Five original mechanics of materials problems in the style of the FE Mechanical exam, each with four answer choices and a full step-by-step solution: axial deformation, bending stress, principal stress, thermal stress and column buckling. Give yourself about 3 minutes per problem.
How to use these problems
Try each problem with a timer and your approved calculator before you open the solution. Keep units on every line; most wrong choices below come from a units slip or the wrong formula variant.
When you miss one, write down which formula you should have recognized, then find it in the FE Reference Handbook so you can find it faster next time.
Keep the mechanics of materials cheat sheet next to you while you study, then practise finding the same formulas in the FE Reference Handbook.
Practice problems
Problem 1: Elongation of a stepped rod
A steel rod has two segments: 0.80 m long with a 20 mm diameter, then 0.50 m long with a 12 mm diameter. It carries a 15 kN axial tension throughout. Take E = 200 GPa (typical for steel). What is the total elongation?
0.19 mm
0.33 mm
0.52 mm
0.62 mm
Show the solution to problem 1
Areas of the two segments.A₁ = π(20)²/4 = 314.2 mm²; A₂ = π(12)²/4 = 113.1 mm²
Elongation of each segment, δ = P·L/(A·E), with E = 200 000 N/mm².δ₁ = 15 000 × 800/(314.2 × 200 000) = 0.191 mm; δ₂ = 15 000 × 500/(113.1 × 200 000) = 0.332 mm
A simply supported beam spans 3.0 m and carries a uniform load of 5.0 kN/m (including its own weight). The cross-section is a solid rectangle 100 mm wide and 200 mm deep. What is the maximum bending stress?
8.4 MPa
2.1 MPa
16.9 MPa
33.8 MPa
Show the solution to problem 2
Maximum moment at mid-span for a uniform load.M = w·L²/8 = 5.0 × 3.0²/8 = 5.625 kN·m
Section modulus of the rectangle.S = b·h²/6 = 100 × 200²/6 = 666 700 mm³
An aluminum bar is fixed between two rigid walls with no gap and no initial stress. It is heated uniformly by 40 °C. Take E = 70 GPa and α = 23 × 10⁻⁶ /°C (typical for aluminum). What is the stress in the bar?
64.4 MPa tension
64.4 MPa compression
0; the bar is free to expand
0.92 MPa compression
Show the solution to problem 4
The walls stop the bar from expanding, so the mechanical strain cancels the thermal strain.ε_thermal = α·ΔT = 23 × 10⁻⁶ × 40 = 920 × 10⁻⁶
The stress needed to push the bar back to its original length.σ = E·α·ΔT = 70 000 MPa × 920 × 10⁻⁶ = 64.4 MPa
The walls push on the bar, so the stress is compressive. The answer does not depend on the bar’s length or area.
A solid round steel column, 30 mm in diameter and 1.5 m long, is pinned at both ends. Take E = 200 GPa (typical for steel). What is the Euler critical buckling load?
8.7 kN
34.9 kN
140 kN
69.8 kN
Show the solution to problem 5
Moment of inertia of the round section.I = π·d⁴/64 = π × 30⁴/64 = 39 760 mm⁴
Pinned at both ends, so the effective-length factor is K = 1.K·L = 1 500 mm
Euler formula. Choice A uses K = 2 and choice C uses K = 0.5.P_cr = π²·E·I/(K·L)² = π² × 200 000 × 39 760 / 1 500² = 34 900 N
Check that Euler applies: the stress at buckling is well below the yield strength of structural steel.σ_cr = 34 900/706.9 = 49 MPa