FE Civil exam practice: practice problems and exam questions

Practice problems for the Fundamentals of Engineering (FE) Civil exam: statics and structures, geotechnics and settlement, hydraulics and open channels, engineering economics and statistics.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

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Topics in this course

  1. Beam support reactions1 point load · 2 point loads · point load + UDLcalculations
  2. Truss: method of jointssymmetric · asymmetric · vertical + horizontal loadcalculations
  3. Shear force & bending momentpoint load · full-span UDL · point load + UDLcalculations
  4. Beam deflectioncantilever, end load · uniform load · superposition and designcalculations
  5. Column bucklingpinned ends · other end conditions · design and critical stresscalculations
  6. Centroid & second moment of areacentroid · second moment of area · bending stresscalculations
  7. Effective stress in soilwater table at surface · water table in the layer · two layerscalculations
  8. Consolidation settlementnormally consolidated clay · overconsolidated clay · time for consolidationcalculations
  9. Hydrostatic forcevertical wall · submerged gate · centre of pressurecalculations
  10. Pipe friction & head lossReynolds number · Darcy–Weisbach head loss · minor losses and total headcalculations
  11. Open-channel flow (Manning)velocity in a channel · flow rate · required slopecalculations
  12. Pumps & pump powerhydraulic power · pump efficiency · system head and powercalculations
  13. Time value of moneyfuture value · compounding and discounting · annuities and loanscalculations
  14. NPV & IRRequal cash flows · uneven cash flows · internal rate of returncalculations
  15. Normal distributionarea in one tail · area between two values · percentilescalculations

Exam tips for FE Civil exam practice

  • Draw the free-body diagram even under time pressure.
  • In Manning’s equation use the hydraulic radius, not the depth.
  • Engineering economics questions are quick points: practise the interest formulas until they are automatic.

Worked examples

Beam support reactionsEasy
P = 55 kNABRA​RB​ = ?2 m2 m

The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RB.

L = 4 mP = 55 kN at 2 m

Find RB (kN)

Show hints
  1. Use static equilibrium: ΣFy = 0 and ΣM = 0. Two unknowns, two equations.
  2. Take moments about A. RA passes through A, so it drops out and you get RB directly.
  3. Total downward load is 55 kN, so RA + RB = 55 kN.
Show the answer and the step-by-step solution

Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.

P = 55 kNABRA​RB​2 m2 m

Free-body diagram

Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.

Equilibrium

Moments about A, counter-clockwise positive. Then vertical force balance.

ΣMA = 0: RB·4 = 55·2
RB = 110/4 = 27.5 kN
ΣFy = 0: RA = 55 − 27.5 = 27.5 kN

Answer

RB = 27.5 kN
Truss: method of jointsMedium
ABBCACACB45 kN6 m1 m2 m

Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.

span = 6 mheight = 2 mB is 1 m from Avertical load = 45 kN

Find FBC (kN)

Show hints
  1. Find the support reactions first from the whole truss.
  2. Go to joint C. Only BC has a vertical component there, so start with ΣFy = 0.
  3. Cy = 7.5 kN.
Show the answer and the step-by-step solution

Method of joints: every pin of a truss is in equilibrium, so ΣFx = 0 and ΣFy = 0 at each joint. Start at a joint with at most two unknown member forces. A negative result means the member is in compression.

ABBCACACB45 kNAy​Cy​

Support reactions

Equilibrium of the whole truss. Moments about A, counter-clockwise positive.

ΣMA = 0: Cy·6 = 45·1 → Cy = 7.5 kN
ΣFy = 0: Ay = 45 − 7.5 = 37.5 kN

Joint C

The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.

sin γ = 2/5.385 = 0.3714, cos γ = 5/5.385 = 0.9285
ΣFy = 0: Cy + FBC·sin γ = 0 → FBC = −7.5/0.3714 = −20.19 kN

Answer

FBC = −20.19 kN (compression)

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Beam reaction calculatorBeam deflection calculatorShear force and bending moment diagram calculatorEuler buckling calculatorPipe head loss calculator (Darcy–Weisbach)Pump power calculatorManning equation calculatorConsolidation settlement calculatorMoment of inertia calculator (second moment of area)Bending stress calculator (σ = My/I)Reynolds number calculatorCentroid calculator (composite shapes and polygons)Truss calculator (method of joints)Z-score and normal probability calculatorNormal distribution calculatorCompound interest calculatorLoan payment calculatorNPV and IRR calculatorROI calculator (return on investment)Payback period calculator (simple and discounted)Density calculator (mass, volume and density)

Related courses

FE Mechanical exam practiceStructural AnalysisHydraulic EngineeringStaticsMechanics of MaterialsGeotechnical EngineeringFluid MechanicsCFA Level I quantitative methods and fixed income practice
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