Mechanical Metallurgy : Response of metals to...

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Mechanical Metallurgy : Response of metals to forces or loads Mechanical assessment of Materials Forming of metals into useful shapes • Structural materials • Machine, aircraft, ship, car etc • Forging, rolling, extrusion, drawing, machining, etc We need to know limiting values of which materials in service can withstand without failure. What is failure? We need to know conditions of load and temperature to minimize the forces that are needed to deform metal without failure.

Transcript of Mechanical Metallurgy : Response of metals to...

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Mechanical Metallurgy : Response of metals to forces or loads

Mechanical assessment of Materials

Forming of metals intouseful shapes

• Structural materials• Machine, aircraft, ship, car etc

• Forging, rolling, extrusion,drawing, machining, etc

We need to know limiting values of which materials in service can withstand without failure.

What is failure?

We need to know conditionsof load and temperature tominimize the forces that areneeded to deform metalwithout failure.

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Strength of materialsStrength of materials deals with relationships between

• external loads which act on some part of a body (member) in equilibrium.

• internal resisting forces

• deformation

• In equilibrium condition, if there are external forces acting on the member,there will be internal forces resistingthe action of the external forces.

• The internal resisting forces are usually expressed by thestress acting over a certain area, so that the internal force is theintegral of the stress times the differential area over which it acts.

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Basic Assumptions

• Continuous:

• Homogeneous:

• Isotropic:

Microscopic scale, metals are made up of an aggregate

of crystal grains having different properties in different

crystallographic directions.

No voids or empty spaces.

Has identical properties at all points.

Has similar properties in all directions or orientation.

However, these crystal grains are very small, and

therefore the properties are homogenous in the

macroscopic scale.

Macroscopic scale, engineering materials such

as steel, cast iron, aluminum seems to be

continuous, homogeneous and isotropic.

micrometers

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Stress - Strain

True stress :

True strain :

(Average strain)

(Average stress, Pa, MPa)

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Measuring Young’s modulus: from linear section of stress-stain diagram

Elastic ModulusHooke’s Law : F = k x ; k: spring constant

E : modulus of elasticity or Young’s modulus (Pa, GPa)E is resistance of a material to elastic deformation. E is a material property (like density)

Also:

• Acoustic Method,

• Resonant Frequency Method

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Bonding Forces

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Elastic ModulusElastic modulus depends on the microstructure and interatomic bonding forces.

Metal [100] [110] [111] E

Aluminum 63.7 72.6 76.1 69

Copper 66.7 130.3 191.1 124

Iron 125.0 210.5 272.7 196

Modulus of Elasticity Values for Several Metals at

Various CrystallographicOrientations (GPa)

Copper

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Ceramics, glasses,: (GPa) Melting Temp. (C)

Diamond (C) 1000

Tungsten Carbide (WC) 450 -650 2870

Silicon Carbide (SiC) 450

Aluminum Oxide (Al2O3) 390 2072

Berylium Oxide (BeO) 380

Magnesium Oxide (MgO) 250

Zirconium Oxide (ZrO) 160 - 241

Mullite (Al6Si2O13) 145

Silicon (Si) 107

Silica glass (SiO2) 94

Soda-lime glass (Na2O - SiO2) 69

Metals:

Tungsten (W) 406 3400

Chromium (Cr) 289 1860

Berylium (Be) 200 - 289

Nickel (Ni) 214

Iron (Fe) 196 1536

Low Alloy Steels 200 - 207

Stainless Steels 190 - 200

Cast Irons 170 - 190

Copper (Cu) 124 1084

Titanium (Ti) 116

Brasses and Bronzes 103 – 124

Aluminum (Al) 69 660

Polymers:

Polyimides 3 - 5

Polyesters 1 - 5

Nylon 2 - 4

Polystryene 3 - 3.4

Polyethylene 0.2 -0.7

Rubbers 0.01-0.1

( Mechanical Behaviour of Engineering Materials:

Metals, Ceramics, Polymers, and Composites)

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MetalsAlloys

GraphiteCeramicsSemicond

Polymers Composites/fibers

E(GPa)

Based on data in Table B2,Callister 7e.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, Ni

Molybdenum

G raphite

Si crystal

Glass -soda

Concrete

Si nitrideAl oxide

PC

Wood( grain)

AFRE( fibers) *

CFRE *

GFRE*

Glass fibers only

Carbon fibers only

A ramid fibers only

Epoxy only

0.4

0.8

2

4

6

10

2 0

4 0

6 08 0

10 0

2 00

6 008 00

10 001200

4 00

Tin

Cu alloys

Tungsten

<100>

<111>

Si carbide

Diamond

PTF E

HDP E

LDPE

PP

Polyester

PSPET

C FRE( fibers) *

G FRE( fibers)*

G FRE(|| fibers)*

A FRE(|| fibers)*

C FRE(|| fibers)*

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Deformations Under Axial Loading

AE

P

EE

• From Hooke’s Law:

• From the definition of strain:

L

• With variations in loading, cross-section or material properties,

i

ii

ii

EA

LP

• Equating and solving for the deformation,

AE

PL

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Resonant Frequency Method

A

l

(Hooke’s Law)

(m: mass)

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Resonant Frequency Method

In our lab we use resonant frequency method in

flexure (bending) mod.

