Question 2 of 8: Strain compatibility and displacements

Question 2: Strain compatibility and 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. A plane strain field with two nonzero constants b and c: $\varepsilon_x=c\!\left(-x^2+\tfrac{7}{3}y^2\right)$, $\varepsilon_y=c\!\left(\tfrac{1}{3}x^2-\tfrac{5}{3}y^2\right)$, $\gamma_{xy}=\tfrac{1}{3}bxy$.

Find. (a) the relationship between b and c for compatibility; (b) the displacements u, v at (5, 2) with c = 4, taking u = v = 0 at the origin.

Approach. Impose the 2-D strain-compatibility equation, then integrate the strain–displacement relations and fix the rigid-body constants from the origin condition.

1. Compatibility. The single 2-D compatibility condition is $$\frac{\partial^2\varepsilon_x}{\partial y^2}+\frac{\partial^2\varepsilon_y}{\partial x^2}=\frac{\partial^2\gamma_{xy}}{\partial x\,\partial y}.$$
2. Substitute. Here $\partial^2\varepsilon_x/\partial y^2=\tfrac{14}{3}c$, $\partial^2\varepsilon_y/\partial x^2=\tfrac{2}{3}c$ and $\partial^2\gamma_{xy}/\partial x\partial y=\tfrac{1}{3}b$, giving $\tfrac{14}{3}c+\tfrac{2}{3}c=\tfrac{1}{3}b$, hence $$\boxed{b=16c}.$$
3. Integrate the strains. From $\varepsilon_x=\partial u/\partial x$ and $\varepsilon_y=\partial v/\partial y$, $$u=c\!\left(-\tfrac{x^3}{3}+\tfrac{7}{3}y^2x\right),\qquad v=c\!\left(\tfrac{1}{3}x^2y-\tfrac{5}{9}y^3\right),$$ the integration functions vanishing because the displacements are zero at the origin and no rigid rotation is admitted (checking $\partial u/\partial y+\partial v/\partial x=\tfrac{16}{3}cxy$ recovers $\gamma_{xy}$ with $b=16c$).
4. Evaluate at (5, 2), c = 4. $$u=c\!\left(-\tfrac{125}{3}+\tfrac{140}{3}\right)=5c=\boxed{20},\qquad v=c\!\left(\tfrac{50}{3}-\tfrac{40}{9}\right)=\tfrac{110}{9}c=\boxed{48.9}.$$

The compatibility check guarantees a single-valued displacement field; the recovered field then reproduces all three strains exactly, confirming the integration constants were chosen correctly. Displacements carry the same length units as the coordinates x and y.