Solution Manual For Materials Selection in Mechanical Design 6th Edition By Michael F. Ashby

Solution Manual For Materials Selection in Mechanical Design 6e By Michael F. Ashby Chapter 1-15

Materials Selection in Mechanical Design 6th edition

Exercises and Solutions

Mike Ashby, June 2024

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Materials Selection in Mechanical Design, 6th edition

Exercises with worked Solutions Contents -------------- PART 1. Materials and Process Selection: the basics -------------Ch 1. Introduction: materials and design

5

Ch 2. Engineering materials and properties

8

Ch 3. Material property charts

20

Ch 4. Material selection – the basics

38

Ch 5. Materials selection – case studies

No exercises

Ch 6. Processes and their effect on properties

61

Ch 7. Processes selection and modelling

69

Ch 8 Additive manufacturing

81

-------------- PART 2. Advanced selection methods -------------Ch 9. Multiple constraints and conflicting objectives Ch 10. Multiple constraints and conflicting objectives – case studies

91 No exercises

Ch 11. Selection of material and shape

109

Ch 12. Material and shape: case studies

No exercises

Ch 13. Designing hybrid materials

122

Ch 14. Hybrids: case studies

No exercises

-------------- PART 3. The wider context -------------Ch 15. Materials and the environment

134

Ch 16. Materials and Sustainable Development

151

Ch 17. The wider context: Materials and Big Issues

156

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Materials Selection in Mechanical Design, 6th edition

Exercises with worked Solutions Ch 1. Introduction: materials and design E1.1 Concepts (working principles). Even if you shut the windows and doors dust settles in rooms like gentle snow. The need: a device to remove dust from household floors. Here are three established concepts:  human-powered bellows pump with cloth filter;  axial-flow electric fan with muslin filter; and  centrifugal fan with centrifugal dust separation and collection.

What other concepts could fill this need? Nothing, at the concept stage, is too far-fetched. Decisions about practicality and cost come later.

Figure 1.12. Vacuum cleaners: a hand-powered cleaner from 1880, the Electrolux cylinder cleaner of 1960 and the Dyson centrifugal cleaner of 2022.

Answer. Here are some concepts for removing dust from floors. The list starts with the three in the figure to illustrate what is wanted 

Entrain dust in airstream created by hand-powered bellows; capture dust from airstream with filter.



Entrain dust in airstream created by axial electric fan; capture dust from airstream with filter.



Entrain dust in airstream created by centrifugal fan; capture dust from airstream with centrifugal separator.

Now some other possibilities.



Sweep the floor with a brush; collect dust in dust pan. It has worked for generations and still does.



Entrain dust in airstream; remove with electrostatic separator. It works for industrial emissions.



Entrain dust in airstream; remove with fine water spray to create a liquid slurry. It’s how oxides of sulphur are removed from power station emissions.



Entrain dust in airstream; remove by passing airstream across wet fabric. It’s what happens when you hang clothes out to dry and a wind stirs up the dust.



Trap dust on adhesive reel-to-reel adhesive tape rolled across the floor. Adhesive tape works well as a way of trapping mosquitos; why not dust?



Pay a cleaning agency to come and take the dust out of your house. They have their own methods.

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E1.2 Function structures. Make a function-structure diagram for the Electrolux cleaner of Figure 1.12.

Answer. The diagram is shown below.

E1.3 Concepts (working principles) for cooling power electronics. Microchips, particularly those for power electronics, get hot. If they get too hot they cease to function. The need: a scheme for removing heat from power microchips. One is sketched in Figure 1.13. Its working principle is that of thermal conduction to fins from which heat is removed by convection.

Figure 1.13 A heat sink

Heat can be removed in four ways (the working principles):    

thermal conduction; convection; exploiting a phase-change; and radiation.

The best solution for the micro-chip may be found by combining two or more of these, such as conduction coupled with convection, as in Figure 1.13. Suggest other combinations (a) for continuous-operation and (b) for short-term, intermittent operation.

Answer. Concepts for continuous or long-term operation. 

Transmission of heat by conduction; removal by natural or fan-assisted convection. (Typical system in personal computer.)



Transmission of heat by conduction; removal by radiation. (Possible system for operation in space.)



Transmission of heat by conduction, heat exchanger to pass heat to working fluid; heat removed from working fluid by convection (a radiator).



Transmission of heat by conduction; heat exchanger to pass heat to working fluid (water, for instance) drawn from renewable source (a river, for example); working fluid then rejected.



Transmission of heat by conduction, active removal by Peltier (thermoelectric) cooling.



Transmission of heat by conduction, active removal by magnetocaloric (MC) cooling.

Concepts for pulsed or short-term operation. 

Conduction of heat into heat sink so massive that the temperature rise is acceptably small.

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Conduction of heat to reserve of phase-change material (ice, paraffins, hydrated salts such as CaCl2.6H2O); the heat is absorbed by the latent heat of melting.



Conduction of heat to sacrificial evaporative coating; the heat is removed by latent heat of vaporization.

E1.4 Function structures. Make a function-structure diagram for the heat sinks.

Answer. The diagram is shown below.

E1.5 Attaching paper sheets. The need: multi-page documents can get muddled if the pages are loose.

Use the internet to explore alternative concepts for attaching a small number of pages together in a way that still allows them to be read. Record, as far as possible, the materials of the attachment and the property that provides the attachment force.

Answer.

Systems that rely on elastic stiffness (thus, elastic modulus) to provide the attachment force.

Wood

Mild steel

Spring steel

Spring steel

Polypropylene

Systems that rely on strength or plasticity to provide the attachment force.

Mild steel

Brass

Brass

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Ch 2. Engineering materials and properties These exercises introduce two useful resources: the data sheets of Appendix A and the Solutions to Standard Problems of Appendix B. When drawing data from Appendix A use the mean values of the ranges listed there.

E2.1 Estimating moduli (1). Young’s modulus E of 6061 grade aluminum is 69 GPa. and its Poisson’s ratio ν is 0.33. What is its shear modulus, G ?

Answer. Inserting the values for E and ν in the central equation (2.1) gives G  25.9 GPa. The measured value is 26 GPa, a difference of less than 1%.

E2.2 Estimating moduli (2). Young’s modulus E for butyl rubber is 0.06 GPa. What is its shear modulus, G ?

Answer. For elastomers,  0.5 Using equation (2.1), G  E / 3  0.02 GPa

E2.3 Sound velocity. The speed of sound, v s , in a solid material is approximately related to its Young’s modulus, E , and density ,  :

vs  E /  The speed of sound in pyrex (Borosilicate) glass is 5610 m/s. Find the density of this glass from Appendix A, Table A3, and use it to estimate the modulus E for the material. How does this compare with the value for E for borosilicate glass listed in Table A3?

Answer. The density  of Borosilicate glass, from Table A3 of Appendix A, is 2250 kg/m . value of Young’s modulus of the glass is 3

The calculated

E  v2s   ( 5610 )2 x2250  7.1 x 1010 Pa  71 GPa The measured value, from Table A3, is 62.5 GPa, a difference of 12%.

E2.4 Deflection of beams (1). A cantilever beam has a length L = 50 mm. It has a rectangular cross-section width b = 5 mm and a thickness t = 1 mm (Figure 2.16). It is made of an aluminum alloy. By how much will the end deflect under an end-load of F = 5 N? Find the equation for the deflection of an end-loaded beam from Appendix B, Table B3, and the (mean) value of Young’s modulus of aluminum alloys from Appendix A, Table A3, and use these to calculate the deflection.

Figure 2.16. A cantilever beam

Answer. The deflection  of a cantilever under an end load F , from Table B3, is bt 3 F L3 with the second moment of area I  12 3E I The mean Young’s modulus E for aluminum, from Appendix A3, is 75 GPa. Inserting the data from the -12 4 question gives a second moment of area I = 0.42x10 m and an end deflection  = 6.7 mm.

 

E2.5 Deflection of beams (2). The wings of a glider are each 4m long. The load-bearing member is the wing spar, a tubular beam running the length of the wing (Figure 2.17). The wing spar in this glider has a diameter of 140 mm and a wall thickness of 6 mm. It is made of an aluminum alloy. In flight the wing spar is loaded in bending with (we will assume) a uniformly distributed force per unit length. By how much will the wing-tip deflect in Figure 2.17. A tubular beam calm flight if the loaded glider weighs 1000 kg? Use data from Appendix A3 for the (mean) value of Young’s modulus of aluminum alloys, the equation for the elastic deflection of a cantilever from Appendix B3 and for the second moment of a beam from Appendix B2.

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Answer. The deflection  of a cantilever under a distributed load F is, from Appendix B, Table B3 F L3 8E I The second moment of area of a thin-walled tube of diameter 2R and wall thickness t (Table B2) is

 

I   R 3 t = 6.5 x 10-6 m4 Each wing spar carries a distributed load of 500kg (4,910 N). The mean Young’s modulus E for aluminum, from Appendix A3, is 75 GPa. Inserting the data from the question results in an end deflection  = 80 mm.

E2.6 Vibration of beams. A cantilever beam with length L = 200 mm, width b = 12 mm and thickness t = 2 mm weighs 38 grams and has a natural vibration frequency of f = 42 hertz (Figure 2.18). What is the modulus of the material of the beam? You will find equations for natural frequencies of vibration and moments of area in Appendix B, Tables B12 and B2.

Answer. From Appendix B, Table B12, the natural frequency of a cantilever beam is f

Figure 2.18. A vibrating beam

3.52 EI EI  0.56 4 2 moL moL4

The mass per unit length of the beam, mo , is mo 

0.038  0.19 kg / m 0.2

And is the second moment of area of the beam, I, is bt 3  8 x 10 12 m4 12 Inverting the equation for f and solving for the modulus gives E = 214 GPa. I

E2.7 Forensic materials science. Exercise E2.6 used a natural vibration frequency to measure the modulus of a beam. Use the information in E2.6 to calculate the density of the beam, then use this and the result of the exercise (the modulus of the beam) as a forensic tool. Scan the Modulus values in Table A3 of Appendix A seeking a match between the calculated modulus and those in the table. What subsets of materials match the calculated modulus? Of these, which subset also matches the calculated density?

Answer. The density (mass/volume) of the beam of exercise E2.6 is

  7920 kg / m3 Its modulus, calculated in exercise E2.6, is

E  214 GPa Only one subset of materials – steels – matches both the density and the modulus. Note that this identification has require only very simple measurement: the dimensions of the beam, its weight and its vibration frequency.

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E2.8 Springs. A helical spring, wound from stainless steel wire with a wire diameter d = 1.5mm, has n = 20 turns of radius R = 10 mm (Figure 2.19). How much will it extend when loaded with a mass M of 1 kg? Assume the shear modulus G of stainless steel to be 3/8 E where E is Young’s modulus. Retrieve this modulus from Appendix A, Table A3, and use the expression for the extension of springs from Appendix B, Table B6.

Answer. From Appendix B, Table B6, the extension u of a helical spring under a force F is

u

Figure 2.19. A helical spring

64 F R3n Gd 4

where R is the radius of the helix, d the diameter of the wire of which the helix is made, n the number of turns and G the shear modulus. In this exercise, the force F is

F  Mg  9.81 N

where g is the acceleration due to gravity. Young’s modulus for stainless steel is 200 GPa, so its shear modulus G  75 GPa. Inserting the data gives a deflection u = 34 mm.

E2.9 Strength from hardness. A titanium alloy has a Brinell hardness of 300. Approximately what is its Vickers hardness and yield strength?

Answer. From Figure 2.6, the Vickers hardness corresponding to a Brinell hardness of 300 is approximately Hv  300 and the corresponding yield strength is approximately 1000 MPa. E2.10 Using yield functions .A titanium alloy pipe of radius r = 20 mm and wall thickness t = 2 mm carries an internal pressure p . The pressure generates a circumferential wall-stress of  1  pr / t and an axial wall-stress

 2  pr / 2t . The yield strength  y of the titanium alloy is 500 MPa. At what pressure will the pipe first yield?

Answer. The von Mises’ yield function describes yield under multi-axial stress

 1  2 2   2  3 2   3  1 2  2 y2 where σ1 , σ 2 , and σ 3 are the principal stresses, positive when tensile;  1 is the greatest or most positive, σ 3 the smallest or least positive and  y is the yield strength in tension. Setting  2   1 / 2 and  3  0 a gives the yield condition

p* 

1  ( 2 / 3 ) y .

The pressure

p*

that just causes first yield is

2 t 2 20 y  500  58 MPa .

3r 3 2

E2.11 Torsion of tubes. A thick-walled tube has an inner radius ri = 10 mm and an outer radius ro = 15 mm (Figure 2.20). It is made from polycarbonate, PC. What is the maximum torque T f that the tube can carry without the onset of yield? Retrieve the (mean) yield strength  y of PC from Appendix A Table A3, the expression for the torque at onset of yield from Appendix B6 and that for the polar moment of a thick-walled tube from Appendix B2 to find out.

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Figure 2.20. A tube in torsion

Answer. The torque at the onset of yield for a thick walled tube is Tf 

K y

with

2 ro

 K  ( ro4  ri4 ) 2

The mean yield strength of  y of PC from Appendix A3 is 65 MPa. Inserting the data from the question gives a -8

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value for the constant K = 6.38 x 10 m and torque at the onset of yield of T f = 138 N.m.

E2.12 Stress concentrations. A round bar, 20 mm in diameter, has a shallow circumferential notch with a depth c = 1 mm and a root radius r = 10 microns (Figure 2.21). The bar is made of a low carbon steel with a yield strength of  y = 250 MPa. It is loaded axially with a nominal stress,  nom (the axial load divided by the un-notched area).

At what value of  nom will yield first commence at the root of the notch? Use the stress concentration estimate of Appendix B, Table B9 to find out.

Figure 2.21. Stress concentrations

Answer. The stress concentration caused by notch of depth c and root radius r is 1/ 2  max c with   2 for tension  1     nom r Yield first starts when  max   y . Inserting the data from the question gives a nominal stress for first yield of

11.9 MPa. Stress concentrations can be very damaging – in this example, a cyclic stress of only  12 MPa will ultimately initiate a fatigue crack at the notch root.

E2.13 Using fracture toughness. The pressure p inside a champagne bottle can reach 5 atmospheres (0.5 r r MPa). The pressure creates a radial stress  r  p and a longitudinal stress  r  p in the wall of the t 2t bottle, where r is the radius of the bottle and t the wall thickness. If the bottle has surface cracks of depth a it will fail by brittle fracture if the maximum principal stress exceeds the value

 fr 

K1c Y a

A long -established vineyard uses bottles with a radius of 50 mm and a wall thickness of 5 mm. Routine handling can generate surface scratches on the bottles deep as a = 1 micron. Is there a risk that the internal pressure will be large enough to cause fracture? Find the value of the fracture toughness, K1c , of soda lime glass from Appendix A, Table A4, (use the mean of the two values) and the value of Y from Appendix B, Table B20 to find out.

The new owner of the vineyard decides to save weight and money by reducing the wall thickness of the bottles to 1 mm. Is there now a risk of fracture?

By how much are these critical pressures reduced if the scratch is particularly severe, meaning a scratch depth of 10 microns?

1/2

Answer. The fracture toughness, K1c , of soda lime glass, from Appendix A, Table A4, is 0.625 MPa.m . The stress required to make a 1 micron surface crack ( a  10 6 m) propagate in glass with a fracture toughness of

K1c  0.625 MPa.m1/2, using equation (2.8) with Y  1.1 , is

 fr 

K1c

Y a



0.625 1.1 1x106 

 321 MPa

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r Equating the maximum principal stress in the bottle wall,  r  p , to  fr gives the maximum pressure the t bottle can contain without fracture t p  321  32.1 MPa r

The bottle is safe. If the wall thickness of the bottle is reduced to 1 mm, the safe pressure falls to 6.7 MPa, barely enough to tolerate a 1 micron scratch.

The scratch depth enters these equations as

a , so the critical pressures in the presence of a 10 micron crack

10  3.2 lower than those for a 1 micron scratch. The thick bottle is still safe. The thin one is

are a factor of unsafe.

E2.14 Thermal stress. An acrylic (PMMA) window is clamped in a thick low carbon steel frame at T = 20 C (Figure 2.22). The temperature falls to T = 20 C, putting the window under tension because the thermal expansion coefficient of PMMA is larger than that of steel. If the window was stressfree at 20C, what stress does it carry at -20 C? Use the result that the bi-axial stress caused by a bi-axial strain difference  is E   1  where E is Young’s modulus for PMMA and Poisson’s ratio   0.33 . You will find data for expansion coefficients in Appendix A, Table A6, and for moduli in Table A3, Appendix A. Use mean values.

Figure 2.22. A clamped pane

Answer. From Appendix A6, the expansion coefficients of PMMA and of low carbon steel are -6

 PMMA = 116x 10 /C

LowC steel = 12.5 x 10-6/C

and

The strain difference caused by difference in thermal expansion,  , when the temperature changes by T is





  PMMA  Low C steel T  104 x106 x 40  4.16 x 103 -3

giving   4.16 x 10 . The modulus of PMMA, from Appendix A3, is E  3.0 GPa. The equation given in the question then predicts a tensile stress in the window of  = 19 MPa.

E2.15 Unstable cracks. The PMMA window described in Exercise 2.14 has a contained crack of length 2a = 0.5 mm. If the maximum tensile stress that is anticipated in the window is  = 20 MPa, will the crack propagate?

Choose an appropriate equation for crack propagation from Appendix B, Table B10, and data for the mean value of the fracture toughness K 1c of PMMA from Appendix A, Table A4, to calculate whether this crack will propagate.

1/2

Answer. The (mean) value of K 1c for PMMA, from table A4, is 1.15 MPa.m . The crack length is small compared with the width of the window, so the appropriate choice of equation describing crack instability is

Y  a  K1c

Y  1.0

with

Inserting the data we find the length of the shortest crack that is just unstable: 2

2a 

2

2  K1c  2  1.15     2.1 x 103 m  2.1mm        20 

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Thus the 0.5mm crack will not propagate.

E2.16 Centrifugal stress. A flywheel with a radius R = 200 mm is designed to spin up to W  8000 rpm (Figure 2.23). It is proposed to make it out of ductile (nodular) cast iron, but the casting shop can guarantee only that it will have no crack-like flaws greater than 2a = 2 mm in length. Use the expression for the maximum stress in a spinning disk in Appendix B, Table B7, the stress intensity at a short, contained, crack from Table B10 and data for cast iron from Appendix A, Tables A2 and A4, to establish if the flywheel is safe. Take Poisson’s ratio  for cast iron to be 0.33.

Figure 2.23. A flywheel

Answer. The maximum tensile stress in a spinning disk of radius R , from Table B7, is

 max 

3   8

  2R 2 .

where   2 W / 60 radians/sec where W is the rotational velocity in rpm.

The stress intensity at a contained crack of length 2a , from Table B10, is

K1  Y max  a

with Y = 1 if the crack is short.

Appendix A gives the mean values of the density  = 7200 kg/m and the fracture toughness K 1c = 38 1/2 MPa.m for nodular cast iron.

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Using these and the data for disk radius R  200mm and crack length 2a = 2 mm, we find the maximum stress the material can support without fracture is K1  max   678 MPa Y a The angular velocity that will generate this stress is

max 

8  max  2380 radians / sec  22 ,700 rpm .

3  R2

The planned rotational speed of 8,000 rpm is safe.

E2.17 Blunt cracks. Liberty ships, thousands of which were constructed during World War II, were welded, not riveted. Many were mistakenly made from a steel that became brittle at the temperatures of the Atlantic in winter.

They had square hatches with sharp corners from which a crack could start and propagate across the deck which, being welded, allowed a continuous path not interrupted by riveted plates. It is said that alert seamen, observing such a crack to start, would seize a power drill and drill a hole at the crack tips, effectively blunting it by reducing the stress concentration. If the crack when the seaman first saw it was 1 m long, and the largest available drill was Figure 2.24 A cracked panel 2r  25 mm in diameter, what is the stress concentration at the end of the crack once the hole was drilled (Figure 2.24)? You will find the equation for the stress at a distance r from the tip of a crack with a length 2a Appendix B, Table B9.

Answer. A narrow slot in a wide plate, with a length 2c and end radius r , concentrates stress locally. This local stress   is greater than the nominal, remote, stress by the factor

l c  1  7.3  nom r where  us a constant, equal to 2 for tensile loading. By drilling out the material at the crack tip the largest stress is that evaluated at the surface of the hole, where r  12.5 mm.

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l 0.5  12  13.6  nom 12.5 x 103

E2.18. Contact forces. A clamp has a hemispherical contact faces of radius R = 10 mm to allow for misorientation (Figure 2.25). It is used to clamp a large lead archaeological treasure for X-ray examination. How much clamping force, F , can be applied without damage (meaning plastic deformation) to the treasured object? You will find the modulus E and yield strength  y of lead alloys in Appendix A, Table A3 (use mean values), and the equation for force required to trigger plastic deformation beneath a spherical indenter in Appendix B, Table B8.

Figure 2.25. Plasticity beneath an indenter

Answer. From Table A3, the modulus of lead alloys is 14 GPa and the yield strength is 11 MPa. The force to cause yielding below a spherical indenter, from Table B8, is

s 

F

 3 y  a2 Immediately before yielding starts, the contact is elastic and the radius of the contact-patch is a where 1/ 3

FR  a  0.7    E  Combining these equations and solving for the clamping force F gives 2

3 R F  98.5    y E  Entering values for the parameters gives a permissible clamping force of 0.067 N. This is an impossibly low clamping force so clamping the object in this way is impractical. A softer clamping system, distributing load over a much larger area, is needed.

 

E2.19. Using specific heat. How much energy is required to heat a 100 mm cube of aluminum from room o temperature (20 C) to its melting point? Retrieve the necessary data from Appendix A, Tables A2, and A5, using mean values of the ranges listed there.

Answer. Data for melting point, Tm , specific heat, C p and density  are listed in Appendix A. The mean values for aluminum are Tm  575 C, C p  925 J/kg.K and   2700 kg/m . The mass of aluminum in the o

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cube is  V  2.7 kg. The energy to heat it though T  555 oC is  V C p T  1.386 MJ E2.20. Steady state heat flow. A room-cooling system pumps cold water at 4 C through a cast aluminum 2 radiator with an area of A  2 m and wall thickness 4 mm. If the temperature of the room is 20 C and convection in the room maintains the outer surface of the radiator at this temperature, how much thermal energy does the system remove in one hour? You will find data for the thermal properties of materials in Appendix A, Tables A5 and A6. Use the means of the values listed there.

Answer. From Appendix A, Table A6 the thermal conductivity of aluminum is 158 W/m/K. The temperature o gradient is dT / dX  16 / 0.004  4 ,000 C /m. The total energy Q that passes across the area A over a time t = 3600 seconds is Q  At q  At 

dT  1.82 x 1010 J  4.6 GJ dX

E2.21. Transient heat flow. A cabin in the Mojavi desert has brick walls 200 mm thick. The temperature during the night falls to 10 C. In the morning, the outer surface of the sun-facing wall of the house quickly warms to 50 14

C. How long will it take for the interior surface to start to get warm? Data for the density, heat capacity and thermal conductivity of brick are listed in Appendix A, Tables A2, A5 and A6. Use mean values of the ranges in the table.

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Answer. From Appendix A, Tables A2, A5 and A6, the density of brick is   1850 kg/m , its heat capacity is C p  800 J/kg.K., and its thermal conductivity is   0.6 W / m.K . Its thermal diffusivity is calculated using

equation (2.14).  a  0.41 x 106 m2 / s  Cp The heat diffusion time, equation (2.15) is

x2  0.20    4.9 x 104 seconds  13.6 hours  6 2a 2 x 0.41 x 10 2

t

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E2.22 Heat exchangers. A power transistor with an area of A of 1 mm dissipates 10 watts. It is attached to a heat sink 2 mm thick, effectively cooled to 30 C on the back face by forced air flow. If the heat sink is made of aluminium nitride, how hot will the microchip be at steady state? You will find equations of heat flow in Appendix B, Table B15, and the thermal properties of aluminium nitride in Appendix A, Tables A5 an A6.

Answer. The thermal conductivity of aluminium nitride, from Table A6, is 170 W/m.K. The basic equation of 2 steady-state heat flux J (J/m .sec) (Table B15), is

J  

dT dx

where  is the thermal conductivity and dT/dx is the temperature gradient. The heat flow, q, away from the transistor is

dT T  1x10 6 x 170 x  8.5 x10 2 T watts 3 dx 2 x10 Equating this to the dissipated power, 10 watts, and solving for T gives q  AJ   A  o

T  118 C Thus the temperature of the transistor is 148 C.

E2.23 Estimating thermal properties. You need a quick estimate of the thermal conductivity  of polyethylene (PE). To find it, you block one end of a PE pipe with a wall thickness of x = 3 mm and diameter of 30 mm and fill it with boiling water while clutching the outside with your other hand. You note that the outer surface of the pipe first becomes appreciably hot at a time t  18 seconds after filling the inside with water. Use this information, plus data for specific heat C p and density  of PE from Appendix A, Tables A2 and A5, to estimate  for PE. How does your result compare with the listed value in Table A7?

Answer. The distance x that heat diffuses in a time t is approximately (equation (2.15) is  x  2 at with a  Cp ( a is the thermal diffusivity). Combining the two equations and solving for  gives

x2  Cp 2t 3 Inserting the data from the question and the mean values  = 950 kg/m and C p = 1850 J/kg/K from

 

Appendix A , we find   0.44 W/m.K. The result given in Appendix A6 for the thermal conductivity of PE is 0.40 – 0.44 W/m.K. E2.24 Choosing dielectrics. The capacitance C of a condenser with two plates each of area A separated by a dielectric of thickness t is

15

C  ro

A t

where  o is the permittivity of free space and  r is the relative dielectric constant of the material between the plates. Select a dielectric by scanning data in Appendix A, Tables A7 and Table A8, (a) to maximize C and (b) to minimize it, for a given A and t.

Answer. (a) Capacitance is maximized by selecting materials with high  r . Appendix A shows that most polymers and ceramics have dielectric constants in the range 2 – 10, the ferroelectric ceramics barium titanate, lead zirconium titanate (PZT) and lead lanthanum zirconium titanate (PLZT) all have dielectric constants greater than 1000. Any one of these would be a good choice. (b) Capacitance is minimized by selection materials with low  r . The non-polar polymers polyethylene, polypropylene and teflon (PTFE) have the lowest values among solid materials. Polymer foams have even lower dielectric constants because they are mainly air (dielectric constant 1)

E2.25 Dielectric losses. Printed circuit boards (PCBs) are made of a rigid dielectric, typically glass-reinforced epoxy or polyimide. When a signal propagates along a conductor on a PCB, it is attenuated by dielectric loss in the dielectric that surrounds it. The loss increases with the loss factor L   r tan(  ) where  r is the dielectric constant and tan δ is the loss tangent. Data for Young’s modulus, dielectric constant and dielectric loss tangent are assembled in the table below. What, typically, are the values of E and L for epoxy and polyimide PCBs? Wearable electronics require that the PCB be flexible, which we will interpret as a low value of Young’s modulus, E. Which materials combine a low loss factor with a low value of Young’s modulus, making them candidates for flexible electronics?

Material

Youngs modulus E (GPa)

Dielectric constant  r (-)

Loss tangent tan δ (-)

Air Alumina, Al2O3 Bakelite, PH Concrete, dry Epoxy, EP Epoxy-glass PCB Glass, borosilicate Glass, soda-lime Paper Polyamide, PA (Nylon) Polyester Polyethylene, PE Polyimide Polyimide-glass PCB Teflon, PTFE Water, distilled Wood, balsa

365 3.8 20 2.75 21.5 62.5 70 3.0 2.9 9.2 0.76 2.4 25.0 0.48 6.0

1.0006 9.0 4.8 4.5 4.4 4.1 3.8 6.0 3.5 3.8 4.0 2.25 4.1 4.4 2.1 77 1.22

0.0 0.0006 0.0002 0.011 0.02 0.022 0.0009 0.02 0.02 0.026 0.005 0.0001 0.015 0.014 0.0002 0.16 0.1

Answer. The table is expanded below to include the values of dielectric loss factor L   r tan(  ) . Epoxy and polyimide-based PCBs have very similar values of E (  25 GPa) and L (0.02).

Two materials, polyethylene and PTFE (Teflon), have exceptionally low modulus and dielectric loss factor. Both are possible candidates for flexible electronics.

16

Material

Youngs modulus E (GPa)

Air Alumina, Al2O3 Bakelite, PH Concrete, dry Epoxy, EP Epoxy-glass PCB Glass, borosilicate Glass,soda-lime Paper Polyamide, PA (Nylon) Polyester Polyethylene, PE Polyimide Polyimide-glass PCB Teflon, PTFE Water, distilled Wood, balsa

Dielectric constant (-)

Loss tangent tan δ (-)

Loss factor  r tan δ (-)

1.0006 9.0 4.8 4.5 4.4 21.5 3.8 6.0 3.5 3.8 4.0 2.25 4.1 25.0 2.1 77 1.22

0.00001 0.0006 0.0002 0.011 0.02 4.1 0.0009 0.02 0.02 0.026 0.005 0.0001 0.015 4.4 0.0002 0.16 0.1

0.00001 0.0054 0.001 0.050 0.088 0.022 0.0034 0.12 0.07 0.10 0.02 0.0002 0.06 0.014 0.0004 12.3 0.122

365 3.8 20 2.75 21.5 62.5 70 3.0 2.9 9.2 0.76 2.4 25.0 0.48 6.0

E2.26 Dielectric heating. A component made of PA46, a general-purpose grade of nylon, is placed in a microwave cavity with a field strength E  2 x 103 V/m and a frequency f  0.5 x 1010 Hz for a time t = 60 seconds at room temperature (20 C). The dielectric loss factor for nylon is exceptionally high at L  4.5 , its 3 density is   1 ,020 kg/m and its heat capacity is C p  1650 J/kg.K. Assuming no heat loss, how hot will the component become?

Answer. The energy dissipated by the field (equation 2.24) is

Q  f E 2 o Lt J/m3 where  o is the permittivity of free space (8.85 x 10 F/m). The energy required to raise the temperature of the PA46 by T is C p  T . Equating this to the dissipated energy Q gives -12

T 

f E 2 o Lt  28 oC Cp 

The component will heat to 28 + 20 = 48 C. E2.27 Piezoelectric actuators. Piezoelectric materials respond to an electric field by change of shape. The strain  induced by a field E is

  d33 E where d33 is the piezoelectric charge coefficient. Its units are pm/V. A sample of soft PZT (lead zirconium titanate) is exposed to a field of 10 MV/m. How large is the resulting strain? You will find the piezoelectric charge coefficient for soft PZT in Appendix A, Table A8. Use a mean value.

Answer. The piezoelectric charge coefficient d 33 for soft PZT is 675 x 10 -3 will induce a strain of   6.75 x 10 = 0.675%.

-12

6

m/V. Thus a field of 10 x 10 V/m

E2.28 Pyroelectric temperature sensing. Remote, non-contact, temperature sensing is possible by focussing radiation from the source onto a pyroelectric material, using the charge that appears on its surface as a measure of temperature. A sensor is required to measure temperature in the range 100 – 200 C. To work in this range the material must have a ferro-electric Curie temperature greater than 200 C (if it doesn’t, it won’t 17

be pyroelectric in the sensing range). The signal is maximized by selecting a material with the largest possible pyroelectric coefficient. Use the data for pyroelectrics in Appendix A, Table A8, to make a selection.

Answer. Lithium niobate has the largest pyroelectric coefficient of those listed in the table but its ferro-electric Curie temperature is too low to allow its use in the range 100 – 200 C. Lithium tantalite best meets the criteria for the sensor described in the question.

