PE Civil (NCEES Principles and Practice of Engineering) — All Questions

For an AADT of 20,000, a K-factor of 0.10, and a directional split D = 0.60, what is the directional design hourly volume?

• a.3,333 veh/hr
• b.1,200 veh/hr✓
• c.12,000 veh/hr
• d.2,000 veh/hr

DDHV = AADT x K x D = 20,000 x 0.10 x 0.60 = 1,200 veh/hr. The D-factor apportions the design-hour volume to the heavier direction of travel, sizing each roadway direction.

In the Highway Capacity Manual, level of service (LOS) for a facility is defined primarily by:

• a.the construction cost per lane-mile
• b.a measure of effectiveness such as density, speed, or delay experienced by users✓
• c.the pavement condition rating
• d.the posted speed limit

LOS grades (A through F) describe operating quality using service measures like density (freeways), control delay (signalized intersections), or average speed. It reflects the driver's experience, not physical or cost attributes.

On an upgrade, heavy trucks are accounted for in capacity analysis by using passenger-car equivalents (PCE) that are:

• a.equal to 1.0 for all vehicles
• b.zero for trucks
• c.greater than 1.0, because a truck occupies more effective road space than a car✓
• d.less than 1.0 on grades

A PCE converts a truck into an equivalent number of passenger cars. On grades, trucks slow down and take up more effective space, so their PCE exceeds 1.0, which reduces the effective capacity of the facility.

A signalized intersection runs a 4-phase cycle with 4 seconds of lost time per phase. What is the total lost time per cycle?

• a.16 s✓
• b.4 s
• c.20 s
• d.8 s

Total lost time L = (lost time per phase) x (number of phases) = 4 x 4 = 16 s. Lost time (startup plus clearance) is unusable green and directly reduces intersection capacity.

A signal phase has a displayed green of 30 s and a yellow of 4 s, with 4 s of lost time. What is the effective green time?

• a.26 s
• b.34 s
• c.30 s✓
• d.38 s

Effective green g = G + Y - lost time = 30 + 4 - 4 = 30 s. Drivers use part of the yellow and lose part of the green to startup; here they happen to offset.

A lane group has a saturation flow rate of 1,900 veh/hr and a green ratio g/C of 0.40. What is its capacity?

• a.1,900 veh/hr
• b.4,750 veh/hr
• c.380 veh/hr
• d.760 veh/hr✓

Capacity c = s x (g/C) = 1,900 x 0.40 = 760 veh/hr. A movement can discharge at the saturation rate only during its effective green, so capacity scales with the green ratio.

A lane group has a demand of 600 veh/hr and a capacity of 760 veh/hr. What is its degree of saturation (v/c ratio)?

• a.0.60
• b.0.79✓
• c.1.27
• d.1.00

Degree of saturation X = v/c = 600/760 = 0.79. Values below 1.0 indicate the movement operates under capacity; as v/c approaches 1.0, delay grows rapidly.

According to delay models such as Webster's, as the degree of saturation (v/c) approaches 1.0, the average control delay:

• a.stays constant
• b.increases sharply toward very large values✓
• c.decreases toward zero
• d.becomes negative

Delay rises nonlinearly and approaches very large values as v/c nears and exceeds 1.0, because arrivals overwhelm the available capacity and queues fail to clear each cycle. This steep rise defines the onset of LOS F.

By the AASHTO fourth-power law for flexible pavements, doubling the axle load increases the pavement damage (load equivalency) by a factor of about:

• a.16✓
• b.4
• c.2
• d.8

Pavement damage grows roughly with the fourth power of axle load, so doubling the load multiplies damage by 2^4 = 16. This is why heavy trucks, not cars, govern pavement design (ESALs).

A rigid pavement differs from a flexible pavement primarily in that it:

• a.uses a Portland cement concrete slab that distributes load through beam (slab) action✓
• b.relies only on asphalt surface friction
• c.has no structural layers
• d.cannot carry truck traffic

A rigid (concrete) pavement spreads wheel loads over a wide area via the flexural stiffness of the slab, so the subgrade sees low pressure. A flexible (asphalt) pavement distributes load through successive granular layers to the subgrade.

