Agricultural and Biosystems Engineers Licensure Examination (ABELE) · Philippines · Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC
Original practice questions written to the published syllabus · not past papersHow members carry load: stress, strain, bending, torsion, deflection and buckling, used in machine and structure design.
In the exams: NCEES FE 9–14 questions; NCEES PE strength of materials; ICAR AIEEA-PG Unit IV
How our questions for Agricultural and Biosystems Engineers Licensure Examination spread across the units of Strength of Materials. Difficulty: 7 easy, 21 medium, 6 hard.
| Stress, strain and Hooke's law | 5 | |
| Thin cylinders and spheres | 5 | |
| Deflection of beams | 4 | |
| Combined stresses and Mohr's circle | 4 | |
| Thermal stresses | 3 | |
| Shear force and bending moment diagrams | 3 | |
| Bending and shear stresses in beams | 3 | |
| Columns: Euler and Rankine formulas | 3 | |
| Elastic constants and their relations | 1 | |
| Torsion of circular shafts | 1 | |
| Strain energy | 1 | |
| Theories of failure | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
For an aluminium alloy used in a sprayer frame, which has no distinct yield point, the yield strength quoted by a supplier is usually the:
D. Stress at 0.2 percent offset strain, drawn parallel to the elastic line
Materials without a sharp yield point are given a proof stress: draw a line parallel to the initial elastic slope from 0.2 percent strain and read the stress where it cuts the curve. Necking begins at the ultimate stress, not yield, and dividing fracture stress by two is not a definition of yield.
Further reading: Gere and Goodno, Mechanics of Materials, ch. 1–2
A 20 mm pin passes through a 12 mm thick plate lug on a trailer chassis and transfers 24 kN. The bearing stress between the pin and the lug is:
C. 100 MPa
Bearing stress uses the projected area d × t: 24,000/(20 × 12) = 100 MPa. P/(πd²/4) = 76.4 MPa is the shear stress in the pin, and P/(πdt) = 31.8 MPa uses the full cylindrical surface, not the projected area. 50 MPa wrongly halves the load as if the pin bore in two lugs (24,000/(2 × 20 × 12)).
Further reading: Gere and Goodno, Mechanics of Materials, ch. 1–2
A 50 mm wide × 150 mm deep timber purlin, placed on edge, carries a bending moment of 3 kN·m. The maximum bending stress is:
C. 16.0 MPa
Z = bd²/6 = 50 × 150²/6 = 187,500 mm³; σ = 3 × 10⁶/187,500 = 16.0 MPa. Laid flat, Z = 150 × 50²/6 and the stress triples to 48.0 MPa. Using bd²/12 as Z doubles the answer. Using bd²/12 as Z gives 32.0 MPa, and converting 3 kN·m with 10⁵ instead of 10⁶ gives 1.6 MPa.
Further reading: Beer and Johnston, Mechanics of Materials, ch. 4 and 6
A simply supported steel beam of 4 m span (E = 200 GPa, I = 20 × 10⁶ mm⁴) carries a uniformly distributed load of 5 kN/m. The midspan deflection is closest to:
D. 4.17 mm
δ = 5wL⁴/(384EI) with w = 5 N/mm: 5 × 5 × 4000⁴/(384 × 200,000 × 20 × 10⁶) = 4.17 mm. wL⁴/(8EI) is the cantilever result; wL⁴/(48EI) mixes up the point-load constant. wL⁴/(8EI) = 40.0 mm, wL⁴/(48EI) = 6.67 mm, and 0.417 mm is a decimal slip.
Further reading: Hibbeler, Mechanics of Materials, ch. 12
A simply supported steel lifting beam of span 3 m (E = 200 GPa, I = 8 × 10⁶ mm⁴) carries 6 kN at midspan. The central deflection is closest to:
A. 2.11 mm
δ = WL³/(48EI) = 6000 × 3000³/(48 × 200,000 × 8 × 10⁶) = 2.11 mm. WL²/(16EI) is the end slope in radians, not a deflection. WL³/(3EI) = 33.8 mm is the cantilever result, 5WL³/(384EI) = 1.318 mm borrows the uniform-load constant, and 0.0021 mm is the answer in metres.
Further reading: Hibbeler, Mechanics of Materials, ch. 12
A steel pipe post of a poly-house frame (outer diameter 60 mm, inner 50 mm, E = 200 GPa) is 3 m long and pinned at both ends. Its Euler critical load is closest to:
C. 72.2 kN
I = π(60⁴ − 50⁴)/64 = 329,376 mm⁴; P_cr = π²EI/L² = π² × 200,000 × 329,376/3000² = 72.2 kN. A quarter of this is the fixed–free value; ignoring the bore overestimates. The fixed–free value is 18.1 kN, fixed–fixed gives 289.0 kN, and a solid 60 mm section gives 139.5 kN.
Further reading: Timoshenko and Gere, Mechanics of Materials; Bansal, Strength of Materials
A mild-steel solid round column, 50 mm in diameter and 2 m long, is pinned at both ends. Using Rankine's formula with σ_c = 320 MPa and a = 1/7500, the crippling load is closest to:
D. 142.4 kN
A = 1963.5 mm², k = d/4 = 12.5 mm, L/k = 160. P = 320 × 1963.5/(1 + 160²/7500) = 142.4 kN. The crushing load σ_cA = 628.3 kN ignores buckling; Euler alone gives 151.4 kN; forgetting to square the slenderness ratio overestimates the load. Forgetting to square L/k gives 615.2 kN.
Further reading: Timoshenko and Gere, Mechanics of Materials; Bansal, Strength of Materials
At a point in a loaded bracket, σ_x = 60 MPa, σ_y = 20 MPa and τ_xy = 15 MPa. The maximum in-plane shear stress is:
C. 25 MPa
τ_max = radius of Mohr's circle = √[((60 − 20)/2)² + 15²] = √(400 + 225) = 25 MPa. Taking only (σ_x − σ_y)/2 = 20 MPa ignores the applied shear. 50 MPa is the diameter of Mohr's circle, not the radius.
Further reading: Popov, Engineering Mechanics of Solids; NCEES FE Reference Handbook
A short 200 mm × 300 mm concrete pedestal for a grain dryer carries 200 kN applied 40 mm off-centre along the 300 mm side. The maximum and minimum compressive stresses are:
D. 6.00 and 0.67 MPa
σ = P/A ± Pe/Z with Z = 200 × 300²/6 about the axis parallel to the 200 mm side: P/A = 3.333 MPa, 6e/b = 0.8, so σ = 3.333(1 ± 0.8) = 6.00 and 0.67 MPa. Using the 200 mm side for b puts e beyond the core and predicts tension (a negative value). 3.33 and 3.33 MPa ignore the eccentricity, and 3.78 and 2.89 MPa drop the 6 in 6e/b.
Further reading: Popov, Engineering Mechanics of Solids; NCEES FE Reference Handbook
A ductile steel has a tensile yield stress of 250 MPa. According to the distortion energy (von Mises) theory, its yield stress in pure shear is closest to:
D. 144 MPa
Von Mises: τ_y = σ_y/√3 = 144 MPa. Tresca gives 125 MPa, about 13% lower; the two theories differ most in pure shear. 177 MPa divides by √2 instead of √3.
Further reading: Popov, Engineering Mechanics of Solids; Shigley-type machine design texts
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