Continuum mechanics
III. Conservation laws and mechanical equilibria
Aleš Janka
office Math 0.107
ales.janka@unifr.ch
http://perso.unifr.ch/ales.janka/mechanics
February 23, 2011, Université de Fribourg
Aleš Janka
III. Conservation laws and equilibria
1. Time-derivative of volume integral in Euler formulation
Scalar field Φ(y, t) (eg. density, concentration)
Volume integral: over current (deformed) domain Ωt :
Z
I=
Φ(y, t)dy
Ωt
Change in I(t): due to two phenomena:
Change of Φ(., t) inside of the fixed domain Ωt over time dt:
Z
Z
∂Φ(y, t)
∂Φ(y, t)
dI1 =
dt dy = dt
dy
∂t
∂t
Ωt
Ωt
Time flow (in- and out-flow) of continuum through ∂Ωt :
Z
Z
Φ ni v i dΓ
dI2 =
Φ n · v dt dΓ = dt
∂Ωt
∂Ωt
By the Divergence theorem:
Z
dI2 = dt
Ωt
Aleš Janka
∂
i
Φv dy
∂y i
III. Conservation laws and equilibria
1. Time-derivative of volume integral in Euler formulation
Time-derivative of volume integral:
Z d
∂Φ
∂
DI
=
(I1 + I2 ) =
+ i Φ v i dy
Dt
dt
∂y
Ωt ∂t
Z
=
i
∂Φ
∂Φ
∂v
+ vi
+ Φ i dy
∂t
∂y i
∂y
DΦ
∂v i
+ Φ i dy
Dt
∂y
Ωt
Z
=
Ωt
Aleš Janka
III. Conservation laws and equilibria
2. Mass conservation in Euler formulation
Application of the formula for DI
Dt : Φ ≡ % is the density:
Z DI
D%
∂v i
≡0=
+ % i dy
for all Ωt ⊂ Ω
Dt
∂y
Ωt Dt
We can tend Ωt to a point to get the point-form of
mass-conservation law (called the continuity equation):
D%
∂v i
∂%
∂
0=
+% i =
+ i (%v i )
Dt
∂y
∂t
∂y
Incompressible continuum: D%
Dt = 0:
∂v i
ie.
ρ i =0
∂y
Continuum at steady state: ∂%
∂t = 0:
∂
i
ρv
=0
ie.
∂y i
Aleš Janka
div(v) = 0
div(ρv) = 0
III. Conservation laws and equilibria
2. Mass conservation in Lagrange formulation
Compare local mass before (Ω0 ) and after (Ωt ) deformation:
Z
Z
%0 (x)dx =
%(y)dy
Ω0
Ωt
∂y i
%(y(x)) · det
dx
∂x j
Ω0
Z
=
Continuity equation:
∂y i
%0 = % · det
∂x j
Aleš Janka
III. Conservation laws and equilibria
3. Stress in Euler formulation: Euler stress tensor
cut the deformed configuration in two
dS0
cutting-plane normal n, ni = n · gi ,
area dS.
~
g3
n0
~
g1
delete one of the two parts (eg. the
upper one)
find a force dF which, applied on a
point in dS, simulates the effects of
the deleted part
dS
g3 = e3
~
g2
e2 =g2
Initial config.
y
e3
e3
e2
e1
dF depends (linearly) on n and dS
dF = τ ij ni gj dS
τ ij is the Euler stress tensor
Aleš Janka
dF
g1 = e1
x(y)
e1
n
III. Conservation laws and equilibria
Deformed config.
e2
3. Stress in Euler formulation: Euler stress tensor
cut the deformed configuration in two
dS0
cutting-plane normal n, ni = n · gi ,
area dS.
~
g3
dS
n0
~
g1
delete one of the two parts (eg. the
upper one)
g3 = e3
e2 =g2
Initial config.
y
e3
e3
e2
e1
dF
g1 = e1
x(y)
find a force dF which, applied on a
point in dS, simulates the effects of
the deleted part
n
~
g2
Deformed config.
e2
e1
dF depends (linearly) on n and dS
dF = τ ij ni gj dS
τ ij is the Euler stress tensor
Aleš Janka
III. Conservation laws and equilibria
3. Stress in Lagrange formulation: 1st Piola-Kirchhoff
cut the deformed configuration in two
cutting-plane normal n, ni = n · gi ,
area dS.
dS0
~
g3
the cutting surface in the initial
configuration has a normal n0 and
area dS0 .
dS
n0
~
g1
g3 = e3
~
g2
e2 =g2
Initial config.