The California Bearing Ratio (CBR) test is used to characterize:

• a.the strength/bearing capacity of subgrade and base materials✓
• b.the concrete compressive strength
• c.the asphalt binder viscosity
• d.the traffic volume

CBR compares a soil's penetration resistance to that of a standard crushed-stone base, giving a strength index used in pavement thickness design. Higher CBR subgrades need thinner pavement structures.

A flexible pavement has an asphalt layer (a1 = 0.44, D1 = 4 in) over a base (a2 = 0.14, D2 = 8 in, drainage coefficient 1.0). What is the structural number SN?

• a.4.40
• b.2.88✓
• c.1.76
• d.12.0

SN = a1·D1 + a2·D2·m2 = 0.44(4) + 0.14(8)(1.0) = 1.76 + 1.12 = 2.88. The structural number sums each layer's coefficient times its thickness (adjusted for drainage) to represent the pavement's total structural capacity.

For an asphalt binder graded PG 64-22, the number 64 represents:

• a.the minimum temperature in Fahrenheit
• b.the maximum pavement design temperature in degrees Celsius the binder can withstand✓
• c.the binder viscosity in poise
• d.the aggregate size in mm

In a Superpave performance grade PG H-L, the first number (64) is the high pavement temperature (deg C) at which the binder resists rutting; the second (-22) is the low temperature at which it resists thermal cracking.

The Marshall stability test is used in the design and evaluation of:

• a.asphalt (hot-mix) concrete mixtures✓
• b.subgrade soils
• c.steel reinforcement
• d.Portland cement concrete

The Marshall method measures the stability and flow of compacted asphalt specimens to select optimum binder content. It is a mix-design test for hot-mix asphalt, not for concrete or soils.

Traffic-control-device standards and signal warrants in the United States are established by:

• a.the National Electrical Code
• b.OSHA regulations
• c.the Manual on Uniform Traffic Control Devices (MUTCD)✓
• d.the Uniform Building Code

The MUTCD (FHWA) defines the standards, warrants, and specifications for signs, signals, and pavement markings so that traffic control is uniform nationwide. Signal installation must satisfy MUTCD warrants.

Compared with stopping sight distance, the passing sight distance required on a two-lane highway is generally:

• a.much smaller
• b.the same
• c.much greater✓
• d.always zero

Passing sight distance must cover the passing vehicle accelerating around a slower vehicle plus a safe clearance from oncoming traffic, so it is several times the stopping sight distance for the same design speed.

A horizontal curve has R = 500 ft and a sight line along a 400-ft arc. What is the required horizontal sightline offset (middle ordinate to a lateral obstruction)? (Use M = R[1 - cos(28.65·S/R)].)

• a.100 ft
• b.78.5 ft
• c.19.6 ft
• d.39.3 ft✓

HSO = R[1 - cos(28.65·S/R)] = 500[1 - cos(28.65 x 400/500)] = 500(1 - cos 22.92 deg) = 500(0.0786) = 39.3 ft. This clearance to obstructions (walls, barriers) preserves sight distance around the curve.

A 4,000-lb vehicle sits on a 5% grade. What is the grade-resistance force acting along the roadway?

• a.20 lb
• b.2,000 lb
• c.200 lb✓
• d.400 lb

Grade resistance = W x G = 4,000 x 0.05 = 200 lb (for small grades, sin ~ tan = G). This longitudinal force must be overcome to climb, and it aids braking on the way down.

On a basic freeway segment with a 70-mph free-flow speed, the approximate per-lane base capacity is:

• a.about 5,000 pc/hr/ln
• b.about 2,400 pc/hr/ln✓
• c.about 800 pc/hr/ln
• d.about 1,000 pc/hr/ln

The Highway Capacity Manual gives a base capacity near 2,400 passenger cars per hour per lane at a 70-mph free-flow speed (dropping to about 2,250 at lower FFS). Capacity is expressed per lane in passenger-car units.

