User:Eml4500.f08.echo.mott/HW7

=HW Report 7=

Element 1: $$ E_1^{(1)} =2 $$ $$ E_2^{(1)} =4 $$ $$ A_1^{(1)}=.5 $$ $$A_2^{(1)} =1.5$$

Element 2: $$ E_1^{(2)}=3$$ $$E_2^{(2)}=7$$ $$A_1^{(2)}=1$$ $$ A_2^{(2)}=3$$

Compute the solution for 2-bar truss with tapered elements

Plot the deformed shape, and also deformed shape for previous 2-bar truss with average E and average A.

Frame element = Truss (bar) element + beam element.
 * where the truss element has axial deformation and the beam element has transverse deformation

(The angle between the 2 elements is constant after deformation)

FBD'S





In general $$d_i^{(e)} \rightarrow f_i^{(e)}$$
 * $$ f$$ is the generated forces
 * $$e = 1,2$$
 * $$i= 1, . . ., 6$$

Rotational Degrees of Freedom: $$ d_3^{(e)}, d_6^{(e)}$$

Bending Moments: $$f_3^{(e)}, f_6^{(e)}$$

Two Dimensional Frame Global Degrees of Freedom



2-Element stiffness matrix $$ \mathbf{k}_{6x6}^{(e)}, e=1,2$$

Global Stiffness Matrix $$ \mathbf{k}_{9x9}= A \mathbf{k}_{6x6}^{(e)}$$

Below is an illustration of the global stiffness matrix for this situation.





For the figure above, all displacements also represent forces and their respective numbers.

$$\mathbf{\tilde k_{6x6}^{(e)}}\mathbf{\tilde d_{6x1}^{(e)}}=\mathbf{\tilde f_{6x1}^{(e)}}$$

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

Note: $$\tilde f_3^{(e)}=f_3^{(e)}, \tilde f_6^{(e)}=f_6^{(e)} $$

The resulting stiffness matrix for the figure above is:

$$\mathbf{\tilde k}_{6x6}=\begin{bmatrix} \frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0 \\ & \frac{12EI}{L^3} & \frac{6EI}{L^2} & 0 & \frac{-12EI}{L^3} & \frac{6EI}{L^2}\\ & & \frac{4EI}{L} & 0 & \frac{-6EI}{L^2} & \frac{2EI}{L} \\ & & & \frac{EA}{L} & 0 & 0 \\ & SYM & & & \frac{12EI}{L^3} & \frac{-6EI}{L^2} \\ & & & & & \frac{4EI}{L}\end{bmatrix}$$

Dimensional Analysis

$$[\tilde d_1]=L=[\tilde d_i]$$ i =1,2,4,5 (L represents a dimension of length)

$$[\tilde d_3]=1=[\tilde d_6]$$ (1 represents rotational and no dimension)

$$\hat{AB}=R \cdot \theta $$

Doing some algebra:

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



$$[\theta ] = \frac{[\hat{AB]}}{[R]}=\frac{L}{L}=1$$

Continuing the dimensional analysis and applying it to collections of matrices:

$$\sigma=E \epsilon \Rightarrow [\sigma]=[E][\epsilon]$$

$$[\epsilon]=\frac{du}{dx}=\frac{L}{L}=1$$

$$[\sigma]=[E]=\frac{F}{L^2}$$

$$[A]=L^2 \mbox{, } [I]=L^4$$

$$[\frac{EA}{L}]=[\tilde k_{11}]=\frac{(F/L^2)(L^2)}{L}=\frac{F}{L}$$

$$[\tilde k_{11} \mbox{ }\tilde d_1]=[\tilde k_{11}][\tilde d_1]=F$$

$$[\tilde k_{23} \mbox{ } \tilde d_3] = [\tilde k_{23}][\tilde d_3]=\frac{[6][E][I]}{[L^2]}$$

Element FD relation in global coordinates from element FD relation in local coordinates. i.e. obtain:

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

$$\mathbf{k}_{6x6}^{(e)} = \mathbf{\tilde T}_{6x6}^{(e)T} \mathbf{\tilde k}_{6x6}^{(e)} \mathbf{\tilde T}_{6x6}^{(e)}$$

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

Putting the above into matrix form:



'''HW7: Solve 2-elem frame system using same data for 2-bar truss system, assuming square cross section. Plot undeformed shape & undeformed shape 2-bar truss & deformed shape 2-elem frame'''



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

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

Where the subscript t in $$f_t$$ signifies that it is transverse

MOTIVATION: Deformed shape of truss element interpolation of transverse displacement $$ v(x) = \tilde{x}$$

