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Applied linear operators and spectral methods/Lecture 1

Linear operators can be thought of as infinite dimensional matrices. Hence we can use well known results from matrix theory when dealing with linear operators. However, we have to be careful. A finite dimensional matrix has an inverse if none of its eigenvalues are zero. For an infinite dimensional matrix, even though all the eigenvectors may be nonzero, we might have a sequence of eigenvalues that tend to zero. There are several other subtleties that we will discuss in the course of this series of lectures.

Let us start off with the basics, i.e., linear vector spaces.

Linear Vector Spaces (S)
Let $$\mathcal{S}$$ be a linear vector space.

Addition and scalar multiplication
Let us first define addition and scalar multiplication in this space. The addition operation acts completely in $$\mathcal{S}$$ while the scalar multiplication operation may involved multiplication either by a real (in $$\mathbb{R}$$) or by a complex number (in $$\mathbb{C}$$). These operations must have the following closure properties:

 * 1) If $$\mathbf{x}, \mathbf{y} \in \mathcal{S}$$ then $$\mathbf{x} + \mathbf{y} \in \mathcal{S}$$.
 * 2) If $$\alpha \in \mathbb{R}$$ (or $$\mathbb{C}$$) and $$\mathbf{x} \in \mathcal{S}$$ then $$\alpha~\mathbf{x} \in \mathcal{S}$$.

And the following laws must hold for addition

 * 1) $$\mathbf{x} + \mathbf{y}$$ = $$\mathbf{y} + \mathbf{x} \qquad$$ Commutative law.
 * 2) $$\mathbf{x} + (\mathbf{y} + \mathbf{z})$$ = $$(\mathbf{x} + \mathbf{y}) + \mathbf{z} \qquad$$ Associative law.
 * 3) $$\exists \mathbf{0} \in \mathcal{S}$$ such that $$\mathbf{0} + \mathbf{x} = \mathbf{x} \quad \forall \mathbf{x} \in \mathcal{S} \qquad $$ Additive identity.
 * 4) $$\forall \mathbf{x} \in \mathcal{S} \quad \exists -\mathbf{x} \in \mathcal{S}$$ such that $$-\mathbf{x} + \mathbf{x} = \mathbf{0} \qquad $$ Additive inverse.

For scalar multiplication we have the properties

 * 1) $$\alpha~(\beta~\mathbf{x}) = (\alpha~\beta)~\mathbf{x}$$.
 * 2) $$(\alpha + \beta)~\mathbf{x} = \alpha~\mathbf{x} + \beta~\mathbf{x}$$.
 * 3) $$\alpha~(\mathbf{x}+\mathbf{y}) = \alpha~\mathbf{x} + \alpha~\mathbf{y}$$.
 * 4) $$\mathbf{1}~\mathbf{x} = \mathbf{x}$$.
 * 5) $$\mathbf{0}~\mathbf{x} = \mathbf{0}$$.

Example 1: n tuples
The $$n$$ tuples $$(x_1, x_2, \dots, x_n)$$ with

\begin{align} (x_1, x_2, \dots, x_n) + (y_1, y_2, \dots, y_n) & = (x_1 + y_1, x_2 + y_2, \dots, x_n + y_n) \\ \alpha~(x_1, x_2, \dots, x_n) & = (\alpha~x_1, \alpha~x_2, \dots, \alpha~x_n) \end{align} $$ form a linear vector space.

Example 2: Matrices
Another example of a linear vector space is the set of $$2 \times 2$$ matrices with addition as usual and scalar multiplication, or more generally $$n\times m$$ matrices.

\alpha\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix} = \begin{bmatrix} \alpha~x_{11} & \alpha~x_{12} \\ \alpha~x_{21} & \alpha~x_{22} \end{bmatrix} $$

Example 3: Polynomials
The space of $$n$$-th order polynomials forms a linear vector space.

p_n = \sum_{j=1}^n \alpha_j~x^j $$

Example 4: Continuous functions
The space of continuous functions, say in $$[0, 1]$$, also forms a linear vector space with addition and scalar multiplication defined as usual.

Linear Dependence
A set of vectors $$\mathbf{x}_1, \mathbf{x}_2, \dots, \mathbf{x}_n \in \mathcal{S}$$ are said to be linearly dependent if $$\exists~ \alpha_1, \alpha_2, \dots, \alpha_n$$ not all zero such that

\alpha~\mathbf{x}_1 + \alpha~\mathbf{x}_2 + \dots + \alpha~\mathbf{x}_n = \mathbf{0} $$ If such a set of constants $$\alpha_1, \alpha_2, \dots, \alpha_n$$ do not exists then the vectors are said to be linearly independent.

Example
Consider the matrices

\boldsymbol{M}_1 = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}, \boldsymbol{M}_2 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \boldsymbol{M}_3 = \begin{bmatrix} 0 & 0 \\ 0 & -1 \end{bmatrix} $$ These are linearly dependent since $$\boldsymbol{M}_1 - \boldsymbol{M}_2 + 2~\boldsymbol{M}_3 = \mathbf{0}$$.

Span
The span of a set of vectors $$(\boldsymbol{T})$$ is the set of all vectors that are linear combinations of the vectors $$\mathbf{x}_i$$. Thus

\text{span}(\boldsymbol{T}) = \{\boldsymbol{T}_1, \boldsymbol{T}_2, \dots, \boldsymbol{T}_n\} $$ where

\boldsymbol{T}_i = \alpha_1~\mathbf{x}_1 + \alpha_2~\mathbf{x}_2 + \dots + \alpha_n~\mathbf{x}_n $$ as $$\alpha_1, \alpha_2, \dots, \alpha_n$$ vary.

Spanning set
If the span = $$\mathcal{S}$$ then $$\boldsymbol{T}$$ is said to be a spanning set.

Basis
If $$\boldsymbol{T}$$ is a spanning set and its elements are linearly independent then we call it a basis for $$\mathcal{S}$$. A vector in $$\mathcal{S}$$ has a unique representation as a linear combination of the basis elements.  why is it unqiue?

Dimension
The dimension of a space $$\mathcal{S}$$ is the number of elements in the basis. This is independent of actual elements that form the basis and is a property of $$\mathcal{S}$$.

Example 1: Vectors in R2
Any two non-collinear vectors $$\mathbb{R}^2$$ is a basis for $$\mathbb{R}^2$$ because any other vector in $$\mathbb{R}^2$$ can be expressed as a linear combination of the two vectors.

Example 2: Matrices
A basis for the linear space of $$2 \times 2$$ matrices is

\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 3 \\ 1 & 1 \end{bmatrix} $$ Note that there is a lot of nonuniqueness in the choice of bases. One important skill that you should develop is to choose the right basis to solve a particular problem.

Example 3: Polynomials
The set $$\{1, x, x^2, \dots, x^n\}$$ is a basis for polynomials of degree $$n$$.

Example 4: The natural basis
A natural basis is the set $$\{ \mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n\}$$ where the $$j$$th entry of $$\mathbf{e}_k$$ is

\delta_{jk} = \begin{cases} 1 & \mbox{for} ~j = k \\ 0 & \mbox{for}~ j \ne k \end{cases} $$ The quantity $$\delta_{jk}$$ is also called the Kronecker delta.

Inner Product Spaces
To give more structure to the idea of a vector space we need concepts such as magnitude and angle. The inner product provides that structure.

The inner product generalizes the concept of an angle and is defined as a function

\langle\bullet,~\bullet\rangle : \mathcal{S}\times\mathcal{S} \rightarrow \mathbb{R} \quad (\text{or}~\mathbb{C}~\text{for a complex vector space}) $$ with the properties

 * 1) $$\langle\mathbf{x},~\mathbf{y}\rangle = \overline{\langle\mathbf{y},~\mathbf{x}\rangle} \qquad$$ overbar indicates complex conjugation.
 * 2) $$\langle\alpha~\mathbf{x},~\mathbf{y}\rangle = \alpha~\langle\mathbf{x},~\mathbf{y}\rangle \quad$$ Linear with respect to scalar multiplication.
 * 3) $$\langle\mathbf{x}+\mathbf{y},~\mathbf{z}\rangle = \langle\mathbf{x},~\mathbf{z}\rangle + \langle\mathbf{y},~\mathbf{z}\rangle \quad$$ Linearity with respect to addition.
 * 4) $$\langle\mathbf{x},~\mathbf{x}\rangle > \mathbf{0}$$ if $$\mathbf{x} \ne 0$$ and $$\langle\mathbf{x},~\mathbf{x}\rangle = \mathbf{0}$$ if and only if $$\mathbf{x} = \mathbf{0}$$.

A vector space with an inner product is called an inner product space.

Example 1:

\langle\mathbf{x},~\beta~\mathbf{y}\rangle = \overline{\langle\beta~\mathbf{y},~\mathbf{x}\rangle} = \overline{\beta}~\overline{\langle\mathbf{y},~\mathbf{x}\rangle} = \overline{\beta}\langle\mathbf{x},~\mathbf{y}\rangle $$

Example 2: Discrete vectors
In $$\mathbb{R}^n$$ with $$\mathbf{x} = \{x_1, x_2, \dots, x_n \}$$ and $$\mathbf{y} = \{y_1, y_2, \dots, y_n \}$$ the Eulidean norm is given by

\langle\mathbf{x},~\mathbf{y}\rangle = \sum_n x_n~y_n $$ With $$\mathbf{x}, \mathbf{y} \in \mathbb{C}^n$$ the standard norm is

\langle\mathbf{x},~\mathbf{y}\rangle = \sum_k x_k~\overline{y_k} $$

Example 3: Continuous functions
For two complex valued continuous functions $$f(x)$$ and $$g(x)$$ in $$[0, 1]$$ we could approximately represent them by their function values at  equally spaced points.

Approximate $$f(x)$$ and $$g(x)$$ by

\begin{align} F & = \{f(x_1), f(x_2), \dots, f(x_n)\} \qquad \text{with} ~x_k = \cfrac{k}{n}\\ G & = \{g(x_1), g(x_2), \dots, g(x_n)\} \qquad \text{with} ~x_k = \cfrac{k}{n} \end{align} $$ With that approximation, a natural norm is

\langle F,~G\rangle = \cfrac{1}{n}~\sum_{k=1}^n f(x_k)~\overline{g(x_k)} $$ Taking the limit as $$n \rightarrow \infty$$ (show this)

\langle f,~g\rangle = \int_0^1 f(x)~\overline{g(x)}~dx $$

If we took  non-equally spaced yet smoothly distributed points we would get

\langle f,~g\rangle = \int_0^1 f(x)~\overline{g(x)}~w(x)~dx $$ where $$w(x) > 0$$ is a smooth weighting function (show this).

There are many other inner products possible. For functions that are not only continuous but also differentiable, a useful norm is

\langle f,~g\rangle = \int_0^1 \left[f(x)~\overline{g(x)} + f^{'}(x)~\overline{g^{'}(x)}\right]~dx $$

We will continue further explorations into linear vector spaces in the next lecture.


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Micromechanics of composites

Welcome to this learning project about !

Learning Project Summary

 * Project code:
 * Suggested Prerequisites:
 * Introduction to Elasticity
 * Continuum mechanics
 * Time investment: 6 months
 * Assessment suggestions:
 * Portal:Enegineering and Technology
 * School:Engineering
 * Department:Mechanical Engineering
 * Stream:Applied Mechanics
 * Level: Second year graduate

Content summary
This course is the micromechanics of composite materials. The purpose is to show you ways in which micromechanics may be used to determine the effective properties of composites.

Goals
This learning project aims to
 * Show you some of the fundamental theorems in the micromechanics of composites.
 * Give you a feel for how the theory can be used to determine the effective properties of composites.
 * Give you an idea about numerical approaches based on the theory.

Learning materials

 * 1) Review of some basic continuum mechanics
 * 2) Conservation of mass
 * 3) Balance of linear momentum
 * 4) Balance of angular momentum
 * 5) Balance of energy
 * 6) Some basic ideas of micromechanics
 * 7) The RVE and governing equations
 * 8) Infinitesimal deformations
 * 9) Average strain in a RVE
 * 10) Average displacement in a RVE
 * 11) Average stress in a RVE
 * 12) Average stress power in a RVE
 * 13) Finite deformations
 * 14) Average deformation gradient in a RVE
 * 15) Average velocity gradient in a RVE
 * 16) Average stress in a RVE
 * 17) Average stress power in a RVE
 * 18) Appendix: Some useful results and proofs
 * 19) Proof 1: Tensor-vector identity - 1
 * 20) Proof 2: Tensor-vector identity - 2
 * 21) Proof 3: Surface and volume integral relation - 1
 * 22) Proof 4: Integral of a cross product
 * 23) Proof 5: Surface and volume integral relation - 2
 * 24) Proof 6: Curl of a gradient - 1
 * 25) Proof 7: Curl of a gradient - 2
 * 26) Proof 8: Relation between axial vector and displacement
 * 27) Proof 9: Relation between axial vector and strain
 * 28) Proof 10: Rigid body motion
 * 29) Proof 11: More tensor identities
 * 30) Proof 12: Relation between volume averaged fields
 * 31) Proof 13: Average stress power identity - Cauchy stress
 * 32) Proof 14: Average stress power identity - 1st P-K stress

Primary texts

 * S. Nemat-Nasser and M. Hori, 1993, Micromechanics: Overall Properties of Heterogeneous Materials, North-Holland.
 * G. W. Milton, 2002, The Theory of Composites, Cambridge University Press.
 * S. Torquato, 2002, Random Heterogeneous Materials, Springer.