E2.29 Magnetostrictive actuation. A magnetostrictive actuator has an active core of length L = 20 mm. It is made of Terfenol D, a giant magnetostrictive alloy of terbium, dysprosium and iron. If a field sufficiently large to saturate the core is applied, by how much will the core change in length? Use the mean value of the data for Terfenol D listed in Appendix A, Table A9, to find out.

-6

Answer. The mean saturation magnetostrictive strain  sat for Terfenol D, from Table A9, is -1,700 x 10 . The negative sign means that the material shrinks when magnetized. Thus the core will change in length by -6

   sat L = 1,700 x 10 x 20 x 10

-3

= 3.4 x 10 m = 34 m -5

E2.30 Material price. It is proposed to replace the ductile (nodular) cast iron casing of a power tool with one with exactly the same dimension molded from nylon. Will the material cost of the nylon casing be greater or less than that made of cast iron? Use data from Appendix A, Table A2, to find out.

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 Cv   C m

Here  is the material density and Cm is the material cost per kg. Data from Table A2 are assembled below, using the means of the ranges.

Material Cast iron Nylon

Density 3 ρ (kg/m ) 7,200 1,100

Price per unit weight Cm ($/kg) 0.40 5.3

Cost per unit vol 3 Cv = ρCm ($/m ) 2,880 5,830

The nylon casing has a greater material cost than that made of cast iron.

E2.31. Carbon footprint. Plastics are sometimes portrayed as environmentally poor because most are synthesized from oil or natural gas. Is this an accurate portrait? Compare the carbon footprint of three of the materials most widely used in household products: polypropylene, aluminium and carbon steel, using data from Appendix A, Tables A11 and A2 (use mean values of the ranges in the tables). Which has the lowest carbon footprint per unit weight? If one material is substituted for another without change of shape or size, the relevant comparison is the carbon footprint per unit volume. On this basis, which material has the lower carbon profile?

Answer. The table lists the carbon footprint per kg, CO2,eq and that per unit volume, ρCO2,eq, where ρ is the density. Carbon steel has the lowest carbon footprint per unit weight, polypropylene the next lowest. If ranked instead by carbon footprint per unit volume, polypropylene has by far the lowest value. Any serious comparison of this type needs a more detailed examination: materials are used in products because they perform a function. If these three materials are candidates, the best choice is that with the lowest carbon footprint per unit of function. We explore ways to do this in later chapters of this book.

18

Material Polypropylene Aluminum Carbon steel

Density, 3) ρ (kg/m 900 2700 7850

Carbon footprint CO2,eq, (kg/kg) 2.9 12 2.3

Carbon footprint 3 ρCO2,eq (kg/m ) 2,610 32,400 18,060

You will find much more about embodied energy and carbon footprint in Chapter 15.

19

Ch 3. Material property charts These 32 exercises help develop fluency in the use of the charts. They are answered by placing selection lines on the appropriate chart and reading off the materials that lie on the appropriate side of the lines. All can be solved by using the printed charts, which you are free to copy for your personal use. If the Ansys Granta Materials Selection software1 is available, it can be used to solve the same exercises.

E3.1 Using the E   chart (1). Use Figure 3.3 to identify one metal and one polymer with a longitudinal wave speed close to 1000 m/s.

Answer. Lead and polypropylene both lie near the contour of longitudinal wave speed of 1000 m/s.

E3.2 Using the E   chart (2). A component is at present made from a brass, a copper alloy. Use the Young’s modulus – Density ( E   ) chart of Figure 3.3 to suggest three other metals that, in the same shape, would be stiffer. “Stiffer” means a higher value of Young’s modulus.

Answer. Metals that are stiffer than brass are listed in the table.

Material Steels Nickel alloys Tungsten alloys

Comment The cheapest stiff, strong structural metal, widely used.

More expensive than steel Refractory (high-melting) and relatively expensive

E3.3 Using the E   chart (3). Use the Young’s modulus – Density ( E   ) chart of Figure 3.3 to identify materials with both a modulus E > 30 GPa and a density  < 2000 kg/m .

3

Answer. There are only two materials on the chart with modulus E > 50 GPa and density  < 2000 kg/m3.

Material Magnesium alloys CFRP – carbon-fiber reinforced plastic

Comment Magnesium is the lightest of all common structural metals – only beryllium is lighter, but it is very expensive and its oxide is toxic.

CFRP is both lighter and stiffer than magnesium. That is one reason it is used for competition cars and bikes.

E3.4 Using the E   chart (4). Use the Young’s Modulus-Density ( E   ) chart of Figure 3.3 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.

1 https://www.ansys.com/en-gb/products/materials/granta-edupack

20

Answer. (a) No metals in Figure 3.3 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, alumina Al203 and (inevitably) diamond.

Material Alumina Al203 Silicon nitride Si3N4 Boron carbide, B4C Silicon carbide, SiC Diamond

Comment Alumina is the most widely used of all technical ceramics (spark plugs, circuit boards…) All ceramics are brittle – they have low values of fracture toughness K 1c and toughness G 1c .

They are, however, exceedingly hard, so are used as cutting tools.

E3.5 Using the E   chart (5). Use the ( E   ) chart of Figure 3.3 to identify metals with both E > 100 GPa 3 and E /  > 0.02 GPa/(kg/m ).

Answer. The chart shows the selection. Draw a line across the chart at E  100 GPa and a second line with a slope parallel to the Guide line for E /  shown at the lower right of the chart, and position it so that it passes through the point E  0.2 GPa,   10 kg / m3 , corresponding to E /  = 0.02 GPa/(kg/m3). The metals that meet both criteria are those that lie above both lines, in the region marked “Search area”. They are listed in the table.

Material Steels Nickel alloys Titanium alloys

Comment Cheap, widely used. Stiff structural material.

More expensive than steel Titanium alloys are very expensive.

E/ = 0.02 GPa/(kg/m3) Search area

E = 100 GPa

21

E3.6 Using the E   chart (6). Use the E /  chart of Figure 3.3 to identify materials with both E > 100 GPa and E 1 / 3 /  > 0.003 (GPa) /(kg/m ). Remember that, on taking logs, the index M  E 1 / 3 /  becomes log( E )  3 log(  )  3 log( M ) 1/3

3

and that this plots as a line of slope 3 on the chart. The Guide line for E 1 / 3 /  at the lower right of the chart shows the slope.

Answer. The chart shows the selection. Draw a line across the chart at E  100 GPa and a second line with a slope parallel to the Guide line for E 1 / 3 /  , and position it so that it passes through the point E = 27 GPa 3 1/3 3 when  = 1000 kg/m , corresponding to E 1 / 3 /  = 0.003 (GPa) /(kg/m ). The materials that meet both criteria are those that lie above both lines, in the region marked “Search area”. They are listed in the table. No metals survive.

Material CFRP Boron carbide, B4C Diamond

Comment Carbon-fiber composites excel in stiffness at low weight.

Boron carbide is exceptionally stiff, hard and light; it is used for body armour.

Diamond just squeezes in

E1/3/ = 0.003 (GPa1/3)/(kg/m3)

Search area

E = 100 GPa

Slope 3

22

E/ = 0.02

E3.7 Using the E   chart (7). Use the E   chart of Figure 3.3 to establish whether woods have a higher specific stiffness E /  than epoxies.

Answer. Draw a line parallel to the E /  Guide line at the lower right of the chart. Slide it upwards until it passes through the tiny bubble for epoxies and observe the positions of the two bubbles for woods, on describing properties parallel to the grain, the other, across the grain. Parallel to the grain, woods have much higher specific stiffness than epoxies. Perpendicular to the grain, woods have about the same value as epoxies.

E/ (GPa/(kg/m ) x 10 3

Material

E3.8

Woods parallel to the grain Woods transverse to the grain

8 - 29 0.7 – 4.0

Epoxies

1.8 – 2.5

3

Using the  f   chart (1). Do titanium alloys have a higher or lower specific strength

(strength/density,  f /  ) than the best steels? This is important when you want strength at low weight (landing gear of aircraft, mountain bikes). Use the  f /  chart of Figure 3.4 to decide.

Answer. The guide line for specific strength,  f /  , on the strength – density chart of Figure 3.4, is shown drawn through the middle of the bubble for titanium alloys. It passes above the elongated bubble for steels.

This means that titanium alloys have a higher specific strength than steels steel, even though the best steels are as strong.

 f /  (MPa/(kg/m3)) x 103

Material Titanium alloys

54 - 270

Steels

25 - 170

Search area

23

E3.9 Using the  f   chart (2). The design of a cycle safety helmet requires a lining that will crush at a stress of 6 – 10 MPa and be as light as possible. Use the  f   chart of Figure 3.4 to select an appropriate class of material to meet this need.

Answer. The required strength range is shown on the chart below – the materials between the two lines yield or crush in the range we want. The lightest are those furthest to the left. Rigid polymer foam, at the upper end of the range of densities in which rigid foams are available, would be a good choice.

Desired range of f = 6 – 10 MPa

E3.10 Using the  f   chart (3). The undercarriage of a cargo plane requires a metal with a compressive strength above 200 MPa and a high bending strength per unit weight, requiring a high value of the index  2f / 3 /  . What metals does the chart of Figure 3.4 suggest?

Answer. Setting a limit of  f = 200 MPa on the strength – density chart of Figure 3.4, then positioning a selection line with a slope of 3/2 corresponding to  2f / 3 /  so that only a few metals remain above it leaves high strength titanium, aluminum and magnesium alloys.

24

Search area = 200 MPa

E 3.11 Using the E   f chart. Use the modulus-strength E   f chart of Figure 3.5 to find materials that have E > 10 GPa and  f > 1,000 MPa. Answer. The limits are shown on the chart. The strongest steels, titanium alloys (Ti-alloys) and carbon fiber reinforced polymers (CFRP) meet these limits.

Material High strength steels High strength nickel alloys Titanium alloys CFRP Diamond

Comment All have  f above 1000 MPa, a very large value

f = 1000 MPa

Search area

E = 10 GPa

25

E3.12 Using the E /    f /  chart. Transport systems require materials that are stiff, strong and light. Use the Specific modulus E /  – Specific strength  f /  chart (Figure 3.6) to identify materials that meet these criteria. Eliminate all ceramics on the grounds that they are brittle. Which material has the best combination of both specific modulus and specific strength? Which metal has the highest specific modulus? Which has the highest specific strength?

Answer. The materials of interest lie in the box at the upper right. The table lists five classes of material that have high values of both indices. Carbon fibre reinforce plastic (CFRP) is best by both criteria. Among metals, aluminum alloys have the highest specific modulus; titanium alloys have the highest specific strength.

Material

E/ (GPa/(kg/m ) x 10

CFRP Titanium alloys Aluminum alloys Magnesium alloys High strength steels

Up to 100 Up to 25 Up to 30 Up to 25 Up to 28

3

3

f/ (MPa/(kg/m3)) x 103 Up to 700 Up to 280 Up to 200 Up to 210 Up to 190 f/ = 0.1 MPa /(kg/m3)

Search area

E/ = 0.01 GPa /(kg/m3)

E3.13 Using the K1c  E chart (1). Do the common polymers polycarbonate, PC, acrylonitrile butadiene styrene, ABS, and polypropylene, PP, have values of fracture toughness, K1c, that are larger or smaller than the engineering ceramic alumina, Al2O3? Are their toughnesses G1c  K12c / E larger or smaller? The K1c –E chart, Figure 3.7, will help.

Answer. Most polymers have values of fracture toughness, K1c, that are about the same that of alumina.

Their toughnesses, G1c  K12c / E , however, are much larger. Even the most brittle of polymers, polystyrene, has a toughness G 1c that is nearly ten times greater than that of alumina. It is this that makes ABS, in particular, the material of choice for the casings of phones and power tools. The values in the table below are read from Chart 3.7, 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.

26

1/2

Material Polycarbonate ABS Polystyrene Alumina Al2O3

2

K1c (MPa.m )

G1c (kJ/m )

2.1 – 4.6 1.2 – 4.3 0.7 – 1.1 3.3 – 4.8

0.5 - 8 1-8 0.25 – 0.7 0.04 – 0.07

Alumina, Al2O3

PC, PP and ABS

E 3.14 Using the K1c  E chart (2). The bumpers of cars have to be tough and light. Compare the use of aluminium and of polypropylene for car bumpers. First read the approximate toughness, G1c for the two materials from the chart of Figure 3.7. Then divide each of these by the density of the material (read from any of the charts with density as an axis or from Appendix A, Table A2) to get a measure of toughness per unit mass. What conclusions can you draw?

Answer. The table lists values. The two materials are about equally tough, but polypropylene, having a lower density, offers more toughness per unit weight. There is, however, more to choosing materials for bumpers that just this. We return to the challenge in later chapters. 

Material

Approx. Toughness, 2 G1c (kJ/m )

Approx. density, (kg/m )

G1c /  x 10 2 3 (kJ/m )/(kg/m )

Aluminum alloys

10

2700

3.7

Polypropylene

10

900

11

3

3

E 3.15 Using the K1c  E chart (3). Use the fracture toughness-modulus chart (Figure 3.7) to find materials that have a fracture toughness K 1c greater than 40 MPa.m 3 contours on Figure 3.7) greater than 10 kJ/m .

27

1/2

and a toughness G1c  K12c / E (shown as

Answer. The chart is reproduced below with the two criteria plotted on it. The table lists the results. Only metals have both high fracture toughness K1c and high toughness G 1c . That is one reason that they are used for pressure vessels (boilers, submarine hulls, gas containers etc).

Material

Comment

High strength steels All have high K 1c and high G 1c

High strength nickel alloys Titanium alloys

Gic = 10 kJ.m2 Search area Kic = 40 MPa.m1/2

E3.16 Using the K1c  E chart (4). The elastic deflection at fracture (“resilience”) of an elastic-brittle solid is proportional to the failure strain,  fr   fr / E , where  fr is the stress that will cause a crack to propagate:

 fr 

K 1c

c

Here K1c is the fracture toughness and c is the length of the longest crack the materials may contain. Thus

 fr

1  K1c    E c E  Materials that can deflect elastically without fracturing are therefore those with large values of K 1c / E . Use 1/2 the K1c –E chart of Figure 3.7 to identify the class of materials with K 1c > 1 MPa.m and high values of K 1c / E . The Design guideline gives the slope of a K 1c / E contour. Draw a line parallel to this one and position it such that all metals are excluded. Which materials have the highest values?

 fr 



Answer. The criteria are plotted on the chart below. 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 1/2 fracture strains. Only polymers have, additionally, K 1c > 1 MPa.m . This makes them attractive for snaptogether parts that must flex without failing.

28

Kic / E =

1.5 x 10-3 m1/2

Search area

Kic = 1 MPa.m1/2

E 3.17

Using the K1c   f chart (1). 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 then 2  4 1  K Ic  p   R   f  where  f is the elastic limit, K1c is the fracture toughness, R is the vessel radius. The pressure is maximized by choosing the material with the greatest value of

M 

K 12c

y

Use the K 1c   f chart of Figure 3.8 to identify three metallic alloys that have particularly high values of M. Answer. A line corresponding to a value K 12c /  y is plotted on the K 1c   f below. It is position so that only a few materials lie above it. All are metals. Those with the highest values are listed in the table.

Material

Comment

Low alloy steels

Traditional material for pressure vessels.

Stainless steels

Used for nuclear pressure vessels.

Copper alloys

Small boilers are made of copper.

Nickel alloys

Reactors for chemical engineering and turbine combustion chambers are made of nickel-based alloys.

29

(Kic)2 / f

Search area

E3.18

Using the K1c   f chart (2). A material is required for the blade of a rotary lawn-mower. Cost is a 1/2

consideration. For safety reasons, the designer specified a minimum fracture toughness K1c > 40 MPa m . for the blade. The other mechanical requirement is for high hardness, H, to minimize blade wear. Hardness, in applications like this one, is related to strength:

H  3 y where  f is the strength (Chapter 2 gives a fuller definition). Use the K 1c  f chart of Figure 3.8 to identify 1/2

three materials that have K1c > 40 MPa m and the highest possible strength. To do this, position a " K1c " 1/2 selection line at 40 MPa m and then adjust a "strength" selection line such that it just admits three candidates. Use the Cost per unit volume chart of Figure 3.21 to rank your selection by material cost, then make a final selection.

Answer. Applying the property limit K 1c > 40 MPa m , and positioning the “strength” line at  f  800 MPa, 1/2

gives three groups of materials. Chart 3.21 identifies the price of each of these per unit volume. The steels are far cheaper than the other two.

Material

Comment

High-strength steels

Traditional material for blades. Price  $500/m

Nickel alloys

Meets the requirements, but more expensive than steel. Price  $150,000/m

3

Titanium alloys

Meets the requirements, but more expensive than steel. Price  $100,000/m

3

30

3

f = 800 MPa

Search area

Kic = 40 MPa.m1/2

E3.19 Using the   E chart (1). Bells ring because they have a low loss (or damping) coefficient,  ; a high damping coefficient gives a dead sound. Use the Loss coefficient – Modulus (   E ) chart of Figure 3.9 to identify material that should make good bells.

Answer. Materials with low loss coefficient, any of which could be used to make a bell, are listed below.

Material

Comment

Copper alloys

The traditional material for bells: bronzes and brasses

Soda glass

Glass makes excellent bells

Silica glass

Even lower damping than soda glass

Ceramics: Al2O3, SiC, Si3N4

Unusual choice, expensive, but should work.

Very high damping materials High damping and high modulus

Very low damping materials

31

E3.20. Using the   E chart (2). A foundation is needed as a mounting for vibration-sensitive measuring equipment. It is suggested that the foundation be made from concrete because of its ability to damp vibration.

Is this assumption justified? Use Figure 3.9 to find out.

Answer. Yes, concrete has a relatively high loss coefficient of 0.02. It is a better choice than – say – cast iron or aluminium. Better might be concrete (for structural strength) with a layer of PE or PTFE or, still better, lead.

E3.21. Using the   E chart (3). The track of a tracked vehicle is made up of hardened steel segments bonded to rubber connectors. When the vehicle is in motion the rubber is subjected to a tension-compression cycle as the track passes over the front and rear drive sprockets. If the vehicle travels at 20 kph, each unit of the rubber is stretched and compressed roughly 1 per second. The thermal conductivity of rubber is extremely low. If the maximum strain  max in the rubber is  0.4 and no heat is lost from it, how hot will the rubber get after 30 minutes? The volume specific heat C p  of the rubber used for the track is 2.5 x 10 J/m .K, its 6

3

modulus E is 2x 10 GPa and its loss coefficient  is 0.5.

-3

Answer. The elastic energy stored in rubber at peak stretch or compression is Umax 

1 5 3 E  max 2  1.6 x 10 J/m 2

The energy dumped into the rubber per cycle is 5

U  2  Umax  5 x 10 J/m

3

giving a temperature rise per cycle of

T 

U o = 0.2 C. Cp

In 30 minutes each rubber pad see 1,800 cycles. If no heat is lost, the temperature rise is o

T30 min  1800 T = 360 C.

The overheating of tank tracks because of heating of this sort is a real problem, requiring the development of low-loss rubbers.

32

E3.22. Using the   E chart (4). Use the Loss coefficient-Modulus (   E ) chart (Figure 3.9) to find metals 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.

Material

E3.23

Comment

Lead alloys

Used to clad buildings to damp sound and vibration

Magnesium alloys

Used to damp vibration in machine tools

Using the    e chart (1). Use the Thermal conductivity – Electrical resistivity (    e ) chart

(Figure 3.10) to find three materials other than diamond with high thermal conductivity,  , and high electrical resistivity,  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 meet the requirements

Silicon carbide, SiC

High ρe

High λ

33

E3.24

Using the    e chart (2). It is easier to measure the electrical resistivity of a metal than to

measure its thermal conductivity. A new alloy has an electrical resistivity  e of 55 .cm . Use the Weidemann-Franz relationship, plotted on Figure 3.10, to estimate its thermal conductivity,  .

Answer. The equation of the line describing the Wiedemann-Franz relationship is

 

2000

e

with  in W/m.K and  e in .cm. Inserting the value for  e gives   36 W/m.K

E3.25 Using the   a chart (1). A novel strontium-silica glass has a thermal conductivity  of 2.1 W/m.K (exceptionally high for a glass). What, approximately, would you expect its thermal diffusivity a to be? Use the empirical relationship between  and a to find out, then examine where it would lie on Figure 3.11.

Answer. The approximate relationship between  and a (equation 3.16 of the text) is

  3 x 106 a (  in W/m.K and a in m /s). Applying it gives a thermal diffusivity for the glass of 0.7 x 10 m /s. The novel glass has values of  and a that lie well above soda-lime glass on Figure 3.11, reproduced below.

2

-6

2

Strontiumsilica glass

E3.26 Using the   a chart (2). A new design of frying pan is made from glass-ceramic 9606, a material with excellent thermal shock resistance originally developed for missile radomes. The pan bottom is 4.2 mm thick. When placed on a gas cooker with a pat of butter in it, it is observed that it takes 7 seconds before the butter starts to melt. Use the rules of thumb of Chapters 2 and 3 to estimate, first, the thermal diffusivity of glass-ceramic 9606, and from this, its thermal conductivity.

34

Answer. If the time t taken for heat to penetrate a distance x into a material, then the thermal diffusivity of the material, from equation (2.15), is approximately

a

x2 -6 2 = 1.26 x 10 m /s 2t

Using the approximate relationship of equation (3.16) relating a and 

  3 x 106 a 2

(  in W/m.K and a in m /s) we estimate the thermal conductivity of glass ceramic 9606 to be   3.8 W/m.K. The measured thermal conductivity of glass ceramic 9606 is 3.4 W/m.K. E3.27 Using the    chart. 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  (to conduct the heat away) and a low expansion coefficient  (to reduce thermal strains), that is, by seeking a window material with a high value of M   /

Use the    chart of Figure 3.12 to identify the best material for an ultra-high powered laser window.

Answer. The chart shows three transparent materials with high  /  : borosilicate glass (Pyrex), silica glass and alumina, which, in single crystal form (sapphire) or ultra-fine-grained form (“Lucalox”) is transparent and hard. Diamond has an exceptionally high value: it has been used for ultra-high-powered laser windows.

Material Borosilicate glass

Comment 6

M = 0.3 x 10 W/m; poor resistance to thermal pulse 10 6

Silica glass

M = 2.3 x 10 W/m; much better than soda glass

Alumina (sapphire)

M = 4.2 x 10 W/m; a little better than silica glass

Diamond

M = 700 x 10 W/m; outstanding

6

6

35

6

E3.28 Using the   E chart. An external panel of a satellite is made of the Grade 9 titanium alloy Ti-3Al-2V, with a yield strength  y of 550 MPa and Poisson’s ratio  of 0.36. The panel heats up when the face of the satellite is directly exposed to the sun and cools when the sun is obscured by the earth. The panel is rigidly constrained around its edges by attachment to the frame of the satellite, which is maintained at a constant temperature in order to stabilize the electrics it contains. If the temperature of the panel can vary from -80 C to +120 C, is there any risk that the resulting thermal stress might cause the panel to yield? Read the value of the thermal stress parameter  E for titanium alloys from Figure 3.13 and use it to find out. (Here  is the thermal expansion coefficient and E is Young’s modulus.)

Answer. The thermal stress  in a biaxially-constrained panel caused by a temperature change T is, from Equation 3.21,

 

E T 1 

The value of the thermal stress parameter  E for titanium alloys, read from Figure 3.13, is almost exactly 1 o MPa/K. Thus a temperature change of 200 C could potentially generate a stress of 313 MPa. This is a large stress but still comfortably below the yield strength of the alloy.

E3.29 Using the Tmax chart. Use the Maximum service temperature (Tmax) chart (Figure 3.14) to find polymers 0 that can be used above 200 C.

Answer. The chart shows just three classes of polymer with maximum service temperatures greater than 200C. They are listed below.

Material

Comment

Polytetrafluorethylene, PTFE

PTFE (Teflon) is used as non-stick coatings for cooking ware, easily surviving the temperatures of baking and frying.

Silicone elastomers

Silicones are polymers with a Si-O-Si chain structure instead of the CC-C chain of polyolefins. They are more stable than carbon-based polymers, but expensive.

Polyether ether ketone (PEEK)

PEEK can be used up to 300°C. It has high tensile and flexural strength. It is used for medical devices, components in car transmissions and aircraft structures.

E3.30 Using the Br  Hc chart. Which class of magnets has the highest energy product? Which has the lowest? Use the chart of Remanent induction and Coercive force, Figure 3.18, to find out, using the product Br.Hc as a metric.

Answer. The diagonal contours on Figure 3.18 show the product Br.Hc. Neodymium-iron-boron and Samariumcobalt magnets have the highest values. Permalloys and soft ferrites have the lowest.

36

E3.31 Using the Magnetostriction chart. If magnetostrictive strain is a linear function of the field H, what strain would appear in a Galfenol rod if it was subjected to a field of 100 A/m? Use data read from the chart of Figure 3.19 to find out.

-6

Answer. The chart suggests that a field of 800 A/m causes Gafenol to strain by 250 x 10 (0.025%). A field of 100 A/m might produce a strain of 1/8 of this, or about 0.003%.

E3.32. Material cost comparison. Window frames are made from extruded aluminum. It is argued that making them instead from extruded PVC would reduce the material cost per frame. If the section shape and thickness of the aluminum and the PVC windows are the same, is the claim justified? The approximate material prices per unit volume is plotted in Figure 3.21. Alternatively, the approximate prices and densities for the two materials can be retrieved from Appendix A. Table A2.

Answer. The volumes of material in the two window frames are the same. Taking density and price values from Table A2: 

3

For aluminum, the material price is C m = 2.6 $/kg and density  = 2,700 kg/m , so the cost per unit volume. C m  = 7.0 x 10 $/m .

3



3

3

For PVC, the material price is C m = 2.6 $/kg and density  = 1450 kg/m , so the cost per unit volume is C m  = 3.8 x 10 $/m .

3

3

The cost of the material for the aluminum window frame is about twice that of the material for the PVC frame.

37

Ch 4. Material selection – the basics These exercises explore Translation and the derivation and plotting of Material Indices.

Translation is the task of re-expressing design requirements in terms that enable material and process selection.

Tackle the exercises by answering the questions in this table. Don’t try to model the behavior at this point (that comes in later exercises). Just think out what the component does, and list the constraints that this imposes on material choice, including processing requirements.

Function



What does component do?

Constraints



What essential conditions must be met?

Objective



What is to be maximized or minimized?

Free variables



What parameters of the problem is the designer free to change?

The distinction between constraints and objectives is important. A constraint is an essential condition that must be met, usually expressed as a limit on a material or process attribute. An objective is a quantity for which an extremum (a maximum or minimum) is sought, frequently cost, mass or volume, but there are others, several of which appear in the exercises below. As an example, a bicycle frame must have a certain stiffness and strength. If it is not stiff and strong enough it will not work, but it is never required to have infinite stiffness or strength. Stiffness and strength are therefore constraints that become limits on modulus, elastic limit and shape. If the bicycle is for sprint racing, it should be as light as possible – if you could make it infinitely light, that would be best of all. Minimizing mass, here, is the objective, perhaps with an upper limit (a constraint) on cost. If instead it is a shopping bike to be sold through supermarkets it should be as cheap as possible – the cheaper it is, the more will be sold. This time minimizing cost is the objective, possible with an upper limit (a constraint) on mass. For most bikes, of course, minimizing mass and cost are both objectives, and then trade-off methods are needed. They come in Chapter 9. For now, use judgement to choose the single most important objective and make all others into constraints.

Two precautionary constrains are worth bearing in mind. Many applications require sufficient fracture toughness for the component to survive mishandling and accidental impact during service. A material like glass might meet the constraints on stiffness, strength, cost and recyclability but would present a risk because its brittleness. The first precautionary constraint is that of “adequate toughness”. This is achieved by requiring that the fracture toughness K 1c  15MPa.m 1 / 2 . Other applications require some ductility, sufficient to allow stress redistribution under loading points, and some ability to bend or shape the material plastically. This is achieved by imposing a second precautionary constraint that the (tensile) ductility  f  2% .

E 4.1 Definitions (1). What is meant by an objective and what by a constraint in the requirements for a design?

How do they differ? How are they used?

Answer. A constraint is an essential condition that must be met, usually expressed as a limit on a material or process attribute. An objective is a quantity for which an extreme value (a maximum or minimum) is sought, frequently cost, mass, volume, or environmental burden, but there are others. Constraints are used for screening to isolate a short list of candidates that can do the job. Objectives are used for ranking to identify the members of the short list that can do the job best.

E 4.2 Definitions (2). What is meant by a free variable? Give examples of free variables.

Answer. Free variables are design parameters that the designer is free to change in order to maximize or minimize the objective. Over-constrained designs have no free variables. More usually, the choice of the material and manufacturing process are free variables, as, often, is one or more dimension. As an example, a panel with prescribed length and width and desired bending stiffness can be made stiffer or less stiff by changing its thickness. The thickness is a free variable because it was not stipulated in the design requirements.

38

E 4.3. Translation (1) You are asked to design a cooking pan for camping. What constraints would you apply in selecting a material for the pan? What objectives would you use to rank the materials that meet the constraints?

Answer.

Function



Constraints

Some possible constraints:  Material of the pan must be non-toxic  Good thermal conductivity 0  Maximum service temperature Tmax > 400 C

Objectives

Cooking pans for camping

 Corrosion resistance in fresh and salt water, weak acids (eg vinegar), weak alkalis (baking soda)  Sufficient stiffness to prevent flexing if lifted when full  Sufficient strength and toughness to survive a drop Some possible objectives:  Maximize life-time of the pan  Minimize the material cost of the pan  Minimize the manufacturing cost of the pan  Maximize consumer appeal of the pan (an issue of industrial design).

E 4.4. Translation (2). Bikes come in many forms, each aimed at a particular sector of the market:  Sprint bikes  Touring bikes  Mountain bikes  Shopping bikes  Children’s bikes  Folding bikes Use your judgement to identify the primary constraints and objective that you would apply in selecting a material for the frame for each bike-type.

Answer. Adequate stiffness and strength are constraints for all six bike types. A frame that is not stiff and strong enough is unsafe. Stiffness and strength are therefore constraints that become limits on modulus, elastic limit and section-shape. If the bicycle is for sprint racing, it should be as light as possible. Minimizing mass, here, is the objective, perhaps with an upper limit (a constraint) on cost. If instead it is a shopping bike to be sold through supermarkets it should be as cheap as possible. This time minimizing cost is the objective, possibly with an upper limit (a constraint) on mass. For most bikes, of course, minimizing mass and cost are both objectives, and then trade-off methods are needed. For now use judgement to choose the single most important objective and make all others into constraints.

Bike type All bikes Sprint bikes Touring bikes Mountain bikes Shopping bikes Children’s bikes Folding bikes

Constraints Adequate stiffness and strength Specified stiffness, upper limit on cost Upper limits on both mass and cost Specified strength and toughness Adequate stiffness and strength Adequate strength Adequate stiffness and strength

Objective Minimize mass Maximize durability Maximize durability Co-minimize cost and mass Minimize cost Minimize volume

E 4.5. Translation (3). A material is required for the windings of an electric heater for a sauna. In use, the o heating coils operate in air at 800 C. What attributes must a coil-material have for this application? List the

39

function and the constraints; set the objective to “minimize material price” and the free variables to “choice of material”.

Answer. o

Function



Windings for sauna heater capable of operating at 800 C in air

Constraints

  

Specified resistivity range (10 – 100 μΩ.cm) Available as wire Sufficient ductility to allow a coil to be wound (  f >5%)



Maximum service temperature Tmax > 800C



Resistance to oxidation: very good

Objective



Minimize material price

Free variables



Choice of material

E 4.6. Translation (4). A material is required to manufacture office scissors. Paper is an abrasive material, and scissors sometimes encounter hard obstacles like staples. List function and constraints; set the objective to “minimize material price” and the free variables to “choice of material”.

