Materials Selection: Case Studies

Transcription

1 MME445: Lecture 21 & 22 Materials Selection: Case Studies A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Learning Objectives Knowledge & Understanding Knowledge of how to use materials property charts and material indices to solve simple material selection problems Skills & Abilities Ability to solve material selection problem using Material Index Values & Attitudes Develop confidence in using Material Indices to solve material selection problem Resources M F Ashby, Materials Selection in Mechanical Design, 4 th Ed., Ch. 06 1

2 Outline of today s lecture Problems in determining material index Case studies in materials selection Materials for ores Materials for table legs Materials for spring Materials for energy-efficient kiln wall E6.1 A material is required for the windings of an electric air-furnace capable of temperatures up to 1000 C. Think out what attributes a material must have if it is to be made into windings and function properly in a furnace. List the function and the constraints; set the objective to minimize material price and the free variables to choice of material. Function Windings for 1000 C electric air furnace Good electrical conductor Capable of being shaped to wire Constraints Sufficient ductility to allow a coil to be wound Maximum service temperature > 1000 C Good resistance to oxidation Objective Minimize material price Free variables Choice of material 2

3 Prob#1 A material is required for a disposable fork for a fast-food chain. List the objective and the constraints that you would see as important in this application. Function Constraints Objective Free variables Disposable cutlery Easy (and thus cheap) to mould Non-toxic Sufficient strength and stiffness (modulus) to withstand loads in use Recyclable or renewable (e.g. wooden cutlery) Ability to be colored (?) Minimize material price Choice of material Prob#2 Trucks rely on compressed air for braking and other power-actuated systems. The air is stored in one or a cluster of cylindrical pressure tanks like that shown here (length L, diameter 2R, hemispherical ends). Most are made of low-carbon steel, and they are heavy. The task is to explore the potential of alternative materials for lighter air tanks, recognizing that there must be a trade-off between mass and cost if it is too expensive, the truck owner will not want it even if it is lighter. Determine the design requirements. Function Constraints Objectives Free variables Air Cylinder for Truck Must not fail by yielding Diameter 2R and length L specified Minimize material cost C Minimize mass m Wall Thickness, t Choice of material 3

4 Prob#3 The elastic deflection at fracture (the resilience ) of an elastic-brittle solid is proportional to the failure strain, e fr = s fr / E, where s fr is the stress that will cause a crack to propagate: s fr = K 1c (pc) 1/2 Determine the materials index for materials that can deflect elastically without fracturing. s fr K 1c e fr = = = E E(pc) 1/2 1 (pc) 1/2 K 1c E Materials that can deflect elastically without fracturing are therefore those with large values of M = K 1c E Prob#4 A material is required for a cheap column with a solid circular cross-section that must support a load F* without buckling. It is to have a height L. Write down an equation for the material cost of the column in terms of its dimensions, the price per kg of the material, C m, and the material density ρ. The cross-section area A is a free variable eliminate it by using the constraint that the buckling load must not the less than F*. Hence read off the index for finding the cheapest column. You will need the equation for the load F* at which a slender column buckles: m = A L r F* C 1 E I L 2 C = m C m = (A L r) C m A C 4pL 2 F* C 1 E 4pL 2 F* C 1 E 1/2 1/2 L r C m = 4pF* L 2 C 1 1/2 r C m E For circular column, I = pr 4 / 4 = A 2 / 4p F* C 1 EI/L 2 = (C 1 E/L 2 ) (A 2 /4p) The material cost of the column is minimized by choosing materials with the largest value of the index M = E r C m 4

5 Materials for oars Mechanically speaking, an oar is a beam, loaded in bending. It must be strong enough to carry, without breaking, the bending moment exerted by the oarsman it must have a stiffness to match the rower s own characteristics it must give the right feel It must be light; the extra weight increases the wetted area of the hull and the drag that goes with it. Thus, an oar is a beam of specified stiffness and minimum weight The material index for a light, stiff beam: There are other obvious constraints: Oars are dropped, and blades sometimes clash. The material must be tough enough to survive this. So brittle materials (having toughness G 1C less than 1 kj/m 2 ) are unacceptable. 5

