User:EML4500.f08.A-team.VandenBerg/HW5

 Comment: This is HW5 located in the user namespace of EML4500.f08.A-team.VandenBerg 06:35, 30 October 2008 (UTC)

Gobal Force Displacement Relationship vs. Principle of Virtual Work
Justification of eliminating rows 1, 2, 5, 6 to obtain $$\mathbf{K}_{_{2x2}}$$ in original two-bar truss:

Global force displacement relation: $$\mathbf{K}_{_{6x6}} \cdot \mathbf{d}_{_{6x1}} = \mathbf{F}_{_{6x1}} \Rightarrow \mathbf{K} \mathbf{d} - \mathbf{F} = \mathbf{0}_{_{6x1}}$$ Eq.(1)

Using Principle of Virtual Work (PVW), (1) becomes: $$\mathbf{W}_{_{6x1}} \cdot \underbrace{(\mathbf{K}\mathbf{d} - \mathbf{F})}_{6x1} = 0_{1x1}$$Eq.(2) for all $$\mathbf{W}_{_{6x1}}$$ which is a "weighting matrix"

Proving Eq.(1) and Eq.(2) are interchangeable:

A)Eq.(1) $$\Rightarrow$$ (2): Trivial

B)Want to show (2) $$\Rightarrow$$ (1):

We must choose different values for $$\mathbf{W}$$

Choice 1:

Select $$\mathbf{W}$$ so that W1=1, W2 = W3 = W4 = W5 = W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{1j} \mathbf{d}_j= \mathbf{F}_1 $$

Choice 2:

Select $$\mathbf{W}$$ so that W1=0, W2 = 1, W3 = W4 = W5 = W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{2j} \mathbf{d}_j= \mathbf{F}_2 $$

Choice 3:

Select $$\mathbf{W}$$ so that W1= W2 = 0, W3 = 1, W4 = W5 = W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{3j} \mathbf{d}_j= \mathbf{F}_3 $$

Choice 4:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = 0, W4 = 1, W5 = W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{4j} \mathbf{d}_j= \mathbf{F}_4 $$

Choice 5:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = W4 = 0, W5 = 1, W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{5j} \mathbf{d}_j= \mathbf{F}_5 $$

Choice 6:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = W4 = W5 = 0, W6=1

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{K}_{6j} \mathbf{d}_j= \mathbf{F}_6 $$

Conclusion : $$\mathbf{K}_{_{6x6}} \mathbf{d}_{_{6x1}} = \mathbf{F}_{_{6x1}}$$

PVW : Accounting for Boundary Conditions
In the original two-bar truss system: d1 = d2 = d5 = d6 = 0.

Weighting coefficients must be "kinematically admissible", meaning they can not violate the boundary conditions, therefore:

$$\Rightarrow$$W1 = W2 = W5 = W6 = 0

These weighting coefficients represent the virtual displacement by calculus of variations

Now,

$$\mathbf{W} \cdot (\mathbf{Kd}-\mathbf{F})$$

Thus

$$= \begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} (\mathbf{\overline{K} \overline{d}- \overline{F} })$$ Eq.(3)

and is true for $$\begin{Bmatrix} W_3 \\ W_4 \end{Bmatrix} $$

With the matrices equating to $$\mathbf{\overline{K}}=\begin{bmatrix} K_{33} && K_{34} \\ K_{43} && K_{44}\end{bmatrix} $$

and:

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

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

being the reduced matrices.

--Eml4500.f08.a-team.robinson 10:47, 3 November 2008 (UTC)

Meeting 25: Wednesday, 22 October 2008. EML4500
No class due to Exam #2

Meeting 26: Monday, 27 October 2008. EML4500
Meeting 26 -- October 27th, 2008

26-1

What was discussed: -Debugging the 2-bar truss Matlab code -Further examination of PVW

Debugging the code:

The code for the two bar truss given to the class by Dr. Vu-Quoc had a slight bug in it since it is a modified version of program in the book for a six bar truss where the Young's Modulus (E) and the area (A) of each element was the same. As this was not the case for the two bar truss, the program had not been corrected to call each different element's properties and thus would not return proper results.

Fixing this code, however, is a fairly simple task and only requires the student to find the function call PlaneTrussResults(e,A) and add a (i) next to the e and the A.

Thus, the homework was given to debug the code, rerun the program, and interpret the results, especially if they differed from the original results. Also another homework assignment was given such that the student should run the code for the six bar truss and interpret those results as well. All Matlab assignments are shown below under the Matlab Section of this report.

After spending the better half of the class on this subject, the discussion returned to grinding through the PVW, which now follows:

26-2

Deriving:

$$\mathbf{k^{(e)}} = \mathbf{T^{(e)^T}}\mathbf{\hat{k}^{(e)}}\mathbf{T^{(e)}}$$

Reference was made to meeting 14 page 3, and also meeting 23 page 3.

