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MATH35021: Reading Week Material
4.1
Experimental determination of elastic constants
I. Simple Extension
II. Simple Shear
T
D
D+∆D
τ
x2
L+∆L
L
x3
γ
000000000000000000000000
111111111111111111111111
111111111111111111111111
000000000000000000000000
000000000000000000000000
111111111111111111111111
x1
T
Figure 4.1: Sketch illustrating the two fundamental experiments for the determination of the elastic
constants.
4.1.1
Experiment I: Simple extension of a thin cylinder
A circular cylinder of diameter D, cross-sectional area A and initial length L is subject to an applied
tension T along its axis which is chosen to lie in the x3 direction. After the load has been applied the
cylinder has length L + ∆L and diameter D + ∆D.
The extension is in the x3 direction and so e33 = ∆L/L ≪ 1. The change in diameter is in the x1 and
x2 directions, so e11 = e22 = ∆D/D. Moreover, the stress is uniaxial (along one axis, x3 ) so τ33 = T /A,
and τij = 0 otherwise.
• Observations:
– Rod extends in the x3 direction in proportion to the applied tension:
τ33 = Ee33 ,
where E is the Elastic (or Young’s) modulus.
(4.1)
– Rod cross-section decreases (contracts in plane) in proportion to its extension in the axial
direction:
e11 = e22 = −ν e33 , where ν is the Poisson ratio.
(4.2)
4.1.2
Experiment II: Simple shear
A block in the x1 -x2 plane is subject to an applied traction, τ , on its upper surface in the positive x1
direction that induces a change in angle γ = 2e12 between two sides. The body is in a state of simple
shear so τ12 = τ and τij = 0 otherwise.
• Observation:
– The change in angle is proportional to the applied traction:
τ = Gγ ⇒ τ12 = G 2e12 ,
4.2
where G is the shear modulus of the material.
(4.3)
Relating experiments (E, ν, G) and theory (λ, µ)
We shall now use the constitutive equations to relate the experimentally measured constants E, ν and
G to the (theoretical) Lamé coefficients λ and µ. The material is homogeneous, isotropic and elastic so
that
τij = λδij ekk + 2µ eij ,
(4.4)
and, alternatively,
eij =
λ
1
τij −
δij τkk .
2µ
2µ(3λ + 2µ)
(4.5)
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Experiment I
In this experiment, τ33 = Ee33 and Θ = τkk = τ33 , because τ11 = τ22 = 0. If we set i = j = 3 in equation
(4.5) we have
1
1
3λ + 2µ − λ
λ
λ
λ+µ
e33 =
τ33 −
τkk =
−
τ33 .
τ33 =
τ33 =
2µ
2µ(3λ + 2µ)
2µ 2µ(3λ + 2µ)
2µ(3λ + 2µ)
µ(3λ + 2µ)
Hence,
1
λ+µ
=
E
µ(3λ + 2µ)
⇒
E=
µ(3λ+2µ)
.
λ+µ
(4.6)
If we substitute i = j = 1 into equation (4.5), we obtain
e11 =
1
λ
λ
τ11 −
τ33 = −
τ33 ,
2µ
2µ(3λ + 2µ)
2µ(3λ + 2µ)
because τ11 = 0.
Using the first experimental observation, τ33 = Ee33 , and the expression (4.6) for E gives
e11 = −
λ
µ(3λ + 2µ)
λ
e33 = −
e33 .
2µ(3λ + 2µ) λ + µ
2(λ + µ)
The second experimental observation is e11 = −ν e33 , so direct comparison yields
ν=
λ
2(λ+µ) .
(4.7)
Note that an identical result is obtained on substitution of i = j = 2 into equation (4.5).
Experiment II
In this experiment, τ12 = 2Ge12 , and using equation (4.4) with i = 1 and j = 2, we have
τ12 = λδ12 ekk + 2µ e12 = 2µ, e12
4.3
⇒
µ = G.
(4.8)
Relations between the elastic constants
Note that only two elastic constants are independent, see equations (4.4, 4.5). The following table lists
the relationships between elastic coefficients for the most commonly used independent pairs.