6 Search Region M 168 Pa ½ / kg m -3 E 1/2 r selection line with slope 2 E 1/2 M 168 Pa ½ / kg m -3 r 10 Materials Selected 1. LD foam 2. Soft and hard woods 3. Bamboo 4. CFRP composite 5. Technical ceramics selection line with slope 2 6

7 After 1st Criterion, 10 Materials Selected 1. LD foam 2. Soft and hard woods 3. Bamboo 4. CFRP composite 5. Technical ceramics These selections must satisfy the other constraint (limit on toughness) G 1C K 1C 2 / E > 1 kj/m 2 K 2 1c E selection line with slope 0.5 Search Region K 1C 2 / E = 1 kj/m 2 The chart shows that, besides LD foam, all ceramics fail to meet the requirement G 1C K 1C 2 / E > 1 kj/m 2 7

8 K 2 1c E selection line with slope 0.5 K 1C 2 / E = 1 kj/m 2 Finally Selected Materials 1. Hard wood (Avg. M = 3950) 2. Soft wood (M = 1500) 3. Bamboo (M = 2020) 4. CFRP (M = 1200) Finally Selected Materials 1. Hard wood (M = 3950) 2. Soft wood (M = 1500) 3. Bamboo (M = 2020) 4. CFRP (M = 1200) So, the recommendation is clear : Make your oars out of hard wood Bamboo would only be suitable for the handle in the shape of a tube and could be combined with a spoon of another material (i.e. shaping requirement) Assuming that the maximum materials index of CFRP is chosen, the oar will be lighter (Wooden oar ~ kg; CFRP oar ~ 3.9 kg) But this is not necessarily true in all cases as the indices overlap this would require the evaluation of individual products to obtain the desired outcome for the design 8

9 Material for table legs A furniture designer wants to design a lightweight table consists of a flat sheet of toughened glass supported on slender, unbraced cylindrical legs The legs must be solid (to make them slender) and as light as possible (to make the table easier to move), and must support the applied design load (the table top and whatever is placed upon it) without buckling and fracture if struck. This is a problem with two objectives: 1. mass is to be minimized, and 2. slenderness maximized There are two constraints: resistance to buckling and resistance to fracture 9

10 Minimizing mass The objective function: subject to the constraint that it supports a load F without buckling The critical elastic buckling load F crit of a column of length L and radius r is given by the Euler s formula: Solving for the free variable, r, and substituting it into the equation for m gives: The material index to maximize: M 1 = E 1/2 /r 5 GPa 1/2 / (Mg/m 3 ) Selected Area E 1/3 /r E 1/2 /r E/r Polymers are out: They are not stiff enough Metals too: They are too heavy 10

11 M 1 5 GPa 1/2 / (Mg/m 3 ) E 1/3 /r E 1/2 /r 11 Materials Selected 1. LD foam 2. Soft and hard woods 3. Bamboo 4. Papers and cardboard 5. CFRP composite 6. Technical ceramics E/r Slenderness Inverting this equation with F crit set equal to F gives an equation for the thinnest leg that will not buckle: The material index: i.e., the thinnest leg is that made of the material with the largest value of the modulus 11

12 M 1 = 5 GPa 1/2 / (Mg/m 3 ) Selected Area M 2 = 100 GPa E 1/3 /r E 1/2 /r E/r Natural materials and foams are out: They are not stiff enough M 1 = 5 GPa 1/2 / (Mg/m 3 ) Selected Area M 2 = 100 GPa E 1/3 /r 7 Materials Selected 1. CFRP composite 2. Technical ceramics E 1/2 /r E/r 12

