1. Materials Selection in Design
The Role of Materials Selection in Design
Exploring relationships - Materials Property Charts
The Materials Selection Process –Design Models
Selecting materials –Materials Indices
Case Studies of Materials Selection using CES
3. Materials selection is a central aspect of design
In many cases materials represent the enabling step
Number of available materials exceeds 100,000…
Concurrent engineering has re-emphasized the role of materials.
The Role of Materials Selection in Design
Why Materials Selection?
New products.
Remain competitive
Factors/Criteria?
Function
Mechanical
Properties Failure
Mode
Manufacturability
Cost
Environmental
Considerations
Materials selection is design-led
Properties of new materials can
suggest new products (optical fiber –
high purity glass).
Optical Fiber
4. The need for a new product can stimulate the development of a
new material
Increase in operational temperature of turbine components. After Schulz et
al, Aero. Sci. Techn.7:2003, p73-80. Y-PSZ (Ytria - Partially Stabilized
Zirconia)
6. http://www.matweb.com/
The goal of design:
“To create products that perform their function effectively, safely,
at acceptable cost”….. What do we need to know about materials
to do this? More than just test data.
7. Materials Data - Organization
The set of properties for a particular material is called the “material
attributes”, which includes both structured and non-structured
information on the material – materials selection involves seeking the
best match between the design requirements and the materials attributes.
8.
9. Interactions
Material
Process
Shape
Functionality
Materials Selection Methodology
•Translate the design requirements into materials specifications. It
should take into consideration the design objectives, constraints and
free variables.
•Screening out of materials that fail the design constraints.
•Ranking the materials by their ability to meet the objectives.
(Material Indices).
•Search for supporting information for the material candidates.
10. Function Objective Constraint
"What does
component do?"
"What is to be maximized
or minimized?"
"What specific requirements
must be met?"
Any engineering
component has
one or more
functions (to
support a load, to
contain a pressure,
to transmit heat,
etc.).
The designer has an
objective (to make it
as cheap as possible,
or as light as possible,
or as safe as possible
or some combination
of these).
The objective must be
achieved subject to
constraints (e.g. the
dimensions are fixed; the
component must carry the
given load without failure, it
should function in a certain
temperature range, etc.
Free variables: What is the designer free to change?
1. Defining the Design requirements
11. 2. List the constraints (e.g. no buckling, high stiffness) of the
problem and develop an equation for them, if possible.
3. Develop an equation of the design objective in terms of functional
requirements, geometry and materials properties (objective
function).
4. Define the unconstrained (free) variables.
5. Substitute the free variable from the constraint equation into the
objective function.
6. Group the variables into three groups, functional requirements (F),
geometry (G) and materials functions (M), to develop the
performance metric (P):
7. Read off the materials index, M, in order to maximize the
performance metric (P).
12. Materials Selection Charts
•The performance metric of a design is limited by the materials.
•Performance metric is a function of multiple properties f(multiple
properties)
•Charts Property 1 versus Property 2 (P1 vs P2)
•It can be plotted for classes and subclasses of materials (Classes:
metals, ceramics, polymers, composites) (Sub-Classes: engineering
ceramics, porous ceramics etc.)
•Combinations of properties are important in evaluating usefulness
of materials.
•Strength to Weight Ratio: σf/ρ
•Stiffness to Weight Ratio: E/ρ
•The properties have ranges
•E(Cu) = few % (purity, texture, etc.)
•Strength of Al2O3 can vary by a factor of 100 due to (porosity,
grain size, heat treatment, etc.)
13. Modulus vs density
• Density depends on
– Atomic weight
– Atom size
– Packing
– Porosity
• Elastic modulus
depends on:
– Bond stiffness
– # bonds per unit area.
Which one to choose?
Depends on the Performance Metrics
Speed of Sound in a solid, ν
( )
ρ
E
v =
14. Material Indices
• Material Indices (MI) are groups of material properties (including
cost) which are useful metrics for comparison of materials
• Better materials have higher MI’s
• The form of the MI depends on the functional requirements (F) and
geometry (G).
