Choosing a material

Every calculation in this course has taken the material as given and asked how big the member must be. The remaining choice is the other one: given a job, what should the thing be made of? It looks like a matter of experience and catalogues, and it turns out to have a clean structure, which Michael Ashby set out systematically in the 1980s and which is now how the question is taught.

Four questions, one of them free

A selection problem has four parts, and separating them is most of the work.

The function is what the component does: carry a tensile load, support a bending moment, store elastic energy, conduct heat while carrying load.

The objective is what is to be minimised or maximised: usually mass, sometimes cost, sometimes volume or embodied carbon.

The constraints are what must not be violated: it must not yield, it must not deflect more than a stated amount, it must fit inside a given envelope, it must survive a given temperature.

The free variable is what may be adjusted to satisfy the constraint, almost always a section dimension.

The method is then mechanical. Write the objective in terms of the geometry and the material. Use the constraint to eliminate the free variable. What is left is a product of a geometric term, a functional term, and a group of material properties. Only that last group depends on the material, so the best material is the one that maximises it, and it is called a material index.

The lightest tie

Take the simplest case. A tie of length $L$ , fixed by the design, must carry a tensile force $F$ without yielding, and its cross-sectional area $A$ is free. Minimise the mass $m = ρ A L$ .

The constraint is $F / A \leq σ_{y}$ , so the smallest area that will do is $A = F / σ_{y}$ . Substituting,

m = ρ L \frac{F}{σ_{y}} = (F) (L) (\frac{ρ}{σ_{y}})

The load and the length are given, so the lightest tie is made of the material with the largest $σ_{y} / ρ$ , the specific strength.

Change one word in the problem and the index changes. Suppose the tie must not stretch more than a stated amount rather than not yield, so the constraint is a stiffness $S = A E / L$ . Then $A = S L / E$ and

m = ρ L \frac{S L}{E} = (S) (L^{2}) (\frac{ρ}{E})

so the lightest stiff tie maximises $E / ρ$ , the specific stiffness. Two apparently similar problems, two different answers, and materials that win one can lose the other badly.

Example. Rank structural steel, aluminium alloy 6082-T6, titanium alloy Ti-6Al-4V and softwood for a stiff tie and for a strong tie, using $ρ$ in megagrams per cubic metre, $E$ in gigapascals and $σ_{y}$ in megapascals.

Material	$ρ$	$E$	$σ_{y}$	$E / ρ$	$σ_{y} / ρ$
Structural steel	7.85	210	275	26.8	35
Aluminium 6082-T6	2.70	70	260	25.9	96
Titanium Ti-6Al-4V	4.43	115	880	26.0	199
Softwood, along grain	0.45	10	40	22.2	89

For stiffness the three metals are indistinguishable, within a few per cent of each other, and softwood is not far behind. That is not a coincidence of these particular alloys: within a class of metals, $E$ and $ρ$ both scale with the strength of the interatomic bonding and the atomic packing, so their ratio is nearly constant. If a design is limited by stiffness in tension, changing from steel to aluminium buys nothing whatever, which is why aircraft skins are not aluminium for stiffness reasons.

For strength the ranking is completely different: titanium beats aluminium beats softwood beats steel, and steel is the worst of the four by a factor of nearly six against titanium.

Now you. A unidirectional carbon fibre laminate has $ρ = 1.60$ , $E = 130$ GPa and a tensile strength of $1000$ MPa. Where does it sit on both indices?

Answer

$E / ρ = 130 / 1.60 = 81$ , three times any of the metals, and $σ / ρ = 1000 / 1.60 = 625$ , three times titanium. It wins both, decisively, which is why it is used where mass is expensive enough to justify the cost and the manufacturing difficulty.

The lightest beam, where the exponent changes

Bending is where the method earns its keep, because the free variable enters the constraint with a different power.

