0.6 Tensors, lagrange methods, and hausdorff measure

Presentation Notes Harry

The Stress Tensor:

Informally, tensors are geometric objects that describe linear relations between vectors, scalars, and other tensors (say, linear maps or dot products). A tensor can be represented by a multi-dimensional array of values and the number of indices needed to specify a component in this array is said to be the order of the tensor.

Multi-dimensional arrays:

1. Scalar [a] - this is a tensor of order 0 because we don't need any indices to specify "a" within a 1x1 array

2. Vector [ai ]i=1,...,n - This is a tensor of order 1 because we need 1 index to specify a component of the vector

3. mxn matrix - this is a tensor of order 2

4. Tensors of higher order can be thought of as multidimensional "boxes"

Notice that we can create a tensor of a new order through the multiplication of tensors. For example, the products of a nxmxp tensor with an px1 vector will yield an nxm matrix.

Definition: The stress force on an object is said to be the force per unit area

So, the stress tensor is an order 2 tensor (a matrix) which takes a unit length direction vector normal to the surface of an object as an input and gives the stress force vector (with respect to the given coordinate system) at this surface as an output.

Then, for our problem, which is in 3 dimensions, the stress tensor is a 3x3 matrix with entries Tij, where i,j=1,2,3.

Imagine a beam and say that we would like to calculate the stress vector at a point, p, in the beam due to a force at the endpoint. Then, given a coordinate system, consider an "epsilon-cube" around p. Given an applied force at the endpoint of the beam (which translates to the same force at p) Tij is the stress vector acting on the ith face in the -jth direction - note that opposite faces have the same index. For an explanation of these properties, see [1].

One can check that in the context of a Michell Truss, the stress tensor will be a rank-one matrix and that the system of forces is in equilibrium if the divergence of the stress tensor is the negative of the applied forces. In other words, admissible trusses satisfy

$\nabla \sigma =-F$

where F is the vector sum of the applied forces. See [5] for an explanation and derivation of this equation.

Definition: Strain is the amount of deformation an object experiences as a result of external forces when compared with its original size and shape.

The strain tensor is operates similarly to the stress tensor, but the entries Tij are replaced by partial derivates of the position of a point, p, with respect to movement in the coordinate directions as a result of external forces. In the context of a Michell Truss, the strain tensor will always be a rank-one matrix as well.

One can investigate the relationship between stress and strain via a stress/strain curve.

Euler-Lagrange Equations

How do we optimize functionals? Consider a functional

$I\left(u\right)={\int }_{a}^{b}f\left(x,u\left(x\right),{u}^{\text{'}}\left(x\right)\right)\phantom{\rule{0.166667em}{0ex}}dx.$

over a region containing a and b.

Let u(x) be a minimizer for I(u). Then u(x) should satisfy the Euler-Lagrange equations as follows:

$\frac{d}{dx}\left[\frac{df\left(x\right)}{d{\mathbf{u}}^{\text{'}}\left(\mathbf{x}\right)}\left(x,\mathbf{u}\left(\mathbf{x}\right),{\mathbf{u}}^{\text{'}}\left(\mathbf{x}\right)\right)\right]\phantom{\rule{4pt}{0ex}}=\frac{df\left(x\right)}{d\mathbf{u}\left(\mathbf{x}\right)}\left(x,\mathbf{u}\left(\mathbf{x}\right),{\mathbf{u}}^{\text{'}}\left(\mathbf{x}\right)\right)$

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