Free PE Civil — Transportation practice problems

Ten real questions from our PE Civil: Transportation bank, drawn from 10 of its 12 study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

4 with figures10 topicsFull worked solutions

1
medium
A. Clear zone, slopes & barrier length-of-need
A rural arterial on a 1,500-ft-radius curve has a design speed of 60 mph and a design-year volume of 6,000 veh/day. For that speed-and-volume cell the agency's clear-zone table (adapted from the Roadside Design Guide) gives 30 ft where the roadside is a fill foreslope of 1V:6H or flatter, 36 ft where it is a fill foreslope of 1V:5H to 1V:4H, and 26 ft where it is a backslope in cut. The agency's curve-adjustment factor for this curve's radius and design speed is $K_{cz}=1.3$ . On the outside of the curve the roadside is a 1V:6H fill foreslope running down to a ditch. The section is drawn schematically (not to scale). The clear-zone distance to be provided on the outside of the curve is most nearly:
Figure — drag or pinch to move around
1. 30 ft
2. 33.8 ft
3. 39 ft
4. 46.8 ft
Show the answer & worked solution
Answer: C.39 ft
The outside roadside is a 1V:6H fill foreslope, so the 1V:6H-or-flatter fill column governs: 30 ft on a tangent. On the outside of the curve that distance is widened by the curve-adjustment factor: $CZ = 30\times1.3 = 39$ ft, measured from the edge of the traveled way. Distractors: stopped at the tangent value and FORGOT the curve adjustment (30 ft); read the backslope (cut) column, which this roadside is not, and widened that (33.8 ft); and read the 1V:5H to 1V:4H fill column, which this roadside is not, and widened that (46.8 ft).
2
medium
A. Signal timing, clearance intervals & warrants
A pedestrian phase is being timed for a marked crosswalk across a four-lane undivided arterial with no median refuge. The crosswalk is 60 ft long, curb to curb. The design uses a walk interval of 7 s and a walking speed of 3.5 ft/s for the pedestrian clearance time. The minimum pedestrian phase — the walk interval plus the pedestrian clearance time (flashing DON'T WALK) — is most nearly:
Figure — drag or pinch to move around
1. 15.6 s
2. 22 s
3. 24.1 s
4. 27 s
Show the answer & worked solution
Answer: C.24.1 s
The pedestrian clearance time must carry a pedestrian who leaves the curb at the end of WALK across the whole crosswalk at the clearance walking speed: FDW $=L/S_p=60/3.5=17.14$ s. The phase is the walk interval plus that clearance: $7+17.14=24.14$ s. Distractors: sized the clearance on HALF the crosswalk, $\tfrac{L/2}{S_p}$ , which only works where a median refuge lets the pedestrian stop (15.57 s); timed the clearance at 4.0 ft/s — the faster speed MUTCD allows only where an extended pushbutton press adds the time — instead of the 3.5 ft/s this crossing is given (22 s); and timed the clearance at 3.0 ft/s — the slower speed MUTCD uses to check the TOTAL crossing time — instead of the 3.5 ft/s this crossing is given (27 s).
3
medium
A. Signs, markings & temporary traffic control (MUTCD)
On a downtown arterial posted at 35 mph, a work zone shifts both through lanes 12 ft to the left; no lane is closed. The agency applies the MUTCD taper criteria: merging taper at least $L$ , shifting taper at least $0.5L$ , shoulder taper at least $0.33L$ , with $L=WS^2/60$ for speeds of 40 mph or less and $L=WS$ for 45 mph or more ( $W$ = offset width, ft; $S$ = speed, mph). The minimum length of the taper at the start of this work zone is most nearly:
Figure — drag or pinch to move around
1. 80.9 ft
2. 123 ft
3. 210 ft
4. 245 ft
Show the answer & worked solution
Answer: B.123 ft
Traffic is moved sideways with no lane lost, so this is a SHIFTING taper, and 35 mph is in the 40-mph-or-less regime: $L=WS^2/60=12(35)^2/60=245$ ft, and the shifting taper is $0.5L=122.5$ ft. Distractors: read the SHOULDER-taper row, $0.33L=80.9$ ft, for a shifting taper; took the right $0.5L$ of the WRONG-regime $L$ ( $WS=420$ ft), giving 210 ft; and read the MERGING-taper row and gave the full $L=245$ ft, although no lane has to merge here.