Other reading materials

 * T. Belytschko, W. K. Liu, and B. Moran.  Nonlinear Finite Elements for Continua and Structures. John Wiley and Sons, Ltd., New York, 2000.
 * J. Bonet and R. D. Wood.  Nonlinear Continuum Mechanics for Finite Element Analysis. Cambridge University Press, 1997.
 * F. Costanzo, G. L. Gray, and P. C. Andia. On the definitions of effective stress and deformation gradient for use in MD: Hill's macro-homogeneity and the virial theorem.  Int. J. Engg. Sci., 43:533--555, 1985. http://dx.doi.org/10.1016/j.ijengsci.2004.12.002
 * P. Chadwick.  Continuum Mechanics: Concise Theory and Problems. George Allen and Unwin Ltd., London, 1976.
 * M. E. Gurtin. The linear theory of elasticity. In C.~Truesdell, editor, Encyclopedia of Physics (Handbuch der Physik), volume VIa/2, pages 1--295. Springer-Verlag, Berlin, 1972.
 * M. E. Gurtin.  An Introduction to Continuum Mechanics. Academic Press, New York, 1981.
 * R. Hill. Elastic properties of reinforced solids : some theoretical principles.  J. Mech. Phys. Solids, 11:357--372, 1963. http://dx.doi.org/10.1016/0022-5096(63)90036-X
 * R. Hill. Theory of mechanical properties of fibre-strengthened materials: I. Elastic behavior.  J. Mech. Phys. Solids, 12:199--212, 1964. http://dx.doi.org/10.1016/0022-5096(64)90019-5
 * R. Hill. On constitutive macro-variables for heterogeneous solids at finite strain.  Proc. Royal Soc. Lond. A, 326:131--147, 1972. http://dx.doi.org/10.1098/rspa.1972.0001
 * R. Hill. On macroscopic effects of heterogeneity in elastoplastic media at finite strain.  Math. Proc. Camb. Phil. Soc, 95:481--495, 1984. http://dx.doi.org/10.1017/S0305004100061818
 * S. Nemat-Nasser. Averaging theorems in finite deformation plasticity.  Mechanics of Materials, 31:493--523, 1999. http://dx.doi.org/10.1016/S0167-6636(98)00073-8
 * S. Nemat-Nasser. Plasticity: A Treatise on Finite Deformation of Heteogeneous Inelastic Materials.  Cambridge University Press, Cambridge, 2004.
 * P. Perzyna. Constitutive equations for thermoinelasticity and instability  phenomena in thermodynamic flow processes.  In Stein E., editor, Progress in Computational Analysis of Inelastic Structures: CISM Courses and Lectures No. 321, pages 1--78. Springer-Verlag-Wien, New York, 1993.
 * W. S. Slaughter.  The Linearized Theory of Elasticity. Birhhauser, Boston, 2002.
 * C. Truesdell and W. Noll.  The Non-linear Field Theories of Mechanics. Springer-Verlag, New York, 1992.
 * T. W. Wright.  The Physics and Mathematics of Adiabatic Shear Bands. Cambridge University Press, Cambridge, UK, 2002.

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title: Micromechanics of composites


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Category:Micromechanics

This category is for easy access to all materials concerned with micromechanics.

title: Category:Micromechanics


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Talk:College Algebra/Real Numbers

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Micromechanics of composites/Conservation of mass

Statement of the balance of mass
The balance of mass can be expressed as: $$  \dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v} = 0 $$ where $$\rho(\mathbf{x},t)$$ is the current mass density, $$\dot{\rho}$$ is the material time derivative of $$\rho$$, and $$\mathbf{v}(\mathbf{x},t)$$ is the velocity of physical particles in the body $$\Omega$$ bounded by the surface $$\partial{\Omega}$$.

Proof
We can show how this relation is derived by recalling that the general equation for the balance of a physical quantity $$f(\mathbf{x},t)$$ is given by

\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~. $$ To derive the equation for the balance of mass, we assume that the physical quantity of interest is the mass density $$\rho(\mathbf{x},t)$$. Since mass is neither created or destroyed, the surface and interior sources are zero, i.e., $$g(\mathbf{x}, t) = h(\mathbf{x},t) = 0$$. Therefore, we have

\cfrac{d}{dt}\left[\int_{\Omega} \rho(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} \rho(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA}~. $$ Let us assume that the volume $$\Omega$$ is a control volume (i.e., it does not change with time). Then the surface $$\partial{\Omega}$$ has a zero velocity ($$u_n = 0$$) and we get

\int_{\Omega} \frac{\partial \rho}{\partial t}~\text{dV} = - \int_{\partial{\Omega}} \rho~(\mathbf{v}\cdot\mathbf{n})~\text{dA}~. $$ Using the divergence theorem

\int_{\Omega} \boldsymbol{\nabla} \bullet \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\cdot\mathbf{n}~\text{dA} $$ we get

\int_{\Omega} \frac{\partial \rho}{\partial t}~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet (\rho~\mathbf{v})~\text{dV}. $$ or,

\int_{\Omega} \left[\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \bullet (\rho~\mathbf{v})\right]~\text{dV} = 0 ~. $$ Since $$\Omega$$ is arbitrary, we must have

\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \bullet (\rho~\mathbf{v}) = 0 ~. $$ Using the identity

\boldsymbol{\nabla} \bullet (\varphi~\mathbf{v}) = \varphi~\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \varphi\cdot\mathbf{v} $$ we have

\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} = 0 ~. $$ Now, the material time derivative of $$\rho$$ is defined as

\dot{\rho} = \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} ~. $$ Therefore,

{ \dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}= 0 ~. } $$


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Micromechanics of composites/Balance of linear momentum

Statement of the balance of linear momentum
The balance of linear momentum can be expressed as: $$ \rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ where $$\rho(\mathbf{x},t)$$ is the mass density, $$\mathbf{v}(\mathbf{x},t)$$ is the velocity, $$\boldsymbol{\sigma}(\mathbf{x},t)$$ is the Cauchy stress, and $$\rho~\mathbf{b}$$ is the body force density.

Proof
Recall the general equation for the balance of a physical quantity

\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~. $$ In this case the physical quantity of interest is the momentum density, i.e., $$f(\mathbf{x},t) = \rho(\mathbf{x},t)~\mathbf{v}(\mathbf{x},t)$$. The source of momentum flux at the surface is the surface traction, i.e., $$g(\mathbf{x},t) = \mathbf{t}$$. The source of momentum inside the body is the body force, i.e., $$h(\mathbf{x},t) = \rho(\mathbf{x},t)~\mathbf{b}(\mathbf{x},t)$$. Therefore, we have

\cfrac{d}{dt}\left[\int_{\Omega} \rho~\mathbf{v}~\text{dV}\right] = \int_{\partial{\Omega}} \rho~\mathbf{v}[u_n - \mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \mathbf{t}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~. $$ The surface tractions are related to the Cauchy stress by

\mathbf{t} = \boldsymbol{\sigma}\cdot\mathbf{n} ~. $$ Therefore,

\cfrac{d}{dt}\left[\int_{\Omega} \rho~\mathbf{v}~\text{dV}\right] = \int_{\partial{\Omega}} \rho~\mathbf{v}[u_n - \mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \boldsymbol{\sigma}\cdot\mathbf{n}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~. $$ Let us assume that $$\Omega$$ is an arbitrary fixed control volume. Then,

\int_{\Omega} \frac{\partial }{\partial t}(\rho~\mathbf{v})~\text{dV} = - \int_{\partial{\Omega}} \rho~\mathbf{v}~(\mathbf{v}\cdot\mathbf{n})~\text{dA} + \int_{\partial{\Omega}} \boldsymbol{\sigma}\cdot\mathbf{n}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~. $$ Now, from the definition of the tensor product we have (for all vectors $$\mathbf{a}$$)

(\mathbf{u}\otimes\mathbf{v})\cdot\mathbf{a} = (\mathbf{a}\cdot\mathbf{v})~\mathbf{u} ~. $$ Therefore,

\int_{\Omega} \frac{\partial }{\partial t}(\rho~\mathbf{v})~\text{dV} = - \int_{\partial{\Omega}} \rho~(\mathbf{v}\otimes\mathbf{v})\cdot\mathbf{n}~\text{dA} + \int_{\partial{\Omega}} \boldsymbol{\sigma}\cdot\mathbf{n}~\text{dA} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} ~. $$ Using the divergence theorem

\int_{\Omega} \boldsymbol{\nabla} \bullet \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\cdot\mathbf{n}~\text{dA} $$ we have

\int_{\Omega} \frac{\partial }{\partial t}(\rho~\mathbf{v})~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet [\rho~(\mathbf{v}\otimes\mathbf{v})]~\text{dV} + \int_{\Omega} \boldsymbol{\nabla} \bullet \boldsymbol{\sigma}~\text{dV} + \int_{\Omega} \rho~\mathbf{b}~\text{dV} $$ or,

\int_{\Omega}\left[ \frac{\partial }{\partial t}(\rho~\mathbf{v}) + \boldsymbol{\nabla} \bullet [(\rho~\mathbf{v})\otimes\mathbf{v})] - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b}\right]~\text{dV} = 0 ~. $$ Since $$\Omega$$ is arbitrary, we have

\frac{\partial }{\partial t}(\rho~\mathbf{v}) + \boldsymbol{\nabla} \bullet [(\rho~\mathbf{v})\otimes\mathbf{v})] - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0~. $$ Using the identity

\boldsymbol{\nabla} \bullet (\mathbf{u}\otimes\mathbf{v}) = (\boldsymbol{\nabla} \bullet \mathbf{v})\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})\cdot\mathbf{v} $$ we get

\frac{\partial \rho}{\partial t}~\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + (\boldsymbol{\nabla} \bullet \mathbf{v})(\rho\mathbf{v}) + \boldsymbol{\nabla} (\rho~\mathbf{v})\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ or,

\left[\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \boldsymbol{\nabla} (\rho~\mathbf{v})\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ Using the identity

\boldsymbol{\nabla} (\varphi~\mathbf{v}) = \varphi~\boldsymbol{\nabla}\mathbf{v} + \mathbf{v}\otimes(\boldsymbol{\nabla} \varphi) $$ we get

\left[\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \left[\rho~\boldsymbol{\nabla}\mathbf{v} + \mathbf{v}\otimes(\boldsymbol{\nabla} \rho)\right]\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ From the definition

(\mathbf{u}\otimes\mathbf{v})\cdot\mathbf{a} = (\mathbf{a}\cdot\mathbf{v})~\mathbf{u} $$ we have

[\mathbf{v}\otimes(\boldsymbol{\nabla} \rho)]\cdot\mathbf{v} = [\mathbf{v}\cdot(\boldsymbol{\nabla} \rho)]~\mathbf{v} ~. $$ Hence,

\left[\frac{\partial \rho}{\partial t} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} +[\mathbf{v}\cdot(\boldsymbol{\nabla} \rho)]~\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 $$ or,

\left[\frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. $$ The material time derivative of $$\rho$$ is defined as

\dot{\rho} = \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} ~. $$ Therefore,

\left[\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}\right]\mathbf{v} + \rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. $$ From the balance of mass, we have

\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v}= 0 ~. $$ Therefore,

\rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. $$ The material time derivative of $$\mathbf{v}$$ is defined as

\dot{\mathbf{v}} = \frac{\partial \mathbf{v}}{\partial t} + \boldsymbol{\nabla} \mathbf{v}\cdot\mathbf{v} ~. $$ Hence,

{ \rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. } $$


text:
Micromechanics of composites/Balance of angular momentum

Statement of the balance of angular momentum
The balance of angular momentum in an inertial frame can be expressed as: $$ \boldsymbol{\sigma}= \boldsymbol{\sigma}^T $$

Proof
We assume that there are no surface couples on $$\partial{\Omega}$$ or body couples in $$\Omega$$. Recall the general balance equation

\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~. $$ In this case, the physical quantity to be conserved the angular momentum density, i.e., $$f = \mathbf{x}\times(\rho~\mathbf{v})$$. The angular momentum source at the surface is then $$g = \mathbf{x}\times\mathbf{t}$$ and the angular momentum source inside the body is $$h = \mathbf{x}\times(\rho~\mathbf{b})$$. The angular momentum and moments are calculated with respect to a fixed origin. Hence we have