E = …

from ASTM E1876

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Resilience, Ur

• Ability of a material to store energy elastically

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

Adapted from Fig. 6.15, Callister 7e.

yyr2

1U @

y

dUr 0

Schematic representation showing how modulus of resilience (corresponding to the shaded area) is determined from thetensile stress-strain behavior of a material.

Ur : Energy per unit volume

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Resilience, Ur

Comparison of stress-strain curves for high and low resilient materials.

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A bar of a material with Young’s modulus, E, length, L, and cross sectional area, A, is subjected to an axial load, P. Derive an expression for strain energy stored in the bar assuming linear elastic deformation.

Energy stored per unit volume.Solution 1:

Multiply by volume, AxLReplace = / E

Solution 2: Energy stored in a spring:

Remember:

Notice = P / A

Example:

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Multi axial stress

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Multi axial stress

Notice : magnitude of stress is negative

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Multi axial stress

Pressurized tank(photo courtesyP.M. Anderson)

z > 0

q

> 0

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Multi axial stress

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Multi axial stress

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Multi axial stress

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Multi axial stress

=

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Plane Stress (z=tyz=txz=0)

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Multiaxial Stress-Strain

G: Shear modulus, Pa, GPa

Nominal tensile strain

Nominal lateral strain

Poisson’s ratio :

Engineering shear strain

for small strains

x

yz

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Poisson’s RatioThe value of Poisson’s ratio, n, can be evaluated for two extreme cases:

the initial and final volumes, V0 and V, are equal

V = V0 [(1 + ε x) (1 + ε y) (1 + ε z)]

Neglecting the cross products of the strains, because they are orders of magnitude smaller than the strains themselves

V = V0 [1 + ε x + ε y + ε z]

(1) when the volume remains constant

Since V=V0 ,

ε x + ε y + ε z= 0 For constant volume deformation.

z

x

y

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For the isotropic case, the two lateral contractions are equal (εx =εy ). Hence,

n = 0.5

2εx = - εz

Remember

For constant volume deformation.

(2) when there is no lateral contraction

Poisson’s Ratio

n = 0

Theoretical boundries are

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Multiaxial Stress-Strain

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Elastic Constants

Hydrostatic or mean stress

Bulk modulus

Volumetric strain

For an isotropic material two elastic constant are required and enough to define elastic behaviour

(D=DV/ V0 )

D =

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Copyright © 2011 Pearson Education South Asia Pte Ltd

Example:A bar made of A-36 steel (E=200 GPa, n=0.32) has the dimensions shown in Fig. 3–22. If an axial force of P = 80 kN is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. What is the change in volume? The material behaves elastically.

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Copyright © 2011 Pearson Education South Asia Pte Ltd

• The normal stress in the bar is

• Given that E = 200 GPa

• The axial elongation of the bar is therefore

Solution

mm/mm 108010200

100.16 6

9

6

E

z

z

(Ans) m120101205.11080 66

z mL

zz

MPa 16Pa 100.1605.01.0

1080 6

3

A

Pz

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Copyright © 2011 Pearson Education South Asia Pte Ltd

• The contraction strains in both the x and y directions are

• The changes in the dimensions of the cross section are

Solution

(Ans) m28.105.0106.25

(Ans) m56.21.0106.25

6

6

yyy

xxx

L

L

(Ans) 106.25- 1080)32.0( - 66 zyx

v

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Copyright © 2011 Pearson Education South Asia Pte Ltd

Solution

To find the change in volume

V = V0 [1 + ε x + ε y + ε z]

zyx

Vo

VoV

66 1028.8 10)6.256.2580(

Vo

VoV

356 1016.21028.8 .)05.0.1.0.5.1( mVoV

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Example: Elastik modulusu E, Poisson oranı n, akma mukavemeti Sy olan aluminyumdan yapılmış L x W x t (Uzunluk X Genişlik X Kalınlık) boyutlarında bir plaka uzunluk yönünde P yükünün etkisi altındadır, genişlik yönünde ise iki ucundan ankstredir (genişlik sabit). Bu koşullar altında uzunluğundaki değişmeyi ve kalınlığındaki azalmayı hesaplayınız.

x

yPP

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A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses x = 12 ksi and z

= 20 ksi.

For E = 10x106 psi and n = 1/3, determine the change in:

a) the length of diameter AB,

b) the length of diameter CD,

c) the thickness of the plate, and

d) the volume of the plate.

Example:

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SOLUTION:

• Apply the generalized Hooke’s Law to find the three components of normal strain.

in./in.10600.1

in./in.10067.1

in./in.10533.0

ksi203

10ksi12

psi1010

1

3

3

3

6

EEE

EEE

EEE

zyxz

zyxy

zyxx

nn

nn

nn

• Find the change in volume

33

333

in75.0151510067.1

/inin10067.1

. DD

D

VV

zyx

3in187.0DV

• Evaluate the deformation components.

in.9in./in.10533.0 3 dxAB

in.9in./in.10600.1 3 dzDC

in.75.0in./in.10067.1 3 tyt

in.108.4 3AB

in.104.14 3DC

in.10800.0 3t