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# Advanced Methods in Materials Selection - PowerPoint PPT Presentation

Advanced Methods in Materials Selection. Conflicting Constraints Lecture 2 & Tutorial 2. IFB Lecture 2: Textbook Chapters 7 & 8. Conflicting Constraints. Fork for a pushbike: what is more important, strength or stiffness?. Stiffness dominated design versus Strength dominated design.

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### Advanced Methods in Materials Selection

• Conflicting Constraints

• Lecture 2 & Tutorial 2

IFB 2012 Conflicting Constraints

IFB 2012 Conflicting Constraints

Fork for a pushbike:

what is more important, strength or stiffness?

• Stiffness dominated design

• versus

• Strength dominated design

IFB 2012 Conflicting Constraints

• Conflicting constraints:

• “most restrictive constraint wins”

• Case study 1: Stiff / strong / light tie rod

• Case study 2: Safe (no yield - no fracture)/ light air tank for a truck

IFB 2012 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

IFB 2012 Conflicting Constraints

Strong tie of length L and minimum mass more than something, must not fail by yielding, by fatigue, by fast-fracture

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

E8.1. 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 =

IFB 2012 Conflicting Constraints

Strong tie of length L and minimum mass more than something, must not fail by yielding, by fatigue, by fast-fracture

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

E8.1. Materials for a strong, light tie-rod Constraint # 2

Equation for constraint on A:

F/A < y (1)

IFB 2012 Conflicting Constraints

• = deflection more than something, must not fail by yielding, by fatigue, by fast-fracture

y = yield strength

• E = elastic modulus

Max. deflection

Must not yield

Competing

performance metrics

E8.1: 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…?

IFB 2012 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  activeconstraint.

Cons to the analytical solution: Solution is not general, as it is a function of L, δ and F. (I.e., the above solution represents only the particular values of L, δ and F used in the calculations.)Lacks visual immediacy.

IFB 2012 Conflicting Constraints

/L= 0.1%:y constraint active (heavier) for short rods

length

Graphical version of the analytical solution (for Aluminium)

/L= 0.1%:E constraint active (heavier) for long rods (rod stretches too much)

IFB 2012 Conflicting Constraints

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)

Cons to the analytical solution (slide #9) (more restrictive constraint) Solution is not general, as it is function of L, δ and F Lacks visual immediacy.

Graphical solution using indices and bubble charts:

• More general/powerful. Changes in L, δ and F are easily visualised

• Allows for a visual while physically based selection.

• Involves all available materials.

• Incorporates geometrical constraints through coupling factors.

IFB 2012 Conflicting Constraints

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/

IFB 2012 Conflicting Constraints

m 1 = m2

E 8.1 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

IFB 2012 Conflicting Constraints

Coupling line for L/  = 1000

E8.1 Tie Rod Graphical solution( /L = 0.1% => L/=1000) Use level 3, exclude ceramics

Active Constraint? 3-D view

Coupling line for L/ = 100

IFB 2012 Conflicting Constraints

m 1 = 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)

IFB 2012 Conflicting Constraints

m1 = m2

Graphical solution (deflection = 1% L/=100)

m1 < m2

Coupling line for L/ = 100

m2 < m1

IFB 2012 Conflicting Constraints

Case Study # 2: Quite Similar to E8.2 (Tute 2), Air cylinder for a truck

Design goal: lighter, safe air cylinders for trucks

IFB 2012 Conflicting Constraints

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

IFB 2012 Conflicting Constraints

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

Eliminate t

transpose

Air cylinder for truck

What is the free variable?

May be either y or f!

IFB 2012 Conflicting Constraints

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.

Equate m1 to m2, and find the coupling factor for given crack size a.

IFB 2012 Conflicting Constraints

= 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

IFB 2012 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

IFB 2012 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

CES Stage 2: Find most restrictive constraint using Material Indices chart

Air cylinder - Conflicting constraints

a = 5 mm intcpt= 8

@ 1/M1= 1

Repeat for a = 5 μm

IFB 2012 Conflicting Constraints

• Most 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 case studies for two conflicting objectives: e.g., weight and cost.

IFB 2012 Conflicting Constraints

Solutions to E8.2

Exercises for Tutorial on multiple constraints

• 8.2 a Coupling factor = 253 for c= 5 µm;

• 8.2 b Coupling factor = 8 for c = 5 mm.

1- Textbook Exercise 8.2, p. 616

2- Exercise on slide # 27

IFB 2012 Conflicting Constraints

There is a typographical error in your textbook, Exercise E8.2,p. 616, 6 lines from the top

IFB 2012 Conflicting Constraints

IFB 2012 Conflicting Constraints

Exercise 2 for Tutorial on multiple constraints

• Hypothetical values to fix the coupling line

IFB 2012 Conflicting Constraints

End of IFB Lecture 2

IFB 2012 Conflicting Constraints