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

What is the degree of curve (arc definition) for a horizontal curve with radius 1,000 ft?

• a.11.46 degrees
• b.5.73 degrees✓
• c.2.86 degrees
• d.1.0 degrees

By the arc definition, D = 5,729.58 / R = 5,729.58 / 1,000 = 5.73 degrees. The degree of curve is the central angle subtending a 100-ft arc; a larger radius gives a flatter (smaller-degree) curve.

Stopping sight distance (SSD) is defined as the sum of the brake-reaction distance and which other component?

• a.The braking (deceleration) distance✓
• b.The roadway grade
• c.The horizontal curve length
• d.The superelevation runoff

SSD = brake-reaction distance (driver perception-reaction time times speed) + braking distance (distance to decelerate to a stop). Both components grow with speed, so SSD increases sharply at higher design speeds.

Using the fundamental traffic-flow relationship, what is the flow rate for a density of 40 veh/mi and a space-mean speed of 60 mph?

• a.2,400 veh/hr✓
• b.1.5 veh/hr
• c.240 veh/hr
• d.100 veh/hr

The fundamental relationship is flow = density x speed: q = k·v = 40 x 60 = 2,400 veh/hr. This links the three primary traffic-stream variables and defines capacity at the flow's peak.

On a horizontal curve, superelevation (banking of the roadway) is provided primarily to counteract what?

• a.The stopping sight distance
• b.The lateral (centripetal) demand of vehicles rounding the curve✓
• c.The vertical grade
• d.The pavement thickness requirement

Superelevation tilts the roadway inward so that a component of the vehicle's weight, together with side friction, supplies the centripetal force needed to turn. This reduces reliance on tire friction and improves safety and comfort on curves.

A crest vertical curve is 200 ft long and connects grades that differ by an algebraic total of 4%. What is the rate of vertical curvature, K?

• a.4
• b.800
• c.50✓
• d.0.02

K = L / A, where L is the curve length in feet and A is the algebraic difference in grades in percent. K = 200 / 4 = 50 ft per percent. K expresses the horizontal distance needed to change the grade by 1% and is checked against sight-distance minimums.

A horizontal curve has a degree of curve of 2 degrees (arc definition). What is its radius?

• a.11,459 ft
• b.1,432.4 ft
• c.2,864.79 ft✓
• d.5,730 ft

By the arc definition, R = 5,729.58 / D = 5,729.58 / 2 = 2,864.79 ft. The degree of curve is the central angle subtending a 100-ft arc; radius is inversely proportional to it.

A horizontal curve has a deflection (intersection) angle of 30 degrees and a degree of curve of 5 degrees (arc definition). What is the length of the curve?

• a.150 ft
• b.600 ft✓
• c.6 ft
• d.6,000 ft

For the arc definition, L = 100·(Delta/D) = 100 x (30/5) = 600 ft. The curve length is proportional to the deflection angle and inversely proportional to the degree of curve.

A horizontal curve has a radius of 500 ft and an intersection angle of 40 degrees. What is the tangent length?

• a.364 ft
• b.420 ft
• c.182 ft✓
• d.146 ft

Tangent length T = R·tan(Delta/2) = 500 x tan(20 deg) = 500 x 0.3640 = 182 ft. The tangent runs from the PC (or PT) to the PI.

For a horizontal curve with R = 500 ft and intersection angle 40 degrees, what is the external distance?

• a.64.2 ft
• b.32.1 ft✓
• c.100 ft
• d.16.0 ft

External distance E = R[sec(Delta/2) - 1] = 500 x (1/cos 20 deg - 1) = 500 x 0.0642 = 32.1 ft. It is the distance from the PI to the middle of the curve.

For a horizontal curve with R = 500 ft and intersection angle 40 degrees, what is the middle ordinate?

• a.60.3 ft
• b.30.2 ft✓
• c.15.1 ft
• d.100 ft

Middle ordinate M = R[1 - cos(Delta/2)] = 500 x (1 - cos 20 deg) = 500 x 0.0603 = 30.2 ft. It is the distance from the midpoint of the long chord to the midpoint of the curve.

For a horizontal curve with R = 500 ft and intersection angle 40 degrees, what is the length of the long chord?

