Civil Engineering PE Exam — Questions and Answers

Question 1: What does the overconsolidation ratio (OCR) represent, and what OCR value defines a normally consolidated soil?

OCR = σ'v0/σ'c; OCR = 1 for normally consolidated
OCR = e0/emin; OCR = 1 for normally consolidated
OCR = σ'c/σ'v0; OCR = 1 for normally consolidated (Correct answer)
OCR = σ'c/σ'v0; OCR > 1 for normally consolidated

Correct answer: OCR = σ'c/σ'v0; OCR = 1 for normally consolidated

OCR = preconsolidation pressure / current effective vertical stress; OCR = 1 means the soil has never experienced stress greater than current.

Question 2: For a sluice gate in a wide rectangular channel with upstream depth y₁ = 2.0 m, gate opening a = 0.3 m, and Cc = 0.62, discharge per unit width assuming negligible velocity of approach is most nearly:

0.74 m²/s
1.17 m²/s (Correct answer)
1.47 m²/s
1.85 m²/s

Correct answer: 1.17 m²/s

q = Cc·a·√(2g·y₁) = 0.62×0.3×√(2×9.81×2.0) = 0.186×6.264 ≈ 1.17 m²/s.

Question 3: Which type of lagoon uses algae to supply oxygen to heterotrophic bacteria for BOD removal?

Polishing lagoon
Facultative lagoon (Correct answer)
Anaerobic lagoon
Aerated lagoon

Correct answer: Facultative lagoon

Facultative lagoons have an aerobic upper zone where algae supply oxygen via photosynthesis and an anaerobic lower zone for sludge digestion.

Question 4: A freeway weaving section has a weaving volume of 1,200 vph and a non-weaving volume of 3,000 vph. The volume ratio (VR) used in HCM weaving analysis is:

0.250
0.400
0.286 (Correct answer)
0.167

Correct answer: 0.286

VR = weaving volume / total volume = 1,200/(1,200+3,000) = 1,200/4,200 ≈ 0.286.

Question 5: A pump is installed in a pipeline to move water from a lower reservoir to a higher reservoir. Which of the following best describes the effect of the pump on the Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)?

An abrupt vertical rise in both the EGL and HGL at the pump's location. (Correct answer)
A rise in the EGL but a drop in the HGL at the pump's location.
An abrupt vertical drop in both the EGL and HGL at the pump's location.
A gradual, steady rise in both the EGL and HGL over the length of the pump.

Correct answer: An abrupt vertical rise in both the EGL and HGL at the pump's location.

A pump adds energy to the fluid in the system. This addition of energy head (pump head, Hp) is represented as a sudden, abrupt vertical rise in both the Energy Grade Line (EGL) and the Hydraulic Grade Line (HGL) at the physical location of the pump. The EGL represents the total energy, and the HGL represents the piezometric head. Since the pump increases the total energy, both lines must jump upwards. The magnitude of the jump is equal to the head added by the pump.

Question 6: Negative skin friction (downdrag) on a deep pile occurs when:

The pile is loaded beyond its axial capacity
The pile end bears directly on bedrock
Surrounding consolidating soil settles more than the pile (Correct answer)
The pile is subjected to uplift tension forces

Correct answer: Surrounding consolidating soil settles more than the pile

Downdrag develops when settling soil moves downward relative to the pile shaft, imposing an additional downward drag force on the pile.

Question 7: The influence line for the force in a diagonal tension member of a Pratt truss shows that the member carries tension when the live load is placed:

Across the entire span
Only directly over the member
In the panel where the member is located and extending to one support (Correct answer)
Only at the joint where the member terminates

Correct answer: In the panel where the member is located and extending to one support

For a diagonal tension member in a Pratt truss, the critical live load position occupies the panel containing the member and extends to the nearest support.

Question 8: A project manager is facing a situation where several key engineering staff are overallocated across multiple concurrent tasks. The project has a fixed, non-negotiable completion date. Which resource optimization technique would be most appropriate to use?

Fast Tracking
Resource Leveling
Resource Smoothing (Correct answer)
Crashing

Correct answer: Resource Smoothing

Resource smoothing is used to optimize resources when the project completion date is fixed. It adjusts activities within their available float to balance resource demand, avoiding peaks and troughs without changing the critical path or extending the project duration. Resource leveling, in contrast, can extend the project schedule to resolve resource over-allocations.

Question 9: A wastewater treatment plant achieves 90% BOD removal. If influent BOD = 250 mg/L, what is the effluent BOD?

25 mg/L (Correct answer)
225 mg/L
10 mg/L
125 mg/L

Correct answer: 25 mg/L

Effluent BOD = influent × (1 − removal efficiency) = 250 × (1 − 0.90) = 250 × 0.10 = 25 mg/L.

