User:Eml4500.f08.gravy.jad/Notes7

$$\ \,$$ $$\  \,$$ $$\  \,$$

Lecture 37

$$\ \left \lbrack \epsilon \right \rbrack = \frac{\left \lbrack du \right \rbrack}{\left \lbrack dx \right \rbrack} = \frac{L}{L}= 1 \,$$

$$\ \left \lbrack \sigma \right \rbrack = \left \lbrack E \right \rbrack = \frac{F}{L^{2}} \,$$

$$\ \left \lbrack A \right \rbrack = L^{2}, \left \lbrack I \right \rbrack = L^{4} \,$$

$$\ \left \lbrack \frac{EA}{L} \right \rbrack = \left \lbrack \tilde{k}_{11} \right \rbrack = \frac {\left ( F/L^2 \right ) \left ( L^2 \right )}{L} = \frac{F}{L} \,$$

$$\ \left \lbrack \tilde{k}_{11} \tilde{d}_{1} \right \rbrack = \left \lbrack \tilde{k}_{11} \right \rbrack \left \lbrack \tilde{d}_{1} \right \rbrack = F \,$$

$$\ \left \lbrack \tilde{k}_{23} \tilde{d}_{3} \right \rbrack = \left \lbrack \tilde{k}_{23} \right \rbrack \left \lbrack \tilde{d}_{2} \right \rbrack = \frac {\left \lbrack 6 \right \rbrack \left \lbrack E \right \rbrack \left \lbrack I \right \rbrack}{\left \lbrack L^2 \right \rbrack} = \frac{1.(F/L^2).L^4}{L^2} = F \,$$

Insert HW verify dimentions for kij and dj

Elem FD rel. in golbal coordinate from element FD rel. in local coordinate:

ie. obtain

$$\ \mathbf{k}^e_{6x6} \mathbf{d}^e_{6x1} = \mathbf{f}^e_{6x1} \,$$ $$\ \mathbf{k}^e_{6x6} = \mathbf{\tilde{T}}^{e^{T}}_{6x6} \mathbf{\tilde{k}}^e_{6x6} \mathbf{\tilde{T}}^{e}_{6x6} \,$$ from $$\ \tilde{\mathbf{k}}^e_{6x6} \tilde{\mathbf{d}}^e_{6x1} = \tilde{\mathbf{f}}^e_{6x1} \,$$

Insert HW7

$$\ \begin{Bmatrix} \tilde{d}_{1} \\ \tilde{d}_{2} \\ \underline{\tilde{d}}_{3} \\ \tilde{d}_{4} \\ \tilde{d}_{5} \\ \underline{\tilde{d}}_{6} \end{Bmatrix}_{6x1} = \underbrace {\begin{bmatrix} \mathbf{\underline{R}}& &0&0&0\\ & & 0&0&0&0\\ 0&0&\mathbf{1}&0&0&0\\ 0&0&0& \mathbf{\underline{R}}& &0\\ 0&0&0& & &0\\ 0&0&0&0&0&\mathbf{1} \end{bmatrix}_{6x6}}_{\tilde{\mathbf{T}}} \begin{Bmatrix} d_1\\d_2\\ \underline{d}_3\\d_4\\d_5\\ \underline{d}_6 \end{Bmatrix}_{6x1} \,$$ where $$\ \underline{d}_3 \,$$ and $$\ \underline{d}_6 \,$$ are rotational components, not displacement components

insert HW7- solve 2elem frame

insert HW7- plot deformed shape on top of previous figure

Derivation of $$\ \mathbf{\tilde{k}}^{(e)} \,$$ from PVW, focusing only on bending effect;

$$\ - \frac{\partial ^2}{\partial x^2} \left ( \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right ) + f_{t} \left ( x \right ) = m \left ( x \right ) \ddot{V} \,$$

$$\ + \frac{\partial ^2}{\partial x^2} \left ( \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right ) + f_{a} \left ( x,t \right ) = m \left ( x \right ) \ddot{u} \,$$

PVW for beams:

$$\ \int_{0}^{L} W\left ( x \right ) \left \lbrack - \frac{\partial^2}{\partial x^2} \left ( \left ( EI \right ) \frac{\partial^2 v}{\partial x} \right ) + f_t - m \ddot v \right \rbrack dx = 0 \,$$ for all possible W(x) (Eq 1)

