Materials Science · Selecting materials for a defined function

Functions, constraints and selection criteria

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

Translate a design need into constraints and an objective.
Separate a mandatory threshold, a ranking objective and a permitted design variable.
Screen material–geometry combinations using a fully stated mechanical model and discrete available sizes.
Reassess a conclusion when a requirement changes or required service evidence is missing.

Begin with the service delivered

A shelf support must carry a specified load within allowed deflection and life, using acceptable space and connections. These are constraints. Reducing mass or cost can then be an objective among feasible options. A material cannot be selected meaningfully before the job is defined.

Suppose the instruction is “make a light shelf support that does not sag.” The function is to hold the shelf and its load in the required position. “Use aluminium” already chooses part of an implementation; it is not a neutral statement of that function. A thick support made from a less stiff material and a thinner support made from a stiffer one may both deliver the same service. What may change must be stated before comparing them.

A constraint is a condition every admitted design must satisfy, such as a maximum displacement at a specified load. An objective orders the admitted designs, such as minimum member mass. A free variable is something this decision is allowed to change: material, section depth, process or number of supports, for example. A fixed dimension is a constraint even if changing it would make the problem easier. These roles belong to the decision, not permanently to a property: mass may be an objective in one brief and a maximum threshold in another.

The following original example uses invented material properties, stress-screen values and exact requirements. It demonstrates a limited design screen, not a tested shelf or a material recommendation. The assumed stress limits are supplied teaching inputs; they are not measured strengths, code allowables or safety factors derived from the references.

Turn an ambiguous request into a checkable brief

Before calculating, resolve what “100 N load” means. Is it the total on a shelf with two supports, or the load on each support? Does it act at an edge or spread along the shelf? Two supports need not share equally if their stiffnesses, locations or connections differ. A number without its force path is insufficient. Likewise, “0.5 mm deflection” must name the point, direction, reference state and loading condition.

For the toy decision, the clarified load is a downward force P = 100 N at the free tip of the single support being evaluated. Its initially straight horizontal rectangular member projects L = 200 mm from an ideal rigid clamp. Its fixed width is b = 20.0 mm. The selectable depth h, measured in the bending direction, must be one of the available stock values 10.0, 15.0 or 20.0 mm. All candidates have the same solid rectangular form and allowed orientation. Changing span, adding supports, rotating a rectangle or substituting a hollow section is outside this decision.

The tip displacement magnitude must be at most δ_lim = 0.500 mm under that load. The peak nominal bending-stress magnitude must not exceed the supplied candidate-specific screen S. The depth may not exceed 20.0 mm. Among combinations satisfying all these conditions in the defined model, minimise member mass m. All dimensions and properties are exact teaching inputs; equality is admitted by the stipulated inequalities, without implying a real-world margin.

Use a uniform, linearly elastic, small-deflection beam model with negligible transverse-shear deformation and no initial curvature or axial force. Only the stated tip force enters it: self-weight, clamp deformation, local connection stresses, creep, repeated loading and environmental changes are omitted. Those omissions define the scope of this illustrative screen. If actual service requires those effects to be controlled, passing this calculation alone does not satisfy the full service brief.

Clarified original teaching decision. These are problem definitions, not performance measured on a product.
Role	Specified item	Why it matters
Function	Support the specified per-support tip load at the required position	A material name alone does not define the service
Fixed conditions	P = 100 N; L = 200 mm; b = 20.0 mm; rigid clamp; rectangular form	Changing one would define a different problem
Mandatory screens	δ ≤ 0.500 mm; nominal \|σ\| ≤ S; h ≤ 20.0 mm	A low objective value cannot compensate for failure
Free variables	Candidate A, B or C and stock h = 10.0, 15.0 or 20.0 mm	Only these combinations may enter this screen
Objective	Minimise mass of the member	Equal connection mass is outside the compared member account

Connect the requirements to quantities the model predicts

Young’s modulus E relates longitudinal stress to small elastic strain; density ρ is mass per solid volume. For a solid rectangular member, m = ρbhL. The newton is a unit of force; a pascal is N/m², and all lengths must use metres in these SI expressions. Mass and stiffness are different: density alone cannot determine the displacement under load.

