PE Civil Practice Questions (2026) — Free Sample Questions with Answers

Answer: Intact, unfissured stiff clay with an unconfined compressive strength of 1.8 tsf, away from traffic and never disturbed: Type A; The same 1.8-tsf clay, but the trench runs beside a highway with heavy truck traffic: Type B; Silty clay with an unconfined compressive strength of 0.8 tsf, unfissured and dry: Type B; Angular crushed-stone-like gravel, dry, with no seepage: Type B; Clean medium sand above the water table: Type C; Clay with an unconfined compressive strength of 1.6 tsf from which water is freely seeping into the trench: Type C

Classify by unconfined compressive strength first (≥ 1.5 tsf A, > 0.5 to < 1.5 tsf B, ≤ 0.5 tsf C), then downgrade for fissuring, vibration, disturbance or seepage; granular sands and rounded gravels are Type C, angular gravel is Type B.

Answer: C

1. Normal durations: A–B–C = 6 + 8 + 6 = 20 days; D–E = 10 + 8 = 18 days.
2. Days 20 → 19: crash B (cheapest on the critical path, $500).
3. Days 19 → 18: B is at its limit; crash A ($800). Both paths are now 18 days.
4. Days 18 → 17: both paths are critical; crash A ($800) and D ($300) = $1,100.
5. Total = 500 + 800 + 1,100 = $2,400.

$1,800: Crashed B for 3 days, ignoring its 1-day crash limit, and then crashed D for 1 day.

$2,100: Shortened only path A–B–C (B 1 day, A 2 days) and missed that path D–E (18 days) then controls.

$2,400: Correct. Crash B 1 day ($500) and A 2 days ($1,600) to bring A–B–C to 17 days; path D–E (18 days) must also lose 1 day at its cheapest rate, D ($300); total $2,400 (least-cost crashing of all critical paths).

$3,600: Crashed only activity C by 3 days; C is the most expensive activity on the path and path D–E still needs 1 day.

Crash the cheapest activity on every critical path one day at a time, and check whether a parallel path becomes critical as the project shortens.

Answer: C

1. Total stress: σ = (8 ft)(110 pcf) + (12 ft)(125 pcf) = 880 + 1,500 = 2,380 psf.
2. Pore pressure at P (12 ft below the GWT): u = (12 ft)(62.4 pcf) = 748.8 psf.
3. Effective stress: σ′ = 2,380 − 748.8 = 1,631 psf ≈ 1,630 psf.

1,130 psf: Computed pore pressure over the full 20 ft depth instead of only the 12 ft below the water table (2,380 − 62.4 × 20).

1,250 psf: Used the buoyant unit weight for the whole 20 ft, including the 8 ft of moist sand above the water table.

1,630 psf: Correct. σ = 8(110) + 12(125) = 2,380 psf; u = 12(62.4) = 749 psf; σ′ = σ − u = 1,631 psf ≈ 1,630 psf (effective stress principle).

2,380 psf: This is the total vertical stress; the pore water pressure was not subtracted.

Effective stress is total stress minus pore pressure, and pore pressure is measured only from the water table down.

Answer: 2.09 in.

1. Mid-clay depth = 15 ft. σ′0 = 10(120) + 5(112 − 62.4) = 1,200 + 248 = 1,448 psf
2. Depth below footing base z = 15 − 3 = 12 ft; Δσ = Q/[(B + z)(L + z)] = 160,000/(20 × 20) = 400 psf
3. Sc = [Cc·H/(1 + e0)] log[(σ′0 + Δσ)/σ′0] = [0.32 × 120 in./1.95] log(1,848/1,448)
4. = 19.692 × 0.1059 = 2.09 in.

For a normally consolidated clay, settlement depends on the log of the final-to-initial effective stress ratio; the 2:1 method spreads the footing load over (B + z)(L + z) at depth z below the base.

