Engineering Materials Vol II (microstructures_ processing_ design) 2nd ed. - M. Ashby_ D. Jones (1999) WW Part 13 pdf

The properties of an alloy yield strength, toughness, oxidation resistance, etc.depend critically on its constitution and on two further features of its structure: the scale nm or µm or

Fig 28.18. Idealized vibration modes of a soundboard The natural frequencies of the modes are

2

1 22

π

where E w|| is the axial modulus and E w⊥ is the radial modulus

In order to estimate the soundboard frequencies, we set d w = 3 mm, l||= 356 mm, and

l⊥ = 93 or 123 mm (Fig 28.20) Violin soundboards are usually made from spruce,

Fig 28.19. A wooden soundboard has both along-grain and across-grain vibrations Although not shown here, both types of vibration have a full set of modes, with n = 1, 2, 3

Fig 28.20. Dimensions of a violin soundboard The dashed outline marks the position of the bass bar –

a wooden “girder” stuck underneath the soundboard in an off-centre position This effectively divides the soundboard into two different across-grain regions.

which typically has E w|| = 11.6 GPa, E w⊥ = 0.71 GPa and ρw = 0.39 Mg m−3 Equations(28.22) and (28.23) then give us the following data

Spruce soundboards have a Young’s modulus anisotropy of about (11.6 GPa/0.71 GPa)

= 16 A replacement material must therefore have a similar anisotropy This ment immediately narrows the choice down to composites (isotropic materials likemetals or polymers will probably sound awful)

require-Because it is fairly cheap, we should begin by looking at GFRP The moduli of glassfibres and resin matrices are 72 GPa and 3 GPa respectively, giving us

Fig 28.21. Modulus anisotropies Ec || /E c⊥ for aligned GFRP and CFRP composites.

1 + −

Having matched f||/f⊥ in this way we must now go on to match the frequenciesthemselves We can see from eqn (28.22) that this requires

Fig 28.22. Sandwich-type sectional composites give a much-improved stiffness-to-mass ratio.

Equation (28.29) then tells us that d c= 2.52 mm

This value for the thickness of the substitute CFRP soundboard sets us a difficult

problem For the mass of the CFRP plate is greater than that of the spruce plate by the

Sectional composites

The solution that has been adopted by makers of composite soundboards is to ate a sandwich structure where a layer of high-quality cardboard is glued between twoidentical layers of CFRP (Fig 28.22) The philosophy of this design modification is toreplace some CFRP by a much lighter material in those regions that contribute least tothe overall stiffness of the section

fabric-The density of the cardboard layer is around 0.2 Mg m−3 To match the mass of thesandwich to that of the spruce soundboard we must then have

or

In order to formulate the criterion for frequency matching we can make the veryreasonable assumption that the CFRP dominates the stiffness of the sandwich section.

We also simplify eqn (28.22) to

l

E I m

π

/

/

(28.34)

where m w = ρw b||l||d w is the total mass of the wooden soundboard Then, because we

want m w = msandwich, frequency matching requires

(3.21d2 − 2.69d1)3 = 4 6 ( ).d2 −d1 (28.38)

This result can be solved numerically to give d1 = 0.63d2; and, using eqn (28.33), we can

then show that d2 = 0.66dw

Conclusions

This design study has shown that it is possible to design a sectional composite thatwill reproduce both the vibrational frequencies and the mass of a traditional woodensoundboard For a soundboard made out of spruce the equivalent composite is asandwich of cardboard glued between two identical layers of aligned CFRP with afibre volume fraction of 0.13 If the wooden soundboard is 3 mm thick the replacementcomposite must be 1.98 mm thick with a cardboard core of 1.25 mm

Background reading

M F Ashby and D R H Jones, Engineering Materials I, 2nd edition, Butterworth-Heinemann,

1996, Chapter 6 (section on composites).

Further reading

C M Hutchins, “The physics of violins”, Scientific American, November 1962.

