The PE Civil Transportation depth exam demands far more than textbook recall—it requires you to apply AASHTO design standards, HCM analysis procedures, and hydraulic equations under real time pressure across 40 depth questions. The problems below mirror the format and rigor of the actual NCEES exam.
These 10 practice problems span the five major topic areas tested on the PE Civil Transportation depth section: traffic engineering, geometric design, signal timing, pavement design, and drainage. Each includes four answer choices and a detailed, step-by-step solution that walks you through every calculation.
Tip: Work each problem on your own before checking the solution. On the real exam you will average roughly 6 minutes per depth question, so hold yourself to that pace here.
Problem 1: Stopping Sight Distance on an Upgrade
A two-lane highway has a design speed of 55 mph and a sustained upgrade of +3%. Using a perception-reaction time of 2.5 seconds and a deceleration rate corresponding to a/g = 0.35, what is most nearly the required stopping sight distance (SSD)?
Show Solution
Answer: B
The AASHTO stopping sight distance formula is:
SSD = 1.47Vt + V² / [30(a/g ± G)]
where V is speed in mph, t is perception-reaction time in seconds, a/g is the deceleration ratio, and G is the grade expressed as a decimal. On an upgrade, gravity assists braking, so we add G in the denominator.
Step 1 — Perception-reaction distance:
d_r = 1.47 × 55 × 2.5 = 202.1 ft
Step 2 — Braking distance:
d_b = 55² / [30 × (0.35 + 0.03)]
d_b = 3,025 / [30 × 0.38]
d_b = 3,025 / 11.4 = 265.4 ft
Step 3 — Total SSD:
SSD = 202.1 + 265.4 = 467.5 ft ≈ 468 ft
Key concept: On an upgrade the grade term is added to the friction factor because gravity helps slow the vehicle. On a downgrade, you would subtract G, which increases the required SSD. Always check the sign convention carefully—this is one of the most common exam mistakes.
Problem 2: Minimum Horizontal Curve Radius
A rural highway has a design speed of 60 mph. The maximum superelevation rate is 0.06 and the side friction factor is 0.12. What is most nearly the minimum radius of a horizontal curve for this highway?
Show Solution
Answer: C
The AASHTO point-mass equilibrium equation for horizontal curves is:
e + f = V² / (15R)
Solving for R:
R = V² / [15(e + f)]
Step 1 — Sum the lateral resistance factors:
e + f = 0.06 + 0.12 = 0.18
Step 2 — Compute the minimum radius:
R = 60² / (15 × 0.18)
R = 3,600 / 2.70
R = 1,333 ft
Key concept: The formula R = V²/(15(e+f)) uses V in mph and gives R in feet. The factor of 15 incorporates the unit conversion and gravitational constant. Remember that higher speeds or lower (e+f) values require larger radii.
Problem 3: Crest Vertical Curve Length Using K-Value
A crest vertical curve connects a +2.0% grade to a −3.0% grade on a highway with a design speed of 50 mph. The required K-value for a crest vertical curve at 50 mph is 84. What is most nearly the minimum length of the vertical curve?
Show Solution
Answer: C
The minimum length of a vertical curve is found from:
L = K × A
where K is the design rate of vertical curvature (ft per percent change in grade) and A is the algebraic difference in grades.
Step 1 — Compute the algebraic difference in grades:
A = |G1 − G2| = |+2.0 − (−3.0)| = |+2.0 + 3.0| = 5.0%
Step 2 — Compute the minimum curve length:
L = 84 × 5.0 = 420 ft
Key concept: Always compute A as the absolute value of the algebraic difference: G1 minus G2. For a crest curve, G1 is positive and G2 is negative (or less positive), so A is their sum in magnitude. The K-value tables in AASHTO’s Green Book (or the CERM) are indexed by design speed and are different for crest versus sag curves.
Problem 4: Level of Service at a Signalized Intersection
A traffic analysis of a signalized intersection yields an average control delay of 42 seconds per vehicle. According to HCM criteria for signalized intersections, what level of service (LOS) does this represent?
Show Solution
Answer: B
The Highway Capacity Manual (HCM) defines LOS for signalized intersections based on average control delay per vehicle:
- LOS A: ≤ 10 s/veh
- LOS B: > 10 and ≤ 20 s/veh
- LOS C: > 20 and ≤ 35 s/veh
- LOS D: > 35 and ≤ 55 s/veh
- LOS E: > 55 and ≤ 80 s/veh
- LOS F: > 80 s/veh
Step 1 — Compare delay to thresholds:
42 s/veh falls in the range > 35 and ≤ 55 s/veh.
