mechanical property of metal

7/28/2019 mechanical property of metal

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Chapter 3: Mechanical Properties of Materials

Why mechanical properties?Why mechanical properties?Why mechanical properties?Why mechanical properties?

Need to design materials that will withstand

applied load and in-service uses for

Bridges for autos and people MEMS devices

1

Space elevator?

skyscrapers

Space exploration


2/55

Chapter 7: Mechanical Properties

ISSUES TO ADDRESS...

Stress and strain: Normalized force and displacements.

Elastic behavior: When loads are small.

Engineering: e= F

i/A

0e= l / l

0

True: T= F

i/A

iT= ln(l

f/ l

0)

'

Chapter 3: Mechanical Properties of Materials

2

Plastic behavior: dislocations andpermanent deformation

Toughness, ductility, resilience, toughness, and hardness:

Define and how do we measure?

Mechanical behavior of the various classes of materials.

Yield Strength: YS

[MPa] (permanent deformation)

UlitmateTensile Strength: TS

[MPa] (fracture)


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Tensile specimen Tensile test machine

Stress & strain Testing

Often 12.8 mm x 60 mm

specimenextensometerAdapted from Fig. 7.2,

Callister & Rethwisch 3e.

4

Other types:-compression: brittle materials (e.g., concrete)

-torsion: cylindrical tubes, shafts.

gauge

length


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Tensile stress, s: Shear stress, t:

Area, A

Ft

Area, A

Ft

Fs

F

Engineering Stress

5

Ft =

FtA

ooriginal areabefore loading

FtF

Fs

=Fs

Ao

Stress has units: N/m2 (or lb/in2 )


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Tensile strain: Lateral (width) strain:

/2

/2

Lowo =

Lo L =L

wo

Engineering Strain

6

Shear strain:/2

/2

/2 -

/2

= tan Strain is always

dimensionless.


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Simple tension: cable

=F

Ao = cross sectionalArea (when unloaded)

FF

Common States of Stress

7

o Simple shear: drive shaft

o =

Fs

A

Note: = M/AcR here.

Ski lift (photo courtesy P.M. Anderson)

M

M Ao

2R

Fs

Ac


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Simple compression:

Ao


8

Canyon Bridge, Los Alamos, NM

Balanced Rock, ArchesNational Park o =

F

A

Note: compressive

structural member ( < 0).

(photo courtesy P.M. Anderson)

(photo courtesy P.M. Anderson)


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Bi-axial tension: Hydrostatic compression:


9

Fish under waterPressurized tank

z > 0

> 0

< 0h

(photo courtesy

P.M. Anderson)

(photo courtesy

P.M. Anderson)


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Pure Tension Pure Compression

e=Fnormal

Ao

e=

l lo

lo

stress

strain

e=EElastic

response

10

Pure Shear

Pure Torsional Shear

e=Fshear

Ao

= tan

stress

strain

e=GElastic

response


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Elastic Deformation

2. Small load

bonds

stretch

1. Initial 3. Unload

return to

initial

11

Elastic means reversible!

F

Linear-

elastic

Non-Linear-

elastic

F


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Plastic Deformation of Metals

1. Initial 2. Small load 3. Unload

planes

still

sheared

bonds

stretch

& planesshear

12

Plastic means permanent!

F

elastic + plastic plastic

elastic

F

linear

elasticlinear

elastic

plastic


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

Hooke's Law: = E

Units: E [GPa] or [psi]

Linear Elasticity

FAxial strain

13

Linear-elastic

E

FWidth strain


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Tangent Modulus is experienced in service.

Secant Modulus is effective modulus at 2% strain.- grey cast iron is also an example

Modulus of polymer changes with time and strain-rate.- must report strain-rate d/dt for polymers.

- must report fracture strain fbefore fracture.

Polymers: Tangent and Secant Modulus

14

%strain

Stress (MPa)initial E

secant E

1 2 3 4 5 ..

tangent E


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Hooke's Law: = E (linear elastic behavior)

Copper sample (305 mm long) is pulled in tension with stress of276

MPa. If deformation is elastic, what is elongation?