S Timoshenko, D H Young and W Weaver, Vibration Problems in Engineering, 4th edition,

Wiley, 1974.

know-is inevitable in a “teach yourself” course which has to accommodate students withdiffering backgrounds Do them anyway The whole thing should take you about

4 hours The material given here broadly parallels the introductory part of the lent text by Hansen and Beiner (referenced below); if you can read German, and want

excel-to learn more, work through this

Phase diagrams are important Whenever materials engineers have to report on theproperties of a metallic alloy, or a ceramic, the first thing they do is reach for the phasediagram It tells them what, at equilibrium, the structure of the alloy or ceramic is Thereal structure may not be the equilibrium one, but equilibrium structure gives a baseline from which other (non-equilibrium) structures can be inferred

Where do you find these summaries-of-structure? All engineering libraries contain:

Source books

M Hansen and K Anderko, Constitution of Binary Alloys, McGraw-Hill, 1958; and supplements,

by R P Elliott, 1965, and F A Shunk, 1969.

J Hansen and F Beiner, Heterogeneous Equilibrium, De Gruyter, 1975.

W Hume-Rothery, J W Christian and W B Pearson, Metallurgical Equilibrium Diagrams,

Insti-tute of Physics, 1952.

E M Levin, C R Robbins and H F McMurdie, Phase Diagrams for Ceramicists, American

Ceramic Society, 1964.

Smithells’ Metals Reference Book, 7th edition, Butterworth-Heinemann, 1992.

ASM Metals Handbook, 10th edition, ASM International, 1990, vol 8.

TEACHING YOURSELF PHASE DIAGRAMS, PART 1 COMPONENTS, PHASES AND STRUCTURES

Definitions are enclosed in boxes and signalled by “DEF.” These you have to learn.

The rest follows in a logical way

DEF A metallic alloy is a mixture of a metal with other

metals or non-metals Ceramics, too, can be mixed

to form alloys

Copper (Cu) and zinc (Zn), when mixed, form the alloy brass Magnesia (MgO) and

alumina (Al2O3) when mixed in equal proportions form spinel Iron (Fe) and carbon (C) mix to give carbon steel.

Components

Alloys are usually made by melting together and mixing the components

DEF The components are the chemical elements which

make up the alloy

In brass the components are Cu and Zn In carbon steel the components are Fe and C.

In spinel, they are Mg, Al and O.

DEF A binary alloy contains two components A ternary

alloy contains three; a quaternary, four, etc.

Symbols

Components are given capital letters: A, B, C or the element symbols Cu, Zn, C

Concentration

An alloy is described by stating the components and their concentrations

DEF The weight % of component A:

WA=

weight of component A weights of all components

(Weight in g)/(atomic or molecular wt in g/mol) = number of mols.

(Number of mols) × (atomic or molecular wt in g/mol) = weight in g

Questions*

1.1 (a) Calculate the concentration in wt% of copper in a brass containing 40 wt% zinc

Concentration of copper, in wt%: WCu= – – – – – – – – – – – – – – – – – – – – – – –(b) 1 kg of an α-brass contains 0.7 kg of Cu and 0.3 kg of Zn

The concentration of copper in α-brass, in wt%: WCu= – – – – – – – – – – – – – –The concentration of zinc in α-brass, in wt%: WZn= – – – – – – – – – – – – – – – –(c) The atomic weight of copper is 63.5 and of zinc 65.4

The concentration of copper in the α-brass, in at%: XCu= – – – – – – – – – – – –The concentration of zinc in the α-brass, in at%: XZn= – – – – – – – – – – – – – –1.2 A special brazing alloy contains 63 wt% of gold (Au) and 37 wt% of nickel (Ni).The atomic weight of Au (197.0) is more than three times that of Ni (58.7) At aglance, which of the two compositions, in at%, is likely to be the right one?

(a) XAu= 0.34, XNi= 0.66

(b) XAu= 0.66, XNi= 0.34

1.3 Your favourite vodka is 100° proof (49 wt% of alcohol) The molecular weight ofwater is 18; that of ethyl alcohol – C2H5OH – is 46 What is the mol% of alcohol inthe vodka?

