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CambridgeTripos-MechanicalBehavior.pdf
Natural Sciences Tripos Part IA
MATERIALS SCIENCE
Course D: Mechanical Behaviour of Materials
Pl view
Plan
i
off a dislocation
di l
ti lloop
Name............................. College..........................
Dr Howard Stone
Lent Term 2013-14
2013 14
IA
DH1
Course D: Mechanical Behaviour of Materials
DH1
Contents
1 Synopsis
3
2 Reading list
2.1 Text books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Web-based resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
5
5
3 Elastic deformation
3.1 Normal stress and normal strain . . . . . . . . .
3.2 Shear stress and shear strain . . . . . . . . . . .
3.3 Elastic deformation . . . . . . . . . . . . . . . . .
3.3.1 Young’s modulus . . . . . . . . . . . . . .
3.3.2 Shear modulus . . . . . . . . . . . . . . .
3.3.3 Poisson’s ratio . . . . . . . . . . . . . . .
3.3.4 Multiaxial stress states . . . . . . . . . . .
3.4 Elastic strain energy . . . . . . . . . . . . . . . .
3.5 Atomic picture of elasticity . . . . . . . . . . . . .
3.6 Thermal expansion . . . . . . . . . . . . . . . . .
3.7 Young’s modulus of composites: Rule of mixtures
3.7.1 Axial modulus . . . . . . . . . . . . . . . .
3.7.2 Transverse modulus . . . . . . . . . . . .
3.8 Materials in structures under stress . . . . . . . .
3.8.1 Deformation of an elastic beam . . . . . .
3.8.2 Second moments of area . . . . . . . . .
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4 Plastic deformation: An introduction to dislocations
4.1 Estimate of the yield stress in a perfect crystal . . . . . . . . . . . . . . . .
4.2 Dislocations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.1 Edge dislocations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.2 Screw dislocations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.3 Mixed dislocations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.4 Dislocation loops . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Motion of dislocations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Shear stress required to move a dislocation . . . . . . . . . . . . . . . . . .
4.5 Force on a dislocation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Slip systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7 Geometry of single crystal slip and Schmid’s law . . . . . . . . . . . . . . .
4.8 Determining the operative slip systems . . . . . . . . . . . . . . . . . . . . .
4.8.1 The OILS rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.9 Geometry as slip proceeds . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.10 The energy associated with a dislocation . . . . . . . . . . . . . . . . . . .
4.11 Dislocation interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.12 Dislocation reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.12.1 Interaction of dislocations on different slip systems: The Lomer lock
4.13 Dislocation generation: Frank-Read sources . . . . . . . . . . . . . . . . . .
4.14 Jogs and kinks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.14.1 The absorption of vacancies/ atoms . . . . . . . . . . . . . . . . . .
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7
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25
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52
DH2
Course D: Mechanical Behaviour of Materials
4.15 Climb and cross slip of dislocations . . . . . .
4.16 Plastic deformation of metallic single crystals
4.16.1 hcp metals . . . . . . . . . . . . . . .
4.16.2 fcc metals . . . . . . . . . . . . . . . .
4.16.3 Polycrystalline metals . . . . . . . . .
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DH2
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53
55
55
56
57
5 Strengthening mechanisms
5.1 Forest hardening . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Dislocation pile ups and the effect of grain size . . . . . . . . . . .
5.3 Solid solution strengthening . . . . . . . . . . . . . . . . . . . . . .
5.3.1 Substitutional solute atoms . . . . . . . . . . . . . . . . . .
5.3.2 Interstitial solute atoms . . . . . . . . . . . . . . . . . . . .
5.4 Precipitate hardening . . . . . . . . . . . . . . . . . . . . . . . . .
5.4.1 Stress required to bow dislocations between precipitates .
5.4.2 Transition from cutting to bowing and maximum hardening .
5.4.3 Changing strengthening mechanisms during age hardening
5.5 Partial dislocations and stacking faults . . . . . . . . . . . . . . . .
5.6 Order hardening . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7 Plastic deformation by cooperative shear - Twinning . . . . . . . .
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58
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76
6 Fracture
6.1 Estimate of ideal fracture stress . . . . . . . . . . . . . . .
6.2 Griffith criterion . . . . . . . . . . . . . . . . . . . . . . . .
6.2.1 The strain energy release rate . . . . . . . . . . .
6.2.2 Fracture energy (crack resistance) . . . . . . . . .
6.2.3 Critical flaw size and ’toughening’ by flaw removal
6.3 Ductile fracture . . . . . . . . . . . . . . . . . . . . . . . .
6.3.1 Effect of plasticity on fracture energy . . . . . . . .
6.3.2 Stress intensity factor . . . . . . . . . . . . . . . .
6.3.3 Uniting the stress and energy approaches . . . . .
6.3.4 Ductile rupture . . . . . . . . . . . . . . . . . . . .
6.4 The Ductile-Brittle Transition Temperature (DBTT) . . . .
6.5 Toughness of composites . . . . . . . . . . . . . . . . . .
6.6 Pressurised pipes . . . . . . . . . . . . . . . . . . . . . .
6.7 Aircraft stresses and materials . . . . . . . . . . . . . . .
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7 Appendix: Observing dislocations
95
8 Glossary
97
DH3
1
Course D: Mechanical Behaviour of Materials
DH3
Synopsis
Elastic deformation
Definitions of stress and strain, both normal and shear. Definitions of elastic deformation,
Young’s modulus, shear modulus, Poisson’s ratio. Multiaxial stress states.
Elastic strain energy. The atomic picture of elasticity: bond stretching. Calculation of Young’s
modulus from binding energy curve for a crystal. Typical values of elastic moduli. Thermal
expansion and bimetallic strips.
Elastic properties of composite materials: axial and transverse moduli using the slab model.
Deformation of an elastic beam. Bending moments. Second moments of area.
Effect of increasing stress on materials: the elastic limit, plastic flow, brittle and ductile fracture.
Plasticity
Why and how does plastic flow occur in crystals? Yield stress for different materials. Frenkel
calculation of theoretical shear strength. Difference between ideal and actual yield stress
explained in terms of dislocations.
Geometry of an edge dislocation in a simple cubic structure. Definitions of line vector, Burgers
vector and slip plane. Screw dislocations, mixed dislocations, dislocation loops.
Observation of dislocations using transmission electron microscopy.
Movement of dislocations. Atom movement at the core of a dislocation due to glide. Concept
of lattice potential (Peierls-Nabarro energy) leading to Peierls-Nabarro stress. Width of a
dislocation determined by bonding type (metallic, covalent, ionic). Expression for Peierls
stress in terms of Burgers vector. Force on a dislocation due to shear stress.
Slip in single crystals. Definition of slip systems. Slip systems for fcc, bcc, hcp metals and
NaCl structures. Geometry of single crystal slip. Schmid factor. Example of calculation of
system with the highest Schmid factor for fcc. The OILS rule. Geometry of a single crystal as
slip proceeds. Sample elongation in single crystal slip. Rotation of the tensile axis.
Dislocation interactions. Repulsion and attraction of dislocations. Energies of dislocations.
Frank’s rule. Formation of a Lomer lock. Dislocation generation (Frank-Read sources). Jogs
and kinks and their consequence on dislocation mobility. Cross slip of screw dislocations.
Climb of edge dislocations. Intersection of propagating dislocations.
Stress-strain curves for single crystals of hcp and fcc metals. Explanations for stages I, II
and III. Duplex slip. Geometrical softening. Differences in polycrystalline metals. Taylor
factor.
DH4
Course D: Mechanical Behaviour of Materials
DH4
Strengthening mechanisms
Factors which control yield stress: Forest hardening; typical values of dislocation density;
dislocation pile ups and the effect of grain size (Hall-Petch relation); interactions between
dislocations and solute atoms. Cottrell atmosphere formation with interstitial solutes. Lüders
bands and the Portevin-Le Chatelier Effect.
Precipitates as obstacles to dislocations. Coherency strains and precipitate cutting. Orowan
bowing. Stress to by-pass obstacles. Age hardening and precipitation sequences. Mechanisms of hardening.
Partial dislocations and stacking faults. Order hardening and anti-phase boundaries.
Deformation twinning as a mechanism of plastic deformation. Twinning elements and the
twinning shear. Factors favouring Twinning. Effects of temperature and strain rate. “Tin Cry”.
Morphology of deformation and annealing twins.
Fracture
Estimate of theoretical cleavage stress. Explanation of difference from observed values in
terms of propagation of pre-existing defects. Griffith criterion for fracture. Strain energy release rates and fracture energies.
Fracture of ductile solids. Concept of stress concentration. Stress intensity factor. Crack
tip plasticity. Fracture toughness. Ductile tearing vs cleavage fracture. Designing against
fast fracture. Pressurised thin walled tubes. Stresses in aircraft fuselages. Toughness of
engineering materials and composites. Ductile/brittle transition temperature.
DH5
2
Course D: Mechanical Behaviour of Materials
DH5
Reading list
2.1
Text books
1. Hull, D. and Bacon, D.J., Introduction to Dislocations, Butterworth, 4th ed., (2001).
2. Hosford, W.F. Mechanical Behaviour of Materials, Cambridge University Press, (2010).
3. Ashby M.F. and Jones D.R.H., Engineering Materials : An Introduction to their Properties and Applications, Pergamon, 2nd ed., (1996).
4. Hull, D. and Clyne, T.W., Introduction to Composite Materials, Cambridge University
Press, (1996).
5. Hamley, I.W., Introduction to Soft Matter, Wiley, (2000).
6. Callister, W.D., Materials Science and Engineering: An Introduction, Wiley, 6th ed.,
(2002).
2.2
Web-based resources
Most of the material associated with the course (handouts, question sheets, practical scripts
etc) can be viewed on the web (www.msm.cam.ac.uk/Teaching) and also downloaded. This
includes model answers, which are released after the work concerned should have been completed. In addition to this text-based material, resources produced within a project based in
the Materials Science Department, called DoITPoMS (Dissemination of Information Technology for the Promotion of Materials Science) are also available (www.doitpoms.ac.uk). These
include libraries of Micrographs, Videos and Teaching and Learning Packages (TLPs). The
following TLPs are relevant to course D:
• Thermal Expansion and the Bimaterial Strip
• Introduction to Dislocations
• Slip in Single Crystals
• Introduction to Mechanical Testing
• Hardness and Indentation
• Brittle Fracture
• Mechanics of Fibre-Reinforced Composites
• Bending and Torsion of Beams
DH6
Course D: Mechanical Behaviour of Materials
DH6
DH7
3
3.1
Course D: Mechanical Behaviour of Materials
DH7
Elastic deformation
Normal stress and normal strain
Consider the effect of forces acting on the surfaces of a cuboid. If a force acts perpendicular to a surface we have a normal stress, denoted by σ. With equal and opposite forces
acting over two opposing faces, as shown in Fig. 1, the cuboid will remain stationary (will not
accelerate).
Area (A)
Force (F)
Force (F)
Stress (σ) =
Force (F)
Area (A)
Figure 1: A cuboid subjected to forces acting perpendicular to two opposite faces generating a normal
stress, σ.
Tensile stress (by convention positive).
Compressive stress (by convention negative).
The stress is defined as the force acting per unit area.
σtrue =
F
A
(3.1)
in which F is the magnitude of the force and A is the area of the face.
This definition of stress is referred to as the true stress.
As the cuboid elongates, conservation of volume would suggest that the area of the cuboid
faces over which the forces act will decrease. To avoid having to account for this effect, it is
often more convenient to define the force as acting over the original area, A◦
σeng =
F
A◦
(3.2)
This is referred to as the nominal stress or engineering stress.
In practice, the difference between the nominal stress/engineering stress and the true stress
is usually small and is often ignored.
DH8
Course D: Mechanical Behaviour of Materials
DH8
The extension of the cuboid under the action of the stress is described by the strain, denoted
by ε. The normal strain is the relative change in linear dimension in the direction of a normal
force. If the cuboid extends by a length δL in the direction of the applied force, from its initial
length of L0 to L1 , as shown in Fig. 2, then the normal strain, ε is defined by
L0
Force (F)
Force (F)
L1
Strain (ε) =
L1 - L0
L0
Figure 2: Extension of a cuboid subjected to forces acting perpendicular to two opposite faces .
δL
(3.3)
L
If the strain is finite, this may be integrated to obtain the true strain or logarithmic strain.
δε =
Z
L1
εtrue =
L0
L1
dL
= ln
L
L0
(3.4)
For small strains, the strain is reasonably approximated by the nominal strain or engineering
strain , which is taken to be
L1 − L0
L1
εeng =
=
−1
(3.5)
L0
L0
The distinction between nominal strain and true strain only becomes important for strains in
excess of a few percent.
Small Strains (up to a few %) → True Stress ∼ Nominal Stress
Large Strains → True Stress > Nominal Stress
DH9
3.2
Course D: Mechanical Behaviour of Materials
DH9
Shear stress and shear strain
If a force acts parallel to the surface of the cuboid we have a shear stress, denoted by τ . As
before, this is considered to act over the surface of the cuboid. To ensure that the cuboid
does not accelerate (rotate) we must define two pairs of opposing forces acting over 4 faces
of the cuboid, as shown in Fig. 3.
Area (A0)
original section
deformed section
rotated deformed section
Δy0
Force (F)
x0
shear stress (τ) =
φ
Force (F)
shear strain =
Area (A0)
Δy0
x0
= tan φ = γ
Figure 3: A cuboid subjected to forces acting parallel to two opposite faces of the cuboid generating
a shear stress, τ .
As before, the shear stress is defined as the force acting per unit area.
τ=
F
A◦
(3.6)
The distortion of the cuboid under the action of the shear stress is shown in Fig. 3. There is
an elongation of the cuboid along one face diagonal accompanied by a contraction along the
other. This distortion can be described by the shear strain, denoted by γ, which is the ratio of
the distances, ∆y◦ and x◦ . The associated angle through which the cuboid has been sheared
is typically referred to as the angle of shear, φ.
∆y◦
x◦
= tan φ
γ =
(3.7)
(3.8)
Normal Stresses & Strains → +ive (tensile) or –ive (compressive) → physically different
Shear Stresses & Strains → +ive or –ive, but depends only on convention used for +ive sense
of reference directions → no physical difference
General stress states more complicated - but can always be resolved into shear stresses and
normal stresses.
Stresses & Strains → Two directions associated with each value. Hence they are NOT vectors
(one direction only).
→ Stresses & strains cannot be resolved as if they were vectors (ie with a single cosine term).
They need two cosine terms - see www.doitpoms.ac.uk/tlplib/tensors/index.php.
DH10
3.3
3.3.1
Course D: Mechanical Behaviour of Materials
DH10
Elastic deformation
Young’s modulus
For an ideal, elastic, isotropic material (no orientational dependence of the material properties) the normal strain, ε, that we measure will be proportional to the applied normal stress, σ
such that
σ = Eε
(3.9)
in which the constant, E, is called the Young’s modulus.
3.3.2
Shear modulus
Similarly, the shear strain, γ, will be proportional to the applied shear stress, τ , such that
τ = Gγ
(3.10)
in which G is the Shear modulus.
It should be noted that G and E are different as they describe the response of a material to
different types of deformation. However, for many materials G ≈ 0.4E.
Young’s modulus is related to Hooke’s law - Force is proportional to displacement
(F = kδL)
Stiffness, k, is an extensive property (depends on the material properties and the shape)
Young’s modulus, E, is an intensive property (depends on the material only) → More generally
useful.
k can be related to E
σ = Eε
F
δL
= E
A
L EA
F =
δL
L
F = kδL
DH11
Course D: Mechanical Behaviour of Materials
DH11
Figure 4: Young’s modulus values for various materials, sub-divided according to type (from: M.F.
Ashby and D.R.H. Jones, Engineering Materials 1, Butterworth-Heinemann, 2001).
DH12
3.3.3
Course D: Mechanical Behaviour of Materials
DH12
Poisson’s ratio
For most materials, the application of a tensile normal stress would be expected to lead to a
contraction in the perpendicular directions, as shown in Fig. 5.
Force (F)
y
Force (F)
z
x
Figure 5: Lateral contraction of a cuboid in directions x and y under the action of a tensile normal
stress applied along direction z.
