Engineering Materials Vol II (microstructures processing design) 2nd ed. - M. Ashby_ D. Jones (1999) Episode 11 ppt

Design with materials 291 be designed to snap together, making assembly fast and cheap. And by accurately sizing the mould, and using pre-coloured polymer, no finishing operations are neces- sary. So great economies of manufacture are possible: polymer parts really can be cheap. But are they inferior? Not necessarily. Polymer densities are low (all are near 1 Mg m −3 ); they are corrosion-resistant; they have abnormally low coefficients of fric- tion; and the low modulus and high strength allows very large elastic deformations. Because of these special properties, polymer parts may be distinctly superior. Composites overcome many of the remaining deficiencies. They are stiff, strong and tough. Their problem lies in their cost: composite components are usually expensive, and they are difficult and expensive to form and join. So, despite their attractive properties, the designer will use them only when the added performance offsets the added expense. New materials are appearing all the time. New polymers with greater stiffness and toughness appear every year; composites are becoming cheaper as the volume of their production increases. Ceramics with enough toughness to be used in conventional design are becoming available, and even in the metals field, which is a slowly devel- oping one, better quality control, and better understanding of alloying, leads to materials with reliably better properties. All of these offer new opportunities to the designer who can frequently redesign an established product, making use of the prop- erties of new materials, to reduce its cost or its size and improve its performance and appearance. Design methodology Books on design often strike the reader as vague and qualitative; there is an implica- tion that the ability to design is like the ability to write music: a gift given to few. And it is true that there is an element of creative thinking (as opposed to logical reasoning or analysis) in good design. But a design methodology can be formulated, and when followed, it will lead to a practical solution to the design problem. Figure 27.1 summarises the methodology for designing a component which must carry load. At the start there are two parallel streams: materials selection and com- ponent design. A tentative material is chosen and data for it are assembled from data sheets like the ones given in this book or from data books (referred to at the end of this chapter). At the same time, a tentative component design is drawn up, able to fill the function (which must be carefully defined at the start); and an approximate stress analysis is carried out to assess the stresses, moments, and stress concentrations to which it will be subjected. The two streams merge in an assessment of the material performance in the tentat- ive design. If the material can bear the loads, moments, concentrated stresses (etc.) without deflecting too much, collapsing or failing in some other way, then the design can proceed. If the material cannot perform adequately, the first iteration takes place: either a new material is chosen, or the component design is changed (or both) to overcome the failing. The next step is a detailed specification of the design and of the material. This may require a detailed stress analysis, analysis of the dynamics of the system, its response 292 Engineering Materials 2 Fig. 27.1. Design methodology. Design with materials 293 to temperature and environment, and a detailed consideration of the appearance and feel (the aesthetics of the product). And it will require better material data: at this point it may be necessary to get detailed material properties from possible suppliers, or to conduct tests yourself. The design is viable only if it can be produced economically. The choice of produc- tion and fabrication method is largely determined by the choice of material. But the production route will also be influenced by the size of the production run, and how the component will be finished and joined to other components; each class of material has its own special problems here; they were discussed in Chapters 14, 19, 24 and 25. The choice of material and production route will, ultimately, determine the price of the product, so a second major iteration may be required if the costing shows the price to be too high. Then a new choice of material or component design, allowing an altern- ative production path, may have to be considered. At this stage a prototype product is produced, and its performance in the market is assessed. If this is satisfactory, full-scale production is