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A13 –Materials Selection in Design
Materials Selection in Design
References
1. Ashby, Michael F., Materials Selection in Design, ButterworthHeinemann, 2nd Edition, 1999.
2. Cambridge Engineering Selector v3.1, Granta Design Limited,
Cambridge, UK, 2000.
Introduction
How does an engineer choose, from a vast menu, the material best
suited to his design purpose? Is it based on experience? Is there a
systematic procedure that can be formulated to make a rational
decision? There is no definitive answer to these questions,
however the procedure can be somewhat approached in a
systematic manner. Ashby's book and the Cambridge Engineering
Selector (CES) software shed some light on the materials selection
decision making process.
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From Ashby: "Materials selection inherently must be based on at
least 5 inter-related criteria:
 Function of structural component
 Materials available and their properties
 Shape and size of structural component
 Process used to manufacture structural component
 Cost and Availability (both of material and process)
Function typically dictates the choice of both material and shape.
Process is influenced by the material selected. Process also
interacts with shape -- the process determines the shape, the size,
the precision and, of course, the cost. The interactions are twoway: specification of shape restricts the choice of material and
process; but equally the specification of process limits the
materials you can use and the shapes they can take. The more
sophisticated the design, the tighter the specifications and the
greater the interaction. The interaction between function, material,
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A13 –Materials Selection in Design
shape and process lies at the heart of the materials selection
process."
Engineering Materials and their Properties
One can generally divide materials into classes, for example:
Metals
Polymers
Ceramics
Composites
Elastomers
Glasses
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When one or more these materials types are combined we obtain a
composite material. In what follows, we will not consider the
metallurgy and chemistry of materials; rather we will focus on the
various properties of the material types that are of importance to
engineers.
Metals have relatively high moduli of elasticity and high strength.
Strength is usually accomplished by alloying and by mechanical
and heat treatment, but they remain ductile, allowing them to be
formed by deformation processes. Typically strength is measured
by the stress at yielding. Tensile and compressive strength is
typically quite close. Ductility of metals may be as low as 2%
(high strength steel) but may be quite high. Metals are subject to
fatigue and typically are the least resistant to corrosion. Some hard
metals may be difficult to machine.
Ceramics and glasses also have high moduli, but, unlike metals,
they are brittle. Their strength in tension means the brittle fracture
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strength; in compression it is the brittle crushing strength, which
about 15 times higher then the tensile strength. Ceramics have no
ductility and therefore have a low tolerance to stress
concentrations. Ductile metals tolerate stress concentration by
deforming inelastically (so that load is redistributed); but ceramics
are unable to do this. Brittle materials tend to have a high scatter
in strength properties. Ceramics are stiff, hard, retain their strength
at high temperatures, are abrasion resistant, and are corrosion
resistant.
Polymers and elastomers are completely different. They have
moduli that are low, roughly 50 times less than those of metals do,
but they can often be nearly as strong as metals. Consequently,
elastic deformations can be very large. They can creep, even at
room temperature, and their properties tend to very greatly with
temperature. Polymers are corrosion resistant. They are easy to
shape through moulding processes.
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Composites can be developed which combine the attractive
properties of the other classes of materials while avoiding some of
their drawbacks. They tend to be light, stiff and strong, and can be
tough. Most readily available composites have a polymer matrix
(usually epoxy or polyester) reinforced by fibers of glass, carbon
or Kevlar. They typically cannot be used above 250C because the
polymer matrix softens. Metal matrix composites can be utilized
at much higher temperatures. Composite components are
expensive and they are relatively difficult to form and join. Thus,
while having attractive properties, the designer will use them only