Example: Hookes Law

For Cu (polycrystalline), E = 110 GPa. FAxial strain

15

=E=E ll

0

l =

l 0E

l =(276MPa)(305mm)

110x10

3

MPa

= 0.77mm

Hookes law involves axial (parallel to applied tensile load) elastic deformation.

FWidth strain


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Mechanical Properties

Recall: Bonding Energy vs distance plots

16

Adapted from Fig. 2.8


tension

compression


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Elastic Behavior

Elasticity of Ceramics

Al O

And Effects of PorosityE= E0(1 - 1.9P + 0.9 P

2)

17

Neither Glass or Alumina experience plastic deformation before fracture!


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Comparison of Elastic Moduli

18

Silicon (single xtal) 120-190 (depends on crystallographic direction)

Glass (pyrex) 70

SiC (fused or sintered) 207-483

Graphite (molded) ~12

High modulus C-fiber 400Carbon Nanotubes ~1000


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Magnesium,

Aluminum

Platinum

Silver, Gold

Tantalum

Zinc, Ti

Steel, Ni

Molybdenum

Si crystal

Glass-soda

Si nitrideAl oxide

Glass fibers only

Carbon fibers only

Aramid fibers only

6080

100

200

600800

10001200

400

Cu alloys

Tungsten

Si carbide

Diamond

*

AFRE(|| fibers)*

CFRE(|| fibers)*

Metals

Alloys

GraphiteCeramics

Semicond

PolymersComposites

/fibers

E(GPa)

Eceramics

> Emetals>> Epolymers

Youngs Modulus, E

19

0.2

8

0.6

1

Graphite

Concrete

PC

Wood( grain)

AFRE( fibers)*

CFRE*

GFRE*

Epoxy only

0.4

0.8

2

4

6

10

20

40 n

PTFE

HDPE

LDPE

PP

Polyester

PSPET

CFRE( fibers)*

GFRE( fibers)*

109 Pa

Based on data in Table B2, Callister 6e.

Composite data based on

reinforced epoxy with 60 vol%

of aligned carbon (CFRE),aramid (AFRE), or glass (GFRE) fibers.


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Poisson's ratio, :

Poisson's ratio,

=

width strain

axial strain =

w /w

l /l =

L

Units: dimensionless

F

L

20

Why does have minus sign?

metals: ~ 0.33

ceramics: ~ 0.25polymers: ~ 0.40

F

Axial strain

Width strain

-


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Limits of the Poisson Ratio

= w/w

l / l= 1

Poisson Ratio has a range 1 1/2

Look at extremes

No change in aspect ratio: w/w = l/l

21

Volume (V = AL) remains constant: V =0.

Hence, V = (L A+A L) = 0. So,

In terms of width, A = w2, then A/A = 2 w w/w2 = 2w/w = L/L.

Hence,

A/A = L/L

= w/w

l / l=

(1

2l / l )

l / l=

1

2

Incompressible solid.

Water (almost).


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Tensile stress is applied along cylindrical brass rod (10 mm diameter).

Poisson ratio is = 0.34 and E = 97 GPa.

Determine load needed for 2.5x103 mm change in diameter if the

deformation is entirely elastic?

F

Example: Poisson Effect

Width strain: (note reduction in diameter)

23

simpletensiontest

x= = . x mm mm = . x

Axial strain: Given Poisson ratio

z= x/ = (2.5x104)/0.34 = +7.35x104

Axial Stress: z = Ez = (97x103 MPa)(7.35x104) = 71.3 MPa.

Required Load: F = zA0 = (71.3 MPa) (5 mm)2 = 5600 N.


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Elastic Shear

modulus, G:

1

G

= G

M

M

simple

Torsion test

Other Elastic Properties

24

modulus, K:

P= -KVVo

V

1-K

Vo

P P

Pressure test:

Init. vol = Vo.

Vol chg. = V


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Simple tension test:

engineering stress,

Elastic+Plastic

at larger stress

Elastic

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

Plastic (Permanent) Deformation

25

engineering strain, p

plastic strain

initially

Adapted from Fig. 7.10 (a),


permanent (plastic)

after load is removed


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Stress where noticeable plastic deformation occurs.

When p = 0.002

Yield Stress, Y

For metals agreed upon 0.2%

P is theproportional limitwhere deviation

from linear behavior occurs.