Question 10: A bid bond submitted with a construction bid typically guarantees that the bidder will:

Maintain the bid price for 90 days after award
Enter into a contract and provide performance/payment bonds if awarded (Correct answer)
Pay all subcontractors and suppliers
Complete the project within the bid price

Correct answer: Enter into a contract and provide performance/payment bonds if awarded

A bid bond ensures the bidder will execute the contract and furnish required performance and payment bonds if their bid is accepted.

Question 11: In a critical path method (CPM) network diagram for a bridge construction project, an activity has an early start (ES) of day 10, a late start (LS) of day 15, an early finish (EF) of day 20, and a late finish (LF) of day 25. What is the total float for this activity?

15 days
5 days (Correct answer)
0 days
10 days

Correct answer: 5 days

Total float is the amount of time that a schedule activity can be delayed without delaying the project finish date. It is calculated as the difference between the late start and early start (LS - ES) or the late finish and early finish (LF - EF). In this case, Float = 15 - 10 = 5 days, or Float = 25 - 20 = 5 days. Activities on the critical path have zero float.

Question 12: The dissolved oxygen (DO) in an aeration basin of an activated sludge system is typically maintained at:

10–12 mg/L
1.0–3.0 mg/L (Correct answer)
0.0–0.5 mg/L (anoxic)
5.0–8.0 mg/L

Correct answer: 1.0–3.0 mg/L

Conventional activated sludge systems are designed to maintain 1.0–3.0 mg/L DO to ensure aerobic conditions without excessive energy consumption.

Question 13: Bishop's simplified method of slices assumes that:

The failure surface is planar
Pore water pressure is negligible
All interslice forces are zero
Interslice shear forces are zero (Correct answer)

Correct answer: Interslice shear forces are zero

Bishop's simplified method neglects interslice shear forces while retaining normal interslice forces, improving accuracy over the ordinary method.

Question 14: When eccentricity of loading on a footing exceeds B/6, the pressure distribution beneath the footing:

Remains uniform but shifts toward the load resultant
Is ignored and replaced with a uniform pressure
Causes immediate punching shear failure by code definition
Develops tensile stress on one side, meaning only part of the footing is in contact with soil (Correct answer)

Correct answer: Develops tensile stress on one side, meaning only part of the footing is in contact with soil

When e > B/6, the resultant falls outside the kern, causing uplift (tension) on the far edge — since soil cannot take tension, the effective footing area reduces.

Question 15: Which ACI 318 provision governs the maximum reinforcement ratio for a tension-controlled singly reinforced beam?

ρmax = 0.75ρb
Net tensile strain εt ≥ 0.004 (Correct answer)
Net tensile strain εt ≥ 0.005
As,max = 0.04bwd

Correct answer: Net tensile strain εt ≥ 0.004

ACI 318-19 replaced the 0.75ρb rule; Section 9.3.3 now requires εt ≥ 0.004 at nominal strength for non-prestressed flexural members.

Question 16: Tension cracks at the crest of a cohesive slope reduce stability primarily because they:

Increase cohesion along the failure surface
Increase the effective normal stress on the failure plane
Reduce the weight of the sliding mass
Can fill with water, adding a hydrostatic driving force (Correct answer)

Correct answer: Can fill with water, adding a hydrostatic driving force

Water-filled tension cracks add a lateral hydrostatic force that increases the driving moment and reduces the factor of safety.

Question 17: A welded connection uses E70XX electrodes. The nominal tensile strength of the weld metal Fexx is:

50 ksi
70 ksi (Correct answer)
90 ksi
60 ksi

Correct answer: 70 ksi

The designation E70XX means the electrode has a minimum tensile strength of 70 ksi.

Question 18: A saturated soil specimen in an unconsolidated-undrained (UU) triaxial test shows zero friction angle (φ = 0). If the unconfined compressive strength (qu) is 80 kPa, what is the undrained shear strength (su)?

40 kPa (Correct answer)
160 kPa
80 kPa
20 kPa

Correct answer: 40 kPa

For φu = 0, su = cu = qu/2 = 80/2 = 40 kPa; the Mohr circle radius equals the shear strength.

Question 19: A traverse line runs from Point A (N 100.0, E 200.0) to Point B (N 240.0, E 350.0). What is the bearing of line AB?

N 47°00' W
S 47°00' E
N 47°00' E (Correct answer)
N 43°36' E

Correct answer: N 47°00' E

Bearing = N arctan(ΔE/ΔN) E = N arctan(150/140) = N arctan(1.071) ≈ N 47°00' E.

Question 20: Flow in a steep channel (S₀ > Sc) with depth greater than critical depth carries which GVF profile classification?

S1 (Correct answer)
S3
S2
M1

Correct answer: S1

An S1 profile occurs on steep slopes when y > yc; depth decreases downstream toward normal depth, with flow remaining supercritical.