Answer: B

1. Ka = tan²(45° − 32°/2) = 0.3073
2. At 8 ft: σ′h = 0.3073(8 × 120) = 295.0 psf. At 20 ft: σ′v = 960 + 12(125 − 62.4) = 1711.2 psf; σ′h = 525.8 psf
3. Soil thrust = ½(295.0)(8) + ½(295.0 + 525.8)(12) = 6,104 lb/ft
4. Water thrust = ½(62.4 × 12)(12) = 4,493 lb/ft
5. Total = 10,597 lb/ft ≈ 10.60 kip/ft

6.10 kip/ft: Included only the effective soil pressure and omitted the water pressure below the water table.

10.60 kip/ft: Correct. Ka = tan²(29°) = 0.3073; soil thrust = 6.10 kip/ft (effective stresses below the GWT); water thrust = ½(62.4)(12)² = 4.49 kip/ft; total = 10.60 kip/ft.

11.87 kip/ft: Applied Ka to the moist unit weight over the full 20 ft and then also added full water pressure, double-counting the water.

13.83 kip/ft: Used the at-rest coefficient Ko = 1 − sin φ′ instead of Ka for a cantilever wall that can yield.

Below the water table, apply the earth pressure coefficient to effective vertical stress only and add the full hydrostatic water pressure separately, because water pressure acts equally in all directions.

Answer: 1.09

1. For slope-parallel seepage with the water table at the surface: FS = [c′ + (γsat − γw)H cos²β tan φ′] / (γsat H sin β cos β)
2. cos²18° = 0.9045; sin 18° cos 18° = 0.2939; tan 28° = 0.5317
3. Resisting: 120 + (124 − 62.4)(12)(0.9045)(0.5317) = 120 + 355.5 = 475.5 psf
4. Driving: (124)(12)(0.2939) = 437.3 psf
5. FS = 475.5/437.3 = 1.09

Slope-parallel seepage cuts the normal effective stress to the buoyant weight while the driving stress still uses the saturated weight, which can roughly halve the dry-condition factor of safety.

Answer: A

1. Mp = FyZx = 50(101)/12 = 420.8 kip-ft; 0.7FySx = 0.7(50)(88.9)/12 = 259.3 kip-ft.
2. Lp = 5.83 ft < Lb = 10 ft < Lr = 16.9 ft: inelastic LTB (AISC 360-16 Eq. F2-2).
3. Mn = 1.0[420.8 − (420.8 − 259.3)(10 − 5.83)/(16.9 − 5.83)] = 420.8 − 60.8 = 360.0 kip-ft ≤ Mp.
4. φbMn = 0.90(360.0) = 324 kip-ft.

324 kip-ft: Correct. Lp < Lb < Lr, so Mn = Mp − (Mp − 0.7FySx)(Lb − Lp)/(Lr − Lp) = 420.8 − 161.5(0.377) = 360 kip-ft; φbMn = 0.9(360) = 324 kip-ft.

360 kip-ft: Omitted the resistance factor φb = 0.90.

379 kip-ft: Used φbMp and ignored lateral-torsional buckling, although Lb > Lp.

421 kip-ft: Reported the nominal plastic moment Mp, with neither φ nor the LTB reduction.

Compare Lb with Lp and Lr to pick the LTB zone; between them, interpolate linearly from Mp to 0.7FySx and multiply by Cb (capped at Mp).

Answer: 53.3 kips

1. By symmetry, R_L0 = R_L4 = 3(20)/2 = 30 kips.
2. Cut through U1–U2, U1–L2 and L1–L2; take moments about L2, where the other two cut members meet.
3. ΣM_L2 (left part): 30(24) − 20(12) − F(9) = 0, so F = 480/9 = 53.3 kips.
4. The top chord is in compression; the magnitude is 53.3 kips.

With the method of sections, take moments about the joint where the other cut members intersect: chord force = panel-point moment ÷ truss depth.

Answer: C

1. Cs = SDS/(R/Ie) = 0.80/8 = 0.100.
2. Upper limit (T ≤ TL): Cs = SD1/[T(R/Ie)] = 0.40/(0.9 × 8) = 0.0556, which governs.
3. Lower limit: 0.044SDSIe = 0.0352 ≥ 0.01, does not govern (S1 < 0.6g, so Eq. 12.8-6 does not apply).
4. V = CsW = 0.0556(3,200) = 178 kips.