Independent pair
λ, µ
λ=
λ
µ=G=
µ
λ, ν
µ, E
E, ν
λ
λ(1−2ν)
2ν
4.3.1
µ(E−2µ)
3µ−E
Eν
(1+ν)(1−2ν)
µ
E=
ν=
µ(3λ+2µ)
λ+µ
(1+ν)(1−2ν)λ
ν
λ
2(λ+µ)
E
E
E
2(1+ν)
ν
E−2µ
2µ
ν
Constitutive equations in terms of the experimentally measurable constants
Using the table above, we can rewrite the equation (4.4) and (4.5) using the Young’s modulus E and
Poisson ratio ν


ν
E 
eij +
(4.9)
δij ekk  .
τij =
1+ν
1 − 2ν |{z}
d
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and


1 
eij =
(1 + ν)τij − ν δij τkk  .
|{z}
E
(4.10)
Θ
For physically realisable materials,
E ≥ 0,
−1 < ν ≤
1
,
2
µ ≥ 0,
3λ + 2µ > 0.
Note that equation (4.10) implies that
d = ekk =
τkk
Θ
((1 + ν) − 3ν) = (1 − 2ν) ,
E
E
and so d = 0, when ν = 1/2. Materials for which ν = 1/2 are thus said to be incompressible. Care
must be taken when considering the limit ν → 1/2 in equation (4.9). In fact, the distinguished limit in
question is ν → 1/2 and d → 0, in such a way that the term d/(1 − 2ν) remains finite.
Chapter 5
The governing equations of linear
elasticity
5.1
Navier–Lamé equations — displacement formulation
The constitutive equation for an isotropic, homogeneous linear elastic body is
τij = λ δij ekk + 2µ eij ,
(5.1)
where λ and µ are the (experimentally determined) Lamé coefficients, τij is the stress tensor and eij is
the strain tensor defined by
1
(5.2)
eij = (ui,j + uj,i ) .
2
Here, ui is the component of the displacement vector, u, in the xi direction. Using equation (5.2) in
equation (5.1) we obtain the an expression for the stresses in terms of the displacements
τij = λ δij uk,k + µ (ui,j + uj,i ) .
(5.3)
The equations of motion for an elastic body are
ρüi = τij,j + Fi ,
(5.4)
where ρ is the density of the body and F is the body force per unit volume.
Now, taking the derivative of equation (5.3) gives
τij,j = λ δij uk,kj + µ (ui,jj + uj,ij ) = λ uk,ki + µ ui,jj + µuj,ji = (λ + µ)uk,ki + µ ui,jj ,
by using the symmetry of the stress tensor and the index-switching property of the Kronecker delta.
Using the last result in equation (5.4), we obtain the Navier–Lamé equations
ρüi = (λ + µ) uk,ki + µ ui,jj + Fi .
The Navier–Lamé equations may be written in the equivalent form
ρü = (λ + µ) grad(div u) + µ ∇2 u + F ,
or even
ρü = (λ + 2µ) grad(div u) − µ curl(curl u) + F ,
by using the vector identity
∇2 u = grad(div u) − curl(curl u).
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(5.5)
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Note
• The general case of a non-zero body force is not very different from the “special” case F = 0
because the Navier–Lamé equations are linear. F acts as an inhomogeneity that can be removed if
a suitable particular solution is found. We seek a function û such that
(λ + µ) ∇∇ · û + µ ∇2 û + F = 0,
irrespective of the boundary conditions. If we then define the displacement field to be u = ũ + û,
then ũ satisfies the homogeneous equation
(λ + µ) ∇∇ · ũ + µ ∇2 ũ = 0,
but the boundary conditions may have changed from the original problem.
5.2
Beltrami–Mitchell equations — stress formulation
The Navier–Lamé equations treat the displacements as the unknowns in elasticity problems; the strains
and stresses are calculated indirectly, once the displacements are known. An alternative approach is to
formulate the set of governing equations so that the stresses are the unknowns and are calculated directly.
