ch06 mech prop summer2016-sent(1)

ME3101-Materials Sc. and Engr.

Dr. Atiqullah. Compiled from: Callister and

others 1

ME3101: Materials Sc. & Engr.

Ch 6: Mech. Properties

How many of these terms do we know?

• Anelasticity - viscoelasticity, polymers

• Elastic recovery

• Engineering Stress/strain – materials testing

• True stress/strain – mfg.

• Poisson’s Ratio – mfg.

• Proportional limit – up to which ..

• Toughness – bumper design for absorbing ? energy

• Resilience (p-168) – spring design- energy absorbing but with no

permanent ..?...

Dr. Mir AtiqullahIncludes materials adopted from Callister text

Tensile and Compressive Deformations

0A

F=σ

Engineering Stress (considers only

the original dimensions)

Engineering Strain (considers only

the original dimensions)

00

0

l

l

l

lli ∆=

−=ε

veA

F−=

−=

0

σ In compression

Poisson’s Ratio: ratio of the lateral and axial strains:

00

00

/)(

/)(

lll

ddd

z

y

z

x

−−

−=−=−=εε

εε

ν

Shear and Torsional Deformations

0

StressShear A

F== τ

J

Tc== τ(Torque) Torsion todue StressShear

GJ

Tl== ϕ(Torque)Torsion todue (rad.)ist Angular tw



others 2

Tensile Tests

Most common among mechanical tests

Standard test methods and samples: ASTM Standard E8 and E8M (metric)

Circular and flat specimens with standard gauge lengths (50 mm or 2”) and areas.

Testing machine usually elongate the specimen at constant rate.

Simultaneously measures load (stress-using load cells) and elongation (strain-

using extensometer).

Engineering stress and engineering strain, from tensile tests are determined by:

0A

F=σ and

00

0

l

l

l

lli ∆=

−=ε

More LATER….

Gage length ?

Compression Tests

Preferred test method if service loads are compressive-e.g. ?

Standard test methods and samples: ASTM Standard E9 and E9M ?

(metric)

Compressive force and stress are NEGATIVE.

Tensile tests are easier to perform and compressive tests rarely reveals

any additional information.

Engineering stress and engineering strain, from compressive tests are

determined by same equations as for tensile tests.

Compressive tests are specially necessary for determining materials

behavior for large strains e.g. in manufacturing (forging, rolling,

bending, stamping,..)

Shear and Torsional Tests

0A

F=τ

Shear caused by parallel forces tending to slide a layer over another.

Shear stress (transverse):

Shear strain γ =tangent of the strain angle θ

Torsion (twisting) produces pure shear

E.g. : drive shaft, vehicle axle, twist drill, screw driver,..

J

Tc=τ



others 3

Geometric Considerations of the Stress State

State of Stress is also a function of the

orientation of the plane along which the

stresses are considered. For a plane at an

angle θ: Resolved stresses are-

+⋅=⋅=

2

2cos1cos2' θ

σθσσ

⋅=⋅=2

2sincossin' θ

σθθστ

o45 is max' θτ when=

F

δ

bonds stretch

return to initial

2

1. Initial 2. Small load 3. Unload

Elastic means reversible!

F

δ

Linear- elastic

Non-Linear-elastic

ATOMIC MODEL OF ELASTIC DEFORMATION

εσ ⋅= E εσ ⋅= E

Consider happens at

the atomic/grain level

Materials properties data:

See Table 6.1 NOW (page 137).

Similarly shear stress and strain are

related by:

γτ ⋅= G γτ ⋅= G

Table 6.1 Page 174 9th ed.

G =~35-40%~ 8.5E6 psi

For a metal/alloy, E=22E6 psi, ν=0.34 Estimate G= ? Use )1(2 ν+= GE



others 4

3

1. Initial 2. Small load 3. Unload

Plastic means permanent!

F

δlinear elastic

linear elastic

δplastic

planes still sheared

F

δelastic + plastic

bonds stretch & planes shear

δplastic

ATOMIC MODEL OF PLASTIC DEFORMATION

Consider what

happens at the

atomic/grain level

7

• Bi-axial tension:2D • Hydrostatic compression:3D

Fish under waterPressurized tank

σz > 0

σθ > 0

σ < 0h

(photo courtesyP.M. Anderson)

(photo courtesyP.M. Anderson)

OTHER COMMON STRESS STATES (2)

• Submarine ? Offshore Pipeline?

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

10

• Hooke's Law:

σ = E ε• Poisson's ratio, ν:

metals: ν ~ 0.30-0.34ceramics: ~0.25polymers: ~0.40 compaction?

