NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersNCEES PE Agricultural and Biological Engineering is a licensing and registration exam run by NCEES (ASABE is the content partner), for candidates in USA. Computer-based, 85 questions, 9.5-hour appointment (8.5 hours of exam). Once a year (next: 28 Oct 2026). Closed book with the electronic NCEES handbook and supplied searchable ASABE standards; SI and US customary units.
Source: ncees.org.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Applied soils and foundations inside facilities and natural-resource questions: bearing capacity and settlement of footings for farm buildings and bins, lateral pressure on retaining and bunker walls, compaction specifications for embankments and ponds, and seepage through small dams. Units are often US customary.
Soil as an engineering material: its phases, flow of water through it, compaction, consolidation, shear strength and stability.
Our 27 NCEES PE Agricultural and Biological Engineering questions in this subject: 7 easy, 14 medium, 6 hard, across 10 units.
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
These 10 were written to NCEES PE Agricultural and Biological Engineering and are not shown on any other exam's page.
A toe filter is to protect a dam base soil with D85 = 0.40 mm and D15 = 0.05 mm. Using Terzaghi's criteria with a factor of 4 (D15 of filter ≤ 4 × D85 of base, and ≥ 4 × D15 of base), which filter D15 is acceptable?
B. 0.8 mm
The retention limit is D15(filter) ≤ 4 × 0.40 = 1.6 mm; the permeability limit is D15(filter) ≥ 4 × 0.05 = 0.2 mm. Only 0.8 mm lies between 0.2 and 1.6 mm. 2.5 mm is too coarse and would let base soil wash through; 0.1 and 0.15 mm are too fine to drain freely. Agencies use refined versions of these rules.
A clay layer 8 m thick (γsat = 19 kN/m³) overlies a sand aquifer whose piezometric level stands 9 m above the clay–sand interface (that is, 1 m above ground level). To what depth can a pit be excavated in the clay before the base is at the point of heaving (factor of safety 1, γw = 9.81 kN/m³)?
D. 3.35 m
Heave begins when the weight of clay left below the pit equals the uplift pressure at the interface: 19 × (8 − d) = 9 × 9.81 = 88.29 kPa. So 8 − d = 4.65 m and d = 3.35 m. 4.65 m is the thickness of clay that must remain, not the depth of the pit. 3.87 m takes the uplift head as the 8 m clay thickness (19(8 − d) = 8 × 9.81); 7.48 m counts only the 1 m of artesian head above ground (19(8 − d) = 9.81).
For a soil with G = 2.70 at a water content of 18%, what is the dry unit weight on the zero-air-voids line (γw = 9.81 kN/m³)?
C. 17.82 kN/m³
γd(ZAV) = Gγw/(1 + wG) = 2.70 × 9.81/(1 + 0.18 × 2.70) = 26.487/1.486 = 17.82 kN/m³. Dividing by (1 + w) confuses it with γd = γ/(1 + w); 16.93 is the 5% air-voids line. 22.45 is Gγw/(1 + w) = 26.487/1.18; 24.83 divides w by G instead of multiplying (26.487/1.0667).
In a sand-replacement test on a pond embankment, 1705 g of calibrated sand (bulk density 1.55 g/cm³) fills the hole from which 2100 g of moist soil at 14% water content was dug. The laboratory maximum dry density is 1.78 g/cm³. What is the relative compaction?
C. 94.1%
Hole volume = 1705/1.55 = 1100 cm³. Bulk density = 2100/1100 = 1.909 g/cm³. Dry density = 1.909/1.14 = 1.675 g/cm³. RC = 1.675/1.78 = 94.1%. Comparing bulk density with maximum dry density gives an impossible 107.3%. 92.2% multiplies by (1 − w) instead of dividing by (1 + w) (1.909 × 0.86/1.78); 106.3% is the ratio inverted (1.78/1.675).
A clay layer 5 m thick has mv = 0.3 m²/MN. What consolidation settlement results from an average effective stress increase of 60 kPa?
