ch21 selection
TRANSCRIPT
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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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 20102
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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 20103
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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 20104
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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 20105
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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 20106
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.
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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)
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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
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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,
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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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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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 201019
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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 201020
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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 201024
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
=
Eliminate the "free" design parameter, A:
Maximize
Light, Stiff Beam
F
=deflection
L bb
specified byapplication
minimize for small m
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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 201025
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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 201026
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.
Eliminate the "free" design parameter, A:
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
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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 201029
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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 201030
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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MSE 280: Introduction to Engineering Materials D.D. Johnson 2004, 2006 201031
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