Chapter 1 Exercises of Chapter I-1: Introduction 1.1 Selection of materials
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Chapter 1 Exercises of Chapter I-1: Introduction
1.1 Selection of materials
Exercise 1.1 Material for a flywheel
This exercise illustrates rational selection of materials. It could have been given as
an example in Sect. 1.1.7 of Volume I.
We wish to design a flywheel to store as much energy as possible given the rotation speed and the thickness. There is no requirement regarding the radius. Which
material should we choose?
1) Give the energy W stored in the flywheel.
Solution
Assume the flywheel to be a disk of constant thickness t
1 2 1M 2 2
Iω =
Rω
2
2 2
M = π R2t ρ
W=
(1.1.1)
I being the moment of inertia of the flywheel, M its mass, R its radius, ω the speed
of rotation and ρ the density of the material.
2) What is the dimensioning factor of a flywheel?
Solution
The centrifugal force should be less than the one which could produce an explosion. This means that the maximum stress should be less than the tensile strength
Rm (provided there is no microcrack present). However, the flywheel stops from
time to time; the maximum stress varies in a cyclic manner between zero and its
maximum value when the rotation velocity is reached. This produces fatigue of
the material. The stress amplitude should be kept below the endurance limit of the
1
2
material (refer to Chapter 6 of Volume II). It can be considered that this endurance
limit is equal to one third to one half of the tensile strength. All being considered,
the dimensioning condition is for the maximum stress not to exceed some fraction
k, including a safety factor, of the tensile strength.
3) Give the location and the value of the maximum stress in the flywheel.
Solution
For a disk, the maximum stresses are located on the axis (refer to mechanics textbooks). The hoop stress and the radial stress are equal there and given by:
σ θ max = σ r max =
3+ν 2 2
R ρω
8
(1.1.2)
4) Write the dimensioning condition
3+ν 2 2
R ρω = kRm
8
(1.1.3)
Being free to choose the dimension of the radius, what is the characteristic ratio of material properties to maximise the stored energy in the flywheel?
Solution
Eliminate R between (1.1.1) and (1.1.3).
W = k2
π
16 ( 3 + ν )
2
t Rm 2
ω2 ρ
(1.1.4)
To maximise W the performance ratio Rm2/ρ should be as high as possible.
5) Indicate the best selection of materials for the flywheel.
Solution
An Ashby map of tensile strength versus density (Fig. 1.1.1) shows that ceramic
materials, composite materials and high strength steels are good candidates. Diamond has a performance ratio equal to about 3x107 in MPa and kg/dm3. That
would be an expensive flywheel! For alumina it is about 2.5x106, 3x105 for high
strength steel, 5x105 for carbon fibre reinforced polymer. For problems in applica-
3
tion, alumina cannot be considered and it turns out that carbon fibre reinforced
polymer is the best choice.
Fig. 1.1.1 Strength-density selection plot
1.2 Mechanical testing
The following exercises aim at a better understanding of mechanical tests described in Sect. 1.3 of Volume I.
Exercise 1.2 Tensile test, Considère's relation
4
We consider a tensile test performed with a cylindrical test piece and we wish to
predict the load versus deformation curves assuming particular constitutive behaviours.
The cross-section of the test piece as a function of time is denoted S(t) and its
length L(t). Assuming constant volume: S(t) L(t) = S(0) L(0), express the relation
between the true strain and the evolution of the cross-section.
a) General case
Assuming a constitutive uniaxial relation σ = σ ( ε , εɺ ) , write the equation relating
the applied load P to the engineering strain denoted η (neglect the elastic deformation).
Find the percentage elongation at maximum force Ag (Considère's relation).
Solution
Give the true strain as a function of the elongation.
ε = log
L
L ( 0)
(1.2.1)
Calculate the evolution of the cross-section.
S (t ) L (0)
=
= exp ( −ε )
S (0) L (t )
(1.2.2)
Calculate the engineering strain
.
η (t ) =
L (t ) − L (0)
= exp ε − 1
L (0)
(1.2.3)
Deduce the evolution of the applied load.
From (1.2.2) and (1.2.3)
P ( t ) = S ( t )σ =
S (0)
σ ( ε , εɺ )
1 + η (t )
(1.2.4)
Calculate, in term of stress, the condition when the maximum force is reached.
This is achieved when dP(η ) = 0
5
σ ( ε , εɺ )
1 dP (η )
1 dσ ( ε , εɺ )
=−
+
2
S ( 0 ) dη
(1 + η ) 1 + η (1 + η ) dε
(1.2.5)
And now from (1.2.5) the Considère relation for dP(η ) = 0
−σ ( Ag ) +
dσ
( Ag ) = 0
dε
(1.2.6)
b) Strain rate independent plasticity
Consider the case of a strain rate independent plastic material. Assume that its
uniaxial behaviour follows the Hollomon relation:
(1.2.7)
σ = σ 0ε n
Determine the percentage elongation at maximum load Ag.
Solution
Equation 1.2.6 yields:
Ag = 100n
(1.2.8)
c) Strain-hardening independent viscous material
Consider the case of a strain-hardening independent viscous material. Assume
that its uniaxial behaviour follows the Norton law
εɺ = (σ K )
M
(1.2.9)
Draw the load versus elongation curve of a material for which M = 4.2
and K = 100 MPa. The cylindrical test piece is 10 mm in diameter and 200 mm
long. The cross-head velocity is 1.2 mm/min.
Solution
Equation 1.2.4 together with (1.2.1) yields:
1M
Lɺ
P (η ) = KA ( 0 )
L ( 0)
1
1 M +1
(1 + η )
(1.2.10)
6
Figure 1.2.1 gives the result in term of P(η)/P(0).
Fig. 1.2.1 Load ratio vs engineering strain for a strain-hardening independent viscous material.
d) Strain-hardening viscous material
To evaluate the percentage elongation at maximum load for a strain-hardening
viscous material, study the stability of a small local perturbation of the test piece
radius.
Solution (Hutchinson and Obrecht 1977)
The test piece of cross-section S0(0) contains at the beginning of the test a local
zone of cross-section S(0) a little smaller than S0(0). Use subscript 0 for the perfect
test piece devoid of any perturbation, that is away from the local perturbation, and
no subscript for the quantities within the perturbation.
Denote ∆S (t ) = S (t ) − S0 (t ) and s (t ) = ∆S (t ) S0 (t ) . These quantities are negative and small. This allows linearisation of the equations. If their absolute values
increase the perturbation is unstable.
Write the quantities ∆σ(t), ∆εɺ and ∆ε in term of s(t).
From (1.2.4) can be written to the first order:
7
∆σ ( t ) S 0 ( t ) + σ 0 ∆S ( t ) = 0
∆σ ( t ) = −σ 0 s ( t )
(1.2.11)
From (1.2.2) can be obtained the strain rate difference:
∆εɺ = −
∆Sɺ ɺ ∆S
+ S0 2 = − sɺ ( t )
S0
S0
(1.2.12)
Integration of the equation yields:
∆ε = s ( 0 ) − s ( t )
(1.2.13)
Find the relation between ∆σ(t), ∆ε and ∆εɺ .
From the uniaxial behaviour:
∂σ
∂σ
∆σ =
∆ε + ɺ ∆εɺ
∂ε 0
∂ε 0
(1.2.14)
Write the differential equation for s(t) and integrate, introducing the quantities
∂σ 1
∂σ εɺ0 ( t )
and µ ( t ) =
ν (t ) =
∂ε 0 σ 0 ( t )
∂εɺ 0 σ 0 ( t )
Elimination of ∆ε, ∆σ and ∆εɺ yields the differential equation:
sɺ +
εɺ0 ( t )
ν (t )
ν ( t ) − 1 s =
εɺ ( t ) s ( 0 )
µ (t )
µ (t ) 0
(1.2.15)
Integration yields:
t
t
t
ν (t )
s ( t ) = s ( 0 ) exp − ∫ h ( t ) dt 1 + ∫ εɺ0 ( t )
exp ∫ h ( t ) dt
0
0
0
µ (t )
(
where h ( t ) =
)
(1.2.16)
εɺ0 ( t )
(ν ( t ) − 1)
µ (t )
Discuss the stability of the perturbation.
The parameter s being negative sɺ is negative when ν < 1; in that case s decreases so that its absolute value increases. This is the case of instability.
Note that the rate of instability increases with the ratio εɺ0 µ .
8
For a strain rate independent material, µ = 0, and then the instability takes place
when ν = 1, the Considère criterion.
εɺ
When ν = 1 , sɺ (ν = 1) = 0 s ( 0 ) , a very small quantity, the more so the higher
µ
the strain rate sensitivity. A high strain rate sensitivity has a stabilising effect.
Necking becomes noticeable for a strain, which can be much higher than the one
given by the Considère criterion. Hutchinson and Obrecht show results of numerical calculations, which illustrate this fact.
Write the solution for a material whose uniaxial behaviour is given by
σ = αε nεɺ m , with m = 1 M .
With this constitutive equation, µ = m and ν = n ε 0 .
Exercise 1.3 Torsion test
Torsion tests are useful for measuring the shear modulus of elasticity and for
the study of the high temperature behaviour of metallic alloys. This is particularly important for optimising hot working conditions. From torsion
tests, finding relations between stress and strain is required. Thus, recording
the torque as a function of the angular rotation must result in a determination
of the corresponding stress, strain and strain rate.
a) Elastic behaviour
Which reasonable assumption can be made about the deformation of the test piece
in a torsion test?
Deduce the strain and the strain rate tensors in term of the angular rotation φ.
Give the stress tensor for isotropic elastic behaviour.
Solution
Assume that the cross-sections remain flat and that diameters remain rectilinear.
This is well verified experimentally.
(i) Give the displacement field, in cylindrical coordinates.
ur = 0 uθ = −rϕ ( z ) = −rz
φ
L
uz = 0
(1.3.1)
where φ is the angle of rotation measured at the extremity of the test piece and L
its length.
Hence the strain and strain rate tensors:
9
0
0
(ε ) = 0 0
φ r
0 −
2L
0
φ r
−
2 L
0
0
0
(εɺ ) = 0 0
0 − φɺ r
2L
0
φɺ r
−
2 L
0
(1.3.2)
The maximum values are found for r = R , the radius of the test piece.
(ii) Find the stress tensor.
The strain is pure shear. The only component of the stress tensor is a shear stress:
σ θ z = τ = 2µε zθ = µφ
r
L
(1.3.3)
µ being the shear modulus of elasticity.
(iii) Give the maximum shear stress.
τR = µ
φ
L
R
(1.3.4)
(iv) Determine the corresponding torque.
C=
∫
R
0
2πr 2τ d r =
∫
R
0
φ
π
φ π
2π µ r 3dr = R 4 µ = R3τ R
L
2
L 2
(1.3.5)
Thus recording φ and the torque as a function of time allows determining the
strain, the strain rate, the maximum shear stress and the shear modulus of elasticity.
b) Strain rate dependent plasticity
Determine the critical torque for plastic deformation to begin. As it is the stressstrain relation which is to be found, determine the equivalent strain as a function
of the angle of rotation φ and the relation between the torque C and the maximum
shear stress τR.
10
Solution
Use the von Mises criterion.
From Eq. 1.3.5, the torque C0 when plasticity begins at the surface of the test
piece is given by:
C0 =
π
2 3
R3 Rp
(1.3.6)
Rp being the yield strength.
(i) Calculate the strain tensor for a large rotation.
Keep the same assumptions than for small deformation. Write the coordinates of
the displaced vector M ′ .
M ′ = M − r (1 − cos ϕ ) e r − r sin ϕ eθ + u ( z ) e z
(1.3.7)
u(z) being the lagrangian displacement in the z direction, that is with respect to the
position of a cross-section which was initially at z.
Deduce the strain tensor.
The strain tensor ∆ij is given by:
2∆ij =
1 1 ∂M ′
∂M ′
(θ k , t )
(θ k , t ) − δij
∂θ j
E i E j ∂θ i
(1.3.8)
where E i = ∂ M ∂θ i and δij is the Kronecker tensor.
From (1.3.7):
∂M ′
= er − (1 − cos ϕ ) e r − sin ϕ eθ = cos ϕ e r − sin ϕ eθ
∂r
1 ∂M ′
= − 1 − (1 − cos ϕ ) eθ + sin ϕ e r = sin ϕ e r + cos ϕ eθ
r ∂θ
∂M ′
r
r
= e z − φ sin ϕ e r − φ cos ϕ eθ + uz′ e z
∂z
L
L
Using Eq 1.3.8, the lagrangian strain rate tensor can be calculated:
(1.3.9)
11
0
0
(∆) = 0 0
φ r
0 −
2L
1
φ
2
0
φ r
−
2L
2
r
2
+ 2u ′ + u′
L
(1.3.10)
There is now an elongation of the test piece if the lagrangian displacement u is
constant, that is a rigid body displacement. For the dilatation ∆zz to be zero, a lagrangian displacement must be introduced, such that:
2
1 r
u ′ + 2u ′ + φ = 0
2 L
2
(1.3.11)
that is:
12
1 r 2
u ′ z = −1 + 1 − φ
2 L
()
φ≤ 2
L
R
(1.3.12)
Plane cross-sections do not remain plane.
Denoting γ = φ r L , calculate the equivalent strain when the test piece length is
kept constant.
2
∆ = ∆ij ∆ij
3
12
=
1
3
(
γ 2+γ2
12
)
(1.3.13)
(ii) Deduce the stresses, neglecting the elastic part.
s33 = σ 33 −
sθz = σ θz
σ 33
=−
3
2γ σ
=−
32∆
2γ2 σ
3 2 ∆
(1.3.14)
where σ is the equivalent stress, a function of the equivalent strain, which must
be determined experimentally.
12
(iii) Deduce a relation between the maximum shear stress τR and the torque.
Integration by parts of Eq. 1.3.5 yields:
C=
2π 3
2π
R τR −
3
3
∫
τR
r 3dτ
(1.3.15)
dτ ∂∆
φ d∆ dτ
dr =
d∆ ∂r
L dγ d∆
(1.3.16)
0
with
dτ =
Differentiate C, given by Eq. 1.3.5, with respect to φ:
R ∂τ
R r ∂∆ dτ
∂C
2π
= 2π ∫
r 2 dr = 2π ∫
r 2dr =
0
0
∂φ
∂φ
L ∂γ d∆
L
∫
R
0
∂∆ dτ 3
r dr
∂γ d∆
(1.3.17)
Now, from (1.3.15), (1.3.16) and (1.3.17):
C=
2π 3
φ ∂C
R τR −
3
3 ∂φ
(1.3.18)
(iv) Find now a relation between the torque and the maximum shear stress
involving the rotation velocity.
The same procedure as the one just used yields:
C=
2π 3
φɺ ∂C
R τR −
3
3 ∂φɺ
(1.3.19)
(v) Explain the procedure to determine, from the torsion test, the relation between
the stress and the strain and the strain rate.
From (1.3.18) and (1.3.19):
τR =
1
∂C ɺ ∂C
+φ ɺ
3C + φ
2πR 3
∂φ
∂φ
(Note that Eq.1.15 in Volume I is mistaken.)
(1.3.20)
13
Perform torsion tests at various rotation velocities or perform a torsion test
changing suddenly the rotation velocity. In each test, record the torque as a function of the angle of rotation. Equation 1.3.20 gives the maximum shear stress.
Equations 1.3.13 and 1.3.14 with γ = γ R = φ R L give the needed relation.
Reference
Hutchinson J W and Obrecht H (1977) Tensile instabilities in strain rate dependent materials,
Proc 4th Int Conf on Fracture Waterloo Canada 1:101-116
14
Chapter 2 Exercises of Chapter I-2: Elastic Behaviour
2.1 Interatomic binding energy in crystals
Exercise 2.1 Cohesive energy, bulk modulus
An interaction potential u per unit mass has the form (cf. Sect. 2.2.3 in Volume I)
u = − Ar − n + Br − m
where m > n
(2.1.1)
By considering the compressibility of a sphere of radius equal to the interatomic distance at rest a, express the bulk modulus k = −Vdp / dV , where p is the
pressure and V the volume, as a function of m, n and the cohesive energy u0 for
the mass density ρ.
Solution
The interatomic distance at rest a is obtained from du / dr = 0 , namely
(du / dr ) r = a = 0
⇒ a = (mB / nA)1/( m− n )
(2.1.2)
and hence
u0 = u (a ) = (1 / n)(n − m) B(mB / nA) − m /( m− n )
(2.1.3)
Let U = Mu where M is the mass of a sphere of radius a; at r = a we have
2
pdV = −(dU / dr )dr ⇒ p = − 1/ (4πa ) (dU / dr )
2
2 2
2
dp / dV = − 1/ (4πa ) (d U / dr ) r = a
(2.1.4)
Since k = −Vdp / dV , we find
k=
ρa2
9
(d 2u / dr 2 )r = a
(2.1.5)
15
Now
(d 2u / dr 2 ) r = a = −n(n + 1) Aa − n− 2 + m(m + 1) Ba − m− 2
= −mn u0 / a 2
(2.1.6)
and hence
k = −(mn / 9) ρ u0
(2.1.7)
showing that the numerical value of k increases with that of u0.
2.2 Anisotropic linear elasticity
The constitutive equations for anisotropic linear elasticity need the use of
fourth-order tensors of moduli or compliances. Due to the symmetry of the
strain and stress tensors, the corresponding 81 components reduce to 21 in
the general case. The number and the value of independent elastic constants
may be still reduced by expressing the possible symmetries of the anisotropy, which can be investigated by use of the properties of sinusoidal wave
propagation and of the necessary conditions of stability of the elastic equilibrium.
Exercise 2.2 Wave propagation in cubic symmetry
Investigate the propagation of plane sinusoidal waves in the [110] direction in a
cubic crystal (cf. Sect. 2.3.5 in Volume I).
Solution
Entering the general form of a sinusoidal wave in the Lamé-Clapeyron equations
(Eq. 2.86 in Volume I) with no body forces other than the inertia forces yields the
classical condition for such waves to exist:
det(Cijkl K j K l − ρω 2 δ ki ) = 0
(2.2.1)
with C the elastic moduli, K the propagation vector, ρ the mass density and ω the
pulsation.
16
In the [110] direction, K1 = K 2 = K / 2, K 3 = 0 ; substituting in Eq.(2.92) of
Sect.2.3.5 of Volume I and using the Voigt notation(2.58) for the elastic moduli
we have
1
2
(C11 + C44 ) K 2 − ρω 2
1
(C12 + C44 ) K 2
2
(C12 + C44 ) K 2
1
(C11 + C44 ) K 2 − ρω 2
2
0
0
1
2
0
0
= 0 (2.2.2)
C44 K 2 − ρω 2
and then:
2
1
1
2
2
2
2
2
4
−
ρω
(
+
)
−
ρω
C
K
C
C
K
( 44
) 2 11 44
− (C12 + C44 ) K = 0
4
(2.2.3)
a) the velocity vTz of transverse waves with displacement parallel to [001] derives from C44 K 2 − ρω 2 = 0 ; it is given by
vTz =
ωT
z
K
=
C44
(2.2.4)
ρ
2
2
b) From 12 (C11 + C44 ) K 2 − ρω 2 = 12 (C12 + C44 ) K 2 it follows that
ω 2 = (1 / 2 ρ ) (C11 + C44 ) K 2 ± (C12 + C44 ) K 2
(2.2.5)
ω12 = ( K 2 / 2 ρ )(C11 + C12 + 2C44 ), ω22 = ( K 2 / 2 ρ )(C11 − C12 )
(2.2.6)
giving the values
The displacement (u1 , v1 ) associated with ω1 obeys
−( K 2 / 2)(C12 + C44 )u1 + ( K 2 / 2)(C11 + C44 )v1 = 0
(2.2.7)
i.e., u1 = v1 . So, these waves are longitudinal and their velocity vL is
vL = (C11 + C12 + 2C44 ) / (2 ρ )
(2.2.8)
17
The displacement (u2 , v2 ) associated with ω2 obeys
( K 2 / 2)(C12 + C44 )(u2 + v2 ) = 0
(2.2.9)
i.e., u2 = −v2 . These are transverse waves with displacement parallel to 110
and their velocity vT is
vT = (C11 − C12 ) / (2 ρ )
(2.2.10)
Thus if the value of ρ is known, the values of all the elastic moduli can be derived from measurements of vTz, vL and vT.
Exercise 2.3 Wave propagation in transverse isotropy
Investigate the propagation of plane sinusoidal waves in a plate normal to x3, constituted with a linear elastic material exhibiting transverse isotropy about the axis
x3. Consider a wave vector K either normal or parallel to the plate. Is it sufficient
for measuring all the elastic moduli CIJ? (cf. Sect. 2.3.5 in Volume I)
Solution
The propagation condition reads: det(Cijkl K j K l − ρω 2 δ ki ) = 0 , with Cijkl given in
Voigt notation by
C11 C12 C13 0
C12 C11 C13 0
C13 C13 C33 0
0
0 C44
0
0
0
0
0
0
0
0
0
that is
0
0
0
0
C44
0
0
0
0
0
0
(1 / 2)(C11 − C12 )
(2.3.1)
18
1
(C11 + C12 ) K1 K 2
2
A − ρω 2
1
(C11 + C12 ) K1 K 2
2
(C13 + C44 ) K1 K 3
with
(C13 + C44 ) K1 K 3
B − ρω 2
(C13 + C44 ) K 2 K 3 =0
C − ρω 2
(C13 + C44 ) K 2 K 3
(2.3.2)
A = C11 K12 + C66 K 22 + C44 K32
2
2
2
B = C11 K 2 + C66 K1 + C44 K3
C = C ( K 2 + K 2 ) + C K 2
44
1
2
33 3
1
with C66 = (C11 − C12 ) .
2
For a wave vector (0, 0, K) normal to the plate, the propagation condition reduces to
C44 K 2 − ρω 2
0
2
0
0
C44 K − ρω
0
2
0
2
0
=0
C33 K − ρω
(2.3.3)
2
which yields v1 = C33 /ρ and v2 = C44 /ρ . The displacement associated with v1
is (0, 0, u30 ), which corresponds to a longitudinal propagation (v1 = vL ) . For v2, we
get u30 = 0 , which corresponds to a transverse propagation (v2 = vT ) .
For a wave vector ( K cos α , K sin α ,0) parallel to the plate, we have
K 2C1 − ρω 2
K 2 (C11 + C22 )sin α cos α
K (C11 + C22 )sin α cos α
K 2C2 − ρω 2
2
0
0
2
0
C44 K − ρω
=0
2
(2.3.4)
C1 = C11 cos α + C66 sin α
2
2
C2 = C11 sin α + C66 cos α
2
with
0
2
which yields v1 = C44 / ρ = vT1 (u10 = u20 = 0) and
K4
K 2C1 − ρω 2 K 2C2 − ρω 2 =
(C11 + C12 ) 2 sin 2 α cos 2 α
4
that is, after reduction
(2.3.5)
19
2
ω 2 C11 + C66 ω 2 C11C66
+
=0
2 −
ρ
ρ2
K2
K
(2.3.6)
and hence
v2 =
C66
ρ
=
The velocity v2 corresponds to
C11 − C12
2ρ
v3 =
C11
ρ
(2.3.7)
u20
cos α
=−
, i.e., a transverse propagation in
u10
sin α
the plane of the plate (v2 = vT2 ) . The velocity v3 corresponds to
u20 sin α
=
, i.e., a
u10 cos α
longitudinal propagation in the plane of the plate (v3 = vL ) . Note that, according
to the property of transverse isotropy, the velocities vL , vT1 and vT2 do not depend on
α. So, the knowledge of ρ and the measurement of these velocities give access to
the moduli C11 , C12 (and then C66 ), C33 and C44 , but not to C13. This modulus can only
be measured from an oblique propagation with respect to the plate plane.
Exercise 2.4 Propagation of sinusoidal longitudinal waves
Show that sinusoidal longitudinal waves cannot propagate in an arbitrary direction α in a linear elastic homogeneous body with arbitrary anisotropy. Consider
then the special case of cubic symmetry.
Solution
The general equation of sinusoidal wave propagation is known to read (see Sect.
2.3.5 in Volume I)
(C
ijkl
K j K l − ρω 2 δik ) uk0 = 0
(2.4.1)
For longitudinal waves along α, we must have uk0 = λ K k with K i = Kα i and
the velocity is vL = ω / K . Equation (2.4.1) then reads
Cijklα jα lα k − ρ vL2α i = 0
(2.4.2)
20
These are three equations for two unknowns (αi), since α is a unit vector, and
one parameter ( ρ vL2 ): there are no solutions in the general case.
Nevertheless, let us suppose that such a propagation is possible in the special
direction α0; the velocity vL0 would then be such that
α i0 ( Cijklα 0j α l0α k0 − ρ (vL0 ) 2 α i0 ) = Cijklα i0α 0j α l0α k0 − ρ (vL0 )2 = 0
(2.4.3)
and then vL0 would be given by
vL0 =
Cijklα i0α 0j α k0α l0
ρ
(2.4.4)
an expression which makes sense since C is definite positive and ρ > 0. For instance, for a cubic symmetry, we would have:
C11
for < 100 >: C α α α α = C = C ⇒ v
ijkl i j k l
1111
11
L <100> =
ρ
1 4
for < 110 >: Cijklα iα jα kα l = (C1111 + 2C1122 + C2222 + 4C1212 )( )
2
2C44 + C11 + C12
and vL<110> =
2ρ
for < 111 >: C α α α α = (3C + 6C + 12C )( 1 ) 4
ijkl i j k l
1111
1122
1212
3
4C44 + C11 + 2C12
and vL<111> =
3ρ
(2.4.5)
We now focus on the special case of cubic symmetry. This can be done in a
more convenient way by using the representation (see Exercise 2.11 below for a
proof of these formulae):
21
C = (C11 − C12 )K ′ + 2C44K ′′ + (C11 + 2C12 )J
1
1
J ijkl = δij δ kl I ijkl = (δik δ jl + δil δ jk ) I = J + K
3
2
K′ = H - J
K ′′ = I - H
K = K ′ + K ′′
(2.4.6)
where all the components of H, in the axes of the cubic symmetry, are zero but
0
0
0
H1111
= H 2222
= H 3333
= 1 . We find
0
H ijkl
α jα lα k = H i0111α13 + H i0222α 23 + H i0333α 33 = α i3
1
1
J ijklα jα lα k = δij δklα jα lα k = α i
3
3
1
1
I ijklα jα lα k = (δik δ jl + δil δ jk )α jα lα k = (α i + α i ) = α i
2
2
(2.4.7)
Hence:
1
′ α jα lα k = α i3 − α i
K ijkl
3
′′ α jα lα k = α i − α i3
K ijkl
(2.4.8)
1
1
Cijklα jα lα k = (C11 − C12 )(α i3 − α i ) + 2C44 (α i − α i3 ) + (C11 + 2C12 )α i
3
3
3
= (C11 − C12 − 2C44 )α i + (2C44 + C12 )α i = ρ vL2α i
(2.4.9)
and then
namely
(C11 − C12 − 2C44 )α13 + (2C44 + C12 )α1 = ρ vL2α1
3
2
(C11 − C12 − 2C44 )α 2 + (2C44 + C12 )α 2 = ρ vLα 2
(C − C − 2C )α 3 + (2C + C )α = ρ v 2α
12
44
3
44
12
3
L 3
11
If no αi vanishes, we must have α12 = α 22 = α 32 =
velocity vL<111> =
4C44 + 2C12 + C11
.
3ρ
(2.4.10)
1
(<111> direction) with the
3
22
If only one αi vanishes, the two others must be equal: this corresponds to the
<110> direction with the velocity vL<110> =
2C44 + C12 + C11
.
2ρ
If two αi vanish, the third one must be equal to 1: this corresponds to the <100>
direction with the velocity vL<100> =
C11
ρ
.
These results fit well with (2.4.5). They are valid only if C11 − C12 − 2C44 ≠ 0 .
When C11 − C12 − 2C44 = 0 , the cubic symmetry is degenerated into isotropy: all
the directions α are equivalent and the longitudinal propagation occurs with the
velocity vLiso =
C11
ρ
=
λ + 2µ
3k + 4 µ
=
, which is a classical result.
ρ
3ρ
Exercise 2.5 Anisotropy in cubic symmetry
For the case of cubic symmetry find the bulk modulus as a function of SIJ; also
Young's modulus in an arbitrary direction u. (cf. Sect. 2.3.2.6 in Volume I)
Solution
(i) Apply a hydrostatic pressure p. We get:
σ ij = − pδij
⇒ ε11 = ε 22 = ε 33 = − p ( S11 + 2S12 )
(2.5.1)
So, the relative volume change ∆V/V is
∆V / V = Tr(ε ) = −3 p( S11 + 2S12 )
(2.5.2)
and then
k =−
pV
1
=
∆V 3( S11 + 2S12 )
(2.5.3)
(ii) Perform a tensile test along the direction u: σ = σ u ⊗ u ⇒ σ ij = σ ui u j . Hence
23
ε11
2
2
2
σ = S11u1 + S12 (u2 + u3 ) etc. (circ. permutation)
2 ε12 = S u u etc. (circ. permutation)
44 1 2
σ
(2.5.4)
and then, with e = ε ij ui u j , we get for the Young modulus E:
1 e
= = S11 + ( S44 − 2 S11 + 2 S12 )(u12u22 + u22u32 + u32u12 )
E σ
(2.5.5)
Exercise 2.6 Elastic anisotropy in hexagonal symmetry
Show that, for linear elasticity, the property of hexagonal symmetry leads to the
same relations between the elastic moduli CIJ as that of transverse isotropy (cf.
Sect. 2.3.2.7 in Volume I). To do that, use the rotation matrices Aij(α) for the rotation angle α about the axis x3 of the hexagonal symmetry and express invariance
with respect to such rotations for α = nπ / 3, n = 1 to 6 .
Solution
According to (2.64), we must have
Cijkl = Aip (α ) Ajq (α ) Akm (α ) Aln (α )C pqmn
cos α
with [ A(α )] = − sin α
0
tem of Fig. 2.6.1.
sin α
cos α
0
(2.6.1)
0
0 α = nπ / 3 in the rectangular coordinate sys1
Fig. 2.6.1 Coordinate system (x3 normal to the plane).
24
In addition to the general symmetry relations Cijkl = C jikl = Cijlk = Cklij , we get,
−1 0 0
for n = 3 , i.e., for [ A(π)] = 0 −1 0 :
0 0 1
C14 = C1123 = A1 p A1q A2 l A3 mC pqlm = −C1123 = 0
C15 = C1131 = A1 p A1q A3l A1mC pqlm = −C1131 = 0
C24 = C2223 = A2 p A2 q A2l A3mC pqlm = −C2223 = 0
C25 = C2231 = A2 p A2 q A3l A1mC pqlm = −C2231 = 0
C34 = C3323 = A3 p A3q A2l A3mC pqlm = −C3323 = 0
(2.6.2)
C35 = C3331 = A3 p A3q A3l A1mC pqlm = −C3331 = 0
C46 = C2312 = A2 p A3 q A1l A2 mC pqlm = −C2312 = 0
C56 = C3112 = A3 p A1q A1l A2 mC pqlm = −C3112 = 0
For n = 1, 2, 4 and 5 , we have:
ε /2
ε′ 3 / 2 0
[ A(nπ / 3)] = −ε ′ 3 / 2 ε / 2 0 with ε , ε ′ = ±1
0
0
1
(2.6.3)
and then
C13 = C1133 = A1 p A1q A3l A3mC pqlm = A1 p A1q C pq 33
= 1 / 4C1133 + 2(εε ′ 3 / 4)C1233 + 3 / 4C2233
C55 = C1313 = A1 p A3q A1l A3mC pqlm = A1 p A1l C p 3l 3
(2.6.4)
= 1 / 4C1313 + 2(εε ′ 3 / 4)C1323 + 3 / 4C2323
With εε ′ = ±1 , this reduces to the four equations:
C13 = 1 / 4C13 + (
C13 = 1 / 4C13 − (
C55 = 1 / 4C55 + (
C55 = 1 / 4C55 − (
3 / 2)C63 + 3 / 4C23
3 / 2)C63 + 3 / 4C23
3 / 2)C54 + 3 / 4C44
3 / 2)C54 + 3 / 4C44
(2.6.5)
25
which leads to
C36 = 0, C13 = C23 , C55 = C44 , C45 = 0.
(2.6.6)
We have also, taking account of the preceding results:
C11 = C1111 = A1 p A1q A1l A1mC pqlm =
= (1 / 16)C11 + ( 3 / 4)εε ′C16 + (3 / 4)C66
+ (3 3 / 4)εε ′C26 + (3 / 8)C12 + (9 / 16)C22
C66 = C1212 = A1 p A2 q A1l A2 mC pqlm =
(2.6.7)
= (3 / 16)C11 − (3 / 8)C12 + (3 / 16)C22 + (1 / 4)C66
+ ( 3 / 4)εε ′(C16 − C26 )
These 4 equations (since εε ′ = ±1 ) reduce to:
C16 = C26 = 0
C11 = C22
C − C = 2C
12
66
11
(2.6.8)
which are identical to the relations deriving from the property of transverse isotropy.
Exercise 2.7 Shear modulus in anisotropic elasticity
1) Find the expression of the shear modulus for given shear plane and direction in
anisotropic elasticity with arbitrary symmetry and specify this expression for various symmetries.
Solution
A uniform shear stress state with the amplitude τ is applied to an anisotropic linear
elastic body along a plane with unit normal n and in the (unit) direction m: the resulting shear strain in the same directions (m, n) has the amplitude γ and the shear
modulus µ is defined by τ = µγ . It depends on (m, n). We define the orientation
symmetric second-order tensor R by
S
R = (m ⊗ n)
1
Rij = (mi n j + m j ni )
2
(2.7.1)
26
We have σ = τ (m ⊗ n + n ⊗ m) = 2τ R , whereas γ is given by
γ = 2m.ε .n = 2ε ij mi n j = ε ij (mi n j + m j ni ) = 2 R : ε
(2.7.2)
so that
1
µ
=
γ 2R : ε 2R : S : σ
=
=
= 4R : S : R
τ
τ
τ
(2.7.3)
Note that we have Tr( R) = 0 and R : R = 1 / 2
In the orthonormal coordinate system (m,n,t) with t = m × n , we have
0 τ 0
0 1/ 2 0
σ = τ 0 0 R = 1 / 2 0 0
0 0 0
0
0 0
(2.7.4)
For an orthotropic symmetry, we have in the Voigt notation and in the axes of
this symmetry
S11
S12
S
[ S ] = 013
0
0
S12
S 22
S23
0
0
0
S13
S 23
S33
0
0
0
0
0
0
S44
0
0
0
0
0
0
S55
0
0
0
0
0
0
S66
(2.7.5)
and then we get
1
4µ
= S11 R112 + S22 R222 + S33 R332 + 2S12 R11 R22 + 2S23 R22 R33 + 2S31 R33 R11
orth
2
23
2
31
(2.7.6)
2
12
+ S44 R + S55 R + S66 R
where the components Rij must be calculated in the axes of the orthotropic symmetry. For a quadratic symmetry with respect to x1 and x2, we would have
1
4µ
= S11 ( R112 + R222 ) + S33 R332 + 2S12 R11 R22 + 2 S13 ( R11 + R22 ) R33
quad
2
23
2
31
2
12
+ S44 ( R + R ) + S66 R
(2.7.7)
27
For transverse isotropy about x3, we have in addition S66 = 2( S11 − S12 ) . We get
1
4µ
= S11 ( R112 + R222 + 2 R122 ) + S33 R332 + 2S12 ( R11 R22 − R122 )
tr. isot
2
33
2
23
(2.7.8)
2
31
− 2S13 R + S44 ( R + R )
where the relation Tr( R) = 0 has been used. If the shear plane is normal to x3, we
have
0
n = (0,0,1) m = (cos α ,sin α ,0) R = 0
cos α
2
0
0
sin α
2
cos α
2
sin α
2
0
(2.7.9)
1
cos 2 α sin 2 α
= S44 (
+
) ⇒ µ = 1 / S44 , which, as expected, is inde4µ
4
4
pendent of α. If n is normal to x3, we can take it parallel to x1 and then
and then
0
cos ϕ
n = (1,0,0) m = (0,cos ϕ ,sin ϕ ) R =
2
sin ϕ
2
and then
cos ϕ
2
0
0
sin ϕ
2
0
0
(2.7.10)
2( S11 − S12 ) 2
= S44 1 +
− 1 cos ϕ .
µ
S44
1
For cubic symmetry, we find, with the “anisotropy parameter” a =
2( S11 − S12 )
S 44
which is greater than 1
1
µ
= S44 {1 + 2(a − 1)( R112 + R222 + R332 )}
cubic
(2.7.11)
28
where the relations Tr( R) = 0 and R : R = 1 / 2 have been used. It is easy to check
that 0 ≤ R112 + R222 + R332 ≤ 1 / 2 , so that we get, since a ≥ 1 :
S44 ≤
1
µ
≤ aS44
cubic
2) A very simple model for a polycrystal with cubic crystals could be constituted
with two cubic crystals in series, subjected to the same shear stress τ, with the
volume fractions c1 and c2 (with c1 + c2 = 1 ). The first crystal is sheared on the system (100 ) [ 011] and the second one on the system (110 ) 1 10 . Express the overall shear strain as the average of the two shear strains and derive the overall
shear modulus. Comment this result.
Solution
We have
τ
1
γ 1 = µ with µ = S44
1
1
1
τ
γ =
with
= aS44
2 µ 2
µ2
(2.7.12)
and hence
γ = c1γ 1 + c2γ 2 = τ (
c1
µ1
+
c2
µ2
)=
τ
µ
(2.7.13)
so that
1
µ
= c1S44 + ac2 S 44 = ( a + c1 (1 − a) ) S44
(2.7.14)
So, when the volume fraction c1 of the first orientation increases from 0 to 1,
the overall shear modulus decreases (here, linearly) from aS44 to S44. This illustrates the fact that the anisotropy induced by a plastic flow and the resulting crystallographic texture can make the overall moduli decrease, even if no damage occurs.
29
Exercise 2.8 Elastic anisotropy in quadratic symmetry
Justify the fact that the compliance matrix in Voigt notation for a linear elastic
material with quadratic symmetry in the symmetry axes is frequently written in the
form (“engineering notation”):
ν
ν
ν
1
− 12
− 13 (or − 31 )
E1
E1
E3
E1
ν
ν
ν
1
− 12
− 13 (or − 31 )
E
E
E3
E1
1
1
ν 13
ν 31
ν 13
ν 31
1
− (or − ) − (or − )
E
E
E
E
E
3
1
3
1
3
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
µ13
0
0
0
1
µ13
0
0
0
0
0
0
1
µ12
(2.8.1)
where the notations E, µ and ν refer to the isotropic Young modulus, shear
modulus and Poisson ratio.
Solution
For an isotropic linear elastic material, the strain response to a tensile test is classically given by
σ
σ = 0
0
1
E
0 0
0 0 ⇒ ε = σ 0
0 0
0
0
−
ν
E
0
whereas the strain response to a simple shear test reads
0
0
ν
−
E
(2.8.2)
30
1
2µ
0
0 τ 0
1
σ = τ 0 0 ⇒ ε = τ
2µ
0 0 0
0
0
0
0
0
0
(2.8.3)
For a quadratic symmetry, the compliance matrix in Voigt notation in the
symmetry axes reads (see Sect. 2.3.2.5 in Volume I)
S11
S12
S
[ S IJ ] = 013
0
0
S12
S11
S13
0
0
0
0
0
0
S44
0
0
S13
S13
S33
0
0
0
0
0
0
0
S44
0
0
0
0
0
0
S66
(2.8.4)
For a tensile test along x1, we have
σ 0 0
S11 0
σ = 0 0 0 ⇒ ε = σ 0 S12
0
0 0
0
0
0
0
S13
(2.8.5)
Comparing with the isotropic case (7.2), we can put
S11 =
1
E1
S12 = −
ν 21
E1
(= −
ν 12
E1
) S13 = −
ν 31
E1
(2.8.6)
For a tensile test along x3, we get
0 0 0
S13
σ = 0 0 0 ⇒ ε =σ 0
0 0 σ
0
0
S13
0
0
0
S33
(2.8.7)
so that we can write
S33 =
1
E3
S13 = −
ν 13
E3
(= −
ν 31
E1
)
(2.8.8)
31
For a shear test on (x1, x2), we have
0
0 τ 0
S
σ = τ 0 0 ⇒ ε = τ 66
2
0 0 0
0
and then we can put S66 =
S44 =
1
µ13
1
µ12
S66
2
0
0
0
0
0
(2.8.9)
. For a shear test on (x1, x3), we find similarly
.
Exercise 2.9 Stability of equilibrium in cubic symmetry
Prove the inequalities (2.85) for the elastic moduli in a cubic system (cf. Sect.
2.3.3 in Volume I).
Solution
Given the definition of CIJ, it suffices to find the conditions for the associated
quadratic form to be positive definite: all the eigenvalues must be positive; therefore all the diagonal sub-determinants must be positive, the conditions for which
are the following:
determinants of order 1: C11 > 0, C44 > 0
2: C112 − C122 > 0
3: (C11 − C12 )2 (C11 + 2C12 ) > 0 ⇒ C11 + 2C12 > 0
with no further conditions for orders 4,5,6.
Exercise 2.10 Stability of equilibrium in transverse isotropy
Find the conditions to be obeyed by the compliances S11 , S13 , S33 , S44 and S66 for
the elastic equilibrium to be stable in transverse isotropy. For given values of
S13 and S33 , draw the stability domain in the plane ( S11 , S66 ) .
Solution
32
Like in Exercise 2.9, the quadratic form σ I S IJ σ J must be positive definite, i.e.,
σ I S IJ σ J > 0, ∀ [σ ] ≠ [ 0] : all the diagonal sub-determinants of SIJ must be positive,
that is:
order 1: S11 > 0, S33 > 0, S44 > 0, S66 > 0
S66 2
S
2
2
2
) = S66 ( S11 − 66 ) > 0
S11 − S12 = S11 − ( S11 −
order 2:
¨
2
4
S S − S 2 > 0
11 33 13
order 3:
S
S
S
S11( S11S33 −S132 ) − ( S11− 66 ) S33 ( S11− 66 ) −S132 + S13 S13 ( S11− 66 ) −S11S13 > 0
2
2
2
S
S
This last condition reduces to S66 ( S11S33 − S132 − 33 66 ) > 0 .
4
Finally, the stability conditions read
S33 > 0, S66 > 0, S44 > 0, S11 >
S66 S132
+
4 S33
(2.10.1)
For given values of S13 and S33, the stability domain is drawn in the (S11, S66)
plane in Fig. 2.10.1.
Fig. 2.10.1 The stability domain lies between the S11 axis and the oblique straight line.
33
2.3 Bounds and estimates for the elastic moduli of heterogeneous
materials
Material scientists and engineers are more and more concerned with the prediction of the overall behaviour and the local response of heterogeneous and
composite materials. We investigate here the simpler case of linear elastic
and thermoelastic heterogeneous materials. For random materials, the RVE
(“representative volume element”) cannot be described completely so that
the effective overall behaviour of the HEM (“homogeneous equivalent medium”) cannot be predicted unambiguously. One can then either make additional assumptions or approximations in order to derive estimates for the
overall behaviour or look for bounds for the overall mechanical properties,
by using variational approaches. Consideration of (either real or fictitious)
eigenstrains or eigenstresses in the initial state is a useful tool for the derivation of both improved estimates and bounds, as illustrated by the various applications which have been made of Eshelby’s solution of the (ellipsoidal)
inclusion problem. This approach can also be applied to heterogeneous
thermoelasticity.
Exercise 2.11: Voigt and Reuss bounds for the elastic moduli of an isotropic
polycrystal with cubic crystals
Find the Voigt ( µ V , k V ) and Reuss ( µ R , k R ) bounds for the shear and bulk moduli
( µ eff , k eff ) of a linear elastic isotropic polycrystal with cubic crystals.
Solution
From (2.135) in Volume I, namely
< c−1 > −1 ≤ (S eff ) −1 = Ceff ≤< c >
(2.11.1)
we know that CV =< c >, S R =< s > .
Note that the scalars cijij , ciijj , sijij and siijj are invariant with respect to any rotation,
which means that they have the same value for all the grains of the polycrystal, so
that we can derive
V
V
< cijij >= cijij = Cijij
, < ciijj >= ciijj = Ciijj
R
R
< sijij >= sijij = Sijij
, < siijj >= siijj = Siijj
(2.11.2)
34
With the notations
A = C11 − C12 , B = C11 + 2C12 , a =
2( S11 − S12 )
S 44
(2.11.3)
we have
cijij = 3C11 + 6C44 = B + 2 A + 3aA
ciijj = 3C11 + 6C12 = 3B
3
1 2
3
sijij = 3S11 + S44 = + +
2
B A Aa
3
siijj = 3S11 + 6 S12 =
B
(2.11.4)
Since the polycrystal is isotropic, CV and S R can be considered as isotropic too,
and written as
1
1
K + V/R J
2 µ V/R
3k
1
1
= (δik δ jl + δil δ jk ) J ijkl = δij δ kl
2
3
CV/R = 2µ V/R K + 3k V/R J S V/R =
with I = J + K
I ijkl
(2.11.5)
so that we get finally
(3a + 2) A
5 Aa
eff
2(2a + 3) ≤ µ ≤
10
B
B
≤ k eff ≤
3
3
(2.11.6)
This last result shows that keff is determined unambiguously and has the value
B/3.
Note: This problem can be solved more straightforwardly by using the representation (2.4.6) given in Exercise 2.4 for cubic symmetry. It uses, in addition to the
unit tensors I, J and K which are especially for isotropy, the new tensors H, K’
and K”, with
1
J ijkl = δij δ kl
3
K = K ′ + K ′′
1
I ijkl = (δik δ jl + δil δ jk ) I = J + K
2
K′ = H - J
K ′′ = I - H
(2.11.7)
35
where all the components of H, in the axes of the cubic symmetry, are zero but
0
0
0
H1111
= H 2222
= H 3333
= 1 . It is easy to prove the following properties:
K′ : K′ = K′
K ′′ : K ′′ = K ′′
H : H = H
K ′ : K ′′ = K ′′ : K ′ = K ′ : J = J : K ′ = K ′′ : J = J : K ′′ = 0
1
1
1
−1
( a′K ′ + a′′K ′′ + bJ ) = a′ K ′ + a′′ K ′′ + b J
Q.F.( a′K ′ + a′′K ′′ + bJ ) > (<)0 ⇔ a′, a′′, b > (<)0
(2.11.8)
where Q.F.(X) denotes the quadratic form associated with X. For instance, in any
system of orthonormal axes turned by the orthogonal matrix [A] from the cubic
axes, we get
0
H ijkl = Aim Ajn Akp Alq H mnpq
= Ai1 Aj1 Ak 1 Al1 + Ai 2 Aj 2 Ak 2 Al 2 + Ai 3 Aj 3 Ak 3 Al 3 (2.11.9)
and then
0
0
H ijmn H mnkl = Aip Ajq Amr Ans Amt Anu Akv Alw H pqrs
H tuvw
0
= Aip Ajq Akv Alw H 0pqrs H tuvw
= Aip Ajq Akv Alw H
0
pqvw
(2.11.10)
= H ijkl
Similarly, we have
1
1
0
Aip Ajq Amr Ans δ mn δ kl H 0pqrs = Aip Ajq δ rs δ kl H pqrs
3
3
1
1
0
= Aip Ajq δkl H pqrr = δij δ kl = J ijkl
3
3
H ijmn J mnkl =
(2.11.11)
We have also H : J = J : H (= J). Consequently
K ′ : K ′ = ( H − J ) : ( H − J ) = H − 2J + J = H − J = K ′
(2.11.12)
and so on for the other equalities in (2.11.8). Using the Voigt notation in the cubic
axes, we have
36
1
0
0
[ I] =
0 0
0 0
0 0
[ K′′] =
0
0 0
1 0 0
1 0
0
0
0 1 0
0 1
0 0 1
[H] =
1/ 2
0
0
1/ 2
0
0
1/ 2
0
0
1 / 3 1 / 3 1 / 3
0
0
0
1 / 3 1 / 3 1 / 3
0
1/ 3 1/ 3 1/ 3
[J] =
1/ 2
0
1/ 2
0
0
1/ 2
0
2 / 3 −1 / 3 −1 / 3
0
−1 / 3 2 / 3 −1 / 3
−1 / 3 −1 / 3 2 / 3
[K] =
1/ 2
0
1/ 2
1 / 2
2 / 3 −1 / 3 −1 / 3
0
−1 / 3 2 / 3 −1 / 3
(2.11.13)
−1 / 3 −1 / 3 2 / 3
[ K ′] =
0
0
0
0
So, if we write c = a′K ′ + a′′K ′′ + bJ , we get from identification in the cubic
axes
2 ′ b
3 a + 3 = C11
b
1
− a′ + = C12
3
3
1
2 a′′ = C44
which yields, as in (2.4.6)
a′ = C11 − C12
⇒ b = C11 + 2C12
a′′ = 2C
44
(2.11.14)
37
c = (C11 − C12 )K ′ + 2C44 K ′′ + (C11 + 2C12 )J
(2.11.15)
and, according to the third line of (2.11.8)
s=
1
1
1
K′ +
K ′′ +
J
C11 − C12
2C44
C11 + 2C12
= ( S11 − S12 )K ′ +
S44
K ′′ + ( S11 + 2S12 )J
2
(2.11.16)
For an isotropic elastic material, we have C11 − C12 = 2C44 and then (2.11.15)
reads
c = (C11 − C12 )(K ′ + K ′′) + (C11 + 2C12 )J
= (C11 − C12 )K + (C11 + 2C12 )J
(2.11.17)
= 2µ K + 3kJ
as we know already.
