Ch21 Selection

8/6/2019 Ch21 Selection

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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 20101

Chapter 21: Materials Selection and Design

For selection, one must establish a link between materials andfunction, with shape and process playing also a possibly

important role (now ignored.) Materials

Attributes: physical,mechanical, thermal,electrical, economic,environmental.

function

shape

process

AREAS OF DESIGN CONCERN

Function- support a load, contain apressure, transmit heat, etc.

What does component do?

Objective- make thing cheaply, light weight,increase safety, etc., or combinations of these.

What is to be maximized or minimized?

Constraints- make thing cheaply, light weight,

increase safety, etc., or combinations of these.What is non-negotiable conditions to be met?

What is negotiable but desired conditions?

See also Materials Selection in Mechanical Design, M. Ashby


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Design & Selection: Materials Indices

Structural elements perform physical functions (carry load or heat, store energy,..),and so they must satisfy certain functional requirements specified by the design,such as specified tensile load, max. heat flux, spring restoring force, etc.

Material index is a combination of materials properties that characterizes thePerformance of a material in a given application.

Performance of a structural element may be specified by thefunctional requirements, the geometry, and the materials properties.

PERFORMANCE:

P[ (Functional needs, F); (Geometric, G); (Material Property, M)]

ForOPTIMUM design, we need to MAXIMIZE (orMINIMIZE) the functional P.

Often easier to MAXIMIZE when plotting!


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Examples of Materials Indices

Function, Objective, and Constraint Index

Tie, minimum weight, stiffness E/

Beam, minimum weight, stiffness E1/2 /

Beam, minimum weight, strength 2/3 /

Beam, minimum cost, stiffness E1/2

/Cm

Beam, minimum cost, strength 2/3 /Cm

Column, minimum cost, buckling load E1/2 /Cm

Spring, min. weight for given energy storage YS 2/E

Thermal insulation, minimum cost, heat flux 1/(Cm)

Electromagnet, maximum field, temperature rise Cp

= thermal cond

Cm =cost/mass

= elec. cond


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Design & Selection: Materials Indices

PERFORMANCE: (assuming separable form) P = f1(F) f2(G) f3(M)

When separable, the optimum subset of materials can be identified

without solving the complete design problem, knowing details of F and G.

There is then enormous simplification and performance can be optimized

by focusing on f3(M), the materials index

S= safety factor should always be included!

Consider only the simplest cases where these factors form a separable equation.


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Price and Availability of Materials

Current Prices on the web(a) : TRENDS-Short term: fluctuations due to supply/demand.

-Long term: prices increase as deposits are depleted.

Materials require energy to process them:- Energy to produce

materials (GJ/ton)AlPETCusteelglasspaper

237 (17)(b)

103 (13)(c)

97 (20)(b)

20(d)

13(e)

9(f)

- Cost of energy used in

processing materials ($/GJ)(g)elect resistancepropanenatural gasoil

2511

98

Yr 2004a http://www.statcan.ca/english/pgdb/economy/primary/prim44.htma http://www.metalprices.comb http://www.automotive.copper.org/recyclability.htmc http://members.aol.com/profchm/escalant.htmld http://www.steel.org.facts/power/energy.htme http://eren.doe.gov/EE/industry_glass.htmlf http://www.aifq.qc.ca/english/industry/energy.html#1g http://www.wren.doe.gov/consumerinfo/rebriefs/cb5.html

Recycling indicated in green.


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Relative Cost (in $) of MaterialsGraphiteCeramicsSemicond

MetalsAlloyComposites

fibe

rsPolymer pl. carbonAuSi wafePETEpoxyNylon 6,0.00.151000

00100002000050000500020001000500200100502010210.Ste

elhigh alloyAl alloysCu alloysMg alloysTi alloysAg alloysPtTungstenAl oxiConcret

DiamondGlass-sodSi carbiSi nitrPCLDPE,HDP

EPPPSPVCAramid fibeCarbon fibeE-glass fibAFRE prepregCFRE prepregG FRE prepregWood

Reference material:-Rolled A36 carbon

steel.

Relative cost fluctuatesless than actual cost overtime.

From Appendix C, Callister,6e.


7/36MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 20107

Materials Selection Examples in Mechanical Designwith Separable Performance Factor

Example 1: Material Index for a Light, Strong, Tie-RodExample 2: Torsionally Stress Shaft (Chpt. 20)Example 3: Safe Pressure Vessel (Chpt. 9)Example 4: Material Index for a Light, Stiff Beam in TensionExample 5: Material Index for a Light, Stiff Beam in Deflection

Example 6: Material Index for a Cheap, Stiff Support ColumnExample 7: Selecting a Slender but strong Table LegExample 8: Elastic Recovery of SpringsExample 9: Optimal Magnet Coil Material (Chpt. 20)

PERFORMANCE: functional needs , geometry, and materials index

P = f1(F) f2(G) f3(M) ---> optimize the material index f3(M).