113 kips: Used the minimum Cs = 0.044SDSIe = 0.0352; the minimum does not govern because the period-limited Cs is larger.

160 kips: Used SD1/(R/Ie) and dropped the period T from the upper-bound equation.

178 kips: Correct. Cs = SDS/(R/Ie) = 0.100 is capped at SD1/[T(R/Ie)] = 0.40/(0.9 × 8) = 0.0556; V = 0.0556(3,200) = 178 kips.

320 kips: Used Cs = SDS/(R/Ie) = 0.100 without checking the upper limit for T ≤ TL.

Always bound Cs between its period-based maximum and its minimum; for moderate-period frames the SD1/T cap usually controls.

Answer: Column C1: Comb. 3: 1.2D + 1.6S + (L or 0.5W); Post P2: Comb. 1: 1.4D; Column C3: Comb. 2: 1.2D + 1.6L + 0.5S; Brace-column C4: Comb. 4: 1.2D + 1.0W + L + 0.5S; Column C5: Comb. 3: 1.2D + 1.6S + (L or 0.5W)

1. Comb. 1: 1.4D. Comb. 2: 1.2D + 1.6L + 0.5S. Comb. 3: 1.2D + 1.6S + max(L, 0.5W). Comb. 4: 1.2D + 1.0W + L + 0.5S.
2. C1: 56.0 / 79.0 / 106.0 / 78.0 kips, so Comb. 3.
3. P2: 84.0 / 81.0 / 80.2 / 78.0 kips, so Comb. 1.
4. C3: 42.0 / 103.0 / 85.6 / 83.0 kips, so Comb. 2.
5. C4: 35.0 / 44.8 / 56.4 / 80.0 kips, so Comb. 4.
6. C5: 28.0 / 45.6 / 77.4 / 72.0 kips, so Comb. 3.

Evaluate every applicable combination: the load with the largest share of the total, given the 1.6 or 1.0 factor it receives as the principal load, usually controls.

Answer: 7–8 AM: 820 major / 160 minor: Meets Condition A; 8–9 AM: 640 major / 130 minor: Meets neither; 11 AM–12 PM: 780 major / 95 minor: Meets Condition B only; 12–1 PM: 520 major / 155 minor: Meets Condition A; 4–5 PM: 910 major / 70 minor: Meets neither; 5–6 PM: 760 major / 140 minor: Meets Condition B only

1. For each hour, check both the major-street and the minor-street threshold of a condition; an hour satisfies a condition only when both are met in that same hour.
2. Warrant 1 is satisfied when either condition is met for any 8 hours of an average day (or the combination of A and B at 80% levels); this table shows only six hours.

Warrant 1 compares paired major- and minor-street volumes hour by hour; a high major-street volume never compensates for a minor-street volume below the threshold.

Answer: 4619 ft

1. Δ = 38°24′ = 38.40°
2. T = R tan(Δ/2) = 1,250 × tan(19.20°) = 435.30 ft
3. L = πRΔ/180 = π × 1,250 × 38.40/180 = 837.76 ft
4. PC = PI − T = 4,216.50 − 435.30 = 3781.20 ft (Sta 37+81.20)
5. PT = PC + L = 3781.20 + 837.76 = 4618.96 ft (Sta 46+18.96) ≈ 4619

Stationing runs along the curve, so PT = PC + L, never PI + T.

Answer: D

1. Sum of predicted crashes over the study period: 2.40 × 3 = 7.20
2. w = 1 / (1 + k × ΣN_predicted) = 1 / (1 + 0.30 × 7.20) = 0.3165
3. N_expected = w × ΣN_predicted + (1 − w) × N_observed = 0.3165 × 7.20 + 0.6835 × 13 = 11.16 crashes (3 yr)
4. Annual expected frequency = 11.16 / 3 = 3.72 crashes/yr

2.40 crashes/yr: Reported the SPF-based predicted frequency, which ignores the site’s observed crash history.

3.01 crashes/yr: Swapped the weights, applying w to the observed crashes and (1 − w) to the predicted crashes.