13 If the legs must be really thin, then the shortlist is reduced to CFRP and ceramics They give legs that weigh the same as the wooden ones but are barely half as thick Table legs are exposed to abuse they get knocked and kicked Common sense suggests that an additional constraint is needed, that of adequate toughness. This can be done using K 1C E chart; it eliminates ceramics, leaving CFRP, although the cost of CFRP may cause the Snr. Ceramics, being brittle, have low values of fracture toughness. Selected Area G c = K 1C 2 / E 1 kj/m 2 Engineering ceramics are out: They are not tough enough K 2 1c E 13

14 G c = K 1C 2 / E = 1 kj/m 2 Finally Selected Material 1. CFRP composite K 2 1c E Material for spring Springs are devices for storing (and releasing) the elastic energy of a deformed material leaf spring Torsion bars and leaf springs are less efficient than axial springs because much of the material is not fully loaded: The material at the neutral axis, for instance, is not loaded at all. axial spring spiral spring The elastic energy stored per unit volume in a block of material stressed uniformly to a stress σ is Axial spring: Leaf spring: torsion spring Torsion spring: 14

15 The spring will be damaged if the stress σ exceeds the yield stress or failure stress σ f ; the constraint is: σ < σ f. Thus the maximum energy density for axial spring is : If mass, rather than volume, matters, then energy stored per unit weight: So, the material index: For max. stored energy per unit volume, we maximise: For max. stored energy per unit mass, we maximise: 15

16 The best choice: high-strength steel Efficient springs can also be made from other suggested materials: CFRP (now used for truck springs) titanium alloys (good but expensive) nylon, PA (children s toys) elastomers Now, let us consider the selection of materials for an efficient light spring 16

17 M = = s f 2 E r s f r E r 2 Metals, because of their high density, are less good than composites, and much less good than elastomers You can store roughly eight times more elastic energy, per unit weight, in a rubber band than in the best spring steel Many additional considerations enter the choice of a material for a spring: Springs for vehicle suspensions must resist fatigue and corrosion engine valve springs must cope with elevated temperatures. Polymers have a relatively high loss factor and dissipate energy when they vibrate; metals, if strongly hardened, do not. Polymers, because they creep, are unsuitable for springs that carry a steady load for long periods of time. 17

18 Material for energy-efficient kiln wall When a kiln is fired, the internal temperature rises quickly from the ambient, T o, to the operating temperature, T i, where it is held for the firing time, t. The energy consumed in the firing time has two contributions: 1. Heat flow/unit area out of the furnace l = thermal conductivity 2. Heat absorbed/unit area by the kiln walls in raising the internal temperature from T o to T i The total energy consumed per unit area is the sum of the two contributions (i.e. heat balance): 18

19 The optimum wall thickness is a balance between low conduction to the environment and internal heat absorption too thin a wall will allow too much heat to flow the environment and too thick a wall will require too much energy to heat up. We find the optimum thickness by differentiating the equation with respect to wall thickness w and equating the result to zero: = Thermal diffusivity Thus, the most energy-efficient kiln wall is the one that only gets really hot on the outside as the firing cycle approaches completion. Eliminating w : q is minimized by choosing a material with a low value of the quantity (lc P r) ½, that is, by maximizing 19

20 Probable candidates: metal foams, ceramic foams, conventional brick and some polymer foam (the last ranked very highly) Polymer foam? Something is wrong here. Have we forgotten something in the way of a constraint? Oxidation resistance?! it is not unusual that kilns are not fired under an inert atmosphere and that the polymer foam will burn during heat up. Real pottery kilns operate near 1000 C, requiring materials with a maximum service temperature above this value. In the limit stage, check off very good oxidation resistance above 500 C and examine the results again. 20

21 It is not generally appreciated that, in an efficiently designed kiln, as much energy goes in heating up the kiln itself as is lost by thermal conduction to the outside environment. It is a mistake to make kiln walls too thick; a little is saved in reduced conduction loss, but more is lost in the greater heat capacity of the kiln itself. That, too, is the reason that foams are good: They have a low thermal conductivity and a low heat capacity. 21

22 Next Class MME445: Lecture 23 Multiple Constraints and Conflicting Objectives 1 22