Question 5 of 8: Strain Compatibility and Displacement Field

Question 5: 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. $\varepsilon_x=c(-1.5x^2+3.5y^2)$, $\varepsilon_y=c(0.5x^2-2.5y^2)$, $\gamma_{xy}=0.5\,b\,x y$.

Find. (a) the $b$–$c$ relation for compatibility; (b) $u(3,5)$ and $v(3,5)$ with $u=v=0$ at the origin.

Approach. Apply the 2-D St. Venant compatibility equation, then integrate the strain–displacement relations, fixing the rigid-body constants from the origin condition.

1. Compatibility (part a). $\dfrac{\partial^2\varepsilon_x}{\partial y^2}+\dfrac{\partial^2\varepsilon_y}{\partial x^2}=\dfrac{\partial^2\gamma_{xy}}{\partial x\,\partial y}$ gives $7c+c=0.5b$, hence $$\boxed{b=16c}.$$
2. Integrate $\varepsilon_x=\partial u/\partial x$. $u=c\!\left(-0.5x^3+3.5xy^2\right)+f(y)$.
3. Integrate $\varepsilon_y=\partial v/\partial y$. $v=c\!\left(0.5x^2y-\tfrac{5}{6}y^3\right)+g(x)$.
4. Enforce the shear strain. $\dfrac{\partial u}{\partial y}+\dfrac{\partial v}{\partial x}=8c\,xy+f'(y)+g'(x)$ must equal $\gamma_{xy}=8c\,xy$, so $f'(y)+g'(x)=0$: both are constants (rigid-body motion). With $u=v=0$ at the origin and no rigid rotation, $f=g=0$.
5. Evaluate at (3,5). $u=c(-0.5\cdot27+3.5\cdot3\cdot25)=249\,c$; $v=c(0.5\cdot9\cdot5-\tfrac{5}{6}\cdot125)=-81.7\,c$: $$\boxed{u(3,5)=249\,c,\qquad v(3,5)=-81.7\,c}$$