User:Eml4500.f08.gravy.mmm/HW7

Element 1: E1(1)=2, E2(1)=4, A1(1)=.5, A2(1)=1.5

Element 2: E1(2)=3, E2(2)=7, A1(2)=1, A2(2)=3

A comparison of the solution for 2-bar truss with tapered elements are shown in a previous report. Modification of 2-bar truss code to accommodate general k(i)

Frame element is a truss bar element(which is axial deformation) and a beam element (which is transverse deformation).

Model frame problem with two elements


Free Body diagrams for the elements



$$d_1^{(1)} \longrightarrow f_1^{(1)}$$

$$ d_2^{(1)} \longrightarrow f_2^{(1)}$$

$$d_3^{(1)} \longrightarrow f_3^{(1)}$$

$$ d_4^{(1)} \longrightarrow f_4^{(1)}$$

$$d_5^{(1)} \longrightarrow f_5^{(1)}$$

$$ d_6^{(1)} \longrightarrow f_6^{(1)}$$

$$d_1^{(2)} \longrightarrow f_1^{(2)}$$

$$d_2^{(2)} \longrightarrow f_2^{(2)}$$

$$d_3^{(2)} \longrightarrow f_3^{(2)}$$

$$d_4^{(2)} \longrightarrow f_4^{(2)}$$

$$ d_5^{(2)} \longrightarrow f_5^{(2)}$$

$$d_6^{(2)} \longrightarrow f_6^{(2)}$$

$$ d_2^{(1)} \longrightarrow f_2^{(1)}$$

$$d_3^{(1)} \longrightarrow f_3^{(1)}$$

$$ d_4^{(1)} \longrightarrow f_4^{(1)}$$

$$d_5^{(1)} \longrightarrow f_5^{(1)}$$

$$d_6^{(1)} \longrightarrow f_6^{(1)}$$

In general, $$d_i^{(e)} \longrightarrow \underbrace{f_i^{(e)}}_{general forces}$$

e=1,2  i=1,...,6

$$\begin{matrix} d_3^{(e)}\\ d_6^{(e)} \end{matrix} \}$$ rotational degrees of freedom $$\longrightarrow \begin{matrix} f_3^{(e)}\\ f_6^{(e)} \end{matrix} \} $$ bending moments

2-D Frame global dofs;



Two element stiffness matrices with $$ \mathbf{k}_{6x6}^{(e)}, e=1,2$$

and the Global stiffness matrix becomes $$ \mathbf{k}_{9x9}= A_{e=1}^{e=2}\mathbf{k}_{6x6}^{(e)}$$



$$ \tilde{\mathbf{k}_{6x6}^{(e)}}\tilde{\mathbf{d}_{6x1}^{(e)}}= \tilde{\mathbf{f}_{6x1}^{(e)}}$$

$$ \tilde{\mathbf{d}^{(e)}}= \begin{bmatrix} d_1^{e} \\ d_2^{e} \\ d_3^{e} \\ d_4^{e} \\ d_5^{e} \\ d_6^{e} \end{bmatrix}$$, $$\tilde{\mathbf{f}^{(e)}}= \begin{bmatrix} f_1^{e} \\ f_2^{e} \\ f_3^{e} \\ f_4^{e} \\ f_5^{e} \\ f_6^{e} \end{bmatrix}$$

Note: $$\tilde{\mathbf{f}_3^{(e)}}= \mathbf{f}_3^{(e)}, \tilde{\mathbf{f}_6^{(e)}}= \mathbf{f}_6^{(e)}$$ moments about $$\tilde{x}=z$$





Dimensional Analysis
$$ \underbrace{\left[\tilde{d_1}\right]}_{dimension of} = \underbrace{L}_{Length}= \underbrace{\left[\tilde{d_1}\right]}_{displacement}$$ i=1,2,4,5

$$\left[\tilde{d_3}\right]= \underbrace{1}_{No dimension}=\underbrace{\left[\tilde{d_6}\right]}_{rotational}$$



$$\underbrace{\widehat{AB}}_{arc lenth}=R \underbrace{\theta}_{radians}$$

$$ \theta=\frac{\widehat{AB}}{R}$$

$$\left[\theta\right]=\frac{\widehat{\left[AB\right]}}{\left[R\right]}=\frac{L}{L}=1$$

$$\sigma=E \epsilon \Rightarrow \left[\sigma\right]=\left[E \right]\underbrace{\left[\epsilon\right]}_1$$

monday 12/8
$$ \tilde{k}_{23}= \frac{6EI}{L^2}= \int_0^L \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2},dx= \frac{dN_2}{dx}(EI)\frac{d^2N_3}{dx^2} \frac{d^2N_2}{dx^2}(EIx)\frac{d^2N_3}{dx^2} \frac{d^2N_2}{dx^2}(EI)\frac{dN_3}{dx} \Bigg|_0^L $$