From the definitions (2.11.7), we find easily the value of the invariants
′ = 2 Kijij = 5 K ijij
′′ = 3 J ijij = 1
K ijij
′
′′ = 0 J iijj = 3
K iijj = 0 K iijj = 0 K iijj
(2.11.18)
which leads to
< cijij >= 2a′ + 3a′′ + b
< c >= 3b
iijj
2 3 1
+ +
a′ a′′ b
3
< siijj >
b
< sijij >=
(2.11.19)
The overall isotropy implies the averages < K ′ > and < K ′′ > to be isotropic.
So, < K ′ > can be put in the form
′ >= 2 = 5α + β
< Kijij
2
< K ′ >= α K + β J with
⇒< K ′ >= K
′
5
< Kiijj >= 0 = 3β
(2.11.20)
3
and then < K ′′ >= K . The derivation of the Voigt and Reuss bounds for the
5
shear and bulk moduli follows straightforwardly.
38
Note that the last line of (2.11.8) would have also made obvious the resolution
of Exercise 2.9.
Exercise 2.12: Voigt and Reuss bounds for an elastic inhomogeneous material
with cubic symmetry and isotropic constituents
eff
Derive the Voigt and Reuss bound for the elastic moduli (C11eff , C12eff , C44
) of an inhomogeneous material with overall cubic symmetry constituted with elastic isotropic components.
Solution
For cubic symmetry, we use the same notations as in Exercise 2.11
A = C11 − C12
B = C11 + 2C12
(2.12.1)
with C = C44 . Note that this leads to the relations
S11 − S12 = A−1
S11 + 2S12 = B −1
S44 = C −1
(2.12.2)
Remind also the results of Exercise 2.11.9 concerning the stability conditions
for cubic symmetry, namely
C11 > 0 C44 > 0 C112 − C122 > 0 C11 + 2C12 > 0
(2.12.3)
which leads to: A > 0, B > 0, C > 0 . For an isotropic material, we would have
(C11 − C12 )iso = 2 µ
(C11 + 2C12 )iso = 3k
(C44 )iso = µ
(2.12.4)
The Voigt bounding inequation reads
E : (< c > −Ceff ) : E ≥ 0, ∀ E
(2.12.5)
According to what precedes, the positivity of (< c > −Ceff ) (Voigt bound) implies the inequalities
2 < µ > − Aeff ≥ 0
eff
3 < k > −B ≥ 0
< µ > −C eff ≥ 0
(2.12.6)
39
whereas the positivity of (< s > −S eff ) (Reuss bound), with help of (2.12.2), leads
to the inequalities
1
−1
eff −1
2 < µ > −( A ) ≥ 0
1
−1
eff −1
< k > −( B ) ≥ 0
3
< µ −1 > −(C eff ) −1 ≥ 0
(2.12.7)
Thus, we obtain the bounds
2 < µ −1 > −1 ≤ Aeff ≤ 2 < µ >
−1
−1
eff
3< k > ≤ B ≤3< k >
−
1
−
1
< µ > ≤ C eff ≤< µ >
(2.12.8)
eff
Voigt and Reuss bounds for C11eff , C12eff , C44
can then be derived easily from their
relations with A, B and C.
Note: the representation (2.4.6) could have been used here again for a more direct
resolution.
Exercise 2.13: Voigt and Reuss bounds for the Young modulus and Poisson
ratio of an isotropic elastic material with isotropic constituents
Show that the laws of mixtures which yield in this case the right values of the
Voigt and Reuss bounds for the shear and bulk moduli are no more valid for the
Young modulus and the Poisson ratio.
Solution
For the Young modulus, we have
E=
9k µ
3k + µ
or
1
1
E −1 = µ −1 + k −1
3
9
(2.13.1)
The Voigt and Reuss bounds for µ eff and k eff can be written as
< µ > −1 ≤ µ eff −1 ≤< µ −1 >
< k > −1 ≤ k eff −1 ≤< k −1 >
(2.13.2)
40
so that we have from (2.13.1)
1
1
1
1
< µ > −1 + < k > −1 ≤ E eff −1 ≤ < µ −1 > + < k −1 > (=< E −1 >)
3
9
3
9
(2.13.3)
and then
−1
1
1
< E −1 > −1 ≤ E eff ≤ < µ > −1 + < k > −1 (≠< E >)
9
3
(2.13.4)
For the Poisson ratio, we start from the general relations for an isotropic material (cf. Eqs (2.82) in Volume I)
E = 2 µ (1 +ν ) k =
E
3(1 − 2ν )
(2.13.5)
and then
k
2(1 +ν ) = 3 (1 − 2ν )
(2.13.6)
µ
which reads also
2(1 +ν )(3 +
µ
k
)=9
(2.13.7)
From the Voigt and Reuss bounds for the shear and bulk moduli, we can derive
the inequalities
1
< k ><
1
µ
≤
>
µ eff
k
eff
≤< µ ><
1
>
k
(2.13.8)
According to (2.13.7), we then have the inequalities
9
9
≤ 1 + ν eff ≤
−1
2 3+ < µ >< k >
2 3+ < k > −1 < µ −1 > −1
(2.13.9)
which yield the searched Voigt and Reuss bounds for ν eff . It can be checked easily that they differ from the predictions of the laws of mixtures, namely < ν > and
< ν-1 >-1, respectively. For example, for a two-phase material with uniform Pois-
41
son ratios (ν1 = ν2 = ν), the Voigt and Reuss bounds for ν eff do not coincide
whereas both laws of mixtures yield the same estimate ν.
Exercise 2.14: Elementary bounds and estimates for an elastic isotropic porous medium
Derive the strain field in a spherical isotropic elastic inhomogeneity embedded in
an infinite isotropic elastic matrix subjected to a uniform strain at infinity from the
solution of the reference problem of a spherical inclusion by using Eshelby’s
method of the “fictitious equivalent inclusion” (see Sect. 2.7.3 of Volume I). Use
this result to derive the Voigt and Reuss bounds, the Hashin-Shtrikman bounds
and the self-consistent estimate for the effective shear modulus of an elastic isotropic porous medium with an elastic isotropic incompressible matrix. Comment
these results at high and low porosity.
Solution
F
We have to choose the (uniform) stress-free strain ε in the inclusion so that the
solutions of the two problems shown in Fig. 2.14.1 are identical. We know the solution of the classical inclusion problem (at right), namely (see Eq. 2.159 in VolI
I
ume I), with e , s the strain and stress (uniform) deviators in the inclusion,
θ I = Tr(ε I ) and E 0 the strain deviator at infinity:
F
I
0
s = 2µ E + ( β − 1)e
⇒
I
0
F
Tr(σ ) = 3k Θ + (α − 1)θ
eI = E 0 + β e F
0
F
θ = Θ + αθ
with α =
1 +ν
2(4 − 5ν )
and β =
.
3(1 −ν )
15(1 −ν )
E
0
E
µ, k
µ, k
µ *, k*
0
=
µ,k
F
ε
Fig. 2.14.1 The equivalent inclusion method
(2.14.1)
42
The solutions of the inclusion and inhomogeneity problems are the same only
if
F
F
*
0
0
µ E + ( β − 1)e = µ E + β e
⇒
0
F
0
F
k Θ + (α − 1)θ = k Θ + αθ
s I = 2µ * e I
I
* I
Tr(σ ) = 3k θ
(2.14.2)
whence the known results (see Eq. 2.165 in Volume I):
I
e =
µ
0
E
*
µ (1 − β ) + µ β
θI =
k
Θ0
*
k (1 − α ) + k α
(2.14.3)
Note that for a spherical cavity ( µ * = k * = 0 ), we get
e Void =
E
0
1− β
θ Void =
Θ0
1− α
(2.14.4)
Voigt and Reuss bounds
From < µ −1 > −1 ≤ µ eff ≤< µ > , we have, for a porosity c and an incompressible
matrix (ν = 1 / 2 )
< µ >= (1 − c) µ
< µ −1 >→ ∞
(2.14.5)
and then
µ Reuss = 0
µ Voigt = (1 − c) µ
(2.14.6)
Hashin-Shtrikman bounds
E0
µ, k
E0
µ, k
µ,k
0
µ,k
F
ε
(1)
(2)
Fig. 2.14.2 Two auxiliary problems for the derivation of the HS+ bound
43
Since the softest phase is made of voids, the lower Hashin-Shtrikman bound is obviously zero: µ HS− = 0 . To derive the upper Hashin-Shtrikman bound, we solve
the two inclusion problems of Fig. 2.14.2:
e1 = E 0
E0
5E
= 0
e2 =
1− β
3
(2.14.7)
since, for ν = 1 / 2 , β = 2 / 5 . With E the macroscopic strain deviator, we must
have
E =< e >= (1 − c)E 0 +
5c
2c
E 0 = (1 + )E 0
3
3
⇒ E0 =
3
E
3 + 2c
(2.14.8)
so that
e1 =
3
E
3 + 2c
e2 =
5
E
3 + 2c
(2.14.9)
The stress tensors are given by
s1 = 2µ e1
⇒ < s >=
s2 = 0
6(1 − c)
µ E = 2µ HS+ < e >
3 + 2c
(2.14.10)
and then
µ HS =
+
3(1 − c)
µ
3 + 2c
(2.14.11)
Self-consistent scheme
E
E
µ SC , k SC
µ SC , k SC
µ,k
0
µ,k
F
ε
(1)
(2)
Fig. 2.14.3 Two auxiliary problems for the derivation of the self-consistent estimate
44
We have to deal with two inclusion problems with the same matrix constituted
with the effective medium ( µ SC , k SC ) and either the sound material ( µ , ∞) or the
void ( 0,0 ) in the spherical inhomogeneity (see Fig. 2.14.3).
We find for the strain deviators
µ SC
E
e1 = SC
µ (1 − β SC ) + µβ SC
e = 1 E
2 1 − β SC
(2.14.12)
with
E = (1 − c)e1 + ce 2
⇒
µ SC (1 − c)
c
+
=1
µ (1 − β SC ) + µβ SC 1 − β SC
SC
(2.14.13)
i.e.,
c
µ SC = µ 1 −
1 − β SC
(2.14.14)
where β SC , which depends on ν SC , has still to be determined (due to the voids,
the effective medium is no longer incompressible). This can be obtained from the
average equation for the volume dilatations:
θ1 = 0 θ 2 =
Θ
1 − α SC
⇒ cθ 2 = c
Θ
1 − α SC
=Θ
⇒ α SC = 1 − c (2.14.15)
From α SC , we derive easily ν SC = (2 − 3c) / (4 − 3c) and then β SC = (2 + c) / 5 ,
so that we have, from (2.14.14)
µ SC =
3(1 − 2c)
µ
3−c
(2.14.16)
Actually, this result does not hold for c > ½ which would correspond to a negative effective shear modulus: beyond this value, according to the self-consistent
model, the porous body has no more any mechanical resistance and the effective
shear modulus vanishes. So, instead of (2.14.16), we have to write
45
µ SC =
3 < 1 − 2c >
µ
3−c
(2.14.17)
where < A > means A as long as A ≥ 0 and 0 when A < 0.
Comments
For c << 1, we get
µ HS+
2c
5c
≈ 1− c −
= 1−
3
3
µ
SC
µ ≈ 1 − 2c + c = 1 − 5c
µ
3
3
(2.14.18)
which shows that the upper Hashin-Shtrikman bound and the self-consistent estimate behave similarly at very low porosity, the latter being always smaller that the
former (no bound violation): this corresponds to the fact that, at low porosity, the
voids are far away from each other with a continuous matrix, which is the situation taken into account by the upper Hashin-Shtrikman bound treatment. At larger
porosity, the voids are still included in a continuous matrix (closed porosity) for
this treatment whereas the self-consistent scheme let both phases (sound material
and voids) play an equivalent morphological role (open porosity), which yields a
lower estimate for the effective shear modulus. This modulus even vanishes for
c ≥ ½. Note that for a cubic lattice of spherical voids with the same radius a, these
4 3
πa π
*
voids can touch each other for c = 3 3 = (≃ 0.524) , which is close to ½.
8a
6
Exercise 2.15: Elementary bounds and estimates for an elastic isotropic twophase material with one rigid phase
Derive the Voigt and Reuss bounds, the Hashin-Shtrikman bounds and the selfconsistent estimate for the effective shear modulus µ eff (or the corresponding
shear compliance η eff = 1/ µ eff ) of a two-phase elastic isotropic material with isotropic and isotropically distributed phases. Phase (1) is supposed to be incompressible and phase (2) to be rigid. Comment these results at low and high volume
fractions c of phase (2).
Solution
Voigt and Reuss bounds
46
From < µ −1 > −1 ≤ µ eff ≤< µ > , we get also < η −1 > −1 ≤ η eff ≤< η > , with η 2 = 0 . So,
we have η eff ≤ (1 − c)η1 or µ eff ≥
µ1
1− c
(= µ Reuss ) and 0 ≤ η eff or µ eff ≤ ∞ (= µ Voigt )
or η Voigt = 0 .
Note that, from < k −1 > −1 ≤ k eff ≤< k >
k
eff
→ ∞ and then ν
eff
with k1 , k2 → ∞ , we have also
= ½ ∀c , for any phase distribution.
Hashin-Shtrikman bounds
We use the same method as in Exercise 2.14 with adequate matrices and inclusions. For the upper bound, the matrix is rigid so that µ HS+ → ∞ (or η HS+ = 0). For
the lower bound, the matrix is made of phase (1) and we get in a spherical inhomogeneity ( µ * , k * ), with the same notations as in Exercise 2.14:
s=
with
β1 = 2 / 5
2µ * µ1
0
E
µ1 (1 − β1 ) + µ * β1
since ν 1 = ½ . So,
s=
0
(2.15.1)
10µ * µ1 0
E
3µ1 + 2µ *
and, when µ * → ∞ ,
0
s → 5µ1 E . The auxiliary strain deviator E is determined from the average strain
equation:
E =< e >= (1 − c)e1 + ce 2
0
with e1 = E , e 2 = 0
(2.15.2)
0
So, we have E = E / (1 − c) and then
2µ1
s1 = 1 − c E
s = 5µ1 E
2 1 − c
(2.15.3)
From S =< s >= 2 µ HS− E , we get finally
µ HS =
−
2 + 3c
µ1
2(1 − c)
(or η HS− =
2(1 − c)
η1 )
2 + 3c
(2.15.4)
Self-consistent scheme
Proceeding like in Exercise 2.14, we find, with β SC = 2 / 5 since ν eff = ½ ∀c :
47
e1 =
5µ SC
E
3µ SC + 2µ1
e2 = 0
(2.15.5)
and then
E = (1 − c)e1 ⇒
5(1 − c) µ SC
=1
3µ SC + 2µ1
(2.15.6)
which yields
µ SC =
2 µ1
2 − 5c
(or η SC = (1 −
5c
)η1 )
2
(2.15.7)
Note that this expression is valid only as long as µ SC ≥ 0 , i.e., for c ≤ 0.4 ; this
means that µ SC → ∞ (or η SC = 0) for c > 0.4 . This could be proved rigorously by
making µ2 tend towards infinity in the second-order equation in µ SC and excluding the negative solution.
Comments
For c << 1, we get from (2.15.4)
η HS
3c
5c
≈ 1− c − = 1−
2
2
η1
−
(2.15.8)
which shows, according to (2.15.7), that the lower Hashin-Shtrikman bound and
the self-consistent estimate behave similarly at very low porosity (with
η HS+ ≤ η SC ≤ η HS− , which ensures that there is no bound violation): this corresponds to the fact that, at low porosity, the rigid phase consists of inclusions embedded in a continuous matrix of phase (1), which is the situation taken into account by the lower Hashin-Shtrikman bound treatment. At larger porosity, the
rigid phase is still embedded in a continuous deformable matrix according to this
treatment whereas the self-consistent scheme let both phases play an equivalent
morphological role, which yields a larger estimate for the effective shear modulus.
From c = 0.4, this estimate is infinite, which corresponds to the formation of a
rigid continuous “skeleton” which makes the whole material rigid itself.
Exercise 2.16: The differential scheme for a two-phase elastic material
The classical self-consistent scheme for a two-phase material is well adapted to
the case of phases which play an equivalent morphological role. For a particle or
48
fibre reinforced composite, one phase is a continuous matrix and the second phase
consists of particles (or fibres) embedded in this matrix and the classical selfconsistent scheme is no more adapted to this morphology. So, this scheme has
been modified in different ways so as to be able to be used for modelling the overall behaviour of such materials. Among the simplest extensions of this scheme let
us quote the “differential (self-consistent) scheme” and the “generalised selfconsistent scheme”. We consider here the first of these schemes; the reader interested in the second one is invited to consult Christensen and Lo (1979). For sake
of simplicity, we restrict ourselves to the case of an isotropic two-phase material
whose phases obey a linear isotropic incompressible elastic behaviour (shear
moduli µ 1 and µ 2 and volume fractions c1 and c2).
The basic idea consists in the incremental introduction of the second phase
made of particles in a matrix which has been homogenised at each step, starting
from the sole continuous phase, and to use the Eshelby-Einstein dilute approximation (see Volume I, Sect. 2.8.2, Eq. 2.191) for this incremental homogenisation
procedure.
1) Derive the expression of µ EE from the tensorial expression of CEE given by
(2.191) for the present case. Check that this result could have been obtained too
from a first-order development of the self-consistent equation (as obtained, e.g.,
from Eq. 2.14.12 and the method used in the Exercises 2.14 and 2.15) for an infinitesimal volume fraction of second phase (this is the reason why the differential
scheme is sometimes also named the “self-consistent” differential scheme).
2) Apply this result to the current situation where the second phase has still the
volume fraction f (with 0 ≤ f ≤ c2 ) and the homogenised matrix has the shear
modulus µ * ( f ) and where f is changed into f+df: what is the resulting variation
of µ * ( f ) ? Derive the corresponding estimate µ DS of the differential scheme by integration of this differential equation.
3) What would be the prediction of the differential scheme for a rigid second
phase? Compare this prediction with the elementary bounds and estimates of Exercise 2.15 and comment on the differences. Why, from a rigorous point of view,
this result cannot be applied directly to a porous medium (where the second phase
is constituted of voids)? What would be the prediction if it was applied nevertheless without modification and what is the rigorous prediction (it will be noticed
that the difference is very small). Compare this prediction with the elementary
bounds and estimates of Exercise 2.14 and comment on the differences.
Solution
1) The tensorial expression of the Eshelby-Einstein estimate is (cf. Eq. 2.191 of
Volume I)
CEE = c mat + f (c2 − c mat ) I + P2mat : (c 2 − c mat )
−1
(2.16.1)
49
which would read with the notations of this exercise
CEE = c1 + f (c 2 − c1 ) I + P21 : (c 2 − c1 )
−1
(2.16.2)
with f infinitesimal. For isotropy, this yields
µ
EE
β
= µ1 + f ( µ 2 − µ1 ) 1 + 1 ( µ2 − µ1 )
µ1
−1
(2.16.3)
Since phase (1) is incompressible, β1 = 2 / 5 and (2.16.3) becomes
µ EE = µ1 1 +
5 f ( µ 2 − µ1 )
3µ1 + 2µ2
(2.16.4)
This means that the introduction of a very low volume fraction dc2 = f of phase
(2) in phase (1) makes the shear modulus µ 1 of this phase be changed by
dµ = µ EE − µ1 =
5µ1 ( µ 2 − µ1 )
dc2
3µ1 + 2µ2
(2.16.5)
To derive the overall shear modulus estimate µ SC according to the classical selfconsistent scheme, we have, from (2.14.12) in Exercise 14
µ SC
E
e1 = SC
µ (1 − β SC ) + µ1β SC
µ SC
e =
E
2 µ SC (1 − β SC ) + µ 2 β SC
(2.16.6)
Since both phases are incompressible, the homogenised material is incompressible too so that β SC = 2 / 5 . So we have
5µ SC
e
=
E
1
3µ SC + 2µ1
SC
e = 5µ
E
2 3µ SC + 2µ 2
and the strain average equation reads
(2.16.7)
50
(1 − c)
5µ SC
5µ SC
+
=1
c
3µ SC + 2 µ1
3µ SC + 2µ 2
(2.16.8)
or
5(1 −c) µ SC (3µ SC + 2µ 2 ) + 5cµ SC (3µ SC + 2µ1 ) = (3µ SC + 2µ1 )(3µ SC + 2µ2 ) (2.16.9)
where µ SC is a function of c. For c infinitesimal, this equation can be differentiated
and yields, with c ≈ dc and µ SC ≈ µ1 + dµ :
−5dcµ1 (3µ1 + 2µ 2 ) + 5[ dµ (3µ1 + 2µ 2 ) + 3µ1dµ ] + ...
...5dc µ1 (5µ1 ) = 3dµ (8µ1 + 2µ 2 )
(2.16.10)
or
10dcµ1 ( µ1 − µ 2 ) + 5(6 µ1 + 2µ2 )dµ = 3(8µ1 + 2 µ2 )dµ
(2.16.11)
5µ1 ( µ1 − µ2 )dc + (3µ1 + 2µ 2 )dµ = 0
(2.16.12)
and then
which is, as expected, the same equation as (2.16.5).
2) Equation (2.16.12) can be used in the current state with µ1 = µ * ( f ) and
dµ = dµ * . As for dc, it is obtained as follows: we start with a volume V containing the volume fV of phase (2); we add the volume αV of phase (2) and the new
( f + α )V
volume fraction of phase (2) is f + df =
. For α = dc (≪ 1) , we get
(1 + α )V
α = dc =
df
1− f
(2.16.13)
so that we have to solve the equation
5µ * ( µ * − µ 2 )
which reads also
df
+ (3µ * + 2µ 2 )dµ * = 0
1− f
(2.16.14)
51
2
df
3µ * + 2 µ
1 *
= − * * 2 dµ * = * − *
dµ
1− f
5µ ( µ − µ 2 )
5µ µ − µ 2
(2.16.15)
This can be easily integrated (through the logarithms) into
1− f = λ
µ2 − µ *
( µ * ) 2/5
(2.16.17)
where the constant λ is determined from the initial condition ( µ * = µ1 for f = 0 ),
i.e.,
λ=
µ12/5
µ 2 − µ1
(2.16.18)
The differential scheme estimate µ DS is finally obtained for f = c2 , that is
1 − c2 =
µ12/5 ( µ2 − µ SD )
µ2 − µ SD µ1
=
( µ2 − µ1 )( µ SD ) 2/5
µ 2 − µ1 µ SD
2/5
(2.16.19)
3) If the second phase is rigid ( µ2 → ∞ ), we get
µ1
1 − c2 → SD
µ
rig
2/5
or
SD
→
µrig
µ1
(1 − c2 )5/2
(2.16.20)
which shows that, at variance with the classical self-consistent estimate, the differential scheme estimate remains finite and only tends to infinity when c2 → 1 :
due to the differential homogenisation procedure, there is no more any possibility
for a continuous rigid skeleton to be formed. On the contrary, when c2 ≪ 1 , these
estimates as well as the Hashin-Shtrikman lower bound behave similarly
5c
( µ ≈ µ1 (1 + 2 ) .
2
When the second phase is constituted of voids, the foregoing calculations are,
from a rigorous point of view, no more valid since neither the second phase nor
the homogenised material is incompressible ( β ≠ 2 / 5 ). If, as a first approximation, this difficulty is omitted, we find from (2.16.19)
52
µ SD µ1
1 − c2 ≈ por SD
µ1 µpor
2/5
or
SD
µpor
≈ µ1 (1 − c2 )5/3
(2.16.21)
This would mean that, now again, the overall shear modulus only vanishes for
c2 = 1 , as expected for a porous material with closed porosity. For c ≪ 1 , we find
5
SD
µpor
≈ µ1 (1 − c2 ) , which conforms with (2.14.18) in Exercise 2.14.
3
An exact calculation would have given
dµ
µ = − (1 −
dk = −
k
(1 −
df
5(3k + 4µ )df
dµ
=−
f )(1 − β ) µ
(1 − f )(9k + 8µ )
⇒
df
dk = − (3k + 4 µ )df
k
f )(1 − α )
4µ (1 − f )
(2.16.22)
and then
dµ
20µ 2
3k − 4µ
=
⇒ µ = µ1
dk k (9k + 8µ )
3k
5/3
(2.16.23)
and finally
3
DS
µ por
µ1
(1 − c2 )6 =
DS 3/5
µpor
2−
µ
1
(2.16.24)
It can be checked that this expression is very close to (2.16.21) and the qualitative comments given before are still valid.
Exercise 2.17: Bounds and estimates for the elastic moduli of an isotropic
polycrystal with cubic crystals
This is an additional question to Exercise 2.11. Apply to the overall shear modulus
of this polycrystal the method proposed in Volume I, Sect.2.8.2, aiming at constructing a whole set of estimates for the overall moduli Cest of heterogeneous materials. Due to the overall isotropy, spheres and an isotropic matrix (µ 0, k0) can be
considered here. Give the expression of µ est as a function µ 0, β 0 (the Eshelby coefficient) and the crystal moduli. Derive the bounds of Voigt and Reuss and of
Hashin and Shtrikman (assuming 2C44 > (C11 –C12)) for µ est and the equation for
53
the self-consistent estimate µ SC. Propose a graphic construction for the derivation
of this estimate.
Solution
According to (2.188) in Volume I, the equation for Cest is
< δcest : (I + P 0 : δc0 )−1 >= 0
(2.17.1)
with P0 for a sphere and an isotropic matrix given by
0
PSph
=
α0
3k
J+
0
β0
3k 0
0
K
α
=
3µ 0
3k 0 + 4µ 0
β0 =
6(k 0 + 2µ 0 )
5(3k 0 + 4µ 0 )
(2.17.2)
and δc0 = c − C0 . Using the representation (2.4.6), we get
δc est = (a′ − 2µ est )K ′ + (a′′ − 2µ est )K ′′ + (b − 3k est )J
−1
β 0 ( a′ − 2 µ 0 )
(I + P : δc ) = 1 +
K ′ + ...
2µ 0
0
0 −1
−1
(2.17.3)
−1
β 0 (a′′ − 2µ 0 )
α 0 (b − 3k 0 )
′′
...1 +
K
+
1 +
J
2µ 0
3k 0
and then the equation
a′ − 2µ est
a′′ − 2µ est
b − 3k est
′
′′
<
K
>
+
<
K
>
+
J = 0 (2.17.4)
β 0 ( a′ − 2 µ 0 )
β 0 (a′′ − 2 µ 0 )
α 0 (b − 3k 0 )
+
+
1+
1
1
2µ 0
2µ 0
3k 0
From Exercise 2.11 (see Eq. 2.11.20), we know that, for an isotropic distribu2
3
tion of cubic crystals, < K ′ >= K and < K ′′ >= K . Consequently we get
5
5
b C11 + 2C12
est
0
est
eff
)
b − 3k = 0, ∀C ⇒ k = k = 3 (=
3
est
3(a′′ − 2µ est )
2(a′ − 2µ )
=0
β 0 ( a′ − 2 µ 0 ) +
β 0 (a′′ − 2µ 0 )
1+
1 +
2µ 0
2µ 0
(2.17.5)
54
The first result in (2.17.5) was already known from Exercise 2.11. The second
equation yields the estimated shear modulus µ est as
µ est =
2 µ 0 (1 − β 0 )(2a′ + 3a′′) + 5a′a′′β 0
20µ 0 (1 − β 0 ) + 2(3a′ + 2a′′) β 0
with β 0 =
2(b + 6 µ 0 )
5(b + 4µ 0 )
(2.17.6)
a′ = C11 − C12 , a′′ = 2C44 , b = C11 + 2C12
As we know from Sect.2.8.2 in Volume I, the bounds of Reuss and Voigt must
be recovered from (2.17.6) when µ 0 tends to zero and infinity, respectively, that is
2
5a′a′′
0
0
est
Reuss
µ → 0 ⇒ β → 5 , µ → 2(3a′ + 2a′′) = µ
µ 0 → ∞ ⇒ β 0 → 3 , µ est → 2a′ + 3a′′ = µ Voigt
5
10
(2.17.7)
It is easy to check that these bounds are the same as those found in Exercise
2.11, taking account of the changes in notations (note that a = a′ / a′′ ).
To derive the lower (upper, resp.) Hashin and Shtrikman bound, we have to
find the value of µ 0 which gives to the quadratic form associated to δc0 = c − C0
the smallest (highest, resp.) possible positive (negative, resp.) value. Since 3k0 = b,
δc 0 = c − C0 = (a′ − 2µ 0 )K ′ + (a′′ − 2 µ 0 )K ′′
(2.17.8)
Since 2C44 > (C11 –C12) (i.e., a” > a’) and taking account of the last property of
(2.11.8) in Exercise 2.11, the lower (upper, resp.) Hashin and Shtrikman bound is
reached when, in (2.17.6), 2 µ 0 = a′ ( 2 µ 0 = a′′ , resp.). Expressing first β 0 in
(2.17.6) as a function of b and µ 0, we get
µ est = f ( µ 0 ) =
8(2a′ + 3a′′)( µ 0 )2 + 3[10a′a′′ + b(2a′ + 3a′′) ] µ 0 + 5a′a′′b
(2.17.9)
80( µ 0 ) 2 + 6 [ 5b + 2(3a′ + 2a′′)] µ 0 + 2b(3a′ + 2a′′)
The Hashin and Shtrikman bounds are found by setting 2 µ 0 = a′ or 2 µ 0 = a′′
in (2.17.9). We find
HS+ a′′(38a′a′′ + 12a′′2 + 16a′b + 9a′′b)
µ =
36a′a′′ + 64a′′2 + 12a′b + 38a′′b
2
µ HS− = a′(42a′a′′ + 8a′ + 6a′b + 19a′′b)
2
24a′a′′ + 76a′ + 42a′b + 8a′′b
(2.17.10)
55
The self-consistent estimate µ SC satisfies the equation µ est = f ( µ est ) where the
function f(µ ) is defined in (2.17.9). This third-order equation reads
µ SC =
8(2a′ + 3a′′)( µ SC ) 2 + 3[10a′a′′ + b(2a′ + 3a′′)] µ SC + 5a′a′′b
80( µ SC )2 + 6 [5b + 2(3a′ + 2a′′)] µ SC + 2b(3a′ + 2a′′)
(2.17.11)
Fig. 2.17.1 Graphical construction for the determination of the self-consistent estimate
If µ est = f ( µ 0 ) is plotted as a function of µ 0 ≥ 0 , the self-consistent estimate is
obtained as the ordinate of the point where the bisector of the first quadrant
crosses this curve. This property suggests an iterative numerical resolution of the
third-order equation (2.17.11) which, among several other equivalent methods
with other initial values, would read (see Fig. 2.17.1)
µ(0) = 0 µ(1) = f (0) µ(2) = f ( µ(1) ) ... µ( n+1) = f ( µ( n ) )
SC
( µ( n ) )
µ = lim
n →∞
(2.17.12)
Exercise 2.18: Elastic and thermoelastic behaviour of an elastic inhomogeneous material with cubic or isotropic symmetry and isotropic constituents
This exercise is a continuation of Exercise 2.12 which is concerned with the investigation of the thermoelastic behaviour of the same material.
1) Consider the special case of identical shear moduli µ i in the phases (i). Show
that the overall behaviour is then isotropic and derive the Hashin and Shtrikman
bounds for the homogenised bulk modulus keff. Comment the obtained result.
2) This result can be corroborated by the construction, according to the procedure
developed in Volume I, Sect.2.8.2, aiming at constructing a whole set of estimates
for the overall moduli Cest of heterogeneous materials. Specify this method for es-
56
timating the sole homogenised bulk modulus keff by considering spherical inhomogeneities with moduli (ki , µi ) embedded in an infinite isotropic matrix with the
moduli (k 0 , µ 0 ) , subjected to the uniform dilatation
Θ0
δ at infinity and calculate
3
the corresponding estimate k . How can we derive from that the Voigt and Reuss
and Hashin and Shtrikman bounds as well as the self-consistent estimate for the
homogenised bulk modulus. What happens when all the shear moduli µ i are identical? Consider the special case of a two-phase material with µ1 ≥ µ 2 .
3) Use these results for proposing the construction of a whole set of estimates for
eff
the overall coefficient of thermal expansion α when these coefficients are isoest
tropic in every phase. Focus on the case of a two-phase material and check that
α eff obeys the relation given by Levin’s theorem. What happens when all the
shear moduli µ i are identical?
Solution
1) When µi = µ , ∀i , (2.12.1) and (2.12.8) give
C11eff − C12eff
=µ
2
(2.18.1)
Ceff = 2 µ K + 3k eff J
(2.18.2)
eff
=
C44
and then
so that the overall behaviour is then isotropic with µ eff = µ and, according to the
Voigt and Reuss bounding, < k −1 > −1 ≤ k eff ≤< k > . Now, the property of local and
global isotropy allows us to use the sharper Hashin and Shtrikman bounding (see
(2.184) in Volume I), namely
α − δk −
−
) −1 >
α + δk +
) −1 >
+
eff
k
k
≤k ≤
α − δk − −1
α + δk + −1
< (1 +
)
>
<
(1
+
) >
k−
k+
with
< k (1 +
< k (1 +
k − = min {ki } , k + = max {ki } ,
−
−
+
δk = k − k , δk = k − k
α− =
+
3k −
3k +
+
=
,
α
3k − + 4µ
3k + + 4 µ
(2.18.3)
57
So, we have
(1 +
α − δk −
k
−
+
(1 +
α δk +
k+
)−1 = (1 + 3
k − k − −1 3k − + 4 µ
) =
3k − + 4 µ
3k + 4µ
3k + + 4 µ
) =
3k + 4µ
(2.18.4)
−1
and then
k HS−
3k − + 4µ
k
)> <
>
−
µ
3
k
+
4
3
k
+
4µ
k
=
=
=
= k HS+ (2.18.5)
1
α − δk − −1
3k − + 4µ
<
>
< (1 +
) >
<
>
3k + 4µ
3k + 4µ
k−
< k (1 +
α − δk −
) −1 >
< k(
since these expressions do not depend on k- (nor on k+) any more. Consequently,
k eff is determined too and reads
k eff =
< k (3k + 4 µ ) −1 >
< (3k + 4 µ ) −1 >
(2.18.6)
where µ is constant and k varies from one phase to another.
2) In the spherical inhomogeneity (i) (see Fig. 2.18.1) the volume dilatation θiest is
given by
1
θiest =
1+
Θ0 =
0
α 0 δki
k0
1
3k 0 + 4µ 0 0
0
Θ
=
Θ
3( k − k 0 )
3ki + 4 µ 0
1+ 0i
3k + 4µ 0
Θ0
δ
3
µ 0 ,k 0
…
µ i , ki
….
(i)
Fig. 2.18.1 Construction of a set of estimates for the overall bulk modulus
(2.18.7)
58
and the mean stress σ mesti is
σ mesti = kiθiest =
ki (3k 0 + 4µ 0 ) 0
Θ
3ki + 4µ 0
(2.18.8)
So, kest is given by
k est =
<σ
<θ
est
m
est
k (3k 0 + 4µ 0 )
k
> <
>
>
3k + 4µ 0
3k + 4 µ 0
=
=
1
3k 0 + 4µ 0
>
<
>
<
>
0
3k + 4 µ 0
3k + 4µ
<
(2.18.9)
since (3k 0 + 4µ 0 ) is constant.
When µ 0 → 0, k est →< k −1 > −1 which is the Reuss bound. When µ 0 → ∞,
k est →< k > which is the Voigt bound. The Hashin and Shtrikman bounds are derived with µ 0 = µ − = min {µi } for the lower bound and µ 0 = µ + = max {µi } for
the upper bound. The self-consistent estimate k SC is obtained for µ 0 = µ SC , the
self-consistent estimate for µ eff .
When all the shear moduli µ i are identical, µ − = µ + = µ and the lower and upper Hashin and Shtrikman bounds for k eff (as well as the self-consistent estimate,
which is intermediate) coincide to the exact value for k eff , namely
k
>
3k + 4µ
=
1
<
>
3k + 4µ
<
k
eff
(2.18.10)
as we already know from (2.18.6).
For a two-phase material, we get, with µ2 < µ1
k est
c1k1
c2 k 2
+
0
0
3
k
+
4
µ
3
k
3k1k2 + 4µ 0 < k >
2 + 4µ
= f (µ 0 ) = 1
=
c1
c2
4µ 0 + 3k1k2 < k −1 >
+
0
0
3k1 + 4µ
3k2 + 4µ
(2.18.11)
The plot of k est = f ( µ 0 ) (see Fig. 2.18.2) yields especially the elementary
bounds and estimates for k eff .
59
Fig. 2.18.2 A set of estimates for the effective bulk modulus of a two-phase material
3) We use the basic equation (see Eqs (2.119) and (2.125) in Volume I) connecting thermoelasticity with elasticity
th
th
E =< B T : ε > ⇒ α
eff
=< BT : α >
(2.18.12)
where B T is the transposed stress concentration tensor of the purely elastic coneff
centration problem. To derive an estimate of α , we use (2.18.12) with an estimate of B T deriving from Question 2. As a matter of fact, due to the isotropy of
α and B T , we have simply
α est =< bmestα > where bmesti =
σ mesti
and Σ m = k est < θ est >
Σm
(2.18.13)
From (2.18.7) and (2.18.8) we get
σ mesti =
ki (3k 0 + 4 µ 0 ) 0
Θ
3ki + 4 µ 0
Σ m = k est <
3k 0 + 4µ 0
> Θ0
3ki + 4µ 0
(2.18.14)
and then
bmesti
ki (3k 0 + 4µ 0 ) 0
ki
ki
Θ
0
0
3ki + 4µ
3ki + 4µ
3ki + 4µ 0
=
=
=
(2.18.15)
0
0
1
k
3k + 4µ
est
est
0
k
<
<
>
>
k <
>Θ
3k + 4µ 0
3k + 4µ 0
3ki + 4 µ 0
So, we get
60
α est =< bmestα >=
< kα (3k + 4 µ 0 ) −1 >
< k (3k + 4µ 0 )−1 >
(2.18.16)
Levin’s theorem (see Eq. 2.127 in Volume I) for a two-phase material leads to
the relation
α eff =< α > +(S eff − < s >) : (s 2 − s1 ) −1 : (α 2 − α 1 )
(2.18.17)
In case of local and global isotropy, this relation reads
α eff =< α > +(
1
1
1 1
− < >)( − ) −1 (α 2 − α1 )
eff
k
k
k2 k1
In order to get consistent estimates for α
eff
(2.18.18)
, we must require this relation to
hold for estimates too, namely
α est =< α > +(
1
1
1 1
− < >)( − ) −1 (α 2 − α1 )
k est
k
k2 k1
(2.18.19)
With k est given by (2.18.11), we get
α est =< α > +(
4 µ 0 + 3k1k2 < k −1 >
1 1
− < k −1 >)( − ) −1 (α 2 − α1 ) (2.18.20)
0
3k1k 2 + 4µ < k >
k2 k1
After some algebra, it can be checked that (2.18.20) and (2.18.16) coincide.
Finally, when all the shear moduli are equal ( µi = µ , ∀i ), we take µ 0 = µ and
obtain the exact value for α eff , namely
α eff =
< kα (3k + 4 µ ) −1 >
< k (3k + 4µ ) −1 >
(2.18.21)
where µ is constant and k and α vary from phase to phase.
Exercise 2.19: Thermal stresses in a two-phase material
When the temperature is uniformly varied in an unloaded heterogeneous material,
for instance during the processing route of a composite material, this gives rise to
an internal thermal stress field which is important to be evaluated in view of a secure further use. As a first approach to this problem, show that these per phase
averaged thermal stresses can be calculated in an elastic two-phase material as a
61
function of the elastic compliances s1 and s 2 , the coefficients of thermal expansion
α 1 and α 2 , the effective coefficient of thermal expansion α eff , the volume fractions c1 and c2 and the temperature variation ∆T. Find a relation involving the efeff
fective compliance S eff in place of α .
Consider the special case when s1 and s 2 , S eff , α 1 and α 2 are isotropic and
discuss the sign of the residual stresses when, as it is more usual in reinforced
composite materials with phase (1) as a matrix and phase (2) made of particles or
fibres, one has α 2 < α1 . Give the solution when S eff is estimated as its Voigt and
Reuss bounds.
Discuss the limits of such a first approach.
Solution
When the material is unloaded, we have
< σ >= 0
(2.19.1)
and, in addition,
ε = ε e + ε th = s : σ + α ∆T
⇓
th
(2.19.2)
eff
< ε >= E = α ∆T =< s : σ > + < α > ∆T
that is
eff
< s : σ >= (α − < α >)∆T
When α
eff
(2.19.3)
is known (or estimated), the two equations (2.19.1) and (2.19.3) al-
low us to calculate (or estimate) the per phase averaged residual stresses
σ 1 and σ 2 in the case of a two-phase material. With c = c2 , these equations read
(1 − c)σ 1 + cσ 2 = 0
eff
(1 − c)s1 : σ 1 + cs 2 : σ 2 = α − (1 − c)α 1 − cα 2 ∆T
When, instead of α
eff
(2.19.4)
, S eff is known (or estimated), Levin’s theorem (see Eq.
2.127 in Volume I or (2.18.17) in Exercise 18) can be used in order to express the
62
relation between α
eff
and S eff since we are dealing with a two-phase material.
This relation reads
α eff =< α > +(S eff − < s >) : (s 2 − s1 ) −1 : (α 2 − α 1 )
(2.19.5)
so that the system (2.19.4) can be transformed into
(1 − c)σ 1 + cσ 2 = 0
(2.19.6)
−1
eff
(1 − c)s1 : σ 1 + cs 2 : σ 2 = S − (1 − c)s1 − cs 2 : (s 2 − s1 ) : (α 2 − α 1 )∆T
or
cσ 2 = −(1 − c)σ 1 = (s 2 − s1 ) −1 : S eff − (1 − c)s1− cs 2 : (s 2 − s1 ) −1 : (α 2 − α 1 )∆T (2.19.7)
In case of isotropy for s1 , s 2 , S eff ,α 1 and α 2 , (2.19.7) reduces to
3(
cσ 2 = −(1 − c)σ 1 =
1
1
− < >)(α 2 − α1 )
k eff
k
∆T δ
1 1
( − )2
k2 k1
(2.19.8)
In this equation, the sign of the residual stresses does not depend on the sign of
(k2 − k1 ) since < k −1 > −1 , which the Reuss bound for k eff , is always smaller than
1
1
k eff , so that eff − < >≤ 0 . So, for ∆T < 0 and α 2 < α1 , the average residual
k
k
stresses are, as expected, compressive in phase (2) and tensile in phase (1). When
1
1
S eff is then estimated as its Reuss bound, Reuss =< > and the per phase average
k
k
residual stresses vanish, which conforms with the underlying uniform stress assumption. When the Voigt bound is used, we get
3(
cσ 2 = −(1 − c)σ 1 =
1
1
− < >)(α 2 − α1 )
<k >
k
∆T δ
1 1
( − )2
k2 k1
= −3c(1 − c)(α 2 − α1 )
k1k2
∆T δ
(1 − c)k1 + ck2
(2.19.9)
63
Roughly speaking, this estimate could be considered as a bound for the thermal
stresses. Nevertheless, one has to keep in mind that this analysis yields estimates
only for the per phase average residual stresses. If these averages are isotropic,
there might exist deviatoric components whose average vanishes but which could
play an important role in local plastification. In addition, the elastic moduli could
depend on temperature, which has not been considered here. Actually, a whole
thermo-elastoviscoplastic of the residual stress field would be necessary for a refined analysis.
2.4 Miscellaneous
Exercise 2.20: Finite Element Method
The principle of the Finite Element Method has been briefly exposed in Sect. 2.4.2
of Volume I. Without going into too much details of the practical implementation
of this method, this exercise aims at studying the simplest way to proceed up to the
complete resolution of an elementary problem.
1) For sake of simplicity, we restrict ourselves to 2D-problems in the (1,2) plane
and triangular 3-node elements. For such a triangle, say 123, calculate the interpolation functions ϕ i = (ϕi1 ,ϕi 2 ) , i = 1,2,3 when a linear interpolation between the
nodes is considered.
Solution
The trial displacement field in the considered triangle, u′( M ) = [u′( x, y ), v′( x, y )] ,
must be a linear function of both the coordinates ( x, y ) and the displacement
components at the nodes of the triangle, that is
u1′( x, y ) = u1′ f1 ( x, y ) + u2′ f 2 ( x, y ) + u3′ f 3 ( x, y )
v1′( x, y ) = v1′ f1 ( x, y ) + v2′ f 2 ( x, y ) + v3′ f 3 ( x, y )
(2.20.1)
with the relations
f1 ( x1 , y1 ) = f 2 ( x2 ,y2 ) = f 3 ( x3 ,y3 ) = 1
(2.20.2)
f1 ( x2 , y2 ) = f1 ( x3 , y3 ) = f 2 ( x1 ,y1 ) = f 2 ( x3 ,y3 ) = f3 ( x1 ,y1 ) = f 3 ( x2 ,y2 ) = 0
So, the linear functions f i ( x, y ) read
64
f1 ( x, y ) = a1 x + b1 y + c1
f 2 ( x, y ) = a2 x + b2 y + c2
f ( x, y ) = a x + b y + c
3
3
3
3
(2.20.3)
where the 9 constants ai , bi , ci are, according to (2.20.2), the solutions of the linear
system of 9 equations
a1 x1 + b1 y1 + c1 = 1
a2 x1 + b2 y1 + c2 = 0
a x + b y + c = 0
3 1 3 1 3
a1 x2 + b1 y2 + c1 = 0
a1 x3 + b1 y3 + c1 = 0
a2 x2 + b2 y2 + c2 = 1
a2 x3 + b2 y3 + c2 = 0
a3 x2 + b3 y2 + c3 = 0
a3 x3 + b3 y3 + c3 = 1
(2.20.4)
From the definition u′( x, y ) = X i ϕ i ( x, y ) , we can write the transposed matrix
T
[ X ]123
as
T
[ X ]123 = [u1′
u3′
v3′ ]
(2.20.5)
0
f3
f2
0
0
f 3
(2.20.6)
v1′ u2′
v2′
f
0
f1 0
T
and the matrix [ϕ ]123 as
T
[ϕ ]123 = 01
f2
with f1, f2 and f3 defined by (2.20.3) and (2.20.4), so that
u ′( x, y )
T
= [ X ]123 .[ϕ ( x, y ) ]123
′
v
(
x
,
y
)
123
[u′( x, y )]123 =
(2.20.7)
2) We consider only plane strain situations in isotropic elasticity (Lamé coefficients λ and µ). The strain and stress matrices are written in the form
T
[ε ]
= ε11 ε 22
2ε12
T
[σ ]
= σ 11 σ 22
2σ 12
(2.20.8)
Calculate the 3x3 matrix [C ]PS of the elastic moduli. Calculate the stiffness
matrix [ R ]123 for one triangle, say 123, with area A123 and thickness e.
