Exercises with Worked Solutions
Selecting Materials (To accompany Lecture Unit 7)
Professor Mike Ashby Department of Engineering University of Cambridge
© M. F. Ashby, 2013 For reproduction guidance see back page This exercise unit is part of a set based on Mike Ashby’s books to help introduce students to materials, processes and rational selection. The Teaching Resources website aims to support teaching of materials-related courses in Design, Engineering and Science. Resources come in various formats and are aimed primarily at undergraduate education.
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Exercises With Worked Solutions – Unit 7
Exercises with Worked Solutions This collection of exercises and solutions has been put together to help you as an instructor choose or develop your own exercises for your students. You may simply want to browse through them for inspiration, or you may use them with your class. We are providing these in Word format so that you may pick and choose the questions you find suitable for your course this year. We have also included variations on a theme so that you can set different questions for different classes. Most of the questions come from or are inspired by the exercises in the following books by Professor Mike Ashby of the University of Cambridge Engineering Department, co-founder of Granta Design. • • •
Materials Selection for Mechanical Design by Design by Michael F. Ashby (ISBN-13: 978-1-85617-663-7) Materials: Engineering, Science, Processing and Design Design by Michael F. Ashby, Hugh Shercliff, and David Cebon (ISBN-13: (ISBN-13: 978-1856178952) Materials and the Environment Environment by Michael F. Ashby (ISBN-13: 978-0-12385971-6)
(Reproduction (Reproduction and copyright information can be found on the t he last page. Please make sure to credit Professor Mike Ashby and Granta Design if you use these questions.) Most of the questions require the use of CES EduPack. However where a topic, such as understanding how to translate design requirements into specific criteria for selection, is important on courses that traditionally use CES EduPack, we have included exercises on these topics too, even though they don’t directly use CES EduPack. (CES EduPack is a materials teaching resource used at 800+ Universities and Colleges worldwide. You can find out all about it here: www.grantadesign.com/education.) www.grantadesign.com/education .) Topics on which there are Exercises with Worked Solutions are 1: Title Materials – classification and properties Elements Design – translating requirements into constraints and objectives Selecting Materials Manufacturing Processes – classification and properties Eco Properties and the Eco Audit Tool Low Carbon Power Systems
Associated Lecture Unit Lecture Units 1 & 2 Lecture Unit 3 Lecture Unit 6 Lecture Unit 7 Lecture Unit 10 Lecture Unit 12 Lecture Unit 14
You can find f ind the other units here: www.grantadesign.com/education/resource www.grantadesign.com/education/resources s If there are other topics upon which it would be good to have a collection of questions, or if you have questions that you have used successfully with your students that you would like to donate to the Materials Education Community, attributed to you, then please contact Granta’s Materials Education Coordinator at
[email protected].
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Exercise units follow the same numbering of the PowerPoint lectures for the same topic www.grantadesign.com/education/resources M.F. Ashby 2013
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Exercises With Worked Solutions – Unit 7
Contents Material Selection Process ....................................................................... .................................................................................................... ............................. 3 Using CES EduPack Graph Stage to select materials .......................................................... 5 Using CES EduPack Limit Stage to select materials .......................................................... 10 Using CES EduPack Tree Stage to select m aterials .......................................................... .......................................................... 15 Materials Selection com bining various Selection Stages in CES EduPack ........................ 16 Using the Results Area of CES EduPack............................................................................ 18 Using Material Property Charts Char ts ................................................................ ........................................................................................... ........................... 20 Material Indices ................................................................................................................... ................................................................................................................... 27 Using Material Indices in Applications................................................................................. 34 An Assortment of Material Selection Examples .................................................................. .................................................................. 38 Economic Factors ......................................................................... ................................................................................................................ ....................................... 65 Longer Questions .............................................................................................................. ................................................................................................................ .. 69
Material Selection Process
Design requirements: Expressed as
Constraints Able to be molded Water and UV resistant Stiff enough Strong enough
Data: Comparison Engine: • • •
Screening Ranking Documentation
Objectives As light as possible As cheap as possible
Materials Attributes Process Attributes Documentation E.g. Density Price Modulus Durability ….
Final Selection In this unit we focus on the central section of the design strategy. We have used translation in the previous unit to establish the design requirements (the section on the left), and CES EduPack provides us with the data (the section on the right). Now, again using CES EduPack, we will learn how to apply the constraints, eliminating materials that cannot meet the design requirements, and rank the ‘survivors’ using the objectives to create a shortlist of candidate materials. CES EduPack offers 3 selection tools: the Graph, Limit, and Tree stages.
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Exercises With Worked Solutions – Unit 7
1. Describe the process you went through the last time you bought an electronic gadget e.g. mobile phone, computer, radio etc. Name 3 constraints you screened on. Name 2 objectives you ranked on. Where did you get your data? What documentation did you read? Answer. Examples may be: Contraints: Size - Fits in my handbag or on the table top. Functionality - Has to be able to do x e.g. receive FM radio or call abroad. Objectives: Cheap as possible Gets highest reviews online Data: Shop labels, shop website, shop assistants, consumer magazines. Documentation: User manuals, online review sites.
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Exercises With Worked Solutions – Unit 7
Using CES EduPack Graph Stage to select materials 1. Video tutorials on the Graph Stage can be accessed by clicking Help > Video Tutorials. 2. Make a bar chart with Fracture toughness on the Y-axis using Level 1, Materials.
Click on “Select”, then select “Edu Level 1” under “MaterialUniverse” and proceed as shown.
Browse
Search
Select
Tools
1. Selection data Edu EduLevel Level11:: Materials Materials 2. Selection Stages Graph
Limit
X-axis Tree
Y-axis
List of properties
Density
Yield strength
Fracture toughness
etc
Label two (or more) materials by clicking on the bars. Label Magnesium alloys (right-click on name in Result window on the left, and click on “Label”). Use the BOX selection tool
to find the five materials with the highest values of
fracture toughness.
Answer. Low alloy steels Nickel alloys Titanium alloys Zinc alloys Stainless steel
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Exercises With Worked Solutions – Unit 7
Nickel
) 100 5 . 0 ^ m . 10 a P M ( s s 1 e n h g u o 0.1 t e r u t c 0.01 a r F
Nickel-chromium alloys Nickel-based superalloys $F%P& epo'y matri' (isotropic) ilicon ca rbide
tainless steel
ranite !o" alloy steel
*thylene +inyl acetate (*,)
#inc die-casting alloys Polyamides (Nylons& P) Polycarbonate (P$)
*po'ies
!imestone
ilicone elastomers (& )
%igid Polymer Foam (!/)
3. Make a bar chart with Price on the Y Axis and the 4 main classes of materials on the x axis. Which materials class has the widest range of values? What is the cheapest material per kilogram? (You can change from metric to imperial units in the options menu.) Under the ‘Select’ mode on the toolbar, click to choose the graph selection stage. The ‘New Graph Stage Wizard’ dialog appears. Set the Y-axis to plot price. For the X-axis, under ‘Category’ click on the button ‘Advanced…’. This will bring up another dialog box which allows you to set a formula for what goes on the X-axis.
Select the ‘Trees’ tab. Double-clicking any folder will insert it into the box above. Whatever goes into this box will thus be reflected on the X-axis.
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Exercises With Worked Solutions – Unit 7
Answer. Ceramics and Glasses have the widest range of prices. Concrete is the cheapest material per mass. 100000 old 10000
) g k 1 / ( e c i r P
1000
luminum nitride
il+er
Polyetherether ketone (P**2)
$eramic 4oam 3itanium alloys
100
10
1
Poly+inylchloride (tpP,$) Ply"ood
0.1
$ast iron& gray
$oncrete
Ceramics and glasses Hybrids: composites, foams, natural materials
Metals and alloys
Polymers and elastomers
4. Which is the cheapest material per unit volume?
Click on this icon to Edit stage
. Click on y-axis and Advanced, then make the
function (Price * Density). Click to the right-hand-side of the highlighted word price and press the * button. Then find density in the property list and double click on it. Once you have [Price]*[Density] in the window click OK twice.
Answer. Concrete and Flexible Polymer Foam (VLD)
5. Make a new bubble chart with Density on the X-axis and Young’s modulus on the Yaxis using Level 1, Materials settings. Browse
Click on “Select” Then select “Edu Level 1” And proceed as shown.
Search
Select
Tools
1. Selection data Edu 22 ::Materials Materials Edu Level Level 1: Materials 2. Selection Stages Graph
Limit
Y-axis
X-axis Tree
List of properties
Density
Young’s modulus
Yield strength
etc
Switch on the envelopes by clicking the icon
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Exercises With Worked Solutions – Unit 7
By what factor are polymers less stiff than metals? Is wood denser or less dense than polyethylene (PE)? Answer. On average polymers are about 50 times less stiff than metals. Polyethylene is denser than woods.
6. Make a new bubble chart with Density on the X-axis and Yield strength on the Y-axis using Level 2, Materials settings. Label two (or more) materials by clicking on the bubbles. Switch on the envelopes by clicking the icon
Do any metals have yield strength less than 10 MPa (2 ksi)? Answer. Yes, Commercially pure Lead, Lead alloys and Tin.
7. With the graph created in the previous exercise, use the BOX selection tool to find materials with yield strength σy > 600 MPa (90 ksi) and a density ρ < 2000 kg/m
3
(120 lbs/cubic ft).
After drawing the box with approximately these values, click on
and then on the
“Selection” tab to adjust the limits.
In materials selection, however, it is more common to use the GRADIENT-LINE selection tool.
. Find the material with the highest “specific strength” ( σy / ρ) by selecting the
line selection tool and entering a slope of 1. Click anywhere on the graph and then click above the line. The materials below the line will be grayed out. Then move the line to the top left of the graph until only one material is left above it. Answer. Both exercises lead to CFRP.
8. Use a Young’s modulus – Density (E- ρ ) chart to identify materials with both a modulus E > 45 GPa and a density ρ < 2000 kg/m 3.
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Exercises With Worked Solutions – Unit 7
Answer. Material
Comment
Cast, and Wrought
Magnesium is the lightest of all common structural metals –
Magnesium alloys
only Beryllium is lighter, but it is very expensive and its oxide is toxic.
CFRP – carbon-fiber
CFRP is both lighter and stiffer than Magnesium. That is one
reinforced plastic
reason it is used for competition cars and bikes.
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Exercises With Worked Solutions – Unit 7 3
11. Use a “Limit” stage to find materials with modulus E > 2 GPa, density < 1000 kg/m and price C m < $3 / kg. Answer. The survivors meeting all 3 constraints are: Hardwood (oak, along grain and across grain), Bamboo, Plywood, Softwood: pine, along grain, Paper and cardboard.
12. Use a “Limit” stage to find materials with maximum service temperature > 200°C, thermal conductivity > 25 W/m. °C, and are good electrical insulators. Answer. Aluminum Nitride, Alumina, Silicon Nitride.