(some from M.F. Ashby)



Example 1: Matls. Index for a Light, Strong, Tie-Rod

A Tie-rod is common mechanical component.

Functional needs: F, L, f

Tie-rod must carry tensile force, F. NO failure. Stress must be less than f. (f=YS, UTS)

L is usually fixed by design, can varyArea A. While strong, need to be lightweight, or low mass.

A = x-areaF

minimize for small m

F

As f

S

-Strength relation: - Mass of rod:

m = r LA

Eliminate the "free" design parameter, A:

m (FS)(L)r

s f

OrMaximize Materials Index: M=s f

rFor light, strong, tie-rod



Example 1: Square Rod (its all the same!)

Carry F without failing; fixed initial length L.

-Strength relation: - Mass of bar:

M=2

Eliminate the "free" design parameter, c:

specified by applicationminimize for small M

Maximize the Materials Performance Index:

M =

(strong, light tension members)

M= (FLS)r

s f

f

S=F

c2

F,



Working with Ashby Plots: Strength (f) versus Density ()

Taken from: Materials Selection forMechanical Design Ed. 2, M. Ashby

Strength can be YS for metals and polymer, compressive strength for ceramics,tear strength for elastomers, and TS for composites, for example.

Strength vs Density Plotif materials optimization isto maximize, e.g,

M = f/

or similar function.


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Where are lines of slope 1 with specified M?

With Materials Index M =f/.

Give some values ofM= 10 Pa/g/m3.

Connect those values by a line. Whatis slope of this log(f) log() line?

Note: slope is log()/log().

Give some values ofM=100 Pa/g/m3.

Why does the M = 100 line have 10times less mass than M =10 line?

From:Materials Selection for Mechanical DesignEd. 2, M. Ashby

Because mass ~ 1/M ~ .

M=100

M=10


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14

All materials along a linehave same performance.

But, M = 30has 1/3 themass ofM = 10, becausethe mass ~ 1/M ~ .

Stren

gth(M

Pa)

Density (Mg/m3)

What are the M values ofthe four lines mark?

- Hint: use values off and that are labeled already.

What are the slopes ofthose four lines?=

- Hint: count decades of rise

and runs, the ratio is slope oflog f /log = (3/2).

=

2 /3

Ashby Plot for Torsionally Stressed Shaft



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Calculating Performance Values and Slopes

Lowest Line: Log-log plot is harder to read, obtain M value with values at tick marks,

e.g. = 0.1 Mg/m3andf = 0.167 MPa or = 0.3 Mg/m3andf = 0.867

MPa.

Performance Value: M = ( f)2/3/ = (0.867 MPa)2/3/0.3 Mg/m3M = 3 (MPa)2/3/Mg/m3.

Slope of3/2: move right2 decades on andup 3 decades on f.

You can verify using (0.867 MPa, 0.3 Mg/m3) and (867 MPa, 30 Mg/m3).

slope =(867) (0.867)

(30) (0.3)=

32

Next Line: P =10 (MPa)2/3 /Mg/m3

=0.1 Mg/m3 and f=1 MPa, M = (1MPa)2/3 /0.1 Mg/m3 = 10 (MPa)2/3/Mg/m3.Again slope of 3/2 is 2 decades on andup 3 decades on f.



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Other factors:-- if by design needs f > 300MPa.

Search area is further limited

(shaded area in plot)

--Rule out ceramics and

glasses: KIc too small.

Details: Strong, Light Torsion Members

Lightest: Carbon fiber reinf. Epoxy (CFRE) member.

Get min. mass by maximizing

materials performance index M:

M =

2 /3

10

()2 /3

/3

Search area



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Considering mass

Maximize:

Details: Strong, Light Torsion Members

Numerical Data:

material

CFRE (vf=0.65)

GFRE (vf=0.65)

Al alloy (2024-T6)Ti alloy (Ti-6Al-4V)4340 steel (oil

quench & temper)

r (Mg/m3)

1.5

2.02.84.47.8

P(MPa)2/3 m3/Mg)

73

52161511

tf(MPa)

1140

1060300525780

M=

2 /3

CRFP are best!



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Overview:Strong & Light Tension/Torsion Members

(adapted from M.F. Ashby,Materials Selection in MechanicalDesign, Butterworth-HeinemannLtd., 1992.)

