Question 4 of 8: Strain compatibility and displacement field

Question 4: Strain compatibility and displacement field (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. $\varepsilon_x=c(-18x^2+42y^2)$, $\varepsilon_y=c(6x^2-30y^2)$, $\gamma_{xy}=6bxy$, with displacements vanishing at the origin.

Find. (a) the constant relation $b(c)$ enforcing 2-D compatibility; (b) $u,v$ at $(3,1)$ with $c=4$.

Approach. Impose the single 2-D compatibility equation to relate $b$ and $c$, then integrate $\varepsilon_x=\partial u/\partial x$ and $\varepsilon_y=\partial v/\partial y$ and fix the integration functions with the shear-strain definition and the origin condition.

1. Compatibility condition. In two dimensions, $$\frac{\partial^2\varepsilon_x}{\partial y^2}+\frac{\partial^2\varepsilon_y}{\partial x^2}=\frac{\partial^2\gamma_{xy}}{\partial x\,\partial y}.$$ Evaluating, $\partial^2\varepsilon_x/\partial y^2=84c$, $\partial^2\varepsilon_y/\partial x^2=12c$, and $\partial^2\gamma_{xy}/\partial x\partial y=6b$, so $$84c+12c=6b\;\Longrightarrow\;\boxed{b=16c}.$$
2. Integrate the normal strains. $$u=\int\varepsilon_x\,dx=c\left(-6x^3+42xy^2\right)+f(y),\qquad v=\int\varepsilon_y\,dy=c\left(6x^2y-10y^3\right)+g(x).$$
3. Fix f and g with the shear strain. $$\gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=84cxy+f'(y)+12cxy+g'(x)=96cxy+f'(y)+g'(x).$$ Since $\gamma_{xy}=6bxy=96cxy$, we need $f'(y)+g'(x)=0$; with no rigid-body motion and zero displacement at the origin, $f=g=0$.
4. Evaluate at (3,1) with c = 4. $$u=4\left(-6(27)+42(3)(1)\right)=4(-162+126)=\boxed{-144},$$ $$v=4\left(6(9)(1)-10(1)\right)=4(54-10)=\boxed{176}.$$ (Displacements are in the same length units as the coordinates; the field is an abstract elasticity exercise.)