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IFB 2012 Materials Selection in Mechanical Design Shape of cross section is kept constant. Only the material changes. INTRODUCTION Materials Selection Without Shape (1/2) Textbook Chapters 5 & 6 IFB 2012 INTRODUCTION Material Indices 1/12 Deriving Materials Indices, Example 1: Material for a stiff, light beam Beam (solid square section) Minimise mass, m: Function Goal 2 m AL b L Constraint Stiffness of the beam, S: Get these equations from the textbook, or pdf file “Useful Solutions” S F δ S CEI L 3 4 I = second moment of area: I b 12 What can be varied ? { • Material choice • Area = b2 12 S L5 m C 1/ 2 To minimise the mass 1/ 2 E m = mass A = cross section L = length = density b = edge length S = stiffness I = second moment of area E = Young’s modulus b2 = m/L b2 = bFree 4 = 12variable SL3/CE (Trade-off variable) Material Index E Chose materials with largest M= 1/ 2 IFB 2012 INTRODUCTION Material Indices Maximise Material Index ! 2/12 Elastic Bending of Beams and Panels; p. 533 or from pdf file “useful solutions” S S S F δ F C 1 EI 3 L CEI L 3 I ?? IFB 2012 INTRODUCTION Material Indices 3/12 Moments of Sections; p 531, or file “Useful Solutions” I b 4 12 IFB 2012 INTRODUCTION Material Indices 4/12 Deriving Materials Indices, Example 2: Material for a stiff, light panel Function Goal Panel, width w and length L specified Minimise mass, m m AL w t L Constraint Stiffness of the panel, S S F S δ CEI L 3 I Free • Material choice. {• Panel thickness t variable..?? What can be varied ? 12 S w 2 m C 1/ 3 2 L 1/ 3 E wt 3 12 Eliminate t m = mass w = width L = length = density t = thickness S = stiffness I = second moment of area E = Young’s modulus Material Index E Chose materials with largest M= IFB 2012 INTRODUCTION Material Indices 1/ 3 5/12 Material Indices for Minimum Mass Function Objective: minimise mass Same Volume Index 1/ ρ E/ ρ Tension (tie) Objective: minimise mass for given stiffness Bending (beam) Bending (panel) To minimise the mass E E 1/2 1/3 /ρ /ρ Maximise Material Index ! IFB 2012 INTRODUCTION Material Indices 6/12 Materials Selection using charts: effect of slope of selection line Exam question: What is the Physics behind Index the different exponents in the Different materials are Indices’ equations? selected, depending on the slope of the selection line E 1/ 3 M E 1/ 2 M E M Selection line slope 1 Selection line slope 2 Selection line slope 3 MECH4301 2011 Lecture 2 Charts 7/12 Demystifying Material Indices This is how the world looks like after you pass the Materials Selection Course IFB 2012 INTRODUCTION Material Indices 8/12 Demystifying Material Indices (beam, elastic bending) Material 1, Mass 1 stiffness S Materials 2, Mass 2 stiffness S 12 S L m 1 C 5 1/ 2 1 1 / 2 E1 2 1 / 2 m1 E2 m2 12 S L m 2 C 5 E 11 / 2 1 1/ 2 2 1 / 2 E2 M1 M2 For given shape, the reduction in mass at constant bending stiffness is given by the reciprocal of the ratio of material indices. Same applies to bending strength. IFB 2012 INTRODUCTION Material Indices 9/12 Example: How good are Mg and Al when it comes to reducing mass? A 10 kg component made of Steel… E m2/m1 = M1/M2 Exam question: Which beam is fatter?? E 1/ 2 E 1/ 3 Mg Steel Same: Which panel is thicker?? E (GPa) (Mg/m3) Tie-rod Beam Panel Equal Volume Steel 210 7.8 10 10 10 10 Al 75 2.7 10 5.9 4.9 3.5 Mg 44 1.7 11 5.1 3.9 2.2 heavier IFB 2012 INTRODUCTION Material Indices lighter 10/12 Comparative weight of panels of equal stiffness The Mg-Li (Steel, Ti, Al and Mg) (Emley, Principles of Mg Technology ) panel is thicker (GPa) (GPa) (Mg/m33) (Mg/m ) Relative weight mass MgLi 44 1 3 Mg 44 1.7 4 Al 75 2.7 5 Ti 115 4.5 7 Steel 210 7.8 10 E E IFB 2012 INTRODUCTION Material Indices 11/12 Example of solution to Tutorial # 1 (Exercise 7.3) IFB 2012 INTRODUCTION Material Indices 12/12 Example of solution to Tutorial # 1 (Exercise 7.3) Derivation of the Material Index: When fully loaded, the beam should not fail, i.e., maximum < * (yield strength) m = lA Solve for A= m/l. The maximum force is Solving for m: 3/2 * I * C m F C l 6 l5/2 3/2 y m 6 m C 2/3 F 2/3 5/3 L * 2/3 ( ) I/ym =A3/2/6 Material Index : M = (*)2/3/. Select using the - chart with a line of slope 1.5, on the upper left corner. 