User:Eml4500.f08.team.guan/hw7

Derivation of the Stiffness Term and Beam Shaping Functions
The motivation of the following derivation is to describe and model the deformed shape of a truss element by using the interpolation of the transverse displacement along with the axial displacement. To begin the derivation, let's recall the second order differential equation that describes the dynamics of an elastic bar. This equation was displayed in the 29.1 lecture. $$\frac{\partial}{\partial{x}}\left[A(x)E(x)\frac{\partial{u}} {\partial{x}}\right]+f(x,t)=m(x)\ddot{u}$$ Here, the term A(x)E(x) is the stiffness matrix, the $$\frac{\partial{u}} {\partial{x}}$$ corresponds to the displacement matrix, and f is the external force exerted on the bar. Applying the above equation to beams and using the principle of virtual work, we get the following equation. $$\int_0^L{w(x)\left[-\frac{\partial^2}{\partial{x^2}}(EI \frac{\partial^2{V}}{\partial{x^2}})+f_t-m\ddot{u}\right]}dx=0$$ for all possible w(x) ....... (1) Next, let's integrate by parts the 1st term. We will call this term alpha. $$\alpha=\int_0^L{w(x)\left[\frac{\partial^2}{\partial{x^2}}(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]dx}$$ where $$\displaystyle s(x)=w(x)$$ and $$r(x)=\frac{\partial}{\partial{x}}\left[(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]$$ and $$r^'(x)=\frac{\partial^2}{\partial{x^2}}\left[(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]$$ Integrating by parts the first time we will get the following. $$=\left[w\frac{\partial}{\partial{x}}((EI) \frac{\partial^2{V}}{\partial{x^2}})\right]\bigg|^L_0-\int_0^L{\frac{dw}{dx}\left[\frac{\partial}{\partial{x}}(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]dx}=\beta_1-\int_0^L{\frac{dw}{dx}\left[\frac{\partial}{\partial{x}}(EI) \frac{\partial^2{V}}{\partial{x^2}}\right]dx}$$ Integrating again by integration by parts, we get the following. $$=\beta_1-\left[\frac{dw}{dx}(EI)\frac{\partial^2{V}}{\partial{x^2}}\right]\bigg|^L_0+ \int_0^L{\frac{d^2w}{dx^2}(EI) \frac{\partial^2{V}}{\partial{x^2}}dx}=\beta_1-\beta_2+ \gamma$$

Thus, in all simplified notation.  $$\displaystyle \alpha=\beta_1-\beta_2+ \gamma$$ Equation 1 above can be written as follows. $$-\beta_1+\beta_2-\gamma+\int_0^L{wf_tdx}-\int_0^L{wm\ddot{V}dx}=0$$ for all possible w(x) The boundary conditions will be evaluated later. Now, let's focus on the stiffness term gamma to derive the beam stiffness matrix and to identify the shape functions for beams. Consider the free body diagram of a beam below. The shaping functions at each node combined with the displacements of the nodes will allow for an equation of the transverse displacement to be written.

 The transverse displacement of the beam can be described by the following equation. $$V(\tilde{x})=N_2(\tilde{x})\tilde{d}_2+N_3(\tilde{x})\tilde{d}_3+N_5(\tilde{x})\tilde{d}_5+N_6(\tilde{x})\tilde{d}_6$$ where the shaping functions are the following. $$N_2(\tilde{x})=1-\frac{3\tilde{x^2}}{L^2}+\frac{2\tilde{x^3}}{L^3}$$ $$N_3(\tilde{x})=\tilde{x}-\frac{2\tilde{x^2}}{L}+\frac{\tilde{x^3}}{L^2}$$ $$N_5(\tilde{x})=\frac{3\tilde{x^2}}{L^2}-\frac{2\tilde{x^3}}{L^3}$$ $$N_6(\tilde{x})=-\frac{\tilde{x^2}}{L}+\frac{\tilde{x^3}}{L^2}$$ Graphically, the shaping functions are represented by the figure below.  Also, recall that the axial displacement described by its corresponding shaping functions and displacements is the following. $$\displaystyle u(\tilde{x})=N_1(\tilde{x})\tilde{d_1}+N_4(\tilde{x})\tilde{d_4}$$