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MECH4301 2007 Lecture # 9 Conflicting Constraints 1/23 Advanced Methods in Materials Selection Conflicting Constraints Lecture 9 & Tutorial 4 Conflicting Objectives Lecture 10 & Tutorial 5

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MECH4301 2007 Lecture # 9 Conflicting Constraints 2/23 Tutorial 4: E7.2 and E7.3 Due Oct 1 Lecture 9: Chapters 9 & 10

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MECH4301 2007 Lecture # 9 Conflicting Constraints 3/23 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

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MECH4301 2007 Lecture # 9 Conflicting Constraints 4/23 Multiple Constraints and Objectives Design with multiple constraints Design with multiple objectives One Objective: the performance metric Rank by performance metric One Constraint Many Constraints One Constraint Many Constraints Multiple Objectives: several performance metrics Trade-off and value function method Rank by most restrictive performance metric Function Combination of methods One Objective: one performance metric Rank by performance metric One Constraint Many Constraints One Constraint Many Constraints Multiple Objectives: several performance metrics Penalty function method Rank by most restrictive performance metric Function Combination of methods Minimise mass Carry force F without yielding, given length Tie rod Simplest case: Design with one objective, meeting a single constraint Or several non-conflicting constraints, such as melting point, corrosion resistance, etc.

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MECH4301 2007 Lecture # 9 Conflicting Constraints 5/23 One Objective: one performance metric Rank by performance metric One Constraint Conflicting Constraints One Constraint Conflicting Constraints Conflicting Objectives: conflicting performance metrics Penalty function method Rank by most restrictive performance metric Function Combination of methods One notch up in complexity: Single objective / Conflicting Constraints 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 Minimise mass No yield & Given deflection No corrosion T max > 100C. The most restrictive constraint determines the performance metric (mass) Tie rod Lecture 10

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MECH4301 2007 Lecture # 9 Conflicting Constraints 6/23 Q7.1. Materials for a stiff, light tie-rod Constraint # 1 Minimise mass m: m = A L (2) Objective Length L is specified Must not stretch more than Constraints Material choice Section area A Free variables Equation for constraint on A: = L = L /E = LF/AE (1) Strong tie of length L and minimum mass L F F Area A Tie-rod Function m = mass A = area L = length = density E= elastic modulus = elastic deflection Performance metric m 1 Eliminate A in (2) using (1): Chose materials with largest M1 =

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MECH4301 2007 Lecture # 9 Conflicting Constraints 7/23 Q7.1. Materials for a strong, light tie-rod Constraint # 2 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 2 Eliminate A in (2) using (1): Chose materials with largest M2 =

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MECH4301 2007 Lecture # 9 Conflicting Constraints 8/23 Q7.1: Conflicting Constraints: Strong /Stiff / Light Tie Rod Rank by the more restrictive of the two, meaning…? Evaluate competing constraints and performance metrics: Max. deflection Must not yield = deflection y = yield strength E = elastic modulus Competing performance metrics Requires stiffer material Stiffness constraint Strength constraint

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MECH4301 2007 Lecture # 9 Conflicting Constraints 9/23 Analytical solution in three steps: Rank by the more restrictive of the constraints 1.Calculate m 1 and m 2 for given L and F 3. Find the smallest of the larger ones The most restrictive constraint requires a larger mass and thus becomes the controlling or active constraint. 2. Find the largest of every pair of m’s

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MECH4301 2007 Lecture # 9 Conflicting Constraints 10/23 Graphical version of the analytical solution (for Aluminium) masslength y constraint active (heavier) E constraint active (heavier ) (long rod stretches too much) /L= 1%: Strength constraint always active Solution for /L= 1% Less demanding E constraint => thinner rod

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MECH4301 2007 Lecture # 9 Conflicting Constraints 11/23 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. 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)

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MECH4301 2007 Lecture # 9 Conflicting Constraints 12/23

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MECH4301 2007 Lecture # 9 Conflicting Constraints 13/23 Graphical solution using Indices and Bubble charts M2 = M1 = make m 1 = m 2 Solve for M1 Straight line, slope = 1 y-intcpt = L/ This is what we know