Solution
In plane strain conditions, we have
65
σ 11= (λ +2µ )ε11+λε 22
σ 22 = λε11+(λ +2 µ )ε 22
σ 12 = 2µε12
σ 11 λ +2µ
⇒ σ 22 = λ
2σ 12 0
λ
λ +2 µ
0
0 ε11
0 ε 22 (2.20.9)
2µ 2ε12
So, the matrix of the elastic moduli in plane strain conditions reads
[C ]PS
λ
λ + 2µ
= λ
λ + 2µ
0
0
0
0
2µ
(2.20.10)
The elementary stiffness matrix [ R ]123 for the element 123 obeys the relation
1
e
T
T
[ X ]123 .[ R ]123 .[ X ]123 = ∫123 [ε ′]123 .[C ]123 .[ε ′]123 dS
2
2
(2.20.11)
T
T
T ∂ϕi1
ε11′ X i ϕ i = X i ε11′ ϕ i = X i
∂x
T
T
T ∂ϕi 2
′ X i ϕ i = X i ε 22
′ ϕi = Xi
with ε 22
∂y
1 T ∂ϕi1 ∂ϕi 2
T
T
+
)
ε12′ X i ϕ i = X i ε12′ ϕ i = X i (
2
∂y
∂x
(2.20.12)
(
)
( )
(
)
( )
(
)
( )
Consequently, we can write
[ε ′]123
=
a1
0
a2
0
a3
0
b1
0
b2
0
b1
2
a1
2
b2
2
a2
2
b3
2
0
b3 .[ X ]123
a3
2
(2.20.13)
and the elementary stiffness matrix for the element 123 reads
T
[ R ]123 = eA123 [ g ] .[C ]PS .[ g ]
with A123 the area of the element 123 and the matrix [ g ] given by
(2.20.14)
66
[g] =
a1
0
a2
0
a3
0
b1
0
b2
0
b1
2
a1
2
b2
2
a2
2
b3
2
0
b3
a3
2
(2.20.15)
3) Find the equivalent nodal forces which are to be concentrated at the concerned
nodes when
a) a uniform density of forces F (q1 , q2 ) is applied inside the element 123
g
b) a linear density of forces T [ (α s + β ) p1 ,(α s + β ) p2 ] is applied on the
boundary edge 12 of length l, with 0 ≤ s ≤ l .
Solution
a) The energetic equivalence is ruled by the relation
∫
T
123
F .u ′ dV = X i ∫ F .ϕ i dV = X i ∫ (q1ϕi1 + q2ϕi 2 )dV = [ X ]123 .[ F ]123 (2.20.16)
123
123
Due to the properties of the interpolation functions f i ( x, y ) (linear variation in
the triangles and 1 or 0 values at the nodes), it is easy to show that
∫
123
f i ( x, y )dV =
eA123
, ∀i
3
(2.20.17)
Consequently, we find
T
[ F ]123 =
eA123
[ q1
3
q2
q1 q2
q1
q2 ]
(2.20.18)
b) Here again, the equivalent nodal forces are defined from the energetic relation
∫
12
g
g
l
T
T .u′ dS = X i ∫ T .ϕ i dS = eX i ∫ (α s+β )( p1ϕi1+ p2ϕi 2 )ds = [ X ]12.[ F ]12 (2.20.19)
12
0
On the boundary edge 12, f 3 = 0, f1 =
l−s
s
, f 2 = . So, one has
l
l
67
l−s
l (α l + 3β )
l
∫0 (α s + β ) l ds =
6
l
s
l
(2
α
l
+
3β )
(α s + β ) ds =
∫0
6
l
(2.20.20)
and then
[ F ]12
(α l + 3β ) p1
(α l + 3β ) p
2
α
β
(2
l
+
3
)
p
el
1
=
6 (2α l + 3β ) p2
0
0
(2.20.21)
4) Show how the foregoing results could be used in order to solve by the finite
element method the problem of an isotropic elastic dam which is addressed in Exercise 8.5 (question 2) from a purely static point of view (see Fig. 2.20.1).
y
0
x
π/4
water
dam
A
B
Fig. 2.20.1 Sketch of the studied dam
This dam can be discretised as indicated in Fig. 2.20.2. The basement 46 is
fixed; the vertical edge 14 suffers the water pressure (linearly increasing with the
depth), 16 is load-free and the dam itself is subjected to its own weight.
Solution
Interpolation functions: using a local coordinate systems, the four triangles are either in the position of triangle (1) (this is the case for (1), (2) and (4)) or in the position of triangle (3). In the first case, one has, with 1 (0, h), 2 (0, 0) and 3 (h, 0)
68
Fig. 2.20.2 Discretised structure: triangles (1) to (4); nodes 1 to 6.
y
f1 ( x, y ) = h
h − ( x + y)
f 2 ( x, y ) =
h
x
f 3 ( x, y ) = h
(2.20.22)
and in the second case, with 1 (0, h), 2 (h, 0) and 3 (h, h)
h− x
f1 ( x, y ) = h
h− y
f 2 ( x, y ) =
h
x+ y−h
f 3 ( x, y ) =
h
(2.20.23)
Stiffness matrix: the [ g ] matrix reads differently for (1), (2), (4) and for (3). According to (2.20.15), we find
[ g ](1),(2),(4)
[ g ](3)
0
0
−1 h
= 0
1h
0
1 h 2 0 −1 h 2
−1 h
0
= 0
0
0
−1 h 2
0
1h
−1 h
0
−1 h 2
0
0
0
−1 h 2
0
1 h 2
0
−1 h
0
1h
0
1 h 2 1 h 2
0
1h
0
(2.20.24)
69
The same remark holds for the elementary stiffness matrices. From (2.20.14),
(2.20.10) and (2.20.24) we find, with A(1) = A(2) = A(3) = A(4) = h 2 2
[ R ](1),(2),(4)
[ R ](3)
0
−µ
−µ
0
µ
µ
λ + 2µ
λ
−λ
−(λ + 2 µ )
0
λ + 3µ
λ+µ
− (λ + 2 µ ) − µ
e
=
λ + 3µ
−λ
−µ
2
sym
λ + 2µ
0
µ
(2.20.25)
λ
0
0
−(λ + 2 µ )
−λ
λ + 2 µ
µ µ
0
−µ
− µ
µ
0
−µ
−µ
e
=
λ + 2µ
−λ
− (λ + 2 µ )
2
sym
λ + 3µ
λ+µ
λ + 3µ
The whole stiffness matrix for the structure is composed by transferring each
component of an elementary matrix at its adequate place in the global matrix;
when several components are transferred at the same place, they have to be added.
Actually, the whole global matrix is not necessary for solving the problem with
the finite element method. In the central final equation (Eq. 2.115 in Volume I),
the submatrix [ R22 ] only is needed, if the first p values of Xi correspond to nodes
where the displacement is given by the boundary conditions. In our case, this
means that the lines and rows of the global matrix corresponding to nodes 4, 5 and
6 where the displacement is forced to vanish can be omitted.
We are then left with the following global [ R22 ] stiffness matrix
0
µ
λ
+
2µ
e
[ R22 ] =
2
sym
−µ
−µ
−λ
−(λ +2µ )
0
λ
2(λ +3µ )
λ +µ
−2(λ +2µ )
2(λ +3µ )
−(λ +µ )
2(λ +3µ )
−(λ +µ )
(2.20.26)
−2 µ
λ +µ
2(λ +3µ )
µ
0
which, in the final equation, will be multiplied by [ X ] transposed of the matrix
given in (2.20.5), since the displacements at 1, 2 and 3 are the sole unknowns.
70
Nodal forces: they are concerned with the weight of the dam and the water pressure. The force density associated with the weight has the form F (q1 , q2 ) studied
in Question 3a, with q1 = 0, q2 = − ρ ′g , with ρ ′ the density of the constitutive
material and g the acceleration of gravity. According to (2.20.18), we find
T
[ F ]123 = −
eh 2 ρ ′g
[ 0 1 0 1 0 1]
6
(2.20.27)
As for the water pressure, the force density as also the form studied in Question
g
3b, namely T [ (2h − y ) ρ g1 ,0] , with ρ the water density. In triangles (1) and (2),
application of (2.20.21) yields
[ F ]12
1
0
eρ gh 2 2
=
6 0
0
0
[ F ]24
4
0
eρ gh 2 5
=
6 0
0
0
(2.20.28)
The global nodal force vector is obtained according to the same process as the
one indicated above for the matrix stiffness. For the sole nodes 1, 2 and 3, we get
ρ
−ρ′
egh 2 6 ρ
[ F2 ] =
6 −3 ρ ′
0
−3ρ ′
(2.20.29)
So, the final equation to be solved reads
0
µ
λ
+
2µ
sym
0
u1′
ρ
−ρ ′
−λ
−(λ+2 µ )
λ
0 v1′
λ+µ
−2(λ+2µ ) −(λ+µ ) u2′ gh 2 6ρ
2(λ+3µ )
. =
(2.20.30)
2(λ+3µ )
−(λ+µ )
−2µ v2′
3 −3ρ ′
0
2(λ+3µ )
λ+µ u3′
2(λ+3µ ) v3′
−3ρ ′
−µ
−µ
µ
71
From the nodal displacements we can then derive the displacement, strain and
stress fields in the whole structure.
Reference
Christensen R.M., Lo K.H. (1979) Solution for effective shear properties in three phase
sphere and cylinder models J. Mech. Phys. Solids, 27: 315-330.
72
Chapter 3 Exercises of Chapter I.3: Elastoplasticity
3.1 Deformation of single crystals
Exercise 3.1 Active slip planes
The active slip planes in a single crystal are dense planes of the crystallographic structure on which the shear stress is larger than the critical value τc.
This is the Schmid and Boas law.
Calculate the tensile stress σp for activation of slip on a plane, whose normal n
makes an angle φ with the tensile axis, in a direction making an angle λ with this
axis.
Solution
Use Cartesian coordinates one direction Ox1 of which coincides with the tensile
axis. Write the stress tensor σij:
σ
(σ ) = 0
0
0 0
0 0
0 0
(3.1.1)
Use the formula: ti = σ ij n j , ti being the stress components on the plane with
unit normal ni:
t1 = σ cos φ t2 = t3 = 0
(3.1.2)
The resolved shear stress is obtained by the projection of the stress ti on the required direction m1 = cos λ
τ = σ cos φ cos λ
(3.1.3)
73
The critical tensile stress corresponds to τ = τ c in (3.1.3):
σ p = τ c cos φ cos λ
(3.1.4)
Find the orientations of the slip plane and slip direction which minimise the critical stress for slip activation.
Solution
Choose the coordinate system such that Ox1 coincides with the tensile axis and
that the unit vector n is perpendicular to the Ox3 axis. The coordinates of this vector are cosφ and sinφ. Let m1, m2, m3 be the coordinates of the unit vector in the
slip direction with m1 = cos λ . The slip direction is perpendicular to n. Hence:
m1 cos φ + m2 sin φ = 0
(3.1.5)
m12 + m22 + m32 = 1
(3.1.6)
Furthermore:
From these two equations is deduced:
cos 2 λ
= 1− m32
sin 2 φ
(3.1.7)
so that the Schmid factor is:
12
mS = cos φ cos λ = cos φ sin φ (1 − m32 )
(3.1.8)
It is maximised for m3 = 0 (thus mS = cos φ sin φ ) and φ = π 4 ( mS = 1 2 ).
Find the relation which the spherical components sinϕ cosθ, sinϕ sinθ, cosϕ, of
the tensile stress direction T must satisfy in order to activate a particular slip system ( hkl ) [ uvw ] in a F.C.C. single crystal, for example (111) 101 .
Solution
Express cosφ and cosλ as a function of h, k, l and u, v, w. Let N be the vector of
components (h, k, l) normal to the slip plane and M the vector of components (u, v,
w) along the slip direction:
74
12
T .N = h sin ϕ cos θ + k sin ϕ sin θ + l cos ϕ = ( h 2 + k 2 + l 2 ) cos φ
(3.1.9)
12
T .M = u sin ϕ cos θ + v sin ϕ sin θ + w cos ϕ = ( u 2 + v 2 + w2 ) cos λ
Equation 3.1.4 yields:
2 12
2 12
σp
( h + k + l ) (u + v + w )
=
(3.1.10)
τ c ( hsinϕ cosθ +ksinϕ sinθ +lcosϕ )( usinϕ cosθ + vsinϕ sinθ + wcosϕ )
2
2
2
2
Note that h, k and l are equal to ±1 , while, one of the coordinates u, v and w
being equal to 0, the other two have the appropriate values ±1 for the slip direction to be perpendicular to the normal to the slip plane. Table 3.1.1 lists the various slip systems in FCC crystals according to this rule. Inverting the + and – signs
for the slip direction merely modifies the orientation. The absolute values of the
cosines only enter the Schmid factor.
Table 3.1.1. Slip systems in FCC crystals
(111)
(11 1 )
(1 11)
(111)
110
101
011
1 10
10 1
01 1
-
-
-
x
x
x
-
x
x
x
-
-
x
-
x
-
x
-
x
x
-
-
-
x
( )
In the particular case of the slip system 111 101 , Eq.3.1.10 yields:
σp
τc
=
6
sin ϕ cos θ + sin ϕ sin θ + cos ϕ 1 − sin ϕ cos θ
(
)(
)
(3.1.11)
Find the tensile direction which minimises the critical stress for slip activation on
the 111 101 slip system.
( )
Solution
(
)
The three vectors of components sin ϕ cos θ ,sin ϕ sin θ ,cos ϕ , (1,1,1) and (-1,
0, 1) must be in the same plane, so that:
75
sin ϕ cos θ sin ϕ sin θ cos ϕ
1
−1
1
0
1
1
= sin ϕ cos θ − 2 sin ϕ sin θ + cos ϕ = 0 (3.1.12)
The second condition for maximising the Schmid factor is:
cos φ =
sin ϕ cos θ + sin ϕ sin θ + cos ϕ
3
=
1
2
(3.1.13)
The solutions of Eqs 3.1.12 and 3.1.13 are:
1 1
cos ϕ = 6 + 2 = 0.908
1
= 0.976
sin θ =
12
7 2− 6
(
(3.1.14)
)
The signs of cosϕ and of sinθ are such that the pole of the tensile axis is in the
reference triangle on the stereographic projection.
For this direction of the tensile axis σ p = 2τ c .
Place the pole of this tensile axis on the stereographic projection.
Solution
On the stereographic projection the distance ρ of the pole to the centre of the base
circle of radius R is equal to Rtan(ϕ/2), ϕ being the angle between the tensile axis
and the direction 001 .This is demonstrated in the following way, x and y being
the coordinates of a pole on the projection plane, r, θ and ϕ the spherical coordinates of a pole on the Ewald sphere of radius R/2:
2x
2y
2R
=
=
R sin ϕ cos θ R sin ϕ sin θ R cos ϕ + 1
(
)
(3.1.15)
Hence:
12
ρ = ( x2 + y2 ) = R
sin ϕ
( cos ϕ + 1)
= R tan
ϕ
2
(3.1.16)
76
( )
The values found for cosϕ and sinθ (Eq. 3.1.14) yield: ρ R = tg ϕ 2 = 0.219
Figure 3.1.1 shows the position of the tensile axis for easiest activation of the
slip system 111 101 .
( )
Fig. 3.1.1. Stereographic projection showing the pole of the tensile axis for the easiest activation
of the (111) plane in the [-101]direction.
Which are the double slip directions?
Solution
( )( )( )
Observing the stereographic projection, the reference triangle 001 , 011 , 111 ,
( )
by two triangles: (001), (011), (111) and (001), (101), (111), within which the
tensile vector activates respectively the (111)101 and the (1 11)011 slip syswithin which the tensile vector activates the 111 101 slip system, is bordered
tems. When the pole of the tensile axis coincides with the lines belonging to two
77
adjacent triangles, the Schmid factors of the corresponding slip systems are equal.
There is then double slip.
Exercise 3.2 Geometrical hardening (Jaoul 2008)
In the course of the plastic deformation by slip of a single crystal loaded by
traction is observed a rotation of the slip planes. The angle of their normal
with the tensile axis increases. The orientation becomes less and less favourable for slip. Hardening occurs, even if the critical shear stress remains constant. This is called geometrical hardening. The exercise is based on the
book of Jaoul, which was reedited in 2008.
In a single crystal, find the relation between the extension e in the tensile direction
and the slip g in a slip direction making initially an angle λ0 with the tensile axis,
along a slip plane whose normal makes initially an angle φ0 with the tensile axis.
Assume free rotation of the specimen.
Using Cartesian coordinates such that axis Ox3 coincides with the tensile axis
and that the slip direction lies in the Ox1, Ox3 plane, find the strain tensor for the
transformation corresponding to the slip g.
Solution
The coordinates of the unit vector along the slip direction are sinλ0, 0, cosλ0 .
The coordinates of the unit vector normal to the slip plane are v1, v2, cosφ0. As it is
perpendicular to the slip direction:
cos φ0 cos λ0
ν 1 = − sin λ
0
1
sin 2 λ0 sin 2 φ0 − cos 2 λ0 cos2 φ0 ) 2
(
v2 =
sin λ0
(3.2.1)
The distance of point M, whose initial coordinates are a1, a2, a3, to the slip
plane is aivi.
Point M is displaced to point M’ whose coordinates are x1, x2, x3. The length
MM’ is gaivi, so that:
x1 = a1 + gai vi sin λ0
x2 = a2
x = a + ga v cos λ
3
3
i i
0
(3.2.2)
78
Hence the gradient tensor (see section A3.1.1 in Volume I) given by the partial
derivatives of xi with respect to ai.
The deformation tensor ∆ij can now be written. As the extension e is in the
Ox3 direction, the ∆33 element only needs to be calculated:
2 ∆33 = g 2 cos 2 φ0 + 2 g cos φ0 sin λ0
(3.2.3)
The extension e is related to this by:
2
2
2
2∆33da32 = dx − da = (1 + e ) − 1 da32
(3.2.4)
Equating the two preceding relations, the slip g can be calculated as the positive root of a second degree equation, yielding:
2
g cos φ0 = (1 + e ) − sin 2 λ0
1
2
− cos λ0
(3.2.5)
The relation between the tensile stress σ and the extension e is now sought. In
order to do that, it is needed to introduce the current angles λ and φ in the course
of deformation. Choosing a point M on the tensile axis with coordinates 0, 0, 1 the
scalar product of the vector OM’ and the unit vector in the slip direction yields
OM’cosλ.
As the coordinates of OM’ are g cosφ0 sinλ0, 0, 1 + g cosφ0 cosλ0 it is found
that:
cos λ =
g cos φ0 + cos λ0
(1 + 2 g cos λ cosφ
0
0
+ g 2 cos 2 φ0 )
1
(3.2.6)
2
and, using the preceding equation (3.2.5) relating g and e:
(1 + e ) 2 − sin 2 λ0
cos λ =
1+ e
1
2
(3.2.7)
The scalar product v.OM’ is equal to OM’cosφ so that:
cos φ =
It can also be written:
cos φ0
(1 + 2 g cosφ
2
2
0 cos λ0 + g cos φ0 )
1
(3.2.8)
2
79
cos φ =
cos φ0
1+ e
(3.2.9)
The tensile flow stress σp is related to the critical shear stress τc by the Schmid
relation found in Exercise 3.1 above:
τ c = σ p cos φ cos λ
(3.2.10)
Equations (3.2.7), (3.2.8) and (3.2.9) yield:
σp =
(1 + e )
2
2
cos φ0 (1 + e ) − sin 2 λ0
1
2
(3.2.11)
τc
It is an increasing function of the extension e, provided
2 sin λ0 ≤ 1 + e , even
when there is no work-hardening (τc is a constant).
Draw the trajectory of the tensile axis on the stereographic projection. Assume
free rotation of the specimen.
Solution
It can be checked that the tensile axis remains in the plane containing the normal
to the slip plane and the slip direction. This is achieved if
cos φ = sin λ . Equations 3.2.7 and 3.2.9 allow checking that this is the case. The
tensile axis rotates, the angle φ increasing, so that its pole on the stereographic
projection moves away from the pole (111) towards the pole 101 .
Starting from the position of the pole of the tensile axis determined in the preceding exercise, it can be seen (Fig. 3.2.1) that this pole, moving towards the pole
101 , meets the border line 001 , 111 (cosθ = sinθ) where double slip starts. If
( )( )
it goes beyond this line, the angle φ keeps increasing, corresponding to a decrease
of the Schmid factor; in the meanwhile the homologous angle φ for the
1 11 011 slip system decreases; this system is then activated; the pole of the
( )
( )
tensile axis moves towards the pole 011 ; the slip system 111 101 is activated again; the pole of the tensile axis moves back to the cosθ = sinθ line. Thus,
the double slip is stabilised along this line; the pole of the tensile axis moves towards the pole 112 .
80
Fig. 3.2.1. Displacement of the pole of the tensile axis in the course of deformation of a FCC
crystal. Single and double slip. The dotted lines show the double slip systems.
3.2 Twinning
A twin is part of a crystal which is related to the parent crystal by an operation of symmetry.
In FCC the twin is the mirror image of the parent crystal across the (111)
plane, called the twin plane.
81
Exercise 3.3 Deformation of a twinned crystal
A twinned FCC crystal, the (111) plane being the twin plane, is compressed in a
direction 134 . Are the deformations of the crystal and the deformation of the
twin compatible?
Solution
Both deformations will be compatible if the compression activates the same slip
systems. The determination of the position of the slip systems in the twin is
needed. The coordinates of a point P being p1, p2, p3, we want to determine the
coordinates p'1, p'2, p'3 of its image P' in the twin.
The equation of the plane (111) is:
x1 + x2 + x3 − 1 = 0
(3.3.1)
The distance d of the point P to the (111) plane is:
d=
p1 + p2 + p3 − 1
(3.3.2)
3
The direction cosines of the normal to (111) being 1
3, 1
3 ,1
3 , the co-
ordinates of the vector PP' are (A, A, A) with A= −2 ( p1+ p2 + p3 − 1) 3.
Therefore, the coordinates of P' are:
p1 − 2 ( p2 + p3 ) + 2
p1′ =
3
p2 − 2 ( p1 + p3 ) + 2
p2′ =
3
p3 − 2 ( p1 + p2 ) + 2
p3′ =
3
(3.3.3)
For a vector OP the image in the twin is O'P' = OP' – OO'. The coordinates of
OO' are (2/3, 2/3, 2/3).
82
The location of the pole of the compression axis on the stereographic projection
is found by noting that it is parallel to the plane (111) and that cos θ = −1
10 , so
that tanθ = 3. Figure 3.3.1 shows this location. It lies in the triangle
( 001) , ( 0 11) , ( 1 11) . In this position it activates the 1 11 101 slip system. The
Schmid factor is:
( )
mS =
( −1 + 3 + 4 )(1 + 4 ) =
26 6
15
13 6
(3.3.4)
The compression axis being parallel to the (111) twin plane remains unchanged
in the twin. It is the same for the vector of the slip direction.
The image P' in the twin of the normal to the slip plane is such that:
p1′ =
p2′ =
p3′ =
1
27
−5
27
1
27
(3.3.5)
Its position is shown in Fig. 3.3.1.The Schmid factor in that case is:
mS′ =
( −1 + 15 + 4 )(1 + 4 ) =
26 6 27
15
13 6
(3.3.6)
It is the same as the one calculated for the parent crystal (Eq. 3.3.4).
Compression in the 134 direction activates as well the 1 11 101 slip sys-
( )
( )
tem in the parent crystal and its image, the 151 101 slip system, in the twin.
The deformation is therefore compatible.
83
Fig. 3.3.1. Positions of the [-1,-3,4] compression axis, P in the parent crystal and P' in the twin
on the stereographic projection.
Exercise 3.4 Twinning and martensitic transformation
The Bogers-Burgers model constitutes an explanation of the martensitic transformation of a FCC structure into a BCC one. Consider in a FCC structure a tetrahedron PVQS of base PVQ the faces of which are {111} planes. Calculate, in
terms of aFCC<112>, the slip of the successive planes parallel to PVQ in the QT direction perpendicular to the PV one (QT is a <112> direction) to transform the
{111} QPS and QVS planes in {110} planes of the BCC structure. What additional shear is needed to obtain a BCC structure?
Solution
Determine the angle between the 111 and the 11 1 directions in the BCC
structure (arctan( 2 2 ), almost 70°). Determine the shear needed to achieve the
transformation of a FCC (111) plane in a BCC (110) plane ( 2 2 − 3 2). Calcu-
84
late, in terms of aFCC<112> the slip of the successive compact {111} plane piled
over the PVQ plane to obtain this shear (almost 1/3(1/6); the summit S moves to
S’; a 1/6 aFCC<112> slip would create a twin).
Observe the way the atoms are piled up in successive planes parallel to the
QPS’ or QVS’ planes and determine the additional slip needed to create a BCC
structure (1/8 aBCC<110>).
3.3 Dislocations
Dislocations are linear defects in crystals. The displacement of dislocations
produces plastic deformation. Their interactions with various obstacles, periodic potential in the crystal itself, other dislocations, grain boundaries and
free surfaces, foreign atoms, increase the force needed to move them; this
explains how yield strengths are modified.
Exercise 3.5 Geometrical properties of dislocations
This exercise illustrates Sect. 3.3.3 of Volume I.
a) Show that at a node where three dislocations meet the sum of the Burgers vectors is zero.
Solution
Draw a Burgers circuit around one of the dislocations; then displace it to surround
the other two. The closure displacement should remain unchanged. Be careful
about the orientation of the lines.
b) Determine the Burgers vector of the dislocation in the bubble raft shown in Fig.
3.13 (p. 175) of Volume I.
Solution
Draw a Burgers circuit as shown in Fig. 3.5.1. Carefully count the same number of
bubbles along each side. The Burgers vector is the closure defect.
85
Fig. 3.5.1. Determination of the Burgers vector b by drawing a Burgers circuit on the bubble raft.
Plastic irreversible deformation stores in crystals geometrically necessary
dislocations. Several examples are given in Volume I (Sect 3.3.3.5).
c) Estimate the density of geometrically necessary dislocations in a thin wall cylinder the radius of which has been increased from r0 to r0 + ∆r by internal pressure.
Solution
The increase of the radius induces a decrease of both the wall thickness and of the
length of the cylinder. As the only component of the stress tensor is the hoop
stress σ θ , the ratio of the longitudinal strain to the circumferential strain is:
εz − σ θ 2
1
=
=−
εθ
σθ
2
(3.5.1)
The volume remaining unchanged, ε r = −1 2 . With e the thickness of the wall
and l its length:
e
l r
= = 0
e0 l0 r
12
(3.5.2)
86
Let a Burgers closed circuit follow the inner and outer circumferences of the
cylinder, these two being joined by two segments across the wall. The variation of
the lengths of these two segments is of second order. The increase of the length Lb
of the circuit is thus:
(
)
∆Lb = 4π r0 + ∆r − 4πr0 = 4π∆r
(3.5.3)
This corresponds to the introduction of n edge dislocations of Burgers vector b
such that nb = 4π∆r. Note that there will be about the same number of positive and
negative edge dislocations because there is no bending of the wall. The corresponding density of geometrically necessary dislocations of length l is:
4π∆r
ρG =
l
b
( 2πrel ) =
2∆r
12
be0 r0 ( r0 + ∆r )
(3.5.4)
d) Find the density of dislocations resulting from the rapid cooling of an Al-SiC
composite from 400°C to room temperature, given that the coefficients of expansion of Al and SiC are 23.10-6 and 5.10-6 respectively and the diameter of the SiC
particles is 10 microns.
Solution
The difference of the coefficients of expansion ∆α results during cooling in the
punching of edge dislocation loops in the aluminium matrix. The relative volume
variation from the injection of a number N of such loops of diameter φ, that of the
SiC particles, is equal to the relative volume variation due to the difference of the
dilatation coefficients:
Nbπφ 2
= 6∆α∆T
4V
(3.5.5)
∆T being the temperature difference.
The density ρG of these geometrically necessary dislocations is then:
ρG =
N πφ 24∆α∆T
=
V
bφ
In the aluminium matrix ρG = 5.7x1013 m-2.
Exercise 3.6 Mobile dislocations
(3.5.6)
87
The displacement of dislocations produces deformations. The average plasS
tic strain rate is given by the equation εɺ p = ∑ γɺ g ( m g ⊗ n g ) , symbol S deg
noting the symmetric part of the tensor and g relates to the various slip systems (Volume I, Sect. 3.5.1.2, p.310, Eq. 3.186). The average strain is given
(
by δε = b ⊗ n
S
) (δA V ), δA being the area swept by the dislocation and V
the volume (Volume I,Sect. 3.3.6.4, p.196, Eq. 3.32b).
a) Find the deformation induced by the climb of an edge dislocation over an area
δA.
Solution
Choose a coordinate system such that Ox3 is along the dislocation line and Ox2
perpendicular to the glide plane, so that the components of the Burgers vector are
(b1, 0, b3) and those of the vector l are (0, 0, 1). The average deformation is given
(
)(
)
by the above equation written: δε ij = bi n j + bj ni δA 2V , where the coordinates
of the Burgers vector reduce to b1, the coordinates of the normal n to the swept
area reduce to n1 and V is the volume of the body. Thus the deformation tensor reduces to
(
δε11 = b1 δA V
)
(3.6.1)
Contrary to dislocation glide, this climb of the dislocation produces a volume
change.
b) Calculate the mean velocity of dislocations in copper to reach a deformation
rate of 10-4 sec-1 in a tensile test, assuming a reasonable mobile dislocation density.
Solution
Remember that the shear rate dγ/dt is given by the relation γɺ = ρ bv , ρ being the
mobile dislocation density, b the Burgers vector and v the average velocity of dislocations. The tensile strain rate Eɺ may be approximately related to the shear rate
γɺ in the grains by the Taylor factor, about 3 for FCC materials (Volume I §3.5.3.6
p.348). For copper the interatomic distance (from Table D1, Appendix D in Volume I) is 2.56x10-10 m. Take 1012 m-2 as a reasonable mobile dislocation density.
The solution is then v = 1.2 micron/s.
88
c) What are the nucleation rates of dislocations Nɺ on each activated slip system in
a copper single crystal of diameter D = 1 cm and length L = 5 cm tested in tension
along the 001 , 011 , 111 or 123 directions with a strain rate of 10-4s-1?
Solution
Find in the first place which are the active slip systems in each case. This is
achieved by calculating the Schmid factor, which is equal to n3b3/b, b being in the
active slip direction. They are listed in Table 3.6.1.
Table 3.6.1. Schmid factors multiplied by √6 for the possible slip systems according to the orientation of the tensile axis. The highest Schmid factors of the active slip systems are in bold characters.
Tensile axis
001
Slip
plane
111
( )
110
101
011
1 10
101
01 1
-
-
-
0
1
1
1
1
0
2/3
2/3
0
16/14
-
011
111
123
001
(11 1 )
-
011
111
123
001
(1 11)
011
111
123
001
011
111
123
( 111)
0
1
1
12/14
0
0
0
0
0
2/3
2/3
4/14
-
10/14
1
6/14
-
1
0
0
0
0
2/3
2/3
1
1
1
0
0
0
0
4/14
-
0
-
-
0
0
4/14
0
-
4/14
-
1
4/14
89
The areas swept by the dislocations are A0/n3, if A0 is the cross-section of the
test piece. Thus
δ Aɺ
V
=
Nɺ A0 n3
A0 L
(3.6.2)
Aɺ
Nɺ
= kn3b3
V
n3 L
(3.6.3)
The strain rate is:
εɺ33 = kn3b3
where k is the number of active slip systems. Hence:
Lεɺ
Nɺ = 33
kb3
(3.6.4)
The results are in Table 3.6.2.
Table 3.6.2. Nucleation rates of dislocations on each active slip system
Tensile axis
001
Number of active
slip systems k
8
Dislocation nucleation rate
Nɺ (s-1)
3.45x103
011
4
9,7x103
111
6
4x103
123
1
25.8x103
Exercise 3.7 Force on a dislocation
The Peach and Köhler formula (Volume I, Sect. 3.3.6.3, p.194, Eq.
3.29): f = b ⋅ σ ∧ l , relates the force f on the dislocation to the Burgers
( )
vector b, the stress tensor and the unit vector along the dislocation line.
a) Show that the glide force is equal to τb and calculate the climb force.
90
Solution
Take a Cartesian coordinate system such that Ox1 lies along the dislocation line
and Ox2 is perpendicular to the glide plane.
The components of the force on the dislocation are found by application of the
Peach and Köhler formula with l1 = 1, l2 = l3 = 0 and, the glide plane of the dislocation being perpendicular to Ox3, b2 = 0. So:
f1 = 0
f 2 = b1σ 12 + b2σ 23 = τ b
f = − (b σ + b σ )
1 13
2 33
3
(3.7.1)
where f3 is the glide force, perpendicular to the dislocation line. The climb force is
equal to – b3σ33 as the edge component is to be considered.
b) Determine the force on a prismatic loop of dislocation, that is a dislocation the
Burgers vector of which is perpendicular to the plane of the loop. How will the
dislocation loop change if gliding is active?
Solution
Let the Ox2 axis be in a direction perpendicular to the plane of the loop; b2 is the
only component of the Burgers vector. The coordinates of the unit vector along the
dislocation line are: sinθ, cosθ, 0. The Peach and Khöler formula gives:
f1 = −bσ 22 cosθ
f 2 = bσ 22 sin θ
f = bσ cos θ − bσ sin θ
3
12
23
(3.7.2)
It is easy to check that the force is perpendicular to the dislocation line. The
glide force f3 is positive if tanθ < σ12/σ23, negative otherwise. Part of the dislocation loop will glide in the positive direction of Ox2, the other part in the negative
direction; the prismatic loop will become an elongated loop, the two long branches
attracting one another to create a dipole.
Exercise 3.8 Jogs
A jog on a dislocation line is a small segment of length b, the Burgers vector
length, perpendicular to the glide plane. Jogs are created when dislocations
91
cross (Volume I, Sect. 3.3.8.1, p.212). Their energy is estimated to be
(1/10)µb3, µ being the shear modulus.
Is there an equilibrium density of jogs along a dislocation line?
Solution
The equilibrium corresponds to a negative free enthalpy G = H − TS.
Let n be the number of jogs per unit length of dislocation. They can occupy N =
1/b positions along the line. Then the enthalpy per unit length is given by:
H=n
µ b3
(3.8.1)
10
with µ = 45,000 MPa and b = 2.5x10-10, H = (n/N) 2.8x10-10 J.
The entropy S is estimated from the number of combinations of N positions n to
n:
n
n
n
n
S = −kN log + 1 − log 1 −
N
N
N
N
(3.8.2)
where k is the Boltzman constant (1.38x10-23 J/K).
S is maximum for n/N = ½; its value is kNlog2. With b = 2.5x10-10 this is equal
to 3.8x10-12 J/K per unit line of dislocation.
Now:
n
n
n
G=H −TS =N 2.8 × 10−10 + kT log
N
N
N
n
n
+1 − log 1 −
N N
(3.8.3)
Stability is achieved if the quantity in bracket, which is negative, is smaller
than −2.8x10-10 /kT = − (2x1013/T) (n/N). At 300 K this is − 6.7x1010(n/N).
This condition is not achieved (except for extremely small values of the jog
concentration).
The conclusion is that jogs are unstable defects.
Exercise 3.9 Local change of the volume near a dislocation
Introduction of dislocations in a material produces but a very small increase
of the overall volume. There are however local changes in the neighbourhood of the dislocation. This constitutes a sound foundation for the assump-
92
tion that plastic deformation takes place at constant volume (Volume I, Sect.
3.3.7.3, p.200).
Find the change in volume in the neighbourhood of a dislocation. Draw the equal
value contours.
Solution
Around a screw dislocation the only component of the strain tensor which is different from 0 is ε θz (Volume I, Sect. 3.3.7.1, p.197). There is no local volume
variation in the neighbourhood of a screw dislocation.
This is not so near an edge dislocation as:
ε rr = ε θθ =
b 1− 2ν sin θ
4π 1 − ν r
(3.9.1)
where b is the Burgers vector, ν the Poisson ratio, θ the polar angle and r the distance to the core. Then the volume variation is twice this value. The contours of
constant volume variation obey equations such that r = Constant x sinθ ; they are
circles in cylindrical coordinates, passing through the dislocation core and with
their centre on the Ox2 coordinate axis.
At each point where there is a positive volume variation corresponds an opposite point where it is negative with the same absolute value. Thus the overall volume does not change.
Exercise 3.10 Interactions between dislocations
A dislocation has a stress field which creates a force on neighbouring dislocations according to the Peach and Köhler equation (Volume I, Sect. 3.3.8,
p.210).
a) Find the equilibrium positions of two parallel edge dislocations on parallel slip
planes.
Solution
Choose (Ox1, Ox3) as the slip plane of the first dislocation with Burgers vector b1,
the dislocation line being oriented in the Ox3 direction. Use the equations of the
stress field of an edge dislocation (Volume I, Sect. 3.3.7.2, p.200, Eq. 3.43a, noting that there is an exponent 2 missing in the denominator) and the Peach and
Köhler equation, which yields:
93
f1 = −
µ (b1 ⋅ b2 )
(
)
8π 1 − ν h
sin 4θ
(3.10.1)
where f1 is the glide force per unit length on the second dislocation with Burgers
vector b2, µ is the shear modulus and ν the Poisson ratio, h is the distance between the slip planes and sinθ the ratio between h and the distance between the
two dislocations.
When the Burgers vectors of the dislocations have the same sign, they attract
each other when -π/4 < θ < π/4, and repel each other otherwise; the stable position
is found for θ = 0. The two dislocations, which must remain in their own slip
plane, form the nucleus of a tilt wall. When the dislocations have opposite signs,
the stable position is found for θ = π/4. The two dislocations form a dipole.
b) What is the resolved shear stress τ needed to destroy a dipole of dislocations in
copper assuming that the distance between the slip planes is equal to 4b.
Solution
To destroy the dipole the applied force τb on the second dislocation must exceed
the maximum attraction force f1 when the distance between the two dislocations
increases. This is found for an angle θ = 3π/8. Hence, from Eq. 10.1:
τb =
µ b2
8π (1− ν )4b
(3.10.2)
For copper, with µ = 45,000 MPa, ν = 0.33 and b = 2.56x10-10 m, τ = 670 MPa.
When the two dislocations are in the same slip plane (h = 0) the stress of the
first dislocation on this plane is (Volume I, Sect. 3.3.7, Eq. 3.35b and 3.43a):
µ ( b1 ⋅ b2 )
for a screw
σ 23 =
2πx1
σ = µ ( b1 ⋅ b2 ) for an edge
12 2π (1 −ν ) x1
c) Calculate the climb force on the second dislocation
Solution
The Peach and Köhler equation yields:
(3.10.3)
94
f2 =
µb2
sin 2 θ ( 2cos 2 θ + 1)
2π (1 −ν ) h
(3.10.4)
The maximum value is found for θ = π/4 and:
f 2max =
µb 2
2π (1 −ν ) h
(3.10.5)
Exercise 3.11 Interaction of a dislocation with an interface
The stress field of a dislocation depends on the elastic moduli of the material
in which it lies. On the other side of the interface, they are different; thus the
stress field is also different there. The problem is to insure the continuity of
the stress and strain fields at the interface. This is achieved by the introduction of image dislocations (Volume I, Sect. 3.3.8.4, p.216).
Find the interaction of a screw dislocation in an isotropic material of shear
modulus µ1 parallel to an interface with an isotropic material of shear modulus
µ2.
Solution
Chose coordinate axes such that Ox1 is perpendicular to the interface, and Ox3
along the dislocation line. The displacements and the stresses, which the dislocation would create on the location of the interface, distant of d from the dislocation,
in an infinite medium would be (Volume I, Sect. 3.3.7.1, p.197):
x
b
arctan 2
2π
d
µ1b x2
σ 13 = −
2π d 2 + x22
u3 =
(3.11.1)
µ1 being the shear modulus of the material in which lies the dislocation.
Notice that the displacement is independent of the shear moduli and, everywhere, is not influenced by the presence of the interface.
The image dislocations must be located either at the same place than the real
dislocation or at a symmetric place with respect to the interface. This is because, if
there are n image dislocations with Burgers vectors ±b , the equality, whatever x2,
95
of the second equation (3.11.1) with the sum of the stresses due to the images,
provides n equations; all the n distances di can be calculated. Now, the equality of
the first equation (3.11.1) with the sum of the displacements due to the images is
respected for one value only of x2. Thus, it is needed that di = ± d .
Designate by βb the Burgers vector of the symmetrical image dislocation assumed to be in the material of modulus µ1, and by γb the Burgers vector of the image dislocation coinciding with the real dislocation and assumed to be in the material of modulus µ2. Matching displacement and stress on both sides of the interface
µ − µ1
2µ1
shows that β = 2
and γ =
.
µ2 + µ1
µ2 + µ1
If the interface is a free surface, µ2 = 0 and β = −1, so that the corresponding
image dislocation attracts the real dislocation. It feels an attractive force towards
the free surface.
In the general case, the force is attractive if µ2 < µ1, repulsive otherwise. The
presence of an oxide layer can prevent the escape of dislocations and harden the
crystal.
Exercise 3.12 Dislocation loop
The energy of a straight dislocation is of the order of 0.5µb2 per unit length,
µ being the shear modulus and b the Burgers vector (Volume I, Sect. 3.3.7.5,
p.202). For a dislocation loop, part of this energy disappears because each
line segment along the line faces an opposite one the Burgers vector of
which is of opposite sign. The energy of a dislocation loop was calculated
8Rc
µ b2 Rc 1
1
(Nabarro 1967, p.75, formula 2.53 a). It is
1+
log 2 ,
2 2 1 − ν
e r0
where Rc is the radius of the loop, ν the Poisson ratio and r0 the core radius
of the order of 10-9 m.
a) Find the shear stress ratio τ/µ required to create ex nihilo a dislocation loop in
copper.
Solution
The work W of the shear stress to expand the loop must be equal to the energy E
of the loop:
96
2
2
R τ b 1 1
1 b Rc 8 Rc
W = π c
= 1 +
log 2
a0 µ a0 2 2 1 −ν a0 a0 e a0
(3.12.1)
With a0 ≃ b and 8 e 2 ≃ 1 , τ/µ reaches a maximum as a function of Rc/a0 for
Rc∗ a0 ≃ 2.7 . Then τ/µ = 0.2. The corresponding energy is 3.4x10-19 J. It is too
large with respect to the thermal energy kT for thermal fluctuations to generate the
loop.
b) Find the force F needed to punch a 10 µm in diameter prismatic dislocation
loop in copper.
Solution
If the punch creates a prismatic loop, the displacement of the force is equal to the
Burgers vector b and the work done is Fb. This must be equal to the energy of the
loop for which the dislocation is of edge character all around the periphery.
Hence:
Fb =
µb2
8R
Rc log 2 c
4 (1 −ν )
e a0
(3.12.2)
The result is F = 2.1x10-4 N.
We neglected the energy of the area 2πRcb created by the punch. With
γ s ≃ 0.1µ b , the extra force needed is 0.36x10-4 N.
c) Find the critical velocity v for a particle of a specific mass ρ of 3,000 kg/m3 to
punch a dislocation loop of 10 µm in diameter in copper.
Solution
( )(
)
Equating the kinetic energy of the particle 1 2 ρ 4π 3 Rc3v 2 to the energy of the
prismatic loop of radius Rc, the critical velocity is found to be 0.26 m/s.
Exercise 3.13 Peierls force
Dislocations tend to lay along Peierls troughs of potential aligned with the
most dense directions of the crystal structure. A force is needed to move a
dislocation from one trough to the next, by the displacement of kinks. This
is the lattice friction (Volume I, Sect. 3.4.2.1, p.225).
97
Calculate the Peierls force τP with respect to the shear modulus for a partial
Shockley dislocation in a FCC structure with a Poisson ratio of 0.3, given that
2π a
2µ
τP =
exp −
, a being the distance between the atomic slip planes.
1− ν
1− ν b
Solution
From the ratio a/b for a partial Shockley dislocation, 1
5x10-3.
2 , the result is τP/µ =
Exercise 3.14 Strain-hardening
Strain-hardening results from the multiplication of dislocations during plastic deformation. The Taylor formula, τ c = αµ b ρD , outlines the effect of
the dislocation density ρD on the critical shear stress τc. (It is reminded that
α is a parameter of the order of 1/3, µ is the shear modulus and b the Burgers vector). This formula results from the density of obstacles to the displacement of gliding dislocations offered by the forest of dislocations (Volume I, Sect. 3.4.2.2, p. 229). The dislocation density includes both the part
due to the multiplication from Frank-Read sources and the part due to geometrically necessary dislocations.
a) Estimate the amount of hardening of an extruded aluminium composite material
due to 10 vol% of rigid inclusions 1 micron in diameter, the drawing ratio being
10%.
Solution
The extrusion introduces geometrically necessary dislocation loops punched from
the rigid inclusions. We first need to determine the displacements to be compensated by dislocations. Let D0 be the initial diameter of the aluminium bar. After
extrusion it becomes D and the volume remaining the same in plastic deformation
the elongation is such that:
LD 2
=1
L0 D02
(3.14.1)
Let ∆D be the decrease of the diameter of the bar after extrusion and ∆L its increase in length.
98
D
∆D
D = 1− D
0
0
2
∆L = D0 − 1
L
0 D
(3.14.2)
If the volume fraction is fV, the mean distance d between the inclusions volume
πφ /6 is:
3
13
π
d =
6
φ
(3.14.3)
f V1 3
The elongation 2u in the extrusion direction of the edges of length d of a cube
containing one inclusion is given by:
D 2
L0
2u
= ∆L = L0 0 − 1
d
D
(3.14.4)
The area decrease of the cube containing one inclusion is such that:
(
2
πD02 2
2
= πD0 − π D0 − ∆D
d
−
d
−
2u
r
4
4
4d 2
(
)
2
)
(3.14.5)
Let Nl be the number of geometrically necessary dislocation loops:
D
N l b = 2 u + ur = d
D0
(
)
D 3 π 1 3 φ D D 3
0 − 1 =
0 − 1 (3.14.6)
D
6 f V1 3 D0 D
The density of geometrically necessary dislocations is:
ρG =
N l πφ
d3
13
( )
= 36π
f
23
V
3
1 D D0
−
1
φ b D0 D
(3.14.7)
With the above values for the parameters ρG = 1.24x1015 m-2.
The critical shear stress being estimated by the Taylor formula, it is found that
τc ≈ 85 MPa.
b) We wish to determine the efficiency of the work of deformation.
99
Assume that the strain-hardening curve for copper can be represented by
σ = 120ε p0.25 in MPa, εp being the plastic deformation.
Find the relation between the density of dislocations ρD and the strain, given
that its value for εp = 0.4 is 1015 per m2.
Solution
Equating the ratio of stresses for a deformation of εp and for a deformation of 0.4
expressed in term of the corresponding dislocation densities, it is found that ρD =
1.58x1015 εp0.5 m-2.
c) Compare the work done in plastic deformation and the stored energy.
Solution
The work done in plastic deformation is:
Wp =
∫
εp
0
σ dε p =
∫
εp
0
120x106 ε p0.25dε p = 9.6x107 ε 1.25
J / m3
p
(3.14.8)
The stored energy Ws is that of the dislocations the total length of which is ρD
m/m3. Hence:
Ws = 0.5µ b2 ρD = 0.5µ b2 1.58x1015 ε p0.5
(3.14.9)
For copper, with µ = 45,000 MPa and b = 2.56x10-10 m, ED = 2.3x106 εp0.5 J/m3.
Hence the efficiency ratio:
0.5
Ws 2.3x106 ε p
0.024
=
= 0.75
Wp 9.7x107 ε 1.25
εp
p
(3.14.10)
This ratio is for instance equal to 0.027 for a deformation of 0.4. This small
value comes from the heat dissipated in plastic deformation.
Exercise 3.15 Depressed modulus of elasticity
A small enough shear stress bends dislocation segments held by anchorage
points without breaking them. As the segments sweep slip plane area, this
produces a deformation. It disappears upon release of the stress as the dislocation segments, because of the line tension, jump back to their original po-
100
sition. The apparent modulus of elasticity, the depressed modulus, is thus
smaller than the true one (Volume I, Sect. 3.4.2.2, p.229).
a) Find the change in the shear modulus of elasticity µ resulting from the reversible movement of dislocations in a FCC material in which the dislocation density is
1012 m-2.
Solution
The relative change of the shear modulus is related to the relative increase of the
shear strain δγ:
δµ
µ
=−
δγ
(3.15.1)
γ
where δγ is related to the total area NδA swept by the N dislocation segments in
volume V by
δγ =
NbδA
V
(3.15.2)
If R is the radius of curvature of the segments of dislocation when the shear
stress is τ, δA is given by:
12
2
L RL L
δA = R arcsin
−
1−
2 2R
2R
2
=
1 L3
16 R
(3.15.3)
where L is the average length of the dislocation segments. If the anchorage points
of the dislocations are dislocation trees, L2 is equal to the dislocation density ρD:
ρD =
NL
V
(3.15.4)
In addition, we have
R=
0.5µ b2
τb
(3.15.5)
τ being the average shear stress on the various active slip planes (Volume I, Eq.
3.50 p. 204).
From these various equations is deduced:
101
δµ
µ
=−
1
8
(3.15.6)
b) Explain why the modulus depression should be greater after polygonisation.
Solution
Let d be the size of a subgrain, assumed to be cubic, and n the total number of dislocation segments of length l between anchoring points. There are V/d3 subgrains,
3V/d3 walls and (3V/d3)(d2/l2) anchoring points. Thus:
( ρ D )P =
nl 3
=
V dl
(3.15.7)
Hence
δγ P =
nb
3bl τ b
3lγ P
9γ
δ AP =
=
= 2 P
V
16d 0.5µ b 2
8d
8d ( ρ D ) P
(3.15.8)
index P referring to the polygonised material.