13. Use a “Limit” stage to find materials with a yield strength greater than 100MPa and 3
density less than 2000 kg/m . List the results. Answer. Cast magnesium alloys, wrought magnesium alloys, and composites CFRP and GFRP.
14. Add two further constraints to the selection of the previous exercise. Require now that the material price be less than $5/kg and the elongation be greater than 5%. Answer. Only the magnesium alloys survive.
15. Use the Limit stage to witness the evolution of materials usage across time. Use a “Limit” stage to select only the materials in usage before 3000BCE. Then produce a graph of Yield Strength vs Young’s Modulus. Copy this plot to a Word document. Repeat this procedure for Years “0”, “1500AD”, “1900AD”, “1950AD” and “2012AD”. Compare the plots obtained.
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Exercises With Worked Solutions – Unit 7
Answer.
Metals and alloys
70008$*
il"er
1000
%old Bron$e
!imestone Hard#ood: oak, along grain ) 100 a P M ( ) t i m i l c 10 i t s a l e ( h t g n 1 e r t s d l e i
Copper
Bamboo oft#ood: pine, along grain !eat(er
%lasses
!ead alloys
oda&lime glass late
'atural materials
%ranite Tin Brick Marble
andstone Commercially pure lead
0.1
oft#ood: pine, across grain
Cork
'on&tec(nical ceramics 0.01 0.001
0.01
0.1
1
10
100
1000
oung6s modulus (Pa)
ear 0
Bron$e
Metals and alloys
Brass
il"er
1000
Cast iron, gray !imestone %old
Hard#ood: oak, along grain ) 100 a P M ( ) t i m i l c 10 i t s a l e ( h t g n 1 e r t s d l e i
Copper
Bamboo oft#ood: pine, along grain !eat(er
%lasses
!ead alloys
oda&lime glass late
'atural materials
%ranite
Marble Commercially pure lead
Brick
andstone
Tin
Cement
0.1
Cork
'on&tec(nical ceramics 0.01 0.001
0.01
0.1
1
10
100
1000
oung6s modulus (Pa)
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Exercises With Worked Solutions – Unit 7
Bron$e
Metals and alloys
9500/
Brass
il"er
1000
Cast iron, gray !imestone %old
Hard#ood: oak, along grain ) a 100 P M ( ) t i m i l c i t 10 s a l e ( h t g n e r 1 t s d l e i
Copper
Bamboo oft#ood: pine, along grain !eat(er
%lasses
!ead alloys
oda&lime glass late
'atural materials
%ranite
Commercially pure lead
Brick
andstone
Paper and cardboard
Marble
Tin
Cement
0.1
Cork
'on&tec(nical ceramics 0.01 0.001
0.01
0.1
1
10
100
1000
oung6s modulus (Pa) teel
Tec(nical ceramics
9:00/ 1000
%lasses
Polymers ) a 100 P M ( ) t i m i l c i t 10 s a l e ( h t g n e r 1 t s d l e i
ilicon carbide
lastomers 'atural rubber )'-+
Tungsten alloys
Cellulose polymers )C*+
ilicon oda&lime glass
'atural materials Borosilicate glass
0.1
'on&tec(nical ceramics 0.01 0.001
0.01
0.1
1
10
100
1000
oung6s modulus (Pa)
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Exercises With Worked Solutions – Unit 7
Metals and alloys
9:50/ 1000
Composites Polymers ) a 100 P M ( ) t i m i l c i t 10 s a l e ( h t g n e r 1 t s d l e i
lastomers
Tec(nical ceramics
'atural materials
0.1
'on&tec(nical ceramics
oams 0.01 0.001
0.01
0.1
1
10
100
1000
oung6s modulus (Pa)
Metals and alloys
;09;/ 1000
Composites Polymers ) a 100 P M ( ) t i m i l c i t 10 s a l e ( h t g n e r 1 t s d l e i
lastomers
Tec(nical ceramics
'atural materials
0.1
'on&tec(nical ceramics
oams 0.01 0.001
0.01
0.1
1
10
100
1000
oung6s modulus (Pa)
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Exercises With Worked Solutions – Unit 7
Using CES EduPack Tree Stage to select materials 16. Video tutorials on the Tree Stage can be accessed by clicking Help > Video Tutorials, or via the link below.
17. This exercise teaches you to find materials that can be molded. In the Level 1 Materials database, in the Tree Stage window, select “ProcessUniverse”, expand “Shaping” in the tree, select Molding and click “Insert”, then “OK”. What are the main classes of materials that can be molded? Answer. Foams, polymers and elastomers. Soda-lime glass is the only member of the Ceramics class that can be molded.
18. This exercise teaches you to find processes that can be used to join steels. For Level 2: Processes – Joining database, in the Tree Stage window, select “MaterialUniverse”, expand “Metals and alloys” in the tree, select Ferrous, and click “Insert”, then “OK”. www.grantadesign.com/education/resources M.F. Ashby 2013
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Exercises With Worked Solutions – Unit 7
Answer. At the Level 2 joining processes database, 17 of the 23 joining processes available can be used to join steels.
19. Make a bar chart of modulus E. Add a tree stage to limit the selection to polymers alone. Which 3 polymers have the highest modulus?
Answer. Several polymers have moduli of about 4 GPa. Based on the mean of the range of moduli, which is the way the software ranks materials to make bar charts, PEEK (Polyetheretherkeytone) has the highest modulus (3.85 GPa). After that come Phenolics Polylactide (PLA) and Acetal (Polyoxymethylane, POM). Taking the maximum of the range of possible values, POM would achive the highest values for Young’s modulus (5GPa) followed by Phenolics and Polyester.
Materials Selection combining various Selection Stages in CES EduPack 20. Video tutorials on Combining Stages can be accessed by clicking Help > Video Tutorials.
21. A material is required for the blade of a rotary lawn mower. Cost is a consideration. For safety reasons, the designer specified a minimum fracture toughness for the blade: it is K 1c > 30 MPa m
1/2
. The other mechanical requirement is for high hardness, H , to
minimize blade wear.
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Exercises With Worked Solutions – Unit 7
Answer. Applying the property limit K 1c > 30 MPam1/2, and reading of materials with high hardness, gives three groups of materials. The steels are far cheaper than the other two (you can see this if you do another bar chart with price on the Y axis). Material
Comment
Steels (low, medium
Traditional material for blades
and high carbon) Stainless steels
Also used for blades, especially for food
Tungsten alloys
Meets the requirements, but more expensive than steel
Nickel superalloys
Used in other applications
Titanium alloys
Meets the requirements, but MUCH more expensive
22. Use a Maximum service temperature (T max ) chart and a Tree stage to find polymers that can be used above 200°C. Answer. The chart shows just four classes of polymer with maximum service temperatures greater than 200°C. They are listed below. Material
Comment
Polytetrafluorethylene,
PTFE (Teflon) is used as non-stick coatings for cooking
PTFE
ware, easily surviving the temperatures of baking and frying.
Silicone elastomers
Silicones are polymers with a Si-O-Si chain structure instead of the C-C-C chain of polyolefins. They are more stable than carbon-based polymers, but expensive.
Polyetheretherketone
PEEK is a high performance thermoplastic and hence
(PEEK)
extremely expensive, limiting it to use in applications where technical performance is paramount.
Phenolics
Their high heat resistance and moldability makes them a good choice for electrical switchgear.
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Exercises With Worked Solutions – Unit 7
23. Plot a modulus-strength E − σ f chart and use a Tree stage to find ferrous metals that have E > 10 GPa and σ f ≥ 1000 MPa. Answer. The strongest ferrous metals (steels) are listed below. Titanium alloys (Ti-alloys), Nickelbased superalloys and carbon fiber reinforced polymers (CFRP) also meet these limits, but they are not ferrous metals! If you eliminate the tree stage, you will get all of these. Material
Comment
High carbon steel
All have σ f above 1000 MPa,
Low alloy steel a very large value
Stainless steel
24. Find Level 2 materials with the following properties: •
Density < 2000 kg/m3
•
Thermal conductivity < 10 W/m.°C
•
Strength (Elastic Limit) > 60 MPa
•
Can be molded
Rank the results by price, using a bar chart. Remember that you can use
to hide
materials that failed to pass all the stages. Answer. (In order of ascending price) PET, PLA, PMMA, POM, Epoxies, PC, PA, PEEK.
Using the Results Area of CES EduPack 25. In MaterialUniverse: Edu Level 1 draw a graph of yield strength vs density. Use a box 3
selection tool to select materials with density between 1,000 and 10,000 Kg/m and a yield strength between 1 and 1000 MPa. Then create a limit stage and rule out materials 0.5
with a fracture toughness below 15 MPa.m . 17 out of 69 materials at level 1 pass all stages. You can see them listed in the bottom left corner of the screen. (a) List the first 3 alphabetically.
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Exercises With Worked Solutions – Unit 7
(b) Now change “Rank by” to “Stage 1: Density” and “Show” to “Pass all Stages”. List the top three i.e. least dense. (c) Now change the “Show” drop down menu to “Pass/Fail Table”. Why did Silicon Carbide fail?
Answer. (a) Aluminum Alloys, Cast Iron, ductile (nodular), Cast Iron, gray (b) Least dense: CFRP, Magnesium Alloys, GFRP (c) Silicon Carbide is not tough enough.
26. In MaterialUniverse: Edu Level 1 draw a graph of yield strength vs. density. Use a gradient line of gradient 1 intercepting the y axis (yield strength) at 1 MPa. Click above the line to select materials of high strength and low density. Then create a limit stage and 0.5
rule out materials with a fracture toughness below 15 MPa.m . 12 out of 69 materials at level 1 pass all stages. You can see them listed in the bottom left corner of the screen. (a) List the first 3 alphabetically. (b) Now change “Rank by” to “Stage 1: Performance Index”. What is the value of the performance index for the material with the biggest performance index? (The performance index in this case is Yield Strength / Density. There is more about performance indices and what they are in a subsequent section) (c) Now change the “Show” drop down menu to “Pass/Fail Table”. Why did Cast Iron, gray fail?
Answer. (a) Aluminum Alloys, Cast Iron, ductile (nodular), CFRP. (b) CFRP – performance index = 0.491 (c) Ratio of yield strength to density for Cast Iron, gray is not high enough.
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Exercises With Worked Solutions – Unit 7
Using Material Property Charts 27. Use a MaterialsUniverse: Edu Level 2 K 1C – E chart to establish: (a) Whether CFRP has higher fracture toughness K 1c than aluminum alloys (b) Whether polypropylene (PP) has a higher toughness G c than age-hardening aluminum alloys (c) Whether polycarbonate (PC) has higher fracture toughness K 1c toughness than glass
Answer. (a) CFRP has lower fracture toughness K 1c than aluminum alloys. (b) Polypropylene (PP) has a similar (slightly higher) toughness G c to age-hardening aluminum alloys. (c) Polycarbonate (PC) is has a much higher fracture toughness K 1c than glass.
