Overview
Energy Formulations of Material Laws
Time, spae, work
Finite Deformation: Speial Cases
Ben M. Jordan
Advaned Elastiity
Harvard University
June 5, 2009
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Outline
Examples
Topis
Denintions
Considè re ondition
Helmholtz free energy
Stability of equilibrium
Time-dependent,
inhomogeneous deformation
Virtual work formulation
Ben M. Jordan
Truss example
Neking example
Coexistent phases example
Critial fore and Gibbs free
energy example
Gibbs free energy of spherial
balloon example
Wave in pre-stressed bar
example
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Denitions
Truss example
Neking Examples
Overview
For small deformations, we applied F=ma in the referene
frame
For large deformations, this is no longer a valid
approximation
If we upgrade our denitions to handle more situations...
...formulate new material models using the free energy
density...
...and link it to the nite element method...
...we an explore all kinds of new phenomena!
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Denitions
Truss example
Neking Examples
Strains
Streth:
Engineering:
Natural:
Lagrange:
λ = Ll
e = l −LL = λ − 1
ε = log ( Ll ) = log (λ )
h i
2
η = 12 Ll − 1 = 12 λ 2 − 1
NOTE: These are all funtions of the streth (λ )
Stresses
True/Cauhy:
Nominal/1st Piola Kirho:
NOTE: Others an be dened as well
Ben M. Jordan
Finite Deformation: Speial Cases
σ = Pa
s = PA
Overview
Energy Formulations of Material Laws
Time, spae, work
Denitions
Truss example
Neking Examples
Inrements and Work-Conjugates
Inrements
Taking the derivates of our strain
denitions w.r.t. λ gives us
δ λ = δLl
δe = δλ
δ ε = δλλ
δη = λδλ
Work-onjugates
Find pairs that give
"inr. of work in ur." = P δ l
AL
"volume in ref."
With nominal stress,
Pδ l = sδ λ = sδ e
AL
With true stress, PaLδ l = σ δ ε
What about Lagrange?
P δ l = δ η (?) = λ δ λ (?)
AL
(?) = S = λs , the 2nd
Work
An inrement of work is given
by P δ l
Ben M. Jordan
Piola-Kirho Stress!
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Denitions
Truss example
Neking Examples
Truss Example
Deformation geometry
√2 2
l +l
λ1 = Ll11 = 1, λ2 = Ll22 , λ3 = Ll33 = √ 12 22
L1 +L2
Note: only unknown here is l2
Material Model: Neo-Hookean
si = µ (λi − λi−2 ), i = 1, 2, 3, s1 = 0
Note: nominal stress form
Fore Balane
Fx : By symmetry, this is 0
Fy : W − 2(os (θ )s3 a3 ) − s2 a2 = 0
Inompressibility
AL = al ⇒ a = ALl
Mixing results in...
W − 2 ll23 s3 A3l3L3
− s2 A2l2L2 = 0,
whih is 1 eqn. for
1 unk.
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Denitions
Truss example
Neking Examples
Neking 1
x-axis: ε , y-axis:P /AK
What is the fore/strain relation for a
power law material, suh as aluminum?
One possible formulation using natural
strain / true stress is:
ε = log (λ ), P = σ a
A power law material model is used,
whih ts the experimental ε → σ urves
well.
σ = K ε N , N = 12 for our example
Inompressibility: AL = al ⇒ a = Ae −ε
Mixing these gives σ = Pa = AeP−ε = K ε N
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Denitions
Truss example
Neking Examples
Neking 2
dP = AK (e −ε N ε N −1 − ε N − e −ε =
dε
a(σ ′ − σ ) = 0 at the maximum
a annot be zero, and thus σ ′ = σ (the
Considè re condition)
Applying our material model, this gives
ε =N
x-axis: ε , y-axis:P /AK
Note that σ ′ is the tangent modulus
Material hardening / geometri softening
Model is invalid after this ondition, as
this is only good for homogenous
deformation
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Denitions
Truss example
Neking Examples
Neking 3
We an also formulate this using s , λ
λ = Ll , P = sA
Using ε = log (λ ) and AL = al ,the power
law material model beomes
N
s = K log (λλ )
Mixing these gives
N
x-axis: λ , y-axis:P /AK
P = KA log (λλ ) = As (λ )
dP = As ′ ⇒ s ′ = 0 (Considè re ondition)
dλ
Taking this derivative from the material
model gives N = log (λ ) ⇔ λ = e N
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Denitions
Truss example
Neking Examples
Neking 4
What about for Neo-Hookean materials?
