User:Eml5526.s11.team04.premchand/HW3

= Problem 3.3 - Direct approach for truss structure =

The following problem is problem 2.3 as taken from Fish and Belytschko's 'A first course in Finite Elements'

Problem Statement
For the truss structure given in Figure 3.1, Nodes A and B are fixed.A force equal to 10N acts in the positive x-direction at node C.Co-ordinates of joints are given in meters.Young's modulus is E=1011Pa and the cross-sectional area for all bars are A=2.10-2m2.


 * 1) Number the elements and nodes.
 * 2) Assemble the global stiffness and force matrix.
 * 3) Partition the system and solve for the nodal displacements.
 * 4) Compute the stresses and reaction forces.

Fig. 3.1 Data for Problem 3

Number the elements and nodes
The nodes are numbered starting with nodes which have prescribed displacements. Numbering in this manner will allow the assembled matrix to be partitioned easily later on.The global node number is given beside the node.While the element number is given in parenthesis beside the elements.

Fig. 3.2

Assembling the global stiffness and force matrix
The following matrix is used for solving the individual element matrix

Element 1 is numbered with global nodes 1 and 4. It is positioned at an angle $$\Phi=90^{0}$$ with respect to positive x axis.So cos $$90^{0}$$=0, sin $$90^{0}$$=1, l(1)=l, A(1)E(1)/l(1)= AE/l

Element 2 is numbered with global nodes 2 and 3. It is positioned at an angle $$\Phi=90^{0}$$ with respect to positive x axis.So cos $$90^{0}$$=0, sin $$90^{0}$$=1, l(2)=l, A(2)E(2)/l(2)= AE/l

Element 3 is numbered with global nodes 4 and 3. It is positioned at an angle $$\Phi=0^{0}$$ with respect to positive x axis.So cos $$0^{0}$$=1, sin $$0^{0}$$=0, l(3)=l, A(3)E(3)/l(3)= AE/l

Element 4 is numbered with global nodes 2 and 4. It is positioned at an angle $$\Phi=135^{0}$$ with respect to positive x axis.So cos $$135^{0}$$=$$-\frac{1}{\sqrt{2}}$$, sin $$135^{0}$$=$$\frac{1}{\sqrt{2}}$$, l(4)=l, A(4)E(4)/l(4)= AE/l

The assembled global stiffness matrix is as follows:

The force,displacement and reaction matrices are as follows:
The global system of equations are as follows:

Partitioning the system and solving for nodal displacements
The global system as shown in equation 3.9 is partitioned after four rows and four columns:

Solving using the above equation, the unknown nodal displacements are

Solving for reactions and stresses
Using the above equation the unknown reaction values are

The stresses in the elements are given by the following expression

For element 1:


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



\mathbf{d}^{2}=l/AE\begin{bmatrix} u_{1x}\\ u_{1y}\\ u_{2x}\\ u_{2y} \end{bmatrix}, \qquad \sigma ^{(2)}=1/A\begin{bmatrix} 0 & -1 &0 &1 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 38.2885\\ 10 \end{bmatrix} = 10/A = 10/(2*10^{-2})= 500N/m^{2}

$$

For element 2:


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



\mathbf{d}^{2}=l/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\\ 48.2885\\ 0 \end{bmatrix} = 0N/m^{2}

$$

For element 3:


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



\mathbf{d}^{2}=l/AE\begin{bmatrix} u_{4x}\\ u_{4y}\\ u_{3x}\\ u_{3y} \end{bmatrix}, \qquad \sigma ^{(2)}=1/A\begin{bmatrix} -1& 0 &1 &0 \end{bmatrix}\begin{bmatrix} 38.2885\\ 10\\ 48.2885\\ 0 \end{bmatrix} = 10/A = 10/(2*10^{-2})= 500N/m^{2}

$$

For element 4:


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



\mathbf{d}^{2}=l/AE\begin{bmatrix} u_{2x}\\ u_{2y}\\ u_{4x}\\ u_{4y} \end{bmatrix}, \qquad \sigma ^{(2)}=1/A\begin{bmatrix} 0.707 & -0.707 &-0.707 &0.707 \end{bmatrix}\begin{bmatrix} 0\\ 0\\ 38.2885\\ 10 \end{bmatrix} = 10/A = -20/(2*10^{-2})= 1000N/m^{2}

$$