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:

Question

[latex]\begin{array}{llll}u_1=0.1 & u_2=0.3 & u_3=0.6 & u_4=0.1 \v_1=0.1 & v_2=0.3 & v_3=0.7 & v_4=0.5\end{array}[/latex]

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.

Accepted Answer

From the first of Eqs (6.96)

[latex]\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} ight\}[/latex] (6.96)

[latex]\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}[/latex]

Adding Eqs (i) and (ii)

[latex]u_1+u_2=2 \alpha_1-2 \alpha_3=0.4 / 10^3[/latex]

i.e.

[latex]\alpha_1-\alpha_3-0.2 / 10^3[/latex] (v)

Adding Eqs (iii) and (iv)

[latex]u_3+u_4=2 \alpha_1+2 \alpha_3=0.7 / 10^3[/latex]

i.e.

[latex]\alpha_1+\alpha_3=0.35 / 10^3[/latex] (vi)

Adding Eqs (v) and (vi)

[latex]\alpha_1=0.275 / 10^3[/latex]

Then from Eq. (v)
[latex]\alpha_3=0.075 / 10^3[/latex]
Now subtracting Eq. (ii) from Eq. (i)

[latex]u_1-u_2=-2 \alpha_2+2 \alpha_4=-0.2 / 10^3[/latex]

i.e.
[latex]\alpha_2-\alpha_4=0.1 / 10^3[/latex] (vii)
Subtracting Eq. (iv) from Eq. (iii)
[latex]u_3-u_4=2 \alpha_2+2 \alpha_4=0.5 / 10^3[/latex]

i.e.
[latex]\alpha_2+\alpha_4=0.25 / 10^3[/latex] (viii)
Now adding Eqs (vii) and (viii)
[latex]2 \alpha_2=0.35 / 10^3[/latex]
whence
[latex]\alpha_2=0.175 / 10^3[/latex]
Then from Eq. (vii)
[latex]\alpha_4=0.075 / 10^3[/latex]

From the second of Eqs (6.96)

[latex]\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}[/latex]

Then, in a similar manner to the above

[latex]\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}[/latex]

Eqs (6.96) are now written

[latex]\begin{aligned}&u_i-(0.275+0.175 x+0.075 y+0.075 x y) imes 10^{-3} \&v_i=(0.4+0.1 x+0.2 y) imes 10^{-3}\end{aligned}[/latex]

Then, from Eqs (6.88)

[latex]\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}[/latex] (6.88)

[latex]\begin{aligned}\varepsilon_x &=(0.175+0.075 y) imes 10^{-3} \\varepsilon_y &=0.2 imes 10^{-3} \\gamma_{x y} &=(0.075+0.075 x+0.1) imes 10^{-3}=(0.175+0.075 x) imes 10^{-3}\end{aligned}[/latex]

At the centre of the element x = y = 0. Then

[latex]\begin{aligned}\varepsilon_x &=0.175 imes 10^{-3} \\varepsilon_y &=0.2 imes 10^{-3} \\gamma_{x y} &=0.175 imes 10^{-3}\end{aligned}[/latex]

so that, from Eqs (6.92)

[latex]\{\sigma\}=\left\{\begin{array}{l}\sigma_x \\sigma_y \ au_{x y}\end{array} ight\}=\frac{E}{1-v^2}\left[\begin{array}{ccc}1 & v & 0 \1 & 1 & 0 \0 & 0 & \frac{1}{2}(1-v)\end{array} ight]\left\{\begin{array}{l}\varepsilon_x \\varepsilon_y \\gamma_{x y}\end{array} ight\}[/latex] (6.92)

[latex]\begin{aligned}\sigma_x &=\frac{200000}{1-0.3^2}(0.175+0.3 imes 0.2) imes 10^{-3}=51.65 \mathrm{~N} / \mathrm{mm}^2 \\sigma_y &=\frac{200000}{1-0.3^2}(0.2+0.3 imes 0.175) imes 10^{-3}=55.49 \mathrm{~N} / \mathrm{mm}^2 \ au_{x y} &=\frac{200000}{2(1+0.3)} imes 0.175 imes 10^{-3}=13.46 \mathrm{~N} / \mathrm{mm}^2\end{aligned}[/latex]

Question p.6.11: A square element 1234, whose corners have coordinates x, y (...

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