Answer.

Function



Blades for office scissors

Constraints

  

High hardness (for wear resistance) Some toughness (to avoid chipping the blade edge) High stiffness to ensure that the blades do not bend in service

Objective



Minimize material price

Free variables



Choice of material

E 4.7. Translation (5). A material is required for a heat exchanger for cooling a small marine engine using o pressurized saline water at 120 C. List function and constraints; set the objective to “minimize material price” and the free variables to “choice of material”.

Answer.

Function



Marine heat exchanger

Constraints

    

Good thermal conductivity Excellent resistance to corrosion by hot salt water o Maximum service temperature > 120 C Sufficient strength to carry the water pressure Sufficient ductility to allow shaping to form tubes

Objective



Minimize material price

Free variables



Choice of material

E 4.8. Translation (6). A material is required for a disposable fork for a fast-food chain that is conscious of its environmental image. List the objective and the constraints that you would see as important in this application.

40

Answer.

Function



Disposable cutlery

Constraints

     

Easy to mold Non-toxic Sufficient strength and stiffness to withstand loads in use Resistant to the mild acids and alkalis that appear in foods Recyclable or renewable (e.g. bioplastic or a natural material) Ability to be colored (?)

Objective



Minimize material price

Free variables



Choice of material

E 4.9. Translation (7). Formulate the constraints and objective you would associate with the choice of material to make the forks of a racing bicycle.

Answer. The loading on the forks is predominantly bending. Both bending stiffness and strength are important in bicycle design. Strength is the most important of the two – the forks must not fail in normal use, which includes riding over curbs and other irregularities holes in the road. If the bicycle is for racing, then the mass is a primary consideration, so the forks should be as light as possible.

Function



Forks for racing bike

Constraints

 

Must not fail under design loads – a constraint on bending strength Dimensions specified

Objective



Minimize weight

Free variables



Choice of material

E4.10 Translation (8). The standard CD comes in a transparent plastic case (known as a “jewel” case) that cracks easily and, if broken, can scratch the CD. Jewel cases are made of injection molded polystyrene, chosen because it is transparent, cheap and easy to mold. A material is sought to make CD cases that do not crack so easily. The case must still be transparent, able to be injection-molded, and able to compete with polystyrene in cost.

Answer. The design requirements express constraints on transparency, moldablity and fracture toughness K 1c (it must be greater than that of polystyrene). Given these, the cheapest material is the best choice.

Function



Improved CD case

Constraints

  

Optically transparent Fracture toughness greater than that of polystyrene Able to be injection-molded

Objective



Minimize material cost

Free variables



Choice of material

41

Figure 4.15. A pivot.

E4.11 Translation (9). Ultra-precise bearings that allow a rocking motion make use of knife-edges or pivots (Figure 4.15). As the bearing rocks, it rolls, translating sideways by a distance that depends on the radius of contact. The further it rolls, the less precise is its positioning, so the smaller the radius of contact R the better.

But the smaller the radius of contact, the greater is the contact pressure (F/A). If this exceeds the hardness H of either face of the bearing, it will be damaged. Elastic deformation is bad too: it flattens the contact, increasing the contact area and the roll. Translate the requirements for the bearing, listing function, constraints, objective and free variable.

Answer. The objective is to minimize the radius of curvature R at the contact. This is limited by the hardness H and modulus E of the materials. The objective is met by choosing materials with maximum hardness H to enable the smallest possible area of contact without damage, and maximum modulus E to minimize elastic flattening of the contact. In a later exercise we return to this problem of two objectives, treating it by trade-off methods. Here we choose maximizing hardness as the primary objective, since deficiency here results in a damaged bearing. We then treat modulus as a constraint, together with the other obvious constraints suggested by the design requirements.

Function



Rocking bearing

Constraints



High modulus, E

Objective



Maximize hardness H of bearing faces

Free variables



Choice of material

Deriving and using material indices. The exercises in this section give practice in deriving indices.

Start each by listing function, constraints, objectives and free variables; without having those straight, you will get in a mess. Then write down an equation for the objective. Consider whether it contains a free variable other than material choice; if it does, identify the constraint that limits it, substitute, and read off the material index.

E4.12 Material indices for elastic beams with differing constraints (1). Show that the best material for a cantilever beam of given length L and given (i.e. fixed) square cross-section (t x t) that will deflect least under a given end load F (Fig. 4.16) is that with the largest value of the index M1 = E, where E is Young's modulus (neglect self-weight). You will need the equations for the deflection  of a cantilever beam with a square cross-section t x t, of length L under an end load F (Appendix B, Section B3):

 

Figure 4.16. A beam with end load.

F L3 3EI

where I is the second moment of area of the beam: I  t 4 / 12 .

Plot the index on a copy of the ( E   ) chart of Figure 3.3 indicating the search area for materials with the highest values M1.

42

Answer. The table lists the design requirements.

Function Constraints Objective Free variables

     

End-loaded cantilever beam Length L specified Section t x t specified End load F specified Minimize the deflection,  Choice of material only

The objective function is an equation for the deflection of the beam. An end-load F produces a deflection  of

 

F L3 3EI

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  F L3   1     4 4    t  E    The quantities F, L and t are all specified. The only free variable is the modulus, E. The deflection  is minimized by choosing a material with the largest value of M1  E

It plots as a horizontal line on the E-ρ chart of Figure 3.3, as shown on the next page. Materials above the line have high values of. E. Pushing the line upwards reduces the number left in the Search Area, identifying the best 1/2  candidates.

M3 = E /orE/ 

M2 = E/

Search area

M1 = E

E4.13 Material indices for elastic beams with differing constraints (2). Show that the best material choice for a cantilever beam of given length L and with a specified section (t x t) that will deflect least under its own self-weight (Fig. 4.17) is that with the largest value of M2 = E/, where  is the density. You will need the equations for the deflection of a cantilever beam with a square cross-

43

Figure 4.17. Self-loaded beam with fixed section t x t.

section t x t (Appendix B, Section B3) under a distributed load f per unit length:

 

1 f L4 8 EI

where I  t 4 / 12 is the second moment of area of the beam. For a self-loaded beam f   A g where  is the 2 density of the material of the beam, A = t its cross-sectional area and g the acceleration due to gravity.

Plot the index on a copy of the ( E   ) chart of Figure 3.3 indicating the search area for materials with the highest values M2.

Answer. The table lists the design requirements.

Function Constraints Objective Free variables

    

Self-loaded cantilever beam Length L specified Section t x t specified Minimize the deflection,  Choice of material only

The beam carries a distributed load, f per unit length, where

f   gt 2 where  is the density of the beam material and g is the acceleration due to gravity. Such a load produces a deflection (Appendix B, Section B3)



1 f L4 8 EI

where I  t 4 / 12 is the second moment of area of the beam. Combing these equations gives the deflection 3gL4    3 f L4   .

4 2 Et 2t 2  E  The quantities t and L are specified. The deflection is minimized by maximizing the index



M2 

E 

The index M2 is plotted on a copy of the E-ρ chart of Figure 3.3, above. The Search Area isolates materials with the largest values of M2.

E4.14 Material indices for elastic beams with differing constraints (3). Show that the material index for the lightest cantilever beam of length L and square section of unspecified area (so the dimension t is a free variable) that will not deflect by more than  * under its own weight (Fig. 4.18) is M3  E /  2 . You will need the equations for the deflection of a cantilever beam with a square cross-section t x t (Appendix B, Section B3) under a distributed load f per unit length:

 

Figure 4.18. Self-loaded beam with free choice of section t x t.

1 f L4 8 EI

where I  t 4 / 12 is the second moment of area of the beam. For a self-loaded beam f   A g where  is the 2 density of the material of the beam, A = t its cross-sectional area and g the acceleration due to gravity.

Plot the index on a copy of the ( E   ) chart of Figure 3.3 indicating the search area for materials with the highest values M3.

44

Answer. The table lists the design requirements.

Function Constraints Objective Free variables

    

Self-loaded cantilever beam Length L specified Maximum deflection,  * , specified Minimize the mass, m Choice of material



Section area A  t 2

The beam deflects under its own weight but the section can be varied to reduce the weight provided the deflection does not exceed  * . The objective function (the quantity to be minimized) is the mass m of the beam

m  t2 L The beam deflects under its own weight but now the section area can be reduced to reduce the load f per unit length

f   gt 2 where  is the density of the beam material and g is the acceleration due to gravity. The deflection of the beam under its own weight must not exceed  *



1 f L4  * 8 EI

where I  t 4 / 12 is the second moment of area of the beam. Eliminating the free variable t (the free variable) from the second equation into the first, gives 3gL5   2  m   2 *  E  The quantities L and  * are given. The mass is minimized by maximizing E M3  2  The index M3 is plotted on a copy of the E-ρ chart of Figure 3.3, above. The Search Area above the M 3 line isolates materials with the largest values of M3.

E4.15 Material indices for elastic beams with differing constraints (4).

Show that the lightest cantilever beam of length L and square section (area free) that will not deflect by more than  * under an end load F (Fig. 4.19) is that made of the material with the largest value of M4  E 1 / 2 /  (neglect self-weight). You will need the equations for the deflection  of a cantilever beam with a square cross-section t x t, of length L under an end load F (Appendix B, Section B3):

 

Figure 4.19. End-loaded beam with free choice of section t x t.

F L3 3EI

where I is the second moment of area of the beam: I  t 4 / 12 .

Plot the index on a copy of the ( E   ) chart of Figure 3.3 indicating the search area for materials with the highest values M4.

Answer. The table lists the design requirements.

45

Function Constraints Objective Free variables

     

End-loaded cantilever beam Length L specified End-load F specified Maximum deflection,  * , specified Minimize the mass, m Choice of material



Section area A  t 2

The section is square, but the dimension t is free. The objective function is

m  t 2L  The deflection

F L3  1    4 4    * t E  Using this to eliminate the free variable, t, gives 1/ 2

 F L5      m  2  1/ 2     E    The quantities F ,  and L are given. The mass is minimized by maximizing

M4 

E1/ 2



From a selection standpoint, M3 and M4 are equivalent.

The selection. The figure shows what the indices look like when plotted on the ( E   ) chart of Figure 3.3 (above). The table lists the top-ranked candidates for each index.

Index

Material choice Metals: tungsten alloys, steels.

Ceramics: SiC, Si3N4, B4C and AlN, but of course all are brittle.

Metals: aluminum, magnesium, titanium alloys and steels all have almost the same value of E / 

High M1 = E

High M2 

E



High M3 

High M4 

Composites: CFRP Ceramics SiC, Si3N4, B4C and AlN

E

2

E1/ 2



Metals: aluminum and magnesium alloys superior to all other metals.

Composites: CFRP excels Ceramics: SiC, Si3N4, B4C and AlN

E4.16 Material index for a light, strong beam. In stiffness-limited applications, elastic deflection is the active constraint. 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 F* and minimum weight (Fig. 4.20). For simplicity, assume the beam to have a solid square cross-section t x t, with free choice of t. You will need the equation for the failure load of a beam (Appendix B, Section B4). It is

46

Figure 4.20. A strong beam with free choice of section t x t..

Ff 

f

I f

ym L Where  f is the yield strength of the materials and y m is the distance between the neutral axis of the beam and its outer filament. I  t 4 / 12  A 2 / 12 is the second moment of the cross-section.

Plot the criterion on a copy of the (  f   ) chart of Figure 3.4 and use it to identify promising candidate materials.

Answer. The table lists the design requirements.

Function

 

Constraints

Beam Length L is specified



Beam must support a bending load Ff* without yield or fracture

Objective

  

Minimize the mass of the beam Cross-section area, A Choice of material

Free variables

The model. The objective is to minimize the mass, giving the objective function m  AL  The failure load Ff is related to the materials strength  f

I f

 Ff* ym L where ym  t / 2 is the distance between the neutral axis of the beam and its outer filament and Ff 

I  t 4 / 12  A 2 / 12 is the second moment of the cross-section. Inverting the equation given in the question leads to an expression for the area A that will support the design load Ff* 2/ 3

 Ff*L   A*   6  f    Substituting A into the objective function gives the mass of the beam that will just support the load Ff* : 2/3 5 / 3 





2/3 f 

  L  ς ρ 

m  6Ff*

The mass is minimized by selecting materials with the largest values of the index ς 2/3 f M5  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  15MPa.m1 / 2 (Almost all metals satisfy this criterion, as does CFRP and GFRP. No ceramics or glasses survive.)

The index M5 is plotted on a copy of the (  f   ) chart of Figure 3.4, below. Materials with high values of M5 are circled.

Material

Comment

CFRP

Exceptionally good, mainly because of its very low density.

Metals: titanium, aluminum and magnesium alloys

Here the light alloys out-perform steel

2/3

M = f /

47

Search area

E4.17 Material cost. A component is at present made from an aluminum alloy. It is suggested that the material cost could be reduced if it was made, instead, from a polymer. To avoid the need to redesign the product, the replacement should have exactly the same dimensions, and thus volume, as the original. Which polymers cost less, per unit volume, than aluminum alloys? Mark up a copy of the Price per unit volume chart (Figure 3.25 of the text) to find out.

Answer. The chart is shown below. The upper limit of the “Polymers” prices is set at the lower limit on the price for aluminum alloys. All the polymers between the lines cost less per unit volume than aluminum alloys. They include Polypropylene (PP), Polyethylene (PE) and Polyvinyl chloride (PVC).

Al-alloys Polymers that are cheaper than Al-alloys

E4.18 Carbon footprint. A component is at present made from an aluminum alloy. It is suggested that its carbon footprint could be reduced if it was made, instead, from a polymer. To avoid the need to redesign the product, the replacement should have exactly the same dimensions, and thus volume, as the original. Which polymers have a lower carbon footprint, per unit volume, than aluminum alloys? Mark up a copy of the Carbon footprint per unit volume chart (Figure 15.6 of the text) to find out.

48

Answer. The chart is shown below. The upper limit of the “Polymers” carbon is set at the lower limit on the carbon for aluminum alloys. The polymers between the lines have lower carbon footprint per unit volume than aluminum alloys. They include all commodity polymers such as Polypropylene (PP), Polyethylene (PE) and Polyvinyl chloride (PVC).

Al-alloys Polymers with lower carbon footprint than Al-alloys

E4.19 Material index for a cheap, stiff column. 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 (Fig.4.21). The objective is to minimise the material cost of the column

C   r 2HCm  Where C m is as measure of the cost of the material per unit weight and  is its density; C m  measures the cost of the material per unit volume. You will need the equation for the load Fcrit at which a slender column buckles (Appendix B, Table B5). It is

Fcrit 

n 2 π 2E I

Figure 4.21. A column.

H2 where n is a constant that depends on the end constraints, E is Young’s modulus and I   r 4 / 4 is the second moment of area of the column (see Appendix B for both). We wish to minimise the materials cost of the column Plot the index on the Modulus – Relative cost chart, Figure 3.22, and identify three of the cheapest materials for a cheap column. Here Cm

Answer. The table summarises the design requirements.

49

Function Constraints Objective Free variables

     

Cylindrical column Height H and load F are specified Column must support a compressive load F without buckling Minimize the material cost of the column Cross-section area, A, and thus the column radius, r Choice of material

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   r 2HCm  * Where C m is the cost/kg of the material of the column. It will buckle elastically if F exceeds the Euler load, Fcrit , given in the question:

Fcrit 

n 2 π 2E I

.

H2 Eliminating the radius r between the two equations, using the definition of I, gives: 1/2

 4F  C ρ C   crit  H 2  m   E 1/2   nπ  The material cost of the column is minimized by choosing materials with the largest value of the index E1 2 Cm ρ . Applying the index M6 to the Modulus – Relative cost chart gives the top-ranked candidates identifies concrete, wood, stone and cast iron as potential candidates. The loading here is compressive, so brittle materials are viable candidates. Since some are also very cheap, they dominate the selection.

M6 

1/2

M6 = E /Cm

Search area

* C is the cost/kg of the processed material, here, the material in the form of a circular rod or column. m

50

4.20 Indices for locally-loaded circular plates.

Aircraft and space structures make use of plates and shells. Derive the material index for a circular plate of radius R carrying a central load F (Fig. 4.22), with a prescribed stiffness S  F /  and of minimum mass. The mid-point deflection  of a circular plate under a load F applied over a small central, circular area is



3 F R2  3   ( 1  2 )   4 Et 3  1  

Figure 4.22. A circular plate.

(Appendix B, Table B7). Here E is Young’s modulus, t is the thickness of the plate and  is Poisson’s ratio. Poisson’s ratio is almost the same for all structural materials and can be treated as a constant.

Answer. . The table summarizes the requirements.

Function



Constraints

 Displacement δ under central load F specified, meaning that the stiffness S  F /  is specified

Stiff circular plate, or



Radius R of plate specified

Objective



Minimize the mass of the plate

Free variables

 

Plate thickness, t Choice of material

The objective is to minimize the mass, m

m   R2 t  where  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, 1/ 3

 3 SR2  t   4  E   

f1  

where f1   is simply a function of Poisson’s ratio,  , and thus a constant. Inserting this into the equation for the mass gives 1/ 3

 3 SR2  m   a2   4    

    E 1 / 3  f  

The lightest plate is that made from a material with a large value of the index

M7 

E1/ 3



The selection. Applying the indices to the ( E   ) chart of Figure 3.3 gives the top-ranked candidates listed below.

Material choice

Index

High M7 

E1/ 3



Natural materials: wood and plywood Composites: CFRP

E4.21 Indices for locally-loaded stiff shells. What is the material index for the hemispherical shell of minimum mass carrying a central load F with a prescribed stiffness S  F /  , as sketched in the figure? The shell has a radius R and a wall

51 Figure 4.23. A spherical shell.

thickness t (Fig. 4.23). The mid-point deflection  of a spherical shell under a load F applied over a small central, circular area, is.

 A

FR Et

2

( 1  2 )

(Appendix B, Table B7). Here A  0.35 is a constant, E is Young’s modulus and  is Poisson’s ratio. Poisson’s ratio is almost the same for all structural materials and can be treated as a constant.

Answer. The table summarizes the requirements.

Function



Stiff hemispherical shell

Constraints

 

Specified stiffness S = F/  under central load F Radius R of shell specified

Objective



Minimize the mass of the shell

Free variables

 

Shell wall thickness, t Choice of material

The objective again is to minimize the mass, m

m  2 R2 t  The deflection  of the shell under the load W, given in the question, is

 A

FR Et 2

( 1  2 )

Inverting this equation gives and writing S  F /  gives 1/2

 SR  t A  f2    E  where f 2   is a function of  , which we treat as a constant. Inserting this into the equation for the mass gives 1/2    m  2 R2  AS R   1 / 2  f2   E



The lightest shell is that made from a material with a largest value of the index

M8 

E1/ 2



The index for the shell differs from that for the plate (Exercise E4.20), 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), making it much stiffer. Singlycurved shells behave like the plate, doubly-curved shells like the hemisphere.

Applying the indices to the ( E   ) chart of Figure 3.3 gives the top-ranked candidates listed below.

Material choice

Index

High M8 

E1/ 2



Metals: aluminum and magnesium alloys superior to all other metals.

Composites: CFRP Ceramics: SiC, Si3N4, B4C and AlN

E4.22 Furnace insulation. A furnace is required to sinter powder-metal parts. It operates continuously at 650 o C while the parts are fed through on a moving belt. You are asked to select a material for furnace insulation to minimize heat loss and thus to make the furnace as energy-efficient as possible. For reasons of space the insulation is limited to a maximum thickness of x  0.2 m. (a) List the function, constraints, objective and free variable.

52

(b) Derive a material index for selecting the material for the insulation.

Answer. The table lists the function, constraints, objective and free variable.

Function

 Insulation for energy-efficient furnace

Constraints

 Maximum service temperature, Tmax > 650 C  Insulation thickness x  0.2 m

Objective

 Minimize heat loss through the insulation

Free variables

 Choice of material

The model. This is a problem involving steady-state heat flow. The heat lost by conduction per unit area of 2 insulation per second, q (J/m .s), is

q 

T x

where  is the thermal conductivity, x the insulation thickness and  T the temperature difference between the interior of the furnace and its surroundings. Both the temperature difference  T and the insulation thickness x are defined by the constraints. The heat loss q is minimized by selection materials with the smallest value of the thermal conductivity  . The material index for the problem is simply M9  

E4.23 Materials for thermal storage (1). A storage heater captures heat over a period of time, then releases it to an air stream when required. Those for domestic heating store solar energy or energy from cheap off-peak electricity and release it slowly during the cold part of the day. What is a good material for the core of a o compact storage material capable of temperatures up to 150 C and a defined storage capacity? Interpret “compact” to mean “of minimum volume". (a) List the function, constraints, objective and free variable. (b) Derive a material index for selecting the material for the heat storage material.

Answer. The table lists the function, constraints, objective and free variable.

Function Constraints Objective Free variables

    

Core for compact storage heater Maximum service temperature, Tmax > 150 C Able to store a thermal energy Q* (joules) Minimize volume of the core Choice of material

The model. When a material is heated from room temperature To to a working temperature T, it absorbs heat Q per unit volume where Q  C p  (T To ) J / m3

and C p is the specific heat of the material of the core (in J/kg.C) and  is its density (in kg/m ). The 3

volume V of material needed to provide a given storage capacity Q* is V 

Q* C p  ( T  To )

The operating temperature T and the storage capacity Q* are specified – they are constraints. The most compact storage heater is one made from a material with a high C p  . The material index for the problem is simply 53

M10  C p  This property combination appears as a set of contours on the (   a ) chart of Figure 3.11, copied 6

3

below. A large number of materials have almost the same high value of M 10, approximately 3 x 10 J/m .

Some of these are very expensive. We need a different objective: that of finding the material that stores the most energy per unit cost. The next exercise explores this.

M10 = Cp 

Search area|

E4.24 Materials for thermal storage (2). The result of Exercise E4.23 was the volume of material needed to store a specified thermal energy. Rework this exercise to minimize the cost of the core instead of its volume.

First use the maximum service temperature chart (Figure 3.14) to eliminate materials that fail the maximum service temperature constraint. Then use the cost per unit volume chart (Figure 3.21) to identify the cheapest candidate materials for the core.

Answer. The table lists the adjusted function, constraints, objective and free variable.

Function Constraints Objective Free variables

    

Core for compact storage heater Maximum service temperature, Tmax > 150 C Able to store a thermal energy Q* (joules) Minimize the cost of the material of the core Choice of material

The model. The volume V of material needed to provide a given storage capacity Q* (from Exercise 4.23) is Q* V  C p  ( T  To ) The conclusion of Exercise 4.23 is that all materials (except foams) have rather similar values of C p  , so the volume V of a storage heater to store Q* joules depends only weakly on the choice of material for the core.

The price, on the other hand, depends strongly on the choice. The maximum service temperature constraint eliminates most polymers and woods. The cost per unit volume chart identifies concrete, stone and brick as the cheapest core materials.

54

55

E4.25 Minimizing thermal inertia. A clamp is required for processing of electronic components. The clamp has a square cross-section of width x and given depth b (Fig. 4.24). 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 – equation (2.15)) is

t

x2 2a

where the thermal diffusivity a   /  C p and  is the thermal conductivity,  the density and C p the specific heat per unit mass.

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 bending moment on the body of the clamp of M  F L , and that the peak stress in the body is given by (Appendix B, Section B15):

xM 2 I where I  b x 3 / 12 is the second moment of area of the body.



Figure 4.24. A clamp.

Answer. The table summarizes the requirements.

Function



C-clamp with 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

The clamp will fail if the stress in it exceeds its elastic limit  f . Equating the peak stress to  f and solving for x gives 1/ 2

 FL   x  6  b f    Inserting this into the equation for the time to reach equilibrium gives F L  1  t3  b  a  f  The time is minimized by choosing materials with large values of the index M  af

In practice, additional constraints on modulus E > 50 GPa (to ensure that the clamp is sufficiently stiff) on fracture toughness K 1c  18MPa.m 1 / 2 (to guard against accidental impact) and on formability will be needed.

E4.26 Springs for trucks. In vehicle suspension design it is desirable to minimize the mass. Select a material and dimensions for a light spring to replace the steel leaf-spring of an existing truck suspension.

The existing leaf-spring is a beam, shown schematically in the figure 4.25. The new spring must have the same length L and stiffness S as the existing one, and must deflect through a maximum safe displacement max without failure. The width b and thickness t are free variables.

56

Figure 4.25. A beam with central load.

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 max and the other on stiffness S. Use the two constraints to fix the values of the free variables.

The mid-point deflection of an elastic beam of length L loaded in three-point bending by a central load F:

1 F L3 48 E I and the deflection at which failure occurs



 max 

1f L 6 tE

2

where I is the second moment of area; for a beam of rectangular section, I  b t 3 / 12 and E and  f are the modulus and failure stress of the material of the beam (see Appendix B.) Plot the index on a copy of the Specific Modulus – Specific Strength (Figure 3.6 of the text) and use it to identify other promising candidates.

Answer. The table lists the requirements.

Function



Leaf spring for truck

Constraints

  

Length L specified Stiffness S specified Maximum displacement  max specified

Objective



Minimize the mass

Free variables

  

Spring thickness t Spring width b Choice of material

The model. The objective function – the quantity to be minimized – is the mass m of the spring: (1) m  bt L  where  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

E bt 3 EI  48 3  4 3 (2)  L L The second constraint is that of a maximum allowable displacement  max without damage to the spring, given in the question as S

F

2 1f L (3) 6 tE Equations (2) and (3) can now be solved for t and b, and these substituted back into (1). The result is  E  2 m  9 S  max L 2     f 

 max 

The mass of the spring is minimized by maximizing the index  2f /  E M12 

 2f E

Additional constraints on fracture toughness K 1c  15MPa.m 1 / 2 (to guard against accidental impact) and on formability will, in practice, be needed.

The selection. The table compares high carbon steel, low alloy steel, titanium alloys and CFRP, ranking them by the value of the index, using data from Appendix A. CFRP is the best choice.

Material

Comment

 2f /  E (kJ/kg) 57

CFRP

3.8

CFRP makes exceptionally good light springs.

Titanium alloys Low alloy steel

1.7 0.36

High carbon steel

0.30

An expensive solution, but one that is lighter than steel.

Less good than CFRP and Titanium alloys but better than carbon steel The standard solution, but one that is heavier than the others above.

The Specific modulus -Specific strength chart is copied below. The index  f / E plots as a contour with a slope of 1, using log scales. The index derived here, M12 

 2f E

plots as a line of slope 2. One M12 contour is plotted in blue, positioned so that only a few materials lie below it – these are the ones with the highest values of M12. It identifies CFRP, steels and titanium alloys (our earlier candidates), but also suggests magnesium alloys and nylons 2

M12 = σf /ρE

58

E4.27 Aperture grills for cathode ray tubes. Long, long ago, in the days of cathode-ray technology, there were two types of cathode ray tube (CRT). In the older technology, colour separation was 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 target and giving a blotchy image. To avoid this, shadow masks were made of Invar, a nickel alloy with a near-zero expansion coefficient between room temperature and 150C̊. A consequence of shadow-mask technology was that the glass screen of the CRT curved inward on all four edges, increasing reflected glare.

Sony’s ‘Trinitron’ technology overcame this problem and allowed greater Figure 4.26. A CRT with aperture grill. brightness by replacing the shadow mask by an aperture grill of fine vertical wires, each about 200 m thick, that allows the intended beam to strike either the red, the green or the blue phosphor to create the image (Fig. 4.26). The glass face of the Trinitron tube was curved in one plane only, reducing glare.

The wires of the aperture grill were pre-tensioned so that they remain taut even when hot – this pre-tension allowed the greater brightness. What index guides the choice of material for the wires? The table summarizes the requirements.

Function Constraints

Objective Free variables

      

Aperture grill for CRT Wire thickness and spacing specified Material must carry pre-tension without failure Electrically conducting to prevent charging Able to be drawn to wire Maximize permitted temperature rise without loss of tension Choice of material

Use the index you derive to rank the following candidate material for this application using data from Appendix A: carbon steel, stainless steel, nickel-chromium alloys and tungsten. All are available as fine wire.

Answer. A thin, taut wire slackens and sags when the strain due to thermal expansion,  th  T

exceeds the elastic strain caused by the pre-tension,  pt ,

 pt 

Here 

 pt

.

E is the thermal expansion coefficient of the wire,  T the temperature rise caused by the electron

beams that strike it, and E its modulus. We wish to maximize the brightness and thus  T .

The tension is limited by the elastic limit of the wire,  f . Inserting this and writing  pt   th

as the condition that the wire remains taut gives

T 

f

.

E To maximize the brightness, maximize f M13  E 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 drawn to wire.

59

The selection. Carbon steel, stainless steel, tungsten and nickel alloys are all available as fine wire. Data for their strength, modulus and expansion coefficient are listed in Appendix A, ranking the survivors by the index M leads to the selection listed below in decreasing order of the index.  f / E (K)

Comment

Tungsten

1.0

Stainless steel

0.32

Carbon steel

0.29

Nickel-chromium alloys

0.26

A logical choice – tungsten has a high melting point and is routinely produced as fine wire Austenitic stainless steels are non-magnetic and available as fine wire – a viable choice.

Carbon steel is ferro-magnetic, so will interact with the scanning magnetic fields – reject.

Many nickel alloys are weakly ferromagnetic – reject for the same reason as carbon steel

Material

Ch 5. Materials selection – case studies This chapter has no exercises.

60

Ch 6. Processes and their effect on properties E 6.1. Exploring processes. Good images of many processes can be found on the internet. Use it to find images for: (a) Plastic film blowing (b) Filament winding (c) Ceramic molding (d) Metal spinning (e) Spot welding (f) Polymer extrusion Answer. Examples of images of the six processes.

Answer.

(a) Plastic film blowing

(b) Filament winding

(c) Ceramic molding

(d) Metal spinning

(e) Spot welding

(f) Polymer extrusion

E 6.2. Making property trajectories (1). Copper and nickel are completely miscible, forming a solid solution at all compositions.

The table lists the thermal conductivity of copper-nickel alloys, ranging from pure copper to pure nickel. Plot a trajectory of thermal conductivity against copper content. How do you explain the shape of the trajectory?

Copper-Nickel alloy 100% copper

T-conductivity (W/m.K) 381

95% copper

22.8

90% copper

20.2

80% copper

22.4

75% copper

25.4

70% copper

31.4

45% copper

47.3

32% copper

66.1

0% copper

88.7

Answer. The trajectory is shown below. Heat, in metals, is conducted by electrons. Pure copper has a high thermal conductivity. The U-shaped trajectory of thermal conductivity against copper content shows how solute increases the electron scattering in copper, driving the conductivity down, rising again as pure nickel is approached. This trajectory is typical of solid solutions in metals.

61

E 6.3. Making property trajectories (2). The table lists the yield strength and fracture toughness of copper-nickel alloys for range of compositions. Plot trajectory of fracture toughness against yield strength, labelling the points with the composition.

Copper-Nickel alloy

Yield strength (MPa)

100% copper

49.7

Fracture toughness 1/2 (MPa.m ) 102

95% copper

85.9

85.8

90% copper

99.3

76.8

80% copper

115

67.5

75% copper

132

67.9

70% copper

145

70.4

45% copper

187

72.9

32% copper

199

79.4

Answer. The trajectory is shown below. The yield strength increases steadily with increasing nickel content. The fracture toughness, by contrast, is at its lowest at about 70% copper, but rises as the composition approaches pure copper or pure nickel.

