Question 24.9: Solve the two-dimensional loaded plate using the CST element......

Considering two CST elements, the finite element model and the elements, and the node numbers are shown in figure . The connectivity table is shown in Table 24.3

Co-ordinate of each nodes are (in mm)

Element no. 1 x_1=60, y_1=0, x_2=60, y_2=30, x_3=0, y_3=0

Element no. 2 x_1=0, y_1=30, x_2=y_2=0, x_3=60, y_3=30

D = material property matrix for plane stress condition

D=\left[\begin{array}{ccc}1 & v & 0 \\v & 1 & 0 \\0 & 0 & \frac{1-v}{2}\end{array}\right] \times \frac{E}{1-v^2}=\frac{70 \times 10^3}{1-0.33^2}\left[\begin{array}{ccc}1 & 0.33 & 0 \\0.33 & 1 & 0 \\0 & 0 & 0.335\end{array}\right]

D=78.555 \times 10^3\left[\begin{array}{ccc}1 & 0.33 & 0 \\0.33 & 1 & 0 \\0 & 0 & 0.335\end{array}\right]

\operatorname{det} J^{(1)}=x_{13} y_{23}-x_{23} y_{13}=60 \times 30-0=1800 \mathrm{~mm}^2

where x_{13}=x_1-x_3=60, y_{23}=y_2-y_3=30

\operatorname{det} J^{(2)}=1800 \mathrm{~mm}^2

Area A=\frac{1}{2}|\operatorname{det} J|=\frac{1}{2} \times 1800=900 \mathrm{~mm}^2

Element strain-displacement matrix is

B^{(1)}=\frac{1}{1800}\left[\begin{array}{cccccc} 30 & 0 & 0 & 0 & -30 & 0 \\0 & -60 & 0 & 60 & 0 & 0 \\-60 & 30 & 60 & 0 & 0 & -30 \end{array}\right]

B^{(2)}=\frac{1}{1800}\left[\begin{array}{cccccc} -30 & 0 & 0 & 0 & 30 & 0 \\0 & 60 & 0 & -60 & 0 & 0 \\60 & -30 & -60 & 0 & 0 & 30 \end{array}\right]

Now performing matrix multiplication, D B^e, we get

The stiffness matrix is obtained from

The boundary conditions for the given problem are Q_5=Q_6=Q_7=Q_8=0 \text { and } Q_1, Q_2, Q_3 and Q_4 which need to be determined. Using the elimination approach, consider the stiffness matrix (assembled) associated with the degrees freedom Q_1, Q_2, Q_3 \text { and } Q_4. The force vector is

\{F\}=\left[\begin{array}{llllllll}0 & 0 & -10,000 & 0 & 0 & 0 & 0 & 0\end{array}\right]^T

The set of equations are

10^5 \times\left[\begin{array}{cccc}4.595 & -2.612 & -2.631 & 1.296 \\-2.612 & 8.513 & 1.316 & -7.855 \\-2.631 & 1.316 & 4.595 & 0 \\1.296 & -7.855 & 0 & 8.513\end{array}\right]\left\{\begin{array}{l}Q_1 \\Q_2 \\Q_3 \\Q_4\end{array}\right\}=\left\{\begin{array}{c}0 \\0 \\-10,000 \\0\end{array}\right\}

Solving for, Q_1, Q_2, Q_3 \text { and } Q_4 we get

Q_1=-0.015 \mathrm{~mm}, Q_2=0.021 \mathrm{~mm}

Q_3=-0.036 \mathrm{~mm}, Q_4=0.022 \mathrm{~mm}

For element 1, the element nodal displacement vector is

q^{(1)}=\left[\begin{array}{lll}-0.015, & 0.021,-0.036, & 0.022,0,0\end{array}\right]^T

Element stress \sigma^{(1)}=D B^{(1)} q

\sigma^{(1)}=\left[\begin{array}{lll}-18.774, & -3.864, & -9.21\end{array}\right]^T ~\mathrm{MPa}

\sigma^{(2)}=\left[\begin{array}{lll}-19.638, & -6.480, & -9.21\end{array}\right]^T ~\mathrm{MPa}