A trapezoidal plate: one Q$4$ element

Consider a linear elasticity problem on the trapezoidal panel domain as shown in the following

figure.

The vertical left edge is fixed. The bottom and the right vertical edges are traction free. Traction

$t_y=-20\frac{N}{m}$ is applied on the top horizontal edge. Material properties are Young's modulus

$E=3{10}^{7}Pa,$ Poisson's ratio $v=0\mathrm{.}3$ and thickness $=1m.$ Plane stress conditions are

considered.

In the Lecture $9,$ the displacement of the trapezoidal panel was solved using one quadrilateral

element $($Q$4).\left[\begin{array}{c}{u}_{x3}\\ u_{y3}\\ u_{x4}\\ u_{y4}\end{array}\right]={10}^{-6}\left[\begin{array}{c}-1\mathrm{.}17\\ -9\mathrm{.}67\\ 2\mathrm{.}67\\ -9\mathrm{.}94\end{array}\right]or,d^e={10}^{-6}\left[\begin{array}{c}0\\ 0\\ 0\\ 0\\ -1\mathrm{.}17\\ -9\mathrm{.}67\\ 2\mathrm{.}67\\ -9\mathrm{.}94\end{array}\right]$

Now you are asked to continue this one$-$element Q$4$ problem to calculate the element nodal

Finite Element Method

stresses using stress extrapolation from the stress at Gauss points.

Hint $($Lecture $10$ video$)$:

Recall the strain$-$displacement matrix at $({}_1,{}_1)=(-\frac{1}{\sqrt[\phantom{2}]{3}},-\frac{1}{\sqrt[\phantom{2}]{3}})$ is given as:

$B^e({}_1,{}_1)=\left[\begin{array}{cccccccc}-0\mathrm{.}44& 0& -0\mathrm{.}06& 0& 0\mathrm{.}12& 0& 0\mathrm{.}38& 0\\ 0& 0\mathrm{.}88& 0& -0\mathrm{.}88& 0& -0\mathrm{.}24& 0& 0\mathrm{.}24\\ 0\mathrm{.}88& -0\mathrm{.}44& -0\mathrm{.}88& -0\mathrm{.}06& -0\mathrm{.}24& 0\mathrm{.}12& 0\mathrm{.}24& 0\mathrm{.}38\end{array}\right]$

Let us compute the strains at the first Gauss point $({}_1,{}_1)=(-\frac{1}{\sqrt[\phantom{2}]{3}},-\frac{1}{\sqrt[\phantom{2}]{3}})$ :

${}^e({}_1,{}_1)=B^e({}_1,{}_1)d^e={10}^{-7}\left[\begin{array}{c}8\mathrm{.}74\\ -0\mathrm{.}65\\ -40\mathrm{.}16\end{array}\right]$

We can compute the stresses at the first Gauss point $({}_1,{}_1)=(-\frac{1}{\sqrt[\phantom{2}]{3}},-\frac{1}{\sqrt[\phantom{2}]{3}})$ via constitutive

relation: The element nodal stresses can be calculated using

$\left[\begin{array}{ccc}{}_{x1}& {}_{y1}& {}_{xy1}\\ {}_{x2}& {}_{y2}& {}_{xy2}\\ {}_{x3}& {}_{y3}& {}_{xy3}\\ {}_{x4}& {}_{y4}& {}_{xy4}\end{array}\right]=\left[\begin{array}{cccc}1+\frac{1}{2}\sqrt[\phantom{2}]{3}& -\frac{1}{2}& 1-\frac{1}{2}\sqrt[\phantom{2}]{3}& -\frac{1}{2}\\ -\frac{1}{2}& 1+\frac{1}{2}\sqrt[\phantom{2}]{3}& -\frac{1}{2}& 1-\frac{1}{2}\sqrt[\phantom{2}]{3}\\ 1-\frac{1}{2}\sqrt[\phantom{2}]{3}& -\frac{1}{2}& 1+\frac{1}{2}\sqrt[\phantom{2}]{3}& -\frac{1}{2}\\ -\frac{1}{2}& 1-\frac{1}{2}\sqrt[\phantom{2}]{3}& -\frac{1}{2}& 1+\frac{1}{2}\sqrt[\phantom{2}]{3}\end{array}\right]\left[\begin{array}{ccc}{}_{x1}^{\text{'}}& {}_{y1}^{\text{'}}& {}_{xy1}^{\text{'}}\\ {}_{x2}^{\text{'}}& {}_{y2}^{\text{'}}& {}_{xy2}^{\text{'}}\\ {}_{x3}^{\text{'}}& {}_{y3}^{\text{'}}& {}_{xy3}^{\text{'}}{?}_{xy4}\\ {}_{x4}^{\text{'}}& {}_{y4}^{\text{'}}& {}_{xy4}^{\text{'}}\end{array}\right]$

If you solve the problem in MATLAB or other software, please post the source codes and

explain how to solve the problem in a short paragraph.