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Strain tensor
E = 1/2 [Φ + Φᵀ + ΦᵀΦ]
can be calculated for any magnitude of strain
E₁₁: stretch of fibers directed along direction 1
E₁₂: change of angle between direction 1 and 2
for normal strain we neglect ΦᵀΦ
Ɛ = Ξ - 1/2 [Φ + Φᵀ] second order term
Ɛᵢⱼ = 1/2 (∂Uᵢ/∂Xⱼ + ∂Uⱼ/∂Xᵢ)
X and X are quite the same, so we can deviate with respect to x or X, it's the same
numerical example:
X₁ - Xₓ, dX₁: we only have displacement in direction
(x₃ = 0)
here x₃ = 0
deformation gradient
F = ∂x₁/∂X₁
Cauchy Green Tensor
E = E1 - E2 = 1/2 [ [1 α 0] [1 α 0] ]T · [ [1 0 0] - [1 0 0] ] [ [0 1 0] - [α 1+α2 0] ]T · [ [0 α 0] - [α 1+α2 0] ] [ [0 0 1] ]
Euler Lagrange Strain Tensor
E2 = 1/2 [ [1 α 0] · [α 0] ]T [ [0 α 0] · [α²] ]
can be neglected because it is small
= small strain
If α is large, so when we have large strain
α² plays a role
therefore, no refers to the elongation of those fibers Em=0 → no stretching of fibers along direction 1
If we have a biological material with fibers directed in direction 2, they start to play a role when strains are large.
Now we want to describe the stress state in a material subjected to large deformation.
Before we define the strain rate that is the derivative of strain components:
elastic energy
W( Εs ), given as the idea of how much energy was put in the material sample when it is subjected to this strain Ε
stress → strain rate → means time derivative of strain
T = N initial
T = N current
cross section Ao
thin in after compression with cross section a
Tz = P / Ao
The practical determination of a during experiments is not easy.
When we do practical experiments we use the initial plate.
when we have F we can go from T to d to T to S
exercise: isotropic material
cross section
x1 - x2 = U1 x3
x2 - x3 = U2 x3
are the same because the material is isotropic
If the material is isotropic we can write
W0 = W0 (I1, I2, I3)
and this means that W0 only depends from invariants example:
W0 = 1/2 (En2 + E122)
this is not an isotropic material
S = 2∂W0/∂E = 2 [ ∂W0/∂I1 ∂I1/∂E + ∂W0/∂I2 ∂I2/∂E + ∂W0/∂I3 ∂I3/∂E ]
depend on the mechanical properties of the material
∂I1/∂E = I
∂I2/∂E = I1I - C
∂I3/∂E = I3 C-1
C1 = 2 ∂W0/∂I1
C2 = 2 ∂W0/∂I2
C3 = 2 ∂W0/∂I3
are functions
Cx (I1; I2; I3)
⟨ S = C1I + C2(I1I - C) + C3I3C-1 ⟩
constitutive eq. for isotropic materials
S22 = 0
C1 + P λ1 = 0, ρ = - CA/λ1
Idrostatic pressure
we put ρ in S11
S11 = CA + (- C1/λ1 ⋅ 1/λ22) = CA(λ - λ/λ3)
stretch (= strain)
when λ1 → ∞, S11 → CA
If we want to go from S to σ we apply the transformation
- σ = 1/λ F S ET
σ11 = S11 λ2
= CA(λ2 - 1/λ)
E = ⎡⎣ λy 0 0 ⎤⎦ ⎡⎣ ΓR 0 0 ⎤⎦ ⎡⎣ λ 0 0 ⎤⎦
⎢ 0 λθ 0 ⎥ ⎢ 0 ΓL 0 ⎥ = ⎢ 0 λ 0 ⎥
⎣ 0 0 λϕ ⎦ ⎣ 0 0 ΓH ⎦ ⎣ 0 0 λ ⎦
We also know that this material is incompressible and so:
J = det(E) = 1
λϕ λr = 1 →
λr = 1/λ2
F = ⎡⎣ 1 0 0 ⎤⎦
⎢ 0 1 0 ⎥
⎣ 0 0 -1/λ2 ⎦
Now we can calculate B or C
→ C = B = F2 = ⎡⎣ λ2 0 0 ⎤⎦
⎢ 0 λ2 0 ⎥
⎣ 0 0 λ-4 ⎦
→ Since F is diagonal → C = B
Now we work with
σ = G B + C2[I B - B2] - P I
Since there are no shear deformations and E is diagonal also the I matrix and so the strain matrix are diagonal. This means that there are no shear strains in this problem, but we have only longitudinal strain, along 3 dir. and so we expect that our material is isotropic and that no shear stresses will be present in our material. So we consider the three direct primary stress components,
σyy = Gθθ = T = C Bθθ + C2(I Bθθ - Bθθ2) - P
σθθ =
σrr = C Brr + C2(I Brr - Brr2) - P
LECTURE 16 - 30-11-2016
Consider now a cylindrical shape.
R, H and L refer to the initial configuration. Then we apply an internal pressure and we consider a boundary condition in when we have σ0, that is the stress every component.
F = 2πRH σ0
This is the total force we are applying to the ends of our cylinder.
The assumption we make is that σ0 is constant along all over the thickness of the membrane. This condition of uniformity of the axial stress is acceptable if the membrane is thin enough and σ0 can be > = 0.
σ0 ≥ 0
Again we have 3 main directions and now the good choice is to use the cylindrical coordinates. If our material is isotropic we can suppose that one we initiate we will have a new configuration.
In the 1st phase the line (9) elongates because we are applying σc and this causes a little reduction of radius. Thus for a given constant σc we start inflating and so part of the structure will have a larger radius and part a smaller radius.
This separation between small and larger radius is faster going along the length of the cylinder.
ANISOTROPIC TISSUE: VASCULAR TISSUE
- anisotropic
- non linear response
Making an uniaxial tensile test we will have a non linear relationship between stress and strain and we have a continuous increase of stiffness of the material during deformation. This does not happen in a rubber like material because we will see a decrease of stiffness. The consequence of this non continuous increase of stiffness is an unstable response like that we had in the balloon and so considering a constitutive eq. which are not of the kind of soft biological tissue.
Moreover applying a stress high enough we have a change of behaviour. However we are interested to describe only the physiological range.
We can also notice that the loading and unloading curves are not overlapped and this means that we have a dissipation mechanism. This means that the energy given back during the unloading is not of the same amount of that given.
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