28. A component is at present made from a brass, a copper alloy. Use a Young’s modulus bar chart to suggest three other metals that, in the same shape, would be stiffer. Answer. Metals that are stiffer than brass are listed below: Material
Comment
Steels
The cheapest stiff, strong structural metal, widely used.
Nickel alloys
More expensive than steel
Tungsten alloys
Refractory (high-melting) and relatively expensive
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Exercises With Worked Solutions – Unit 7
29. Use a Young’s Modulus-Density ( E- ρ ) chart to find (a) metals that are stiffer and less dense than steels and (b) materials (not just metals) that are both stiffer and less dense than steel. Answer. (a) No engineering metals are both stiffer and less dense than steel, though nickel alloys come close. (b) Several ceramics qualify: Boron carbide B4C, silicon carbide SiC, silicon nitride Si3N4, Aluminum nitride AlN, and Alumina Al 203. Material
Comment
Alumina Al203
Alumina is the most widely used of all technical ceramics (spark plugs, circuit boards…) All ceramics are brittle – they
Silicon nitride Si 3N4
have low values of fracture toughness K 1c and toughness
Boron carbide, B4C
G1c .
Silicon carbide, SiC Aluminum nitride
30. Bells ring because they have a low loss (or damping) coefficient; a high damping gives a dead sound. Use a loss coefficient – modulus chart to identify materials that would make good bells. Answer. Materials with a low loss coefficient, any of which could be used to make bells, are listed below: Material
Comment
Copper alloys
The traditional material for bells: bronzes and brasses
Glass, silica, SiO 2
Glasses make excellent bells
Ceramics: Al2O3, SiC
Unusual choice as they are expensive, but should work.
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Exercises With Worked Solutions – Unit 7
31. Use a loss coefficient – modulus chart to find materials with the highest possible damping. Answer. Lead alloys have very high damping – they are used to clad buildings to deaden sound and vibration. Magnesium alloys also have high damping: they are used to dampen vibration in machine tools. Most of the foams, polymers and natural materials also have high damping. Material
Comment
Lead alloys
Used to clad buildings to damp sound and vibration
Magnesium alloys
Used to damp vibration in machine tools
32. Use a Thermal conductivity-Electrical resistivity ( λ − ρ e ) chart to find three materials with high thermal conductivity, λ , and high electrical resistivity, ρ e . Answer. Aluminum nitride is the best choice. The next best are alumina and silicon carbide. Material
Comment
Aluminum nitride, AlN
Favored material for heat sinks requiring this combination of properties
Alumina, Al2O3 Both also meet the requirements Silicon carbide, SiC
33. Make a E – ρ chart. Apply a selection line of slope 1, corresponding to the index E / ρ positioning the line such that six materials are left above it. Which are they and what families do they belong to?
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Exercises With Worked Solutions – Unit 7
Answer. The six that have the highest index E / ρ are alumina, aluminum nitride, boron carbide, CFRP, silicon carbide, and silicon nitride. They come from just two families: technical ceramics and composites.
34. Use a E- ρ chart to establish whether woods have a higher specific stiffness E/ ρ than epoxies. Answer. Parallel to the grain, woods have much higher specific stiffness than epoxies. Perpendicular to the grain, woods have on average about the same value as epoxies. 3
3
Material
E/ ρ (GPa / (kg/m )) x 10
Woods parallel to the grain
8 - 29
Woods transverse to the grain
0.7 – 4.0
Epoxies
1.8 – 2.5
35. Use a E- ρ chart to identify metals with both E > 100 GPa and specific stiffness E/ ρ > 0.02 GPa/(kg/m3). Answer. Material
Comment
Steels
Cheap, widely used. Stiff structural material.
Nickel alloys
More expensive than steel
Titanium alloys
Titanium alloys are very expensive.
Tungsten alloys
Refractory, heavy and expensive materials
36. The speed of longitudinal waves in a material is proportional to E / ρ . Plot contours of this quantity onto a copy of a E – ρ chart, allowing you to read off approximate values for any material on the chart. Which metals have the about the same sound velocity as steel? Does sound move faster in titanium or glass?
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Exercises With Worked Solutions – Unit 7
Answer. Tungsten, titanium, nickel, aluminum, magnesium and steel all have about the same value of E / ρ , and thus similar sound velocities. The sound of velocity in glass is a little higher than that in titanium.
37. Use a E – ρ chart to identify materials with both E > 100 GPa and E 1 / 3 / ρ > 0.003 (GPa)
1/3 /(kg/m3). Remember that, on taking logs, the index M = E 1 / 3 / ρ
becomes
Log ( E ) = 3 Log ( ρ ) + 3 Log ( M )
and that this plots as a line of slope 3 on the chart, passing through the point E = 27GPa when ρ = 1000 kg/m3 in the units on the chart. Answer. Material
Comment
CFRP
Carbon-fiber composites excel in stiffness at low weight.
Boron carbide, B4C
Boron carbide is exceptionally stiff, hard and light; it is used for body armor.
38. Do titanium alloys have a higher or lower specific strength (strength/density, σ f / ρ ) than the best steels? This is important when you want strength at low weight (landing gear of aircraft, mountain bikes). Use a σ f / ρ chart to decide. Answer. 3
Material
σ ρ σf / ρ ρ (MPa / (kg/m )) x 10
Titanium alloys
60 – 262
Steels
32 – 191
-3
39. Are the fracture toughness, K 1c , of the common polymers polycarbonate, ABS, or polystyrene larger or smaller than the engineering ceramic alumina, Al 2O3? Are their toughness G1c = K 12c / E larger or smaller? Use a graph of K 1c–E.
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Exercises With Worked Solutions – Unit 7
Answer. Most polymers have lower fracture toughness, K 1c , than alumina. Their toughness, 2 G1c = K 1c / E , however, are much larger. Even the most brittle of polymers, polystyrene,
has a toughness G1c that is a few times greater than that of alumina. The values in the table are read from the graph, using the K 1c axis to read off values of fracture toughness, and the G1c = K 12c / E contours to read off values of toughness.
1/2
2
Material
K 1c ( MPa.m )
G 1c ( kJ/m )
Polycarbonate
2.1 – 4.6
2.0 – 9.6
ABS
1.2 – 4.3
0.8 – 10.5
Polystyrene
0.7 – 1.1
0.25 – 0.8
Alumina Al2O3
3.3 – 4.8
0.03 – 0.06
40. Use the fracture toughness-modulus chart to find materials that have a fracture toughness K 1c greater than 100 MPa.m and a toughness G1c = K 12c / E greater than 1/2
2
10 kJ/m . Answer. The table lists the results. Only metals have both high fracture toughness K 1c and high toughness G1c . That is one reason that they are used for pressure vessels (boilers, submarine hulls, gas containers etc). Material Low alloy steels
2
K 1c /E (kJ/m )
0.93 – 190
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Stainless steels
19.3 – 113
Nickel
31.1 – 59.5
Nickel-chromium alloys
30.4 – 57.8
Nickel superalloys
21.4 – 65.1
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Material Indices 41. What is meant by a material index? (Hint: In the Help menu of the CES EduPack there is a section called Selection Methodology.) Answer. The performance of a component or product is limited by the properties of the materials of which it is made. Sometimes it is limited by a single property, sometimes by a combination of them. The property or property-group that measures the performance of a material for a given design is called its material index. Performance is maximized by seeking the materials with the most extreme (biggest or smallest, usually the smallest) value of the index.
42. What property would you rank on if you wanted a material that is a good thermal insulator? Answer. Thermal Conductivity.
43. What property would you rank on if you wanted to make the heaviest paperweight in the world? Answer. Density.
Strong tie of length L and minimum mass
F Area A
F L
44. Consider a tie rod with length L defined. It must carry a prescribed tensile load F without failure (a constraint) and at the same time be as light a possible (an objective). The area the force acts on is A. It must not yield so we know the yield strength of the material must be greater than F/A. σy > F/A
The mass of the tie rod is equal to the density multiplied by the volume. Mass = ρV = ρLA
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We want to minimise the mass so we need to know the mass in terms of the materials properties and the fixed values. σy > F/A can be rearranged to F/ σy > A
Therefore
Mass = ρLA = ρL F/ σy
(a) Given that F and L are fixed what combination of material properties needs to be maximised in order to get the lightest tie-rod? Use the advanced function in the graph stage wizard to plot this combination of properties as a function on the y-axis of a bar chart for materials at Edu Level 1. (b) What are the three best performing materials by this index? (Video tutorials on Plotting Charts can be accessed by clicking Help > Video Tutorials.) Performance indices only deal with the information you have used in order to create them. Their results always need to be checked by common sense and other information you have about your application. (c) Why might the third best material be a bad choice for certain applications? Answer. (a) σy / ρ (b)
(c) Silicon Carbide does not have a very high fracture toughness. Therefore it might not be a good choice in any industrial application where the tie rod might get knocked about. www.grantadesign.com/education/resources M.F. Ashby 2013
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45. The most usual mode of loading of engineering structures is bending: wing-spars of aircraft, ceiling and floor joists of buildings, golf club shafts, oars, skis….all these structures carry bending moments; they are beams. The requirement here is for a beam of specified stiffness and minimum mass. A beams bending stiffness depends on its shape and its young’s modulus. 3
2
3
S= CEI/L = CEA /12L
S = Bending Stiffness, m = mass, A = area, L = length, ρ = density, E = Young’s modulus 4
2
I = second moment of area (I = b /12 = A /12) C = constant (here, 48)
Stiff beam of length L and minimum mass
b
L
Square section, area A = b2
If we want to minimise mass we need an equation for it in terms of the materials properties and fixed values. Mass = Volume * Density = AL * ρ Area can be found from the stiffness equation and substituted into the mass equation. (a) What is the Mass in terms of Stiffness, Length, Density, Young’s Modulus and constants? If the stiffness and shape of the beam are specified and become fixed then separate the mass equation into fixed values and materials properties. (b) What combination of materials properties needs to be maximised to get the lightest beam? Most performance indices are of the form α
m=
P1
P2
Where m = performance index. P1 = a materials property and P2 = another material property. Taking logarithm gives: Log(P1) = 1/ α Log(P2) + Log (m) www.grantadesign.com/education/resources M.F. Ashby 2013
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For a line on a linear graph y=ax+c, a is the gradient of the graph. Compare this equation to the Log equation before. (c) What gradient would you set for a line representing your performance index?