62

E 6.4. Using property trajectories. An aluminium alloy is sought for a demanding application. For safety reasons the 1/2 designer has specified a minimum fracture toughness of 30 MPa.m and the highest possible strength. Use the property trajectory of Figure 6.18 to select an alloy and a heat treatment that meets the requirements. Use the internet to find out more about the alloy and the heat treatment that you select.

1/2

Answer. Figure 6.18 is copied, below. The horizontal red line is set at a fracture toughness of 30 MPa.m , only materials above this line are acceptable. The vertical red line has been pushed as far as possible to the right, leaving only one alloy above it. The alloy that best meets the criteria is Al7075 in the T7352 condition.

30 MPa.m1/2

Many websites provide information about aluminium alloys. Typical of the descriptions of Al 7075 is: “Aluminum alloy 7075 is an aluminium alloy with zinc (5.5%), magnesium (2.5%) and copper (2.5%). It is the strongest of the aluminium alloys, with a strength comparable to many steels, has good fatigue strength and average machinability, but has less resistance to corrosion than many other Al alloys.”

Information about tempers for aluminum alloys is available on EngineersEdge website: http://www.engineersedge.com/aluminum_tempers.htm on which we find: “T7352: Solution heat treated and specially artificially aged. Applies to 7075 alloy forgings which have both compression-stress relief and special aging to make the material resistant to stress-corrosion.”

E 6.5. Choosing optimal processing conditions. An approximate measure of the energy absorbed in plastic deformation of a metal is the product  y  f of its yield strength  y and its elongation to fracture  f . These two properties for copper and copper-nickel alloys are the axes of Figure 6.16 of the text. Plot contours of  y  f onto a copy of this figure and use them to identify the optimal processing to maximize energy absorption capacity.

Answer. Figure 6.16 is reproduced below with contours of energy  y  f shown in red. Copper nickel solutionstrengthened alloys with a composition in the range 30 – 40 % copper, 60 – 70% nickel absorb the most. If alloying is not an option, then pure copper in the 1/4 hard condition is the best choice.

63

50

100

20 10 5 Energy 3 MJ/m

E6.6. You are asked to suggest a material for a light casing for transporting a sensitive measuring device. Foamed ABS is suggested as a possible choice. Estimate the value of Young’s modulus for ABS when foamed to 10% of its solid density. Retrieve the value of the modulus for solid ABS from Appendix A, Table A3, using the mean value of the range listed there.

Answer. The mean value of Young’s modulus for solid ABS, retrieved from Appendix A, is 2.0 GPa. Figure 6.27 of the text demonstrates that the modulus of foams scales with the square of the relative density, that is that 2

   E  Es    s  where E s and  s are the modulus and density of the solid material, and E and  are those of the foam. If the ABS is foamed to a density that is 10% that of the solid, the modulus is expected to fall to 0.02 GPa. E6.7. A component made from AISI 1137 steel is to be heat-treated to give it a yield strength of 1000 MPa. What tempering temperature would you choose? What fracture toughness range would the steel be expected to have after the heat treatment? Use the trajectory of Figure 6.21 to find out.

Answer. Tempering at 425 C gives a yield strength of 1000 MPa and maximizes fracture toughness

K1c range: 1/2 23 – 40 MPa.m

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E6.8. A stainless steel is required for a pressure vessel. A yield strength exceeding 1000 MPa is required. For safety, the steel should have a fracture toughness exceeding 80 MPa.m1/2. What grades of stainless steel, in what state of thermos-mechanical treatment, would meet the requirements. Use the trajectory of Figure 6.22 to find out.

Answer. The requirements are met by AISI 450 precipitation-hardened stainless steels in the H1000, H950 and H900 states of heat treatment.

E6.9. Polypropylene has a Youngs modulus of 1.3 GPa and a yield strength of 35 MPa. The low modulus is a handicap in some applications. What cheap filler could be blended into the polypropylene to increase its modulus to 3 GPa without its strength falling below 30 MPa? How much filler would be required? Possible fillers are calcium carbonate (the cheapest), talc (a little more expensive) and chopped glass fiber (the most expensive). Use the trajectory of Figure 6.24 to find the cheapest option.

Answer. The requirements on modulus and strength are met by 40% talc or 10% of glass fiber. Talc is the cheaper option.

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E6.10. The effect of porosity on modulus E and strength  f of alumina (Al2O3) ceramic is plotted in the trajectory of Figure 6.26. On the log scales used here, the relationship is almost linear. How are the two related? Express the result in the form  f  AE n where n in and integer exponent. What are the units of A in your equation?

Answer. The trajectory, annotated, is shown below. The straight line corresponds to the (approximate) relationship

 f ; 2.3 x 106 E 2 (both E and  f in GPa). The strength is much more sensitive to porosity than the modulus. The units of the -1

constant A are GPa .

Log (Modulus) 2.0

2.3

2.5

2.7 2.7

2.3

2.0

Log (Strength)

1.7

1.3

2

E6.11. The Ansys Granta EduPack software contains sets of records for making Process-Property charts like those shown in Section 6.4 of this chapter. If this tool is available, use these record-sets to plot the process-property trajectory for (a) Thermal conductivity vs electrical resistivity for solution hardened copper as a function of solute concentration. (b) Vickers hardness vs elongation for medium carbon steel as function of tempering temperature. (c) Heat exchangers require materials with high thermal conductivity and – if pressurized – high strength.

Explore the process-property trajectory for the Thermal conductivity vs Yield strength for aluminum grades to find the best combination for heat exchangers. (d) Do fillers change the thermal properties of polymers? Explore the process property trajectories for the Thermal conductivity and the Expansion coefficient for filled polypropylene to find out.

Answer. The process-property profiles are shown below

2 https://www.ansys.com/products/materials/granta-edupack

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Increasing solute concentration

(a) Solute drives down thermal conductivity and drives up resistivity. The two are inversely proportional

Increasing tempering temperature

(b) Tempering martensite reduces hardness but increases elongation (ductility); the higher the tempering temperature the greater is the effect

67

Increasing cold work

(c) Wrought non age-hardening (1000 series) Al alloys offer the best combination of high thermal conductivity and high strength.

Increasing cold work

(d) All fillers increase the thermal conductivity and reduce the thermal expansion of polypropylene. The effect is large; the more the filler, the greater is the effect.

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Ch 7. Processes selection and modelling E7.1. Selecting shaping processes (1). How would you make the rocker arm of Figure 7.19? It is to be made from and aluminium-silicon alloy. The shape is “3D solid”. The expected production run (batch size) is 2,000. It weighs about 1kg.

Surface roughness and tolerance are not critical because the bores and the boss will be machined to final size. Present the results in the format of Table 7.3 of the text.

Answer. The first three columns of the constraint-matrix, below, show the principal selection steps. Of the eleven processes capable of shaping nonferrous metals, four meet the primary constraints. The last column comments on issues of quality.

Compatible with nonferrous metal (Figure 7.3) Sand casting Die casting Investment casting Centrifugal casting

Compatibility with shape (Figure 7.4)

Compatibility with batch size (Figure 7.6)

Quality comment Surface roughness irrelevant; low die cost High precision, good surface finish, expensive dies

Fails Fails Good mechanical properties; requires much subsequent machining

Forging Extrusion Sheet forming Powder methods Electro-machining Machining from solid Additive manufacture

Figure 7.19. A rocker arm

Fails Fails Fails Fails High precision possible; much waste generated Fails

E7.2. Selecting shaping processes (2). A complex camera body is to be made of a magnesium alloy with an initial production run of 10,000 units. Surface finish and tolerance must be high. Which processes can make it? Use the matrices of Figures 7.3, 7.4 and 7.6 to identify potential candidates. Present the results in the format of Table 7.3 of the text.

Answer. The material is a non-ferrous metal. The shape is a 3-D hollow. The planned batch size is 10,000 units.

The first three columns of the constraint-matrix, below, show the principal selection steps. Of the eleven processes capable of shaping non-ferrous metals, four meet the primary constraints. The last column comments on issues of quality.

Compatible with nonferrous metal (Figure 7.3) Sand casting Investment casting Low pressure casting Forging Extrusion Sheet forming Powder methods Electro-machining Machining Additive manufacture

Compatibility with shape (Figure 7.4)

Compatibility with batch size (Figure 7.6)

Quality comment Surface roughness rules out sand casting Capable of precise shape with good finish Less precise than pressure die casting

Fails Fails Fails Fails Fails The complex shape makes machining difficult Fails

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E7.3 Selecting shaping processes (3). A car roof-box is to be made of glass-fiber reinforced polyester in a batch of 10,000. It is made up of two separate shell-like components, one for the main shell, the other for the lid. What processes are available to shape them? Use the matrices of Figures 7.3, 7.4 and 7.6 to identify potential candidates.

Answer. Of the four processes listed in the matrices, three meet the requirements. Filament winding appears because it can be used to make certain dished shapes, but it does not lend itself well to this shape because it is not axisymmetric. Thermosetting polyester can be shaped by two-part resin transfer or vacuum bag molding.

Both give a good surface finish and high quality if entrapped bubbles are avoided.

Compatible with nonferrous metal (Figure 7.3) Resin transfer molding Filament winding Lay-up methods Vacuum-bag molding

Compatibility with shape (Figure 7.4)

Compatibility with batch size (Figure 7.6)

Quality comment Meets the requirements Meets requirements but awkward for this shape

Fails Meets the requirements

E7.4 Casting slender shapes -surface tension limited. Metal foams can be made by first making an open-cell polymer foam, then embedding this in plaster, burning out the polymer then forcing metal under pressure into the resulting mold to replicate the foam (Figure 7.20). If the cell walls of the polymer foam are cylindrical with a diameter of is 0.5 mm, what pressure will be needed to replicated it using a zinc die-casting alloy? The 2 surface tension of the zinc alloy is is 0.76 J/m .

Answer. The pressure required to overcome surface tension and force the metal into a cylindrical channel of diameter 2r is p  2 / r . The overpressure needed to force the zinc alloy into a cylinder of radius 0.25 mm is 2 2 2 x 0.76 p   6080 N / m  6.08 kPa  3 r 0.25x10

Figure 7.20. A metal foam made by replicating a polymer foam.

This is a little less than 0.1 bar.

E7.5 Casting slender shapes – viscosity limited. (a) A cylinder-head casting has fins with a thickness w = 2mm and length L = 20mm., extensive in the third dimension. How long will it take to fill the fins with molten -3 aluminum under a pressure p of 5 kPa if the viscosity of the aluminum is  = 1.4 x 10 Pa.s? o

(b) Molten glass at 1000 C has a viscosity of 17 Pa.s. If glass were cast into the fin under the same pressure of 5 kPa, how long would it take to fill it?

Answer. (a)The partly filled fin is shown in the figure.

The volume per second entering the channel of width w is given by equation (7.2,a).

dV pw 3  dt 12 x Where  is the viscosity of the fluid x is the length of plate in contact with the flowing metal. If the front advances by dx , a volume dV  w.dx must enter the channel per unit depth, so

dx 1 dV pw 2   dt w dt 12 x 70

Integrating from x  0 to x  L gives the time to fill the channel as 2

6  L    p w  Inserting data for aluminum gives a filling time of 0.17 sec. t

o

3

(b) Inserting data for molten glass at 1000 C gives a filling time of 2 x 10 s = 33 mins. A much higher pressure is required to cast glass that to cast aluminum.

E7.6 Consequences of friction hills. A copper ingot is compressed between two anvils. If the yield strength  y

of copper is 80 MPa and the coefficient of friction  at the interface between the copper and the anvil is 0.1, what is the maximum forming pressure at the center of the ingot when the aspect ratio 2w / h is 20:1?

Answer. The forming pressure across the ingot, is given by equation (7.3) of the text:  2 w  x  p  1.15  y exp  h  Evaluating this at the mid-point of the ingot where x  0 gives

p  8.5 y  680 MPa

Compressing a copper ingot

E7.7 Powder compaction. A cylindrical magnet is to be made from Alnico (an iron-aluminum-nickel-cobalt alloy) by powder compaction and sintering. The aspect ratio of the cylinder, h / 2r , is 2. The coefficient of friction  of the powder with the die wall, if unlubricated, is 0.5. By what factor will the compaction pressure have fallen at the mid plane of the cylinder because of die-wall friction? If, instead, a lubricant is mixed with the powder (it will burn out during sintering), reducing the coefficient of friction to 0.05, by how much will the compaction pressure have fallen at the mid plane?

Answer. The pressure drop in the powder with distance x from the compression platen is given by equation (7.4) of the text: 2 x p( x )  po exp 

r

With a coefficient of friction of   0.5 , the pressure drop at the point x  h / 2 is p h  exp    exp 4  po r A coefficient of friction of 0.5 results in a large pressure drop from p to only 0.14 p at the mid plane. Lubrication, reducing  to 0.05, reduces this drop to 0.82 p.

Compacting a powder

E7.8 Elevator control quadrant. The quadrant of Figure 7.21 is part of the control system for the wing-elevator of a commercial aircraft. It is to be made of a light alloy (aluminum or magnesium based). It weighs about 5 kg. The minimum section thickness is 5 mm, and – apart from the bearing surfaces – the requirements on surface finish and precision are not strict: surface finish  10 µm and precision  0.5 mm. The bearing surfaces require a surface finish  1 µm and a precision  0.05 mm. A production run of 100 is planned. (a) Itemize the function and constraints, leave the objective blank and enter “Choice of process” for the free variable.

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Figure 7.21. Elevator control quadrant

(b) Use the charts of figures 7.3, 7.4 and 7.6 to identify processes to shape the quadrant.

Answer. (a) The function and constraints are listed in the table.

Function



Constraints

      

Objective Free variable

Shape an elevator quadrant Material class: aluminum or magnesium alloy Shape class: 3D-hollow Mass: 5 kg Minimum section: 5 mm Tolerance: 0.5 mm, (0.05mm on bearing surfaces) Surface roughness: 10 µm (1 µm on bearing surfaces) Planned batch size of 100

Technical constraints

Quality constraints Economic constraint



Choice of process

(b) Applying the first two constraints and the last, using the charts of Figures 7.3, 7.4 and 7.6 leads to the following picture.

Compatible with non-ferrous metal (Figure 7.3) Sand casting Die casting Investment casting Centrifugal casting Forging Extrusion Sheet forming Powder methods Electro-machining Machining from solid Additive manufacture

Compatibility with shape (Figure 7.4)

Compatibility with batch size (Figure 7.6)

Quality comment Surface roughness irrelevant; low die cost

Fails Allows very intricate shapes Fails Fails Fails Fails Fails Extremely slow and expensive High precision possible; much waste generated A possibility – see Chapter 8 for details

Five processes survive. Machining or electro-machining from solid, for such a complex shape, presents difficulties: multiple jigging, difficult access for tooling. Sand and investment casting and additive manufacture can cope with the complexity. One of these looks like the best choice. A documentation search confirms that investment casting is extremely versatile, both for material and shape.

E7.9. Casing for an electric plug. A plug has a casing, pins, connectors, a cable clamp, fasteners, and, in some plugs, a fuse (Figure 7.22). How would you make the two-part insulating casing? For safety reasons, it is to be made of a thermosetting plastic such as phenolic. Each part weighs about 30 6 grams. A batch size of 2 x 10 is planned. The required tolerance of 0.3 mm and surface roughness of 1 m must be achieved without using secondary operations. (a) Itemize the function and constraints, leave the objective blank and enter “Choice of process” for the free variable. (b) Use the charts of figures 7.3, 7.4 and 7.6 to identify processes to shape the casing.

Answer. (a) The table lists the function and constraints.

72

Figure 7.22. Electrical plug

Function



Shape an electric plug casing

Constraints

      

Material class: thermosetting polymer Shape class: 3D-solid Mass: 0.03 kg Minimum section: 2 mm Tolerance: 0.3 mm Surface roughness: 1µm 6 Batch size of 2 x10

Objective Free variables

Technical constraints

Quality constraints Economic constraint

-

Choice of process

(b) Applying the first two and the last of the constraints using the charts of Figures 7.3, 7.4 and 7.6 leads to the following picture.

Compatible with thermosetting polymers (Figure 7.3) Machining Injection molding Compression molding Polymer casting Resin transfer molding Additive manufacture

Compatibility with shape (Figure 7.4)

Compatibility with batch size (Figure 7.6) Fails

Quality comment

Meets requirement for finish and tolerance Meets requirement for finish and tolerance Fails Meets requirements for finish and tolerance Fails

I n j e c ti o n

molding, compression molding and resin transfer molding meet the requirements. To get further we need to explore documentation and examine cost in more detail.

E7.10 Ceramic valves for taps. Few things are more irritating than a dripping tap. Taps drip because the rubber washer is worn or the brass seat is pitted by corrosion. Ceramics have good wear resistance and excellent corrosion resistance in both pure and salt water. Many household taps now use ceramic valves.

Figure 7.23 shows how they work. A ceramic valve consists of two disks mounted one above the other, spring-loaded so that their faces are in contact. Each disk has a diameter of 20 mm, a thickness of 3 mm and weighs about 10 grams. In order to seal well, the mating surfaces of the two disks must be flat and smooth, requiring high levels of precision and surface finish: typically, a tolerance < 0.02 mm and a surface roughness < 0.1 µm. The outer face of each has a slot that registers it o and allows the upper disc to be rotated through 90 (1/4 turn). In the “off” position the holes in the upper disc are blanked off by the solid part of the lower one; in the “on” position the holes are aligned. A 5 6 production run of 10 –10 is envisaged. The task is to select a process to make the ceramic disks.

Figure 7.23. Tap with ceramic valve

(a) List the function and constraints, leave the objective blank and enter “Choice of process” for the free variable. (b) Use the charts of figures 7.3, 7.4 and 7.6 to identify processes to shape the casing

Answer. (a) The table lists the function and constraints.

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Function

 Shape a ceramic valve

Constraints

      

Material class: technical ceramic Shape class: prismatic Mass: 0.01 kg Minimum section: 3 mm Tolerance: 0.02 mm Surface roughness: 0.1µm 5 6 Planned batch size of 10 –10

Objective

--

Free variables

 Choice of process

Technical requirements

Quality requirements Economic requirement

(b) Applying the constraints leaves two possible candidates: powder methods and machining (some ceramics can be machined). The tolerance and roughness requirements for the mating faces require a subsequent grinding, lapping or polishing step.

Compatible with ceramics (Figure 7.3) Powder methods Electro-machining Machining

Compatibility with shape (Figure 7.4)

Compatibility with batch size (Figure 7.6)

Quality comment Will need finishing treatment for the faces

Fails Limits material choice to machinable ceramics

Grinding, lapping, polishing

Routinely used to finish ceramics

E7.11 Shaping plastic bottles. Polyethylene bottles (Figure 7.24) are used to contain fluids as different as milk and engine oil. A typical 1 liter polyethylene bottle weighs about 30 grams and has a wall thickness of about 0.8 mm. The shape is 3-D hollow. The batch size is large (1,000,000 bottles). Which process could be used to make them? (a) List the function and constraints, leave the objective blank and enter “Choice of process” for the free variable.

Figure 7.24. PE bottle

(b) Use the charts of figures 7.3, 7.4 and 7.6 to identify processes to shape the casing.

Answer. (a) The table lists the function and constraints.

Function



Shape a polyethylene bottle

Constraints

      

Material class: Polyethylene (or thermoplastic) Shape class: 3-D hollow Mass: 0.02 – 0.04 kg Minimum section: 0.7 - 1 mm Tolerance: 1 mm Surface roughness: 10 µm 6 Planned batch size of > 10

Objective Free variables

Technical constraints

Quality constraints Economic constraint

-

Choice of process

(b) Injection molding, blow molding meet the requirements. Although 3-D hollow shapes can be made by injection molding, it does not lend itself well to closed shapes like bottles, which could only be produced by molding two halves separately and then joining. We conclude that blow molding is the preferred process.

Compatible with

Compatibility

Compatibility

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Quality comment

polymers (Figure 7.3) Machining Injection molding Blow molding Compression molding Rotation molding Thermo-forming Polymer casting Additive manufacture

with shape (Figure 7.4)

with batch size (Figure 7.6) Fails Meets technical constraints Meets technical constraints

Fails Fails Fails Fails Fails

Figure 7.25. Car hood.

E7.12 Car hood. As weight-saving assumes greater importance in automobile design, the replacement of steel parts with polymer-composite substitutes becomes increasingly attractive. Weight can be saved by replacing a steel hood with one made from a thermosetting composites such as glass-fiber/polyester or carbon fiber/epoxy. The weight of the hood depends on the car model: a typical composite hood weighs 8 10 kg. The shape is a dished-sheet and the requirements on tolerance and roughness are 1 mm and 2 m, respectively. A production run of 100,000 is envisaged. (a) List the function and constraints, leave the objective blank and enter “Choice of process” for the free variable. (b) Use the charts of figures 7.3, 7.4 and 7.6 to identify processes to shape the casing.

Answer. (a) The table lists the function and constraints.

Function



Shape a car hood

Constraints

      

Material class: Thermosetting composite Shape class: Dished sheet Mass: 8 – 10 kg Minimum section: 3 mm Tolerance: 1 mm Surface roughness: 2 µm Planned batch size of 100,000

Objective Free variables

Technical constraints

Quality constraints Economic constraint

-

Choice of process

(b) Applying the constraints leaves resin transfer molding as the prime choice.

Compatible with nonferrous metal (Figure 7.3) Resin transfer molding Filament winding Lay-up methods Vacuum-bag molding

Compatibility with shape

Compatibility with batch size

(Figure 7.4)

(Figure 7.6)

Quality comment Meets the requirements

Fails Fails Fails

E7.13 Cost. A process is needed to mold the wide-necked ABS containers show in Figure 7.26. It is suggested that they might be blow-molded, injection molded or rotation molded. The table lists approximate values for the parameters that enter the cost model. Use these to assess the cheapest 6 process choice for a batch size of 10 . Make a plot of cost per unit against batch size, n, to identify the best choice using equation (7.15) of the text: C 

 C mCm 1 C  t   c  C&oh   C pp   1  f  n n& Ltwrof 

Figure 7.26. ABS containers.

75

Parameters Material cost, mCm/(1 – f) ($)

Basic overhead Coh ($ per hour) Capital write-off time twrof (years) Load factor, L Dedicated tooling cost, Ct ($) Capital cost, Cc ($) Production rate, n&(per hour) Post-processing cost, Cpp ($/unit)

Blow molding 0.1 100

Injection molding 0.1 100

Rotation molding 0.1 100

5 0.5 3000 16,000 270

5 0.5 10,000 160,000 240

5 0.5 5000 6,800 8

0

0

0

Answer.

The output of the cost model for the three

processes is show below. Blow molding is the least expensive for all but the smallest batch sizes.

E7.14. Modelling process cost. Create an EXCEL spread sheet to evaluate and plot the cost equation, equation (7.15) of the text:  C mCm 1 C C   t   c  C&oh   C pp   1  f  n n& Ltwrof  The parameters and their units are defined in the table.

The figure below suggests a structure.    

Parameters Mass of component, m Material cost, Cm Scrap fraction, f Basic overhead Coh

Capital write-off time twrof Load factor, L Dedicated tooling cost, Ct Capital cost, Cc Production rate, n& Post-processing cost, Cpp

Units kg $/kg $ per hour Years $ $ Units per hour

$ per unit The heading names the process, here, Sand casting.

Block A1-A4 defines the component for which the cost will be calculated (here, 1 connecting rod).

Block B1- C5 allows entry of the parameters characterizing the process system.

Block H1-L6 does the same for the parameters describing the component that the process will make.

Overheads are the aggregated costs of the energy, labor, information cost (R&D, royalties, licenses), rent and administration, pro-rated to give an hourly rate, C&oh , spread across the output of the company. The basic elements that enter C&oh are entered in block B7 – C12, and summed and divided

by the production rate n&in line B13 -C13 to evaluate C&oh . 

That sets up values for all the parameters that appear in equation (7.15). The calculation proper starts on Line 15, which sets up the steps in batch size for which a cost will be calculated, spaced in an approximately logarithmic way. The individual terms of equation (7.15) are evaluated in lines 18 - 23 and summed for each batch size in line 26. The batch-sizes are repeated in line 25 to make plotting easier.

Check the proper running of the spread sheet by using it to plot the unit cost of sand-casting a connecting rod using the input data shown in the spread sheet below.

76

Suggested structure for the spread sheet, with data for sand casting of con rods.

Answer. The figure shows the predicted cost per unit if made by sand casting. The cost of the mold (Ct) is low, so producing just one or two castings cost only a few hundred dollars. But the process is slow (producing only 2 per hour), so the unit cost does not diminish much at higher batch sizes.

E7.15. Modelling process cost. Use the spread sheet created in Exercise E7.14 to plot the cost of making the same con rod by forging, entering the data show in the table below. Leave the inputs for the overhead rate, C&oh unchanged at $100/hr. What accounts for the high unit cost at batch sizes below 100? Why does the curve flatten out at about $10 per unit instead of continuing to fall with increasing batch size?

Process system parameters – forging Capital cost of equipment C c ($)

Value 55,000

Rate of production, n&(units/hr) Load factor L (-) Write off time, twrof (years)

200 0.5 5

Project related parameters – cast iron con-rod

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Value

Final mass of component, m (kg) Cost of material C m ($/kg)

1.0 5.0

Loss factor, f (-) Tooling - Cost of die, C t ($)

0.3 8,500

Post processing cost, C pp ($/unit)

1.0

Answer. The figures shows the spread sheet and the predicted cost per unit if made by forging. At batch sizes below 300, the fixed costs (tooling) dominate. Above 3000, variable costs, particularly materials, dominate

At batch sizes below 300, the fixed costs (tooling) dominate

78

Above 3000, variable costs, particularly materials, dominate

The next four exercises require the use of the Ansys Granta EduPack 3 Materials Selection software.

E7.16 Selecting a joining process (1). Use the process selection capability of the Ansys Granta EduPack to select a joining process to meet the following requirements.

Function



Create a permanent butt joint between steel plates

Constraints

    

Material class: carbon steel Joint geometry: butt joint Section thickness: 8 mm Permanent Watertight

Objective

--

Free variables



Technical constraints

Choice of process

Answer. Applying the EduPack to the problem gives the selection shown in the below.

Selection using EduPack Level 2 Brazing Friction welding Friction-stir welding Gas – metal arc (MIG) welding Gas-tungsten arc (TIG) welding Manual metal arc (MMA) welding Oxyacetylene welding Power (electron, laser) beam welding Ultrasonic welding

Comment

There are many ways to make a permanent watertight joint in steel.

E7.17 Selecting a joining process (2). Use the process selection capability of the Ansys Granta EduPack to select a joining process to meet the following requirements.

Function

 Create a watertight, demountable lap joint between glass and polymer

Constraints

    

Objective Free variables

Material class: glass and polymers Joint geometry: lap joint Section thickness: 4 mm Demountable Watertight

Technical constraints

-

Choice of process

Answer. Applying the EduPack to the problem gives the selection shown in the below.

Selection using EduPack Level 2

Comment

Snap fit Threaded fasteners 3 Ansys Granta EduPack (2023) https://www.ansys.com/en-gb/products/materials/granta-edupack

79

Flexible adhesives Rigid adhesives

All are practical

E7.18 Selecting surface treatment processes (1). Use the process selection capability of the Ansys Granta EduPack to select a joining process to meet the following requirements.

Function

 Increase the surface hardness and wear resistance of a high carbon steel component

Constraints

 

Objective

--

Free variables



Material class: carbon steel Increase surface hardness and wear resistance

Technical constraints

Choice of process

Answer. Applying the EduPack to the problem gives the selection shown in the below.

Selection using EduPack Level 2 Carburizing and carbonitriding Electroless plating Electroplating Induction and flame hardening Laser surface hardening and melting Metal flame spray Nitriding Vapor metallizing (PVD)

Comment There are many ways to increase the surface hardness of carbon steel. A search for documentation is required to narrow the choice further.

E7.19 Selecting surface treatment processes (2). Use the process selection capability of the Ansys Granta EduPack to select a joining process to meet the following requirements.

Function



Apply color and pattern to the curved surface of a polymer molding

Constraints

  

Material class: thermoplastic Purpose of treatment: aesthetics, color Curved surface coverage: good or very good

Objective

--

Free variables



Technical constraints

Choice of process

Answer. Applying EduPack to the problem gives the selection shown in the below.

Selection using EduPack Level 2 Cubic printing Electroless plating Electrophoretic and Autophoretic painting Organic solvent-based painting Pad printing Vitreous enameling Water-based painting

Comment All are practical. A search for documentation is required to narrow the choice further.

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Ch. 8 Additive manufacturing 8.11 Exercises E8.1. Use the internet to explore headlines about AM in the contemporary press, meaning this year. What perceptions of the potential for AM emerge most strongly?

Answer. Here are four headlines from 2023, the year this book was written. They reflect the perceived importance of AM for defense and sustainable manufacturing, and the great increase in the scale of AM products that are being produced in that year.

“Manufacturing advances are key to national security. Advances in manufacturing capabilities are critical to the Defense Department's overall national security strategy, a senior Pentagon technology official said today.

DOD launched America Makes in 2012 to bring together industry, academia, government and workforce to accelerate the adoption of additive manufacturing in defense-related manufacturing.”

4 US Department of Defense,.15 August, 2023 “Manufacturing Innovation. The Ministry of Defence (MOD) has set up the first framework for additive manufacturing, teaming up with industry to unlock the benefits of Additive Manufacturing. It will work with industry on tasks of increasing complexity, all focused around additively manufacture defence inventory parts to fit these to in-service platforms.”

5 UK Ministry of Defence (MOD), 6 April 2023 “By its very nature, additive manufacturing is considered a sustainable technology. Because additive technologies deposit material in a structured way instead of removing it, as in some other processes such as machining, 3D printing materials are used very efficiently“ 6 Tct Magazine, 20 October 2023 “Space startup forced to call off launch of world's first 3D-printed rocket. Relativity Space was set to launch its Terran 1 booster Wednesday from Florida’s Cape Canaveral Space Force Station, but technical difficulties at the launchpad forced the company to call off the attempt. Relativity’s booster is the first to be made almost entirely with 3D printing. The test flight was being closely watched because, if successful, it would represent a milestone for the space industry.”

7 NBC news, 8 March 2023 E8.2. When is AM appropriate? And which process? (1) The most expensive of the Rolls Royce range of vehicles is the Phantom, starting at around $500,000. Production averages 700 per year. But you don’t buy a Phantom, you commission it: almost every aspect of the interior of the car can be customized to meet the taste of the purchaser. What are the opportunities for Additive manufacturing here? Is AM used in building Phantoms today? Use the internet to find out Answer. Additive manufacturing is well suited to mass customization: small volume production of many similar but not identical parts in plastics and metals. Facia components, heating ducts, drink holders, lighting fixtures, storage containers, all require expensive tooling if made by conventional methods. They are economic when many identical units are planned, but not when the production volume is small. AM allows small volume production without the tooling cost.

4 https://www.defense.gov/News/News-Stories/Article/Article/3493769/tech-official-says-manufacturing-advances-are-key-to-nationalsecurity/ 5 https://des.mod.uk/first-framework-for-additive-manufacturing-awarding-industry-contracts/ 6 https://www.tctmagazine.com/ 7 https://www.nbcnews.com/science/space/worlds-first-3d-printed-rocket-set-make-inaugural-flight-rcna73868

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Control of color and texture of polymer parts is practical with fused material extrusion (MEX) and material jetting (MJT). Laser powder bed melting (PBF) and laser engineered net shape (DED) methods produce metallic parts from the range of metals listed in Table 8.1.

The BMW Group (makers of Rolls Royce cars) has been using AM components in series production since 2012.