PVW For Beams:

$$\displaystyle\int_0^L w(x) \left [ - \frac{\partial ^2}{\partial x^2} \left ((EI)\frac{ \partial ^2 v}{\partial x^2} \right ) + f_t - m \ddot v \right ]dx$$


 * For all possible $$w(x)$$


 * $$ w(x)= \tilde{x}$$


 * $$d(x)=0$$

Integration by parts of 1st term : (use p. 30-1 as reference)

$$\displaystyle\int_0^L \underbrace{w(x)}_{s(x)} \frac{\partial ^2}{\partial x^2} \left [(EI)\frac{\partial ^2 v}{\partial x^2} \right ]dx$$

$$\underbrace{\frac{\partial}{\partial x}\underbrace{ \left (\frac{\partial}{\partial x} (EI)\frac{\partial ^2 v}{\partial x^2} \right )}_{r(x)}}_{ r\prime(x)}$$

$$= \underbrace {w \frac{\partial}{ \partial x} { (EI) \frac{\partial ^2v}{\partial x^2} }  }_{\beta _1}|_0^L-\displaystyle\int_0^L \underbrace{\frac{dw}{dx}}_{s \prime (x)} \underbrace{\frac{\partial}{\partial x}{(EI) \frac{\partial v}{ \partial x^2}  }  dx}_{r(x)}$$

$$= \beta_1 - \left [ \underbrace{\frac{dw}{dx}(EI) \frac{\partial ^2v}{\partial x^2}}_{\beta_2} \right ]|_0^L+ \underbrace{\displaystyle\int_0^L \frac{d^2w}{dx^2}(EI) \frac{\partial ^2v}{\partial x^2}}_{ \gamma}$$

Note the symmetry.

Eq (1) becomes:

$$=- \beta_1 + \beta_2 - \gamma + \displaystyle\int_0^L wf_t \, dx - \displaystyle\int_0^L wm\dot v \, dx$$

for all possible w(x)

Focus on the stiffness term for now to derive the beam stiffness matrix and to identify the beam shape functions.



Recall: $$u(\tilde x)=N_1(\tilde x) \tilde d_1+N_4(\tilde x) \tilde d_4 \mbox{       (Equation 1)}$$

$$V(\tilde x)=N_2(\tilde x)\tilde d_2 + N_3(\tilde x) \tilde d_3 + N_5 (\tilde x) \tilde d_5 + N_6(\tilde x)\tilde d_6 \mbox{     (Equation 2)}$$



$$N_2(\tilde x) = 1 - \frac{3\tilde x^2}{L^2}+\frac{2\tilde x^3}{L^3} \mbox{ :  } \tilde d_2$$

$$N_3(\tilde x) = \tilde x - \frac{2\tilde x^2}{L}+\frac{\tilde x^3}{L^2} \mbox{ :  } \tilde d_3$$

$$N_5(\tilde x) = \frac{3 \tilde x^2}{L^2} - \frac{2\tilde x^3}{L^3} \mbox{ :  } \tilde d_5$$

$$N_6(\tilde x) = \frac{-\tilde x^2}{L} - \frac{\tilde x^3}{L^2} \mbox{ :  } \tilde d_6$$



$$\tilde{ \mathbf {d} }^{(e)} = \tilde{ \mathbf {T} }^{(e)} + \mathbf {d}^{(e)}$$



The element displacement matrix $$\mathbf {d}^{(e)}$$ is known.

Computing $$u(\tilde{x})$$ and $$v(\tilde{x})$$ :

$$ \mathbf{u}(\tilde{x}) = u(\tilde{x}) \hat{\tilde{i}} + v(\tilde{x}) \hat{\tilde{j}}$$

$$ = u_x(\tilde{x}) \hat{i} + u_y(\tilde{x}) \hat{j}$$

Equation 1 presented above can be used to solve for $$v(\tilde{x})$$ and Equation 2 can be used to solve for $$u(\tilde{x})$$.