For an isotropic material the strains in the two directions perpendicular to the tensile normal
stress will be the same. We define the relationship between the tensile normal stress (acting
along direction z) and the normal strains (in directions x and y) to be
εx = εy = −νεz = −ν
σz
E
(3.11)
in which ν is the Poisson’s ratio.
Poisson’s ration is dimensionless.
It can take values between -1 and 0.5.
The conservation of volume would suggest that Poisson’s ratio should be 0.5 (K=∞).
In practice, Poisson’s ratios between 0.2 and 0.5 are common (most metals have Poisson’s
ratios around 0.3)
DH13
3.3.4
Course D: Mechanical Behaviour of Materials
DH13
Multiaxial stress states
If an elastic, isotropic material is subjected to normal stresses in three perpendicular directions, σx , σy and σz , as shown in Fig. 6, the resultant normal strains in the directions, x, y and
z will be given by
εx =
εy =
εz =
1
(σx − ν (σy + σz ))
E
1
(σy − ν (σx + σz ))
E
1
(σz − ν (σx + σy ))
E
(3.12)
(3.13)
(3.14)
σy
σx
σz
σz
y
σx
z
x
σy
Figure 6: A cuboid subjected to three perpendicular normal stresses.
Some special cases:
• Plane stress - One of the normal stresses is zero (e.g. stretching a thin sheet.)
• Plane strain - One of the normal strains is zero
• Hydrostatic stress - All three normal stresses have the same magnitude.
DH14
3.4
Course D: Mechanical Behaviour of Materials
DH14
Elastic strain energy
When a body is elastically deformed the work done is stored as elastic strain energy.
If we consider a cylinder extended by a length δL under the action of opposing forces, F , the
work done on the cylinder is F δL.
L
δL
F
F
A
Figure 7: A cylinder extended under the action of opposing forces.
From our definition of normal stress, σ = F/A. Hence, F = σA
Similarly, from our definition of normal strain, δε = δL/L. Hence, δL = Lδε
Hence the work done on the cylinder is
work done = F δL = (σA) (Lδε)
(3.15)
Assuming the volume of the cylinder, V = AL is invariant and substituting σ = Eε, Equation 3.15 may be integrated between 0 and εmax
Z εmax
V Eε dε
(3.16)
work done =
0
=
1
V Eε2max
2
(3.17)
Hence
the work done per unit volume under a normal stress is 12 Eε2
This energy is stored as potential energy in strained interatomic bonds and is also referred to
as the strain energy density.
It can be similarly shown that
the work done per unit volume under a shear stress is 12 Gγ 2
As σ = Eε, elastic strain energy can also be written as 21 σε.
This is the area under a σ − ε curve.
The equivalent for shear strain energy is 21 τ γ.
DH15
3.5
Course D: Mechanical Behaviour of Materials
DH15
Atomic picture of elasticity
The variation of potential energy, U , with distance, r, between two atoms, and the resultant
force, is shown schematically in Fig. 8.
Energy, U
Inter-atomic
spacing, r
ro
F (=dU/dr)
Figure 8: Variation of potential energy with inter-atomic spacing.
At very large distances the potential energy is approximately zero. As the atoms approach
one another the potential energy falls, reaching a minimum of Umin at a distance of r◦ , after
which the potential energy rises, ultimately becoming positive at small distances.
This variation in potential energy with distance can be reasonably approximated by the
Lennard - Jones potential , ULJ
r 6 r◦ 12
◦
ULJ = Umin
−2
(3.18)
r
r
As the atoms are displaced from their equilibrium separation of r◦ they will experience a
restoring force.
F =
dU
dr
(3.19)
For small displacements from r◦ the variation in the potential energy is approximately linear
(see Fig. 8). Hence, the force will be approximately proportional to the displacement - this is
the origin of Hooke’s law. We can therefore estimate the Young’s modulus from the LennardJones potential.
DH16
Course D: Mechanical Behaviour of Materials
DH16
Consider a simple cubic crystal. If we stretched the crystal parallel to [100] we will be stretching a square array of atomic bonds as shown in Fig. 9
σ
σ
z
y
x
Figure 9: A simple cubic crystal - consider the square array of bonds identified by the dotted lines.
Prior to loading, we may reasonably expect the atoms to be r◦ apart. The area occupied by
each bond in the plane perpendicular to the applied normal stress is r◦2 .
The stress on each bond, σ, is therefore,
σ=
F
1 dU
= 2
2
r◦
r◦ dr
(3.20)
By definition (from Eq. 3.3), the strain in the bond, ε, is
δε =
δr
r◦
(3.21)
The Young’s modulus is therefore
dσ dr
dσ
=
=
E=
dε
dr dε
!
1 d2 U (r◦ )
r◦2 dr2 r◦
(3.22)
The Young’s modulus can therefore be obtained from the interatomic potential by
E=
1 d2 U r◦ dr2 r◦
(3.23)
DH17
3.6
Course D: Mechanical Behaviour of Materials
DH17
Thermal expansion
“Temperature becomes a quantity definable either in terms of macroscopic thermodynamic
quantities such as heat and work, or, with equal validity and identical results, in terms of a
quantity which characterised the energy distribution among the particles in a system.” (Quinn,
T. J. Temperature 1990 Academic Press ISBN 0-12-569681-7)
The temperature of any system, T , can be related to the mean kinetic energies of the atoms
it contains according to
3
1
kb T = h mv 2 i
(3.24)
2
2
where kb is Boltzmann’s constant
At higher temperatures, the greater kinetic energy of the atoms will allow them to move further
away from their equilibrium positions. The asymmetry of the potential energy vs distance
curve results in the mean position increasing for oscillations of higher energy. This is shown
schematically in Fig. 10. This give rise to thermal expansion.
Energy, U
Inter-atomic
spacing, r
Thermal
Expansion
rhigh T
rlow T
ro
Figure 10: Variation of potential energy with inter-atomic spacing, illustrating the effect of temperature
on the mean atomic position.
Macroscopically we define the relationship between a temperature change ∆T and the accompanying increase in thermal strain ε to be
εthermal = α∆T
(3.25)
where α is the coefficient of thermal expansion or thermal expansivity.
In general, materials with stronger interatomic bonds (e.g. ceramics) show low coefficients of
thermal expansion as well as high Young’s moduli. The converse is true of weakly bonded
materials (e.g. polymers). Examples of the temperature dependence of the coefficients of
thermal expansion of a range of materials are given in Fig. 11
DH18
Course D: Mechanical Behaviour of Materials
DH18
linear coefficient of thermal expansion, α ([µstrain]/°C)
3
10
polyester
2
10
epoxy
aluminium
1
titanium
10
SiC
Al2O3
0
10
SiO2
−1
10
0
50
100
150
200
250
300
350
400
450
500
temperature/°C
Figure 11: Temperature dependence of the coefficients of thermal expansion of selected materials.
DH19
3.7
Course D: Mechanical Behaviour of Materials
DH19
Young’s modulus of composites: Rule of mixtures
Composites
Metals typically combine both high strength and damage tolerance. However, they have a
number of drawbacks. In particular, they are often quite dense and may corrode. Ceramics
and polymers are generally lighter and more corrosion-resistant. Regretably, ceramics and
polymers do not have much damage tolerance. However, excellent properties can be obtained by combining different materials to produce composites, usually with one constituent
in fibre form.
Ceramics and polymers are often much stronger when produced as fine fibres. For ceramics,
this strengthening arises from removal of flaws. This is discussed in greater detail in Section 6.
In polymers it is achieved by molecular alignment. Three main types of fibre are used in
composites: glass, carbon and aramid (aromatic polyamides, such as KevlarTM ). These fibres
are embedded in a matrix - usually a polymer and commonly a thermosetting resin, such as
an epoxy or a polyester. In some types of composite, the fibres are oriented randomly within
a plane, while in others the material is made up of a stack of differently-oriented “plies”, each
containing an aligned set of parallel fibres.
Stiffness of Composites
The materials used for the fibres and matrices of composites typically have very different
Young’s moduli. As such, the composite will exhibit highly anisotropic elastic behaviour depending upon whether it is loaded parallel or perpendicular to the fibres. The Young’s moduli
in these two cases can be reasonably predicted using a simple ‘slab’ model and the rule of
mixtures. These are shown in Fig. 12.
σ
1-Vf
σm
σ
σf
(a)
(b)
Vf
σ
1-Vf
Vf
(c)
σm
σf
σ
Figure 12: A unidirectional long fibre composite material and simplified ’slab’ models for (b) axial
deformation and (c) transverse deformation.
DH20
3.7.1
Course D: Mechanical Behaviour of Materials
DH20
Axial modulus
The axial modulus may be obtained assuming the ‘slabs’ shown in Fig. 12(b) are subjected
to equal strain. This is known as the Voigt Model.
Using the subscripts, c, f and m to refer to the composite, fibres and matrix respectively.
The stress experienced by the composite will be equal to the sum of the stresses experienced
by the fibres and matrix weighted by their area/volume fractions
σc = σf Vf + σm Vm
(3.26)
σc = σf Vf + σm (1 − Vf )
(3.27)
Ec εc = Ef εf Vf + Em εm (1 − Vf )
(3.28)
Given that σ = Eε
Using the equal strain assumption, εc = εf = εm
Ec = Ef Vf + Em (1 − Vf )
(3.29)
Hence, the axial stiffness of a unidirectional fibre composite is
Ec = Ef Vf + Em (1 − Vf )
Axial - Rule of mixtures provides a reasonable estimate of the composite stiffness
(3.30)
DH21
3.7.2
Course D: Mechanical Behaviour of Materials
DH21
Transverse modulus
The transverse modulus may be obtained assuming the ’slabs’ shown in Fig. 12(c) are subjected to equal stress. This is known as the Reuss Model.
The strain experienced by the composite will be equal to the sum of the strains experienced
by the fibres and matrix weighted by their area/volume fractions
Given that ε = σ/E
εc = εf Vf + εm Vm
(3.31)
εc = εf Vf + εm (1 − Vf )
(3.32)
σf
σc
σm
=
(1 − Vf )
Vf +
Ec
Ef
Em
(3.33)
Using the equal stress assumption, σc = σf = σm
1
Ec
1
Ec
=
=
1
1
(1 − Vf )
Vf +
Ef
Em
Em Vf + Ef (1 − Vf )
Ef Em
(3.34)
(3.35)
Hence, the transverse stiffness of a unidirectional fibre composite is
Ec =
Ef Em
Em Vf + Ef (1 − Vf )
(3.36)
Transverse modulus / Inverse rule of mixtures is substantially lower that the axial value.
Critically, it is not very accurate as parts of the matrix are shielded from the applied stress by
being “in parallel” with the fibres
DH22
3.8
Course D: Mechanical Behaviour of Materials
DH22
Materials in structures under stress
The effective design of structures capable of withstanding mechanical loads involves calculations of the stresses and strains as well as an understanding of how the materials will deform
and fail. Today, the stresses that develop in engineering structures may be calculated using
well established and powerful numerical models. These may be used without deep background knowledge. However, materials science is still a relatively new discipline and there
have been, and continue to be, many examples of structures failing because of a poor choice
of material, an inadequate understanding of material behaviour, or simply failing to understand the stresses and strains generated in the structure.
3.8.1
Deformation of an elastic beam
Consider a uniform beam bent under a bending moment, M , as shown in Fig. 13. In bending,
the top surface has become longer (in tension) whilst the bottom surface has become shorter
(in compression). Along the mid thickness of the beam, the length is unchanged - this is
termed the neutral axis.
l
y
neutral
axis
M
M
R
ε
θ
y
θ
Figure 13: Strains induced during bending of a beam by the application of a moment, M .
The axial strain at a distance of y from the neutral axis is given by
εaxial =
extension
(R + y)θ − Rθ
y
=
=
original length
Rθ
R
(3.37)
DH23
Course D: Mechanical Behaviour of Materials
DH23
Hence, the axial stress is
Ey
(3.38)
R
The force on a strip of thickness dy and width b at a distance y from the neutral axis is
Ey
F = σA =
(b · dy)
(3.39)
R
σaxial = Eεaxial =
The bending moment about the neutral axis associated with this force is
Ey
Bending moment = y
(b · dy)
R
(3.40)
Hence, the total bending moment is
E
M=
R
Z
h/2
y 2 bdy
(3.41)
−h/2
R h/2
For any section, we can define the second moment of area, I = −h/2 y 2 bdy. Hence the total
bending moment may be written in terms of the Young’s modulus, E, the second moment of
area, I and the radius of curvature, R, or the curvature, κ = 1/R
M=
EI
= κEI
R
The product EI is termed the beam stiffness or sometimes as flexural rigidity.
(3.42)
DH24
3.8.2
Course D: Mechanical Behaviour of Materials
DH24
Second moments of area
For a general shape, the second moment of area about a neutral axis is defined as
Z
I=
y 2 b{y}dy
(3.43)
section
Large values of I will give a stiff beam. This is most readily achieved with sections that have
the greatest proportion of their sections at large distances from the neutral axis, hence the
use of I-beams, box- and tube- sections.
The formulae giving the second moment of area for a wide variety of common shapes are
given in many books.
h
w
3
I = wh
12
D
4
I = πD
64
d
D
4 4
I= π(D -d )
64
DH25
4
Course D: Mechanical Behaviour of Materials
DH25
Plastic deformation: An introduction to dislocations
If a stress below the yield stress is applied, the material deforms elastically, i.e. the material
is able to return to its original shape after the stress is removed. With larger stresses, unrecoverable or plastic deformation occurs, i.e. it yields. The stresses at which materials yield
vary widely. They are typically in the range 30-1000 MPa for most engineering materials but
can be greater than 1 GPa or as low as 1 MPa.
Figure 14: Yield stresses for various materials, sub-divided according to type (from: M.F. Ashby and
D.R.H. Jones, Engineering Materials 1, Butterworth-Heinemann, 2001).
DH26
4.1
Course D: Mechanical Behaviour of Materials
DH26
Estimate of the yield stress in a perfect crystal
Plastic deformation usually occurs by atomic planes sliding over each other, under the influence of shear stresses. Note that when a normal stress is applied, shear stresses are
generated on planes inclined at an angle to the stress axis, as shown in Fig. 15.
σ
τmax≈ σ/2
τcrit
Figure 15: Shear stresses generated in a material subjected to a normal stress and the associated
slip of atomic planes over one another under the action of the shear stresses.
To estimate the yield stress of a perfect crystal, consider two close-packed planes with separation h and interatomic spacing b sliding over one another by a distance of u under the action
of a shear stress, τ , as shown in Fig. 16. We can also imagine that, for small values of u, the
deformation will be elastic.
We need to understand how τ varies with u. It is clearly going to be periodic with the distance
moved, u, and will be zero when u = 0, u = b/2, u = b, etc.
u=0
τ
u = 0.5b
τ
u=b
τ
h
τ
b
τ
τ
Figure 16: Movement of a close packed plane past another close packed plane from positions, u = 0
to u = b.
DH27
Course D: Mechanical Behaviour of Materials
DH27
If we assume it is sinusoidal then we can write
τ = C sin
2πu
b
(4.44)
in which C is a constant.
From the relationship between shear stress and shear strain given in Eqn. 3.10, τ = Gγ.
In this case, the shear strain is
(4.45)
γ = u/h
From Equation 4.44, for small u/b
2πu
2π (hγ)
τ ≈C
=C
=
b
b
2πCh
b
γ
(4.46)
Hence,
Gb
2πh
(4.47)
2πu
Gb
sin
2πh
b
(4.48)
C=
and Equation 4.44 becomes
τ=
For slip to occur we must apply a critical shear stress, τcrit , that will be sufficient to overcome
the maximum resistance that will be encountered. Hence,
τcrit =
For close packed spheres,
Gb
2πh
(4.49)
b
2
G
G
= √ and hence, τcrit = √ ≈
h
5
3
π 3
Note that for many materials G ≈ 0.4E. This is ∼1000 larger than the shear stresses actually
required to deform real materials!
DH28
4.2
4.2.1
Course D: Mechanical Behaviour of Materials
DH28
Dislocations
Edge dislocations
A dislocation is a line defect in a crystal in which the atoms are systematically displaced from
their ideal positions. Arguably the simplest dislocation to visualise is an edge dislocation. An
example of an edge dislocation is shown in Fig. 17. This can be simply considered as an
extra half plane of atoms in an otherwise perfect crystal structure. The line of the dislocation
can easily be seen at the base of the extra half plane, going into the page.