In general, $$\tilde{k}_{ij}=\int_0^L \frac{d^2N_i}{dx^2}(EI)\frac{d^2N_3}{dx^2}dx$$ i,j=2,3,5,6

Elastodynamics (trusses, frames, 2-D, 3-D, elasticity)
A model problem can be found in a previous report Here

Discrete Principle of Virtual work
$$ \mathbf{\bar{w}} \cdot [ \mathbf{\bar{M}} \mathbf{\ddot{\bar{d}}} + \mathbf{\bar{K}} \mathbf{\bar{d}}- \mathbf{\bar{F}} ] =0$$

for all $$\mathbf{\bar{w}}$$

The bars mean the matrix is reduced which is to say the boundary conditions have already been applied

$$\Rightarrow $$ $$\mathbf{\bar{M}} \mathbf{\ddot{\bar{d}}} + \mathbf{\bar{K}} \mathbf{\bar{d}}=  \mathbf{\bar{F}}(t)$$ equation 1 $$\mathbf{\bar{d}}(0) = \mathbf{\bar{d}}_0$$ $$\mathbf{\dot{\bar{d}}}(0) = \mathbf{\bar{v}}_0$$

Complete ordinary differential equation (ODEs) (2nd order in time)+ initial conditions governing the elastodynamics of discretized continuous problem (MDOF)

Solving equation 1:

1) Consider unforced vibration problem

$$ \mathbf{\bar{M}}_{nxn} \mathbf{\ddot{v}}_{nx1} + \mathbf{\bar{K}}_{nxn} \mathbf{v}_{nx1}= \mathbf{0}_{nx1}$$   unforced equation 2

assume $$v(t)=(sin \omega t) \mathbf{\phi}_{nx1}$$ where $$\mathbf{\phi}$$ is not time dependent

$$ \mathbf{\ddot{\bar{v}}}=-\omega^2sin\omega t \phi$$

$$=\omega^2sin\omega t \mathbf{\bar{M}} \phi + sin\omega t \mathbf{\bar{K}} \phi = \mathbf{0}$$

$$\Rightarrow \mathbf{\bar{K}} \phi=\omega^2\mathbf{\bar{M}}\phi$$ equation 3 Which is the Generalized eval. Problem

$$ \mathbf{Ax}=\lambda \mathbf{Bx}$$ with $$\lambda$$ as the eval. Which is the General Form

Standard eval. problem: $$ \mathbf{Ax}=\lambda \mathbf{x}$$ with B = I with I as the Identity Matrix

The Identity Matrix I=$$\begin{bmatrix} 1 & 0\\0&1 \end{bmatrix}$$ which has ones on the diagonal and zeros everywhere else.

$$\lambda =\omega^2$$ and ($$\lambda_i,\phi_i$$) are eigenpairs with i=1,...,n

mode i $$\Rightarrow \begin{cases} v_1(t)=(sin\omega_1t)\phi \\ i=1,\cdots,n \end{cases}$$

2)Model superposition method:

Orthogonal properties of eignepairs:

$$\underbrace{\mathbf{\phi}_i^T}_{1xn} \underbrace{\mathbf{\bar{M}}}_{nxn} \underbrace{\mathbf{\phi}_i}_{nx1} = \underbrace{\delta_{ij}}_{Kronecker \delta}= \begin{cases} 1,& \mbox{if } i=j \\  0,  &\mbox{if } i=j \end{cases}$$

Mass orthogonal of eigenvectors

Using equation 2 and equation 3

$$\mathbf{\bar{M}}\omega_j=\lambda_j^{-1}\mathbf{\bar{K}}\omega_j$$

$$

\underbrace{\phi_i^T\mathbf{\bar{M}}}_{\delta_{ij}}\phi_j=\lambda_j^{-1}\phi_i^T\mathbf{\bar{K}}\phi_j$$

$$\Rightarrow \phi_i^T\mathbf{\bar{K}}\phi_j=\lambda_j \delta_{ij}$$

From equation 1 we get: $$\mathbf{\bar{d}}=\sum_{i=1}^n \xi_i(t)\phi_i$$

$$ \mathbf{\bar{M}} \underbrace{(\sum_{j}\ddot{\xi}_j \phi_j)}_{\mathbf{\ddot{d}}} + \mathbf{\bar{K}}\underbrace{(\sum_{j}\xi_j \phi_j)}_{\mathbf{\bar{d}}}= \mathbf{F}$$

$$\sum_j\ddot{\xi}_j\underbrace{(\phi_i^T\mathbf{\bar{M}}\phi_j)}_{\delta_{ij}} + \sum_j\xi_j \underbrace{(\phi_i^T\mathbf{\bar{K}}\phi_j)}_{\lambda_i\delta_{ij}}= \phi_i^T \mathbf{F}$$

$$\Rightarrow \ddot{\xi}_i+\lambda_i\xi_i=\phi_i^T F$$ i=1,...n

Pictures