established. But the designer’s role does not end at this point. Continuous analysis of the performance of a compon- ent usually reveals weaknesses or ways in which it could be improved or made more cheaply. And there is always scope for further innovation: for a radically new design, or for a radical change in the material which the component is made from. Successful designs evolve continuously, and only in this way does the product retain a competit- ive position in the market place. Further reading (a) Design G. Pahl and W. Beitz, Engineering Design, The Design Council, 1984. V. Papanek, Design for the Real World, Random House, 1971. (b) Metals ASM Metals Handbook, 8th edition, American Society for Metals, 1973. Smithells’ Metals Reference Book, 7th edition, Butterworth-Heinemann, 1992. (c) Ceramics W. E. C. Creyke, I. E. J. Sainsbury, and R. Morrell, Design with Non-Ductile Materials, Applied Science Publishers, 1982. D. W. Richardson, Modern Ceramic Engineering, Marcel Dekker, 1982. (d) Polymers DuPont Design Handbooks, DuPont de Nemours and Co., Polymer Products Department, Wilmington, Delaware 19898, USA, 1981. ICI Technical Services Notes, ICI Plastics Division, Engineering Plastics Group, Welwyn Garden City, Herts., England, 1981. 294 Engineering Materials 2 (e) Materials selection J. A. Charles and F. A. A. Crane, Selection and Use of Engineering Materials, 2nd edition, Butterworth- Heinemann, 1989. M. F. Ashby, Materials Selection in Mechanical Design, Pergamon, 1992. M. F. Ashby and D. Cebon, Case Studies in Materials Selection, Granta Design, 1996. Problems 27.1 You have been asked to prepare an outline design for the pressure hull of a deep- sea submersible vehicle capable of descending to the bottom of the Mariana Trench in the Pacific Ocean. The external pressure at this depth is approximately 100 MPa, and the design pressure is to be taken as 200 MPa. The pressure hull is to have the form of a thin-walled sphere with a specified radius r of 1 m and a uniform thickness t. The sphere can fail in one of two ways: external-pressure buckling at a pressure p b given by pE t r b .,=       03 2 where E is Young’s modulus; yield or compressive failure at a pressure p f given by p t r ff ,=       2 σ where σ f is the yield stress or the compressive failure stress as appropriate. The basic design requirement is that the pressure hull shall have the min- imum possible mass compatible with surviving the design pressure. By eliminating t from the equations, show that the minimum mass of the hull is given by the expressions mrp E b b ., . . =       22 9 305 05 ρ for external-pressure buckling, and mrp ff f ,=       2 3 π ρ σ for yield or brittle compressive failure. Hence obtain a merit index to meet the design requirement for each of the two failure mechanisms. [You may assume that the surface area of the sphere is 4 π r 2 .] Answers: E 0.5 / ρ for external-pressure buckling; σ f / ρ for yield or brittle compressive failure. 27.2 For each material listed in the following table, calculate the minimum mass and wall thickness of the pressure hull of Problem 27.1 for both failure mechanisms at the design pressure. Design with materials 295 Material E (GPa) s f (MPa) Density, r (kg m − 3 ) Alumina 390 5000 3900 Glass 70 2000 2600 Alloy steel 210 2000 7800 Titanium alloy 120 1200 4700 Aluminium alloy 70 500 2700 Hence determine the limiting failure mechanism for each material. [Hint: this is the failure mechanism which gives the larger of the two values of t.] What is the optimum material for the pressure hull? What are the mass, wall thickness and limiting failure mechanism of the optimum pressure hull? Answers: Material m b (tonne) t b (mm) m f (tonne) t f (mm) Limiting failure mechanism Alumina 2.02 41 0.98 20 Buckling Glass 3.18 97 1.63 50 Buckling Alloy steel 5.51 56 4.90 50 Buckling Titanium alloy 4.39 74 4.92 83 Yielding Aluminium alloy 3.30 97 6.79 200 Yielding The optimum material is alumina, with a mass of 2.02 tonne, a wall thickness of 41 mm and a limiting failure mechanism of external-pressure buckling. 