Figure 17: Schematic representation of an edge dislocation in a simple cubic material.
If a dislocation is present in a crystal, slip can occur by the progressive movement of the
dislocation. This motion is analogous to the movement of a caterpillar, as shown in Fig. 18.
A better analogy may be moving a carpet by moving a ruck across it rather than trying to
drag the whole carpet. As only a few bonds are being distorted at any point in time by
the progressive motion of the dislocation, the shear stress required to move a dislocation is
significantly lower than that required to shear the whole crystal at once.
Figure 18: Movement of an edge dislocation under the action of a shear stress
An animation of the glide motion of an edge dislocation can be seen in a DoITPoMS TLP at
www.doitpoms.ac.uk/tlplib/dislocations/dislocation glide.php.
DH29
Course D: Mechanical Behaviour of Materials
DH29
Geometry of a dislocation
For any dislocation we can define a dislocation line vector, l. In an edge dislocation this
corresponds to the base of the extra half plane of atoms. Perhaps more usefully, we can
define the dislocation line vector to be the line that separates the slipped from the unslipped
regions of the crystal.
Note that the dislocation line cannot end within the crystal (but it can form loops).
The lattice displacement caused by the dislocation can be described by its Burgers vector, b.
The Burgers vector may be obtained by constructing a Burgers circuit around the dislocation
core. This is shown schematically in Fig. 19.
(a)
(b)
b A
E A
E
D
B
D
B
C
C
Figure 19: A Burgers circuit around an edge dislocation in a simple cubic crystal. The Burgers vector,
b is defined by the closure failure, EA.
The Burgers vector is obtained by making integer steps between lattice points in the positive
and negative directions as you go around the dislocation core in a right-handed sense. If
the same number of steps in the same directions is performed in a perfect crystal you will
NOT end up back where you started. This leads to a closure failure. The vector defining the
−→
closure failure, EA defines the Burgers vector, b.
For an edge dislocation, the Burgers vector is perpendicular to its line vector.
A dislocation can only slip in the plane containing both its Burgers vector and line vector. This
is termed the slip plane. In the case of an edge dislocation, as the Burgers and line vectors
are perpendicular to one another there is only ever one plane in which an edge dislocation
can glide.
Course D: Mechanical Behaviour of Materials
b
DH30
Figure 20: Plan view of an edge dislocation in a simple cubic crystal.
DH30
DH31
4.2.2
Course D: Mechanical Behaviour of Materials
DH31
Screw dislocations
There are other types of dislocation. The other simple one is the screw dislocation, see
Fig. 21. A screw dislocation can be considered as a helical spiral defect in the crystal. The
axis of the helix defines the associated dislocation line vector, l, as well as the associated
Burgers vector, b (i.e. b k l).
(a)
(b)
A b
G
E
D
A
D
F
G
E
F
B
B
C
C
b
Figure 21: (a) A screw dislocation in a simple cubic crystal. The Burgers circuit is shown. (b) Burgers
circuit in a perfect crystal indicating the Burgers vector associated with the screw dislocation (GA).
Figure 22: Plan view of a screw dislocation in a simple cubic crystal.
Critically, because the Burgers vector and line vector of screw dislocations are parallel, screw
dislocations may glide on a family of possible slip planes. In the same way that the movement of an edge dislocation caused shear of the crystal, (see Fig. 18), passage of a screw
dislocation will shear a crystal.
DH32
4.2.3
Course D: Mechanical Behaviour of Materials
DH32
Mixed dislocations
Dislocations may also be intermediate between edge and screw and are termed mixed dislocations. The relative orientations of the line and Burgers vectors of edge, screw and mixed
dislocations are shown in Fig. 23
l
slipped
l
unslipped
slipped
l
unslipped
b
slipped
b
edge
screw
(a)
(b)
unslipped
b
mixed
(c)
Figure 23: Relative orientations of (a) edge, (b) screw and (c) mixed dislocations.
4.2.4
Dislocation loops
It is also possible to have a dislocation loop, as shown in Fig. 24. As before, the dislocation line
separates the slipped from the unslipped crystal. The Burgers vector is the same all around
the dislocation loop but as the line vector changes from being parallel to the Burgers vector
to being perpendicular to the Burgers vector so the character of the dislocation changes from
screw to mixed to edge and back again.
mixed
unslipped
edge
mixed
slipped
unslipped
l
screw
screw
b
mixed
edge
mixed
Figure 24: A dislocation loop. The Burgers vector is constant all around the loop whilst the character
changes between screw, mixed and edge.
DH33
Course D: Mechanical Behaviour of Materials
Figure 25: A plan view of a dislocation loop in a simple cubic crystal.
DH33
DH34
4.3
Course D: Mechanical Behaviour of Materials
DH34
Motion of dislocations
The motion of an edge and a screw dislocation are shown in Fig. 26. Note that in both cases
the dislocation moves perpendicular to the dislocation line vector and the resultant slip of
the crystal is equal to the Burgers vector of the dislocation. However, in the case of an edge
dislocation the dislocation line lies perpendicular to the applied shear stress, whilst for a screw
vector the dislocation line lies parallel to the applied shear stress
τ
(a)
τ
τ
(b)
τ
(c)
d
pe
p
i
l
S
ed
ipp
l
s
n
U
Figure 26: Schematic representations of the motion of (a) an edge dislocation and (b) a screw dislocation under an applied shear stress leading to shear of the crystal (c).
Note that for the crystal structure to be recovered after the passage of a dislocation the
Burgers vector MUST correspond to a vector between lattice points. These are termed perfect
dislocations. Other dislocations, not corresponding to lattice vectors, may exist but these will
not recover the original crystal structure in their wake.
DH35
Course D: Mechanical Behaviour of Materials
DH35
Motion of dislocation loops
Under the action of a shear stress applied parallel to the Burgers vector a dislocation loop,
like that shown in Fig. 24, will expand (or contract) in all directions. This can be understood by
considering that the edge segments of the dislocation loop will move out in the direction of the
applied shear stress whilst the screw line segments will move out sideways, perpendicular to
the applied shear stress. Note that the top and bottom (and left and right) segments move in
opposite directions because, whilst they have the same Burgers vector defined with reference
to the direction of the dislocation line marked on the diagram, they would have Burgers vectors
of opposite sense if Burgers circuits were drawn around them in the same direction.
edge
unslipped
slipped
screw
unslipped
screw
b
edge
Figure 27: Schematic representations of the motion of a dislocation loop.
DH36
4.4
Course D: Mechanical Behaviour of Materials
DH36
Shear stress required to move a dislocation
As a dislocation is moved, bonds are required to break and reform. As such, a shear stress
sufficient to overcome the associated energy barrier must be applied. This must move a
dislocation beyond its highest energy configuration.
Figure 28: (a) lowest and (b) highest energy configurations of an edge dislocation moving in a simple
cubic crystal.
The energy barrier to move a dislocation from one position to the next is called the Peierl’sNabarro energy and the associated shear stress is called the Peierl’s-Nabarro stress, τP .
The Peierl’s-Nabarro stress cannot be easily calculated as it is very sensitive to the nature
and directionality of the interatomic forces. However, a simple model gives
2πw
τP ≈ 3G exp −
(4.50)
b
where w is the dislocation width. This function is strongly dependent upon w/b.
The width of a dislocation is the distance over which atoms are significantly displaced from
their perfect crystal positions. This can be taken to be the region in which the atoms are
displaced by & b/4. Note that at the core of the dislocation, the atoms are displaced by b/2.
Schematic illustrations of wide and narrow edge dislocations are shown in Fig. 29.
0.25b
w
0.25b
0.25b
w
0.25b
Figure 29: Atomic disregistry in (a) wide and (b) narrow edge dislocations.
DH37
4.5
Course D: Mechanical Behaviour of Materials
DH37
Force on a dislocation
The force acting on a dislocation as a result of an applied shear stress may be obtained by
considering the work done in moving the dislocation - see Fig. 30
F
Figure 30: Movement of an edge dislocation with Burgers vector, b, under the action of an applied
shear stress.
If an edge dislocation of length, L experiences a force acting per unit length, F , then the total
force acting over the dislocation will be F L.
The work done in moving the dislocation through the crystal a distance d is therefore F Ld.
This must be equal to the work done by the shear stress, τ , which acts over the entire area
of the slip plane, Ld. Thus the work done in shearing the crystal a distance b is τ Ldb.
Hence,
τ Ldb = F Ld
(4.51)
Therefore, the force per unit length acting on a dislocation as a result of an applied shear
stress is
F = τb
Applies to all types of dislocation
NB dot product → = τ b only if τ acts k to b
Force always acts normal to the dislocation line (⊥ to l)
(4.52)
DH38
4.6
Course D: Mechanical Behaviour of Materials
DH38
Slip systems
If we deform a single crystal of a material, we usually find that it deforms on a specific set of
parallel planes and in a specific set of directions within these planes.
The combination of the plane in which slip occurs (slip plane) and the direction (slip direction)
together make up the slip system.
The slip system that operates in a given material is governed by the crystal structure and
the nature of the interatomic bonds. These are most commonly, but not always, the closest
packed planes and the closest directions within those planes.
For example, in a cubic close-packed metal, slip invariably occurs on the close-packed planes,
{111}, and in the close-packed directions in those planes, h11̄0i.
Table 1: Slip systems in some common materials
Material
fcc metals
Slip systems
{111}h11̄0i
bcc metals
{110}h11̄1i
(others occur at high T)
C, Si, Ge
(diamond structure)
{111}h11̄0i
NaCl*
{110}h11̄0i
CsCl*
{110}h001i
hcp metals
{001}h100i
(others can also occur)
* and other ionic crystals with this structure
Figure 31: Slip systems and Burgers vectors of common crystal structures
Burgers vector of a (perfect) dislocation
→ must be a lattice vector
Other defects may exist but these lead to defects
DH39
4.7
Course D: Mechanical Behaviour of Materials
DH39
Geometry of single crystal slip and Schmid’s law
If a stress is applied at an arbitrary angle to a single crystal, as shown in Fig. 32, each
available slip system will experience a resolved shear stress acting in the associated slip
plane in the slip direction.
Area (A)
slip plane
normal (n)
Force (F)
φ
λ
slip
direction
Force (F)
Figure 32: Geometry of slip during tensile testing of a single crystal.
If the normal to the slip plane makes an angle φ with the tensile axis then the area of the
plane is A/ cos φ.
If the slip direction is at an angle of λ to the tensile axis then the resolved component of the
applied force, F , parallel to the slip direction is F cos λ.
The resolved component of the shear stress on the slip plane acting parallel to the slip direction, τR , is therefore
τR =
=
F cos λ
A/ cos φ
F
cos φ cos λ
A
(4.53)
(4.54)
For a given material we find experimentally that the value of τR at which slip occurs is constant. This value is called the critical resolved shear stress, τc . This is Schmid’s law. The
quantity cos φ cos λ is called the Schmid factor.
The relationship between the yield stress, σy , at which plastic deformation initiates and the
critical resolved shear stress is therefore
τc = σy cos φ cos λ
(4.55)
Schmid factor → Note the two cosine terms
(Both the area of the slip plane & the component of force in the slip direction change as
different slip systems are considered)
DH40
4.8
Course D: Mechanical Behaviour of Materials
DH40
Determining the operative slip systems
During deformation, all of the available slip systems will experience a resolved shear stress.
As the critical resolved shear stress is the same for all slip systems, so we would expect the
one with the greatest Schmid factor to yield first. If we know the orientation of the tensile axis
we can calculate the Schmid factor of all of the slip systems and identify the largest Schmid
factor. This is effective but time consuming.
For bcc and fcc structures there is a simple method that identifies the slip system with the
greatest Schmid factor very easily, called the OILS rule.
4.8.1
The OILS rule
For fcc metals, the slip systems are {111}h11̄0i.
For bcc metals, the slip systems are {110}h11̄1i
Using the OILS rule:
1. Write down the indices of the tensile axis [U V W ].
2. Ignoring the signs, identify the Highest, Intermediate and Lowest valued indices.
3. The slip direction in a fcc crystal is the h110i direction with zero in the position of the
Intermediate index and the signs of the other two preserved.
or
The slip plane in a bcc crystal is the {110} plane with zero in the position of the
Intermediate index and the signs of the other two preserved from the tensile axis.
4. The slip plane in a fcc crystal is the {111} plane with the sign of the Lowest valued index
reversed and the signs of the other two preserved from the tensile axis.
or
The slip direction in a bcc crystal is the h111i direction with the sign of the Lowest valued
index reversed and the signs of the other two preserved.
The mnemonic OILS stands for zerO Intermediate, Lowest Sign.
Example of OILS rule: If the tensile axis is parallel to the [2̄14] direction of a fcc crystal, the
sequence is [ILH]. As the slip systems for fcc metals are {111}h11̄0i
zerO Intermediate gives the slip direction as being [011]
and Lowest Sign reversed gives the slip plane as being (1̄1̄1)
Hence, in this case, the slip system with the largest Schmid factor is (1̄1̄1)[011].
DH41
Course D: Mechanical Behaviour of Materials
DH41
Schmid factor calculation for the example above
cos φ can be obtained from the dot product of the normal to the slip plane and the tensile
axis.
   
1̄
2̄
 1̄  ·  1 
1
4
5
√ √
cos φ =
= √ = 0.630
(4.56)
3 21
63
Similarly cos λ can be obtained from the dot product of the slip direction and the tensile
axis.
   
0
2̄
 1 · 1 
1
4
5
√ √
cos λ =
= √ = 0.772
(4.57)
2 21
42
Hence, in this case cos φ cos λ = 0.486.
√
Note that the largest possible value of the Schmid factor is 0.5 (cos φ = cos λ = 1/ 2).
DH42
4.9
Course D: Mechanical Behaviour of Materials
DH42
Geometry as slip proceeds
Slip on a single system tends to cause lateral displacement (as well as axial extension). If
the grips used to hold the sample are aligned so as to prevent this, then the tensile axis will
rotate towards the slip direction, as shown in Fig. 33.
(a)
(c)
(b)
(d)
bending
λ0
λ1
slip
direction
λ1
rotation
bending
Figure 33: Deformation of a single crystal with one operative slip system (a) before loading, (b) during
loading, (c) allowing rotation at the grips, (d) with fixed grips.
The spacing between the planes remains constant through the deformation and the number
of planes is conserved. Hence, l cos φ will remain constant as deformation proceeds. This is
shown schematically in Fig. 34. Hence,
l◦ cos φ◦ = l1 cos φ1
(4.58)
Similarly, if we consider the section of the sample that contains both the slip direction and the
tensile axis, by the same argument l cos(90◦ − λ) = l sin λ is constant. Hence,
l◦ sin λ◦ = l1 sin λ1
(4.59)
As slip proceeds and the slip direction rotates towards the tensile axis, the resolved shear
stress on all available slip systems will change.
It is often more convenient to refer to this movement from the reference frame of the sample
and consider the tensile axis rotating towards the slip direction.
The direction of the tensile axis relative to the crystal lattice vectors as slip proceeds can also
be obtained by adding multiples of the slip direction to the tensile axis indices. Using the
example we had earlier
TA = [2̄14] + n[011]
(4.60)
DH43
Course D: Mechanical Behaviour of Materials
slip plane
normal
l
tensile
axis
DH43
tensile
axis
l
λ
slip
direction
Figure 34: Geometry as slip proceeds showing the number of planes remains constant.
This continues until two of the indices have the same value (ignoring their signs). In the example we are working with this will occur when the tensile axis reaches [2̄25]. At this point two
slip systems now have the same Schmid factor and slip proceeds on both of them. The new
slip system can also be obtained by the OILS rule and, in our example, is (111)[1̄01].
The subsequent reorientation of the tensile axis will now proceed by adding equal multiples of
both slip directions to the tensile axis indices. Again, using the example we had earlier
TA = [2̄25] + n[011] + n[1̄01]
(4.61)
As can be seen from this equation, the tensile axis will approach a h112i type direction, but
will never quite get there. For deformation along a h112i type direction the rotations induced
by the two slip systems exactly cancel.
DH44
4.10
Course D: Mechanical Behaviour of Materials
DH44
The energy associated with a dislocation
The energy associated with the elastic distortion of the crystal around a dislocation can be
most readily estimated by examining a screw dislocation. Consider the annulus of material
around a screw dislocation (see Fig. 35(b)). If this annulus is unwrapped we can see that the
original cuboid of material has been sheared.
