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Advanced Methods in Materials Selection

Advanced Methods in Materials Selection. Conflicting Constraints Lecture 9 & Tutorial 4 Conflicting Objectives Lecture 10 & Tutorial 5. Lecture 9: Chapters 9 & 10. Tutorial 4: E7.2 and E7.3 Due Oct 1. Outline Lecture 9. Conflicting constraints:

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Advanced Methods in Materials Selection

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  1. Advanced Methods in Materials Selection • Conflicting Constraints • Lecture 9 & Tutorial 4 • Conflicting Objectives • Lecture 10 & Tutorial 5 MECH4301 2007 Lecture # 9 Conflicting Constraints

  2. Lecture 9: Chapters 9 & 10 Tutorial 4: E7.2 and E7.3 Due Oct 1 MECH4301 2007 Lecture # 9 Conflicting Constraints

  3. Outline Lecture 9 • Conflicting constraints: • “most restrictive constraint wins” • Case study 1: Stiff / safe / light column • Case study 2: Safe (no yield - no fracture)/ light air tank for a truck MECH4301 2007 Lecture # 9 Conflicting Constraints

  4. Tie rod Function Function Carry force F without yielding, given length Minimise mass One Objective: one performance metric One Objective: the performance metric Multiple Objectives: several performance metrics Multiple Objectives: several performance metrics One Constraint One Constraint Many Constraints Many Constraints Many Constraints Many Constraints One Constraint One Constraint Rank by performance metric Rank by performance metric Rank by most restrictive performance metric Rank by most restrictive performance metric Combination of methods Combination of methods Trade-off and value function method Penalty function method Or several non-conflicting constraints, such as melting point, corrosion resistance, etc. Multiple Constraints and Objectives • Design with multiple constraints • • Design with multiple objectives Simplest case: Design with one objective, meeting a single constraint MECH4301 2007 Lecture # 9 Conflicting Constraints

  5. Tie rod Minimise mass No yield & Given deflection No corrosion Tmax > 100C. Function One Objective: one performance metric Conflicting Objectives: conflicting performance metrics One Constraint ConflictingConstraints Conflicting Constraints One Constraint The most restrictive constraint determines the performance metric (mass) Rank by performance metric Rank by most restrictive performance metric Combination of methods Penalty function method 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 One notch up in complexity: Single objective / Conflicting Constraints Lecture 10 MECH4301 2007 Lecture # 9 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 Q7.1. 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 = MECH4301 2007 Lecture # 9 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 Q7.1. Materials for a strong, light tie-rod Constraint # 2 Equation for constraint on A: F/A < y (1) MECH4301 2007 Lecture # 9 Conflicting Constraints

  8. = deflection y = yield strength • E = elastic modulus Max. deflection Must not yield Competing performance metrics Q7.1: 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…? MECH4301 2007 Lecture # 9 Conflicting Constraints

  9. Analytical solution in three steps: Rank by the more restrictive of the constraints • Calculate m1 and m2 for given L and F 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 and thus becomes the controlling or active constraint. MECH4301 2007 Lecture # 9 Conflicting Constraints

  10. mass y constraint active (heavier) length Graphical version of the analytical solution (for Aluminium) E constraint active (heavier) (long rod stretches too much) /L= 1%:Strength constraint always active Less demanding E constraint => thinner rod Solution for/L= 1% MECH4301 2007 Lecture # 9 Conflicting Constraints

  11. Pros to the graphical analytical solution : it makes explicit the dependence on L and L/.Cons: it is specific to the material considered (requires a dedicated graph per material) Graphical solution using indices and bubble charts. More general/powerful. Allows for a visual while physically based selection. Involves all available materials. Incorporates geometrical constraints through coupling factors. MECH4301 2007 Lecture # 9 Conflicting Constraints

  12. MECH4301 2007 Lecture # 9 Conflicting Constraints

  13. 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/ MECH4301 2007 Lecture # 9 Conflicting Constraints

  14. MECH4301 2007 Lecture # 9 Conflicting Constraints

  15. Materials for High-Performance Con-Rods MECH4301 2007 Lecture # 9 Conflicting Constraints

  16. MECH4301 2007 Lecture # 9 Conflicting Constraints

  17. MECH4301 2007 Lecture # 9 Conflicting Constraints

  18. High F/L2 LowF/L2 MECH4301 2007 Lecture # 9 Conflicting Constraints

  19. MECH4301 2007 Lecture # 9 Conflicting Constraints

  20. MECH4301 2007 Lecture # 9 Conflicting Constraints

  21. m1 = m2 E 7.1 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 MECH4301 2007 Lecture # 9 Conflicting Constraints

  22. Coupling line for L/ = 1000 E7.1 Tie Rod Graphical solution( /L = 0.1% L/=1000) Use level 3, exclude ceramics Active Constraint? 3-D view Coupling line for L/ = 100 MECH4301 2007 Lecture # 9 Conflicting Constraints

  23. 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) MECH4301 2007 Lecture # 9 Conflicting Constraints

  24. lighter m1 = m2 Graphical solution (deflection = 1% L/=100) m1 < m2 Coupling line for L/ = 100 m2 < m1 MECH4301 2007 Lecture # 9 Conflicting Constraints

  25. Compressed air tank Case Study # 2: Quite Similar to E7.2, Air cylinder for a truck Design goal: lighter, safe air cylinders for trucks MECH4301 2007 Lecture # 9 Conflicting Constraints

  26. 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 MECH4301 2007 Lecture # 9 Conflicting Constraints

  27. 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 May be either y or f Eliminate t transpose Air cylinder for truck What is the free variable? MECH4301 2007 Lecture # 9 Conflicting Constraints

  28. 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. Rank by the more restrictive of the two MECH4301 2007 Lecture # 9 Conflicting Constraints

  29. 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 MECH4301 2007 Lecture # 9 Conflicting Constraints

  30. 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 MECH4301 2007 Lecture # 9 Conflicting Constraints

  31. Summary • Real 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 air cylinder problem again for two conflicting objectives: e.g., weight and cost. End of Lecture 9 MECH4301 2007 Lecture # 9 Conflicting Constraints

  32. There is a typographical error in textbook, Exercise E7.2,p. 581, 8 lines from the bottom It reads: It should read: MECH4301 2007 Lecture # 9 Conflicting Constraints

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