Gradient = ? )
1
P ( g o L
Log (P2)
Plot a graph of logarithmic graph of the two properties in the performance index you just derived. (In the graph wizard a log plot is the default.) Line selection tool
Line selection Enter slope OK
? Cancel
Plot a line on the graph using the gradient line icon in the toolbar and choosing the gradient you just worked out. Click anywhere on the graph for the line to be drawn. (d) Which side of the line are the best materials by your index? Sense check this. Pick the property you are most comfortable with. In this case perhaps Density. For a light tie rod we need low Density, so click on the graph on the side of the line that has the lowest values of density. You’ll see that bubble on the other side go grey and are ruled out of the selection. You can now hover over the line until the cursor changes and then click and drag the line to reduce the number of coloured bubbles.
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(e) What are the top 3 materials in Edu Level 1 for your performance index? (You can rank by performance index in the results window. Note that it will use average values of the properties to do this.) Check this by deleting this stage (right click on the stage in the selection stages window and choose delete), and creating a bar chart with your performance index as a function on the y-axis. (The video tutorial Plotting Charts has a section on how to do this.) (f) Are the materials the same?
Answer. 1 / 2
12 L5 S * m = C (a)
ρ 1 / 2 E
1/2
1/2
(b) ρ /E needs to be minimised to get the least mass. Therefore E / ρ needs to be maximised. (c) & (d) & (e)
Gradient = 2
(f)
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46. Find the table of performance indices in the help menu of the CES EduPack and look up the value to minimise for a stiffness limited, minimum mass, Cylinder with Internal Pressure. Answer. ρ /E
47. Find the booklet Useful Solutions to Standard Problems from the website www.grantadesign.com/education/resources (you need to register for the open access password). Use the Elastic Bending of Beams section to write an equation for the stiffness of an I beam with fixed ends and a central load in terms of Young’s Modulus and the shape of the beam. Answer.
1 3 b 192E h t (1 + 3 ) 6 h S= 3
L
Where S = Stiffness, E = Young’s Modulus, h = height, L = Length, b = width across I and t= thickness of I.
48. The objective in selecting a material for a panel of given in-plane dimensions for the lidcasing of an ultra-thin portable computer is that of minimizing the panel thickness h while
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meeting a constraint on bending stiffness, S* to prevent damage to the screen. What is the appropriate material index? Answer. The index is that for a stiffness limited panel of minimum volume: minimize 1 / E 1/3 , where E is Young’s modulus (or, more strictly, the flexural modulus, equal to E for isotropic
materials but not for those, like fiber composites, that are not isotropic).
49. Derive the material index for a torsion bar with a solid circular section. The length L and the stiffness S* are specified, and the torsion bar is to be as light as possible. Follow the steps for a beam bent under load, but replace the bending stiffness S* = Fδ by the torsional stiffness S* = T / θ. Answer. The material index is M = G 1/2 / ρ where G is the shear modulus and ρ is the density.
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Using Material Indices in Applications (Use CES EduPack Level 1 unless otherwise indicated.) 50. Plot the index for a light, stiff panel on a Modulus-Density chart, positioning the line such that eight materials are left above it, excluding ceramics because of their brittleness. What material classes do they belong to? Answer. 1/3
The index for selecting materials for light stiff panels is ρ /E . The materials that have the lowest (best) values of the index belong to the classes •
Woods
•
Polymer foams
•
Composites (CFRP)
•
Technical ceramics (Boron Carbide)
Polymer foams have the problem that their modulus is low, so although they are light, the panel has to be thick to achieve much stiffness. It is possible to apply a constraint on thickness (it translates into a lower limit for modulus); applying it then removes any materials that can only meet the stiffness constraint if they are too thick. CFRP and Boron carbide are both the ‘top of their class’, but both very expensive, and only about half as good as bamboo.
51. A material is required for a cheap column with a solid circular cross-section that must support a load F* without buckling. It is to have a height L . Write down an equation for the material cost of the column in terms of its dimensions, the price per kg of the material, C m , and the material density ρ. The cross-section area A is a free variable – eliminate it
by using the constraint that the buckling load must not the less than F*. Hence read off the index for finding the cheapest column.
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Answer. Plot a graph of modulus E plotted against material cost per unit volume C m ρ . A line 1/2
corresponding to the material index = E / C m ρ is positioned on it so that only a few materials remain above it. These are the materials with the most attractive value of the index: concrete, brick, stone, cast iron and wood.
52. The window through which the beam emerges from a high-powered laser must obviously be transparent to light. Even then, some of the energy of the beam is absorbed in the window and can cause it to heat and crack. This problem is minimized by choosing a window material with a high thermal conductivity λ (to conduct the heat away) and a low expansion coefficient α (to reduce thermal strains), that is, by seeking a window material with a high value of M = λ / α . Use a limit stage to choose transparent and optical quality materials and a α − λ chart to identify the best material for an ultra-high powered laser window. Use the Level 3 database.
Answer. Looking at the top 8 transparent materials with high λ / α : Diamond, Titanium Silicate, Alumina, which, is in one of three transparent forms, single crystal form (sapphire), ultrafine grained form (“Lucalox”), or as a Bio-ceramic, and Silica.. At Level 3 of CES EduPack, Diamond has the highest λ / α . Diamond, has an exceptionally high value: it has been used for ultra high-powered laser windows. It is however very expensive!
53. The elastic deflection at fracture (the “resilience”) of an elastic-brittle solid is proportional to the failure strain, ε fr = σ fr / E , where σ fr is the stress that will cause a crack to propagate:
σ fr =
K 1c
π c
Here K 1c is the fracture toughness and c is the length of the longest crack the materials may contain. Thus
ε fr =
K 1c π c E 1
Materials that can deflect elastically without fracturing are therefore those with large values of K 1c / E . Use a K 1c –E chart to identify the class of materials with K 1c > 1 MPa.m
1/2
and high values of K 1c / E .
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Answer. Polymers and foams both have large K 1c / E allowing them to flex without fracturing. They have much higher values K 1c / E than metals or ceramics, and thus large fracture strains. Only polymers have, additionally, K 1c > 1 MPa.m1/2. This makes them attractive for snap-together parts that must flex without failing. Leather has a particularly high value of resilience, K 1c / E .
54. One criterion for design of a safe pressure vessel is that it should leak before it breaks: the leak can be detected and the pressure released. This is achieved by designing the vessel to tolerate a crack of length equal to the thickness t of the pressure vessel wall, without failing by fast fracture. The safe pressure p is thus
4 1 K Ic π R σ f 2
p ≤
where σ f is the elastic limit, K 1c is the fracture toughness, and R is the vessel radius. The pressure is maximized by choosing the material with the greatest value of
M =
K 12c
σ y
Use a K 1c − σ y chart to identify three alloys that have particularly high values of M.
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Answer. Alloys with high values of K 12c / σ y , read from the chart, are listed below: Material Nickel alloys
2
K 1c / σ y
7.95 – 126
Reactors for chemical engineering and turbine combustion chambers are made of nickel based alloys.
Copper alloys
4.72 – 103
Small boilers are made of copper.
Tin
20.9 – 92.3
Used for storage tanks for chemical solutions, but its main use is for steel coatings and alloying with copper.
Stainless steels
7.49 – 67.9
Used for nuclear pressure vessels.
Zinc alloys
0.52 – 53.3
Used for corrosion resistance and for its castability.
Low alloy steels
0.25 – 51.8
Traditional material for pressure vessels.
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An Assortment of Material Selection Examples This section combines the efforts of the previous two sections; in here students will have to derive material indices, then apply them to select materials using the CES EduPack. They can be equivalently solved with materials property charts.
55. Aperture grills for cathode ray tubes. There are two types of cathode ray tube (CRT). In the older technology, color separation is achieved by using a shadow mask : a thin metal plate with a grid of holes that allow only the correct beam to strike a red, green or blue phosphor. A shadow mask can heat up and distort at high brightness levels (‘doming’), causing the beams to miss their targets, and giving a blotchy image. To avoid this, shadow masks are made of Invar, a nickel alloy with a near-zero expansion coefficient between room temperature and 150˚C. It is a consequence of shadow-mask technology that the glass screen of the CRT curves inward on all four edges, increasing the probability of reflected glare. Sony’s ‘Trinitron’ technology overcame this problem and allowed greater brightness by replacing the shadow mask by an aperture grill of fine vertical wires, each about 200 µm in thickness, that allows the intended beam to strike either the red, the green or the blue phosphor to create the image. The glass face of the Trinitron tube was curved in one plane only, reducing glare. The wires of the aperture grill are tightly stretched, so that they remain taut even when hot – it is this tension that allows the greater brightness. What index guides the choice of material to make them? The table summarizes the requirements.
Function
Aperture grill for CRT
Constraints
Wire thickness and spacing specified Must carry pre-tension without failure Electrically conducting to prevent charging Able to be drawn to wire
Objective
Maximize permitted temperature rise without loss of tension
Free variables
Choice of material
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Answer. A thin, taut wire slackens and sags when the strain due to thermal expansion, ε th = α ∆T
exceeds the elastic strain caused by the pre-tension, σ ε pt = . E
Here α is the thermal expansion coefficient of the wire, ∆T the temperature rise caused by the electron beams that strike it, σ the tensile pre-stress in the wire and E its modulus. We wish to maximize the brightness, and thus ∆T . The tension is limited by the elastic limit of the wire, σ f . Inserting this and writing ε pt ≥ ε th as the condition that the wire remains taut gives
∆T =
σ f E α
The result could hardly be simpler. To maximize the brightness, maximize M =
σ f E α
.
There is a second requirement. The wires must conduct, otherwise they would charge up, distorting the image. We therefore require, also, that the material be a good electrical conductor and that it is capable of being rolled or drawn to wire.
Applying the constraints listed in the table above to the CES EduPack Level 1 or 2 database and ranking the survivors by the index M leads to the selection listed below.
Material High Carbon steel
Comment Carbon steel is ferro-magnetic, so will interact with the scanning magnetic fields – reject.
Commercially pure
Extracting titanium from its oxide is difficult making it an
Titanium
expensive option
Tungsten alloys
A logical choice – tungsten has a high melting point and is routinely produced as fine wire
Nickel-based
Many nickel alloys are weakly ferromagnetic – reject for the
alloys
same reason as carbon steel
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The final choice, using this database, is stainless steel or tungsten. If the selection is repeated using the more detailed Level 3 database and “All materials”, not just “All Bulk Materials”, carbon fiber emerges as a potential candidate. Carbon fiber of the desired diameter is available; it conducts well, both electrically and thermally, it has high strength and – best of all – it has almost zero thermal expansion along the fiber direction.