Question 21: In a PERT analysis, an activity has an optimistic duration of 4 days, a most likely duration of 7 days, and a pessimistic duration of 16 days. What is the expected duration?

7.0 days
7.5 days
9.0 days
8.0 days (Correct answer)

Correct answer: 8.0 days

PERT expected duration = (O + 4M + P)/6 = (4 + 4×7 + 16)/6 = (4 + 28 + 16)/6 = 48/6 = 8 days.

Question 22: A spillway has discharge coefficient Cd = 0.85, head H = 1.5 m, and length L = 10 m. The discharge over the spillway is most nearly:

83 m³/s
99 m³/s
116 m³/s
66 m³/s (Correct answer)

Correct answer: 66 m³/s

Q = Cd(2/3)√(2g)·L·H^(3/2) = 0.85×(2/3)×4.429×10×1.837 ≈ 66 m³/s.

Question 23: For a spirally reinforced column, the minimum spiral reinforcement ratio ρs per ACI 318 is:

0.45(Ag/Ac - 1)(f'c/fyt) (Correct answer)
0.12f'c/fyt
0.25(Ag/Ac)(f'c/fyt)
0.01(Ag/Ac)(fy/f'c)

Correct answer: 0.45(Ag/Ac - 1)(f'c/fyt)

ACI 318 Section 25.7.3.3 requires ρs ≥ 0.45(Ag/Ac − 1)(f'c/fyt), ensuring the spiral replaces the lost capacity of the shell.

Question 24: The modified Proctor compaction test (ASTM D1557) delivers approximately how many times more compactive energy than the standard Proctor (ASTM D698)?

2.5×
4.5× (Correct answer)
6.0×
1.5×

Correct answer: 4.5×

The modified Proctor uses a 10-lb hammer with an 18-in drop in 5 layers, delivering roughly 4.5 times the energy of the standard Proctor.

Question 25: The traffic intensity (ρ) in a D/D/1 queuing model where arrival rate λ = 800 veh/h and service rate μ = 1,000 veh/h is:

0.50
0.80 (Correct answer)
1.00
1.25

Correct answer: 0.80

Traffic intensity ρ = λ/μ = 800/1,000 = 0.80; since ρ < 1, the queue is stable.

Question 26: The primary purpose of a pile load test (ASTM D1143) is to:

Measure pile installation energy during driving
Verify pile capacity and measure load-settlement behavior at the specific site (Correct answer)
Determine soil stratification and SPT N-values
Evaluate corrosion potential of the pile material

Correct answer: Verify pile capacity and measure load-settlement behavior at the specific site

Static pile load tests confirm that the installed pile achieves the required capacity and document the load-settlement curve for design verification.

Question 27: The design of a left-turn bay requires a storage length based on the number of vehicles expected per cycle. If a cycle is 90 s, the arrival rate is 120 vph, and average vehicle length + spacing is 25 ft, the required storage length is:

100 ft
125 ft
50 ft
75 ft (Correct answer)

Correct answer: 75 ft

Vehicles per cycle = 120×(90/3600) = 3 vehicles; storage = 3×25 = 75 ft.

Question 28: In the analysis of a space (3D) truss, the determinacy condition is m + r = 3j. A space truss has 30 members, 6 reactions, and 12 joints. The truss is:

Statically determinate (Correct answer)
Statically indeterminate to 1st degree
A mechanism
Statically indeterminate to 3rd degree

Correct answer: Statically determinate

For a 3D truss: m + r = 30 + 6 = 36 = 3j = 3(12) = 36, confirming the truss is statically determinate.

Question 29: The balanced reinforcement ratio ρb for a singly reinforced beam with f'c = 4000 psi and fy = 60,000 psi is approximately:

0.0320
0.0180
0.0285
0.0214 (Correct answer)

Correct answer: 0.0214

ρb = (0.85β1 f'c/fy) × (87,000/(87,000+fy)) = (0.85×0.85×4/60) × (87/147) ≈ 0.0285 × 0.592 ≈ 0.0214.

Question 30: A contractor submits a bid of $2,400,000 for a project. If the contractor's overhead is 12% and profit is 8% of the total bid, what is the estimated direct cost?

$1,800,000
$1,680,000
$2,016,000
$1,920,000 (Correct answer)

Correct answer: $1,920,000

Direct cost = $2,400,000 × (1 - 0.12 - 0.08) = $2,400,000 × 0.80 = $1,920,000.

Question 31: The Equivalent Single Axle Load (ESAL) for a tandem axle group weighing 34,000 lb using the AASHTO Load Equivalency Factor (LEF) tables is approximately:

1.11 (Correct answer)
0.62
2.44
3.27

Correct answer: 1.11

Per AASHTO tables, a 34,000 lb tandem axle group has an LEF of approximately 1.11 ESALs for a standard pavement SN.