Mol% of alcohol: XC H OH

2 5 = – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

1.4 An alloy consists of XA at% of A with an atomic weight aA, and XB at% of B with an

atomic weight of aB Derive an equation for the concentration of A in wt% Bysymmetry, write down the equation for the concentration of B in wt%

(a) a single solid solution;

(b) two separated, essentially pure, components;

(c) two separated solid solutions;

(d) a chemical compound, together with a solid solution

* Answers are given at the end of each section But don’t look at them until you have done your best to

answer all the questions in a given group.

How can you tell which form you have got? By examining the microstructure To do

this, the alloy is cut to expose a flat surface which is then polished, first with ively finer grades of emery paper, and then with diamond pastes (on rotating feltdiscs) until it reflects like a brass doorknob Finally, the polished surface is etched,usually in a weak acid or alkali, to reveal the microstructure – the pattern of grainsand phases; brass doorknobs often show this pattern, etched by the salts from sweatyhands Grain boundaries show up because the etch attacks them preferentially Theetch also attacks the crystals, leaving densely packed crystallographic planes exposed;light is reflected from these planes, so some grains appear light and others dark,depending on whether the light is reflected in the direction in which you are looking.Phases can be distinguished, too, because the phase boundaries etch, and becausemany etches are designed to attack one phase more than another, giving a contrastdifference between phases

success-The Al–11 wt% Si casting alloy is typical of (b): the Si separates out as fine needles(≈ 1 µm diameter) of essentially pure Si in a matrix of pure Al The Cd–60 wt% Zn alloy typifies (c): it consists of a zinc-rich phase of Zn with 0.1 wt% Cd dissolved in it plus

a cadmium-rich phase of Cd with 0.8 wt% Zn dissolved in it Finally, slow-cooled

DEF All parts of an alloy with the same physical and

chemical properties and the same composition are

parts of a single phase.

The Al–Si, Cd–Zn and Al–Cu alloys are all made up of two phases

Questions

1.6 You heat pure copper At 1083°C it starts to melt While it is melting, solid and liquidcopper co-exist Using the definition above, are one or two phases present? – – –Why? – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

1.7 Three components A, B and C of an alloy dissolve completely when liquid buthave no mutual solubility when solid They do not form any chemical compounds.How many phases, and of what compositions, do you think would appear in thesolid state?

Phases – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –Compositions – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

The constitution of an alloy

DEF The constitution of an alloy is described by:

(a) The phases present

(b) The weight fraction of each phase

(c) The composition of each phase

The properties of an alloy (yield strength, toughness, oxidation resistance, etc.)depend critically on its constitution and on two further features of its structure: the

scale (nm or µm or mm) and shape (round, or rod-like, or plate-like) of the phases, not

described by the constitution The constitution, and the scale and shape of the phases,depend on the thermal treatment that the material has had

EXAMPLE

The alloy aluminium–4 wt% copper forms the basis of the 2000 series (Duralumin, orDural for short) It melts at about 650°C At 500°C, solid Al dissolves as much as 4 wt%

of Cu completely At 20°C its equilibrium solubility is only 0.1 wt% Cu If the material

is slowly cooled from 500°C to 20°C, 4 wt% − 0.1 wt% = 3.9 wt% copper separates outfrom the aluminium as large lumps of a new phase: not pure copper, but of thecompound CuAl2 If, instead, the material is quenched (cooled very rapidly, often bydropping it into cold water) from 500°C to 20°C, there is not time for the dissolvedcopper atoms to move together, by diffusion, to form CuAl2, and the alloy remains asolid solution

At room temperature, diffusion is so slow that the alloy just stays like this, frozen as

a single phase But if you heat it up just a little – to 160°C, for example – and hold itthere (“ageing”), the copper starts to diffuse together to form an enormous number ofvery tiny (nm) plate-like particles, of composition roughly CuAl2 On recooling toroom temperature, this new structure is again frozen in

The yield strength and toughness of Dural differ enormously in these three tions (slow-cooled, quenched, and quenched and aged); the last gives the highest yieldand lowest toughness because the tiny particles obstruct dislocations very effectively

condi-It is important to be able to describe the constitution and structure of an alloyquickly and accurately So do the following, even if they seem obvious

