ENSC 1004 Engineering Materials
Student feedback is instrumental to the progressive
improvement of this unit. So I encourage you to do the
SELT (Student Experience of Learning and Teaching)
for this unit.
Be critical but constructive, tell us what has worked
well and what needs improvement!
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Associate Professor Ali Karrech
Room: ENCM 1.40
Phone: 6488 3523, email: ali.karrech@uwa.edu.au
Materials were prepared by Professor Hong Yang
Important Notices
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The final exam will be on Tuesday 14 June 2022 at 9am for
regular students.
You will find it available online (on LMS). This exam will not
be invigilated.
3ENSC 1004 Engineering Materials
Final Revision
W12 – L24 Revision Part I
W12 – L25 Revision Part II
T3: Thermal properties
• Heat capacity
• Thermal expansion
• Thermal conductivity
• Thermal stresses
Key topic areas in Chapters 4-7:
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T1: Engineering materials in three basic types
• Basic classifications dictated by intrinsic bonding types
• General physical, mechanical and deteriorative properties resulted from bonding nature
• Typical mechanical behaviour of three basic types
• Typical deteriorative behaviour of three basic types
• Typical applications of three basic types, more specifically into engineering alloys
• Life cycle of materials – sustainable utilization and environmental/societal impacts
T2: Mechanical deformation and failure of materials
• Foundational knowledge: terminology, concept, tensile deformation behaviour, tensile
mechanical properties, bonding nature and crystal structure basis. (assume established)
• Failure by plastic yielding – design by yield strength
• Failure by brittle fracture with existing cracks – design by fracture toughness
• Failure by fatigue under cyclic loading – design by fatigue life
T1: Engineering materials in three basic types
You should be able to extend and add depth to what was summarized in mid-
semester exam revision to provide an overall view on typical behaviour and properties
of three basic types of materials.
• It may be easier to list in a table form
• It may be easier to compare and contrasting – be clear of your references
• It is essential that you are using appropriate terminology to describe
• It is essential you are able to use clearly labelled schematic diagram to demonstrate
• It is important you establish clarity in magnitudes when comparing properties
• It is expected that you are able to select right materials for intended applications
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Metals (and alloys)
• steels & cast irons
• nonferrous – Cu, Al, Ti &
refractory
Ceramics
• crystalline ceramics
• glasses
Polymers
• thermoplastics
• thermosets
• elastomers
Bonding Type Entirely primary:
metallic + covalent
Entirely primary:
ionic + covalent
Primary + Secondary:
• covalent within molecules
• secondary between molecules
Bonding Energy
(Strength)
Moderate High Low between molecules, although
high within molecules
Structures • mostly highly crystalline
• BCC, FCC and HCP
• containing dislocations
• mostly crystalline except
glasses
• containing coiled long chains
which entangle
• semicrystalline to amorphous
Melting temperature Moderate High Low
Electrical conductivity High Low Low
Thermal conductivity High Low Low
Chemical activity High Low Low
Young’s modulus
• Perhaps demonstrate by − curve
• Ensure you are aware of the magnitude comparison
• Link properties to bonding nature and crystal structures
• Link to typical applications
Yield strength
Ultimate tensile strength
Ductility
Modulus of resilience
Toughness
Deteriorative Behaviour High tendency of corrosion Generally corrosion resistant
Typical Applications • List 2-3 typical applications for each type of materials and sub-classes of engineering alloys
• Able to select right materials for specific applications
Recycle and Disposal • Understand technical aspects of materials’ life cycle
• More importantly, appreciate the impact of materials’ life cycle on environment and society
Physical Properties
Mechanical Properties
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Metals (and alloys)
• steels & cast irons
• nonferrous – Cu, Al, Ti &
refractory
Ceramics
• crystalline ceramics
• glasses
Polymers
• thermoplastics
• thermosets
• elastomers
Bonding Type Entirely primary:
metallic + covalent
Entirely primary:
ionic + covalent
Primary + Secondary:
• covalent within molecules
• secondary between molecules
Bonding Energy
(Strength)
Moderate High Low between molecules, although
high within molecules