( δµ µ )P
δµ µ
=
9
d 2 ( ρD )P
(3.15.9)
Figure 3.15.1 shows an abacus giving this ratio as a function of the subgrain
size and of the dislocation density.
Exercise 3.16 Kinematic hardening
Strain-hardening corresponds to an increase of the dislocation density. Dislocations form clusters and then cells within which the dislocation density is
much smaller than in the walls. The behaviour then corresponds to kinematic hardening. The composite model of Mughrabi explains this (Volume
I, Sect. 3.4.2.2, p.237). It considers the material to be a composite of two
materials, the cells and the walls. The shear stress in each is given by the
Taylor formula (Eq.3.100).
In copper subjected to cyclic work-hardening the volume fraction of dislocation
walls is estimated to be 5%, within which the dislocation density would be about
1016 m-2, against 1013 m-2 in the interior of the cells. Draw the cyclic hardening
102
curve, assuming absence of isotropic hardening, and estimate the amount of
stored energy when the total strain amplitude is 0.01 after load removal (for copper µ = 45,000 MPa, ν = 0.33 and b = 2.56x10-10 m).
Fig. 3.15.1. Dislocation density and subgrain size yielding a certain ratio of the depressed
modulus of a polygonised material to that of the material with a uniform distribution of dislocations.
Solution
Apply the composite model of Mughrabi in which the shear stress is the average
of the one of the cell interior τc and the one of the walls τw, each one given by the
Taylor formula.
103
τ = (1 3) µb 1013 = 4 MPa
c
16
τ
w = (1 3) µb 10 = 128 MPa
(3.16.1)
The shear stress of the composite is the average:
τ = 0.95τ c + 0.05τ w = 10 MPa
(3.16.2)
Figure 3.16.1 shows the behaviour.
For the stored energy is the energy of the dislocations: 0.5µb2 J/m:
WS = 0.95 ( ρ D )c + 0.05 ( ρD ) w µ b 2 ≃ 8.4x105 Jm −3
(3.16.3)
This stored energy can be compared to the work W = τγ of the shear stress,
which is about 1.4x105 J/m3. The difference is the heat produced.
Exercise 3.17 Interaction of solute atoms with dislocations
Solute atoms interact with dislocations owing to two effects: size effect and
modulus effect. The first one comes from the local change of volume and
distortion, which a solute atom creates, interacting with the stress field of the
dislocations. The second one is of second order (Volume I, Sect. 3.4.3.1,
p.249).
a) Calculate the approximate interaction energy of an atom of zinc with an edge
dislocation in copper (for copper µ = 45,000 MPa, ν = 0.33 and b = 2.56x10-10 m,
and for zinc the interatomic distance is 2.67x10-10 m).
Solution
A zinc atom in substitution in the copper crystal has a larger volume than the solvent. It interacts with the hydrostatic component of the dislocation stress field.
Neglect any modulus effect. The hydrostatic stress in the neighbourhood of an
edge dislocation (Volume I, Eq. 3.110) is, r and θ being the polar coordinates:
σm = −
1+ ν µ b
sin θ
1− ν 3π r
(3.17.1)
104
Fig. 3.16.1. Cyclic behaviour of copper containing dislocation cells according to the model of
Mughrabi.
The relative size increase due to a zinc atom is η = ∆a0/a0 = 0.043. (a0 = b/2 is
the radius of the copper atom).
The volume increase is: ∆V = 4πa03η = 1.13x10-30.
The interaction energy, for θ = 0, is:
U = −σ m ∆V = 2.74x10−30
1
J
r
(3.17.2)
At a distance r = 2b, U = 5.3x10-21 J (0.03 eV) per solute atom.
b) Find the approximate interaction energy of a carbon atom with a dislocation in
α iron.
105
Solution
A carbon atom occupies the middle of the edge of the cubic iron cell. The length
of this edge is a = 2.866x10-10 m. The insertion of the carbon atom increases it to
(2.482 + 1.544)x10-10 m. The resulting strain is log (4.026/2.866) = 0.339. The
elongations in the two other perpendicular directions being zero, the relative volume increase is 0.339. The volume increase is 0.339 a3 = 6x10-30 m3. The interaction energy U = − σm∆V with an edge dislocation, calculated at a distance 2b, is
deduced from Eq. 3.17.1 applied to iron. It yields U = 6x10-20 J (0.4 eV) per atom.
There is no volume interaction with a screw dislocation the stress field of
which being of zero hydrostatic stress. However, the distortion introduced by the
carbon atom interacts with the shear components of the stress field.
We calculate the coordinates of the displacement vector u in the coordinates of
the screw dislocation the Burgers vector b of which lies on the Ox3 axis (Fig.
3.17.1). The vector u is collinear with A/A and u = 1.16x10-10 m. Hence:
2
1
u=
e1 +
e3 u
3
3
(3.17.3)
The normal stress in the direction of u is:
σn =
2 2
2 2 µ b x2
2 µ b sin θ
σ 13 = −
=−
3
3 2π x12 + x2 2
3 π r
The coordinates of the carbon atom in position C1 are:
(3.17.4)
x1 = 0,
x2 = − 3a 2 2 , x3 = 0 (Fig.3.17.1).
In order to calculate the interaction energy U, we need to know the intensity of
the force being displaced by u. It is equal to the average normal stress over a small
area s2 multiplied by this same area. As σn varies as 1/x2, its average value over a
small area remains equal to σn. It is therefore difficult to decide which value to
choose for s. If we arbitrarily decide s2 to be equal to the cross-section of the carbon atom, we get:
U≃
4 u 2
µb rC = 2x10−20 J ( 0.13eV )
9
a
(3.17.5)
This is a stronger interaction energy than for edge dislocations.
Note that the interaction energy of the dislocation with the carbon atom situated
in the centre of the cell face (position C2 in Fig. 3.17.1) is zero because in that case
x2 = 0 so that σ13 = 0. This is also found to be the case for the carbon atom situated
106
in position C3 when the calculation of the stress in the direction of the displacement shows that it is zero.
Fig. 3.17.1. Positions of carbon atoms inserted in a BCC cell and direction A of the displacement
due to the insertion.
107
c) Find the carbon concentration needed to saturate all the Cottrell’s atmospheres
in α Fe containing a dislocation density of 1014m-2, assuming that the saturation
means that all the sites C1 closest to the dislocation line are occupied.
Solution
There are four carbon atoms in position C1 per cell along the dislocation line. The
number of cells along the dislocation line per cubic meter is equal to ρ D
(a 3) .
3
The number of cells per cubic metre is equal to 1/a and there are 2 Fe atoms per
cell. For the saturation of the Cottrell atmosphere the atomic concentration of C
atoms is equal to 2 3ρD a 2 = 2.8x10−5
d) What is the shear stress needed for an edge dislocation to escape from a saturated carbon atmosphere in α Fe?
Solution
Assume that the carbon atoms in the Cottrell atmosphere lie at a distance a below
the dislocation line. The evolution of the interaction energy U per unit length
when the dislocation glides away from the atmosphere is
U =−
1 1+ ν
a
µ∆V 2
3π 1 − ν
x1 + a 2
(3.17.6)
where ∆V is the increase of the atomic volume in inserting a carbon atom and x1 is
the distance away from the original position of the dislocation. The stress to pull
2 1 +ν
ax1
away the dislocation is τ =
µ∆V
. It goes through a maximum
2
2
3π 1 −ν
( x + a2 )
1
3 1 + ν µ∆V
, that is 6,100 MPa! It is a very high value; the escape of a
8π 1 − ν ba 2
loop would be easier.
Following Nabarro (Nabarro 1967) in the calculation of such a loop escape, the
loop is represented by two “jogs” of length x1 and a free straight line of length w.
τe =
The various components of the energy UL are the interaction with the atmosphere, the increase of the line tension and the work done. The equation yielding
the extremums as a function of the distance x1 is
108
∂U L
2 Awax1
µb 2
=
+
− τ bw = 0
∂x1 b ( x 2 + a 2 ) 2
π
1
(3.17.7)
1 1 +ν
A=
µ b∆V
3π 1 −ν
⇒
Setting x1 = atanφ, the energy goes through a minimum followed by a maxi-
3 3 A
. (τe corresponds to φ = π/6 and
8 a 2 b2
τ c ≃ 10,000 MPa . For plastic deformation to be possible new unanchored dislocations must be created.
Minimising with respect to w the activation energy, UL*, for the escape of the
loop for values of φ close to π/6 yields:
mum if w is large and τ ≤ τ e =
w* =
*
L
U =
µb
2π τ e − τ
(
4
3π
)
µ b a (1− τ τ )
(3.17.8)
12
2
e
e) Determine the precise trajectory of carbon atoms diffusing towards a screw dislocation in α Fe (c0 = 0.001). Deduce the kinetics of the creation of the Cottrell
atmosphere. Achieve the computation for temperatures of 25, 100 and 200°C,
given the diffusion coefficient of carbon D = 2x10-6 exp (-29.93/RT) m2/s (activation energy in kcal/mole).
Solution
If U is the interaction energy of the carbon atoms, their velocity is v =
D(−gradU)/kT . From (3.17.4):
2
3π
2
=
3π
v=
µb
D
πrC 2u
( e1 sin 2θ − e2 cos 2θ )
r2
kT
µb 2 D
πrC u
( er sin θ − eθ cosθ )
r2
kT
(3.17.9)
(It is reminded that rC is the radius of the carbon atom 0.772x10-10 m and u =
1.16x10-10 m.
It follows that the trajectories are circles (Fig.3.17.2). They bring the carbon atoms to the dislocation atmosphere.
109
Fig. 3.17.2. Trajectories of carbon atoms towards the dislocation.
From (3.17.9) the velocity along the circle of radius r0 is:
v = −r0
( )=
d 2θ
dt
2 µb 2 D
1 D
πrC u
= 4.15x10−29 2
2
3π r
kT
r kT
(3.17.10)
With r = 2r0cosθ, the integration of (3.17.10) yields the time t(r0) for a carbon
atom to travel along the entire circle:
()
t r0 =
3
1 kT r0
kT
= 2x1030 r03
2
µ
b
D
D
πr
u
2
C
12π 2
(3.17.11)
During this time an area S = πr02 has been emptied of carbon atoms.
An area increment dS is swept during time dt:
dS = πdr02 =
2π 0.5x10−30 D
3
kT
23
()
t (r )
dt r0
13
(3.17.12)
0
This brings a number dN of carbon atoms to a unit length of dislocation equal
to:
110
2c 2π 0.5x10−30 D
dN = 30
kT
a 3
23
()
t (r )
dt r0
13
(3.17.13)
0
The number of carbon atoms needed to saturate the unit length of dislocation is:
N=
4 3
a
(3.17.14)
Integration of (3.17.13) yields the time t needed to create a Cottrell atmosphere:
t=
7x10−22 kT
c03 2 D
(3.17.15)
The results are in Table 3.17.1.
Table 3.17.1. Approximate times for the creation of a Cottrell atmosphere
Temperature (K)
300
373
473
Time (s) (hours) (min) (days)
1.4x107 3830
160
2160
36
6
Exercise 3.18 Piobert-Lüders band
When dislocations are anchored by Cottrell atmosphere, for plastic deformation to occur fresh dislocations must be created. This avalanche in a localised zone of concentrated stress results in the formation of bands of localised
strain, Piobert-Lüders bands (Volume I, Sect. 3.4.3.1, p.256, Fig. 3.80)
Find the angle made by a Piobert-Lüders band with the tensile axis in a thin sheet.
Solution
The band, which deforms plastically, is in plane stress and at the same time in
plane strain in its own direction. If axis Ox'1, making an angle φ with the axis Ox1,
1
is along the band and Ox'2 perpendicular to it, Eε11′ = 0 = σ 11′ − σ 22
′ . Figure
2
111
3.18.1 shows, on the Mohr circle, the positions which these stresses must therefore
occupy. So, cos2φ = 1/3.
Fig. 3.18.1. Piobert-Lüders band in a thin sheet and Mohr circle.
112
Exercise 3.19 GP zone hardening
Guinier-Preston (GP) zones are formed during the first stage of precipitation
of solute atoms. They are clusters of solutes which remain on the lattice sites
of the solvent atoms. A classical example is that of aluminium copper alloys.
(Volume I, Sect. 3.4.3.2, p.260)
Figure 3.89 of Volume I (p. 264) shows a GP zone in a 1.7at.% Al-Cu alloy, which
Karlik and Jouffrey (Karlik M and Jouffrey B (1996) Guinier-Preston platelets interaction with dislocations. Study at the atomic level. J Phys III 6:825-829) observed by high resolution transmission electron microscopy. They demonstrated
that it was a 100% copper monolayer disk on {100} planes of the aluminium crystallographic lattice. The figure shows that a dislocation sheared the GP zone.
Evaluate the hardening resulting from the precipitation of the GP zones in the
1.7at% Cu aluminium alloy.
Solution
The dislocations glide in {111} planes with [110] Burgers vector. Two kinds of
interactions with the GP zones can be considered: elastic interaction and chemical
interaction. The first one can be considered as the interaction of a gliding dislocation with the solute atoms disk represented by a dislocation loop of Burgers vector
ηb, η being the misfit atomic parameter equal to 0.11 for the smaller Cu atoms in
Al. A rough approximation consists in applying the equation giving the shear
stress needed for crossing the forest of dislocations:
τc =
1
µηb ρD
4
(3.19.1)
The dislocation density is the total length of dislocations in a unit volume:
ρ D = N disk 2πr
(3.19.2)
Ndisk being the number of disks of radius r in a unit volume.
If the atomic concentration of precipitated copper atoms is c and a the parameter of FCC aluminium:
c
Hence:
4
2πr 2
= N disk 2
3
a
a
(3.19.3)
113
ρD =
4
c
ar
(3.19.4)
Combining (3.19.1) and (3.19.4), with for aluminium µ = 2.6x1010, a = 4x10-10,
b = a/21/2 = 2.8x10-10, η = 0.11 and r = 5x10-9, as measured on Fig. 3.89 of Volume I:
τc =
1 µη b
c = 2.8x108 c
2 ar
(3.19.5)
If c = 0.017, the critical shear stress is 36 MPa, a value of a factor 3 too low
with respect to measurement of the yield strength. Moreover, the concentration of
precipitated carbon is lower than 0.017. Elastic interaction is insufficient to account for the critical shear stress.
If the slip plane of the dislocation is 1 11 , the Burgers vectors are 011 ,
( )
( )
which is parallel to the 100 plane of the GP zone, 10 1 and 110 ,which are
inclined to this plane.
When a 011 dislocation penetrates in a GP zone, it replaces a Cu−Cu bond
by a Al−Al one. The difference of the bond energies is small. The GP zone offers
little resistance to the crossing of such dislocations.
When a 10 1 or a 110 dislocation crosses a GP zone, one Cu−Cu bond
and one Al−Al bond are replaced by two Al−Cu bonds. This involves an energy
∆H = 2hAlCu − hCuCu − hAlAl . This is evaluated by Karlik and Jouffrey to be 1.72 eV
= 2.7x10-19 J.
Per unit area of the glide plane, the disks which cross it are in a volume
4r 2 . The number of disks per unit volume being given by (3.19.3) and as there
are 2/3 effective disks, the number of effective disks crossing the glide plane is
16c 3 2 πar . A gliding dislocation zigzags between the crossing points with a
small "amplitude" x for a "period" y, xy being the area per crossing point:
xy =
3 2πar
16c
(3.19.6)
The average length of a crossed disk is πr/2. So, the average pinning energy is
πr
H=
∆H .
2b
The gain of energy owing to the pinning action of the crossed GP zones is:
114
E1 =
H
πr
=
∆H
y 2by
(3.19.7)
The increase of the line energy is:
12
E2 =
2
2
µb 2 ( x + y ) − y
2
y
=
µb2 x
4 y
2
(3.19.8)
Minimising E2 – E1 yields the zigzag shape:
xc3 =
π (xy )r∆H
(3.19.9)
2µ b3
2
3
c
y =
( )
2 xy µ b3
(3.19.10)
πr∆H
It can be checked that xc/yc = 0.024 is indeed a small quantity.
According to the hypothesis of Friedel (1964), escaping from a pinning point
the dislocation sweeps the area (xy), so that:
τ cb ( xy ) = yc ( E1 − 2 E2 ) =
πr ∆H
4b
(3.19.11)
Using (3.19.6) the critical shear stress is:
τc =
4c∆H
3 2ab2
= 140 MPa
(3.19.12)
Exercise 3.20 Irradiation hardening by cavities
The vacancies produced in stainless steel by irradiation condense to create
cavities. The result is swelling as well as hardening.
Estimate the yield strength increase by a volume fraction fv of cavities of radius r.
Solution
115
The hardening when a cavity is crossed by a dislocation comes from the elimination of a segment of this dislocation and hence of the corresponding energy. In
reality the problem is more complicated as the stress concentration around the cavity plays also a role. We look only for the hardening coming from the annihilation.
Take the origin of the coordinates at the centre of the cavity (Fig.3.20.1), y being in the direction of the displacement of the dislocation. The coordinates of a
point on the surface of the cavity are rcosθ and rsinθ. Let L be the mean distance
between cavities centres. With t the line tension, the radius of curvature of the dislocation is t/τb = (µb2/2)/τb, µ being the shear modulus, b the Burgers vector and τ
the shear stress. Thus the angle ϕ of the dislocation with the x axis where it meets
the cavity is such that:
(
sin ϕ = L − 2r cos θ
Fig.3.20.1. Dislocation crossing a cavity
)µτb
(3.20.1)
116
The work done by the force on the dislocation where it meets the surface of the
cavity when it moves forward a distance d(rsinθ) is equal to the variation of the
line energy (µb2/2)d(rcosθ). Hence:
t cos ϕ sin θ − t sin ϕ cos θ = −t sin θ
(3.20.2)
This equation shows that the angle ϕ is equal to 2θ. The dislocation arc between
the cavities becomes unstable (the Frank-Read condition is met) for θ = π/4.
From (3.20.1), the evolution of the angle θ as a function of the shear stress is
such that:
τ µb =
sin 2θ
L − 2r cos θ
(3.20.3)
This shows that the dislocation finds no resistance until it crosses the cavity by
the middle and that the shear stress must be increased until θ = π/4. At that point:
τ c µb =
1
(3.20.4)
L−r 2
It remains to evaluate L. According to the Friedel hypothesis the area swept by
a dislocation until it meets another obstacle is equal to the average area per obstacle on the glide plane s. This is related to the volume fraction fv:
f v = fs = πr 2 s
(3.20.5)
The Friedel hypothesis yields:
s=
2
πr 2 πRc π µ b
=
=
fv
2
2 2τ c
2
(3.20.6)
Rearranging, it is found that:
fv b
τc
=
µ 2 2r
(3.20.7)
Note that the volume fraction has a limit, 0.9, when the cavities are close
packed.
117
The cavities are not all crossed by the equatorial plane, so that the radius should
be replaced by the average value 2r/π.
Assuming then for instance r/b = 10 and fv = 0.4, the critical shear stress is
equal to 0.035µ.
3.4 Composite materials
Exercise 3.21 Composite wire
A 2 mm diameter steel wire, coated with aluminium to give an external diameter
of 2.5 mm, is subjected to traction. Which metal yields first? What is the yield
strength of the composite wire? What is its modulus of elasticity? (The Young
moduli E are 210,000 MPa and 70,000 MPa and the yield strength 200 MPa and
70 MPa respectively for the steel and for aluminium; they both have the same
Poisson ratio).
Solution
Considering that the tensile strain ε of the two metals is the same, the stress in the
steel part is equal to 210,000 ε MPa and to 70,000 ε MPa in the aluminium part.
Thus the yield strength is reached for ε = 1/1,000 for the aluminium part and for ε
= 1/1,050 for the steel one, which yields first.
The cross-section of the steel part is π mm2 and that of the aluminium one is
0.562π mm2. The average stress in the composite wire for a strain ε is:
Ecompε =
1
0.562
200,000ε +
70,000ε = 153, 200ε
2
2.5 4
2.52 4
(3.21.1)
Ecomp being the Young modulus of the composite wire.
When the yield strength is reached in the steel part, the stress in the composite
is 153,200/1050 = 146 MPa. To reach the yield strength in the aluminium part, a
further strain is needed: 1/1,000 – 1/1,050. It corresponds to a stress increase equal
to:
1 0.562
1
∆σ =
−
2 70,000 = 1.5 MPa
1,000 1,050 2.5 4
(3.21.2)
Assuming that there is no strain-hardening in the steel, the yield strength of the
composite wire is 147.5 MPa.
118
Exercise 3.22 Fibre reinforced composites
The fracture strength of a fibre reinforced composite is the larger of its
strength at fracture of the reinforcing fibres or of the remaining strength of
the matrix when the fibres are broken. If the volume fraction of reinforcing
fibres is too low, the fracture strength of the composite is lower than the one
of the matrix with no reinforcement (Volume I, Sect. 3.4.6.2, p.299).
The shear lag model (Volume I, Sect. 3.4.6.3, p.300) shows that the tensile
stress along the length of the fibres is maximum in their middle. The load
transfer is the higher when the matrix is fully plastified around the fibres.
The tensile stress in the fibres then increases linearly up to the maximum at
the middle, where its ratio to the shear yield strength of the matrix is equal
to the ratio of the fibres length to their radius.
a) A composite material consists of 40% by volume of high strength steel wires (Rp
= 1,050 MPa, A% = 0.15) aligned in an aluminium matrix for which the strain
hardening curve is σ = 270 ε0.2 MPa. Find the fracture strength of the composite.
Solution
Neglecting the strain-hardening of the high resistance steel before fracture, the
tensile stress σ in the composite is:
σ = 0.4 × 1,050 + 0.6 × 270ε 0.2 MPa
(3.22.1)
When the steel wires fracture the tensile strain is 0.15 and the corresponding
stress, according to Eq. 3.22.1, is 531 MPa.
Increasing further the strain, the stress comes from the aluminium matrix only:
σ = 0.6 × 270ε 0.2 MPa
(3.22.2)
The Considere condition gives the strain at maximum load: 0.2. According to
Eq. 3.22.2 the corresponding stress is 117 MPa.
The fracture strength of the composite corresponds to the breaking of the steel
wires: 531 MPa.
b) A boron fibres-aluminium composite has the following characteristics: unidirectional reinforcement; fibres: volume fraction 50%, length 100 mm, diameter
0.1 mm, breaking strength 3,000 MPa, Young modulus 415,000 MPa; matrix:
yield strength 70 MPa, flow strength upon fibres breaking 70 MPa, breaking
strength 200 MPa. Calculate the critical length for load transfer and the critical
volume fraction.
119
Solution
According to the shear lag model the critical length lcf of the reinforcing fibres of
radius rf is such that the maximum tensile stress in the middle of their length is
equal to their fracture resistance Rmf:
Rmf = k m
lcf
rf
= R mp
lcf
2r f
(3.22.3)
where km and Rpm are respectively the yield strengths of the matrix in shear and in
tensile loading.
With the given values lcf = 0.1x3,000/70 = 4.3 mm.
The critical volume fraction fvc is such that:
σ R = f vc 3,000 + (1 − f vc ) 70 = 200
(3.22.4)
Hence fvc = 4.4 %.
c) How does the critical length for load transfer vary with temperature?
Solution
The yield strength of the matrix decreases when the temperature is increased.
Hence, Eq.3.22.3 shows that the critical length increases.
3.5 Continuum crystalline plasticity
Continuum crystalline plasticity combines the macroscopic approach of continuum plasticity with the crystalline structure of the considered materials.
The plastic deformation is considered as resulting from glide occurring on
easy glide systems when their resolved shear stress reaches a critical value.
This critical shear stress evolves with strain-hardening which generally depends on the plastic activity of all the slip systems.
Exercise 3.23 Latent hardening tests
120
Some aspects of the strain-hardening properties of single crystals can be investigated through “latent hardening tests” (see Sect. 3.5.1.2.d2 in Volume I): a primary specimen is pre-deformed by single slip (l) and from this are cut several secondary samples at different orientations. Each of these is then deformed again
according to some system (g) for which the new critical resolved shear stress
(CRSS) will depend on the short-range interactions between (l) and (g).
The studied FCC crystal is supposed to obey the Schmid law with equal initial
CRSSs τ 0 and the strain-hardening law
τ cg = τ 0 + ∑ H gl γ l
(γ l > 0)
(3.23.1)
l
where τ cg is the CRSS on system (g), γ l the plastic glide on the active system (l)
and H gl a strain-hardening matrix to be determined experimentally. The initial
sample is a plate normal to 1 11 .
1) A tensile test is performed on this plate in the 123 direction. Find the resolved shear stress on the 12 easy glide systems {111} 110 , which are denoted
according to the classical notation
A : ( 111) B : (111) C : ( 1 11) D : (1 11)
1: [ 011] 2 : 0 11 3 : [101] 4 : 101 5 : 110 6 : [110]
Which is the primary system? Write the yield stress σ 0 as a function of τ 0 .
2) This tensile test is performed up to the axial plastic strain ε1p and the axial
stress σ 1 . Write, by use of (3.23.1), the self-hardening modulus on the primary
system as a function of σ 0 ,σ 1 and ε1p .
3) A secondary sample is then cut from the initial plate in the direction 3 12 , at
an angle of 60° with the 123 direction, and then submitted to a tensile test. The
measured yield stress is σ 2 . Calculate the resolved shear on the easy glide systems. Which would be the active system in case of an isotropic strain-hardening
matrix (i.e., H gl = H , ∀g , l )? What would be the H value? Actually, it is observed
that, as long as ε1p < ε cp , the active system is the primary one but that, for ε1p > ε cp ,
the active system is the secondary one, namely C5. What are the consequences of
this observation on the components of the H gl matrix which can be determined or
bounded?
Solution
1) Let k be the unit vector along the tensile axis. The stress tensor then reads
121
(3.23.2)
σ =σk ⊗k
where a ⊗ b denotes the tensorial product between the vectors a and b. The resolved shear stress τ on any slip system (n, m) is given by
τ = m.σ .n = σ (k .m)(k .n)
(3.23.3)
With k along 123 , we get in the cubic axes
k=
1
σ
(−1, 2,3) ⇒ τ = (−m1 + 2m2 + 3m3 )(−n1 + 2n2 + 3n3 )
14
14
(3.23.4)
It follows that, on the 12 easy glide systems {111} 110 , the resolved shear
stress is
14
14
14
14
6τ A 2 = 6σ
14 6τ A3 = 12σ 14 6τ A6 = 6σ
6τ B2 = 4σ
14 6τ B4 = 16σ
14 6τ B5 = 12σ
6τ C1 = 10σ 14 6τ C3 = 4σ
14 6τ C5 = 6σ
6τ D1 = 0
14 6τ D6 = 0
14 6τ D4 = 0
(3.23.5)
Thus, the primary system is B4, with
τ B4 =
8
7 6
σ
⇒ σ0 =
7 6
τ0
8
(3.23.6)
2) As long as the primary system B4 is the only active one and neglecting the
change of orientation of the cubic lattice with respect to the tensile axis, we have
on the system B4
τ=
8
σ
7 6
σ : ε p = τγ = σε p ⇒ γ =
7 6 p
ε
8
(3.23.6)
and then, according to (3.23.1)
τ=
8
7 6
σ = τ 0 + H B4B4γ =
8
7 6
σ 0 + H B4B4
7 6 p
ε
8
(3.23.7)
122
The self-hardening modulus H B4B4 can then be deduced from the primary tensile test by the relation
H B4B4 =
32 σ 1 − σ 0
147 ε1p
(3.23.8)
1
3) We still use (3.23.3) but now with k along 3 12 , that is k =
(−3, −1, 2) .
14
We find
τ=
σ
14
(−3m1 − m2 + 2m3 )(−3n1 − n2 + 2n3 )
(3.23.9)
The resulting resolved shear stresses on the 12 systems {111} 110 is then now
14
14
14
14
6τ A 2 = 12σ
14 6τ A3 = −4σ
14 6τ A 6 = −16σ
6τ B2 = −6σ 14 6τ B4 = −10σ 14 6τ B5 = −4σ
6τ C1 = 6σ
14 6τ C3 = −6σ
14 6τ C5 = 12σ
6τ D1 = 0
14 6τ D4 = 0
14 6τ D6 = 0
(3.23.10)
In case of an isotropic strain-hardening matrix, the initial CRSS for this secondary test would be the same on the 12 systems and given by
τ cg = τ 0 + H
7 6 p
ε1
8
∀g
(3.23.11)
and the active system would be the primary one, namely A6, (say ( 111) 1 10 to
take the sign into account) with
σ2 =
7 6 A6 7 6
7 6 p
τc =
ε1
τ 0 + H
8
8
8
(3.23.12)
Thus, H would be equal to
8
σ −τ
8 7 6 2 0 32 σ 2 − σ 0
H=
=
p
ε1p
7 6
147 ε1
(3.23.13)
123
Since H would also be equal to HB4B4, given by (3.23.8), this proves that σ2
would be equal to σ1, which is consistent with the assumption of isotropic strainhardening.
If σ 2 ≠ σ 1 , which is supposed from now on, this means that the strainhardening is not isotropic. Let then Rg be the Schmid factor on system (g) (as deduced from (3.23.10), with a positive sign). Since A6 is observed to be the only
active system in the secondary test as long as ε1p < ε cp , this means that, on any system (g), one has
ε1p < ε cp ⇒ R gσ 2 ≤ τ 0 + H gB4
7 6 p
ε1
8
(3.23.14)
where the equality only holds for (g) = A6. This leads to the value of H A6B4 ,
namely
H A6B4
8
σ −τ
8 7 6 2 0 32 σ 2 − σ 0
=
=
p
ε1p
7 6
147 ε1
( for ε
p
1
< ε cp )
(3.23.15)
and to lower bounds derived from (3.23.14) for H gB4 , g ≠ A6 . For (g) = C5,
which is observed as active when ε1p ≥ ε cp , we have, with R C5 =
ε1p ≥ ε cp ⇒
6
7 6 p
σ 2 = τ 0 + H C5B4
εc
7
8
6
7
(3.23.16)
or
H C5B4 =
8 3σ 2 − 4σ 0
ε cp
147
(3.23.17)
Exercise 3.24 Pencil glide
In most BCC crystals, plastic slip occurs along a <111> direction on planes including this direction ({110}, {112}, {123}... or even any plane in zone with
<111>); this gives rise to what is called “pencil glide”: the plastic deformation
resembles that of a stack of hexagonal shaped pencils (see Volume I, Sect. 3.3.2.1
and Sect 3.3.5.1). We aim here at expressing a plastic criterion according to
124
which, for a given stress state σ , plastic glide occurs in the <111> direction on
the (non necessarily crystalline) plane in zone with it which suffers the maximum
resolved shear stress when it reaches the critical value τ 0 . We then apply this criterion to the uniform stress state which has, in the cubic axes, the matrix
0 0 0
(σ ij ) = 0 p τ
0 τ p 2
p≥0
(3.24.1)
and we compare the corresponding elastic domain in the ( τ , p ) half-plane with
the one deriving with the Tresca criterion with the critical maximum shear stress
τ 0 . Finally, we derive the form of the plastic strain rate tensor εɺ p , up to a multiplicative scalar factor λɺ > 0 .
Solution
We have to calculate max n (T (n).m ) over all the unit vectors n normal to m, for a
given slip direction m, with T (n) = σ .n . Since we have
T (n).m = n.σ .m = m.σ .n = T (m).n
(3.24.2)
max n (T (n).m ) = max n (T (m).n ) = τ (m)
(3.24.3)
it follows that
where τ (m) is the shear stress on the plane normal to m . This shear stress is
given by
2
2
τ (m) 2 = T (m) − (T (m).m ) = TT
i i − ( Tk mk )
= σ ij m jσ ik mk − (σ ij mi m j )
2
2
(3.24.4)
In the cubic axes, for the <111> directions, we have
mi = ±
and then
1
ε
= i
3
3
with ε i = ±1, ∀i
(3.24.5)
125
1
3
τ (m) 2 = σ ijσ ik ε jε k −
2
1
(σ ijε jε j )
9
(3.24.6)
The considered plastic criterion, which is a special case of the Schmid law, implies that
max <111> τ = ±τ 0
⇒ max <111> τ 2 = τ 02
(3.24.7)
that is
2
1
1
max <111> σ ijσ ik ε j ε k − (σ ij ε j ε j ) = τ 02
9
3
(3.24.8)
The maximum has to be searched over the four <111> directions, namely
(A) : [111]
(B) : 11 1
(C) : 1 11
(D) : 1 1 1
ε1 = 1 ε 2 = 1
ε3 = 1
ε1 = 1 ε 2 = 1
ε 3 = −1
ε1 = 1 ε 2 = −1 ε 3 = 1
(3.24.9)
ε1 = 1 ε 2 = −1 ε 3 = −1
For the stress state (3.24.1), we have
0
σ ijε j pε 2 + τε 3
1
τε 2 + pε 3
2
2
1
2
σ ij ε jσ ik ε k = ( pε 2 + τε 3 ) + τε 2 + pε 3
2
5
⇒
(3.24.10)
= p 2 + 2τ 2 + 3ε 2ε 3 pτ
4
3
σ ij ε i ε j = p + 2ε 2ε 3τ
2
and then the criterion (3.24.8) reads
2
1 5 2
1 3
2
max <111> p + 2τ + 3ε 2ε 3 pτ − p + 2ε 2ε 3τ = τ 02
9 2
3 4
(3.24.11)
1
2
1
max <111> p 2 + ε 2ε 3 pτ + τ 2 = τ 02
3
9
6
(3.24.12)
or
126
For ε 2ε 3 > 0 ( ε 2ε 3 < 0 , resp.) i.e. for (A) and (D) (for (B) and (C), resp.), the
maximum is reached for τ ≥ 0 ( τ ≤ 0 , resp.) and reads
2
2
2
(A) and (D), τ ≥ 0 ⇒ 3 p + 6 pτ + 4τ = 18τ 0
2
2
2
(B) and (C), τ ≤ 0 ⇒ 3 p − 6 pτ + 4τ = 18τ 0
(3.24.13)
In the ( τ , p ) half-plane ( p ≥ 0 ), the yield locus consists of two portions of ellipses centred at the origin and symmetric with respect to the p half-axis, according to Fig. 3.24.1
Fig. 3.24.1 Yield locus for pencil glide in the (τ, p) half plane
Since the Tresca criterion relies on a critical value τ 0 of the maximum shear
stress, which is always larger than any resolved shear stress, the Tresca yield locus
would necessarily be interior to the one in Fig. 3.24.1, which relies on the same
critical value τ 0 of some resolved shear stress. In other words, the plastic flow
would occur earlier according to Tresca’s criterion than according to the criterion
studied here.
We know (see Volume I, Sect. 3.5.2.3) that the Schmid law associated with a
deformation mode by slip leads to the principle of maximum plastic work. This
principle leads itself to the normality rule, which can thus be used here. The global
load function f (σ ) reads
f (σ ) = 3 p 2 ± 6 pτ + 4τ 2 − 18τ 02
or, for every system
(3.24.14)
127
f (σ )(ABCD) = 6σ ijσ ik ε j ε k − 2(σ ij ε i ε j ) 2 − 18τ 02
(3.24.15)
The normality rule reads
∂f
εɺ = fij λɺ λɺ > 0 fij =
∂σ ij
S
p
ij
where α
S
(3.24.16)
denotes the symmetric part of α . With (3.24.15), we get for every ac-
tive system
f ij (ABCD) = 6(σ ik ε j ε k + σ jk ε iε k ) − 4σ kl ε k ε l ε iε j
(3.24.17)
which leads in the cubic axes to the matrix
−6 p − 8ε 2ε 3τ
( fij (ABCD) ) = −2ε1ε 3τ
−3ε ε p − 2ε ε τ
1 3
1 2
−2ε1ε 3τ
6 p + 4ε 2ε 3τ
3ε 2ε 3 p + 4τ
−3ε1ε 3 p − 2ε1ε 2τ
3ε 2ε 3 p + 4τ
(3.24.18)
4ε 2ε 3τ
and then to the deformation rates
−3 p − 2τ
−6 p − 8τ −2τ
(εɺ A ) = λɺA −2τ
6 p + 4τ 3 p + 4τ
−3 p − 2τ 3 p + 4τ 4τ
3 p − 2τ
−6 p + 8τ 2τ
p
(εɺ B ) = λɺB 2τ
6 p − 4τ
−3 p + 4τ
3 p − 2τ
−3 p + 4τ −4τ
p
−6 p + 8τ
p
ɺ
(εɺ C ) = λC −2τ
−3 p + 2τ
−6 p − 8τ
p
(εɺ D ) = λɺD 2τ
3 p + 2τ
−2τ
−3 p + 2τ
6 p − 4τ
−3 p + 4τ
−3 p + 4τ −4τ
2τ
3 p + 2τ
6 p + 4τ 3 p + 4τ
3 p + 4τ 4τ
(3.24.19)
p
where it can be checked that the condition σ : εɺ > 0 is satisfied with positive val-
ues for all the parameters λɺ . Assuming that two systems which suffer the same resolved shear stress have the same shear strain rate, we find finally
128
0
0
−12 p − 16τ
p
ɺ
ɺ
ɺ
ɺ
0
12 p + 8τ 6 p + 8τ
τ > 0 ⇒ λA = λD = λ ⇒ (εɺ ) = λ
0
6 p + 8τ
8τ
(3.24.20)
0
0
−12 p +16τ
τ < 0 ⇒ λɺ = λɺ = λɺ′ ⇒ (εɺ p ) =λɺ′
0
12 p − 8τ −6 p +8τ
B
C
0
−6 p + 8τ
−8τ
Exercise 3.25 Card glide
A hexagonal crystal is plastically deformed by single slip occurring on the sole
basal plane, with unit normal c, along the direction corresponding to the maximum resolved shear stress τ on the basal plane, when it reaches the critical value
τ0 (“card glide”).
1) Write the corresponding plastic criterion for a uniform stress state σ. Consider
the special case of a simple shear stress T in the plane (x1, x2), along x2, with the c
axis referred by the polar angles θ and ϕ in the rectangular axes (x1, x2, x3).
2) Find the yield stress Tc(θ) and the slip direction m(θ) when the c axis is such
p
that sin ϕ = 1 / 3, cos ϕ = 2 2 / 3. Find the plastic strain rate tensor εɺ for a given
slip rate γɺ .
Solution
1) The maximum resolved shear stress on a given plane is the shear stress. On the
basal plane, it is given by
τ = σ .c − (c.σ .c) c
(3.25.1)
The plastic criterion then reads
f (σ ) = σ .c − (c.σ .c) c − τ 0 = 0
(3.25.2)
The considered stress state has the matrix
0 0
(σ ) = 0 0
0 T
0
T
0
(3.25.3)
129
and the shear vector τ is
sin θ cos ϕ
c sin θ sin ϕ
cos θ
0
σ .c T cosθ
T sin θ sin ϕ
c.σ .c = 2T sin θ cos θ sin ϕ
(3.25.4)
−2T sin 2 θ cos θ sin ϕ cos ϕ
⇒ τ T cosθ − 2T sin 2 θ cos θ sin 2 ϕ
2
T sin θ sin ϕ (1 − 2 cos θ )
The plastic criterion then reads
4sin 4 θ cos 2 θ sin 2 ϕ cos 2 ϕ + cos 2 θ (1 − 2sin 2 θ sin 2 ϕ ) 2 ...
2
τ =T
= τ 0 (3.25.5)
2
2
2
2
... + sin θ sin ϕ (1 − 2cos θ )
2
2
2) With sin ϕ = 1 / 3, cos ϕ = 2 2 / 3, we get
2
32 4
2
2 2 1 2
2
2
τ = T sin θ cos θ + cos θ 1 − sin θ + sin θ (1 − 2cos 2θ )
9
9
81
2
2
(3.25.6)
2
T2
=
1 + 2cos 2 θ ) = τ 02
(
9
and then
Tc (θ ) =
3τ 0
1 + 2cos 2 θ
(3.25.7)
So, Tc (θ ) increases from τ 0 to 3τ 0 when θ goes from 0 to π/2. The slip direction m can be calculated as
130
2
−4 2 sin θ cos θ
3 1 + 2cos 2 θ
cos θ ( 9 − 2sin 2 θ )
τ
3τ
m= =
=
τ T (1 + 2cos 2 θ ) 3 (1 + 2cos 2 θ )
2
sin θ (1 − 2cos θ )
2
3 (1 + 2cos θ )
(3.25.8)
For a single slip on the system (c, m), we have
S
εɺ p = γɺ ( m ⊗ c )
(3.25.9)
Using (3.25.4) and (3.25.9), we find
(ε )
p
32sin 3 θ cos θ 2 2 sin θ cos θ ( 9 − 4sin 2 θ ) 2 2 sin 2 θ (1 − 4sin 2 θ )
−
9 (1 + 2cos 2 θ )
3 (1 + 2cos 2 θ )
9 (1 + 2cos 2 θ )
2sin θ cos θ ( 9 − 2sin 2 θ )
1 + 2cos 2 θ
= Sym
2
3
9 (1 + 2cos θ )
2
2sin θ cos θ (1 − 2cos θ )
Sym
Sym
1 + 2cos 2 θ
(3.25.10)
Note that this result could have been obtained as well by use of the normality
rule and of the plastic potential f (σ ) given by (3.25.2) through the relation
∂f
εɺ = γɺ
∂σ
p
like in Exercise 3.24.
S
(3.25.11)
131
3.6 Macroscopic Formulation of Plastic Behaviour
Such a formulation is based on a plastic criterion in the initial state and its
evolution with the plastic flow (the yield function, including a law of workhardening) and on a law ruling the plastic flow. The initial plastic criterion
may be isotropic, such as the Tresca and von Mises criteria, but it becomes
generally anisotropic with an increasing plastic deformation, such as it occurs during the classical operations of metal forming (cold rolling, deepdrawing…) due to the formation of a crystallographic and morphologic texture in polycrystals. The work-hardening law may itself be isotropic or not,
such as the kinematic hardening or any combination of isotropic and kinematic hardening. The description of the plastic flow is made easier when the
material (then named a “standard material”) obeys Hill’s principle of maximum plastic work, which is the case for polycrystals whose grains deform
only by plastic slip according to the Schmid law: thanks to the resulting
normality and convexity rules, the yield function determines the plastic flow
almost completely, up to a scalar function which can be identified from a
simple test. Particular models, such as the Prandtl-Reuss and the Prager
models, derived from these properties, are frequently used for a first approximation of the actual plastic behaviour under investigation.
Exercise 3.26 Non-isochoric plastic flow
We want to study the elastoplastic behaviour of a metallic material which has
been processed according to a powder metallurgy route (sintering, hot isostatic
pressing…), responsible for some residual porosity and then for a non-isochoric
plastic flow. This porosity can be considered as distributed isotropically and the
original powder as isotropic. Show why a plastic criterion described by the load
function
f (σ ) = A(Trσ ) 2 + BTr( s : s ) − C
( = Aσ
2
kk
+ Bsij sij − C )
A, B, C positive
(3.26.1)
where s is the stress deviator could be an adequate criterion for such a material.
Derive the yield stress σ 0 for a tensile test and τ 0 for a simple shear test. What is
the yield surface in the space of the principal stresses
(σ 1 , σ 2 , σ 3 )
when
A = B / 3 ?. What are the conditions for the plastic flow to occur? Assuming the
132
principle of the maximum plastic work to be valid, express the components of an
increment dε ijp of plastic strain for a tensile test. Check that there is a plastic volume variation. To characterise the volume change during the plastic flow, derive
the corresponding “plastic Poisson ratio” and check that it is different from ½.
Solution
The proposed criterion only uses symmetric invariants of the stress tensor σ . So,
it is isotropic, as expected for an isotropic material. It depends on the stress deviator but also on Trσ and so on the hydrostatic pressure, as expected for a porous
material. So, (3.26.1) can be an adequate yield function for the considered material.
For a tensile test, we have
σ 0
(σ ) = 0
0
2
3σ0
0 0
0 0 ⇒ (s) = 0
0 0
0
0
−
σ0
3
0
0
0
σ
− 0
3
(3.26.2)
and then (3.26.1) yields
2B 2
3C
A+
σ 0 = C ⇒ σ 0 =
3
3A + 2B
(3.26.3)
For a simple shear test, we have
τ 0
(σ ) = 0
0
0
−τ 0
0
0
C
0 ⇒ 2 Bτ 02 = C ⇒ τ 0 =
2B
0
(3.26.4)
1
With sij = σ ij − σ kk δij , we have
3
1
1
sij sij = σ ij − σ kk δij σ ij − σ kk δij
3
3
2 2 1 2
1
= σ ijσ ij − σ kk + σ kk = σ ijσ ij − σ kk2
3
3
3
(3.26.5)
133
and then
f (σ ) = Aσ kk2 + Bsij sij − C
B
= A − σ kk2 + Bσ ijσ ij − C
3
B
2
= A − (σ 1 + σ 2 + σ 3 ) + B (σ 12 + σ 22 + σ 32 ) − C
3
(3.26.6)
When A = B / 3 , the equation of the yield surface in the space of the principal
stresses reduces to
σ 12 + σ 22 + σ 32 =
C
B
(3.26.7)
which is the equation of the sphere of radius C / B centred at the origin.
The conditions for the plastic flow to occur read
B 2
f (σ ) = A − 3 σ kk + Bσ ijσ ij − C = 0
∂f : dσ = 2 Bσ + 2 A − B σ δ dσ > 0
ij
kk ij ij
∂σ
3
(3.26.8)
B 2
A − 3 σ kk + Bσ ijσ ij = C
2 Bσ dσ + 2 A − B σ dσ > 0
ij
ij
kk ll
3
(3.26.9)
or
If the material obeys the principle of maximum plastic work, the normality rule
implies a plastic strain increment of the form
dε ijp =
∂f
dλ , dλ > 0
∂σ ij
(3.26.10)
that is
B
dε ijp = 2 Bσ ij + A − σ kk δij dλ , dλ > 0
3
(3.26.11)
134
For a tensile test, this gives
( dε )
p
2B
A+ 3
= 2σ 0 dλ 0
0
0
A−
0
B
A−
3
0
B
3
0
(3.26.12)
The associated relative volume variation is given by
( ) = dε
Tr dε
p
p
kk
= 6 Aσ 0 dλ > 0
(3.26.13)
since A > 0,dλ > 0 .
The corresponding “plastic Poisson ratio” is
νp =−
p
dε 22
α −3
B − 3A
B
=
=
, α = >0
p
dε11 2 B + 3 A 2α + 3
A
(3.26.14)
It increases from −1 to 1/ 2 when α goes from 0 to + ∞ and vanishes for α = 3
(i.e., for A = B / 3 ).
Exercise 3.27 Cold rolling induced plastic anisotropy
1) A metallic thin sheet has been cold-rolled in such a way that it has acquired an
induced plastic anisotropy prior to further forming steps such as stamping or deep
drawing. The corresponding yield function is expressed as
f (σ , K ) = σ : B : σ − K 2
(3.27.1)
where B is a positive fourth-order tensor (i.e., such that σ : B : σ ≥ 0, ∀σ ) and K
is a hardening parameter. What are the minimal symmetries of B? Show that the
Voigt notation can be used for it as well as it can be for elastic compliances S. In
all what follows, we assume that the studied plastic anisotropy obeys a transverse
isotropy around the axis x3 normal to the sheet. Using the results of Exercise 2.10
concerning the stability of elastic equilibrium in transverse isotropy, find the form
of the BIJ matrix and the domain of the possible values for its components.
135
Solution
Since we have
σ ij = σ ji and σ ij Bijklσ kl = σ kl Bklijσ ij
(3.27.2)
Bijkl = B jikl = Bijlk = Bklij
(3.27.3)
it follows that
So, we can use the Voigt notation as we do for elastic compliances S. The
transverse isotropy obeyed by B leads in addition, as known for S in this case, to a
matrix BIJ of the form
B11
( BIJ ) =
B12
B11
Sym.
B13
B13
B33
0
0
0
B44
0
0
0
0
B44
0
0
0
0
0
2( B11 − B12 )
(3.27.4)
Furthermore, the positivity of B leads, as shown for S in Exercise 2.10, to the
properties
B33 ≥ 0 B66 ≥ 0 B44 ≥ 0 B11 ≥
B66 B132
+
4 B33
(3.27.5)
2) The yield function is now supposed not to depend on the hydrostatic pressure.
Show that this implies that the yield function can be written as
2f (σ , K ) = F (σ 22 − σ 33 ) 2 + G (σ 33 − σ 11 ) 2 + H (σ 11 − σ 22 )2 + ...
2
... + 2 Lσ 23
+ 2 M σ 312 + 2 Nσ 122 − 2 K 2
(3.27.6)
where G, M and N can be expressed as functions of F, H and L. Show what are
these relations and, assuming that H > F, what is the variation domain of F, H
and L.
