Question 5 of 8: Strain Compatibility and Recovery of Displacements

Question 5: Strain Compatibility and Recovery of Displacements (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The plane strain field $$\varepsilon_x=c(-18x^2+42y^2),\quad \varepsilon_y=c(6x^2-30y^2),\quad \gamma_{xy}=6bxy.$$

Find. (a) the constant relation making the field compatible; (b) the displacement field (u, v) up to rigid-body motion.

Approach. Impose the 2-D compatibility equation to relate b and c; then integrate the strain–displacement relations, using the shear-strain equation to fix the integration functions.

1. Compatibility. The single 2-D compatibility condition is $$\frac{\partial^2\varepsilon_x}{\partial y^2}+\frac{\partial^2\varepsilon_y}{\partial x^2}=\frac{\partial^2\gamma_{xy}}{\partial x\,\partial y}.$$ Evaluating: $84c+12c=6b$, hence $$\boxed{b=16c.}$$
2. Integrate ε_x. Since $\varepsilon_x=\partial u/\partial x$, $$u=\int c(-18x^2+42y^2)\,dx=c(-6x^3+42xy^2)+f(y).$$
3. Integrate ε_y. Since $\varepsilon_y=\partial v/\partial y$, $$v=\int c(6x^2-30y^2)\,dy=c(6x^2y-10y^3)+g(x).$$
4. Apply the shear relation. $\gamma_{xy}=\partial u/\partial y+\partial v/\partial x = 84cxy+f'(y)+12cxy+g'(x)=96cxy+f'(y)+g'(x)$. With $b=16c$, $\gamma_{xy}=96cxy$, so $f'(y)+g'(x)=0$; each side is a constant $\omega$ (rigid-body rotation), giving $f(y)=-\omega y+u_0$, $g(x)=\omega x+v_0$.
5. Displacement field. Discarding rigid-body terms, $$\boxed{u=6c\,x(7y^2-3x^2),\qquad v=2c\,y(3x^2-5y^2).}$$