Free FE Civil Practice Problems with Solutions [2026]

Correct answer: C. 3.74 m²

Reference: FE Reference Handbook 10.6, Mathematics — Circular Segment

Above half full, the segment formula measures the wrong region on purpose - the AIR at the crown is the segment, and the water is everything else

The air segment first: φ = 2 arccos[(r − d)/r] = 2 arccos(0.65/1.2) = 1.997 rad, so A = r²(φ − sin φ)/2 = 1.44(1.997 − 0.911)/2 = 0.78 m²

The flow area is the complement: A = πr² − 0.78 = 4.52 − 0.78 = 3.74 m²

The 0.55-m dimension measures down from the CROWN, not up from the invert - reading which region the segment formula is computing is the whole judgment, and the segment itself is one of the printed answers waiting for whoever skips it.

Correct answer: D. $112,000

Reference: FE Reference Handbook 10.6, Engineering Probability and Statistics — Least Squares; Linear Regression and Goodness of Fit

The intercept comes at the end of a slope-first chain, and both sums of squares need their correction terms before the slope exists.

Sxy = Σxy − (Σx)(Σy)/n = 2,338.0 − (42)(399.0)/6 = −455.0

Sxx = Σx² − (Σx)²/n = 364 − (42)²/6 = 70.0

b = Sxy/Sxx = −455.0/70.0 = −6.50 ($1000s per year)

The means are x̄ = 42/6 = 7.0 yr and ȳ = 399.0/6 = 66.5 ($1000s).

â = ȳ − b·x̄ = 66.5 − (−6.50)(7.0) = 112.0 ($1000s)

The prices were tallied in $1000s, so the intercept converts to $112,000 — the fitted price at age zero. The negative slope makes the subtraction in â = ȳ − b·x̄ add magnitude, which is why the intercept lands well above the mean price.

Correct answer: A. 3.0 ft

Reference: FE Reference Handbook 10.6, Mathematics — Quadratic Equation

The gravel fixes the border area. Converting the volume, 8 yd³ × 27 ft³/yd³ = 216 ft³, and spreading it 6 in. (0.5 ft) deep gives 216/0.5 = 432 ft² of border.

With border width x on all four sides, the bordered footprint is (40 + 2x) by (25 + 2x), so the border area is (40 + 2x)(25 + 2x) − 40(25) = 4x² + 130x

Setting that equal to 432 ft²: 4x² + 130x − 432 = 0 → 2x² + 65x − 216 = 0

The quadratic formula gives x = [−65 + √(65² + 4(2)(216))]/(2 · 2) = (−65 + 77.16)/4 = 3.04 ft

The formula's other root is negative, and a border cannot have a negative width, so that root is rejected on physical grounds. The width is most nearly 3.0 ft.

Correct answer: D. No contract exists - the conditional reply was a counteroffer that ended the original offer

Reference: NCEES FE Civil exam specification — Ethics and Professional Practice (contract-law elements); general knowledge item — no FE Reference Handbook coverage

An acceptance must mirror the offer's terms. The owner's reply changed the price, and a reply that changes the terms is not an acceptance at all — it is a counteroffer, and a counteroffer TERMINATES the offer it answers.

From that moment the $180,000 offer was dead. The owner's later message could not accept what no longer existed; it stands only as a fresh offer to the contractor, who never accepted it. Silence bound nobody in either direction — a party's failure to respond to a counteroffer forms nothing.

Correct answer: C. $13,580

Reference: FE Reference Handbook 10.6, Engineering Economics — Interest Rate Tables

Bring both deposits to time zero:

P = 50,000 + 20,000(P/F, 6%, 3) = 50,000 + 20,000(0.8396) = $66,792

The A/P factor prices an annuity whose first payment falls ONE period after the valuation point, so a series that begins at the end of year 5 must be valued at the end of year 4:

V₄ = 66,792(F/P, 6%, 4) = 66,792(1.2625) = $84,325

A = 84,325(A/P, 6%, 8) = 84,325(0.16104) = $13,580

Rolling the account forward to year 5 — the date the first withdrawal is actually taken — overshoots by one period of interest, because that valuation point demands a beginning-of-period series the tables do not price.

Correct answer: B. Member b

Reference: FE Reference Handbook 10.6, Statics — Plane Truss: Method of Joints

Each reaction is half the 60-kN total load: R = 30 kN.