Take a beam of given length $L$ , simply supported, carrying a given load, and required to have a stiffness $S$ , meaning a load per unit deflection. Let the section be square of side $b$ , free to vary. From the deflection lesson, the stiffness of any standard case has the form $S = C E I / L^{3}$ with $C$ a number depending only on how it is loaded and supported. For a square section, $I = b^{4} / 12$ , so

b^{4} = \frac{12 S L^{3}}{C E}

The mass is $m = ρ b^{2} L$ , and $b^{2}$ is the square root of the expression above, so

m = ρ L \sqrt{\frac{12 S L^{3}}{C E}} = \sqrt{\frac{12 S}{C}} L^{5 / 2} (\frac{ρ}{E^{1 / 2}})

The material group is now $ρ / E^{1 / 2}$ , so the lightest stiff beam maximises

M = \frac{E^{1 / 2}}{ρ}

The square root is the whole story. It appears because the free variable can grow the section outwards, and $I$ grows faster than the area does, so a light material is allowed to be bulky and still win. Repeating the argument with a strength constraint, where $σ = M_{b} / Z$ and $Z = b^{3} / 6$ , gives $b \propto σ^{- 1 / 3}$ and a mass proportional to $ρ / σ^{2 / 3}$ , so the lightest strong beam maximises $σ^{2 / 3} / ρ$ . For a panel of fixed width where only the thickness is free, the same argument gives $E^{1 / 3} / ρ$ .

Recomputing the table with the beam indices changes the answer completely.

Material	$E^{1 / 2} / ρ$	$σ^{2 / 3} / ρ$
Structural steel	1.85	5.4
Aluminium 6082-T6	3.10	15.1
Titanium Ti-6Al-4V	2.42	20.7
Softwood, along grain	7.03	26.0
Carbon fibre laminate	7.13	62.5
Concrete	2.28	4.5

Wood is now the best of the traditional materials for a light stiff beam, ahead of steel by a factor of nearly four and level with carbon fibre. That is not a quirk of the arithmetic: it is why aircraft were built of spruce until the 1930s, why model aeroplanes are balsa, why floor joists are timber, and why a wooden oar is not a steel one. Nature reached the same conclusion, which is why a tree trunk and a bird's bone are both light, hollow and stiff rather than dense and solid.

The comparison in kilograms

Example. A simply supported beam of square section spans $3$ m and must not deflect more than $10$ mm under a central point load of $2$ kN. Find the section size and mass in structural steel, and in softwood with $E = 10$ GPa.

The deflection is $W L^{3} / 48 E I$ , so $I = W L^{3} / (48 E δ)$ . In steel, $I = 2000 \times 3000^{3} / (48 \times 210 000 \times 10) = 5.36 \times 10^{5}$ mm⁴. With $I = b^{4} / 12$ , $b = (12 \times 5.36 \times 10^{5})^{1 / 4} = 50.4$ mm, so the mass is $7850 \times (0.0504)^{2} \times 3 = 59.7$ kg.

In softwood, $I = 2000 \times 3000^{3} / (48 \times 10 000 \times 10) = 1.125 \times 10^{7}$ mm⁴, giving $b = 107.8$ mm and a mass of $450 \times (0.1078)^{2} \times 3 = 15.7$ kg.

The wooden beam is more than twice as deep and less than a third of the weight, and the ratio $59.7 / 15.7 = 3.8$ is exactly the ratio of the two indices, $7.03 / 1.85$ . The index predicted the answer without any of the arithmetic.

Now you. Repeat for aluminium, with $E = 70$ GPa and $ρ = 2700$ kg m⁻³.

Answer

$I = 2000 \times 3000^{3} / (48 \times 70 000 \times 10) = 1.607 \times 10^{6}$ mm⁴, so $b = 66.3$ mm and the mass is $2700 \times (0.0663)^{2} \times 3 = 35.6$ kg. The ratio to steel, $59.7 / 35.6 = 1.68$ , again matches the ratio of indices, $3.10 / 1.85$ .

Why steel is still everywhere

If steel loses on nearly every index in the tables above, its dominance needs explaining, and the explanation is that mass was the wrong objective for most structures.

