Meeg309 Chapter 5 Used Finally 2011

8/3/2019 Meeg309 Chapter 5 Used Finally 2011

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Chapter 6 - 1

ISSUES TO ADDRESS...

Stress and strain: What are they and why arethey used instead of load and deformation?

Elastic behavior: When loads are small, how muchdeformation occurs? What materials deform least?

Plastic behavior: At what point does permanentdeformation occur? What materials are most

resistant to permanent deformation? Toughness and ductility: What are they and how

do we measure them?

Chapter 5:Mechanical Properties of

Materials


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Chapter 6 -

Mechanical Properties of Metals

How do metals respond to external loads?

Stress and Strain

Tension

Compression

Shear

Torsion

Elastic deformation

Plastic Deformation

Yield Strength

Tensile Strength

Ductility

Toughness

Hardness

Chapter 6 Outline


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Chapter 6 - 3

Learning Objectives

Introduce the basic concepts associated with mechanical

properties of materials Review some of the basic testing procedures that engineers

use to evaluate many of these properties.


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Chapter 6 - 4

Often materials are subject to forces (loads) whenthey are used. It is important to know howmaterials deform (elongate, compress, twist) or

break as a function of applied load, time,temperature, and other conditions.


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Chapter 6 -5

(1) Important to understand capabilities andlimitations of materials: Failure is caused by a lack of fundamental

understanding of materials, their properties, andfailure modes.

Technological Importance:Why Study Properties and Failure of Materials ?


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Chapter 6 -

6

(2) An understanding of materials properties help us todesign better components, parts, devices, etc.Why and how do materials fail? Can we prevent failure? How do we make metals stronger?

Why do materials behave differently under dynamic loadscompared to static loads? How do we select the right material for the job?

(3) Its interesting and helps to make you a moreinformed person

Why study Properties and Failure of Materials ?


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Chapter 6 -

plastic (permanent) deformation of a

bridge

deformation led to eventual collapse

Tacoma Narrows suspension bridge,near Puget Sound, failed on at 11 amNov. 7, 1940, after only having beenopen for traffic a few months

Failure by Plastic Deformation

De Havilland Comet, first commercial jetaircraft, had five major crashes in 1952 -54 period

caused by fatigue cracks initiated at

square windows, driven by cabinpressurization and depressurization


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Chapter 6 -8

Catastrophic Failure - examples


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Chapter 6 -9

Catastrophic Failure - examples


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Chapter 6 -

10

Types of Loading

Tensile Compressive

Shear Torsion


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Chapter 6 - 11

Elastic means reversible!

Elastic Deformation

2. Small load

F

d

bondsstretch

1. Initial 3. Unload

return to

initial

F

d

Linear-

elastic

Non-Linear-elastic


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Chapter 6 - 12

Plastic means permanent!

Plastic Deformation (Metals)

F

d

linearelastic

linearelastic

dplastic

1. Initial 2. Small load 3. Unload

planesstillsheared

F

delastic + plastic

bondsstretch& planesshear

dplastic


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Chapter 6 - 13

Stress has units:N/m2 or lb

f/in2

Engineering Stress

Shear stress, t:

Area, Ao

Ft

Ft

Fs

F

F

Fs

t =Fs

Ao

Tensile stress, s:

original areabefore loading

s= FtAo

2f

2mNor

inlb=

Area, Ao

Ft

Ft


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Chapter 6 - 14

Simple tension: cable

Note: t = M/AcRhere.

Common States of Stress

os=

F

A

o

t =Fs

A

ss

M

M Ao

2R

FsAc

Torsion (a form of shear): drive shaftSki lift (photo courtesyP.M. Anderson)

Ao= cross sectional

area (when unloaded)

FF


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Chapter 6 - 15

(photo courtesy P.M. Anderson)Canyon Bridge, Los Alamos, NM

os= F

A

Simple compression:

Note: compressivestructure member(s < 0 here).(photo courtesy P.M. Anderson)

OTHER COMMON STRESS STATES (i)

Ao

Balanced Rock, ArchesNational Park


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Chapter 6 - 16

Tensile strain: Lateral strain:

Strain is alwaysdimensionless.

Engineering Strain

Shear strain:

q

90

90 - qy

x qg = x/y= tan

e =d

Lo

Adapted from Fig. 6.1(a) and (c), Callister & Rethwisch 8e.

d/2

Lowo

-deL =

L

wo

dL/2


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Chapter 6 - 17

Linear Elastic Properties

Modulus of Elasticity, E:(also known as Young's modulus)

Hooke's Law:

s = Ee s

Linear-elastic

Ee

F

Fsimpletensiontest


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Chapter 6 -

18

Nonlinear elastic behaviorIn some materials (many polymers, concrete...), elastic deformation is notlinear, but it is still reversible.

