introduction to engineering materials engr2000 chapter...

Introduction to Engineering MaterialsENGR2000

Chapter 6: Mechanical Properties of Metals

Dr. Coates

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6.2 Concepts of Stress and Strain

• Principal ways in which a load may be applied– Tension– Compression– Shear– Torsional




Fracture & Failure

• Fracture occurs when a structural componentseparates into two or more pieces.– fracture represents failure of the component

• Failure is the inability of a component to performits desired function.– failure may occur prior to fracture– e.g. a car axle that bends when you drive over a

pothole




Tension Tests

• The specimen is deformed to fracture with agradually increasing tensile load…




Engineering Stress & Engineering Strain

0

0

0

0

:strain gEngineerin

:stress gEngineerin

lll

lΔlε

AF




Compression Tests

• Similar to the tensile tests• Force is compressive• Specimen contracts along direction of the stress




Engineering Stress & Engineering Strain

0

:strain gEngineerin

:stress gEngineerin

0

0

0

0

lll

lΔlε

AF




Shear and Torsional Tests

smallfor tan

:strainShear

:stressShear

0

hw

AF




Shear and Torsional Tests

f

Tf

: torqueapplied toduestrain Shear

: torqueapplied todue stressShear




Deformation

• All materials undergo changes in dimensions inresponse to mechanical forces. This phenomenonis called deformation.– If the material reverts back to its original size and

shape upon removal of the load, the deformation issaid to be elastic deformation.

– If application and removal of the load results in apermanent change in size or shape, the deformation issaid to be plastic deformation.




Elastic Deformation6.3 Stress-Strain Behavior

εE

E

:elasticity of Modulus

:law sHooke'

G

:modulusShear




Modulus of Elasticity

• Young’s modulus• Elastic modulus• Stiffness• Material’s resistance to elastic deformation

– greater the elastic modulus, the stiffer the material orsmaller the elastic strain that results from theapplication of a given stress.




Elastic modulus of metals, ceramics & polymers

• Ceramics– high elastic modulus– diamond, graphite (covalent bonds)

• Metals– high elastic modulus

• Polymers– low elastic modulus– weak secondary bonds between chains

Increasing E




Tangent or Secant Modulus

• Materials with non-linear elastic behavior– Gray cast iron,– concrete,– many polymers




Elastic Modulus on an Atomic Scale

• Elastic strain– small changes in inter-atomic spacing– stretching of inter-atomic bonds

• Elastic modulus– resistance to separation of adjacent atoms– inter-atomic bonding forces




Review2.5 Bonding Forces and Energies

length. bond theasknown is apart. are atoms two theof centers the-

0:mequilibriuAt

:forcenet The

0

0

rr

FFF

FFF

RAN

RAN




Elastic Modulus on an Atomic Scale

??0

WhydrdFtorelatedbeEmightHow

r




Effect of Temperature on the Elastic Modulus




Example 6.1

• A piece of Cu originally 305 mm long is pulled intension with a stress of 276 Mpa. If thedeformation is entirely elastic, what will be theresultant elongation?




Example 6.1

?110

:6.1 Table Using276305

:Given

0

lGpaE

MPamml

Cu

mmElσΔl

lΔlEEεσ

77.0

:law sHooke' Using

0

0




6.4 Anelasticity- time dependent elastic behavior

• Time dependent elastic strain component.• Elastic deformation continues after the stress

application.• Upon load release, finite time is required for

complete recovery.




Anelasticity of metals, ceramics & polymers

• Polymers– Visco-elastic behavior

• Metals

Increasinganelasticity




6.5 Elastic Properties of Materials




Poisson’s Ratio

35.025.0:metalsmost For

change net volume no50.0

25.041

0

:constant) material (a ratio sPoisson'

max

ltheoretica

z

y

z

x

09-13-10




Elastic Properties of Materials

12

:)directions allin properties (same materials isotropicFor ,,

:)properties (material constants Elastic

EG

GE




Example 6.2

• A tensile stress is to be applied along the long axisof a cylindrical brass rod that has a diameter of 10mm. Determine the magnitude of the loadrequired to produce a 2.5 x 10-3 mm change indiameter if the deformation is entirely elastic.




?34.0

97:6.1 Table Using

105.2

10:Given

30

F

GpaE

mmdmmd

brass

brass




NdF

MPaE

dd

z

xz

x

56004

:force theCompute3.71

:law sHooke' Using

1035.7

:ratio sPoisson' theUsing

105.2

:strain theCompute

20

4

4

0




Plastic Deformation

• Elastic deformation– Up to strains of 0.005

• Plastic deformation– Hooke’s law is no longer valid– Non-recoverable or permanent deformation– Phenomenon of yielding




6.6 Tensile Properties

fractureat strain fracture)at (stressstrength fracture

strain uniform stress gengineerin maximumstrength tensileultimate

strainpoint yieldlimit elastic the toingcorrespond stressstrength yield

f

f

u

uts

yp

ys

σ

σ




Yielding and Yield Strength

• Gradual elastic-plastic transition• Proportional limit (P)

– Initial departure from linearity– Point of yielding

• Yield strength– Determined using a 0.002 strain

offset method




Yielding and Yield Strength

• Steels & other materials• Yield point phenomenon• Yield strength

– Average stress associatedwith the lower yield point




Elastic deformation, plastic deformation, necking & fracture

necking




Example 6.3

• From the tensile stress-strain behavior for thebrass specimen shown, determine the following– The modulus of elasticity– The yield strength at a strain offset of 0.002– The maximum load that can be sustained by a

cylindrical specimen having an original diameter of12.8 mm

– The change in length of a specimen originally 250 mmlong that is subjected to a tensile stress of 345 Mpa.