A tip force produces a bending moment of magnitude M(x) = P(L − x) at a section x from the clamp. Moment measures the turning effect of a force about that section. The largest magnitude is PL at the clamp. In the small-deflection beam model, a plane section rotates while remaining plane, and its longitudinal strain changes linearly with distance y from the central neutral axis. Integrating the corresponding linear-elastic stresses to supply M introduces the second moment of area I = ∫y² dA. For width b and depth h, integrating b y² from −h/2 to h/2 gives I = bh³/12, in m⁴. It is a geometric quantity, not a mass moment of inertia.

Let v(x) be downward displacement, so curvature in this sign convention is approximately vʺ(x) = P(L − x)/(EI). The prime means a derivative with respect to position; curvature is the rate at which slope changes. A rigid clamp gives v(0) = 0 and vʹ(0) = 0. Integrating once gives vʹ(x) = P(Lx − x²/2)/(EI); again gives v(x) = P(Lx²/2 − x³/6)/(EI). At the free tip, δ = v(L) = PL³/(3EI) = 4PL³/(Ebh³). The units are (N m³)/(N m²) = m. A different load position or support condition needs its own relation.

The same section model gives longitudinal stress σ(x,y) = −M(x)y/I with a sign set by the chosen axes. Its largest magnitude for this rectangular tip-loaded member is σ_max = PL(h/2)/I = 6PL/(bh²), in Pa. Opposite outer surfaces have opposite signs. For this toy screen S bounds both signs of that nominal bending stress. It is not an additional law for crack growth, local clamp failure or transverse shear.

Both δ and σ_max fall when h increases, but mass rises in direct proportion to h. The inequalities imply h ≥ [4PL³/(Ebδ_lim)]^(1/3) and h ≥ [6PL/(bS)]^(1/2). A cube root is the positive number whose cube equals its argument; a square root has the corresponding square property. The required depth is the larger bound, followed by selection of an actually available stock size. A real valued bound such as 16.6 mm does not create that stock size.

This derivation also shows why a ranking index cannot be borrowed independently of the geometry freedoms. Here b is fixed and only h changes. With only the displacement condition active and continuously adjustable h, the minimum-mass expression contains ρ/E^(1/3). If geometrically similar square sections scale in both width and depth, the dependence differs. The complete discrete, multi-constraint screen below is more direct than using an index outside its assumptions.

Original idealised rigid-clamp rectangular cantilever with a downward 100 newton force at its free tip, fixed span 200 millimetres and fixed width 20 millimetres. The upper arrangement is schematic; selectable stock depth h is not the drawn vertical scale. The lower plot evaluates the stated small-deflection beam equation at 201 positions for each of two A designs. Horizontal position runs from zero at the clamp to 200 millimetres at the tip; downward displacement magnitude runs from zero to 0.8 millimetres. A15 is dashed orange with a square tip marker and reaches 0.677249 millimetres; A20 is solid blue with a circular tip marker and reaches 0.285714 millimetres. The specified 0.500 millimetre limit applies at the tip, so A15 fails and A20 passes that displacement screen. Both curves have zero displacement and slope at the rigid clamp. This is a calculated model, not a measured deformed shape or a complete qualification; nominal stress, stock availability, mass and omitted connection or service effects require their own checks.

Screen the designs before choosing the lightest

Use A: E = 70.0 GPa, ρ = 2700 kg/m³, S = 120 MPa; B: E = 20.0 GPa, ρ = 1200 kg/m³, S = 30.0 MPa; C: E = 200 GPa, ρ = 7800 kg/m³, S = 160 MPa. A gigapascal is 10⁹ Pa and a megapascal 10⁶ Pa. Labels A, B and C are fictional candidates; similar-looking numbers do not identify a material grade, processing state or reliable service property.

For A at h = 15.0 mm, m = 2700(0.0200)(0.0150)(0.200) = 0.162 kg. Its I is 5.625 × 10⁻⁹ m⁴, so δ = 100(0.200)³/[3(70.0 × 10⁹)(5.625 × 10⁻⁹)] = 0.677249 mm. The peak nominal stress is 26.6667 MPa, comfortably below its supplied S, but displacement exceeds 0.500 mm. Passing the stress screen cannot compensate for that failed constraint.