Solution
Since the yield function is independent of the hydrostatic pressure p, we must have
136
(σ + pδ) : B : (σ + pδ) = σ + : B : σ , ∀σ , ∀p
(3.27.7)
2 pB jjklσ kl + p 2 B jjll = 0, ∀σ , ∀p
(3.27.8)
B jjkl = 0
(3.27.9)
B11 + B12 + B13 = 0
2 B13 + B33 = 0
(3.27.10)
that is
and then
These 9 equations reduce to
which implies also, since B66 = 2( B11 − B12 ) , the relation
B11 =
B66 B132
+
4 B33
(3.27.11)
Referring to (3.27.5), this is associated to the fact that B is still positive, but
not definite positive since we can have σ : B : σ = 0 with σ ≠ 0 , as soon as
σ = pδ :
( p δ) : B : ( p δ) = 0, ∀p
(3.27.12)
So, the BIJ matrix only depends on three independent parameters such as B12,
B33 and B44, since we have
B11 = − B12 +
B33
B
, B13 = − 33 , B66 = B33 − 4 B12
2
2
Now, the yield function can be written as
(3.27.13)
137
f (σ , K ) = (σ 11 σ 22 σ 33 σ 23 σ 31 σ 12 ) ...
B11
...
B12
B11
B13
B13
0
0
0
0
B33
0
0
B44
0
Sym.
0 σ 11
0 σ 22
0 σ 33
2
−K
0 σ 23
0 σ 31
B66 σ 12
B44
(3.27.14)
with the relations (3.27.13). Keeping B12, B33 and B44 as independent parameters, it
reads
B
2
f (σ , K ) = 33 − B12 (σ 112 + σ 22
) + 2 B12σ 11σ 22 − B33σ11σ 33 − B33σ 22σ 33...
2
... + B33σ 332 + B44 (σ 232 + σ 312 ) + ( B33 − 4 B12 )σ 122 − K 2
B
B
2
= − B12 (σ 11 − σ 22 ) + 33 (σ 22 − σ 33 ) 2 + 33 (σ 33 − σ 11 )2 + B44σ 23
...
2
2
... + B44σ 312 + ( B33 − 4 B12 )σ 122 − K 2
(3.27.15)
2
This expression conforms with (3.27.6) if we put
F = B33
G = B33
⇒ G=F
H = −2 B12
M =L
L = B44
N = F + 2H
M = B44
with
N = B33 − 4 B12
H > F > 0, L > 0
(3.27.16)
So, in what follows, we use the expression
2f (σ , K ) = F (σ 22 − σ 33 ) 2 + F (σ 33 − σ 11 ) 2 + H (σ 11 − σ 22 )2 + ...
2
... + 2 Lσ 23
+ 2 Lσ 312 + 2( F + 2 H )σ 122 − 2 K 2
(3.27.17)
3) Check that the tensile yield stress in any direction of the plane (x1, x2) making
the angle α with x1 does not depend on α. We consider now plane stress states in
the plane (x1, x2) of the form
σ1 0
(σ ) = 0 σ 2
0 0
0
0
0
(3.27.18)
Draw the corresponding yield surface in a plane (σ 1 , σ 2 ) after having shown
that such a description makes sense. How does this surface evolve when the
138
strain-hardening is isotropic? How do the parameters H and F evolve in this
case?
Solution
The stress matrix in the axes (x1, x2, x3) reads
cos 2 α
(σ ) = σ 0 sin α cos α
0
sin α cos α
2
sin α
0
0
0
0
(3.27.19)
The tensile yield stress σ 0 is reached when f (σ , K ) vanishes, namely
F (sin 4 α + cos 4 α ) + H ( cos 2 α − sin 2 α )2 ...
2
σ
= 2K
2
2
... + 2( F + 2 H )sin α cos α
2
0
(3.27.20)
or
σ 02 ( F + 2 H ) = 2 K 2 ⇒ σ 0 =
K 2
, ∀α
F+H
(3.27.21)
which, as expected from the property of transverse isotropy, is independent of α.
For the considered plane stress state, the plastic domain is reached when
2
F (σ 12 + σ 22 ) + H (σ 1 − σ 2 ) = 2 K 2
⇒ ( F + H ) (σ 12 + σ 22 ) − 2 H σ 1σ 2 = 2 K 2
(3.27.22)
Due to the transverse isotropy, this criterion only depends on the principal
stresses σ1 and σ2 and the elastic domain can be drawn in the plane (σ 1 , σ 2 ) . It is
limited by an ellipse centred at the origin and with its axes making the angle π/4
with the axes x1 and x2. For isotropic strain-hardening, this ellipse is dilated by a
homothetic transformation centred at the origin during the plastic flow, which implies that the ratio H/F remains constant.
4) The studied material is now supposed to be standard (i.e., to obey Hill’s principle of maximum plastic work). Derive from this property the value of the Lankford
coefficient defined as the ratio r = εɺ2p / εɺ3p of the lateral and normal plastic strain
rates during a tensile test along the x1 axis. Derive also the response to a balanced
139
biaxial tension ( σ 1 = σ 2 > 0 ). We then define an equivalent plane plastic strain
rate by
1/2
p
p
p
εɺequ
= ( εɺαβ
εɺβα
)
α , β = 1, 2
(3.27.23)
p
Compare the values of the parameter εɺ3p εɺequ
in the same hardening state for
simple tension and balanced biaxial tension.
Finally, discuss, from what precedes, the deep-drawing ability of such a thin
sheet by comparing pure drawing ( σ 2 ≃ 0 ) and pure stretching ( σ 1 ≃ σ 2 > 0 ) deformation modes by taking both yield stress levels and risks of tearing into account.
Solution
From (3.27.17) written in the principal axes, the normality rule implies
∂f
εɺ = λɺ
∂σ ij
p
ij
S
(λɺ > 0) with
∂f
= H (σ 1 − σ 2 ) − F (σ 3 − σ 1 )
∂σ 1
∂f
= F (σ 2 − σ 3 ) − H (σ 1 − σ 2 ) (3.27.24)
∂σ 2
∂f
= − F (σ 2 − σ 3 ) + F (σ 3 − σ 1 )
∂σ 3
In simple tension, we then have
εɺ1p = λɺ ( H + F )σ
p
εɺ2 = −λɺ H σ
εɺ p = −λɺ Fσ
3
⇒ r=
εɺ2p H
= >1
εɺ3p F
(3.27.25)
and then
p
ɺ H 2 + ( H + F )2
εɺequ
= λσ
εɺ3p
p
equ
εɺ
=
ST
F
2
H + (H + F )
2
=
1
2
H H
+ 1 +
F
F
(3.27.26)
2
140
In balanced biaxial tension ( σ 1 = σ 2 > 0 ), we get
εɺ1p = λɺ Fσ
p ɺ
εɺ2 = λ Fσ
εɺ p = −2λɺ Fσ
3
(3.27.27)
and then
p
ɺ F 2
εɺequ
= λσ
εɺ3p
p
equ
εɺ
= 2
(3.27.28)
BBT
Since we have H F > 1 , it follows that
εɺ3p
p
εɺequ
<
ST
εɺ3p
p
εɺequ
(3.27.29)
BBT
This proves that the thickness reduction, and then the risk of tearing, is larger
in balanced biaxial tension than in simple tension. At the same time, the yield
stress is larger too ( K 2 / F instead of K 2 / F + H ); the difference increases with H F . Consequently, the deep-drawing ability is better in simple tension and, more generally, in pure drawing than in pure stretching and this is favoured by larger values of H F , the Lankford coefficient, i.e., by a stronger coldrolling induced plastic anisotropy.
3.7 Plastic response of structures
In the general case, the computation of the plastic response of structures
needs the resolution of a difficult boundary value problem due to the
multibranched and incremental nature of the constitutive equations (see Exercise 3.29 below). Thanks to adequate variational approaches, the problems
are mainly solved numerically. Nevertheless, the problem is easier in some
situations, which are mainly considered in the following: when the loading
is proportional, a direct integration is possible, which transforms a “flow
problem” into a “deformation problem”, so that plasticity is no more very
far from nonlinear elasticity (Exercise 3.28); for rigid perfectly plastic
141
standard material, a limit yield analysis can be developed by use of variational principles and of rigorous or approximate powerful specific methods
such as the slip lines technique and the method of rigid blocks, respectively
(Exercise 3.30).
Exercise 3.28 Cylindrical thin tube under proportional loading
A circular cylindrical thin tube is constituted with an elastoplastic standard material obeying the von Mises criterion with isotropic hardening. It is subjected to the
stress field
0
(σ )rθ z = 0
0
p
0 τ
0
τ
p/2
p>0
(3.28.1)
written in cylindrical polar coordinates (r, θ, z), with z along the tube axis.
1) What is the considered loading? Find, in the half-plane (τ, p), the initial yield
surface (let σ 0 be the initial tensile yield stress).
2) Why can the plastic flow law be written as
p
dε = g( J 2 ) s d J 2
(3.28.2)
where s is the stress deviator and J 2 its second symmetric invariant? Find the
function g( J 2 ) by using the fact that, for this material, the tensile curve equation
reads
σ = σ0 + A ε p
(3.28.3)
3) The tube is now subjected to a monotonic proportional loading. Give the corresponding relation between the equivalent stress and strain ( σ and ε ). Find the
plastic strain tensor ε
p
for such a loading path. Consider the special case when
p
τ = p and find ε for the final value τ = p = 2σ 0 .
Solution
1) The cylindrical tube is subjected to the internal pressure p and to the axial couple of torsion C = 2πRtτ where R is the mean radius of the cross-section and t is
142
the (small with respect to R) thickness of this section. The stress deviator in cylindrical polar coordinates is
− p/2
( s ) = 0
0
0
0
p/2 τ
τ 0
(3.28.4)
So we have
J2 =
1
1 p2 p2
p2
+ 2τ 2 ) =
+τ 2
sij sij = ( +
2
2 4
4
4
In simple tension, J 2 =
σ2
3
(3.28.5)
. So, the von Mises criterion reads
σ2
p2
+τ 2 = 0
4
3
or
p 2 + 4τ 2 =
4σ 02
3
(3.28.6)
In the (τ, p) half-plane (p > 0), this is the equation of a semi-ellipse symmetrical with respect to the axes (see Fig. 3.28.1).
Fig. 3.28.1 Initial yield surface in the (τ, p) half-plane
2) The von Mises yield function with isotropic hardening reads
143
f (σ , Y ) = J 2 − K 2
(3.28.7)
with Y the hardening parameters.
For a standard material (i.e., obeying Hill’s maximum plastic work principle),
the normality rule implies
dε ijp = g(σ , Y ) f ij f kl dσ kl
f ij =
∂f
= sij
∂σ ij
(3.28.8)
So, with f kl dσ kl = skl dσ kl = skl ds kl = dJ 2 , we get
dε ijp = g(σ , Y ) sij dJ 2
(3.28.9)
where g(σ , Y ) must be a hydrostatic pressure independent isotropic function of
σ and where the hardening parameters Y reduce to K2, i.e., the value of J2 during
the plastic flow. Thus, we can write
p
dε = g( J 2 , J 3 ) s dJ 2
for J 2 = K 2 and dJ 2 > 0
(3.28.10)
Like for the von Mises criterion, we can then neglect the influence of the third
invariant J3 and we adopt (3.28.2) as the plastic flow law. The flow function
g( J 2 ) can be deduced from a tensile test as indicated in Sect. 3.5.2.4 of Volume I
(Eqs 3.239 and 3.240). The tensile test curve ε p = ϕ (σ ) is given here, from
(3.28.3), by
ϕ (σ ) =
(σ − σ 0 )
2
σ ≥ σ0
(3.28.11)
2(σ − σ 0 )
A2
(3.28.12)
A2
and then we have
g (σ ) = ϕ ′(σ ) =
Since, in a tensile test, σ = 3J 2 , we get
g(J2 ) =
2( 3J 2 − σ 0 )
A2
(3.28.13)
144
3) For a monotonic proportional loading path, we know (Sect. 3.5.3.2 of Volume
I) that the relation between the equivalent stress and plastic strain is the same as
the one between the axial stress and plastic strain in a tensile test, that is
ε p = ϕ (σ ) =
with ε p =
(σ − σ 0 ) 2
A2
if σ > σ 0
(3.28.14)
2 p p
ε ij ε ij . In addition, we have
3
ε ijp =
3sij
2σ
ϕ (σ ) =
3sij (σ − σ 0 ) 2
2 A2σ
if σ > σ 0
(3.28.15)
When τ = p , the loading path is proportional (see Fig. 3.28.1) and the foregoing results may be used. We have
− p/2
( s ) = 0
0
0
p/2
p
0
3 p2
p 15
p ⇒ σ =
+ 3τ 2 =
4
2
0
(3.28.16)
so that we get
2
3 p 15
2
ε =
− σ 0 2
s
2 2
A p 15
p
(3.28.17)
or
(ε )
p
0 0
2 −1/2
3 p 15
= 2
− σ 0 0 1/2 1
A 5 2
0
1 0
(3.28.18)
For p = 2σ 0 , we get
(ε )
p
−1/2 0 0
2(8 15 − 15)σ 02
=
0 1/2 1
5 A2
0
1 0
(3.28.19)
145
Exercise 3.29 Deep-drawing analysis
We consider a situation of deep-drawing of a thin metallic sheet similar to the one
studied in Exercise 3.27 but here attention is focussed on macroscopic structural
aspects whereas the material behaviour is somewhat simplified (no anisotropy, no
strain-hardening). We restrict ourselves to the initial stage of the forming of a circular cylindrical flat-bottom can with the radius a, starting from a circular blank
with the radius R and a thickness t much smaller than R, and to the mechanical
analysis of the flange, that is this part of the blank which is located between the
die and the blank-holder and is not yet in contact with the punch (see Fig. 3.29.1)
Fig. 3.29.1 Scheme of the deep drawing forming of a flat-bottom thin sheet
The flange is supposed to deform by isotropic pure drawing (the French notion
of isotropic “rétreint”), a deformation mode according to which the exterior circular edge as well as any material circle centred on the symmetry axis z remain
circular while its radius r, with a ≤ r ≤ R , is decreasing. Its initial state is supposed to be natural; the constitutive material obeys an isotropic elastic-perfectly
plastic (no strain-hardening) behaviour, the von Mises criterion and the principle
maximum plastic work. The friction action of the blank-holder and the associated
pressure are neglected, so that both faces of the flange and its exterior edge (r =
R) are free of forces. The action of the punch and of the rest of the can on the
flange is schematised as a density of prescribed forces Tg on the section r = a of
the form
g
T = − p er
r =a
( p ≥ 0)
(3.29.1)
where p, which is progressively increasing from 0, can be considered as the loading parameter of the studied structure. The stress state in the flange is plane in
146
both the elastic and the plastic regimes (i.e., the only non-zero stress components
are σ rr ,σ rθ and σ θθ and they depend on r and θ only).
1) Preliminary analysis of the stress and strain fields
a) Write the equilibrium equations and the boundary conditions on r = a and on
r = R . From now on, we suppose that σ rr ,σ rθ and σ θθ do not depend on θ. What
does this assumption imply?
b) What is the relation between σ rr , σ θθ and the tensile yield stress σ 0 inside a
plastic zone? Draw the cross-section of the yield surface by the plane σ zz = 0 in
the space of the principal stresses.
c) In this plane, which part of the yield boundary corresponds to a “retreint”
p
mode of plastic flow implying εɺrrp ≥ 0 and εɺθθ
≤ 0 ? When does the flange thickness
decrease and when does it increase by plastic flow?
Solution
a) The equilibrium equations read (see Volume I, Appendix F.2)
∂σ rr ∂σ rθ σ rr − σ θθ
=0
∂r + r ∂θ +
r
∂σ rθ + ∂σ θθ + 2σ rθ = 0
r ∂θ
r
∂r
(3.29.2)
The boundary conditions are
σ rr ( R) = σ rθ ( R ) = 0
σ rr (a) = p σ rθ (a) = 0
(3.29.3)
If σ rr ,σ rθ and σ θθ do not depend on θ, we get
dσ rr σ rr − σ θθ
=0
dr +
r
dσ rθ + 2σ rθ = 0 ⇒ σ = A
rθ
dr
r
r2
with
A
A
= 2 =0
2
a
R
(3.29.4)
and then σ rθ = 0 . The first equation (3.29.4) reads also
σ θθ =
with σ rr ( R) = 0 σ rr (a) = p .
d(rσ rr )
dr
(3.29.5)
147
b) The von Mises criterion reads
J2 =
1
1
s : s = σ 02
2
3
(3.29.6)
with
2σ rr − σ θθ
2σ − σ rr
σ + σ θθ
, sθθ = θθ
, szz = − rr
srr =
3
3
3
srz = szθ = srθ = 0
(3.29.7)
and then
J2 =
1 2
1
σ rr − σ rrσ θθ + σ θθ2 ) = σ 02
(
3
3
(3.29.8)
In the ( σ rr ,σ θθ ) plane, this is the equation of an ellipse centred at the origin
with the principal axis inclined at 45° with the coordinate axes (Fig. 3.29.2).
p
c) The plastic flow obeys the normality rule εɺ = λɺ s (λɺ ≥ 0) :
(εɺ )
p
p λɺ
εɺrr = ( 2σ rr − σ θθ )
3
p λɺ
= εɺθθ = ( 2σ θθ − σ rr )
3
p λɺ
εɺzz = (σ rr + σ θθ )
3
(3.29.9)
148
Fig. 3.29.2 Cross-section of the yield surface by the plane ( σ rr , σ θθ )
p
The conditions εɺrrp ≥ 0 and εɺθθ
≤ 0 imply 2σ rr ≥ σ θθ and σ rr ≥ 2σ θθ respectively, so that the admissible part of the ellipse is restricted to the arc ACB (see
Fig. 3.29.2) with
2σ
σ 2σ σ
σ
σ
A − 0 , − 0 , C 0 , − 0 , B 0 , 0
3
3
3
3
3 3
(3.29.10)
In addition, we have εɺzzp ≥ 0 (plastic thickness increase) on AC and εɺzzp ≤ 0
(plastic thickness decrease) on CB.
2) Elastic regime
This regime is defined by an elastic response at any point of the flange. Let E be
the Young modulus and ν the Poisson ratio.
a) Show that the non-zero stress components must have the form
σ rr =
k
+l
r2
σ θθ = −
k
+l
r2
(3.29.11)
where the constants k and l have to be derived from the boundary conditions. Give
the expression of the radial displacement ur(r).
149
ur ( R )
and find the maximum value of p,
R
pc say, for which this relation is valid. Find the location of the incipient plastic
flow in the flange.
b) Give the relation between p and δ = −
Solution
a) The strain compatibility implies
dur σ rr −νσ θθ
ε rr = dr =
E
−
σ
νσ
u
rr
ε = r = θθ
θθ r
E
⇒ ε rr =
d ( rε θθ )
dr
(3.29.12)
so that we have
d
( r (σ θθ −νσ rr ) ) = σ rr −νσ θθ
dr
with σ θθ =
d ( rσ rr )
dr
(3.29.13)
which leads to
r2
d 2σ rr
dσ
k
k
+ 3r rr = 0 ⇒ σ rr = 2 + l σ θθ = − 2 + l
2
dr
dr
r
r
(3.29.14)
The boundary conditions are
σ rr ( R) = 0 ⇒
σ (a) = p ⇒
rr
k
+l = 0
pa 2 R 2
R2
⇒ k= 2
k
R − a2
+
l
=
p
a2
l=−
pa 2
R − a2
2
(3.29.15)
So, we find
σ rr =
pa 2 ( R 2 − r 2 )
r 2 (R2 − a2 )
σ θθ =
− pa 2 ( R 2 + r 2 )
r 2 (R2 − a2 )
(3.29.16)
and then
ur = rε θθ =
r (σ θθ −νσ rr )
pa 2
=−
( (1 +ν ) R 2 + (1 −ν )r 2 ) (3.29.17)
E
E ( R 2 − a 2 )r
150
b) From (3.29.17) we find
δ =−
ur ( R )
2a 2
=
p or
R
E(R2 − a2 )
p=
E(R2 − a2 )
δ
2a 2
(3.29.18)
This relation is no more valid as soon as any plastic flow occurs in the flange.
This happens when
2
max r∈[ a , R ] (σ rr2 − σ rrσ θθ + σ θθ
) = σ 02
(3.29.19)
With help of (3.29.16) we find
σ rr2 − σ rrσ θθ + σ θθ2 =
p2a4
(R
2
−a
2 2
)
3R 4
1 + 4
r
(3.29.20)
This expression has its maximum value when r ∈ [ a, R ] at r = a and this
maximum equals σ 02 when p reaches the critical value pc given by
pc =
σ 0 (R2 − a2 )
3R 4 + a 4
2σ 0 a 2
δc =
4
4
E 3R + a
(3.29.21)
3) Elastoplastic regime
In this regime, let us consider that the flange is divided into two zones: an elastoplastic one, defined by a ≤ r ≤ ρ , and a still elastic one, with ρ ≤ r ≤ R . We
have to find the relation between p and ρ or between δ and ρ.
a) elastic zone: the general form of the equations (3.29.11) is still valid but k and l
take values different from (3.29.15). Find these values and the expression of δ as
functions of ρ.
b) plastic zone:
- kinematics: how can the kinematic compatibility be assessed in the plastic
zone and at its junction with the elastic zone?
- statics: write the equations to be satisfied by σ rr and σ θθ in the plastic zone.
Show that σ θθ can be expressed as a function of σ rr and then that σ rr is the solution of a first-order nonlinear differential equation. Solve this equation thanks to
the variable change
σ rr =
2σ 0
π
sin(ϕ + )
6
3
(3.29.22)
151
where ϕ (r , p ) has to be determined. Show that one can choose ϕ (r , p) = 0 when
σ rr = −σ θθ = σ 0 / 3 and that, for the “rétreint” mode, ϕ (r , p ) may be supposed
π
π
and + only. What are the conditions to be satisfied at
3
3
r = a and r = ρ ? Propose a computation method to derive ρ and δ . Show that
to vary between −
the existence of a solution needs the condition p ≤ 2σ 0 / 3 to be satisfied. Is this
condition sufficient to ensure the existence of a solution? Comment on the variations of ρ and δ with p.
Solution
a) The boundary condition at r = R gives
σ rr ( R) =
k
+ l = 0 ⇒ k = −lR 2
R2
(3.29.23)
so that we get
σ rr = l 1 −
R2
r2
σ θθ = l 1 +
R2
r2
(3.29.24)
The plastic criterion is satisfied at r = ρ , namely
σ rr2 ( ρ ) − σ rr ( ρ )σ θθ ( ρ ) + σ θθ2 ( ρ ) = l 2 1 +
3R 4
2
= σ0
ρ4
(3.29.25)
In order to conform with the “rétreint” deformation mode (see Question 1.c and
the conditions 2σ rr ≥ σ θθ and σ rr ≥ 2σ θθ ) we have to choose
3R 4
l = −σ 0 1 + 4
ρ
−1/2
=−
σ0ρ 2
ρ 4 + 3R 4
(3.29.26)
so that we get
σ rr =
σ0ρ 2
R2
2 − 1
ρ 4 + 3R 4 r
Finally we have
σ θθ = −
σ 0ρ 2
R2
2 + 1
4
4
r
ρ + 3R
(3.29.27)
152
δ =−
ur ( R )
σ ( R ) −νσ rr ( R)
2σ 0 ρ 2
2l
= − θθ
=− =
R
E
E E ρ 4 + 3R 4
(3.29.28)
b) The kinematic compatibility is a condition which has to be applied to the total
strain rate field εɺ (r ) and not to the elastic and plastic parts separately. We have in
the plastic zone
ν
e 1 +ν
εɺ = E σɺ − E Tr(σɺ ) δ
εɺ p = λɺ s
λɺ ≥ 0
(3.29.29)
where the time derivatives of ε , σ and λ may be understood as derivatives with
respect to the loading parameter p as well. In addition, one has to prescribe the
continuity of the associated displacement field at r = ρ .
The stress field has to satisfy the equilibrium equations, the plastic criterion,
the boundary condition at r = a and the continuity condition at r = ρ .
d
σ θθ = dr ( rσ rr )
σ rr (a) = p
σ rr2 − σ rrσ θθ + σ θθ2 = σ 02
σ rr ( ρ ) =
σ 0 (R2 − ρ 2 )
(3.29.30)
ρ 4 + 3R 4
The first two equations yield, with the notation σ rr = y and y′ = dy/dr
y 2 + ryy′ + r 2 y′2 = σ 02
(3.29.31)
which is a first-order nonlinear differential equation for y (r ) . In order to solve
this equation, it is more convenient to use a variable change suggested by a parametrisation of the elliptic yield surface of Fig. 3.29.2, namely
y = σ rr =
2σ 0
π
sin ϕ +
6
3
(3.29.32)
It is easy to check that, for the three points A, B and C of Fig. 3.29.2, one has
π
3
ϕ A = − ,ϕC = 0,ϕ B =
π
3
(3.29.33)
153
so that the “rétreint” mode regime is fully described by the interval −
π
π
≤ϕ ≤ .
3
3
With this new variable ϕ (r , p ) , (3.29.32) becomes
π
π
π
π
3
sin 2 (ϕ + ) + r sin(ϕ + )cos(ϕ + )ϕ ′ + r 2 cos 2 (ϕ + )ϕ ′2 =
6
6
6
6
4
(3.29.34)
or equivalently
π
π
π
r cos ϕ + 6 ϕ ′ + cos ϕ r cos ϕ + 6 ϕ ′ + sin ϕ − 6 = 0
(3.29.35)
It is easy to check that forcing the second term of this equation to be zero leads
to the same equation as that obtained by doing the same on the first term, but leads
to values of ϕ larger than π/3. Actually, the “rétreint” mode corresponds to
π
π
π
r cos ϕ + ϕ ′ + cos ϕ = 0 − ≤ ϕ ≤
6
3
3
(3.29.36)
This equation may be rewritten as
sin ϕ
2
2
3−
ϕ ′ + = 0 ⇒ r cos ϕ exp ϕ 3 = cst
cos
r
ϕ
(
)
(3.29.37)
The constant can be determined from the third equation of (3.29.30), i.e.,
(
)
(3.29.38)
2σ 0
π
sin ϕ (a ) + = p
6
3
(3.29.39)
cst = a 2 cos ϕ (a)exp ϕ (a ) 3
where ϕ (a) is given by
σ rr (a ) =
Note that this is possible only if
p≤
2σ 0
3
(3.29.40)
Combining this condition with the yield equation (3.29.21), we must have
154
σ 0 (R2 − a2 )
4
3R + a
4
≤ p≤
2σ 0
3
(3.29.41)
Finally, the last equation (3.29.30) yields
σ rr ( ρ ) =
2σ 0
π σ (R2 − ρ 2 )
sin ϕ ( ρ ) + = 0
6
3
ρ 4 + 3R 4
(3.29.42)
This allows the derivation of ρ since ϕ ( ρ ) is given, from (3.29.37) and
(3.29.38), by
(
)
(
ρ 2 cos ϕ ( ρ )exp ϕ ( ρ ) 3 = a 2 cos ϕ (a)exp ϕ (a) 3
)
(3.29.43)
where ϕ (a) is given by (3.29.39). From ρ ( p) , we get δ ( p ) by (3.29.28).
Note that the conditions (3.29.41) are necessary but not sufficient for the solution to exist. We must have in addition the property ρ ≤ R , which yields a condition on ϕ ( ρ ) , and then on ϕ (a) , and then on p: this condition may be more restrictive than (3.29.41).
The variations of ϕ (a, p) , ϕ ( ρ , p) and δ ( p ) with p are given by
dϕ (a, p)
=
dp
3
π
2σ 0 cos ϕ (a, p) +
6
with ϕ (a, p) ≤
π
dϕ ( a , p )
⇒
≥ 0 (3.29.44)
3
dp
2
4
4 3/2
dρ a (3R + ρ ) exp 3 (ϕ (a) − ϕ ( ρ ) )
=
≥0
dp
6σ 0 ρ R 4 ( R 2 − ρ 2 )
Since
(3.29.45)
∂ϕ ( ρ , p)
∂ϕ ( ρ , p)
≤ 0 , this leads to
≤ 0 . Finally, we find, as expected
∂ρ
∂p
2
dδ 2a exp 3 (ϕ (a) − ϕ ( ρ ) )
=
≥0
dp
E(R2 − ρ 2 )
(3.29.46)
4) Limit analysis
Finally, we investigate briefly the plastic breakdown of the structure. Show that a
complete (i.e., ρ = R ) plastic forming of the flange can only occur when the ratio
R/a does not exceed a critical value; give a numerical approximation of this value
155
and draw an indicative plot of the relation p(δ ) when R/a is either lower or larger than this critical value. What is the maximum value of p in both cases? Comment qualitatively on the deformation mode in both cases. What practical recommendation can be given from what precedes in order to favour the deep-drawing
of a can by diminishing the risk of an early tearing of the flange?
Solution
When ρ → R,ϕ ( ρ ) → −
R2
π
according to (3.29.42) and then (3.29.43) yields
6
π 3
3
2
exp −
= a cos ϕ (a )exp ϕ (a ) 3
2
6
(
)
(3.29.47)
The right member in this equation obeys, since ϕ (a) ≤ π / 3
(
)
a 2 cos ϕ (a) exp ϕ (a) 3 ≤
π 3
a2
exp
2
3
(3.29.48)
So, (3.29.47) needs the condition
R2
π 3 a2
π 3
3
exp −
≤ exp
2
6 2
3
(3.29.49)
to be fulfilled, i.e.,
π 3
R2
1
exp
≤
⇒
2
a
3
2
When this condition is satisfied, ρ = R, δ =
R
≤ 2.96
a
σ0
and p = pmax ≤
(3.29.50)
2σ 0
. When it is
3
E
2σ 0
not, p reaches the value
for ρ max < R . Then we have ϕ (a) = π / 3 and ρ max
3
is given by
156
2
ρ max
cos ϕ ( ρ )exp
with
(
)
3ϕ ( ρ ) =
cos ϕ ( ρ ) =
3π
a2
exp
2
3
R2 3
(3.29.51)
4
3R 4 + ρ max
which would make it possible to derive ρ max (< R ) and then δ max (< σ 0 /E ) .
Both cases are schematically illustrated in Fig. 3.29.3.
Fig. 3.29.3 Schematic illustration of the (p, δ) relation according to the value of R/a
In the first case (R/a < 2.96), the flange is fully converted into the can wall. In
the second case (R/a > 2.96), there remains in the flange an elastic ring which
cannot follow the plastic flow of the internal part of the flange: hence the risk of
ductile tearing at the boundary of the plastic zone. So, in order to favour a deeper
drawing, it can be recommended to limit the initial flange radius R at values
smaller than three times the punch radius.
Exercise 3.30 Method of rigid blocks compared with the slip lines technique
As indicated in Sect 3.5.3.4 of Volume I, the method of rigid blocks is a convenient and simple method to build up a kinematic approach to limit analysis. The
material is still considered as a standard, rigid-perfectly plastic one. The method
consists in envisaging the flow as the blocks sliding relative to one another (it is
an incipient flow): we then have a velocity field in a rigid solid with discontinuities at the interfaces between moving blocks, tangential to the interfaces. The rela-
157
tive movements of the blocks must be compatible one with another; this can be ensured by the hodograph method (Sect. 3.5.3.5c in Volume I): for a plane problem,
we associate with the physical plane a velocity plane (the hodograph plane), in
which m(u, v) is the image of a physical point M (x, y) at which the velocity components are (u, v); if O is the origin of the hodograph plane Om = v( M ) .
The plastic power developed over an interface element dS with the velocity
discontinuity ∆v is k ∆vdS (k is the flow limit in simple shear). At a metal-tool
interface we assume a friction law of the Tresca, or boundary layer, type in which
the frictional shear τi is independent of the normal stress: τ i = mk with 0 ≤ m ≤ 1
( m = 0 corresponds to perfect lubrication and m = 1 to “stick friction”). If vf is
the relative slip velocity at the interface element dS over which friction is acting,
this is tangential to dS and the friction power over the element is mk vf dS .
1) Apply this method to the same problem of a flat punch acting on a semi-infinite
body (plane strain, isotropic criterion, stick friction) treated in Sect 3.5.3.4 of
Volume I from a static point of view. Use for that the 5 blocks scheme indicated in
Fig. 3.30.1 (note that the triangles are right-angled isosceles ones) and then optimise this treatment with respect to the depth h, which is a/2 in Fig. 3.30.1, of the
plastic zone. Compare the result with the one derived from the slip lines method.
Fig. 3.30.1. Punch on a semi-infinite body: method of rigid blocks
Solution
In the hodograph plane (for example, 32 is parallel to AD, the seat of the velocity discontinuity 32 between blocks 3 and 2), we get the diagram of Fig. 3.30.2.
158
Fig. 3.30.2 Hodograph analysis (stick friction)
The plastic power dissipated at the interfaces between the blocks per unit
thickness, assuming no slip under the punch, is:
over AD and A′D′ : 2( a 2 )V0 2 k = 2kV0 a
over AC and A′C′ : 2(a 2 )(V0 / 2 )k = kV0 a
⇒ total : 6kV0 a
over DC and D′C′ : 2kV0 a
over BC and B′C′ : 2(a 2 )(V / 2 )k = kV a
0
0
(3.30.1)
Hence, according to the upper bound theorem
FV0 ≤ 6kV0 a, ∀V0
⇒ F ≤ 6ka
(3.30.2)
Note that this upper bound is larger than the one derived from the slip lines
method (see Sect. 3.5.3.5 in Volume I), namely ka (2 + π) . The estimate can be
improved by optimising with respect to the depth of the plastic zone; thus if we
assume that the same flow scheme applies over a deformed zone of variable depth
h, where in the above we have fixed h = a/2 , we find for the plastic power P(h)
P( h ) = 2kV0 (
a2
+ 2h )
h
(3.30.3)
This upper bound for the limit force is minimum for h = a / 2 and it is equal to
4ka 2 ; it is still larger than the bound obtained by the method of slip lines.
159
2) We treat the same problem for conditions of perfect lubrication (no friction) between the punch and the body. Does the static approach developed in the former
case and referred to in question 1) apply now without change? Use the method of
rigid blocks according to the flow scheme of Fig. 3.30.3 (all the triangles are
right-angled isosceles triangles) and find an upper bound for F. Use then the
method of slip lines by considering two Prandtl fans with opening π/2 at A and A’
and homogeneous fields on both sides. Comment on these results.
Fig. 3.30.3 Punch on a semi-infinite body (no friction): method of rigid blocks
Solution
The static approach developed for the case of stick friction under the punch treat
the interface as a principal surface so that the shear stress vanishes, as it does
when there is no friction; so, the associated lower bounds for F are still valid in
the latter case and thus we have
F ≥ ka (2 + π )
(3.30.4)
The hodograph analysis of Fig. 3.30.3 leads to the diagram of Fig. 3.30.4
Fig. 3.30.4 Hodograph analysis (no friction)
160
So, the plastic power dissipated on the interfaces is
over BA′ and BA : 0 (no friction)
over BC and BC′ : 2(a / 2 2 )(V0 2 )k = kV0 a
over AC and A′C′ : 2( a / 2 2 )(V 2 )k = kV a
0
0
⇒ total : 6kV0 a (3.30.5)
′
′
a
V
k
kV
a
over
DC
and
D
C
:
2(
/
2)2
=
2
0
0
over AD and A′D′ : 2(a / 2 2 )V 2 k = kV a
0
0
over ED and E′D′ : 2(a / 2 2 )V0 2 k = kV0 a
and we find
FV0 ≤ 6kV0 a, ∀V0
⇒ F ≤ 6ka
(3.30.6)
The slip lines under A (and A’) are suggested in Fig. 3.30.5
Fig. 3.30.5 Sketch of slip lines under A and A’ (no friction)
On a line α, we have: p + 2kθ = cst
On Ax, we get: θ = π / 2 σ 1 = 0 σ 2 = −2k
p=k
On A’A, we find: θ = 0 σ 2 = − pAA′ − k
Consequently, we have
pAA′ = k + k π
σ 2 = − k (1 + π) − k = −k (2 + π)
(3.30.7)
and then
FV0 ≤ −σ 2 aV0
⇒ F ≤ ka (2 + π)
(3.30.8)
This upper bound is lower (and then better) than the one found in 3.30.6. In
addition, it is equal to the lower bound (3.30.4). So, it gives the exact value of the
limit punching force.
161
References
Friedel J (1964) Dislocations. Pergamon Press
Jaoul B (2008) Étude de la plasticité et application aux métaux. Mines Paris Tech Paris
Nabarro F R N (1967) Theory of dislocations. Clarendon, Oxford.
162
Chapter 4 Exercises of Chapter I.4: Elastoviscoplasticity
High temperature (T/TM > 0.4) deformation of crystalline materials involves
diffusion flow as well as dislocation motion. Creep is a high temperature
phenomenon in which a constant load results in continuing permanent deformation. Upon application of a fixed load, transient creep (or primary
creep), characterised by a decreasing creep rate, is observed. After a certain
time (strain), the strain rate decreases to a characteristic constant and minimum value called the steady-state creep rate. This stage is followed by a
progressive increase of the creep rate corresponding to tertiary creep leading
to failure. Constitutive equations have been developed to describe these
three stages, in particular steady creep. All relate the creep rate to stress and
diffusivity appropriate to the particular creep mechanism. Additionally, diffusion-controlled creep mechanisms explicitly relate creep rate to the material grain size. The creep rate increases with decreasing grain size for volume-diffusion-controlled (Nabarro-Herring) creep, and depends even more
strongly on grain size when mass transport occurs along grain boundary
(Coble creep). This creep mechanism is prevalent in small grain-sized materials, such as ceramics.
4.1 Mechanisms of creep deformation
Exercise 4.1 Coble creep
The basis of Coble creep is presented in Volume I, p. 402. This model was proposed for creep in polycrystals in which the strain rate is controlled by diffusion,
not through the grains but along the grain boundaries, as it is assumed in the
Herring-Nabarro creep (Volume I, p. 400). The original paper was published by
Coble (1963).
Let us consider a spherical grain of radius R submitted to a vertical uniaxial
stress, σ. The driving force for diffusional creep was shown by Nabarro and Herring to result from a change in equilibrium vacancy concentration with stress.
Their result can be reduced to a simple expression (see Volume I, Eq. 4.73) for
the change in vacancy concentration.
∆C = C0σ Ω / kT
(4.1.1)
163
where Ω ≈ b3 , b being the Burgers vector and C0 is the equilibrium vacancy concentration at temperature T in a stress-free crystal; σ is the component of the local stress normal to the boundary. For a spherical grain the stress direction defines the polar axis. The vacancy concentration at the equator is C0, and the
vacancy concentration at the poles changes by the amount given by Eq.4.1.1. In
this figure the applied stress is vertical. It is assumed that excess vacancies created at the pole diffuse to the equator exchanging atoms, which produces an elongation. Since each half of the sphere acts independently, we need only to consider
one hemisphere.
a) Indicate schematically the path of vacancies and atoms
Solution
Figure 4.1.1 is a sketch showing the vacancies and the atoms fluxes.
A
vacancies
S1
S2
atoms
α
B
Fig. 4.1.1. Sketch illustrating the motion of vacancies and atoms in the Coble creep mechanism
b) The hemisphere is divided into two parts, one part corresponding to the source
of vacancies (pole A), the other one corresponding to the sink (equator B). To
simplify, the concentration of vacancies in each part is assumed to be uniform.
The sphere keeps a constant volume; the sink surface must be equal to the source
surface.
Show that the corresponding parallel is obtained for α = π / 3
Solution
The areas S1 and S2 can be easily calculated:
164
dS1 = 2πR sin α Rdα
(
S1 = 2πR 2 1 − cos α
)
2
S2 = 2πR cos α
(4.1.1)
S1 = S2 ⇒ α = π / 3
(
)
c) The mean gradient of the concentration of vacancies is therefore ∆C / πR / 2 .
The maximum value of this gradient occurs at α = π / 3 and can be written as:
1 dC
∆C
=N
R dθ α = π / 3
πR / 2
(4.1.2)
where N is a proportionality factor equal to 2.15 (See Coble 1963).
Calculate the flux of vacancies per second.
Solution
If δGB is the thickness of the grain boundary; the area, S, crossed by the vacancies
at α = π / 3 is given by:
( )
S = δ GB 2πR sin π / 3 = π 3Rδ GB
(4.1.3)
The number of vacancies crossing this area is given by:
J = DL N
∆C
π 3Rδ GB = 7.45DLδ GB ∆C
πR / 2
(
)
(4.1.4)
d) Calculate the corresponding strain rate
Solution
The volume change associated with the flux of vacancies:
J Ω = πR 2
dR
= 7.45DLδ GB ∆CΩ
dt
(4.1.5)
The mean strain rate is given by:
εɺ =
1 dR J Ω 7.45DLδ GB ∆CΩ
=
=
R dt πR3
πR 3
(4.1.6)
165
The vacancy diffusion coefficient, DL is related to the grain boundary diffusion
coefficient, DGB, by DGB = DL C0 Ω. The strain rate can thus be expressed as (refer
to Eq (4.60a and b) in Volume I):
εɺ =
7.45σ DGBδ GBΩ
πR3kT
(4.1.7)
When R is considered as one half of the grain diameter, one obtains
εɺ = α
DGBδ GBσΩ
d 3 kT
(4.1.8)
where α is a numerical factor.
This expression is similar to the Herring- Nabarro creep law except for the d-1/3
variation instead of d-1/2.
Exercise 4.2 Enhanced role of diffusion creep in ceramics
Diffusion creep with n = 1 (n =1/M in Eq. 4.5 in Volume I) is a common behaviour
in ceramics. Give two reasons why the role of diffusion is enhanced in ceramics
compared to metals (Cannon and Langdon, 1988)
Solution
(i) The grain boundary mobility in ceramics is lower than in metals. It is thus easier to stabilise, and maintain, a very fine grain structure. Whereas most metals
(and especially pure metals) have grain sizes of 50-100 µm or larger, many ceramics have stable grain sizes of the order of 1-10 µm. As indicated in figures 4.33,
4.34, 4.35, and 4.36 (Volume I), diffusion creep with n = 1 extends to higher values of σ/µ when the grain size is reduced (see Eq. 4.76 in Volume I) and at the
same time the creep rates also become faster.
(ii) Diffusion creep becomes important in many ceramics because of the preferential enhancement of diffusion of one of the ionic species along the gain boundaries. The faster lattice diffusion coefficient for the cation usually dominates in
creep diffusion creep regime.
Exercise 4.3 Diffusion creep in alumina
Using the creep deformation map for alumina (grain size =10 µm) given in Fig.
4.46 in Volume I, determine:
166
a) The shear stress (in MPa) that will give rise to a creep rate of 10-10/s at 1400°C
b) The creep mechanism under these conditions.
c) The stress at which power law creep becomes dominant in the temperature
range between 1,400°C and 1,500°C.
Hint: Melting point, TM = 2,320K
Burgers vector, b = 4.76 x 10-10 m
Shear modulus at 300K, µ0 = 1.55x105 MPa
T dµ
Temperature dependence of modulus, M
= −0.35
µ 0 dT
Solution
a)T/TM = 1,673/2,320 = 0.72
1,373
= − 0.321x105 (MPa)
∆µ = − 0.35x1.55x105x
2,320
µ (1,400°C)= 1.23x105 (MPa)
b) Figure 4.46a (Volume I) shows that σ is about 0.10 MPa. This corresponds to
diffusion (Coble) creep.
c) The stress is about 300 MPa
4.2 Mechanical models
Exercise 4.4 Stress relaxation
A bolt made of steel is used to assemble two plates which are maintained at
550°C. Creep tests at this temperature showed that the steady creep law can be
written as εɺ = Bσ 3 , where B is a constant, such as for σ = 27.5 MPa, εɺ =
2.80x10-8/hour
This bolt has been initially loaded to a stress σ0 of 70 MPa. Calculate the
tightening stress remaining after one year of operation.
Solution
One neglects the part of the relaxed stress due to primary creep. This assumption
is not necessarily justified (see Exercise 4.5) because of the low values of the
creep strain involved in creep relaxation. So, B = 1.35 x 10-12 (h x MPa)-1.
167
During stress relaxation the elastic strain, ε e , is converted into a creep strain,
ε vp , such as:
dε vp
dt
=−
dε e
1 dσ
=−
= Bσ n
dt
E dt
(4.4.1)
1
1
−
σ n−1 σ 0n−1
(4.4.2)
After integrating one obtains:
t=
1
BE n − 1
( )
The numerical application leads to σ = 16.3 MPa
Exercise 4.5 Creep under variable stress
In many situations one has to deal with variable stresses (see Exercise 4.4). Creep
tests were performed at 700°C on one heat of 316 stainless steel (see Calvet et al.,
1972). The specimens were loaded continuously at constant stress rate such that σ
= Kt. The strains which were obtained were limited so that the deformation behaviour could be described using only primary creep law.
a) Two assumptions are usually made when the creep stress is variable, as shown
in Fig. 4.10 (Volume I): (i) time hardening, (ii) strain hardening. Indicate the
most appropriate assumption.
Solution
As in primary creep there is an increase in dislocation density, one is tempted
to say that the strain-hardening assumption is better than the time-hardening rule.
b) Tests at constant load showed that the primary creep law could be written as:
εɺ = At − pσ n
(4.5.1)
Show that this law can also be expressed as:
εɺ = Bε − qσ n′
Relate B to A and p; q to p and n' to n and p.
Solution
(4.5.2)
168
Integration of (4.5.1) leads to:
1
1
1 − p 1− p 1− p
⇒t =
ε
Aσ n
1 1− p
ε = Aσ
t
1− p
n
(4.5.3)
The integration of (4.5.2) yields:
1
ε = (q + 1)Bσ n′ t q+1
(4.5.4)
Eliminating t between (4.5.3) and (4.5.4) and using (4.5.2) one obtains:
B= A
1
1− p
p
(1 − p )− 1− p
⇒q=
p
1− p
and
n′ =
n
1− p
(4.5.5)
c) Tests at constant increasing stress, σ = Kt were performed. The prediction of
the strain with time can be made using either the time-hardening (Eq. 4.5.1) or the
strain-hardening (Eq. 4.5.2) rules.
Calculate as a function of time the evolution of the creep strains εTH and εSH
corresponding to the two rules of time- and strain-hardening and the time to reach
a given strain as a function of K.
Solution
As the strain hardening assumption predicts a higher strain rate the relative position of the curves calculated with the two extreme assumptions is the following
(Fig.4.5.1):
Time to reach
a given strain
b)
Creep strain
a)
SH
TH
Time
(a)
K
(b)
Fig. 4.5.1. (a) Schematic evolution of creep strain with time in stress increasing creep tests.
Comparison between strain hardening (SH) and time hardening (TH) rules; (b) Variation of time
to reach a given strain as a function of K parameter
169
More quantitatively, εTH and εSH can be calculated as follows:
Time-hardening: integration of Eq.4.5.1 gives:
εTH =
AK n n− p+1
t
n − p +1
(4.5.6)
Strain-hardening: integration of Eq.4.5.2 gives:
1
ε SH
B q + 1 K n'
( )
=
n' + 1
(
)
q +1 n' +1
1
A
t q +1 =
K nt n− p +1
1− p
p
( n − p + 1) (1 − p )
(4.5.7)
The ratio εTH/εSH is given by:
p
ε SH n − p + 1
=
≥1
ε TH 1 − p
(4.5.8)
because p < 1 and n > 0. Hence, the evolutions of the creep strain as a function of
time and of the time to reach a given strain as a function of K are shown in Fig.
4.5.1.
The results of the tests on 316 stainless steel corresponding to two values of K
(Calvet et al., 1972) are shown below (Fig.4.5.2) where it is observed that the SH
rule gives the best predictions.
Fig. 4.5.2. Creep curves under increasing stress – Steel AISI 316 − θ = 700°C. Comparison between experiments and calculations; (a) K = 0.13 MPa/h; (b) K = 0.3 MPa/h (Adapted from
Calvet et al., 1972)
170
Exercise 4.6 Relationship between creep ductility and time to failure
accounting for tertiary creep
This exercise is largely based on a publication by Phaniraj et al (2003). It deals
not only with Chapter 4 in Volume I, but also Chapter 6 in Volume II.
The shape of the creep curves in many materials is shown in Fig. 4.6.1. In many
cases primary creep strain, εp, is relatively insignificant and is neglected in this
sv
exercise. The material deforms at constant rate, εɺ , in the stationary regime and
then at an increasing rate in the tertiary regime.
