User:Eml4500.f08.ateam.nobrega/Homework Report Seven

=Lecture Notes=

Frame Element
A frame element is the combination of a truss, or bar element, and a beam element. The truss element corresponds to the axial deformation experienced by the frame and the beam element corresponds to the transverse deformation. A model of a frame with two elements bound together by a rigid connection is shown below.



A rigid connection requires that the angle between two elements remain constant even after deformation. The free body diagrams of each element are presented below.



The general forces associated with the above elements are presented in the following form and include forces and bending moments.

fie $$\rightarrow$$ Generalized forces

Where e is equal to 1 or 2 and i can range from 1 to 6.

Looking at the above free body diagrams d3e and d6e are rotational terms. Correspondingly, f3e and f6e are the bending moments.

A diagram of the frame global degrees of freedom are shown below.



Here d3,6,9 are the rotational degrees of freedom and the rest displacements. For this frame the 2 element stiffness matrix Ke is a 6x6 matrix where e can be equal to 1 or 2. Therefore the global stiffness matrix, K, becomes a 9x9 matrix where,


 * $$ K = A_{e=1}^{e=2}K^{(e)}$$






 * $$ \mathbf \tilde{f}^{(e)} = \mathbf \tilde{k}^{(e)} \mathbf \tilde{d}^{(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} $$

The displacement matrix includes both displacements and rotaions.


 * $$ \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: $$ \tilde{f_{3}}^{e}= f_{3}^{e} $$ and $$ \tilde{f_{6}}^{e}= f_{6}^{e} $$. These terms are moments not forces. (Moments about the $$ \tilde{z}=z $$ axis.)



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

Therfore, the matrix multiplication is:

$$ \begin{bmatrix} \tilde{f_{1}}^{e}\\ \tilde{f_{2}}^{e}\\ \tilde{f_{3}}^{e}\\ \tilde{f_{4}}^{e}\\ \tilde{f_{5}}^{e}\\ \tilde{f_{6}}^{e} \end{bmatrix} =    \begin{bmatrix} \frac{EA}{L} & 0 & 0 & -\frac{EA}{L} & 0 & 0\\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & -\frac{12EI}{L^{3}} & \frac{6EI}{L^{2}}\\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{2EI}{L} \\ -\frac{EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0\\ 0 & -\frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & -\frac{6EI}{L^{2}}\\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & -\frac{6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} \begin{bmatrix} \tilde{d_{1}}^{e}\\ \tilde{d_{2}}^{e}\\ \tilde{d_{3}}^{e}\\ \tilde{d_{4}}^{e}\\ \tilde{d_{5}}^{e}\\ \tilde{d_{6}}^{e} \end{bmatrix} $$

Dimensional Analysis
$$ [\tilde{d}_{1}]=L=[d_{i}^{N}] $$ where i=1,2,3,4,5,6

The rotational terms in the $$ \mathbf\tilde{d} $$ matrix are dimensionless. Therfore,

$$ [\tilde{d}_{3}]=1 $$

$$ AB=R\theta $$

Where AB is the arch length and theta is in radians.

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

$$ \sigma =E\varepsilon

[\sigma ]=[E][\varepsilon]

$$

Where, $$ [\varepsilon]=1 $$.

Why is the dimension of strain equal to 1? The modulous of Elasticity (E) and stress (σ) have the same units; for example, Pascal can be in SI, metric units, or psi, in English units. Therefore, strain (ε) is dimensionless.

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

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

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

$$ [k_{11}d_1]=[k_{11}][d_1]=F $$

$$ [k_{23}d_3]=[k_{23}][d_3]=[\frac{[6][E][I]}{L^2]}]=\frac{((I)\frac{F}{L^2}(L^4))}{L^2}=F $$

The FD relationship is derived in the global coordinate system from the element force displacement relationship. That is:

k (e)6x6 d (e)6x1= f (e)6x1 k (e)6x6= T (e)T6x6 k (e)6x6 T (e)6x6

from k (e)6x6 d (e)6x1= f (e)6x1



\begin{bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\\ \end{bmatrix} =    \begin{bmatrix} R & R & 0 & 0 & 0 & 0\\ R & R & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\     0 & 0 & 0 & R & R & 0\\ 0 & 0 & 0 & R & R & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\    \end{bmatrix} \begin{bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\\ \end{bmatrix} $$

Where d is a 6x1 matrix and the transformation matrix T is a 6x6 matrix.

