User:Eml4500.f08.gravy.jad/Notes5.g

 See my comments below. Please don't remove these comments in case you amend your work; just add a comment in these comment boxes to say what you did. Eml4500.f08 12:56, 10 November 2008 (UTC) The 2-bar truss code was corrected and the error was explained. This is a link to the orginal page.Orginal and here is the differences between the pages Differences Eml4500.f08.gravy.mmm 00:26, 12 November 2008 (UTC)

FEA Notes and Homework 5

Principal of virtual work
Justification of eliminating rows 1,2,5,6 to obtain K2x2 in 2-bar truss:

as described in an earlier report Elimination of dofs

FD relation $$\mathbf{K}_{2X2} \mathbf{d}_{6x1}= \mathbf{F}_{6x1} $$

becomes

$$\mathbf{Kd-F}= \mathbf{0}_{6x1}$$  (Equation 1)

The Principal of virtual work gives the equation :$$ \mathbf{W}_{6x1} \cdot \overbrace{( \mathbf{Kd-F}))}^{6x1} = 0_{1x1}$$(Equation 2)

Is true for all W values Where $$\mathbf{W}_{6x1}$$ is the weighting matrix.

So (Equation 1)$$\iff$$ (Equation 2)

Proof:

A)(Equation 1)$$\Rightarrow$$ (Equation 2) is trivial

B)To prove (Equation 1)$$\Rightarrow$$ (Equation 2) we must choose different values for W

Choice 1: Select 'W' such that W1=1, W2=W3=W4=W5=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 1[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j= \mathbf{F}_1 $$ (First Equation)

Choice 2: Select 'W' such that W2=1, W1=W3=W4=W5=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 1[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j= \mathbf{F}_2 $$ (Second Equation)

Choice 3: Select 'W' such that W3=1, W1=W2=W4=W5=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 1 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j= \mathbf{F}_3 $$ (Third Equation)

Choice 4: Select 'W' such that W4=1, W1=W2=W3=W5=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 1 & 0 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 1 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j= \mathbf{F}_4 $$ (Fourth Equation)

Choice 5: Select 'W' such that W5=1, W1=W2=W3=W4=W6=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 1 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 0 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j= \mathbf{F}_5 $$ (Fifth Equation)

Choice 6: Select 'W' such that W6=1, W1=W2=W3=W4=W5=0 so

$$\mathbf{W^T}= \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$

$$ \mathbf{W} \cdot ( \mathbf{Kd-F} ) = 0[ \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j- \mathbf{F}_1] + 0[ \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j- \mathbf{F}_2] + 0 [ \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j- \mathbf{F}_3] + 0 [ \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j- \mathbf{F}_4] + 0 [ \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j- \mathbf{F}_5] + 1 [ \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j- \mathbf{F}_6]  $$

so the result becomes  $$\sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j= \mathbf{F}_6 $$ (Sixth Equation)

Hence the equation becomes $$\mathbf{Kd}= \mathbf{F} $$ which is Equation 1 = For the Principal of virtual work when you account for the boundary conditions d1 =  d2 =  d5=  d6=0 The $$\mathbf{K}$$  matrix becomes the  $$\mathbf{\overline{K}}$$  matrix.

The weighing coefficients must be "kinetically admissible" which means they cannot violate the boundary conditions so W1 = W2 =  W5=  W6=0  where the weighing coefficients represent the virtual displacement.

The $$\overline{ \mathbf{K}}$$ is created as described in an eariler report Elimination of dofs

$$\mathbf{W}_{6x1} \cdot (

\underbrace{\mathbf{Kd}}_{\Downarrow}-\mathbf{F}) $$

$$\begin{bmatrix} && \\&& \end{bmatrix}_{6x2} -\begin{Bmatrix} \\ \\ \end{Bmatrix}_{2x1}

$$

$$= \begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} (\mathbf{\overline{K}_{2x2} \overline{d}_{2x1}- \overline{F}_{2x1} })$$ (Equation 3)

for all values of $$\begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} $$

with $$\mathbf{\overline{K}}=\begin{bmatrix} K_{33} && K_{34} \\ K_{43} && K_{44}\end{bmatrix} $$

$$\mathbf{\overline{d}}= \begin{Bmatrix}d_{3} \\ d_{4} \end{Bmatrix} $$

$$ \overline{F } = \begin{Bmatrix}F_{3} \\ F_{4} \end{Bmatrix} $$

Fixing the bug in the 2-bar truss code
There was a bug in the code solving the 2-bar truss solved in and earlier report 2-bar truss. The code was originally written for a 6-bar truss and was modified for the 2-bar truss. Unfortunately an error was created. The 2-bar truss had different values for the E's and A's for the elements,but the 6-bar truss had the same E's and A's for all of the elements. This was fixed by substituting the e and A in the following for loop with e(i) and A(i):