Question 32: A steel plate girder with a very slender web uses transverse stiffeners to develop post-buckling tension field action. This behavior is described by:

St. Venant torsion resisting applied shear
Reduced effective width method for web compression
Euler column theory applied to web panels
Wagner's tension field theory, where diagonal tension carries shear after web buckling (Correct answer)

Correct answer: Wagner's tension field theory, where diagonal tension carries shear after web buckling

Tension field action (Wagner theory) allows the web to carry shear through diagonal tensile stresses after the web has buckled, significantly increasing shear capacity.

Question 33: A rectangular primary sedimentation basin at a water treatment plant is 100 ft long, 30 ft wide, and has a water depth of 12 ft. If the plant flow rate is 3.0 MGD (Million Gallons per Day), what is the surface overflow rate?

83,333 gal/day
1,000 gpd/ft² (Correct answer)
1,250 gpd/ft²
83 gpd/ft²

Correct answer: 1,000 gpd/ft²

The surface overflow rate (SOR) is calculated by dividing the flow rate by the surface area of the basin. First, calculate the surface area: Area = Length × Width = 100 ft × 30 ft = 3,000 ft². Next, convert the flow rate to gallons per day: Flow = 3.0 MGD = 3,000,000 gpd. Finally, calculate the SOR: SOR = Flow / Area = 3,000,000 gpd / 3,000 ft² = 1,000 gpd/ft². The depth of the basin is not used in this calculation.

Question 34: When was this university's course established?

1717
1818
1838 (Correct answer)
1840

Correct answer: 1838

This question refers to the establishment of the Civil Engineering and Mining Class at King's College London. This significant program, which contributed to the formalization of engineering education in the UK, was established in 1838.

Question 35: According to HCM, the peak hour factor (PHF) for a road with a peak hour volume of 1,200 vph and a peak 15-minute volume of 350 vehicles is:

0.91
0.86 (Correct answer)
0.78
0.82

Correct answer: 0.86

PHF = Peak Hour Volume / (4 × Peak 15-min Volume) = 1200 / (4 × 350) = 1200/1400 ≈ 0.86.

Question 36: A roundabout entry has a flow of 400 vph and circulating flow of 600 vph. Using FHWA capacity formula (entry capacity ≈ 1,130 − 0.7 × circulating flow), the entry capacity is closest to:

840 vph
710 vph (Correct answer)
530 vph
900 vph

Correct answer: 710 vph

Entry capacity = 1,130 − 0.7 × 600 = 1,130 − 420 = 710 vph, indicating the entry is within capacity.

Question 37: Which estimating method uses historical cost data from similar completed projects as its primary basis?

Parametric estimating
Bottom-up estimating
Analogous estimating (Correct answer)
Three-point estimating

Correct answer: Analogous estimating

Analogous estimating relies on actual cost data from previous similar projects to estimate future project costs.

Question 38: The minimum passing sight distance (PSD) for a design speed of 55 mph on a two-lane highway per AASHTO is approximately:

800 ft
2,000 ft
1,500 ft (Correct answer)
1,000 ft

Correct answer: 1,500 ft

AASHTO specifies a minimum PSD of approximately 1,500 ft for a 55 mph design speed on two-lane highways.

Question 39: The factor of safety for slope stability is defined as the ratio of:

Resisting forces to driving forces (Correct answer)
Driving forces to resisting forces
Cohesion to friction angle
Normal stress to shear stress

Correct answer: Resisting forces to driving forces

Factor of safety equals resisting moment (or force) divided by driving moment (or force), with FS > 1 indicating a stable slope.

Question 40: The MUTCD requires that a 'School Speed Limit' sign be used only when the speed limit applies during specific times. Which beacon type is typically used with school speed limit assemblies?

Flashing yellow LED beacon (Correct answer)
Red flashing beacon
White strobe beacon
Green steady beacon

Correct answer: Flashing yellow LED beacon

MUTCD recommends flashing yellow LED beacons with school speed limit signs to alert drivers when the reduced speed is in effect.

Question 41: The dimensionless time factor Tv in Terzaghi's consolidation theory is defined as:

Tv = cv / (t × Hdr²)
Tv = cv × t / Hdr² (Correct answer)
Tv = Hdr² / (cv × t)
Tv = t / (cv × Hdr)

Correct answer: Tv = cv × t / Hdr²

Tv = cv·t/Hdr², where cv is the coefficient of consolidation and Hdr is the length of the drainage path.

Question 42: For a doubly reinforced concrete beam, the compression steel stress f's is typically determined by:

Setting f's = 0.85f'c
Using strain compatibility with εs' = (c - d')/c × 0.003 (Correct answer)
Using f's = 0.5fy as an ACI simplification
Assuming f's = fy always

Correct answer: Using strain compatibility with εs' = (c - d')/c × 0.003

Strain compatibility requires εs' = 0.003(c − d')/c, then f's = Es×εs' ≤ fy to verify whether compression steel yields.

