Mechanical Properties

Chapter 7

Stress and Strain, Elastic Deformation

Chapter 7

• 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 do dislocationscause permanent deformation? What materials aremost resistant to permanent deformation?

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

• Ceramic Materials: What special provisions/tests aremade for ceramic materials?

Issues to address…

Loading Forms

• Simple tension: cable

Note: = M/AcR here.

Ao = cross sectional area (when unloaded)

FF

o F

A

o

FsA

M

M Ao

2R

FsAc

• Torsion (a form of shear): drive shaft Ski lift

Common States of Stress

Canyon Bridge, Los Alamos, NM

• Simple compression:

Ao

Balanced Rock, Arches National Park o

FA

Note: compressivestructure member(s < 0 here).

(photo courtesy P.M. Anderson)

(photo courtesy P.M. Anderson)

Other Common Stress States_1

• Bi-axial tension: • Hydrostatic compression:

Fish under waterPressurized tank

z > 0

> 0

s < 0h

(photo courtesy P.M. Anderson) (photo courtesy P.M. Anderson)

Other Common Stress States_2

Elastic means reversible!

1. Initial 2. Small load 3. Unload

F

bonds stretch

return to initial

F

Linear-elastic

Non-Linear-elasticCf. Anelasticity

Elastic Deformation

Stress has units:N/m2 or lbf/in2

Area, A

Ft

Ft

Fs

F

F

Fs

=Fs

Ao

original area before loading

Area, A

F t

F t

=FtAo 2

f2m

Nor

in

lb=

Engineering Stress

• Tensile strain: • Lateral strain:

• Shear strain:

Strain is alwaysdimensionless.

90º

90º - y

x = x/y = tan

Lo

L Lwo

/2

L/2

Lowo

Engineering Strain

(2) Tensile strain

00

0i ==εl

lΔ

lll

-

(1) Tensile stress

oAF

σ = psi = 0.006895 MPa 1 MPa = 1 MN/m2

(3) Shear stress

(4) Shear strain

θtan=hx

γ =

+ tensile- compressive

Engineering Stress & Strain

oAF

= τ

22cos+1

=cos=' 2θ

σθσσ

22sin

=cossin='θ

σθθστ

Area : A/cos

90-

Geometric Considerations of Stress State

• Typical tensile test machine

specimenextensometer

• Typical tensile specimen

gauge length

Stress-Strain Testing

Region I Region II Region III

Stress & Strain Curves

Yield Point Phenomena

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

• Hooke's Law:

= E

Linear-elastic

E

F

Fsimple tension test

Linear Elastic Properties

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

Elastic Deformation_1

The stress is linearly proportional to strainσ = EεThe slope of curve : Modulus of elasticity

Young's modulus ( Typical metals ~ 109 psi )

(1) Deformation is recoverable.(2) Deformation is linear. ( Hooke's law )(3) Modulus determined by atomic bonding. (atom type and crystal structure)

):G(= modulus shear γ τ

Elastic Deformation_2

G

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

46

10

20

406080

100

200

600800

10001200

400

Tin

Cu alloys

Tungsten

Si carbide

Diamond

PTFE

HDPE

LDPE

PP

Polyester

PSPET

CFRE( fibers)*

GFRE( fibers)*

GFRE(|| fibers)*

AFRE(|| fibers)*

CFRE(|| fibers)*

MetalsAlloys

GraphiteCeramicsSemicond

Polymers Composites/fibers

E(GPa)

Eceramics > Emetals >> Epolymers

Based on data in Table B2, Callister 6e.Composite data based onreinforced epoxy with 60 vol%of alignedcarbon (CFRE),aramid (AFRE), orglass (GFRE)fibers.

Young’s Moduli: Comparison

109 Pa

Non-elastic Behaviors

• Poisson's ratio, :

: dimensionless

L

-

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

Poisson’s Ratio_1

εε

= L-v

Poisson's ratio is determined by two opposing factor :(1) Conservation of volume(2) Conservation of bond angles

rigid body ν= 0 diamond ν= 0.2soft body ν= 0.5 natural rubber ν= 0.49

most structural materials ν≈1/3

strainaxialstrainlateral

=

ν -

z

y

z

x

ZyxZZ

εε

=εε

ν=

νε=ε= εEε=σ

--∴

-

compression

elongation

Poisson’s Ratio_2

• Elastic Shearmodulus, G:

G

= G

simpletorsiontest

M

M

• Special relations for isotropic materials:

2(1 +ν)E

G =3(1 - 2ν)

EK =

• Elastic Bulkmodulus, K:

Pressure test: Vol. init = VoVol. chg = ∆V

P

P PK

P

∆V

K Vo

Other Elastic Properties

oVV

=PΔ

-

• Simple tension:

FLoEAo

L Fw o

EAo

• Material, geometric, and loading parameters all contribute to deflection.

• Larger elastic moduli minimize elastic deflection.

F

Ao/2

L/2

Lowo

• Simple torsion:

2MLoro

4GM = moment = angle of twist

2ro

Lo

Useful Linear Elastic Relationships

- Time dependent elastic deformation

- Elastic deformation continue after the stress application,

and finite time is required for complete recovery after unloading.

- Viscoelastic behavior of polymers, mud.