At the support joint, the end diagonal must supply the full 30-kN vertical reaction. With panel geometry 1.5 m horizontal and 2.0 m vertical over a 2.5-m diagonal length, its force is 30(2.5/2.0) = 37.5 kN in compression.

Sweeping inward joint by joint, the chord forces grow toward midspan as the moment builds: the mid top chord carries 45.0 kN in compression, and the mid bottom chord carries 45.0 kN in tension.

The governing compressive member is therefore the mid top chord at 45.0 kN — larger than the end diagonal that dominates the first joint solved.

In a truss the chords resist the global bending moment through the force couple across the truss depth, so chord forces peak where the moment peaks — at midspan — while diagonals peak at the supports where the shear peaks. Ranking members means sweeping past the first joint and keeping tension and compression separate as the signs emerge.

Correct answer: D. 5.59 m

Reference: FE Reference Handbook 10.6, Statics — Systems of Forces (Centroids of Masses, Areas, Lengths, and Volumes)

Replace each load block by its resultant at its own centroid:

Uniform: W₁ = 2(4) = 8 kN at x = 2 m → Triangular: W₂ = ½(6)(3) = 9 kN at x = 4 + ⅔(6) = 8 m → Point: 5 kN at x = 7 m

x̄ = ΣWx/ΣW = [8(2) + 9(8) + 5(7)]/22 = 123/22 = 5.59 m

The triangular block's centroid sits two-thirds of the way toward its DEEP end — the load accumulates where the intensity is, not at the block's geometric middle — and that judgment alone separates three of the four choices.

Correct answer: D. 136.9 mm

Reference: FE Reference Handbook 10.6, Dynamics — Potential Energy (Elastic Potential Energy)

From release to maximum compression the block is momentarily at rest at both ends, so the gravitational potential energy lost equals the elastic energy stored. The block falls through the 0.5-m gap PLUS the compression x itself: mg(h + x) = kx²/2

12(9.81)(0.5 + x) = 8,000x²/2 → 4,000x² − 117.72x − 58.86 = 0

x = [117.72 + √(117.72² + 4(4,000)(58.86))]/(2 × 4,000) = 0.1369 m

x = 136.9 mm

Two habits fail here. Dropping the "+ x" term — as if the block stopped falling when it touched the spring — undershoots by about 15 mm. And no static or suddenly-applied formula applies: this is a free fall through half a metre, so the spring must absorb far more than 2×(static deflection).

Correct answer: C. 13.4 ft

Reference: FE Reference Handbook 10.6, Mechanics of Materials — Shearing Force and Bending Moment Sign Conventions

Because dM/dx = V, the bending moment reaches its maximum at the section where the shear passes through zero.

ΣMB = 0: RA(30) = (2 × 18)(21) + 8(6) → RA = 804/30 = 26.8 kips

Within the loaded length the shear falls at the rate of the load intensity: V(x) = 26.8 − 2x = 0 → x = 13.4 ft

The zero crossing lies inside the 18-ft loaded length, and just beyond that length the shear is already negative (26.8 − 36 = −9.2 kips), so the crossing at 13.4 ft is the only one and the moment peaks there. The governing section sits where the shear diagram crosses zero — a point fixed by the left reaction and the load intensity.

Correct answer: B. 980 mm

Reference: FE Reference Handbook 10.6, Mechanics of Materials — Uniaxial Loading and Deformation

Series segments carry the SAME 60-kN force and their elongations add. The steel spends its share of the limit first: δs = PL/AE = 60,000(1,200)/(500 × 200,000) = 0.72 mm

The aluminum gets what remains: δa ≤ 2.4 − 0.72 = 1.68 mm

Solving δ = PL/AE for the length, with the ALUMINUM modulus: La = δaAEa/P = 1.68(500)(70,000)/60,000

La = 980 mm

The budget arithmetic is the discrimination — the steel's 0.72 mm is spent whether or not it is counted, and skipping the deduction over-lengthens the aluminum by 43%. Millimetre for millimetre, the softer aluminum consumes the remaining allowance nearly three times as fast as steel would.