Question 43: An engineer is calculating the required yellow interval for a signalized intersection approach. The design speed is 40 mph, the driver perception-reaction time is 1.0 second, and the comfortable deceleration rate is 10 ft/s². The approach is on a +2% (uphill) grade. Using the standard kinematic equation, what is the minimum required yellow interval?

3.8 s (Correct answer)
2.8 s
4.5 s
5.1 s

Correct answer: 3.8 s

The formula for the yellow interval (Y) is: Y = t + V / (2a + 2Gg), where t = perception-reaction time, V = approach speed, a = deceleration rate, G = grade, and g = acceleration due to gravity (32.2 ft/s²). First, convert speed: V = 40 mph * 1.467 (ft/s)/mph ≈ 58.7 ft/s. Then, plug in the values: Y = 1.0 s + 58.7 ft/s / (2 * 10 ft/s² + 2 * 0.02 * 32.2 ft/s²). Y = 1.0 + 58.7 / (20 + 1.288) = 1.0 + 58.7 / 21.288 = 1.0 + 2.76 s ≈ 3.76 s. The closest answer is 3.8 s.

Question 44: Which ASCE 7 load combination governs for uplift on a roof during a wind event?

0.9D + 1.0W (Correct answer)
1.2D + 1.0E
1.4D
1.2D + 1.6W

Correct answer: 0.9D + 1.0W

The combination 0.9D + 1.0W uses minimum dead load to maximize net uplift, which governs when wind causes upward (negative) pressure on a roof.

Question 45: The NRCS Curve Number (CN) method estimates:

Groundwater recharge volume
Evapotranspiration losses only
Direct runoff depth from storm rainfall depth (Correct answer)
Peak discharge rate only

Correct answer: Direct runoff depth from storm rainfall depth

The CN method calculates direct runoff Q from rainfall P using Q = (P − 0.2S)²/(P + 0.8S), where S is the potential maximum retention based on CN.

Question 46: The Schedule Performance Index (SPI) for a project is 0.75. This means the project is:

75% complete
25% ahead of schedule
Progressing at 75% of the planned rate (Correct answer)
25% over budget

Correct answer: Progressing at 75% of the planned rate

SPI = EV/PV = 0.75 means for every $1.00 of work planned, only $0.75 worth of work has been accomplished.

Question 47: The AASHTO flexible pavement design uses the Structural Number (SN). If layer coefficients are a1=0.44, a2=0.14, a3=0.11, drainage coefficients m2=m3=1.0, and layer thicknesses D1=4 in, D2=6 in, D3=8 in, the SN is:

4.12
2.96
3.48 (Correct answer)
5.20

Correct answer: 3.48

SN = a1D1 + a2D2m2 + a3D3m3 = 0.44×4 + 0.14×6×1.0 + 0.11×8×1.0 = 1.76 + 0.84 + 0.88 = 3.48.

Question 48: In a continuous beam analyzed by the three-moment equation (Clapeyron's theorem), the equation relates:

Moments at three consecutive supports to the applied loads and spans (Correct answer)
Deflections at midspan of adjacent bays
Shear forces at three consecutive sections
Reactions at three consecutive supports

Correct answer: Moments at three consecutive supports to the applied loads and spans

The three-moment equation (Clapeyron's) relates the bending moments at three consecutive supports M_A, M_B, M_C in terms of span lengths, loads, and EI values.

Question 49: In steel design, the shear factor of safety in ASD is typically set at approximately:

1.50 (Fv = 0.45Fy)
2.5 (Fv = 0.33Fy)
1.67 (Fv = 0.40Fy) (Correct answer)
2.0 (Fv = 0.50Fy)

Correct answer: 1.67 (Fv = 0.40Fy)

ASD allowable shear stress is Fv = 0.40Fy, corresponding to a safety factor of about 1.67 against shear yielding.

Question 50: A sand has a maximum void ratio of 0.82 and minimum void ratio of 0.45. A field sample has a void ratio of 0.60. What is the relative density (Dr)?

26.8%
73.2%
59.5% (Correct answer)
40.5%

Correct answer: 59.5%

Dr = (emax − e)/(emax − emin) × 100 = (0.82 − 0.60)/(0.82 − 0.45) × 100 = 0.22/0.37 × 100 ≈ 59.5%.

Question 51: Biochemical oxygen demand (BOD) measures:

Dissolved oxygen concentration in a receiving stream
Chemical oxygen demand of inorganic compounds
Oxygen consumed by microorganisms decomposing organic matter (Correct answer)
Total organic carbon concentration in wastewater

Correct answer: Oxygen consumed by microorganisms decomposing organic matter

BOD quantifies dissolved oxygen consumed by biological processes decomposing organic matter, typically measured over 5 days at 20°C (BOD₅).

