Nonlinear finite elements/Axial bar strong form

Axially loaded bar : Strong form
In this case, we consider an  infinitesimal slice of the bar and perform a balance of forces on that slice (see Figure 1).

A balance of forces gives

A \boldsymbol{\sigma} = \mathbf{q}(\mathbf{x})~\Delta x + A (\boldsymbol{\sigma} + \Delta\boldsymbol{\sigma}) \qquad \implies \qquad A \left(\cfrac{(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}) - \boldsymbol{\sigma}}{\Delta x}\right) + \mathbf{q}(\mathbf{x}) = 0 ~. $$

Taking the limit as $$\Delta x \rightarrow 0$$, we get

A \lim_{\Delta x \rightarrow 0} \left(\cfrac{(\boldsymbol{\sigma}+\Delta\boldsymbol{\sigma}) - \boldsymbol{\sigma}}{\Delta x}\right) + \mathbf{q}(\mathbf{x}) = 0 \qquad \implies \qquad {    A \cfrac{d\boldsymbol{\sigma}}{dx} + \mathbf{q}(\mathbf{x}) = 0 ~. } $$

This is a differential equation written in terms of stress. We want to express it in terms of displacements. How do we do this?

First, we convert the stresses into strains using the constitutive equation ($$\boldsymbol{\sigma} = E \boldsymbol{\varepsilon}$$).

{    AE \cfrac{d\boldsymbol{\varepsilon}}{dx} + \mathbf{q}(\mathbf{x}) = 0 ~. } $$

Then, we convert the strains into displacements using the  strain-displacement relations:

\boldsymbol{\varepsilon} := \lim_{l\rightarrow 0}\cfrac{\Delta l}{l} \qquad \implies \qquad \boldsymbol{\varepsilon} = \lim_{\Delta x \rightarrow 0} \cfrac{(\mathbf{u} + \Delta\mathbf{u}) - \mathbf{u}}{\Delta x}    \equiv {\boldsymbol{\varepsilon} = \cfrac{d\mathbf{u}}{dx}} ~. $$

The differential equation in terms of the displacement $$\mathbf{u}$$ is

{    AE \cfrac{d^2\mathbf{u}}{dx^2} + \mathbf{q}(\mathbf{x}) = 0 ~. } $$ Since $$\mathbf{q}(\mathbf{x}) = a \mathbf{x}$$, we have an  inhomogeneous ordinary differential equation for the displacements in the bar.

{    AE \cfrac{d^2\mathbf{u}}{dx^2} + a \mathbf{x} = 0 ~. } $$

If we can find the displacements in the bar, then we can solve for the strains and therefore the stresses. But to do that, we have to solve the  governing differential equation.

To get a unique solution, we need to provide  boundary conditions.

In this case, these are Since the ODE is in terms of $$\mathbf{u}$$, the BCs must also be in terms of $$\mathbf{u}$$. But we have a force at one end. This force has to be converted into an equivalent displacement. How?
 * 1) At $$\mathbf{x} = 0$$ (wall), $$\mathbf{u} = 0$$. This is an  essential BC.
 * 2) At $$\mathbf{x} = L$$ (end), $$\mathbf{f} = \mathbf{R}$$. This is a  natural BC.

\mathbf{f} = A \boldsymbol{\sigma} = A E \boldsymbol{\varepsilon} = A E \cfrac{d\mathbf{u}}{dx} ~. $$

Therefore, the BCs become

The problem can then be stated as
 * 1) At $$\mathbf{x} = 0$$ (wall), $$\mathbf{u} = 0$$.
 * 2) At $$\mathbf{x} = L$$ (end), $$A E \cfrac{d\mathbf{u}}{dx} = \mathbf{R}$$.

$$   \begin{align} \text{Find}~ & \mathbf{u}(\mathbf{x})~\text{such that it satisfies} \\ & AE \cfrac{d^2\mathbf{u}}{dx^2} = - a \mathbf{x} \\ & \mathbf{u}(0) = 0 \\ & \left.\cfrac{d\mathbf{u}}{dx}\right|_{\mathbf{x} = L} = \cfrac{\mathbf{R}}{AE} \end{align} $$

The differential formulation is also called the  strong form of the problem.

Analytical Solution
The analytical solution strategy is as follows:
 * 1) Set right hand side to zero and solve ( Homogeneous solution).
 * 2) Find  one solution with RHS = $$-a \mathbf{x}$$ ( Particular solution).
 * 3) General solution = homogeneous + particular.
 * 4) Apply BCs.

Homogeneous solution

 * The homogeneous ODE is

AE \cfrac{d^2\mathbf{u}}{dx^2} = 0~. $$
 * Integrate twice to get the homogeneous solution

A E \mathbf{u} = C_1 \mathbf{x} + C_2 \quad \implies \quad \mathbf{u}(\mathbf{x}) = \cfrac{C_1 \mathbf{x} + C_2}{A E} ~. $$

Particular solution

 * The ODE is

AE \cfrac{d^2\mathbf{u}}{dx^2} = -a \mathbf{x} ~. $$
 * Integrate twice to get the particular solution (and assume that the constants of integration are zero)

A E \mathbf{u} = -\cfrac{a \mathbf{x}^3}{6} \quad \implies \quad \mathbf{u}(\mathbf{x}) = -\cfrac{a \mathbf{x}^3}{6 A E} ~. $$

General solution

 * Add the homogeneous and particular parts to get

\mathbf{u}(\mathbf{x}) = \cfrac{-a \mathbf{x}^3 + D_1 \mathbf{x} + D_2}{6AE} ~. $$

Apply BCs

 * At $$\mathbf{x} = 0$$, $$\mathbf{u} = 0$$. Therefore, $$D_2 = 0$$.
 * At $$\mathbf{x} = L$$,

\cfrac{d\mathbf{u}}{dx} = \cfrac{-3aL^2 + D_1}{6AE} = \cfrac{R}{AE} \quad \implies \quad D_1 = 6R + 3aL^2 ~. $$

Solution

 * The displacement field in the bar is

{       \mathbf{u}(\mathbf{x}) = \cfrac{-a \mathbf{x}^3 + (6\mathbf{R} + 3aL^2)\mathbf{x}}{6AE} ~. }     $$
 * The strain in the bar is

{       \boldsymbol{\varepsilon}(\mathbf{x}) = \cfrac{-a \mathbf{x}^2 + (2\mathbf{R} + aL^2)}{2AE} ~. }     $$
 * The stress in the bar is

\boldsymbol{\sigma}(\mathbf{x}) = \cfrac{-a \mathbf{x}^2 + (2\mathbf{R} + aL^2)}{2A} ~. $$ A plot of this solution is shown in Figure 2. All known quantities have been taken to be 1 for the plot.