ALL 17 CHAPTERS COVERE
Exercises with worked solutions
,Contents
E1. Introduction to exercises
E2 Material evolution in products (Chapter
1) E3. Devising concepts (Chapter 2)
E4. Using material properties (Chapter 3)
E5. Using material selection charts (Chapter 4)
E6. Translation: constraints and objectives (Chapters 5
and 6) E7. Deriving and using material indices (Chapters
5 and 6)
E8. Multiple constraints and objectives (Chapters 7
and 8) E 9. Selecting material and shape (Chapters
9 and 10) E10. Hybrid materials (Chapters 11 and
12)
E11. Selecting processes (Chapters 13 and
14) E12. Materials and the environment
(Chapter 15)
,E1 Introduction to exercises
These exercises are designed to develop facility in selecting 3. A request for a selection based on one material index alone (such
materials, processes and shape, and in devising hybrid materials when
no monolithic material completely meets the design requirements. as M E / ) is correctly answered by listing the subset of
Each exercise is accompanied by a worked solution. They are materials that maximize this index. But a request for a selection of
organized into the twelve sections listed on the first page. materials for a component – a wing spar, for instance (which is a
light, stiff beam, for which the index is M E / ) – requires
The early exercises are easy. Those that follow lead the reader
more: some materials with high E / such as silicon carbide,
through the use of material properties and simple solutions to
mechanics problems, drawing on data and results contained in are unsuitable for obvious reasons. It is a poor answer that
Appendices A and B; the use of material property charts; techniques for ignores common sense and experience and fails to add further
the translation of design requirement to identify constraints and constraints to incorporate them. Students should be encouraged
objectives; the derivation of indices, screening and ranking, multi- to discuss the implications of their selection and to suggest further
objective optimization; coupled choice of material and shape; devising selection stages.
hybrids; and the choice of materials to meet environmental criteria.
The best way to use the charts that are a feature of the book is to
Three important points. make clean copies (or down-load them from
http://www.grantadesign.com ) on which you can draw, try out
1. Selection problems are open-ended and, generally, under-
alternative selection criteria, write comments and so forth. Although
specified; there is seldom a single, correct answer. The proper the book itself is copyrighted, the reader is authorized to make copies
answer is sensible translation of the design requirements into
of the charts and to reproduce these, with proper reference to their
material constraints and objectives, applied to give a short-list of
source, as he or she wishes.
potential candidates with commentary suggesting what supporting
information would be needed to narrow the choice further.
2. The positioning of selection-lines on charts is a matter of All the materials selection problems can be solved using the CES
judgement. The goal is to place the lines such that they leave an EduPack software, which is particularly effective when multiple criteria
adequately large "short list" of candidates (aim for 4 or so), drawn, and unusual indices are involved.
if possible, from more than one class of material.
www.grantadesign.com/education/resources -2-
,E2 Material evolution in products (Chapter 1)
E 2.1. Use Google to research the history and uses of one of the
following materials
Tin
Glass
Cement
Titanium
Carbon fiber
Present the result as a short report of about 100 - 200 words (roughly
half a page).
Specimen answer: tin. Tin (symbol Sn), a silver-white metal, has a long
history. It was traded in the civilisations of the Mediterranean as early
as 1500 BC (the Old Testament of the Christian bible contains many
references to it). Its importance at that time lay in its ability to harden
copper to give bronze (copper containing about 10% tin), the key
material for weapons, tools and statuary of the Bronze age (1500 BC –
500 BC). Today tin is still used to make bronze, for solders and as a
corrosion resistant coating on steel sheet (“tin plate” ) for food and drink
containers – a “tinnie”, to an Australian, is a can of beer. Plate glass is
made by floating molten glass on a bed of liquid tin (the Pilkington
process). Thin deposits of tin compounds on glass give transparent,
electrically conducting coatings used for frost-free windshields and for
panel lighting.
E2.2 Research, at the level of the mini case studies in this chapter, the
evolution of material use in
Writing implements (charcoal, “lead” (graphite), quill pens, steel
nib pens, gold plus osmium pens, ball points..)