Plot the deformed shape of the figure above. Derivation of k can be found using the Principal of Virtual Work (PVW), focusing only on bending effects leads to

$$ \frac{d^2}{dx^2}(EI\frac{d^2v}{dx^2})-f_t (x)=m(x)\ddot{v} $$

Shape Functions and the Principle of Virtual Work
Motivation: deformed shape of truss, interpertation of transverse displacements $$ v(\tilde x) $$ element.

Principle of Virtual Work: Beams $$ \int\limits_{0}^{L} w(\tilde x) \left \{ \frac{-\partial^2 }{\partial x^2}\left [(EI) \frac{\partial^2 v}{\partial x^2}\right ] + f_{t} - m \ddot v \right \} dx = 0 \qquad \qquad (1) \qquad$$ for all possible $$ w(\tilde x) $$

Integration by Parts of 1st Term: $$ \alpha = \int\limits_{0}^{L} w(\tilde x) \frac{-\partial^2 }{\partial x^2}\left [(EI) \frac{\partial^2 v}{\partial x^2}\right ] dx = 0 \qquad \qquad  \qquad$$ $$ \frac{\partial}{\partial x} \left \{\frac{\partial}{\partial x} \left [(EI) \frac{\partial^2 v}{\partial x^2}\right ] \right \} $$ $$ = \left \{w \frac{\partial}{\partial x} \left [(EI) \frac{\partial^2 v}{\partial x^2} \right ] \right \}_{0}^{L} - \int\limits_{0}^{L} \frac{\partial w}{\partial x} \frac{\partial}{\partial x} \left \{(EI) \frac{\partial^2 v}{\partial x^2}\right \} dx  $$ Note: Where $$ \left \{w \frac{\partial}{\partial x} \left [(EI) \frac{\partial^2 v}{\partial x^2} \right ] \right \}_{0}^{L} $$ will be noted as $$ \beta_{1} $$ $$ \beta_{1} - \left [\frac{dw}{dx}(EI)\frac{\partial^2 v}{\partial x^x} \right ]_{0}^{L} + \int\limits_{0}^{L} \frac{d^2w}{dx^2}(EI)\frac{\partial^2v}{\partial x^2}dx$$ Note: Where $$ \left [\frac{dw}{dx}(EI)\frac{\partial^2 v}{\partial x^x} \right ]_{0}^{L} $$ will be noted as $$ \beta_{2} $$ and $$ \int\limits_{0}^{L} \frac{d^2w}{dx^2}(EI)\frac{\partial^2v}{\partial x^2}dx $$ will be noted as $$ \gamma $$ Note: Also notice the symmetry in the expression above Therefore Equation (1) becomes: $$ -\beta_{1} + \beta_{2} - \gamma + \int\limits_{0}^{L} wf_{t}dx-\int\limits_{0}^{L}wm \ddot v dx $$, for all possible $$ w(x) $$

Now let us focus on the stiffness term $$ \gamma $$ for now to derive the beam stiffness matrix and to identify the beam shape functions.



$$ v(\tilde x) = N_{2}(\tilde x)\tilde d_{2} + N_{3}(\tilde x)\tilde d_{3} + N_{5}(\tilde x)\tilde d_{5} + N_{6}(\tilde x)\tilde d_{6} $$

Recall: $$ u(\tilde x) = N_{1}(\tilde x)\tilde d_{1} + N_{4}(\tilde x)\tilde d_{4} $$