-

tensile stress,

Y

26

Note: for 2 in. sample

= 0.002 = z/z

z = 0.004 in

Start at 0.2% strain (for most metals). Draw line parallel to elastic curve (slope of E).

Y is value of stress where dotted line crosses

stress-strain curve (dashed line).

Eng. strain, p = 0.002

Elastic

recovery

P


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Yield-point phenomenon occurs when elastic plastic

transition is abrupt.

Yield Points and YS

No offset method required.

In steels, this effect is seen when

dislocations start to move and unbind

27

For steels, take the avg. stress

of lower yield point since less

sensitive to testing methods.

.

Lower yield point taken as Y.

Jagged curve at lower yield point occurs

when solute binds dislocation and

dislocation unbinding again, until work-hardening begins to occur.


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3 different types of behavior

Stress-Strain in Polymers

Brittle

plastic

For plastic polymers: YS at maximum stress just

after elastic region.

TS is stress at fracture!

28

Highly elastic

Highly elastic polymers:

Elongate to as much as 1000% (e.g. silly putty).

7 MPa < E < 4 GPa 3 order of magnitude!

TS(max) ~ 100 MPa some metal alloys up to 4 GPa


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Graphite/

Ceramics/Semicond

Metals/

Alloys

Composites/

fibersPolymers

,y(MP

a)

ure,

occursbefore

yield.

300

400

500600700

1000

2000

Al (6061)ag

hr

Steel (1020)cdSteel (4140)a

Steel (4140)qt

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

Mo (pure)Cu (71500)cw

ure,

trixcomposite

s,since

ccursbeforey

ield.

y(ceramics)

>>y(metals)

>> y(polymers)

Compare Yield Stress, YS

29

Yield

strength

PVC

Hardtome

as

sinceinte

nsion,fractureusuall

Nylon 6,6

LDPE

70

20

40

6050

100

10

30

Tin (pure)

Al (6061)a

Cu (71500)hrTa (pure)

Ti (pure)a

Hardtom

ea

inceramicmatrixandepoxym

intension,fractureusua

lly

HDPEPP

humid

dry

PC

PET

oom va ues

Based on data in Table B4,

Callister 6e.

a = annealed

hr = hot rolled

ag = aged

cd = cold drawncw = cold worked

qt = quenched & tempered


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Maximum possible engineering stress in tension.

(Ultimate) Tensile Strength, TS

y F= fracture orultimate

strength

eering

TS

ress

30

Metals: occurs when necking starts.

Ceramics: occurs when crack propagation starts.

Polymers: occurs when polymer backbones are

aligned and about to break.

strain

Typical response of a metalNeck acts

as stress

concentratorengi

s

engineering strain


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Metals: Tensile Strength, vTS

For Metals: max. stress in tension when necking starts, which

is the metals work-hardening tendencies vis--vis those that

initiate instabilities.

dF= 0 Maximum eng. Stress (at necking)

dF= 0 = dA +Ad

dT =

dAi

31

decreased force due to

decrease in gage diameter

Increased force due to

increase in applied stress

At the point where these two competing changes in forceequal, there is permanent neck.

Determined by slope of true stress - true strain curve

T iFractionalIncrease in

Flow stress

fractionaldecrease

in load-

bearing

area

dT

T

= dA

i

Ai

=dl

i

li

dT

d

T

dT

=T

IfT=K(

T)n, then n=

T

n = strain-hardening coefficient


32/55

Room T values

TS(ceram)

~TS(met)

~ TS(comp)

>> TS(poly)

Compare Tensile Strength, TSGraphite/

Ceramics/Semicond

Metals/

Alloys

Composites/

fibersPolymers

TS

(M

Pa)

200

300

1000

Al (6061) ag

Cu (71500) hr

Ta (pure)Ti (pure) a

Steel (1020)

Steel (4140) a

Steel (4140) qt

Ti (5Al-2.5Sn) a

W (pure)

Cu (71500) cw

2000

3000

5000

Al oxide

Diamond

Si nitride

GFRE (|| fiber)

C FRE (|| fiber)

A FRE (|| fiber)

