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

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

MECH4301 2007 Lecture # 9 Conflicting Constraints 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

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. MECH4301 2007 Lecture # 9 Conflicting Constraints

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) MECH4301 2007 Lecture # 9 Conflicting Constraints

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 = MECH4301 2007 Lecture # 9 Conflicting Constraints

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 = MECH4301 2007 Lecture # 9 Conflicting Constraints

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…? MECH4301 2007 Lecture # 9 Conflicting Constraints

MECH4301 2007 Lecture # 9 Conflicting Constraints 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

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

MECH4301 2007 Lecture # 9 Conflicting Constraints 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

MECH4301 2007 Lecture # 9 Conflicting Constraints

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/ MECH4301 2007 Lecture # 9 Conflicting Constraints

MECH4301 2007 Lecture # 9 Conflicting Constraints

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

MECH4301 2007 Lecture # 9 Conflicting Constraints

MECH4301 2007 Lecture # 9 Conflicting Constraints

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

MECH4301 2007 Lecture # 9 Conflicting Constraints

MECH4301 2007 Lecture # 9 Conflicting Constraints

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 MECH4301 2007 Lecture # 9 Conflicting Constraints

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 MECH4301 2007 Lecture # 9 Conflicting Constraints

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) MECH4301 2007 Lecture # 9 Conflicting Constraints

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

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

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 MECH4301 2007 Lecture # 9 Conflicting Constraints

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 May be either y or f Eliminate t transpose MECH4301 2007 Lecture # 9 Conflicting Constraints

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

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 MECH4301 2007 Lecture # 9 Conflicting Constraints

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

MECH4301 2007 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. End of Lecture 9 MECH4301 2007 Lecture # 9 Conflicting Constraints

MECH4301 2007 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: MECH4301 2007 Lecture # 9 Conflicting Constraints