We can then use the constitutive law to determine the strains and hence the displacements. There is a
potential problem, however, the displacements cannot be uniquely recovered from the strain field unless
the strain compatibility conditions are satisfied.
eij,kl + ekl,ij − ekj,il − eil,kj = 0.
(5.6)
The inverse form of the linear constitutive law (5.1) is
eij =
1
λ
τij −
δij τkk ,
2µ
2µ(3λ + 2µ)
and using the engineering (experimental) constants E and ν, we obtain a slightly neater expression
E
ν
eij = τij −
δij Θ.
1+ν
1+ν
(5.7)
Substituting the expression (5.7) into the strain compatibility equations (5.6) yields
i
ν h
δij Θ,kl + δkl Θ,ij − δkj Θ,il − δil Θ,kj = 0.
τij,kl + τkl,ij − τkj,il − τil,kj −
1+ν
All the terms in the above expression are second derivatives of τij , so that the equations are all satisfied
if the stress is a linear function of position, in which case every term is zero. Remember that only six
of these 81 equations are linearly independent, so we are free to combine equations by setting k = l and
summing over the repeated index to obtain nine equations, of which six are still independent through the
symmetry of the stress tensor,
τij,kk + τkk,ij − τkj,ik − τik,kj −
ν
[δij Θ,kk + δkk Θ,ij − δkj Θ,ik − δik Θ,kj ] = 0.
1+ν
Using the fact that τkk = Θ, δkk = 3 and the index-switching property of the Kronecker delta, the
equations become
ν
[δij Θ,kk + 3Θ,ij − Θ,ij − Θ,ij ] = 0,
1+ν
ν
[δij Θ,kk + Θ,ij ] = 0,
⇒ τij,kk + Θ,ij − τkj,ik − τik,kj −
1+ν
ν
1
Θ,ij −
Θ,kk δij − τkj,ik − τik,kj = 0.
⇒ τij,kk +
1+ν
1+ν
The equation (5.8) can be written in the alternative form
τij,kk + Θ,ij − τkj,ik − τik,kj −
∇2 τij +
ν
1
Θ,ij −
δij ∇2 Θ = τkj,ik + τik,kj .
1+ν
1+ν
(5.8)
(5.9)
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Differentiating the equations of static equilibrium τij,j + Fi = 0, we obtain τij,jk + Fi,k = 0, which can
be used to replace the terms on the right-hand side of equation (5.9).
∇2 τij +
1
ν
Θ,ij −
δij ∇2 Θ = −Fj,i − Fi,j .
1+ν
1+ν
(5.10)
Setting i = j in equation (5.10) gives
∇2 τii +
1
ν
Θ,ii −
3∇2 Θ = −2Fi,i
1+ν
1+ν
⇒
⇒ ∇2 Θ +
1
3ν
∇2 Θ −
∇2 Θ = −2Fi,i
1+ν
1+ν
1−ν 2
∇ Θ = −Fi,i ;
1+ν
(5.11)
and on substitution of (5.11) into equation (5.10) we finally obtain the Beltrami–Mitchell (compatibility)
equations
1
ν
∇2τij + 1+ν
(5.12)
Θ,ij + 1−ν
δij Fk,k + Fi,j + Fj,i = 0.
These are stress compatibility equations and we note that in unsteady problems, we must add the inertia
force −ρüi to Fi .
The Beltrami–Mitchell equations are six linear partial differential equations for the six independent
components of the stress tensor. As one might expect, these equations are (very) useful if the problem
has stress boundary conditions. The equations are less useful if displacement boundary conditions are
specified.
Notes
• The equations of equilibrium are automatically satisfied by construction.
• The strains eij can be recovered from the constitutive equation.
• Once the strain field is known, the displacements can be obtained by integrating
eij =
1
(ui,j + uj,i ) .
2
Remember that the constants of integration represent rigid-body motions.
• The above integration is possible because we started from the strain compatibility equations and,
therefore, if the Beltrami–Mitchell equations are satisfied so are the strain compatibility equations;
and hence, the body will be continuous.
