User:EML5526.S11.Team3.vnarayanan/Homework3

Problem 3.2
For the spring system given in the figure below, a. Number the elements and nodes. b. Assemble the global stiffness and force matrix. c. Partition the system and solve for the nodal displacements. d. Compute the reaction forces.

3.2(b) : Assembled Global Stiffness matrix and Force matrix
The stiffness matrices of individual elements are assembled according to the connectivity table to get the global stiffness matrix. Connectivity table is a one which shows the nodes associated with each element.

Element stiffness matrix is given by,

Connectivity table :

$$\begin{array}{|c|c||c|}  Element No. & Local Node 1 & Local Node 2 \\ \hline 1&1&4\\ 2&2&4\\ 3&1&3\\ 4&3&4\\ \end{array}$$

Assembled global stiffness matrix,

Since the nodes 1 and 2 are fixed, displacements at these nodes are 0. But there will be reaction forces acting at these nodes, in order to maintain static equilibrium of the system. A force of 50 N acts at node 4 and at all other nodes forces are 0. Therefore, the global force, displacement and reaction matrices are as follows,

3.2(c): Partition of the system and Nodal displacements
The system can be expressed in the equation form as follows,

The above system of equations can be partitioned as,

where,

$$ \left[\textbf{K}_E \right] = \begin{bmatrix} k_1+k_3 &0 \\ 0 &k_2  \\ \end{bmatrix} \;\;\left[\textbf{K}_{EF} \right] = \begin{bmatrix} &-k_3 &-k_1 \\ &0 &-k_2 \\ \end{bmatrix}\; \; \left[\textbf{K}^{T}_{EF} \right] = \begin{bmatrix} k_3 &0  \\ -k_1 &-k_2\\ \end{bmatrix}\; \; \left[\textbf{K}_{F} \right] = \begin{bmatrix} &k_3+k_4 &-k_4 \\ &-k_4 &k_1+k_2+k_4\\ \end{bmatrix} $$

$$ \left[\textbf{d}_{E} \right] = \begin{bmatrix} 0\\ 0\\ \end{bmatrix}\; \; \left[\textbf{d}_{F} \right] = \begin{bmatrix} u_3\\ u_4\\ \end{bmatrix}\; \; \left\{\textbf{r}_{E} \right\} = \begin{Bmatrix} r_1\\ r_2\\ \end{Bmatrix}\; \; \left\{\textbf{f}_{F} \right\} = \begin{Bmatrix} 0\\ f_4\\ \end{Bmatrix}\;$$

The reduced system of equations is given by,

Therefore,

Substituting,  $$\;\;k_1 = 3k, k_2 = k, k_3 = k, k_4 = 2k \; and \; f_4 = 50\;\;$$, we get,

$$ \begin{bmatrix} &3k &-2k \\ &-2k &6k\\ \end{bmatrix}\;\begin{bmatrix} u_3\\ u_4\\ \end{bmatrix}\; \; = \begin{Bmatrix} 0\\ 50\\ \end{Bmatrix}\; $$

$$ \Rightarrow \;\; $$ $$3ku_3 - 2ku_4 = 0;$$

$$ -2ku_3 + 6ku_4 = 50 $$

Solving the two equations, we get,

3.2(d): Reaction Forces
The reaction forces are found using the equation,

Therefore, $$ \begin{bmatrix} &-k_3 &-k_1 \\ &0 &-k_2 \\ \end{bmatrix}\; \begin{bmatrix} u_3\\ u_4\\ \end{bmatrix}\; \; = \begin{Bmatrix} r_1\\ r_2\\ \end{Bmatrix}\; $$

Substituting,  $$\;\;k_1 = 3k, k_2 = k, k_3 = k, u_3 = \frac{50}{7k}\;\; u_4 = \frac{75}{7k} \;\;$$,

we get,

$$ \begin{bmatrix} &-k &-3k \\ &0 &-k\\ \end{bmatrix}\;\begin{bmatrix} 50/7k\\ 75/7k\\ \end{bmatrix}\; \; = \begin{Bmatrix} r_1\\ r_2\\ \end{Bmatrix}\; $$

$$ \Rightarrow \;\; $$ $$\;\; \frac{-50}{7}-\frac{225}{7}= r_1\;\; ;$$ $$ \frac{-75}{7} = r_2$$

Problem 3.4
$$ \mathbf{\alpha } = b_i\left [ a_2'b_j'+a_2b_j'' \right ]$$

$$\mathbf{\beta } = b_j\left [ a_2'b_i'+a_2b_i'' \right ] $$

Where,

$$\displaystyle b_i(x) = \cos(ix)$$

$$b_j(x) = \cos(jx)\;; i\;\neq\;j$$

Show that $$ \;\; \mathbf{\alpha }\;\neq\; \mathbf{\beta }$$

Solution
$$\displaystyle b_i(x) = \cos(ix)$$

$$\displaystyle b_j(x) = \cos(jx)$$

$$b_i'(x) = -i\sin(ix)\; ; \;b_i''(x) = -i^2\cos(ix) $$

$$b_j'(x) = -j\sin(jx)\; ; \;b_j''(x) = -j^2\cos(jx) $$

Therefore,

$$\mathbf{\alpha } = \cos(ix)\left [ a_2'(-j\sin(jx))+a_2(-j^2\cos(jx))\right] $$

$$\mathbf{\alpha } = \cos(ix)\left [ -ja_2'\sin(jx)-j^2a_2\cos(jx)\right] $$

Therefore,

$$\mathbf{\beta } = \cos(jx)\left [ a_2'(-i\sin(ix))+a_2(-i^2\cos(ix))\right] $$

$$\mathbf{\beta } = \cos(jx)\left [ -ia_2'\sin(ix)-i^2a_2\cos(ix)\right] $$

From, Equations (3.4.2) and (3.4.4), it is evident that

If $$\;\; a_2 \neq a_2' \neq 0 \;\; then,$$

Let us consider a specific example, say $$i = 2 \; and \; j = 1$$

Therefore,

$$\displaystyle b_2(x) = \cos(2x)$$

$$\displaystyle b_1(x) = \cos(x)$$

$$b_2'(x) = -2\sin(2x)\; ; \;b_2''(x) = -4\cos(2x) $$

$$b_j'(x) = -\sin(x)\; ; \;b_j''(x) = -\cos(x) $$

$$\mathbf{\alpha } = b_2\left [ a_2'b_1'+a_2b_1'' \right ]$$

Therefore,

$$\mathbf{\alpha } = \cos(2x)\left [ a_2'(-\sin(x))+a_2(-\cos(x))\right] $$

$$\mathbf{\alpha } = \cos(2x)\left [ -a_2'\sin(x)-a_2\cos(x)\right] $$

$$\mathbf{\beta} = b_1\left [ a_2'b_2+a_2b_2'' \right ]$$

Therefore,

$$\mathbf{\beta } = \cos(x)\left [ a_2'(-2\sin(x))+a_2(-4\cos(2x))\right] $$

$$\mathbf{\beta } = \cos(jx)\left [ -2a_2'\sin(2x)-4a_2\cos(2x)\right] $$

From Equations (3.4.6) and (3.4.7), it is clear that,