1. IFB 2012 INTRODUCTION Material Indices1/12 IFB 2012 Materials Selection in Mechanical Design INTRODUCTION Materials Selection Without Shape (1/2) Textbook Chapters 5 & 6 Shape of cross section is kept constant. Only the material changes.

2. b 2 = Free variable (Trade-off variable) IFB 2012 INTRODUCTION Material Indices2/12 Deriving Materials Indices, Example 1: Material for a stiff, light beam Chose materials with largest M= Material Index Material choice Area = b 2 What can be varied ? { I = second moment of area: Beam (solid square section) Function m = mass A = cross section L = length  = density b = edge length S = stiffness I = second moment of area E = Young’s modulus Stiffness of the beam, S: Constraint Minimise mass, m: Goal b 2 = m/L  b 4 = 12 SL 3 /CE To minimise the mass Maximise Material Index ! Get these equations from the textbook, or pdf file “Useful Solutions”

3. IFB 2012 INTRODUCTION Material Indices3/12 Elastic Bending of Beams and Panels; p. 533 or from pdf file “useful solutions”

4. IFB 2012 INTRODUCTION Material Indices4/12 Moments of Sections; p 531, or file “Useful Solutions”

5. IFB 2012 INTRODUCTION Material Indices5/12 m = mass w = width L = length  = density t = thickness S = stiffness I = second moment of area E = Young’s modulus Panel, width w and length L specified Stiffness of the panel, S Minimise mass, m Chose materials with largest M= Deriving Materials Indices, Example 2: Material for a stiff, light panel Material Index Function Goal Constraint Material choice. Panel thickness t What can be varied ? { Eliminate t Free variable..??

6. IFB 2012 INTRODUCTION Material Indices6/12 Material Indices for Minimum Mass Function Index Same Volume Tension (tie) Bending (beam) Bending (panel) Objective: minimise mass for given stiffness Objective: minimise mass To minimise the mass Maximise Material Index !

7. MECH4301 2011 Lecture 2 Charts7/12 Materials Selection using charts: effect of slope of selection line Index Selection line slope 2 Selection line slope 1 Selection line slope 3 Different materials are selected, depending on the slope of the selection line Exam question: What is the Physics behind the different exponents in the Indices’ equations?

8. IFB 2012 INTRODUCTION Material Indices8/12 Demystifying Material Indices This is how the world looks like after you pass the Materials Selection Course

9. IFB 2012 INTRODUCTION Material Indices9/12 Demystifying Material Indices (beam, elastic bending) 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. Material 1, Mass 1 stiffness S Materials 2, Mass 2 stiffness S

10. IFB 2012 INTRODUCTION Material Indices10/12 Example: How good are Mg and Al when it comes to reducing mass? E (GPa)  ( Mg/m 3 ) Tie-rodBeamPanel Equal Volume Steel2107.810 Al752.7105.94.93.5 Mg441.7115.13.92.2 A 10 kg component made of Steel… heavier lighter m 2 /m 1 = M 1 /M 2 Exam question: Which beam is fatter?? Same: Which panel is thicker??

11. Comparative weight of panels of equal stiffness (Steel, Ti, Al and Mg) (Emley, Principles of Mg Technology ) IFB 2012 INTRODUCTION Material Indices11/12 E (GPa)  ( Mg/m 3 ) Relative weight MgLi4413 Mg441.74 Al752.75 Ti1154.57 Steel2107.810 E (GPa)  ( Mg/m 3 ) Relative mass MgLi4413 Mg441.74 Al752.75 Ti1154.57 Steel2107.810 The Mg-Li panel is thicker

12. IFB 2012 INTRODUCTION Material Indices12/12 Example of solution to Tutorial # 1 (Exercise 7.3)

13. 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 I/y m =A 3/2 /6 2011 Lecture 3 Material Indices13/12 Solving for m: Select using the  -  chart with a line of slope 1.5, on the upper left corner. Material Index : M = (  * ) 2/3 / . Example of solution to Tutorial # 1 (Exercise 7.3)

14. 2011 Lecture 3 Material Indices14/12 NameX-AxisY-AxisStage 1: Index CFRP, epoxy matrix1500 - 1600550 - 10500.05375 Wood, typical along grain600 - 80060 - 1000.02626 Flexible Polymer Foam16 - 350.24 - 0.850.02488 Magnesium alloys1740 - 1950185 - 4750.02414 Polyamides (Nylons, PA)1120 - 114090 - 1650.02175 Rigid Polymer Foam (LD)36 - 700.45 - 2.250.02 Silicon carbide3100 - 3210400 - 6100.01981 GFRP, epoxy matrix1750 - 1970138 - 2410.01732 Titanium alloys4400 - 4800300 - 16250.01713 Bamboo600 - 80036 - 450.01695 Flexible Polymer Foam (LD)38 - 700.24 - 2.350.01602 Alumina3800 - 3980350 - 5880.01518 Rigid Polymer Foam (MD)78 - 1650.65 - 5.10.01314 Stainless steel7600 - 8100480 - 22400.01306 Polyester1040 - 140041.4 - 89.60.01283 Low alloy steel7800 - 7900550 - 17600.0126 High carbon steel7800 - 7900550 - 16400.01261 Flexible Polymer Foam70 - 1150.43 - 2.950.01207 Aluminum alloys2500 - 290058 - 5500.01178 Copy the results from CES 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! Select using the  -  chart with a line of slope 1.5, upper left corner. Sort the materials by their Index

