Question p.6.11: A square element 1234, whose corners have coordinates x, y (...
A square element 1234, whose corners have coordinates x, y (in metres) of
(−1, −1), (1, −1), (1, 1), and (−1, 1), respectively, was used in a plane stress finite element analysis. The following nodal displacements (mm) were obtained:
If Young’s modulus E = 200 000 N/mm² and Poisson’s ratio ν = 0.3, calculate the stresses at the centre of the element.
From the first of Eqs (6.96)
\left.\begin{array}{l}u(x, y)=\alpha_1+\alpha_2 x+\alpha_3 y+\alpha_4 x y \\v(x, y)=\alpha_5+\alpha_6 x+\alpha_7 y+\alpha_8 x y\end{array}\right\} (6.96)
\begin{aligned}&u_1=\alpha_1-\alpha_2-\alpha_3+\alpha_4=0.1 / 10^3 (i) \\&u_2=\alpha_1+\alpha_2-\alpha_3-\alpha_4=0.3 / 10^3 (ii) \\&u_3=\alpha_1+\alpha_2+\alpha_3+\alpha_4=0.6 / 10^3 (iii) \\&u_4=\alpha_1-\alpha_2+\alpha_3-\alpha_4=0.1 / 10^3 (iv)\end{aligned}Adding Eqs (i) and (ii)
u_1+u_2=2 \alpha_1-2 \alpha_3=0.4 / 10^3i.e.
\alpha_1-\alpha_3-0.2 / 10^3 (v)
Adding Eqs (iii) and (iv)
u_3+u_4=2 \alpha_1+2 \alpha_3=0.7 / 10^3i.e.
\alpha_1+\alpha_3=0.35 / 10^3 (vi)
Adding Eqs (v) and (vi)
\alpha_1=0.275 / 10^3Then from Eq. (v)
\alpha_3=0.075 / 10^3
Now subtracting Eq. (ii) from Eq. (i)
i.e.
\alpha_2-\alpha_4=0.1 / 10^3 (vii)
Subtracting Eq. (iv) from Eq. (iii)
u_3-u_4=2 \alpha_2+2 \alpha_4=0.5 / 10^3
i.e.
\alpha_2+\alpha_4=0.25 / 10^3 (viii)
Now adding Eqs (vii) and (viii)
2 \alpha_2=0.35 / 10^3
whence
\alpha_2=0.175 / 10^3
Then from Eq. (vii)
\alpha_4=0.075 / 10^3
From the second of Eqs (6.96)
\begin{aligned}&v_1=\alpha_5-\alpha_6-\alpha_7+\alpha_8=0.1 / 10^3 (ix) \\&v_2=\alpha_5+\alpha_6-\alpha_7-\alpha_8=0.3 / 10^3 (x)\\&v_3=\alpha_5+\alpha_6+\alpha_7+\alpha_8=0.7 / 10^3 (xi) \\&v_4=\alpha_5-\alpha_6+\alpha_7-\alpha_8=0.5 / 10^3 (xii)\end{aligned}Then, in a similar manner to the above
\begin{aligned}&\alpha_5=0.4 / 10^3 \\&\alpha_7=0.2 / 10^3 \\&\alpha_6=0.1 / 10^3 \\&\alpha_8=0\end{aligned}Eqs (6.96) are now written
\begin{aligned}&u_i-(0.275+0.175 x+0.075 y+0.075 x y) \times 10^{-3} \\&v_i=(0.4+0.1 x+0.2 y) \times 10^{-3}\end{aligned}Then, from Eqs (6.88)
\varepsilon_x=\frac{\partial u}{\partial x} \quad \varepsilon_y=\frac{\partial v}{\partial y} \quad \gamma_{x y}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x} (6.88)
At the centre of the element x = y = 0. Then
\begin{aligned}\varepsilon_x &=0.175 \times 10^{-3} \\\varepsilon_y &=0.2 \times 10^{-3} \\\gamma_{x y} &=0.175 \times 10^{-3}\end{aligned}so that, from Eqs (6.92)
\{\sigma\}=\left\{\begin{array}{l}\sigma_x \\\sigma_y \\\tau_{x y}\end{array}\right\}=\frac{E}{1-v^2}\left[\begin{array}{ccc}1 & v & 0 \\1 & 1 & 0 \\0 & 0 & \frac{1}{2}(1-v)\end{array}\right]\left\{\begin{array}{l}\varepsilon_x \\\varepsilon_y \\\gamma_{x y}\end{array}\right\} (6.92)