What a mechanics of materials viva tests
A viva on mechanics of materials asks you to explain how loads produce stresses, strains and deformation, and to defend the assumptions behind your calculations. The examiner may give you a short problem, ask you to sketch a force or stress diagram, or change a condition partway through your explanation. There is no single international viva format or universal marking scheme for this subject; follow the guidance for your course or assessment from your institution.
The most reliable answer has a visible sequence:
- State the model. Identify the member, supports, loads, dimensions and material assumptions.
- Set up equilibrium or compatibility. Draw a free-body diagram where it helps, and state the relevant boundary conditions.
- Choose and apply a relation. Define symbols, show units and keep the calculation in a clear order.
- Interpret the result. Say where the maximum occurs, what the sign means, and whether the result is plausible.
- Name the limitation. Explain what could invalidate the simple model, such as yielding, a stress concentration or large deformation.
Do not start with a formula before deciding what the question is asking. A correct number with no load path, assumptions or interpretation is difficult to assess. If you do not know a value, state what information is missing rather than silently assuming it.
The examiner strip in MySummaries can be set up like this:
This strip represents a practice examiner persona, not a published rubric for every engineering viva. It keeps the practice focused on the reasoning that can be observed in a spoken answer.
Turn your course material into viva stations
Start from the equations and examples in your own lecture notes, problem sheets and laboratory material. Group related ideas into a revision board: axial loading and thermal strain; beam shear and bending; torsion; combined loading; and failure or design checks. Include the sign conventions and assumptions used in your course, since notation and conventions can vary.
For each topic, make a station with an opening problem and at least one follow-up. The opening problem should be solvable aloud in a few minutes. The follow-up should test whether you understand the model, not whether you can repeat the same calculation. For example, after finding the extension of a bar, ask what changes if the bar is heated while restrained. After finding beam stress, ask where the neutral axis is and what a notch might do to the local stress.
A set of stations on this topic ends up looking like this:
The table is a way to plan practice, not a statement that every course uses these stations. Use the actual dimensions, material properties and notation from your own coursework when you practise. In a live answer, ask for a missing value if the result depends on it.
Station 1: axial stress and extension
This station tests whether you can distinguish stress, strain and extension, and whether you can keep units consistent. Before calculating, identify the bar as a uniform, prismatic member under an axial tensile force. The simple elastic relations are average normal stress (\sigma=P/A), strain (\epsilon=\sigma/E), and extension (\delta=PL/(AE)). They assume small deformation, uniform material behaviour and no significant stress concentration.
The recorded answer below has a sound result but does not make the unit conversion or model sufficiently explicit. The practice feedback reflects platform criteria, not a universal exam rubric.
A recorded attempt at the axial-bar station looks like this:
Examiner
A 2 m steel bar has area 400 mm² and carries a tensile force of 40 kN. Find its average axial stress and elastic extension, then state your assumptions. Take E = 200 GPa.
Identified axial tension but did not clearly state that the bar is uniform and remains elastic.
ImproveName the uniform, prismatic, linear-elastic bar model before using the equations.
The stress and extension are correct, but the conversion from mm² and GPa was not shown.
ImproveShow 40 kN / 400 mm² = 100 N/mm² = 100 MPa, then calculate extension.
Recognised tensile stress but did not connect the result to strain or the deformation assumption.
ImproveState that strain is 100 MPa / 200,000 MPa = 0.0005 and the extension is 1.0 mm.
The answer did not say what could make the elastic calculation unsuitable.
ImproveMention yielding, non-uniform area or a local stress concentration as reasons to refine the check.
A strong answerI will model this as a uniform prismatic bar in uniaxial tension, with linear-elastic behaviour and small strain. The average stress is P/A = 40,000 N / 400 mm² = 100 N/mm², or 100 MPa. With E = 200,000 MPa, the strain is σ/E = 0.0005, so over 2,000 mm the extension is 1.0 mm. This is an average-stress calculation; a shoulder, hole or yielding would require a more detailed assessment.
That is a MySummaries station, filled with mechanics of materials material. Yours is written from your own notes. Start free
The key unit step is that 1 N/mm² equals 1 MPa. With the given modulus, 200 GPa is 200,000 MPa, so strain is dimensionless and the extension calculation can be performed in millimetres. If the examiner changes the material or asks about a restrained temperature increase, do not reuse the answer unchanged: thermal strain and the restraint condition now matter. For a fully restrained, uniform bar in a simple elastic model, the magnitude of thermal stress is related to (E\alpha\Delta T); its sign is compressive for a temperature rise that the bar cannot accommodate.
Listen back for whether you said what “average” means. The equation (P/A) does not resolve a local peak at a notch or a contact surface. That distinction is often more valuable than adding an extra decimal place.
Station 2: bending stress in a simply supported beam
This station checks equilibrium, the bending-moment diagram and the flexure relation. The dimensions are given so you can calculate rather than recite a general rule. Say which way the beam bends before assigning tension and compression: for a simply supported beam under a downward midspan load, the central sagging region puts the top fibres in compression and the bottom fibres in tension.
A recorded attempt at the beam station looks like this:
Examiner
A simply supported 4 m beam carries a 12 kN point load at midspan. Its rectangular section is 100 mm wide and 200 mm deep, with the depth vertical. Find the maximum bending stress and describe the stress distribution.
Found equal 6 kN reactions and the 12 kN·m central moment, with limited explanation.
ImproveState the symmetry and use Mmax = PL/4 for a central point load on a simply supported span.
The 18 MPa result is correct, but the second moment of area was not shown.
ImproveGive I = bh³/12 = 6.67 × 10⁻⁵ m⁴ and use σ = Mc/I with c = 0.1 m.
Described a linear variation but did not identify the neutral axis or which face is in tension.