Therefore, the intersection operates at LOS D.
Key concept: Memorize the HCM delay thresholds for signalized intersections—they are tested frequently. Note that the LOS criteria differ for unsignalized intersections (which use different delay ranges) and for freeway segments (which use density in pc/mi/ln). Do not mix up the tables.
Problem 5: Webster’s Optimal Signal Cycle Length
A signalized intersection operates with three phases. The total lost time per cycle is 12 seconds. The critical flow ratios for the three phases are 0.30, 0.25, and 0.20. Using Webster’s formula, what is most nearly the optimal cycle length?
Show Solution
Answer: C
Webster’s formula for the optimal cycle length that minimizes total intersection delay is:
C_o = (1.5L + 5) / (1 − Y)
where L is the total lost time per cycle and Y is the sum of the critical-phase flow ratios.
Step 1 — Compute Y (sum of critical flow ratios):
Y = 0.30 + 0.25 + 0.20 = 0.75
Step 2 — Substitute into Webster’s formula:
C_o = (1.5 × 12 + 5) / (1 − 0.75)
C_o = (18 + 5) / 0.25
C_o = 23 / 0.25 = 92 s
Key concept: Webster’s formula is only valid when Y < 1.0. As Y approaches 1.0, the optimal cycle length grows toward infinity, indicating the intersection cannot handle the demand. Also note that L typically equals the number of phases multiplied by the lost time per phase (e.g., 3 phases × 4 s = 12 s).
Problem 6: Required Superelevation Rate
A horizontal curve on a two-lane highway has a radius of 1,200 ft and a design speed of 60 mph. If the side friction factor at this speed is 0.11, what is most nearly the required superelevation rate?
Show Solution
Answer: B
From the AASHTO horizontal curve equation:
e + f = V² / (15R)
Solve for the superelevation rate e:
e = V² / (15R) − f
Step 1 — Compute V²/(15R):
V² / (15R) = 60² / (15 × 1,200) = 3,600 / 18,000 = 0.20
Step 2 — Subtract the friction factor:
e = 0.20 − 0.11 = 0.09
A superelevation rate of 9% is within the typical maximum of 8–12% used by most state DOTs. In practice, you would round to the nearest standard increment (typically 2% increments) and verify it does not exceed the agency’s maximum allowable superelevation rate.
Key concept: This problem is the inverse of Problem 2. Given the radius, you solve for the required superelevation. On the PE exam, always double-check that your computed e does not exceed the maximum superelevation for the roadway class (e_max is commonly 0.06, 0.08, 0.10, or 0.12 depending on the agency and road type).
Problem 7: Peak Runoff Using the Rational Method
A 30-acre commercial drainage basin has a runoff coefficient of 0.65. The rainfall intensity for the design storm at the time of concentration is 4.2 in/hr. Using the Rational Method, what is most nearly the peak runoff?
Show Solution
Answer: C
The Rational Method equation is:
Q = C × i × A
where Q is peak runoff in cfs, C is the dimensionless runoff coefficient, i is rainfall intensity in inches per hour, and A is drainage area in acres. When using these units, Q comes out directly in cubic feet per second (the conversion factor of 1.008 is close enough to 1.0 that it is omitted by convention).
Step 1 — Substitute the given values:
Q = 0.65 × 4.2 × 30
Step 2 — Calculate:
Q = 0.65 × 4.2 = 2.73
Q = 2.73 × 30 = 81.9 cfs
Key concept: The Rational Method is valid for small drainage areas (typically under 200 acres). The critical design rainfall intensity corresponds to a storm duration equal to the time of concentration. A common exam mistake is to forget that “Q = CiA” already gives cfs when i is in in/hr and A is in acres—no additional unit conversion is needed.
Problem 8: Flow Capacity Using Manning’s Equation
A 24-inch-diameter concrete storm drain pipe (Manning’s n = 0.013) is laid on a slope of 0.5%. If the pipe is flowing full, what is most nearly the discharge capacity?
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Answer: B
Manning’s equation for open-channel flow (US customary units) is:
Q = (1.49 / n) × A × R2/3 × S1/2
where n is the roughness coefficient, A is the cross-sectional flow area (ft²), R is the hydraulic radius (ft), and S is the slope (ft/ft).
