Study guides

Mechanics of Materials

3 guides for this subject — how to prepare, what to practise, and how to know it has gone in.

A practice question

Choose an answer and it is marked, with the explanation. From Mechanics of Materials Practice Questions: Worked Answers.

Question 11 mark

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.

A single-best-answer item tests calculation of direct axial stress.

A spoken answer, marked

An examiner's question, answered out loud and marked criterion by criterion. From Mechanics of Materials Viva Questions: Practice Guide.

Oral — Axial extension of a uniform barMarked

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.

2:283:00Mark answer
78%Axial extension of a uniform bar — marked78/100 · Sound calculation; make the setup explicit · 2:28 spoken of 3:00
Model and assumptions15/20

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.

Calculation and units17/20

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.

Interpretation16/20

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.

Limits of the model14/20

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.

Start studying mechanics of materials

Yours are written from your own notes, and the free plan needs no payment details.

Guides

Studying mechanics of materials? Build your mechanics of materials board from your own notes — the cards, papers and lectures are written from it. Free plan: 20 cards, a paper, a lecture and a viva to try.

Start on MySummaries

Other subjects