13.Materials Selection F08

7/26/2019 13.Materials Selection F08

1/15

MSE 280: Introduction to Engineering Materials

Reading: Callister Ch. 21Materials Selection and Design

MSE280 2007, 2008 Moonsub Shim

Materials

Selection

and

Design

For selection, one must establish a link between materials andfunction, with shape and processing also playing possiblyimportant roles (ignored for now)

Materials

Attributes: physical,

function

AREAS OF DESIGN CONCERN

Function- support a load, contain amechanical, thermal,

electrical, economic,

environmental.shape

process

pressure, transmit heat, etc.What does the component do?

Objective- make things cheaply, light weight,increase safety, etc., or combinations of these.What needs to be maximized or minimized?

Constraints- make things within given budget,


, ,etc., or combinations of these.What are non-negotiable conditions to be

met?

What are negotiable but desired

conditions? Following Materials Selection in MechanicalDesign, M. Ashby Modified from D. Johnson


2/15

Design & Selection: Materials Indices

Material index (performance index) is a combination of materialsproperties that characterizes the Performance of a material in a givenapplication.

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

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

For OPTIMUM design, we need to MAXIMIZE or MINIMIZE thePerformance.


Consider only the simplest cases where these factors form a separable

equation. Performance = f1(F) f2(G) f3(M)

From D. Johnson

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

Cm =cost/mass


Spring, minimum weight for given energy storage YS2/E

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

Electromagnet, maximum field, temperature rise Cp

=thermal cond

=elec. cond

Modified from D. Johnson

Why different dependences? linear vs. square root etc.


3/15

Example 1: Material 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 thanf. (f=YS, UTS)

L is usually fixed by design, can varyArea A.

A = x-area

FL

e strong, nee s to e g twe g t, or ow mass.

Ff

-Strength relation:Tie-Rod has to be able to withstand applied tensile load

fA

FApplied tensile stress =

Including safety factor:


- Mass of rod: m=LA

- Eliminate the "free" design parameter,A:

A = m/L SLm

F f

/

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

minimize for small mm (FS)(L)

f

- Rearrange:

** Com are to Performance = f1 F f2 G f3 M

Functional Needs (how muchapplied load and safety factor)

Geometricalparameter

Material properties(1/Performance index)

Fixed by service requirements!


Or Maximize the

Performance Index:

fP=

Light, strong, tie-rod


4/15

Example 1B: square rod (its all the same!)

Carry F without failing; fixed initial length L.

-Strength relation: - Mass of bar:

2Lcm=

f =F

F,

Eliminate the "free" design parameter, c:

minimize for small Mf

FLSm

)(=

S c

cc


Maximize the Materials Performance Index:

f

P=(strong, light tension members)

From D. Johnson

Ashby Plots: Strength (f ) versus Density ()

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

fP= or Pf =

-

Pf logloglog +=

- To select materials, consider aminimum performance index(e.g. P = 10 Pa/g/m3)

With P = 10 and = 0.1, = 1


Taken from:Materials Selection for

Mechanical Design Ed. 2, M. Ashby

With P = 10 and = 1, f= 10

With P = 10 and = 10, f= 100

All materials lying on this line canperform equally well.

P = 10 Pa/g/m3


5/15

Ashby Plots: Strength (f ) versus Density ()

Pf

logloglog +=

P = 10 Pa/g/m3

P = 100 Pa/g/m3

Notice that for the samestrength, P =100 material willrequire 10 times less mass thanP = 10 material.e.g. for f= 100 MPa


Taken from:Materials Selection for

Mechanical Design Ed. 2, M. Ashbyf

FLSm

)(=

then:MPa

mMgFLSmP

100

/1)(

3

100== MPamMg

FLSmP100

/10)(

3

10==

10010 10 == = PP mm

and

Bar must not lengthen by more than under force F; must have initial length L.