Watering cans (wood – galvanized iron – polypropylene)
Bicycles (wood – bamboo – steel, aluminum, magnesium,
titanium – CFRP)
Small boat building (wood – aluminum – GFRP)
Book binding (Wood – leather – cardboard – vinyl)
www.grantadesign.com/education/resources -3-
,E3 Devising concepts (Chapter 2)
These two examples illustrate the way in which concepts are Concept Embodiment
generated. The left-hand part of each diagram describes a physical
principle by which the need might be met; the right-hand part
elaborates, suggesting how the principle might be used.
Electric fan pulling air stream through
paper or cloth filter in portable unit
C1 Entrain in air stream
and filter
Central pump and filter linked to
E3.1 Concepts and embodiments for dust removers. We met the need rooms by ducting
for a “device to remove household dust” in Chapter 1, with
examples of established solutions. Now it is time for more creative
thinking. Devise as many concepts to meet this need as you can. Centrifugal turbo-fan with surrounding
Nothing, at the concept stage, is too far-fetched; decisions about dust collector, in portable unit
practicality and cost come later, at the detailed stage. So think C2 Entrain in air
along the lines of Figure 2.2 of the main text and list concepts and stream,
Central turbo-fan with centrifugal
outline embodiments as block diagrams like this: dust collector linked to rooms by ducting
Centrifugal turbo-fan, injected water
Concept Embodiment spray to wash air stream, in portable unit
C3 Entrain in air
stream,
Central turbo-fan , water spray to wash
Electric fan pulling air stream through air stream, linked to rooms by ducting
paper or cloth filter in portable unit
C1 Entrain in air stream
and filter Axial fan drawing air stream between
Central pump and filter linked to
rooms by ducting charged plates, in portable unit
C4 Entrain in air
stream, trap
Central fan with electrostatic collector,
linked to rooms by ducting
C5 Trap dust on Reel-to-reel single sided adhesive tape
Answer. Design problems are open-ended; there are always adhesive strip running over flexible pressure-pad
alternative solutions. Here are some for dust removers.
www.grantadesign.com/education/resources -4-
,E3.2 Cooling power electronics. Microchips, particularly those for
power electronics, get hot. If they get too hot they cease to
Concept Embodiment
function. The need: a scheme for removing heat from power
microchips. Devise concepts to meet the need and sketch an
embodiment of one of them, laying out your ideas in the way
Massive, with sufficient heat capacity to absorb heat
suggested in exercise E3.1. over work cycle without significant increase in
C1 Conduction
Answer. Four working principles are listed below: thermal conduction, Compact, requiring back-up by coupling to convection,
convection by heat transfer to a fluid medium, evaporation evaporation or radiation
exploiting the latent heat of evaporation of a fluid, and radiation,
best achieved with a surface with high emissivity. Free convection not requiring fan or pump.
The best solutions may be found by combining two of these:
conduction coupled with convection (as in the sketch – an often- C2 Convection
used combination) or radiation coupled with evaporation (a Forced convection with fan or pump
possibility for short-life space structures).
Unconfined, such as a continuous spray of volatile fluid
C3 Evaporation
Confined, utilising heat-pipe technology
Radiation to ambient, using high emissivity coatings
C4 Radiation
Radiation to cooled surface, from high emissivity
surface to highly absorbent surface
Sketch.
www.grantadesign.com/education/resources -5-
,E4 Using material properties (Chapter 3) E4.3 A thick-walled tube has an inner radius ri = 10 mm and an outer
These exercises introduce the reader to 2 useful resources: the data radius ro = 15 mm. It is made from polycarbonate, PC. What is the
sheets of Appendix A and the Solutions to Standard Problems of maximum torque that the tube can carry without the onset of yield?