when the added performance justifies the added cost.
Some important definitions for material properties
Elastic modulus (units: psi, MPa) - the slope of the linear-elastic
part of the stress-strain curve.
 Young's modulus, E, describes tension or compression.
 The shear modulus, G, describes shear loading.
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 Poisson's ratio, , is dimensionless and is the negative ratio of the
lateral strain to the axial strain in axial loading.
 Accurate moduli are often measured dynamically by exciting the
natural vibrations of a beam or wire, or by measuring the speed
of sound waves in the material.
Strength,  f (units: psi, MPa)
 For metals, the strength  f is identified by the 0.2% offset yield
strength,  y .
 For polymers, the strength  f is identified as the stress  y at
which the stress-strain curve becomes significantly nonlinear;
typically a strain of 1%.
 Strength for ceramics and glasses depends strongly on the mode
of loading - in tension strength means the fracture strength  tf
while in compression it means the crushing strength  cf which is
10 to 15 times larger.
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For metals, yield under multiaxial loads are related to that in
simple tension by a yield function; for example the vonMises yield
function:
(1   2 )2  ( 2   3 )2  ( 3  1 ) 2  2 2f
Ultimate Tensile Strength,  u (units: psi, MPa)
The nominal stress at which a round bar of the material, loaded in
tension, separates. For brittle materials (ceramics, glasses and
brittle polymers) it is the same as the failure strength in tension.
For metals, ductile polymers and most composites, it larger than
the strength  f by a factor of 1.1 to 3 because of the work
hardening, or, for composites because of load transfer to the
reinforcing fibers.
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Resilience, R (units J / m3 )
The maximum energy stored elastically without any
damage to the material, and which is released again
on unloading, i.e., the area under the elastic portion
of the stress-strain curve.
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Hardness, H (units: psi, MPa)
A measure of its strength. It is measured by pressing a pointed
diamond or hardened steel ball into the surface of the material;
defined as the indenter force divided by the projected area of the
indent.
Toughness, Gc (units: kJ / m 2 ) and fracture toughness, K c (units:
psi in1/ 2 , MPa m1/ 2 )
A measure of the resistance of the material to the propagation of a
crack. The fracture toughness is measured by loading a sample in
tension that contains a deliberately introduced crack of length 2c
(which is perpendicular to load), and the determining the tensile
stress  c at which the crack propagates. Fracture toughness is
c
Kc2
defined by K c  Y
, and the toughness is Gc 
, where
E (1   )
c
Y is a geometric factor, near 1, which depends on the sample
geometry.
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Loss coefficient,  (dimensionless)
A measure of the degree to which a material dissipates energy in
cyclic loading. Essentially, the ratio of energy dissipated to the
elastic energy (for a given stress that the material is loaded to).
Related to the damping capacity of a material (how much damping
a material has). If the loss coefficient is zero, there is no damping.
Depends on the frequency of the loading.
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Materials Selection Charts
Mechanical, thermal and other properties for materials may be
displayed in a variety of ways. What is needed is a way to
compare materials in a useful way for properties that are important
for the design problem under consideration.
For example,
 If we want a structure to stiff but light, then we want to choose a
material that has a high stiffness (E) to density () ratio.
 If we want a structure to be strong but light, then we want to
choose a material that has a high strength ( f ) to density ()
ratio.
 If we want a structure that is tough (resistant to crack formation
or propagation) and light, then we want to choose a material that
has a high fracture toughness ( K IC ) to density () ratio.
Here are some charts from Ashby and the Cambridge Engineering
Selector v3.1 software (CES).
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Materials Selection - the basics
Lets take a look at the basics of material selection. First we need
to define the concept of material indices. The design of any
structural element is specified by three things: the functional
requirement (F), the geometry (G) and the properties of the
material of which it is made (M). The performance (P) is
described functionally by an equation of the form:
 Functional
  Geometric   Material

