Choosing a Material Without Guessing

"Which material is strongest?" has an answer and it is almost never useful. Strength alone ignores weight, cost, temperature, corrosion, how the part will be made, and what failure mode you are actually guarding against.

The right question is narrower and has three parts: what does this part have to do, what is fixed, and what are you trying to minimise? A tie rod carrying a set load, with a fixed length, minimising mass, is a different problem from a beam carrying a set deflection, and they rank materials differently.

Key idea: Selection is an optimisation with a stated objective and stated constraints, not a lookup. Once the problem is phrased that way, a single combined number, the material index, ranks every candidate correctly, and the arithmetic that produces it is short.

Take the simplest case: a tie of length $L$ that must carry force $F$ without yielding, minimising mass. Cross-section $A$ is free.

The constraint is $F/A \leq \sigma_y$, so the smallest usable area is $A = F/\sigma_y$. Substitute into the mass:

$m = A L \rho = \frac{F L \rho}{\sigma_y} = (F L)\left(\frac{\rho}{\sigma_y}\right)$

The first bracket is fixed by the problem. The second is all material. Minimising mass therefore means maximising

$M = \frac{\sigma_y}{\rho}$

which is specific strength. Repeat the derivation for a beam in bending, where $A$ enters the second moment of area rather than linearly, and you get $\sigma_y^{2/3}/\rho$ instead. Same method, different exponent, and the exponent changes the winner.

In practice: The exponent comes from the geometry, so a material that wins for a tie can lose for a beam. This is why "the best material for aerospace" is not a well-formed statement, and why a real selection is done per component.

Steel is the strongest of these in absolute terms and the worst per kilogram. Oak beats it. Titanium is four times better, which is why it appears in landing gear and engine mounts despite costing many times more per kilogram.

Unidirectional carbon fibre composite is another factor of five beyond titanium, and that number carries a large caveat: it applies along the fibres only. Across them the same material is weaker than the polymer holding it together.

Do the same calculation for stiffness per unit mass, $E/\rho$, and something odd appears.

Steel, titanium and aluminium are within four percent of each other. They differ by a factor of three in stiffness and by exactly the same factor in density, and the two cancel.

Key idea: Swapping between structural metals cannot make a stiffness-limited part lighter. It is not that the gain is small: there is no gain. If your part is stiffness-limited you must change the geometry, put the material further from the neutral axis, or leave metals entirely. Composites are the only entry in the table that breaks the pattern, and that is the actual reason they took over aerospace primary structure.

With indices defined, selection becomes a filter and a sort. The constraints eliminate; the objective ranks.

MATERIALS = [
    # name,             E(GPa), sigma_y(MPa), rho(kg/m3), Tmax(C), cost(/kg)
    ("structural steel",  210,  400,  7850,  450,   1.0),
    ("6061-T6 aluminium",  70,  310,  2700,  150,   3.5),
    ("Ti-6Al-4V",         116,  900,  4430,  400,  35.0),
    ("CFRP (UD)",         150, 1500,  1600,  120,  60.0),
]

def select(materials, min_strength, max_temp, index):
    ok = [m for m in materials if m[2] >= min_strength and m[4] >= max_temp]
    return sorted(ok, key=index, reverse=True)

# lightest tie carrying its load, must survive 200 C
for m in select(MATERIALS, 250, 200, lambda m: m[2] / m[3]):
    print(f"{m[0]:<20} index {m[2] / m[3]:6.3f}   cost/kg {m[5]}")

The 200 degree requirement removes aluminium and CFRP before ranking begins. That ordering matters: constraints are absolute and cannot be traded against the objective, so apply them first and rank only what survives.

A composite puts strong, stiff, brittle fibres in a weak, tough matrix. The fibres carry the load; the matrix transfers it between them, holds their alignment, and protects them. Neither constituent is useful alone.

The reason it works is Griffith's size effect from the fracture lesson: a thin fibre can contain only a small flaw, so it approaches its theoretical strength. Bundling many of them also means a broken fibre sheds load into its neighbours instead of failing the part.

Gotcha: A composite's published strength is along the fibres. A laminate with plies at several angles has a fraction of that number in every direction, because only some plies carry load in each. Quoting the unidirectional figure for a real part overstates it substantially.

Key idea: The index ranks materials on one objective. Every reason in that list is a different objective, and real selection is the intersection of several. A material that wins decisively on the metric you optimised and loses on inspectability, repair, or failure warning is a material you have not finished evaluating.

Whether a material creeps depends not on the absolute temperature but on the homologous temperature, the ratio $T/T_m$ in kelvin. The rule of thumb is that creep becomes a design concern above about $0.4\,T_m$.

Gotcha: Solder creeps at room temperature. Tin melts at 232 degrees Celsius, so a solder joint sitting on a desk is already above 0.4 of its absolute melting point, and a joint under constant mechanical load will slowly deform for its whole life. This is why solder is specified for electrical connection and never as a structural fastening, and why lead roofing visibly ripples after decades.

A material that passes every mechanical check can still fail because of what surrounds it, and these mechanisms produce failures far below any static strength.

• Galvanic corrosion. Two different metals in electrical contact with an electrolyte between them form a battery, and the less noble one dissolves. Steel fasteners in an aluminium structure, or carbon fibre against aluminium, are the classic pairings.
• Stress corrosion cracking. A specific alloy in a specific chemical environment under sustained tensile stress cracks at a stress it would otherwise carry indefinitely. Austenitic stainless steel in chlorides is the textbook case, and the alloy is otherwise excellent.
• Hydrogen embrittlement. Hydrogen from plating, pickling or cathodic protection diffuses into high-strength steel and drops its toughness sharply. It is a delayed failure, appearing hours or days after loading.
• Fretting. Two surfaces rubbing microscopically at a joint break through the oxide layer and initiate fatigue cracks far earlier than the fatigue calculation predicted.

In practice: Each of these is triggered by a combination, material plus environment plus stress, and none appears in a datasheet's strength figures. Checking them is a separate pass over the design.

Everything in this course collapses into a sequence, and the order is what stops you optimising the wrong number.

1. State the function, the constraints, and the objective. What it must do, what cannot change, what you are minimising. Vague answers here produce a defensible-looking selection of the wrong material.
2. Screen on constraints. Temperature, environment, minimum toughness, forbidden materials. These are absolute and eliminate candidates outright.
3. Rank on the index. Derive it from the geometry rather than reaching for specific strength by default, since the exponent changes the winner.
4. Check every failure mode separately. Yield, buckling, fracture with a realistic assumed flaw, fatigue over the service cycles, creep at temperature, corrosion in the actual environment. A part is only as good as its worst mode.
5. Check that it can be made. Process determines achievable geometry, achievable tolerance, and the defect population you inherit. A casting and a forging of the same alloy are not the same material for fatigue purposes.
6. Then look at cost, including processing, joining, inspection and eventual repair, rather than price per kilogram.

Key idea: The index tells you what to try. Steps 4 and 5 are what stop you shipping it.