Shape Functions: $$ N_{2}(\tilde x) = 1 - \frac{3\tilde x^3}{L^2} + \frac{2\tilde x^3}{L^3} => \tilde d_{2} $$ $$ N_{3}(\tilde x) = \tilde x - \frac{2\tilde x^2}{L} + \frac{\tilde x^3}{L^2} => \tilde d_{3} $$ $$ N_{5}(\tilde x) = \frac{3\tilde x^2}{L^2} - \frac{2\tilde x^3}{L^3} => \tilde d_{5} $$ $$ N_{6}(\tilde x) = \frac{-\tilde x^2}{L} + \frac{\tilde x^3}{L^2} => \tilde d_{6} $$

Shape Functions Continued


From Lecture 37-3,


 * $$ \tilde{d}_{6x1}^{(e)} = \tilde{T}_{6x6}^{(e)} \tilde{d}_{6x1}^{(e)}$$

Where $$ \tilde{d}^{(e)} $$ is known after solving the Finite Element System.

To compute $$ u(\tilde{x}), v(\tilde{x})$$,




 * $$ u(\tilde{x}) = U(\tilde{x}) \overrightarrow{ \tilde{i}} + V(\tilde{x}) \overrightarrow{ \tilde{j}} $$


 * $$ U(\tilde{x}) = U_x(\tilde{x}) \overrightarrow{ \tilde{i}} + U_y(\tilde{x}) \overrightarrow{ \tilde{j}} $$

To compute $$ U_x(\tilde{x}) $$ and $$ U_y(\tilde{x}) $$



\begin{bmatrix} U_x(\tilde{x})\\ U_y(\tilde{x}) \end{bmatrix} = R^{T} \begin{bmatrix} U(\tilde{x})\\ V(\tilde{x}) \end{bmatrix} $$

Where $$ U(\tilde{x}) $$ and $$ V(\tilde{x}) $$ are found



\begin{bmatrix} U(\tilde{x})\\ V(\tilde{x}) \end{bmatrix} = \begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix} \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} $$



\mathbb{N} = \begin{bmatrix} N_1 & 0 & 0 & N_4 & 0 & 0 \\ 0 & N_2 & N_3 & 0 & N_5 & N_6 \end{bmatrix} $$

Therefore,



\begin{bmatrix} U_x(\tilde{x})\\ U_y(\tilde{x}) \end{bmatrix} = R^T \mathbb{N} \tilde{T}^{(e)} d^{(e)} $$

Note: Equation 2 from the previous lecture P 39-2 was used for dimensional analysis.

[u]=L
 * P 31-4 [N1]=[N4]=1

Using Equation 2 determine that [N1][d1]=[N4][d4]

[v]=L
 * [N2][d2]=L

Thus


 * [N3][d3]=L

Derivation of the beam shape functions N2, N3, N5, and N6 were previously determined in Lecture 37-4, without the use of ft (distributed transmitted load) and the inertia force.


 * $$ m \ddot{v} (static) =\frac{d^2}{dx^2}(EI\frac{d^2x}{dx^2})$$

Consider further constant values for E & I, the equation simplifies down to $$\frac{d^4V}{dx^4}=0$$

When the above equation is integrated four times, four constant values are determined: C0, C1, C2, and C4.

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

To obtain N2(x), use $$ V(0)=1 $$ $$ V(L)=0 $$

$$ \frac{dV(0)}{dx}=0 $$

$$ \frac{dV(L)}{dx}=0 $$

Use the boundary conditions listed above to solve for C0 and C3.