Question 52: The slenderness ratio klu/r below which slenderness effects may be neglected for a braced (non-sway) reinforced concrete column is:

34 − 12(M1/M2) (Correct answer)
40
50
22

Correct answer: 34 − 12(M1/M2)

ACI 318 Section 6.2.5 allows slenderness to be neglected for braced columns when klu/r < 34 − 12(M1/M2), where M1/M2 is the end moment ratio.

Question 53: A truss bridge deck girder delivers concentrated loads only at panel points. This is a critical requirement because:

Loads applied between joints would cause bending in chord members, violating the two-force member assumption (Correct answer)
Diagonal members cannot carry transverse loads
Panel points are the strongest locations in the chord
Bridge codes prohibit mid-panel loading

Correct answer: Loads applied between joints would cause bending in chord members, violating the two-force member assumption

Classical truss analysis assumes members are two-force members (axial only); loading between joints introduces bending moments that the analysis ignores.

Question 54: The stopping sight distance (SSD) on a level road for a design speed of 60 mph, assuming perception-reaction time of 2.5 s and deceleration of 11.2 ft/s², is closest to:

645 ft
720 ft
490 ft
570 ft (Correct answer)

Correct answer: 570 ft

SSD = 1.47×60×2.5 + (88)²/(2×11.2) = 220.5 + 345.7 ≈ 566 ft, rounded to 570 ft per AASHTO.

Question 55: The critical chain project management method differs from CPM primarily because it:

Eliminates the need for a network diagram
Accounts for resource constraints and uses buffers rather than individual activity float (Correct answer)
Requires a fully resourced schedule before sequencing activities
Uses three-point duration estimates for all activities

Correct answer: Accounts for resource constraints and uses buffers rather than individual activity float

Critical chain focuses on resource-constrained scheduling and uses project buffers and feeding buffers instead of padding individual activity durations.

Question 56: The standard step method for GVF computation differs from the direct step method primarily because the standard step:

Only applies to prismatic channels while the direct step applies to non-prismatic channels
Specifies distance increments and iterates to find the corresponding depth (Correct answer)
Specifies depth increments and directly solves for distance
Uses the momentum equation while the direct step uses the energy equation

Correct answer: Specifies distance increments and iterates to find the corresponding depth

The standard step method fixes the distance increment and iterates to find depth at each section; it applies to non-prismatic channels where cross-section geometry varies.

Question 57: A 1-inch diameter A325 bolt in single shear has a nominal shear strength. The AISC nominal shear stress for A325 bolts in a bearing-type connection is approximately:

30 ksi
48 ksi (Correct answer)
60 ksi
20 ksi

Correct answer: 48 ksi

AISC Table J3.2 lists Fnv = 48 ksi for A325 bolts in bearing-type connections.

Question 58: Which of the following factors would most likely lead to a DECREASE in the coefficient of permeability (k) of a sandy soil?

A decrease in the average particle size (D10) (Correct answer)
An increase in the void ratio
An increase in the degree of saturation from 85% to 100%
An increase in the temperature of the permeating water

Correct answer: A decrease in the average particle size (D10)

The coefficient of permeability (k) is highly dependent on the size of the void spaces through which water flows. A decrease in the average particle size means the void spaces become smaller and more tortuous, significantly reducing the ease with which water can pass through, thus decreasing permeability. Conversely, a higher void ratio, full saturation (which eliminates air blockages), and higher water temperature (which reduces viscosity) all tend to increase the coefficient of permeability.

Question 59: Net allowable bearing capacity is calculated as:

qu minus the footing self-weight
(qu − q) / FS, where q is the overburden stress at footing depth (Correct answer)
qu times the factor of safety
qu divided by a factor of safety

Correct answer: (qu − q) / FS, where q is the overburden stress at footing depth

Net allowable bearing capacity = (qu − q)/FS, subtracting the existing overburden pressure before applying the safety factor.

Question 60: Torsional effects in a reinforced concrete beam can be neglected per ACI 318 when the factored torsional moment Tu is less than:

φ(√f'c/4)(Acp²/Pcp)
φ(0.85f'c)(Acp/Pcp)
φ(√f'c/12)(Acp²/Pcp) (Correct answer)
φ(√f'c)(Acp²/Pcp)

Correct answer: φ(√f'c/12)(Acp²/Pcp)

ACI 318 Section 9.5.4.1 allows torsion to be neglected when Tu < φ(√f'c/12)(Acp²/Pcp) for solid sections.

Question 61: In AISC Chapter F, which condition defines a compact section regarding the compression flange?