8 More than 10,000 additively-manufactured parts have been built into the Rolls-Royce Phantoms to date .

E8.3 When is AM appropriate? And which process? (2) You are approached by a company with an idea for a light-weight plastic support for mobile phones to allow hands-free reading of text messages. The company’s designer has made sketches and CAD renderings, but the CEO wants reassurance that the support feels and looks right, and is stable when carrying a phone. The CEO requests a prototype, using the same plastic (ABS) as the planned production model. Does this sound like a job for additive manufacturing? If it is, what system would be most appropriate?

Answer. The requirements – a single prototype, small in size, made from a thermoplastic readily available for AM use – match the capabilities of AM well. Table 8.1 of the text suggests Fused material extrusion (MEX) or Powder bed fusion (PBF) as possible processes. The first, MEX, is practical with low cost, desk-top systems, and is well-tried for projects of this sort. It’s the best choice.

E8.4 When is AM appropriate? And which process (3)? A typical bicycle frame has 14 parts that are assembled by brazing, welding or with fasteners. The cost of assembly is significant. A standard street bike is optimized for cost; weight is a consideration, but not the decisive one. If customization is wanted, the purchaser of a street bike does if for themselves. Production runs are high. The material – mild steel – is readily available. By contrast, a bike for a professional all-terrain competitor, today, is customized to the body shape of the competitor. The material of choice is titanium alloy Ti-6AL-4V. The section shape and internal structure is optimized to minimize weight with constraints on strength and stiffness. It can be complex.

Compare the relative merits of AM and standard commercial manufacturing methods (CM) for the production of street bikes and competition bikes. Assign a value between 1 (poorly suited for AM) and 5 (highly valued suited for AM) using a copy of the pro-template of Figure 8.12. Do you think that this would be provide a helpful startingAM Value assessment point for a Street bike Competition bike further Design attribute Desired batch size Shape complexity Part integration desirable Lightweighting a priority Customization valued

Value

Rank

Value

Rank

discussion with the designers about the design of their bikes?

8 https://www.press.bmwgroup.com/global/article/detail/T0261924EN/series-components-made-by-3d-printers:-bmw-group-expandsuse-of-additive-manufacturing-processes?language=en

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Figure 8.12. Blank target plot

Answer. The table lists estimated values for the AM-suited design attributes. The resulting target plots are shown below. What conclusions? AM could be a valid choice for highly customized competition bikes. It offers little for street bikes unless a special edition (a centennial model, for example) was planned.

Design attribute Desired batch size Shape complexity Part integration desirable Lightweighting a priority Customization valued

Street bike Value Rank Large 1 Moderate 2 Yes 3 No 1 No 1

Competition bike Value Rank Very small 5 High 5 Yes 3 Yes 5 Yes 5

E8.5 Multicolor Additive manufacturing. You are approached by a sculptor intrigued by the idea of using AM to make solid or hollow works of art from plastic or ceramic. Control of color and surface texture are important. Which AM technology would you suggest? Use the internet to gather additional information.

Answer. Making multi-color plastic parts by conventional manufacturing is complex, requiring either a multistep molding process (e.g., over-molding) or post-processing of unicolor parts (e.g., screen printing or manual coloring). Several of the AM technologies shown in Figure 8.2 can do it. Material extrusion (MEX, shown as the cover picture of this chapter) can combine feeds to build multi-colored parts. Multi jetting (MJT) allows controlled texture and color gradients. Binder jetting (BJT) produces full-color objects from gypsum. Mcor’s sheet lamination (SHL) process does the same with colored papers.

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https://3faktur.com/en/full-color-3d-printing-technology-overview/ https://www.sculpteo.com/blog/2017/03/22/multicolor-3d-printing-how-does-it-work/

E8.6. Cost components Direct Energy Depostion, (DED) systems, typified by the Optomec LENS 450, build metal structures by direct energy deposition from powder feedstock. The equipment costs $350,000, deposits metal at 0.15 kg/hr and consumes 12 kW of electrical power while running. One full-time operator can manage four sets of LENS 450 systems at any one time. 

If metal powder costs $35/kg and the build efficiency is 80%, meaning that 20% of the powder is lost, what is the material cost per kg of build?  If electrical power costs $0.2 per kW.hr, what is the energy cost per kg of build?  If labor plus overheads is costed at $80/hr, what the contribution of the direct cost labor to the cost per kg of build?  If capital write-off time is 5 years, and the equipment is productive for 8 hours per day for 250 days per year, what is its contribution to the cost per kg of build?

Surveying the results, what are the main cost drivers in building a 1 kg structure using a LENS system?

Answer.  

A 20% material loss means that powder costing $35/(1.0-0.2) = $43.8 is required for a 1 kg build At a build rate of 0.15 kg/hr, it takes 6.7 hours for a 1kg build, consuming 80 kW.hr. At $0.2/kW.hr, the cost of the electrical power required for a 1 kg build is $16.  An overhead rate of $80/hr, shared between four systems, contributes 6.7 x 80/4 = $133 to the cost of a 1 kg build  The reported load factor (8 hrs/day, 250 days/yr, 5 yrs) means the equipment is productive for 10,000 hours over its life time. The cost of the capital investment of $350,000, pro-rated over productive life, is $35 per hour. It contributes 6.7 x 35 = $235 to the cost per kg of the build The big contributions to the cost of a 1 kg build are time (the overhead of $133/kg) and capital ($235/kg). The cost of material and energy are much smaller.

E8.7. Cost components of AM by material extrusion. Material extrusion (MEX) systems, typified by Cincinnati Inc SAAM, build polymer or polymer-based composite parts. The system extrudes material at 0.07 kg/hr, consumes 0.25 kW, and costs $10,000. It uses a nozzle diameter of 0.4 mm and builds in layers 0.2 mm high, defining the resolution and surface roughness. 

If PLA thread costs $8/kg and the build efficiency is 90%, meaning that 10% of the PLA is lost, what is the material cost per kg of building a PLA part?  If electrical power costs $0.2 per kW.hr, what is the energy cost per kg of build?  If labor plus overheads is costed at $60/hr and 10% of one operator’s time is assigned to the system while building, what is its contribution to the cost per kg of build?  If capital write-off time is 5 years, and the equipment is productive for 8 hours per day for 250 days per year, what is its contribution to the cost per kg of build?

Surveying the results, what are the main cost drivers in building a 1 kg structure using PLA extrusion?

Answer.    

A 10% material loss means that PLA thread costing $8/(1.0-0.1) = $9 is required for a 1 kg build At a build rate of 0.07 kg/hr, it takes 14.3 hours for a 1kg build, consuming 3.6 kW.hr. At $0.2/kW.hr, the cost of the electrical power required for a 1 kg build is $0.7.

An overhead rate of $60/hr contributes 0.1 x 14.3 x 60 = $86 to the cost of a 1 kg build The reported load factor (8 hrs/day, 250 days/yr, 5 yrs) means the equipment is productive for 10,000 hours over its life time. The cost of the capital investment of $10,000, pro-rated over productive life, is $1 per hour. It contributes 14.3 x 1 = $14.3 to the cost per kg of the build.

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The biggest contribution to the cost of a 1 kg build is time (the overhead of $86/kg), responsible for 78% of the total cost. System development to increase the extrusion rate and to make the system semi-autonomous, requiring less operator time per build, will, in the future, reduce this contribution.

E8.8. The table summarizes the design features that best match the capabilities of Additive and Conventional manufacturing. Propose two products that meet as many features listed for CM on left, and two that do the same for those of AM on the right, as possible.

For Conventional Manufacture (CM) More precise than AM Less expensive feedstock than AM Greater choice of materials Economic at large production volume Economic for simple shapes Easily machined materials Fast production rate Centralized manufacture No size limitations

For Additive manufacturing (AM) Economic at low production volumes Shape complexity and part integration Ease of customization Little or no expensive tooling No need to hold large stocks Can be cheaper when material costs high Can be cheaper when machining costs high Enables localised, distributed manufacture Enables extreme light-weight design

Answer.

Conventional manufacturing (1). High volume aluminum appliance housing weighing 0.25kg, with near-uniform wall thickness, designed for easy removal from a 2-part mold (meaning no re-entrants or internal cavities and with appropriate chamfer). Note the expensive toolsteel die.

Chosen process: pressure die casting.

Conventional manufacturing (2). Roughly 60 billion Lego bricks are made each year. They are made from ABS (acrylonitrile-butadiene-styrene), a non-toxic thermoplastic that can be colored and has good wear properties. The bricks have a simple shape that can be molded in a twopart mold.

Die casting (source: indiamart.com)

Chosen process: injection molding.

Additive manufacturing (1). A design for a light-weight bicycle saddle for competition use envisages an innovative design inspired by the All-Wings Falcon Road MTB Saddle shown on the right. The proposal is to customize the dimensions and weight of each saddle to match those of of individual riders. It is to be made from and ABS-elastomer blend. It is anticipated that no two saddles with be identical.

Leggo bricks

Chosen process: material extrusion (MEX)

Additive manufacturing (2). Roughly 60% of the population of a developed country wear glasses, most of the over 50, who have become used to paying over $100 for frames. Some find that standard frames suit them, but others have problems because of the shape of their faces, particularly their noses. The minimum order quantity for an injection-molded frame is about 500, making the process uneconomic if used to meet individual needs. There is a market for customized frames, based on laser scans of the face, to optimally position the lenses and give a comfortable fit.

Chosen process: DED – laser sintering of PA 2200, a standard material for additive manufacture.

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Innovative bike saddle

Customized glasses

E8.9. Complexity, batch size and cost. Figure 8.6 of the text showed how manufacturing cost changes with complexity and production volume for conventional and additive manufacturing. The two images in Figure 8.13 are both heat exchangers. The first, a cast iron house radiator, is made up of 10 simple units clamped together and is produced in large numbers. The second is, an Inconel 718 rocket nozzle, with internal cooling channels (sectioned here) and is at present a prototype. Position each on Figure 8.6, identifying the manufacturing path that best suits it.

Figure 8.13. Heat exchangers.

Answer. The two heat exchangers are positioned on the cost/batch-size/complexity plots below, identifying the best manufacturing class. Simple (here, repeated) shapes and large production runs is most economically done by conventional casting.

Prototype production of complex shapes is most economically done by additive manufacture.

E8.10. Modelling AM process cost: DED. Create an EXCEL spread sheet to evaluate and plot the cost equation, equation (8.1) of the text:

C AM 

mC fdsk

1  f 



 C STC m  Cc    C&oh   C post n D& Ltwrof 

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The Excel spread sheet of Chapter 7, explained in Exercise 7.18, can be adapted to do this. Use it (or start from scratch if it is not available) to create the spread sheet shown Figure 8.14.

The heading names the process, here, Direct energy deposition, DED. Block A1-A3 defines the component for which the cost will be calculated. Block B1- C5 allows entry of the parameters characterizing the AM system. Block H1-L6 does the same for the parameters describing the component that the process will make.

Overheads are the aggregated costs of the energy, labor, information cost (R&D, royalties, licenses), rent and administration, pro-rated to give an hourly rate, C& oh , spread across the output of the company. The basic & elements that enter Coh are entered in Block B7 – C12, and summed and divided by the production rate n& in line B13 -C13 to evaluate C& oh .

That sets up values for all the parameters that appear in equation (8.1). The calculation proper starts on Line 15, which sets up the steps in batch size for which a cost will be calculated, spaced in an approximately logarithmic way. The individual terms of equation (8.1) are evaluated in lines 18 - 23 and summed for each batch size in line 26. The batch-sizes are repeated in line 25 to make plotting easier.

The input data are summarized in Table E8.5, below. Enter the data in this table and plot the unit cost of the component, made by DED, as a function of the batch size n. Inspect the spread sheet to answer the following questions.   

Which term of the model dominates the unit cost at a batch size of 3?

Which dominates the unit cost at a batch size of 300?

If DED is to be made more competitive, what should be the priority in its further development?

Figure 8.14 The EXCEL spread sheet

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Table E8.5. Input for the cost model Process system parameters Capital cost of equipment C c ($)

100,000

Build rate D& (kg/hr), Load factor L (-) Capital write-off time twrof (yrs)

0.5 0.5 5

Project related parameters Final mass of component, m (kg) Material cost C fdsk ($/kg)

1 25.0

Loss factor, f (-) Cost of developing the STC file, C STC ($)

0.2 1,000

Post-processing cost C pp ($/kg)

10

Components of overhead Cost of labor, C lab ($/person/hr)

70

Operators per system, Nop (-)

1

Cost of energy, Cenergy ($/kW.hr)

0.2

Power required by the system, P (kW) Indirect overheads (heating, lighting, admin) C.indr ($/hr)

10 28

Direct Energy Deposition (DED)

Answer. The output of the model looks like this. The cost of the STC file is small compared with the tooling cost of most conventional processes, but the long processing time and higher material cost gives a higher plateau.

At a batch size of 3, the cost of developing the STC file dominates.

At a batch size of 300, the overheads dominate, because the process is slow.

Making DED more competitive requires that it be made faster, or that multiple copies of the component can be made during the same build.

E8.11. Modelling AM process cost: Material extrusion (MEX). A design for an eco-friendly bicycle light includes a plastic casing, estimated to weigh 100 grams, to be made of PLA feedstock costing $8/kg. To make sure that the casing is compatible with the contents and with external fixtures, the designer commissions a prototype, to be made by material extrusion (MEX) using a Cincinnati Inc SAAM system. This model extrudes material at 0.07 kg/hr, consumes 0.25 kW, and costs $10,000. Overheads are costed at $50 per employee per hour; managing the system requires 10% of one employee’s time. The input data are assembled in Table E8.6, below. Enter these in the EXCEL spreadsheet of Exercise 8.10 to show how the part-cost changes with batch size.

What is the cost of the single prototype? The target price for this part, when in production, is $2. Is MEX a viable production process for producing the part?

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Table E8.6. Inputs for MEX the cost model Process system parameters Capital cost of equipment C c ($)

AM FDM 10,000

Build rate D& (kg/hr), Load factor L (-) Capital write-off time twrof (yrs)

0.07 0.23 5

Project related parameters Final mass of component, m (kg) Material cost C fdsk ($/kg)

0.1 8.0

Loss factor, f (-) Cost of developing the STC file, C STC ($)

0.1 400

Post-processing cost C pp ($/kg)

0

Components of overhead Cost of labor, C lab ($/person/hr)

50

Operators per system, Nop (-)

0.1

Cost of energy, C energy ($/kW.hr)

0.2

Power required by the system, P (kW) Indirect overheads (heating, lighting, admin) C.indr ($/hr)

0.25 8

Answer. The Excel spread sheet and a plot of its output is reproduced below. The cost of a single prototype is $421 – manageable as reassurance that the design is compatible with its mating parts. The cost drops with increasing production numbers, flattening out at $21 per unit, largely determined by the cost of time reflected in the overhead rate. This exceeds the target cost for the product when in production, so a conventional process better suited for mass production (injection molding, for example) must be sought for the next phase of the design.

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Ch 9. Multiple constraints and conflicting objectives Over-constrained problems are normal in materials selection. Often it is just a case of applying each constraint in succession, retaining only those solutions that meet them all. But when constraints are used to eliminate free variables in an objective function, (as discussed in Section 9.2 of the text) the “active constraint” method must be used. The first six exercises in this section illustrate problems with multiple constraints. Those that follow concern multiple objectives, trade-off methods and penalty functions (Section 9.3 of the text).

E9.1 Multiple constraints: safe, cheap pipe-work. Many industrial systems involve pipes, some enormous (the oil industry), others smaller (domestic gas and water pipes). Almost all must contain an internal pressure and do so, for safety reasons, without yielding (material yield strength  y ) or failing by fast fracture (material fracture toughness K1c ). When the length of pipe-work is large, as it often is, the material cost becomes significant. What is the best material for pipe-work that is safe (no failure) and as cheap as possible when the largest detectable flaw in the pipe wall is (a) 10 microns in length and (b) 10 mm in length?

Figure 9.12. A pipe containing a pressure

Here are the steps. Identify the tensile stress in the wall of a pipe (radius R and wall thickness t) when the pressure difference across the pipe wall is p (Figure 9.12). Write an expression for the cost of unit length of the pipe (radius R and wall thickness t) in terms of its dimension and the density  and cost per kg of the material C m . This is the objective function, the quantity to be minimised. Express the condition that the stress in the wall be (a) less than the yield strength of the material and (b) less than its fracture stress if the largest flaw it contains has a length c. Treat the wall thickness t as the free variable – it must be chosen so that the wall does not fail. Substitute each constraint into the objective function, eliminating t, to get two expressions for the cost, one if yielding is the failure mechanism and one if fracture is the mechanism. Identify the two material indices M1 and M2 associated with the two failure modes. Identify the coupling equation that links them when the two costs are equal.

The requirements are summarized below.

Table 9.2. Requirements for safe low-cost pipework Function Constraints

Objective Free variables

      

Safe low-cost pipe work Pipe diameter 2R and pressure difference p specified Must not fail by yielding Must not fail by fast fracture.

Minimize cost per unit length, C, of the pipe Wall thickness, t Choice of material

The table lists the properties of materials that are used for pipes. Enter the data into an Excel spread sheet. Use it to evaluate the two material indices M1 and M2 and plot them as a min-max diagram to identify the cheapest material when the largest detectable flaw the pipe wall is (a) 10 microns in length and (b) 1 mm in length.

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Table 9.3. Data for materials that are used for pipes Cost Cm ($/kg) 6.1 0.36 0.76 2.4 3.4 3.9 1.6 1.4 2.6 0.05

Materials for pipes Copper 1220 Cast iron, grey Mild steel 2024 Aluminum alloy 304 Stainless steel Lead Polyethylene (PE) Polypropylene (PP) Polyvinyl chloride (PVC) Concrete

Density ρ (kg/m^3) 8940 7,200 7,850 2,770 7,950 10,500 950 900 1,450 2,450

Yield strength σy (MPa) 292 290 300 286 743 10 24 29 44 2

Fr. toughness K1c (MPa.m^0.5) 47 15 70 39 110 12 2 3.8 3.3 0.4

Answer. The pipe is cylindrical with a radius R, and that the wall thickness, t (the free variable), small compared with R. Then the tensile stress in the pipe wall is pR  t The objective function is the cost per unit length of the pipe: C 2 R t C m  where Cm is the cost per kg of the material. The first constraint is that the pipe should not yield – that is, that the tensile stress in the wall should not exceed y. The second is that it should not fail by fast fracture; this requires that this stress should be less than K 1c /  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. Using each constraint in turn to substitute for the free variable t in the objective function leads to the performance equations and indices laid out below.

Objective

Constraints Yield constraint ς 

p R t

 y

Performance equation

Index

C ρ  C1  2 π  p.R 2   m   ς f 

C  M1  m

1/ 2

C  M2  m K1c

y

Substitute for t

C 2 R t C m 

Fracture constraint

ς

pR t



K1c c

 c C 2  2 πp  R 2  π   t

 Cm ρ     K1c 

Substitute for t

The coupling equation is found by equating m1 to m2, giving a relationship between M1 and M2:

M 1  ( πc)1/2 M 2 The position of the coupling line depends on the detection limit, c1 for cracks, through the term ( c) .

The figure shows the appropriate chart with two coupling lines, one for c = 10 mm and the other for c = 10 m.

The data are assembled in the table.

1/2

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Materials for pipes

Cost Cm ($/kg)

Density ρ (kg/m^3)

Yield strength σy (MPa)

Fr. toughness K1c (MPa.m^0.5)

M1  Cm  /  y

M2  Cm  / K1c

Copper 1220 Grey cast iron Mild steel 2024 Aluminum alloy 304 Stainless steel Lead Polyethylene (PE) Polypropylene (PP) Polyvinyl chloride (PVC) Concrete

6.1 0.36 0.76 2.4 3.4 3.9 1.6 1.4 2.6 0.05

8,940 7,200 7,850 2,770 7,950 10,500 950 900 1,450 2,450

292 290 300 286 743 10 24 29 44 2

47 15 70 39 110 12 2 3.8 3.3 0.4

185 8.9 19.8 22.7 35.9 4095 63.3 43.4 85.6 61.2

1152 172.8 85.2 166.9 242.8 3412 760.0 331.5 1142 306.2

The best choice is the material that has the lowest value of the larger of C1 and C2 – the min-max condition.

The best way to find it is to plot the two indices against each other as in the figure below. Two coupling lines are shown, one for c = 10 mm and the other for c = 10 m. The best choice of material is that captured in the box with its corner furthest down the coupling line. When the tolerable flaw size is large, the best choice is mild steel. When it can be guaranteed to be much smaller (10 m), grey cast iron is a cheaper option.

E9.2 Multiple constraints: a light, stiff, strong tie (Part 1) (Figure 9.13).

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  (implying a constraint on stiffness, F /  ). The table summarizes the requirements.

Figure 9.13. A tie.

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Table 9.4. Requirements for the tie-rod Function



Tie rod

Constraints

     

Must not fail by yielding under force F Must have specified stiffness, F/ Length L and axial load F specified Minimize mass m Section area A Choice of material

Objective Free variables

Follow the methods of Chapter 4 to 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 and the coupling equation are (in order):    L      and and M1  M2    y    E  E y   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





EA L

F   E

M1 

    m2  L F  y   

M2 

m1  L2

 E

(1)

mAL Substitute for A

Strength constraint

F  y A

 y

(2)

The symbols have their usual meanings: A = area, L= length,   density, F/ =stiffness, E = Young’s modulus,

 y = yield strength or elastic limit. The coupling equation is found by equating m1 to m2, giving    L      y    E    defining the coupling constant Cc = L/.

or

M2 

94

L



M1

E9.3 Multiple constraints: a light, stiff, strong tie (Part 2). Figure 9.14 is a chart with the two indices M1 and M2 L of exercise E9.2 as axes9. The two are related by the coupling line M2  M1 . Materials appear as white  bubbles, some of which are labeled. Use it to identify candidate materials for the tie (i) when the permitted -3 -2 elastic strain  /L = 10 and (ii) when the permitted strain  /L = 10

Figure 9.14. A chart with the two indices M1 and M2 of exercise E9.2 as axes

Answer. The chart below shows the positions of the coupling line when the permitted strain  /L = 10 and -3

-2 % max  m1 ,m2  that when the permitted strain  /L = 10 . The materials that give a tie with the lowest mass m

meet both constraints are those lying furthest down the appropriate coupling line: CFRP when the permitted strain is 1%, and technical ceramics boron carbide, silicon carbide or alumina when the permitted strain is only 0.1%.

Technical ceramics are available as fibers, but may not be a good bet for a tie rod because of their incipient brittleness (low fracture toughness). To get a more realistic selection for the permitted strain of 0.1%, the constraint has been relaxed, moving the selection box back up the coupling line to the position shown in red. It now encloses low alloy steel, titanium alloys and – again – CFRP.

9 The Ansys-Granta Edupack software allows the construction of charts with axes that are combinations of properties, like

those of  / E and  /  y shown here, the drawing of coupling lines and of a selection box to identify the optimum choice of material.

95

E9.4 Multiple constraints: a cheap column that must not buckle or crush (Part 1) (Figure 9.15). 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 (material compression strength  c , MPa); tall slender columns buckle 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, the first using the constraints that the column must not crush, the second that it must not buckle. The table summarizes the needs.

Table 9.5. Requirements for the column Function Constraints

Objective Free variables

      

Column Must not fail by compressive crushing Must not buckle Height H and compressive load F specified.

Minimize material cost C Diameter D Choice of material

Show that the material indices for selecting the cheapest material for the column for each failure criterion are C   C ρ  and M1   m  M2   m    E 1/2   c 

Figure 9.15. A column

where C m is the material cost per kg,  the material density,  c its crushing strength and E its modulus, and that the coupling equation linking M1 and M2 when the two costs are equal is

M2 

1/ 2

π1/2  F    2  H2 

 M1

Proceed as follows (1) Write an expression for the material cost of the column – its mass times its cost per unit mass, Cm. 96

(2) Express the two constraints as equations, the first expressing the condition that the load must not exceed the crushing load, the second expressing the condition that the load must not exceed the elastic buckling load (3) Use the constraints to substitute for the free variable, D. (4) Read off the group of material properties that minimize the equation for the cost.

(5) Equate the cost when limited by one failure mode to that when limited by the other to find the coupling constant linking M1 and M2.

Answer. We seek materials for a light column with circular section. The column must neither buckle nor crush under a design load F. The material cost of the column, C, (the quantity to be minimized) is



D 2HC m  4 where D is the diameter (the free variable) and H the height of the column, Cm is the cost per kg of the material and  is its density. The column must not crush, requiring that F  ςc πD 2 / 4 C





where c is the compressive strength of the material. Nor must it buckle, requiring that the load F be less than the Euler buckling load (Appendix B, Table B5) F

π 2 EI H2

Here E is Young’s modulus and I is the second moment of area, which, for a circular cross-section (Appendix B, Table B2), is π I  D4 64 The subsequent steps in the derivation of the performance equations are laid out below.

Objective

Constraint Substitute for D

Crushing constraint

C

 4

Ff 

 D2 4

c

Performance equation

Index

C   C1  F H m   c 

Cm 

c

D 2H C m  Substitute for D

Buckling constraint

F 

π 2 EI H

2



π3 D4 E 64 H

2

C2 

2 π

1/2

C ρ  F 1/2H 2  m   E 1/2 

Cm ρ E 1/2

C   C ρ The first performance equation contains the index M1   m  , the second, the index M2   m  .  E 1/2   c  The cost of the column that must not fail by either mode is the greater of these two costs. This is a min-max

~

problem: we seek the material with the lowest (min) cost C which itself is the larger (max) of C1 and C2.

The coupling equation is found by equating C1 to C2, giving

π 1/2  F   M2   2  H 2 

1/ 2

 M1

97

E9.5 Multiple constraints: a cheap column that must not buckle or crush ( Part 2). Data for seven possible candidate materials for the column are listed in the Table. Use these to identify candidate materials when F = 0.1 MN and H = 3m. Ceramics are admissible here, because they have high strength in compression.

Table E9.6. Data for candidate materials for the column Material Wood (spruce) Brick Granite Concrete Cast iron Structural steel Al-alloy 6061

Density  3 (kg/m ) 700 2100 2600 2300 7150 7850 2700

Cost/kg Cm ($/kg) 0.5 0.35 0.6 0.08 0.25 0.4 1.2

Modulus E (MPa) 10,000 22,000 20,000 20,000 130,000 210,000 69,000

Compression strength  c (MPa) 25 95 150 25 200 300 150

Answer. The two performance equations are evaluated in the Table for a column of height H = 3m, carrying a load F = 0.1 MN. The table also lists C~  max( C ,C ). The cheapest choice is concrete.

1

Material Wood (spruce) Brick Granite Concrete Cast iron Structural steel Al-alloy 6061

2

Density  3 (kg/m )

Cost/kg Cm ($/kg)

Modulus E (MPa)

Compression Strength c (MPa)

C1 $

C2 $

C%

700 2100 2600 2300 7150 7850 2700

0.5 0.35 0.6 0.08 0.25 0.4 1.2

10,000 22,000 20,000 20,000 130,000 210,000 69,000

25 95 150 25 200 300 150

4.2 2.3 3.1 2.2 2.6 3.0 6.5

11.2 16.1 35.0 4.7 16.1 21.8 39.5

11.2 16.1 35.0 4.7 16.1 21.8 39.5

$

E9.6 Multiple constraints: a cheap column that must not buckle or crush ( Part 3). Figure 9.16 shows a material chart with the two indices of Exercise E9.3 as axes. Identify and plot coupling lines for selecting materials for a column with F = 1 MN and H = 3m (the same conditions as above), and for a second column with F = 10 MN and H = 1m.

98

Figure 9.16. A chart with the two indices of Exercise E9.3 as axes

Answer. The coupling equation is found by equating C1 to C2 giving 1/ 2

π 1/2  F     M1 2  H 2  2 It contains the structural loading coefficient F/H . Two positions for the coupling line are shown, one 2 2 corresponding to a low value of F/H = 0.011 MPa (F = 0.1 MN, H = 3 m) and to a high one F/H = 10 MPa (F = 10 MN, H = 1 m), with associated solutions. Remember that, since E and  c are measured in MPa, the load F 2 must be expressed in units of MN. Concrete is the cheapest choice for the smaller value of F/H ; cast iron or 2 steel is the cheapest choice for larger values of F/H .

M2 

E9.7 Multple objectives: light, cheap pipe-work (Part 1). Exercise E9.1 explored the material selection for a pipe that must contain an internal pressure without yielding or failing by fast fracture (two constraints) and is as cheap as possible (one objective). This exercise explores selection for a pipe that must contain an internal pressure without yielding (one constraint) and is both as light and as cheap as 99

Figure 9.17. A pipe containing a pressure

possible (two objectives). The requirements are summarized in Table 9.7

Table 9.7. Requirements for light, low-cost pipework Function Constraints Objectives Free variables

      

Safe, light, low-cost, pipe work Pipe diameter 2R and pressure difference p specified Must not fail by yielding Minimize mass per unit length, m, of the pipe Minimize cost per unit length, C, of the pipe Wall thickness, t Choice of material

Here are the steps. Identify the tensile stress in the wall of a pipe (radius R and wall thickness t) when the pressure difference across the pipe wall is p (Figure 9.17). Express the constraint that the stress must be less than the yield strength of the material of the pipe. Write an expression for the mass of unit length of the pipe (radius R and wall thickness t) in terms of its dimension and the density  . Write and expression for the material cost of unit length of the pipe – it is just the mass per unit length multiplied by of the material C m .

These are the two objective functions, both to be minimized. Treat the wall thickness t as the free variable – it must be chosen so that the wall does not fail by yielding. Substitute the constraint into the objective functions, eliminating t, to get one expression for the mass and one for the cost. Identify the two material indices M1 and M2 that they contain.

The table lists the properties of materials that are used for pipes. Enter the data into an Excel spread sheet. Use it to evaluate the two material indices M1 and M2 and plot them as a trade-off diagram. Which materials lie closest to the trade-off surface?

Table 9.8 Properties of materials for pipes Materials for pipes

Cost Cm ($/kg)

Density ρ (kg/m^3)

Yield strength σy (MPa)

Copper 1220 Grey cast iron Nodular cast iron Mild steel 2024 Aluminum alloy 304 Stainless steel Lead Polyethylene (PE) Polypropylene (PP) Polyvinyl chloride (PVC) Concrete

6.1 0.36 0.40 0.76 2.4 3.4 3.9 1.6 1.4 2.6 0.05

8,940 7,200 7,200 7,850 2,770 7,950 10,500 950 900 1,450 2,450

292 290 400 300 286 743 10 24 29 44 2

Answer. The pipe is cylindrical with a radius R, and that the wall thickness, t (the free variable), small compared with R. Then the tensile stress in the pipe wall is pR  t There are two objective functions. The first is the mass per unit length, m, to be minimized. m 2 R t  where  is the density of the material. The second is the cost per unit length of the pipe: C 2 R t C m 

100

The constraint is that the pipe should not yield – that is, that the tensile stress in the wall should not exceed y.

Using the constraint to substitute for the free variable t in each objective in turn leads to the performance equations and indices laid out below.

Objectives Mass

Constraint

Performance equation

Index

M1 

Substitute for t

ρ m  2 π  p.R 2     ςy 

C ρ  C  2 π  p.R 2   m   ςy 

C  M1  m

m 2 R t 

Yield constraint ς 

p R t

 y

 y

Substitute for t

Cost C 2 R t Cm 

y

The two indices are evaluated in the table and plotted in the accompanying figure. The trade-off line is sketched in by hand. Remember that it must be convex downward, and is the lower envelope for the materials plotted on the chart. Two materials lie on the trade-off line: the aluminum alloy and nodular cast iron. Two others lie close: stainless steel and grey cast iron.

