IFB 2012
Materials Selection in Mechanical Design
Efficient?
Lecture 1
Materials and Shape:
Materials for efficient
structures
“Efficient” = use
least amount of
material for
given stiffness
or strength.
Extruded shapes
Textbook Chapters 9 and 10
IFB 2012 Lecture 1 Shape Factors
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Shape and Mechanical Efficiency
• Section shape becomes
important when materials
are loaded in bending, in
torsion, or are used as
slender columns.
Is shape important
for tie rods?
• Examples of “Shape”:
• Shapes to which a
material can be formed
are limited by the
material itself.
Shapes from: http://www.efunda.com/math/areas/RolledSteelBeamsS.cfm
IFB 2012 Lecture 1 Shape Factors
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Certain materials can
only be made with
certain shapes: what is
the best material/shape
combination (for each
loading mode) ?
Extruded shapes
IFB 2012 Lecture 1 Shape Factors
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Shape efficiency: bending stiffness
pp.bh
248-249
3
I 
2
Define a standard reference section: a solid square, area A
=
b
12
S 
Io 
b
4

C 1 EI
3
Area A is
constant
2
12


L
Define shape factor
for
elastic bending,
measuring efficiency, as
Shaped
sections
Neutral
reference
section
F
Ao
12
Area Ao = b2
modulus E
unchanged
b
b
So 
C E I0
L
S 
3
e 
S
So

E I
E Io

I
Io
 12
I
2
Ao
 12
L
I
A
C E I
2
IFB 2012 Lecture 1 Shape Factors
3
Ao = A
4/24
bending stiffness
e 
S
So

EI
E Io

I
Io
 12
I
A
2
A shaped beam of shape factor for
elastic bending, e = 10, is 10 times
stiffer than a solid square section
beam of similar cross section area.
IFB 2012 Lecture 1 Shape Factors
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Properties of the shape factor
e = 2
Rectangular
Sections
The shape factor
is dimensionless
-- a pure number.

It characterises
shape, regardless
of size.

I-sections
 e  10
Circular tubes
 e  15
Increasing size at constant shape = constant SF
These sections are φe times stiffer in bending than a solid
square section of the same cross-sectional area

IFB 2012 Lecture 1 Shape Factors
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Shape efficiency: bending strength
p. 254
Define a standard reference section: a solid square, area A = b2
Neutral
reference
section
yield strength
unchanged
Area A is
constant
factor for the
onset of
plasticity
b
M
f 0
Io
ym

b
3
6
3/2

M
b
A0
Z 
6
 I  *
  0   Z 0
 ym 
f 
*
Define shape
Area A = b2
Zo 

M
M
I
f
fo

*
Z o
*

Z
Zo
IFB 2012 Lecture 1 Shape Factors
*
 Z
(failure), yI
measuring
efficiency, as
m
 I  *
  Z 
M f  
 ym 
Z
 I 

 
 ym 
section modulus , Z 
y max
*
f
6
*
Z
A
3/2
A = Ao
7/24
*
bending strength
f 
M
M
f
fo

Z
Z o
*
*

Z
Zo
6
Z
A
3/2
A shaped beam of shape factor for
bending strength, f = 10, is 10 times
stronger than a solid square section
beam of similar cross section area.
IFB 2012 Lecture 1 Shape Factors
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Tabulation of shape factors (elastic bending) p. 252
Section shape
h
Area A
m
bh
Second
moment I, m 4
Elastic shape
factor
b h3
12
h
b
 3
a b
4
3a
b
b
2a
ab
e 
I
Io
I
 bh   1  h
 12 2  12 
  2 2
A
 12   b h  b
3
Second moment of
section, I
A2 = Ao2
2b
t
2ri
2ro
 (ro2  ri2 )
 2r t
 4 4
(ro  ri )
4
  r3 t
t
h
2 t ( h  b)
(h, b  t )
1 3
b
h t (1  3 )
6
h
3 r
 
 t
(r  t )
e 
I
Io
 12
I
A
r t
3
2
 12
( 2  rt )
2

3r
 t

r
t
1 h (1  3b / h )
2 t (1  b / h )2
(h , b  t
b
t
hi
ho
b (h o  h i )
 2bt
(h , b  t )
b
b 3
(h o  h 3i )
12
1
 b t h o2
2
t
h
2t
2 t ( h  b)
(h, b  t )
1 3
b
h t (1  3 )
6
h
3 h o2
2 bt
(h , b  t )
1 h (1  3b / h )
2 t (1  b / h )2
(h , b  t )
b
IFB 2012 Lecture 1 Shape Factors
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Comparison of shapes done so far for given
material (E, y) and constant cross section area, A.
Interesting, but not very useful.
This is a case of Material Substitution at constant structural
stiffness or strength, allowing for differences in shape

How to compare different materials and
Example: compare Steel scaffoldings
different shapes at:
with Bamboo scaffoldings

Constant structural stiffness, S ?