Fig. 4.6.1. Schematic creep curve showing negligible primary creep strain εp, tMGD as time to
reach Monkman-Grant ductility, εt limiting tertiary creep strain and λ as damage tolerance factor
(Phaniraj et al., 2003)
According to continuum damage mechanics (CDM) (see Sect. 8.4.4, Volume
II) damage is treated as an internal variable, D, and for uniaxial stress conditions
it can be expressed by two coupled differential equations:
m
(4.6.1)
k
k
Dɺ = Dɺ 0 (σ / σ 0 ) 1 / (1 − D )
(4.6.2)
m
εɺ = εɺ0 (σ / σ 0 ) 1 / (1 − D )
It is assumed that the exponents m and k are constant, independent of the applied stress, σ.
a) Find the evolution of the damage parameter D and of the creep strain as a function of the reduced time: t/tr, tr being the end of life time. Show that, integrating
(4.6.2), the time to fracture tr and the evolution of damage can be determined:
Show that integration of (4.6.2) yields:
171
k
tr = 1 / ( k + 1) Dɺ 0 (σ / σ 0 )
(4.6.3)
corresponding to D = 1, and:
1
t k +1
D = 1 − 1 −
tr
(4.6.4)
Solution
Integration of (4.6.2) yields:
1
k +1
1 − D = 1 − ( k + 1) Dɺ 0 (σ / σ 0 ) t
k
with tr =
1
ɺ
( k + 1) D0 (σ / σ 0 )k
1
t k +1
= 1 −
tr
when D = 1
(4.6.5a)
(4.6.5b)
b) Show that, integrating (4.6.1) and using (4.6.5b), one obtains the evolution of
the creep strain:
t
ε / εr = 1 − 1 −
tr
k +1− m
k +1
(4.6.6)
with
εɺ
1
m−k
ε r = ɺ0
(σ / σ 0 )
D0 k − m + 1
Solution
Substituting Eq. 4.6.5a in Eq. 4.6.1, and integrating one obtains:
(4.6.7)
172
ε = εɺ0 (σ / σ 0 )
m
k − m +1
k +1
t k +1
tr
1 − 1 −
k − m + 1 tr
(4.6.8)
The variation of strain with time is given by:
ε = εɺ0 (σ / σ 0 )
m
k −m +1
k +1
t k +1
tr
1 − 1 −
k − m + 1 tr
(4.6.9)
The strain to failure is given by:
m
ε r = εɺ0 (σ / σ 0 ) tr
εɺ
k +1
1
m−k
= ɺ0
(σ / σ 0 )
k − m + 1 D0 k − m + 1
(4.6.10)
Thus one obtains:
t
ε
= 1 − 1 −
εr
tr
k − m+1
k +1
(4.6.11)
g
c) The Monkman-Grant law states that the product ( εɺ sv ) tr is a constant denoted
MGD. The exponent g does not differ much from 1, and in this exercise it is assumed that g = 1. The ratio λ = ε r εɺ sv tr measures the relative importance of the
tertiary creep strain. The time to reach the Monkman-Grant ductility is denoted
tMGD.
Calculate the ratio λ in terms of the reduced time tMGD/tr.
Solution
The ratio λ = ε r εɺ sv tr measures the relative importance of the tertiary creep strain.
It can simply be expressed as:
λ=
εr
εɺ svtr
=
k +1
k − m +1
Denoting the Monkman-Grant strain εMGD, Eq. 4.6.5 yields:
(4.6.12)
173
t
ε
1 − MGD = 1 − MGD
εr
tr
1λ
(4.6.13)
As εt = εr − εMGD, the tertiary ductility is such that:
ε t tMGD
= 1−
ε r
tr
1/ λ
(4.6.14)
This expression is only valid when D is negligible, in the stationary regime.
Equation 4.6.11 can thus be written as:
ε sv
t
= 1 − 1 −
εr
tr
1/ λ
(4.6.15)
where λ is assumed to be a constant independent of the applied stress.
d) Show that the time to reach the Monkman-Grant ductility (MGD) can simply be
related to the value of λ which measures the part of the ductility associated with
tertiary creep. Comment the expression obtained between tMGD /tr and
λ = ε r εɺsvtr . MGD is defined as MGD = εɺsvtr . (See Fig. 4.6.1)
Solution
Equation. 4.6.11 indicates that:
ε
t
1−
= 1 −
ε r tr
1/ λ
(4.6.16)
So, the ductility due to tertiary creep, εt, can simply be written as (see (4.6.14))
ε t tMGD
= 1−
ε r
tr
1/ λ
with ε t = ( λ − 1) εɺ sv tr and ε r = λεɺ sv tr
Using the definition of λ, Eq. 4.6.17 can be rewritten as follows:
(4.6.17)
(4.6.18)
174
( λ − 1) εɺsvtr
λεɺsvtr
1/ λ
t
= 1 − MGD
tr
(4.6.19)
One deduces that:
tMGD
tr
λ
λ − 1
= 1−
= constant
λ
(4.6.20)
The variation of tMGD / tr as a function of λ is shown in Fig. 4.6.2 where it is
observed that this ratio reaches a value which is almost constant as soon as the
damage tolerance factor λ is sufficiently large ≥ 5 . It can easily be shown that
( )
this constant value is equal to (1 − 1 / e ) ≈ 0.63 .
Fig. 4.6.2. Variation of tMGD/tr as a function of λ in two steels (adapted from Phaniraj et al 2003)
4.3 Miscellaneous
Exercise 4.7 Bending of creeping bars and application of deformationmechanisms maps
The bending behaviour of a prismatic bar is investigated in the frame of classical
assumptions used in the study of the strength of materials (small displacement, a
planar section remains plane; the stress state is assumed to be uniaxial). The
length of the bar is L; the bar is aligned along the x axis, the y axis being vertical.
175
This bar is submitted to a bending moment, M (Fig. 4.7.1). The viscoplastic law
which is used is the simple Norton law:
σ
εɺv =
K
n
(4.7.1)
a) Rectangular beam (width b, height h)
Show that at a given time, t, the profile of the stress σ xx = σ can be written as:
2y
h
σ = σM
1/ n
(4.7.2)
where σ M is the maximum stress in the beam at time t and σ is the stress along
the x axis across the beam section. What are the simplifying assumptions made to
obtain this expression? Remember that assuming that a section remains planar
leads to:
ε = εM
2y
h
or εɺ = εɺM
2y
h
(4.7.3)
Calculate the maximum stress σ M as a function of the applied bending moment, M. Introducing the inertia moment, I, used in elasticity, show that the stress
can be expressed as:
σ =
M h
2I
( )
f y, n
(4.7.4)
Check that the function, f, is equal to 1 for y = h/2 when n = 1. Indicate qualitatively the shape of this function f, as a function of y for different values of exponent
n.
An “equivalent” inertia moment (IC) can be defined for creep deformation. IC
can be defined such that:
σM =
M h
2 IC
(4.7.5)
Derive the expression relating the curvature to the bending moment, M, and the
time, t. Express IC as a function of the elastic inertia I.
176
Solution
If the elastic strain is neglected, one can write:
εɺ = εɺv =
n
σ
K
(4.7.6)
This implies that:
σ = σM
2y
h
1/ n
(4.7.7)
The stress is maximum at the free surface and is null in the centre of the beam.
The situation is similar to that observed in elasticity.
The bending moment can be written as:
M = 2b
∫
h/ 2
0
σ ydy
(4.7.8)
Using (4.7.1) and integrating, one obtains:
σM =
M h 2n + 1
2I
3n
where I is the “elastic” inertia moment of the beam (I = bh3/12)
Fig. 4.7.1. Sketch showing a bending bar
(4.7.9)
177
Fig. 4.7.2. . Stress distribution in bending of a rectangular beam for different values of the exponent n. The elastic stress distribution is the same as that given for n = 1
One obtains:
M h 2y
σ =
2I h
1/ n
2n + 1
3n
(4.7.10)
The function f (y, n) is:
f =
2n + 1 2 y
3n h
1/ n
(4.7.11)
The ratio between the bending stress and the maximum elastic strain, as a function of the distance from the centre line (2y/h) is shown in Fig. 4.7.2 for various
values of the exponent n. The stress redistribution associated with creep deformation is relatively independent of the value of exponent n when it is larger than 1. It
can be noted that when n tends to infinity, the maximum stress at the surface tends
to 2/3 the stress calculated in elasticity.
Equation 4.7.9 shows that
IC =
3n
I
2n + 1
(4.7.12)
178
b) Circular beam and deformation-mechanisms maps
One considers the case where n =1. The corresponding creep mechanisms are diffusional ones (Coble or Herring-Nabarro creep: see Sect. 4.3.2.2 in Volume I).
Creep of lead water pipes. Analysis of sagging pipes. Use of the deformation
map for antimonial lead (Figures 4.7.3a and b. See Ashby and Jones 1980).
Since pre-Roman times lead has been used for water conduits and roof coverings. It has now been largely eliminated for these usages One distinguishes “soft
lead”, typically 99.5% Pb containing some impurities such as antimony, Sb, and
“hard lead” containing 1 to 6% Sb. The earliest pipes were cast which produced
large grain-sized (about 1mm) material. Pipes produced by plastic deformation
had a smaller grain size (about 50µm) due to recrystallisation. Figures 4.7.3a and
b indicate that the reduction in grain size from 1 mm down to 50 µm produces a
large acceleration in creep strain rate at low stresses, as expected from Eqs. 4.71
and 4.76 in Volume I. The melting point of pure lead is 327°C (600 K). The addition of antimony lowers the melting point of lead to 525 K (eutectic temperature),
such as at room temperature T/TM is between 0.50 and 0.57.
Two examples of creeping lead pipes were examined by Ashby and Jones: one
is an external drain pipe, 75 years old, the other one is an interior hot-water pipe
of about the same age (Figures 4.7.4 and 4.7.5).The maximum strain in the rectangular section drain pipe occurs in the lower surface of the pipe. This strain was
calculated from the radius of curvature, the depth of the sag, h, and the height, b,
of the pipe. These measurements were made from Fig. 4.7.5 showing that the drain
pipe has crept under its own weight. The shear strain rate for small sags (h<b)
can be calculated as γɺ = 4 3bh / l 2t .
Calculate the corresponding strain rate from this expression. The stress varies
with position along the pipe; expressed as an equivalent shear stress it has a
maximum value of 3x10-5µ. Is the observed strain consistent with that predicted
from Fig. 4.7.3? What would have happened if the grain size had been 1mm?
Solution
Fig. 4.7.3. Deformation mechanisms of antimonial lead with a grain size of 50 µm ((a) at left)
and 1 mm ((b), at right)(Ashby and Jones 1980)
179
Fig. 4.7.4. Creep of lead pipes (Ashby and Jones 1980).
Fig. 4.7.5. Dimensions of drain and hot water pipe (Ashby and Jones 1980).
One finds γɺ ≈ 6 x10−12 s −1 which corresponds to a very low strain rate. This
strain rate is compatible with the results reported in Fig. 4.7.3a (d = 50 µm) for an
applied stress of 3.5x10-5µ.
An increase of the grain size by a factor of 1,000/50 = 20 would have produced
a strain rate of about 10-15 due to Coble creep. However it is difficult to conclude
unambiguously from the examination of Fig. 4.7.3a. This figure shows that, for a
stress of 3.5x10-5 µ, and a grain size of 1 mm, diffusion flow and power-law creep
are overlapping. The positions corresponding to the drain pipe and the hot water
pipe are indicated in Fig. 4.7.3.
180
References
Ashby MF, Jones DRH (1980) Engineering Materials 1. An introduction to their properties and
Applications. Pergamon Press and applications of deformation-mechanism maps. See website: engineering.dartmouth.edu/defmech/chapter_19.htm
Calvet JN, Le Bret P, Besnault R, Lehman D, Blanc B, Roulliay R, Castaing A (1972) Etude des
déformations de fluage sous sollicitations complexes d’un acier inoxydable austénitique
Z6CND18-12 (AISI 316) – Fluage sous charge progressive à haute température – Fluage sous
contrainte thermique. Note technique RMA (72) 507
Cannon WR, Langdon TG (1988) Review-Creep of ceramics. Part II: an examination of flow
mechanisms. J. Mater Sci 23:1-20
Coble RL (1963) A model for boundary diffusion controlled creep in polycrystalline materials.
Journal of Applied Physics 34: 1679-1682
Phaniraj C, Choudary BK, Bhanu Sankara K, Raj B (2003) Relationship between time to reach
181
Chapter 5 Exercises of Chapter I.5: Viscoelasticity
5.1 One-dimensional responses in linear viscoelasticity
The coupling of elasticity and viscosity, which is the characteristic feature of
viscoelasticity, is responsible for the hereditary nature of this constitutive
behaviour and for the specific delayed responses to various loadings: creep
and retardation, relaxation, stress- and strain-recovery... One-dimensional
linear viscoelasticity, which obeys the Boltzmann superposition principle, is
fully characterised by only one scalar function (the creep or the relaxation
function) of two time variables, and it can be described by an integral relation between the force and displacement variables. This relation reduces to a
convolution product when non-ageing linear viscoelasticity is considered, so
that the Laplace-Carson transform can be used for an easier resolution and
for a conversion of the viscoelastic problem into an elastic one: the creep or
the relaxation function only depends then on one time variable. The method
of the spectral representation shows that all the constitutive behaviour characteristics are contained in the relaxation or retardation spectrum; when this
spectrum reduces to discrete single lines, the constitutive equation reduces
to a differential equation and can be represented by series or parallel assemblages of the Kelvin and Maxwell models. The Dynamic Mechanical Analysis (DMA) may be used to get the complex modulus from an experimental
investigation and then to a complete description of the 1-D constitutive behaviour.
Exercise 5.1 Elementary rheological models
Calculate the constitutive differential equations for series and for parallel assemblages of one Kelvin model and one Maxwell model. For that, use both a direct
method and the Laplace-Carson transform technique. Check that the short and
long term creep and relaxation responses coincide with the intuitive predictions
which can be made from the spring and dashpot representation.
Solution
1) Burgers model (series assemblage):
182
For the Maxwell model, the differential equation is known to read (see Sect.
5.1.3.3 in Volume I), with obvious notations
qɺM =
Qɺ M QM
+
EM η M
(5.1.1)
For the Kelvin model, it is
(5.1.2)
QK = EK qK + η K qɺK
For a series assemblage of these models, we have
QM = QK = Q
q = qM + qK
qɺ = qɺM + qɺK
(5.1.3)
and then, from (5.1.2)
Qɺ = EK (qɺ − qɺM ) + η K (qɺɺ − qɺɺM )
ɺɺ
Qɺ
Q
Q
Qɺ
= EK qɺ −
+
+
+ ηK qɺɺ −
EM η M
EM η M
(5.1.4)
which may also be written as
EK qɺ + η K qɺɺ =
E
η
η ɺɺ
Q + 1 + K + K Qɺ + K Q
ηM
η
E
EM
M
M
EK
(5.1.5)
This result could have been derived straightforwardly by using the fact that, for
a series assemblage, the resulting creep function is the sum of the creep functions
of each element. According to Sect. 5.1.3.3, we get for the Laplace-Carson transformed resulting creep function
f *( p ) =
q *( p )
1
1
1
( E + ηM p )( EK + ηK p ) + EMηM p (5.1.6)
=
+
+
= M
*
Q ( p ) EM η M p EK + η K p
EMη M p ( EK + η K p )
After reduction, it is easy to show that the second-order polynomial relation
with respect to p which derives from (5.1.6) admits (5.1.5) as its original function
of time.
From the spring and dashpot representation of this model, it is obvious that it
exhibits instantaneous elasticity (due to the unconstrained spring of the Maxwell
model), unlimited creep (due to the unconstrained dashpot of the Maxwell model)
and complete relaxation (due to the unconstrained dashpot of the Maxwell model).
183
This is corroborated by the behaviour of f*(p) and r*(p) given by (5.1.6) or derived
from it for p → 0 (i.e., t → ∞) and p → ∞ (i.e., t → 0+ ) :
1
f * ( p) = lim f (t ) =
(instantaneous elasticity)
lim
p →∞
t →0 +
EM
f * ( p) = limf (t ) = +∞ (unlimited creep)
lim
p →0
t →∞
r * ( p) = lim r(t ) = 0 (complete relaxation)
lim
t →∞
p →0
(5.1.7)
2) Parallel assemblage
We now have
qM = qK = q
Q = QM + QK
Qɺ = Qɺ M + Qɺ K
(5.1.8)
From (5.1.1), we have
(Qɺ − Qɺ K ) (Q − QK )
+
EM
ηM
ɺ
Q − ( EK qɺ + η K qɺɺ) Q − ( EK q + η K qɺ )
=
+
ηM
EM
qɺ =
(5.1.9)
which may also be written as
E
η
q + 1 + K + K
ηM
EM η M
EK
ηK
Q
Qɺ
qɺɺ =
+
qɺ +
η M EM
EM
(5.1.10)
Like for the series assemblage, we can also use the Laplace-Carson transform
technique with the fact that, for a parallel assemblage, the resulting relaxation
function is the sum of the relaxation functions of each element. Thus, according to
Sect. 5.1.3.3, we get for the Laplace-Carson transformed resulting relaxation function
r *( p) =
=
Q*( p )
E η p
= EK + η K p + M M
*
q ( p)
EM + η M p
( EM + ηM p )( EK + ηK p ) + EMηM p
( EM + η M p )
which leads to the second-order polynomial relation with respect to p
(5.1.11)
184
( EM +ηM p ) Q*( p )= EM EK +( EMηK +EKηM +EMηM ) p+ηMηK p 2 q*( p)
(5.1.12)
It is easy to check that (5.1.12) admits the differential equation (5.1.10) as its
original function of time.
From the spring and dashpot representation, it is obvious that this model does
not exhibit instantaneous elasticity (due to the dashpot of the Kelvin model), has
limited creep (due to the spring of the Kelvin model) and cannot suffer a relaxation test (due to the dashpot of the Kelvin model, which cannot deform instantaneously). This is corroborated by the behaviour of r*(p) and f*(p) given by (5.1.11)
or derived from it for p → 0 (i.e., t → ∞) and p → ∞ (i.e., t → 0+ ) :
lim f * ( p) = lim f (t ) = 0 (no instantaneous elasticity)
t →0 +
p →∞
1
f * ( p) = limf (t ) =
(limited creep)
lim
p →0
t →∞
E
K
lim r* ( p) = lim r(t ) = ∞ (relaxation test impossible)
t →0
p →∞
(5.1.13)
Exercise 5.2 Creep recovery and relaxation
1) Preliminary creep recovery tests performed on a linear non-ageing viscoelastic
material described by the parameters Q (load) and q (deformation) have shown
that there remains a permanent strain qp proportional to the load Q0 and to the
time interval τ during which this load has been imposed, say q p = AQ0τ , where A
is a constant. Show from that property that the creep flow is unlimited with a finite
limit creep rate and that the relaxation is complete.
2) A test at constant strain rate qɺ0 is performed on the same material. Show that
Q tends toward a finite limit value Q∞ and derive the relation between Q∞ and qp
(question 1), for fixed values of Q0, τ and qɺ0 .This test is performed up to t = θ
and the load is then suppressed. Show that q then tends toward the permanent
θ
strain qp′ given by qp′ = A∫ Q (t )dt where A is the constant of question 1
0
Solution
1) A creep recovery test is described by the relations
Q (t ) = Q0 [ Y(t ) − Y(t − τ ) ]
q (t ) = Q0 [ f (t ) − f (t − τ ) ]
⇒ *
*
*
*
Q ( p ) = Q0 [1 − exp(− pτ )]
q ( p ) = f ( p )Q ( p )
(5.2.1)
185
where Y(t) is the (Heaviside) unit step function and f(t) is the creep function. We
have
q p = lim q(t ) = lim q* ( p) = lim f * ( p)Q* ( p ) = lim f * ( p) .0
p →0
t →∞
p→0
(5.2.2)
p →0
Since q p (= AQ0τ ) is finite and non zero, this means that f * ( p) → ∞ when
p → 0 , i.e., that f (t ) → ∞ when t → ∞ : so, the creep flow is unlimited. Its time
derivative for t → ∞ can be evaluated as follows
*
( fɺ )
p q *( p )
= p f * ( p ) − f (0+ ) = *
− pf (0+ )
Q ( p)
(5.2.3)
*
()
and, since f(0+) is necessarily finite, pf (0+ ) → 0 when p → 0 . So, fɺ
like
behaves
p q *( p )
when p → 0 and we get
Q* ( p )
p q *( p )
p q *( p )
lim fɺ (t ) = lim fɺ * ( p) = lim *
= lim
t →∞
p →0
p →0
p
→
0
Q ( p)
Q0 [1 − exp(− pτ ) ]
(5.2.4)
Now, we have
lim
p→0
1 − exp(− pτ )
=τ
p
lim q* ( p ) = lim q(t ) = qp = AQ0τ
p →0
t →∞
(5.2.5)
and then
lim fɺ (t ) =
t →∞
AQ0τ
=A
Q0τ
(5.2.6)
So, when p → 0 , we have
pf * ( p ) ≈ A, f * ( p) ≈
A *
1
p
, r ( p) = *
≈ , lim r(t ) = lim r* ( p ) = 0
p
f ( p ) A t →∞
p →0
which means that the relaxation is complete.
2) We impose q (t ) = qɺ0tY(t ) . For the Laplace-Carson transforms, we get
(5.2.7)
186
q* ( p) =
qɺ0
p
⇒ Q* ( p ) = r * ( p ) q * ( p ) =
qɺ0 r* ( p )
p
(5.2.8)
The limit values for t → ∞ derive from
qɺ0 r* ( p) qɺ0
=
= Q∞
p →0
p
A
lim Q (t ) = lim Q* ( p) = lim
p →0
t →∞
since we know that
(5.2.9)
r* ( p) 1
≈
when p → 0 . With qp = AQ0τ , we get
p
A
qp =
Q0
qɺ0τ
Q∞
(5.2.10)
The general relation
t
q (t ) = Q (t )f (0) + ∫ Q (τ )f ′(t − τ )dτ
0
(5.2.11)
reads, for t > θ ,
t >θ
θ
⇒ q(t ) = ∫ Q (τ )f ′(t − τ ) dτ
0
(5.2.12)
So, when t → ∞, f ′(t − τ ) → A and then
θ
lim q(t ) = A∫ Q(τ ) dτ = qp′
t →∞
0
(5.2.13)
Exercise 5.3 Tensile and relaxation tests
A tensile test at constant axial strain rate qɺ0 is performed on a linear non-ageing
viscoelastic material described by the parameters Q (load) and q (deformation).
The load response has the form
Q(t ) = qɺ0 α (1 − exp(− at ) ) + β
where α, β and a are constants obtained experimentally.
(5.3.1)
187
1) Find the differential equation obeyed by this material. Calculate the creep and
relaxation functions and give the stiffness and viscosity characteristics of two
equivalent rheological models which can be used to represent this viscoelastic behaviour.
2) The tensile test is performed up to time t0 and then followed by a relaxation test.
Calculate Q(t) for t ≥ t0 .
Solution
1) Using the Laplace-Carson transforms of q(t) and Q(t), we get
q(t ) = qɺ0tY(t ) ⇒ Q (t ) = qɺ0 α (1 − exp(−at ) ) + β
*
qɺ0
p
⇒ Q* ( p) = qɺ0 (α + β ) − qɺ0α
q ( p) =
p
p+a
(5.3.2)
Thus, the relation between q* ( p ) and Q* ( p) may read
Q* = (α + β ) pq* − α
p2 *
q
p+a
(5.3.3)
or
pQ* + aQ* = β p 2 q* + a(α + β ) pq*
(5.3.4)
whose original is the differential equation
Qɺ + aQ = β qɺɺ + a (α + β )qɺ
(5.3.5)
From (5.3.4), we can write
( p + a)Q* = p ( β p + a(α + β ) ) q*
and then
(5.3.6)
188
α
1
*
q
p
+
a
β (α + β )
α +β
f * ( p) = * =
=
+
α
Q
p ( β p + a (α + β ) )
p
p + a(1 + )
β
*
p ( β p + a(α + β ) )
*
Q
p
= β p + aα
r ( p) = * =
q
p+a
p+a
(5.3.7)
The original creep and relaxation functions then read
1
α
α
tY(t ) +
1 − exp −a (1 + )t Y(t )
f (t ) =
2
α +β
β
a (α + β )
r* ( p) = βδ(t ) + aα exp −at Y(t )
( )
(5.3.8)
Referring to the creep functions of a Kelvin model (EK, ηK) and of a dashpot
(ηD), we conclude that the studied viscoelastic behaviour can be represented by a
series assemblage of the dashpot ( η D = α + β ) and the Kelvin model defined by
a
β
the parameters ( EK = (α + β ) 2 , η K = (α + β ) ). Equivalently, referring to the
α
α
relaxation functions of a Maxwell model (EM, ηM) and of a dashpot (ηD’), we conclude that the studied viscoelastic behaviour can be represented as well by a parallel assemblage of the dashpot ( η D′ = β ) and the Maxwell model defined by the
parameters ( EM = aα , η M = α ).
2) The whole deformation path is described by the relation
q (t ) = qɺ0 [ tY(t ) − (t − t0 )Y(t − t0 ) ]
(5.3.9)
which transforms into
q* ( p) =
qɺ0
1 − exp ( − pt0 )
p
(5.3.10)
Thus we have
Q* ( p) = r* ( p)q* ( p) = qɺ0 ( β +
aα
) (1 − exp(− pt0 )
p+a
(5.3.11)
The original function Q(t) has then the form
Q(t ) = qɺ0 [ Φ (t ) − Φ (t − t0 ) ]
(5.3.12)
189
with Φ (t ) defined by its Laplace-Carson transform Φ * ( p ) given by
aα
p
= α + β −α
p+a
p+a
(5.3.13)
Φ (t ) = (α + β )Y(t ) − α exp(−at )Y(t )
(5.3.14)
Φ* ( p ) = β +
So, we find
and then we get for t ≥ t0
Q(t ) = qɺ0α exp(−at ) [ exp(at0 ) − 1]
(5.3.15)
Thus we note that there is a load discontinuity at t = t0 with
+
Q (t0 ) = qɺ0α [1 − exp(−at0 ) ]
−
Q (t0 ) = qɺ0α [1 − exp(−at0 ) ] + β qɺ0
⇒ ∆Q (t0 ) = − β qɺ0
(5.3.16)
Exercise 5.4 Complex modulus
1) A preliminary approach to the modelling of the 1-D response of a linear nonageing viscoelastic material uses a parallel assemblage of one Kelvin model and
one Maxwell model, as studied in Exercise 5.1. Using the solution of this exercise
(part 2) for the relaxation function, show that the parameters EK, ηK, EM and ηM
of this model can be identified from the sole following information obtained from
the normalised response Q(t )/qɺ0 to the stepwise deformation rate qɺ0 Y(t ) : value
and time derivative at t = 0+ , ordinate at t = 0 and slope of the asymptote for
t →∞.
2) In order to get an improved modelling of the observed response, we consider
now the parallel assemblage of one Kelvin model ( E∞ ,η0 ) and n Maxwell models
( Ei ,ηi , i = 1 to n) and we compare the corresponding complex modulus to the experimental one. Calculate this modulus as a function of the parameters E∞ ,η0 , Ei ,
the relaxation times τ i = ηi /Ei and the pulsation ω.
3) A further improvement consists in the consideration of a continuous relaxation
spectrum g(τ ) . Calculate the corresponding complex modulus and find its limits
for ω → 0 and ω → ∞ . Draw qualitatively the associated Cole and Cole diagram. Find the parameters EK, ηK, EM and ηM of a scheme with only one Maxwell
190
model (see question 1) which would lead to the best approximation to this diagram when, at once, ω → 0 and ω → ∞ .
Solution
1) From (5.1.11) in Exercise 5.1, we get
r *( p) =
E η p
Q*( p )
= EK + η K p + M M
*
q ( p)
EM + η M p
(5.4.1)
The normalised response Q(t )/qɺ0 to the stepwise deformation rate qɺ0 Y(t ) can
be derived from its Laplace-Carson transform through the relations
q* ( p ) =
qɺ0 Q* ( p)
=
p r* ( p)
⇒
EMη M
Q * r * EK
= =
+ ηK +
qɺ0
p
p
EM + η M p
(5.4.2)
The limit values for t → 0, ∞ are obtained from the limits of the LaplaceCarson transforms for p → ∞,0 , namely
Q*
EK + EM
Q (t )
,
≈ η K + ( EK + EM )t
t →0+ ⇔ p →∞ ⇒ ɺ ≈ η K +
q
p
qɺ0
0
(5.4.3)
*
t →∞ ⇔ p →0 ⇒ Q ≈ η + η + EK , Q(t ) ≈ η + η + E t
K
M
K
M
K
qɺ0
p
qɺ0
Consequently, the four model parameters EK, ηK, EM and ηM can be obtained
Q (t )
from the experimental curve
using the values ηK and ( η K + ηM ) and the
qɺ0
slopes ( EK + EM ) and EK.
2) The Laplace-Carson transform of the relaxation function now reads
r *( p) = E∞ + η0 p + ∑
i
Eiηi p
Ei + ηi p
(5.4.4)
and the complex modulus r *(iω ) , with i 2 = −1 , is given by
r *(iω ) = E∞ + iη0ω + ∑
i
iEiηiω
Ei + iηiω
This can also be written in the classical form, with τ i = ηi /Ei :
(5.4.5)
191
Eiω 2τ i2
′
G
(
)
=
E
+
ω
∑i 1 + ω 2τ 2
∞
i
r *(iω ) = G′(ω ) + iG′′(ω ) with
τ
E
G ′′(ω ) = ω η +
i i
0 ∑
2 2
i 1+ ω τi
(5.4.6)
3) For a continuous relaxation spectrum g(τ ) , with τ varying from 0 to ∞, the
Maxwell models with a relaxation time lying between τ and ( τ + dτ ) have the
equivalent stiffness dE = g(τ )dτ . Consequently, by analogy with (5.4.6), we get
r *(iω ) = E∞ + ω 2 ∫
∞
0
∞ τ g(τ )
τ 2 g(τ )
dτ + iω η 0 + ∫
dτ
2 2
2 2
0
1+ ω τ
1+ ω τ
(5.4.7)
When ω → 0 , we have
∞
∞
r *(iω )ω ≈0 ≈ E∞ + iω η0 + ∫ τ g(τ )dτ + ω 2 ∫ τ 2 g(τ )dτ + ...
0
0
(5.4.8)
whereas, when ω → ∞ ,
∞
r *(iω )ω →∞ ≈ E∞ + ∫ g(τ )dτ + iη0ω + ...
0
(5.4.9)
So, the associated Cole and Cole diagram qualitatively looks like suggested in
Fig. 5.4.1. For the simple model of Question 1, the complex modulus would read,
according to (5.4.6) with only one Maxwell model
r1*(iω ) = EK +
EMω 2τ M2
E τ
+ iω η K + M 2M 2
1 + ω 2τ M2
1
+
ω τM
(5.4.10)
Fig. 5.4.1 Qualitative sketch of the Cole and Cole diagram for a continuous relaxation spectrum
192
At very low and very high frequencies, we have
*
η M2 2
ω + ...
r1 (iω )ω ≈0 ≈ EK + iω (η K + η M ) +
EM
r*(iω )
ω →∞ ≈ EK + EM + iη Kω + ...
1
(5.4.11)
So, the best approximation to (5.4.7) by a parallel assemblage of one Kelvin
model and one Maxwell model when, at once, ω → 0 and ω → ∞ , is obtained
when the following conditions are fulfilled:
E = E η + η = η + ∞τ g(τ )dτ
K
M
0
∞
∫0
K
∞
EK + EM = E∞ + ∫ g(τ )dτ η K = η0
0
(5.4.12)
EK = E∞ η K = η0
∞
∞
EM = ∫0 g(τ )dτ η M = ∫0 τ g(τ )dτ
(5.4.13)
that is when
5.2 Local viscoelastic constitutive equations and response of
viscoelastic structures
The Boltzmann superposition principle can still be used to derive the local
expression of the constitutive equations for linear viscoelastity. Their general form is unchanged, except that the scalar functions must be changed
into fourth-order tensors. Advantage may be taken of the assumption of nonageing and of the use of the Laplace-Carson transform to reduce viscoelastic
to elastic structural design through the correspondence theorem. This approach may also be used to derive estimates for the overall properties of heterogeneous viscoelastic materials (see Exercise 5.8).
Exercise 5.5 Identification of constitutive equations
193
An experimental investigation is performed in view of identifying the local constitutive equation of an isotropic linear non ageing viscoelastic material in isothermal conditions. This equation can be written (see Eq. 5.110 in Volume I)
σ (t ) = λ (t ) ⊙ Tr ε (t ) δ + 2µ (t ) ⊙ ε (t )
E (t ) ⊙ ε (t ) = [1 + ν (t )] ⊙ σ (t ) −ν (t ) ⊙ Tr σ (t ) δ
(5.5.1)
Two tests (simple shearing and hydrostatic pressure) are conducted in order to
identify the functions E(t) and ν(t).
1) Simple shear: in its principal axes, the stress tensor σ (t ) reads
0
σ (t ) 0
(σ (t ) ) = 0 −σ (t ) 0
0
0
0
(5.5.2)
Show that this test can be characterised by only two parameters, namely
Q(t ) = σ (t ) for the load and q(t) for the deformation. What will be q(t)? Express
*
the Laplace-Carson transform rSS
( p ) of the relaxation function for this test and
for these parameters as a function of E *( p) and ν *( p) . A “dynamic” investigation (sinusoidal vibration q = q0 cos ω t with the stationary response with the amplitude Q0 , the same pulsation ω and the phase lag ϕ (ω ) ) yields the relations
Q0
Bω
with B > 0
R(ω ) = q = ( A2 + ω 2 )1/2
0
π
tan ϕ (ω ) = A
with A > 0, 0 ≤ ϕ ≤
ω
4
(5.5.3)
Calculate the complex modulus in the form G′(ω ) + iG′′(ω ) . Check that it has
the same form as the one associated to a parallel assemblage of one Kelvin model
(EK, ηK) and one Maxwell model (EM, ηM) and find the associated relaxation function rSS (t ) .
2) Hydrostatic pressure: the stress field σ (t ) = − P(t )δ (with P > 0 ) is uniform.
Find the deformation parameter q(t) which can be associated with the load pa*
( p ) of the rerameter Q(t ) = − P(t ) . Express the Laplace-Carson transform rHP
laxation function for this test and for these parameters as a function of E *( p) and
ν *( p) . This test is performed at constant deformation rate, i.e., qɺ = qɺ0Y (t ) and
the response is
194
Q(t ) = qɺ0 (Ct + D )
C >0 D>0
(5.5.4)
*
Calculate rHP
( p ) as a function of C and D and find a rheological model consistent with this relaxation function.
3) Application: Calculate E *( p) and ν *( p) and comment qualitatively on their
variations. A horizontal cantilever beam, with a square cross-section, made of this
material, is fixed at one end and subjected at the other end to a constant vertical
force F parallel to two edges of the cross-section. Calculate the vertical displacement v(t) at the free end of the beam and find a rheological model yielding the
same response for the parameters (F, v).
Solution
1) Simple shear: the deformation work variation per unit volume must read
dW = Qdq = σ ij dε ij
→ q = ε11 − ε 22
= σ d(ε11 − ε 22 )
(5.5.5)
Application of the Laplace-Carson transform of (5.5.1) yields
1 +ν * * ν * *
1 +ν * *
*
σ 11 − * (σ 11 + σ 22
σ
)=
*
2(1 + ν * ) *
E
E
E*
*
⇒
q
=
σ
E*
1 +ν * * ν * *
1 +ν * *
*
σ
σ
σ
σ
=
−
(
+
)
=
−
22
11
22
E*
E*
E*
ε11* =
*
ε 22
(5.5.6)
With Q* = σ * , the corresponding (transformed) relaxation function is then
*
rSS
( p) =
Q* ( p )
E* ( p)
=
= µ * ( p)
*
*
q ( p) 2 (1 +ν ( p) )
(5.5.7)
*
From R(ω) and tan ϕ , the complex modulus rSS
(iω ) can be derived by the relations
*
rSS
(iω ) = G′(ω ) + iG ′′(ω ) = R(ω ) ( cos ϕ (ω ) + isin ϕ (ω ) )
= R(ω ) cos ϕ (ω ) (1 + i tan ϕ (ω ) )
Since 0 ≤ ϕ ≤
π
, one has simply
4
(5.5.8)
195
cos ϕ =
1
2
1 + tan ϕ
1
=
1+
A
2
ω2
=
ω
(A
2
1/2
+ ω2 )
(5.5.9)
and then
*
rSS
(iω ) =
Bω
ω
A
1 + i
( A2 + ω 2 )1/2 ( A2 + ω 2 )1/2
ω
(5.5.10)
Bω 2
ABω
= 2
+i 2
A + ω2
A + ω2
As already shown in Exercise 4 (see Eq. 5.4.6), the complex modulus for a parallel assemblage of one Kelvin model and one Maxwell model reads
*
rK/
/ M (iω ) = EK +
EMω 2τ M2
E τ
+ iω ηK + M 2M 2 with τ M = η M /EM (5.5.11)
2 2
1+ ω τM
1+ ω τM
So, (5.5.10) and (5.5.11) coincide if
EK = 0 η K = 0 EM = B η M = B / A
(5.5.12)
which means that one single Maxwell model is needed to model the studied response in simple shear. The associated relaxation function reads
rSS (t ) = EM exp(−
EM
ηM
t ) = B exp(− At )
(5.5.13)
2) Hydrostatic pressure: the deformation work variation per unit volume must read
dW = Qdq = σ ij dε ij
→ q = ε kk
= − Pdε kk
(5.5.14)
From (5.5.1), we get
E *( p) ε kk* ( p) = 1 − 2ν * ( p) σ kk* ( p) = −3 1 − 2ν * ( p ) P* ( p)
(5.5.15)
E *( p) q* ( p) = 3 1 − 2ν * ( p) Q* ( p)
(5.5.16)
and then
196
or
*
rHP
( p) =
Q* ( p )
E * ( p)
=
q* ( p) 3 1 − 2ν * ( p)
(5.5.17)
With qɺ = qɺ0Y (t ) and Q(t ) = qɺ0 (Ct + D) , we find
E * ( p)
*
⇒
r
(
p
)
=
C
+
Dp
=
C
HP
Q* = qɺ0 ( + D )
3 1 − 2ν * ( p)
p
pq* = qɺ0
(5.5.18)
i.e., the transformed relaxation function of the Kelvin model ( EK = C ,η K = D ).
3) Application: from (5.5.7) and (5.5.17), we get
*
*
*
3rHP
− 2rSS
ν ( p) =
*
*
2 ( rSS
+ 3rHP
)
* *
9rSS rHP
*
E ( p) = r* + 3r*
SS
HP
(5.5.19)
From (5.5.13) and (5.5.18), namely
Bp
*
rSS ( p) =
p
+A
r* ( p) = C + Dp
HP
(5.5.20)
3(C + Dp )( p + A) − 2 Bp
*
ν ( p) = 2 ( 3(C + Dp)( p + A) + Bp )
9 Bp(C + Dp )
E * ( p) =
3(C + Dp)( p + A) + Bp
(5.5.21)
E *( p) and ν *( p) are given by
Without trying to perform the complete inversion of these Laplace-Carson
transformed functions, we can at least easily derive the limit values of the original
functions and their time derivatives when t → 0+ and t → ∞ from the limit values of ν * ( p ), p ν * ( p) −ν (0+ ) , E * ( p) and p E * ( p) − E (0+ ) when p → ∞ and
p → 0 , respectively. We find that ν (t ) starts from ½ at t = 0 and decreases
197
(ν ′(0+ ) = − B / 2 D < 0 ) whereas it tends to the asymptotic value ½ when t → ∞ .
As for E (t ) , it starts from 3B and decreases ( E ′(0+ ) = − B( B + 3 AD ) / D < 0 ) and
it tends to 0 when t → ∞ .
The problem of the cantilever beam can be solved by using the elastic solution
and the correspondence theorem (see Sect. 5.3.2.1 in Volume I). For linear isotropic elasticity, the displacement at the free end of the beam vel is known to be
vel =
Fl 3
3EI
with I =
a4
12
(5.5.22)
where E is the Young modulus, l is the beam length, I is the principal moment of
inertia and a is the edge length of the square cross-section. Consequently, the
Laplace-Carson transformed viscoelastic solution v(t ) will be
v* =
F *l 3
3E * I
with I =
a4
12
(5.5.23)
With F constant, we find
v* =
4 Fl 3 1
a4 E*
⇒ v(t ) =
4 Fl 3
1
J (t ) with J *( p) = *
4
a
E ( p)
(5.5.24)
According to (5.5.21), we have
3(C + Dp)( p + A) + Bp
9 Bp(C + Dp )
1
A
1
=
+
+
3B 3Bp 9 D( p + C )
D
(5.5.25)
1
A
1
C
+
t+
1 − exp − t
3B 3B 9C
D
(5.5.26)
J * ( p) =
and then
J (t ) =
and hence v(t ) by (5.5.24).
198
Obviously, J (t ) is the creep function of a Burgers model, i.e. of the series as3B
semblage of the Maxwell model ( EM = 3B, η M =
) and of the Kelvin model
A
( EK = 9C , ηK = 9 D ).
Exercise 5.6 Responses of a circular thin tube
A circular thin tube (height h, mean radius R0, thickness e ≪ R0 ), open at its ends,
is made of a linear isotropic non ageing viscoelastic material.
1) This tube is subjected to a uniform internal pressure P, considered as the load
parameter Q for this test. What should be the associated deformation parameter
q? The creep function for this test fIP relating Q and q is found to read
f IP (t ) = a + b t
(5.6.1)
Why must the constants a and b be positive? Does the studied behaviour exhibit
for this test instantaneous elasticity? limited or unlimited creep flow? complete or
partial relaxation? complete or partial creep recovery? (NB: remind that
α *
(t )
π
1
! =
2
2
= α ! p −α
(0 ≤ α ≤ 1) )
(5.6.2)
Show that such a response is not consistent with a finite and discrete relaxation
spectrum.
The stress state which results from the uniform internal pressure applied to
this circular cylindrical open tube reads, in cylindrical polar coordinates (r, θ, z)
(σ )
( r ,θ , z )
0 0
= 0 Σ
0 0
0
0
0
(5.6.3)
Prove that Σ = PR0 /e and derive from that the relation between the LaplaceCarson transformed creep (or relaxation) function f IP* ( p) (or rIP* ( p) ) and the
transformed viscoelastic Young modulus E * ( p).
2) The same tube is subjected to axial sinusoidal vibrations with the pulsation ω.
Calculate the associated complex compliance
J * (iω ) = 1 / E * (iω ) = E (ω )exp ( iϕ (ω ) )
−1
(5.6.4)
199
Show that the Cole and Cole diagram for the complex modulus E * (iω ) is a
circular arc starting at the origin for ω = 0 .
3) A static tensile test is performed on the same tube. It is observed that, during
this test, the lateral shortening ε 22 is always three times smaller that the axial
strain ε 11 . Deduce from this property and from what precedes the constituve equations of the studied material.
Solution
1) The increment of strain energy by unit volume reads
dW = P
d(πR 2 h)
dR
=P
= Pdq
2πRhe
e
⇒q=
R − R0
e
(5.6.5)
The creep function must be positive and non decreasing. Consequently a and b
must be positive. Since f (0+ ) = a ≠ 0, the response exhibits instantaneous elasticity and since f (t ) → ∞ when t → ∞, the creep flow is unlimited. The LaplaceCarson transformed creep function reads
f IP* ( p) = a + b
π
1
= *
2 p rIP ( p)
(5.6.6)
A relaxation test exhibits no singularity at t = 0+ (since lim rIP* ( p ) = 1/a ≪ ∞ ).
p →∞
*
IP
In addition, the relaxation is complete (since lim r ( p ) = 0 ). A creep recovery test
p →0
is ruled by the following equations
P(t ) = P0 [ Y(t ) − Y(t − τ )] ⇒ P* ( p) = P0 [1 − exp(− pτ )]
b π
q* ( p) = f IP* ( p) P* ( p ) = P0 a +
[1 − exp(− pτ ) ]
2 p
(5.6.7)
b π
lim q(t ) = lim q* ( p) = lim P0 pτ a +
=0
t →∞
p →0
p →0
2 p
So, the creep recovery is complete. For a viscoelastic body exhibiting instantaneous elasticity and unlimited creep flow defined by a finite and discrete relaxation spectrum, the transformed creep function reads (see Sect. 5.1.3.4 in Volume I)
200
f * ( p) =
1
1
Js
+
+ ∑s
E0 η ∞ p
1 + pτ s
(5.6.8)
with E0 ≠ 0 and η∞ ≪ ∞ . The creep recovery response is then
lim q(t ) = lim q* ( p ) = lim P0 pτ f * ( p) = P0
p →0
t →∞
p →0
τ
η∞
(5.6.9)
which means that the recovery is only partial. Consequently, we can state that the
studied behaviour cannot be described by a finite and discrete relaxation spectrum.
Using (5.6.3), we can write the strain energy density (5.6.5) in the form
dW = P
dR
dR
= σ : dε = Σ
e
R0
⇒ Σ=P
R0
e
(5.6.10)
The transformed constitutive equations yield
1 + ν * * ν * * Σ * R0 P*
Σ − *Σ = * =
E*
E
E
e E*
*
R − R0
R r
e
=
= q* = 0 IP* q*
R0
R0
e E
*
=
ε θθ
(5.6.11)
so that we find
rIP* ( p) =
1
e2
e2 1
= 2 E * ( p) = 2 *
f ( p ) R0
R0 J ( p)
*
IP
(5.6.12)
2) According to (5.6.12), the complex compliance reads
J * (iω ) =
e2 *
e2
b π
f (iω ) = 2 a +
2 IP
R0
R0
2 iω
(5.6.13)
For physical reasons (namely the positivity of G’(ω) and G”(ω), which will be
1 1− i
checked later on), we have to choose
=
instead of its opposite and
i
2
(5.6.13) becomes
J * (iω ) =
e2
b π
b π
a+
−i
2
R0
2 2ω
2 2ω
(5.6.14)
201
If ψ is the phase lag for J * (iω ) , we have
ϕ = −ψ
b π
2
2ω
⇒ tan ϕ = − tanψ =
b π
a+
2 2ω
(5.6.15)
From this expression, several simplifications can be derived, namely
b π
a tan ϕ
=
2 2ω 1 − tan ϕ
E = E * (iω ) =
⇒
J = J * (iω ) =
ae 2 1 + tan 2 ϕ
ae 2
=
R02 (1 − tan ϕ )
R02 (cos ϕ − sin ϕ )
R02
(cos ϕ − sin ϕ )
ae 2
R02
(cos ϕ − sin ϕ )(cos ϕ + isin ϕ )
ae 2
R02
′
ω
G
(
)
cos ϕ (cos ϕ − sin ϕ )
=
R2
ae 2
⇒ G ′2 + G ′′2 − 02 (G ′ − G ′′) = 0
2
ae
G ′′(ω ) = R0 sin ϕ (cos ϕ − sin ϕ )
2
ae
E * (iω ) =
(5.6.16)
The last relation between G’(ω) and G”( ω) is the equation of the Cole and Cole
curve (with the restriction G′ ≥ 0, G′′ ≥ 0 ): it is the equation of an arc or circle,
centred at (
R02
R02
,
−
) , starting from the origin when ω → 0 and tending to
2ae 2 2ae 2
R02
,0) when ω → ∞ .
ae 2
3) The described property implies the relation
(
1
3
1
3
*
ε 22
= −ν *ε11* = − ε11* ⇒ ν * ( p) = = ν (t )
(5.6.17)
The constitutive equations for the studied material then read
4 *
1
*
*
*
E ( p)ε ( p) = 3 σ ( p) − 3 Tr σ ( p) δ
or
4
1
ε (t ) = J (t ) ⊙ σ (t ) − J (t ) ⊙ Tr σ (t ) δ
3
3
(5.6.18)
202
with
J * ( p) =
1
e2
e2
π
= 2 f IP* ( p ) = 2 a + b
E ( p) R0
R0
2 p
*
(5.6.19)
5.3 Miscellaneous
Exercise 5.7 Ageing viscoelasticity
A concrete slab has been poured on a horizontal rigid substratum with conditions
of perfect bonding at the interface. The slab, with the thickness 0.25m, is considered as infinite in the horizontal directions x2 and x3 and there are no exterior
forces. When it is not subjected to stresses, the concrete material suffers at age u a
SH
shrinkage isotropic contraction ε (u ) = −a(u )δ . It obeys an isotropic ageing linear viscoelastic behaviour. In Table 5.7.1 are given, for several values of the age
u
- the delayed Young relaxation function E(t,u) for t = 200 days
- the instantaneous Young relaxation function E(u,u)
Table 5.7.1 Observation data
u days
a(u) mm/m
E(200,u) MPa
E(u,u) MPa
0
0
5,000
20,000
7
0.1
10,000
32,000
28
0.2
15,000
38,000
77
0.3
18,000
39,000
163
0.4
22,000
40,000
200
0.425
40,000
40,000
We aim at predicting the risk of cracking, given that the tensile strength of the
used concrete is about 4 MPa. It is assumed that
- the rigid substratum forces the resulting displacement field to reduce to
its vertical component, depending on x1 only, say v1 ( x1 ).