Appendix B. Retrieve the (mean) yield strength y of PC from Appendix A5, the
expression for the torque at onset of yield from Appendix B6 and that for
E4.1 A cantilever beam has a length L = 50 mm, a width b = 5 mm and the polar moment of a thick walled tube from Appendix B2 to find out.
a thickness t = 1 mm. It is made of an aluminum alloy. By how much
will the end deflect under and end-load of 5 N? Use data from
Appendix A4 for the (mean) value of Young’s modulus of aluminum Answer. The torque at the onset of yield for a thick walled tube is
alloys, the equation for the elastic deflection of a cantilever from
Appendix B3 and for the second moment of a beam from Appendix B2 K 4 4
y
to find out. Tf with K ( ro r
2
ro )2 i
Answer. The deflection of a cantilever under an end load F is, from The mean yield strength of y of PC from Appendix A5 is 65 MPa.
Appendix B, Inserting the data from the question gives a torque at the onset of yield of
F L3 with
bt3 T f = 138 N.m.
I
3E I 12
The mean Young’s modulus E for aluminum, from Appendix A4, is 75
GPa. Inserting the data from the question results in an end deflection E4.4 A round bar, 20 mm in diameter, has a shallow circumferential
= 6.7 mm. notch with a depth c = 1 mm with a root radius r = 10 microns. The bar
is made of a low carbon steel with a yield strength of y = 250 MPa. It is
loaded axially with a nominal stress, (the axial load divided by the
E4.2 A spring, wound from stainless steel wire with a wire diameter
nom will yield first commence at the
d = 1mm, has n = 20 turns of radius R = 10 mm. How much will it
un-notched area). At what value of nom
extend when loaded with a mass P of 1 kg? Assume the shear root of the notch? Use the stress concentration estimate of Appendix B9
modulus G of stainless steel to be 3/8 E where E is Young’s to find out.
modulus, retrieve this from Appendix A4, and use the expression for
the extension of springs from Appendix B6 to find out. Answer. The stress concentration caused by notch of depth c and root
radius r is
Answer. The extension u of a spring under a force F Pg = 9.81 N 1/ 2
m ⎛ c⎞
ax
1 ⎜ ⎟ with for tension
2
(here g is the acceleration due to gravity) is n ⎝ r⎠
om
www.grantadesign.com/education/resources -6-
, 64 F R 3 Yield first starts when max y . Inserting the data from the
n question
u gives a nominal stress for first yield of 11.9 MPa. Stress concentrations
Gd4 can be very damaging – in this example, a cyclic stress of only 12 MPa
will ultimately initiate a fatigue crack at the notch root.
Young’s modulus for stainless steel is 200 GPa, so shear modulus
G 76 GPa. Inserting the data gives a deflection u = 10.4 mm.
www.grantadesign.com/education/resources -7-
,E4.5 An acrylic (PMMA) window is clamped in a low carbon steel Answer. The crack length is small compared with the width of the
frame at T = 20 C. The temperature falls to T = -20 C, putting the window, so the appropriate choice of equation describing crack
window under tension because the thermal expansion coefficient of instability is
PMMA is larger than that of steel. If the window was stress-free at
20C, what stress does it carry at -20 C? Use the result that the bi-axial C a with C 1.0
K1
c
stress caused by a bi-axial strain difference
Inserting the data we find the length of the shortest crack that is just
is
E unstable:
1 2
⎛⎞
where E is Young’s modulus for PMMA and Poisson’s ratio 0.33 2a 2 ⎜ = 2.1 mm, using 1.15
. K1c K
⎟
⎜ 1c PMMA
You will find data for expansion coefficients in Table A7, and for moduli
in Table B5. Use mean values. MPa.m
⎟
⎝
⎠
1/2
Thus the 0.5mm crack will not propagate.
