Chapter 6.1-6.2, 6.6:

Mechanical Properties of Metals

EM321: Lesson 8b

Before class, pass out photocopy of Stress-Strain plot

• Stress and Strain: What are they and why are they 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 dislocations cause permanent deformation? What materials are most resistant to permanent deformation?

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

Important Topics: Mechanical Properties

F

bonds stretch

return to initial

1. Initial 2. Small load 3. Unload

Elastic means reversible!

F

Linear- elastic

Non-Linear-elastic

Elastic Deformation

1. Initial 2. Small load 3. Unload

Plastic means permanent!

F

linear elastic

linear elastic

plastic

planes still sheared

F

elastic + plastic

bonds stretch & planes shear

plastic

Plastic Deformation (Metals)

Tensile (Normal) stress, s Shear stress, t

Area, A

Ft

Ft

s FtAo

original area before loading

Area, A

Ft

Ft

Fs

F

F

Fs

t FsAo

Stress has units:

Pa =N/m2 or psi = lb/in2

Engineering Stress

Shear stress causes a shape

change

6

• Simple tension: cable

t

Common States of Stress

os F

A

ot

FsA

ss

M

M A o

2R

FsA c

• Torsion (a form of shear): drive shaftSki lift

Ao = cross sectional

area (when unloaded)

FF

Canyon Bridge, Los Alamos, NM

os F

A

• Simple compression:

Note: compressivestructure member(s < 0 here).

A o

Balanced Rock, Arches National Park

Other Common Stress States (i)

• Bi-axial tension: • Hydrostatic compression:

Pressurized tank

s < 0h

Other Common Stress States (ii)

Fish under water

sz > 0

sq

> 0

• Tensile strain, ε: • Lateral strain, εL:

Strain is alwaysdimensionless.

Engineering Strain

• Shear strain, g:

q

90º

90º - qy

xqg = D x/y = tan

e Lo

-eL L

wo

d /2

Lowo

dL /2

Typical tensile specimen

• Other types of tests: --Compression: brittle materials (i.e. concrete) --Torsion: cylindrical tubes, shafts.

Typical tensile test machine

load cell

extensometerspecimen

moving cross head

Stress-Strain Testing: The Tensile Test

gauge length

(portion of sample with reduced cross section)=

Tensile Test Output: Stress-Strain Curve• The tensile test gives an output of Force, F, versus Deflection, d

• Convert this output to Stress versus Strain:– Divide Force,F, by Initial Cross Sectional Area, Ao, to get _______, σ

– Divide Deflection, , d by Initial Length, Lo, to get ________, ε

• Example Stress-Strain curve:– Note: inner graph is a zoomed in

picture of the beginning of the curve

Group Question: Why do we typically use Stress vs. Strain plot, rather than Force vs. Deflection plot?

Stress

Strain

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

• Hooke's Law:

s = E e

• Poisson's ratio, n :

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

en - Le

F

Fsimple tension test

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

Linear Elastic Properties, E and n s

Linear-

elastic

Ee

eL

e-n

Example: Finding E• Modulus of Elasticity, E, of a material is equal to the slope of

the initial, linear section of the Stress-Strain plot.

• What is the Modulus of Elasticity

of the material shown in the

Stress-Strain plot that was handed

out (Fig 6.21)?

E ≈ 200 * 106 Pa / 0.001 = 200 * 109 Pa

More Comments on Elastic Deformation

• The Modulus of Elasticity, E, is the stiffness of the material, or resistance to elastic deformation.

• E is an important design parameter• Elastic deformation is non-permanent

– Spring like return to original shape

• Values of E are a direct consequence of bonding type– Slope of Stress-Strain plot (which is proportional to the Modulus of

Elasticity) depends on bond strength of the metal

Even More Comments About Elasticity• It is assumed elastic deformation is

time independent• Also assumed that as load is released,

the strain is totally recovered.• In most engineering materials, there is

a time dependent elastic strain component. – Elastic deformation will continue after the

stress application, and upon release some time is required for complete recovery.

– Time dependent behavior is called “anelasticity”

– Usually negligible for metals, is significant for some polymers, where it is called “visco-elasticity”

• Elastic Shear Modulus, G:t

Ggt = G g

Other Elastic Properties

simpletorsiontest

M

M

• Special relations for isotropic materials:

2(1 + n)

EG =

3(1 - 2n)

EK =

• Elastic Bulk Modulus, K:

Pressure Test:

Init. Vol =Vo. Vol Chg. = DV

P

P PP = -KDVVo

P

DVK Vo

Isotropic means that the material behaves the same in all directions

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

<100>

<111>

Si carbide

Diamond

PTFE

HDPE

LDPE

PP

Polyester

PSPET

CFRE( fibers)*

GFRE( fibers)*

GFRE(|| fibers)*

AFRE(|| fibers)*

CFRE(|| fibers)*

MetalsAlloys

GraphiteCeramicsSemicond

PolymersComposites

/fibers

E(GPa)