y
e3
e1
recopy dF to the initial configuration!!
e3
e2
e1
Deformed config.
e2
↑ Euler formulation ↑
dF depends (linearly) on n0 and dS0
dF = σ ij n0i gj dS0
1st Piola-Kirchhoff stress tensor σ ij
Aleš Janka
dF
g1 = e1
x(y)
find a force dF which, applied on a
point in dS, simulates the effects of
the deleted (eg. upper) part
n
III. Conservation laws and equilibria
3. Stress in Lagrange formulation: 1st Piola-Kirchhoff
cut the deformed configuration in two
cutting-plane normal n, ni = n · gi ,
area dS.
dS0
g3 = e3 n
0
the cutting surface in the initial
configuration has a normal n0 and
area dS0 .
dS
^
g3
n
dF
e2 =g2
dF
^
g1 = e1
g2
^
g1
x(y)
Initial config.
find a force dF which, applied on a
point in dS, simulates the effects of
the deleted (eg. upper) part
y
e3
e3
e2
e1
Deformed config.
e2
e1
↑ Lagrange formulation ↑
recopy dF to the initial configuration!!
dF depends (linearly) on n0 and dS0
dF = σ ij n0i gj dS0
1st Piola-Kirchhoff stress tensor σ ij
Aleš Janka
III. Conservation laws and equilibria
3. Stress in Lagrange formulation: 2nd Piola-Kirchhoff
cut the deformed configuration in two
cutting-plane normal n, ni = n · gi ,
area dS.
dS0
g3 = e3 n
0
cutting surface in initial config.:
normal n0 and area dS0 .
dS
^
g3
n
dF
e2 =g2
^
g1 = e1
find a force dF which, applied on a
point in dS, simulates the effects of
the deleted (eg. upper) part
g2
^
g1
x(y)
Initial config.
y
e3
transform dF to dF0 in initial config.:
i
j ∂x
dF0 = dF
gi
∂y j
dF0 depends (linearly) on n0 and dS0
e1
e3
e2
e1
Deformed config.
e2
↑ 1st Piola-Kirchhoff ↑
dF0 = T ij n0i gj dS0
2nd Piola-Kirchhoff stress tensor T ij
Aleš Janka
dF
III. Conservation laws and equilibria
3. Stress in Lagrange formulation: 2nd Piola-Kirchhoff
cut the deformed configuration in two
cutting-plane normal n, ni = n · gi ,
area dS.
dS0
g3 = e3 n
0
cutting surface in initial config.:
normal n0 and area dS0 .
dS
dF0
^
g3
n
e2 =g2
^
g1 = e1
find a force dF which, applied on a
point in dS, simulates the effects of
the deleted (eg. upper) part
g2
^
g1
x(y)
Initial config.
y
e3
transform dF to dF0 in initial config.:
i
j ∂x
dF0 = dF
gi
∂y j
dF0 depends (linearly) on n0 and dS0
e3
e2
e1
Deformed config.
e2
e1
↑ 2nd Piola-Kirchhoff ↑
dF0 = T ij n0i gj dS0
2nd Piola-Kirchhoff stress tensor T ij
Aleš Janka
III. Conservation laws and equilibria
3. Stress: Cauchy vs. 1st/2nd Piola-Kirchhoff tensors
How to transform areas dS0 → dS?: Nanson’s relation:
dV0
dx
∂x `
dSj = J ·
dS0`
∂y j
dS
Initial config.
dS0
dy
dV
i
∂y
with J = det
.
∂x k
x
e3
e1
y (x)
e2
Deformed config.
Relation between Euler and 1st Piola-Kirchhoff:
dF
i
∂x `
= τ nj dS = τ dSj = τ J
dS0`
∂y j
= σ i` n0` dS0 = σ i` dS0`
ij
ij
ij
∂x `
σ =τ J
∂y j
i`
Aleš Janka
dF
ij
III. Conservation laws and equilibria
3. Stress: Cauchy vs. 1st/2nd Piola-Kirchhoff tensors
How to transform areas dS0 → dS?: Nanson’s relation:
dV0
dx
∂x `
dS0`
dSj = J ·
∂y j
dS
Initial config.
dS0
dy
dV
∂y i
.
with J = det
∂x k
x
e3
y (x)
e2
e1
Deformed config.