Definitions of E

s/e = tangent modulus at s2

s/e = secant modulus between originand s1


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Chapter 6 - 19

Linear Elastic Properties

Modulus of elasticity is defined as the ratio of stress to strain when

deformation is totally elastic

It is a measure of the stiffness of a material: the higher the modulus ofelasticity of a material, the stiffer the material

A plot of stress against strain for elastic deformation is linear and the

slope of this linear plot gives the modulus of elasticity For some materials such as gray cast iron, concrete and many

polymers, the initial elastic deformation is not linear.

Where the initial elastic modulus deformation is not linear, tangent orsecant modulus can be used to determine the modulus of elasticity


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Chapter 6 - 20

Anelasticity

The following assumptions are generally made with regard to elasticity of

materials:a. The deformation is time independent, i.e. an applied stress produces

an instantaneous elastic strain that remains constant as long as thestress is maintained

b. The strain is completely recovered after removal of the stress

However, elastic deformation will continue for a finite time and onremoving the applied stress, a finite time is required for a completerecovery of the strain

This time-dependent elastic behavior is referred to as an anelasticity

For metals, the anelastic components are negligible but for polymer its

magnitude is significant


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Chapter 6 -

Example:

A piece of metal originally 305 mm long is pulled in tension

with a stress of 276 MPa. If the deformation is entirely elasticwhat will be its elongation? E = 110 GPa

Solutions Lo = 305mm = 0.305 m,s = 270 x 106 Pa

L = ?E = 110 x 109 Pa

3

9

6

10x2.50910x110

10x276

E

-===

0.305mx10x2.509LLL

L

3

o

o

-===

= 0.765 mm


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Chapter 6 -

Tensile loads

Lf LO

s

ex

ey The ratio of lateral strain (in

direction perpendicular to theapplied stress) to axial strain (indirection parallel to the appliedstress) is called Poissons ratio.

For many metals and alloys,Poissons ratio lies between

0.25 and 0.35

For isotropic materials,

E = 2G(1 + )

y

x-ratio,sPoisson'e

e=

Poisson's ratio,


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Chapter 6 - 23

Poisson's ratio, Poisson's ratio, :

Units:E: [GPa] or [psi]: dimensionless

> 0.50 density increases

< 0.50 density decreases(voids form)

eL

e

-

e=- L

e

metals: ~ 0.33

ceramics: ~ 0.25

polymers: ~ 0.40

Example


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Chapter 6 -

Example

A tensile stress is to be applied along the long axis of a cylindrical brass rodthat has a diameter of 10 mm. Determine the magnitude of the load requiredto produce a 2.5 x 10-3 mm change in diameter if the deformation is entirely

elastic. Ebrass = 97 GPa and brass = 0.34

y

x

P

P

Do = 10 mm, D = 2.5 x 10-3 mm

E = 97 x 109 Pa u = 0.34

Solution

44

y

4

y

x 10x.353734.0

10x2.5

10x2.50.34

strainaxial

strainlateral

---

==-

-=-=-= ye

43

o

10x2.510

10x2.5DD -

-

-=-=-== xlateral ee

Pa10x71.32310xx7.35310x97 649 == -es E

NALoad o 56024

)01.0(x10x71.323

26 =

==

s


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Chapter 6 - 25

Mechanical Properties

Slope of stress strain plot (which is

proportional to the elastic modulus) dependson bond strength of metal

Adapted from Fig. 6.7,Callister & Rethwisch 8e.


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Chapter 6 - 26

Metals

Alloys

GraphiteCeramics

Semicond

PolymersComposites

/fibers

E(GPa)

Based on data in Table B.2,Callister & Rethwisch 8e.Composite data based onreinforced epoxy with 60 vol%of alignedcarbon (CFRE),aramid (AFRE), or

glass (GFRE)fibers.