p  f 
,
,


  (13.1)
Requirements,
F
Parameters,
G
Properties,
M
 
 


 f ( F , G, M )
The quantity p describes some aspect of the performance of the
component: its mass, or volume, or cost, or life, etc. Optimum
design is the selection of the material and geometry that maximize
or minimize p according to its desirability or otherwise.
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In many cases, the three groups of parameters are separable so that
p can be written as:
(13.2)
p  f1 ( F ) f 2 (G ) f3 ( M )
When the groups are separable, the optimum choice of material
becomes independent of the other details of the design, i.e., it is the
same for all geometries, G, and for all the values of the functional
requirement, F. The optimum subset of materials can now be
identified without solving the complete design problem, i.e.,
without considering or even know all the details of F and G . The
function f3 ( M ) is called the material efficient coefficient, or
material index. Lets take a look at an example to see how this
works.
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Example 1: Material index for a light, strong axial bar (rod)
We want to design a bar of length L to carry a tensile force F
without failure; and to be of minimum mass. Thus, maximizing
performance means minimizing the mass while still carrying the
load F safely. Function, objective and constraints may be listed as:
Function:
Objective:
Constraints:
Axial rod
Minimize the mass
(a) Length ( L ) specified
(b) Support tensile load F without material failure
We need an equation describing the quantity to be maximized or
minimized. This is the mass m of the rod. This equation, called
the objective function, is given by:
m  AL 
(13.3)
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where A is the cross-sectional area of the rod and  is the density
of the material out of which it is made. The length L and force F
are specified and are therefore fixed; the cross-sectional area A is
free to choose.
We could obviously reduce the mass by reducing the crosssectional area A, but there is a constraint; the area must be
sufficient to carry the load and not fail, i.e.,
F
 f
A
(13.4)
where  f is the failure strength. Eliminating A from the last two
equations gives:
  
m  ( F )( L) 
(13.5)
  f 


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Notice that the first term contains the specified load F while the
second term contains the specified length L . The last term
contains the material properties. Hence, the lightest bar which will
carry F safely is that made of the material with smallest value of
 /  f . [Note: we should be including the safety factor SF here
so that (13.4) becomes F / A   f / SF . However, if the same
safety factor is used for each material in a problem, its value does
not enter into the material selection.]
It might be easier, or more natural, to ask what must be maximized
in order to maximize performance. We therefore invert the
material properties in (13.5) and define the material index M as
f
M

(13.6)
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The lightest bar that will safely carry the load F without failing
is that with the largest value of the material index M . This index
is sometimes called the specific strength.
How do we determine the candidate materials with the best
f
f
(largest)
ratio? We use the
chart in Fig. 4.4 from Ashby,


or generate the chart using the CES software.
Note: The material index for stiff, light bar is similarly obtained as
the largest value for the following material index M .
M
We now use the
E

E

chart in Fig. 4.3 from Ashby (or CES).
(13.7)
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Example 2: Material index for a light, stiff simply supported
beam
Consider a simply supported beam of length L , square crosssection (bxb), and subjected to a transverse force F at mid-span.
We want to design a beam which must meet a constraint on its
stiffness S , i.e., it must not deflect more than  under the load F .
F
Function:
Objective:
Constraints:
Beam
Minimize the mass
L/2
L/2
(a) Length ( L ) specified
(b) Support bending load F without deflecting too
much
What does the term stiffness mean? Recall that for a cantilever
FL3
beam with a load F at its end, the deflection is given by  
3EI
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which can be written as
3EI
S
F
. The term S 
3EI
is called

L
L
the stiffness and is similar to a stiffness coefficient in a finite
F
element analysis. Hence, for the cantilevered beam: S  .

3
3
For the problem at hand (simply supported beam with point load at
the center), beam theory gives the maximum deflection (at the
center of the beam) as:
FL3
F


48 EI S
where S 
48EI
3
(13.8)
="stiffness" of the simply supported beam (for a
L
point load at the center). The constraint equation than requires that
S
F


48EI
3
L
(13.9)
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The moment of inertia is given by:
(base)(height )3 b 4 A2
I


12
12 12
(13.10)
Note that the length ( L ) is specified and the stiffness S is specified
by equation (13.9). The area A is free to be determined.
The mass of the beam (objective function) is given by:
m  AL 
(13.11)
The mass can be reduced by reducing the area, but only so far that
the stiffness constraint [equation (13.9)] is still met.
Substituting I from equation (13.10) into (13.9) gives
F 4 EA2
S  3

L
(13.12)
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Substituting A from equation (13.12) into equation (13.11) gives:
1/ 2
 S 
3   
m
L  1/ 2 

 4L 
E 
 f1 ( F ) f 2 (G ) f3 ( M )
 
(13.13)
Note that we have separated the design problem into the three
parameters: function (F), geometry (G) and material (M). The
best materials for a light, stiff beam are those which maximize the
material index M:
E1/ 2
(13.14)
M

It will turn out that the above result is valid for beams with any
support condition and with any type of bending load location or
E
E1/ 2
distribution. We now use the
guideline in the chart in Fig.