A at 20.0 mm gives 0.216 kg, 0.285714 mm and 15.0 MPa, passing both mechanical screens. C at 15.0 mm gives 0.468 kg, 0.237037 mm and 26.6667 MPa, also passing. A20 therefore has lower member mass among these two admitted designs. C is stiffer as a material, but the objective is not maximum modulus. B20 is lighter still, 0.0960 kg, yet its 1.000 mm displacement excludes it.

The table and monotonicity account for all nine allowed combinations. A10 fails displacement because the deeper A15 already fails. Every allowed B fails because B20, its deepest permitted option, already fails. C10 fails, while C20 passes but is heavier than the passing C15. Thus the minimum-mass admitted stock choice is A20 under this particular brief. No out-of-range large-deflection prediction for a thinner B is needed to reach that conclusion.

Original calculated combinations under the clarified tip-load model. The suffix is stock depth in millimetres. Passing applies only to the stated model and supplied screens.
Design	Member mass / kg	δ / mm	σ_max / MPa	Screen result
A15	0.162	0.677249	26.6667	Fails displacement
A20	0.216	0.285714	15.0000	Passes supplied screens
B20	0.0960	1.000000	15.0000	Fails displacement
C10	0.312	0.800000	60.0000	Fails displacement
C15	0.468	0.237037	26.6667	Passes supplied screens
C20	0.624	0.100000	15.0000	Passes, heavier than C15

Original fictional mass versus tip-displacement plot for six labelled stock combinations. Horizontal mass runs from zero to 0.7 kilograms; vertical downward tip displacement magnitude runs from zero to 1.1 millimetres. A uses orange triangles, B a green circle and C blue squares; suffixes are stock depths in millimetres. A15 is at 0.162 kilograms and 0.677249 millimetres; A20 at 0.216 and 0.285714; B20 at 0.0960 and 1.000000; C10 at 0.312 and 0.800000; C15 at 0.468 and 0.237037; C20 at 0.624 and 0.100000. All plotted combinations pass their separately stipulated nominal bending-stress screen. The dashed violet original tip limit is 0.500 millimetres; its shaded lower band contains the displacement-admitted points. A20 is their lowest-mass option. A tighter dotted limit at 0.250 millimetres excludes A20, leaving C15 as the lightest admitted point. Discrete points are not connected because arbitrary intermediate stock sizes are unavailable. A10 and smaller B options are excluded by monotonic displacement screening; C20 is heavier than passing C15. No real material rating, statistical uncertainty, connection evidence or service-life qualification is supplied.

A changed requirement can change the selected combination

For A with the original 0.500 mm limit, the displacement bound is h ≥ 16.5965 mm and the stress bound is h ≥ 7.07107 mm. Displacement governs the continuous lower bound. The next allowed stock depth is 20.0 mm. For C the corresponding bounds are 11.6961 and 6.12372 mm, so stock 15.0 mm suffices. Optimising a continuous depth and choosing among discrete stock sizes are different decisions.

Now require δ ≤ 0.250 mm at the same tip load while preserving every other condition. A20 gives 0.285714 mm and is excluded; no shallower allowed A can repair it. B remains excluded. C15 gives 0.237037 mm and still passes, so it is the lightest admitted combination in the revised screen. The earlier choice did not become an intrinsically bad material. The requirement changed.

Alternatively, keep 0.500 mm but double the supplied tip force to 200 N. The linear model doubles δ and nominal bending stress without changing member mass. A20 then gives 0.571429 mm and fails; C15 gives 0.474074 mm and 53.3333 MPa, passing its supplied screens. This transfer is conditional on the same elastic model; it is not a rule that a real overloaded support retains the same response.

A continuous ranking alone also misses availability. If A is available only at 10.0 and 15.0 mm while C retains all three sizes, A has no admitted option under the original brief even though its continuously sized ideal member could be light. The missing stock depth is a constraint, not a small penalty that a favourable density can cancel. Geometry, manufacturability and joining continues that separate practical question.