(a)
(b)
l
(c)
r
b
b
dr
l
dr
2πr
Figure 35: Lattice distortion around a screw dislocation.
The associated shear strain of the annulus, γ, is
γ=
b
2π r
(4.62)
The volume of material in this annulus, V , is
V = 2π r l dr
(4.63)
The shear strain energy per unit volume is 12 Gγ 2 . Hence the strain energy in this element,
dU , is
1
b 2
dU = G
(2π r l dr)
(4.64)
2
2πr
The total strain energy can be obtained by integrating this expression between r = r◦ , the
radius at which the strain is too great to be treated as elastic, and some outer limit, r = r∞ .
It is also necessary to make allowance for the core energy (the energy associated with the
region for which r < r◦ ).
Z
Gb2 l r∞ dr
U =
+ core energy
(4.65)
4π r◦ r
Gb2 l
r∞
=
ln
+ core energy
(4.66)
4π
r◦
The challenge is to choose suitable values for r∞ , r◦ and the core energy.
DH45
Course D: Mechanical Behaviour of Materials
DH45
Clearly as r◦ approaches zero, Hooke’s law would suggest that the stress would become
infinite. However, the actual stress will remain finite (as dU/dr will remain finite). It has been
suggested that taking a limit of integration of r◦ = b/4 provides a slight overestimate of the
actually stress and hence the strain energy density in the region immediately above r◦ that is
approximately equal to the core energy (see Fig. 36).
real
Figure 36: Strain energy density close to the dislocation core.
Similarly, r∞ cannot be larger than the crystal. In practice taking a value equal to half the
distance between dislocations is considered reasonable.
With such assumptions, the overall result is
1
U ≈ Gb2 l
2
(4.67)
A similar result can be obtained for an edge dislocation, but is more difficult to derive. Importantly, the energy of an edge dislocation is always higher than that of a screw dislocation.
It is often more useful to express the strain energy per unit length of dislocation, Λ.
1
Λ ≈ Gb2
2
(4.68)
DH46
4.11
Course D: Mechanical Behaviour of Materials
DH46
Dislocation interactions
The distortion induced in a crystal by the presence of a dislocation leads to local tensile,
compressive and shear stresses.
A screw dislocation only gives rise to localised shear stresses. As we saw earlier, the shear
strain associated with a screw dislocation is γ = b/2πr (from Eq. 4.62). Hence, the local
shear stress acting along the axis of the dislocation is
Gb
2πr
in which r is the radial distance from the dislocation core.
(4.69)
τ = Gγ =
The local stress field around an edge dislocation is much more complex and contains tensile, compressive and shear stresses. However, by just considering the hydrostatic tensile
and compressive stresses around an edge dislocation it is easy to visualise the local stress
fields. Specifically, the extra half plane of atoms above the slip plane puts this region into
compression whilst the region below the slip plane goes into tension.
The stress fields associated with dislocations interact. The resulting stress at any point will
then be the sum of the stress contributions from each dislocation. As the elastic strain energy
is proportional to the square of the local strain, it is energetically favourable for the stress
fields to configure themselves to minimise this strain.
As a result, two dislocations of the same sign on the same slip plane will repel one another,
whilst two dislocations of opposite sign on the same slip plane will attract one another. Similarly, dislocations of the same sign moving on different slip planes may be attracted to one
another, and possibly form dislocation arrays.
(a)
(c)
C
C
T
T
(b)
(d)
C
C
T
T
C
T
C
C
C
T
T
T
Figure 37: Interactions between the hydrostatic stress fields surrounding (a) two dislocations of the
same sign moving on the same slip plane, (b) two dislocations of opposite sign moving on the same
slip plane, (c) two dislocations of the same sign moving on different slip planes, (d) two dislocations of
opposite sign moving on different slip planes.
Dislocation arrays may form that minimise the overall energy, as shown in Fig. 38.
DH47
Course D: Mechanical Behaviour of Materials
(a)
DH47
(b)
Figure 38: (a) original dislocation configuration. (b) dislocation array.
Figure 39: Dislocation arrays in a single-slip-oriented copper single crystal (from: Lepisto et al.,
Materials Science and Engineering, 81 (1986) 457-463).
DH48
4.12
Course D: Mechanical Behaviour of Materials
DH48
Dislocation reactions
If two dislocations combine, the new dislocation will have a Burgers vector, b3 , which will be
the sum of the Burgers of the two dislocations from which it formed, b1 and b2 .
The line vector of the new dislocation, l3 , will be the intersection of the slip planes of the two
original dislocations.
The plane in which the new dislocation may slip will be the plane containing both its Burgers
vector, b3 , and its line vector, l3 .
For such a dislocation reaction to occur, however, it must be energetically favourable. As the
energy associated with a dislocation is proportional to the square of the magnitude of the
Burgers vector, dislocations will combine if
b23 < b21 + b22
(4.70)
This is known as Frank’s rule.
Consider the example of slip in the fcc single crystal we considered earlier.
The two slips systems that became active when the tensile axis rotated to [2̄25] were
(1̄1̄1)[011] and (111)[1̄01].
a
a
[011] and b2 = [1̄01].
2
2
If dislocations with these two Burgers vectors were to combine then the resultant dislocation
would have a Burgers vector
The associated Burgers vectors are therefore b1 =
b3 = b1 + b2 =
a
a
a
[011] + [1̄01] = [1̄12]
2
2
2
(4.71)
The square of the magnitude of the Burgers vectors of the two original dislocations are both
a2 /2, whilst the square of the magnitude of the Burgers vectors of the resultant dislocation
would be 3a2 /2
In this case, b23 > b21 + b22 , so we do not expect these dislocations to combine.
DH49
4.12.1
Course D: Mechanical Behaviour of Materials
DH49
Interaction of dislocations on different slip systems: The Lomer lock
Dislocation interactions can have profound effects upon the plastic deformation behaviour of
materials.
Consider the example of the interaction of the dislocations moving on the following two slip
systems in a fcc metal. (1̄11)[101] and (111)[1̄10].
Figure 40: Dislocation interaction leading to the formation of a Lomer lock.
a
a
[101] and b2 = [1̄10].
2
2
If these dislocations combine then the resultant dislocation would have a Burgers vector
The associated Burgers vectors are b1 =
b3 = b1 + b2 =
a
a
a
[101] + [1̄10] = [011]
2
2
2
(4.72)
In this case, the square of the magnitude of the Burgers vectors of the two original dislocations
and the resultant dislocation are all equal to a2 /2. We therefore expect these dislocations to
combine as b23 < b21 + b22 .
The line vector of the new dislocation, l3 , must lie at the intersection of both slip planes, i.e.
along [01̄1] (or [011̄]), which is perpendicular to b3 (you can obtain this result using the Weiss
Zone law). The new dislocation is therefore an edge dislocation and can glide in the plane
containing b3 and l3 , i.e. (100).
Unfortunately, (100) is not a plane on which slip occurs in fcc metals. As such this dislocation
will not move - it is sessile and will block the movement of further dislocations on both slip
planes.
This type of sessile dislocation is called a Lomer lock.
DH50
4.13
Course D: Mechanical Behaviour of Materials
DH50
Dislocation generation: Frank-Read sources
Once a dislocation has moved across a crystal and generated a step on the surface of the
crystal it is no longer available to participate in plastic deformation. The gross plastic deformation we see in metals therefore requires sources of dislocations. In practice, dislocations
can be created at free surfaces, grain boundaries and within grains.
The most famous type of dislocation source is a Frank-Read source - named after the two
people who postulated its existence. This is shown in Fig. 41
τb
Figure 41: A schematic representation of the operation of a Frank-Read source
Imagine a dislocation with Burgers vector, b, of initial length, L, pinned at both ends. With
no applied shear stress, the dislocation will lie in a straight line, as this minimises its elastic
strain energy.
When a shear stress, τ , is applied, the dislocation will experience a force (F = τ b) normal
to the dislocation line. As the dislocation is pinned at both ends it will bow out, balancing the
line tension (i.e. dislocation energy) and the force due to the applied shear stress.
As the shear stress and hence the force on the dislocation is increased, it will eventually
become unstable and bow outwards until it forms a loop. When the two segments at the
bottom of the loop touch one another they annihilate each other as their line vectors are
defined in opposite directions, implying that their Burgers vectors will also be opposite.
Provided the shear stress is maintained, the dislocation loop formed may continue to propagate outwards whilst the remaining segment between the two pinning points may go on to
produce further dislocations.
An example of a Frank-Read source in silicon is shown in Fig. 42.
Figure 42: A Frank-Read source in Si (from: W.G. Dash, Dislocations and Mechanical Properties of
Crystals, ed. J.C. Fisher, Wiley New York, 1957.)
DH51
4.14
Course D: Mechanical Behaviour of Materials
DH51
Jogs and kinks
Dislocations are rarely straight and may not lie entirely in a single slip plane. Steps which lie
in the same slip plane are termed kinks.
(a)
(b)
Figure 43: Schematic illustration of kinks in (a) an edge dislocation, and (b) a screw dislocation.
Steps in the dislocation from one slip plane to another are termed jogs.
(a)
(b)
Figure 44: Schematic illustration of a jog in (a) an edge dislocation, and (b) a screw dislocation.
As kinks lie in the same slip plane they do not inhibit the movement of the dislocation. In fact
they may assist its motion, as atoms or vacancies diffusing to them may enable the dislocation
to move at stresses below the critical resolved shear stress!
Although, in the case of the screw dislocation, the kink will have some edge character the
screw dislocation may now be confined to a single slip plane.
Kinks and jogs are formed by:
• The absorption of vacancies/ atoms
• Intersection of propagating dislocations
DH52
Course D: Mechanical Behaviour of Materials
4.14.1
DH52
The absorption of vacancies/ atoms
The absorption of vacancies or atoms into a dislocation may lead to the formation of kinks or
jogs. This is a thermally activated process.
Similarly, if kinks or jogs are already present along the dislocation, they may get larger or
smaller (and possibly annihilate) as a result of this process. The mobility of atoms in most
materials is sufficient that an equilibrium exchange of atoms is set up between the slipped
and unslipped crystals either side of the dislocation.
If it is energetically favourable for one process to dominate, for example under an applied
stress, the kink or jog may move.
This is most readily visualised with the migration of kinks in the extra half plane of an edge
dislocation.
(a)
(b)
Figure 45: Movement of a kink in an edge dislocation by the transfer of vacancies/ atoms.
Spontaneous creation of double kinks by thermal motion
A dislocation may move under an applied shear stress lower than the critical resolved shear
stress by the spontaneous formation and growth of double kinks. As this is a thermally activated process it is favoured at higher temperatures.
(a)
(b)
Figure 46: Formation and propagation of double kinks by thermal motion of atoms.
DH53
4.15
Course D: Mechanical Behaviour of Materials
DH53
Climb and cross slip of dislocations
If dislocation glide on a given slip plane is inhibited by an obstacle it may be possible for the
dislocation to migrate onto another slip plane and continue gliding from there. Two important
processes by which this can occur are: cross slip of screw dislocations and climb of edge
dislocations.
Climb of edge dislocations
As the Burgers vectors and line vectors of edge dislocations are perpendicular to one another,
they are confined to a single slip plane and, as such, it is not possible for edge dislocations to
undergo cross slip.
Edge dislocations may, however, move onto other slip planes to avoid obstacles or to arrange themselves into energetically favourable configurations by the process of dislocation
climb.
If a vacancy in the crystal structure migrates to the dislocation core, the segment of the
dislocation where the vacancy is absorbed will rise by one atomic spacing. This is termed
positive climb.
It is also possible for extra atoms to diffuse to the edge dislocation core, moving the dislocation
core downward. This is termed negative climb.
(a)
(b)
(c)
Obstacle
Figure 47: Positive climb of a dislocation to bypass an obstacle.
Note that a lot of vacancies/ extra atoms may be required to raise or lower an entire edge dislocation to the next slip plane. The diffusion processes required for either positive or negative
climb are thermally activated and therefore occur more readily at high temperatures.
DH54
Course D: Mechanical Behaviour of Materials
DH54
Cross slip of screw dislocations
As the Burgers vector and line vector of screw dislocations are parallel, they may glide on the
family of crystallographically related planes that contain these vectors. This enables these
dislocations to change the slip system in which they move to detour past obstacles.
Consider a screw dislocation with Burgers vector a/2[11̄0] propagating along (111) as shown
in Fig. 48. It will continue to slip on this slip plane until the local stress field changes so that
motion on the (111̄) becomes preferred. At which point it will begin gliding on this new slip
system.
Double cross slip may also be possible, as shown in Fig. 48(d). By this process, screw
dislocations may move past obstacles in their initial slip plane.
[110]
(111)
b
(111)
(a)
(b)
(c)
(d)
Figure 48: Schematic illustrations of cross slip in a fcc metal. (a) A screw dislocation is gliding along
(111). (b) The dislocation may begin slipping on (111̄) if it is favourable for it to do so. (c) The dislocation
may continue to propagate along (111̄). (d) Double cross slip back onto a (111) plane may also be
possible.
DH55
Course D: Mechanical Behaviour of Materials
4.16
DH55
Plastic deformation of metallic single crystals
If we conduct a tensile test on a metallic single crystal we expect to see the normal strain
increase approximately linearly with applied normal stress, according to Hooke’s law, until the
stress reaches the yield stress, σy . At this stress, the resolved shear stress on the slip system
with the largest Schmid factor will be sufficient for dislocations to move.
As the crystal plastically deforms, the tensile axis rotates towards the slip direction, with a
progressive change in the Schmid factor.
4.16.1
hcp metals
During the deformation of hcp metals, only one slip system will operate. The critical resolved
shear stress will not vary during deformation, as shown in Fig. 49. However, as the tensile axis
Initial
elastic
strain
Plastic deformation
Figure 49: Shear stress vs shear strain typically observed for hcp metal single crystals.
rotates towards the slip direction, its Schmid factor will change. Depending upon the initial
orientation of the crystal the Schmid factor may decrease (Fig. 50 - A) or it may first increase
before decreasing (Fig. 50 - B). As a result, for some initial orientations, the tensile normal
stress required to continue plastic deformation may actually decrease after yield (Fig. 50 - B).
This effect is referred to as geometric softening.
A
Initial
elastic
strain
B
Plastic deformation
Figure 50: Normal stress vs strain observed for hcp metal single crystals. A - decreasing Schmid
factor. B - initially increasing Schmid factor (Geometric softening).
DH56
4.16.2
Course D: Mechanical Behaviour of Materials
DH56
fcc metals
During the deformation of fcc metals, one or two slip systems will operate, depending upon
the orientation of the crystal. The critical resolved shear stress will vary during deformation,
as shown below.
Stage
Stage
III
II
Initial
elastic
strain
Stage
I
Plastic deformation
Figure 51: Shear stress vs shear strain typically observed for fcc metal single crystals.
Stage I: One operative slip system. The tensile axis rotates towards the slip direction. The
critical resolved shear stress remains constant but the Schmid factors on all slip systems
change. The extent of Stage I is determined by the initial orientation of the crystal. Stage I is
often referred to as easy glide.
Stage II: Two operative slip systems (duplex slip). The tensile axis has rotated into a position
in which two slip systems share the largest Schmid factor. Dislocations moving on these slip
systems interact with each other producing jogs, locks and pile ups. This leads to a rapid
increase in the critical resolved shear stress required to move further dislocations on these
slip systems. The increase in strength with plastic strain that arises by this process is referred
to as work hardening or strain hardening .
Stage III: The resolved shear stress becomes sufficient to activate other slip systems and
allow dislocations to bypass obstacles. The transition to stage III correlates with stacking fault
energy in fcc metals as cross-slip occurs more readily with higher stacking fault energies and
therefore the transition in such materials occurs at lower applied stresses.
DH57
4.16.3
Course D: Mechanical Behaviour of Materials
DH57
Polycrystalline metals
In a polycrystalline metal, the orientation of the individual grains and hence also their Schmid
factors are different. The stress-strain curve (Fig. 52) shows no Stage I deformation behaviour
as duplex slip and work hardening initiates after different plastic strains in each grain. As a
result, the stress-strain curve shows a continuous work hardening after yield.