εν = − Lε

εL

ε

1-ν

F

Fsimple tension test

σ

Linear- elastic

1

E

ε

Units:

E: [GPa] or [psi]

ν: ?

LINEAR ELASTIC PROPERTIES

Materials properties data:

See Table 6.1 AGAIN (page 137) 6th ed. P-118).

Verify the range of ν values.



others 5

• Elastic Shear

modulus, G:

11

τ

1

Gγ

τ = G γ

• Elastic Bulk

modulus, K:

P= -K∆VVo

P

∆V

1-K

Vo

• Special relations for isotropic materials:

P

P P

M

M

G =

E

2(1+ ν) K =

E

3(1− 2ν)

simple

torsion

test

pressure

test:

Init. Vol. =Vo.

Vol. chg.= ∆V

OTHER ELASTIC PROPERTIES

120.2

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

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)*

MetalsAlloys

GraphiteCeramics

Semicond

PolymersComposites

/fibers

E(GPa)

Eceramics

> Emetals

>> Epolymers

109 Pa

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

• Simple tension:

13

δ = FLo

EAo

δL

= −νFwo

EAo

δ/2

δ/2

δL/2δL/2

Lowo

F

Ao

• Simple torsion:

M=moment α =angle of twist

2ro

Lo

• Material, geometric, and loading parameters all

contribute to deflection.

• Larger elastic moduli minimize elastic deflection.

USEFUL LINEAR ELASTIC RELATIONS

JG

LM

⋅⋅

= 0α



others 6

Example.

Rod diameter 10 mm undergoes a tensile loading

Calc the load necessary to produce a 2.5E-3 mm change in

diameter within elastic region.

units? 45.20

−−=∆

= Ed

dxε

νε

ε xz −=

G =

E

2(1+ ν)

Ezεσ =

0AF ⋅= σ

PLASTIC (PERMANENT) DEFORMATION

Elastic limit= ?

Max strain at Elastic limit= ? e.g. 0.005 or 0.5%

Plastic deformation ~ breaking the bond with old neighbors

and building new bonds next…

In crystalline solids deformation ~ slip between adjacent

crystal planes.

Noncrystalline solids (also liquids)- deformation by

viscous flow of material. Material is not an organized

array of atoms, to begin with.

14

• Simple tension test:

(at lower temperatures, T < Tmelt/3)

PLASTIC (PERMANENT) DEFORMATION

tensile stress, σ

engineering strain, ε

Elastic initially

Elastic+Plastic at larger stress

permanent (plastic) after load is removed

εp

plastic strain Elastic strain/

recovery



others 7

15

• Stress at which noticeable plastic deformation has

occurred.when εp = 0.002 (standard)Other standards exist: e.g. 0.4% etc.Other values may also be used.tensile stress, σ

engineering strain, ε

σy

εp = 0.002

YIELD STRENGTH, σy

Partial plastic deformation

No distinct point where yield starts

16

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibers

Polymers

Yie

ld s

tre

ng

th,

σy

(M

Pa

)

PVC

Ha

rd t

o m

ea

su

re,

s

inc

e i

n t

en

sio

n,

fra

ctu

re u

su

all

y o

cc

urs

be

fore

yie

ld.

Nylon 6,6

LDPE

70

20

40

6050

100

10

30

200

300

400

500600700

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

su

re,

in c

era

mic

ma

trix

an

d e

po

xy

ma

trix

co

mp

os

ite

s,

sin

ce

in

te

ns

ion

, fr

ac

ture

us

ua

lly

oc

cu

rs b

efo

re y

ield

.

HDPEPP

humid

dry

PC

PET

¨

Room T values

σy(ceramics)

>>σy(metals)

>> σy(polymers)

Based on data in Table B4,Callister 6e.

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

YIELD STRENGTH: COMPARISON

17

• Maximum possible engineering stress in tension.

• 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

strain

en

gin

ee

rin

g

str

es

s

TS

Typical response of a metal Practice

Example 6.3



others 8

18

Room T valuesSi crystal

<100>

Graphite/ Ceramics/ Semicond

Metals/ Alloys

Composites/ fibers

Polymers

Te

ns

ile

str

en

gth

, T

S (

MP

a)

PVC

Nylon 6,6

10

100

200

300

1000

Al (6061)a

Al (6061)ag

Cu (71500)hr

Ta (pure)Ti (pure)a

Steel (1020)

Steel (4140)a

Steel (4140)qt

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

Cu (71500)cw

LDPE

PP

PC PET

20

3040

2000

3000

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 & tempered

AFRE, GFRE, & CFRE =aramid, glass, & carbonfiber-reinforced epoxycomposites, with 60 vol%fibers.