5.3
The steady, constant-body-force case
If the body force is a constant and there are no unsteady terms, then all of the derivatives Fi,j will be
zero. The Beltrami–Mitchell equations become
∇2 τij +
1
Θ,ij = 0,
1+ν
and from equation (5.11) we have
∇2 Θ = 0.
Thus, Θ is a harmonic function: a function that satisfies Laplace’s equation. Recall that Θ = (3λ + 2µ) d,
which implies that
∇2 d = 0,
and so d = ekk = uk,k is also a harmonic function. If we apply the operator ∇2 to the (steady) Navier–
Lamé equations (5.5) we obtain
(λ + µ) uk,kijj + µ ui,kkjj = 0.
MATH35021: Reading Week Material
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The first term may be written as (uk,kjj ),i and because uk,k is harmonic then uk,kjj = 0 and hence the
first term is zero. Thus,
ui,kkjj = 0, or rather ∇4 u = 0,
and so the displacement field is a solution of the biharmonic equation. Furthermore, by using the definition
of the strain tensor (5.2) and the expression for the stress tensor in terms of the displacements (5.3), it
follows that
∇4 τij = 0, and ∇4 eij = 0;
the stress and strain components are both biharmonic functions.
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5.4
Alternative coordinate systems:
5.4.1
Governing Equations in Cylindrical Polar Coordinates
• x1 = x = r cos θ, x2 = y = r sin θ, x3 = z = z.
u = (ur , uθ , uz ),
e = (eij ),
τ = (τij ),
where i, j = r, θ, z.
• Vector calculus:
∂f
∂f
1 ∂f
1 ∂(rur ) 1 ∂uθ
∂uz
θ̂ +
r̂ +
ẑ,
div u =
+
+
,
∂r
r ∂θ
∂z
r ∂r
r ∂θ
∂z
∂ur
∂uθ
∂uz
1 ∂(ruθ ) 1 ∂ur
1 ∂uz
r̂ +
ẑ.
θ̂ +
−
−
−
curl u =
r ∂θ
∂z
∂z
∂r
r ∂r
r ∂θ
grad f =
• Stress-strain relations have the same form as in Cartesian coordinates:
τij = λδij div u + 2µeij ,
i, j = r, θ, z.
• Stress-displacement relations:
τrr
∂ur
= λ div u + 2µ
,
∂r
τθθ = λ div u + 2µ
τθr
∂uθ
uθ
1 ∂ur
τrθ
=
=
−
+
,
µ
µ
∂r
r
r ∂θ
ur
1 ∂uθ
+
r ∂θ
r
τrz
τzr
∂uz
∂ur
=
=
+
,
µ
µ
∂r
∂z
,
τzz = λ div u + 2µ
∂uz
,
∂z
τθz
τzθ
1 ∂uz
∂uθ
=
=
+
.
µ
µ
r ∂θ
∂z
• Strain-displacement relations:
err =
2erθ = 2eθr =
∂ur
,
∂r
eθθ =
∂uθ
uθ
1 ∂ur
−
+
,
∂r
r
r ∂θ
1 ∂uθ
ur
+ ,
r ∂θ
r
2erz = 2ezr =
ezz =
∂ur
∂uz
+
,
∂z
∂r
∂uz
,
∂z
2ezθ = 2eθz =
1 ∂uz
∂uθ
+
.
r ∂θ
∂z
• Equilibrium equations (statics): for the displacement formulation, use Navier’s equation,
(λ + 2µ) grad div u − µ curl curl u + F = 0,
whereas for the stress formulation, use
1 ∂τrθ
∂τrz
τrr − τθθ
∂τrr
+
+
+
+ Fr
∂r
r ∂θ
∂z
r
1 ∂τθθ
∂τθz
2
∂τrθ
+
+
+ τrθ + Fθ
∂r
r ∂θ
∂z
r
∂τrz
1 ∂τθz
∂τzz
1
+
+
+ τrz + Fz
∂r
r ∂θ
∂z
r
=
0
=
0
=
0.
