User:EML4500.F08.RAMROD.B/Homework 5

The following is a FBD of an infinitesimal section area of the elastic bar with varying properties.



The bottom force that is not labeled in the picture can be represented as the following equation;

$$\displaystyle A(x) \partial x \rho(x) \frac{\partial^2 u}{\partial t^2}$$

Within this equation $$ A(x) \partial x \rho(x)$$ is equal to $$\displaystyle m(x)$$ which has the units of $$ \frac{mass}{length}$$. Now we proceed to sum the forces in the x-direction and the results are as follows;

$$\displaystyle \sum{F_x}= -N(x,t) + N(x+dx,t) + f(x,t)dx - m(x)\ddot{u} = 0$$

where $$\ddot{u}$$ is equal to the acceleration as given in Newton's notation. Now we perform a Taylor Series expansion on this equation but we will be ignoring the higher order terms. This is performed below.

$$\displaystyle \sum{F_x}= \frac{\partial N}{\partial x}(x,t) + (Higher Order Terms ) + f(x,t) - m(x)\ddot{u}$$  $$\displaystyle f(x+dx) = f(x) + \frac{df}{dx}(x)dx + \frac{1}{2}\frac{d^2f}{dx^2} +.....$$  $$\displaystyle \frac{\partial N}{\partial x} + f = m\ddot{u}$$

The second equation is part of the expansion while the last equation is known as the equation of motion for the elastic bar which ignores the higher order terms of the expansion. The equation of motion is otherwise abbreviated as $$\displaystyle EOM$$. An expression for the force $$N(x,t)$$ can be obtained and the results are found below.

$$\displaystyle N(x,t) = A(x)\sigma(x) = A(x) E(x) \varepsilon (x,t) = A(x) E(x) \frac{\partial u}{\partial x}(x,t)$$

Putting this expression into the $$EOM$$ yields the partial differential equation of motion (PDE of Motion) for the elastic bar.

$$\displaystyle \frac{\partial }{\partial x}\left[ A(x) E(x) \frac{\partial u}{\partial x}\right] + f(x,t) = m(x) \ddot{u}$$

In order to solve this equation two boundary conditions along with two initial conditions are needed. The two initial conditions are initial displacement and initial velocity. For the boundary conditions, two cases need to be considered. They are depicted below.



For case a, the two boundary conditions are as follows;

$$\displaystyle u(0,t) = u(L,t) = 0$$

$$\displaystyle \frac{\partial u}{\partial x}(L,t) = \frac{F(t)}{A(L)E(L)}$$

For case two the boundary conditions are as follows;

$$\displaystyle u(0,t) = o$$

$$\displaystyle N(L,t) = F(t)$$

The initial conditions are specific to different problems