On a basic freeway segment, as the free-flow speed increases, the per-lane capacity generally:

• a.increases✓
• b.becomes negative
• c.decreases to zero
• d.stays exactly the same

Higher free-flow speeds (better geometry, wider lanes, fewer interruptions) support slightly higher per-lane capacities in the HCM freeway model. The relationship is positive though the gain is modest.

Level of service F on a facility corresponds to conditions where:

• a.demand exceeds capacity (v/c > 1) and flow breaks down with queuing✓
• b.the road is empty
• c.only trucks are present
• d.traffic flows freely at high speed

LOS F represents breakdown: arriving demand exceeds capacity, queues form and grow, and speeds collapse. It is the failing condition that capacity improvements aim to avoid.

A bottleneck serves 800 veh/hr while 1,000 veh/hr arrive for 1 hour (deterministic queuing). What is the maximum queue length at the end of that hour?

• a.125 veh
• b.1,800 veh
• c.200 veh✓
• d.1,000 veh

For deterministic queuing, queue growth = (arrival rate - service rate) x time = (1,000 - 800) x 1 = 200 vehicles. The queue accumulates at the difference between demand and capacity.

A parabolic vertical curve 400 ft long begins (BVC) at elevation 100.00 ft on a +2% grade and ends on a -2% grade. What is the curve elevation at 200 ft from the BVC?

• a.100.0 ft
• b.104.0 ft
• c.102.0 ft✓
• d.98.0 ft

Elevation = BVC + g1·x + [(g2 - g1)/(2L)]·x^2 = 100 + 0.02(200) + [(-0.04)/(800)](200^2) = 100 + 4 - 2 = 102.0 ft. The parabolic offset (-2 ft) pulls the point below the tangent elevation of 104 ft.

A vertical curve 400 ft (4 stations) long joins a +2% grade to a -2% grade. What is the rate of grade change per station?

• a.-4 %/station
• b.-0.01 %/station
• c.-1 %/station✓
• d.+1 %/station

Rate of change r = (g2 - g1)/L = (-2 - 2)/4 stations = -1 percent per station. The negative sign indicates a crest curve; the magnitude sets how quickly the grade transitions.

A reverse curve consists of:

• a.a single curve with a spiral
• b.two curves turning the same direction with different radii
• c.two circular curves of opposite direction (turning opposite ways) joined at a common tangent point✓
• d.a straight tangent only

A reverse curve has two arcs curving in opposite directions, one right and one left, meeting at a common point. A compound curve, by contrast, has two same-direction arcs of different radii.

The primary purpose of a spiral (transition) curve between a tangent and a circular curve is to:

• a.gradually introduce curvature and superelevation for a smooth, comfortable transition✓
• b.eliminate the need for superelevation
• c.shorten the overall alignment
• d.increase the design speed automatically

A spiral provides a continuously changing radius so lateral acceleration and superelevation build up gradually rather than abruptly at the PC. This improves driver comfort, safety, and the natural steering path.

On a horizontal curve, the superelevation runoff length is the distance required to:

• a.provide passing sight distance
• b.change the horizontal radius to zero
• c.stop a vehicle from the design speed
• d.transition the pavement cross-slope from zero (adverse crown removed) to full superelevation✓

Superelevation runoff is the length over which the outside lane rotates from a level (or crown-removed) cross-section to the full banked cross-slope of the curve. It is set to keep the edge-of-pavement grade change gradual.

When selecting a design vehicle for intersection turning templates, the largest turning radius is required by:

• a.a passenger car (P)
• b.a bicycle
• c.a large tractor-semitrailer (e.g., WB-67)✓
• d.a single-unit truck (SU)

Larger, longer design vehicles like the WB-67 tractor-semitrailer track wide and need the largest turning paths. Designing curb returns and turning lanes for the largest expected vehicle ensures smaller vehicles are accommodated too.

A vehicle accelerates uniformly from rest to 60 mph (88 ft/s) in 10 seconds. What is its acceleration?

• a.8.8 ft/s^2✓
• b.6.0 ft/s^2
• c.88 ft/s^2
• d.0.88 ft/s^2

a = (v - v0)/t = (88 - 0)/10 = 8.8 ft/s^2 (60 mph = 88 ft/s). Uniform acceleration is the change in velocity divided by the time interval.

