Chapter 6: Mechanical Properties - Konkukccdjko.konkuk.ac.kr/upload/sub0503/ch06.pdfmaterial, having the effect of ... extensometer specimen • Typical tensile specimen ... - The

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 much deformation occurs? What materials deform least?

• Plastic behavior: At what point does permanent deformation occur? What materials are most resistant to permanent deformation?

• Toughness and ductility: What are they and howdo we measure them?

Chapter 6: Mechanical Properties

Chapter 6 - 2

Stress and Strain

(a) Tensile Load – elongation and (+)strain

(b) Compressive load –contraction and (-)strain

(c) Shear stress – shear strain (γ)

(d) Torque - Torsional deformation

Chapter 6 - 3

• Simple tension: ex. cable

Common States of Stress

• Torsion(a form of shear): ex. drive shaft

Ski lift (photo courtesy P.M. Anderson)

FF

- Shear stress is a function the applied torque T

- Shear strain is related to the angle of twist, ϕ

Chapter 6 - 4

Common States of Stress

- Shear Force is when opposingforces are applied across the material, having the effect of cutting it in two. The material would require shear strength to resist this force.

- Scissors cut material by applying a local shear stress at the cutting location which exceeds the material's shear strength.

• Shear stress: Bench shears, Scissors,

Chapter 6 - 5

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

• Simple compression:

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

OTHER COMMON STRESS STATES (i)

Balanced Rock, Arches National Park

Chapter 6 - 6

• Bi-axial tension: • Hydrostatic compression:

Pressurized tank

< 0h

(photo courtesyP.M. Anderson)

(photo courtesyP.M. Anderson)

OTHER COMMON STRESS STATES (ii)

Fish under water

z > 0

> 0

Chapter 6 - 7

Stress has units:N/m2 (Pa) or lbf /in2 (psi)

Engineering Stress• Shear stress, :

Area, Ao

Ft

Ft

Fs

F

F

Fs

= Fs

Ao

• Tensile stress, :

original area before loading

=Ft

Ao2f

2m

Nor

in

lb=

Area, Ao

Ft

Ft acts tangentialto the surface

acts perpendicularto the cross-section

Chapter 6 - 8

• Tensile strain (): • Lateral strain ( ):

Strain is alwaysdimensionless.

Engineering Strain

• Shear strain ():

90º

90º - y

x = x/y = tan

Lo

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

/2

Lowo

L L

wo

L/2

L

Chapter 6 - 9

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 from Fig. 6.2,Callister & Rethwisch 8e.

gauge length

→ to measure changes in the length of an object

Chapter 6 - 10

Stress-Strain Testing• Universal Testing Machine (UTM) : INSTRON

http://www.youtube.com/watch?v=CkSsRLUeRQIhttp://www.youtube.com/watch?v=qDqc-HAxNrQ

Chapter 6 - 11

Linear Elastic Properties

• Modulus of Elasticity, E: (also known as Young's modulus)the slope of the linear portion of the stress-strain curve, it is usually specific to each material; a constant, known value

• Hooke's Law:

= E

Linear-elastic

E

F

Fsimple tension test

SI unit for the E is Gpa :

E= F/(A

1GPa = 109N/m2

cf. psi (lb/ in2)kgf/cm2

Elastic deformation is not permanent; when the load is removed, the part returns to its original shape and dimensions.

Chapter 6 -

Shear Modulus

y

x

12

= x/y = tan Shear stress() to shear strain():

= G ,

G is Shear Modulus (Units: N/m2)

90º

90º - y

x

Chapter 6 -

- Ratio of lateral to axial strain calledPoisson's ratio, .

- (Tension) shrink laterally (Compression) bulge.

Poisson's ratio,

13

- dimensionless.

- Sign: lateral strain opposite to longitudinal strain

- Theoretical value: for isotropic material: 0.25

- (Typical value) 0.24 - 0.30 (Maximum value) 0.50

L

Lateral strain

Tensile strain

x

z

y

z

Chapter 6 - 14

Chapter 6 - 15

Poisson's ratio,

• Poisson's ratio, :

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

> 0.50 density increases

< 0.50 density decreases (voids form)

L

-

L

metals: ~ 0.33ceramics: ~ 0.25polymers: ~ 0.40

Lateral strain

Tensile strain

Chapter 6 -

Elastic Modulus, Poisson’s Ratioand Shear Modulus

- For isotropic material:

E = 2G(1+) G ~ 0.4E

- Single crystals are usually elastically anisotropic

- Elastic behavior varies with crystallographic direction.