MPa

GPa

E

ys 250:strength Yield

8.9300016.0

0150

:modulus Elastic




NdF

MPa

Fmmd

uts

uts

900,574

450:strength tensile theUsing

?8.12

:Given

20

max

max

0




mmll

lMPamml

15:elongation theCompute

06.0:A)(point strain theDetermine

?345250

:Given

0

0

09-12-11




Ductility

• Measure of the degree of plastic deformation thathas been sustained at fracture.

• Brittle materials– Little or no plastic deformation

prior to fracture• Ductile materials

– Significant plastic deformationprior to fracture




Ductility

region. necked theof area sectional-cross final theis and area sectional-cross original theis where

100%

:areain reduction Percent fracture.at length theis and

length original theis where

100100%

:elongationPercent

0

0

0

0

0

0

f

f

f

ff

AA

AAA

RA

ll

lll

EL




Ductility

• Why do we need to know the ductility ofmaterials?– The degree to which a structure will deform plastically

prior to fracture (design & safety measures)– The degree of allowable deformation during fabrication




Mechanical properties of metals

• Depend on– prior deformation,– presence of impurities &– heat treatments




Engineering stress-strain behavior for Fe at different temperatures

brittle

ductile




Resilience

• Capacity of a material to absorb energy when it isdeformed elastically and then, upon unloading, tohave this energy recovered.

• Modulus of resilience– Strain energy per unit volume required to stress a

material from an unloaded state up to the point ofyielding.




Resilience

EE

U

dU

yyy

yyr

r

y

22121

:region elasticlinear a Assuming

:resilience of Modulus

2

0




Resilience

• Materials used in spring applications– High resilience– High yield strength– Low elastic modulus

09-17-10




Toughness

• Measure of the ability of a material to absorbenergy up to fracture.

• Area under the stress-strain curve up to the pointof fracture.




Toughness

f

dU

0

:Toughness




Toughness

• Ductile materials• Brittle materials

Increasingtoughness




Example

• Which material has the highest elastic modulus?• Which material has the highest ductility?• Which material has the highest toughness?• Which material does not exhibit

any significant plastic deformationprior to fracture?




6.7 True Stress and Strain

• Decline in stress necessary to continuedeformation past the maximum point M.

• Is the material getting weaker?





• Short-coming in the engineering stress-straincurve– Decreasing cross-sectional area not considered





0

lln

:strain True area. ousinstantane theis where

:stress True

0 ll

ldlε

AAF

lt

i

it





11ln

:necking Prior to

:change volumeno Assuming

00

t

t

ii lAlA





• Necking introduces a complex stress state…




Strain-hardening exponent

t.coefficienstrength theis andexponent hardening-strain theis where

:necking ofpoint thetondeformatio plastic ofonset theFrom

Kn

K ntt




Example 6.4

• A cylindrical specimen of steel having an originaldiameter of 12.8 mm is tensile tested to fractureand found to have an engineering fracture stress of460 Mpa. If its cross-sectional diameter atfracture is 10.7 mm, determine– The ductility in terms of percent reduction in area– The true stress at fracture




??%

7.10

4608.12

:Given

0

tf

f

f

RAmmdMPammd

%30100

100%

:Ductility

20

220

0

0

ddd

AAA

RA

f

f




MPaAF

NAF

AFMPa

ftf

f

f

660

:fractureat stress True

200,59:load applied The

460

:fractureat stress gEngineerin

0

0




Example 6.5

• Compute the strain-hardening exponent n for analloy in which a true stress of 415 Mpa produces atrue strain of 0.10. Assume a value of 1035 Mpafor K.




Example 6.5

?1035

10.0415

:Given

nMPaK

MPa

t

t

40.0

:exponent hardening-Strain

n

K ntt




6.10 Hardness

• Measure of a materials resistance to localizedplastic deformation (small dent or scratch)




Rockwell Hardness Tests

• A hardness number is determined– Difference in depth of penetration

from an initial minor loadfollowed by a major load

• Combinations of indenters &loads are used




Hardness Testing Techniques




Minor load=10 kg- used for thick specimens




Minor load = 3 kg- thin specimens




Correlations between hardness & tensile strength

• Both are measures ofresistance to plasticdeformation




Property Variability & Design/Safety Factors

• Reading assignment– Section 6.11 on variability of material properties– Example 6.6




6.12 Design/Safety Factors

dys

c

cd

N

N

:criteriaselection Materialfactor.design theis 1'

load)max estimated theof basis (on the stress calculated theis '

:stressDesign




6.12 Design/Safety Factors

0.42.1:safety ofFactor

safety. offactor theis 1and material theofstrength yield theis

:stress or working stress Safe

N

N

Nys

ysw




Design Example 6.1

• A tensile-testing apparatus is to be constructedthat must withstand a maximum load of 220 kN.The design calls for two cylindrical support posts,each of which is to support half of the max load.Furthermore, plain carbon steel ground andpolished shafting rounds are to be used; theminimum yield and tensile strengths of this alloyare 310 Mpa and 565 Mpa respectively. Specify asuitable diameter for these support posts.




5:safety offactor a Assume

?565

310

11022021:Given

max

N

dMPaMPa

kNkNF

uts

ys

mmd

dF

MPaN

w

ysw

5.474/

:requireddiameter Minimum

62

:stress Working

0

20

max




Design problem 6.D1

• A large tower is to be supported by a series ofsteel wires; it is estimated that the load on eachwire will be 13,300 N. Determine the minimumrequired wire diameter, assuming a factor of safetyof 2 and a yield strength of 860 Mpa for the steel.




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