User:Eml4500.f08.Ateam.philbin/HW5

Homework 5

Mtg. 28 10-31-08

Example of a Composite Material
Some composite materials are examples of how the modulus of elasticity can be a function of x. One example is reinforced concrete walls. The modulus of the concrete is significantly different than that of the reinforcing steel. In order to accurately model this material the modulus would change as one moved through each material.



source: Rebar

Another material that has a modulus that changes with x is an aluminum composite panel which consists of two aluminum sheets sandwiched around a non-aluminum material.

source: (Aluminium_composite_panel)

Partial Differential Method


A very small slice of the beam shall be considered of width dx. The force balance in the x direction gives the following result.

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

$$=\frac{\partial {N}}{\partial{x}}(x,t)dx + higher\ order\ terms + f(x,t)dx-m(x)\ddot{u}dx\quad (Eqn. 1)$$

These higher order terms come from the Taylor Series expansion, shown in the general form here.

$$f(x+dx)=f(x)+\frac{df(x)}{dx}dx+\frac{1}{2}\frac{d^2f(x)}{dx^2}dx^2...$$

Where all the $$dx^2...$$ and higher terms are neglected due to their small effect on the result. After those simplifications are made, Eqn. 1 can be written in the following form.

$$\frac{\partial{N}}{\partial{x}}+f(x,t)=m(x)\ddot{u}\quad (Eqn. 2)$$

In this form, it is easy to see that this is the equation of motion. In order to get to a more usable form, N(x,t) needs to be rewritten into its constituent components.

$$N(x,t)=A(x)\ \sigma(x,t)$$

$$\sigma(x,t)=E(x)\ \epsilon(x,t)$$

$$\epsilon(x,t)=\frac{\partial{u}}{\partial{x}}(x,t)$$

Using these relationships, N(x,t) can be written as follows.

$$N(x,t)=[A(x)\ E(x)\ \frac{\partial{u}}{\partial{x}}(x,t)]$$

Now Eqn. 1 can be rewritten in the final form.

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

This is the final form of the equation of motion that will be used to solve finite element problems with the partial derivative method.

Initial and Boundary Conditions
In order to solve these problems, certain aspects of the conditions must be known. For the following case, two boundary conditions are known.

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

In another example, two initial conditions are known.



$$u(0,t)=0\ and\ N(L,t)=A(L)\sigma(L,t)=F(t)$$

3-bar Space Truss Statics


The three bar space truss was found to be statically determinate. There are three unknowns and three equations to use, summing the forces in x, y, and z directions. By using these three equations, the following relationships are made.

$$\sum{F_{x}}= F_{1}l^{(1)} + F_{2}l^{(2)} = 0$$

$$\sum{F_{y}}= F_{1}m^{(1)} + F_{2}m^{(2)} = -P$$

$$\sum{F_{z}}= F_{1}n^{(1)} + F_{2}n^{(2)} + F_{3}n^{(3)} = 0$$

Plugging in P = 20,000 N, gives the following forces along each element.

$$F_{1} = 20375\ N$$

$$F_{2} = 13124\ N$$

$$F_{3} = -23149\ N$$

In order to solve for the axial stress and strain, the following equations will be used.

$$\sigma_{n} = \frac{F_{n}}{A_{n}}$$

$$\epsilon_{n} = \frac{\sigma_{n}}{E_{n}}$$

Plugging in the given values for the Young's Modulus and area for each element yields these values. (Sigma is expressed in MPa and epsilon is expressed in mm.)

$$\sigma_{1} = \frac{20375 N}{200\ mm^2}= 101.875\qquad \qquad \epsilon_{1} = \frac{\sigma_{1}}{200\ GPa}= 0.5094$$

$$\sigma_{2} = \frac{13214 N}{200\ mm^2}= 66.07\qquad \qquad \ \ \ \epsilon_{2} = \frac{\sigma_{2}}{200\ GPa}= 0.33035$$

$$\sigma_{3} = \frac{-23149 N}{600\ mm^2}= -38.58\qquad \qquad \epsilon_{3} = \frac{\sigma_{3}}{200\ GPa}= -0.1929$$

These values for the displacement of each element matches the results from the MATLAB confirming the method.