User:EML4500.f08.delta 6.ramirez/110308

PVW (Continuous case) Dynamics of Elastic Bar

PDE:

$$ \frac{d}{dx}[ (EA)\frac{\partial u}{\partial x} ] + \mathit{f} = m\ddot{u} $$ (1)

Note: Equation (1) is known as the discrete equation of motion (EOM)

The first term in equation (1) as seen below,

$$ \frac{d}{dx}[ (EA)\frac{\partial u}{\partial x} ] $$

will give rise to

$$ -Kd $$

Likewise, the second term,

$$ \mathit{f} = m\ddot{u} $$

will give rise to

$$ +F = M\ddot{d} $$

where M represents the mass matrix.

Thus,

$$ -Kd + F = M\ddot{d} \Rightarrow M\ddot{d} + Kd = F $$ (2)

Note: Equation (2) is a multi-degree of freedom equation

Goal: Derive (2) from (1)

Review: Associate a single degree of freedom system to a multi-degree of freedom system.



Note: Equation from single degree of freedom system looks very similar to the multi-degree of freedom system

Now deriving (2) from (1) results in equation (3),

$$ \int_{x = 0}^{x = L}{w(x)\left\{\frac{d}{dx} [ (EA) \frac{\partial u}{\partial x}] + f - m\ddot{u}\right\}dx = 0} $$

for all possible w(x), where w(x) represents the weighting function.

Note: (1) → (2) is trivial, but (3) → (1) is not trivial

Further, equation (3) can be rewritten as,

$$ \int w(x) g(x) dx = 0 $$

for all possible w(x)

Since equation (3) holds for all w(x), select w(x) = g(x). Thus, equation (3) becomes,

$$ \int g(x)^2 dx = 0 $$

Since g(x)2 > 0, therefore, g(x) = 0.