4.3 from Ashby (or CES) to determine the best candidate materials.
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Example 3: Material index for a light, strong simply
supported beam
In stiffness-limited applications, it is elastic deflection that is the
active constraint, i.e., deflection limits performance. In strengthlimited applications, deflection is acceptable provided the
component does not fail, i.e., strength is the active constraint.
Consider the selection of a simply supported beam (square crosssection) for a strength-limited application. The dimensions are as
in the previous case. The design
F
requirements are summarized by:
Function:
Objective:
Constraints:
Beam
L/2
L/2
Minimize the mass
(a) Length ( L ) specified
(b) Support bending load F without failing by
yield or fracture
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The objective function is still on mass [equation (13.11)] but the
constraint is now that of strength, i.e., the beam must the support
the load without failing. The bending moment is a maximum at
the center and is equal to M  FL / 4 . The stress at the top surface
( y  ymax ) is given by   ( Mymax ) / I  ( FLymax ) /(4 I ) . Hence F
is given by F  (4 I ) /( Lymax ) . The failure load F f occurs when
   f , or
I f
(13.15)
F f  C2
ymax L
where C2 is a constant depending upon support conditions and
load application/distribution, and ymax is the distance between the
neutral (centroidal) axis of the beam and its outer most fiber. Note
that for the simply supported beam with point load at the center,
C2  4 (as derived above) and ymax  b / 2 (half the height). Using
equation (13.15) and equation (13.10) to eliminate A from the
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objective function in equation (13.11) gives the mass of the beam
that will support the load F f :
 6Ff 
m
 C L2 
 2 
2/3
  
L  2/3 
 f 


3
(13.16)
Note that  f is typically  y for ductile metals. The mass is now
minimized by selecting materials that maximize the material index
M:
2
 3/
f
M

(13.17)
As stated before, the design requirement is characterized by:
function, an objective and constraints. What is the difference
between constraints and an objective? A constraint is a feature of
the design that must be met at a specified level (for example,
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deflection or stiffness). An objective is a feature for which a
maximum or minimum is sought (mass in the last few cases).
The objective function is sometimes not easy to choose because
there may be many options. For example, the objective function
might be cost, it might be corrosion resistance, it might be elastic
energy storage (for a spring), it might be thermal efficiency for an
insulation system, and the list goes on.
We note from these three cases, that the satisfaction of the
objective function requires choosing materials where is a ratio is
maximized. Plotting these stiffness to mass (weight) or strength to
mass ratios for broad classes of materials allows one to very
quickly see which materials are "better." Note also that, like the
last example, it is often not a simple ratio like  f /  but
2
something more complex like  3/
f /  , or for the deflection-
limited case E1/ 2 /  .
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Example 4: Material index for a cheap, stiff column
We desire the cheapest cylindrical column of length L and
diameter 2r that will safely support a compressive load F without
buckling. The requirements are:
Function:
Column
Objective:
Minimize the cost
Constraints: (a) Length ( L ) specified
(b) Support compressive force F without buckling
The objective function is the cost C defined by
C  AL Cm
(13.18)
where Cm is the cost/kg of the material of the cylindrical column
(we consider the cost of the processed material only and ignore
fabrication and other costs). A long slender column will buckle
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elastically if F exceeds the Euler critical load Fcrit . The solution is
safe if
 2 EI
F  Fcrit  n 2
(13.19)
L
where n is a constant that depends on the end conditions (n=1 for
pinned-pinned condition, n=4 for clamped-clamped condition).
For the cylindrical cross-section , the moment of inertia is
I   r 4 / 4  A2 /(4 )
(13.20)
where A is the cross-sectional area. Note that the load F and the
length L are specified; the free variable is the cross-sectional area
 EA2
A. Substituting (13.20) into (13.19) gives F  Fcrit  n
.
2
4L
Substituting A from (13.18) into this last result gives
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1/ 2
 4 
C 

