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2 - Ashby Method

2.3 - Materials Selection for best performance

Outline

Resources: M. F. Ashby, Materials Selection in Mechanical Design Butterworth Heinemann, 1999

Chapters 5 and 6 The Cambridge Material Selector (CES) software -- Granta Design, Cambridge

(www.grantadesign.com)

Deriving performance indices

Performance maximizing criteria

Selection with multiple constraints

Materials selection and function

Design involves choosing a material, process and part shape to perform some function.

Function dictates the choice of both materials and shape. In many cases materials choice is not directly dependent of

shape.

Performance Indices

Minimum cost

Minimumweight

Maximum energy storage

Minimum environ. impact

FUNCTION

OBJECTIVE

CONSTRAINTS

INDEX

Tie

Beam

Shaft

Column

Mechanical,Thermal,

Electrical...

Stiffnessspecified

Strengthspecified

Fatigue limit

Geometryspecified

=

y

M

Minimise this!

Each combination ofFunctionObjectiveConstraintFree variable

Has a characterising material index

Performance Indices

Performance Indices

Performance Indices

Deriving Performance Indices: Procedure

Identify the attribute to be maximized or minimized (weight, cost, energy, stiffness, strength, safety, environmental damage, etc.).

Develop an equation for this attribute in terms of the functional requirements, the geometry and the material properties (the objective function).

Identify the free (unspecified) variables. Identify the constraints; rank them in order of

importance.

Develop equations for the constraints (no yield; no fracture; no buckling, maximum heat capacity, cost below target, etc.).

Substitute for the free variables from the constraints into the objective function.

Group the variables into three groups: functional requirements, F, geometry, G, and material properties, M, thus, we can write:

p = f1(F) f2 (G) f3 (M) Read off the performance index, expressed as a quantity f3 (M),

to be minimized or maximized.

Deriving Performance Indices: Procedure

Deriving Performance Indices: Light, strong tie

Strong tie of length L and minimum mass

L

FF

Area A

m = massA = areaL = length = density

= yield strengthy


Minimize mass, m , of a solid cylindrical tie rod of length L, which carries a tensile force F with safety factor Sf. The mass is given by:

m = A L

where A is the area of the cross-section and is the density. This is called the Objective Function

The length L and force F are specified; the radius r is free.

The section must, however, be sufficient to carry the tensile load F, requiring that:

F / A = f / Sf

where f is the failure strength.


Eliminating A between these two equations gives:

m = ( Sf F ) ( L ) { / f } or Note the form of this result.

The first bracket contains the functional requirementthat the specified load is safely supported.

The second bracket contains the specified geometry(the length of the tie).

The last bracket contains the material properties.

=

yFLm


We want to minimize the performance, m , while meeting the functional and geometric requirements.This means we want the smallest value of { / f } or the largest value of

M = f /

Ashby defines this latter quantity as the performance index.

Tie-rod

Minimise mass m


L

FF

Area A

Length L is specified Must not fail under load F Adequate fracture toughness

Material choice Section area A

Function

Objective

Constraints

Free variables


= yield strengthy

STEP 1Identify function, constraints, objective and free variables


Tie-rod

Minimise mass m:m = A L (1)


L

FF

Area A



Function

Objective

Constraints

Free variables


= yield strengthy

STEP 2Define equation for objective -- the performance equation


Tie-rod



L

FF

Area A



Function

Objective

Constraints

Free variables


= yield strengthy

STEP 3If the performance equationcontains a free variable other than material, identify the constraint that limits it

Equation for constraint on A: F/A < y (2)


Tie-rod



L

FF

Area A


Function

Objective

Constraints

Free variables


= yield strengthy


STEP 4Use this constraint to eliminate the free variable in performance equation

Material choice Section area A;

eliminate in (1) using (2):

=

yFLm


Tie-rod



L

FF

Area A


Function

Objective

Constraints

Free variables


= yield strengthy

Equation for constraint on A: F/A < y (2) STEP 5

Read off the combination of material properties that maximise performance



=

yFLm


Tie-rod



L

FF

Area A


Function

Objective

Constraints

Free variables


= yield strengthy


PERFORMANCEINDEX



=

yFLm Chose materials with smallest

y


Deriving Performance Indices: Light, stiff tie

Tie-rod


=

ESLm 2

Stiff tie of length L and minimum mass

L

FF

Area A

Material choice Section area A; eliminate in (1) using (2):

Function

Objective

Constraints

Free variables

Chose materials with smallest

E

m = massA = areaL = length = densityS = stiffnessE = Youngs Modulus

Stiffness of the tie S:

LAES = (2)

Performance Indices for weight: Tie

Function Stiffness Strength

Tension (tie)

Bending (beam)

Bending (panel)

Cost, CmDensity, Modulus, EStrength, yEndurance limit, eThermal conductivity, T- expansion coefficient,

the Physicists view of materials, e.g.Material properties --

the Engineers view of materialsMaterial indices --

/E y/1/2

/E 2/3y/

Objective: minimise mass

Minimise these!

1/3/E 1/2y/

Deriving Performance Indices: Light, stiff beam

m = massA = areaL = length = densityb = edge lengthS = stiffnessI = second moment of areaE = Youngs Modulus

Beam (solid square section).