• a.684 ft
• b.342 ft✓
• c.383 ft
• d.171 ft

Long chord LC = 2R·sin(Delta/2) = 2 x 500 x sin 20 deg = 1,000 x 0.3420 = 342 ft. It connects the PC to the PT directly.

A curve is designed for 60 mph on a road with a maximum side-friction factor of 0.12 and a radius of 1,500 ft. What superelevation rate is required? (Use e + f = V^2/(15R), V in mph, R in ft.)

• a.0.12
• b.0.28
• c.0.16
• d.0.04✓

e + f = V^2/(15R) = 60^2/(15 x 1,500) = 3,600/22,500 = 0.16. Then e = 0.16 - f = 0.16 - 0.12 = 0.04 (4%). Side friction carries part of the lateral demand; the rest is met by banking.

What is the minimum radius for a 70-mph curve with maximum superelevation e = 0.08 and side-friction factor f = 0.10? (Use R = V^2/[15(e + f)].)

• a.907 ft
• b.2,722 ft
• c.272 ft
• d.1,815 ft✓

R_min = V^2/[15(e + f)] = 70^2/[15(0.08 + 0.10)] = 4,900/2.70 = 1,815 ft. The minimum radius drops as the allowable superelevation and side friction increase.

A driver travels at 60 mph with a perception-reaction time of 2.5 s and a braking friction factor of 0.35 on a level road. What is the stopping sight distance? (Use SSD = 1.47Vt + V^2/[30f].)

• a.784 ft
• b.220 ft
• c.343 ft
• d.563 ft✓

SSD = brake-reaction distance + braking distance = 1.47 x 60 x 2.5 + 60^2/(30 x 0.35) = 220.5 + 342.9 = 563 ft. Using only one component (220 or 343) is the common error.

At 50 mph with a perception-reaction time of 2.5 s, what is the brake-reaction (perception-reaction) distance?

• a.367 ft
• b.184 ft✓
• c.110 ft
• d.73.5 ft

Reaction distance d = 1.47·V·t = 1.47 x 50 x 2.5 = 183.75 ft (the 1.47 converts mph to ft/s). The vehicle travels this far before braking even begins.

A vehicle at 55 mph brakes on a 3% downgrade with a friction factor of 0.35. What is the braking distance? (Use d = V^2/[30(f + G)], G negative downhill.)

• a.275 ft
• b.288 ft
• c.350 ft
• d.315 ft✓

d = V^2/[30(f + G)] = 55^2/[30(0.35 - 0.03)] = 3,025/9.6 = 315 ft. A downgrade reduces the effective deceleration, lengthening the braking distance relative to the level value (288 ft).

A crest vertical curve must provide a stopping sight distance of 500 ft with an algebraic grade change of A = 5%. Assuming S < L, what curve length is required? (Use L = A·S^2/2158.)

• a.579 ft✓
• b.290 ft
• c.158 ft
• d.1,158 ft

For S < L on a crest curve, L = A·S^2/2158 = 5 x 500^2/2158 = 1,250,000/2158 = 579 ft. The 2158 constant embeds the standard driver eye height and object height.

A sag vertical curve must provide 400 ft of headlight sight distance with A = 4%. Assuming S < L, what curve length is required? (Use L = A·S^2/[400 + 3.5S].)

• a.200 ft
• b.178 ft
• c.356 ft✓
• d.711 ft

For a sag curve controlled by headlight distance (S < L), L = A·S^2/(400 + 3.5S) = 4 x 400^2/(400 + 1,400) = 640,000/1,800 = 356 ft. The 400 and 3.5S terms model the headlight height and 1-degree upward beam.

A vertical curve is designed with a rate of vertical curvature K = 158 for an algebraic grade change of A = 5%. What is the required curve length?

• a.395 ft
• b.163 ft
• c.31.6 ft
• d.790 ft✓

L = K·A = 158 x 5 = 790 ft. K is the horizontal distance needed to change the grade by 1 percent, so multiplying by the total grade change gives the length.

A crest vertical curve 500 ft long joins a +3% grade to a -2% grade. How far from the beginning of the curve (BVC) is the high point?