Correct answer: C. 0.43

Reference: FE Reference Handbook 10.6, Materials Science/Structure of Matter — Lever Rule (Binary Phase Diagrams)

At 850°C the 40% B alloy lies inside the two-phase α + L field, so its mass is shared between solid α with the tie line's solidus-end composition and liquid with its liquidus-end composition. Reading the tie line at 850°C: xα = 20% B and xL = 55% B.

mass fraction solid = (xL − x)/(xL − xα) = (55 − 40)/(55 − 20) = 15/35 = 0.43

The lever rule weighs each phase by the arm of the tie line opposite to it: the solid fraction takes the distance from the overall composition out to the liquidus end, divided by the full tie-line length between the two phase compositions.

Correct answer: C. 3.5 × 10⁵ cycles

Reference: FE Reference Handbook 10.6, Materials Science/Structure of Matter — Properties of Materials

The Basquin relation for fully reversed high-cycle fatigue is N = (σᵣ/A)^(1/B), a power law between stress amplitude and life.

Fitting the two tests: B = ln(250/350)/ln(10⁶/10⁴) = −0.3365/4.6052 = −0.0731

Life at 270 MPa from the 350-MPa point: N = 10⁴ × (270/350)^(1/−0.0731) = 10⁴ × (0.7714)^(−13.7) = 10⁴ × 34.9 = 3.5 × 10⁵ cycles

A power law is a straight line only on log-log axes — a 20-MPa change in amplitude moves the life by half an order of magnitude, which is why the linear readings miss so badly.

Correct answer: D. 7.83 ft

Reference: FE Reference Handbook 10.6, Fluid Mechanics — Forces on Submerged Surfaces and the Center of Pressure (with Statics — Equilibrium Requirements)

With the surface below the gate top, the wetted panel is the water depth d itself, so the resultant force and its location both move with the unknown — depth drives the pressure prism, the center of pressure, and the hinge arm at once.

F = γ·h_c·A = (62.4)(d/2)(4.0·d) = 124.8 d²

For a rectangle wetted from the surface, the center of pressure sits at 2d/3 below the surface → arm above the bottom hinge = d − 2d/3 = d/3

M_hinge = F(d/3) = 124.8 d²(d/3) = 41.6 d³

The gate is on the verge of opening when the hydrostatic moment reaches the hinge capacity:

41.6 d³ = 20,000 → d³ = 480.8 → d = 7.83 ft

The result respects the setup: 7.83 ft is below the 9.0-ft gate top, so the wetted-depth assumption stands, and the 350-lb gate weight acts in the plane of the vertical gate straight through the hinge axis — zero moment arm, so it never enters the equilibrium. A depth-reverse like this cubes the unknown because the resultant grows as d² while its arm above the hinge grows as d — the center of pressure of a triangular prism sits a third of the depth above the base, and that geometry is the whole moment computation.

Correct answer: C. 3.85 m/s

Reference: FE Reference Handbook 10.6, Fluid Mechanics — Pitot Tubes

The manometer converts the stagnation-static difference into a mercury column, and the mercury displaces WATER, so only the density difference drives the deflection: Δp = (SG − 1)γwh = 12.6(9,810)(0.060) = 7,420 Pa

The stagnation relation gives the velocity: V = √(2Δp/ρ) = √(2 × 7,420/1,000)

V = 3.85 m/s

The (SG − 1) is where this item is won: the 60 mm of mercury is opposed by 60 mm of water on the other limb, and forgetting that buyback inflates the driving pressure by 8% — a small-looking slip that survives to the option list. The bare deflection is not a water head either; it must pass through the manometer equation before it means anything to the probe.

Correct answer: A. 827 ft²

Reference: FE Reference Handbook 10.6, Mathematics — Mensuration of Areas and Volumes

The fill height is the grade difference:

h = 851.0 − 842.0 = 9.0 ft

Each side slope projects horizontally by its ratio times the height: 2(9.0) = 18 ft on the left and 3(9.0) = 27 ft on the right, so the base of the fill is 76 + 18 + 27 = 121 ft

Gross trapezoid between the finished top and the ground: A = ½(76 + 121)(9.0) = 886.5 ft²

Triangular median swale cut out of the top: A = ½(24)(5.0) = 60 ft²

Net embankment area = 886.5 − 60 = 826.5 ft² ≈ 827 ft²

Decomposing an irregular section into simple shapes is the working method for every end-area computation: build the largest simple figure the outline allows, then correct it with the pieces that are not fill. The swale is air, not embankment, so its triangle carries a minus sign — keeping track of which decomposed pieces add material and which remove it is the whole bookkeeping of the method.