A. 90 mm
S = mv·Δσ′·H = 0.3 × 10⁻³ m²/kN × 60 kPa × 5 m = 0.090 m = 90 mm. The key step is converting m²/MN to m²/kN (divide by 1000); 18 mm forgets the layer thickness. 9 mm and 900 mm are factor-of-ten slips in converting m²/MN to m²/kN.
A smooth vertical wall 6 m high retains dry, level cohesionless backfill with γ = 18 kN/m³ and φ = 30°. What is the Rankine active thrust per metre and where does it act?
B. 108 kN/m at 2 m above the base
Ka = (1 − sin30°)/(1 + sin30°) = 1/3. Pa = ½KaγH² = 0.5 × (1/3) × 18 × 36 = 108 kN/m, acting at H/3 = 2 m above the base because the pressure diagram is triangular. 972 uses Kp = 3.
A smooth vertical wall 6 m high retains level sand with φ = 30°. The water table is 2 m below the top of the backfill; the sand weighs 17 kN/m³ above it and 20 kN/m³ (saturated) below it, and there is no drainage. With γw = 9.81 kN/m³, what is the total horizontal thrust (soil plus water) per metre?
D. 162.3 kN/m
Ka = 1/3. Effective vertical stress: 17 × 2 = 34 kPa at 2 m; 34 + (20 − 9.81) × 4 = 74.76 kPa at 6 m. Active pressure: 11.33 kPa at 2 m and 24.92 kPa at 6 m. Soil thrust = ½ × 11.33 × 2 + 11.33 × 4 + ½ × (24.92 − 11.33) × 4 = 83.8 kN/m. Water pressure at the base = 9.81 × 4 = 39.24 kPa, thrust = ½ × 39.24 × 4 = 78.5 kN/m, with K = 1 for water. Total = 162.3 kN/m. 83.8 omits the water; 110.0 applies Ka to the saturated unit weight and forgets the water separately. 120.0 kN/m applies ½Ka·γsat·H² over the full height and ignores the water (½ × ⅓ × 20 × 36).
A rigid, propped culvert wall retains normally consolidated sand (φ′ = 30°, γ = 18 kN/m³, dry). Using Jaky's formula, what is the lateral earth pressure at 4 m depth?
A. 36 kPa
A wall that cannot move carries at-rest pressure. K0 = 1 − sinφ′ = 0.5, so σh = 0.5 × 18 × 4 = 36 kPa. 24 kPa is the active value, which needs the wall to move away from the soil. 216 kPa uses the passive coefficient Kp = 3 (3 × 18 × 4), and 72 kPa takes K = 1 (18 × 4).
A strip footing 2 m wide is founded at 1 m depth in saturated clay (cu = 50 kPa, φu = 0, γ = 18 kN/m³). Using Terzaghi's factors for φ = 0 (Nc = 5.7, Nq = 1, Nγ = 0), what is the net ultimate bearing capacity?
C. 285 kPa
qu = cNc + γDf·Nq = 50 × 5.7 + 18 × 1 × 1 = 303 kPa (gross). Net ultimate = qu − γDf = 285 kPa. 303 kPa is gross; 5.14cu = 257 uses the Prandtl value, not Terzaghi's; the 1.3 factor applies only to square and circular footings. 370 kPa applies the 1.3 square-footing factor (1.3 × 50 × 5.7).
A 0.3 m plate on sand settles 10 mm under a given pressure. Using Terzaghi and Peck's relation (widths in metres), what settlement is expected under a 2.0 m wide footing at the same pressure?
C. 30.2 mm
Sf = Sp[Bf(Bp + 0.3)/(Bp(Bf + 0.3))]² = 10 × [2.0 × 0.6/(0.3 × 2.3)]² = 10 × (1.739)² = 30.2 mm. Scaling in proportion to width gives 66.7 mm, far too much for sand; forgetting to square gives 17.4 mm.
This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
Do these early in revision so they are fast on the day.
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