F,

- Stiffness relation: - Mass of bar:

Example 2: Material Index for a Light, Stiff Beam in Tension

L

cc

L

LE

c

F =

2( = E) m=Lc2

Eliminate the "free" design parameter, c:

EL

FLm

=

2


Maximize the Performance Index:

EP=

specified by application m n m ze or sma m

(stiff, light tension members)

From D. Johnson


6/15

Example 3: Material Index for a Light , Stiff Beam in Deflection

Bendingiscommonmodeofloading,e.g.,

golfclubs,wingspars,floorjoists.

Bar with initial length L must not deflect by more than under force F.

F

=deflectionL

bb

A = b2

- Stiffness relation: - Mass of bar:

ALm=

Eliminate the "free" design parameter,A:

=

2

3

4

3

44 A

LF

b

LFE

2

2

=

L

mAor

2

25

4m

LFE


2/1E

P=

If only beam height can change (not A), then P = (E1/3/) (Car door)If only beam width can change (not A), then P= (E/)

Maximize

Light, Stiff Beam

application

minimize for small m

2/1

5

4 E

FLm

Light, Stiff Plate E/(width change only)

Performance of Square Beam (deflection) vs. Fixed Height or Width

Light, Stiff Beam E1/2/(cross sectional area change)

Light, Stiff Panel E1/3/(height change only)

MSE280 2007, 2008 Moonsub ShimFrom D. Johnson


7/15

Example 4: Energy efficient car design

One way to improve fuel efficiency is to reduce mass ofthe car.

yp ca we g s r u on n a car Steel 71% (body panels) Cast iron 15% (engine, gear box, axle) Rubber (tires, hoses) Rest: glass, zinc, copper, aluminum, polymers.

Largest reduction may come from body panels


ompare erent can ate mater a s or o ypanel that can reduce weight withoutcompromising yield strength.


L

b d L and b must remain constant (only dcan change).

- Flexural strength relation: - Mass of panel:

Lbdm =2

2

3

bd

FLf=

Lbd

=

- Eliminate free design parameter:2

3

=

LbFL


m m

=

2/1

3

2

3

f

bFLm

Minimize

or maximizeperformance index:

=

2/1fP


8/15


=

2/1f

P

(Mg/m3) y(MPa) y1/2/Mild steel 7.8 220 1.9

High str. steel 7.8 500 2.9

Al alloys

Steel Al alloy 2.7 193 5.1

GFRP 1.8 75 4.8

- Glass Fiber Reinforced Polymer: Cancreep at T above 60oC.

- Al alloy: best performance but limited


P = 2

strength and higher cost of production.

- High strength steel: least amount ofchanges w.r.t. manufacturing changesfrom mild steel but also least inperformance improvement.

Example 5: Torsionally stressed shaft

Establish a criterion for selection of light and strong solid cylindrical shaftthat is subjected to torsional stress.

1. Required strength


2. Mass3. Other considerations (e.g. materials cost)

Goal: choose material to maximize strength with minimummass and cost


9/15


- Strength requirement:Shear stress at radius r

- Mass of the shaft:

Mt= applied

torque3

2

r

Mt

=

3

2

r

Mtf

=

Lr

m

V

m2

==


- Eliminate free design parameter:

L

mr=

3

2

=

L

m

Mtf

= 3/13/2


32

f

t

Functional needs (how muchapplied load should alsoinclude safety factor)

Geometricalparameter

Material properties(1/Performance index)

** Compare to:


requirements!

3/2

f

Choose material to minimize or maximize

3/2f

Performance = f1(F) f2(G) f3(M)


10/15

Materials selection chart(Ashby plot)

=

3/2fP


log2

3log

2

3log += P

f

Strength vs Density

Lets say we want to

constrain our search to

materials with P > 10.

300MPa

P = 10

t ona constra nts may

be added, e.g.

-Require minimum

strength:f> 300 MPa.

-Rule out brittle

materials no ceramics.


Search area is now

limited to the shaded

area (minus ceramics) in

the Ashby plot.


11/15

Other factors:--require f> 300MPa.

Maximize the Performance Index: P =

f2 /3

Details: Strong, Light Torsion

Members

>10

--Rule out ceramics and glasses: KIc too small.

