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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 MECH Lecture # 9 Conflicting Constraints

2 MECH4301 2008 Lecture # 9 Conflicting Constraints
Lecture 9: Chapters 9 & 10 Conflicting constraints: “most restrictive constraint wins” Case study 1: Stiff / strong / light column Case study 2: Safe (no yield - no fracture)/ light air tank for a truck MECH Lecture # 9 Conflicting Constraints

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

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

5 Q7.1. Materials for a stiff, light tie-rod Constraint # 1
Strong tie of length L and minimum mass L F Area A Tie-rod Function Length L is specified Must not stretch more than  Constraints m = mass A = area L = length  = density E= elastic modulus  = elastic deflection Equation for constraint on A:  = L = L/E = LF/AE (1) Minimise mass m: m = A L  (2) Objective Material choice Section area A Free variables Eliminate A in (2) using (1): Performance metric m1 Chose materials with largest M1 = MECH Lecture # 9 Conflicting Constraints

6 Q7.1. Materials for a strong, light tie-rod Constraint # 2
Strong tie of length L and minimum mass L F Area A Tie-rod Function m = mass A = area L = length  = density = yield strength Length L is specified Must not fail under load F Constraints Equation for constraint on A: F/A < y (1) Eliminate A in (2) using (1): Minimise mass m: m = A L  (2) Objective (Goal) Material choice Section area A Free variables Performance metric m2 Chose materials with largest M2 = MECH Lecture # 9 Conflicting Constraints

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

8 MECH4301 2008 Lecture # 9 Conflicting Constraints
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 and thus becomes the controlling or active constraint. Cons to the analytical solution: Solution is not general, as it depends on F and L Lacks visual immediacy. MECH Lecture # 9 Conflicting Constraints

9 Graphical version of the analytical solution (for Aluminium)
/L= 1%: E constraint active (heavier) for long rods (rod stretches too much) mass /L= 1%: Strength constraint always active /L= 0.1%: y constraint active (heavier) for short rods Less demanding E constraint => thinner rod mE Solution for /L= 1% MECH Lecture # 9 Conflicting Constraints length

10 MECH4301 2008 Lecture # 9 Conflicting Constraints
Cons to the analytical solution (slide #9) (more restrictive constraint) Solution is not general, as it depends on F and L Lacks visual immediacy. 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) => 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. MECH Lecture # 9 Conflicting Constraints

11 Graphical solution using Indices and Bubble charts
M2 = M1 = This is what we know make m1 = m2 Solve for M1 Straight line, slope = 1 y-intcpt = L/ MECH Lecture # 9 Conflicting Constraints

12 E 7.1 Tie Rod Graphical solution (/L = 1% => L/ = 100) Use level 3, exclude ceramics
m1 < m2 Simultaneously Maximise M1 and M2 m1 = m2 Coupling line for L/ = 100 m2 < m1 MECH Lecture # 9 Conflicting Constraints

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

14 3-D view of the interacting constraints
m1 m1 > m2 m2 m2 > m1 m1 = m2 Locus of coupling line depends on coupling factor 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) MECH Lecture # 9 Conflicting Constraints

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

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

17 Case study: Air cylinder for truck
Density  Yield strength y Fracture toughness K1c Pressure p 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 +1000C Wall thickness, t; choice of material Conflicting constraints lead to competing performance metrics MECH Lecture # 9 Conflicting Constraints

18 Vol of material in cylinder wall
Air cylinder for truck t L 2R Density  Yield strength y Fracture toughness K1c Pressure p What is the free variable? Aspect ratio,  Objective: mass Vol of material in cylinder wall Stress in cylinder wall Failure stress Safety factor Eliminate t transpose May be either y or f ! MECH Lecture # 9 Conflicting Constraints

19 Air cylinder : graphical solution using CES charts
CES Stage 1; apply simple (non conflicting) constraints: working temp up to 1000C, resist organic solvents etc. CES Stage 2: evaluate conflicting performance metrics: Must not yield: Must not fracture S = safety factor a = crack length y = yield strength K1c = Fracture toughness Competing performance metrics for minimum mass Equate m1 to m2, and find the coupling factor for given crack size a. MECH Lecture # 9 Conflicting Constraints

20 Air cylinder - Simple (non- conflicting) constraints
Corrosion resistance in organic solvents CES Stage 1: Impose constraints on corrosion in organic solvents Impose constraint on maximum working temperature Select above this line Max service temp = 373 K (1000C) Corrosion resistance MECH Lecture # 9 Conflicting Constraints

21 MECH4301 2008 Lecture # 9 Conflicting Constraints
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 MECH Lecture # 9 Conflicting Constraints

22 Air cylinder - Conflicting constraints
CES Stage 2: Find most restrictive constraint using Material Indices chart Air cylinder - Conflicting constraints a = 5 mm intcpt= 8 @ 1/M1= 1 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 Repeat for a = 5 μm MECH Lecture # 9 Conflicting Constraints

23 MECH4301 2008 Lecture # 9 Conflicting Constraints
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. MECH Lecture # 9 Conflicting Constraints

24 MECH4301 2008 Lecture # 9 Conflicting Constraints
Solutions to E7.2 and E7.3 7.2 a Coupling factor = 253 for c= 5 µm; 7.2 b Coupling factor = 8 for c = 5 mm. 7.3 X- axis (E) multiplied by 1000 to convert GPa to MPa. 7.3a Coupling factor = 0.14 7.3b Coupling factor = OR: (0.14-1= 7.04) and ( =704). MECH Lecture # 9 Conflicting Constraints

25 MECH4301 2008 Lecture # 9 Conflicting Constraints
There is a typographical error in textbook, Exercise E7.2, p. 581, 8 lines from the bottom It reads: It should read: End of Lecture 9 MECH Lecture # 9 Conflicting Constraints


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