4
medium
A. Equal-tangent parabola geometry & elevations
On a rural highway cresting a ridge, an equal-tangent vertical curve joins a $+3\%$ grade to a $-2\%$ grade over $L=600$ ft. Its PVI is at Sta 48+00.00, elevation 853.20 ft. The finished roadway elevation at Sta 47+80.00 is most nearly:
Figure — drag or pinch to move around
1. 838.58 ft
2. 849.33 ft
3. 852.60 ft
4. 858.33 ft
Show the answer & worked solution
Answer: B.849.33 ft
The parabola is written from the PVC, so locate it first: PVC $=$ Sta 48+00.00 $-\,L/2=$ Sta 45+00.00, at elevation $y_{PVC}=853.20 - 9.00=844.20$ ft, and the station in question lies $x=280$ ft beyond it. With $a=\dfrac{g_2-g_1}{2L}=\dfrac{(-0.02)-(0.03)}{2(600)}=-4.167\times10^{-5}$ ft $^{-1}$ , $y=y_{PVC}+g_1x+ax^2=844.20 + 8.40 - 3.27=849.33$ ft. Distractors: used the PVI elevation as the PVC elevation, skipping the back-up along $g_1$ over $L/2$ (858.33 ft); stopped on the incoming tangent, DROPPING the parabolic term (852.60 ft); and located the PVC a FULL curve length behind the PVI instead of $L/2$ , so $x=580$ ft from a PVC at 835.20 ft (838.58 ft).
5
medium
A. Uninterrupted flow, capacity & level of service
On an urban freeway, the peak-direction hourly volume is $V=4{,}400$ veh/h on $N=3$ lanes, with a peak-hour factor PHF $=0.90$ . Trucks make up 8% of the traffic, and the passenger-car equivalent for trucks on this segment is $E_T=2.5$ . The demand flow rate per lane, $v_p$ , is most nearly:
1. 1455 pc/h/ln
2. 1630 pc/h/ln
3. 1825 pc/h/ln
4. 1956 pc/h/ln
Show the answer & worked solution
Answer: C.1825 pc/h/ln
Heavy-vehicle factor: $f_{HV}=1/[1+P_T(E_T-1)]=1/[1+0.08(2.5-1)]=0.893$ . Demand flow rate per lane: $v_p=\dfrac{V}{PHF\cdot N\cdot f_{HV}}=\dfrac{4{,}400}{0.90(3)(0.893)}=1825$ pc/h/ln. Distractors: put $E_T$ in the bracket where $(E_T-1)$ belongs, $f_{HV}=1/[1+0.08(2.5)]=0.833$ , counting each truck's own passenger-car space as extra (1956); LEFT OUT $f_{HV}$ , giving a rate in mixed vehicles rather than passenger cars (1630); and MULTIPLIED by $f_{HV}$ instead of dividing (1455 — the wrong direction, since $f_{HV}<1$ must raise $v_p$ ).
6
medium
A. Soil properties, classification, CBR & compaction
A field density test on a lean-clay roadway subgrade lift gives a total (moist) unit weight of $\gamma_t=105.5$ pcf at a water content of $w=12.5\%$ . The standard Proctor test on the same soil gives a maximum dry unit weight of $\gamma_{d,\max}=108.0$ pcf at an optimum water content of $w_{opt}=16.5\%$ , and the specification requires 92% relative compaction. The relative compaction achieved is most nearly:
1. 83.8%
2. 86.8%
3. 94.4%
4. 97.7%
Show the answer & worked solution
Answer: B.86.8%
First the field DRY unit weight, with the lift's own water content (NCEES PE Civil Reference Handbook §3.9.1): $\gamma_d=\dfrac{\gamma_t}{1+w}=\dfrac{105.5}{1+0.125}=93.78$ pcf. Then relative compaction (§3.9.2): $RC=\dfrac{\gamma_{d,\text{field}}}{\gamma_{d,\max}}\times100=\dfrac{93.78}{108.0}\times100=86.8\%$ , which falls short of the 92% required. The lift was placed 4.0 points dry of optimum; the optimum water content describes the laboratory curve and never enters the ratio. Distractors: dried the field weight with the OPTIMUM water content, $105.5/1.165$ , instead of the lift's own $w$ , 83.8%; divided by the REQUIRED density, $92\%\times\gamma_{d,\max}$ , instead of by the Proctor maximum, 94.4% (that ratio measures the lift against the spec, not its relative compaction); and never removed the water, putting the TOTAL unit weight over the dry maximum, $105.5/108.0$ , 97.7%.
7
medium
A. Circular curve geometry & stationing
On a mountain two-lane highway, the PC of a simple circular curve is at Sta 24+63.90 and its PI is at Sta 28+00.00, where the tangents deflect $\Delta=70^\circ$ . The length of the curve is most nearly:
1. 672.2 ft
2. 586.4 ft
3. 550.6 ft
4. 410.6 ft
Show the answer & worked solution
Answer: B.586.4 ft
The PC lies on the back tangent, so the station difference is the tangent distance: Sta 28+00.00 − Sta 24+63.90 gives $T=336.10$ ft. From $T=R\tan(\Delta/2)$ (NCEES PE Civil Reference Handbook 5.2.1), $R=\dfrac{336.10}{\tan 35^\circ}=480$ ft, and the length of curve is $L=R\Delta\dfrac{\pi}{180}=480(70)\dfrac{\pi}{180}=586.4$ ft. Check: it falls between the long chord 550.6 ft and the tangent path $2T=672.2$ ft, as an arc must. Distractors: reported the LONG CHORD $C=2R\sin(\Delta/2)=550.63$ ft, the straight line from PC to PT, instead of the arc; measured PC $\to$ PI $\to$ PT along both TANGENTS, $2T=672.2$ ft, the path the curve replaces; and used the tangent distance as if it were the RADIUS, $T\Delta\tfrac{\pi}{180}=410.62$ ft.