A vehicle traveling 60 mph (88 ft/s) decelerates at 15 ft/s^2. What braking distance is needed to stop?

• a.516 ft
• b.129 ft
• c.88 ft
• d.258 ft✓

Using v^2 = v0^2 - 2a·d with final v = 0: d = v0^2/(2a) = 88^2/(2 x 15) = 7,744/30 = 258 ft. Braking distance grows with the square of speed.

The standard AASHTO perception-reaction time used for stopping sight distance design is:

• a.1.0 second
• b.2.5 seconds✓
• c.5.0 seconds
• d.0.5 second

AASHTO uses a 2.5-second brake-reaction (perception-reaction) time for stopping sight distance, which accommodates most drivers under most conditions. It combines the time to perceive, identify, decide, and initiate braking.

Skid marks 100 ft long are measured on a surface with a friction factor of 0.70. What was the approximate initial speed? (Use v = sqrt(30·f·d), v in mph, d in ft.)

• a.2,100 mph
• b.30 mph
• c.45.8 mph✓
• d.21 mph

v = sqrt(30·f·d) = sqrt(30 x 0.70 x 100) = sqrt(2,100) = 45.8 mph. This accident-reconstruction formula estimates speed from skid length and pavement friction.

For a fixed design vehicle and roadway, increasing the design speed generally requires the maximum allowable grade to be:

• a.lower (flatter)✓
• b.unchanged
• c.exactly zero
• d.higher (steeper)

Higher design speeds call for flatter maximum grades so that heavy vehicles can maintain speed on upgrades and stopping distances remain adequate on downgrades. AASHTO tables reduce allowable grade as design speed rises.

A roadway direction must carry a directional design hour volume of 1,800 veh/hr, and each lane provides a service flow of 900 veh/hr at the target LOS. How many lanes are needed in that direction?

• a.1 lane
• b.2 lanes✓
• c.4 lanes
• d.3 lanes

Number of lanes = DDHV / per-lane service flow = 1,800/900 = 2 lanes. Dividing the directional demand by the acceptable per-lane service volume gives the lanes required to hold the target level of service.

An approach has a design speed of 30 mph (44 ft/s), a driver perception-reaction time of 1.0 s, and a comfortable deceleration of 10 ft/s^2 on a level grade. What is the minimum yellow (change) interval? (Use y = t + V/[2a].)

• a.1.0 s
• b.2.2 s
• c.4.4 s
• d.3.2 s✓

Yellow change interval y = t + V/(2a) = 1.0 + 44/(2 x 10) = 1.0 + 2.2 = 3.2 s. It gives a driver time to react and either stop or clear the stop line, avoiding the dilemma zone.

A borrow soil has a bank (in-place) volume of 500 cubic yards and a shrinkage of 10%. What compacted fill volume will it produce?

• a.550 cubic yards
• b.405 cubic yards
• c.450 cubic yards✓
• d.500 cubic yards

Compacted volume = bank volume x (1 - shrinkage) = 500 x (1 - 0.10) = 450 CY. Shrinkage reflects densification of soil in a compacted embankment and governs how much borrow is required for a given fill.

An embankment requires 1,800 compacted cubic yards. If the borrow soil has a shrinkage of 10%, what bank volume must be excavated?

• a.1,800 cubic yards
• b.2,000 cubic yards✓
• c.1,980 cubic yards
• d.1,620 cubic yards

Bank volume = compacted volume / (1 - shrinkage) = 1,800 / 0.90 = 2,000 CY. You must excavate more bank material than the finished fill because compaction reduces the volume.

A soil has a bank volume of 2,000 cubic yards and a swell of 25%. What is the loose (hauled) volume?

• a.2,250 cubic yards
• b.1,600 cubic yards
• c.2,500 cubic yards✓
• d.2,000 cubic yards

Loose volume = bank volume x (1 + swell) = 2,000 x 1.25 = 2,500 CY. Excavation loosens soil so it occupies more volume; swell governs the number of truck loads to haul.

A soil swells 25% when excavated. What is its load factor?