Answer. The strain difference caused by difference in thermal
expansion, , when the temperature changes by T is E4.7 A flywheel with a radius R = 200 mm is designed to spin up to
800 of cast iron, but the casting shop can guarantee only that it will have
PM 0 no crack-like flaws greater
MA rpm
T . It
Low C steel
is
pro
pos
ed
to
ma
ke
it
out
www.grantadesign.com/education/resources -8-
, THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
www.grantadesign.com/education/resources -9-
Exercises with worked solutions
,Contents
E1. Introduction to exercises
E2 Material evolution in products (Chapter
1) E3. Devising concepts (Chapter 2)
E4. Using material properties (Chapter 3)
E5. Using material selection charts (Chapter 4)
E6. Translation: constraints and objectives (Chapters 5
and 6) E7. Deriving and using material indices (Chapters
5 and 6)
E8. Multiple constraints and objectives (Chapters 7
and 8) E 9. Selecting material and shape (Chapters
9 and 10) E10. Hybrid materials (Chapters 11 and
12)
E11. Selecting processes (Chapters 13 and
14) E12. Materials and the environment
(Chapter 15)
,E1 Introduction to exercises
These exercises are designed to develop facility in selecting 3. A request for a selection based on one material index alone (such
materials, processes and shape, and in devising hybrid materials when
no monolithic material completely meets the design requirements. as M E / ) is correctly answered by listing the subset of
Each exercise is accompanied by a worked solution. They are materials that maximize this index. But a request for a selection of
organized into the twelve sections listed on the first page. materials for a component – a wing spar, for instance (which is a
light, stiff beam, for which the index is M E / ) – requires
The early exercises are easy. Those that follow lead the reader
more: some materials with high E / such as silicon carbide,
through the use of material properties and simple solutions to
mechanics problems, drawing on data and results contained in are unsuitable for obvious reasons. It is a poor answer that
Appendices A and B; the use of material property charts; techniques for ignores common sense and experience and fails to add further
the translation of design requirement to identify constraints and constraints to incorporate them. Students should be encouraged
objectives; the derivation of indices, screening and ranking, multi- to discuss the implications of their selection and to suggest further
objective optimization; coupled choice of material and shape; devising selection stages.
hybrids; and the choice of materials to meet environmental criteria.
The best way to use the charts that are a feature of the book is to
Three important points. make clean copies (or down-load them from
http://www.grantadesign.com ) on which you can draw, try out
1. Selection problems are open-ended and, generally, under-
alternative selection criteria, write comments and so forth. Although
specified; there is seldom a single, correct answer. The proper the book itself is copyrighted, the reader is authorized to make copies
answer is sensible translation of the design requirements into
of the charts and to reproduce these, with proper reference to their
material constraints and objectives, applied to give a short-list of
source, as he or she wishes.
potential candidates with commentary suggesting what supporting
information would be needed to narrow the choice further.
2. The positioning of selection-lines on charts is a matter of All the materials selection problems can be solved using the CES
judgement. The goal is to place the lines such that they leave an EduPack software, which is particularly effective when multiple criteria
adequately large "short list" of candidates (aim for 4 or so), drawn, and unusual indices are involved.
if possible, from more than one class of material.
www.grantadesign.com/education/resources -2-
,E2 Material evolution in products (Chapter 1)
E 2.1. Use Google to research the history and uses of one of the
following materials
Tin
Glass
Cement
Titanium
Carbon fiber
Present the result as a short report of about 100 - 200 words (roughly
half a page).
Specimen answer: tin. Tin (symbol Sn), a silver-white metal, has a long
history. It was traded in the civilisations of the Mediterranean as early
as 1500 BC (the Old Testament of the Christian bible contains many
references to it). Its importance at that time lay in its ability to harden
copper to give bronze (copper containing about 10% tin), the key
material for weapons, tools and statuary of the Bronze age (1500 BC –
500 BC). Today tin is still used to make bronze, for solders and as a
corrosion resistant coating on steel sheet (“tin plate” ) for food and drink
containers – a “tinnie”, to an Australian, is a can of beer. Plate glass is
made by floating molten glass on a bed of liquid tin (the Pilkington
process). Thin deposits of tin compounds on glass give transparent,
electrically conducting coatings used for frost-free windshields and for
panel lighting.
E2.2 Research, at the level of the mini case studies in this chapter, the
evolution of material use in
Writing implements (charcoal, “lead” (graphite), quill pens, steel
nib pens, gold plus osmium pens, ball points..)