E-glass fib

C fibersAramid fib

32

Based on data in Table B4,


Si crystal

Tensile

strength

,

PVC

Nylon 6,6

10

100Al (6061) a

LDPE

PP

PC PET

20

3040

Graphite

Concrete

Glass-soda

HDPE

wood ( fiber)

wood(|| fiber)

1

GFRE ( fiber)C FRE ( fiber)A FRE( fiber)


33/55

Example for Metals: Determine E, YS, and TS

Stress-Strain for Brass Youngs Modulus, E (bond stretch)

GPaMPa

E 8.9300016.0

)0150(

12

12 =

=

=

0ffset Yield-Stress, YS (plastic deformation)

YS= 250 MPa

33

Max. Load from Tensile Strength TS

Fmax =TSA0 =TSd0

2

2

= 450MPa12.8x103m

2

2

= 57,900N

Change in length atPoint A, l = l0

Gage is 250 mm (10 in) in length and 12.8 mm

(0.505 in) in diameter.

Subject to tensile stress of 345 MPa (50 ksi)

l = l0 = (0.06)250 mm = 15 mm


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Most metals are ductile at RT and above, but can become brittle at low T

bcc Fe

Temperature matters (see Failure)

34

cup-and-cone fracture in Al brittle fracture in mild steel


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36/55

Energy to break a unit volume of material,

or absorb energy to fracture.

Approximate as area under the stress-strain curve.

Toughness

E ngineering

tensile

small toughness (ceramics)

large toughness (metals)

36

very small toughness

(unreinforced polymers)

Engineering tensile strain,

,

UT= do

f

Brittle fracture: elastic energy

Ductile fracture: elastic + plastic energy


37/55


brittle polymer

plastic

elastomer

37

elastic moduli

less than for metals Adapted from Fig. 7.22,Callister & Rethwisch 3e.

Fracture strengths of polymers ~ 10% of those for metals.

Deformation strains for polymers > 1000%.

for most metals, deformation strains < 10%.


38/55

Decreasing T...

-- increases E

-- increases TS

-- decreases %EL

Increasing

Influence ofTand Strain Rate on Thermoplastics

40

60

80 4C

20C

(MPa)

Plots for

semicrystalline

PMMA (Plexiglas)

38

strain rate...-- same effectsas decreasing T.

Adapted from Fig. 7.24, Callister & Rethwisch 3e. (Fig. 7.24 is from T.S. Carswell

and J.K. Nason, 'Effect of Environmental Conditions on the Mechanical Properties

of Organic Plastics", Symposium on Plastics, American Society for Testing and

Materials, Philadelphia, PA, 1944.)

20

00 0.1 0.2 0.3

60Cto 1.3


39/55

Necking appears along entire

sample after YS!


Mechanism unlike metals, necking due

to alignment of crystallites.

Load vertical

39

Align crystalline sections by

straightening chains in the

amorphous sections

After YS, necking

proceeds by

unraveling; hence,

neck propagates,

unlike in metals!See Chpt 8


40/55

Time-dependent deformation in Polymers

Stress relaxation test:

-strain in tension to and hold.

- observe decrease instress with time.

tensile test

Large decrease in Erfor T> Tg.

(amorphous

polystyrene)

Fig. 7.28, Callister &

Rethwisch 3e.

(Fig. 7.28 from A.V.Tobolsky, Properties

and Structures of

Polymers, Wiley and

Sons, Inc., 1960.)

103

101

10-1

105

rigid solid

(small relax)

transitionregion

Er(10 s)

in MPa

viscous liquid

40

Representative Tg values (in C):

PE (low density)

PE (high density)PVC

PS

PC

-110

- 90+ 87

+100

+150Selected values from Table 11.3, Callister & Rethwisch 3e.

o

r

ttE

=

)()(

Relaxation modulus:

time

straino

(t)

10-

60 100 140 180 T(C)Tg

(large relax)


41/55

Wh T S i ?


42/55

Why use True Strain?

Up to YS, there is volume change due to Poisson Effect!

In a metal, from YS and TS, there is plastic deformation, as dislocations move

atoms by slip, but V=0 (volume is constant).

A0l 0 = Ail i

t= ln

li

l0

lnli l

0+ l

0

l0

= ln(1+)

42

Sum of incremental strain does

NOT equal total strain!

Test length Eng. Eng.