• a.0.80✓
• b.1.00
• c.1.25
• d.0.75

Load factor = 1 / (1 + swell) = 1 / 1.25 = 0.80. Multiplying a loose (truck) volume by the load factor converts it back to bank volume, the basis for pay quantities.

A project must haul 2,500 loose cubic yards of soil in trucks that carry 25 loose cubic yards each. How many truck loads are required?

• a.100 loads✓
• b.50 loads
• c.80 loads
• d.125 loads

Number of loads = total loose volume / truck capacity = 2,500 / 25 = 100 loads. Hauling calculations must use loose (not bank) volume because trucks carry loosened, swelled material.

A front-end loader has a 3-cubic-yard bucket and completes one load cycle in 30 seconds. At 100% efficiency, what is its production rate?

• a.120 CY/hr
• b.720 CY/hr
• c.360 CY/hr✓
• d.180 CY/hr

Cycles per hour = 3,600 s / 30 s = 120; production = 120 x 3 CY = 360 CY/hr. Production is bucket volume times cycles per hour, before applying an efficiency factor.

Applying a 50-minute working hour (efficiency factor 50/60), what is the realistic production of a loader whose ideal output is 360 CY/hr?

• a.300 CY/hr✓
• b.432 CY/hr
• c.250 CY/hr
• d.360 CY/hr

Realistic production = ideal x efficiency = 360 x (50/60) = 300 CY/hr. The 50-minute-hour factor accounts for normal job and management delays within each clock hour.

A haul truck has a total cycle time of 24 minutes, and the loader fills it in 6 minutes. How many trucks are needed to keep the loader working continuously?

• a.6 trucks
• b.4 trucks✓
• c.3 trucks
• d.5 trucks

Number of trucks = truck cycle time / load time = 24 / 6 = 4 trucks. Matching the fleet to the loader's pace avoids the loader idling or trucks queuing.

A haul truck spends 6 min loading, 4 min hauling, 2 min dumping, and 3 min returning. What is its total cycle time?

• a.12 min
• b.18 min
• c.15 min✓
• d.20 min

Cycle time is the sum of all activity times: 6 + 4 + 2 + 3 = 15 min. The full cycle, not just haul time, sets how many trips a truck makes per hour.

A 20-CY truck has a cycle time of 15 minutes. What is its hauling production rate?

• a.20 CY/hr
• b.100 CY/hr
• c.80 CY/hr✓
• d.60 CY/hr

Trips per hour = 60 / 15 = 4; production = 4 x 20 = 80 CY/hr. Truck production equals capacity times trips per hour, driven by the complete cycle time.

In a CPM forward pass, an activity has an early start (ES) of day 5 and a duration of 8 days. What is its early finish (EF)?

• a.3
• b.13✓
• c.8
• d.40

Early finish = early start + duration = 5 + 8 = 13. The forward pass propagates ES and EF from project start to compute the earliest completion.

An activity has an early start of day 6 and a late start of day 10. What is its total float?

• a.6
• b.4✓
• c.0
• d.16

Total float = late start - early start = 10 - 6 = 4 days (equivalently LF - EF). It is the delay an activity can absorb without extending the project.

An activity finishes at early finish day 10, and its only successor has an early start of day 12. What is the activity's free float?

• a.12
• b.10
• c.2✓
• d.22

Free float = (earliest ES of successors) - EF = 12 - 10 = 2 days. Free float is the delay available without disturbing the early start of any following activity.

In CPM, the critical path is best defined as:

• a.the path containing the most activities
• b.the shortest path between start and finish
• c.the continuous path whose activities carry the greatest amount of total float in the network
• d.the longest continuous path of activities through the network✓

The critical path is the longest-duration path through the network; it sets the minimum project duration, and its activities have zero total float. Any delay on it delays the project.

A network has three paths from start to finish: A-B-C = 12 days, A-D-C = 15 days, and A-E = 9 days. What is the project duration?

• a.36 days
• b.15 days✓
• c.12 days
• d.9 days

Project duration equals the longest path (the critical path): A-D-C = 15 days. Paths are not summed; the network finishes only when its longest chain completes.