User:Eml4500.f08.bike.viale/homework report 7

= Mtg 41 =

Use above boundary conditions to solve for $$c_0$$ through $$c_3$$

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

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

$$V'(x) = C_1 + 2C_2x + 3C_3x^2$$

$$V'(0) = C_1 = 0$$

(2) $$V'(L) = 2C_2L +3C_3L^2 = 0$$

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

(1) $$\Rightarrow 0 = 1 + C_2L^2 + (-\frac{2}{3}\frac{C_2}{L}) - L^3$$

Where the second, third, and fourth terms combine to be:

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

This is similar to the result for $$N_2$$ found previously.

For $$N_3$$

Boundary Condition

$$V(0) = V(L) = 0$$

$$\tilde{d}_3$$ rotational

$$V'(0) = 1, V'(L) = 0$$

For $$N_5$$:

$$V(0) = 0, V(L) = 1$$

$$V'(0) = 0, V'(L) = 0$$

For $$N_2$$:

$$V(0) = V(L) = 0$$

$$V'(0) = 0, V'(L) = 1$$

$$\tilde{d}_6$$ rotational

The plots for $$N_5 and N_6$$ where previously provided.

Now we must derive the coefficients in $$\mathbf{\tilde{k}}$$ ( the element stiffness matrix)

The coefficients with EA are already complete, however the ones with EI still need to be computed.

$$\tilde{k}_{22} = \frac{12EI}{L^3} = \int_{0}^{L}{\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}dx}$$

=Mtg 41=

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

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)

Analyzing the Modal problem from lecture 31 using discrete Principle of Virtual Work is as follows:

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

In the above equation, the boundary conditions are already applied and the K matrix represents the reduced stiffness matrix.

This equation is true for all $$\mathbf{\bar{w}}$$.

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

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

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

The above equations will be referred to as (1) for the rest of this discussion.

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

Now, solving the method for solving equation (1) is as follows:

1) Consider the unforced vibrations problem:

$$\mathbf{\bar{M}}_{nxn}\mathbf{\ddot{v}}_{nx1} + \mathbf{\bar{K}}_{nxn}\mathbf{v}_{nx1} = \mathbf{0}_{nx1}$$

This equation is equal to zero becuase it is unforced. This will be referred to as equation (2) from now on.

Assume:

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

Where the phi matrix is not time dependent. This solution gives the oscillating motions that have been seen in the FEA animations that have been shown.

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,

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

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

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

Where lambda is an eigenvalue.

A standard eigenvalue problem is given when the B matrix is equal to the identity matrix and the equation becomes:

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

This means that

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

Also the eigenpairs can be represented as:

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

with i = 1 through n

Now the 'ith' mode for the animation can be represented as:

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

for i = 1 through n

Now step 2 in the solution is as follows:

2) Modal superposition method:

By using the orthogonal properties of the eigenpairs we can equate:

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

This delta is referred to as the Kronecker delta.

This reduction is possible due to the mass orthogonality of the eigenvector.

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

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

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

where the left hand side is equal to the Kronecker delta, therefore:

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

$$\mathbf{\bar{d}}_{nx1}(t) = \sum{\zeta _i(t)\mathbf{\phi }_{inx1}}$$

This means 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 itself. 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. This means that this equation can be written as an ordinary differential equation as follows:

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

with i = 1 through n