eng1050lecture4 s1 2014

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www.monash.edu.au Department of Materials Engineering ENG1050/MCD4220 Engineering Materials Lecture 4: Measuring material properties

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Page 1: ENG1050Lecture4 S1 2014

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www.monash.edu.au

Department of Materials Engineering

ENG1050/MCD4220 Engineering Materials

Lecture 4: Measuring material

properties

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Department of Materials Engineering

Objectives

1.  Appreciate the influence of atomic structure, bonding and nano/microstructures have on some physical properties;

2. Have an understanding of different materials responses to forces and stresses

3. Have an understanding of the basic mechanical properties, principally elastic modulus and yield stress, and be ab

to use these as design criteria

4. Be familiar with processes occurring during plastic deformation and to draw upon these concepts in order to knowhow to strengthen the material

5. Know how to tailor the mechanical properties of a polymeric material using control over crystallinity and the glass transition

6. Understand the role of composite materials in engineering, and their responses to applied stresses

7. Understand the processes involved during fracture and have a broad understanding of how fracture can be avoided byappropriate selection of materials and design

8. Have a basic understanding of the thermal, electrical and magnetic properties of materials in terms of the atomic and electrcharacteristics of materials and to use these criteria for material selection

9. Understand the processes of corrosion and degradation in the environment and to draw upon these to increase the lifetimethrough appropriate protection and material selection

10. Be able to select an appropriate material for a given application based on the above points

11. Appreciate the socio-political and sustainability issues influencing material selection, commonly experienced as aprofessional engineer

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Department of Materials Engineering

Typical assessment

On completion you should be able to

• Calculate elastic modulus, yield strength, proof stress,tensile strength and ductility

• Define work hardening

• Describe qualitatively and quantitatively the stress-strain

behaviour of a metal (including work hardening)

• Define onset of necking and estimate the necking strain

• Define uniform elongation

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Department of Materials Engineering

 Nominal Stress (N/mm2 or MPa)

 Nominal Strain

Yield stress

 Plastic region

Stress - strain graph (nominal/engineering)

• The stress - strain graph is

independent of dimensions

• The yield stress is the stress

beyond which the original

dimensions of the material are

not recovered on unloading.

ExamplesContext Detail Application

Units:

Stress - 1 Pascal = 1 Pa = 1 N/m2)

1 MPa = 1 x 106 Pa = 1 x 106 N/m2 

= 1 N /10-6 m2 = 1 N/mm2 

Strain has no units (dimensionless quan

 Elastic region

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Department of Materials Engineering

What properties are we interested in?

 Nominal stress (n)

 Nominal strain (n)

Slope =Young’s modulus (E) = / 

YS

ExamplesContext Detail Application

 = E - Hooke’s Law (valid fo

linear Elastic region)

E (= /) –  Young’s Modulus (E

Modulus)

YS  –  Yield Stress (Yield Streng

Higher Elastic Modulus

Lower Elastic Modulus

R G

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Department of Materials Engineering

What properties are we interested in?

Plastic strain after fracture (Ductility)

TS

YS

ExamplesContext Detail Application

 = E - Hooke’s Law (valid fo

linear Elastic region)

E (= /) –  Young’s Modulus (E

Modulus)

YS  –  Yield Stress (Yield Streng

TS  –  Tensile strength (also call

Ultimate Tensile Strength - UT

Elastic strain

recovered

Fracture point

 Nominal stress (n)

 Nominal strain (n)

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Department of Materials Engineering

What properties are we interested in?

0.2% strain

(0.002)

PS

TS

YS

ExamplesContext Detail Application

 = E - Hooke’s Law (valid fo

linear Elastic region)

E (= /) –  Young’s Modulus (E

Modulus)

YS  –  Yield Stress (Yield Streng

TS  –  Tensile strength (also call

Ultimate Tensile Strength - UT

PS –  Proof Stress (also called o

yield strength)(0.2 for 0.2% strain)

 Nominal stress (n)

 Nominal strain (n

)

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Department of Materials Engineering

What properties are we interested in?