Eceramics > Emetals >> Epolymers

109 Pa

Modulus of Elasticity: Comparison

Simple Tensile Test:

tensile stress, s

engineering strain, e

Elastic initially

Elastic+Plastic at larger stress

permanent (plastic) after load is removed

epplastic strain

Plastic (Permanent) Deformation

• Proportional Limit = Stress at which plastic deformation first occurs

• Yield Strength = Stress at which noticeable plastic deformation has occurred.

tensile stress, s

engineering strain, e

sy

ep = 0.002

Yield Strength, sy or Sy

Since it is sometimes difficult to determine the initial departure from linear behavior, the convention is to offset strain by 0.002, draw a

parallel line to the F-displacement curve and mark the intercept point.

This is the “yield strength”.

The Yield Point Phenomenon

• Most materials do not exhibit a distinct yield point (Fig. a)– Use 0.002 Strain offset method

to calculate

• Some materials exhibit 2 distinct yield points (Fig. b)– Stress drops dramatically at

Upper Yield Point– Stress is relatively constant for

a while at Lower Yield Point

Group Question: For materials (e.g. some steels) that exhibit the yield point phenomenon, should the upper or lower yield point be used? Why?

Lower Yield point should be used. Better defined. Stable for a long time. Insensitive to testing procedure. More conservative (because it is lower value).

Example: Finding Yield Strength

• Use the 0.002 Strain offset method to find the Yield Strength of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)?

Sy ≈ 400 MPa

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibersPolymers

Yield

str

ength

, sy (M

Pa)

PVC

Ha

rd t

o m

ea

sure,

si

nce

in t

ensi

on, fr

act

ure

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

Ha

rd t

o m

ea

sure

, in

cera

mic

matr

ix a

nd e

pox

y m

atr

ix c

om

posi

tes,

sin

ce in

tensi

on, fr

act

ure

usu

ally

occ

urs

befo

re y

ield

.HDPEPP

humid

dryPC

PET

¨ Room Temp values

y(ceramics) >> y(metals) >> y(polymers)

Yield Strength: Comparison

Ultimate Tensile Strength, SUt or ST or TS

y

strain

Typical response of a metal

SF = fracture

strength

Neck – acts as stress concentrator

eng

inee

ring

SUt s

tres

s

engineering strain

Ultimate Tensile Strength = Maximum stress on engineering stress-strain curve.

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

Typical response of a metal

Example: Finding Ultimate Tensile Strength

• Find the Ultimate Tensile Strength of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)?

SUt ≈ 500 MPa

Room Temp values

Si crystal<100>

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibersPolymers

Tensi

le s

trength

, TS (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)

Ultimate Tensile Strength: Comparison

Deformation Terms• Ductility—Measure of the degree of plastic

deformation that has been sustained at fracture• Brittle—A material that sustains very little to no plastic

deformation prior to fracture• Curves for brittle and ductile behavior:

Percent Elongation = Plastic tensile strain at failure

Engineering tensile strain, e

Engineering tensile stress, s

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

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

Another ductility measure = Percent Reduction of Area

100% xA

AARA

o

fo -

Lo LfAo Af

%EL

L f - LoLo

x100

Ductility: Percent Elongation

Note: Typically, Percent Elongation ≠ Percent Reduction of Area

Toughness = Energy to break a unit volume of material - Approximated 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

Engineering tensile stress, s

small toughness (ceramics)

large toughness (metals)

Resilience, Ur

Resilience = Ability of a material to store energy – Energy stored best in elastic region– Approximated as the area of the elastic region

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

yyr2

1U es@

e

es y dUr 0

Example: Finding Ductility, Toughness and Resilience

• Find the Ductility (%EL), Toughness and Resilience of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)?

%EL ≈ 0.185 = 18.5%

Toughness ≈ 500 MPa * 0.185 – 0.5 * 150 Mpa * 0.02 – 0.5 * 130 MPa * 0.065Toughness ≈ 86.8 Mpa

Resilience ≈ 0.5 * 400 MPa * 0.0038Resilience ≈ 0.76 MPa

Elastic Strain Recovery

Str

ess

Strain

3. Reapplyload

2. Unload

D

Elastic strainrecovery

1. Load

syo

syi

Conclusion• Stress and Strain• Elastic and Plastic Behavior• Mechanical Properties found from Stress-Strain Curve

– Modulus of Elasticity– Yield Strength– Ultimate Tensile Strength– Ductility– Toughness– Resilience

Example Stress-Strain Curve (Callister Fig. 6.21)