User:Eml5526.s11.team6.tupsakhare/hwk3

=HOMEWORK 3 -Problem 3.3=

Given
Referred from - Pg 37, Reference (1)

Consider the truss shown in the following FIGURE 3.3.1(PLEASE INSERT PROBLEM STATEMENT FIG HERE)

-Nodes A and B are fixed.

-A force F = 10 N acts in the positive x- direction at node C. Coordinates of the joints are given in meters.

-Young's modulus E = 1011 Pa

-Cross sectional area for all bars, A = '''2. 10-2 m2'''

Find
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 stresses & reactions

Number the Element and Nodes
Node and element numbering is done in the following fashion. Refer to the FIGURE 3.3.2(PLEASE INSERT NUMBERERD FIGURE HERE) as given below-

Assemble the Global Stiffness and Force Matrix
We use the Following formula Referred from - Pg 30, Reference (1) for Elemental Stiffness Matrix of each element,

$$\displaystyle (Eq. 3.1) $$

Where, $${k^{e}} = \frac{l^{e}}$$ and the superscript e denotes the element number. Now in this case as seen from the given data, A and E for all the 4 elements have same but distinct values. So we can eliminate the superscript e for A and E of each element.

Element 1

 * Element 1 goes from Node 1 to Node 3


 * Orientation : 90 degrees to the Positive X-direction, $$\phi^{(1)} =90^{o}$$ . Hence, $$cos(90^{o})=0,sin(90^{o})=1$$ and $${{l}^{(1)}}=1$$

Using Eq. 3.3.1 for Element 1,

Element 2

 * Element 2 goes from Node 2 to Node 4


 * Orientation : 90 degrees to the Positive X-direction, $$\phi^{(2)} =90^{o}$$ . Hence, $$cos(90^{o})=0,sin(90^{o})=1$$ and $${{l}^{(2)}}=1$$

Using Eq. 3.3.1 for Element 2,

Element 3

 * Element 3 goes from Node 2 to Node 3


 * Orientation : 135 degrees to the Positive X-direction, $$\phi^{(3)} =135^{o}$$ . Hence, $$cos(135^{o})=\frac{-1}{\sqrt{2}},sin(135^{o})=\frac{1}{\sqrt{2}}$$ and $$l^{(3)}=\sqrt{2}$$

Using Eq. 3.3.1 for Element 3,

Element 4

 * Element 4 goes from Node 3 to Node 4


 * Orientation : 0 degrees to the Positive X-direction,$$\phi^{(4)} =0^{o}$$, $$cos(0^{o})=1,sin(0^{o})=0$$, $$l^{(4)}=1$$,

Using Eq. 3.3.1 for Element 4,

GLOBAL STIFFNESS MATRIX
Using above Elemental Matrices, we construct the Global Stiffness Matrix for the given Truss System as follows,

FORCE MATRIX
The force matrix contains reactions and externally applied force components at various nodes. Let Rx = Reaction at node in x -direction, Ry = Reaction at node in y -direction, subscripts denote the node number Also node 3 is free to move in both x- and y- direction and is free of any external forces. And node 4 is free to move in both x- and y- direction and has an external force of 10 N applied in the x- direction as per the given data in the problem statement.

The Force Matrix is assembled as follows -

NODAL DISPLACEMENT MATRIX
The noddal displacement matrix contains displacement components at various nodes. Let ux = Displacement at node in x -direction, uy = Displacement at node in y -direction, subscripts denote the node number Nodes 1 and 2 are fixed so they have all the components of displacement as zeros (0). Also node 3 and 4 are free to move in both x- and y- direction.

The Nodal Displacement Matrix is assembled as follows -

Partition the system and solve for the Nodal Displacements.
Now, using the Global stiffness matrix (K), Force matrix (F) and the Nodal displacement matrix (d), we can write the Global System as follows,

$$\displaystyle (Eq.3.3.2) $$

As the displacements for first two nodes are prescribed, we partition the systems after four rows and columns.

System Partition
Also we use the following equation from - Pg 23, 'A first course in finite elements' - Fish & Belytschko -2007 and compare our global system with it to get the nodal reactions and elemental stresses using $$\displaystyle (Eq.3.3.2) $$ & $$\displaystyle (Eq.3.3.3)  $$ -

$$\displaystyle (Eq.3.3.3) $$

On comparison then we get the following partitioned matrices -

Nodal Displacements
We solve the following system of equations to get the nodal displacement matrix,

But $$\bar{d}_{E}$$ is a Zero Matrix, which will yield, $$d_{F}=K_{F}^{-1}.f_{F}$$ Solving using the above equation, the unknown nodal displacements are

$$\displaystyle (Eq.3.3.4) $$ -

Nodal reactions
Using the above equation the unknown reaction values are

$$\displaystyle (Eq.3.3.6) $$

Elemental Stresses
The stresses in the elements are given by the following expression from page 34 of the reference (1)

For element 1:


 * $$ cos90^{\circ} = 0,\quad sin90^{\circ}=1, $$



\mathbf{d}^{(3)}=l^{(3)}/AE\begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{2x}\\ u_{2y} \end{bmatrix}, \qquad \sigma ^{(3)}=1/A\begin{bmatrix} 0 & -1 &0 &1 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 38.2843\\ 10 \end{bmatrix} = 500N/m^{2}

$$

For element 2:


 * $$ cos135^{\circ} = -0.707,\quad sin135^{\circ}=0.707, $$



\mathbf{d}^{(2)}=l^{(2)}/AE\begin{bmatrix} u_{2x}\\ u_{2y}\\ u_{3x}\\ u_{3y} \end{bmatrix}, \qquad \sigma ^{(2)}=1/A\begin{bmatrix} 0 & -1 &-0 &1 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 48.2843\\ 0 \end{bmatrix} = 0 N/m^{2}

$$

For element 3: 


 * $$ cos90^{\circ} = 0,\quad sin90^{\circ}=1, $$



\mathbf{d}^{(1)}=l^{(1)}/AE\begin{bmatrix} u_{2x}\\ u_{2y}\\ u_{4x}\\ u_{4y} \end{bmatrix}, \qquad \sigma ^{(1)}=1/A\begin{bmatrix} 0.5 & -0.5 &-0.5 &0.5 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 38.2843\\ 10 \end{bmatrix} = -1000N/m^{2}

$$

For element 4:


 * $$ cos0^{\circ} = 1,\quad sin0^{\circ}=0, $$



\mathbf{d}^{(4)}=l^{(4)}/AE\begin{bmatrix} u_{3x}\\ u_{3y}\\ u_{4x}\\ u_{4y} \end{bmatrix}, \qquad \sigma ^{(4)}=1/A\begin{bmatrix} -1& 0 &1 &0 \end{bmatrix}\begin{bmatrix} 38.2843\\ 1\\ 48.2843\\ 0 \end{bmatrix} = 10/(2*10^{-2})= 500N/m^{2}

$$