NCEES FE Other Disciplines · USA · NCEES
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
NCEES FE Other Disciplines
9–14 of 110 questions. Short numericals solved with the NCEES Reference Handbook: axial stress and deformation, thermal stress, shear and moment diagrams, bending and transverse shear stress, torsion of circular shafts, beam deflection by superposition from handbook tables, Euler buckling with effective length, thin-walled pressure vessels, Mohr's circle and principal stresses, and failure criteria. Speed with units and knowing where each formula sits in the handbook decide the score.
How our questions for NCEES FE Other Disciplines spread across the units of Strength of Materials. Difficulty: 18 easy, 32 medium, 13 hard.
| Stress, strain and Hooke's law | 8 | |
| Shear force and bending moment diagrams | 6 | |
| Columns: Euler and Rankine formulas | 6 | |
| Combined stresses and Mohr's circle | 6 | |
| Theories of failure | 6 | |
| Thermal stresses | 5 | |
| Bending and shear stresses in beams | 5 | |
| Thin cylinders and spheres | 5 | |
| Elastic constants and their relations | 4 | |
| Torsion of circular shafts | 4 | |
| Deflection of beams | 4 | |
| Strain energy | 4 |
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 20 m steel irrigation pipe (α = 12 × 10⁻⁶ per °C, E = 200 GPa) is laid with a 3 mm expansion gap and then heats by 35 °C, with the far end anchored. The stress in the pipe is:
A. 54 MPa
Free expansion = 12 × 10⁻⁶ × 20,000 × 35 = 8.4 mm. The gap absorbs 3 mm, so 5.4 mm is blocked: σ = 200,000 × 5.4/20,000 = 54 MPa. Ignoring the gap gives 84 MPa. 114 MPa adds the gap to the free expansion instead of subtracting it (11.4 mm).
Further reading: Gere and Goodno, Mechanics of Materials, ch. 2
Along a beam, the maximum bending moment occurs at the section where:
C. The shear force is zero or changes sign
Because dM/dx = V, the bending moment is stationary where V = 0 or changes sign. The shear force is usually largest at the supports, where the moment of a simply supported beam is zero.
Further reading: Hibbeler, Mechanics of Materials, ch. 6
For a column fixed at the base and free at the top, the effective length used in Euler's formula is:
C. 2L
The fixed–free column buckles like half of a pinned–pinned column of length 2L, so L_e = 2L and the critical load is a quarter of the pinned–pinned value. L/2 is fixed–fixed; L/√2 is fixed–pinned.
Further reading: Timoshenko and Gere, Mechanics of Materials; Bansal, Strength of Materials
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 slender column is pinned at one end and fixed at the other. Compared with the same column pinned at both ends, its Euler load is about:
B. Two times
Fixed–pinned L_e = L/√2, so P_cr ∝ 1/L_e² = 2/L². 1.41 = √2 is the ratio of lengths, not loads.
Further reading: Timoshenko and Gere, Mechanics of Materials; Bansal, Strength of Materials
On a principal plane, the shear stress is:
D. Zero
Principal planes are defined as the planes of zero shear; on them the normal stress is an extreme. Maximum shear planes lie at 45° to them.
Further reading: Popov, Engineering Mechanics of Solids; NCEES FE Reference Handbook
A 2 m steel tie (area 500 mm², E = 200 GPa) carries a gradually applied load of 50 kN. The strain energy stored is:
B. 25 J
U = P²L/(2AE) = 50,000² × 2000/(2 × 500 × 200,000) = 25,000 N·mm = 25 J. Forgetting the ½ gives 50 J (a suddenly applied load would double the stress and give 100 J); not converting N·mm to J gives 25,000.
Further reading: Timoshenko and Gere, Mechanics of Materials; Rajput, Strength of Materials
A steel shear block (G = 80 GPa) is uniformly stressed in shear to 60 MPa. The strain energy stored per unit volume is:
B. 22.5 kJ/m³
u = τ²/(2G) = 60²/(2 × 80,000) = 0.0225 N·mm/mm³ = 22.5 kJ/m³. Using E instead of G gives 9.0 kJ/m³. Omitting the ½ gives 45.0 kJ/m³, and the solid-shaft factor τ²/(4G) gives 11.25 kJ/m³.
Further reading: Timoshenko and Gere, Mechanics of Materials; Rajput, Strength of Materials
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