User:Eml4500.f08.ateam.boggs.t/Lecture 29

The Principle of Virtual Work: A Continuation of the Dynamics of an Elastic Bar
For the initial conditions, at t = 0, prescribe,


 * $$ U(x,t=0) = \overline{U}(x) $$       The known functional displacement
 * $$ \dot{U}(x,t=0) = \overline{V}(x) $$       The known functional velocity

To create an equation for a multiple degree of freedom system, start with


 * $$ \frac{d}{dz}[EA \frac{du}{dx}] + f = m \ddot{u} $$    Equation (1)

This becomes,


 * $$ -kd + f = m \ddot{d} $$

or


 * $$ m \ddot{d} + kd = f $$    Equation (2)

To derive Equation 2 from 1 move all the terms to one side and take the integral with respect to x, producing,


 * $$ \int_{x=0}^{x=L} w(x)(\frac{d}{dx}[EA \frac{du}{dx}] + f - m \ddot{u}) \,dx = 0 $$    Equation (3)

This is true for all w(x), the weighting function. It should be noted that producing Equation 3 from Equation 1 is a trivial solution, but producing Equation 1 from Equation 3 is not. Equation 3 can be re-written as,


 * $$ \int_0^L w(x)g(x) \,dx = 0 $$

Here g(x) is substituted into Equation 3 for


 * $$ \frac{d}{dx}[EA \frac{du}{dx}] + f - m \ddot{u} $$

Because 3 holds true for all w(x), select w(x) that equals g(x). Doing so allows Equation 3 to become:


 * $$ \int_0^L g^2(x) \,dx = 0 $$

for $$ g(x) \ge 0 $$