Constant failure moment, Mf ?
IFB 2012 Lecture 1 Shape Factors
10/24
Indices that include shape (1): minimise mass at constant stiffness allowing
for changes in shape p. 265-270
F
F
L
Function
Beam (shaped section).
Objective
Minimise mass, m, where:
Area A
m  AL
Constraint
Bending stiffness of the beam S:
S 
CEI
L
3
Trick to bring the Shape Factor in ?
e 
I
Io
 12
 A 
 I e

 12 
I
A
2
2

A
m
L
Eliminating A from the eq. for the mass gives:
 12 S L 

m  
C


5
1/ 2



  E 1 / 2
 e




L
m = mass
A = area
L = length
 = density
b = edge length
S = stiffness
I = second moment of area
E = Youngs Modulus
Shape factor part of
the material index
  e E 

 

1/ 2
Chose materials with largest
IFB 2012 Lecture 1 Shape Factors
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



Indices that include shape (2): minimise mass at constant strength p . 311
F
Function
Beam (shaped section).
Objective
Minimise mass, m, where:
F
L
Area A
m  AL
Constraint
Bending strength of the beam Mf:
M
f

I
  Z
*
*
ym
Trick to bring the Shape Factor in ?
f 
Z
Z
 6
Zo
A
3/2
 Z 
fA
6
3/2

A
m
L
m = mass
A = area
L = length
 = density
Mf = bending strength
I = moment of section
E = Young’s Modulus
Z = section modulus
L
Eliminating A from the equation for m gives:
m  6 M
f

2/3
L
2/3


f

   * 
 f




* 2/3
Chose materials with largest
Shape factor part of
the material index
IFB 2012 Lecture 1 Shape Factors
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



From Introduction: Demystifying Material Indices (elastic
bending)
Given shape,
Same shape,
Material 2, same S
Material 1, given S
 12 S L 

m 1  
C


5
1/ 2
 1
 1 / 2
 E1



 2
  1 / 2
m1
 E2
m2
 12 S L 

m 2  
C


5
  E 11 / 2
 
  1
1/ 2
 2
 1 / 2
 E2



 M1
 
 M2
For given shape, the reduction in mass at
constant bending stiffness is determined by
the reciprocal of the ratio of material indices.
Same conclusion applies to bending strength.
IFB 2012 Lecture 1 Shape Factors
13/24
Demystifying Shape Factors (elastic bending)
Shaped to φe, same
material, same S
F
Square beam,
mo , given S
L
 12 S L 

mo  


C


5
1/ 2
ms
mo
F
L
 1
 1/ 2
E
 1





1
 
1/ 2
(

E
)

1
 12 S L 

m s  
C


5
  E11 / 2

 
 1

1

  1/ 2

1/ 2

1

  E 1 / 2
 e 1




EXAM QUESTION Is
the cross section area
constant when going
from mo to ms?
Shaping (material fixed) at constant bending
stiffness reduces the mass of the
component in proportion to e-1/2 .
Optimum approach: simultaneously
maximise both M and .
IFB 2012 Lecture 1 Shape Factors
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Demystifying Shape Factors
F
Square beam,
mo , Mf
m o  6 M
(failure of beams)
Shaped to φf,
same material,
same Mf
L
f 
ms
mo
2/3
L

m s  6 M
 
* 2/3
 
 




* 2/3








*


 f


2/3
f


1


  2/3
f

2/3
F
L
L


f


* 2/3
EXAM QUESTION:
Is the cross section
area constant when
going from mo to ms?
Shaping (material fixed) at constant
bending strength reduces the mass of the
component in proportion to f-2/3.
Optimum approach: simultaneously
maximise both M and .
IFB 2012 Lecture 1 Shape Factors
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Practical examples of material-shape combinations
F
• Materials for stiff beams of minimum weight
• Fixed shape (e fixed): choose materials with greatest E
• Shape e a variable: choose materials with greatest
Material
, Mg/m3
E, GPa
e,max
E
1/ 2
/



/
E  /ρ
1/2
e, max
E  /ρ
1/2
e, max
1020 Steel
7.85
205
65
1.8
14.7
6061 Al
2.70
70
44
3.1
20.4
GFRP
1.75
28
39
2.9
18.9
0.9
13
8
4
11.4
Wood (oak)
1/ 2 L
Same shape for all (up to e = 8): wood is best
Maximum shape factor (e = e,max): Al-alloy is best
Steel recovers some performance through high e,max
See textbook pp. 266 and 268 for more examples.
IFB 2012 Lecture 1 Shape Factors
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Tute #3:
p.269-270
Note that  e E 
 e E 1 / 2
1/ 2


e
 /e
E /  e 
1/ 2

s
new material with
Silicon Carbide
Alumina
Boron Carbide
Silicon
Tungsten Carbides
Steels
Nickel alloys
(GPa)
modulus
Young’s
Young's Modulus
(typical) (GPa)
Al alloys
100
Al: e = 44
 C
Zinc alloys
Titanium
Lead alloys
Concrete
Plywood
We call this
“dragging the
material’s label”
PET
PVC
PUR
PE
PTFE
PP
1
Rigid Polymer Foams
0.1
EVA
Silicone
1/ 2