ExamplesContext Detail Application

 Nominal stress

 Nominal strain

Slope represents

the work

hardening rate

(d/d, units MPa)

TS

YS

Uniform elongation

Uniform strain

Fracture strain

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Department of Materials Engineering

Yield strength (elastic limit) of various materials

ExamplesContext Detail Application

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Department of Materials Engineering

ExamplesContext Detail Application

Elongation at fracture of various materials

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Department of Materials Engineering

0

50

100

150

200

250

300

0 0.05 0.1 0.15 0.2 0.25 0.3

Strain

 

ExamplesContext Detail Application

Example question

• Calculate the 0.2% pr

stress, and ultimatetensile strength for th

alloy.

   S   t  r  e  s  s   (   N   /  m  m   2   o

  r   M   P  a   )

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Department of Materials Engineering

0

50

100

150

200

250

300

0 0.05 0.1 0.15 0.2 0.25 0.3

Strain

 

ExamplesContext Detail Application

0.002 (0.2%)

140MPa

Proof stress = 140M

240MPa

UTS = 240MPa

Example question

• Calculate the 0.2% pr

stress, and ultimate

tensile strength for th

alloy.

   S   t  r  e  s  s   (   N   /  m  m   2   o

  r   M   P  a   )

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Department of Materials Engineering

How does the stress-strain curve relate to can manufac

• Recall that

 –for the body we require a metal that is strong but will deformsufficiently in order to shape the can,

 –for the end we want a metal that is strong, but will open on

demand

• How do we translate these requirements into material

properties?

ExamplesContext Detail Application

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Department of Materials Engineering

Can manufacture

• When we talk about a ‘strong’ can body, which property are

we referring to?

Yield/Proof Stress

• When we talk about ‘deforming sufficiently’, what feature of

the stress-strain curve are we referring to?

Uniform elongation

• When we talk about ‘opening on demand’, what features of

the stress-strain curve is this related to?

Ductility

ExamplesContext Detail Application

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Department of Materials Engineering

Stress-strain curve (True versus Engineering)

• Since volume is constant,

L0 A0 = LA

 A = (L0A0 /L)

• True stress, t = F/A = F/(L0A0 /L)

= FL/L0 A0 = (F/A0)(L/

= n ((L0 + L)/L0)

t = n (1 + n) > n 

• Earlier we found that

t = ln(1+n) < n 

•  Actual (true) stress keeps increasing

final failure

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Department of Materials Engineering

• The true stress – true strain curve ca

be characterised by the relation

  t  = A(  t  )n

• The constant, n , which is characteris

of the material, is called the work

hardening exponent

• Work hardening may be described a

increase in yield stress ( Y  ) due toplastic deformation – it is a measure

storage of plastic energy (higher wor

hardening exponent, n , means great

ability to store plastic energy)

Work hardening

ExamplesContext Detail  Application

d   t 

d   t 

True Stress - True Strain

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Department of Materials Engineering

Work hardening

y2

y1

y0

t1

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Department of Materials Engineering

Ductile vs brittle materials

ExamplesContext Detail Application

• Ductility may be expressed

quantitatively as either percentag

elongation or percentage reductio

area.

• Toughness may be defined as th

amount of work done (per unit

volume) in breaking the material.

•  Area under the stress-strain curv

a measure of toughness.

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Department of Materials Engineering

Necking

• Concentration of deformation in a

small region, leading to the

formation of a “neck”

• Occurs when the load-carrying

ability of the metal is no longer

able to support the stress

increase due to the reduction incross-sectional area

• To understand the phenomenon

we must think in terms of true

stress

Fractu

occur

the ne

forme

Deformation

Concentrated

  Uniform    at the  neck

Localized deformation of a ductile material during

a tensile test produces a necked region.

n

n

M

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Department of Materials Engineering

Neck formation

Microscopic phenomenon

• As strain increases, material workhardens

• As work hardening proceeds the

material is able to support a highe

stress

Macroscopic phenomenon

•Specimen has constant volume

•As length increases, cross-

sectional area decreases

•As area decreases, stress

increases

There is a feedback loop between the micro and the macro phenomena

When the increase in stress overtakes the increase in strength, plastic

instability and necking occurs

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Department of Materials Engineering

Onset of necking (Mathematically...)