56. Material index for a light, strong beam. In stiffness-limited applications, it is elastic deflection that is the active constraint: it limits performance. In strength-limited applications, deflection is acceptable provided the component does not fail; strength is the active constraint. Derive the material index for selecting materials for a beam of length L, specified strength and minimum weight, where the beam is secured at one end as in a cantilever. For simplicity, assume the beam to have a solid square cross-section t x t. You will need the equation for the failure load of a beam. It is F f =
I σ f y m L
where y m is the distance between the neutral axis of the beam and its outer most surface and I = t 4 / 12 = A 2 / 12 is the second moment of the cross-section. y m = t/2
The table itemizes the design requirements
Neutral Axis
Function
Beam
Constraints
Length L is specified Beam must support a bending load F without yield or fracture
Objective
Minimize the mass of the beam
Free variables
Cross-section area, A Choice of material
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Example light stiff beam – an oar. Answer. The objective is to minimize the mass, giving the objective function m = A L ρ
Inverting the equation given in the question leads to an expression for the area A that will support the design load F.
F L A = 6 σ f
2 / 3
Substituting A into the objective function gives the mass of the beam that will just support the load F f :
m = (6 F )
2/3 5 / 3
L
ρ 2/3 σ f
The mass is minimized by selecting materials with the largest values of the index 2/3 f
σ
M =
ρ
If the cantilever is part of a mechanical system it is important that it have sufficient fracture toughness to survive accidental impact loads. For this we add the requirement of adequate toughness: K 1c > 15 MPa .m 1 / 2
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The selection. Applying the constraint on K 1c and ranking by the index M using the CES EduPack Level 1 or 2 database gives the top-ranked candidates listed below. Material
Comment
Carbon Fiber Reinforced
Exceptionally good, mainly because of its very low
Plastic (CFRP)
density.
Metals: titanium, aluminum
Here the light alloys out-perform steel
and magnesium alloys
57. Indices for stiff plates and shells Aircraft and space structures make use of plates and shells. The index depends on the configuration. Here you are asked to derive the material index for (a) a circular plate of radius a carrying a central load W with a prescribed stiffness S = W / δ and of minimum mass, and
(b) a hemispherical shell of radius a carrying a central load W with a prescribed stiffness S = W / δ and of minimum mass, as shown in the figures.
Use the two results listed below for the mid-point deflection δ of a plate or spherical shell under a load W applied over a small central, circular area. 3 W a
Circular plate:
δ =
Hemispherical shell
δ = A
2
4 π E t 3 W a 2
2
3 + ν 1 + ν
( 1 −ν )
2
( 1 − ν )
E t
in which A ≈ 0.35 is a constant. Here E is Young’s modulus, t is the thickness of the plate or shell and ν is Poisson’s ratio. Poisson’s ratio is almost the same for all structural materials and can be treated as a constant. The table summarizes the requirements.
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Function
Stiff circular plate, or Stiff hemispherical shell
Constraints
Stiffness S under central load W specified Radius a of plate or shell specified
Objective
Minimize the mass of the plate or shell
Free variables
Plate or shell thickness, t Choice of material
Answer. (a) The plate. The objective is to minimize the mass, 2
m = π a t ρ
where ρ is the density of the material of which the plate is made. The thickness t is free, but must be sufficient to meet the constraint on stiffness. Inverting the first equation in the question gives, for the plate,
3 S a 2 t = 4 π E
1 / 3
f 1 (ν )
where f 1 (ν ) is simply a function of ν , and thus a constant. Inserting this into the equation for the mass gives
3 S a 2 m = π a 2 4 π
1 / 3
ρ 1 / 3 f (ν ) E
The lightest plate is that made from a material with a large value of the index 1 / 3
M 1 =
E
ρ
(b) The hemispherical shell. The objective again is to minimize the mass, m = 2 π a 2 t ρ
Inverting the second equation in the question gives, for the shell,
S a t = A E
1 / 2
f 2 (ν )
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where f 2 (ν ) as before is a function of ν , and thus a constant. (Possions ratio does not vary very much between materials). Inserting this into the equation for the mass gives
ρ
m = 2 π a 2 ( A S a )1 / 2 f (ν ) E 1 / 2
The lightest shell is that made from a material with a large value of the index 1 / 2
M 2 =
E
ρ
The index for the shell differs from that for the plate, requiring a different choice of material. This is because the flat plate, when loaded, deforms by bending. The hemispherical shell, by contrast, carries membrane stresses (tension and compression in the plane of the shell wall), and because of this is much stiffer. Singly-curved shells behave like the plate, doubly-curved shells like the hemisphere.
The selection. Applying the three indices to the CES EduPack Level 1 or 2 database gives the top-ranked candidates listed below. Material
Comment Natural materials: wood and plywood
1 / 3
High M 1 =
High M 2 =
E
ρ
Composites: CFRP
1 / 2
Metals: aluminum and magnesium alloys superior
ρ
to all other metals.
E
Composites: CFRP. Ceramics: SiC, Si 3N4, B4C and AlN.
58. The C-clamp.
The clamp has a square cross-section of width x and given depth b.
It is essential that the clamp have low thermal inertia so that it reaches temperature quickly. The time t it takes a component of thickness x to reach thermal equilibrium when the temperature is suddenly changed (a transient heat flow problem) is t ≈
x
2
2a
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The time to reach thermal equilibrium is reduced by making the section x thinner, but it must not be so thin that it fails in service. Use this constraint to eliminate x in the equation above, thereby deriving a material index for the clamp. Use the fact that the clamping force F creates a moment on the body of the clamp of M = F L , and that the peak stress in the body is given by
σ =
x M 2 I
where I = b x 3 / 12 is the second moment of area of the body. The table summarizes the requirements. Function
C-clamp of low thermal inertia
Constraints
Depth b specified Must carry clamping load F without failure
Objective
Minimize time to reach thermal equilibrium
Free variables
Width of clamp body, x Choice of material
Answer. The clamp will fail if the stress in it exceeds its elastic limit σ f . Equating the peak stress to σ f and solving for x gives
F L x = 6 b σ f
1 / 2
Inserting this into the equation for the time to reach equilibrium gives
t = 3
F L 1 b a σ f
The time is minimized by choosing materials with large values of the index
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M = a σ f . Additional constraints on modulus E > 50 GPa (to ensure that the clamp is sufficiently stiff) on fracture toughness K 1c > 18 MPa.m 1 / 2 (to guard against accidental impact) and on formability will, in practice, be needed. The selection. Applying the constraint on K 1c and formability, and ranking by the index M using the CES EduPack Level 1 or 2 database gives the top-ranked candidates listed
below. Material Aluminum alloys
Comment The obvious candidate – good thermal conductor, adequately stiff and strong, and easy to work.
Copper alloys
Here the high thermal diffusivity of copper is dominating the selection.
59. Springs for trucks. In vehicle suspension design it is desirable to minimize the mass of all components. You have been asked to select a material and dimensions for a light spring to replace the steel leaf-spring of an existing truck suspension. The existing leafspring is a beam, shown schematically in the figure. The new spring must have the same length L and stiffness S as the existing one, and must deflect through a maximum safe displacement δ max without failure. The width b and thickness t are free variables.
Derive a material index for the selection of a material for this application. Note that this is a problem with two free variables: b and t ; and there are two constraints, one on safe deflection δ max and the other on stiffness S . Use the two constraints to fix free variables.
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The table catalogs the requirements. Function
•
Leaf spring for truck
Constraints
• •
Length L specified Stiffness S specified
•
Objective
•
Maximum displacement Minimize the mass
Free variables
• • •
Spring thickness t Spring width b Choice of material
δ max
specified
Stiffnes S= F/ δ You will need the equation for the mid-point deflection of an elastic beam of length L loaded in three-point bending by a central load F: 3
δ =
1 F L
48 E I
and that for the deflection at which failure occurs
δ max =
2 1 σ f L
6
t E
where I is the second moment of area; for a beam of rectangular section, I = b t 3 / 12 and E and σ f are the modulus and failure stress of the material of the beam.
Answer. The objective function – the quantity to be minimized – is the mass m of the spring: m = b t L ρ
(1)
where ρ is its density. The length L is fixed. The dimensions b and t are free. There are two constraints. The first is a required stiffness, S . From the first equation given in the question 3
E b t EI = 48 = 4 S = δ L3 L3 F
(2)
The second constraint is that of a maximum allowable displacement δ max without damage to the spring, given in the question as 2
δ max =
1 σ f L 6
t E
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Equations (2) and (3) can now be solved for t and b , and these substituted back into (1). The result is
ρ E 2 m = 9 S δ max L 2 σ f The mass of the spring is minimized by maximizing the index
M =
2 σ f
ρ E
Additional constraints on fracture toughness K 1c > 15 MPa .m 1 / 2 (to guard against accidental impact) and on formability will, in practice, be needed.
The selection. Applying the constraint on K 1c and formability, and ranking by the index M using the CES EduPack Level 1 or 2 database gives the top-ranked candidates listed
below. Material
Comment
Elastomers (rubber)
Oops! We have missed a constraint here. Elastomers excel as light springs, but the constraint on thickness t and depth b in this application translates via equation (2) into an additional constraint on modulus: E > SL3 / 4bt 3 .
Titanium alloys
An expensive solution, but one that is lighter than steel.
CFRP
CFRP makes exceptionally good light springs.
High carbon steel
The standard solution, but one that is heavier than the others above.
60. Fin for a rocket.
A tube-launched rocket has stabilizing fins at its rear. During launch o
the fins experience hot gas at T g = 1700 C for a time t = 0.3 seconds. It is important that the fins survive launch without surface melting. Suggest a material index for selecting a material for the fins. The table summarizes the requirements.
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Exercises With Worked Solutions – Unit 7
Function
High heat transfer rocket fins
Constraints
All dimensions specified o
Must not suffer surface melting during exposure to gas at 1700 C for 0.3 seconds Objective
Minimize the surface temperature rise during firing Maximize the melting point of the material
Free variables
Choice of material
This is tricky. Heat enters the surface of the fin by transfer from the gas. If the heat transfer coefficient is h, the heat flux per unit area is q = h ( T g − T s )
where T s is the surface temperature of the fin – the critical quantity we wish to minimize. Heat is diffused into the fin surface by thermal conduction. If the heating time is small compared with the characteristic time for heat to diffuse through the fin, a quasi steady-state exists in which the surface temperature adjusts itself such that the heat entering from the gas is equal to that diffusing inwards by conduction. This second is equal to q = λ
(T s − T i ) x
where λ is the thermal conductivity, T i is the temperature of the (cold) interior of the fin, and x is a characteristic heat-diffusion length. When the heating time is short (as here) the thermal front, after a time t , has penetrated a distance x ≈ (2 a t )1 / 2
where a = λ / ρ C p is the thermal diffusivity. Substituting this value of x in the previous equation gives
(
q = λ ρ C p
)1 / 2
(T s − T i ) 2t
where ρ is the density and C p the specific heat of the material of the fin. Proceed by equating the two equations for q, solving for the surface temperature T s to give the objective function. Read off the combination of properties that minimizes T s ; it is the index for the problem.