$$ V(0)=1=C_0 $$

$$ V(L)=1+C_1L+C_2L^2+C_3L^3 $$

$$ \frac{dV(x)}{dx}=C_1+2C_2x+3C_3x^2 $$

$$ \frac{dV(0)}{dx}=C_1=0 $$

$$ \frac{dV(L)}{dx}=2C_2L+3C_3L^2=0 $$

$$ C_3=\frac{-2C_2}{3L} $$

$$ 0=1+C_2L^2+\frac{-2C_2L^2}{3} $$

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

$$ C_3=\frac{2}{L^3} $$

For N3: Boundary Conditions are

$$V(0)=V(L)=0$$ $$\frac{dV(0)}{dx}=0$$ $$\frac{dV(L)}{dx}=0$$

For N5: Boundary Conditions are

$$V(0)=0$$ $$V(L)=1$$ $$\frac{dV(0)}{dx}=0$$ $$\frac{dV(L)}{dx}=0$$

For N6: Boundary Conditions are

$$V(0)=0$$ $$V(L)=0$$ $$\frac{dV(0)}{dx}=0$$ $$\frac{dV(L)}{dx}=L$$

See Lecture 39-1 for the plots of N5 and N6

Derivation of the coefficients of K (element stiffness matrix) coefficients with EA: Done coefficients with EI: To be Done $$k_{22}=\frac{12EI}{L^2}=\int_{0}^{L}\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}dx$$

$$ \tilde k_{23} = \frac {6EI}{L^2} = \int\limits_{0}^{L} \frac{d^2N_{2}}{dx^2}(EI) \frac{d^2N_{3}}{dx^2}dx$$

In general, $$ \tilde k_{ij} = \int\limits_{0}^{L} \frac{d^2N_{i}}{dx^2}(EI) \frac{d^2N_{j}}{dx^2}dx $$, where $$ i,j = 2,3,5,6 $$

Structural Dynamics
Elastodynamics (truss, frames, 2D & 3D elasticity)

Discrete Principle of Virtual Work:

Note: Bold Face implies that boundary conditions are already applied; Therefore the expression below is the reduced version.

$$ \mathbf w \cdot \left [\mathbf M \mathbf \ddot d + \mathbf K \mathbf d - \mathbf F \right ] = 0 $$, for all  $$  \mathbf w $$ Equation (1): $$ \mathbf M \mathbf \ddot d + \mathbf K \mathbf d = \mathbf F_{t} $$ $$ \mathbf d(0) = \mathbf d_{0} $$ $$ \mathbf \dot d_{0} = \mathbf v_{0} $$ Note: Complete ordinary differential equations and initial conditions governing the elastodynamics of the discretized problem.

Solving Equation (1):

1. Consider unforced vibrations problem

$$ \mathbf M \underline \ddot v + \mathbf K \underline v = \underline 0 $$

Assume:

$$ \underline v(t)_{nx1} = (sin(wt)) \phi_{nx1(not line dependent)} $$

Then:

$$ \underline \ddot v = w^2sin(wt) \underline \phi $$

Then:

$$ -w^2sin(wt) \mathbf M \phi + sin(wt) \mathbf K \phi = \underline 0 $$

Then giving us the generalized eigen value problem below:

$$ \mathbf K \phi = w^2 \mathbf M \phi $$

Which we can compare to the general form below:

$$ \underline A \underline x = \lambda \underline B \underline x $$, where $$ \lambda $$ is the eigen value

Where in your standard eigen value problem $$ \underline A \underline x = \lambda \underline B \underline x $$, where $$ ( \underline B = \underline I ) $$, where $$ \underline I $$ is your identity matrix shown below:

$$ \underline I = \begin{bmatrix} 1 &   &   & 0 \\   &. &  &   \\   &   & . &   \\ 0 &   &   & 1 \end{bmatrix} $$

Where:

$$ \lambda = w^2 $$  eigen value

and,

$$ (\lambda_{i}, \underline \phi_{i}) $$  eigen pairs, where i = 1,. . ., n

where,

Mode $$ i => animation => \underline V_{i}(t) - (sin(w_{i}t))d_{i} $$

2. Model Superposition Method

Orthogonal problem of eigen pairs:

$$ \phi_{i}^T \mathbf M \phi_{j} = \delta_{ij} = 1 $$ if $$ i=j $$

or,

$$ \phi_{i}^T \mathbf M \phi_{j} = \delta_{ij} = 0 $$ if $$ i $$ does not equal $$ j $$

Mass orthogonal of eigenvalue.

$$ \mathbf M \phi_{j} = \lambda_{j}^{-1} \mathbf K \phi_{j} $$

$$ \phi_{i}^T \mathbf M \phi_{j} = \lambda_{j}^{-1} \phi_{j}^{-1} \mathbf K \phi_{j} $$

