How to use these practice questions
These ten single-best-answer questions cover common mechanics of materials calculations and concepts. They are not tied to a particular university or professional examination; check your course outline for its exact topic coverage and conventions.
Answer each question as it appears, then compare your reasoning with the worked explanation. Keep units consistent, write down the governing relationship before substituting values, and check whether the result is a stress, force, moment or displacement. Those habits help catch errors that a correct formula alone will not prevent.
1. Axial stress
This question tests direct stress from an axial force.
A MySummaries practice item on axial stress looks like this:
A steel bar carries an axial tensile load of 20 kN. Its uniform cross-sectional area is 200 mm². What is the average normal stress?
Average normal stress is σ = P/A. Convert 20 kN to 20,000 N: σ = 20,000/200 = 100 N/mm² = 100 MPa. The strongest distractor is 10 MPa, which comes from a factor-of-ten error in dividing the load by the area.
Since 1 N/mm² equals 1 MPa, the calculation gives 100 MPa. Tensile stress is positive under the usual sign convention; compressive stress is negative.
2. Axial elongation
This question tests elastic extension and consistent use of length units.
A MySummaries item on axial deformation uses the following values:
A uniform bar is 2.0 m long, has an area of 300 mm² and is loaded in tension by 30 kN. Take Young's modulus as 200 GPa. Assuming linear-elastic behaviour, what is its elongation?
For a uniform bar, δ = PL/(AE). Use P = 30,000 N, L = 2,000 mm, A = 300 mm² and E = 200,000 N/mm². This gives δ = 1 mm. The strongest distractor is 10 mm, usually caused by a conversion error when changing metres to millimetres or gigapascals to N/mm².
The unit conversions are the main trap: 200 GPa is 200,000 N/mm², and 2 m is 2,000 mm. Extension increases with load and length, and decreases with area and elastic modulus.
3. Fully restrained thermal expansion
This question tests thermal stress when a member cannot expand freely.
A MySummaries thermal-stress question presents a restrained steel member:
A steel bar is fully restrained against axial movement and its temperature rises by 50°C. Take E = 200 GPa and the coefficient of thermal expansion α = 12 × 10⁻⁶/°C. What thermal stress develops?
A free temperature rise would produce strain αΔT = 12 × 10⁻⁶ × 50 = 0.0006. Full restraint produces stress of magnitude EαΔT = 200,000 MPa × 0.0006 = 120 MPa. The stress is compressive because the bar is prevented from expanding. The strongest distractor is 120 MPa tension: it has the right magnitude but the wrong sign.
A uniform temperature change does not create stress in a freely expanding bar. Restraint changes that: the support reaction prevents the thermal strain and creates compression on heating.
4. Torsion of a solid circular shaft
This question tests maximum elastic shear stress in a solid shaft.
A MySummaries torsion item asks for the outer-surface stress:
A solid circular shaft of diameter 40 mm carries a torque of 0.50 kN·m. What is the maximum shear stress at its outer surface?
For a solid circular shaft, τmax = 16T/(πd³). Convert 0.50 kN·m to 500,000 N·mm. Substitution gives τmax = 500,000 × 16/(π × 40³) = 39.8 N/mm² = 39.8 MPa. The strongest distractor is 79.6 MPa, which is twice the correct result and can follow from using an incorrect diameter or section relation.
The cubic dependence on diameter matters: for the same torque, a modest increase in shaft diameter substantially reduces maximum shear stress.
5. Maximum bending moment
This question tests the bending-moment diagram for a simply supported beam with a central point load.
A MySummaries beam question tests the peak moment, not the support reactions alone:
A simply supported beam has a 4 m span and carries a 12 kN point load at midspan. What is the maximum bending moment?
For a simply supported beam with a central point load, Mmax = PL/4. Thus Mmax = 12 × 4/4 = 12 kN·m. The strongest distractor is 24 kN·m, which results from using PL/2, the product of one support reaction and the full span rather than the distance to midspan.
The reactions are 6 kN each. The moment rises linearly from a support to the centre, where it reaches 6 kN × 2 m = 12 kN·m.
6. Flexural stress
This question tests the flexure formula and the position of the point where stress is required.
A MySummaries bending-stress item supplies the section property directly:
At a beam section, the bending moment is 6 kN·m. The second moment of area is 8.0 × 10⁶ mm⁴, and the outer-fibre distance from the neutral axis is 50 mm. What is the magnitude of the outer-fibre bending stress?