27.3 Briefly describe the processing route which you would specify for making the pressure hull of Problem 27.2 from each of the materials listed in the table. Com- ment on any particular problems which might be encountered. [You may assume that the detailed design will call for a number of apertures in the wall of the pressure hull.] 296 Engineering Materials 2 Chapter 28 Case studies in design 1. D ESIGNING WITH METALS : CONVEYOR DRUMS FOR AN IRON ORE TERINAL Introduction The conveyor belt is one of the most efficient devices available for moving goods over short distances. Billions of tons of minerals, foodstuffs and consumer goods are handled in this way every year. Figure 28.1 shows the essentials of a typical conveyor system. The following data are typical of the largest conveyors, which are used for handling coal, iron ore and other heavy minerals. Capacity: 5000 tonne h −1 Belt speed: 4 m s −1 Belt tension: 5 tonne Motor rating: 250 k W Belt section: 1.5 m wide × 11 mm thick Distance between centres of tail drum and drive drum: 200 m Fig. 28.1. Schematic of a typical conveyor system. Because the belt tends to sag between the support rollers it must be kept under a constant tension T . This is done by hanging a large weight on the tension drum. The drive is supplied by coupling a large electric motor to the shaft of the drive drum via a suitable gearbox and overload clutch. Case studies in design 297 It is important that conveyor systems of this size are designed to operate continuously for long periods with minimum “down-time” for routine maintenance: the unsched- uled breakdown of a single unit in an integrated plant could lead to a total loss of production. Large conveyors include a number of critical components which are designed and built essentially as “one-offs” for a particular installation: it is doubly important to check these at the design stage because a failure here could lead to a damagingly long down-time while a harassed technical manager phones the length of the country looking for fabrication shops with manoeuvrable capacity, and steel merchants with the right sections in stock. Tail drum design The tail drum (Fig. 28.1) is a good example of a critical component. Figure 28.2 shows the general arrangement of the drum in its working environment and Fig. 28.3 shows a detailed design proposal. We begin our design check by looking at the stresses in the shaft. The maximum stress comes at the surface of the shaft next to the shaft-plate weld (Fig. 28.4). We can calculate the maximum stress from the standard formula σ max = Mc I (28.1) where the bending moment M is given by M = Fx (28.2) and the second moment of area of the shaft is given by I c .= π 4 4 (28.3) Using values of F = 5000 × 9.81 N, x = 380 mm, and c = 75 mm, we get a value for σ max of 56 MPa. This stress is only a quarter of the yield stress of a typical structural steel, and the shaft therefore has an ample factor of safety against failure by plastic overload. Fig. 28.2. Close-up of the tail drum. The belt tension applies a uniformly distributed sideways loading to the drum. 298 Engineering Materials 2 Fig. 28.3. Cross-section through the tail drum. All dimensions are in mm. We have assumed a belt tension of 5 tonnes, giving a total loading of 10 tonnes. Fig. 28.4. Shaft-plate detail. The second failure mode to consider is fatigue. The drum will revolve about once every second, and each part of the shaft surface will go alternately into tension and compression. The maximum fatigue stress range (of 2 × 56 = 112 MPa) is, however, only a quarter of the fatigue limit for structural steel (Fig. 28.5); and the shaft should therefore last indefinitely. But what about the welds? There are in fact a number of reasons for expecting them to have fatigue properties that are poorer than those of the parent steel (see Table 28.1). Figure 28.6 shows the fatigue properties of structural steel welds. The fatigue limit stress range of 120 MPa for the best class of weld is a good deal less than the limiting range of 440 MPa for the parent steel (Fig. 28.5). And the worst class of weld has a limiting range of only 32 MPa! Case studies in design 299 Fig. 28.5. Fatigue data for a typical structural steel in dry air. Note that, if the fatigue stress range is less than 440 MPa (the fatigue limit ) the component should last indefinitely. The data relate to a fatigue stress cycle with a zero mean stress, which is what we have in the case of our tail drum. The shaft-plate weld can be identified as a class E/F weld with a limiting stress range of 69 to 55 MPa. This is a good deal less than the stress range of 112 MPa experienced by the shaft. We thus have the curious situation where a weld which is merely an attachment to the shaft has weakened it so much that it will only last for about 2 × 10 6 cycles – or 1 month of operation. The obvious way of solving this problem is to remove the attachment weld from the surface of the shaft. Figure 28.7 shows one way of doing this. Gives stress concentration. In the case of butt welds this can be removed by grinding back the weld until flush with the parent plates. Grinding marks must be parallel to loading direction otherwise they can initiate fatigue cracks. Helps initiate fatigue cracks. Improve finish by grinding. Weld liable to fatigue even