Correct answer: B. 21.6 ft

Reference: FE Reference Handbook 10.6, Mathematics — Trigonometry

The vertical component of the slope distance is s·cos(z) = 285.60 × cos 85°30′ = 22.4 ft

The line runs from the instrument's trunnion axis to the prism, so the ground-to-ground difference restores both offsets: ΔElev = 22.4 + hi − ht = 22.4 + 5.2 − 6.0 = 21.6 ft

The instrument height lifts the line's origin above the station mark and the prism floats above the surveyed point — both offsets must be restored to move from line-of-sight geometry to ground-to-ground elevation.

Correct answer: B. 14.2 ft³/sec

Reference: FE Reference Handbook 10.6, Civil Engineering — Weir Formulas (with Orifices)

Both openings sit in series, so the basin can only release what the tighter one passes

Riser weir: Q = CwLH^3/2 = 3.33(6.0)(2.0)^1.5 = 56.5 ft³/sec

Barrel orifice: Q = CdA√(2gh) = 0.60(1.2)√(64.4 × 6.0) = 14.2 ft³/sec

The orifice governs: 14.2 ft³/sec — water pours over the crest faster than the barrel can carry it away, so the riser simply runs full.

Correct answer: A. 15,100 gal/day

Reference: FE Reference Handbook 10.6, Civil Engineering — Darcy's Law

The hydraulic gradient comes from piezometric heads, not depths to water, so each water-table elevation is computed first from its ground surface.

Head at the first well: h₁ = 741.0 − 18.6 = 722.4 ft

Head at the second well: h₂ = 752.5 − 34.6 = 717.9 ft

Even though the ground rises toward the second well, the water table there is lower, so flow moves from the first well toward the second: i = (722.4 − 717.9)/1,500 = 0.0030

The discharge crosses the saturated section of the aquifer: A = 40 × 800 = 32,000 ft²

Darcy's law gives the volumetric discharge directly from the superficial velocity — the porosity plays no part in Q: Q = KiA = 21 × 0.0030 × 32,000 = 2,016 ft³/day

Converting, Q = 2,016 × 7.48 = 15,080 gal/day.

Correct answer: C. 17.6 m

Reference: FE Reference Handbook 10.6, Environmental Engineering — Clarifier

Each criterion demands its own surface area. Overflow rate: A = Q/vo = 6,800/28 = 243 m²

Residence time: V = Qθ = 6,800 × (2.5/24) = 708 m³, so A = V/depth = 708/3.5 = 202 m²

The LARGER area governs - both criteria are minimums, and only the 243-m² tank satisfies the pair: D = √(4 × 243/π) = 17.6 m

The 2.5 hours must become a day fraction before it meets the per-day flow, and a tank sized to the smaller area passes one criterion while silently failing the other.

Correct answer: B. 25 kN·m

Reference: FE Reference Handbook 10.6, Civil Engineering — Elementary Statically Indeterminate Structures by Force Method of Analysis

Releasing the center support and restoring compatibility (the force method) gives the center-support moment of a two-equal-span beam under uniform load:

M_B = −wL²/8 = −10(6)²/8 = −45 kN·m

The end reaction follows: R_A = wL/2 − |M_B|/L = 30 − 7.5 = 22.5 kN

Shear vanishes at x = R_A/w = 2.25 m, where the span moment peaks:

M_max = R_A²/(2w) = 22.5²/20 = 25 kN·m

Continuity trades moment between the span and the support: the positive peak drops to 9/16 of the simple-span value while the support takes the negative maximum — and the LARGEST moment in the beam is actually the 45 kN·m over the support, not in the span. Reading which moment a design check needs is half the analysis.

Correct answer: A. 96 kips

Reference: FE Reference Handbook 10.6, Civil Engineering — Tension Members (Limit States and Available Strengths)

For net-area computation each deduction is the nominal hole diameter plus 1/16 in., and the nominal hole is itself the bolt diameter plus 1/16 in., so each hole removes 3/4 + 1/8 = 7/8 in. of width.

Rupture is checked on the weakest chain across the plate, so every potential failure path must be enumerated — including the chain that zigzags through ALL THREE bolts.