Step 1 — Convert diameter to feet and compute geometry for a full pipe:
D = 24 in = 2.0 ft
A = πD² / 4 = π(2.0)² / 4 = π = 3.1416 ft²
For a circular pipe flowing full, the hydraulic radius is:
R = D / 4 = 2.0 / 4 = 0.50 ft
Step 2 — Compute R2/3 and S1/2:
R2/3 = 0.502/3 = 0.6300
S1/2 = 0.0051/2 = 0.07071
Step 3 — Substitute into Manning’s equation:
Q = (1.49 / 0.013) × 3.1416 × 0.6300 × 0.07071
Q = 114.6 × 3.1416 × 0.6300 × 0.07071
Q = 114.6 × 3.1416 = 360.1
Q = 360.1 × 0.6300 = 226.9
Q = 226.9 × 0.07071 = 16.0 cfs
Key concept: For a circular pipe flowing full, R = D/4. The factor 1.49 is used for US customary units; if working in SI, use 1.00 instead. Also remember that maximum discharge in a partially full circular pipe actually occurs at about 93% full, not 100%—a favorite exam trick question.
Problem 9: AASHTO Flexible Pavement Structural Number
A flexible pavement section consists of 3.5 inches of hot-mix asphalt (HMA) surface (layer coefficient a1 = 0.44), 8.0 inches of granular base (a2 = 0.14, drainage coefficient m2 = 1.0), and 10.0 inches of granular subbase (a3 = 0.11, drainage coefficient m3 = 0.8). What is most nearly the structural number (SN) of this pavement?
Show Solution
Answer: B
The AASHTO 1993 flexible pavement design method uses the structural number equation:
SN = a1D1 + a2D2m2 + a3D3m3
where ai is the layer coefficient, Di is the layer thickness in inches, and mi is the drainage coefficient for unbound layers.
Step 1 — HMA surface layer (no drainage coefficient for bound layers):
a1D1 = 0.44 × 3.5 = 1.54
Step 2 — Granular base layer:
a2D2m2 = 0.14 × 8.0 × 1.0 = 1.12
Step 3 — Granular subbase layer:
a3D3m3 = 0.11 × 10.0 × 0.8 = 0.88
Step 4 — Sum the contributions:
SN = 1.54 + 1.12 + 0.88 = 3.54
Key concept: The drainage coefficient m applies only to unbound (granular) layers, not to asphalt or portland cement concrete. A value of m = 1.0 indicates good drainage conditions; values below 1.0 penalize the structural contribution of that layer for poor drainage. The PE exam may give you a required SN and ask you to determine the minimum layer thickness—solve each layer from the top down.
Problem 10: Intersection Sight Distance at a Stop-Controlled Intersection
At a two-way stop-controlled intersection, a vehicle on the minor street must turn left onto the major street. The design speed on the major street is 40 mph. The departure sight triangle requires a time gap of 7.5 seconds for a left turn from stop. What is most nearly the required intersection sight distance (ISD) along the major street?
Show Solution
Answer: B
The AASHTO intersection sight distance for a vehicle departing from a stop is based on the distance a vehicle on the major road travels during the required gap time:
ISD = 1.47 × Vmajor × tgap
where Vmajor is the major-street design speed in mph, tgap is the required time gap in seconds, and 1.47 converts mph to ft/s.
Step 1 — Substitute the given values:
ISD = 1.47 × 40 × 7.5
Step 2 — Calculate:
ISD = 1.47 × 40 = 58.8 ft/s
ISD = 58.8 × 7.5 = 441 ft
A clear sight triangle of at least 441 ft must be maintained along the major street in both directions from the minor-street stop bar.
Key concept: The time gap tgap depends on the maneuver: 6.5 s for a right turn or through movement from a stop, 7.5 s for a left turn across one lane of opposing traffic. Add 0.5 s for each additional lane the vehicle must cross. These values are tabulated in AASHTO’s A Policy on Geometric Design of Highways and Streets (the “Green Book”).
How to Use These Problems Effectively
Simply reading solutions is not enough—active practice is what builds exam-day fluency. Here are three strategies to get the most out of this problem set:
- Simulate exam conditions. Work all 10 problems in one sitting with only the PE Civil Reference Handbook open. Give yourself 60 minutes total.
- Diagnose your weaknesses. If you missed the Manning’s or Rational Method problems, drill more hydraulics. If the geometric design problems tripped you up, revisit the AASHTO formulas until you can set them up from memory.
- Redo problems you got wrong. Wait 48 hours, then attempt the missed problems again without looking at the solutions. Repetition builds the automaticity you need under time pressure.
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