Structures • mostly highly crystalline
• BCC, FCC and HCP
• containing dislocations
• mostly crystalline except
glasses
• containing coiled long chains
which entangle
• semicrystalline to amorphous
Melting temperature Moderate High Low
Electrical conductivity High Low Low
Thermal conductivity High Low Low
Chemical activity High Low Low
Young’s modulus
• Perhaps demonstrate by − curve
• Ensure you are aware of the magnitude comparison
• Link properties to bonding nature and crystal structures
• Link to typical applications
Yield strength
Ultimate tensile strength
Ductility
Modulus of resilience
Toughness
Deteriorative Behaviour High tendency of corrosion Generally corrosion resistant
Typical Applications • List 2-3 typical applications for each type of materials and sub-classes of engineering alloys
• Able to select right materials for specific applications
Recycle and Disposal • Understand technical aspects of materials’ life cycle
• More importantly, appreciate the impact of materials’ life cycle on environment and society 7
T2: Mechanical deformation and failure of materials
Understanding the basics – you are expected to be able to do the following:
• Use appropriate engineering terminology to describe mechanical behaviour of materials
and mechanical properties used to describe such behaviour.
• Deferential applied load (, ) vs materials’ response (deformation; direct ∆, ; reactive∆, $%&'(%$)
• Determine , , $%&'(%$ given , ∆, ∆ and vice versa
• Name and determine tensile mechanical properties, including E, ), *+,, ductility,
modulus of resilience and toughness, given a - curve.
• Use schematic - curve to highlight tensile deformation characteristics of three basic
types of engineering materials.
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Able to solve problems associated brittle fracture failure:
• Ductile fracture occurs with plastic yielding, which absorbs large energy, blunts crack tips,
slows down crack propagation, and warns the imminent failure.
• Brittle facture, on the other hand, is accompanied with little or no plastic deformation,
crack tips are able to raise local stress to above fracture strength, causing cleavage of
interatomic bonds, resulting in rapid crack propagation, and fast failure.
• Fracture toughness - is a material’s property, which measures the material’s resistance
against brittle fracture when it contains pre-existing cracks.
- stress intensity factor = Yσ
- largest crack; surface crack length = ; internal crack length = 2
- brittle facture criteria: when ≥ -, brittle facture will occur
• Scenarios associated with solving brittle fracture problems:
- given - and , able to work out maximum allowed σ to avoid brittle facture
- given - and σ, able to work outmaximum allowed to avoid brittle facture
- given σ and , able to select right material (i.e. -) to avoid brittle facture
T2: Mechanical deformation and failure of materials
Able to solve problems associated failure by plastic yielding:
• Yield strength should be the design parameter for work load.
• Appreciate the need to consider and apply design safety factor.
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T2: Mechanical deformation and failure of materials
Able to solve problems associated fatigue failure:
• Two basic fatigue behaviours of materials: fatigue limit vs
no fatigue limit
• On top of basic concepts: number of cycles to failure,
fatigue life; stress amplitude, fatigue strength, fatigue limit
• Understand the S-N curve, and how the curve would
change under different loading conditions.
• Given the S-N curve of a material, able to determine the
fatigue limit or fatigue strength for a required fatigue life.
• Know how to work out the stress amplitude (S) given the
cyclic loading parameters for reversed or repeated stress
loading mode.
• Know how to work out maximum and minimum
stresses/loads given other cyclic loading parameters such
as %, ., ( and .
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4340 Steel 316 Stainless Steel 7075 Al Alloy Ti-6Al-4V Ti Alloy) (MPa) 862 310 505 830*+, (MPa) 1280 620 572 900 (GPa) 207 193 71 114 0.30 0.30 0.33 0.34/- ( ) 50 112 24 55
Final exam Question 2020
You have been tasked with selecting the material for manufacturing spokes for a bicycle
wheel. The spokes will be made from a cylindrical wire of one of the alloys listed in the
following table:
The spokes will be tensioned to 1200 N (this is the nominal load condition). The stress should
be no more than 50% of the yield strength of the material under the nominal load. Apply the
following design constraints sequentially to eliminate materials from consideration as you go.
