Engineering Materials vol 2 Part 14 docx

316 Engineering Materials 2 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 ρ w = 0.39 Mg m −3 . Equations (28.22) and (28.23) then give us the following data. f || (s −1 ) ≈ 59 236 531 944 1475 2124 2891 etc. ! 217 868 1953 3472 etc. f ⊥ (s −1 ) ≈ @ 124 496 1116 1984 3100 etc. For all the crudity of our calculations, these results show clearly that wooden violin soundboards have an impressive number of natural frequencies. Replacement materials 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 require- ment immediately narrows the choice down to composites (isotropic materials like metals or polymers will probably sound awful). Because it is fairly cheap, we should begin by looking at GFRP. The moduli of glass fibres and resin matrices are 72 GPa and 3 GPa respectively, giving us E c|| = 72V f + 3(1 − V f ) (28.24) Case studies in design 317 Fig. 28.21. Modulus anisotropies Ec || / E c ⊥ for aligned GFRP and CFRP composites. and E c⊥ = VV ff 72 1 3 1 .+ −       − (28.25) Now, as Fig. 28.21 shows, E c|| /E c⊥ can never be greater than 6.6 for GFRP; and this is far too small to give a useful soundboard. We must therefore hope that CFRP can give us the required anisotropy. The modu- lus of type-1 carbon fibres is 390 GPa along the fibre axis (although it is only 12 GPa at right angles to this). So E c|| = 390V f + 3(1 − V f ) (28.26) and E c⊥ = VV ff 12 1 3 1 .+ −       − (28.27) This gives a maximum ratio of 43, which is more than enough. In fact, to bring the anisotropy down to the target figure of 16 we need to reduce the volume fraction of fibres to only 0.13! Having matched f || /f ⊥ in this way we must now go on to match the frequencies themselves. We can see from eqn. (28.22) that this requires n l Ebd bd n l Ebd bd ww ww cc cc 2 2 3 12 2 2 3 12 212212 π ρ π ρ || || || || || || || ||             =             // (28.28) which tells us that the composite plate must have a thickness of 318 Engineering Materials 2 Fig. 28.22. Sandwich-type sectional composites give a much-improved stiffness-to-mass ratio. d c = d w E E wc cw || || ρ ρ       12/ . (28.29) Now we already know three of the terms in this equation: d w is 3 mm, E w|| is 11.6 GPa and ρ w is 0.39 Mg m −3 . The last two unknowns, E c|| and ρ c , can be calculated very easily. When V f = 0.13 eqn. (28.24) tells us that E c|| = 53 GPa. ρ c can be found from the simple rule of mixtures for density, with ρ c = (0.13 × 1.9 + 0.87 × 1.15) Mg m −3 = 1.25 Mg m −3 . (28.30) 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 factor 252 125 3039 269 3 3 . . . mm Mg m mm Mg m × × = − − (28.31) This excessive mass is quite unacceptable – the CFRP plate will need far too much energy to get it vibrating. But how can we reduce the mass without shifting the vibration frequency? Sectional composites The solution that has been adopted by makers of composite soundboards is to fabric- ate a sandwich structure where a layer of high-quality cardboard is glued between two identical layers of CFRP (Fig. 28.22). The philosophy of this design modification is to replace some CFRP by a much lighter material in those regions that contribute least to the overall stiffness of the section. The density of the cardboard layer is around 0.2 Mg m −3 . To match the mass of the sandwich to that of the spruce soundboard we must then have 0.39d w = 1.25(d 2 − d 1 ) + 0.2d 1 (28.32) or d w = 3.21d 2 − 2.69d 1 . (28.33) Case studies in design 319 In order to formulate the criterion for frequency matching we can make the very reasonable assumption that the CFRP dominates the stiffness of the sandwich section. We also simplify eqn. (28.22) to fn l EI m ww w || || || =             2 32 12 2 π / / (28.34) where m w = ρ w b || l || d w is the total mass of the wooden soundboard. Then, because we want m w = m sandwich , frequency matching requires E w|| I w = E c|| I c (28.35) or E bd Eb dd w w c|| || || || ( ) . 3 2 3 1 3 12 12 = − (28.36) We know that E c|| /E w|| = 53/11.6 = 4.6 so that eqn. (28.36) reduces to ddd w 3 2 3 1 3 46 .