USA · NCEES · licensing
Computer-based, 110 questions, 6-hour appointment (5 h 20 min exam).
The usual first step (EIT/EI) for agricultural and biological engineers. Specifications effective from the July 2020 exams. Fluid mechanics is the heaviest area.
Official source: https://ncees.org/wp-content/uploads/FE-Other-Disciplines-CBT-specs.pdf
Original questions written to this exam's published syllabus and pattern. Not past papers.
1. What is the rank of the matrix [[1, 2, 3], [2, 4, 6], [1, 1, 1]] (rows listed in order)?
B. 2
Row 2 is exactly twice row 1, so R2 − 2R1 gives a zero row. Rows 1 and 3 are not multiples of each other, so two independent rows remain and the rank is 2. Answering 3 assumes a 3×3 matrix is always full rank; its determinant is 0, so it is not. Rank 1 would need row 3 to be a multiple of row 1 as well.
Engineering Mathematics · Matrices, determinants and systems of linear equations
2. Six grain samples from one bin give moisture contents (per cent, wet basis) of 12.1, 13.4, 11.8, 12.9, 13.6 and 12.4. What is the sample standard deviation?
C. 0.721
Mean = 76.2/6 = 12.7000. Sum of squared deviations = 2.6000. Sample variance = 2.6000/(n − 1) = 2.6000/5 = 0.5200, so s = 0.721. Dividing by n = 6 gives the population SD 0.658; 0.520 is the variance, not the SD; 0.600 is the mean absolute deviation.
Probability, Statistics & Data Analysis · Measures of central tendency and dispersion
3. A well water contains, per litre, 162 mg Ca(HCO3)2, 120 mg MgSO4, 111 mg CaCl2 and 58.5 mg NaCl. What are its temporary and permanent hardness, as mg/L CaCO3?
C. Temporary 100, permanent 200
Convert with mass × 50/equivalent weight: Ca(HCO3)2 162 × 50/81 = 100 (temporary); MgSO4 120 × 50/60 = 100 and CaCl2 111 × 50/55.5 = 100 (permanent, total 200). NaCl causes no hardness, so adding 50 for it is wrong; 162 and 231 are raw salt masses, not CaCO3 equivalents.
Engineering Physics & Chemistry · Water chemistry: hardness, pH and treatment
4. Two chains pull on a tree stump from one ring with forces of 6 kN and 4 kN; the angle between the chains is 120 degrees. What is the magnitude of the resultant pull on the stump?
D. 5.29 kN
R = sqrt(P^2 + Q^2 + 2PQ cos theta) = sqrt(36 + 16 + 2 x 6 x 4 x cos 120) = sqrt(28) = 5.29 kN. Using +cos 60 instead of cos 120 gives 8.72 kN; treating the chains as perpendicular gives 7.21 kN; adding arithmetically gives 10 kN, which is only right for chains pulling in the same direction.
Engineering Mechanics (Statics & Dynamics) · Force systems and resultants
5. A trailer hitch pin of 20 mm diameter connects a drawbar clevis and carries a pull of 30 kN in double shear. The average shear stress in the pin is closest to:
A. 47.7 MPa
Double shear means two planes carry the load. A = π(20)²/4 = 314.2 mm²; τ = 30,000/(2 × 314.2) = 47.7 MPa. Using one plane gives 95.5 MPa (single-shear slip); using πd² as the area gives 30,000/(2 × 1256.6) = 11.9 MPa, and d² instead of πd²/4 gives 30,000/(2 × 400) = 37.5 MPa.
Strength of Materials · Stress, strain and Hooke's law
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