Watering cans (wood – galvanized iron – polypropylene)
Bicycles (wood – bamboo – steel, aluminum, magnesium,
titanium – CFRP)
Small boat building (wood – aluminum – GFRP)
Book binding (Wood – leather – cardboard – vinyl)
www.grantadesign.com/education/resources -3-
,E3 Devising concepts (Chapter 2)
These two examples illustrate the way in which concepts are Concept Embodiment
generated. The left-hand part of each diagram describes a physical
principle by which the need might be met; the right-hand part
elaborates, suggesting how the principle might be used.
Electric fan pulling air stream through
paper or cloth filter in portable unit
C1 Entrain in air stream
and filter
Central pump and filter linked to
E3.1 Concepts and embodiments for dust removers. We met the need rooms by ducting
for a “device to remove household dust” in Chapter 1, with
examples of established solutions. Now it is time for more creative
thinking. Devise as many concepts to meet this need as you can. Centrifugal turbo-fan with surrounding
Nothing, at the concept stage, is too far-fetched; decisions about dust collector, in portable unit
practicality and cost come later, at the detailed stage. So think C2 Entrain in air
along the lines of Figure 2.2 of the main text and list concepts and stream,
Central turbo-fan with centrifugal
outline embodiments as block diagrams like this: dust collector linked to rooms by ducting
Centrifugal turbo-fan, injected water
Concept Embodiment spray to wash air stream, in portable unit
C3 Entrain in air
stream,
Central turbo-fan , water spray to wash
Electric fan pulling air stream through air stream, linked to rooms by ducting
paper or cloth filter in portable unit
C1 Entrain in air stream
and filter Axial fan drawing air stream between
Central pump and filter linked to
rooms by ducting charged plates, in portable unit
C4 Entrain in air
stream, trap
Central fan with electrostatic collector,
linked to rooms by ducting
C5 Trap dust on Reel-to-reel single sided adhesive tape
Answer. Design problems are open-ended; there are always adhesive strip running over flexible pressure-pad
alternative solutions. Here are some for dust removers.
www.grantadesign.com/education/resources -4-
,E3.2 Cooling power electronics. Microchips, particularly those for
power electronics, get hot. If they get too hot they cease to
Concept Embodiment
function. The need: a scheme for removing heat from power
microchips. Devise concepts to meet the need and sketch an
embodiment of one of them, laying out your ideas in the way
Massive, with sufficient heat capacity to absorb heat
suggested in exercise E3.1. over work cycle without significant increase in
C1 Conduction
Answer. Four working principles are listed below: thermal conduction, Compact, requiring back-up by coupling to convection,
convection by heat transfer to a fluid medium, evaporation evaporation or radiation
exploiting the latent heat of evaporation of a fluid, and radiation,
best achieved with a surface with high emissivity. Free convection not requiring fan or pump.
The best solutions may be found by combining two of these:
conduction coupled with convection (as in the sketch – an often- C2 Convection
used combination) or radiation coupled with evaporation (a Forced convection with fan or pump
possibility for short-life space structures).
Unconfined, such as a continuous spray of volatile fluid
C3 Evaporation
Confined, utilising heat-pipe technology
Radiation to ambient, using high emissivity coatings
C4 Radiation
Radiation to cooled surface, from high emissivity
surface to highly absorbent surface
Sketch.
www.grantadesign.com/education/resources -5-
,E4 Using material properties (Chapter 3) E4.3 A thick-walled tube has an inner radius ri = 10 mm and an outer
These exercises introduce the reader to 2 useful resources: the data radius ro = 15 mm. It is made from polycarbonate, PC. What is the
sheets of Appendix A and the Solutions to Standard Problems of maximum torque that the tube can carry without the onset of yield?
