FE Civil Transportation Engineering practice problems
6 free FE Civil Transportation Engineering practice problems with answers and full worked solutions. Transportation Engineering carries 9–14 of the 110 questions on the FE Civil exam (NCEES specification effective July 2020). Pick an answer to check it, then read the worked solution and why each wrong choice is wrong.
6 worked Transportation Engineering problems
Question 1 of 6
A circular curve has a radius of 1650 ft and a central angle of 66 degrees. Its length is most nearly:
Plan view of a circular curve of radius 1650 ft with a central angle of 66 degrees, showing the PC, PI and PT. Show the answer and the reasoning
Why C
Given: R = 1650 ft, central angle = 66 degrees. Relation: L = R x delta, with delta in radians. Substitute: delta = 66(pi/180) = 1.15192 rad; L = 1650(1.15192). Result: L = 1901 ft. Check: D = 5729.58/1650 = 3.4725 degrees, and 100(66)/D gives the same length.
Why A is tempting
Computed the tangent length rather than the arc length.
Why B is tempting
Computed the long chord, which is always shorter than the arc.
Why D is tempting
Doubled the arc by using delta over 180 with the full circumference.
Source FHWA geometric design and pavement guidance — Geometric design and pavement structural design
Question 2 of 6
For a given traffic loading and subgrade, which pavement type generally requires the least surface thickness?
Show the answer and the reasoning
Why C
A rigid pavement is stiff enough to act as a slab in bending, so a wheel load is distributed over a large area of subgrade and the required surface thickness is comparatively small. A flexible pavement transmits load through a much narrower cone of pressure, so it needs greater combined thickness to hold the subgrade stress down.
Why A is tempting
Flexibility is exactly what makes an asphalt pavement spread load over a smaller area, so it needs more thickness, not less, for the same subgrade stress.
Why B is tempting
Compaction temperature is a construction and emissions consideration. It has nothing to do with how thick the layer must be for a given traffic loading.
Why D is tempting
A granular base is far less stiff than a concrete slab and spreads load over a correspondingly smaller area, so the total section ends up thicker.
Source FHWA geometric design and pavement guidance — Geometric design and pavement structural design
Question 3 of 6
A lane carries 1,500 veh/hr. The average time headway between vehicles is most nearly:
Show the answer and the reasoning
Why D
Given: q = 1500 veh/hr. Relation: headway = 3600/q, the inverse of flow. Substitute: h = 3600/1500. Result: h = 2.40 s.
Why A is tempting
Used the minutes in an hour rather than the seconds.
Why B is tempting
Inverted the relation.
Why C is tempting
Halved the headway, which corresponds to twice the stated flow rate.
Source FHWA geometric design and pavement guidance — Geometric design and pavement structural design
Question 4 of 6
A crosswalk is 82 ft long. Using a walking speed of 3.5 ft/s and a start-up interval of 3.2 s, the total pedestrian crossing time is most nearly:
Show the answer and the reasoning
Why B
Given: crosswalk 82 ft, walking speed 3.5 ft/s, start-up 3.2 s. Relation: t = start-up + W/speed. Substitute: t = 3.2 + 82/3.5 = 3.2 + 23.43. Result: t = 26.6 s.
Why A is tempting
Subtracted the start-up interval.
Why C is tempting
Multiplied by the start-up time rather than adding it.
Why D is tempting
Multiplied by the walking speed rather than dividing by it.
Source FHWA geometric design and pavement guidance — Geometric design and pavement structural design
Question 5 of 6
A road currently carries 36,000 veh/day. Traffic is forecast to grow at 2.0% per year, compounded. The volume in 20 years is most nearly:
Show the answer and the reasoning
Why D
Given: ADT = 36,000 veh/day, i = 0.02, n = 20 years. Relation: ADT_n = ADT(1 + i)^n. Substitute: 36000(1 + 0.02)^20 = 36000(1.4859). Result: 5.349e+4 veh/day.
Why A is tempting
Gave only the increase, and computed it as simple growth at that.
Why B is tempting
Discounted backwards instead of projecting forwards.
Why C is tempting
Applied simple rather than compound growth. Compounding gives a larger figure over a long horizon.
Source FHWA geometric design and pavement guidance — Geometric design and pavement structural design
Question 6 of 6
A circular highway curve has a radius of 550 ft. Using the arc definition, its degree of curve is most nearly:
Show the answer and the reasoning
Why B
Given: R = 550 ft, arc definition (100-ft arc). Relation: D = 5729.58/R, where 5729.58 = 100(180/pi). Substitute: D = 5729.58/550. Result: D = 10.4 degrees.
Why A is tempting
Used the 20-metre arc definition of degree of curve.
Why C is tempting
Used a 200-ft arc. Degree of curve is defined on a 100-ft station.
Why D is tempting
Divided the radius by 3.281, as if it had been given in metres.
Source FHWA geometric design and pavement guidance — Geometric design and pavement structural design
Going further
Frequently asked questions
How many Transportation Engineering questions are on the FE Civil exam?
Between 9 and 14 of the 110 questions, according to the NCEES FE Civil specification effective July 2020.
Are these real NCEES exam questions?
No. NCEES does not release its exam questions. These are original FE-style problems written to the NCEES specification and the FE Reference Handbook; every calculated key is checked by two independent solution methods.
Are these Transportation Engineering practice problems free?
Yes. All 6 problems on this page, with answers and full worked solutions, are free and need no account.
Check where you stand, free
The quick diagnostic is eight FE Civil questions in about ten minutes, with no account. The free mock is a full 110-question, 5 hour 20 minute FE Civil paper with a worked solution for every question.