Straight transverse section (one hole): wn = 8 − 0.875 = 7.125 in.

Two-hole staggered chain (one diagonal): wn = 8 − 2(0.875) + (2)²/[4(4)] = 6.50 in.

Three-hole zigzag through both gage lines (two diagonals): wn = 8 − 3(0.875) + 2(2)²/[4(4)] = 5.375 + 0.50 = 5.875 in. → governs

An = 5.875(0.375) = 2.203 in². Every element of a flat plate is connected at the splice, so the effective net area carries no shear-lag reduction: Ae = An = 2.203 in².

φPn = 0.75 Fu Ae = 0.75(58)(2.203) = 95.8 kips, most nearly 96 kips

The s²/4g term credits the extra length of each diagonal ligament, and a chain may step back and forth between gage lines: three deducted holes with only two stagger credits tear less material than either shorter path here. The governing net section is always the minimum over all chains — a staggered pattern is never safe to judge by any single path.

Correct answer: A. 0.94 mm upward

Reference: FE Reference Handbook 10.6, Civil Engineering — Truss Deflection by Unit Load Method

Δ = Σ fᵢ(ΔL)ᵢ, where fᵢ comes from a unit load at the joint in the direction sought, and (ΔL)ᵢ = αLᵢ(ΔT)ᵢ for a heated member. Only DE changes length, so the sum has one term.

With a unit downward load at B, the top chord force is f_DE = −2/3 (compression).

(ΔL)_DE = (11.7 × 10⁻⁶)(4,000)(30) = 1.40 mm (expansion)

Δ_B = (−2/3)(1.40) = −0.94 mm — negative, meaning opposite the unit load: 0.94 mm upward.

A heated top chord lengthens while the bottom chord holds, so the truss arches — the same reason sun on a bridge deck's top surface cambers it upward. The unit-load force translates the one member's length change into joint movement, and its sign carries the direction automatically.

Correct answer: D. 8.0

Reference: FE Reference Handbook 10.6, Civil Engineering — Soil Classification

From the curve, the diameters at 60% and 10% passing are D60 = 0.60 mm and D10 = 0.075 mm.

Cu = D60/D10 = 0.60/0.075 = 8.0

The uniformity coefficient spans the curve between its 60% and 10% ordinates — the larger diameter over the smaller, so the ratio always exceeds one, and a value below one flags an inverted read. The 30% ordinate belongs to the OTHER gradation coefficient, and a curve read one gridline off at the top swaps the answer for the D50 impostor.

Correct answer: C. 1.92

Reference: FE Reference Handbook 10.6, Civil Engineering — Retaining Walls

The active thrust follows directly from the given coefficient: P = ½ Ka γ h² = ½ × ⅓ × 18 × 3² = 27 kN/m, acting at h/3 = 1.0 m above the base

Overturning moment about the toe: Mot = 27 × 1.0 = 27 kN·m/m

The wall resists with its weight: W = 24 × 3 × 1.2 = 86.4 kN/m at the section centroid, 0.6 m from the toe, giving Mr = 51.8 kN·m/m

FS = Mr/Mot = 51.8/27 = 1.92

Both arms are measured from the same toe - the weight acts through the centroid at half the base width, and the triangular pressure resultant sits at the lower third point of the retained height.

Correct answer: A. 0.6

Reference: FE Reference Handbook 10.6, Civil Engineering — Coefficient of concavity (or curvature)

The exhibit's ordinate is cumulative percent RETAINED, while every defining D-size is a percent-FINER diameter. Sieve results are tallied as mass caught on each screen, so the two scales are complements: percent finer = 100 − percent retained. D10 therefore reads where the curve crosses the 90% retained line, D30 at 70% retained, and D60 at 40% retained.

From those three crossings: D10 = 0.2 mm, D30 = 0.6 mm, D60 = 3.0 mm.

Cc = (D30)²/(D10 × D60) = (0.6)²/[(0.2)(3.0)] = 0.36/0.60 = 0.6

The middle diameter is the one that enters squared, because the coefficient measures how the center of the gradation bends between its two anchor sizes. With Cc below 1 this soil's curve dips under the smooth well-graded shape between D10 and D60 — a gap-tending gradation despite its wide size range — and that judgment starts with reading the axis for what it actually plots before pulling a single number off the curve.

