Engineering & Technology · Material Selection, Degradation & Failure

Property maps, performance indices and material-selection reasoning

A lightweight tie must hold two points apart while stretching only a little under tension. One candidate has a high elastic modulus but is dense; another is less stiff and lighter. Comparing modulus alone does not answer which tie can meet the same requirement with less mass.

Material Structure & Properties and Engineering Design Methods & Trade-offs supply stress, strain, modulus and requirements. This chapter recalls each relationship before using it. Calculations use multiplication, division and simple algebra; the tie example can be read first as a worked substitution. All candidate properties, component histories, profiles and impact values are original teaching assumptions, not supplier data, experiments, safe-life assessments or production specifications.

Back to the chapter

After reading, you can

Derive a material index from a specific task and recognise what a property map leaves out.

Translate the task into a comparison

First state function, constraints, objective and free choices. Here the function is carrying a steady axial tensile force. The length, force and allowable extension are fixed. We want to minimise mass and may choose cross-sectional area. This is a deliberately simple elastic model of a straight uniform tie, with no bending, joint flexibility or time-dependent deformation.

Screen requirements that cannot be traded away: adequate strength, compatible environment, available dimensions and a feasible manufacturing route. An excellent mass score cannot compensate for failing one of these. A property value must also belong to a stated grade, condition, direction and temperature; a family label such as “polymer” is not a complete data record.

Find the combination that the task rewards

Let F be force, L length, A area, E elastic modulus and δ extension. For our uniform linear-elastic tie, δ = FL/(AE). To keep extension below δmax, the area must be at least FL/(Eδmax). Mass is density ρ multiplied by volume AL. Substitution gives minimum model mass m = ρFL²/(Eδmax).

With F, L and δmax fixed, less ρ/E means less model mass. Equivalently, a larger E/ρ is favourable. This combination is a performance index: a property combination derived for a particular objective and set of free variables. Change the task to bending or restrict the available thickness, and the useful index may change.

A: density 2000 kg/m³, modulus 40 GPa. B: 5000, 100. C: 1500, 50. Linear axes: density 0–6000 kg/m³, modulus 0–120 GPa. A and B have E/ρ = 20 million m²/s²; C approximately 33 million. These are invented values, not real material grades.

Read a map as a way to ask better questions

Our map contains three imaginary candidates, not handbook data. A: E = 40 GPa, ρ = 2000 kg/m³; B: 100 GPa, 5000 kg/m³; C: 50 GPa, 1500 kg/m³. A and B share E/ρ = 20 million m²/s². C has about 33 million m²/s² and lies above their constant-index line. The axes are linear; many published selection maps instead use logarithmic axes.

This map screens only the stated relationship. It cannot say whether C has adequate strength, tolerates moisture, creeps during service, can be joined or exists in the required shape. Real material data often occupy ranges rather than single points. Keep promising candidates for further evaluation, with their data sources and conditions attached.

Compare equal-function ties

Use constructed F = 1000 N, L = 1 m and δmax = 0.001 m. Candidate A needs A = 1000/(40 × 10⁹ × 0.001) = 0.000025 m², or 25 mm². Its model mass is 2000 × 0.000025 × 1 = 0.050 kg.

B needs 10 mm² and also has mass 0.050 kg. Higher modulus alone has not reduced the mass. C needs 20 mm² and gives 0.030 kg, a 40% reduction relative to 0.050 kg.

C’s nominal tensile stress would be 1000 N/20 mm² = 50 MPa. No strength value has been supplied, so we cannot approve C. The result is a stiffness-based shortlist, conditional on the model and on other checks.

Try the reasoning

Why do A and B give the same minimum model mass?
Show the answer and reasoning
Their E/ρ ratios are equal. B needs less area but its greater density exactly offsets that reduction in this model.
Does C’s favourable index prove that it is suitable?
Show the answer and reasoning
No. Strength, durability, manufacture, joints and other non-negotiable requirements still need evidence.
Would the same index automatically rank a thin panel against bending?
Show the answer and reasoning
No. Different geometry and allowable changes produce different relationships. Derive the index from the new task.

What to carry forward

A property index answers a defined optimisation question. A map helps shortlist candidates; product performance still needs a complete design and evidence.

References

Teaching resources and research supporting this lesson’s explanations and examples.