1.8 In the example above:

(a) How many phases are present at 500°C? – – – – – – – – – – – – – – – – – – – – – –(b) How many phases after slow cooling to 20°C? – – – – – – – – – – – – – – – – – – –(c) How many phases after quenching to 20°C? – – – – – – – – – – – – – – – – – – – –(d) How many phases after quenching and ageing? – – – – – – – – – – – – – – – – – –1.9 An alloy of 120 g of lead (Pb) and 80 g of tin (Sn) is melted and cast At 100°C, twophases are found There is 126.3 g of the lead-rich phase and 73.7 g of the tin-rich

phase It is known that the lead-rich phase contains WPb = 95% of lead The stitution of the alloy at room temperature is described by:

con-(a) Number of phases – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –(b) Weight% of lead-rich phase – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –Weight% of tin-rich phase – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

(c) Composition of lead-rich phase, in wt%: WPb= – – – – – – – – – – – – – – – – – – –

equilibrium constitution for an alloy, to which it tends.

DEF A sample has its equilibrium constitution when, at a given,

constant temperature T and pressure p, there is no further

tendency for its constitution to change with time Thisconstitution is the stable one

Alloys can exist in non-equilibrium states – the Al–Cu example was an illustration.But it is always useful to know the equilibrium constitution It gives a sort of base-linefor the constitution of the real alloy, and the likely non-equilibrium constitutions canoften be deduced from it

State variables

Ten different samples with the same composition, held at the same T and p, have the

same equilibrium constitution Ten samples each of different composition, or each

held at different T or p values, have ten different equilibrium constitutions.

DEF The independent constitution variables or state variables are T, p and composition.

EXAMPLE FOR THE AL-CU ALLOY (DESCRIBED ON PAGE 311):

Values of the state variables Equilibrium constitution

They are equilibrium constitutions because they are the ones reached by very slow

cooling; slow cooling gives time for equilibrium to be reached

Certain thermodynamic relations exist between the state variables In general for a

binary alloy we choose p, T and XB (the at% of component B) as the

independ-ent variables – though presindepend-ently we shall drop p The volume V and the composition

XA (= 1 − XB) are then determined: they are the dependent variables Of course, the weight percentages WA and WB can be used instead

Equilibrium constitution (or phase) diagrams

The equilibrium constitution of an alloy can be determined experimentally by lography and thermal analysis (described later) If the pressure is held constant at

metal-1 atm., then the independent variables which control the constitution of a binary alloy

are T and XB or WB

DEF An equilibrium-constitution diagram or equilibrium

diagram for short (or, shorter still, phase diagram),

is a diagram with T and XB (or WB) as axes It showsthe results of experiments which measure the

equilibrium constitution at each T and XB (or WB)

Figure A1.1 shows a phase diagram for the lead–tin system (the range of alloysobtained by mixing lead and tin, which includes soft solders) The horizontal axis is

composition X (at%) below and W (wt%) above The vertical axis is temperature

Fig A1.1.

in °C The diagram is divided into fields: regions in which the number of phases is

con-stant In the unshaded fields the equilibrium constitution is single phase: liquid (above),

or tin containing a little dissolved lead (left), or lead containing a little dissolved tin(right) In the shaded fields the equilibrium constitution has two phases: liquid plussolid Sn, or liquid plus solid Pb, or solid Pb mixed with solid Sn (each containing alittle of the other in solution)

DEF The diagram shows the equilibrium constitution for

all the binary alloys that can be made of lead and tin,

in all possible proportions, or, in short, for the

lead–tin system.

A binary system is a system with two components.

A ternary system is a system with three components.

The constitution point

The state variables define a point on the diagram: the “constitution point” If this point

is given, then the equilibrium number of phases can be read off So, too, can theircomposition and the quantity of each phase – but that comes later So the diagram tellsyou the entire constitution of any given alloy, at equilibrium Refer back to the defini-

tion of constitution (p 311) and check that this is so.

Questions

1.10 Figure A1.2 shows the Pb–Sn diagram again, but without shading

(a) What is the composition and temperature (the state variables) of point 1?Composition – – – – – – – – – – at% Pb and – – – – – – – – – – at% Sn

Temperature – – – – – – – – – – °C