Relation between Euler and 2nd Piola-Kirchhoff:
dF0j
∂x j
∂x j km ∂x `
k
=
dF =
τ J
dS0`
∂y m
∂y k
∂y k
= T `j dS0`
T
`j
=τ
km
∂x j ∂x `
J
∂y k ∂y m
Aleš Janka
III. Conservation laws and equilibria
3. Stress: Cauchy vs. 1st/2nd Piola-Kirchhoff tensors
How to transform areas dS0 → dS?: Nanson’s relation:
dV0
dx
∂x `
dSj = J ·
dS0`
∂y j
dS
Initial config.
dS0
dy
dV
i
∂y
with J = det
.
∂x k
x
e3
e1
y (x)
e2
Deformed config.
Relation between 1st and 2nd Piola-Kirchhoff:
From relations to Euler stress:
T
`j
=τ
km
∂x j ∂x `
J
∂y k ∂y m
T
`j
and
=σ
Aleš Janka
k`
σ
k`
=τ
km
∂x `
J
∂y m
∂x j
∂y k
III. Conservation laws and equilibria
4. Force equilibria: balance of force and of momentum
Body loads: forces on internal points: electro-magnetic field,
gravity, acceleration
dF i = f i % dV − % ai dV
where f i is the body force per unit mass, ai is acceleration
and % is density
Surface (contact) loads: pressure and drag on the surface:
can be expressed by
τ ij nj
Equilibrium of forces and of momentum of forces (F × x).
Euler formulation is more natural for force equilibria than
Lagrange formulation
Aleš Janka
III. Conservation laws and equilibria
4. Force equilibria: Euler formulation
Z
ω
% f i dV +
Z
∂ω
τ ij nj dS = 0
for any ω ⊂ Ωt
By the Divergence theorem:
Z ij
∂τ
%f i +
dV = 0
j
∂y
ω
We can tend ω ⊂ Ωt to one point to get:
∂ ij
τ + %f i = 0
j
∂y
in the deformed domain Ωt .
Aleš Janka
III. Conservation laws and equilibria
4. Force equilibria: Lagrange formulation
Using 1st Piola-Kirchhoff stress: and body forces per unit mass
of initial configuration:
dF0i = f0i %0 dV0
Force equilibrium with 1st Piola-Kirchhoff stress:
∂ ij
i
σ
+
%
f
=0
0
0
∂x j
in the initial configuration Ω0 .
Using 2nd Piola-Kirchhoff stress: transform σ ij to T k` :
∂
∂x `
∂y k j`
T
∂x j
k
∂
∂u
j`
k
k
+ %0 f0k =
T
+
δ
+
%
f
=0
0
0
j
∂x j
∂x `
Aleš Janka
III. Conservation laws and equilibria
4. Force equilibria: momentum conditions
Euler formulation: momentum equilibrium about the origin:
Z
Z
y × % f i gi dV +
y × τ ij nj gi dS = 0
ω
∂ω
Levi-Civita symbol:

 +1 if (i, j, k) is an even permutation of (1, 2, 3)
−1 if (i, j, k) is an odd permutation of (1, 2, 3)
ijk =

0 if i = j or i = k or j = k
Properties of the Levi-Civita symbol:
a×b=c
ci = εijk aj b k .
⇔
Hence for moment equilibrium:
Z
Z
εijk y i % f j dV +
εijk y i τ j` n` dS = 0
ω
∂ω
Aleš Janka
III. Conservation laws and equilibria
4. Force equilibria: momentum conditions
By the Divergence theorem:
Z
Z
εijk y i % f j dV +
ω
ω
∂ i j`
εijk y τ
dS = 0
∂y `
Work-out the derivative in the second term to get:
j`
Z
Z
∂τ
+ % f j dV + εijk δ`i τ j` dV = 0
εijk y i
`
∂y
ω
ω
As the red term is 0 by component-wise conditions, momentum
conditions reduce to
εijk τ ji = 0
∀k = 1, 2, 3.
By definition of the Levi-Civita symbol, this means:
τ 12 = τ 21
,
τ 13 = τ 31
,
τ 23 = τ 32
Euler and 2nd Piola-Kirchhoff stress tensors are symmetric!
Aleš Janka
III. Conservation laws and equilibria
5. Transformation of tensors Lagrange ↔ Euler
Deformation gradient:
∂y i
i
Fj =
∂xj
i
F−1 j
∂x i
=
∂yj
∂y i
J = det
∂xj
Strain tensors: Lagrange ↔ Euler
εk` = Fki · Eij · F`j
Stress tensors: Lagrange (2nd Piola-Kirchhoff stress) ↔ Euler
j
i
T ij = J · F−1 k · τ km · F−1 m
Aleš Janka
III. Conservation laws and equilibria