Materials for pipes

Cost Cm ($/kg)

Density ρ (kg/m^3)

Yield strength σy (MPa)

M1   / y

M2  Cm  /  y

Copper 1220 Grey cast iron Nodular cast iron Mild steel 2024 Aluminum alloy 304 Stainless steel Lead Polyethylene (PE) Polypropylene (PP) Polyvinyl chloride (PVC) Concrete

6.1 0.36 0.40 0.76 2.4 3.4 3.9 1.6 1.4 2.6 0.05

8,940 7,200 7,200 7,850 2,770 7,950 10,500 950 900 1,450 2,450

292 290 400 300 286 743 10 24 29 44 2

31 25 18 26 10 11 1050 40 31 33 1225

186 9 7 20 23 36 4095 63 43 86 61

Trade-off plot for mass and cost of pipe-work 101

E9.8 Multple objectives: light, cheap pipe-work (Part 2). Create a penalty function Z (equation 9.10 of the text), combining the two performance equations of Exercise 9.7, mass m and cost C, choosing the cost in $ as the unit for Z. To do this, we write the penalty function as

C ρ  ρ Z  C   m  2 π  p.R 2   m   2 π  p.R 2  .   ςy   ςy  The exchange constant  then has units of $/kg, measuring the value (in $) of weight saving (in kg). The factor

2 π  p.R 2 appears in both terms and contains no material properties, so we can get rid of it by defining a normalize penalty function Z Z*  2 π  p.R 2

The penalty function now becomes C ρ  ρ Z*   m    .   M2   M1  ςy   ςy  We seek the materials with the lowest values of Z*. M1 and M2 are the axes of the trade of plot created in Exercise 9.7. The best material choice, for a given value of  , is the one nearest the point where Z* is tangent to the trade-off surface (Figure 9.9).

Answer. Here is the trade-off plot for the pipe materials, with contours of constant Z* for three values of α added. They are curved because the axes of the plot are logarithmic (on linear scales, they are straight lines).

The value of Z* is increased until the contour just touches the trade-off surface. The material nearest the contact point is the best choice. When mass is not a problem (α = 0.05 $/kg), cast iron is the best choice. But if the pipe work has to be transported over large distances, or is part of a transport system, mass becomes a concern (α = 20 $/kg). Then more expensive, but lighter, choices (aluminum alloy, stainless steel) are better choices.

Trade-off plot for mass and cost of pipe-work with Z*-contours for different exchange constants, α 102

E9.9 Conflicting objectives: an air cylinder for a truck – Part 1: analysis. Trucks use 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 in Figure 9.18. The tank has length L, diameter 2R, and hemispherical ends. Most are made of lowcarbon 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. Table 9.9 summarizes the design requirements.

Figure 9.18. An air cylinder Table 9.9. Requirements for light, cheap, air cylinder Function Constraints Objectives Free variables

      

Air cylinder for truck Must not fail by yielding Diameter 2R and length L specified, so the ratio Q = 2R/L is fixed.

Minimize mass m Minimize material cost C Wall thickness, t Choice of material

Show that the mass m and material cost C of the tank relative to one made of low-carbon steel (mo, Co) are given by

m      yo     mo   y   o 

and

C  Cm     yo     Co   y   Cmo o 

where  is the density, y the yield strength and Cm the cost per kg of the material, and the subscript “o” indicates values for mild steel.

Answer. The mass m of the tank is

m  2 RLt  4 R2 t   2 RLt  ( 1  Q) where Q, the aspect ratio 2R/L, is fixed by the design. This is the first objective – the quantity to be minimized.

The stress in the wall of the tank caused by the pressure p must not exceed  y , the yield strength of the material of the tank wall, giving the constraint pR    y t Substituting for t, the free variable, gives     m  2 R 2 L p ( 1  Q )  y    The material cost C is simply the mass m times the cost per kg of the material, Cm, giving the second objective 103

C   C  C m m  2 R 2 L p ( 1  Q )  m   y    The steps are summarized below.

Objectives

Constraints

Performance equation

m  2 RL t ( 1  Q)

Index

   m  2 R2 Lp ( 1  Q)    y   

M1 

C   C  2 R2 Lp ( 1  Q)  m   y   

M2 

Substitute for t p R Strength constraint   y t

C  Cm m

 y

Cm 

y

Dividing m by the value for steel, mo, and C by the value for steel, Co, gives the mass and cost relative to that of steel tank as m      yo  C  Cm     yo  and        mo   y   o  Co   y   Cmo o 

E9.10 Conflicting objectives: an air cylinder for a truck – Part 2, material choice. Explore the trade-off between relative cost and relative mass for the air cylinder of Exercise E9.9, considering the replacement of a mild steel tank with one made, first, of low alloy steel, and, second, one made of filament-wound CFRP. The relevant material properties are listed in the table below.

Define a relative penalty function m C Z*   *  mo Co where  is a relative exchange constant, and evaluate Z for  = 1 and for  = 100. *

*

*

*

Table 9.10. Materials and properties for the air cylinder Material

Density  3 (kg/m )

Mild steel

7850

Yield strength c (MPa) 314

Price per /kg Cm ($/kg)

Low alloy steel

7850

775

0.85

CFRP

1550

760

42.1

0.55

*

Answer. To get further we need a relative penalty function, Z : m C Z*   *  mo C o This is evaluated for Low alloy steel and for CFRP in the table below. For  = 1 (meaning that weight carries a * * low cost-premium) – low alloy steel has by far the lowest Z . But for  = 100 (meaning that weight carriers a * large cost-premium), CFRP has the lowest Z . *

Density  3 (kg/m )

Yield strength c (MPa)

Price per /kg Cm ($/kg)

α =1

α = 100

Mild steel

7850

314

0.55

2

101

Low alloy steel

7850

775

0.85

1.03

45.6

CFRP

1550

760

42.1

5.2

13.4

Material

104

*

*

Z, *

Z, *

E9.11 Conflicting objectives: an air cylinder for a truck – Part 3: the graphical method. Figure 9.19, below, is a chart with axes of m/mo and C/Co derived in Exercise E9.9. Mild steel (here labelled “Low carbon steel”) lies at the co-ordinates (1,1) and is colored black. Sketch a trade-off surface – remember that it is simply a lower * envelope to the data, with no mathematical function associated with it. Then plot contours of Z (see Exercise * * E9.8) that are approximately tangent to the trade-off surface for  = 1 and for  = 100. They are plots of the linear relationship *

m 1 C  mo * Co

but appear as curves because of the log scales. What selections do these suggest?

Figure 9.19. Cost and mass relative to low carbon steel Answer. The figure shows the trade-off surface. Materials on or near this surface have best combinations of * mass and cost. Several are better low-carbon steel. Two contours of Z that just touch the trade-off line are * * shown, one for  = 1, the other for  = 100 – they are curved because of the logarithmic axes.

The first, that for  = 1, identifies high carbon and low alloy steels as good choices. This is because their * higher strength allows a thinner tank wall. The contour for  = 100 touches near CFRP, aluminum and magnesium alloys – if weight saving is very highly valued, these become attractive solutions. *

105

E9.12 Conflicting objectives: low-cost insulating walls. 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, , 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. Table 9.10 summarizes the design requirements.

Table 9.10. Requirements for low-cost insulating walls.

Function Constraint Objectives Free variables

    

Foam for panel-wall insulation Panel wall thickness specified.

Minimize foam thermal conductivity,  Maximize foam stiffness, meaning Young’s modulus, E Choice of material

Figure 9.20 shows the thermal conductivity  of foams plotted against their elastic compliance 1 / E (the reciprocal of Young’s moduli E, since we must express the objectives in a form that requires minimization). The 3 numbers in brackets are the 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.

106

Figure 9.20. Thermal conductivity and elastic compliance of foams

Answer. The steps in making a reasoned choice are as follows (1) Sketch the trade-off surface: the lower envelope of the  - 1/E chart, as shown below. The foams that lie on or near the surface are a better choice than those the lie far above it. This already eliminates a large number of foams and identifies the family from which a choice should be made. Metal foams are not a good choice; ceramic foams an attractive choice only when high stiffness is needed.

(2) To go further, construct a penalty function: 1 Z  1   2   E  Z is to be minimized, so  1 is a measure of the value associated with reducing heat flow;  2 a measure of the value associated with reducing core compliance. Rearranging the equations gives  1 Z   2  1 1  E  If the axes were linear, this equation would be that of a family of straight, parallel, lines on the  vs. 1/E diagram, of slope  2 /  1 , each line corresponding to a value of Z /  1 . In fact the scales are logarithmic, and that leads instead to a set of curved lines. They are sketched for  2 /  1 = 0.01 (meaning that thermal insulation is considered very important, and stiffness less important) and for  2 /  1 = 100 (meaning the opposite). The foam nearest the point at which a penalty line is tangent to the trade-off surface is the best 3 choice. In the first example, PVC or ABS foams with a density of about 0.1 Mg/m are the best. 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 are viable.

107

E9.13 Approximate values for exchange constants (1). In Britain, a typical family car covers 150,000 km over its life and consumes on average 5.6 liters of petrol per 100 km. Petrol in Britain costs £1.5 ($1.95) per liter, so the life cost for fuel is £12,600 ($15,372). The typical car, when loaded, weighs 1400 kg. If the petrol consumption is directly proportional to the weight, what is the value of reducing the weight of the car by 1 kg?



Answer. The reduction in life fuel-cost if the weight is reduced by 1 kg is

12600  9.0 £ / kg 1400

($11/kg)

E9.14 Approximate values for exchange constants (2). In the US a typical family car lasts 12 years, covers 200,000 miles (322,000 km) over its life and delivers, on average 25.4 miles per US gallon (10.75 km/liter).

Petrol in the US costs $ 0.98 per liter, so the life cost for fuel is $29,354. The typical family car, when loaded, weighs 2315 kg. If the petrol consumption is directly proportional to the weight, what is the value of reducing the weight of the car by 1 kg?

Answer. The reduction in life fuel-cost if the weight is reduced by 1 kg is  

29 ,354  12.7 $ / kg .

2 ,315

E9.15 Approximate values for exchange constants (3). A Routemaster – a London bus (Figure 9.21) – weighs 12 tonnes when empty, 18 tonnes when full. It does, on average, 6.6 miles (11 km) per imperial gallon of diesel, which cost £9.76 per imperial gallon (2.15 $/liter). The bus travels an average of 57,000 km per year and has a life of 20 years. Assuming that the average load factor is 50% (meaning that the bus weighs 15 tonnes on average) and that the fuel consumption is directly proportional to the weight, what is the value of weight saving for a Routemaster? (Data from London Transport.)

Figure 9.21. A Routemaster bus

Answer. The average Routemaster covers 1.14 million kilometers over its life, consuming 104,000 gallons of diesel, costing £ 1.01 million ($1.23 million). If its weight is reduced by 1 kg the cost saving would be 1 ,010 ,000    67.3 £ / kg ( 82$ / kg) . This exchange constant  is roughly six times that for a car, making 15 ,000 the use of light alloys more attractive in a bus or truck than a car.

Ch 10. Multiple constraints and conflicting objectives – case studies This chapter has no exercises.

Ch 11. Selection of material and shape These exercises cover the derivation of shape factors, of indices that combine material and shape, and the use of the 4-quadrant chart arrays to explore material and shape combinations. For this last purpose it is useful to have clean copies of the chart arrays of Figures 11.9 and 11.12. Like the material property charts, they can be copied for educational purposes without restriction of copyright.

E11.1 Stiffness shape factors for box sections and tubes (Figure 11.17).

Evaluate the shape factor  B for stiffness-limited design in bending of a e

square box section of outer edge-length h = 100mm and wall thickness t = 108

Figure 11.17

3mm. Is this box shape more efficient (meaning stiffer for the same mass) than a tube made of the same material with diameter 2r = 100mm and wall thickness t = 3.82 mm (giving it the same mass per unit length, m/L)? Treat both as thin-walled shapes. Use the expressions in Table 11.3 of the text for the shape factor  B . e

Answer. The shape-efficiency factor for elastic bending for the square box section (from Table 11.3) is 1h  Be   16.7 2t That for the tube is 3r  Be   12.5  t The box is more efficient than the tube of the same m/L, and both are much stiffer in bending than a solid square section of the same area (and thus mass per unit length).

E11.2 Strength shape factors for box-sections and tubes (Figure 11.17 again). Evaluate the shape factor Bf for strength-limited design in bending of a square box section of outer edge-length h = 100mm and wall thickness t = 3mm. Is this box shape more efficient (meaning stronger for the same mass) than tube made of the same material with diameter 2r = 100mm and wall thickness t = 3.82 mm (giving it the same mass per unit length, m/L)? Treat both as thin-walled shapes.

Use the expressions given in Table 11.3 of the text for the shape factors Bf .

Answer. The shape-efficiency factor for bending failure for the square box section (from Table 11.3) is

Bf 

h  5.8 t

That for the tube is

 Bf 

3 2

r  4.33 t

The box is more efficient than the tube, and both are roughly 5 times stronger in bending than a solid square section of the same area (and thus mass per unit length).

E11.3 Deriving shape factors for stiffness-limited design (Figure 11.18).

Derive the expression for the shape-efficiency factor  Be for stiffness-limited design for a circular tube with outer radius 5t and wall thickness t, loaded in bending (Figure 11.18). Do not assume that the thin-wall approximation is valid.

Figure 11.18

Answer. The shape factor for elastic bending of a tubular beam (left-hand figure) is defined by 12 I  Be  2 A where I is the second moment of area and A is the area. The second moment of area, I, of the tube about the axis shown as a dotted line is

I

 4

5 t 4   4t 4  92.25 t 4

Its cross-section area is

4

A   5 t 2   4 t 2  9  t 2 from which 109

 Be 

12 I

 4.35

A2

E11.4 Deriving shape factors for stiffness-limited design (Figure 11.19). Derive the expression for the shape-efficiency factor  Be for stiffness-limited design for a channel section of thickness t, overall flange width 5t and overall depth 10 t, bent about its major axis (the dash-dot line in Figure 11.19). Do not assume that the thin-wall approximation is valid.

Figure 11.19 Answer. The second moment of area, I, of the channel about the axis shown as a dotted line is

I

10t 3 .5t  8t 3 .4t  246 t 4 12

12 2

Its cross-section area is A  18 t , from which

Be 

12 I A2

 9.11

E11.5 Deriving shape factors for stiffness-limited design (Figure 11.20). Derive the expression for the shape-efficiency factor  Be for stiffness-limited design for a square box section of wall thickness t, height and width h1 = 10 t bent about its major axis (the dash-dot in Figure 11.20). Do not assume that the thin-wall approximation is

and line valid.

Figure 11.20

Answer. The second moment of area, I, of the square box-section (right-hand figure) about the axis shown as a dotted line is h1 / 2

I2 

0

h2 / 2

y 2 h1 dy  2 





0

y 2 h2 dy

1 4 2  h1  h24  th13 12 3 and the cross-section area is A  h12  h22  4 h1 t

Assembling these results, and inserting h1 = 10t gives 12 I 1 h Be  2  1  5 2 t A E11.6 Deriving shape factors for strength-limited design. Determine the value of the shape-efficiency factor Bf for strength limited design in bending using the dimensions shown on the diagrams (a) For the tube-section shown in Figure 11.21 (a) (b) For the box section shown in Figure 11.21 (b) and (c) For the channel section shown in Figure 11.21 (c).

The expression for Bf for the first two sections can be read from Table 11.3. You will have to derive the expression for the third.

110

Figure 11.21

Answer. (a) The shape factor for strength-limited design in bending for a tube, read from Table 11.3, is

Bf 

3 2

r  2.68 t

(b) The shape factor for strength-limited design in bending for the box, read from Table 11.3, is h Bf   3.16 t (c) The second moment of area, I, of the channel about the axis shown as a dotted line is

I

10 t 3 .5t  8 t 3 .4 t  246 t 4

12 12 The section modulus Z is I 246 t 4 Z   49 t 3 ymax 5t 2

Its cross-section area is A  18 t , from which

Bf 

6Z

 3.85

A3 / 2

The channel is stronger in bending than the tube or the box because more of the material is distributed far from the neutral axis.

E11.7 Deriving indices that include shape. A beam of length L, loaded in bending, must support a specified bending moment M without failing and be as light as possible. Show that to minimize the mass of the beam per unit length, m/L, one should select a material and a section-shape to maximize the quantity

   M f B

2/3

f



where  f is the failure stress and  the density of the material of the beam, and Bf is the shape-efficiency factor for failure in bending.

Answer. The beam of length L, loaded in bending, must support a specified load F without failing and be as light as possible. Its mass m is (1) m  AL  where A is the area of its cross section. Failure occurs if the load exceeds the moment M  Z f (2) where Z is the section modulus . Replacing Z by the shape-factor Bf  6 Z / A 3 / 2 (equation 11.10 of the text) gives

111

f

M 

6

Bf A 3 / 2

(3)

Substituting this into equation (1) for the mass gives  2/3 m  6 M 

L

   f B

(4)

2/3

f

The best material-and-shape combination is that with the greatest value of the index  f  f 2/3 M  B 





E11.8 Calculating shape factors from stiffness data. The elastic shape factor measures the gain in stiffness by shaping, relative to a solid square section of the same crosssection area and thus mass per unit length. Shape factors can be determined by experiment by measuring the stiffness and mass of a structure and using these to calculate

 Be by inverting equation 11.19 of the text, solving for  Be .

Figure 11.22

Apply this approach to calculate the shape factor  B from the following experimental data, measured on an aluminum alloy beam (Figure 11.22) loaded in 3-point bending (for which C1 = 48 – see Appendix B, Section B3) using the data shown in the table. e

Table 11.8. Beam properties Attribute Bending stiffness SB Mass/unit length Beam length L Beam material

Value 5 7.2 x 10 N/m m/L

Material density  Material modulus E

1 kg/m 1m 6061 aluminum alloy 2670 kg/m

3

69 GPa

Answer. Equation 11.19 of the text relates the mass m of a beam of length L and prescribed bending stiffness

S B to the shape factor  Be :  12 SB   m     C1 

1/ 2

    L5 / 2   e E 1 / 2  B  

 

where C1 is a constant that depends on the distribution of load on the beam (here, 48) and E and density  are the modulus and density of the material of which it is made. Inverting the equation gives  12 L5 S B    2    Be    C m 2  E    1  Inserting the data given in the question into the equation for  B given in the example gives e

Be  18.6 E11.9 Calculating shape factors from stiffness data. A steel truss bridge shown in Figure 11.23 has a span L and is simply supported at both ends. It weighs m tonnes. As a 112

Figure 11.23

rule of thumb, bridges are designed with a stiffness SB such that the central deflection  of a span under its self-weight is less than 1/300 of the length L (thus SB  300mg / L where g is the acceleration due to gravity, 9.81 m/s ). Use this information to calculate a minimum value for the shape factor  B of the three steel truss 3 bridge spans listed in the table. Take the density  of steel to be 7900 kg/m and its modulus E to be 205 GPa. The constant C1 = 384/5 = 76.8 for uniformly distributed load (Appendix B, Section B3). e

2

Span L (m)

Mass m (tonnes)

Royal Albert bridge, Tamar, Saltash UK (1857)

139

1060

Carquinez Strait bridge, California (1927)

132

650

Chesapeake Bay bridge, Maryland USA (1952)

146

850

Bridge and construction date*

Table 11.9. properties

Bridge

*Data from Bridges Handbook, Chen and Duan (2014)

Answer. The shape factor is given in terms of m, L and SB by inverting equation 11.19 of the text:

 12 L5 S B    2     C m 2  E    1   Inserting the data given in the question together with SB  300 mg / L gives

Be  

 3600L4 g    2     C 1 m  E    

Be  

Inserting C1 = 384/5 (Appendix B, Section B3) and the data from the table gives the results shown below.

F = mg 6 (10 N)

Bridge and construction date* Royal Albert bridge, Tamar, Saltash UK (1857) Carquinez Strait bridge, California (1927) Chesapeake Bay bridge, Maryland USA (1952)

10.6 6.5 8.5

A m / L 2

(m ) 0.97 0.62 0.74

 Be 49.3 65.4 74.8

The results show that “structured structures” (trusses, rib-stiffened structures) can offer very large shape efficiencies. Note how the efficiency has grown over time.

E11.10 Deriving indices for bending. A beam, loaded in bending, must support a specified bending moment M* without failing and be as light as possible. Section shape is a variable, and “failure” here means the first onset of plasticity. Derive the material index. The table summarizes the requirements.

Function Constraints Objective Free variables

     

Light weight beam Specified failure moment M* Length L specified Minimize mass m Choice of material Section shape and scale

113

Table 11.10 requirements for the beam

Design

Answer. The mass m of a beam of length L and section area A is m  AL

where  is the density of the material of which the beam is made. Failure occurs if the bending moment exceeds the value M  Z f f

Replacing Z by the shape-factor B (equation (11.10) of the text)gives  f f 3/2 M   A 6 B Substituting this into equation (1) for the mass of the beam gives   2/3    m  6 M* L 2/3   f   B  f  The best material-and-shape combination is that with the greatest value of the index

 

M 





Bf  f  2 / 3 

E11.11 Deriving indices for torsion (Figure 11.24). A shaft of length L, loaded in torsion, must support a specified torque T* without failing and be as cheap as possible. Section shape is a variable and “failure” again means the first onset of plasticity. Derive the material index. The table summarizes the requirements.

Figure 11.24

Table 11.11. Design requirements for the torsion bar Function Constraints Objective Free variables

 Cheap torsion bar  Specified failure torque T*  Length L specified  Minimize material cost C  Choice of material  Section shape and scale

Answer. The material cost C of a shaft of length L and section area A is .

C  ALCm 

where Cm is the cost per kg and ρ is the density of the material of which the shaft is made. Failure occurs if the torque T exceeds the moment T  Q f where  f is the shear stress at which plasticity first starts.

Replacing Q by the shape-factor Tf using equation (11.13) of the text gives

T 

f

Tf A 3 / 2

4.8 Substituting this into equation (1) for the material cost of the shaft gives   2/3 C  m  4.8 T * L  m 2/3  f   T  f  Approximating  f by  f / 2 , we find that the material-and-shape combination with the lowest associated





 

material cost is that with the greatest value of the index

M

Tf  f  2 / 3 Cm 

114

E11.12 Material and shape to suppress buckling (Figure 11.25). The figure shows a concept for a lightweight display stand. The stalk must support a mass m of 100 kg, to be placed on its upper surface at a height h, without failing by elastic buckling. It is to be made of stock tubing and must be as light as possible. Use the methods of Chapter 11 to derive a material index for the tubular material of the stalk of the stand that includes the shape of the section, described by the shape factor

 Be 

12 I A2

where I is the second moment of area and A is the section area of the stalk. The table summarizes the requirements.

Figure 11.25

Table 11.12 Design requirements for the column  Light weight column  Specified buckling load F  Height h specified  Minimize mass m  Choice of material  Section shape and scale

Function Constraints Objective Free variables

Answer. The objective function is the mass. The mass m of the tubular stalk is (1) m  Ah  where  is the density and A the section area of the tube. The constraint is that the stalk should not buckle.

The buckling load for a slender column of height h (Appendix B, Table B5) is Fcr 

n2 π 2 EI



n 2 π 2 E Be A 2 12 h2

(2) h2 where E is Young’s modulus, I is the second moment of area of the tube and n is a constant that depends on the end constraint (n = 1/2 in this instance). Substituting for A from equation (2) into equation (1) gives

m

2 3 h2 π

 F 1/2

ρ (EBe ) 1/2

The best choice of tubing for the stalk is that with the greatest value of the index M 

( E Be ) 1/2 ρ

E11.13 Selection of material and shape to suppress buckling. Cylindrical tubing is available from stock in the following materials and sizes (Table 11.13). Use this information and the material index to identify the best stock material for the stalk of the stand shown in Figure 11.25.

Material Aluminum alloys Steel Copper alloys Polycarbonate (PC) Various woods

Modulus E (GPa) 69 210 120 3.0 7 - 12

Tube radius r 25 mm 30 mm 20 mm 20 mm 40 mm

Table 11.13 Materials available for the stalk 115

Wall thickness/tube radius, t/r 0.07 to 0.25 0.045 to 0.1 0.075 to 0.1 0.15 to 0.3 Solid circular sections only

Answer. As we saw in the previous exercise, the best choice of tubing for the stalk is that with the greatest ( E Be ) 1/2 . The bending moment of inertia, I for a thin walled tube with radius r and wall thickness t ρ (Table 11.2) is: 3 I=r t The area A of the section is 2 rt, giving a shape factor 12 I 3 r Be  2   t A

value of

(Note that, for a solid section, B  0.95  1 ). The selection can be made by evaluating and comparing the e

quantity M  ( EB )

e 1/2

/  for each of the materials and sections, as shown

Material

E (GPa)

 (kg/m3)

 Be

M (GPa /Mg/m )

Aluminum alloys Steel Copper alloys Polycarbonate (PC) Various woods

71 210 120 3 8-15

2700 7900 8900 1200 500 - 800

4 - 14 10 - 21 10 - 13 3.3 - 6.7 1

6.2 - 11.7 5.8 - 8.2 3.9 – 4.4 2.6 – 3.7 3.5 – 7.7

1/2

3

Aluminum offers the highest index and thus the lightest stalk. Steel is the next best choice, then wood.

116

E11.14 Microscopic shape: tube arrays (Figure 11.26). Calculate the gain in bending efficiency,  Be , when a solid is formed into small, thin-walled tubes of radius r and wall thickness t that are then stacked and bonded into a large array, part of which is shown in the figure. Let the solid of which the tubes are made have modulus Es and density  s . Express the result in terms of r and t. If the tubes have a diameter 2r of 2mm and a wall thickness t of 0.1 mm, what is the gain in bending efficiency?

Figure 11.26 Answer. Equation (11.38) of the text defines the gain in bending efficiency,  Be , of a prismatic array about an axis normal to that of the array as

 Be



EI E sIs



s 

Where  s is the density of the solid in the array and  is that of the array itself. The unit cell of the structure is shown as a red hexagon on the right-hand part of the figure. The area As of solid in the unit cell, is

As  6  r t and that of the hexagonal unit cell is

A  6 r 2 tan( 60 ) the relative density of the array,  /  s , (assuming r >> t) is As  t   1.8 2 s r 6 tan( 60 ) r Thus the bending efficiency is EI  r  Be   s  0.56 E s Is  t

If r = 1 mm and t = 0.1 mm, the gain in efficiency is  Be  5.6 .

E11.15 The structural efficiency of foamed panels (Figure 11.27). Calculate the change in structural efficiency for both bending stiffness and strength when a solid flat panel of unit area and thickness t is foamed to give a foam panel of unit area and thickness h, at constant mass. The modulus E and strength  f of foams scale with relative density  /  s as 2

3/2

      E    E s  f     f ,s and  s   s  where E,  f and  are the modulus, strength and density of the foam and E s ,  f ,s and  s those of the solid panel. (The derivation of these equations is given in Chapter 13.)

117 Figure 11.27

Answer. The panel, initially solid and of thickness t, is foamed to a height h at constant area. The density falls from  s to

t h

  s

and the modulus and strength fall from Es to  f ,s 2

3/2

t  t  E  Es   and  f   f ,s   h   h The bending stiffness per unit area changes from Ss to S, so that EI S h 1  Be      /  s  So E sIs t 3

3

(using Is  t / 12 and I  h / 12 per unit width). Foaming imparts a shape-efficiency factor for stiffness equal to the reciprocal of the relative density.

2 2 Following the same procedure for strength (using Z s  t / 6 and Z  h / 6 per unit width) gives 2 3/2 Z f h 1 h  t   Bf         Z s  f ,s t h t       / s 1 / 2

The shape-efficiency factor for strength is equal to the reciprocal of the square root of the relative density.

E11.16 Use of the four segment chart for stiffness-limited design. Use the 4-segment chart for stiffness5 limited design of Figure 11.9 of the text to compare the mass per unit length, m/L, of a section with EI = 10 2 Nm made from (a) structural steel with a shape factor  B of 20, modulus E = 210 GPa and density  = 7900 kg/m e

3

e (b) carbon fiber reinforced plastic with a shape factor  B of 10, modulus E = 70 GPa and density  = 3

1600 kg/m .

Answer. The results appear on the chart below. The CFRP beam with EI = 10 Nm and  B = 10 weighs about 5

2

e

2.5 kg/m; the steel beam with the same stiffness weighs about twice as much (5kg/m) despite having a more efficient shape(  B = 20). e

118

E11.17 Numerical comparison for stiffness-limited design. Show, by direct calculation, that the conclusion of Exercise E11.16 – that the CFRP beam with EI = 10 Nm and  B = 10 weighs less than the steel beam with 5

e

2

the same  B = 20 – is consistent with the idea that to minimize mass for a given stiffness one should maximize e

E * /  * with E  E /  B and    /  B . *

e

Answer. Direct calculation of

The higher the value of

*

e

* e * e E * /  * with E  E /  B and    /  B is shown in the table.

Material

E *  E / Be

Steel CFRP

(N/m ) 9 10.5 x 10 9 7.0 x 10

 *   /  Be

2

3

(kg/m ) 395 160

E* / * 3 (N/m ) /(kg/m ) 259 523 2 1/2

E * /  * the lighter is the beam. The ranking is the same as that arrived at in Exercise

E11.16.

119

E11.18 Use of the four segment chart for strength. Use the 4-segment chart for strength-limited design of 4 Figure 11.12 of the text to compare the mass per unit length, m/L, of a section with Z  f = 10 Nm (where Z is the section modulus) made from

(a) 6061 grade aluminum alloy with a shape factor Bf of 3, strength  f = 200 MPa and density  = 3

2700 kg/m , and

(b) a titanium alloy with a shape factor Bf of 10, strength  f = 480 MPa and density  = 4420 kg/m .

3

Answer. The results appear on the chart below. The titanium alloy beam with Z  f = 10 Nm and Bf of 10 is 4

much lighter ( m / L  2.5 kg/m) than the aluminium alloy beam of the same strength but the lesser shape efficiency Bf of 3 ( m / L  2.5 kg/m).

120

E11.19 Numerical comparison for stiffness-limited design. Show, by direct calculation, that the conclusion of 4 Exercise E11.18 – that the titanium alloy beam with Z  f = 10 Nm and Bf of 10 is much lighter than the 6061 aluminium alloy beam of the same strength and Bf – is consistent with the idea that to minimize mass for a



2 / 3 .

f given stiffness one should maximize B  f





Answer. Direct calculation of  Bf  f



2/3

f

Material

The higher the value of   Bf  f  

2/3

 3

(MPa)

(kg/m )

200 480

2700 4420

6061 Al alloy Ti alloy



/  gives the values shown in the table.

/

Bf

   f B

2/3

f

/

2/3

3

(MPa) /(kg/m ) 3 10

-2

2.6 x 10 -2 6.4 x 10

the lighter is the beam. The ranking is the same as that found in

Exercise E11.18.

121

Ch 12. Material and shape: case studies This chapter has no exercises.

Ch 13. Designing hybrid materials The examples in this section relate to the design of hybrid material. The first four illustrate the use of the key concepts of Section 13.3. The next four involve the use of bounds for evaluating the potential of composite systems and of hybrid design to fill holes in property space. The remaining exercises make use of the charts for natural materials and raise a challenge: to explore the potential of hybridizing two very different materials.

E 13.1. Packing fraction for non-directional bonding. The BCC and diamond cubic structures of Table 13.1 are stable only when the bonds linking the atoms are directional. When bonding is non-directional (when spheres are allowed to pack in the earths gravitational field, for example) packing can be loose-random, dense-random or close packed. Plot the packing fraction P against coordination number C (Table 13.2) for non-directional packings. Is there a correlation?

