User:EML4500.F08.RAMROD.B/Homework 7

=Motivation=

The motivation for this section is to find the deformed shape of the truss element through interpolation of the transverse displacement $$\displaystyle v(s) = v(\tilde{x})$$. We will first look at the Principal of Virtual Work for beams;

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

For the first term we must integrate by parts. This is done below.

$$\displaystyle \alpha := \int_{0}^{L}{w(\tilde{x}) \left[-\frac{\partial ^2}{\partial x^2}\left((EI)\frac{\partial^2v}{\partial x^2} \right)\right]}$$

$$\displaystyle \alpha := \frac{\partial }{\partial x}\left[ \frac{\partial }{\partial x}\left((EI)\frac{\partial^2v}{\partial x^2} \right)\right]$$

What lies inside the square brackets will be designated as $$\displaystyle r(x)$$ and the whole expression is designated as $$\displaystyle r'(x)$$. Performing the integration by parts yields;

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

The first term (non integral term) will be denoted by $$\displaystyle \beta_1$$, the first derivative in the interal will be represented by $$\displaystyle s'(x)$$, and as your can see $$\displaystyle r(x)$$ is imbedded in the integral. Performing another integration by parts yields the following;

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

The term after $$\displaystyle \beta_1 $$ will be denoted $$\displaystyle \beta_2$$ and the whole integral term will be denoted as $$\displaystyle \gamma$$. It is important to note that this new equation is symmetric. Inputting the new variables the new equation of the Principal of Virtual Work for beams is as follows;

$$\displaystyle -\beta_1 + \beta_2 - \gamma +\int_{0}^{L}{w(\tilde{x})f_t dx} - \int_{0}^{L}{Wm\ddot{u}dx} = 0$$ for all possible $$\displaystyle w(x)$$.

=Beam Stiffness Matrix=

Let us now focus on the stiffness term $$\displaystyle \gamma$$ so we can derive the beam stiffness matrix and identify the beam shape.



We need to plot the following eqaution;

$$\displaystyle v(\tilde{x}) = N_2(\tilde{x})\tilde{d_2} + N_3(\tilde{x})\tilde{d_3} + N_5(\tilde{x})\tilde{d_5} + N_6(\tilde{x})\tilde{d_6}$$

We recall that

$$\displaystyle u(\tilde{x}) = N_1(\tilde{x})\tilde{d_1} +N_4(\tilde{x})\tilde{d_4}$$





The equations for $$\displaystyle N_2(\tilde{x}),N_3(\tilde{x}),N_5(\tilde{x}),N_6(\tilde{x})$$ can now be found from these drawings and are shown below.

$$\displaystyle N_2(\tilde{x}) = 1 - \frac{3\tilde{x}^2}{L^2} + \frac{2\tilde{x}^3}{L^3}$$, $$\displaystyle \tilde{d_2}$$

$$\displaystyle N_3(\tilde{x}) = \tilde{x} - \frac{2\tilde{x}^2}{L} + \frac{\tilde{x}^2}{L^2}$$, $$\displaystyle \tilde{d_3}$$

$$\displaystyle N_5(\tilde{x}) = \frac{3\tilde{x}^2}{L^2} - \frac{2\tilde{x}^3}{L^3}$$, $$\displaystyle \tilde{d_5}$$

$$\displaystyle N_6(\tilde{x}) = -\frac{\tilde{x}^2}{L} + \frac{\tilde{x}^3}{L^2}$$, $$\displaystyle \tilde{d_6}$$