Numerical Data:materialCFRE (vf=0.65)GFRE (vf=0.65)

Al alloy (2024-T6)Ti alloy (Ti-6Al-4V)

(Mg/m3)1.52.02.84.4

P (MPa)2/3m3/Mg)73521615

f(MPa)11401060300525


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

quench & temper)

.

Data from Table 6.6, Callister 6e.

From D. Johnson

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)

AlPETCu

237 (17)(b)

103 (13)(c)

97 (20)(b)

- Cost of energy used inprocessing materials ($/GJ)(g)

elect resistancepropanenatural gas

25119


glasspaper

20(d)13(e)

9(f)

a 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.

From D. Johnson


12/15

Relative Cost (in $) of MaterialsGraphite/

Ceramics/

Semicond

Metals/

Alloys

Composites/

fibersPolymers

Au

100000

10000

20000

50000

PtDiamond $=

$/kg

($ /kg)ref material

ativeCost

($)

Si wafer

Nylon 6,6

5000

2000

1000

500

200

100

50

20

10Al alloys

Cu alloys

Mg alloys

Ti alloys

Ag alloys

Tungsten

Al oxide

Si carbide

Si nitride

PC

Aramid fibersCarbon fibers

AFRE prepreg

CFRE prepreg

GFRE prepreg

Reference material:-Rolled A36 carbon steel.

Relative cost fluctuates lessthan actual cost over time.


Rel

pl. carbon

PETEpoxy

0.05

0.1

5

2

1

0.5

Steel

high alloy

Concrete

Glass-sodaLDPE,HDPEPPPS

PVCE-glass fibers

Wood

Based on data in AppendixC, Callister, 6e.

From D. Johnson

Details: Strong, Low-Cost Torsion

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

Numerical Data:material P MPa 2/3m3/M $/P x100

CFRE (vf=0.65)GFRE (vf=0.65)

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

quench & temper)

8040151105

7352161511

1127693

74846

Data from Table 6.7, Callister 6e.


owes cos : s ee o quenc emper

Need to consider machining, joining costs also.

From D. Johnson


13/15

Example 6: Safe Pressure Vessel

Spherical gas/fluid tankunder pressure p

Recall Design Example from Failure Section of the course

Two possible designs for safety:A) Plastic distort ion before leaking (i.e. the mechanical deformation

before leak occurs).- Calculate relative maximum critical crack length where plastic

Circumferential wall stress:

t

pr

2=


deformation occurs before catastrophic crack propagation for 1040Steel, Ti alloy and Stainless steel.

B) Leak-before-break (e.g. to prevent pressure build-up leading toexplosion). Achieved when ac = t (i.e. complete opening before crackpropagation).- Calculate the relative maximum pressure for same 3 materials as A).


- Yield strength - Fracture toughness

A) Plast ic distor tion before leaking

=cy

Critical stress for crack propagation

ccIc

- Sub iny for c: cayYIc

K

2

Larger the tolerable cracksize the better.


-solve for ac:2

1

y

Icc

K

Ya

Therefore, maximize thisquantity.

y

IcKP

=1


14/15


Pressure on the wall:- Fracture toughness

B) Leak-before-break

tpr2

=cIc

aYK = or 2prt=

Since leak-before-break occurs when ac = t:

tYKIc =sub in for t

2

prYKIc=

-Vessel should not

yield from the stress

2

2

=Y

K

pr

Ic

2

2

= K

Ic


-solve for p:higher the tolerablepressure the better.Therefore, maximizethis quantity.

y

IcKP

2

2=

on the wall:

y

IcK

rYp

2

2

2

Ypr

y

Yield-before-break

P1 = KIc/y

Leak-before-breakSteels

Cu-alloys


P2 = (KIc)2/y

Thin wall, strongP3 = yse.g. Require a minimum

strength of 100 MPa

Al-alloys

P1=0.5 m1/2

P2=10 m


P3=100 MPa Large pressure vesselsare always made of steel.


15/15

Summary

Performance index and Ashby Plot.Combination of properties to choose optimum

materials to satisfy two or more needs.

Specific examples: Strong and light tie-rod

Stiff and light beam (tension)

Stiff and light (deflection)


Strong (torsion), light and cheap shafts

Pressure vessel

13.Materials Selection F08

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