Integrate by Parts of 1 st term:

$$\ \alpha = \int_{0}^{L} \underbrace{W \left ( x \right )}_{s \left ( x \right )} \frac{\partial^2}{\partial x^2} \left \lbrace \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right \rbrace dx \,$$ where $$\ \underbrace{\frac{\partial}{\partial x} \left ( \frac{\partial}{\partial x} \left \lbrace \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right \rbrace \right )}_{r'\left ( x \right )} \,$$

$$\ = \left \lbrack W \frac{\partial}{\partial x} \left \lbrace \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right \rbrace \right \rbrack _{0}^{L} - \int _{0}^{L} \underbrace{\frac{dW}{dx}}_{s' \left ( x \right )} \underbrace{ \frac{\partial}{\partial x} \left \lbrace \left ( EI \right ) \frac{\partial ^2 v}{\partial x^2} \right \rbrace}_{r \left ( x \right )} dx \,$$

$$\ = \beta_1 - \underbrace{\left \lbrack \frac{dW}{dx} \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} \right \rbrack_{0}^{L}}_{\beta_2} + \underbrace{\int_{0}^{L} \frac{d^2 W}{dx^2} \left ( EI \right ) \frac{\partial^2 v}{\partial x^2} dx}_{\gamma} \,$$

Equation (1) becomes: $$\ \beta_1 - \beta_2 - \gamma + \int_{0}^{L} W f_t dx - \int_{0}^{L} W m \ddot{v} dx = 0 \,$$ for all possible W(x)

Now focusing on the stiffness term $$\ \gamma \,$$ for the time being to derive the beam stiffness matrix and to identify the beam shape functions:

insert pic

$$\ v \left ( \tilde{x} \right ) = N_2 \left ( \tilde{x} \right ) \tilde{d}_2 + N_3 \left ( \tilde{x} \right ) \tilde{d}_3 + N_5 \left ( \tilde{x} \right ) \tilde{d}_5 + N_6 \left ( \tilde{x} \right ) \tilde{d}_6 \,$$ Equation 2

Recall: $$\ u \left ( \tilde{x} \right ) = N_1 \left ( \tilde{x} \right ) \tilde{d}_1 + N_4 \left ( \tilde{x} \right ) \tilde{d}_4 \,$$ Equation 4

insert pics

In accordance with the book, the following equations represent the above picure:

$$\ N_2 \left ( \tilde{x} \right ) = 1 - \frac{3 \tilde{x}^2}{L^2} + \frac{2 \tilde{x}^3}{L^3} \,$$ for $$\ \tilde{d}_2 \,$$

$$\ N_3 \left ( \tilde{x} \right ) = \tilde{x} - \frac{2 \tilde{x}^2}{L} + \frac{ \tilde{x}^3}{L^2} \,$$ for $$\ \tilde{d}_3 \,$$

$$\ N_5 \left ( \tilde{x} \right ) = \frac{3 \tilde{x}^2}{L^2} - \frac{2 \tilde{x}^3}{L^3} \,$$ for $$\ \tilde{d}_5 \,$$

$$\ N_6 \left ( \tilde{x} \right ) = - \frac{ \tilde{x}^2}{L} + \frac{ \tilde{x}^3}{L^2} \,$$ for $$\ \tilde{d}_6 \,$$

insert pictures

As determined previously in an above section:

$$\ \tilde{\mathbf{d}}^{(e)}_{6x1} = \tilde{\mathbf{T}}^{(e)}_{6x6} \mathbf{d}^{(e)}_{6x1} \,$$ where $$\ \mathbf{d}^{(e)}_{6x1} \,$$ is known after solving the Finite Element Systems.

The only thing left is to compute $$\ u \left ( \tilde{x} \right ) \,$$ and $$\ v \left ( \tilde{x} \right ) \,$$

insert picture

where $$\ \mathbf{u} \left ( \tilde{x} \right ) = u \left ( \tilde{x} \right ) \vec{\tilde{i}} + v \left ( \tilde{x} \right ) \vec{\tilde{j}} \,$$

Now we compute $$\ u \left ( \tilde{x} \right ) \,$$, $$\ v \left ( \tilde{x} \right ) \,$$ using Equations (1) and (2) as stated previously.

Computing $$\ u_x \left ( \tilde{x} \right ), u_y \left ( \tilde{x} \right ) \,$$ from $$\ u \left ( \tilde{x} \right ), v \left ( \tilde{x} \right ) \,$$

$$\ \begin{Bmatrix} u_x \left ( \tilde{x} \right ) \\ u_y \left ( \tilde{x} \right ) \end{Bmatrix} = \mathbf{R}^T \begin{Bmatrix} u \left ( \tilde{x} \right ) \\ v \left ( \tilde{x} \right ) \end{Bmatrix} \,$$

$$\ \begin{Bmatrix} u_x \left ( \tilde{x} \right ) \\ u_y \left ( \tilde{x} \right ) \end{Bmatrix} = \underbrace_{\mathbb{N} \left ( \tilde{x} \right )} \underbrace{\begin{Bmatrix} \tilde{d}_1^{(e)} \\ \vdots \\ \tilde{d}_6^{(e)} \end{Bmatrix}}_{\mathbf{\tilde{d}}^{(e)}} \,$$ Equation (2)

$$\ \begin{Bmatrix} u_x \left ( \tilde{x} \right ) \\ u_y \left ( \tilde{x} \right ) \end{Bmatrix} = \underbrace{\mathbf{R}^T}_{2x2} \mathbb{N} \underbrace{\left ( \tilde{x} \right )}_{2x6} \underbrace{\mathbf{\tilde{T}}^{(e)}}_{6x6} \underbrace{\mathbf{d}^{(e)}}_{6x1} \,$$ Equation (3)