Recall FD relation with axial dof's, or q(e).

$$\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} = \mathbf{p^{(e)}}$$

From this week get:

$$\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{p^{(e)}} = \mathbf{0}$$ which we label Eqn. (1)

Using the PVW:

$$\mathbf{\hat{W}} \cdot \left ( \mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{p^{(e)}} \right ) = \mathbf{0}$$ which we label Eqn. (2)

Which is true for all $$\mathbf{\hat{W}}$$

Earlier it was shown that a similar set of equations where equivalent, and thus by the same logic these equations [(1) and (2)] are also equivalent. Thus:

Proving Eq.(1) and Eq.(2) are interchangeable:

A)Eq.(1) $$\Rightarrow$$ (2): Trivial

B)Want to show (2) $$\Rightarrow$$ (1):

We must choose different values for $$\mathbf{W}$$

Choice 1:

Select $$\mathbf{W}$$ so that W1=1, W2 = W3 = W4 = W5 = W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{1j} \mathbf{q}_j= \mathbf{p}_1 $$

Choice 2:

Select $$\mathbf{W}$$ so that W1=0, W2 = 1, W3 = W4 = W5 = W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{2j} \mathbf{q}_j= \mathbf{p}_2 $$

Choice 3:

Select $$\mathbf{W}$$ so that W1= W2 = 0, W3 = 1, W4 = W5 = W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{3j} \mathbf{q}_j= \mathbf{p}_3 $$

Choice 4:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = 0, W4 = 1, W5 = W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{4j} \mathbf{q}_j= \mathbf{p}_4 $$

Choice 5:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = W4 = 0, W5 = 1, W6=0

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{5j} \mathbf{q}_j= \mathbf{p}_5 $$

Choice 6:

Select $$\mathbf{W}$$ so that W1= W2 = W3 = W4 = W5 = 0, W6=1

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

and

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

Thus, the result becomes

$$\Rightarrow \sum^6_{j=1} \mathbf{\hat{k}}_{6j} \mathbf{q}_j= \mathbf{p}_6 $$

Conclusion : $$\mathbf{\hat{k}}_{_{6x6}} \mathbf{q}_{_{6x1}} = \mathbf{p}_{_{6x1}}$$

Recall also that: $$\mathbf{q^{(e)}} = \mathbf{T^{(e)}}\mathbf{d^{(e)}}$$     Eqn. (3)

is interchangeable with:

$$\mathbf{\hat{W}} = \mathbf{T^{(e)}}\mathbf{W}$$    Eqn. (4)

Meeting 27: Wednesday, 29 October 2008. EML4500
$$ \hat{\underline{w}}_{2x1} {=}$$ virtual axial displacement, corresponding to $$ q^{(e)}_{2x1} $$

$$ \hat{\underline{w}}_{2x1} {=}$$ virtual axial displacement, corresponding to $$ q^{(e)}_{2x1} $$

$$ \underline{w}_{4x1} {=} $$ virtual displacement in global coordinate system, corresponding to $$ d^{(e)}_{4x1} $$

Replace Equations (3) and (4) into (2):

$$ \hat{\underline{w}}_{2x1} \cdot (\hat{\underline{k}}^{(e)} \underline{q}^{(e)} - \underline{p}^{(e)}) {=} 0_{1x1} $$ (2)

$$ \Rightarrow (\underline{T}^{(e)}\underline{w}) \cdot [\hat{\underline{k}}^{(e)}(\underline{T}^{(e)}\underline{d}^{(e)}) - \underline{p}^{(e)}] {=} 0 $$ for all $$ w_{4x1} $$ (5)

Recall: $$ (\underline{A}\underline{B})^T {=} \underline{B}^T\underline{A}^T $$ (6)

Recall: $$ \underline{a}_{nx1} \cdot \underline{b}_{nx1} {=} \underbrace{ \underline{a}^T_{1xn}\underline{b}_{nx1} }_{1x1 (scalar)} $$ (7)

Apply (7) and (6) in (5):

$$ \underbrace{(\underline{T}^{(e)} \underline{w})^T}_{(7)}[\underline{\hat{k}}^{(e)}(\underline{T}^{(e)}\underline{d}^{(e)}) - \underline{p}^{(e)}] {=} 0_{1x1} $$ for all $$ \underline{w}_{4x1} $$

$$ \Rightarrow \underbrace{\underline{w}^T\underline{T}^{(e)T}}_{(6)}[-] {=} 0 $$

$$ \Rightarrow \underbrace{\underline{w}}_{(7)} \cdot [(\underbrace{\underline{T}^{(e)T} \underline{\hat{k}}^{(e)} \underline{T}^{(e)} \underline{d}^{(e)}}_{\underline{k^{(e)}}}) - \underbrace{\underline{T}^{(e)T} \underline{p}^{(e)}}_{\underline{f}^{(e)}}] {=} 0 $$ for all $$ \underline{w}_{4x1} $$

$$ \Rightarrow \underline{w} \cdot [\underline{k}^{(e)}\underline{d}^{(e)} - \underline{f}^{(e)}] {=} 0 $$ for all $$ \underline{w} $$

$$ \Rightarrow \underline{k}^{(e)}\underline{d}^{(e)} {=} \underline{f}^{(e)} $$

So far, only discrete cases have been discussed. Now continuous cases will be discussed.

