User:Eml4500.f08.team.figuerrez/hw7/meeting41

Meeting 41 - Derivation Elastodynamics (Combining FEA & Vibrations)
$$\tilde{K_{23}}=\frac{6EI}{L^{2}}=\int_{0}^{L}\frac{\partial ^2N_{2}}{\partial x^{2}}(EI)\frac{\partial ^2N_{3}}{\partial x^{2}}dx$$

In general, $$\tilde{K_{ij}}=\int_{0}^{L}\frac{\partial ^2N_{i}}{\partial x^{2}}(EI)\frac{\partial ^2N_{j}}{\partial x^{2}}dx$$, where $${\boldsymbol{i,j=2,3,5,6}}$$

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

From meeting 31: Model problem

Discrete PVW (Principle of Virtual Work):

$${\boldsymbol{w\left[M\ddot{d}+KJ-F\right]=0}}$$

Note: The boundary conditions are already applied to w, and K is the reduced stiffness matrix. The equation then becomes:

$${\boldsymbol{M\ddot{d}+Kd=F(t)}}$$

$${\boldsymbol{d(0)=d_{0}}}$$, where 0 is unforced

$${\boldsymbol{\dot{d}(0)=v_{0}}}$$

Complete ordinary differential equations (ODEs)

2nd order in time plus inital conditions governing the elastodynamics of discrete-fixed cont. problems (MDoF)

Solving Equation (1):

1.) Consider unforced vibration problem.

$$M_{nxn}\ddot{v}_{nx1}+K_{nxn}v_{nx1}=0_{nx1}$$

Now, assume that $${\boldsymbol{v(t)_{nx1}=(sinwt)_{nx1}\phi}}$$, where $$\phi$$ is not time dependent.

Then, taking the double derivative of the function yields:

$${\boldsymbol{\ddot{v}=-w^{2}sinwt\phi}}$$

$${\boldsymbol{-w^{2}sinwtM\phi+sinwtK\phi=0}}$$

$${\boldsymbol{K\phi=w^{2}M\phi}}$$

Generalized Eigen value problem:

$${\boldsymbol{Ax = \lambda Bx}}$$, where lambda is an eigenvalue.

standard eigenvalue problem: $${\boldsymbol{Ax = \lambda x}}$$

This means that:

$${\boldsymbol{\lambda = \omega ^{2}}}$$ is an eigenvalue and $${\boldsymbol{(\lambda _i,\mathbf{\phi _i})}}$$ is the eigenpairs with i = 1 through n.

The ith mode animation can be represented as:

$${\boldsymbol{v_i(t)=(sinw_it)\phi _i}}$$, where i = 1 through n

2.) Model superposition method (using orthogonal properties of the eigenpairs):

$$\phi _{i}^{T} M \phi = \delta _{ij} = 1$$, if $$i=j$$

$$\phi _{i}^{T} M \phi = \delta _{ij} = 0$$, if $$i\neq j$$

Where $${\boldsymbol{\delta _ij}}$$ 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) yields:

$${\boldsymbol{M \phi _{j} = \lambda _{j}^{-1} \cdot K \phi _{j}}}$$

$${\boldsymbol{\phi _{i}^{T} M \phi _{j} = \lambda _{j}^{-1} \phi _{i}^{T} K \phi _{j}}}$$, where $${\boldsymbol{\phi _{i}^{T} M \phi _{j}}}$$ is equal to the Kronecker delta, therefore:

$${\boldsymbol{\phi _{i}^{T} K \phi _{j} = \lambda _{j} \delta _{ij}}}$$

$${\boldsymbol{d_{nx1}(t) = \sum{\zeta _{i}(t) \phi _{inx1}}}}$$

Equation (1) can then be written as:

$${\boldsymbol{M(\sum_{j}^{} \ddot{\zeta }_{j} \phi _{j}) + K(\sum_{j}^{} \zeta _{j} \phi _{j}) = F}}$$

$${\boldsymbol{\sum_{j}^{} \ddot{\zeta}_{j}(\phi _{i}^{T} M \phi _{j}) + \sum_{j}^{} \zeta _{j} \phi _{i}^{T} K \phi _{j} = \phi _{i}^{T} F}}$$

$${\boldsymbol{\ddot{\zeta } + \lambda _{i} \zeta _{i} = \phi _{i}^{T} F}}$$, with i = 1 through n.