Can compute $$u_x(\tilde{x}) $$ and $$u_y(\tilde{x})$$ from $$u(\tilde{x})$$ and $$v(\tilde{x})$$ :

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

Also,

$$\begin{Bmatrix} u(\tilde{x}) \\ v(\tilde{x}) \end{Bmatrix} = \overbrace{\begin{pmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{pmatrix}}^{\mathbb{N}} \overbrace{\begin{Bmatrix} {\tilde{d_1}}^{(e)} \\ \vdots \\ {\tilde{d_6}}^{(e)} \end{Bmatrix}}^{{\tilde{\mathbf d}}^{(e)}}$$

The equations presented can be combined to form the following:

$$\begin{Bmatrix} u_x \\ u_y \end{Bmatrix} = {\mathbf R}^T \ \mathbb{N}(\tilde x) \ {\tilde{\mathbf T}}^{(e)} \ {\tilde{\mathbf d}}^{(e)} $$

 Note:  Dimensional Analysis

$$[u_{ }^{ }]=L$$

Previously: $$[N_1^{ }]= [N_4]=1$$

Equation 2 previously: $$ [N_1]^{} \, [ \tilde {d_1}] + [N_4]^{} \, [ \tilde {d_4}]$$


 * $$N_1^{ } = 1$$


 * $$ \tilde{d_1} = L$$


 * $$N_4^{ } = 1$$


 * $$ \tilde{d_4}= L$$

$$[v_{ }^{ }]=L$$

$$[N_2]\,[\tilde{d_2}]=L$$


 * $$ N_2^{ }=1$$


 * $$ \tilde{d_2}= L$$ (translational displacement)

$$ [N_3] \, [\tilde{d_3}] = L $$


 * $$ N_3^{ } = L$$


 * $$ \tilde{d_3}=1$$ (rotational displacement)

 End Note 

Derivation of beam shape functions (shown earlier)

$$ N_2, N_3, N_5, N_6 $$

Recall: Governing PDE for beams, Eq (1), without $$f_t$$ (distributed transverse load) and without inertia force.

$$ m \ddot {v} \mbox{(static case)}:$$

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

Further, consider constant EI values

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

Integrate 4 times. End up with 4-constants : $$ C_0 - C_3$$

$$ \rightarrow V(x)= C_0 + C_1x^1 + C_2x^2 +C_3x^3$$

To obtain $$N_2(x)$$ $$(\tilde{x} =x for simplicity)$$

$$ V(0)=1, \, V(L)=0, \, V\prime(0) = v\prime (L) = 0$$

Use the above boundary conditions to solve for $$C_0 - C_3$$

$$V(0_{ }^{ }) =1 =C_0$$

$$V(L)=1 +C_1 \cdot L + C_2 \cdot L^2 + C_3 \cdot L^3 =0$$

$$ V^{\prime} (x) = C_1 +2C_2x+3C_3x^2$$

$$V^{\prime} (0) = C_1=0$$

$$(1) V^{\prime}(L)=2C_2 \cdot L + 3 C_3 \cdot L^2 = 0$$

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

$$ (1) \rightarrow 0= 1+ \underbrace{ C_2 \cdot L^2 + \left (- \frac{2}{3} \, \frac{C_2}{L} \right ) L^3}_{C_2 \cdot L^2[\underbrace {1- \frac{2}{3}}_{\frac{1}{3}}]}$$

$$ C_2= \frac{-3}{L^2}$$

$$ C_3 = -\frac{2}{3} \, \frac{1}{L} \left ( -\frac{3}{L^2} \right ) = \frac{2}{L^3}$$

Compute with expressions for $$N_2$$

Boundary Conditions

For $$\mathbf{N_3}$$:

$$ V(0)=V(L)=0, \, V^{\prime}(0)=1, \, V^{\prime}(L)=0$$


 * $$ \tilde {d_3}, \, \mbox{rotational}$$

For $$\mathbf{N_5}$$:

$$ V(0)=0, \, V(L)=1, \, V^{\prime}(0)=0, \, V^{\prime}(L)=0$$


 * $$ \tilde {d_5}, \, \mbox{ displacement}$$

For $$ \mathbf {N_6}$$:

$$ V(0) = V(L)=0, \, V^{\prime}(0)=0, \, V^{\prime}(L)=1$$


 * $$ \tilde{d_6}, \, \mbox{rotational}$$

Plots for $$ N_5, \, N_6$$ were shown previously.