Motivational model problem: Elastic bar with varying A(x), E(x), subjected to varying axial load (distributed) and concentrated load and inertia force (dynamics). Concentrated load and inertia force are time dependent.

EML4500.f08.A-team.kirley 04:22, 30 October 2008 (UTC)

Meeting 28: Friday, 31 October 2008. EML4500


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

$$\Sigma$$ Fx = $$ \frac{\delta N}{\delta x} $$ dx + h.o.t. + f(x,t) dx - m(x) ,math> \ddot{u}, dx neglect h.o.t.

(h.o.t. = higher order terms)

Recall Taylor series expansion:

f(x+dx) = f(x) + $$ \frac{df(x)}{dx} $$ dx + $$ \underbrace{\frac{1}{2} \frac{d^2f(x)}{dx^2} dx^2 + ...}_{h.o.t.} $$

Eq. (1) $$ \Rightarrow \frac{\delta N}{\delta x} + f = m \ddot{u} $$ (2)

Equation of motion (EOM)

$$ N(x,t) = A(x) \underbrace{\sigma (x,t)}_{E(x) \underbrace{\epsilon (x,t)}_{\frac{\delta u}{\delta x} (x,t)}} $$ (3)

Constitutive relation

(3) in (2) yields

$$ \frac{\delta}{\delta x} [A(x)E(x) \frac{\delta u}{\delta x}] + f(x,t) = m(x) \underbrace{\ddot{u}}_{\frac{\delta^2 u}{\delta t^2}} $$

Need 2 b.c.s (2nd order derivative w.r.t. x)

2 initial conditions (2nd order derivative w.r.t. t)

(initial displacement, initial velocity)



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

$$ \Rightarrow \frac{\delta u}{\delta x} (L,t) = \frac{F(t)}{A(L)E(L)} $$



1) u(0,t) = 0

2)

$$ \underbrace{N(L,t)}_{A(t)\underbrace{\sigma (L,t)}_{E(L) \underbrace{\epsilon (L,t)}_{\frac{\delta u}{\delta x} (L,t)}}} = F(t)$$

Initial condition at t=0, prescribe

u(x,t=0) = $$\overline{u}$$(x) known function (displacement)

$$ \frac{\delta u}{\delta t} $$ (x,t=0) = $$ \dot{u} $$ (x,t=0) = $$ \overline{\nu} $$ (x)

known function velocity

Matlab and Related Homework
Here is the debugged code for the 2 bar truss:

It can be seen that there are 4 less values in the results matrix, which is appropriate for the 2 bar truss system. Previously the program had printed the correct values, and then spit out garbage for the final four since the code was bugged.

6 Bar Truss from pg. 226
This is the original 6 bar truss code.

Below is a plot of the deformed and un-deformed 6 bar truss. The values for the deformation have been magnified by a factor of 5000 to make the deformation of the truss glaringly obvious.

The following is the code used to create this masterpiece.

Next is a modified code for the 6 bar truss, this time with varying values for the Young's Modulus (E) for each of the elements of the truss

Following is the plot for the modified 6 bar truss. This plot has also had the deformation values magnified by a factor of 5000 for glaring obviousness.

The following is the code used to create the plot.

3 Bar Space Truss From pg 230
Below is the Matlab code from the book that calculates the results for the 3 bar space truss.

Next up are the plots of the truss from several viewpoints of the graph.

This one is from the vantage point of (-2,-2,3)

And now looking at the XY plane:

And looking the XZ plane:

And finally facing the YZ plane:

Below is the code used to generate the graph

Unfortunately, this 3 bar space truss happens to be statically determinate. Personally I'd rather not be bothered to show you how I came up with that other than there are enough equations to cover all of the unknowns. But I suppose I'll prove it out for your benefit (and my own, I suppose) with a section titled....