8
medium
A. Highway hydrology & runoff
An interchange ramp catchment drains 2 acres of asphalt ramp pavement ( $C=0.9$ ) and 6 acres of turf infield ( $C=0.2$ ). The design rainfall intensity for the catchment's time of concentration is $I=4.2$ in/hr. By the rational method, the peak discharge at the outlet is most nearly:
1. 7.56 cfs
2. 12.6 cfs
3. 18.5 cfs
4. 30.2 cfs
Show the answer & worked solution
Answer: B.12.6 cfs
Rational formula with an area-weighted runoff coefficient (NCEES PE Civil Reference Handbook §6.5.2.1): $C_w=\dfrac{0.9(2)+0.2(6)}{8}=0.375$ , which sits between the two covers' coefficients and nearer the one with more area; then $Q=C_wIA=0.375(4.2)(8)=12.6$ cfs, taking 1 acre-in./hr as 1 cfs. Distractors: took the PLAIN average of the two coefficients, $(0.9+0.2)/2=0.55$ , ignoring the areas, $18.48$ cfs; let the pavement's $C=0.9$ govern the whole 8 ac, $30.24$ cfs; and counted only the 2 paved acres, as if the turf infield contributed nothing, $7.56$ cfs.
9
medium
A. Intersection sight distance & roundabout geometry
A driver stopped on a minor-road approach will turn left onto a two-lane highway with a design speed of 45 mph. The driver's eye is 20.0 ft from the center of the near highway lane. Looking left, the driver's line of sight just clears the corner of a building, which is 14.0 ft from that lane center and 120 ft from the driver's eye, measured parallel to the highway. For a vehicle approaching in the near lane at the design speed, the time gap the available sight distance provides is most nearly:
1. 1.81 s
2. 2.59 s
3. 6.05 s
4. 8.89 s
Show the answer & worked solution
Answer: C.6.05 s
The sight line from the eye past the corner reaches the near-lane center a distance $a$ up the highway. With the eye $b=20.0$ ft from the lane center and the corner $p=14.0$ ft from it and $q=120$ ft along the highway, similar triangles give $\dfrac{a}{b}=\dfrac{q}{b-p}$ , so $a=\dfrac{120(20.0)}{20.0-14.0}=400$ ft. At 45 mph $=45(1.47)=66.2$ ft/s, the time gap is $t=\dfrac{400}{66.2}=6.05$ s. That is less than the 7.5 s the AASHTO time-gap method gives a passenger car turning left from a stop onto a two-lane road (a clear sight distance of $1.47(45)(7.5)=496$ ft), so this sight line is deficient: the obstruction would have to be cleared. Distractors: stopped at the obstruction, using the corner's own 120 ft along the highway as the sight distance instead of extending the sight line to the approaching car's path (1.81 s); set up the similar triangles with the corner's offset from the LANE, $a=qb/p=171$ ft, where the corner's offset from the driver's eye, $b-p$ , belongs (2.59 s); and divided the 400 ft by the speed in mph, dropping the $1.47$ ft/s per mph conversion (8.89 s).
10
medium
A. Traffic loading (ESALs) & pavement structural design
A four-lane divided arterial carries a first-year ADT of 12,000 veh/day, 10% of it trucks averaging 0.45 ESAL per truck. Traffic is expected to grow 3% per year over a 20-year design period. Take the directional distribution factor $D_D=0.5$ and the lane distribution factor $D_L=0.9$ . The design-lane ESALs for the design period are most nearly:
1. 1,320,000 ESALs
2. 1,770,000 ESALs
3. 2,380,000 ESALs
4. 2,650,000 ESALs
Show the answer & worked solution
Answer: C.2,380,000 ESALs
Trucks: $12{,}000(0.1)=1{,}200$ per day, so the first-year ESALs (both directions) are $1{,}200(0.45)(365)=197{,}100$ . Growing 3% a year for 20 years, they accumulate by the uniform-series compound-amount factor, $G=(F/A)=\dfrac{(1+0.03)^{20}-1}{0.03}=26.8704$ (NCEES PE Civil Reference Handbook §1.7 and its factor table; this is the growth factor $G$ in the handbook's ESAL accumulation equation), so the design period carries $197{,}100(26.8704)=5{,}296{,}156$ ESALs in both directions. In the design lane (§5.1.6): $\text{ESAL}_{DL}=D_D\,D_L\,\text{ESAL}=0.5(0.9)(5{,}296{,}156)=2{,}383{,}270$ , which is 2,380,000 ESALs. Distractors: left out the lane distribution factor $D_L$ (2,650,000 ESALs); multiplied by the 20 years with NO growth, $G=n$ (1,770,000 ESALs); and read the PRESENT-worth column $(P/A)=14.88$ for the growth factor; the ESALs accumulate to the END of the period, so it is $(F/A)$ (1,320,000 ESALs).

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