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Advanced Methods in Materials Selection

Advanced Methods in Materials Selection. Conflicting Constraints Lecture 2 & Tutorial 2. IFB Lecture 2: Textbook Chapters 7 & 8. Conflicting Constraints. Fork for a pushbike: what is more important, strength or stiffness?. Stiffness dominated design versus Strength dominated design.

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Advanced Methods in Materials Selection

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  1. Advanced Methods in Materials Selection • Conflicting Constraints • Lecture 2 & Tutorial 2 IFB 2012 Conflicting Constraints

  2. IFB Lecture 2: Textbook Chapters 7 & 8 IFB 2012 Conflicting Constraints

  3. Conflicting Constraints Fork for a pushbike: what is more important, strength or stiffness? • Stiffness dominated design • versus • Strength dominated design IFB 2012 Conflicting Constraints

  4. Outline Lecture 2 • Conflicting constraints: • “most restrictive constraint wins” • Case study 1: Stiff / strong / light tie rod • Case study 2: Safe (no yield - no fracture)/ light air tank for a truck IFB 2012 Conflicting Constraints

  5. Most designs are over-constrained: “Should not deflect more than something, must not fail by yielding, by fatigue, by fast-fracture …” more constraints than free variables IFB 2012 Conflicting Constraints

  6. Strong tie of length L and minimum mass Function Tie-rod F F Area A L Minimise mass m: m = A L  (2) Objective • Length L is specified • Must not stretch more than  Constraints Free variables • Material choice • Section area A Performance metric m1 E8.1. Materials for a stiff, light tie-rod Constraint # 1 m = mass A = area L = length  = density E= elastic modulus  = elastic deflection Equation for constraint on A:  = L = L/E = LF/AE (1) Eliminate A in (2) using (1): Chose materials with largest M1 = IFB 2012 Conflicting Constraints

  7. Strong tie of length L and minimum mass Function Tie-rod F F Area A L Objective (Goal) Minimise mass m: m = A L  (2) m = mass A = area L = length  = density = yield strength • Length L is specified • Must not fail under load F Constraints Eliminate A in (2) using (1): Free variables • Material choice • Section area A Chose materials with largest M2 = Performance metric m2 E8.1. Materials for a strong, light tie-rod Constraint # 2 Equation for constraint on A: F/A < y (1) IFB 2012 Conflicting Constraints

  8. = deflection y = yield strength • E = elastic modulus Max. deflection Must not yield Competing performance metrics E8.1: Conflicting Constraints: Strong /Stiff / Light Tie Rod Requires stiffer material Evaluate competing constraints and performance metrics: Stiffness constraint Strength constraint Rank by the more restrictive of the two, meaning…? IFB 2012 Conflicting Constraints

  9. Analytical solution in three steps: Rank by the more restrictive of the constraints 1. Calculate m1 and m2 for given L (1 m) and F (10 kN) 2. Find the largest of every pair of m’s 3. Find the smallest of the larger ones The most restrictive constraint requires a larger mass  activeconstraint. Cons to the analytical solution: Solution is not general, as it is a function of L, δ and F. (I.e., the above solution represents only the particular values of L, δ and F used in the calculations.)Lacks visual immediacy. IFB 2012 Conflicting Constraints

  10. mass /L= 0.1%:y constraint active (heavier) for short rods length Graphical version of the analytical solution (for Aluminium) /L= 0.1%:E constraint active (heavier) for long rods (rod stretches too much) IFB 2012 Conflicting Constraints

  11. Pros to the graphical-analytical solution (Slide #10): it makes explicit the dependence on L and L/.Cons: it is specific to the material considered (requires a dedicated graph per material) Cons to the analytical solution (slide #9) (more restrictive constraint) Solution is not general, as it is function of L, δ and F Lacks visual immediacy. Graphical solution using indices and bubble charts: • More general/powerful. Changes in L, δ and F are easily visualised • Allows for a visual while physically based selection. • Involves all available materials. • Incorporates geometrical constraints through coupling factors. IFB 2012 Conflicting Constraints

  12. M1 = M2= Graphicalsolution using Indices and Bubble charts This is what we know make m1= m2 Solve for M1 Straight line, slope = 1 y-intcpt = L/ IFB 2012 Conflicting Constraints

  13. m1 = m2 E 8.1 Tie Rod Graphical solution (/L = 1% => L/ = 100) Use level 3, exclude ceramics Simultaneously Maximise M1 and M2 m1 < m2 Coupling line for L/ = 100 m2 < m1 IFB 2012 Conflicting Constraints

