User:Eml4500.f08.Ateam.paquette/Homework6

Lecture Notes 29, November 3rd, 2008
The initial conditions for the system at t = 0 can be defined below.

$$ u\left(x,t=0\right) = \overline{u}(x) $$ for the given function of x, a known displacement function,

and

$$\frac{\partial u}{\partial t}(x,t=0)={\ddot{u}}(x,t=0)=\overline{v}(x) $$ for a known velocity function.

The principle of virtual work (continuous) of dynamics of the elastic bar can be shown as a partial differential equation:

$$\frac{\partial}{\partial x}\left[(EA)\frac {\partial u}{\partial x}\right]+ f = m{\ddot{u}} $$ (1)

The Discrete Equation of Motion can then be represented as

$$-\underline{K}\underline{d} + \underline{F} = \underline{M}\underline{\ddot{d}}$$

or $$\underline{M}\underline{\ddot{d}} + \underline{K}\underline{d} = \underline{F} $$ (2)

Where $$\mathbf\underline{M}$$ is a mass matrix.

$$\frac{\partial }{\partial x}\left[\left(EA \right)\frac{\partial }{\partial x} \right] = -Kd$$

$$f = F$$

$$m\ddot{u} = M\ddot{d} $$

Note that equation (2) is the equation of motion for a system with multiple degrees of freedom. A system with a single degree of freedom will be shown in equation 3.

For a single degree of freedom



$$\underline{m}\underline{\ddot{d}} + \underline{k}\underline{d} = \underline{F}$$

All of these equations have been derived with the help of the following integral:

$$\int_{x=0}^{x=L} W(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right] + f - m\ddot{u}}\right)dx = 0 $$ (3)

for all possible $$\mathbf {W(x)}$$, where $$\mathbf {W(x)}$$ is the weighting coefficient. As seen before, going from equation 1 to equation 3 is trivial but it is important to know how to get from equation 3 to equation 1. Equation 3 can be rewritten as:

$$\int_{x=0}^{x=L} w(x)g(x)dx = 0$$ for all $$\mathbf {w(x)}$$

Since equation 3 holds true for all values of $$\mathbf{w(x)}$$ by the principle of virtual work, we can choose $$ \mathbf {w(x) = g(x)}$$. This changes equation 3 to

$$\int_{x=0}^{x=L} g^2(x)dx = 0$$

Integrating by Parts
Integrating by parts is needed when trying to integrate two functions multiplied together. The formula for integrating by parts can be seen below. The symbols 'r' and 's' will represent the two different functions of x in this example.

$$ \left(rs\right)^'=r^'s+rs^'$$

Where: $$ r^' = \frac{dr}{dx} $$ and $$ s^'=\frac{ds}{dx} $$

Integrating both sides forms the equation:

$$\int \left(rs\right)^' = \int r^'s + \int rs^'$$

This leads to the final equation of:

$$ \int r^'s = rs - \int rs^'$$

Now to continue with the Principle of Virtual Work, recall equation 3:

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

For the first term, let $$r(x) = (EA){\frac{\partial u}{\partial x}}$$ and $${s\left(x\right) = W\left(x\right)}$$

After integrating by parts, the equation yields:

$$ \int_{x=0}^{x=L} W(x)\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]dx = \left[W(EA)\frac{\partial u}{\partial x}\right]_{x=0}^{x=L} - \int_{x=0}^{x=L}\frac{\partial W}{\partial x}(EA)\frac{\partial u}{\partial x}dx$$

Where the minus sign on the right side of the equation relates to the - Kd in the step between equations 1 and 2 above.

Replacing the x's with either L or 0, sets the right side equal to:

$$ = W(L)(EA)(L)\frac{\partial u}{\partial x}(L,t) - W(0)(EA)(0)\frac{\partial u}{\partial x}(0,t) - \int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx $$

Now consider the model equation with two boundary conditions:



At x = 0, select W(x) such that $$W\left(0\right) = 0$$ or so that it is kinematically admissible.

Motivation: Discrete PVW applied to equation below:

$$ \mathbf{{W}_{6x1}\left(\left[K \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1} - {F}_{6x1}\right)=0_{1x1}}$$

Where $$\left(\left[K \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1} - {F}_{6x1}\right)$$ becomes a 6x1 matrix and $$ \mathbf{F}^{T}=\left[F_{1}F_{2}F_{3}F_{4}F_{5}F_{6}\right] $$ where F1, F2, F5, and F6 are unknown reactions.

Note that W can be selected arbitrarily, but for simpliciity select W such that W1 = W2 = W5 = W6 = 0. This is so all equations involving unknown reactions can be eliminated, or rows 1,2,5 and 6 are eliminated. Equation 1 shown below can then be formed.

$$ \mathbf{\overline{K}_{2x2}\overline{d}_{2x1}=\overline{F}_{2x1}} $$

Where $$\mathbf\overline{d} = \begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}$$ and $$\mathbf\overline{F} = \begin{bmatrix}F_{3} \\ F_{4}\end{bmatrix}$$.

Note that $$ \mathbf{\overline{W}\left(\overline{K}\overline{d}-\overline{F}\right)}=0 $$ (2) is actually the step before equation 1.

$$ W(L)F(t) - \int_{0}^{L}\frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx+\int w(x)[f - m\ddot{u}]dx=0 $$

This becomes the final equation:

$$ \int_{0}^{L}w(m\ddot{u})dx+\int_{0}^{L} \frac{\partial w}{\partial x}(EA)\frac{\partial u}{\partial x}dx

= W(L)F(t) + \int_{0}^{L}wfdx $$

These equations are for all $$ W\left(x\right)$$ such that $$ W\left(0\right) = 0 $$

Two cases of PVW


For the rest of this section, we will be focusing on element i in the above figures.

Assume the displacement for u(x) for $$x_i \le x \le x_{i+1}$$. (i.e., x $$\epsilon [x_i, x_{i+1}]$$)

The motivation for the linear interpolation of u(x) is shown below using a 2-bar truss.



The deformed shape is a straight line. This means there was an implicit assumption of the linear interpolation of the displacement between two nodes. Consider the case where there is only axial displacement, or zero transverse displacements. The question that needs to be solved is:

Express u(x) in terms of di = u(xi) and di+1 = u(xi+1) as a linear function in x.

The answer to this question is defined by linear interpolation, or:

$$u(x) = N_i\left(x\right)d_i + N_{i+1}\left(x\right)d_{i+1}$$

where $$N_i\left(x\right)$$ and $$N_{i+1}\left(x\right)$$ are linear functions of x.

$$N_i\left(x\right) = $$

$$N_{i+1}\left(x\right) = \frac {x - x_{i}}{x_{i+1} - x_{i}}$$



Continuing PVW to Discrete PVW
Lagrangian interpolation The Motivation for the form of $$\mathbf{N_i(x)}$$ and $$\mathbf{N_{i+1}(x)}$$: 1) $$\mathbf{N_i(x)}$$ and $$\mathbf{N_{i+1}(x)}$$ are linear, straight lines, thus any linear combination of $$\mathbf{N_i(x)}$$ and $$\mathbf{N_{i+1}(x)}$$ are also linear, and in particular, the expression for u(x).