User:Eml4500.f08.team.dwyer.jd/HW5 Oct 31 (Meeting 28)

Introduction to Finite Element Analysis using Differential Equations
Looking back at meeting 27, we take: $$\sum{F_x}=0= -N(x,t) +N(x +dx,t) + f(x,t)dx - m(x) \ddot{u}dx$$

$$= \frac{\partial N}{\partial x}(x,t)dx + h.o.t. + f(x,t)dx + m(x)\ddot{u}dx\;\;(1)$$

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

Expanding

$$-N(x,t)+N(x+dx,t)$$ to $$-N(x,t)+N(x,t)+N(dx,t)=N(dx,t)$$

Equation 1

$$0=\frac{\partial N}{\partial x}(x,t)dx+h.o.t.+f(x,t)dx-m(x)\ddot{u}dx$$

Where H.O.T. stands for higher order terms which can be neglected by recalling the Taylor series expansion:

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

Therefore Equation 1 can be re-written as Equation 2- The Equation of Motion for an Elastic Bar

$$\frac{\partial N}{\partial x}+f=m\ddot{u}$$

$$N(x,t)=A(x)\sigma(x)\;\;(3)$$ is the constructive relation where,

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

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

PDE of motion

(3) in (2) yields: $$\frac{\partial}{\partial x}[A(x)E(x)\frac{\partial u}{\partial x}] + f(x,t) = m(x) \ddot{u}\;\;(4)$$ where, $$\ddot{u} = \frac{{\partial^2}u}{\partial t^2}$$

Single Degree of Freedom Example


A single degree of freedom (SDOF) situation is shown above. The equation of motion is: $$m\ddot{d}+kd=F$$


 * 2 initial conditions are needed since it is a 2nd order derivative.

Boundary Conditions

 * Need 2 b.c.'s ( 2nd order derivation with respect to x)
 * Need 2 initial conditions (2nd derivative with respect to t) - initial displacement, initial velocity

2 initial conditions(initial displacement, initial velocity)

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

(2nd derivative with respect to t)

1) $$u(o,t) = 0$$

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

$$N(L,t)=A(L)\sigma(L,t)$$ where $$(\sigma(L,t) = E(L)\epsilon(L,t)$$ where $$\epsilon(x,t)=\frac{\partial U}{\partial x}(x,t)$$