\cfrac{d}{dt}\left[\int_{\Omega} \mathbf{x}\times(\rho~\mathbf{v})~\text{dV}\right] = \int_{\partial{\Omega}} [\mathbf{x}\times(\rho~\mathbf{v})] [u_n - \mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \mathbf{x}\times\mathbf{t}~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ Assuming that $$\Omega$$ is a control volume, we have

\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\partial{\Omega}} [\mathbf{x}\times(\rho~\mathbf{v})][\mathbf{v}\cdot\mathbf{n}]~\text{dA} + \int_{\partial{\Omega}} \mathbf{x}\times\mathbf{t}~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ Using the definition of a tensor product we can write

[\mathbf{x}\times(\rho~\mathbf{v})][\mathbf{v}\cdot\mathbf{n}] = [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]\cdot\mathbf{n} ~. $$ Also, $$\mathbf{t} = \boldsymbol{\sigma}\cdot\mathbf{n}$$. Therefore we have

\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\partial{\Omega}} [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]\cdot\mathbf{n} ~\text{dA} + \int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ Using the divergence theorem, we get

\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]~\text{dV} + \int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ To convert the surface integral in the above equation into a volume integral, it is convenient to use index notation. Thus,

\left[\int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA}\right]_i = \int_{\partial{\Omega}} e_{ijk}~x_j~\sigma_{kl}~n_l~\text{dA}= \int_{\partial{\Omega}} A_{il}~n_l~\text{dA}= \int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} $$ where $$[~]_i$$ represents the $$i$$-th component of the vector. Using the divergence theorem

\int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \boldsymbol{\nabla} \bullet \boldsymbol{A}~\text{dV} = \int_{\Omega} \frac{\partial A_{il}}{\partial x_l}~\text{dV} = \int_{\Omega} \frac{\partial }{\partial x_l}(e_{ijk}~x_j~\sigma_{kl})~\text{dV}~. $$ Differentiating,

\int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \left[ e_{ijk}~\delta_{jl}~\sigma_{kl} + e_{ijk}~x_j~\frac{\partial \sigma_{kl}}{\partial x_l}\right]~\text{dV} = \int_{\Omega} \left[ e_{ijk}~\sigma_{kj} + e_{ijk}~x_j~\frac{\partial \sigma_{kl}}{\partial x_l}\right]~\text{dV} = \int_{\Omega} \left[ e_{ijk}~\sigma_{kj} + e_{ijk}~x_j~[\boldsymbol{\nabla} \bullet \boldsymbol{\sigma}]_k\right]~\text{dV} ~. $$ Expressed in direct tensor notation,

\int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \left[ [\mathcal{E}:\boldsymbol{\sigma}^T]_i + [\mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})]_i\right]~\text{dV} $$ where $$\mathcal{E}$$ is the third-order permutation tensor. Therefore,

\left[\int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA}\right]_i = = \int_{\Omega} \left[ [\mathcal{E}:\boldsymbol{\sigma}^T]_i + [\mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})]_i\right]~\text{dV} $$ or,

\int_{\partial{\Omega}} \mathbf{x}\times(\boldsymbol{\sigma}\cdot\mathbf{n})~\text{dA} = = \int_{\Omega} \left[ \mathcal{E}:\boldsymbol{\sigma}^T + \mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})\right]~\text{dV} ~. $$ The balance of angular momentum can then be written as

\int_{\Omega} \mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right]~\text{dV} = - \int_{\Omega} \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}]~\text{dV} + \int_{\Omega} \left[ \mathcal{E}:\boldsymbol{\sigma}^T + \mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})\right]~\text{dV} + \int_{\Omega} \mathbf{x}\times(\rho~\mathbf{b})~\text{dV} ~. $$ Since $$\Omega$$ is an arbitrary volume, we have

\mathbf{x}\times\left[\cfrac{\partial}{\partial t}(\rho~\mathbf{v})\right] = - \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] + \mathcal{E}:\boldsymbol{\sigma}^T + \mathbf{x}\times(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma}) + \mathbf{x}\times(\rho~\mathbf{b}) $$ or,

{\mathbf{x}}\times {\left[\frac{\partial }{\partial t}(\rho~\mathbf{v}) - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - \boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] + \mathcal{E}:\boldsymbol{\sigma}^T ~. $$ Using the identity,

\boldsymbol{\nabla} \bullet (\mathbf{u}\otimes\mathbf{v}) = (\boldsymbol{\nabla} \bullet \mathbf{v})\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})\cdot\mathbf{v} $$ we get

\boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] = (\boldsymbol{\nabla} \bullet \mathbf{v})[\mathbf{x}\times(\rho~\mathbf{v})] + (\boldsymbol{\nabla} [\mathbf{x}\times(\rho~\mathbf{v})])\cdot\mathbf{v} ~. $$ The second term on the right can be further simplified using index notation as follows.

\begin{align} \left[(\boldsymbol{\nabla} [\mathbf{x}\times(\rho~\mathbf{v})])\cdot\mathbf{v}\right]_i = \left[(\boldsymbol{\nabla} [\rho~(\mathbf{x}\times\mathbf{v})])\cdot\mathbf{v}\right]_i & = \frac{\partial }{\partial x_l}(\rho~e_{ijk}~x_j~v_k)~v_l \\ & = e_{ijk}\left[ \frac{\partial \rho}{\partial x_l}~x_j~v_k~v_l+ \rho~\frac{\partial x_j}{\partial x_l}~v_k~v_l + \rho~x_j~\frac{\partial v_k}{\partial x_l}~v_l\right] \\ & = (e_{ijk}~x_j~v_k)~\left(\frac{\partial \rho}{\partial x_l}~v_l\right)+ \rho~(e_{ijk}~\delta_{jl}~v_k~v_l) + e_{ijk}~x_j~\left(\rho~\frac{\partial v_k}{\partial x_l}~v_l\right) \\ & = [(\mathbf{x}\times\mathbf{v})(\boldsymbol{\nabla} \rho\cdot\mathbf{v}) + \rho~\mathbf{v}\times\mathbf{v} + \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})]_i \\ & = [(\mathbf{x}\times\mathbf{v})(\boldsymbol{\nabla} \rho\cdot\mathbf{v}) + \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})]_i ~. \end{align} $$ Therefore we can write

\boldsymbol{\nabla} \bullet [[\mathbf{x}\times(\rho~\mathbf{v})]\otimes\mathbf{v}] = (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) + (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})] ~. $$ The balance of angular momentum then takes the form

{\mathbf{x}}\times {\left[\frac{\partial }{\partial t}(\rho~\mathbf{v}) - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) - \mathbf{x}\times(\rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T $$ or,

{\mathbf{x}}\times {\left[\frac{\partial }{\partial t}(\rho~\mathbf{v}) + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T $$ or,

{\mathbf{x}}\times {\left[\rho\frac{\partial \mathbf{v}}{\partial t} + \frac{\partial \rho}{\partial t}~\mathbf{v} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T $$ The material time derivative of $$\mathbf{v}$$ is defined as

\dot{\mathbf{v}} = \frac{\partial \mathbf{v}}{\partial t} + \boldsymbol{\nabla} \mathbf{v}\cdot\mathbf{v} ~. $$ Therefore,

{\mathbf{x}}\times {\left[\rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} \right]} = - \mathbf{x}\times\cfrac{\partial \rho}{\partial t}~\mathbf{v} + - (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) - (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) + \mathcal{E}:\boldsymbol{\sigma}^T ~. $$ Also, from the conservation of linear momentum

\rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} = 0 ~. $$ Hence,

\begin{align} 0 & = \mathbf{x}\times\cfrac{\partial\rho}{\partial t}~\mathbf{v} + (\rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times~\mathbf{v}) + (\boldsymbol{\nabla} \rho\cdot\mathbf{v})(\mathbf{x}\times\mathbf{v}) - \mathcal{E}:\boldsymbol{\sigma}^T \\ & = \left(\frac{\partial \rho}{\partial t} + \rho\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} \right)(\mathbf{x}\times\mathbf{v}) - \mathcal{E}:\boldsymbol{\sigma}^T ~. \end{align} $$ The material time derivative of $$\rho$$ is defined as

\dot{\rho} = \frac{\partial \rho}{\partial t} + \boldsymbol{\nabla} \rho\cdot\mathbf{v} ~. $$ Hence,

(\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v})(\mathbf{x}\times\mathbf{v}) - \mathcal{E}:\boldsymbol{\sigma}^T = 0 ~. $$ From the balance of mass

\dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v} = 0 ~. $$ Therefore,

\mathcal{E}:\boldsymbol{\sigma}^T = 0 ~. $$ In index notation,

e_{ijk}~\sigma_{kj} = 0 ~. $$ Expanding out, we get

\sigma_{12} - \sigma_{21} = 0 ~; \sigma_{23} - \sigma_{32} = 0 ~; \sigma_{31} - \sigma_{13} = 0 ~. $$ Hence,

{ \boldsymbol{\sigma} = \boldsymbol{\sigma}^T } $$


text:
Micromechanics of composites/Balance of energy

Statement of the balance of energy
The balance of energy can be expressed as: $$ \rho~\dot{e} - \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{v}) + \boldsymbol{\nabla} \bullet   \mathbf{q} - \rho~s= 0 $$ where $$\rho(\mathbf{x},t)$$ is the mass density, $$e(\mathbf{x},t)$$ is the internal energy per unit mass, $$\boldsymbol{\sigma}(\mathbf{x},t)$$ is the Cauchy stress, $$\mathbf{v}(\mathbf{x},t)$$ is the particle velocity, $$\mathbf{q}$$ is the heat flux vector, and $$s$$ is the rate at which energy is generated by sources inside the volume (per unit mass).

Proof
Recall the general balance equation

\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~. $$ In this case, the physical quantity to be conserved the total energy density which is the sum of the internal energy density and the kinetic energy density, i.e., $$f = \rho~e + 1/2~\rho~|\mathbf{v}\cdot\mathbf{v}|$$. The energy source at the surface is a sum of the rate of work done by the applied tractions and the rate of heat leaving the volume (per unit area), i.e, $$g = \mathbf{v}\cdot\mathbf{t} - \mathbf{q}\cdot\mathbf{n}$$ where $$\mathbf{n}$$ is the outward unit normal to the surface. The energy source inside the body is the sum of the rate of work done by the body forces and the rate of energy generated by internal sources, i.e., $$h = \mathbf{v}\cdot(\rho\mathbf{b}) + \rho~s$$.

Hence we have

\cfrac{d}{dt}\left[\int_{\Omega} \rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)~\text{dV}\right] = \int_{\partial{\Omega}}\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right) (u_n - \mathbf{v}\cdot\mathbf{n})~\text{dA} + \int_{\partial{\Omega}} (\mathbf{v}\cdot\mathbf{t} - \mathbf{q}\cdot\mathbf{n})~\text{dA} + \int_{\Omega} \rho~(\mathbf{v}\cdot\mathbf{b} + s)~\text{dV} ~. $$ Let $$\Omega$$ be a control volume that does not change with time. Then we get

\int_{\Omega} \frac{\partial }{\partial t}\left[\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\right]~\text{dV} = -\int_{\partial{\Omega}}\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right) (\mathbf{v}\cdot\mathbf{n})~\text{dA} + \int_{\partial{\Omega}} (\mathbf{v}\cdot\mathbf{t} - \mathbf{q}\cdot\mathbf{n})~\text{dA} + \int_{\Omega} \rho~(\mathbf{v}\cdot\mathbf{b} + s)~\text{dV} ~. $$ Using the relation $$\mathbf{t} = \boldsymbol{\sigma}\cdot\mathbf{n}$$, the identity $$\mathbf{v}\cdot(\boldsymbol{\sigma}\cdot\mathbf{n}) = (\boldsymbol{\sigma}^T\cdot\mathbf{v})\cdot\mathbf{n}$$, and invoking the symmetry of the stress tensor, we get

\int_{\Omega} \frac{\partial }{\partial t}\left[\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\right]~\text{dV} = -\int_{\partial{\Omega}}\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right) (\mathbf{v}\cdot\mathbf{n})~\text{dA} + \int_{\partial{\Omega}} (\boldsymbol{\sigma}\cdot\mathbf{v} - \mathbf{q})\cdot\mathbf{n}~\text{dA} + \int_{\Omega} \rho~(\mathbf{v}\cdot\mathbf{b} + s)~\text{dV} ~. $$ We now apply the divergence theorem to the surface integrals to get

\int_{\Omega} \frac{\partial }{\partial t}\left[\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\right]~\text{dV} = -\int_{\Omega} \boldsymbol{\nabla} \bullet \left[\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\mathbf{v}\right]~\text{dV} +\int_{\Omega} \boldsymbol{\nabla} \bullet (\boldsymbol{\sigma}\cdot\mathbf{v})~\text{dV} - \int_{\Omega} \boldsymbol{\nabla} \bullet \mathbf{q}~\text{dV} +\int_{\Omega} \rho~(\mathbf{v}\cdot\mathbf{b} + s)~\text{dV} ~. $$ Since $$\Omega$$ is arbitrary, we have

