User:Eml4500.f08.team.guan/hw6117

The Continuous FEM and PVW continued
The following table compares the different terms of the FEM equations of the continuous and discrete methods: inertia, stiffness, and the applied force terms.
 * {| border=1 class="wikitable" style="margin: 1em auto 1em auto"


 * style="text-align: center;" |Continuous setting (PVW)
 * style="text-align: center;" |Discrete setting (PVW)


 * style="text-align: center;" |inertia: $$\displaystyle \alpha=\int_0^Lwm \ddot {u}dx$$
 * style="text-align: center;" |inertia: $$\displaystyle \bar {w} \cdot (\bar{M}\bar{d})$$
 * style="text-align: center;" |inertia: $$\displaystyle \bar {w} \cdot (\bar{M}\bar{d})$$


 * style="text-align: center;" |stiffness term: $$\displaystyle \beta=\int_0^L\frac{dw}{dx}(EA)\frac{\partial{u}}{\partial{x}}dx$$
 * style="text-align: center;" |stiffness term: $$\displaystyle \bar {w} \cdot (\bar{K}\bar{d})$$
 * style="text-align: center;" |stiffness term: $$\displaystyle \bar {w} \cdot (\bar{K}\bar{d})$$


 * style="text-align: center;" |applied force term: $$\displaystyle \gamma=w(L)F(t)+\int_0^L{wfdx}$$
 * style="text-align: center;" |applied force term: $$\displaystyle \bar {w} \cdot \bar{F}$$
 * style="text-align: center;" |applied force term: $$\displaystyle \bar {w} \cdot \bar{F}$$


 * style="text-align: center;" |for all $$\displaystyle w(x)$$ such that $$\displaystyle w(0)=0$$, the boundary condition.
 * style="text-align: center;" |for all $$\bar{w}$$, the boundary conditions eliminated.
 * style="text-align: center;" |for all $$\bar{w}$$, the boundary conditions eliminated.


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Now, the stiffness term of the continuous finite element method will be examined by discretizing the clamped bar at its axis. It will be discretized into "n,nodes" nodes and "n,nodes-1" elements. The nodes need not be equidistant from each other and consequently the elements will not need to be the same length. The displacements will be only in the direction of the axis and will be denoted by d. No transverse forces will be taken into consideration. The following figure will describe the system being analyzed.  u(x) is the value of the axial displacement and is linearly interpolated along the element i. Assume displacement u(x) for $$x_i \le x \le x_{i+1}$$ (i.e. x $$\in [x_i,x_{i+1}]$$). The symbol $$\in$$ means "belongs to" or "is an element of a set". The local nodes are labeled in order of increasing x. x tilda is the incremented x coordinate where x tilda equals x plus x_i. Motivation for the linear interpolation of u(x) of the 2 bar truss is as follows. Let's consider an element in the figure below. The dotted line is the undeformed truss and the solid line is the deformed truss. The deformed shape is a straight line. In other words, there was an implicit assumption of linear interpolation of displacement between 2 nodes.  This is true considering the case where there are only axial displacements (i.e. zero transverse displacements). (31.2)    The question is how to express u(x) in terms of $$d_i=u(x_i)$$ as a linear function in x (i.e. using linear interpolation). Let u(x) be written in the following linear form. $$\displaystyle u(x)=N_i(x)d_i + N_{i+1}d_{i+1}$$ where $$N_i$$ and $$N_{i+1}$$ are linear functions in the x domain.  Writing the equation for N_i+1, we have the following. $$N_{i+1}(x)=\frac{x-x_i}{x_{i+1}-x_i}$$ Checking this to be the correct equation, we plug in x_i in the equation to get zero and plug in x_i+1 to get one. This parallels the figure so the equation is correct. The equation for N_i is given in the collapsible box below.