Initial
elastic
strain
Plastic deformation
Figure 52: Normal stress vs strain typically observed for polycrystalline fcc metals.
For randomly orientated grains the average value of the Schmid factor is ∼1/3. This average value is referred to as the Taylor factor. Hence for polycrystalline materials it might be
expected that
σy ≈ 3τc
(4.73)
In practice, the yield stress of polycrystalline metals is often much higher than this due to
effect of grain boundaries. In a polycrystal, the deformation of each individual grain has to be
compatible with that of its neighbours – i.e. there is a strong constraint effect. Multiple slip
is normally required from the outset in virtually all grains in order to satisfy this requirement,
and substantially higher stresses are needed for yielding and plasticity, compared with single
crystals.
DH58
5
5.1
Course D: Mechanical Behaviour of Materials
DH58
Strengthening mechanisms
Forest hardening
The dislocation density, ρ, is a measure of the total dislocation line length per unit volume or,
equivalently, the number of dislocations intersecting unit area. It has units of m–2 .
The average spacing between dislocations, L, is related to the dislocation density. If we
assume we have a simple cubic array of dislocations, then an area of L2 has 1 dislocation
passing through it. Hence, ρ = 1/L2
With higher dislocation densities, dislocation motion becomes increasingly difficult due to the
presence of sessile dislocations blocking the easy glide of glissile dislocations. A greater
shear stress is therefore required to push dislocations through the material resulting in higher
yield strengths. This effect is known as forest hardening.
Figure 53: TEM micrograph of a creep deformed nickel alloy showing regions with high and low
dislocation densities.
Dislocation densities in metals range from ∼ 1010 m–2 in annealed metals to ∼ 1016 m–2 in
cold worked metals. These correspond to spacings of ∼ 10 µm to 10 nm.
DH59
5.2
Course D: Mechanical Behaviour of Materials
DH59
Dislocation pile ups and the effect of grain size
Dislocations will move along a slip plane until they encounter an obstacle. A grain boundary
forms such an obstacle as the dislocation cannot pass from one grain to the next. Instead the
grain must exert a sufficient stress on the neighbouring grain to initiate slip in it. Dislocations
travelling on the same slip plane will therefore accumulate at grain boundaries, forming a
dislocation pile-up, see Fig. 54.
Similar dislocations travelling on the same slip plane will repel each other. If there are n
dislocations in a pile up, the stress at the obstacle will be n times the applied stress.
B
A
Figure 54: Dislocation pile up a a grain boundary.
The effect of grain size on yield strength can now be understood. For macroscopic yielding
of the metal, slip must be initiated in all grains. If slip in a favourably orientated grain (A) is
underway a dislocation pile-up must be established that is capable of exerting sufficient stress
on the neighbouring grain (B) to initiate yield. With a large grain size it is possible to reach
the necessary stress by forming a long pile-up. If the grain size is small only shorter pile-ups
can be supported and hence a greater stress is required.
Experimentally, the tensile yield stress, σy , is seen to be related to the diameter of the grains,
d by the Hall-Petch relation.
k
σy = σ◦ + √
d
in which, σ◦ is the intrinsic yield stress and k is a constant for the material.
Neighbouring grains:
→ constraint on deformation (“back-stress” on dislocations)
→ yield stresses much higher than for single crystals
→ finer grain size usually gives higher yield stress
(5.74)
DH60
5.3
Course D: Mechanical Behaviour of Materials
DH60
Solid solution strengthening
The effect of solute atoms can best be understood by considering the interaction between the
stress fields they produce and those of the dislocations with which they interact. Substitutional
and interstitial solute atoms act differently and must be considered separately.
vacancy
substitutional
defects
interstitial
defect
Figure 55: Lattice distortions associated with point defects
5.3.1
Substitutional solute atoms
If the substitutional solute atom is larger than the atoms in the crystal it resides in it will
introduce a spherically symmetric compressive stress field. If the solute atom is smaller, it will
introduce a spherically symmetric tensile stress field.
As the stress fields induced by either large or small solute atoms contain no shear stress
component, they will not interact with screw dislocations, because the stress fields of screw
dislocations involve shear only.
Substitutional solute atoms may, however, interact with the stress fields generated by edge
dislocations. It is favourable for the solute atoms to position themselves such that the overall
strain energy is minimised.
DH61
Course D: Mechanical Behaviour of Materials
DH61
Larger solute atoms will position themselves in the tensile field of the dislocation (below the
extra half plane of an edge dislocation - see Fig. 56(a)). Smaller solute atoms will position
themselves in the compressive field (at the end of the extra half plane of an edge dislocation
- see Fig. 56(b)). In practice, dislocations are much more mobile than substitutional solute
atoms. As such, the presence of solute atoms in the slip plane can be considered on the
basis of their relative size and where they sit relative to the gliding dislocation.
(a)
(b)
Figure 56: (a) Larger solute atoms in the tensile field of an edge dislocation. (b) Smaller solute atoms
in the compressive field of an edge dislocation.
For example, if a small solute atom sits just above the slip plane of an edge dislocation,
it will attract the dislocation as this will form an energetically favourable configuration (see
Fig. 56(b)). Similarly, if it was located on the other side of the slip plane it would be expected
to repel the dislocation.
As it is energetically favourable for the dislocation to be in certain arrangements with substitutional solute atoms, the dislocation will spend more time in these energetically favourable
configurations. Overall this results in dislocation motion being retarded with a greater shear
stress being required to move dislocations.
DH62
5.3.2
Course D: Mechanical Behaviour of Materials
DH62
Interstitial solute atoms
Interstitial solute atoms typically have a stronger effect than substitutional solute atoms. This
may be attributed to the larger strains they typically generate and, more importantly, they
may also produce asymmetric distortions of the crystal structure. The best known example is
carbon atoms in α-iron.
Octahedral
interstice
a
Figure 57: Carbon atom occupying an octahedral interstice in α-iron.
The octahedral interstice occupied by the carbon atom has two iron atoms
√ at a distance of
a/2 from the centre of the interstice and four iron atoms at a distance of a/ 2 from the centre
of the interstice. If the atom in the interstice has a diameter greater than a − 2rFe it will induce
an asymmetric distortion of the lattice.
Such asymmetric distortions can interact with both edge and screw dislocations.
The comparative effectiveness of solid solution strengthening with substitutional and interstitial additions is shown in Fig. 58
C,N
150
Δσy (MPa)
Si
100
Mn
50
0
Mo
Ni
0
1.0
2.0
% alloying element
Figure 58: Comparative effect of solid solution strengthening in α-iron. (from: F.B. Pickering and T.
Gladman, ISI Special Report 81, Iron and Steel Inst., London, p109 (1963)).
DH63
Course D: Mechanical Behaviour of Materials
DH63
Interstitial diffusion rates can be very high (see Fig. 59). In steels carbon atoms may therefore
migrate to dislocations very quickly (in a few hours at room temperature, a few seconds at
150◦ C, or virtually instantaneously at 300◦ C).
Diffusion distance (µm)
100
10
300˚C
1
150˚C
0.1
20˚C
0.01
0.001
0
10
1
10
2
10
3
10
4
10
5
10
Time (s)
Figure 59: Diffusion distances for (interstitial) carbon in α-iron.
These rapid diffusion rates can allow carbon atoms to locate themselves in every crosssectional plane containing the dislocation (see Fig. 60). These accumulations of interstitial
carbon atoms at dislocations are said to form carbon atmospheres or Cottrell atmospheres.
Figure 60: Schematic depiction of the formation of a carbon atmosphere along the length of a dislocation in a steel.
Cottrell atmospheres can also have a profound effect upon the yield behaviour as a function
of temperature.
DH64
Course D: Mechanical Behaviour of Materials
DH64
Lüders bands
The stress-strain curve of a low carbon steel is shown schematically in Fig. 61.
σ
upper yield stress
lower yield stress
Luders bands
unyielded
yielded
ε
Figure 61: Schematic illustration of the occurrence of Lüders bands associated with a localised drop
in yield stress, observed with low-carbon steels.
The steel first starts to deform plastically at the upper yield stress as the dislocations escape
from their Cottrell atmospheres.
During a tensile test, stress concentrations towards the ends of the sample will initiate yield
in those regions. As the dislocations in these regions are no longer pinned by their Cottrell
atmospheres, the local yield stress will now be lower than that of the rest of the sample.
Plastic deformation is therefore concentrated in these regions. However, as they deform they
work harden and the local yield stress will again rise.
Plastic deformation may continue at the lower yield stress by the movement of the boundary
between the yielding and unyielding regions of the sample. These boundaries, or Lüders
bands typically lie at ∼50◦ to the tensile axis and move together towards the centre of the
sample. When they meet, continued plastic deformation will lead to work hardening and a
progressive increase in the yield stress.
DH65
Course D: Mechanical Behaviour of Materials
DH65
Portevin-Le Chatelier effect
As the temperature increases, the shape of the stress-strain curves changes markedly as the
mobility of carbon atoms increases.
At high temperatures (∼300◦ C) carbon atoms are sufficiently mobile to move with the dislocations. As such, they no longer inhibit dislocation motion and provide no strengthening
effect.
At intermediate temperatures (∼150◦ C) the carbon atoms and dislocations move at comparable speeds. This enables dislocations to repeatedly escape their Cottrell atmospheres and for
them to reform. This produces serrations in the stress-strain curve. This behaviour is known
as the Portevin–Le Chatelier effect.
σ
room temperature
150 ˚C
300 ˚C
ε
Figure 62: Schematic stress-strain plots for low carbon steel, showing the Portevin-Le Chatelier effect
with increasing test temperature
DH66
5.4
Course D: Mechanical Behaviour of Materials
DH66
Precipitate hardening
A highly effective method of restricting dislocation motion is through the use of precipitates in
the crystal matrix. Small coherent precipitates may be sheared by the passage of a dislocation.
(a)
(b)
matrix
b
(c)
r
precipitate
Figure 63: Shearing of a coherent precipitate by a dislocation. (a) the dislocation approaches the
precipitate. (b) the precipitate has passed part way through the precipitate. (c) the dislocation has
exited the precipitate, leaving it sheared by the Burgers vector.
For precipitate shearing in this fashion, the increase in shear stress required for dislocation
√
motion, ∆τ , is proportional to the square root of the precipitate radius, i.e. ∆τ ∝ r.
Larger precipitates will normally have a different crystal structure from the matrix. This prevents dislocations from the matrix passing directly into the precipitate. However, they may
bow around the precipitate, leaving a dislocation loop around the precipitate (Fig. 64). This is
know as Orowan bowing.
(a)
(b)
(c)
(d)
(e)
Figure 64: Schematic depiction of Orowan bowing of dislocations around a pair of large, unshearable
precipitates leaving dislocation loops around them.
DH67
5.4.1
Course D: Mechanical Behaviour of Materials
DH67
Stress required to bow dislocations between precipitates
Consider a dislocation bowing between two precipitates separated by a distance, L (Fig. 65).
The peak curvature of the dislocation line, and hence also the peak energy, will occur when
the dislocation forms a semi-circle between the precipitates.
R
dθ
dl
θ
r
L
Figure 65: Force on a short segment of a dislocation to bow it between two precipitates.
For a segment of the dislocation of length, dl, the force acting normal to the segment is F dl.
Thus the resolved component of the force acting in the vertical direction is
Z π
Z π
L
FL
F sin θdl =
F sin θ
dθ =
[− cos θ]π0 = F L = τ bL
(5.75)
2
2
0
0
This force is balanced by the line tension of the dislocation in the downwards direction at each
precipitate. The line tension is equivalent to the energy per unit length of the dislocation at
each precipitate and is given by Λ = Gb2 /2. Thus,
2
Gb
(5.76)
τ bL = 2
2
The bowing stress or Orowan stress required to bow a dislocation between precipitates separated by a distance, L, is therefore
τ=
Gb
L
(5.77)
Note that as the Orowan stress is inversely proportional to the precipitate spacing we need
very fine arrays of precipitates to provide significant strengthening.
DH68
5.4.2
Course D: Mechanical Behaviour of Materials
DH68
Transition from cutting to bowing and maximum hardening
If the precipitates are very small (and coherent) they may be sheared readily providing little hardening. As their radii increases (and number density decreases for constant volume
√
fraction), so the resistance to dislocation cutting will increase with r. Eventually it becomes
easier for the dislocations to bow between the precipitates rather than cut them. Further increases in precipitate radii will result in a decrease in the critical resolved shear stress with
1/r. A peak strength will be seen at the transition from precipitate cutting to Orowan bowing
(Fig. 66).
Δτc
Cutting
Bowing
r
Figure 66: Change in the critical resolved shear stress as a function of precipitate radius in precipitate
hardened alloys showing the transition from cutting to bowing mechanisms.
DH69
5.4.3
Course D: Mechanical Behaviour of Materials
DH69
Changing strengthening mechanisms during age hardening
In nucleating and growing precipitates in an alloy from a supersaturated solid solution, several factors will contribute to the strength of the alloy and these will change with time (see
Fig. 67).
• Solute hardening
At short ageing times the majority of the strength of the alloy will come from solid solution strengthening in the supersaturated solid solution. This will decrease with time as
solute is rejected from the matrix to form the precipitates.
• Coherency strains
The coherent and semi-coherent precipitates that form in the early stages of ageing will
distort the lattice in which they sit. This induces coherency strains in the matrix which
extend the range of influence of the precipitates and inhibit dislocation motion. At this
stage the coherent precipitates may be readily sheared by dislocations.
• Precipitate cutting
As the coherent and semi-coherent precipitates become larger significantly larger stresses
are required to shear the precipitates.
• Orowan bowing
As the precipitates become large and the distance between precipitates increases, dislocation bypass by bowing becomes possible. This becomes easier as the precipitate
array becomes coarser (over-ageing).
yield strength, σy
overall yield stress
precipitate
cutting
solution
strengthening
precipitate
bypass
(Orowan
coherency
strengthening
aging time
Figure 67: Origins of changes in yield stress of an age hardenable alloy as a function of heat treatment
time.
DH70
Course D: Mechanical Behaviour of Materials
DH70
Precipitate hardening in Al-Cu alloys: If we quench an Al-Cu alloy from the single phase
field we obtain a supersaturated solid solution. Ageing the alloy will first lead to coherent
Gunier-Preston (GP) zones, then coherent θ”, then semi coherent θ0 and finally incoherent
precipitates θ. As the precipitates age their average size and spacing will increase. This will
be accompanied an initial increase in yield stress with ageing time followed by a decrease in
yield stress with longer ageing times, see Fig. 68.
Figure 68: Hardness of Al-Cu alloys (related to yield stress) as a function of heat treatment time at
two different temperatures. (from: J. Silcock, et al., Inst. Metals, 82, 239, 1953-54.)
• Higher peak strength at lower T
• Longer to reach peak strength at lower T (note the log scale)
• Sometimes get multiple peaks (optimum precipitate size of a given type)
DH71
Course D: Mechanical Behaviour of Materials
DH71
Images of the precipitates in an Al-Cu alloy are shown in Fig. 69.
Figure 69: TEM micrographs of Al-Cu alloys at different stages of ageing:
GP zones (from: R.B. Nicholson & J. Nutting, Phil. Mag., 3, 531, 1958), θ” (from: R.B. Nicholson et al., J. Inst.
Metals., 87, 431, 1958-59), θ0 (from: G.C. Weatherly & R.B. Nicholson, Phil. Mag., 17, 813, 1968), θ (from: G.A.
Chadwick, Metallography of phase transformations, Butterworths, London, 1973).
The nature of the precipitates also has a pronounced effect upon the rate of strain hardening
(see Fig. 70). Whilst GP zones may provide significant strengthening, once slip starts the
σ
Aged to
peak hardness
Underaged to produce GP zones
Overaged
Solid solution
Pure Al
ε
Figure 70: Schematic illustrations of the stress-strain curves of Al-Cu alloys after different stages of
ageing.
precipitates may be repeatedly sheared by dislocations and the yield stress may actually
decrease. As dislocations may pass through the entire crystals relatively unhindered the rate
of strain hardening is low.
In the peak hardened and overaged states, dislocations must bow around precipitates. The
increase in dislocation debris (loops, tangles, etc) left by the process inhibits the passage of
further dislocations and the material strain hardens rapidly.