Make success measurable

State load cases, environment, duration, manufacture and maintenance. Separate required thresholds from preferences. If requirements conflict, expose the conflict rather than silently dropping one to make a preferred material win.

A requirement is incomplete if “acceptable life” has no duration, load history, environment or observable failure definition. Low displacement in one instantaneous elastic calculation does not establish ten-year creep, repeated-load durability or corrosion resistance. A connection assumed rigid here also needs evidence if its motion or failure matters to actual service. Missing required information is unresolved; it must not be entered as a passing score or replaced with a favourable property from a different test condition.

For example, two specimens can share E, density and the same initial nominal stress screen while differing in time-dependent deformation or crack populations. The present calculation cannot rank their long-term performance because those mechanisms and observations were not supplied. A relevant next test would use the required load history, temperature, geometry and connections and measure the actual displacement or failure endpoint over an appropriate duration. An unrelated short tensile strength cannot supply that missing evidence.

Keep the record of the function, clarified requirements, fixed conditions, variable freedoms, data provenance, model and exclusions alongside the chosen candidate. If none survives, report that the supplied design space is infeasible under the stated screens. Seek an authorised change of requirements or design freedoms, additional evidence or a new concept; do not quietly erase the limiting condition. Service conditions and reliable property data develops the service question, while Trade-offs, uncertainty and transparent decisions develops how to compare objectives and uncertain evidence.

Worked examples

Work through the case

A candidate is light but exceeds the allowable deflection. Is it the best feasible option?

Deflection is a required constraint.
The candidate fails that requirement in its current design.
Change geometry or consider another option.

It is not feasible as specified, regardless of its low mass.

Resolve what the load statement actually requires

A brief says a shelf carries 100 N and uses two supports. A designer immediately applies 50 N to each cantilever. What must be clarified before that screen is justified?

The 100 N might mean total shelf load, per-support load or a particular load case. Clarify the intended definition first.
Even a total 100 N does not prove equal reactions: load position, support locations and the stiffness/connection model matter.
The deflection requirement must identify its observation point and whether shelf deformation and connection motion are included.
For the worked toy screen the brief explicitly assigns 100 N at each individual free tip; use that condition rather than silently halving it.

Resolve load allocation, location, support model and the displacement measurand. Equal sharing requires justification.

A low mass and passed stress limit do not establish feasibility

Apply all supplied screens to A15 and compare it with A20. Which change makes the second design admitted?

Use h = 0.0150 m: I = 0.0200(0.0150)³/12 = 5.625 × 10⁻⁹ m⁴.
The tip relation gives δ = 0.677249 mm, above 0.500 mm; nominal stress is 26.6667 MPa, below 120 MPa. Mass is 0.162 kg.
At h = 0.0200 m, I grows by (20/15)³. Deflection falls by (15/20)³ to 0.285714 mm.
Stress falls by (15/20)² to 15.0 MPa, while mass rises by 20/15 to 0.216 kg. Both now satisfy their separate screens.

A15 fails displacement; A20 passes the supplied model screens after the permitted depth increase. The mass increase purchases a different geometric response.

Convert a calculated bound into an allowed candidate

Use candidate A and the original brief. Find both continuous lower bounds on h and the lowest-mass stock option. Can 16.5965 mm be ordered merely because the calculation gives it?

The displacement bound is [4(100)(0.200)³/(70.0 × 10⁹ × 0.0200 × 0.000500)]^(1/3) = 0.0165965 m.
The stress bound is [6(100)(0.200)/(0.0200 × 120 × 10⁶)]^(1/2) = 0.00707107 m.
The larger bound is 16.5965 mm. It is below the maximum 20.0 mm, but neither 10.0 nor 15.0 mm reaches it.
The smallest allowed stock size is 20.0 mm, giving 0.216 kg. A newly custom-made section would change the allowed design/manufacturing space.

Select A20 under the original stock restriction. The continuous minimum is not an available design by itself.

Re-screen when a requirement becomes tighter

Change only the deflection limit from 0.500 to 0.250 mm. Compare the former A20 choice and C15.