- the sole non zero components of the stress field are σ 22 and σ 33 .
- the Poisson ratio is constant and equal to 0.2.
1) Draw a qualitative plot of the time variations of the relaxation functions E(t,u)
for several values of u.
2) Use first an elastic approximation of the concrete behaviour to estimate the
tensile stress σ 22 for t = 200 days with the elastic Young modulus given by
203
E (u, u ) for u = 0, 7, 28, 77, 163 and 200 days. What can be concluded from that
concerning the risk of crack formation?
3) Going back to the actual ageing viscoelastic behaviour, give the theoretical expression of σ 22 (t ) and find numerically an estimate of it for t ≤ 200 days by use
of a stepwise approximation of a (u ) . Comment on the influence of viscosity on the
risk of cracking.
Solution
1) See Fig. 5.7.1.
E (t , u )
E (u, u )
E (t , u4 )
E (t , u3 )
E (t , u2 )
0
u1
u2
u3
u4
E (t , u1 )
u, t
Fig. 5.7.1 Qualitative plot of the time variations of E (t, u) for several values of u
2) According to an elastic treatment, we would have, with ε ije the elastic strain
ε ij = ε ijSH + ε ije
= −a(u )δij +
1 +ν
ν
σ ij − σ kk δij
E
E
(5.7.1)
By symmetry, σ kk = σ 22 + σ 33 = 2σ 22 . From (5.7.1) and ε 22 = 0 , we get
ε 22 = 0 = −a(u ) +
and then
1 −ν
σ 22
E
(5.7.2)
204
σ 22 =
E
a(u )
1 −ν
(5.7.3)
At t = 200 days , we find
σ 22 (200) =
.425x10−3
E = .53,125x10−3 E
.8
(5.7.4)
With E = E (u, u ) for u = 0, 7, 28, 77, 163 and 200 days, we get
σ 22 (200) = 10.6, 17.0, 20.2, 20.7, 21.3 and 21.3 MPa , respectively
(5.7.5)
So, with σ 22 (200) ≫ 4 MPa , an elastic treatment would unambiguously predict
the formation of shrinkage induced cracks.
3) According to the general expression (5.101) in Volume I and taking the shrinkage deformation and the isotropy into account, we can write, with σ (0+ ) = 0
t
t
t
0
0
0
∫ E (t , u)d (ε (u) + a(u)δ ) = ∫ (1 +ν (t , u ) ) d (σ (u ) ) − ∫ ν (t , u )d ( Trσ (u )δ ) (5.7.6)
Since ν (t , u ) is constant, this relation reads
t
∫ E (t , u)d (ε (u ) + a(u )δ ) = (1 +ν )σ (t ) −ν Trσ (t )δ
(5.7.7)
0
As before in 2), σ kk = σ 22 + σ 33 = 2σ 22 and ε 22 = 0 so that (5.7.7) yields
σ 22 (t ) =
1
1 −ν
t
∫ E (t , u )da(u )
(5.7.8)
0
Table 5.7.2 Stepwise approximation for a numerical integration of (5.7.8)
∆u1 = 7days
∆a1 = 10
−4
E1 = 104
∆u2 = 21days
∆a1 = 10
∆u3 = 49days
−4
E2 = 1.5x104
∆a1 = 10
∆u4 = 86days
−4
∆a1 = 10
E3 = 1.8x104
−4
E4 = 2.2x104
∆u5 = 37days
∆a1 = .25x10−4
E2 = 4x104
This integral can be approximated numerically by
σ 22 (t ) ≃
1
1 −ν
∑
5
i =1
Ei ∆ai
(5.7.9)
205
with the stepwise approximation of Table 5.7.2. This treatment yields an estimate
of σ 22 (t ) for different values of t, namely
σ 22 (7days) ≃ 1.25MPa
σ (28days) ≃ 3.125MPa
22
σ 22 (77days) ≃ 5.375MPa
σ (163days) ≃ 8.125MPa
22
σ 22 (200days) ≃ 9.375MPa
(5.7.10)
These results show that the risks of crack formation are lowered by the action
of viscosity but that these risks remain present (here after about 40 days) due to
the ageing behaviour of concrete.
Exercise 5.8 Viscoelastic two-phase material
An isotropic two-phase viscoelastic material is constituted of two isotropic incompressible non-ageing linear viscoelastic maxwellian phases (i), with the volume
fractions c1 and c2, which obey the constitutive equations
eɺ =
sɺ
Ei
+
s
ηi
(5.8.1)
where e and s are the strain and stress deviators, respectively. Due to a particulate morphology of the composite, the Mori-Tanaka model is used to get an estimate of the overall behaviour, with phase (2) considered as a spherical inclusion
in an infinite matrix made of phase (1).
Show that the effective behaviour can be described by one scalar creep function f MT (t ) . Prove that f MT* ( p ) is a rational fraction in p and, without trying to
derive f MT (t ) explicitly, give the conditions for it to describe the creep response
of a parallel assemblage of one Kelvin model (EK, ηK) and one Maxwell model
(EM, ηM). Derive the corresponding transformed effective relaxation function
r MT* ( p) and show that the associated relaxation spectrum reduces to two single
lines. Comment qualitatively on the differences between the relaxation spectra as
predicted by the self-consistent scheme (see Sect. 5.3.3.3 in Volume I) and the
Mori-Tanaka model.
Solution
206
Due to the constitutive behaviour of both phases, the overall behaviour of the twophase material is necessarily isotropic incompressible non-ageing linear viscoelastic and obeys the constitutive equations
S (t ) = 2µ MT (t ) ⊙ E (t ) or
*
*
S ( p ) = 2µ MT* ( p )E ( p)
(5.8.2)
where S (t ) and E (t ) are the macroscopic stress and strain deviators. These equations show that the effective behaviour can be described by one scalar creep function f MT (t ) whose Laplace-Carson transform is given by
f MT* ( p) =
1
1
=
MT*
r ( p) 2µ ( p)
MT*
(5.8.3)
The elastic Mori-Tanaka estimate of the shear modulus is known to be (see Eq.
2.194), with f = c2 :
µ MT = µ1
5 [ (1 − f ) µ1 + f µ 2 ] + 2(1 − f )( µ 2 − µ1 )
5µ1 + 2(1 − f )( µ 2 − µ1 )
(2 + 3 f )( µ 2 − µ1 )
= µ1 1 +
5µ1 + 2(1 − f )( µ2 − µ1 )
(5.8.4)
According to the correspondence theorem, we can then write for the viscoelastic problem
µ
MT*
5 (1 − f ) µ1* ( p ) + f µ 2* ( p ) + 2(1 − f ) ( µ2* ( p) − µ1* ( p ) )
( p) = µ ( p)
*
5µ1 ( p) + 2(1 − f ) ( µ2* ( p) − µ1* ( p) )
*
1
5µ1* ( p) + 2(1 − f ) ( µ2* ( p) − µ1* ( p) )
1
f MT* ( p) = *
*
*
*
*
2µ1 ( p ) 5 (1 − f ) µ1 ( p) + f µ 2 ( p) + 2(1 − f ) ( µ2 ( p) − µ1 ( p ) )
(5.8.5)
or
µ2* ( p)
µ1* ( p)
1
f MT* ( p) = *
µ * ( p)
2µ1 ( p )
3(1 − f ) + (2 + 3 f ) 2*
µ1 ( p)
(3 + 2 f ) + 2(1 − f )
with, from (5.8.1)
(5.8.6)
207
*
s ( p)
1
1 *
*
e ( p) = +
s ( p) = *
2 µi ( p )
Ei ηi p
(5.8.7)
and then
1
1
1
= +
2µ ( p ) Ei ηi p
*
i
or 2µi* ( p ) =
Eiηi p
Ei + ηi p
⇓
(5.8.8)
*
2
*
1
µ ( p) E2η 2 E1 + η1 p
=
µ ( p) E1η1 E2 + η2 p
We then have
f MT* ( p) =
=
( E1 + η1 p ) [ (3 + 2 f ) E1η1 ( E2 + η 2 p) + 2(1 − f ) E2η 2 ( E1 + η1 p)]
E1η1 p [3(1 − f ) E1η1 ( E2 + η2 p) + (2 + 3 f ) E2η 2 ( E1 + η1 p )]
(η1 p + E1 )( Ap + B )
E1η1 p(Cp + D)
(5.8.9)
E1 BC ( E1C − η1 D)( AD − BC )
A η1 AD
η1 ACD
=
+
1 +
D
E1C
p
p+
C
This result proves first that f MT* ( p ) is a rational fraction in p. After decomposition into simple elements, it happens then to read as the sum of two contributions: the two first terms in the brackets express the creep behaviour of a Maxwell
model (EM, ηM) with
EM =
E1C
ηD
ηM = 1
A
B
iff
C
≥ 0 and
A
D
≥0
B
(5.8.10)
It is easy to check in (5.8.9) that both these conditions are satisfied, since
f ≤1.
The third term in the brackets expresses the creep behaviour of a Kelvin model
(EK, ηK) with
208
E1η1CD 2
E
=
K
( E1C − η1 D )( AD − BC )
E1η1C 2 D
η =
K
( E1C − η1 D )( AD − BC )
iff
( E1C − η1 D )( AD − BC ) ≥ 0
(5.8.11)
This condition is fulfilled too since
A =η1η 2 [ (3 + 2 f ) E1 + 2(1 − f ) E2 ]
B =E1 E2 [ (3 + 2 f )η1 + 2(1 − f )η 2 ] E1C−η1 D = 3(1 − f ) E1η1 ( E1η 2 − E2η1 )
⇒
(5.8.12)
C =η1η 2 [3(1 − f ) E1 + (2 + 3 f ) E2 ] AD−BC = 25 fE1 E2η1η 2 ( E1η 2 − E2η1 )
D =E1 E2 [3(1 − f )η1 + (2 + 3 f )η 2 ]
From (5.8.9), the Laplace-Carson transformed relaxation function reads
D
p+
E
η
p
(
Cp
+
D
)
E
C
C
1 1
r MT* ( p ) =
= 1 p
(η1 p + E1 )( Ap + B)
A ( p + E1 )( p + B )
A
η1
(5.8.13)
After a decomposition of this rational fraction into simple elements, we find
A( E C − η D ) p η ( AD − BC ) p
E
1
1
1
r MT* ( p ) =
+ 1
E1
B
A( E1 A − η1 B)
p+
p+
η1
A
(5.8.14)
We know that E1/η1 and B/ A are positive and ( E1C − η1 D ) and ( AD − BC )
are given by (5.8.12); In addition we have
E1 A −η1 B = E1η1η 2 [ (3 + 2 f ) E1+ 2(1 − f ) E2 ] − E1 E2η1 [ (3 + 2 f )η1+ 2(1 − f )η 2 ]
= (3 + 2 f ) E1η1 ( E1η 2 − E2η1 )
(5.8.15)
Consequently we get
E1C − η1 D 3(1 − f ) E1η1 ( E1η 2 − E2η1 ) 3(1 − f )
E A − η B = (3 + 2 f ) E η ( E η − E η ) = (3 + 2 f )
1
1
1 1
1 2
2 1
AD
−
BC
25
f
E
E
η
η
(
E
η
−
E
25 f E2η 2
1 2 1 2
1 2
2η1 )
=
=
E1 A − η1 B (3 + 2 f ) E1η1 ( E1η2 − E2η1 )
3+ 2 f
(5.8.16)
209
and (5.8.14) reads simply
r MT* ( p ) =
3(1 − f ) E1 p
25 f E1 E2η1η 2 p
+
3 + 2 f p + E1
(3 + 2 f ) A p + B
A
η1
(5.8.17)
The original relaxation function r MT (t ) is then
r MT (t ) =
E 25 f E1 E2η1η 2
3(1 − f ) E1
B
exp − 1 t +
exp − t
η
3+ 2 f
(3
+
2
f
)
A
A
1
(5.8.18)
This is the relaxation function of a parallel assemblage of two Maxwell models
with the relaxation times τ 1 = η1 E1 and τ 2 = A B and with the amplitudes
3(1 − f ) E1
25 f E1 E2η1η2
G1 =
and G2 =
, respectively, which corresponds to a re3+ 2 f
(3 + 2 f ) A
laxation spectrum reducing to two single lines. The prominent morphological role
of the matrix according to the Mori-Tanaka model is expressed by the fact that one
of these relaxation times, τ1, is the one of the matrix, with an amplitude which
only depends of the elastic modulus of the matrix, E1 (and, of course, of the volume fractions). These properties strongly differ from those which would have derived from the use of the self-consistent scheme, which makes both phases play
analogous morphological roles and then forces the associated relaxation spectrum
to be continuous (see Sect. 5.3.3.3 in Volume I). This is also reflected by the fact
that the “long range memory effect”, which reduces to the additional presence of a
Kelvin-type response (5.8.11) to the basic Maxwellian one according to the MoriTanaka model, is much more effective (see the square root in Eq. 5.133 in Volume
I) when the self-consistent scheme is used.
210
Chapter 6 Exercises of Appendix I-A (Annex 1): Atomic and
Molecular Structures
Exercise 6.1 Binding energy
Calculate the repulsion potential of NaCl, given the lattice parameter 3.96x10-10m
and the binding energy 777.9 kJ/mole.
Solution
The repulsion potential is the difference between the binding energy and the Coulomb attractive potential, which is:
Uc = N
e2
1
M
4πε0
r0
(6.1.1)
where N is the Avogadro number 6x1023, e the charge of the electron 1.6x10-19 C,
M the Madelung constant 1.75 for NaCl structure and 4πε0 = 1.112 C2/Jm. The
distance between ions r0 is 3.96x10-10/√2 = 2.8x10-10 m. Thus the Coulomb attractive potential is equal to −863 kJ/mole and the repulsive potential is -778 + 863 =
85 kJ/mole.
Exercise 6.2 Miller indices
Which {110} planes contain the direction [111]?
Solution
(1 10) (10 1 ) (01 1 ) for which hu + kv + lw = 0
Exercise 6.3 Miller indices
Is the direction [123] in the plane (111)?
211
Solution
No: 1x1 + 2x1 + 3x1 ≠ 0
Is the direction [123] in the plane (11 1 ) ?
Solution
Yes: 1x1 + 2x1 + 3x(−1) = 0
Exercise 6.4 Densest planes
Find the densest planes and directions in FCC, BCC and CPH structures.
Solution
In FCC, the closest atom to the one at the origin of coordinates is at ½a, ½a, 0.
Therefore the densest directions are <110>. 110 , 10 1 and 0 11 directions
are perpendicular to the 111 one. Thus the (111) plane, which contains three
densest directions, is a densest plane. The densest planes are {111} planes.
In BCC, the closest atom to the one at the origin of coordinates is at ½a, ½a,
½a. Therefore the densest directions are <111>, 111 and 1 11 directions are
( )
perpendicular to the 110 one. Thus the 110 plane, which contains two densest directions, is a densest plane. The densest planes are {110} planes.
(
)
In CPH, the densest directions < 10 10 > are in the basal plane 0001 , which is
the densest plane.
Exercise 6.5 FCC, BCC and CPH structures
a) For the three structures FCC, BCC and CPH, what is the number of atoms in
the unit cell?
Solution
FCC: 8 atoms at the cube corners, each belonging to 8 unit cells + 6 atoms at the
face centres, each belonging to 2 cells = 4 atoms per unit cell.
BCC: 8 atoms at the cube corners, each belonging to 8 unit cells + 1 atom at the
cell centre = 2 atoms per unit cell.
212
CPH: Twice 6 atoms at the hexagons summits, each belonging to 6 unit cells +
twice 1 atom at the hexagon centre, each belonging to 2 cells + 3 atoms within the
cell = 6 atoms per unit cell. (We consider really super unit cells).
b) Find the ratio of the atomic volume to the unit cell volume.
Solution
FCC: In the unit cube of side a, along a diagonal of length a 2 there are 2 close
packed atoms; their radius being a 2 4 , the atomic volume is π 2a 3 24 . With
4 atoms per unit cell, the sought ratio is π 2 6 = 0.74.
BCC: In the unit cube of side a, along a diagonal of length a 3 there are 2 close
packed atoms; their radius being a 3 4 , the atomic volume is π 3a3 16 . With 2
atoms per unit cell, the sought ratio is π 3 8 = 0.68.
CPH: close packed like FCC.
c) Find the dimension of the insertion site.
Solution
FCC: The insertion site is at ¼a, ¼a, ¼a.The distance to the centres of the surrounding atoms is a 3 16 . The radius of the insertion site is therefore
(
)
3 4 − 2 4 a = 0.079a . Its ratio to the atomic radius is 0.225.
BCC: The insertion site at the centre of a cube face (or similarly at the centre of a
side) is not symmetrical (Fig.6.5.1). Its small axis is, r being the atomic radius:
(
) (
a 2−r = 1 2− 3 4 a = 2
Fig.6.5.1. Insertion site in BCC
)
3 − 1 r = 0.154r
213
CPH: Close packed: same as FCC.
d) What are the reciprocal lattices?
Solution
FCC: The reciprocal lattice is BCC with a*. a = 1
BCC: The reciprocal lattice is FCC with a*. a = 1
CPH: The reciprocal lattice is hexagonal rotated through 30° about the c axis with
c* = 2π/c and a∗ = 4π a 3
e) How many (a) nearest (b) second nearest neighbours are there?
Solution
FCC: (a) 12 nearest neighbours (b) 12 second nearest neighbours.
BCC: (a) 8 nearest neighbours (b) 6 second nearest neighbours
CPH: closed packed as FCC
Exercise 6.6 The c/a ratio in hexagonal structures
What is the value of the c/a ratio for close packing in the hexagonal system?
Solution
The height of the Thomson tetrahedron is a
3 . Therefore c a = 2
3.
Exercise 6.7 Angle between directions
Find the angle between the directions [123] and [110] as well as between [111]
and [122] in cubic systems.
Solution
If θ is the angle between [123] and [110], we have
cos θ =
and so θ = 55.46°.
1⋅1 + 2 ⋅1 + 3⋅ 0
12
12
(1 + 4 + 9) (1 + 1)
=
3
28
(6.7.1)
214
If θ is the angle between [111] and [122], we find
cos θ =
1x1 + 1x2 + 1x2
12
12
(1 + 1 + 1) (1 + 4 + 4)
=
5
27
(6.7.2)
and so θ = 15.79°
Exercise 6.8 Planes in zone
Find the condition that must be satisfied by the indices h, k, l, for planes in zone
whose direction is [uvw].
Solution
hu + kv + lw = 0
(6.8.1)
Exercise 6.9 Orientation of slip lines
Using the stereographic projection, find the possible orientation of the grains on
which the slip lines of Fig.A1.30 (Volume I, p. 540) are seen.
Solution
Figure A1.30a shows single slip lines. Aluminium being FCC, they correspond to
the activation of a (111) slip plane. The number of possible orientations of the
grain is infinite.
Figure A1.30b shows two families of slip line making an angle of 73°
(Fig.6.9.1 below). Now two (111) planes make an angle arcos(1/3) = 70.5°. It can
be assumed that the slip lines result from the activation of two (111) slip planes
perpendicular to the surface on which the slip lines are seen. The tensile stress acting on the main slip system increased this angle, explaining the difference between 70.5° and 73°. The surface is perpendicular to a [110] direction.
215
Fig. 6.9.1. The two families of slip lines make an angle of 73°, slightly larger than arcos (1/3),
owing to tensile deformation.
Exercise 6.10 Laue spots
Explain the shape of the Laue spots in Fig.A1.31 (Volume I).
Solution
The spots are elongated. This corresponds to a continuous distribution of diffraction angles denoting a strain-hardened crystal.
Exercise 6.11 Laue transmission
Explain the Laue transmission photograph of a steel sheet in Fig.A1.32 (Volume
I).
Solution
216
Each ring corresponds to diffraction by a family of (hkl) planes. They are not continuous owing to a limited number of grains in the annealed sheet.
Exercise 6.12 Diffraction spectrum
Interpret the diffraction spectrum of Fig.A1.33 (Volume I).
Solution
The factor Q = h 2 + k 2 + l 2 = ( 4a 2 λ 2 ) sin 2 θ (refer to p.522, Volume I) can be
found from the approximate diffraction angles of the various rays. Table 6.12.1
shows that they can be interpreted as the diffraction spectrum of a BCC structure.
The lattice parameter is deduced given the wave length of Cu Kα radiation: λ =
1.6x10-10 m.
Table 6.12.1. Interpretation of the diffraction spectrum given in figure A1.33 (Volume I)
Angle 2θ (°)
40
58
75
88
101
115
132
155
sin2θ
0.117
0.235
0.371
0.483
0.595
0.711
0.834
0.953
sin2θ/sin22θ
1
2
3.17
4.13
5.08
6.07
7.12
8.14
Q
2
4
6
8
10
12
14
16
{hkl}
{110}
{200}
{112}
{220}
{310}
{222}
{123}
{400}
a (10-10m)
3.30
3.23
3.22
3.26
3.28
3.29
3.28
3.28
The average value of the lattice parameter is 3.27x10-10m. Elements Ta and Nb
with a lattice parameter of 3.30x10-10m could be the diffracting elements. It can be
checked that this lattice parameter gives diffraction angles 2θ which fit well the
ones on the figure A1.33 (Volume I).
Exercise 6.13 Internal stress measurements
a) Find the change in the diffraction angles for a polycrystal subjected to a tension
E/105, where E is the Young modulus.
Solution
From the Bragg relation:
217
2d sin θ = nλ
(6.13.1)
we have
σ
E
=
∆d
∆θ
=−
d
tan θ
(6.13.2)
Hence ∆θ = −10−5 tan θ .
b) Show how X-ray diffraction can be used to measure an applied tension.
Solution
Fig.6.13.1. X-ray diffraction for measuring the tensile stress.
Measure the diffraction angle θ for {hkl} at an angle φ (Fig.6.13.1). Then:
∆d
σ
= (1 +ν ) sin 2 φ − 1
d
E
(6.13.3)
The operation can be repeated for various angles φ. Plotting ∆d/d, from (6.13.3)
as a function of sin2φ yields σ/E.
Exercise 6.14 Parity rules
Prove the parity rules for the indices of reflecting planes for BCC and FCC
structures.
Solution
218
The structure factor for atom hxj, kyj, lzj is: Fj ( hkl ) = e
(
2 πi hx j + ky j + lz j
)
. The reflection
is extinct if this factor is equal to 0.
For BCC:
Fj ( hkl ) = e
=e
(
(
2 πi hx j + ky j + lz j
2 πi hx j + ky j + lz j
)
1 + e
)
+e
πi ( h + k + l )
(
) (
) (
)
2 πi h x j +1 2 + k y j +1 2 + l z j +1 2
(6.14.1)
Hence the reflection is extinct if:
e
πi ( h + k + l )
= −1
(6.14.2)
implying (h + k + l) is odd. The sum of the indices for reflecting planes is even for
the BCC structure {110}, {200}, {112}, etc.
For FCC the reflection is extinct if h + k +2l, h + 2k +l, 2h + k +l are odd. Thus
h, k and l are of different parities. The indices of the reflecting planes are of the
same parities for the FCC structure {111}, {200}, {220} etc.
219
Chapter 7 Exercises of Appendix I-B (Annex 2): Phase
Transformations
Phase transformations in materials play a key role on their mechanical properties. The basis of the study of phase transformations is briefly introduced
in Appendix B of Volume I. A number of exercises (A2.11.1 to 9) are proposed at the end of this Appendix. The solutions of these exercises are given
below. Initial numbers of these exercises are indicated in brackets
Exercise 7.1 (A2.11.1) Equilibrium diagram. Purification by zone melting
A beam of an alloy AB, whose equilibrium diagram is given below, is placed in a
furnace which is moved slowly so that a moving liquid region, of length l, is
formed from the left to the right (Fig.7.1.1). Use the phase diagram (Fig. 7.1.2) to
explain the redistribution of the solute B between the liquid and solid phases. Can
the operation be repeated? If so, in what circumstances? This method is used to
purify Ga and Si semiconductors and to produce certain high-purity metals.
Fig.7.1.1 Principle of zone refining.
Solution
In a binary alloy in which the addition of the element B lowers the melting temperature of solute A, the phase diagram shows the existence of a different solubility of the element B between the solid and the liquid phases (Figs 7.1.2 a and b).
This difference in concentration is defined with the partitioning coefficient, k, of
element B in A such as k = CS/CL. In Fig. 7.1.2a, k <1, while in Fig. 7.1.2b, k > 1.
220
Fig. 7.1.2 Phase diagrams (a) k<1; (b) k>1.
It can easily be shown that with a diagram such as that of Fig. 7.1.2a, it is possible to purify the solid phase when moving the melting device from the left to the
right. This can be shown by writing the mass balance during a small translation of
the melted zone by dx. The melted zone gives rise to the formation of a small slice
of solid dx with a concentration CS such as CS dx of impurities is lost (for a bar
with a unit section) as indicated in Fig. 7.1.3. But the melting of another slice of
thickness dx in the initial solid located on the right leads to an increase of impurity
by C0 dx
Fig. 7.1.3 Mass balance.
The solute equilibrium is such that:
C0dx − CS dx = ldCL
(7.1.1)
Using the relation between CS and CL (k = CS/CL), and the boundary condition:
x = 0, CS = kC0, Eq. 7.1.1 can easily be integrated to give:
kx
CS ( x ) = C0 1 + ( k − 1) exp −
l
(7.1.2)
221
The variation of the ratio CS/C0 is schematically shown in Fig. 7.1.4 for different values of the partitioning factor, k. These curves show a horizontal asymptote
CS/C0 = 1.
Fig. 7.1.4. Theoretical profiles of concentration along a bar as a function of the value of the partitioning coefficient, k.
In (7.1.2) the length of the bar does not appear because we have assumed that
the bar is infinitively long. The asymptotical regime is reached more rapidly when
k increases and when the length of the melted zone, l , is small.
In practice zone melting with multisource is used in order to accelerate the
process of purification (Fig. 7.1.5).
Fig. 7.1.5 Zone refining with muti-sources.
Exercise 7.2 (A2.11.2) Steel microstructures
Steels of various carbon contents are cooled slowly from the austenite state. Fig.
A2.52 (Volume I) gives micrographs of the structures. What is the carbon content
of each? Use the Fe-Fe3C part of the iron-carbon equilibrium diagram to justify
your answers.
Solution
222
Remind the lever rule. Using this rule, the composition of the four steels can be
given approximately.
Steel a contains some pearlite (~ 10%). This means that the carbon content is
larger than 0.02 (see Fig. A.26). Using the lever rule, this leads to C~ 0.08.
Steel b contains a larger amount of pearlite (~ 0.40). This means that C ~ 0.33.
Steel c is fully pearlitic. This indicates that C ~ 0.80.
Steel d contains some intergranular cementite. This steel is slightly hypereutectoid, i.e. C > 0.80
Exercise 7.3 (A2.11.3) Martensitic transformation
Two steels with 12% Cr and 5% and 7% Ni respectively are thermally treated.
They are quenched from 1,000°C: what can be predicted from the Fe-Cr-Ni equilibrium phase diagram given in Fig. A2.53 (Volume I)? (martensite laths are
formed in both cases). They are then subjected to a repeated annealing: 1h at
700°C – cooling in air – cooling to –196°C – 5h at 500°C, after which the residual austenite content is measured at ambient temperature. The data are given in
the table. Explain what has happened; would you expect an increase in the nickel
content in the austenite? Electron microscopy shows that the γ phase is distributed
in bands: how do you explain this? Figure A2.54 (Volume I) gives the results of
tensile tests to determine the yield strength of the two specimens: how do you explain the rapid variation of this with temperature?
Solution
Figure A2.53 (Volume I) shows that at 1,000°C these materials are fully austenitic. Quenching produces lath martensite with eventually some retained austenite.
The equation giving MS page 583 (Volume I) predicts that MS(12Cr−5Ni) =
272°C, MS(12Cr−7Ni) = 238°C.
Annealing at 500°C will produce a two-phase microstructure: austenite reverted
from tempered martensite (or ferrite) + ferrite (or tempered martensite). The austenite will be enriched in nickel (which is a γ former). This reverted austenite can
partly be transformed into martensite during cooling, but residual austenite will
remain present. This austenite is metastable and will be transformed into martensite (strain induced phase transformation) during a tensile test. This explains the
anomalous variation of the yield strength of these materials when tested at 0°C
and below.
Exercise 7.4 (A2.11.4) Steel microstructures
Various heat treatments have been applied to a Cr-Mo steel (0.39C, 1.5Cr,
0.5Mo), the TTC curve for which is given in Fig. A2.55 (Volume I). This steel is
223
much used in mechanical engineering for items such as gears. Why is such care
taken over the conditions for austenitising (maintaining in the γ phase) and the resulting grain size of the austenite? Four structures resulting from various cooling
rates are shown in the figure. Identify these microstructures.
Solution
Low austenitising temperatures are used to avoid a detrimental growth of the austenite grain size. The material with a hardness of 634 DPH (DPH: diamond pyramid hardness is the same as Vickers hardness (HV)) was obtained with a high
cooling rate (~ 25°C/s). The microstructure is fully martensitic. The material with
a hardness of 327 DPH was obtained with a cooling rate of about 1°C/s. The CCT
diagram (see Sect. A.2.6.2, Volume I) shows that the microstructure is fully
bainitic. The material with a hardness of 281 DPH was obtained with a cooling
rate of about 10°C/min: it contains some ferrite (white phase) in a bainite matrix.
The material corresponding to a hardness of 227 DPH was slowly cooled at about
40°C/h. It contains a significant amount of pro-eutectoid ferrite (white phase).
Exercise 7.5 (A2.11.5) Surface treatment: Cementation
Carbon cementation: Cementation is a thermo-chemical treatment, the aim of
which is to increase the carbon content of the sample at the surface. This enrichment is achieved by keeping the surface in contact with a carbon-donating material in powder, paste or liquid form. The diffusion process is always followed by
the thermal treatment of quenching: why is this? The quenching creates residual
compressive stresses in the surface: why?
Suppose the operation takes place in the gaseous phase (Fig. A2.56, Volume
I); show that the solution to Fick’s second equation is:
x
C − C0
= erf
CS − C0
2 Dt
(7.5.1)
2 x
exp −u 2 du , with C0 the initial carbon content of the steel
π 0
and CS the saturation value for the γ phase at the temperature of the treatment.
Figure A2.56 (Volume I) gives carbon concentration profiles for a steel with
C0 = 0.15%, CS = 1.3% at 925°C. Use these to find the diffusion coefficient of carbon in the γ phase and compare your value with that given in the tables in Sect.
A2.3.2 (Table A2.4, Volume I). What do you think of the result? Why is it necessary to conduct this process at a temperature above that of the eutectoid?
where erf ( x ) =
Solution
∫
( )
224
Quenching has to be applied for the formation of martensite.
The formation of martensite occurs preferentially at the free surface (which is
cooler than the bulk). As this phase transformation is accompanied by an expansion (a few %, see p. 580, Volume I) this volume change produces compressive
residual stresses at the free surface, which is beneficial for the fatigue properties of
the components. Modeling of these residual stresses is complex. In particular the
“transformation plasticity” (see Sect. 3.4.4.4 in Volume I) must be taken into account.
See that the expression:
x
C − C0
= erf
CS − C0
2 Dt
(7.5.2)
is demonstrated in classical books on diffusion (see for example Crank 1956).
Numerical calculations must be made to identify the diffusion coefficient from
the results reported in Fig.A2.56 (Volume I).
Simplified calculations can be made using an approximation for the erf function which is valid for small distances, x from the free surface and for large times,
t, i.e.:
1/2
1 x2
C − C0
≃ 1 − exp −
π Dt
CS − C0
(7.5.3)
Taking, for instance x = 0.5 mm, one obtains with Eq. 7.5.2, at t = 2 hours, D ~
1.20x10-10 m2/s; t = 4 hours, D ~ 0.30x10-10 m2/s.
Using Table A2.4 (Volume I, p. 558), one obtains at 925°C (1,198 K), D = 0.28
10-10 m2/s, which is a value very close to those derived from the carbon profiles
shown in Fig. A2.56 (Volume I).
This cementation heat-treatment must be performed in the γ field (see Fe−C
diagram in Fig. A2.6 in Volume I) where the solubility of carbon is sufficiently
high. This diagram shows that, at 925°C, the material is in the γ field.
Exercise 7.6 (A2.11.6) Solidification
Solidification of a Fe-C-Ni alloy has resulted in segregation of the additive elements C, Ni. The distances over which these segregations occur depend on the
conditions during the solidification and on the growth of dendrites. Taking the
data in Table A2.4 (Volume I) concerning the diffusion coefficients for these elements in Feγ , show that there is a possibility of homogenising this alloy by holding
it in the austenitic phase for a long time. What quicker methods can you suggest
225
for the process, bearing in mind that band structures such as those shown in Fig.
A2.37 (Volume I), are to be avoided?
Solution
The dendrites spacing is between 0.1 mm and 1 mm in many solidification processes.
A heat treatment of homogenisation is efficient if the alloying elements (C, Ni)
have enough time to diffuse over these distances. This heat treatment has to be
performed at elevated temperature. However the temperature must be below the
local melting point which is, most often, decreased by the presence of alloying
elements. See Fe-C phase diagram (Fig. A2.6, Volume I) and Fe-Ni diagram (Fig.
A2.7, Volume I).
A simple relationship can be used to estimate the diffusion distance, x:
x=
Dt
(7.6.1)
where D is the diffusion coefficient and t, the time for homogenisation. Assuming
that the heat treatment is made at 1,200°C (1,473K) and using the values of the
diffusion coefficients for C and Ni given in Table A2.4 (Volume I), it is found
that:
- For x = 0.1 mm, tNi = 30 hours, and tC = 26s. Carbon which is an interstitial
element is known to diffuse must faster than Ni which a substitutional element in
Fe. The inhomogeneities can therefore be largely reduced with a heat treatment
lasting about one day.
- This is not the case when x = 1mm. In these conditions it is found that tNi =
2,800 hours (which is clearly not realistic) and tC = 0.8 hour.
One way to reduce the times of homogenisation is to predeform the material to
reduce the mean distance between the dendrites. This predeformation must be
made at elevated temperature (1,000 – 1,100°C). The existence of bands can be related to some segregation.
Exercise 7.7 (A2.11.7) Hardening by precipitation and coalescence of the
precipitates in a ferritic stainless steel
Experiments have been made with steels of the following percentage composition:
(a) Fe – 19.6Cr – 2.03Ni – 0.97Al
(b) Fe – 19.6Cr – 4.15Ni – 1.87Al
1) Explain the choice of these compositions, knowing that the hardening phase
produced by the precipitation has the composition NiAl (structure B2, CSCl, see
Appendix A, Volume I). Is it normal for these steels to have a BCC structure?
2) Samples of the steels have been held at 1,150°C for 1 h; what structure will
they have? After this they are cooled quickly and then aged at various tempera-
226
tures. The curves of Fig. A2.57 (Volume I) give the variations of hardness with
time of ageing and the micrographs of Fig. A2.58 (Volume I) show the precipitation of the NiAl phase.
a) Is the form of the hardness curves what you would expect?
b) Comment on the legend on the micrographs. Why is there a diffraction spot
of structure of the type (001) in the [001] section?
Solution
1) These materials containing about 20%Cr are stainless. The presence of Ni and
Al leads to the precipitation of the ordered B2 (NiAl) phase. Electron diffraction
shows the existence of this phase after ageing: an extra spot (100) is observed
(Fig. A2.58d in Volume I). These precipitates are ordered particles whose lattice
parameter is close to that of the matrix and they are in cube-cube orientation relationship with the matrix: (100)matrix // (100)precipitate, and [001]matrix // [001]precipitate
The materials have a BCC structure as indicated by the Schaeffler diagram.
This is due to the fact that Cr and Al are strong ferrite stabilizers.
2) At 1,150°C, the materials keep a BCC structure and Ni and Al atoms are in
solid solution in a high chromium ferrite.
Figure A2.57 (Volume I) shows a typical behaviour for precipitation hardening
with the existence of a peak. This suggests that at the early beginning of ageing
small precipitates are sheared by dislocations with a Burgers vector ½ <110>. The
dislocations are then grouped in pairs separated by a stacking fault (see Fig. 3.93
in Volume 1). After the maximum of the hardness, the particles are bigger and
they are by-passed by the Orowan process with the formation of dislocations loops
left around the precipitates (see Fig. 3.90 in Volume 1).
Exercise 7.8 (A2.11.8) Precipitation
Figure A2.59 (Volume I) shows the variation of precipitate size (d) with time at
various temperatures, suggesting a growth law of the form:
A=
64γ S DCE∞ Ω 2
9kT
(7.8.1)
Deduce the value of the apparent activation energy Qd for diffusion of Al and
Ni in α−iron (Fe−20Cr), assuming that γS and Ω are independent of temperature.
The variation of the solubility has been determined – how could this be done? –
and the results tabulated; discuss the values found for Qd.
Solution
227
The growth law:
d 3 − d 03 = A ( t − t0 ) with A =
64γ S DCE∞ Ω 2
9kT
(7.8.2)
corresponds to the LSW theory (see Sect. A2.5.3.2 in Volume I). In this expression t0 is the time at which the system enters the coalescence regime, when the
mean particle size is d0, CE∞ is the concentration of the solute at infinity, γS is the
surface energy between the precipitate and the matrix, D is the diffusion coefficient, Ω is the molar volume of the species entering into the particles.
The results shown in Fig. A2.59 (Volume I) indicate that this law is well
obeyed in the alloys presented in Exercise 7.7. The concentration CE∞ is slightly
dependent on temperature. The activation energy of diffusion can be determined,
assuming that γS and Ω are independent of temperature. This equation can be written as:
log
AT
Q
= cste −
C∞
kT
(7.8.3)
The values of A can be determined from the results of Fig. A2.59 (Volume I),
knowing the value of t0 (for more details, see Taillard, Pineau, 1982).
Fig. 7.8.1. Coarsening of NiAl precipitates in a Fe-Cr ferritic steel (Taillard, Pineau, 1982).
228
The representation of Eq. 7.8.3 is shown in Fig. 7.8.1 where a straight line can
be drawn through the data points. The slope of this line leads to a value of
Q = 302 ± 81KJmol−1 . This value is compatible with the diffusion activation energies of the elements present in the alloys which are given in Table 7.8.1. This table
also includes the estimated values for the diffusion coefficients of aluminium,
nickel, chromium and iron in α−Fe. From this table it can be concluded that
chromium probably does not intervene as the element controlling the coarsening
rate but it is impossible to distinguish between the other elements, although the
apparent energy Q is closer to that of Al.
Table 7.8.1. Values of the effective diffusion coefficients in Fe − 19.6Cr − 4.15Ni − 1.8Al and
DAl, DNi, DCr and DFe in α−Fe (Taillard, Pineau, 1982).
Temperature
(K)
DCr (m2s-1)
DFe (m2s-1)
DNi (m2s-1)
DAl (m2s-1)
D (m2s-1)
823
5.0 x 10-22
2.3 x 10-20
3.6 x 10-20
9.3 x 10-18
1.7 x 10-19
873
8.9 x 10
-21
-19
-19
-17
3.9 x 10-18
923
1.1 x 10-20
1.9 x 10-16
7.2 x 10-17
1.9 x 10
1.2 x 10-18
2.8 x 10
1.8 x 10-18
4.6 x 10
Exercise 7.9 (A2.11.9) Effect of applied stress on the morphology of
precipitates in nickel-based alloys
As Fig. A2.60 (Volume I) shows, the morphology of single-crystal materials containing a large volume fraction of precipitates (~ 50−60%) is considerably
changed by the application of stress. In this case (alloy Udimet 700), tension leads
to a rafting process and a redistribution of precipitates, initially cubic and distributed isotropically, into layers perpendicular to the direction of the stress. In
contrast to this, compression rearranges them into rods or needles aligned parallel to the stress. These changes have an important effect on the creep behaviour of
the material.
Give a qualitative explanation of these changes, bearing in mind the following:
- the precipitates are coherent with the matrix
- any departure from coherence, measured by the relative difference in the values of the parameters for the γ’ phase and the matrix, δ = (aγ’ − aγ)/aγ, can be positive, negative or zero.
- the elastic constants, Young’s modulus in particular, are different fro the precipitates and the matrix.
Further information can be found in:
Pineau A (1976) Influence of uniaxial stress on the morphology of coherent
precipitates during coarsening-elastic energy considerations. Acta Metall 24:559564
Nabarro FRN (1996) Rafting in superalloys. Metall Mater Trans 27A:513-530
229
Solution
The most advanced nickel-based alloys used for the fabrication of turbine blades
are single crystalline with the <001> axis parallel to the longitudinal direction of
the blades. These materials contain a very high volume fraction (~60%) of γ’ precipitates (Ni3Al, L12 structure, see Figure 1.41 in Volume I). These particles have
initially a cubic shape. The blades are submitted to complex thermo-mechanical
loads, in particular an axial load due to the centrifugal force. As these materials
are working at very elevated temperatures (~1,000°C), the precipitates are not stable. They grow and they can rearrange themselves to adopt a plate-like morphology (Fig. A2.60a in Volume I) or a needle shape (Fig. A.2.60b in Volume I), depending on the orientation of the applied stress and the misfit parameter δ = (aγ’ –
am)/am, where aγ’ and am are the lattice parameters of the precipitates and that of
the matrix, respectively. The stress free misfit is a pure dilatation. This partly explains why the particles adopt initially a cube shape in the stress free conditions.
Many studies have been devoted to the ageing of these materials under an applied stress. The rearrangement of the particles in the presence of an applied stress
is named “rafting” in the literature (see e.g. Nabarro, 1996). In this exercise an attempt is made to explain qualitatively these shape changes in terms of the elastic
theory of inclusions (see Sect.2.7 in Volume 1).
The matrix and the precipitates are assumed to be elastically isotropic. Equation 2.174 in Volume I can be extended to include the inhomogeneity effect
( Em ≠ Ep ) , where E
m
and Ep are the Young modulus of the matrix and that of the
precipitates. The total energy per unit volume, ξ t (ξ =Wst /V in Eq.2.174 ) can be
written as the sum of three terms (see Pineau, 1967):
ξt = ξincl + ξint + ξinh
(7.9.1)
where:
1
2
ξincl = − σ ij I eij T
ξint = −σ A e33T
1
2
ξinh = − σ A e33T
∗
(a )
(b)
(7.9.2)
(c)
In this expression, σA is the applied stress (σA > 0 for a tensile stress; σA< 0 for
a compressive stress; eijT* is the transformation strain for the inclusion effect
∗
∗
∗
∗
(
)
e11T = e22T = e33T = aɺɺ and eijT i ≠ j = 0 ; σ ijI is the stress inside the inclusion in
230
T
is the equivalent strain for an inhomogeneous
the absence of applied stress; e33
( Ep ≠ Em ) inclusion. This equivalence was defined by Eshelby (1961).
The inhomogeneity effect (ξinh ) is only due to the fact that, in the absence of
coherency misfit, the particles have elastic constants different from those of the
matrix. The inclusion effect (ξincl ) is related to the misfit. The interaction effect
(ξint )
is related to the fact that the precipitation occurs under an applied stress.
After some lengthy calculations it can be shown that:
2
σ
σ
= A A + B A + C
2
Em aɺɺ
Em aɺɺ
Em aɺɺ
ξt
(7.9.3)
In this equation, the coefficients A, B and C are dependent on the particle shape
and on the ratio Ep/Em.
The principle of the calculations is to determine the shape (sphere, plates perpendicular to the applied stress, or needles aligned with the stress axis) which
minimizes the value of the total strain energy given by Eq. 7.9.3.
It should be realised that, due to the difference in elastic moduli between the
matrix and the precipitates, anisotropy in the misfit is produced, as schematically
shown in Fig. 7.9.1 where the deformations of the matrix (solid lines) and that of
the precipitates (dotted lines) are indicated. Let us consider for instance the case
where σA > 0, δ > 0 and Ep < Em. Figure 7.9.1b shows that under a tensile stress the
deformation which has to be applied to the inclusion to be relocated within the
matrix is larger along the stress axis than in the other directions. The situation is
thus similar to that corresponding to a particle located in a matrix without any apT
T
plied stress but such as: e11
=e22
T
T
and e33
>e11
. Inclusion theory shows that in
these conditions the strain energy for a plate perpendicular to the stress axis is
lower than that of a cube. Other qualitative variations for the shape of the particles
based on this energy analysis are shown in Figs 7.9.1c, d and e. In particular the
needle-like morphology observed in Fig. A2.60 is in qualitative agreement with
the predicted shape indicated in Fig. 7.9.1d.
During coarsening the variation in the surface energy has also to be considered.
When the shape change occurs by the agglomeration of particles, the surface energy is always decreased when initially independent cube-shaped particles are
transformed into needles or plates.
The observations reported in Fig. A2.60 were made by Tien and Copley (1971)
on single crystals of Udimet 700 in which δ = +2 x10-4.
231
Fig. 7.9.1 Stress-induced coarsening of precipitates in nickel-based superalloys. Sketch showing
the influence of the orientation of the applied stress and of the difference in Young modulus of
the matrix and the precipitates on the relative shape change of both phases.
References
Crank J (1956) The mathematics of diffusion. Oxford Science Publications, Oxford University
Press.
Eshelby JD (1961) Elastic inclusions and inhomogeneities in Progress in solid mechanics II. In:
Sneddon I.N., Hill R. (ed)),vol 3, North-Holland, Amsterdam, 87-140
Nabarro FRN (1996) Rafting in superalloys. Metall Mater Trans 27A:513-530
Pineau A (1976) Influence of uniaxial stress on the morphology of coherent precipitates during
coarsening-elastic energy considerations. Acta Metall 24:559-564
Taillard R, Pineau A (1982) The precipitation of the intermetallic compound NiAl in Fe19Wt%Cr alloys. Mater Sci Eng 54:209-219
Tien JK, Copley SM (1971) The effect of uniaxial stress on the periodic morphology of coherent
gamma prime precipitates in nickel-base superalloy crystals. Metall Trans 2:215-219
232
Chapter 8 Exercises of Appendix I-C (Annex 3): Continuum
Mechanics: Basic Concepts and Equations
8.1 Deformations
Exercise 8.1 Calculation of strain fields in Cartesian coordinates
Calculate the strain tensors for the following transformations T, first for finite
transformations and second assuming small perturbations. Interpret these transformations and find the principal directions and strains for the case of small perturbations.
In orthonormal Cartesians T is defined by
1)
x = a + λa
y = b + µb
z = c +ν c
2)
x = a
y = b + 2γ a
z = c
(8.1.1)
Solution
1) We have
dx = (1 + λ )da
d x = dy = (1 + µ )db
dz = (1 +ν )dc
da
d a = db
dc
(8.1.2)
and then
2
2
2∆ij dai da j = d x − d a = λ (2 + λ )da 2 + µ (2 + µ )db 2 + ν (2 + ν )dc 2
(8.1.3)
Equating coefficients, we find for the Green-Lagrange strain tensor
0
0
λ (1 + λ 2)
0
µ (1 + µ 2)
(∆) = 0
0
0
ν (1 + ν 2)
(8.1.4)
233
For small perturbations, we get
0
0
µ 0
0 ν
λ
(ε ) = 0
0
(8.1.5)
So, the principal directions coincide with the coordinate system and λ is the
dilatation in the direction e1, etc.
2) Similarly, we find
dx = da
d x = dy = 2γ da + db
dz = dc
da
d a = db
dc
(8.1.6)
and then
2
2
2∆ij dai da j = d x − d a = 4γ 2 da 2 + 4γ dadb
(8.1.7)
So, the Green-Lagrange strain tensor is
2γ 2
(∆) = γ
0
0
0 0
0 0
γ
(8.1.8)
which for small perturbations reduces to
0 γ 0
0 0
0 0 0
(ε ) = γ
(8.1.9)
This is a shear strain fied; the principal strains are ±γ and 0 and the principal
directions e1 + e 2 , e1 − e2 , e3 .
Exercise 8.2 Calculation of strain fields in cylindrical polars
In cylindrical polars (e r , eθ , e z ) , with M = M ( r,θ , z ) , T takes M to M' with
234
1)
2)
M ′ = M + u( r )e r + w( z )e z
M ′ = M + v ( r )e θ
3)
M ′ = M + aze θ
4)
M′= M +
(8.2.1)
bθ
ez
2π
Calculate the corresponding strain tensors, first for finite transformations and
second assuming small perturbations. Interpret these transformations and find the
principal directions and strains for the case of small perturbations.