DH72
5.5
Course D: Mechanical Behaviour of Materials
DH72
Partial dislocations and stacking faults
In practice, it is often energetically favourable for dislocations to dissociate into dislocations
with smaller Burgers vectors.
Consider the (111) plane of an fcc metal. The crystal structure is restored after the passage
of a perfect dislocation that shifts atoms in a B position to the next B position. The Burgers
vector required to do this is shown by b1 = a/2[1̄01] in the figure below.
A
A
b2
B
A
C
b3
b1
A
C
A
C
B
A
C
Figure 71: The (111) plane of an fcc metal. Atoms in the A layer are drawn. The B and C layer atomic
positions are marked.
An alternative path between B positions is to first shift an atom in a B position to a C position
and then on to the next B position. This requires the passage of two smaller dislocations that
are not lattice vectors, and are shown by b2 = a/6[2̄11] and b3 = a/6[1̄1̄2]. These are referred
to as partial dislocations.
In an fcc metal these partial dislocations will have Burgers vectors of a/6h112̄i
After passage of the first dislocation with Burgers vector b2 , the stacking sequence will
change from
A B C A B C A B C
to
A B C A C A B C A
This generates a stacking fault. The passage of the second dislocation with Burgers vector
b3 will restore the original structure. An illustration of a stacking fault formed between two
partial dislocations is shown in Fig. 72.
Consider the dissociation of the dislocation b1 into the two partial dislocations, b2 and b3 .
a
a
a
[1̄01] → [2̄11] + [1̄1̄2]
2
6
6
(5.78)
The energies of these dislocations are proportional to
a2
a2 a2
a2
→
+
=
2
6
6
3
(5.79)
Hence, the dissociation of the dislocation into partial dislocations is energetically favourable.
DH73
Course D: Mechanical Behaviour of Materials
DH73
Figure 72: A stacking fault in a fcc (or hcp) metal.
Once a perfect dislocation has dissociated into two partial dislocations, the elastic stress
fields will cause the partial dislocations to repel each other, generating an extended stacking
fault between them. As there is an energy penalty associated with forming the stacking fault
(stacking fault energy). The width of the stacking fault will be limited by the condition where
Monday, 23 January 12
the net repulsive force (or reduction in elastic strain energy) between the partial dislocations
equals the total stacking fault energy.
Hence, if the stacking fault energy (per unit area) is low, the two partial dislocations may be
well separated. If the stacking fault energy is high, then these two partial dislocations will be
close together.
The stacking fault energy has a strong influence on the ease with which cross-slip may occur
in fcc metals. A perfect screw dislocation may cross-slip readily. However, as partial dislocations have both screw and edge components they may not cross-slip directly. Instead they
must recombine before cross-slip can occur. As a consequence, the onset of Stage III is
delayed in metals with low stacking faults energies (e.g. brass and stainless steels) and they
will work-harden rapidly.
Table 2: Stacking fault energies of selected fcc metals (in mJ m−2 ).
Ag
16
Al
166
Au
32
Cu
45
Ni
125
Pd
180
Pt
322
Rh
750
Ir
300
DH74
5.6
Course D: Mechanical Behaviour of Materials
DH74
Order hardening
For those compounds that show an transition between ordered and disordered (related) structures an abrupt change in strength is often seen. For example, in β brass a large change in
the critical resolved shear stress is observed around the order-disorder transition at 470◦ C.
τc (MPa)
Long range order
0.6
0.5
0.2 0
5
4
3
2
1
400
440
Tc 480
Temperature ˚C
520
Figure 73: Effect of order/disorder on the critical resolved shear stress of β brass [from: N. Brown
in Mechanical Properties of Intermetallic Compounds, J.H. Westbrook (ed), John Wiley and Sons,
(1960)]
In bcc metals, like high temperature β brass, slip would be expected on the {110}h11̄1i slip
systems. The associated Burgers vectors are a/2h11̄1i. However, these are not lattice vectors in the CsCl structure, like low temperature β brass. The passage of such a dislocation
through a crystal with the CsCl structure would therefore generate a layer with energetically
unfavourable like-like bonds.
This energetically unfavourable layer is called an anti-phase boundary and is a 2D planar
defect.
The passage of a second a/2h11̄1i dislocation following the first one will restore the original
structure. The two dislocations (and their anti-phase boundary) collectively form a superdislocation and the individual dislocations are often referred to as superpartial dislocations.
Superdislocations can be incredibly effective in strengthening materials at high temperature
(more on this in Course F).
Antiphase boundaries are therefore associated with dislocations at either end in a similar
way that stacking faults are associated with partial dislocations, as described previously. As
before, if the energy associated with the antiphase boundary is low these two dislocations
DH75
Course D: Mechanical Behaviour of Materials
DH75
[001]
[110]
Figure 74: (11̄0) plan view of the CsCl structure showing the formation of an anti-phase boundary
after a displacement of a/2[111].
may be well separated. If the energy associated with the antiphase boundary is high, these
two dislocations will need to be close together.
Note that in the CsCl structure, the shortest vectors between lattice points are ah110i. These
are much longer than a/2h11̄1i dislocations and therefore have a much greater energy associated with them.
DH76
5.7
Course D: Mechanical Behaviour of Materials
DH76
Plastic deformation by cooperative shear - Twinning
Whilst dislocation motion is the predominant deformation mechanism in most metals at room
temperature, plastic deformation may also occur by other mechanisms.
Another important deformation mechanism is the co-operative (simultaneous) shearing of
successive atomic planes. If the crystal structure formed by this process differs from the
original structure it is termed a martensitic phase transformation. If the crystal structure
formed is the same as the original, but in a different orientation, it is termed deformation
twinning or mechanical twinning.
It is called twinning because the crystal either side of the twin boundary are mirror images of
each other.
Deformation twinning occurs through a combination of a twinning plane and a twinning direction. These are analogous to the slip plane and slip direction that define deformation by
dislocation slip.
Table 3: The twinning planes and twinning directions seen in selected crystal structures.
Crystal structure
bcc
bct
fcc
Twinning plane
{112}
{011}, {102}, etc
{111}
Twinning direction
h111̄i
h100i, h2̄01i, etc
h112̄i
As with other shear processes, the shear associated with twinning (twinning shear) is simply
the displacement parallel to the plane divided by the interplanar separation.
Slip
→ Lattice unaffected
→ can generate large shear displacements on a single plane (by repeated passage of dislocations)
Deformation Twinning
→ can generate similar type of displacement, but homogeneously distributed
→ elastic strain due to constraint from surrounding material (in polycrystals)
DH77
Course D: Mechanical Behaviour of Materials
DH77
Twinning in fcc metals
Fcc metals rarely undergo significant deformation twinning, except under specific conditions
(e.g. Cu or Ag-Au alloys at cryogenic temperatures and alloys with low stacking fault energies, such as TWIP steels and Ni-Co alloys). However, the twinning process in fcc metals is
perhaps the easiest to visualise. It requires the shear of each successive {111} plane. This
is shown schematically in Fig. 75 below.
{111} in fcc
Layer A
Layer B
Layer C
(a/6)[112]
A
C
B
A
C
B
A
C
B
A
twin
B
A
C
B
A
C
C
B
A
C
B
A
C
B
A
A
C parent
B
A
B twin
C
A
C
B parent
A
parent
[001]
[110]
[112]
Figure 75: Stacking sequence changes during shearing of {111} planes in the fcc structure to form a
twin.
Each successive
atomic plane is displaced by a/6h112̄i. The magnitude of this displacement
√
is δ = a/ 6. Hence, the twinning shear in fcc metals is
√
δ
a/ 6
1
= √ =√
(5.80)
Twinning shear =
d111
a/ 3
2
DH78
Course D: Mechanical Behaviour of Materials
DH78
Effect of temperature: Deformation twinning is athermal (does not require thermal activation). This is in contrast to dislocation glide, which does show temperature dependence,
particularly if cross-slip or climb are required. Deformation twinning is therefore favoured
at lower temperatures. The temperature dependence of dislocation glide and deformation
twinning are shown in Fig. 76 below.
τcrit
Ambient T
Twinning
Slip
(high strain rate)
Slip
(low strain rate)
T
Figure 76: Effect of temperature and strain rate on dislocation slip and deformation twinning.
Effect of strain rate: Deformation twinning typically requires higher shear stresses than
dislocation glide (at room temperature). However, it occurs very quickly, propagating at the
speed of sound in the material. As such, deformation twinning may become favoured at
higher strain rates.
Effect of crystal structure: Deformation twinning is more likely to occur in crystal systems
in which limited slip systems are available (e.g. hcp metals) or where dislocation mobility is
impaired.
DH79
Course D: Mechanical Behaviour of Materials
DH79
Morphology of twins
Deformation twins typically display a lenticular (lens like) shape as it is necessary to maintain
strain compatibility at grain boundaries. In contrast, annealing twins formed naturally at high
temperature in the absence of stress typically display straight interfaces. Such twins are
common in materials with very low twin boundary energies.
Annealing Twins
Deformation Twins
Figure 77: Schematic representation of the morphologies of annealing and deformation twins.
Figure 78: Annealing twins in an α brass (DoITPoMS micrograph library number 430)
Figure 79: Deformation twins in commercially pure
Ti. (Courtesy of N.G. Jones)
All twins crystallographically-related to parent grain
Annealing twins
• formed by reconstructive (diffusional) processes during heat treatment (often as recrystallisation occurs)
• boundaries usually straight (fully coherent)
• common in materials with very low twin boundary energy
Deformation twins
• formed by cooperative shear, accommodating imposed shear strain
• often lens-shaped, since anchored to grain boundaries at ends
DH80
6
Course D: Mechanical Behaviour of Materials
DH80
Fracture
The catastrophic extension of cracks throughout the thickness of a component is termed
fracture. In brittle materials, where no plastic deformation occurs, brittle fracture is to be
expected. However, it is also possible for materials that exhibit plastic deformation to fail in
a similar catastrophic fashion by ductile fracture. In both cases, understanding the circumstances under which fracture may occur is therefore critical and has been addressed through
the field of fracture mechanics.
In brittle materials, the path of the fracture surface may pass straight across grain boundaries in transgranular fracture or follow the path of grain boundaries in intergranular fracture.
Examples of the fracture surfaces from these two modes of failure are shown below.
Figure 80: Example of a transgranular fracture surface on a notched low carbon ferritic steel (DoITPoMS micrograph library number 143).
Figure 81: Example of a intergranular fracture surface on HY100 forging steel (DoITPoMS micrograph
library number 146).
DH81
6.1
Course D: Mechanical Behaviour of Materials
DH81
Estimate of ideal fracture stress
The ideal fracture stress of a material can be estimated by considering the interatomic forces
between atoms.
If the atomic bonds across the surfaces that are being pulled apart are separated by a distance of r◦ , there will be A/r◦2 bonds crossing the area, A. The force on each bond is therefore
F = σr◦2
(6.81)
Given that the force on a bond may be related to the potential energy of the bond, U , by
dU
dr
(6.82)
1 dU
r◦2 dr
(6.83)
F =
The stress across the bond is
σ=
The Young’s modulus is related to the potential energy of the bond by
1 d2 U E=
r◦ dr2 r◦
(6.84)
Using the Lennard-Jones potential, the Young’s modulus can be approximated by
E=
72Umin
r◦3
(6.85)
Hence,
Er◦3
(6.86)
72
Substituting this expression for Umin into the Lennard-Jones potential, the stress can be obtained
E r◦ 7 r◦ 13
σ=
−
(6.87)
6
r
r
Umin =
This has a maximum value at
r = r◦
13
7
1
6
(6.88)
Substituting this value into the equation for the stress, the maximum stress is therefore
σmax =
7
169
7
13
1
6
E = 0.0373E
(6.89)
For a material with a Young’s modulus of 300 GPa, a value typical of many ceramics, the expected fracture stress is therefore ∼ 11.2 GPa. In practice is it almost impossible to approach
such theoretical fracture stresses due to the presence of cracks, holes or flaws in the material.
We will examine this effect in the next section.
DH82
6.2
Course D: Mechanical Behaviour of Materials
DH82
Griffith criterion
Alan Arnold Griffith, one of the pioneers of fracture mechanics, used energetic considerations
to determine when a crack present in a material under stress will grow. For such a crack to
grow, sufficient energy must be supplied to allow the crack to extend. This energy can come
from the stored elastic strain energy in the material or work done by the applied stress.
Following this approach, consider a plate subjected to an applied stress, σ, with a pre-existing
crack of length 2c at its centre, as shown in Fig. 82. As the crack extends by a length, dc,
at each end it relaxes the elastically strained material around it. This is shown by the dotted
region.
σ0
dc
c
σ0
Figure 82: Stress-free region shielded by a crack from the applied load.
The strain energy per unit volume in an elastically stressed material is given by
σ2
Eε2
= ◦
2
2E
(6.90)
The strain energy released by the extension of the crack by dc at each end is the product of
this expression and the increase in the stress-free volume.
The shape of the stress-free region is not well-defined, and the stress was not uniform within
it before the crack advanced, but taking the relieved area to be twice that of the circle having
the crack as the diameter gives a fair approximation. Thus, for a plate of thickness, t, the
energy released during incremental crack advance under a stress of σ◦ is given by
dW =
σ◦2
2σ 2 πc t dc
2 (2πc t dc) = ◦
2E
E
(6.91)
DH83
6.2.1
Course D: Mechanical Behaviour of Materials
DH83
The strain energy release rate
The concept of stored elastic strain energy being released as the crack advances is central to
fracture mechanics. The strain energy release rate (crack driving force) is usually given the
symbol G (not to be confused with shear modulus or Gibbs free energy!). It is a ‘rate’ with
respect to the creation of new crack area (and so has units of J m–2 ) and does not relate to
time in any way. It follows that
2 2σ◦2 πc t dc /E
dW
σ◦ c
G=
=
=π
new crack area
(2t dc)
E
(6.92)
The value of the constant (π in this case) is not well defined. It depends on specimen geometry, crack shape/orientation and loading conditions. In any event, the approximation used
for the stress-free volume is simplistic. However, the dependence of G on (σ◦2 c/E) is more
general and has important consequences.
G∼
σ◦2 c
E
The strain energy release rate is therefore:
* Larger for larger cracks (flaws)
* More energy released (per unit of new crack area) when longer cracks propagate
* Larger under larger applied stresses
6.2.2
Fracture energy (crack resistance)
In order for crack propagation to be possible, the strain energy release rate must be greater
than or equal to the rate of energy absorption (expressed as energy per unit area of crack).
This energy requirement is sometimes known as the Griffith criterion. For a brittle material this
fracture energy is simply given by 2γ (where γ is the surface energy, with the factor of 2 arising
because there are two new surfaces created when a crack forms). It can be considered as
a critical strain energy release rate, Gc . It is a material property. It is sometimes termed the
crack resistance or work of fracture. The fracture strength can therefore be written as
G ≥ Gc = 2γ
(6.93)
Therefore, combining this inequality with the expression for the strain energy release rate we
obtained earlier
2 σ◦ c
π
≥ Gc = 2γ
(6.94)
E
Rearranging this expression, the fracture stress, σ∗ , is
r
σ∗ =
2γE
πc
(6.95)
DH84
6.2.3
Course D: Mechanical Behaviour of Materials
DH84
Critical flaw size and ’toughening’ by flaw removal
If a component is to be subjected to a particular stress level in service, then the concept
emerges of a critical flaw size, c∗ ,which must be present in order for fracture to occur
c∗ ≈
2γE
πσ◦2
(6.96)
Fracture stress (MPa)
Fig. 83 shows predicted values of the fracture stress for Al2 O3 obtained using this equation
and data from the databook (Gc = 0.05 kJ m−2 , E = 390 GPa). Note that the tensile strength
of Al2 O3 given in the databook of 500 MPa corresponds to a critical crack size of 0.025 mm .
6
4
2
100
6
4
2
10
0.01
0.1
1
10
100
Critical crack length (µm)
Figure 83: Fracture stress of Al2 O3 as a function of crack size.
If high stress levels (∼several hundred MPa) are to be sustained, then all sizeable flaws (&
1 µm) must be removed. This can often be done by various kinds of surface treatment, some
of which put the surface into compression (close up cracks). However, it should be noted that,
while these procedures can raise the maximum tolerable stress, they do not actually change
the material toughness (fracture energy), which is still equal to 2γ (for a brittle material).
