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8/12/2019 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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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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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
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?
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