ImproveState zero longitudinal bending stress at the neutral axis, increasing linearly to the outer fibres; bottom is in tension for sagging.
Did not qualify the use of elementary beam theory or distinguish bending from shear stress.
ImproveState the beam-theory assumptions and note that shear stress is a separate distribution.
A strong answerThe reactions are 6 kN each by symmetry, so the maximum sagging moment at midspan is PL/4 = 12 kN·m. For the rectangular section, I = bh³/12 = 0.1 × 0.2³/12 = 6.67 × 10⁻⁵ m⁴, and the outer-fibre distance is 0.1 m. The flexure relation gives σmax = Mc/I = 18 MPa. Longitudinal bending stress varies linearly through the depth, is zero at the neutral axis, and is tensile at the bottom for this sagging case. This uses elementary beam theory; transverse shear stress is a different stress distribution.
In the calculation, keep all lengths in metres: (M=12{,}000) N·m, (c=0.1) m and (I=6.67\times10^{-5}) m⁴. The resulting outer-fibre stress is 18 MPa. The section’s maximum transverse shear stress is not found from (Mc/I); for a rectangular section it has a parabolic distribution and is greatest at the neutral axis. If asked to calculate it, identify the shear force at the section and use the appropriate relation for the section rather than confusing it with bending stress.
Practise drawing a quick bending-moment sketch as you speak. For this loading, the moment is zero at the supports and reaches its maximum at midspan. If the load position changes, symmetry no longer gives equal reactions automatically; return to equilibrium instead of relying on the memorised result.
The transcript view can isolate one phrase that cost detail in the beam answer:
The reactions are six kilonewtons each, and the maximum moment is twelve kilonewton-metres. The second moment of area is the rectangle formula. I get eighteen megapascals at the outside of the beam. The top is in tension.
The second moment of area is the rectangle formula.
The phrase names a formula but does not show which dimension is cubed or what value follows from the given orientation.
Say: For a rectangle with the 200 mm depth vertical, I = bh³/12 = 0.1 × 0.2³/12 = 6.67 × 10⁻⁵ m⁴.The last phrase is marked as a knowledge gap because the sign is reversed for the stated sagging beam: the top is in compression and the bottom is in tension. When reviewing a transcript, separate missing detail from a wrong concept. One needs a fuller explanation; the other needs correction before it becomes a fluent habit.
Station 3: torsion of a solid circular shaft
This station combines a polar second moment of area, shear stress and angle of twist. State that the shaft is solid and circular, and that the calculation assumes uniform elastic torsion away from load-introduction effects. A useful order is to find (J), calculate the outer-surface shear stress, then calculate twist. Do not confuse the polar second moment (J) with the bending second moment (I).
A recorded attempt at the torsion station looks like this:
Examiner
A 1.2 m long solid circular shaft of diameter 40 mm transmits 600 N·m. With shear modulus 80 GPa, estimate the maximum shear stress and angle of twist.
Used the solid-shaft expression for J and obtained a consistent value.
ImproveState J = πd⁴/32 and make the diameter-to-radius distinction explicit.
Calculated a value close to 47.7 MPa at the outer surface.
ImproveShow τmax = Tc/J with c = d/2 = 0.02 m and consistent SI units.
Found about 0.0358 rad but did not initially convert it to degrees or interpret it.
ImproveUse θ = TL/(JG), then state 0.0358 rad, approximately 2.05°.
Recognised elastic torsion but did not mention local effects or the radial stress distribution.
ImproveSay shear stress rises linearly from zero at the centre to its maximum at the surface in the ideal solid shaft.
A strong answerFor a solid circular shaft, J = πd⁴/32 = 2.51 × 10⁻⁷ m⁴, with d = 0.04 m. The maximum shear stress occurs at the outer radius c = 0.02 m, so τmax = Tc/J = 600 × 0.02 /(2.51 × 10⁻⁷) ≈ 47.7 MPa. The elastic angle of twist is θ = TL/(JG) = 600 × 1.2 /(2.51 × 10⁻⁷ × 80 × 10⁹) ≈ 0.0358 rad, or 2.05°. In the ideal shaft, shear stress varies linearly from zero at the centre to the maximum at the surface. This assumes a uniform elastic shaft and does not include stress concentrations or other local effects.
Check the units before deciding whether the result is plausible: torque is in N·m, length in metres, (G=80\times10^9) Pa, and (J) in m⁴. The maximum shear stress is about 47.7 MPa and the twist is about 0.0358 radians, or 2.05 degrees. If the shaft were hollow, the expression for (J) would change. If a keyway, shoulder or other geometric discontinuity were included, the nominal torsion formula would not by itself describe the local peak stress.
A repeatable practice routine
For each station, record one uninterrupted attempt before reading the feedback. Aim to speak the setup, calculation and interpretation—not just the final result. Then review the recording and mark only specific weaknesses: a missing assumption, an incorrect sign, an unstated unit conversion, or a conclusion that is not supported by the calculation.
Repeat the station later with one changed condition. For the bar, change the area or ask about restraint during heating. For the beam, move the point load away from midspan and recalculate reactions. For the shaft, change the diameter or ask how a hollow section alters (J). A changed condition tests whether you understand the governing relation rather than remember a worked answer.
Keep a short error log with three columns: what you said, why it was incomplete or wrong, and the exact sentence or calculation you will use next time. Retest weak stations until you can explain each step without rushing. Use your course’s own sign convention, permitted formula sheet and material data; where the question omits a necessary property, ask for it or state a clearly labelled assumption.
How MySummaries helps
Build a board from your mechanics notes, problem sheets and photographed working, then turn its sections into spoken stations. Recorded viva practice lets you compare your calculation, assumptions and interpretation with feedback and revisit the points that cost marks. Start with your own material at MySummaries.