User:EML4500.f08.Ateam.Miller/homework7

Dimensional Analysis of Displacements
Axial Displacements

$$\left[u \right]=L$$

$$\left[N_{1} \right]= \left[N_{4} \right]=1$$

$$\left[N_{1} \right]\left[\tilde{d_{1}} \right]+\left[N_{4} \right]\left[\tilde{d_{4}} \right]$$

Transverse Displacements

$$\left[v \right]=L$$

$$\left[N_{2} \right]\left[\tilde{d_{2}} \right]=L$$

$$\left[N_{3} \right]\left[\tilde{d_{3}} \right]=L$$

Derivation of Beam Shape Functions
$$N_2,\ N_3,\ N_5,\ N_6$$

Refer to plots shown above to derive the beam shape functions.

The governing partial differential equation (PDE) for beams is as follows.

$$\frac{\delta ^{2}}{\delta x^{2}}\left\{\left(EI \right) \frac{\delta^{2}v }{\delta x^{2}}\right\}=0$$

Considering EI to be constant reveals the following.

$$\frac{\delta ^{4}}{\delta x^{4}}v=0$$

By integrating this four times we obtain four constants in a relation for V(x).

$$V\,(x) = c_0+c_1x^1+c_2x^2+c_3x^3$$

In order to obtain $$N_2(x)$$ where $$\tilde{x}\equiv x$$ the following boundary conditions must be considered to solve for the four constants determined above.

$$V\,(0)=1$$

$$V\,(L)=0$$

$$V\,'(0)=V\,'(L)=0$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=1=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=1$$

$$V'(0)=0=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=0$$

$$V(L)=0=1+0+c_2(L)^2+c_3(L)^3=1+c_2L^2+c_3L^3\Rightarrow (1)$$

$$V'(L)=0=0+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2\Rightarrow (2)$$

From equation (2),

$$c_3=-\frac{2}{3}\frac{c_2}{L}$$

Now, inserting c3 from above into equation (1) yields

$$0=1+c_2L^2+\left( -\frac{2}{3}\frac{c_2}{L}\right)L^3\Rightarrow c_2=-\frac{3}{L^2}$$

and $$c_3=-\frac{2}{3}\frac{1}{L}\left({-\frac{3}{L^2}} \right)=\frac{2}{L^3}$$

Now, inserting c0, c1, c2, and c3 into V(x),

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

at the N2 boundary conditions. This expression is the same as the N2(x) found previously.

Likewise for $$N_3$$ the following boundary conditions are considered.

$$V\,(0)=0$$

$$V\,(L)=0$$

$$V\,'(0)=1$$

$$V\,'(L)=0$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=0=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=0$$

$$V'(0)=1=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=1$$

$$V(L)=0=1+0+c_2(L)^2+c_3(L)^3=1+c_2L^2+c_3L^3\Rightarrow (1)$$

$$V'(L)=0=1+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2\Rightarrow (2)$$

Using MATLAB to solve the system of 2 equations and 2 unknowns for equations (1) and (2),

$$c_2=\frac{-2}{L}$$

$$c_3=\frac{1}{L^2}$$

Now, inserting c0, c1, c2, and c3 into V(x),

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

at the N3 boundary conditions. This expression is the same as the N3(x) found previously.

Likewise for $$N_5$$ the following boundary conditions are considered.

$$V\,(0)=0$$

$$V\,(L)=1$$

$$V\,'(0)=0$$

$$V\,'(L)=0$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=0=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=0$$

$$V'(0)=0=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=0$$

$$V(L)=1=0+0+c_2(L)^2+c_3(L)^3=c_2L^2+c_3L^3=1\Rightarrow (1)$$

$$V'(L)=0=0+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2=0\Rightarrow (2)$$

Using MATLAB to solve the system of 2 equations and 2 unknowns for equations (1) and (2),

$$c_2=\frac{3}{L^2}$$

$$c_3=\frac{-2}{L^3}$$

Now, inserting c0, c1, c2, and c3 into V(x),

$$V(x)=\frac{3}{L^2}x^2-\frac{2}{L^3}x^3$$

at the N5 boundary conditions. This expression is the same as the N5(x) found previously.

Likewise for $$N_6$$ the following boundary conditions are considered.

$$V\,(0)=0$$

$$V\,(L)=0$$

$$V\,'(0)=0$$

$$V\,'(L)=1$$

Evaluating V(x) at the boundary conditions yields

$$V(0)=0=c_0+c_1(0)+c_2(0)^2+c_3(0)^3=c_0\Rightarrow c_0=0$$

$$V'(0)=0=c_1+2c_2(0)+3c_3(0)^2=c_1\Rightarrow c_1=0$$

$$V(L)=0=0+0+c_2(L)^2+c_3(L)^3=c_2L^2+c_3L^3=0\Rightarrow (1)$$

$$V'(L)=0=0+2c_2(L)+3c_3(L)^2=2c_2L+3c_3L^2=0\Rightarrow (2)$$

Using MATLAB to solve the system of 2 equations and 2 unknowns for equations (1) and (2),

$$c_2=\frac{-1}{L}$$

$$c_3=\frac{1}{L^2}$$

Now, inserting c0, c1, c2, and c3 into V(x),

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

at the N6 boundary conditions. This expression is the same as the N6(x) found previously.

Element Stiffness Matrix Coefficient Derivation
The coefficients containing EA have already been derived. This leaves the terms containing an EI component. These coefficients are derived here.

$$\tilde{k_{22}}=\frac{12EI}{L^{3}}=\int_{0}^{L}{\frac{d^{2}N_{2}}{dx^{2}}(EI)\frac{d^{2}N_{2}}{dx^{2}}dx}$$

$$\tilde{k}_{23}=\frac{6EI}{L^2}= \int_{0}^{L} \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}\, dx$$

Or to write this in a more generalized form:

$$\tilde{k}_{ij}= \int_{0}^{L} \frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}\, dx\quad i,j=2,3,5,6$$