TENSILE STRENGTH: COMPARISON

• Plastic tensile strain at failure:

19

Engineering tensile strain, ε

Engineering tensile stress, σ

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

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

• Another ductility measure:% reduction in area: 100% xA

AARA

o

fo −=

• Note: %RA and %EL are often comparable.

--Reason: crystal slip does not change material volume.

--%RA > %EL possible if internal voids form in neck.

Lo LfAo

Af

%EL =

L f − Lo

Lo

x100

Adapted from Fig. 6.13, Callister 6e.

DUCTILITY, %EL

Resilience

Defn: Capacity of a material to absorb energy when it is deformed

elastically, which releases this energy upon unloading.

Modulus of ResilienceEE

Uyy

yyyr22

1

2

12σσ

σεσ =

==

Resilient material � high in ?

Resilient material � low in ?

Yield strength

Elasticity

Application� ? springssprings



others 9

• Energy to break a unit volume of material

• Approximate by the area under the stress-strain

curve.

20

smaller toughness- unreinforced polymers

Engineering tensile strain, ε

Engineering tensile stress, σ

smaller toughness (ceramics)

larger toughness (metals, PMCs)

TOUGHNESS

Test Procedures:

Notch/Impact Toughness (Charpy or Izod methods)

Fracture Toughness: Resistance to the growth of an existing crack.

Open p-260 eq. 8.4 and P- 262 Table 8.1 KIC Values.

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

--resistance to plastic deformation or cracking in compression.--better wear properties.

--higher E or stiffness.

21

e.g., 10mm 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

HARDNESS

Principle of

hardness tests

CBN

Why Hardness Tests so Common?

• Simple and inexpensive- no special preparation of

workpiece/specimen. Testers are relatively inexpensive.

• (Usually) test is non destructive. Only a small indentation

is left. Microhardness tests leave an indentation, which is

so small, it requires a microscope to locate and measure

hardness.

• Other mechanical properties may be estimated from

hardness values. e.g. tensile strength: see figure 6.19

Hardness test is performed more frequently than other

methods, because:



others 10

Hardness vs tensile strength

Source : Metals Handbook

Some linear relationship:

TS(Mpa) = 3.45X HB

TS(Psi) = 500 X HB

HB-Brinnel hardness number.

Can we set up such a Linear

relationship between TS and

HRC? Why? Why Not?

HRC scale is not linear..

Standard Hardness Measuring Methods

Brinell Test: Oldest system, uses hardened steel ball upto 3000 kg of load.

Rockwell Test: Most common, Uses diamond cone and various loads (60, 100,

150 kg). Minor load 10 kg.

Indenters: Diamond ‘brale’, steel balls( 1/16, 1/8 inch diam.).

Superficial Rockwell Test: Similar to Rockwell but using lower loads (15, 30,

and 45 kg), Uses same indenters. Minor load 3 kg.

Knoop and Vickers Microhardness tests: uses very small diamond tip,

produces almost invisible indentation, which is later measured by a cross-hair

under microscope. Loads are much smaller between 1 and 1000 grams. These

methods are suitable for selective hardness measurements in a small region of

a specimen. Knoop test (uses elongated diamond indenter) is suitable for

ceramic materials. Microhardness test is suitable for ALL type of materials.

Brinell Tester-applications

The Brinell test is generally used for bulk metal hardness

measurements - the impression is larger than that of the

Vickers or Rockwell test and this is useful as it averages out

any local heterogeneity and is affected less by surface

roughness.

80 HRB =?



others 11

Brinell Hardness Test-Limitations

However, because of the large ball diameter the test cannot be

used to determine the hardness variations in a welded joint for

which the Vickers test is preferred.

Very hard metals, over 450BHN may also cause the ball to

deform resulting in an inaccurate reading.

To overcome this limitation a tungsten carbide ball is used

instead of the hardened steel ball but there is also a hardness

limit of 600BHN with this indentor.

Rockwell Tester

A modern Rockwell

Hardness tester ( Dream)

Specimen

Indenter

80 HRC =?

60 HR45T =?

A Rockwell Hardness

tester

Micro hardness Testers

There are two types of indenters, a square

base pyramid shaped diamond for testing in

a Vickers tester and a narrow rhombus

shaped indenter for a Knoop tester.

Vickers indentation

250 HK =?200 VHN =?



others 12

Summary Hardness Testing Methods.

Errors in Hardness Testing

� Flatness and surface finish -flatness is most important - a maximum angle of

approximately ± 1° would be regarded as acceptable.

� To achieve the required flatness tolerance and surface finish surface grinding

or machining may be necessary.