  14. Coupling line for L/ = 1000 E8.1 Tie Rod Graphical solution( /L = 0.1% => L/=1000) Use level 3, exclude ceramics Active Constraint? 3-D view Coupling line for L/ = 100 IFB 2012 Conflicting Constraints

  15. m1 = m2 lighter 3-D view of the interacting constraints m1 m2 m1 > m2 m2 > m1 Locus of coupling line depends on coupling factor • m1 = m2 on the coupling line. • The closer to the bottom corner, the lighter the component. • Away from the coupling line, one of the constraints is active (larger m) IFB 2012 Conflicting Constraints

  16. lighter m1 = m2 Graphical solution (deflection = 1% L/=100) m1 < m2 Coupling line for L/ = 100 m2 < m1 IFB 2012 Conflicting Constraints

  17. Compressed air tank Case Study # 2: Quite Similar to E8.2 (Tute 2), Air cylinder for a truck Design goal: lighter, safe air cylinders for trucks IFB 2012 Conflicting Constraints

  18. t Density  Yield strength y Fracture toughness K1c Pressure p 2R FunctionPressure vessel Objective Minimise mass ConstraintsDimensions L, R, pressure p, given Safety: must not fail by yielding Safety: must not fail by fast fracture Must not corrode in water or oil Working temperature -50 to +1000C Wall thickness, t; choice of material L Conflicting constraints lead to competing performance metrics Free variables Case study: Air cylinder for truck IFB 2012 Conflicting Constraints

  19. t Density  Yield strength y Fracture toughness K1c Pressure p 2R L Aspect ratio,  Vol of material in cylinder wall Objective: mass Failure stress Stress in cylinder wall Safety factor Eliminate t transpose Air cylinder for truck What is the free variable? May be either y or f! IFB 2012 Conflicting Constraints

  20. CES Stage 2: evaluate conflicting performance metrics: S = safety factor a = crack length y = yield strength K1c = Fracture toughness Must not yield: Must not fracture Competing performance metrics for minimum mass Air cylinder : graphical solution using CES charts CES Stage 1; apply simple (non conflicting) constraints: working temp up to 1000C, resist organic solvents etc. Equate m1 to m2, and find the coupling factor for given crack size a. IFB 2012 Conflicting Constraints

  21. Max service temp = 373 K (1000C) Corrosion resistance in organic solvents Corrosion resistance Air cylinder - Simple (non- conflicting) constraints • CES Stage 1: • Impose constraints on corrosion in organic solvents • Impose constraint on maximum working temperature Select above this line IFB 2012 Conflicting Constraints

  22. CES, Stage 2: Equate m1 to m2, and find the coupling factor for given crack size a. = = for a crack a = 5 mm, the coupling factor is 1/√(3.14*0.005) = 1/0.12 = 8 IFB 2012 Conflicting Constraints

  23. Results so far: • Epoxy/carbon fibre composites • Epoxy/glass fibre composites • Low alloy steels • Titanium alloys • Wrought aluminium alloy • Wrought austenitic stainless steels • Wrought precipitation hardened stainless steels Lighter this way CES Stage 2: Find most restrictive constraint using Material Indices chart Air cylinder - Conflicting constraints a = 5 mm intcpt= 8 @ 1/M1= 1 Repeat for a = 5 μm IFB 2012 Conflicting Constraints

  24. Summary • Most designs are over-constrained and many have multiple objectives • Method of maximum restrictiveness copes with conflicting multiple constraints • Analytical method useful but depends on the particular conditions set and lacks the visual power of the graphical method • Graphical method produces a more general solution • Next lecture will solve case studies for two conflicting objectives: e.g., weight and cost. IFB 2012 Conflicting Constraints

  25. Solutions to E8.2 Exercises for Tutorial on multiple constraints • 8.2 a Coupling factor = 253 for c= 5 µm; • 8.2 b Coupling factor = 8 for c = 5 mm. 1- Textbook Exercise 8.2, p. 616 2- Exercise on slide # 27 IFB 2012 Conflicting Constraints

  26. There is a typographical error in your textbook, Exercise E8.2,p. 616, 6 lines from the top It reads: It should read: IFB 2012 Conflicting Constraints

  27. Exercise 2 for Tutorial on multiple constraints IFB 2012 Conflicting Constraints

  28. Exercise 2 for Tutorial on multiple constraints • Hypothetical values to fix the coupling line IFB 2012 Conflicting Constraints

  29. End of IFB Lecture 2 IFB 2012 Conflicting Constraints

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