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 constraintsm1
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