• a.500 ft
• b.250 ft
• c.200 ft
• d.300 ft✓

The turning point is at x = g1·L/(g1 - g2) = 3 x 500/(3 - (-2)) = 1,500/5 = 300 ft from the BVC. The high (or low) point is where the curve's slope reaches zero.

In the fundamental traffic-flow relationship q = k·v, the speed term v refers to:

• a.the space-mean speed✓
• b.the maximum speed
• c.the posted speed limit
• d.the time-mean speed

The flow-density-speed identity q = k·v uses the space-mean speed (harmonic mean over a length of road). The time-mean speed (measured at a point) is always slightly higher and does not satisfy the identity exactly.

A roadway carries a flow of 1,800 veh/hr at a space-mean speed of 45 mph. What is the traffic density?

• a.25 veh/mi
• b.40 veh/mi✓
• c.81,000 veh/mi
• d.400 veh/mi

From q = k·v, density k = q/v = 1,800/45 = 40 veh/mi. Density is vehicles per unit length; multiplying flow by speed (81,000) is the classic mistake.

At a density of 40 veh/mi, what is the average spacing between vehicles?

• a.66 ft
• b.264 ft
• c.40 ft
• d.132 ft✓

Average spacing = 5,280 ft/mi / density = 5,280/40 = 132 ft per vehicle. Spacing (center-to-center distance) is the reciprocal of density expressed in feet.

A lane carries 1,200 veh/hr. What is the average time headway between vehicles?

• a.0.33 s
• b.1,200 s
• c.30 s
• d.3 s✓

Average headway = 3,600 s/hr / flow = 3,600/1,200 = 3 s. Time headway is the reciprocal of flow rate; capacity conditions correspond to the minimum safe headway.

Under the Greenshields linear model with a free-flow speed of 60 mph and a jam density of 200 veh/mi, what is the maximum flow (capacity)?

• a.1,500 veh/hr
• b.6,000 veh/hr
• c.12,000 veh/hr
• d.3,000 veh/hr✓

For Greenshields, q_max = v_f·k_j/4 = 60 x 200/4 = 3,000 veh/hr. Capacity occurs at half the free-flow speed and half the jam density.

Under the Greenshields model with a free-flow speed of 60 mph, at what speed does maximum flow (capacity) occur?

• a.45 mph
• b.30 mph✓
• c.60 mph
• d.15 mph

In the Greenshields model, capacity occurs at the optimum speed v_o = v_f/2 = 60/2 = 30 mph (and density k_j/2). Flow falls off on either side of this point.

At the jam density on a highway, the speed and flow are respectively:

• a.maximum and maximum
• b.both zero✓
• c.zero and maximum
• d.maximum and zero

At jam density vehicles are bumper-to-bumper and stopped, so speed is zero; with zero speed the flow q = k·v is also zero. Maximum flow occurs at an intermediate density, not at jam.

A location has an hourly volume of 1,800 veh and a peak 15-minute count of 500 veh. What is the peak-hour factor?

• a.0.90✓
• b.0.45
• c.1.11
• d.0.72

PHF = hourly volume / (4 x peak 15-min volume) = 1,800/(4 x 500) = 1,800/2,000 = 0.90. A PHF near 1.0 indicates uniform flow; lower values indicate sharper peaking.

An intersection approach has an hourly volume of 1,800 veh and a peak-hour factor of 0.90. What is the equivalent peak flow rate?

• a.500 veh/hr
• b.1,620 veh/hr
• c.1,800 veh/hr
• d.2,000 veh/hr✓

Flow rate = hourly volume / PHF = 1,800/0.90 = 2,000 veh/hr. Dividing by the PHF converts a full-hour count into the flow rate sustained during the peak 15 minutes, which controls capacity analysis.

A highway has an AADT of 20,000 veh/day and a K-factor of 0.10. What is the design hourly volume?

• a.20,000 veh/hr
• b.2,000 veh/hr✓
• c.1,000 veh/hr
• d.200,000 veh/hr

DHV = AADT x K = 20,000 x 0.10 = 2,000 veh/hr. The K-factor is the proportion of daily traffic occurring in the design (typically 30th-highest) hour.

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.