15. Materials indices are specific functions derived from design
equations that involve only materials properties that can be used in
conjunction with materials selection charts
•e.g. strong, light tie rod in tension–minimize ρ/σy
•e.g. stiff, light beam in bending –minimize ρ/E1/2
•e.g. stiff, light panel in bending -minimize ρ/E1/3
Materials Indices
Derivation of MI’s
The derivation of the MI will be illustrated by examples:
16. Example 1: Strong and light tie-rod
Function
Objective
Constraints
Free Variables
Tie-rod
Minimize mass ρ
AL
m =
The length (L) is specified
Must not fail under load
Must have adequate fracture toughness
y
A
F
σ
≤
Materials choice
Section Area (A) – eliminate using above equations
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
=
y
FL
AL
m
σ
ρ
ρ Minimize mass , hence, choose
materials with smallest
y
σ
ρ
17. Example 2: Stiff and light beam
Function
Objective
Constraints
Free Variables
Beam of solid square
section
Minimize mass ρ
L
b
m 2
=
The beam must be stiff,
i.e. small deflection (C is
a constant) 3
4
1
3
3
_
)
(
48
L
Eb
C
L
CEI
S
Stiffness
Stiffness
F
EI
FL
Max
Max
=
=
=
= δ
δ
Materials choice
Dimension b – eliminate using above equations
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
2
1
2
1
1
5
2
1
1
3
E
C
SL
L
E
C
SL
m
ρ
ρ Minimize mass , choose
materials with smallest 2
1
E
ρ
18. Example 3: Stiff, light panel
Function
Objective
Constraints
Free Variables
Panel with given
width (w) and
length (L)
Minimize mass
ρ
twL
m =
The panel must be stiff,
i.e. small deflection (C is
a constant)
Materials choice
Dimension t – eliminate using above equations
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
3
1
3
1
2
6
3
1
3
E
C
w
SL
wL
CEw
SL
m
ρ
ρ Minimize mass , choose
materials with smallest 3
1
E
ρ
F
t = thickness
3
3
_
)
(
L
CEwt
S
Stiffness
Stiffness
F Max
=
= δ
25. Standards
and codes
(ISO 14000)
Supplier
information
Sector-specific
approval
(FDA, MilSpec)
z Supporting information,
general: what else do you need to know?
FE modules
The nature of material data
z Supporting information,
specific: what is the experience with the
material?
Design
guide
lines
Established
applications
Failure
analyses
Case
studies
“Structured” and “Unstructured” data
Handbooks,
data sheets
Reports, papers,
the Web
z Non-numeric: properties measured by
yes - no (Boolean) or
poor-average-good type (Rankings)
Can extrude?
Good or bad
in sea water?
z Numeric:
properties measured by
numbers:
density, modulus, cost
…other properties
U it 1 F 1 8
27. Summary: Material Indices
• A method is necessary for translating design requirements into a
prescription for a material
• Modulus-Density charts
– Reveal a method of using lines of constant
to allow selection of materials for minimum weight and
deflection-limited design.
• Material Index
– Combination of material properties which characterize
performance in a given application.
• Performance of a material:
3
,
2
,
1
1
=
n
E n
ρ
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
M
stics
Characteri
Material
G
Parameters
Geometeric
F
Needs
Functional
f
p
,
,
,
,
,
)
(
)
(
)
( 3
2
1 M
f
G
f
F
f
p =
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45. Case 1: Materials for Table legs
Design a slender, light table legs that will support
the applied design load and will not fracture if
struck.
Function
Objective
Constraints
Free Variables
Column, supporting compressive loads.
