User:Eml4500.f08.Ateam.paquette/Homework7

Lecture Notes 36, November 19, 2008
The element force displacement in local coordinates is represented by:

$$\mathbf{\tilde{k}}^{(e)}_{6x6}\mathbf{\tilde{d}}^{(e)}_{6x1} = \mathbf{\tilde{f}}^{(e)}_{6x1}$$

Where the local element stiffness matrix is:

$$\mathbf{\tilde{k}} = \begin{bmatrix} \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{2EI}{L} \\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L}\end{bmatrix}$$

note that $$\tilde{f_3}^{(e)} = f_3^{(e)}$$ and $$\tilde{f_6}^{(e)} = f_6^{(e)}$$ because of the moment about the z axis.

Dimensional Analysis
When doing a dimensional analysis, the following is the notation that will be used to represent the dimension of:

$$\left[\tilde{d_i}\right]$$ where i = 1,2,4,5

Where $$\left[\tilde{d_1}\right] = L$$ and $$\left[\tilde{d_3}\right] = 1 = \left[\tilde{d_6}\right]$$

PICTURE

From the previous figure, R can be found to help calculate the arc length, AB : $$\widehat{AB} = R\theta$$

$$\theta$$ should be in radians so it can be used in the next equations.

$$[\theta] = \frac{[AB]}{[R]} = \frac{L}{L} = 1$$

Now stress and strain can be found which will help verify the values of the stiffness matrix.

$$[\sigma] = [E][\epsilon]$$

Lecture Notes 38, November 24, 2008
Motivation: The deform shape of the tress element and the interpolation of the transverse displacements v(s), where $$s = \tilde{x}$$.

Principle of Virtual Work for Beams
The PVW for beams can be defined as:

$$ \int_{0}^{L}{w(\tilde{x})}\left[-\frac{\partial ^2}{\partial x^2} \left((EI)\frac{\partial ^2v}{\partial x^2}\right) + f_t - m\ddot{v}\right]dx = 0$$ (1) for all possible w(x)

Begin the integration by parts for the first term:

$$\alpha = \int_{0}^{L}{w(\tilde{x})}\frac{\partial^2}{\partial x^2}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}dx$$

where $$r(x) = \left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}$$

and where $$r'(x) = \frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right)$$

This equation becomes: $$\alpha = \left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right]^L_0 - \int_{0}^{L}\frac{dw}{dx}\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}dx$$

where $$s'(x) = \frac{\partial w}{\partial x}$$

and $$\beta_1 = \left[w\frac{\partial}{\partial x}\left\{(EI)\frac{\partial^2v}{\partial x^2}\right\}\right]^L_0$$.

After integrating by parts again, $$\alpha$$ becomes:

$$ = \beta_1 - \left[\frac{dw}{dx}\left(EI\right)\frac{\partial ^2v}{\partial x^2}\right]^L_0 + \int_{0}^{L}{\frac{\partial^2w}{\partial x^2}\left(EI\right)\frac{\partial ^2v}{\partial x^2}}dx$$

where $$\beta_2 = \left[\frac{dw}{dx}(EI)\frac{\partial ^2v}{\partial x^2}\right]^L_0$$

and the last integral can be replaced with the symbol $$\gamma$$. Note that this integral is symmetric. Equation (1) can now be rewritten and shortened to:

$$ - \beta_1 + \beta_2 - \gamma + \int_{0}^{L}{wf_tdx} - \int_{0}^{L}{wm\ddot{v}dx} = 0$$ for all possible w(x).

Stiffness term $$\gamma$$
The focus is now on the stiffness term $$\gamma$$ to derive the beam stiffness matrix and to identify the beam shape function.

PICTURE 38.1

For this beam, the equation for $$v(\tilde{x})$$ can be seen below.

$$ v(\tilde{x}) = N_2(\tilde{x})\tilde{d}_2 + N_3(\tilde{x})\tilde{d}_3 + N_5(\tilde{x})\tilde{d}_5 + N_6(\tilde{x})\tilde{d}_6$$

Recall that $$u(\tilde{x}) = N_1(\tilde{x})\tilde{d}_1 + N_4(\tilde{x})\tilde{d}_4$$

PICTURE 38.2

$$N_2(\tilde{x}) = 1 - \frac{3\tilde{x}^2}{L^2} + \frac{2\tilde{x}^3}{L^3}\,\,\,\,\,\,\,\tilde{d}_2 $$

$$N_3(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\,\,\,\,\,\,\,\tilde{d}_3 $$

$$N_5(\tilde{x}) = \frac{3\tilde{x}^2}{L^2} - \frac{2\tilde{x}^3}{L^3}\,\,\,\,\,\,\,\tilde{d}_5$$

$$N_6(\tilde{x}) = -\frac{\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}\,\,\,\,\,\,\,\tilde{d}_6$$