Now lets note: how Equ (2), Dimensional Analysis

$$\ \lbrack u \rbrack = L \,$$, previously we also noted that $$\ \lbrack N_1 \rbrack = \lbrack N_4 \rbrack = 1 \,$$, and $$\ \underbrace{\lbrack N_1 \rbrack}_{=1} \underbrace{\lbrack \tilde{d}_1 \rbrack}_{= L} + \underbrace{\lbrack N_4 \rbrack}_{= 1} \underbrace{\lbrack \tilde{d}_4 \rbrack}_{= L} \,$$

also

$$\ \lbrack v \rbrack = L, \underbrace{\lbrack N_2 \rbrack}_{=1} \underbrace{\lbrack \tilde{d}_2 \rbrack}_{= L} = L \,$$(displacement, transverse)

The same can be said with the third component: $$\ \underbrace{\lbrack N_2 \rbrack}_{=1} \underbrace{\lbrack \tilde{d}_2 \rbrack}_{= L} = L \,$$

Insert HW for N5 and N6

Derive of beam shape functions $$\ N_2, N_3, N_5, N_6 \,$$ as previously shown.

Recall: Governing PDE for beams

Equ (1), without ft (distributed transversal load) and without inertia force, $$\ m \ddot{v} \,$$ (static case = 0) :

$$\ \frac{\partial ^4}{\partial x^2} \begin{Bmatrix} \left ( EI \right ) \frac{\partial ^2 v}{\partial x^2} \end{Bmatrix} = 0 \,$$

Further consider constant EI: $$\ \frac{\partial ^4}{\partial x^4} v = 0 \,$$ integrate fout times and get C0, ..., C3.

Thus $$\ \Rightarrow v \left ( x \right ) = C_0 + C_1 x^1 + C_2 x^2 + C_3 x^3 \,$$

To obtain $$\ N_2 \left ( x \right ) \,$$ ( $$\ \tilde{x} \cong x \,$$ ) for simplicity

$$\ v \left ( 0 \right ) = 1, v \left ( L \right ) = 0, v' \left ( L \right ) = 0, v' \left ( L \right ) = 0 \,$$

These are the boundary conditions to use to solve for C0, ..., C3.

$$\ v \left ( 0 \right ) = 1 = C_0 \,$$

$$\ v \left ( L \right ) = 1 + C_1 \cdot L + C_2 \cdot L^2 + C_3 \cdot L^3 = 0 \,$$ Equation (1)

$$\ v' \left ( x \right ) = C_1 + 2 C_2 \cdot x + 3 C_3 \cdot L^2 = 0 \,$$

$$\ v' \left ( 0 \right ) = C_1 = 0 \,$$

$$\ v' \left ( L \right ) = 2 C_2 L^2 + 3 C_3 L^3 = 0 \,$$

$$\ (2) \Rightarrow C_3 = - \frac{2}{3} \frac{C_2}{L} \,$$

$$\ (1) \Rightarrow 1 + C_2 L^2 - \frac{2}{3} \frac{C_2}{L} (L^3) = 1 + \frac{1}{3} C_2 L^2 = 0 \,$$

$$\ C_2 = \frac{-3}{L^2} \,$$ and $$\ C_3 = \frac{-2}{3} \frac{1}{L} ( \frac{-3}{L^2} ) = \frac{2}{L^3} \,$$ Compare with expression for N2 previously.

For N3: Boundary Conditions

$$\ v \left ( 0 \right ) = 0, v \left ( L \right ) = 0, v' \left ( L \right ) = 1, v' \left ( L \right ) = 0 \,$$

For N5: Boundary Conditions

$$\ v \left ( 0 \right ) = 0, v \left ( L \right ) = 1, v' \left ( L \right ) = 1, v' \left ( L \right ) = 0 \,$$

For N6: Boundary Conditions

$$\ v \left ( 0 \right ) = 0, v \left ( L \right ) = 0, v' \left ( L \right ) = 1, v' \left ( L \right ) = 1 \,$$

(See previous plots for N5 and N6.)

Derive coefficients in $$\ \tilde{\mathbf{k}} \,$$ (element stiffness matrix) coefficients with EA: Done

Derive coefficients in $$\ \tilde{\mathbf{k}} \,$$ (element stiffness matrix) coefficients with EI: To be done

$$\ \tilde{k}_{22} = \frac{12EI}{L^3} = \int_{0}^{L} \frac{d^2 N_2}{dx^2} (EI) \frac{d^2 N_2}{dx^2} dx \,$$