Increasing Mfor strong

tensionmembers

Increasing Mfor strongtorsion members

0.1 1 10 30

1

10

102

103

104

Density,(Mg/m3)

Strength, f (MPa)

slope

=1

0.1

Metalalloys

Steels

Ceramics

PMCs

Polymers

|| grain

grainwood

Cermets

slope=3/2


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Considering

(Cost/mass)*massMaximize:

M = 2/3/Cm

22x103

13x103

11x103

9x103

1x103

M=

2 /3

Minimize Cost: Cost Index ~ m$~ $/M (since m ~ 1/M)

materialCFRE (vf=0.65)GFRE (vf=0.65)Al alloy (2024-T6)Ti alloy (Ti-6Al-4V)4340 steel (oil quench & temper)

$

80

40151105

M (MPa)2/3 m3/Mg)7352161511

($/M)x100112769374846

Lowest cost: 4340 steel (oil quench & temper)

Maximize:

Need to consider machining, joining costs also.


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Example 3: Safe Pressure VesselUses info from leak-before-fail example.

Design requirements

Function: contain pressure, pObjective: maximum safetyConstraints: (a) must yield before break

(b) must leak before break(c) t small: reduces mass and cost

Choose t so that at working pressure, p, the stress is less than ys .

Check (by x-ray, ultrasonics, etc.) that no cracks > 2ac are present:

Stress to active crack propagation is

Safety (also use safety factor, S) achieved for stress less than this, but greatersafety obtained requiring no cracks propagate even if = ys (stably deform).

This condition ( = ys /S) yields

=

ac2

2

2

So, M1 = KIc/sys

p R

tt

2a

= 2


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Safe Pressure Vessel (cont)

Tolerable crack size is maximized by choosing largest

Large pressure vessels cannot always be tested for cracks and stresstesting is impractical. Cracks grow over time by corrosion or cyclic loading(cannot be determined by one measurement at start of service).

Leak-before-fail criterion (leaks can be detected over lifetime)

Wall thickness designed to contain pressure w/o yielding, so

Two equations solved for maximum pressure gives

Wall thickness must be thin for lightness and economy. Thinnest wall has largest yield stress, so

M1 = KIc/ys

=

t 2

M2 = (KIc)2/ys

M3 = ys

Note largest M1 and M2 for smallest ys . FOOLISH for pressure vessel.

S f P V l ( t)


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Safe Pressure Vessel (cont)

Yield-before-breakM1 = KIc/ys

Leak-before-breakM2 = (KIc)

2/ys

Thin wall, strongM3 = ys

Large pressure vessels arealways made of steel.

Models are made of Cu,for resistance to corrosion.Check that M2 favors steel.

M3=100 MPaeliminates Al.

SteelsCu-alloys

Al-alloys

SEAR

CH

M3=100 MPa

M1=0.6 m1/2


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Bar must not lengthen by more than d

under force F; must have initial length L.

Maximize the Materials Index:

F,

- Stiffness relation: - Mass of bar:

F

c2=

Hookes Law = E

m =2

Eliminate the "free" design parameter, c:

m =FL

2

d

r

E

M=

specified by applicationminimize for small m

(stiff, light tension members)

Example 4: Material Index for a Light, Stiff Beam in Tension

Example 5: Material Index for a Light Stiff Beam in Deflection


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Example 5: Material Index for a Light, Stiff Beam in Deflection

Bending is common mode of loading, e.g.,

golf clubs, wing spars, floor joists, ships.

m

121

1/2

(3

)

1/2

M=

1/ 2

If only beam height can change (not A), then M= (E1/3 /) (Car door) I ~ b3w

If only beam width can change (not A), then M= (E/)

Bar with initial length L must not deflect by morethan d under force F.

- Stiffness relation: - Mass of bar:

F

1

3 =

1

34

12

=

1

32

12

m = 2

=


Maximize

Light, Stiff Beam

F

=deflection

L bb

specified byapplication

minimize for small m


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Light, Stiff Plate E/

Light, Stiff Beam E1/2 /

Light, Stiff Panel E1/3 /

Performance of Square Beam vs. Fixed Height or Width


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Example 6: Material Index for a Cheap, Stiff Support Column

L

Buckledeflection

Radius, rA slender column of fixed intial length L uses

less material than a fat one; but must not be so

slender than it buckles under load F.

F

= 2

2

Load less than Euler Load.N given by end constraint on column.

C4

np

1/2 F

L2

1/2L3

Cmr

E1/2

E1/2

Cm

C=

=

- No buckling relation: - Cost objective:

Cm is the cost/kg of (usuallyprocessed) material.


Maximize

Cheap, Stiff Beam

specified byapplication minimize for small m


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Performance of Stiff but Cost Effective Beam

With cost considered,now polymers andmetals area useful!