when applied stress cycle is wholly compressive . Reduce residual stresses by stress relieving, hammering or shot peening. Help initiate fatigue cracks. Critical welds must be tested non-destructively and defects must be gouged out. Sharp changes in mechanical properties give local stress concentrations. Table 28.1. Weld characteristics giving adverse fatigue properties Characteristic Comments Change in section at weld bead. Poor surface finish of weld bead. Contain tensile residual stresses which are usually as large as the yield stress. Often contain defects (hydrogen cracks, slag inclusions, stop–start marks). Large differences in microstructure between parent metal, heat-affected zone and weld bead. 300 Engineering Materials 2 Fig. 28.6. Fatigue data for welded joints in clean air. The class given to a weld depends critically on the weld detail and the loading direction. Classes B and C must be free from cracks and must be ground flush with the surface to remove stress concentrations. These conditions are rarely met in practice, and most welds used in general construction have comparatively poor fatigue properties. Fig. 28.7. Modification to remove attachment weld from surface of shaft. The collar can be pressed on to the shaft and secured with a feather key, but we must remember that the keyway will weaken the shaft. [...]... 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,... direct load Of course, when two of the spokes straddle the vertical 312 Engineering Materials 2 Fig 28.16 Tentative design for a self-sprung wheel The spokes and the rim both deflect The design ensures that the sum of the deflections is constant and share the load, the deflection would be only one-half this amount (1.3 mm) The cross-section of the rim could then be sized to deflect as a beam between spokes,... Violin soundboards are usually made from spruce, 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 Ew|| = 11. 6 GPa, Ew⊥ = 0.71 GPa and ρw = 0.39 Mg m−3 Equations... is the couple which must be applied at the mid span of the shaft to make (dy/dx)x=0 = 0; and MB is the couple that is needed to cause the out-of-plane deflection of the end discs Using the standard beam-bending formula of d2 y M = dx 2 EI (28.5) 302 Engineering Materials 2 we can write dy  dy   =  −  B dx A dx Setting lm M ∫ EI dx (28.6) 0 dy  dy   = 0, and M (0 ഛ x ഛ 1 m) = MA then gives ... (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... therefore has both along-grain and across-grain vibrations (Fig 28.19) The frequencies of these vibrations are then  π   Ew || I w  f|| = n 2  2     2l ||   ρw b|| dw  1/2 (28.22) Case studies in design 315 Fig 28.18 Idealized vibration modes of a soundboard The natural frequencies of the modes are proportional to n2 Fig 28.19 A wooden soundboard has both along-grain and across-grain vibrations... in the tyre and the radius of the tyre cross-section, R This force creates bending stresses in the rim section which can be minimised by keeping the rim height H as small as possible: the left-hand design of Fig 28.15 is poor; the right-hand design is better Radial ribbing, added as shown, further stiffens the rim without substantially affecting the cross-section wall thickness Innovative design Polymers... 236 531 944 1475 2124 2891 etc ! 217 868 1953 3472 etc −1 f⊥(s ) ≈ @ 124 496 111 6 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... match the mass of the sandwich to that of the spruce soundboard we must then have 0.39dw = 1.25(d2 − d1) + 0.2d1 (28.32) dw = 3.21d2 − 2.69d1 (28.33) or Fig 28.22 Sandwich-type sectional composites give a much-improved stiffness-to-mass ratio 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... radius R The general rule for Fig 28 .11 Changes of section cause distortion, sink-marks and internal stresses in moulded plastic parts Case studies in design Fig 28.12 Example of good and bad design with polymers Fig 28.13 A sharp change of section induces stress concentrations The local stress can be many times greater than the nominal stress 309 310 Engineering Materials 2 estimating stress concentrations . applied at the mid span of the shaft to make (dy/dx) x=0 = 0; and M B is the couple that is needed to cause the out-of-plane deflection of the end discs. Using the standard beam-bending formula. Engineering Materials, 2nd edition, Butterworth- Heinemann, 1989. M. F. Ashby, Materials Selection in Mechanical Design, Pergamon, 1992. M. F. Ashby and D. Cebon, Case Studies in Materials Selection,. rule of mould design is to aim for a uniform section throughout the component. During mould- ing, hot polymer is injected or pressed into the mould. Solidification proceeds from the outside in. Abrupt