λ ≤ λp where λ = h/tw ≤ 2.24√(E/Fy)
λ ≤ λr where λ = bf/2tf ≤ 1.0√(E/Fy)
λ ≤ λp where λ = bf/2tf ≤ 0.38√(E/Fy) (Correct answer)
No flange slenderness limit applies to rolled W-shapes

Correct answer: λ ≤ λp where λ = bf/2tf ≤ 0.38√(E/Fy)

A compact flange requires bf/(2tf) ≤ 0.38√(E/Fy), allowing the section to reach the plastic moment before local buckling.

Question 62: Who was the first female civil engineer to earn a degree?

Nora Blatch (Correct answer)
Patricia Jacobs
Nelly Craig
Sharon Dufresne

Correct answer: Nora Blatch

Nora Blatch (later Nora Blatch de Forest and Nora Blatch Barney) is recognized as the first woman to earn a civil engineering degree. She graduated from Cornell University in 1905, marking a significant milestone for women in engineering and paving the way for future generations of female civil engineers.

Question 63: An NPDES permit under the Clean Water Act regulates:

Point source discharges of pollutants to navigable waters of the US (Correct answer)
Nonpoint source pollution from agricultural fields
Atmospheric deposition of nitrogen to water bodies
Underground injection of industrial wastes

Correct answer: Point source discharges of pollutants to navigable waters of the US

NPDES requires permits for any discrete point source discharging pollutants into waters of the United States, setting effluent limits and monitoring requirements.

Question 64: In designing a high-strength bolted slip-critical connection, the design slip resistance per bolt depends on:

Bolt diameter and grade only
Weld size and electrode classification
Class of faying surface, number of slip planes, and minimum bolt pretension (Correct answer)
Bearing strength of the connected material

Correct answer: Class of faying surface, number of slip planes, and minimum bolt pretension

Slip resistance = μ × Du × hf × Tb × ns, where μ is the mean slip coefficient for the surface class, Tb is pretension, and ns is the number of slip planes.

Question 65: A drilled shaft is constructed in a deep clay layer. The alpha (α) method is used to estimate skin friction. If Su = 1,200 psf and α = 0.55, the unit skin friction is:

660 psf (Correct answer)
540 psf
720 psf
840 psf

Correct answer: 660 psf

Unit skin friction = α × Su = 0.55 × 1,200 = 660 psf using the alpha (adhesion factor) method.

Question 66: Which of the following is classified as a systematic error in tape surveying?

Transposing digits when recording a field measurement
Steel tape length error due to temperature variation from the calibration temperature (Correct answer)
Random wind vibration shifting the tape during measurement
Misidentifying a survey monument in the field

Correct answer: Steel tape length error due to temperature variation from the calibration temperature

Temperature-induced tape length change is systematic because it follows a predictable formula (αLΔT) and can be fully corrected.

Question 67: A two-lane highway has a directional split of 60/40 and a PHF of 0.90. The directional design hourly volume (DDHV) based on an AADT of 12,000 vpd and a K-factor of 0.10 is:

800 vph
648 vph
540 vph
720 vph (Correct answer)

Correct answer: 720 vph

DDHV = AADT × K × D = 12,000 × 0.10 × 0.60 = 720 vph.

Question 68: What is the mean cell residence time (MCRT or SRT) if the aeration basin volume is 500,000 gallons, MLSS = 2,500 mg/L, daily sludge wasting = 5,000 lb/day, and effluent TSS is negligible?

~4.2 days
~6.3 days
~2.1 days (Correct answer)
~8.4 days

Correct answer: ~2.1 days

Mass in basin = 500,000 gal × 2,500 mg/L × 8.34×10⁻⁶ lb·L/(mg·gal) ≈ 10,425 lb; SRT = 10,425/5,000 ≈ 2.1 days.

Question 69: Darcy's Law for groundwater flow through a porous medium is expressed as:

Q = A / (K × i)
Q = K × A / i
Q = i / (K × A)
Q = K × i × A (Correct answer)

Correct answer: Q = K × i × A

Darcy's Law states Q = KiA, where K is hydraulic conductivity (ft/day or cm/s), i is the hydraulic gradient (Δh/ΔL), and A is the cross-sectional area.

Question 70: According to ACI 318, the minimum clear cover for reinforcement in a concrete beam exposed to weather (deicing salts) is:

2.5 in (Correct answer)
2.0 in
1.5 in
3.0 in

Correct answer: 2.5 in

ACI 318 Table 20.6.1.3 requires 2.5 in clear cover for #6 through #11 bars exposed to weather (deicing salts).

Question 71: A construction manager uses the S-curve on a project. What does the S-curve represent?

Activity relationships in a network diagram
Cumulative planned versus actual cost or work over time (Correct answer)
Resource histogram showing daily labor needs
Risk probability plotted against impact

Correct answer: Cumulative planned versus actual cost or work over time

An S-curve plots cumulative cost or work over time, typically forming an S-shape due to slow start, rapid mid-project progress, and tapering completion.