It’s also worth noting that certain conditions, such as a corrosive environment or a repeatedly
cycled load (fatigue), can cause cracks to elongate progressively with time, even if they are
initially below the critical length for fast (unstable) fracture. Clearly this may eventually lead
to unstable (energetically favoured) crack growth. Fortunately, many materials are not brittle,
and so can sustain relatively high stress levels without fracturing, even in the presence of
large flaws.
Can apparently “toughen” by reducing the size of the largest flaw present.
As most flaws are likely to be found on the surface
→ polish surface
→ put surface into compression (eg “shot peen”, quench surface of glass plates) (BUT,
doesn’t raise the fracture energy → that is real “toughening” of material)
DH85
6.3
Course D: Mechanical Behaviour of Materials
DH85
Ductile fracture
In the calculations in the preceding section we did not consider the effect of stress concentrations around the crack tip.
Figure 84: Schematic representation of the ‘lines of force’
that lead to stress concentrations around crack tips.
Inglis examined this mathematically, focussing on the crack tip region, and derived the following expression for the peak stress, σmax , in terms of the applied stress, σ◦ .
r c
(6.97)
σmax = σ0 1 + 2
r
where 2c is the crack length (c for a surface crack) and r is the radius of curvature at the tip.
(It follows that the stress concentration factor for a circular hole is 3.) The stress distribution
ahead of a crack tip is shown schematically in Fig. 85.
σ0
(a)
(b)
c
x
y
2c
r
σmax
σ0
σ0
Figure 85: Stress concentration at a crack tip: (a) loading of a flat plate containing an ellipsoidal crack
and a surface crack and (b) schematic stress distribution in the vicinity of the ellipsoidal crack.
In ductile materials, the high stresses in the vicinity of the crack tip may result in local plastic
deformation producing a zone of plasticity. This is shown in Fig. 86. This local plasticity
limits the peak stress to approximately the yield stress of the material, σy , and increases
DH86
Course D: Mechanical Behaviour of Materials
DH86
the radius of curvature, r, at the crack tip and thereby also reduces the stress concentration
effect. Critically, this local plastic deformation requires work to be done, most of which is
dissipated as heat. This demands a much higher strain energy release rate to propagate the
crack.
σ0
X
σ0
X
Y
σ0
σ
σ0
σ
σmax
X
Y
σ0
Y
σmax = σy
X
σ0
Y
Figure 86: Crack tip shapes and stress distributions for (a) brittle and (b) ductile materials.
Stress concentration at crack tip
•
•
•
•
stimulates plasticity
blunts the crack tip
reduces the peak stress (to ∼ σY )
increases energy absorption
DH87
6.3.1
Course D: Mechanical Behaviour of Materials
DH87
Effect of plasticity on fracture energy
The Griffith criterion is only truly valid for brittle materials. However, Irwin proposed (1948)
that the Griffith condition could be modified to include the plastic work done during crack
advance, γp
G ≥ Gc = 2 (γ + γp )
(6.98)
As before, we can write
π
σ◦2 c
E
≥ Gc = 2 (γ + γp )
which can be rearranged to give the associated fracture stress
r
2 (γ + γp ) E
σ◦ =
πc
(6.99)
(6.100)
The extra energy required by γp make a big difference. Gc for metals (∼ 1-100 kJ m−2 ) Gc
brittle materials (ceramics, glasses etc) (∼ 1-100 J m−2 )
6.3.2
Stress intensity factor
In most cases, the Inglis stress concentration equation (Eqn. 6.97) can be approximated
by
r r
c
c
σmax = σ0 1 + 2
≈ 2σ◦
(6.101)
r
r
Since a critical stress is needed at the crack tip to open up the atomic planes, it follows that
crack propagation is expected when
√
σ◦ c ≥ critical value
(6.102)
where the critical value is expected to be constant for a given material, but to vary between
√
materials (σmax r is constant for a given material). In the 1950’s, Irwin proposed the concept
of a stress intensity factor, K, such that
√
K = σ◦ πc
(6.103)
√
The stress intensity factor (units of MPa m) scales with the level of stress at the crack tip.
Fracture is expected when K reaches a critical value, Kc , the critical stress intensity factor,
which is often termed the fracture toughness.
K ≥ Kc {crack tip conditions} → analogous to G ≥ Gc {global energy balance}
DH88
6.3.3
Course D: Mechanical Behaviour of Materials
DH88
Uniting the stress and energy approaches
There are clearly parallels between K reaching a critical value, Kc , and G reaching a critical
value of Gc . The magnitude of K can be considered to represent the crack driving force,
analogous to G. Consider again the Griffiths energy criterion
2 σ◦ c
G ≥ Gc , i.e. π
=G
(6.104)
E
Therefore,
p
√
σ◦ πc∗ ≥ EGc = Kc
It can be seen that this is similar in form to Eqn. 6.103. It follows that
p
√
K = EG
and
Kc = EGc
(6.105)
(6.106)
Kc → “fracture toughness” (critical stress intensity factor)
Gc → “fracture energy” (critical strain energy release rate)
While it is reassuring to be able to treat fracture from both stress and energy viewpoints, it is
not immediately apparent what advantages are conferred by using a stress intensity criterion,
rather than an energy-based one. However, in practice it is possible to establish K values,
and corresponding Kc values, for various loading and specimen geometries, whereas this is
not really possible with an energy-based approach. (The 3-D stress state at the crack tip, and
hence the size and shape of the plastic zone, can be affected by specimen thickness and
width.) Moreover, rates of sub-critical crack growth (progressive advance of a crack, due to a
corrosive environment or cyclic loading) can often be predicted from the stress intensity factor,
since it is directly related to conditions at the crack tip, whereas the strain energy release rate
is a more global parameter.
DH89
6.3.4
Course D: Mechanical Behaviour of Materials
DH89
Ductile rupture
The preceding section suggests that the Griffith criterion may be applied to ductile materials
and that they should be as sensitive to flaws as brittle materials, although they would have
higher fracture stresses for a given crack size. In practice, failure stresses are not systematically higher for ductile materials, although they certainly fracture less readily and require
much more energy input. Furthermore, ductile materials show little or no sensitivity to initial
flaw size. The differences between these modes of fracture, which are illustrated in Fig. 87
below are clearly important.
Figure 87: Micrographs of fractured specimens, and schematic depictions, of different modes of failure, showing (a) ductile rupture, (b) moderately ductile fracture and (c) brittle fracture.
Very ductile materials often fail by ductile rupture (progressive necking down to a point), with
little or no crack propagation as such. The failure stress in such cases can only be predicted
by analysis of the plastic flow and Eqn. 6.100 is
irrelevant. In practice, such highly ductile materials are too soft to be useful for most purposes.
However, fracture of engineering metals is commonly preceded by significant amounts of plastic flow and some necking, (Fig. 87(b)). Clearly,
use of Eqn. 6.100 requires care in such cases. It
should be noted that, even if the fracture stress
is not very high, a good toughness will ensure
that components are durable and robust under a
range of service conditions. An example of a high
magnification image of fracture surface following Figure 88: Example of a fracture surface
on aluminium (DoITPoMS micrograph library
ductile rupture is shown in Fig. 88.
number 83).
Often get extensive macroscopic plastic flow before fracture → “Necking”
True stress at fracture of the neck may be high, but the nominal stress (apparent strength) is
often quite low
DH90
6.4
Course D: Mechanical Behaviour of Materials
DH90
The Ductile-Brittle Transition Temperature (DBTT)
Toughness is sensitive to plastic flow characteristics, which can change with T . As such,
fracture behaviour can change on heating or cooling. In general, heating leads to increased
plasticity and hence an improved toughness. There is thus less concern about fracture at
higher T , and more about excessive plastic flow. Conversely, as T is reduced, fracture becomes of greater concern.
There is a broad relationship between strain rate and T . Increasing the strain rate is often
similar to reducing T , since both tend to inhibit dislocation motion. Hence, fracture is often
more likely if a higher strain rate is imposed (although relatively large changes in strain rate
may be needed to generate the same effect as a relatively small change in T ).
It is conventional when investigating the onset of brittleness that a fairly high strain rate is used
to reflect the most demanding conditions likely to be encountered in service, e.g. by impacting
a specimen with a pendulum. This also provides a convenient method of characterising the
toughness, since the recovery height of the pendulum after fracturing the specimen gives an
indication of the energy absorbed, which may be related to the specimen toughness.
Figure 89: Impact energy as a function of temperature for several materials, some of which exhibit
ductile to brittle transitions on cooling.
The data shown in Fig. 89 illustrate how the toughness (measured by impact testing) of a few
materials varies with T in the range from 100◦ C to –150◦ C. An fcc metal such as Cu remains
tough down to very low T , whereas an hcp metal like Zn, in which dislocation motion occurs
by basal slip, may be quite brittle even at ambient T . Interest centres mainly on steel (and
other bcc metals). Dislocations are less mobile in bcc metals than in fcc and are more prone
to pinning by interstitials. Grain boundary embrittlement by segregation of impurities, such as
P and S, can also occur in steels. The transition T may be around ambient, or not far below,
so that service conditions may be encountered (e.g. in polar regions) in which the toughness
is much lower than expected. Steels have, however, been developed with good toughness
down to very low T .
DH91
6.5
Course D: Mechanical Behaviour of Materials
DH91
Toughness of composites
In composite materials, both the matrix and reinforcing fibres may be brittle but the composite
may still exhibit considerable toughness. A component such as a vaulting pole will acquire
many, relatively large, surface defects, so the fact that it can sustain such high stresses without fracturing indicates that, as a material, the composite has a high toughness. The main
energy-absorbing mechanism raising the toughness of fibre composites is the pulling of fibres
out of their sockets in the matrix during crack advance - see Fig. 90.
Figure 90: (a) Schematic of a crack passing through an aligned fibre composite, showing interfacial
debonding and fibre pull-out, and (b) a SEM micrograph of a composite fracture surface.
Consider a fibre with a remaining embedded length of x being pulled out an increment of
distance dx. The associated work is given by the product of the force acting on the fibre and
the distance it moves
dW = (2πrx) τi∗ dx
(6.107)
where τi∗ is the interfacial shear stress, taken as constant along the length of the fibre. The
work done in pulling this fibre out completely is therefore given by
Z
∆W =
x◦
2πrxτi∗ dx = πrx2◦ τi∗
(6.108)
0
where x◦ is the pull-out length. The number of fibres per unit sectional area, M , is related to
the fibre volume fraction, f , and the fibre radius, r, by
M=
so the pull-out work of fracture, Gcp , is given by
f
πr2
(6.109)
DH92
Course D: Mechanical Behaviour of Materials
Gcp = M ∆W =
f πrx2◦ τi∗
= 4f s2 rτi∗
πr2
DH92
(6.110)
where s is the pull-out aspect ratio (x◦ /2r). Since there will be a range of values, an (RMS)
average value should be used. The process can make large contributions to the fracture
energy. For example, with f =0.5, s=20, r=5 µm and τi∗ =20 MPa we get a fracture toughness
of ∼80 kJ m-2 , a value typical of a tough metal.
It might be expected that continuous fibres would break in the crack plane, where the stress
is highest, so there would be no pull-out. However, since the fibres are brittle, flaw-sensitive
materials, they will often break some distance from the crack plane, where there is a flaw,
even though the stress is somewhat lower there. Pull-out aspect ratios of the order of 2050 (e.g. pull-out lengths of several hundred µm for a 10 µm diameter fibre) are commonly
observed - see Fig. 90(b).
It may be necessary to carefully control certain fibre and interfacial characteristics in order to
maximise the pull-out toughness.
DH93
6.6
Course D: Mechanical Behaviour of Materials
DH93
Pressurised pipes
A common and important structural application where fast fracture should be avoided is in
pressurised pipes. The material from which they are fabricated should be sufficiently tough
that flaws and cracks can be detected before they cause fast fracture.
Consider a cylindrical pipe of length, L, and radius, r, with an internal pressure, P , as shown
in Fig. 91.
Figure 91: Derivation of the stresses in the wall of an internally pressurised cylinder.
The force associated with the hoop stress, σθ , in the pipe wall must balance the force from
the internal pressure acting over the projected area across the pipe.
σθ (2tL) = P (2rL)
(6.111)
Hence the hoop stress in a pressurised pipe is
σθ =
Pr
t
(6.112)
Similarly, the force associated with the axial stress, σz , in the pipe wall must balance the force
from the internal pressure acting over the projected area along the pipe.
σz (2πrt) = P πr2
(6.113)
Hence the axial stress in a pressurised pipe is
σz =
Pr
2t
It can be seen from these expressions that the hoop stress is twice the axial stress.
Many components & structures in this condition
•
•
•
•
•
Boilers, pressure vessels etc
Hoses, pipelines etc
Many biological structures (blood vessels etc)
Submarines (external pressure → compressive stresses)
Aircraft & spacecraft
(6.114)
DH94
6.7
Course D: Mechanical Behaviour of Materials
DH94
Aircraft stresses and materials
Aircraft are subject to both bending moments and pressurisation stresses. During flight, wings
experience bending moments, generating compressive stresses in the top surface and tensile
stresses in the underside. (During landing, these stresses are reversed.) The bending moment, and hence the stress levels, are highest close to the fuselage. Air regulations require
that the cabin pressure should be at least ∼0.8 bar. At typical long-haul operating heights,
atmospheric pressure is ∼0.2 bar, so the over-pressure ∼0.6 bar. Resultant stresses in the
fuselage skin clearly depend on cabin diameter and skin thickness, but often reach significant
levels. Parts of the fuselage are also subject to bending moments, since it is being supported
where it is attached to the wings, but gravity acts on the parts fore and aft of this region. A
schematic depiction of stress distributions during flight is shown in Fig. 92.
Figure 92: Schematic depiction of the stress distributions in a large aircraft during flight at high altitude
NB Airframes are made of either Al alloy or composite material
→ Both have low density, but neither has a very high fracture toughness
→ Must design to avoid both fast fracture and fatigue crack growth
Create circular flaw (with needle, diameter 100 µm) → c∗ < 1 µm (if rubber ideally brittle) →
fast fracture predicted?
NB Sticky tape → constrains relaxation of rubber → reduces energy release rate
DH95
7
Course D: Mechanical Behaviour of Materials
DH95
Appendix: Observing dislocations
This is non-examinable and has been included for reference only.
High resolution transmission electron microscopy
With modern transmission electron microscopes, it is possible to image the atoms in the
vicinity of the dislocation core directly. From such images it is possible to determine the
Burgers vector of the dislocation from a Burgers circuit.
However, imaging dislocations in this way is challenging as it requires the most advanced
electron microscopes and extremely careful sample preparation.
A useful and effective method of determining the character of dislocations is through diffraction contrast.
Diffraction contrast imaging of dislocations
As shown in Fig. 93, the local distortion of the lattice planes in the vicinity of a dislocation
core may result in it fulfilling diffraction criteria. The diffraction of electrons away from the
transmitted beam in such regions will show the dislocations as dark lines in bright field images.
e-
Strong transmitted beam away
from the dislocation core
e-
Strong diffracted beam in the
vicinty of the dislocation core
Figure 93: Schematic illustration showing the origin of diffraction contrast in transmission electron
microscopy in the vicinity of dislocation cores.
Similarly, if a dark field image is taken using the diffraction spots from the crystal, the region
in the vicinity of the dislocation will not diffract at the same angle as the surrounding crystal
and will again appear dark in the image.
However, as those lattice planes with normals perpendicular to the Burgers vector of the
dislocation remain undistorted, they will not diffract at an angle different to the perfect crystal.
If a bright field image is formed from one of these spots, the dislocation will be invisible.
DH96
Course D: Mechanical Behaviour of Materials
DH96
We can therefore use this phenomenon to determine the Burgers vectors of dislocations by
taking bright field images from a selection of spots with different reciprocal lattice vectors and
seeing in which ones the dislocation lines are invisible. When they are invisible
g·b=0
(7.115)
in which g is the reciprocal lattice vector of the diffraction spot used to generate the bright
field image and b is the Burgers vector of the invisible dislocation line segment.
An example of a g · b analysis is shown in Fig. 94.
Figure 94: A series of transmission electron images of intersecting dislocation segments. The operative strongly diffracting beam g is shown on each image. The Burgers vectors of the various segments
have been indexed using the g · b criterion.