16

Chapter 6 - 17

Non-linear Elastic Behavior

- Elastic portion of the stress-strain curve is not linear. ex. Gray cast iron, concrete

& many polymer

- It is not possible to determine a modulus of elasticity

- Either tangent or secant modulus is normally used

http://www.youtube.com/watch?v=clLEhdbQ6Ug

Chapter 6 - 18

Mechanical Properties : atomic scale

• Slope of stress strain plot (which is proportional to the elastic modulus) depends on bond strength of metal:

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

→ elastic modulus is proportional to the slope of each curve at the equil. interaction separation

E∝

Chapter 6 - 19

Mechanical Properties- The modulus of elasticity diminishes with increasing temperature

Chapter 6 -20

• Have assumed elastic deformation is time independent

(applied stress produces instantaneous strain)

• Elastic deformation takes time; can continue even after loadrelease. This behavior is known as anelasticity.

(→ time dependence of elastic deformation)

• Small effect in metals; can be significant for polymers(visco-elastic behavior).

Anelasticity (visco-elastic behavior)

Chapter 6 - 21

MetalsAlloys

GraphiteCeramicsSemicond

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), orglass (GFRE)fibers.

Young’s Moduli: Comparison

109 Pa

0.2

8

0.6

1

Magnesium,Aluminum

Platinum

Silver, Gold

Tantalum

Zinc, Ti

Steel, NiMolybdenum

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

20

40

6080

100

200

600800

10001200

400

Tin

Cu alloys

Tungsten

<100>

<111>

Si carbide

Diamond

PTFE

HDPE

LDPE

PP

Polyester

PSPET

CFRE( fibers) *

GFRE( fibers)*

GFRE(|| fibers)*

AFRE(|| fibers)*

CFRE(|| fibers)*

Chapter 6 - 22

• Simple tension:

FLo

EAo

L Fw o

EAo

• Material, geometric, and loading parameters allcontribute to deflection.

• Larger elastic moduli minimize elastic deflection.

Useful Linear Elastic Relationships

F

Ao/2

L/2

Lowo

• Simple torsion:

2MLo

ro4G

M = moment = angle of twist

2ro

Lo

Chapter 6 - 23

Elastic means reversible!

Elastic Deformation2. Small load

F

bonds stretch

1. Initial 3. Unload

return to initial

- Reversible for small strains

- Stress removed → material returns tooriginal size

F

Linear- elastic

Non-Linear-elastic

Chapter 6 - 24

Plastic Deformation (Metals)1. Initial 2. Small load 3. Unload

planesstill sheared

F

elastic + plastic

bonds stretch & planes shear

plastic

F

linear elastic

linear elastic

plastic

• stress not proportional to strain

• deformation is not reversible : permanent/plastic

• deformation occurs by breaking and re-arrangement of atomic bonds (crystalline materials by motion of defects)

→ Permanent deformation for metals is accomplished by means of a process called slip, which involves the motion of dislocations.

Chapter 6 - 25

(at lower temperatures, i.e. T < Tmelt/3)Plastic (Permanent) Deformation

• Simple tension test:

Engineeringstress,

engineering strain,

Elastic+Plastic at larger stress

p

plastic strain

Elastic initially

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

permanent (plastic) after load is removed

- After removal of the stress, the large number of atoms that have relocated, do not return to original position.

- plastic deformationcorresponds to the breaking of bonds with original atom neighbors and then reforming bondswith new neighbors.

Chapter 6 - 26

(at lower temperatures, i.e. T < Tmelt/3)Plastic (Permanent) Deformation

• Most structures are designed to ensure that only elastic deformationresults when stress is applied.

• A structure that has plastically deformed, or experienced a permanent change in shape, may not be capable of functioning as intended.

• Yield strength is usually more important than tensile strength. Once yield stress has been passed, structure has deformed beyond acceptable limits.

Necked region in a fractured sample

http://www.youtube.com/watch?v=hD_NJaZIpT0

Chapter 6 - 27

• Stress at which noticeable plastic deformation hasoccurred.

when p = 0.002

Yield Strength, y

y = yield strength

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

tensile stress,

engineering strain,

y

p = 0.002

Elastic Plastic

Yield Point : strain deviates from being proportional to stress (Proportional Limit)

P - Yield point (P) is the gradual elastic to plastic transition.