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MECH4301 2007 Lecture # 9 Conflicting Constraints 14/23

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MECH4301 2007 Lecture # 9 Conflicting Constraints 15/23 Materials for High-Performance Con-Rods

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MECH4301 2007 Lecture # 9 Conflicting Constraints 16/23

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MECH4301 2007 Lecture # 9 Conflicting Constraints 17/23

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MECH4301 2007 Lecture # 9 Conflicting Constraints 18/23 High F/L 2 LowF/L 2

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MECH4301 2007 Lecture # 9 Conflicting Constraints 19/23

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MECH4301 2007 Lecture # 9 Conflicting Constraints 20/23

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MECH4301 2007 Lecture # 9 Conflicting Constraints 21/23 E 7.1 tie rod Graphical solution ( /L = 1% L/ = 100) Use level 3, exclude ceramics Simultaneously Maximise M1 and M2 m 1 = m 2 m 1 < m 2 m 2 < m 1 Coupling line for L/ = 100

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MECH4301 2007 Lecture # 9 Conflicting Constraints 22/23 E7.1 Tie Rod Graphical solution ( /L = 0.1% L/ =1000) Use level 3, exclude ceramics Coupling line for L/ = 1000 Coupling line for L/ = 100 Active Constraint? 3-D view

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MECH4301 2007 Lecture # 9 Conflicting Constraints 23/23 3-D view of the interacting constraints m 1 = m 2 m 1 > m 2 m 2 > m 1 m2m2 m1m1 lighter 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) Locus of coupling line depends on coupling factor

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MECH4301 2007 Lecture # 9 Conflicting Constraints 24/23 Graphical solution (deflection = 1% L/ =100) Coupling line for L/ = 100 m 1 = m 2 m 2 < m 1 m 1 < m 2 lighter

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MECH4301 2007 Lecture # 9 Conflicting Constraints 25/23 Case Study # 2: Quite Similar to E7.2, Air cylinder for a truck Design goal: lighter, safe air cylinders for trucks Compressed air tank

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MECH4301 2007 Lecture # 9 Conflicting Constraints 26/23 Case study: Air cylinder for truck Free variables Function Pressure vessel Objective Minimise mass Constraints Dimensions 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 +100 0 C Wall thickness, t; choice of material t L 2R Density Yield strength y Fracture toughness K 1c Pressure p Conflicting constraints lead to competing performance metrics

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MECH4301 2007 Lecture # 9 Conflicting Constraints 27/23 Air cylinder for truck t L 2R Density Yield strength y Fracture toughness K 1c Pressure p Objective: mass Vol of material in cylinder wall Eliminate t Stress in cylinder wall Failure stress Safety factor Aspect ratio, May be either y or f transpose What is the free variable?

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MECH4301 2007 Lecture # 9 Conflicting Constraints 28/23 Air cylinder : graphical solution using CES charts CES Stage 1; apply simple (non conflicting) constraints: working temp up to 100 0 C, resist organic solvents etc. Rank by the more restrictive of the two CES Stage 2: evaluate conflicting performance metrics: Must not yield: Must not fracture S = safety factor a = crack length y = yield strength K 1c = Fracture toughness Competing performance metrics for minimum mass

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MECH4301 2007 Lecture # 9 Conflicting Constraints 29/23 Air cylinder - Simple (non- conflicting) constraints CES Stage 1: Impose constraints on corrosion in organic solvents Impose constraint on maximum working temperature Max service temp = 373 K (100 0 C) Corrosion resistance in organic solvents Corrosion resistance Select above this line

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MECH4301 2007 Lecture # 9 Conflicting Constraints 30/23 Air cylinder - Conflicting constraints CES Stage 2: Find most restrictive constraint using Material Indices chart 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

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MECH4301 2007 Lecture # 9 Conflicting Constraints 31/23 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 End of Lecture 9 Graphical method produces a more general solution Next lecture will solve air cylinder problem again for two conflicting objectives: e.g., weight and cost.

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MECH4301 2007 Lecture # 9 Conflicting Constraints 32/23 There is a typographical error in textbook, Exercise E7.2, p. 581, 8 lines from the bottom It reads: It should read:

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