Copper alloys
Mg alloys
CFRP
GFRP
Wood
10
0.01
E
Bamboo
   /e
s
 /e
Al: e = 1
1000
E  E /e
1e-003
Cork
Material substitution
at constant structural stiffness
allowing for differences in cross sectional
Flexible Polymer Foams
Polyisoprene
Polyurethane
shape/size
to increase
the structural efficiency
Butyl Rubber
Neoprene
1e-004
0.01
0.1
1
10
Density (typical) (Mg/m^3)
3
Density
(Mg/m
)
IFB 2012
Lecture 1 Shape
Factors
17/24
Dragging the material labels in CES  shaping at constant stiffness
Unshaped
Steel SF =1
1000
Drag the
labels
along lines
of slope 1
Selection line
of slope 2
Unshaped
Aluminium
0
steel SF=1
Unshaped
Bamboo SF= 1
100
Bamboo SF=1
Al SF =1
Young's Modulus (GPa)
Shaped
steel
Shaped
Bamboo
SF=65
SF=5.6
Shaped
aluminium
SF = 44
10
Shaping makes Steel
beams competitive
with Al beams and
Bamboo cane
Bamboo SF =5.6
Steel SF = 65
1
Al Sf=44
1
0.1
10
Density (Mg/m^3)
IFB 2012 Lecture 1 Shape Factors
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Dragging the material’s label in CES  shaping at constant strength
Note that

f


*


2/3
f

*

2/3
f
2

 / f
2
 /  
 /  
*


2/3
2
f
new material with
2
f
*s
  / f
*
2
   / f
s
2
steel  f  1
2
steel  f  100
2

*2 /3
Material substitution at constant structural strength
allowing for differences in cross sectional shape/size
to increase the structural efficiency

IFB 2012 Lecture 1 Shape Factors
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Dragging the material labels in CES  shaping at constant strength
Selection line of
slope 1.5
selection line slope 1.5
steel SF =1
1000
Shaped
Bamboo SF=2
(SF)2=4
Al Sf =1
Tensile Strength (MPa)
Bamboo SF =1
Shaped Steel
SF=7; (SF)2=49
100
bamboo SF = 2 SF^2=4
Shaped Aluminium
SF=10; (SF)2=100
Shaping makes
Steel beams
competitive with Al
beams and
Bamboo cane
steels SF=7 SF^2=49
10
Al 2024, SF=10 SF^2=100
0.01
0.1
1
10
Density (Mg/m^3)
IFB 2012 Lecture 1 Shape Factors
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
Steel, Al and
Bamboo scaffoldings

IFB 2012 Lecture 1 Shape Factors
Shaping allows
you to choose.
Use what is more
mass-efficient,
convenient,
cheap, and, of
course, available.
21/24
Exam questions:
Shaping at constant cross section A increases the
bending stiffness or strength by  at constant mass.
This stems from the definition of shape factor
e = S/So= I/Io
f = M/Mo = Z/Zo
Dragging the material label in the CES charts is
equivalent to shaping at constant bending stiffness
or strength, so the mass is reduced by 1/e1/2
(stiffness) or by 1/f2/3 (strength).
Dragging the material label along a line of slope 1 keeps the ratio E/ρ = E*/ρ* constant (* = shaped).
Shaping sacrifices the stiffness in tension (tie rod) in favour of the bending stiffness (beam), thus
increasing the mass efficiency of the section.
IFB 2012 Lecture 1 Shape Factors
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Really scary bamboo scaffoldings
http://www.flickr.com/photos/photocapy/41857542/
IFB 2012 Lecture 1 Shape Factors
23/24
-Tutorial 1, Materials and Shape.
Solve in this order: (4 Exercises)
E9.1 (p. 623)
CASE STUDY 10.2 (p.279)
CASE STUDY 10.4 (p. 284)
E9.8 (p. 627)
Notes and Hints for E9.1 and CS10.4:
E9.1 does not require the use of charts.
CS 10.4: follow the procedure of case study 10.2;
create a CES chart and analyse the effect of shaping
on the position of the bubbles (Do that by dragging
the materials’ labels.)
IFB 2012 Lecture 1 Shape Factors
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End of Lecture 1
IFB 2012 Lecture 1 Shape Factors
IFB 2012
25/24
Shape and mode of loading
Standard structural members
Loading:
tension/compression
Area A matters,
not shape
Both, Area A and
shape IXX, IYY matter
Loading: bending
Both, Area A and
shape J matter
Loading: torsion
Both, Area A and
shape Imin matter
Loading: axial
compression
IFB 2012 Lecture 1 Shape Factors
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Examples of Materials Indices including shape
p. 278
Buckling: Same as
elastic bending
IFB 2012 Lecture 1 Shape Factors
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Shape factors for twisting and buckling
p. 252/253
Elastic twisting
T 
ST
K

S To
K 0  0 . 14 A
Ko
2

 T  7 . 14
K
A
2
Failure under torsion
 fT 
Q
Q
 4 .8
Qo
A
3/2
Buckling
e 
S
So

E I
E Io
 12
I
A
2
IFB 2012 Lecture 1 Shape Factors
Same as elastic
bending
28/24