- Yield strength is a measure of resistance to plastic deformation

Chapter 6 - 28

Yield Strength, y

→ σy = average stress at the lower yield point

Strain Strain

Str

ess

Str

ess

0.005

σy σy→ σy = Stress required to produce some amount of strain (e.g., ε= 0.005)

• Nonlinear elastic deformation • Some low-carbon steels with yield point phenomenon

Chapter 6 - 29

Room temperaturevalues

Based on data in Table B.4,Callister & Rethwisch 8e.a = annealedhr = hot rolledag = agedcd = cold drawncw = cold workedqt = quenched & tempered

Yield Strength : ComparisonGraphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibers

PolymersY

ield

str

engt

h, y

(MP

a)

PVC

Har

d to

mea

sure

,

sinc

e in

tens

ion,

frac

ture

usu

ally

occ

urs

befo

re y

ield

.

Nylon 6,6

LDPE

70

20

40

6050

100

10

30

200

300

400500600700

1000

2000

Tin (pure)

Al (6061) a

Al (6061) ag

Cu (71500) hrTa (pure)Ti (pure) aSteel (1020) hr

Steel (1020) cdSteel (4140) a

Steel (4140) qt

Ti (5Al-2.5Sn) aW (pure)

Mo (pure)Cu (71500) cw

Har

d to

mea

sure

, in

cer

amic

mat

rix a

nd e

poxy

mat

rix c

ompo

site

s, s

ince

in te

nsio

n, fr

actu

re u

sual

ly o

ccur

s be

fore

yie

ld.

HDPEPP

humid

dryPC

PET

¨

Chapter 6 - 30

Tensile Strength, TS

• Metals: occurs when noticeable necking starts.• Polymers: occurs when polymer backbone chains are

aligned and about to break.

Adapted from Fig. 6.11, Callister & Rethwisch 8e.

y

strain

Typical response of a metal

F = fracture or

ultimate

strength

Neck – acts as stress concentrator

engi

neer

ing

TS

stre

ss

engineering strain

• Maximum stress on engineering stress-strain curve.

Chapter 6 - 31

• After yielding, the stress necessary to continue plastic deformation in metals increases to a maximum point (M) and then decreases to the eventual fracture point (F).• All deformation up to the maximum stress is uniform throughout the tensile sample. • However, at max stress, a small constriction or neck begins to form.• Subsequent deformation will be confined to this neck area.• Fracture strength corresponds to the stress at fracture.

Region between M and F:• Metals: occurs when noticeable necking starts.• Ceramics: occurs when crack propagation starts.• Polymers: occurs when polymer backbones are aligned and about to break.

Tensile Strength, TS

Chapter 6 - 32

Elastic Strain Recovery (during plastic deformation)


Str

ess

Strain

3. Reapplyload

2. Unload

D

Elastic strainrecovery

1. Load

yo

yi

- Deformed plastically,stress released, materialhas permanent strain.

- If stress is reapplied,material again respondselastically at the beginningup to a new yield point thatis higher than the originalyield point.

(Strain hardening)

- Elastic strain beforereaching the yield point iscalled elastic strainrecovery.

Chapter 6 -

Stress-Strain Diagram

Strain ( ) (L/Lo)

41

2

3

5

Elastic Region

PlasticRegion

StrainHardening Fracture

ultimatetensile strength

Elastic regionslope =Young’s (elastic) modulusyield strength

Plastic regionultimate tensile strengthstrain hardeningfracture

necking

yieldstrength

UTS

y

εEσ

ε

σE

12

y

ε ε

σE

Chapter 6 -

Stress-Strain Diagram (cont)

• Elastic Region (Point 1 –2)- The material will return to itsoriginal shape after the material is unloaded(like a rubber band).

- The stress is linearly proportionalto the strain in this region.

εEσ σ : Stress(psi)E : Elastic modulus (Young’s Modulus) (psi)ε : Strain (in/in)

- Point 2 : Yield Strength : a point where permanent deformation occurs. (If it is passed, the material will no longer return to its original length.)

ε

σE or

Strain ( )

41

2

3

5

Elastic Region

PlasticRegion

StrainHardening

Fracture

UTS

y

Chapter 6 -

• Strain Hardening

- If the material is loaded again

from Point 4, the curve will

follow back to Point 3 with the

same Elastic Modulus (slope).