� The correct load must be applied and to achieve this there must be no friction

in the loading system otherwise the impression will be smaller than expected

- regular maintenance and calibration of the machine is therefore essential.

� The condition of the indentor is crucial - whilst the Vickers diamond is

unlikely to deteriorate with use unless it is damaged or loosened in its

mounting by clumsy handling, the Brinell ball will deform over a period of

time and inaccurate readings will result. This deterioration will be accelerated

if a large proportion of the work is on hard materials.

� The length of time that the load is applied is important and must be controlled

to some uniform value.

Errors in Hardness Testing(contd.)

� The specimen dimensions are important - if the test piece is too

thin the hardness of the specimen table will affect the result. As a

rule of thumb the specimen thickness should be ten times the depth

of the impression for the Brinell test and twice that of the Vickers

diagonal.

� If the impression is too close to the specimen edge then low

hardness values will be recorded - again as a rule the impression

should be some 4 to 5 times the impression diameter from any free

edge.

� Performing hardness testing on cylindrical surfaces eg pipes and

tubes, the radius of curvature will affect the indentation shape and

can lead to errors. It may be necessary to apply a correction factor

- this is covered in, ISO 6507 Part 1.



others 13

Errors in Hardness Testing(contd.)

� The specimen table should be rigidly supported and must be

in good condition - burrs or raised edges beneath the

sample will give low readings.

� Impact loading must be avoided. It is very easy to force the

indentor into the specimen surface when raising the table

into position. This can strain the equipment and damage the

indentor.

� Operator (or student) training is crucial and

� Regular validation or calibration is essential if hardness rest

results are to be accurate and reproducible.

(material compiled from various sources)

True Stress and True Strain

Defn: True Stress= applied load divided by instantaneous area.

Defn: True Stress= applied load divided by instantaneous area.

0

lnl

lT =ε

i

TA

F=σ

)1( εσσ +=T)1ln( εε +=T

For some metals/alloys, true stress-strain relationship during plastic deformation

meaning after yielding and until necking begins, is approximated by:

n

TT K εσ ⋅=n=strain hardening exponent.

(0.12-0.5) See Table 6.4 page 171

NOW

What is the value of n for 2024-

T3 Al ?

• What is the equation

relating engr. Stress and

engr strain?

• An increase in σy due to plastic deformation.

• Curve fit to the stress-strain response –yield to necking :

σ

ε

large hardening

small hardening

un

loa

d

relo

ad

σy 0

σy 1

σT = C εT( )n

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

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

HARDENING

Elastic Recovery:

Return to original shape/size (slope)

New yield point

Who said it

works?



others 14

Table 6.4 : n and K values for some alloys (page-171)

n

TT K εσ ⋅=

True vs. Engineering Stress/Strain Curves

M� tensile strength

For Some metals and alloys,

Approximate true stress/strain curve between yield point and max

strength (M), is given by: n

TT Kεσ =

Table of n and K: see previous page

Class Exercise

I: True Stress 500 MPa (72,50 psi) � true strain 0.16 (unit?)

K=825 MPa (120,000 psi)

II: If true stress is 600 MPa ( 87,000 psi) � true strain = ?

n

TT K εσ ⋅=

Solution strategy:

1. For part I – calculate n. [take log, solve

for n.]

2. In part II use this value of n to calculate

true strain.



others 15

Another Example #6.47 page 212

• True Stress 60 ksi at true strain 0.15

• True Stress 70 ksi at true strain 0.25

• True stress = 65 ksi At true strain 0.20 – is it true? Why?

• Strategy:

– Set up 2 equations

– Sove for K and n.

– Use these K, n to calculate True stress at true strain 0.21.

• Design uncertainties mean we do not push the limit.

• Factor of safety, N

23

σworking =

σy

N

Often N is between

1.2 and 4, but depends

on application

• Ex: Calculate a diameter, d, to ensure that yield does

not occur in the 1045 carbon steel rod below. Use a

factor of safety of 5.

1045 plain carbon steel: σy=310MPa

TS=565MPa

F = 220,000N

d

Lo σworking =

σy

N

220,000N

π d2 /4

/5=310 MPa

DESIGN OR SAFETY FACTORS

d= ?

Key Terms this chapter

� Anelasticity

� Elastic limit

� Elastic recovery

� Engineering Stress VS True Stress

� Engineering Strain VS True Strain

� True stress True strain ..

� Hardness Test Types – Brinell, Rockwell, Micro(Vickers,

knoop), Superficial (rockwell), Durometer..(for ____)

� Rockwell Scales: A, B, C, ....many.

� Typical hardness for CR steels HRC 35-40 HR lower.

� Resilience – useful property for design of ___?