Stress, strain, torsion, bending and buckling. These problems are heavily weighted on the exam and reward clean units.
Choose any section below to get a free practice problem. Work it with the FE Reference Handbook open, tap your answer, then check the solution.
Normal stress is force over area, σ = P/A; shear stress is τ = V/A.
A 10 mm pin in double shear carries 12 kN. What is the shear stress?
Two shear planes: A = 2(78.54) = 157.1 mm2. τ = 12,000/157.1 = 76.4 MPa.
Common trap: (D) uses single shear.
Mechanics of MaterialsStress and StrainPoisson’s RatioThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
A free bar heated by ΔT grows δ = αLΔT.
A 1 m diameter tank with 10 mm walls holds 2 MPa. What is the hoop stress?
σh = pD/2t = 2(1000)/20 = 100 MPa.
Common trap: (A) is the longitudinal stress.
Thermal DeformationsThin-Walled Pressure VesselsThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
A shaft under torque T has shear stress τ = Tr/J, largest at the outer surface.
A 40 mm shaft, 2 m long (G = 80 GPa), carries 1 kN·m. What is the angle of twist?
J = 251,300 mm4. φ = TL/GJ = (106)(2000)/(80,000 × 251,300) = 0.0995 rad = 5.70°.
Common trap: (A) slips a decimal. Keep every length in mm and every stress in MPa (N/mm2).
Torsionangle of twistpolar momentThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Shear and moment diagrams show where a beam works hardest.
A 45 kN·m moment acts on a 150 × 300 mm rectangular beam. What is the maximum bending stress?
S = bh2/6 = 150(90,000)/6 = 2.25 × 106 mm3. σ = 45 × 106/2.25 × 106 = 20 MPa.
Common trap: (D) uses bh2/12 for the section modulus; that’s half of the correct S = bh2/6.
BeamsShearing Force and Bending MomentBeam DeflectionThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Long slender columns fail by buckling, not crushing.
A 3 m steel column (I = 2 × 106 mm4, E = 200 GPa) is fixed at the base and free at the top (K = 2). What is its buckling load?
Pinned at both ends it would buckle at π2EI/L2 = 439 kN. Pcr ∝ 1/(KL)2, so K = 2 divides that by 4: 110 kN.
Common trap: (C) multiplies by 4.
ColumnsEulerMohr’s CircleThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
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