(1) Considering aerodynamics and handling in a cross-wind, spoke diameter should not
exceed 2.4 mm; which materials are possible?
(2) It is preferred that the spokes fail by yielding rather than brittle fracture; impact with loose
road debris may lead to spoke surface damage to depth of 0.75 mm, you may assume the
geometric parameter Y is equal to 1.0 for fracture toughness calculations; which materials
are possible?
(3) Under the nominal load condition, the spokes should stretch by no more than 0.3% over
their unloaded length (~300 mm), which materials are possible?
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4340 Steel (Steel) 316 stainless steel (SS) 7075 Al Alloy (Al) Ti-6Al-4V Ti Alloy (Ti)
= 2 2×1200 ×862 = 1.88 m = 2
2×1200 ×862 = 3.09 m = 2
2×1200 ×862 = 2.46 m = 2
2×1200 ×862 = 1.92 m
Given:
• 4 candidate alloys and their properties as tabulated
• Loading conditions: nominal load = 1200 ; applied stress: = 0.5)
• Requirements for the spokes:
(1) diameter of spoke ≤ 2.4
(2) no brittle fracture before plastic yielding, surface crack = 0.75 , = 1
(3) ≤ 0.3%; = 300
(1) To Solve: select alloy(s) meeting diameter constraint of ≤ 2.4
Let be the cross-sectional area of the spoke, thus = = 140 → = 4 = 2 2) → () = 2 2())()
1 = 1 0
Answer:
• Only the Steel and Ti alloys meet the diameter constraint of ≤ 2.4 , under the
specified applied load conditions.
• The other 2 alloys need not be considered passed this point.
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4340 Steel (Steel) Ti-6Al-4V Ti Alloy (Ti) = ) 7.5×1012= 862× 7.5×1012 = 41.8 < /- = 50
= ) 7.5×1012= 830× 7.5×1012 = 40.3 < /- = 55
(2) To Solve: select alloy(s) to ensure no brittle facture before plastic yielding
Since plastic yielding occurs when applied stress % reaches the yield strength ) of the
material, this constraint demands the stress intensity factor = % not to exceed
fracture toughness of the alloy at % = ). Given = 1 and = 0.75 = 7.5×1012,
we have = ) 7.5×1012
Answer:
• Both the Steel and Ti alloys have a fracture toughness above the stress intensity factor and
are thus expected to fail by plastic yielding rather than by brittle fracture.
Given:
• 4 candidate alloys and their properties as tabulated
• Loading conditions: nominal load = 1200 ; applied stress: = 0.5)
• Requirements for the spokes:
(1) diameter of spoke ≤ 2.4
(2) no brittle fracture before plastic yielding, surface crack = 0.75 , = 1
(3) ≤ 0.3%; = 300
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4340 Steel (Steel) Ti-6Al-4V Ti Alloy (Ti) = 0.5×862 207×103 = 2.1×1013= 0.21% < 0.3% = 0.5×830 114×103 = 3.6×1013= 0.36% > 0.3%
(3) To Solve: select alloy(s) to meet strain constraint of ≤ 0.3%
Under nominal load condition of = 0.5) , Hooke’s law applies. Thus the strain
experienced by the alloy can be calculated as = 0.5!
Final Answer:
• The 4340 steel is the material of choice, as it satisfied all the constraints.
• The minimum diameter of the spoke is 1.88 mm.
Given:
• 4 candidate alloys and their properties as tabulated
• Loading conditions: nominal load = 1200 ; applied stress: = 0.5)
• Requirements for the spokes:
(1) diameter of spoke ≤ 2.4
(2) no brittle fracture before plastic yielding, surface crack = 0.75 , = 1
(3) ≤ 0.3%; = 300