( ).=− (28.37) Finally, combining eqns. (28.33) and (28.37) gives us (3.21d 2 − 2.69d 1 ) 3 = 46 2 3 1 3 .( ).dd− (28.38) This result can be solved numerically to give d 1 = 0.63d 2 ; and, using eqn. (28.33), we can then show that d 2 = 0.66d w . Conclusions This design study has shown that it is possible to design a sectional composite that will reproduce both the vibrational frequencies and the mass of a traditional wooden soundboard. For a soundboard made out of spruce the equivalent composite is a sandwich of cardboard glued between two identical layers of aligned CFRP with a fibre volume fraction of 0.13. If the wooden soundboard is 3 mm thick the replacement composite 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. 320 Engineering Materials 2 Appendix 1 Teaching yourself phase diagrams To the student If you work through this course, doing the problems, you will get a working know- ledge of what a phase diagram (or equilibrium diagram) means, and how to use it. Don’t rush it: learn the definitions and meditate a little over the diagrams themselves, checking yourself as you go. Some parts (the definitions, for instance) are pretty con- centrated stuff. Others (some of the problems, perhaps) may strike you as trivial. That is inevitable in a “teach yourself” course which has to accommodate students with differing backgrounds. Do them anyway. The whole thing should take you about 4 hours. The material given here broadly parallels the introductory part of the excel- lent text by Hansen and Beiner (referenced below); if you can read German, and want to learn more, work through this. Phase diagrams are important. Whenever materials engineers have to report on the properties of a metallic alloy, or a ceramic, the first thing they do is reach for the phase diagram. It tells them what, at equilibrium, the structure of the alloy or ceramic is. The real structure may not be the equilibrium one, but equilibrium structure gives a base line 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. T EACHING 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. Teaching yourself phase diagrams 321 Alloys 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 (Al 2 O 3 ) 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: W A = weight of component A weights of all components∑ × 100 The atom (or mol) % of component A: X A = number of atoms (or mols) of component A number of atoms (or mols) of all components∑ × 100 322 Engineering Materials 2 (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%: W Cu = ––––––––––––––––––––––– (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%: W Cu = –––––––––––––– The concentration of zinc in α -brass, in wt%: W Zn = –––––––––––––––– (c) The atomic weight of copper is 63.5 and of zinc 65.4. The concentration of copper in the α -brass, in at%: X Cu = ––––––––––– – The concentration of zinc in the α -brass, in at%: X Zn = ––––––––––––– – 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 a glance, which of the two compositions, in at%, is likely to be the right one? (a) X Au = 0.34, X Ni = 0.66. (b) X Au = 0.66, X Ni = 0.34. 1.3 Your favourite vodka is 100° proof (49 wt% of alcohol). The molecular weight of water is 18; that of ethyl alcohol – C 2 H 5 OH – is 46. What is the mol% of alcohol in the vodka? Mol% of alcohol: X C HOH 25 = –––––––––––––––––––––––––––––––– 1.4 An alloy consists of X A at% of A with an atomic weight a A , and X B at% of B with an atomic weight of a B . Derive an equation for the concentration of A in wt%. By symmetry, write down the equation for the concentration of B in wt%. Structure Alloys are usually made by melting the components and mixing them together while liquid, though you can make them by depositing the components from the vapour, or by diffusing solids into each other. No matter how you make it, a binary alloy can take one of four forms: (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. Teaching yourself phase diagrams 323 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 success- ively finer grades of emery paper, and then with diamond pastes (on rotating felt discs) 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 grains and phases; brass doorknobs often show this pattern, etched by the salts from sweaty hands. Grain boundaries show up because the etch attacks them preferentially. The etch 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 because many etches are designed to attack one phase more than another, giving a contrast difference between phases. 