 The above array "results" was incomplete; it showed only the first column of this array. The reason was because this team used the incomplete code PlaneTrussResults from Team Delta_6; see my comments in "The best of HW5". Eml4500.f08 12:59, 10 November 2008 (UTC)

The code was actually not incomplete, but the way it was written would not work in the code used for the two-bar truss. The following modifactions were made to PlaneTrussResults.m to make it work in the two-bar truss code:

This line was added to the end of the code: results = [eps, sigma, force];

And the "[eps, sigma, force]" in the first line of code was changed to "results". This gave the desired values for the results variable in the two-bar truss code, and the fixed results were changed in the code above. Eml4500.f08.gravy.sms 9:52, 10 November 2008 (UTC)

Principal of virtual work with axial dofs
The Principal of virtual work can also be expressed with the equation $$q^{(1)}_{2x1}= \underline{T}_{2x4} d^{(i)}_{4x1}$$

This is much like how FEA course grades are calculated, $$Grade= \alpha_o \cdot HW grade + \sum^3_{i=1} Exam_i $$

Deriving $$\mathbf{K}^{(e)}_{4x4}= \mathbf{T}^{(e)T}_{4x2} \mathbf{\hat{K}}^{(e)}_{2x2} \mathbf{T}^{(e)}_{2x4} $$

Remember from previous reports [http://en.wikiversity.org/w/index.php?title=User:Eml4500.f08.gravy.jad/Notes3.g&oldid=348085#Similarly_for_node_2_:_HW. Axial dofs equation]and Force Displacement

That:

$$\mathbf{\hat{K}}^{(e)}_{2x2} \mathbf{q}^{(e)}_{2x1}= \mathbf{p}^{(e)}_{2x1} \Longrightarrow \mathbf{\hat{K}}^{(e)} \mathbf{q}^{(e)}- \mathbf{p}^{(e)}= \mathbf{0}^{(1)}_{2x2}$$ (Equation 1)

$$ \mathbf{\hat{W}}_{2x1} \cdot \underbrace{(\mathbf{\hat{K}}^{(e)} \mathbf{q}^{(e)}- \mathbf{p}^{(e)}}_{2x1)}= \mathbf{0}_{1x1} $$ (Equation 2)

We showed (Equation 1)$$\iff$$ (Equation 2) earlier in this report

Recall: $$\mathbf{q}^{(e)}_{2x1}= \mathbf{T}^{(e)}_{2x4} \mathbf{d}^{(e)}_{4x1} $$

Similarly

$$ \mathbf{\hat{W}}^{(e)}_{2x1}= \mathbf{T}^{(e)}_{2x4} \mathbf{W}^{(e)}_{4x1} $$

$$\ \mathbf{\hat{W}}_{2x1} = \,$$virtual axial displacement, corresponding $$\ \mathbf{q}^{e}_{2x1} \,$$

$$\ \mathbf{W}_{4x1} = \,$$virtual displacement in global coordinate system, with corresponding $$\ \mathbf{d}^{e}_{4x1} \,$$

Now if you replace (Equation 3) and (Equation 4) into (Equation 2), the end result is:

$$\ \left ( \mathbf{T}^{(e)} \mathbf{W} \right ) \cdot \left\lbrack \hat{\mathbf{k}}^{(e)} \left ( \mathbf{T}^{(e)} \mathbf{d}^{(e)} \right ) - \mathbf{P}^{(e)} \right\rbrack = 0 \,$$for all $$\ \mathbf{W}_{4x1} \,$$ (Equation 5)

Recall: $$\ \left ( \mathbf{A}_{pxq} \mathbf{B}_{qxr} \right )^{T} = \mathbf{B}^{T} \mathbf{A}^{T} \,$$ (Equation 6)

As an Ex:

$$\ \mathbf{A}_{2x3} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}_{2x3} \,$$ $$\ \mathbf{B}_{3x3} = \begin{bmatrix} 7 & 8 & 9 \\ 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}_{3x3} \,$$

Proof of equation $$( \mathbf{T}^{(e)}W)\cdot [\mathbf{\hat{k}^{(e)}q^{(e)} - p^{(e)}}] = 0 $$
Show that the following equation is true:

$$( \mathbf{T}^{(e)}W)\cdot [\mathbf{\hat{k}^{(e)}q^{(e)} - p^{(e)}}] = 0 $$

Continuous case with Partial Differential Equations(PDE)
So far, we are dealing with the discrete case [not continuous][matrices], up until now...