\frac{\partial }{\partial t}\left[\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\right] = - \boldsymbol{\nabla} \bullet \left[\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\mathbf{v}\right] + \boldsymbol{\nabla} \bullet (\boldsymbol{\sigma}\cdot\mathbf{v}) - \boldsymbol{\nabla} \bullet \mathbf{q} + \rho~(\mathbf{v}\cdot\mathbf{b} + s)~. $$ Expanding out the left hand side, we have

\begin{align} \frac{\partial }{\partial t}\left[\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\right] & = \frac{\partial \rho}{\partial t}\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right) + \rho~\left(\frac{\partial e}{\partial t} + \frac{1}{2}~\frac{\partial }{\partial t}(\mathbf{v}\cdot\mathbf{v})\right) \\ & = \frac{\partial \rho}{\partial t}\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right) + \rho~\frac{\partial e}{\partial t} + \rho~\frac{\partial \mathbf{v}}{\partial t}\cdot\mathbf{v} ~. \end{align} $$ For the first term on the right hand side, we use the identity $$\boldsymbol{\nabla} \bullet (\varphi~\mathbf{v}) = \varphi~\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \varphi\cdot\mathbf{v}$$ to get

\begin{align} \boldsymbol{\nabla} \bullet \left[\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\mathbf{v}\right] & = \rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)~\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \left[\rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\right]\cdot\mathbf{v} \\ & = \rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)~\boldsymbol{\nabla} \bullet \mathbf{v} + \left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\boldsymbol{\nabla} \rho\cdot\mathbf{v} + \rho~\boldsymbol{\nabla} \left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\cdot\mathbf{v} \\ & = \rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)~\boldsymbol{\nabla} \bullet \mathbf{v} + \left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\boldsymbol{\nabla} \rho\cdot\mathbf{v} + \rho~\boldsymbol{\nabla} e\cdot{\mathbf{v}} + \frac{1}{2}~\rho~\boldsymbol{\nabla} (\mathbf{v}\cdot\mathbf{v})\cdot\mathbf{v} \\ & = \rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)~\boldsymbol{\nabla} \bullet \mathbf{v} + \left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\boldsymbol{\nabla} \rho\cdot\mathbf{v} + \rho~\boldsymbol{\nabla} e\cdot{\mathbf{v}} + \rho~(\boldsymbol{\nabla}\mathbf{v}^T\cdot\mathbf{v})\cdot\mathbf{v} \\ & = \rho~\left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)~\boldsymbol{\nabla} \bullet \mathbf{v} + \left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right)\boldsymbol{\nabla} \rho\cdot\mathbf{v} + \rho~\boldsymbol{\nabla} e\cdot{\mathbf{v}} + \rho~(\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v})\cdot\mathbf{v} ~. \end{align} $$ For the second term on the right we use the identity $$\boldsymbol{\nabla} \bullet (\boldsymbol{S}^T\cdot\mathbf{v}) = \boldsymbol{S}:\boldsymbol{\nabla}\mathbf{v} + (\boldsymbol{\nabla} \bullet \boldsymbol{S})\cdot\mathbf{v}$$ and the symmetry of the Cauchy stress tensor to get

\boldsymbol{\nabla} \bullet (\boldsymbol{\sigma}\cdot\mathbf{v}) = \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + (\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})\cdot\mathbf{v} ~. $$ After collecting terms and rearranging, we get

\left(\frac{\partial \rho}{\partial t}+ \rho~\boldsymbol{\nabla} \bullet \mathbf{v} + \boldsymbol{\nabla} \rho\cdot\mathbf{v}\right) \left(e + \frac{1}{2}~\mathbf{v}\cdot\mathbf{v}\right) + \left(\rho~\frac{\partial \mathbf{v}}{\partial t} + \rho~\boldsymbol{\nabla}\mathbf{v}\cdot\mathbf{v} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b}\right)\cdot\mathbf{v} + \rho~\left(\frac{\partial e}{\partial t} + \boldsymbol{\nabla} e\cdot{\mathbf{v}}\right) + - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \bullet \mathbf{q} - \rho~s = 0 ~. $$ Applying the balance of mass to the first term and the balance of linear momentum to the second term, and using the material time derivative of the internal energy

\dot{e} = \frac{\partial e}{\partial t} + \boldsymbol{\nabla} e\cdot\mathbf{v} $$ we get the final form of the balance of energy:

{ \rho~\dot{e} - \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} + \boldsymbol{\nabla} \bullet \mathbf{q} - \rho~s = 0 ~. } $$


text:
Micromechanics of composites/Introduction

The main aim of micromechanics is to find the properties of a continuum material point based on the microstructure. We assume that the microstructure can be described as a continuum even though it is heterogeneous and may contain voids.

The properties at a material point are computed using an averaging procedure. Many such averaging procedures exist, some of which assume the existence of a representative volume element (RVE). A RVE is usually chosen such that it represents adequately the local microstructure of the macroscopic scale continuum (in a statistical sense).

We follow the following principle from Hill (1972):

Experimental determinations of mechanical behaviour rest ultimately on measured loads or mean displacements over pairs of opposite faces of a representative cube.  Macro-variables intended for constitutive laws should thus be capable of definition in terms of surface data alone, ''either directly or indirectly. It is not necessary, by any means, that macro-variables so defined should be unweighted volume averages of their macroscopic counterparts. However, variables that do have this special property are naturally the easiest to handle analytically in the transition between levels. Accordingly, we approach the construction of macro-variables by first identifying some relevant averages that depend uniquely on surface data.''

In this section, the main focus is on defining average quantities that may be used to describe the macroscopic constitutive behavior of composites. The average quantities are then shown to depend solely on boundary data. Both infinitesimal and finite deformations are considered.

Governing Equations
The equations that govern the motion of a solid include the balance laws for mass, momentum, and energy. Kinematic equations and constitutive relations are needed to complete the system of equations. Physical restrictions on the form of the constitutive relations are imposed by an entropy inequality that expresses the second law of thermodynamics in mathematical form.

The balance laws express the idea that the rate of change of a quantity (mass, momentum, energy) in a volume must arise from three causes:

 * 1) the physical quantity itself flows through the surface that bounds the volume,
 * 2) there is a source of the physical quantity on the surface of the volume, or/and,
 * 3) there is a source of the physical quantity inside the volume.

Let $$\Omega$$ be an RVE and let $$\partial{\Omega}$$ be the surface of the RVE. Let the position vector of a point in space be given by $$\mathbf{x}$$. We can write

\mathbf{x} = x_1~\mathbf{e}_1 + x_2~\mathbf{e}_2 + x_3~\mathbf{e}_3 \qquad \implies \qquad \mathbf{x} = x_i~\mathbf{e}_i $$ where $$(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)$$ are the unit vectors that define a Cartesian coordinate system.

Let $$f(\mathbf{x},t)$$ be the physical quantity that is flowing through the RVE. Let $$g(\mathbf{x},t)$$ be sources on the surface of the RVE and let $$h(\mathbf{x},t)$$ be sources inside the RVE. Let $$\mathbf{n}(\mathbf{x},t)$$ be the outward unit normal to the surface $$\partial{\Omega}$$. Let $$\mathbf{v}(\mathbf{x},t)$$ be the velocity of the physical particles that carry the physical quantity that is flowing. Also, let the speed at which the bounding surface $$\partial{\Omega}$$ is moving be $$u_n$$ (in the direction $$\mathbf{n}$$).

Then, balance laws can be expressed in the general form

\cfrac{d}{dt}\left[\int_{\Omega} f(\mathbf{x},t)~\text{dV}\right] = \int_{\partial{\Omega}} f(\mathbf{x},t)[u_n(\mathbf{x},t) - \mathbf{v}(\mathbf{x},t)\cdot\mathbf{n}(\mathbf{x},t)]~\text{dA} + \int_{\partial{\Omega}} g(\mathbf{x},t)~\text{dA} + \int_{\Omega} h(\mathbf{x},t)~\text{dV} ~. $$ Note that the functions $$f(\mathbf{x},t)$$, $$g(\mathbf{x},t)$$, and $$h(\mathbf{x},t)$$ can be scalar valued, vector valued, or tensor valued - depending on the physical quantity that the balance equation deals with.

The balance laws of mass, momentum, and energy can be written as:

\begin{align} \dot{\rho} + \rho~\boldsymbol{\nabla} \bullet \mathbf{v} & = 0 & & \qquad\text{Balance of Mass} \\ \rho~\dot{\mathbf{v}} - \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} - \rho~\mathbf{b} & = 0 & & \qquad\text{Balance of Linear Momentum} \\ \boldsymbol{\sigma} & = \boldsymbol{\sigma}^T & & \qquad\text{Balance of Angular Momentum} \\ \rho~\dot{e} - \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{v}) + \boldsymbol{\nabla} \bullet \mathbf{q} - \rho~s & = 0 & & \qquad\text{Balance of Energy.} \end{align} $$

In the above equations $$\rho(\mathbf{x},t)$$ is the mass density (current), $$\dot{\rho}$$ is the material time derivative of $$\rho$$, $$\mathbf{v}(\mathbf{x},t)$$ is the particle velocity, $$\dot{\mathbf{v}}$$ is the material time derivative of $$\mathbf{v}$$, $$\boldsymbol{\sigma}(\mathbf{x},t)$$ is the Cauchy stress tensor, $$\mathbf{b}(\mathbf{x},t)$$ is the body force density, $$e(\mathbf{x},t)$$ is the internal energy per unit mass, $$\dot{e}$$ is the material time derivative of $$e$$, $$\mathbf{q}(\mathbf{x},t)$$ is the heat flux vector, and $$s(\mathbf{x},t)$$ is an energy source per unit mass.

The gradient and divergence operators are defined such that

\boldsymbol{\nabla} \mathbf{v} = \sum_{i,j = 1}^3 \frac{\partial v_i}{\partial x_j}\mathbf{e}_i\otimes\mathbf{e}_j = v_{i,j}\mathbf{e}_i\otimes\mathbf{e}_j ~; \boldsymbol{\nabla} \bullet \mathbf{v} =\sum_{i=1}^3 \frac{\partial v_i}{\partial x_i} = v_{i,i} ~; \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} = \sum_{i,j=1}^3 \frac{\partial \sigma_{ij}}{\partial x_j}~\mathbf{e}_i = \sigma_{ij,j}~\mathbf{e}_i ~. $$

The contraction operation is given as

\boldsymbol{A}:\boldsymbol{B} = \sum_{i,j=1}^3 A_{ij}~B_{ij} = A_{ij}~B_{ij} ~. $$


text:
Micromechanics of composites/Infinitesimal deformations

Infinitesimal Deformations
Let us first consider only infinitesimal strains and linear elasticity. If we assume that the RVE is small enough, we can neglect inertial and body forces. In addition, if the RVE is in equilibrium, conservation of mass automatically holds. Then the equations that govern the motion of the RVE can be written as:

{ \begin{align} \boldsymbol{\varepsilon} & = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T) & & \qquad \text{Strain-Displacement Relations} \\ \boldsymbol{\sigma} & = \boldsymbol{\mathsf{C}}:\boldsymbol{\varepsilon} & & \qquad \text{Stress-Strain Relations} \\ \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} & = 0& & \qquad \text{Balance of Linear Momentum} \\ \boldsymbol{\sigma} & = \boldsymbol{\sigma}^T& & \qquad \text{Balance of Angular Momentum} \\ \rho~\dot{e} & = \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{v}) - \boldsymbol{\nabla} \bullet \mathbf{q} + \rho~s & & \qquad\text{Balance of Energy.} \\ \end{align} } $$ In the above equations, $$\boldsymbol{\varepsilon}(\mathbf{x})$$ is the strain tensor (small strain), $$\mathbf{u}(\mathbf{x})$$ is the displacement vector, and $$\boldsymbol{\mathsf{C}}(\mathbf{x})$$ is the fourth-order tensor of elastic moduli at the point $$\mathbf{x} \in \Omega$$.

To get a unique solution of the governing equations, we need boundary conditions on $$\partial{\Omega}$$. These boundary conditions may be in the form of applied tractions:

\boldsymbol{\sigma}\bullet\mathbf{n} = \bar{\mathbf{t}} ~;\qquad \sigma_{ij}~n_j = \bar{t_i} $$ where $$\mathbf{n}(\mathbf{x})$$ is the outward normal vector to the surface $$\partial{\Omega}$$ and $$\bar{\mathbf{t}}(\mathbf{x})$$ is the applied traction.

Alternatively, the boundary conditions may be in the form of applied displacements:

\mathbf{u}= \bar{\mathbf{u}} ~;\qquad u_i = \bar{u_i} $$ where $$\bar{\mathbf{u}}(\mathbf{x})$$ is the applied displacement.

We usually assume that the portions of the boundary on which tractions and displacements are applied are nonoverlapping, i.e., $$\partial{\Omega} = \partial{\Omega}_t \cup \partial{\Omega}_u$$ and $$\partial{\Omega}_t \cap \partial{\Omega}_u = \varnothing$$.