Solution
1) Using the same method as in Exercise 8.1, we can write
∂M
∂M
∂M
dr +
dθ +
dz
∂r
∂θ
∂z
= e r dr + re θdθ + e z dz
dM =
With
(8.2.2)
der
de
= e θ and θ = − e r , we find from (8.2.1)
dθ
dθ
d M ′ = d M + u′( r )e r d r + u( r )e θdθ + w′( z )e z d z
(8.2.3)
= [1 + u′( r )] e r d r + [ r + u( r )] e θdθ + [1 + w′( z ) ] ez d z
Thus we get
2
2
2
2
2
d M ′ − d M = [1 + u′] dr 2 + [ r + u ] dθ 2 + [1 + w′] dz 2 − ...
... − dr 2 − r 2dθ 2 − dz 2
(8.2.4)
= 2u′ + u′ dr + 2ur + u dθ + 2 w′ + w′ dz
2
2
2
2
2
2
According to (A3.8) in Volume I, this is to be identified with
2
2
d M ′ − d M = 2 E i E j ∆ij dθi dθ j
(8.2.5)
where (θi ) = ( r ,θ , z ) and ( E i ) = (1, r ,1) . The non zero components of ∆ are then
easily found to be
235
∆rr = u′( r ) +
u′( r ) 2
2
∆θθ =
u( r ) u( r ) 2
+
r
2r 2
∆zz = w′( z ) +
w′( z )2
2
(8.2.6)
We could have as well used the general intrinsic definition of the GreenLagrange tensor, namely
T
2∆ = F .F − δ
(8.2.7)
which leads to (A3.7) in Volume I, that is
2∆ij =
1 ∂M ′
∂M ′
(θ k , t ).
(θ k , t ) − δij
∂θi
E i E j ∂θ j
(8.2.8)
with (θi ) = ( r ,θ , z ) and ( E i ) = (1, r ,1) in cylindrical polars. We then find, with
use of (8.2.3)
2∆rr = (1 + u′)2 − 1 = 2u′ + u′2
1
u2
2
2∆θθ = 2 ( r + u ) − 1 = 2ru + 2
r
r
2∆zz = (1 + w′)2 − 1 = 2 w′ + w′2
(8.2.9)
and all other components vanish. Hence
u′2 ( r )
u′( r ) +
2
∆ =
0
0
( )
0
u( r ) u 2 ( r )
+
r
2r 2
0
0
w′2 ( z )
w′( z ) +
2
0
(8.2.10)
and for small perturbations
u′( r )
ε = 0
0
()
0
u( r )
r
0
0
0
w′( z )
(8.2.11)
236
So, the principal directions coincide with the coordinate system (e r , eθ , ez ) and
the non zero components are the principal strains.
2) Instead of (8.2.3), dM ′ is now given by
d M ′ = d M − v ( r )e r d θ + v′( r )e θdr
= [e r + v′( r )e θ ] d r + [ re θ − v( r )e r ] dθ + e z d z
(8.2.12)
and then we have
∂M ′
∂r = e r + v′( r )e θ
∂M ′
= re θ − v ( r )e r
∂θ
∂M ′
∂z = e z
(8.2.13)
so that, according to (8.2.8), we find
2
2∆rr = v′ ( r )
v2 (r)
2∆θθ = 2
r
v( r )
2∆rθ = 2∆θr = − r + v′( r )
(8.2.14)
and all other components vanish. Hence
v′2 ( r )
2
v ( r ) v′( r )
∆ = −
+
2
2r
0
( )
and for small perturbations
−
v( r ) v′( r )
+
2r
2
v2 (r )
+
2r 2
0
0
0
0
(8.2.15)
237
0
v ( r ) v′( r )
ε = −
+
2r
2
0
−
v ( r ) v′( r )
+
2r
2
()
0
0
0
0
0
(8.2.16)
v( r ) v′( r )
This represents a pure shear. The principal strains are ±
−
and 0;
2
2r
the principal axes e r + e θ , e r − eθ , e z . Such a transformation can be achieved, for
example, by first preventing the internal surface of a tube from moving by soldering it to a rigid cylinder and then applying a torque over the external surface.
3) The increment dM ′ is given by
d M ′ = d M − aze r d θ + ae θdz
= e r d r + [ re θ − aze r ] dθ + [ae θ + e z ] d z
(8.2.17)
and then
∂M ′
= er
∂r
∂M ′
= re θ − aze r
∂θ
∂M ′
= ae θ + e z
∂z
(8.2.18)
From (8.2.8), we find
az
2∆rθ = 2∆θr = −
r
2∆ = 2∆ = a
θz
zθ
a2 z2
r2
2
2∆zz = a
2∆θθ =
(8.2.19)
and all other components vanish. Hence; the strain tensor is
0
az
∆ = −
2r
0
( )
az
2r
a2 z2
2r 2
a
2
−
0
a
2
a2
2
For small perturbations ( a ≪ 1 ), this becomes
(8.2.20)
238
0
az
ε = −
2r
0
−
()
0
a
2
0
az
2r
0
a
2
(8.2.21)
a
r 2+ z2 .
2r
4) Using the same method as in the foregoing questions, we find
whose eigenvalues are 0 and ±
bθ
ez
2π
b
= e r d r + re θ +
e z dθ + e z d z
2π
dM ′ = dM +
∂M ′
= er
∂r
∂M ′
b
= re θ +
ez
∂θ
2π
0
0
b2
∆ = 0
8π 2 r 2
b
0
4 πr
( )
0
ε = 0
0
()
0
0
b
4 πr
0
b
4 πr
0
0
b
4 πr
0
∂M ′
= ez
∂z
(8.2.22)
(8.2.23)
(8.2.24)
(8.2.25)
As we already know from Sect. 3.3.7.1 in Volume I, this strain field is associated with a screw dislocation.
239
Exercise 8.3 Strain compatibility (small perturbations)
1) Show that the tensor field
(α ( x, y ) ) = ex + fy
ax + by ex + fy
cx + dy
(8.3.1)
where a, b, c, d, e and f are constants can represent a plane strain tensor field and
deduce the components of the associated displacement vector field.
2) In an orthonormal fixed system (e1 , e 2 , e 3 ) , the non-zero components of a strain
field are ε12 = ε 21 = g( x1 , x2 , x3 ) . Find the general form of g( x1 , x2 , x3 ) .
3) The given unit vectors n and m are such that the base (n, m, e 3 ) is orthonormal
and fixed. The strain tensor has the form ε = h( x1 , x2 , x3 )(n ⊗ m + m ⊗ n ) . Find the
most general form of the function h( x1 , x2 , x3 ) .
Solution
1) For a plane problem the compatibility equations reduce to
2
∂ 2ε xy
∂x∂y
=
2
∂ 2ε xx ∂ ε yy
+
∂y 2
∂x 2
(8.3.2)
This involves only second order partial derivatives; since α ( x, y ) is affine with
respect to x and y, it satisfies the equation (8.3.2) identically. For the corresponding displacement ( u ( x, y ), v( x, y ) ) we have
∂u
a
= ax + by ⇒ u = x 2 + bxy + h( y )
ε xx =
∂x
2
∂v
d 2
= cx + dy ⇒ v = y + cxy + g( x)
ε yy =
∂y
2
∂u ∂v
+
= bx + cy + h′( y ) + g′( x)
2ε xy =
∂y ∂x
(8.3.3)
Hence, according to (8.3.1), we must have
bx + cy + h ′( y ) + g′( x) = 2ex + 2 fy
or
bx − 2ex + g′( x) = 2 fy − cy − h′( y ) = −G
(8.3.4)
240
where G is a constant, since it is the common value of two functions depending
only on x and y, respectively. So, we find
b 2
g( x) = (e − 2 ) x − Gx + B
h( y ) = ( f − c ) y 2 + Gy + A
2
(8.3.5)
where A and B are constants and the displacement vector field is
a 2
c 2
u ( x, y ) = 2 x + bxy + ( f − 2 ) y + Gy + A
v( x, y ) = d y 2 + cxy + (e − b ) x 2 − Gx + B
2
2
(8.3.6)
Note that this displacement field is defined to within an arbitrary rigid body
displacement OM ∧ G + K where G = (0,0, G ) and K = ( A, B,0).
2) The compatibility equations give
ε12,13 = ε12,23 = ε12,33 = 0
ε12,12 = 0
(8.3.7)
The three first equations imply
g( x1 , x2 , x3 ) = ax3 + b( x1 , x2 )
(8.3.8)
where a is a constant. Then from the fourth equation we get
g( x1 , x2 , x3 ) = ax3 + b( x1 ) + c( x2 )
(8.3.9)
3) In the base (n, m, e 3 ) , with coordinates n,m,x3 the only non-vanishing components of the strain tensor are ε nm = ε mn = h( x1 , x2 , x3 ). So, according to (8.3.9) in
question 2), we must have
ε nm = ε mn = ax3 + b(n) + c(m)
(8.3.10)
Therefore, changing the axes and expressing n and m in terms of x1 and x2, we
obtain
h( x1 , x2 , x3 ) = ax3 + b [ n( x1 , x2 )] + c [ m( x1 , x2 )]
(8.3.11)
241
which is the most general expression for glide on the slip system (n, m).
Exercise 8.4 Deformation of curves and surfaces
1) A wire in the form of a curve in 3-dimensional space, for which we take the arc
length s as parameter, is subjected to the transformation
M ( s ) → M ′( s ) = M ( s ) + u ( s )
(8.4.1)
where u(s) is a vector field in R3. Determine the associated strain tensor.
2) A canvas sheet acting as a roof is held firmly by its four sides. As a result of a
fall of snow it is subjected to a displacement field u such that
M ( x, y,0) → M ′( x, y + v( x), w( x, y ))
(8.4.2)
Find the associated strain tensor.
3) We consider a straight beam which in its undeformed state is a (long) cylinder
with generators parallel to Ox and has a plane of symmetry containing Ox. We
take for Ox the inertial axis of the beam, also called the mid fibre. We seek to express the strain at any point of the beam in terms of the quantities that describe
the change of shape of the mid fibre only, with the following assumptions:
H1: a point M on the mid fibre is displaced to M' such that
M ( x,0,0) → M ′ [ x + u ( x), v( x),0]
(8.4.3)
H2: plane cross-sections of the beam are not deformed and remain normal to
the axis (the plane of symmetry is conserved).
(a) Find the strain tensor ε ( M ) for the axis Ox.
(b) Find an expression for the displacement of an arbitrary point P( x, y,0)
and the associated strain tensor ε ( P ) : for simplicity, u' and v' can be taken as
negligible relative to 1.
Solution
With t the unit tangent vector at s, we write
dM
2
2
d M = ds ds = tds ⇒ d M = ds
d M ′ = tds + du ds ⇒ d M ′2 = 1 + 2t. du + ( du ) 2 ds 2
ds
ds
ds
(8.4.4)
242
from which we find the single non zero component of the strain tensor to be
∆tt ( s ) = t.
du 1 du 2
+ ( )
ds 2 ds
(8.4.5)
du
.
ds
2) With i , j and k the unit vectors of the orthonormal base, we have
For small perturbations this becomes ε tt ( s ) = t.
d M = dxi + dy j ⇒ d M 2 = dx 2 + dy 2
∂w
∂w
dy k
d M ′ = dxi + [ dy + v′( x)dx ] j + dx +
∂
∂y
x
2
⇒ d M ′2 = dx 2 + dy + v′( x)dx 2 + ∂w dx + ∂w dy
[
]
∂y
∂x
(8.4.6)
and then
′2 ∂w 2
v + ( ∂x )
1
( ∆ ) = 2 ∂w ∂w
v′ + ∂x ∂y
∂w ∂w
∂x ∂y
∂w 2
( )
∂y
v′ +
(8.4.7)
a second-order tensor in R2
3) (a) On the axis Ox, we have
d M = dxi ⇒ d M 2 = dx 2
2
2
2
2
2
d M ′ = (1 + u′)dxi + v′dx j ⇒ d M ′ = (1 + u′) dx + v′ dx
(8.4.8)
and the single non-zero component of the strain tensor is ∆xx given by
2
2
2∆ xx ( M )dx 2 = d M ′ − d M = (2u′ + u ′2 + v′2 )dx 2
1
⇒ ∆ xx ( M ) = u ′ + (u′2 + v′2 )
2
(8.4.9)
(b) The vector OP( x, y,0) is displaced to OP′ = OM + MM ′ + yn , with n the unit
vector normal to the tangent vector t : (1 + u′, v′,0) , that is
243
n=
k ∧t
1
:
(−v′,1 + u′,0)
k ∧t
(1 + u′)2 + v′2
(8.4.10)
Therefore we have
OP′ : ( x + u, v,0) +
y
(1 + u′) 2 + v′2
(−v′,1 + u ′,0)
(8.4.11)
2
With the assumptions made, y does not vary and then dP = dx 2 . The exact
2
calculation of dP′ is made difficult by the need to evaluate (d/dx)n(x); but taking
u' and v' to be small relative to 1 gives the simplified expression
P′ : ( x + u − yv′, y + v,0) ⇒
dP ′
: (1 + u′ − yv′′, v′,0)
dx
(8.4.12)
So
dP′ 2
= (1 + u′ − yv′′) 2 + v′2 = (1 + u′) 2 + v′2 − 2 yv′′(1 + u ′) + y 2 v′′2
dx
(8.4.13)
1
1 2 2
2
2
′
′′
′
′′
⇒ ∆xx ( P) = 2dx 2 (dP − dP ) = ∆xx ( M ) − yv (1 + u ) + 2 y v
Neglecting the terms of higher order of smallness we have
ε xx ( P) = ε xx ( M ) − yv′′
(8.4.14)
where, to the approximation we are making, v'' is the change in curvature, here the
curvature of the beam in its final state. The linearised strain component tensor
ε xx ( P) at an arbitrry point with ordinate y is
ε xx ( P) = u′ − yv′′
8.2 Stresses
Exercise 8.5 Statically admissible stress fields
(8.4.15)
244
1) For homogeneous plates acted on by forces in their own planes, find in each
case a uniform statically admissible stress state and give the associatedMohr representation:
(a) a rectangular plate ABCD with shear τ applied at each edge and compression p at a pair of opposite edges.
(b) a parallelogram plate ABCD, with acute internal angle (π/2 − α ) , with
compression p1 and p2 parallel to the opposite edges.
2) The wall of a dam has the form of a prism whose section is a right-angled triangle OAB (see Fig. 8.5.1) with OA, facing up-stream, vertical, OB, facing downstream inclined at 45° to the vertical and AB fixed. The density of the wall material is ρ, that of the water in the dam, which is full, is ϖ (atmospheric pressure
can be neglected). Show that the equilibrium equations for the dam are satisfied
by a stress field as follows:
ax + by ex + fy
cx + dy
(σ ) = ex + fy
(8.5.1)
where a, b, c, d, e and f are constants to be calculated.
y
0
x
π/4
water
dam
A
B
Fig. 8.5.1 Sketch of the studied dam
3) A semi-infinite, heavy homogeneous body is subjected to atmospheric pressure
at the surface; investigate the stress field and show that there are an unlimited
number of statically admissible fields.
Solution
1) (a) The plane stress tensor has the form
σ xx σ xy
xy σ yy
(σ ) = σ
On AB n : (1,0) and σ .n = σ xx i + σ xy j = −τ j .
(8.5.2)
245
On BC n : (0,1) and σ .n = σ xy i + σ yy j = −τ i − p j .
Hence a statically admissible tensor is
0 τ
p
(σ ) = − τ
(8.5.3)
a uniform tensor field which clearly satisfies the equilibrium equations without a
body force term. It is easily shown that the boundary conditions are indeed satisfied on the edges CD ( n : (−1,0)) and DA ( n : (0, −1)). The associated Mohr representation is indicated in Fig. 8.5.2.
τ
τ
-p
-p/2
0
σ
Fig. 8.5.2 Mohr representation for question 1(a) of Exercise 8.5.
(b) We reason as before with, on AB, n AB : (cos α , − sin α ) and on CB, n CB : (0,1) .
Hence
σ xx σ xy cos α − p1
σ = − p1 cos α
=
with
σ
σ
yy − sin α
0
τ = p1 sin α
xy
σ xx σ xy 0 − p2 sin α
σ = − p2 cos α
with
=
τ = − p2 sin α
σ xy σ yy 1 − p2 cos α
(8.5.4)
Looking for a uniform stress field, we have from (8.5.4)
σ xx cos α − σ xy sin α = − p1
σ xy cos α − σ yy sin α = 0
σ = − p sin α
σ yy = − p2 cos α
2
xy
arriving finally at the following (uniform) statically admissible stress field
(8.5.5)
246
p1 + p2 sin 2 α
−
(σ ) =
cos α
− p sin α
2
− p2 sin α
− p2 cos α
(8.5.6)
The Mohr circle, which is centred on the σ axis, must go, according to (8.5.4),
through the points
σ = − p1 cos α
AB
τ = p1 sin α
σ = − p2 cos α
CB
τ = − p2 sin α
(8.5.7)
Hence the diagram in Fig. 8.5.3.
Fig. 8.5.3 Mohr representation for question 1(b) of Exercise 8.5.
2) From the equilibrium equations we get
σ xx , x + σ xy, y = 0 ⇒ a + f = 0
σ xy, x + σ yy, y − ρ g = 0 ⇒ e + d − ρ g = 0
(8.5.8)
with g the gravitational acceleration.
The boundary conditions on OB ( x = − y ) are
and on OA ( x = 0 )
1
(σ xx + σ xy ) = 0 ⇒ a − b + e − f = 0
2
1
(σ xy + σ yy ) = 0 ⇒ e − f + c − d = 0
2
(8.5.9)
247
−σ xx = −by = −ϖy ⇒ b = ϖ
−σ xy = − fy = 0 ⇒ f = 0
(8.5.10)
From the 6 equations (8.5.8-10), the considered stress field is statically admissible if
a = f = 0 b = e = ϖ d = ρ g − ϖ c = ρ g − 2ϖ
(8.5.11)
3) With Oy in the direction of the upward vertical the problem is invariant with respect to x and z, so the stress field is a function of y only. The equilibrium equations are
∂σ xy ∂y = 0 ∂σ yy ∂y − ρ g = 0 ∂σ yz ∂y = 0
(8.5.12)
whence
σ xy = a
σ yz = b
σ yy = ρ gy + c
(8.5.13)
The conditions at the boundary y = 0 are
σ xy = 0 σ yy = − p0 σ yz = 0 ⇒ a = b = 0 c = − p0
(8.5.14)
Thus the equilibrium equations enable us to determine σ xy , σ yy and σ yz
uniquely, but only these components. From the nature of the problem it seems reasonable to infer that σ xz = 0 , but this still leaves σ xx and σ zz arbitrary until the
constitutive equations for the medium have been specified.
8.3 Problems in linear elasticity
Exercise 8.6 Notches and cracks
It is known that, near the origin of a crack in an elastic medium that is linear, homogeneous and isotropic, the stress field σij, (and therefore the deformation field
εij) has a singularity of the form r-½: for example, in the symmetric crack opening
mode (Mode I)
248
σ ij (r ,θ ) =
KI
f ij (θ )
r
(8.6.1)
where KI is the stress-intensity factor for Mode I. This exercise is to prove the existence of such a singularity, considering the crack as a speci al case of a notch
(see also Chapter 2, Volume II).
We assume that perturbations are small and that at a point M (r ,θ ) near to the
origin of the notch of angle 2ϕ symmetrical with respect to the x1 axis (see Fig.
8.6.1) the displacement field u ( M ) has the form ui = r α fi (θ ) in the cylindrical
coordinate system (er, eθ), with uz = 0 in plane strain.
Fig. 8.6.1 Coordinates and geometry of the notch
1) Express the strain ε (r ,θ ) and stress σ (r ,θ ) tensors at point M as functions of
f, g and α.
Solution
In cylindrical polars we have in plane strain conditions (see Appendix F in Volume I)
ur,r
(ε ) = 1
(ru + u − u )
θ, r
r,θ
θ
2r
1
(ruθ,r + ur,θ − uθ )
2r
1
(uθ,θ + ur )
r
(8.6.2)
249
With ur = r α f (θ ) and uθ = r α g(θ ) , we find
1
1
αf
(α − 1)g + f ′
2
2
(ε (r ,θ ) ) = rα −1 1
(α − 1)g + 1 f ′
′
g
+
f
2
2
(8.6.3)
Using the constitutive equations we get
1 0 0
σ = λε kk δ + 2µε = λ r α −1 ( (α + 1)f + g′ ) 0 1 0 + ...
0 0 1
1
1
αf
(α − 1)g + f ′ 0
2
2
1
α −1 1
... + 2µ r
(α − 1)g + f ′
g′ + f
0
2
2
0
0
0
(8.6.4)
that is
(σ ( r , θ ) ) = r
α −1
(λ + 2µ )α f + λ (f + g′)
µ ( f ′ + (α − 1)g )
λα f + (λ + 2 µ )(f + g′)
µ ( f ′ + (α − 1)g )
(8.6.5)
with σ rz = σ θz = 0 and σ zz = λε kk = λ r α −1 ( (α + 1)f + g′ ) .
2) From the equilibrium equations for the stresses find a system of two linear differential equations satisfied by the functions f and g. Deduce from that the differential equation, independent of the elastic constants, for f(θ) and find the general
solution. Find f and g for the particular case of a load that is symmetrical with respect to the notch, i.e. to the x1 axis.
Solution
Expanding the equilibrium equation in cylindrical polars (see Appendix F in Volume I for the expression of divσ ), we find
250
∂σ r r 1 ∂σ r θ ∂σ rz σ r r − σ θθ
+
+
+
+ ρ fr = 0
r ∂θ
∂z
r
∂r
σ
∂σ r θ 1 ∂σ θθ ∂σ z θ
+
+
+ 2 rθ + ρ fθ = 0
r ∂θ
∂z
r
∂r
∂σ z r 1 ∂σ z θ ∂σ z z σ z r
+
+
+
+ ρ fz = 0
r ∂θ
∂z
r
∂r
(8.6.6)
from which in this case, with (8.6.5)
2
µ f ′′ + [ (α − 1)(λ + µ ) − 2µ ] g′ + (α − 1)(λ + 2µ )f + = 0
2
(λ + 2µ )g′′ + [α (λ + µ ) + λ + 3µ ] f ′ + (α − 1) µ g = 0
(8.6.7)
which do not depend on r, or, with use of E and ν instead of λ and µ
α − 3 + 4ν
(α 2 − 1)(1 −ν )
′′
′
f
+
g
+
2
f =0
1 − 2ν
1 − 2ν
2(1 −ν ) g′′ + α + 3 − 4ν f ′ + (α 2 − 1)g = 0
1 − 2ν
1 − 2ν
(8.6.8)
which do not depend on E.
Elimination of g between the equations (8.6.8) for f and g gives the following
fourth-order homogneous differential equation for f(θ):
f (4) + 2(α 2 + 1)f ′′ + (α 2 − 1)2 f = 0
(8.6.9)
which does not involve any of the elastic constants. Putting f (θ ) = exp ( kθ ) we
have the characteristic equation
k 4 + 2(α 2 + 1)k 2 + (α 2 − 1) 2 = 0
(8.6.10)
with roots ± (α + 1)i and ± (α − 1)i where i 2 = −1 ; the general solution is therefore
f (θ ) = a1 cos(α + 1)θ + a2 cos(α − 1)θ + a3 sin(α + 1)θ + a4 sin(α − 1)θ (8.6.11)
For symmetric loads the displacements must be symmetric, with
u (r ,θ ) = u (r , −θ ) ; f is therefore an even function of θ, so the sine terms vanish,
giving
f (θ ) = a1 cos(α + 1)θ + a2 cos(α − 1)θ
(8.6.12)
251
Substituting this in the first of the equations (8.6.8) gives an equation for g'
which after simplifying is
g′(θ ) = − a1 (α + 1)cos(α + 1)θ − a2 (α − 1)
α + 3 − 4ν
cos(α − 1)θ
α − 3 + 4ν
(8.6.13)
Integrating, with g'(0) = 0, we have
g(θ ) = −a1 sin(α + 1)θ − a2
α + 3 − 4ν
sin(α − 1)θ
α − 3 + 4ν
(8.6.14)
3) The equations found for the functions f and g have the trivial solution
f = g = 0 (a1 = a2 = 0) . Using the fact that the stress vector is zero at the edges of
the notch, find the condition for a non-trivial solution to exist. Deduce that this solution can be either regular or singular and that for a crack we have α = 1/2 .
Solution
The condition that the edges of the notch are free of stress is expressed by stating
that the stress vector components are zero for θ = ±ϕ . Taking account of the parity of f and g this gives, for θ = ϕ
σ rθ θ =ϕ = 0 ⇒ f ′(ϕ ) + (α − 1)g(ϕ ) = 0
σ θθ θ =ϕ = 0 ⇒ [ (α + 1)λ + 2µ ] f (ϕ ) + (λ + 2µ )g′(ϕ ) = 0
(8.6.15)
f ′(ϕ ) + (α − 1)g(ϕ ) = 0
(1 −ν + αν )f (ϕ ) + (1 −ν )g′(ϕ ) = 0
(8.6.16)
that is
Substituting (8.6.12) and (8.6.14) for f and g we get two linear homogeneous
equations for a1 and a2, namely
(α − 1)sin(α − 1)ϕ
=0
a1 sin(α + 1)ϕ + a2
α − 3 + 4ν
−a1α (1 − 2ν )cos(α + 1)ϕ + a2 (1 −ν + αν + A)cos(α − 1)ϕ = 0
with A = −
(1 −ν )(α − 1)(α + 3 − 4ν )
α − 3 + 4ν
(8.6.17)
252
The condition for the existence of a non-zero solution for a1 and a2 is that the
determinant of the coefficients vanishes, that is
sin(α + 1)ϕ
−α (1 − 2ν )cos(α + 1)ϕ
(α − 1)sin(α − 1)ϕ
=0
α − 3 + 4ν
(1 −ν + αν + A)cos(α − 1)ϕ
(8.6.18)
with A given in (8.6.18). This equation reduces to
(α + 1) cos(α − 1)ϕ sin(α + 1)ϕ − (α − 1) cos(α + 1)ϕ sin(α − 1)ϕ = 0
(8.6.19)
sin 2αϕ + α sin 2ϕ = 0
(8.6.20)
or
This can be solved graphically by studying the intersection, for any fixed value
of ϕ lying between 0 and π, of the straight line y = −(sin 2ϕ ) x and the sine wave
y = sin(2ϕ x) with x > 0 ; there are two cases to be considered.
(a) 0 < ϕ < π 2 , i.e. π 2ϕ > 1
This corresponds to a salient, not a notch. The construction for the graphical solution shows that in this case the positive abscissa α0 of the intersection is larger
than π 2ϕ > 1 . Since the radial variation of the stress and strain solution is rα-1, it
follows that this remains finite at the origin and there is no singularity.
(b) π 2 < ϕ < π , i.e. π 2ϕ < 1 and sin 2ϕ < 0
The solution is now such that α 0 < π 2ϕ < 1 , so that ( α − 1 ) is negative and
there is a singularity at the apex of the notch. For a crack ( ϕ = π ), sin 2ϕ = 0 , giving the root α 0 = π 2π = 1 2 . The strains and stresses then have a singularity of
the order r −1 2 and the displacements are like r 1 2 .
Exercise 8.7 The Green tensor for the elastic isotropic infinite medium
The problem (see Sect. 2.7.1 in Volume I) is to find the displacement field u(M) in
an infinite linearly elastic isotropic medium due to a unit point force F applied at
a point P, that is to prove Eq. 2.154 in Volume I. As we know, since the system is
linear, the Green tensor G ( M , P ) for the infinite homogeneous medium, with zero
displacement at infinity, obeys the relation
u ( M ) = G ( M , P ).F ( P )
(8.7.1)
253
With this, the principle of superposition can be used again to construct the solution for any distribution of applied forces.
1) Show that the Navier equations (A3.17) in Appendix C of Volume I are satisfied identically by a displacement field of the form
u = ∇2 B −
1
grad(div B)
2(1 −ν )
(8.7.2)
where ν is the Poisson ratio and B satisfies an equation of the form
∇4 =
ρ
f
µ
(8.7.3)
with ρ, µ the density and elastic shear modulus of the medium.
Solution
In rectangular Cartesians, the Navier equations read
ui ,kk +
1
ρ
uk ,ki + f i = 0
1 − 2ν
µ
(8.7.4)
According to (8.7.2), we have
ui = Bi ,kk − α Bk ,ki
α=
1
2(1 −ν )
(8.7.5)
and then
ui ,kk = Bi ,kkll − α Bk ,ikll
uk ,k = Bk ,kll − α Bk ,kll = (1 − 2ν )α Bk ,kll
(8.7.6)
so that (8.7.4) becomes
Bi ,kkll − α Bk ,ikll + α Bk ,ikll = Bi ,kkll = −
ρ
f
µ i
or ∇ 4 B = −
ρ
f
µ
(8.7.7)
2) Find the field B due to a point force F applied at the origin; deduce the corresponding dispacement field u and the Green tensor, assuming an infinite medium
and zero displacement at infinity.
254
Solution
The body forces f must be taken in the form
ρ f = F δ(r )
(8.7.8)
where δ(r) is the Dirac function and r is the distance of a point from the origin; so
the equation to be solved is
∇4 B = −
F
µ
δ( r )
(8.7.9)
We can use for that a classical result of the theory of Coulomb potentials,
namely
∇ 2 (1 r ) = −4πδ(r )
(8.7.10)
Now
r,k =
xk
r
r,kl =
δ kl xk xl
− 3
r
r
⇒ ∇ 2 r = r,kk =
2
r
(8.7.11)
whence
2
∇ 4 r = ∇ 2 ( ) = −8πδ(r )
r
(8.7.12)
So
∇4 B = −
F
µ
δ(r ) =
F 4
r
∇ r ⇒ ∇4 B −
F=0
8πµ
8
µ
π
(8.7.13)
from which
B=
r
F+K
8πµ
(8.7.14)
where K is any biharmonic vector field (i.e., satisfying ∇ 4 K = 0 ). Taking account
that u depends on the second derivatives of B and vanishes at infinity, one can
show that it suffices to take K = 0, giving
255
B=
r
F
8πµ
(8.7.15)
From (8.7.5) and (8.7.15) we find
Bi ,kk =
Fi
4πµ r
Bk ,k =
xk Fk
8πµ r
⇒ Bk ,ki =
Fi
x xF
− k i 3k
8πµ r 8πµ r
(8.7.16)
and then
ui =
xi x j
xi x j
1
1
(2 − α )δij + α 2 Fj =
(3 − 4ν )δij + 2 Fj (8.7.17)
8πµ r
r
16πµ (1 −ν )r
r
So, referring to (8.7.1), we have proved that
Gij ( M , O) =
xi x j
1
(3 − 4ν )δij + 2
16πµ (1 −ν )r
r
(8.7.18)
If now we put e = ( x − x′ ) x − x′ , we have, for a unit force applied at P( x′)
Gij ( M , P ) = Gij ( x, x′) = Gij ( x − x′) =
(3 − 4ν )δij + ei e j
16πµ (1 −ν ) x − x′
(8.7.19)
which is Eq. 2.154 in Volume I.
Exercise 8.8 Spherical inhomogeneity
The problem is to find directly, i.e. without use of the general Eshelby’s solution of
the ellipsoidal inhomogeneity problem, the strain state in a spherical elastic isotropic inhomogeneity (I) with properties (µI, νI) and perfet bonding at the interface to an infinite elastic isotropic medium with properties (µm, νm) which is subjected to a pure shear γ at infinity. We solve this by assuming a general form for
the displacement field and determining the details by setting up the Navier equations with the boundary conditions at the interface and at infinity. We use spherical co-ordinates with origin at the centre of the sphere.
1) Write the Navier equations for the problem, assuming no body forces.
Solution
The Navier equations in the absence of body forces read
256
(λ + µ )grad ( div(u ) ) + µ∆(u ) = 0
(8.8.1)
Using spherical polars, we have, according to Appendix F of Volume I (Sect. F.3)
1
u
1
uθ
divu = ur,r + uθ,θ + 2 r +
uϕ,ϕ +
r
r r sin θ
r tan θ
(8.8.2)
(divu ),ϕ
1
eϕ
grad(divu ) = (divu ),r e r + (divu ),θ eθ +
r
r sin θ
(8.8.3)
1
u
1
uθ
X = grad(divu ) = (ur,r + uθ,θ + 2 r +
uϕ,ϕ +
), r e r
r
r r sin θ
r tan θ
uθ
1
1
u
1
+ (ur,r + uθ,θ + 2 r +
uϕ,ϕ +
),θ eθ
r
r
r r sin θ
r tan θ
uθ
u
1
1
1
+
(ur,r + uθ,θ + 2 r +
uϕ,ϕ +
),ϕ eϕ
r sin θ
r
r r sin θ
r tan θ
(8.8.4)
and then
We find
u
u ru − uϕ,ϕ ruθ,r − uθ
1
1
X = (ur,rr + uθ,θ r − 2 uθ,θ + 2 r,r − 2 2r + ϕ,2ϕ r
+ 2
) er
r
r
r
r
r sin θ
r tan θ
u
u cos θ
u
u
uθ
1
1
+ (ur,rθ + uθ,θθ + 2 r,θ + ϕ,ϕθ − ϕ,ϕ 2 + θ,θ −
) eθ (8.8.5)
r
r
r
r sin θ
r sin θ
r tan θ r sin 2 θ
u
u
1
1
1
+
(ur,rϕ + uθ,θϕ + 2 r,ϕ +
uϕ,ϕϕ + θ,ϕ ) eϕ
r sin θ
r
r
r sin θ
r tan θ
On the other hand, we have from Sect F.3 in Volume I (note that in the second
term of the component on eϕ , sin 2θ must be changed into sin θ , as follows)
2u
2u 2(uθ sin θ ),θ
Y = ∆u = ∆ur − 2r −
− 2 ϕ,ϕ e r
2
r
r sin θ
r sin θ
2u cos θ
2u
u
+ ∆uθ + 2r,θ − 2 θ 2 − 2ϕ,ϕ 2 eθ
r
r sin θ
r sin θ
2u
u
2u cos θ
+ ∆uϕ + 2 r,ϕ − 2 ϕ 2 + 2θ,ϕ 2 eϕ
r sin θ r sin θ
r sin θ
(8.8.6)
257
with ∆ur , ∆uθ , ∆uϕ to be calculated according to the relation
∆f = f ,rr +
2 f ,r
r
+
f,θθ
r
2
+
f,θ
2
r tan θ
+
f ,ϕϕ
(8.8.7)
2
r sin 2 θ
Thus, the Navier equations will read in this case
(λ + µ ) X + µ Y = 0
or
X + (1 − 2ν )Y = 0
(8.8.8)
with X and Y given by (8.8.5) and (8.8.6), respectively.
2) The boundary conditions at infinity are as follows:
1
ux∞ = γ x
2
1
uy∞ = − γ y
2
uz∞ = 0
(8.8.9)
or in spherical polars
ur∞ =
γr
2
sin 2θ cos 2ϕ
uθ∞ =
γr
2
sin θ cos θ cos 2ϕ
uϕ∞ = −
γr
2
sin θ sin 2ϕ (8.8.10)
Look for the solution in the form
ur = U r (r )sin 2θ cos2ϕ
uθ = U θ (r )sinθ cosθ cos2ϕ
uϕ = U ϕ (r )sinθ sin2ϕ (8.8.11)
Solution
To determine these functions we first substitute (8.8.11) into the Navier equations.
This gives the following equations:
2
2
cos 2ϕ sin θ (r U r′′ + 2rU r′ − 2U r − 3rU θ′ + 3U θ )
er
r 2 +2(rU ′ − U ) + 2(rU ′ − U )
ϕ
ϕ
θ
θ
2sin θ cos θ cos 2ϕ ′
+
rU + 2U r − 3U θ eθ
r
r2
2sin 2ϕ 2
− 2
sin θ (rU r′ + 2U r − 3U θ ) + 2(U θ + U ϕ ) eϕ
r sin θ
X=
and
(8.8.12)
258
u
2ur,r ur,θθ
u
+ 2 + 2 r,θ + 2 r,ϕϕ2
ur,rr +
r
r
r tan θ r sin θ
Y =
er
2uϕ,ϕ
2ur 2(uθ sin θ ),θ
− 2
− 2 −
r 2 sin θ
r sin θ
r
u
2
u
u
u
θ, r
+ θ,2θθ + 2 θ,θ + 2 θ,ϕϕ2
uθ,rr +
θ
θ
r
r
r
tan
r
sin
eθ
+
2uϕ,ϕ cos θ
2ur,θ
uθ
+ 2 − 2 2 − 2 2
r sin θ
r sin θ
r
2uϕ,r uϕ,θθ
u
u
+ 2 + 2 ϕ,θ + 2 ϕ,ϕϕ2
uϕ,rr +
r
r
r tan θ r sin θ
+
eϕ
uϕ
2uθ,ϕ cos θ
2ur,ϕ
−
+ 2 2
+ 2
r sin θ
r sin θ r 2 sin 2 θ
(8.8.13)
that is
cos 2ϕ
sin 2θ (r 2U r′′+ 2rU r′− 8U r + 6U θ ) − 4(U θ + U ϕ ) e r
r2
4(U θ + U ϕ )
cosθ cos 2ϕ
2
+
sin θ (r U θ′′+ 2rU θ′+ 4U r − 6U θ ) −
eθ (8.8.14)
2
r
sin θ
Y=
+
(
)
4(U θ + U ϕ )
sin θ sin 2ϕ 2 ′′
r U ϕ + 2rU ϕ′− 2U ϕ − 4U r + 4U θ −
eϕ
2
r
sin 2 θ
Now, we set separately to zero those expressions that depend on θ and those
that do not. We are then left with 3 equations which read
U θ + U ϕ = 0
2
2(1 −ν )(r U r′′+ 2rU r′ ) − 2(5 − 8ν )U r − 3rU θ′+ 3(3 − 4ν )U θ = 0
2
(1 − 2ν )(r U θ′′ + 2rU θ′ ) + 2rU r′ + 4(1 −ν )(2U r − 3U θ ) = 0
(8.8.15)
We now look for the solutions as linear functions of r, r3, r-2 and r-4, in the following form:
3 A3 5 − 4ν A4
6ν
3
U r = A1r − 1 − 2ν A2 r + r 4 + 1 − 2ν r 2
U = −U = A r − 7 − 4ν A r 3 − 2 A3 + 2 A4
θ
ϕ
1
2
1 − 2ν
r4
r2
(8.8.16)
259
where ν is either νI or νm according as r <a or r >a, the radius of the inclusion,
and the constants Ai have different values in the inclusion and the matrix. Let us
call them Ai in (I) and Bi in the matrix. Now
(i) in the inclusion (r ≤ a) there is no singularity at r = 0 ; so A3 = A4 = 0
(ii) in the matrix the conditions at infinity give U r (∞) = U θ (∞) ≈ γ r/2 ; so
B1 = γ /2 and B2 = 0 .
The remaining coefficients A1, A2, B3 and B4 are determined.by stating continuity of the diplacement and the stress vector across the matrix-inclusion interface:
(i) displacement: U r (a− ) = U r (a+ ) and U θ (a− ) = U θ (a+ ) , giving
6ν I
3B3 5 − 4ν m B4
γ
3
A1a − 1 − 2ν A2 a = 2 a + a 4 + 1 − 2ν a 2
I
m
A a − 7 − 4ν I A a 3 = γ a − 2 B3 + 2 B4
1 1 − 2ν I 2
2
a4
a2
(8.8.17)
(ii) stress vector: σ rr (a− ) = σ rr (a+ ), σ rθ (a− ) = σ rθ (a+ ), σ rϕ (a− ) = σ rϕ (a+ ) with
2µ
σ rr = 1 − 2ν ( (1 −ν )ε rr + ν (ε θθ + ε ϕϕ ) )
= 2µ sin 2 θ cos 2ϕ (1 −ν )U ′ + ν (2U − 3U )
r
r
θ
1 − 2ν
r
σ
µε
=
2
rθ
rθ
′ 1
= µ sin θ cos θ cos 2ϕ U θ + r (2U r − U θ )
σ rϕ = 2µε rϕ
1
= − µ sin θ sin 2ϕ U θ′ + (2U r − U θ )
r
(8.8.18)
from which, after simplification, we have the two further equations:
γ
B
3ν I
5 −ν m B4
A2 a 2 = µ m − 12 53 − 2
µ I A1 +
1
−
2
2
a
1
− 2ν m a3
ν
I
µ A − 7 + 2ν I A a 2 = µ γ + 8 B3 + 2 1 + ν m B4
m
I 1 1 − 2ν 2
a5
1 − 2ν m a 3
I
2
Solving (8.8.17) and (8.8.19), we find
(8.8.19)
260
15(1 −ν m ) µm
γ
A1 =
A2 = 0
2(4
−
5
ν
)
µ
+
(7
−
5
ν
)
µ
m
I
m
m 2
B
3( µ m − µ I )
γ
3
5 =−
2 ( 2(4 − 5ν m ) µ I + (7 − 5ν m ) µm ) 2
a
B
5(1 − 2ν m )( µm − µ I )
γ
34 =
a
2
2(4
5
)
(7
5
)
−
+
−
ν
µ
ν
µ
(
)
m
I
m
m 2
(8.8.20)
The vanishing of A2 implies that in the inclusion U r = U θ = −U ϕ = A1r ; hence
ur (I) =
2 A1
γ
uθ (I) =
ur∞
2 A1
γ
uθ∞
uϕ (I) =
2 A1
γ
uϕ∞
(8.8.21)
Thus the strain (and therefore the stress) is uniform in the inhomogeneity; the
tensor is proportional to that of the prescribed shear strain. Putting
2 4 − 5ν m
15 1 −ν m
(8.8.22)
µm
γ
µm + β ( µ I − µm ) 2
(8.8.23)
β=
we get
A1 =
and hence, for the strain deviators
eI =
Since a deviatoric strain E
µm
∞
E
µm + β ( µ I − µm )
∞
(8.8.24)
can be decomposed into two orthogonal shear
∞
strains, (8.8.24) still holds for arbitrary E : it is the Eshelby relation for a spherical inhomogeneity (see Sect. 2.7.3.4 in Volume I).
8.4 Miscellaneous
Exercise 8.9 Tensile test at finite strain
261
A circular cylindrical rod (initial length l0, initial radius r0) deforms uniformly,
without volume change, under the axial load Q. In a Cartesian rectangular coordinate system with 1 as the rod axis, any point of the rod with the initial coordinates (X1, X2, X3) moves to (x1, x2, x3) with
x1 =
l
X1
l0
x2 =
r
X2
r0
x3 =
r
X3
r0
(8.9.1)
Let A = (l − l0 ) l0 be the corresponding axial elongation.
1) Using the polar decomposition of the transformation gradient F of a finite
transformation T (see Sect. 1.4.3 in Volume I), give the general relation between
the Lagrangian (Green-Lagrange) and Eulerian (Almansi-Euler) strain tensors
and the symmetric part S of F . Show that with ε defined by ε = S − δ , one has
1
2
−1
L = Fɺ .F . Express all the foregoing tensors for the studied tensile test. Compare
∆ = ε + ε 2 . Express the Almansi-Euler strain rate tensor d as a function of
l
dλ
l0
λ
the numerical values of A, ∆11, ε11, A11 and the axial logarithmic strain e = ∫
for A = 50% .
Solution
The polar decomposition of F reads
F = U .S
with
S=S
T
T
and U = U
−1
(8.9.2)
From d x = F .d X , d x′ = F .d X ′ , one has
T
T
2
d x.d x′ = d X .F .F .d X ′ = d X .S .U .U .S .d X ′ = d X .S .d X ′
(8.9.3)
So, for d x = d x′ and d X = d X ′ , we get
2
2
ds = d X .S .d X
2
ds0 = d X .d X
2
⇒ ds 2 − ds02 = d X .( S − δ).d X = 2d X .∆.d X
(8.9.4)
2
and then the relation between the Green-Lagrange ( ∆ ) and the Cauchy ( S ) strain
tensors, namely
262
∆=
1 2
S −δ
2
(
)
(8.9.5)
With ε = S − δ , we find
∆=
1
1 2
(δ + ε ) 2 − δ ) = ε + ε
(
2
2
(8.9.6)
−1
−1
Similarly, for a Eulerian description, we have d X = F .d x, d X ′ = F .d x′ with
F
−1
−1
= S .U
T
and then
d X .d X ′ = d x.F
−1T
−1
−1
.F .d x′
= d x.F .F
−1T
−1
T
−1
−2
.d x′ = d x.S .U .U .S .d X ′ = d x.S .d x′
(8.9.7)
Hence, for d x = d x′ and d X = d X ′
ds 2 = d x.d x
−2
⇒ ds 2 − ds02 = d x.(δ − S ).d x = 2d x.Α.d x
2
−2
ds0 = d x.S d x
(8.9.8)
with the Almansi-Euler strain tensor Α given by
1
2
Α = (δ − S −2 )
(8.9.9)
From (8.9.4), the Lagrangian strain rate tensor is simply ∆ɺ , whereas, from
(8.9.8), we find for a Eulerian description
d
d
−1
(d x) = ( F .d X ) = Fɺ .d X = Fɺ .F .d x = L.d x
dt
dt
(8.9.10)
d
d
d
T
(d x.d x) = (d x).d x + d x. (d x) = d x.( L + L ).d x = 2d x.d .d x
dt
dt
dt
(8.9.11)
and then
Hence, the Eulerian strain rate tensor d is given by
263
1
T
d = (L + L )
2
with
L = Fɺ .F
−1
(8.9.12)
For the problem under investigation, we find
l l0
F
=
( ) 0
0
0
r r0
0
0
0 = (S ) ⇒ U = δ
r r0
(8.9.13)
and
l
0
0
−1
l
0
r
(ε ) = ( S − δ ) = 0 r − 1 0
0
r
0
− 1
0
r0
(8.9.14)
Note that, since there is no volume change, we must have
J = det( F ) =
lr 2
=1
l0 r02
(8.9.15)
From (8.9.13) we get
(S )
2
l 2 l02
= 0
0
0
2
2
0
r r
0
0
0
r 2 r02
(8.9.16)
and then
l2
0
0
2 −1
l0
1 2
1
r2
(∆ ) = 2 S − δ = 2 0 r 2 −1 0
0
2
r
0
0
−
1
r02
(
)
(8.9.17)
264
For a Eulerian description, we find
(S )
−2
l02 l 2
= 0
0
0
2
0
r2
r
0
l02
1 − 2
l
0
1
0 ⇒ ( A) = 0
2
r02 r 2
0
0
1−
r02
r2
0
0
0 (8.9.18)
r02
1− 2
r
As for the strain rate tensors, we find for the Lagrangian description
llɺ
2
l0
( ∆ɺ ) == 0
0
rrɺ
r02
0
0
rɺ r0
(F )
0
0
0
0
rrɺ
r02
(8.9.19)
and for the Eulerian one
lɺ l0
ɺ
(F ) = 0
0
0
rɺ r0
0
−1
l0 l
= 0
0
0
r0 r
0
0
0
r0 r
(8.9.20)
and then, due to the symmetry
( L ) = ( Fɺ .F )
−1
lɺ l
= 0
0
0
rɺ r
0
0
0 = (d )
rɺ r
(8.9.21)
For A = 50% , we have l l0 = 1.5 and then ∆11 = .625 , ε11 = .5 , Α11 = .278 and
e = log1.5 = .405 .
2) Let C = Q S0 the nominal axial stress, with S0 the initial cross-section area.
Calculate in the same axes system the components of the Cauchy stress tensor σ ,
referred to the current configuration, and those of the Piola-Khirchhoff II stress
tensor Π , referred to the initial configuration (see Sect. 1.4.3 in Volume I), as
265
functions of C and A. Compare, for A = 50% , the numerical values of σ 11 C and
Π11 C.
Solution
The Cauchy stress tensor for this tensile test reads
Q S
(σ ) = 0
0
0 0
0 0
0 0
(8.9.22)
with, according to (8.9.15)
σ 11 =
r2
Q Q S0
l
=
= C 02 = C = C (1 + A)
S S0 S
r
l0
(8.9.23)
The Lagrangian Piola-Khirchhoff II stress tensor Π is linked with the Cauchy
stress tensor by
Π = J F −1.σ .( F T )−1
(8.9.24)
Ql02
2
Sl
(Π ) = 0
0
(8.9.25)
which gives, with J = 1
0 0
0 0
0 0
Hence we have
Π11 =
Q S0 l02 Q l l02
C
=
=
2
2
S0 S l
S0 l0 l
1+ A
(8.9.26)
Thus, comparing (8.9.23) and (8.9.26), we find for A = 50% σ 11 C = 1.5 and
Π11 C = 1 1.5 = .667.
