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.
Machinery systems: using ASABE D497 draft and traction relationships to size tractors and implements, drawbar and PTO power, ballast and slip.
How soil resists tools and supports wheels: draft, slip, tractive efficiency, weight transfer and stability.
Our 16 NCEES PE Agricultural and Biological Engineering questions in this subject: 0 easy, 11 medium, 5 hard, across 9 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 tractor rated at 45 kW is to pull a mouldboard plough in clay. The farmer wants the widest plough possible. Which reasoning is sound?
C. Estimate draft for local soil, divide drawbar power available at efficient slip by speed, then choose width with a margin
Width must be set from drawbar power actually available (engine power × transmission efficiency × tractive efficiency, at a sustainable load) and the local draft. Sandy soil draft is lower, not higher, than clay, and draft is not set by implement mass.
A bias-ply tyre (b = 0.45 m, d = 1.6 m, deflection/section height δ/h = 0.2) carries 18 kN on soil with CI = 800 kPa. What is the Brixius mobility number Bn?
B. 34.7
Bn = (CI·b·d/W)(1 + 5δ/h)/(1 + 3b/d) = (32.0) × 2.0/1.8438 = 34.7. Leaving out both correction factors gives 32.0 (the Wismer-Luth Cn). 64.0 leaves out the 1 + 3b/d factor; 17.4 leaves out the 1 + 5δ/h factor.
For the same tyre (Bn = 34.7) at 12 % slip, use GTR = 0.88(1 − e^(−0.1Bn))(1 − e^(−7.5s)) + 0.04 and MRR = 1/Bn + 0.04 + 0.5s/√Bn to find the net traction ratio.
A. 0.467
GTR = 0.88 × (1 − e^(−3.471)) × (1 − e^(−0.90)) + 0.04 = 0.546. MRR = 1/34.71 + 0.04 + 0.5 × 0.12/√34.71 = 0.079. NTR = 0.546 − 0.079 = 0.467. Forgetting the +0.04 in GTR gives 0.427. 0.625 adds MRR instead of subtracting it.
A national test report gives maximum drawbar power on a concrete track. Why will a farmer rarely achieve it in the field on tilled soil?
A. Tractive efficiency on tilled soil is lower than on concrete because of greater sinkage, motion resistance and slip
Test tracks give comparable conditions across tractors, but on tilled soil more of the axle power is lost to rolling resistance and slip, so field drawbar power is lower.
What insertion rate does the ASABE procedure specify for the soil cone penetrometer?
A. About 30 mm/s
ASABE EP542 specifies a steady insertion rate of about 30 mm/s (about 1829 mm/min). Readings taken much faster or slower are not comparable.
A two-wheel-drive tractor must develop 18 kN net pull. On this soil at the target slip, the drive tyres can achieve a net traction ratio of 0.45. What dynamic load must the drive wheels carry?
C. 40.0 kN
W_dyn = NT/NTR = 18/0.45 = 40.0 kN. Multiplying gives 8.1 kN, far too little. Part of this load comes from weight transfer; ballast provides the rest. 32.7 kN divides by (1 − 0.45); 0.4 kN reads 0.45 as 45.
A contractor wants to reduce subsoil compaction from a heavy grain cart during a wet harvest. Which measure addresses subsoil compaction most directly?
D. Reduce the load per axle, for example by a smaller cart or more axles
Lower pressure helps the topsoil but subsoil stress follows total axle load. More axles or smaller loads reduce subsoil stress. Staying off wet soil is also important.
Dual wheels on a tractor, compared with singles at the same load and pressure, typically
C. reduce contact pressure and rutting but spread stress over a wider zone
Duals roughly double contact area, lowering surface pressure and rut depth, but the total load is the same and stress bulbs overlap at depth, spreading moderate stress more widely.
A tyre's load-inflation table gives a minimum pressure for the axle load. Why should a field engineer not go below it to gain traction?
A. Excess deflection overstresses the carcass and sidewalls and can cause failure or rim slip
Traction usually keeps improving as pressure falls, but the tyre is only designed to deflect so far at a given load and speed; beyond that, cords and sidewalls are damaged and the tyre may rotate on the rim.
In mixed control of a tractor hydraulic lift, the system
A. responds partly to draft and partly to implement position, limiting depth variation while still controlling load
Mixed control blends draft sensing with position sensing, so the implement neither plunges when draft falls nor holds a fixed depth regardless of load.
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.
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