User:Eml4500.f08.delta 6.krueger/Notes Monday November 24th

Motivation: deformed shape of truss element and Interpolation of transverse displacement v(s), where s=$$tilde{x}$$. Recall the governing partial differential equation for the bar element is: $$ +\frac{d^{2}}{dx^{2}}[ (EA)\frac{\partial u}{\partial x} ] + \mathit{f_{a}(x,t)} = m(x)\ddot{u} $$ The positive sign in front of the equation is to emphasize the difference between the above equation and the governing equation for a distributed transversal load on a bar element. The first term, $$\frac{d}{dx^{2}}[ (EA)\frac{\partial u}{\partial x} ]$$, is a second order differential equation with respect to x. The second term ,$$\mathit{f_{a}(x,t)}$$, refers to an axial distributed load. Where the subscript a is to distinguish the axial load equation from the transversal load equation. The load may be changing with time. Therefore, the load may be a function of time as well as distance (x).

 Principle of Virtual Work for Beams: 

The derivation is the same as the derivation of the principle of virtual work for the dynamics of elastic bar (refer to homework 6, PVW of dynamics of elastic bar) with one difference. The difference is that integration by parts has to be performed twice instead of once. Eqn (1)

$$ \int_{0}^{L}{w(\tilde{x})\left\{\frac{d^{2}}{dx^{2}} [ (EI) \frac{\partial^{2} u}{\partial x^{2}}] + f_{t} - m\ddot{v}\right\}dx = 0} $$

For all possible w(x), where $$f_{t}$$ is the transversal distributed load. First the first term will be integrated by parts. For simplicity the minus sign will be dropped and the first term will be defined as $$\alpha$$:

$$\alpha:= \int_{0}^{L}{w(\tilde{x})\frac{d^{2}}{dx^{2}} [ (EI) \frac{\partial^{2} u}{\partial x^{2}}]dx} $$ Where $$\tilde{x}=s(\tilde{x})$$. The position (r) is: $$r = \frac{d}{dx} [ (EI) \frac{\partial^{2} v}{\partial x^{2}}] $$ and the velocity ($$ \dot{r}$$) is: $$\dot{r} = \frac{d}{dx}(\frac{d}{dx} [ (EI) \frac{\partial^{2} v}{\partial x^{2}}]) $$

Now the argument x is omitted for simplicity and the equation is integrated.

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

Now again for simplicity $$\beta_{1}$$ is defined as:

$$\beta_{1} = w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}$$

Now Integration by parts once more.

$$=\beta_{1} - \left[w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}\right]_{0}^{L} + \int_{0}^{L}\frac{d^{2}w}{dx^{2}}\frac{\partial }{\partial x}\left\{(EI)\frac{\partial^{2} v}{\partial x^{2}}\right\}dx$$

Now $$\beta_{2}$$ and $$\gamma$$ are defined as:

$$\beta_{2} = \left[w\frac{\partial }{\partial x}\left\{(EI) \frac{\partial^{2} v}{\partial x^{2}}\right\}\right]_{0}^{L}$$ and  $$\gamma = \int_{0}^{L}\frac{d^{2}w}{dx^{2}}\frac{\partial }{\partial x}\left\{(EI)\frac{\partial^{2} v}{\partial x^{2}}\right\}dx$$

Eqn (1) now becomes:

$$-\beta_{1}+\beta_{2}-\gamma+\int_{0}^{L}wf_{t}dx-\int_{0}^{L}wm\ddot{v}dx=0$$

For all possible $$w(\tilde{x})$$

Now look at the stiffness term $$\gamma$$ to derive the beam stiffness matrix and to identify the beam shape functions.

The axial displacements $$\tilde{d_{1}}$$ and $$\tilde{d_{4}}$$ are ignored because the focus is on the beam.

The transverse displacement ($$v(\tilde{x})$$) is: $$v(\tilde{x})=N_{2}\tilde{d_{2}} + N_{3}\tilde{d_{3}} + N_{5}\tilde{d_{5}} + N_{6}\tilde{d_{6}}$$. Recall: $$u(\tilde{x})= N_{1}(\tilde{x})\tilde{d_{1}} + N_{4}(tilde{x})\tilde{d_{4}}$$





The rotation of $$N_{2}(\tilde{x})$$ is equal to zero. The slope is therefore equal to zero. This is true at the ends of the beam. The expressions for the shape functions are:

$$N_{2}(\tilde{x})=1-\frac{3\tilde{x^{2}}}{L^{2}}+\frac{2\tilde{x^{3}}}{L^{3}}$$

$$N_{3}(\tilde{x})=\tilde{x}-\frac{2\tilde{x^{2}}}{L}+\frac{\tilde{x^{3}}}{L^{2}}$$

$$N_{5}(\tilde{x})=\frac{3\tilde{x^{2}}}{L^{2}}-\frac{2\tilde{x^{3}}}{L^{3}}$$

$$N_{6}(\tilde{x})=-\frac{\tilde{x^{2}}}{L}+\frac{\tilde{x^{3}}}{L^{2}}$$

Observe that when the shape function is multiplied by the displacement (d) it must give a displacement.