User:Eml4500.f08.team.figuerrez/hw7/meeting40

Meeting 40 - Derivation & Application Of Shape Functions
Equation (2) from meeting (39), we will be using dimensional analysis:

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

From meeting (31): $$\left[ N_1 \right]=\left[ N_4 \right]=1$$

Therefore equation (2) from meeting (39) becomes:

$$\left[ N_1 \right]\left[ \tilde{d}_1 \right]+\left[ N_4 \right]\left[ \tilde{d}_4 \right]$$

Converting the given values equates to:

$${\boldsymbol{1 * L}} + {\boldsymbol{1 * L}}$$

$$\left[ V \right]=Length=L$$

$$\left[ N_2 \right]\left[ \tilde{d}_2 \right]=L$$

Converting the given values equates to:

$${\boldsymbol{1 * L}} = {\boldsymbol{L}}$$    (displacement, transverse)

$$\left[ N_3 \right]\left[ \tilde{d}_3 \right]=L$$

Converting the given values equates to:

$${\boldsymbol{L * L}} = {\boldsymbol{L}}$$    (rotational)

Now we solve the derivation of beam shape functions: N2, N3, N5, N6  (from meeting 38)

Recall: Governing PDE for beams

From meeting (37), equation (1), without ft (distributed trans. load) and without inertia force $$m\ddot{v}$$ (static case):

$$\frac{\partial^{2}}{\partial x^{2}}\left[(EA)\frac{\partial^{2}v}{\partial x^{2}}\right]=0$$

Further more, consider the constant EI:

$$\frac{\partial^{4}}{\partial x^{4}}V=0$$, where you integrate four times to get four constants: C0, C1, C2, C3

Solving for v equates to: $${\boldsymbol{v(x)=C_{0}+C_{1}x^{1}+C_{2}x^{2}+C_{3}x^{3}}}$$

To obtain N2(x)    ($$\tilde{x}=x$$ for simplicity), use the boundary conditions:

$${\boldsymbol{v(0)=1, v(L)=0}}$$

$${\boldsymbol{v'(0)=v'(L)=0}}$$

Use the above boundary conditions to solve for: C0, C1, C2, C3

$${\boldsymbol{v(0)=C_{0}}}$$

$${\boldsymbol{v(L)=1+C_{1}L+C_{2}L^{2}+C_{3}L^{3}}}$$    (1)

$${\boldsymbol{v'(L)=C_{1}+2C_{2}x+3C_{3}x^{2}}}$$

$${\boldsymbol{v'(L)=C_{1}=0}}$$

$${\boldsymbol{v'(L)=2C_{2}L+3C_{3}L^{2}=0}}$$    (2)

Solving for C3 from equation (2) yields:

$$C_{3}=-\frac{2}{3}\frac{C_{2}}{L}$$

Plugging C3 back into equation (1) yields:

$$0=1+C_{2}L^{2}+\left[-\frac{2}{3}\frac{C_{2}}{L}\right]L^{3}$$

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

$$C_{3}=-\frac{2}{3}\frac{1}{L}\frac{-3}{L^{2}}=\frac{2}{L^{3}}$$

To obtain N3(x), use the boundary conditions:

$${\boldsymbol{v(0)=v(L)=0}}$$

$${\boldsymbol{v'(0)=1, v'(L)=0}}$$

To obtain N5(x), use the boundary conditions:

$$\tilde{d}_{5}=displacement$$

$${\boldsymbol{v(0)=0, v(L)=L}}$$

$${\boldsymbol{v'(0)=0, v'(L)=0}}$$

To obtain N6(x), use the boundary conditions:

$$\tilde{d}_{6}=rotationals$$

$${\boldsymbol{v(0)=v(L)=0}}$$

$${\boldsymbol{v'(0)=0, v'(L)=1}}$$

Derivation of coefficient in $$\tilde{K}$$ (element stiffness matrix)

Coefficient with EA: D

Coefficient with EI yields:

$$K_{22}=\frac{12EI}{L^{3}}=\int_{0}^{L}\frac{\partial ^2N_{2}}{\partial x^{2}}(EI)\frac{\partial ^2N_{2}}{\partial x^{2}}dx$$