffic and highway engineers are continually engaged in working to

ensure that highway crash rates can be reduced.

a) What Causes Transportation Crashes? Explain in brief.

b) Compare between road crashes at night and road crashes during the

day.

Question 2:

A curve of radius 75 m with 0.08 super elevation is located at a section of an

existing rural highway , which has been constructed on a highway of +3%

gradients. The maximum speed allowed on the curve is 45 km/h and the

driver must decelerate to safely enter the curve with deceleration rate 3.5 m/s2

Maximum allowable speed on highway = 90 km/h.

Letter height of road sign = 4"

Gross vehicle weight = 1500 kg

Frontal cross-sectional area = 2 m2

Assume that a driver can read a road sign within his or her area of vision at

distance of (10 m) for each inch of letter height.

a) Determine the minimum distance from the curve that road sign should be

located in forming acute turning on the highway.

b) Calculate the total resistance force on the vehicle as it traverses the

horizontal curve .

c) Determine the total horsepower on the curve the vehicle must produce .

Useful Formula

𝑆𝑆𝐷 = 0.278 𝑢 𝑡 +𝑢1

2 − 𝑢22

254(𝑎𝑔

± 𝐺) 𝑅𝑎 = 0.5

(0.077 𝜌𝐶𝐷 𝐴 𝑢2)

𝑔

6-01×7= rvC ,0.012 = rsC )W2u rv+ 0.077 C rs =( C rR

𝑅𝑐 = 0.5(0.077 𝑢2𝑊)

𝑔𝑅 𝑃 =

2.73 𝛴𝑅 𝑢

746

𝑅 = 𝑢2

127(𝑒+𝑓𝑠)

Highways & Transportation (ECIV4333)

The Islamic University of Gaza Civil Engineering Department 1st Semester ,NOV 2016 (B)

Time Allowed 30 minutes

Instructors:

Dr. Yahya R. Sarraj T.A :Hesham A. Hanoona

MIDTERM EXAM

الرقم الجامعي اسم الطالب

Notes Mark Question No.

1

2

total

Question 1:

Traffic and highway engineers are continually engaged in working to

ensure that highway crash rates can be reduced.

a) Describe in brief how serious road crashes are to the world.

b) What is meant by traffic calming?

Question 2:

A horizontal curve of a radius R is located on a highway (leveled) that has a

design speed of 100 km/h. A warning sign is to be located before the curve to

warn the drivers to reduce their speed before entering the curve. Assuming

that 85th percentile of drivers can see the sign from 40 m distance and the

distance between the sign and the curve is 120m.

Determine:

a) The speed of the car on the curve.

b) The radius of the curve, assuming a super elevation of 6%. (fs = 0.12)

c) The additional horsepower needed at the curve for a car weight 1100 kg

and has a frontal cross section of 4 m2.

Useful Formula

Rr =( Crs + 0.077 Crv u2)W Crs = 0.012 , Crv = 7×10-6

Highways & Transportation (ECIV4333)

The Islamic University of Gaza Civil Engineering Department

1st Semester, NOV 2016 (C) Time Allowed 30 minutes

: Instructors

Dr. Yahya R. Sarraj T.A :Hesham A. Hanoona

MIDTERM EXAM

الرقم الجامعي اسم الطالب

Notes Mark Question No.

1

2

total

Question 1:

The table below compares road fatality rates in Gaza Strip with other

countries in 2007 according to the World Health Organization.

Country/region Gaza Strip UK Qatar Lebanon World

average

Fatalities / 100,000 people 8 6 24 2 19

a) Explain the above table.

b) Making use of this table, discuss the best method to compare between

the highway safety conditions in different countries.

Question 2:

A level horizontal curve of radius R with 0.08 super elevation is located at the

end of s section of existing rural highway , which has gradient equal to +3%.

All vehicles must stop at the intersection as shown in the figure below .

Assume that a driver sees the stop sign immediately on entering the curve .

Driver must decelerate with deceleration rate 3 m/s2

Design speed on highway =70 km/h.

The maximum speed allowed on the curve is 60 km/h.

Gross vehicle weight = 1800 kg.

Frontal cross-sectional area of vehicle = 3 m2

a) Find the minimum radius of curve (R).(fs=0.12)

b) Find the length of section on highway from the beginning of curve to the

intersection (A to B).

c) Calculate the resistance forces on the vehicle as it moves on highway with

the design speed.

Useful Formula

𝑆𝑆𝐷 = 0.278 𝑢 𝑡 +𝑢1

2 − 𝑢22

254(𝑎𝑔

± 𝐺) 𝑅𝑎 = 0.5

(0.077 𝜌𝐶𝐷 𝐴 𝑢2)

𝑔

6-01×7= rvC ,0.012 = rsC )W2u rv+ 0.077 C rs =( C rR

𝑅𝑐 = 0.5(0.077 𝑢2𝑊)

𝑔𝑅 𝑃 =

2.73 𝛴𝑅 𝑢

746

𝑅 = 𝑢2

127(𝑒+𝑓𝑠)

Highways & Transportation (ECIV4333)

The Islamic University of Gaza Civil Engineering Department 1st Semester, NOV 2016 (D)

Time Allowed 30 minutes

Instructors:

Dr. Yahya R. Sarraj T.A :Hesham A. Hanoona

MIDTERM EXAM

الرقم الجامعي اسم الطالب

Notes Mark Question No.

1

2

total

Question 1:

The table below compares road fatality rates in Gaza Strip with other

countries in 2007 according to the World Health Organization.

Country/region Gaza Strip UK Qatar Lebanon World

average

Fatalities / 100,000 people 8 6 24 2 19

a) Explain the above table.

b) Making use of this table, discuss the best method to compare between

the highway safety conditions in different countries.

Question 2:

A level horizontal curve of radius R with 0.06 super elevation is located at the

end of s section of existing rural highway , which has gradient equal to +2%.

All vehicles must stop at the intersection as shown in the figure below .

Assume that a driver sees the stop sign immediately on entering the curve .

Driver must decelerate with deceleration rate 2.5 m/s2

Speed limit on highway =80 km/h.

The maximum speed allowed on the curve is 55 km/h.

Gross vehicle weight = 1600 kg.

Frontal cross-sectional area of vehicle = 4 m2

a) Find the minimum radius of curve (R).(fs=0.13)

b) Find the length of section on highway from the beginning of curve to the

intersection (A to B).

c) Calculate the resistance forces on the vehicle as it moves on highway with

the speed limit.

Useful Formula

𝑆𝑆𝐷 = 0.278 𝑢 𝑡 +𝑢1

2 − 𝑢22

254(𝑎𝑔

± 𝐺) 𝑅𝑎 = 0.5

(0.077 𝜌𝐶𝐷 𝐴 𝑢2)

𝑔

6-01×7= rvC ,0.012 = rsC )W2u rv+ 0.077 C rs =( C rR

𝑅𝑐 = 0.5(0.077 𝑢2𝑊)

𝑔𝑅 𝑃 =

2.73 𝛴𝑅 𝑢

746

𝑅 = 𝑢2

127(𝑒+𝑓𝑠)