Correct answer: B. 79.7 ft

Reference: FE Reference Handbook 10.6, Civil Engineering — Horizontal Curves

The printed long-chord relation recovers the radius first: LC = 2R sin(Δ/2) → R = 860/[2 sin 21°] = 1,200 ft

The printed middle-ordinate relation then closes it: M = R[1 − cos(Δ/2)] = 1,200(1 − cos 21°) = 79.7 ft

Both formulas run on the half-angle — the chord and its ordinate belong to the half-triangles the curve's bisector cuts — and the middle ordinate hangs from the CHORD toward the arc, while its near-twin, the external distance, reaches from the PI. The two answer different setback questions, and only one of them is asked.

Correct answer: B. 4,310 pc/h

Reference: FE Reference Handbook 10.6, Civil Engineering — Level of Service (LOS) (Peak Hour Factor)

LOS C holds while density stays at or below the printed threshold of 26 pc/mi/ln.

The flow identity q = kv converts that ceiling to a per-lane flow RATE: q = 26 × 60 = 1,560 pc/h/ln, or 4,680 pc/h across three lanes — but that is a peak 15-minute rate, not an hourly volume.

The hour can only average PHF times its worst quarter: V = 0.92 × 4,680 = 4,310 pc/h

Both table reads are load-bearing — the LOS C row (not D), and the direction of the PHF adjustment: peaking always shrinks what a full hour can serve.

Correct answer: B. 11 in.

Reference: FE Reference Handbook 10.6, Civil Engineering — AASHTO Structural Number Equation

The structural number sums the layered contributions: SN = a₁D₁ + a₂D₂m₂ + a₃D₃m₃

The surface and subbase are fixed: a₁D₁ = 0.44(4.0) = 1.76 and a₃D₃m₃ = 0.11(8.0)(1.0) = 0.88

The base must supply the remainder: a₂D₂m₂ = 4.0 − 1.76 − 0.88 = 1.36

D₂ = 1.36/(0.14 × 0.9) = 1.36/0.126 = 10.8 in.

The drainage coefficient discounts the base layer's contribution, so it lives in the divisor when the equation runs backward — a wet-side m below 1.0 always demands MORE thickness, and any manipulation that yields less has put m on the wrong side.

Correct answer: B. 5.4 ft

Reference: NCEES FE Civil exam specification — Construction Engineering (interpretation of engineering drawings); general knowledge item — no FE Reference Handbook coverage

Cover is measured from the finished ground surface down to the outside top — the crown — of the pipe, so three reads off the sheet chain together: the pipe grade, the ground elevation at the station, and the barrel dimensions from the callout.

A profile sheet states pipe position by INVERT, because grade control is set on the flowline. Station 11+40 lies 140 ft beyond MH-1 at 10+00, so the invert there comes by interpolation down the 1.5% grade: inv El. = 425.60 − 0.015(140) = 425.60 − 2.10 = 423.50 ft

The invert is the INSIDE bottom of the pipe. From it, the outside crown sits one inside diameter plus one wall thickness higher — the callout's 24-in. bore plus a (30 − 24)/2 = 3-in. wall: crown El. = 423.50 + 2.00 + 0.25 = 425.75 ft

Cover = ground − crown = 431.10 − 425.75 = 5.35 ft, most nearly 5.4 ft

Adding the full 30-in. outside diameter to the invert counts the bottom wall too, and that wall sits BELOW the flowline the invert already marks. The rim elevations on the same sheet describe the casting heights at the two structures, not the surface along the run — the spot elevation at 11+40 is the one that counts.

Correct answer: C. $1,000

Reference: FE Reference Handbook 10.6, Civil Engineering — CPM Precedence Relationships

Crashing shortens the project only while the crashed activity lies on EVERY critical path - and the critical path moves as durations change.

Start: B-D = 10 days governs (A-C = 9).

Day 1: crash the cheapest B-D activity - B at $400. Duration 9; now BOTH chains are critical at 9 days.

Day 2: both chains must shorten together. Cheapest on B-D is B again ($400); cheapest on A-C is A ($200). Day-2 cost $600.

Total: $400 + $600 = $1,000 for an 8-day project.

Re-identifying the critical path after every crash day is the discipline: once the chains tie, a day of project compression must be purchased on every critical chain simultaneously.