Use σ = My/I. Convert 6 kN·m to 6 × 10⁶ N·mm. Then σ = (6 × 10⁶ × 50)/(8.0 × 10⁶) = 37.5 N/mm² = 37.5 MPa. The strongest distractor is 75 MPa, which doubles the result, often through a misplaced factor of two when interpreting the neutral-axis distance.
Stress is zero at the neutral axis under the elementary beam model and reaches its largest magnitude at the outer fibres. The sign depends on which side is in tension for the stated moment direction.
7. Maximum transverse shear in a rectangular section
This question tests the non-uniform shear-stress distribution across a rectangular cross-section.
A MySummaries shear question distinguishes maximum stress from average stress:
A rectangular beam section has area 1,000 mm² and carries a transverse shear force of 20 kN. What is the maximum shear stress, using the elementary elastic distribution for a rectangle?
The average shear stress is V/A = 20,000/1,000 = 20 MPa. For a rectangular section, the maximum at the neutral axis is 3V/(2A), or 1.5 times the average: 30 MPa. The strongest distractor is 20 MPa; it is the average value, not the maximum. The distribution falls to zero at the top and bottom surfaces.
A frequent mistake is applying V/A when asked for the maximum in a rectangle. First identify whether the question asks for average or peak shear stress.
8. Euler buckling
This question tests elastic critical load for a pin-ended column.
A MySummaries stability question makes the end condition explicit:
An ideal pin-ended column has length 2,000 mm, second moment of area 8.0 × 10⁶ mm⁴ and Young's modulus 200,000 N/mm². Using Euler theory, what is its critical buckling load?
For a pin-ended column, Pcr = π²EI/L². Substitution gives π² × 200,000 × 8.0 × 10⁶/(2,000²) = 3.95 × 10⁶ N, or 3.95 MN. The strongest distractor is 0.395 MN, a factor-of-ten error in evaluating the powers of ten. Euler theory assumes an ideal slender column and elastic behaviour.
The effective length is the actual length for ideal pin-ended supports. Other end conditions change effective length, so do not use this result without checking the support condition.
9. Principal stresses
This question tests principal stresses in plane stress using the stress transformation relation.
A MySummaries principal-stress item includes both normal and shear components:
At a point in plane stress, σx = 80 MPa, σy = 20 MPa and τxy = 30 MPa. What are the two in-plane principal stresses?
The centre of Mohr's circle is (σx + σy)/2 = 50 MPa. Its radius is √[((σx − σy)/2)² + τxy²] = √(30² + 30²) = 42.4 MPa. The principal stresses are 50 ± 42.4, or 92.4 MPa and 7.6 MPa. The strongest distractor is 80 MPa and 20 MPa, which ignores the effect of shear stress.
The principal planes have zero in-plane shear stress. The pair of principal stresses is centred on the average normal stress; the shear component determines the radius of the circle.
10. Factor of safety against yielding
This question tests the simple ratio of yield strength to a stated equivalent stress.
A MySummaries final question checks whether the ratio is set up in the correct direction:
A ductile component has a yield strength of 250 MPa. Its maximum von Mises stress under the stated load is 100 MPa. Using factor of safety = yield strength / equivalent stress, what is the factor of safety against yielding?
The factor of safety is 250/100 = 2.5. The strongest distractor is 0.40, which reverses the ratio. This result is based on the specified von Mises stress and yield strength; it does not by itself establish safety against fatigue, fracture, buckling or other failure modes.
A factor of safety is dimensionless. State the failure mode and the stress measure used, rather than treating one ratio as a complete assessment of every possible failure.
Review the errors, not just the score
After checking your answers, sort errors by cause: unit conversion, wrong section formula, sign convention, or misunderstanding what quantity was requested. Rework a missed calculation from the governing equation without looking at the answer, then make a short recall prompt for any definition or relationship you could not state.
A MySummaries weak-topic view can make the next revision target explicit:
Use the weakest section to choose a focused follow-up, but revisit stronger topics as well. In particular, check that a formula remains clear when the geometry or boundary condition changes.
If the same point costs you marks repeatedly, turn it into a single-question prompt and practise retrieving it without cues. A MySummaries remediation tray could hold a reminder like this:
You lost this mark twice: for a rectangular section, what is the maximum transverse shear stress in terms of V and area A?
The answer is 3V/(2A), at the neutral axis. Keep the prompt narrow: one card should test one relationship, not an entire chapter.
How MySummaries helps
MySummaries lets you build a revision board from your own lecture notes, slides or problem sheets, then practise retrieval with flashcards and written questions generated from that material. For this subject, that can help you keep your course's notation and assumptions alongside the worked calculations. Open MySummaries.