Minimize mass and
maximize slenderness
Specified length,
Must not buckle
Must not fracture if struck
Diameter of the legs
Choice of materials
mass:
Maximum elastic buckling load:
Solving for r
The weight is minimized by selecting materials
with the greatest value of the materials index:
46. Inverting equation (2) gives and equation
for the thinnest legs which will not
buckle:
to yield the second materials index (maximize):
Set M1to be minimum of 5 and
M2to be greater than 100 (an
arbitrary choice –it can be
modified later if a wider
choice of materials to be
screened is desired). Candidate
materials include some
ceramics, CFRP
•engineering ceramics are not
tough –legs are subjected to
abuse and this makes them a
bad selection for this
application
Selection = CFRP
must consult designer wrt cost
-expensive
47. Case 2: Materials for Flywheels
Flywheels are rotating devices that store
rotational energy in applications such as
automotive transmissions. An efficient
flywheel stores maximum energy per unit
volume/mass at a specified angular
velocity.
Function
Objective
The kinetic energy the device can the device can store is limited store
is limited by the material by the material strength.
Flywheel for energy storage.
Maximize kinetic energy per unit mass.
Mass of the disc
Kinetic energy (J is the mass moment of
inertia)
2
2
1
Jw
KE =
For a solid round disc J
around its rotation axis
2
2
1
mR
J = 2
2
4
1
w
mR
KE =
48. Constraints
Free Variables
The quantity to be maximized is the energy per unit
mass
2
2
4
1
w
R
m
KE
=
The outer radius is fixed.
It must not burst.
It must have adequate toughness (crack tolerance)
Choice of materials
The maximum radial stress (principal
stress) is given by the equation: 2
8
3 2
2
2
2
,
R
R
Max
r
ρω
ρω
υ
σ ≅
+
=
The stress must not exceed the yield
stress: ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
ρ
σ y
m
KE
2
1
Hence, the material index to maximize is:
ρ
σ y
M =
49. The choices are some composites (CFRP), some engineering ceramics
and high strength Ti and Al alloys
•engineering ceramics eliminated due to lack of toughness
•further selection must be made on the basis of cost and energy storage
capacity for specific materials
–e.g. CFRP can store 400kJ/kg
50. Case 3: Materials for Passive Solar Heating
A simple way of storing solar energy for
residential heating is by heating the walls
during the day and transferring heat to the
interior via forced convection at night.
Need to diffuse heat from the outer to inner
surface in 12h. For architectural reasons,
the wall thickness (W) cannot exceed 0.5m
Function
Objective
Heat storage medium
Maximize thermal energy storage per unit material cost.
Constraints
Free Variables
Heat diffusion time through wall time (t) ~12h
Wall thickness w<0.5m
Working temperature TMax ~ 100oC
Wall thickness w.
Choice of materials
What material will maximize the thermal energy captured by the wall
while retaining the required heat diffusion time of up to 12h?
51. For a wall of thickness w, the heat (Q) per unit
area of wall heated through ΔT is given by:
T
c
w
Q pΔ
= ρ
For the heat diffusion distance in time t:
where Δ is the thermal diffusivity, λ is the
thermal conductivity and ρ is the density
Dt
w 2
=
p
c
D
ρ
λ
=
⎟
⎠
⎞
⎜
⎝
⎛
Δ
=
Δ
=
D
T
t
T
D
c
t
Q p
λ
ρ 2
2 2
1
The heat capacity of the wall is maximized by choosing a
material with a high value of:
D
λ
The restriction on the wall thickness (w) and diffusion
time (t) yield the constraint:
s
m
t
w
D
2
6
2
10
3
2
−
×
≤
≤
52.
53. Cost must be a significant consideration in this selection because the
application is for housing, where cost is always a significant factor.
Taking cost into consideration, the most likely choice is concrete, with
stone and brick as alternatives.
54. M.F. Ashby, “Materials Selection in Mechanical Design, 2ndEd.”
•multiple sources listed in appendices of the book
•materials selection charts in lecture Appendix
Other references :
•ASM Metals Handbook
•Perry’s Chemical Engineering Handbook
•CRC Handbook of Mathematics and Physics
•ASM Handbook of Ceramics and Composites
Literature Resources