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. A plane strain field $\varepsilon_x=c(-3x^2+7y^2)$, $\varepsilon_y=c(x^2-5y^2)$, $\gamma_{xy}=bxy$, with the displacement fixed to zero at the origin.

Find. (a) the $b$–$c$ relation from compatibility; (b) $u,v$ as functions of $c$, evaluated at (5, 2).

Approach. Impose the 2-D compatibility equation on the given strains, then integrate the strain–displacement relations, fixing the rigid-body constants with $u(0,0)=v(0,0)=0$.

1. (a) Compatibility. The plane condition is $\dfrac{\partial^2\varepsilon_x}{\partial y^2}+\dfrac{\partial^2\varepsilon_y}{\partial x^2}=\dfrac{\partial^2\gamma_{xy}}{\partial x\,\partial y}$. Here $14c+2c=b$, giving $$\boxed{b=16c.}$$
2. Integrate the normal strains. From $\varepsilon_x=\partial u/\partial x$ and $\varepsilon_y=\partial v/\partial y$, $u=c(-x^3+7xy^2)+f(y)$ and $v=c(x^2y-\tfrac{5}{3}y^3)+g(x)$.
3. Enforce the shear strain. $\gamma_{xy}=\partial u/\partial y+\partial v/\partial x=16cxy+f'(y)+g'(x)$. Matching $\gamma_{xy}=16cxy$ requires $f'(y)+g'(x)=0$, i.e. both are constants; with no rigid rotation and $u(0,0)=v(0,0)=0$, $f=g=0$.
4. (b) Evaluate at (5, 2). $u=c(-125+7\cdot5\cdot4)=15c$ and $v=c(25\cdot2-\tfrac{5}{3}\cdot8)=\tfrac{110}{3}c$: $$\boxed{u(5,2)=15c,\qquad v(5,2)=\tfrac{110}{3}c\approx36.7c.}$$