- The material now has a higher

yield strength of Point 3.


Strain ( )

41

2

3

5

Elastic Region

PlasticRegion

StrainHardening

Fracture

UTS

y

- Raising the yield strength by permanently straining the material is called Strain Hardening; an increase in σy due to plastic deformation.

Chapter 6 -

• Tensile Strength (Point 3)

- The largest value of stress on the diagram is called Tensile Strength(TS) or Ultimate TensileStrength(UTS)

- It is the maximum stress whichthe material can support without breaking.


Strain ( )

41

2

3

5

Elastic Region

PlasticRegion

StrainHardening

Fracture

UTS

y

• Fracture (Point 5)

- If the material is stretched beyond Point 3, the stress decreases as necking and non-uniform deformation occur.

- Fracture will finally occur at Point 5.

Chapter 6 - 37

Tensile Strength: Comparison

Si crystal<100>

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibers

PolymersTe

nsile

stre

ngth

, TS

(MP

a)

PVC

Nylon 6,6

10

100

200300

1000

Al (6061) a

Al (6061) agCu (71500) hr

Ta (pure)Ti (pure) aSteel (1020)

Steel (4140) a

Steel (4140) qt

Ti (5Al-2.5Sn) aW (pure)

Cu (71500) cw

LDPE

PP

PC PET

20

3040

20003000

5000

Graphite

Al oxide

Concrete

Diamond

Glass-soda

Si nitride

HDPE

wood ( fiber)

wood(|| fiber)

1

GFRE(|| fiber)

GFRE( fiber)

CFRE(|| fiber)

CFRE( fiber)

AFRE(|| fiber)

AFRE( fiber)

E-glass fib

C fibersAramid fib

Based on data in Table B.4,Callister & Rethwisch 8e.a = annealedhr = hot rolledag = agedcd = cold drawncw = cold workedqt = quenched & temperedAFRE, GFRE, & CFRE =aramid, glass, & carbonfiber-reinforced epoxycomposites, with 60 vol%fibers.

Room temperaturevalues

Chapter 6 - 38

Ductility

- A measure of the degree of plastic deformation that has been sustained at fracture.

- A metal that experiences very little or no plastic deformation upon fracture is termed brittle.

- expressed quantitatively as either percent elongation(%EL) or percent reduction in area(%RA)

Chapter 6 - 39

• Plastic deformation that has been sustained at fracture

Ductility

• Another ductility measure: 100xA

AARA%

o

fo -=

x 100L

LLEL%

o

of

Lf

Ao AfLo


Engineering tensile strain,

Engineering tensile stress,

smaller %EL(brittle if %EL<5% )

larger %EL(ductile if %EL>5%)

→ A material that suffers very little plastic deformation is brittle.

Chapter 6 - 40

Ductility

- The magnitudes of both yield and tensile strengths decline with increasing temperature

- Ductility increases with temperature. (ex. Fe)

Chapter 6 - 41

--brittle response (aligned chain, cross linked & networked case)--plastic response (semi-crystalline case)

Stress-Strain Behavior: Elastomers

3 different responses:

A – brittle failure

B – plastic failure

C - highly elastic (elastomer)

Chapter 6 - 42

• The ability to absorb energy up to fracture : Energy to break a unit volume of material (energy per unit volume of material, e.g. J/m3 ).

• Approximate by the area under the stress-strain curve.• A “tough” material has strength and ductility.

Toughness

Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy


small toughness (unreinforced polymers)

Engineering tensile strain,

Engineering tensile stress,

small toughness (ceramics)

large toughness (metals)

Chapter 6 - 43

Resilience, Ur• Ability of a material to store energy

– Energy stored best in elastic region

– Resilient materials have high yield strengths and low moduli of elasticity; ex. spring applications.

If we assume a linear stress-strain curve this simplifies to


yyr 21U y2

1 y

E

y

2E2

y dUr 0

Chapter 6 - 44

True Stress & Strain- The decline in the stress

past the maximum(M) seems to indicate that the metal is becoming weaker. However, it is increasing in strength.

- However, cross-sectional area is decreasing rapidly within the neck region, where deformation is occurring.

iT AF

- It is more meaningful to use a true stress-true strain scheme.