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 Al–4 wt% Cu is typical of (d) (p. 311). Questions 1.5 List the compositions of the alloy and the phases mentioned above. wt% Cd wt% Zn Cadmium-zinc alloy Zinc-rich phase Cadmium-rich phase Phases 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 liquid copper co-exist. Using the definition above, are one or two phases present? – –– Why? –––––––––––––––––––––––––––––––––––––––––––– – 324 Engineering Materials 2 1.7 Three components A, B and C of an alloy dissolve completely when liquid but have 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 the solid 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. E XAMPLE The alloy aluminium–4 wt% copper forms the basis of the 2000 series (Duralumin, or Dural 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 out from the aluminium as large lumps of a new phase: not pure copper, but of the compound CuAl 2 . If, instead, the material is quenched (cooled very rapidly, often by dropping it into cold water) from 500°C to 20°C, there is not time for the dissolved copper atoms to move together, by diffusion, to form CuAl 2 , and the alloy remains a solid 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 it there (“ageing”), the copper starts to diffuse together to form an enormous number of very tiny (nm) plate-like particles, of composition roughly CuAl 2 . On recooling to room temperature, this new structure is again frozen in. The yield strength and toughness of Dural differ enormously in these three condi- tions (slow-cooled, quenched, and quenched and aged); the last gives the highest yield and lowest toughness because the tiny particles obstruct dislocations very effectively. It is important to be able to describe the constitution and structure of an alloy quickly and accurately. So do the following, even if they seem obvious. Teaching yourself phase diagrams 325 Questions 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, two phases 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 W Pb = 95% of lead. The con- stitution of the alloy at room temperature is described by: (a) Number of phases ––––––––––––––––––––––––––––––––––– (b) Weight% of lead-rich phase –––––––––––––––––––––––––––––– Weight% of tin-rich phase ––––––––––––––––––––––––––––––– (c) Composition of lead-rich phase, in wt%: W Pb = ––––––––––––––––––– W Sn = ––––––––––––––––––– (d) Composition of tin-rich phase, in wt%: W Pb = –––––––––––––––––––– W Sn = –––––––––––––––––––– Equilibrium constitution The Al–4 wt% Cu alloy of the example can exist at 20°C in three different states. Only one – the slowly cooled one – is its equilibrium state, though given enough time the others would ultimately reach the same state. At a given temperature, then, there is an 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. This constitution 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-line for the constitution of the real alloy, and the likely non-equilibrium constitutions can often 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. [...]... 