Giving us:

=> $$ \phi_{i}^T \mathbf K \phi_{j} = \lambda_{j} \delta_{ij} $$

$$ \mathbf d(t) = \sum_{i=1}^{n} \xi_{i}(t) \phi_{i} $$

$$ \mathbf M (\sum_{j} \xi_{j} \underline \phi_{j}) + \mathbf K(\sum_{j} \xi_{j} \underline \phi_{j}) = \underline F $$

Where,

$$ \sum_{j} \xi_{j} \underline \phi_{j} = \mathbf \ddot d $$

$$ \sum_{j} \xi_{j} \underline \phi_{j} = \mathbf \dot d $$

$$ \sum_{j} \xi_{j}(\phi_{i}^{T} \mathbf M \phi_{j}) + \sum_{j} \xi_{j} \phi_{i}^{T} \mathbf K \phi_{j} = \phi_{i}^{T}F $$

Finally Giving:

$$ => \ddot \xi_{i} + \lambda_{i} \xi_{i} = \phi_{i}^T \underline F $$, where $$ i=1,....,n $$

=Wikiversity vs E-learning=

=MATLAB=

Two-Bar Tapered Truss
The class was presented with a two-bar truss with tapered elements. The material properties are applied at the nodes instead of an average value applied over the length of the element. The general set up for the problem is presented below.





The class was asked to compute the solution to the two-bar truss with tapered elements and to plot and compare the deformed shape to that of a two-bar truss with average material properties applied across the entire length of the element.

Two Bar Truss General Vs. Average K Matrices with Undeformed and Deformed Shapes


In the image above the red line represents the original “undeformed” shape of the Two Bar truss system. The green line represents the deformed shape of the truss system using the “general” stiffness matrix, and the blue line represents the deformed shape of the truss system using the “average” stiffness matrix.

Electric Pylon as a Frame
The electric pylon structure from before is now analyzed as a frame. The following procedures are to be completed:


 * 1) Update Matlab code to accommodate frame elements
 * 2) Plot the undeformed and deformed structure
 * 3) Identify the frame element(s) with the highest nodal bending moment and shear force
 * 4) Is the frame model statically determinate?
 * 5) Find the lowest 3 eigenpairs
 * 6) Plot the eigenvectors

Recall the properties of the pylon:

Young's Modulus: 200 GPa

Moment of inertia: 1.33 x 10-8m4

Cross-sectional area: 4 x 10-4m2

Additional Code Required
Length.m ReduceM.m ReduceS.m PlaneTrussElement.m NodalSoln.m PlaneTrussResults.m

PlaneFrameResults.m

2) Plot of Undeformed and Deformed Truss and Frame Structures


The red represents the Pylon as a frame approximation while the cyan is representative of the truss modeling.

Code output
Per the Matlab output:

max_moment = 6.2872 max_momentele = 81 max_shear = 5.4826 max_shearele = 81

The highest moment is 6.29 N-m and occurs in element 81. The highest shear stress (transverse) is 5.48Pa and also occurs in element 81.

4) Statically determinate?
This problem is not statically determinate. The two-force member assumption does not hold and therefore there are more unknowns than available equations.

5) Lowest Theee Eigenpairs
Per the Matlab output, the lowest three eigenpairs are:

lam = 132.86      2471.2       2942.6

6) Plot of the Three Corresponding Eigenvectors






=Contributing Members= Eml4500.f08.ateam.boggs.t 15:21, 9 December 2008 (UTC)

Eml4500.f08.ateam.nobrega 14:50, 9 December 2008 (UTC)

Eml4500.f08.ateam.mcnally 15:49, 9 December 2008 (UTC)

Eml4500.f08.ateam.rivero 15:50, 9 December 2008 (UTC)

Eml4500.f08.ateam.carr 17:16, 9 December 2008 (UTC)

Eml4500.f08.ateam.shah 20:43, 9 December 2008 (UTC)