User:Eml4500.f08.a-team.robinson

Homework 3 Contributions

Homework 4 Contributions

Using Global Free Body Relationships to Compute Reactions
Using force-displacement relations from meeting 10:

(6x2 matrix)x(2x1 mat.)=(6x1 mat.):

$$\begin{bmatrix} K_{13} &  K_{14} \\ K_{23} &  K_{24} \\ K_{33} &  K_{34} \\ K_{43} &  K_{44} \\ K_{53} &  K_{54} \\ K_{63} &  K_{64} \\ \end{bmatrix} \begin{Bmatrix} d_{3}\\ d_{4}\\ \end{Bmatrix} = \begin{Bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}\\ \end{Bmatrix} $$

Only necessary to do computations for rows 1,2,5,6 to get F1,F2,F5,F6 (reactions)

What "the Loop" is
Clarifying the loop between the F.E.M. and statics:

For two-bar truss system:

FEM --> Compute displacement --> Compute reactions

Statics --> Compute reactions --> Compute displacement

The step in statics from computing reactions to computing displacement is what we define as "closing the loop"

Closing the Loop
FBD's of two-force bodies (elem.'s 1,2):

Figure 8: Axes Forces in Elem. 1

Figure 9: Axes Forces in Elem. 2

By statics, reactions known and therefore the member forces: P1(1), P2(2)

Computing axial displ. dof's (amount of extension of bars):

$$q2^1 = \frac{P1^1}{k^1}=\frac{P2^1}{k^1} $$, $$q1^2 = \frac{-P2^2}{k^2}$$

Fixed node at 1 & 2 means: q1^2 = q2^2 = 0

To back out from above results the displ. dof's of node 2:



Infinitesimal displacement
Using same two-bar picture from meeting 15:



AC = \frac{|P_{1}^{(1)}|}{k^{(1)}} =\frac{5.1243}{3/4} = 6.8324 $$



AB = \frac{|P_{1}^{(2)}|}{k^{(2)}} =\frac{6.276}{5} = 1.2552 $$

Next, must solve for (x, y) coordinates of B & C:



X_{B} = AB \cdot \cos{(135)} = 1.2552\cos{(135)} = -.88756 $$



Y_{B} = AB \cdot \sin{(135)} = 1.2552\sin{(135)} = .88756 $$



X_{C} = AC \cdot \cos{(30)} = 6.8324\cos{(30)} = 5.91703 $$



Y_{C} = AC \cdot \sin{(30)} = 6.8324\sin{(30)} = 3.4162 $$

Other unknowns are XD and YD. You need an equation for lines AB & BC





PQ = (PQ)i^{~} = (X-X_{P})i + (Y-Y_{P})j $$



(PQ)i^{~} = PQ[ \cos{\theta}i + \sin{\theta}j ] $$

Therefore



X - X_{P} = PQ\cdot \cos{\theta} $$



Y - Y_{P} = PQ\cdot \sin{\theta} $$



\frac {Y-Y_{P}}{X-X_{P}} = \tan{\theta} $$



Y-Y{P} = (\tan{\theta})\cdot(X-X_{P}) $$

Equation for a line perpendicular to above line, passing P:



Y-Y_{P} = \tan{(\theta + \frac{\pi}{2})}(X-X{P}) $$

By definition:

AD = d3i + d4j

Meeting 21: Monday, 13 October 2008. EML4500
Eigenvalue problem: $$\mathbf{kv} = \lambda \mathbf{v}$$

Let $$\left\{\mathbf{u_1}, \mathbf{u_2}, \mathbf{u_3}, \mathbf{u_4} \right\}$$ be the pure eigenvectors corresponding to the 4 zero eigenvalues:

$$\mathbf{ku_i} = 0\mathbf{u_i} = \mathbf{0}\left(i = 1,2,3,4 \right)$$

Linear combination of $$\left\{u_i, i = 1,2,3,4 \right\} $$

$$\sum_{i=1}^{4}{\alpha_{i_{1x1}}\mathbf{u}_{i_{6x1}}=:\mathbf{W}_{_{6x1}}}=\mathbf{0}_{_{1x1}}\cdot \mathbf{W}_{_{6x1}}$$