Anelasticity

Deformation is plastic by dislocation movements which are not recoverable.

ultimate tensile strength : 50MPa ~3,000MPa

work hardening ( iron wire once bended )

P : proportional limit

σy : yield strength

Plastic deformation

Plastic Deformation

Elastic Strain Recovery

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

• Simple tension test:

engineering stress,

engineering strain,

Elastic+Plastic at larger stress

permanent (plastic) after load is removed

pplastic strain

Elastic initially

Plastic (permanent) Deformation

• Stress at which noticeable plastic deformation has occurred.

when p = 0.002 (0.2%)

y = yield strength

Note: for 2 inch sample

= 0.002 = z/z

z = 0.004 in

tensile stress,

engineering strain,

y

p = 0.002

Yield Strength, σy

Room T values

a = annealedhr = hot rolledag = agedcd = cold drawncw = cold workedqt = quenched & tempered

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibersPolymers

Yiel

d st

reng

th,

y

(MP

a)

PVC

Har

d to

mea

sure

, si

nce

in te

nsio

n, fr

actu

re u

sual

ly o

ccur

s be

fore

yie

ld.

Nylon 6,6

LDPE

70

20

40

6050

100

10

30

200

300400500600700

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

¨

Yield Strength: Comparison

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

aligned and about to break.* Cermics: ??

y

strain

Typical response of a metal

F = fracture or ultimate strength

Neck – acts as stress concentrator

engi

neer

ing

TSst

ress

engineering strain

• Maximum stress on engineering stress-strain curve.

Tensile Strength, TS

Room T valuesSi crystal

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibersPolymers

Te

nsi

le s

tre

ng

th, T

S (M

Pa

)

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 TS(ceram)

~TS(met)

~ TS(comp) >> TS(poly)

Based on data in Table B4,Callister 6e.a = annealedhr = hot rolledag = agedcd = cold drawncw = cold workedqt = quenched & temperedAFRE, GFRE, & CFRE =aramid, glass, & carbonfiber-reinforced epoxycomposites, with 60 vol%fibers.

Tensile Strength: Comparison

• Plastic tensile strain at failure:

• Another ductility measure: 100xA

AARA%o

fo -=

x 100L

LLEL%

oof

Engineering tensile strain,

Engineering tensile stress,

smaller %EL

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

LfAo AfLo

Ductility, %EL

Brittle and Ductile

• Ability of a material to store energy – Energy stored best in elastic region

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

Adapted from Fig. 6.15, Callister 7e.

∫y

0rd=U

εεσ

Resilience, Ur

yyr εσ21

U

• Energy to break a unit volume of material• Approximate by the area under the stress-strain curve

Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy

very small toughness(unreinforced polymers)

Engineering tensile strain,

Engineering tensile stress, s

small toughness (ceramics)

large toughness (metals)

Toughness

iT AF=σ

( )oiT ln=ε ll( )( )+ε1ln=ε+ε1=σσ

T

T

Ai : instantaneous cross-sectional area

00ii =Aif A ll

)+ε1(ln= ε, )+ε1(=σσ TT

Note: S.A. changes when sample stretched

• True stress

• True Strain

True Stress & Strain

)+ε(ln=)(ln=)(ln=d

=ε0

T llΔl+l

ll

ll

0

0

0

l

l∫

• Curve fit to the stress-strain response:

T = K T( )

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 y due to plastic deformation.

large hardening

small hardeningy0

y1

Hardening

• Room T behavior is usually elastic, with brittle failure.• 3-Point Bend Testing often used.

--tensile tests are difficult for brittle materials.

Measuring Elastic Modulus

FL/2 L/2

= midpoint deflection

cross section

R

b

d

rect. circ.

Location of max tension

• Flexural strength:

rect. fs m

fail 1.5FmaxL

bd2

FmaxL

R3

xF

Fmax

max

• Typ. values:Material fs(MPa) E(GPa)

Si nitrideSi carbideAl oxideglass (soda)

700-1000550-860275-550

69

30043039069

Measuring Strength

• Determine elastic modulus according to:

E

F

L3

4bd3

F

L3

12R4

rect. cross

section

circ. cross

section

Fx

linear-elastic behavior

F

slope =

Measure of a materials resistance to localizedplastic deformation

(1) simple and inexpensive

(2) nondestructive

Depend on type of indenter

- Rockwell Hardness

- Brinell Hardness

- Knoop and Vickers Hardness

Hardness_1

• 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 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

Hardness_2

• Rockwell– No major sample damage– Each scale runs to 130 but only useful in range 20-100. – 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

Hardness Measurement_1

Hardness Measurement_2

• Elastic modulus is material property• Critical properties depend largely on sample flaws (defects, etc.).

Large sample to sample variability. • Statistics

– Mean

– Standard Deviation

where n is the number of data points

Variability in Material Properties

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

Nσ

=σ yworking

Often N isbetween1.2 and 4

• Example: Calculate a diameter, d, to ensure thatyield does not occur in the 1045 carbon steel rod below. Use a factor of safety of 5.

( )4/dN000,220

2π5

Nσ

=σ yworking 1045 plain carbon steel: = 310 MPa

TS = 565 MPa

F = 220,000N

d

L o

d = 0.067 m = 6.7 cm

Design or Safety Factors

• 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).

• Plastic behavior: This permanent deformationbehavior occurs when the tensile (or compressive)uniaxial stress reaches sy.

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

• Ductility: The plastic strain at failure.

Summary