Youngs Moduli: Comparison

109 Pa

0.2

8

0.6

1

Magnesium,

Aluminum

Platinum

Silver, Gold

Tantalum

Zinc, Ti

Steel, Ni

Molybdenum

Graphite

Si crystal

Glass-soda

Concrete

Si nitrideAl oxide

PC

Wood( grain)

AFRE( fibers) *

CFRE*

GFRE*

Glass fibers only

Carbon fibers only

Aramid fibers only

Epoxy only

0.4

0.8

2

4

6

10

2 0

4 0

6 08 010 0

200

600800

10 001200

400

Tin

Cu alloys

Tungsten

Si carbide

Diamond

PTFE

HDPE

LDPE

PP

Polyester

PSPET

CFRE( fibers) *

GFRE( fibers)*

GFRE(|| fibers)*

AFRE(|| fibers)*

CFRE(|| fibers)*


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Chapter 6 -

27

Stress-Strain Behavior(Tension)

Elastic Plastic

Stre

ss

Strain

Elastic deformation

Reversible:

( For small strains)

Stress removed

material returns tooriginal size

Plastic deformation

Irreversible:

Stress removed material does notreturn to original dimensions.


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Chapter 6 - 28

Stress at which noticeableplastic deformation hasoccurred.

when ep = 0.002

Yield Strength, sy

sy= yield strength

Note: for 2 inch sample

e = 0.002 = z/z

z= 0.004 in

Adapted from Fig. 6.10(a),Callister & Rethwisch 8e.

tensile stress,s

engineering strain, e

sy

ep= 0.002


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Chapter 6 -

March 07, 2008

29

1:

Example 1: Convert the change in length data in Table 1 to engineering stress

and strain and plot a stress-strain curve.


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Chapter 6 -30

Example 1: SOLUTION

below


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Chapter 6 -31

(c)2003 Brooks/Co le, a division of Thomson Learning, Inc. Thomson Learning

is a trademark used herein under license.

The stress-strain curve for an aluminum alloy fromTable 1


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Chapter 6 -32


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Chapter 6 - 33

Stress-Strain Testing

Typical tensile test

machine

Adapted from Fig. 6.3, Callister & Rethwisch 8e. (Fig. 6.3 is taken from H.W.Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials,

Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.)

specimenextensometer

Typical tensile

specimen

Adapted fromFig. 6.2,

Callister &Rethwisch 8e.

gauge

length


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Chapter 6 -34

A unidirectional force is

applied to a specimen in

the tensile test by means

of the moveable

crosshead. The cross-

head movement can be

performed using screws

or a hydraulic

mechanism


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Chapter 6 - 35

Experimental Tests:


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Chapter 6 -

36

Experimental Tests:Test Specimen Geometry, Facilities

& Set-Up

Tension Torsion Load Frame

Optical Strain Measurement System

Specimen with Random Patterns

D = 25.38

r= 6.35RAD

+

2

11

3

D = 25.382

1

1

3

90.00o0.50

r= 3.00 0.03

Mild Notch

Sharp Notch


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Chapter 6 -37

The Tensile Test: Use of the Stress-Strain

Diagram

Load - The force applied to a material during testing.

Strain gage or Extensometer - A device used for measuring

change in length and hence strain. Engineering stress - The applied load, or force, divided by

the original cross-sectional area of the material.

Engineering strain - The amount that a material deforms per

unit length in a tensile test.


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Chapter 6 - 38

Tensile Strength, TS

Metals: occurs when noticeable necking starts. Polymers: occurs when polymer backbonechains are

aligned and about to break.


sy

strain

Typical response of a metal

F= fracture or

ultimate

strength

Neck actsas stressconcentrator

enginee

ring

TS

stres

s

engineering strain

Maximum stress on engineering stress-strain curve.


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Chapter 6 -

The maximum stress on the stress-strain curve of a materialrepresents its tensile strength

Deformation up to the maximum stress value is uniform

throughout the gauge length of the tensile specimen

At the maximum stress, necking occurs at some point andall subsequent deformation is confined to that point untilfracture occurs

The fracture strength corresponds to the stress level, atwhich failure occurs

Yield strength is normally used for design purpose instead oftensile strength because a material will undergo a large

amount of undesirable plastic deformation at maximumstress


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Chapter 6 -40

Stress-Strain Curve


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Chapter 6 -41

Important Design Considerations

D tilit


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Chapter 6 - 42

Plastic tensile strain at failure:

Ductility

Another ductility measure: 100xA

AARA%

o

fo-

=

x 100

LLL

EL%o

of -=

LfAo AfLoAdapted from Fig. 6.13,Callister & Rethwisch 8e.

Engineering tensile strain, e

Engineeringtensilestress, s

smaller %EL

larger %EL

D tilit


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Chapter 6 -

100xEL%

o

of

L

LL -=

Ductility is a measure of plastic deformation that a materialcan sustain before fracture

It is expressed either as percent elongation to fracture or aspercent reduction in area

When the fracture strain is low and less than 5 %, thematerial is said to be brittle

Ductility


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Chapter 6 -

Toughness is a measure of materials ability to absorb

energy up to fracture It is a measure of a materials resistance to fracture when a

crack is present

For low strain rate condition, toughness is given by the areaunder the stress-strain curve measured up to fracture point

For high strain condition and when a notch is present, notchtoughness is assessed using impact test

A tough material must have high strength and ductility

Toughness


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Chapter 6 - 45

Energy to break a unit volume of material

Approximate by the area under the stress-strain curve.