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 328 Engineering Materials 2 Fig A1 .2 (b) Mark the constitution point for a Pb–70 at% Sn alloy at 25 0°C onto Fig A1 .2 What does the alloy consist of at 25 0°C?... – – – – – Answers to questions: part 1 1.1 (a) WCu = 60% (b) WCu = 70%, WZn = 30% (c) XCu = 71%, XZn = 29 % 1 .2 (a) is the correct composition 1.3 Your vodka contains 27 mol% of alcohol 1.4 WA = aA X A a A X A + aB X B WB = aB X B a A X A + aB X B 1.5 wt% Cd Cadmium–zinc alloy Zinc-rich phase Cadmium-rich phase wt% Zn 40 0.1 99 .2 60 99.9 0.8 330 Engineering Materials 2 1.6 Two phases: liquid and solid... (d) 1 2 1 2 1.9 (a) (b) (c) (d) 2 63%, 37% 95%, 5% 0%, 100% 1.10 (a) 50%, 50%, 300°C (See Fig A1.4.) Fig A1.4 (b) Liquid; 1 (c) Liquid plus lead-rich solid; 2 (d) 24 0°C, 183°C At point 2: liquid plus solid (Pb) At point 3: two solids, (Sn) and (Pb) (e) XPb = 80%, T = 20 0°C 1 phase Lead-rich solid 155°C Two phases: lead-rich solid (Pb) and tin-rich solid (Sn) 1.11 Yes (see definition, on p 308) 1. 12 (See... – – – – – – – – – – – – – – Fig A1 .14 338 Engineering Materials 2 The composition of phase 1 is – – – – – – – – – – – – – – – – – – – – – – – – – – – – – The composition of phase 2 is – – – – – – – – – – – – – – – – – – – – – – – – – – – – – (d) Indicate with arrows on the figure the lines along which: 1 The composition of phase 1 moves 2 The composition of phase 2 moves The overall composition of... cooling curve Questions 2. 4 Construct the general shape of a cooling curve for iron, starting at 3000°C, and ending at 0°C, using the data given in problem 2. 3 (see Fig A1.10) Fig A1.10 2. 5 Which phases do you expect at each of the following constitution points, for the lead–tin system? (a) (b) (c) (d) XPb = 40%, T WPb = 15%, T WSn = 10%, T WPb = 35%, T = = = = 175°C 20 0°C 20 0°C 20 0°C –––––––––––––––––––––––––––––––––... 326 Engineering Materials 2 DEF The independent constitution variables or state variables are T, p and composition EXAMPLE FOR THE AL-CU ALLOY (DESCRIBED ON PAGE Values of the state variables 311): Equilibrium constitution (a) T = p = WAl = WCu = 500°C 1 atm 96% 4% 5 4 6 4 7 → Single-phase solid solution of copper in aluminium (b) T = p = WAl = WCu = 20 °C 1 atm 96% 4% 5 4 6 4... from the surface to the centre, reaching a peak of around 2 GPa Radioactive decay causes the centre to have a temperature of about 30°C, but at the surface the temperature is below −100°C Assuming a linear temperature gradient from the surface to the centre, which phases of ice would be found in Ganymede? Fig A1. 12 336 Engineering Materials 2 Binary systems What defines the constitution of an alloy?... co-exist at a point: the triple point 3 32 Engineering Materials 2 Fig A1.6 The behaviour at constant p is given by a horizontal cut through the diagram The solid melts at Tm and vaporises at Tv The phase diagram at constant pressure is a line (shown on the right) along which the span of stability of each phase is marked, as shown in Fig A1.7 Fig A1.7 Questions 2. 1 List the phases shown on the T–p diagram... A lead–tin alloy with composition WPb = 80% is held at a temperature T (a) At T = 28 0°C which is the dominant phase? (b) At T = 20 0°C which is the dominant phase? Indicate by arrows on Fig A1.17 the changes in the compositions of the liquid phase and the solid phase as the alloy is cooled from 28 0°C to 20 0°C Fig A1.17 2. 10 The alloy is cooled to 150°C (a) How many phases are present? – – – – – – – –... phases shown on the T–p diagram (Fig A1.7) – – – –––––––––––––– –––––––––––––––––––– 2. 2 If the pressure is increased, does the melting point of the material of the diagram increase – – – – – – – – – , decrease – – – – – – – – – or stay constant – – – – – – – – –? 2. 3 At 1 atmosphere, iron melts at 1536°C and boils at 28 60°C When it solidifies (a phase change), it does so in the b.c.c crystal structure . 0.39 Mg m −3 . Equations (28 .22 ) and (28 .23 ) then give us the following data. f || (s −1 ) ≈ 59 23 6 531 944 147 5 21 24 28 91 etc. ! 21 7 868 1953 34 72 etc. f ⊥ (s −1 ) ≈ @ 124 496 1116 1984 3100 etc. For. n l Ebd bd n l Ebd bd ww ww cc cc 2 2 3 12 2 2 3 12 2 122 12 π ρ π ρ || || || || || || || ||             =             // (28 .28 ) which tells us that the composite plate must have a thickness of 318 Engineering. around 0 .2 Mg m −3 . To match the mass of the sandwich to that of the spruce soundboard we must then have 0.39d w = 1 .25 (d 2 − d 1 ) + 0.2d 1 (28 . 32) or d w = 3 .21 d 2 − 2. 69d 1 . (28 .33) Case