Where $$=:$$ means "equal by definition"

$$\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \equiv \mathbf{W} $$ (equal by definition)

$$ \alpha_i \Rightarrow $$(real numbers)

W is also an eigenvector corresponding to a zero eigenvalue:

$$ \mathbf{kW} = \mathbf{k}\left(\sum_{i=1}^{4}{\alpha_i\mathbf{u}_i} \right) $$

$$= \sum_{i=1}^{4}{\alpha_i\left(\mathbf{ku_i} \right)} = \mathbf{0} = 0 \times \mathbf{W}$$

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.

Integration by Parts: r(x), s(x)
$$ (rs)' = r's + rs' $$

$$ r' = \frac{dr}{dx} \quad, \quad s' = \frac{ds}{dx}$$

$$ \underbrace{\int(rs)'}_{rs} = \int r's + \int rs' $$

$$\Rightarrow \int r's = rs - \int rs' $$

Recall continuous PVW (Eqn. (3) from meeting 29)


 * 1st term: $$ \qquad r(x) = (EA)\frac{\partial u}{\partial x} $$
 * 2nd term: $$ \qquad s(x) = W(x) $$

By integration by parts:

$$ \int\limits_{x=0}^{x=L} \underbrace{W(x)}_{s} \frac{\partial }{\partial x} \underbrace{\left [(EA) \frac{\partial u}{\partial x}\right ]}_{r} dx = \left [W (EA) \frac{\partial u}{\partial x}  \right]_{x=0}^{x=L} - \int\limits_{0}^{L}\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx $$

$$ \Rightarrow W(L) \underbrace{(EA)(L)\frac{\partial u}{\partial x}(L,t)}_{F=N(L,t)} - W(0) \underbrace{(EA)(0)\frac{\partial u}{\partial x}(0,t)}_{N(0,t)}- \int\limits_{0}^{L}\frac{dW}{dx}(EA)\frac{\partial u}{\partial x}dx $$

Model Problem
Image

At x = 0, select W(x) so that W(0) = 0, meaning the system is kinematically admissible.

Motivation: Discrete PVW applied to the equation shown below:

$$ \mathbf{W}_{6x1}\left(\left[ \mathbf{K} \right]_{6x2}\begin{Bmatrix} d_{3} \\ d_{4}\end{Bmatrix}_{2x1}-\mathbf{F}_{6x1}\right)=\mathbf{0}_{1x1} $$ for all $$ \mathbf{W} $$

where $$ \mathbf{F}^{T}=\left[F_{1 }F_{2} F_{3} F_{4} F_{5} F_{6}\right] $$

$$ F_{3}, F_{4} $$ are known reactions and $$ F_{1}, F_{2}, F_{5}, F_{6} $$ are unknown reactions.

Since $$ \mathbf{W} $$ can be selected arbitrarily, select $$ \mathbf{W} $$ such that $$ W_{1}=W_{2}=W_{5}=W_{6}=0 $$ so to eliminate equations involving unknown reactions $$\Rightarrow$$ This eliminates rows 1, 2, 5, and 6. The result is shown below:

$$ \mathbf{K}_{2x2}\mathbf{d}_{2x1}=\mathbf{F}_{2x1} \Rightarrow$$ Equation (1)

Note: $$ \mathbf{W}\left(\mathbf{K}\mathbf{d}-\mathbf{F}\right)=0 $$ for all $$ \mathbf{W} $$

Back to the continuous principle of virtual work

The unknown reaction is:

$$ N(0,t)=(EA)(0)\frac{\partial u}{\partial x}(0,t) $$

Due to continuous PVW, the equation then becomes:

$$ W\left(L\right)F(t)-\int_{0}^{L}\frac{\partial W}{\partial x}(EA)\frac{\partial u}{\partial x}dx+\int W(x)\left[f-m\ddot{u}\right]dx=0 $$ for all $$ W\left(x\right) $$ such that $$ W\left(0\right)=0 $$