If we need to solve the energy equation, we also have to specify heat flux or specified temperature boundary conditions on the RVE.


text:
Micromechanics of composites/Average strain in a RVE

Average Strain in a RVE
The average strain tensor is defined as

{ \langle \boldsymbol{\varepsilon} \rangle := \frac{1}{2}\left(\langle \boldsymbol{\nabla} \mathbf{u} \rangle + \langle \boldsymbol{\nabla} \mathbf{u} \rangle^T\right) } $$ where the average displacement gradient is

{ \langle \boldsymbol{\nabla} \mathbf{u} \rangle = \cfrac{1}{V}\int_{\Omega} \boldsymbol{\nabla}\mathbf{u}~\text{dV} ~. } $$

We would like to find the relation between the average strain in a RVE and the applied displacements at the boundary of the RVE. To do that, recall the relation (see Appendix)

\int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA} $$ where $$\mathbf{v}$$ is a vector field on $$\Omega$$ and $$\mathbf{n}$$ is the normal to $$\partial{\Omega}$$. Using this relation, we get

\langle \boldsymbol{\nabla} \mathbf{u} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}} \mathbf{u}\otimes\mathbf{n}~\text{dA} ~. $$ Hence,

\langle \boldsymbol{\nabla} \mathbf{u} \rangle^T = \cfrac{1}{V}\int_{\partial{\Omega}} \mathbf{n}\otimes\mathbf{u}~\text{dA} ~. $$ Plugging these into the definition of average strain, we get

{ \langle \boldsymbol{\varepsilon} \rangle := \cfrac{1}{2V} \int_{\partial{\Omega}} (\mathbf{u}\otimes\mathbf{n} + \mathbf{n}\otimes\mathbf{u})~\text{dA} ~. } $$ This implies that '''the average strain is completely defined in terms of the applied displacements at the boundary! ''' Also, the average strain tensor is symmetric by virtue of its definition.

We can define the average rotation tensor (which represents an infinitesimal rotation) in an analogous manner. The rotation tensor is given by

\boldsymbol{\omega} = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} - \boldsymbol{\nabla}\mathbf{u}^T) ~. $$ Therefore, the average rotation can be defined as

\langle \boldsymbol{\omega} \rangle := \frac{1}{2}\left(\langle \boldsymbol{\nabla} \mathbf{u} \rangle - \langle \boldsymbol{\nabla} \mathbf{u} \rangle^T\right) ~. $$ In terms of the applied boundary displacements,

\langle \boldsymbol{\omega} \rangle = \cfrac{1}{2V} \int_{\partial{\Omega}} (\mathbf{u}\otimes\mathbf{n} - \mathbf{n}\otimes\mathbf{u})~\text{dA} ~. $$

The effect of rigid body motions on the average strain
Let us consider a rigid body displacement given by (see Appendix)

\mathbf{u}(\mathbf{x}) = \mathbf{c} + \boldsymbol{\omega}\cdot\mathbf{x} $$ where $$\mathbf{c}$$ is a constant translation and $$\boldsymbol{\omega}$$ is a second-order skew symmetric tensor representing an infinitesimal rotation. Then,

\langle \boldsymbol{\nabla} \mathbf{u} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}} \mathbf{u}\otimes\mathbf{n}~\text{dA} = \cfrac{1}{V}\int_{\partial{\Omega}} (\mathbf{c} + \boldsymbol{\omega}\cdot\mathbf{x})\otimes\mathbf{n}~\text{dA} = \cfrac{1}{V}\int_{\partial{\Omega}} \mathbf{c}\otimes\mathbf{n}~\text{dA} + \cfrac{1}{V}\int_{\partial{\Omega}} (\boldsymbol{\omega}\cdot\mathbf{x})\otimes\mathbf{n}~\text{dA} ~. $$ Recall that

(\boldsymbol{A}\cdot\mathbf{b})\otimes\mathbf{c} = \boldsymbol{A}\cdot(\mathbf{b}\otimes\mathbf{c}) $$ where $$\boldsymbol{A}$$ is a second-order tensor and $$\mathbf{b}$$ and $$\mathbf{c}$$ are vectors. Therefore,

\langle \boldsymbol{\nabla} \mathbf{u} \rangle = \mathbf{c}\otimes\left(\cfrac{1}{V}\int_{\partial{\Omega}} \mathbf{n}~\text{dA}\right) + \boldsymbol{\omega}\cdot\left(\cfrac{1}{V}\int_{\partial{\Omega}} \mathbf{x}\otimes\mathbf{n}~\text{dA}\right) ~. $$ From the divergence theorem,

\int_{\Omega} \boldsymbol{\nabla} \bullet \boldsymbol{A}~\text{dV} = \int_{\partial{\Omega}} \boldsymbol{A}\cdot\mathbf{n}~\text{dA} $$ where $$\boldsymbol{A}$$ is a second-order tensor field and $$\mathbf{n}$$ is the unit outward normal vector to $$\partial{\Omega}$$. Hence,

\int_{\partial{\Omega}} \mathbf{n}~\text{dA} = \int_{\partial{\Omega}} \boldsymbol{\mathit{1}}\cdot\mathbf{n}~\text{dA} = \int_{\Omega} \boldsymbol{\nabla} \bullet \boldsymbol{\mathit{1}}~\text{dV} = \mathbf{0} ~. $$ We also have (see appendix),

\int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA} $$ where $$\mathbf{v}$$ is a vector and $$\mathbf{n}$$ is the unit outward normal to $$\partial{\Omega}$$. Therefore,

\int_{\partial{\Omega}} \mathbf{x}\otimes\mathbf{n}~\text{dA} = \int_{\Omega} \boldsymbol{\nabla} \mathbf{x}~\text{dV} = \int_{\Omega} \boldsymbol{\mathit{1}}~\text{dV} = V~\boldsymbol{\mathit{1}} ~. $$ We then have

\langle \boldsymbol{\nabla} \mathbf{u} \rangle = \mathbf{c}\otimes\mathbf{0} + \boldsymbol{\omega}\cdot\boldsymbol{\mathit{1}} = \boldsymbol{\omega} ~. $$ Since $$\boldsymbol{\omega}$$ is a skew-symmetric second-order tensor we have

\boldsymbol{\omega} = - \boldsymbol{\omega}^T ~. $$ Therefore,

\langle \boldsymbol{\varepsilon} \rangle = \frac{1}{2}\left(\langle \boldsymbol{\nabla} \mathbf{u} \rangle + \langle \boldsymbol{\nabla} \mathbf{u} \rangle^T\right) = \frac{1}{2}(\boldsymbol{\omega} + \boldsymbol{\omega}^T) = \mathbf{0} ~. $$ Hence, the average strain is not affected by rigid body motions. However, for simplicity, we assume that the displacement field in a RVE does not contain any rigid body motions.


text:
User:Aftabs

Aftab Ahmed Siddiqui

This is just an intro page. Once I am familer with the formatting and style I will definately going to add more about my self and other things related to my field that is computer networks.

Brief Introduction:

I have completed my Bachelors in Computer Science from University of Karachi, Pakistan and for last 2 years I have been working in the field of core network operations in an Internet Service Provider. I am also doing my Masters in the field of Computer Networks. My presence here may help someone and definately going to help me to groom in the field.

In case someone wants to contact me: aftab.siddiqui@gmail.com

title: User:Aftabs


text:
Micromechanics of composites/Average displacement in a RVE

Average Displacement in a RVE
The average displacement in a RVE may be defined as

{ \langle\mathbf{u}\rangle := \cfrac{1}{V}\int_{\Omega} \mathbf{u}(\mathbf{x})~\text{dV} ~. } $$ We would like to find the relation between the average displacement in a RVE and the applied displacements at the boundary of the RVE. To do that, recall the identity

\boldsymbol{\nabla} \bullet (\mathbf{v}\otimes\boldsymbol{w}) = \mathbf{v}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{w}) + (\boldsymbol{\nabla}\mathbf{v})\cdot\boldsymbol{w} $$ where $$\mathbf{v}$$ and $$\boldsymbol{w}$$ are two vector fields.

If we choose $$\mathbf{v}$$ such that $$\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{\mathit{1}}$$ in the above identity, then we can get an equation for $$\boldsymbol{w}$$, i.e.,

\boldsymbol{\nabla} \bullet (\mathbf{v}\otimes\boldsymbol{w}) = \mathbf{v}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{w}) + \boldsymbol{\mathit{1}}\cdot\boldsymbol{w} = \mathbf{v}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{w}) + \boldsymbol{w} ~. $$ Now, $$\boldsymbol{\nabla}\mathbf{v} = \boldsymbol{\mathit{1}}$$ if $$\mathbf{v} = \mathbf{x}$$. Therefore,

\boldsymbol{\nabla} \bullet (\mathbf{x}\otimes\boldsymbol{w}) = \mathbf{x}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{w}) + \boldsymbol{w}\qquad\implies\qquad \boldsymbol{w} =\boldsymbol{\nabla} \bullet (\mathbf{x}\otimes\boldsymbol{w}) - \mathbf{x}\cdot(\boldsymbol{\nabla} \bullet \boldsymbol{w})~. $$ Using the above in the expression for the average displacement, we have

\langle\mathbf{u}\rangle = \cfrac{1}{V}\int_{\Omega} [\boldsymbol{\nabla} \bullet (\mathbf{x}\otimes\mathbf{u}) - \mathbf{x}\cdot(\boldsymbol{\nabla} \bullet \mathbf{u})]~\text{dV} ~. $$ Applying the divergence theorem to the first term on the right, we get

{ \langle\mathbf{u}\rangle = \cfrac{1}{V}\int_{\partial{\Omega}}(\mathbf{x}\otimes\mathbf{u})\cdot\mathbf{n}~\text{dV} - \cfrac{1}{V}\int_{\Omega}\mathbf{x}\cdot(\boldsymbol{\nabla} \bullet \mathbf{u})~\text{dV} ~. } $$ There are two terms in the above expression: the first is a boundary term while the second requires information from the interior of the body. Hence, in general, the  average displacement of a RVE cannot be determined solely on the basis of boundary displacements.

Incompressible materials
In the material is incompressible, the balance of mass gives us

\boldsymbol{\nabla} \bullet \mathbf{u} = 0 ~. $$ In that case,

\langle\mathbf{u}\rangle = \cfrac{1}{V}\int_{\partial{\Omega}}(\mathbf{x}\otimes\mathbf{u})\cdot\mathbf{n}~\text{dV} ~. $$ It's only in this special case that the average displacement in the RVE can be expressed in terms of boundary displacements.


text:
Micromechanics of composites/Average stress in a RVE

Average Stress in a RVE
Let the average stress in the RVE be defined as

{ \langle \boldsymbol{\sigma} \rangle := \cfrac{1}{V}~\int_{\Omega} \boldsymbol{\sigma}(\mathbf{x})~\text{dV} } $$ where $$V$$ is the volume of $$\Omega$$.

We would like to find out the relation between the average stress in a RVE and the applied tractions on the boundary of the RVE. To do that, recall the relation (see Appendix)

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} ~. $$ If we choose $$\mathbf{v}$$ such that $$\boldsymbol{\nabla} \mathbf{v} = \boldsymbol{\mathit{1}}$$, we have

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\mathit{1}}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} = \int_{\Omega} \boldsymbol{S}~\text{dV} + \int_{\Omega} \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)~\text{dV} ~. $$ Therefore,

\int_{\Omega} \boldsymbol{S}~\text{dV} =- \int_{\Omega} \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)~\text{dV} + \int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA} ~. $$ If we choose $$\boldsymbol{S}$$ to be the stress tensor $$\boldsymbol{\sigma}$$, and involve the symmetry of the stress tensor, we get

\int_{\Omega} \boldsymbol{\sigma}~\text{dV} =- \int_{\Omega} \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})~\text{dV} + \int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{\sigma}\bullet\mathbf{n})~\text{dA} ~. $$ Now, the divergence of the stress is zero (from the conservation of linear momentum). Therefore,

\int_{\Omega} \boldsymbol{\sigma}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{\sigma}\bullet\mathbf{n})~\text{dA} ~. $$ Using the traction boundary condition, we have

\int_{\Omega} \boldsymbol{\sigma}~\text{dV} = \int_{\partial{\Omega}} \mathbf{v}\otimes\bar{\mathbf{t}}~\text{dA} ~. $$ Now $$\boldsymbol{\nabla} \mathbf{v} = \boldsymbol{\mathit{1}}$$ if $$\mathbf{v} = \mathbf{x}$$. Therefore, we have

\int_{\Omega} \boldsymbol{\sigma}~\text{dV} = \int_{\partial{\Omega}} \mathbf{x}\otimes\bar{\mathbf{t}}~\text{dA} ~. $$ Hence the average stress is given by

{ \langle \boldsymbol{\sigma} \rangle := \cfrac{1}{V}~\int_{\partial{\Omega}} \mathbf{x}\otimes\bar{\mathbf{t}}~\text{dA} ~. } $$ This implies that the average stress is completely determined by the applied tractions!