15. IFB 2012 INTRODUCTION Material Indices15/12 The End Introduction

16. IFB 2012 INTRODUCTION Material Indices16/12 The CES software: Demonstration

17. IFB 2012 INTRODUCTION Material Indices17/12 Organising information: the MATERIALS TREE Kingdom Materials Family Ceramics & glasses Metals & alloys Polymers & elastomers Hybrids Class Steels Cu-alloys Al-alloys Ti-alloys Ni-alloys Zn-alloys Member 1000 2000 3000 4000 5000 6000 7000 8000 A material record Attributes Density Mechanical props. Thermal props. Electrical props. Optical props. Corrosion props. Supporting information -- specific -- general Density Mechanical props. Thermal props. Electrical props. Optical props. Corrosion props. Supporting information -- specific -- general

18. IFB 2012 INTRODUCTION Material Indices18/12 CES : the 3 levels Level 2 enough for most exercises 3400

19. IFB 2012 INTRODUCTION Material Indices19/12 Chart created with the CES software (level 1, 60 materials) Edensity

20. IFB 2012 INTRODUCTION Material Indices20/12 Chart created with the CES software (level 3, ~3400 materials ) Edensity

21. IFB 2012 INTRODUCTION Material Indices21/12 Ranking Materials using Charts 1 2 3 E  metals ceramics composites & polymers foams Selection line for tie rods Selection line for beams Selection line for panels 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. Selection corner Important: Read textbook pp.93-95: Summary and Conclusions to Ch. 4, Properties of charts.

22. IFB 2012 INTRODUCTION Material Indices22/12 mass =  xV price [c ] = $/kg Total cost C = c x mass = c  V Total cost C   c [$/m 3 ] Function Index Tension (tie) Bending (beam) Bending (panel) Material Indices for Minimum Cost? Goal: minimise cost Performance metric = cost per given stiffness To minimise the cost Maximise Material Index ! Same for embodied energy Q =  q, etc. [  c ] = [$/m 3 ]  “price density”

23. IFB 2012 INTRODUCTION Material Indices23/12 Comparative stiffness of panels of equal weight (Steel, Ti, Al and Mg) (Emley, Principles of Mg Technology ) E (GPa)  ( Mg/m 3 ) Relative stiffness MgLi44123 Mg441.719 Al752.78 Ti1154.53 Steel2107.81

24. IFB 2012 INTRODUCTION Material Indices24/12 Material for a stiff tie-rod of minimum mass Minimise mass m: m = A L  Length L is specified Must not deflect more than  under load F Material Cross section area A Equation for constraint on  :  ≤ L  = L  /E = L F/A E Chose materials with largest M = Material Index Constraints Goal What can be varied to meet the goal ? Performance metric: mass { { A = LF/E  m = mass  = density E = Young’s modulus  = deflection Tie-rod Function To minimise the mass Maximise Material Index ! A = Free variable ; or Trade-off variable

25. IFB 2012 INTRODUCTION Material Indices 25/12 Materials for a strong, light beam m = mass A = area L = length  = density M f = bending strength I = second moment of area E = Youngs Modulus Z = section modulus Beam (shaped section). Bending strength of the beam M f : Combining the equations to eliminate A gives: Chose materials with largest M = Minimise mass, m, where: Function Objective Constraint Area A To minimise the mass Maximise Material Index !

26. IFB 2012 INTRODUCTION Material Indices26/12 Failure of Beams; p. 535

27. IFB 2012 INTRODUCTION Material Indices27/12 Moments of Sections; p 531

28. IFB 2012 INTRODUCTION Material Indices28/12 Materials for a strong, light tie-rod Minimise mass m: m = A L  (2) Objective (Goal) Length L is specified Must not fail under load F Constraints Material choice Section area A. Free variables Equation for constraint on A: F/A <  y (1) Strong tie of length L and minimum mass L F F Area A Tie-rod Function m = mass A = area L = length  = density = yield strength Performance metric m Chose materials with largest M = Eliminate A in (2) using (1): To minimise the mass Maximise Material Index !

29. IFB 2012 INTRODUCTION Material Indices29/12 Example Objective: minimise mass Performance metric = mass Function Stiffness Strength Tension (tie) Bending (beam) Bending (panel) 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 Either is a material index

30. IFB 2012 INTRODUCTION Material Indices30/12 Material Indices Each combination of Function Constraint Objective Free variable has a characterising material index Maximise this! I NDEX Maximise this!

31. IFB 2012 INTRODUCTION Material Indices31/12 The End Introduction