The final equation can be written as:

$$ \int_{0}^{L}W\left(m\ddot{u}\right)dx+\int_{0}^{L} \frac{\partial W}{\partial x}(EA)\frac{\partial u}{\partial x}dx=W(L)F(t)+\int_{0}^{L}Wfdx $$ for all $$ W\left(x\right) $$ such that $$ W\left(0\right)=0 $$

Verifying Dimensions and Element Force Displacement Relations
Using the definition shown above, the following dimensional analysis shows the rotational degrees of freedom are dimensionless. $$\left[\theta \right]=\frac{\left[AB \right]}{\left[R \right]}=\frac{L}{L}=1$$

A dimensional analysis can also be done on other important parameters:

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

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

$$ \left[A\right] = L^2, [I] = L^4 $$

$$ \left[\frac{EA}{L}\right] = [\tilde{k}_{11}] = \frac{\frac{F}{L^2}{L^2}}{L} = \frac{F}{L} $$

$$ \left[\tilde{k}_{11}\tilde{d_1}\right] = [\tilde{k}_{11}][\tilde{d_1}] = F $$

$$ \left[\tilde{k}_{23}\tilde{d_3}\right] = [\tilde{k}_{23}][\tilde{d_3}] $$ where $$ \tilde{d_3} = 1 $$

$$ = \frac{\left[6\right][E][I]}{L^2} $$

$$ = \frac{\left(1\right)\left(\frac{F}{L^2}\right)(L^4)}{L^2} = F $$

Verifying Dimensions of $$ \tilde{k}_{ij} $$ and $$ \tilde{d}_{j} $$ for $$ i $$ and $$ j $$ = 1,...,6:
The matrix $$ \tilde{k_{ij}}\tilde{d_j} $$ is shown below

$$ \begin{bmatrix}\frac{EA}{L} & 0 & 0 & \frac{-EA}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{4EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{2EI}{L} \\ \frac{-EA}{L} & 0 & 0 & \frac{EA}{L} & 0 & 0 \\ 0 & \frac{-12EI}{L^{3}} & \frac{-6EI}{L^{2}} & 0 & \frac{12EI}{L^{3}} & \frac{-6EI}{L^{2}} \\ 0 & \frac{6EI}{L^{2}} & \frac{2EI}{L} & 0 & \frac{-6EI}{L^{2}} & \frac{4EI}{L} \end{bmatrix} $$

The dimension of each term in the $$ \tilde{k_{ij}}\tilde{d_j} $$ matrix was calculated just as the above terms were in the previous notes and the resulting matrix is found to be:

$$ \begin{bmatrix}\frac{F}{L} & 0 & 0 & \frac{-F}{L} & 0 & 0 \\ 0 & \frac{F}{L} & F & 0 & \frac{-F}{L} & F \\ 0 & F & FL & 0 & -F & FL \\ \frac{-F}{L} & 0 & 0 & \frac{F}{L} & 0 & 0 \\ 0 & \frac{-F}{L} & -F & 0 & \frac{F}{L} & -F \\ 0 & F & FL & 0 & -F & FL \end{bmatrix} $$

The element force displacement relation in global coordinates from the element force displacement relation in local coordinates yields:

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

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

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

$$ \begin{bmatrix} \tilde{d}_{1}\\ \tilde{d}_{2}\\ \tilde{d}_{3}\\ \tilde{d}_{4}\\ \tilde{d}_{5}\\ \tilde{d}_{6} \end{bmatrix} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0 & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & l^{(e)} & m^{(e)} & 0 \\ 0 & 0 & 0 & -m^{(e)} & l^{(e)} & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6} \end{bmatrix} $$

The derivation of $$ \tilde{K}^{(e)} $$ from PVW, focusing only on bending effects:

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

Vibrational Analysis Using FEA
$$\hat{k}_{23} = \frac{6EI}{L^2} = \int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_3}{dx^2}dx}$$