2011 Lecture 3 Material Indices 13/12 Select using the - chart with a line of slope 1.5, upper left corner. Titanium alloys Stainless steel CFRP, epoxy matrix (isotropic) 1000 Silicon carbide Magnesium alloys Magnesium alloys Polyamides (Nylons, PA) Wood, typical along grain Tensile strength (MPa) 100 Bamboo Rigid Polymer Foam (HD) Aluminum alloys GFRP, epoxy matrix (isotropic) Copy the results from CES Sort the materials by their Index Name CFRP, epoxy matrix Wood, typical along grain Flexible Polymer Foam Magnesium alloys Polyamides (Nylons, PA) X-Axis 1500 - 1600 600 - 800 16 - 35 1740 - 1950 1120 - 1140 Y-Axis 550 - 1050 60 - 100 0.24 - 0.85 185 - 475 90 - 165 Stage 1: Index 0.05375 0.02626 0.02488 0.02414 0.02175 Rigid Polymer Foam (LD) Silicon carbide GFRP, epoxy matrix Titanium alloys Bamboo Flexible Polymer Foam (LD) 36 - 70 3100 - 3210 1750 - 1970 4400 - 4800 600 - 800 38 - 70 0.45 - 2.25 400 - 610 138 - 241 300 - 1625 36 - 45 0.24 - 2.35 0.02 0.01981 0.01732 0.01713 0.01695 0.01602 Alumina Rigid Polymer Foam (MD) Stainless steel Polyester Low alloy steel High carbon steel Flexible Polymer Foam Aluminum alloys 3800 - 3980 78 - 165 7600 - 8100 1040 - 1400 7800 - 7900 7800 - 7900 70 - 115 2500 - 2900 350 - 588 0.65 - 5.1 480 - 2240 41.4 - 89.6 550 - 1760 550 - 1640 0.43 - 2.95 58 - 550 0.01518 0.01314 0.01306 0.01283 0.0126 0.01261 0.01207 0.01178 10 Rigid Polymer Foam (MD) Flexible Polymer Foam (LD) Selection line gradient1.5 1 100 1000 10000 Density (kg/m^3) Conclusions to the chart/table: Composites, timber are the best materials. Al, Mg and steels are good competitors. Foams perform generally well, due to their low density. However, if made out of foams, the beams will be rather fat/big! 2011 Lecture 3 Material Indices 14/12 The End Introduction IFB 2012 INTRODUCTION Material Indices 15/12 The CES software: Demonstration IFB 2012 INTRODUCTION Material Indices 16/12 Organising information: the MATERIALS TREE Kingdom Family • Ceramics & glasses Materials • Metals & alloys Class Steels Cu-alloys Al-alloys Ni-alloys Zn-alloys • Hybrids Attributes Density Ti-alloys • Polymers & elastomers Member 1000 2000 3000 4000 5000 6000 7000 8000 Mechanical props. Thermal props. Electrical props. Optical props. Corrosion props. Supporting information -- specific -- general A material record IFB 2012 INTRODUCTION Material Indices 17/12 CES : the 3 levels 3400 Level 2 enough for most exercises IFB 2012 INTRODUCTION Material Indices 18/12 Chart created with the CES software (level 1, 60 materials) Alumina CFRP, epoxy matrix (isotropic) E 100 Tungsten carbides Silicon Tungsten alloys Wood, typical along grain Brick Zinc alloys Rigid Polymer Foam (HD) 10 Young's modulus (GPa) Rigid Polymer Foam (MD) Wood, typical across grain Concrete Lead alloys 1 Rigid Polymer Foam (LD) Polyvinylchloride (tpPVC) Leather 0.1 Cork 0.01 Silicone elastomers 1e-3 Polychloroprene (Neoprene, CR) Flexible Polymer Foam (VLD) 100 Density (kg/m^3) 1000 IFB 2012 INTRODUCTION Material Indices 10000 density 19/12 Chart created with the CES software (level 3, ~3400 materials) Diamond 1000 Palm (0.35) 100 E Balsa (l) (ld) 10 Polymethacrylimide Foam: Young's modulus (GPa) Tin-Lead 63-37 Solder 1 Metal Impregnated Carbon Melamine Foam Leather 0.1 Silicone elastomer (Shore A40) 0.01 Silicone elastomer 1e-3 Ethylene-Propylene Rubber (EPM) 1e-4 Polyurethane Foam 1e-5 10 100 Density (kg/m^3) 1000 IFB 2012 INTRODUCTION