User:Eml4500.f08.gravy.jad/Notes5

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Lecture 27

$$\ \hat{\mathbf{w}}_{2x1} = \,$$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 Eqs (3) and (4) into Eq (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 $$\ \mathbf{w}_{4x1} \,$$ [This is Eq (5)]

Recall: $$\ \left ( \mathbf{A}_{pxq} \mathbf{B}_{qxr} \right )^{T} = \mathbf{B}^{T} \mathbf{A}^{T} \,$$ [This is Eq 6]

As an Ex:

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

(insert hw5 verify equation 6)

Recall: $$\ \mathbf{a}_{nx1} \cdot \mathbf{b}_{nx1} = \mathbf{a}^{T}_{1xn} \mathbf{b}_{nx1} = \sharp _{1x1} \,$$ where # equals some number, a scalar [Eq 7]

Now, apply (7) and (6) to (5), to reveal: $$\ \left ( \mathbf{T}^{(e)} \mathbf{w} \right )^{T} \left \lbrack \hat{\mathbf{k}}^{(e)} \left ( \mathbf{T}^{(e)} \mathbf{d}^{(e)} \right ) - \mathbf{P}^{(e)} \right \rbrack =0_{1x1} \,$$for all $$\ \mathbf{w}_{4x1} \,$$

$$\ \Rightarrow \left ( \mathbf{w}^{T} {\mathbf{T}^{(e)}}^{T} \right ) \left \lbrack \hat{\mathbf{k}}^{(e)} \left ( \mathbf{T}^{(e)} \mathbf{d}^{(e)} \right ) - \mathbf{P}^{(e)} \right \rbrack =0 \,$$

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

$$\ \Rightarrow \mathbf{w} \cdot \left \lbrack \mathbf{k}^{(e)} \mathbf{d}^{(e)} - \mathbf{f}^{(e)} \right \rbrack =0 \,$$for all $$\ \mathbf{w}_{4x1} \,$$

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

insert hw5 proof

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]



HW5 literature search



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

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

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

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

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

Eq 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