Question 72: Which of the following is the primary purpose of creating a Work Breakdown Structure (WBS) in project planning?

To identify and quantify all potential project risks and their impacts.
To assign specific tasks to individual team members.
To define the total scope of the project by breaking it down into smaller, manageable components. (Correct answer)
To sequence all project activities chronologically.

Correct answer: To define the total scope of the project by breaking it down into smaller, manageable components.

The Work Breakdown Structure (WBS) is a foundational project management tool used to define and organize the total scope of a project. It is a hierarchical decomposition of the project into smaller, more manageable deliverables or work packages. The WBS ensures that all required work is identified and serves as the basis for subsequent planning activities like scheduling, cost estimating, and resource allocation.

Question 73: In situ air sparging remediates contaminated groundwater by:

Installing a permeable reactive barrier to intercept the plume
Adding chemical oxidants to destroy contaminants in place
Injecting air below the water table to volatilize dissolved contaminants for SVE capture (Correct answer)
Pumping groundwater to the surface and treating it above grade

Correct answer: Injecting air below the water table to volatilize dissolved contaminants for SVE capture

Air sparging injects compressed air into the saturated zone, stripping volatile organic compounds into the vapor phase where soil vapor extraction (SVE) collects them.

Question 74: A 20-ft × 20-ft raft foundation on soft clay settles 2 inches at the center. To reduce this settlement by 50%, an engineer could:

Replace 6 ft of soil with compacted structural fill to reduce stress on the compressible layer (Correct answer)
Increase column spacing to redistribute load
Add waterproofing membrane beneath the raft
Increase the raft thickness by 50%

Correct answer: Replace 6 ft of soil with compacted structural fill to reduce stress on the compressible layer

Partial soil replacement (surcharge reduction) decreases net stress on the compressible stratum, directly reducing consolidation settlement.

Question 75: A steel column with KL/r = 80 and Fy = 50 ksi falls into which AISC buckling category?

Elastic buckling (KL/r > 113)
Post-buckling reserve applies
No buckling concern (KL/r < 25)
Inelastic buckling (KL/r < 4.71√(E/Fy) ≈ 113) (Correct answer)

Correct answer: Inelastic buckling (KL/r < 4.71√(E/Fy) ≈ 113)

For Fy = 50 ksi, the transition slenderness is 4.71√(E/Fy) = 4.71√(29000/50) ≈ 113; KL/r = 80 is in the inelastic buckling range.

Question 76: The minimum acceptable factor of safety against sliding for a retaining wall is typically:

1.5 (Correct answer)
3.0
1.0
2.0

Correct answer: 1.5

AASHTO and most codes require FS ≥ 1.5 against sliding, computed as total horizontal resisting forces divided by total active horizontal thrust.

Question 77: A closed traverse has a linear misclosure of 0.42 ft and a total perimeter of 1,260 ft. What is the precision ratio?

1:3,000 (Correct answer)
1:1,500
1:4,000
1:2,000

Correct answer: 1:3,000

Precision = 1:(perimeter/misclosure) = 1:(1,260/0.42) = 1:3,000.

Question 78: When preparing a construction cost estimate, which factor is included in the general conditions (indirect costs) rather than direct costs?

Structural steel material
Concrete placement labor
Project superintendent salary (Correct answer)
Subcontractor work

Correct answer: Project superintendent salary

Project superintendent salary is an indirect (general conditions) cost, while direct costs include materials, labor, and subcontracted work tied to specific work items.

Question 79: In water treatment, what is the purpose of breakpoint chlorination?

To add enough chlorine to oxidize all ammonia-nitrogen and achieve free chlorine residual (Correct answer)
To reduce chlorine to below 0.2 mg/L for taste control
To form chloramines for distribution system residual
To lower the pH before coagulation

Correct answer: To add enough chlorine to oxidize all ammonia-nitrogen and achieve free chlorine residual

At the breakpoint, all chloramines and organic chlorine compounds are destroyed and any additional chlorine appears as free residual (HOCl/OCl⁻).

Question 80: A project has Total Float = 0 and Free Float = 0 on the critical path. If an activity on the critical path is delayed by 5 days, the project completion date will be delayed by:

Less than 5 days
5 days (Correct answer)
More than 5 days
0 days

Correct answer: 5 days

Any delay on the critical path directly extends the project completion date by the same duration since there is no float available.

Civil Engineering PE Exam

The NCEES PE Civil exam is a computer-based licensure exam for civil engineers. It consists of 80 questions across a chosen specialty area (Construction, Geotechnical, Structural, Transportation, or Water Resources & Environmental). Examinees have 9 hours to complete the exam. The pass rate ranges from 49–65% depending on specialty. Effective April 2024, the standalone breadth section was eliminated — all questions are now specialty-depth focused.

Questions

540 min

Time Limit

70%

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