Symmetry of the average stress and the effect of rigid body translation
Let us now assume that the applied tractions are self equilibrating. Then the resultant forces and moments due to the applied tractions vanish and we have

\int_{\partial{\Omega}} \bar{\mathbf{t}}~\text{dA} = \mathbf{0} \qquad\text{and}\qquad \int_{\partial{\Omega}} \mathbf{x}\times\bar{\mathbf{t}}~\text{dA} = \mathbf{0} ~. $$

From the moment balance equation above we can show that (see Appendix)

\int_{\partial{\Omega}} \mathbf{x}\otimes\bar{\mathbf{t}}~\text{dA} = \int_{\partial{\Omega}} \bar{\mathbf{t}}\otimes\mathbf{x}~\text{dA} ~. $$ Therefore the average stress tensor $$\langle \boldsymbol{\sigma} \rangle$$ is symmetric if the applied tractions are self equilibrated.

Now, if we translate the body by a constant amount $$\mathbf{u}_0$$ (rigid body translation), we get

\bar{\langle \boldsymbol{\sigma} \rangle} = \cfrac{1}{V}~\int_{\partial{\Omega}} (\mathbf{x}+\mathbf{u}_0)\otimes\bar{\mathbf{t}}~\text{dA} = \cfrac{1}{V}~\int_{\partial{\Omega}} [\mathbf{x}\otimes\bar{\mathbf{t}} + \mathbf{u}_0\otimes\bar{\mathbf{t}}]~\text{dA}~. $$ or

\bar{\langle \boldsymbol{\sigma} \rangle} = \cfrac{1}{V}\left[ \int_{\partial{\Omega}} \mathbf{x}\otimes\bar{\mathbf{t}}~\text{dA} + \mathbf{u}_0\otimes\int_{\partial{\Omega}} \bar{\mathbf{t}}~\text{dA}\right] = \langle \boldsymbol{\sigma} \rangle $$ Therefore, the average stress is not affected by a rigid body translation only if the applied tractions are self equilibrated.

We can conclude that  the average stress $$\langle \boldsymbol{\sigma} \rangle$$ is an acceptable measure of stress in a RVE if the applied tractions are self equilibrated.


text:
Micromechanics of composites/Average stress power in a RVE

Average Stress Power in a RVE
The equation for the balance of energy is

\rho~\dot{e} - \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{v}) + \boldsymbol{\nabla} \bullet \mathbf{q} - \rho~s = 0~. $$ If the absence of heat flux or heat sources in the RVE, the equation reduces to

\rho~\dot{e} = \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{v})~. $$ The quantity on the right is the stress power density and is a measure of the internal energy density of the material.

The average stress power in a RVE is defined as

{ \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle := \cfrac{1}{V}\int_{\Omega} \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v}~\text{dV} ~. } $$ Note that the quantities $$\boldsymbol{\sigma}$$ and $$\boldsymbol{\nabla}\mathbf{v}$$ need not be related in the general case.

The average velocity gradient $$\langle\boldsymbol{\nabla}\mathbf{v} \rangle$$ is defined as

{ \langle\boldsymbol{\nabla}\mathbf{v} \rangle := \cfrac{1}{V}\int_{\Omega} \boldsymbol{\nabla}\mathbf{v}~\text{dV} ~. } $$ To get an expression for the average stress power in terms of the boundary conditions, we use the identity

\boldsymbol{\nabla} \bullet (\boldsymbol{S}^T\cdot\mathbf{v}) = \boldsymbol{S}:\boldsymbol{\nabla}\mathbf{v} + (\boldsymbol{\nabla} \bullet \boldsymbol{S})\cdot\mathbf{v} $$ to get

\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\Omega} \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v}~\text{dV} = \cfrac{1}{V}\int_{\Omega} \left[\boldsymbol{\nabla} \bullet (\boldsymbol{\sigma}^T\cdot\mathbf{v}) - (\boldsymbol{\nabla} \bullet \boldsymbol{\sigma})\cdot\mathbf{v}\right]~\text{dV} ~. $$ Using the balance of linear momentum ($$\boldsymbol{\nabla} \bullet \boldsymbol{\sigma} = 0$$), we get

\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\Omega}\boldsymbol{\nabla} \bullet (\boldsymbol{\sigma}^T\cdot\mathbf{v})~\text{dV} ~. $$ Using the divergence theorem, we have

\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}}(\boldsymbol{\sigma}^T\cdot\mathbf{v})\cdot\mathbf{n}~\text{dV} = \cfrac{1}{V}\int_{\partial{\Omega}}(\boldsymbol{\sigma}^T\cdot\mathbf{v})\cdot\mathbf{n}~\text{dV} = \cfrac{1}{V}\int_{\partial{\Omega}}(\boldsymbol{\sigma}\cdot\mathbf{n})\cdot\mathbf{v}~\text{dV} ~. $$ Now, the surface traction is given by $$\bar{\mathbf{t}} = \boldsymbol{\sigma}\cdot\mathbf{n}$$. Therefore,

{ \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}}\bar{\mathbf{t}}\cdot\mathbf{v}~\text{dV} ~. } $$ {\scriptsize } In micromechanics, it is of interest to see how the average stress power of a RVE is related to the product of the average stress $$\langle \boldsymbol{\sigma} \rangle$$ and the average velocity gradient $$\langle\boldsymbol{\nabla}\mathbf{v} \rangle$$. While homogenizing a RVE, we would ideally like to have

\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle ~. $$ However, this is not true in general. We can show that  if the gradient of the velocity is a symmetric tensor (i.e., there is no spin), then (see Appendix for proof)

{ \langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle - \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V} \int_{\partial{\Omega}} [\mathbf{v} - \langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}]\cdot [(\boldsymbol{\sigma} - \langle \boldsymbol{\sigma} \rangle)\cdot\mathbf{n}]~\text{dA} ~. } $$

We can arrive at $$\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle$$ if either of the following conditions is met on the boundary $$\partial{\Omega}$$:

 * 1) $$\mathbf{v} = \langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x}$$ ~.
 * 2) $$\boldsymbol{\sigma}\cdot\mathbf{n} = \langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n}$$ ~.

Linear boundary velocities
If the prescribed velocities on $$\partial{\Omega}$$ are a linear function of $$\mathbf{x}$$, then we can write

\mathbf{v}(\mathbf{x}) = \boldsymbol{G}\cdot\mathbf{x} \qquad \qquad \forall ~\mathbf{x} \in \partial{\Omega} $$ where $$\boldsymbol{G}$$ is a constant second-order tensor.

From the divergence theorem

\int_{\Omega} \boldsymbol{\nabla} \mathbf{a}~\text{dV} = \int_{\partial{\Omega}} \mathbf{a}\otimes\mathbf{n}~\text{dA} ~. $$ Therefore,

\langle\boldsymbol{\nabla}\mathbf{v} \rangle = \cfrac{1}{V}\int_{\Omega} \boldsymbol{\nabla} \mathbf{v}~\text{dV} = \cfrac{1}{V}\int_{\partial{\Omega}} \mathbf{v}\otimes\mathbf{n}~\text{dA}~. $$ Hence, on the boundary

\mathbf{v} - \langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x} = \boldsymbol{G}\cdot\mathbf{x} - \left[\cfrac{1}{V}\int_{\partial{\Omega}} (\boldsymbol{G}\cdot\mathbf{x})\otimes\mathbf{n}~\text{dA}\right]\cdot\mathbf{x} $$ Using the identity (see Appendix)

(\boldsymbol{A}\cdot\mathbf{a})\otimes\mathbf{b} = \boldsymbol{A}\cdot(\mathbf{a}\otimes\mathbf{b}) $$ and since $$\boldsymbol{G}$$ is constant, we get

\mathbf{v} - \langle\boldsymbol{\nabla}\mathbf{v} \rangle^T\cdot\mathbf{x} = \boldsymbol{G}\cdot\mathbf{x} - \left[\boldsymbol{G}\cdot\left(\cfrac{1}{V} \int_{\partial{\Omega}}\mathbf{x}\otimes\mathbf{n}~\text{dA}\right) \right]\cdot\mathbf{x} ~. $$ From the divergence theorem,

\int_{\partial{\Omega}} \mathbf{x}\otimes\mathbf{n}~\text{dA} = \int_{\Omega} \boldsymbol{\nabla} \mathbf{x}~\text{dV} = \int_{\Omega} \boldsymbol{\mathit{1}}~\text{dV} = V~\boldsymbol{\mathit{1}}~. $$ Therefore,

\mathbf{v} - \langle\boldsymbol{\nabla}\mathbf{v} \rangle\cdot\mathbf{x} = \boldsymbol{G}\cdot\mathbf{x} - (\boldsymbol{G}\cdot\boldsymbol{\mathit{1}})\cdot\mathbf{x} = \boldsymbol{G}\cdot\mathbf{x} - \boldsymbol{G}\cdot\mathbf{x} = \mathbf{0} \qquad \implies \qquad {\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle}~. $$

Uniform boundary tractions
If the prescribed tractions on the boundary $$\partial{\Omega}$$ are uniform, they can be expressed in terms of a constant symmetric second-order tensor $$\boldsymbol{G}$$ through the relation

\bar{t}(\mathbf{x}) = \boldsymbol{G}\cdot\mathbf{n}(\mathbf{x}) \qquad\qquad \forall~\mathbf{x} \in \partial{\Omega}~. $$ The tractions are related to the stresses at the boundary of the RVE by $$ \bar{t} = \boldsymbol{\sigma}\cdot\mathbf{n}$$.

The average stress in the RVE is given by

\langle \boldsymbol{\sigma} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}} \mathbf{x}\otimes\bar{t}~\text{dA} = \cfrac{1}{V}\int_{\partial{\Omega}} \mathbf{x}\otimes(\boldsymbol{G}\cdot\mathbf{n})~\text{dA} ~. $$ Using the identity $$\mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T$$ (see Appendix), we have

\langle \boldsymbol{\sigma} \rangle = \cfrac{1}{V}\int_{\partial{\Omega}} (\mathbf{x}\otimes\mathbf{n})\cdot\boldsymbol{G}^T~\text{dA} ~. $$ Since $$\boldsymbol{G}$$ is constant and symmetric, we have

\langle \boldsymbol{\sigma} \rangle = \left(\cfrac{1}{V}\int_{\partial{\Omega}}\mathbf{x}\otimes\mathbf{n}~\text{dA}\right) \cdot\boldsymbol{G}~. $$ Applying the divergence theorem,

\langle \boldsymbol{\sigma} \rangle = \left(\cfrac{1}{V}\int_{\Omega}\boldsymbol{\nabla} \mathbf{x}~\text{dV}\right) \cdot\boldsymbol{G} = \boldsymbol{\mathit{1}}\cdot\boldsymbol{G} = \boldsymbol{G} ~. $$ Therefore,

\boldsymbol{\sigma}\cdot\mathbf{n} - \langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n} = \bar{t} - \boldsymbol{G}\cdot\mathbf{n} = \mathbf{0} \qquad \implies \qquad {\langle \boldsymbol{\sigma}:\boldsymbol{\nabla}\mathbf{v} \rangle = \langle \boldsymbol{\sigma} \rangle:\langle\boldsymbol{\nabla}\mathbf{v} \rangle}~. $$

Remark
Recall that for small deformations, the displacement gradient $$\boldsymbol{\nabla}\mathbf{u}$$ can be expressed as

\boldsymbol{\nabla}\mathbf{u} = \boldsymbol{\varepsilon} + \boldsymbol{\omega} ~. $$ For small deformations, the time derivative of $$\boldsymbol{\nabla}\mathbf{u}$$ gives us the velocity gradient $$\boldsymbol{\nabla}\mathbf{v}$$, i.e.,

\boldsymbol{\nabla}\mathbf{v} = \dot{\boldsymbol{\varepsilon}} + \dot{\boldsymbol{\omega}} ~. $$ If $$\boldsymbol{\omega} = 0$$, we get

\boldsymbol{\nabla}\mathbf{v} = \dot{\boldsymbol{\varepsilon}} ~. $$ Hence, for small strains and in the absence of rigid body rotations, the stress power density is given by $$\boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}$$. Then the average stress power is defined as

{     \langle \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}\rangle := \cfrac{1}{V}\int_{\Omega} \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}~\text{dV} ~. }   $$ and the average strain rate is defined as

{     \langle \dot{\boldsymbol{\varepsilon}} \rangle := \cfrac{1}{V}\int_{\Omega} \dot{\boldsymbol{\varepsilon}}~\text{dV} ~. }   $$ In terms of the surface tractions and the applied boundary velocities, we have

{     \langle \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}\rangle = \cfrac{1}{V}\int_{\partial{\Omega}}\bar{\mathbf{t}}\cdot\dot{\mathbf{u}}~\text{dV} ~. }   $$ For small strains and no rotation, the stress-power difference relation becomes

{     \langle \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}\rangle - \langle \boldsymbol{\sigma} \rangle:\langle \dot{\boldsymbol{\varepsilon}} \rangle = \cfrac{1}{V} \int_{\partial{\Omega}} [\dot{\mathbf{u}} - \langle\boldsymbol{\nabla} \dot{\mathbf{u}} \rangle\cdot\mathbf{x}]\cdot [(\boldsymbol{\sigma} - \langle \boldsymbol{\sigma} \rangle)\cdot\mathbf{n}]~\text{dA} ~. }   $$ We can arrive at $$\langle \boldsymbol{\sigma}:\dot{\boldsymbol{\varepsilon}}\rangle = \langle \boldsymbol{\sigma} \rangle:\langle \dot{\boldsymbol{\varepsilon}} \rangle$$ if either of the following conditions is met on the boundary $$\partial{\Omega}$$: We can also show in an identical manner that
 * 1) $$\dot{\mathbf{u}} = \langle\boldsymbol{\nabla} \dot{\mathbf{u}} \rangle\cdot\mathbf{x} \qquad\implies\qquad$$ Linear boundary velocity field.
 * 2) $$\boldsymbol{\sigma}\cdot\mathbf{n} = \langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n} \qquad\implies\qquad$$ Uniform boundary tractions.