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Exercises With Worked Solutions – Unit 7
The selection is made by seeking materials with large values of the index and with a high melting point, T m . If the CES software is available, make a chart with these two as axes and identify materials with high values of the index that also have high melting points. Answer. Equating the equations and solving for T s gives
T s =
h T g
t +
λ ρ C p T i
h
t +
λ ρ C p
When t = 0 the surface temperature T s = T i and the fin is completely cold. When t is large, T s = T g and the surface temperature is equal to the gas temperature. For a given t , T s is minimized by maximizing λ ρ C p . The first index is therefore M 1 = λ ρ C p
and index that often appears in problems involving transient heat flow. Melting is also made less likely by choosing a material with a high melting point T m . The second index is therefore M 2 = T m
The selection. The figure shows the two indices, using the CES EduPack Level 2 database. The top-ranked candidates are listed below. Material
Comment
Silicon carbide,
Silicon carbide out-performs all metals except tungsten, and is
SiC
much lighter. If used, its brittleness would have to be reckoned with.
Copper alloys
The exceptional thermal conductivity of copper is dominating here – it is able to conduct heat away from the surface quickly, limiting the surface heating.
Aluminum alloys
An attractive choice, since Al-alloys are also light
Aluminum –
This metal matrix composite has almost the thermal conductivity
Silicon carbide
of aluminum and is stiffer and stronger.
MMC
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Exercises With Worked Solutions – Unit 7
Handling Multiple Constraints and Objectives
The first two case studies in this section illustrate the problem of material selection with multiple constraints. Each constraint would lead to its own material index and separate performance equation. It can then be observed which equation is the limiting factor (“active constraint”) in trying to achieve the objective. However by plotting a chart equating the material indices it would then be possible to see which material indices would be the active constraint and which materials should then be selected as the specifications of the design, or functional requirements, are changed. The remaining two concern multiple objectives and trade-off methods. . When a problem has two objectives – minimizing both mass and cost of a component, for instance – a conflict arises: the cheapest solution is not necessarily the lightest and vice versa. The best combination is sought by constructing a trade-off plot using mass as one axis, and cost as the other. To get further we need a penalty function with an exchange constant describing the penalty associated with unit increase in mass, or, equivalently, the value associated with a unit decrease. The best solutions are found where the line defined by this equation is tangential to the trade-off surface.
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Case Study on a Light, Stiff, Strong Tie (Multiple constraints) 61. A tie, of length L loaded in tension, is to support a load F , at minimum weight without failing (implying a constraint on strength) or extending elastically by more than δ (implying a constraint on stiffness, F / δ ). The table summarizes the requirements.
Function
Tie rod
Constraints
Must not fail by yielding under force F Must have specified stiffness, F/ δ Length L and axial load F specified
Objective
Minimize mass m
Free variables
Section area A Choice of material
Establish two performance equations for the mass, one for each constraint, from which two material indices and one coupling equation linking them are derived. Show that the two indices are M 1 =
ρ E
and
M 2 =
ρ σ y
and that a minimum is sought for both. Use CES EduPack to produce a graph, which has the indices as axes, to identify candidate materials for the tie when (i)
-3
-2
δ /L = 10 and (ii) δ /L = 10 .
Remember y=ax Log(y)=Log(a)+Log(x).
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Exercises With Worked Solutions – Unit 7
Answer. The derivation of performance equations and the indices they contain is laid out here:
Objective
Constraints
Performance equation
Index
Substitute for A
Stiffness constraint
F
δ
=
2 F ρ
E A
m1 = L
L
δ E
M 1 =
ρ σ y
M 2 =
ρ E
(1)
m = A L ρ Substitute for A
Strength constraint
m 2 = L F
F = σ y A
ρ (2) σ y
(The symbols have their usual meanings: A = area, L= length, ρ = density, F/ δ =stiffness, E = Young’s modulus, σ y = yield strength or elastic limit.)
The coupling equation is found by equating m 1 to m 2, giving
ρ L ρ = σ y δ E defining the coupling constant C c = L/ δ. The chart below shows the positions of the 3
coupling line when L/ δ = 100 and when L/ δ = 10 (corresponding to the required values of δ /L in the question) and the materials that are the best choice for each.
Coupling lines
δ
= 10 - 3
δ
L
= 10 - 2
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Exercises With Worked Solutions – Unit 7
Coupling
Material choice
Comment
Ceramics: boron carbide, silicon
These materials are available as fibers as
carbide
well as bulk.
Composites: CFRP; after that,
If ductility and toughness are also
Ti, Al and Mg alloys
required, the metals are the best choice.
condition L/ δ = 100
L/ δ = 1000
The use of ceramics for a tie, which must carry tension, is normally ruled out by their low fracture toughness – even a small flaw can lead to brittle failure. But in the form of fibers both boron carbide and silicon carbide are used as reinforcement in composites, where they are loaded in tension, and their stiffness and strength at low weight are exploited. The CES EduPack software allows the construction of charts with axes that are combinations of properties, like those of ρ / E and ρ / σ y shown here. You can also add a line and specify its intercept with the y-axis and gradient using the section tab in the dialogue box that opens on clicking the properties icon. However you cannot add more than one line at a time in the software.
Case Study on a Light, Safe Pressure Vessel 62. When a pressure vessel has to be mobile; its weight becomes important. Aircraft bodies, rocket casings and liquid-natural gas containers are examples; they must be light, and at the same time they must be safe, and that means that they must not fail by yielding or by fast fracture. What are the best materials for their construction? The table summarizes the requirements.
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Exercises With Worked Solutions – Unit 7
Function
Pressure vessel
Constraints
Must not fail by yielding Must not fail by fast fracture. Diameter 2R and pressure difference ∆ p specified
Objective
Minimize mass m
Free variables
Wall thickness, t Choice of material
(a) Write, first, a performance equation for the mass m of the pressure vessel. Assume, for simplicity, that it is spherical, of specified radius R , and that the wall thickness, t (the free variable) is small compared with R. Then the tensile stress in the wall is σ =
∆ p R 2 t
where ∆ p , the pressure difference across this wall, is fixed by the design. The first constraint is that the vessel should not yield – that is, that the tensile stress in the wall should not exceed σ y . The second is that it should not fail by fast fracture; this requires that the wall-stress be less than K 1c / π c , where K 1c is the fracture toughness of the material of which the pressure vessel is made and c is the length of the longest crack that the wall might contain. (b) Use each of these in turn to eliminate t in the equation for m ; use the results to identify two material indices, M 1 =
ρ σ y
and
M 2 =
ρ K 1c
and a coupling relation between them. It contains the crack length, c . The figure shows the chart you will need with the two material indices as axes. (c) Plot the coupling equation onto this figure for two values of c : one of 5 mm, the other 5 µm. Identify the lightest candidate materials for the vessel for each case. You can add a line and specify its intercept with the y-axis and gradient using the section tab in the dialogue box that opens on clicking the properties icon. However you cannot
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Exercises With Worked Solutions – Unit 7
add more than one line at a time in the software. Remember y=ax+c Log(ax) = Log(a) +
Log(x)
Answer. 2
(a) The objective function is the mass of the pressure vessel:
m = 4 π R t ρ
(b) The tensile stress in the wall of a thin-walled pressure vessel is:
σ =
∆ p R 2 t
Equating this first to the yield strength σ y , then to the fracture strength K 1c / π c and substituting for t in the objective function leads to the performance equations and indices laid out below:
Objective
Constraints
No Yield:
Performance equation σ =
Substitute for t
∆ pR 2 t
≤ σ y
3
σ f
m1 = 2 π ∆ p.R ⋅
ρ
Index
M 1 =
ρ (1) σ y
2
m = 4 π R t ρ
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Exercises With Worked Solutions – Unit 7
No Fracture: σ =
∆ pR 2t
≤
K 1 C
π c
ρ (2) M 2 = K 1c K 1c
m 2 = 2 π ∆ p ⋅ R 3 (π c )1 / 2
ρ
Substitute for t
The coupling equation is found by equating m 1 = m 2 , giving a relationship between M 1 and M 2:
M 1 = ( π c)
1/2
M 2
(c) The position of the coupling line depends on the detection limit , c 1 for cracks, through 1/2
the term ( π c) . The figure on the next page shows the appropriate chart with two coupling lines, one for c = 5 mm and the other for c = 5µm. The resulting selection is summarized in the table: Coupling condition
Crack length c ≤ 5 mm
Material choice
Comment
Titanium alloys
These are the standard materials for
Aluminium alloys
pressure vessels. Steels appear, despite their high density, because their toughness and strength are so high
( π c = 0.125 ) Steels CFRP Crack length c ≤ 5 µ m
Silicon carbide
Ceramics, potentially, are attractive structural materials, but the difficult of fabricating and maintaining them with no
( π c = 3.96 x 10 −3 )
Silicon nitride
flaws greater than 5 µm is enormous
Alumina
Coupling line, c = 5 microns Coupling line, c = 5 mm
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In large engineering structures it is difficult to ensure that there are no cracks of length greater than 1 mm; then the tough engineering alloys based on steel, aluminum and titanium are the safe choice. In the field of MEMS (micro electro-mechanical systems), in which films of micron-thickness are deposited on substrates, etched to shapes and then loaded in various ways, it is possible – even with brittle ceramics – to make components with no flaws greater than 1 µm in size. In this regime, the second selection given above has relevance.
Case Study on an Air Cylinder for a Truck 63. Trucks rely on compressed air for braking and other power-actuated systems. The air is stored in one or a cluster of cylindrical pressure tanks like that shown here (length L, diameter 2R , hemispherical ends). Most are made of low-carbon steel, and they are heavy. The task: to explore the potential of alternative materials for lighter air tanks, recognizing that there must be a trade-off between mass and cost – if it is too expensive, the truck owner will not want it even if it is lighter. The table summarizes the design requirements.
Function
Air Cylinder for Truck
Constraints
Must not fail by yielding Diameter 2R and length L specified, so the ratio Q = 2R/L is fixed.
Objectives
Minimize material cost C Minimize mass m
Free variables
Wall Thickness, t Choice of material
(a) Show that the mass and material cost of the tank relative to one made of low-carbon steel are given by
ρ σ y ,o = mo σ y ρ o m
and
ρ σ y ,o = Cm C o σ y C m ,o ρ o C
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Exercises With Worked Solutions – Unit 7
where ρ is the density, σ y the yield strength and C m the cost per kg of the material, and the subscript “o” indicates values for mild steel. Remember that the stress in the wall of a pressure vessel, where t<
m mo
+
C C o
*
*
*
*
where α is a relative exchange constant, and evaluate Z for α = 1 and for α = 100.
