Question p.6.12: A rectangular element used in plane stress analysis has corn...
A rectangular element used in plane stress analysis has corners whose coordinates in metres referred to an Oxy axes system are 1(−2, −1), 2(2, −1), 3(2, 1), 4(−2, 1). The displacements of the corners (in metres) are
If Young’s modulus is 200 000 N/mm² and Poisson’s ratio is 0.3 calculate the strains at the centre of the element.
Suitable displacement functions are:
\begin{aligned}&u=\alpha_1+\alpha_2 x+\alpha_3 y+\alpha_4 x y \\&v=\alpha_5+\alpha_6 x+\alpha_7 y+\alpha_8 x y\end{aligned}Then
\begin{aligned}&u_1=\alpha_1-2 \alpha_2-\alpha_3+2 \alpha_4=0.001 (i) \\&u_2=\alpha_1+2 \alpha_2-\alpha_3-2 \alpha_4=0.003 (ii)\\&u_3=\alpha_1+2 \alpha_2+\alpha_3+2 \alpha_4=-0.003 (iii) \\&u_4=\alpha_1-2 \alpha_2+\alpha_3-2 \alpha_4=0 (iv)\end{aligned}Subtracting Eq. (ii) from Eq. (i)
\alpha_2-\alpha_4 = 0.0005 (v)
Subtracting Eq. (iv) from Eq. (iii)
\alpha_2+\alpha_4 = −0.00075 (vi)
Subtracting Eq. (vi) from Eq. (v)
\alpha_4= −0.000625
Then, from either of Eqs (v) or (vi)
\alpha_2 = −0.000125
Adding Eqs (i) and (ii)
\alpha_1-\alpha_3= 0.002 (vii)
Adding Eqs (iii) and (iv)
\alpha_1+\alpha_3 = −0.0015 (viii)
Adding Eqs (vii) and (viii)
\alpha_1 = 0.00025
Then from either of Eqs (vii) or (viii)
\alpha_3 = −0.00175
Similarly
Then
\begin{aligned}&u_i=0.00025-0.00125 x-0.00175 y-0.000625 x y \\&v_i=-0.001+0.00025 x+0.002 y-0.00025 x y\end{aligned}From Eqs (1.18) and (1.20)
\left.\begin{array}{rl}\varepsilon_x & =\frac{\partial u}{\partial x} \\\varepsilon_y & =\frac{\partial v}{\partial y} \\\varepsilon_z & =\frac{\partial w}{\partial z}\end{array}\right\} (1.18)
\left.\begin{array}{l}\gamma_{x z}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z} \\\gamma_{x y}=\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y} \\\gamma_{y z}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}\end{array}\right\} (1.20)
\begin{aligned}\varepsilon_x &=\frac{\partial u}{\partial x}=-0.000125-0.000625 y \\\varepsilon_y &=\frac{\partial v}{\partial y}=0.002-0.00025 x \\\gamma_{x y} &=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=-0.0015-0.000625 x-0.00025 y\end{aligned}At the centre of the element where x = y = 0
\varepsilon_x=-0.000125 \quad \varepsilon_y=0.002 \quad \gamma_{x y}=-0.0015