{     \langle \boldsymbol{\sigma}:\boldsymbol{\varepsilon}\rangle = \cfrac{1}{V}\int_{\partial{\Omega}}\bar{\mathbf{t}}\cdot\mathbf{u}~\text{dV} ~. }   $$ and that, when $$\boldsymbol{\nabla}\mathbf{u}$$ is symmetric,

{     \langle \boldsymbol{\sigma}:\boldsymbol{\varepsilon}\rangle - \langle \boldsymbol{\sigma} \rangle:\langle \boldsymbol{\varepsilon} \rangle = \cfrac{1}{V} \int_{\partial{\Omega}} [\mathbf{u} - \langle \boldsymbol{\nabla} \mathbf{u} \rangle\cdot\mathbf{x}]\cdot [(\boldsymbol{\sigma} - \langle \boldsymbol{\sigma} \rangle)\cdot\mathbf{n}]~\text{dA} ~. }   $$ In this case, we can arrive at the relation $$\langle \boldsymbol{\sigma}:\boldsymbol{\varepsilon}\rangle = \langle \boldsymbol{\sigma} \rangle:\langle \boldsymbol{\varepsilon} \rangle$$ if either of the following conditions is met at the boundary:
 * 1) $$\mathbf{u} = \langle \boldsymbol{\nabla} \mathbf{u} \rangle\cdot\mathbf{x} \qquad\implies\qquad$$ Linear boundary displacement field.
 * 2) $$\boldsymbol{\sigma}\cdot\mathbf{n} = \langle \boldsymbol{\sigma} \rangle\cdot\mathbf{n} \qquad\implies\qquad$$ Uniform boundary tractions.


text: User talk:Dkcreatto

title: User talk:Dkcreatto


text:
Micromechanics of composites/Finite deformations

Finite deformations
If a RVE undergoes finite deformations (i.e., large strains and large rotations), then we have to make a distinction between the initial and deformed configuration. Let us assume that the deformation can be described by a map

\mathbf{x} = \boldsymbol{\varphi}(\mathbf{X}) = \mathbf{x}(\mathbf{X}) $$ where $$\mathbf{X}$$ is the position of a point in the RVE in the initial configuration and $$\mathbf{x}$$ is the location of the same point in the deformed configuration.

The deformation gradient is given by

\boldsymbol{F} = \frac{\partial \mathbf{x}}{\partial \mathbf{X}} = \boldsymbol{\nabla}_0~ \mathbf{x} ~. $$

If we assume that the RVE is small enough, we can neglect inertial and body forces. Then the equations that govern the motion of the RVE can be written with respect to the reference configuration as

{ \begin{align} \boldsymbol{C} & = \boldsymbol{F}^T\cdot\boldsymbol{F} & & \qquad \text{Strain-deformation Relations} \\ \boldsymbol{P} & = \hat{\boldsymbol{P}}(\boldsymbol{C}) & & \qquad \text{Stress-Strain Relations} \\ \rho~\det(\boldsymbol{F}) & = \rho_0 & &\qquad \text{Balance of Mass} \\ \boldsymbol{\nabla}_0 \bullet \boldsymbol{P}^T & = 0& & \qquad \text{Balance of Linear Momentum} \\ \boldsymbol{F}\cdot\boldsymbol{P} & = \boldsymbol{P}^T\cdot\boldsymbol{F}^T& & \qquad \text{Balance of Angular Momentum} \\ \rho_0~\dot{e} & = \boldsymbol{P}^T:\dot{\boldsymbol{F}} - \boldsymbol{\nabla}_0 \bullet \mathbf{q} + \rho_0~s & & \qquad\text{Balance of Energy.} \\ \end{align} } $$ In the above $$\boldsymbol{C}$$ is the right Cauchy-Green deformation tensor, $$\boldsymbol{P}$$ is the first Piola-Kirchhoff stress tensor, and $$rho_0$$ is the mass density in the reference configuration. The first Piola-Kirchhoff stress tensor is related to the Cauchy stress tensor by

\boldsymbol{P} = \det(\boldsymbol{F})~\boldsymbol{F}^{-1}\cdot\boldsymbol{\sigma} ~. $$ The gradient and divergence operators are defined such that

\boldsymbol{\nabla}_0~ \mathbf{v} = \sum_{i,j = 1}^3 \frac{\partial v_i}{\partial X_j}\boldsymbol{E}_i\otimes\boldsymbol{E}_j = v_{i,j}\boldsymbol{E}_i\otimes\boldsymbol{E}_j ~; \boldsymbol{\nabla}_0 \bullet \mathbf{v} =\sum_{i=1}^3 \frac{\partial v_i}{\partial X_i} = v_{i,i} ~; \boldsymbol{\nabla}_0 \bullet \boldsymbol{S} = \sum_{i,j=1}^3 \frac{\partial S_{ij}}{\partial X_j}~\boldsymbol{E}_i = S_{ij,j}~\boldsymbol{E}_i $$ where $$\mathbf{v}$$ is a vector field, $$BS$$ is a second-order tensor field, and $$\boldsymbol{E}_i$$ are the components of an orthonormal basis in the reference configuration.

With respect to the deformed configuration, the governing equations are

{ \begin{align} \boldsymbol{b} & = \boldsymbol{F}\cdot\boldsymbol{F}^T & & \qquad \text{Strain-deformation Relations} \\ \boldsymbol{\sigma} & = \hat{\boldsymbol{\sigma}}(\boldsymbol{b}) & & \qquad \text{Stress-Strain Relations} \\ \rho~\det(\boldsymbol{F}) & = \rho_0 & &\qquad \text{Balance of Mass} \\ \boldsymbol{\nabla} \bullet \boldsymbol{\sigma} & = 0& & \qquad \text{Balance of Linear Momentum} \\ \boldsymbol{\sigma} & = \boldsymbol{\sigma}^T& & \qquad \text{Balance of Angular Momentum} \\ \rho~\dot{e} & = \boldsymbol{\sigma}:(\boldsymbol{\nabla}\mathbf{v}) - \boldsymbol{\nabla} \bullet \mathbf{q} + \rho~s & & \qquad\text{Balance of Energy.} \\ \end{align} } $$ Here, $$\boldsymbol{b}$$ is the left Cauchy-Green deformation tensor, $$\boldsymbol{\sigma}$$ is the Cauchy stress, and $$\rho$$ is the mass density in the deformed configuration. The gradient and divergence operators are defined such that

\boldsymbol{\nabla} \mathbf{v} = \sum_{i,j = 1}^3 \frac{\partial v_i}{\partial x_j}\mathbf{e}_i\otimes\mathbf{e}_j = v_{i,j}\mathbf{e}_i\otimes\mathbf{e}_j ~; \boldsymbol{\nabla} \bullet \mathbf{v} =\sum_{i=1}^3 \frac{\partial v_i}{\partial x_i} = v_{i,i} ~; \boldsymbol{\nabla} \bullet \boldsymbol{S} = \sum_{i,j=1}^3 \frac{\partial S_{ij}}{\partial x_j}~\mathbf{e}_i = S_{ij,j}~\mathbf{e}_i ~. $$

It is convenient to express all (unweighted) volume average quantities for finite deformation in terms of integrals over the reference volume ($$\Omega_0$$) and the reference surface ($$\partial{\Omega}_0$$).

Note that the strain measures used for finite deformation contain products of the deformation gradient. For example,

\boldsymbol{C} = \boldsymbol{F}^T\cdot\boldsymbol{F} ~. $$ A volume average of $$\boldsymbol{C}$$ may be defined in two ways:

\langle\boldsymbol{C}\rangle := \cfrac{1}{V_0} \int_{\Omega_0} \boldsymbol{C}~\text{dV} = \cfrac{1}{V_0} \int_{\Omega_0} \boldsymbol{F}^T\cdot\boldsymbol{F}~\text{dV} \qquad\text{or}\qquad \overline{\boldsymbol{C}} := \left(\cfrac{1}{V_0} \int_{\Omega_0} \boldsymbol{F}^T~\text{dV}\right)\cdot \left(\cfrac{1}{V_0} \int_{\Omega_0} \boldsymbol{F}^T~\text{dV}\right) = \langle \boldsymbol{F}\rangle^T\cdot\langle \boldsymbol{F}\rangle ~. $$ The choice of the definition of a macroscopic average quantity is based on physical considerations. Ideally, { such quantities are chosen such that their unweighted volume averages are completely defined by the surface data.}Unweighted average quantities that satisfy these requirements are the deformation gradient $$\boldsymbol{F}$$, its rate $$\dot{\boldsymbol{F}}$$, the first Piola-Kirchhoff stress $$\boldsymbol{P}$$, and its rate $$\dot{\boldsymbol{P}}$$.


text:
Micromechanics of composites/Average deformation gradient in a RVE

Average deformation gradient in a RVE
The average deformation gradient is defined as

{ \langle \boldsymbol{F}\rangle := \cfrac{1}{V_0}\int_{\Omega_0} \boldsymbol{F}~\text{dV} } $$ where $$V_0$$ is the volume in the reference configuration.

We can express the average deformation gradient in terms of surface quantities by using the divergence theorem. Thus,

\langle \boldsymbol{F}\rangle = \cfrac{1}{V_0}\int_{\Omega_0} \boldsymbol{F}~\text{dV} = \cfrac{1}{V_0}\int_{\Omega_0} \boldsymbol{\nabla}_0~ \mathbf{x}~\text{dV} = \cfrac{1}{V_0} \int_{\partial{\Omega}_0} \mathbf{x}\otimes\mathbf{N}~\text{dA} = \cfrac{1}{V_0} \int_{\partial{\Omega}_0} (\mathbf{X}+\mathbf{u})\otimes\mathbf{N}~\text{dA} $$ where $$\mathbf{N}$$ is the unit outward normal to the reference surface $$\partial{\Omega}_0$$ and $$\mathbf{u}(\mathbf{X}) = \mathbf{x} - \mathbf{X}$$ is the displacement.

The surface integral can be converted into an integral over the deformed surface using Nanson's formula for areas:

\text{d}\mathbf{a} = \det(\boldsymbol{F})~\boldsymbol{F}^{-T}~\text{d}\mathbf{A}\qquad\equiv\qquad \mathbf{n}~\text{da} = \det(\boldsymbol{F})~\boldsymbol{F}^{-T}\cdot\mathbf{N}~\text{dA} \quad \implies \quad \cfrac{1}{\det{\boldsymbol{F}}}~\boldsymbol{F}^T\cdot\mathbf{n}~\text{da} = \mathbf{N}~\text{dA} $$ where $$\text{da}$$ is an element of area on the deformed surface, $$\mathbf{n}$$ is the outward normal to the deformed surface, and $$\text{dA}$$ is an element of area on the reference surface.