3.21 crashes/yr: Computed the weight with one year of predicted crashes (2.40) instead of the 3-year sum (7.20).

3.72 crashes/yr: Correct. ΣN_pred = 2.40 × 3 = 7.20; w = 1/(1 + 0.30 × 7.20) = 0.3165; N_exp = 0.3165(7.20) + (1 − 0.3165)(13) = 11.16 crashes in 3 yr = 3.72 crashes/yr.

EB blends the SPF prediction and the observed count; the weight uses the predicted crashes summed over all study years, which corrects for regression to the mean.

Answer: 34.2 cfs

1. Segment 2: t = 900 ft/(1.5 ft/sec × 60 sec/min) = 10.0 min.
2. Segment 3: t = 1,500 ft/(5.0 ft/sec × 60 sec/min) = 5.0 min.
3. Tc = 10.0 + 10.0 + 5.0 = 25.0 min.
4. Interpolate between 20 min (4.2 in./hr) and 30 min (3.4 in./hr): i = 4.2 − (5/10)(0.8) = 3.8 in./hr.
5. Q = CiA = 0.60 × 3.8 × 15 = 34.2 cfs.

Time of concentration is the sum of the travel times (length ÷ velocity) of the sheet, shallow concentrated and channel or pipe segments along the hydraulically most remote path; the rational-method intensity is read for a duration equal to Tc.

Answer: C

1. y_c = (q²/g)^(1/3) = (3.0²/9.81)^(1/3) = 0.972 m: 0.40 m < y_c (supercritical chute); 1.60 m > y_c (subcritical outlet). A jump must occur.
2. Fr₁ = q/(y₁√(gy₁)) = 3.0/(0.40√(9.81 × 0.40)) = 3.79.
3. Sequent depth y₂ = (y₁/2)[√(1 + 8Fr₁²) − 1] = 1.95 m.
4. Compare with the downstream normal depth: 1.60 m < 1.95 m, so the tailwater is too low to hold the jump at the break. The flow stays supercritical on an M3 curve and jumps on the mild reach.

Steep reach, upper portion: Not here. The jump moves onto the steep reach only if the downstream depth exceeds the sequent depth of the chute flow; in that case it would sit near the break (S1 curve), not high on the chute.

Steep reach, just upstream of the slope break: Not here. A jump is forced up onto the chute only when the mild-reach depth (1.60 m) exceeds the sequent depth of the 0.40-m chute flow, 1.95 m. It does not, so the tailwater cannot push the jump upstream.

Mild reach, downstream of the slope break: Correct. Critical depth y_c = (q²/g)^(1/3) = 0.97 m, so the chute flow (0.40 m) is supercritical and the mild-reach flow (1.60 m) is subcritical. The sequent depth of the chute flow is 1.95 m (Fr₁ = 3.79). Because 1.60 m < 1.95 m, the supercritical flow continues past the break along an M3 curve until its depth reaches the sequent of 1.60 m (about 0.54 m), and the jump forms on the mild reach.

Locate a jump at a steep-to-mild break by comparing the downstream normal depth with the sequent depth of the upstream supercritical flow: lower tailwater pushes the jump downstream (M3 then jump), higher tailwater drives it up the chute (S1 curve).

Answer: A

1. NPSHA = h_atm − h_vp ± z_s − h_f,s, with z_s negative for a suction lift.
2. NPSHA = 33.0 − 0.8 − 12.0 − 2.6.
3. NPSHA = 17.6 ft.

17.6 ft: Correct. NPSHA = h_atm − h_vp − z_s − h_f,s = 33.0 − 0.8 − 12 − 2.6 = 17.6 ft.

18.4 ft: Omitted the vapor-pressure head; NPSHA is measured above the vapor pressure.

20.2 ft: Omitted the 2.6 ft of suction-line losses.

41.6 ft: Added the 12-ft static lift as if the water surface were above the pump (flooded suction); it is a suction lift and must be subtracted.

NPSHA is the absolute suction head above vapor pressure; a suction lift and suction-line losses both reduce it, and it must exceed the pump's NPSHR to avoid cavitation.