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Example 7: Select a Slender but strong Table Leg

For attractiveness, legs must be solid (to be thin) and light as possible (to make

table easy to move). Legs must support table top and load without buckling.

What material would you recommend to Luigi?

m 4

1/22

1/2

r=4

3

1/4

1/2

1

1/4

M2= E

M

1

= 1/2

2 indicesto meet

m =2

- Critical Elastic Load: - Mass of leg:

F 2

2 = 3

4

42

Eliminate the "free" design parameter, R:

Maximize

For slenderness, get R forCritical Load:

(Example from Ashby.)

Luigi Tavolina (furniture designer) conceives of a lightweight table ofsimplicity, with a flat toughened glass top on slender, unbraced, cylindrical legs.

1/2


29/36


Material indices: M1

= 1/2

and M2= E

Wood is good choice.

So is composite CFRP (higher E). Ceramic meets stated design

goals, but are brittle

E1/2 / guideline (slope of 2)

M2 = E =100 GPa

M1= 6 (GPa)1/2 /(Mg/m3)

Example 8: Elastic Recovery of Springs


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Example 8: Elastic Recovery of SpringsRecall from Hookes Law and Resilience, Uel = 2/2E.

We wish to maximize this, but the spring willl be damage if > ys . Uel = ys

2/2E

(Torsion bars and lead spring are less efficient than axial springs because some of thematerial is not fully loaded, for instance, the neutral axis it is not loaded at all!)

F

F/2F/2

Deflection,d

Can show that Uel = (ys2/E)/18

Addition constraint can be added.

Ifin-service, a spring under goes deflection of d under force F, then ys2/E has to

be high enough to avoid permanent set (a high resilience!).

So spring materials are heavily SS-strengtheningand work-hardening(e.g, cold-rolled single-phase brass or bronze), SS- + precipitation-strengthening (spring steel).

Annealingany spring material removes work-hardening, or cause precipitation tocoarsen, reducing YS and making materials useless as a spring!


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High magnetic fields permit study(2) of:

- electron energy levels,- conditions for superconductivity- conversion of insulators into conductors.

Largest Example:

- short pulse of 800,000 gauss(Earth's magnetic field: ~ 0.5 Gauss)

Technical Challenges:- Intense resistive heatingcan melt the coil.- Lorentz stress can exceed the material strength.

Goal: Select an optimal coil material.(1) Based on discussions with Greg Boebinger, Dwight Rickel, and James Sims, National HighMagnetic Field Lab (NHMFL), Los Alamos National Labs, NM (April, 2002).(2) See G. Boebinger, Al Passner, and Joze Bevk, "Building World Record Magnets", ScientificAmerican, pp. 58-66, June 1995, for more information.

Pulsed magneticcapable of 600,000gauss field during 20msperiod.

Fractured magnet coil.(Photos from NHMFL,Los Alamos National Labs,NM (Apr. 2002) by P.M.Anderson)

Example 9: Optimal Magnet Coil Material


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Applied magnetic field, H:

H = N I/L

Lorentz "hoop" stress: Resistive heating: (adiabatic)

R

= (

)

temp increaseduring current

pulse of DtMagnetic fieldout of plane.

elect. resistivity

specific heat

currentN turns toL = length of eachtur

Force

length=

Lorentz Stress & Heating


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Relative cost of coil:

$ = $M

Eliminate M from the stress & heating equations:

Applied magnetic field:

H = N I/L

--Stress requirement

specified by application

Cost Index C1:maximize for

large H2/$

specified by application

Cost Index C2:maximize for

large Ht1/2 /$

--Heating requirement

H

2

1

H 2

$

122

Magnet Coil: Cost Index

I di F A C il M t i l


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From Appendices B and C,Callister 6e:

Material1020 steel (an)1100 Al (an)7075 Al (T6)11000 Cu (an)17200 Be-Cu (st)71500 Cu-Ni (hr)Pt

Ag (an)Ni 200units

f

39590572220475380145

170462MPa

d

7.852.712.808.898.258.9421.5

10.58.89g/cm3

$

0.812.313.47.9

51.412.91.8e4

27131.4

--

cv

486904960385420380132

235456J/kg-K

e

1.600.290.520.170.573.751.06

0.150.95

-m3

P1

50332042558437

1652

f/ d

P2

2 2115531

19


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Material costs fluctuate but rise over long term as:- rich deposits are depleted,- energy costs increase.

Recycled materials reduce energy use significantly. Materials are selected based on:

- performance orcost indices.

Examples:- design of minimum mass, maximum strength of:

shafts under torsion, bars under tension, plates under bending,

- selection to optimize more than one property: leg slenderness and mass.

pressure vessel safety. material for a magnet coil.

SUMMARY

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Ch21 Selection

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