User:Eml4500.f08.team.Bengtson/teamHW5

Meeting 28: Friday, 31st Oct 08


This is the full body diagram of a composite material with force acting on it(shows the different reactions a displacement)

The equation of force in the x direction for the above diagram is

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

$$ =\frac{\partial N}{\partial x}(x,t)dx + HOT+f(x,t)dx-M\ddot{u}dx $$
 * HOT = Higher order term

The HOT is neglected.

The dx can be factored out of all the terms leaving

$$ =\frac{\partial N}{\partial x}(x,t) + HOT+f(x,t)-M\ddot{u} $$

Review of Taylor Series

This is to show why the HOT are $$ f(x+dx)=f(x) + \frac{\partial F(x)}{\partial x}dx + \frac{1}{2}\frac{\partial^2 f(x)}{\partial x^2}dx^2 +..... $$

After the first two terms, the terms are considered HOT and as such in this class be neglected.

eq(1)$$\Rightarrow \frac{\partial N}{\partial x}+f=m\ddot{u}$$ (2)

The above is the equation of motion.

$$ N(x,t) = A(x) \sigma (x,t) $$(3)

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

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

We substitute (3) into (2)

This yields a new equation representative of the equation of motion $$ \frac{\partial }{\partial x}[A(x) E(x) \frac{\partial u}{\partial x}]+ F(x,t) = m(x)\ddot{u} $$

Please Note:

$$ \ddot{u}=\frac{\partial^2 u}{\partial x^2}$$

It is found that in order to solve this equation we need two boundary conditions and two initial conditions.

These can be two initial conditions, such as the initial displacement or initial velocity.

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

This is the initial conditions of a bar that is fixed at both ends.

Shown as:



2nd condition

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

$$ N(L,t)=A(L) \sigma (L,t) (3) $$

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

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

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

The velocity of the bar at length L at time t of the bar below.