Formulate this using s , λ
λ = Ll , P = sA
Material model beomes s = µ (λ − λ −2 )
P = µ A(λ − λ −2)
dP = µ A 1 + 2 , 1 + 2 = 0
dλ
λ3
λ3
(Considè re ondition)
√
x-axis: λ , y-axis:P /Aµ
This is satised when λ = 3 −2, whih
is not valid.
Conlusion: no neking in Neo-Hookean
materials.
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Helmholtz free energy
Gibbs free energy
Examples
Helmholtz free energy 1
θ is temp., l is length
Given F (l , θ ),δ F = ∂∂Fl δ l + ∂∂ Fθ δ θ = P δ l + ηδ θ
Assuming that temperate hange is negligible, i.e. adiobati, and
thus onsider δ F = P δ l
Reall P = sA and l = λ L ⇒ δALF = PALδ l
δ W = s δ λ ⇒ s (λ ) = dW
d λ , where W is the energy density
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Helmholtz free energy
Gibbs free energy
Examples
Helmholtz free energy 2
Using our material models, we an write them in terms of energy
density
For neo-Hookean, s (λ ) = µ (λ − λ −2 ) = dW
dλ Integrating gives W (λ ) = µ2 λ 2 + 2λ −1 − ,W (1) = 0 ⇒ = 3
N
For power law, s = K log (λλ ) = dW
dλ
Integrating gives W (λ ) = NK+1 log (λ )N +1
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Helmholtz free energy
Gibbs free energy
Examples
Stability of equilibrium
If fore is stati, then objet may equilibrate. If so, what is λ in this
state? Is it stable?
Gibbs free energy = potential energy=Helmholtz free energy - work
done by external fore
G = ALW (λ ) − PL · (λ − 1) = F (λ ) − P · (l − L)
Using thermodynamis, we an show that the equilibrium state is
reahed when G is minimized.
Evaluating G at λ + δ λ by expanding a Taylor series about λ
(x0 = λ , x − x0 = δ λ )
G (λ + δ λ ) = G (λ ) + G ′(λ )(δ λ ) + G 2(λ ) (δ λ )2 + ...
Setting G ′ (λ ) = ALW ′ − PL = 0 ⇒ W ′ = PA = s (stress is reovered)
This is a stable equillibrium if G ′′ (λ ) > 0 i.e. if W ′′ > 0
NOTE: Reall from neking that W ′′ = s ′ = 0 where maximal fore is
′′
ahieved.
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Helmholtz free energy
Gibbs free energy
Examples
Coexistent phases example
Mixed phases → non-onvex energy
density → material model not one to one.
L′ + L′′ = L, l ′ + l ′′ = l , λ ′ L′ + λ ′′L′′ = l
Problem: For some nononvex W (λ ),nd
the unstable stress regime.
′
dW
st = Wλ ′′′′−−W
λ ′ = d λ is the tangent line that
passes through the points λ ′ and λ ′′
Maxwell's rule:
dW = s at λ ′ and λ ′′
t
Rdλλ′′
λ ′ sd λ = 0
The area under the urve is equal
F = W (λ ′ )AL′ + W (λ ′′ )AL′′
Summary: In equilibrium the phases
separate, but if metastable, they oexist.
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Helmholtz free energy
Gibbs free energy
Examples
Gibbs example 1
Problem: For a power law material, determine P and plot
G (λ ) for value around it.
