Materials Science · Selecting materials for a defined function

Property combinations and performance indices

Useful background: Measuring and checking material claims; Elasticity, plasticity and tensile strength; Processing routes and microstructural evolution. Each worked case explains the quantities it uses.

Back to the chapter

After reading, you can

Derive a property combination from a defined design model.
Calculate each candidate’s required area and mass when both stress and extension are limited.
Identify the active constraint and explain why the preferred candidate can change with the extension limit.

The governing equation suggests the index

For a uniform tie carrying force F over length L, mass is ρAL. If allowable tensile stress σ limits area, A must be at least F/σ. Minimum ideal mass is therefore ρLF/σ. Maximising σ/ρ helps this particular design.

A performance index collects material properties that determine an objective within a specified design model. Start with the function, the requirements that must be met, the quantity to minimise and the dimensions that can change. Here the function is to carry a prescribed static tensile force. The objective is minimum mass, and the uniform cross-sectional area is the freely chosen dimension. A material with a larger index is preferred only while that derivation’s requirements and free dimensions apply.

Use F for tensile force in newtons (N), L for initial length in metres (m), A for initial cross-sectional area in square metres (m²), and ρ for density in kilograms per cubic metre (kg/m³). Then m = ρAL has units kg. Engineering stress is F/A, in pascals (Pa = N/m²). Let σ_allow be a supplied allowable tensile stress for the specified conditions. Requiring F/A ≤ σ_allow gives A ≥ F/σ_allow.

Because density and length are positive, adding area increases mass. If this stress requirement is the only area constraint, choose its boundary A = F/σ_allow. Substitution gives m_min = ρLF/σ_allow = LF/(σ_allow/ρ). For the same F and L, increasing σ_allow/ρ decreases minimum mass. This is the original tie derivation with the allowable explicitly named; an unexplained tensile maximum is not automatically an allowable.

A different constraint gives a different answer

For a uniform linear-elastic tie, extension δ = FL/(AE). If a maximum extension δmax governs, A ≥ FL/(Eδmax), giving minimum mass ρFL²/(Eδmax); this model therefore favours high E/ρ. Bending, buckling or fixed dimensions create other combinations. An index is a compact result of assumptions, not an intrinsic ranking of all possible materials.

In δ = FL/(AE), δ is the extension in metres and E is Young’s modulus in pascals. The formula assumes small, uniform, uniaxial linear-elastic deformation. A permitted maximum extension δ_max is positive. Rearranging δ ≤ δ_max gives A ≥ FL/(Eδ_max). If only extension limits area, substituting its boundary into m = ρAL gives m_min = ρFL²/(Eδ_max). For identical F, L and δ_max, maximise E/ρ.

The two indices have the same dimensions: (N/m²)/(kg/m³) = N·m/kg = m²/s². Equal dimensions do not mean they answer the same question. σ_allow/ρ describes the supplied stress requirement; E/ρ describes the extension requirement. Neither should be multiplied or averaged into an unexplained overall score.

Keep the original doubling example below within its stress-only case. Once both requirements apply, the area must satisfy both lower bounds: A_req = max(F/σ_allow, FL/(Eδ_max)). This maximum chooses the larger required area. The active constraint is the requirement reached at equality; the other is satisfied with some margin, or both are active when their area bounds tie.

Two candidates for the same tie

Consider an invented uniform tie with F = 10,000 N and L = 1.00 m. The table supplies hypothetical candidate properties, including allowables that are inputs to this exercise rather than real product recommendations. Both materials are assumed linear elastic over the calculated stress range. Area can vary continuously; equality at a supplied bound is allowed in this ideal comparison.

For candidate A the stress floor is 10,000/(200 × 10⁶) = 0.000050 m² = 50 mm². Candidate B needs at least 10,000/(250 × 10⁶) = 40 mm². The conversion is 1 mm² = 10⁻⁶ m². If extension were unrestricted, masses would be 2,000 × 1 × 50 × 10⁻⁶ = 0.100 kg and 8,000 × 1 × 40 × 10⁻⁶ = 0.320 kg. A wins this stress-only comparison.