The design responses have not changed: A20 remains 0.285714 mm and C15 remains 0.237037 mm.
A20 exceeds the new threshold; smaller A sizes deflect more. B20 also fails, so no allowed B helps.
C15 still passes displacement and its nominal 26.6667 MPa remains below the supplied 160 MPa screen.
C20 passes but weighs 0.624 kg, more than C15 at 0.468 kg. C10 fails.

C15 is now the minimum-mass admitted stock design. Reusing the earlier material ranking would miss a changed feasible set.

Try the reasoning

Can a design objective be “minimise mass” without dimensions or loads?
Show the answer and reasoning
It is incomplete.
Constraints define a meaningful feasible design.
Why list failure modes?
Show the answer and reasoning
They identify relevant constraints and data.
One strength value does not cover every mode.
Keep the original 0.500 mm brief but remove A20 from the allowed stock list. What is the lightest admitted remaining combination?
Show the answer and reasoning
C15 at 0.468 kg.
A15 already fails displacement, so A10 cannot pass. All B sizes fail because B20 fails. C10 fails, and C20 is heavier than passing C15. An unavailable size cannot be restored by a favourable ranking score.
Under the same linear model, apply 200 N at the tip. Compute δ and nominal stress for A20 and C15, retaining the original displacement limit.
Show the answer and reasoning
A20: 0.571429 mm and 30.0 MPa, displacement fail. C15: 0.474074 mm and 53.3333 MPa, passes supplied screens.
The governing equations are proportional to P for fixed E and geometry. Passing stress alone does not rescue A20. This calculation retains the stated elastic/no-other-load assumptions; it supplies no overload qualification.
Increase only B’s nominal stress screen from 30 to 300 MPa. Does B20 become feasible?
Show the answer and reasoning
No. Its displacement remains 1.000 mm, above 0.500 mm.
Changing S changes the stress inequality, not E or I. A strength number cannot alter the displacement calculated from the stipulated stiffness model.
A proposal keeps A15 but increases width from 20 to 30 mm. What happens to mass and model displacement, and is it an allowed repair?
Show the answer and reasoning
Mass becomes 0.243 kg and displacement becomes about 0.451499 mm; the proposal violates the fixed-width condition.
At fixed depth, I and mass are proportional to width, so the factors are 1.5 and 1/1.5. Passing the mechanical screen does not make a changed fixed dimension part of the original allowed space.
A20 deflects less than A15 even though both use exactly the same E. Has the material modulus increased?
Show the answer and reasoning
No; the geometric second moment increased.
Depth changes I as h³. The same material model can produce different component stiffnesses, so comparing supports requires their dimensions and boundary conditions.
A20 passes the instantaneous screen. A new mandatory requirement limits displacement after ten years at a different temperature. No time-dependent data are supplied. Should it be recorded as passed, failed, or unresolved?
Show the answer and reasoning
Unresolved for that new requirement.
The model contains no time-dependent constitutive law, temperature-dependent inputs or validated accelerated-test relation. Neither its instantaneous pass nor a missing record establishes the ten-year outcome.
Keep the 100 N tip load but require δ ≤ 0.0500 mm with all original stock choices. Does any candidate survive?
Show the answer and reasoning
No candidate survives the supplied displacement screen.
The deepest C20 is the stiffest available combination and gives 0.100000 mm, already above the threshold. A20 and B20 give larger values, and reducing depth only increases them. Do not select the least bad failed option as feasible.
Under the original constraints, an owner changes the objective from minimum member mass to minimum delivered cost. Are A20 and its cost ranking already determined by this lesson’s data?
Show the answer and reasoning
The mechanical feasible set is unchanged, but the cost ranking is unresolved.
No purchase, processing, connection, quantity or delivery-cost model was provided. A density or mass ranking is not a delivered-cost calculation. Clarify the new objective and gather the needed evidence while retaining all mandatory screens.

What to carry forward

Define the function and constraints first.
Rank only feasible alternatives.
State what each requirement measures, where, under which load and with which evidence.
A free variable may be changed only within the permitted design space; a calculated continuous optimum need not be an available stock option.
A model-feasible combination is conditional on its model and inputs. Missing service evidence is unresolved, not a silent pass.

References

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

Massachusetts Institute of Technology3 sources