User:Eml4500.f08.bike.bernal

Fri Dec 5

Now, a dimensional analysis will be done to the axial as well as the transverse displacements. Axial Displacements

$$ \left[u \right] =L $$ $$ \left[N_1 \right] =[N_4]=1 $$ $$ [\tilde{d}^{(e)}_1]= [\tilde{d}^{(e)}_4]= 1$$ Therefore, $$ [N_1][\tilde{d}^{(e)}_1]= [N_4][\tilde{d}^{(e)}_4]= L $$

Transverse Displacements

$$ \left[v \right] =L $$ The shape functions for the transverse displacements are shown below. $$\left[N_2 \right]=\left[N_3 \right]=\left[N_5 \right]=\left[N_6 \right]=1$$ The displacements corresponding rotation are shown below, which are dimensionless.

$$ [\tilde{d}^{(e)}_3]= [\tilde{d}^{(e)}_6]= 1$$

The rest of the displacements are in the $$ \tilde{Y}$$ direction, and those are shown below.

$$ [\tilde{d}^{(e)}_2]= [\tilde{d}^{(e)}_5]= 1$$

Therefore,

$$ [N_2][\tilde{d}^{(e)}_2]= [N_4][\tilde{d}^{(e)}_5]= L $$

$$ [N_1][\tilde{d}^{(e)}_3]= [N_4][\tilde{d}^{(e)}_6]= 1 $$

Derivation of Beam Shape Functions
First, the governing PDE for beams is analyzed. This PDE is considered without a distributed transverse load and also without an inertia force.

$$ m \ddot{v} $$(static case) is shown below.

$$\frac{\partial^{2} }{\partial x^{2}} = \begin{Bmatrix} (EI)\frac{\partial^{2}v }{\partial x^{2}} \end{Bmatrix}=0$$

Also, let's consider a constant EI, and the final equation is shown below.

$$\frac{\partial^{4} }{\partial x^{4}}=0$$

Solving for this fourth degree derivative, the following equation is obtained.

$$ v(x)= C_0 + C_{1}x+ C_{3}x^{2}+C_{3}x^{3}$$

Now, to get $$ N_2(x) $$, and since it is in the axial direction, $$\tilde{x} = x$$.

This results in the following values using the boundary conditions.

$$ v(0)=1$$ $$ v(L)=0$$ $$ v'(0)=0$$ $$ v'(L)=0$$