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Question 1Geotechnical Engineering

The Taylor stability chart is used to analyze:

  • A.Bearing capacity of strip footings and shallow foundations per Terzaghi theory
  • B.Liquefaction potential and cyclic behavior of saturated granular soils
  • C.Factor of safety for circular failure surfaces in cohesive slopes
  • D.Lateral earth pressure on retaining walls per Coulomb and Rankine theory
Show solution

Answer · C

Taylor stability chart: gives stability number N = c/(γH×FS) as function of slope angle and depth factor. Used for total stress analysis (φ=0, undrained). More rigorous: Bishop's simplified method, Spencer's method, Morgenstern-Price. Critical failure surface: minimum FS circle. Swedish circle: simpler but conservative.

Question 2Hydraulics & Hydrology

Drainage area A=25 acres, C=0.60, rainfall intensity i=4.2 in/hr. Calculate peak runoff Q = CiA.

  • A.48 cfs
  • B.63 cfs
  • C.91 cfs
  • D.35 cfs
Show solution

Answer · B

Q = 0.60 × 4.2 × 25 = 63 cfs. Rational Method assumes: uniform rainfall, peak runoff occurs when entire area contributes (time of concentration tc). Good for small urban areas <200 acres. For larger areas, use unit hydrograph or HEC-HMS.

Question 3Means & Methods

Why is fly ash commonly added to structural concrete?

  • A.To greatly increase early strength per ACI 232R and meet initial load requirements
  • B.To reduce heat of hydration and permeability while improving later-age strength
  • C.To automatically entrain air bubbles and improve freeze-thaw durability
  • D.To accelerate set time and increase workability for rapid construction
Show solution

Answer · B

Fly ash reacts pozzolanically with calcium hydroxide, reducing heat and permeability while enhancing long-term strength, though early strength gain is slower. Air entrainment requires an air-entraining agent, not fly ash.

Question 4Project Planning

Resource leveling is primarily used to:

  • A.Shorten the project duration
  • B.Smooth resource usage and reduce peaks
  • C.Eliminate float from noncritical activities
  • D.Lock activities to specific dates
Show solution

Answer · B

Resource leveling shifts noncritical work within available float to achieve a more even resource histogram. It may lengthen the schedule if resources are limited. Note: This type of problem requires multi-step analysis, careful unit tracking, and application of the relevant engineering/management standard. Always verify against the applicable code or reference before finalizing.

Question 5Soil Mechanics

Bjerrum correction for vane shear: VST gives su=55kPa, PI=40, Bjerrum mu=0.88. Design su?

  • A.su_design=48.4kPa
  • B.su_design=62.5kPa (over-corrected)
  • C.su_design=55kPa (no correction)
  • D.su_design=38.5kPa (excessive correction)
Show solution

Answer · A

su_design=mu*su_VST=0.88*55=48.4kPa. Answer A. The Bjerrum correction accounts for rate effects (vane test is faster than field loading) and anisotropy. For PI=40: mu~0.85-0.90. Always apply correction when using VST data for stability calculations.

Question 6Structural Engineering

A high-strength bolt (A325) in a bearing-type connection fails by:

  • A.Tension in bolt only, limited by 0.75FyAb per AISC M27 specification
  • B.Bolt tension induced by prying action under eccentric combined shear-tension loads
  • C.Shear through bolt shank or bearing/tearout of connected material
  • D.Corrosion failure causing material loss in high-strength bolt shank
Show solution

Answer · C

Bearing-type connections must check: (1) bolt shear (Rn = 0.6FuAbNv per LRFD), (2) bearing on connected material (Rn = 2.4FudbT), and (3) tearout (Rn = 1.2FuLcT). The failure mode with minimum capacity governs. Prying is critical only in slip-critical or tension-dominated cases, not routine bearing connections.

Question 7Structural Mechanics

Tied column 14x14in (e_min=1.0in), 8-#7 bars=4.8in2, f'c=4ksi, fy=60ksi. Nominal axial capacity Po and phi*Pn,max.