At δ_max = 1.00 mm, extension instead requires 250 mm² for A and 50 mm² for B. Both exceed their respective stress floors. The resulting minimum masses are 0.500 kg and 0.400 kg, so B wins. Its density is four times larger, but its required area is five times smaller. Actual stress is 40 MPa for A and 200 MPa for B, each below its supplied allowable.

Relax the extension limit to 2.00 mm. A now requires 125 mm² for extension and 50 mm² for stress, so choose 125 mm² and obtain 0.250 kg. B requires 25 mm² for extension and 40 mm² for stress, so choose 40 mm² and obtain 0.320 kg. B’s actual extension is 1.25 mm, less than the permitted 2.00 mm. A wins the combined comparison even though B has the larger E/ρ.

Invented candidate inputs and independently calculated minimum areas and masses. Same force, length and ideal tensile conditions throughout.
Quantity	Candidate A	Candidate B
Density ρ (kg/m³)	2,000	8,000
Young’s modulus E (GPa)	40	200
Allowable stress σ_allow (MPa)	200	250
σ_allow/ρ (m²/s²)	100,000	31,250
E/ρ (m²/s²)	20,000,000	25,000,000
Stress-only area / mass	50 mm² / 0.100 kg	40 mm² / 0.320 kg
At δ_max = 1.00 mm: area / mass	250 mm² / 0.500 kg	50 mm² / 0.400 kg
At δ_max = 2.00 mm: area / mass	125 mm² / 0.250 kg	40 mm² / 0.320 kg

Read where the governing requirement changes

For each candidate, the two area bounds are equal when F/σ_allow = FL/(Eδ_max). Cancelling the common positive force gives δ_max = Lσ_allow/E. The transition is 5.00 mm for A and 1.25 mm for B. Below its transition, extension controls the minimum area; above it, the stress floor controls. At the transition both constraints are active.

The figure plots the mass calculated from the maximum of both bounds, not two separate mass estimates to choose between. Its horizontal portions are stress-controlled mass floors. Its falling portions occur because allowing more extension reduces the elastic area requirement. All curve points refer to the same force, length and candidate properties; the horizontal axis changes only the permitted extension.

Between 1.25 mm and 5.00 mm, B stays at 0.320 kg while A has m_A = (0.500 kg·mm)/δ_max when δ_max is expressed in mm. Set 0.500/δ_max = 0.320 to find the ranking crossover at δ_max = 1.5625 mm. A then uses 160 mm² and B uses 40 mm², giving equal masses. Their actual extensions differ: A reaches 1.5625 mm, whereas B extends only 1.25 mm. A common permitted maximum does not require every candidate to reach it.

These are component-model results, not universal material rankings. A fixed area removes the freedom used in the derivation; a compression member may buckle; bending, joints, manufacturing limits, variable temperature, fatigue, corrosion, self-weight or nonuniform stress require additional equations or constraints. Real allowables need evidence for the service conditions. The ideal calculations here neither supply them nor approve a component.

Invented uniform tensile tie: force 10,000 N, length 1 m. Candidate A has density 2,000 kg/m³, E 40 GPa and allowable stress 200 MPa; B has density 8,000 kg/m³, E 200 GPa and allowable stress 250 MPa. Minimum area is the larger of the stress and extension bounds. Horizontal axis is permitted extension from 0.5 to 6 mm; vertical axis is minimum feasible mass from zero to 1.1 kg. Solid blue A falls as 0.500 divided by the extension limit in mm until 5 mm, then remains at 0.100 kg. Dashed orange B falls as 0.400 divided by the extension limit in mm until 1.25 mm, then remains at 0.320 kg. The curves cross at 1.5625 mm and 0.320 kg. Square markers indicate the transition where each candidate changes active constraint. A hollow circle indicates equal mass, not equal actual extension. Listed exact values supplement the schematic line interpolation. Properties and allowables are hypothetical; this is not measured material performance.

Worked examples

Work through the case

For fixed F and L, an ideal tie material doubles σ/ρ. How does its minimum mass change?

Mass is inversely proportional to σ/ρ.
Doubling the ratio halves the modelled mass.
Retain the static tensile assumptions.

It halves in this model.

Check the selected areas

For the invented 10,000 N, 1 m tie with δ_max = 2 mm, choose an area for each candidate and check both requirements.