• Stress boundary conditions: these are when t is prescribed. We have, from ti = n̂j τij ,
tr
=
n̂r τrr + n̂θ τrθ + n̂z τrz
tθ
tz
=
=
n̂r τrθ + n̂θ τθθ + n̂z τθz
n̂r τrz + n̂θ τθz + n̂z τzz
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MATH35021: Reading Week Material
5.4.2
Governing Equations in Spherical Polar Coordinates
• x1 = x = r sin θ cos φ, x2 = y = r sin θ sin φ, x3 = z = r cos θ.
u = (ur , uθ , uφ ),
e = (eij ),
• Vector calculus:
grad f =
τ = (τij ),
where i, j = r, θ, φ.
∂f
1 ∂f
1 ∂f
θ̂ +
φ̂,
r̂ +
∂r
r ∂θ
r sin θ ∂φ
∂ 2
∂
∂
(r sin θ ur ) +
(r sin θ uθ ) +
(ruφ ) ,
∂r
∂θ
∂φ
r̂
rθ̂ r sin θ φ̂ ∂
1
∂
∂
.
curl u = 2
∂φ
r sin θ ∂r ∂θ
ur ruθ r sin φ uφ 1
r2 sin θ
div u =
• Stress-strain relations have the same form as in Cartesian coordinates:
τij = λδij div u + 2µeij ,
i, j = r, θ, φ.
• Stress-displacement relations:
τrr = λ div u + 2µ
τφφ
2µ
= λ div u +
r
∂ur
,
∂r
τθθ = λ div u +
1 ∂uφ
+ ur + uθ cot θ ,
sin θ ∂φ
τrφ
τφr
1 ∂ur
∂uφ
uφ
=
=
+
−
,
µ
µ
r sin θ ∂φ
∂r
r
2µ
r
∂uθ
+ ur ,
∂θ
τrθ
τθr
∂uθ
uθ
1 ∂ur
=
=
−
+
,
µ
µ
∂r
r
r ∂θ
τθφ
τφθ
1 ∂uθ
1 ∂uφ
uφ cot θ
=
=
+
−
.
µ
µ
r sin θ ∂φ
r ∂θ
r
• Strain-displacement relations:
err =
∂ur
,
∂r
2erθ = 2eθr =
eθθ =
1 ∂uθ
ur
+ ,
r ∂θ
r
∂uθ
uθ
1 ∂ur
−
+
,
∂r
r
r ∂θ
2eφθ = 2eθφ =
eφφ =
1 ∂uφ
ur
uθ cot θ
+
+
,
r sin θ ∂φ
r
r
2erφ = 2eφr =
1 ∂ur
∂uφ
uφ
+
−
,
r sin θ ∂φ
∂r
r
1 ∂uθ
1 ∂uφ
uφ cot θ
+
−
.
r sin θ ∂φ
r ∂θ
r
• Equilibrium equations (statics): for the displacement formulation, use Navier’s equation,
(λ + 2µ) grad div u − µ curl curl u + F = 0,
whereas for the stress formulation, use
1 ∂τrθ
1 ∂τrφ
2τrr − τθθ − τφφ + cot θ τrθ
∂τrr
+
+
+
+ Fr
∂r
r ∂θ
r sin θ ∂φ
r
∂τrθ
1 ∂τθθ
1 ∂τθφ
3τrθ + (τθθ − τφφ ) cot θ
+
+
+
+ Fθ
∂r
r ∂θ
r sin θ ∂φ
r
1 ∂τθφ
1 ∂τφφ
3τrφ + 2τθφ cot θ
∂τrφ
+
+
+
+ Fφ
∂r
r ∂θ
r sin θ ∂φ
r
=
0
=
0
=
0.
• Stress boundary conditions: these are when t is prescribed. We have, from ti = n̂j τij ,
tr
tθ
=
=
n̂r τrr + n̂θ τrθ + n̂φ τrφ
n̂r τrθ + n̂θ τθθ + n̂φ τθφ
tφ
=
n̂r τrφ + n̂θ τθφ + n̂φ τφφ