In general,

$$\hat{k}_{ij} = \int_{0}^{L}{\frac{d^2N_i}{dx^2}(EI)\frac{d^2N_j}{dx^2}dx}$$

With i,j = 2,3,5,6

Elastodynamics (trusses, 2-D frames, 3-D elasticity)

Using modal problem from lecture 31 using discrete Principle of Virtual Works:

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

In the above equation, the boundary conditions are already applied

The following equation is true for all $$\mathbf{\bar{w}}$$:

$$\mathbf{\bar{M}}\mathbf{\ddot{\bar{d}}} + \mathbf{\bar{k}}\mathbf{\bar{d}} = \mathbf{\bar{F}}(t)$$

where $$\mathbf{\bar{d}}(0) = \mathbf{\bar{d}}_0$$

and $$\mathbf{\dot{\bar{d}}}(0) = \mathbf{\bar{V}}_0$$

The above will be referred to as (1)

These are the complete ordinary differential equations (ODEs), second order in time, and initial conditions governing the elastodynamics of the discretized continuous problem with multiple degrees of freedom.

Solving equation (1):

1) Consider the unforced vibrations problem:
$$\mathbf{\bar{M}}_{nxn}\mathbf{\ddot{v}}_{nx1} + \mathbf{\bar{K}}_{nxn}\mathbf{v}_{nx1} = \mathbf{0}_{nx1}$$

Unforced meaning the equation is equal to zero. This will be referred to as equation (2)

Assume: $$\mathbf{v}(t)_{nx1} = (sinwt)\mathbf{\phi _{nx1}}$$

Where the phi matrix is not time dependent

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

This means equation (2) becomes:

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

Therefore:

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

Which is the generalized eigenvalue problem, as it is of the form:

$$\mathbf{A}\mathbf{x} = \lambda \mathbf{B}\mathbf{x}$$

With lambda as the eigenvalue.

A standard eigenvalue problem: $$\mathbf{A}\mathbf{x} = \lambda \mathbf{x}$$ is given when the B matrix is equal to the identity matrix.

This means that

$$\lambda =\omega ^2$$ is an eigenvalue.

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

with i = 1 through n

Now the mode i for the animation can be represented as:

$$\mathbf{v}_i(t) = (sinw_it)\mathbf{\phi _i}$$

for i = 1 through n

2) Modal superposition method:
Using orthogonal properties of the eigenpairs:

$$\mathbf{\phi _i}^T_{1xn}\mathbf{\bar{M}}_{nxn}\mathbf{\phi }_{nx1} = \delta _{ij} = \begin{cases} & \text{1 if } i=j \\ & \text{0 if } i\neq j \end{cases}$$

This delta is the Kronecker delta.

Mass orthogonality of the eigenvector.

Now, applying this to Eq (1) and (2) gives:

$$\mathbf{\bar{M}}\mathbf{\phi }_j = \lambda _j \cdot \mathbf{\bar{k}}\mathbf{\phi }_j$$

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

Therefore:

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

Equation (1) can be written as:

$$\mathbf{\bar{M}}(\sum_{j}^{}{\ddot{\zeta }_j\mathbf{\phi }_j}) + \mathbf{\bar{K}}(\sum_{j}^{}{\zeta _j\mathbf{\phi }_j}) = \mathbf{F}$$

In the above equation, the first term in parenthesis is equal to the second derivative of the d matrix, while the second term in parenthesis is equal to the d matrix.

We can also write:

$$\sum_{j}^{}{\ddot{\zeta }_j(\mathbf{\phi }_i^T\mathbf{\bar{M}}\mathbf{\phi }_j} + \sum_{j}^{}{\zeta _j(\mathbf{\phi }_i^T\mathbf{\bar{K}}\mathbf{\phi }_j} = \mathbf{\phi }_i^T\mathbf{F}$$

The first term in parenthesis in this equation is equal to the Kronecker delta, while the second term in parenthesis is equal to an eigenvalue times the Kronecker delta.

Finally,

$$\Rightarrow \ddot{\zeta } + \lambda _i\zeta _i = \mathbf{\phi }_i^T\mathbf{F}$$

with i = 1 through n