  • A.Po = 938 k; φPn,max = 488 k
  • B.Po = 938 k; φPn,max = 610 k (0.80 factor omitted)
  • C.Po = 681 k; φPn,max = 354 k (Ag = 12×14 in² error)
  • D.Po = 1,173 k; φPn,max = 610 k (f'c = 5 ksi entered)
Show solution

Answer · A

Po = 0.85f'c(Ag−Ast) + fy×Ast. Ag = 196 in². Ast = 8×0.60 = 4.80 in². Po = 0.85×4×191.2 + 60×4.80 = 650.1 + 288 = 938 k. φPn,max = 0.65 × 0.80 × 938 = 488 k → Answer A. Distractor B omits 0.80: 0.65×938 = 610 k. Distractor C uses 12-in width: Ag=168 in², Po=3.4×163.2+288=843 k; φPn=0.52×843=438 k. Distractor D uses f'c=5 ksi: Po=0.85×5×191.2+288=1,100 k; φPn=0.52×1,100=572 k — illustrates why reading f'c from drawings carefully matters.

Question 8Transportation Engineering

The AASHTO pavement design equation uses the concept of Equivalent Single Axle Load (ESAL). An 18-kip (80-kN) single axle has an ESAL of:

  • A.0.1
  • B.10
  • C.1.0
  • D.2.0
Show solution

Answer · C

The 18-kip (80-kN) single axle = 1 ESAL by definition. ESAL = (axle load/18 kips)^4 (approximately, load equivalency factor). A 9-kip axle = (9/18)^4 = 0.0625 ESAL. A 30-kip axle = (30/18)^4 = 7.7 ESAL. Heavier axles cause disproportionately more damage (4th power law). Design traffic = sum of ESALs over design life.

Question 9Water Resources Engineering

The Modified Puls method for reservoir routing is based on:

  • A.Manning's equation for open channel flow and velocity-depth relationships in reservoirs
  • B.Muskingum method combined with dynamic wave routing for channel reaches
  • C.Rational method applied directly to reservoir peak outflow calculations
  • D.Continuity equation relating inflow, outflow, and storage
Show solution

Answer · D

Modified Puls method applies continuity (ΔS/Δt = I − O) with a storage-indication curve relating (2S/Δt + O) to outflow O. This discretized form solves I₁ + I₂)/2 = (2S₁/Δt − O₁)/2 + (2S₂/Δt + O₂)/2 iteratively. Manning's equation governs spillway/outlet discharge, not the routing algorithm itself.

Question 10Geotechnical Engineering

The Rock Quality Designation (RQD) is determined by:

  • A.Schmidt hammer rebound test to assess rock surface hardness in geotechnical exploration
  • B.Rock classification from outcrop mapping and weathering assessment using GSI method
  • C.Point load strength test as specified in ASTM D5731 to evaluate intact rock strength
  • D.Percentage of intact core pieces ≥ 100 mm (4 in) from borehole drilling
Show solution

Answer · D

RQD = Σ(lengths of pieces ≥ 100 mm) / total core length × 100%. ASTM D6032 defines RQD from borehole core. Classification: 90-100% excellent, 75-90% good, 50-75% fair, 25-50% poor, <25% very poor. Used in RMR and Q-system for rock mass characterization and tunnel support design. Low RQD indicates more jointing and reduced strength.

Question 11Hydraulics & Hydrology

Pump must deliver 850 GPM against 65 ft total dynamic head (TDH). Calculate water horsepower WHP = Q×TDH/(3,960).

  • A.14.0 HP
  • B.18.5 HP
  • C.9.2 HP
  • D.22.0 HP
Show solution

Answer · A

WHP = (850×65)/3,960 = 55,250/3,960 = 13.95 HP ≈ 14.0 HP. Brake horsepower BHP = WHP/pump_efficiency. If pump is 75% efficient: BHP = 14.0/0.75 = 18.7 HP. Motor sized with service factor, typically select 20 HP motor.

Question 12Means & Methods

A Job Hazard Analysis (JHA) performed each morning primarily:

  • A.Serves as the primary replacement for OSHA mandatory safety training on construction sites
  • B.Identifies task-specific hazards and control measures before work begins
  • C.Eliminates the need for PPE by thoroughly documenting all potential hazards in advance
  • D.Sets daily production targets and scheduling priorities for crew workflow management
Show solution

Answer · B

JHAs (or JSAs) are task-level hazard reviews performed before work begins each day to identify hazards, assign control measures, and specify required PPE. They complement (not replace) formal OSHA training (A), work alongside PPE requirements (C), and focus on safety rather than scheduling (D). This daily ritual ensures crews start with situational awareness and documented controls.

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