 n 
1/ 2
F
 2
L 
3  Cm 
L  1/ 2
E



(13.21)
As before, we obtain the functional, geometry and material
parameters. The cost of the column is minimized by choosing
materials with largest value of the material index given by:
E1/ 2
M
Cm 
(13.22)
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Table 6.6 Procedure for deriving material indices (from Ashby)
Step
Action
1
Define the design requirements:
(a) Function: what does the component do?
(b) Objective: what is to be maximized or minimized?
(c) Constraints: essential requirements which must be met:
stiffness, strength, corrosion resistance, forming
characteristics
2
Develop an equation for the objective in terms of the
functional requirements, the geometry and the material
properties (the objective function).
3
Identify the free (unspecified) variables.
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4
Develop equations for the constraints (no yield; no fracture;
no buckling, etc.).
5
Substitute for the free variables from the constraint
equations into the objective function.
6
Group the variables into three groups: functional
requirements (F), geometry (G), and material properties (M);
thus
Performance characteristic  f1 ( F ) f 2 (G ) f3 ( M )
or
7
Performance characteristic  f1 ( F ) f 2 (G ) f3 ( M )
Read off the material index, expressed as a quantity M,
which optimizes the performance characteristic.
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Table 5.7 Examples of material indices (from Ashby)
Function, Objective and Constraint
Bar, minimum weight, stiffness prescribed
Beam, minimum weight, stiffness prescribed
Index
E

E1/ 2

Beam, minimum weight, strength prescribed
 2y / 3

Beam, minimum cost, stiffness prescribed
E1/ 2
Cm 
Beam, minimum cost, strength prescribed
 2y / 3
Cm 
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Column, minimum cost, buckling load prescribed
E1/ 2
Cm 
Spring, minimum weight for given energy storage
 2y
E
Thermal insulation, minimum cost, heat flux prescribed
1
 Cm 
Electromagnet, maximum field, temperature rise prescribed  C p 
(  = density; E = Young's modulus;  y = elastic limit;
Cm = cost/kg;  = thermal conductivity;  electrical conductivity;
C p = specific heat)
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In the material selection charts shown previously (from CES), note
that the various material properties are plotted on log-log scales.
The reason for this is as follows. For a condition like
E1/ 2

C
where C is a constant; we can take the log of each side to obtain
1 log E  log 
2
or
 log C
log E  2log   2log C
When plotted as log E vs. log  (or E vs.  on log-log scale), this
equation represents a family of straight parallel lines with a slope
of 2 and an intercept on the log E -axis of 2log C ; and each line
corresponds to a value of the constant C. These lines are referred
to as selection guide lines in CES. Any material falling on a given
straight line will have equal values of E1/ 2 /  , i.e., be of equal
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"goodness" in satisfying a material index for stiffness to weight
ratio. Selecting a higher curve (greater C) in the family of curves
is equivalent to selecting a family of materials with higher stiffness
to weight ratio.
There are many go-no go limits that may limit the values of
specific properties. For example, in Example 4 (column buckling),
if the diameter is constrained to 2r *, this will require a material
with a modulus greater than [found by inverting equation (13.19)]
4 FL2
E*  3
n (r*)4
Property limits will plot as horizontal or vertical lines on material
selection charts. The restriction on r* leads to a lower bound for E
then given by the equation above. It might also be a design
requirement that the column diameter lie within certain limits (for
example, the column diameter must satisfy r1  r  r2 ). In this
case, we would have both upper and lower limits on the diameter
and thus upper and lower limits on the modulus E.