True stress; the load divided by the actual cross-sectional area(Ai) of the specimen at that load;

- True strain; calculated using actual and not original dimensions, defined by

oiT ln

Chapter 6 - 45

True Stress & Strain• If no volume change occurs during deformation ;

• Engineering stress and strain are related according to :

1ln

1

T

T


The true stress necessary to sustain increasing strain continues to rise past the tensile point M’

00 AA ii

iT AF

oiT ln

00 AA ii

Valid only to the onset of necking

Chapter 6 -46

In an undeformedthermoplastic polymer tensile sample,

(a) the polymer chains are randomly oriented.

(b)When a stress is applied, a neck develops as chains become aligned locally. The neck continues to grow until the chains in the entire gage length have aligned.

(c) The strength of the polymer is increased

Chapter 6 - 47

Stress-Strain Results for Steel Sample

Chapter 6 -

Example 1. Convert the change in length data in the table to engineering stress and strain and plot a stress-strain curve.

Chapter 6 -

Example 1 SOLUTION

Chapter 6 -

Example 2. Compare engineering stress and strain with true stress and strain for the aluminum alloy in Example 1 at (a) the maximum load. The diameter at maximum load is 0.497 in. and at fracture is 0.398 in.

Example 3 SOLUTION

Chapter 6 - 51

Hardness• Resistance to permanently indenting the surface.• Large hardness means:

-- resistance to plastic deformation or cracking incompression.

-- better wear properties.

e.g., 10 mm sphere

apply known force(1 to 1000g)

measure size of indent after removing load

dDSmaller indents mean larger hardness.

increasing hardness

most plastics

brasses Al alloys

easy to machine steels file hard

cutting tools

nitrided steels diamond

Chapter 6 - 52

Hardness: MeasurementTable 6.5

Chapter 6 - 53

Conversion of Hardness Scales

Also see: ASTM E140 - 07 Volume 03.01Standard Hardness Conversion Tables for Metals Relationship Among Brinell Hardness, Vickers Hardness, Rockwell Hardness, Superficial Hardness, Knoop Hardness, and Scleroscope Hardness

Chapter 6 - 54

Correlation between Hardness and Tensile Strength

• Both hardness and tensile strength are indicators of a metal’s resistance to plastic deformation.

• For cast iron, steel and brass, the two are roughly proportional.

• Tensile strength (psi) = 500*BHR

Chapter 6 - 55

Variability in Material Properties

• Elastic modulus is material property

• Measured material properties are not exact quantities.

ex. A number of identical tensile samples give a variety of modulus, yield strength, and tensile strength.

• A number of factors lead to uncertainties in measure data;

ex. Test method, variation in specimen fabrication procedures, operator bias, apparatus calibration, inhomogeneities, and lot difference

• Scatter and variability of materials properties are inevitable and must be dealt with appropriately.

• “What is the probability of failure of this alloy under these given circumstances” instead of asking the question, “What is the fracture strength of this ally?”

Chapter 6 - 56

Variability in Material Properties

• Specify a typical value and degree of dispersion;

• Statistics

– Mean

– Standard Deviation

s n

xi x 2n1

1

2

n

xx n

n

where n is the number of data points

Chapter 6 - 57

• Design uncertainties mean we do not push the limit.• Factor of safety, N

Ny

working

Often N isbetween1.2 and 4

• Example: Calculate a diameter, d, to ensure that yield doesnot occur in the 1045 carbon steel rod below. Use a factor of safety of 5.

Design or Safety Factors

220,000N

d2 / 4 5

Ny

working

1045 plain

carbon steel: y = 310 MPa

TS = 565 MPa

F = 220,000N

d

Lo

d = 0.067 m = 6.7 cm

Chapter 6 - 58

• 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 alarge elastic modulus (E or 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 y.

Chapter 6 - 59

• Concept Check : 6.1-6.4

• Example problem : 6.1~6.6

• Design example : 6.1

• Questions and Problems :6.4~6.8, 6.10, 6.11, 6.13, 6.15~6.18, 6.20, 6.22~6.28, 6.35, 6.36, 6.39, 6.49, 6.50, 6.54, 6.57, 6D1

Problems

Chapter 6: Mechanical Properties - Konkukccdjko.konkuk.ac.kr/upload/sub0503/ch06.pdfmaterial, having the effect of ... extensometer specimen • Typical tensile specimen ... - The

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