The conservation of mass gives us

J := \det(\boldsymbol{F}) = \cfrac{\rho_0}{\rho} = \cfrac{V}{V_0}~. $$ Therefore,

\mathbf{x}\otimes\mathbf{N}~\text{dA} = \mathbf{x}\otimes(\mathbf{N}~\text{dA}) = \mathbf{x}\otimes\left(\cfrac{V_0}{V}~\boldsymbol{F}^T\cdot\mathbf{n}~\text{da}\right) = \left(\cfrac{V_0}{V}\right)~\mathbf{x}\otimes(\boldsymbol{F}^T\cdot\mathbf{n})~\text{da} $$ Plugging into the surface integral, we have

\langle \boldsymbol{F} \rangle = \cfrac{1}{V_0} \int_{\partial \Omega_0} \mathbf{x}\otimes\mathbf{N}~\text{dA} = \cfrac{1}{V_0} \int_{\partial \Omega} \left[ \left(\cfrac{V_0}{V}\right)~\mathbf{x}\otimes(\boldsymbol{F}^T\cdot\mathbf{n}) \right]~\text{da} = \cfrac{1}{V} \int_{\partial \Omega } \mathbf{x}\otimes(\boldsymbol{F}^T\cdot\mathbf{n})~\text{da} ~. $$ Using the identity $$\mathbf{a}\otimes(\boldsymbol{A}\cdot\mathbf{b}) = (\mathbf{a}\otimes\mathbf{b})\cdot\boldsymbol{A}^T$$ (see Appendix), we get

\langle \boldsymbol{F}\rangle = \cfrac{1}{V} \int_{\partial{\Omega}} (\mathbf{x}\otimes\mathbf{n})\cdot\boldsymbol{F}~\text{da} ~. $$ Therefore, the average deformation gradient in surface integral form can be written as

{ \langle \boldsymbol{F}\rangle = \cfrac{1}{V_0} \int_{\partial{\Omega}_0} \mathbf{x}\otimes\mathbf{N}~\text{dA} = \cfrac{1}{V} \int_{\partial{\Omega}} (\mathbf{x}\otimes\mathbf{n})\cdot\boldsymbol{F}~\text{da} ~. } $$

Note that there are three more conditions to be satisfied for the average deformation gradient to behave like a macro variable, i.e.,

\det\langle \boldsymbol{F}\rangle > 0 ~; \langle \boldsymbol{F}\rangle^{-1} = \langle \boldsymbol{F}^{-1 \rangle} ~; V = V_0 \det\langle \boldsymbol{F}\rangle ~. $$ These considerations and their detailed exploration can be found in Costanzo et al.(2005).


text:
Micromechanics of composites/Average velocity gradient in a RVE

Average velocity gradient in a RVE
The time rate of the deformation gradient is given by

\dot{\boldsymbol{F}} = \frac{\partial }{\partial t}[\boldsymbol{F}(\mathbf{X}, t)] = \frac{\partial }{\partial t}\left(\frac{\partial }{\partial \mathbf{X}}[\mathbf{x}(\mathbf{X},t)]\right) = \frac{\partial }{\partial \mathbf{X}}\left(\frac{\partial }{\partial t}[\mathbf{x}(\mathbf{X},t)]\right) = \frac{\partial \dot{\mathbf{x}}}{\partial \mathbf{X}} = \boldsymbol{\nabla}_0~ \dot{\mathbf{x}} ~. $$ The average time rate of the deformation gradient is defined as

{ \langle \dot{\boldsymbol{F}} \rangle := \cfrac{1}{V_0}\int_{\Omega_0} \dot{\boldsymbol{F}}~\text{dV} ~. } $$ Following the same procedure as in the previous section, we can show that

{ \langle \dot{\boldsymbol{F}} \rangle = \cfrac{1}{V_0} \int_{\partial{\Omega}_0} \dot{\mathbf{x}}\otimes\mathbf{N}~\text{dA} = \cfrac{1}{V} \int_{\partial{\Omega}} (\dot{\mathbf{x}}\otimes\mathbf{n})\cdot\boldsymbol{F}~\text{da} ~. } $$ The velocity gradient ($$\boldsymbol{l}$$) is given by

\boldsymbol{l} = \boldsymbol{\nabla}\mathbf{v} = \dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1} $$ where $$\mathbf{v}(\mathbf{x})$$ is the velocity.

The average velocity gradient in a RVE is defined as

{ \overline{\boldsymbol{l}} := \langle \dot{\boldsymbol{F}} \rangle\cdot\langle \boldsymbol{F}\rangle^{-1} ~. } $$ Note that $$\overline{\boldsymbol{l}} = \langle\boldsymbol{l}\rangle = \langle\dot{\boldsymbol{F}}\cdot\boldsymbol{F}^{-1}\rangle$$ only if $$\boldsymbol{F} = \boldsymbol{\mathit{1}}$$.


text:
Micromechanics of composites/Average stress in a RVE with finite strain

Average stress in a RVE
The average nominal (first Piola-Kirchhoff ) stress is defined as

{ \langle \boldsymbol{P} \rangle = \cfrac{1}{V_0} \int_{\Omega_0} \boldsymbol{P}~\text{dV} ~. } $$ Recall the relation (see Appendix)

\int_{\partial{\Omega}} \mathbf{v}\otimes(\boldsymbol{S}^T\bullet\mathbf{n})~\text{dA} = \int_{\Omega} [\boldsymbol{\nabla} \mathbf{v}\cdot\boldsymbol{S} + \mathbf{v}\otimes(\boldsymbol{\nabla} \bullet \boldsymbol{S}^T)]~\text{dV} ~. $$ In the above equation, let the volume integral be over $$\Omega_0$$ and let the surface integral be over $$\partial{\Omega}_0$$. Let the unit outward normal to $$\partial{\Omega}_0$$ be $$\mathbf{N}$$. Let the gradient and divergence operations be with respect to the reference configuration. Also, let $$\mathbf{v} \rightarrow \mathbf{X}$$ and let $$\boldsymbol{S} \rightarrow \boldsymbol{P}$$. Then we have

\int_{\partial{\Omega}_0} \mathbf{X}\otimes(\boldsymbol{P}^T\bullet\mathbf{N})~\text{dA} = \int_{\Omega_0} [\boldsymbol{\nabla}_0~ \mathbf{X}\cdot\boldsymbol{P} + \mathbf{X}\otimes(\boldsymbol{\nabla}_0 \bullet \boldsymbol{P}^T)]~\text{dV} = \int_{\Omega_0} [\boldsymbol{\mathit{1}}\cdot\boldsymbol{P} + \mathbf{X}\otimes(\boldsymbol{\nabla}_0 \bullet \boldsymbol{P}^T)]~\text{dV} = \int_{\Omega_0} [\boldsymbol{P} + \mathbf{X}\otimes(\boldsymbol{\nabla}_0 \bullet \boldsymbol{P}^T)]~\text{dV} ~. $$ If we assume that there are  no inertial forces or body forces, then $$\boldsymbol{\nabla}_0 \bullet \boldsymbol{P}^T = 0$$ (from the conservation of linear momentum), and we have

\int_{\partial{\Omega}_0} \mathbf{X}\otimes(\boldsymbol{P}^T\bullet\mathbf{N})~\text{dA} = \int_{\Omega_0} \boldsymbol{P}~\text{dV} = V_0~\langle \boldsymbol{P} \rangle ~. $$ Let $$\bar{\mathbf{T}}$$ be a  self equilibrating traction that is applied to the RVE, i.e., it does not lead to any inertial forces. Then, Cauchy's law states that $$\bar{\mathbf{T}} = \boldsymbol{P}^T\cdot\mathbf{N}$$ on $$\partial{\Omega}_0$$. Hence we get

{ \langle \boldsymbol{P} \rangle = \cfrac{1}{V_0} \int_{\partial{\Omega}_0} \mathbf{X}\otimes\bar{\mathbf{T}}~\text{dA} ~. } $$ Given the above, the average Cauchy stress in the RVE is defined as

{ \langle \overline{\boldsymbol{\sigma}} \rangle := \cfrac{1}{\det\langle \boldsymbol{F}\rangle}~\langle \boldsymbol{F}\rangle\cdot\langle \boldsymbol{P} \rangle ~. } $$ Note that, in general, $$\langle \overline{\boldsymbol{\sigma}} \rangle \ne \langle \boldsymbol{\sigma} \rangle$$.

The Kirchhoff stress is defined as $$\boldsymbol{\tau} := \det\boldsymbol{F}~\boldsymbol{\sigma}$$. The average Kirchhoff stress in the RVE is defined as

{ \langle \overline{\boldsymbol{\tau}} \rangle := \det\langle \boldsymbol{F}\rangle~\langle \overline{\boldsymbol{\sigma}} \rangle = \langle \boldsymbol{F}\rangle\cdot\langle \boldsymbol{P} \rangle ~. } $$ In general, $$\langle \overline{\boldsymbol{\tau}} \rangle \ne \langle \boldsymbol{\tau} \rangle$$.


text:
Power Generation

Welcome to the Home page for Electrical Power Generation

In this course we will discuss Methods used to Generate Electricity ( i.e Types of Electric Power Stations ). At the end of this course the student is expected to be familiar with the Principles of Electrical Generation and their application in real life.

You are advised to read carefully through given material and attempt all quiz/questionaires in this course and dont forget to follow necessary safety code that will be outlined in this course. We hope you will enjoy this major course in Electrical Engineering.

Course Outcomes
At the end of this course the student is expected to be comfortable with the following:
 * 1) Basics of Energy Conversion
 * 2) Sources of Energy
 * 3) Types of Power stations ( Construction & Operation )
 * 4) Variable Load on Power Stations
 * 5) Comparison of Different Power stations.

Prerequisites
The following courses are considered important for the student to be at ease with material that will be covered in this course:
 * 1) Electrical Engineering Fundamentals
 * 2) Alternating Current Circuits

Resources
Wikiversity has a respectable body of content related to this course, you are advised to use the resources wisely to achieve your success. This course includes a formula sheet that can help you remember important Formulae and principles.Do not rely on the formula sheet entirely without understanding course material.

Lessons

 * /Introduction/
 * /Steam Power/
 * /Hydro Power/

 * /Diesel Power/
 * /Nuclear Power/
 * /Gas Turbine Power/

 * /Variable Load/
 * /Comparison/
 * /Formulae/

Organization
This course consists of:
 * 8 Lessons → [ 8 weeks ]
 * 1 Quiz Test → [ 1 Day ]

All the above Span about 8 weeks. Plan your time wisely, by starting a journal and track your progress. If you work consistently and keep to your planned Study Time, you will Do Very well in this course.

Here are some of the resources available to Help you understand and make better used of this personal Learning Experience:
 * History of free learning
 * Learning by doing
 * How to use wiki technology as a free learner

title: Power Generation


text:
Markerless Tracking/Related Work

This page sumarizes related work for the research project Markerless Tracking.

Recognition
[http://mi.informatik.uni-siegen.de/ Blanz et. al.] showed how to use computer generated images of faces to recognize real ones. This work is based on a morphable face model which was presented before.

Not jet reviewd:

Tracking
Motion Capturing could greatly improve with analysis-by-synthesis as stated by Moeslund and Granum.

Not jet reviewd:

Theory
Not jet reviewd:

Commercial
Organic Motion's solution to Motion Capturing was presented at SIGGRAPH 2007 (demo video) and is the first commercial available product to track all bodyparts including bones, geometry and textures. Their website claims that "Tschesnok created a thinking system which looks at people in a manner very similar to the way the brain process human vision." . Obviously it is not quite clear how this works. Hopefully their pending patent will be available soon.

Interdisciplinary
As briefly described in neuro-scientists found evidence that the nervous system use probabilistic population codecs.

title: Markerless Tracking/Related Work


text: Category:Technical Writing Level 1

title: Category:Technical Writing Level 1


text:
Technical writing/Tools/Making Wiki Tables

How to edit Wiki

Wikiversity Main Page

Why and When to use Tables
Tables are an excellent way to organize large amounts of information in an easy to read way.

Table Syntax
The following code will build a basic table.

Tables always begin with {| and end with |}

In the first line we set the widths of the borders and the space between cells.

{| border="1" cellpadding="1" cellspacing="1"

On the second line we enter the title of the table.

|+ 2006/07 Premier League Table

The following lines create the column headings, when creating the column headings we use a "!" instead of a pipe.

|-    !Position !Club Name !Points

Now we create the rows and the information that goes in them. Each line starts with a pipe and each row of information is separated by a |-.

|-  |1  ||  Manchester United  ||  89 |-  |2  ||  Chelsea  ||  83 |-  |3  ||  Liverpool  ||  68 |-  |4  ||  Arsenal  ||  68 |-  |5  ||  Tottenham  ||  60 |-  |6  ||  Bolton  ||  59

Notice that each cell is split by a double pipe ||. Usually there would be no space between the pipes, here the space is there to make reading the code easier.

This will create a basic table, which looks like this.

Tips and Tricks
Table width:

In the first line of the table it is possible to define the table's width, for example:

{| width=1000pt border="1" cellpadding="1" cellspacing="1"

Column width:

To define the column width just add a percentage of the overall width to the column heading lines, as follows:

! width=25% | Position ! width=50% | Club Name ! width=25% | Points

The table now looks like this:

Common Problems

 * Missing pipes
 * Too many spaces
 * Not ending the table with |}

Exercise 1 Adding a Table to a Wiki

 * 1) Create a page
 * 2) Add a table with 2 columns and five rows
 * 3) Fill in the table with the names of five people and their favorite drink
 * 4) Save your page
 * 5) Debug the table if needed


text:
User:AlistairReece/Technical writing

Back to Technical Writing

title: User:AlistairReece/Technical writing


text:
Technical writing/Alistair Reece

Back to Technical Writing

title: Technical writing/Alistair Reece


text:
User:Est~enwikiversity

My Interests are:Game design, micro and nanoelectronics.

title: User:Est~enwikiversity


text:
User:Joe.lawson

Joseph Lawosn

Age: 22

Born: 7th, December, 1984

Occupation: Entrepreneur

As a student of Georgia Gwinnett College I plan on pursuing and completing a degree in Pychology specializing in Teen Therapy.

title: User:Joe.lawson