DH97
8
Course D: Mechanical Behaviour of Materials
DH97
Glossary
Terms are listed alphabetically. Words in italics provide cross-references to other entries in
the glossary.
Beam stiffness (Σ) Product of Young’s modulus and moment of inertia, characterising the
resistance to deflection when subjected to a bending moment.
Bending moment (M ) Turning moment generated in a beam by a set of applied forces. The
bending moment is balanced at each point along the beam by the moment of the internal
stresses.
Brittle fracture Fracture not involving gross plastic flow, although some local plastic deformation may occur at the tip of the crack.
Burgers vector (b) Vector giving magnitude and direction of lattice displacement generated
by passage of a dislocation. It is a lattice vector for a perfect dislocation.
Climb Movement of an edge dislocation by absorption or emission of a vacancy.
Coefficient of thermal expansion (α) See thermal expansivity .
Critical resolved shear stress (τc ) Value of resolved shear stress at which slip occurs on a
specified slip system.
Cross-slip Movement of a screw dislocation from one permissible slip plane on to another,
usually in order to bypass an obstacle.
Curvature (κ) Reciprocal of the radius of curvature adopted by a beam subject to a bending
moment. Also equal to the through-thickness gradient of strain in the beam.
Deformation twinning Mode of plastic deformation involving co-operative shear of atomic
arrays into a new orientation of the same crystal structure (reflecting the parent orientation
across the twin plane). Also termed mechanical twinning.
Dislocation line Boundary between slipped and unslipped regions of a crystal.
Dispersion Strengthening Raising of yield stress via obstacles to dislocation glide that are
thermally stable.
Ductile fracture Fracture involving gross plastic flow.
Edge dislocation Dislocation with a Burgers vector normal to the dislocation line.
Engineering strain (ε) Normal strain, given by ratio of change in length to original length.
Also called nominal strain. See also true strain.
Engineering stress (σ) Normal stress, given by ratio of applied force to original sectional
area. Also called nominal stress. See also true stress.
Fracture energy (Gc ) Energy per unit area required to extend a crack. Also termed the critical
strain energy release rate.
Fracture toughness (Kc ) Critical value of the stress intensity factor, characterising the resistance of a material to crack propagation.
DH98
Course D: Mechanical Behaviour of Materials
DH98
Frank’s rule A dislocation reaction will occur only if the energy of the product dislocation(s)
is less than that of the reacting dislocation(s). The energy of a dislocation is proportional to
the square of its Burgers vector.
Frank-Read source A mechanism by which dislocations can multiply.
Glide Motion of a dislocation on its slip plane by small co-operative movements of atoms
close to the dislocation core.
Glissile dislocation A dislocation which is able to glide.
Griffith criterion Energy-based condition for fracture.
Hooke’s Law Stress is proportional to strain during elastic deformation.
Jog A step in a dislocation that does not lie in the slip plane.
Kink A step in a dislocation that lies in the slip plane.
Martensitic transformation Phase transformation occurring via co-operative shear of atomic
arrays into a new crystal structure.
Mechanical twinning See deformation twinning.
Mixed dislocation Dislocation in which the Burgers vector is neither parallel nor perpendicular to the dislocation line.
Neutral axis (of a beam) Axis (strictly, a plane) parallel to the length of a beam, along which
there is no change in length on bending.
Nominal strain (ε) See engineering strain .
Nominal stress (σ) See engineering stress .
Normal strain (ε) Deformation in which change in length is parallel to original length.
Normal stress (σ) Stress induced by a force acting normal to the sectional area to which it is
applied.
OILS rule Non-graphical method to find the slip system with the highest Schmid factor in a
cubic crystal.
Orowan bowing The bowing of dislocations between precipitates that cannot be cut.
Partial dislocation A dislocation for which the Burgers vector is a smaller than a lattice
vector.
Perfect dislocation A dislocation for which the Burgers vector is a lattice vector.
Primary slip system Slip system which first becomes active. Normally the one with the
highest Schmid factor.
Resolved shear stress (τR ) Component of shear stress acting on a slip plane and parallel to
a slip direction in that plane.
Schmid factor (cos φ cos λ) Geometrical factor relating resolved stress to normal stress along
the tensile axis in a single crystal under tension or compression.
Schmid’s law Slip initiates at a critical value of resolved shear stress.
DH99
Course D: Mechanical Behaviour of Materials
DH99
Screw dislocation Dislocation with a Burgers vector parallel to its dislocation line.
Second moment of area (I) Parameter dependent on sectional shape, characterising the
resistance a beam offers to deflection under an applied bending moment. Also called the
moment of inertia .
Sessile dislocation A dislocation which is unable to glide.
Shear modulus (G) Constant of proportionality between shear stress and shear strain.
Shear strain (γ) Distortional deformation (an angle) arising from a shear displacement.
Shear stress (τ ) Stress induced by a force acting parallel to the sectional area to which it is
acting.
Slip system [UVW ] (hkl) Combination of slip plane (hkl) and slip direction [UVW ], which lies
in the (hkl) plane.
Solution strengthening Increase in yield stress of a material through the addition of solute
atoms, caused by the interaction of these solute atoms with dislocations.
Stacking fault A two-dimensional defect in which the sequence of stacking of atomic planes
is interrupted.
Strain energy release rate (G) Elastic strain energy released per unit of created crack
area.
Strain hardening see work hardening.
Stress intensity factor (K) Parameter characterising the crack driving force, in terms of
applied stress level and crack length.
Surface energy (γ) Energy per unit area associated with a free surface.
Thermal expansivity (α) Ratio of change in length to original length, per unit change in
temperature. Also known as coefficient of thermal expansion.
True strain (ε) Ratio of change in length to current length. May differ from the nominal, or
engineering, strain.
True stress (σ) Ratio of force to current sectional area over which it is acting. May differ from
the nominal, or engineering, stress.
Twinning plane Plane within which shear takes place during deformation twinning, which
also forms a mirror plane between parent and twin structures).
Work hardening Increase in plastic flow stress with strain. Associated with an increase
in dislocation density, formation of entanglements, locks etc. Also known as strain hardening.
Yield point Point (stress level = σy ) at which plastic deformation first occurs.
Young’s modulus (E) Constant of proportionality between normal stress and normal strain
.
DH100
Course D: Mechanical Behaviour of Materials
DQ1
Question Sheet 13
1. A titanium wire of circular section has a diameter of 200 µm. What would cause it to
exhibit the greater elongation:
(a) suspending a weight of 1 kg from it or
(b) heating it up by 300◦ C?
(Property data are supplied in the Data Book and the coefficient of thermal expansion
can be obtained from DH18. State any assumptions.)
2. The potential energy, U , between two carbon atoms joined by a single covalent bond
and separated by a distance, x, can be approximated by
U =−
A
B
+ 8
2
x
x
where A = 9.0 × 10-39 J m2 and B = 3.0 × 10-98 J m8 . Calculate the equilibrium separation of the atoms.
If solid carbon had a simple cubic structure (primitive cubic lattice, with one atom per
lattice point) estimate the Young’s modulus, E.
For diamond, E is actually ∼1200 GPa, while for graphite it is 27 GPa (averaged over
all directions). Compare these values with the value you have estimated for the simple
cubic structure, and discuss the reasons for any discrepancies.
Polyethylene ([-CH2 -]n ) contains long hydrocarbon chains but exhibits a much lower
value of E, typically 0.2 to 0.7 GPa, why is this?
3. A composite (60% aligned glass fibres (7 µm diameter) in an epoxy matrix) is in the form
of thin sheets. Using equal strain and equal stress models respectively, calculate the
axial and transverse Young’s moduli.
A “cross-ply” is produced by bonding together several sheets, each with its fibre direction
normal to that of its neighbours. Using an equal strain assumption, calculate the Young’s
modulus of such a material, parallel to either of the two fibre directions.
(Young’s moduli: for glass, E = 76 GPa: for epoxy, E = 3 GPa)
4. A cantilever is loaded with a static force, F , at its tip, as shown in the figure below.
L
x
z
δ
F
Obtain an expression for the deflection, z, as a function of position along the cantilever,
x. You may take the relationship between the curvature, (1/R), and the deflection, z, to
be 1/R = d2 z/dx2 .
DH101
Course D: Mechanical Behaviour of Materials
DQ1
Hence show that the deflection at the tip of the cantilever, δ, is
δ=
F L3
3EI
At what position does the cantilever experience the largest tensile stress? At this position, what stress will be experienced by a solid, square sectioned bar of thickness,
b?
DH102
Course D: Mechanical Behaviour of Materials
DQ2
Question Sheet 14
1. Determine whether the following dislocations in aluminium are edge, screw or mixed.
Identify in each case the slip plane(s) on which the dislocation will be able to glide.
(Al has an fcc crystal structure and its slip system is {111}h11̄0i.)
Burgers vector, b, k to
Line vector k to
(a)
[11̄0]
[1̄10]
(b)
[101]
[112]
(c)
[11̄0]
[1̄1̄2]
2. Aluminium (fcc) slips on {111} planes in h11̄0i directions. List all the physically distinct slip planes in this structure, and for each plane list all the physically distinct slip
directions. How many physically distinct slip systems are there?
A single crystal of aluminium is subjected to tensile load along [123]. For each of the distinct slip systems, evaluate the Schmid factor (cos φ cos λ), and hence determine which
slip system will operate first. Show that the OILS rule leads to the same prediction.
3. A cylindrical single crystal of copper (fcc) is tested in tension along the axis parallel to
[12̄5]. Which slip system has the highest Schmid factor?
As the specimen is deformed, how does the orientation of the tensile axis change?
Deformation is continued until the Schmid factors for two slip systems become equal.
What is the orientation of the tensile axis at this point, and what are the two systems?
4. A bar of polycrystalline aluminium is plastically deformed in tension, to a strain of 10%. It
is found that the flow stress (yield stress) is initially 50 MPa, rising linearly with increasing
strain to 150 MPa at 10% (plastic) strain. Estimate the mechanical work done on the bar
per unit volume (= area under stress-strain plot).
While this plastic deformation is taking place, the dislocation density in the bar increases
by a factor of 10, from 1013 to 1014 m−2 . Using the standard expression for the stored
elastic strain energy per unit length of dislocation line, and assuming that it applies to all
types of dislocation, estimate the increase in stored energy per unit volume in the bar,
associated with this increase in dislocation density. Explain why the mechanical work
done is much greater than the energy associated with the increased dislocation density
and describe the fate of the ‘lost’ energy.
(Al is fcc, with a = 0.405 nm and shear modulus G = 28 GPa)
DH103
Course D: Mechanical Behaviour of Materials
DQ3
Question Sheet 15
1. The following refer to dislocations in a fcc metal. In each case, determine whether the
dislocation will move under the action of the stress, and if so, in what direction will its
line move?
(a) a screw dislocation with b k [110] subjected to a shear stress along [110] on (11̄1).
(b) a screw dislocation with b k [110] subjected to a shear stress along [1̄12] on (11̄1).
(c) an edge dislocation with b k [01̄1] subjected to a shear stress along [01̄1] on (111).
2. The peak yield stress, σy , of an age hardening Al-Cu alloy is 430 MPa. When heavily
over-aged (precipitates too widely spaced to affect dislocations), σy falls to 100 MPa.
Assuming that the difference between these two yield stress values corresponds to the
Orowan bowing stress, estimate the spacing between precipitate surfaces in the peakaged alloy.
In the peak-aged condition, the precipitate morphology is thin cylindrical discs (thickness ∼20% of diameter). Estimate the volume fraction of precipitates present, assuming that they are mono-sized, with a disc diameter of 20 nm, and distributed in a simple
cubic array
(The average peak shear stress on slip planes in a polycrystal (with randomly-oriented
grains) is 1/3 (= Taylor factor) of the applied uniaxial stress. Al is fcc, with a = 0.405 nm
and shear modulus G = 28 GPa)
3. During deformation of an aluminium alloy, a dislocation with Burgers vector b1 = a2 [1̄10]
moving on (111) interacts with a second dislocation with Burgers vector b2 = 2a [101̄]
moving on (11̄1). Using Frank’s rule, show that it is energetically favourable for these
dislocations to add. Describe the character of the resultant dislocation. What is the
significance of the new slip system in terms of continued plastic deformation?
4. Outline the factors favouring deformation twinning during imposed plastic straining. Calculate the twinning shear (= displacement of an atom in the twinning direction / distance
of the atom from the twin-parent boundary plane) for the fcc structure.
If the same shear strain were to be generated by dislocation glide (in a h11̄0i direction), with slip occurring on ({111}-type) planes spaced 1 µm apart, estimate how many
dislocations would need to glide across each slip plane.
(Assume a ≈ 0.4 nm.)
5. The following table lists the values of the tensile yield stress for four copper-based materials. In each case, explain briefly the dominant mechanism which gives rise to the
value of the yield stress.
(a) 99.99% Cu, single crystal, σy = 10 MPa
(b) 99.99% Cu, polycrystalline, annealed, grain size 10 µm, σy = 60 MPa
(c) material (b) after deformation to 50% plastic strain, σy = 325 MPa
(d) Cu - 2 at.% Be, polycrystalline, annealed, grain size 10 µm, σy = 185 MPa
DH104
Course D: Mechanical Behaviour of Materials
DQ4
Question Sheet 16
1. Explain why the strengths of bulk ceramics and inorganic glasses are much lower in
tension than in compression, and outline methods by which the tensile strengths of
brittle materials may be increased.
A material with tensile yield stress, σy , Young’s modulus, E, and toughness, Gc contains
an internal crack of half-length, a. What condition must be obeyed for the material to
yield in tension rather than failing in a brittle manner?
Use this condition and information from the databook to rank the following materials in
order of increasing flaw tolerance: polyethylene, high strength steel, aluminium alloy,
and alumina (Al2 O3 ).
2. A small stone weighing 30 g, thrown up from the tyre of a car, strikes the windscreen of
a following car with an impact velocity of 50 km h−1 . Calculate its kinetic energy.
The windscreen, which has an area of 2 m2 and is 8 mm thick, fractures throughout
in a mosaic of through-thickness cracks, creating segments approximating to squares
of side 5 mm. If the surface energy, γ, of the glass is ∼2 J m−2 , estimate the energy
associated with the newly-created surface area.
Your calculations should indicate that the impact energy is insufficient to create the observed new surface area, even if the fracture is ideally brittle. It’s suggested that the
extra energy could have been supplied by release of residual stresses in the glass,
which were deliberately created during manufacture by quenching the surface layers
from high temperature with air jets (in order to “toughen” the glass). This process imparts a residual compressive stress of ∼30 MPa into 1 mm-thick surface layers on both
sides. Taking the Young’s modulus, E, of the glass to be ∼70 GPa, estimate the associated stored energy (= 1/2 stress × strain × volume) and check whether it is sufficient to
account for the observed fracture behaviour.
Comment on the nature of “toughening” of windscreens achieved in this way.
3. Explain carefully why the Griffith fracture criterion cannot be used to obtain the failure
stress of a typical engineering metal, such as mild steel, being subjected to a conventional tensile test. What information would be needed in order for this tensile strength
to be estimated by analysing the fracture behaviour?
4. The air pressure in a large aircraft is 0.8 atm (∼0.08 MPa) and it flies at an altitude where
atmospheric pressure is 0.2 atm. The fuselage approximates to a cylinder of diameter
∼7 m and skin thickness ∼3 mm, with a length behind the wings of ∼40 m. The weight
distribution in the rear half of the plane, taking account of lift provided by the tailplane,
is equivalent to cantilever loading with a weight of 25 tonnes (25,000 kg) acting on the
end of the fuselage. Where is the peak tensile stress in the fuselage and what is its
orientation and magnitude?
An Air Marshall on the plane inadvertently fires his weapon. Despite prior assurances
that this is impossible, the bullet penetrates the fuselage skin, leaving a jagged hole
∼10 mm in diameter (in the region of peak tensile stress). Neglecting any effects of
depressurisation, will major disaster follow in the form of rapid crack propagation around
DH105
Course D: Mechanical Behaviour of Materials
DQ4
the fuselage? As cabin pressure drops, does the likelihood of such fracture increase or
decrease?
(The fuselage√is made of an aluminium alloy (2024), which has a fracture toughness,
Kc ≈ 40 MPa m. The moment of area, I, of the section of a hollow, thin-walled cylinder,
diameter D, wall thickness t, is approximately πD3 t/8)
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