Toughness

Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy


very small toughness(unreinforced polymers)

Engineering tensile strain, e

Engineeringtensile

stress, s

small toughness (ceramics)

large toughness (metals)

R ili U


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Chapter 6 -

Resilience of a material is a measure of its capacity to absorb energywhen it elastically deformed and have this energy recovered on

removing the applied load. Modulus of resilience, Ur is the strain energy per unit volume required to

stress a material from an unloaded state up to the yield point

Modulus of resilience, Ur, is the area under the stress-strain curve takento the yield point

=y

0

r dU

e

es

Assuming a linear elastic regionyesy2

1r U =

sy and ey are yield stress and yield strain respectively

Substituting for strain

2E

E

U

2

yy

21

y21

r =

== yy ses

Thus resilient materials are those having high yieldstrength and low moduli of elasticity

Resilience, Ur


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Chapter 6 - 47

Resilience, Ur

Ability of a material to store energy

Energy stored best in elastic region

If we assume a linearstress-strain curve thissimplifies to


yyr2

1U es@

e

es= y dUr0


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Chapter 6 - 48

Elastic Strain Recovery


Stress

Strain

3. Reapplyload

2. Unload

D

Elastic strainrecovery

1. Load

syo

syi


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Chapter 6 - 49

Hardness Resistance to permanently indenting the surface.

Large hardness means:-- resistance to plastic deformation or cracking in

compression.-- better wear properties.

e.g.,10 mm sphere

apply known force measure size

of indent afterremoving load

dDSmaller indentsmean largerhardness.

increasing hardness

mostplastics

brassesAl alloys

easy to machinesteels file hard

cuttingtools

nitridedsteels diamond


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Chapter 6 - 50

Hardness: Measurement

Rockwell

No major sample damage

Each scale runs to 130 but only useful in range20-100.

Minor load 10 kg Major load 60 (A), 100 (B) & 150 (C) kg

A = diamond, B = 1/16 in. ball, C = diamond

HB = Brinell Hardness TS(psia) = 500 x HB

TS(MPa) = 3.45 x HB


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Chapter 6 - 51

Hardness: MeasurementTable 6.5


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Chapter 6 - 52

Hardening

Curve fit to the stress-strain response:

sT =K eT( )n

true stress (F/A) true strain: ln(L/Lo)

hardening exponent:n= 0.15 (some steels)to n= 0.5 (some coppers)

An increase in sydue to plastic deformation.s

e

large hardening

small hardeningsy0

sy1


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Chapter 6 - 53

Variability in Material Properties

Elastic modulus is material property

Critical properties depend largely on sample flaws(defects, etc.). Large sample to sample variability.

Statistics

Mean

Standard Deviation s=

n

xi -

x

( )

2

n-1

1

2

nxx n

n

=

where nis the number of data points


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Chapter 6 - 54

Design uncertainties mean we do not push the limit.

Factor of safety, N

N

y

working

s=s

Often Nisbetween1.2 and 4

Example: Calculate a diameter, d, to ensure that yield does

not occur in the 1045 carbon steel rod below. Use afactor of safety of 5.

Design or Safety Factors

220,000N

d2/ 4( )

5N

y

working

s=s 1045 plain

carbon steel:

sy= 310 MPaTS= 565 MPa

F= 220,000N

d

Lo

d= 0.067 m = 6.7 cm


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Chapter 6 - 55

Stress and strain: These are size-independentmeasures of load and displacement, respectively.

Elastic behavior: This reversible behavior oftenshows a linear relation between stress and strain.To minimize deformation, select a material with a

large elastic modulus (Eor G).

Toughness: The energy needed to break a unitvolume of material.

Ductility: The plastic strain at failure.

Summary

Plastic behavior: This permanent deformationbehavior occurs when the tensile (or compressive)

uniaxial stress reaches sy.

S


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56

Summary

Anelasticity

Ductility Elastic deformation

Elastic recovery

Engineering strain

Engineering stress

Hardness Modulus of elasticity

Plastic deformation

Poissons ratio

Proportional limit

Shear Tensile strength

Toughness

Yielding

Yield strength

Make sure you understand language and concepts:

Meeg309 Chapter 5 Used Finally 2011

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