NOW, the continuous case: (PDEs)

$$\ \Rightarrow \,$$Motivational Model Problem: Elastic bar with varying A(x), E(x), subjected to varying axial load (distributed) + concentrated load + inirtia force(dynamics) [the distributed and concentrated loads are both time dependent]





$$\ \sum F_{x} = 0 = -N(x,t) + N(x + dx, t) + f(x,t)dx - m(x)dx \ddot{u} \,$$

$$\ \sum F_{x} = \frac{dN}{dt} (x,t)dx + H.O.T + f(x,t)dx - m(x)dx \ddot{u} \,$$ (Equation 1)

Recall the Taylor series exp: $$\ f(x+dx) = f(x) + \frac{df}{dx}dx+ \frac{1}{2} \frac{d^{2}f}{dx^{2}}(x)dx+... \,$$from the third term onwards are are H.O.Ts = Higher Order Terms

(Equation 1) $$\ \Rightarrow \frac{dN}{dx} + f = m \ddot{u} \,$$ Equation of motion (EOM) (Equation 2)

$$\ N(x,t) = A(x) \sigma(x,t) \,$$where $$\ \sigma(x,t) = E(x) \epsilon(x,t) \,$$where $$\ \epsilon(x,t) = \frac{du}{dx}(x,t) \,$$ (Equation 3)

Combining (Equation 2) and (Equation 3) yields: $$\ \frac{d}{dx} \left \lbrack A(x)E(x) \frac{du}{dx} + f(x,t) = m(x) \ddot{u} \right \rbrack \,$$where $$\ \ddot{u} = \frac{d^{2}u}{dt^{2}} \,$$ (Equation 4)

(Equation 4) describes the PDE of motion.

Now, we need two boundary conditions (2nd order derivative with respect to x), AND two initial conditions (2nd derivative with respect to t) [initial displacement and initial velocity]



$$\ u(0,t) = 0 = u(L,t) \,$$



1)$$\ u(0,t)=0 \,$$, 2)$$\ N(L,t) = F(t) \,$$; $$\ A(L) \sigma (L,t) = A(L)E(L) \epsilon (L,t) \,$$where $$\ \epsilon (L,t) = \frac{du}{dx} (x,t) \,$$

$$\ \Rightarrow \frac{du}{dx} (L,t) = \frac{F(t)}{A(L)E(L)} \,$$@ x=l

Initial conditions: @$$\ t=0, u(x,0) = u(x) \,$$known function displacement, and $$\ \frac{du}{dx}(x,0) = \dot{u}(x,0)= \overline{V}(x) \,$$known function velocity

FEM Analysis 3-Bar Space Truss Using MATLAB
The following Matlab code was taken from the 3-bar space truss system found in the textbook on page 230. The code uses three other functions: SpaceTrussElement.m, NodalSoln.m, and SpaceTrussResults.m. These functions were obtained from the online resources located at the book's website here.

This space truss is found to be statically determinate. So taking the sum of the forces in the x,y and z direction yields the same reactions as as the FEA method. Which are

R1= 20375

R2= 13214

R3= -23148

Contributing Team Members
The following students contributed to this report:

Jessica Dyer Eml4500.f08.gravy.jad 09:04, 7 November 2008 (UTC)

Mikal McFarlane Eml4500.f08.gravy.mmm 15:21, 5 November 2008 (UTC)

Justin Tennant Eml4500.f08.gravy.JMT 22:28, 6 November 2008 (UTC)

Patrick Clayton Eml4500.f08.gravy.psc 06:48, 7 November 2008 (UTC)

Sean Sullivan Eml4500.f08.gravy.sms 10:45, 7 November 2008 (UTC)

Brian Eggeman Eml4500.f08.gravy.bje 15:59, 7 November 2008 (UTC)