User:Eml4500.f08.Lulz.js/HW5

Justification of Reduction of Stiffness Matrix for 2-bar Truss Problem
From the general force displacement relationship, the following relation can be derived:


 * $$\mathbf{kd}= \mathbf{F}$$ => $$\mathbf{kd - F}= \mathbf{0}\;\;(1)$$

Using this relation, the Principle of Virtual Work can be expressed as follows:


 * $$\mathbf{W}\cdot\mathbf{(kd - F)} = \mathbf{0}$$ for all $$\mathbf{W}\;\;(2)$$

Where $$\mathbf{W}\;$$ refers to the weighting matrix. It is useful to show the equivalency of the above two expressions. While is is trivial to show how equation 2 comes from equation 1, it is more challenging to show how equation 1 comes from equation 2. In order to prove that this is the case, consider the following weighting coefficients:


 * $$W_{1} = 1\;$$, $$\;W_{2}=...=W_{6} = 0$$

Which can be expressed in transposed form as:


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

Applying this choice for $$\mathbf{W}$$ to equation 2 yields the following result:


 * $$\mathbf{W}\cdot\mathbf{(kd - F)} = [1\cdot\sum_{j=1}^{6} (k_{1j}d_{j} - F_{1}) + 0\cdot\sum_{j=1}^{6} (k_{2j}d_{j} - F_{2}) + ... + 0\cdot\sum_{j=1}^{6} (k_{6j}d_{j} - F_{6})] = 0$$

It is easily shown that the result of the above expression is:


 * $$\sum_{j=1}^{6} k_{1j}d_{j} = F_{1}$$

Consider another set of weighting coefficients and resulting transposed form:


 * $$W_{1} = 0\;$$, $$W_{2} = 0\;$$ , $$\;W_{3}=...=W_{6} = 0$$


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

As before, applying this matrix to equation 2 yields a result of $$\sum_{j=1}^{6} k_{2j}d_{j} - F_{2} = 0$$. Thus, it is apparent that for any choice of weighting matrix $$\mathbf{W}$$, the result is equivalent to $$\mathbf{kd - F}= \mathbf{0}$$, which is equation 1.

Returning to the specific 2-bar truss problem, accounting for the boundary conditions allows the removal of the following global displacements:


 * $$d_{1}=d_{2}=d_{5}=d_{6}=0\;$$

The weighting coefficients must be "kinematically admissable" - this means that it must not violate the boundary conditions. Thus, the following is a result of the above expression:


 * $$W_{1}=W_{2}=W_{5}=W_{6}=0\;$$

This leads to the conclusion that the weighting coefficients are essentially virtual displacements. As a result of the previous 2 relations, equation 2 can now be rewritten as:


 * $$\begin{Bmatrix}W_{3}\\W_{4}\end{Bmatrix}\cdot\mathbf{(\bar{k}\bar{d} - \bar{F})} = 0$$ for all $$\begin{Bmatrix}W_{3}\\W_{4}\end{Bmatrix}$$

Where $$\mathbf{\bar{k}}\;$$, $$\mathbf{\bar{d}}\;$$, and $$\mathbf{\bar{F}}$$ are defined as follows:


 * $$\mathbf{\bar{k}} = \begin{bmatrix}k_{33}&k_{34}\\k_{43}&k_{44}\end{bmatrix}$$


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


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

An example of weighting is the calculation of a student's final grade in this course, which is given as follows: $$course\;grade = \alpha_{0} + HW Grade + \Sigma(\alpha Exam Grade)$$

Derivation of $$\mathbf{k^{(e)}} = \mathbf{T^{(e)^{T}}}\mathbf{\hat{k}^{(e)}}\mathbf{T^{(e)}}$$
Recall the force-displacement relationship adjusted for axial degrees of freedom $$\mathbf{q^{(e)}}$$:


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

This relationship can be rearranged to the following:


 * $$\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{P^{(e)}} = \mathbf{0}\;\;(1)$$

Applying the Principle of Virtual Work results in the following:


 * $$\mathbf{\hat{W}}\cdot(\mathbf{\hat{k}^{(e)}}\mathbf{q^{(e)}} - \mathbf{P^{(e)}}) = \mathbf{0}$$ for all $$\mathbf{\hat{W}}$$

It has previously been shown that equations 1 and 2 are equivalent. Recalling the following two relations:


 * $$\mathbf{q^{(e)}} = \mathbf{T^{(e)}}\mathbf{d^{(e)}}\;\;(3)$$


 * $$\mathbf{\hat{W}^{(e)}} = \mathbf{T^{(e)}}\mathbf{W^{(e)}}\;\;(4)$$

$$\mathbf{\hat{W}}$$ refers to virtual axis displacements, corresponding to $$\mathbf{q^{(e)}}$$.

$$\mathbf{W}$$ refers to virtual displacements in global coordinate system, corresponding to $$\mathbf{d^{(e)}}$$.