Material
Yield strength σc
Price per /kg Cm
(kg/m )
(MPa)
($/kg)
Mild steel
7850
314
0.702
Low alloy steel
7850
775
0.937
CFRP
1550
760
42.1
Density ρ 3
(c) Below is a chart with axes of m/m o and C/C o . Mild steel (here labelled “Low carbon steel”) lies at the co-ordinates (1,1). *
Sketch a trade-off surface and plot contours of Z that are approximately tangent to *
*
the trade-off surface for α = 1 and for α = 100. What selections do these suggest? ) ) t i m i l c i t s a 10 l e ( h t g n e r t s d l e 1 i 1 ? 9 7 ( > ) 0 0.1 5 = < 1 y t i s n e / (0.01
Non age-hardening "rought l-alloys #inc die-casting alloys
!ead alloys
!o" carbon steel $opper $ast iron& gray
Medium carbon steel
Nickel @igh carbon ste el
!o" alloy st eel
P**2 $F%P
Nickel-based superalloys
ge-hardening "rought l-alloys 3itanium alloys Arought magnesium alloys
0.01
0.1
1
10
100
1000
( /ensity 1 <=50 ) > ( 79? 1 ield strength (elastic limit) ) > ( Price 1 0.<0; )
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Exercises With Worked Solutions – Unit 7
Answer. (a) The mass m of the tank is
m = (2π R L t + 4π R 2 t ) ρ = (2π R L t (1 + Q)) ρ
where Q , the aspect ratio 2R/L, is fixed by the design requirements. The stress in the wall of the tank caused by the pressure p must not exceed σ y , which is the yield strength of the material of the tank wall, meaning that σ =
p R t
≤ σ y
Substituting for t , the free variable, gives
m =
ρ σ y
R 2 L p ( 1 + Q) 2π
The material cost C is simply the mass m times the cost per kg of the material, C m, giving
C m ρ σ y
C = C m m = 2π R L p ( 1 + Q ) 2
from which the mass and cost relative to that of a low-carbon steel (subscript o) tank are
ρ σ y ,o = σ y ρ o mo m
ρ σ y ,o . = Cm σ y C m ,o ρ o C o C
and
(b) To get further we need a penalty function: The relative penalty function * * Z = α
m mo
+
C C o
This is evaluated for Low alloy steel and for CFRP in the table below, for α * = 1 (meaning *
that weight carries a low cost premium) – Low alloy steel has by far the lowest Z . But *
when it is evaluated for α = 100 (meaning that weight carriers a large cost premium), CFRP has the lowest Z * . (c) The figure shows the trade-off surface. Materials on or near this surface have attractive combinations of mass and cost. Several are better than low-carbon steel. Two contours of Z * that just touch the trade-off line are shown, one for α * = 1, the other for α * = 100 – they are curved because of the logarithmic axes. (Note for CES EduPack enthusiasts: The trade surface is not dictated by an equation, but is in effect a wiggly line created by the location of the material bubbles on the chart. This cannot be plotted using the CES EduPack. You can however copy and paste the charts into WORD and add any lines or curves you want to demonstrate your point. The Z* lines can be added (one at a time) to the charts in CES EduPack by first adding them in the linear plot and then changing the graph properties to a log plot.)
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Exercises With Worked Solutions – Unit 7
*
The first, for α = 1 identifies higher strength steels as good choices. This is because *
their higher strength allows a thinner tank wall. The contour for α = 100 touches near CFRP, aluminum and magnesium alloys – if weight saving is very highly valued, these become attractive solutions.
Material
Density ρ
Yield strength
3
*
Price per /kg
*
Z,
Z,
(kg/m )
σc (MPa)
Mild steel
7850
314
0.702
2
101
Low alloy steel
7850
775
0.937
0.946
41.0
CFRP
1550
760
42.1
4.9
12.9
) ) t i m i l c i t s a 10 l e ( h t g n e r t s d l e 1 i 1 ? 9 7 ( > ) 0 0.1 5 = < 1 y t i s n e / (0.01
Cm ($/kg)
*
α
= 1
*
α
= 100
Non age-hardening "rought l-alloys #inc die-cas ting alloys
!ead alloys
!o" carbon steel $opper $ast iron& gray
Medium carbon steel
Nickel @igh carbon steel
!o" alloy steel
P**2 $F%P
Nickel-based supe ralloys
ge-hardening "rought l-alloys 3itanium alloys Arought magnesium alloys
Z* with α* = 1 0.01
0.1
1
Z* with α* = 100 10
100
1000
( /ensity 1 <=50 ) > ( 79? 1 ield strength (elastic limit) ) > ( Price 1 0.<0; )
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Case Study on Insulating Walls for Freezers 64. Freezers and refrigerated trucks have panel-walls that provide thermal insulation, and at the same time are stiff, strong and light (stiffness to suppress vibration, strength to tolerate rough usage). To achieve this, the panels are usually of sandwich construction, with two skins of steel, aluminum or GFRP (providing the strength) separated by, and bonded to, a low density insulating core. In choosing the core we seek to minimize thermal conductivity, λ, and at the same time to maximize stiffness, because this allows thinner steel faces, and thus a lighter panel, while still maintaining the overall panel stiffness. The table summarizes the design requirements.
Function
Foam for panel-wall insulation
Constraints
Panel wall thickness specified
Objective
Minimize foam thermal conductivity,
λ
Maximise foam stiffness, i.e. Young’s modulus, E Free variables
Choice of material
The graph below shows the thermal conductivity λ of foams plotted against their elastic compliance 1 / E (the reciprocal of their Young’s moduli E , since we must express the objectives in a form that requires minimization). The numbers in brackets are the 3
densities of the foams in Mg/m . The foams with the lowest thermal conductivity are the least stiff; the stiffest have the highest conductivity. Explain the reasoning you would use to select a foam for the truck panel using a penalty function.
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Exercises With Worked Solutions – Unit 7
Answer. The steps in making a reasoned choice are as follows: 1. Sketch the trade-off surface: the low λ vs. low 1/E envelope of the data, as shown below. The foams that lie on or near the surface are a better choice than those far from it. This already eliminates a large number of foams and identifies the family from which a choice should be made. Note that most metal foams are not a good choice; only if the highest stiffness is wanted is the metal foam Aluminium-SiC (1.0) an attractive choice. 2. Construct a penalty function:
1 E
Z = α 1 λ + α 2
Z is to be minimized, so α 1 is a measure of the value associated with reducing heat flow; α 2 a measure of the value associated with reducing core compliance. Rearranging the equations gives λ =
Z
α 1
−
α 2 1 α 1 E
If the axes were linear, this equation would be that of a family of straight, parallel, lines on the λ vs. 1/E diagram, of slope −α 2 / α 1 , each line corresponding to a value of Z / α 1 . In fact the scales are logarithmic, and that leads instead to a set of curved lines. One such line is sketched below for values α 2 / α 1 = 0.01 (meaning that thermal insulation is considered very important, and stiffness less important) and for α 2 / α 1 = 100 (meaning the opposite). The foam nearest the point at which the
penalty lines are tangent to the trade-off surface is the best choice. In the first 3
example PVC foam with a density of about 0.1 Mg/m is the best choice, but in the second a ceramic or even a metal foam is a better choice. Ceramic foams are brittle. This probably rules them out for the truck body because is exposed to impact loads. But in other applications ceramic foams – particularly glass foams – are viable.
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Exercises With Worked Solutions – Unit 7
Ceramic foams
Metal foams
Penalty line for α1 / α2 = 100 Polymer foams Penalty line for α1 / α2 = 0.1
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Economic Factors In design, and materials selection, the science dominates the constraints and considerations, but engineers and materials scientists cannot work in isolation from the world around them. There are environmental, social, and political factors as well, but one of the most common factors to consider when designing a product has to be the price of it relative to the budget of its target audience. In this section are examples with minimizing cost as the main objective. (See Unit 6 for questions where the environment is considered.)
65. (a) Use the Young’s modulus-Price (E – C m) chart to find the cheapest materials with a modulus, E, greater than 100 GPa. (b) Use the Strength-Price ( σ f − C m ) chart to find the cheapest materials with a strength, σ f , above 100MPa. Answer. (a) The two cheap classes of material that meet the constraints are cast irons and carbon steels. (b) Cast irons and steels are again the best choice. It is because of their high stiffness and strength at low cost that they are so widely used.
66. Plot a Modulus-Price ( E – C m) chart to find, from among the materials that appear on it, (a) The cheapest material with a modulus greater than 1 GPa (b) The cheapest metal (c) The cheapest polymer (d) whether magnesium alloys are more or less expensive than aluminum alloys (e) whether PEEK (a high performance engineering polymer) is more or less expensive than PTFE Answer. (a) concrete; (b) cast iron; (c) polypropylene; (d) magnesium alloys are more expensive than aluminum alloys; (e) PEEK is much more expensive than PTFE, which itself is an expensive polymer.
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67. It is proposed to replace the cast iron casing of a power tool with one with precisel y the same dimensions molded from nylon. Will the material cost of the nylon casing be greater or less than that made of cast iron? Answer. If the dimensions of the cast iron and nylon cases are the same, the volume of material required to make them are equal. Thus the cheaper option is the one with the lower material cost per unit volume C v , where C v = ρ C m ( ρ is the material density and C m the material cost per kg). Surprisingly, the nylon casing has a lower material cost than that made of cast iron.
68. A material is required for a tensile tie to link the front and back walls of a barn to stabilize both. It must meet a constraint on stiffness and be as cheap as possible. To be safe the 1/2
material of the tie must have a fracture toughness K 1c > 18 MPa.m . The relevant material index is M = E / C m ρ . (E is the elastic modulus, ρ the material density and C m the material cost per kg).
Construct a chart of E plotted against C m ρ . Add the constraint of adequate fracture 1/2
toughness, meaning K 1c > 18 MPa.m , using a “Limit” stage. Then plot an appropriate selection line on the chart and report the 3 materials that are the best choices for the tie. Answer. All 3 are plain-carbon steels: High carbon steel, Medium carbon steel, Low carbon steel.
69. Columns support compressive loads: the legs of a table; the pillars of the Parthenon. Derive the index for selecting materials for the cheapest cylindrical column of specified height, H, that will safely support a load F without buckling elastically. You will need the equation for the load F crit at which a slender column buckles. It is
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Exercises With Worked Solutions – Unit 7
F crit =
n
2
π
2
E I
H 2
where n is a constant that depends on the end constraints and I = π r 4 / 4 = A 2 / 4π is the second moment of area of the column. The table lists the requirements.
Function
Cylindrical column
Constraints
Length L is specified Column must support a compressive load F without buckling
Objective
Minimize the material cost of the column
Free variables
Cross-section area, A Choice of material
Answer. A slender column uses less material than a fat one, and thus is cheaper; but it must not be so slender that it will buckle under the design load, F . The objective function is the cost C = A H ρ C m
Where C m is the cost/kg of the material of the column. It will buckle elastically if F exceeds the Euler load, F crit , given in the question. Eliminating A between the two equations, using the definition of I , gives:
4 F C ≥ nπ
1/2
C ρ H 2 m 1/2 E
The material cost of the column is minimized by choosing materials with the largest value of the index 12
M =
E
C m ρ
The loading here is compressive, so brittle materials are viable candidates. Since some are also very cheap, they dominate the selection. Applying the index M to the CES EduPack Level 1 or 2 database gives the top-ranked candidates listed below.