Candidate A: A_stress = 10,000/(200 × 10⁶) = 50 × 10⁻⁶ m². A_extension = 10,000 × 1/(40 × 10⁹ × 0.002) = 125 × 10⁻⁶ m². Choose the larger, 125 mm².
A’s stress is 10,000/(125 × 10⁻⁶) = 80 MPa ≤ 200 MPa. Its extension is 10,000/(125 × 10⁻⁶ × 40 × 10⁹) = 0.002 m. Its mass is 2,000 × 125 × 10⁻⁶ × 1 = 0.250 kg.
Candidate B: A_stress = 40 mm² and A_extension = 25 mm². Choose 40 mm². Its stress is 250 MPa and extension is 10,000/(40 × 10⁻⁶ × 200 × 10⁹) = 0.00125 m = 1.25 mm. Both meet their limits.
B’s mass is 8,000 × 40 × 10⁻⁶ × 1 = 0.320 kg. Compare only after making each candidate feasible: A is lighter by 0.070 kg.

Choose A at 125 mm² for a modelled minimum mass of 0.250 kg; extension controls A, while allowable stress controls B.

Transfer to a tighter requirement

Keep the same force, length and candidates, but require δ_max = 0.80 mm. Which has the lower feasible mass?

A_extension = 10,000/(40 × 10⁹ × 0.00080) = 312.5 mm², larger than its 50 mm² stress floor. Its mass is 0.625 kg.
B_extension = 10,000/(200 × 10⁹ × 0.00080) = 62.5 mm², larger than its 40 mm² stress floor. Its mass is 0.500 kg.
The resulting stresses are 32 MPa and 160 MPa, both below the respective allowables. Both extend by 0.80 mm, because extension controls both choices.
Now E/ρ gives the appropriate ordering: B’s value is 25/20 = 1.25 times A’s, so its extension-limited mass is 1/1.25 = 0.80 times A’s.

B is lighter at 0.500 kg. Tightening the permitted extension changed which comparison determines the winner.

Try the reasoning

Can the tie index be assumed optimal for a buckling column?
Show the answer and reasoning
No.
A compression member can become unstable before reaching the supplied tensile allowable. Buckling introduces stiffness, length, support and cross-sectional geometry through different equations. Derive the objective and constraints again for that model.
Why use allowable rather than an unexplained best strength?
Show the answer and reasoning
The design criterion needs a justified service value.
A reported maximum may describe another test, direction, temperature or failure criterion. The allowable must be justified for the specified conditions; in this lesson it is explicitly supplied rather than inferred from a headline value.
At δ_max = 2 mm a calculation chooses A’s smaller bound, 50 mm², to minimise mass. Which requirement fails, and why must the maximum be used?
Show the answer and reasoning
The extension would be 5 mm, exceeding 2 mm.
At 50 mm² the stress is 200 MPa, meeting its allowable, but FL/(AE) = 10,000/(50 × 10⁻⁶ × 40 × 10⁹) = 0.005 m. An area must be at least each bound simultaneously. The minimum feasible choice is their maximum, 125 mm².
At the 1.5625 mm ranking crossover, do the equal 0.320 kg masses imply equal areas or equal actual extensions?
Show the answer and reasoning
No. A uses 160 mm² and extends 1.5625 mm; B uses 40 mm² and extends 1.25 mm.
Mass also contains density: A’s density is one quarter of B’s, so four times the area yields equal mass at fixed length. B is stress-controlled and has extension margin. The requirement is an inequality, not an instruction to make both actual extensions equal.
Double F to 20,000 N, leaving L = 1 m, δ_max = 2 mm and the supplied properties unchanged. Find the new areas and masses. Do the transition extension limits change?
Show the answer and reasoning
A: 250 mm² and 0.500 kg. B: 80 mm² and 0.640 kg. The transitions remain 5.00 mm and 1.25 mm.
Both area bounds are proportional to force, so their maximum and the minimum mass double. The stresses and actual extensions at the new areas stay as before. Force cancels from the transition equation δ_max = Lσ_allow/E; this conclusion depends on keeping the ideal properties and constraints unchanged.

What to carry forward

Derive indices from constraints.
Carry the assumptions with the comparison.
For simultaneous lower area bounds, use their maximum and check feasibility before comparing masses.
An active constraint and the preferred candidate can change when a service requirement changes.

References

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

Massachusetts Institute of Technology3 sources