Placing equations 3 and 4 into equation 2 yields the following:


 * $$(\mathbf{T^{(e)}}\mathbf{W})\cdot[\mathbf{\hat{k}^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}}) - \mathbf{P^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}\;\;(5)$$

Recall the following two vector identities:


 * $$(\mathbf{A}\cdot\mathbf{B})^{T} = \mathbf{B}^{T}\mathbf{A}^{T}\;\;(6)$$


 * $$\mathbf{a}\cdot\mathbf{b} = \mathbf{a}^{T}\mathbf{b}\;\;(7)$$

Applying equations 7 and 8 into equation 5 yields the following derivation:

$$(\mathbf{T^{(e)}}\mathbf{W})^{T}[\mathbf{\hat{k}^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}}) - \mathbf{P^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}$$

$$\mathbf{T^{(e)}}^{T}\mathbf{W}^{T}[\mathbf{\hat{k}^{(e)}}(\mathbf{T^{(e)}}\mathbf{d^{(e)}}) - \mathbf{P^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}$$

$$\mathbf{W}^{T}[\left(\mathbf{T^{(e)}}^{T}\mathbf{\hat{k}^{(e)}}\mathbf{T^{(e)}}\right)\mathbf{d^{(e)}} - \left\{\mathbf{T^{(e)}}^{T}\mathbf{P^{(e)}}\right\}] = \mathbf{0}$$

In the last expression, the term inside the parentheses is equivalent to $$\mathbf{k^{(e)}}$$, and the term inside the curly brackets is equivalent to $$\mathbf{f^{(e)}}$$. Making these substitutions yields the following result:

$$\mathbf{W^{T}}[\mathbf{k^{(e)}}\mathbf{d^{(e)}} - \mathbf{f^{(e)}}] = \mathbf{0}$$ for all $$\mathbf{W}$$.

Continuous Case - Motivational Model Problem


Consider the elastic bar shown in Figure ##. The bar has a varying cross-section and Young's modulus, and is subject to a time-dependent loading condition consisting of a distributed, varying axial load as well as a concentrated load, signified by $$f(x,t)\;$$ and $$P(x)\;$$ respectively. A section of the bar is shown in Figure ##.

The product $$A(x)\rho (x)\;$$ can be expressed as $$m(x)\cdot dx\;$$, where $$m(x)\;$$ is the mass per unit length. The term $$\frac{\partial{^{2}u}}{\partial{x^{2}}}$$ refers to the acceleration of the deforming bar, and can be represented as $$\ddot{u}$$.

The sum of forces on the section can be written as follows:


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


 * $$\Sigma F_{x} = 0 = \frac{\partial{N(x,t)}}{\partial x}dx + f(x,t)dx - m(x)\ddot{u}dx + h.o.t\;\;(1)$$

"h.o.t" refers to the higher order terms.

Recall the general definition of the Taylor Series expansion, $$f(x+dx) = f(x) + \frac{\partial{f(x)}}{\partial x}dx + \frac{1}{2}\frac{\partial{^{2}x}}{\partial{x}^{2}} + ...$$, where $$\frac{1}{2}\frac{\partial{^{2}x}}{\partial{x}^{2}} + ...$$ are the higher order terms and are typically neglected. Dividing out the $$dx\;$$ terms and applying this concept to equation 1 yields equation 2, the equation of motion for the elastic bar:


 * $$\frac{\partial{N}}{\partial x} + f = m\ddot{u}\;\;(2)$$

The constitutive relation for the bar can be given as equation 3:


 * $$N(x,t) = A(x)\sigma (x,t)\;\;(3)$$

The $$\sigma (x,t)\;$$ term can be broken down as follows:


 * $$\sigma (x,t)= E(x)\epsilon (x,t)\;$$


 * $$\epsilon (x,t)= \frac{\partial{u}}{\partial{x}}(x,t)\;$$

The partial differential equation of motion for the elastic bar can be given as equation 4, which is the result of substituting equation 3 and resulting breakdown of terms into equation 2:


 * $$\frac{\partial}{\partial x}[A(x)E(x)\frac{\partial{u}}{\partial{x}}] + f(X,t) = m(x)\ddot{u}\;\;(4)$$

In order to manipulate this equation, boundary and initial conditions are required. Since the equation is of 2nd order in both x and t, two boundary and two initial conditions are necessary. The initial conditions consist of an initial displacement and velocity, given as $$[init. vel., init. disp]\;$$. The boundary conditions depend on the type of system being analyzed. Figure ## shows a bar of length L with both ends constrained. The boundary conditions for this system are:


 * $$u(0,t) = 0 = u(L,t)\;$$

Figure ## again shows a bar of length L, but with one end free and subject to force F. The first boundary condition, as with the constrained case, is:


 * $$u(0,t) = 0\;\;(bc\;1)$$

The second boundary condition can be derived using the following line of thought, similar to that used to derive equation 4:


 * $$N(L,t) = F(t)\;$$


 * $$N(L,t) = A(L)\sigma (L,t)\;$$


 * $$\sigma (L,t) = E(L)\epsilon (L,t)\;$$


 * $$\epsilon (L,t) = \frac{\partial u}{\partial x} (L,t)$$

Using the above relations, the second boundary condition for this case can be written as follows:


 * $$\frac{\partial u(L,t)}{\partial x} = \frac{F(t)}{A(L)E(L)}\;\;(bc\;2)$$