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Exercises With Worked Solutions – Unit 7
Material
Comment
Ceramics: brick, concrete,
The low cost and fairly high modulus makes these
cement and stone
the top-ranked candidates.
Wood
Exceptional stiffness parallel to the grain, and cheap.
Carbon steel, cast iron
Steel out-performs all other materials when strength at low cost is sought.
* C m is the cost/kg of the processed material, here, the material in the form of a circular rod or column.
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Longer Questions 70. Elastic beams with differing constraints. Start each of the four parts of this problem by listing the function, the objective and the constraints. You will need the equations for the deflection of a cantilever beam with a square cross-section t x t. The two that matter are that for the deflection δ of a beam of length L under an end load F: 3
δ =
F L
3 EI
(1)
and that for the deflection of a beam under a distributed load f per unit length: 4
δ =
1 f L
8 E I
(2)
where I = t 4 / 12 . For a self-loaded beam f = ρ A g where ρ is the density of the material of the beam, A its cross-sectional area and g the acceleration due to gravity. (a) Show that the best material for a cantilever beam of given length L and given square cross-section (t x t ) that will deflect least under a given end load F is that with the largest value of the index M = E , where E is Young's modulus (neglect self-weight). (Figure a) (b) Show that the best material choice for a cantilever beam of given length L and with a given section (t x t ) that will deflect least under its own self-weight is that with the largest value of M = E/ ρ , where ρ is the density. (Figure b) (c) Show that the material index for the lightest cantilever beam of length L and square section (not given, i.e., the area is a free variable) that will not deflect by more than δ under its own weight is M = E / ρ 2 . (Figure c) (d) Show that the lightest cantilever beam of length L and square section (area free) that will not deflect by more than δ under an end load F is that made of the material with the largest value of M = E 1 / 2 / ρ (neglect self weight). (Figure d)
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Exercises With Worked Solutions – Unit 7
Answer. The point of this problem is that the material index depends on the mode of loading, on the geometric constraints and on the design objective. (a) The table lists the design requirements for part (a) of the problem. Function
End-loaded cantilever beam
Constraints
Length L specified Section t x t specified End load F specified
Objective
Minimize the deflection, δ
Free variables
Choice of material only
The objective function is an equation for the deflection of the beam. An end-load F produces a deflection δ of 3
δ =
F L
3 EI
where E is the modulus of the beam material and I = t 4 / 12 is the second moment of the area, so that the deflection becomes 3
δ = 4
F L 1 t 4
E
The magnitude of the load F and the dimensions L and t are all given. The deflection δ is minimized by maximizing M1 = E.
(b) The design requirements for part (b) are listed below Function
Self-loaded cantilever beam
Constraints
Length L specified Section t x t specified
Objective
Minimize the deflection, δ
Free variables
Choice of material only
The beam carries a distributed load, f per unit length, where f = ρ g t 2
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Exercises With Worked Solutions – Unit 7
where ρ is the density of the beam material and g is the acceleration due to gravity. Such a load produces a deflection 4
δ =
3 f L
2 E t 4
4
3 g L ρ
=
2
2 t
E
(the objective function). As before, t and L are given. The deflection is hence minimized by maximizing M 2 =
E
ρ
(c) The design requirements for part (c) are listed below: Function
Self-loaded cantilever beam
Constraints
Length L specified Maximum deflection, δ , specified
Objective
Minimize the mass, m
Free variables
Choice of material Section area A = t 2
The beam deflects under its own weight but now the section can be varied to reduce the weight provided the deflection does not exceed δ , as in the figure. The objective function (the quantity to be minimized) is the mass m of the beam m = t 2 L ρ
Using equation (2) above and putting in values for f and I in terms of t and then substituting for t (the free variable) into the mass equation, gives
m =
5 2 3 g L ρ
2 δ E
The quantities L and δ are given. The mass is minimized by maximizing M 3 =
E
ρ 2
(d) The design requirements for part (d) are listed below
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Exercises With Worked Solutions – Unit 7
Function
End-loaded cantilever beam
Constraints
Length L specified End-load F specified Maximum deflection, δ , specified
Objective
Minimize the mass, m
Free variables
Choice of material Section area A = t 2
The section is square, but the dimension t is free. The objective function is 2
m = t L ρ
The deflection is, as in part (a) 3
δ = 4
F L 1 4
t
E
Using this to eliminate the free variable, t, gives
F L5 m = 2 δ
1 / 2
ρ 1 / 2 E
The quantities F, δ and L are given. The mass is minimized by maximizing 1 / 2
M 4 =
E
ρ
From a selection standpoint, M3 and M4 are equivalent. The selection. Applying the three indices to the CES EduPack Level 1 or 2 database gives the top-ranked candidates listed below.
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Exercises With Worked Solutions – Unit 7
Index
Material choice Metals: tungsten alloys, nickel alloys, steels.
High M1 = E
Ceramics: SiC, Si 3N4, B4C, Al2O3 and AlN, but of course all are brittle. Metals: aluminum, magnesium, nickel and titanium alloys and
High M 2 =
steels all have almost the same value of E / ρ
E
ρ
Composites: CFRP All technical ceramics
High M 3 =
High M 4 =
Metals: aluminum and magnesium alloys superior to all other
E
ρ 2
metals.
1 / 2
Composites: CFRP excels
ρ
Ceramics: SiC, Si 3N4, B4C, Al2O3 and AlN
E
Case Study on a Cheap Column that must not Buckle or Crush 71. The best choice of material for a light strong column depends on its aspect ratio: the ratio of its height H to its diameter D. This is because short, fat columns fail by crushing; tall slender columns buckle instead. instead. Derive two performance equations for the material cost of a column of solid circular section and specified height H, designed to support a load F large compared to its self-load, one using the constraints that the column must not crush, the other that it must not buckle. The table below summarizes the needs:
Function
Column
Constraints
Must not fail by compressive crushing Must not buckle Height H and compressive load F specified.
Objective
Minimize material cost C
Free variables
Diameter D Choice of material
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Exercises With Worked Solutions – Unit 7
(a) Proceed as follows: •
Write down down an expression for the material cost of the column – its mass times its cost per unit mass, C m m.
•
Express the two constraints as equations, and use them to substitute for the free variable, D , to find the cost of the column that will just support the load without failing by either mechanism. (Hint: Remember the open access booklet Useful Solutions for Standard Problems contains Second Moments of Area and Buckling equations.)
•
Identify the material indices M 1 and M 2 2 that enter the two equations for the mass, showing that they are
C ρ M 1 = m σ c
and
C ρ M 2 = m E 1/2
where C m is the material cost per kg, ρ the material density, σ c its crushing strength and E its modulus.
(b) Data for six possible candidates for the column are listed in the Table. Use these to to 5
identify candidate materials when F = = 10 N and H = = 3m. Ceramics are admissible here, because they have high strength in compression.
Data for candidate materials for the column Material
Density ρ
Cost/kg
3
(kg/m )
Cm ($/kg)
Modulus E
Compression
(MPa)
strength, σ c (MPa)
Wood (spruce)
700
0.5
10,000
25
Brick
2100
0.35
22,000
95
Granite
2600
0.6
20,000
150
Poured concrete
2300
0.08
20,000
13
Cast iron
7150
0.25
130,000
200
Structural steel
7850
0.4
210,000
300
Al-alloy 6061
2700
1.2
69,000
150
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Exercises With Worked Solutions – Unit 7
(c) Below is a material chart with the two indices as axes. Identify and plot plot coupling lines for 5
selecting materials for a column with F = = 10 N and H = = 3m (the same conditions as 3
above), and for a second column with F = = 10 N and H = = 20m.
Answer. This exercise illustrates the method of solving over-constrained problems. This one concerns materials for a light column with circular section which must neither buckle nor crush under a design load F . The cost, C , is to be minimized: C =
π 4
2
D H C m ρ
where D is the diameter (the free variable) and H the height of the column, C m m is the cost per kg of the material and ρ is is its density. The column must not crush, requiring that 4 F π D
≤ σ c
2
where σ c c is the compressive strength. Nor must it buckle: 2
≤ F ≤
π
EI 2
H
The right-hand side is the Euler buckling load in which E is Young’s modulus. The second moment of area for a circular column is
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Exercises With Worked Solutions – Unit 7
I =
π
64
D 4
The subsequent steps in the derivation of performance equations are laid out below: Objective
Constraint
Performance equation
Substitute for D
Crushing constraint
C =
π 4
F f ≤
π D 2
C ρ C 1 = F H m
σ c
4
σ c
(1)
2
D H C m ρ Substitute for D
Buckling constraint F ≤
π
3
2
EI 2
H
=
π
4
D E 2
64 H
The performance equations contain the indices
2
C 2 =
1/2
2 C m ρ H (2) E 1/2
1/2
F
π
C ρ C ρ M 1 = m and M 2 = m . σ c E 1/2 ~
This is a min-max problem: we seek the material with the lowest (min) cost C which itself is the larger (max) of C 1 and C 2 . The two performance equations are evaluated in the ~
table, which also lists C = max ( C 1 , C 2 ). for a column of height H = 3m, carrying a load F 5
= 10 N. The cheapest choice is concrete.
Material
Density ρ 3
(kg/m )
Cost/kg
Modulus
Compression
Cm ($/kg)
E (MPa)
strength σc
C1
C2
~ C
$
$
$
(MPa) Wood (spruce)
700
0.5
10,000
25
4.2
11.2
11.2
Brick
2100
0.35
22,000
95
2.3
16.1
16.1
Granite
2600
0.6
20,000
150
3.1
35.0
35.0
Poured concrete
2300
0.08
20,000
13
4.3
4.7
4.7
Cast iron
7150
0.25
130,000
200
2.6
16.1
16.1
Structural steel
7850
0.4
210,000
300
3.0
21.8
21.8
Al-alloy 6061
2700
1.2
69,000
150
6.5
39.5
39.5
The coupling equation is found by equating C 1 to C 2 giving
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Exercises With Worked Solutions – Unit 7
1/2
M 2 =
π
F ⋅ 2 H 2
1 / 2
⋅ M 1 2
It contains the structural loading coefficient F/H . Two positions for the coupling line are 2
2
5
shown, one corresponding to a low value of F/H = 0.011 MN/m (F = 10 N, H = 3 m) and to a high one F/H = 2.5 MN/m2 (F = 107 N, H = 2 m), with associated solutions. 2
Remember that, since E and σ c are measured in MPa, the load F must be expressed in units of MN.
Coupling lines
Large F/H
Small F/H
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