0-1-2-3 0-3

0 2.00

1 2.20 0.1

2 2.42 0.1

3 2.662 0.1 0.662/2.0

TOTAL 0.3 0.331

t= ln

2.2

2.0+ ln

2.42

2.20+ ln

2.662

2.42= ln

2.662

2.00

Eng.

Strain

True

StrainSum of incremental strain does

equal total strain.


43/55

Example:

A cylindrical specimen of steel having an original diameter of 12.8

mm (0.505 in.) is tensile tested to fracture and found to have an

engineering fracture strength of 460 MPa (67,000 psi). If its cross-

sectional diameter at fracture is 10.7 mm (0.422 in.), determine:

(a) The ductility in terms of percent reduction in area(b) The true stress at fracture


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Hardness


45/55

Resistance to permanently indenting the surface.

Large hardness means:--resistance to plastic deformation or cracking in compression.

--better wear properties.

e.g.,apply known force(1 to 1000g)

measure sizeof indent after

Hardness

45

mm sp ere

removing load

dDSmaller indentsmean largerhardness.

increasing hardness

mostplastics

brassesAl alloys

easy to machinesteels file hard

cuttingtools

nitridedsteels diamond

Adapted from Fig. 7.18.


46/55

Hardness: Measurement

Rockwell

No major sample damage

Each scale runs to 130 (useful in range 20-100). Minor load 10 kg

Major load 60 (A), 100 (B) & 150 (C) kg

46

A = diamond, B = 1/16 in. ball, C = diamond

HB = Brinell Hardness

TS (psia) = 500 x HB

TS (MPa) = 3.45 x HB

H d M t


47/55

Hardness: Measurement

47

F ti


48/55

Under fluctuating / cyclic stresses, failure can occur atconsiderably lower loads than tensile or yield strengths of

material under a static load.Causes 90% of all failures of metallic structures (bridges,aircraft, machine components, etc.)

Fatigue

Failure under fluctuating stress

48

Fatigue failure is brittle-like (relatively little plastic deformation)

- even in normally ductile materials. Thus sudden and

catastrophic!

Fatigue failure has three stages:

crack initiation in areas of stress concentration

(near stress raisers)

incremental crack propagation

catastrophic failure

F i C li S


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Fatigue: Cyclic StressesCyclic stresses characterized by maximum, minimum and mean stress, the range of stress, the

stress amplitude, and the stress ratio

Mean stress m = (max + min) / 2

Range of stress r = (max - min)

Stress amplitude a = r/2 = (max - min) / 2

Stress ratio R = min / max

49

Convention: tensile stresses positive

compressive stresses negative

Creep


50/55

pTime-dependent and permanent

deformation of materials subjected to a constant load at hightemperature (> 0.4 Tm).

Examples: turbine blades, steam generators.

Creep testing:

50 Creep testing

Furnace

Stages of creep


51/55

S ages o c eep

51

1. Instantaneous deformation, mainly elastic.

2. Primary/transient creep. Slope of strain vs. time decreases with time:

work-hardening

3. Secondary/steady-state creep. Rate of straining constant: work-hardening

and recovery.

4. Tertiary. Rapidly accelerating strain rate up to failure: formation of internal

cracks, voids, grain boundary separation, necking, etc.


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Mechanisms of Creep


53/55

Mechanisms of Creep

Different mechanisms act in different materials and underdifferent loading and temperature conditions:

Stress-assisted vacancy diffusion

Grain boundary diffusion

Grain boundary sliding

Dislocation motion

53

Different mechanisms different n, Qc.

Grain boundary diffusion Dislocation glide and climb

Typical Forming Operations


54/55

Typical Forming Operations

rollingforging

54

Wire drawing extrusion Deep drawing

die

Stretch formingbending

shearing

Summary


55/55

Stress and strain: These are size-independent measures of load and

displacement, respectively.

Elastic behavior: This reversible behavior often shows a linear

relation between stress and strain.

To minimize deformation, select a material with a large elastic

modulus (E or G).

Summary

55

Plastic behavior: This permanent deformation

behavior occurs when the tensile (or compressive)

uniaxial stress reaches sy.

Toughness: The energy needed to break a unit

volume of material. Ductility: The plastic strain at failure.

mechanical property of metal

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