Mechanics of biological structures (seconda parte) - Prof. Vena

Name: Mechanics of biological structures (seconda parte) - Prof. Vena
Brand: Skuola.net
Price: 7.99 EUR
Availability: InStock
Rating: 3.0 (1 reviews)
Author: Ing_bio

Revisionato il 17/05/2026

di Ing_bio

Publisher

Vota 3,0/5 (1)

Contenuto verificato e approvato dal Team di Esperti di Skuola.net

Appunti chiari e completi dell'interno corso di "Mechanics of biological structures" tenuto dal prof. Vena basati su appunti personali del publisher presi alle lezioni del prof. …

Esame Mechanics of Biological Structures

Facoltà Ingegneria dei sistemi

Dal corso del Prof. Vena Pasquale

Università Politecnico di Milano

A.A. 2016-2017

101 pagine

1 download

Appunto

Scarica

Estratto del documento

Strain tensor

Strain tensor can be calculated for any magnitude of strain:

Ξ = 1/2 [Φ + Φ_transpose + Φ_transposeΦ]
E₁₁ ➔ stretch of fibers directed along direction 1
E₁₂ ➔ change of angle between direction 1 and 2

For normal strain, we neglect Φ_transposeΦ:

Ξ ≈ Ξ - 1/2 [Φ + Φ_transpose] _{second order term}

Ξ_ij = 1/2 (∂U_i/∂X_j + ∂U_j/∂X_i)

X and X are quite the same, so we can deviate with respect to X or X; it's the same.

Numerical example

X₁ = X₁ + αX₁ we only have displacement in direction 1 α > 0

X₂ = X₂

X₃ = X₃ ➔ here X₂ = 0

Deformation gradient F:

1	α	0
0	1	0
0	0	1

Strain tensor calculation

ε = 1/2 [Φ + Φ_transpose + Φ_transposeΦ] 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 Φ_transposeΦ:

ε ➔ ξ - 1/2 [Φ + Φ_transpose] _{second order term}

ε_ij = 1/2 (∂u_i/∂x_j + ∂u_j/∂x_i)

X and X are quite the same, so we can deviate with respect to x or X; it's the same.

Numerical example

x₁ = X₁ + αX₁ _d>0 we only have displacement in direction 1

x₂ = X₂

x₃ = X₃ here, ξ₂ = 0

Deformation gradient F:

1	α	0
0	1	0
0	0	1

Green tensor

Green Tensor E:

E = E₁ - E = [α₀¹]- [n d 0]...[0 α 0]^{small strain}

Lagrange Strain Tensor E = ₁ / ₂ (E - I) = ₁ / ₂[0 α 0] can be neglected because it is small. If α is large, no way we have large strain α² plays a role.

α² E₂₂ ➔ no refers to the elongation of those fibers. E₄₄ = 0 ➔ no stretching of fibers along direction α. If we have a biological material with fibers directed in direction Z, they start to play a role when strains are large.

Stress state in large deformations

Now we want to describe the stress state in a material subjected to large deformation before we define the strain rate, which is the derivative of strain components:

σ = ∂U / ∂ε

Elastic energy W(ε) gives us the idea of how much energy we put into a material sample when it is subjected to this strain ε

σ = ∂W / ∂ε

ε̇ strain rate ➔ measures time derivative of strain

ξ = 1/2 ( ∂M_i/∂x_j + ∂M_j/∂x_i )

DD = 1/2 ( L + L^T ) where L = ∂M_i/∂x_j velocity gradient

Strain rate in small strain problem: D 2D = ( L+L^T )

F → EE = ∂x/∂X so F = ∂E/∂t

∂x/∂X = ∂x/∂XE = FE_t - (F^T E_t) - F^T E + E^T E = 2F^T D

EE = 1/2 C

It comes from ξ = 1/2 ( C - I ), we make the deviate = F^T D E

We use these results to define the stress. In large strain problems, we have more than one definition of stress.

σ̇ - ∂W/∂ξ = σ + ∂W/∂ξ

W : energy rate per unit volume

ρW : energy rate per unit mass

x x _{mean product that} ẇ = gives a scalar as result

ρ ẇ = ρ ẇ = σ_u D_u

When we define w(ξ), we need also to define ẇ̇ and then ρ ẇ

From energy considerations, we give other definitions:

ρ_o is density in current configurations. Now we define ρ in the initial configuration as ρ_o

ρ_o ẇ = ρ_o - ρ_o ẇ = ρ_o : D ρ = ρ_o ρ = ρ _o ρ = ρ = ρ_o = ρ_o = &ratio;o ρ _o = 0_op= dmso ρ ẇ = σ : D : ρ - σ : D :σ ρ σI = I

σ Cauchy stress Kirchhoff Strain Tensor

First Piola-Kirchhoff stress tensor

^{T _o}_i&suppro;&o;_m is a non symmetric stress tensor model for this reason modal for this reason F ⊃ energy based definition of stress

Anteprima

Vedrai una selezione di 22 pagine su 101