Transportation Engineering
9–14 of 110 itemsNCEES draws between 9 and 14 of the 110 from this area on any legal paper. Its outline lists 5 sub-topics; the bank holds 53 verified templates against them, which expand to 1,438 questions.
Sub-topics, from the NCEES outline
- AGeometric design (e.g., streets, highways, intersections)
- BPavement system design (e.g., thickness, subgrade, drainage, rehabilitation)
- CTraffic capacity and flow theory
- DTraffic control devices
- ETransportation planning (e.g., travel forecast modeling, safety, trip generation)
5 worked problems
Real questions from the bank, at the depth the exam asks them. Pick an answer first: the reasoning opens when you do, along with why each wrong answer is tempting.
Question 1 of 5
A sag vertical curve joins a -4.5% downgrade to a 2.5% upgrade. The headlight sight distance required is 700 ft and is less than the curve length. The minimum curve length is most nearly:
Sag vertical curve joining a -4.5 percent downgrade to a 2.5 percent upgrade, controlled by the headlight beam reaching 700 ft ahead. Show the answer and the reasoning
Why B
Given: g1 = -4.5%, g2 = 2.5%, headlight sight distance S = 700 ft, S < L. Relation: L = AS^2/(400 + 3.5S), from a 2.0-ft headlight and a 1-degree upward beam. Substitute: A = 7.0%; L = 7.0(700)^2/(400 + 3.5(700)) = 3430000/2850.0. Result: L = 1204 ft. Check: L exceeds S = 700 ft, so the S < L form was the right one.
Why A is tempting
Added the grades rather than taking their algebraic difference.
Why C is tempting
Used the crest curve expression. A sag curve at night is limited by how far the headlights reach.
Why D is tempting
Left out the sight distance term in the denominator entirely.
Source FHWA geometric design and pavement guidance — Sag vertical curve, headlight criterion
Question 2 of 5
Fifteen-minute counts during the peak hour are 220, 165, 198 and 198 vehicles. The peak hour factor is most nearly:
Show the answer and the reasoning
Why A
Given: 15-min counts 220, 165, 198, 198; the peak interval is 220. Relation: PHF = hourly volume/(4 x peak 15-min volume). Substitute: PHF = 781/(4 x 220) = 781/880. Result: PHF = 0.887.
Why B is tempting
Inverted the ratio. The peak hour factor is never greater than one.
Why C is tempting
Used the lowest fifteen-minute count instead of the peak one.
Why D is tempting
Did not multiply the peak fifteen-minute count by four. The comparison is against the hourly rate that interval implies.
Source Traffic flow theory, standard relations — Peak hour factor
Question 3 of 5
An intersection approach has a speed of 55 mph on a -3% grade. Using a perception-reaction time of 1 s and a deceleration rate of 10 ft/s^2, the yellow change interval is most nearly:
Show the answer and the reasoning
Why C
Given: V = 55 mph, t = 1 s, a = 10 ft/s^2, G = -0.03. Relation: y = t + 1.47V/(2a + 2gG), with G signed (negative downhill). Substitute: y = 1 + 1.47(55)/(2(10) + 64.4(-0.03)) = 1 + 80.85/18.068. Result: y = 5.47 s.
Why A is tempting
Left out the perception-reaction time.
Why B is tempting
Ignored the grade entirely.
Why D is tempting
Omitted the factor of two in the denominator.
Source FHWA geometric design and pavement guidance — Geometric design and pavement structural design
Question 4 of 5
A highway carries 8,000 vehicles per day, 10% of them trucks with a truck factor of 1.5. The growth factor over the design period is 24.3, the directional distribution is 0.5 and the lane distribution is 0.9. The design lane ESALs are most nearly:
Show the answer and the reasoning
Why A
Given: ADT = 8,000, trucks = 0.1, TF = 1.5, GF = 24.3, DD = 0.5, LD = 0.9. Relation: ESAL = ADT x %trucks x TF x 365 x GF x DD x LD. Substitute: ESAL = 8000(0.1)(1.5)(365)(24.3)(0.5)(0.9). Result: ESAL = 4.79e+6 in the design lane.
Why B is tempting
Left out the lane distribution factor.
Why C is tempting
Left out the directional and lane distribution factors. Only the traffic in the design lane and direction counts.
Why D is tempting
Applied the truck factor to all vehicles. Passenger cars contribute almost no pavement damage.
Source FHWA geometric design and pavement guidance — Equivalent single axle load computation
Question 5 of 5
An intersection with 45,000 entering vehicles per day recorded 36 crashes over 3 years. The crash rate per million entering vehicles is most nearly:
Show the answer and the reasoning
Why C
Given: 36 crashes, 45,000 entering veh/day, 3 years. Relation: R = crashes x 10^6/(365 x years x entering ADT). Substitute: exposure = 365(3)(45000) = 4.928e+7; R = 36(10^6)/4.928e+7. Result: R = 0.731 crashes per million entering vehicles.
Why A is tempting
Used a hundred-thousand-vehicle base rather than a million.
Why B is tempting
Doubled the entering volume, counting both directions of an ADT that is already two-way.
Why D is tempting
Counted a single year of exposure rather than the full study period.
Source FHWA geometric design and pavement guidance — Geometric design and pavement structural design
Four mistakes that cost the question
Each of these lands on an answer that is offered, so it costs the question outright. The minutes are our estimate of the time each one burns on top, against an average of 2.9 minutes a question (320 minutes for 110).
Costs the question and about 4 min
Using the wrong vertical curve formula
Crest curves are governed by stopping sight distance with driver eye and object heights; sag curves by headlight sight distance. Each has one formula for sight distance shorter than the curve and another for longer. Using one without checking the result against the curve length, or a crest formula on a sag, gives a length that is offered.
Costs the question and about 2 min
Multiplying by the peak hour factor instead of dividing
The peak hour factor is the hourly volume over four times the peak 15-minute count, so it is at most 1. The design flow rate is the hourly volume divided by it, which is larger. Multiplying gives a lower flow, and that is the distractor.
Costs the question and about 3 min
Dropping the grade, or its sign, from braking distance
Braking distance on a grade is V² over 30 times (a/32.2 ± G), with V in mph: a downgrade lengthens it, an upgrade shortens it. The same grade term appears in the yellow change interval. Ignoring the grade, or using its sign the wrong way round, gives answers a few percent apart, all offered.
Costs the question and about 2 min
Using the wrong crash rate denominator
Intersection crash rates are per million entering vehicles; segment rates are per hundred million vehicle-miles, so they also take the length. Both need the count of days, 365 times the years of record. Leaving out a term changes the answer by a factor the choices are built around.
Questions about Transportation Engineering
What transportation topics does the FE Civil exam cover?
Five sub-topics: geometric design of streets, highways and intersections; pavement design; traffic capacity and flow theory; traffic control devices; and transportation planning, including trip generation, forecasting and safety.
Do I need the Highway Capacity Manual?
Not as a book. Traffic flow questions use flow equals density times speed, the Greenshields model, the peak hour factor and simple capacity and signal timing calculations, all of which are in the FE Reference Handbook.
Which pavement design method appears?
The AASHTO flexible pavement method: the structural number is the sum of layer coefficients times layer thicknesses, with drainage coefficients on the unbound layers, and traffic is expressed in 18-kip equivalent single axle loads.
How do stopping sight distance questions work?
Stopping sight distance is the distance travelled during perception and reaction, 1.47 times speed in mph times the reaction time, plus the braking distance, adjusted for grade. It then sets the minimum length of a crest vertical curve.
How many study hours this area is worth · Using the handbook under the clock