Stiffness of the beam S:

I is the second moment of area:

Material choice. Edge length b. Combining the equations gives:

3LIECS =

12bI

4=

= 2/1

2/15

ECLS12

m

== LbLAm 2


2/1E

b

b

L

F

Minimise mass, m, where:

Function

Objective

Constraint

Free variables

Deriving Performance Indices: Light, strong beam

m = massA = areaL = length = densityb = edge lengthI = second moment of areay = yield strength

Beam (solid square section).

Must not fail under load F


Material choice. Edge length b. Combining the equations gives:

=

> 3y b3FL

Ib/2M

12bI

4=

( ) ( )

= 2/3

y

2/35/3

3FLm

== LbLAm 2


2/3

y

b

b

L

F

Minimise mass, m, where:

Function

Objective

Constraint

Free variables

Performance Indices for weight: Beam


Tension (tie)

Bending (beam)

Bending (panel)




/E y/1/2

/E 2/3y/


Minimise these!

1/3/E 1/2y/

Deriving Performance Indices: Light, stiff panel

m = massw = widthL = length = densityt = thicknessS = stiffnessI = second moment of areaE = Youngs Modulus

Panel with given width w and length L

Stiffness of the panel S:


3LIECS =

12twI3

=

tw

= 3/1

23/12

EL

CwS12

m

L

F

== LtwLAm


3/1E

Material choice. Panel thickness t. Combining the equations gives:

Minimise mass, m, where

Function

Objective

Constraint

Free variables

Deriving Performance Indices: Light, strong panel

m = massw = widthL = length = densityt = thicknessI = second moment of areay = yield strength

Panel with given width w and length L

Must not fail under load F


12twI3

=

tw

( ) ( )

= 1/2

y

3/21/2

L3Fwm

L

F

== LtwLAm

Material choice. Panel thickness t. Combining the equations gives:

Minimise mass, m, where

Function

Objective

Constraint

Free variables

=

> 2y wt3FL

It/2M


1/2

y

Performance Indices for weight: Panel


Tension (tie)Bending (beam)Bending (panel)




/E y/1/2

/E 2/3y/


Minimise these!

1/3/E 1/2y/

Minimum cost

Minimumweight



FUNCTION

OBJECTIVE

CONSTRAINTS

INDEX

Tie

Beam

Shaft

Column

Mechanical,Thermal,

Electrical...

Stiffnessspecified

Strengthspecified

Fatigue limit

Geometryspecified

= 2/1E

M

Minimise this!

Optimised selection using charts

b

b

L

F


Index 1/2EM =

22 M/E =

( ) ( ) ( )MLog2Log2ELog =

Contours of constantM are lines of slope 2

on an E- chart

CE 2/1

=

0.1

10

1

100

Metals

Polymers

Elastomers

Woods

Composites

Foams0.01

1000

1000.1 1 10Density (Mg/m3)

Young

s m

odu

lus

E, (G

Pa)

Ceramics

2


Index 1/2EM =

22 M/E =


0.1

10

1

100

Metals

Polymers

Elastomers

Woods

Composites

Foams

1000

1000.1 1 10

Density (Mg/m3)

Young

s m

odu

lus

E, (G

Pa)

Ceramics

2

y = a x + b

= Log() = -1 0 1


on an E- chart

y = Log (E)

x = Log ()


Index 1/2EM =

22 M/E =


0.1

10

1

100

Metals

Polymers

Elastomers

Woods

Composites

Foams

1000

1000.1 1 10

Density (Mg/m3)

Young

s m

odu

lus

E, (G

Pa)

Ceramics

2

y = a x + b

=

b = 1-2Log(M) =1M = 10-1/2 = 0.31

y = Log (E)

x = Log ()

Log() = -1 0 1


Index 1/2EM =

22 M/E =


0.1

10

1

100

Metals

Polymers

Elastomers

Woods

Composites

Foams

1000

1000.1 1 10

Density (Mg/m3)

Young

s m

odu

lus

E, (G

Pa)

Ceramics

2y = a x + b

=

b = 2-2Log(M) =2M = 10-1 = 0.1

Minimising M

y = Log (E)

x = Log ()

Log() = -1 0 1

Minimum cost

Minimumweight



FUNCTION

OBJECTIVE

CONSTRAINTS

INDEX Stiffnessspecified

Strengthspecified

Fatigue limit

Geometryspecified

( )[ ]EfM ,=

Minimise this!

Each combination ofFunctionObjectiveConstraintFree variable

Has a characterising material index

Performance Indices for weight: Stiffness



Tension (tie)Bending (beam)Bending (panel)




/E y/1/2

/E 2/3y/


Minimise these!

1/3/E 1/2y/


Index 1/2EM =

22 M/E =



on an E- chart

CE

=

CE 2/1

=

CE 3/1

=

0.1

10

1

100

Metals

Polymers

Elastomers

Woods

Composites

Foams0.01

1000

1000.1 1 10Density (Mg/m3)

Young

s m

odu

lus

E, (G

Pa)

Ceramics

12 3

Performance Indices for weight

Deriving Performance Indices for Cost and Energy

To minimize Cost use the indices for minimum weight, replacing density by C, where C is the cost per kg.

To minimize Energy use the indices for minimum weight, replacing density by q, where q is the energy content per kg.

M = / f M = C / f

M = / f M = q / f

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