Material Indices 10000 density 20/12 Ranking Materials using Charts Selection line for beams Selection line for tie rods ceramics Selection corner Selection line for panels E 1 E E C C E 1/ 3 foams 3 1/ 2 2 C metals composites & polymers One very significant conclusion from this course, so far: For beams and panels, materials with very low density are more important than for tie-rods. This is why foams are not used for tie rods, but are preferred for beams and more so for flat panels. Important: Read textbook pp.93-95: Summary and Conclusions to Ch. 4, Properties of charts. IFB 2012 INTRODUCTION Material Indices 21/12 Material Indices for Minimum Cost? Same for embodied energy Q = q, etc. mass = xV price [c ] = $/kg Total cost C = c x mass = c V Total cost C c [$/m3] [ c ] = [$/m3] “price density” Goal: minimise cost Performance metric = cost per given stiffness Function Index Tension (tie) E/ cρ Bending (beam) E 1/2 / cρ Bending (panel) E 1/3 / cρ To minimise the cost Maximise Material Index ! IFB 2012 INTRODUCTION Material Indices 22/12 Comparative stiffness of panels of equal weight (Steel, Ti, Al and Mg) (Emley, Principles of Mg Technology) Relative stiffness E (GPa) (Mg/m3) MgLi 44 1 23 Mg 44 1.7 19 Al 75 2.7 8 Ti 115 4.5 3 Steel 210 7.8 1 IFB 2012 INTRODUCTION Material Indices 23/12 Material for a stiff tie-rod of minimum mass Function Constraints Tie-rod { • Length L is specified • Must not deflect more than under load F m = mass = density E = Young’s modulus = deflection Equation for constraint on : ≤ L = L /E = L F/A E Goal A = LF/E Minimise mass m: m = AL What can be varied to meet the goal ? Performance metric: mass { Material Index • Material • Cross section area A L F m 2 E E Chose materials with largest M = IFB 2012 INTRODUCTION Material Indices To minimise the mass A = Free variable ; or Trade-off variable 24/12 Maximise Material Index ! Materials for a strong, light beam Function Beam (shaped section). Objective Minimise mass, m, where: Area A m AL Constraint Bending strength of the beam Mf: M I f Z * ym Ab 2 * Z bh 8 2 b 3 m = mass A = area L = length = density Mf = bending strength I = second moment of area E = Youngs Modulus Z = section modulus 8 m L Combining the equations to eliminate A gives: m 6 F f 2/3 5/3 L * 2/3 * 2/3 Chose materials with largest M = IFB 2012 INTRODUCTION Material Indices To minimise the mass Maximise Material Index ! 25/12 Failure of Beams; p. 535 IFB 2012 INTRODUCTION Material Indices 26/12 Moments of Sections; p 531 IFB 2012 INTRODUCTION Material Indices 27/12 Materials for a strong, light tie-rod Strong tie of length L and minimum mass Function Tie-rod F F Area A L Constraints • Length L is specified • Must not fail under load F Equation for constraint on A: F/A < y (1) Objective (Goal) Minimise mass m: m = AL Free variables • Material choice • Section area A. Performance metric m m FL y m = mass A = area L = length = density y= yield strength (2) Eliminate A in (2) using (1): Chose materials with largest M = IFB 2012 INTRODUCTION Material Indices To minimise the mass y 28/12 Maximise Material Index ! Material Indices An objective defines a performance metric: e.g. mass or cost. The equation for the performance metric contains material properties. Sometimes a single property Sometimes a combination Function Example Objective: minimise mass Performance metric = mass Either is a material index Stiffness Tension (tie) E/ ρ Bending (beam) E 1/2 /ρ Bending (panel) E 1/3 /ρ Strength y /ρ IFB 2012 INTRODUCTION Material Indices 2/3 σy /ρ 1/2 σy /ρ 29/12 Material Indices Each combination of Function Constraint Objective Free variable has a characterising material index Maximise this! INDEX E M 1/ 2 INDEX y M Maximise this! IFB 2012 INTRODUCTION Material Indices 30/12 The End Introduction IFB 2012 INTRODUCTION Material Indices 31/12