Thank God For Vector Calculus

First off, let's observe the problem and notice that there are three reaction forces (which from here on out will be know generically as R1, R2, and R3. Also notice that there is a single applied force on this system, which will be known as P.  Next we can divide up each of the three reaction forces into three partial reactions each in the conventional $$\hat{i}$$,  $$\hat{j}$$, and  $$\hat{k}$$ unit vector directions.  Algebraically this looks like:

$$\mathbf{R_1} = \mathbf{R_{1x}\hat{i}} + \mathbf{R_{1y}\hat{j}} + \mathbf{R_{1z}\hat{k}}$$

$$\mathbf{R_2} = \mathbf{R_{2x}\hat{i}} + \mathbf{R_{2y}\hat{j}} + \mathbf{R_{2z}\hat{k}}$$

$$\mathbf{R_3} = \mathbf{R_{3x}\hat{i}} + \mathbf{R_{3y}\hat{j}} + \mathbf{R_{3z}\hat{k}}$$

With a bit of observation of the system, it is seen that $$\mathbf{R_{3x}\hat{i}}$$ and $$\mathbf{R_{3y}\hat{j}}$$ drop out to zero since there are no forces that can be applied in those directions on this element. This is not a necessary step, but greatly simplifies the calculations. Next up we must define each of the partial reactions in order to fully setup the problem. With some knowledge of vectors from calculus, it is quite simple to derive the each of the partial reactions, as seen below:

$$\mathbf{R_{1x}} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1}$$

$$\mathbf{R_{1y}} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1}$$

$$\mathbf{R_{1z}} = \mathbf{R_1}\sin{\phi_1}$$

$$\mathbf{R_{2x}} = \mathbf{R_2}\cos{\phi_2}\cos{\theta_2}$$

$$\mathbf{R_{2y}} = \mathbf{R_2}\cos{\phi_2}\sin{\theta_2}$$

$$\mathbf{R_{2z}} = \mathbf{R_2}\sin{\phi_2}$$

$$\mathbf{R_{3z}} = \mathbf{R_3}$$

Note that R3 does not have any x or y direction values since it begins at the origin and extends in the positive z direction. Next thing to be done is to calculate the theta's and phi's in order to finish setting up the problem. Before that happens however, we must the length between node 1 and the origin, which I'll generically call "a" and the length between node 2 and the origin, which I'll generically call "b". Solving for these values is as easy as using the Pythagorean Theorem, such as follows:

$$a = \sqrt{0.96^2 + 1.92^2} = 2.146625$$

$$b = \sqrt{1.44^2 + 1.44^2} = 2.036468$$

Now we are in a position to calculate all of the angles that we will need. Thus,

$$\tan{\theta_1} = \frac{1.92}{0.96}$$

$$\tan{\theta_2} = \frac{1.44}{-1.44}$$

$$\tan{\phi_1} = \frac{2}{2.146625}$$

$$\tan{\phi_2} = \frac{2}{2.036468}$$

Therefore,

$$\theta_1 = 63.435^\circ$$

$$\theta_2 = -45^\circ$$

$$\phi_1 = 42.975^\circ$$

$$\phi_2 = 44.482^\circ$$

Setting up the 3 summation of the force equations we get:

$$\mathbf{F_x} = \mathbf{R_1}\cos{\phi_1}\cos{\theta_1} + \mathbf{R_2}\cos{\phi_2}\cos{\theta_2} = 0$$

$$\mathbf{F_y} = \mathbf{R_1}\cos{\phi_1}\sin{\theta_1} + \mathbf{R_2}\cos{\phi_2}\sin{\theta_2} - \mathbf{P} = 0$$

$$\mathbf{F_z} = \mathbf{R_1\sin{\phi_1}} + \mathbf{R_2\sin{\phi_2}} + \mathbf{R_3} = 0$$

And after a small bit of rearranging and plugging in (or plugging and using a bit of linear algebra) which I'll leave up to the reader, you will get:

$$\mathbf{R_1} = -20374.65877$$

$$\mathbf{R_2} = -13214.4037$$

$$\mathbf{R_3} = 23148.11723$$

These values match up nearly perfectly to the Matlab's code, and are really only off by about a few hundreths of a point (due to rounding of the decimals on the lengths and the angles). However, you may notice that the values seem to have the opposite signs than what the code gives. You must remember that the code solves for the axial forces in the element, and while the absolute values of these numbers are the same, the signs are not. Simply put, the reactions with negative (-) signs are in tension and conventionally listed as positive. Also, the reaction with the positive (+) sign is in compression, which is conventionally listed as negative.

Contributions
--EML4500.f08.A-team.kirley 04:22, 30 October 2008 (UTC)

--Eml4500.f08.a-team.robinson 10:47, 3 November 2008 (UTC)

--EML4500.f08.A-team.melvin 18:22, 6 November 2008 (UTC)

--EML4500.f08.A-team.rieth 04:25, 7 November 2008 (UTC)

--EML4500.f08.A-team.morford 20:06, 7 November 2008 (UTC)

--EML4500.f08.A-team.VandenBerg 21:57, 7 November 2008 (UTC)