Derive coefficients in $$\mathbf{\tilde{k}}$$ (element stiffness matrix)

Coefficient with EA: DONE

Coefficient with EI: TO BE DONE

$$ \tilde{k}_{22} = \frac{12EI}{L^2}= \displaystyle\int_0^L \frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}$$

$$\tilde k_{23}=\frac{6EI}{L^2}=\int_0^L \frac{d^2 N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}\,dx$$

In general: $$\tilde k_{ij}=\int_0^L \frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}\,dx$$ for: i,j=2,3,5,6

Elastodynamics (trusses, frames, 2-D & 3-D Elasticity)

Model problem

Discrete PVW (Boundary conditions already applied):

$$\mathbf{\bar w} \cdot \left [ \mathbf{\bar M \ddot{\bar d}} + \mathbf{\bar K \bar d} - \mathbf{\bar F}\right ]=0$$

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

Defining Equation 1:

$$ \Rightarrow \mathbf{\bar M \ddot{\bar d}} + \mathbf{\bar K \bar d} = \mathbf{\bar F(t)}$$ $$\mathbf{\bar d}(0)=\mathbf{\bar d_O}$$ $$\mathbf{\dot{\bar d}}(0)=\mathbf{\bar V_O}$$

Complete ordinary differential equations (ODEs) (2nd order in time) and initial conditions governing the elastodynamic of discretized continuous problems (MDOF).

Solving Equation 1:

1) Consider unforced vibrations problem:

$$\mathbf{\bar M_{nxn} \ddot v_{nx1} + \bar K_{nxn} v_{nx1} = 0_{nx1}}$$ Where 0 represents unforced

Assume $$\mathbf{v}(t)=(sin\omega t) \phi$$

The individual $$\phi$$ means the equation is not time dependent

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

The generalized evaluation problem is:

$$\Rightarrow \mathbf{\bar K \phi}=\omega ^2 \mathbf{\bar M \phi}$$

The general form is:

$$\mathbf{Ax}=\lambda \mathbf{Bx}$$

Standard evaluation problem: $$\mathbf{Ax}=\lambda \mathbf{x}$$

where B is set equal to the identity matrix: $$\mathbf{B}=\mathbf{I}$$

$$\mathbf{I}=\begin{bmatrix} 1 & & 0 \\ & \ddots & \\ 0 & & 1\end{bmatrix}$$

$$\lambda =\omega ^2$$ eval

$$( \lambda _i, \mathbf{\phi}_i)$$ eigenpairs

i = 1, ..., n

The animation is dictated by: Mode i $$\Rightarrow \mathbf{V}_i(t)=(sin(\omega _i t))\mathbf{\phi}_i$$ i=1, ..., n

2) Model superposition method:

Orthogonal properties of eigenpairs:

$$\mathbf{\phi}_i^T \mathbf{\bar M \phi}_j=\partial _{ij}= \begin{cases}1 & \mbox{if } i=j \\ 0 & i \ne j\end{cases}$$

This equation represents Kronecker delta

Mass orthogonal of eigenvector:

$$\mathbf{\bar M \phi}_j=\lambda _j \mathbf{\bar K \phi}_j $$

$$\mathbf{\phi}_i^T \mathbf{\bar M \phi}_j=\lambda _j \mathbf{\phi}_i^T \mathbf{\bar K \phi}_j$$

$$\Rightarrow \mathbf{\phi}_i^T \mathbf{\bar K \phi}_j=\frac{1}{\lambda}_j \partial _{ij}$$

Equation 1: $$\mathbf{\bar d}(t)_{nx1}=\sum_{i=1}^n \zeta _{i(1x1)}(t)\phi _{i(nx1)}$$

$$\mathbf{\bar M} \underbrace { \left (\sum_j \ddot{\zeta}_j \mathbf{\phi}_j \right )}_{\mathbf{\ddot{\bar d}}} + \mathbf{\bar K} \underbrace{\left (\sum_j \zeta_j\mathbf{\phi}_j \right )}_{\mathbf{\bar d}}=\mathbf{F}$$

$$\sum_j \ddot{\zeta}_j \underbrace{ \left ( \mathbf{\phi}_i^T \mathbf{\bar M \phi}_j \right ) }_{\partial _{ij}} + \sum_j \zeta _j \underbrace{ \left ( \mathbf{\phi}_i^T \mathbf{\bar K \phi}_j  \right ) }_{\lambda_j \partial_{ij}} = \mathbf{\phi}_i^T\mathbf{F}$$

$$\Rightarrow \ddot{\zeta}_i+\lambda_i \zeta_i=\phi_i^T \mathbf{F}$$ $$i=1,...,n$$

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 * Use university secure password to login. No extra passwords to learn and you trust that no one will get on your page.
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 * No way to create your own content.
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Contributing Team Members
Chris Mott 05:16, 9 December 2008 (UTC)

Bradley LaCroix 05:58, 9 December 2008 (UTC)

Brandon Sell 07:15, 9 December 2008 (UTC)

Jason Bruce 14:57, 9 December 2008 (UTC)

Charles Marshall 15:41, 9 December 2008 (UTC)

Chris Sell 19:33, 9 December 2008 (UTC)