Change the objective to cost and redo the derivation. Nothing in the algebra changes except that the mass $ρ$ is replaced by the cost per unit volume $C_{m} ρ$ , where $C_{m}$ is the cost per kilogram.

Example. Structural steel costs roughly $1$ unit of currency per kilogram and a carbon fibre laminate roughly $50$ . Compare them for a light stiff beam when the objective is cost rather than mass.

On mass, the indices from the table are $1.85$ for steel and $7.13$ for the laminate, so the laminate wins by a factor of $3.9$ . On cost the index becomes $E^{1 / 2} / (C_{m} ρ)$ , giving $1.85 / 1 = 1.85$ for steel and $7.13 / 50 = 0.14$ for the laminate. Steel now wins by a factor of $13$ , and the two materials have swapped places by a wide margin in both directions.

For a building, where the structure is a modest fraction of the total cost and its mass is carried by the ground rather than by an engine, cost is the right objective and steel and concrete win. For an aircraft, where every kilogram costs fuel for thirty years, mass is the right objective and the laminate wins. Both answers come from the same method with a different objective, which is the point.

Now you. A bridge girder and a bicycle frame are both loaded mainly in bending. Why do they end up in different materials?

Answer

The objectives differ. The girder is sized by cost per unit of bending stiffness and strength, and its own mass is carried by foundations, so steel or concrete wins. The frame is sized by mass, because the rider carries it and accelerates it constantly, so aluminium, titanium or carbon fibre wins despite costing many times more per kilogram.

What no index knows

The tables above use one number per property, and every one of them hides something.

Toughness and fracture. An index built on yield strength assumes the material yields. Carbon fibre laminates and ceramics do not: they fail by delamination or by fast fracture from a flaw, so their useful design stress is set by damage tolerance rather than by strength, and a laminate that has been struck by a dropped tool may have lost half its compressive strength with nothing visible on the surface.

Fatigue. For anything cycled, the relevant strength is the endurance limit at the required number of cycles and at the actual detail, not the tensile strength. For welded steel that can be under a third of the yield stress, and it depends on the weld geometry rather than on the steel grade, so the whole index calculation is being done with the wrong number.

Directionality. Wood's spectacular beam index is measured along the grain. Across it, both stiffness and strength fall by a factor of ten to thirty, and its strength in shear parallel to the grain is a few megapascals. A laminate's properties depend on the ply layup and can be tuned, which also means they can be got wrong.

Variability. Steel arrives with a guaranteed minimum yield strength. Timber's strength varies by a factor of two between pieces of the same grade, which is why timber design uses characteristic values at the fifth percentile and large partial factors. Concrete is made on site.

Everything that is not a mechanical property. Fire: steel loses about half its strength by $550$ degrees Celsius and needs protection, timber chars predictably and can outlast it. Corrosion, and the maintenance it implies over a hundred year life. Whether the material can be joined at all: welding, bolting, gluing and casting each rule out some materials entirely. Whether anyone can make the shape. And, increasingly, embodied carbon, which ranks materials differently again and by a wide margin.

An index narrows a field of thousands to a handful. Choosing among that handful is engineering judgement informed by all of the above, and the honest statement is that the calculation is the beginning of the decision rather than the end of it.

What has been built

The course started with a bridge that does nothing, and the chain from there was short. Equilibrium gives the reactions and, joint by joint, every force in a truss. Cutting a beam open turns those into shear and bending moment along the span. Dividing by the area that carries them gives stress, which a tensile test can finally judge. Letting the material strain supplies the equations statics lacked, and the yield point says where the straight line stops. Torsion and bending show that how the area is arranged matters more than how much of it there is, shear explains the web, and integrating the curvature says how far the thing moves. Compression turns out to have a failure mode with no stress in it at all, several actions at once need a criterion rather than a comparison, and the material itself is chosen by an index that falls out of writing down the mass.

That is enough to say why things stand up, and enough to size a real member with a stated margin. What it does not include is time: fatigue, creep, corrosion, and the slow loss of everything the first calculation assumed. Structures fail far more often from those than from an arithmetic error in a bending moment.