Appendix B. Retrieve the (mean) yield strength y of PC from Appendix A5, the
expression for the torque at onset of yield from Appendix B6 and that for
E4.1 A cantilever beam has a length L = 50 mm, a width b = 5 mm and the polar moment of a thick walled tube from Appendix B2 to find out.
a thickness t = 1 mm. It is made of an aluminum alloy. By how much
will the end deflect under and end-load of 5 N? Use data from
Appendix A4 for the (mean) value of Young’s modulus of aluminum Answer. The torque at the onset of yield for a thick walled tube is
alloys, the equation for the elastic deflection of a cantilever from
Appendix B3 and for the second moment of a beam from Appendix B2 K 4 4
y
to find out. Tf with K ( ro r
2
ro )2 i
Answer. The deflection of a cantilever under an end load F is, from The mean yield strength of y of PC from Appendix A5 is 65 MPa.
Appendix B, Inserting the data from the question gives a torque at the onset of yield of
F L3 with
bt3 T f = 138 N.m.
I
3E I 12
The mean Young’s modulus E for aluminum, from Appendix A4, is 75
GPa. Inserting the data from the question results in an end deflection E4.4 A round bar, 20 mm in diameter, has a shallow circumferential
= 6.7 mm. notch with a depth c = 1 mm with a root radius r = 10 microns. The bar
is made of a low carbon steel with a yield strength of y = 250 MPa. It is
loaded axially with a nominal stress, (the axial load divided by the
E4.2 A spring, wound from stainless steel wire with a wire diameter
nom will yield first commence at the
d = 1mm, has n = 20 turns of radius R = 10 mm. How much will it
un-notched area). At what value of nom
extend when loaded with a mass P of 1 kg? Assume the shear root of the notch? Use the stress concentration estimate of Appendix B9
modulus G of stainless steel to be 3/8 E where E is Young’s to find out.
modulus, retrieve this from Appendix A4, and use the expression for
the extension of springs from Appendix B6 to find out. Answer. The stress concentration caused by notch of depth c and root
radius r is
Answer. The extension u of a spring under a force F Pg = 9.81 N 1/ 2
m ⎛ c⎞
ax
1 ⎜ ⎟ with for tension
2
(here g is the acceleration due to gravity) is n ⎝ r⎠
om
www.grantadesign.com/education/resources -6-
, 64 F R 3 Yield first starts when max y . Inserting the data from the
n question
u gives a nominal stress for first yield of 11.9 MPa. Stress concentrations
Gd4 can be very damaging – in this example, a cyclic stress of only 12 MPa
will ultimately initiate a fatigue crack at the notch root.
Young’s modulus for stainless steel is 200 GPa, so shear modulus
G 76 GPa. Inserting the data gives a deflection u = 10.4 mm.
www.grantadesign.com/education/resources -7-
,E4.5 An acrylic (PMMA) window is clamped in a low carbon steel Answer. The crack length is small compared with the width of the
frame at T = 20 C. The temperature falls to T = -20 C, putting the window, so the appropriate choice of equation describing crack
window under tension because the thermal expansion coefficient of instability is
PMMA is larger than that of steel. If the window was stress-free at
20C, what stress does it carry at -20 C? Use the result that the bi-axial C a with C 1.0
K1
c
stress caused by a bi-axial strain difference
Inserting the data we find the length of the shortest crack that is just
is
E unstable:
1 2
⎛⎞
where E is Young’s modulus for PMMA and Poisson’s ratio 0.33 2a 2 ⎜ = 2.1 mm, using 1.15
. K1c K
⎟
⎜ 1c PMMA
You will find data for expansion coefficients in Table A7, and for moduli
in Table B5. Use mean values. MPa.m
⎟
⎝
⎠
1/2
Thus the 0.5mm crack will not propagate.
Answer. The strain difference caused by difference in thermal
expansion, , when the temperature changes by T is E4.7 A flywheel with a radius R = 200 mm is designed to spin up to
800 of cast iron, but the casting shop can guarantee only that it will have
PM 0 no crack-like flaws greater
MA rpm
T . It
Low C steel
is
pro
pos
ed
to
ma
ke
it
out
www.grantadesign.com/education/resources -8-
, THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
www.grantadesign.com/education/resources -9-