Answer. The figure shows the correlation It is approximated by the equation P  0.025C  0.45

E 13.2. Chopped fiber composites. Nylon 66 has a tensile strength  m of 55 MPa. It is to be reinforced with 12 micron diameter chopped S-glass fibers with a tensile strength  ts , f of 4750 MPa. What is the maximum length  of fiber for useful strengthening?

Answer. The stress in the fiber when the composite is loaded beyond the yield strength of the matrix rises to (Section 13.3):   fiber  m 2r The fiber will break if this stress exceeds its tensile strength, so nothing is gained by using fibers of length greater than  ts , f   2 r  1036 m  1mm m

E 13.3. Rigid lattices (Figure 13.40). The condition that a 3-dimensional pin-jointed frame (meaning one that is hinged at its corners) made up of b struts and j frictionless joints to be both statically and kinematically determinate is M b  3j  6  0

122

(the Maxwell stability criterion). The unit cell of the lattice shown in Figure 13.40 has 12 bars that are unique to the cell and24 bars that are shared between 2 cells, so there are b 24 bars per cell. It has 6 joints each shared between 2 cells and 8 joints each shared between 4 cells, so j = 6 joints per cell. Apply Maxwell’s stability criterion (equation 13.3) to decide if this structure is (a) A mechanism (b) An optimally rigid structure, free of self-stresses (c) An over-constrained structure, with potential for self-stress.

Figure 13.40. The lattice and its unit cell

Answer. Maxwell’s stability criterion for 3-D structures requires that M b  3j  6  0 where the equality sign describes optimal stability and the inequality describes over-constraint. Inserting the data we find: M  24  3 x6  6  0 The structure is optimally stable.

E 13.4. Conducting polymers. Conducting carbon nanotubes with and aspect ratio of  = L/d = 200 (where L is the nanotube length and d its diameter) are dispersed in a polylactide matrix. Polylactide is an electrical insulator. If the dispersion is random (not an easy thing to achieve) what volume fraction of nanotubes would be required to first allow conduction?

Answer. The percolation threshold at which a continuous contacting path forms for randomly dispersed rods 1/ 2 is f *  fc /  where fc  0.2

is the percolation threshold for randomly dispersed spheres. We expect

first conduction at a volume fraction f *  0.2 / 200  0.014 , or about 1.5%.

In practice the measured

threshold is considerably higher (Wu, D. et al, (2010) Polymer Physics Volume 48, pp. 479–489) because the nanotubes tend to stick together to form a floc and the dispersion is not random.

E 13.5 Concepts for light, stiff composites (1). Figure 13.41 is a chart for exploring stiff composites with light alloy or polymer matrices. Three matrix materials are shown, highlighted in red. They are surrounded by a large number of possible reinforcing fibers (suffix f), particulates (suffix p) and whiskers (suffix w). A construction like that of Figure 13.11 of the text allows the potential of any given matrix-reinforcement combination to be assessed. The criteria of excellence (the indices E /  , E

1/ 2

/  and E 1 / 3 /  for light,

stiff structures) are shown; they increase in value towards the top left. Use the chart to compare the performance of a titanium-matrix composite reinforced with (a) zirconium carbide, ZrC, particles (b) Saffil alumina fibers and (c) Nicalon silicon carbide fibers.

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Keep it simple: use equations 13.4 and 13.5 to calculate the density and upper and lower bounds for the modulus at a volume fraction of f = 0.5 and plot these points. Then sketch arcs from the matrix to the reinforcement to pass through them. In making your judgement, assume that f = 0.5 is the maximum practical reinforcement level.

Figure 13.41. The moduli and densities of light alloys and fibers

Answer. The figure below shows the result. The guide lines have been moved to pass through titanium, allowing the performance of each composite system to be assessed. Upper and lower bounds for the modulus of the three proposed composites (equations 13.5a and 13.5b) are plotted by evaluatin E~ , E~ and % for a u

L

volume fraction f  0.5 , the sketching arcs from titanium to the chosen reinforcement that pass through these points. A vertical, broken, line through each of the bound-envelopes marks the f = 0.5 point. It marks a cut-off: volume fractions of reinforcement about 0.5 are not realizable.

The Ti-ZrC composite system offers little or no gain in performance because the composite trajectory lies almost parallel to the guide-lines. By contrast, Ti-Saffil composites offer gains. Best of the lot is Ti-Nicalon.

These composites have a trajectory that lies almost normal to the guide-lines, offering the greatest increase in all three criteria of excellent.

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E13.6 Concepts for light, stiff composites (2). Use the Modulus–Density composite design chart of Figure 13.41 to explore the relative potential of (a) magnesium–E-glass-fiber composites and (b) magnesium -beryllium composites for light, stiff structures.

Answer. The results are shown on the Answer-figure of Exercise E13.5. Magnesium–E-glass composites offer no gain in performance – in fact they are less good than unreinforced magnesium when the criteria of excellence E 1 / 2 /  and E 1 / 3 /  are applied. Magnesium–beryllium composites offer considerable gains.

E13.7 Concepts for composites with tailored thermal properties (Figure 13.42).

The chart below allows exploration of components for composites with desired combinations of thermal conductivity and thermal expansion, using a light alloy or a polymer matrix. Three possible matrices –aluminum, magnesium and titanium – are shown in red. The materials shown in green are available as fibers (f), whiskers (w) and particles (p). A construction like that of Figure 13.14 allows the potential of any given matrix-reinforcement combination to be assessed. One criterion of excellence (the index for materials to minimize thermal distortion,  /  ) is shown; it increases in value towards the bottom right. Use the chart to compare the performance of a magnesium AZ63 alloy-matrix composite reinforced with (a) alloy steel fibers, (b) silicon 125

carbide fibers, SiC (f) and (c) diamond-structured carbon particles. Keep it simple: use equation 13.10 to 13.13 to calculate the upper and lower bounds for  and  at a volume fraction of f = 0.5 and plot these points.

Then sketch curves linking matrix to reinforcement to pass through the outermost of the points. In making your judgement, assume that f = 0.5 is the maximum practical reinforcement level.

Answer. The figure below shows the result. Mg-alloy steel composites perform less well than magnesium itself. By contrast, Mg-SiC composites offer gains (and are currently available). Even bigger gains lie with Mgdiamond composites. These are not as outrageous as they might sound: industrial diamond in particulate form is produced in large quantities for cutting tools, rock drills, grinding and polishing. Using it for a reinforcement in a composite for very high-performance applications is perfectly practical.

E 13.8

Hybrids with exceptional combinations of stiffness and damping. The loss coefficient  measures

the fraction of the elastic energy that is dissipated during a load-unload cycle. The Loss-coefficient – Modulus (   E ) chart of Figure 3.9 is populated only along one diagonal band. Monolithic materials with low E have high  , those with high E have low  . The challenge here is to devise hybrids to fill the holes, with the following applications in mind.

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(a) Sheet steel (as used in car body panels, for instance) is prone to lightly damped flexural vibration. Devise a hybrid sheet that combines the high stiffness, E, of steel with high loss coefficient,  . (b) High loss coefficient means that energy is dissipated on mechanical cycling. This energy appears as heat, sometimes with undesirable consequences. Devise a hybrid with low modulus E and low loss coefficient, .

Answer. The figure shows two simple solutions.

High modulus with high damping is achieved by coating the steel with high-loss polymer or elastomeric skins to give a hybrid as suggested by the sketch at the top right. When the sheet vibrates – a bending mode – the outer high-loss layers suffer the greatest strain and dissipate energy, damping the vibration. The in-plane and flexural moduli remain those of the steel sheet.

Low modulus and low damping can be achieved by internal shape, as sketched in the lower left. A bendingdominated cellular structure has a modulus E%given by equation 13.18 of the text: 2

   E% E s    s 

where  is the density of the cell-wall material, and Es and  s are the modulus and density of the cellular structure. The loss coefficient of the cellular structure,  s , is the same as that of the material from which it is made, provided deformation remains elastic. Thus a low-loss aluminum alloy with a bending-dominated cellular structure of relative density  /  s = 0.1 has a has a modulus that is 100 times lower than that of aluminum, with the same value of  . It lies at the point marked on the figure.

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E 13.9

Sandwich panels. An aircraft quality sandwich panel has the characteristics listed in the table.

(a) Use the data and equations 13.30, 13.31 and 13.32 of the text to calculate the equivalent density, flexural modulus and strength. (b) Plot these on a copy of a Modulus – Density, a Strength – Density and Modulus – Strength chart (Figures 3.3, 3.4 and 3.5). Do the flexural properties of the panel lie in a region of property space not filled by monolithic materials?

Data for glass fiber / aluminum honeycomb sandwich panel Face material Core material

0.38 mm glass fiber / epoxy 3 3.2 mm cell, 97.8 kg/m , 5052 Alu honeycomb

Panel weight per unit area, ma Panel length, L Panel width, b

2.65 kg/m 510 mm 51 mm

Panel thickness, d Flexural stiffness, EI

10.0 mm 2 67 N.m

Failure moment, Mf

160 Nm

2

~ Answer. (a) The equivalent density ~ , equation (13.30) is 

ma  265 kg / m 3 d

12 EI The equivalent modulus E~ , equation (13.31) is E~   15.8 GPa bd 3 4M f  The equivalent strength  f , equation (13.32), is ~ f   125 MPa bd 2 (b) See the three figures below. Yes, the flexural properties of the panel lie in areas of property space not occupied by monolithic materials.

128

129

E 13.10 Natural hybrids that are light and stiff. Plot aluminum alloys, steels, CFRP and GFRP onto a copy of the E   chart for natural materials (Figure 13.34), where E is Young’s modulus and  is the density. How do they compare with natural materials, using the flexural stiffness index E

1/ 2

/  as criterion of excellence?

The table lists the necessary data.

Material Aluminum alloy Steel CFRP GFRP

Young’s modulus E (GPa)

Density  (kg/m3)

74 210 100 21

2700 7850 1550 1850

Answer. The comparison of the four materials with those of nature is shown in the chart below. The criterion of excellence for bending stiffness at minimum weight, E

1/ 2

/  , is highlighted as red broken lines. The higher

the value of this criterion for a material, the better is its performance as a light, stiff beam. Natural materials are extraordinarily efficient by this measure. CFRP is comparable with the best of them, but the other engineering materials perform less well than those of nature.

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E 13.11 Natural hybrids that are light and strong. Plot aluminum alloys, steels, CFRP and GFRP onto a copy of the strength-density chart for natural materials (Fig. 13.35). How do they compare with natural materials, using the flexural strength index  2f / 3 /  (where  f is the failure stress) as criteria of excellence. The table lists the necessary data.

Material Aluminum alloy Steel CFRP GFRP

Density  (kg/m3)

Strength,  f (MPa)

2700 7850 1550 1850

335 700 760 182

Answer. The comparison of the four engineering materials with those of nature is shown in the chart below.

The criterion of excellence for bending strength at minimum weight,  2f / 3 /  , is highlighted as red broken lines. The greater the value of this criterion for a material, the better is its performance as a light, strong beam.

Natural materials are extraordinarily efficient by this measure. CFRP is comparable with the best of them, but the other engineering materials perform less well than those of nature.

E 13.12 Natural hybrids that act as springs. The table lists the moduli and strengths of materials that are used for springs. Plot these onto a copy of the E   f chart for natural materials of Figure 13.36, and compare their

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energy-storing performance with that of natural materials, using the  2f / E as the criterion of choice. Here

 f is the failure stress, E is Young’s modulus and  is the density.

Material

Modulus E (GPa)

Strength  f (MPa)

Density  (kg/m )

Spring steel

206

1100

7850

Copper-2%Beryllium

130

980

8250

CFRP, filament wound

68

760

1580

3

Answer. The comparison is shown on the chart below. Contours of the criterion of excellence,  2f / E , are highlighted in red. The greater the value of this criterion for a material, the better is its performance as a spring. The best man-made spring materials have a performance that is comparable with that of antler, but falls short of that offered by silks of various types. If the criterion for light springs,  2f /  E is used instead, the natural materials rank even more highly.

E 1 3 .

1 3 F i n d i n g a s u bstitute for bone. Find an engineering material that most closely resembles the longitudinal strength of compact bone (cortical bone (L)) in its strength and density characteristics by plotting data for this material, listed in the table below, onto a copy of the Strength – Density chart for engineering materials (Figure 3.4).

Material

Density  (kg/m3)

Strength,  f (MPa)

Cortical bone

1950

130

132

Answer. The modulus and density of compact bone are plotted on a copy of the strength-density chart, below. These properties are matched closely by those of magnesium and glass fiber reinforced epoxy, GFRP. These materials work well for external braces. For prostheses, bio-compatibility is essential. This leads to the used of cobalt and titanium-based alloys even though the match of mechanical properties is poor. Efforts are underway to develop a bio-compatible polyethylene/hydroxyapatite composites with a near perfect match of properties. Hydroxyapatite is a natural ceramic that is found in real bone – it appears on the charts for natural materials.

E13.14 Creativity : what could you do with X? The same materials appear on all the charts of Chapter 3. These can be used as the starting point for “what if…?” exercises. As a challenge, use any chart or combination of charts to explore what might be possible by hybridizing any pair of the materials listed below, in any configuration you care to choose. (a) Cement (b) Wood (c) Polypropylene (d) Steel (e) Copper.

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Ch 14. Hybrids: case studies This chapter has no exercises.

Ch 15. Materials and the environment E15.1 Functional units. Choose a functional unit for an E-LCA of the products in the table below, defining the unit to be used in making the inventory. Think of the basic need filled by the product and express it as “per…what?”.        

Washing machine Air conditioner Lighting Office space Paint A smartphone Copper mining Packaging for retail sale of milk

Answer. The table lists the suggested functional units

Product

Functional unit

Washing machines

per 5 kg load of washing

Air conditioners

per unit volume of cooled space per unit time (per cool m per year)

Lighting

per lumen per year (the lumen is a measure of light intensity)

Office space

per m per year (accommodation is characterized per unit area, not per unit volume)

Paint

per m protected for 10 years

Smartphone

per minute of use-time

Copper mining

per kg of metallic copper

Packaging for milk

per liter

3

2 2

E15.2 System boundaries (1). What, in the context of life-cycle assessment, is meant by “system boundaries”?

How are they set?

Answer. The system boundary should normally enclose all four phases of product life: material production, product manufacture, product use and product disposal. For the E-LCA to meet ISO standards, the practitioner must define the system boundary and state which aspects of each phase of life are included within it and which remain outside.

A partial E-LCA, setting the system boundary around only one phase of life, is sometimes useful. The embodied energy and carbon footprint of a material are examples. They are found by treating the material as a product, enclosing only the material production within the system boundary.

E15.3 System boundaries (2). Between 2010 and 2015, a number of European car-makers installed “defeat” software in their vehicles to conceal the real level of exhaust emissions. Where would you place 134

the foreground system boundary on Figure 15.3 of the text if the goal of the assessment was to compare emissions from different makes of car?

Answer. The foreground is now the cars and their emissions. The background is not entirely irrelevant – the nature of the fuel (petrol, diesel) and additives in petrol affect the exhaust emissions.

E15.4 System boundaries (3). An energy company supplying diesel fuel for vehicles proposes to increase production by drilling new wells. Shareholders ask for an environmental comparison of its supply chain with that of competing companies. Where would you place the foreground system on Figure 15.3 of the text to make this comparison?

Answer. The foreground is the entire production process, including the drilling of the well. The vehicles and their emissions are not relevant unless the company produces a dirty fuel (one with aboveaverage sulphur content, for instance), when they become relevant.

E15.5 System boundaries (4). Figure 15.20 shows the steps in providing milk from cow to consumer. It involves milking, bottling, transport to point of sale, refrigeration and final consumption. The point of sale is a supermarket. The milk is provided by independent dairy farmers and the bottling is carried out by a subcontractor. The rest of the supply chain is under the direct control of the supermarket itself. As the manager of the supermarket, what will you choose as the functional unit? Where will you place the system boundary for an environmental life cycle assessment of provision of bottled milk to your customers?

Figure 15.20 Steps in providing milk to consumer

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Answer. The proper functional unit is “per liter of refrigerated milk provided to the customer”. If consumers are concerned about poor environmental practice anywhere along the supply chain of the products they buy, they will hold the retailer responsible. To the supermarket manager, the entire supply chain falls within the foreground system even though part of it is not under their direct control.

E15.6 System boundaries (5). As part of due diligence, the supermarket selling the milk shown the exercise E15.5 requires the CEO of the milk-bottling company to provide an environmental assessment of the bottling process. Where should the CEO place the foreground system to conduct this assessment?

Answer. The requirement is met if the CEO limits the foreground assessment to bottle-making and bottling. The milk and the subsequent delivery of filled bottles to the supermarket become the background system.

E15.7 System boundaries (6). The environmental officer of the supermarket described in Exercise E5.5 wishes to commission an E-LCA of the parts of the supply chain under its direct control. What should be included in the foreground system for this assessment?

Answer. The supermarket has direct control of transport from the bottling plant, refrigeration for a length of time not exceeding the sell-by date, facilitating sale to the consumer and perhaps recovering empty bottles for recycling.

E15.8 Inventory compilation: PE and glass bottles. Supplier A supplies milk in 1 liter disposable polyethylene bottles or in glass bottles that can be reused. A 1 liter polyethylene milk bottle, when empty, weighs 36 grams. An equivalent glass milk bottle weighs 300 grams. Compare the material carbon footprints, and molding carbon footprints of a PE and a glass milk bottle. The eco-properties of the two materials are listed here10 Polyethylene

Soda-lime glass

Primary production Embodied energy, MJ/kg Carbon footprint, kgCO2,eq/kg Water usage, l/kg

75 1.9 58

11 0.76 14

Recycling Embodied energy, MJ/kg Carbon footprint, kgCO2,eq/kg Recycle fraction, %

27 0.94 8.4

8.7 0.53 24

Molding Energy. MJ/kg Carbon footprint, kgCO2,eq/kg

20.5 1.3

9.5 0.62

Bottle washing Energy per bottle per wash, MJ Carbon per bottle per wash, kg

0.4 0.03

(a) What is the appropriate functional unit to use when assessing the emissions associated with the bottles? (b) Which bottle has the lower carbon footprint if made from virgin materials?

10 Data from Appendix A, Tables A11 and A12

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(c) Which has the lower carbon footprint if made from 100% recycled material? (d) If the polyethylene bottle is used only once but the glass bottle is returned and reused five times before it is broken or discarded, does the conclusion change?

Answer. (a) The appropriate functional unit is not “carbon footprint per bottle” but “carbon footprint per liter of milk delivered”. (b) and (c) Multiplying the carbon footprint per kg by the bottle-weight for each material and summing the contributions of the material itself and the molding process

100% virgin material

36g PE bottle 300g glass bottle

Carbon footprint, kgCO2,eq 0.12 0.41

100% recycled material

36g PE bottle 300g glass bottle

0.08 0.35

Bottle

The carbon footprint of glass and glass-molding, per kg, are almost half those of polyethylene, but the much heavier glass bottle, whether virgin or recycled, has a carbon footprint that is factor of three greater than the PE bottle.

(d) If the PE bottle is used only once and the glass bottle is reused five times, the glass bottle has delivered five times more milk. To do this, it must be washed and sterilized four times. The carbon footprint per liter of milk delivered then becomes 0.41/5 + 4 x 0.03 = 0.20 kg per liter of milk delivered if the bottle was made from virgin glass, (or 0.19 kg if the bottle was made of recycled glass).

E15.9. Embodied energy. What is meant by embodied energy per kilogram of a metal? Why does it differ from the free energy of formation of the oxide, carbonate or sulfide from the which the metal was extracted?

Answer. The embodied energy of a metal the energy needed to extract and refine it. To convert an oxide, sulfide, or carbonate ore to metal, the free energy of formation of those compounds must, of course, be provided. The embodied energy includes this, but it also includes the energy to mine, transport and concentrate the ore, and the energy “loss” that results from the inefficiencies of the processes.

E15.10 Ideal and real embodied energy. Iron is made by the reduction of iron oxide, Fe 2O3, with carbon.

Aluminum is made by the electro-chemical reduction of Bauxite, basically Al 2O3. The enthalpy of oxidation of iron to iron oxide Fe2O3 is 7.4 MJ/kg. That of aluminum to its oxide Al2O3 is 30.4 MJ/kg. These are the lower limits for the energy required to extract these metals. Compare these energies with the embodied energies of cast iron and of aluminum, retrieved from Appendix A, Table A11. What conclusions can you draw?

Answer. The table shows the data. The embodied energies are larger, by a factor of 4 to 6, than the enthalpy of oxidation. This is because of the inherent irreversibility of the reduction process, heat and other energy losses, and the energy required for mining transporting and concentrating the ores before reduction.

Material Cast iron Aluminum

Enthalpy of oxidation (MJ/kg) 7.4 30.4

Embodied energy (MJ/kg) from Table A11 30 190

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Ratio actual/ideal embodied energy 4.0 6.3

E 15.11 Embodied energies. (a) Window frames are made from extruded aluminum. It is argued that making them instead from extruded PVC would give a product with a lower embodied energy and carbon footprint. If the section shape and thickness of the aluminum and the PVC windows are the same and both are made from virgin material, is the claim justified? Embodied energies and densities for the two materials are listed in the data sheets of Appendix A, Tables A2 and A11. (b) Products like window frames are usually made from recycled aluminum. If the embodied energy of recycled aluminum is 30 MJ/kg, does the conclusion change?

Answer. (a) The volumes of material in the two window frames are the same. The table lists the mean values of the properties from the data sheets of Appendix A, the embodied energy per unit volume, H m 

Mean density ρ 3 (kg/m ) from Table A2

Embodied energy Hm (MJ/kg) from Table A11

Embodied energy 3) H m  (GJ/m

Aluminum PVC

2,700 1,450

513 94

Recycled aluminum

2,700

190 65 From Table A12 33

Material

89

The aluminum window frame has an embodied energy that is larger, by a factor of 6, than the one made of PVC. (b) If the frame is made from 100% recycled aluminum (bottom row of the table), the aluminum and the PVC frames have about the same embodied energy.

The ratio of CO2 footprint / Embodied energy for the two materials is nearly the same (it is plotted in the answer to Exercise E15.1) so the conclusion is the same when based on CO2 footprint.

E15.12. Getting a feel for embodied energies. Aluminum is made by the electrolysis of bauxite. The embodied energy of virgin aluminum is approximately 190 MJ/kg. What else could you do with this much electrical energy? Explore (a) How much water would this much energy bring to the boil from an initial temperature of 20C (the specific heat of water is 4.18 kJ/kg.K)? (b) How far would it drive an electric car if the energy needed to propel the car is 5 MJ/km?

Answer. (a) The specific heat of water is 4.18 kJ/kg.K so it takes 0.33 MJ to heat 1kg (or 1 liter) of water from room temperature (20C) to its boiling point. You could boil 568 liters of water with the energy it takes to make just 1 kg of aluminum. (b) The electric car could be driven 190/5= 38 km with the energy it takes to make just 1 kg of aluminum.

The message here is that embodied energies of materials are large. environmental reasons for selecting materials with low embodied energy.

There are good economic and

E15.13 Extraction efficiency of copper. Copper, atomic weight 59 kg/kmol, can be extracted from natural copper oxide, CuO. The free energy of formation of CuO is 134 kJ/mol. This much energy must be provided to 138

liberate one mol of copper from CuO. How does this compare with the reported embodied energy of copper, 59 MJ/kg (Appendix A, Table A11)?

Answer. The “ideal” energy per kg to reduce CuO to copper is H

134 x 1000  2,380 kJ / kg  2.3MJ / kg 59

This is 4% of the embodied energy. The difference arises because of the additional energy needed to mine and concentrate the CuO (the typical mining grade is 4%), and the energy lost as low-grade heat during the refining.

E15.14 Carbon footprints. Rank the three common commodity materials Low carbon steel, Aluminum alloy and Polypropylene by carbon footprint per unit weight (kgCO2,eq/kg), and carbon footprint per unit volume 3 (kgCO2,eq/m ), using data from Appendix A, Table A2 for density and Table A10 for carbon footprint..

Answer. The table shows the data and the calculated information. Steel has the lowest carbon footprint, CO2 per unit weight. Polyethylene has by far the lowest carbon footprint per unit volume, CO2  .

Material

Carbon footprint CO2 (kgCO2,eq/kg)

Low carbon steel Aluminum alloys Polypropylene

2.5 12 2.9

Mean Density  3

(kg/m ) 7850 2700 900

Carbon footprint 3 CO2 .  (kgCO2,eq/m ) 19,630 32,400 2,610

E15.15 Carbon footprint and embodied energy of materials.. Make a bar chart of CO2 footprint divided by Embodied energy, using data from the data sheets of Appendix A, Table A11, for      

Cement Low carbon steel Copper Aluminum alloys PVC and Softwood

Which material has the highest ratio? Why?

Answer. The figure shows ratio of carbon footprint to embodied energy. For most materials the ratio is in the range 0.05 - 0.07 kg/MJ, allowing an estimate of one if the other is known. The value for cement, however, is far higher (0.17). This is because making cement (a component of concrete) involves the calcining of calcium carbonate, CaCO3, driving off carbon dioxide, CO2, to leave lime, CaO. This CO2 contribution adds to that derived from the fuel that is used to heat the kiln to give the high total.

Note the median value of this ratio, about 0.06 kg/MJ, is almost constant, useful when estimating carbon footprints from embodied energies and vice versa.

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E15.16. Carbon footprints. (a) A range of office furniture includes a chunky hardwood table weighing 18 kg and a much lighter table with a 3.0 kg aluminum frame and a 3.0 kg glass top. Which of the two tables has the lower carbon footprint? Use data from the data sheets of Appendix A, Tables A11, and A12. (b) If the alumina-glass table is made from recycled aluminum with a carbon footprint that is 14% that of virgin aluminum, and recycled glass with a carbon footprint that is 70% that of virgin glass, does the conclusion change?

Answer. (a) Appendix A provides data for carbon footprint Material Hardwood Aluminum alloy Glass

Carbon footprint, virgin material (kgco2,eq/kg) 0.6* 12 0.76

Carbon footprint, recycled material (kgco2,eq/kg) 2.6 0.53

* The embodied energy of wood arises because of harvesting, cutting and kiln drying.

The total carbon footprint of the hardwood table is 18 x 0.6 = 10.8 kgco2,eq Total carbon footprint of aluminum/glass table is 3.0 x 12 + 3 x 0.76 =38.3 kgco2,eq The aluminum-glass table a carbon footprint that is three times that of the hardwood table, despite only weighing one third as much. (b) Replacing virgin with recycled aluminum and glass reduces carbon footprint of the aluminum/glass table to 3.0 x 2.6 + 3 x 0.53= 9.4 kgco2,eq The carbon footprint of the aluminum/glass table is now comparable with that of the hardwood table.

E15.17 Minimizing embodied energy. Use the E  Hp  chart of Figure 15.14 to find the polymer with a modulus E greater than 1 GPa and the lowest embodied energy per unit volume.

140

Answer. The construction is shown below. The polymer with a modulus greater than 1 GPa and the lowest embodied energy is polypropylene, PP, ringed on the figure.

E15.18. Comparing products by embodied energy. A maker of polypropylene (PP) garden furniture is concerned that the competition is stealing market share by claiming that the “traditional” material for garden furniture, cast iron, is less energy and CO2 intensive than the PP and much more durable. A typical PP chair weighs 1.6 kg; one made of cast iron weighs 7.5 kg. (a) What is sensible functional unit for a garden chair? (b) Use the data for these two materials in Appendix A, Table A11, to find out who is right. Are the differences significant if the data for embodied energy are only accurate to  25%? (c) If the PP chair lasts 5 years and the cast iron chair lasts 25 years, does the conclusion change?

Answer. (a) Functional unit. When the life-times of competing products differ, the functional unit must recognize that the energy and carbon release during manufacture should be scaled accordingly. An appropriate functional unit is “provision of garden seating, per year”. (b)The table lists mean values of embodied energy and carbon footprint, per kg, for the two materials. The last two columns show the values per chair. If the difference in lifetime is ignored, the polypropylene chair wins – it has lower embodied energy and much lower carbon footprint than the cast iron chair.

Material Cast iron Polypropylene

Embodied energy* MJ/kg 29 80

*

CO2

kgCO2,eq /kg 2.1 2.9

141

Embodied energy MJ/chair 217 128

CO2 kg CO2,eq /chair 15.7 4.6

(c) If the cast iron chair lasts for 25 years and the polypropylene chair for only 5, the proper measure is the energy and carbon per year of functional life. The cast iron chair wins.

E15.19 Materials with low carbon footprint. Identical casings for a power tool could be die-cast in aluminum or molded in ABS or polyester GFRP. Use the carbon footprint per unit volume bar-chart of Figure 15.6 to decide which choice minimizes the carbon footprint of the casing, assuming the same volume of material is used for each casing.

3

Answer. The carbon footprint /m chart of Figure 15.6 is reproduced below, with the materials circled. ABS has the lowest carbon footprint per unit volume. For a casing using the same volume of material, ABS emits only one sixth as much carbon as (virgin) aluminium. GFRP lies in between. If recycled aluminum, carbon footprint 3 7.2 tonne/m , is used instead, the aluminum casing has a lower footprint than GFRP but is still higher than ABS

E15.20 Disposable knives and forks (Fig. 15.21). Disposable knives and forks are ordered by an environmentally-conscious pizza-house.

The shape of each (and thus the length, L, width, b, and profile) are fixed, but the thickness, t, is free: it is chosen to give enough bending-stiffness and strength to impale the pre-cut pizza without excessive flexure or failure. The pizzeria-proprietor wishes to project a green image by minimizing the energy-content of his throw-away tableware, which could be molded from polystyrene (PS), stamped from aluminum sheet or thermo-formed from wood ply.

Figure 15.21 The fork

Formulate the design requirements for the material to make the fork. Derive, or look up, an appropriate material index for selecting materials for low embodied-energy forks subject (a) to a stiffness constraint, and (b) to a strength constraint. Model the eating implement as a beam of fixed length L and width b, but with a thickness t that is free, loaded in bending, as in the figure. The objective-function is the volume of material in the fork times its energy content, Hm  , per unit volume ( Hm is the embodied energy per kg, and  the density). The limit on flexure imposes a stiffness constraint (Appendix B, Section B3). Use this information to develop the index.

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Flexure, in cutlery, is an inconvenience. Failure – whether by plastic deformation or by fracture – is more serious: it causes loss-of-function; it might even cause hunger. Repeat the analysis, deriving an index when a required strength is the constraint.

Answer. This is a straightforward application of the method illustrated in the text by the derivation of equations (4.15) and (4.16); the only difference is that energy content, mHm, (where Hm is the embodied energy per kg of the material) rather than mass, m, is minimized. The free variable is the thickness t of the shaft of the fork; all other dimensions are fixed. There are two alternative constraints, first, that the fork should not flex too much, second, that it should not fail.

Function



Environmentally friendly disposable forks

Constraints

   

Length L specified Width b specified Stiffness S specified, or Failure load F is specified

Objective



Minimize the embodied energy mHm

Free variables

 

Shaft thickness t Choice of material

The resulting indices are

 E 1 / 2 and M   f  M 

2/ 3

1

Hm 

2

Hm 

The selection can be implemented using Figures 15.14 and 15.15 of the text. A selection line is plotted on each passing through Polystyrene (PS), picking up the relevant slope from the Guidelines on the lower right of each chart. The three materials are ringed on both charts. Aluminum lies a little above PS, wood laminate far above it on the stiffness chart – it has a slightly lower embodied energy for the same bending stiffness as PS. But on the strength chart aluminum alloys, even the strongest, lie below PS; wood remains far above it. Wood is the best choice for both stiffness and strength a low embodied energy.