At the onset of necking (highest point on Stress-strain curve) dF  = 0 ….

From the definition of stress  F  =   t A

  dF  =   t dA + A d   t  = 0

  - (dA/A) = (d  t  /   

t ) ….

Volume does not change. Therefore  AL = constant

  AdL + LdA = 0 ….

  - (dA/A) = (dL/L) = d  t   ….

From Eqn 2 and Eqn 4, (d  t  /   

t ) = d  

t  

  (d   t  /d  t ) = t   …..E

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Department of Materials Engineering

• On the nominal stress-strain curv

necking commences at the highes

• On the true stress-true strain curv

necking occurs when:

Onset of necking (Graphically…) 

ExamplesContext Detail  Application

 Nominal stress

 Nominal strain

TS

Note that the units of the two curves are the same. 

Gradient of the true 

stress - strain curve

True  stress - straincurve

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Department of Materials Engineering

Est imate   of necking strain

We know that the plastic part of the curve can be

described by:

ExamplesContext Detail Application

At necking

We also know that at necking:

  t  = A(  t  )n 

Consequently =nA(  t  )n-1 

Therefore nA(  t  )n-1 = A(  t  )

n = (  

t  )

n

 /(  

t  )

n-1

 = 

t  

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Department of Materials Engineering

Definitions• Elastic modulus; Young’s modulus: 

 –The slope of the elastic (linear) region of the stress-strain curve.

• Proof stress (offfset yield strength), PS

 or 0.2 (units MPa):

 –The stress measured at the intersection of the engineering stress-strain cuand a line drawn parallel to the elastic portion of the curve, offset by a stra0.2% (or 0.002)

• Tensile strength; ultimate tensile strength (units MPa):

 –The maximum stress measured on the engineering stress strain curve

• Work hardening rate (d/d, units MPa) –The rate of change of stress with strain in the plastic region of the stress s

curve.

 –Strictly speaking, measured only on the true stress-strain curve.

ExamplesContext Detail Application

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Department of Materials Engineering

• Ductility

 –The permanent (plastic) strain after fracture, measured by reassemblingthe fractured  specimen or by construction on the stress-strain curve of a

line from the point of fracture parallel to the elastic region until it intersectswith the strain axis (zero stress).

• Fracture strain

 –The sum of elastic and plastic strain at the instant before fracture

 – Always greater than ductility (compare the definitions).

• Uniform elongation

 –The strain that can be achieved before necking (ie, with uniform thinningof a sample).

 –On an engineering stress-strain curve this is coincident with the strain atwhich the ultimate tensile strength is found.

ExamplesContext Detail Application

Definitions

S

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Department of Materials Engineering

Summary

• Measuring mechanical properties

• Definitions of mechanical properties

• Onset of plastic instability form a mathematical point of view

- why does this happen?

• Uniform elongation

• Work hardening

• Use of mechanical properties to sort material suitability for

purpose

R di d t d ti

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Department of Materials Engineering

Reading and study questions

•  Read Callister excerpts (pdf document online): Chapter 7

(pages 186 – 211), including example problems 7.1, 7.2,

7.3, 7.4, 7.5

• http://aluminium.matter.org.uk 

f

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Department of Materials Engineering

Determine the mechanical properties of this material

R l t ti

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Department of Materials Engineering

Relevant properties

• Elastic modulus

• 0.2% Proof stress (yield strength)

• Tensile strength

• Uniform elongation

• Work hardening exponentHint: work hardening exponent is equal to true strain at the onset of

necking

• Ductility

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• Two copper specimens with the same original dimensions have been plasticalldeformed under different tensile stresses. The figures below show their newdimensions after stretching. If these two specimens are tested under tensilestresses again, which one is expected to yield at a lower stress? Briefly explaiyour reason.

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Rapid feedback

• What aspect of the unit so far do you understand?

• What aspect do you need assistance with?