Preliminary¶

Reference

Mechanics Lecture Notes Part III: Foundations of Continuum Mechanics, pa.kelly@auckland.ac.nz.

Tensors¶

A tensor of order zero is simply another name for a scalar $α$ .

A first-order tensor is simply another name for a vector $u u$ .

We use uppercase bold-face Latin letters to denote second order tensor.

A second-order tensor $T T$ may be defined as an operator that acts on a vector $u u$ generating another vector $v v$ , such that

T T (u u) = v v

which is a linear operator.

dyad(tensor product)

the tensor product of two vectors $u u$ and $v v$

u u \otimes v v

is defined by

(u u \otimes v v) w w = u u (v v \cdot w w)

Properties

(i)

(u u \otimes v v) (w w \otimes x x) = (v v \cdot w w) (u u \otimes x x)

cause

\begin{aligned} (u u \otimes v v) (w w \otimes x x) y y & = (u u \otimes v v) (x x \cdot y y) w w \\ = (x x \cdot y y) (u u \otimes v v) w w \\ = (x x \cdot y y) (v v \cdot w w) u u \\ = (v v \cdot w w) (x x \cdot y y) u u \\ = (v v \cdot w w) (u u \otimes x x) y y \end{aligned}

(ii)

u u (v v \otimes w w) = (u u \cdot v v) w w

cause

\begin{aligned} (y y \otimes u u) (v v \otimes w w) & = (u u \cdot v v) (y y \otimes w w) \\ = y y \otimes [(u u \cdot v v) w w] \end{aligned}

Some example

Projection Tensor $(e e \otimes e e)$

So

(e e \otimes e e) u u = (e e \cdot u u) e e

is the vector projection of $u u$ on $e e$ , denoted by $P P$ .

A dyadic is a linear combination of dyads (with scalar coefficients).

In the following discussion, we can treat $T T$ as a matrix.

Cartesian Tensors¶

A second order tensor and the the vector it operates on can be described in terms of Cartesian components.

Example

Identity tensor/(or unit tensor).

I I = \sum_{i = 1}^{d} e e_{i} \otimes e e_{i}

cause it follows

\begin{aligned} I I u u & = \sum_{i = 1}^{d} (e e_{i} \otimes e e_{i}) u u \\ = \sum_{i = 1}^{d} (e e_{i} \cdot u u) e e_{i} \\ = \sum_{i = 1}^{d} u_{i} e e_{i} \\ = u u \end{aligned}

Or identity tensor can be written as

I I = \sum_{i, j = 1}^{d} δ_{i j} (e e_{i} \otimes e e_{j})

Second order tensor as a Dyadic

Every second order tensor can always be written as a dyadic involving the Cartesian base vectors $e e_{i}$ , that is, if we denote

E E_{i} = T T (e e_{i})

then

T T = \sum_{i = 1}^{d} (E E_{i} \otimes e e_{i})

Proof

\begin{aligned} b b & = T T (a a) \\ = T T (\sum_{i = 1}^{d} a_{i} e e_{i}) \\ = \sum_{i = 1}^{d} a_{i} T T (e e_{i}) \end{aligned}

Denote $T T (e e_{i})$ to be $E E_{i}$ , then

\begin{aligned} b b & = \sum_{i = 1}^{d} a_{i} E E_{i} \\ = \sum_{i = 1}^{d} (a a \cdot e e_{i}) E E_{i} \\ = \sum_{i = 1}^{d} (E E_{i} \otimes e e_{i}) a a \end{aligned}

So

T T = \sum_{i = 1}^{d} (E E_{i} \otimes e e_{i})

If we write $E E_{i}$ with base vectors like

E E_{i} = \sum_{j = 1}^{d} E_{i j} e e_{j}, i = 1, \dots, d

Then

\begin{aligned} (E E_{i} \otimes e e_{i}) & = (\sum_{j = 1}^{d} E_{i j} e e_{j}) \otimes e e_{i} \\ = \sum_{j = 1}^{d} E_{i j} (e e_{j} \otimes e e_{i}), i = 1, \dots, d \end{aligned}

Thus

T T = \sum_{i = 1}^{d} \sum_{j = 1}^{d} E_{i j} (e e_{j} \otimes e e_{i})

Introduce 9 scalars $T_{i j} = E_{j i}$ , then

\begin{aligned} T T & = \sum_{i = 1}^{d} \sum_{j = 1}^{d} T_{j i} (e e_{j} \otimes e e_{i}) \\ = \sum_{j = 1}^{d} \sum_{i = 1}^{d} T_{j i} (e e_{j} \otimes e e_{i}) switch summation turn \\ = \sum_{i = 1}^{d} \sum_{j = 1}^{d} T_{i j} (e e_{i} \otimes e e_{j}) switch i and j \end{aligned}

We can see that 9 dyads ${e e_{i} \otimes e e_{j}}_{i, j = 1}^{3}$ forms a basis for the space of second order tensors.

Recall that

\begin{aligned} T_{i j} & = E_{j i} \\ = E E_{j} \cdot e e_{i} \\ = T (e e_{j}) \cdot (e e_{i}) \\ = (e e_{i}) \cdot T (e e_{j}) \end{aligned}

So we can get the component of a tensor by the above way.

Cauchy Stress Tensor¶

The traction vector, the limiting value of the ratio of force over area, that is,

t t^{n n} = lim_{Δ_{s} \to 0} \frac{Δ F}{Δ S}

where $n n$ denotes normal vector to the surface.

The stress $σ σ$ , a second order tensor which maps $n n$ onto $t t$

t t = σ σ n n

If we consider a coordinate system with base vectors $e e_{i}$ , then $σ σ = \sum_{i, j = 1}^{d} (σ_{i j} e e_{i} \otimes e e_{j})$

σ σ = (σ_{i j})

So

t_{i} e_{i} e_{i} = \sum_{j = 1}^{3} σ_{i j} n_{j} e e_{i}

For example,

σ σ e e_{j} = \sum_{i = 1}^{3} σ_{i j} e e_{i}

which denotes the summation of the $j$ th column of matrix $σ σ$ .

So the components $σ_{11}, σ_{21}, σ_{31}$ of the stress tensor are the three components of the traction vector which acts on the plane with normal $e e_{1}$ .

Hamilton Operator¶

First we want to introduce the operator

\nabla = i i \frac{\partial}{\partial x} + j j \frac{\partial}{\partial y} + k k \frac{\partial}{\partial z}

then

\nabla f = i i \frac{\partial f}{\partial x} + j j \frac{\partial f}{\partial y} + k k \frac{\partial f}{\partial z} = grad f

we also have inner product

\begin{aligned} \nabla \cdot a a & = (i i \frac{\partial}{\partial x} + j j \frac{\partial}{\partial y} + k k \frac{\partial}{\partial z}) \cdot (P i i + Q j j + R k k) \\ = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \\ = div a a \end{aligned}

and cross product

\begin{aligned} \nabla \times a a & = (i i \frac{\partial}{\partial x} + j j \frac{\partial}{\partial y} + k k \frac{\partial}{\partial z}) \times (P i i + Q j j + R k k) \\ = | \begin{array}{ccc} i i & j j & k k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{array} | \\ = (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) i i + (\frac{\partial R}{\partial x} - \frac{\partial P}{\partial z}) j j + (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) k k \\ = rot a a \end{aligned}

Then Gauss Formula can be expressed by

\iint_{\partial Ω} a a d S S = ∭_{Ω} \nabla \cdot a a d V

Stokes Formula can be expressed by

\int_{\partial Σ} a a d s s = \iint_{Σ} (\nabla \times a a) \cdot d S S