Power law in terms of nominal stress: s = Kλ log (λ )N
N −1
N
Considè re ondition: s ′ = 0 ⇒ ddsλ = K (log (λ ) λ 2N −log (λ ) = 0
λ = 1 or λ = e N
N
N
P = kKA log (λλ ) = sA ⇒ P = KAN
eN
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Helmholtz free energy
Gibbs free energy
Examples
Gibbs example 2
Let's examine the behaviour
around PC of G (λ )
Reall W (λ ) = NK+1 log (λ )N +1
G (λ ) = ALW (λ ) − PL · (λ − 1)
For the above G(lambda), the
fore is less than the ritial
fore, and for small lambda
near 1, the helmholtz free
energy is less than the work
being done.
x-axis: λ , y-axis:G (λ ), P < P
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Helmholtz free energy
Gibbs free energy
Examples
Gibbs example 3
G (λ ) = ALW (λ ) − PL · (λ − 1)
For Pmore, the work being
done is always greater than
the Helmholz free energy.
x-axis: λ , y-axis:G (λ ), P > P
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Helmholtz free energy
Gibbs free energy
Examples
Spherial balloon example 1
Problem: Disuss Gibbs free energy of
balloon
G = 4π r 2 HW − p 34 π r 3
Using three ingredients:
Def Geom: λ1 = λ2 = 22ππRr = Rr , λ3 = Hh
Inompressibility:
4π R 2 H = 4π r 2 h ⇒ λ3 = λ1−2
Fore balane: σ3 ≈ 0, σ1 = σ2 = Pr
2h ,
(biaxial state)
Consider a half sphere:
2π rhσ1 = π r 2 p
σ = (σ1 , σ2 , 0) and add hydrostati
pressure to get (0, 0, −σ1 )
Material model (true stress):
−σ1 = µ · (λ 2 − λ −1 )
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Helmholtz free energy
Gibbs free energy
Examples
Spherial balloon example 2
Result: p = −R2H (λ1−7 − λ1−1 )
We an nd ritial pressure from here, and
plot G (λ )as before.
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Time-dependent, inhomogeneous deformation
Now we move to a more ompliated model and onsider time
and spae variation
4 ingredients:
,t )−x (X ,t )
(X ,t )
Def Geom:
λ (X , t ) = x (X +dXdX
= ∂ xdX
Cons. of Mass:
ρ (X ) only.
Cons. of Momentum (Fore Balane):
s (X , t )A(X ))dX + B (X , t )A(X )dX = ρ (X ) ∂
∂
∂X (
Mat. Model:
Ben M. Jordan
2 x (X ,t )
∂t
Finite Deformation: Speial Cases
A(X )dX
s = g (λ )
Overview
Energy Formulations of Material Laws
Time, spae, work
Wave in pre-stressed bar example
Using the ingredients:
Def. Geom: x (X , t ) = λ0 X + u (X , t ),
λ = ∂∂Xx = λ0 + ∂∂Xu
Mat. Model: Expand s = g (λ ) using TS to
nd:
s ≈ g (λ0 ) + g ′(λ0 )(λ − λ0) + 12 g ′′(λ0 )(λ −
λ0 )2 + ...
2
Cons. of Momentum: ∂∂X s (X , t ) = ρ ∂ x∂(tX2 ,t )
By keeping up to linear terms from TS, we an mix
ingredients to nd:
g ′ (λ0 ) ∂ 2 u
∂ 2u
=
2
ρ ∂ t2
∂t
q ′
= g (λ0 )
ρ
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Virtual work formulation
Dene virtual displaement: δ x = δ x (X )
Virtual streth: δ λ = ∂ (∂δXλ )
Using onservation of momentum, we an write an expression
for virtual work:
As δ x |XX21 +
R X2 ∂
X1 ∂ X (sA)δ xdX
R
Integrating by parts gives us: XX12 As ∂∂X (δ x )dX
For an arbitrary segment, s δ λ = s ∂∂X (δ x ) is the virtual work
done by all fores on the segment.
This is the basis for the nite element method.
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
Researh
Biomaterials are ompliated, inhomogenous, anisotropi
materials
Tissues, in partiular, require onsideration of speial material
models
Tissues also grow and this addition of mass and volume
hange must be onsidered
My ongoing work will onsider various models for growth
Appliations to limb, root, and ell, and embryo growth.
Ben M. Jordan
Finite Deformation: Speial Cases
Overview
Energy Formulations of Material Laws
Time, spae, work
The end
Thank you all for a great semester.
Ben M. Jordan
Finite Deformation: Speial Cases