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E15.21 Eco-audit (1): a wheeled bin. The bin shown sketched in Figure 15.12, top right, is manufactured in Poland and imported by road to a distribution point in the UK, a distance of 1,750 km, by 26 tonne (3 axle) truck. The bin is molded from high density polyethylene (HDPE) with butyl-rubber wheels and weighs 15 kg. Its guaranteed life is 3 years, after which it could be recycled but in practice it is combusted with energy recovery because of shortage of recycling capacity. The bin uses no energy during its life. The energy fingerprint over life is shown in Figure 15.12, top row right.. What changes could improve it?

Answer. The energy profile is dominated, on the left, by that of the material. The contribution of transport is small. Combustion with heat recovery captures a little of the embodied energy of the polymer.

The framing of the analysis in the question is misleading. The functional unit used there is “one wheeled bin with a life of 3 years”, but the 3-year limit is an unnecessary restriction. A better choice is “one wheeled bin, per year of active life”. Increasing the life to 6 years reduces the contributions per year by a factor of 2.

Incineration at end of life recovers little energy but releases much carbon. PE can be recycled. The embodied energy of virgin PE is 75 MJ/kg. That of recycled PE is 27 MJ/kg, a factor of 3 less. If the original bin was made of PE with a 50% recycled content, the large material contribution at start of life is reduced from 1.2 to 0.8 GJ. So the recommendation: encourage the retention of the bin in use for as long as possible (perhaps by increasing the guarantee period), and recycle it when it is retired.

E15.22 Eco-audit (2): a.jug kettle. A 2 kW jug kettle sketched in figure 15.12, second row left, is designed in Britain, manufactured in Shangai China, shipped from Shanghai to Los Angeles and transported by road in a 4axel (32 tonne) truck to Chicago where it is sold. The kettle has a life of 3 years before the style-conscience owner replaces it, during which time it is used, on average, three times a day to boil water, taking 2 minutes per boil. At the end of life the materials of the kettle are recycled. The energy audit shows the energies required for each phase of life. What steps would you take to reduce them?

Answer. The largest contribution to the energy fingerprint of the kettle is that of the use-phase. If energy is to be saved, the kettle must be used less or better insulated, retaining the heat in water left inside – a stainless stell thermos flask keeps 0.5 liters of water hot for up to 20 hours.. The energy and carbon fingerprints of many energy-using products – hair dryers, mixers, refrigerators, dishwashers – are dominated by the use-phase, as in this example.

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E15.23 Eco-audit (3): a family car. The energy fingerprint of a small gasoline-powered family car is sketched in figure 15.12, third row left. Its mass is 1000 kg, using materials with a total embodied energy of 70 GJ. A further 27 GJ is required for the manufacturing phase. The car is manufactured in Germany, shipped by very large container ship from Bremen to New York (6460 km) and delivered to the US show room in Chicago by 40 tonne (6 axle) truck (1270 km). The car has a useful life of 10 years, and will be driven on average 20,000 km per year consuming 2 MJ/km. We assume that collection and shredding at end of life consumes 0.5 GJ but recovers materials with an energy-credit of 25 GJ per vehicle. The bar chart shows the contribution to energy over life.

What steps would do most to reduced life-energy requirement?

Answer. The use-phase of the vehicle remains the dominant contribution to life-energy consumption.

Material choice should focus first on reducing vehicle mass since it is this that most directly correlates with fuel consumption. It is essential to then check that the consequent change in embodied energy of the material choice has not negated the gain resulting from the lower mass.

E15.24 Eco-audit (4): a smartphone11. It takes almost 1 GJ to extract, refine and process the materials needed to make a smartphone. Typically, a phone consumes 3.8 kW.hr of electrical energy per year to charge it. Most smartphones are manufactured in China and air freighted from there to the rest of the world. Typical phone contracts last for 24-months, after which a new contract, with a new handset included, is set up. This rapid turnround creates a waste-management problem. Recycling is possible but practiced only at a low level at present.

The resulting energy fingerprint is shown in the bottom row of Figure 15.12. How could the energy profile of the phone be improved?

Answer. Materials and manufacture consume more energy than all the other phases of life combined.

Figure 15.13 offers guidance on what to do about it.

First and most obvious: extend the product life. In the past the rapid evolution of the technology has made 2-year replacement attractive, but this is slowing, making longer life practical for the user, but constraining the sale of new phones. The other option is to make the product more modular, allowing components (such as the battery) to be replaced, and the reuse or restoration of components.

11 Ercam, M. et al (2016) “Life cycle assessment of a smartphone” Conference: ICT for Sustainability https://www.researchgate.net/publication/308986891_Life_Cycle_Assessment_of_a_Smartphone

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E15.25 Materials for low embodied energy structures. Show that the index for selecting materials for a strong panel with the dimensions shown in the Figure 15.22, loaded in bending, with minimum embodied energy content, is that with the largest value of

 y1 / 2 M  Hm 

Figure 15.22 A panel

where Hm is the embodied energy of the material,  its density and  y its yield strength. The constraint is that *

the failure load F must exceed a chosen value F where: F  C2

I y hL

 F*

where C2 is a constant and I is the second moment of area of the panel, I 

b h3 (Appendix B, Table B4).

12

The table summarizes the requirements

Function Constraint Objective Free variables

    

Panel, loaded in bending Must carry load F* without yielding Minimize the embodied energy of the panel Maximize energy absorbed per unit mass Choice of material

Answer. The objective function for the embodied energy of the panel, H , is the volume Lbh times the embodied energy of the material per unit volume, Hm 

:

H  bhL  Hm *

Its failure load F must be at least F where I y * F  C2

hL

 F

Here C 2 is a constant that depends only on the distribution of the loads and I is the second moment of area, which, for a rectangular section, is b h3 12 We can reduce the energy H by reducing h , but only so far that the strength constraint is still met. Combining the last two equations and solving for h gives I 

1. / 2

 12 F * L   h  C 2b y    Using this to eliminate h in the objective function gives 1/2

 12 F* L3 b  H    C 2  

 Hm  y1 / 2

146

*

The quantities F , L , b and C 2 are all specified; the only freedom of choice left is that of the material. The best materials for a strong panel with the lowest embodied energy are those with the largest values of

M 

 y1 / 2 Hm 

E15.26. Energy-efficient materials for transport. The maker of a small electric vehicle plans to make bumpers out of a molded thermoplastic. Which index is the one to guide this selection if the aim is to maximize the range of the vehicle for a given battery storage capacity? Plot it on the appropriate chart taken from the text and identify promising thermoplastics.

Answer. The use-energy of any vehicle, whether gasoline or electric powered, increases with its mass. The bumper contributes to this mass. Maximizing range for a given battery capacity means minimizing mass of passive components like bumpers. The function of the bumper is to sustain bending loads without failure. The required index (equation 15.5 of the text) is

 y2 / 3  This is plotted on the strength – density chart below, picking up the slope of 1.5 from the guide line at the lower right. The selection line is positioned such that a few polymers remain above it – they are the best choice, shown enlarged at the lower right. They are:  Polycarbonate, PC  Polyamide (nylon) PA and  Polyetheretherketone PEEK In practice bumpers are made of blends of Polycarbonate with other polymers such as polypropylene or polyamide, some filled with calcium carbonate or glass fiber, or of fiber reinforced thermosets such as polyester.

PEEK is too expensive for application such as this.

M

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E15.27. Energy-efficient floor joists: embodied energy and shape. Floor joists are beams loaded in bending (Fig. 15.23). They can be made of wood, of steel, or of steel-reinforced concrete, with the shape factors listed below. For a given bending stiffness and strength, which of these carries the lowest production-energy burden? The relevant data, drawn from the tables of Appendix A, are listed.

Figure 15.23. A plain and a shaped section

(a) Start with stiffness. Locate from equation 11.20 of the text the material index for stiffness-limited, shaped beams of minimum mass. Adapt this to make the index for stiffness-limited, shaped beams of embodied energy by multiplying density  by the embodied energy per kg, Hm . Use the modified index to rank the three beams. (b) Repeat the procedure, this time for strength, creating the appropriate index for strength-limited shaped beams at minimum energy content by adapting equation 11.28.

What do you conclude about the relative energy-penalty of design with wood and with steel?

Strength

Material

Density  kg/m3

Modulus E GPa

 f MPa

Energy Hm MJ/kg

Soft wood Reinforced concrete Steel

700 2900 79000

10 35 210

40 10 200

7.5 5 30

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Shape factor  Be 2 2 15

Shape factor

Bf 1.4 1.4 4

Answer. (a) The index for an energy-efficient beam with a specified stiffness, modified to include shape, is

Be E   M 

1/ 2

1

Hm  e

where E is the modulus,  the density, Hm the energy content per unit weight, and  B is the shape factor for stiffness controlled design. The first Table shows values of M1 for the wood, the concrete and the shaped steel beam. Wood wins.

Hm 

Material

E (GPa)

(GJ/m )

Wood Reinforced concrete Steel

10 35 210

5.25 14.5 238

3

 Be

M1 3 (GPa /GJ/m )

2 2 15

0.86 0.57 0.23

1/2

(b) The index for an energy efficient beam of prescribed strength is

Bf  f  M 

2/ 3

2

Hm 

where Bf is the shape factor for failure in bending. The next table shows that, as before, wood is the most efficient material.

Material Wood Reinforced concrete Steel

f (MPa) 40 10 200

Hm  3

(GJ/m ) 5.25 14.5 238

Bf

M2 2/3 3 (MPa /GJ/m )

1.4 1.4 4

2.8 0.4 0.36

E15.28 Power systems: energy pay-back time. The table below shows data for the main materials used in a wind turbine (Figure 15.24). It lists the approximate mass used per kW installed and the embodied energy per kg of these materials.

Bill of materials Aluminum CFRP Concrete Copper GFRP Steel Neodymium (magnets) Plastics

(a)

Mass per installed unit of power capacity (kg/kW) 2 7.5 500 1.5 7.5 120 0.012 5

Embodied energy of materials (MJ/kg) 200 480 1.8 65 160 25 85 80

Figure 15.24 A wind turbine

Use these data to estimate the total embodied energy of the materials in a 1 MW wind turbine.

149

(b) The capacity factor for a typical land-based wind turbine is 20%. (Capacity factor is the equivalent fraction of the time that the turbine is operating at its rated capacity; thus a 1MW turbine produces, on average, the equivalent of 0.2 MW of electrical power). How long will it take a 1MW turbine to return the embodied energy of the materials of which it is made?

Answer. (a) The embodied energies of the components are listed and summed in the last column of the table below. The embodied energy per kW of nominal generating capacity is 9,589 MJ/kW. Thus the embodied energy of the materials in a 1 6 MW turbine is 9.589 x 10 MJ. Bill of materials Aluminum CFRP Concrete Copper GFRP Steel Neodymium (magnets) Plastics

Mass per installed unit of power capacity (kg/kW) 2 7.5 500 1.5 7.5 120 0.012 5

Embodied energy of materials (MJ/kg) 200 480 1.8 65 160 25 85 80 Total

Mass x Embodied energy (MJ/kW) 400 3,600 900 67 1.200 3000 1 400 9,598

(b) A 1 MW turbine running at a capacity factor of 0.2 produces, when averaged over time, 0.2 MW (that 6 is, 0.2 MJ/sec) of electrical power. In one year it produces 0.2 x 3600 x 24 x 365 = 6.3 x 10 MJ. Thus the time it 6 6 takes for the turbine to return the embodied energy contained in its structure is 9.6 x 10 / 6.3 x 10 = 1.5 years.

150

Ch 16. Materials and Sustainable Development E16.1 Exploring the UN Sustainability goals (1). In 2015, the United Nations 2030 Agenda for Sustainable Development identified 17 Goals12,13, shown in Fig 16.16. It was adopted by all 193 member states of the United Nations. Goal 12, Responsible Consumption and Production, most directly addresses materials and their uses. Summarize the aims of Goal 12 in plain language. You’ll find the targets starting on page 26 of this rather wordy document: https://sustainabledevelopment.un.org/post2015/transformingourworld/publication

Figure 16.18. The UN Sustainable Development Goals for 2030.

Answer. SDG 12, Responsible consumption and production, focusses on conserving resources and protecting the environment. Paraphrased, its aims are, by 2030, to • Implement the 10-Year Framework of Programs on Sustainable Consumption and Production • Encourage the efficient use of natural resources.

• Reduce food waste in the supply chain and at the retail and consumer levels.

• Reduce the release of chemicals to air, water and soil.

• Encourage sustainable procurement and reduce waste by prevention, recycling and reuse.

• Promote lifestyles inharmony with nature, and sustainable tourism that respects local culture.

• Rationalize fossil-fuel subsidies, making allowance for impacts this might have on developing countries.

It would be interesting to explore how AI tools like ChatGPT or similar respond to each of these challenges.

E16.2. Exploring the UN Sustainability Goals (2). Paraphrase the aims of Goal 8, Decent Work and Economic Growth, set out in the UN 2030 Agenda for Sustainable Development. It is accessible from page 23 of https://sustainabledevelopment.un.org/post2015/transformingourworld/publication

Answer. SDG 8, Decent Work and Economic Growth, is about combining economic growth with fair employment and environmental protection. Paraphrased, its aims are, by 2030, to  Sustain at least 7% GDP growth per annum in the least developed countries.  Increase productivity through innovation and promote growth of small and medium-sized enterprises.  Improve resource efficiency and decouple economic growth from environmental degradation.  Achieve full employment and equal pay for work of equal value.  Reduce the proportion of youth not in employment, education or training by 2020.

12 https://unstats.un.org/sdgs/report/2019/Overview/ 13 https://www.un.org/sustainabledevelopment/sustainable-development-goals/

151

   

Eradicate forced and child labour, modern slavery and human trafficking by 2025.

Promote sustainable tourism that creates jobs.

Expand access to banking, insurance and financial services for all.

Increase Aid for Trade support for developing countries.

E16.3 Progress in achieving the UN Sustainable Development Goals (SDGs). The UN SDGs set ambitions targets for global change towards a more sustainable world order. Will they be met by the target date of 2030?

Try an internet search for responses, limiting it to recognized reliable sources.

Answer.

“Even before the pandemic, the world was off track to fulfil the SDGs. Projections suggested that 6 per cent of the global population would still be living in extreme poverty in 2030, missing the target of ending poverty. The global maternal mortality rate declined by 38 per cent between 2000 and 2017 but this is less than half the rate needed to achieve the 2030 target.”

UN Department of Economic and Social Affairs, 2020 https://www.un.org/development/desa/dpad/publication/un-desa-policy-brief-81-impact-of-covid-19-on-sdgprogress-a-statistical-perspective/

“How much progress is being made on the UN’s Sustainable Development Goals? The Sustainable Development Goals Report 2022 painted a worrying picture of progress in almost all areas, but the 2023 update delivers an even graver warning: "The promises enshrined in the SDGs are in peril," the UN states. This year marks the halfway point to the 2030 Agenda deadline, increasing global urgency to ignite progress.”

World Economic Forum, Sept 11, 2023. https://www.weforum.org/agenda/2023/09/un-sustainable-development-goals-progress-report/ It should be remembered that the UN SDGs are extremely ambitious, aimed at inspiring efforts rather than setting immutable deadlines.

E16.4. Life cycle assessments. What do E-LCAs and S-LCAs have in common? In what ways are they different?

Answer. S-LCAs and E-LCAs have much in common. Both use the ISO framework (definition of goals and scope, inventory assembly, impact assessment and interpretation and impact on areas of principal importance, AoPs).

The main differences are the impact categories and the AoPs. 



The E-LCA assesses the environmental impacts of product life. Examples of its impact categories are climate change, resource depletion and fresh water availability. AoPs include human health, ecosystem health and resource availability. the S-LCA assesses its social and socio-economic impacts. Examples of its impact categories are equal opportunity, fair salary and working hours, indigenous rights and freedom of speech. AoPs include human and civil rights, human wellbeing and dignity.

E16.5. Life cycle costing (LCC). (a) What is life cycle costing? (b) A car costs $40,000, which could instead have been invested at 4%/year. The value of the car depreciates at 15% per year. It consumes 10 liters of gasoline per 100 km (23 mpg). Routine maintenance, taxes and insurance average $2,200 per year. If gasoline costs $1.6/liter ($4.00 per US gallon), and the car is driven 32,000 km (20,000 miles) per year, what is the cost of ownership over 10 years?

Answer. (a) Section 16.6 of this Chapter gives this definition: 152

Life-cycle costing (LCC) is the estimation of the cost of owning an asset over its entire life. It includes:  The acquisition cost and cost of delivery and installation;  The interest payments on the capital used to buy the asset (if borrowed), or the loss of interest that could have been earned by investing it (if self-financed);  Operating costs, including consumables, energy, water and servicing;  End-of-life costs and residual value that could be recovered by sale or reclaim of a deposit;  Environmental costs such as carbon tax and the cost of internalizing those that are at present externalized.

10

(b) The capital cost, discounted over 10 years, is $40,000 x (1 + 0.04) = $59,000 10 The residual value of the car after 10 years is $40,000 x (1 – 0.15) = $7,900 The running cost, not including fuel = $22,000* Fuel cost. if driven 20,000 miles per year for 10 years = $34,800* Total 10 year life cost = (59,000 – 7,900) + 22,000 + 34,800 = $107,600

This is 2.6 times the initial capital cost of the car. *These are future costs and should be discounted appropriately – See Ashby (2022), Chapter 7 for the way to do this.

E16.6 Externalised costs. What is meant by an “externalized” cost in an environmental context? Numerous web-sites carry good but wordy explanations. Put together one that is as simple and clear as you can.

Answer. Externalised costs are costs you generate but don’t pay. Industrial emissions, for example, damage health and property by polluting air and water, cause acid rain, ozone layer depletion and, ultimately climate change. When the creator accepts responsibility for the problem and pays to fix it, the environmental cost is said to be internalized. When the damage is global, it is difficult to establish who is-responsible. The environmental cost becomes a burden on society as a whole and is said to be externalized. Much current legislation aims to internalize environmental costs that, at present, are externalized.

E16.7. The triple bottom line. What is meant by the Triple Bottom Line in economics? What are its key features? Explore the 2020 Investopedia article by Will Kenton and Janet Berry-Johnson (link below) to assemble a brief response. https://www.investopedia.com/terms/t/triple-bottom-line.asp

Answer. Triple bottom line (3BL) thinking, in economics, advocates that companies should give as much weight to social and environmental concerns as they do to profits. The basic idea is that, instead of one bottom line, there should be three: profit, people, and the planet. A 3BL analysis assesses a corporation's commitment to be socially responsible and minimize its impact on the environment over time. Without doing so, it cannot account for the full cost of doing business.

E16.8. Areas of Protection. “Well-being and Dignity” is a social Area of Protection, along with Fair Working Conditions and Observation of Civil and Human rights. It is sometimes argued that Well-being is the most important AoP. But how do you measure it? The two links below describe two ways of assessing it. How, briefly, do they do it?

The first is the Gallup-Healthways Global Well-being index: http://info.healthways.com/hs-fs/hub/162029/file-1634508606pdf/WBI2013/Gallup-Healthways_State_of_Global_Well-Being_vFINAL.pdf

The second is the UN Human Development Index, plotted in Figure 16.5: http://hdr.undp.org/en/content/humandevelopment-index-hdi

Answer. The Gallup-Healthways Global Well-Being Index is a worldwide barometer of individuals’ perceptions of their own well-being. It is based on a questionnaire exploring individual sense of purpose, social integration, financial security, sense of community and physical health.

153

The UN Human Development Index (HDI) combines data for life expectancy, education and national literacy, and standard of living (measured by the Gross National Income) to give a single ranking between zero (undeveloped) and 100 (highly developed) of the population of a country.

E16.9. Supply chain due diligence (1). A company producing bronze bearings has, till now, sourced tin from China, the world largest producer. The price of tin doubled in early 2021 as the electronics industry ramped up demand, creating a supply-side risk. The CEO of the company dealt with this by exploring alternative supply channels from other tin-producing nations: Malaysia, Indonesia and Bolivia. Part of the task was to anticipate the social risk of dealing with producers in those countries. The table ranks the three countries by six social impact categories. Color-code the results in a way that the CEO can grasp quickly.

Ranking 1 (least good) to 100 (best)

Impact category

Malaysia

Indonesia

Bolivia

Forced or child labor Minimum wage Political, civil rights Voice, accountability UN HDI Satisfaction with life

55 49 51 34 71 58

75 36 51 51 39 54

61 61 84 60 74 21

Area of protection Fair working conditions Human and civil rights Wellbeing and dignity

Answer. The data suggest that the minimum wage in both Malaysia and is low – it may be necessary to check that the selected supplier pays a fair wage and that workers have an adequate voice in the way the operation is run. The UN Human Development Index for Indonesia and Bolivia are low, implying lower than average competencies in the work force. Bolivia scores more highly one category: satisfaction with life.

The data are color-coded in the table below. Rankings above 50 are highlighted in green, those below 50 but above 25 in yellow, to suggest where further investigation and possible additional training may be needed.

None are below 25, the bottom quartile. Overall, Bolivia has 5 greens. The other two have only 4.

Ranking 1 (least good) to 100 (best)

Impact category

Malaysia

Indonesia

Bolivia

Forced or child labor Minimum wage Political, civil rights Voice, accountability UN HDI Satisfaction with life

55 49 51 34 71 58

75 36 51 51 39 54

53 63 40 47 38 64

Area of protection Fair working conditions Human and civil rights Wellbeing and dignity

E16.10. Supply chain due diligence (2). Most electric vehicles (EVs), today, have lithium-ion batteries. Cobalt, in the form of the mixed oxide LiCoO2, is a key element in these batteries. The supply chain for cobalt starts at the mine level. Most cobalt comes from one nation; there are Nation 1000 tonnes/yr significant contributions from two more (see table). A maker of EVs D.R. Congo 95 wants assurance that the supply chain of the batteries for their cars Australia 7.7 Philippines 4.7 meets ethical standards. Table 16.Y lists social rankings for the three source-nations involved. Color-code the rankings to help with presenting the data for subsequent discussion.

Use green for rankings above 50%, amber for those between 25% and 50%, and red for those below 25%.

Given the results, what steps would you recommend the EV should take to establish that the cobalt they us is ethically sourced?

154

Table 16.7 Social rankings for nations supplying cobalt.

Ranking 1 (least good) to 100 (best)

Impact category

D R Congo

Australia

Philippines

Area of protection

Child / forced labor Minimum wage Political, civil rights Voice, accountability UN HDI Satisfaction with life

3 0 34 12 8 11

100 100 100 94 95 95

63 34 67 51 41 41

Fair working conditions Human and civil rights Well-being and dignity

Answer. The color-coding is shown here. The Democratic Republic of the Congo lags behind other nations in five of the six social categories listed there, flagging significant social risk. Due diligence requires that this is recognized and dealt with.

Ranking 1 (least good) to 100 (best)

Impact category

D R Congo

Australia

Philippines

Area of protection

Child / forced labor Minimum wage Political, civil rights Voice, accountability UN HDI Satisfaction with life

3 1 34 12 8 11

100 100 100 94 95 95

63 34 67 51 41 41

Fair working conditions Human and civil rights Well-being and dignity

One solution is to establish a working partnership with one or more of the Congolese producers and to set standards for workplace management. Another is to source the cobalt from somewhere else. Replacing the Congo by either of the other two significantly reduces social risk.

155

Ch 17. The wider context: Materials and Big Issues E17.1. Big hitters – Agriculture and forestry. Agriculture and forestry make large negative contributions to more than one of the big issues identified in this chapter. Which are they? Roughly how much do they contribute to each? Examine figures 17.2 to 17.7 to find out.

Answer. Figures 17.2 and 17.6 indicate that agriculture and forestry (including deforestation) are responsible for in the release about 24% of global greenhouse gasses, they are implicated in about 45% of species loss per year, and are the origin of about 1% of the global waste stream.

E17.2. Big hitters – Electrical power generation. What fraction of global carbon emissions come from electricity generation? Use the Figure 17.2 to find out.

Answer. Figures 17.2 shows that electricity generation is responsible for about 27% of global greenhouse gasses emissions per year.

E17.3. Big hitters – Building and construction. Building and construction make large negative contributions to more than one of the big issues identified in this chapter. Which are they?

Answer. Figures 17.2, 17.3 and 17.7 show that concrete for construction is responsible for about 7% of the global greenhouse gasses each year. About 70% of material consumption, and about 20% of the waste. Figure 17.2 also shows that heating and cooling of buildings is responsible for a further 11% of greenhouse gas emissions.

E17.4. Big hitters – Materials (1). Which materials make the largest contribution to global CO2 emissions?

Answer. Figures 17.2 suggests that making steel and cement for concrete are the biggest players, each contributing about 6.5% of the global total.

E17.5. Big hitters – Materials (2). Does the production of materials play a significant role in causing species loss?

Answer. Figures 17.6 suggests that material production contributes to habitat loss (mining, rubber, bio-polymer production) and deforestation and unsustainable use of land and water (toxic emissions and water use from mining) E17.6. Big hitters – Materials (3). What is the single biggest source of waste?

Answer. Figures 17.7 indicates the mining, quarrying and construction waste exceeds that from all other sources. Material extraction is involved in much of it.

E17.7. Material production. Primary material production has a large carbon footprint. The table lists the world primary production (not including recycled material) per year and the carbon footprint for nine widely used materials. Use these values to calculate the global greenhouse gas emission associated with the production of each material.

Material Blast-furnace steel Concrete Primary aluminum Paper and cardboard Polypropylene (PP) Soda-lime glass Polyethylene (PE) Polyethylene terephthalate (PET) Natural rubber

World production (million tonnes/year) 2,160 41,000 64 400 62 110 40 19 13

156

Carbon footprint (kg/kg) 2.3 0.12 14 1.2 2.9 0.76 1.9 2.7 2

Answer. The last column shows the product of the world production and carbon footprint. The results are plotted in the figure below. Primary steel and concrete production make particularly large contributions.

Material Blast-furnace steel Concrete Primary aluminum Paper and cardboard Polypropylene (PP) Soda-lime glass Polyethylene (PE) Polyethylene terephthalate (PET) Natural rubber

World production (million tonnes/year) 2,160 41,000 64 400 62 110 40 19 13

Carbon footprint (kg/kg) 2.3 0.12 14 1.2 2.9 0.76 1.9 2.7 2

Green house gas emissions (million tonnes per year) 5,900 4,920 896 480 180 84 76 51 26

E17.8. Supply risk for materials of hard disk drives. The installed global data storage capacity will reach 175 zettabytes in 202514 , generating about 50 million end-of-life hard-disk drives (HDDs) per year. For security reasons, most are shredded, even when still functional15. The table lists approximate materials intensity of a hard disk drive (adapted from European Commission 2020). Mark up a copy of Figure 17.5, supply chain risk, to indicate these elements. Use your judgement to assess the risk level for this technology.

Technology

Hard disk drive

Element Co Pt Nd (a LREE) Cr Ru

Tonnes per zettabyte 1.0 0.4 420 0.1 0.4

Answer. The marked-up copy of Figure 17.5 is shown below, highlighting elements used in hard disk drives.

Four have supply risk estimates ranging from moderate to very high. The materials intensity for neodymium, a heavy rare earth (HREE), is large and the risk very high. There is reason for concern here.

14A zettabyte is 1021 bytes. It takes one billion, 1TB hard drives to store one zettabyte of data..

15Walzberg, J. (2022) “Hard-disk drive circularity” Resources, Conservation and Recycling, vol. 178, 106102 https://www.sciencedirect.com/science/article/abs/pii/S0921344921007102

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E17.9. Supply risk for materials for RAM. The table lists approximate materials intensity of magnetic RAM (adapted from European Commission 2020). Mark up a copy of Figure 17.5, supply chain risk, to indicate these elements. Use your judgement to assess the risk level for this technology.

Technology

Magnetic RAM

Element Al Co Ni W Ta Ir Mn

Tonnes per zettabyte 0.1 0.55 2.6 7.7 7.7 2.0 8.0

Answer. The marked-up copy of Figure 17.5 is shown below, highlighting elements in magnetic RAM. One element, cobalt, is ranked as having high supply risk. The rest are low or very low. The overall risk is lower than that for hard disk drives (Exercise E17.8).

E17.10. Batteries for e-transport. The table compares the current EU consumption of five elements used in lithium-ion batteries with the anticipated demand for electric cars in 2030. Mark up a copy of Figure 17.5, supply chain risk, to show where these elements lie on the supply-risk scale. Use this information plus that in the table to assess where problems might lie.

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Material Cobalt Lithium Nickel Manganese Graphite

Current EU consumption in all applications (Thousand tonnes per year) 30 6 500 4,000 250

Anticipated demand for etransport by 2030 (Thousand tonnes per year) 67 51 280 63 500

https://rmis.jrc.ec.europa.eu/uploads/CRMs_for_Strategic_Technologies_and_Sectors_in_the_EU_2020.pdf

Answer. The marked-up copy of Figure 17.5 is shown below, highlighting elements in lithium-ion batteries.

Cobalt and graphite are of greatest concern: the anticipated demand exceeds current supply and both lie high on the risk-scale. Future demand for Lithium exceeds current supply, but the current supply risk is low. The current consumption of manganese in all applications greatly exceeds that anticipated from batteries – no problem here.

E17.11. Stock dynamics and circularity. Ink cartridges for bubble jet printers last about 4 months before they are discarded or returned to the maker. The UK consumes 65 million cartridges per year of which 15% are returned for reconditioning and refilling, saving 60% on materials and energy. The sale of print cartridges is increasing at 13% per year. What fraction of the present supply of cartridges is met by the reconditioned cartridges? What fraction would be met of all cartridges were returned for reconditioning?

Answer. If consumption yesterday was C 0 and has grown at rc % per year, today’s consumption is

rc t ) 100 where t is the lifetime of the product. The quantity recovered is R  f C0 where f is the fraction collected at end of life. The fractional contribution of yesterday’s scrap to today’s consumption is then C  C 0 exp(

r R 13  f exp( c t )  0.15 exp( 0.33)  0.14 C 100 100 Reconditioning meets 14% of current demand. If, instead, all cartridges were returned (f = 100%) and all were undamaged, they would meet 96% of current demand.

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E17.12. Decarbonising the construction industry. Construction consumes materials, generates waste and releases greenhouse gases on a large scale. A recent report* from the UK Royal Academy of Engineering explores ways to reduce its impact. Scan the report and summarise the steps it suggests. *https://www.raeng.org.uk/publications/reports/decarbonising-construction-building-a-new-net-zero Answer. The report identifies four broad areas for improvement.

Product outcomes. The construction industry does not, at present, prioritize whole-life carbon (the sum of that arising from initial construction, heating, cooling, lighting, maintenance over life, and final decommissioning.)

It should.

Design and specification. Design decisions made in the early stages of a building project ‘lock-in’ both the embodied and operational carbon for the life of the building. Establishing the availability of low carbon materials and specifying them at the design stage is essential.

Construction and reuse. The construction sector at present operates a linear materials economy. This is unsustainable. Extraction and waste can be minimized by reusing materials either for their original purpose or to fulfil a different function.

Changes to procurement. Procurers, such as government, must change their approach to procurement to reflect whole life carbon performance. Procurers can use their purchasing power to stipulate carbon performance, set out in both public and private contracts.

E17.13. Project: data-centers. Explore the energy and materials demands of data-centers in greater depth than is done in the text. The following links are good starting points. https://blog.izm.fraunhofer.de/the-hunger-of-data-centers-for-critical-raw-materials-and-how-we-counter-it/ https://datacentersustainability.org/data-centers-and-critical-raw-materials/

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